Precalculus: Graphs & Models

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Graphs and Models John W. Coburn St. Louis Community College at Florissant Valley

J.D. Herdlick St. Louis Community College at Meramec-Kirkwood

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TM

PRECALCULUS: GRAPHS AND MODELS Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 4 3 2 1 ISBN 978–0–07–351953–1 MHID 0–07–351953–7 ISBN 978–0–07–723052–4 (Annotated Instructor’s Edition) MHID 0–07–723052–3 Vice President, Editor-in-Chief: Marty Lange Vice President, EDP: Kimberly Meriwether David Vice-President New Product Launches: Michael Lange Editorial Director: Stewart K. Mattson Sponsoring Editor: John R. Osgood Developmental Editor: Eve L. Lipton Marketing Manager: Kevin M. Ernzen Senior Project Manager: Vicki Krug

Buyer II: Sherry L. Kane Senior Media Project Manager: Sandra M. Schnee Senior Designer: Laurie B. Janssen Cover Image: © Georgette Douwma and Sami Sarkis / Gettyimages Senior Photo Research Coordinator: John C. Leland Compositor: Aptara®, Inc. Typeface: 10.5/12 Times Roman Printer: R. R. Donnelley

Chapter 1 Opener/p.1: © Neil Beer/Getty Images/RF; p. 2/left: © PhotoAlto/RF; p. 2/middle: © Brand X Pictures/PunchStock/RF; p. 2/right: © Lars Niki/RF; p. 48: © Royalty-Free/CORBIS; p. 77: NASA/RF; p. 90: © Tom Grill/Corbis/RF; p. 100: © Lourens Smak/Alamy/RF. Chapter 2 Opener/p. 105: © Getty Images, Inc./ RF; p. 106: © Siede Preis/Getty Images/RF; p. 107: © The McGraw-Hill Companies, Inc./Ken Cavanagh Photographer; p. 145: © PhotoLink/Getty Images/RF; p. 164: © Alan and Sandy Carey/Getty Images/RF; p. 175: Courtesy John Coburn; p. 182: © Royalty-Free/CORBIS. Chapter 3 Opener/p. 203: © Christian Pondella Photography; p. 233: © Photodisc Collection/Getty Images/RF; p. 251: © Photodisc Collection/Getty Images/RF; p. 252: Courtesy NASA; p. 260: © Steve Cole/ Getty Images/RF; p. 274: © U.S. Fish & Wildlife Service/Tracy Brooks/RF; p. 282: © Royalty-Free/CORBIS; p. 283: © The McGraw-Hill Companies, Inc./Barry Barker, photographer; p. 284: © Royalty-Free/CORBIS; p. 289/top: © Patrick Clark/Getty Images/RF; p. 289/bottom: © Digital Vision/PunchStock/RF; p. 290/left: © Goodshoot/PunchStock/RF; p. 290/right: © Royalty-Free/CORBIS; p. 291: © Charles Smith/CORBIS/RF. Chapter 4 Opener/p. 307: © Royalty-Free/CORBIS; p. 316: © Adalberto Rios Szalay/Sexto Sol/Getty Images/RF; p. 319, 335, 370, 394: © Royalty-Free/CORBIS. Chapter 5 Opener/p. 409: © Comstock Images/RF; p. 431/left: © Geostock/Getty Images/RF; p. 431/right: © Lawrence M. Sawyer/Getty Images/RF; p. 440: © U.S. Geological Survery; p. 441: © Lars Niki/RF; p. 445: © Medioimages/Superstock/RF; p. 455/top-left: © Ingram Publishing/age Fotostock/RF; p. 455/bottom-left: © Keith Brofsky/Getty Images/RF; p. 455/top-right: © The McGraw-Hill Companies, Inc./Andrew Resek, photographer; p. 455/bottom-right: © McGraw-Hill Higher Education/Carlyn Iverson, photographer; p. 467: © Stock Trek/Getty Images/RF; p. 484: Courtesy Dawn Bercier; p. 501: © CMCD/Getty Images/RF; p. 503: © John A. Rizzo/Getty Images/RF. Chapter 6 Opener/p. 509: © Digital Vision/RF; p. 521: © Jules Frazier/Getty Images/RF; p. 524: © Karl Weatherly/Getty Images/RF; p. 525: © Dynamic GraphicsGroup/PunchStock/RF; p. 526: © Michael Fay/Getty Images/RF; p. 558: © Royalty-Free/CORBIS; p. 589: © Digital Vision/Punchstock/RF; p. 608: © Royalty-Free/CORBIS; p. 625: © Steve Cole Getty Images/RF; p. 638: © Royalty-Free/CORBIS. Chapter 7 Opener/p. 653: © NPS Photo by William S. Keller/RF; p. 730: © John Wang/Getty Images/RF. Chapter 8 Opener/p. 745: © Royalty-Free/CORBIS. Chapter 9 Opener/p. 837/Blackberry phone: © The McGraw-Hill Companies, Inc./Lars A. Niki photographer; p. 837/Ipod: © The McGraw-Hill Companies, Inc; p. 837/MP3 with headphones: © Don Farrall/Getty Images/RF; p. 837/Group of various cell phones: © Stockbyte/ Getty Images/RF; p. 846: © I. Rozenbaum/F. Cirou/Photo Alto/RF; p. 851: © 2009 Jupiterimages Corporation/RF; p. 852: © Royalty-Free/CORBIS; p. 863: © Creatas/ Punchstock/RF; p. 886: © Royalty-Free/CORBIS; p. 940: © Steve Cole/Getty Images/RF; p. 948: © F. Shussler/PhotoLink/Getty Images/RF. Chapter 10 Opener/ p. 961: © Mark Downey/Getty Images/RF; p. 992: © Brand X Pictures/PunchStock/RF; p. 993: © Digital Vision/Getty Images/RF; p. 996: © H. Wiesenhofer/PhotoLink/ Getty Images/RF; p. 1004: © Jim Wehtje/Getty Images/RF; p. 1005/left: © Getty Images/RF; p. 1005:/right © Creatas/Punchstock/RF; p. 1014: © Ryan McVay/Getty Images/RF; p. 1017: © The McGraw-Hill Companies, Inc./Jill Braaten. Photographer; p. 1061: © PhotoLink/Getty Images/RF. Chapter 11 Opener/p. 1077: © Digital Vision/RF; p. 1097: © Royalty-Free/CORBIS; p. 1110: © Anderson Ross/Getty Images/RF; p. 1141: © Royalty-Free/CORBIS. Chapter 12 Opener/p. 1169: © Royalty-Free/CORBIS. Appendix A p. A-7: © Photodisc/Getty Images/RF; p. A-23: © Royalty-Free/CORBIS; p. A-78: © Glen Allison/Getty Images/RF.

Library of Congress Cataloging-in-Publication Data Coburn, John W. Precalculus : graphs and models / John W. Coburn, J. D. Herdlick. p. cm. Includes index. ISBN 978–0–07–351953–1 — ISBN 0–07–351953–7 (hard copy : alk. paper) 1. Functions— Graphic methods—Textbooks. 2. Trigonometry—Graphic methods—Textbooks. I. Herdlick, John D. II. Title. QA331.3.C6325 2012 515—dc22 2010047030 www.mhhe.com

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Brief Contents Preface vi Index of Applications

1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 CHAPTER 9 C H A P T E R 10 C H A P T E R 11 C H A P T E R 12 CHAPTER

xxxv

Relations, Functions, and Graphs

1

More on Functions 105 Quadratic Functions and Operations on Functions

203

Polynomial and Rational Functions 307 Exponential and Logarithmic Functions 409 An Introduction to Trigonometric Functions 509 Trigonometric Identities, Inverses, and Equations 653 Applications of Trigonometry

745

Systems of Equations and Inequalities

837

Analytic Geometry and the Conic Sections 961 Additional Topics in Algebra 1077 Bridges to Calculus: An Introduction to Limits 1169

Appendix A

A Review of Basic Concepts and Skills A-1

Appendix B

Proof Positive-A Selection of Proofs from Precalculus A-84

Appendix C

More on Synthetic Division A-89

Appendix D

Reduced Row-Echelon Form and More on Matrices A-91

Appendix E

The Equation of a Conic A-93

Appendix F

Families of Polar Curves A-95 Student Answer Appendix (SE only)

SA-1

Instructor Answer Appendix (AIE only) Index

IA-1

I-1

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About the Authors John Coburn

John Coburn grew up in the Hawaiian Islands, the seventh of sixteen children. He received his Associate of Arts degree in 1977 from Windward Community College, where he graduated with honors. In 1979 he earned a Bachelor’s Degree in Education from the University of Hawaii. After working in the business world for a number of years, he returned to teaching, accepting a position in high school mathematics where he was recognized as Teacher of the Year (1987). Soon afterward, the decision was made to seek a Master's Degree, which he received two years later from the University of Oklahoma. John is now a full professor at the Florissant Valley campus of St. Louis Community College. During his tenure there he has received numerous nominations as an outstanding teacher by the local chapter of Phi Theta Kappa, two nominations to Who’s Who Among America’s Teachers, and was recognized as Post Secondary Teacher of the Year in 2004 by the Mathematics Educators of Greater St. Louis (MEGSL). He has made numerous presentations and local, state, and national conferences on a wide variety of topics and maintains memberships in several mathematics organizations. Some of John’s other interests include body surfing, snorkeling, and beach combing whenever he gets the chance. He is also an avid gamer, enjoying numerous board, card, and party games. His other loves include his family, music, athletics, composition, and the wild outdoors.

J.D. Herdlick

J.D. Herdlick was born and raised in St. Louis, Missouri, very near the Mississippi river. In 1992, he received his bachelor’s degree in mathematics from Santa Clara University (Santa Clara, California). After completing his master’s in mathematics at Washington University (St. Louis, Missouri) in 1994, he felt called to serve as both a campus minister and an aid worker for a number of years in the United States and Honduras. He later returned to education and spent one year teaching high school mathematics, followed by an appointment at Washington University as visiting lecturer, a position he held until 2006. Simultaneously teaching as an adjunct professor at the Meramec campus of St. Louis Community College, he eventually joined the department full time in 2001. While at Santa Clara University, he became a member of the honorary societies Phi Beta Kappa, Pi Mu Epsilon, and Sigma Xi under the tutelage of David Logothetti, Gerald Alexanderson, and Paul Halmos. In addition to the Dean’s Award for Teaching Excellence at Washington University, J.D. has received numerous awards and accolades for his teaching at St. Louis Community College. Outside of the office and classroom, he is likely to be found in the water, on the water, and sometimes above the water, as a passionate wakeboarder and kiteboarder. It is here, in the water and wind, that he finds his inspiration for writing. J.D. and his family currently split their time between the United States and Argentina.

Dedication With boundless gratitude, we dedicate this work to the special people in our lives. To our children, whom we hope were joyfully oblivious to the time, sacrifice, and perseverance required; and to our wives, who were well acquainted with every minute of it.

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About the Cover Most coral reefs in the world are 7000–9000 years old, but new reefs can fully develop in as few as 20 years. In addition to being home to over 4000 species of tropical or reef fish, coral reefs are immensely beneficial to humans and must be carefully preserved. They buffer coastal regions from strong waves and storms, provide millions of people with food and jobs, and prompt advances in modern medicine. Similar to the ancient reefs, a course in Precalculus is based on thousands of years of mathematical curiosity, insight, and wisdom. In this one short course, we study a wealth of important concepts that have taken centuries to mature. Just as the variety of fish in the sea rely on the coral reefs to survive, students in a Precalculus course rely on mastery of this bedrock of concepts to successfully pursue more advanced courses, as well as their career goals.

From the Authors

nges. From the ion has seen some enormo us cha cat edu tics ma the ma s, ade dec In the last two to online homework and and the adv ent of the Intern et, ors ulat calc ng phi gra of n ctio elen ting. intr odu s ago, the cha nges hav e been unr ade dec ut abo am dre only ld cou visual sup plement s we nce tea ching re a combined 40 yea rs of exp erie sha k dlic Her . J.D and urn Cob n Tog eth er, Joh e dev elop ed a wea lth of ors and oth er technologies, and hav ulat calc ng phi gra h wit ulus calc pre endeav or. firs tha nd exp erience related to the con ver sat iona l style and Models text, we hav e combined the and phs Gra ck rdli /He urn Cob one of In the , wit h this dep th of exp erience. As for wn kno are s text our t tha s the wea lth of application y see functions think visually, to a poin t where the ts den stu help to out set we ls, our primary goa ediately lead to a of gra phs, wit h attr ibutes that imm 2 ily fam a of one as 4x – x = , the nat ure of like f(x) ior, zer oes, solu tions to ineq ualities hav -be end ms, imu min and ums discussion of ma xim an equation that es in con text — instead of mer ely ibut attr se the of tion lica app the le the roots, and the scr een of a calculat or. And whi on ph gra a g etin rpr inte by or off ers much must be solv ed by factor ing nal drudgery, we believe our text atio put com e som eve reli y ma ors gra phing calculat gra phical met hods, wit h ison of algebra ic met hods ver sus par com e -sid -by side ple sim a nua lly. n more tha checking answer s to wor k don e ma ply sim n tha role ant ific sign re mo the calculat or playing a sible wit h pap er and investigate far bey ond what’s pos and k wor to d use are ors ulat Gra phing calc age more applications, and e more tru e-to-lif e equations, eng solv to d use gy nolo tech the h wit text is built on pencil, the end we believe you’ll see this In t. res inte of ns stio que l ntia explore more substa accent uates the visual and dynamic excursion that a ers off t tha one yet als, ent strong fundam use in all areas of their solv ing acumen that studen ts will blem pro and g nnin pla nal atio aniz l tool for the org Gra phs and Models text as an idea ck rdli /He urn Cob the er off we lives. To this end —John Coburn and J.D. Her dlick tics. tea ching and lear ning of mathema

v

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Making Connections . . . Precalculus tends to be a challenging course for many students. They may not see the connections that Precalculus has to their life or why it is so critical that they succeed in this course. Others may enter into this course underprepared or improperly placed and with very little motivation. Instructors are faced with several challenges as well. They are given the task of improving pass rates and student retention while ensuring the students are adequately prepared for more advanced courses, as a Precalculus course attracts a very diverse audience, with a wide variety of career goals and a large range of prerequisite skills. The goal of this textbook series is to provide both students and instructors with tools to address these challenges, so that both can experience greater success in Precalculus. For instance, the comprehensive exercise sets have a range of difficulty that provides very strong support for weaker students, while advanced students are challenged to reach even further. The rest of this preface further explains the tools that John Coburn, J.D. Herdlick, and McGraw-Hill have developed and how they can be used to connect students to Precalculus and connect instructors to their students.

The Coburn/Herdlick Precalculus Series provides you with strong tools to achieve better outcomes in your Precalculus course as follows:

vi

▶

Making Connections Visually, Symbolically, Numerically, and Verbally

▶

Better Student Preparedness Through Superior Course Management

▶

Increased Student Engagement

▶

Solid Skill Development

▶

Strong Mathematical Connections

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▶

Making Connections Visually, Symbolically, Numerically, and Verbally

In writing their Graphs and Models series, the Coburn/Herdlick team took great care to help students think visually by relating a basic graph to an algebraic equation at every opportunity. This empowers students to see the “Why?” behind many algebraic rules and properties, and offers solid preparation for the connections they’ll need to make in future courses which often depend on these visual skills. ▶

Better Student Preparedness Through Superior Course Management

McGraw-Hill is proud to offer instructors a choice of course management options to accompany Coburn/ Herdlick. If you prefer to assign text-specific problems in a brand new, robust online homework system that contains stepped out and guided solutions for all questions, Connect Math Hosted by ALEKS may be for you. Or perhaps you prefer the diagnostic nature and artificial intelligence engine that is the driving force behind our ALEKS 360 Course product, a true online learning environment, which has been expanded to contain hundreds of new College Algebra & Precalculus topics. We encourage you to take a closer look at each product on preface pages x through xiii and to consult your McGraw-Hill sales representative to setup a demonstration. ▶

Increased Student Engagement

There are many texts that claim they “engage” students, but only the Coburn/Herdlick Series has carefully studied and implemented features and options that make it truly possible. From the on-line support, to the textbook design and a wealth of quality applications, students will remain engaged throughout their studies. ▶

Solid Skill Development

The Coburn/Herdlick series intentionally relates the examples to the exercise sets so there is a strong connection between what students are learning while working through the examples in each section and the homework exercises that they complete. This development of strong mechanical skills is followed closely by a careful development of problem solving skills, with the use of interesting and engaging applications that have been carefully chosen with regard to difficulty and the skills currently under study. There is also an abundance of exercise types to choose from to ensure that homework challenges a wide variety of skills. Furthermore, John and J.D. reconnect students to earlier chapter material with Mid-Chapter Checks; students have praised these exercises for helping them understand what key concepts require additional practice. ▶

Strong Mathematical Connections

John Coburn and J.D. Herdlick’s experience in the classroom and their strong connections to how students comprehend the material are evident in their writing style. This is demonstrated by the way they provide a tight weave from topic to topic and foster an environment that doesn’t just focus on procedures but illustrates the big picture, which is something that so often is sacrificed in this course. Moreover, they employ a clear and supportive writing style, providing the students with a tool they can depend on when the teacher is not available, when they miss a day of class, or simply when working on their own.

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Making Connections . . . Visually, Symbolically, Numerically, and Verbally , the concre te and numer ic “It is widely known that for studen ts to grow stronger algebr aically entations. In this transit ion experiences from their past must give way to more symbolic repres visual connections and verbal from numer ic, to symbolic, to algebr aic thinking, the importance of of rich concep ts or subtle ideas, connections is too often overlooked. To reach a deep understanding concep t or idea using the terms studen ts must develop the ability to menta lly “see” and discuss the seeing the connections that and names needed to describ e it accurately. Only then can they begin . A large part of this involves exist between each new concep t, and concep ts that are already known they’re able to see functio ns like helping our studen ts to begin thinking visuall y, to a point where ical attribu tes that immediately f(x) = x2 – 4x as only one of a large family of functions, with graph , solutions to inequa lities, the lead to a discussion of maximums and minimums, end-behavior, zeroes t. And while it’s important for nature of the roots, and the application of these attribu tes in contex , and that the intersection of students to see that zeroes are x-intercepts and x-intercepts are zeroes g these graphs, these should not two graphs provides a simulta neous solution to the equations formin tions, investigations, connections, remain the sole focus of the tool. Graphing calculators allow explora and we should use the technology and visualizations far beyond what’s possible with paper and pencil, more true-to-life equations, to aid the development of these menta l-visua l skills, in addition to solving ns involving real data, domain engaging more applications, and explor ing the more substa ntial questio tables, and other questions of and range, anticipated graphical behavior, additional uses of lists and be successful in these endeavors.” interest. We believe this text offers instructors the tools they need to —The Authors

EXAMPLE 1

䊲

䊳

Solve for x and check your answer: log x log 1x 32 1. 䊲

Algebraic Solution

log x log 1x 32 1 log 3x 1x 32 4 1 x2 3x 101 x2 3x 10 0 1x 52 1x 22 0 x 5 or x 2

“I think there is a good balance between technology

Solving a Logarithmic Equation

original equation product property exponential form, distribute x set equal to 0 factor result

and paper/pencil techniques. I particularly like how the technology portion does not take the place of paper/pencil, but instead supplements it. I think a lot of departments will like that.

Graphical Solution

Using the intersection-ofgraphs method, we enter Y1 log X log1X 32 and Y2 1. From the domain we know x 7 0, indicating the solution will occur in QI. After graphing both functions using the window shown, the intersection method shows the only solution is x 2.

3

”

0

—Daniel Brock, Arkansas State University-Beebe

5

▶ Graphical Examples show students how

3

Check: The “solution” x 5 is outside the domain and is ignored. For x 2, log x log1x 32 1 original equation log 2 log12 32 1 substitute 2 for x log 2 log 5 1 simplify log12 # 52 1 product property log 10 1 Property I

the calculator can be used to supplement their understanding of a problem. EXAMPLE 1A

You could also use a calculator to verify log 2 log 5 1 directly. roug g 14 Now try Exercises 7 through

Precalculus: Graphs and Models textbook the best approach ever to the teaching of Precalculus with the inclusion of graphing calculator.

”

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—Alvio Dominguez, Miami-Dade College-Wolfson

Solving an Equation Graphically 1 Solve the equation 21x 32 7 x 2 using 2 a graphing calculator.

䊳

Solution

“I have certainly found the Coburn/Herdlick’s

䊳

䊳

Begin by entering the left-hand expression as Y1 and the right-hand expression as Y2 (Figure 1.74). To find points of intersection, press 2nd TRACE (CALC) and select option 5:intersect, which automatically places you on the graphing window, and asks you to identify the “First curve?.” As discussed, pressing three times in succession will identify each graph, bypass the “Guess?” option, then find and display the point of intersection (Figure 1.75). Here the point of intersection 10 is (2, 3), showing the solution to this equation is x 2 (for which both expressions equal 3). This can be verified by direct substitution or by using the TABLE feature. ENTER

Figure 1.74

Figure 1.75 10

10

10

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▶ Calculator Explanations incorporate the

calculator without sacrificing content.

Figure 3.2 Most graphing calculators are programmed to work with imaginary and complex numbers, though for some models the calculator must be placed in complex number mode. After pressing the MODE key (located to the right of the 2nd option key), the screen shown in Figure 3.2 appears and we use the arrow keys to access “a bi” and active this mode (by pressing ). Once active, we can validate our previo previous statements about imaginary numbers (Figure gure 3.3 3.3), as well as verify our previous calculations like those in Examples 3(a), 3(d), and an d 4(a) (F (Figure 3.4). Note the imaginary unit i is the 2nd option for the decimal point. ENTER

“The technology (graphing calculator) explanations and illustrations are superb. The level of detail is valuable; even an experienced user (myself) learned some new techniques and “tricks” in reading through the text. The text frequently references use of the calculator—yet without sacrificing rigor or mathematical integrity.

Figure 3.3

Figure 3.4

”

—Light Bryant, Arizona Western College

Figure 4.4A

To help illustrate the Intermediate Value Theorem, many graphing calculators offer a useful feature called split screen viewing, that enables us to view a table of values and the graph of a function at the same time. To illustrate, enter the function y x3 9x 6 (from Example 6) as Y1 on the Y= screen, then set the viewing window as shown in Figure 4.4. Set your table in AUTO mode with ¢Tbl 1, then press the MODE key (see Figure 4.4A) and notice the second-to-last entry on this screen reads: Full for full screen viewing, Horiz for splitting the screen horizontally with the graph above a reduced home screen, and G-T, which represents Graph-Table and splits the screen vertically. In the G-T mode, the graph appears on the left and the table of values on the right. Navigate the cursor to the G-T mode and press . Pressing the GRAPH key at this point should give you a screen similar to Figure 4.5. Scrolling downward shows the function also changes sign between x 2 and x 3. For more on this idea, see Exercises 31 and 32. werful yet simple As a final note, while the intermediate value theorem is a powerful tool, it must be used with care. For example, given p1x2 x4 10x2 5, p112 7 0 lly, and p112 7 0, seeming to indicate that no zeroes exist in the intervall (1, 1). Actual Actually, there are two zeroes, as seen in Figure 4.6. ENTER

in every section. I have been using TI calculators for 15 years and I learned a few new tricks while reading this book.

Figure 4.6

Figure 4.5

“The authors give very good uses of the calculator

25

”

B. You’ve just seen how we can use the intermediate value theorem to identify intervals containing a polynomial zero

5

5

—George Hurlburt, Corning Community College

10

▶ Technology Applications show

students how technology can be used to help apply lessons from the classroom to real life.

“I think that the graphing examples, explanations,

and problems are perfect for the average college algebra student who has never touched a graphing calculator. . . . . I think this book would be great to actually have in front of the students.

”

—Dale Duke, Oklahoma City Community College

Use Newton’s law of cooling to complete Exercises 75 and 76: T(x) ⫽ TR ⫹ (T0 ⫺ TR)ekx.

75. Cold party drinks: Janae was late getting ready for the party, and the liters of soft drinks she bought were still at room temperature (73°F) with guests due to arrive in 15 min. If she puts these in her freezer at 10°F, will the drinks be cold enough (35°F) for her guests? Assume k ⬇ 0.031. 76. Warm party drinks: Newton’s law of cooling applies equally well if the “cooling is negative,” meaning the object is taken from a colder medium and placed in a warmer one. If a can of soft drink is taken from a 35°F cooler and placed in a room where the temperature is 75°F, how long will it take the drink to warm to 65°F? Assume k ⬇ 0.031. Photochromat Photochromatic sunglasses: Sunglasses that darken in sunlight sunl unligh ghtt (p (photo (photochromatic sunglasses) contain millions of mole m ole lecules le lec cu of a substance known as silver halide. The molecules molecules m mo o are ttransparent indoors in the absence of ultraviolent (UV) (U light. Outdoors, UV light from the sun causes the mol molecules to change shape, darkening the lenses in respo response to the intensity of the UV light. For certain lenses, the function T1x2 0.85x models the transparency of the lenses (as a percentage) based on a UV index x. Find Fi the transparency (to the nearest percent), if the lenses are exposed to 77 li ht with a UV index of 7 (a high exposure). 77. sunlight 78. sunlight with a UV index of 5.5 (a moderate exposure)

80. Use a trial-and-error process and a graphing calculator to determine the UV index when the lenses are 50% transparent. Modeling inflation: Assuming the rate of inflation is 5% per year, the predicted price of an item can be modeled by the function P1t2 P0 11.052 t, where P0 represents the initial price of the item and t is in years. Use this information to solve Exercises 81 and 82. 81. What will the price of a new car be in the year 2015, if it cost $20,000 in the year 2010? 82. What will the price of a gallon of milk be in the year 2015, if it cost $3.95 in the year 2010? Round to the nearest cent. Modeling radioactive decay: The half-life of a radioactive substance is the time required for half an initial amount of the substance to disappear through decay. The amount of the substance remaining is given t by the formula Q1t2 Q0 1 12 2 h, where h is the half-life, t represents the elapsed time, and Q(t) represents the amount that remains (t and h must have the same unit of time). Use this information to solve Exercises 83 and 84. 83. Some isotopes of the substance known as thorium have a half-life of only 8 min. (a) If 64 grams are initially present, how many grams (g) of the substance remain after 24 min? (b) How many minutes until only 1 gram (g) of the substance remains?

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Connect Math Hosted by ALEKS Corporation is an exciting, new assignment and assessment platform combining the strengths of McGraw-Hill Higher Education and ALEKS Corporation. Connect Math Hosted by ALEKS is the first platform on the market to combine an artificially-intelligent, diagnostic assessment with an intuitive ehomework platform designed to meet your needs. Connect Math Hosted by ALEKS Corporation is the culmination of a one-of-a-kind market development process involving math full-time and adjunct Math faculty at every step of the process. This process enables us to provide you with a solution that best meets your needs. Connect Math Hosted by ALEKS Corporation is built by Math educators for Math educators!

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You want a way to identify the strengths and weaknesses of your class at the beginning of the term rather than after the first exam.

Integrated ALEKS® Assessment ▶ This artificially-intelligent (AI), diagnostic assessment identifies precisely what a student knows and is ready to learn next. ▶ Detailed assessment reports provide instructors with specific information about where students are struggling most. ▶ This AI-driven assessmentt is the only one of its kind in an online homework platform.

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Built by Math Educators for Math Educators 3

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Efficient Assignment Navigation ▶ Students have access to immediate feedback and help while working through assignments. ▶ Students have direct access ess to a media-rich eBook forr easy referencing. ▶ Students can view detailed, ed, step-by-step solutions written by instructors who teach the course, providing a unique solution on to each and every exercise. e

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Students can easily monitor and track their progress on a given assignment.

Y want a more intuitive and efficient assignment creation process You because of your busy schedule. b

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Flexible Instructor Gradebook ▶ Based on instructor feedback, Connect Math Hosted by ALEKS Corp’s straightforward design creates an intuitive, visually pleasing grade management environment. ▶ Assignment types are color-coded for easy viewing. ▶ The gradebook allows instructors the flexibility to import and export additional grades. Instructors have the ability to drop grades as well as assign extra credit.

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Built by Math Educators for Math Educators 7

Y want algorithmic content that was developed by math faculty to You ensure the content is pedagogically sound and accurate. e

Digital Content Development Story The development of McGraw-Hill’s Connect Math Hosted by ALEKS Corp. content involved collaboration between McGraw-Hill, experienced instructors, and ALEKS, a company known for its high-quality digital content. The result of this process, outlined below, is accurate content created with your students in mind. It is available in a simple-to-use interface with all the functionality tools needed to manage your course. 1. McGraw-Hill selected experienced instructors to work as Digital Contributors. 2. The Digital Contributors selected the textbook exercises to be included in the algorithmic content to ensure appropriate coverage of the textbook content. 3. The Digital Contributors created detailed, stepped-out solutions for use in the Guided Solution and Show Me features. 4. The Digital Contributors provided detailed instructions for authoring the algorithm specific to each exercise to maintain the original intent and integrity of each unique exercise. 5. Each algorithm was reviewed by the Contributor, went through a detailed quality control process by ALEKS Corporation, and was copyedited prior to being posted live.

Connect Math Hosted by ALEKS Corp. Built by Math Educators for Math Educators Lead Digital Contributors

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Amy Naughten

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Robust Graphing Features: ALEKS Precalculus provides more graphing coverage and includes a built-in graphing calculator, an adaptive, open-response environment, and realistic answer input tools to ensure student mastery. The ALEKS Graphing Calculator is accessible via the Student Module and can be turned on or off by the instructor.

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. . .T hrough Superior Course Management New Instructor Module Features for Precalculus Help Students Achieve Success While Saving Instructor Time ALEKS includes an Instructor Module with powerful, assignment-driven features and extensive content flexibility to simplify course management so instructors spend less time with administrative tasks and more time directing student learning. The ALEKS Instructor Module also includes two new features that further simplify course management and provide content flexibility: Partial Credit on Assignments and Supplementary Textbook Integration Topic Coverage.

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Increased Student Engagement . . . Through g Meaningful g Applications pp a con nec tion bet wee n the req uires that student s exp erie nce Ma king mat hematics mea ning ful is the result of a pow erf ul on the wor ld they live in. This text act imp its and y, stud y the s atic ing close ties to the mat hem qua lity, and greates t inte res t, hav est high the of s tion lica app vide commit men t to pro larly ma de an eff ort to sup ply ed leve ls of diff icul ty. We par ticu itor mon lly efu car h wit and les, assignm ent s, and exa mp illus trations, incl uded as hom ework lass in-c for d use be to y ntit qua these in suf ficient ir sup ply premat urely. Ma ny and test s, wit hou t exhaus ting the zes quiz of n ctio stru con the in d cur ious, eve n emp loye nces, wit h oth ers com ing from a erie exp rse dive own our of n bor atics in app lications wer e nts of life, and to see the mat hem eve ay ryd eve the on e seiz to ch tools, wit h visiona ry folly that ena bles one tial libr ary of ref ere nce and resear stan sub a by ted por sup e wer se the backgr oun d. The Aut hor s nts, and modern tren ds. —The an eye toward hist ory, cur ren t eve

▶ Chapter Openers highlight Chapter Connections, an interesting

application exercise from the chapter, and provide a list of other real-world connections to give context for students who wonder how math relates to them.

“I think the book has very modern applications and quite a few of them. The calculator instructions are very well done.”

CHAPTER CONNECTIONS

—Nezam Iraniparast, Western Kentucky University

More on Functions CHAPTER OUTLINE

2.5 Piecewise-Defined Functions 245

power in watts and v is the wind velocity in miles per hour. While the formula enables us to predict the power generated for a given wind speed, the graph offers a visual representation of this relationship, where we note a rapid growth in power output as the wind speed increases. This application appears as Exercise 107 in Section 2.2.

2.6 Variation: The Toolbox Functions in Action 259

Check out these other real-world connections:

2.1 Analyzing the Graph of a Function 188

▶ Examples throughout the text feature word problems, providing

students with a starting point for how to solve these types of problems in their exercise sets.

Viewing a function in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For example, the power generated by a wind turbine is often modeled 8v3 by the function P 1v2 , where P is the 125

2.2 The Toolbox Functions and Transformations 202 2.3 Absolute Value Functions, Equations, and Inequalities 218

2.4 Basic Rational Functions and Power Functions; More on the Domain 230

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“ The students always want to know ‘When am I ever going to have

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Analyzing the Path of a Projectile (Section 2.1, Exercise 57) Altitude of the Jet Stream (Section 2.3, Exercise 61) Amusement Arcades (Section 2.5, Exercise 42) Volume of Phone Calls (Section 2.6, Exercise 55)

to use algebra anyway?’ Now it will not be hard for them to see for themselves some REAL ways. —Sally Haas, Angelina College

187

”

EXAMPLE 2

▶ Application Exercises at the end of each section are the hallmark of

the Coburn series. Never contrived, always creative, and born out of the author’s life and experiences, each application tells a story and appeals to a variety of teaching styles, disciplines, backgrounds, and interests. The authors have ensured that the applications reflect the most common majors of precalculus students.

“ The amount of technology is great, as are the applications. The quality of the applications is better than my current text.” —Daniel Russow, Arizona Western College–Yuma

▶ M Math th iin A Action ti A Applets, l t llocated t d online, li enable bl students t d t tto work k

collaboratively as they manipulate applets that apply mathematical concepts in real-world contexts. xvi

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Identifying Functions Two relations named f and g are given; f is pointwise-defined (stated as a set of ordered pairs), while g is given as a set of plotted points. Determine whether each is a function. f: 13, 02, 11, 42, 12, 52, 14, 22, 13, 22, 13, 62, 10, 12, (4, 5), and (6, 1)

Solution

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The relation f is not a function, since 3 is paired with two different outputs: 13, 02 and 13, 22 . The relation g shown in the figure is a function. Each input corresponds to exactly one output, otherwise one point would be directly above the other and have the same first coordinate.

g

5

y (0, 5)

(4, 2) (3, 1)

(2, 1) 5

5

x

(4, 1) (1, 3) 5

Now try Exercises 11 through 18 䊳

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Through Timely Examples mp les that set the sta ge to overstate the imp ortance of exa t icul diff be ld wou it , tics ma the was too In ma falt ered due to an exa mp le that e hav s nce erie exp l iona cat edu for lear ning. No t a few a car efu l and a dist rac ting result. In this ser ies, had or ce, uen seq of out fit, r on diff icul t, a poo ely and clea r, wit h a direct foc us tim e wer t tha les mp exa ct sele to deliber ate eff ort was ma de link pre vious e, they wer e fur the r designed to sibl pos e her ryw Eve d. han at l the concep t or skil e. As a tra ined educat or gro undwor k for concep ts to com the lay to and s, idea t ren cur to seq uence of concep ts ore it’s ever asked, and a tim ely bef en oft is n stio que a wer ans knows, the bes t tim e to new idea simply the way in this regard, ma king each long a go can les mp exa ed uct car efu lly constr ity of a studen t grows successfu l, the mathematica l matur en Wh . step d ate icip ant n eve cal, nex t logi . —The Aut hor s it was just sup posed to be that way in unn otic ed incr ements, as tho ugh

“ The authors have succeeded with numerous

calculator examples with easy-to-use instructions to follow along. I truly enjoy seeing plenty of calculator examples throughout the text!!

▶ Side by side graphical and algebraic solutions illustrate the

difference between problem-solving methods, emphasize the connections between algebraic and graphical information, and enable students to understand why one method might be preferable to another for any given problem.

”

—David Bosworth, Huchinson Community College

▶ Titles have been added to examples to

highlight relevant learning objectives and reinforce the importance of speaking mathematically using vocabulary.

EXAMPLE 8

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Analytical Solution

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Solve the inequality x2 6x 9.

▶ Annotations located to the right of the

solution sequence help the student recognize which property or procedure is being applied. ▶ “Now Try” boxes immediately following

examples guide students to specific matched exercises at the end of the section, helping them identify exactly which homework problems coincide with each discussed concept.

Solving a Quadratic Inequality

WORTHY OF NOTE

Begin by writing the inequality in standard form: x2 6x 9 0. Note this is equivalent to g1x2 0 for g1x2 x2 6x 9. Since a 6 0, the graph of g will open downward. The factored form is g1x2 1x 32 2, showing 3 is a zero and a repeated root. Using the x-axis, we plot the point (3, 0) and visualize a parabola opening downward through this point. Figure 3.29 shows the graph is below the x-axis (outputs are negative) for all values of x except x 3. But since this is a less than or equal to inequality, the solution is x 僆⺢.

Since x 3 was a zero of multiplicity 2, the graph “bounced off” the x-axis at this point, with no change of sign for g. The graph is entirely below the x-axis, except at the vertex (3, 0).

Graphical Solution

Figure 3.29 1

0

1

2

3

4

5

6

7

x

a0 䊳

The complete graph of g shown in Figure 3.30 confirms the analytical solution (using the zeroes method). For the intervals of the domain shown in red: 1q, 32 ´ 13, q2 , the graph of g is below the x-axis 3g1x2 6 04 . The point (3, 0) is on the x-axis 3 g132 04 . As with the analytical solution, the solution to this “less than or equal to” inequality is all real numbers. A calculator check of the original inequality is shown in Figure 3.31. Figure 3.30

Figure 3.31

y

10

2

2

6

x 2

8

g(x)

“The modeling and regression

examples in this text are excellent, and the instructions for using the graphing calculator to investigate these types of problems are great.

3 8

Now try Exercises 121 thro through 132

”

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—Allison Sutton, Austin Community College

“ The examples support the exercises which is very important. The chapter is very well written and is easy to read and understand.

”

—Joseph Lloyd Harris, Gulf Coast Community College

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Solid Skill Development . . . Through g Exercises s. The exe rcise sup por t of eac h sec tion’s ma in idea in es rcis exe of lth wea a d ude We hav e incl t for wea ker stu den ts, e, in an eff ort to pro vide sup por car at gre h wit ed uct str con e of wer set s n fur the r. The qua ntit y and qua lity eve ch rea to ts den stu ed anc adv whi le cha llenging mo re ies to guide eff ort s, and num ero us opp ort unit r’s che tea a for t por sup ong str exe rcis es off ers ing idea s. to illus tra te imp ortant pro blem solv and ions ulat calc t icul diff h oug stu den ts thr —The Aut hor s

Mid-Chapter Checks

MID-CHAPTER CHECK

Mid-Chapter Checks provide students with a good stopping place to assess their knowledge before moving on to the second half of the chapter.

4. Write the equation of the function that has the same graph of f 1x2 2x, shifted left 4 units and up 2 units.

1. Determine whether the following function is even, 冟x冟 odd, or neither. f 1x2 x2 4x

5. For the graph given, (a) identify the function family, (b) describe or identify the end-behavior, inflection point, and x- and y-intercepts, (c) determine the domain and range and

2. Use a graphing calculator to find the maximum and minimum values of f 1x2 1.91x4 2.3x3 2.2x 5.12 . Round to the nearest hundredth. 3 Use interval notation to identify the interval(s)

End-of-Section Exercise Sets

Exercise 5 y 5

f(x)

5

5 x

2.2 EXERCISES

▶ Concepts and Vocabulary exercises to help students

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CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

recall and retain important terms.

1. After a vertical , points on the graph are farther from the x-axis. After a vertical , points on the graph are closer to the x-axis.

2. Transformations that change only the location of a graph and not its shape or form, include and .

4. The inflection point of f 1x2 21x 42 3 11 is at and the end-behavior is , .

3. The vertex of h1x2 31x 52 9 is at and the graph opens . 2

5 Gi

▶ Developing Your Skills exercises to provide

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practice of relevant concepts just learned with increasing levels of difficulty.

th

h f

lf

ti

f ( ) di

/

/E l i

h th

7. f 1x2 x2 4x 5

2

3i

15. r 1x2 3 14 x 3 16. f 1x2 2 1x 1 4 5

y

5

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r(x)

8. g1x2 x2 2x

y

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5 x

5

5

5 x

5

f(x)

5 x

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y

y

5

5 x

17. g1x2 2 14 x

formulas and applications bring forward some interesting ideas and problems that are more in depth. These would help hold the students’ interest in the topic.

hift f f 1 2

DEVELOPING YOUR SKILLS

By carefully inspecting each graph given, (a) identify the function family; (b) describe or identify the end-behavior, vertex, intervals where the function is increasing or decreasing, maximum or minimum value(s) and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

“ The sections in the assignments headed working with

6 Di

18. h1x2 2 1x 1 4

y

5

y

g(x) h(x)

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5

5

—Sherri Rankin, Huchinson Community College

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WORKING WITH FORMULAS

61. Discriminant of the reduced cubic x3 ⴙ px ⴙ q ⴝ 0: D ⴝ ⴚ14p3 ⴙ 27q2 2 The discriminant of a cubic equation is less well known than that of the quadratic, but serves the same purpose. The discriminant of the reduced cubic is given by the formula shown, where p is the linear coefficient and q is the constant term. If D 7 0, there will be three real and distinct roots. If D 0, there are still three real roots, but one is a repeated root (multiplicity two). If D 6 0, there are one real and two complex roots. Suppose we wish to study the family of cubic equations where q p 1. a. Verify the resulting discriminant is D 14p3 27p2 54p 272. b. Determine the values of p and q for which this family of equations has a repeated real root. In other words, solve the equation 14p3 27p2 54p 272 0 using the rational zeroes theorem and synthetic division to write D in completely factored form.

▶ Working with Formulas exercises to demonstrate

contextual applications of well-known formulas. ▶ Extending the Concept exercises that require

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communication of topics, synthesis of related concepts, and the use of higher-order thinking skills.

EXTENDING THE CONCEPT

59. Use the general solutions from the quadratic formula to show that the average value of the x-intercepts is b . Explain/Discuss why the result is valid even if 2a the roots are complex. b 2b2 4ac

▶ Maintaining Your Skills exercises that address

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skills from previous sections to help students retain previously learning knowledge.

b 2b2 4ac

62. Referring to Exercise 39, discuss the nature (real or complex, rational or irrational) and number of zeroes (0, 1, or 2) given by the vertex/intercept formula if (a) a and k have like signs, (b) a and k k have unlike signs, (c) k is zero, (d) the ratio a is positive and a perfect square and (e) the

MAINTAINING YOUR SKILLS

37. (1.3) Is the graph shown here, the graph of a function? Discuss why or why not.

38. (R.2/R.3) Determine the area of the figure shown 1A LW, A ␲r2 2.

18 cm 24 cm

39. (1.5) Solve for r: A P Prt

“The exercise sets are plentiful. I like having many to

choose from when assigning homework. When there are only one or two exercises of a particular type, it’s hard for the students to get the practice they need.

”

—Sarah Jackson, Pratt Community College

xviii

40

S l

f

(if if

ibl )

“ There seems to be a good selection of easy, moderate, and difficult problems in the exercises.”

—Ed Gallo, Sinclair Community College

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End-of-Chapter Review Material Exercises located at the end of the chapter provide students with the tools they need to prepare for a quiz or test. Each chapter features the following: ▶

Making Connections matching exercises are groups of problems where students must identify graphs based on an equation or description. This feature helps students make the connection between graphical and algebraic information while it enhances students’ ability to read and interpret graphical data.

“ Not only was the algebra rigorously treated, but it

was reinforced throughout the chapters with the MidChapter Check and the Chapter Review and Tests.

”

—Mark Crawford, Waubonsee Community College

MAKING CONNECTIONS Making M ki Connections: C ti G Graphically, hi ll Symbolically, S b li ll Numerically, N i ll and d Verbally V b ll Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

5 x

5

▶

Chapter Summary and Concept Reviews that present key concepts with corresponding exercises by section in a format easily y used by y students.

5 x

“

”

▶

Practice Tests that give students the opportunity to check their knowledge and prepare for classroom quizzes, tests, and other assessments.

▶

Cumulative Reviews that are presented at the end of each chapter help students retain previously learned skills and concepts by revisiting important ideas from earlier chapters (starting with Chapter 2).

▶

Graphing Calculator icons appear next to exercises where important concepts can be supported by the use of graphing technology.

5 x

5

1 1. ____ y x 1 3 2. ____ y x 1

y

(h)

5

5 x

5

5

5

5 x

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y

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(f)

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(c)

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The problem sets are really magnificent. I deeply enjoy and appreciate the many problems that incorporate telescopes, astronomy, reflector design, nuclear cooling tower profiles, charged particle trajectories, and other such examples from science, technology, and engineering. —Light Bryant, Arizona Western College

y

(b)

5

5

5 x

5

5

9. ____ f 132 4, f 112 0

10. ____ f 142 3, f 142 3

SUMMARY AND CONCEPT REVIEW SECTION 1.1 SE

Rectangular Coordinates; Graphing Circles and Other Relations

KEY CONCEPTS KE • A relation is a collection of ordered pairs (x, y) and can be stated as a set or in equation form. • As a set of ordered pairs, we say the relation is pointwise-defined. The domain of the relation is the set of all first coordinates, and the range is the set of all corresponding second coordinates. • A relation can be expressed in mapping notation x S y, indicating an element from the domain is mapped to (corresponds to or is associated with) an element from the range. • The graph of a relation in equation form is the set of all ordered pairs (x, y) that satisfy the equation. We plot a sufficient number of points and connect them with a straight line or smooth curve, depending on the pattern formed. • The x- and y-variables of linear equations and their graphs have implied exponents of 1. • With a relation entered on the Y= screen, a graphing calculator can provide a table of ordered pairs and the related graph. x1 x2 y1 y2 , b. • The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is a 2 2

• The distance between the points (x1, y1) and (x2, y2) is d 21x2 x1 2 2 1y2 y1 2 2. • The equation of a circle centered at (h, k) with radius r is 1x h2 2 1y k2 2 r2. EXERCISES 1. Represent the relation in mapping notation, then state the domain and range. 517, 32, 14, 22, 15, 12, 17, 02, 13, 22, 10, 826 2

Homework Selection Guide A list of suggested homework exercises has been provided for each section of the text (Annotated Instructor’s Edition only). This feature may prove especially useful for departments that encourage consistency among many sections, or those having a large adjunct population. The feature was also designed as a convenience to instructors, enabling them to develop an inventory of exercises that is more in tune with the course as they like to teach it. The guide provides prescreened and preselected p assignments at four different levels: Core, Standard, Extended, and In Depth. 8 10 • Core: These assignments go right to the heart of the material, HOMEWORK SELECTION GUIDE offering a minimal selection of exercises that cover the primary concepts and solution strategies of the section, along with a small selection of the best applications. • Standard: The assignments at this level include the Core exercises, while providing for additional practice without excessive drill. A wider assortment of the possible variations on a theme are included, as well as a greater variety of applications. • Extended: Assignments from the Extended category expand on the Standard exercises to include more applications, as well as some conceptual or theory-based questions. Exercises may include selected items from the Concepts and Vocabulary, Working with Formulas, and the Extending the Concept categories of the exercise sets. • In Depth: The In Depth assignments represent a more comprehensive look at the material from each section, while attempting to keep the assignment manageable for students. These include a selection of the most popular and highest-quality exercises from each category of the exercise set, with an additional emphasis on Maintaining Your Skills. Additional answers can be found in the Instructor Answer Appendix.

Core: 7–91 every other odd, 95–101 odd (26 Exercises) Standard: 1–4, 7–83 every other odd, 85–92 all, 95–101 odd (36 Exercises)

13 3

19 2

Extended: 1–4, 7–31 every other odd, 35–38 all, 39–79 every other odd, 85–92 all, 95–101 odd, 106, 109 (39 Exercises) In Depth: 1–4, 7–31 every other odd, 35–38 all, 39–83 every other odd, 85–92 all, 95, 96, 98, 99, 100, 101, 105, 106, 109 (44 Exercises)

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Strong Mathmatical Connections . . . Through a Conversational Writing Style featur es of a mathematics text, While examples and applications are arguably the most promin ent togeth er. It may be true that it’s the readability and writing style of the author s that bind them when looking for an example some studen ts don’t read the text, and that others open the text only for those studen ts that do (read similar to the exercise they’re working on. But when they do and g concep ts in a form and the text), it’s important they have a text that “speak s to them,” relatin style of this text will help draw at a level they understand and can relate to. We feel the writing and bringing them back a second studen ts in and keep their interest, becoming a positiv e experience begin to see the true value and third time, until it becomes habitual. At this point studen ts might g with any other form of of their text, as it becomes a resour ce for learning on equal footin —The Authors direction. supplementa l instruction. This text represents our best effort s in this

Conversational Writing Style John and J.D.’s experience in the classroom and their strong connections to how students comprehend the material are evident in their writing style. They use a conversational and supportive writing style, providing the students with a tool they can depend on when the teacher is not available, when they miss a day of class, or simply when working on their own. The effort they have put into the writing is representative of John Coburn’s unofficial mantra: “If you want more students to reach the top, you gotta put a few more rungs on the ladder.”

“Coburn strikes a good balance between

providing all of the important information necessary for a certain topic without going too deep.

”

—Barry Monk, Macon State College

“I think the authors have done an excellent job

of interweaving the formal explanations with the ‘plain talk’ descriptions, illustrating with meaningful examples and applications.

”

—Ken Gamber, Hutchinson Community College

Through Student Involvement How do you design a student-friendly textbook? We decided to get students involved by hosting two separate focus groups. During these sessions we asked students to advise us on how they use their books, what pedagogical elements are useful, which elements are distracting and not useful, as well as general feedback on page layout. During this process there were times when we thought, “Now why hasn’t anyone ever thought of that before?” Clearly these student focus groups were invaluable. Taking direct student feedback and incorporating what is feasible and doesn’t detract from instructor use of the text is the best way to design a truly student-friendly text. The next two pages will highlight what we learned from students so you can see for yourself how their feedback played an important role in the development of the Coburn/Herdlick series.

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5.2

Students said that Learning Objectives should clearly define the goals of each section.

Exponential Functions

LEARNING OBJECTIVES In Section 5.2 you will see how we can:

A. Evaluate an exponential

Demographics is the statistical study of human populations. In this section, we introduce the family of exponential functions, which are widely used to model population growth or decline with additional applications in science, engineering, and many other fields. As with other functions, we begin with a study of the graph and its characteristics.

function

A. Evaluating Exponential Functions

B. Graph general exponential functions

C. Graph base-e exponential functions D. Solve exponential equations and applications

In the boomtowns of the old west, it was not uncommon for a town to double in size every year (at least for a time) as the lure of gold drew more and more people westward. When this type of growth is modeled using mathematics, exponents play a lead role. Suppose the town of Goldsboro h d 1000 id t h ld fi t di d

Examples are “boxed” so students can clearly see where they begin and end. Examples are called out in the margins so they are easy for students to spot.

EXAMPLE 4

䊳

Graphing Exponential xponential Functions Using Transformations i transformations f i Graph F1x2 2xx11 2 using off the basic function f 1x2 2x (not by simply plotting points). Clearly state what transformations are applied.

Solution

䊳

Students asked for Check Points throughout each section to alert them when a specific learning objective has been covered and to reinforce the use of correct mathematical terms.

(1, 3) y2

4

4

x

To help sketch a more accurate graph, the point (3, 6) can be used: F132 6. Now try Exercises 15 through 30

䊳

Students told us they liked when the examples were linked to the exercises.

Described by students as one of the most useful features in a math text, Caution Boxes signal a student to stop and take note in order to avoid mistakes in problem solving.

CAUTION

䊳

S d Students told ld us that h the h color l red d should h ld only be used for things that are really important. Also, anything significant should be included in the body of the text; marginal readings imply optional.

Because students spend a lot of time in the exercise section of a text, they said that a white background is hard on their eyes . . . so we used a soft, off-white color for the background.

(0, 2.5)

(3, 6)

F102 21021 2 21 2 1 2 2 2.5 B. You’ve just seen how we can graph general exponential functions

Students said having a lot of icons was confusing. The graphing calculator is the only icon used in the exercise sets; no unnecessary icons are used.

The graph of F is that of the basic function f 1x2 2x with a horizontal shift 1 unit right and a vertical shift 2 units up. With this in mind the horizontal asymptote also shifts from y 0 to y 2 and (0, 1) shifts to (1, 3). The y-intercept of F is at (0, 2.5):

y F(x) = 2x is shifted 1 unit right 2 units up

For equations like those in Example 1, be careful not to treat the absolute value bars as simple grouping symbols. The equation 51x 72 2 13 has only the solution x 10, and “misses” the second solution since it yields x 7 3 in simplified form. The equation 5冟x 7冟 2 13 simplifies to 冟x 7冟 3 and there are actually two solutions. Also note that 5冟x 7冟冟5x 35冟!

Students told us that directions should be in bold so they are easily distinguishable from the problems.

䊳

APPLICATIONS

Use the information given to build a linear equation model, then use the equation to respond. For exercises 71 to 74, develop both an algebraic and a graphical solution.

71. Business depreciation: A business purchases a copier for $8500 and anticipates it will depreciate in value $1250 per year. a. What is the copier’s value after 4 yr of use? b. How many years will it take for this copier’s value to decrease to $2250? 72. Baseball card value: After purchasing an autographed baseball card for $85, its value increases by $1.50 per year. a. What is the card’s value 7 yr after purchase? b. How many years will it take for this card’s value to reach $100?

74. Gas mileage: When empty, a large dump-truck gets about 15 mi per gallon. It is estimated that for each 3 tons of cargo it hauls, gas mileage decreases by 34 mi per gallon. a. If 10 tons of cargo is being carried, what is the truck’s mileage? b. If the truck’s mileage is down to 10 mi per gallon, how much weight is it carrying? 75. Parallel/nonparallel roads: Aberville is 38 mi north and 12 mi west of Boschertown, with a straight “farm and machinery” road (FM 1960) connecting the two cities. In the next county, Crownsburg is 30 mi north and 9.5 mi west of Dower, and these cities are likewise connected by a straight road (FM 830). If the two roads continued indefinitely in both directions, would they intersect at some point?

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Connections to Calculus. . . What’s This Feature All About?

metersfallen

Calculus is often described as the study of change, Figure 1 Figure 2 motion, and accumulation. While there are two main d 0 divisions (differential calculus and integral calculus), both depend on a single, fundamental idea—the use Average velocity 10 120 of successive approximations of increasing accuracy = 78.4 m − 4.9 m to find an exact result. 20 4 sec − 1 sec 100 Differential calculus allows us to quantify the = 24.5 m rate of change in a quantity at a specific instant, using 30 80 sec a tool called the derivative. For instance, if an object 40 rolls off of an elevated surface (Figure 1), the equation 60 2 d ⫽ 4.9t gives the distance the object has fallen (in 50 meters) after t seconds. Due to gravity, the velocity of 40 the object increases the further it falls. Since d = 4.9t 2 60 20 velocity is computed as the distance traveled divided D by the time in motion aor R ⫽ b , we can easily 70 T t 1 2 3 4 5 6 find the average velocity of the object for any time 80 seconds interval (Figure 2). The derivative concept uses a series of time intervals that become “infinitely small,” and ultimately result in a precise formula (called the limit) for the instantaneous velocity of the object at any time t. On the other hand, integral calculus develops a tool called Figure 3 Figure 4 the integral, which can find a total accumulation over an infinite number of very small intervals. For instance, we can approximate the volume of the solid shown in Figure 3 using a series of thin cylinders or disks (V ⫽ ␲r 2 h) that become more numerous and ever thinner, while continuing to fit snugly within the frame of the original solid (Figure 4). As the height of each disk becomes infinitely small, and the number of disks becomes infinitely large, we find these approximations tend toward a precise formula (called the limit) for the exact volume of the solid. It’s important to note that in each case, the tools supporting both concepts are algebraic in nature, and involve (1) algebraically rewriting an expression in a form that allows the tools of calculus to be applied, or (2) simplifying the result of such an application. This means your success will likely be measured in direct proportion to your mastery of the algebraic concepts we study in this course, many of which will form the “connections” we emphasize in this feature. So more specifically—this feature is designed to point out connections between the algebraic concepts you’re currently studying, and the calculus concepts you’ll soon encounter. It’s not intended as an exhaustive coverage of all connections to be found in each chapter, but simply a hint or gleaning of how the concepts you’re studying now, are connected to the calculus concepts to come. The Authors

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Coburn’s Precalculus Series College Algebra: Graphs & Models, First Edition A Review of Basic Concepts and Skills ◆ Functions and Graphs ◆ Relations; More on Functions ◆ Quadratic Functions and Operations on Functions ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Systems of Equations and Inequalities ◆ Matrices and Matrix Applications ◆ Analytic Geometry and the Conic Sections ◆ Additional Topics in Algebra

Precalculus: Graphs & Models, First Edition Functions and Graphs ◆ Relations; More on Functions ◆ Quadratic Functions and Operations on Functions ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Introduction to Trigonometry ◆ trigonometric Identities, Inverses, and Equations ◆ Applications of Trigonometry ◆ Systems of Equations and Inequalities; Matrices ◆ Analytic Geometry; Polar and parametric Equations ◆ Sequences, Series, Counting, and Probability ◆ Bridges to Calculus—An Introduction to Limits

College Algebra Second Edition Review ◆ Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Systems of Equations and Inequalities ◆ Matrices ◆ Geometry and Conic Sections ◆ Additional Topics in Algebra MHID 0-07-351941-3, ISBN 978-0-07-351941-8

College Algebra Essentials Second Edition Review ◆ Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Systems of Equations and Inequalities MHID 0-07-351968-5, ISBN 978-0-07-351968-5

Algebra and Trigonometry Second Edition Review ◆ Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Trigonometric Functions ◆ Trigonometric Identities, Inverses, and Equations ◆ Applications of Trigonometry ◆ Systems of Equations and Inequalities ◆ Matrices ◆ Geometry and Conic Sections ◆ Additional Topics in Algebra MHID 0-07-351952-9, ISBN 978-0-07-351952-4

Precalculus Second Edition Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Trigonometric Functions ◆ Trigonometric Identities, Inverses, and Equations ◆ Applications of Trigonometry ◆ Systems of Equations and Inequalities ◆ Matrices ◆ Geometry and Conic Sections ◆ Additional Topics in Algebra ◆ Limits MHID 0-07-351942-1, ISBN 978-0-07-351942-5

Trigonometry Second Edition Introduction to Trigonometry ◆ Right Triangles and Static Trigonometry ◆ Radian Measure and Dynamic Trigonometry ◆ Trigonometric Graphs and Models ◆ Trigonometric Identities ◆ Inverse Functions and Trigonometric Equations ◆ Applications of Trigonometry ◆ Trigonometric Connections to Algebra MHID 0-07-351948-0, ISBN 978-0-07-351948-7

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Making Connections . . . Through 360º Development McGraw-Hill’s 360° Development Process is an ongoing, never-ending, market-oriented approach to building accurate and innovative print and digital products. It is dedicated to continual large-scale and incremental improvement driven by multiple customer feedback loops and checkpoints. This process is initiated during the early planning stages of our new products, intensifies during the development and production stages, and then begins again on publication, in anticipation of the next edition. A key principle in the development of any mathematics text is its ability to adapt to teaching specifications in a

universal way. The only way to do so is by contacting those universal voices—and learning from their suggestions. We are confident that our book has the most current content the industry has to offer, thus pushing our desire for accuracy to the highest standard possible. In order to accomplish this, we have moved through an arduous road to production. Extensive and open-minded advice is critical in the production of a superior text. By investing in this extensive endeavor, McGraw-Hill delivers to you a product suite that has been created, refined, tested, and validated to be a successful tool in your course.

Student Focus Groups Two student focus groups were held at Illinois State University and Southeastern Louisiana University to engage students in the development process and provide feedback as to how the design of a textbook impacts homework and study habits in the College Algebra, Precalculus, and Trigonometry course areas. Francisco Arceo, Illinois State University Candace Banos, Southeastern Louisiana University Dave Cepko, Illinois State University Andrea Connell, Illinois State University Nicholas Curtis, Southeastern Louisiana University M. D. “Boots” Feltenberger, Southeastern Louisiana University Regina Foreman, Southeastern Louisiana University Ashley Lae, Southeastern Louisiana University Brian Lau, Illinois State University Daniel Nathan Mielneczek, Illinois State University Mingaile Orakauskaite, Illinois State University Todd Michael Rapnikas, Illinois State University Bethany Rollet, Illinois State University Teddy Schrishuhn, Illinois State University

Josh Schultz, Illinois State University Jessica Smith, Southeastern Louisiana University Andy Thurman, Illinois State University Ashley Youngblood, Southeastern Louisiana University

Digital Contributors Jeremy Coffelt, Blinn College Vanessa Coffelt, Blinn College Vickie Flanders, Baton Rouge Community College Anne Marie Mosher, Saint Louis Community CollegeFlorissant Valley

Special Thanks Sherry Meier, Illinois State University Rebecca Muller, Southeastern Louisiana University Anne Schmidt, Illinois State University

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Kristen Stoley, Blinn College David Ray, University of Tennessee-Martin Stephen Toner, Victor Valley Community College Paul Vroman, Saint Louis Community College-Florissant Valley

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Making Connections . . . Developmental Editing The manuscript has been impacted by numerous developmental reviewers who edited for clarity and consistency. Efforts resulted in cutting length from the manuscript, while retaining a conversational and casual narrative style. Editorial work also ensured the positive visual impact of art and photo placement. Chapter Reviews and Manuscript Reviews Teachers and academics from across the country reviewed the current edition text, the proposed table of contents, and first-draft manuscript to give feedback on reworked narrative, design changes, pedagogical enhancements, and organizational changes. This feedback was summarized by the book team and used to guide the direction of the second-draft manuscript. Betty Anderson, Howard Community College David Bosworth, Hutchinson Community College Daniel Brock, Arkansas State University-Beebe Barry Brunson, Western Kentucky University Light Bryant, Arizona Western College Brenda Burns-Williams, North Carolina State University-Raleigh Charles Cooper, Hutchinson Community College Mark Crawford, Waubonsee Community College Joseph Demaio, Kennesaw State University Alvio Dominguez, Miami-Dade College-Wolfson Dale Duke, Oklahoma City Community College Frank Edwards, Southeastern Louisiana University Caleb Emmons, Pacific University Mike Everett, Santa Ana College Maggie Flint, Northeast State Technical Community College Ed Gallo, Sinclair Community College Ken Gamber, Hutchinson Community College David Gurney, Southeastern Louisiana University

Sally Haas, Angelina College Ben Hill, Lane Community College Jody Hinson, Cape Fear Community College Lynda Hollingsworth, Northwest Missouri State University George Hurlburt, Corning Community College Sarah Jackson, Pratt Community College Laud Kwaku, Owens Community College Kathryn Lavelle, Westchester Community College Joseph Lloyd Harris, Gulf Coast Community College Austin Lovenstein, Pulaski Technical College Rodolfo Maglio, Northeastern Illinois University Barry Monk, Macon State College Camille Moreno, Cosumnes River College Anne Marie Mosher, Saint Louis Community College-Florissant Valley Lilia Orlova, Nassau Community College Susan Pfeifer, Butler Community College Sherri Rankin, Hutchinson Community College Daniel Russow, Arizona Western College-Yuma Rose Shirey, College of the Mainland Joy Shurley, Abraham Baldwin Agricultural College Sean Simpson, Westchester Community College Pam Stogsdill, Bossier Parish Community College Allison Sutton, Austin Community College Linda Tremer, Three Rivers Community Collge Dahlia Vu, Santa Ana College Jackie Wing, Angelina College

Acknowledgments We first want to express a deep appreciation for the guidance, comments, and suggestions offered by all reviewers of the manuscript. We have once again found their collegial exchange of ideas and experience very refreshing and instructive, and always helping to create a better learning tool for our students. Vicki Krug has continued to display an uncanny ability to bring innumerable pieces from all directions into a unified whole, in addition to providing spiritual support during some extremely trying times; Patricia Steele’s skill as a copy editor is as sharp as ever, and her attention to detail continues to pay great dividends; which helps pay the debt we owe Katie White, Michelle Flomenhoft, Christina Lane, and Eve Lipton for their useful suggestions, infinite patience, tireless efforts, and art-counting eyes, which helped in bringing the manuscript to completion. We must also thank John Osgood for his ready wit, creative energies, and ability to step into the flow without missing a beat; Laurie Janssen and our magnificent

design team, and Dawn Bercier whose influence on this project remains strong although she has moved on, as it was her indefatigable spirit that kept the ship on course through trial and tempest, and her ski-jumper’s vision that brought J.D. on board. In truth, our hats are off to all the fine people at McGraw-Hill for their continuing support and belief in this series. A final word of thanks must go to Rick Armstrong, whose depth of knowledge, experience, and mathematical connections seems endless; Anne Marie Mosher for her contributions to various features of the text, Mitch Levy for his consultation on the exercise sets, Stephen Toner for his work on the videos, Jon Booze and his team for their work on the test bank, Cindy Trimble for her invaluable ability to catch what everyone else misses; and to Rick Pescarino, Kelly Ballard, John Elliot, Jim Frost, Barb Kurt, Lillian Seese, Nate Wilson, and all of our colleagues at St. Louis Community College, whose friendship, encouragement, and love of mathematics makes going to work each day a joy.

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Making Connections . . . Through Supplements *All online supplements are available through the book’s website: www.mhhe.com/coburn.

Instructor Supplements • Computerized Test Bank Online: Utilizing Brownstone Diploma® algorithm-based testing software enables users to create customized exams quickly. • Instructor’s Solutions Manual: Provides comprehensive, worked-out solutions to all exercises in the text. • Annotated Instructor’s Edition: Contains all answers to exercises in the text, which are printed in a second color, adjacent to corresponding exercises, for ease of use by the instructor.

Student Supplements • Student Solutions Manual provides comprehensive, worked-out solutions to all of the odd-numbered exercises. • Graphing Calculator Manual includes detailed instructions for using calculators to solve problems throughout the text. Written by the authors to accompany their text, it is designed to match and supplement the text. • Videos • Interactive video lectures are provided for each section in the text, which explain to the students how to do key problem types, as well as highlighting common mistakes to avoid. • Exercise videos provide step-by-step instruction for the key exercises which students will most wish to see worked out. • Graphing calculator videos help students master the most essential calculator skills used in the college algebra course. • The videos are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design.

Connect Math™ Hosted by ALEKS® www.connectmath.com Connect Math Hosted by ALEKS is an exciting, new assessment and assignment platform combining the strengths of McGraw-Hill Higher Education and ALEKS Corporation. Connect Math Hosted by ALEKS is the first platform on the market to combine an artificial-intelligent, diagnostic

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assessment with an intuitive ehomework platform designed to meet your needs. Connect Math Hosted by ALEKS is the culmination of a one-of-a-kind market development process involving math full-time faculty members and adjuncts at every step of the process. This process enables us to provide you with an end product that better meets your needs. Connect Math Hosted by ALEKS is built by mathematicians educators for mathematicians educators!

www.aleks.com ALEKS (Assessment and LEarning in Knowledge Spaces) is a dynamic online learning system for mathematics education, available over the Web 24/7. ALEKS assesses students, accurately determines their knowledge, and then guides them to the material that they are most ready to learn. With a variety of reports, Textbook Integration Plus, quizzes, and homework assignment capabilities, ALEKS offers flexibility and ease of use for instructors. • ALEKS uses artificial intelligence to determine exactly what each student knows and is ready to learn. ALEKS remediates student gaps and provides highly efficient learning and improved learning outcomes • ALEKS is a comprehensive curriculum that aligns with syllabi or specified textbooks. Used in conjunction with McGraw-Hill texts, students also receive links to textspecific videos, multimedia tutorials, and textbook pages. • ALEKS offers a dynamic classroom management system that enables instructors to monitor and direct student progress toward mastery of course objectives.

ALEKS Prep/Remediation: • Helps instructors meet the challenge of remediating underprepared or improperly placed students. • Assesses students on their prerequisite knowledge needed for the course they are entering (i.e., Calculus students are tested on Precalculus knowledge) and prescribes a unique and efficient learning path specifically to address their strengths and weaknesses. • Students can address prerequisite knowledge gaps outside of class freeing the instructor to use class time pursuing course outcomes.

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Making Connections . . . McGraw-Hill Higher Education and Blackboard® have teamed up. Blackboard, the Web-based course-management system, has partnered with McGraw-Hill to better allow students and faculty to use online materials and activities to complement face-to-face teaching. Blackboard features exciting social learning and teaching tools that foster more logical, visually impactful and active learning opportunities for students. You’ll transform your closed-door classrooms into communities where students remain connected to their educational experience 24 hours a day. This partnership allows you and your students access to McGraw-Hill’s Connect™ and Create™ right from within your Blackboard course—all with one single sign-on. Not only do you get single sign-on with Connect and Create, you also get deep integration of McGraw-Hill content and content engines right in Blackboard. Whether you’re choosing a book for your course or building Connect assignments, all the tools you need are right where you want them—inside of Blackboard. Gradebooks are now seamless. When a student completes an integrated Connect assignment, the grade for that assignment automatically (and instantly) feeds your Blackboard grade center. McGraw-Hill and Blackboard can now offer you easy access to industry leading technology and content, whether your campus hosts it, or we do. Be sure to ask your local McGraw-Hill representative for details.

TEGRITY—tegritycampus.mhhe.com McGraw-Hill Tegrity Campus™ is a service that makes class time available all the time by automatically capturing every lecture in a searchable format for students to review when they study and complete assignments. With a simple one-click start and stop process, you capture all computer screens and corresponding audio. Students replay any part of any class with easy-to-use browser-based viewing on a PC or Mac. Educators know that the more students can see, hear, and experience class resources, the better they learn. With Tegrity, students quickly recall key moments by using Tegrity’s unique search feature. This search helps students efficiently find what they need, when they need it across an entire semester of class recordings. Help turn all your students’ study time into learning moments immediately supported by your lecture. To learn more about Tegrity watch a 2-minute Flash demo at tegritycampus.mhhe.com.

Electronic Books: If you or your students are ready for an alternative version of the traditional textbook, McGraw-Hill eBooks offer a cheaper and eco-friendly alternative to traditional textbooks. By purchasing eBooks from McGraw-Hill, students can save as much as 50% on selected titles delivered on the most advanced eBook platform available. Contact your McGraw-Hill sales representative to discuss eBook packaging options.

Create: Craft your teaching resources to match the way you teach! With McGraw-Hill Create, www.mcgrawhillcreate.com, you can easily rearrange chapters, combine material from other content sources, and quickly upload content you have written like your course syllabus or teaching notes. Find the content you need in Create by searching through thousands of leading McGraw-Hill textbooks. Arrange your book to fit your teaching style. Create even allows you to personalize your book’s appearance by selecting the cover and adding your name, school, and course information. Order a Create book and you’ll receive a complimentary print review copy in 3–5 business days or a complimentary electronic review copy (eComp) via email in minutes. Go to www.mcgrawhillcreate.com today and register to experience how McGraw-Hill Create empowers you to teach your students your way. xxvii

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Contents Preface vi Index of Applications

CHAPTER

1

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Relations, Functions, and Graphs

1

1.1 Rectangular Coordinates; Graphing Circles and Other Relations 2 1.2 Linear Equations and Rates of Change 19 1.3 Functions, Function Notation, and the Graph of a Function 33 Mid-Chapter Check 48 Reinforcing Basic Concepts: Finding the Domain and Range of a Relation from Its Graph 48

1.4 Linear Functions, Special Forms, and More on Rates of Change 50 1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving 64 1.6 Linear Function Models and Real Data 79 Making Connections 93 Summary and Concept Review 94 Practice Test 99 Strengthening Core Skills: The Various Forms of a Linear Equation 100 Calculator Exploration and Discovery: Evaluating Expressions and Looking for Patterns 101 Connections to Calculus: Tangent Lines 103

CHAPTER

2

More on Functions 105 2.1 Analyzing the Graph of a Function 106 2.2 The Toolbox Functions and Transformations 120 2.3 Absolute Value Functions, Equations, and Inequalities 136 Mid-Chapter Check 146 Reinforcing Basic Concepts: Using Distance to Understand Absolute Value Equations and Inequalities 147

2.4 Basic Rational Functions and Power Functions; More on the Domain 148 2.5 Piecewise-Defined Functions 163 2.6 Variation: The Toolbox Functions in Action 177 Making Connections 188 Summary and Concept Review 189 Practice Test 193 Calculator Exploration and Discovery: Studying Joint Variations 195 Strengthening Core Skills: Variation and Power Functions: y ⴝ kxp 196 Cumulative Review: Chapters 1–2 197 Connections to Calculus: Solving Various Types of Equations; Absolute Value Inequalities and Delta/Epsilon Form 199

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CHAPTER

3

Quadratic Functions and Operations on Functions

203

3.1 Complex Numbers 204 3.2 Solving Quadratic Equations and Inequalities 214 3.3 Quadratic Functions and Applications 235 Mid-Chapter Check 249 Reinforcing Basic Concepts: An Alternative Method for Checking Solutions to Quadratic Equations 249

3.4 Quadratic Models; More on Rates of Change 250 3.5 The Algebra of Functions 262 3.6 The Composition of Functions and the Difference Quotient 274 Making Connections 292 Summary and Concept Review 292 Practice Test 297 Calculator Exploration and Discovery: Residuals, Correlation Coefficients, and Goodness of Fit 298 Strengthening Core Skills: Base Functions and Quadratic Graphs 300 Cumulative Review: Chapters 1–3 301 Connections to Calculus: Rates of Change and the Difference Quotient;

Transformations and the Area Under a Curve

CHAPTER

4

303

Polynomial and Rational Functions 307 4.1 Synthetic Division; the Remainder and Factor Theorems 308 4.2 The Zeroes of Polynomial Functions 320 4.3 Graphing Polynomial Functions 337 Mid-Chapter Check 354 Reinforcing Basic Concepts: Approximating Real Zeroes

355

4.4 Graphing Rational Functions 356 4.5 Additional Insights into Rational Functions 371 4.6 Polynomial and Rational Inequalities 385 Making Connections 396 Summary and Concept Review 396 Practice Test 400 Calculator Exploration and Discovery: Complex Zeroes, Repeated Zeroes, and Inequalities 401 Strengthening Core Skills: Solving Inequalities Using the Push Principle 402 Cumulative Review: Chapters 1–4 403 Connections to Calculus: Graphing Techniques 405

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CHAPTER

5

Exponential and Logarithmic Functions 409 5.1 5.2 5.3 5.4

One-to-One and Inverse Functions 410 Exponential Functions 422 Logarithms and Logarithmic Functions 433 Properties of Logarithms 446 Mid-Chapter Check 456 Reinforcing Basic Concepts: Understanding Properties of Logarithms 457 5.5 Solving Exponential and Logarithmic Equations 457 5.6 Applications from Business, Finance, and Science 469 5.7 Exponential, Logarithmic, and Logistic Equation Models 482 Making Connections 495 Summary and Concept Review 496 Practice Test 501 Calculator Exploration and Discovery: Investigating Logistic Equations 502 Strengthening Core Skills: The HerdBurn Scale — What’s Hot and What’s Not 503 Cumulative Review: Chapters 1–5 504 Connections to Calculus: Properties of Logarithms; Area Functions; Expressions Involving ex 505

CHAPTER

6

An Introduction to Trigonometric Functions 509 6.1 6.2 6.3 6.4

6.5 6.6 6.7 6.8

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Contents

Angle Measure, Special Triangles, and Special Angles 510 Unit Circles and the Trigonometry of Real Numbers 527 Graphs of the Sine and Cosine Functions 542 Graphs of the Cosecant, Secant, Tangent, and Cotangent Functions 561 Mid-Chapter Check 577 Reinforcing Basic Concepts: Trigonometry of the Real Numbers and the Wrapping Function 577 Transformations and Applications of Trigonometric Graphs 578 The Trigonometry of Right Triangles 595 Trigonometry and the Coordinate Plane 610 Trigonometric Equation Models 622 Making Connections 633 Summary and Concept Review 634 Practice Test 643 Calculator Exploration and Discovery: Variable Amplitudes and Modeling the Tides 645 Strengthening Core Skills: Standard Angles, Reference Angles, and the Trig Functions 646 Cumulative Review: Chapters 1–6 648 Connections to Calculus: Right Triangle Relationships; Converting from Rectangular Coordinates to Trigonometric (Polar) Form 650

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CHAPTER

7

Trigonometric Identities, Inverses, and Equations 653 7.1 7.2 7.3 7.4

Fundamental Identities and Families of Identities 654 More on Verifying Identities 661 The Sum and Difference Identities 669 The Double-Angle, Half-Angle and Product-to-Sum Identities 680 Mid-Chapter Check 693 Reinforcing Basic Concepts: Identities—Connections and Relationships 693 7.5 The Inverse Trig Functions and Their Applications 695 7.6 Solving Basic Trig Equations 711 7.7 General Trig Equations and Applications 721 Making Connections 732 Summary and Concept Review 733 Practice Test 737 Calculator Exploration and Discovery: Seeing the Beats as the Beats Go On 739 Strengthening Core Skills: Trigonometric Equations and Inequalities 739 Cumulative Review: Chapters 1–7 741 Connections to Calculus: Simplifying Expressions Using a Trigonometric Substitution; Trigonometric Identities and Equations 743

CHAPTER

8

Applications of Trigonometry

745

8.1 Oblique Triangles and the Law of Sines 746 8.2 The Law of Cosines; the Area of a Triangle 759 8.3 Vectors and Vector Diagrams 771 Mid-Chapter Check 786 Reinforcing Basic Concepts: Scaled Drawings and the Laws of Sine and Cosine 786

8.4 Vectors Applications and the Dot Product 787 8.5 Complex Numbers in Trigonometric Form 802 8.6 De Moivre’s Theorem and the Theorem on nth Roots 813 Making Connections 821 Summary and Concept Review 822 Practice Test 826 Calculator Exploration and Discovery: Investigating Projectile Motion 828 Strengthening Core Skills: Vectors and Static Equilibrium 828 Cumulative Review: Chapters 1–8 829 Connections to Calculus: Trigonometry and Problem Solving; Vectors in Three Dimensions 832

Contents

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CHAPTER

9

Systems of Equations and Inequalities 9.1 9.2 9.3 9.4

9.5 9.6 9.7 9.8

CHAPTER

10

10.5 10.6 10.7 10.8

Contents

Linear Systems in Two Variables with Applications 838 Linear Systems in Three Variables with Applications 853 Systems of Inequalities and Linear Programming 865 Partial Fraction Decomposition 879 Mid-Chapter Check 891 Reinforcing Basic Concepts: Window Size and Graphing Technology 892 Solving Linear Systems Using Matrices and Row Operations 893 The Algebra of Matrices 905 Solving Linear Systems Using Matrix Equations 917 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More 933 Making Connections 947 Summary and Concept Review 948 Practice Test 953 Calculator Exploration and Discovery: Cramer’s Rule 954 Strengthening Core Skills: Augmented Matrices and Matrix Inverses 955 Cumulative Review: Chapters 1–9 956 Connections to Calculus: More on Partial Fraction Decomposition; The Geometry of Vectors and Determinants 958

Analytical Geometry and the Conic Sections 961 10.1 10.2 10.3 10.4

xxxii

837

A Brief Introduction to Analytic Geometry 962 The Circle and the Ellipse 969 The Hyperbola 984 The Analytic Parabola 997 Mid-Chapter Check 1006 Reinforcing Basic Concepts: More on Completing the Square 1006 Nonlinear Systems of Equations and Inequalities 1007 Polar Coordinates, Equations, and Graphs 1018 More on the Conic Sections: Rotation of Axes and Polar Form 1035 Parametric Equations and Graphs 1051 Making Connections 1064 Summary and Concept Review 1064 Practice Test 1069 Calculator Exploration and Discovery: Elongation and Eccentricity 1070 Strengthening Core Skills: Ellipses and Hyperbolas with Rational/Irrational Values of a and b 1071 Cumulative Review: Chapters 1–10 1072 Connections to Calculus: Polar Graphs and Instantaneous Rates of Change; Systems of Polar Equations 1073

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CHAPTER

11

Additional Topics in Algebra 1077 11.1 11.2 11.3 11.4

Sequences and Series 1078 Arithmetic Sequences 1089 Geometric Sequences 1098 Mathematical Induction 1112 Mid-Chapter Check 1119 Reinforcing Basic Concepts: Applications of Summation 1119 11.5 Counting Techniques 1120 11.6 Introduction to Probability 1132 11.7 The Binomial Theorem 1145 Making Connections 1153 Summary and Concept Review 1154 Practice Test 1158 Calculator Exploration and Discovery: Infinite Series, Finite Results 1160 Strengthening Core Skills: Probability, Quick-Counting, and Card Games 1161 Cumulative Review: Chapters 1–11 1162 Connections to Calculus: Applications of Summation 1165

CHAPTER

12

Bridges to Calculus: An Introduction to Limits 1169 12.1 An Introduction to Limits Using Tables and Graphs 1170 12.2 The Properties of Limits 1180 Mid-Chapter Check 1190

12.3 Continuity and More on Limits 1191 12.4 Applications of Limits: Instantaneous Rates of Change and the Area Under a Curve 1203 Making Connections 1215 Summary and Concept Review 1216 Practice Test 1218 Calculator Exploration and Discovery: Technology and the Area Under a Curve 1219 Cumulative Review: Chapters 1–12 1220

Contents

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Appendix A

A Review of Basic Concepts and Skills

A-1 Algebraic Expressions and the Properties of Real Numbers A-1 Exponents, Scientific Notation, and a Review of Polynomials A-10 Solving Linear Equations and Inequalities A-24 Factoring Polynomials and Solving Polynomial Equations by Factoring A-38 A.5 Rational Expressions and Equations A-52 A.6 Radicals, Rational Exponents, and Radical Equations A-64 Overview of Appendix A A-80 Practice Test A-82

A.1 A.2 A.3 A.4

Appendix B

Proof Positive—A Selection of Proofs from Precalculus

Appendix C

More on Synthetic Division

Appendix D

Reduced Row-Echelon Form and More on Matrices

Appendix E

The Equation of a Conic

A-93

Appendix F

Families of Polar Curves

A-95

A-89

Student Answer Appendix (SE only)

SA-1

Instructor Answer Appendix (AIE only) Index

xxxiv

Contents

I-1

A-84

IA-1

A-91

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Index of Applications ANATOMY/PHYSIOLOGY age of child related to circumference of head, 484 average height related to average weight, 30 distance in Plow yoga position from shoulders to toes, 643–644 height versus male shoe size, 91, 162 height versus wingspan of human body, 90 ideal weight for adult males, 46 length of radial bone when height of person is known, A–37 neck and waist measurements of human body, 92

ARCHITECTURE area of trapezoidal window, A–32–A–33 area of walkway around fountain, A–38 distance between fountains in the Ellipse, 978–979 distances among points on mosaic zodiac floor circle, 540 elliptical fireplace with glass doors, 981 elliptical garden with fountains, 981 Gateway Arch dimensions of crosssections, 524 height of Burj Dubai building, 607–608 height of center of elliptical bridge arches, 981 height of CNN Tower, 607 height of Eiffel Tower, 607 height of indoor waterfall, 648 height of Petronas Tower I, 607 height of St. Louis Arch, 1220 height of Stratosphere Tower in Las Vegas, 661 height of Washington Monument, 621, 786 height of window washers above ground, 603–604 heights of Eiffel Tower and Chrysler Building, 1063 heights of tallest buildings, 948 number of stairs and building height, 183 sight distance from external elevator, 1207–1208 sum of heights of Eiffel and CNN Towers, 852 surface area of Khufu’s pyramid, 823 velocities of ascending and descending elevators, 609 viewing angle of Petronas Towers, 741

viewing angle of Sears Tower and its antenna, 732 volume of cement needed for circular walkway, A–38 whispering gallery height and width, 983 width of Hall of Mirrors at Versailles, A–36 window height above ground, 603–604

special effects lighting in art gallery, 834 thickness of book and number of pages, 183 ticket pricing, 245, 721 viewing angle for paintings at art show, 709 work done in moving piano across stage, 1219

ART/FINE ARTS/THEATER

BIOLOGY/ZOOLOGY

actors’ ages, A–36 angular and linear velocities of camera cart on tracks, 643 area of border of tablecloth, 982 auction prices of three paintings, 863 auditorium seating capacity, 1095, 1097 classical music fund drive, 915–916 demand for popular songs, 290–291 distance from seat to bottom of movie screen, 669 distance from seat to top of movie screen, 668 drive-in movie theater viewing angles, 704–705 elliptical exhibit hall with stands, 981 falling/diving speeds of Star Trek characters, 99 frequencies of musical notes, 492 heights of favorite cartoon characters, 930 jewelry composed of gold alloy mixture, 844–845 Leonardo da Vinci’s human body diagram, 90 logo for engineering firm, 720 movie theater revenue, 294 music store clearance sale, 946 number of tickets sold, 259, 850 original price of baseball cards, 931 playing times of Mozart’s arias, 930 playing times of Rolling Stones songs, 930 purchase prices of art objects, 901 purchase prices of rare books, 904 rare books arrangement on shelves, 1159 revenue generated by movies, 1072 seating arrangement possibilities for small number of persons, 1123, 1128, 1131 sight angle of center seat in movie theater, 835 sizes of soft drinks sold at theater, 926–927 song selecting by band for contest, 1130

animal diets in zoo, 932 animal population maximums and minimums, 580, 591 animal territories, 1034 average birth rate of animal species, 1087 bacterial population growth, 1111 biorhythm potential cycles, 629 box turtle longevity, 1160 cell size and cell division, 500 chicken production in U.S., 491 chimpanzee’s probability of spelling a word, 1144 circadian rhythm of human body temperature, 629 concentration of chemical in bloodstream of large animal, 401 crawling speed of insect, 306 crop duster’s flying speed, 608 deep-sea fishing depths, 145 deer and predators, 161 deer population estimates, 1098 dimensions of fish tank, 1018 elk herd repopulation, A–63 fish and shark populations, 161 fish length to weight relationship, A–78 flight path of scavenger bird, 1030 fluctuating sizes of animal populations, 624 fruit fly population exponential growth, 475–476 gestation periods of selected mammals, 162, 863 hawk species repopulation, 1154 heights of tree and of baboons in tree, 607 human life expectancy, 31 insect infestation maximums and minimums, 406–407 lion and hyena populations, 290 motion detectors at watering holes, 756 prairie dog population growth rate, 500 predator/prey concentrations, 273, 481, 494

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pricing pet care products, 852 rabbit populations, 481, 488 raccoon and mosquito populations, 290 rodent and wolf populations, 284–285, 494 sinusoidal population of mosquitoes varied by humidity, 591 sinusoidal population of porcupines varied by solar cycle, 591 species/area relationship, 162, 455 speed of racing pigeons and of wind, 851 stocking a lake, 467, 830, 956, 1088 temperature and cricket chirps, 32 temperature and insect population, 352 vector angles between flying geese and ornithologist and car, 835 volume of an egg, 186 wave motion of sharks and fish, 559 weight and age of dog, 501 weight and daily food intake of nonpasserine birds, 157–158 whale weight and length, 162 wingspan and weight of selected birds, 162 wingspan of each of three birds, 863 wolf population preservation, 1088 year when two squirrel populations were closest in size, 1221 yeast culture growth, 486

BUSINESS/ECONOMICS advertising results, 174, 463, 467, 488 amount of crude oil imported to U.S., 174 annual profit, 1119–1120 annuities, 472–474, 479–480 automated filling of cereal boxes, 146 billboard viewing angle from highway, 709 blue-book value, 1088 break-even costs, 267–268, 272, 846, 851, 1013, 1016, 1017 business losses of a company, 48 car rental charges, 46 coffee sales fluctuating with weather, 731 coin value appreciation, 1159 commission sales, 27 committee/management composition possibilities, 1129, 1131, 1157 cost, revenue and profit, 233–234, 272 cost of undeveloped lot, 770, 830 cost to package, load, install each type of pool table, 916 credit card use, 92

xxxvi

decline of newspaper publishing, 174 depreciation, 31, 57–58, 62, 77, 428, 431, 467, 497, 501, 1018, 1110, 1159, 1203 e-mail address standardization, 1131 estimated price, 954 exponential growth, 431 gas mileage, 31, 46 GDP per capita in selected years, 88, 89 home selling prices within a range, 88, 90 home value growth, 456 hourly customer count in restaurant, 353 inflation rate, 432, 1088, 1111 law firm choices to attend forum, 1141 length of shipping cartons, 76 manufacturing cost, 370, 377–378, 382 market equilibrium, 847 market equilibrium point, 847, 851–852, 948, 954 marketing budgets, 445 marketing strategy and sales volume, 441, 463, 467 market share, 463–464 maximum profit, 239, 245–246, 872, 877–878, 949–950 maximum revenue, 248, 877, 891 military expenditures in U.S., 175 minimal shipping costs, 873–874, 878 minimum wage, A–9 monthly earnings of employee, 48 natural gas pricing, 175 number of books shipped per box, A–51 number of circuit boards assembled and length of time working, 29, 468, 489 number of units/services required to break even, 851 numbers of units manufactured to maximize revenue, 891, 954 online sales of pet supplies, 251 overtime wages, 175 patent applications, 90 patents issued, 91 percent of full acre being purchased, 770 plumbing service charges, 46–47 postage rates by ounce, 176 price, demand and supply, 1072 price and demand, 185 production cost, 246 production level to minimize cost, 401 profit and loss, 117 profit growth, 320 profit of new company, 301, 404 quality control tests, 145

randomly choosing corporate office location, 1140 rate of change for CD sales, 253–254 rate of growth change for Starbucks, 295 repetitive task learning curve, 468, 489 resource allocation, 930–931 retail sales revenue, 243 revenue and costs of space travel, 270 revenue generated by shirt sizes, 916 revenue growth, 431 sale price increase to return to original price, 1063 sales goals, 1097 sales of hybrid cars, 92 sales volume rate of change, 298 seasonal income, 730 seasonal sales/revenue, 730, 737, 738 selling price to maximize revenue on an item, 741 shelf size to number of cans stocked, 456 sinking funds, 474 start-up costs, 493 supply and demand, 186, 368, 847, 851–852, 948, 954, 1017 theft of precious metals in production line, 944–945 T-shirt production in each of two plants, 915 used vehicle sales, 269 vehicle tires sold by three retail outlets, 915 volumes of raw materials needed to manufacture drums, 944 volunteers arrangements to replace managers, 1129 wage increases, 1087 wages and hours worked, 183

CHEMISTRY acid solution concentrations, 891 chlorine dissipation from swimming pool, 1110–1111 freezing time, 467 froth height of carbonated beverages, 490 household cleaner solution, 850 mixture/solution concentrations, 73–74, 77–78, 369, 844, 863 pH levels, 443, 445, 453, 455 venting landfill gases, 1111 volume of raw materials needed to manufacture combustibles, 953 water temperature mixture, 99

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COMMUNICATION cable television subscriptions, 489 cell phone charges, 176 cell phone subscribers, 234 earnings of current and previous years for video company, A–82 flagpole height, 512–513 late DVD rental returns, 1152 length of cable securing radio tower, A–78 MP3 market research, 847, 851 number of phone calls between two cities, 186 phone service calling plans, 175 profit from sales of Wi-Fi phones, 296 radar detection, 17, 753, 756 radio broadcast range, 17–18 sales of Apple iPhones, 100 signal from tall radio tower, 1006 television programming possibilities, 1131 tracking wait time for cable service installation, 1159 TV repair costs, A–9 walkie-talkie range, 826

COMPUTERS animation on computer game, 1097 compact disk circumference, 541 consultant salary, 145 cost of repair service call, A–8 elastic rebound of animated dropping ball, 1111 encryption of messages, 941–942, 945, 953 households with Internet connections, 62 internet revenue, 295 internet selling, 186 memory card assembly rate by robot, 1162 probability of home computer ownership, 1143 randomly-generated numbers, 1142 rebate amount in incentive packages for internet service, 909–910 sales of Apple iPhones, 100 storage space on a hard drive, A–15 upgrading source of data traffic, 939 year cable installations will be greater than 30,000, 905

CONSTRUCTION/ MANUFACTURING area of a Norman window, 943 area of circular sidewalk, 63 area of house and of lot, 379 box manufacturing, 1017 building heights measured by angles of elevation, 574, 633 cable angles to steady a radio tower, 679 cable lengths for trolley car, 482 cable winch turning angles, 541 composition of crew to complete small job, 1157 condominiums on triangular lot, 766 cost of a new home and size of the house, 29 cost of copper tubing, 186 cost of elliptical mirror and its frame, 1050 deflection of beam when force applied, 394 dimensions of cylindrical tank, 1017 dimensions of deck, 1017 drill bit efficiency, 1107 drywall sheets required, A–38 elliptical hole in roof for plumbing vent, 982 equipment aging, 1110 fencing area, 247 generator failure probability, 1140 height of vertical support to give exact degrees of incline, 641 instantaneous velocity of falling wrench, 1206–1207 largest number of shingle packs and nail buckets lifted onto roof, A–36 length of base of water tower, 1018 length of ladder leaning against building, A–75 length of roof rafters, 757 light bulb defectiveness, 1160 maximum load supported by a post, 468 measuring depth of a well, 273 number of screws to manufacture to maximize revenue, 877 parabolic flashlight reflector, 1004, 1005 power tool rental cost, A–37 profit per contract for home improvement items, 915 propeller manufacturing, 1034 ramp angle of incline, 799 rate of production for specialized part, 1214

rolling new sod over selected areas, 540–541 roof slope, 30, A–9 rotations of ceiling fan blades, 620, 621 safe weight supported by horizontal beam, 185, 187, 195 sand and cement mixtures to maximize profit, 878 sewer line slope, 30 sheep pens dimensions, 247 shopping complex distance from avenue, 603 sizes of rejected ball bearings in bin, 946 speed of a winch, 525 speed of pallet pulled by winch, 633 swimming pool dimensions, 1017 ventilation of home, 445 winch-drum radian angle turned while lifting, 636 window areas and shapes, 247 wiring an apartment by pulling wire from spool, 541

CRIMINAL JUSTICE/ LEGAL STUDIES area taped by surveillance camera, 770 cost to seize illegal drugs, A–62 estimating time of death, 466 federal law enforcement officers employed for selected years, 89 parabolic sound receivers for private investigators, 1005 prison population increases, 62

DEMOGRAPHICS ages of child prodigies, 891 animal population maximums and minimums, 580, 591 arrangement in photographs of multiplebirth families, 1129 basketball salaries, 492 cable television subscriptions, 489 cell phone use, 493 centenarian population increases, 483 children and AIDS cases, 494 credit card use, 92 debit card use, 491 debt-per-capita of U.S., A–21 decrease in percentage of smokers, 32, 88 family farms with milk cows, 490 federal law enforcement officers employed for selected years, 89 female physicians for selected years, 32 females/males in workforce, 91

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fertility rates of women in U.S., 260 growth rates of children, 494 households with internet connections, 62 incomes over $200,000, 455 instantaneous rate of change of population, 1213 logistics and population size, 890 longest-living human, 176 low birth weight of baby and age of mother, 493 multiple births increasing, 175 number of post offices in U.S., 490 population density of urban areas, 367 population growth, 466, 1110–1111 population growth in space colony, 1110 population of coastal areas, 491 prison population increases, 62 registered podcast growth, 649 research and development expenditures, 492 sets of triplets born in U.S., 504 students joining clubs by gender and class, 916 telephone calls per capita, 490 telephone opinion poll possibilities, 1152 tourist population, 319 union workers chosen to be interviewed, 1160 veterans in civilian life, 491 volunteer enlistments in military by age, 912–913

EDUCATION/TEACHING college loans from alumni contributions, 850 cost of college tuition and fees, 32 course schedule possibilities, 1128 exam scores needed to keep scholarship, A–37 grade point averages, 370 grade possibilities at college, 1128 guessing randomly on multiple choice test, 1152 IQ of each of three persons, 904 language retention, 369 memory retention, 445, A–63 number of children homeschooled, 297 number of credit hours taught at community college, 1156 quiz grade versus study time, 491 scholarship award possibilities, 1131 score needed on last exam to earn 80 average, A–31

xxxviii

sight angle for drivers approaching school, 835 students joining clubs by gender and class, 916 test averages, 370 textbook committee membership possibilities, 1127 true-false quiz probability, 1144 typing speed of student, A–63 value of doing homework, 98 vector from top of light pole to corner of playground, 834 ways for children to line up for lunch, 1130 ways to choose students to attend seminar/conference, 1130, 1136

ENGINEERING Civil minimum bid for boring tunnel through mountain, 768 tunnel length through mountain, 768 vector from top of light pole to corner of playground, 834 Electrical AC circuits, 678, 811 alternating current, 560 current in circuit, 812 electrical resistance and wire dimensions, 161, 186, 878, 983, A–9 impedance, 213, 608, 809, 812, 820 parallel circuits, 812 phase angle between current and voltage, 809 temperature and resistance, 394–395 voltage in a circuit, 213, 811–812 voltage supplied to Japan, 812 Mechanical angles between hands of clock, 621, 691 area cleaned by windshield wiper, 610 car jack and equilibrium, 678 center-pivot irrigation wheel velocities, 526 coordinates of angle in multijointed arm, 827 industrial spotlights, 1005 loss of length in pencils through use, 1221 machine gears and size, 692 position of engine piston, 630 radius and height of cylindrical vents, 1016 straight-line and angle distances of carts on circular tracks, 834

wind-powered energy, 30, 89, 134, 135, 186, 194, 353, 421 wind-power kite height, 514

ENVIRONMENTAL STUDIES acres of each crop planted to maximize profits, 877 area of circular forest fire, 289 area of oil spill, 284 barrels of toxic waste in storage, 1119, 1163 city park dimensions, A–7 cost of removing pollutants, 155–156, 161, 368, 401, 1208–1209, A–63 deer and predators, 161 distance of forest fire to tower, 757, 826 draining of water reservoir, 1156 drinking water pollution testing, 1152 fish versus shark populations, 161 gold mining and depletion of resources, 490 hauling capacity of hazardous waste to maximize revenue, 877, 997 lawn dimensions, A–9 ounces of platinum mined over time, 501 pest control cost, A–8 recycling costs, 161, 368 sulphur dioxide emissions in U.S., 741 tomato production and watering amounts, 259–260 total area of Southeast Asia’s Coral Triangle, 827 velocity of debris from strip mine explosion, 1219 venting landfill gases, 1111 volume of garbage in landfill and number of garbage trucks dumping, 29 volume of grain storage silo, A–38 waste product monthly maximums and miniumums, 742 water leaking from reservoir dam, 1214 water supply in reservoir, 352 wind-power kite height, 514

FINANCE amount invested in bonds with different interests, 850 amounts borrowed at three interest rates, 861–862, 865 amounts invested in each interest rate, 853, 904, 931, 954 annuities, 472–474, 479–480, 830, 957 balance of payments, 352, 401

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bequests to charity, 1097 checking account balance, A–37 coin denominations in collection container, 850, 931, 949 coin denominations purchased, 850 coin value appreciation, 1159 compound interest, 288, 470–472, 478–481, 500, 526, 742, 1109, 1163, 1190 credit card payments, A–31–A–32 currency conversion, 289 debt load, 319 deposits required monthly to meet final goal, 500, 501 doubling time for interest, 443, 470–471 federal investment in military over time, 270 federal surplus of U.S., 118 fixed interest, 905 government deficits, 335 guns versus butter spending, 877 interest rates, 117 international trade balance, 234 investing in coins, 62 investment growth, 479 investment tripling rate, 445 mortgage payments, 480, A–23 national debt increase, 198 opening price of stock, A–63 payday loans, 469 percentage of families owning stocks, 174 possible ways to divide monetary gift, 877 research and development dividends, 270 retirement investments, 869, 931 return on investments, 945 shares traded on New York Stock Exchange, 88 simple interest, 31, 185, 469, 478, 500 sinking funds, 474 stock prices over time, A–63 stock purchases, 1084–1085 stock value, 145 value of gold coins, 863–864 value of investment over time, 347

GEOGRAPHY/GEOLOGY area and angle measurements of Bermuda Triangle, 827 area of center-pivot irrigated crop, 526 area of Nile River Delta, 772 area of Yukon Territory, 772 atmospheric pressure and altitude, 440, 464–465, 499

avalanche conditions, 709 contour map and length of trail up mountain, 525 depth of channel cut by river through canyon, 407–408 depth of tidal pool, 957 dimensions of a tract of land, 1017 distance across base of volcano, 762–763 distance from Four Corners USA to point on Kansas border, 643 distance north of equator of Las Vegas, 577 distance of Invercargill, NZ, from equator, 560 distance of kayak from glacier, 602 distance of viewer’s house from Petronas Tower I, 607 earthquake intensity, 440, 444, 456 earthquake magnitude, 439, 498 earthquakes and elastic rebound, 630–631 fluid mechanics, 678 height of a mountain peak, 758, 827, 830, 957 height of canyon rim, 607 height of cliff, 522 height of Mount McKinley, 741 height of Mount Rushmore, 638 height of tree on mountainside, 822 highest and lowest points on each continent, A–38 land area of Tahiti and Tonga, 852, 954 length of Panama Canal, A–37 lengths of each of two rivers, 76 location of stranded boat using polar coordinates, 1018–1019 ocean depths and temperatures, 163, 493 pathways around a pond, 776 radius of Earth at latitude of Beijing, 609 rainbow height, 608 river velocities, 162 seasonal ice thickness on lake, 730 seasonal water temperature in mountain stream, 730 tide height predictions, 47 vertical and horizontal changes on contour map, 608 viewing distance changes with increasing elevations, 157, 256, 290 water depth and pressure, 197, 492 water level in lake, 31 weight of diamonds extracted from mine, 488

width of a canyon, 758 width of Africa at equator, 524 width of large rock formation, 1072 work done by arctic explorer dragging sled to haul supplies, 800, 825

HISTORY boom town population growth, 1111 years Declaration of Independence signed and Civil War ended, 852 years each of three documents signed, 863 years each of three wars ended, 863

MATHEMATICS angle between two vectors of box, 833–834 angle complement and supplement, 511 angle measurements of triangle, 904 angle of depression, 602, 607 angle of diagonal of parallelepiped with base, 731 angle of elevation, 607, 633, 644, 741 angle of inclination for selected materials sliding down inclines, 575 angle of rotation, 603, 607, 620, 621, 692 angles formed at each vertex of triangle on geoboard, 769 angles formed by bread sliced diagonally, 641 angles formed by radii of three tangent circles, 786 angular and linear velocities, 521 arc length, 516, 1097, 1105–1106 area of circle, 289, 421, A–21 area of circular ripple over time, 296 area of circular sector, 517 area of circular segment, 731, 737 area of circular sidewalk, 63 area of ellipse, 981 area of equilateral triangle, 184 area of first quadrant triangle, 381 area of nonright triangle, 764–765 area of parallelogram, 620, 943 area of polygon inscribed around circle, 575 area of polygon inscribed in circle, 558 area of print on page within standard border dimensions, 383 area of rectangle, 876, A–37 area of rectangle inscribed in semicircle, 834 area of right parabolic segment, 1004 area of trapezoidal window, A–32–A–33

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area of triangle, 46, 738, 770, 876, A–37 area of triangle using Heron’s formula, 770, 1180 area of wall illuminated by circle and by ellipse, 982 areas of small triangles within large triangle, 384 average rate of change, 254–256, 260–261 center of circle, 197 circumference of circle, 90 circumscribed triangles, 758 colored ball sequences withdrawn from bag, 1129, 1143, 1159 combination lock sequence possibilities, 1121–1122, 1128, 1157 consecutive integers, 77 coterminal angles, 515 degrees/minutes/seconds and decimal degrees conversion, 511 diagonal of cube, 609 diagonal of rectangular parallelepiped, 609 diagonal of rectangular prism, A–83 diameter of circle circumscribed around triangle, 758 dimensions of box if only volume is known, 335 dimensions of open box, 318, 1014 dimensions of rectangular solid, 863 dimensions of right triangle, 904 focal chord of hyperbola, 996 horizontal and vertical components of vector, 775 icon-choosing probability, 1160 inscribed circle area, 17 inscribed triangle area, 17 lateral surface area of cone, A–79 lateral surface area of frustum, A–79 length of apothem of hexagon, A–36 length of chord of circumscribed triangle, 758 length of diagonal of parallelepiped, 731 length of diagonals within cube, 610 length of hypotenuse of triangle, 197 length of laser pen measured by beam of light and angle of pen, 594 length of rectangle, 197 length of side of triangle, 176 letter rearrangements possibilities in words, 1129 letters/numbers written on slips of paper to be randomly drawn, 1143, 1157

xl

line segment measuring, 541 magnitude of a diagonal, 784 maximum area included by several fenced pens, 242, 247, 336 multiple arc swings distances, 1110 nonacute angles, 620 number combination possibilities, 1128 number of sides needed for polygon inscribed in circle, 1170–1171, 1177 number of vertices of dodecahedron, A–36 original dimensions of cube after slice removed, 334 perimeter of ellipse, 981 perimeter of hexagon circumscribed by circle, 769 perimeter of legal-size paper, 954 perimeter of pentagon circumscribed by circle, 769 perimeter of polygon inscribed around circle, 575 perimeter of trapezoid, 731, 768 perimeter of triangle, 197, 198, 384 perimeter of triangle on geoboard, 769 Pick’s theorem to calculate area of triangle, 46 points on a unit circle, 529, 540 possibilities for food group arrangements in table settings, 1159 probability of drawing the perfect number, 1144 Pythagorean theorem, 668 radian and degree conversion, 518 radian angle, 636 radius of circle, 197, 289 radius of circular ripple over time, 296 radius of circumscribed circle, 758 radius of sphere, 420 rearrangement possibilities of digits in numbers, 1129 remote access door opener number possibilities, 1128 rewriting an expression using negative exponents, A–23 security code PIN possibilities, 1122, 1221 simultaneous calculation of perimeter and area, 915 single digit randomly chosen from digit 10, 1140 sum of consecutive cubes, 395 sum of consecutive squares, 395 surface area of box, 382

surface area of cube, 184 surface area of cylinder, 76, 119, 233, 271, 381, A–51 surface area of cylinder when only volume is known, 383 surface area of open cylinder, 384 surface area of sphere, 184 surface area of spherical cap, 384 tangent functions, 630 thickness of paper folded several times, 1111 time of day expressed metrically, A–9 triangles formed by three rods attached at pivot points, 757, 769, 826 volume of balloon, 261 volume of composite figure, A–38 volume of cone, 421, 730, 876 volume of conical shell, A–51 volume of cube, 46 volume of cylinder, 46, 90, 119, 730, 876 volume of cylindrical shell, A–51 volume of equipoise cylinder, 417 volume of paraboloid, 504 volume of parallelepiped, 960 volume of prism, 944 volume of pyramid, 943 volume of rectangular box, A–51 volume of rolling snowball, 298 volume of sphere, 134 volume of sphere circumscribed by a cylinder, 78 volume of spherical cap, 649 volume of spherical shell, A–51 width of rectangle, 197

MEDICINE/NURSING/ NUTRITION/DIETETICS/HEALTH absorption rate of drug, 467 age of child related to circumference of head, 484 amounts of each nut in mixture to maximum profit, 872–873 angles formed by bread sliced diagonally, 641 angles of jointed light in dentist’s office, 785 animal diets in zoo, 932 average height related to average weight, 30 bacterial growth, 431, 481, 1111 calories allotted for lunch on a diet, A–36 calories of each nutrient provided daily for geriatric patient, 864

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coffee blends to maximize profits, 877 concentration of medication in bloodstream, 368, A–62 connections between weather and mood, 631 cost of milk, A–8 cost of types of meat per pound, 957 costs of ordering food from selected restaurants, 916 doctors chosen randomly to visit hospitals, 1144 fertility rates of women in U.S., 260 food service supply inventories, 938 freezing time, 467 genetics and fruit fly population growths, 475–476 grams of each of three fats in soup, 864 growth weight of fetus, 260 heart rate during exercise routine, 731 hodophobics randomly chosen to receive therapy, 1144 human life expectancy, 31 ideal weight for adult males, 46 instantaneous rate of change for bacteria in human body, 1213 measures of grain in each bundle received by bakery, 945 menu items possible in restaurant, 1128 military physical conditioning regimen, 1121 milkfat mixture, 850 milligrams of painkiller in bloodstream, A–22 mixture concentrations, 77–78 pH levels, 443, 445, 453, 455 placement of kidney stone from lithotripter, 981 pounds of each coffee bean type used each week, 946 prescription drugs sales per selected years, 62 range of weights lifted to prevent heart disease, A–36 rolling pin speed, 525 sandwiches needed to maximize revenue, 877 snowcone dimensions, 709 spread of disease after outbreak, 489 tracking wait time after making doctors appointments, 1158 training diet, 932 vegetarians randomly chosen from dietetics class, 1141

ways to make hamburgers and fruit trays, 1130 ways to select recipes for competition, 1130 weight loss over time, 489 yeast culture growth, 486

METEOROLOGY altitude and atmospheric pressure, 445, 464–465, 499 altitude and temperature, 420, 467 altitude of jet stream, 145 altitudes of fighter pilot training, 354 annual rainfall effect on number of cattle per acre, 100 atmospheric pressure and altitude, 440, 445, 464–465, 499 atmospheric pressure and vacuum strength, 1110 atmospheric temperature and altitude, 56–57, 420, 444–445 average monthly rainfall for Reno and Cheyenne, 632, 639 boiling temperature of water and altitude, 30, 89 connections between weather and mood, 631 daily water usage in arid city, 644 days in year with 10.5 hours of daylight, 1006 discharge rate of river, 316, 631, 726–727, 730 distance between sides of nuclear cooling towers, 996 earthquake intensity, 440, 444, 456 earthquake magnitude, 439, 498 earthquake range, 17 focus of parabolic radio antenna dish, 1002 gnomon shadow lengths at equator, 625–626 height of ocean floor, 331 hours of daylight by month, 124, 632 hours of daylight in selected cities, 591, 592, 594, 632 location of storm, 993 monthly temperatures for selected U.S. cities, 629, 631, 642, 644 ocean temperatures, 163, 493–494 ocean wave height, 558 parabolic sound receivers, 1005 record low temperatures for Denver, 738 seasonal precipitation in Wyoming and Washington, 627

seasonal temperatures in North Dakota, 626 sinusoidal temperature patterns at selected countries, 579, 590 speed of Caribbean Current, 850 speed of flowing water, 525 temperature and altitude, 420 temperature and atmospheric pressure, 444–445 temperature changes during daylight hours, 649 temperature drop, 261 temperature fluctuation over time, 559, 579, 1097 tidal depths of water in bay, 644 tide heights over time, 559 tsunami height and wavelength, 558 tubular fluid flow, A–51 vectors for three vessels mapping ocean floor, 782 water depth and pressure, 197, 492 wind powered energy, 30, 89, 134, 135, 186, 194, 353, 421 wind-power kite height, 514 wind velocity, 845–848

PHYSICS/ASTRONOMY/ PLANETARY STUDIES absorption rates of fabric, 194, 493, 501 acceleration due to gravity on ball rolling down incline, 135, 720 angle made by light source and point on surface, 576 angle of illuminance, 834 angle of projection of projectile at maximum height, 1063 angular and linear velocities of Ganymede, 526 angular and linear velocities of Venus around Sun, 1162 aphelion and perihelion of Mars, 1069 area of a lune on the surface of Earth, 576 attraction between particles, A–23 average speed of UFO between two cities, 759 brightness of a star, 444 carbon-14 dating, 481 catapults and projectiles, 117, 246 Celsius to Fahrenheit temperature conversion, 46, 850 closest and farthest distances between two planets, 752–753, 756

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collision of two particles moving in a medium, 1063 days required for selected planets to orbit Sun, A–79 densities of objects floating on or sinking in water, 335 distance ball rolls due to gravity, 135 distance between cities measured by radio waves on satellite, 768 distance between planets when arrived at focal chords simultaneously, 1049 distance from satellite to Earth horizon, 710 distance of Earth to Mars, 77 distance of image reflected from spherical lens, 574 distance of Mars to Sun, 195 distances traveled by bouncing rubber ball, 1145 distances traveled by swinging pendulum, 1105–1106, 1110, 1159 distance traveled by light beam over time, 588 distance traveled by planets orbiting the Sun, 541 eccentricities of planetary orbits, 1044–1045, 1049 elastic rebound of dropped balls, 1111 electrical constant between charged particles, 186 electron motion, 1062 elliptical orbit of Halley’s comet, 1069 elliptical orbit of planet, 1058 fluctuating size of Mars polar ice cap, 623 focal chord for orbit of planet, 1046, 1068 force acting on object on inclined plane, 708, 790, 799, 825, 830 forces acting on a point, 772, 780 gravitational force between two objects, 160, 187 gravity and speed of falling object, 420 half-life of radioactive substances, 432, 477, 481, 499, 504 harmonic motion of float bobbing on waves, 593 harmonic motion of oscillating pendulum, 593 harmonic motion of sound waves, 585–586, 593 harmonic motion of stretched spring, 584, 592 heat flow around circumference of pipe, 708 height of bounced ball, 294

xlii

height of falling object, 259, 285 height of light source to provide maximum illumination, 202, 834 height of projectile, 233, 240, 246, 260, 294 heights and times of rocket, 228–229, 246 hydrostatics and surface tension, 558 illumination of book surface, 606 illumination of Moon surface, 631 index of refraction for selected mediums, 720 instantaneous velocity and heights of model rockets, 1213 instantaneous velocity of bowling ball dropped from building, 1213 intensity of light, 161, 187, 669, A–23 intensity of sound, 187 kinetic energy and velocity, 185 kinetic energy of planets, 559 latitudinal miles between two cities, 741 length of pendulum and its period, 299 light refraction angles, 720 linear velocity of Neptune, 526 linear velocity of turning wheel, 741 locations of blips on radar screen, 618 low-orbit space travel packages, 710 magnetic attractive force and distance, 648 magnetic field on Mars, 609 maximum range of projectile, 687, 691 minimum distance between like particles with common charge, 996 motion detectors, 524 motionless moments of particle floating in turbulence, 200, 202 noise level, 446 orbital velocity of planets, 982 orbits of an inner and an outer planet, 1062 orbits of satellites around Earth, 192 oscillation of stretched spring, 145 parabolic radio wave receivers, 1005 particle motion, 1062 path of comet, 992 perihelion, aphelion, period of comets, 1050 perihelion of asteroid Ceres, 1049 perihelions of planetary orbits, 1045, 1049 period of pendulum, 186 photochromatic sunglasses, 432 planetary orbits, 162, 493, 982

potential energy of planets, 559 pressure of gas in closed container, 181, A–9 pressure on eardrums, 678 projected image height, 184, 290, 420 projectile motion, 796, 801, 825, 828 pulling object up frictionless ramp, 997 radioactive decay, 432, 477, 481, 488, 499, 504, 577 radius and surface area of supernova, 289 radius of Jupiter, A–82 randomly choosing first speaker at space conference, 1139 range of projectile, 719, 742, 799 rate of change of projectile velocity, 255 revenue and costs of space travel, 270 ring tones on telephone, 560 rocket testing, 352–353 satellite orbiting distance north or south of Equator, 629 seasonal size of Antarctica ice sheet, 629 shadow length changes over time, 589, 594, 625 solar furnace, 1005 sound intensity, 444 spaceship velocity, 467 space-time relationship, A–51 speed of sound, 62, 691 speed of sound and temperature/ altitude, 691 standing waves, 692 strokes of hand pump to create vacuum, 1110 submarine depth, 145 temperature of pizza over time, 428–429 temperatures of cool and warm drinks, 432 temperatures of Earth’s atmosphere, 348 tension on rope supporting heavy object between two points, 829 thermal conductivity, 931 time for liquid stain to spread, 194, 493, 501 time for projectile/dropped object to hit ground, 134, 184, 294, 956, A–78 time required for rocket to turn around, 252 time required for satellite to reach Jupiter from Earth, A–21 touch-tone phone sounds, 687–688, 691 vectors for three forces in opposite directions, 782

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velocity of a bullet, 160 velocity of falling object, 261, 271, 285–286 velocity of fluid flowing from tank, 134 velocity of particle in positive direction while floating in turbulence, 390–391 volume of bathtub draining over time, 295 wavelengths of visible light, 559 waves traveling along a string, 678 weight of astronaut on moon, 186 weight of object above surface of Earth, 185 wind powered energy, 30, 89, 134, 135, 186, 194, 353, 421 work done along distance of object being moved, 790–791, 799

POLITICS amounts spent on defense and domestic improvements, 877 federal income tax rate and income, 853 retiree randomly chosen to be interviewed, 1144 Supreme Court justices, 30 veteran randomly chosen to be interviewed, 1141–1142 voter randomly chosen to be interviewed, 1138 women in U.S. Congress, 88

SOCIAL SCIENCES/HUMAN SERVICES choosing soldiers for reconnaissance team, 1129 cost of picking up trash along highways, 192 daily water usage in arid city, 644 dimensions of envelopes, A–52 disarming a bomb probabilities, 1144 distances between each vertex for search-and-rescue team, 830 eating out popularity, 62 energy rationing in U.S., 174 European shoe sizes, A–37 Express Mail cost for selected years, 88 heating and cooling subsidies for lowincome families, A–37 height of ladder leaning against building, 522 home location near river and golf course, 1050 home value growth, 456 international shoe sizes, 289 lawn service charge, 851

library fines on overdue materials, 496 number of customers in shopping mall, 397 number of new books published in U.S., 298 paper sizes, A–52 per capita spending on police protection, 170 phone service calling plans, 175 physical training for military recruits, 145 plumbing service charges, 46–47 postage costs, A–9 postage rate for large envelopes, 171 probability of choosing teenager from family of five children, 1133 recycling program costs, 161, 368 selecting books to read on vacation, 1130 sizes of Slushies sold at convenience store, 930 speed of lawn mower, 648 telephone area codes possibilities, 1131 time required to pick up trash along highway, 186 tracking wait time for customer service, 1142 two men pulling on mule, 782 two tractors pulling at stump, 782 watering a lawn, 524, 648 water rationing in southwestern U.S., 175 ways to dress for work, 1157 wild stallion roped by cowhands, 798 work done by mule plowing field, 799 work done to mow lawn, 800

SPORTS/LEISURE admission charged by age at Water World, 175 amusement park attendance, 319 amusement park revenue, 838 angles between each marker in orienteering meet, 822–823 angular and linear velocities of bowling ball, 635 anxiety level of Olympic skater, 407–408 archery competition, 1061 archery target hits probabilities, 1160 area of elliptical race track, 982 areas of football and soccer fields, A–36 attendance at state park depending on weather, 731 average weight of football offensive line, A–37

awarding medals in sprinting competition, 1130 barrel races, 541 baseball card value, 31 baseball height while falling, 954 baseball hits probabilities, 1152 baseball home run angle and velocity, 1061 basketball championship final score, 904 basketball free throw possibilities, 1121, 1150–1151 basketball team formation possibilities, 1130 blanket toss competition, 246–247 bowling scores needed to obtain average, 99 bystander’s possibility of being hit by flying model airplane, 822 cable lengths for trolley car, 482 card drawing probabilities, 1141 chess tournament place finishing possibilities, 1129 circus human cannonball height from ground and distance from net, 680, 801, 1057 coin flipping possibilities, 145, 1128, 1130, 1139, 1142, 1145, 1152 dartboard hits probabilities, 1143 depth of scuba diver over time, 400 descent distances of spelunkers, 76 dice rolling probabilities, 1130, 1132–1133, 1135–1136, 1139–1142, 1157 differences in speed and distance of adult and kid bicycles, 526 dimensions of flag, 1017 dimensions of sail, 1017 distance of boat to lighthouse, 644 distance of golf ball to hole, 560, 709 distance of thrown softball, A–9 distance run while training for marathon, 1211–1212, A–36 distance traveled by a seat on a Ferris wheel, 721 diving speeds, 99 domino randomly drawn from bag, 1140 elliptical billiard table, 983 exponential decay of pitcher’s mound, 431 finishing a foot race possibilities, 1124 Five Card Stud probability of hands dealt, 1161–1162 football field goal kick angle and velocity, 1062 football throwing competition, 1061

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game show contestants randomly chosen to start final round, 1140 game spinner angles turned through, 620 game spinners and probability, 431 golf swing arm lengths and angles, 719 height of canyon rim, 607 height of high-wire acrobat and angles of elevation, 826 height of kicked football/soccerball, 240–241, 864 height of lacrosse long pass, 827 height of thrown baseball, 801 heights of rides at amusement park, 365 high diver entry into water, 620 high jump Olympic records, 91 high-wire walking on incline, 608 horse race place finishing possibilities, 1129, 1157 hot air ballooning, 162 instantaneous velocity of carabiner dropped by rock climber, 1213 javelin throw angle of release, 799 jogging distance with increased speed, 305 jogging speed, 77 jumping frog distances jumped, 249 kiteboarding and wind speed, 146 kite height, A–78 length of boat and drag resistance, 335 linear velocity of passenger on Ferris wheel, 1111 lottery winning possibilities, 1126 magnitude and angle of tug-of-war rope, 798 maximum height of jai alai thrown ball, 956 maximum height of shot arrow and time airborne, 864 maximum number of registrants for 5-km race, 259 minimum altitude of plane flying over target, 996 minimum altitude of stunt plane, 996 motorcycle jumps, 247 number of swimmers in pool over time, A–23 number of words formed from a single word, 1157 Olympic 400-meter swimming competition, 119 origami angles, 692 parabolic sound receivers at sports events, 1005

xliv

pitcher’s mound loss of height, 431 pool balls randomly chosen in game, 1141–1142 probability of bowler rolling strikes, 1158 probability of completing tandem bicycle trip, 1159 projectile components of kicked football, 785 projectile components of shot arrow, 785, 796–797 randomly choosing little league coach to speak first, 1139 remaining distance for climber to reach top of rim, 607 revenue for selected months of Water World, 297 riding a round-a-bout, 524 roller coaster design, 720 runner first to finish line, 214, 320 running times in 400-meter race, 47, 118 scores in Trumps game, 852 Scrabble game letter possibilities, 1125, 1129 shot-put angle of rotation of thrower, 644 single outcome probability of card drawn from deck, 1133, 1135–1140, 1157 size of guided tour groups and start-up times needed, 259 SkeeBall machine programming, 175 snowboarder heights, 247 soccer kick for winning goal, 1069 soccer shooting angle for goal, 710 soccer starting roster randomly chosen, 1140 soccer team starting line-up possibilities, 1131 speed of bicycle converted from rpm to mph, 521 speed of river current, 849, 850 speed of rower in still water, 849, 850 spinner outcomes possibilities, 1128, 1130, 1134, 1139, 1140, 1142, 1159 sports promotions, 445 super ball elastic rebound, 1111 swimming race and current in river, 824 target hits probability for markswoman, 1143 tennis court dimensions, 234 tension and angle of rope holding runaway steer, 827 thrown baseball reaching catcher without bouncing, 1004

throws at moving targets made by sports players, 757, 826 Tic-Tac-Toe ending board possibilities, 1132 timed swimming trials, A–37 time for thrown/dropped object to hit ground, 134, 184, 294, 956 tolerances for sport balls, 145 training diet, 932 training time for sprinter, 187 trampoline flips and belly-flops, 620 treadmill angle of incline during workout, 731 unicycle racing speed, 525 vectors for family bowling at three lanes, 782 velocity of Gravity Drum ride, 524 velocity of soccer ball kicked straight up, 1204–1206 volume of water in swimming pool, 319 waterski jump angle from lake surface, 709 ways to choose cats in pet show, 1157 ways to commit crime in Clue game, 1131 winning score in table tennis tournament, 198 work done to give wheelbarrow rides to kids, 800 work done to pull bus in tough-man contest, 800 work done to pull sled on level surface, 799

TRANSPORTATION acceleration and velocity of car, 62 aerial distance between planes after five hours, 769 aircraft N-number possibilities, 1131 airplane speed with no wind, 845–846 airport moving walkway speed, 850 altitude of helicopter, 522 angle between plane making food drop and observation post, 836 angle of ground searchlight tracking jet plane, 835 angle of tow lines pulling barge along river bank, 827 angle of tow lines pulling yacht into port, 824 apparent height of building seen while traveling toward it, 630 bicycle sales growth, 489

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biking time, 73 billboard design, 770 billboard viewing angle from highway, 709 car repair costs, A–9 celebrity fans waiting at arrival gates, A–36 coordinates of plane and of town, 1163 cost of gasoline, A–8 cost per gallon of grades of gasoline, 892 course and speed of ship in crosscurrent, 785, 957, 1051 course heading and speed of airplane flying through crosswind, 781, 785, 801, 827, 830, 1221 cruise ship speed, 850 cruising speed of airliner, A–9 dimensions of a trailer, 1017 distance between cities, 519–520, 524, 756–760, 768, 786 distance from start when one vehicle overtakes another, 77 distance of aircraft carrier from home port and hours since departure, 29 distance of car wreck from Eiffel Tower, 607 distance of ship to closest observation post, 812 distance of two boats from building, 641 distance on map to destination, 99 distance traveled based on speed and fuel capacity, 182 drag force and speed of car, 295 driving time, 73, 77, 301, 403 fines for speeding, 420 flight heading between cities, 769 flight path of rescue plane, 1034 flow of traffic per minute, 352 fuel consumption of competing car manufacturers, 831 fuel economy for selected years, 500 gas mileage, 46

groundspeed and direction of airplane, 983 height of stacks on cruise liner, A–7 helicopter’s altitude from ground, A–7 highway sign erected on steep hillside, 786 horsepower of vehicle engines, 504 hydrofoil service engine failure probability, 1135 length of bridge over river/lake, 607, 635, 669 length of flights to several cities along route, 904 lengths of suspension bridges, 77 license plate possibilities, 1128, 1130, 1159 locating ship/airplane using radar, 996 minimizing distribution distance, 272–273 minimizing transportation costs, 878 moving van rental costs, A–37 nautical distance between boats after ten hours, 769 odometer reading during emissions testing, 621 parabolic car headlights, 1004 parallel/nonparallel roads, 31 parking lot dimensions, A–8 perpendicular/nonperpendicular course headings, 31 price per gallon of gasoline, 850 probability of unclogged route for fire truck, 1140 ramp to loading dock, 523 rate of climb of aircraft, 30 revenue and costs of space travel, 270 round-trip average speed, 394 rowing distance to shore plus running distance to house, 199–200 runway length, 768 runway takeoff distance, 445

sight angle for drivers approaching school, 835 sight angle of road sign on pole, 832–833 speed of car from skid marks, A–78 speed of Caribbean Current, 850 speed of scooter with wheels of known radius, 997 speed of vehicle after selected seconds, 99 speeds of two vehicles traveling at right angles, 526 stopping distance of car based on skid marks, 134 stopping distance of car based on speed, 186 time for boats traveling in opposite directions to be certain distance apart, 214 time for one vehicle to overtake another, 77 time required for coasting boat to stop, 194 total cost for car rental, A–2 tow trucks winching van from ditch, 784 traffic volume, 145, 352 train speed at road crossing, 608 tugboats attempting to free barge stuck on sandbar, 772, 780 tunnel clearance, 1016 two tugboats docking large ship, 784 vectors for three boats traveling in opposite directions, 782 velocity and fuel economy, 291 walking speed of person walking against airport walkway, 850 weight of truck being winched up inclined ramp, 799 width of fighter plane wing along its length, 1155 width of sign over highway, 607 work done to push stranded car, 799

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Precalculus—

CHAPTER CONNECTIONS

Relations, Functions, and Graphs CHAPTER OUTLINE 1.1 Rectangular Coordinates; Graphing Circles

Viewing relations and functions in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For instance, while many business are aware that Internet use is increasing with time, they are ver y interested in the rate of growth, in order to prepare and develop related goods and ser vices. 䊳

This application appears as Exercise 109 in Section 1.4.

and Other Relations 2

1.2 Linear Equations and Rates of Change 19 1.3 Functions, Function Notation, and the Graph of a Function 33

1.4 Linear Functions, Special Forms, and More on Rates of Change 50

1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving 64

1.6 Linear Function Models and Real Data 79

Connections to Calculus

A solid understanding of lines and their equations is fundamental to a study of differential calculus. The Connections to Calculus feature for Chapter 1 reviews these essential concepts and skills, and provides an opportunity for practice in the context of a future calculus course. 1

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Precalculus—

1.1



A. Express a relation in

In everyday life, we encounter a large variety of relationships. For instance, the time it takes us to get to work is related to our average speed; the monthly cost of heating a home is related to the average outdoor temperature; and in many cases, the amount of our charitable giving is related to changes in the cost of living. In each case we say that a relation exists between the two quantities.

mapping notation and ordered pair form B. Graph a relation C. Graph relations on a calculator D. Develop the equation and graph of a circle using the distance and midpoint formulas

A. Relations, Mapping Notation, and Ordered Pairs

Consumer spending (dollars per year)

Figure 1.1 In the most general sense, a relation is simply a correspondence between two sets. Relations can be repreP B sented in many different ways and may even be very Missy April 12 “unmathematical,” like the one shown in Figure 1.1 Jeff Nov 11 between a set of people and the set of their correspondAngie Sept 10 ing birthdays. If P represents the set of people and B Megan Nov 28 represents the set of birthdays, we say that elements of Mackenzie May 7 Michael P correspond to elements of B, or the birthday relation April 14 Mitchell maps elements of P to elements of B. Using what is called mapping notation, we might simply write P S B. From a purely practical standpoint, we note that while it is possible for two different people to share the same birthday, it is quite impossible for the same person to have two different birthdays. Later, this observation will help us mark the difference between a relation and special kind of relation called a function. Figure 1.2 The bar graph in Figure 1.2 is also an 500 example of a relation. In the graph, each year is related to annual consumer spend($411-est) 400 ($375) ing per person on cable and satellite television. As an alternative to mapping or a bar ($281) 300 graph, this relation could also be repre($234) sented using ordered pairs. For example, ($192) 200 the ordered pair (5, 234) would indicate that in 2005, spending per person on 100 cable and satellite TV in the United States averaged $234. When a relation is 3 11 5 7 9 represented using ordered pairs, we say Year (0 → 2000) Source: 2009 Statistical Abstract of the United States, the relation is pointwise-defined. Table 1089 (some figures are estimates) Over a long period of time, we could collect many ordered pairs of the form (t, s), where consumer spending s depends on the time t. For this reason we often call the second coordinate of an ordered pair (in this case s) the dependent variable, with the first coordinate designated as the independent variable. The set of all first coordinates is called the domain of the relation. The set of all second coordinates is called the range. 1–2

2

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Precalculus—

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3

Section 1.1 Rectangular Coordinates; Graphing Circles and Other Relations

EXAMPLE 1

䊳

Expressing a Relation as a Mapping and as a Pointwise-Defined Relation Represent the relation from Figure 1.2 in mapping notation and as a pointwise-defined relation, then state its domain and range.

Solution

䊳

A. You’ve just seen how we can express a relation in mapping notation and ordered pair form

Let t represent the year and s represent consumer spending. The mapping t S s gives the diagram shown. As a pointwisedefined relation we have (3, 192), (5, 234), (7, 281), (9, 375), and (11, 411). The domain is the set {3, 5, 7, 9, 11}; the range is {192, 234, 281, 375, 411}.

t

s

3 5 7 9 11

192 234 281 375 411

Now try Exercises 7 through 12

䊳

For more on this relation, see Exercise 93.

B. The Graph of a Relation Table 1.1 y ⴝ x ⴚ 1 x

y

4

5

2

3

0

1

2

1

4

3

Table 1.2 x ⴝ 円 y円 x

y

2

2

1

1

0

0

1

1

2

2

Relations can also be stated in equation form. The equation y x 1 expresses a relation where each y-value is one less than the corresponding x-value (see Table 1.1). The equation x 冟y冟 expresses a relation where each x-value corresponds to the absolute value of y (see Table 1.2). In each case, the relation is the set of all ordered pairs (x, y) that create a true statement when substituted, and a few ordered pair solutions are shown in the tables for each equation. Relations can be expressed graphically using a recFigure 1.3 tangular coordinate system. It consists of a horizontal y 5 number line (the x-axis) and a vertical number line (the 4 y-axis) intersecting at their zero marks. The point of inter3 QII QI section is called the origin. The x- and y-axes create a 2 flat, two-dimensional surface called the xy-plane and 1 divide the plane into four regions called quadrants. 5 4 3 2 1 1 2 3 4 5 x 1 These are labeled using a capital “Q” (for quadrant) and 2 the Roman numerals I through IV, beginning in the QIII QIV 3 upper right and moving counterclockwise (Figure 1.3). 4 The grid lines shown denote the integer values on each 5 axis and further divide the plane into a coordinate grid, where every point in the plane corresponds to an ordered Figure 1.4 pair. Since a point at the origin has not moved along either y 5 axis, it has coordinates (0, 0). To plot a point (x, y) means we place a dot at its location in the xy-plane. A few of the (4, 3) ordered pairs from y x 1 are plotted in Figure 1.4, where a noticeable pattern emerges—the points seem to (2, 1) lie along a straight line. 5 x If a relation is pointwise-defined, the graph of the 5 (0, 1) relation is simply the plotted points. The graph of a re(2, 3) lation in equation form, such as y x 1, is the set of (4, 5) all ordered pairs (x, y) that are solutions (make the 5 equation true). Solutions to an Equation in Two Variables 1. If substituting x a and y b results in a true equation, the ordered pair (a, b) is a solution and on the graph of the relation. 2. If the ordered pair (a, b) is on the graph of a relation, it is a solution (substituting x a and y b will result in a true equation).

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Precalculus—

4

1–4

CHAPTER 1 Relations, Functions, and Graphs

We generally use only a few select points to determine the shape of a graph, then draw a straight line or smooth curve through these points, as indicated by any patterns formed. EXAMPLE 2

䊳

Graphing Relations Graph the relations y x 1 and x 冟y冟 using the ordered pairs given in Tables 1.1 and 1.2.

Solution

䊳

For y x 1, we plot the points then connect them with a straight line (Figure 1.5). For x 冟y冟, the plotted points form a V-shaped graph made up of two half lines (Figure 1.6). Figure 1.5 5

Figure 1.6 y

y yx1

5

x y

(4, 3) (2, 2)

(2, 1) (0, 0) 5

5

x

5

5

(2, 3)

x

(2, 2)

(0, 1)

5

5

(4, 5)

Now try Exercises 13 through 16 WORTHY OF NOTE As the graphs in Example 2 indicate, arrowheads are used where appropriate to indicate the infinite extension of a graph.

䊳

While we used only a few points to graph the relations in Example 2, they are actually made up of an infinite number of ordered pairs that satisfy each equation, including those that might be rational or irrational. This understanding is an important part of reading and interpreting graphs, and is illustrated for you in Figures 1.7 through 1.10. Figure 1.7

Figure 1.8

y x 1: selected integer values

y x 1: selected rational values

y

y

5

5

5

5

x

5

5

5

5

Figure 1.9

Figure 1.10

y x 1: selected real number values

y x 1: all real number values

y

y

5

5

5

5

5

x

x

5

5

5

x

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Precalculus—

1–5

5


Since there are an infinite number of ordered pairs forming the graph of y x 1, the domain cannot be given in list form. Here we note x can be any real number and write D: x 僆⺢. Likewise, y can be any real number and for the range we have R: y 僆⺢. All of these points together make these graphs continuous, which for our purposes means you can draw the entire graph without lifting your pencil from the paper. Actually, a majority of graphs cannot be drawn using only a straight line or directed line segments. In these cases, we rely on a “sufficient number” of points to outline the basic shape of the graph, then connect the points with a smooth curve. As your experience with graphing increases, this “sufficient number of points” tends to get smaller as you learn to anticipate what the graph of a given relation should look like. In particular, for the linear graph in Figure 1.5 we notice that both the x- and y-variables have an implied exponent of 1. This is in fact a characteristic of linear equations and graphs. In Example 3 we’ll notice that if the exponent on one of the variables is 2 (either x or y is squared ) while the other exponent is 1, the result is a graph called a parabola. If the x-term is squared (Example 3a) the parabola is oriented vertically, as in Figure 1.11, and its highest or lowest point is called the vertex. If the y-term is squared (Example 3c), the parabola is oriented horizontally, as in Figure 1.13, and the leftmost or rightmost point is the vertex. The graphs and equations of other relations likewise have certain identifying characteristics. See Exercises 85 through 92. EXAMPLE 3

䊳

Graphing Relations Graph the following relations by completing the tables given. Then use the graph to state the domain and range of the relation. a. y x2 2x b. y 29 x2 c. x y2

Solution

䊳

For each relation, we use each x-input in turn to determine the related y-output(s), if they exist. Results can be entered in a table and the ordered pairs used to assist in drawing a complete graph. Figure 1.11 a. y ⴝ x2 2x y

x

y

(x, y) Ordered Pairs

4

24

(4, 24)

3

15

(3, 15)

2

8

(2, 8)

1

3

(1, 3)

0

(0, 0)

1

0 1

(1, 1)

2

0

(2, 0)

3

3

(3, 3)

4

8

(4, 8)

(4, 8)

(2, 8) y x2 2x

5

(1, 3)

(3, 3)

(0, 0)

(2, 0)

5

5 2

x

(1, 1)

The resulting vertical parabola is shown in Figure 1.11. Although (4, 24) and (3, 15) cannot be plotted here, the arrowheads indicate an infinite extension of the graph, which will include these points. This “infinite extension” in the upward direction shows there is no largest y-value (the graph becomes infinitely “tall”). Since the smallest possible y-value is 1 [from the vertex (1, 1)], the range is y 1. However, this extension also continues forever in the outward direction as well (the graph gets wider and wider). This means the x-value of all possible ordered pairs could vary from negative to positive infinity, and the domain is all real numbers. We then have D: x 僆⺢ and R: y 1.

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y ⴝ 29 ⴚ x2

b.

Figure 1.12

x

y


4

not real

—

3

0

(3, 0)

2

15

(2, 15)

1

212

(1, 212)

0

3

(0, 3)

1

212

(1, 212)

2

15

(2, 15)

3

0

(3, 0)

4

not real

—

y 兹9 x2

y 5

(1, 2兹2) (2, 兹5)

(0, 3) (1, 2兹2) (2, 兹5)

(3, 0)

(3, 0)

5

5

x

5

The result is the graph of a semicircle (Figure 1.12). The points with irrational coordinates were graphed by estimating their location. Note that when x 6 3 or x 7 3, the relation y 29 x2 does not represent a real number and no points can be graphed. Also note that no arrowheads are used since the graph terminates at (3, 0) and (3, 0). These observations and the graph itself show that for this relation, D: 3 x 3, and R: 0 x 3. c. Similar to x 冟y冟, the relation x y2 is defined only for x 0 since y2 is always nonnegative (1 y2 has no real solutions). In addition, we reason that each positive x-value will correspond to two y-values. For example, given x 4, (4, 2) and (4, 2) are both solutions to x y2. x ⴝ y2

Figure 1.13 y

x

y


2

not real

—

1

not real

—

0

(0, 0)

0 1

1, 1

(1, 1) and (1, 1)

2

12, 12

(2, 12) and (2, 12)

3

13, 13

(3, 13) and (3, 13)

4

2, 2

(4, 2) and (4, 2)

5

x y2 (4, 2)

(2, 兹2) (0, 0) 5

5

5

x

(2, 兹2) (4, 2)

This relation is a horizontal parabola, with a vertex at (0, 0) (Figure 1.13). The graph begins at x 0 and extends infinitely to the right, showing the domain is x 0. Similar to Example 3a, this “infinite extension” also extends in both the upward and downward directions and the y-value of all possible ordered pairs could vary from negative to positive infinity. We then have D: x 0 and R: y 僆⺢. B. You’ve just seen how we can graph relations


䊳

C. Graphing Relations on a Calculator For relations given in equation form, the TABLE feature of a graphing calculator can be used to compute ordered pairs, and the GRAPH feature to draw the related graph. To use these features, we first solve the equation for the variable y (write y in terms of x), then enter the right-hand expression on the calculator’s Y= (equation editor) screen.

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7

We can then select either the GRAPH feature, or set-up, create, and use the TABLE feature. We’ll illustrate here using the relation ⫺2x ⫹ y ⫽ 3. 1. Solve for y in terms of x. ⫺2x ⫹ y ⫽ 3 given equation y ⫽ 2x ⫹ 3 add 2x to each side 2. Enter the equation. Press the Y= key to access the equation editor, then enter 2x ⫹ 3 as Y1 (see Figure 1.14). The calculator automatically highlights the equal sign, showing that equation Y1 is now active. If there are other equations on the screen, you can either them or deactivate them by moving the cursor to overlay the equal sign and pressing . 3. Use the TABLE or GRAPH . To set up the table, we use the keystrokes 2nd (TBLSET). For this exercise, we’ll put the table in the “Indpnt: Auto Ask” mode, which will have the calculator automatically generate the input and output values. In this mode, we can tell the calculator where to start the inputs (we chose TblStart 3), and have the calculator produce the input values using any increment desired (we choose Tbl 1). See Figure 1.15A. Access the table using 2nd GRAPH (TABLE), and the table resulting from this setup is shown in Figure 1.15B. Notice that all ordered pairs satisfy the equation y ⫽ 2x ⫹ 3, or “y is twice x increased by 3.”

Figure 1.14

CLEAR

ENTER

Figure 1.15A

WINDOW

Figure 1.15B

Since much of our graphical work is centered at (0, 0) on the coordinate grid, the calculator’s default settings for the standard viewing are [⫺10, 10] for both x and y (Figure 1.16). The Xscl and Yscl values give the scale used on each axis, and indicate here that each “tick mark” will be 1 unit apart. To graph the line in this window, we can use the ZOOM key and select 6:ZStandard (Figure 1.17), which resets the window to these default settings and automatically graphs the line (Figure 1.18). WINDOW

Figure 1.16

Figure 1.18

Figure 1.17

10

⫺10

10

⫺10

In addition to using the calculator’s TABLE feature to find ordered pairs for a given graph, we can also use the calculator’s TRACE feature. As the name implies, this feature allows us to “trace” along the graph by moving a cursor to the left and right using the arrow keys. The calculator displays the coordinates of the cursor’s location each time it moves. After pressing the TRACE key, the marker appears automatically and as you move it to the left or right, the current coordinates are shown at the bottom

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of the screen (Figure 1.19). While not very “pretty,” (⫺0.8510638, 1.2978723) is a point on this line (rounded to seven decimal places) and satisfies its equation. The calculator is displaying these decimal values because the viewing screen is exactly 95 pixels wide, 47 pixels to the left of the y-axis, and 47 pixels to the right. This means that each time you press the left or right arrow, the x-value changes by 1/47th—which is not a nice round number. To have the calculator TRACE through “friendlier” values, we can use the ZOOM 4:ZDecimal feature, which sets Xmin ⫽ ⫺4.7 and Xmax ⫽ 4.7, or ZOOM 8:ZInteger, which sets Xmin ⫽ ⫺47 and Xmax ⫽ 47. Let’s use the ZOOM 4:ZDecimal option here, noting the calculator automatically regraphs the line. Pressing the TRACE key once again and moving the marker shows that more “friendly” ordered pair solutions are displayed (Figure 1.20). Other methods for finding a friendly window are discussed later in this section. EXAMPLE 4

䊳

Figure 1.19 10

⫺10

10

⫺10

Figure 1.20 3.1

⫺4.7

4.7

⫺3.1

Graphing a Relation Using Technology Use a calculator to graph 2x ⫹ 3y ⫽ ⫺6. Then use the TABLE feature to determine the value of y when x ⫽ 0, and the value of x when y ⫽ 0. Write each result in ordered pair form.

Solution

䊳

We begin by solving the equation for y, so we can enter it on the given equation 2x ⫹ 3y ⫽ ⫺6 3y ⫽ ⫺2x ⫺ 6 subtract 2x (isolate the y-term) ⫺2 x ⫺ 2 divide by 3 y⫽ 3

Y=

screen.

⫺2 x ⫺ 2 on the Y= screen and using ZOOM 6:ZStandard produces 3 the graph shown. Using the TABLE and scrolling as needed, shows that when x ⫽ 0, y ⫽ ⫺2, and when y ⫽ 0, x ⫽ ⫺3. As ordered pairs we have 10, ⫺22 and 1⫺3, 02 . Entering y ⫽

10

⫺10

C. You’ve just seen how we can graph a relation using a calculator

10

⫺10


䊳

D. The Equation and Graph of a Circle Using the midpoint and distance formulas, we can develop the equation of another important relation, that of a circle. As the name suggests, the midpoint of a line segment is located halfway between the endpoints. On a standard number line, the midpoint of the line segment with endpoints 1 and 5 is 3, but more important, note that 3 is the

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6 15 3. This observation 2 2 can be extended to find the midpoint between any two points (x1, y1) and (x2, y2) in the xy-plane. We simply find the average distance between the x-coordinates and the average distance between the y-coordinates. average distance (from zero) of 1 unit and 5 units:

The Midpoint Formula

Given any line segment with endpoints P1 1x1, y1 2 and P2 1x2, y2 2 , the midpoint M is given by M: a

x1 x2 y1 y2 , b 2 2

The midpoint formula can be used in many different ways. Here we’ll use it to find the coordinates of the center of a circle. 䊳

EXAMPLE 5

䊳

Solution

Using the Midpoint Formula

The diameter of a circle has endpoints at P1 13, 22 and P2 15, 42 . Use the midpoint formula to find the coordinates of the center, then plot this point. x1 x2 y1 y2 , b 2 2 3 5 2 4 , b M: a 2 2 2 2 M: a , b 11, 12 2 2

y 5

Midpoint: a

P2

(1, 1) 5

5

x

P1 5

The center is at (1, 1), which we graph directly on the diameter as shown. Now try Exercises 29 through 38 䊳

The Distance Formula

Figure 1.21 y

c

(x2, y2)

b

x

a

(x1, y1)

(x2, y1)

P2

P1

a 兩 x2 x1兩

b 兩 y2 y1兩

d

In addition to a line segment’s midpoint, we are often interested in the length of the segment. For any two points (x1, y1) and (x2, y2) not lying on a horizontal or vertical line, a right triangle can be formed as in Figure 1.21. Regardless of the triangle’s orientation, the length of side a (the horizontal segment or base of the triangle) will have length 冟x2 x1冟 units, with side b (the vertical segment or height) having length 冟y2 y1冟 units. From the Pythagorean theorem (Appendix A.6), we see that c2 a2 b2 corresponds to c2 1 冟x2 x1冟 2 2 1 冟y2 y1冟 2 2. By taking the square root of both sides we obtain the length of the hypotenuse, which is identical to the distance between these two points: c 21x2 x1 2 2 1y2 y1 2 2. The result is called the distance formula, although it’s most often written using d for distance, rather than c. Note the absolute value bars are dropped from the formula, since the square of any quantity is always nonnegative. This also means that either point can be used as the initial point in the computation. The Distance Formula

Given any two points P1 1x1, y1 2 and P2 1x2, y2 2, the straight line distance d between them is

d 21x2 x1 2 2 1y2 y1 2 2

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EXAMPLE 6

䊳

Using the Distance Formula Use the distance formula to find the diameter of the circle from Example 5.

Solution

䊳

For 1x1, y1 2 13, 22 and 1x2, y2 2 15, 42, the distance formula gives d

21x2 x1 2 2 1y2 y1 2 2 2 3 5 132 4 2 34 122 4 2 282 62 1100 10

The diameter of the circle is 10 units long. Now try Exercises 39 through 48

䊳

A circle can be defined as the set of all points in a plane that are a fixed distance called the radius, from a fixed point called the center. Since the definition involves distance, we can construct the general equation of a circle using the distance formula. Assume the center has coordinates (h, k), and let (x, y) represent any point on the graph. The distance between these points is equal to the radius r, and the distance formula yields: 21x h2 2 1y k2 2 r. Squaring both sides gives the equation of a circle in standard form: 1x h2 2 1y k2 2 r2. The Equation of a Circle A circle of radius r with center at (h, k) has the equation 1x h2 2 1y k2 2 r2

If h 0 and k 0, the circle is centered at (0, 0) and the graph is a central circle with equation x2 y2 r2. At other values for h or k, the center is at (h, k) with no change in the radius. Note that an open dot is used for the center, as it’s actually a point of reference and not a part of the graph.

y

Circle with center at (h, k) r

k

(x, y)

(h, k)

Central circle

(x h)2 (y k)2 r2 r

(x, y)

(0, 0)

h

x

x2 y2 r2

EXAMPLE 7

䊳

Finding the Equation of a Circle in Standard Form

Solution

䊳

Since the center is at (0, 1) we have h 0, k 1, and r 4. Using the standard form 1x h2 2 1y k2 2 r2 we obtain

Find the equation of a circle with center 10, 1) and radius 4. 1x 02 2 3y 112 4 2 42 x2 1y 12 2 16

substitute 0 for h, 1 for k, and 4 for r simplify

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11

The graph of x2 1y 12 2 16 is shown in the figure. y (0, 3) Circle

r4 (4, 1)

Center: (0, 1) Radius: r 4 Diameter: 2r 8

x (4, 1)

(0, 1)

(0, 5)


䊳

The graph of a circle can be obtained by first identifying the coordinates of the center and the length of the radius from the equation in standard form. After plotting the center point, we count a distance of r units left and right of center in the horizontal direction, and up and down from center in the vertical direction, obtaining four points on the circle. Neatly graph a circle containing these four points. EXAMPLE 8

䊳

Graphing a Circle

Solution

䊳

Comparing the given equation with the standard form, we find the center is at 12, 32 and the radius is r 213 ⬇ 3.5.

Graph the circle represented by 1x 22 2 1y 32 2 12. Clearly label the center and radius.

standard form

S

S

S

1x h2 2 1y k2 2 r2

1x 22 2 1y 32 2 12 h 2 k 3 h2 k 3

given equation

r2 12 r 112 2 13 ⬇ 3.5

radius must be positive

Plot the center 12, 32 and count approximately 3.5 units in the horizontal and vertical directions. Complete the circle by freehand drawing or using a compass. The graph shown is obtained. y Some coordinates are approximate

Circle (2, 0.5) x

r ~ 3.5 (1.5, 3)

(2, 3)

(5.5, 3)

Center: (2, 3) Radius: r 2兹3 Endpoints of horizontal diameter (2 2兹3, 3) and (2 2兹3, 3) Endpoints of vertical diameter (2, 3 2兹3) and (2, 3 2兹3)

(2, 6.5)


䊳

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In Example 8, note the equation is composed of binomial squares in both x and y. By expanding the binomials and collecting like terms, we can write the equation of the circle in general form:

WORTHY OF NOTE After writing the equation in standard form, it is possible to end up with a constant that is zero or negative. In the first case, the graph is a single point. In the second case, no graph is possible since roots of the equation will be complex numbers. These are called degenerate cases. See Exercise 105.

EXAMPLE 9

1x 22 2 1y 32 2 12

x 4x 4 y 6y 9 12 x2 y2 4x 6y 1 0 2

2

standard form expand binomials combine like terms—general form

For future reference, observe the general form contains a sum of second-degree terms in x and y, and that both terms have the same coefficient (in this case, “1”). Since this form of the equation was derived by squaring binomials, it seems reasonable to assume we can go back to the standard form by creating binomial squares in x and y. This is accomplished by completing the square. 䊳

Finding the Center and Radius of a Circle Find the center and radius of the circle with equation x2 y2 2x 4y 4 0. Then sketch its graph and label the center and radius.

Solution

䊳

To find the center and radius, we complete the square in both x and y. x2 y2 2x 4y 4 0 1x2 2x __ 2 1y2 4y __ 2 4 1x2 2x 12 1y2 4y 42 4 1 4 adds 1 to left side

given equation group x-terms and y-terms; add 4 complete each binomial square

add 1 4 to right side

adds 4 to left side

1x 12 2 1y 22 2 9

factor and simplify

The center is at 11, 22 and the radius is r 19 3. (1, 5)

(4, 2)

y

r3 (1, 2)

(2, 2) Circle Center: (1, 2) Radius: r 3

x

(1, 1)


EXAMPLE 10

䊳

䊳

Applying the Equation of a Circle To aid in a study of nocturnal animals, some naturalists install a motion detector near a popular watering hole. The device has a range of 10 m in any direction. Assume the water hole has coordinates (0, 0) and the device is placed at 12, 12 . a. Write the equation of the circle that models the maximum effective range of the device. b. Use the distance formula to determine if the device will detect a badger that is approaching the water and is now at coordinates 111, 52 .

y 5

10

5

x

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Solution

䊳

13

a. Since the device is at (2, ⫺1) and the radius (or reach) of detection is 10 m, any movement in the interior of the circle defined by 1x ⫺ 22 2 ⫹ 1y ⫹ 12 2 ⫽ 102 will be detected. b. Using the points (2, ⫺1) and (11, ⫺5) in the distance formula yields: d⫽ ⫽ ⫽ ⫽ ⫽

21x2 ⫺ x1 2 2 ⫹ 1y2 ⫺ y1 2 2 2111 ⫺ 22 2 ⫹ 3⫺5 ⫺ 1⫺12 4 2 292 ⫹ 1⫺42 2 181 ⫹ 16 197 ⬇ 9.85

distance formula substitute given values simplify compute squares result

Since 9.85 6 10, the badger is within range of the device and will be detected. Now try Exercises 95 through 100

䊳

When using a graphing calculator to study circles, it’s important to note that most standard viewing windows have the x- and y-values preset at [⫺10, 10] even though the calculator screen is not square. This tends to compress the y-values and give a skewed image of the graph. If the circle appears oval in shape, use ZOOM 5:ZSquare to obtain the correct perspective. Graphing calculators can produce the graph of a circle in various ways, and the choice of method simply depends on what you’d like to accomplish. To simply view the graph or compare two circular graphs, the DRAW command is used. From the home screen press: 2nd PRGM (DRAW) 9:Circle(. This generates the “Circle(” command, with the left parentheses indicating we need to supply three inputs, separated by commas. These inputs are the x-coordinate of the center, the y-coordinate of the center, and the radius of the circle. For the circle defined by the equation 1x ⫺ 32 2 ⫹ 1y ⫹ 22 2 ⫽ 49, we know the center is at 13, ⫺22 and the radius is 7 units. The resulting command and graph are shown in Figures 1.22 and 1.23. While the DRAW command will graph any circle, we are unable to use the TRACE or CALC commands to interact with the graph. To make these features available, we must first solve for x in terms of y, as we did previously (the 1:ClrDraw command is used to clear the graph). Consider the relation x2 ⫹ y2 ⫽ 25, which we know is the equation of a circle centered at (0, 0) with radius r ⫽ 5. x2 ⫹ y2 ⫽ 25

original equation

y ⫽ 25 ⫺ x y ⫽ ⫾ 225 ⫺ x2 2

2

isolate y 2 solve for y

Note that we can separate this result into two parts, enabling the calculator to graph Y1 ⫽ 225 ⫺ x2 (giving the “upper half” of the circle), and Y2 ⫽ ⫺ 225 ⫺ x2 (giving the “lower half”). Enter these on the Y= screen (note that Y2 ⫽ ⫺Y1 can be used instead of reentering the entire expression: VARS ). If we graph Y1 and Y2 on the standard screen, the result appears more oval than circular (Figure 1.24). Using the ZOOM 5:ZSquare option, the tick marks become equally spaced on both axes (Figure 1.25). ENTER

Figure 1.22, 1.23 10

⫺15.2

15.2

⫺10

Figure 1.24

Figure 1.25

10

10

⫺10

10

⫺10

⫺15.2

15.2

⫺10

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Although it is a much improved graph, the circle does not appear “closed” as the calculator lacks sufficient pixels to show the proper curvature. A second alternative is to manually set a “friendly” window. Using Xmin 9.4, Xmax 9.4, Ymin 6.2, and Ymax 6.2 will generate a better graph due to the number of pixels available. Note that we can jump between the upper and lower halves of the circle using the up or down arrows. See Exercises 101 and 102.

D. You’ve just seen how we can develop the equation and graph of a circle using the distance and midpoint formulas

1.1 EXERCISES 䊳



䊳

1. If a relation is defined by a set of ordered pairs, the domain is the set of all ________ components, the range is the set of all ________ components.

4. For x2 y2 25, the center of the circle is at ________ and the length of the radius is ________ units. The graph is called a ________ circle.

2. For the equation y x 5 and the ordered pair (x, y), x is referred to as the input or ________ variable, while y is called the ________ or dependent variable.

5. Discuss/Explain how to find the center and radius of the circle defined by the equation x2 y2 6x 7. How would this circle differ from the one defined by x2 y2 6y 7?

3. A circle is defined as the set of all points that are an equal distance, called the ________, from a given point, called the ________.

6. In Example 3(b) we graphed the semicircle defined by y 29 x2. Discuss how you would obtain the equation of the full circle from this equation, and how the two equations are related.


Represent each relation in mapping notation, then state the domain and range.

GPA

7.

9. {(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)}

4.00 3.75 3.50 3.25 3.00 2.75 2.50 2.25 2.00 0

10. {(2, 4), (3, 5), (1, 3), (4, 5), (2, 3)} 11. {(4, 0), (1, 5), (2, 4), (4, 2), (3, 3)} 12. {(1, 1), (0, 4), (2, 5), (3, 4), (2, 3)}

1

2

3

4

Complete each table using the given equation. For Exercises 15, 16, 21, and 22, each input may correspond to two outputs (be sure to find both if they exist). Use these points to graph the relation. For Exercises 17 through 24, also state the domain and range.

5

Year in college

Efficiency rating

8.

95 90 85 80 75 70 65 60 55 0

State the domain and range of each pointwise-defined relation.

2 13. y x 1 3 x

1

2

3

4

Month

5

6

y

5 14. y x 3 4 x

6

8

3

4

0

0

3

4

6

8

8

10

y

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15. x 2 冟y冟

16. 冟y 1冟 x

x

x

y

Use a graphing calculator to graph the following relations. Then use the TABLE feature to determine the value of y when x ⴝ 0, and the value(s) of x when y ⴝ 0, and write the results in ordered pair form.

y

2

0

0

1

1

3

3

5

26. x 2y 6

6

6

27. y x2 4x

7

7

28. y x2 2x 3

17. y x2 1 x

25. 2x 5y 10

18. y x2 3

y

x

y

Find the midpoint of each segment with the given endpoints.

3

2

29. (1, 8), (5, 6)

2

1

0

0

30. (5, 6), (6, 8)

2

1

31. (4.5, 9.2), (3.1, 9.8)

3

2

32. (5.2, 7.1), (6.3, 7.1)

4

3

19. y 225 x2 x

20. y 2169 x2

y

x

y

4

12

3

5

0

0

2

3

3

5

4

12

21. x 1 y

2

x

y

1 0

2

1

1.25

2

1

7

9

2

2

1

1

0

0

1

4

2

7

3

36.

1 2 3 4 5 x

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

1 2 3 4 5 x

Find the center of each circle with the diameter shown.

37.

24. y 1x 12 3 x

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

y

4

y

35.

x

5

x

Find the midpoint of each segment.

22. y 2 x 2

3

3 1 3 5 34. a , b, a , b 4 3 8 6

2

10

23. y 2x 1

1 3 1 2 33. a , b, a , b 5 3 10 4

y

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

1 2 3 4 5 x

38.

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

1 2 3 4 5 x

39. Use the distance formula to find the length of the line segment in Exercise 35. 40. Use the distance formula to find the length of the line segment in Exercise 36. 41. Use the distance formula to find the length of the diameter of the circle in Exercise 37. 42. Use the distance formula to find the length of the diameter of the circle in Exercise 38.

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In Exercises 43 to 48, three points that form the vertices of a triangle are given. Use the distance formula to determine if any of the triangles are right triangles (the three sides satisfy the Pythagorean Theorem a2 ⴙ b2 ⴝ c2).

Write each equation in standard form to find the center and radius of the circle. Then sketch the graph.

73. x2 y2 10x 12y 4 0 74. x2 y2 6x 8y 6 0

43. (3, 7), (2, 2), (5, 5)

75. x2 y2 10x 4y 4 0

44. (7, 0), (1, 0), (7, 4)

76. x2 y2 6x 4y 12 0

45. (4, 3), (7, 1), (3, 2)

77. x2 y2 6y 5 0

46. (5, 2), (0, 3), (4, 4)

78. x2 y2 8x 12 0

47. (3, 2), (1, 5), (6, 4)

79. x2 y2 4x 10y 18 0

48. (0, 0), (5, 2), (2, 5)

80. x2 y2 8x 14y 47 0

Find the equation of a circle satisfying the conditions given, then sketch its graph.

81. x2 y2 14x 12 0 82. x2 y2 22y 5 0

49. center (0, 0), radius 3

83. 2x2 2y2 12x 20y 4 0


84. 3x2 3y2 24x 18y 3 0

51. center (5, 0), radius 13 52. center (0, 4), radius 15 53. center (4, 3), radius 2 54. center (3, 8), radius 9 55. center (7, 4), radius 17 56. center (2, 5), radius 16 57. center (1, 2), diameter 6

In this section we looked at characteristics of equations that generated linear graphs, and graphs of parabolas and circles. Use this information and ordered pairs of your choosing to match the eight graphs given with their corresponding equation (two of the equations given have no matching graph).

a. y x2 6x c. x2 y 9

b. x2 1y 32 2 36 d. 3x 4y 12

3 x4 2

f. 1x 12 2 1y 22 2 49

58. center (2, 3), diameter 10

e. y

59. center (4, 5), diameter 4 13

g. 1x 32 2 y2 16 h. 1x 12 2 1y 22 2 9

60. center (5, 1), diameter 4 15 61. center at (7, 1), graph contains the point (1, 7) 62. center at (8, 3), graph contains the point (3, 15)

i. 4x 3y 12

85.

63. center at (3, 4), graph contains the point (7, 9) 64. center at (5, 2), graph contains the point (1, 3)

ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

65. diameter has endpoints (5, 1) and (5, 7) 66. diameter has endpoints (2, 3) and (8, 3) Identify the center and radius of each circle, then graph. Also state the domain and range of the relation.

67. 1x 22 2 1y 32 2 4 68. 1x 52 2 1y 12 2 9

69. 1x 12 2 1y 22 2 12 70. 1x 72 2 1y 42 2 20 71. 1x 42 2 y2 81 72. x2 1y 32 2 49

y 10 8 6 4 2

87.

86.

y

2 4 6 8 10 x

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x

10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

j. 6x y x2 9

88.

2 4 6 8 10 x

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x

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89.

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

䊳

2 4 6 8 10 x

90.

91.

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x

92.

2 4 6 8 10 x

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x


93. Spending on Cable and Satellite TV: s ⴝ 29t ⴙ 96

94. Radius of a circumscribed circle: r ⴝ

The data from Example 1 is closely modeled by the formula shown, where t represents the year (t 0 corresponds to the year 2000) and s represents the average amount spent per person, per year in the United States. (a) List five ordered pairs for this relation using t 3, 5, 7, 9, 11. Does the model give a good approximation of the actual data? (b) According to the model, what will be the average amount spent on cable and satellite TV in the year 2013? (c) According to the model, in what year will annual spending surpass $500? (d) Use the table to graph this relation by hand. 䊳

17


The radius r of a circle circumscribed around a square is found by using the formula given, where A is the area of the square. Solve the formula for A and use the result to find the area of the square shown.

A B2 y

(5, 0) x

APPLICATIONS

95. Radar detection: A luxury liner is located at map coordinates (5, 12) and has a radar system with a range of 25 nautical miles in any direction. (a) Write the equation of the circle that models the range of the ship’s radar, and (b) use the distance formula to determine if the radar can pick up the liner’s sister ship located at coordinates (15, 36). 97. Inscribed circle: Find the equation for both the red and blue circles, then find the area of the region shaded in blue.

y

(2, 0) x

99. Radio broadcast range: Two radio stations may not use the same frequency if their broadcast areas overlap. Suppose station KXRQ has a broadcast area bounded by x2 y2 8x 6y 0 and WLRT has a broadcast area bounded by x2 y2 10x 4y 0. Graph the circle representing each broadcast area on the same grid to determine if both stations may broadcast on the same frequency.

96. Earthquake range: The epicenter (point of origin) of a large earthquake was located at map coordinates (3, 7), with the quake being felt up to 12 mi away. (a) Write the equation of the circle that models the range of the earthquake’s effect. (b) Use the distance formula to determine if a person living at coordinates (13, 1) would have felt the quake. 98. Inscribed triangle: The area of an equilateral triangle inscribed in a circle is given 313 2 r, by the formula A 4 where r is the radius of the circle. Find the area of the equilateral triangle shown.

y (3, 4)

x

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100. Radio broadcast range: The emergency radio broadcast system is designed to alert the population by relaying an emergency signal to all points of the country. A signal is sent from a station whose broadcast area is bounded by x2 ⫹ y2 ⫽ 2500 (x and y in miles) and the signal is picked up and relayed by a transmitter with range 1x ⫺ 202 2 ⫹ 1y ⫺ 302 2 ⫽ 900. Graph the circle representing each broadcast area on the same grid to determine the greatest distance from the original station that this signal can be received. Be sure to scale the axes appropriately.

䊳

101. Graph the circle defined by x2 ⫹ y2 ⫽ 36 using a friendly window, then use the TRACE feature to find the value of y when x ⫽ 3.6. Now find the value of y when x ⫽ 4.8. Explain why the values seem “interchangeable.” 102. Graph the circle defined by 1x ⫺ 32 2 ⫹ y2 ⫽ 16 using a friendly window, then use the TRACE feature to find the value of the y-intercepts. Show you get the same intercept by computation.


103. Although we use the word “domain” extensively in mathematics, it is also commonly seen in literature and heard in everyday conversation. Using a collegelevel dictionary, look up and write out the various meanings of the word, noting how closely the definitions given are related to its mathematical use. 104. Consider the following statement, then determine whether it is true or false and discuss why. A graph will exhibit some form of symmetry if, given a point that is h units from the x-axis, k units from the y-axis, and d units from the origin, there is a second point on the graph that is a like distance from the origin and each axis.

䊳

1–18


105. When completing the square to find the center and radius of a circle, we sometimes encounter a value for r2 that is negative or zero. These are called degenerate cases. If r2 6 0, no circle is possible, while if r2 ⫽ 0, the “graph” of the circle is simply the point (h, k). Find the center and radius of the following circles (if possible). a. x2 ⫹ y2 ⫺ 12x ⫹ 4y ⫹ 40 ⫽ 0 b. x2 ⫹ y2 ⫺ 2x ⫺ 8y ⫺ 8 ⫽ 0 c. x2 ⫹ y2 ⫺ 6x ⫺ 10y ⫹ 35 ⫽ 0


106. (Appendix A.2) Evaluate/Simplify the following expressions. a.

x2x5 x3

b. 33 ⫹ 32 ⫹ 31 ⫹ 30 ⫹ 3⫺1 c. 125⫺3 1

2

d. 273 e. (2m3n)2

f. 15x2 0 ⫹ 5x0

107. (Appendix A.3) Solve the following equation. x 1 5 ⫹ ⫽ 3 4 6 108. (Appendix A.4) Solve x2 ⫺ 27 ⫽ 6x by factoring. 109. (Appendix A.6) Solve 1 ⫺ 1n ⫹ 3 ⫽ ⫺n and check solutions by substitution. If a solution is extraneous, so state.

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Linear Equations and Rates of Change


A. Graph linear equations

B.

C. D. E.

using the intercept method Find the slope of a line and interpret it as a rate of change Graph horizontal and vertical lines Identify parallel and perpendicular lines Apply linear equations in context

In preparation for sketching graphs of other equations, we’ll first look more closely at the characteristics of linear graphs. While linear graphs are fairly simple models, they have many substantive and meaningful applications. For instance, most of us are aware that satellite and cable TV have been increasFigure 1.26 ing in popularity since they were first 500 introduced. A close look at Figure 1.2 from ($411-est) 400 Section 1.1 reveals that spending on these ($375) forms of entertainment increased from ($281) 300 $192 per person per year in 2003 to $281 ($234) in 2007 (Figure 1.26). From an investor’s ($192) 200 or a producer’s point of view, there is a very high interest in the questions, “How 100 fast are sales increasing? Can this relationship be modeled mathematically to help 3 11 5 7 9 predict sales in future years?” Answers to Year (0 → 2000) these and other questions are precisely Source: 2009 Statistical Abstract of the United States, what our study in this section is all about. Table 1089 (some figures are estimates) Consumer spending (dollars per year)

1.2

A. The Graph of a Linear Equation A linear equation can be identified using these three tests: 1. the exponent on any variable is one, 2. no variable occurs in a denominator, and 3. no two variables are multiplied together. The equation 3y 9 is a linear equation in one variable, while 2x 3y 12 and y 23 x 4 are linear equations in two variables. In general, we have the following definition: Linear Equations A linear equation is one that can be written in the form ax by c where a, b, and c are real numbers, with a and b not simultaneously equal to zero. As in Section 1.1, the most basic method for graphing a line is to simply plot a few points, then draw a straight line through the points. EXAMPLE 1

䊳

Graphing a Linear Equation in Two Variables Graph the equation 3x 2y 4 by plotting points.

Solution

WORTHY OF NOTE If you cannot draw a straight line through the plotted points, a computational error has been made. All points satisfying a linear equation lie on a straight line.

1–19

䊳

Selecting x 2, x 0, x 1, and x 4 as inputs, we compute the related outputs and enter the ordered pairs in a table. The result is x input

y output

2

5

0

2

1

1 2

4

4

y

(2, 5)

5

(0, 2)

(x, y) ordered pairs 12, 52

(1, q) 5

5

10, 22 11, 12 2

14, 42

x

(4, 4) 5

Now try Exercises 7 through 12 䊳 19

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Notice that the line in Example 1 crosses the y-axis at (0, 2), and this point is called the y-intercept of the line. In general, y-intercepts have the form (0, y). Although difficult to see graphically, substituting 0 for y and solving for x shows this line crosses the x-axis at ( 43, 0) and this point is called the x-intercept. In general, x-intercepts have the form (x, 0). The x- and y-intercepts are usually easier to calculate than other points (since y 0 or x 0, respectively) and we often graph linear equations using only these two points. This is called the intercept method for graphing linear equations. The Intercept Method 1. Substitute 0 for x and solve for y. This will give the y-intercept (0, y). 2. Substitute 0 for y and solve for x. This will give the x-intercept (x, 0). 3. Plot the intercepts and use them to graph a straight line.

EXAMPLE 2

䊳

Graphing Lines Using the Intercept Method Graph 3x 2y 9 using the intercept method.

Solution

䊳

Substitute 0 for x (y-intercept) 3102 2y 9 2y 9 9 y 2 9 a0, b 2

Substitute 0 for y (x-intercept) 3x 2102 9 3x 9 x3 (3, 0)

5

y 3x 2y 9

冢0, t 冣

(3, 0) 5

5

x

5

A. You’ve just seen how we can graph linear equations using the intercept method


䊳

B. The Slope of a Line and Rates of Change After the x- and y-intercepts, we next consider the slope of a line. We see applications of this concept in many diverse areas, including the grade of a highway (trucking), the pitch of a roof (carpentry), the climb of an airplane (flying), the drainage of a field (landscaping), and the slope of a mountain (parks Figure 1.27 y and recreation). While the general concept is an (x2, y2) intuitive one, we seek to quantify the concept (as- y2 sign it a numeric value) for purposes of comparison and decision making. In each of the preceding y2 y1 examples (grade, pitch, climb, etc.), slope is a rise measure of “steepness,” as defined by the ratio vertical change (x1, y1) . Using a line segment through horizontal change y1 arbitrary points P1 1x1, y1 2 and P2 1x2, y2 2 , x2 x1 run we can create the right triangle shown in Figx ure 1.27 to help us quantify this relationship. The x2 x1 figure illustrates that the vertical change or the

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Section 1.2 Linear Equations and Rates of Change

21

change in y (also called the rise) is simply the difference in y-coordinates: y2 y1. The horizontal change or change in x (also called the run) is the difference in x-coordinates: x2 x1. In algebra, we typically use the letter “m” to represent slope, y y change in y giving m x22 x11 as the change in x. The result is called the slope formula. WORTHY OF NOTE

The Slope Formula

Given two points P1 1x1, y1 2 and P2 1x2, y2 2 , the slope of the nonvertical line through P1 and P2 is

While the original reason that “m” was chosen for slope is uncertain, some have speculated that it was because in French, the verb for “to climb” is monter. Others say it could be due to the “modulus of slope,” the word modulus meaning a numeric measure of a given property, in this case the inclination of a line.

y2 y1 x2 x1 where x2 x1. m

Actually, the slope value does much more than quantify the slope of a line, it expresses a rate of change between the quantities measured along each axis. In ¢y change in y applications of slope, the ratio change in x is symbolized as ¢x . The symbol ¢ is the Greek letter delta and has come to represent a change in some quantity, and the ¢y notation m ¢x is read, “slope is equal to the change in y over the change in x.” Interpreting slope as a rate of change has many significant applications in college algebra and beyond. EXAMPLE 3

䊳

Using the Slope Formula ¢y Find the slope of the line through the given points, then use m to find an ¢x additional point on the line. a. (2, 1) and (8, 4) b. (2, 6) and (4, 2)

Solution

䊳

a. For P1 12, 12 and P2 18, 42 , b. y2 y1 m x2 x1 41 82 3 1 6 2 The slope of this line is 12. ¢y 1 Using , we note that y ¢x 2 increases 1 unit (the y-value is positive), as x increases 2 units. Since (8, 4) is known to be on the line, the point 18 2, 4 12 110, 52 must also be on the line.

For P1 12, 62 and P2 14, 22, y2 y1 m x2 x1 26 4 122 4 2 6 3 The slope of this line is 2 3 . ¢y 2 Using , we note that y ¢x 3 decreases 2 units (the y-value is negative), as x increases 3 units. Since (4, 2) is known to be on the line, the point 14 3, 2 22 17, 02 must also be on the line. Now try Exercises 33 through 40 䊳

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䊳

CAUTION

When using the slope formula, try to avoid these common errors. 1. The order that the x- and y-coordinates are subtracted must be consistent, y ⫺ y y ⫺ y since x22 ⫺ x11 ⫽ x21 ⫺ x12. 2. The vertical change (involving the y-values) always occurs in the numerator: y2 ⫺ y1 x2 ⫺ x1 x2 ⫺ x1 ⫽ y2 ⫺ y1 . 3. When x1 or y1 is negative, use parentheses when substituting into the formula to prevent confusing the negative sign with the subtraction operation.

EXAMPLE 4

䊳

Interpreting the Slope Formula as a Rate of Change Jimmy works on the assembly line for an auto parts remanufacturing company. By 9:00 A.M. his group has assembled 29 carburetors. By 12:00 noon, they have completed 87 carburetors. Assuming the relationship is linear, find the slope of the line and discuss its meaning in this context.

Solution WORTHY OF NOTE Actually, the assignment of (t1, c1) to (9, 29) and (t2, c2) to (12, 87) was arbitrary. The slope ratio will be the same as long as the order of subtraction is the same. In other words, if we reverse this assignment and use 1t1, c1 2 ⫽ 112, 872 and 1t2, c2 2 ⫽ 19, 292 , we have ⫺ 87 ⫺58 58 m ⫽ 29 9 ⫺ 12 ⫽ ⫺3 ⫽ 3 .

䊳

First write the information as ordered pairs using c to represent the carburetors assembled and t to represent time. This gives 1t1, c1 2 ⫽ 19, 292 and 1t2, c2 2 ⫽ 112, 872. The slope formula then gives: c2 ⫺ c1 ¢c 87 ⫺ 29 ⫽ ⫽ ¢t t2 ⫺ t1 12 ⫺ 9 58 ⫽ or 19.3 3 carburetors assembled Here the slope ratio measures , and we see that Jimmy’s group can hours assemble 58 carburetors every 3 hr, or about 1913 carburetors per hour. Now try Exercises 41 through 44 䊳

Positive and Negative Slope If you’ve ever traveled by air, you’ve likely heard the announcement, “Ladies and gentlemen, please return to your seats and fasten your seat belts as we begin our descent.” For a time, the descent of the airplane follows a linear path, but the slope of the line is negative since the altitude of the plane is decreasing. Positive and negative slopes, as well as the rate of change they represent, are important characteristics of linear graphs. In Example 3(a), the slope was a positive number (m 7 0) and the line will slope upward from left to right since the y-values are increasing. If m 6 0 as in Example 3(b), the slope of the line is negative and the line slopes downward as you move left to right since y-values are decreasing.

m ⬎ 0, positive slope y-values increase from left to right

m ⬍ 0, negative slope y-values decrease from left to right

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23


EXAMPLE 5

䊳

Applying Slope as a Rate of Change in Altitude At a horizontal distance of 10 mi after take-off, an airline pilot receives instructions to decrease altitude from their current level of 20,000 ft. A short time later, they are 17.5 mi from the airport at an altitude of 10,000 ft. Find the slope ratio for the descent of the plane and discuss its meaning in this context. Recall that 1 mi 5280 ft.

Solution

䊳

Let a represent the altitude of the plane and d its horizontal distance from the airport. Converting all measures to feet, we have 1d1, a1 2 152,800, 20,0002 and 1d2, a2 2 192,400, 10,0002 , giving 10,000 20,000 a2 a1 ¢a ¢d d2 d1 92,400 52,800 10,000 25 39,600 99

¢altitude Since this slope ratio measures ¢distance , we note the plane is decreasing 25 ft in altitude for every 99 ft it travels horizontally.

B. You’ve just seen how we can find the slope of a line and interpret it as a rate of change


䊳

C. Horizontal Lines and Vertical Lines Horizontal and vertical lines have a number of important applications, from finding the boundaries of a given graph (the domain and range), to performing certain tests on nonlinear graphs. To better understand them, consider that in one dimension, the graph of x 2 is a single point (Figure 1.28), indicating a Figure 1.28 location on the number line 2 units from zero x2 in the positive direction. In two dimensions, the 5 4 3 2 1 0 1 2 3 4 5 equation x 2 represents all points with an x-coordinate of 2. A few of these are graphed in Figure 1.29, but since there are an infinite number, we end up with a solid vertical line whose equation is x 2 (Figure 1.30). Figure 1.29

Figure 1.30

y 5

y (2, 5)

5

x2

(2, 3) (2, 1) 5

(2, 1)

WORTHY OF NOTE If we write the equation x 2 in the form ax by c, the equation becomes x 0y 2, since the original equation has no y-variable. Notice that regardless of the value chosen for y, x will always be 2 and we end up with the set of ordered pairs (2, y), which gives us a vertical line.

5

x

5

5

x

(2, 3) 5

5

The same idea can be applied to horizontal lines. In two dimensions, the equation y 4 represents all points with a y-coordinate of positive 4, and there are an infinite number of these as well. The result is a solid horizontal line whose equation is y 4. See Exercises 49 through 54. Horizontal Lines

Vertical Lines

The equation of a horizontal line is yk where (0, k) is the y-intercept.

The equation of a vertical line is xh where (h, 0) is the x-intercept.

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So far, the slope formula has only been applied to lines that were nonhorizontal or nonvertical. So what is the slope of a horizontal line? On an intuitive level, we expect that a perfectly level highway would have an incline or slope of zero. In general, for any two points on a horizontal line, y2 y1 and y2 y1 0, giving a slope of m x2 0 x1 0. For any two points on a vertical line, x2 x1 and x2 x1 0, making the slope ratio undefined: m

y y 2

Figure 1.31 For any horizontal line, y2 ⴝ y1

1

0

(see Figures 1.31 and 1.32).

Figure 1.32 For any vertical line, x2 ⴝ x1

y

(x1, y1)

y

y y2 y1 x x x 2 1 y1 y1 x x 2 1

(x2, y2)

y y2 y1 x x 2 1 x y2 y1 x x 1 1 y2 y1 x 0 undefined

(x2, y2)

0 x x 2 1

x

0 (x1, y1)

The slope of any horizontal line is zero.

The slope of any vertical line is undefined.

Calculating Slopes The federal minimum wage remained constant from 1997 through 2006. However, the buying power (in 1996 dollars) of these wage earners fell each year due to inflation (see Table 1.3). This decrease in buying power is approximated by the red line shown. a. Using the data or graph, find the slope of the line segment representing the minimum wage. b. Select two points on the line representing buying power to approximate the slope of the line segment, and explain what it means in this context. Table 1.3 Time t (years)

5.15

Minimum wage w

Buying power p

1997

5.15

5.03

1998

5.15

4.96

1999

5.15

4.85

2000

5.15

4.69

2001

5.15

4.56

2002

5.15

4.49

4.15

2003

5.15

4.39

4.05

2004

5.15

4.28

2005

5.15

4.14

2006

5.15

4.04

5.05 4.95 4.85 4.75 4.65 4.55 4.45 4.35 4.25

97 19 98 19 9 20 9 00 20 01 20 02 20 03 20 04 20 0 20 5 06

䊳

The Slope of a Vertical Line

19

EXAMPLE 6

The Slope of a Horizontal Line

Wages/Buying power

24

Time in years

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Solution

䊳

WORTHY OF NOTE In the context of lines, try to avoid saying that a horizontal line has “no slope,” since it’s unclear whether a slope of zero or an undefined slope is intended.

a. Since the minimum wage did not increase or decrease from 1997 to 2006, the line segment has slope m 0. b. The points (1997, 5.03) and (2006, 4.04) from the table appear to be on or close to the line drawn. For buying power p and time t, the slope formula yields: ¢p p2 p1 ¢t t2 t1 4.04 5.03 2006 1997 0.11 0.99 9 1 The buying power of a minimum wage worker decreased by 11¢ per year during this time period.

C. You’ve just seen how we can graph horizontal and vertical lines

Now try Exercises 55 and 56

䊳

D. Parallel and Perpendicular Lines Two lines in the same plane that never intersect are called parallel lines. When we place these lines on the coordinate grid, we find that “never intersect” is equivalent to saying “the lines have equal slopes but different y-intercepts.” In Figure 1.33, notice the rise ¢y and run of each line is identical, and that by counting ¢x both lines have slope m 34. Figure 1.33

y 5

Generic plane L 1

Run L2

L1 Run

Rise

L2

Rise

5

5

x

5

Coordinate plane

Parallel Lines Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively. 1. If m1 m2, then L1 is parallel to L2. 2. If L1 is parallel to L2, then m1 m2. In symbols, we write L1||L2. Any two vertical lines (undefined slope) are parallel.

EXAMPLE 7A

䊳

Determining Whether Two Lines Are Parallel Teladango Park has been mapped out on a rectangular coordinate system, with a ranger station at (0, 0). Brendan and Kapi are at coordinates 124, 182 and have set a direct course for the pond at (11, 10). Caden and Kymani are at (27, 1) and are heading straight to the lookout tower at (2, 21). Are they hiking on parallel or nonparallel courses?

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1–26


Solution

䊳

To respond, we compute the slope of each trek across the park. For Brendan and Kapi: m

y2 y1 x2 x1 10 1182

For Caden and Kymani: m

y2 y1 x2 x1

21 1 2 1272 4 20 25 5

11 1242 28 4 35 5

Since the slopes are equal, the two groups are hiking on parallel courses. Two lines in the same plane that intersect at right angles are called perpendicular lines. Using the coordinate grid, we note that intersect at right angles suggests that their slopes are negative reciprocals. While certainly not a proof, notice in Figure 1.34, the 4 rise 3 ratio rise run for L1 is 3 and the ratio run for L2 is 4 . Alternatively, we can say their slopes have a product of ⴚ1, since m1 # m2 1 implies m1 m12. Figure 1.34

Generic plane

y

L1

5

L1

Run

Rise Rise Run 5

5

L2

x

L2 5

Coordinate plane

Perpendicular Lines Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively. 1. If m1 # m2 1, then L1 is perpendicular to L2. 2. If L1 is perpendicular to L2, then m1 # m2 1. In symbols we write L1 ⬜ L2. Any vertical line (undefined slope) is perpendicular to any horizontal line (slope m 0). We can easily find the slope of a line perpendicular to a second line whose slope is known or can be found—just find the reciprocal and make it negative. For a line with slope m1 37, any line perpendicular to it will have a slope of m2 73. For m1 5, the slope of any line perpendicular would be m2 15. EXAMPLE 7B

䊳

Determining Whether Two Lines Are Perpendicular

The three points P1 15, 12, P2 13, 22 , and P3 13, 22 form the vertices of a triangle. Use these points to draw the triangle, then use the slope formula to determine if they form a right triangle.

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27


Solution

䊳

Figure 1.35

For a right triangle to be formed, two of the lines through these points must be perpendicular (forming a right angle). From Figure 1.35, it appears a right triangle is formed, but we must verify that two of the sides are actually perpendicular. Using the slope formula, we have: For P1 and P2

For P1 and P3

2 1 m1 35 3 3 2 2

21 m2 3 5 1 8

y 5

P1

P3 5

x

5

P2

5

For P2 and P3 m3

2 122

3 3 4 2 6 3

Since m1 # m3 1, the triangle has a right angle and must be a right triangle.

D. You’ve just seen how we can identify parallel and perpendicular lines


䊳

E. Applications of Linear Equations The graph of a linear equation can be used to help solve many applied problems. If the numbers you’re working with are either very small or very large, scale the axes appropriately. This can be done by letting each tic mark represent a smaller or larger unit so the data points given will fit on the grid. Also, many applications use only nonnegative values and although points with negative coordinates may be used to graph a line, only ordered pairs in QI can be meaningfully interpreted. EXAMPLE 8

䊳

Applying a Linear Equation Model — Commission Sales Use the information given to create a linear equation model in two variables, then graph the line and answer the question posed: A salesperson gets a daily $20 meal allowance plus $7.50 for every item she sells. How many sales are needed for a daily income of $125? Verify your answer by graphing the line on a calculator and using the

䊳

feature.

Let x represent sales and y represent income. This gives

verbal model: Daily income (y) equals $7.5 per sale 1x2 $20 for meals y equation model: y 7.5x 20

Using x 0 and x 10, we find (0, 20) and (10, 95) are points on this line and these are used to sketch the graph. From the graph, it appears that 14 sales are needed to generate a daily income of $125.00.

(10, 95)

100

Since daily income is given as $125, we substitute 125 for y and solve for x. 125 7.5x 20 105 7.5x 14 x

y 7.5x 20

150

Income

Algebraic Solution

TRACE

50

(0, 20) 0

2

4

6

8 10 12 14 16

Sales substitute 125 for y subtract 20 divide by 7.5

x

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Graphical Solution

䊳

Begin by entering the equation y 7.5x 20 on the Y= screen, recognizing that in this context, both the input and output values must be positive. Reasoning the 10 sales will net $95 (less than $125) and 20 sales will net $170 (more than $125), we set the viewing as shown in Figure 1.36. We can then GRAPH the equation and use the TRACE feature to estimate the number of sales needed. The result shows that income is close to $125 when x is close to 14 (Figure 1.37). In addition to letting us trace along a graph, the TRACE option enables us to evaluate the equation at specific points. Simply entering the number “14” causes the calculator to accept 14 as the desired input (Figure 1.38), and after pressing , it verifies that (14, 125) is indeed a point on the graph (Figure 1.39).

Figure 1.36

WINDOW

Figure 1.37 200

20

0

ENTER

0

Figure 1.38

E. You’ve just seen how we can apply linear equations in context

Figure 1.39


䊳

1.2 EXERCISES 䊳



1. To find the x-intercept of a line, substitute ______ for y and solve for x. To find the y-intercept, substitute _________ for x and solve for y.

4. The slope of a horizontal line is _______, the slope of a vertical line is _______, and the slopes of two parallel lines are ______.

2. The slope formula is m ______ ______, and indicates a rate of change between the x- and y-variables.

5. Discuss/Explain If m1 2.1 and m2 2.01, will the lines intersect? If m1 23 and m2 23 , are the lines perpendicular?

3. If m 6 0, the slope of the line is ______ and the line slopes _______ from left to right.

6. Discuss/Explain the relationship between the slope formula, the Pythagorean theorem, and the distance formula. Include several illustrations.

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Create a table of values for each equation and sketch the graph.

x

y

3 9. y ⫽ x ⫹ 4 2 x

8. ⫺3x ⫹ 5y ⫽ 10 x

y

36. (⫺3, ⫺1), (0, 7)

37. (1, ⫺8), (⫺3, 7)

38. (⫺5, 5), (0, ⫺5)

39. (⫺3, 6), (4, 2)

40. (⫺2, ⫺4), (⫺3, ⫺1)

41. The graph shown models the relationship between the cost of a new home and the size of the home in square feet. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the cost of a 3000 ft2 home. Exercise 41

5 10. y ⫽ x ⫺ 3 3

y

x

Exercise 42

500

y

1200 960

Volume (m3)

7. 2x ⫹ 3y ⫽ 6

35. (10, 3), (4, ⫺5)

Cost ($1000s)

䊳

29


250

720 480 240

0

1

2

3

4

5

0

ft2 (1000s)

Graph the following equations using the intercept method. Plot a third point as a check.

13. 3x ⫹ y ⫽ 6

14. ⫺2x ⫹ y ⫽ 12

15. 5y ⫺ x ⫽ 5

16. ⫺4y ⫹ x ⫽ 8

17. ⫺5x ⫹ 2y ⫽ 6

18. 3y ⫹ 4x ⫽ 9

19. 2x ⫺ 5y ⫽ 4

20. ⫺6x ⫹ 4y ⫽ 8

21. 2x ⫹ 3y ⫽ ⫺12

22. ⫺3x ⫺ 2y ⫽ 6

1 23. y ⫽ ⫺ x 2

24. y ⫽

25. y ⫺ 25 ⫽ 50x

26. y ⫹ 30 ⫽ 60x

2 27. y ⫽ ⫺ x ⫺ 2 5

3 28. y ⫽ x ⫹ 2 4

29. 2y ⫺ 3x ⫽ 0

30. y ⫹ 3x ⫽ 0

31. 3y ⫹ 4x ⫽ 12

32. ⫺2x ⫹ 5y ⫽ 8

2 x 3

43. The graph shown models the relationship between the distance of an aircraft carrier from its home port and the number of hours since departure. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the distance from port after 8.25 hours. Exercise 43 300

33. (3, 5), (4, 6)

34. (⫺2, 3), (5, 8)

500

150

0

Compute the slope of the line through the given points, ¢y then graph the line and use m ⴝ ¢x to find two additional points on the line. Answers may vary.

Exercise 44

Circuit boards

12. If you completed Exercise 10, verify that 37 (⫺1.5, ⫺5.5) and 1 11 2 , 6 2 also satisfy the equation given. Do these points appear to be on the graph you sketched?

100

42. The graph shown models the relationship between the volume of garbage that is dumped in a landfill and the number of commercial garbage trucks that enter the site. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the number of trucks entering the site daily if 1000 m3 of garbage is dumped per day.

Distance (mi)

11. If you completed Exercise 9, verify that (⫺3, ⫺0.5) and (12, 19 4 ) also satisfy the equation given. Do these points appear to be on the graph you sketched?

50

Trucks

10

Hours

20

250

0

5

10

Hours

44. The graph shown models the relationship between the number of circuit boards that have been assembled at a factory and the number of hours since starting time. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate how many hours the factory has been running if 225 circuit boards have been assembled.

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45. Height and weight: While there are many exceptions, numerous studies have shown a close relationship between an average height and average weight. Suppose a person 70 in. tall weighs 165 lb, while a person 64 in. tall weighs 142 lb. Assuming the relationship is linear, (a) find the slope of the line and discuss its meaning in this context and (b) determine how many pounds are added for each inch of height. 46. Rate of climb: Shortly after takeoff, a plane increases altitude at a constant (linear) rate. In 5 min the altitude is 10,000 ft. Fifteen minutes after takeoff, the plane has reached its cruising altitude of 32,000 ft. (a) Find the slope of the line and discuss its meaning in this context and (b) determine how long it takes the plane to climb from 12,200 ft to 25,400 ft. 47. Sewer line slope: Fascinated at how quickly the plumber was working, Ryan watched with great interest as the new sewer line was laid from the house to the main line, a distance of 48 ft. At the edge of the house, the sewer line was 6 in. under ground. If the plumber tied in to the main line at a depth of 18 in., what is the slope of the (sewer) line? What does this slope indicate? 48. Slope (pitch) of a roof: A contractor goes to a lumber yard to purchase some trusses (the triangular frames) for the roof of a house. Many sizes are available, so the contractor takes some measurements to ensure the roof will have the desired slope. In one case, the height of the truss (base to ridge) was 4 ft, with a width of 24 ft (eave to eave). Find the slope of the roof if these trusses are used. What does this slope indicate? Graph each line using two or three ordered pairs that satisfy the equation.

55. Supreme Court justices: The table given shows the total number of justices j sitting on the Supreme Court of the United States for selected time periods t (in decades), along with the number of nonmale, nonwhite justices n for the same years. (a) Use the data to graph the linear relationship between t and j, then determine the slope of the line and discuss its meaning in this context. (b) Use the data to graph the linear relationship between t and n, then determine the slope of the line and discuss its meaning. Exercise 55 Time t (1960 S 0)

Justices j

Nonwhite, nonmale n

0

9

0

10

9

1

20

9

2

30

9

3

40

9

4

50

9

5

56. Boiling temperature: The table shown gives the boiling temperature t of water as related to the altitude h. Use the data to graph the linear relationship between h and t, then determine the slope of the line and discuss its meaning in this context. Exercise 56 Altitude h (ft)

Boiling Temperature t (ⴗF)

0

212.0

1000

210.2

2000

208.4

3000

206.6 204.8

49. x 3

50. y 4

4000

51. x 2

52. y 2

5000

203.0

6000

201.2

Write the equation for each line L1 and L2 shown. Specifically state their point of intersection.

53.

y

54.

L1

L1

L2

4 2 ⴚ4

ⴚ2

2 ⴚ2 ⴚ4

4

x

ⴚ4

ⴚ2

y 5 4 3 2 1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

L2 2

4

x

Two points on L1 and two points on L2 are given. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicular, or neither.

57. L1: (2, 0) and (0, 6) L2: (1, 8) and (0, 5)

58. L1: (1, 10) and (1, 7) L2: (0, 3) and (1, 5)

59. L1: (3, 4) and (0, 1) 60. L1: (6, 2) and (8, 2) L2: (5, 1) and (3, 0) L2: (0, 0) and (4, 4) 61. L1: (6, 3) and (8, 7) L2: (7, 2) and (6, 0)

62. L1: (5, 1) and (4, 4) L2: (4, 7) and (8, 10)

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31

In Exercises 63 to 68, three points that form the vertices of a triangle are given. Use the points to draw the triangle, then use the slope formula to determine if any of the triangles are right triangles. Also see Exercises 43–48 in Section 1.1.

63. (3, 7), (2, 2), (5, 5)

66. (5, 2), (0, 3), (4, 4)

64. (7, 0), (1, 0), (7, 4)

67. (3, 2), (1, 5), (6, 4)

65. (4, 3), (7, 1), (3,2)

68. (0, 0), (5, 2), (2, 5)

䊳


69. Human life expectancy: L ⴝ 0.15T ⴙ 73.7 In the United States, the average life expectancy has been steadily increasing over the years due to better living conditions and improved medical care. This relationship is modeled by the formula shown, where L is the average life expectancy and T is number of years since 1980. (a) What was the life expectancy in the year 2010? (b) In what year will average life expectancy reach 79 yr?

䊳

70. Interest earnings: 100I ⴝ 35,000T If $5000 dollars is invested in an account paying 7% simple interest, the amount of interest earned is given by the formula shown, where I is the interest and T is the time in years. Begin by solving the formula for I. (a) How much interest is earned in 5 yr? (b) How much is earned in 10 yr? (c) Use the two points (5 yr, interest) and (10 yr, interest) to calculate the slope of this line. What do you notice?

APPLICATIONS

Use the information given to build a linear equation model, then use the equation to respond. For exercises 71 to 74, develop both an algebraic and a graphical solution.

71. Business depreciation: A business purchases a copier for $8500 and anticipates it will depreciate in value $1250 per year. a. What is the copier’s value after 4 yr of use? b. How many years will it take for this copier’s value to decrease to $2250? 72. Baseball card value: After purchasing an autographed baseball card for $85, its value increases by $1.50 per year. a. What is the card’s value 7 yr after purchase? b. How many years will it take for this card’s value to reach $100? 73. Water level: During a long drought, the water level in a local lake decreased at a rate of 3 in. per month. The water level before the drought was 300 in. a. What was the water level after 9 months of drought? b. How many months will it take for the water level to decrease to 20 ft?

74. Gas mileage: When empty, a large dump-truck gets about 15 mi per gallon. It is estimated that for each 3 tons of cargo it hauls, gas mileage decreases by 34 mi per gallon. a. If 10 tons of cargo is being carried, what is the truck’s mileage? b. If the truck’s mileage is down to 10 mi per gallon, how much weight is it carrying? 75. Parallel/nonparallel roads: Aberville is 38 mi north and 12 mi west of Boschertown, with a straight “farm and machinery” road (FM 1960) connecting the two cities. In the next county, Crownsburg is 30 mi north and 9.5 mi west of Dower, and these cities are likewise connected by a straight road (FM 830). If the two roads continued indefinitely in both directions, would they intersect at some point? 76. Perpendicular/nonperpendicular course headings: Two shrimp trawlers depart Charleston Harbor at the same time. One heads for the shrimping grounds located 12 mi north and 3 mi east of the harbor. The other heads for a point 2 mi south and 8 mi east of the harbor. Assuming the harbor is at (0, 0), are the routes of the trawlers perpendicular? If so, how far apart are the boats when they reach their destinations (to the nearest one-tenth mi)?

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77. Cost of college: For the years 2000 to 2008, the cost of tuition and fees per semester (in constant dollars) at a public 4-yr college can be approximated by the equation y 386x 3500, where y represents the cost in dollars and x 0 represents the year 2000. Use the equation to find: (a) the cost of tuition and fees in 2010 and (b) the year this cost will exceed $9000. Source: The College Board

78. Female physicians: In 1960 only about 7% of physicians were female. Soon after, this percentage began to grow dramatically. For the years 1990 to 2000, the percentage of physicians that were female can be approximated by the equation y 0.6x 18.1, where y represents the percentage (as a whole number) and x 0 represents the year 1990. Use the equation to find: (a) the percentage of physicians that were female in 2000 and (b) the projected year this percentage would have exceeded 30%.

79. Decrease in smokers: For the years 1990 to 2000, the percentage of the U.S. adult population who were smokers can be approximated by the equation y 13 25 x 28.7, where y represents the percentage of smokers (as a whole number) and x 0 represents 1990. Use the equation to find: (a) the percentage of adults who smoked in the year 2005 and (b) the year the percentage of smokers is projected to fall below 15%. Source: WebMD

80. Temperature and cricket chirps: Biologists have found a strong relationship between temperature and the number of times a cricket chirps. This is modeled by the equation T 14N 40, where N is the number of times the cricket chirps per minute and T is the temperature in Fahrenheit. Use the equation to find: (a) the outdoor temperature if the cricket is chirping 48 times per minute and (b) the number of times a cricket chirps if the temperature is 70°.

Source: American Journal of Public Health

䊳


81. If the lines 4y 2x 5 and 3y ax 2 are perpendicular, what is the value of a? 82. Let m1, m2, m3, and m4 be the slopes of lines L1, L2, L3, and L4, respectively. Which of the following statements is true? a. m4 6 m1 6 m3 6 m2 y L2 L m 6 m 6 m 6 m b. 3 1 2 4 1 L3 c. m3 6 m4 6 m2 6 m1 L4 x d. m1 6 m3 6 m4 6 m2 e. m1 6 m4 6 m3 6 m2

䊳

83. An arithmetic sequence is a sequence of numbers where each successive term is found by adding a fixed constant, called the common difference d, to the preceding term. For instance 3, 7, 11, 15, . . . is an arithmetic sequence with d 4. The formula for the “nth term” tn of an arithmetic sequence is a linear equation of the form tn t1 1n 12d, where d is the common difference and t1 is the first term of the sequence. Use the equation to find the term specified for each sequence. a. 2, 9, 16, 23, 30, . . . ; 21st term b. 7, 4, 1, 2, 5, . . . ; 31st term c. 5.10, 5.25, 5.40, 5.55, . . . ; 27th term 9 d. 32, 94, 3, 15 4 , 2 , . . . ; 17th term


84. (1.1) Name the center and radius of the circle defined by 1x 32 2 1y 42 2 169 85. (Appendix A.6) Compute the sum and product indicated: a. 120 3 145 15 b. 13 152 13 252 86. (Appendix A.4) Solve the equation by factoring, then check the result(s) using substitution: 12x2 44x 45 0

87. (Appendix A.5) Factor the following polynomials completely: a. x3 3x2 4x 12 b. x2 23x 24 c. x3 125

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Functions, Function Notation, and the Graph of a Function


A. Distinguish the graph of a function from that of a relation B. Determine the domain and range of a function C. Use function notation and evaluate functions D. Read and interpret information given graphically

In this section we introduce one of the most central ideas in mathematics—the concept of a function. Functions can model the cause-and-effect relationship that is so important to using mathematics as a decision-making tool. In addition, the study will help to unify and expand on many ideas that are already familiar.

A. Functions and Relations There is a special type of relation that merits further attention. A function is a relation where each element of the domain corresponds to exactly one element of the range. In other words, for each first coordinate or input value, there is only one possible second coordinate or output. Functions A function is a relation that pairs each element from the domain with exactly one element from the range. If the relation is defined by a mapping, we need only check that each element of the domain is mapped to exactly one element of the range. This is indeed the case for the mapping P S B from Figure 1.1 (page 2), where we saw that each person corresponded to only one birthday, and that it was impossible for one person to be born on two different days. For the relation x ⫽ 冟y冟 shown in Figure 1.6 (page 4), each element of the domain except zero is paired with more than one element of the range. The relation x ⫽ 冟y冟 is not a function.

EXAMPLE 1

䊳

Determining Whether a Relation is a Function Three different relations are given in mapping notation below. Determine whether each relation is a function. a. b. c.

Solution

䊳

Person

Room

Marie Pesky Bo Johnny Rick Annie Reece

270 268 274 276 272 282

Pet

Weight (lb)

Fido

450 550 2 40 8 3

Bossy Silver Frisky Polly

War

Year

Civil War

1963

World War I

1950

World War II

1939

Korean War

1917

Vietnam War

1861

Relation (a) is a function, since each person corresponds to exactly one room. This relation pairs math professors with their respective office numbers. Notice that while two people can be in one office, it is impossible for one person to physically be in two different offices. Relation (b) is not a function, since we cannot tell whether Polly the Parrot weighs 2 lb or 3 lb (one element of the domain is mapped to two elements of the range). Relation (c) is a function, where each major war is paired with the year it began. Now try Exercises 7 through 10 䊳

1–33

33

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If the relation is pointwise-defined or given as a set of individual and distinct plotted points, we need only check that no two points have the same first coordinate with a different second coordinate. This gives rise to an alternative definition for a function. Functions (Alternate Definition) A function is a set of ordered pairs (x, y), in which each first component is paired with only one second component.

EXAMPLE 2

䊳

Identifying Functions Two relations named f and g are given; f is pointwise-defined (stated as a set of ordered pairs), while g is given as a set of plotted points. Determine whether each is a function. f: 1⫺3, 02, 11, 42, 12, ⫺52, 14, 22, 1⫺3, ⫺22, 13, 62, 10, ⫺12, (4, ⫺5), and (6, 1)

Solution

䊳

The relation f is not a function, since ⫺3 is paired with two different outputs: 1⫺3, 02 and 1⫺3, ⫺22 .

g

5

y (0, 5)

(⫺4, 2)

The relation g shown in the figure is a function. Each input corresponds to exactly one output, otherwise one point would be directly above the other and have the same first coordinate.

(3, 1)

(⫺2, 1) ⫺5

5

x

(4, ⫺1) (⫺1, ⫺3) ⫺5

Now try Exercises 11 through 18 䊳 The graphs of y ⫽ x ⫺ 1 and x ⫽ 冟y冟 from Section 1.1 offer additional insight into the definition of a function. Figure 1.40 shows the line y ⫽ x ⫺ 1 with emphasis on the plotted points (4, 3) and 1⫺3, ⫺42. The vertical movement shown from the x-axis to a point on the graph illustrates the pairing of a given x-value with one related y-value. Note the vertical line shows only one related y-value (x ⫽ 4 is paired with only y ⫽ 3). Figure 1.41 gives the graph of x ⫽ 冟y冟, highlighting the points (4, 4) and (4, ⫺4). The vertical movement shown here branches in two directions, associating one x-value with more than one y-value. This shows the relation y ⫽ x ⫺ 1 is also a function, while the relation x ⫽ 冟y冟 is not. Figure 1.41

Figure 1.40 5

y y⫽x⫺1

y

x ⫽ ⱍyⱍ (4, 4)

5

(4, 3) (2, 2) (0, 0) ⫺5

5

x

⫺5

5

x

(2, ⫺2) (⫺3, ⫺4)

⫺5

⫺5

(4, ⫺4)

This “vertical connection” of a location on the x-axis to a point on the graph can be generalized into a vertical line test for functions.

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35

Section 1.3 Functions, Function Notation, and the Graph of a Function

Vertical Line Test A given graph is the graph of a function, if and only if every vertical line intersects the graph in at most one point. Applying the test to the graph in Figure 1.40 helps to illustrate that the graph of any nonvertical line must be the graph of a function, as is the graph of any pointwisedefined relation where no x-coordinate is repeated. Compare the relations f and g from Example 2. 䊳

EXAMPLE 3

Using the Vertical Line Test Use the vertical line test to determine if any of the relations shown (from Section 1.1) are functions.

䊳

Solution

Visualize a vertical line on each coordinate grid (shown in solid blue), then mentally shift the line to the left and right as shown in Figures 1.42, 1.43, and 1.44 (dashed lines). In Figures 1.42 and 1.43, every vertical line intersects the graph only once, indicating both y ⫽ x2 ⫺ 2x and y ⫽ 29 ⫺ x2 are functions. In Figure 1.44, a vertical line intersects the graph twice for any x 7 0 [for instance, both (4, 2) and 14, ⫺22 are on the graph]. The relation x ⫽ y2 is not a function.

Figure 1.42

Figure 1.43

y (4, 8)

(⫺2, 8) y ⫽ x2 ⫺ 2x

5

Figure 1.44 y

y y ⫽ 兹9 ⫺ x2 (0, 3)

5

x ⫽ y2 (4, 2)

(2, 兹2)

5

(⫺1, 3)

(⫺3, 0)

(3, 3)

(0, 0)

(3, 0)

⫺5

5

x

⫺5

5

(2, 0)

(0, 0) ⫺5

5

(1, ⫺1)

⫺2

x ⫺5

⫺5

x

(2, ⫺兹2) (4, ⫺2)


EXAMPLE 4

䊳

Using the Vertical Line Test Use a table of values to graph the relations defined by a. y ⫽ 冟x冟 b. y ⫽ 1x, then use the vertical line test to determine whether each relation is a function.

Solution

䊳

WORTHY OF NOTE For relations and functions, a good way to view the distinction is to consider a mail carrier. It is possible for the carrier to put more than one letter into the same mailbox (more than one x going to the same y), but quite impossible for the carrier to place the same letter in two different boxes (one x going to two y’s).

a. For y ⫽ 冟x冟, using input values from x ⫽ ⫺4 to x ⫽ 4 produces the following table and graph (Figure 1.45). Note the result is a V-shaped graph that “opens upward.” The point (0, 0) of this absolute value graph is called the vertex. Since any vertical line will intersect the graph in at most one point, this is the graph of a function.

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CHAPTER 1 Relations, Functions, and Graphs y ⴝ 円 x円 x

y ⴝ 円 x円

⫺4

4

⫺3

3

⫺2

2

⫺1

1

0

0

1

1

2

2

3

3

4

4

Figure 1.45 y 5

⫺5

5

x

⫺5

b. For y ⫽ 1x, values less than zero do not produce a real number, so our graph actually begins at (0, 0) (see Figure 1.46). Completing the table for nonnegative values produces the graph shown, which appears to rise to the right and remains in the first quadrant. Since any vertical line will intersect this graph in at most one place, y ⫽ 1x is also a function. Figure 1.46 y

y ⴝ 1x x

5

y ⴝ 1x

0

0

1

1

2

12 ⬇ 1.4

3

13 ⬇ 1.7

4

⫺5

5

x

2 ⫺5

A. You’ve just seen how we can distinguish the graph of a function from that of a relation


B. The Domain and Range of a Function Vertical Boundary Lines and the Domain

WORTHY OF NOTE On a number line, some texts will use an open dot “º” to mark the location of an endpoint that is not included, and a closed dot “•” for an included endpoint.

In addition to its use as a graphical test for functions, a vertical line can help determine the domain of a function from its graph. For the graph of y ⫽ 1x (Figure 1.46), a vertical line will not intersect the graph until x ⫽ 0, and then will intersect the graph for all values x ⱖ 0 (showing the function is defined for these values). These vertical boundary lines indicate the domain is x ⱖ 0. Instead of using a simple inequality to write the domain and range, we will often use (1) a form of set notation, (2) a number line graph, or (3) interval notation. Interval notation is a symbolic way of indicating a selected interval of the real numbers. When a number acts as the boundary point for an interval (also called an endpoint), we use a left bracket “[” or a right bracket “]” to indicate inclusion of the endpoint. If the boundary point is not included, we use a left parenthesis “(” or right parenthesis “).”

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EXAMPLE 5

䊳

37

Using Notation to State the Domain and Range Model the given phrase using the correct inequality symbol. Then state the result in set notation, graphically, and in interval notation: “The set of real numbers greater than or equal to 1.”

Solution

䊳

WORTHY OF NOTE Since infinity is really a concept and not a number, it is never included (using a bracket) as an endpoint for an interval.

Let n represent the number: n ⱖ 1. • Set notation: 5n |n ⱖ 16 [ • Graph: ⫺2 ⫺1

0

1

2

3

4

• Interval notation: n 僆 3 1, q 2

5

Now try Exercises 35 through 50 䊳 The “僆” symbol says the number n is an element of the set or interval given. The “ q ” symbol represents positive infinity and indicates the interval continues forever to the right. Note that the endpoints of an interval must occur in the same order as on the number line (smaller value on the left; larger value on the right). A short summary of other possibilities is given here for any real number x. Many variations are possible. Conditions (a ⬍ b) x is greater than k x is less than or equal to k x is less than b and greater than a x is less than b and greater than or equal to a x is less than a or x is greater than b

Set Notation 5x |x 7 k6 5x |x ⱕ k6

5x |a 6 x 6 b6 5x |a ⱕ x 6 b6 5x |x 6 a or x 7 b6

Number Line

Interval Notation x 僆 1k, q 2

) k

x 僆 1⫺q, k4

[ k

)

)

a

b

[

)

a

b

x 僆 1a, b2 x 僆 3 a, b2

)

)

a

b

x 僆 1⫺q, a2 ´ 1b, q2

For the graph of y ⫽ 冟x冟 (Figure 1.45), a vertical line will intersect the graph (or its infinite extension) for all values of x, and the domain is x 僆 1⫺q, q 2 . Using vertical lines in this way also affirms the domain of y ⫽ x ⫺ 1 (Section 1.1, Figure 1.5) is x 僆 1⫺q, q 2 while the domain of the relation x ⫽ 冟y冟 (Section 1.1, Figure 1.6) is x 僆 30, q 2 .

Range and Horizontal Boundary Lines The range of a relation can be found using a horizontal “boundary line,” since it will associate a value on the y-axis with a point on the graph (if it exists). Simply visualize a horizontal line and move the line up or down until you determine the graph will always intersect the line, or will no longer intersect the line. This will give you the boundaries of the range. Mentally applying this idea to the graph of y ⫽ 1x (Figure 1.46) shows the range is y 僆 3 0, q 2. Although shaped very differently, a horizontal boundary line shows the range of y ⫽ 冟x冟 (Figure 1.45) is also y 僆 30, q 2. EXAMPLE 6

䊳

Determining the Domain and Range of a Function Use a table of values to graph the functions defined by 3 a. y ⫽ x2 b. y ⫽ 1 x Then use boundary lines to determine the domain and range of each.

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Solution

䊳

a. For y ⫽ x2, it seems convenient to use inputs from x ⫽ ⫺3 to x ⫽ 3, producing the following table and graph. Note the result is a basic parabola that “opens upward” (both ends point in the positive y direction), with a vertex at (0, 0). Figure 1.47 shows a vertical line will intersect the graph or its extension anywhere it is placed. The domain is x 僆 1⫺q, q 2 . Figure 1.48 shows a horizontal line will intersect the graph only for values of y that are greater than or equal to 0. The range is y 僆 30, q 2 . Figure 1.47

Squaring Function x

y ⴝ x2

⫺3

9

⫺2

4

⫺1

1

0

0

1

1

2

4

3

9

5

Figure 1.48

y y ⫽ x2 5

⫺5

5

x

y y ⫽ x2

⫺5

⫺5

5

x

⫺5

3 b. For y ⫽ 1x, we select points that are perfect cubes where possible, then a few others to round out the graph. The resulting table and graph are shown. Notice there is a “pivot point” at (0, 0) called a point of inflection, and the ends of the graph point in opposite directions. Figure 1.49 shows a vertical line will intersect the graph or its extension anywhere it is placed. Figure 1.50 shows a horizontal line will likewise always intersect the graph. The domain is x 僆 1⫺q, q 2 , and the range is y 僆 1⫺q, q 2 .

Figure 1.49

Cube Root Function x

3 yⴝ 1 x

⫺8

⫺2

⫺4

⬇ ⫺1.6

⫺1

⫺1

0

0

1

1

4

⬇ 1.6

8

2

Figure 1.50 3

5

⫺10

y y ⫽ 兹x

5

10

⫺5

x

⫺10

3 y y ⫽ 兹x

10

x

⫺5


Implied Domains When stated in equation form, the domain of a function is implicitly given by the expression used to define it, since the expression will dictate what input values are allowed. The implied domain is the set of all real numbers for which the function represents a real number. If the function involves a rational expression, the domain will exclude any input that causes a denominator of zero, since division by zero is undefined. If the function involves a square root expression, the domain will exclude inputs that create a negative radicand, since 1A represents a real number only when A ⱖ 0.

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EXAMPLE 7

Solution

䊳

䊳

Figure 1.51

Y1 ⫽ 1X ⫺ 12/1X2 ⫺ 92

39

Determining Implied Domains State the domain of each function using interval notation. 3 a. y ⫽ b. y ⫽ 12x ⫹ 3 x⫹2 x⫺1 c. y ⫽ 2 d. y ⫽ x2 ⫺ 5x ⫹ 7 x ⫺9 a. By inspection, we note an x-value of ⫺2 results in a zero denominator and must be excluded. The domain is x 僆 1⫺q, ⫺22 ´ 1⫺2, q2. b. Since the radicand must be nonnegative, we solve the inequality 2x ⫹ 3 ⱖ 0, ⫺3 giving x ⱖ ⫺3 2 . The domain is x 僆 3 2 , q 2. c. To prevent division by zero, inputs of ⫺3 and 3 must be excluded (set x2 ⫺ 9 ⫽ 0 and solve by factoring). The domain is x 僆 1⫺q, ⫺32 ´ 1⫺3, 32 ´ 13, q 2 . Note that x ⫽ 1 is in the domain X⫺1 0 ⫽ 0 is defined. See Figure 1.51, where Y1 ⫽ 2 since ⫺8 . X ⫺9 d. Since squaring a number and multiplying a number by a constant are defined for all real numbers, the domain is x 僆 1⫺q, q 2. Now try Exercises 63 through 80 䊳

EXAMPLE 8

䊳

Determining Implied Domains Determine the domain of each function: 2x 14x ⫹ 5 7 7 a. For y ⫽ , we must have ⱖ 0 (for the radicand) and x ⫹ 3 ⫽ 0 Ax ⫹ 3 x⫹3 (for the denominator). Since the numerator is always positive, we need x ⫹ 3 7 0, which gives x 7 ⫺3. The domain is x 僆 1⫺3, q 2 . a. y ⫽

Solution

䊳

7 Ax ⫹ 3

b. y ⫽

2x , we must have 4x ⫹ 5 ⱖ 0 and 14x ⫹ 5 ⫽ 0. This shows 14x ⫹ 5 ⫺5 we need 4x ⫹ 5 7 0, so x 7 ⫺5 4 . The domain is x 僆 1 4 , q 2 .

b. For y ⫽ B. You’ve just seen how we can determine the domain and range of a function


C. Function Notation Figure 1.52 x

Input f Sequence of operations on x as defined by f

Output

y ⫽ f (x)

In our study of functions, you’ve likely noticed that the relationship between input and output values is an important one. To highlight this fact, think of a function as a simple machine, which can process inputs using a stated sequence of operations, then deliver a single output. The inputs are x-values, a program we’ll name f performs the operations on x, and y is the resulting output (see Figure 1.52). Once again we see that “the value of y depends on the value of x,” or simply “y is a function of x.” Notationally, we write “y is a function of x” as y ⫽ f 1x2 using function notation. You are already familiar with letting a variable represent a number. Here we do something quite different, as the letter f is used to represent a sequence of operations to be performed on x. Consider x x the function y ⫽ ⫹ 1, which we’ll now write as f 1x2 ⫽ ⫹ 1 3since y ⫽ f 1x24 . 2 2

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In words the function says, “divide inputs by 2, then add 1.” To evaluate the function at x ⫽ 4 (Figure 1.53) we have: input 4

Figure 1.53

input 4

S

S

f 1x2 ⫽

4

x ⫹1 2 4 f 142 ⫽ ⫹ 1 2 ⫽2⫹1 ⫽3

Input f Divide inputs by 2 then add 1: 4 +1 2

Output

3

Function notation enables us to summarize the three most important aspects of a function using a single expression, as shown in Figure 1.54. Figure 1.54 Output value y = f(x)

f(x) Sequence of operations to perform on the input

Input value

Instead of saying, “. . . when x ⫽ 4, the value of the function is 3,” we simply say “f of 4 is 3,” or write f 142 ⫽ 3. Note that the ordered pair (4, 3) is equivalent to (4, f (4)). 䊳

CAUTION

EXAMPLE 9

Solution

䊳

䊳

Although f(x) is the favored notation for a “function of x,” other letters can also be used. For example, g(x) and h(x) also denote functions of x, where g and h represent different sequences of operations on the x-inputs. It is also important to remember that these represent function values and not the product of two variables: f1x2 ⫽ f # 1x2 .

Evaluating a Function

Given f 1x2 ⫽ ⫺2x2 ⫹ 4x, find 7 a. f 1⫺22 b. f a b 2 a.

f 1x2 ⫽ ⫺2x2 ⫹ 4x f 1⫺22 ⫽ ⫺21⫺22 2 ⫹ 41⫺22 ⫽ ⫺8 ⫹ 1⫺82 ⫽ ⫺16

c. f 1x2 ⫽ ⫺2x2 ⫹ 4x f 12a2 ⫽ ⫺212a2 2 ⫹ 412a2 ⫽ ⫺214a2 2 ⫹ 8a ⫽ ⫺8a2 ⫹ 8a

c. f 12a2 b.

d.

d. f 1a ⫹ 12

f 1x2 ⫽ ⫺2x2 ⫹ 4x 7 7 2 7 f a b ⫽ ⫺2 a b ⫹ 4 a b 2 2 2 ⫺21 ⫺49 ⫹ 14 ⫽ or ⫺10.5 ⫽ 2 2

f 1x2 ⫽ ⫺2x2 ⫹ 4x f 1a ⫹ 12 ⫽ ⫺21a ⫹ 12 2 ⫹ 41a ⫹ 12 ⫽ ⫺21a2 ⫹ 2a ⫹ 12 ⫹ 4a ⫹ 4 ⫽ ⫺2a2 ⫺ 4a ⫺ 2 ⫹ 4a ⫹ 4 ⫽ ⫺2a2 ⫹ 2 Now try Exercises 87 through 102 䊳

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41


A graphing calculator can evaluate the function Y1 ⫽ ⫺2X2 ⫹ 4X using the TABLE feature, the TRACE feature, or function notation (on the home screen). The first two have been illustrated previously. To use function notation, we access the function names using the VARS key and right arrow to select Y-VARS (Figure 1.55). The 1:Function option is the default, so pressing will enable us to make our choice (Figure 1.56). In this case, we selected 1:Y1, which the calculator then places on the home screen, enabling us to enclose the desired input value in parentheses (function notation). Pressing ENTER completes the evaluation (Figure 1.57), which verifies the result from Example 9(b).

Figure 1.55

ENTER

Figure 1.56

Figure 1.57

C. You’ve just seen how we can use function notation and evaluate functions

D. Reading and Interpreting Information Given Graphically Graphs are an important part of studying functions, and learning to read and interpret them correctly is a high priority. A graph highlights and emphasizes the all-important input/output relationship that defines a function. In this study, we hope to firmly establish that the following statements are synonymous: 1. 2. 3. 4. EXAMPLE 10A

䊳

f 1⫺22 ⫽ 5 1⫺2, f 1⫺22 2 ⫽ 1⫺2, 52 1⫺2, 52 is on the graph of f, and when x ⫽ ⫺2, f 1x2 ⫽ 5

Reading a Graph For the functions f and g whose graphs are shown in Figures 1.58 and 1.59 a. State the domain of the function. b. Evaluate the function at x ⫽ 2. c. Determine the value(s) of x for which y ⫽ 3. d. State the range of the function. Figure 1.58 y

4

3

3

2

2

1

1 1

2

g

5

4

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

䊳

y

f

5

Solution

Figure 1.59

3

4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

⫺2

⫺2

⫺3

⫺3

1

2

3

4

5

x

For f, a. The graph is a continuous line segment with endpoints at (⫺4, ⫺3) and (5, 3), so we state the domain in interval notation. Using a vertical boundary line we note the smallest input is ⫺4 and the largest is 5. The domain is x 僆 3⫺4, 54 .

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b. The graph shows an input of x ⫽ 2 corresponds to y ⫽ 1: f (2) ⫽ 1 since (2, 1) is a point on the graph. c. For f (x) ⫽ 3 (or y ⫽ 3) the input value must be x ⫽ 5 since (5, 3) is the point on the graph. d. Using a horizontal boundary line, the smallest output value is ⫺3 and the largest is 3. The range is y 僆 3⫺3, 34 . For g, a. Since g is given as a set of plotted points, we state the domain as the set of first coordinates: D: 5⫺4, ⫺2, 0, 2, 46 . b. An input of x ⫽ 2 corresponds to y ⫽ 2: g(2) ⫽ 2 since (2, 2) is on the graph. c. For g(x) ⫽ 3 (or y ⫽ 3) the input value must be x ⫽ 4, since (4, 3) is a point on the graph. d. The range is the set of all second coordinates: R: 5⫺1, 0, 1, 2, 36.

EXAMPLE 10B

Solution

䊳

䊳

Reading a Graph

Use the graph of f 1x2 given to answer the following questions: a. What is the value of f 1⫺22 ? (⫺2, 4) b. What value(s) of x satisfy f 1x2 ⫽ 1?

y 5

f a. The notation f 1⫺22 says to find the value (0, 1) of the function f when x ⫽ ⫺2. Expressed (⫺3, 1) graphically, we go to x ⫽ ⫺2 and locate the ⫺5 corresponding point on the graph (blue arrows). Here we find that f 1⫺22 ⫽ 4. b. For f 1x2 ⫽ 1, we’re looking for x-inputs that result in an output of y ⫽ 1 3since y ⫽ f 1x2 4 . ⫺5 From the graph, we note there are two points with a y-coordinate of 1, namely, (⫺3, 1) and (0, 1). This shows f 1⫺32 ⫽ 1, f 102 ⫽ 1, and the required x-values are x ⫽ ⫺3 and x ⫽ 0.

5

x

Now try Exercises 103 through 108 䊳 In many applications involving functions, the domain and range can be determined by the context or situation given. EXAMPLE 11

䊳

Determining the Domain and Range from the Context Paul’s 2009 Voyager has a 20-gal tank and gets 18 mpg. The number of miles he can drive (his range) depends on how much gas is in the tank. As a function we have M1g2 ⫽ 18g, where M(g) represents the total distance in miles and g represents the gallons of gas in the tank (see graph). Find the domain and range.

Solution

D. You’ve just seen how we can read and interpret information given graphically

䊳

M 600 480

(20, 360) 360 240

120 Begin evaluating at x ⫽ 0, since the tank cannot (0, 0) hold less than zero gallons. With an empty tank, the 0 10 20 (minimum) range is M102 ⫽ 18102 or 0 miles. On a full tank, the maximum range is M1202 ⫽ 181202 or 360 miles. As shown in the graph, the domain is g 僆 [0, 20] and the corresponding range is M(g) 僆 [0, 360].

g


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43


1.3 EXERCISES 䊳



䊳

1. If a relation is given in ordered pair form, we state the domain by listing all of the coordinates in a set.

2. A relation is a function if each element of the is paired with element of the range.

3. The set of output values for a function is called the of the function.

4. Write using function notation: The function f evaluated at 3 is negative 5:

5. Discuss/Explain why the relation y ⫽ x2 is a function, while the relation x ⫽ y2 is not. Justify your response using graphs, ordered pairs, and so on.

6. Discuss/Explain the process of finding the domain and range of a function given its graph, using vertical and horizontal boundary lines. Include a few illustrative examples.


Determine whether the mappings shown represent functions or nonfunctions. If a nonfunction, explain how the definition of a function is violated.

7.

8.

9.

Woman

Country

Indira Gandhi Clara Barton Margaret Thatcher Maria Montessori Susan B. Anthony

Britain

Book

Author

Hawaii Roots Shogun 20,000 Leagues Under the Sea Where the Red Fern Grows

Rawls

Basketball star

10.

Country

Language

Canada Japan Brazil Tahiti Ecuador

Japanese Spanish French Portuguese English

U.S. Italy India

Verne Haley Clavell Michener Reported height

Determine whether the relations indicated represent functions or nonfunctions. If the relation is a nonfunction, explain how the definition of a function is violated.

11. (⫺3, 0), (1, 4), (2, ⫺5), (4, 2), (⫺5, 6), (3, 6), (0, ⫺1), (4, ⫺5), and (6, 1) 12. (⫺7, ⫺5), (⫺5, 3), (4, 0), (⫺3, ⫺5), (1, ⫺6), (0, 9), (2, ⫺8), (3, ⫺2), and (⫺5, 7) 13. (9, ⫺10), (⫺7, 6), (6, ⫺10), (4, ⫺1), (2, ⫺2), (1, 8), (0, ⫺2), (⫺2, ⫺7), and (⫺6, 4) 14. (1, ⫺81), (⫺2, 64), (⫺3, 49), (5, ⫺36), (⫺8, 25), (13, ⫺16), (⫺21, 9), (34, ⫺4), and (⫺55, 1) 15.

7'1" 6'6" 6'7" 6'9" 7'2"

y (⫺3, 5)

(2, 4)

(⫺3, 4)

Air Jordan The Mailman The Doctor The Iceman The Shaq

16.

y 5

(⫺1, 1)

5

(3, 4) (1, 3)

(4, 2)

(⫺5, 0) ⫺5

5 x

⫺5

5 x

(0, ⫺2) (5, ⫺3)

(⫺4, ⫺2) ⫺5

(1, ⫺4)

⫺5

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17.

18.

y 5

29.

y 5

(3, 4) (⫺2, 3)

(1, 1)

⫺5

5 x

⫺5

20.

y

⫺5

5 x

22.

y 5

Graph each relation using a table, then use the vertical line test to determine if the relation is a function.

31. y ⫽ x

32. y ⫽ 1 x 3

34. x ⫽ 冟y ⫺ 2冟

Use an inequality to write a mathematical model for each statement, then write the relation in interval notation.

35. To qualify for a secretarial position, a person must type at least 45 words per minute.

⫺5

⫺5

⫺5

33. y ⫽ 1x ⫹ 22 2

y 5

⫺5

5 x

5 x

⫺5

⫺5

Determine whether or not the relations given represent a function. If not, explain how the definition of a function is violated. 5

⫺5

5 x

(3, ⫺2)

(⫺1, ⫺4)

(4, ⫺5)

⫺5

⫺5

5 x

(⫺5, ⫺2)

(⫺2, ⫺4)

21.

y 5

(3, 3)

(1, 2) (⫺5, 1)

19.

30.

y 5

(⫺3, 4)

36. The balance in a checking account must remain above $1000 or a fee is charged.

y 5

37. To bake properly, a turkey must be kept between the temperatures of 250° and 450°. ⫺5

⫺5

5 x

24.

y 5

⫺5

5

⫺5

5 x

Graph each inequality on a number line, then write the relation in interval notation.

y

⫺5

25.

38. To fly effectively, the airliner must cruise at or between altitudes of 30,000 and 35,000 ft.

⫺5

⫺5

23.

5 x

5 x

⫺5

26.

y 5

39. p 6 3

40. x 7 ⫺2

41. m ⱕ 5

42. n ⱖ ⫺4

43. x ⫽ 1

44. x ⫽ ⫺3

45. 5 7 x 7 2

46. ⫺3 6 p ⱕ 4

Write the domain illustrated on each graph in set notation and interval notation.

y 5

47. ⫺5

⫺5

5 x

5 x

48. 49.

⫺5

⫺5

50. 27.

28.

y 5

⫺5

5 x

⫺5

y

0

1

2

3

)

⫺3 ⫺2 ⫺1

0

[

1

2

3

[

⫺3 ⫺2 ⫺1

0

1

2

3

)

[

⫺3 ⫺2 ⫺1

0

1

2

3

4

Determine whether or not the relations indicated represent functions, then determine the domain and range of each.

5

⫺5

[

⫺3 ⫺2 ⫺1

5 x

51.

52.

y 5

y 5

⫺5 ⫺5

5 x

⫺5

⫺5

5 x

⫺5

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53.

54.

y 5

⫺5

y 5

⫺5

5 x

5 x

⫺5

55.

⫺5

56.

y 5

⫺5

y 5

⫺5

5 x

5 x

⫺5

57.

⫺5

58.

y 5

⫺5

5 x

⫺5

59.

60.

y

⫺5

5 x

62.

y

⫺5

5 x

x x ⫺ 3x ⫺ 10

78. y2 ⫽

x⫺4 x ⫹ 2x ⫺ 15

79. y ⫽

1x ⫺ 2 2x ⫺ 5

80. y ⫽

1x ⫹ 1 3x ⫹ 2

2

2

81. h1x2 ⫽

⫺2 1x ⫹ 4

82.

83. g1x2 ⫽

⫺4 A3 ⫺ x

84. p1x2 ⫽

⫺7 15 ⫺ x

85. r 1x2 ⫽

2x ⫺ 1 13x ⫺ 7

86. s1x2 ⫽

x2 ⫺ 4 111 ⫺ 2x

f 1x2 ⫽

y 5

⫺5

5 x

⫺5

1 87. f 1x2 ⫽ x ⫹ 3 2

2 88. f 1x2 ⫽ x ⫺ 5 3

90. f 1x2 ⫽ 2x2 ⫹ 3x

91. h1x2 ⫽

3 x

93. h1x2 ⫽

5冟x冟 x

2 x2 4冟x冟 94. h1x2 ⫽ x 92. h1x2 ⫽

⫺5

95. g1r2 ⫽ 2␲r

96. g1r2 ⫽ 2␲rh

97. g1r2 ⫽ ␲r

98. g1r2 ⫽ ␲r2h

2

Determine the value of p(5), p1 32 2, p(3a), and p(a ⴚ 1), then simplify.

99. p1x2 ⫽ 12x ⫹ 3

100. p1x2 ⫽ 14x ⫺ 1

3x ⫺ 5 x2 2

Determine the domain of the following functions, and write your response in interval notation.

3 63. f 1x2 ⫽ x⫺5

⫺2 64. g1x2 ⫽ 3⫹x

65. h1a2 ⫽ 13a ⫹ 5 66. p1a2 ⫽ 15a ⫺ 2 67. v1x2 ⫽ 69. u ⫽ 71. y ⫽

x⫹2 x2 ⫺ 25

5 Ax ⫺ 2

Determine the value of g(4), g 1 32 2, g(2c), and g(c ⴙ 3), then simplify.

⫺5

5

77. y1 ⫽

Determine the value of h(3), h1ⴚ23 2 , h(3a), and h(a ⫺ 2), then simplify.

y

⫺5

61.

76. y ⫽ 冟x ⫺ 2冟 ⫹ 3

5

⫺5

5 x

75. y ⫽ 2冟x冟 ⫹ 1

89. f 1x2 ⫽ 3x2 ⫺ 4x

⫺5

5

74. s ⫽ t2 ⫺ 3t ⫺ 10

For Exercises 87 through 102, determine the value of f 1ⴚ62, f 1 32 2, f 12c2, and f 1c ⴙ 12 , then simplify. Verify results using a graphing calculator where possible.

y 5

⫺5

5 x

73. m ⫽ n2 ⫺ 3n ⫺ 10

68. w1x2 ⫽

v⫺5 v2 ⫺ 18

70. p ⫽

17 x ⫹ 123 25

72. y ⫽

x⫺4 x2 ⫺ 49

101. p1x2 ⫽

102. p1x2 ⫽

2x2 ⫹ 3 x2

Use the graph of each function given to (a) state the domain, (b) state the range, (c) evaluate f (2), and (d) find the value(s) x for which f 1x2 ⴝ k (k a constant). Assume all results are integer-valued.

103. k ⫽ 4

104. k ⫽ 3 y

y

5

5

q⫹7 q2 ⫺ 12 11 x ⫺ 89 19

⫺5

5 x

⫺5

⫺5

5 x

⫺5

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105. k ⫽ 1

106. k ⫽ ⫺3

107. k ⫽ 2

y

⫺5

5

5 x

⫺5

⫺5

y 5

5

5 x

⫺5

⫺5

5 x

⫺5

⫺5

5 x

⫺5


9 109. Ideal weight for males: W1H2 ⴝ H ⴚ 151 2 The ideal weight for an adult male can be modeled by the function shown, where W is his weight in pounds and H is his height in inches. (a) Find the ideal weight for a male who is 75 in. tall. (b) If I am 72 in. tall and weigh 210 lb, how much weight should I lose? 5 110. Celsius to Fahrenheit conversions: C ⴝ 1F ⴚ 322 9 The relationship between Fahrenheit degrees and degrees Celsius is modeled by the function shown. (a) What is the Celsius temperature if °F ⫽ 41? (b) Use the formula to solve for F in terms of C, then substitute the result from part (a). What do you notice? 䊳

108. k ⫽ ⫺1 y

y

5

䊳

1–46


1 111. Pick’s theorem: A ⴝ B ⴙ I ⴚ 1 2 Pick’s theorem is an interesting yet little known formula for computing the area of a polygon drawn in the Cartesian coordinate system. The formula can be applied as long as the vertices of the polygon are lattice points (both x and y are integers). If B represents the number of lattice points lying directly on the boundary of the polygon (including the vertices), and I represents the number of points in the interior, the area of the polygon is given by the formula shown. Use some graph paper to carefully draw a triangle with vertices at 1⫺3, 12 , (3, 9), and (7, 6), then use Pick’s theorem to compute the triangle’s area.

APPLICATIONS

112. Gas mileage: John’s old ’87 LeBaron has a 15-gal gas tank and gets 23 mpg. The number of miles he can drive is a function of how much gas is in the tank. (a) Write this relationship in equation form and (b) determine the domain and range of the function in this context. 113. Gas mileage: Jackie has a gas-powered model boat with a 5-oz gas tank. The boat will run for 2.5 min on each ounce. The number of minutes she can operate the boat is a function of how much gas is in the tank. (a) Write this relationship in equation form and (b) determine the domain and range of the function in this context. 114. Volume of a cube: The volume of a cube depends on the length of the sides. In other words, volume is a function of the sides: V1s2 ⫽ s3. (a) In practical terms, what is the domain of this function? (b) Evaluate V(6.25) and (c) evaluate the function for s ⫽ 2x2. 115. Volume of a cylinder: For a fixed radius of 10 cm, the volume of a cylinder depends on its height. In other words, volume is a function of height:

V1h2 ⫽ 100␲h. (a) In practical terms, what is the domain of this function? (b) Evaluate V(7.5) and 8 (c) evaluate the function for h ⫽ . ␲ 116. Rental charges: Temporary Transportation Inc. rents cars (local rentals only) for a flat fee of $19.50 and an hourly charge of $12.50. This means that cost is a function of the hours the car is rented plus the flat fee. (a) Write this relationship in equation form; (b) find the cost if the car is rented for 3.5 hr; (c) determine how long the car was rented if the bill came to $119.75; and (d) determine the domain and range of the function in this context, if your budget limits you to paying a maximum of $150 for the rental. 117. Cost of a service call: Paul’s Plumbing charges a flat fee of $50 per service call plus an hourly rate of $42.50. This means that cost is a function of the hours the job takes to complete plus the flat fee. (a) Write this relationship in equation form; (b) find the cost of a service call that takes 212 hr; (c) find the number of hours the job took if the charge came to $262.50; and (d) determine the

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118. Predicting tides: The graph shown approximates the height of the tides at Fair Haven, New Brunswick, for a 12-hr period. (a) Is this the graph of a function? Why? (b) Approximately what time did high tide occur? (c) How high is the tide at 6 P.M.? (d) What time(s) will the tide be 2.5 m? 5

Meters

3 2 1

5

7

1.0

9

11 1 A.M.

6

8

10

12 2 A.M.

4

Time

3

Time


Distance in meters

120. A father challenges his son to a 400-m race, depicted in the graph shown here. 400 300 200 100 0

10

20

30

40

50

60

70

80

Time in seconds Father:

Son:

a. Who won and what was the approximate winning time? b. Approximately how many meters behind was the second place finisher? c. Estimate the number of seconds the father was in the lead in this race. d. How many times during the race were the father and son tied?

121. Sketch the graph of f 1x2 ⫽ x, then discuss how you could use this graph to obtain the graph of F1x2 ⫽ 冟x冟 without computing additional points. 冟x冟 What would the graph of g1x2 ⫽ look like? x 122. Sketch the graph of f 1x2 ⫽ x2 ⫺ 4, then discuss how you could use this graph to obtain the graph of F1x2 ⫽ 冟x2 ⫺ 4冟 without computing additional points. 冟x2 ⫺ 4冟 Determine what the graph of g1x2 ⫽ 2 would x ⫺4 look like. 123. If the equation of a function is given, the domain is implicitly defined by input values that generate real-valued outputs. But unless the graph is given or can be easily sketched, we must attempt to find the range analytically by solving for x in terms of y. We should note that sometimes this is an easy task, while at other times it is virtually impossible and we must rely on other methods. For the following functions, determine the implicit domain and find the range by solving for x in terms of y. a. y ⫽ xx

䊳

0.5

4 P.M.

4

3 P.M.

119. Predicting tides: The graph shown approximates the height of the tides at Apia, Western Samoa, for a 12-hr period. (a) Is this the graph of a function? Why? (b) Approximately what time did low tide occur? (c) How high is the tide at 2 A.M.? (d) What time(s) will the tide be 0.7 m? Meters

domain and range of the function in this context, if your insurance company has agreed to pay for all charges over $500 for the service call.

䊳

47


⫺ 3 ⫹ 2

b. y ⫽ x2 ⫺ 3


124. (1.1) Find the equation of a circle whose center is 14, ⫺12 with a radius of 5. Then graph the circle. 125. (Appendix A.6) Compute the sum and product indicated: a. 124 ⫹ 6 154 ⫺ 16 b. 12 ⫹ 132 12 ⫺ 132

126. (Appendix A.4) Solve the equation by factoring, then check the result(s) using substitution: 3x2 ⫺ 4x ⫽ 7. 127. (Appendix A.4) Factor the following polynomials completely: a. x3 ⫺ 3x2 ⫺ 25x ⫹ 75 b. 2x2 ⫺ 13x ⫺ 24 c. 8x3 ⫺ 125

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1–48


MID-CHAPTER CHECK Exercises 5 and 6 y L1

5

L2

2. Find the slope of the line passing through the given points: 1⫺3, 82 and 14, ⫺102 . 3. In 2009, Data.com lost $2 million. In 2010, they lost $0.5 million. Will the slope of the line through these points be positive or negative? Why? Calculate the slope. Were you correct? Write the slope as a unit rate and explain what it means in this context.

⫺5

5 x

⫺5

Exercises 7 and 8 y 5

h(x)

⫺5

5 x

⫺5

4. To earn some spending money, Sahara takes a job in a ski shop working primarily with her specialty—snowboards. She is paid a monthly salary of $950 plus a commission of $7.50 for each snowboard she sells. (a) Write a function that models her monthly earnings E. (b) Use a graphing calculator to determine her income if she sells 20, 30, or 40 snowboards in one month. (c) Use the results of parts a and b to set an appropriate viewing window and graph the line. (d) Use the TRACE feature to determine the number of snowboards that must be sold for Sahara’s monthly income to top $1300.

6. Write the equation for line L2 shown. Is this the graph of a function? Discuss why or why not. 7. For the graph of function h(x) shown, (a) determine the value of h(2); (b) state the domain; (c) determine the value(s) of x for which h1x2 ⫽ ⫺3; and (d) state the range. 8. Judging from the appearance of the graph alone, compare the rate of change (slope) from x ⫽ 1 to x ⫽ 2 to the rate of change from x ⫽ 4 to x ⫽ 5. Which rate of change is larger? How is that demonstrated graphically? Exercise 9 F 9. Compute the slope of the line F(p) shown, and explain what it means as a rate of change in this context. Then use the slope to predict the fox population when the pheasant population P is 13,000. Pheasant population (1000s) Fox population (in 100s)

1. Sketch the graph of the line 4x ⫺ 3y ⫽ 12. Plot and label at least three points.

10

9 8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

7

8

9 10

10. State the domain and range for each function below. y y a. b. 5

5

⫺5

5 x

⫺5

5 x

⫺5

⫺5

y

c.

5

⫺5

5. Write the equation for line L1 shown. Is this the graph of a function? Discuss why or why not.

5 x

⫺5

REINFORCING BASIC CONCEPTS Finding the Domain and Range of a Relation from Its Graph The concepts of domain and range are an important and fundamental part of working with relations and functions. In this chapter, we learned to determine the domain of any relation from its graph using a “vertical boundary line,” and the range by using a “horizontal boundary line.” These approaches to finding the domain and range can be combined into a single step by envisioning a rectangle drawn around or about the graph. If the entire graph can be “bounded” within the rectangle, the domain and range can be based on the rectangle’s related length and width. If it’s impossible to bound the graph in a particular direction, the related x- or y-values continue infinitely. Consider the graph in Figure 1.60. This is the graph of an ellipse (Section 8.2), and a rectangle that bounds the graph in all directions is shown in Figure 1.61.

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Reinforcing Basic Concepts

Figure 1.60

Figure 1.61

y

y

10

10

8

8

6

6

4

4

2

2

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

2

4

6

8 10

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

⫺4

⫺4

⫺6

⫺6

⫺8

⫺8

⫺10

⫺10

2

4

6

8 10

x

The rectangle extends from x ⫽ ⫺3 to x ⫽ 9 in the horizontal direction, and from y ⫽ 1 to y ⫽ 7 in the vertical direction. The domain of this relation is x 僆 3⫺3, 9 4 and the range is y 僆 31, 7 4 . The graph in Figure 1.62 is a parabola, and no matter how large we draw the rectangle, an infinite extension of the graph will extend beyond its boundaries in the left and right directions, and in the upward direction (Figure 1.63). Figure 1.62 Figure 1.63 y

y

10

10

8

8

6

6

4

4

2

2

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

2

4

6

8 10

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

⫺4

⫺4

⫺6

⫺6

⫺8

⫺8

⫺10

⫺10

2

4

6

8 10

x

The domain of this relation is x 僆 1⫺q, q 2 and the range is y 僆 3⫺6, q 2 . Finally, the graph in Figure 1.64 is the graph of a square root function, and a rectangle can be drawn that bounds the graph below and to the left, but not above or to the right (Figure 1.65). Figure 1.64 Figure 1.65 y

y

10

10

8

8

6

6

4

4

2

2

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

2

4

6

8 10

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

⫺4

⫺4

⫺6

⫺6

⫺8

⫺8

⫺10

⫺10

2

4

6

8 10

x

The domain of this relation is x 僆 3 ⫺7, q 2 and the range is y 僆 3 ⫺5, q 2 . Use this approach to find the domain and range of the following relations and functions. Exercise 1:

Exercise 2:

y 10 8 6 4 2 ⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2 4 6 8 10 x

Exercise 4:

y 10 8 6 4 2

10 8 6 4 2 2 4 6 8 10 x

Exercise 3:

y

⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

y 10 8 6 4 2

2 4 6 8 10 x

⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2 4 6 8 10 x

49

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Linear Functions, Special Forms, and More on Rates of Change


A. Write a linear equation in slope-intercept form and function form B. Use slope-intercept form to graph linear equations C. Write a linear equation in point-slope form D. Apply the slope-intercept form and point-slope form in context

EXAMPLE 1

䊳

The concept of slope is an important part of mathematics, because it gives us a way to measure and compare change. The value of an automobile changes with time, the circumference of a circle increases as the radius increases, and the tension in a spring grows the more it is stretched. The real world is filled with examples of how one change affects another, and slope helps us understand how these changes are related.

A. Linear Equations, Slope-Intercept Form and Function Form In Section 1.2, we learned that a linear equation is one that can be written in the form ax by c. Solving for y in a linear equation offers distinct advantages to understanding linear graphs and their applications.

Solving for y in a Linear Equation Solve 2y 6x 4 for y, then evaluate at x 4, x 0, and x 13.

Solution

䊳

2y 6x 4 2y 6x 4 y 3x 2

given equation add 6x divide by 2

Since the coefficients are integers, evaluate the function mentally. Inputs are multiplied by 3, then increased by 2, yielding the ordered pairs (4, 14), (0, 2), and 113, 12 . Now try Exercises 7 through 12

䊳

This form of the equation (where y has been written in terms of x) enables us to quickly identify what operations are performed on x in order to obtain y. Once again, for y 3x 2: multiply inputs by 3, then add 2. EXAMPLE 2

䊳

Solving for y in a Linear Equation Solve the linear equation 3y 2x 6 for y, then identify the new coefficient of x and the constant term.

Solution

䊳

3y 2x 6 3y 2x 6 2 y x2 3

given equation add 2x divide by 3

The coefficient of x is 23 and the constant term is 2. Now try Exercises 13 through 18

䊳

WORTHY OF NOTE In Example 2, the final form can be written y 23 x 2 as shown (inputs are multiplied by two-thirds, then increased by 2), or written as y 2x 3 2 (inputs are multiplied by two, the result divided by 3 and this amount increased by 2). The two forms are equivalent.

50

When the coefficient of x is rational, it’s helpful to select inputs that are multiples of the denominator if the context or application requires us to evaluate the equation. This enables us to perform most operations mentally. For y 23x 2, possible inputs might be x 9, 6, 0, 3, 6, and so on. See Exercises 19 through 24. In Section 1.2, linear equations were graphed using the intercept method. When the equation is written with y in terms of x, we notice a powerful connection between the graph and its equation—one that highlights the primary characteristics of a linear graph. 1–50

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Section 1.4 Linear Functions, Special Forms, and More on Rates of Change

EXAMPLE 3

䊳

Noting Relationships between an Equation and Its Graph Find the intercepts of 4x 5y 20 and use them to graph the line. Then, a. Use the intercepts to calculate the slope of the line, then identify the y-intercept. b. Write the equation with y in terms of x and compare the calculated slope and y-intercept to the equation in this form. Comment on what you notice.

Solution

䊳

Substituting 0 for x in 4x 5y 20, we find the y-intercept is 10, 42. Substituting 0 for y gives an x-intercept of 15, 02 . The graph is displayed here. a. The y-intercept is 10, 42 and by calculation or ¢y , the slope is m 4 counting 5 [from the ¢x intercept 15, 02 we count down 4, giving ¢y 4, and right 5, giving ¢x 5, to arrive at the intercept 10, 42 ]. b. Solving for y: given equation 4x 5y 20 5y 4x 20 subtract 4x 4 y x 4 divide by 5 5

y 5 4 3 2

(5, 0)

1

5 4 3 2 1 1

4

1

2

3

4

5

x

2 3

5

4

(0, 4)

5

The slope value seems to be the coefficient of x, while the y-intercept is the constant term. Now try Exercises 25 through 30

䊳

After solving a linear equation for y, an input of x 0 causes the “x-term” to become zero, so the y-intercept automatically involves the constant term. As Example 3 illustrates, we can also identify the slope of the line—it is the coefficient of x. In general, a linear equation of the form y mx b is said to be in slope-intercept form, since the slope of the line is m and the y-intercept is (0, b). Slope-Intercept Form For a nonvertical line whose equation is y mx b, the slope of the line is m and the y-intercept is (0, b). Solving a linear equation for y in terms of x is sometimes called writing the equation in function form, as this form clearly highlights what operations are performed on the input value in order to obtain the output (see Example 1). In other words, this form plainly shows that “y depends on x,” or “y is a function of x,” and that the equations y mx b and f 1x2 mx b are equivalent. Linear Functions A linear function is one of the form

f 1x2 mx b,

where m and b are real numbers. Note that if m 0, the result is a constant function f 1x2 b. If m 1 and b 0, the result is f 1x2 x, called the identity function.

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EXAMPLE 4

䊳

Finding the Function Form of a Linear Equation Write each equation in both slope-intercept form and function form. Then identify the slope and y-intercept of the line. a. 3x 2y 9 b. y x 5 c. 2y x

Solution

䊳

A. You’ve just seen how we can write a linear equation in slope-intercept form and function form

a. 3x 2y 9

b. y x 5

2y 3x 9

y x 5

3 9 y x 2 2 3 9 f 1x2 x 2 2 3 9 m ,b 2 2 9 y-intercept a0, b 2

y 1x 5 f 1x2 1x 5 m 1, b 5

c. 2y x x y 2 1 y x 2 1 f 1x2 x 2 1 m ,b0 2

y-intercept (0, 5)

y-intercept (0, 0)


䊳

Note that we can analytically develop the slope-intercept form of a line using the slope formula. Figure 1.66 shows the graph of a general line through the point (x, y) with a y-intercept of (0, b). Using these points in the slope formula, we have Figure 1.66

y2 y1 m x2 x1

y 5

(x, y)

5

5

x

(0, b)

yb m x0 yb m x y b mx y mx b

5

slope formula

substitute: (0, b) for (x1, y1), (x, y) for (x2, y2)

simplify multiply by x add b to both sides

This approach confirms the relationship between the graphical characteristics of a line and its slope-intercept form. Specifically, for any linear equation written in the form y mx b, the slope must be m and the y-intercept is (0, b).

B. Slope-Intercept Form and the Graph of a Line If the slope and y-intercept of a linear equation are known or can be found, we can construct its equation by substituting these values directly into the slope-intercept form y mx b. EXAMPLE 5

䊳

y

Finding the Equation of a Line from Its Graph

5

Find the slope-intercept equation of the line shown.

Solution

䊳

Using 13, 22 and 11, 22 in the slope formula, ¢y or by simply counting , the slope is m 42 or 21. ¢x By inspection we see the y-intercept is (0, 4). Substituting 21 for m and 4 for b in the slopeintercept form we obtain the equation y 2x 4.

(1, 2)

5

5

x

(3, 2) 5


䊳

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53


Actually, if the slope is known and we have any point (x, y) on the line, we can still construct the equation since the given point must satisfy the equation of the line. In this case, we’re treating y mx b as a simple formula, solving for b after substituting known values for m, x, and y. EXAMPLE 6

䊳

Using y ⫽ mx ⫹ b as a Formula Find the slope-intercept equation of a line that has slope m 45 and contains 15, 22. Verify results on a graphing calculator.

Solution

䊳

10

Use y mx b as a “formula,” with m 45, x 5, and y 2. y mx b 2 45 152 b 2 4 b 6b

10

slope-intercept form

10

substitute 45 for m, 5 for x, and 2 for y simplify 10 (5, 2) is on the line

solve for b

The equation of the line is y 6. After entering the equation on the Y= screen of a graphing calculator, we can evaluate x 5 on the home screen, or use the TRACE feature. See the figures provided. 4 5x


䊳

Writing a linear equation in slope-intercept form enables us to draw its graph with a minimum of effort, since we can easily locate the y-intercept and a second point using ¢y ¢y 2 . For instance, the rate of change indicates that counting down 2 and ¢x ¢x 3 right 3 from a known point will locate another point on this line. EXAMPLE 7

䊳

Graphing a Line Using Slope-Intercept Form and the Rate of Change Write 3y 5x 9 in slope-intercept form, then graph the line using the y-intercept and the rate of change (slope).

Solution

WORTHY OF NOTE Noting the fraction 35 is equal to 5 3 , we could also begin at (0, 3) and ¢y 5 count (down 5 and left 3) ¢x 3 to find an additional point on the line: 13, 22 . Also, for any ¢y a , note negative slope ¢x b a a a . b b b

䊳

3y 5x 9 3y 5x 9 y 53x 3

y fx 3 y

Run 3

given equation isolate y term Rise 5

divide by 3

The slope is m 53 and the y-intercept is (0, 3). ¢y 5 (up 5 and Plot the y-intercept, then use ¢x 3 right 3 — shown in blue) to find another point on the line (shown in red). Finish by drawing a line through these points.

(3, 8)

y f x (0, 3)

5

5

x

2


䊳

For a discussion of what graphing method might be most efficient for a given linear equation, see Exercises 103 and 114.

Parallel and Perpendicular Lines From Section 1.2 we know parallel lines have equal slopes: m1 m2, and perpendicular 1 lines have slopes with a product of 1: m1 # m2 1 or m1 . In some applications, m2 we need to find the equation of a second line parallel or perpendicular to a given line, through a given point. Using the slope-intercept form makes this a simple four-step process.

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Finding the Equation of a Line Parallel or Perpendicular to a Given Line 1. Identify the slope m1 of the given line. 2. Find the slope m2 of the new line using the parallel or perpendicular relationship. 3. Use m2 with the point (x, y) in the “formula” y ⫽ mx ⫹ b and solve for b. 4. The desired equation will be y ⫽ m2 x ⫹ b.

EXAMPLE 8

䊳

Finding the Equation of a Parallel Line

Solution

䊳

Begin by writing the equation in slope-intercept form to identify the slope.

Find the slope-intercept equation of a line that goes through 1⫺6, ⫺12 and is parallel to 2x ⫹ 3y ⫽ 6. 2x ⫹ 3y ⫽ 6 3y ⫽ ⫺2x ⫹ 6 y ⫽ ⫺2 3 x ⫹ 2

given line isolate y-term result

The original line has slope m1 ⫽ ⫺2 3 and this will also be the slope of any line parallel to it. Using m2 ⫽ ⫺2 3 with 1x, y2 S 1⫺6, ⫺12 we have y ⫽ mx ⫹ b ⫺2 1⫺62 ⫹ b ⫺1 ⫽ 3 ⫺1 ⫽ 4 ⫹ b ⫺5 ⫽ b

The equation of the new line is y ⫽

⫺2 3 x

slope-intercept form substitute ⫺2 3 for m , ⫺6 for x, and ⫺1 for y simplify solve for b

⫺ 5. Now try Exercises 63 through 76

䊳

31 Graphing the lines from Example 8 as Y1 and Y2 on a graphing calculator, we note the lines do appear to be parallel (they actually must be since they have identical slopes). Using the 47 ZOOM 8:ZInteger feature of the calculator, we ⫺47 can quickly verify that Y2 indeed contains the point (⫺6, ⫺1). For any nonlinear graph, a straight line ⫺31 drawn through two points on the graph is called a secant line. The slope of a secant line, and lines parallel and perpendicular to this line, play fundamental roles in the further development of the rate-of-change concept.

EXAMPLE 9

䊳

Finding Equations for Parallel and Perpendicular Lines A secant line is drawn using the points (⫺4, 0) and (2, ⫺2) on the graph of the function shown. Find the equation of a line that is a. parallel to the secant line through (⫺1, ⫺4). b. perpendicular to the secant line through (⫺1, ⫺4).

Solution

䊳

¢y Either by using the slope formula or counting , we find the secant line has slope ¢x ⫺2 ⫺1 m⫽ . ⫽ 6 3

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WORTHY OF NOTE The word “secant” comes from the Latin word secare, meaning “to cut.” Hence a secant line is one that cuts through a graph, as opposed to a tangent line, which touches the graph at only one point.

a. For the parallel line through (1, 4), m2 y mx b 1 112 b 4 3 12 1 b 3 3 13 b 3

1 . 3

y 5

slope-intercept form substitute 1 3 for m , 1 for x, and 4 for y

5

(1, 4) result

x

5

x

5

13 1 . x 3 3

b. For the perpendicular line through (1, 4), m2 3. y mx b 4 3112 b 4 3 b 1 b

5

simplify 14 12 32

The equation of the parallel line (in blue) is y

y 5

slope-intercept form substitute 3 for m, 1 for x, and 4 for y simplify

5

result

The equation of the perpendicular line (in yellow) is y 3x 1. B. You’ve just seen how we can use the slope-intercept form to graph linear equations

55


(1, 4)

5


䊳

C. Linear Equations in Point-Slope Form As an alternative to using y mx b, we can find the equation of the line using the y2 y1 m, and the fact that the slope of a line is constant. For a given slope formula x2 x1 slope m, we can let (x1, y1) represent a given point on the line and (x, y) represent any y y1 m. Isolating the “y” terms other point on the line, and the formula becomes x x1 on one side gives a new form for the equation of a line, called the point-slope form: y y1 m x x1 1x x1 2 y y1 a b m1x x1 2 x x1 1 y y1 m1x x1 2

slope formula

multiply both sides by (x x1) simplify S point-slope form

The Point-Slope Form of a Linear Equation

For a nonvertical line whose equation is y y1 m1x x1 2 , the slope of the line is m and (x1, y1) is a point on the line.

While using y mx b (as in Example 6) may appear to be easier, both the slope-intercept form and point-slope form have their own advantages and it will help to be familiar with both.

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EXAMPLE 10

䊳

Using y y1 m(x x1) as a Formula Find the equation of the line in point-slope form, if m 23 and (3, 3) is on the line. Then graph the line. y

Solution

䊳

C. You’ve just seen how we can write a linear equation in point-slope form

y y1 m1x x1 2 2 y 132 3 x 132 4 3 2 y 3 1x 32 3

5

point-slope form

y 3 s (x 3)

substitute 23 for m; (3, 3) for (x1, y1) simplify, point-slope form

¢y 2 To graph the line, plot (3, 3) and use ¢x 3 to find additional points on the line.

x3

5

5

x

y2 (3, 3) 5


䊳

D. Applications of Linear Equations As a mathematical tool, linear equations rank among the most common, powerful, and versatile. In all cases, it’s important to remember that slope represents a rate of change. ¢y The notation m literally means the quantity measured along the y-axis, is chang¢x ing with respect to changes in the quantity measured along the x-axis. EXAMPLE 11

䊳

Relating Temperature to Altitude In meteorological studies, atmospheric temperature depends on the altitude according to the formula T1h2 3.5h 58.5, where T(h) represents the approximate Fahrenheit temperature at height h (in thousands of feet, 0 h 36). a. Interpret the meaning of the slope in this context. b. Determine the temperature at an altitude of 12,000 ft. c. If the temperature is 8°F what is the approximate altitude?

Algebraic Solution

䊳

3.5 ¢T , ¢h 1 meaning the temperature drops 3.5°F for every 1000-ft increase in altitude. b. Since height is in thousands, use h 12. a. Notice that h is the input variable and T is the output. This shows

T1h2 3.5h 58.5 T1122 3.51122 58.5 16.5

Technology Solution

original formula substitute 12 for h result

䊳

At a height of 12,000 ft, the temperature is about 16.5°.

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c. Replacing T(h) with 8 and solving gives

Algebraic Solution

䊳

T1h2 3.5h 58.5 8 3.5h 58.5 66.5 3.5h 19 h

original formula substitute 8 for T(h) subtract 58.5 divide by 3.5

The temperature is about 8°F at a height of 19 1000 19,000 ft.

Graphical Solution

䊳

Since we’re given 0 h 36, we can set Xmin 0 and Xmax 40. At ground level 1x 02 , the formula gives a temperature of 58.5°, while at h 36, we have T1362 67.5. This shows appropriate settings for the range would be Ymin 50 and Ymax 50 (see figure). After setting Y1 3.5X 58.5, we press TRACE and move the cursor until we find an output value near 8, which occurs when X is near 19. To check, we input 19 for x and the calculator displays an output of 8, which corresponds with the algebraic result (at 19,000 ft, the temperature is 8°F). 50

0

40

50


䊳

In many applications, outputs that are integer or rational values are rare, making it difficult to use the TRACE feature alone to find an exact solution. In the Section 1.5, we’ll develop additional ways that graphs and technology can be used to solve equations. In some applications, the relationship is known to be linear but only a few points on the line are given. In this case, we can use two of the known data points to calculate the slope, then the point-slope form to find an equation model. One such application is linear depreciation, as when a government allows businesses to depreciate vehicles and equipment over time (the less a piece of equipment is worth, the less you pay in taxes). EXAMPLE 12A

䊳

Using Point-Slope Form to Find a Function Model Five years after purchase, the auditor of a newspaper company estimates the value of their printing press is $60,000. Eight years after its purchase, the value of the press had depreciated to $42,000. Find a linear equation that models this depreciation and discuss the slope and y-intercept in context.

Solution

䊳

Since the value of the press depends on time, the ordered pairs have the form (time, value) or (t, v) where time is the input, and value is the output. This means the ordered pairs are (5, 60,000) and (8, 42,000). v2 v1 t2 t1 42,000 60,000 85 18,000 6000 3 1

m

slope formula 1t1, v1 2 15, 60,0002; 1t2, v2 2 18, 42,0002 simplify and reduce

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6000 ¢value , indicating the printing press loses ¢time 1 $6000 in value with each passing year. The slope of the line is

WORTHY OF NOTE Actually, it doesn’t matter which of the two points are used in Example 12A. Once the point (5, 60,000) is plotted, a constant slope of m 6000 will “drive” the line through (8, 42,000). If we first graph (8, 42,000), the same slope would “drive” the line through (5, 60,000). Convince yourself by reworking the problem using the other point.

v v1 m1t t1 2 v 60,000 60001t 52 v 60,000 6000t 30,000 v 6000t 90,000

point-slope form substitute 6000 for m; (5, 60,000) for (t1, v1) simplify solve for v

The depreciation equation is v1t2 6000t 90,000. The v-intercept (0, 90,000) indicates the original value (cost) of the equipment was $90,000. Once the depreciation equation is found, it represents the (time, value) relationship for all future (and intermediate) ages of the press. In other words, we can now predict the value of the press for any given year. However, note that some equation models are valid for only a set period of time, and each model should be used with care.

EXAMPLE 12B

䊳

Using a Function Model to Gather Information From Example 12A, a. How much will the press be worth after 11 yr? b. How many years until the value of the equipment is $9000? c. Is this function model valid for t 18 yr (why or why not)?

Solution

䊳

a. Find the value v when t 11: v1t2 6000t 90,000 v1112 60001112 90,000 24,000

equation model substitute 11 for t result (11, 24,000)

After 11 yr, the printing press will only be worth $24,000. b. “. . . value is $9000” means v1t2 9000: v1t2 9000 6000t 90,000 9000 6000t 81,000 t 13.5

D. You’ve just seen how we can apply the slopeintercept form and point-slope form in context

value at time t substitute 6000t 90,000 for v (t ) subtract 90,000 divide by 6000

After 13.5 yr, the printing press will be worth $9000. c. Since substituting 18 for t gives a negative quantity, the function model is not valid for t 18. In the current context, the model is only valid while v 0 and solving 6000t 90,000 0 shows the domain of the function in this context is t 30, 15 4 . Now try Exercises 107 through 112

䊳

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1.4 EXERCISES 䊳



7 x 3, the slope is 4 and the y-intercept is .

1. For the equation y

3. Line 1 has a slope of 0.4. The slope of any line perpendicular to line 1 is . 5. Discuss/Explain how to graph a line using only the slope and a point on the line (no equations).

䊳

¢cost indicates the ¢time changing in response to changes in

2. The notation

is .

4. The equation y y1 m1x x1 2 is called the form of a line.

6. Given m 35 and 15, 62 is on the line. Compare and contrast finding the equation of the line using y mx b versus y y1 m1x x1 2.


Solve each equation for y and evaluate the result using x 5, x 2, x 0, x 1, and x 3.

7. 4x 5y 10

8. 3y 2x 9

9. 0.4x 0.2y 1.4 10. 0.2x 0.7y 2.1 11.

1 3x

1 5y

1

12.

1 7y

1 3x

2

For each equation, solve for y and identify the new coefficient of x and new constant term.

13. 6x 3y 9

14. 9y 4x 18

15. 0.5x 0.3y 2.1 16. 0.7x 0.6y 2.4 17. 65x 17y 47

18.

7 12 y

4 15 x 76

Write each equation in slope-intercept form (solve for y) and function form, then identify the slope and y-intercept.

31. 2x 3y 6

32. 4y 3x 12

33. 5x 4y 20

34. y 2x 4

35. x 3y

36. 2x 5y

37. 3x 4y 12 0

38. 5y 3x 20 0

For Exercises 39 to 50, use the slope-intercept form to state the equation of each line. Verify your solutions to Exercises 45 to 47 using a graphing calculator.

39.

Evaluate each equation by selecting three inputs that will result in integer values. Then graph each line.

19. y 43x 5

20. y 54x 1

21. y 32x 2

22. y 25x 3

23. y 16x 4

24. y 13x 3

Find the x- and y-intercepts for each line, then (a) use these two points to calculate the slope of the line, (b) write the equation with y in terms of x (solve for y) and (c) compare the calculated slope and y-intercept to the equation from part (b). Comment on what you notice.

25. 3x 4y 12

26. 3y 2x 6

27. 2x 5y 10

28. 2x 3y 9

29. 4x 5y 15

30. 5y 6x 25

40.

y 5 4 3 2 1 54321 1 2 (3, 1) 3 4 5

41.

(5, 5)

(3, 3) (0, 1) 1 2 3 4 5 x

54321 1 2 3 4 5

y

(1, 0)

5 4 3 (0, 3) 2 1

54321 1 2 (2, 3) 3 4 5

y 5 4 (0, 3) 3 2 1

1 2 3 4 5 x

42. m 2; y-intercept 10, 32 43. m 3; y-intercept (0, 2)

44. m 3 2 ; y-intercept 10, 42

45. m 4; 13, 22 is on the line

(5, 1)

1 2 3 4 5 x

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46. m 2; 15, 32 is on the line 47. m

3 2 ;

48.

y

50.

14, 72 is on the line 49.

10,000

y 1500

8000

1200

6000

900

4000

600

2000

300 12

Write the equations in slope-intercept form and state whether the lines are parallel, perpendicular, or neither.

14

16

18

20 x

26

28

30

32

71. 4y 5x 8 5y 4x 15

72. 3y 2x 6 2x 3y 6

73. 2x 5y 20 4x 3y 18

74. 4x 6y 12 2x 3y 6

75. 3x 4y 12 6x 8y 2

76. 5y 11x 135 11y 5x 77

34 x

A secant line is one that intersects a graph at two or more points. For each graph given, find an equation of the line (a) parallel and (b) perpendicular to the secant line, through the point indicated.

y 2000 1600 1200 800

77.

400

78.

y 5

y 5

(1, 3) 8

10

12

14

16 x

Write each equation in slope-intercept form, then use the rate of change (slope) and y-intercept to graph the line.

51. 3x 5y 20

52. 2y x 4

53. 2x 3y 15

54. 3x 2y 4

5

55. y 23x 3

56. y 52x 1

57. y 1 3 x 2

58. y 4 5 x 2

59. y 2x 5

60. y 3x 4

61. y 12x 3

62. y 3 2 x 2

(2, 4)

5

79.

5 x

5

80.

y 5

Graph each linear equation using the y-intercept and rate of change (slope) determined from each equation.

5

5 x

y 5

(1, 3)

5

5

5 x

5

81.

5 x

5

82.

y 5

(1, 2.5)

y 5

(1, 3)

Find the equation of the line using the information given. Write answers in slope-intercept form.

63. parallel to 2x 5y 10, through the point 15, 22 64. parallel to 6x 9y 27, through the point 13, 52

65. perpendicular to 5y 3x 9, through the point 16, 32 66. perpendicular to x 4y 7, through the point 15, 32 67. parallel to 12x 5y 65, through the point 12, 12 68. parallel to 15y 8x 50, through the point 13, 42 69. parallel to y 3, through the point (2, 5)

70. perpendicular to y 3 through the point (2, 5)

5

5 x

5

5 x

(0, 2) 5

5

Find the equation of the line in point-slope form, then graph the line.

83. m 2; P1 12, 52

84. m 1; P1 12, 32

85. P1 13, 42, P2 111, 12 86. P1 11, 62, P2 15, 12

87. m 0.5; P1 11.8, 3.12

88. m 1.5; P1 10.75, 0.1252

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Find the equation of the line in point-slope form, and state the meaning of the slope in context — what information is the slope giving us?

89.

90.

y Typewriters in service (in ten thousands)

Income (in thousands)

y 10 9 8 7 6 5 4 3 2 1 0

x

1 2 3 4 5 6 7 8 9

10 9 8 7 6 5 4 3 2 1 0

Student’s final grade (%) (includes extra credit)

x Hours of television per day 0

93.

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

60 40 20

1

2

3

4

Rainfall per month (in inches)

5

x

y C

y D

x y F

x y G

x

x y H

x

x

95. While driving today, I got stopped by a state trooper. After she warned me to slow down, I continued on my way.

x Independent investors (1000s) 1 2 3 4 5 6 7 8 9 10

y Eggs per hen per week

Cattle raised per acre

80

y E

x

10 9 8 7 6 5 4 3 2 1 0

94.

y 100

0

x

1 2 3 4 5 6 7 8 9

y Online brokerage houses

92.

y

y B

x

Year (1990 → 0)

100 90 80 70 60 50 40 30 20 10

Using the concept of slope, match each description with the graph that best illustrates it. Assume time is scaled on the horizontal axes, and height, speed, or distance from the origin (as the case may be) is scaled on the vertical axis. y A

Sales (in thousands)

91.

61


96. After hitting the ball, I began trotting around the bases shouting, “Ooh, ooh, ooh!” When I saw it wasn’t a home run, I began sprinting. 97. At first I ran at a steady pace, then I got tired and walked the rest of the way.

10 8 6

98. While on my daily walk, I had to run for a while when I was chased by a stray dog.

4 2 0

60

65

70

75

80

x

Temperature in °F

99. I climbed up a tree, then I jumped out. 100. I steadily swam laps at the pool yesterday. 101. I walked toward the candy machine, stared at it for a while then changed my mind and walked back. 102. For practice, the girls’ track team did a series of 25-m sprints, with a brief rest in between.

䊳


103. General linear equation: ax by c The general equation of a line is shown here, where a, b, and c are real numbers, with a and b not simultaneously zero. Solve the equation for y and note the slope (coefficient of x) and y-intercept (constant term). Use these to find the slope and y-intercept of the following lines, without solving for y or computing points. a. 3x 4y 8 b. 2x 5y 15 c. 5x 6y 12 d. 3y 5x 9

104. Intercept-Intercept form of a linear y x equation: 1 h k The x- and y-intercepts of a line can also be found by writing the equation in the form shown (with the equation set equal to 1). The x-intercept will be (h, 0) and the y-intercept will be (0, k). Find the x- and y-intercepts of the following lines using this method. How is the slope of each line related to the values of h and k? a. 2x 5y 10 b. 3x 4y 12 c. 5x 4y 8

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1–62


APPLICATIONS

105. Speed of sound: The speed of sound as it travels through the air depends on the temperature of the air according to the function V 35T 331, where V represents the velocity of the sound waves in meters per second (m/s), at a temperature of T° Celsius. (a) Interpret the meaning of the slope and y-intercept in this context. (b) Determine the speed of sound at a temperature of 20°C. (c) If the speed of sound is measured at 361 m/s, what is the temperature of the air? 106. Acceleration: A driver going down a straight highway is traveling 60 ft/sec (about 41 mph) on cruise control, when he begins accelerating at a rate of 5.2 ft/sec2. The final velocity of the car is given by V 26 5 t 60, where V is the velocity at time t. (a) Interpret the meaning of the slope and y-intercept in this context. (b) Determine the velocity of the car after 9.4 seconds. (c) If the car is traveling at 100 ft/sec, for how long did it accelerate? 107. Investing in coins: The purchase of a “collector’s item” is often made in hopes the item will increase in value. In 1998, Mark purchased a 1909-S VDB Lincoln Cent (in fair condition) for $150. By the year 2004, its value had grown to $190. (a) Use the relation (time since purchase, value) with t 0 corresponding to 1998 to find a linear equation modeling the value of the coin. (b) Discuss what the slope and y-intercept indicate in this context. (c) How much was the penny worth in 2009? (d) How many years after purchase will the penny’s value exceed $250? (e) If the penny is now worth $170, how many years has Mark owned the penny? 108. Depreciation: Once a piece of equipment is put into service, its value begins to depreciate. A business purchases some computer equipment for $18,500. At the end of a 2-yr period, the value of the equipment has decreased to $11,500. (a) Use the relation (time since purchase, value) to find a linear equation modeling the value of the equipment. (b) Discuss what the slope and y-intercept indicate in this context. (c) What is the equipment’s value after 4 yr? (d) How many years after purchase will the value decrease to $6000? (e) Generally, companies will sell used equipment while it still has value and use the funds to purchase new equipment. According to the function, how many years will it take this equipment to depreciate in value to $1000?

109. Internet connections: The number of households that are hooked up to the Internet (homes that are online) has been increasing steadily in recent years. In 1995, approximately 9 million homes were online. By 2001 this figure had climbed to about 51 million. (a) Use the relation (year, homes online) with t 0 corresponding to 1995 to find an equation model for the number of homes online. (b) Discuss what the slope indicates in this context. (c) According to this model, in what year did the first homes begin to come online? (d) If the rate of change stays constant, how many households were on the Internet in 2006? (e) How many years after 1995 will there be over 100 million households connected? (f) If there are 115 million households connected, what year is it? Source: 2004 Statistical Abstract of the United States, Table 965

110. Prescription drugs: Retail sales of prescription drugs have been increasing steadily in recent years. In 1995, retail sales hit $72 billion. By the year 2000, sales had grown to about $146 billion. (a) Use the relation (year, retail sales of prescription drugs) with t 0 corresponding to 1995 to find a linear equation modeling the growth of retail sales. (b) Discuss what the slope indicates in this context. (c) According to this model, in what year will sales reach $250 billion? (d) According to the model, what was the value of retail prescription drug sales in 2005? (e) How many years after 1995 will retail sales exceed $279 billion? (f) If yearly sales totaled $294 billion, what year is it? Source: 2004 Statistical Abstract of the United States, Table 122

111. Prison population: In 1990, the number of persons sentenced and serving time in state and federal institutions was approximately 740,000. By the year 2000, this figure had grown to nearly 1,320,000. (a) Find a linear function with t 0 corresponding to 1990 that models this data, (b) discuss the slope ratio in context, and (c) use the equation to estimate the prison population in 2010 if this trend continues. Source: Bureau of Justice Statistics at www.ojp.usdoj.gov/bjs

112. Eating out: In 1990, Americans bought an average of 143 meals per year at restaurants. This phenomenon continued to grow in popularity and in the year 2000, the average reached 170 meals per year. (a) Find a linear function with t 0 corresponding to 1990 that models this growth, (b) discuss the slope ratio in context, and (c) use the equation to estimate the average number of times an American will eat at a restaurant in 2010 if the trend continues. Source: The NPD Group, Inc., National Eating Trends, 2002

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113. Locate and read the following article. Then turn in a one-page summary. “Linear Function Saves Carpenter’s Time,” Richard Crouse, Mathematics Teacher, Volume 83, Number 5, May 1990: pp. 400–401. 114. The general form of a linear equation is ax by c, where a and b are not simultaneously zero. (a) Find the x- and y-intercepts using the general form (substitute 0 for x, then 0 for y). Based on what you see, when does the intercept method work most efficiently? (b) Find the slope and y-intercept using the general form (solve for y). Based on what you see, when does the slopeintercept method work most efficiently?

115. Match the correct graph to the conditions stated for m and b. There are more choices than graphs. a. m 6 0, b 6 0 b. m 7 0, b 6 0 c. m 6 0, b 7 0 d. m 7 0, b 7 0 e. m 0, b 7 0 f. m 6 0, b 0 g. m 7 0, b 0 h. m 0, b 6 0 (1)

y

(2)

y

x

(4)

y

y

(3)

x

(5)

x

䊳

63


y

x

y

(6)

x

x


116. (1.3) Determine the domain: a. y 12x 5 5 b. y 2x 5 117. (Appendix A.6) Simply without the use of a calculator. 2 a. 273 b. 281x2 118. (Appendix A.3) Three equations follow. One is an identity, another is a contradiction, and a third has a solution. State which is which. 21x 52 13 1 9 7 2x 21x 42 13 1 9 7 2x 21x 52 13 1 9 7 2x

119. (Appendix A.2) Compute the area of the circular sidewalk shown here 1A ␲r2 2 . Use your calculator’s value of ␲ and round the answer (only) to hundredths. 10 yd

8 yd

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Solving Equations and Inequalities Graphically; Formulas and Problem Solving


A. Solve equations

B.

C. D.

E.

graphically using the intersection-of-graphs method Solve equations graphically using the x-intercept/zeroes method Solve linear inequalities graphically Solve for a specified variable in a formula or literal equation Use a problem-solving guide to solve various problem types

In this section, we’ll build on many of the ideas developed in Appendix A.3 (Solving Linear Equations and Inequalities), as we learn to manipulate formulas and employ certain problem-solving strategies. We will also extend our understanding of graphical solutions to a point where they can be applied to virtually any family of functions.

A. Solving Equations Graphically Using the Intersect Method For some background on why a graphical solution is effective, consider the equation 2x 9 31x 12 2. By definition, an equation is a statement that two expressions are equal for some value of the variable (Appendix A.3). To highlight this fact, the expressions 2x 9 and 31x 12 2 are evaluated independently for selected integers in Tables 1.4 and 1.5. Table 1.4

Table 1.5

x

2x ⫺ 9

x

⫺3(x ⫺ 1) ⫺ 2

3

15

3

10

2

13

2

7

1

11

1

4

0

9

0

1

1

7

1

2

2

5

2

5

3

3

3

8

Note the two expressions are equal (the equation is true) only when the input is x 2. Solving equations graphically is a simple extension of this observation. By treating the expression on the left as the independent function Y1, we have Y1 2X 9 and the related linear graph will contain all ordered pairs shown in Table 1.4 (see Figure 1.67). Doing the same for the right-hand expression yields Y2 31X 12 2, and its related graph will likewise contain all ordered pairs shown in the Table 1.5 (see Figure 1.68).

f

∂

2x 9 31x 12 2 Y1

Y2

The solution is then found where Y1 Y2, or in other words, at the point where these two lines intersect (if it exists). See Figure 1.69. Most graphing calculators have an intersect feature that can quickly find the point(s) where two graphs intersect. On many calculators, we access this ability using the sequence 2nd TRACE (CALC) and selecting option 5:intersect (Figure 1.70).

10

10

64

10

10

10

10

Figure 1.69

Figure 1.68

Figure 1.67

10

10

10

10

10

10

1–64

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Section 1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving

Figure 1.71 Figure 1.70

65

Figure 1.72 10

10

10

10

10

10

10

10

Because the calculator can work with up to 10 Figure 1.73 expressions at once, it will ask you to identify 10 each graph you want to work with—even when there are only two. A marker is displayed on each graph in turn, and named in the upper left 10 corner of the window (Figure 1.71). You can 10 select a graph by pressing , or bypass a graph by pressing one of the arrow keys. For situations involving multiple graphs or multiple 10 solutions, the calculator offers a “GUESS?” option that enables you to specify the approximate location of the solution you’re interested in (Figure 1.72). For now, we’ll simply press two times in succession to identify each graph, and a third time to bypass the “GUESS?” option. The calculator then finds and displays the point of intersection (Figure 1.73). Be sure to check the settings on your viewing window before you begin, and if the point of intersection is not visible, try ZOOM 3:Zoom Out or other windowresizing features to help locate it. ENTER

ENTER

EXAMPLE 1A

䊳

Solving an Equation Graphically 1 Solve the equation 21x 32 7 x 2 using 2 a graphing calculator.

Solution

䊳

Begin by entering the left-hand expression as Y1 and the right-hand expression as Y2 (Figure 1.74). To find points of intersection, press 2nd TRACE (CALC) and select option 5:intersect, which automatically places you on the graphing window, and asks you to identify the “First curve?.” As discussed, pressing three times in succession will identify each graph, bypass the “Guess?” option, then find and display the point of intersection (Figure 1.75). Here the point of intersection 10 is (2, 3), showing the solution to this equation is x 2 (for which both expressions equal 3). This can be verified by direct substitution or by using the TABLE feature. ENTER

Figure 1.74

Figure 1.75 10

10

10

This method of solving equations is called the Intersection-of-Graphs method, and can be applied to many different equation types.

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Intersection-of-Graphs Method for Solving Equations For any equation of the form f(x) g(x), 1. Assign f (x) as Y1 and g (x) as Y2. 2. Graph both function and identify any point(s) of intersection, if they exist. The x-coordinate of all such points is a solution to the equation. Recall that in the solution of linear equations, we sometimes encounter equations that are identities (infinitely many solutions) or contradictions (no solutions). These possibilities also have graphical representations, and appear as coincident lines and parallel lines respectively. These possibilities are illustrated in Figure 1.76. Figure 1.76 y

y

y

x

x

One point of intersection (unique solution)

EXAMPLE 1B

䊳

Infinitely many points of intersection (identity)

x

No point of intersection (contradiction)

Solving an Equation Graphically Solve 0.75x 2 0.511 1.5x2 3 using a graphing calculator.

Solution

䊳

With 0.75X 2 as Y1 and 0.511 1.5X2 3 as Y2, we use the 2nd TRACE (CALC) option and select 5:intersect. The graphs appear to be parallel lines (Figure 1.77), and after pressing three times we obtain the error message shown (Figure 1.78), confirming there are no solutions. ENTER

Figure 1.78

Figure 1.77 10

10

A. You’ve just seen how we can solve equations using the Intersection-of-Graphs method

10

10


䊳

B. Solving Equations Graphically Using the x-Intercept/Zeroes Method The intersection-of-graphs method works extremely well when the graphs of f(x) and g(x) (Y1 and Y2) are simple and “well-behaved.” Later in this course, we encounter a number of graphs that are more complex, and it will help to develop alternative methods for solving graphically. Recall that two equations are equivalent if they have the same solution set. For instance, the equations 2x 6 and 2x 6 0 are equivalent (since x 3 is a solution to both), as are 3x 1 x 5 and 2x 6 0 (since

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x 3 is a solution to both). Applying the intersection-of-graphs method to the last two equivalent equations, gives

Y2

2x 6 0 5

Y1

and

f

The intersection-of-graphs method focuses on a point of intersection (a, b), and names x a as the solution. The zeroes method focuses on the input x r, for which the output is 0 3 Y1 1r2 0 4 . All such values r are called the zeroes of the function. Much more will be said about functions and their zeroes in later chapters.

f

WORTHY OF NOTE

f

3x 1 x 5

Y1

Y2

The intersection method will work equally well in both cases, but the equation on the right has only one variable expression, and will produce a single (visible) graph (since Y2 0 is simply the x-axis). Note that here we seek an input value that will result in an output of 0. In other words, all solutions will have the form (x, 0), which is the x-intercept of the graph. For this reason, the method is alternatively called the zeroes method or the x-intercept method. The method employs the approach shown above, in which the equation f 1x2 g1x2 is rewritten as f 1x2 g1x2 0, with f 1x2 g1x2 assigned as Y1. Zeroes/x-Intercept Method for Solving Equations

For any equation of the form f 1x2 g1x2, 1. Rewrite the equation as f 1x2 g1x2 0. 2. Assign f 1x2 g1x2 as Y1. 3. Graph the resulting function and identify any x-intercepts, if they exist. Any x-intercept(s) of the graph will be a solution to the equation. To locate the zero (x-intercept) for 2x 6 0 on a graphing calculator, enter 2X 6 for Y1 and use the 2:zero option found on the same menu as the 5:intercept option (Figure 1.79). Since some equations have more than one zero, the 2:zero option will ask you to “narrow down” the interval it should search, even though there is only one zero here. It does this by asking for a “Left Bound?”, a “Right Bound?”, and a “GUESS?” (the Guess? option can once again be bypassed). You can enter these bounds by tracing along the graph or by inputting a chosen value, then pressing (note how the calculator posts a marker at each Figure 1.79 bound). Figure 1.80 shows we entered x 0 as the left bound and x 4 as the right, and the calculator will search for the x-intercept in this interval (note that in general, the cursor will be either above or below the x-axis for the left bound, but must be on the opposite side of the x-axis for the right bound). Pressing once more bypasses the Guess? option and locates the x-intercept at (3, 0). The solution is x 3 (Figure 1.81). ENTER

ENTER

Figure 1.80

Figure 1.81 10

10

10

10

10

10

10

10

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EXAMPLE 2

䊳

Solving an Equation Using the Zeroes Method Solve the equation 41x 32 6 2x 3 using the zeroes method.

Solution

䊳

Figure 1.82 As given, we have f 1x2 41x 32 6 10 and g1x2 2x 3. Rewriting the equation as f 1x2 g1x2 0 gives 41x 32 6 12x 32 0, where the expression for g(x) is parenthesized to ensure 10 10 the equations remain equivalent. Entering 41X 32 6 12X 32 as Y1 and pressing 2nd TRACE (CALC) 2:zero produces the screen shown in Figure 1.82, with the 10 calculator requesting a left bound. We can input any x-value that is obviously to the left of the x-intercept, or move the cursor to any position left of the x-intercept and press (we input x 4, see Figure 1.83). The calculator then asks for a right bound and as before we can input any x-value obviously to the right, or simply move the cursor to any location on the opposite side of the x-axis and press (we chose x 2, see Figure 1.84). After bypassing the Guess? option (press once again), the calculator locates the x-intercept at (3.5, 0), and the solution to the original equation is x 3.5 (Figure 1.85). ENTER

ENTER

ENTER

Figure 1.83

Figure 1.84

10

10

10

10

B. You’ve just seen how we can solve equations using the x-intercept/zeroes method

Figure 1.85 10

10

10

10

10

10

10

10


䊳

C. Solving Linear Inequalities Graphically The intersection-of-graphs method can also be applied to solve linear inequalities. The point of intersection simply becomes one of the boundary points for the solution interval, and is included or excluded depending on the inequality given. For the inequality f 1x2 7 g1x2 written as Y1 7 Y2, it becomes clear the inequality is true for all inputs x where the outputs for Y1 are greater than the outputs for Y2, meaning the graph of f(x) is above the graph of g(x). A similar statement can be made for f 1x2 6 g1x2 written as Y1 6 Y2. Intersection-of-Graphs Method for Solving Inequalities

For any inequality of the form f 1x2 7 g1x2 , 1. Assign f(x) as Y1 and g(x) as Y2. 2. Graph both functions and identify any point(s) of intersection, if they exist. The solution set is all real numbers x for which the graph of Y1 is above the graph of Y2. For strict inequalities, the boundary of the solution interval is not included. A similar process is used for the inequalities f 1x2 g1x2 , f 1x2 6 g1x2 , and f 1x2 g1x2 . Note that we can actually draw the graphs of Y1 and Y2 differently (one more bold than the

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other) to clearly tell them apart in the viewing window. This is done on the Y= screen, by moving the cursor to the far left of the current function and pressing until a bold line appears. From the default setting, pressing one time produces this result. ENTER

ENTER

EXAMPLE 3

䊳

Solving an Inequality Using the Intersection-of-Graphs Method Solve 0.513 x2 5 2x 4 using the intersection-of-graphs method.

Solution

䊳

Figure 1.86

To assist with the clarity of the solution, we set the calculator to graph Y2 using a bolder line than Y1 (Figure 1.86). With 0.513 X2 5 as Y1 and 2X 4 as Y2, we use the 2nd TRACE (CALC) option and select 5:intersect. Pressing three times serves to identify both graphs, bypass the “Guess?” option, and display the point of intersection 13, 22 (Figure 1.87). Since the graph of Y1 is below the graph of Y2 1Y1 Y2 2 for all values of x to the right of 13, 22 , x 3 is the left boundary, with 10 the interval extending to positive infinity. Due to the less than or equal to inequality, we include x 3 and the solution interval is x 僆 33, q 2 . ENTER

C. You’ve just seen how we can solve linear inequalities graphically

Figure 1.87 10

10

10


䊳

D. Solving for a Specified Variable in Literal Equations A formula is an equation that models a known relationship between two or more quantities. A literal equation is simply one that has two or more variables. Formulas are a type of literal equation, but not every literal equation is a formula. For example, the formula A P PRT models the growth of money in an account earning simple interest, where A represents the total amount accumulated, P is the initial deposit, R is the annual interest rate, and T is the number of years the money is left on deposit. To describe A P PRT, we might say the formula has been “solved for A” or that “A is written in terms of P, R, and T.” In some cases, before using a formula it may be convenient to solve for one of the other variables, say P. In this case, P is called the object variable. EXAMPLE 4

䊳

Solving for Specified Variable Given A P PRT, write P in terms of A, R, and T (solve for P).

Solution

䊳

Since the object variable occurs in more than one term, we first apply the distributive property. A P PRT A P11 RT2 P11 RT2 A 1 RT 11 RT2 A P 1 RT

focus on P — the object variable factor out P solve for P [divide by (1 RT )]

result


䊳

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We solve literal equations for a specified variable using the same methods we used for other equations and formulas. Remember that it’s good practice to focus on the object variable to help guide you through the solution process, as again shown in Example 5. EXAMPLE 5

䊳

Solving for a Specified Variable Given 2x 3y 15, write y in terms of x (solve for y).

Solution

䊳

WORTHY OF NOTE In Example 5, notice that in the second step we wrote the subtraction of 2x as 2x 15 instead of 15 2x. For reasons that will become clearer as we continue our study, we generally write variable terms before constant terms.

2x 3y 15 3y 2x 15 1 13y2 13 12x 152 3 y 2 3 x 5

focus on the object variable subtract 2x (isolate y-term) multiply by 13 (solve for y ) simplify and distribute


䊳

Literal Equations and General Solutions Solving literal equations for a specified variable can help us develop the general solution for an entire family of equations. This is demonstrated here for the family of linear equations written in the form ax b c. A side-by-side comparison with a specific linear equation demonstrates that identical ideas are used. Specific Equation 2x 3 15 2x 15 3 x

15 3 2

Literal Equation focus on object variable

ax b c ax c b

subtract constant

x

divide by coefficient

cb a

Of course the solution on the left would be written as x 6 and checked in the original equation. On the right we now have a general formula for all equations of the form ax b c. EXAMPLE 6

䊳

Solving Equations of the Form ax ⫹ b ⫽ c Using a General Formula Solve 6x 1 25 using the formula just developed, and check your solution in the original equation.

Solution

䊳

For this equation, a 6, b 1, and c 25, giving x

cb a 25 112

24 6 4

6

→

Check:

6x 1 25 6142 1 25 24 1 25 25 25 ✓ Now try Exercises 55 through 60

D. You’ve just seen how we can solve for a specified variable in a formula or literal equation

䊳

Developing a general solution for the linear equation ax b c seems to have little practical use. But in Section 3.2 we’ll use this idea to develop a general solution for quadratic equations, a result with much greater significance.

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71

E. Using a Problem-Solving Guide Becoming a good problem solver is an evolutionary process. Over time and with continued effort, your problem-solving skills grow, as will your ability to solve a wider range of applications. Most good problem solvers develop the following characteristics: • A positive attitude • A mastery of basic facts • Strong mental arithmetic skills

• Good mental-visual skills • Good estimation skills • A willingness to persevere

These characteristics form a solid basis for applying what we call the Problem-Solving Guide, which simply organizes the basic elements of good problem solving. Using this guide will help save you from two common stumbling blocks—indecision and not knowing where to start. Problem-Solving Guide • Gather and organize information. Read the problem several times, forming a mental picture as you read. Highlight key phrases. List given information, including any related formulas. Clearly identify what you are asked to find. • Make the problem visual. Draw and label a diagram or create a table of values, as appropriate. This will help you see how different parts of the problem fit together. • Develop an equation model. Assign a variable to represent what you are asked to find and build any related expressions referred to in the problem. Write an equation model based on the relationships given in the problem. Carefully reread the problem to double-check your equation model. • Use the model and given information to solve the problem. Substitute given values, then simplify and solve. State the answer in sentence form, and check that the answer is reasonable. Include any units of measure indicated.

General Modeling Exercises Translating word phrases into symbols is an important part of building equations from information given in paragraph form. Sometimes the variable occurs more than once in the equation, because two different items in the same exercise are related. If the relationship involves a comparison of size, we often use line segments or bar graphs to model the relative sizes. EXAMPLE 7

䊳

Solving an Application Using the Problem-Solving Guide The largest state in the United States is Alaska (AK), which covers an area that is 230 square miles (mi2) more than 500 times that of the smallest state, Rhode Island (RI). If they have a combined area of 616,460 mi2, how many square miles does each cover?

䊳

Combined area is 616,460 mi2, AK covers 230 more than 500 times the area of RI.

gather and organize information highlight any key phrases

230

…

Solution

make the problem visual

500 times

Rhode Island’s area R

Alaska

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Let R represent the area of Rhode Island. Then 500R 230 represents Alaska’s area.

assign a variable build related expressions

Rhode Island’s area Alaska’s area Total R 1500R 2302 616,460 501R 616,230 R 1230

write the equation model combine like terms, subtract 230 divide by 501

2

Rhode Island covers an area of 1230 mi , while Alaska covers an area of 500112302 230 615,230 mi2. Now try Exercises 63 through 68

䊳

Consecutive Integer Exercises Exercises involving consecutive integers offer excellent practice in assigning variables to unknown quantities, building related expressions, and the problem-solving process in general. We sometimes work with consecutive odd integers or consecutive even integers as well. EXAMPLE 8

䊳

Solving a Problem Involving Consecutive Odd Integers The sum of three consecutive odd integers is 69. What are the integers?

Solution

䊳

The sum of three consecutive odd integers . . . 2

2

4 3 2 1

odd

WORTHY OF NOTE The number line illustration in Example 8 shows that consecutive odd integers are two units apart and the related expressions were built accordingly: n, n 2, n 4, and so on. In particular, we cannot use n, n 1, n 3, . . . because n and n 1 are not two units apart. If we know the exercise involves even integers instead, the same model is used, since even integers are also two units apart. For consecutive integers, the labels are n, n 1, n 2, and so on.

odd

2

2

gather/organize information highlight any key phrases 2 make the problem visual

0

1

odd

2

3

odd

4

n n1 n2 n3 n4

odd

odd

odd

Let n represent the smallest consecutive odd integer, then n 2 represents the second odd integer and 1n 22 2 n 4 represents the third.

In words: first second third odd integer 69 n 1n 22 1n 42 69 3n 6 69 3n 63 n 21


write the equation model equation model combine like terms subtract 6 divide by 3

The odd integers are n 21, n 2 23, and n 4 25. 21 23 25 69 ✓ Now try Exercises 69 through 72

䊳

Uniform Motion (Distance, Rate, Time) Exercises Uniform motion problems have many variations, and it’s important to draw a good diagram when you get started. Recall that if speed is constant, the distance traveled is equal to the rate of speed multiplied by the time in motion: D RT.

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EXAMPLE 9

䊳

Solving a Problem Involving Uniform Motion I live 260 mi from a popular mountain retreat. On my way there to do some mountain biking, my car had engine trouble — forcing me to bike the rest of the way. If I drove 2 hr longer than I biked and averaged 60 miles per hour driving and 10 miles per hour biking, how many hours did I spend pedaling to the resort?

Solution

䊳

The sum of the two distances must be 260 mi. The rates are given, and the driving time is 2 hr more than biking time.

Home

gather/organize information highlight any key phrases make the problem visual

Driving

Biking

D1 RT

D2 rt

Resort

D1 D2 Total distance 260 miles

Let t represent the biking time, then T t 2 represents time spent driving. D1 D2 260 RT rt 260 601t 22 10t 260 70t 120 260 70t 140 t2

assign a variable build related expressions write the equation model RT D1, rt D2 substitute t 2 for T, 60 for R, 10 for r distribute and combine like terms subtract 120 divide by 70

I rode my bike for t 2 hr, after driving t 2 4 hr. Now try Exercises 73 through 76

䊳

Exercises Involving Mixtures Mixture problems offer another opportunity to refine our problem-solving skills while using many elements from the problem-solving guide. They also lend themselves to a very useful mental-visual image and have many practical applications. EXAMPLE 10

䊳

Solving an Application Involving Mixtures As a nasal decongestant, doctors sometimes prescribe saline solutions with a concentration between 6% and 20%. In “the old days,” pharmacists had to create different mixtures, but only needed to stock these concentrations, since any percentage in between could be obtained using a mixture. An order comes in for a 15% solution. How many milliliters (mL) of the 20% solution must be mixed with 10 mL of the 6% solution to obtain the desired 15% solution? Provide both an algebraic solution and a graphical solution.

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Algebraic Solution

䊳

Only 6% and 20% concentrations are available; mix some 20% solution with 10 mL of the 6% solution. (See Figure 1.88.)

gather/organize information highlight any key phrases

Figure 1.88 20% solution

6% solution ? mL

10 mL make the problem visual

(10 ?) mL 15% solution

Let x represent the amount of 20% solution, then 10 x represents the total amount of 15% solution. 1st quantity times its concentration

1010.062 0.6

2nd quantity times its concentration

x10.22 0.2x 0.2x 0.05x x


1st2nd quantity times desired concentration

110 x2 10.152 1.5 0.15x 0.9 0.15x 0.9 18

write equation model distribute/simplify subtract 0.6 subtract 0.15x divide by 0.05

To obtain a 15% solution, 18 mL of the 20% solution must be mixed with 10 mL of the 6% solution.

Graphical Solution

䊳

WORTHY OF NOTE For mixture exercises, an estimate assuming equal amounts of each liquid can be helpful. For example, assume we use 10 mL of the 6% solution and 10 mL of the 20% solution. The final concentration would be halfway in between, 6 20 13%. This is too low a 2 concentration (we need a 15% solution), so we know that more than 10 mL of the stronger (20%) solution must be used.

Although both methods work equally well, here we elect to use the intersection-of-graphs method and enter 1010.062 X10.22 as Y1 and 110 X2 10.152 as Y2. Virtually all graphical solutions require a careful study of the context to set the viewing window prior to graphing. If 10 mL of liquid were used from each concentration, we would have 20 mL of a 13% solution (see Worthy of Note), so more of the stronger solution is needed. This shows that an appropriate Xmax might be close to 30. If all 30 mL were used, the output would be 30 10.152 4.5, so an appropriate Ymax might be around 6 (see Figure 1.89). Using 2nd TRACE (CALC) 5:Intersect and pressing three times gives 0 (18, 4.2) as the point of intersection, showing x 18 mL of the stronger solution must be used (Figure 1.90).

Figure 1.89

Figure 1.90 6

ENTER

30

0

E. You’ve just seen how we can use the problemsolving guide to solve various problem types


䊳

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75

1.5 EXERCISES 䊳


Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. When using the method, one side of an equation is entered as and the other side as on a graphing calculator. The of the point of is the solution of the equation. 2. To solve a linear inequality using the intersectionof-graphs method, first find the point of . The of this point is a boundary value of the solution interval and if the inequality is not strict, this value is in the solution. 3. A(n) equation is an equation having more unknowns.

䊳

or

4. For the equation S 2␲r2 2␲rh, we can say that S is written in terms of and . 5. Discuss/Explain the similarities and differences between the intersection and zeroes methods for solving equations. How can the zeroes method be applied to solving linear inequalities? Give examples in your discussion. 6. Discuss/Explain each of the four basic parts of the problem-solving guide. Include a solved example in your discussion.


Solve the following equations using a graphing calculator and the intersection-of-graphs method. For Exercises 7 and 8, carefully sketch the graphs you designate as Y1 and Y2 by hand before using your calculator.

7. 3x 7 21x 12 10 8. 2x 1 21x 32 1 9. 0.8x 0.4 0.2512 0.4x2 2.8 10. 0.5x 2.5 0.7513 0.2x2 0.4 11. x 13x 12 0.514x 62 2

12. 3x 14 x2 0.216 10x2 5.2 13. 13x 2 3 x 9 14.

4 5 x

16.

1 3 1x

8 65x

15. 12 1x 42 10 x 12 12x2

62 5 x 16 23x2

Solve the following equations using a graphing calculator and the x-intercept/zeroes method. Compare your results for Exercises 17 and 18 to those of Exercises 7 and 8.

21. 1.51x 42 2.5 3x 3.5 22. 0.813x 12 0.2 2x 3.8 23. 21x 22 1 x 11 x2

24. 312x 12 1 2x 11 4x2

25. 3x 10.7x 1.22 211.1x 0.62 0.1x

26. 3x 210.2x 1.42 410.8x 0.72 0.2x Solve the following inequalities using a graphing calculator and the intersection-of-graphs method. Compare your results for Exercises 27 and 28 to those of Exercises 7 and 8.

27. 3x 7 7 21x 12 10 28. 2x 1 7 21x 32 1

29. 2x 13 x2 215 2x2 7

30. 413x 52 2 312 4x2 24 31. 0.31x 22 1.1 6 0.2x 3 32. 0.2514 x2 1 6 1 0.5x 33. 31x 12 1 x 41x 12

17. 3x 7 21x 12 10

34. 1.112 x2 0.2 7 510.1 0.2x2 0.1x

18. 2x 1 21x 32 1

35. 211.5x 1.12 0.1x 4x 0.313x 42

19. 213 2x2 5 3x 3

36. 41x 12 2x 7 6 21x 1.52

20. 313 x2 4 2x 5

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Solve for the specified variable in each formula or literal equation.

37. P C CM for C (retail) 38. S P PD for P (retail) 39. C 2␲r for r (geometry) 40. V LWH for W (geometry) P1V1 P2V2 41. for T2 (science) T1 T2 42.

P1 C 2 for P2 (communication) P2 d

43. V 43␲r2h for h (geometry) 44. V 13␲r2h for h (geometry) a1 an 45. Sn na b for n (sequences) 2 46. A

h1b1 b2 2 2

for h (geometry)

47. S B 12PS for P (geometry)

䊳

48. s 12gt2 vt for g (physics) 49. Ax By C for y 50. 2x 3y 6 for y 51. 56x 38y 2 for y 52. 23x 79y 12 for y

53. y 3 4 5 1x 102 for y

54. y 4 2 15 1x 102 for y

The following equations are given in ax ⴙ b ⴝ c form. Solve by identifying the value of a, b, and c, then using cⴚb the formula x ⴝ . a

55. 3x 2 19 56. 7x 5 47 57. 6x 1 33 58. 4x 9 43 59. 7x 13 27 60. 3x 4 25


61. Surface area of a cylinder: SA ⴝ 2␲r2 ⴙ 2␲rh The surface area of a cylinder is given by the formula shown, where h is the height of the cylinder and r is the radius of the base. Find the height of a cylinder that has a radius of 8 cm and a surface area of 1256 cm2. Use ␲ ⬇ 3.14.

䊳

1–76


62. Using the equation-solving process for Exercise 61 as a model, solve the formula SA 2␲r2 2␲rh for h.

APPLICATIONS

Solve by building an equation model and using the problem-solving guidelines as needed. Check all answers using a graphing calculator. General Modeling Exercises

63. Two spelunkers (cave explorers) were exploring different branches of an underground cavern. The first was able to descend 198 ft farther than twice the second. If the first spelunker descended 1218 ft, how far was the second spelunker able to descend? 64. The area near the joining of the Tigris and Euphrates Rivers (in modern Iraq) has often been called the Cradle of Civilization, since the area has evidence of many ancient cultures. The length of the Euphrates River exceeds that of the Tigris by 620 mi. If they have a combined length of 2880 mi, how long is each river?

65. U.S. postal regulations require that a package Girth can have a maximum combined length and girth (distance around) L of 108 in. A shipping H carton is constructed so that it has a width of W 14 in., a height of 12 in., and can be cut or folded to various lengths. What is the maximum length that can be used? Source: www.USPS.com

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66. Hi-Tech Home Improvements buys a fleet of identical trucks that cost $32,750 each. The company is allowed to depreciate the value of their trucks for tax purposes by $5250 per year. If company policies dictate that older trucks must be sold once their value declines to $6500, approximately how many years will they keep these trucks? 67. The longest suspension bridge in the world is the Akashi Kaikyo (Japan) with a length of 6532 feet. Japan is also home to the Shimotsui Straight bridge. The Akashi Kaikyo bridge is 364 ft more than twice the length of the Shimotsui bridge. How long is the Shimotsui bridge? Source: www.guinnessworldrecords.com

68. The Mars rover Spirit landed on January 3, 2004. Just over 1 yr later, on January 14, 2005, the Huygens probe landed on Titan (one of Saturn’s moons). At their closest approach, the distance from the Earth to Saturn is 29 million mi more than 21 times the distance from the Earth to Mars. If the distance to Saturn is 743 million mi, what is the distance to Mars?

77

Uniform Motion Exercises 73. At 9:00 A.M., Linda leaves work on a business trip, gets on the interstate, and sets her cruise control at 60 mph. At 9:30 A.M., Bruce notices she’s left her briefcase and cell phone, and immediately starts after her driving 75 mph. At what time will Bruce catch up with Linda? 74. A plane flying at 300 mph has a 3-hr head start on a “chase plane,” which has a speed of 800 mph. How far from the airport will the chase plane overtake the first plane? 75. Jeff had a job interview in a nearby city 72 mi away. On the first leg of the trip he drove an average of 30 mph through a long construction zone, but was able to drive 60 mph after passing through this zone. If driving time for the trip was 112 hr, how long was he driving in the construction zone? 76. At a high-school cross-country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi long and Jared finished in 2 hr, how far did he jog at the faster pace? Mixture Exercises Give the total amount of the mix that results and the percent concentration or worth of the mix.

77. Two quarts of 100% orange juice are mixed with 2 quarts of water (0% juice). 78. Ten pints of a 40% acid are combined with 10 pints of an 80% acid. Consecutive Integer Exercises 69. Find two consecutive even integers such that the sum of twice the smaller integer plus the larger integer is one hundred forty-six. 70. When the smaller of two consecutive integers is added to three times the larger, the result is fiftyone. Find the smaller integer. 71. Seven times the first of two consecutive odd integers is equal to five times the second. Find each integer. 72. Find three consecutive even integers where the sum of triple the first and twice the second is eight more than four times the third.

79. Eight pounds of premium coffee beans worth $2.50 per pound are mixed with 8 lb of standard beans worth $1.10 per pound. 80. A rancher mixes 50 lb of a custom feed blend costing $1.80 per pound, with 50 lb of cheap cottonseed worth $0.60 per pound. Solve each application of the mixture concept.

81. To help sell more of a lower grade meat, a butcher mixes some premium ground beef worth $3.10/lb, with 8 lb of lower grade ground beef worth $2.05/lb. If the result was an intermediate grade of ground beef worth $2.68/lb, how much premium ground beef was used?

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82. Knowing that the camping/hiking season has arrived, a nutrition outlet is mixing GORP (Good Old Raisins and Peanuts) for the anticipated customers. How many pounds of peanuts worth $1.29/lb, should be mixed with 20 lb of deluxe raisins worth $1.89/lb, to obtain a mix that will sell for $1.49/lb?

䊳

83. How many pounds of walnuts at 84¢/lb should be mixed with 20 lb of pecans at $1.20/lb to give a mixture worth $1.04/lb? 84. How many pounds of cheese worth 81¢/lb must be mixed with 10 lb cheese worth $1.29/lb to make a mixture worth $1.11/lb?


85. Look up and read the following article. Then turn in a one page summary. “Don’t Give Up!,” William H. Kraus, Mathematics Teacher, Volume 86, Number 2, February 1993: pages 110–112. 86. A chemist has four solutions of a very rare and expensive chemical that are 15% acid (cost $120 per ounce), 20% acid (cost $180 per ounce), 35% acid (cost $280 per ounce) and 45% acid (cost $359 per ounce). She requires 200 oz of a 29% acid solution. Find the combination of any two of these concentrations that will minimize the total cost of the mix. 87. P, Q, R, S, T, and U represent numbers. The arrows in the figure show the sum of the two or three numbers added in the indicated direction

䊳

1–78


(Example: Q T 23). Find P Q R S T U. P

Q

26

S

30 40

R

T 19

U 23

34

88. Given a sphere circumscribed by a cylinder, verify the volume of the sphere is 23 that of the cylinder.


1 2 3 2 x x2 x 2x 90. (1.4) Solve for y, then state the slope and y-intercept of the line: 6x 7y 42

89. (Appendix A.5) Solve for x:

91. (Appendix A.4) Factor each expression: a. 4x2 9 b. x3 27

92. (1.3) Given g1x2 x2 3x 10, evaluate g1 13 2, g122, and g152

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1.6

Linear Function Models and Real Data

LEARNING OBJECTIVES

Collecting and analyzing data is a tremendously important mathematical endeavor, having applications throughout business, industry, science, and government. The link between classroom mathematics and real-world mathematics is called a regression, in which we attempt to find an equation that will act as a model for the raw data. In this section, we focus on situations where the data is best modeled by a linear function.

In Section 1.6 you will see how we can:

A. Draw a scatterplot and

B.

C.

D.

E.

identify positive and negative associations Use a scatterplot to identify linear and nonlinear associations Use a scatterplot to identify strong and weak correlations Find a linear function that models the relationships observed in a set of data Use linear regression to find the line of best fit

A. Scatterplots and Positive/Negative Associations In this section, we continue our study of ordered pairs and functions, but this time using data collected from various sources or from observed real-world relationships. You can hardly pick up a newspaper or magazine without noticing it contains a large volume of data presented in graphs, charts, and tables. In addition, there are many simple experiments or activities that enable you to collect your own data. We begin analyzing the collected data using a scatterplot, which is simply a graph of all of the ordered pairs in a data set. Often, real data (sometimes called raw data) is not very “well behaved” and the points may be somewhat scattered—the reason for the name.

Positive and Negative Associations Earlier we noted that lines with positive slope rise from left to right, while lines with negative slope fall from left to right. We can extend this idea to the data from a scatterplot. The data points in Example 1A seem to rise as you move from left to right, with larger input values generally resulting in larger outputs. In this case, we say there is a positive association between the variables. If the data seems to decrease or fall as you move left to right, we say there is a negative association.

EXAMPLE 1A

䊳

Drawing a Scatterplot and Observing Associations The ratio of the federal debt to the total population is known as the per capita debt. The per capita debt of the United States is shown in the table for the odd-numbered years from 1997 to 2007. Draw a scatterplot of the data and state whether the association is positive, negative, or cannot be determined. Source: Data from the Bureau of Public Debt at www.publicdebt.treas.gov

Per Capita Debt ($1000s)

30

1997

20.0

28

1999

20.7

2001

20.5

2003

23.3

2005

27.6

2007

30.4

Debt ($1000s)

Year

26 24 22 20 1997 1999 2001 2003 2005 2007

Year

Solution

1–79

䊳

Since the amount of debt depends on the year, year is the input x and per capita debt is the output y. Scale the x-axis from 1997 to 2007 and the y-axis from 20 to 30 to comfortably fit the data (the “squiggly lines,” near the 20 and 1997 in the graph are used to show that some initial values have been skipped). The graph indicates a positive association between the variables, meaning the debt is generally increasing as time goes on. 79

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EXAMPLE 1B

䊳

Drawing a Scatterplot and Observing Associations A cup of coffee is placed on a table and allowed to cool. The temperature of the coffee is measured every 10 min and the data are shown in the table. Draw the scatterplot and state whether the association is positive, negative, or cannot be determined. Temperature (ºF)

0

110

10

89

20

76

30

72

40

71

120 110

Temp (°F)

Elapsed Time (minutes)

100 90 80 70 0

Solution

䊳

A. You’ve just seen how we can draw a scatterplot and identify positive and negative associations

5 10 15 20 25 30 35 40

Time (minutes)

Since temperature depends on cooling time, time is the input x and temperature is the output y. Scale the x-axis from 0 to 40 and the y-axis from 70 to 110 to comfortably fit the data. As you see in the figure, there is a negative association between the variables, meaning the temperature decreases over time. Now try Exercises 7 and 8 䊳

B. Scatterplots and Linear/Nonlinear Associations The data in Example 1A had a positive association, while the association in Example 1B was negative. But the data from these examples differ in another important way. In Example 1A, the data seem to cluster about an imaginary line. This indicates a linear equation model might be a good approximation for the data, and we say there is a linear association between the variables. The data in Example 1B could not accurately be modeled using a straight line, and we say the variables time and cooling temperature exhibit a nonlinear association.

䊳

Drawing a Scatterplot and Observing Associations A college professor tracked her annual salary for 2002 to 2009 and the data are shown in the table. Draw the scatterplot and determine if there is a linear or nonlinear association between the variables. Also state whether the association is positive, negative, or cannot be determined.

Year

Salary ($1000s)

2002

30.5

2003

31

2004

32

2005

33.2

2006

35.5

2007

39.5

2008

45.5

2009

52

55 50

Salary ($1000s)

EXAMPLE 2

45

Appears nonlinear

40 35 30

2002

2004

2006

2008

2010

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Section 1.6 Linear Function Models and Real Data

Solution

䊳

B. You’ve just seen how we can use a scatterplot to identify linear and nonlinear associations

Since salary earned depends on a given year, year is the input x and salary is the output y. Scale the x-axis from 2002 to 2010, and the y-axis from 30 to 55 to comfortably fit the data. A line doesn’t seem to model the data very well, and the association appears to be nonlinear. The data rises from left to right, indicating a positive association between the variables. This makes good sense, since we expect our salaries to increase over time. Now try Exercises 9 and 10 䊳

C. Identifying Strong and Weak Correlations Using Figures 1.91 and 1.92 shown, we can make one additional observation regarding the data in a scatterplot. While both associations shown appear linear, the data in Figure 1.91 seems to cluster more tightly about an imaginary straight line than the data in Figure 1.92. Figure 1.91

Figure 1.92 y

y 10

10

5

5

0

5

x

10

0

5

10

x

We refer to this “clustering” as the “goodness of fit,” or in statistical terms, the strength of the correlation. To quantify this fit we use a measure called the correlation coefficient r, which tells whether the association is positive or negative: r 7 0 or r 6 0, and quantifies the strength of the association: 0r 0 ⱕ 100% . Actually, the coefficient is given in decimal form, making 0r 0 ⱕ 1. If the data points form a perfectly straight line, we say the strength of the correlation is either ⫺1 or 1, depending on the association. If the data points appear clustered about the line, but are scattered on either side of it, the strength of the correlation falls somewhere between ⫺1 and 1, depending on how tightly/loosely they’re scattered. This is summarized in Figure 1.93. Figure 1.93 Perfect negative correlation Strong negative correlation ⫺1.00

Moderate negative correlation

No correlation Weak negative correlation

Weak positive correlation 0

Moderate positive correlation

Perfect positive correlation Strong positive correlation ⫹1.00

The following scatterplots help to further illustrate this idea. Figure 1.94 shows a linear and negative association between the value of a car and the age of a car, with a strong correlation. Figure 1.95 shows there is no apparent association between family income and the number of children, and Figure 1.96 appears to show a linear and positive association between a man’s height and weight, with a weak correlation.

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Figure 1.94

Figure 1.96

Age of auto

Male weights

Family income

Figure 1.95

Value of auto

Number of children

Male heights

Until we develop a more accurate method of calculating a numerical value for this correlation, the best we can do are these broad generalizations: weak correlation, strong correlation, or no correlation. EXAMPLE 3A

䊳

High School and College GPAs Many colleges use a student’s high school GPA as a possible indication of their future college GPA. Use the data from Table 1.6 (high school/college GPA) to draw a scatterplot. Then a. Sketch a line that seems to approximate the data, meaning it has the same general direction while passing through the observed “center” of the data. b. State whether the association is positive, negative, or cannot be determined. c. Decide whether the correlation is weak or strong. Table 1.6

Solution

䊳

EXAMPLE 3B

䊳

High School GPA

College GPA

1.8

1.8

2.2

2.3

2.8

2.5

3.2

2.9

3.4

3.6

3.8

3.9

4.0 3.5

College GPA

82

3.0 2.5 2.0 1.5

1.5

2.0

2.5

3.0

3.5

4.0

High School GPA

a. A line approximating the data set as a whole is shown in the figure. b. Since the line has positive slope, there is a positive association between a student’s high school GPA and their GPA in college. c. The correlation appears strong.

Natural Gas Consumption The amount of natural gas consumed by homes and offices varies with the season, with the highest consumption occurring in the winter months. Use the data from Table 1.7 (outdoor temperature/gas consumed) to draw a scatterplot. Then a. Sketch an estimated line of best fit. b. State whether the association is positive, negative, or cannot be determined. c. Decide whether the correlation is weak or strong.

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Table 1.7 Gas Consumed (cubic feet)

30

800

40

620

50

570

60

400

70

290

80

220

800

Gas consumption (ft3)

Outdoor Temperature (ºF)

600

400

200

0

30

40

50

60

70

80

Outdoor temperature (⬚F)

Solution

䊳

C. You’ve just seen how we can use a scatterplot to identify strong and weak correlations

a. We again use appropriate scales and sketch a line that seems to model the data (see figure). b. There is a negative association between temperature and the amount of natural gas consumed. c. The correlation appears to be strong. Now try Exercises 11 and 12

䊳

D. Linear Functions That Model Relationships Observed in a Set of Data Finding a linear function model for a set of data involves visually estimating and sketching a line that appears to best “fit” the data. This means answers will vary slightly, but a good, usable model can often be obtained. To find the function, we select two points on this imaginary line and use either the slope-intercept form or the pointslope formula to construct the function. Points on this estimated line but not actually in the data set can still be used to help determine the function.

EXAMPLE 4

䊳

Finding a Linear Function to Model the Relationship Between GPAs Use the scatterplot from Example 3A to find a function model for the line a college might use to project an incoming student’s future GPA.

Solution

䊳

Any two points on or near the estimated best-fit line can be used to help determine the linear function (see the figure in Example 3A). For the slope, it’s best to pick two points that are some distance apart, as this tends to improve the accuracy of the model. It appears (1.8, 1.8) and (3.8, 3.9) are both on the line, giving y2 ⫺ y1 x2 ⫺ x1 3.9 ⫺ 1.8 ⫽ 3.8 ⫺ 1.8 ⫽ 1.05 y ⫺ y1 ⫽ m1x ⫺ x1 2 y ⫺ 1.8 ⫽ 1.051x ⫺ 1.82 y ⫺ 1.8 ⫽ 1.05x ⫺ 1.89 y ⫽ 1.05x ⫺ 0.09 m⫽

slope formula

substitute (x2, y2) for S (3.8, 3.9), (x1, y1) for S (1.8, 1.8) slope point-slope form substitute 1.05 for m, (1.8, 1.8) for (x1, y1) distribute add 1.8 (solve for y )

One possible function model for this data is f 1x2 ⫽ 1.05x ⫺ 0.09. Slightly different functions may be obtained, depending on the points chosen. Now try Exercises 13 through 22

䊳

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WORTHY OF NOTE Sometimes it helps to draw a straight line on an overhead transparency, then lay it over the scatterplot. By shifting the transparency up and down, and rotating it left and right, the line can more accurately be placed so that it’s centered among and through the data.

EXAMPLE 5

䊳

The function from Example 4 predicts that a student with a high school GPA of 3.2 will have a college GPA of almost 3.3: f 13.22 ⫽ 1.0513.22 ⫺ 0.09 ⬇ 3.3, yet the data gives an actual value of only 2.9. When working with data and function models, we should expect some variation when the two are compared, especially if the correlation is weak. Applications of data analysis can be found in virtually all fields of study. In Example 5 we apply these ideas to an Olympic swimming event.

Finding a Linear Function to Model the Relationship (Year, Gold Medal Times) The men’s 400-m freestyle times (gold medal times — to the nearest second) for the 1976 through 2008 Olympics are given in Table 1.8 (1900 S 0). Let the year be the input x, and winning race time be the output y. Based on the data, draw a scatterplot and answer the following questions. a. Does the association appear linear or nonlinear? b. Is the association positive or negative? c. Classify the correlation as weak or strong. d. Find a function model that approximates the data, then use it to predict the winning time for the 2012 Olympics. Table 1.8 Year (x) (1900 S 0)

Time ( y) (sec)

76

232

80

231

84

231

88

227

92

225

96

228

100

221

104

223

108

223

242

Solution

䊳

234

Time (sec)

84

226

218

210 76

84

92

100

108

Year

Begin by choosing appropriate scales for the axes. The x-axis (year) could be scaled from 76 to 112, and the y-axis (swim time) from 210 to 246. This will allow for a “frame” around the data. After plotting the points, we obtain the scatterplot shown in the figure. a. The association appears to be linear. b. The association is negative, showing that finishing times tend to decrease over time. c. There is a moderate to strong correlation. d. The points (76, 232) and (104, 223) appear to be on a line approximating the data, and we select these to develop our equation model. y2 ⫺ y1 x2 ⫺ x1 223 ⫺ 232 ⫽ 104 ⫺ 76 ⬇ ⫺0.32 y ⫺ 232 ⫽ ⫺0.321x ⫺ 762 y ⫺ 232 ⫽ ⫺0.32x ⫹ 24.32 y ⫽ ⫺0.32x ⫹ 256.32 m⫽

slope formula 1x1, y1 2 S 176, 2322 , 1x2, y2 2 S 1104, 2232 slope (rounded to tenths) point-slope form distribute add 232 (solve for y )

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85

One model for this data is y ⫽ ⫺0.32x ⫹ 256.32. Based on this model, the predicted time for the 2012 Olympics would be f 1x2 ⫽ ⫺0.32x ⫹ 256.32 f 11122 ⫽ ⫺0.3211122 ⫹ 256.32 ⫽ 220.48

function model substitute 112 for x (2012) result

In 2012 the winning time is projected to be about 220.5 sec. Now try Exercises 23 and 24

D. You’ve just seen how we can find a linear function that models relationships observed in a set of data

䊳

As a reminder, great care should be taken when using equation models obtained from real data. It would be foolish to assume that in the year 2700, swim times for the 400-m freestyle would be near 0 sec—even though that’s what the model predicts for x ⫽ 800. Most function models are limited by numerous constraining factors, and data collected over a much longer period of time might even be better approximated using a nonlinear model.

E. Linear Regression and the Line of Best Fit There is actually a sophisticated method for calculating the equation of a line that best fits a data set, called the regression line. The method minimizes the vertical distance between all data points and the line itself, making it the unique line of best fit. Most graphing calculators have the ability to perform this calculation quickly. The process involves these steps: (1) clearing old data, (2) entering new data, (3) displaying the data, (4) calculating the regression line, and (5) displaying and using the regression line. We’ll illustrate by finding the regression line for the data shown in Table 1.8 in Example 5, which gives the men’s 400-m freestyle gold medal times (in seconds) for the 1976 through the 2008 Olympics, with 1900S0.

Step 1: Clear Old Data To prepare for the new data, we first clear out any old data. Press the STAT key and select option 4:ClrList. This places the ClrList command on the home screen. We tell the calculator which lists to clear by pressing 2nd 1 to indicate List1 (L1), then enter a comma using the , key, and continue entering other lists we want to clear: 2nd 2nd 2 , 3 will clear List1 (L1), List2 (L2), and List3 (L3). ENTER

Step 2: Enter New Data Press the STAT key and select option 1:Edit. Move the cursor to the first position of List1, and simply enter the data from the first column of Table 1.8 in order: 76 80 84 , and so on. Then use the right arrow to navigate to List2, and enter the data from the second column: 232 231 231 , and so on. When finished, you should obtain the screen shown in Figure 1.97.

Figure 1.97

Step 3: Display the Data

Figure 1.98

ENTER

ENTER

ENTER

ENTER

WORTHY OF NOTE As a rule of thumb, the tick marks for Xscl can be set by mentally 冟Xmax冟 ⫹ 冟Xmin冟 estimating and 10 using a convenient number in the neighborhood of the result (the same goes for Yscl). As an alternative to manually setting the window, the ZOOM 9:ZoomStat feature can be used.

ENTER

ENTER

With the data held in these lists, we can now display the related ordered pairs on the coordinate grid. First press the Y= key and any existing equations. Y= Then press 2nd to access the “STATPLOTS” screen. With the cursor on 1:Plot1, press and be sure the options shown in Figure 1.98 are highlighted. If you need to make any changes, navigate the cursor to CLEAR

ENTER

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the desired option and press . Note the data in L1 ranges from 76 to 108, while the data in L2 ranges from 221 to 232. This means an appropriate viewing window might be [70, 120] for the x-values, and [210, 240] for the y-values. Press the key and set up the window accordingly. After you’re finished, pressing the GRAPH key should produce the graph shown in Figure 1.99.

Figure 1.99

ENTER

WINDOW

240

70

120

Step 4: Calculate the Regression Equation

210

To have the calculator compute the regression equation, press the STAT and keys to move the cursor over to the CALC options (see Figure 1.100). Since it appears the data is best modeled by a linear equation, we choose option 4:LinReg(ax ⴙ b). Pressing the number 4 places this option on the home screen, and pressing computes the values of a and b (the calculator automatically uses the values in L1 and L2 unless instructed otherwise). Rounded to hundredths, the linear regression model is y ⫽ ⫺0.33x ⫹ 257.06 (Figure 1.101).

Figure 1.100

ENTER

Figure 1.101

Step 5: Display and Use the Results Although graphing calculators have the ability to paste the regression equation directly into Y1 on the Y= screen, for now we’ll enter Y1 ⫽ ⫺0.33x ⫹ 257.06 by hand. Afterward, pressing the GRAPH key will plot the data points (if Plot1 is still active) and graph the line. Your display screen should now look like the one in Figure 1.102. The regression line is the best estimator for the set of data as a whole, but there will still be some difference between the values it generates and the values from the set of raw data (the output in Figure 1.102 shows the estimated time for the 2000 Olympics was about 224 sec, when actually it was the year Ian Thorpe of Australia set a world record of 221 sec).

EXAMPLE 6

䊳

Figure 1.102 240

70

䊳

210

Using Regression to Model Employee Performance Riverside Electronics reviews employee performance semiannually, and awards increases in their hourly rate of pay based on the review. The table shows Thomas’ hourly wage for the last 4 yr (eight reviews). Find the regression equation for the data and use it to project his hourly wage for the year 2011, after his fourteenth review.

Solution

120

Following the prescribed sequence produces the equation y ⫽ 0.48x ⫹ 9.09. For x ⫽ 14 we obtain y ⫽ 0.481142 ⫹ 9.09 or a wage of $15.81. According to this model, Thomas will be earning $15.81 per hour in 2011.

Review (x)

Wage ( y)

(2004) 1

$9.58

2

$9.75

(2005) 3

$10.54

4

$11.41

(2006) 5

$11.60

6

$11.91

(2007) 7

$12.11

8

$13.02


䊳

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87

With each linear regression, the calculator can be set to compute a correlation coefficient that is a measure of how well the equation fits the data (see Subsection C). To 0 display this “r-value” use 2nd (CATALOG) and activate DiagnosticOn. Figure 1.103 shows a scatterplot with perfect negative correlation 1r ⫽ ⫺12 and notice all data points are on the line. Figure 1.104 shows a strong positive correlation 1r ⬇ 0.982 of the data from Example 6. See Exercise 35. Figure 1.103

Figure 1.104

E. You’ve just seen how we can use a linear regression to find the line of best fit

1.6 EXERCISES 䊳



1. When the ordered pairs from a set of data are plotted on a coordinate grid, the result is called a .

2. If the data points seem to form a curved pattern or if no pattern is apparent, the data is said to have a association.

3. If the data points seems to cluster along an imaginary line, the data is said to have a association.

4. If the pattern of data points seems to increase as they are viewed left to right, the data is said to have a association.

5. Compare/Contrast: One scatterplot is linear, with a weak and positive association. Another is linear, with a strong and negative association. Give a written description of each scatterplot.

6. Discuss/Explain how this is possible: Working from the same scatterplot, Demetrius obtained the equation y ⫽ ⫺0.64x ⫹ 44 for his equation model, while Jessie got the equation y ⫽ ⫺0.59x ⫹ 42.

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1–88


DEVELOPING YOUR SKILLS 7. For mail with a high priority, x “Express Mail” offers next day 1981 delivery by 12:00 noon to most 1985 destinations, 365 days of the 1988 year. The service was first offered by the U.S. Postal 1991 Service in the early 1980s and 1995 has been growing in use ever 1999 since. The cost of the service 2002 (in cents) for selected years is shown in the table. (a) Draw a 2010 scatterplot of the data, then (b) decide if the association is positive, negative, or cannot be determined.

y 935 1075 1200 1395 1500 1575 1785 1830

Source: 2004 Statistical Abstract of the United States; USPS.com

8. After the Surgeon General’s x y first warning in 1964, 1965 42.4 cigarette consumption began a 1974 37.1 steady decline as advertising was banned from television 1979 33.5 and radio, and public 1985 29.9 awareness of the dangers of 1990 25.3 cigarette smoking grew. The 1995 24.6 percentage of the U.S. adult population who considered 2000 23.1 themselves smokers is shown 2002 22.4 in the table for selected years. 2005 16.9 (a) Draw a scatterplot of the data, then (b) decide if the association is positive, negative, or cannot be determined. Source: 1998 Wall Street Journal Almanac and 2009 Statistical Abstract of the United States, Table 1299

9. Since the 1970s women have x y made tremendous gains in the 1972 32 political arena, with more and 1978 46 more female candidates running 1984 65 for, and winning seats in the U.S. Senate and U.S. Congress. 1992 106 The number of women 1998 121 candidates for the U.S. Congress 2004 141 is shown in the table for selected years. (a) Draw a scatterplot of the data, (b) decide if the association is linear or nonlinear and (c) if the association is positive, negative, or cannot be determined. Source: Center for American Women and Politics at www.cawp.rutgers.edu/Facts3.html

10. The number of shares traded on the New York Stock Exchange experienced dramatic change in the 1990s as more and more individual investors gained access to the stock market via the Internet

and online brokerage houses. The volume is shown in the table for 2002, and the odd numbered years from 1991 to 2001 (in billions of shares). (a) Draw a scatterplot of the data, (b) decide if the association is linear or nonlinear, and (c) if the association is positive, negative, or cannot be determined.

x

y

1991

46

1993

67

1995

88

1997

134

1999

206

2001

311

2002

369

Source: 2000 and 2004 Statistical Abstract of the United States, Table 1202

The data sets in Exercises 11 and 12 are known to be linear.

11. The total value of the goods x and services produced by a (1970 S 0) y nation is called its gross 0 5.1 domestic product or GDP. The 5 7.6 GDP per capita is the ratio of 10 12.3 the GDP for a given year to the 15 17.7 population that year, and is one of many indicators of 20 23.3 economic health. The GDP per 25 27.7 capita (in $1000s) for the 30 35.0 United States is shown in the 33 37.8 table for selected years. (a) Draw a scatterplot using scales that appropriately fit the data, then sketch an estimated line of best fit, (b) decide if the association is positive or negative, then (c) decide whether the correlation is weak or strong. Source: 2004 Statistical Abstract of the United States, Tables 2 and 641

12. Real estate brokers carefully Price Sales track sales of new homes 130’s 126 looking for trends in location, 150’s 95 price, size, and other factors. 170’s 103 The table relates the average selling price within a price 190’s 75 range (homes in the $120,000 210’s 44 to $140,000 range are 230’s 59 represented by the $130,000 250’s 21 figure), to the number of new homes sold by Homestead Realty in 2004. (a) Draw a scatterplot using scales that appropriately fit the data, then sketch an estimated line of best fit, (b) decide if the association is positive or negative, then (c) decide whether the correlation is weak or strong.

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For the scatterplots given: (a) Arrange them in order from the weakest to the strongest correlation, (b) sketch a line that seems to approximate the data, (c) state whether the association is positive, negative, or cannot be determined, and (d) choose two points on (or near) the line and use them to approximate its slope (rounded to one decimal place).

13. A.

y 60 55 50 45 40 35 30 25

B.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

C.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

D.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

14. A.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

B.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

C.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

For the scatterplots given, (a) determine whether a linear or nonlinear model would seem more appropriate. (b) Determine if the association is positive or negative. (c) Classify the correlation as weak or strong. (d) If linear, sketch a line that seems to approximate the data and choose two points on the line and use them to approximate its slope.

15.

y 60 55 50 45 40 35 30 25

16.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

17.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

0 1 2 3 4 5 6 7 8 9 x

18.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

20.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

21. In most areas of the country, x y law enforcement has become (1990→0) (1000s) a major concern. The number 3 68.8 of law enforcement officers 6 74.5 employed by the federal 8 83.1 government and having the authority to carry firearms 10 88.5 and make arrests is shown in 14 93.4 the table for selected years. (a) Draw a scatterplot using scales that appropriately fit the data and sketch an estimated line of best fit and (b) decide if the association is positive or negative. (c) Choose two points on or near the estimated line of best fit, and use them to find a function model and predict the number of federal law enforcement officers in 1995 and the projected number for 2011. Answers may vary. Source: U.S. Bureau of Justice, Statistics at www.ojp.usdoj.gov/bjs/fedle.htm

0 1 2 3 4 5 6 7 8 9 x

D.

19.

89

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

22. Due to atmospheric pressure, x y the temperature at which ⫺1000 213.8 water will boil varies 0 212.0 predictably with the altitude. Using special equipment 1000 210.2 designed to duplicate 2000 208.4 atmospheric pressure, a lab 3000 206.5 experiment is set up to study 4000 204.7 this relationship for altitudes 5000 202.9 up to 8000 ft. The set of data 6000 201.0 collected is shown in the table, with the boiling temperature y 7000 199.2 in degrees Fahrenheit, 8000 197.4 depending on the altitude x in feet. (a) Draw a scatterplot using scales that appropriately fit the data and sketch an estimated line of best fit, (b) decide if the association is positive or negative. (c) Choose two points on or near the estimated line of best fit, and use them to find a function model and predict the boiling point of water on the summit of Mt. Hood in Washington State (11,239 ft height), and along the shore of the Dead Sea (approximately 1312 ft below sea level). Answers may vary. 23. For the data given in Exercise 11 (Gross Domestic Product per Capita), choose two points on or near the line you sketched and use them to find a function model for the data. Based on this model, what is the projected GDP per capita for the year 2010?

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24. For the data given in Exercise 12 (Sales by Real Estate Brokers), choose two points on or near the line you sketched and use them to find a function

䊳

model for the data. Based on this model, how many sales can be expected for homes costing $275,000? $300,000?


25. Circumference of a Circle: C ⴝ 2␲r: The formula for the circumference of a circle can be written as a function of C in terms of r: C1r2 ⫽ 2␲r. (a) Set up a table of values for r ⫽ 1 through 6 and draw a scatterplot of the data. (b) Is the association positive or negative? Why? (c) What can you say about the strength of the correlation? (d) Sketch a line that “approximates” the data. What can you say about the slope of this line? 26. Volume of a Cylinder: V ⴝ ␲r2h: As part of a project, students cut a long piece of PVC pipe with a diameter of 10 cm into sections that are 5, 10, 15, 20, and 25 cm long. The bottom of each is then made watertight and each section is filled to the 䊳

1–90


brim with water. The volume Height Volume is then measured using a flask (cm) (cm3) marked in cm3 and the results 5 380 collected into the table shown. (a) Draw a scatterplot 10 800 of the data. (b) Is the 15 1190 association positive or 20 1550 negative? Why? (c) What can 25 1955 you say about the strength of the correlation? (d) Would the correlation here be stronger or weaker than the correlation in Exercise 25? Why? (e) Run a linear regression to verify your response.

APPLICATIONS

Use the regression capabilities of a graphing calculator to complete Exercises 27 through 34.

27. Height versus wingspan: Leonardo da Vinci’s famous diagram is an illustration of how the human body comes in predictable proportions. One such comparison is a person’s height to their wingspan (the maximum distance from the outstretched tip of one middle finger to the other). Careful measurements were taken on eight students and the set of data is shown here. Using the data, (a) draw the scatterplot; (b) determine whether the association is linear or nonlinear; (c) determine whether the association is positive or negative; and (d) find the regression equation and use it to predict the wingspan of a student with a height of 65 in. Height (x)

Wingspan ( y)

28. Patent applications: Every year the U.S. Patent and Trademark Office (USPTO) receives thousands of applications from scientists and inventors. The table given shows the number of applications received for the odd years from 1993 to 2003 (1990 S 0). Use the data to (a) draw the scatterplot; (b) determine whether the association is linear or nonlinear; (c) determine whether the association is positive or negative; and (d) find the regression equation and use it to predict the number of applications that will be received in 2011. Source: United States Patent and Trademark Office at www.uspto.gov/web

Year (1990 S 0)

Applications (1000s)

61

60.5

3

188.0

61.5

62.5

5

236.7

54.5

54.5

7

237.0

73

71.5

9

278.3

67.5

66

11

344.7

51

50.75

13

355.4

57.5

54

52

51.5

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29. Patents issued: An Year Patents increase in the (1990 S 0) (1000s) number of patent 3 107.3 applications (see Exercise 28), typically 5 114.2 brings an increase in 7 122.9 the number of patents 9 159.2 issued, though many 11 187.8 applications are denied due to 13 189.6 improper filing, lack of scientific support, and other reasons. The table given shows the number of patents issued for the odd years from 1993 to 2003 (1999 S 0). Use the data to (a) draw the scatterplot; (b) determine whether the association is linear or nonlinear; (c) determine whether the association is positive or negative; and (d) find the regression equation and use it to predict the number of applications that will be approved in 2011. Which is increasing faster, the number of patent applications or the number of patents issued? How can you tell for sure? Source: United States Patent and Trademark Office at www.uspto.gov/web

30. High jump records: In the Year Height sport of track and field, the (1900S 0) in. high jumper is an unusual 0 75 athlete. They seem to defy gravity as they launch their 12 76 bodies over the high bar. 24 78 The winning height at the 36 80 summer Olympics (to the 56 84 nearest unit) has steadily increased over time, as 68 88 shown in the table for 80 93 selected years. Using the 88 94 data, (a) draw the 92 92 scatterplot, (b) determine whether the association 96 94 is linear or nonlinear, 100 93 (c) determine whether the 104 association is positive or 108 negative, and (d) find the regression equation using t ⫽ 0 corresponding to 1900 and predict the winning height for the 2004 and 2008 Olympics. How close did the model come to the actual heights? Source: athens2004.com

31. Females/males in the workforce: Over the last 4 decades, the percentage of the female population in the workforce has been increasing at a fairly steady rate. At the same time, the percentage of the male population in the workforce has been declining. The set of data is shown in the tables. Using the data, (a) draw scatterplots for both data sets, (b) determine whether the associations are linear or nonlinear, (c) determine whether the associations are positive or negative, and (d) determine if the percentage of females in the workforce is increasing faster than the percentage of males is decreasing. Discuss/Explain how you can tell for sure. Source: 1998 Wall Street Journal Almanac, p. 316

Exercise 31 (women)

Exercise 31 (men)

Year (x) (1950 S 0)

Percent

Year (x) (1950 S 0)

Percent

5

36

5

85

10

38

10

83

15

39

15

81

20

43

20

80

25

46

25

78

30

52

30

77

35

55

35

76

40

58

40

76

45

59

45

75

50

60

50

73

32. Height versus male shoe Height Shoe Size size: While it seems 66 8 reasonable that taller people should have larger 69 10 feet, there is actually a 72 9 wide variation in the 75 14 relationship between 74 12 height and shoe size. The data in the table show the 73 10.5 height (in inches) 71 10 compared to the shoe size 69.5 11.5 worn for a random sample 66.5 8.5 of 12 male chemistry students. Using the data, 73 11 (a) draw the scatterplot, 75 14 (b) determine whether the 65.5 9 association is linear or nonlinear, (c) determine whether the association is positive or negative, and (d) find the regression equation and use it to predict the shoe size of a man 80 in. tall and another that is 60 in. tall. Note that the heights of these two men fall outside of the range of our data set (see comment after Example 5 on page 85).

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33. Plastic money: The total x amount of business (1990 S 0) y transacted using credit 1 481 cards has been changing 2 539 rapidly over the last 15 to 20 years. The total 4 731 volume (in billions of 7 1080 dollars) is shown in the 8 1157 table for selected years. 9 1291 (a) Use a graphing calculator to draw a 10 1458 scatterplot of the data 12 1638 and decide whether the association is linear or nonlinear. (b) Calculate a regression equation with x ⫽ 0 corresponding to 1990 and display the scatterplot and graph on the same screen. (c) According to the equation model, how many billions of dollars were transacted in 2003? How much will be transacted in the year 2011? Source: Statistical Abstract of the United States, various years

34. Sales of hybrid Year Hybrid Sales cars: Since their (2000 S 0) (in thousands) mass introduction 2 35 near the turn of the century, the 3 48 sales of hybrid 4 88 cars in the United 5 200 States grew 6 250 steadily until late 2007, when 7 352 the price of 8 313 gasoline began 9 292 showing signs of weakening and eventually dipped below $3.00/gal. Estimates for the annual sales of hybrid cars are given in the table for the years 2002 through 2009 12000 S 02 . (a) Use a graphing calculator to draw a scatterplot of the data and decide if the association is linear or nonlinear. (b) If linear, calculate a regression model for the data and display the scatterplot and data on the same screen. (c) Assuming that sales of hybrid cars recover, how many hybrids does the model project will be sold in the year 2012? Source: http://www.hybridcar.com

䊳


35. It can be very misleading to x y rely on the correlation 50 67 coefficient alone when 100 125 selecting a regression model. 150 145 To illustrate, (a) run a linear regression on the data set 200 275 given (without doing a 250 370 scatterplot), and note the 300 550 strength of the correlation 350 600 (the correlation coefficient). (b) Now run a quadratic regression ( STAT CALC 5:QuadReg) and note the strength of the correlation. (c) What do you notice? What factors other than the correlation coefficient must be taken into account when choosing a form of regression?

36. In his book Gulliver’s Travels, Jonathan Swift describes how the Lilliputians were able to measure Gulliver for new clothes, even though he was a giant compared to them. According to the text, “Then they measured my right thumb, and desired no more . . . for by mathematical computation, once around the thumb is twice around the wrist, and so on to the neck and waist.” Is it true that once around the neck is twice around the waist? Find at least 10 willing subjects and take measurements of their necks and waists in millimeters. Arrange the data in ordered pair form (circumference of neck, circumference of waist). Draw the scatterplot for this data. Does the association appear to be linear? Find the equation of the best fit line for this set of data. What is the slope of this line? Is the slope near m ⫽ 2?

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Making Connections


37. (1.3) Is the graph shown here, the graph of a function? Discuss why or why not.

38. (Appendix A.2/A.3) Determine the area of 18 cm the figure shown 2 1A ⫽ LW, A ⫽ ␲r 2.

24 cm

39. (1.5) Solve for r: A ⫽ P ⫹ Prt 40. (Appendix A.3) Solve for w (if possible): ⫺216w2 ⫹ 52 ⫺ 1 ⫽ 7w ⫺ 413w2 ⫹ 12

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

⫺5

y

(b)

5

⫺5

5 x

y

5 x

⫺5

5 x

⫺5

1 1. ____ y ⫽ x ⫹ 1 3 2. ____ y ⫽ ⫺x ⫹ 1 3. ____ m 7 0, b 6 0 4. ____ x ⫽ ⫺1

⫺5

5 x

5 x

⫺5

y

(g)

5

⫺5

5

⫺5

y

(f)

5

⫺5

⫺5

5 x

y

(d)

5

⫺5

⫺5

(e)

y

(c)

5

⫺5

y

(h)

5

5 x

5

⫺5

⫺5

5 x

⫺5

9. ____ f 1⫺32 ⫽ 4, f 112 ⫽ 0

10. ____ f 1⫺42 ⫽ 3, f 142 ⫽ 3

11. ____ f 1x2 ⱖ 0 for x 僆 3⫺3, q2 12. ____ x ⫽ 3

5. ____ y ⫽ ⫺2

13. ____ f 1x2 ⱕ 0 for x 僆 3 1, q 2

6. ____ m 6 0, b 6 0

14. ____ m is zero

7. ____ m ⫽ ⫺2

15. ____ function is increasing, y-intercept is negative

8. ____ m ⫽

2 3

16. ____ function is decreasing, y-intercept is negative

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1–94

SUMMARY AND CONCEPT REVIEW SECTION 1.1


KEY CONCEPTS • A relation is a collection of ordered pairs (x, y) and can be stated as a set or in equation form. • As a set of ordered pairs, we say the relation is pointwise-defined. The domain of the relation is the set of all first coordinates, and the range is the set of all corresponding second coordinates. • A relation can be expressed in mapping notation x S y, indicating an element from the domain is mapped to (corresponds to or is associated with) an element from the range. • The graph of a relation in equation form is the set of all ordered pairs (x, y) that satisfy the equation. We plot a sufficient number of points and connect them with a straight line or smooth curve, depending on the pattern formed. • The x- and y-variables of linear equations and their graphs have implied exponents of 1. • With a relation entered on the Y= screen, a graphing calculator can provide a table of ordered pairs and the related graph. x1 ⫹ x2 y1 ⫹ y2 , b. • The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is a 2 2

• The distance between the points (x1, y1) and (x2, y2) is d ⫽ 21x2 ⫺ x1 2 2 ⫹ 1y2 ⫺ y1 2 2. • The equation of a circle centered at (h, k) with radius r is 1x ⫺ h2 2 ⫹ 1y ⫺ k2 2 ⫽ r2.

EXERCISES 1. Represent the relation in mapping notation, then state the domain and range. 51⫺7, 32, 1⫺4, ⫺22, 15, 12, 1⫺7, 02, 13, ⫺22, 10, 82 6 2. Graph the relation y ⫽ 225 ⫺ x2 by completing the table, then state the domain and range of the relation. x ⫺5 ⫺4 ⫺2 0 2 4 5

y

3. Use a graphing calculator to graph the relation 5x ⫹ 3y ⫽ ⫺15. Then use the TABLE feature to determine the value of y when x ⫽ 0, and the value(s) of x when y ⫽ 0, and write the results in ordered pair form. Mr. Northeast and Mr. Southwest live in Coordinate County and are good friends. Mr. Northeast lives at 19 East and 25 North or (19, 25), while Mr. Southwest lives at 14 West and 31 South or (⫺14, ⫺31). If the streets in Coordinate County are laid out in one mile squares, 4. Use the distance formula to find how far apart they live. 5. If they agree to meet halfway between their homes, what are the coordinates of their meeting place? 6. Sketch the graph of x2 ⫹ y2 ⫽ 16. 7. Sketch the graph of x2 ⫹ y2 ⫹ 6x ⫹ 4y ⫹ 9 ⫽ 0. 8. Find an equation of the circle whose diameter has the endpoints (⫺3, 0) and (0, 4).

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Summary and Concept Review

95

Linear Equations and Rates of Change

SECTION 1.2

KEY CONCEPTS • A linear equation can be written in the form ax ⫹ by ⫽ c, where a and b are not simultaneously equal to 0. y2 ⫺ y1 • The slope of the line through (x1, y1) and (x2, y2) is m ⫽ x ⫺ x , where x1 ⫽ x2. 2 1 ¢y vertical change change in y rise ⫽ ⫽ . • Other designations for slope are m ⫽ run ⫽ change in x ¢x horizontal change • Lines with positive slope 1m 7 02 rise from left to right; lines with negative slope (m 6 0) fall from left to right. • The equation of a horizontal line is y ⫽ k; the slope is m ⫽ 0. • The equation of a vertical line is x ⫽ h; the slope is undefined. • Lines can be graphed using the intercept method. First determine (x, 0) (substitute 0 for y and solve for x), then (0, y) (substitute 0 for x and solve for y). Then draw a straight line through these points. Parallel lines have equal slopes (m1 ⫽ m2); perpendicular lines have slopes that are negative reciprocals • 1 (m1 ⫽ ⫺ or m1 # m2 ⫽ ⫺1). m2 EXERCISES ¢y rise ⫽ 9. Plot the points and determine the slope, then use the ratio to find an additional point on the line: run ¢x a. 1⫺4, 32 and 15, ⫺22 and b. (3, 4) and 1⫺6, 12. 10. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicular, or neither: a. L1: 1⫺2, 02 and (0, 6); L2: (1, 8) and (0, 5) b. L1: (1, 10) and 1⫺1, 72 : L2: 1⫺2, ⫺12 and 11, ⫺32 11. Graph each equation by plotting points: (a) y ⫽ 3x ⫺ 2 (b) y ⫽ ⫺32x ⫹ 1. 12. Find the intercepts for each line and sketch the graph: (a) 2x ⫹ 3y ⫽ 6 (b) y ⫽ 43x ⫺ 2. 13. Identify each line as either horizontal, vertical, or neither, and graph each line. a. x ⫽ 5 b. y ⫽ ⫺4 c. 2y ⫹ x ⫽ 5 14. Determine if the triangle with the vertices given is a right triangle: 1⫺5, ⫺42, (7, 2), (0, 16). 15. Find the slope and y-intercept of the line shown and discuss the slope ratio in this context. Hawk population (100s)

10

y

8 6 4 2 2

4

6

x

8

Rodent population (1000s)

SECTION 1.3

Functions, Function Notation, and the Graph of a Function

KEY CONCEPTS • A function is a relation, rule, or equation that pairs each element from the domain with exactly one element of the range. • The vertical line test says that if every vertical line crosses the graph of a relation in at most one point, the relation is a function. • The domain and range can be stated using set notation, graphed on a number line, or expressed using interval notation.

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1–96


• On a graph, vertical boundary lines can be used to identify the domain, or the set of “allowable inputs” for a • • • •

function. On a graph, horizontal boundary lines can be used to identify the range, or the set of y-values (outputs) generated by the function. When a function is stated as an equation, the implied domain is the set of x-values that yield real number outputs. x-values that cause a denominator of zero or that cause the radicand of a square root expression to be negative must be excluded from the domain. The phrase “y is a function of x,” is written as y ⫽ f 1x2 . This notation enables us to summarize the three most important aspects of a function with a single expression (input, sequence of operations, output).

EXERCISES 16. State the implied domain of each function: a. f 1x2 ⫽ 24x ⫹ 5

b. g1x2 ⫽

17. Determine h1⫺22, h1⫺23 2 , and h(3a) for h1x2 ⫽ 2x2 ⫺ 3x.

x⫺4 x ⫺x⫺6 2

18. Determine if the mapping given represents a function. If not, explain how the definition of a function is violated. Mythological deities

Primary concern

Apollo Jupiter Ares Neptune Mercury Venus Ceres Mars

messenger war craftsman love and beauty music and healing oceans all things agriculture

19. For the graph of each function shown, (a) state the domain and range, (b) find the value of f (2), and (c) determine the value(s) of x for which f 1x2 ⫽ 1. I.

II.

y 5

⫺5

5 x

⫺5

SECTION 1.4

III.

y 5

⫺5

5 x

⫺5

y 5

⫺5

5 x

⫺5

Linear Functions, Special Forms, and More on Rates of Change

KEY CONCEPTS • The equation of a nonvertical line in slope-intercept form is y ⫽ mx ⫹ b or f 1x2 ⫽ mx ⫹ b. The slope of the line is m and the y-intercept is (0, b). ¢y to • To graph a line given its equation in slope-intercept form, plot the y-intercept, then use the slope ratio m ⫽ ¢x find a second point, and draw a line through these points. • If the slope m and a point (x1, y1) on the line are known, the equation of the line can be written in point-slope form: y ⫺ y1 ⫽ m1x ⫺ x1 2 .

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97


• A secant line is the straight line drawn through two points on a nonlinear graph. ¢y literally means the quantity measured along the y-axis is changing with respect to changes ¢x in the quantity measured along the x-axis.

• The notation m ⫽

27. For the graph given, (a) find the equation of the line in point-slope form, (b) use the equation to predict the x- and y-intercepts, (c) write the equation in slope-intercept form, and (d) find y when x ⫽ 20, and the value of x for which y ⫽ 15.

SECTION 1.5

Exercise 26 W Wolf population (100s)

EXERCISES 20. Write each equation in slope-intercept form, then identify the slope and y-intercept. a. 4x ⫹ 3y ⫺ 12 ⫽ 0 b. 5x ⫺ 3y ⫽ 15 21. Graph each equation using the slope and y-intercept. a. f 1x2 ⫽ ⫺23 x ⫹ 1 b. h1x2 ⫽ 52 x ⫺ 3 22. Graph the line with the given slope through the given point. a. m ⫽ 23; 11, 42 b. m ⫽ ⫺12; 1⫺2, 32 23. What are the equations of the horizontal line and the vertical line passing through 1⫺2, 52? Which line is the point (7, 5) on? 24. Find the equation of the line passing through (1, 2) and 1⫺3, 52. Write your final answer in slope-intercept form. 25. Find the equation for the line that is parallel to 4x ⫺ 3y ⫽ 12 and passes through the point (3, 4). Write your final answer in slope-intercept form. 26. Determine the slope and y-intercept of the line shown. Then write the equation of ¢W the line in slope-intercept form and interpret the slope ratio m ⫽ in the context ¢R of this exercise.

10 8 6 4 2

2

4

6

8

10

R

Rabbit population (100s)

Exercise 27 y 100 80 60 40 20 0

2

4

6

8

10 x

Solving Equations and Inequalities Graphically; Formulas and Problem Solving

KEY CONCEPTS • To use the intersection-of-graphs method for solving equations, assign the left-hand expression as Y1 and the right-hand as Y2. The solution(s) of the original equation are the x-coordinate(s) of the point(s) of intersection of the graphs of Y1 and Y2. • When an equation is written in the form h1x2 ⫽ 0, the solutions can be found using the x-Intercept/Zeroes method. Assign h(x) as Y1, and find the x-intercepts of its graph. • Linear inequalities can be solved by first applying the intersection-of-graphs method to identify the boundary value of the solution interval. Next, the solution is determined by a careful observation of the relative positions of the graphs (is Y1 above or below Y2) and the given inequality. • To solve formulas for a specified variable, focus on the object variable and apply properties of equality to write this variable in terms of all others. • The basic elements of good problem solving include: 1. Gathering and organizing information 2. Making the problem visual 3. Developing an equation model 4. Using the model to solve the application For a complete review, see the problem-solving guide on page 71.

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EXERCISES 28. Solve the following equation using the intersection-of-graphs method. 31x ⫺ 22 ⫹ 10 ⫽ 16 ⫺ 213 ⫺ 2x2 29. Solve the following equation using the x-intercept/zeroes method. 21x ⫺ 12 ⫹ 32 ⫽ 51 25x ⫹ 15 2 ⫺ 32 30. Solve the following inequality using the intersection-of-graphs method. 31x ⫹ 22 ⫺ 2.2 6 ⫺2 ⫹ 410.2 ⫺ 0.5x2 Solve for the specified variable in each formula or literal equation. 31. V ⫽ ␲r2h for h 32. P ⫽ 2L ⫹ 2W for L 33. ax ⫹ b ⫽ c for x

34. 2x ⫺ 3y ⫽ 6 for y

Use the problem-solving guidelines (page 71) to solve the following applications. 35. At a large family reunion, two kegs of lemonade are available. One is 2% sugar (too sour) and the second is 7% sugar (too sweet). How many gallons of the 2% keg, must be mixed with 12 gallons of the 7% keg to get a 5% mix? 36. A rectangular window with a width of 3 ft and a height of 4 ft is topped by a semi-circular window. Find the total area of the window. 37. Two cyclists start from the same location and ride in opposite directions, one riding at 15 mph and the other at 18 mph. If their radio phones have a range of 22 mi, how many minutes will they be able to communicate?

SECTION 1.6

Linear Function Models and Real Data

KEY CONCEPTS • A scatterplot is the graph of all the ordered pairs in a real data set. • When drawing a scatterplot, be sure to scale the axes to comfortably fit the data. • If larger inputs tend to produce larger output values, we say there is a positive association. • If larger inputs tend to produce smaller output values, we say there is a negative association. • If the data seem to cluster around an imaginary line, we say there is a linear association between the variables. • If the data clearly cannot be approximated by a straight line, we say the variables exhibit a nonlinear association (or sometimes no association). • The correlation coefficient r measures how tightly a set of data points cluster about an imaginary curve. The strength of the correlation is given as a value between ⫺1 and 1. Measures close to ⫺1 or 1 indicate a very strong correlation. Measures close to 0 indicate a very weak correlation. • We can attempt to model linear data sets using an estimated line of best fit. • A regression line minimizes the vertical distance between all data points and the graph itself, making it the unique line of best fit. EXERCISES Exercise 38 38. To determine the value of doing homework, a student in college algebra collects data x y on the time spent by classmates on their homework in preparation for a quiz. Her data (min study) (quiz score) is entered in the table shown. (a) Use a graphing calculator to draw a scatterplot of 45 70 the data. (b) Does the association appear linear or nonlinear? (c) Is the association 30 63 positive or negative? 10 59 39. If the association in Exercise 38 is linear, (a) use a graphing calculator to find a linear 20 67 function that models the relation (study time, grade), then (b) graph the data and the line 60 73 on the same screen. (c) Does the correlation appear weak or strong? 70 85 40. According to the function model from Exercise 39, what grade can I expect if I study 90 82 for 120 minutes? 75

90

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99

Practice Test

PRACTICE TEST

2. How much water that is 102°F must be mixed with 25 gal of water at 91°F, so that the resulting temperature of the water will be 97°F.

11. My partner and I are at coordinates 1⫺20, 152 on a map. If our destination is at coordinates 135, ⫺122 , (a) what are the coordinates of the rest station located halfway to our destination? (b) How far away is our destination? Assume that each unit is 1 mi. 12. Write the equations for lines L1 and L2 shown. y L1

3. To make the bowling team, Jacques needs a threegame average of 160. If he bowled 141 and 162 for the first two games, what score S must be obtained in the third game so that his average is at least 160? 4. In the 2009 movie Star Trek (Chris Pine, Zachary Quinto, Zoy Zaldana, Eric Bana), Sulu falls off of the drill platform without a parachute, and Kirk dives off the platform to save him. To slow his fall, Sulu uses a spread-eagle tactic, while Kirk keeps his body straight and arms at his side, to maximize his falling speed. If Sulu is falling at a rate of 180 ft/sec, while Kirk is falling at 250 ft/sec, how long would it take Kirk to reach Sulu, if it took Kirk a full 2 sec to react and dive after Sulu? 5. Two relations here are functions and two are not. Identify the nonfunctions (justify your response). a. x ⫽ y2 ⫹ 2y b. y ⫽ 25 ⫺ 2x c. 冟y冟 ⫹ 1 ⫽ x d. y ⫽ x2 ⫹ 2x 6. Determine whether the lines are parallel, perpendicular, or neither: L1: 2x ⫹ 5y ⫽ ⫺15 and L2: y ⫽ 25 x ⫹ 7. 7. Graph the line using the slope and y-intercept: x ⫹ 4y ⫽ 8 8. Find the center and radius of the circle defined by x2 ⫹ y2 ⫺ 4x ⫹ 6y ⫺ 3 ⫽ 0, then sketch its graph. 9. After 2 sec, a car is traveling 20 mph. After 5 sec, its speed is 40 mph. Assuming the relationship is linear, find the velocity equation and use it to determine the speed of the car after 9 sec. 10. Find the equation of the line parallel to 6x ⫹ 5y ⫽ 3, containing the point 12, ⫺22 . Answer in slopeintercept form.

5

L2

⫺5

5 x

⫺5

13. State the domain and range for the relations shown on graphs 13(a) and 13(b). Exercise 13(a)

Exercise 13(b)

y

y

5

⫺4

5

6 x

⫺5

⫺4

6 x

⫺5

W(h) 14. For the linear function shown, a. Determine the value of W(24) from the graph. b. What input h will give an output of W1h2 ⫽ 375? c. Find a linear function for Hours worked the graph. d. What does the slope indicate in this context? e. State the domain and range of h. 2 ⫺ x2 15. Given f 1x2 ⫽ , evaluate and simplify: x2 a. f 1 23 2 b. f 1a ⫹ 32 500 400

Wages earned

1. Solve each equation. 2 a. ⫺ x ⫺ 5 ⫽ 7 ⫺ 1x ⫹ 32 3 b. ⫺5.7 ⫹ 3.1x ⫽ 14.5 ⫺ 41x ⫹ 1.52 c. P ⫽ C ⫹ k C; for C d. P ⫽ 2L ⫹ 2W; for W

300 200 100

0

8

16

24

32

40

h

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Exercise 16 16. In 2007, there were 3.3 million Apple iPhones sold worldwide. By 2009, this figure had jumped to approximately 30.3 million [Source: http://brainstormtech.blogs. fortune.cnn.com/2009/03/12/]. Assume that for a time, this growth could be modeled by a linear function. (a) Determine ¢sales the rate of change , and ¢time (b) interpret it in this context. Then use the rate of change to (c) approximate the number of sales in 2008, and what the projected sales would be for 2010 and 2011.

17. Solve the following equations using the x-intercept/zeroes method. 1 a. 2x ⫹ a4 ⫺ xb ⫽ ⫺120 ⫹ x2 3 b. 210.7x ⫺ 1.32 ⫹ 2.6 ⫽ 2x ⫺ 310.2x ⫺ 22 18. Solve the following inequalities using the intersection-of-graphs method. a. 3x ⫺ 15 ⫺ x2 ⱖ 215 ⫺ x2 ⫹ 3 b. 210.75x ⫺ 12 6 0.7 ⫹ 0.513x ⫺ 12

Exercise 19 19. To study how annual rainfall affects the ability Rainfall Cattle to attain certain levels of (in.) per Acre livestock production, a 12 2 local university collects 16 3 data on the average 19 7 annual rainfall for a 23 9 particular area and 28 11 compares this to the average number of 32 22 free-ranging cattle per 37 23 acre for ranchers in that 40 26 area. The data collected are shown in the table. (a) Use a graphing calculator to draw a scatterplot of the data. (b) Does the association appear linear or nonlinear? (c) Is the association positive or negative?

20. If the association in Exercise 19 is linear, (a) use a graphing calculator to find a linear function that models the relation (rainfall, cattle per acre), (b) use the function to find the number of cattle per acre that might be possible for an area receiving 50 in. of rainfall per year, and (c) state whether the correlation is weak or strong.

STRENGTHENING CORE SKILLS The Various Forms of a Linear Equation Learning mathematics is very much like the construction of a skyscraper. The final height of the skyscraper ultimately depends on the strength of the foundation and quality of the frame supporting each new floor as it is built. Our previous work with linear functions and their graphs, while having a number of useful applications, is actually the foundation on which much of our future work is built. For this reason, it’s important you gain a certain fluency with linear functions and relationships — even to a point where things come to you effortlessly and automatically. As noted mathematician Henri Lebesque once said, “An idea reaches its maximum level of usefulness only when you understand it so well that it seems like you have always known it. You then become incapable of seeing the idea as anything but a trivial and immediate result.” These formulas and concepts, while simple, have an endless number of significant and substantial applications. Forms and Formulas

slope formula

point-slope form

slope-intercept form

standard form

y2 ⫺ y1 x2 ⫺ x1 given any two points on the line

y ⫺ y1 ⫽ m1x ⫺ x1 2

y ⫽ mx ⫹ b

Ax ⫹ By ⫽ C

given slope m and any point 1x1, y1 2

given slope m and y-intercept (0, b)

A, B, and C are integers (used in linear systems)

m⫽

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Calculator Exploration and Discovery

101

Characteristics of Lines

y-intercept

x-intercept

increasing

decreasing

(0, y) let x ⫽ 0, solve for y

(x, 0) let y ⫽ 0, solve for x

m 7 0 line slants upward from left to right

m 6 0 line slants downward form left to right

intersecting

parallel

perpendicular

dependent

m1 ⫽ m2

m1 ⫽ m2, b1 ⫽ b2 lines do not intersect

m1m2 ⫽ ⫺1 lines intersect at right angles

m1 ⫽ m2, b1 ⫽ b2 lines intersect at all points

horizontal

vertical

identity

y⫽k horizontal line through k

x⫽h vertical line through h

y⫽x the input value identifies the output

Relationships between Lines

lines intersect at one point Special Lines

Use the formulas and concepts reviewed here to complete the following exercises. For the two points given: (a) compute the slope of the line through the points and state whether the line is increasing or decreasing, (b) find the equation of the line in point-slope form, then write the equation in slope-intercept form, and (c) find the x- and y-intercepts and graph the line. Exercise 1: P1(0, 5)

P2 (6, 7)

Exercise 2: P1(3, 2)

P2 (0, 9)

Exercise 3: P1(3, 2)

P2 (9, 5)

Exercise 4: P1 1⫺5, ⫺42

P2 (3, 2)

Exercise 6: P1 12, ⫺72

P2 1⫺8, ⫺22

Exercise 5: P1 1⫺2, 52

P2 16, ⫺12

CALCULATOR EXPLORATION AND DISCOVERY Evaluating Expressions and Looking for Patterns These “explorations” are designed to explore the full potential of a graphing calculator, as well as to use this potential to investigate patterns and discover connections that might otherwise be overlooked. In this exploration and discovery, we point out the various ways an expression can be evaluated on a graphing calculator. Some ways seem easier, faster, and/or better than others, but each has advantages and disadvantages depending on the task at hand, and it will help to be aware of them all for future use. One way to evaluate an expression is to use the TABLE feature of a graphing calculator, with the expression (TBLSET) screen entered as Y1 on the Y= screen. If you want the calculator to generate inputs, use the 2nd to indicate a starting value (TblStartⴝ) and an increment value (⌬Tbl ⴝ), and set the calculator in Indpnt: AUTO ASK mode (to input specific values, the calculator should be in Indpnt: AUTO ASK mode). After pressing 2nd GRAPH (TABLE), the calculator shows the corresponding Figure 1.105 input and output values. Expressions can also be evaluated on the home screen for a single value or a series of values. Enter the expression ⫺34x ⫹ 5 on the Y= screen (see Figure 1.105) and use 2nd MODE (QUIT) to get back to the home screen. To evaluate this expression, (Y-VARS), and use the first option 1:Function . This access Y1 using VARS brings us to a submenu where any of the equations Y1 through Y0 (actually Y10) can be accessed. Since the default setting is the one we need (1:Y1), simply press and Y1 appears on the home screen. To evaluate a single input, simply enclose it in WINDOW

ENTER

ENTER

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parentheses. To evaluate more than one input, enter the numbers as a set of values with the set enclosed in parentheses. In Figure 1.106, Y1 has been evaluated for x ⫽ ⫺4, then simultaneously for x ⫽ ⫺4, ⫺2, 0, and 2. A third way to evaluate expressions is using a list, with the desired inputs entered in List 1 (L1), then List 2 (L2) defined in terms of L1. For example, L2 ⫽ ⫺34L1 ⫹ 5 will return the same values for inputs of ⫺4, ⫺2, 0, and 2 seen previously on the home screen (remember to clear the lists first). Lists are accessed by pressing STAT 1:Edit. Enter the numbers ⫺4, ⫺2, 0, and 2 in L1, then use the right arrow to move to L2. It is important to note that you next press the up arrow key so that the cursor overlies L2. The bottom of the screen now reads “L2 ⫽ ” and the calculator is waiting for us to define L2. After entering L2 ⫽ ⫺34L1 ⫹ 5 (see Figure 1.107) and pressing we obtain the same outputs as before (see Figure 1.108). The advantage of using the “list” method is that we can further explore or experiment with the output values in a search for patterns.

Figure 1.106

Figure 1.107

ENTER

Exercise 1: Evaluate the expression 0.2L1 ⫹ 3 on the list screen, using consecutive integer inputs from ⫺6 to 6 inclusive. What do you notice about the outputs? Exercise 2: Evaluate the expression 12L1 ⫺ 19.1 on the list screen, using consecutive integer inputs from ⫺6 to 6 inclusive. We suspect there is a pattern to the output values, but this time the pattern is very difficult to see. On the home screen, compute the difference between a few successive outputs from L2 [for example, L2112 ⫺ L212)]. What do you notice?

Figure 1.108

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CONNECTIONS TO CALCULUS Understanding and internalizing concepts related to linear functions is one of the main objectives of Chapter 1 and its Strengthening Core Skills feature. The ability to quickly and correctly write the equation of a line given sufficient information has a number of substantial applications in the calculus sequence. ln calculus, we will extend our understanding of secant lines to help understand lines drawn tangent to a curve and further to lines and planes drawn tangent to three-dimensional surfaces. These relationships will lead to significant and meaningful applications in topography, meteorology, engineering, and many other areas.

Tangent Lines In Section 1.4 we learned how to write the equation of a line given its slope and any point on the line. ln that section, Example 6 uses the slope-intercept form y mx b as a formula, while Example 10 applies the point-slope form y y1 m(x x1). In calculus, we regularly use both forms to write the equations of secant lines and tongent lines. Consider the graph of f 1x2 x2 2x 3, a parabola that opens upward, with y-intercept (0, 3) and vertex at (1, 4). Using the tools of calculus, we can show that the function g1x2 2x 2 gives the slope of any line drawn tangent to this graph at a given x. For example, to find the slope of the line tangent to this curve when x 3, we evaluate g(x) at x 3, and find the slope will be g(3) 4. Note how this information is used in Example 1. EXAMPLE 1

䊳

Finding the Equation of a Tangent Line

Solution

䊳

To begin, we first determine the slope of the tangent line. For x 0, we have

Find an equation of the line drawn tangent to the graph of f 1x2 x2 2x 3 at x 0, and write the result in slope-intercept form. g1x2 2x 2

given equation and slope of tangent line

g102 2102 2

substitute 0 for x

2 y x 2 2x 3

y 5 4 3

y 2x 3

2 1

5 4 3 2 1 1 2

(0, 3) 3 4 5

1

2

3

4

5

x

result

The slope of the line drawn tangent to this curve at x 0, is m 2. However, recall that to find the equation of this line we also need a point on the line. Using x 0 once again, we note (0, 3) is the y-intercept of the graph of f(x) and is also a point on the tangent line. Using the point (0, 3) and the slope m 2,we have y y1 m1x x1 2 y 132 21x 02 y 3 2x y 2x 3

point-slope form substitute 2 for m, (o, 3) for (x1, y 1) simplify slope-intercept form

In the figure shown, note the graph of y 2x 3 is tangent to the graph of y x2 2x 3 at the point (0, 3). Now try Exercises 1 through 8

䊳

Similar to our work here with tangent lines, many applications of advanced mathematics require that we find the equation of a line drawn perpendicular to this tangent line. In this case, the line is called a normal to the curve at point (x, y). See Exercises 9 through 12.

1–103

103

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Connections to Calculus Exercises For Exercises 1 through 4, the function g(x) ⴝ 2x ⴙ 4 gives the slope of any line drawn tangent to the graph of f(x) ⴝ x2 ⴙ 4x ⴚ 5 at a given x. Find an equation of the line drawn tangent to the graph of f(x) at the following x-values, and write the result in slope-intercept form. 1. x 4 3. x 2

2. x 0 4. x 1

For Exercises 5 through 8, the function v(x) ⴝ 6x2 ⴚ 6x ⴚ 36 gives the slope of any line drawn tangent to the graph of s1x2 ⴝ 2x3 ⴚ 3x2 ⴚ 36x at given x. Find an equation of the line drawn tangent to the graph of s(x) at the following x-values, and write the result in slope-intercept form. 5. x 2 7. x 1

6. x 3 8. x 4

12.

y x 4

y 5 4 3

(1, 3)

2 1 5 4 3 2 1 1

1

2

3

4

4 5

13. A theorem from elementry geometary states, “A line drawn tangent to a circle is perpendicular to the radius at the point of tangency.” Find the equation of the tangent line shown in the figure. y 4 3 2 1 5 4 3 2 1

4

1

3

1

4 1

2

3

4

5

(1, 2)

5

y

y 5

5

4

y 0.5x 3

3 2

2

1

(3, 1.5)

5 4 3 2 1 1

1

2

3

4

5

x

2

3

3

4

4

5

5

y 5 4 3 2

(1, 1)

5 4 3 2 1 1

2

y 2x 1

1

5 4 3 2 1 1 2 3 4 5

x

14. Find the equation of the line containing the perpendicular bisector of the line segment with endpoints (2, 3) and (4, 1) shown.

4

1

5

x

3

3

4

5

5 4 3 2 1 1

4

3

2

2

10.

2

(0.5, 1)

y 2x

3

11.

5

y 5

2

x

3

For Exercises 9 through 12, find the equation of the normal to each tangent shown, or the equation of the tangent line for each normal given. 9.

5

2

1

2

3

4

5

x

1

2

3

4

5

x

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CHAPTER CONNECTIONS

More on Functions CHAPTER OUTLINE 2.1 Analyzing the Graph of a Function 106 2.2 The Toolbox Functions and Transformations 120 2.3 Absolute Value Functions, Equations, and Inequalities 136

2.4 Basic Rational Functions and Power Functions; More on the Domain 148

2.5 Piecewise-Defined Functions 163 2.6 Variation: The Toolbox Functions in Action 177

Viewing a function in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For example, the power generated by a wind turbine is often modeled 8v3 by the function P 1v2 ⫽ , where P is the 125 power in watts and v is the wind velocity in miles per hour. While the formula enables us to predict the power generated for a given wind speed, the graph offers a visual representation of this relationship, where we note a rapid growth in power output as the wind speed increases. 䊳


The foundation and study of calculus involves using absolute value inequalities to analyze very small differences. The Connections to Calculus for Chapter 2 expands on the notation and language used in Connections this analysis, and explores the need to solve a broad range of equation types. to Calculus

105

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2.1

Analyzing the Graph of a Function

LEARNING OBJECTIVES

In this section, we’ll consolidate and refine many of the ideas we’ve encountered related to functions. When functions and graphs are applied as real-world models, we create numeric and visual representations that enable an informed response to questions involving maximum efficiency, positive returns, increasing costs, and other relationships that can have a great impact on our lives.

In Section 2.1 you will see how we can

A. Determine whether a

B.

C.

D.

E.

function is even, odd, or neither Determine intervals where a function is positive or negative Determine where a function is increasing or decreasing Identify the maximum and minimum values of a function Locate local maximum and minimum values using a graphing calculator

A. Graphs and Symmetry While the domain and range of a function will remain dominant themes in our study, for the moment we turn our attention to other characteristics of a function’s graph. We begin with the concept of symmetry. Figure 2.1

Symmetry with Respect to the y-Axis

Consider the graph of f 1x2 x 4x shown in Figure 2.1, where the portion of the graph to the left of the y-axis appears to be a mirror image of the portion to the right. A function is symmetric to the y-axis if, given any point (x, y) on the graph, the point 1x, y2 is also on the graph. We note that 11, 32 is on the graph, as is 11, 32, and that 12, 02 is an x-intercept of the graph, as is (2, 0). Functions that are symmetric with respect to the y-axis are also known as even functions and in general we have: 4

2

5

y f(x) x4 4x2 (2.2, ~4)

(2.2, ~4)

(2, 0)

(2, 0)

5

5

x

(1, 3) 5 (1, 3)

Even Functions: y-Axis Symmetry A function f is an even function if and only if, for each point (x, y) on the graph of f, the point (x, y) is also on the graph. In function notation f 1x2 f 1x2

Symmetry can be a great help in graphing new functions, enabling us to plot fewer points and to complete the graph using properties of symmetry. EXAMPLE 1

䊳

Graphing an Even Function Using Symmetry a. The function g(x) in Figure 2.2 (shown in solid blue) is known to be even. Draw the complete graph. 2 Figure 2.2 b. Show that h1x2 x3 is an even function using y the arbitrary value x k [show h1k2 h1k2 ], 5 then sketch the complete graph using h(0), g(x) h(1), h(8), and y-axis symmetry. (1, 2)

Solution

106

䊳

a. To complete the graph of g (see Figure 2.2) use the points (4, 1), (2, 3), (1, 2), and y-axis symmetry to find additional points. The corresponding ordered pairs are (4, 1), (2, 3), and (1, 2), which we use to help draw a “mirror image” of the partial graph given.

(1, 2)

(4, 1)

(4, 1) 5 x

5

(2, 3)

(2, 3) 5

2–2

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107

Section 2.1 Analyzing the Graph of a Function 2

b. To prove that h1x2 x3 is an even function, we must show h1k2 h1k2 for any constant k. 2 1 After writing x3 as 3x2 4 3 , we have: h1k2 ⱨ h1k2

2

2 1k2 ⱨ 2 1k2 3

WORTHY OF NOTE The proof can also1be demonstrated 2 by writing x 3 as 1x 3 2 2, and you are asked to complete this proof in Exercise 69.

2

3

y 5

(8, 4)

first step of proof

3 1k2 4 ⱨ 3 1k2 4 2

1 3

Figure 2.3

1 3

h(x)

(8, 4)

evaluate h (k ) and h(k ) (1, 1)

2

radical form result: 1k2 2 k2

3 2 3 2 2 k 2 k ✓

(1, 1)

10

(0, 0)

Using h102 0, h112 1, and h182 4 with y-axis symmetry produces the graph shown in Figure 2.3.

10

x

5


䊳

Symmetry with Respect to the Origin Another common form of symmetry is known as symmetry to the origin. As the name implies, the graph is somehow “centered” at (0, 0). This form of symmetry is easy to see for closed figures with their center at (0, 0), like certain polygons, circles, and ellipses (these will exhibit both y-axis symmetry and symmetry with respect to the origin). Note the relation graphed in Figure 2.4 contains the points (3, 3) and (3, 3), along with (1, 4) and (1, 4). But the function f (x) in Figure 2.5 also contains these points and is, in the same sense, symmetric to the origin (the paired points are on opposite sides of the x- and y-axes, and a like distance from the origin). Figure 2.4

Figure 2.5

y

y

5

5

(1, 4)

(1, 4)

(3, 3)

(3, 3)

5

5

x

f(x)

5

5

(3, 3) (1, 4)

(3, 3) (1, 4)

5

x

5

Functions symmetric to the origin are known as odd functions and in general we have: Odd Functions: Symmetry About the Origin A function f is an odd function if and only if, for each point (x, y) on the graph of f, the point (x, y) is also on the graph. In function notation f 1x2 f 1x2

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CHAPTER 2 More on Functions

EXAMPLE 2

䊳

Graphing an Odd Function Using Symmetry a. In Figure 2.6, the function g(x) given (shown in solid blue) is known to be odd. Draw the complete graph. b. Show that h1x2 x3 4x is an odd function using the arbitrary value x k [show h1x2 h1x2 ], then sketch the graph using h122 , h112 , h(0), and odd symmetry.

Solution

䊳

a. To complete the graph of g, use the points (6, 3), (4, 0), and (2, 2) and odd symmetry to find additional points. The corresponding ordered pairs are (6, 3), (4, 0), and (2, 2), which we use to help draw a “mirror image” of the partial graph given (see Figure 2.6). Figure 2.6

Figure 2.7

y

y

10

5

(1, 3)

g(x) (6, 3)

(2, 2) (4, 0)

10

h(x)

(2, 0) x (6, 3) 10

(4, 0)

(2, 2)

5

(2, 0) (0, 0)

5

x

(1, 3) 10

5

b. To prove that h1x2 x3 4x is an odd function, we must show that h1k2 h1k2. h1k2 ⱨ h1k2 1k2 41k2 ⱨ 3 k3 4k 4 k3 4k k3 4k ✓ 3

Using h122 0, h112 3, and h102 0 with symmetry about the origin produces the graph shown in Figure 2.7. Now try Exercises 13 through 24 A. You’ve just seen how we can determine whether a function is even, odd, or neither

䊳

Finally, some relations also exhibit a third form of symmetry, that of symmetry to the x-axis. If the graph of a circle is centered at the origin, the graph has both odd and even symmetry, and is also symmetric to the x-axis. Note that if a graph exhibits x-axis symmetry, it cannot be the graph of a function.

B. Intervals Where a Function Is Positive or Negative

Consider the graph of f 1x2 x2 4 shown in Figure 2.8, which has x-intercepts at (2, 0) and (2, 0). As in Section 1.5, the x-intercepts have the form (x, 0) and are called the zeroes of the function (the x-input causes an output of 0). Just as zero on the number line separates negative numbers from positive numbers, the zeroes of a function that crosses the x-axis separate x-intervals where a function is negative from x-intervals where the function is positive. Noting that outputs (y-values) are positive in Quadrants I and II, f 1x2 7 0 in intervals where its graph is above the x-axis. Conversely, f 1x2 6 0

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Section 2.1 Analyzing the Graph of a Function

in x-intervals where its graph is below the x-axis. To illustrate, compare the graph of f in Figure 2.8, with that of g in Figure 2.9. Figure 2.8 5

(2, 0)

Figure 2.9

y f(x) x2 4

5

y g(x) (x 4)2

(2, 0)

5

5

x

3

(4, 0)

x

5

(0, 4) 5

The graph of f is a parabola, with x-intercepts of (2, 0) and (2, 0). Using our previous observations, we note f 1x2 0 for x 1q, 24 ´ 3 2, q2 since the graph is above the x-axis, and f 1x2 6 0 for x 12, 22 . The graph of g is also a parabola, but is entirely above or on the x-axis, showing g1x2 0 for x ⺢. The difference is that zeroes coming from factors of the form (x r) (with degree 1) allow the graph to cross the x-axis. The zeroes of f came from 1x 22 1x 22 0. Zeroes that come from factors of the form 1x r2 2 (with degree 2) cause the graph to “bounce” off the x-axis (intersect without crossing) since all outputs must be nonnegative. The zero of g came from 1x 42 2 0.

WORTHY OF NOTE These observations form the basis for studying polynomials of higher degree in Chapter 4, where we extend the idea to factors of the form 1x r2 n in a study of roots of multiplicity.

EXAMPLE 3

5

䊳

Solving an Inequality Using a Graph Use the graph of g1x2 x3 2x2 4x 8 given to solve the inequalities a. g1x2 0 b. g1x2 6 0

Solution

䊳

From the graph, the zeroes of g (x-intercepts) occur at (2, 0) and (2, 0). a. For g1x2 0, the graph must be on or above the x-axis, meaning the solution is x 32, q 2 . b. For g1x2 6 0, the graph must be below the x-axis, and the solution is x 1q, 22 . As we might have anticipated from the graph, factoring by grouping gives g1x2 1x 221x 22 2, with the graph crossing the x-axis at 2, and bouncing off the x-axis (intersects without crossing) at x 2.

y (0, 8) g(x) 5

5

5

x

2


䊳

Even if the function is not a polynomial, the zeroes can still be used to find x-intervals where the function is positive or negative.

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EXAMPLE 4

Solution

䊳

䊳

Solving an Inequality Using a Graph

y

For the graph of r 1x2 1x 1 2 shown, solve a. r 1x2 0 b. r 1x2 7 0 a. The only zero of r is at (3, 0). The graph is on or below the x-axis for x 31, 3 4 , so r 1x2 0 in this interval. b. The graph is above the x-axis for x 13, q 2 , and r 1x2 7 0 in this interval.

10

r(x) 10

10

10

Now try Exercises 29 through 32 B. You’ve just seen how we can determine intervals where a function is positive or negative

x

䊳

This study of inequalities shows how the graphical solutions studied in Section 1.5 are easily extended to the graph of a general function. It also strengthens the foundation for the graphical solutions studied throughout this text.

C. Intervals Where a Function Is Increasing or Decreasing In our study of linear graphs, we said a graph was increasing if it “rose” when viewed from left to right. More generally, we say the graph of a function is increasing on a given interval if larger and larger x-values produce larger and larger y-values. This suggests the following tests for intervals where a function is increasing or decreasing. Increasing and Decreasing Functions Given an interval I that is a subset of the domain, with x1 and x2 in I and x2 7 x1, 1. A function is increasing on I if f 1x2 2 7 f 1x1 2 for all x1 and x2 in I (larger inputs produce larger outputs). 2. A function is decreasing on I if f 1x2 2 6 f 1x1 2 for all x1 and x2 in I (larger inputs produce smaller outputs). 3. A function is constant on I if f 1x2 2 f 1x1 2 for all x1 and x2 in I (larger inputs produce identical outputs).

f(x)

f (x) is increasing on I

f(x2)

f (x)

f (x) is decreasing on I

f(x) is constant on I

f (x)

f (x1)

f(x1)

f (x2)

f (x2)

f (x1)

f (x1)

f (x1) x1 Interval I

x2

x2 x1 and f (x2) f (x1) for all x I graph rises when viewed from left to right

x

x1 Interval I

x2

x2 x1 and f (x2) f (x1) for all x I graph falls when viewed from left to right

f(x2)

f(x1)

f (x2) x

x1 Interval I

x2

x

x2 x1 and f(x2) f(x1) for all x I graph is level when viewed from left to right

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Consider the graph of f 1x2 x2 4x 5 given in Figure 2.10. Since the parabola opens downward with the vertex at (2, 9), the function must increase until it reaches this peak at x 2, and decrease thereafter. Notationally we’ll write this as f 1x2c for x 1q, 22 and f 1x2T for x 12, q 2. Using the interval 13, 22 shown below the figure, we see that any larger input value from the interval will indeed produce a larger output value, and f 1x2c on the interval. For instance, 1 7 2

x2 7 x1

and

and

f 112 7 f 122 8 7 7

Figure 2.10 10

y f(x) x2 4x 5 (2, 9) (0, 5)

(1, 0)

(5, 0)

5

5

x

10

x (3, 2)

f 1x2 2 7 f 1x1 2

A calculator check is shown in the figure. Note the outputs are increasing until x 2, then they begin decreasing. EXAMPLE 5

䊳

Finding Intervals Where a Function Is Increasing or Decreasing

y 5

Use the graph of v(x) given to name the interval(s) where v is increasing, decreasing, or constant.

Solution

䊳

From left to right, the graph of v increases until leveling off at (2, 2), then it remains constant until reaching (1, 2). The graph then increases once again until reaching a peak at (3, 5) and decreases thereafter. The result is v1x2c for x 1q, 22 ´ 11, 32, v1x2T for x 13, q 2, and v(x) is constant for x 12, 12 .

v(x)

5

5

x

5


䊳

WORTHY OF NOTE Questions about the behavior of a function are asked with respect to the y outputs: is the function positive, is the function increasing, etc. Due to the input/output, cause/effect nature of functions, the response is given in terms of x, that is, what is causing outputs to be positive, or to be increasing.

EXAMPLE 6

Notice the graph of f in Figure 2.10 and the graph of v in Example 5 have something in common. It appears that both the far left and far right branches of each graph point downward (in the negative y-direction). We say that the end-behavior of both graphs is identical, which is the term used to describe what happens to a graph as 冟x冟 becomes very large. For x 7 0, we say a graph is, “up on the right” or “down on the right,” depending on the direction the “end” is pointing. For x 6 0, we say the graph is “up on the left” or “down on the left,” as the case may be. 䊳

Describing the End-Behavior of a Graph

The graph of f 1x2 x 3x is shown. Use the graph to name intervals where f is increasing or decreasing, and comment on the end-behavior of the graph.

y 5

3

Solution

C. You’ve just seen how we can determine where a function is increasing or decreasing

䊳

From the graph we observe that f 1x2c for x 1q, 12 ´ 11, q 2 , and f 1x2T for x 11, 12 . The end-behavior of the graph is down on the left, and up on the right (down/up).

f(x) x2 3x

5

5

x

5


䊳

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D. Maximum and Minimum Values The y-coordinate of the vertex of a parabola that opens downward, and the y-coordinate of “peaks” from other graphs are called maximum values. A global maximum (also called an absolute maximum) names the largest y-value over the entire domain. A local maximum (also called a relative maximum) gives the largest range value in a specified interval; and an endpoint maximum can occur at an endpoint of the domain. The same can be said for any corresponding minimum values. We will soon develop the ability to locate maximum and minimum values for quadratic and other functions. In future courses, methods are developed to help locate maximum and minimum values for almost any function. For now, our work will rely chiefly on a function’s graph. EXAMPLE 7

䊳

Analyzing Characteristics of a Graph Analyze the graph of function f shown in Figure 2.11. Include specific mention of a. domain and range, b. intervals where f is increasing or decreasing, c. maximum (max) and minimum (min) values, d. intervals where f 1x2 0 and f 1x2 6 0, and e. whether the function is even, odd, or neither.

Solution

D. You’ve just seen how we can identify the maximum and minimum values of a function

䊳

a. Using vertical and horizontal boundary lines show the domain is x ⺢, with a range of: y 1q, 74 . b. f 1x2c for x 1q, 32 ´ 11, 52 shown in blue in Figure 2.12, and f 1x2T for x 13, 12 ´ 15, q 2 as shown in red. c. From part (b) we find that y 5 at (3, 5) and y 7 at (5, 7) are local maximums, with a local minimum of y 1 at (1, 1). The point (5, 7) is also a global maximum (there is no global minimum). d. f 1x2 0 for x 36, 84 ; f 1x2 6 0 for x 1q, 62 ´ 18, q 2 e. The function is neither even nor odd.

Figure 2.11 y 10

(5, 7) f(x)

(3, 5)

(1, 1) 10

10

x

10

Figure 2.12 y 10

(5, 7) (3, 5) (6, 0)

(1, 1)

10

(8, 0) 10 x

10


䊳

E. Locating Maximum and Minimum Values Using Technology In Section 1.5, we used the 2nd TRACE (CALC) 2:zero option of a graphing calculator to locate the zeroes/x-intercepts of a function. The maximum or minimum values of a function are located in much the same way. To illustrate, enter the function y x3 3x 2 as Y1 on the Y= screen, then Figure 2.13 graph it in the window shown, where x 3 4, 44 5 and y 35, 5 4 . As seen in Figure 2.13, it appears a local maximum occurs at x 1 and a local minimum at x 1. To actually find the local maximum, we access the 2nd TRACE 4 4 (CALC) 4:maximum option, which returns you to the graph and asks for a Left Bound?, a Right Bound?, and a Guess? as before. Here, we entered a left bound of “3,” a right bound 5

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Figure 2.14 of “0” and bypassed the guess option by pressing a third time (the calculator again sets 5 the “䉴” and “䉳” markers to show the bounds chosen). The cursor will then be located at the local maximum in your selected interval, with 4 the coordinates displayed at the bottom of the 4 screen (Figure 2.14). Due to the algorithm the calculator uses to find these values, a decimal number very close to the expected value is 5 sometimes displayed, even if the actual value is an integer (in Figure 2.14, 0.9999997 is displayed instead of 1). To check, we evaluate f 112 and find the local maximum is indeed 0. ENTER

EXAMPLE 8

䊳

Locating Local Maximum and Minimum Values on a Graphing Calculator Find the maximum and minimum values of f 1x2

Solution

䊳

1 4 1x 8x2 72 . 2

1 4 1X 8X2 72 as Y1 on the Y= screen, and graph the 2 function in the ZOOM 6:ZStandard window. To locate the leftmost minimum value, we access the 2nd TRACE (CALC) 3:minimum option, and enter a left bound of “4,” and a right bound of “1” (Figure 2.15). After pressing once more, the cursor is located at the minimum in the interval we selected, and we find that a local minimum of 4.5 occurs at x 2 (Figure 2.16). Repeating these steps using the appropriate options shows a local maximum of y 3.5 occurs at x 0, and a second local minimum of y 4.5 occurs at x 2. Note that y 4.5 is also a global minimum. Begin by entering

ENTER

Figure 2.15

Figure 2.16 10

10

10

E. You’ve just seen how we can locate local maximum and minimum values using a graphing calculator

10

10

10

10

10


䊳

The ideas presented here can be applied to functions of all kinds, including rational functions, piecewise-defined functions, step functions, and so on. There is a wide variety of applications in Exercises 57 through 64.

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2.1 EXERCISES 䊳



1. The graph of a polynomial will cross through the x-axis at zeroes of factors of degree 1, and off the x-axis at the zeroes from linear factors of degree 2.

3. If f 1x2 2 7 f 1x1 2 for x1 6 x2 for all x in a given interval, the function is in the interval.

5. Discuss/Explain the following statement and give an example of the conclusion it makes. “If a function f is decreasing to the left of (c, f (c)) and increasing to the right of (c, f (c)), then f (c) is either a local or a global minimum.” 䊳

2. If f 1x2 f 1x2 for all x in the domain, we say that f is an function and symmetric to the axis. If f 1x2 f 1x2 , the function is and symmetric to the . 4. If f 1c2 f 1x2 for all x in a specified interval, we say that f (c) is a local for this interval.

6. Without referring to notes or textbook, list as many features/attributes as you can that are related to analyzing the graph of a function. Include details on how to locate or determine each attribute.


The following functions are known to be even. Complete each graph using symmetry.

7.

8.

y 5

5

5 x

3 15. f 1x2 4 1 xx

1 16. g1x2 x3 6x 2 1 17. p1x2 3x3 5x2 1 18. q1x2 x x

y 10

10

10 x

Determine whether the following functions are even, odd, or neither.

10

5

Determine whether the following functions are even: f 1k2 f 1k2 .

9. f 1x2 7冟 x 冟 3x2 5 10. p1x2 2x4 6x 1

1 11. g1x2 x4 5x2 1 3

1 冟x冟 x2 The following functions are known to be odd. Complete each graph using symmetry. 13.

12. q1x2

14.

y 10

Determine whether the following functions are odd: f 1k2 f 1k2 .

19. w1x2 x3 x2

3 20. q1x2 x2 3冟x冟 4

1 3 21. p1x2 2 1 x x3 4

22. g1x2 x3 7x

23. v1x2 x3 3冟x冟

Use the graphs given to solve the inequalities indicated. Write all answers in interval notation.

25. f 1x2 x3 3x2 x 3; f 1x2 0

y 10

y

5

10

10 x

10

10 x 5

10

24. f 1x2 x4 7x2 30

5 x

10 5

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26. f 1x2 x3 2x2 4x 8; f 1x2 7 0

32. g1x2 1x 12 3 1; g1x2 6 0 y

y

5

5 5

5 x

g(x) 5

5 x

1

27. f 1x2 x4 2x2 1; f 1x2 7 0 y

5

5

5 x

5

Name the interval(s) where the following functions are increasing, decreasing, or constant. Write answers using interval notation. Assume all endpoints have integer values.

33. y V1x2

34. y H1x2 y

y

10

5

28. f 1x2 x3 2x2 4x 8; f 1x2 0 y

1

5

10

10 x

5

5 x

H(x)

5

5 x 10

35. y f 1x2

5

5

36. y g1x2 y

y

10 10

3 29. p1x2 1 x 1 1; p1x2 0 y

f(x)

8

g(x)

6

10

5

10 x 4 2

10 5

2

4

6

8

x

10

5 x

p(x)

For Exercises 37 through 40, determine (a) interval(s) where the function is increasing, decreasing or constant, and (b) comment on the end-behavior.

5

30. q1x2 1x 1 2; q1x2 7 0 y

37. p1x2 0.51x 22 3

3 38. q1x2 1 x1

y

5

y

5

5

(0, 4) q(x) 5

5 x

(2, 0)

5

31. f 1x2 1x 12 1; f 1x2 0 3

y

5

(1, 0)

5

5 x

5

5

39. y f 1x2

5 x

(0, 1)

5

40. y g1x2 y

y 10 5

5

f(x)

5 x

10

5 5

10 x

5 x 3

10

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For Exercises 41 through 48, determine the following (answer in interval notation as appropriate): (a) domain and range of the function; (b) zeroes of the function; (c) interval(s) where the function is greater than or equal to zero, or less than or equal to zero; (d) interval(s) where the function is increasing, decreasing, or constant; and (e) location of any local max or min value(s).

42. y f 1x2

41. y H1x2 5

y (2, 5)

45. y Y1

46. y Y2 y

y

5

5

5

5

5 x

5 x

5

5

47. p1x2 1x 32 3 1 48. q1x2 冟x 5冟 3

y 5

y

y

10 10

(1, 0)

(3.5, 0)

(3, 0)

5

5 x

5

8

5 x

6

10 5 (0, 5)

10 x

4

5 2

43. y g1x2

44. y h1x2

10

y

y

5 5 x

g(x) 2

5

x

2

4

6

8

10

x

Use a graphing calculator to find the maximum and minimum values of the following functions. Round answers to nearest hundredth when necessary.

5

5

2

5

3 3 6 1x 5x2 6x2 50. y 1x3 4x2 3x2 4 5 51. y 0.0016x5 0.12x3 2x 49. y

52. y 0.01x5 0.03x4 0.25x3 0.75x2 54. y x2 2x 3 2

53. y x 24 x 䊳


55. Conic sections—hyperbola: y 13 24x2 36 y While the conic sections are 5 not covered in detail until f(x) later in the course, we’ve already developed a number 5 5 x of tools that will help us understand these relations and their graphs. The 5 equation here gives the “upper branches” of a hyperbola, as shown in the figure. Find the following by analyzing the equation: (a) the domain and range; (b) the zeroes of the relation; (c) interval(s) where y is increasing or decreasing; (d) whether the relation is even, odd, or neither, and (e) solve for x in terms of y.

56. Trigonometric graphs: y sin1x2 and y cos1x2 The trigonometric functions are also studied at some future time, but we can apply the same tools to analyze the graphs of these functions as well. The graphs of y sin x and y cos x are given, graphed over the interval x 3360°, 360°4 . Use them to find (a) the range of the functions; (b) the zeroes of the functions; (c) interval(s) where y is increasing/decreasing; (d) location of minimum/maximum values; and (e) whether each relation is even, odd, or neither. y

y

(90, 1)

1

1

y cos x

y sin x

(90, 0) 360 270 180

90

90

1

180

270

360 x

360 270 180

90

90

1

180

270

360 x

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APPLICATIONS c. d. e. f. g. h.

Height (feet)

57. Catapults and projectiles: Catapults have a long and interesting history that dates back to ancient times, when they were used to launch javelins, rocks, and other projectiles. The diagram given illustrates the path of the projectile after release, which follows a parabolic arc. Use the graph to determine the following: 80 70 60 50 40 30

20

60

100

140

180

220

260

Distance (feet)

a. State the domain and range of the projectile. b. What is the maximum height of the projectile? c. How far from the catapult did the projectile reach its maximum height? d. Did the projectile clear the castle wall, which was 40 ft high and 210 ft away? e. On what interval was the height of the projectile increasing? f. On what interval was the height of the projectile decreasing? P (millions of dollars)

58. Profit and loss: The profit of DeBartolo Construction Inc. is illustrated by the graph shown. Use the graph to t (years since 1990) estimate the point(s) or the interval(s) for which the profit P was: a. increasing b. decreasing 16 12 8 4 0 4 8

1 2 3 4 5 6 7 8 9 10

constant a maximum a minimum positive negative zero

59. Functions and rational exponents: The graph of 2 f 1x2 x3 1 is shown. Use the graph to find: a. domain and range of the function b. zeroes of the function c. interval(s) where f 1x2 0 or f 1x2 6 0 d. interval(s) where f (x) is increasing, decreasing, or constant e. location of any max or min value(s) Exercise 59

Exercise 60 y

y 5

5

(1, 0) (1, 0) 5

(0, 1)

5

(3, 0) 5 x

(3, 0) (0, 1)

5

5 x

5

60. Analyzing a graph: Given h1x2 冟x2 4冟 5, whose graph is shown, use the graph to find: a. domain and range of the function b. zeroes of the function c. interval(s) where h1x2 0 or h1x2 6 0 d. interval(s) where f(x) is increasing, decreasing, or constant e. location of any max or min value(s)

61. Analyzing interest rates: The graph shown approximates the average annual interest rates I on 30-yr fixed mortgages, rounded to the nearest 14 % . Use the graph to estimate the following (write all answers in interval notation). a. domain and range b. interval(s) where I(t) is increasing, decreasing, or constant c. location of any global maximum or d. the one-year period with the greatest rate of increase and minimum values the one-year period with the greatest rate of decrease Source: 2009 Statistical Abstract of the United States, Table 1157 16

Mortgage rate

14 12 10 8 6 4 2 0

t

83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 00 01 02 03 04 05 06 07 08 09

Year (1983 → 83)

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62. Analyzing the surplus S: The following graph approximates the federal surplus S of the United States. Use the graph to estimate the following. Write answers in interval notation and estimate all surplus values to the nearest $10 billion. a. the domain and range b. interval(s) where S(t) is increasing, decreasing, or constant c. the location of any global maximum and minimum values d. the one-year period with the greatest rate of increase, and the one-year period with the greatest rate of decrease S(t): Federal Surplus (in billions)

Source: 2009 Statistical Abstract of the United States, Table 451 400 200 0 200 400 600 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108

t

Year (1980 → 80)

64. Constructing a graph: Draw a continuous function g that has the following characteristics, then state the zeroes and the location of all maximum and minimum values. [Hint: Write them as (c, g(c)).] a. Domain: x 1q, 8 4 Range: y 3 6, q2 b. g102 4.5; g162 0 c. g1x2c for x 16, 32 ´ 16, 82 g1x2T for x 1q, 62 ´ 13, 62 d. g1x2 0 for x 1q, 9 4 ´ 33, 8 4 g1x2 6 0 for x 19, 32

63. Constructing a graph: Draw a continuous function f that has the following characteristics, then state the zeroes and the location of all maximum and minimum values. [Hint: Write them as (c, f (c)).] a. Domain: x 110, q2 Range: y 16, q2 b. f 102 0; f 142 0 c. f 1x2c for x 110, 62 ´ 12, 22 ´ 14, q 2 f 1x2T for x 16, 22 ´ 12, 42 d. f 1x2 0 for x 3 8, 44 ´ 3 0, q 2 f 1x2 6 0 for x 1q, 82 ´ 14, 02 䊳

EXTENDING THE CONCEPT Exercise 65

65. Does the function shown have a maximum value? Does it have a minimum value? Discuss/explain/justify why or why not.

y 5

Distance (meters)

66. The graph drawn here depicts a 400-m race between a mother and her daughter. Analyze the graph to answer questions (a) through (f). a. Who wins the race, the mother or daughter? b. By approximately how many meters? c. By approximately how many seconds? Exercise 66 Mother Daughter d. Who was leading at t 40 seconds? 400 e. During the race, how many seconds was 300 the daughter in the lead? f. During the race, how many seconds was 200 the mother in the lead?

5

5 x

5

100

10

20

30

40

50

Time (seconds)

60

70

80

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67. The graph drawn here depicts the last 75 sec of the competition between Ian Thorpe (Australia) and Massimiliano Rosolino (Italy) in the men’s 400-m freestyle at the 2000 Olympics, where a new Olympic record was set. a. Who was in the lead at 180 sec? 210 sec? b. In the last 50 m, how many times were they tied, and when did the ties occur? c. About how many seconds did Rosolino have the lead? d. Which swimmer won the race? e. By approximately how many seconds? f. Use the graph to approximate the new Olympic record set in the year 2000. Thorpe

Rosolino

Distance (meters)

400

350

300

250

150

155

160

165

170

175

180

185

190

195

200

205

210

215

220

225

Time (seconds)

68. Draw the graph of a general function f (x) that has a local maximum at (a, f (a)) and a local minimum at (b, f (b)) but with f 1a2 6 f 1b2 . 䊳

2

69. Verify that h1x2 ⫽ x3 is an even function, by first rewriting h as h1x2 ⫽ 1x3 2 2. 1


70. (Appendix A.4) Solve the given quadratic equation by factoring: x2 ⫺ 8x ⫺ 20 ⫽ 0.

71. (Appendix A.5) Find the (a) sum and (b) product of the 3 3 rational expressions and . x⫹2 2⫺x

72. (1.4) Write the equation of the line shown, in the form y ⫽ mx ⫹ b.

73. (Appendix A.2) Find the surface area and volume of the cylinder shown 1SA ⫽ 2␲r 2 ⫹ ␲r 2h, V ⫽ ␲r 2h2 .

y

36 cm

5

12 cm ⫺5

5 x

⫺5

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The Toolbox Functions and Transformations


A. Identify basic

B. C.

D.

E.

characteristics of the toolbox functions Apply vertical/horizontal shifts of a basic graph Apply vertical/horizontal reflections of a basic graph Apply vertical stretches and compressions of a basic graph Apply transformations on a general function f (x )

Many applications of mathematics require that we select a function known to fit the context, or build a function model from the information supplied. So far we’ve looked at linear functions. Here we’ll introduce the absolute value, squaring, square root, cubing, and cube root functions. Together these are the six toolbox functions, so called because they give us a variety of “tools” to model the real world (see Section 2.6). In the same way a study of arithmetic depends heavily on the multiplication table, a study of algebra and mathematical modeling depends (in large part) on a solid working knowledge of these functions. More will be said about each function in later sections.

A. The Toolbox Functions While we can accurately graph a line using only two points, most functions require more points to show all of the graph’s important features. However, our work is greatly simplified in that each function belongs to a function family, in which all graphs from a given family share the characteristics of one basic graph, called the parent function. This means the number of points required for graphing will quickly decrease as we start anticipating what the graph of a given function should look like. The parent functions and their identifying characteristics are summarized here.

The Toolbox Functions Identity function

Absolute value function y

y

5

x

f(x) x

3

3

2

2

1

1

0

0

0

1

1

1

1

2

2

2

2

3

3

3

3

x

f (x) x

3

3

2

2

1

1

0

f(x) x 5

5

x

5

5

Square root function y

y

5

f (x) 1x

f(x) x2

x

3

9

2

2

4

1

1

1

0

0

0

0

1

1

1

1

2

1.41

2

4

3

1.73

9

4

2

x

3

120

x

Domain: x (q, q), Range: y [0, q) Symmetry: even Decreasing: x (q, 0); Increasing: x (0, q ) End-behavior: up on the left/up on the right Vertex at (0, 0)

Domain: x (q, q), Range: y (q, q) Symmetry: odd Increasing: x (q, q) End-behavior: down on the left/up on the right

Squaring function

5

5

x

Domain: x (q, q), Range: y [0, q) Symmetry: even Decreasing: x (q, 0); Increasing: x (0, q) End-behavior: up on the left/up on the right Vertex at (0, 0)

5

5

x

Domain: x [0, q), Range: y [0, q) Symmetry: neither even nor odd Increasing: x (0, q) End-behavior: up on the right Initial point at (0, 0)

2–16

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Section 2.2 The Toolbox Functions and Transformations

Cubing function

Cube root function y

y 10

x

f (x) 1 x

27

27

3

2

8

8

2

1

1

1

1

0

0

0

0

1

1

1

1

2

8

8

2

3

27

27

3

x

f (x) x

3

3

5

x

5

3

f(x) 3 x

10

10

x

5

Domain: x (q, q), Range: y (q, q) Symmetry: odd Increasing: x (q, q) End-behavior: down on the left/up on the right Point of inflection at (0, 0)

Domain: x (q, q), Range: y (q, q) Symmetry: odd Increasing: x (q, q) End-behavior: down on the left/up on the right Point of inflection at (0, 0)

In applications of the toolbox functions, the parent graph may be “morphed” and/or shifted from its original position, yet the graph will still retain its basic shape and features. The result is called a transformation of the parent graph. EXAMPLE 1

Solution

䊳

䊳

Identifying the Characteristics of a Transformed Graph The graph of f 1x2 x2 2x 3 is given. Use the graph to identify each of the features or characteristics indicated. a. function family b. domain and range c. vertex d. max or min value(s) e. intervals where f is increasing or decreasing f. end-behavior g. x- and y-intercept(s) a. b. c. d. e. f. g.

y 5

5

5

x

5

The graph is a parabola, from the squaring function family. domain: x 1q, q 2 ; range: y 3 4, q 2 vertex: (1, 4) minimum value y 4 at (1, 4) decreasing: x 1q, 12, increasing: x 11, q 2 end-behavior: up/up y-intercept: (0, 3); x-intercepts: (1, 0) and (3, 0) Now try Exercises 7 through 34

A. You’ve just seen how we can identify basic characteristics of the toolbox functions

䊳

Note that for Example 1(f), we can algebraically verify the x-intercepts by substituting 0 for f(x) and solving the equation by factoring. This gives 0 1x 121x 32 , with solutions x 1 and x 3. It’s also worth noting that while the parabola is no longer symmetric to the y-axis, it is symmetric to the vertical line x 1. This line is called the axis of symmetry for the parabola, and for a vertical parabola, it will always be a vertical line that goes through the vertex.

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B. Vertical and Horizontal Shifts As we study specific transformations of a graph, try to develop a global view as the transformations can be applied to any function. When these are applied to the toolbox functions, we rely on characteristic features of the parent function to assist in completing the transformed graph.

Vertical Translations We’ll first investigate vertical translations or vertical shifts of the toolbox functions, using the absolute value function to illustrate. EXAMPLE 2

䊳

Graphing Vertical Translations

Solution

䊳

A table of values for all three functions is given, with the corresponding graphs shown in the figure.

Construct a table of values for f 1x2 x, g1x2 x 1, and h1x2 x 3 and graph the functions on the same coordinate grid. Then discuss what you observe.

x

f (x) x

g(x) x 1

h(x) x 3

3

3

4

0

2

2

3

1

1

1

2

2

0

0

1

3

1

1

2

2

2

2

3

1

3

3

4

0

(3, 4)5

y g(x) x 1

(3, 3) (3, 0)

1

f(x) x

5

5

x

h(x) x 3 5

Note that outputs of g(x) are one more than the outputs of f (x), and that each point on the graph of f has been shifted upward 1 unit to form the graph of g. Similarly, each point on the graph of f has been shifted downward 3 units to form the graph of h, since h1x2 f 1x2 3. Now try Exercises 35 through 42

䊳

We describe the transformations in Example 2 as a vertical shift or vertical translation of a basic graph. The graph of g is the graph of f shifted up 1 unit, and the graph of h, is the graph of f shifted down 3 units. In general, we have the following: Vertical Translations of a Basic Graph

Given k 7 0 and any function whose graph is determined by y f 1x2 , 1. The graph of y f 1x2 k is the graph of f(x) shifted upward k units. 2. The graph of y f 1x2 k is the graph of f(x) shifted downward k units. Graphing calculators are wonderful tools for exploring graphical transformations. To emphasize that a given graph is being shifted vertically as in 3 Example 2, try entering 1 X as Y1 on the Y= screen, then Y2 Y1 2 and Y3 Y1 3 (Figure 2.17 — recall the Y-variables are accessed using VARS (Y-VARS) ). Using the Y-variables in this way enables us to study identical transformations on a variety of graphs, simply by changing the function in Y1. ENTER

Figure 2.17

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Using a window size of x 35, 54 and y 3 5, 54 for the cube root function, produces the graphs shown in Figure 2.18, which demonstrate the cube root graph has been shifted upward 2 units (Y2), and downward 3 units (Y3). Try this exploration again using Y1 1X.

Figure 2.18 5 Y2

5

5 Y3

5

Horizontal Translations

The graph of a parent function can also be shifted left or right. This happens when we alter the inputs to the basic function, as opposed to adding or subtracting something to the function itself. For Y1 x2 2 note that we first square inputs, then add 2, which results in a vertical shift. For Y2 1x 22 2, we add 2 to x prior to squaring and since the input values are affected, we might anticipate the graph will shift along the x-axis—horizontally. EXAMPLE 3

䊳

Graphing Horizontal Translations

Solution

䊳

Both f and g belong to the quadratic family and their graphs are parabolas. A table of values is shown along with the corresponding graphs.

Construct a table of values for f 1x2 x2 and g1x2 1x 22 2, then graph the functions on the same grid and discuss what you observe.

x

f (x) x2

y

g(x) (x 2)2

3

9

1

2

4

0

1

1

1

0

0

4

1

1

9

2

4

16

3

9

25

9 8

(3, 9)

(1, 9)

7

f(x) x2

6 5

(0, 4)

4

(2, 4)

3

g(x) (x 2)2

2 1

5 4 3 2 1 1

1

2

3

4

5

x

It is apparent the graphs of g and f are identical, but the graph of g has been shifted horizontally 2 units left. Now try Exercises 43 through 46

䊳

We describe the transformation in Example 3 as a horizontal shift or horizontal translation of a basic graph. The graph of g is the graph of f, shifted 2 units to the left. Once again it seems reasonable that since input values were altered, the shift must be horizontal rather than vertical. From this example, we also learn the direction of the shift is opposite the sign: y 1x 22 2 is 2 units to the left of y x2. Although it may seem counterintuitive, the shift opposite the sign can be “seen” by locating the new x-intercept, which in this case is also the vertex. Substituting 0 for y gives 0 1x 22 2 with x 2, as shown in the graph. In general, we have Horizontal Translations of a Basic Graph

Given h 7 0 and any function whose graph is determined by y f 1x2 , 1. The graph of y f 1x h2 is the graph of f(x) shifted to the left h units. 2. The graph of y f 1x h2 is the graph of f(x) shifted to the right h units.

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Figure 2.19

To explore horizontal translations on a graphing calculator, we input a basic function in Y1 and indicate how we want the inputs altered in Y2 and Y3. Here we’ll enter X3 as Y1 on the Y= screen, then Y2 Y1 1X 52 and Y3 Y1 1X 72 (Figure 2.19). Note how this 10 duplicates the definition and notation for horizontal shifts in the orange box. Based on what we saw in Example 3, we expect the graph of y x3 will first be shifted 5 units left (Y2), then 7 units right (Y3). This in confirmed in Figure 2.20. Try this exploration again using Y1 abs1X2.

Figure 2.20 Y3

10

10

10

EXAMPLE 4

䊳

Graphing Horizontal Translations Sketch the graphs of g1x2 x 2 and h1x2 1x 3 using a horizontal shift of the parent function and a few characteristic points (not a table of values).

Solution

䊳

The graph of g1x2 x 2 (Figure 2.21) is the absolute value function shifted 2 units to the right (shift the vertex and two other points from y x 2 . The graph of h1x2 1x 3 (Figure 2.22) is a square root function, shifted 3 units to the left (shift the initial point and one or two points from y 1x). Figure 2.21 5

Figure 2.22

y g(x) x 2

y h(x) x 3

(1, 3)

5

(6, 3)

(5, 3) (1, 2) 5

Vertex

(2, 0)

5

x 4

B. You’ve just seen how we can perform vertical/ horizontal shifts of a basic graph

(3, 0)

5

x


䊳

C. Vertical and Horizontal Reflections The next transformation we investigate is called a vertical reflection, in which we compare the function Y1 f 1x2 with the negative of the function: Y2 f 1x2 .

Vertical Reflections EXAMPLE 5

䊳

Graphing Vertical Reflections Construct a table of values for Y1 x2 and Y2 x2, then graph the functions on the same grid and discuss what you observe.

Solution

䊳

A table of values is given for both functions, along with the corresponding graphs.

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Section 2.2 The Toolbox Functions and Transformations y 5

x

Y1 x2

Y2 x2

2

4

4

1

1

1

0

0

0

1

1

1

2

4

4

Y1 x2

(2, 4)

5 4 3 2 1

Y2 x2

1

2

3

4

5

x

(2, 4) 5

As you might have anticipated, the outputs for f and g differ only in sign. Each output is a reflection of the other, being an equal distance from the x-axis but on opposite sides. Now try Exercises 51 and 52

䊳

The vertical reflection in Example 5 is called a reflection across the x-axis. In general, Vertical Reflections of a Basic Graph

For any function y f 1x2 , the graph of y f 1x2 is the graph of f(x) reflected across the x-axis.

To view vertical reflections on a graphing calculator, we simply define Y2 Y1, as seen here 3 using 1 X as Y1 (Figure 2.23). As in Section 1.5, we can have the calculator graph Y2 using a bolder line, to easily distinguish between the original graph and its reflection (Figure 2.24). To aid in the viewing, we have set a window size of x 35, 5 4 and y 3 3, 34 . Try this exploration again using Y1 X2 4.

Figure 2.23

Figure 2.24 3

5

5

3

Horizontal Reflections It’s also possible for a graph to be reflected horizontally across the y-axis. Just as we noted that f (x) versus f 1x2 resulted in a vertical reflection, f(x) versus f 1x2 results in a horizontal reflection. EXAMPLE 6

䊳

Graphing a Horizontal Reflection

Solution

䊳

A table of values is given here, along with the corresponding graphs.

Construct a table of values for f 1x2 1x and g1x2 1x, then graph the functions on the same coordinate grid and discuss what you observe.

x

f(x) 1x

g(x) 1x

4

not real

2

2

not real

12 1.41

1

not real

1

0

0

0

1

1

not real

2

12 1.41

not real

4

2

not real

y (4, 2)

(4, 2) 2

g(x) x

f(x) x

1

5 4 3 2 1 1 2

1

2

3

4

5

x

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The graph of g is the same as the graph of f, but it has been reflected across the y-axis. A study of the domain shows why — f represents a real number only for nonnegative inputs, so its graph occurs to the right of the y-axis, while g represents a real number for nonpositive inputs, so its graph occurs to the left. Now try Exercises 53 and 54

䊳

The transformation in Example 6 is called a horizontal reflection of a basic graph. In general, Horizontal Reflections of a Basic Graph

For any function y f 1x2 , the graph of y f 1x2 is the graph of f(x) reflected across the y-axis.

C. You’ve just seen how we can apply vertical/horizontal reflections of a basic graph

D. Vertically Stretching/Compressing a Basic Graph As the words “stretching” and “compressing” imply, the graph of a basic function can also become elongated or flattened after certain transformations are applied. However, even these transformations preserve the key characteristics of the graph. EXAMPLE 7

䊳

Stretching and Compressing a Basic Graph

Solution

䊳

A table of values is given for all three functions, along with the corresponding graphs.

Construct a table of values for f 1x2 x2, g1x2 3x2, and h1x2 13x2, then graph the functions on the same grid and discuss what you observe.

x

f (x) x2

g(x) 3x2

h(x) 13 x2

3

9

27

3

2

4

12

4 3

1

1

3

1 3

0

0

0

0

1

1

3

1 3

2

4

12

4 3

3

9

27

3

y g(x) 3x2

(2, 12) (2, 4)

f(x) x2

10

h(x) ax2 (2, d) 5 4 3 2 1

1

2

3

4

5

x

4

The outputs of g are triple those of f, making these outputs farther from the x-axis and stretching g upward (making the graph more narrow). The outputs of h are one-third those of f, and the graph of h is compressed downward, with its outputs closer to the x-axis (making the graph wider). Now try Exercises 55 through 62

䊳

WORTHY OF NOTE In a study of trigonometry, you’ll find that a basic graph can also be stretched or compressed horizontally, a phenomenon known as frequency variations.

The transformations in Example 7 are called vertical stretches or compressions of a basic graph. Notice that while the outputs are increased or decreased by a constant factor (making the graph appear more narrow or more wide), the domain of the function remains unchanged. In general,

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Stretches and Compressions of a Basic Graph

For any function y f 1x2 , the graph of y af 1x2 is 1. the graph of f(x) stretched vertically if a 7 1, 2. the graph of f(x) compressed vertically if 0 6 a 6 1. Figure 2.25

Figure 2.26

To use a graphing calculator in a study of stretches and compressions, we simply define Y2 and Y3 as constant multiples of Y1 (Figure 2.25). As seen in Example 7, if a 7 1 the graph will be stretched vertically, if 0 6 a 6 1, the graph will be vertically compressed. This is further illus- 0 trated here using Y1 1X, with Y2 2Y1 and Y3 0.5Y1. Since the domain of y 1x is restricted to nonnegative values, a window size of x 30, 10 4 and y 3 1, 7 4 was used (Figure 2.26). Try this exploration again using Y1 abs1X2 4.

D. You’ve just seen how we can apply vertical stretches and compressions of a basic graph

7

10

1

E. Transformations of a General Function If more than one transformation is applied to a basic graph, it’s helpful to use the following sequence for graphing the new function. General Transformations of a Basic Graph

Given a function y f 1x2 , the graph of y af 1x h2 k can be obtained by applying the following sequence of transformations: 1. horizontal shifts 2. reflections 3. stretches/compressions 4. vertical shifts We generally use a few characteristic points to track the transformations involved, then draw the transformed graph through the new location of these points. EXAMPLE 8

䊳

Graphing Functions Using Transformations Use transformations of a parent function to sketch the graphs of 3 a. g1x2 1x 22 2 3 b. h1x2 2 1 x21

Solution

a. The graph of g is a parabola, shifted left 2 units, reflected across the x-axis, and shifted up 3 units. This sequence of transformations is shown in Figures 2.27 through 2.29. Note that since the graph has been shifted 2 units left and 3 units up, the vertex of the parabola has likewise shifted from (0, 0) to 12, 32 .

䊳

Figure 2.27 y (x

Figure 2.28

y

2)2

(4, 4)

5

y x2

5

Figure 2.29

y y (x 2)2

5

y g(x) ⫽ ⫺(x ⫹ 2)2 ⫹ 3

(⫺2, 3)

(0, 4)

(2, 0) 5

(2, 0) Vertex

5

Shifted left 2 units

5

x

5

5

x

⫺5

(⫺4, ⫺1) (4, 4)

5

(0, 4)

Reflected across the x-axis

(0, ⫺1)

⫺5

Shifted up 3 units

5

x

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b. The graph of h is a cube root function, shifted right 2, stretched by a factor of 2, then shifted down 1. This sequence is shown in Figures 2.30 through 2.32 and illustrate how the inflection point has shifted from (0, 0) to 12, 12 . Figure 2.30 y 5

Figure 2.31

3

y x 2

5

Figure 2.32

3 y y 2x 2

5

3 y h(x) 2x 21

(3, 2) (3, 1) (2, 0) 6 x Inflection (1, 1)

4

(2, 0) 6

x

4

(2, 1)

(1, 2)

6

x

(1, 3) 5

5

5

Shifted right 2

(3, 1)

4

Shifted down 1

Stretched by a factor of 2


䊳

It’s important to note that the transformations can actually be applied to any function, even those that are new and unfamiliar. Consider the following pattern: Parent Function

Transformation of Parent Function y 21x 32 2 1

quadratic: y x2

absolute value: y 0 x 0

y 2 0 x 3 0 1

3 y 21 x31

cube root: y 1x 3

general: y f 1x2

y 2f 1x 32 1

In each case, the transformation involves a horizontal shift 3 units right, a vertical reflection, a vertical stretch, and a vertical shift up 1. Since the shifts are the same regardless of the initial function, we can generalize the results to any function f(x).

WORTHY OF NOTE Since the shape of the initial graph does not change when translations or reflections are applied, these are called rigid transformations. Stretches and compressions of a basic graph are called nonrigid transformations, as the graph is distended in some way.

vertical reflections, vertical stretches and compressions

S

y af 1x h2 k S

y f 1x2

Transformed Function S

General Function

horizontal shift h units, opposite direction of sign

vertical shift k units, same direction as sign

Also bear in mind that the graph will be reflected across the y-axis (horizontally) if x is replaced with x. This process is illustrated in Example 9 for selected transformations. Remember — if the graph of a function is shifted, the individual points on the graph are likewise shifted.

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EXAMPLE 9

䊳

Graphing Transformations of a General Function

Solution

䊳

For g, the graph of f is (1) shifted horizontally 1 unit left (Figure 2.34), (2) reflected across the x-axis (Figure 2.35), and (3) shifted vertically 2 units down (Figure 2.36). The final result is that in Figure 2.36.

Given the graph of f(x) shown in Figure 2.33, graph g1x2 f 1x 12 2.

Figure 2.34

Figure 2.33 y

y

5

5

(2, 3)

(3, 3)

f (x)

(0, 0) 5

5

x

5

(1, 0)

(2, 3)

5

x

5

x

(1, 3)

5

5

Figure 2.36

Figure 2.35 y

y

5

5

(1, 3) (1, 1) g (x)

(1, 0) 5

5

x

5

(3, 2) (1, 2)

(5, 2) (3, 3) 5

(3, 5)

5


䊳

As noted in Example 9, these shifts and transformation are often combined— particularly when the toolbox functions are used as real-world models (Section 2.6). On a graphing calculator we again define Y1 as needed, then define Y2 as any desired combination of shifts, stretches, and/or reflections. For Y1 X2, we’ll define Y2 as 2 Y1 1X 52 3 (Figure 2.37), and expect that the graph of Y2 will be that of Y1 shifted left 5 units, reflected across the x-axis, stretched vertically, and shifted up three units. This shows the new vertex should be at 15, 32 , which is confirmed in Figure 2.38 along with the other transformations. Figure 2.38 Figure 2.37

10

10

10

10

Try this exploration again using Y1 abs1X2 .

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Using the general equation y af 1x h2 k, we can identify the vertex, initial point, or inflection point of any toolbox function and sketch its graph. Given the graph of a toolbox function, we can likewise identify these points and reconstruct its equation. We first identify the function family and the location (h, k) of any characteristic point. By selecting one other point (x, y) on the graph, we then use the general equation as a formula (substituting h, k, and the x- and y-values of the second point) to solve for a and complete the equation. EXAMPLE 10

䊳

Writing the Equation of a Function Given Its Graph Find the equation of the function f(x) shown in the figure.

Solution

䊳

The function f belongs to the absolute value family. The vertex (h, k) is at (1, 2). For an additional point, choose the x-intercept (3, 0) and work as follows: y ax h k 0 a 132 1 2

E. You’ve just seen how we can apply transformations on a general function f(x)

0 4a 2 2 4a 1 a 2

general equation (function is shifted right and up) substitute 1 for h and 2 for k, substitute 3 for x and 0 for y simplify

y 5

f(x) 5

5

x

subtract 2 5

solve for a

The equation for f is y 12 0 x 1 0 2. Now try Exercises 97 through 102

䊳

2.2 EXERCISES 䊳



1. After a vertical , points on the graph are farther from the x-axis. After a vertical , points on the graph are closer to the x-axis. 3. The vertex of h1x2 31x 52 2 9 is at and the graph opens . 5. Given the graph of a general function f (x), discuss/ explain how the graph of F1x2 2f 1x 12 3 can be obtained. If (0, 5), (6, 7), and 19, 42 are on the graph of f, where do they end up on the graph of F?

2. Transformations that change only the location of a graph and not its shape or form, include and .

4. The inflection point of f 1x2 21x 42 3 11 is at and the end-behavior is , .

6. Discuss/Explain why the shift of f 1x2 x2 3 is a vertical shift of 3 units in the positive direction, while the shift of g1x2 1x 32 2 is a horizontal shift 3 units in the negative direction. Include several examples along with a table of values for each.

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131



By carefully inspecting each graph given, (a) identify the function family; (b) describe or identify the end-behavior, vertex, intervals where the function is increasing or decreasing, maximum or minimum value(s) and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

7. f 1x2 x2 4x

15. r 1x2 314 x 3 16. f 1x2 21x 1 4 y

y

5

5

⫺5

5 x

8. g1x2 x2 2x

y

⫺5

⫺5

⫺5

y

5 x

f(x)

r(x)

5

5

17. g1x2 2 14 x

18. h1x2 21x 1 4

y ⫺5

5 x

⫺5

y 5

5

5 x

g(x) h(x)

⫺5

⫺5

9. p1x2 x2 2x 3

⫺5

5 x

⫺5

5 x

10. q1x2 x2 2x 8

y

⫺5

⫺5

y 10

5

⫺5

5 x

⫺10

10 x

⫺10

⫺5

11. f 1x2 x2 4x 5

12. g1x2 x2 6x 5

y

For each graph given, (a) identify the function family; (b) describe or identify the end-behavior, vertex, intervals where the function is increasing or decreasing, maximum or minimum value(s) and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

19. p1x2 2x 1 4

y

5

10

10

20. q1x2 3x 2 3

y

y

5

q(x)

⫺10

10 x

⫺10

10 x

⫺5

For each graph given, (a) identify the function family; (b) describe or identify the end-behavior, initial point, intervals where the function is increasing or decreasing, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

13. p1x2 2 1x 4 2

5 x

⫺5

⫺5

⫺10

⫺10

p(x)

5 x

⫺5

21. r 1x2 2x 1 6 22. f 1x2 3x 2 6 y

y 4

6

r(x) ⫺5 ⫺5

5 x

5 x

f(x)

14. q1x2 2 1x 4 2 ⫺6

⫺4

y

y

5

5

23. g1x2 3x 6

p(x)

24. h1x2 2x 1

y

y 6

6 ⫺5

5 x

⫺5

5 x

q(x)

g(x) ⫺5

h(x)

⫺5 ⫺5

5 x

⫺4

⫺5

5 x

⫺4

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For each graph given, (a) identify the function family; (b) describe or identify the end-behavior, inflection point, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values. Be sure to note the scaling of each axis.

25. f 1x2 1x 12 3

26. g1x2 1x 12 3

y 5

f(x)

⫺5

5 x

5 x

27. h1x2 x3 1

3 28. p1x2 2x 1

y

⫺5

5 x

5 x

⫺5

29. q1x2 2x 1 1 3

30. r 1x2 2x 1 1 3

y

y

⫺5

⫺5

5 x

q(x)

⫺5

5 x

r(x)

y 5

32.

f(x)

42. t1x2 0 x 0 3

g1x2 1x 4

45. Y1 0 x 0 , Y2 0 x 4 0

H1x2 1x 42 3

Sketch each graph by hand using transformations of a parent function (without a table of values).

47. p1x2 1x 32 2

48. q1x2 1x 1

51. g1x2 0 x 0

52. j1x2 1x

3 53. f 1x2 2 x

3 50. f 1x2 1 x2

54. g1x2 1x2 3

Use a graphing calculator to graph the functions given in the same window. Comment on what you observe.

⫺5

For Exercises 31–34, identify and state the characteristic features of each graph, including (as applicable) the function family, end-behavior, vertex, axis of symmetry, point of inflection, initial point, maximum and minimum value(s), x- and y-intercepts, and the domain and range.

31.

40. g1x2 1x 4

q1x2 1x 52 2

49. h1x2 x 3

5

5

q1x2 x2 7, r 1x2 x2 3


46. h1x2 x3, ⫺5

3 3 g1x2 2 x 3, h1x2 2 x4

39. f 1x2 x3 2

44. f 1x2 1x,

p(x) h(x) ⫺5

h1x2 1x 3

37. p1x2 x, q1x2 x 5, r 1x2 x 2

43. p1x2 x2,

y 5

5

3 36. f 1x2 2 x,

41. h1x2 x2 3

⫺5

⫺5

g1x2 1x 2,

Sketch each graph by hand using transformations of a parent function (without a table of values).

g(x)

⫺5

35. f 1x2 1x,

38. p1x2 x2,

y

5


y 5

55. p1x2 x2,

q1x2 3x2, r 1x2 15x2

56. f 1x2 1x, g1x2 41x,

h1x2 14 1x

57. Y1 0 x 0 , Y2 3 0 x 0 , Y3 13 0 x 0 58. u1x2 x3,

v1x2 8x3,

w1x2 15x3

g(x)

Sketch each graph by hand using transformations of a parent function (without a table of values). ⫺5

⫺5

5 x

5 x

3 59. f 1x2 4 2 x

61. p1x2 13x3 y 5

⫺5

34.

f(x)

5 x

⫺5

62. q1x2 34 1x

⫺5

⫺5

33.

60. g1x2 2 0x 0

y 5

⫺5

Use the characteristics of each function family to match a given function to its corresponding graph. The graphs are not scaled — make your selection based on a careful comparison.

g(x)

5 x

⫺5

63. f 1x2 12x3

64. f 1x2 2 3 x 2

3 65. f 1x2 1x 32 2 2 66. f 1x2 1 x11

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67. f 1x2 x 4 1

68. f 1x2 1x 6

71. f 1x2 1x 42 2 3

72. f 1x2 x 2 5

Graph each function using shifts of a parent function and a few characteristic points. Clearly state and indicate the transformations used and identify the location of all vertices, initial points, and/or inflection points.

69. f 1x2 1x 6 1 70. f 1x2 x 1 73. f 1x2 1x 3 1 y a.

74. f 1x2 1x 32 2 5 y b.

x

x

75. f 1x2 1x 2 1

76. g1x2 1x 3 2

79. p1x2 1x 32 3 1

80. q1x2 1x 22 3 1

83. f 1x2 x 3 2

84. g1x2 x 4 2

77. h1x2 1x 32 2 2 78. H1x2 1x 22 2 5 3 81. s1x2 1 x12

3 82. t1x2 1 x31

85. h1x2 21x 12 2 3 86. H1x2 12x 2 3 c.

d.

y

3 87. p1x2 13 1x 22 3 1 88. q1x2 41 x12

y

89. u1x2 2 1x 1 3 90. v1x2 3 1x 2 1 x

x

91. h1x2 15 1x 32 2 1

92. H1x2 2x 3 4

Apply the transformations indicated for the graph of the general functions given.

e.

f.

y

93.

y

y 5

94.

f(x)

y 5

g(x)

(⫺1, 4) (⫺4, 4)

(3, 2)

(⫺1, 2)

x

x

⫺5

⫺5

5 x

5 x

(⫺4, ⫺2) ⫺5

⫺5

g.

h.

y

y

a. f 1x 22 b. f 1x2 3 c. 12 f 1x 12 d. f 1x2 1

x

x

95. i.

y

j.

y 5

(2, ⫺2)

a. b. c. d. 96.

h(x)

g1x2 2 g1x2 3 2g1x 12 1 2 g1x 12 2 y 5

y

(⫺1, 3)

(2, 0)

(⫺1, 0) ⫺5

x

5 x

⫺5

y

l.

x

⫺5

y

x

a. b. c. d.

(1, ⫺3)

(2, ⫺4)

h1x2 3 h1x 22 h1x 22 1 1 4 h1x2 5

5 x

(⫺2, 0)

x (⫺4, ⫺4)

k.

H(x)

⫺5

a. b. c. d.

H1x 32 H1x2 1 2H1x 32 1 3 H1x 22 1

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Use the graph given and the points indicated to determine the equation of the function shown using the general form y af(x h) k.

97.

98.

y 5

99.

y (⫺5, 6)

y

5

(6, 4.5)

5

p(x) g(x)

(2, 0) ⫺5

5 x

f(x)

⫺5

5 x

⫺3(⫺3, 0)

(0, ⫺4)

⫺5

100.

⫺4

⫺3

(0, ⫺4)

101.

y (⫺4, 5) 5

x

5

102.

y 5

y (3, 7)

7

(1, 4) f(x)

h(x)

r(x) ⫺4

5

x

⫺8

(5, ⫺1)

⫺5

䊳

⫺3 ⫺5

7 x ⫺3

(0, ⫺2)


103. Volume of a sphere: V(r) 43␲r3 The volume of a sphere is given by the function shown, where V(r) is the volume in cubic units and r is the radius. Note this function belongs to the cubic family of functions. (a) Approximate the value of 43␲ to one decimal place, then graph the function on the interval [0, 3]. (b) From your graph, estimate the volume of a sphere with radius 2.5 in., then compute the actual volume. Are the results close? (c) For V 43 ␲r3, solve for r in terms of V.

䊳

2 x

(⫺4, 0)

104. Fluid motion: V(h) 4 1h 20 Suppose the velocity of a fluid flowing from an open tank (no top) through an opening in its side is given by the function shown, where V(h) is the velocity of the fluid (in feet per second) at water height h (in feet). Note this function belongs to the square root family of functions. An open tank is 25 ft deep and filled to the brim with fluid. (a) Use a table of values to graph the 25 ft function on the interval [0, 25]. (b) From your graph, estimate the velocity of the fluid when the water level is 7 ft, then find the actual velocity. Are the answers close? (c) If the fluid velocity is 5 ft/sec, how high is the water in the tank?

APPLICATIONS

105. Gravity, distance, time: After being released, the time it takes an object to fall x ft is given by the function T1x2 14 1x, where T(x) is in seconds. (a) Describe the transformation applied to obtain the graph of T from the graph of y 1x, then sketch the graph of T for x 30, 100 4 . (b) How long would it take an object to hit the ground if it were dropped from a height of 81 ft? 106. Stopping distance: In certain weather conditions, accident investigators will use the function v1x2 4.91x to estimate the speed of a car (in miles per hour) that has been involved in an accident, based on the length of the skid marks x (in feet). (a) Describe the transformation applied to

obtain the graph of v from the graph of y 1x, then sketch the graph of v for x 3 0, 4004. (b) If the skid marks were 225 ft long, how fast was the car traveling? Is this point on your graph? 107. Wind power: The power P generated by a certain 8 3 v wind turbine is given by the function P1v2 125 where P(v) is the power in watts at wind velocity v (in miles per hour). (a) Describe the transformation applied to obtain the graph of P from the graph of y v3, then sketch the graph of P for v 30, 25 4 (scale the axes appropriately). (b) How much power is being generated when the wind is blowing at 15 mph?

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108. Wind power: If the power P (in watts) being generated by a wind turbine is known, the velocity of the wind can be determined using the function 3 v1P2 52 2 P. (a) Describe the transformation applied to obtain the graph of v from the graph of 3 y 2 P, then sketch the graph of v for P 3 0, 512 4 (scale the axes appropriately). (b) How fast is the wind blowing if 343W of power is being generated? Is this point on your graph? 109. Distance rolled due to gravity: The distance a ball rolls down an inclined plane is given by the function d1t2 2t2, where d(t) represents the distance in feet after t sec. (a) Describe the transformation applied to obtain the graph of d from the graph

䊳

of y t2, then sketch the graph of d for t 30, 3 4. (b) How far has the ball rolled after 2.5 sec? 110. Acceleration due to gravity: The velocity of a steel ball bearing as it rolls down an inclined plane is given by the function v1t2 4t, where v(t) represents the velocity in feet per second after t sec. (a) Describe the transformation applied to obtain the graph of v from the graph of y t, then sketch the graph of v for t 30, 34 . (b) What is the velocity of the ball bearing after 2.5 sec? Is this point on your graph?


111. Carefully graph the functions f 1x2 x and g1x2 2 1x on the same coordinate grid. From the graph, in what interval is the graph of g(x) above the graph of f (x)? Pick a number (call it h) from this interval and substitute it in both functions. Is g1h2 7 f 1h2? In what interval is the graph of g(x) below the graph of f (x)? Pick a number from this interval (call it k) and substitute it in both functions. Is g1k2 6 f 1k2?

䊳

135

112. Sketch the graph of f 1x2 2x 3 8 using transformations of the parent function, then determine the area of the region in quadrant I that is beneath the graph and bounded by the vertical lines x 0 and x 6.

113. Sketch the graph of f 1x2 x2 4, then sketch the graph of F1x2 x2 4 using your intuition and the meaning of absolute value (not a table of values). What happens to the graph?


114. (1.1) Find the distance between the points 113, 92 and 17, 122, and the slope of the line containing these points. 2x2 3x 5x 2

115. (Appendix A.2) Find the perimeter of the figure shown.

5x 2x2 3x 5

1 1 7 2 116. (1.5) Solve for x: x x . 3 4 2 12 117. (2.1) Without graphing, state intervals where f 1x2c and f 1x2T for f 1x2 1x 42 2 3.

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Absolute Value Functions, Equations, and Inequalities While the equations x 1 5 and 冟x 1冟 5 are similar in many respects, note the first has only the solution x 4, while either x 4 or x 6 will satisfy the second. The fact there are two solutions shouldn’t surprise us, as it’s a natural result of how absolute value is defined.


A. Solve absolute value equations

A. Solving Absolute Value Equations

B. Solve “less than” absolute value inequalities C. Solve “greater than” absolute value inequalities D. Solve absolute value equations and inequalities graphically E. Solve applications involving absolute value

The absolute value of a number x can be thought of as its distance from zero on the number line, regardless of direction. This means 冟x冟 4 will have two solutions, since there are two numbers that are four units from zero: x 4 and x 4 (see Figure 2.39). Exactly 4 units from zero

Figure 2.39

⫺5 ⫺4

Exactly 4 units from zero ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

This basic idea can be extended to include situations where the quantity within absolute value bars is an algebraic expression, and suggests the following property. Property of Absolute Value Equations If X represents an algebraic expression and k is a positive real number,

WORTHY OF NOTE Note if k 6 0, the equation 冟X冟 k has no solutions since the absolute value of any quantity is always positive or zero. On a related note, we can verify that if k 0, the equation 冟X冟 0 has only the solution X 0.

then 冟X冟 k implies X k or X k As the statement of this property suggests, it can only be applied after the absolute value expression has been isolated on one side.

EXAMPLE 1

䊳

Solving an Absolute Value Equation Solve: 5冟x 7冟 2 13.

Solution

䊳

Begin by isolating the absolute value expression. 5冟x 7冟 2 13 original equation 5冟x 7冟 15 subtract 2 冟x 7冟 3 divide by 5 (simplified form) Now consider x 7 as the variable expression “X” in the property of absolute value equations, giving or x 7 3 x 7 3 apply the property of absolute value equations x4 or x 10 add 7 Substituting into the original equation verifies the solution set is {4, 10}. Now try Exercises 7 through 18

CAUTION

136

䊳

䊳

For equations like those in Example 1, be careful not to treat the absolute value bars as simple grouping symbols. The equation 51x 72 2 13 has only the solution x 10, and “misses” the second solution since it yields x 7 3 in simplified form. The equation 5冟x 7冟 2 13 simplifies to 冟x 7冟 3 and there are actually two solutions. Also note that 5冟x 7冟冟5x 35冟!

2–32

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Section 2.3 Absolute Value Functions, Equations, and Inequalities

137

If an equation has more than one solution as in Example 1, they cannot be simultaneously stored using the, X,T,␪,n key to perform a calculator check (in function or “Func” mode, this is the variable X). While there are other ways to “get around” this (using Y 1 on the home screen, using a TABLE in ASK mode, enclosing the solutions in braces as in {4, 10}, etc.), we can also store solutions using the ALPHA keys. To illustrate, we’ll place the solution x 4 in storage location A, using 4 STO ALPHA MATH (A). Using this “ STO ALPHA ” sequence we’ll next place the solution x 10 in storage location B (Figure 2.40). We can then check both solutions in turn. Note that after we check the first solution, we can recall the expression using 2nd and simply change the A to B (Figure 2.41). ENTER

Figure 2.40

Figure 2.41

Absolute value equations come in many different forms. Always begin by isolating the absolute value expression, then apply the property of absolute value equations to solve.

EXAMPLE 2

䊳

Solving an Absolute Value Equation Solve:

Solution

䊳

2 ` 5 x ` 9 8. 3 2 `5 x ` 9 8 3 2 ` 5 x ` 17 3 2 5 x 17 3 2 x 22 3 x 33

Check WORTHY OF NOTE As illustrated in both Examples 1 and 2, the property we use to solve absolute value equations can only be applied after the absolute value term has been isolated. As you will see, the same is true for the properties used to solve absolute value inequalities.

䊳

2 For x 33: ` 5 1332 ` 3 | 5 21112 | 05 22 0 0 17 0 17

original equation

add 9

or or or

98 98 98 98 98 8 8✓

2 5 x 17 3 2 x 12 3 x 18

apply the property of absolute value equations subtract 5 multiply by 32

2 1182 ` 9 8 3 |5 2162 | 9 8 0 5 12 0 9 8 0 17 0 9 8 17 9 8 8 8✓

For x 18: ` 5

Both solutions check. The solution set is 518, 336.


䊳

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For some equations, it’s helpful to apply the multiplicative property of absolute value: Multiplicative Property of Absolute Value If A and B represent algebraic expressions, then 冟AB冟冟A冟冟B冟. Note that if A 1 the property says 冟1 # B冟冟1冟冟B冟冟B冟. More generally the property is applied where A is any constant.

EXAMPLE 3

䊳

Solution

䊳

Solving Equations Using the Multiplicative Property of Absolute Value Solve: 冟2x冟 5 13. 冟2x冟 5 13 冟2x冟 8 冟2冟冟x冟 8 2冟x冟 8 冟x冟 4 x 4 or x 4

original equation subtract 5 apply multiplicative property of absolute value simplify divide by 2 apply property of absolute value equations

Both solutions check. The solution set is 54, 46. Now try Exercises 23 and 24

䊳

In some instances, we have one absolute value quantity equal to another, as in 冟A冟冟B冟. From this equation, four possible solutions are immediately apparent: (1) A B

(2) A B

(3) A B

(4) A B

However, basic properties of equality show that equations (1) and (4) are equivalent, as are equations (2) and (3), meaning all solutions can be found using only equations (1) and (2).

EXAMPLE 4

䊳

Solving Absolute Value Equations with Two Absolute Value Expressions Solve the equation 冟2x 7冟冟x 1冟.

Solution

䊳

This equation has the form 冟A冟冟B冟, where A 2x 7 and B x 1. From our previous discussion, all solutions can be found using A B and A B. AB 2x 7 x 1 2x x 8 x 8

solution template substitute subtract 7 subtract x

A B 2x 7 1x 12 2x 7 x 1 3x 6 x 2

solution template substitute distribute add x, subtract 7 divide by 3

The solutions are x 8 and x 2. Verify the solutions by substituting them into the original equation. A. You’ve just seen how we can solve absolute value equations


䊳

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139

B. Solving “Less Than” Absolute Value Inequalities Absolute value inequalities can be solved using the basic concept underlying the property of absolute value equalities. Whereas the equation 冟x冟 4 asks for all numbers x whose distance from zero is equal to 4, the inequality 冟x冟 6 4 asks for all numbers x whose distance from zero is less than 4. Distance from zero is less than 4

Figure 2.42

)

⫺5 ⫺4

⫺3 ⫺2 ⫺1

) 0

1

2

3

4

5

As Figure 2.42 illustrates, the solutions are x 7 4 and x 6 4, which can be written as the joint inequality 4 6 x 6 4. This idea can likewise be extended to include the absolute value of an algebraic expression X as follows. Property I: Absolute Value Inequalities (Less Than) If X represents an algebraic expression and k is a positive real number, then 冟X冟 6 k implies k 6 X 6 k Property I can also be applied when the “” symbol is used. Also notice that if k 6 0, the solution is the empty set since the absolute value of any quantity is always positive or zero.

EXAMPLE 5

Solution

䊳

䊳

Solving “Less Than” Absolute Value Inequalities Solve the inequalities: 冟3x 2冟 a. 1 4 冟3x 2冟 a. 1 4 冟3x 2冟 4 4 3x 2 4 6 3x 2 2 2 x 3

b. 冟2x 7冟 6 5 original inequality multiply by 4 apply Property I subtract 2 from all three parts divide all three parts by 3

The solution interval is 3 2, 23 4.

b. 冟2x 7冟 6 5

original inequality

Since the absolute value of any quantity is always positive or zero, the solution for this inequality is the empty set: { }. Now try Exercises 27 through 38

䊳

As with the inequalities from Section 1.5, solutions to absolute value inequalities can be checked using a test value. For Example 5(a), substituting x 0 from the solution interval yields: 1 1✓ 2

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2–36


In addition to checking absolute value inequalities using a test value, the TABLE feature of a graphing calculator can be used, alone or in conjunction with a relational test. Relational tests have the calculator return a “1” if a given statement is true, and a “0” otherwise. To illustrate, consider the inequality 2冟x 3冟 1 5. Enter the expression on the left as Y1, recalling the “abs(” notation is accessed in the MATH menu: MATH (NUM) “1:abs(” (note this option gives only the left parenthesis, you must supply the right). We can then simply inspect the Y1 column of the TABLE to find outputs that are less than or equal to 5. To use a relational test, we enter Y1 5 as Y2 (Figure 2.43), with the “less than or equal to” symbol accessed using 2nd MATH 6:ⱕ. Now the calculator will automatically check the truth of the statement for any value of x (but note we are only checking integer values), and display the result in the Y2 column of the TABLE (Figure 2.44). After scrolling through the table, both approaches show that 2冟x 3冟 1 5 for x 僆 [1, 5]. ENTER

Figure 2.43

Figure 2.44

B. You’ve just seen how we can solve “less than” absolute value inequalities

C. Solving “Greater Than” Absolute Value Inequalities For “greater than” inequalities, consider 冟x冟 7 4. Now we’re asked to find all numbers x whose distance from zero is greater than 4. As Figure 2.45 shows, solutions are found in the interval to the left of 4, or to the right of 4. The fact the intervals are disjoint (disconnected) is reflected in this graph, in the inequalities x 6 4 or x 7 4, as well as the interval notation x 僆 1q, 42 ´ 14, q 2. Distance from zero is greater than 4

Figure 2.45

)

⫺7 ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

Distance from zero is greater than 4

) 0

1

2

3

4

5

6

7

As before, we can extend this idea to include algebraic expressions, as follows: Property II: Absolute Value Inequalities (Greater Than) If X represents an algebraic expression and k is a positive real number, then 冟X冟 7 k implies X 6 k or X 7 k

EXAMPLE 6

䊳

Solving “Greater Than” Absolute Value Inequalities Solve the inequalities: 1 x a. ` 3 ` 6 2 3 2

b. 冟5x 2冟

3 2

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Solution

䊳

141

a. Note the exercise is given as a less than inequality, but as we multiply both sides by 3, we must reverse the inequality symbol. x 1 ` 3 ` 6 2 3 2 x `3 ` 7 6 2 x x or 3 7 6 3 6 6 2 2 x x 6 9 or 7 3 2 2 or x 7 6 x 6 18

original inequality multiply by ⴚ3, reverse the symbol

apply Property II

subtract 3 multiply by 2

Property II yields the disjoint intervals x 僆 1q, 182 ´ 16, q2 as the solution. )

⫺30 ⫺24 ⫺18 ⫺12 ⫺6

) 0

6

12

18

24

30

3 original inequality 2 Since the absolute value of any quantity is always positive or zero, the solution for this inequality is all real numbers: x 僆⺢.

b. 冟5x 2冟


䊳

A calculator check is shown for part (a) in Figures 2.46 through 2.48. Figure 2.46

Figure 2.47

Figure 2.48

This helps to verify the solution interval is x 僆 1q, 182 ´ 16, q 2 . Due to the nature of absolute value functions, there are times when an absolute value relation cannot be satisfied. For instance the equation 冟x 4冟 2 has no solutions, as the left-hand expression will always represent a nonnegative value. The inequality 冟2x 3冟 6 1 has no solutions for the same reason. On the other hand, the inequality 冟9 x冟 0 is true for all real numbers, since any value substituted for x will result in a nonnegative value. We can generalize many of these special cases as follows.

C. You’ve just seen how we can solve “greater than” absolute value inequalities

Absolute Value Functions — Special Cases Given k is a positive real number and A represents an algebraic expression, 冟A冟 k 冟A冟 6 k 冟A冟 7 k has no solutions

has no solutions

is true for all real numbers

See Exercises 51 through 54. CAUTION

䊳

Be sure you note the difference between the individual solutions of an absolute value equation, and the solution intervals that often result from solving absolute value inequalities. The solution 52, 56 indicates that both x 2 and x 5 are solutions, while the solution 32, 52 indicates that all numbers between 2 and 5, including 2, are solutions.

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D. Solving Absolute Value Equations and Inequalities Graphically The concepts studied in Section 1.5 (solving linear equations and inequalities graphically) are easily extended to other kinds of relations. Essentially, we treat each expression forming the equation or inequality as a separate function, then graph both functions to find points of intersection (equations) or where one graph is above or Figure 2.49

Figure 2.50

3.1

4.7

3.1

4.7

4.7

3.1

4.7

3.1

below the other (inequalities). For 2冟x 1冟 3 6 2, enter the expression 2冟X 1冟 3 as Y1 on the Y= screen, and 2 as Y2. Using ZOOM 4:ZDecimal produces the graph shown in Figure 2.49. Using 2nd TRACE (CALC) 5:intersect, we find the graphs intersect at x 1.5 and x 3.5 (Figure 2.50), and the graph of Y1 is above the graph of Y2 in this interval. Since this is a “less than” inequality, the solutions are outside of this interval, which gives x 僆 1q, 1.52 ´ 13.5, q2 as the solution interval. Note that the zeroes/x-intercept method could also have been used. EXAMPLE 7

䊳

Solving Absolute Equations and Inequalities Graphically 1 For f 1x2 2.5冟x 2冟 8 and g1x2 x 3, solve 2 a. f 1x2 g1x2 b. f 1x2 g1x2 c. f 1x2 7 g1x2

Solution

䊳

a. With f 1x2 2.5冟x 2冟 8 as Y1 and 1 g1x2 x 3 as Y2 (set to graph in bold), 2 using 2nd TRACE (CALC) 5:intersect 10 shows the graphs intersect 1Y1 Y2 2 at x 0 and x 5 (see figure). These are 1 the solutions to 2.5冟x 2冟 8 x 3. 2

10

10

10

b. The graph of Y1 is below the graph of Y2 1Y1 6 Y2 2 between these points of 1 intersection, so the solution interval for 2.5冟x 2冟 8 x 3 is x 僆 [0, 5]. 2 c. The graph of Y1 is above the graph of Y2 1Y1 7 Y2 2 outside this interval, 1 giving a solution of x 僆 1q, 02 ´ 15, q 2 for 2.5冟x 2冟 8 7 x 3. 2 D. You’ve just seen how we can solve absolute value equations and inequalities graphically


䊳

E. Applications Involving Absolute Value Applications of absolute value often involve finding a range of values for which a given statement is true. Many times, the equation or inequality used must be modeled after a given description or from given information, as in Example 8.

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EXAMPLE 8

䊳

143

Solving Applications Involving Absolute Value Inequalities For new cars, the number of miles per gallon (mpg) a car will get is heavily dependent on whether it is used mainly for short trips and city driving, or primarily on the highway for longer trips. For a certain car, the number of miles per gallon that a driver can expect varies by no more than 6.5 mpg above or below its field tested average of 28.4 mpg. What range of mileage values can a driver expect for this car?

Solution

䊳

Field tested average: 28.4 mpg mileage varies by no more than 6.5 mpg ⫺6.5

gather information highlight key phrases

⫹6.5

28.4

make the problem visual

Let m represent the miles per gallon a driver can expect. Then the difference between m and 28.4 can be no more than 6.5, or 冟m 28.4冟 6.5. 冟m 28.4冟 6.5 6.5 m 28.4 6.5 21.9 m 34.9

assign a variable write an equation model equation model apply Property I add 28.4 to all three parts

The mileage that a driver can expect ranges from a low of 21.9 mpg to a high of 34.9 mpg. E. You’ve just seen how we can solve applications involving absolute value


䊳

2.3 EXERCISES 䊳


Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. When multiplying or dividing by a negative quantity, we the inequality symbol to maintain a true statement.

2. To write an absolute value equation or inequality in simplified form, we the absolute value expression on one side.

3. The absolute value equation 冟2x 3冟 7 is true when 2x 3 or when 2x 3 .

4. The absolute value inequality 冟3x 6冟 6 12 is true when 3x 6 7 and 3x 6 6

Describe the solution set for each inequality (assume k > 0). Justify your answer.

5. 冟ax b冟 6 k

6. 冟ax b冟 7 k

.

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Solve each absolute value equation. Write the solution in set notation. For Exercises 7 to 18, verify solutions by substituting into the original equation. For Exercises 19–26 verify solutions using a calculator.

7. 2冟m 1冟 7 3 8. 3冟n 5冟 14 2 9. 3冟x 5冟 6 15 10. 2冟y 3冟 4 14 11. 2冟4v 5冟 6.5 10.3 12. 7冟2w 5冟 6.3 11.2 13. 冟7p 3冟 6 5 14. 冟3q 4冟 3 5 15. 2冟b冟 3 4 16. 3冟c冟 5 6 17. 2冟3x冟 17 5 18. 5冟2y冟 14 6 19. 3 `

w 4 ` 1 4 2

20. 2 ` 3

v ` 1 5 3

21. 8.7冟p 7.5冟 26.6 8.2

35.

冟5v 1冟 8 6 9 4

36.

冟3w 2冟 6 6 8 2

37. `

1 7 4x 5 ` 3 2 6

38. `

2y 3 3 15 ` 4 8 16

39. 冟n 3冟 7 7 40. 冟m 1冟 7 5 41. 2冟w冟 5 11 42. 5冟v冟 3 23 43. 44.

冟q冟 2 冟p冟 5

5 1 6 3

3 9 2 4

45. 3冟5 7d冟 9 15 46. 5冟2c 7冟 1 11 47. 2 6 ` 3m

1 4 ` 5 5

3 5 2n ` 4 4

22. 5.3冟q 9.2冟 6.7 43.8

48. 4 `

23. 8.7冟2.5x冟 26.6 8.2

49. 4冟5 2h冟 9 7 11

24. 5.3冟1.25n冟 6.7 43.8

50. 3冟7 2k冟 11 7 10

25. 冟x 2冟冟3x 4冟

51. 3.9冟4q 5冟 8.7 22.5

26. 冟2x 1冟冟x 3冟

52. 0.9冟2p 7冟 16.11 10.89

Solve each absolute value inequality. Write solutions in interval notation. Check solutions by back substitution, or using a calculator.

27. 3冟p 4冟 5 6 8 28. 5冟q 2冟 7 8 29. 3 冟m冟 2 7 4 30. 2冟n冟 3 7 7 31. 冟3b 11冟 6 9 32. 冟2c 3冟 5 6 1 33. 冟4 3z冟 12 6 7 34. 冟2 3u冟 5 4

53. 冟4z 9冟 6 4 54. 冟5u 3冟 8 7 6 Use the intersect command on a graphing calculator and the given functions to solve (a) f 1x2 g1x2 , (b) f 1x2 g1x2 , and (c) f 1x2 6 g1x2 .

55. f 1x2 冟x 3冟 2, g1x2 12x 2

56. f 1x2 冟x 2冟 1, g1x2 32x 9

57. f 1x2 0.5冟x 3冟 1, g1x2 2冟x 1冟 5 58. f 1x2 2冟x 3冟 2, g1x2 冟x 4冟 6

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59. Spring Oscillation: 冟d x冟 L A weight attached to a spring hangs at rest a distance of x in. off the ground. If the weight is pulled down (stretched) a distance of L inches and released, the weight begins to bounce and its distance d off the ground must satisfy the indicated formula. (a) If x equals 4 ft and the spring is stretched 3 in. and released, solve the inequality to find what distances from the ground the weight will oscillate between. (b) Solve for x in terms of L and d. 䊳

145

60. A “Fair” Coin: `

h 50 ` 6 1.645 5

If we flipped a coin 100 times, we expect “heads” to come up about 50 times if the coin is “fair.” In a study of probability, it can be shown that the number of heads h that appears in such an experiment should satisfy the given inequality to be considered “fair.” (a) Solve this inequality for h. (b) If you flipped a coin 100 times and obtained 40 heads, is the coin “fair”?

APPLICATIONS

Solve each application of absolute value.

61. Altitude of jet stream: To take advantage of the jet stream, an airplane must fly at a height h (in feet) that satisfies the inequality 冟h 35,050冟 2550. Solve the inequality and determine if an altitude of 34,000 ft will place the plane in the jet stream. 62. Quality control tests: In order to satisfy quality control, the marble columns a company produces must earn a stress test score S that satisfies the inequality 冟S 17,750冟 275. Solve the inequality and determine if a score of 17,500 is in the passing range. 63. Submarine depth: The sonar operator on a submarine detects an old World War II submarine net and must decide to detour over or under the net. The computer gives him a depth model 冟d 394冟 20 7 164, where d is the depth in feet that represents safe passage. At what depth should the submarine travel to go under or over the net? Answer using simple inequalities. 64. Optimal fishing depth: When deep-sea fishing, the optimal depths d (in feet) for catching a certain type of fish satisfy the inequality 28冟d 350冟 1400 6 0. Find the range of depths that offer the best fishing. Answer using simple inequalities. For Exercises 65 through 68, (a) develop a model that uses an absolute value inequality, and (b) solve.

65. Stock value: My stock in MMM Corporation fluctuated a great deal in 2009, but never by more than $3.35 from its current value. If the stock is worth $37.58 today, what was its range in 2009?

66. Traffic studies: On a given day, the volume of traffic at a busy intersection averages 726 cars per hour (cph). During rush hour the volume is much higher, during “off hours” much lower. Find the range of this volume if it never varies by more than 235 cph from the average. 67. Physical training for recruits: For all recruits in the 3rd Armored Battalion, the average number of sit-ups is 125. For an individual recruit, the amount varies by no more than 23 sit-ups from the battalion average. Find the range of sit-ups for this battalion. 68. Computer consultant salaries: The national average salary for a computer consultant is $53,336. For a large computer firm, the salaries offered to their employees vary by no more than $11,994 from this national average. Find the range of salaries offered by this company. 69. Tolerances for sport balls: According to the official rules for golf, baseball, pool, and bowling, (a) golf balls must be within 0.03 mm of d 42.7 mm, (b) baseballs must be within 1.01 mm of d 73.78 mm, (c) billiard balls must be within 0.127 mm of d 57.150 mm, and (d) bowling balls must be within 12.05 mm of d 2171.05 mm. Write each statement using an absolute value inequality, then (e) determine which sport gives the least width of interval b for the diameter tolerance t at average value of the ball.

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70. Automated packaging: The machines that fill boxes of breakfast cereal are programmed to fill each box within a certain tolerance. If the box is overfilled, the company loses money. If it is underfilled, it is considered unsuitable for sale. 䊳

Suppose that boxes marked “14 ounces” of cereal must be filled to within 0.1 oz. Find the acceptable range of weights for this cereal.


71. Determine the value or values (if any) that will make the equation or inequality true. x a. 冟x冟 x 8 b. 冟x 2冟 2 c. x 冟x冟 x 冟x冟 d. 冟x 3冟 6x e. 冟2x 1冟 x 3 72. The equation 冟5 2x冟冟3 2x冟 has only one solution. Find it and explain why there is only one. 73. In many cases, it can be helpful to view the solutions to absolute value equations and inequalities as follows. For any algebraic expression X and positive

䊳

2–42


constant k, the equation 冟X冟 k has solutions X k and X k, since the absolute value of either quantity on the left will indeed yield the positive constant k. Likewise, 冟X冟 6 k has solutions X 6 k and X 6 k. Note the inequality symbol has not been reversed as yet, but will naturally be reversed as part of the solution process. Solve the following equations or inequalities using this idea. a. 冟x 3冟 5 b. 冟x 7冟 7 4 c. 3冟x 2冟 12 d. 3冟x 4冟 7 11


74. (Appendix A.4) Factor the expression completely: 18x3 21x2 60x. 76. (Appendix A.6) Simplify

1

by rationalizing the 3 23 denominator. State the result in exact form and approximate form (to hundredths).

75. (1.5) Solve V2

2W for ␳ (physics). C␳A

77. (Appendix A.3) Solve the inequality, then write the solution set in interval notation: 312x 52 7 21x 12 7.

MID-CHAPTER CHECK 1. Determine whether the following function is even, 冟x冟 odd, or neither. f 1x2 x2 4x 2. Use a graphing calculator to find the maximum and minimum values of f 1x2 1.91x4 2.3x3 2.2x 5.12 . Round to the nearest hundredth. 3. Use interval notation to identify the interval(s) where the function from Exercise 2 is increasing, decreasing, or constant. Round to the nearest hundredth.

4. Write the equation of the function that has the same graph of f 1x2 2x, shifted left 4 units and up 2 units. 5. For the graph given, (a) identify the function family, (b) describe or identify the end-behavior, inflection point, and x- and y-intercepts, (c) determine the domain and range, and (d) determine the value of k if f 1k2 2.5. Assume required features have integer values.

Exercise 5 y 5

f(x)

⫺5

5 x

⫺5

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6. Use a graphing calculator to graph the given functions in the same window and comment on what you observe. p1x2 1x 32 2

r1x2 12 1x 32 2

147

9. Solve the following absolute value inequalities. Write solutions in interval notation. a. 3.1冟d 2冟 1.1 7.3 冟1 y冟 11 2 7 b. 3 2 c. 5冟k 2冟 3 6 4

q1x2 1x 32 2

7. Solve the following absolute value equations. Write the solution in set notation. 2 11 a. 冟d 5冟 1 7 b. 5 冟s 3冟 3 2

10. Kiteboarding: With the correct sized kite, a person can kiteboard when the wind is blowing at a speed w (in mph) that satisfies the inequality 冟w 17冟 9. Solve the inequality and determine if a person can kiteboard with a windspeed of (a) 5 mph? (b) 12 mph?

8. Solve the following absolute value inequalities. Write solutions in interval notation. x a. 3冟q 4冟 2 6 10 b. ` 2 ` 5 5 3

REINFORCING BASIC CONCEPTS Using Distance to Understand Absolute Value Equations and Inequalities For any two numbers a and b on the number line, the distance between a and b can be written 冟a b冟 or 冟b a冟. In exactly the same way, the equation 冟x 3冟 4 can be read, “the distance between 3 and an unknown number is equal to 4.” The advantage of reading it in this way (instead of “the absolute value of x minus 3 is 4”), is that a much clearer visualization is formed, giving a constant reminder there are two solutions. In diagram form we have Figure 2.51. Distance between 3 and x is 4. ⫺5 ⫺4 ⫺3 ⫺2

Figure 2.51

4 units ⫺1

0

1

4 units 2

3

4

5

Distance between 3 and x is 4. 6

7

8

9

From this we note the solutions are x 1 and x 7. In the case of an inequality such as 冟x 2冟 3, we rewrite the inequality as 冟x 122 冟 3 and read it, “the distance between 2 and an unknown number is less than or equal to 3.” With some practice, visualizing this relationship mentally enables a quick statement of the solution: x 僆 35, 14 . In diagram form we have Figure 2.52. Distance between 2 and x is less than or equal to 3. 8 7 6

Figure 2.52

3 units

Distance between 2 and x is less than or equal to 3.

3 units

5 4 3 2 1

0

1

2

3

4

5

6

Equations and inequalities where the coefficient of x is not 1 still lend themselves to this form of conceptual understanding. For 冟2x 1冟 3 we read, “the distance between 1 and twice an unknown number is greater than or equal to 3.” On the number line (Figure 2.53), the number 3 units to the right of 1 is 4, and the number 3 units to the left of 1 is 2. Distance between 1 and 2x is greater than or equal to 3.

Figure 2.53

3 units

6 5 4 3

ⴚ2 1

0

3 units 1

2

3

Distance between 1 and 2x is greater than or equal to 3 4

5

6

7

8

For 2x 2, x 1, and for 2x 4, x 2, and the solution set is x 僆 1q, 14 ´ 32, q 2. Attempt to solve the following equations and inequalities by visualizing a number line. Check all results algebraically. Exercise 1: 冟x 2冟 5

Exercise 2: 冟x 1冟 4

Exercise 3: 冟2x 3冟 5

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Basic Rational Functions and Power Functions; More on the Domain


A. Graph basic rational functions, identify vertical and horizontal asymptotes, and describe end-behavior B. Use transformations to graph basic rational functions and write the equation for a given graph C. Graph basic power functions and state their domains D. Solve applications involving basic rational and power functions

In this section, we introduce two new kinds of relations, rational functions and power functions. While we’ve already studied a variety of functions, we still lack the ability to model a large number of important situations. For example, functions that model the amount of medication remaining in the bloodstream over time, the relationship between altitude and weightlessness, and the equations modeling planetary motion come from these two families.

A. Rational Functions and Asymptotes Just as a rational number is the ratio of two integers, a rational function is the ratio of two polynomials. In general, Rational Functions A rational function V(x) is one of the form V1x2

p1x2 d1x2

,

where p and d are polynomials and d1x2 0. The domain of V(x) is all real numbers, except the zeroes of d. The simplest rational functions are the reciprocal function y 1x and the reciprocal square function y x12, as both have a constant numerator and a single term in the denominator. Since division by zero is undefined, the domain of both excludes x 0. A preliminary study of these two functions will provide a strong foundation for our study of general rational functions in Chapter 4.

The Reciprocal Function: y ⴝ

1 x

The reciprocal function takes any input (other than zero) and gives its reciprocal as the output. This means large inputs produce small outputs and vice versa. A table of values (Table 2.1) and the resulting graph (Figure 2.54) are shown. Table 2.1

148

Figure 2.54

x

y

x

y

1000

1/1000

1/1000

1000

5

1/5

1/3

3

4

1/4

1/2

2

3

1/3

1

1

2

1/2

2

1/2

1

1

3

1/3

1/2

2

4

1/4

1/3

3

5

1/5

1/1000

1000

1000

1/1000

0

undefined

y 3

冢a, 3冣

y

1 x

2

(1, 1)

冢3, a冣

1

5

冢3, a冣 5

冢5, Q冣 (1, 1)

冢 a, 3冣

冢5, Q冣 x

1 2 3

2–44

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149

Section 2.4 Basic Rational Functions and Power Functions; More on the Domain

WORTHY OF NOTE The notation used for graphical behavior always begins by describing what is happening to the x-values, and the resulting effect on the y-values. Using Figure 2.55, visualize that for a point (x, y) on the graph of y 1x , as x gets larger, y must become smaller, particularly since their product must always be 1 1y 1x 1 xy 12 .

Table 2.1 and Figure 2.54 reveal some interesting features. First, the graph passes the vertical line test, verifying y 1x is indeed a function. Second, since division by zero is undefined, there can be no corresponding point on the graph, creating a break at x 0. In line with our definition of rational functions, the domain is x 僆 1q, 02 ´ 10, q 2 . Third, this is an odd function, with a “branch” of the graph in the first quadrant and one in the third quadrant, as the reciprocal of any input maintains its sign. Finally, we note in QI that as x becomes an infinitely large positive number, y becomes infinitely small and closer to zero. It seems convenient to symbolize this endbehavior using the following notation:

Figure 2.55

as x S q,

y

yS0

as x becomes an infinitely large positive number

Graphically, the curve becomes very close to, or approaches the x-axis. We also note that as x approaches zero from the right, y becomes an infinitely large positive number: as x S 0 , y S q . Note a superscript or sign is used to indicate the direction of the approach, meaning from the positive side (right) or from the negative side (left).

y x

x

EXAMPLE 1

y approaches 0

䊳

Describing the End-Behavior of Rational Functions For y 1x in QIII (Figure 2.54), a. Describe the end-behavior of the graph. b. Describe what happens as x approaches zero.

Solution

䊳

Similar to the graph’s behavior in QI, we have a. In words: As x becomes an infinitely large negative number, y approaches zero. In notation: As x S q , y S 0. b. In words: As x approaches zero from the left, y becomes an infinitely large negative number. In notation: As x S 0 , y S q . Now try Exercises 7 and 8

The Reciprocal Square Function: y ⴝ

䊳

1 x2

From our previous work, we anticipate this graph will also have a break at x 0. But since the square of any negative number is positive, the branches of the reciprocal square function are both above the x-axis. Note the result is the graph of an even function. See Table 2.2 and Figure 2.56. Table 2.2

Figure 2.56

x

y

x

y

1000

1/1,000,000

1/1000

1,000,000

5

1/25

1/3

9

4

1/16

1/2

4

3

1/9

1

1

2

1/4

2

1/4

1

1

3

1/9

1/2

4

4

1/16

1/3

9

5

1/25

1/1000

1,000,000

1000

1/1,000,000

0

undefined

y x12

y 3

(1, 1)

冢5,

1

1 25 冣 5

2

冢3, 19冣

(1, 1)

冢3, 19 冣

冢5, 5

1 2 3

1 25 冣

x

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Similar to y 1x , large positive inputs generate small, positive outputs: as x S q, y S 0. This is one indication of asymptotic behavior in the horizontal direction, and we say the line y 0 (the x-axis) is a horizontal asymptote for the reciprocal and reciprocal square functions. In general, Horizontal Asymptotes Given a constant k, the line y k is a horizontal asymptote for V if, as x increases or decreases without bound, V(x) approaches k: as x S q, V1x2 S k

or

as x S q, V1x2 S k

As shown in Figures 2.57 and 2.58, asymptotes are represented graphically as dashed lines that seem to “guide” the branches of the graph. Figure 2.57 shows a horizontal asymptote at y 1, which suggests the graph of f(x) is the graph of y 1x shifted up 1 unit. Figure 2.58 shows a horizontal asymptote at y 2, which suggests the graph of g(x) is the graph of y x12 shifted down 2 units. Figure 2.57 y 3

f(x)

1 x

Figure 2.58 1

y 3

2

2

y1

1

5

EXAMPLE 2

䊳

g(x) x12 2

5

x

1

5

5

1

1

2

2

3

3

x

y 2

Describing the End-Behavior of Rational Functions For the graph in Figure 2.58, use mathematical notation to a. Describe the end-behavior of the graph and name the horizontal asymptote. b. Describe what happens as x approaches zero.

Solution

䊳

a. as x S q, g1x2 S 2, as x S q, g1x2 S 2,

b. as x S 0 , g1x2 S q , as x S 0 , g1x2 S q

y 2 is a horizontal asymptote Now try Exercises 9 and 10

䊳

While the graphical view of Example 2(a) (Figure 2.58) makes these concepts believable, a numerical view of this end-behavior can be even more compelling. Try entering x12 2 as Y1 on the Y= screen, then go to the TABLE feature 1TblStart 3, ¢Tbl 1; Figure 2.59). Scrolling in either direction shows that as 冟x冟 becomes very large, Y1 becomes closer and closer to 2, but will never be equal to 2 (Figure 2.60). Figure 2.59

Figure 2.60

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151

From Example 2(b), we note that as x becomes smaller and close to 0, g becomes very large and increases without bound. This is one indication of asymptotic behavior in the vertical direction, and we say the line x 0 (the y-axis) is a vertical asymptote for g (x 0 is also a vertical asymptote for f in Figure 2.57). In general, Vertical Asymptotes Given a constant h, the vertical line x h is a vertical asymptote for a function V if, as x approaches h, V(x) increases or decreases without bound: as x S h , V1x2 S q

or

as x S h , V1x2 S q

Here is a brief summary: Reciprocal Function f 1x2

A. You’ve just seen how we can graph basic rational functions, identify vertical and horizontal asymptotes, and describe end-behavior

1 x Domain: x 僆 1q, 02 ´ 10, q 2 Range: y 僆 1q, 02 ´ 10, q2 Horizontal asymptote: y 0 Vertical asymptote: x 0

Reciprocal Quadratic Function 1 x2 Domain: x 僆 1q, 02 ´ 10, q2 Range: y 僆 10, q 2 Horizontal asymptote: y 0 Vertical asymptote: x 0 g1x2

B. Using Asymptotes to Graph Basic Rational Functions Identifying these asymptotes is useful because the graphs of y 1x and y x12 can be transformed in exactly the same way as the toolbox functions. When their graphs shift — the vertical and horizontal asymptotes shift with them and can be used as guides to redraw the graph. In shifted form, a k for the reciprocal function, and f 1x2 xh a k for the reciprocal square function. g1x2 1x h2 2 When horizontal and/or vertical shifts are applied to simple rational functions, we first apply them to the asymptotes, then calculate the x- and y-intercepts as before. An additional point or two can be computed as needed to round out the graph. EXAMPLE 3

䊳

Graphing Transformations of the Reciprocal Function 1 1 using transformations of the parent function. x2 1 The graph of g is the same as that of y , but shifted 2 units right and 1 unit upward. x This means the vertical asymptote is also shifted 2 units right, and the horizontal 1 asymptote is shifted 1 unit up. The y-intercept is g102 . For the x-intercept: 2 1 1 substitute 0 for g (x ) 0 x2 1 1 subtract 1 x2 11x 22 1 multiply by 1x 22 x1 solve Sketch the graph of g1x2

Solution

y 5

䊳

x2

4 3

y1 5

(0, 0.5) (1, 0)

2 1 1 2 3 4 5

5

x

The x-intercept is (1, 0). Knowing the graph is from the reciprocal function family and shifting the asymptotes and intercepts yields the graph shown. Now try Exercises 11 through 26

䊳

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These ideas can be “used in reverse” to determine the equation of a basic rational function from its given graph, as in Example 4. EXAMPLE 4

䊳

Writing the Equation of a Basic Rational Function, Given Its Graph Identify the function family for the graph given, then use the graph to write the equation of the function in “shifted form.” Assume 冟a冟 1.

Solution

䊳

The graph appears to be from the reciprocal square family, and has been shifted 2 units right (the vertical asymptote is at x 2), and 1 unit down (the horizontal asymptote is at y 1). From y x12, we obtain f 1x2 1x 1 22 2 1 as the shifted form.

y 6

6

6 x

6

Now try Exercises 27 through 38 B. You’ve just seen how we can use asymptotes and transformations to graph basic rational functions and write the equation for a given graph

䊳

Using the definition of negative exponents, the basic reciprocal and reciprocal square functions can be written as y x1 and y x2, respectively. In this form, we note that these functions also belong to a family of functions known as the power functions (see Exercise 80).

C. Graphs of Basic Power Functions Italian physicist and astronomer Galileo Galilei (1564–1642) made numerous contributions to astronomy, physics, and other fields. But perhaps he is best known for his experiments with gravity, in which he dropped objects of different weights from the Leaning Tower of Pisa. Due in large part to his work, we know that the velocity of an object after it has fallen a certain distance is v 12gs, where g is the acceleration due to gravity (32 ft/sec2), s is the distance in feet the object has fallen, and v is the velocity of the object in feet per second (see Exercise 71). As you will see, this is an example of a formula that uses a power function. From previous coursework or a review of radicals and rational exponents (Ap1 1 3 pendix A.6), we know that 1x can be written1 as x2, and 1 x1as x3, enabling us to write these functions in exponential form: f 1x2 x2 and g1x2 x3. In this form, we see that these actually belong to a larger family of functions, where x is raised to some power, called the power functions. Power Functions and Root Functions For any constant real number p and variable x, functions of the form f 1x2 x p

are called power functions in x. If p is of the form

1 for integers n 2, the functions n

f 1x2 xn 3 f 1x2 1 x 1

are called root functions in x.

n

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1

3

5 y x2, y x4, y x3, y 1 x, and y x2 are all power functions, The functions 1 5 4 but only y x and y 1 x are also root functions. Initially we will focus on power functions where p 7 0.

EXAMPLE 5

䊳

Comparing the Graphs of Power Functions

Use a graphing calculator to graph the power functions f 1x2 x4, g1x2 x3, 3 1 2 h1x2 x , p1x2 x , and q1x2 x2 in the standard viewing window. Make an observation in QI regarding the effect of the 7exponent on each function, then 1 discuss what the graphs of y x6 and y x2 would look like. 1

Solution

䊳

First we enter the functions in sequence as Y1 through Y5 on the Y= screen (Figure 2.61). Using ZOOM 6:ZStandard produces the graphs shown in Figure 2.62. Narrowing the window to focus on QI (Figure 2.63: x 僆 3 4, 104, y 僆 34, 10 4 ), we quickly see that for x 1, larger values of p cause the graph of y x p to increase at a faster rate, and smaller values at a slower rate. In other words 1 1 (for x 1), since 6 , the graph of 6 4 1 y x6 would increase slower and appear 1 to be “under” the graph of Y1 X4. 7 7 Since 7 2, the graph of y x2 would 2 increase faster and appear to be “more narrow” than the graph of Y5 X2 (verify this).

2

Figure 2.61, 2.62

10

10

10

10

Figure 2.63 10 Y5

Y4

Y3 Y2 Y1

4

10

4


䊳

The Domain of a Power Function In addition to the observations made in Example 5, we can make other important notes, particularly regarding the domains of power functions. When the exponent on a power m function is a rational number 7 0 in simplest form, it appears the domain is all real n 2 1 numbers if n 2 is odd, as seen in the graphs of g1x2 x3 , h1x2 x1 x1 , and 2 q1x2 x x1. If n is an even1 number, the domain is all nonnegative real numbers as 3 seen in the graphs of f 1x2 x4 and p1x2 x2. Further exploration will show that if p is irrational, as in y x␲, the domain is also all nonnegative real numbers and we have the following:

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The Domain of a Power Function

Given a power function f 1x2 x p with p 7 0. m 1. If p is a rational number in simplest form, n a. the domain of f is all real numbers if n is odd: x 僆 1q, q 2 , b. the domain of f is all nonnegative real numbers if n is even: x 僆 3 0, q 2 . 2. If p is an irrational number, the domain of f is all nonnegative real numbers: x 僆 30, q 2 . Further confirmation of statement 1 can be found by recalling the graphs of 1 1 3 y 1x x2 and y 1 x x3 from Section 2.2 (Figures 2.64 and 2.65). Figure 2.64 y

Figure 2.65

f(x) ⫽ 兹x

5

5

(note n is even) (9, 3) (4, 2) (6, 2.4)

(0, 0) ⫺1

(note n is odd)

(8, 2)

(1, 1)

(0, 0) 4

8

3 y g(x) ⫽ 兹x

x

⫺8

⫺4

(1, 1) 4

8

x

(⫺1, ⫺1) (⫺8, ⫺2) ⫺5

⫺5

Domain: x 僆 30, q2 Range: y 僆 3 0, q 2

EXAMPLE 6

䊳

Domain: x 僆 3 q, q2 Range: y 僆 3 q, q 2

Determining the Domains of Power Functions State the domain of the following power functions, and identity whether each is also a root function. 4 1 2 8 a. f 1x2 x5 b. g1x2 x10 c. h1x2 1 x d. q1x2 x3 e. r 1x2 x1 5

Solution

䊳

a. Since n is odd, the domain of f is all real numbers; f is not a root function. b. Since n is even, the domain of g is x 僆冤0, q 2 ; g is a root function. 1 c. In exponential form h1x2 x8. Since n is even, the domain of h is x 僆冤0, q 2 ; h is a root function. d. Since n is odd, the domain of q is all real numbers; q is not a root function e. Since p is irrational, the domain of r is x 僆冤0, q 2 ; r is not a root function Now try Exercises 49 through 58

䊳

Transformations of Power and Root Functions As we saw in Section 2.2 (Toolbox Functions and Transformations), the graphs of the 3 root functions y 1x and y 1 x can be transformed using shifts, stretches, reflections, and so on. In Example 8(b) (Section 2.2) we noted the graph of 3 3 h1x2 2 1 x 2 1 was the graph of y 1 x shifted 2 units right, stretched by a factor of 2, and shifted 1 unit down. Graphs of other power functions can be transformed in exactly the same way.

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EXAMPLE 7

䊳

Graphing Transformations of Power Functions Based on our previous observations, 2 3 a. Determine the domain of f 1x2 x3 and g1x2 x2 , then verify by graphing them on a graphing calculator. 2 3 b. Next, discuss what the graphs of F1x2 1x 22 3 3 and G1x2 x2 2 will look like, then graph each on a graphing calculator to verify.

Solution

䊳

m

a. Both f and g are power functions of the form y x n . For f, n is odd so its domain is all real numbers. For g, n is even and the domain is x 僆 3 0, q 2 . Their graphs support this conclusion (Figures 2.66 and 2.67). Figure 2.66

Figure 2.67

10

10

10

10

10

10

10

10

b. The graph of F will be the same as the graph of f, but shifted two units right and three units down, moving the vertex to 12, 32 . The graph of G will be the same as the graph of g, but reflected across the x-axis, and shifted 2 units up (Figures 2.68 and 2.69). Figure 2.69

Figure 2.68 10

10

C. You’ve just seen how we can graph basic power functions and state their domains

10

10

10

10

10

10


䊳

D. Applications of Rational and Power Functions These new functions have a variety of interesting and significant applications in the real world. Examples 8 through 10 provide a small sample, and there are a number of additional applications in the Exercise Set. In many applications, the coefficients may be rather large, and the axes should be scaled accordingly. EXAMPLE 8

䊳

Modeling the Cost to Remove Waste For a large urban-centered county, the cost to remove chemical waste and other 18,000 180, pollutants from a local river is given by the function C1p2 p 100 where C( p) represents the cost (in thousands of dollars) to remove p percent of the pollutants.

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a. Find the cost to remove 25%, 50%, and 75% of the pollutants and comment on the results. b. Graph the function using an appropriate scale. c. Use mathematical notation to state what happens as the county attempts to remove 100% of the pollutants.

Solution

䊳

a. We evaluate the function as indicated, finding that C1252 60, C1502 180, and C1752 540. The cost is escalating rapidly. The change from 25% to 50% brought a $120,000 increase, but the change from 50% to 75% brought a $360,000 increase! C(p) b. From the context, we need only graph the x ⫽ 100 1200 portion from 0 p 6 100. For the C-intercept we substitute p 0 and find C102 0, which 900 seems reasonable as 0% would be removed (75, 540) 600 if $0 were spent. We also note there must be a vertical asymptote at x 100, since this 300 x-value causes a denominator of 0. Using (25, 60) (50, 180) p this information and the points from part (a) 100 50 75 25 produces the graph shown. c. As the percentage of pollutants removed y ⫽ ⫺180 approaches 100%, the cost of the cleanup skyrockets. Using notation: as p S 100 , C S q . Now try Exercises 65 through 70

䊳

While not obvious at first, the function C(p) in Example 8 is from the family of 1 reciprocal functions y . A closer inspection shows it has the form x 18,000 1 a k S 180, showing the graph of y is shifted right y x xh x 100 100 units, reflected across the x-axis, stretched by a factor of 18,000 and shifted 180 units down (the horizontal asymptote is y 180). As sometimes occurs in real-world applications, portions of the graph were ignored due to the context. To see the full graph, we reason that the second branch occurs on the opposite side of the vertical and horizontal asymptotes, and set the window as shown in Figure 2.70. After entering C(p) as Y1 on the Y= screen and pressing GRAPH , the full graph appears as shown in Figure 2.71 (for effect, the vertical and horizontal asymptotes were drawn separately using the 2nd PRGM (DRAW) options). Figure 2.71 Figure 2.70

2000

200

0

2000

Next, we’ll use a root function to model the distance to the horizon from a given height.

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EXAMPLE 9

䊳

The Distance to the Horizon On a clear day, the distance a person can see from a certain height (the distance to the horizon) is closely approximated by the root function d1h2 3.571h, where d(h) represents the viewing distance (in kilometers) from a height of h meters above sea level. a. To the nearest kilometer, how far can a person see when standing on the observation level of the John Hancock building in Chicago, Illinois, about 335 m high? b. To the nearest meter, how high is the observer’s eyes, if the viewing distance is 130 km?

Solution

䊳

a. Substituting 335 for h we have d1h2 3.57 1h d13352 3.57 1335 ⬇ 65.34

original function substitute 335 for h result

On a clear day, a person can see about 65 kilometers. b. We substitute 130 for d(h): d1h2 130 36.415 1326.052

3.57 1h 3.57 1h ⬇ 1h ⬇h

original function substitute 130 for d(h) divide by 3.57 square both sides

If the distance to the horizon is 130 km, the observer’s eyes are at a height of approximately 1326 m. Check the answer to part (b) by solving graphically. Now try Exercises 71 through 74

䊳

One area where power functions and modeling with regression are used extensively is allometric studies. This area of inquiry studies the relative growth of a part of an animal in relation to the growth of the whole, like the wingspan of a bird compared to its weight, or the daily food intake of a mammal or bird compared to its size.

EXAMPLE 10

䊳

Modeling the Food Requirements of Certain Bird Species To study the relationship between the weight of a nonpasserine bird and its daily food intake, the data shown in the table was collected (nonpasserine: nonsinging, nonperching birds). a. On a graphing calculator, enter the data in L1 and L2, then set an appropriate window to view a scatterplot of the data. Does a power regression STAT CALC, A:PwrReg seem appropriate?

Average weight (g)

Daily food intake (g)

Common pigeon

350

25

Ring-necked duck

725

50

Ring-necked pheasant

1400

70

Canadian goose

4525

165

White swan

9075

240

Bird

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b. Use a graphing calculator to find an equation model using a power regression on the data, and enter the equation in Y1 (round values to three decimal places). c. Use the equation to estimate the daily food intake required by a barn owl (470 g), and a gray-headed albatross (6800 g). d. Use the intersection of graphs method to find the weight of a Great-Spotted Kiwi, given the daily food requirement is 130 g.

Solution

䊳

a. After entering the weights in L1 and Figure 2.72 food intake in L2, we set a window that 300 will comfortably fit the data. Using x 僆 30, 10,000 4 and y 僆 330, 300 4 produces the scatterplot shown (Figure 2.72). The data does not appear 0 10,000 linear, and based on our work in Example 5, a power function seems appropriate. 30 b. To access the power regression option, use STAT (CALC) A:PwrReg. To Figure 2.73 three decimal places the equation for Y1 would be 0.493 X0.685 (Figure 2.73). c. For the barn owl, x 470 and we find the estimated food requirement is about 33.4 g per day (Figure 2.74). For the grayheaded albatross x 6800 and the model estimates about 208.0 g of food daily is required. d. Here we’re given the food intake of the Great-Spotted Kiwi (the output value), and want to know what input value (weight) was used. Entering Y2 130, we’ll attempt to find where the graphs of Y1 and Y2 intersect (it will help to deactivate Plot1 on the Y= screen, so that only the graphs of Y1 and Y2 appear). Using 2nd TRACE (CALC) 5:Intersect shows the graphs intersect at about (3423.3, 130) (Figure 2.75), indicating the average weight of a Great-Spotted Kiwi is near 3423.3 g (about 7.5 lb). Figure 2.75 Figure 2.74

300

10,000

0

D. You’ve just seen how we can solve applications involving basic rational and power functions

30


䊳

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159

2.4 EXERCISES 䊳



1. Write the following in notational form. As x becomes an infinitely large negative number, y approaches 2. 1 2, 1x 32 2 a asymptote occurs at x 3 and a horizontal asymptote at .

䊳

2. For any constant k, the notation “as 冟 x 冟 S q, y S k ” is an indication of a asymptote, while “x S k, 冟y冟 S q ” indicates a asymptote. 1 has branches in Quadrants I x 1 and III. The graph of Y2 has branches in x Quadrants and .

3. Given the function g1x2

4. The graph of Y1

5. Discuss/Explain how and why the range of the reciprocal function differs from the range of the reciprocal quadratic function. In the reciprocal quadratic function, all range values are positive.

6. If the graphs of Y1

1 1 and Y2 2 were drawn x x on the same grid, where would they intersect? In what interval(s) is Y1 7 Y2?


For each graph given, (a) use mathematical notation to describe the end-behavior of each graph and (b) describe what happens as x approaches 1.

1 2 x1

7. V1x2

8. v1x2

1 2 x1

y

y

6 5 4 3 2 1 ⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4

3 2 1

1 2 3 4 5 6 x

1 x ⫺5⫺4⫺3⫺2⫺1 ⫺1 1 2 3 4 5 ⫺2 ⫺3 ⫺4 ⫺5 ⫺6 ⫺7

For each graph given, (a) use mathematical notation to describe the end-behavior of each graph, (b) name the horizontal asymptote, and (c) describe what happens as x approaches ⴚ2.

9. Q1x2

1 1 1 10. q1x2 2 2 1x 22 1x 22 2 y

⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4

y

5 4 3 2 1

6 5 4 3 2 1 1 2 3 x

⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

Sketch the graph of each function using transformations of the parent function (not by plotting points). Clearly state the transformations used, and label the horizontal and vertical asymptotes as well as the x- and y-intercepts (if they exist). Also state the domain and range of each function.

11. f 1x2

12. g1x2

13.

14.

15. 17. 19. 21. 23. 25.

1 2 3 x

1 1 x 1 h1x2 x2 1 g1x2 x2 1 f 1x2 1 x2 1 h1x2 1x 12 2 1 g1x2 1x 22 2 1 f 1x2 2 2 x 1 h1x2 1 1x 22 2

16. 18. 20. 22. 24. 26.

1 2 x 1 f 1x2 x3 1 h1x2 2 x 1 g1x2 2 x3 1 f 1x2 1x 52 2 1 h1x2 2 2 x 1 g1x2 2 3 x 1 g1x2 2 1x 12 2

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Identify the parent function for each graph given, then use the graph to construct the equation of the function in shifted form. Assume |a| ⴝ 1. y

27. S(x)

⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5 ⫺6

y

28.

4 3 2 1

5 4 3 2 1

1 2 3 4 x

⫺7⫺6 ⫺5⫺4⫺3⫺2⫺1 ⫺1

1 2 3 x

⫺2 ⫺3 ⫺4 ⫺5

s(x)

For each pair of functions given, state which function increases faster for x ⬎ 1, then use the INTERSECT command of a graphing calculator to find where (a) f(x) ⫽ g(x), (b) f(x) ⬎ g(x), and (c) f(x) ⬍ g(x).

39. f 1x2 x2, g1x2 x3

40. f 1x2 x4, g1x2 x5

43. f 1x2 x , g1x2 x

44. f 1x2 x 4, g1x2 x2

41. f 1x2 x4, g1x2 x2 2 3

42. f 1x2 x3, g1x2 x5

4 5

7

6 3 5 4 45. f 1x2 1 x, g1x2 1 x 46. f 1x2 1 x, g1x2 1 x

3 2 4 3 47. f 1x2 2 x , g1x2 x4 48. f 1x2 x2, g1x2 2 x 5

29.

3 2 1

y 4 3 2 1

Q(x)

⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5 ⫺6 ⫺7

31.

30.

y

1 2 3 4 x

⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5 ⫺6

y 2 1 ⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 v(x) ⫺3 ⫺4 ⫺5 ⫺6 ⫺7 ⫺8

1 2 3 x

q(x)

y 2 1 ⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 w(x) ⫺3 ⫺4 ⫺5 ⫺6 ⫺7 ⫺8

1 2 3 x

Use the graph shown to Exercises 33 through 38 y complete each statement using 10 the direction/approach notation.

33. As x S q, y ______. 34. As x S q, y ______.

⫺10

10 x

35. As x S 1, y ______.

49. f 1x2 x8 7

50. g1x2 x7

51. h1x2 x5

6

52. q1x2 x6

7 53. r1x2 1 x

54. s1x2 x6

6

5

1

Using the functions from Exercises 49–54, identify which of the following are defined and which are not. Do not use a calculator or evaluate.

55. a. f 122

b. f(2)

56. a. h(0.3)

b. h10.32 c. q(0.3)

d. q(0.3)

57. a. h11.22

b. r172

d. s(0)

7 58. a. f a b 8

8 b. ga b c. q11.92 7

c. g122

36. As x S 1 , y ______.

⫺10

37. The line x 1 is a vertical asymptote, since: as x S ____, y S ____. 38. The line y 2 is a horizontal asymptote, since: as x S ____, y S ____.

c. s1␲2

d. g122

d. q(0)

Compare and discuss the graphs of the following functions. Verify your answer by graphing both on a graphing calculator.

59. f 1x2 x8; F1x2 1x 12 8 2 7

䊳

3

State the domain of the following functions.

1 2 3 4 5 6 x

32.

3

7

60. g1x2 x7; G1x2 1x 32 7 2 8

8

6

6

61. p1x2 x5; P1x2 1x 22 5 5

5

62. q1x2 x6; Q1x2 2x6 5


63. Gravitational attraction: F ⴝ

km1m2 2

d The gravitational force F between two objects with masses m1 and m2 depends on the distance d between them and some constant k. (a) If the masses of the two objects are constant while the distance between them gets larger and larger, what happens to F? (b) Let m1 and m2 equal 1 mass unit with k 1 as well, and investigate using a table of values. What family does this function belong to? (c) Solve for m2 in terms of k, m1, d and F.

mⴙM 22gh m For centuries, the velocity v of a bullet of mass m has been found using a device called a ballistic pendulum. In one such device, a bullet is fired into a stationary block of wood of mass M, suspended from the end of a pendulum. The height h the pendulum swings after impact is measured, and the approximate velocity of the bullet can then be calculated using g 9.8 m/sec2 (acceleration due to gravity). When a .22-caliber bullet of mass 2.6 g is fired into a wood block of mass 400 g, their combined mass swings to a height of 0.23 m. To the nearest meter per second, find the velocity of the bullet the moment it struck the wood.

64. Velocity of a bullet: v ⴝ

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APPLICATIONS

65. Deer and predators: By banding deer over a period of 10 yr, a capture-and-release project determines the number of deer per square mile in the Mark Twain National Forest can be modeled by 75 the function D1p2 , where p is the number of p predators present and D is the number of deer. Use this model to answer the following. a. As the number of predators increases, what will happen to the population of deer? Evaluate the function at D(1), D(3), and D(5) to verify. b. What happens to the deer population if the number of predators becomes very large? c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the deer population if the number of predators becomes very small (less than 1 per square mile). 66. Balance of nature: A marine biology research group finds that in a certain reef area, the number of fish present depends on the number of sharks in the area. The relationship can be modeled by the 20,000 , where F(s) is the fish function F1s2 s population when s sharks are present. a. As the number of sharks increases, what will happen to the population of fish? Evaluate the function at F(10), F(50), and F(200) to verify. b. What happens to the fish population if the number of sharks becomes very large? c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the fish population if the number of sharks becomes very small. 67. Intensity of light: The intensity I of a light source depends on the distance of the observer from the source. If the intensity is 100 W/m2 at a distance of 5 m, the relationship can be modeled by the 2500 function I1d2 2 . Use the model to answer the d following. a. As the distance from the lightbulb increases, what happens to the intensity of the light? Evaluate the function at I(5), I(10), and I(15) to verify. b. If the intensity is increasing, is the observer moving away or toward the light source?

c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the intensity if the distance from the lightbulb becomes very small. 68. Electrical resistance: The resistance R (in ohms) to the flow of electricity is related to the length of the wire and its gauge (diameter in fractions of an inch). For a certain wire with fixed length, this relationship can be modeled by the function 0.2 R1d2 2 , where R(d) represents the resistance in d a wire with diameter d. a. As the diameter of the wire increases, what happens to the resistance? Evaluate the function at R(0.05), R(0.25), and R(0.5) to verify. b. If the resistance is increasing, is the diameter of the wire getting larger or smaller? c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the resistance in the wire as the diameter gets larger and larger. 69. Pollutant removal: For a certain coal-burning power plant, the cost to remove pollutants from plant emissions can be modeled by 8000 80, where C(p) represents the C1p2 p 100 cost (in thousands of dollars) to remove p percent of the pollutants. (a) Find the cost to remove 20%, 50%, and 80% of the pollutants, then comment on the results; (b) graph the function using an appropriate scale; and (c) use mathematical notation to state what happens if the power company attempts to remove 100% of the pollutants. 70. City-wide recycling: A large city has initiated a new recycling effort, and wants to distribute recycling bins for use in separating various recyclable materials. City planners anticipate the cost of the program can be modeled by the 22,000 220, where C(p) function C1p2 p 100 represents the cost (in $10,000) to distribute the bins to p percent of the population. (a) Find the cost to distribute bins to 25%, 50%, and 75% of the population, then comment on the results; (b) graph the function using an appropriate scale; and (c) use mathematical notation to state what happens if the city attempts to give recycling bins to 100% of the population.

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71. Hot air ballooning: If air resistance is neglected, the velocity (in ft/s) of a falling object can be closely approximated by the function V1s2 8 1s, where s is the distance the object has fallen (in feet). A balloonist suddenly finds it necessary to release some ballast in order to quickly gain altitude. (a) If she were flying at an altitude of 1000 ft, with what velocity will the ballast strike the ground? (b) If the ballast strikes the ground with a velocity of 225 ft/sec, what was the altitude of the balloon? 72. River velocities: The ability of a river or stream to move sand, dirt, or other particles depends on the size of the particle and the velocity of the river. This relationship can be used to approximate the velocity (in mph) of the river using the function V1d2 1.77 1d, where d is the diameter (in inches) of the particle being moved. (a) If a creek can move a particle of diameter 0.095 in., how fast is it moving? (b) What is the largest particle that can be moved by a stream flowing 1.1 mph? 73. Shoe sizes: Although there may be some notable exceptions, the size of shoe worn by the average man is related to his height. This relationship is 3 modeled by the function S1h2 0.75h2, where h is the person’s height in feet and S is the U.S. shoe size. (a) Approximate Denzel Washington’s shoe size given he is 6 ft, 0 in. tall. (b) Approximate Dustin Hoffman’s height given his shoe size is 9.5. 74. Whale weight: For a certain species of whale, the relationship between the length of the whale and the weight of the 27whale can be modeled by the function W1l2 0.03l 11, where l is the length of the whale in meters and W is the weight of the whale in metric tons (1 metric ton ⬇ 2205 pounds). (a) Estimate the weight of a newborn calf that is 6 m long. (b) At 81 metric tons, how long is an average adult? 75. Gestation periods: The data shown in the table can be used to study the relationship between the weight of mammal and its length of pregnancy. Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to three decimal places). Use the equation to estimate (c) the length of pregnancy of a racoon (15.5 kg) and (d) the weight of a fox, given the length of pregnancy is 52 days. Average Weight (kg)

Gestation (days)

Rat

0.4

24

Rabbit

3.5

50

Armadillo

6.0

51

Coyote

13.1

62

Dog

24.0

64

Mammal

76. Bird wingspans: The data in the table explores the relationship between a bird’s weight and its wingspan. Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to three decimal places). Use the equation to estimate (c) the wingspan of a Bald Eagle (16 lb) and (d) the weight of a Bobwhite Quail with a wingspan of 0.9 ft. Weight Wingspan (lb) (ft)

Bird Golden Eagle

10.5

6.5

Horned Owl

3.1

2.6

Peregrine Falcon

3.3

4.0

Whooping Crane

17.0

7.5

1.5

2.0

Raven

77. Species-area relationship: To study the relationship between the number of species of birds on islands in the Caribbean, the data shown in the table was collected. Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to three decimal places). Use the equation to estimate (c) the number of species of birds on Andros (2300 mi2) and (d) the area of Cuba, given there are 98 such species. Island

Area (mi2)

Great Inagua

Species

600

16

Trinidad

2000

41

Puerto Rico

3400

47

Jamaica Hispaniola

4500

38

30,000

82

78. Planetary orbits: The table shown gives the time required for the first five planets to make one complete revolution around the Sun (in years), along with the average orbital radius of the planet in astronomical units (1 AU 92.96 million miles). Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to four decimal places). Use the equation to estimate (c) the average orbital radius of Saturn, given it orbits the Sun every 29.46 yr, and (d) estimate how many years it takes Uranus to orbit the Sun, given it has an average orbital radius of 19.2 AU. Planet

Years

Radius

Mercury

0.24

0.39

Venus

0.62

0.72

Earth

1.00

1.00

Mars Jupiter

1.88

1.52

11.86

5.20

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Section 2.5 Piecewise-Defined Functions


79. Consider the graph of f 1x2

1 once again, and the x x by f(x) rectangles mentioned in the Worthy of Note on page 149. Calculate the area of each rectangle formed for x 僆 51, 2, 3, 4, 5, 66 . What do 1 you notice? Repeat the exercise for g1x2 2 and x the x by g(x) rectangles. Can you detect the pattern formed here?

80. All of the power functions presented in this section had positive exponents, but the definition of these types of functions does allow for negative exponents as well. In addition to the reciprocal and reciprocal square functions (y x1 and y x2), these types

䊳

163

of power functions have Depth Temp significant applications. For (meters) (°C) example, the temperature of 125 13.0 ocean water depends on several 250 9.0 factors, including salinity, 500 6.0 latitude, depth, and density. 750 5.0 However, between depths of 125 m and 2000 m, ocean 1000 4.4 temperatures are relatively 1250 3.8 predictable, as indicated by the 1500 3.1 data shown for tropical oceans 1750 2.8 in the table. Use a graphing 2000 2.5 calculator to find the power regression model and use it to estimate the water temperature at a depth of 2850 m.


81. (1.4) Solve the equation for y, then sketch its graph using the slope/intercept method: 2x 3y 15. 82. (1.3) Using a scale from 1 (lousy) to 10 (great), Charlie gave the following ratings: {(The Beatles, 9.5), (The Stones, 9.6), (The Who, 9.5), (Queen, 9.2), (The Monkees, 6.1), (CCR, 9.5), (Aerosmith, 9.2), (Lynyrd Skynyrd, 9.0), (The Eagles, 9.3), (Led

2.5

Zeppelin, 9.4), (The Stones, 9.8)}. Is the relation (group, rating) as given, also a function? State why or why not. 83. (1.5) Solve for c: E mc2. 84. (2.3) Use a graphing calculator to solve 冟x 2冟 1 2冟x 1冟 3.

Piecewise-Defined Functions


A. State the equation, domain, and range of a piecewise-defined function from its graph B. Graph functions that are piecewise-defined C. Solve applications involving piecewisedefined functions

Most of the functions we’ve studied thus far have been smooth and continuous. Although “smooth” and “continuous” are defined more formally in advanced courses, for our purposes smooth simply means the graph has no sharp turns or jagged edges, and continuous means you can draw the entire graph without lifting your pencil. In this section, we study a special class of functions, called piecewisedefined functions, whose graphs may be various combinations of smooth/not smooth and continuous/not continuous. The absolute value function is one example (see Exercise 31). Such functions have a tremendous number of applications in the real world.

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A. The Domain of a Piecewise-Defined Function For the years 1990 to 2000, the American bald eagle remained on the nation’s endangered species list, although the number of breeding pairs was growing slowly. After 2000, the population of eagles grew at a much faster rate, and they were removed from the list soon afterward. From Table 2.3 and plotted points modeling this growth (see Figure 2.76), we observe that a linear model would fit the period from 1992 to 2000 very well, but a line with greater slope would be needed for the years 2000 to 2006 and (perhaps) beyond. Figure 2.76 10,000

Bald eagle breeding pairs

9,000

Table 2.3 Year (1990 S 0)

Bald Eagle Breeding Pairs

Year (1990 S 0)

Bald Eagle Breeding Pairs

2

3700

10

6500

4

4400

12

7600

6

5100

14

8700

8

5700

16

9800

8,000 7,000 6,000 5,000 4,000 3,000

Source: www.fws.gov/midwest/eagle/population 0

2

4

6

8

10

12

14

16

18

t (years since 1990)

WORTHY OF NOTE For the years 1992 to 2000, we can estimate the growth in breeding pairs ¢pairs ¢time using the points (2, 3700) and (10, 6500) in the slope formula. The result is 350 1 , or 350 pairs per year. For 2000 to 2006, using (10, 6500) and (16, 9800) shows the rate of growth is significantly larger: ¢pairs 550 ¢years 1 or 550 pairs per year.

The combination of these two lines would be a single function that modeled the population of breeding pairs from 1990 to 2006, but it would be defined in two pieces. This is an example of a piecewise-defined function. The notation for these functions is a large “left brace” indicating the equations it groups are part of a single function. Using selected data points and techniques from Section 1.4, we find equations that could represent each piece are p1t2 350t 3000 for 0 t 10 and p1t2 550t 1000 for t 7 10, where p(t) is the number of breeding pairs in year t. The complete function is then written: function name

function pieces

350t 3000, p1t2 e 550t 1000,

domain of each piece

2 t 10 t 7 10

In Figure 2.76, note that we indicated the exclusion of t 10 from the second piece of the function using an open half-circle.

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EXAMPLE 1

䊳

Writing the Equation and Domain of a Piecewise-Defined Function The linear piece of the function shown has an equation of y 2x 10. The equation of the quadratic piece is y x2 9x 14. 10 a. Use the correct notation to write them as a 8 single piecewise-defined function and state the domain of each piece by inspecting the graph. 6 b. State the range of the function.

Solution

䊳

y

f(x)

4

a. From the graph we note the linear portion is defined between 0 and 3, with these endpoints 2 included as indicated by the closed dots. The domain here is 0 x 3. The quadratic 0 portion begins at x 3 but does not include 3, as indicated by the half-circle notation. The equation is function name

f 1x2 e

function pieces

2x 10, x2 9x 14,

(3, 4)

2

4

6

8

x

10

domain

0x3 3 6 x7

b. The largest y-value is 10 and the smallest is zero. The range is y 僆 30, 104 .

A. You’ve just seen how we can state the equation, domain, and range of a piecewise-defined function from its graph


䊳

Piecewise-defined functions can be composed of more than two pieces, and can involve functions of many kinds.

B. Graphing Piecewise-Defined Functions As with other functions, piecewise-defined functions can be graphed by simply plotting points. Careful attention must be paid to the domain of each piece, both to evaluate the function correctly and to consider the inclusion/exclusion of endpoints. In addition, try to keep the transformations of a basic function in mind, as this will often help graph the function more efficiently.

EXAMPLE 2

䊳

Graphing a Piecewise-Defined Function Evaluate the piecewise-defined function by noting the effective domain of each piece, then graph by plotting these points and using your knowledge of basic functions. h1x2 e

Solution

䊳

x 2, 2 1x 1 1,

5 x 6 1 x 1

The first piece of h is a line with negative slope, while the second is a transformed square root function. Using the endpoints of each domain specified and a few additional points, we obtain the following: For h1x2 x 2, 5 x 6 1, x

h(x)

For h1x2 2 1x 1 1, x 1, x

h(x)

5

3

1

1

3

1

0

1

1

(1)

3

3

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After plotting the points from the first piece, we connect them with a line segment noting the left endpoint is included, while the right endpoint is not (indicated using a semicircle around the point). Then we plot the points from the second piece and draw a square root graph, noting the left endpoint here is included, and the graph rises to the right. From the graph we note the complete domain of h is x 僆 3 5, q 2 , and the range is y 僆 3 1, q 2 .

h(x) 5

h(x) x 2 h(x) 2 x 1 1 5

5

x

5


䊳

Most graphing calculators are able to graph piecewise-defined functions. Consider Example 3.

EXAMPLE 3

Solution

䊳

䊳

Graphing a Piecewise-Defined Function Using Technology x 5, 5 x 6 2 Graph the function f 1x2 e on a graphing calculator 2 1x 42 3, x 2 and evaluate f (2). Figure 2.77 10 Both “pieces” are well known—the first is a line with slope m 1 and y-intercept (0, 5). The second is a parabola that opens upward, shifted 4 units to the right and 3 units up. If we attempt to graph 10 f(x) using Y1 X 5 and Y2 1X 42 2 3 10 as they stand, the resulting graph may be difficult to analyze because the pieces overlap and intersect (Figure 2.77). To graph the functions 10 we must indicate the domain for each piece, separated by a slash and enclosed in parentheses. Figure 2.78 For instance, for the first piece we enter Y1 X 5/1X 5 and X 6 22 , and for the second, Y2 1X 42 2 3 1X 22 (Figure 2.78). The slash looks like (is) the division symbol, but in this context, the calculator interprets it as a means of separating the function from the domain. The inequality symbols are accessed using the 2nd MATH (TEST) keys. As shown for Y , compound 1 inequalities must be entered in two parts, using the logical connector “and”: 2nd MATH (LOGIC) 1:and. The graph is shown in Figure 2.79, where we see the function is linear for x 僆 [5, 2) and quadratic for x 僆 [2, q ). Using the 2nd GRAPH (TABLE) feature reveals the calculator will give an ERR: (ERROR) message for inputs outside the domains of Y1 and Y2, and we see that f is defined for x 2 only for Y2: f 122 7 (Figure 2.80). Figure 2.79 Figure 2.80 10

10

10

10


䊳

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As an alternative to plotting points, we can graph each piece of the function using transformations of a basic graph, then erase those parts that are outside of the corresponding domain. Repeat this procedure for each piece of the function. One interesting and highly instructive aspect of these functions is the opportunity to investigate restrictions on their domain and the ranges that result.

Piecewise and Continuous Functions EXAMPLE 4

䊳

Graphing a Piecewise-Defined Function Graph the function and state its domain and range: f 1x2 e

Solution

䊳

1x 32 2 12, 3,

0 6 x6 x 7 6

The first piece of f is a basic parabola, shifted three units right, reflected across the x-axis (opening downward), and shifted 12 units up. The vertex is at (3, 12) and the axis of symmetry is x 3, producing the following graphs. 1. Graph first piece of f (Figure 2.81)

2. Erase portion outside domain. of 0 6 x 6 (Figure 2.82).

Figure 2.81

Figure 2.82 y

y 12

y (x 3)2 12

12

10

10

8

8

6

6

4

4

2

2

1

1 2 3 4 5 6 7 8 9 10

x

y (x 3)2 12

1

1 2 3 4 5 6 7 8 9 10

x

The second function is simply a horizontal line through (0, 3). 3. Graph second piece of f (Figure 2.83).

4. Erase portion outside domain of x 7 6 (Figure 2.84).

Figure 2.83

Figure 2.84

y 12

y y (x 3)2 12

12

10

10

8

8

6

6

4

y3

4

2 1

f (x)

2

1 2 3 4 5 6 7 8 9 10

x

1

1 2 3 4 5 6 7 8 9 10

x

The domain of f is x 僆 10, q 2, and the corresponding range is y 僆 3 3, 124. Now try Exercises 15 through 18

䊳

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Piecewise and Discontinuous Functions Notice that although the function in Example 4 was piecewise-defined, the graph was actually continuous—we could draw the entire graph without lifting our pencil. Piecewise graphs also come in the discontinuous variety, which makes the domain and range issues all the more important.

EXAMPLE 5

䊳

Graphing a Discontinuous Piecewise-Defined Function Graph g(x) and state the domain and range: g1x2 e

Solution

䊳

12x 6, x 6 10,

0x4 4 6 x9

The first piece of g is a line, with y-intercept (0, 6) and slope 1. Graph first piece of g (Figure 2.85)

12.

¢y ¢x

2. Erase portion outside domain. of 0 x 4 (Figure 2.86).

Figure 2.85

Figure 2.86

y

y

10

10

8

8

6

6

y qx 6

4

4

2

2

1

2

3

4

5

6

7

8

9 10

y qx 6

x

1

2

3

4

5

6

7

8

9 10

x

The second is an absolute value function, shifted right 6 units, reflected across the x-axis, then shifted up 10 units. WORTHY OF NOTE As you graph piecewise-defined functions, keep in mind that they are functions and the end result must pass the vertical line test. This is especially important when we are drawing each piece as a complete graph, then erasing portions outside the effective domain.

3. Graph second piece of g (Figure 2.87).

4. Erase portion outside domain of 4 6 x 9 (Figure 2.88).

Figure 2.87

Figure 2.88

y x 6 10

y

y

10

10

8

8

6

6

4

4

2

2

1

2

3

4

5

6

7

8

9 10

x

g(x)

1

2

3

4

5

6

7

8

9 10

x

Note that the left endpoint of the absolute value portion is not included (this piece is not defined at x 4), signified by the open dot. The result is a discontinuous graph, as there is no way to draw the graph other than by “jumping” the pencil from where one piece ends to where the next begins. Using a vertical boundary line, we note the domain of g includes all values between 0 and 9 inclusive: x 僆 30, 9 4. Using a horizontal boundary line shows the smallest y-value is 4 and the largest is 10, but no range values exist between 6 and 7. The range is y 僆 34, 6 4 ´ 37, 10 4. Now try Exercises 19 and 20

䊳

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EXAMPLE 6

䊳

Graphing a Discontinuous Function The given piecewise-defined function is not continuous. Graph h(x) to see why, then comment on what could be done to make it continuous. x2 4 , h1x2 • x 2 1,

Solution

䊳

x2 x2

The first piece of h is unfamiliar to us, so we elect to graph it by plotting points, noting x 2 is outside the domain. This produces the table shown. After connecting the points, the graph turns out to be a straight line, but with no corresponding y-value for x 2. This leaves a “hole” in the graph at (2, 4), as designated by the open dot (see Figure 2.89). Figure 2.89

Figure 2.90 y

y

WORTHY OF NOTE The discontinuity illustrated here is called a removable discontinuity, as the discontinuity can be removed by redefining a single point on the function. Note that after factoring the first piece, the denominator is a factor of the numerator, and writing the result in lowest terms gives h1x2 1x x221x2 22 x 2, x 2. This is precisely the equation of the line in Figure 2.89 3y x 2 4 .

x

h(x)

4

2

2

0

0

2

2

—

4

6

5

5

5

5

x

5

5

x

5

5

The second piece is pointwise-defined, and its graph is simply the point (2, 1) shown in Figure 2.90. It’s interesting to note that while the domain of h is all real numbers (h is defined at all points), the range is y 僆 1q, 42 ´ 14, q 2 as the function never takes on the value y 4. In order for h to be continuous, we would need to redefine the second piece as y 4 when x 2. Now try Exercises 21 through 26

䊳

To develop these concepts more fully, it will help to practice finding the equation of a piecewise-defined function given its graph, a process similar to that of Example 10 in Section 2.2.

EXAMPLE 7

䊳

Determining the Equation of a Piecewise-Defined Function y

Determine the equation of the piecewise-defined function shown, including the domain for each piece.

Solution

䊳

By counting ¢y ¢x from (2, 5) to (1, 1), we find the linear portion has slope m 2, and the y-intercept must be (0, 1). The equation of the line is y 2x 1. The second piece appears to be a parabola with vertex (h, k) at (3, 5). Using this vertex with the point (1, 1) in the general form y a1x h2 2 k gives y a1x h2 k 1 a11 32 2 5 4 a122 2 4 4a 1 a 2

5

4

6

5

general form, parabola is shifted right and up substitute 1 for x, 1 for y, 3 for h, 5 for k simplify; subtract 5 122 2 4 divide by 4

x

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The equation of the parabola is y 1x 32 2 5. Considering the domains shown in the figure, the equation of this piecewise-defined function must be p1x2 e

2x 1, 1x 32 2 5,

B. You’ve just seen how we can graph functions that are piecewise-defined

2 x 6 1 x1


䊳

C. Applications of Piecewise-Defined Functions The number of applications for piecewise-defined functions is practically limitless. It is actually fairly rare for a single function to accurately model a situation over a long period of time. Laws change, spending habits change, and technology can bring abrupt alterations in many areas of our lives. To accurately model these changes often requires a piecewise-defined function.

EXAMPLE 8

䊳

Modeling with a Piecewise-Defined Function For the first half of the twentieth century, per capita spending on police protection can be modeled by S1t2 0.54t 12, where S(t) represents per capita spending on police protection in year t (1900 corresponds to year 0). After 1950, perhaps due to the growth of American cities, this spending greatly increased: S1t2 3.65t 144. Write these as a piecewise-defined function S(t), state the domain for each piece, then graph the function. According to this model, how much was spent (per capita) on police protection in 2000 and 2010? How much will be spent in 2014? Source: Data taken from the Statistical Abstract of the United States for various years.

Solution

䊳

function name

S1t2 e

function pieces

effective domain

0.54t 12, 3.65t 144,

0 t 50 t 7 50

Since both pieces are linear, we can graph each part using two points. For the first function, S102 12 and S1502 39. For the second function S1502 39 and S1802 148. The graph for each piece is shown in the figure. Evaluating S at t 100: S1t2 3.65t 144 S11002 3.6511002 144 365 144 221

S(t) 240 200 160

(80, 148)

120 80 40 0

(50, 39) 10 20 30 40 50 60 70 80 90 100 110 t (1900 → 0)

About $221 per capita was spent on police protection in the year 2000. For 2010, the model indicates that $257.50 per capita was spent: S11102 257.5. By 2014, this function projects the amount spent will grow to S11142 272.1 or $272.10 per capita. Now try Exercises 33 through 44

䊳

Step Functions The last group of piecewise-defined functions we’ll explore are the step functions, so called because the pieces of the function form a series of horizontal steps. These functions find frequent application in the way consumers are charged for services, and have several applications in number theory. Perhaps the most common is called the greatest integer function, though recently its alternative name, floor function, has gained popularity (see Figure 2.91). This is in large part due to an improvement in notation

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and as a better contrast to ceiling functions. The floor function of a real number x, denoted f 1x2 :x ; or Œ x œ (we will use the first), is the largest integer less than or equal to x. For instance, :5.9 ; 5, : 7; 7, and :3.4 ; 4. In contrast, the ceiling function C1x2 <x = is the smallest integer greater than or equal to x, meaning < 5.9 = 6, 0 (only y is positive)

QI x > 0, y > 0 (both x and y are positive)

sin is positive

All functions are positive

tan is positive

cos is positive

QIII x < 0, y < 0 (both x and y are negative)

EXAMPLE 5

䊳

QIV x > 0, y < 0 (only x is positive)

Evaluating Trig Functions for a Rotation ␪

y q

5␲ . Evaluate the six trig functions for ␪ 4

Solution

䊳

5␲ A rotation of terminates in QIII, so 4 5␲ ␲ ␲ . The associated point is ␪r 4 4 12 12 a , b since x 6 0 and y 6 0 in QIII. 2 2

x

5 4

2`

r d

√22 , √22 3 2

x

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CHAPTER 6 An Introduction to Trigonometric Functions

This yields cosa

5␲ 12 b 4 2

Noting the reciprocal of seca

C. You’ve just seen how we can define the six trig functions in terms of a point on the unit circle

sina

5␲ 12 b 4 2

tana

5␲ b1 4

12 is 12 after rationalizing, we have 2

5␲ b 12 4

csca

5␲ b 12 4

cota

5␲ b1 4


D. The Trigonometry of Real Numbers Figure 6.41 Defining the trig functions in terms of a point on the y unit circle is precisely what we needed to work with 3 s 4 them as functions of real numbers. This is because √22 , √22 when r 1 and ␪ is in radians, the length of the subtended arc is numerically the same as the measure of sr d 3 4 the angle: s 112␪ 1 s ␪! This means we can view any function of ␪ as a like function of arc 1x length s, where s 僆⺢ (see the Reinforcing Basic Concepts feature following Section 6.4). As a compromise the variable t is commonly used, with t representing either the amount of rotation or the length of the arc. As such we will assume t is a unitless quantity, although there are other reasons 3␲ for this assumption. In Figure 6.41, a rotation of ␪ is subtended by an arc length 4 ␲ 3␲ of s (about 2.356 units). The reference angle for ␪ is , which we will now 4 4 refer to as a reference arc. As you work through the remaining examples and the exercises that follow, it will often help to draw a quick sketch similar to that in Figure 6.41 to determine the quadrant of the terminal side, the reference arc, and the sign of each function.

EXAMPLE 6

䊳

Evaluating Trig Functions for a Real Number t Evaluate the six trig functions for the given value of t. 11␲ 3␲ a. t b. t 6 2

Solution

䊳

Figure 6.42 y q

11

t 6

x

tr k

√32 , 12 3 2

2`

11␲ , the arc terminates in QIV where x 7 0 and y 6 0. The reference 6 11/␲ ␲ and from our previous work we know the corresponding arc is 2␲ 6 6 13 1 , b. See Figure 6.42. This gives point (x, y) is a 2 2

a. For t

cosa

11␲ 13 b 6 2

sina

11␲ 1 b 6 2

tana

11␲ 13 b 6 3

seca

2 13 11␲ b 6 3

csca

11␲ b 2 6

cota

11␲ b 13 6

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Section 6.2 Unit Circles and the Trigonometry of Real Numbers

3␲ is a quadrantal angle and the associated point is 10, 12. 2 See Figure 6.43. This yields

Figure 6.43

b. t

y q

t 3 2`

(0, 1)

3␲ b0 2 3␲ seca b undefined 2

cosa

2

535

x

3␲ b 1 2 3␲ csca b 1 2

3␲ b undefined 2 3␲ cota b 0 2

sina

tana

3 2

Now try Exercises 41 through 44 䊳 As Example 6(b) indicates, as functions of a real number the concept of domain comes into play. From their definition it is apparent there are no restrictions on the domains of cosine and sine, but the domains of the other functions must be restricted to exclude division by zero. For functions with x in the denominator, we cast out the ␲ odd multiples of , since the x-coordinate of the related quadrantal points is zero: 2 ␲ 3␲ S 10, 12, S 10, 12, and so on. The excluded values can be stated as 2 2 ␲ t ␲k for all integers k. For functions with y in the denominator, we cast out all 2 multiples of ␲ 1t ␲k for all integers k) since the y-coordinate of these points is zero: 0 S 11, 02, ␲ S 11, 02, 2␲ S 11, 02, and so on. The Domains of the Trig Functions as Functions of a Real Number For t 僆⺢ and k 僆⺪, the domains of the trig functions are: cos t x t僆⺢

sin t y t僆⺢

1 sec t ; x 0 x ␲ t ␲k 2

1 csc t ; y 0 y t ␲k

y ;x0 x ␲ t ␲k 2 x cot t ; y 0 y t ␲k

tan t

For a given point (x, y) on the unit circle associated with the real number t, the value of each function at t can still be determined even if t is unknown. EXAMPLE 7

䊳

Finding Function Values Given a Point on the Unit Circle 24 Given 1 7 25 , 25 2 is a point on the unit circle corresponding to a real number t, find the value of all six trig functions of t.

Solution D. You’ve just seen how we can define the six trig functions in terms of a real number t

䊳

24 Using the definitions from the previous box we have cos t 7 25 , sin t 25 , and sin t 24 25 tan t cos t 7 . The values of the reciprocal functions are then sec t 7 , 25 7 csc t 24, and cot t 24 .


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E. Finding a Real Number t Whose Function Value Is Known Figure 6.44 In Example 7, we were able to determine the values of the trig functions even (0, 1) y 12 , √32 though t was unknown. In many cases, however, we need to find the value of t. √22 , √22 For instance, what is the value of t given √32 , 12 13 cos t with t in QII? Exercises 2 d u k (1, 0) of this type fall into two broad catex gories: (1) you recognize the given number as one of the special values: 1 12 13 13 , , , 13, 1 f ; e 0, , or 2 2 2 3 (2) you don’t. If you recognize a special value, you can often name the real number t after a careful consideration of the related quadrant and required sign. The diagram in Figure 6.44 reviews these special ␲ values for 0 t but remember—all other special values can be found using 2 reference arcs and the symmetry of the circle.

EXAMPLE 8

䊳

Finding t for Given Values and Conditions Find the value of t that corresponds to the given function values. 12 a. cos t b. tan t 13; t in QIII ; t in QII 2

Solution

䊳

a. The cosine function is negative in QII and QIII, where x 6 0. We recognize 12 as a standard value for sine and cosine, related to certain multiples of 2 3␲ ␲ t . In QII, we have t . 4 4 b. The tangent function is positive in QI and QIII, where x and y have like signs. We recognize 13 as a standard value for tangent and cotangent, related to ␲ 8␲ 4␲ a b. certain multiples of t . For tangent in QIII, we have t 6 3 6 Now try Exercises 61 through 84 䊳 If the given function value is not one of the special values, properties of the inverse trigonometric functions must be used to find the associated value of t. The inverse functions are fully developed in Section 7.5. Using radian measure and the unit circle is much more than a simple convenience to trigonometry and its applications. Whether the unit is 1 cm, 1 m, 1 km, or even 1 light-year, using 1 unit designations serves to simplify a great many practical applications, including those involving the arc length formula, s r␪. See Exercises 95 through 102. The following table summarizes the relationship between a special arc t (t in QI) and the value of each trig function at t. Due to the frequent use of these relationships, students are encouraged to commit them to memory.

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E. You’ve just seen how we can find the real number t corresponding to given values of sin t, cos t, and tan t

t

sin t

cos t

tan t

csc t

sec t

cot t

0

0

1

0

undefined

1

undefined

␲ 6

1 2

13 2

13 1 ⫽ 3 13

2

2 213 ⫽ 3 13

13

␲ 4

12 2

12 2

1

12

12

1

␲ 3

13 2

1 2

13

2 2 13 ⫽ 3 13

2

13 1 ⫽ 3 13

␲ 2

1

0

undefined

1

undefined

0

6.2 EXERCISES 䊳



1. A central circle is symmetric to the _________ axis, the _________ axis and to the _________.

䊳

5 2. Since 1 13 , ⫺12 13 2 is on the unit circle, the point _________ in QII is also on the circle.

3. On the unit circle, cos t ⫽ _________, sin t ⫽ 1 _________, and tan t ⫽ __; while ⫽ _______, x x 1 ⫽ _________, and ⫽ ________. y y

4. On a unit circle with ␪ in radians, the length of a(n) _________ is numerically the same as the measure of the _________, since for s ⫽ r ␪, s ⫽ ␪ when r ⫽ 1.

5. Discuss/Explain how knowing only one point on the unit circle, actually gives the location of four points. Why is this helpful to a study of the circular functions?

6. A student is asked to find t using a calculator, given sin t ⬇ 0.5592 with t in QII. The answer submitted is t ⫽ sin⫺1 0.5592 ⬇ 34°. Discuss/Explain why this answer is not correct. What is the correct response?


Given the point is on the unit circle, complete the ordered pair (x, y) for the quadrant indicated. Answer in radical form as needed.

7. 1x, ⫺0.82; QIII

8. 1⫺0.6, y2; QII

5 9. a , yb; QIV 13

8 10. ax, ⫺ b; QIV 17

11. a

111 , yb; QI 6

13. a⫺

111 , yb; QII 4

12. ax, ⫺ 14. ax,

113 b; QIII 7

16 b; QI 5

Given the point is on the unit circle, complete the ordered pair (x, y) for the quadrant indicated. Round results to four decimal places. Check your answer using 2nd TRACE (CALC) with a graph of the unit circle.

15. 1x, ⫺0.21372 ; QIII

16. (0.9909, y); QIV

17. (x, 0.1198); QII

18. (0.5449, y); QI

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Verify the point given is on the unit circle, then use symmetry to find three more points on the circle. Coordinates for Exercises 19 to 22 are exact, coordinates for Exercises 23 to 26 are approximate.

19. a 21. a

13 1 , b 2 2

111 5 , b 6 6

23. (0.3325, 0.9431)

25. 10.9937, 0.11212

20. a

17 3 , b 4 4

22. a

13 16 , b 3 3

24. 10.7707, 0.63722

26. 10.2029, 0.97922

␲ ␲ ␲ : : triangle with a hypotenuse of length 6 3 2 1 13 1 to verify that a , b is a point on the unit circle. 2 2

27. Use a

28. Use the results from Exercise 27 to find three additional points on the circle and name the quadrant of each point. Find the reference angle associated with each rotation, then find the associated point (x, y) on the unit circle.

29. ␪

5␲ 4

31. ␪

5␲ 6

30. ␪

5␲ 3

32. ␪

7␲ 4

33. ␪

11␲ 4

34. ␪

11␲ 3

35. ␪

25␲ 6

36. ␪

39␲ 4

Without the use of a calculator, state the exact value of the trig functions for the given angles. A diagram may help.

␲ 37. a. sina b 4 5␲ c. sina b 4 9␲ e. sina b 4 5␲ g. sina b 4 ␲ 38. a. tana b 3 4␲ c. tana b 3

3␲ b 4 7␲ d. sina b 4 ␲ f. sina b 4 11␲ b h. sina 4 b. sina

2␲ b 3 5␲ d. tana b 3 b. tana

␲ f. tana b 3 10␲ b h. tana 3

7␲ b 3 4␲ g. tana b 3

e. tana

39. a. cos ␲ ␲ c. cosa b 2

b. cos 0 3␲ d. cosa b 2

40. a. sin ␲ ␲ c. sina b 2

b. sin 0 3␲ d. sina b 2

Use the symmetry of the circle and reference arcs as needed to state the exact value of the trig functions for the given real numbers without the use of a calculator. A diagram may help.

␲ 41. a. cosa b 6 7␲ c. cosa b 6 13␲ b e. cosa 6 5␲ g. cosa b 6

5␲ b 6 11␲ b d. cosa 6 ␲ f. cosa b 6 23␲ b h. cosa 6

b. cosa

␲ 42. a. csca b 6 7␲ c. csca b 6 13␲ b e. csca 6 11␲ b g. csca 6

5␲ b 6 11␲ b d. csca 6 ␲ f. csca b 6 17␲ b h. csca 6 b. csca

43. a. tan ␲ ␲ c. tana b 2

b. tan 0 3␲ d. tana b 2

44. a. cot ␲ ␲ c. cota b 2

b. cot 0 3␲ d. cota b 2

Given (x, y) is a point on the unit circle corresponding to t, find the value of all six trig functions of t.

45.

46.

y

y

(0.8, 0.6) t

(1, 0) x

t

(1, 0) x

15 , 8 17 17

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47.

Without using a calculator, find the value of t in [0, 2␲) that corresponds to the following functions.

y

t

61. sin t

(1, 0) x

13 ; t in QII 2

1 62. cos t ; t in QIV 2

5 , 12 13 13

48.

y

63. cos t

5 , √11 6 6

1 64. sin t ; t in QIV 2

(1, 0)

t

23 ; t in QIII 2

x

65. tan t 13; t in QII 2 121 b 49. a , 5 5

17 3 , b 50. a 4 4

1 212 b 51. a , 3 3

2 16 1 , b 52. a 5 5

On the unit circle, the real number t can represent either the amount of rotation or the length of the arc when we associate t with a point (x, y) on the circle. In the circle diagram shown, the real number t in radians is marked off along the circumference. For Exercises 53 through 60, name the quadrant in which t terminates and use the figure to estimate function values to one decimal place (use a straightedge). Check results using a calculator. Exercises 53 to 60 q

2.0

y

1.5 1.0

2.5 0.5 3.0

0 x

1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1

6.0 3.5 5.5 4.0 4.5 3 2

5.0

66. sec t 2; t in QIII 67. sin t 1 68. cos t 1 Without using a calculator, find the two values of t (where possible) in [0, 2␲) that make each equation true.

2 13

69. sec t 12

70. csc t

71. tan t undefined

72. csc t undefined

73. cos t 75. sin t 0

1 35,

12 2

45 2

74. sin t

12 2

76. cos t 1

77. Given is a point on the unit circle that corresponds to t. Find the coordinates of the point corresponding to (a) t and (b) t ␲. 7 24 78. Given 125 , 25 2 is a point on the unit circle that corresponds to t. Find the coordinates of the point corresponding to (a) t ␲ and (b) t ␲.

Find an additional value of t in [0, 2␲) that makes the equation true.

79. sin 0.8 0.7174 80. cos 2.12 0.5220 81. cos 4.5 0.2108

53. sin 0.75

54. cos 2.75

82. sin 5.23 0.8690

55. cos 5.5

56. sin 4.0

83. tan 0.4 0.4228

57. tan 0.8

58. sec 3.75

84. sec 5.7 1.1980

59. csc 2.0

60. cot 0.5

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85. From Pythagorean triples to points on the unit x y circle: 1x, y, r2 S a , , 1b r r While not strictly a “formula,” dividing a Pythagorean triple by r is a simple algorithm for rewriting any Pythagorean triple as a triple with hypotenuse 1. This enables us to identify certain points on the unit circle, and to evaluate the six trig functions of the related acute angle. Rewrite each x y triple as a triple with hypotenuse 1, verify a , b is r r a point on the unit circle, and evaluate the six trig functions using this point. a. (5, 12, 13) c. (12, 35, 37)

䊳

6–32


b. (7, 24, 25) d. (9, 40, 41)

␲ 86. The sine and cosine of 12k ⴙ 12 ; k 僆⺪ 4 In the solution to Example 8(a), we mentioned 12 were standard values for sine and cosine, 2 ␲ “related to certain multiples of .” Actually, we 4 ␲ meant “odd multiples of .” The odd multiples of 4 ␲ are given by the “formula” shown, where k is 4 ␲ any integer. (a) What multiples of are generated 4 by k 3, 2, 1, 0, 1, 2, 3? (b) Find similar formulas for Example 8(b), where 13 is a standard value for tangent and cotangent, “related to certain ␲ multiples of .” 6

APPLICATIONS

Unit circle points: In Exercises 23 through 26, four decimal approximations of unit circle points were given. Find such unit circle points that are on the terminal side of the following angles in standard position. (Hint: Use the definitions of the trig functions.)

87. ␪ 2

88. ␪ 1

89. ␪ 5

90. ␪ 4

Mosaic design: The floor of a local museum contains a mosaic zodiac circle as shown. The circle is 1 m ( 100 cm) in radius, with all 12 zodiac signs represented by small, evenly spaced bronze disks on the circumference. Exercises 91 to 94

91. If the disk representing Libra is 96.6 cm to the right of and 25.9 cm above the center of the zodiac, what is the distance d (see illustration) from Libra’s disk to Pisces’ disk? What is the distance from Libra’s disk to Virgo’s disk?

P

S

US RI

CO I RP

O

AQ UA

CA

SAGITA RIU S

CORN PRI

RA

PISCE S

LIB

d

VIRG

L

S RU

O

I ES AR

92. If the disk representing Taurus is 70.7 cm to the left of and 70.7 cm below the center of the zodiac, what is the distance from Taurus’ disk to Leo’s disk? What is the distance from Taurus’ disk to Scorpio’s disk? 93. Using a special triangle, identify (in relation to the center of the zodiac) the location of the point P (see illustration) where the steel rod separating Scorpio and Sagittarius intersects the circumference. Round to the nearest tenth of a centimeter. 94. Using a special triangle, identify (in relation to the center of the zodiac) the location of the point where the steel rod separating Aquarius and Pisces intersects the circumference. Round to the nearest tenth of a centimeter.

EO

TA U CER CAN

GEMI

NI

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95. Laying new sod: When new sod is laid, a heavy roller is used to press the sod down to ensure good contact with the ground beneath. The 1 ft radius of the roller is 1 ft. (a) Through what angle (in radians) has the roller turned after being pulled across 5 ft of yard? (b) What angle must the roller turn through to press a length of 30 ft? 96. Cable winch: A large winch with a radius of 1 ft winds in 3 ft of cable. (a) Through what angle (in radians) has it turned? (b) What angle must it turn through in order to winch in 12.5 ft of cable?

541


Exercise 96

97. Wiring an apartment: In the wiring of an apartment complex, electrical wire is being pulled from a spool with radius 1 decimeter (1 dm ⫽ 10 cm). (a) What length (in decimeters) is removed as the spool turns through 5 rad? (b) How many decimeters are removed in one complete turn 1t ⫽ 2␲2 of the spool? 98. Barrel races: In the barrel races popular at some family reunions, contestants stand on a hard rubber barrel with a radius of 1 cubit (1 cubit ⫽ 18 in.), and try to “walk the barrel” from the start line to the finish line without falling. (a) What distance (in cubits) is traveled as the barrel is walked through an angle of 4.5 rad? (b) If the race is 25 cubits long, through what angle will the winning barrel walker walk the barrel?

Interplanetary measurement: In the year 1905, astronomers began using astronomical units or AU to study the distances between the celestial bodies of our solar system. One AU represents the average distance between the Earth and the Sun, which is about 93 million miles. Pluto is roughly 39.24 AU from the Sun. Exercise 102 102. Verifying s ⴝ r␪: On a protractor, carefully measure the distance from the middle of the protractor’s eye to the edge of the eye protractor along the 1 unit 0° mark, to the nearest half-millimeter. Call this length “1 unit.” Then use a ruler to draw a straight line on a blank sheet of paper, and with the protractor on edge, start the zero degree mark at one end of the line, carefully roll the protractor until it reaches 1 radian 157.3°2 , and mark this spot. Now measure the length of the line segment created. Is it very close to 1 “unit” long? 20 160

100 80

110 70

12 60 0

13 50 0

10 170

90 90

0 180

80 100

170 10

180 0

101. Compact disk circumference: A standard compact disk has a radius of 6 cm. Call this length “1 unit.” Mark a starting point on any large surface, then carefully roll the compact disk along this line without slippage, through one full revolution (2␲ rad) and mark this spot. Take an accurate measurement of the resulting line segment. Is the result close to 2␲ “units” (2␲ ⫻ 6 cm)?

70 110

160 20

100. If you include the dwarf planet Pluto, Jupiter is the middle (fifth of nine) planet from the Sun. Suppose astronomers had decided to use its average distance from the Sun as 1 AU. In this case, 1 AU would be 480 million miles. If Jupiter travels through an angle of 4 rad about the Sun, (a) what distance in the “new” astronomical units (AU) has it traveled? (b) How many of the new AU does it take to complete one-half an orbit about the Sun? (c) What distance in the new AU is the dwarf planet Pluto from the Sun?

60 0 12

1500 3

3 1500

4 14 0 0

50 0 13

0 14 0 4

99. If the Earth travels through an angle of 2.5 rad about the Sun, (a) what distance in astronomical units (AU) has it traveled? (b) How many AU does it take for one complete orbit around the Sun?

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103. In this section, we discussed the domain of the circular functions, but said very little about their range. Review the concepts presented here and determine the range of y cos t and y sin t. In other words, what are the smallest and largest output values we can expect?

Use the radian grid given with Exercises 53–60 to answer Exercises 105 and 106.

105. Given cos12t2 0.6 with the terminal side of the arc in QII, (a) what is the value of 2t? (b) What quadrant is t in? (c) What is the value of cos t? (d) Does cos12t2 2cos t?

sin t , what can you say about the cos t range of the tangent function?

104. Since tan t

䊳

6–34


106. Given sin12t2 0.8 with the terminal side of the arc in QIII, (a) what is the value of 2t? (b) What quadrant is t in? (c) What is the value of sin t? (d) Does sin12t2 2sin t?


107. (1.3) Given the points (3, 4) and (5, 2) find a. the distance between them b. the midpoint between them c. the slope of the line through them.

109. (2.3, R.6) Solve each equation: a. 2x 1 3 7 b. 2 1x 1 3 7 110. (4.2) Use the rational zeroes theorem to solve the equation completely, given x 3 is one root.

108. (5.4) Use a calculator to find the value of each expression, then explain the results. a. log 2 log 5 ______ b. log 20 log 2 ______

6.3

x4 x3 3x2 3x 18 0

Graphs of the Sine and Cosine Functions


A. Graph f (t) sin t using special values and symmetry B. Graph f (t) cos t using special values and symmetry C. Graph sine and cosine functions with various amplitudes and periods D. Write the equation for a given graph

As with the graphs of other functions, trigonometric graphs contribute a great deal toward the understanding of each function and its applications. For now, our primary interest is the general shape of each basic graph and some of the transformations that can be applied. We will also learn to analyze each graph, and to capitalize on the features that enable us to apply the functions as real-world models.

A. Graphing f (t) ⴝ sin t Consider the following table of values (Table 6.1) for sin t and the special angles in QI. Table 6.1 t

0

␲ 6

␲ 4

␲ 3

␲ 2

sin t

0

1 2

12 2

13 2

1

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Section 6.3 Graphs of the Sine and Cosine Functions

␲ to ␲ 2 (QII), special values taken from the unit circle show sine values are decreasing from 1 to 0, but through the same output values as in QI. See Figures 6.45 through 6.47. Observe that in this interval, sine values are increasing from 0 to 1. From

Figure 6.45

冢 12 , √32 冣

y (0, 1)

Figure 6.46

Figure 6.47

y (0, 1)

y (0, 1)

冢√22 , √22 冣

冢√32 , 12 冣 3␲ 4

2␲ 3

(1, 0) x

(1, 0)

sin a

5␲ 6

(1, 0) x

(1, 0)

23 2␲ b 3 2

sin a

(1, 0) x

(1, 0)

sin a

22 3␲ b 4 2

5␲ 1 b 6 2

With this information we can extend our table of values through ␲, noting that sin ␲ 0 (see Table 6.2). Table 6.2 t

0

␲ 6

␲ 4

␲ 3

␲ 2

2␲ 3

3␲ 4

5␲ 6

␲

sin t

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

Using the symmetry of the circle and the fact that y is negative in QIII and QIV, we can complete the table for values between ␲ and 2␲. EXAMPLE 1

䊳

Finding Function Values Using Symmetry Use the symmetry of the unit circle and reference arcs of special values to complete Table 6.3. Recall that y is negative in QIII and QIV. Table 6.3 t

␲

7␲ 6

5␲ 4

4␲ 3

3␲ 2

5␲ 3

7␲ 4

11␲ 6

2␲

sin t

Solution

䊳

12 ␲ depending on , sin t 4 2 1 ␲ the quadrant of the terminal side. Similarly, for any reference arc of , sin t , 6 2 ␲ 13 while any reference arc of will give sin t . The completed table is shown 3 2 in Table 6.4. Symmetry shows that for any odd multiple of t

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Table 6.4 t

␲

7␲ 6

sin t

0

⫺

1 2

5␲ 4 ⫺

4␲ 3

12 2

⫺

3␲ 2

13 2

5␲ 3 ⫺

⫺1

11␲ 6

7␲ 4

13 2

⫺

12 2

⫺

1 2

2␲ 0


䊳

13 1 12 ⫽ 0.5, 0.71, and 0.87, we plot these points and 2 2 2 connect them with a smooth curve to graph y ⫽ sin t in the interval 30, 2␲4. The first five plotted points are labeled in Figure 6.48. Noting that

Figure 6.48

␲, 6

0.5

␲4 , 0.71

␲3 , 0.87

sin t

␲2 , 1

1

ng asi cre De

rea

si

ng

0.5

Inc

544

(0, 0)

␲ 2

␲

3␲ 2

2␲

t

⫺0.5 ⫺1

These approximate values can be quickly generated using the table feature of a graphing calculator set in radian MODE . Begin by noticing the least common ␲ ␲ ␲ ␲ denominator of the standard values , , , and is 12. After entering Y1 ⫽ sin X, 6 4 3 2 we use this observation and the keystrokes 2nd (TBLSET) to set the table as shown in Figure 6.49. Note after pressing , the calculator automatically replaces ␲ the exact value with a decimal approximation (0.261...). Pressing the keys 2nd GRAPH 12 (TABLE) results in the table shown in Figure 6.50, and we quickly recognize the Figure 6.48 decimal approximations in the Y1 column. Exercises 9 and 10 explore the two additional values that appear in this column. WINDOW

ENTER

Figure 6.49

Figure 6.50

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Expanding the table from 2␲ to 4␲ using reference arcs and the unit circle 13␲ ␲ b sina b since shows that function values begin to repeat. For example, sina 6 6 ␲ 9␲ ␲ ␲ ␪r , sina b sina b since ␪r , and so on. Functions that cycle through 6 4 4 4 a set pattern of values are said to be periodic functions. Using the down arrow to scroll through the table in Figure 6.50, while observing the graph in Figure 6.48, helps highlight these concepts. Periodic Functions A function f is said to be periodic if there is a positive number P such that f 1t P2 f 1t2

for all t in the domain. The smallest number P for which this occurs is called the period of f. For the sine function note sin t sin1t 2␲2, as in sina

␲ 13␲ b sina 2␲b 6 6

9␲ ␲ b sina 2␲b, with the idea extending to all other real numbers t: 4 4 sin t sin 1t 2␲k2 for all integers k. The sine function is periodic with period P 2␲. Although we initially focused on positive values of t in 3 0, 2␲4, t 6 0 and k 6 0 are certainly possibilities and we note the graph of f 1t2 sin t extends infinitely in both directions (see Figure 6.51). and sina

Figure 6.51 ␲

2 , 1

f 1

f(t) sin t

0.5

␲

4␲ 3

␲ 3

2␲ 3

␲

2 , 1

␲ 3

0.5

␲

2␲ 3

4␲ 3

t

1

To see even more of this graph, we can use a graphing calculator and a preset ZOOM option that automatically sets a window size convenient to many trigonometric graphs. The resulting after pressing ZOOM 7:ZTrig is shown in Figure 6.52 for a calculator MODE set in radian . Pressing GRAPH with Y1 sin X displays the graph of f 1t2 sin t in this predefined window (see Figure 6.53). WINDOW

Figure 6.53

Figure 6.52

4

2

2

4

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Finally, the table, graph, and unit circle all confirm that the range of f 1t2 sin t is 3 1, 14 , and that f 1t2 sin t is an odd function. In particular, the graph ␲ ␲ shows sina b sina b, and the unit circle 2 2 shows sin t y, and sin1t2 y (see Figure 6.54), from which we obtain sin1t2 sin t by substitution. As a handy reference, the following box summarizes the main characteristics of f 1t2 sin t.

Figure 6.54 y (0, 1)

y sin t ( x, y) t

(1, 0)

(1, 0)

x

t (0, 1)

(x, y)

Characteristics of f(t) ⴝ sin t For all real numbers t and integers k, Domain 1q, q 2

Range 3 1, 14

Period 2␲

Symmetry odd

Maximum value sin t 1 ␲ at t 2␲k 2

Minimum value sin t 1 3␲ 2␲k at t 2

Decreasing ␲ 3␲ a , b 2 2

Zeroes

sin1t2 sin t Increasing ␲ 3␲ a0, b ´ a , 2␲b 2 2

EXAMPLE 2

䊳

Using the Period of sin t to Find Function Values

Use the characteristics of f 1t2 sin t to match the given value of t to the correct value of sin t. ␲ ␲ 17␲ 11␲ a. t 8␲ b. t c. t d. t 21␲ e. t 4 6 2 2 I. sin t 1

Solution

䊳

t k␲

II. sin t

1 2

III. sin t 1

IV. sin t

12 2

V. sin t 0

␲ ␲ 8␲b sin , the correct match is (IV). 4 4 ␲ ␲ Since sina b sin , the correct match is (II). 6 6 ␲ ␲ 17␲ Since sina b sina 8␲b sin , the correct match is (I). 2 2 2 Since sin 121␲2 sin 1␲ 20␲2 sin ␲, the correct match is (V). 3␲ 3␲ 11␲ Since sina b sina 4␲b sina b, the correct match is (III). 2 2 2

a. Since sina b. c. d. e.


䊳

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Many of the transformations applied to algebraic graphs can also be applied to trigonometric graphs. These transformations may stretch, reflect, or translate the graph, but it will still retain its basic shape. In numerous applications, it will help if you’re able to draw a quick, accurate sketch of the transformations involving f 1t2 sin t. To assist this effort, we’ll begin with the interval 30, 2␲4 , combine the characteristics just listed with some simple geometry, and offer the following four-step process. Step I:

Draw the y-axis, mark zero halfway up, with 1 and 1 an equal distance from this zero. Then draw an extended t-axis and tick mark 2␲ to the extreme right (Figure 6.55).

Figure 6.55 y 1

0 2

t

3 2

2

t

3 2

2

t

1

Step II:

On the t-axis, mark halfway between 0 and 2␲ and label it “␲,” mark halfway between ␲ on either side ␲ 3␲ . and label the marks and 2 2 Halfway between these you can draw additional tick marks to repre␲ sent the remaining multiples of 4 (Figure 6.56).

Step III: Next, lightly draw a rectangular frame, which we’ll call the reference rectangle, P 2␲ units wide and 2 units tall, centered on the t-axis and with the y-axis along the left side (Figure 6.57).

Figure 6.56 y 1

0 2

1

Figure 6.57 y 1

0 2

1

Figure 6.58 Step IV: Knowing y sin t is positive and y increasing in QI, that the range is 1 3 1, 14, that the zeroes are 0, ␲, and 2␲, and that maximum and minimum values occur halfway between 0 3 the zeroes (since there is no horizon2 t 2 2 tal shift), we can draw a reliable graph of y sin t by partitioning the 1 rectangle into four equal parts to locate these values (note bold tick-marks). We will call this partitioning of the reference rectangle the rule of fourths, since we are then scaling P the t-axis in increments of (Figure 6.58). 4

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EXAMPLE 3

䊳

Graphing y ⴝ sin t Using a Reference Rectangle ␲ 3␲ d. Use steps I through IV to draw a sketch of y sin t for the interval c , 2 2

Solution

䊳

Start by completing steps I and II, then extend the t-axis to include 2␲. Beginning at t 0, draw a reference rectangle 2␲ units wide and 2 units tall, centered on the t-axis. After applying the rule of fourths, we note the zeroes occur at t 0, t ␲, and t 2␲, while the max/min values fall halfway between them ␲ 3␲ at t and t (see Figure 6.59). Plot these points and connect them with a 2 2 smooth, dashed curve. This is the primary period of the sine curve. Figure 6.59 y 1

2

3

2

2

2

3 2

2

t

1

Using the periodic nature of the sine function, we can also graph the sine curve on the interval 32␲, 0 4 , as shown in Figure 6.60. The rule of fourths again helps to locate the zeroes and max/min values (note the bold tick-marks) over this interval. Figure 6.60 y 1

2

3

2

2

2

3 2

2

t

1

␲ 3␲ d , we simply highlight the graph in this For the graph of y sin t in c , 2 2 interval using a solid curve, as shown in Figure 6.61. Figure 6.61

WORTHY OF NOTE

y

␲ 3␲ d is Since the interval c , 2 2 P 2␲ units wide, this section of the graph can generate the entire graph of y sin t.

1

y sin t 2

3

2

2

2

3 2

2

t

1


䊳

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Figure 6.62

We can verify the results of Example 3 by changing the settings from the ␲ predefined ZOOM 7:ZTrig to Xmin , 2 3␲ Xmax , Ymin 2, and Ymax 2. 2 The GRAPH of Y1 sin X with these window settings is shown in Figure 6.62.

2

WINDOW

A. You’ve just seen how we can graph f (t) ⴝ sin t using special values and symmetry

2

3 2

2

B. Graphing f(t) ⴝ cos t

With the graph of f 1t2 sin t established, sketching the graph of f 1t2 cos t is a very natural next step. First, note that when t 0, cos t 1 so the graph of y cos t 1 13 b, will begin at 10, 12 in the interval 3 0, 2␲ 4. Second, we’ve seen a , 2 2 12 13 1 12 a , b and a , b are all points on the unit circle since they satisfy 2 2 2 2 x2 y2 1. Since cos t x and sin t y, the Pythagorean identity 1 cos2t sin2t 1 can be obtained by direct substitution. This means if sin t , 2 13 then cos t and vice versa, with the signs taken from the appropriate quadrant. 2 Similarly, if sin t 0, then cos t 1 and vice versa. The table of values for cosine then becomes a simple variation of the table for sine, as shown in Table 6.5 for t 僆 30, ␲ 4. Table 6.5 t

0

␲ 6

␲ 4

␲ 3

␲ 2

2␲ 3

3␲ 4

5␲ 6

␲

sin t

0

1 0.5 2

12 0.71 2

13 0.87 2

1

13 0.87 2

12 0.71 2

1 0.5 2

0

cos t

1

13 ⬇ 0.87 2

12 ⬇ 0.71 2

1 ⴝ 0.5 2

0

1 ⴚ ⴝ ⴚ0.5 2

ⴚ

13 ⬇ ⴚ0.87 2

ⴚ1

The same values can be taken from the unit circle, but this view requires much less effort and easily extends to values of t in 3 ␲, 2␲ 4 . Using the points from Table 6.5 and its extension through 3␲, 2␲ 4 , we can draw the graph of y cos t in 30, 2␲ 4 and identify where the function is increasing and decreasing in this interval. See Figure 6.63. Figure 6.63 cos t 1

D

2

0 0.5

1

g sin rea ec

0.5

6 , 0.87 4 , 0.71 3 , 0.5 2 , 0 2

ng

As with the sine function, a graphing calculator can quickly produce a table of points on the graph of the cosine function. With the independent variable in the table set to Ask mode, we generate the following special values, which also appear in Table 6.5 and Figure 6.63.

12 ⬇ ⴚ0.71 2

Inc rea si

WORTHY OF NOTE

ⴚ

3 2

2

t

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The function is decreasing for t in 10, ␲2, and increasing for t in 1␲, 2␲2. The end ␲ result appears to be the graph of y sin t shifted to the left units, a fact more easily 2 ␲ seen if we extend the graph to as shown. This is in fact the case, and 2 is a relationship we will later prove in Chapter 7. Like y sin t, the function y cos t is periodic with period P 2␲, with the graph extending infinitely in both directions. Finally, we note that cosine is an even function, meaning cos1t2 cos t for all ␲ ␲ t in the domain. For instance, cos a b cos a b 0 (see Figure 6.63). Here is a 2 2 summary of important characteristics of the cosine function. Characteristics of f (t) ⴝ cos t For all real numbers t and integers k,

EXAMPLE 4

䊳

Domain 1q, q 2

Range 3 1, 14

Period 2␲

Symmetry even cos1t2 cos t

Maximum value cos t 1 at t 2␲k

Minimum value cos t 1 at t ␲ 2␲k

Increasing

Decreasing

1␲, 2␲2

10, ␲2

Zeroes ␲ t ␲k 2

Graphing y ⴝ cos t Using a Reference Rectangle ␲ 3␲ d and use a graphing calculator to Draw a sketch of y cos t for t in c , 2 2 check your graph.

Solution

䊳

As with the graph of y sin t, begin by completing steps I and II, then extend the t-axis to include 2␲. Beginning at t 0, draw a reference rectangle 2␲ units wide and 2 units tall, centered on the t-axis. After applying the rule of fourths, we ␲ 3␲ note the zeroes occur at t and t with the max/min values at t 0, t ␲, 2 2 and t 2␲. Plot these points and connect them with a smooth, dashed curve (see Figure 6.64). This is the primary period of the cosine curve. Figure 6.64 y 1

2

3

2

2

2

1

3 2

2

t

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Using the periodic nature of the cosine function, we can also graph the cosine curve on the interval 3 ⫺2␲, 04 , as shown in Figure 6.65. The rule of fourths again helps to locate the zeroes and max/min values (note the bold tick-marks) over this interval.

WORTHY OF NOTE We also could have graphed the cosine curve in this interval using the table

t

cos t

␲ ⫺ 2

0

0

1

␲ 2

0

␲

⫺1

3␲ 2

0

Figure 6.65 y 1

2␲

3␲

2

␲

␲

␲

␲ 2

2

3␲ 2

2␲

t

1

␲ 3␲ d , we simply highlight the graph in this For the graph of y ⫽ cos t in c⫺ , 2 2 interval using a solid curve, as shown in Figure 6.66.

and connecting these points with a smooth curve.

Figure 6.66 y 1

y cos t

2␲

3␲

2

␲

␲

␲

␲ 2

2

3␲ 2

2␲

t

1

Check

䊳

With the

WINDOW

settings shown, the graphs are identical.

2

B. You’ve just seen how we can graph f (t) ⴝ cos t using special values and symmetry

␲ 2

3␲ 2

2


䊳

C. Graphing y ⴝ A sin(Bt) and y ⴝ A cos(Bt)

WORTHY OF NOTE Because the functions sin t and cos t take on both positive and negative values, we can use the graphs of y ⫽ 冟 A sin t 冟 and y ⫽ 冟 A cos t 冟 to emphasize the role of amplitude 冟 A 冟 as the maximum displacement. See Exercises 17 through 20.

In many applications, trig functions have maximum and minimum values other than 1 and ⫺1, and periods other than 2␲. For instance, in tropical regions the daily maximum and minimum temperatures may vary by no more than 20°, while for desert regions this difference may be 40° or more. This variation is modeled by the amplitude of the sine and cosine functions.

Amplitude and the Coefficient A (assume B ⴝ 1) For functions of the form y ⫽ A sin t and y ⫽ A cos t, let M represent the Maximum M⫹m value and m the minimum value of the functions. Then the quantity gives the 2

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Mm gives the amplitude of the function. 2 Amplitude is the maximum displacement from the average value in the positive or negative direction. It is represented by A, with A playing a role similar to that seen for algebraic graphs 3 Af 1t2 vertically stretches or compresses the graph of f, and reflects it across the t-axis if A 6 0 4. Graphs of the form y sin t (and y cos t) can quickly be sketched with any amplitude by noting (1) the zeroes of the function remain fixed since sin t 0 implies A sin t 0, and (2) the maximum and minimum values are A and A, respectively, since sin t 1 or 1 implies A sin t A or A. Note this implies the reference rectangle will be 2A units tall and P units wide. Connecting the points that result with a smooth curve will complete the graph. average value of the function, while

EXAMPLE 5

䊳

Graphing y ⴝ A sin t Where A ⴝ 1

Solution

䊳

With an amplitude of A 4, the reference rectangle will be 2142 8 units tall, by 2␲ units wide. Using the rule of fourths, the zeroes are still t 0, t ␲, and t 2␲, with the max/min values spaced equally between. The maximum value is 3␲ ␲ 4 sina b 4112 4, with a minimum value of 4 sina b 4112 4. 2 2 Connecting these points with a “sine curve” gives the graph shown 1y sin t is also shown for comparison).

Draw a sketch of y 4 sin t in the interval 3 0, 2␲ 4.

4

y 4 sin t Zeroes remain fixed 2

y sin t

3 2

2

t

4


䊳

Figure 6.67 In the graph for Example 5, we note the 4 zeroes of y A sin t remained fixed for A 1 and A 4. Additionally, it appears the max/min values occur at the same t-values. We can use a graphing calculator to 2 further investigate these observations. On the 2 1 Y= screen, enter Y1 sin X, Y2 sin X, 2 Y3 2 sin X, and Y4 4 sin X, then use 4 ZOOM 7:ZTrig to graph the functions (note Y 2 and Y4 were graphed in Example 5). As you see in Figure 6.67, each graph “holds on” to the zeroes, while rising to the expected amplitude A halfway between these zeroes.

Period and the Coefficient B While basic sine and cosine functions have a period of 2␲, in many applications the period may be very long (tsunamis) or very short (electromagnetic waves). For the equations y A sin1Bt2 and y A cos1Bt2, the period depends on the value of B. To see why, consider the function y cos12t2 and Table 6.6. Multiplying input values by 2 means each cycle will be completed twice as fast. The table shows that y cos12t2

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completes a full cycle in 30, ␲ 4, giving a period of P ␲. Figure 6.68 verifies these observations, with the graphs of Y1 cos1X2 and Y2 cos12X2 displayed over the interval 30, 2␲ 4 . Figure 6.68

Table 6.6

2

t

0

␲ 4

␲ 2

3␲ 4

␲

2t

0

␲ 2

␲

3␲ 2

2␲

cos (2t)

1

0

1

0

1

2

0

2

Dividing input values by 2 (or multiplying by 12 2 will cause the function to complete a cycle only half as fast, doubling the time required to complete a full cycle. Table 6.7 shows y cos1 12 t2 completes only one-half cycle in 2␲. Figure 6.69 shows the graphs of Y1 cos1X2 and Y2 cos1 11/22X2 . Figure 6.69

Table 6.7

2

(values in blue are approximate)

2

0

t

0

␲ 4

␲ 2

3␲ 4

␲

5␲ 4

3␲ 2

7␲ 4

2␲

1 t 2

0

␲ 8

␲ 4

3␲ 8

␲ 2

5␲ 8

3␲ 4

7␲ 8

␲

1 cos a tb 2

1

0.92

12 2

0.38

0

0.38

0.92

1

12 2

2

Figure 6.70 The graphs of y cos t, y cos12t2, 1 y and y cos1 2 t2 shown in Figure 6.70 y cos(2t) y cos t 1 clearly illustrate this relationship and how the value of B affects the period of a graph. To find the period for arbitrary values of 2 3 2␲ B, the formula P is used. Note for B 1 2␲ y cos12t2, B 2 and P ␲, as 2 2␲ 1 1 4␲. shown. For y cos a tb, B , and P 2 2 1/2

y cos 12 t

4

t

Period Formula for Sine and Cosine For B a real number and functions y A sin1Bt2 and y A cos1Bt2, P

2␲ . B

To sketch these functions for periods other than 2␲, we still use a reference rectangle of height 2A and length P, then break the enclosed t-axis into four equal parts to help draw the graph. In general, if the period is “very large” one full cycle is appropriate for the graph. If the period is “very small,” graph at least two cycles. Note the value of B in Example 6 includes a factor of ␲. This actually happens quite frequently in applications of the trig functions.

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EXAMPLE 6

䊳

Graphing y ⴝ A cos(Bt), Where A, B ⴝ 1

Solution

䊳

The amplitude is A 2, so the reference rectangle will be 2122 4 units high. Since A 6 0, the graph will be vertically reflected across the t-axis. The period is 2␲ P 5 (note the factors of ␲ reduce to 1), so the reference rectangle will 0.4␲ be 5 units in length. Breaking the t-axis into four parts within the frame (rule of fourths) gives 1 14 25 54 units, indicating that we should scale the t-axis in multiples 10 of 14. Note the zeroes occur at 54 and 15 4 , with a maximum value at 4 . In cases where the ␲ factor reduces, we scale the t-axis as a “standard” number line, and estimate the location of multiples of ␲. For practical reasons, we first draw the unreflected graph (shown in blue) for guidance in drawing the reflected graph, which is then extended to fit the given interval.

Draw a sketch of y 2 cos10.4␲t2 for t in 3␲, 2␲ 4 .

y y 2cos(0.4t)

2

3

2

1

2

1

1

2

3

4

5

6

t

1 2

y 2 cos(0.4t)


C. You’ve just seen how we can graph sine and cosine functions with various amplitudes and periods

Figure 6.71 While graphing calculators can quickly 2 and easily graph functions over a given interval, understanding the analytical solution presented remains essential. A true test of effective calculator use comes when the amplitude or period is 1 a very large or a very small number. For in- 1 stance, the tone you hear while pressing “5” on your telephone is actually a combination of the tones modeled by Y1 sin 32␲17702X4 and 2 Y2 sin 3 2␲113362X4 . Graphing these functions requires a careful analysis of the period, Figure 6.72 otherwise the graph can appear garbled, mis2 leading, or difficult to read (try graphing Y1 on the ZOOM 7:ZTrig screen). Firstnote for Y1, 1 2␲ A 1 and P or . With a period 1 1 2␲17702 770 770 this short, even graphing the function from 770 Xmin 1 to Xmax 1 gives a distorted graph (see Figure 6.71). Because of the calculated period of this function, we set Xmin 2 to 1/770, Xmax to 1/770 and Xscl to (1/770)/10. This gives the graph in Figure 6.72, which can then be used to investigate characteristics of the function. See Exercises 39 and 40.

D. Writing Equations from Graphs Mathematical concepts are best reinforced by working with them in both “forward and reverse.” Where graphs are concerned, this means we should attempt to find the equation of a given graph, rather than only using an equation to sketch the graph. Exercises

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of this type require that you become very familiar with the graph’s basic characteristics and how each is expressed as part of the equation. EXAMPLE 7

䊳

Determining the Equation of a Given Graph The graph shown here is of the form y A sin1Bt2. Find the values of A and B. y 2

y A sin(Bt)

⫺

␲ 2

␲ 2

␲

3␲ 2

2␲

t

2

Solution

䊳

3␲ By inspection, the graph has an amplitude of A 2 and a period of P . 2 2␲ 3␲ , substituting To find B we used the period formula P for P and solving. 冟 B冟 2 2␲ 冟B冟 3␲ 2␲ 2 B 3␲B 4␲ 4 B 3 P

D. You’ve just seen how we can write the equation for a given graph

period formula

substitute

3␲ for P; B 7 0 2

multiply by 2B solve for B

The result is B 43, which gives us the equation y 2 sin1 43t2. Now try Exercises 41 through 60

䊳

There are a number of interesting applications of this “graph to equation” process in the exercise set. See Exercises 63 to 74.

6.3 EXERCISES 䊳



1. For the sine function, output values are ________ ␲ in the interval c 0, d . 2

2. For the cosine function, output values are ________ ␲ in the interval c 0, d . 2

3. For the sine and cosine functions, the domain is ________ and the range is ________.

4. The amplitude of sine and cosine is defined to be the maximum ________ from the ________ value in the positive and negative directions.

5. Discuss/Explain how the values generated by the unit circle can be used to graph the function f 1t2 sin t. Be sure to include the domain and range of this function in your discussion.

6. Discuss/Describe the four-step process outlined in this section for the graphing of basic trig functions. Include a worked-out example and a detailed explanation.

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DEVELOPING YOUR SKILLS For the following, (a) simplify the given values, (b) use a calculator to evaluate the sine of any nonstandard value, and (c) compare your results to Figure 6.50 on page 544.

7. Use the symmetry of the unit circle and reference arcs of standard values to complete a table of values for y ⫽ cos t in the interval t 僆 3 ␲, 2␲ 4 .

8. Use the standard values for y ⫽ cos t for t 僆 3 ␲, 2␲4 to create a table of values for y ⫽ sin t on the same interval.

9.

1␲ 2␲ 3␲ , , 12 12 12

10.

4␲ 5␲ 6␲ , , 12 12 12

Use the characteristics of f 1t2 ⴝ sin t to match the given value of t to the correct value of sin t.

11. a. t ⫽

␲ ⫹ 10␲ 6

b. t ⫽ ⫺

I. sin t ⫽ 0 12. a. t ⫽

II. sin t ⫽

␲ ⫺ 12␲ 4

I. sin t ⫽ ⫺

␲ 4

b. t ⫽

1 2

c. t ⫽

1 2

III. sin t ⫽ 1

11␲ 6

II. sin t ⫽ ⫺

⫺15␲ 4

c. t ⫽ 12 2

23␲ 2

III. sin t ⫽ 0

d. t ⫽ 13␲ IV. sin t ⫽

12 2

d. t ⫽ ⫺19␲ IV. sin t ⫽

12 2

e. t ⫽

21␲ 2

V. sin t ⫽ ⫺ e. t ⫽ ⫺

12 2

25␲ 4

V. sin t ⫽ ⫺1

Use steps I through IV given in this section to draw a sketch of each graph. Verify your results with a graphing calculator.

13. y ⫽ sin t for t 僆 c ⫺

3␲ ␲ , d 2 2

14. y ⫽ sin t for t 僆 3⫺␲, ␲ 4

␲ 15. y ⫽ cos t for t 僆 c ⫺ , 2␲ d 2

␲ 5␲ 16. y ⫽ cos t for t 僆 c ⫺ , d 2 2

Earlier we defined the amplitude of the sine and cosine functions as the maximum displacement from the average value. When the average value is 0 (no vertical translation), this “maximum displacement” can more clearly be seen using the concept of absolute value. The graphs shown in Exercises 17 to 20 are of the form y ⴝ 円 A sin t 円 or y ⴝ 円 A cos t 円, where A is the amplitude of the function. Use the graphs to state the value of A.

17.

18.

4

⫺2␲

6

⫺2␲

2␲

⫺6

⫺4

19.

20.

1

⫺2␲

2␲

⫺1

2␲

1

⫺2␲

2␲

⫺1

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557


e.

Use a reference rectangle and the rule of fourths to draw an accurate sketch of the following functions through two complete cycles—one where t > 0, and one where t ⬍ 0. Clearly state the amplitude and period as you begin.

f.

y 4

2

2 2

0 2

3 2

2

22. y 4 sin t

23. y 2 cos t

24. y 3 cos t

1 25. y sin t 2

3 26. y sin t 4

g.

2

h.

2

29. y 0.8 cos 12t2 1 31. f 1t2 4 cosa tb 2

35. y 4 sina

36. y 2.5 cosa

37. f 1t2 2 sin1256␲t2

49.

2␲ tb 5

39. Graph the tone Y2 sin 3 2␲113362X4 and find its value at X 0.00025. Identify the settings of your “friendly” viewing window.

Clearly state the amplitude and period of each function, then match it with the corresponding graph.

43. y 3 sin12t2

44. y 3 cos12t2

45. f 1t2

46. g1t2

3 cos 10.4t2 4

47. y 4 sin 1144␲t2 a.

b.

y

1

0

2

3

4

5 t

c.

2 0

4

1 72

1 48

1 36

5 t 144

t

4

8

0

2

3 8

5 t 8

0 0.2

1

51.

y 0.4

4

2

3 4

t

2

3

4

t

0.4

52.

y 6

0

1

2

3

4

y 1.6

5 t

0

6

1.6

The graphs shown are of the form y ⴝ A sin1Bt2 . Use the characteristics illustrated for each graph to determine its equation.

54.

y

y

300 10 t

200 t

y

55.

2

3

4

5

56.

y

1 22

0.2 0.2 t

y 4

0 2 4

1 144

1 72

1 48

1 36

y

6 t

2 1 144

2

0

d.

y

3 4

0.2

1

2

4

2

7 cos 10.8t2 4

1

2

2

4

0

50.

y

53.

1

1

t

1

1

48. y 4 cos 172␲t2

2

y 2

2

0.5

40. Graph the function Y3 950 sin(0.005X) and find the value at X 550. Identify the settings of your “friendly” viewing window.

42. y 2 sin14t2

0.5

38. g1t2 3 cos 1184␲t2

41. y 2 cos14t2

t

2

The graphs shown are of the form y ⴝ A cos(Bt). Use the characteristics illustrated for each graph to determine its equation.

34. g1t2 5 cos18␲t2

5␲ tb 3

3 4

2

3 32. y 3 cosa tb 4

33. f 1t2 3 sin 14␲t2

2

4

0

30. y 1.7 sin 14t2

3 2

1

1

28. y cos 12t2

4

y

1

27. y sin 12t2

2

0

t

4

21. y 3 sin t

y 4

5 t 144

t 3

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Match each graph to its equation, then graphically estimate the points of intersection. Confirm or contradict your estimate(s) by substituting the values into the given equations using a calculator.

57. y cos x; y sin x

58. y cos x; y sin12x2

y

y

y

y

1

2

2

0.5

0.5

1

1

0

2

3 2

2 x

1

0 0.5

2

3 2

2 x

1

0 1 2

2

3 2

2 x

0 1

1 2

1

3 2

2 x

2


61. Area of a regular polygon inscribed in a circle: nr2 2␲ Aⴝ sin a b n 2 Exercise 61 The formula shown gives the area of a regular r polygon inscribed in a circle, where n is the number of sides 1n 32 and r is the radius of the circle. Given r 10 cm, a. What is the area of the circle? b. What is the area of the polygon when n 4? Find the length of the sides of the polygon using two different methods. c. Calculate the area of the polygon for n 10, 20, 30, and 100. What do you notice?

䊳

60. y 2 cos12␲x2; y 2 sin1␲x2

1

0.5

䊳

59. y 2 cos x; y 2 sin13x2

62. Hydrostatics, surface tension, and contact 2␥ cos ␪ angles: y ⴝ kr The height that a liquid will rise in a capillary tube is given by the formula shown, Capillary y where r is the radius of the Tube tube, ␪ is the contact angle of the liquid (the meniscus), Liquid is the surface tension of the liquid-vapor film, and k is a constant that depends on the weight-density of the liquid. How high will the liquid rise given that the surface tension 0.2706, the tube has radius r 0.2 cm, the contact angle ␪ 22.5°, and k 1.25?

APPLICATIONS

Tidal waves: Tsunamis, also known as tidal waves, are ocean waves produced by earthquakes or other upheavals in the Earth’s crust and can move through the water undetected for hundreds of miles at great speed. While traveling in the open ocean, these waves can be represented by a sine graph with a very long wavelength (period) and a very small amplitude. Tsunami waves only attain a monstrous size as they approach the shore, and represent a very different phenomenon than the ocean swells created by heavy winds over an extended period of time. 63. A graph modeling a Height in feet tsunami wave is given in 2 1 the figure. (a) What is the height of the tsunami 1 20 40 60 80 100Miles 2 wave (from crest to trough)? Note that h 0 is considered the level of a calm ocean. (b) What is the tsunami’s wavelength (period)? (c) Find an equation for this wave.

64. A heavy wind is kicking up ocean swells approximately 10 ft high (from crest to trough), with wavelengths (periods) of 250 ft. (a) Find an equation that models these swells. (b) Graph the equation. (c) Determine the height of a wave measured 200 ft from the trough of the previous wave.

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Sinusoidal models: The sine and cosine functions are of great importance to meteorological studies, as when modeling the temperature based on the time of day, the illumination of the moon as it goes through its phases, or even the prediction of tidal motion.

Kinetic energy: The kinetic energy a planet possesses as it orbits the Sun can be modeled by a cosine function. When the planet is at its apogee (greatest distance from the Sun), its kinetic energy is at its lowest point as it slows down and “turns around” to head back toward the Sun. The kinetic energy is at its highest when the planet “whips around the Sun” to begin a new orbit.

Sinusoidal movements: Many animals exhibit a wavelike motion in their movements, as in the tail of a shark as it swims in a straight line or the wingtips of a large bird in flight. Such movements can be modeled by a sine or cosine function and will vary depending on the animal’s size, speed, and other factors. Distance 67. The graph shown models in inches 20 the position of a shark’s 10 t sec tail at time t, as measured 10 2 3 4 5 1 to the left (negative) and 20 right (positive) of a straight line along its length. (a) Use the graph to determine the related equation. (b) Is the tail to the right, left, or at center when t 6.5 sec? How far? (c) Would you say the shark is “swimming leisurely,” or “chasing its prey”? Justify your answer.

50 25 0

75 50 25 0

12 24 36 48 60 72 84 96

12 24 36 48 60 72 84 96

t days

t days

70. The potential energy of the planet is the antipode of its kinetic energy, meaning when kinetic energy is at 100%, the potential energy is 0%, and when kinetic energy is at 0% the potential energy is at 100%. (a) How is the graph of the kinetic energy related to the graph of the potential energy? In other words, what transformation could be applied to the kinetic energy graph to obtain the potential energy graph? (b) If the kinetic energy is at 62.5% and increasing, what can be said about the potential energy in the planet’s orbit at this time? Visible light: One of the narrowest bands in the electromagnetic spectrum is the region involving visible light. The wavelengths (periods) of visible light vary from 400 nanometers (purple/violet colors) to 700 nanometers (bright red). The approximate wavelengths of the other colors are shown in the diagram. Violet

68. The State Fish of Hawaii is the humuhumunukunukuapua’a, a small colorful fish found abundantly in coastal waters. Suppose the tail motion of an adult fish is modeled by the equation d1t2 sin115␲t2 with d(t) representing the position of the fish’s tail at time t, as measured in inches to the left (negative) or right (positive) of a straight line along its length. (a) Graph the equation over two periods. (b) Is the tail to the left or right of center at t 2.7 sec? How far? (c) Would you say this fish is “swimming leisurely,” or “running for cover”? Justify your answer.

75

Percent of KE

␲ 66. The equation y 7 sina tb models the height of 6 the tide along a certain coastal area, as compared to average sea level. Assuming t 0 is midnight, (a) graph this function over a 12-hr period. (b) What will the height of the tide be at 5 A.M.? (c) Is the tide rising or falling at this time?

69. Two graphs are given here. (a) Which of the graphs could represent the kinetic energy of a planet orbiting the Sun if the planet is at its perigee (closest distance to the Sun) when t 0? (b) For what value(s) of t does this planet possess 62.5% of its maximum kinetic energy with the kinetic energy increasing? (c) What is the orbital period of this planet? a. 100 b. 100 Percent of KE

Temperature 65. The graph given shows deviation 4 the deviation from the 2 average daily t 0 temperature for the hours 4 8 12 16 20 24 of a given day, with 2 t 0 corresponding to 4 6 A.M. (a) Use the graph to determine the related equation. (b) Use the equation to find the deviation at t 11 (5 P.M.) and confirm that this point is on the graph. (c) If the average temperature for this day was 72°, what was the temperature at midnight?

559

400

Blue

Green

500

Yellow Orange

600

Red

700

71. The equations for the colors in this spectrum have 2␲ the form y sin1t2, where gives the length of the sine wave. (a) What color is represented by the ␲ tb? (b) What color is equation y sina 240 ␲ tb? represented by the equation y sina 310

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72. Name the color represented by each of the graphs (a) and (b) and write the related equation. a. 1 y t (nanometers) 0

300

600

900

1200

73. Find an equation of the household current modeled by the graph, then use the equation to determine I when t 0.045 sec. Verify that the resulting ordered pair is on the graph.

1

b.

y 1

t (nanometers) 0

300

600

900

Alternating current: Surprisingly, even characteristics of the electric current supplied to your home can be modeled by sine or cosine functions. For alternating current (AC), the amount of current I (in amps) at time t can be modeled by I A sin1␻t2, where A represents the maximum current that is produced, and ␻ is related to the frequency at which the generators turn to produce the current.

1200

1

Exercise 73 Current I 30 15

t sec 15

1 50

1 25

3 50

2 25

1 10

30

74. If the voltage produced by an AC circuit is modeled by the equation E 155 sin1120␲t2, (a) what is the period and amplitude of the related graph? (b) What voltage is produced when t 0.2? 䊳


75. For y A sin1Bx2 and y A cos1Bx2, the Mm gives the average value of the 2 function, where M and m represent the maximum and minimum values, respectively. What was the average value of every function graphed in this section? Compute a table of values for y 2 sin t 3, and note its maximum and minimum values. What is the average value of this function? What transformation has been applied to change the average value of the function? Can you name the average value of y 2 cos t 1 by inspection? expression

䊳

2␲ B came from, consider that if B 1, the graph of y sin1Bt2 sin11t2 completes one cycle from 1t 0 to 1t 2␲. If B 1, y sin1Bt2 completes one cycle from Bt 0 to Bt 2␲. Discuss how this observation validates the period formula.

76. To understand where the period formula P

77. The tone you hear when pressing the digit “9” on your telephone is actually a combination of two separate tones, which can be modeled by the functions f 1t2 sin 3 2␲18522t 4 and g1t2 sin 3 2␲114772t4. Which of the two functions has the shorter period? By carefully scaling the axes, graph the function having the shorter period using the steps I through IV discussed in this section.


78. (6.2) Given sin 1.12 0.9, find an additional value of t in 30, 2␲2 that makes the equation sin t 0.9 true.

80. (6.1) Invercargill, New Zealand, is at 46° 14¿ 24– south latitude. If the Earth has a radius of 3960 mi, how far is Invercargill from the equator?

79. (6.1) Use a special triangle to calculate the distance from the ball to the pin on the seventh hole, given the ball is in a straight line with the 100-yd plate, as shown in the figure.

81. (3.1) Given z1 1 i and z2 2 5i, compute the following: a. z1 z2 b. z1 z2 c. z1z2 z2 d. z1

Exercise 79

100 yd 60 100 yd

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A. Graph the functions

B.

C.

D.

E.

y A csc1Bt2 and y A sec1Bt2 Graph y tan t using asymptotes, zeroes, and sin t the ratio cos t Graph y cot t using asymptotes, zeroes, and cos t the ratio sin t Identify and discuss important characteristics of y tan t and y cot t Graph y A tan 1Bt 2 and y A cot 1Bt 2 with various values of A and B

EXAMPLE 1

䊳

Unlike sine and cosine, the cosecant, secant, tangent, and cotangent functions have no maximum or minimum values over their domains. However, it is precisely this unique feature that adds to their value as mathematical models. Collectively, these six trig functions give scientists the tools they need to study, explore, and investigate a wide range of phenomena, extending our understanding of the world around us.

A. Graphs of y ⴝ A csc(Bt) and y ⴝ A sec(Bt) From our earlier work, we know that y sin t and y csc t are reciprocal functions: 1 1 csc t . Likewise, we have sec t . The graphs of these reciprocal funcsin t cos t tions follow quite naturally from the graphs of y A sin1Bt2 and y A cos1Bt2 by using these observations: 1. You cannot divide by zero. 2. The reciprocal of a very small number ( | x | 6 1) is a very large number (and vice versa). 3. The reciprocal of 1 is 1. Just as with rational functions, division by zero creates a vertical asymptote, so the 1 graph of y csc t will have a vertical asymptote at every point where sin t sin t 0. This occurs at t ␲k, where k is an integer 1p , 2␲, ␲, 0, ␲, 2␲, p2. The table shown in Figure 6.73 shows that as x S ␲, Y1 sin x approaches zero, 1 Figure 6.73 csc X becomes infinitely large. while Y2 Y1 Further, when sin1Bt2 1, csc1Bt2 1 since the reciprocals of 1 and 1 are still 1 and 1, respectively. Finally, due to observation 2, the graph of the cosecant function will be increasing when the sine function is decreasing, and decreasing when the sine function is increasing. Considering the case where A 1, we can graph y csc1Bt2 by first drawing a sketch of y sin1Bt2 . Then using the previous observations and a few well-known values, we can accurately complete the graph. From this 2␲ graph, we discover that the period of the cosecant function is also and that y csc1Bt2 B is an odd function. Graphing y ⴝ A csc(Bt) for A ⴝ 1 Graph the function y csc t over the interval 3 0, 4␲4 .

Solution

6–53

䊳

Begin by sketching the function y sin t, using a standard reference rectangle 2␲ 2␲ 2A 2112 2 units high by 2␲ B 1 1 units in length. Since csc t , the sin t graph of y csc t will be asymptotic at the zeroes of y sin t 1t 0, ␲, and 2␲). As in Section 6.3, we can then extend the graph

t

sin t

csc t

0

0

1 S undefined 0

␲ 6

1 0.5 2

2 2 1

␲ 4

12 0.71 2

2 1.41 12

␲ 3

13 0.87 2

2 1.15 13

␲ 2

1

1

561

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into the interval 3 2␲, 4␲ 4 by reproducing the graph from 30, 2␲ 4 . A partial table and the resulting graph are shown. y csc t

y • Vertical asymptotes where sin(Bt) is zero • When sin(Bt) 1, csc(Bt) 1 • Output values are reciprocated

2 2

1.41

√2 2

0.71

2 1

1

y sin t ␲ 4

␲ 2

␲

3␲ 2

5␲ 2

2␲

3␲

7␲ 2

4␲

t

2


䊳

With Y1 and Y2 entered as shown in Figure 6.74, a graphing calculator set in radian MODE will confirm the results of Example 1 (Figure 6.75).

WORTHY OF NOTE Due to technological limitations, some older graphing calculators may draw what appears to be the asymptotes of y csc t. It is important to realize these are not part of the graph.


3

4␲

0

3

Similar to Example 1, the graph of y sec t

1 will have vertical asymptotes cos t

␲ 3␲ and b. Note that if A 1, we would then use the graph of 2 2 y A cos1Bt2 to graph y A sec1Bt2 , as in Example 2.

where cos t 0 at

EXAMPLE 2

䊳

Graphing y ⴝ A sec(Bt) for A, B ⴝ 1 ␲ Graph the function y 3 sec a tb over the interval 3 2, 64 and use a graphing 2 calculator to check your graph.

Solution

䊳

Begin by sketching the function ␲ y 3 cos a tb using a reference rectangle 2 2␲ 2␲ 4 units long 6 units high by B ␲/2 as a guide. Within the rectangle, each special feature will be 1 unit apart (rule of fourths), with the asymptotes occurring at the ␲ zeroes of y 3 cos a tb 1t 1 and t 32 . 2 We then extend the graph to cover the interval 32, 6 4 by reproducing the

t

3 cos a

␲ tb 2

3 sec a

␲ tb 2

0

3

3

1 3

13 3# 2.60 2

2 3# 3.46 13

1 2

3#

2 3 1

12 2.12 2

3#

1 1.5 2 0

3#

2 4.24 12 3

#26 1

1 S undefined 0

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Section 6.4 Graphs of the Cosecant, Secant, Tangent, and Cotangent Functions

appropriate sections of the graph. A partial table and the resulting graph are shown. y 3 sec ␲2 t

y

• Vertical asymptotes where cos(Bt) is zero • When A cos(Bt) A, A sec(Bt) A

Check

䊳

y 3 cos ␲2 t

1 2

1

1

0.5

1

2

3

4

6

5

t

With Y1 and Y2 entered as shown, the graphs match. ✓ 10

2

6

10


䊳

2␲ , and that B y A sec1Bt2 is an even function. The most important characteristics of the cosecant and secant functions are summarized in the following box. Note that for these functions, there is no discussion of amplitude, and no mention is made of their zeroes since neither graph intersects the t-axis.

From the graph, we discover that the period of the secant function is also A. You’ve just seen how we can graph the functions y ⴝ A csc(Bt) and y ⴝ A sec(Bt)

Characteristics of f(t) ⴝ csc t and f(t) ⴝ sec t For all real numbers t and integers k, y ⴝ csc t

y ⴝ sec t

Domain

t k␲

Range

Asymptotes

1q, 1 4 ´ 3 1, q2

t k␲

Period 2␲

Domain

t

␲ ␲k 2

Symmetry odd csc1t2 csc t

Range

1q, 14 ´ 31, q 2 Period 2␲

Asymptotes

t

␲ ␲k 2

Symmetry even sec1t2 sec t

B. The Graph of y ⴝ tan t Like the secant and cosecant functions, tangent is defined in terms of a ratio, creating asymptotic behavior at the zeroes of the denominator. In terms of the unit circle, y ␲ ␲ tan t , which means in 3␲, 2␲ 4 , vertical asymptotes occur at t , t , x 2 2

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Figure 6.76

3␲ , since the x-coordinate on the unit circle is zero (see Figure 6.76). We further 2 note tan t 0 when the y-coordinate is zero, so the function will have t-intercepts at t ␲, 0, ␲, and 2␲ in the same interval. This produces the following framework for graphing the tangent function shown in Figure 6.77. and

y (0, 1) (x, y) t (1, 0)

(0, 0)

(1, 0)

Figure 6.77

x tan t 4

(0, 1) y tan t x

Asymptotes at odd multiples of

␲ 2

␲

t-intercepts at integer multiples of ␲

2

␲

␲

␲ 2

2

2␲

3␲ 2

t

2 4

Figure 6.78

Knowing the graph must go through these zeroes and approach the asymptotes, we are left with determining the direction of the approach. This can be discovered by noting that in QI, the y-coordinates of points on the unit circle start at 0 and increase, while y the x-values start at 1 and decrease. This means the ratio defining tan t is increasing, x ␲ and in fact becomes infinitely large as t gets very close to . (Figure 6.78). Using the 2 ␲ ␲ additional points provided by tan a b 1 and tan a b 1, we find the graph 4 4 ␲ ␲ of tan t is increasing throughout the interval a , b and that the function has a 2 2 period of ␲. We also note y tan t is an odd function (symmetric about the origin), since tan1t2 tan t as evidenced by the two points just computed. The completed graph is shown in Figure 6.79 with the primary interval in red. Figure 6.79 tan t 4

␲4 , 1

2

␲ ␲

4 , 1

␲

␲ 2

2

␲

3␲ 2

2␲

t

2 4 ␲

␲

y The graph can also be developed by noting sin t y, cos t x, and tan t . x sin t This gives tan t by direct substitution and we can quickly complete a table of cos t values for tan t, as shown in Example 3.

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EXAMPLE 3

䊳

Constructing a Table of Values for f (t) ⴝ tan t y Complete Table 6.8 shown for tan t using the values given for sin t and cos t, x then graph the function by plotting points.

Solution

䊳

The completed table is shown here. Table 6.8

t

0

␲ 6

␲ 4

␲ 3

␲ 2

2␲ 3

3␲ 4

5␲ 6

␲

sin t y

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

cos t x

1

13 2

12 2

1 2

0

y x

0

1 0.58 13

1

13 1.7

undefined

13

tan t

1 2

12 2

1

13 2

1

1 13

0

For the noninteger values of x and y, the “twos will cancel” each time we compute y . This means we can simply list the ratio of numerators. The resulting values are x shown in red in Table 6.8, along with the corresponding plotted points. The graph shown in Figure 6.80 was completed using symmetry and the previous observations. Figure 6.80 ␲

6 , 0.58

f (t) 4

␲

3 , 1.7

y tan t

2 ␲

2

␲

4 , 1

␲ 2

2

␲

3␲

t 4

3␲

4

3

2␲

␲

6

5␲

2

, 1

, 0.58

, 1.7

␲


䊳

Additional values can be found using a calculator as needed. For future use and reference, it will help to recognize the approximate decimal equivalent of all special 1 0.58. See values and radian angles. In particular, note that 13 1.73 and 13 Exercises 17 through 22. While we could easily use

Y1 tan X on a graphing calculator to generate the graph shown in Figure 6.80, we would miss an opportunity to reinforce the previous observations using sin x the ratio definition tan x . To begin, enter cos x Y1 Y1 sin X, Y2 cos X, and Y3 , as shown in Y2 Figure 6.81 [recall that function variables are accessed using VARS (Y-VARS) (1:Function)]. ENTER

Figure 6.81

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Note that Y2 has been disabled by overlaying the cursor on the equal sign and pressing . Pressing ZOOM 7:ZTrig at this point produces the screen shown in Figure 6.82, where we note that tan x is zero everywhere that sin x is zero. Similarly, disabling Y1 while Y2 is enabled emphasizes that the asymptotes of y tan x occur everywhere cos x is zero (see Figure 6.83). ENTER

Figure 6.82

Figure 6.83 4

4

2␲

B. You’ve just seen how we can graph y ⴝ tan t using asymptotes, zeroes, and the sin t ratio cos t

2␲

2␲

2␲

4

4

C. The Graph of y ⴝ cot t Since the cotangent function is also defined in terms of a ratio, it too displays asymptotic behavior at the zeroes of the denominator, with t-intercepts at the zeroes of the x numerator. Like the tangent function, cot t can be written in terms of cos t x y cos t , and the graph obtained by plotting points. and sin t y: cot t sin t EXAMPLE 4

Constructing a Table of Values for f (t) ⴝ cot t

䊳

x for t in 3 0, ␲ 4 using its ratio relationship with cos t y and sin t. Use the results to graph the function for t in 1␲, 2␲2. Complete Table 6.9 for cot t

Solution

䊳

The completed table is shown here, with the computed values displayed in red. Table 6.9

t

0

␲ 6

␲ 4

␲ 3

␲ 2

2␲ 3

3␲ 4

5␲ 6

␲

sin t y

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

cos t x

1

13 2

12 2

1 2

0

undefined

13

1

1 13

0

cot t

x y

1 2

1 13

12 2

1

13 2

13

1 undefined

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In this interval, the cotangent function has asymptotes at 0 and ␲ since y 0 at these ␲ points, and has a t-intercept at since x 0. The graph shown in Figure 6.84 was 2 completed using the period of y cot t, P ␲. Note that due to how it is defined, the function y cot t is a decreasing function. Figure 6.84 cot t 4 2

␲

␲ 2

2

␲ 2

␲

3␲ 2

2␲

t

4 ␲

␲

␲


C. You’ve just seen how we can graph y ⴝ cot t using asymptotes, zeroes, and the cos t ratio sin t

䊳

Figure 6.85 Because the cotangent (and tangent) of integer ␲ multiples of will always be undefined, 1, 0, or 1, 4 these values are often used to generate the graph cos X of the function. By entering Y1 and setting sin X ␲ TblStart ␲ and ¢Tbl , we can quickly 4 generate these standard values (see Figure 6.85). Note the five points calculated in Figure 6.85 have been colored blue in Figure 6.84.

D. Characteristics of y ⴝ tan t and y ⴝ cot t The most important characteristics of the tangent and cotangent functions are summarized in the following box. There is no discussion of amplitude, maximum, or minimum values, since maximum or minimum values do not exist. For future use and reference, perhaps the most significant characteristic distinguishing tan t from cot t is that tan t increases, while cot t decreases over their respective domains. Also note that due to symmetry, the zeroes of each function are always located halfway between the asymptotes. Characteristics of f(t) ⴝ tan t and f(t) ⴝ cot t For all real numbers t and integers k, y ⴝ tan t Domain

␲ ␲k 2 Period ␲

t

y ⴝ cot t Range

Asymptotes

1q, q2

␲ ␲k 2 Symmetry odd tan1t2 tan t

Behavior increasing

t

Domain

Range

Asymptotes

t k␲

1q, q 2

t k␲

Period ␲

Behavior decreasing

Symmetry odd cot1t2 cot t

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EXAMPLE 5

䊳

Using the Period of f(t) ⴝ tan t to Find Additional Points ␲ 1 7␲ 13␲ 5␲ , what can you say about tan a b, tan a b, and tan a b? Given tan a b 6 6 6 6 13

Solution

䊳

␲ 7␲ ␲ by a multiple of ␲: tan a b tan a ␲b, 6 6 6 5␲ 13␲ ␲ ␲ tana b tana 2␲b, and tana b tana ␲b. Since the period of 6 6 6 6 1 . the tangent function is P ␲, all of these expressions have a value of 13 Each value of t differs from


D. You’ve just seen how we can identify and discuss important characteristics of y ⴝ tan t and y ⴝ cot t

䊳

Since the tangent function is more common than the cotangent, many needed calculations will first be done using the tangent function and its properties, then ␲ reciprocated. For instance, to evaluate cot a b we reason that cot t is an odd 6 ␲ ␲ function, so cot a b cot a b. Since cotangent is the reciprocal of tangent and 6 6 ␲ ␲ 1 tan a b , cot a b 13. See Exercises 31 and 32. 6 6 13

E. Graphing y ⴝ A tan(Bt) and y ⴝ A cot(Bt) The Coefficient A: Vertical Stretches and Compressions For the tangent and cotangent functions, the role of coefficient A is best seen through an analogy from basic algebra (the concept of amplitude is foreign to these functions). Consider the graph of y x3 (Figure 6.86). Comparing the parent function y x3 with functions y Ax3, the graph is stretched vertically if A 7 1 (see Figure 6.87) and compressed if 0 6 A 6 1. In the latter case the graph becomes very “flat” near the zeroes, as shown in Figure 6.88. Figure 6.86

Figure 6.87

Figure 6.88

y x3 y

y 4x3; A 4 y

y 14 x3; A 14 y

x

x

x

While cubic functions are not asymptotic, they are a good illustration of A’s effect on the tangent and cotangent functions. Fractional values of A 1 A 6 12 compress the graph, flattening it out near its zeroes. Numerically, this is because a fractional part of ␲ a small quantity is an even smaller quantity. For instance, compare tan a b with 6 1 ␲ ␲ ␲ 1 tan a b. Rounded to two decimal places, tan a b 0.57, while tan a b 0.14, 4 6 6 4 6 so the graph must be “nearer the t-axis” at this value.

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EXAMPLE 6

䊳

569

Comparing the Graph of f(t) ⴝ tan t and g(t) ⴝ A tan t Draw a “comparative sketch” of y tan t and y 14 tan t on the same axis and discuss similarities and differences. Use the interval 3␲, 2␲ 4 .

Solution

䊳

Both graphs will maintain their essential features (zeroes, asymptotes, period, increasing, and so on). However, the graph of y 14 tan t is vertically compressed, causing it to flatten out near its zeroes and changing how the graph approaches its asymptotes in each interval. See Figure 6.89. The table feature of a graphing calculator 1 with Y1 tan X and Y2 a b tan 1X2 reinforces these observations (Figure 6.90). 4 Figure 6.90

Figure 6.89 y 4

y tan t y 14 tan t

2

␲

␲ 2

␲ 2

␲

3␲ 2

2␲

t

2 4


䊳

The Coefficient B: The Period of Tangent and Cotangent

WORTHY OF NOTE It may be easier to interpret the phrase “twice as fast” as 2P ␲ and “one-half as fast” as 12 P ␲. In each case, solving for P gives the correct interval for the period of the new function.

Like the other trig functions, the value of B has a material impact on the period of the function, and with the same effect. The graph of y cot12t2 completes a cycle twice 1 ␲ as fast as y cot t aP versus P ␲b, while y cot a tb completes a cycle 2 2 one-half as fast 1P 2␲ versus P ␲2. This reasoning leads us to a period formula for tangent and cotangent, namely, ␲ P , where B is the coefficient of the input variable. B Similar to the four-step process used to graph sine and cosine functions, we can ␲ graph tangent and cotangent functions using a rectangle P units in length and 2A B units high, centered on the primary interval. After dividing the length of the rectangle into fourths, the t-intercept will always be the halfway point, with y-values of A occuring at the 14 and 34 marks. See Example 7.

EXAMPLE 7

䊳

Graphing y ⴝ A cot(Bt) for A, B ⴝ 1

Solution

䊳

For y 3 cot12t2, A 3, which results in a vertical stretch, and B 2, which ␲ gives a period of . The function is still undefined (asymptotic) at t 0 and then 2 ␲ at all integer multiples of P . We also know the graph is decreasing, with 2 ␲ 3␲ zeroes of the function halfway between the asymptotes. The inputs t and t 8 8

Sketch the graph of y 3 cot12t2 over the interval 3␲, ␲ 4 .

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3␲ 1 3 ␲ ␲ and marks between 0 and b yield the points a , 3b and a , 3b, which 4 4 2 8 8 we’ll use along with the period and symmetry of the function to complete the graph in Figure 6.91. ␲ A graphing calculator check is shown in Figure 6.92 (note 0.392 . 8

a the

Figure 6.92

Figure 6.91

6

y y 3 cot(2t) 6 3 ␲

␲ 2

␲8 , 3 ␲

␲ 2

␲

␲

t

6

, 3 3␲ 8

6


䊳

As with the trig functions from Section 6.3, it is possible to determine the equation of a tangent or cotangent function from a given graph. Where previously we used the amplitude, period, and max/min values to obtain our equation, here we first determine the period of the function by calculating the “distance” between asymptotes, then choose any convenient point on the graph (other than a t-intercept) and substitute the x- and y- values into the general equation to solve for A. EXAMPLE 8

䊳

Constructing the Equation for a Given Graph Find the equation of the graph shown in Figure 6.93, given it’s of the form y A tan1Bt2. Check your answer with a graphing calculator. Figure 6.93 y A tan(Bt)

y 3 2 1 ␲

2␲ 3

␲ 3

1 2 3

Solution

䊳

␲ 3

␲

2␲ 3

␲, 2

t

2

␲ ␲ and t , we find the 3 3 ␲ ␲ 2␲ 2␲ . To find the value of B we substitute period is P a b for P in 3 3 3 3 3 ␲ 3 P and find B (verify). This gives the equation y A tan a tb. B 2 2 Using the primary interval and the asymptotes at t

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571

␲ ␲ To find A, we take the point a , 2b shown, and use t with y 2 to 2 2 solve for A: 3 y A tan a tb 2 3 ␲ 2 A tan c a b a b d 2 2 3␲ 2 A tan a b 4 2 A 3␲ tan a b 4 2

substitute

substitute 2 for y and

䊳

With the

solve for A

result; tan a

3␲ b 1 4

Figure 6.94 3

set to match the given graph ␲ aXmin ␲, Xmax ␲, Xscl , 6 WINDOW

Ymin 3, Ymax 3, and Yscl 1b,

E. You’ve just seen how we can graph y ⴝ A tan(Bt) and y ⴝ A cot(Bt) with various values of A and B

␲ for t 2

multiply

The equation of the graph is y 2 tan1 32t2.

Check

3 for B 2

␲

the calculator produces the GRAPH shown in Figure 6.94. The graphs match. ✓

␲

3


䊳

6.4 EXERCISES 䊳



1. The period of y tan t and y cot t is ________. To find the period of y tan1Bt2 and y cot1Bt2, the formula _________ is used.

2. The function y tan t is _________ everywhere it is defined. The function y cot t is _________ everywhere it is defined.

3. Tan t and cot t are _________ functions, so 11␲ f 1t2 _________. If tan a b 0.268, 12 11␲ b _________. then tan a 12

4. The asymptotes of y _______ are located at odd ␲ multiples of . The asymptotes of y ________ 2 are located at integer multiples of ␲.

5. Discuss/Explain (a) how you could obtain a table of values for y sec t in 3 0, 2␲ 4 from the values for y cos t, and (b) how you could obtain a table 3 of values for y csc a tb t in 3 0, 2␲ 4 , given the 2 3 values for y sin a tb. 2

6. Explain/Discuss how the zeroes of y sin t and y cos t are related to the graphs of y tan t and y cot t. How can these relationships help graph functions of the form y A tan1Bt2 and y A cot1Bt2 ?

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Draw the graph of each function by first sketching the related sine and cosine graphs, and applying the observations from this section. Confirm your graph using a graphing calculator.

7. y ⫽ csc12t2

8. y ⫽ csc1␲t2

1 9. y ⫽ sec a tb 2

10. y ⫽ sec12␲t2

11. y ⫽ 3 csc t

20. State the value of each expression without the use of a calculator. ␲ a. cot a b b. tan ␲ 2 5␲ 5␲ c. tan a⫺ b d. cot a⫺ b 4 6

12. g1t2 ⫽ 2 csc14t2 14. f 1t2 ⫽ 3 sec12t2

13. y ⫽ 2 sec t

Use the values given for sin t and cos t to complete the tables.

15. t

␲

7␲ 6

sin t ⫽ y

0

⫺

cos t ⫽ x

⫺1

tan t ⫽

⫺

5␲ 4

1 2

⫺

12 2

13 2

⫺

12 2

4␲ 3 ⫺

13 2 1 2

3␲ 2 ⫺1 0

y x

16. t

3␲ 2

sin t ⫽ y

⫺1

cos t ⫽ x

0

tan t ⫽

5␲ 3 ⫺

11␲ 6

7␲ 4

13 2

⫺

12 2

12 2

1 2

2␲

1 2

0

13 2

1

⫺

19. State the value of each expression without the use of a calculator. ␲ ␲ a. tan a⫺ b b. cot a b 4 6 3␲ ␲ c. cot a b d. tan a b 4 3

y x

21. State the value of t without the use of a calculator, given t 僆 3 0, 2␲2 terminates in the quadrant indicated. a. tan t ⫽ ⫺1, t in QIV b. cot t ⫽ 13, t in QIII 1 , t in QIV c. cot t ⫽ ⫺ 13 d. tan t ⫽ ⫺1, t in QII 22. State the value of t without the use of a calculator, given t 僆 30, 2␲2 terminates in the quadrant indicated. a. cot t ⫽ 1, t in QI b. tan t ⫽ ⫺ 13, t in QII 1 , t in QI c. tan t ⫽ 13 d. cot t ⫽ 1, t in QIII Use the values given for sin t and cos t to complete the tables.

17. Without reference to a text or calculator, attempt to name the decimal equivalent of the following values to one decimal place. ␲ 2

␲ 4

␲ 6

12

12 2

2 13

18. Without reference to a text or calculator, attempt to name the decimal equivalent of the following values to one decimal place. ␲ 3

␲

3␲ 2

13

13 2

1 13

23. t

␲

7␲ 6

sin t ⫽ y

0

⫺

cos t ⫽ x

⫺1

cot t ⫽

x y

⫺

5␲ 4

1 2

⫺

12 2

13 2

⫺

12 2

4␲ 3 ⫺

3␲ 2

13 2

⫺1

1 2

0

⫺

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24.

Graph each function over the interval indicated, noting the period, asymptotes, zeroes, and values of A and B. t

3␲ 2

sin t y

1

cos t x cot t

0

5␲ 3

13 2

7␲ 4

1 2

12 2

12 2

11␲ 6

2␲

1 2

0

13 2

1

x y

11␲ 25. Given t is a solution to tan t 7.6, use the 24 period of the function to name three additional solutions. Check your answer using a calculator. 7␲ is a solution to cot t 0.77, use the 24 period of the function to name three additional solutions. Check your answer using a calculator.

26. Given t

27. Given t 1.5 is a solution to cot t 0.07, use the period of the function to name three additional solutions. Check your answers using a calculator.

␲ ␲ 37. y tan 12t2; c , d 2 2 1 38. y tan a tb; 34␲, 4␲ 4 4 ␲ ␲ 39. y cot 14t2; c , d 4 4 1 40. y cot a tb; 3 2␲, 2␲ 4 2 ␲ ␲ 41. y 2 tan14t2; c , d 4 4 1 42. y 4 tan a tb; 3 2␲, 2␲ 4 2 1 43. y 5 cot a tb; 3 3␲, 3␲ 4 3 44. y

␲ ␲ 1 cot12t2; c , d 2 2 2

28. Given t 1.25 is a solution to tan t 3, use the period of the function to name three additional solutions. Check your answers using a calculator.

1 1 45. y 3 tan12␲t2; c , d 2 2

Verify the value shown for t is a solution to the equation given, then use the period of the function to name all real roots. Check two of these roots on a calculator.

47. f 1t2 2 cot1␲t2; 31, 1 4

29. t

␲ ; tan t 0.3249 10

30. t

␲ ; tan t 0.1989 16

31. t

␲ ; cot t 2 13 12

32. t

5␲ ; cot t 2 13 12

␲ 46. y 4 tana tb; 32, 2 4 2

48. p1t2

Clearly state the period of each function, then match it with the corresponding graph.

1 49. y 2 csca tb 2

1 50. y 2 seca tb 4

51. y sec18␲t2

52. y csc112␲t2

a.

Graph each function over the interval indicated, noting the period, asymptotes, zeroes, and value of A. Include a comparative sketch of y ⴝ tan t or y ⴝ cot t as indicated.

1 tan t; 3 2␲, 2␲4 2

y 4 2

0

c.

2␲

4␲

6␲

8␲

t

0

2

2

4

4

d.

y 4

35. h1t2 3 cot t; 3 2␲, 2␲4

2

1 cot t; 3 2␲, 2␲ 4 4

2

36. r1t2

b.

y 4 2

33. f 1t2 2 tan t; 32␲, 2␲ 4 34. g1t2

1 ␲ cota tb; 34, 44 2 4

0

4

2␲

4␲

6␲

8␲

1 6

1 4

1 3

t

y 4 2

1 12

1 6

1 4

1 3

5 12

t

0 2 4

1 12

5 12

t

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Find the equation of each graph, given it is of the form y ⴝ A csc(Bt). Check with a graphing calculator.

53.

6–66


54.

y 8

Find the equation of each graph, given it is of the form y ⴝ A cot(Bt). Check with a graphing calculator.

57.

y

14 , 2√3

y

0.8

4 4

0.4 2 5

1 5

0 4

3 5

4 5

1

t

2␲

0

4␲

t

6␲

0.4

8

3

2

1

0.8

2

3

t

1 2

t

4

Find the equation of each graph, given it is of the form y ⴝ A tan(Bt). Check with a graphing calculator.

55.

1

58.

y 4

y

9 , √3 1

9 ␲

2 , 3

2␲

␲

3␲ 2

␲ 2

␲ 2

␲

1 2

t

1 3

1 6

1 6

1 3

4

9

56.

␲ 3␲ and t are solutions to 8 8 cot13t2 tan t, use a graphing calculator to find two additional solutions in 30, 2␲4 .

59. Given that t

y 2 1 ␲ 2

␲ 3

␲

␲ 6

1

12 , 2

䊳

1

␲ 6

2

␲ 3

␲ 2

t

60. Given t 16 is a solution to tan12␲t2 cot1␲t2, use a graphing calculator to find two additional solutions in 31, 14 .

WORKING WITH FORMULAS d cot u ⴚ cot v The height h of a tall structure can be computed using two angles of elevation measured some distance apart along a straight line with the object. This height is given by the formula shown, where d is the distance between the two points from which angles u and v were measured. Find the height h of a building if u 40°, v 65°, and d 100 ft.

61. The height of an object calculated from a distance: h ⴝ

h u

h sⴚk The equation shown is used to help locate the position of an image reflected by a spherical lens, where s is the distance of the object from the lens along a horizontal axis, ␪ ␪ is the angle of elevation from this axis, h is the altitude of the right triangle P indicated, and k is distance from the lens to the foot of altitude h. Find the Object ␲ distance k given h 3 mm, ␪ , and that the object is 24 mm from the lens. 24

v xd

d

62. Position of an image reflected from a spherical lens: tan ␪ ⴝ

Lens h P Reflected image s

k

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APPLICATIONS

63. Circumscribed polygons: The perimeter of a regular polygon circumscribed about a ␲ circle of radius r is given by P 2nr tana b, where n is the number of sides 1n 32 and n r is the radius of the circle. Given r 10 cm, a. What is the circumference of the circle? b. What is the perimeter of the polygon when n 4? Why? c. Calculate the perimeter of the polygon for n 10, 20, 30, and 100. What do you notice? See Exercise 61 from Section 6.3.

Exercises 63 and 64

r

64. Circumscribed polygons: The area of a regular polygon circumscribed about a circle of ␲ radius r is given by A nr2tana b, where n is the number of sides 1n 32 and r is the radius of the circle. n Given r 10 cm, a. What is the area of the circle? b. What is the area of the polygon when n 4? Why? c. Calculate the area of the polygon for n 10, 20, 30, and 100. What do you notice? Coefficients of friction: Pulling someone on a sled is much easier during the winter than in the summer, due to a phenomenon known as the coefficient of friction. The friction between the sled’s skids and the snow is much lower than the friction between the skids and the dry ground or pavement. Basically, the coefficient of friction is defined by the relationship ␮ tan ␪, where ␪ is the angle at which a block composed of one material will slide down an inclined plane made of another material, with a constant velocity. Coefficients of friction have been established experimentally for many materials and a short list is shown here.

Material

Coefficient

steel on steel

0.74

copper on glass

0.53

glass on glass

0.94

copper on steel

0.68

wood on wood

0.5

65. Graph the function ␮ tan ␪, with ␪ in degrees over the interval 30°, 60° 4 and use the intersection-of-graphs method to estimate solutions to the following. Confirm or contradict your estimates using a calculator. a. A block of copper is placed on a sheet of steel, which is slowly inclined. Is the block of copper moving when the angle of inclination is 30°? At what angle of inclination will the copper block be moving with a constant velocity down the incline? b. A block of copper is placed on a sheet of cast-iron. As the cast-iron sheet is slowly inclined, the copper block begins sliding at a constant velocity when the angle of inclination is approximately 46.5°. What is the coefficient of friction for copper on cast-iron? c. Why do you suppose coefficients of friction greater than ␮ 2.5 are extremely rare? Give an example of two materials that likely have a high ␮-value. 5␲ d and use the intersection-of-graphs 12 method to estimate solutions to the following. Confirm or contradict your estimates using a calculator. a. A block of glass is placed on a sheet of glass, which is slowly inclined. Is the block of glass moving when ␲ the angle of inclination is ? What is the smallest angle of inclination for which the glass block will be 4 moving with a constant velocity down the incline (rounded to four decimal places)? b. A block of Teflon is placed on a sheet of steel. As the steel sheet is slowly inclined, the Teflon block begins sliding at a constant velocity when the angle of inclination is approximately 0.04. What is the coefficient of friction for Teflon on steel? c. Why do you suppose coefficients of friction less than ␮ 0.04 are extremely rare for two solid materials? Give an example of two materials that likely have a very low ␮ value.

66. Graph the function ␮ tan ␪ with ␪ in radians over the interval c 0,

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67. Tangent lines: The actual definition of the word tangent comes from the Latin tangere, meaning “to touch.” In mathematics, a tangent line touches the graph of a circle at only one point and function values for tan ␪ are obtained from the length of the line segment tangent to a unit circle. a. What is the length of the line segment when ␪ ⫽ 80°? b. If the line segment is 16.35 units long, use the intersection-of-graphs method to find the value of ␪. c. Can the line segment ever be greater than 100 units long? Why or why not? d. How does your answer to (c) relate to the asymptotic behavior of the graph? 68. Illumination: The angle ␣ made by a light source and a point on a horizontal I surface can be found using the formula csc ␣ ⫽ 2 , where E is the rE illuminance (in lumens/m2) at the point, I is the intensity of the light source in lumens, and r is the distance in meters from the light source to the point. Use the graph from Example 1 to help determine the angle ␣ given I ⫽ 700 lumens, E ⫽ 55 lumens/m2, and the flashlight is held so that the distance r is 3 m. 䊳

tan ␪ ␪ 1

Intensity I

r

Illuminance E ␣

EXTENDING THE CONCEPT ⫺1 ⫹ 15 has long been thought to be the most pleasing ratio in art and architecture. It is 2 commonly believed that many forms of ancient architecture were constructed using this ratio as a guide. The ratio actually turns up in some surprising places, far removed from its original inception as a line segment cut in “mean and extreme” ratio. Given x ⫽ 0.6662394325, try to find a connection between y ⫽ cos x, y ⫽ tan x, y ⫽ sin x, and the golden ratio.

69. The golden ratio

70. Determine the slope of the line drawn through the parabola (called a secant line) in Figure I. Use the same method (any two points on the line) to calculate the slope of the line drawn tangent to the parabola in Figure II. Compare your calculations to the tangent of the angles ␣ and ␤ that each line makes with the x-axis. What can you conclude? Write a formula for the point/slope equation of a line using tan ␪ instead of m. Figure II

Figure I y

y 10

10

5

5

␣ 5

䊳

␤ 10

x

5

10

x


71. (6.1) A lune is a section of surface area on a sphere, which is subtended by an angle ␪ at the circumference. For ␪ in radians, the surface area of a lune is A ⫽ 2r2␪, where r is the radius of the sphere. Find the area of a lune on the surface of the Earth which is subtended by an angle of 15°. Assume the radius of the Earth is 6373 km.

␪ r

72. (4.4/4.5) Find the y-intercept, x-intercept(s), and all asymptotes of each function, but do not graph. 3x2 ⫺ 9x x⫹1 x2 ⫺ 1 t1x2 ⫽ a. h1x2 ⫽ b. c. p1x2 ⫽ x⫹2 2x2 ⫺ 8 x2 ⫺ 4x

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74. (5.5) The radioactive element potassium-42 is sometimes used as a tracer in certain biological experiments, and its decay can be modeled by the formula Q1t2 Q0e0.055t, where Q(t) is the amount that remains after t hours. If 15 grams (g) of potassium-42 are initially present, how many hours until only 10 g remain?

73. (6.2) State the points on the unit circle that ␲ ␲ 3␲ 3␲ , , and 2␲. correspond to t 0, , , ␲, 4 2 4 2 ␲ What is the value of tan a b? Why? 2

MID-CHAPTER CHECK 6. For the point on the unit circle in Exercise 5, use the intersection-of-graphs method to find the related angle t in both degrees (to tenths) and radians (to ten-thousandths).

1. The city of Las Vegas, Nevada, is located at 36° 06¿ 36– north latitude, 115° 04¿ 48– west longitude. (a) Convert both measures to decimal degrees. (b) If the radius of Exercise 2 the Earth is 3960 mi, how y far north of the equator is 86 cm Las Vegas? 2. Find the angle subtended by the arc shown in the figure, then determine the area of the sector.

7. Name the location of the asymptotes and graph ␲ y 3 tan a tb for t 僆 32␲, 2␲4. 2

␪ 20 cm

x

3. Evaluate without using a calculator: (a) cot 60° and (b) sin a

7␲ b. 4

␲ 4. Evaluate using a calculator: (a) sec a b and 12 (b) tan 83.6°. Exercise 5 y 5. Complete the ordered pair indicated on the unit circle in the figure and find the value of all six trigonometric t 1 functions at this point.

8. Clearly state the amplitude and period, then sketch ␲ the graph: y 3 cos a tb. 2 9. On a unit circle, if arc t has length 5.94, (a) in what quadrant does it terminate? (b) What is its reference arc? (c) Of sin t, cos t, and tan t, which are negative for this value of t? Exercise 10 10. For the graph given here, y (a) clearly state the f(t) amplitude and period; (b) find an equation of t the graph; (c) graphically find f 1␲2 and then confirm/contradict your estimation using a calculator. 8

4

0

␲ 4

␲ 2

3␲ 4

␲

5␲ 4

3␲ 2

4 8

x

√53 , y

REINFORCING BASIC CONCEPTS Trigonometry of the Real Numbers and the Wrapping Function The trigonometric functions are sometimes discussed in terms of what is called a wrapping function, in which the real number line is literally wrapped around the unit circle. This approach can help illustrate how the trig functions can be seen as functions of the real numbers, and apart from any reference to a right triangle. Figure 6.95 shows (1) a unit circle with the location of certain points on the circumference clearly marked and (2) a blue number line that has been marked in ␲ multiples of to coincide with the length of the special arcs (integers are shown in red). Figure 6.96 shows this same 12 blue number line wrapped counterclockwise around the unit circle in the positive direction. Note how the resulting dia␲ 12 12 ␲ 12 , b on the unit circle: cos gram confirms that an arc of length t is associated with the point a and 4 2 2 4 2

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␲ 12 5␲ 13 1 5␲ 13 5␲ 1 ; while an arc of length of t is associated with the point a , b: cos and sin . 4 2 6 2 2 6 2 6 2 Use this information to complete the exercises given. Figure 6.95 Figure 6.96 sin

1 √3 2, 2 √2 √2 2, 2 √3 1 2, 2

(0, 1) y

(1, 0)

1 , √3 2 2 √2, √2 2 2 √3 , 1 2 2 ␲ 0 45 4

7␲ 2␲ 12 3␲ 3 2 4

x␲ 12

1 ␲ 6

␲ 4

␲ 3

2 5␲ ␲ 12 2

3

7␲ 2␲ 3␲ 5␲ 11␲ ␲ 12 3 4 6 12

1. What is the ordered pair associated with an arc length of t 2. What arc length t is associated with the ordered pair a

t

5␲ 6 11␲ 12 3 ␲

␲ 2

y

5␲ 12 ␲ 3 1

45

␲ 4

␲ 6

␲ 12

␲ 4

0 x

2␲ ? What is the value of cos t? sin t? 3

13 1 , b? Is cos t positive or negative? Why? 2 2

3. If we continued to wrap this number line all the way around the circle, in what quadrant would an arc length of 11␲ terminate? Would sin t be positive or negative? t 6 4. Suppose we wrapped a number line with negative values clockwise around the unit circle. In what quadrant would 5␲ an arc length of t terminate? What is cos t? sin t? What positive rotation terminates at the same point? 3

6.5

Transformations and Applications of Trigonometric Graphs


A. Apply vertical translations in context

From your algebra experience, you may remember beginning with a study of linear graphs, then moving on to quadratic graphs and their characteristics. By combining and extending the knowledge you gained, you were able to investigate and understand a variety of polynomial graphs—along with some powerful applications. A study of trigonometry follows a similar pattern, and by “combining and extending” our understanding of the basic trig graphs, we’ll look at some powerful applications in this section.

B. Apply horizontal translations in context C. Solve applications involving harmonic motion D. Apply vertical and horizontal translations to cosecant, secant, tangent, and cotangent E. Solve applications involving the tangent, cotangent, secant, and cosecant functions

A. Vertical Translations: y ⴝ A sin(Bt) ⴙ D and y ⴝ A cos(Bt) ⴙ D On any given day, outdoor temperatures tend to follow a Figure 6.97 sinusoidal pattern, or a pattern that can be modeled by a C sine function. As the sun rises, the morning temperature 15 begins to warm and rise until reaching its high in the late afternoon, then begins to cool during the early evening and nighttime hours until falling to its nighttime low just prior 6 12 18 24 t to sunrise. Next morning, the cycle begins again. In the northern latitudes where the winters are very cold, it’s not 15 unreasonable to assume an average daily temperature of 0°C 132°F2, and a temperature graph in degrees Celsius that looks like the one in Figure 6.97. For the moment, we’ll assume that t 0 corresponds to 12:00 noon. Note 2␲ ␲ that A 15 and P 24, yielding 24 or B . B 12

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If you live in a more temperate area, the daily temperatures still follow a sinusoidal pattern, but the average temperature could be much higher. This is an example of a vertical shift, and is the role D plays in the equation y A sin1Bt2 D. All other aspects of a graph remain the same; it is simply shifted D units up if D 7 0 and D units down Mm if D 6 0. As in Section 6.3, for maximum value M and minimum value m, 2 Mm gives the amplitude A of a sine curve, while gives the average value D. 2 EXAMPLE 1

䊳

Modeling Temperature Using a Sine Function On a fine day in Galveston, Texas, the high temperature might be about 85°F with an overnight low of 61°F. a. Find a sinusoidal equation model for the daily temperature. b. Sketch the graph. c. Approximate what time(s) of day the temperature is 65°F. Assume t 0 corresponds to 12:00 noon.

Solution

䊳

␲ , and the equation model 12 Mm ␲ 85 61 , we find the will have the form y A sin a tb D. Using 12 2 2 85 61 12. The resulting average value D 73, with amplitude A 2 ␲ equation is y 12 sin a tb 73. 12 b. To sketch the graph, use a reference rectangle 2A 24 units tall and P 24 units wide, along with the rule of fourths to locate zeroes and max/min values (see Figure 6.98). Then lightly sketch a sine curve through these points and ␲ within the rectangle as shown. This is the graph of y 12 sin a tb 0. 12 Using an appropriate scale, shift the rectangle and plotted points vertically upward 73 units and carefully draw the finished graph through the points and within the rectangle (see Figure 6.99). a. We first note the period is still P 24, so B

Figure 6.98

Figure 6.99

F

90

12

␲

y 12 sin 12 t

6

85

t (hours) 0

F

␲

y 12 sin 12 t 73

80 75

6

12

18

Average value

24

6

70

12

65 60

(c) t (hours)

0 6

12

18

24

␲ tb 73. Note the broken-line notation 12 “ ” in Figure 6.99 indicates that certain values along an axis are unused (in this case, we skipped 0° to 60°2, and we began scaling the axis with the values needed. This gives the graph of y 12 sina

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Figure 6.100

c. As indicated in Figure 6.99, the temperature hits 65° twice, at about 15 and 21 hr after 12:00 noon, or at 3:00 A.M. and 9:00 A.M. This can be verified by setting ␲ Y1 12 sina Xb 73, 12 then going to the home screen and computing Y1(15) and Y1(21) as shown in Figure 6.100.

WORTHY OF NOTE Recall from Section 2.2 that transformations of any function y f 1x2 remain consistent regardless of the function f used. For the sine function, the transformation y af 1x h2 k is more commonly written y A sin1t C2 D, and A gives a vertical stretch or compression, C is a horizontal shift opposite the sign, and D is a vertical shift, as seen in Example 1.


䊳

Sinusoidal graphs actually include both sine and cosine graphs, the difference being that sine graphs begin at the average value, while cosine graphs begin at the maximum value. Sometimes it’s more advantageous to use one over the other, but equivalent forms can easily be found. In Example 2, a cosine function is used to model an animal population that fluctuates sinusoidally due to changes in food supplies. EXAMPLE 2

䊳

Modeling Population Fluctuations Using a Cosine Function The population of a certain animal species can be modeled by the function ␲ P1t2 1200 cosa tb 9000, where P(t) represents the population in year t. 5 Use the model to a. Find the period of the function. b. Graph the function over one period. c. Find the maximum and minimum values. d. Use a graphing calculator to determine the number of years the population is less than 8000.

Solution

䊳

␲ 2␲ , the period is P 10, meaning the population of this 5 ␲/5 species fluctuates over a 10-yr cycle. b. Use a reference rectangle (2A 2400 by P 10 units) and the rule of fourths to locate zeroes and max/min values, then sketch the unshifted graph ␲ y 1200 cosa tb. With P 10, these important points occur at t 0, 5 2.5, 5, 7.5, and 10 (see Figure 6.101). Shift this graph upward 9000 units (using an appropriate scale) to obtain the graph of P(t) shown in Figure 6.102. a. Since B

Figure 6.102

Figure 6.101 P 1500

P

␲

10,500

y 1200 cos 5 t

␲

P(t) 1200 cos 5 t 9000

10,000

1000

9500

500

t (years)

9000

10

8500

Average value

0 500

2

4

6

8

1000

8000

1500

7500

t (years)

0 2

4

6

8

10

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c. The maximum value is 9000 ⫹ 1200 ⫽ 10,200 and the minimum value is 9000 ⫺ 1200 ⫽ 7800. d. Begin by entering Y1 ⫽ 1200 cos a␲

X b ⫹ 9000 and Y2 ⫽ 8000 in a graphing 5 calculator. With the window settings identified in Figure 6.103, pressing GRAPH confirms the graph we produced in Figure 6.102. The intersection of Y1 and Y2 can be located using the 2nd TRACE (CALC) 5:intersect feature. After twice, the identifying Y1 and Y2 as the intersecting curves by pressing calculator asks us to make a guess. A casual observation of the graph indicates the leftmost point of intersection occurs near 4, so pressing 4 will produce the screen shown in Figure 6.104. With a minor adjustment, this process determines the rightmost point of intersection occurs when x ⬇ 5.93. We can now determine that the population drops below 8000 animals for approximately 5.93 ⫺ 4.07 ⫽ 1.86 yr. ENTER

ENTER

Figure 6.103

Figure 6.104

10,500

10,500

10

0

A. You’ve just seen how we can apply vertical translations in context

10

0

7000

7000


䊳

B. Horizontal Translations: y ⴝ A sin(Bt ⴙ C) ⴙ D and y ⴝ A cos(Bt ⴙ C) ⴙ D In some cases, scientists would rather “benchmark” their study of sinusoidal phenomena by placing the average value at t ⫽ 0 instead of a maximum value (as in Example 2), or by placing the maximum or minimum value at t ⫽ 0 instead of the average value (as in Example 1). Rather than make additional studies or recompute using available data, we can simply shift these graphs using a horizontal translation. To help understand how, consider the graph of Y1 ⫽ X2. The graph is a parabola that opens upward with a vertex at the origin. Comparing this function with Y2 ⫽ 1X ⫺ 32 2 and Y3 ⫽ 1X ⫹ 32 2, we note Y2 is simply the parent graph shifted 3 units right (Figure 6.105), and Y3 is the parent graph shifted 3 units left or “opposite the sign” (Figure 6.106). While quadratic functions have no maximum value if A ⬎ 0, these graphs are a good reminder of how a basic graph can be horizontally shifted. We simply replace the independent variable x with 1x ⫾ h2 or t with 1t ⫾ h2, where h is the desired shift and the sign is chosen depending on the direction of the shift. Figure 6.106

Figure 6.105 5

⫺3

5

6

⫺1

⫺6

3

⫺1

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EXAMPLE 3

䊳

Investigating Horizontal Shifts of Trigonometric Graphs Use a horizontal translation to shift the graph from Example 2 so that the average population begins at t ⫽ 0. Verify the result on a graphing calculator, then find a sine function that gives the same graph as the shifted cosine function.

Solution

䊳

␲ For P1t2 ⫽ 1200 cosa tb ⫹ 9000 from 5 Example 2, the average value occurs at t ⫽ 2.5, and by symmetry at t ⫽ ⫺2.5. For the average value to occur at t ⫽ 0, we shift the graph to the right 2.5 units. Replacing t with 1t ⫺ 2.52

11,000

⫺1

10

␲ gives P1t2 ⫽ 1200 cos c 1t ⫺ 2.52 d ⫹ 9000. 5 7000 A graphing calculator shows the desired result is obtained (see figure). The new graph appears to be a sine function with the same ␲ amplitude and period, and in fact the equation is y ⫽ 1200 sina tb ⫹ 9000. 5 Now try Exercises 21 and 22

䊳

␲ Equations like P1t2 ⫽ 1200 cos c 1t ⫺ 2.52 d ⫹ 9000 from Example 3 are said to 5 be written in shifted form, since changes to the input variable are seen directly and we can easily tell the magnitude and direction of the shift. To obtain the standard form ␲ ␲ we distribute the value of B: P1t2 ⫽ 1200 cos a t ⫺ b ⫹ 9000. In general, the 5 2 standard form of a sinusoidal equation (using either a cosine or sine function) is written y ⫽ A sin1Bt ⫾ C2 ⫹ D, with the shifted form found by factoring out B from Bt ⫾ C: y ⫽ A sin1Bt ⫾ C2 ⫹ D S y ⫽ A sin c B at ⫾

C bd ⫹ D B

In either case, C gives what is known as the phase angle of the function, and is used in a study of AC circuits and other areas, to discuss how far a given function is C “out of phase” with a reference function. In the latter case, is simply the horizontal B shift (or phase shift) of the function and gives the magnitude and direction of this shift (opposite the sign). Characteristics of Sinusoidal Equations Transformations of the graph of y ⫽ sin t can be written as y ⫽ A sin1Bt2, where 1. 冟A冟 gives the amplitude of the graph, or the maximum displacement from the average value. 2␲ 2. B is related to the period P of the graph according to the ratio P ⫽ . 冟B冟 Translations of y ⫽ A sin1Bt2 can be written as follows:

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WORTHY OF NOTE

Standard form

It’s important that you don’t confuse the standard form with the shifted form. Each has a place and purpose, but the horizontal shift can be identified only by focusing on the change in an independent variable. Even though the equations y ⫽ 41x ⫹ 32 2 and y ⫽ 12x ⫹ 62 2 are equivalent, only the first explicitly shows that y ⫽ 4x2 has been shifted three units left. Likewise y ⫽ sin321t ⫹ 32 4 and y ⫽ sin12t ⫹ 62 are equivalent, but only the first explicitly gives the horizontal shift (three units left). Applications involving a horizontal shift come in an infinite variety, and the shifts are generally not uniform or standard.

EXAMPLE 4

583

䊳

Shifted form C y ⫽ A sin1Bt ⫾ C2 ⫹ D y ⫽ A sin c B at ⫾ b d ⫹ D B C 3. In either case, C is called the phase angle of the graph, while ⫾ gives the B magnitude and direction of the horizontal shift (opposite the given sign). 4. D gives the vertical shift of the graph, and the location of the average value. The shift will be in the same direction as the given sign. Knowing where each cycle begins and ends is a helpful part of sketching a graph of the equation model. The primary interval for a sinusoidal graph can be found by solving the inequality 0 ⱕ Bt ⫾ C 6 2␲, with the reference rectangle and rule of fourths giving the zeroes, max/min values, and a sketch of the graph in this interval. The graph can then be extended in either direction, and shifted vertically as needed.

Analyzing the Transformation of a Trig Function Identify the amplitude, period, horizontal shift, vertical shift (average value), and endpoints of the primary interval. ␲ 3␲ b⫹6 y ⫽ 2.5 sin a t ⫹ 4 4

Solution

䊳

The equation gives an amplitude of 冟 A冟 ⫽ 2.5, with an average value of D ⫽ 6. The maximum value will be y ⫽ 2.5112 ⫹ 6 ⫽ 8.5, with a minimum of 2␲ ␲ y ⫽ 2.51⫺12 ⫹ 6 ⫽ 3.5. With B ⫽ , the period is P ⫽ ⫽ 8. To find the 4 ␲/4 ␲ ␲ 3␲ b⫽ horizontal shift, we factor out to write the equation in shifted form: a t ⫹ 4 4 4 ␲ 1t ⫹ 32. The horizontal shift is 3 units left. For the endpoints of the primary 4 ␲ interval we solve 0 ⱕ 1t ⫹ 32 6 2␲, which gives ⫺3 ⱕ t 6 5. 4 Now try Exercises 23 through 34

䊳

Figure 6.107 ␲ The analysis of y ⫽ 2.5 sin c 1t ⫹ 32 d ⫹ 6 10 4 from Example 4 can be verified on a graphing calculator. Enter the function as Y1 on the Y= screen and set an appropriate window 5.2 size using the information gathered. Pressing ⫺3 the TRACE key and ⫺3 gives the average value y ⫽ 6 as output. Repeating this for x ⫽ 5 shows one complete cycle has been completed 0 (Figure 6.107). To help gain a better understanding of sinusoidal functions, their graphs, and the role the coefficients A, B, C, and D play, it’s often helpful to reconstruct the equation of a given graph. ENTER

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EXAMPLE 5

䊳

Determining an Equation of a Trig Function from Its Graph Determine an equation of the given graph using a sine function.

Solution

䊳

From the graph it is apparent the maximum value is 300, with a minimum of 50. This gives a value 300 50 300 50 175 for D and 125 of 2 2 for A. The graph completes one cycle from t 2 ␲ to t 18, showing P 18 2 16 and B . 8 The average value first occurs at t 2, so the basic graph has been shifted to the right 2 units. ␲ The equation is y 125 sin c 1t 22 d 175. 8

y 350 300 250 200 150 100 50 0

4

8

12

16

20

24


䊳

␲ 1X 22 d 175 from Example 5, the 8 table feature of a calculator can provide a convincing Figure 6.108 check of the solution. After recognizing the average, maximum, and minimum values of the given graph occur at regular intervals (as is true for any sinusoidal function), setting up a table to start at 2 with step size of 4 units will produce the 2nd GRAPH (TABLE) shown in Figure 6.108. Note the screen shown contains the coordinates of all seven points indicated in the graph for this example. In addition to the

B. You’ve just seen how we can apply horizontal translations in context

t

GRAPH

of Y1 125 sin c

C. Simple Harmonic Motion: y ⴝ A sin(Bt) or y ⴝ A cos(Bt) The periodic motion of springs, tides, sound, and other phenomena all exhibit what is known as harmonic motion, which can be modeled using sinusoidal functions.

Harmonic Models—Springs Consider a spring hanging from a beam with a weight attached to one end. When the weight is at rest, we say it is in equilibrium, or has zero displacement from center. See Figure 6.109. Stretching the spring and then releasing it causes the weight to “bounce up and down,” with its displacement from center neatly modeled over time by a sine wave (see Figure 6.110). For objects in harmonic motion (there are other harmonic models), the input variable t is always a time unit (seconds, minutes, days, etc.), so in addition to the period of the sinusoid, we are very interested in its frequency—the number of cycles it completes per unit time. Since the period gives the time required to complete one cycle, the frequency B 1 . f is given by f P 2␲

Figure 6.109 At rest

Stretched

Released

4

4

4

2

2

2

0

0

0

2

2

2

4

4

4

Figure 6.110 Harmonic motion Displacement (cm) 4

t (seconds) 0 0.5

4

1.0

1.5

2.0

2.5

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EXAMPLE 6

䊳

585

Applications of Sine and Cosine: Harmonic Motion For the harmonic motion of the weight modeled by the sinusoid in Figure 6.110, a. Find an equation of the form y A cos1Bt2 . b. Determine the frequency. c. Use the equation to find the position of the weight at t 1.8 sec.

Solution

䊳

a. By inspection the graph has an amplitude A 3 and a period P 2. After 2␲ substitution into P , we obtain B ␲ and the equation y 3 cos1␲t2. B 1 b. Frequency is the reciprocal of the period so f , showing one-half a cycle is 2 completed each second (as the graph indicates). c. Evaluating the model at t 1.8 gives y 3 cos 3␲11.82 4 2.43, meaning the weight is 2.43 cm below the equilibrium point at this time. Now try Exercises 55 through 58

䊳

Harmonic Models—Sound Waves A second example of harmonic motion is the production of sound. For the purposes of this study, we’ll look at musical notes. The vibration of matter produces a pressure wave or sound energy, which in turn vibrates the eardrum. Through the intricate structure of the middle ear, this sound energy is converted into mechanical energy and sent to the inner ear where it is converted to nerve impulses and transmitted to the brain. If the sound wave has a high frequency, the eardrum vibrates with greater frequency, which the brain interprets as a “high-pitched” sound. The intensity of the sound wave can also be transmitted to the brain via these mechanisms, and if the arriving sound wave has a high amplitude, the eardrum vibrates more forcefully and the sound is interpreted as “loud” by the brain. These characteristics are neatly modeled using y A sin1Bt2 . For the moment we will focus on the frequency, keeping the amplitude constant at A 1. The musical note known as A4 or “the A above middle C” is produced with a frequency of 440 vibrations per second, or 440 hertz (Hz) (this is the note most often used in the tuning of pianos and other musical instruments). For any given note, the same note one octave higher will have double the frequency, and the same note one octave 1 lower will have one-half the frequency. In addition, with f the value of P 1 B 2␲a b can always be expressed as B 2␲f, so A4 has the equation P y sin 344012␲t2 4 (after rearranging the factors). The same note one octave lower is A3 and has the equation y sin 322012␲t2 4 , Figure 6.111 with one-half the frequency. To draw the representative graphs, we must scale the t-axis in A4 y sin[440(2␲t)] y very small increments (seconds 103) A3 y sin[220(2␲t)] 1 1 0.0023 for A4, and since P 440 t (sec 103) 1 P 0.0045 for A3. Both are graphed 0 220 1 2 3 4 5 in Figure 6.111, where we see that the higher note completes two cycles in the same inter- 1 val that the lower note completes one.

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EXAMPLE 7

䊳

Applications of Sine and Cosine: Sound Frequencies The table here gives the frequencies for three octaves of the 12 “chromatic” notes with frequencies between 110 Hz and 840 Hz. Two of the 36 notes are graphed in the figure. Which two? y 1

0

y1 sin[ f (2␲t)] y2 sin[ f (2␲t)] t (sec

103)

1.0 2.0 3.0 4.0 5.0 6.0 7.0 1

Solution

䊳

C. You’ve just seen how we can solve applications involving harmonic motion

Frequency by Octave Note

Octave 2

Octave 3

Octave 4

A

110.00

220.00

440.00

A#

116.54

233.08

466.16

B

123.48

246.96

493.92

C

130.82

261.64

523.28

C#

138.60

277.20

554.40

D

146.84

293.68

587.36

D#

155.56

311.12

622.24

E

164.82

329.24

659.28

F

174.62

349.24

698.48

F#

185.00

370.00

740.00

G

196.00

392.00

784.00

G#

207.66

415.32

830.64

Since amplitudes are equal, the only differences are the frequency and period of the notes. It appears that y1 has a period of about 0.004 sec, giving a frequency of 1 ⫽ 250 Hz—very likely a B3 (in bold). The graph of y2 has a period of about 0.004 1 0.006, for a frequency of ⬇ 167 Hz—probably an E2 (also in bold). 0.006 Now try Exercises 59 through 62

䊳

D. Vertical and Horizontal Translations of Other Trig Functions In Section 6.4, we used the graphs of y ⫽ A sin1Bt2 and y ⫽ A cos1Bt2 to help graph y ⫽ A csc1Bt2 and y ⫽ A sec1Bt2 , respectively. If a vertical or horizontal translation is being applied to a cosecant/secant function, we simply apply the same translations to the underlying sine/cosine function as an aid to graphing the function at hand. EXAMPLE 8

䊳

Graphing y ⴝ A sec1Bt ⴙ C2 ⴙ D ␲ ␲ Draw a sketch of y ⫽ 2 sec a t ⫺ b ⫺ 3 for t in [0, 9]. 3 2

Solution

䊳

␲ ␲ To graph this secant function, consider the function y ⫽ 2 cos a t ⫺ b ⫺ 3. In 3 2 3 ␲ shifted form, we have y ⫽ 2 cos c at ⫺ b d ⫺ 3. With an amplitude of 0 A 0 ⫽ 2 3 2 2␲ ⫽ 6, the reference rectangle will be 4 units high and 6 units and period of P ⫽ ␲/3 wide. The rule of fourths shows the zeroes and max/min values of the unshifted 3 9 3 graph will occur at 0, , 3, , and 6. Applying a horizontal shift of ⫹ units gives 2 2 2

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15 3 9 the values , 3, , 6, and , which we combine with the reference rectangle to 2 2 2 produce the cosine and secant graphs shown next. Note the vertical asymptotes of the secant function occur at the zeroes of the cosine function (Figure 6.112). By shifting these results 3 units downward (centered on the line y 3), we ␲ ␲ obtain the graph of y 2 sec a t b 3 (Figure 6.113). 3 2 Figure 6.113

Figure 6.112 ␲ 3t

y 2 sec

y

␲

␲ 2

y

y 2 sec 3 t

␲ 2

5

t

3

1

4 3 2

1

1

2

1

2

3

4

6

7

8

9

3 1 2 3

1

2

3

4

5

6

7

8

9

t ␲

y 2 cos 3 t

4 ␲ 2

4

5

␲

y 2 cos 3 t

6

␲ 2

3

7


䊳

To horizontally and/or vertically translate the graphs of y A tan1Bt2 and y A cot1Bt2, we once again apply the translations to the basic graph. Recall for these two functions: ␲ ␲ 1. The primary interval of y tan t is c , d , while that of y cot t is 30, ␲4 . 2 2 2. y tan t increases while y cot t decreases on their respective domains. EXAMPLE 9

䊳

Graphing y ⴝ A tan1Bt ⴙ C2 Draw a sketch of y 2.5 tan10.5t 1.52 over two periods.

Figure 6.114 y y 2.5 tan[0.5(t)] 10

Solution

WORTHY OF NOTE The reference rectangle shown is 5 units high by 2 units wide.

䊳

In shifted form, this equation is y 2.5 tan 3 0.51t 32 4 . With A 2.5 and B 0.5, the graph is vertically stretched with a period of 2␲ (see Figure 6.114). Note that t ␲ and ␲ gives the location of the asymptotes for the unshifted graph, ␲ ␲ while t and gives 2.5 and 2.5, 2 2 respectively. Shifting the graph 3 units left produces the one shown in Figure 6.115 (␲ 3 6.14, ␲ 3 4.57, 0 3 3, 2 ␲ 3 1.43, and ␲ 3 0.14). 2

8 6 4 2 10 8 6 4 2 2

2

4

6

8 10

t

4 6 8 10

Figure 6.115 y 2.5 tan(0.5t 1.5)

y 10 8 6 4 2 10 8 6 4 2 2

2

4

6

8 10

t

4 6

D. You’ve just seen how we can apply vertical and horizontal translations to cosecant, secant, tangent, and cotangent functions

8 10


䊳

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E. Applications of the Remaining Trig Functions We end this section with two examples of how tangent, cotangent, secant, and cosecant functions can be applied. Numerous others can be found in the exercise set. EXAMPLE 10

䊳

Applications of y ⴝ A tan1Bt2 : Modeling the Movement of a Light Beam One evening, in port during a Semester at Sea, Richard is debating a project choice for his Precalculus class. Looking out his porthole, he notices a revolving light turning at a constant speed near the corner of a long warehouse. The light throws its beam along the length of the warehouse, then disappears into the air, and then returns time and time again. Suddenly—Richard has his project. He notes the time it takes the beam to traverse the warehouse wall is very close to 4 sec, and in the morning he measures the wall’s length at 127.26 m. His project? Modeling the distance of the beam from the corner of the warehouse as a function of time using a tangent function. Can you help?

Solution

䊳

The equation model will have the form D1t2 A tan1Bt2, where D(t) is the distance (in meters) of the beam from the corner after t sec. The distance along the wall is measured in positive values so we’re using only 12 the period of the function, giving 12P 4 (the beam “disappears” at t 4) so P 8. Substitution in the ␲ ␲ period formula gives B and the equation D A tan a tb. 8 8 Knowing the beam travels 127.26 m in about 4 sec (when it disappears into infinity), we’ll use t 3.9 and D 127.26 in order to solve for A and complete our equation model (see note following this example). ␲ A tan a tb D 8 ␲ A tan c 13.92 d 127.26 8 127.26 A ␲ tan c 13.92 d 8 5

equation model

substitute 127.26 for D and 3.9 for t

solve for A

result

One equation approximating the distance of the beam from the corner of the ␲ warehouse is D1t2 5 tan a tb. 8 Now try Exercises 63 through 66

䊳

For Example 10, we should note the choice of 3.9 for t was arbitrary, and while we obtained an “acceptable” model, different values of A would be generated for other choices. For instance, t 3.95 gives A 2.5, while t 3.99 gives A 0.5. The true value of A depends on the distance of the light from the corner of the warehouse wall. In any case, it’s interesting to note that at t 2 sec (one-half the time it takes the beam to disappear), the beam has traveled only 5 m from the corner of the building: ␲ D122 5 tan a b 5 m. Although the light is rotating at a constant angular speed, 4 the speed of the beam along the wall increases dramatically as t gets close to 4 sec.

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EXAMPLE 11

䊳

Applications of y ⴝ A csc1Bt2 : Modeling the Length of a Shadow During the long winter months in southern Michigan, Daniel begins planning for his new solar-powered water heater. He needs to choose a panel location on his roof, and is primarily concerned with the moving shadows of the trees throughout the day. To this end, he begins studying the changing shadow length of a tree in front of his house. At sunrise, the shadow is too long to measure. As the morning progresses, the shadow decreases in length, until at midday it measures its shortest length of 7 ft. As the day moves on, the shadow increases in length, until at sunset it is again too long to measure. If on one particular day sunrise occurs at 6:00 A.M. and sunset at 6:00 P.M.: a. Use the cosecant function to model the tree’s shadow length. b. Approximate the shadow length at 10:00 A.M.

Solution

䊳

a. It appears the function L1t2 A csc1Bt2 will model the shadow length, since the shadow decreases from an “infinitely large” length to a minimum length, then returns to an infinite length. Let t 0 correspond to 6:00 A.M. and note Daniel’s observations over the 12 hr of daylight. Since the length of the shadow is measured in positive values, we need only 12 the period of the ␲ function, giving 12P 12 (so P 24). The period formula gives B and 12 ␲ the equation model is L1t2 A csc a tb. We solve for A to complete the 12 model, using L 7 when t 6 (noon). ␲ tb 12 ␲ 7 A csc c 162 d 12 7 A ␲ csc a b 2 7 1 7

L1t2 A csc a

equation model

substitute 7 for L and 6 for t

solve for A;

6␲ ␲ 12 2

␲ csc a b 1 2 result

The equation modeling the shadow length of the tree is L1t2 7 csc a

E. You’ve just seen how we can solve applications involving the tangent, cotangent, secant, and cosecant functions

␲ tb. 12

␲ 2 13 b. Evaluating the model at t 4 (10 A.M.) gives L 7 csc a b 7 a b 8.08, 3 3 meaning the shadow is approximately 8 ft, 1 in. long at 10 A.M. Now try Exercises 67 and 68

䊳

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6.5 EXERCISES 䊳



䊳

1. A sinusoidal wave is one that can be modeled by functions of the form or .

2. Given h, k > 0, the graph of y sin x k is the graph of y sin x shifted k units. The graph of y sin1x h2 is the graph of y sin x shifted h units.

3. To find the primary interval of a sinusoidal graph, solve the inequality .

4. Given the period P, the frequency is given the frequency f, the value of B is

5. Explain/Discuss the difference between the standard form of a sinusoidal equation, and the shifted form. How do you obtain one from the other? For what benefit?

6. Write out a step-by-step procedure for sketching 1 ␲ the graph of y 30 sina t b 10. Include 2 2 use of the reference rectangle, primary interval, zeroes, max/mins, and so on. Be complete and thorough.

, and .


Use the graphs given to (a) state the amplitude A and period P of the function; (b) estimate the value at x ⴝ 14; and (c) estimate the interval in [0, P] where f(x) 20.

7.

8.

f (x) 50

6

12

18

24

30

5

50

10 15 20 25

30

x

50

Use the graphs given to (a) state the amplitude A and period P of the function; (b) estimate the value at x ⴝ 2; and (c) estimate the interval in [0, P], where f(x) < ⴚ100.

9.

10.

f (x) 250

3 3 9 3 15 9 21 6 27 4 2 4 4 2 4 4

f (x) 125

x

250

1

2

3

4

5

6

7

8 x

125

Use the information given to write a sinusoidal equation 2␲ and sketch its graph. Use B ⴝ . P

11. Max: 100, min: 20, P 30 12. Max: 95, min: 40, P 24

14. Max: 12,000, min: 6500, P 10 Use the information given to write a sinusoidal equation, sketch its graph, and answer the question posed.

f (x) 50

x

13. Max: 20, min: 4, P 360

15. In Geneva, Switzerland, the daily temperature in January ranges from an average high of 39°F to an average low of 29°F. (a) Find a sinusoidal equation model for the daily temperature; (b) sketch the graph; and (c) approximate the time(s) each January day the temperature reaches the freezing point (32°F). Assume t 0 corresponds to noon. 16. In Nairobi, Kenya, the daily temperature in January ranges from an average high of 77°F to an average low of 58°F. (a) Find a sinusoidal equation model for the daily temperature; (b) sketch the graph; and (c) approximate the time(s) each January day the temperature reaches a comfortable 72°F. Assume t 0 corresponds to noon. 17. In Oslo, Norway, the number of hours of daylight reaches a low of 6 hr in January, and a high of nearly 18.8 hr in July. (a) Find a sinusoidal equation model for the number of daylight hours each month; (b) sketch the graph; and (c) approximate the number of days each year there are more than 15 hr of daylight. Use 1 month 30.5 days. Assume t 0 corresponds to January 1. Source: www.visitnorway.com/templates.

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18. In Vancouver, British Columbia, the number of hours of daylight reaches a low of 8.3 hr in January, and a high of nearly 16.2 hr in July. (a) Find a sinusoidal equation model for the number of daylight hours each month; (b) sketch the graph; and (c) approximate the number of days each year there are more than 15 hr of daylight. Use 1 month 30.5 days. Assume t 0 corresponds to January 1. Source: www.bcpassport.com/vital/temp.

19. Recent studies seem to indicate the population of North American porcupine (Erethizon dorsatum) varies sinusoidally with the solar (sunspot) cycle due to its effects on Earth’s ecosystems. Suppose the population of this species in a certain locality is modeled by the 2␲ function P1t2 250 cosa tb 950, where P(t) 11 represents the population of porcupines in year t. Use the model to (a) find the period of the function; (b) graph the function over one period; (c) find the maximum and minimum values; and (d) estimate the number of years the population is less than 740 animals. Source: Ilya Klvana, McGill University (Montreal), Master of Science thesis paper, November 2002.

20. The population of mosquitoes in a given area is primarily influenced by precipitation, humidity, and temperature. In tropical regions, these tend to fluctuate sinusoidally in the course of a year. Using trap counts and statistical projections, fairly accurate estimates of a mosquito population can be obtained. Suppose the population in a certain region was modeled by the function ␲ P1t2 50 cos a tb 950, where P(t) was the 26 mosquito population (in thousands) in week t of the year. Use the model to (a) find the period of the function; (b) graph the function over one period; (c) find the maximum and minimum population values; and (d) estimate the number of weeks the population is less than 915,000. 21. Use a horizontal translation to shift the graph from Exercise 19 so that the average population of the North American porcupine begins at t 0. Verify results on a graphing calculator, then find a sine function that gives the same graph as the shifted cosine function. 22. Use a horizontal translation to shift the graph from Exercise 20 so that the average population of mosquitoes begins at t 0. Verify results on a graphing calculator, then find a sine function that gives the same graph as the shifted cosine function.

Identify the amplitude (A), period (P), horizontal shift (HS), vertical shift (VS), and endpoints of the primary interval (PI) for each function given.

23. y 120 sin c

␲ 1t 62 d 12

24. y 560 sin c

␲ 1t 42 d 4

␲ ␲ 25. h1t2 sin a t b 6 3

26. r1t2 sin a

␲ ␲ 27. y sin a t b 4 6

␲ 5␲ 28. y sin a t b 3 12

29. f 1t2 24.5 sin c

␲ 1t 2.52 d 15.5 10

30. g1t2 40.6 sin c

␲ 1t 42 d 13.4 6

2␲ ␲ t b 10 5

␲ 5␲ 31. g1t2 28 sin a t b 92 6 12 32. f 1t2 90 sin a

␲ ␲ t b 120 10 5

␲ ␲ 33. y 2500 sin a t b 3150 4 12 34. y 1450 sin a

3␲ ␲ t b 2050 4 8

Find an equation of the graph given. Write answers in the form y ⴝ A sin1Bt ⴙ C2 ⴙ D and check with a graphing calculator.

35.

0

37.

36.

y 700 600 500 400 300 200 100 6

12

18

t 24

0

38.

y 20 18 16 14 12 10 8 0

39.

25

50

75

100

t 125

0

40.

t 90

180

270

360

t 8

16

24

32

y 140 120 100 80 60 40 20

y 12 10 8 6 4 2 0

y 140 120 100 80 60 40 20

t 6

12

18

24

30

36

y 6000 5000 4000 3000 2000 1000 0

t 6

12

18

24

30

36

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Sketch one complete period of each function.

␲ 41. f 1t2 25 sin c 1t 22 d 55 4 42. g1t2 24.5 sin c

␲ 1t 2.52 d 15.5 10

43. h1t2 3 sin 14t ␲2

␲ 44. p1t2 2 sin a3t b 2 Sketch the following functions over the indicated intervals.

3␲ 1 45. y 5 sec c at b d 2; 3 3␲, 6␲ 4 3 2

47. y 0.7 csc a␲t

␲ b 1.2; 31.25, 1.754 4

␲ ␲ 48. y 1.3 sec a t b 1.6; 3 2, 64 3 6 Sketch two complete periods of each function.

␲ 49. y 0.5 tan c 1t 22 d 4 ␲ 50. y 1.5 cot c 1t 12 d 2 51. y 10 cot12t 12 52. y 8 tan13t 22

1 46. y 3 csc c 1t ␲2 d 1; 32␲, 4␲ 4 2 䊳


53. The relationship between the coefficient B, the frequency f, and the period P In many applications of trigonometric functions, the equation y A sin1Bt2 is written as y A sin 3 12␲f 2t 4 , where B 2␲f. Justify the new 1 2␲ equation using f and P . In other words, P B explain how A sin(Bt) becomes A sin 3 12␲f 2t 4 , as though you were trying to help another student with the ideas involved.

䊳

54. Number of daylight hours: 2␲ K 1t ⴚ 792 d ⴙ 12 D1t2 ⴝ sin c 2 365 The number of daylight hours for a particular day of the year is modeled by the formula given, where D(t) is the number of daylight hours on day t of the year and K is a constant related to the total variation of daylight hours, latitude of the location, and other factors. For the city of Reykjavik, Iceland, K 17, while for Detroit, Michigan, K 6. How many hours of daylight will each city receive on June 30 (the 182nd day of the year)?

APPLICATIONS

55. Harmonic motion: A weight on the end of a spring is oscillating in harmonic motion. The equation model for the oscillations is ␲ d1t2 6 sina tb, where d is the 2 distance (in centimeters) from the equilibrium point in t sec. a. What is the period of the motion? What is the frequency of the motion? b. What is the displacement from equilibrium at t 2.5? Is the weight moving toward the equilibrium point or away from equilibrium at this time?

c. What is the displacement from equilibrium at t 3.5? Is the weight moving toward the equilibrium point or away from equilibrium at this time? d. How far does the weight move between t 1 and t 1.5 sec? What is the average velocity for this interval? Do you expect a greater or lesser velocity for t 1.75 to t 2? Explain why.

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56. Harmonic motion: The bob on the end of a 24-in. pendulum is oscillating in harmonic motion. The equation model for the oscillations is d1t2 20 cos14t2 , where d is the distance (in inches) from the equilibrium point, t sec after being released from d d one side. a. What is the period of the motion? What is the frequency of the motion? b. What is the displacement from equilibrium at t 0.25 sec? Is the weight moving toward the equilibrium point or away from equilibrium at this time? c. What is the displacement from equilibrium at t 1.3 sec? Is the weight moving toward the equilibrium point or away from equilibrium at this time? d. How far does the bob move between t 0.25 and t 0.35 sec? What is its average velocity for this interval? Do you expect a greater velocity for the interval t 0.55 to t 0.6? Explain why. 57. Harmonic motion: A simple pendulum 36 in. in length is oscillating in harmonic motion. The bob at the end of the pendulum swings through an arc of 30 in. (from the far left to the far right, or onehalf cycle) in about 0.8 sec. What is the equation model for this harmonic motion? 58. Harmonic motion: As part of a study of wave motion, the motion of a float is observed as a series of uniform ripples of water move beneath it. By careful observation, it is noted that the float bobs up and down through a distance of 2.5 cm every 1 sec. What is the equation model for this 3 harmonic motion? 59. Sound waves: Two of the musical notes from the chart on page 586 are graphed in the figure. Use the graphs given to determine which two. y

y2 sin[ f (2␲t)]

1

t (sec 103) 0 2

1

4

y1 sin[ f (2␲t)]

6

8

10

60. Sound waves: Two chromatic notes not on the chart from page 586 are graphed in the figure. Use the graphs and the discussion regarding octaves to determine which two. Note the scale of the t-axis has been changed to hundredths of a second. y 1

y2 sin[ f (2␲t)] t (sec 102)

0 0.4

0.8

1.2

1.6

2.0

y1 sin[ f (2␲t)] 1

Sound waves: Use the chart on page 586 to write the equation for each note in the form y ⴝ sin[ f(2␲t)] and clearly state the period of each note.

61. notes D2 and G3

62. the notes A4 and C2

Tangent function data models: Model the data in Exercises 63 and 64 using the function y A tan(Bx). State the period of the function, the location of the asymptotes, the value of A, and name the point (x, y) used to calculate A (answers may vary). Use your equation model to evaluate the function at x 2 and x 2. What observations can you make? Also see Exercise 73. 63.

Input

Output

Input

Output

6

q

1

1.4

5

20

2

3

4

9.7

3

5.2

3

5.2

4

9.7

2

3

5

20

1

1.4

6

q

0

0

64. Input Input

Output Input Output

3

q

2.5 2 1.5 1

Output Input

Output

0.5

6.4

91.3

1

13.7

44.3

1.5

23.7

23.7

2

44.3

13.7

2.5

91.3

0.5

6.4

3

q

0

0

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Laser Light

Exercise 65 65. As part of a lab setup, a laser pen is made to swivel on a large protractor as illustrated in the figure. For their lab project, students are asked to take the Distance ␪ instrument to one end of (degrees) (cm) a long hallway and 0 0 measure the distance of 10 2.1 the projected beam relative to the angle the 20 4.4 pen is being held, and 30 6.9 collect the data in a 40 10.1 table. Use the data to 50 14.3 find a function of the 60 20.8 form y A tan1B␪2. 70 33.0 State the period of the function, the location of 80 68.1 the asymptotes, the 89 687.5 value of A, and name the point (␪, y) you used to calculate A (answers may vary). Based on the result, can you approximate the length of the laser pen? Note that in degrees, the 180° . period formula for tangent is P B

66. Use the equation model obtained in Exercise 65 to compare the values given by the equation with the actual data. As a percentage, what was the largest deviation between the two? 67. Shadow length: At high noon, a flagpole in Oslo, Norway, casts a 10-m-long shadow during the month of January. Using information from Exercise 17, (a) find a cosecant function that models the shadow length, and (b) use the model to find the length of the shadow at 2:00 P.M. 䊳

6–86


68. Shadow length: At high noon, the “Living Shangri-La” skyscraper in Vancouver, British Columbia, casts a 15-m-long shadow during the month of June. Given there are 16 hr of daylight that month, (a) find a cosecant function that models the shadow length, and (b) use the model to find the length of the shadow at 7:30 A.M. Daylight hours model: Solve using a graphing calculator and the formula given in Exercise 54. 69. For the city of Caracas, Venezuela, K 1.3, while for Tokyo, Japan, K 4.8. a. How many hours of daylight will each city receive on January 15th (the 15th day of the year)? b. Graph the equations modeling the hours of daylight on the same screen. Then determine (i) what days of the year these two cities will have the same number of hours of daylight, and (ii) the number of days each year that each city receives 11.5 hr or less of daylight. 70. For the city of Houston, Texas, K 3.8, while for Pocatello, Idaho, K 6.2. a. How many hours of daylight will each city receive on December 15 (the 349th day of the year)? b. Graph the equations modeling the hours of daylight on the same screen. Then determine (i) how many days each year Pocatello receives more daylight than Houston, and (ii) the number of days each year that each city receives 13.5 hr or more of daylight.


71. The formulas we use in mathematics can sometimes seem very mysterious. We know they “work,” and we can graph and evaluate them — but where did they come from? Consider the formula for the number of daylight hours from Exercise 54: 2␲ K 1t 792 d 12. D1t2 sin c 2 365 a. We know that the addition of 12 represents a vertical shift, but what does a vertical shift of 12 mean in this context?

b. We also know the factor 1t 792 represents a phase shift of 79 to the right. But what does a horizontal (phase) shift of 79 mean in this context? K c. Finally, the coefficient represents a change 2 in amplitude, but what does a change of amplitude mean in this context? Why is the coefficient bigger for the northern latitudes?

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Section 6.6 The Trigonometry of Right Triangles

72. Use a graphing calculator to graph the equation 3x f 1x2 ⫽ ⫺ 2 sin12x2 ⫺ 1.5. 2 a. Determine the interval between each peak of the graph. What do you notice? 3x ⫺ 1.5 on the same screen b. Graph g1x2 ⫽ 2 and comment on what you observe. c. What would the graph of 3x f 1x2 ⫽ ⫺ ⫹ 2 sin12x2 ⫹ 1.5 look like? 2 What is the x-intercept? 73. Rework Exercises 63 and 64, obtaining a new equation for the data using a different ordered pair to compute the value of A. What do you notice? Try yet another ordered pair and calculate A once again for another equation Y2. Complete a table of values

䊳

595

using the given inputs, with the outputs of the three equations generated (original, Y1, and Y2). Does any one equation seem to model the data better than the others? Are all of the equation models “acceptable”? Please comment. 74. Regarding Example 10, we can use the standard distance/rate/time formula D ⫽ RT to compute the average velocity of the beam of light along the wall D in any interval of time: R ⫽ . For example, using T ␲ D1t2 ⫽ 5 tana tb, the average velocity in the 8 D122 ⫺ D102 ⫽ 2.5 m/sec. interval [0, 2] is 2⫺0 Calculate the average velocity of the beam in the time intervals [2, 3], [3, 3.5], and [3.5, 3.8] sec. What do you notice? How would the average velocity of the beam in the interval [3.9, 3.99] sec compare?


75. (6.1) In what quadrant does the arc t ⫽ 3.7 terminate? What is the reference arc?

76. (3.2) Given f 1x2 ⫽ ⫺31x ⫹ 12 2 ⫺ 4, name the vertex and solve the inequality f 1x2 7 0.

6.6


A. Find values of the six

C. D.

E.

78. (6.3/6.4) Sketch the graph of (a) y ⫽ cos t in the interval [0, 2␲) and (b) y ⫽ tan t in the interval ␲ 3␲ b. a⫺ , 2 2

The Trigonometry of Right Triangles

LEARNING OBJECTIVES

B.

77. (3.1) Compute the sum, difference, product, and quotient of ⫺1 ⫹ i 15 and ⫺1 ⫺ i 15.

trigonometric functions from their ratio definitions Solve a right triangle given one angle and one side Solve a right triangle given two sides Use cofunctions and complements to write equivalent expressions Solve applications of right triangles

Over a long period of time, what began as a study of chord lengths by Hipparchus, Ptolemy, Aryabhata, and others became a systematic application of the ratios of the sides of a right triangle. In this section, we develop the sine, cosine, and tangent functions from a right triangle perspective, and explore certain relationships that exist between them. This view of the trig functions also leads to a number of significant applications.

A. Trigonometric Ratios and Their Values In Section 6.1, we looked at applications involving 45-45-90 and 30-60-90 triangles, using the fixed ratios that exist between their sides. To apply this concept more generally using other right triangles, each side is given a specific name using its location relative to a specified angle. For the 30-60-90 triangle in Figure 6.116(a), the side opposite (opp) and the side adjacent (adj) are named with respect to the 30° angle, with the hypotenuse (hyp) always across from the right angle. Likewise for the 45-45-90 triangle in Figure 6.116(b).

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Figure 6.116 (a) hyp 2x

60 opp x

30 adj √3x

for 45

(b)

for 30 opp 1 2 hyp

hyp √2x

adj √3 2 hyp

45

opp x

45 adj x

opp 1 √3 3 adj √3

opp 1 √2 2 hyp √2 adj 1 √2 2 hyp √2 opp adj

1 1

Using these designations to define the various trig ratios, we can now develop a systematic method for applying them. Note that the x’s “cancel” in each ratio, reminding us the ratios are independent of the triangle’s size (if two triangles are similar, the ratio of corresponding sides is constant). Ancient mathematicians were able to find values for the ratios corresponding to any acute angle in a right triangle, and realized that naming each ratio would be opp opp adj S sine, S cosine, and S tangent. Since helpful. These names are hyp hyp adj each ratio depends on the measure of an acute angle ␪, they are often referred to as functions of an acute angle and written in function form. sine ␪ ⫽

opp hyp

cosine ␪ ⫽

adj hyp

tangent ␪ ⫽

opp adj

hyp opp , also play a signifinstead of opp hyp icant role in this view of trigonometry, and are likewise given names: The reciprocal of these ratios, for example,

cosecant ␪ ⫽

hyp opp

secant ␪ ⫽

hyp adj

cotangent ␪ ⫽

adj opp

The definitions hold regardless of the triangle’s orientation or which of the acute angles is used. As before, each function name is written in abbreviated form as sin ␪, cos ␪, tan ␪, csc ␪, sec ␪, and cot ␪ respectively. Note that based on these designations, we have the following reciprocal relationships:

WORTHY OF NOTE Over the years, a number of memory tools have been invented to help students recall these ratios correctly. One such tool is the acronym SOH CAH TOA, from the first letter of the function and the corresponding ratio. It is often recited as, “Sit On a Horse, Canter Away Hurriedly, To Other Adventures.” Try making up a memory tool of your own.

sin ␪ ⫽

1 csc ␪

cos ␪ ⫽

1 sec ␪

tan ␪ ⫽

1 cot ␪

csc ␪ ⫽

1 sin ␪

sec ␪ ⫽

1 cos ␪

cot ␪ ⫽

1 tan ␪

In general: Trigonometric Functions of an Acute Angle a c b cos ␣ ⫽ c a tan ␣ ⫽ b

␤

sin ␣ ⫽

c ␣ b

b c a cos ␤ ⫽ c b tan ␤ ⫽ a sin ␤ ⫽

a

Now that these ratios have been formally named, we can state values of all six functions given sufficient information about a right triangle.

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EXAMPLE 1

䊳

Finding Function Values Using a Right Triangle Given sin ␪ ⫽ 47, find the values of the remaining trig functions.

Solution

䊳

opp 4 ⫽ , we draw a triangle with a side of 4 units opposite a designated 7 hyp angle ␪, and label a hypotenuse of 7 (see the figure). Using the Pythagorean theorem we find the length of the adjacent side: adj ⫽ 272 ⫺ 42 ⫽ 133. The ratios are

For sin ␪ ⫽

133 7 7 sec ␪ ⫽ 133

4 7 7 csc ␪ ⫽ 4

4 133 133 cot ␪ ⫽ 4

cos ␪ ⫽

sin ␪ ⫽

tan ␪ ⫽

7

4

␪ adj

Now try Exercises 7 through 12 A. You’ve just seen how we can find values of the six trigonometric functions from their ratio definitions

䊳

Note that due to the properties of similar triangles, identical results would be 2 8 ⫽ 47 ⫽ 14 ⫽ 16 obtained using any ratio of sides that is equal to 74. In other words, 3.5 28 and so on, will all give the same value for sin ␪.

B. Solving Right Triangles Given One Angle and One Side 60⬚

hyp 2x

opp x

Example 1 gave values of the trig functions for an unknown angle ␪. Using the special triangles, we can state the value of each trig function for 30°, 45°, and 60° based on the related ratio (see Table 6.10). These values are used extensively in a study of trigonometry and must be committed to memory.

30⬚

Table 6.10

adj √3x hyp √2x

45⬚

opp x

␪

sin ␪

cos ␪

tan ␪

csc ␪

sec ␪

cot ␪

30°

1 2

13 2

1 13 ⫽ 3 13

2

2 213 ⫽ 3 13

13

45°

12 2

12 2

1

12

12

1

60°

13 2

1 2

13

2 2 13 ⫽ 3 13

2

1 13 ⫽ 3 13

45⬚ adj x

To solve a right triangle means to find the measure of all three angles and all three sides. This is accomplished using a combination of the Pythagorean theorem, the properties of triangles, and the trigonometric ratios. We will adopt the convention of naming each angle with a capital letter at the vertex or using a Greek letter on the interior. Each side is labeled using the related lowercase letter from the angle opposite. The complete solution should be organized in table form as in Example 2. Note the quantities shown in bold were given, and the remaining values were found using the relationships mentioned.

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EXAMPLE 2

䊳

Solving a Right Triangle Solve the triangle shown in the figure.

Solution

䊳

Applying the sine ratio (since the side opposite 30° is given), we have: sin 30° 17.9 c c sin 30° 17.9 17.9 c sin 30° 35.8 sin 30°

sin 30°

opposite hypotenuse

multiply by c

c

17.9

divide by sin 30°

1 2

opp . hyp

B

30

C

b

result

Using the Pythagorean theorem shows b ⬇ 31, and since ⬔A and ⬔B are complements, B 60°. Note the results would have been identical if the special ratios from the 30-60-90 triangle were applied. The hypotenuse is twice the shorter side: c 2117.92 35.8, and the longer side is 1 3 times the shorter: b 17.91 132 ⬇ 31.

Angles

A

Sides

A 30ⴗ

a 17.9

B 60°

b ⬇ 31

C 90ⴗ

c 35.8


䊳

Prior to the widespread availability of handheld calculators, tables of values were used to find sin ␪, cos ␪, and tan ␪ for nonstandard angles. Table 6.11 shows the sine of 49° 30¿ is approximately 0.7604. Table 6.11 sin ␪ ␪

0ⴕ

10ⴕ

20ⴕ

30ⴕ

45ⴗ

0.7071

0.7092

0.7112

0.7133

46

0.7193

0.7214

0.7234

0.7254

47

0.7314

0.7333

0.7353

0.7373

48

0.7431

0.7451

0.7470

0.7490

49

0.7547

0.7566

0.7585

" 0.7604

"

598

Today these trig values are programmed into your calculator and we can retrieve them with the push of a button (or two). To find the sine of 49° 30¿ , recall the degree and minute symbols are often found in the 2nd APPS (ANGLE) menu. A calculator in degree MODE can produce the screen shown in Figure 6.117 and provide an even more accurate approximation of sin 49° 30¿ than the table.

Figure 6.117

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EXAMPLE 3

䊳

Solving a Right Triangle Solve the triangle shown in the figure.

Solution

Figure 6.119

䊳

We know ⬔B 58° since A and B are complements (Figure 6.118). We can find length b using the tangent function: Figure 6.118 24 opp B tan 32° tan 32° adj b multiply by b b tan 32° 24 c 24 mm 24 b divide by tan 32° tan 32° 32 A C b ⬇ 38.41 mm result We can find the length c by simply applying the Pythagorean theorem, as shown in the third line of Figure 6.119. Alternatively, we could find c by using another trig ratio and a known angle. 24 opp sin 32° For side c: sin 32° hyp c multiply by c c sin 32° 24 Angles Sides 24 c divide by sin 32° A 32ⴗ a 24 sin 32° B 58° b ⬇ 38.41 ⬇ 45.29 mm result C 90ⴗ

The complete solution is shown in the table.

c ⬇ 45.29


䊳

When solving a right triangle, any combination of the known triangle relationships can be employed: 1. 2. 3. 4.

B. You’ve just seen how we can solve a right triangle given one angle and one side

The angles in a triangle must sum to 180°: A B C 180°. The sides of a right triangle are related by the Pythagorean theorem: a2 b2 c2. The 30-60-90 and 45-45-90 special triangles. The six trigonometric functions of an acute angle.

However, the resulting equation must have only one unknown or it cannot be used. For the triangle shown in Figure 6.120, we cannot begin with the Pythagorean theorem since sides a and b are unknown, and tan 51° is unusable for the same reason. Since the b hypotenuse is given, we could begin with cos 51° and solve 152 a for b, or with sin 51° and solve for a, then work out a 152 complete solution. Verify that a ⬇ 118.13 ft and b ⬇ 95.66 ft.

Figure 6.120 B

a

152 ft

A 51

b

C

C. Solving Right Triangles Given Two Sides In Section 6.2, we used our experience with the special values to find an angle given only the value of a trigonometric function. Unfortunately, if we did not have (or recognize) a special value, we could not determine the angle. In times past, the partial table for sin ␪ given earlier was also used to find an angle whose sine was known, meaning if sin ␪ ⬇ 0.7604, then ␪ must be 49.5° (see the last line of Table 6.11). The modern

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notation for “an angle whose sine is known” is ␪ sin1x or ␪ arcsin x, where x is the known value for sin ␪. The values for the acute angles ␪ sin1x, ␪ cos1x, and ␪ tan1x are also programmed into your calculator and are generally accessed using the 2nd key with the related SIN , COS , or TAN key. With these we are completely equipped to find all six measures of a right triangle, given at least one side and any two other measures. 䊳

CAUTION

EXAMPLE 4

䊳

When working with the inverse trig functions, be sure to use the inverse trig keys, and not the reciprocal key x-1 . For instance, using special triangles, cos1 10.52 60°. 1 Compare this with 3cos10.52 4 1 sec 0.5. cos 0.5 Both calculations are done in degree MODE and shown in the figure.

Solving a Right Triangle ␤

Solve the triangle given in the figure.

Solution

C. You’ve just seen how we can solve a right triangle given two sides

䊳

Since the hypotenuse is unknown, we cannot begin with the sine or cosine ratios. The opposite and adjacent 17 sides for ␣ are known, so we use tan ␣. For tan ␣ 25 1 17 we find ␣ tan a b ⬇ 34.2° [verify that 25 17 tan134.2°2 ⬇ 0.6795992982 ⬇ 4 . Since ␣ and ␤ 25 are complements, ␤ ⬇ 90° 34.2° 55.8°. The Pythagorean theorem shows the hypotenuse is about 30.23 m (verify).

c

17 m

␣ 25 m

Angles ␣ ⬇ 34.2°

Sides a 17

␤ ⬇ 55.8°

b 25

␥ 90ⴗ

c ⬇ 30.23


䊳

D. Using Cofunctions and Complements to Write Equivalent Expressions WORTHY OF NOTE The word cosine is actually a shortened form of the words “complement of sine,” a designation suggested by Edmund Gunter around 1620 since the sine of an angle is equal to the cosine of its complement 3sine1␪2 cosine190° ␪2 4 .

For the right triangle in Figure 6.121, ⬔␣ and ⬔␤ are compleFigure 6.121 ments since the sum of the three angles must be 180°. The com␤ plementary angles in a right triangle have a unique relationship c that is often used. Specifically, since ␣ ␤ 90°, a a a ␤ 90° ␣. Note that sin ␣ and cos ␤ . This ␣ c c b means sin ␣ cos ␤ or sin ␣ cos190° ␣2 by substitution. In words, “The sine of an angle is equal to the cosine of its complement.” For this reason sine and cosine are called cofunctions (hence the name cosine), as are secant/cosecant, and tangent/cotangent. As a test, we use a calculator to check the statement sin 52.3° cos190° 52.3°2 sin 52.3° ⱨ cos 37.7° 0.791223533 0.791223533 ✓

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601

To verify the cofunction relationship for sec ␪ and csc ␪, recall their reciprocal relationship to cosine and sine, respectively.

Figure 6.122

sec 52.3° ⱨ csc 37.7° 1 1 ⱨ cos 52.3° sin 37.7° 1.635250666 ⫽ 1.635250666 ✓ The cofunction relationship for tan ␪ and cot ␪ is similarly verified in Figure 6.122. Summary of Cofunctions sine and cosine sin ␪ ⫽ cos190° ⫺ ␪2 cos ␪ ⫽ sin190° ⫺ ␪2

tangent and cotangent tan ␪ ⫽ cot190° ⫺ ␪2 cot ␪ ⫽ tan190° ⫺ ␪2

secant and cosecant sec ␪ ⫽ csc190° ⫺ ␪2 csc ␪ ⫽ sec190° ⫺ ␪2

For use in Example 5 and elsewhere in the text, note the expression tan215° is simply a more convenient way of writing 1tan 15°2 2. EXAMPLE 5

䊳

Applying the Cofunction Relationship Given cot 75° ⫽ 2 ⫺ 13 in exact form, find the exact value of tan215° using a cofunction. Check the result using a calculator.

Solution

Check

䊳

䊳

Using cot 75° ⫽ tan190° ⫺ 75°2 ⫽ tan 15° gives tan2 15° ⫽ cot2 75° ⫽ 12 ⫺ 132 2 ⫽ 4 ⫺ 413 ⫹ 3 ⫽ 7 ⫺ 4 13

cofunction; square both sides substitute known value square as indicated result

To clearly calculate the square of the tangent, note we first evaluate tan 15° and then square the Answer. As the screen shows, the approximate forms of tan215° and 7 ⫺ 4 13 are equal. ✓

D. You’ve just seen how we can use cofunctions and complements to write equivalent expressions


䊳

E. Applications of Right Triangles While the name seems self-descriptive, in more formal terms an angle of elevation is defined to be the acute angle formed by a horizontal line of orientation (parallel to level ground) and the line of sight (see Figure 6.123). An angle of depression is likewise defined but involves a line of sight that is below the horizontal line of orientation (Figure 6.124). Figure 6.123

Figure 6.124 Line of orientation

t

h sig

f eo Lin S angle of elevation Line of orientation

S angle of depression Lin

eo

fs

igh

t

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Angles of elevation and depression make distance and length computations of all sizes a relatively easy matter and are extensively used by surveyors, engineers, astronomers, and even the casual observer who is familiar with the basics of trigonometry. EXAMPLE 6

䊳

Applying Angles of Depression From the edge of the Perito Moreno glaciar in Patagonia, Argentina, Gary notices a kayak on the lake 30 m below. If the angle of depression at that point is 35°, what is the distance d from the kayak to the glacier?

Solution

䊳

From the diagram, we note that d is the side opposite a 55° angle (the complement of the 35° angle). Since the adjacent side is known, we use the tangent function to solve for d.

35

30 m

d

d 30 30 tan 55° d 42.8 ⬇ d tan 55°

tan 55°

opp adj

multiply by 30 (exact form) result (approximate form)

The kayak is approximately 42.8 m (141 ft) from the glacier. Now try Exercises 71 through 74

䊳

Closely related to angles of elevation/depression are acute angles of rotation from a fixed line of orientation to a fixed line of sight. In this case, the movement is simply horizontal rather than vertical. Land surveyors use a special type of measure called bearings, where they indicate the acute angle the line of sight makes with a due north or due south line of reference. For instance, Figure 6.125(a) shows a bearing of N 60° E and Figure 6.125(b) shows a bearing of S 40° W. Figure 6.125 N

N

N 60 E 60 W

E

W

E 40 S 40 W

S

S

(a)

(b)

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EXAMPLE 7

䊳

603

Applying Angles of Rotation A city building code requires a new shopping complex to be built at least 150 ft from an avenue that runs due east and west. Using a modern theodolite set up on the avenue, a surveyor finds the tip of the shopping complex closest to the avenue lies 200 ft away on a bearing of N 41°17¿15– W. How far from the avenue is this tip of the building?

Solution

䊳

To find the distance d, we first convert the given angle to decimal degrees, then find its complement. For 41° 17¿ 15– we have 3 41 1711/602 115/36002 4° 41.2875°. The complement of this angle is 90° 41.2875° 48.7125°. Since we need the side opposite this angle and the hypotenuse is known, we use a sine function to solve for d.

N 200 ft

d N 4117'15" W 48.7125

d 200 200 sin 48.7125° d 150.28 ⬇ d sin 48.7125°

sin 48.7125°

opp hyp

multiply by 200 (exact form) result (approximate form)

The shopping complex is approximately 150.28 ft from the avenue, just barely in compliance with the city code. Now try Exercises 75 and 76

䊳

In their widest and most beneficial use, the trig functions of acute angles are used with other problem-solving skills, such as drawing a diagram, labeling unknowns, working the solution out in stages, and so on. Example 8 serves to illustrate some of these combinations. EXAMPLE 8

䊳

Applying Angles of Elevation and Depression From his hotel room window on the sixth floor, Singh notices some window washers high above him on the hotel across the street. Curious as to their height above ground, he quickly estimates the buildings are 50 ft apart, the angle of elevation to the workers is about 80°, and the angle of depression to the base of the hotel is about 50°. a. How high above ground is the window of Singh’s hotel room? b. How high above ground are the workers?

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Solution

䊳

a. Begin by drawing a diagram of the situation (see figure). To find the height of the window we’ll use the tangent ratio, since the adjacent side of the angle is known, and the opposite side is the height we desire. For the height h1:

(not to scale) h2

h1 50 50 tan 50° h1 59.6 ⬇ h1 tan 50°

tan 50°

opp adj

solve for h1

result 1tan 50° ⬇ 1.19182

The window is approximately 59.6 ft above ground. b. For the height h2:

80° 50° h1

h2 50 50 tan 80° h2 283.6 ⬇ h2 tan 80°

tan 80°

opp adj

solve for h2

result 1tan 80° ⬇ 5.67132

The workers are approximately 283.6 59.6 343.2 ft above ground.

x


䊳

50 ft

There are a number of additional interesting applications in the Exercise Set. E. You’ve just seen how we can solve applications of right triangles

6.6 EXERCISES 䊳



䊳

1. The phrase, “an angle whose tangent is known,” is written notationally as _________.

7 2. Given sin ␪ 24 , csc ␪ _________ because _________ they are .

3. The sine of an angle is the ratio of the _________ side to the _________.

4. The cosine of an angle is the ratio of the _________ side to the _________.

5. Discuss/Explain exactly what is meant when you are asked to “solve a triangle.” Include an illustrative example.

6. Given an acute angle and the length of the adjacent leg, which four (of the six) trig functions could be used to begin solving the triangle?


Use the function value given to determine the values of the other five trig functions of the acute angle ␪. Answer in exact form (a diagram will help).

7. cos ␪

5 13

8. sin ␪

20 29

9. tan ␪

84 13

10. sec ␪

53 45

11. cot ␪

2 11

12. cos ␪

2 3

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Precalculus—

6–97 Solve each triangle using trig functions of an acute angle ␪. Give a complete answer (in table form) using exact values.

13.

Use a calculator to find the value of each expression, rounded to four decimal places.

14. B

420 ft

C b

a

196 cm

B

C

b

60

16.

C

B 45

a

c

c

a

45 A

B

C

81.9 m

Solve the triangles shown and write answers in table form. Round sides to the nearest 100th of a unit. Verify that angles sum to 180ⴗ and that the three sides satisfy (approximately) the Pythagorean theorem.

17.

24. cos 72°

25. tan 40°

26. cot 57.3°

27. sec 40.9°

28. csc 39°

29. sin 65°

30. tan 84.1°

Use a calculator to find the acute angle whose corresponding ratio is given. Round to the nearest 10th of a degree. For Exercises 31 through 38, use Exercises 23 through 30 to answer.

30 A

9.9 mm

A

23. sin 27°

c

A

15.

605


31. sin A 0.4540

32. cos B 0.3090

33. tan ␪ 0.8391

34. cot A 0.6420

35. sec B 1.3230

36. csc ␤ 1.5890

37. sin A 0.9063

38. tan B 9.6768

39. tan ␣ 0.9896

40. cos ␣ 0.7408

41. sin ␣ 0.3453

42. tan ␣ 3.1336

Select an appropriate function to find the angle indicated (round to 10ths of a degree).

43.

44.

18. B

B

6m

c

14 m

89 in.

45.

22

46.

A

b

A 49

19.

20.

C

b

B c

14 in. 18 m

c

C

15 in.

18.7 cm

B

5 mi

6.2 mi

58 238 ft

a

5.6 mi

19.5 cm A

A

C

C

b

b

51 A

47.

48. A

21.

625 mm

C

B

22.

B 20 mm

c

b

42 mm

207 yd

28 c

65

a

A A

45.8 m

C

B 221 yd

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Draw a right triangle ABC as shown, using the information given. Then select an appropriate ratio to find the side indicated. Round to the nearest 100th.

Exercises 49 to 54 B c A

a

b

49. ⬔A 25° c 52 mm find side a

50. ⬔B 55° b 31 ft find side c

51. ⬔A 32° a 1.9 mi find side b

52. ⬔B 29.6° c 9.5 yd find side a

53. ⬔A 62.3° b 82.5 furlongs find side c

54. ⬔B 12.5° a 32.8 km find side b

C

Use a calculator to evaluate each pair of functions and comment on what you notice.

55. sin 25°, cos 65°

Based on your observations in Exercises 55 to 58, fill in the blank so that the functions given are equal.

59. sin 47°, cos ___

60. cos ___, sin 12°

61. cot 69°, tan ___

62. csc 17°, sec ___

Complete the following tables without referring to the text or using a calculator.

63. ␪ ⴝ 30ⴗ

sin ␪

cos ␪

tan ␪

sin(90ⴗ ⴚ ␪)

cos(90ⴗ ⴚ ␪)

tan(90ⴗ ⴚ ␪)

csc ␪

sec ␪

cot ␪

␪ ⴝ 45ⴗ

sin ␪

cos ␪

tan ␪

sin(90ⴗ ⴚ ␪)

cos(90ⴗ ⴚ ␪)

tan(90ⴗ ⴚ ␪)

csc ␪

sec ␪

cot ␪

64.

Evaluate the following expressions without a calculator, using the cofunction relationship and the following exact forms: sec 75ⴗ ⴝ 16 ⴙ 12; tan 75ⴗ ⴝ 2 ⴙ 13.

56. sin 57°, cos 33° 57. tan 5°, cot 85° 58. sec 40°, csc 50°

䊳

6–98


65. 16 csc 15°

66. csc2 15°

67. cot2 15°

68. 13 cot 15°


69. The sine of an angle between two sides of a 2A triangle: sin ␪ ⴝ ab If the area A and two sides a and b of a triangle are known, the sine of the angle between the two sides is given by the formula shown. Find the angle ␪ for the triangle shown given A ⬇ 38.9, and use it to solve the triangle. (Hint: Apply the same concept to angle ␤ or .)

17

8

24

70. Illumination of a surface: E ⴝ

I cos ␪ d2

The illumination E of a surface by a light source is a measure of the luminous flux per unit area that reaches the surface. The value of E [in lumens (lm)

90 cd (about 75 W) per square foot] is given by the formula shown, where d is the distance from the light source (in feet), I is the intensity of the light [in candelas (cd)], and ␪ is the angle the light source makes with the vertical. For 65° reading a book, an illumination E of at least 18 lm/ft2 is recommended. Assuming the open book is lying on a horizontal surface, how far away should a light source be placed if it has an intensity of 90 cd (about 75 W) and the light flux makes an angle of 65° with the book’s surface (i.e., ␪ 25°)?

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607

APPLICATIONS

71. Angle of elevation: For a person standing 100 m from the center of the base of the Eiffel Tower, the angle of elevation to the top of the tower is 71.6°. How tall is the Eiffel Tower? 72. Angle of elevation: In 2001, the tallest building in the world was the Petronas Tower I in Kuala Lumpur, Malaysia. For a person standing 25.9 ft from the base of the tower, the angle of elevation to the top of the tower is 89°. How tall is the Petronas tower? 73. Angle of depression: A person standing near the top of the Eiffel Tower notices a car wreck some distance from the tower. If the angle of depression from the person’s eyes to the wreck is 32°, how far away is the accident from the base of the tower? See Exercise 71. 74. Angle of depression: A person standing on the top of the Petronas Tower I looks out across the city and pinpoints her residence. If the angle of depression from the person’s eyes to her home is 5°, how far away (in feet and in miles) is the residence from the base of the tower? See Exercise 72. 75. Angles of rotation: From a point 110 ft due south of the eastern end of a planned east/west bridge spanning the Illinois river, a surveyor notes the western end lies on a bearing of N 38°35¿15– W. To the nearest inch, how long will the bridge be?

N

110 ft

78. Observing wildlife: From her elevated observation post 300 ft away, a naturalist spots a troop of baboons high up in a tree. Using the small transit attached to her telescope, she finds the angle of depression to the bottom of this tree is 14°, while the angle of elevation to the top of the tree is 25°. The angle of elevation to the troop of baboons is 21°. Use this information to find (a) the height of the observation post, (b) the height of the baboons’ tree, and (c) the height of the baboons above ground. 79. Angle of elevation: The tallest free-standing tower in the world is the CNN Tower in Toronto, Canada. The tower includes a rotating restaurant high above the ground. From a distance of 500 ft the angle of elevation to the pinnacle of the tower is 74.6°. The angle of elevation to the restaurant from the same vantage point is 66.5°. How tall is the CNN 66.5 Tower? How far below the 74.6 pinnacle of the tower is the restaurant located? 500 ft

76. Angles of rotation: A large sign spans an east/west highway. From a point 35 m due west of the southern base of the sign, a surveyor finds the northern base lies on a bearing of N 67°11¿42– E. To the nearest centimeter, how wide is the sign?

N 35 m

77. Height of a climber: A local Outdoors Club has just hiked to the south rim of a large canyon, when they spot a climber attempting to scale the taller northern face. Knowing the distance between the sheer walls of the northern and southern faces of the canyon is approximately 175 yd, they attempt to compute the distance remaining for the climbers to reach the top of the northern rim. Using a homemade transit, they sight an angle of depression of 55° to the bottom of the north face, and angles of elevation of 24° and 30° to the climbers and top of the northern rim respectively. (a) How high is the southern rim of the 30 canyon? 24 (b) How high is 55 the northern rim? (c) How much farther 175 yd until the climber reaches the top?

80. Angle of elevation: In January 2009, Burj Dubai unofficially captured the record as the world’s tallest building, according to the Council on Tall Buildings and Urban Habitat (Source: www.ctbuh.org). Measured at a point 159 m from its base, the angle of elevation to the top of the spire is 79°. From a

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distance of about 134 m, the angle of elevation to the top of the roof is also 79°. How tall is Burj Dubai from street level to the top of the spire? How tall is the spire itself? 81. Crop duster’s speed: While standing near the edge of a farmer’s field, Johnny watches a crop duster dust the farmer’s field for insect control. 50 ft Curious as to the plane’s speed during each drop, Johnny attempts an estimate using the angle of rotation from one end of the field to the other, while standing 50 ft from one corner. Using a stopwatch he finds the plane makes each pass in 2.35 sec. If the angle of rotation was 83°, how fast (in miles per hour) is the plane flying as it applies the insecticide? 82. Train speed: While driving to their next gig, Josh and the boys get stuck in a line of cars at a railroad crossing as the gates go down. As the sleek, speedy express train approaches, Josh decides to pass the time estimating its speed. He spots a large oak tree beside the track some distance away, and figures the angle of rotation from the crossing to the tree is about 80°. If their car is 60 ft from the crossing and it takes the train 3 sec to reach the tree, how fast is the train moving in miles per hour? Alternating current: In AC (alternating current) applications, Z the relationship between measures known as the impedance (Z), resistance (R), and the phase R angle 1␪2 can be demonstrated using a right triangle. Both the resistance and the impedance are measured in ohms 12 .

83. Find the impedance Z if the phase angle ␪ is 34°, and the resistance R is 320 . 84. Find the phase angle ␪ if the impedance Z is 420 , and the resistance R is 290 . 85. Contour maps: In the figure shown, the contour interval is 175 m (each concentric line represents an increase of 175 m in elevation), and the scale of horizontal distances is 1 cm 500 m. (a) Find the vertical change from A to B (the increase in elevation); (b) use a proportion to find

A

B

6–100

the horizontal change between points A and B if the measured distance on the map is 2.4 cm; and (c) draw the corresponding right triangle and use it to estimate the length of the trail up the mountain side that connects A and B, then use trig to compute the approximate angle of incline as the hiker climbs from point A to point B. 86. Contour maps: In the figure shown, the contour interval is 150 m (each concentric line represents an increase of 150 m in elevation), and the B scale of horizontal distances is 1 cm 250 m. (a) Find A the vertical change from A to B (the increase in elevation); (b) use a proportion to find the horizontal change between points A and B if the measured distance on the map is 4.5 cm; and (c) draw the corresponding right triangle and use it to estimate the length of the trail up the mountain side that connects A and B, then use trig to compute the approximate angle of incline as the hiker climbs from point A to point B. 87. Height of a rainbow: While visiting the Lapahoehoe Memorial on the island of Hawaii, Bruce and Carma see a spectacularly vivid rainbow arching over the bay. Bruce speculates the rainbow is 500 ft away, while Carma estimates the angle of elevation to the highest point of the rainbow is about 42°. What was the approximate height of the rainbow? 88. High-wire walking: As part of a circus act, a highwire walker not only “walks the wire,” she walks a wire that is set at an incline of 10° to the horizontal! If the length of the (inclined) wire is 25.39 m, (a) how much higher is the wire set at the destination pole than at the departure pole? (b) How far apart are the poles?

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89. Diagonal of a cube: A d 35 cm cubical box has a diagonal measure of x 35 cm. (a) Find the ␪ x dimensions of the box x and (b) the angle ␪ that the diagonal makes at the lower corner of the box.

䊳

609


90. Diagonal of a rectangular parallelepiped: A h ␪ rectangular box has a width 50 of 50 cm and a length of 70 cm 70 cm. (a) Find the height h that ensures the diagonal across the middle of the box will be 90 cm and (b) the angle ␪ that the diagonal makes at the lower corner of the box.


91. One of the challenges facing any terrestrial exploration of Mars is its weak and erratic magnetic field. An unmanned rover begins a critical mission at landing site A where magnetic north lies on a bearing of N 30° W of true (polar) north (see figure). The rover departs in the magnetic direction of S 20° E and travels straight for 1.4 km to site B. At this point, magnetic north lies on a bearing of N 45° E of true north. The rover turns to magnetic direction of N 5° W and travels 2 km straight to site C. At C, magnetic north lies on a bearing of N 10° W of true north. In what magnetic direction should the rover head to return to site A? 92. As of the year 2009, the Bailong elevator outside of Zhangjiajie, China, was the highest exterior glass elevator in the world. While Christine was descending in the red car, she noticed Simon ascending in the yellow car, below her at an angle of depression of 50°. Seven seconds later, Simon was above her at an angle of elevation of 50°. If the cars have the same velocity ascending and descending, and the horizontal distance between the two cars is 47 ft, how fast do the cars travel? 93. The radius of the Earth at the equator (0° N latitude) is approximately 3960 mi. Beijing, China, is located at 39.5° N latitude, 116° E longitude. Philadelphia, Pennsylvania, is located at the same latitude, but at 75° W longitude. (a) Use the diagram given and a cofunction relationship to find the radius r of the Earth (parallel to the equator) at this latitude; (b) use the arc length formula to compute the shortest distance between these two cities along this latitude; and (c) if the supersonic Concorde flew a direct flight between Beijing and Philadelphia along this latitude, approximate the flight time assuming a cruising speed of 1250 mph. Note: The shortest distance is actually traversed by heading northward, using the arc of a “great circle” that goes through these two cities.

Exercise 91

True north Magnetic north

30

A

Exercise 92

Exercise 93 North Pole h Rh

␪N

r R ␪ R

South Pole

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94. (3.2) Solve by factoring: a. g2 9g 0 b. g2 9 0 c. g2 9g 10 0 d. g2 9g 10 0 e. g3 9g2 10g 90 0 95. (2.1) For the graph of T(x) 3 given, (a) name the local maximums and minimums, 5 (b) the zeroes of T, 3 (c) intervals where T1x2T and T1x2c, and (d) intervals where T1x2 7 0 and T1x2 6 0.

6.7

96. (6.1) The armature for the rear windshield wiper has a length of 24 in., with a rubber wiper blade that is 20 in. long. What area of my rear windshield is cleaned as the armature swings back-and-forth through an angle of 110°?

y T(x)

5

x

97. (6.1) The boxes used to ship some washing machines are perfect cubes with edges measuring 38 in. Use a special triangle to find the length of the diagonal d of one side, and the length of the interior diagonal D (through the middle of the box).

D

d

Trigonometry and the Coordinate Plane


A. Define the trigonometric functions using the coordinates of a point in QI B. Use reference angles to evaluate the trig functions for any angle C. Solve applications using the trig functions of any angle

This section tends to bridge the study of static trigonometry and the angles of a right triangle, with the study of dynamic trigonometry and the unit circle. This is accomplished by noting that the domain of the trig functions (unlike a triangle point of view) need not be restricted to acute angles. We’ll soon see that the domain can be extended to include trig functions of any angle, a view that greatly facilitates our work in Chapter 8, where many applications involve angles greater than 90°.

A. Trigonometric Ratios and the Point P(x, y) Regardless of where a right triangle is situated or how Figure 6.126 it is oriented, each trig function can be defined as a y given ratio of sides with respect to a given angle. In this light, consider a 30-60-90 triangle placed in the first quadrant with the 30° angle at the origin and the (5√3, 5) longer side along the x-axis. From our previous review 5 of similar triangles, the trig ratios will have the same 60 10 value regardless of the triangle’s size so for conven5 ience, we’ll use a hypotenuse of 10 giving sides of 5, 513, and 10. From the diagram in Figure 6.126 we 30 10 x 5√3 note the point (x, y) marking the vertex of the 60° angle has coordinates (5 13, 5). Further, the diagram shows that sin 30°, cos 30°, and tan 30° can all be expressed in terms of triangle side y opp adj 5 513 x 1sine2, 1cosine2, lengths or these coordinates since r r hyp 10 hyp 10 y opp 5 1tangent2, where r is the distance from (x, y) to the origin. Each and x adj 5 13 13 1 , result reduces to the more familiar values seen earlier: sin 30° , cos 30° 2 2

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Section 6.7 Trigonometry and the Coordinate Plane

1 13 . This suggests we can define the six trig functions in terms 3 13 of x, y, and r, where r 2x2 y2. Consider that the slope of the line coincident with the hypotenuse is rise 13 5 , and since the line goes through the origin its equation must be run 3 5 13 13 y x. Any point (x, y) on this line will be at the 60° vertex of a right triangle 3 formed by drawing a perpendicular line from the point (x, y) to the x-axis. As Example 1 shows, we obtain the special values for sin 30°, cos 30°, and tan 30° regardless of the point chosen. and tan 30°

EXAMPLE 1

䊳

Evaluating Trig Functions Using x, y, and r y

13 x, 3 then draw the corresponding right triangle and evaluate sin 30°, cos 30°, and tan 30°. Pick an arbitrary point in QI that satisfies y

Solution

䊳

The coefficient of x has a denominator of 3, so we choose a multiple of 3 for convenience. For x 6 13 162 2 13. As seen in the figure, we have y 3

10

y √3 x 3 (6, 2√3) 4√3

10

2√3

30 6

30

10

x

10

the point (6, 2 13) is on the line and at the vertex of the 60° angle. Using the triangle and evaluating the trig functions at 30°, we obtain: y y x 213 6 213 cos 30° sin 30° tan 30° r r x 6 4 13 4 13 13 1 6 13 13 2 2 3 4 13 13 Now try Exercises 7 and 8

䊳

In general, consider any two points (x1, y1) and (x2, y2) on an arbitrary line y kx, at corresponding distances r1 and r2 from the origin (Figure 6.127). Because the triangles y1 y2 x1 x2 , , and so on, and we conclude that the formed are similar, we have x1 x2 r1 r2 values of the trig functions are indeed independent of the point chosen. Viewing the trig functions in terms of x, y, and r produces significant results. In 13 x from Example 1 also extends into QIII, and Figure 6.128, we note the line y 3 Figure 6.127

Figure 6.128

y

y 10

(x2, y2)

␪ y kx

(x1, y1) r1 y1 ␪ x1 x2

y

y2 x

2√3

6 30 4√3

(6, 2√3)

210

(6, 2√3) 10

x

√3 x 3

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creates another 30° angle whose vertex is at the origin (since vertical angles are equal). The sine, cosine, and tangent functions can still be evaluated for this angle, but in QIII both x and y are negative. If we consider the angle in QIII to be a positive rotation of 210° 1180° 30°2, we can evaluate the trig functions using the values of x, y, and r from any point on the terminal side, since these are fixed by the 30° angle created and are the same as those in QI except for their sign: y 2 13 r 4 13 1 2

x 6 r 4 13 13 2

sin 210°

y 213 x 6 13 3

tan 210°

cos 210°

For any rotation ␪ and a point (x, y) on the terminal side, the distance r can be found using r 2x2 y2 and the six trig functions likewise evaluated. Note that evaluating them correctly depends on the quadrant of the terminal side, since this will dictate the signs for x and y. Students are strongly encouraged to make these quadrant and sign observations the first step in any solution process. In summary, we have Trigonometric Functions of Any Angle Given P(x, y) is any point on the terminal side of angle ␪ in standard position, with r 2x2 y2 1r 7 02 the distance from the origin to (x, y). The six trigonometric functions of ␪ are y y x tan ␪ sin ␪ cos ␪ r r x x0 csc ␪

r y

sec ␪

y0

EXAMPLE 2

䊳

x y y0

r x

cot ␪

x0

Evaluating Trig Functions Given the Terminal Side is on y ⴝ mx Given that P(x, y) is a point on the terminal side of angle ␪ in standard position, find the values of sin ␪ and cos ␪, if a. The terminal side is in QII and coincident with the line y 12 5 x, b. The terminal side is in QIV and coincident with the line y 12 5 x.

Solution

䊳

a. Select any convenient point in QII that satisfies this equation. We select x 5 since x is negative in QII, which gives y 12 and the point (5, 12). Solving for r gives r 2152 2 1122 2 13. The ratios are sin ␪

y 12 r 13

cos ␪

5 x r 13

b. In QIV we select x 10 since x is positive in QIV, giving y 24 and the point (10, 24). Solving for r gives r 21102 2 1242 2 26. The ratios are y 24 r 26 12 13

sin ␪

x 10 r 26 5 13

cos ␪


䊳

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613

In Example 2, note the ratios are the same in QII and QIV except for their sign. We will soon use this observation to great advantage. EXAMPLE 3

䊳

Evaluating Trig Functions Given a Point P Find the values of the six trigonometric functions given P1⫺5, 52 is on the terminal side of angle ␪ in standard position.

Solution

䊳

For P1⫺5, 52 we have x 6 0 and y 7 0 so the terminal side is in QII. Solving for r yields r ⫽ 21⫺52 2 ⫹ 152 2 ⫽ 150 ⫽ 5 12. For x ⫽ ⫺5, y ⫽ 5, and r ⫽ 5 12, we obtain y y x 5 ⫺5 5 sin ␪ ⫽ ⫽ cos ␪ ⫽ ⫽ tan ␪ ⫽ ⫽ r r x ⫺5 5 12 5 12 12 12 ⫽ ⫽⫺ ⫽ ⫺1 2 2 The remaining functions can be evaluated using reciprocals. csc ␪ ⫽

2 ⫽ 12 12

sec ␪ ⫽ ⫺

2 ⫽ ⫺ 12 12

cot ␪ ⫽ ⫺1

Note the connection between these results and the special values for ␪ ⫽ 45°. Now try Exercises 13 through 28

䊳

Graphing calculators offer a number of features that can assist this approach to a study of the trig functions. On many calculators, the keystrokes 2nd APPS (ANGLE) will bring up a menu with options 1 through 4 which are basically used for angle conversions. Of interest to us here are options 5 and 6, which can be used to determine the radius r (option 5) or the angle (option 6) related to a given point (x, y). For (5, ⫺5) from Example 3, the home screen and press 2nd APPS (ANGLE) 5:R 䊳 Pr(, which will place the option on the home screen. This feature supplies the left parenthesis of the ordered pair, and you simply complete it: 5:R 䊳 Pr(–5, 5). As shown in Figure 6.129, the calculator returns 7.07…, the decimal approximation of 512. To find the related angle, it is assumed that ␪ is in standard position and (x, y) is on the terminal side. Pressing 2nd APPS (ANGLE) 6:R 䊳 P␪( and completing the ordered pair as before shows the corresponding angle is 135⬚ (Figure 6.129). Note this is a QII angle as expected, since x 6 0 and y 7 0, and we can check the results of Example 3 by evaluating the trig functions of 135° (see Figure 6.130). See Exercises 29 and 30. CLEAR

Figure 6.129

Figure 6.130

Figure 6.131 y (0, b) 180⬚ (⫺a, 0)

(a, 0) 90⬚

(0, ⫺b)

x

Now that we’ve defined the trig functions in terms of ratios involving x, y, and r, the question arises as to their value at the quadrantal angles. For 90° and 270°, any point on the terminal side of the angle has an x-value of zero, meaning tan 90°, sec 90°, tan 270°, and sec 270° are all undefined since x ⫽ 0 is in the denominator. Similarly, at 180° and 360°, the y-value of any point on the terminal side is zero, so cot 180°, csc 180°, cot 360°, and csc 360° are likewise undefined (see Figure 6.131).

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䊳

EXAMPLE 4

Evaluating the Trig Functions for ␪ ⴝ 90ºk, k an Integer Evaluate the six trig functions for ␪ 270°.

䊳

Solution

Here, ␪ is the quadrantal angle whose terminal side separates QIII and QIV. Since the evaluation is independent of the point chosen on this side, we choose (0, 1) for convenience, giving r 1. For x 0, y 1, and r 1 we obtain 1 0 1 cos 270° tan 270° 1 0 1undefined2 1 1 0 The remaining ratios can be evaluated using reciprocals. sin 270°

csc 270° 1

sec 270°

1 1undefined2 0

cot 270°

0 0 1


䊳

Results for the quadrantal angles are summarized in Table 6.12. Table 6.12 ␪

A. You’ve just seen how we can define the trigonometric functions using the coordinates of a point in QI

0°/360° S 11, 02

y sin ␪ ⴝ r 0

90° S 10, 12

180° S 11, 02 270° S 10, 12

cos ␪ ⴝ

x r

tan ␪ ⴝ

1

0

1

0

0

1

1

0

y x

csc ␪ ⴝ

r y

sec ␪ ⴝ

r x

cot ␪ ⴝ

x y

undefined

1

undefined

undefined

1

undefined

0

0

undefined

1

undefined

undefined

1

undefined

0

B. Reference Angles and the Trig Functions of Any Angle Recall that for any angle ␪ in standard position, the acute angle ␪r formed by the terminal side and the nearest x-axis is called the reference angle. Several examples of this definition are illustrated in Figures 6.132 through 6.135 for ␪ 7 0° (note that if 0° 6 ␪ 6 90°, then ␪r ␪). Figure 6.132

Figure 6.133

y

Figure 6.134

y

5

Figure 6.135 y

y

5

5

5

(x, y) (x, y) r

5

5

x

r

5

5

x

5

r

5

x

r

5

5

(x, y) (x, y) 5

90 180 r 180

5

180 270 r 180

5

270 360 r 360

5

360 450 r 360

x

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EXAMPLE 5

䊳

Finding Reference Angles Determine the reference angle for a. 315° b. 150° c. 239°

Solution

䊳

d. 425°

Begin by mentally visualizing each angle and the quadrant where it terminates. a. 315° is a QIV angle: b. 150° is a QII angle: ␪r 360° 315° 45° ␪r 180° 150° 30° c. 239° is a QIII angle: d. 425° is a QI angle: ␪r 239° 180° 59° ␪r 425° 360° 65° Now try Exercises 33 through 44

䊳

The reference angles from Examples 5(a) and 5(b) were special angles, which means we automatically know the absolute values of the trig ratios using ␪r. The best way to remember the signs of the trig functions is to keep Figure 6.136 in mind that sine is associated with y, cosine with x, and tangent with both x and y (r is always positive). In addiQuadrant II Quadrant I tion, there are several mnemonic devices (memory tools) Sine All to assist you. One is to use the first letter of the function is positive are positive that is positive in each quadrant and create a catchy acronym. For instance ASTC S All Students Take Cosine Tangent Classes (see Figure 6.136). Note that a trig function and its is positive is positive reciprocal function will always have the same sign. See Quadrant IV Quadrant III Exercises 45 through 48. EXAMPLE 6

䊳

Evaluating Trig Functions Using ␪r Use a reference angle to evaluate sin ␪, cos ␪, and tan ␪ for ␪ 315°.

Solution

䊳

The terminal side is in QIV where cosine is positive while sine and tangent are negative. With ␪r 45°, we have: sin 315° ⴚsin ␪r sin 45° 22 2

QIV: sin ␪ 6 0

y

315

r 45

x

replace ␪r with 45° sin 45°

12 2

Similarly, the values for cosine and tangent are cos 315° ⴙcos ␪r tan 315° ⴚtan ␪r cos 45° tan 45° 22 1 2 Now try Exercises 49 through 56

䊳

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EXAMPLE 7

䊳

Finding Function Values Using a Quadrant and Sign Analysis 5 and cos ␪ 6 0, find the value of the other ratios. 13 Always begin with a quadrant and sign analysis: sin ␪ is positive in QI and QII, while cos ␪ is negative in QII and QIII. Both conditions are satisfied in QII only. For r ⫽ 13 and y ⫽ 5, the Pythagorean theorem shows x ⫽ ⫾ 2132 ⫺ 52 ⫽ ⫾ 1144 ⫽ ⫾12. ⫺12 5 With ␪ in QII, x ⫽ ⫺12 and this gives cos ␪ ⫽ and tan ␪ ⫽ . The 13 ⫺12 ⫺12 13 13 reciprocal values are csc ␪ ⫽ , sec ␪ ⫽ , and cot ␪ ⫽ . 5 ⫺12 5 Given sin ␪ ⫽

Solution

䊳


䊳

In our everyday experience, there are many Figure 6.137 actions and activities where angles greater than or y equal to 360° are applied. Some common instances are 8 a professional basketball player who “does a three ⫽ 510⬚ sixty” (360°) while going to the hoop, a diver who ⫽ 150⬚ completes a “two-and-a-half” (900°) off the high 30⬚ board, and a skater who executes a perfect triple axel 8 x (312 turns or 1260°). As these examples suggest, angles ⫽ ⫺210⬚ greater than 360° must still terminate on a quadrantal axis, or in one of the four quadrants, allowing a reference angle to be found and the functions to be evaluated for any angle regardless of size. Figure 6.137 illustrates that ␣ ⫽ 150°, ␤ ⫽ ⫺210°, and ␪ ⫽ 510° are all coterminal, with each having a reference angle of 30°. EXAMPLE 8

䊳

Evaluating Trig Functions of Any Angle

Solution

䊳

The three angles are coterminal in QII where sine is positive while cosine and tangent are negative. With ␪r ⫽ 30°, we have:

Evaluate sin 150°, cos 1⫺210°2, and tan 510°.

sin 150° ⫽ ⴙsin ␪r ⫽ sin 30° 1 ⫽ 2

cos 1⫺210°2 ⫽ ⴚcos ␪r ⫽ ⫺cos 30° 23 ⫽⫺ 2

tan 510° ⫽ ⴚtan ␪r ⫽ ⫺tan 30° 23 ⫽⫺ 3


B. You’ve just seen how we can use reference angles to evaluate the trig functions for any angle

䊳

Since 360° is one full rotation, all angles ␪ ⫹ 360°k will be coterminal for any integer k. For angles with a very large magnitude, we can find the quadrant of the terminal side by adding or subtracting as many integer multiples of 360° as needed until ␪ 6 360°. 1908° ⫽ 5.3 and 1908° ⫺ 360°152 ⫽ 108°. This angle is in QII with For ␣ ⫽ 1908°, 360° ␪r ⫽ 72°. See Exercises 77 through 92.

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C. Applications of the Trig Functions of Any Angle One of the most basic uses of coterminal angles is determining all values of ␪ that satisfy a stated relationship. For example, by now you are aware that 1 if sin ␪ (positive one-half), then ␪ 30° or 2 ␪ 150° (see Figure 6.138). But this is also true for all angles coterminal with these two, and we would write the solutions as ␪ 30° 360°k and ␪ 150° 360°k for all integers k.

Figure 6.138 y 8

5

150 r 30

30 x

EXAMPLE 9

䊳

Finding All Angles that Satisfy a Given Equation Find all angles satisfying the relationship given. Answer in degrees. 12 a. cos ␪ b. tan ␪ 1.3764 2

Solution

WORTHY OF NOTE Since reference angles are acute, we find them by using the sin1, cos1, or tan1 keys with positive values.

䊳

12 , we reason 2 ␪r 45° and two solutions are ␪ 135° from QII (see Example 3) and ␪ 225° from QIII. For all values of ␪ satisfying the relationship, we have ␪ 135° 360°k and ␪ 225° 360°k. See Figure 6.139. b. Tangent is negative in QII and QIV. For 1.3764 we find ␪r using a calculator: 2nd TAN (tan1) 1.3764 shows tan1 11.37642 ⬇ 54, so ␪r 54°. Two solutions are ␪ 180° 54° 126° from QII, and in QIV ␪ 360° 54° 306°. The result is ␪ 126° 360°k and ␪ 306° 360°k. Note these can be combined into the single statement ␪ 126° 180°k. See Figure 6.140. a. Cosine is negative in QII and QIII. Recognizing cos 45°

ENTER

Figure 6.139

Figure 6.140

y 8

y √2

cos 2

8

tan 1.3764

5

r 45 r 45

135

r 54 x

126 r 54

x


䊳

We close this section with an additional application of the concepts related to trigonometric functions of any angle.

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EXAMPLE 10

䊳

Applications of Coterminal Angles: Location on Radar A radar operator calls the captain over to her screen saying, “Sir, we have an unidentified aircraft at bearing N 20 E (a standard 70° rotation). I think it’s a UFO.” The captain asks, “What makes you think so?” To which the sailor replies, “Because it’s at 5000 ft and not moving!” Name all angles for which the UFO causes a “blip” to occur on the radar screen.

Solution C. You’ve just seen how we can solve applications using the trig functions of any angle

䊳

y

Blip!

70 x

Since radar typically sweeps out a 360° angle, a blip will occur on the screen for all angles ␪ 70° 360°k, where k is an integer. Now try Exercises 103 through 108

䊳

6.7 EXERCISES 䊳

CONCEPTS SQAND VOCABULARY


䊳

1. An angle is in standard position if its vertex is at the _______ and the initial side is along the ________.

2. A(n) ________ angle is one where the ________ side is coincident with one of the coordinate axes.

3. Angles formed by a counterclockwise rotation are _________ angles. Angles formed by a ________ rotation are negative angles.

4. For any angle ␪, its reference angle ␪r is the positive ________ angle formed by the ________ side and the nearest x-axis.

5. Discuss the similarities and differences between the trigonometry of right triangles and the trigonometry of any angle.

6. Let T(x) represent any one of the six basic trig functions. Explain why the equation T1x2 k will always have exactly two solutions in [0, 2␲) if x is not a quadrantal angle.

DEVELOPING YOUR SKILLS 7. Draw a 30-60-90 triangle with the 60° angle at the origin and the short side along the positive x-axis. Determine the slope and equation of the line coincident with the hypotenuse, then pick any point on this line and evaluate sin 60°, cos 60°, and tan 60°. Comment on what you notice. 8. Draw a 45-45-90 triangle with a 45° angle at the origin and one side along the positive x-axis. Determine the slope and equation of the line coincident with the hypotenuse, then pick any point on this line and evaluate sin 45°, cos 45°, and tan 45°. Comment on what you notice.

Graph each linear equation and state the quadrants it traverses. Then pick one point on the line from each quadrant and evaluate the functions sin ␪, cos ␪ and tan ␪ using these points.

3 9. y x 4 11. y

13 x 3

10. y

5 x 12

12. y

13 x 2

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Find the values of the six trigonometric functions given P(x, y) is on the terminal side of angle ␪, with ␪ in standard position.

13. (8, 15)

14. (7, 24)

15. (⫺20, 21)

16. (⫺3, ⫺1)

17. (7.5, ⫺7.5)

18. (9, ⫺9)

19. (4 13, 4)

20. (⫺6, 613)

21. (2, 8)

22. (6, ⫺15)

23. (⫺3.75, ⫺2.5)

24. (6.75, 9)

5 2 25. a⫺ , b 9 3

7 3 26. a , ⫺ b 4 16

1 15 27. a , ⫺ b 4 2

28. a⫺

13 22 , b 5 25

29. Use the R 䊳 Pr( feature of a graphing calculator to find the radius corresponding to the point (–5, 5 13). 30. Use the R 䊳 P␪( feature of a graphing calculator to find the angle corresponding to (⫺28, ⫺45), then evaluate sin ␪, cos ␪, and tan ␪ for this angle. Compare each result to the values given by y y x sin ␪ ⫽ , cos ␪ ⫽ , and tan ␪ ⫽ . r r x 31. Evaluate the six trig functions in terms of x, y, and r for ␪ ⫽ 90°. 32. Evaluate the six trig functions in terms of x, y, and r for ␪ ⫽ 180°. Name the reference angle ␪r for the angle ␪ given.

33. ␪ ⫽ 120°

34. ␪ ⫽ 210°

35. ␪ ⫽ 135°

36. ␪ ⫽ 315°

37. ␪ ⫽ ⫺45°

38. ␪ ⫽ ⫺240°

39. ␪ ⫽ 112°

40. ␪ ⫽ 179°

41. ␪ ⫽ 500°

42. ␪ ⫽ 750°

43. ␪ ⫽ ⫺168.4°

44. ␪ ⫽ ⫺328.2°

State the quadrant of the terminal side of ␪, using the information given.

45. sin ␪ 7 0, cos ␪ 6 0 46. cos ␪ 6 0, tan ␪ 6 0 47. tan ␪ 6 0, sin ␪ 7 0 48. sec ␪ 7 0, tan ␪ 7 0

619

Find the exact values of sin ␪, cos ␪, and tan ␪ using reference angles.

49. ␪ ⫽ 330°

50. ␪ ⫽ 390°

51. ␪ ⫽ ⫺45°

52. ␪ ⫽ ⫺120°

53. ␪ ⫽ 240°

54. ␪ ⫽ 315°

55. ␪ ⫽ ⫺150°

56. ␪ ⫽ ⫺210°

For the information given, find the values of x, y, and r. Clearly indicate the quadrant of the terminal side of ␪, then state the values of the six trig functions of ␪.

4 5 and sin ␪ 6 0

58. tan ␪ ⫽ ⫺

59. csc ␪ ⫽ ⫺

37 35 and tan ␪ 7 0

60. sin ␪ ⫽ ⫺

61. csc ␪ ⫽ 3 and cos ␪ 7 0

62. csc ␪ ⫽ ⫺2 and cos ␪ 7 0

57. cos ␪ ⫽

7 8 and sec ␪ 6 0

63. sin ␪ ⫽ ⫺

12 5 and cos ␪ 7 0

20 29 and cot ␪ 6 0

5 12 and sin ␪ 6 0

64. cos ␪ ⫽

Find two positive and two negative angles that are coterminal with the angle given. Answers will vary.

65. 52°

66. 12°

67. 87.5°

68. 22.8°

69. 225°

70. 175°

71. ⫺107°

72. ⫺215°

Evaluate in exact form as indicated.

73. sin 120°, cos ⫺240°, tan 480° 74. sin 225°, cos 585°, tan ⫺495° 75. sin ⫺30°, cos ⫺390°, tan ⫺690° 76. sin 210°, cos 570°, tan ⫺150° Find the exact values of sin ␪, cos ␪, and tan ␪ using reference angles.

77. ␪ ⫽ 600°

78. ␪ ⫽ 480°

79. ␪ ⫽ ⫺840°

80. ␪ ⫽ ⫺930°

81. ␪ ⫽ 570°

82. ␪ ⫽ 495°

83. ␪ ⫽ ⫺1230°

84. ␪ ⫽ 3270°

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For each exercise, state the quadrant of the terminal side and the sign of the function in that quadrant. Then evaluate the expression using a calculator. Round to four decimal places.

85. sin 719°

86. cos 528°

87. tan ⫺419°

88. sec ⫺621°

89. csc 681°

90. tan 995°

91. cos 805°

92. sin 772°

䊳


93. The area of a parallelogram: A ⴝ ab sin ␪ The area of a parallelogram is given by the formula shown, where a and b are the lengths of the sides and ␪ is the angle between them. Use the formula to complete the following: (a) find the area of a parallelogram with sides a ⫽ 9 and b ⫽ 21 given ␪ ⫽ 50°. (b) What is the smallest integer value of ␪ where the area is greater than 150 units2? (c) State what happens when ␪ ⫽ 90°. (d) How can you find the area of a triangle using this formula?

䊳

94. The angle between two intersecting lines: m2 ⴚ m1 tan ␪ ⴝ 1 ⴙ m2m1 Given line 1 and line 2 with slopes m1 and m2, respectively, the angle between the two lines is given by the formula shown. Find the angle ␪ if the equation of line 1 is y1 ⫽ 34x ⫹ 2 and line 2 has equation y2 ⫽ ⫺23x ⫹ 5.

APPLICATIONS

Find all angles satisfying the stated relationship. For standard angles, express your answer in exact form. For nonstandard values, use a calculator and round function values to tenths.

95. cos ␪ ⫽

1 2

97. sin ␪ ⫽ ⫺

96. sin ␪ ⫽ 13 2

99. sin ␪ ⫽ 0.8754 101. tan ␪ ⫽ ⫺2.3512

12 2

98. tan ␪ ⫽ ⫺

13 1

100. cos ␪ ⫽ 0.2378

make over three complete, counterclockwise rotations, with the blade stopping at the 8 o’clock position. What angle ␪ did the blade turn through? Name all angles that are coterminal with ␪. 105. High dives: As part of a diving competition, David executes a perfect reverse two-and-a-half flip. Does he enter the water feet first or head first? Through what angle did he turn from takeoff until the moment he entered the water? Exercise 105

102. cos ␪ ⫽ ⫺0.0562

103. Nonacute angles: At a Exercise 103 recent carnival, one of the games on the midway was played using a large spinner that turns clockwise. On Jorge’s spin the number 25 began at the 12 o’clock (top/center) position, returned to this position five times during the spin and stopped at the 3 o’clock position. What angle ␪ did the spinner spin through? Name all angles that are coterminal with ␪. 104. Nonacute angles: One of the four blades on a ceiling fan has a decal on it and begins at a designated “12 o’clock” position. Turning the switch on and then immediately off, causes the blade to

106. Gymnastics: While working out on a trampoline, Charlene does three complete, forward flips and then belly-flops on the trampoline before returning to the upright position. What angle did she turn through from the start of this maneuver to the moment she belly-flops?

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107. Spiral of Archimedes: The graph shown is called the spiral of Archimedes. Through what angle ␪ has the spiral turned, given the spiral terminates at 16, 22 as indicated?

Exercise 107 y 10

10

0

10 x

(6, 2) 10

䊳

621


108. Involute of a circle: The graph shown is called the involute of a circle. Through what angle ␪ has the involute turned, given the graph terminates at 14, 3.52 as indicated?

Exercise 108 y 10

10

10 x

0

(4, 3.5) 10


109. In an elementary study of trigonometry, the hands of a clock are often studied because of the angle relationship that exists between the hands. For example, at 3 o’clock, the angle between the two hands is a right angle and measures 90°. a. What is the angle between the two hands at 1 o’clock? 2 o’clock? Explain why. b. What is the angle between the two hands at 6:30? 7:00? 7:30? Explain why. c. Name four times at which the hands will form a 45° angle. Exercise 110 y 110. In the diagram shown, the indicated ray is of arbitrary length. (a) Through what additional 5 C angle ␣ would the ray have to be rotated to create triangle ABC? (b) What will be the length of side AC once the triangle is complete? 111. Referring to Exercise 104, suppose the fan blade has a radius of 20 in. and is turning at a rate of 12 revolutions per second. (a) Find the angle the blade turns through in 3 sec. (b) Find the circumference of the circle traced out by the tip of the blade. (c) Find the total distance traveled by the blade tip in 10 sec. (d) Find the speed, in miles per hour, that the tip of the blade is traveling.

䊳

B 5

A

(3, 2)

5 x

5


112. (6.1) For emissions testing, automobiles are held stationary while a heavy roller installed in the floor allows the wheels to turn freely. If the large wheels of a customized pickup have a radius of 18 in. and are turning at 300 revolutions per minute, what speed is the odometer of the truck reading in miles per hour?

114. (6.6) Jazon is standing 117 ft from the base of the Washington Monument in Washington, D.C. If his eyes are 5 ft above level ground and he must hold his head at a 78° angle from horizontal to see the top of the monument (the angle of elevation is 78º), estimate the height of the monument. Answer to the nearest tenth of a foot.

113. (5.4) Solve for t. Answer in both exact and approximate form:

115. (1.4) Find an equation of the line perpendicular to 4x 5y 15 that contains the point 14, 32.

250 150e0.05t 202.

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Trigonometric Equation Models


A. Create a trigonometric model from critical points or data B. Create a sinusoidal model from data using regression

In the most common use of the word, a cycle is any series of events or operations that occur in a predictable pattern and return to a starting point. This includes things as diverse as the wash cycle on a washing machine and the powers of i. There are a number of common events that occur in sinusoidal cycles, or events that can be modeled by a sine wave. As in Section 6.5, these include monthly average temperatures, monthly average daylight hours, and harmonic motion, among many others. Less well-known applications include alternating current, biorhythm theory, and animal populations that fluctuate over a known period of years. In this section, we develop two methods for creating a sinusoidal model. The first uses information about the critical points (where the cycle reaches its maximum or minimum values); the second involves computing the equation of best fit (a regression equation) for a set of data.

A. Critical Points and Sinusoidal Models Although future courses will define them more precisely, we will consider critical points to be inputs where a function attains a minimum or maximum value. If an event or phenomenon is known to behave sinusoidally (regularly fluctuating between a maximum and minimum), we can create an acceptable model of the form y A sin1Bx C2 D given these critical points (x, y) and the period. For instance, 2␲ , we find many weather patterns have a period of 12 months. Using the formula P B ␲ 2␲ B and substituting 12 for P gives B (always the case for phenomena with P 6 a 12-month cycle). The maximum value of A sin1Bx C2 D will always occur when sin1Bx C2 1, and the minimum at sin1Bx C2 1, giving this system of equations: max value M A112 D and min value m A112 D. Solving the Mm Mm system for A and D gives A and D as before. To find C, assume 2 2 the maximum and minimum values occur at (x2, M) and (x1, m), respectively. We can substitute the values computed for A, B, and D in y A sin1Bx C2 D, along with either (x2, M) or (x1, m), and solve for C. Using the minimum value (x1, m), where x x1 and y m, we have y A sin1Bx C2 D m A sin1Bx1 C2 D mD sin1Bx1 C2 A

sinusoidal equation model substitute m for y and x1 for x isolate sine function

Fortunately, for sine models constructed from critical points we have yD mD S , which is always equal to 1 (see Exercise 33). This gives a simple A A 3␲ 3␲ Bx1 C or C Bx1. result for C, since 1 sin1Bx1 C2 leads to 2 2 See Exercises 7 through 12 for practice with these ideas.

622

6–114

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Section 6.8 Trigonometric Equation Models

EXAMPLE 1

䊳

623

Developing a Model for Polar Ice Cap Extent from Critical Points When the Spirit and Odyssey Rovers landed on Mars (January 2004), there was a renewed public interest in studying the planet. Of particular interest were the polar ice caps, which are now thought to hold frozen water, especially the northern cap. The Martian ice caps expand and contract with the seasons, just as they do here on Earth but there are about 687 days in a Martian year, making each Martian “month” just over 57 days long (1 Martian day 1 Earth day). At its smallest size, the northern ice cap covers an area of roughly 0.17 million square miles. At the height of winter, the cap covers about 3.7 million square miles (an area about the size of the 50 United States). Suppose these occur at the beginning of month 4 1x ⫽ 42 and month 10 1x ⫽ 102 respectively. a. Use this information to create a sinusoidal model of the form f 1x2 ⫽ A sin1Bx ⫹ C2 ⫹ D. b. Use the model to predict the area of the ice cap in the eighth Martian month. c. Use a graphing calculator to determine the number of months the cap covers less than 1 million mi2.

Solution

䊳

␲ . The maximum 6 and minimum points are (10, 3.7) and (4, 0.17). Using this information, 3.7 ⫹ 0.17 3.7 ⫺ 0.17 D⫽ ⫽ 1.935 and A ⫽ ⫽ 1.765. Using 2 2 3␲ ␲ 5␲ 3␲ C⫽ ⫺ Bx1 gives C ⫽ ⫺ 142 ⫽ . The equation model is 2 2 6 6 ␲ 5␲ b ⫹ 1.935, where f (x) represents millions of square f 1x2 ⫽ 1.765 sin a x ⫹ 6 6 miles in month x. b. For the size of the cap in month 8 we evaluate the function at x ⫽ 8. ␲ 5␲ d ⫹ 1.935 substitute 8 for x f 182 ⫽ 1.765 sin c 182 ⫹ 6 6 ⫽ 2.8175 result

a. Assuming a “12-month” weather pattern, P ⫽ 12 and B ⫽

In month 8, the polar ice cap will cover about 2,817,500 mi2. c. Of the many options available, we opt to solve by locating the points where ␲ 5␲ Y1 ⫽ 1.765 sin a X ⫹ b ⫹ 1.935 and Y2 ⫽ 1 intersect. After entering the 6 6 functions on the Y= screen, we set x 僆 30, 12 4 and y 僆 3 ⫺2, 54 for a window with a frame around the output values. 5 Press 2nd TRACE (CALC) 5:intersect to find the intersection points. To four decimal places they occur at x ⫽ 2.0663 and x ⫽ 5.9337. The ice cap at the 0 12 northern pole of Mars has an area of less than 1 million mi2 from early in the second month to late in the fifth month. The second intersection is shown in ⫺2 the figure. Now try Exercises 19 and 20

䊳

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While this form of “equation building” can’t match the accuracy of a regression model (computed from a larger set of data), it does lend insight as to how sinusoidal functions work. The equation will always contain the maximum and minimum values, and using the period of the phenomena, we can create a smooth sine wave that “fills in the blanks” between these critical points. EXAMPLE 2

䊳

Developing a Model of Wildlife Population from Critical Points Naturalists have found that many animal populations, such as the arctic lynx, some species of fox, and certain rabbit breeds, tend to fluctuate sinusoidally over 10-year periods. Suppose that an extended study of a lynx population began in 2000, and in the third year of the study, the population had fallen to a minimum of 2500. In the eighth year the population hit a maximum of 9500. a. Use this information to create a sinusoidal model of the form P1x2 ⫽ A sin1Bx ⫹ C2 ⫹ D. b. Use the model to predict the lynx population in the year 2006. c. Use a graphing calculator to determine the number of years the lynx population is above 8000 in a 10-year period.

Solution

䊳

2␲ ␲ ⫽ . Using 2000 as year zero, the minimum 10 5 and maximum populations occur at (3, 2500) and (8, 9500). From the information 9500 ⫺ 2500 9500 ⫹ 2500 ⫽ 6000, and A ⫽ ⫽ 3500. Using the given, D ⫽ 2 2 3␲ ␲ 9␲ ⫺ 132 ⫽ , giving an equation model of minimum value we have C ⫽ 2 5 10 ␲ 9␲ b ⫹ 6000, where P(x) represents the lynx P1x2 ⫽ 3500 sin a x ⫹ 5 10 population in year x. b. For the population in 2006 we evaluate the function at x ⫽ 6.

a. Since P ⫽ 10, we have B ⫽

␲ 9␲ b ⫹ 6000 P1x2 ⫽ 3500 sin a x ⫹ 5 10 9␲ ␲ d ⫹ 6000 P162 ⫽ 3500 sin c 162 ⫹ 5 10 ⬇ 7082 In 2006, the lynx population was about 7082. c. Using a graphing calculator and the functions ␲ 9␲ Y1 ⫽ 3500 sin a X ⫹ b ⫹ 6000 and 5 10 Y2 ⫽ 8000, we attempt to find points of intersection. Enter the functions (press Y= ) and set a viewing window (we used x 僆 30, 12 4 and y 僆 30, 10,000 4 ). Press 0 2nd TRACE (CALC) 5:intersect to find where Y1 and Y2 intersect. To four decimal places this occurs at x ⫽ 6.4681 and x ⫽ 9.5319. The lynx population exceeded 8000 for roughly 3 years. The first intersection is shown.

sinusoidal function model

substitute 6 for x result

10,000

12

0


䊳

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This type of equation building isn’t limited to the sine function. In fact, there are many situations where a sine model cannot be applied. Consider the length of the shadows cast by a flagpole or radio tower as the Sun makes its way across the sky. The shadow’s length follows a regular pattern (shortening then lengthening) and “always returns to a starting point,” yet when the Sun is low in the sky the shadow becomes (theoretically) infinitely long, unlike the output values from a sine function. If we consider signed shadow lengths at a latitude where the Sun passes directly overhead (in contrast with Example 11 in Section 6.5), an equation involving tan x might provide a good model. We’ll attempt to model the data using y ⫽ A tan1Bx ⫾ C2, with the D-term absent since a vertical shift in this context has no meaning. Recall that the C ␲ period of the tangent function is P ⫽ and that ⫾ gives the magnitude and direc冟B冟 B tion of the horizontal shift, in a direction opposite the sign.

EXAMPLE 3

䊳

The data given tracks the length Hour of Length Hour of of a gnomon’s shadow for the the Day (cm) the Day 12 daylight hours at a certain location q 0 7 near the equator (positive and 1 29.9 8 negative values indicate lengths before noon and after noon 2 13.9 9 respectively). Assume t ⫽ 0 3 8.0 10 represents 6:00 A.M. 4 4.6 11 a. Use the data to find an equation 5 2.1 12 model of the form L1t2 ⫽ A tan1Bt ⫾ C2. 6 0 b. Graph the function and scatterplot. c. Find the shadow’s length at 4:30 P.M. d. If the shadow is 6.1 cm long, what time in the morning is it?

WORTHY OF NOTE A gnomon is the protruding feature of a sundial, casting the shadow used to estimate the time of day (see photo).

Solution

Using Data to Develop a Function Model for Shadow Length

䊳

Length (cm) ⫺2.1 ⫺4.6 ⫺8.0 ⫺13.9 ⫺29.9 ⫺q

a. We begin by noting this phenomenon has a period of P ⫽ 12. Using the period ␲ ␲ ␲ formula for tangent we solve for B: P ⫽ gives 12 ⫽ , so B ⫽ . Since we B B 12 want (6, 0) to be the “center” of the function [instead of (0, 0)], we desire a C ␲ horizontal shift 6 units to the right. Using the ratio awith B ⫽ b gives B 12 12C ␲ ⫺6 ⫽ so C ⫽ ⫺ . To find A we use the equation built so far: ␲ 2 ␲ ␲ L1t2 ⫽ A tan a t ⫺ b, and any data point to solve for A. Using (3, 8) 12 2 we obtain 8 ⫽ A tan a ␲ 8 ⫽ A tan a⫺ b 4 ⫺8 ⫽ A

␲ ␲ 132 ⫺ b: 12 2 Figure 6.141

simplify ␲ solve for A: tan a⫺ b ⫽ ⫺1 4

The equation model is ␲ ␲ L1t2 ⫽ ⫺8 tan a t ⫺ b. 12 2 b. The scatterplot and graph are shown in Figure 6.141.

40

⫺1.2

13.2

⫺40

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c. 4:30 P.M. indicates t ⫽ 10.5. Evaluating L(10.5) gives ␲ ␲ t⫺ b 12 2 ␲ ␲ L110.52 ⫽ ⫺8 tan c 110.52 ⫺ d 12 2 3␲ ⫽ ⫺8 tan a b 8 ⬇ ⫺19.31 L1t2 ⫽ ⫺8 tan a

A. You’ve just seen how we can create a trigonometric model from critical points or data

function model

substitute 10.5 for t

Figure 6.142

simplify

30

result

At 4:30 P.M., the shadow has a length of 冟⫺19.31冟 ⫽ 19.31 cm. d. Substituting 6.1 for L(t) and solving for t graphically gives the graph shown in Figure 6.142, where we note the day is about 3.5 hr old—it is about 9:30 A.M.

0

12

⫺30


䊳

B. Data and Sinusoidal Regression Most graphing calculators are programmed to handle numerous forms of polynomial and nonpolynomial regression, including sinusoidal regression. The sequence of steps used is the same regardless of the form chosen. Exercises 13 through 16 offer further practice with regression fundamentals. Example 4 illustrates their use in context. EXAMPLE 4

䊳

Calculating a Regression Equation for Seasonal Temperatures The data shown give the record high Month Temp. Month Temp. temperature for selected months in (Jan S 1) (°F) (Jan S 1) (°F) Bismarck, North Dakota. a. Use the data to draw a scatterplot, 1 63 9 105 then find a sinusoidal regression 3 81 11 79 model and graph both on the same 5 98 12 65 screen. 7 109 b. Use the equation model to estimate the record high temperatures for months 2, 6, and 8. c. Determine what month gives the largest difference between the actual data and the computed results. Source: NOAA Comparative Climate Data 2004.

Solution

䊳

a. Entering the data and running the regression (in radian mode) results in the coefficients shown in Figure 6.143. After entering the equation in Y1 and pressing ZOOM 9:Zoom Stat we obtain the graph shown in Figure 6.144 (indicated window settings have been rounded). Figure 6.144 Figure 6.143

117

0

13

55

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b. Using x ⫽ 2, x ⫽ 6, and x ⫽ 8 as inputs, the equation model projects record high temperatures of 68.5°, 108.0°, and 108.1°, respectively. to evaluate c. In the header of L3, use Y1(L1) the regression model using the inputs from L1, and place the results in L3. Entering L2 ⫺ L3 in the header of L4 gives the results shown in Figure 6.145 and we note the largest difference occurs in September—about 4°.

Figure 6.145

ENTER


䊳

Weather patterns differ a great deal depending on the locality. For example, the annual precipitation received by Seattle, Washington, far exceeds that received by Cheyenne, Wyoming. Our final example compares the two amounts and notes an interesting fact about the relationship. EXAMPLE 5

䊳

Calculating a Regression Model for Seasonal Precipitation The average monthly precipitation (in inches) for Cheyenne, Wyoming, and Seattle, Washington, are shown in the table. a. Use the data to find a sinusoidal regression model for the average monthly precipitation in each city. Enter or paste the equation for Cheyenne in Y1 and the equation for Seattle in Y2. b. Graph both equations on the same screen (without the scatterplots) and use TRACE or 2nd TRACE (CALC) 5:intersect to help estimate the number of months Cheyenne receives more precipitation than Seattle.

Month WY WA (Jan. S 1) Precip. Precip.

Source: NOAA Comparative Climate Data 2004.

Solution

䊳

1

0.45

5.13

2

0.44

4.18

3

1.05

3.75

4

1.55

2.59

5

2.48

1.77

6

2.12

1.49

7

2.26

0.79

8

1.82

1.02

9

1.43

1.63

a. Setting the calculator in Float 0 1 2 3 4 5 6 7 8 9 10 0.75 3.19 MODE and running sinusoidal regressions gives 11 0.64 5.90 the equations shown in Figure 6.146. 12 0.46 5.62 b. Both graphs are shown in Figure 6.147. Using the TRACE feature, we find the graphs intersect at approximately (4.7, 2.0) and (8.4, 1.7). Cheyenne receives more precipitation than Seattle for about 8.4 ⫺ 4.7 ⫽ 3.7 months of the year. Figure 6.147 Figure 6.146

7

0

B. You’ve just seen how we can create a sinusoidal model from data using regression

13

0


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6.8 EXERCISES 䊳



1. For y A sin1Bx C2 D, the maximum value occurs when _________ 1, leaving y ________.

2. For y A sin1Bx C2 D, the minimum value occurs when _________ 1, leaving y _________.

3. For y A sin 1Bx C2 D, the value of C can be

4. Any phenomenon with sinusoidal behavior regularly fluctuates between a ________ and a ________ value.

found using C Bx, where x is a _________ point.

5. Discuss/Expand on Example 4. Why does the equation calculated from the maximum and minimum values differ from the regression equation, even though both use the same set of data?

䊳

6. Discuss/Explain: (1) How is data concerning average rainfall related to the average discharge rate of rivers? (2) How is data concerning average daily temperature related to water demand? Are both sinusoidal with the same period? Are there other associations you can think of?


Find a sinusoidal equation for the information as given.

7. minimum value at (9, 25); maximum value at (3, 75); period: 12 min

10. minimum value at (3, 3.6); maximum value at (7, 12); period: 8 hr

8. minimum value at (4.5, 35); maximum value at (1.5, 121); period: 6 yr

11. minimum value at (5, 279); maximum value at (11, 1285); period: 12 yr

9. minimum value at (15, 3); maximum value at (3, 7.5); period: 24 hr

12. minimum value at (6, 8280); maximum value at (22, 23,126); period: 32 yr

Data and sinusoidal regression models: For the following sets of data (a) find a sinusoidal regression equation using your calculator; (b) construct an equation manually using the period and maximum/minimum values; and (c) graph both on the same screen, then use a TABLE to find the largest difference between output values.

13.

14.

15.

16.

Output

Month (Jan. 1)

Output

Month (Jan. 1)

Output

15

1

179

1

16

1

86

41

4

201

2

19

2

96

7

69

7

195

3

21

3

99

10

91

10

172

4

22

4

95

13

100

13

145

5

21

5

83

16

90

16

120

6

19

6

72

19

63

19

100

7

16

7

56

22

29

22

103

8

13

8

48

25

5

25

124

9

11

9

43

28

2

28

160

10

10

10

49

18

31

188

11

11

11

58

12

13

12

73

Output

Day of Month

1 4

Day of Month

31

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17. Orbiting Distance North or South of the Equator: D(t) A cos(Bt) Unless a satellite is placed in a strict equatorial orbit, its distance north or south of the equator will vary according to the sinusoidal model shown, where D(t) is the distance t minutes after entering orbit. Negative values indicate it is south of the equator, and the distance D is actually a twodimensional distance, as seen from a vantage point in outer space. The value of B depends on the speed of the satellite and the time it takes to complete one orbit, while 冟A冟 represents the maximum distance from the equator. (a) Find the equation model for a

䊳

629


satellite whose maximum distance north of the equator is 2000 miles and that completes one orbit every 2 hours (P ⫽ 120). (b) Is the satellite north or south of the equator 257 min after entering orbit? How far north or south? 18. Biorhythm Theory: P(d) 50 sin(Bd) 50 Advocates of biorhythm theory believe that human beings are influenced by certain biological cycles that begin at birth, have different periods, and continue throughout life. The classical cycles and their periods are: physical potential (23 days), emotional potential (28 days), and intellectual potential (33 days). On any given day of life, the percent of potential in these three areas is purported to be modeled by the function shown, where P(d) is the percent of available potential on day d of life. Find the value of B for each of the physical, emotional, and intellectual potentials and use it to see what the theory has to say about your potential today. Use day d ⫽ 365.251age2 ⫹ days since last birthday.

APPLICATIONS

19. Record monthly temperatures: The U.S. National Oceanic and Atmospheric Administration (NOAA) keeps temperature records for most major U.S. cities. For Phoenix, Arizona, they list an average high temperature of 65.0°F for the month of January (month 1) and an average high temperature of 104.2°F for July (month 7). Assuming January and July are the coolest and warmest months of the year, (a) build a sinusoidal function model for temperatures in Phoenix, and (b) use the model to find the average high temperature in September. (c) If a person has a tremendous aversion to temperatures over 95°, during what months should they plan to vacation elsewhere? 20. Seasonal size of polar ice caps: Much like the polar ice cap on Mars, the sea ice that surrounds the continent of Antarctica (the Earth’s southern polar cap) varies seasonally, from about 8 million mi2 in September to about 1 million mi2 in March. Use this information to (a) build a sinusoidal equation that models the advance and retreat of the sea ice, and (b) determine the size of the ice sheet in May. (c) Find the months of the year that the sea ice covers more than 6.75 million mi2. 21. Body temperature cycles: A phenomenon is said to be circadian if it occurs in 24-hr cycles. A person’s body temperature is circadian, since there are normally small, sinusoidal variations in body temperature from a low of 98.2°F to a high of 99°F throughout a 24-hr day. Use this information to (a) build the circadian equation for a person’s body temperature, given t ⫽ 0 corresponds to midnight and that a person usually reaches their minimum temperature at 5 A.M.; (b) find the time(s) during a day when a person reaches “normal” body temperature 198.6°2; and (c) find the number of hours each day that body temperature is 98.4°F or less.

Exercise 20

Ice caps mininum maximum

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25. Distance and Distance Height apparent height: Traveled (mi) (cm) While driving toward 0 0 a Midwestern town 3 1 on a long, flat stretch 6 1.8 of highway, I decide to pass the time by 9 2.8 measuring the 12 4.2 apparent height of 15 6.3 the tallest building in 18 10 the downtown area 21 21 as I approach. At the 24 q time the idea occurred to me, the buildings were barely visible. Three miles later I hold a 30-cm ruler up to my eyes at arm’s length, and the apparent height of the tallest building is 1 cm. After three more miles the apparent height is 1.8 cm. Measurements are taken every 3 mi until I reach town and are shown in the table (assume I was 24 mi from the parking garage when I began this activity). (a) Use the data to come up with a tangent function model of the building’s apparent height after traveling a distance of x mi closer. (b) What was the apparent height of the building after I had driven 19 mi? (c) How many miles had I driven when the apparent height of the building took up all 30 cm of my ruler?

22. Position of engine piston: For an internal combustion engine, the position of a piston in the cylinder can be modeled by a sinusoidal function. For a particular engine size and idle speed, the piston head is 0 in. from the top of the cylinder (the minimum value) when t 0 at the beginning of the intake stroke, and reaches a maximum distance of 4 in. from the top of the cylinder (the maximum 1 value) when t 48 sec at the beginning of the compression stroke. Following the compression 2 2, the exhaust stroke is the power stroke 1t 48 3 4 2, after stroke 1t 48 2, and the intake stroke 1t 48 which it all begins again. Given the period of a four-stroke engine under these conditions is 1 P 24 second, (a) find the sinusoidal equation modeling the position of the piston, and (b) find the distance of the piston from the top of the cylinder at t 19 sec. Which stroke is the engine in at this moment? Intake 0 1 2 3 4

Compression

0 1 2 3 4

0 1 2 3 4

0 1 2 3 4

26. Earthquakes and elastic rebound: The theory of elastic rebound has been used by seismologists to study the cause of earthquakes. As seen in the figure, the Earth’s crust is stretched to a breaking point by the slow movement of one tectonic plate in a direction opposite the other along a fault line, and when the rock snaps—each half violently rebounds to its original alignment causing the Earth to quake.

Data and tangent functions: Use the data given to find an equation model of the form f(x) A tan(Bx C ). Then graph the function and scatterplot to help find (a) the output for x 2.5, and (b) the value of x where f (x) 16.

23.

x

y

x

y

0

q

7

1.4

1

20

8

3

2

9.7

9

5.2

3

5.2

10

9.7

4

3

11

20

5

1.4

12

q

6

0

Stress line Start

After years of movement

Elastic rebound

POP!

24.

x

y

x

y

0

q

7

6.4

1

91.3

8

13.7

2

44.3

9

23.7

3

23.7

10

44.3

4

13.7

11

91.3

5

6.4

12

q

6

0

(0, 0)

Fault line

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x

y

⫺4.5

⫺61

1

2.1

⫺4

⫺26

2

6.8

⫺3

⫺14.8

3

15.3

⫺2

⫺7.2

4

25.4

⫺1

⫺1.9

4.5

59

0

x

y

0

Suppose the misalignment of these plates through the stress and twist of crustal movement can be modeled by a tangent graph, where x represents the horizontal distance from the original stress line, and y represents the vertical distance from the fault line. Assume a “period” of 10.2 m. (a) Use the data from the table to come up with a trigonometric model of the deformed stress line. (b) At a point 4.8 m along the fault line, what is the distance to the deformed stress line (moving parallel to the original stress line)? (c) At what point along the fault line is the vertical distance to the deformed stress line 50 m? 27. Record monthly Month High temperatures: The (Jan. S 1) Temp. (°F) highest temperature of 2 76 record for the even 4 89 months of the year are 6 98 given in the table for the city of Pittsburgh, 8 100 Pennsylvania. (a) Use 10 87 the data to draw a 12 74 scatterplot, then find a sinusoidal regression model and graph both on the same screen. (b) Use the equation to estimate the record high temperatures for the odd-numbered months. (c) What month shows the largest difference between the actual data and the computed results? Source: 2004 Statistical Abstract of the United States, Table 378.

28. River discharge rate: Month Rate The average discharge (Jan. S 1) (m3/sec) rate of the Alabama 1 1569 River is given in 3 1781 the table for the 5 1333 odd-numbered months of the year. (a) Use 7 401 the data to draw a 9 261 scatterplot, then find a 11 678 sinusoidal regression model and graph both on the same screen. (b) Use the equation to estimate the flow rate for the even-

numbered months. (c) Use the graph and equation to estimate the number of days per year the flow rate is below 500 m3/sec. Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

29. Illumination of the moon’s surface: The table given indicates the percent of the Moon that is illuminated for the days of a particular month, at a given latitude. (a) Use a graphing calculator to find a sinusoidal regression model. (b) Use the model to determine what percent of the Moon is illuminated on day 20. (c) Use the maximum and minimum values with the period and an appropriate horizontal shift to create your own model of the data. How do the values for A, B, C, and D compare? Day

% Illum.

Day

% Illum.

1

28

19

34

4

55

22

9

7

82

25

0

10

99

28

9

13

94

31

30

16

68

30. Connections between weather and mood: The mood of persons with SAD syndrome (seasonal affective disorder) often depends on the weather. Victims of SAD are typically more despondent in rainy weather than when the sun is out, and more comfortable in the daylight hours than at night. The table shows the average number of daylight hours for Vancouver, British Columbia, for 12 months of a year. (a) Use a calculator to find a sinusoidal regression model. (b) Use the model to estimate the number of days per year (use 1 month ⬇ 30.5 days) with more than 14 hr of daylight. (c) Use the maximum and minimum values with the period and an appropriate horizontal shift to create a model of the data. How do the values for A, B, C, and D compare? Source: Vancouver Climate at www.bcpassport.com/vital.

Month

Hours

Month

Hours

1

8.3

7

16.2

2

9.4

8

15.1

3

11.0

9

13.5

4

12.9

10

11.7

5

14.6

11

9.9

6

15.9

12

8.5

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31. Average monthly rainfall: The average monthly rainfall (in inches) for Reno, Nevada, is shown in the table. (a) Use the data to find a sinusoidal regression model for the monthly rainfall. (b) Graph this equation model and the rainfall equation model for Cheyenne, Wyoming (from Example 5), on the same screen, and estimate the number of months that Reno gets more rainfall than Cheyenne. Source: NOAA Comparative Climate Data 2004.

䊳

6–124


Month (Jan S 1)

Reno Rainfall

Month (Jan S 1)

Reno Rainfall

1

1.06

7

0.24

2

1.06

8

0.27

3

0.86

9

0.45

4

0.35

10

0.42

5

0.62

11

0.80

6

0.47

12

0.88

32. Hours of Month TX MN daylight by (Jan S 1) Sunlight Sunlight month: The 1 10.4 9.1 number of 2 11.2 10.4 daylight hours 3 12.0 11.8 per month (as measured on the 4 12.9 13.5 15th of each 5 14.4 16.2 month) is shown 6 14.1 15.7 in the table for 7 13.9 15.2 the cities of 8 13.3 14.2 Beaumont, 9 12.4 12.6 Texas, and Minneapolis, 10 11.5 11.0 Minnesota. (a) 11 10.7 9.6 Use the data to 12 10.2 8.7 find a sinusoidal regression model of the daylight hours for each city. (b) Graph both equations on the same screen and use the graphs to estimate the number of days each year that Beaumont receives more daylight than Minneapolis (use 1 month ⫽ 30.5 days). Source: www.encarta.msn.com/media_701500905/ Hours_of_Daylight_by_Latitude.html.


y⫺D m⫺D S is equal A A M⫹m M⫺m to ⫺1. Then verify this relationship in general by substituting for A and for D. 2 2

33. For the equations from Examples 1 and 2, use the minimum value (x, m) to show that

34. A dampening factor is any function whose product with a sinusoidal function causes a systematic reduction in amplitude. In the graph shown, the function y ⫽ sin13x2 has been dampened by the linear function 1 y ⫽ ⫺ x ⫹ 2 over the interval 3⫺2␲, 2␲ 4 . Notice that the peaks of the sine graph are points on the graph of 4 this line. The table gives approximate points of intersection for y ⫽ sin13x2 with another dampening factor. Use the regression capabilities of a graphing calculator to find the approximate equation of the dampening factor. (Hint: It is not linear.)

4

⫺2␲

2␲

⫺4

x

y

x

y

␲ 6

2.12

13␲ 6

1.03

5␲ 6

1.58

17␲ 6

1.02

3␲ 2

1.22

7␲ 2

1.18

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Making Connections


35. (2.4) State the domains of the following functions: 4 a. f 1x2 x1.7 b. g1x2 x5 9 c. h1x2 2x3

36. (5.4) Use properties of logarithms to write the following as a single term.

37. (6.1) The barrel on a winch has a radius of 3 in. and is turning at 30 rpm. As the barrel turns it winds in a cable that is pulling a heavy pallet across the warehouse floor. How fast is the pallet moving in feet per minute?

38. (6.6) Clarke is standing between two tall buildings. The angle of elevation to the top of the building to her north is 60°, while the angle of elevation to the top of the building to her south is 70. If she is 400 m from the base of the northern building and 200 m from the southern, which one is taller?

4 log 2

1 log 25 2

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs A through H are given. Match the characteristics given in 1 through 16 to one of the eight graphs. y

(a)

y

(b)

2

⫺2

4

2 x

y

2␲

2␲ t

2

4

2␲

2␲ t

2␲ t

2␲ t 3

4

4

9. ____ period 4

1. ____ period ␲

␲ 10. ____ f 1t2 cos a tb 2

2. ____ amplitude 2 3. ____ f 1t2 sec at

y

(h)

4

2␲

2␲

5

y

(g)

5

2␲ t

4 t

2

y

(f)

2

3

2␲

4

4 t

y

(d)

2

4

⫺2

(e)

y

(c)

4

␲ b1 2

␲ 11. ____ f 1t2 2 tan a tb 2

4. ____ f 1t2 2 sin t

1 13 b is on the graph 12. ____ The point a , 2 2

5. ____ minimum of 1 occurs at t ␲

13. ____ t k␲, shifted up 1 unit

6. ____ x y 1 2

2

7. ____ f 1t2 sin at 8. ____ period 2

␲ b 2

14. ____ f 1t2 csc t 1

15. ____ f 1t2 2 cot at 16. ____ f 1t2 T for t ⺢

␲ b 2

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Angle Measure, Special Triangles, and Special Angles

KEY CONCEPTS • An angle is defined as the joining of two rays at a common endpoint called the vertex. • An angle in standard position has its vertex at the origin and its initial side on the positive x-axis. • Two angles in standard position are coterminal if they have the same terminal side. • A counterclockwise rotation gives a positive angle, a clockwise rotation gives a negative angle. 1 of a full revolution. One (1) radian is the measure of a central angle • One 11°2 degree is defined to be 360 subtended by an arc equal in length to the radius. • Degrees can be divided into a smaller unit called minutes: 1° 60¿; minutes can be divided into a smaller unit called seconds: 1¿ 60–. This implies 1° 3600–. • Two angles are complementary if they sum to 90° and supplementary if they sum to 180°. • Properties of triangles: (I) the sum of the angles is 180°; (II) the combined length of any two sides must exceed that of the third side and; (III) larger angles are opposite longer sides. • Given two triangles, if all three corresponding angles are equal, the triangles are said to be similar. If two triangles are similar, then corresponding sides are in proportion. • In a 45-45-90 triangle, the sides are in the proportion 1x: 1x: 12x. • In a 30-60-90 triangle, the sides are in the proportion 1x: 13x: 2x. • The formula for arc length: s r␪, ␪ in radians. 1 • The formula for the area of a circular sector: A r2 ␪, ␪ in radians. 2 ␲ 180° ; for radians to degrees, multiply by . • To convert degree measure to radians, multiply by ␲ 180° ␲ ␲ ␲ ␲ • Special angle conversions: 30° , 45° , 60° , 90° . 6 4 3 2 • A location north or south of the equator is given in degrees latitude; a location east or west of the Greenwich Meridian is given in degrees longitude. ␪ • Angular velocity is a rate of rotation per unit time: ␻ . t ␪r • Linear velocity is a change in position per unit time: V or V r␻. t EXERCISES 1. Convert 147° 36¿ 48– to decimal degrees. 2. Convert 32.87° to degrees, minutes, and seconds.

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635

3. All of the right triangles given are similar. Find the dimensions of the largest triangle. Exercise 3

Exercise 4

16.875 60

3 6

4

d

40

0y

d

4. Use special angles/special triangles to find the length of the bridge needed to cross the lake shown in the figure. 5. Convert to degrees:

2␲ . 3

6. Convert to radians: 210°.

7. Find the arc length if r 5 and ␪ 57°.

8. Evaluate without using a calculator: 7␲ sin a b. 6

Find the angle, radius, arc length, and/or area as needed, until all values are known. y y 9. 10. 11. 96 in.

152 m2

s 1.7

2.3 15 cm

y

x

r

x

8

x

12. With great effort, 5-year-old Mackenzie has just rolled her bowling ball down the lane, and it is traveling painfully slow. So slow, in fact, that you can count the number of revolutions the ball makes using the finger holes as a reference. (a) If the ball is rolling at 1.5 revolutions per second, what is the angular velocity? (b) If the ball’s radius is 5 in., what is its linear velocity in feet per second? (c) If the distance to the first pin is 60 feet and the ball is true, how many seconds until it hits?

SECTION 6.2

Unit Circles and the Trigonometry of Real Numbers

KEY CONCEPTS • A central unit circle is a circle with radius 1 unit having its center at the origin. • A central circle is symmetric to both axes and the origin. This means that if (a, b) is a point on the circle, then 1a, b2, 1a, b2 , and 1a, b2 are also on the circle and satisfy the equation of the circle. • On the unit circle with ␪ in radians, the length of a subtended arc is numerically the same as the subtended angle, making the arc a “circular number line” and associating any given rotation with a unique real number. • A reference angle is defined to be the acute angle formed by the terminal side of a given angle and the x-axis. For functions of a real number we refer to a reference arc rather than a reference angle.

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• For any real number t and a point on the unit circle associated with t, we have: cos t x

sin t y

y x x0

tan t

1 x x0

1 y y0

sec t

csc t

x y y0

cot t

• Given the specific value of any function, the related real number t or angle ␪ can be found using a special reference arc/angle, or the sin1, cos1, or tan1 features of a calculator.

EXERCISES 113 , yb is on the unit circle, find y if the point is in QIV, then use the symmetry of the circle to locate 13. Given a 7 three other points. 3 17 b is on the unit circle, find the value of all six trig functions of t without the use of a 14. Given a , 4 4 calculator. 2 . 13 16. Use a calculator to find the value of t that corresponds to the situation described: cos t 0.7641 with t in QII. 15. Without using a calculator, find two values in [0, 2␲) that make the equation true: csc t

17. A crane used for lifting heavy equipment has a winch-drum with a 1-yd radius. (a) If 59 ft of cable has been wound in while lifting some equipment to the roof-top of a building, what radian angle has the drum turned through? (b) What angle must the drum turn through to wind in 75 ft of cable?

Graphs of the Sine and Cosine Functions

SECTION 6.3

KEY CONCEPTS • Graphing sine and cosine functions using the special values from the unit circle results in a periodic, wavelike graph with domain 1q, q 2. 6 , 0.87

4 , 0.71

cos t

3 ,

6 , 0.5 0.5

2 , 0

2 , 1

ing 2

3 2

2

t

(0, 0)

0.5

0.5

1

1

Incr easi

0.5

reas

(0, 0)

3 , 0.87

1

Dec

0.5

sin t

ng

1

4 , 0.71

2

3 2

2

t

• The characteristics of each graph play a vital role in their contextual application, and these are summarized on pages 546 and 550. • The amplitude of a sine or cosine graph is the maximum displacement from the average value. For y A sin1Bt2 and y A cos1Bt2, the amplitude is A. • The period of a periodic function is the smallest interval required to complete one cycle. 2␲ For y A sin1Bt2 and y A cos1Bt2, P gives the period. B

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637

• If A 7 1, the graph is vertically stretched, if 0 6 A 6 1 the graph is vertically compressed, and if A 6 0 the graph is reflected across the x-axis. • If B 7 1, the graph is horizontally compressed (the period is smaller/shorter); if B 6 1 the graph is horizontally stretched (the period is larger/longer). 2␲ units wide, B centered on the x-axis, then use the rule of fourths to locate zeroes and max/min values. Connect these points with a smooth curve.

• To graph y A sin1Bt2 or A cos (Bt), draw a reference rectangle 2A units high and P

EXERCISES Use a reference rectangle and the rule of fourths to draw an accurate sketch of the following functions through at least one full period. Clearly state the amplitude and period as you begin. 18. y 3 sin t

19. y cos12t2

21. f 1t2 2 cos14␲t2 23. The given graph is of the form y A sin1Bt2 . Determine the equation of the graph.

22. g1t2 3 sin1398␲t2 24. Referring to the chart of colors visible in the electromagnetic spectrum (page 559), what color is

y 1 0.5

(0, 0) 0.5 1

SECTION 6.4

20. y 1.7 sin14t2

6

3

2

2 3

5 t 6

represented by the equation y sin a By y sin a

␲ tb? 270

␲ tb? 320

Graphs of the Cosecant, Secant, Tangent, and Cotangent Functions

KEY CONCEPTS 1 will be asymptotic everywhere cos t 0, increasing where cos t is decreasing, and cos t decreasing where cos t is increasing. 1 The graph of y csc t will be asymptotic everywhere sin t 0, increasing where sin t is decreasing, and sin t decreasing where sin t is increasing. y Since tan t is defined in terms of the ratio , the graph will be asymptotic everywhere x 0 x ␲ 2 (0, 1) on the unit circle, meaning all odd multiples of . (x, y) 2 x 0 Since cot t is defined in terms of the ratio , the graph will be asymptotic everywhere y 0 (1, 0) (1, 0) x y on the unit circle, meaning all integer multiples of ␲. The graph of y tan t is increasing everywhere it is defined; the graph of y cot t is 3 (0, 1) 2 decreasing everywhere it is defined. The characteristics of each graph play vital roles in their contextual application, and these are summarized on pages 563 and 567.

• The graph of y sec t

• •

• • •

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• For the more general tangent and cotangent graphs y A tan1Bt2 and y A cot1Bt2, if A 7 1, the graph is vertically stretched, if 0 6 A 6 1 the graph is vertically compressed, and if A 6 0 the graph is reflected across the x-axis. • If B 7 1, the graph is horizontally compressed (the period is smaller/shorter); if B 6 1 the graph is horizontally stretched (the period is larger/longer). ␲ • To graph y A tan 1Bt2, note A tan (Bt) is zero at t 0. Compute the period P and draw asymptotes a B P distance of on either side of the y-axis. Plot zeroes halfway between the asymptotes and use symmetry to 2 complete the graph. ␲ • To graph y A cot 1Bt2, note it is asymptotic at t 0. Compute the period P and draw asymptotes a B distance P on either side of the y-axis. Plot zeroes halfway between the asymptotes and use symmetry to complete the graph.

EXERCISES 25. Use a reference rectangle and the rule of fourths to draw an accurate sketch of Exercise 26 y 3 sec t through at least one full period. Clearly state the period as you begin. y 8 26. The given graph is of the form y A csc 1Bt2. Determine the equation of the 6 graph. 4 2 27. State the value of each expression without the aid of a calculator: (0, 0) 1 2 2 1 4 3 3 3 ␲ 7␲ 4 a. tana b b. cot a b 6 3 4 8 28. State the value of each expression without the aid of a calculator, given that t terminates in QII. 1 a. tan1 1 132 b. cot1a b 13 1 29. Graph y 6 tan a tb in the interval [2␲, 2␲]. 2 1 30. Graph y cot 12␲t2 in the interval [1, 1]. 2 31. Use the period of y cot t to name three additional solutions to cot t 0.0208, given t 1.55 is a solution. Many solutions are possible. 32. Given t 0.4444 is a solution to cot1t 2.1, use an analysis of signs and quadrants to name an additional solution in [0, 2␲). d 33. Find the approximate height of Mount Rushmore, using h and the values shown. cot u cot v

h (not to scale) v 40

u 25 144 m

5 3

t

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639

Transformations and Applications of Trigonometric Graphs

SECTION 6.5

KEY CONCEPTS • Many everyday phenomena follow a sinusoidal pattern, or a pattern that can be modeled by a sine or cosine function (e.g., daily temperatures, hours of daylight, and more). • To obtain accurate equation models of sinusoidal phenomena, vertical and horizontal shifts of a basic function are used. • The equation y A sin1Bt C2 D is called the standard form of a general sinusoid. The equation C y A sin c B at b d D is called the shifted form of a general sinusoid. B In either form, D represents the average value of the function and a vertical shift D units upward if D 7 0, • Mm Mm D, A. D units downward if D 6 0. For a maximum value M and minimum value m, 2 2 C • The shifted form y A sin c Bat b d D enables us to quickly identify the horizontal shift of the function: B C units in a direction opposite the given sign. B • To graph a shifted sinusoid, locate the primary interval by solving 0 Bt C 6 2␲, then use a reference rectangle along with the rule of fourths to sketch the graph in this interval. The graph can then be extended as needed, then shifted vertically D units. • One basic application of sinusoidal graphs involves phenomena in harmonic motion, or motion that can be modeled by functions of the form y A sin1Bt2 or y A cos1Bt2 (with no horizontal or vertical shift). • If the period P and critical points (X, M) and (x, m) of a sinusoidal function are known, a model of the form y A sin 1Bt C2 D can be obtained: B

2␲ P

A

Mm 2

D

Mm 2

C

3␲ Bx 2

EXERCISES For each equation given, (a) identify/clearly state the amplitude, period, horizontal shift, and vertical shift; then (b) graph the equation using the primary interval, a reference rectangle, and rule of fourths. ␲ 34. y 240 sin c 1t 32 d 520 6

␲ 3␲ b 6.4 35. y 3.2 cos a t 4 2

For each graph given, identify the amplitude, period, horizontal shift, and vertical shift, and give the equation of the graph. 36. 350 y 37. 210 y 300

180

250

150

200

120

150

90

100

60

50

30

0 6

12

18

24 t

0

4

2

3 4

t

38. Monthly precipitation in Cheyenne, Wyoming, can be modeled by a sine function, by using the average precipitation for June (2.26 in.) as a maximum (actually slightly higher in May), and the average precipitation for December (0.44 in.) as a minimum. Assume t 0 corresponds to March. (a) Use the information to construct a sinusoidal model, and (b) use the model to estimate the inches of precipitation Cheyenne receives in August 1t 52 and September 1t 62. Source: 2004 Statistical Abstract of the United States, Table 380.

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The Trigonometry of Right Triangles

SECTION 6.6

KEY CONCEPTS • The sides of a right triangle can be named relative to their location with respect to a given angle. B

B hypotenuse

A

hypotenuse

side opposite A

C

side adjacent

side adjacent C

side opposite

• The ratios of two sides with respect to a given angle are named as follows: sin ␣

opp hyp

cos ␣

adj hyp

tan ␣

opp adj

• The reciprocal of the ratios above play a vital role and are likewise given special names: hyp opp 1 csc ␣ sin ␣

csc ␣

hyp adj 1 sec ␣ cos ␣

sec ␣

adj opp 1 cot ␣ tan ␣ cot ␣

• Each function of ␣ is equal to the cofunction of its complement. For instance, the complement of sine is cosine and sin ␣ cos190° ␣2.

• To solve a right triangle means to apply any combination of the trig functions, along with the triangle properties, until all sides and all angles are known.

• An angle of elevation is the angle formed by a horizontal line of sight (parallel to level ground) and the true line of sight. An angle of depression is likewise formed, but with the line of sight below the line of orientation.

EXERCISES 39. Use a calculator to solve for A: a. cos 37° A b. cos A 0.4340

40. Rewrite each expression in terms of a cofunction. a. tan 57.4° b. sin119° 30¿ 15– 2

Solve each triangle. Round angles to the nearest tenth and sides to the nearest hundredth. 41. 42. B B

20 m c

A

89 in.

49 b

C

C

c

21 m

A

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43. Josephine is to weld a vertical support to a 20-m ramp so that the incline is exactly 15°. What is the height h of the support that must be used? 44. From the observation deck of a seaside building 480 m high, Armando sees two fishing boats in the distance. The angle of depression to the nearer boat is 63.5°, while for the boat farther away the angle is 45°. (a) How far out to sea is the nearer boat? (b) How far apart are the two boats?

20 m

641

h

15

45 63.5 480 m

45. A slice of bread is roughly 14 cm by 10 cm. If the slice is cut diagonally in half, what acute angles are formed?

SECTION 6.7


KEY CONCEPTS • In standard position, the terminal sides of 0°, 90°, 180°, 270°, and 360° angles coincide with one of the axes and are called quadrantal angles. • By placing a right triangle in the coordinate plane with one acute angle at the origin and one side along the x-axis, we note the trig functions can be defined in terms of a point P(x, y) on the hypotenuse. • Given P(x, y) is any point on the terminal side of an angle ␪ in standard position. Then r 2x2 y2 is the distance from the origin to this point. The six trigonometric functions of ␪ are defined as y r r x csc ␪ sec ␪ cot ␪ x x y y x0 y0 x0 y0 A reference angle ␪r is defined to be the acute angle formed by the terminal side of a given angle ␪ and the x-axis. Reference angles can be used to evaluate the trig functions of any nonquadrantal angle, since the values are fixed by the ratio of sides and the signs are dictated by the quadrant of the terminal side. If the value of a trig function and the quadrant of the terminal side are known, the related angle ␪ can be found using a reference arc/angle, or the sin1, cos1, or tan1 features of a calculator. If ␪ is a solution to sin ␪ k, then ␪ 360°k is also a solution for any integer k. sin ␪

• • • •

y r

cos ␪

x r

tan ␪

EXERCISES 46. Find two positive angles and two negative angles that are coterminal with ␪ 207°. 47. Name the reference angle for the angles given: ␪ 152°, ␪ 521°, ␪ 210° 48. Find the value of the six trigonometric functions, given P(x, y) is on the terminal side of angle ␪ in standard position. a. P112, 352 b. 112, 182 49. Find the values of x, y, and r using the information given, and state the quadrant of the terminal side of ␪. Then state the values of the six trig functions of ␪. 12 4 a. cos ␪ ; sin ␪ 6 0 b. tan ␪ ; cos ␪ 7 0 5 5 50. Find all angles satisfying the stated relationship. For standard angles, express your answer in exact form. For nonstandard angles, use a calculator and round to the nearest tenth. 13 a. tan ␪ 1 b. cos ␪ c. tan ␪ 4.0108 d. sin ␪ 0.4540 2

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SECTION 6.8

Trigonometric Equation Models

KEY CONCEPTS • If the period P and critical points (X, M) and (x, m) of a sinusoidal function are known, a model of the form y A sin1Bt C2 D can be obtained: B

2␲ P

A

Mm 2

D

Mm 2

C

3␲ Bx 2

• If an event or phenomenon is known to be sinusoidal, a graphing calculator can be used to find an equation model if at least four data points are given.

EXERCISES For the following sets of data, (a) find a sinusoidal regression equation using your calculator; (b) construct an equation manually using the period and maximum/minimum values; and (c) graph both on the same screen, then use a TABLE to find the largest difference between output values. 51. 52. Day of Year

Output

Day of Year

Output

Day of Month

Output

Day of Month

Output

1

4430

184

90

1

69

19

98

31

3480

214

320

4

78

22

92

62

3050

245

930

7

84

25

85

92

1890

275

2490

10

91

28

76

123

1070

306

4200

13

96

31

67

153

790

336

4450

16

100

53. The highest temperature on record for the even months of the year are given in the table for the city of Juneau, Alaska. (a) Use a graphing calculator to find a sinusoidal regression model. (b) Use the equation to estimate the record high temperature for the month of July. (c) Compare the actual data to the results produced by the regression model. What comments can you make about the accuracy of the model? Source: 2004 Statistical Abstract of the United States, Table 378.

Month (Jan S 1)

High Temp(ºF)

2

57

4

72

6

86

8

83

10

61

12

54

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Practice Test

643

PRACTICE TEST 1. Name the reference angle of each angle given. a. 225° b. 510° c.

7␲ 6

d.

25␲ 3

2. Find two negative angles and two positive angles that are coterminal with ␪ 30°. Many solutions are possible. 3. Convert from DMS to decimal degrees or decimal degrees to DMS as indicated. a. 100° 45¿ 18– to decimal degrees b. 48.2125° to DMS

4. Four Corners USA is the point at which Utah, Colorado, Arizona, and New Mexico meet. The southern border of Colorado, the western border of Kansas, and the point P where Colorado, Nebraska, and Kansas meet, very nearly approximates a 30-60-90 triangle. If the western border of Kansas is 215 mi long, (a) what is the distance from Four Corners USA to point P? (b) How long is Colorado’s southern border?

Colorado

Nebraska P Kansas

Utah Arizona

New Mexico

5. Complete the table from memory using exact values. If a function is undefined, so state. t

sin t

cos t

tan t

csc t

sec t

cot t

0 2␲ 3 7␲ 6 5␲ 4 5␲ 3 13␲ 6

2 and tan ␪ 6 0, find the value of the 5 other five trig functions of ␪.

6. Given cos ␪

1 212 b is a point on the unit circle, 7. Verify that a , 3 3 then find the value of all six trig functions associated with this point. 8. In order to take pictures of a dance troupe as it performs, a camera crew rides in a cart on tracks that trace a circular arc. The radius of the arc is 75 ft, and from end to end the cart sweeps out an angle of 172.5° in 20 seconds. Use this information to find (a) the length of the track in feet and inches, (b) the angular velocity of the cart, and (c) the linear velocity of the cart in both ft/sec and mph.

9. Solve the triangle shown. Answer in table form. Exercise 9 A

C 15.0 cm

57 B

10. The “plow” is a yoga position in which a person lying on their back brings their feet up, over, and behind their head and touches them to the floor. If distance from hip to shoulder (at the right angle) is 57 cm and from hip to toes is 88 cm, find the

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distance from shoulders to toes and the angle formed at the hips. Exercise 10 Hips Legs

Torso

Arms

Head

15. The lowest temperature on record for the even months of the year are given in the table for the city of Denver, Colorado. (a) Use a graphing calculator to find a sinusoidal regression model and (b) use the equation to estimate the record low temperature for the odd numbered months. Source: 2004 Statistical Abstract of the United States, Table 379.

Toes

Month (Jan S 1)

Low Temp (ºF)

2

30

4

2

6

30

11. While doing some night fishing, you round a peninsula and a tall lighthouse comes into view. Taking a sighting, you find the angle of elevation to the top of the lighthouse is 25°. If the lighthouse is known to be 27 m tall, how far from the lighthouse are you? Exercise 11

25

12. Find the value of t 僆 3 0, 2␲4 satisfying the conditions given. 1 a. sin t , t in QIII 2 2 13 , t in QIV 3

c. tan t 1, t in QII 13. In arid communities, daily water usage can often be approximated using a sinusoidal model. Suppose water consumption in the city of Caliente del Sol reaches a maximum of 525,000 gallons in the heat of the day, with a minimum usage of 157,000 gallons in the cool of the night. Assume t 0 corresponds to 6:00 A.M. (a) Use the information to construct a sinusoidal model, and (b) use the model to approximate water usage at 4:00 P.M. and 4:00 A.M. 14. State the domain, range, period, and amplitude (if it exists), then graph the function over 1 period. ␲ a. y 2 sin a tb b. y sec t 5 c. y 2 tan 13t2

41 3

12

25

16. State the amplitude, period, horizontal shift, vertical shift, and endpoints of the primary interval. Then sketch the graph using a reference rectangle and the rule of fourths: 27 m

b. sec t

8 10

y 12 sin a3t

␲ b 19. 4

17. An athlete throwing the shot-put begins his first attempt facing due east, completes three and onehalf turns and launches the shot facing due west. What angle did his body turn through? 18. State the domain, range, and period, then sketch the graph in 3 0, 2␲2. 1 a. y tan12t2 b. y cot a tb 2 19. Due to tidal motions, the depth of water in Brentwood Bay varies sinusoidally as shown in the diagram, where time is in hours and depth is in feet. Find an equation that models the depth of water at time t. Exercise 19 Depth (ft) 20 16 12 8 4

Time (hours) 0

4

8

12

16

20

24

20. Find the value of t satisfying the given conditions. a. sin t 0.7568; t in QIII b. sec t 1.5; t in QII

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CALCULATOR EXPLORATION AND DISCOVERY Variable Amplitudes and Modeling the Tides Tidal motion is often too complex to be modeled by a single sine function. In this Exploration and Discovery, we’ll look at a method that combines two sine functions to help model a tidal motion with variable amplitude. In the process, we’ll use much of what we know about the amplitude, horizontal shifts, and vertical shifts of a sine function, helping to reinforce these important concepts and broaden our understanding about how they can be applied. The graph in Figure 6.148 shows three days of tidal motion for Davis Inlet, Canada. As you can see, the amplitude of the graph varies, and there is no single sine function that can serve as a model. However, notice that the amplitude varies predictably, and that the high tides and low tides can independently be modeled by sine functions. To simplify our exploration, we will use the assumption that tides have an exact 24-hr period (close, but no), that variations between high and low tides takes place every 12 hr (again close but not exactly true), and the variation between the “low-high” (1.9 m) and the “high-high” (2.4 m) is uniform. A similar assumption is made for the low tides. The result is the graph in Figure 6.149. Figure 6.149

Figure 6.148 Height (m)

Height (m) 2.4

2.3 1.9

2

2.4

2.4

2.4 2.0

1.9

1.9

2

2.4 1.9

1.9

1

1

0.9 0 Midnight

0.8

0.7 Noon

Midnight

Noon

0.9

0.7

0.7

Midnight

0.6 Noon

t

0 Midnight

Midnight

0.9

0.7

0.9

0.7

0.7 t

Noon

Midnight

Noon

Midnight

Noon

Midnight

First consider the high tides, which vary from a maximum of 2.4 to a minimum of 1.9. Using the ideas from 2.4 ⫹ 1.9 2.4 ⫺ 1.9 ⫽ 0.25 and D ⫽ ⫽ 2.15. With a period Section 6.7 to construct an equation model gives A ⫽ 2 2 ␲ of P ⫽ 24 hr we obtain the equation Y1 ⫽ 0.25 sin a xb ⫹ 2.15. Using 0.9 and 0.7 as the maximum and minimum 12 ␲ xb ⫹ 0.8 (verify this). Graphing these two functions 12 over a 24-hr period yields the graph in Figure 6.150, where we note the high and low values are correct, but the two functions are in phase with each other. As can be determined from Figure 6.149, we want the high tide model to start at the average value and decrease, and the low tide equation model to start at high-low and decrease. Replacing x with x ⫺ 12 in Y1 and x with x ⫹ 6 in Y2 accomplishes this result (see Figure 6.151). Now comes the fun part! Since Y1 represents the low/high maximum values for high tide, and Y2 represents the low/high minimum values for low tide, low tides, similar calculations yield the equation Y2 ⫽ 0.1 sin a

Figure 6.150

Figure 6.151

3

3

0

24

0

0

24

0

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the amplitude and average value for the tidal motion at Davis Inlet are A

Y1 Y2 Y1 Y2 and D ! By entering 2 2

Y1 Y2 Y1 Y2 , the equation for the tidal motion (with its variable amplitude) will have the form and Y4 2 2 Y5 Y3 sin1Bx C2 Y4, where the value of B and C must be determined. The key here is to note there is only a ␲ 12-hr difference between the changes in amplitude, so P 12 (instead of 24) and B for this function. Also, 6 from the graph (Figure 6.149) we note the tidal motion begins at a minimum and increases, indicating a shift of 3 units to the right is required. Replacing x with x 3 gives the equation modeling these tides, and the final equation ␲ is Y5 Y3 sin c 1X 32 d Y4. Figure 6.152 gives a screen shot of Y1, Y2, and Y5 in the interval [0, 24]. The tidal 6 graph from Figure 6.149 is shown in Figure 6.153 with Y1 and Y2 superimposed on it. Y3

Figure 6.152

Figure 6.153

3

Height (m)

2.4

2.4 1.9

2

2.4 Y1

1.9

1.9

Y5 0

24

1

0.9 0 Midnight

0.7

0.9

0.7

0.9

0.7

Y2 t

Noon

Midnight

Noon

Midnight

Noon

Midnight

0

Exercise 1: The website www.tides.com/tcpred.htm offers both tide and current predictions for various locations around the world, in both numeric and graphical form. In addition, data for the “two” high tides and “two” low tides are clearly highlighted. Select a coastal area where tidal motion is similar to that of Davis Inlet, and repeat this exercise. Compare your model to the actual data given on the website. How good was the fit?

STRENGTHENING CORE SKILLS Standard Angles, Reference Angles, and the Trig Functions A review of the main ideas discussed in this chapter indicates there are four of what might be called “core skills.” These are skills that (a) play a fundamental part in the acquisition of concepts, (b) hold the overall structure together as we move from concept to concept, and (c) are ones we return to again and again throughout our study. The first of these is (1) knowing the standard angles and standard values. These values are “standard” because no estimation, interpolation, or special methods are required to name their values, and each can be expressed as a single factor. This gives them a great advantage in that further conceptual development can take place Figure 6.154 without the main points being obscured by large expressions or decimal y approximations. Knowing the values of the trig functions for each standard angle sin r 12 2 will serve you very well throughout this study. Know the chart on page 537 and the (√3, 1) 30 30 (√3, 1) r r ideas that led to it. 2 2 The standard angles/values brought us to the trigonometry of any angle, forming a strong bridge to the second core skill: (2) using reference angles to determine the 2 2 x values of the trig functions in each quadrant. For review, a 30-60-90 triangle will 2 2 always have sides that are in the proportion 1x: 13x: 2x, regardless of its size. This 30 r 30 (√3, 1) r means for any angle ␪, where ␪r 30°, sin ␪ 12 or sin ␪ 12 since the ratio is (√3, 1) fixed but the sign depends on the quadrant of ␪: sin 30° 12 [QI], sin 150° 12 2 [QII], sin 210° 12 [QIII], sin 330° 12 [QII], and so on (see Figure 6.154).

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647

Figure 6.155 In turn, the reference angles led us to a third core skill, helping us realize that if ␪ y 13 was not a quadrantal angle, (3) equations like cos1␪2 must have two solutions 2 cos r √3 2 2 in 30, 360°2. From the standard angles and standard values we learn to recognize that (√3, 1) 210 13 2 , ␪r 30°, which will occur as a reference angle in the two for cos ␪ 150 r 30 2 2 2 x quadrants where cosine is negative, QII and QIII. The solutions in 3 0, 360°2 are r 30 2 ␪ 150° and ␪ 210° (see Figure 6.155). (√3, 1) Of necessity, this brings us to the fourth core skill, (4) effective use of a calculator. The standard angles are a wonderful vehicle for introducing the basic ideas of 2 trigonometry, and actually occur quite frequently in real-world applications. But by far, most of the values we encounter will be nonstandard values where ␪r must be found using a calculator. However, once ␪r is found, the reason and reckoning inherent in these ideas can be directly applied. The Summary and Concept Review Exercises, as well as the Practice Test offer ample opportunities to refine these skills, so that they will serve you well in future chapters as we continue our attempts to explain and understand the world around us in mathematical terms.

Exercise 1: Fill in the table from memory. t

0

␲ 6

␲ 4

␲ 3

␲ 2

2␲ 3

3␲ 4

5␲ 6

␲

7␲ 6

5␲ 4

sin t y cos t x tan t

y x

Exercise 2: Solve each equation in 3 0, 2␲2 without the use of a calculator.

Exercise 3: Solve each equation in 30, 2␲2 using a calculator and rounding answers to four decimal places.

a. 2 sin t 13 0

a. 16 sin t 2 1

b. 312 cos t 4 1

b. 312 cos t 12 0 1 1 c. 3 tan t 2 4 d. 2 sec t 5

c. 13 tan t 2 1 d. 12 sec t 1 3

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CUMULATIVE REVIEW CHAPTERS 1–6 1. Solve the inequality: 2x 1 3 6 5 2. Find the domain of the function: y 2x 2x 15 80 3. Given that tan ␪ , draw a right triangle that 39 corresponds to this ratio, then use the Pythagorean theorem to find the length of the missing side. Finally, find the two acute angles. 2

4. Without a calculator, what values in 30, 2␲2 make the equation true: sin t

13 ? 2

3 17 b is a point on the unit circle 5. Given a , 4 4 corresponding to t, find all six trig functions of t. State the domain and range of each function: 6. y f 1x2

7. a. f 1x2 12x 3 2x b. g1x2 2 x 49

y 5

5

5 x

f(x) 5

8. y T1x2 x

T(x)

0

7

1

5

2

3

3

1

4

1

5

3

6

5

9. Analyze the graph of the function in Exercise 6, including: (a) local maximum and minimum values; (b) intervals where f 1x2 0 and f 1x2 6 0; (c) intervals where f is increasing or decreasing; and (d) any symmetry noted. Assume the features you are to describe have integer values. 10. The attractive force that exists between two magnets varies inversely as the square of the distance between them. If the attractive force is 1.5 newtons (N) at a distance of 10 cm, how close are the magnets when the attractive force reaches 5 N?

11. The world’s tallest indoor waterfall is in Detroit, Michigan, in the lobby of the International Center Building. Standing 66 ft from the base of the falls, the angle of elevation to the top of the waterfall is 60°. How tall is the waterfall? 12. It’s a warm, lazy Saturday and Hank is watching a county maintenance crew mow the park across the street. He notices the mower takes 29 sec to pass through 77° of rotation from one end of the park to the other. If the corner of the park is 60 ft directly across the street from his house, (a) how wide is the park? (b) How fast (in mph) does the mower travel as it cuts the grass? 13. Graph using transformations of a parent function: 1 f 1x2 2. x1 14. Graph using transformations of a parent function: g1x2 ex1 2.

15. Find f 1␪2 for all six trig functions, given the point P19, 402 is a point on the terminal side of the angle. Then find the angle ␪ in degrees, rounded to tenths. 16. Given t 5.37, (a) in what quadrant does the arc terminate? (b) What is the reference arc? (c) Find the value of sin t rounded to four decimal places. 17. A jet-stream water sprinkler shoots water a distance of 15 m and turns back-and-forth through an angle of t 1.2 rad. (a) What is the length of the arc that the sprinkler reaches? (b) What is the area in m2 of the yard that is watered? 18. Determine the equation of graph shown given it is of the form y A tan1Bt2. y 4 2 3

4

2

4

4 2

2

3 4

t

4

8 , 1

19. Determine the equation of the graph shown given it is of the form y A sin1Bt C2 D. y 2 1

0 1 2

4

2

3 4

x

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Cumulative Review Chapters 1–6

20. In London, the average temperatures on a summer day range from a high of 72°F to a low of 56°F. Use this information to write a sinusoidal equation model, assuming the low temperature occurs at 6:00 A.M. Clearly state the amplitude, average value, period, and horizontal shift. 21. The graph of a function f (x) is given. Sketch the graph of f 1 1x2. y 5

5

5 x

5

22. The volume of a spherical cap is given by ␲h2 13r h2. Solve for r in terms of V and h. V 3 h r

Exercises 26 through 30 require the use of a graphing calculator 26. Determine the minimum value of the function f 1x2 0.5x4 3x3 e. Round your answers to one decimal place. 27. Use the TABLE feature of a calculator to find the unknown coordinates of the following QII points on the unit circle. Round your answers to two decimal places when necessary.

x

y

0.2 0.4 0.6 0.8

28. In 2006, the Podcast Alley directory had cataloged about 26,000 different podcasts with over a million episodes. In 2010, there were over 77,000 different podcasts in this directory. This exponential growth can be modeled by the equation p 19.8211.312 t, where p is the number of podcasts (in thousands) registered in a given year t (t 0 corresponds to 2005). According to this model, when did the number of registered podcasts exceed (a) 50,000 and (b) 100,000? 29. Solve the equation and round your answer to two decimal places. 2x2 27x 62 11 ln x

23. Find the slope and y-intercept: 3x 4y 8. 24. Solve by factoring: 4x3 8x2 9x 18 0. 25. At what interest rate will $1000 grow to $2275 in 12 yr if compounded continuously?

30. Use the R S P␪( feature of a graphing calculator to find the angle corresponding to the point (12, 32), then evaluate tan ␪. Compare your result to y tan ␪ . x

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CONNECTIONS TO CALCULUS While right triangles have a number of meaningful applications as a problem-solving tool, they can also help to rewrite certain expressions in preparation for the tools of calculus, and introduce us to an alternative method for graphing relations and functions using polar coordinates.

Right Triangle Relationships Drawing a diagram to visualize relationships and develop information is an important element of good problem solving. This is no less true in calculus, where it is often a fundamental part of understanding the question being asked. As a precursor to applications involving trig substitutions, we’ll illustrate how right triangle diagrams are used to rewrite trigonometric functions of ␪ as algebraic functions of x. EXAMPLE 1

Solution

䊳

䊳

Using Right Triangle Diagrams to Rewrite Trig Expressions Use the equation x ⫽ 5 sin ␪ and a right triangle diagram to write 5 x cos ␪, tan ␪, sec ␪, and csc ␪ as functions of x. ␪ x Using x ⫽ 5 sin ␪, we obtain sin ␪ ⫽ . From our work in Chapter 6, adjacent side 5 opp we know the right triangle definition of sin ␪ is , and we draw a triangle with side x oriented hyp opposite an angle ␪, and label a hypotenuse of 5 (see figure). To find an expression for the adjacent side, we use the Pythagorean theorem: 1adj2 2 ⫹ x2 ⫽ 52 1adj2 2 ⫽ 25 ⫺ x2 adj ⫽ 225 ⫺ x2

Pythagorean theorem isolate term result (length must be positive); ⫺5 6 x 6 5

Using this triangle and the standard definition of the remaining trig functions, we x 5 225 ⫺ x2 5 find cos ␪ ⫽ and , tan ␪ ⫽ , sec ␪ ⫽ , . csc ␪ ⫽ x 5 225 ⫺ x2 225 ⫺ x2 Now try Exercises 1 through 4

EXAMPLE 2

䊳

Using Right Triangle Diagrams to Rewrite Trig Expressions Find expressions for tan ␪ and csc ␪, given sec ␪ ⫽

Solution

䊳

䊳

With sec ␪ ⫽

2u2 ⫹ 144 . u

冪u2 ⫹ 144

hyp , we draw a right triangle diagram as in Example 1, with a adj

opp

␪ u

opp hypotenuse of 2u2 ⫹ 144 and a side u adjacent to angle ␪. For tan ␪ ⫽ and adj hyp csc ␪ ⫽ , we use the Pythagorean theorem to find an expression for the opposite side: opp 1opp2 2 ⫹ u2 ⫽ 1 2u2 ⫹ 1442 2 1opp2 2 ⫹ u2 ⫽ u2 ⫹ 144 1opp2 2 ⫽ 144 opp ⫽ 12

Pythagorean theorem square radical subtract u2 result (length must be positive)

With an opposite side of 12 units, the figure shows tan ␪ ⫽

12 2u2 ⫹ 144 and csc ␪ ⫽ . u 12 Now try Exercises 5 and 6

䊳

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Converting from Rectangular Coordinates to Trigonometric (Polar) Form y x Using the equations cos ␪ ⫽ , sin ␪ ⫽ , and r ⫽ 2x2 ⫹ y2 from Section 6.7, we r r can derive x ⫽ r cos ␪, y ⫽ r sin ␪, and x2 ⫹ y2 ⫽ r2. With these equations, we can rewrite functions of x given in rectangular form as equations of ␪ in trigonometric (polar) form. In Chapter 10 we’ll see how this offers us certain advantages. Note that while algebraic equations are often written with y in terms of x, polar equations are written with r in terms of ␪. EXAMPLE 3

䊳

Converting from Rectangular to Trigonometric Form Rewrite the equation 2x ⫹ 3y ⫽ 6 in trigonometric form using the substitutions indicated and solving for r. Note the given equation is that of a line with x-intercept (3, 0) and y-intercept (0, 2).

Solution

䊳

Using x ⫽ r cos ␪ and y ⫽ r sin ␪ we proceed as follows: 2x ⫹ 3y ⫽ 6 21r cos ␪2 ⫹ 31r sin ␪2 ⫽ 6

given substitute r cos ␪ for x, r sin ␪ for y

r 32 cos ␪ ⫹ 3 sin ␪ 4 ⫽ 6

factor out r

6 r⫽ 2 cos ␪ ⫹ 3 sin ␪

solve for r

This is the equation of the same line, but in trigonometric form. Now try Exercises 7 through 10

䊳

While it is somewhat simplistic (there are other subtleties involved), we can verify the equation obtained in Example 3 produces the same line as 2x ⫹ 3y ⫽ 6 by evaluating the equation at ␪ ⫽ 0° to find a point on the positive x-axis, ␪ ⫽ 90° to find a point of the positive y-axis, and ␪ ⫽ 135° to find a point in QII. For ␪ ⫽ 0°:

r⫽

6 2 cos 0° ⫹ 3 sin 0°

For ␪ ⫽ 90°:

6 6 ⫽ 2 2112 ⫹ 3102 ⫽3

⫽

For ␪ ⫽ 135°:

r⫽

r⫽

6 2 cos 90° ⫹ 3 sin 90°

6 6 ⫽ 3 2102 ⫹ 3112 ⫽2 ⫽

y

6 2 cos 135° ⫹ 3 sin 135°

10 8 6

6

6 ⫽ ⫽ 12 12 12 2 a⫺ b⫹3a b 2 2 2 ⫽

12 ⫽ 6 12 ⬇ 8.5 12

Using a distance r from the origin at each ␪ given, we note that all three points are on the line 2x ⫹ 3y ⫽ 6, as shown in the figure.

4

6冪2

2

⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2

135⬚

4

6

8 10

x

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EXAMPLE 4

䊳

Converting from Polar to Rectangular Form Rewrite the equation r ⫽ 4 cos ␪ in rectangular form using the relationships x ⫽ r cos ␪, y ⫽ r sin ␪, and/or x2 ⫹ y2 ⫽ r2. Identify the resulting equation.

Solution

䊳

x From x ⫽ r cos ␪ we have cos ␪ ⫽ . Substituting into the given equation we have r r ⫽ 4 cos ␪ x r ⫽ 4a b r 2 r ⫽ 4x 2 x ⫹ y2 ⫽ 4x x2 ⫺ 4x ⫹ y2 ⫽ 0 2 1x ⫺ 4x ⫹ 42 ⫹ y2 ⫽ 4 1x ⫺ 22 2 ⫹ y2 ⫽ 4

given substitute

x for cos ␪ r

multiply by r substitute x 2 ⫹ y 2 for r 2 set equal to 0 complete the square in x standard form

The result shows r ⫽ 4 cos ␪ is a trigonometric form for the equation of a circle with center at (2, 0) and radius r ⫽ 2. Now try Exercises 11 through 14

䊳

Connections to Calculus Exercises Use the diagram given to find the remaining side of the triangle. Then write the six trig functions as functions of x. 1.

2.

hyp 4

x⫺2

冪2x2 ⫹ 8

␪

␪ 冪x2 ⫹ 8x

adj

Use the equation given and a sketch of the corresponding right triangle diagram to write the remaining five trig functions as functions of x. 3. x ⫽ 4 tan ␪

4. x ⫽ 5 sec ␪

5. Find expressions for cot ␪ and sec ␪, given 2u2 ⫹ 169 . csc ␪ ⫽ u

6. Find expressions for sin ␪ and cos ␪, given 2 15 . cot ␪ ⫽ x

Use x ⴝ r cos ␪, y ⴝ r sin ␪, and x2 ⴙ y2 ⴝ r2 to rewrite the expressions in trigonometric form. 7. y ⫽ 2 9. y ⫽ ⫺2x ⫹ 3

1 8. y ⫽ x2 4

10. 1x ⫺ 22 2 ⫹ y2 ⫽ 4

Use x ⴝ r cos ␪, y ⴝ r sin ␪ and x2 ⴙ y2 ⴝ r2 to rewrite the expressions in rectangular form, then identify the equation as that of a line, circle, vertical parabola, or horizontal parabola. 11. r ⫽ 5 sin ␪ 13. r 13 cos ␪ ⫺ 2 sin ␪2 ⫽ 6

12. r ⫽

4 1 ⫹ sin ␪

14. r ⫽

4 2 ⫺ 2 cos ␪

y aHint: sin ␪ ⫽ .b r x aHint: cos ␪ ⫽ .b r

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CHAPTER CONNECTIONS

Trigonometric Identities, Inverses, and Equations CHAPTER OUTLINE 7.1 Fundamental Identities and Families of Identities 654 7.2 More on Verifying Identities 661 7.3 The Sum and Difference Identities 669 7.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities 680

7.5 The Inverse Trig Functions and Their Applications 695 7.6 Solving Basic Trig Equations 711

This chapter will unify much of what we’ve learned so far, and lead us to some intriguing, sophisticated, and surprising applications of trigonometry. Defining the trig functions helped us study a number of new relationships not possible using algebra alone. Their graphs gave us insights into how the functions were related to each other, and enabled a study of periodic phenomena. We will now use identities to simplify complex expressions and show how trig functions often work together to model natural events. One such “event” is a river’s seasonal discharge rate, which tends to be greater during the annual snow melt. In this chapter, we’ll learn how to predict the discharge rate during specific months of the year, information of great value to fisheries, oceanographers, and other scientists. 䊳

This application appears as Exercises 53 and 54 in Section 7.7

7.7 General Trig Equations and Applications 721

Trigonometric equations, identities, and substitutions also play a vital role in a study of calculus, helping to simplify complex expressions, or rewrite an expression in a form more suitable for the tools of calculus. These connections are explored in the Connections to Calculus feature following Chapter 7. Connections to Calculus 653

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CHAPTER 7 Trigonometric Identities, Inverses, and Equations

7.1

Fundamental Identities and Families of Identities

LEARNING OBJECTIVES

In this section, we begin laying the foundation necessary to work with identities successfully. The cornerstone of this effort is a healthy respect for the fundamental identities and vital role they play. Students are strongly encouraged to do more than memorize them—they should be internalized, meaning they must become a natural and instinctive part of your core mathematical knowledge.


A. Use fundamental identities to help understand and recognize identity “families” B. Verify other identities using the fundamental identities and basic algebra skills C. Use fundamental identities to express a given trig function in terms of the other five

A. Fundamental Identities and Identity Families An identity is an equation that is true for all elements in the domain. In trigonometry, some identities result directly from the way the functions are defined. For instance, the reciprocal relationships we first saw in Section 6.2 followed directly from their definitions. We call identities of this type fundamental identities. Successfully working with other identities will depend a great deal on your mastery of these fundamental types. For convenience, the definitions of the trig functions are reviewed here, followed by the fundamental identities that result. Given point P(x, y) is on the terminal side of angle ␪ in standard position, with r ⫽ 2x2 ⫹ y2 the distance from the origin to (x, y), we have y r r csc ␪ ⫽ ; y ⫽ 0 y

x r r sec ␪ ⫽ ; x ⫽ 0 x

cos ␪ ⫽

sin ␪ ⫽

y tan ␪ ⫽ ; x ⫽ 0 x x cot ␪ ⫽ ; y ⫽ 0 y

Fundamental Trigonometric Identities Reciprocal identities

Ratio identities

sin ␪ cos ␪ sec ␪ tan ␪ ⫽ csc ␪ cos ␪ cot ␪ ⫽ sin ␪

1 csc ␪ 1 cos ␪ ⫽ sec ␪ 1 tan ␪ ⫽ cot ␪ sin ␪ ⫽

EXAMPLE 1

䊳

Pythagorean identities

tan ␪ ⫽

cos2␪ ⫹ sin2␪ ⫽ 1 1 ⫹ tan2␪ ⫽ sec2␪ cot2␪ ⫹ 1 ⫽ csc2␪

Proving a Fundamental Identity Use the coordinate definitions of the trigonometric functions to prove the identity cos2␪ ⫹ sin2␪ ⫽ 1.

Solution

WORTHY OF NOTE The Pythagorean identities are used extensively in future courses. See Example 1 of the Connections to Calculus feature at the end of this chapter.

䊳

We begin with the left-hand side. y 2 x 2 cos2␪ ⫹ sin2␪ ⫽ a b ⫹ a b r r 2 2 y x ⫽ 2⫹ 2 r r 2 x ⫹ y2 ⫽ r2

substitute

y x for cos ␪, for sin ␪ r r

square terms

add terms

Noting that r ⫽ 2x2 ⫹ y2 implies r2 ⫽ x2 ⫹ y2, we have ⫽

r2 ⫽1 r2

substitute x 2 ⫹ y 2 for r 2, simplfy

Now try Exercises 7 through 10 654

䊳

7–2

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655

The fundamental identities seem to naturally separate themselves into the three groups or families listed, with each group having additional relationships that can be inferred from the definitions. For instance, since sin ␪ is the reciprocal of csc ␪, csc ␪ must be the reciprocal of sin ␪. Similar statements can be made regarding cos ␪ and sec ␪ as well as tan ␪ and cot ␪. Recognizing these additional “family members” enlarges the number of identities you can work with, and will help you use them more effectively. In particular, since they are reciprocals: sin ␪ csc ␪ ⫽ 1, cos ␪ sec ␪ ⫽ 1, and tan ␪ cot ␪ ⫽ 1. See Exercises 11 and 12. EXAMPLE 2

䊳

Identifying Families of Identities Starting with cos2␪ ⫹ sin2␪ ⫽ 1, use algebra to write four additional identities that belong to the Pythagorean family.

Solution

䊳

cos2 ␪ ⫹ sin2 ␪ ⫽ 1 1) sin2␪ ⫽ 1 ⫺ cos2␪ 2) sin ␪ ⫽ ⫾21 ⫺ cos2␪ cos2␪ ⫹ sin2␪ ⫽ 1 3) cos2␪ ⫽ 1 ⫺ sin2␪ 4) cos ␪ ⫽ ⫾21 ⫺ sin2␪

original identity subtract cos2␪ take square root original identity subtract sin2␪ take square root

For the identities involving a radical, the choice of sign will depend on the quadrant of the terminal side. Now try Exercises 13 and 14

A. You’ve just seen how we can use fundamental identities to help understand and recognize identity “families”

䊳

The fact that each new equation in Example 2 represents an identity gives us more options when attempting to verify or prove more complex identities. For instance, since cos2␪ ⫽ 1 ⫺ sin2␪, we can replace cos2␪ with 1 ⫺ sin2␪, or replace 1 ⫺ sin2␪ with cos2␪, any time they occur in an expression. Note there are many other members of this family, since similar steps can be performed on the other Pythagorean identities. In fact, each of the fundamental identities can be similarly rewritten and there are a variety of exercises at the end of this section for practice.

B. Verifying an Identity Using Algebra Note that we cannot prove an equation is an identity by repeatedly substituting input values and obtaining a true equation. This would be an infinite exercise and we might easily miss a value or even a range of values for which the equation is false. Instead we attempt to rewrite one side of the equation until we obtain a match with the other side, so there can be no doubt. As hinted at earlier, this is done using basic algebra skills combined with the fundamental identities and the substitution principle. For now we’ll focus on verifying identities by using algebra. In Section 7.2 we’ll introduce some guidelines and ideas that will help you verify a wider range of identities.

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EXAMPLE 3

䊳

Using Algebra to Help Verify an Identity Use the distributive property to verify that sin ␪1csc ␪ sin ␪2 cos2␪ is an identity.

Solution

䊳

Use the distributive property to simplify the left-hand side. sin ␪ 1csc ␪ sin ␪2 sin ␪ csc ␪ sin2␪ 1 sin2␪ cos2␪

distribute substitute 1 for sin ␪ csc ␪ 1 sin2␪ cos2␪

Since we were able to transform the left-hand side into a duplicate of the right, there can be no doubt the original equation is an identity. Now try Exercises 15 through 24

䊳

Often we must factor an expression, rather than multiply, to begin the verification process. EXAMPLE 4

䊳

Using Algebra to Help Verify an Identity Verify that 1 cot2␣ sec2␣ cot2␣ is an identity.

Solution

䊳

The left side is as simple as it gets. The terms on the right side have a common factor and we begin there. cot2␣ sec2␣ cot2␣ cot2␣ 1sec2␣ 12 cot2␣ tan2␣ 1cot ␣ tan ␣2 2 12 1

factor out cot2␣ substitute tan2␣ for sec2␣ 1 power property of exponents cot ␣ tan ␣ 1


䊳

Examples 3 and 4 show you can begin the verification process on either the left or right side of the equation, whichever seems more convenient. Example 5 shows how the special products 1A B21A B2 A2 B2 and/or 1A B2 2 A2 2AB B2 can be used in the verification process. EXAMPLE 5

䊳

Using a Special Product to Help Verify an Identity Use a special product and fundamental identities to verify that 1sin ␤ cos ␤2 2 1 ⴚ 2 sin ␤ cos ␤ is an identity.

Solution

䊳

Begin by squaring the left-hand side, in hopes of using a Pythagorean identity. 1sin ␤ cos ␤2 2 sin2␤ 2 sin ␤ cos ␤ cos2␤ cos2␤ sin2␤ 2 sin ␤ cos ␤ 1 ⴚ 2 sin ␤ cos ␤

binomial square rewrite terms substitute 1 for cos2␤ sin2␤


B. You’ve just seen how we can verify other identities using the fundamental identities and basic algebra skills

䊳

Another common method used to verify identities is simplification by combining sin2␪ A C AD BC . For sec ␪ cos ␪, the right-hand terms, using the model B D BD cos ␪ sin2␪ cos2␪ 1 sec ␪. See Exercises 39 side immediately becomes , which gives cos ␪ cos ␪ through 44.

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C. Writing One Function in Terms of Another Any one of the six trigonometric functions can be written in terms of any of the other functions using fundamental identities. The process involved offers practice in working with identities, highlights how each function is related to the other, and has practical applications in verifying more complex identities. EXAMPLE 6

䊳

Writing One Trig Function in Terms of Another Write the function cos ␪ in terms of the tangent function.

Solution

䊳

Begin by noting these functions share “common ground” via sec ␪, since 1 . Starting with sec2␪, sec2␪ ⫽ 1 ⫹ tan2␪ and cos ␪ ⫽ sec ␪ sec2␪ ⫽ 1 ⫹ tan2␪ sec ␪ ⫽ ⫾21 ⫹ tan2␪

Pythagorean identity square roots

We can now substitute ⫾21 ⫹ tan2␪ for sec ␪ in cos ␪ ⫽ cos ␪ ⫽

1 ⫾21 ⫹ tan ␪ 2

1 . sec ␪

substitute ⫾21 ⫹ tan2␪ for sec ␪

Note we have written cos ␪ in terms of the tangent function. Now try Exercises 45 through 50

WORTHY OF NOTE Although identities are valid where both expressions are defined, this does not preclude a difference in the domains of each function. For example, the result of Example 6 is indeed an identity, even though the ␲ left side is defined at while the 2 right side is not.

EXAMPLE 7

䊳

Example 6 also reminds us of a very important point—the sign we choose for the final answer is dependent on the terminal side of the angle. If the terminal side is in QI or QIV we chose the positive sign since cos ␪ 7 0 in those quadrants. If the angle terminates in QII or QIII, the final answer is negative since cos ␪ 6 0 in those quadrants. Similar to our work in Section 6.7, given the value of cot ␪ and the quadrant of ␪, the fundamental identities enable us to find the value of the other five functions at ␪. In fact, this is generally true for any given trig function and angle ␪. 䊳

Using a Known Value and Quadrant Analysis to Find Other Function Values ⫺9 with the terminal side of ␪ in QIV, find the value of the other 40 five functions of ␪. Use a calculator to check your answer. Given cot ␪ ⫽

Solution

䊳

40 follows immediately, since cotangent and tangent 9 are reciprocals. The value of sec ␪ can be found using sec2␪ ⫽ 1 ⫹ tan2␪. The function value tan ␪ ⫽ ⫺

sec2␪ ⫽ 1 ⫹ tan2 ␪ 40 2 ⫽ 1 ⫹ a⫺ b 9 81 1600 ⫽ ⫹ 81 81 1681 sec2␪ ⫽ 81 41 sec ␪ ⫽ ⫾ 9

Pythagorean identity substitute ⫺ square ⫺

40 for tan ␪ 9

40 81 , substitute for 1 9 81

combine terms

take square roots

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Since sec ␪ is positive for a terminal side in QIV, we have sec ␪

41 . 9

9 (reciprocal identities), and we find 41 sin ␪ 40 sin ␪ using sin2 ␪ 1 cos2␪ or the ratio identity tan ␪ (verify). 41 cos ␪ 41 This result and another reciprocal identity gives us our final value, csc ␪ . 40 This automatically gives cos ␪

Check

䊳

As in Example 9 of Section 6.7, we find ␪r using 2nd TAN (TANⴚ1) 40 ⴜ 9 ) , which shows ␪r ⬇ 1.3495 (Figure 7.1). Since the terminal side of ␪ is in QIV, one possible value for ␪ is 2␲ ␪r. Note in Figure 7.1, the 2nd (–) (ANS) feature was used to compute ␪, which we then stored as X. In Figure 7.2, we verify that 40 9 40 tan ␪ , cos ␪ , and sin ␪ . 9 41 41 ENTER

Figure 7.1

C. You’ve just seen how we can use fundamental identities to express a given trig function in terms of the other five

Figure 7.2


䊳

7.1 EXERCISES 䊳



1. Three fundamental ratio identities are ? ? ? tan ␪ , tan ␪ . , and cot ␪ cos ␪ csc ␪ sin ␪

4. An is an equation that is true for all elements in the . To show an equation is an identity, we employ basic algebra skills combined with the identities and the substitution principle.

2. The three fundamental reciprocal identities are sin ␪ 1/ , cos ␪ 1/ , and tan ␪ 1/ . From these, we can infer three additional reciprocal relationships: csc ␪ 1/ sec ␪ 1/ , and cot ␪ 1/ .

5. Use the pattern

3. Starting with the Pythagorean identity cos2 ␪ sin2 ␪ 1, the identity 1 tan2␪ sec2 ␪ can be derived by dividing both sides by . Alternatively, dividing both sides of this equation by sin2 ␪, we obtain the identity .

,

A C AD BC to add the B D BD following terms, and comment on this process versus “finding a common denominator.” sin ␪ cos ␪ sin ␪ sec ␪

6. Name at least four algebraic skills that are used with the fundamental identities in order to rewrite a trigonometric expression. Use algebra to quickly rewrite 1sin ␪ cos ␪2 2.

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659


Use the definitions of the trigonometric functions to prove the following fundamental identities.

7. 1 tan2␪ sec2␪

8. cot2␪ 1 csc2␪ cos ␪ 10. cot ␪ sin ␪

sin ␪ 9. tan ␪ cos ␪

Starting with the ratio identity given, use substitution and fundamental identities to write four new identities belonging to the ratio family. Answers may vary.

11. tan ␪

sin ␪ cos ␪

12. cot ␪

cos ␪ sin ␪

Starting with the Pythagorean identity given, use algebra to write four additional identities belonging to the Pythagorean family. Answers may vary.

13. 1 tan2␪ sec2␪

14. cot2␪ 1 csc2␪

Verify the equation is an identity using multiplication and fundamental identities.

15. sin ␪ cot ␪ cos ␪

16. cos ␪ tan ␪ sin ␪

17. sec ␪ cot ␪ csc ␪

18. csc2␪ tan2␪ sec2␪

2

2

2

32.

cos ␪ cot ␪ cos ␪ sin ␪ cot ␪ cot2␪

Verify the equation is an identity using special products and fundamental identities.

33.

1sin ␪ cos ␪2 2

34.

11 tan ␪2 2

cos ␪

sec ␪

sec ␪ 2 sin ␪

sec ␪ 2 sin ␪

35. 11 sin ␪211 sin ␪2 cos2␪

36. 1sec ␪ 12 1sec ␪ 12 tan2␪ 37.

1csc ␪ cot ␪21csc ␪ cot ␪2

cot ␪

38.

1sec ␪ tan ␪21sec ␪ tan ␪2

sin ␪

tan ␪

csc ␪

Verify the equation is an identity using fundamental A C AD ⴞ BC identities and ⴞ ⴝ to combine terms. B D BD

19. cos ␪ 1sec ␪ cos ␪2 sin ␪

39.

20. tan ␪ 1cot ␪ tan ␪2 sec2␪

sin ␪ cos2␪ csc ␪ sin ␪ 1

21. sin ␪ 1csc ␪ sin ␪2 cos2␪

40.

22. cot ␪ 1tan ␪ cot ␪2 csc2␪

tan2␣ sec ␣ cos ␣ sec ␣ 1

23. tan ␪ 1csc ␪ cot ␪2 sec ␪ 1

41.

sin ␪ sin ␪ 1 tan ␪ csc ␪ cos ␪ cot ␪

24. cot ␪ 1sec ␪ tan ␪2 csc ␪ 1

42.

cos ␪ cos ␪ 1 cot ␪ sec ␪ sin ␪ tan ␪

43.

csc ␪ sec ␪ tan ␪ sin ␪ sec ␪

44.

csc ␪ sec ␪ cot ␪ cos ␪ csc ␪

2

Verify the equation is an identity using factoring and fundamental identities.

25. tan2␪ csc2␪ tan2␪ 1 26. sin2␪ cot2␪ sin2␪ 1 27.

sin ␪ cos ␪ sin ␪ tan ␪ cos ␪ cos2␪

28.

sin ␪ cos ␪ cos ␪ cot ␪ sin ␪ sin2␪

1 sin ␪ 29. sec ␪ cos ␪ cos ␪ sin ␪

Write the given function entirely in terms of the second function indicated.

45. tan ␪ in terms of sin ␪ 46. tan ␪ in terms of sec ␪ 47. sec ␪ in terms of cot ␪

1 cos ␪ csc ␪ sin ␪ cos ␪ sin ␪

48. sec ␪ in terms of sin ␪

30. 31.

sin ␪ tan ␪ sin ␪ cos ␪ tan ␪ tan2␪

50. cot ␪ in terms of csc ␪

49. cot ␪ in terms of sin ␪

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For the given trig function f (␪) and the quadrant in which ␪ terminates, state the value of the other five trig functions. Use a calculator to verify your answers.

51. cos ␪

20 with ␪ in QII 29

56. csc ␪

7 with ␪ in QII x 7 with ␪ in QIII 13

52. sin ␪

12 with ␪ in QII 37

57. sin ␪

53. tan ␪

15 with ␪ in QIII 8

58. cos ␪

54. sec ␪

5 with ␪ in QIV 3

9 59. sec ␪ with ␪ in QII 7

55. cot ␪

x with ␪ in QI 5

60. cot ␪

䊳

11 with ␪ in QIV 2


61. The versine function: V ⴝ 2 sin2␪ For centuries, the haversine formula has been used in navigation to calculate the nautical distance between any two points on the surface of the Earth. One part of the formula requires the calculation of V, where ␪ is half the difference of latitudes between the two points. Use a fundamental identity to express V in terms of cosine.

䊳

23 with ␪ in QIV 25

nx2 cos1 180ⴗ n 2 b 180ⴗ 4 sin1 n 2 The area of a regular polygon is given by the formula shown, where n represents the number of sides and x is the length of each side. a. Rewrite the formula in terms of a single trig function. b. Verify the formula for a square with sides of 8 m given the point (2, 2) is on the terminal side of a 45° angle in standard position.

62. Area of a regular polygon: A ⴝ a

APPLICATIONS

Writing a given expression in an alternative form is a skill used at all levels of mathematics. In addition to standard factoring skills, it is often helpful to decompose a power into smaller powers (as in writing A3 as A # A2).

63. Show that cos3␪ can be written as cos ␪11 sin2␪2 . 64. Show that tan3␪ can be written as tan ␪1sec2␪ 12 . 65. Show that tan ␪ tan3␪ can be written as tan ␪1sec2␪2 . 66. Show that cot3␪ can be written as cot ␪1csc2␪ 12 . 67. Show tan2␪ sec ␪ 4 tan2␪ can be factored into 1sec ␪ 42 1sec ␪ 121sec ␪ 12 .

68. Show 2 sin2␪ cos ␪ 13 sin2 ␪ can be factored into 11 cos ␪2 11 cos ␪2 12 cos ␪ 132 .

69. Show cos2␪ sin ␪ cos2␪ can be factored into 111 sin ␪211 sin ␪2 2. 70. Show 2 cot2␪ csc ␪ 2 12 cot2␪ can be factored into 21csc ␪ 1221csc ␪ 12 1csc ␪ 12 . 71. Angle of intersection: At their point of intersection, the angle ␪ between any two nonparallel lines satisfies the relationship 1m2 m1 2cos ␪ sin ␪ m1m2sin ␪, where m1 and m2 represent the slopes of the two lines. Rewrite the equation in terms of a single trig function. 72. Angle of intersection: Use the result of Exercise 71 to find the tangent of the angle between the lines 2 7 Y1 x 3 and Y2 x 1. 5 3 73. Angle of intersection: Use the result of Exercise 71 to find the tangent of the angle between the lines Y1 3x 1 and Y2 2x 7.

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74. The word tangent literally means “to touch,” which in mathematics we take to mean touches in only and exactly one point. In the figure, the circle has a radius of 1 and the vertical line is “tangent” to the circle at the x-axis. The figure can be used to verify the Pythagorean identity for sine and cosine, as well as the ratio identity for tangent. Discuss/Explain how.

Exercise 74 y sin ␪

3

tan ␪

1

75. Simplify 2 sin ␪ 13 sin ␪ 2 sin ␪ 13 sin ␪ using factoring and fundamental identities. 4

䊳

661

Section 7.2 More on Verifying Identities

2

␪ cos ␪

x


76. (5.5) Solve for x: 2351

78. (4.2) Use the rational zeroes theorem and other “tools” to find all zeroes of the function f 1x2 2x4 9x3 4x2 36x 16.

2500 1 e1.015x

77. (6.6) Standing 265 ft from the base of the Strastosphere Tower in Las Vegas, Nevada, the angle of elevation to the top of the tower is about 77°. Approximate the height of the tower to the nearest foot.

7.2

79. (6.3) Use a reference rectangle and the rule of fourths to sketch the graph of y 2 sin12t2 for t in [0, 2␲).

More on Verifying Identities


A. Identify and use identities

In Section 7.1, our primary goal was to illustrate how basic algebra skills can be used to rewrite trigonometric expressions and verify simple identities. In this section, we’ll sharpen and refine these skills so they can be applied more generally, as we develop the ability to verify a much wider range of identities.

due to symmetry

B. Verify general identities C. Use counterexamples and contradictions to show an equation is not an identity

A. Identities Due to Symmetry The symmetry of the unit circle and the wrapping function presented in Chapter 6’s Reinforcing Basic Skills feature, lead us directly to the final group of fundamental identities. Given t 7 0, consider the points on the unit circle associated with t and t, as shown in Figure 7.3. From our definitions of the trig functions, sin t y and sin1t2 y, and we recognize sine is an odd function: sin1t2 sin t. The remaining identities due to symmetry can similarly be developed, and are shown here with the complete family of fundamental identities.

Figure 7.3 1 (x, y) t

y x y

t (x, y)

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WORTHY OF NOTE

Fundamental Trigonometric Identities

The identities due to symmetry are sometimes referred to as the even/odd properties. These properties can help express the cofunction identities in shifted form. For example, ␲ sin t cos a tb 2 cos c at cos at

␲ bd 2

␲ b. 2

EXAMPLE 1

䊳

Reciprocal identities

1 csc t 1 cos t sec t 1 tan t cot t sin t

Ratio identities

Pythagorean identities

sin t cos t sec t tan t csc t cos t cot t sin t

cos2t sin2t 1

sin1t2 sin t

1 tan2t sec2t

cos1t2 cos t

cot2t 1 csc2t

tan1t2 tan t

tan t

Identities due to symmetry

Using Symmetry to Verify an Identity Verify the identity: (1 tan t)2 sec2t ⴙ 2 tan(ⴚt)

Solution

䊳

Begin by squaring the left-hand side, in hopes of using a Pythagorean identity. 11 tan t2 2 1 2 tan t tan2t 1 tan2t 2 tan t sec2t 2 tan t

binomial square rewrite terms substitute sec2t for 1 tan2t

At this point, we appear to be off by two signs, but quickly recall that tangent is an odd function and tan t tan(t). By writing sec2t 2 tan t as sec2t 2(tan t), we can complete the verification. A. You’ve just seen how we can identify and use identities due to symmetry

sec2t 21tan t2 sec2t ⴙ 2 tan1ⴚt 2

rewrite expression to obtain tan t

substitute tan 1t 2 for tan t


䊳

B. Verifying Identities We’re now ready to put these ideas and the ideas from Section 7.1 to work for us. When verifying identities we attempt to mold, change, or rewrite one side of the equality until we obtain a match with the other side. What follows is a collection of the ideas and methods we’ve observed so far, which we’ll call the Guidelines for Verifying Identities. But remember, there really is no right place to start. Think things over for a moment, then attempt a substitution, simplification, or operation and see where it leads. If you hit a dead end, that’s okay! Just back up and try something else. Guidelines for Verifying Identities WORTHY OF NOTE When verifying identities, it is actually permissible to work on each side of the equality independently, in the effort to create a “match.” But properties of equality can never be used, since we cannot assume an equality exists.

1. As a general rule, work on only one side of the identity. • We cannot assume the equation is true, so properties of equality cannot be applied. • We verify the identity by changing the form of one side until we get a match with the other. 2. Work with the more complex side, as it is easier to reduce/simplify than to “build.” 3. If an expression contains more than one term, it is often helpful to combine A C AD BC terms using . B D BD 4. Converting all functions to sines and cosines can be helpful. 5. Apply other algebra skills as appropriate: distribute, factor, multiply by a conjugate, and so on. 6. Know the fundamental identities inside out, upside down, and backward — they are the key!

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663

Note how these ideas are employed in Examples 2 through 5, particularly the frequent use of fundamental identities. EXAMPLE 2

䊳

Verifying an Identity Verify the identity: sin2␪ tan2␪ tan2␪ sin2␪.

Solution

䊳

As a general rule, the side with the greater number of terms or the side with rational terms is considered “more complex,” so we begin with the right-hand side. tan2␪ sin2␪

sin2␪ sin2␪ 2 cos ␪ sin2␪ # 1 sin2␪ 1 cos2␪ sin2␪ sec2␪ sin2␪ sin2␪ 1sec2␪ 12 sin2␪ tan2␪

substitute

sin2␪ cos2␪

for tan2␪

decompose rational term substitute sec2␪ for factor out sin2␪

1 cos2␪

substitute tan2␪ for sec2 ␪ 1


䊳

In the first step of Example 2, we converted all functions to sines and cosines. Due to their use in the ratio identities, this often leads to compound fractions that we’ll need to simplify to complete a proof. EXAMPLE 3

䊳

Verifying an Identity by Simplifying Compound Fractions Verify the identity

Solution

䊳

1 cot x cos x sec x csc x

Beginning with the left-hand side, we’ll use the reciprocal and ratio identities to express all functions in terms of sines and cosines. cos x cos x substitute for cot x, 1 sin x 1 cot x sin x 1 1 for sec x, for csc x sec x csc x 1 1 cos x sin x cos x sin x cos x a1 b1cos x sin x2 multiply numerator and sin x denominator by cos x sin x (the LCD for all fractions) 1 1 a b1cos x sin x2 cos x sin x

1121cos x sin x2 a a

cos x b1cos x sin x2 sin x

1 1 b1 cos x sin x2 a b1cos x sin x2 cos x sin x cos x sin x cos 2x sin x cos x cos x 1sin x cos x2 1sin x cos x2 cos x

distribute

simplify

factor out cos x in numerator simplify


䊳

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Examples 2 and 3 involved factoring out a common expression. Just as often, we’ll need to multiply numerators and denominators by a common expression, as in Example 4. EXAMPLE 4

䊳

Verifying an Identity by Multiplying Conjugates Verify the identity:

Solution

䊳

1 ⴚ cos t cos t . 1 sec t tan2t

Both sides of the identity have a single term and one is really no more complex than the other. As a matter of choice we begin with the left side. Noting the denominator on the left has the term sec t, with a corresponding term of tan2t on the right, we reason that multiplication by a conjugate might be productive. cos t 1 sec t cos t a ba b 1 sec t 1 sec t 1 sec t cos t 1 1 sec2t cos t 1 tan2t 1 ⴚ cos t tan2t

multiply numerator and denominator by the conjugate distribute: cos t sec t 1, 1A B 21A B 2 A 2 B 2 substitute tan2t for 1 sec2t 11 tan2t sec2t 1 1 sec2t tan2t 2 multiply above and below by 1


䊳

Example 4 highlights the need to be very familiar with families of identities. To replace 1 sec2t, we had to use tan2t, not simply tan2t, since the related Pythagorean identity is 1 tan2t sec2t. As noted in the Guidelines, combining rational terms is often helpful. At this point, C AD BC A students are encouraged to work with the pattern as a means of B D BD combing rational terms quickly and efficiently. EXAMPLE 5

䊳

Verifying an Identity by Combining Terms Verify the identity:

Solution

䊳

sec x sin x tan2x ⴙ cos2x . sec x sin x tan x

We begin with the left-hand side. sec x sin x sec2x sin2x sec x sin x sin x sec x 11 tan2x2 11 cos2x2 sin x 1 a ba b cos x 1 tan2x ⴙ cos2x tan x

combine terms:

A C AD BC B D BD

substitute 1 tan2x for sec2x, 1 cos2x for sin2x,

1 for sec x cos x

simplify numerator, substitute tan x for

sin x cos x


B. You’ve just seen how we can verify general identities

䊳

Identities come in an infinite variety and it would be impossible to illustrate all variations. The general ideas and skills presented should prepare you to verify any of those given in the Exercise Set, as well as those you encounter in your future studies. See Exercises 35 through 58.

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665

C. Showing an Equation Is Not an Identity To show an equation is not an identity, we need only find a single value for which the functions involved are defined but the equation is false. This can often be done by trial and error, or even by inspection. To illustrate the process, we’ll use two common misconceptions that arise in working with identities. EXAMPLE 6

䊳

Showing an Equation Is Not an Identity Show the equations given are not identities. a. sin12x2 ⫽ 2 sin x b. cos1␣ ⫹ ␤2 ⫽ cos ␣ ⫹ cos ␤

Solution

䊳

a. The assumption here seems to be that we can factor out the coefficient from the argument. By inspection we note the amplitude of sin(2x) is A ⫽ 1, while the amplitude of 2 sin x is A ⫽ 2. This means they cannot possibly be equal for all values of x, although they are equal for integer multiples of ␲. For instance, substituting ␲ for x shows the left- and right-hand sides of the equation are equal (see Figure 7.4). However, Figure 7.5 shows the two sides of the equation ␲ are not equal when x ⫽ . This equation is not an identity. 6 Figure 7.4

Figure 7.5

b. The assumption here is that we can distribute function values. This is similar to saying 1x ⫹ 4 ⫽ 1x ⫹ 2, a statement obviously false for all values except x ⫽ 0. Here we’ll substitute convenient values to prove the equation false, 3␲ ␲ namely, ␣ ⫽ and ␤ ⫽ . 4 4 cos a

3␲ 3␲ ␲ ␲ ⫹ b ⫽ cos a b ⫹ cos a b 4 4 4 4 12 12 ⫹ cos ␲ ⫽ ⫺ 2 2 ⫺1 ⫽ 0

substitute

3␲ ␲ for ␣ and for ␤ 4 4

simplify result is false


䊳

Many times, a graphical test can be used to help determine if an equation is an identity. While not fool-proof, seeing if the graphs appear identical can either suggest the identity is true, or definitely show it is not. When testing identities, it helps to the left-hand side of the equation as Y1 on the Y= screen, and Figure 7.6 the right-hand side as Y2. We can then test whether 4 an identity relationship might exist by graphing both relations, and noting whether two graphs or a single graph appears on the GRAPH screen. Consider the equation 1 ⫺ cos2x ⫽ sin2x. 2␲ ⫺2␲ Y1 ⫽ 1 ⫺ cos1X2 2 After entering and Y2 ⫽ sin1X2 2, we obtain the graph shown in Figure 7.6 using the calculator’s ZOOM 7:ZTrig feature. ⫺4 ENTER

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Since only one graph appears in this window, it seems that an identity relationship likely exists between Y1 and Y2. This can be further supported (but still not proven) by entering Y3 Y2 1 to vertically translate the graph of Y2 up 1 unit (be sure to deactivate Y2) and observing the GRAPH (see Figure 7.7). Of course the calculator’s TABLE feature could also be used on Y1 and Y2. To graphically investigate the equation sin12x2 2 sin x, GRAPH Y1 sin12X2 and Y2 2 sin1X2 and note the existence of two distinct graphs (Figure 7.8). The resulting graphs confirm our solution to Example 6(a): sin12x2 2 sin x —even though sin12x2 2 sin x for integer multiples of ␲ (the points of intersection). See Exercises 65 through 68. Figure 7.7

Figure 7.8

4

4

2␲

C. You’ve just seen how we can use counterexamples and contradictions to show an equation is not an identity

2␲

2␲

2␲

4

4

7.2 EXERCISES 䊳



䊳

1. The identities sin x tan x sin1x2 tan x and cos1x2 cot x cos x cot x are examples of due to .

2. To show an equation is not an identity, we must find at least input value where both sides of the equation are defined, but results in a equation.

3. To verify an identity, always begin with the more expression, since it is easier to than to .

4. Converting all terms to functions of may help verify an identity.

5. Discuss/Explain why you must not add, subtract, multiply, or divide both sides of the equation when verifying identities.

6. Discuss/Explain the difference between operating on both sides of an equation (see Exercise 5) and working on each side independently.


Verify that the following equations are identities.

7. 11 sin x2 3 1 sin1x2 4 cos2x

8. 1sec x 12 3sec1x2 14 tan2 x 9. sin2 1x2 cos2x 1

10. 1 cot2 1x2 csc2x 11. 12.

1 sin1x2

cos x cos1x2 sin x 1 cos1x2

sin x cos x sin 1x2

sec x csc x

and

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13. cos2x tan2x 1 cos2x 14. sin2x cot2x 1 sin2x 15. tan x cot x sec x csc x 16. cot x cos x csc x sin x

35.

sec2x tan2x 1 cot2x

36.

csc2x cot2x 1 tan2x

37. sin2x 1cot2x csc2x2 sin2x

38. cos2x 1tan2x sec2x2 cos2x 39. cos x cot x sin x csc x

17.

cos x csc x sin x tan x

18.

sin x sec x cos x cot x

41.

1 sin x 1tan x sec x2 2 1 sin x

19.

sec x sin x cot x tan x

42.

1 cos x 1csc x cot x2 2 1 cos x

20.

csc x cos x cot x tan x

43.

cos x sin x cos x sin x 1 tan x 1 tan x

21.

sin x csc x cos2x csc x

44.

sin x cos x 1 cot x 1 cot x sin x cos x

22.

cos x sec x sin2x sec x

45.

tan2x cot2x csc x sec x tan x cot x

23.

1 tan x sec x csc x sin x

46.

cot x tan x sin x cos x cot2x tan2x

24.

1 cot x csc x sec x cos x

47.

cot x 1 sin2x cot x tan x

25.

cos ␪ sec ␪ tan ␪ 1 sin ␪

48.

tan x 1 cos2x cot x tan x

26.

sin ␪ csc ␪ cot ␪ 1 cos ␪

49.

sec4x tan4x 1 sec2x tan2x

27.

1 sin x cos x cos x 1 sin x

51.

cos4x sin4x 2 sec2x cos2x

28.

sin x 1 cos x sin x 1 cos x

52.

sin4x cos4x 2 csc2x sin2x

29.

csc x cos x cot2x sin2x cos x csc x cot x

30.

1 1 csc2x sec2x 2 cos x sin2x

sin x sin x 31. 2 tan2x 1 sin x 1 sin x

40. sin x tan x cos x sec x

53. 1sec x tan x2 2 54. 1csc x cot x2 2

50.

csc4x cot4x 1 csc2x cot2x

1sin x 12 2 cos2x

1cos x 12 2 sin2x

55.

cos x sin x csc x sec x cos x cos x sec x sin x sin x

32.

cos x cos x 2 cot2x 1 cos x 1 cos x

56.

sin x cos x sec x csc x sin x cos x csc x cos x sin x

33.

cot x cot x 2 sec x 1 csc x 1 csc x

57.

34.

tan x tan x 2 csc x 1 sec x 1 sec x

sin x cos x sin4x cos4x 3 3 1 sin x cos x sin x cos x

58.

sin4x cos4x sin x cos x 3 3 1 sin x cos x sin x cos x

667

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Show that the following equations are not identities.

59. sin a␪ ⫹

␲ ␲ b ⫽ sin ␪ ⫹ sin a b 3 3

␲ ␲ 60. cos a b ⫹ cos ␪ ⫽ cos a ⫹ ␪b 4 4

Determine which of the following are not identities by using a calculator to compare the graphs of the left- and right-hand sides of each equation.

65.

1 ⫺ sin2␪ ⫽ cos ␪ cos ␪

66. cos12x2 ⫽ 1 ⫺ 2 sin2x

61. cos 12␪2 ⫽ 2 cos ␪ 62. tan 12␪2 ⫽ 2 tan ␪ tan ␪ ␪ 63. tan a b ⫽ 4 tan 4

64. cos2␪ ⫺ sin2␪ ⫽ ⫺1 䊳

7–16


67.

1 ⫹ sin x cos x ⫽ cos x 1 ⫹ sin x

68.

cos x ⫽ sec x ⫺ tan x 1 ⫺ sin x


69. Distance to top of movie screen: d2 ⴝ 120 ⴙ x cos ␪2 2 ⴙ 120 ⴚ x sin ␪2 2 At a theater, the optimum viewing angle depends on a number of factors, like the height of the screen, the incline of the auditorium, the location of a seat, the height of your eyes while seated, and so on. One of the measures needed to find the “best” seat is the distance from your eyes to the top of the screen. For a theater with the dimensions shown, this distance is given by the formula here (x is the diagonal distance from the horizontal floor to your seat). (a) Show the formula is equivalent to d 2 ⫽ 800 ⫹ 40x 1cos ␪ ⫺ sin ␪2 ⫹ x2. (b) Find the distance d if ␪ ⫽ 18° and you are sitting in the eighth row with the rows spaced 3 ft apart. 2

c sin A sin B 70. The area of triangle ABC: A ⴝ 2 sin C

(not to scale)

d 20 ft

␣ D x 3 ft

␪

20 ft C

If one side and three angles of a triangle are known, its area can be computed using this formula, where side c is opposite angle C. Find the area of the triangle shown in the diagram.

75⬚

A

䊳

3 ft

45⬚ 20 cm

60⬚ B

APPLICATIONS

71. Pythagorean theorem: For the triangle shown, (a) find an expression for the length of the hypotenuse in terms of tan x and cot x, then determine the length of the hypotenuse when x ⫽ 1.5 rad. (b) Show the expression you found in part (a) is equivalent to h ⫽ 1csc x sec x and recompute the length of h the hypotenuse using this √cot x expression. Did the answers match? √tan x

72. Pythagorean theorem: For the triangle shown, (a) find an expression for the area of the triangle in terms of cot x and cos x, then determine its area ␲ given x ⫽ . (b) Show the expression you found in 6 1 part (a) is equivalent to A ⫽ 1csc x ⫺ sin x2 2 Exercise 72 and recompute the area using this expression. Did the cos x answers match? cot x

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73. Viewing distance: Referring to Exercise 69, find a formula for D — the distance from a patron’s eyes to the bottom of the movie screen. Simplify the result using a Pythagorean identity, then find the value of D for the same seat described in part (b) of Exercise 69. 74. Viewing angle: Referring to Exercises 69 and 73, once d and D are known, the viewing angle ␣ (the angle subtended by the movie screen and the viewer’s eyes) can be found using the formula d2 D2 202 . Find the value of cos ␣ cos ␣ 2dD 䊳

for the same seat described in part (b) of Exercise 69. 75. Intensity of light: In a study of the luminous intensity of light, the expression I1 cos ␪ sin ␣ can occur. 21I1cos ␪2 2 1I2sin ␪2 2 Simplify the equation for the moment I1 I2. 76. Intensity of light: Referring to Exercise 75, find the angle ␪ given I1 I2 and ␣ 60°.


77. Verify the identity

䊳

669

Section 7.3 The Sum and Difference Identities

sin6x cos6x 1 sin2x cos2x. sin4x cos4x

78. Use factoring to show the equation is an identity: sin4x 2 sin2x cos2x cos4x 1.


79. (4.4) Graph the rational function given. x1 h1x2 2 x 4 27 3 , b is a point on the unit 4 4 circle, then state the values of sin t, cos t, and tan t associated with this point.

80. (6.2) Verify that a

7.3

81. (6.6) Use an appropriate trig ratio to find the length of the bridge needed to cross the lake shown in the figure.

Exercise 82 400 yd 62

d

82. (2.3) Graph using transformations of a toolbox function: f 1x2 2冟x 3冟 6

The Sum and Difference Identities


A. Develop and use sum and difference identities for cosine B. Use the cofunction identities to develop the sum and difference identities for sine and tangent C. Use the sum and difference identities to verify other identities

The sum and difference formulas for sine and cosine have a long and ancient history. Originally developed to help study the motion of celestial bodies, they were used centuries later to develop more complex concepts, such as the derivatives of the trig functions, complex number theory, and the study of wave motion in different mediums. These identities are also used to find exact results (in radical form) for many nonstandard angles, a result of great importance to the ancient astronomers and still of notable mathematical significance today.

A. The Sum and Difference Identities for Cosine On the unit circle with center C, consider the point A on the terminal side of angle ␣, and point B on the terminal side of angle ␤, as shown in Figure 7.9. Since r 1, the coordinates of A and B are 1cos ␣, sin ␣2 and 1cos ␤, sin ␤2, respectively. Using the

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Figure 7.9

distance formula, we find that segment AB is equal to

AB ⫽ 21cos ␣ ⫺ cos ␤2 2 ⫹ 1sin ␣ ⫺ sin ␤2 2

(cos , sin ) A

1

⫽ 2cos2␣ ⫺ 2 cos ␣ cos ␤ ⫹ cos2␤ ⫹ sin2␣ ⫺ 2 sin ␣ sin ␤ ⫹ sin2␤

(cos , sin ) B

binomial squares

⫽ 21cos2␣ ⫹ sin2␣2 ⫹ 1cos2␤ ⫹ sin2␤2 ⫺ 2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ regroup ⫽ 22 ⫺ 2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤

cos2u ⫹ sin2u ⫽ 1

With no loss of generality, we can rotate sector ACB clockwise, until side CB coincides with the x-axis. This creates new coordinates of (1, 0) for B, and new coordinates of 1cos1␣ ⫺ ␤2, sin1␣ ⫺ ␤2 2 for A, but the distance AB remains unchanged! (See Figure 7.10.) Recomputing the distance gives

C

Figure 7.10

AB ⫽ 2 3cos1␣ ⫺ ␤2 ⫺ 1 4 2 ⫹ 3sin1␣ ⫺ ␤2 ⫺ 0 4 2

⫽ 2cos2 1␣ ⫺ ␤2 ⫺ 2 cos1␣ ⫺ ␤2 ⫹ 1 ⫹ sin2 1␣ ⫺ ␤2

(cos( ), sin( )) A 1 C

⫽ 23 cos2 1␣ ⫺ ␤2 ⫹ sin2 1␣ ⫺ ␤2 4 ⫺ 2 cos1␣ ⫺ ␤2 ⫹ 1 ⫽ 12 ⫺ 2 cos1␣ ⫺ ␤2

(1, 0)

Since both expressions represent the same distance, we can set them equal to each other and solve for cos1␣ ⫺ ␤2.

B

12 ⫺ 2 cos1␣ ⫺ ␤2 ⫽ 12 ⫺ 2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ 2 ⫺ 2 cos1␣ ⫺ ␤2 ⫽ 2 ⫺ 2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ ⫺2 cos1␣ ⫺ ␤2 ⫽ ⫺2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ cos1␣ ⫺ ␤2 ⫽ cos ␣ cos ␤ ⫹ sin ␣ sin ␤

AB ⫽ AB property of radicals subtract 2 divide both sides by ⫺2

The result is called the difference identity for cosine. The sum identity for cosine follows immediately, by substituting ⫺␤ for ␤. cos1␣ ⫺ ␤2 ⫽ cos ␣ cos ␤ ⫹ sin ␣ sin ␤

cos1␣ ⫺ 3 ⫺␤4 2 ⫽ cos ␣ cos1⫺␤2 ⫹ sin ␣ sin1⫺␤2 cos1␣ ⫹ ␤2 ⫽ cos ␣ cos ␤ ⫺ sin ␣ sin ␤

difference identity substitute ⫺␤ for ␤ cos1⫺␤2 ⫽ cos ␤; sin1⫺␤2 ⫽ ⫺sin ␤

The sum and difference identities can be used to find exact values for the trig functions of certain angles (values written in nondecimal form using radicals), simplify expressions, and to establish additional identities. EXAMPLE 1

䊳

Finding Exact Values for Non-Standard Angles Use the sum and difference identities for cosine to find exact values for a. cos 15° ⫽ cos145° ⫺ 30°2 b. cos 75° ⫽ cos145° ⫹ 30°2 Check results on a calculator in degree MODE .

Solution

䊳

Each involves a direct application of the related identity, and uses special values. a.

Figure 7.11

cos1␣ ⫺ ␤2 ⫽ cos ␣ cos ␤ ⫹ sin ␣ sin ␤ cos145° ⫺ 30°2 ⫽ cos 45° cos 30° ⫹ sin 45° sin 30° 13 12 1 12 ba b⫹a ba b ⫽a 2 2 2 2 16 ⫹ 12 cos 15° ⫽ 4 See Figure 7.11.

difference identity ␣ ⫽ 45°, ␤ ⫽ 30° standard values

combine terms

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Figure 7.12

b.

cos1␣ ⫹ ␤2 ⫽ cos ␣ cos ␤ ⫺ sin ␣ sin ␤ cos145° ⫹ 30°2 ⫽ cos 45° cos 30° ⫺ sin 45° sin 30° 13 12 1 12 ba b⫺a ba b ⫽a 2 2 2 2 16 ⫺ 12 cos 75° ⫽ 4

671

sum identity ␣ ⫽ 45°, ␤ ⫽ 30° standard values

combine terms

See Figure 7.12. Now try Exercises 7 through 12

䊳

CAUTION

䊳

Be sure you clearly understand how these identities work. In particular, note that 1 13 cos160° ⫹ 30°2 ⫽ cos 60° ⫹ cos 30° a0 ⫽ ⫹ b and in general f 1a ⫹ b2 ⫽ f 1a2 ⫹ f 1b2. 2 2

These identities are listed here using the “⫾” and “⫿ ” notation to avoid needless repetition. In their application, use both upper symbols or both lower symbols, with the order depending on whether you’re evaluating the cosine of a sum or difference of two angles. As with the other identities, these can be rewritten to form other members of the identity family. One such version is used in Example 2 to consolidate a larger expression. The Sum and Difference Identities for Cosine cosine family: cos1␣ ⫾ ␤2 ⫽ cos ␣ cos ␤ ⫿ sin ␣ sin ␤ cos ␣ cos ␤ ⫿ sin ␣ sin ␤ ⫽ cos1␣ ⫾ ␤2

EXAMPLE 2

䊳

functions repeat, signs alternate can be used to expand or contract

Using a Sum/Difference Identity to Simplify an Expression Write as a single expression in cosine and evaluate: cos 57° cos 78° ⫺ sin 57° sin 78°

Solution

䊳

Since the functions repeat and are expressed as a difference, we use the sum identity for cosine to rewrite the difference as a single expression. cos ␣ cos ␤ ⫺ sin ␣ sin ␤ ⫽ cos1␣ ⫹ ␤2 cos 57° cos 78° ⫺ sin 57° sin 78° ⫽ cos157° ⫹ 78°2 ⫽ cos 135° The expression is equal to cos 135° ⫽ ⫺

sum identity for cosine ␣ ⫽ 57°, ␤ ⫽ 78° 57° ⫹ 78° ⫽ 135°

12 . See the figure. 2 Now try Exercises 13 through 16

䊳

The sum and difference identities can be used to evaluate the cosine of the sum of two angles, even when they are not adjacent or expressed in terms of cosine.

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EXAMPLE 3

䊳

Computing the Cosine of a Sum 5 Given sin ␣ ⫽ 13 with the terminal side of ␣ in QI, and tan ␤ ⫽ ⫺24 7 with the terminal side of ␤ in QII. Compute the value of cos 1␣ ⫹ ␤2.

Solution Figure 7.13 y 13 12

䊳

To use the sum formula we need the value of cos ␣, sin ␣, cos ␤, and sin ␤. Using the given information about the quadrants along with the Pythagorean theorem, we draw the triangles shown in Figures 7.13 and 7.14, yielding the values that follow. cos ␣ ⫽

5 x

52 ⫹ 122 ⫽ 132

Figure 7.14

5 7 24 12 1QI2, sin ␣ ⫽ 1given2, cos ␤ ⫽ ⫺ 1QII2, and sin ␤ ⫽ 1QII2 13 13 25 25

Using cos 1␣ ⫹ ␤2 ⫽ cos ␣ cos ␤ ⫺ sin ␣ sin ␤ gives this result: 5 12 7 24 b a⫺ b ⫺ a b a b 13 25 13 25 120 84 ⫺ ⫽⫺ 325 325 204 ⫽⫺ 325

cos 1␣ ⫹ ␤2 ⫽ a

y

25 24 ⫺7

x

(⫺7)2


⫹ 242 ⫽ 252

A. You’ve just seen how we can develop and use sum and difference identities for cosine

䊳

To verify the result of Example 3, we can use the inverse trigonometric functions and our knowledge of reference angles to approximate the values of ␣ and ␤. In the first line of Figure 7.15, we find the QI value of ␣ is Figure 7.15 approximately 22.62°, and store it in memory location A using STO ALPHA MATH (A). Noting the terminal side of ␤ is in QII, we compute its value 1⬇106.26°2 as shown in the second calculation and store it in memory location B. For the final step of the verification, we evaluate cos 1A ⫹ B2 in fractional ⫺204 form and obtain , as in Example 3. 325

B. The Sum and Difference Identities for Sine and Tangent The cofunction identities were introduced in Section 6.6, using the complementary ␲ angles in a right triangle. In this section we’ll verify that cos a ⫺ ␪b ⫽ sin ␪ and 2 ␲ sin a ⫺ ␪b ⫽ cos ␪. For the first, we use the difference identity for cosine to obtain 2 cos a

␲ ␲ ␲ ⫺ ␪b ⫽ cos cos ␪ ⫹ sin sin ␪ 2 2 2 ⫽ 102 cos ␪ ⫹ 112 sin ␪ ⫽ sin ␪

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For the second, we use cos a

673

␲ ␲ ⫺ ␪b ⫽ sin ␪, and replace ␪ with the real number ⫺ t. 2 2

This gives cos a cos a

␲ ⫺ ␪b ⫽ sin ␪ 2

cofunction identity for cosine

␲ ␲ ␲ ⫺ c ⫺ t d b ⫽ sin a ⫺ tb 2 2 2 ␲ cos t ⫽ sin a ⫺ tb 2

replace ␪ with result, note c

␲ ⫺t 2

␲ ␲ ⫺ a ⫺ tb d ⫽ t 2 2

␲ ⫺ tb ⫽ cos t for any 2 real number t. Both identities can be written in terms of the real number t. See Exercises 19 through 24. This establishes the cofunction relationship for sine: sin a

The Cofunction Identities cos a

␲ ⫺ tb ⫽ sin t 2

sin a

␲ ⫺ tb ⫽ cos t 2

The sum and difference identities for sine can easily be developed using cofunction ␲ identities. Since sin t ⫽ cos a ⫺ tb, we need only rename t as the sum 1␣ ⫹ ␤2 or the 2 difference 1␣ ⫺ ␤2 and work from there. sin t ⫽ cos a sin 1␣ ⫹ ␤2 ⫽ cos c

␲ ⫺ tb 2

cofunction identity

␲ ⫺ 1␣ ⫹ ␤2 d 2

substitute 1␣ ⫹ ␤2 for t

⫽ cos c a

␲ ⫺ ␣b ⫺ ␤ d 2

regroup argument

␲ ␲ ⫺ ␣b cos ␤ ⫹ sin a ⫺ ␣b sin ␤ 2 2 sin 1␣ ⫹ ␤2 ⫽ sin ␣ cos ␤ ⫹ cos ␣ sin ␤ ⫽ cos a

apply difference identity for cosine result

The difference identity for sine is likewise developed. The sum and difference identities for tangent can be derived using ratio identities and their derivation is left as an exercise (see Exercise 78). The Sum and Difference Identities for Sine and Tangent sine family: sin 1␣ ⫾ ␤2 ⫽ sin ␣ cos ␤ ⫾ cos ␣ sin ␤ sin ␣ cos ␤ ⫾ cos ␣ sin ␤ ⫽ sin 1␣ ⫾ ␤2 tan ␣ ⫾ tan ␤ 1 ⫾ tan ␣ tan ␤ tan ␣ ⫾ tan ␤ ⫽ tan 1␣ ⫾ ␤2 1 ⫿ tan ␣ tan ␤

tangent family: tan 1␣ ⫾ ␤2 ⫽

functions alternate, signs repeat can be used to expand or contract signs match original in numerator, signs alternate in denominator can be used to expand or contract

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EXAMPLE 4A

Solution

䊳

Simplifying Expressions Using Sum/Difference Identities

䊳

Since the functions in each term alternate and the expression is written as a sum, we use the sum identity for sine:

Write as a single expression in sine: sin 12t2 cos t ⫹ cos 12t2 sin t. sin ␣ cos ␤ ⫹ cos ␣ sin ␤ ⫽ sin 1␣ ⫹ ␤2 sin12t2 cos t ⫹ cos 12t2 sin t ⫽ sin 12t ⫹ t2 ⫽ sin(3t) The expression is equal to sin(3t).

sum identity for sine substitute 2t for ␣ and t for ␤ simplify

With Y1 ⫽ sin 12X2 cos 1X 2 ⫹ cos12X2 sin 1X2 and Y2 ⫽ sin 13X2 , the TABLE feature of a calculator set in radian MODE provides strong support for the result of Example 4A (see Figure 7.16).

EXAMPLE 4B

䊳

Figure 7.16

Simplifying Expressions Using Sum/Difference Identities Use the sum or difference identity for tangent to find the exact value of tan

Solution

䊳

11␲ must be the sum or difference of two standard 12 2␲ ␲ 11␲ ⫽ ⫹ . This gives angles. A casual inspection reveals 12 3 4 Since an exact value is requested,

tan ␣ ⫹ tan ␤ 1 ⫺ tan ␣ tan ␤ 2␲ ␲ tan a b ⫹ tan a b 3 4 2␲ ␲ tan a ⫹ b⫽ 2␲ ␲ 3 4 1 ⫺ tan a b tan a b 3 4 ⫺ 13 ⫹ 1 ⫽ 1 ⫺ 1⫺ 132112 1 ⫺ 13 ⫽ 1 ⫹ 13 tan 1␣ ⫹ ␤2 ⫽

B. You’ve just seen how we can use the cofunction identities to develop the sum and difference identities for sine and tangent

11␲ . 12

sum identity for tangent

␣⫽

2␲ ␲ ,␤⫽ 3 4

tan a

2␲ ␲ b ⫽ ⫺ 13, tan a b ⫽ 1 3 4

simplify expression


䊳

C. Verifying Other Identities Once the sum and difference identities are established, we can simply add these to the tools we use to verify other identities. EXAMPLE 5

䊳

Verifying an Identity Verify that tan a␪ ⫺

␲ tan ␪ ⴚ 1 b⫽ is an identity. 4 tan ␪ ⴙ 1

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Solution

䊳

675

Using a direct application of the difference formula for tangent we obtain ␲ 4 ␲ 1 ⫹ tan ␪ tan 4 tan ␪ ⫺ 1 tan ␪ ⴚ 1 ⫽ ⫽ 1 ⫹ tan ␪ tan ␪ ⴙ 1

␲ tan a␪ ⫺ b ⫽ 4

tan ␪ ⫺ tan

␣ ⫽ ␪, ␤ ⫽

␲ 4

␲ tan a b ⫽ 1 4


EXAMPLE 6

Solution

䊳

Verifying an Identity

䊳

Using the sum and difference formulas for sine we obtain

䊳

Verify that sin 1␣ ⫹ ␤2 sin 1␣ ⫺ ␤2 ⫽ sin2␣ ⴚ sin2␤ is an identity. sin 1␣ ⫹ ␤2 sin 1␣ ⫺ ␤2 ⫽ 1sin ␣ cos ␤ ⫹ cos ␣ sin ␤21sin ␣ cos ␤ ⫺ cos ␣ sin ␤2 1A ⫹ B21A ⫺ B2 ⫽ A 2 ⫺ B 2 ⫽ sin2␣ cos2␤ ⫺ cos2␣ sin2␤ use cos2x ⫽ 1 ⫺ sin2x to write the ⫽ sin2␣ 11 ⫺ sin2␤2 ⫺ 11 ⫺ sin2␣2 sin2␤ expression solely in terms of sine

⫽ sin ␣ ⫺ sin ␣ sin ␤ ⫺ sin ␤ ⫹ sin ␣ sin ␤ ⫽ sin2␣ ⴚ sin2␤ 2

C. You’ve just seen how we can use the sum and difference identities to verify other identities

2

2

2

2

2

distribute simplify


7.3 EXERCISES 䊳



1. Since tan 45° ⫹ tan 60° 7 1, we know tan 45° ⫹ tan 60° ⫽ tan 105° is since tan ␪ 6 0 in .

2. To find an exact value for tan 105°, use the sum identity for tangent with ␣ ⫽ and ␤ ⫽ .

3. For the cosine sum/difference identities, the functions in each term, with the sign between them.

4. For the sine sum/difference identities, the functions in each term, with the sign between them.

5. Discuss/Explain how we know the exact value for 2␲ ␲ 11␲ ⫽ cos a ⫹ b will be negative, prior cos 12 3 4 to applying any identity.

6. Discuss/Explain why tan1␣ ⫺ ␤2 ⫽

sin1␣ ⫺ ␤2

cos1␤ ⫺ ␣2 is an identity, even though the arguments of cosine have been reversed. Then verify the identity.

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7–24



Find the exact value of the expression given using a sum or difference identity. Some simplifications may involve using symmetry and the formulas for negatives. Check results on a calculator.

7. cos 105° 9. cos a

7␲ b 12

8. cos 135° 10. cosa⫺

5␲ b 12

Use sum/difference identities to verify that both expressions give the same result. Check results on a calculator.

11. a. cos145° ⫹ 30°2 12. a. cos a

␲ ␲ ⫺ b 6 4

b. cos1120° ⫺ 45°2 b. cos a

␲ ␲ ⫺ b 4 3

Rewrite as a single expression in cosine.

13. cos17␪2 cos12␪2 ⫹ sin17␪2 sin12␪2 ␪ ␪ ␪ ␪ 14. cos a b cos a b ⫺ sin a b sin a b 3 6 3 6 Find the exact value of the given expressions. Check results on a calculator.

15. cos 183° cos 153° ⫹ sin 183° sin 153° 16. cos a

7␲ 5␲ 7␲ 5␲ b cos a b ⫺ sin a b sin a b 36 36 36 36

4 17. For sin ␣ ⫽ ⫺ with terminal side of ␣ in QIV and 5 5 tan ␤ ⫽ ⫺ with terminal side of ␤ in QII, find 12 cos1␣ ⫹ ␤2. 112 with terminal side of ␣ in QII and 113 89 sec ␤ ⫽ ⫺ with terminal side of ␤ in QII, find 39 cos1␣ ⫺ ␤2.

18. For sin ␣ ⫽

Use a cofunction identity to write an equivalent expression.

19. cos 57°

20. sin 18°

21. tan a

22. sec a

5␲ b 12

␲ 23. sin a ⫺ ␪b 6

␲ b 10

␲ 24. cos a ⫹ ␪b 3

Rewrite as a single expression.

25. sin13x2 cos15x2 ⫹ cos13x2 sin15x2 x x x x 26. sin a b cos a b ⫺ cos a b sin a b 2 3 2 3 27.

tan15␪2 ⫺ tan12␪2 1 ⫹ tan15␪2 tan12␪2

x x tan a b ⫹ tan a b 2 8 28. x x 1 ⫺ tan a b tan a b 2 8 Find the exact value of the given expressions.

29. sin 137° cos 47° ⫺ cos 137° sin 47° 30. sin a

11␲ 5␲ 11␲ 5␲ b cos a b ⫹ cos a b sin a b 24 24 24 24

11␲ 4␲ b ⫺ tan a b 21 21 31. 11␲ 4␲ 1 ⫹ tan a b tan a b 21 21 tan a

3␲ ␲ b ⫹ tan a b 20 10 32. 3␲ ␲ 1 ⫺ tan a b tan a b 20 10 tan a

7 with terminal side of ␣ in QII 25 15 and cot ␤ ⫽ with terminal side of ␤ in QIII, find 8 a. sin1␣ ⫹ ␤2 b. tan1␣ ⫹ ␤2

33. For cos ␣ ⫽ ⫺

29 with terminal side of ␣ in QI and 20 12 cos ␤ ⫽ ⫺ with terminal side of ␤ in QII, find 37 a. sin1␣ ⫺ ␤2 b. tan1␣ ⫺ ␤2

34. For csc ␣ ⫽

Find the exact value of the expression given using a sum or difference identity. Some simplifications may involve using symmetry and the formulas for negatives.

35. sin 105°

36. sin1⫺75°2

37. sin a

38. sin a

5␲ b 12

11␲ b 12

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40. tana

41. tan 75°

42. tana⫺

44. a. sina

␲ ␲ ⫺ b 3 4

b. sin1135° ⫺ 120°2 b. sina

Exercise 52

␲ b 12

Use sum/difference identities to verify that both expressions give the same result.

43. a. sin145° ⫺ 30°2

Exercise 51

2␲ b 3

39. tan 150°

␲ ␲ ⫺ b 4 6

45. Find sin 255° given 150° ⫹ 105° ⫽ 255°. See Exercises 7 and 35. 5␲ 19␲ 19␲ b given 2␲ ⫺ ⫽ . See 12 12 12 Exercises 10 and 37.

46. Find cos a

15 A

30⬚

12

28 49. Given ␣ and ␤ are obtuse angles with sin ␣ ⫽ 53 13 and cos ␤ ⫽ ⫺ , find 85 a. sin1␣ ⫺ ␤2 b. cos1␣ ⫹ ␤2 c. tan1␣ ⫺ ␤2 60 50. Given ␣ and ␤ are obtuse angles with tan ␣ ⫽ ⫺ 11 35 and sin ␤ ⫽ , find 37 a. sin1␣ ⫺ ␤2 b. cos1␣ ⫹ ␤2 c. tan1␣ ⫺ ␤2 51. Use the diagram indicated to compute the following: a. sin A b. cos A c. tan A

45⬚

8

15

52. Use the diagram indicated to compute the following: a. sin ␤ b. cos ␤ c. tan ␤ 53. For the figure indicated, show that ␪ ⫽ ␣ ⫹ ␤ and compute the following: a. sin ␪ b. cos ␪ c. tan ␪ Exercise 53

12 47. Given ␣ and ␤ are acute angles with sin ␣ ⫽ 13 35 and tan ␤ ⫽ , find 12 a. sin1␣ ⫹ ␤2 b. cos1␣ ⫺ ␤2 c. tan1␣ ⫹ ␤2 8 48. Given ␣ and ␤ are acute angles with cos ␣ ⫽ 17 25 and sec ␤ ⫽ , find 7 a. sin1␣ ⫹ ␤2 b. cos1␣ ⫺ ␤2 c. tan1␣ ⫹ ␤2

5

45

24

32

28

54. For the figure indicated, show that ␪ ⫽ ␣ ⫹ ␤ and compute the following: a. sin ␪ b. cos ␪ c. tan ␪ Exercise 54 8

6

5

12

Verify each identity.

55. sin1␲ ⫺ ␣2 ⫽ sin ␣

56. cos1␲ ⫺ ␣2 ⫽ ⫺cos ␣

57. cos ax ⫹

␲ 22 b⫽ 1cos x ⫺ sin x2 4 2

58. sin ax ⫹

␲ 12 b⫽ 1sin x ⫹ cos x2 4 2

59. tan ax ⫹

␲ 1 ⫹ tan x b⫽ 4 1 ⫺ tan x

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60. tan ax ⫺

␲ tan x ⫺ 1 b⫽ 4 tan x ⫹ 1

61. cos1␣ ⫹ ␤2 ⫹ cos1␣ ⫺ ␤2 ⫽ 2 cos ␣ cos ␤ 62. sin1␣ ⫹ ␤2 ⫹ sin1␣ ⫺ ␤2 ⫽ 2 sin ␣ sin ␤ 63. cos12t2 ⫽ cos2t ⫺ sin2t 64. sin12t2 ⫽ 2 sin t cos t 65. sin13t2 ⫽ ⫺4 sin3t ⫹ 3 sin t 䊳

7–26


66. cos13t2 ⫽ 4 cos3t ⫺ 3 cos t 67. Use a difference identity to show cos ax ⫺

12 ␲ b⫽ 1cos x ⫹ sin x2. 4 2

68. Use sum/difference identities to show sin ax ⫹

␲ ␲ b ⫹ sin ax ⫺ b ⫽ 12 sin x. 4 4

WORKING WITH FORMULAS Wk tan1 p ⴚ ␪2 c The force required to maintain equilibrium when a screw jack is used can be modeled by the formula shown, where p is the pitch angle of the screw, W is the weight of the load, ␪ is the angle of friction, with k and c being constants related to a particular jack. Simplify the formula using ␲ ␲ the difference formula for tangent given p ⫽ and ␪ ⫽ . 6 4

69. Force and equilibrium: F ⴝ

n2 n1 Brewster’s law of optics states that when unpolarized light strikes a dielectric surface, the transmitted light rays and the reflected light rays are perpendicular to each other. The proof of Brewster’s law involves the expression ␲ n1 sin ␪p ⫽ n2 sin a ⫺ ␪p b. Use the difference identity for sine to verify that this expression leads to 2 Brewster’s law.

70. Brewster’s law of reflection: tan ␪p ⴝ

䊳

APPLICATIONS

71. AC circuits: In a study of AC circuits, the equation cos s cos t sometimes arises. Use a sum R⫽ ␻C sin1s ⫹ t2 identity and algebra to show this equation is 1 equivalent to R ⫽ . vC1tan s ⫹ tan t2 72. Fluid mechanics: In studies of fluid mechanics, the equation ␥1V1sin ␣ ⫽ ␥2V2sin1␣ ⫺ ␤2 sometimes arises. Use a difference identity to show that if ␥1V1 ⫽ ␥2V2, the equation is equivalent to cos ␤ ⫺ cot ␣ sin ␤ ⫽ 1. 73. Art and mathematics: When working in two-point geometric perspective, artists must scale their work to fit on the paper or canvas they are using. In A tan ␪ doing so, the equation ⫽ arises. B tan190° ⫺ ␪2 Rewrite the expression on the right in terms of sine and cosine, then use the difference identities to A show the equation can be rewritten as ⫽ tan2 ␪. B

74. Traveling waves: If two waves of the same frequency, velocity, and amplitude are traveling along a string in opposite directions, they can be represented by the equations Y1 ⫽ A sin1kx ⫺ vt2 and Y2 ⫽ A sin1kx ⫹ vt2 . Use the sum and difference formulas for sine to show the result YR ⫽ Y1 ⫹ Y2 of these waves can be expressed as YR ⫽ 2A sin1kx2cos1vt2. 75. Pressure on the eardrum: If a frequency generator is placed a certain distance from the ear, the pressure on the eardrum can be modeled by the function P1 1t2 ⫽ A sin12␲ft2, where f is the frequency and t is the time in seconds. If a second frequency generator with identical settings is placed slightly closer to the ear, its pressure on the eardrum could be represented by P2 1t2 ⫽ A sin12␲ft ⫹ C2 , ␲ where C is a constant. Show that if C ⫽ , the 2 total pressure on the eardrum 3P1 1t2 ⫹ P2 1t2 4 is P1t2 ⫽ A 3sin12␲ft2 ⫹ cos12␲ft2 4 .

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76. Angle between two cables: Two cables used to steady a radio tower are attached to the tower at heights of 5 ft and 35 ft, with both secured to a stake 12 ft from the tower (see figure). Find the value of cos ␪, where ␪ is the angle between the upper and lower cables.

Exercise 76

77. Difference quotient: Given f 1x2 ⫽ sin x, show that the difference quotient f 1x ⫹ h2 ⫺ f 1x2 cos h ⫺ 1 sin h b ⫹ cos x a b. results in the expression sin x a h h h

35 ft

78. Difference identity: Derive the difference identity for tangent using sin1␣ ⫺ ␤2 . (Hint: After applying the difference identities, tan1␣ ⫺ ␤2 ⫽ cos1␣ ⫺ ␤2 divide the numerator and denominator by cos ␣ cos ␤.)

䊳

5 ft

12 ft


A family of identities called the angle reduction formulas will be of use in our study of complex numbers and other areas. These formulas use the period of a function to reduce large angles to an angle in [0, 360°) or [0, 2␲) having an equivalent function value: (1) cos1t ⫹ 2␲k2 ⫽ cos t; (2) sin1t ⫹ 2␲k2 ⫽ sin t. Use the reduction formulas to find values for the following functions (note the formulas can also be expressed in degrees).

83. An alternative method of proving the difference formula for cosine uses a unit circle and the fact that equal arcs are subtended by equal chords (D ⫽ d in the diagram). Using a combination of algebra, the distance formula, and a Pythagorean identity, show that cos1␣ ⫺ ␤2 ⫽ cos ␣ cos ␤ ⫹ sin ␣ sin ␤ (start by computing D2 and d 2). Then discuss/explain how the sum identity can be found using the fact that ␤ ⫽ ⫺1⫺␤2.

81. sin a

41␲ b 6

82. sin 2385°

84. Verify the Pythagorean theorem for each right triangle in the diagram, then discuss/explain how the diagram offers a proof of the sum identities for sine and cosine. Exercise 84 A

Exercise 83

1

(cos( ), sin( )) (cos , sin )

sin B sin A

(cos , sin ) cos B

D

B

d

(1, 0)

sin B

A cos B cos A

cos B sin A

91␲ b 6

sin B cos A

80. cos a

79. cos 1665°

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85. (6.5) State the period of the functions given: ␲ ␲ a. y ⫽ 3 sin a x ⫺ b 8 3 ␲ b. y ⫽ 4 tan a2x ⫹ b 4

87. (6.6) Clarence the Clown is about to be shot from a circus cannon to a safety net on the other side of the main tent. If the cannon is 30 ft long and must be aimed at 40° for Clarence to hit the net, the end of the cannon must be how high from ground level?

86. (2.5) Graph the piecewise-defined function given: 3 x 6 ⫺1 2 ⫺1 ⱕ x ⱕ 1 f 1x2 ⫽ • x x x 7 1

88. (1.4) Find the equation of the line parallel to 2x ⫹ 5y ⫽ ⫺10, containing the point (5, ⫺2). Write your answer in standard form.

7.4

The Double-Angle, Half-Angle, and Product-to-Sum Identities


A. Derive and use the doubleangle identities for cosine, tangent, and sine B. Develop and use the power reduction and half-angle identities C. Derive and use the product-to-sum and sumto-product identities D. Solve applications using identities

The derivation of the sum and difference identities in Section 7.3 was a “watershed event” in the study of identities. By making various substitutions, they lead us very naturally to many new identity families, giving us a heightened ability to simplify expressions, solve equations, find exact values, and model real-world phenomena. In fact, many of the identities are applied in very practical ways, as in a study of projectile motion and the conic sections (Chapter 10). In addition, one of the most profound principles discovered in the eighteenth and nineteenth centuries was that electricity, light, and sound could all be studied using sinusoidal waves. These waves often interact with each other, creating the phenomena known as reflection, diffraction, superposition, interference, standing waves, and others. The product-to-sum and sum-to-product identities play a fundamental role in the investigation and study of these phenomena.

A. The Double-Angle Identities The double-angle identities for sine, cosine, and tangent can all be derived using the related sum identities with two equal angles 1␣ ⫽ ␤2. We’ll illustrate the process here for the cosine of twice an angle. cos1␣ ⫹ ␤2 ⫽ cos ␣ cos ␤ ⫺ sin ␣ sin ␤ cos1␣ ⫹ ␣2 ⫽ cos ␣ cos ␣ ⫺ sin ␣ sin ␣ cos12␣2 ⫽ cos2␣ ⫺ sin2␣

sum identity for cosine assume ␣ ⫽ ␤ and substitute ␣ for ␤ simplify — double-angle identity for cosine

Using the Pythagorean identity cos2␣ ⫹ sin2␣ ⫽ 1, we can easily find two additional members of this family, which are often quite useful. For cos2␣ ⫽ 1 ⫺ sin2␣ we have cos12␣2 ⫽ cos2␣ ⫺ sin2␣

⫽ 11 ⫺ sin ␣2 ⫺ sin ␣ cos12␣2 ⫽ 1 ⫺ 2 sin2␣ 2

2

double-angle identity for cosine substitute 1 ⫺ sin2␣ for cos2␣ double-angle in terms of sine

Using sin ␣ ⫽ 1 ⫺ cos ␣ we obtain an additional form: 2

2

cos12␣2 ⫽ cos2␣ ⫺ sin2␣

⫽ cos ␣ ⫺ 11 ⫺ cos ␣2 cos12␣2 ⫽ 2 cos2␣ ⫺ 1 2

2

double-angle identity for cosine substitute 1 ⫺ cos2␣ for sin2␣ double-angle in terms of cosine

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Section 7.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities

681

The derivations of sin12␣2 and tan12␣2 are likewise developed and are asked for in Exercise 105. The double-angle identities are collected here for your convenience. The Double-Angle Identities cosine: cos12␣2 ⫽ cos2␣ ⫺ sin2␣

sine: sin12␣2 ⫽ 2 sin ␣ cos ␣

⫽ 1 ⫺ 2 sin ␣ 2

⫽ 2 cos2␣ ⫺ 1 tangent:

EXAMPLE 1

䊳

tan12␣2 ⫽

2 tan ␣ 1 ⫺ tan2␣

Using a Double-Angle Identity to Find Function Values 5 Given sin ␣ ⫽ , find the value of cos12␣2. 8

Solution

䊳

Using the double-angle identity for cosine interms of sine, we find cos12␣2 ⫽ 1 ⫺ 2 sin2␣ 5 2 ⫽ 1 ⫺ 2a b 8 25 ⫽1⫺ 32 7 ⫽ 32

double-angle in terms of sine substitute

5 for sin ␣ 8

5 2 25 2a b ⫽ 8 32 result

7 5 If sin ␣ ⫽ , then cos12␣2 ⫽ . A calculator check is shown in the figure. 8 32 Now try Exercises 7 through 20

䊳

Like the fundamental identities, the double-angle identities can be used to verify or develop others. In Example 2, we explore one of many multiple-angle identities, verifying that cos13␪2 can be rewritten as 4 cos3␪ ⫺ 3 cos ␪ (in terms of powers of cos ␪). EXAMPLE 2

䊳

Verifying a Multiple Angle Identity Verify that cos13␪2 ⫽ 4 cos3␪ ⫺ 3 cos ␪ is an identity.

Solution

䊳

Use the sum identity for cosine, with ␣ ⫽ 2␪ and ␤ ⫽ ␪. Note that our goal is an expression using cosines only, with no multiple angles. cos1␣ ⫹ ␤2 ⫽ cos ␣ cos ␤ ⫺ sin ␣ sin ␤ cos12␪ ⫹ ␪2 ⫽ cos12␪2 cos ␪ ⫺ sin12␪2sin ␪ cos13␪2 ⫽ 12 cos2␪ ⫺ 12 cos ␪ ⫺ 12 sin ␪ cos ␪2 sin ␪ ⫽ 2 cos3␪ ⫺ cos ␪ ⫺ 2 cos ␪ sin2␪ ⫽ 2 cos3␪ ⫺ cos ␪ ⫺ 2 cos ␪11 ⫺ cos2 ␪2 ⫽ 2 cos3 ␪ ⫺ cos ␪ ⫺ 2 cos ␪ ⫹ 2 cos3␪ ⫽ 4 cos3␪ ⫺ 3 cos ␪

sum identity for cosine substitute 2␪ for ␣ and ␪ for ␤

substitute for cos 12␪2 and sin12␪2 multiply substitute 1 ⫺ cos2␪ for sin2␪ multiply combine terms


䊳

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EXAMPLE 3

䊳

Using a Double-Angle Formula to Find Exact Values Find the exact value of sin 22.5° cos 22.5°.

Solution

䊳

A product of sines and cosines having the same argument hints at the double-angle identity for sine. Using sin12␣2 ⫽ 2 sin ␣ cos ␣ and dividing by 2 gives sin ␣ cos ␣ ⫽ sin 22.5° cos 22.5° ⫽

sin12␣2 2 sin 32122.5°2 4

2 sin 45° ⫽ 2 12 b a 12 2 ⫽ ⫽ 2 4

A. You’ve just seen how we can derive and use the double-angle identities for cosine, tangent, and sine

double-angle identity for sine

replace ␣ with 22.5°

multiply

sin 45° ⫽

12 2


䊳

B. The Power Reduction and Half-Angle Identities Expressions having a trigonometric function raised to a power occur quite frequently in various applications. We can rewrite even powers of these trig functions in terms of an expression containing only cosine to the power 1, using what are called the power reduction identities. In calculus, this becomes an indispensible tool, making expressions easier to use and evaluate. It can legitimately be argued that the power reduction identities are actually members of the double-angle family, as all three are a direct consequence. To find identities for cos2␣ and sin2␣, we solve the related double-angle identity involving cos12␣2 . 1 ⫺ 2 sin2␣ ⫽ cos12␣2 ⫺2 sin ␣ ⫽ cos12␣2 ⫺ 1 1 ⫺ cos12␣2 sin2␣ ⫽ 2 2

cos 12␣2 in terms of sine subtract 1, then divide by ⫺2 power reduction identity for sine

Using the same approach for cos2␣ gives cos2␣ ⫽

1 ⫹ cos12␣2 2

. The identity

2 tan ␣ (see Exercise 106), but in this 1 ⫺ tan2␣ 1 ⫺ cos12␣2 sin2␣ . . The result is case it’s easier to use the identity tan2␣ ⫽ 2 1 ⫹ cos12␣2 cos ␣ for tan2␣ can be derived from tan12␣2 ⫽

The Power Reduction Identities cos2␣ ⫽

EXAMPLE 4

䊳

1 ⫹ cos12␣2 2

sin2␣ ⫽

1 ⫺ cos12␣2 2

tan2␣ ⫽

1 ⫺ cos12␣2 1 ⫹ cos12␣2

Using a Power Reduction Formula Write 8 sin4x in terms of an expression containing only cosines to the power 1 and use the GRAPH feature of a calculator to support your result.

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Solution

䊳

8 sin4x ⫽ 81sin2x2 2 1 ⫺ cos12x2 2 d ⫽ 8c 2 ⫽ 2 31 ⫺ 2 cos12x2 ⫹ cos2 12x2 4 1 ⫹ cos14x2 ⫽ 2 c 1 ⫺ 2 cos12x2 ⫹ d 2 ⫽ 2 ⫺ 4 cos12x2 ⫹ 1 ⫹ cos14x2 ⫽ 3 ⫺ 4 cos12x2 ⫹ cos14x2

683

original expression substitute

1 ⫺ cos 12x2

for sin2x

1 ⫹ cos 14x2

for cos2(2x)

2

multiply substitute

2

multiply result

To support our result, we enter Y1 ⫽ 8 sin(X)^4 and Y2 ⫽ 3 ⫺ 4 cos12X2 ⫹cos14X2 in our calculator. Recognizing Y1 can only take on nonnegative values less than or ⫺2␲ equal to 8, we set the window as shown in the figure. Anticipating that only one graph will be seen (since Y1 and Y2 should be coincident), we can vertically shift Y2 down 4 units and graph Y3 ⫽ Y2 ⫺ 4. The figure then helps support that 8 sin4x ⫽ 3 ⫺ 4 cos12x2 ⫹ cos14x2 .

10

2␲

⫺6


䊳

The half-angle identities follow directly from the power reduction identities, using 1 ⫹ cos12␣2 algebra and a simple change of variable. For cos2␣ ⫽ , we first take square 2 1 ⫹ cos12␣2 u roots and obtain cos ␣ ⫽ ⫾ . Using the substitution u ⫽ 2␣ gives ␣ ⫽ , B 2 2 and making these substitutions results in the half-angle identity for cosine: u 1 ⫹ cos u cos a b ⫽ ⫾ , where the radical’s sign depends on the quadrant in which 2 A 2 u u 1 ⫺ cos u , terminates. Using the same substitution for sine gives sin a b ⫽ ⫾ 2 2 A 2 u 1 ⫺ cos u u . In the case of tan a b, we can and for the tangent identity, tan a b ⫽ ⫾ 2 A 1 ⫹ cos u 2 actually develop identities that are free of radicals by rationalizing the denominator or numerator. We’ll illustrate the former, leaving the latter as an exercise (see Exercise 104). 11 ⫺ cos u211 ⫺ cos u2 u tan a b ⫽ ⫾ 2 B 11 ⫹ cos u211 ⫺ cos u2 11 ⫺ cos u2 2

B 1 ⫺ cos2u

rewrite

B

11 ⫺ cos u2 2

Pythagorean identity

⫽⫾ ⫽⫾

multiply by the conjugate

⫽⫾`

sin2u

1 ⫺ cos u ` sin u

2x 2 ⫽ 冟x冟

u Since 1 ⫺ cos u 7 0 and sin u has the same sign as tan a b for all u in its domain, 2 u 1 ⫺ cos u the relationship can simply be written tan a b ⫽ . 2 sin u

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The Half-Angle Identities u 1 ⫹ cos u cos a b ⫽ ⫾ 2 A 2

u 1 ⫺ cos u sin a b ⫽ ⫾ 2 A 2

1 ⫺ cos u u tan a b ⫽ 2 sin u

EXAMPLE 5

䊳

u 1 ⫺ cos u tan a b ⫽ ⫾ 2 A 1 ⫹ cos u

sin u u tan a b ⫽ 2 1 ⫹ cos u

Using Half-Angle Formulas to Find Exact Values Use the half-angle identities to find exact values for (a) sin 15° and (b) tan 15°.

Solution

䊳

Noting that 15° is one-half the standard angle 30°, we can find each value by applying the respective half-angle identity with u ⫽ 30° in Quadrant I. 30° 1 ⫺ cos 30° b⫽ 2 B 2 13 1⫺ 2 ⫽ R 2 22 ⫺ 13 sin 15° ⫽ 2

a. sin a

30° 1 ⫺ cos 30° b⫽ 2 sin 30° 13 1⫺ 2 tan 15° ⫽ ⫽ 2 ⫺ 13 1 2

b. tan a

Note the verification of part (a) in the figure. Part (b) can be similarly checked with a calculator in degree MODE . Now try Exercises 37 through 48

EXAMPLE 6

䊳

Using Half-Angle Formulas to Find Exact Values For cos ␪ ⫽ ⫺

Solution

B. You’ve just seen how we can develop and use the power reduction and half-angle identities

䊳

䊳

␪ ␪ 7 and ␪ in QIII, find exact values of sin a b and cos a b. 25 2 2

␲ 3␲ ␪ ␪ 3␲ , we know must be in QII S 6 6 2 2 2 2 4 ␪ ␪ and we choose our signs accordingly: sin a b 7 0 and cos a b 6 0. 2 2 With ␪ in QIII S ␲ 6 ␪ 6

␪ 1 ⫺ cos ␪ sin a b ⫽ 2 A 2 7 1 ⫺ a⫺ b 25 ⫽ Q 2 16 4 ⫽ ⫽ A 25 5

␪ 1 ⫹ cos ␪ cos a b ⫽ ⫺ 2 A 2 7 1 ⫹ a⫺ b 25 ⫽⫺ Q 2 3 9 ⫽⫺ ⫽⫺ 5 A 25 Now try Exercises 49 through 64

䊳

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C. The Product-to-Sum Identities As mentioned in the introduction, the product-to-sum and sum-to-product identities are of immense importance to the study of any phenomenon that travels in waves, like light and sound. In fact, the tones you hear as you dial a telephone are actually the sum of two sound waves interacting with each other. Each derivation of a product-to-sum identity is very similar (see Exercise 107), and we illustrate by deriving the identity for cos ␣ cos ␤. Beginning with the sum and difference identities for cosine, we have cos ␣ cos ␤ ⫹ sin ␣ sin ␤ ⫽ cos1␣ ⫺ ␤2

cosine of a difference

⫹ cos ␣ cos ␤ ⫺ sin ␣ sin ␤ ⫽ cos1␣ ⫹ ␤2

cosine of a sum

2 cos ␣ cos ␤ ⫽ cos1␣ ⫺ ␤2 ⫹ cos1␣ ⫹ ␤2 cos ␣ cos ␤ ⫽

1 3cos1␣ ⫺ ␤2 ⫹ cos1␣ ⫹ ␤2 4 2

combine equations divide by 2

In addition to wave phenomenon, the identities from this family are very useful in a study of calculus and are listed here. The Product-to-Sum Identities 1 3 cos1␣ ⫺ ␤2 ⫹ cos1␣ ⫹ ␤2 4 2 1 sin ␣ cos ␤ ⫽ 3 sin1␣ ⫹ ␤2 ⫹ sin1␣ ⫺ ␤2 4 2

cos ␣ cos ␤ ⫽

EXAMPLE 7

䊳

1 3cos1␣ ⫺ ␤2 ⫺ cos1␣ ⫹ ␤2 4 2 1 cos ␣ sin ␤ ⫽ 3sin1␣ ⫹ ␤2 ⫺ sin1␣ ⫺ ␤2 4 2 sin ␣ sin ␤ ⫽

Rewriting a Product as an Equivalent Sum Using Identities Write the product 2 cos(27t) cos(15t) as the sum of two cosine functions.

Solution

䊳

This is a direct application of the product-to-sum identity, with ␣ ⫽ 27t and ␤ ⫽ 15t. 1 3cos1␣ ⫺ ␤2 ⫹ cos1␣ ⫹ ␤2 4 2 1 2 cos127t2cos115t2 ⫽ 2a b 3cos127t ⫺ 15t2 ⫹ cos127t ⫹ 15t2 4 2 ⫽ cos112t2 ⫹ cos142t2 cos ␣ cos ␤ ⫽

product-to-sum identity

substitute result

Now try Exercises 65 through 74 䊳 There are times we find it necessary to “work in the other direction,” writing a sum of two trig functions as a product. This family of identities can be derived from the product-to-sum identities using a change of variable. We’ll illustrate the process for sin u ⫹ sin v. You are asked for the derivation of cos u ⫹ cos v in Exercise 108. To begin, we use 2␣ ⫽ u ⫹ v and 2␤ ⫽ u ⫺ v. This creates the sum 2␣ ⫹ 2␤ ⫽ 2u and the difference 2␣ ⫺ 2␤ ⫽ 2v, yielding ␣ ⫹ ␤ ⫽ u and ␣ ⫺ ␤ ⫽ v, respectively. u⫹v u⫺v , which all Dividing the original expressions by 2 gives ␣ ⫽ and ␤ ⫽ 2 2 together make the derivation a matter of direct substitution. Using these values in any product-to-sum identity gives the related sum-to-product, as shown here.

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sin ␣ cos ␤ ⫽ sin a 2 sin a

1 3sin1␣ ⫹ ␤2 ⫹ sin1␣ ⫺ ␤2 4 2

u⫹v u⫺v 1 b cos a b ⫽ 1sin u ⫹ sin v2 2 2 2 u⫹v u⫺v b cos a b ⫽ sin u ⫹ sin v 2 2

product-to-sum identity (sum of sines)

u⫺v u⫹v for ␣, for ␤, 2 2 substitute u for ␣ ⫹ ␤ and v for ␣ ⫺ ␤ substitute

multiply by 2

The sum-to-product identities follow. The Sum-to-Product Identities u⫹v u⫺v b cos a b 2 2 u⫹v u⫺v sin u ⫺ sin v ⫽ 2 cos a b sin a b 2 2

cos u ⫹ cos v ⫽ 2 cos a

EXAMPLE 8

䊳

u⫹v u⫺v b cos a b 2 2 u⫹v u⫺v cos u ⫺ cos v ⫽ ⫺2 sin a b sin a b 2 2 sin u ⫹ sin v ⫽ 2 sin a

Rewriting a Sum as an Equivalent Product Using Identities Given y1 ⫽ sin112␲t2 and y2 ⫽ sin110␲t2, express y1 ⫹ y2 as a product of trigonometric functions.

Solution

䊳

This is a direct application of the sum-to-product identity sin u ⫹ sin v, with u ⫽ 12␲t and v ⫽ 10␲t. sin u ⫹ sin v ⫽ 2 sina sin112␲t2 ⫹ sin110␲t2 ⫽ 2 sina

u⫺v u⫹v b cos a b 2 2

sum-to-product identity

12␲t ⫺ 10␲t 12␲t ⫹ 10␲t b cosa b 2 2

substitute 12␲t for u and 10␲t for v

⫽ 2 sin111␲t2 cos1␲t2 C. You’ve just seen how we can derive and use the product-to-sum and sum-toproduct identities

substitute

Now try Exercises 75 through 84 䊳 For a mixed variety of identities, see Exercises 85 through 102.

D. Applications of Identities In more advanced mathematics courses, rewriting an expression using identities enables the extension or completion of a task that would otherwise be very difficult (or even impossible). In addition, there are a number of practical applications in the physical sciences.

Projectile Motion A projectile is any object that is thrown, shot, kicked, dropped, or otherwise given an initial velocity, but lacking a continuing source of propulsion. If air resistance is ignored, the range of the projectile depends only on its initial velocity v and the angle ␪ at which it is propelled. This phenomenon is modeled by the function r 1␪2 ⴝ

1 2 v sin ␪ cos ␪ 16

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EXAMPLE 9

䊳

687

Using Identities to Solve an Application a. Use an identity to show r1␪2 ⫽

1 2 v sin ␪ cos ␪ is equivalent to 16

r 1␪2 ⫽

1 2 v sin12␪2. 32 b. If the projectile is thrown with an initial velocity of v ⫽ 96 ft/sec, how far will it travel if ␪ ⫽ 15°? c. From the result of part (a), determine what angle ␪ will give the maximum range for the projectile.

Solution

䊳

a. Note that we can use a double-angle identity if we rewrite the coefficient. 1 1 Writing as 2 a b and commuting the factors gives 16 32 1 1 r 1␪2 ⫽ a b v2 12 sin ␪ cos ␪2 ⫽ a bv2sin12␪2. 32 32 1 b. With v ⫽ 96 ft/sec and ␪ ⫽ 15°, the formula gives r 115°2 ⫽ a b1962 2 sin 30°. 32 Evaluating the result shows the projectile travels a horizontal distance of 144 ft.

c. For any initial velocity v, r 1␪2 will be maximized when sin12␪2 is a maximum. This occurs when sin12␪2 ⫽ 1, meaning 2␪ ⫽ 90° and ␪ ⫽ 45°. The maximum range is achieved when the projectile is released at an angle of 45°. For verification we’ll assume an initial velocity of 96 ft/sec and 1 enter the function r 1␪2 ⫽ 1962 2sin12␪2 ⫽ 32 288 sin12␪2 as Y1. With an amplitude of 288 and results confined to the first quadrant, we set an appropriate window, graph the function, and use the 2nd TRACE (CALC) 4:maximum feature. As shown in the figure, the max occurs at ␪ ⫽ 45°.

350

0

90

0

Now try Exercises 111 and 112 䊳

Sound Waves Each tone you hear on a touch-tone phone is actually the combination of precisely two sound waves with different frequencies (frequency f is defined as B f⫽ ). This is why the tones you hear sound identical, regardless of what phone 2␲ you use. The sum-to-product and product-to-sum formulas help us to understand, study, and use sound in very powerful and practical ways, like sending faxes and using other electronic media. EXAMPLE 10

䊳

Using an Identity to Solve an Application On a touch-tone phone, the sound created by pressing 5 is produced by combining a sound wave with frequency 1336 cycles/sec, with another wave having frequency 770 cycles/sec. Their respective equations are y1 ⫽ cos12␲ 1336t2 and y2 ⫽ cos 12␲ 770t2, with the resultant wave being y ⫽ y1 ⫹ y2 or y ⫽ cos12672␲t2 ⫹ cos 11540␲t2. Rewrite this sum as a product.

1

2

3

697 cps

4

5

6

770 cps

7

8

9

852 cps

*

0

#

941 cps

1209 cps

1477 cps

1336 cps

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Solution

䊳

This is a direct application of a sum-to-product identity, with u ⫽ 2672␲t and v ⫽ 1540␲t. Computing one-half the sum/difference of u and v gives 2672␲t ⫺ 1540␲t 2672␲t ⫹ 1540␲t ⫽ 2106␲t and ⫽ 566␲t. 2 2 u⫹v u⫺v b cos a b sum-to-product identity cos u ⫹ cos v ⫽ 2 cos a 2 2 substitute 2672␲t for u cos12672␲t2 ⫹ cos11540␲t2 ⫽ 2 cos12106␲t2 cos1566␲t2 and 1540␲t for v Now try Exercises 113 through 116 䊳

D. You’ve just seen how we can solve applications using identities

Note we can identify the button pressed when the wave is written as a sum. If we have only the resulting wave (written as a product), the product-to-sum formula must be used to identify which button was pressed. Additional applications requiring the use of identities can be found in Exercises 117 through 122.

7.4 EXERCISES 䊳



1. The double-angle identities can be derived using the identities with ␣ ⫽ ␤. For cos12␪2 we expand cos1␣ ⫹ ␤2 using . 3. Multiple-angle identities can be derived using the sum and difference identities. For sin (3x) use ⫹ sin ( ). 5. Explain/Discuss how the three different identities u for tan a b are related. Verify that 2 sin u 1 ⫺ cos u ⫽ . sin u 1 ⫹ cos u 䊳

2. If ␪ is in QIII then 180° 6 ␪ 6 270° and

␪ 6 . 2 4. For the half-angle identities the sign preceding the u radical depends on the in which . 2 7 6. In Example 6, we were given cos ␪ ⫽ ⫺ and ␪ 25 in QIII. Discuss how the result would differ if we stipulate that ␪ is in QII instead. in

since

6


Find exact values for sin(2␪), cos(2␪), and tan(2␪) using the information given. Check results on a calculator.

7. sin ␪ ⫽

5 ; ␪ in QII 13

9. cos ␪ ⫽ ⫺ 11. tan ␪ ⫽

8. cos ␪ ⫽ ⫺

21 ; ␪ in QII 29

9 63 ; ␪ in QII 10. sin ␪ ⫽ ⫺ ; ␪ in QIII 41 65

13 ; ␪ in QIII 84

12. sec ␪ ⫽

53 ; ␪ in QI 28

␪ must be 2

13. sin ␪ ⫽

48 ; cos ␪ 6 0 73

14. cos ␪ ⫽ ⫺

8 ; tan ␪ 7 0 17

5 15. csc ␪ ⫽ ; sec ␪ 6 0 3 16. cot ␪ ⫽ ⫺

80 ; cos ␪ 7 0 39

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Find exact values for sin ␪, cos ␪, and tan ␪ using the information given.

17. sin12␪2 ⫽

18. sin12␪2 ⫽ ⫺

Use the results of Exercises 37–40 and a half-angle identity to find the exact value.

45. sin 11.25°

24 ; 2␪ in QII 25

47. sin a

240 ; 2␪ in QIII 289

120 20. cos12␪2 ⫽ ; 2␪ in QIV 169

49.

21. Verify the following identity:

51.

sin13␪2 ⫽ 3 sin ␪ ⫺ 4 sin ␪ 3

22. Verify the following identity: cos14␪2 ⫽ 8 cos4␪ ⫺ 8 cos2␪ ⫹ 1 Use a double-angle identity to find exact values for the following expressions.

23. cos 75° sin 75°

24. cos215° ⫺ sin215°

␲ 25. 1 ⫺ 2 sin2 a b 8

␲ 26. 2 cos2 a b ⫺ 1 12 28.

␲ 2 2 tan1 12

␲ 1 ⫺ tan2 1 12 2

29. Use a double-angle identity to rewrite 9 sin (3x) cos (3x) as a single function. [Hint: 9 ⫽ 92 122 .] 30. Use a double-angle identity to rewrite 2.5 ⫺ 5 sin2x as a single term. [Hint: Factor out a constant.] Rewrite in terms of an expression containing only cosines to the power 1. Verify result with a calculator.

31. sin2x cos2x

32. sin4x cos2x

33. 3 cos4x

34. cos4x sin4x

35. 2 sin6x

36. 4 cos6x

Use a half-angle identity to find exact values for sin ␪, cos ␪, and tan ␪ for the given value of ␪. Check results on a calculator.

37. ␪ ⫽ 22.5° 39. ␪ ⫽

␲ 12

41. ␪ ⫽ 67.5° 43. ␪ ⫽

3␲ 8

46. tan 37.5°

␲ b 24

48. cos a

5␲ b 24

Use a half-angle identity to rewrite each expression as a single, nonradical function.

41 19. cos12␪2 ⫽ ⫺ ; 2␪ in QII 841

2 tan 22.5° 27. 1 ⫺ tan222.5°

689

38. ␪ ⫽ 75° 40. ␪ ⫽

5␲ 12

42. ␪ ⫽ 112.5° 44. ␪ ⫽

11␲ 12

53.

1 ⫹ cos 30° A 2

50.

1 ⫺ cos 14␪2

B 1 ⫹ cos 14␪2

52.

sin 12x2

1 ⫹ cos 12x2

54.

1 ⫺ cos 45° A 2

1 ⫹ cos 16␪2

B 1 ⫺ cos 16␪2 1 ⫺ cos 16x2 sin 16x2

␪ ␪ ␪ Find exact values for sin a b, cos a b, and tan a b using 2 2 2 the information given.

55. sin ␪ ⫽

12 ; ␪ is obtuse 13

56. cos ␪ ⫽ ⫺

8 ; ␪ is obtuse 17

4 57. cos ␪ ⫽ ⫺ ; ␪ in QII 5 58. sin ␪ ⫽ ⫺

7 ; ␪ in QIII 25

59. tan ␪ ⫽ ⫺

35 ; ␪ in QII 12

60. sec ␪ ⫽ ⫺

65 ; ␪ in QIII 33

61. sin ␪ ⫽

15 ; ␪ is acute 113

62. cos ␪ ⫽

48 ; ␪ is acute 73

63. cot ␪ ⫽

3␲ 21 ;␲ 6 ␪ 6 20 2

64. csc ␪ ⫽

41 ␲ ; 6 ␪ 6 ␲ 9 2

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Write each product as a sum using the product-to-sum identities.

65. sin1⫺4␪2 sin18␪2

66. cos115␣2 sin 1⫺3␣2

7t 3t 67. 2 cos a b sin a b 2 2

5t 9t 68. 2 sin a b sin a b 2 2

70. 2 cos12150␲t2 cos1268␲t2

7␲ ␲ b cos a b 8 8

72. 2 cos 105° cos 165° 74. sin a

7␲ ␲ b sin a⫺ b 12 12

Write each sum as a product using the sum-to-product identities.

75. cos19h2 ⫹ cos14h2 77. sin a

11x 5x b ⫺ sin a b 8 8

76. sin114k2 ⫹ sin141k2 78. cos a

79. cos1697␲t2 ⫹ cos 11447␲t2

⫽ cot2␪ ⫺ 1

92. csc2␪ ⫺ 2 ⫽

cos12␪2 sin2␪

2 cot ␪ ⫺ tan ␪

7x 5x b ⫺ cos a b 6 6

96. csc12x2 ⫽

1 csc x sec x 2

x x 98. 1 ⫺ 2 sin2a b ⫽ cos a b 4 2

99. 1 ⫺ sin2 12␪2 ⫽ 1 ⫺ 4 sin2␪ ⫹ 4 sin4␪ x 100. 2 cos2 a b ⫺ 1 ⫽ cos x 2

Find the exact value using sum-to-product identities.

102.

82. cos 285° ⫺ cos 195°

sin12␪2

x x 97. cos2 a b ⫺ sin2 a b ⫽ cos x 2 2

101. 81. cos 75° ⫹ cos 15°

2 cos12␪2

95. tan x ⫹ cot x ⫽ 2 csc12x2

80. cos1852␲t2 ⫹ cos11209␲t2

sin1120␲t2 ⫹ sin180␲t2 cos1120␲t2 ⫺ cos180␲t2

⫽ ⫺cot120␲t2

sin m ⫹ sin n m⫹n ⫽ tan a b cos m ⫹ cos n 2

103. Show sin2␣ ⫹ 11 ⫺ cos ␣2 2 ⫽ c 2 sin a

17␲ 13␲ 83. sin a b ⫺ sin a b 12 12 84. sin a

sin ␪ 2

94. cot ␪ ⫺ tan ␪ ⫽

Find the exact value using product-to-sum identities.

73. sin a

cos12␪2

93. tan12␪2 ⫽

69. 2 cos11979␲t2 cos1439␲t2

71. 2 cos 15° sin 135°

91.

␣ 2 bd . 2

u 1 ⫺ cos u 104. Show that tan a b ⫽ ⫾ is equivalent 2 A 1 ⫹ cos u

11␲ 7␲ b ⫹ sin a b 12 12

to

Verify the following identities.

sin u by rationalizing the numerator. 1 ⫹ cos u

85.

2 sin x cos x ⫽ tan12x2 cos2x ⫺ sin2x

105. Derive the identity for sin12␣2 and tan12␣2 using

86.

1 ⫺ 2 sin x ⫽ cot12x2 2 sin x cos x

106. Derive the identity for tan2 1␣2 using

sin1␣ ⫹ ␤2 and tan1␣ ⫹ ␤2, where ␣ ⫽ ␤.

2

87. 1sin x ⫹ cos x2 2 ⫽ 1 ⫹ sin12x2

tan12␣2 ⫽

. Hint: Solve for tan2␣ and 1 ⫺ tan2 1␣2 work in terms of sines and cosines.

88. 1sin2x ⫺ 12 2 ⫽ sin4x ⫹ cos12x2 89. cos18␪2 ⫽ cos2 14␪2 ⫺ sin2 14␪2

2 tan1␣2

107. Derive the product-to-sum identity for sin ␣ sin ␤.

90. sin14x2 ⫽ 4 sin x cos x11 ⫺ 2 sin x2 2

108. Derive the sum-to-product identity for cos u ⫹ cos v.

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691


␪ 109. Supersonic speeds, the sound barrier, and Mach numbers: M ⴝ csc a b 2 The speed of sound varies with temperature and altitude. At 32°F, sound travels about 742 mi/hr at sea level. A jet-plane flying faster than the speed of sound (called supersonic speed) has “broken the sound barrier.” The plane projects three-dimensional sound waves about the nose of the craft that form the shape of a cone. The cone intersects the Earth along a hyperbolic path, with a sonic boom being heard by anyone along this path. The ratio of the plane’s speed to the speed of sound is called its Mach number M, meaning a plane flying at M ⫽ 3.2 is traveling 3.2 times the speed of sound. This Mach number can be determined using the formula given here, where ␪ is the vertex angle of the cone described. For the following exercises, use the formula to find M or ␪ as required. For parts (a) and (b), answer in exact form (using a half-angle identity) and approximate form. a. ␪ ⫽ 30° b. ␪ ⫽ 45° c. M ⫽ 2 110. Malus’s law: I ⴝ I0 cos2␪ When a beam of plane-polarized light with intensity I0 hits an analyzer, the intensity I of the transmitted beam of light can be found using the formula shown, where ␪ is the angle formed between the transmission axes of the polarizer and the analyzer. Find the intensity of the beam when ␪ ⫽ 15° and I0 ⫽ 300 candelas (cd). Answer in exact form (using a power reduction identity) and approximate form. 䊳

APPLICATIONS

Range of a projectile: Exercises 111 and 112 refer to Example 9. In Example 9, we noted that the range of a projectile was maximized at ␪ ⫽ 45°. If ␪ 7 45° or ␪ 6 45°, the projectile falls short of its maximum potential distance. In Exercises 111 and 112 assume that the projectile has an initial velocity of 96 ft/sec. 111. Compute how many feet short of maximum the projectile falls if (a) ␪ ⫽ 22.5° and (b) ␪ ⫽ 67.5°. Answer in both exact and approximate form. 112. Use a calculator to compute how many feet short of maximum the projectile falls if (a) ␪ ⫽ 40° and ␪ ⫽ 50° and (b) ␪ ⫽ 37.5° and ␪ ⫽ 52.5°. Do you see a pattern? Discuss/explain what you notice and experiment with other values to confirm your observations. Touch-tone phones: The diagram given in Example 10 shows the various frequencies used to create the tones for a touch-tone phone. Use a sum-to-product identity to write the resultant wave when the following numbers are pressed. 113. 3 114. 8

One button is randomly pressed and the resultant wave is modeled by y(t) shown. Use a product-to-sum identity to write the expression as a sum and determine the button pressed. 115. y1t2 ⫽ 2 cos12150␲t2 cos1268␲t2 116. y1t2 ⫽ 2 cos11906␲t2 cos1512␲t2 117. Clock angles: Kirkland City has a large clock atop city hall, with a minute hand that is 3 ft d long. Claire and Monica independently attempt to devise a function that will track the distance between the tip of the minute hand at t minutes between the hours, and the tip of the minute hand when it is in the vertical position as shown. Claire finds the function ␲t d1t2 ⫽ ` 6 sin a b ` , while Monica devises 60 ␲t d1t2 ⫽ 18 c 1 ⫺ cos a b d . Use the identities A 30 from this section to show the functions are equivalent.

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21.6 cm

118. Origami: The ␪ Japanese art of L origami involves the repeated folding of a single piece of paper to create various art forms. When the 28 cm upper right corner of a rectangular 21.6-cm by 28-cm piece of paper is folded down until the corner is flush with the other side, the length L of 10.8 the fold is related to the angle ␪ by L ⫽ . sin ␪ cos2␪ 21.6 sec ␪ (a) Show this is equivalent to L ⫽ , sin12␪2 (b) find the length of the fold if ␪ ⫽ 30°, and (c) find the angle ␪ if L ⫽ 28.8 cm. 119. Machine gears: A machine part involves two gears. The first has a radius of 2 cm and the second a radius of 1 cm, so the smaller gear ␪ 2 cm turns twice as fast as the larger gear. Let ␪ represent the angle of rotation in the larger gear, measured from a vertical and downward 2␪ 1 cm h starting position. Let P be a P point on the circumference of the smaller gear, starting at the vertical and downward position. Four engineers working on an improved design for this component devise

䊳

7–40


functions that track the height of point P above the horizontal plane shown, for a rotation of ␪° by the larger gear. The functions they develop are: Engineer A: f 1␪2 ⫽ sin12␪ ⫺ 90°2 ⫹ 1; Engineer B: g 1␪2 ⫽ 2 sin2␪; Engineer C: k1␪2 ⫽ 1 ⫹ sin2 ␪ ⫺ cos2 ␪; and Engineer D: h1␪2 ⫽ 1 ⫺ cos 12␪2. Use any of the identities you’ve learned so far to show these four functions are equivalent.

120. Working with identities: Compute the value of sin 15° two ways, first using the half-angle identity for sine, and second using the difference identity for sine. (a) Find a decimal approximation for each to show the results are equivalent and (b) verify algebraically that they are equivalent. (Hint: Square both sides.) 121. Working with identities: Compute the value of cos 15° two ways, first using the half-angle identity for cosine, and second using the difference identity for cosine. (a) Find a decimal approximation for each to show the results are equivalent and (b) verify algebraically that they are equivalent. (Hint: Square both sides.) 122. Standing waves: A clapotic (or standing) wave is formed when a wave strikes and reflects off a seawall or other immovable object. Against one particular seawall, the standing wave that forms can be modeled by summing the incoming wave represented by the equation yi ⫽ 2 sin11.1x ⫺ 0.6t2 with the outgoing wave represented by the equation yo ⫽ 2 sin11.1x ⫹ 0.6t2 . Use a sum-to-product identity to express the resulting wave y ⫽ yi ⫹ yo as a product.


123. Can you find three distinct, real numbers whose sum is equal to their product? A little known fact from trigonometry stipulates that for any triangle, the sum of the tangents of the angles is equal to the products of their tangents. Use a calculator to test this statement, recalling the three angles must sum to 180°. Our website at www.mhhe.com/coburn shows a method that enables you to verify the statement using tangents that are all rational values.

22 ⫹ 12 ⫹ 1 2 ⫹ 1 3 that cos 3.75° is equal to . Verify the result 2 using a calculator, then use the patterns noted to write the value of cos 1.875° in closed form (also verify this result). As ␪ becomes very small, what appears to be happening to the value of cos ␪?

Exercise 124

1 ␪ 2

1

␪ cos ␪

sin ␪

124. A proof without words: From elementary geometry we have the following: (a) an angle inscribed in a semicircle is a right angle; and (b) the measure of an inscribed angle (vertex on the circumference) is one-half the measure of its intercepted arc. Discuss/explain how the unit-circle x sin x diagram offers a proof that tan a b ⫽ . Be detailed and thorough. 2 1 ⫹ cos x 125. Using ␪ ⫽ 30° and repeatedly applying the half-angle identity for cosine, show

s

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693



126. (4.2) Use the rational roots theorem to find all zeroes of x4 ⫹ x3 ⫺ 8x2 ⫺ 6x ⫹ 12 ⫽ 0. 127. (6.1) The hypotenuse of a certain right triangle is twice the shortest side. Solve the triangle. 63 128. (6.2) Verify that 1 16 65 , 65 2 is on the unit circle, then find tan ␪ and sec ␪ to verify 1 ⫹ tan2␪ ⫽ sec2␪.

129. (6.5) Write the equation of the function graphed in terms of a sine function of the form y ⫽ A sin1Bx ⫹ C2 ⫹ D.

y 3 2 1 ⫺1 ⫺2 ⫺3

␲

2␲ x

MID-CHAPTER CHECK 1. Verify the identity using a multiplication: sin x 1csc x ⫺ sin x2 ⫽ cos2x 2. Verify the identity by factoring: cos2x ⫺ cot2x ⫽ ⫺cos2x cot2x 3. Verify the identity by combining terms: 2 sin x cos x ⫺ ⫽ cos x sin x sec x csc x 4. Show the equation given is not an identity. 1 ⫹ sec2x ⫽ tan2x 5. Verify each identity. sin3x ⫹ cos3x a. ⫽ 1 ⫺ sin x cos x sin x ⫹ cos x 1 ⫹ sec x 1 ⫹ cos x ⫽0 b. ⫺ csc x cot x 6. Verify each identity. sec2x ⫺ tan2x ⫽ cos2x a. sec2x cot x ⫺ tan x ⫽ cos2x ⫺ sin2x b. csc x sec x

56 7. Given ␣ and ␤ are obtuse angles with sin ␣ ⫽ 65 80 and tan ␤ ⫽ ⫺ , find 39 a. sin1␣ ⫺ ␤2 b. cos 1␣ ⫹ ␤2 c. tan1␣ ⫺ ␤2 8. Use the diagram shown to compute sin A, cos A, and tan A.

Exercise 8 ␪

60⬚ A

48

9. Given 15 17 and ␪ in QII, find

cos ␪ ⫽ ⫺

14

␪ ␪ exact values of sin a b and cos a b. 2 2 7 with ␣ in QIII, find the value 25 of sin12␣2, cos12␣2 , and tan12␣2 .

10. Given sin ␣ ⫽ ⫺

REINFORCING BASIC CONCEPTS Identities—Connections and Relationships It is a well-known fact that information is retained longer and used more effectively when it is organized, sequential, and connected. In this feature, we attempt to do just that with our study of identities. In flowchart form we’ll show that the entire range of identities has only two tiers, and that the fundamental identities and the sum and difference identities are really the keys to the entire range of identities. Beginning with the right triangle definition of sine, cosine, and tangent, the reciprocal identities and ratio identities are more semantic (word related) than mathematical, and the Pythagorean identities follow naturally from the properties of right triangles. These form the first tier.

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Basic Definitions sin ␪ ⫽

opp hyp

cos ␪ ⫽

adj hyp

tan ␪ ⫽

opp adj

Fundamental Identities defined

Reciprocal Identities

defined

derived

Ratio Identities sin ␪ ⫽ tan ␪ cos ␪ cos ␪ ⫽ cot ␪ sin ␪ sec ␪ ⫽ tan ␪ csc ␪

1 ⫽ csc ␪ sin ␪ 1 ⫽ sec ␪ cos ␪ 1 ⫽ cot ␪ tan ␪

Pythagorean Identities cos2␪ ⫹ sin2␪ ⫽ 1 1 ⫹ tan2␪ ⫽ sec2␪

cot2␪ ⫹ 1 ⫽ csc2␪

(divide by cos ␪)

(divide by sin2␪)

2

The reciprocal and ratio identities are actually defined, while the Pythagorean identities are derived from these two families. In addition, the identity cos2␪ ⫹ sin2␪ ⫽ 1 is the only Pythagorean identity we actually need to memorize; the other two follow by division of cos2␪ and sin2␪ as indicated. In virtually the same way, the sum and difference identities for sine and cosine are the only identities that need to be memorized, as all other identities in the second tier flow from these. Sum/Difference Identities

e

cos 1␣ ⫾ ␤2 ⫽ cos ␣ cos ␤ ⫿ sin ␣ sin ␤ sin 1␣ ⫾ ␤2 ⫽ sin ␣ cos ␤ ⫾ cos ␣ sin ␤

Double-Angle Identities Power Reduction Identities use ␣ ⫽ ␤ solve for cos2␣, sin2␣ in in sum identities related cos12␣2 identity

Half-Angle Identities solve for cos ␣, sin ␣ and use ␣ ⫽ u/2 in the power reduction identities

Product-to-Sum Identities combine various sum/difference identities

sin 12␣2 ⫽ 2 sin ␣ cos ␣

cos2␣ ⫽

1 ⫹ cos 12␣2 2

u 1 ⫹ cos u cos a b ⫽ ⫾ 2 A 2

see Section 7.4

cos 12␣2 ⫽ cos2␣ ⫺ sin2␣

sin2␣ ⫽

1 ⫺ cos 12␣2 2

u 1 ⫺ cos u sin a b ⫽ ⫾ 2 A 2

see Section 7.4

cos 12␣2 ⫽ 2 cos2␣ ⫺ 1 (use sin2␣ ⫽ 1 ⫺ cos2␣)

cos 12␣2 ⫽ 1 ⫺ 2 sin2␣ (use cos2␣ ⫽ 1 ⫺ sin2␣)

Exercise 1: Starting with the identity cos2␣ ⫹ sin2␣ ⫽ 1, derive the other two Pythagorean identities.

Exercise 2: Starting with the identity cos 1␣ ⫾ ␤2 ⫽ cos ␣ cos ␤ ⫿ sin ␣ sin ␤, derive the double-angle identities for cosine.

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The Inverse Trig Functions and Their Applications


A. Find and graph the inverse

B.

C.

D.

E.

sine function and evaluate related expressions Find and graph the inverse cosine and tangent functions and evaluate related expressions Apply the definition and notation of inverse trig functions to simplify compositions Find and graph inverse functions for sec x, csc x, and cot x Solve applications involving inverse functions

While we usually associate the number ␲ with the features of a circle, it also occurs in some “interesting” places, such as the study of normal (bell) curves, Bessel functions, Stirling’s formula, Fourier series, Laplace transforms, and infinite series. In much the same way, the trigonometric functions are surprisingly versatile, finding their way into a study of complex numbers and vectors, the simplification of algebraic expressions, and finding the area under certain curves—applications that are hugely important in a continuing study of mathematics. As you’ll see, a study of the inverse trig functions helps support these fascinating applications.

A. The Inverse Sine Function In Section 5.1 we established that only one-to-one functions have an inverse. All six trig functions fail the horizontal line test and are not one-to-one as given. However, by suitably restricting the domain, a one-to-one function can be defined that makes finding an inverse possible. For the sine function, it seems natural to choose the interval ␲ ␲ c , d since it is centrally located and the sine function attains all possible range 2 2 values in this interval. A graph of y sin x is shown in Figure 7.17, with the portion corresponding to this interval colored in red. Note the range is still 3 1, 1 4 (Figure 7.18).

Figure 7.17 2

2

y sin x

1

2

2

1

2

Figure 7.18

y

Figure 7.19

y

y sin x

1

3 2

2

x

2

1

2 , 1

1

1

2

y y sin1x

2

y sin x

1

2,

1

2

x

2 , 1

2 2 ,

1

1

2

x

1 1

2

yx

We can obtain an implicit equation for the inverse of y sin x by interchanging x- and y-values, obtaining x sin y. By accepted convention, the explicit form of the inverse sine function is written y sin1x or y arcsin x. Since domain and range values have been interchanged, the domain of y sin1x is 3 1, 1 4 and the range is ␲ ␲ c , d . The graph of y sin1x can be found by reflecting the portion in red across 2 2 the line y x and using the endpoints of the domain and range (see Figure 7.19). The Inverse Sine Function ␲ ␲ For y sin x with domain c , d 2 2 and range 31, 14 , the inverse sine function is

2

7–43

1, 2

1

y ⴝ sinⴚ1x or y ⴝ arcsin x, ␲ ␲ with domain 3 1, 14 and range c , d . 2 2 ⴚ1 y ⴝ sin x if and only if sin y ⴝ x

y y sin1x

1 2

1

2

x

1

1, 2

2

695

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From the implicit form x sin y, we learn to interpret the inverse function as, “y is the number or angle whose sine is x.” Learning to read and interpret the explicit form in this way will be helpful. y ⴝ sinⴚ1x 3 x ⴝ sin y EXAMPLE 1

䊳

x ⴝ sin y 3 y ⴝ sinⴚ1x

Evaluating y ⴝ sinⴚ1x Using Special Values Evaluate the inverse sine function for the values given: 13 1 a. y sin1a b. y arcsin a b c. y sin1 122 b 2 2

Solution

䊳

␲ ␲ For x in 31, 1 4 and y in c , d , 2 2 13 13 a. y sin1a b: y is the number or angle whose sine is 2 2 13 ␲ 13 1 sin y , so sin1a b . 2 2 3 1 1 b. y arcsin a b: y is the arc or angle whose sine is 2 2 1 1 ␲ 1 sin y , so arcsin a b . 2 2 6 c. y sin1 122: y is the number or angle whose sine is 2 1 sin y 2. Since 2 is not in 3 1, 14, sin1 122 is undefined.

Table 7.1


x

sin x

␲2

1

␲3

13 2

␲4

12 2

␲6

12

0

0

␲ 6

1 2

␲ 4

12 2

␲ 3

13 2

␲ 2

1

EXAMPLE 2

䊳

1 13 and sin y 2 2 ␲ ␲ each have an infinite number of solutions, but only one solution in c , d . 2 2 1 13 , 1, and so on b, y sin1 x can be When x is one of the standard values a0, , 2 2 evaluated by reading a standard table “in reverse.” For y arcsin 112, we locate the number 1 in the right-hand column of Table 7.1, and note the “number or angle ␲ whose sine is 1,” is . If x is between 1 and 1 but is not a standard value, we can 2 use the sin1 function on a calculator, which is most often the 2nd or INV function for SIN . In Examples 1(a) and 1(b), note that the equations sin y

䊳

Evaluating y ⴝ sinⴚ1x Using a Calculator Evaluate each inverse sine function twice. First in radians rounded to four decimal places, then in degrees to the nearest tenth. a. y sin10.8492 b. y arcsin 10.23172

Solution

䊳

For x in 3 1, 14 , we evaluate y sin1x. a. y sin10.8492: With the calculator in radian MODE , use the keystrokes 2nd SIN 0.8492 ) . We find sin1 10.84922 1.0145 radians. In degree MODE , the same sequence of keystrokes gives sin1 10.84922 58.1° (note that 1.0145 rad 58.1°2 . ENTER

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Section 7.5 The Inverse Trig Functions and Their Applications

WORTHY OF NOTE The sin⫺1x notation for the inverse sine function is a carryover from the f ⫺1 1x2 notation for a general inverse function, and likewise has nothing to do with the reciprocal of the function. The arcsin x notation derives from our work in radians on the unit circle, where y ⫽ arcsin x can be interpreted as “y is an arc whose sine is x.”

697

b. y ⫽ arcsin1⫺0.23172: In radian MODE , we find sin⫺1 1⫺0.23172 ⬇ ⫺0.2338 rad. In degree MODE , sin⫺1 1⫺0.23172 ⬇ ⫺13.4°. Now try Exercises 13 through 16

䊳

From our work in Section 5.1, we know that if f and g are inverses, 1 f ⴰ g21x2 ⫽ x and 1g ⴰ f 21x2 ⫽ x. This suggests the following properties. Inverse Function Properties for Sine For f 1x2 ⫽ sin x and g1x2 ⫽ sin⫺1x:

I. 1 f ⴰ g21x2 ⫽ sin1sin⫺1x2 ⫽ x for x in 3⫺1, 1 4 and

␲ ␲ II. 1g ⴰ f 21x2 ⫽ sin⫺1 1sin x2 ⫽ x for x in c ⫺ , d 2 2

EXAMPLE 3

䊳

Evaluating Expressions Using Inverse Function Properties Evaluate each expression and verify the result on a calculator. 1 ␲ a. sin c sin⫺1a b d b. arcsin c sina b d c. sin⫺1 1sin 150°2 2 4

Solution

䊳

1 1 1 a. sin c sin⫺1a b d ⫽ , since is in 3 ⫺1, 14 Property I 2 2 2 ␲ ␲ ␲ ␲ ␲ b. arcsin c sin a b d ⫽ , since is in c ⫺ , d Property II 4 4 4 2 2 ⫺1 c. sin 1sin 150°2 ⫽ 150°, since 150° is not in 3 ⫺90°, 90° 4. This doesn’t mean the expression cannot be evaluated, only that we cannot use Property II. Since sin 150° ⫽ sin 30°, sin⫺1 1sin 150°2 ⫽ sin⫺1 1sin 30°2 ⫽ 30°. The calculator verification for each is shown in Figures 7.20 and 7.21. Note ␲ ⬇ 0.7854. 4 Figure 7.20 Parts (a) and (b)

Figure 7.21 Part (c)


䊳

The domain and range concepts at play in Example 3(c) can be further explored with the TABLE feature of a calculator. Begin by using the TBLSET screen ( 2nd ) to set TblStart ⫽ 90 with ¢Tbl ⫽ ⫺30. After placing the calculator in degree MODE , go to the screen and input Y1 ⫽ sin X, Y2 ⫽ sin⫺1X, and Y3 ⫽ Y2 1Y1 2 Y= (the composition Y2 ⴰ Y1). Then disable Y2 so that only Y1 and Y3 will be displayed (Figure 7.22). Note the inputs are standard angles, the outputs in Y1 are the (expected) WINDOW

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standard values, and the outputs in Y3 return the original standard angles. Now scroll upward until 180 is at the top of the X column (Figure 7.23), and note that Y3 continues to return standard angles from the interval 3 90°, 90° 4 , including Example 3(c)’s result: sin1 1sin 150°2 150°. This is a powerful reminder that the inverse function properties cannot always be used when working with inverse trigonometric functions. Figure 7.22

Figure 7.23

A. You’ve just seen how we can find and graph the inverse sine function and evaluate related expressions

B. The Inverse Cosine and Inverse Tangent Functions Like the sine function, the cosine function is not one-to-one and its domain must also be restricted to develop an inverse function. For convenience we choose the interval x 僆 3 0, ␲ 4 since it is again somewhat central and takes on all of its range values in this interval. A graph of the cosine function, with the interval corresponding to this interval shown in red, is given in Figure 7.24. Note the range is still 3 1, 14 (Figure 7.25). Figure 7.24

Figure 7.25

y

Figure 7.26

y

3

y

(1, )

3

3

y cos1(x) 2

y cos(x)

2

2

3 2

1

2

x

2

2

(0, 1)

(0, 1)

2

2

1

y cos(x)

x (, 1)

2

(1, 0) 1

y cos(x)

x

(, 1)

For the implicit equation of inverse cosine, y cos x becomes x cos y, with the corresponding explicit forms being y cos1x or y arccos x. By reflecting the graph of y cos x across the line y x, we obtain the graph of y cos1 x shown in Figure 7.26. The Inverse Cosine Function

For y cos x with domain 3 0, ␲ 4 and range 3 1, 14 , the inverse cosine function is y ⴝ cosⴚ1x or y ⴝ arccos x

with domain 31, 14 and range 3 0, ␲ 4.

y ⴝ cosⴚ1x if and only if cos y ⴝ x

EXAMPLE 4

䊳

y

(1, ) 3

y cos1(x)

2 1

2

1

Evaluating y ⴝ cosⴚ1x Using Special Values Evaluate the inverse cosine for the values given: 13 a. y cos1 102 b. y arccos a b 2

c. y cos1 1␲2

(1, 0)

x

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Solution

䊳

699

For x in 31, 14 and y in 3 0, ␲ 4 , a. y cos1 102: y is the number or angle whose cosine is 0 1 cos y 0. ␲ This shows cos1 102 . 2 13 b. y arccos a b: y is the arc or angle whose cosine is 2 13 13 13 5␲ 1 cos y . This shows arccos a b . 2 2 2 6 c. y cos1 1␲2: y is the number or angle whose cosine is ␲ 1 cos y ␲. Since ␲ 僆 31, 14 , cos1 1␲2 is undefined. Attempting to evaluate cos1 1␲2 on a calculator will produce the error message shown in the figure. Now try Exercises 25 through 34

䊳

Knowing that y cos x and y cos1x are inverse functions enables us to state inverse function properties similar to those for sine. Inverse Function Properties for Cosine

For f 1x2 cos x and g1x2 cos1x: I. 1 f ⴰ g21x2 cos1cos1x2 x for x in 31, 1 4 II. 1g ⴰ f 21x2 cos

and

1

EXAMPLE 5

䊳

1cos x2 x for x in 30, ␲4

Evaluating Expressions Using Inverse Function Properties Evaluate each expression. a. cos 3 cos1 10.732 4

Solution

䊳

b. arccos c cos a

␲ bd 12

c. cos1 c cos a

a. cos 3cos1 10.732 4 0.73, since 0.73 is in 3 1, 14 Property I ␲ ␲ ␲ b. arccos c cos a b d , since is in 30, ␲ 4 Property II 12 12 12 4␲ 4␲ 4␲ , since c. cos1 c cos a b d is not in 30, ␲ 4. 3 3 3 This expression cannot be evaluated using Property II. Since

4␲ bd 3

cos a

4␲ 2␲ 4␲ 2␲ 2␲ b cos a b, cos1 c cos a b d cos1 c cos a b d . 3 3 3 3 3 The results can also be verified using a calculator. Now try Exercises 35 through 42

䊳

For the tangent function, we likewise restrict the domain to obtain a one-to-one ␲ ␲ function, with the most common choice being a , b. The corresponding range is ⺢. 2 2 The implicit equation for the inverse tangent function is x tan y with the explicit forms y tan1x or y arctan x. With the domain and range interchanged, the domain ␲ ␲ ␲ ␲ of y tan1x is ⺢, and the range is a , b. The graph of y tan x for x in a , b 2 2 2 2

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is shown in red (Figure 7.27), with the inverse function y tan1 x shown in blue (Figure 7.28). Figure 7.27 y tan x

Figure 7.28 y

y 3

3 2 1

␲

冢 ␲4 , 1冣

␲ 2

␲

冢 4 , 1冣

2

y tan1x

1

␲

x

␲

1 2

冢1,

␲

冢1, 4 冣 ␲

3

3

The Inverse Tangent Function

Inverse Function Properties for Tangent

␲ ␲ For y tan x with domain a , b and 2 2 range ⺢, the inverse tangent function is

For f 1x2 tan x and g1x2 tan1x:

y ⴝ tanⴚ1x or y ⴝ arctan x, ␲ ␲ with domain ⺢ and range a , b. 2 2 y ⴝ tanⴚ1x if and only if tan y ⴝ x

EXAMPLE 6

䊳

x

1 ␲2 ␲4 冣 2

I. 1 f ⴰ g21x2 tan1tan1x2 x for x in ⺢ and

␲ ␲ II. 1g ⴰ f 21x2 tan1 1tan x2 x for x in a , b. 2 2

Evaluating Expressions Involving Inverse Tangent Evaluate each expression. a. tan1 1 132 b. arctan 3 tan10.892 4

Solution

䊳

B. You’ve just seen how we can find and graph the inverse cosine and tangent functions and evaluate related expressions

␲ ␲ For x in ⺢ and y in a , b, 2 2 ␲ ␲ a. tan1 1 132 , since tan a b 13 3 3 ␲ ␲ b. arctan 3tan10.892 4 0.89, since 0.89 is in a , b 2 2

Property II


䊳

C. Using the Inverse Trig Functions to Evaluate Compositions 1 In the context of angle measure, the expression y sin1a b represents an angle— 2 1 the angle y whose sine is . It seems natural to ask, “What happens if we take the 2 1 tangent of this angle?” In other words, what does the expression tan c sin1a b d 2 ␲ mean? Similarly, if y cos a b represents a real number between 1 and 1, how do 3 ␲ 1 we compute sin c cos a b d ? Expressions like these occur in many fields of study. 3

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EXAMPLE 7

䊳

Simplifying Expressions Involving Inverse Trig Functions Simplify each expression: 1 a. tan c arcsin a⫺ b d 2

Solution

䊳

701

␲ b. sin⫺1 c cos a b d 3

1 ␲ ␲ 1 a. In Example 1 we found arcsin a⫺ b ⫽ ⫺ . Substituting ⫺ for arcsin a⫺ b 2 6 6 2 13 13 ␲ 1 gives tan a⫺ b ⫽ ⫺ , showing tan c arcsin a⫺ b d ⫽ ⫺ . 6 3 2 3 1 ␲ ␲ b. For sin⫺1 c cos a b d , we begin with the inner function cos a b ⫽ . Substituting 3 3 2 1 ␲ 1 for cos a b gives sin⫺1a b. With the appropriate checks satisfied we have 2 3 2 ␲ ␲ 1 ␲ sin⫺1a b ⫽ , showing sin⫺1 c cos a b d ⫽ . 2 6 3 6 Now try Exercises 53 through 64

䊳

If the argument is not a special value and we need the Figure 7.29 answer in exact form, we can draw the triangle described by the inner expression using the definition of the trigono17 metric functions as ratios. In other words, for either y or 8 8 ⫺1 ␪ ⫽ sin a b, we draw a triangle with hypotenuse 17 17 ␪ adj and side 8 opposite ␪ to model the statement, “an angle opp 8 ⫽ whose sine is ” (see Figure 7.29). Using the Pythagorean theorem, we find 17 hyp the adjacent side is 15 and can now name any of the other trig functions. EXAMPLE 8

䊳

Using a Diagram to Evaluate an Expression Involving Inverse Trig Functions Evaluate the expression tan c sin⫺1a⫺

Solution Figure 7.30 ␪

15 17

⫺8 (15, ⫺8)

䊳

8 bd. 17

The expression tan c sin⫺1a⫺

8 8 b d is equivalent to tan ␪, where ␪ ⫽ sin⫺1a⫺ b 17 17 8 ␲ ␲ with ␪ in c ⫺ , d (QIV or QI). For sin ␪ ⫽ ⫺ 1sin ␪ 6 02, ␪ must be in 2 2 17 Figure 7.31 QIII or QIV. To satisfy both, ␪ must be in 8 QIV. From Figure 7.30 we note tan ␪ ⫽ ⫺ , 15 8 8 ⫺1 showing tan c sin a⫺ b d ⫽ ⫺ . 17 15 A calculator check is shown in Figure 7.31. Now try Exercises 65 through 72

䊳

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These ideas apply even when one side of the triangle is unknown. In other words, x b, since “␪ is an angle whose we can still draw a triangle for ␪ cos1a 2 2x 16 adj x .” cosine is hyp 2x2 16 EXAMPLE 9

䊳

Using a Diagram to Evaluate an Expression Involving Inverse Trig Functions Evaluate the expression tan c cos1a

x

2x 16 function is defined for the expression given.

Solution

䊳

Rewrite tan c cos1a ␪ cos1a

x 2x 16 2

2

b d . Assume x 7 0 and the inverse

b d as tan ␪, where

√x2 16

opp

b. Draw a triangle with 2x2 16 x side x adjacent to ␪ and a hypotenuse of 2 2x 16 (see the figure). The Pythagorean theorem gives x2 opp2 1 2x2 162 2, which leads to opp2 1x2 162 x2 x 4 giving opp 116 4. This shows tan ␪ tan c cos1a bd . 2 x 2x 16

C. You’ve just seen how we can apply the definition and notation of inverse trig functions to simplify compositions

x


䊳

D. The Inverse Functions for Secant, Cosecant, and Cotangent As with the other functions, we restrict the domains of the secant, cosecant, and cotangent functions to obtain one-to-one functions that are invertible (an inverse can be found). Once again the choice is arbitrary, and because some domains are easier to work with than others, these restrictions are not necessarily used uniformly in subsequent mathematics courses. For y sec x, we’ve chosen the “most intuitive” restriction, one that seems more centrally located (nearer the origin). The graph of y sec x is reproduced here, along with its inverse function (see Figures 7.32 and 7.33). The domain, range, and graphs of the functions y csc1x and y cot1x are given in Figures 7.34 and 7.35.

Figure 7.32

Figure 7.33 y ⴝ sec ⴚ1x

y ⴝ sec x

y

y 3

y sec1x

y sec x x僆

[0, 2 )

傼

( 2 , ]

y 僆 (, 1] 傼 [1, )

3

2

x 僆 (, 1] 傼 [1, )

2

1

y 僆 [0, 2 ) 傼 ( 2 , ]

1

x

x

1

1

2

2

3

3

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703


Figure 7.34

Figure 7.35 y cot1x y

y csc1x y 2

y csc1x

y cot1x

x 僆 (⬁, 1] 傼 [1, ⬁) y僆

[ 2 ,

0) 傼 (0,

2

x 僆 (⬁, ⬁)

] 2

y 僆 (0 , ) 1

1

x 1

x

1

2

2

The functions y sec1x, y csc1x, and y cot1x can be evaluated by noting their relationships to y cos1x, y sin1x, and y tan1x, respectively. For y sec1x, we have sec y x 1 1 sec y x 1 cos y x

WORTHY OF NOTE While the domains of y cot1x and y tan1x both include all real numbers, evaluating cot1x using 1 tan1a b involves the restriction x x 0. To maintain consistency, ␲ the equation cot1x tan1x 2 is often used. The graph of ␲ y tan1x is that of 2 y tan1x reflected across the ␲ x-axis and shifted units up, with 2 the result identical to the graph of y cot1x.

EXAMPLE 10

1 y cos1a b x 1 sec1x cos1a b x

definition of inverse function property of reciprocals reciprocal ratio rewrite using inverse function notation

substitute sec1x for y

1 In other words, to find the value of y sec1x, evaluate y cos1a b, x 1. x 1 Similarly, the expression csc1x can be evaluated using sin1a b, x 1. The x expression cot1x can likewise be evaluated using an inverse tangent function: 1 cot1x tan1a b. x

䊳

Evaluating an Inverse Trig Function Evaluate using a calculator only if necessary: 2 ␲ a. sec1a b. cot1a b b 12 13

Solution

䊳

2 13 b, we evaluate cos1a b. 2 13 Since this is a standard value, no calculator is needed and the result is 30°. ␲ 12 b. For cot1a b, find tan1a b on a calculator: ␲ 12 ␲ 12 cot1a b tan1a b 1.3147. ␲ 12 a. From our previous discussion, for sec1a

Now try Exercises 77 through 86 A summary of the highlights from this section follows.

䊳

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Summary of Inverse Function Properties and Compositions 1. For sin x and sin1x, sin1sin1x2 x, for any x in the interval 3 1, 14 ; ␲ ␲ sin1 1sin x2 x, for any x in the interval c , d 2 2 1 3. For tan x and tan x, tan1tan1x2 x, for any real number x; ␲ ␲ tan1 1tan x2 x, for any x in the interval a , b 2 2

2. For cos x and cos1x, cos1cos1x2 x, for any x in the interval 31, 1 4 ; cos1 1cos x2 x, for any x in the interval 30, ␲4 1 4. To evaluate sec1x, use cos1a b, x 1; x 1 csc1x, use sin1a b, x 1; x ␲ 1 cot x, use tan1x, for all real numbers x 2

Our calculators can lend some insight to the varied domain restrictions the inverse trig functions demand, simply by graphing y f 1g1x2 2 . The result should yield the identity function y x for the domain specified. With the calculator in radian MODE and a ZOOM 4:ZDecimal window, compare the graphs of Y1 sin1sin1X2 and Y2 sin1 1sin X2 shown in Figures 7.36 and 7.37, respectively. A casual observation verifies property 1: sin1sin1x2 x for any x in the interval 31, 14 and sin1 1sin x2 x for any x in the ␲ ␲ interval c , d . See Exercises 87 through 90 for verifications of properties 2 and 3. 2 2 Figure 7.36

Figure 7.37

3.1

4.7

D. You’ve just seen how we can find and graph inverse functions for sec x, csc x, and cot x

3.1

4.7

4.7

3.1

4.7

3.1

E. Applications of Inverse Trig Functions We close this section with one example of the many ways that inverse functions can be applied. EXAMPLE 11

䊳

Using Inverse Trig Functions to Find Viewing Angles Believe it or not, the drive-in Figure 7.38 movie theaters that were so popular in the 1950s are making a comeback! If you arrive early, you can park in 30 ft one of the coveted “center spots,” but if you arrive late, you might have to park very close and strain your neck to 10 ft watch the movie. Surprisingly, the maximum viewing angle x (not the most comfortable viewing angle in this case) is actually very close to the front. Assume the base of a 30-ft screen is 10 ft above eye level (see Figure 7.38).

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705

a. Use the inverse function concept to find expressions for angle ␣ and angle ␤. b. Use the result of part (a) to find an expression for the viewing angle ␪. c. Use a calculator to find the viewing angle ␪ (to tenths of a degree) for distances of 15, 25, 35, and 45 ft, then to determine the distance x (to tenths of a foot) that maximizes the viewing angle.

Solution

E. You’ve just seen how we can solve applications involving inverse functions

䊳

a. The side opposite ␤ is 10 ft, and we want to know x — the adjacent side. This 10 10 suggests we use tan ␤ , giving ␤ tan1a b. In the same way, we find x x 40 that ␣ tan1a b. x b. From the diagram we note that ␪ ␣ ␤, and substituting for ␣ and ␤ 40 10 directly gives ␪ tan1a b tan1a b. x x 40 10 c. After we enter Y1 tan1a b tan1a b, a graphing calculator in degree X X Figure 7.39 MODE gives approximate viewing angles of 50 35.8°, 36.2°, 32.9°, and 29.1°, for x 15, 25, 35, and 45 ft, respectively. From this data, we note the distance x that makes ␪ a maximum must be between 15 and 0 100 35 ft, and using 2nd TRACE (CALC) 4:maximum shows ␪ is a maximum of 36.9° at a distance of 20 ft from the screen (see Figure 7.39). 0


䊳

7.5 EXERCISES 䊳



1. All six trigonometric functions fail the test and therefore are not -to-

.

2. The two most common ways of writing the inverse function for y sin x are and .

3. The domain for the inverse sine function is and the range is .

4. The domain for the inverse cosine function is and the range is .

5. Most calculators do not have a key for evaluating an expression like sec15. Explain how it is done using the COS key.

6. Discuss/Explain what is meant by the implicit form of an inverse function and the explicit form. Give algebraic and trigonometric examples.

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The tables here show values of sin ␪, cos ␪, and tan ␪ for ␪ 僆 [ⴚ180ⴗ to 210ⴗ]. The restricted domain used to develop the inverse functions is shaded. Use the information from these tables to complete the exercises that follow. y ⴝ cos ␪

y ⴝ sin ␪ ␪

sin ␪

␪

sin ␪

␪

cos ␪

␪

cos ␪

␪

tan ␪

␪

tan ␪

30°

1 2

180°

1

30°

13 2

180°

0

30°

13 3

1 2

60°

13 2

150°

13 2

60°

1 2

150°

13 3

60°

13

13 2

90°

1

120°

1 2

90°

0

120°

13

90°

—

120°

13 2

90°

0

120°

1 2

90°

—

120°

13

13 2

150°

1 2

60°

1 2

150°

13 2

60°

13

150°

1 2

180°

0

30°

13 2

180°

30°

13 3

180°

0

210°

0

1

210°

0

210°

13 3

180°

0

150°

120°

90° 60°

1

y ⴝ tan ␪

30°

0

0

1 2

Use the preceding tables to fill in each blank (principal values only).

7.

sin 0 0

sin10

␲ sin a b 6 sin a

1 ␲ arcsin a b 2 6

5␲ 1 b 6 2

␲ sin a b 1 2

8.

1 2 13 sin 120° 2 sin 30°

sin 160°2 sin 180°

1 sin1a b 2 1

sin

112

11. sin11

12 b 2

13 13 arcsina b 2 2 arcsin 0 0°

10. arcsin a

1

13 2

13 b 2

1 12. arcsin a b 2

0

13 3

Evaluate using a calculator. Answer in radians to the nearest ten-thousandth and in degrees to the nearest tenth.

13. arcsin 0.8892

7 14. arcsin a b 8

15. sin1a

16. sin1a

1 b 17

1 15 b 2

Evaluate each expression, keeping the domain and range of each function in mind. Check results using a calculator.

17. sin c sin1a

1 sin1 2 13 sin1a b 2

Evaluate without the aid of calculators or tables, keeping the domain and range of each function in mind. Answer in radians.

9. sin1a

12 bd 2

18. sin c arcsin a

13 bd 2

␲ 19. arcsin c sin a b d 3

20. sin1 1sin 30°2

21. sin1 1sin 135°2

22. arcsin c sin a

23. sin1sin1 0.82052

3 24. sin c arcsin a b d 5

2␲ bd 3

Use the tables given prior to Exercise 7 to fill in each blank (principal values only).

25.

cos11

cos 0 1 ␲ cos a b 6 cos 120° cos ␲ 1

arccos a 1 2

13 ␲ b 2 6

1 arccos a b 2 cos1 112

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26.

707


cos160°2

1 2

␲ 13 cos a b 6 2 cos1120°2 cos 12␲2 1

44.

1 cos1a b 2 cos1a

tan 1150°2

13 b 2

13 3

tan ␲ 0 tan 120° 13

1 arccos a b 120° 2

␲ tan a b 4

cos11

tan1a

13 b 3

tan10

arctan 1132 arctan 1

␲ 4

Evaluate without the aid of calculators or tables. Evaluate without the aid of calculators or tables. Answer in radians.

1 27. cos1a b 2

28. arccos a

29. cos1 112

13 b 2

30. arccos (0)

Evaluate using a calculator. Answer in radians to the nearest ten-thousandth, degrees to the nearest tenth.

4 32. arccos a b 7

31. arccos 0.1352 33. cos1a

15 b 3

34. cos1a

16 1 b 5

Evaluate each expression, keeping the domain and range of each function in mind. Check results using a calculator.

␲ 35. arccos c cos a b d 4

36. cos1 1cos 60°2

37. cos1cos1 0.55602

38. cos c arccos a

39. cos c cos1a 41. cos1 c cos a

8 bd 17

42. arccos 1cos 315.8°2

Use the tables presented before Exercise 7 to fill in each blank.

43.

tan 0 0 ␲ tan a b 3 tan 30° ␲ tan a b 3

13 3

Evaluate using a calculator. Answer in radians to the nearest ten-thousandth and in degrees to the nearest tenth.

49. tan1 12.052

50. tan1 10.32672

51. arctan a

52. arctan1 162

53. sin1 c cos a

2␲ bd 3

55. tan c arccos a

13 bd 2

12 bd 2

59. arccos3sin130°2 4

␲ 54. cos1 c sin a b d 3 1 56. sec c arcsin a b d 2 1 58. cot c cos1a b d 2 60. arcsin1cos 135°2

Explain why the following expressions are not defined.

61. tan1sin112

62. cot(arccos 1)

␲ 63. sin1 c csc a b d 4

64. cos1 c sec a

2␲ bd 3

Use the diagrams below to write the value of: (a) sin ␪, (b) cos ␪, and (c) tan ␪.

65.

66.

0.5

␲ 3

13 b 3

tan1 1 132

29 b 21

Simplify each expression without using a calculator.

tan 0

arctan a

46. arctan112 48. tan10

47. arctan1 132

1

arctan1132

13 b 3

57. csc c sin1a

12 13 b d 40. cos c arccos a bd 2 2

5␲ bd 4

45. tan1a

0.4

10 10√3

0.3

67.

√x 2 36

6

20

x

68. √100 9x2

10

3x

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Precalculus—

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Evaluate each expression by drawing a right triangle and labeling the sides.

7 69. sin c cos1a b d 25 71. sin c tan1a

15 bd 2

73. cot c arcsina

3x bd 5

75. cos c sin1a

72. tan c cos1a

123 bd 12

74. tan c arcseca

5 bd 2x

x

Use the tables given prior to Exercise 7 to help fill in each blank. sec11

sec 0 1 ␲ seca b 3 sec130°2 sec1␲2

䊳

␲ arcsec 2 3 2 13

78.

sec160°2 2 2 7␲ seca b 6 13

11 70. cos c sin1a b d 61

bd 212 x2 29 x2 76. tan c sec1a bd x

77.

7–56


arcseca

2 b 13

sec1 112 ␲

arcsec 2 arcseca

sec1360°2 1

2 b 13

arcsec 1 sec12 60°

sec160°2

Evaluate using a calculator in degree necessary.

only as

MODE

79. arccsc 2

80. csc1a

2 b 13

81. cot1 13

82. arccot112

83. arcsec 5.789

84. cot1a

85. sec1 17

86. arccsc 2.9875

17 b 2

Use the graphing feature of a calculator to determine the interval where the following functions are equivalent to the identity function y ⴝ x. If necessary, use the TRACE or 2nd TRACE (CALC) features to determine whether or not endpoints should be included.

88. Y cos1 1cos x2

87. Y cos1cos1x2

90. Y tan1 1tan x2

89. Y tan1tan1x2


91. The force normal to an object on an inclined plane: FN ⴝ mg cos ␪ When an object is on an inclined plane, the normal force is the force acting perpendicular to the plane and away from the force of gravity, and is measured in a unit called newtons (N). The magnitude of this force depends on the angle of incline of the plane according to the formula above, where m is the mass of the object in kilograms and g is the force of gravity (9.8 m/sec2). Given m 225 g, find (a) FN for ␪ 15° and ␪ 45° and (b) ␪ for FN 1 N and FN 2 N. 92. Heat flow on a cylindrical pipe: T ⴝ 1T0 ⴚ TR 2 sin a

FN 225

FN g

kg

5k

22

g

b ⴙ TR; y 0 2x2 ⴙ y2 When a circular pipe is exposed to a fan-driven source of heat, the temperature of the air reaching the pipe is greatest at the point nearest to the source (see diagram). As you move around the circumference of the pipe away from the source, the temperature of the air reaching the pipe gradually decreases. One possible model of this phenomenon is given by the formula shown, where T is the temperature of the air at a point (x, y) on the circumference of a pipe with outer radius r 2x2 y2, T0 is the temperature of the air at the source, and TR is the surrounding room temperature. Assuming T0 220°F, TR 72° and r 5 cm: (a) Find the temperature of the air at the points (0, 5), (3, 4), (4, 3), (4.58, 2), and (4.9, 1). (b) Why is the temperature decreasing for this sequence of points? (c) Simplify the formula using r 5 and use it to find two points on the pipe’s circumference where the temperature of the air is 113°.

g

y

Fan Heat source y

r x

Pipe x2 y 2 r 2

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Precalculus—

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709

APPLICATIONS

93. Snowcone dimensions: Made in the Shade Snowcones sells a colossal size cone that uses a conical cup holding 20 oz of ice and liquid. The cup is 20 cm tall and has a radius of 5.35 cm. Find the angle ␪ formed by a cross-section of the cup.

Exercise 93 5.35 cm

20 cm

97. Viewing angles for advertising: A 25-ft-wide billboard is erected perpendicular to a straight highway, with the closer edge 50 ft away (see figure). Assume the advertisement on the billboard is most easily read when the viewing angle is 10.5° or more. (a) Use inverse functions to find an expression for the viewing angle ␪. (b) Use a calculator to help determine the distance d (to tenths of a foot) for which the viewing angle is greater than 10.5°. (c) What distance d maximizes this viewing angle? Exercise 97

94. Avalanche conditions: Exercise 94 Winter avalanches occur for many reasons, one being the slope of the mountain. Avalanches seem to occur most often for slopes between 35° and 60° 2000 ft (snow gradually slides off steeper slopes). The slopes at a local ski resort have an 2559 ft average rise of 2000 ft for each horizontal run of 2559 ft. Is this resort prone to avalanches? Find the angle ␪ and respond. Exercise 95 95. Distance to hole: A popular story on the PGA Tour has Gerry H Yang, Tiger Woods’ teammate at Stanford and occasional caddie, using the Pythagorean 150 yd theorem to find the distance Tiger needed to reach a particular hole. Suppose you M Marker notice a marker in the B 48 yd ground stating that the straight-line distance from the marker to the hole (H) is 150 yd. If your ball B is 48 yd from the marker (M) and angle BMH is a right angle, determine the angle ␪ and your straight-line distance from the hole.

96. Ski jumps: At a waterskiing contest on a large lake, skiers use a ramp rising out of the water that is 30 ft long and 10 ft high at the high end. What angle ␪ does the ramp make with the lake?

Exercise 96

d

50 ft

25 ft

98. Viewing angles at an art show: At an art show, a painting 2.5 ft in height is hung on a wall so that its base is 1.5 ft above the eye level of an average viewer (see figure). (a) Use inverse functions to find expressions for angles ␣ and ␤. (b) Use the result to find an expression for the viewing angle ␪. (c) Use a calculator to help determine the distance x (to tenths of a foot) that maximizes this viewing angle. Exercise 98

2.5 ft

1.5 ft x

10 ft

30 ft

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7–58

CHAPTER 7 Trigonometric Identities, Inverses, and Equations 24 ft

99. Shooting angles and shots on goal: A soccer player is on a breakaway and is dribbling just inside the right sideline toward the opposing goal (see figure). As the defense closes in, she has just a few seconds to decide when to shoot. (a) Use inverse functions to find an expression for the shooting angle ␪. (b) Use a calculator to help determine the distance d (to tenths of a foot) that will maximize the shooting angle for the dimensions shown.

70 ft

(goal area) (penalty area)

␪

d

(sideline)

䊳


Consider a satellite orbiting at an altitude of x mi above Earth. The distance d from the satellite to the horizon and the length s of the corresponding arc of Earth are shown in the diagram.

100. To find the distance d we use the formula d ⫽ 22rx ⫹ x2. (a) Show how this formula was developed using the Pythagorean theorem. (b) Find a formula for the angle ␪ in terms of r and x, then a formula for the arc length s. 101. If Earth has a radius of 3960 mi and the satellite is orbiting at an altitude of 150 mi, (a) what is the measure of angle ␪? (b) how much longer is d than s?

䊳

x

d s

Earth r ␪ r Center of the Earth

MAINTAINING YOUR SKILLS 104. (4.6) Solve the inequality f 1x2 ⱕ 0 using zeroes and end-behavior given f 1x2 ⫽ x3 ⫺ 9x.

102. (7.4) Use the triangle given with a double-angle identity to find the exact value of sin12␪2. ␣

␪ √85

89

6

39

␤ 7

80

103. (7.3) Use the triangle given with a sum identity to find the exact value of sin1␣ ⫹ ␤2.

105. (1.4) In 2000, Space Tourists Inc. sold 28 low-orbit travel packages. By 2005, yearly sales of the loworbit package had grown to 105. Assuming the growth is linear, (a) find the equation that models this growth 12000 S t ⫽ 02 , (b) discuss the meaning of the slope in this context, and (c) use the equation to project the number of packages that were sold in 2010.

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Precalculus—

7.6

Solving Basic Trig Equations


A. Use a graph to gain information about principal roots, roots in [ 0, 2␲ ), and roots in ⺢ B. Use inverse functions to solve trig equations for the principal root C. Solve trig equations for roots in [ 0, 2␲ ) or [ 0, 360° ) D. Solve trig equations for roots in ⺢

In this section, we’ll take the elements of basic equation solving and use them to help solve trig equations, or equations containing trigonometric functions. All of the algebraic techniques previously used can be applied to these equations, including the properties of equality and all forms of factoring (common terms, difference of squares, etc.). As with polynomial equations, we continue to be concerned with the number of solutions as well as with the solutions themselves, but there is one major difference. There is no “algebra” that can transform a function like sin x 12 into x solution. For that we rely on the inverse trig functions from Section 7.5.

A. The Principal Root, Roots in [0, 2␲), and Real Roots In a study of polynomial equations, making a connection between the degree of an equation, its graph, and its possible roots, helped give insights as to the number, location, and nature of the roots. Similarly, keeping graphs of basic trig functions in mind helps you gain information regarding the solution(s) to trig equations. When solving trig equations, we refer to the solution found using sin1, cos1, and tan1 as the principal root. You will alternatively be asked to find (1) the principal root, (2) solutions in 3 0, 2␲2 or 3 0°, 360°2, and (3) solutions from the set of real numbers ⺢. For convenience, graphs of the basic sine, cosine, and tangent functions are repeated in Figures 7.40 through 7.42. Take a mental snapshot of them and keep them close at hand.

Figure 7.40

Figure 7.41

y 1

Figure 7.42 y

y y sin x

1

y cos x

3

y tan x

2 2

1

2

x

2

1

2

1

x 2

1

2

x

2 3

EXAMPLE 1

䊳

Visualizing Solutions Graphically Consider the equation sin x 23. Using a graph of y sin x and y 23, a. state the quadrant of the principal root. b. state the number of roots in 30, 2␲2 and their quadrants. c. comment on the number of real roots.

Solution

WORTHY OF NOTE ␲ b as 2 Quadrant I or QI, regardless of whether we’re discussing the unit circle or the graph of the function. In Example 1(b), the solutions correspond to those found in QI and QII on the unit circle, where sin x is also positive. Note that we refer to a0,

䊳

y We begin by drawing a quick sketch of 2 y sin x y sin x and y 3, noting that solutions will 1 ys occur where the graphs intersect. a. The sketch shows the principal root 2 2 x ␲ occurs between 0 and in QI. 2 1 b. For 3 0, 2␲2 we note the graphs intersect twice and there will be two solutions in this interval, one in QI and one in QII. c. Since the graphs of y sin x and y 23 extend infinitely in both directions, they will intersect an infinite number of times — but at regular intervals! Once a root is found, adding integer multiples of 2␲ (the period of sine) to this root will give the location of additional roots.

Now try Exercises 7 through 10 7–59

䊳

711

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7–60


When this process is applied to the equation tan x 2, the graph shows the principal root occurs ␲ between and 0 in QIV (see Figure 7.43). In the 2 interval 30, 2␲2 the graphs intersect twice, in QII and QIV where tan x is negative (graphically—below the x-axis). As in Example 1, the graphs continue infinitely and will intersect an infinite number of times—but again at regular intervals! Once a root is found, adding integer multiples of ␲ (the period of tangent) to this root will give the location of other roots.

A. You’ve just seen how we can use a graph to gain information about principal roots, roots in [0, 2␲), and roots in ⺢

Figure 7.43 y 3

y tan x

2 1 2

y 2

1

2

x

2 3

B. Inverse Functions and Principal Roots To solve equations having a single variable term, the basic goal is to isolate the variable term and apply the inverse function or operation. This is true for algebraic equations like 2x 1 0, 2 1x 1 0, or 2x2 1 0, and for trig equations like 2 sin x 1 0. In each case we would add 1 to both sides, divide by 2, then apply the appropriate inverse. When the inverse trig functions are applied, the result is only the principal root and other solutions may exist depending on the interval under consideration. EXAMPLE 2

䊳

Finding Principal Roots Find the principal root of 13 tan x 1 0.

Solution B. You’ve just seen how we can use inverse functions to solve trig equations for the principal root

Table 7.2

䊳

We begin by isolating the variable term, then apply the inverse function. 13 tan x 1 0

given equation

1 tan x 13 tan1 1tan x2 tan1a x

␲ 6

add 1 and divide by 13

1 b 13

apply inverse tangent to both sides

␪

sin ␪

cos ␪

0 ␲ 6

0 1 2

1 13 2

␲ 4

12 2

12 2

␲ 3 ␲ 2

13 2

1 2

1

0

2␲ 3

13 2

3␲ 4

12 2

12 2

C. Solving Trig Equations for Roots in [0, 2␲) or [0ⴗ, 360ⴗ)

5␲ 6

1 2

13 2

␲

0

To find multiple solutions to a trig equation, we simply take the reference angle of the principal root, and use this angle to find all solutions within a specified range. A mental image of the graph still guides us, and the standard table of values (also held in memory) allows for a quick solution to many equations.

1 2

1

result (exact form)


䊳

Equations like the one in Example 2 demonstrate the need to be very familiar with the functions of special angles. They are frequently used in equations and applications to ensure results don’t get so messy they obscure the main ideas. For convenience, the values of sin ␪ and cos ␪ are repeated in Table 7.2 for ␪ 僆 30, ␲ 4 . Using symmetry and the appropriate sign, the table can easily be extended to all values in 30, 2␲2 . Using the reciprocal and ratio relationships, values for the other trig functions can also be found.

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713

Section 7.6 Solving Basic Trig Equations

EXAMPLE 3

䊳

Finding Solutions in [0, 2␲)

Solution

䊳

Isolate the variable term, then apply the inverse function.

For 2 cos ␪ 12 0, find all solutions in 3 0, 2␲2. 2 cos ␪ 12 0

given equation

12 cos ␪ 2

subtract 12 and divide by 2

cos1 1cos ␪2 cos1a ␪

12 b 2

3␲ 4

apply inverse cosine to both sides

result

3␲ ␲ as the principal root, we know ␪r . 4 4 Since cos x is negative in QII and QIII, the second 5␲ . The second solution could also have solution is 4 been found from memory, recognition, or symmetry on the unit circle. Our (mental) graph verifies these are the only solutions in 30, 2␲2 .

WORTHY OF NOTE

With

Note how the graph of a trig function displays the information regarding quadrants. From the graph of y cos x we “read” that cosine is negative in QII and QIII [the lower “hump” of the graph is below the x-axis in 1␲/2, 3␲/22 ] and positive in QI and QIV [the graph is above the x-axis in the intervals 10, ␲/22 and 13␲/2, 2␲2 ].

1

2

y y cos x

2

x

√2

1

y 2

Now try Exercises 29 through 34 EXAMPLE 4

䊳

Finding Solutions in [0, 2␲)

Solution

䊳

As with the other equations having a single variable term, we try to isolate this term or attempt a solution by factoring.

䊳

For tan2x 1 0, find all solutions in 30, 2␲2 .

tan2x 1 0

given equation

2tan x 11 tan x 1 2

add 1 to both sides and take square roots result

The algebra gives tan x 1 or tan x 1 and we solve each equation independently. 1

tan

tan x 1 1tan x2 tan1 112 ␲ x 4

1

tan

tan x 1 1tan x2 tan1 112 ␲ x 4

apply inverse tangent principal roots

␲ y is in the 3 4 specified interval. With tan x positive in QI and 2 ␲ 5␲ y1 1 . While x is QIII, a second solution is 4 4 not in the interval, we still use it as a reference y 1 to identify the angles in QII and QIV (for 1 tan x 12 and find the solutions 2 7␲ 3␲ x . The four solutions are and 3 4 4 ␲ 3␲ 5␲ 7␲ x , , , which is supported by the graph shown. , and 4 4 4 4 Of the principal roots, only x

y tan x

2


x

䊳

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For any trig function that is not equal to a standard value, we can use a calculator to approximate the principal root or leave the result in exact form, and apply the same ideas to this root to find all solutions in the interval. EXAMPLE 5

䊳

Finding Solutions in [0ⴗ, 360ⴗ)

Solution

䊳

Use a u-substitution to simplify the equation and help select an appropriate strategy. For u cos ␪, the equation becomes 3u2 u 2 0 and factoring seems the best approach. The factored form is 1u 12 13u 22 0, with solutions u 1 and u 23. Re-substituting cos ␪ for u gives

Find all solutions in 30°, 360°2 for 3 cos2␪ cos ␪ 2 0.

cos ␪ 1 cos1 1cos ␪2 cos1 112 ␪ 180° Figure 7.44 1

360

180

y y cos ␪ ys 180

1

360

␪

y 1

2 3

equations from factored form

2 cos1 1cos ␪2 cos1a b 3 ␪ ⬇ 48.2°

apply inverse cosine principal roots

Both principal roots are in the specified interval. The first is quadrantal, the second was found using a calculator and is approximately 48.2°. With cos ␪ positive in QI and QIV, a second solution is 1360 48.22° 311.8°. The three solutions seen in Figure 7.44 are 48.2°, 180°, and 311.8° although only ␪ 180° is exact. With the calculator still in degree MODE , the solutions ␪ 180° and ␪ ⬇ 311.8° are verified in Figure 7.45. While we may believe the second calculation is not exactly zero due to round-off error, a more satisfactory check can be obtained by storing the result of 360 cos1 1 23 2 as X, and using 3 cos1X2 2 cos1X2 2, as shown in Figure 7.46. Figure 7.45

C. You’ve just seen how we can solve trig equations for roots in [0, 2␲) or [0, 360ⴗ)

cos ␪

Figure 7.46


䊳

D. Solving Trig Equations for All Real Roots (⺢) As we noted, the intersections of a trig function with a horizontal line occur at regular, predictable intervals. This makes finding solutions from the set of real numbers a simple matter of extending the solutions we found in 3 0, 2␲2 or 30°, 360°2. To illustrate, consider the 3␲ solutions to Example 3. For 2 cos ␪ 12 0, we found the solutions ␪ and 4 5␲ ␪ . For solutions in ⺢, we note the “predictable interval” between roots is identi4 cal to the period of the function. This means all real solutions will be represented by 5␲ 3␲ ␪ 2␲k and ␪ 2␲k, k 僆⺪ (k is an integer). Both are illustrated in 4 4 Figures 7.47 and 7.48 with the primary solution indicated with a ✸.

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Precalculus—

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715


Figure 7.47 y

f 2k

Figure 7.48

1

y

h 2k

y cos

y cos x

1

43 2

etc.

etc. 2

s 4

EXAMPLE 6

䊳

2 h

Finding Solutions in ⺢ Find all real solutions to 13 tan x 1 0.

Solution

䊳

43 2

2 3 4

etc. 2 z 4

z 4

z

2

2 f

2 h

m 4

y

y tan x

etc. In Example 2 we found the ␲ principal root was x . Since 3 6 the tangent function has a period of ␲, adding integer multiples of ␲ to this root will identify all solutions: ␲ x ␲k, k 僆⺪, as illustrated here. 6

2 3 4

etc.

2 f

3

l

k

'

m

etc. 2

3

y ⬇ 0.58

x

3


䊳

These fundamental ideas can be extended to many different situations. When asked to find all real solutions, be sure you find all roots in a stipulated interval before naming solutions by applying the period of the function. For instance, cos x 0 has two solutions ␲ 3␲ d , which we can quickly extend to find all real roots. in 30, 2␲2 c x and x 2 2 ␲ But using x cos10 or a calculator limits us to the single (principal) root x , and 2 3␲ . Note that solutions involving multiples of we’d miss all solutions stemming from 2 an angle (or fractional parts of an angle) should likewise be “handled with care,” as in Example 7. EXAMPLE 7

䊳

Finding Solutions in ⺢ Find all real solutions to 2 sin12x2 cos x cos x 0.

Solution

䊳

Since we have a common factor of cos x, we begin by rewriting the equation as cos x 32 sin12x2 1 4 0 and solve using the zero factor property. The resulting equations are cos x 0 and 2 sin12x2 1 0 S sin12x2 12. cos x 0

sin12x2

1 2

equations from factored form

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7–64


WORTHY OF NOTE When solving trig equations that involve arguments other than a single variable, a u-substitution is sometimes used. For Example 7, substituting u for 2x gives the 1 equation sin u , making it “easier 2 ␲ 1 to see” that u (since is a 6 2 special value), and therefore ␲ ␲ 2x and x . 6 12

In 3 0, 2␲2, cos x 0 has solutions x

␲ ␲ 3␲ and x , giving x 2␲k and 2 2 2

3␲ 2␲k as solutions in ⺢. Note these can actually be combined and written 2 ␲ 1 as x ␲k, k 僆⺪. For sin12x2 we know that sin u is positive in QI and 2 2 ␲ 1 ␲ QII, and the reference angle for sin u is . This yields the solutions u (QI) 2 6 6 5␲ ␲ 5␲ . Since we seek all real and u (QII), or in this case 2x and 2x 6 6 6 roots, we first extend each solution by 2␲k before dividing by 2, otherwise multiple

x

solutions would be overlooked. ␲ 2␲k 6 ␲ x ␲k 12

2x

5␲ 2␲k 6 5␲ x ␲k 12

2x

solutions from sin (2x) 12 ; k 僆⺪

divide by 2


䊳

In the process of solving trig equations, we sometimes employ fundamental identities to help simplify an equation, or to make factoring or some other method possible. EXAMPLE 8

䊳

Solving Trig Equations Using an Identity Find all real solutions for cos12x2 sin2x 3 cos x 1.

Solution

䊳

With a mixture of functions, exponents, and arguments, the equation is almost impossible to solve as it stands. But we can eliminate the sine function using the identity cos12x2 cos2x sin2x, leaving a quadratic equation in cos x. cos12x2 sin2 x 3 cos x 1 cos2x sin2x sin2x 3 cos x 1 cos2 x 3 cos x 1 cos2 x 3 cos x 1 0

given equation substitute cos2x sin2x for cos (2x) combine like terms subtract 1

Let’s substitute u for cos x to give us a simpler view of the equation. This gives u2 3u 1 0, which is clearly not factorable over the integers. Using the quadratic formula with a 1, b 3, and c 1 gives u

3 2132 2 4112112 2112 3 113 2

quadratic formula in u

simplified

To four decimal places we have u ⬇ 3.3028 and u ⬇ 0.3028. To answer in terms of the original variable we re-substitute cos x for u, realizing that cos x ⬇ 3.3028 has no solution, so solutions in 30, 2␲2 must be provided by cos x ⬇ 0.3028 and will occur in QII and QIII. The primary solutions are x cos1 10.30282 ⬇ 1.8784 and 2␲ 1.8784 ⬇ 4.4048 rounded to four decimal places, so all real solutions are given by x ⬇ 1.8784 2␲k and x ⬇ 4.4048 2␲k. Now try Exercises 67 through 82

䊳

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The TABLE feature of a calculator in radian MODE can partially (yet convincingly) verify the solutions to Example 8. On the Y= screen, enter Y1 4.4048 2␲X. This expression is then used in place of X when the left-hand side of the original equation is entered as Y2 (see Figure 7.49). After using the TBLSET screen to set TblStart 0 and ¢Tbl 1, the keystrokes 2nd GRAPH (TABLE) will produce the table shown in Figure 7.50. Note the values in the Y2 column are all very close to 1 (the right-hand side of the original equation), implying the Y1 values are solutions. Figure 7.49

Figure 7.50

D. You’ve just seen how we can solve trig equations for roots in ⺢

7.6 EXERCISES 䊳


Fill in each blank with the appropriate word or phrase. Carefully reread the section if necessary.

1. For simple equations, a mental graph will tell us the quadrant of the root, the number of roots in , and show a pattern for all roots. 12 the principal root is 2 solutions in 3 0, 2␲2 are and and an expression for all real roots is and ; k 僆⺪.

3. For sin x

5. Discuss/Explain/Illustrate why tan x

, ,

3 and 4

3 have two solutions in 3 0, 2␲2, even 4 though the period of y tan x is ␲, while the period of y cos x is 2␲.

cos x

2. Solving trig equations is similar to solving algebraic equations, in that we first the variable term, then apply the appropriate function. 4. For tan x 1, the principal root is solutions in 30, 2␲2 are and and an expression for all real roots is

, , .

1 has four solutions in 2 3 0, 2␲2 . Explain how these solutions can be viewed as the vertices of a square inscribed in the unit circle.

6. The equation sin2x

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DEVELOPING YOUR SKILLS 3 7. For the equation sin x and the graphs of 4 3 y sin x and y given, state (a) the quadrant 4 of the principal root and (b) the number of roots in 3 0, 2␲2. Exercise 7 1

2

Exercise 8

y y sin x

y y cos x

1

2

2 x

y 1

34

3

yE

2 x

1

8. For the equation cos x and the graphs of y cos x and y 34 given, state (a) the quadrant of the principal root and (b) the number of roots in 3 0, 2␲2. 3 4

9. Given the graph y tan x shown here, draw the horizontal line y 1.5 and then for tan x 1.5, state (a) the quadrant of the principal root and (b) the number of roots in 3 0, 2␲2. Exercise 9

y

3

3

2

2

1

1

2 x

1

2 3 21 2

2

2

3

3

cos ␪

0

0

1

␲ 6

1 2

␲ 4

12 2

␲ 3

13 2

␲ 2

0

␲ 2

1

3␲ 4

2␲ 3

13 2

␲

5␲ 6

1 2

␲

0

7␲ 6

tan ␪

sin ␪

sin ␪

5␲ 4

1 2

13 2

tan ␪

12 2

1

12 2

3␲ 2

0

7␲ 4

12 2

2␲

1

␲ between 0 and 4 2␲, with the values for cos ␪ given. Complete the table without a calculator or references using your knowledge of the unit circle, the signs of the trig functions in each quadrant, memory/recognition, sin ␪ tan ␪ , and so on. cos ␪

12. The table shows ␪ in multiples of

y sec x

2

␪

␪

4␲ 3

cos ␪

Exercise 12

0

Exercise 10

y y tan x

2

Exercise 11

3 2 x 2

10. Given the graph of y sec x shown, draw the horizontal line y 54 and then for sec x 54, state (a) the quadrant of the principal root and (b) the number of roots in 3 0, 2␲2.

␲ 11. The table shows ␪ in multiples of between 0 and 6 4␲ , with the values for sin ␪ given. Complete the 3 table without a calculator or references using your knowledge of the unit circle, the signs of the trig functions in each quadrant, memory/recognition, sin ␪ , and so on. tan ␪ cos ␪

Find the principal root of each equation.

13. 2 cos x 12

14. 2 sin x 1

15. 4 sin x 2 12

16. 4 cos x 213

17. 13 tan x 1

18. 2 13 tan x 2

19. 2 13 sin x 3

20. 312 csc x 6

21. 6 cos x 6

22. 4 sec x 8

7 7 cos x 8 16 25. 2 4 sin ␪

5 5 24. sin x 3 6 26. ␲ tan x 0

27. 513 10 cos ␪

28. 4 13 4 tan ␪

23.

Find all solutions in [0, 2␲).

29. 9 sin x 3.5 1

30. 6.2 cos x 4 7.1

31. 8 tan x 7 13 13 32.

1 3 7 sec x 2 4 4

33.

2 5 3 cot x 3 6 2

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34. 110 sin x 55 13 35. 4 cos2x 3

63. 12 cos x sin12x2 3 cos x 0

36. 4 sin x 1

37. 7 tan x 21

64. 13 sin x tan12x2 sin x 0

38. 3 sec x 6

39. 4 csc x 8

40. 6 13 cos x 313

41. 4 12 sin x 4 12

2

2

2

42.

2

2

2

2 4 5 cos2x 3 6 3

Solve the following equations by factoring. State all solutions in [0°, 360°). Round to one decimal place if the result is not a standard value.

65. cos13x2 csc 12x2 2 cos13x2 0

66. 13 sin12x2 sec12x2 2 sin12x2 0 Solve each equation using a calculator and inverse trig functions to determine the principal root (not by graphing). Clearly state (a) the principal root and (b) all real roots.

67. 3 cos x 1

68. 5 sin x 2

43. 3 cos2␪ 14 cos ␪ 5 0

69. 12 sec x 3 7

70. 13 csc x 2 11

44. 6 tan2␪ 213 tan ␪ 0

71.

45. 2 cos x sin x cos x 0 46. 2 sin2x 7 sin x 4 47. sec2x 6 sec x 16 48. 2 cos3x cos2x 0 49. 4 sin2x 1 0 50. 4 cos x 3 0 2

Find all real solutions. Note that identities are not required to solve these exercises.

1 1 sin12␪2 2 3

72.

2 1 cos12␪2 5 4

73. 5 cos12␪2 1 0 74. 6 sin12␪2 3 2 Solve the following equations using an identity. State all real solutions in radians using the exact form where possible and rounded to four decimal places if the result is not a standard value.

75. cos2x sin2x

1 2

51. 2 sin x 12

52. 2 cos x 1

53. 4 cos x 212

54. 4 sin x 2 13

76. 4 sin2x 4 cos2x 2 13

55. 13 tan x 13

56. 213 tan x 2

57. 6 cos12x2 3

58. 2 sin13x2 12

1 1 77. 2 cos a xb cos x 2 sin a xb sin x 1 2 2

59. 13 tan12x2 13 60. 2 13 tan13x2 6 1 61. 2 13 cos a xb 2 13 3 1 62. 8 sin a xb 4 13 2 䊳

719

78. 12 sin12x2cos13x2 12 sin13x2cos12x2 1 79. 1cos ␪ sin ␪2 2 1 80. 1cos ␪ sin ␪2 2 2

81. cos12␪2 2 sin2␪ 3 sin ␪ 0 82. 3 sin12␪2 cos2 12␪2 1 0

WORKING WITH FORMULAS 5 2 v sin12␪2 49 The distance a projectile travels is called its range and is modeled by the formula shown, where R is the range in meters, v is the initial velocity in meters per second, and ␪ is the angle of release. Two friends are standing 16 m apart playing catch. If the first throw has an initial velocity of 15 m/sec, what two angles will ensure the ball travels the 16 m between the friends?

83. Range of a projectile: R ⴝ

84. Fine-tuning a golf swing: (club head to shoulder)2 ⴝ (club length)2 ⴙ (arm length)2 ⴚ 2 (club length)(arm length)cos ␪ A golf pro is taking specific measurements on a client’s swing to help improve her game. If the angle ␪ is too small, the ball is hit late and “too thin” (you top the ball). If ␪ is too large, the ball is hit early and “too fat” (you scoop the ball). Approximate the angle ␪ formed by the club and the extended (left) arm using the given measurements and formula shown.

Exercise 84 37 in.

39 in.

27 in.

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APPLICATIONS

Acceleration due to gravity: When a steel ball is

released down an inclined plane, the rate of the ball’s acceleration depends on the angle of incline. The acceleration can be approximated by the formula A1␪2 9.8 sin ␪, where ␪ is in degrees and the acceleration is measured in meters per second/per second. To the nearest tenth of a degree, 85. What angle produces an acceleration of 0 m/sec2 when the ball is released? Explain why this is reasonable. 86. What angle produces an acceleration of 9.8 m/sec2? What does this tell you about the acceleration due to gravity? 87. What angle produces an acceleration of 5 m/sec2? Will the angle be larger or smaller for an acceleration of 4.5 m/sec2? 88. Will an angle producing an acceleration of 2.5 m/sec2 be one-half the angle required for an acceleration of 5 m/sec2? Explore and discuss. Snell’s law states that Exercises 89 to 92 when a ray of light light incidence reflection passes from one medium into another, the sine of the angle of incidence ␣ varies directly with the sine of refraction new medium the angle of refraction ␤ sin() k sin() (see the figure). This phenomenon is modeled by the formula sin ␣ k sin ␤, where k is called the index of refraction. Note the angle ␪ is the angle at which the light strikes the surface, so that ␣ 90° ␪. Use this information to work Exercises 89 to 92. 89. A ray of light passes from air into water, striking the water at an angle of 55°. Find the angle of incidence ␣ and the angle of refraction ␤, if the index of refraction for water is k 1.33. 90. A ray of light passes from air into a diamond, striking the surface at an angle of 75°. Find the angle of incidence ␣ and the angle of refraction ␤, if the index of refraction for a diamond is k 2.42. 91. Find the index of refraction for ethyl alcohol if a beam of light strikes the surface of this medium at an angle of 40° and produces an angle of refraction ␤ 34.3°. Use this index to find the angle of incidence if a second beam of light created an angle of refraction measuring 15°.

92. Find the index of refraction for rutile (a type of mineral) if a beam of light strikes the surface of this medium at an angle of 30° and produces an angle of refraction ␤ 18.7°. Use this index to find the angle of incidence if a second beam of light created an angle of refraction measuring 10°. 93. Roller coaster design: As part of a science fair project, Hadra builds a scale model of a roller

Loading platform

1 coaster using the equation y 5 sin a xb 7, 2 where y is the height of the model in inches and x is the distance from the “loading platform” in inches. (a) How high is the platform? (b) What distances from the platform does the model attain a height of 9.5 in.? 94. Company logo: Part of the logo for an engineering firm was modeled by a cosine function. The logo was then manufactured in steel and installed on the entrance marquee of the home office. The position and size of the logo is modeled by the function y 9 cos x 15, where y is the height of the graph above the base of the marquee in inches and x represents the distance from the edge of the marquee. Assume the graph begins flush with the edge. (a) How far above the base is the beginning of the cosine graph? (b) What distances from the edge does the graph attain a height of 19.5 in.?

International Engineering

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Section 7.7 General Trig Equations and Applications


95. Find all real solutions to 5 cos x x x in two ways. First use a calculator with Y1 5 cos X X and Y2 X to determine the regular intervals between points of intersection. Second, simplify by adding x to both sides, and draw a quick sketch of the result to locate x-intercepts. Explain why both methods give the same result, even though the first presents you with a very different graph.

䊳

721

96. Once the fundamental ideas of solving a given family of equations are understood and practiced, a student usually begins to generalize them—making the numbers or symbols used in the equation irrelevant. (a) Use the inverse sine function to find the principal root of y A sin1Bx C2 D, by solving for x in terms of y, A, B, C, and D. (b) Solve the following equation using the techniques addressed in this section, and then using the “formula” from part (a): ␲ 1 5 2 sin a x b 3. Do the results agree? 2 4


97. (3.1) Use a substitution to show that x 2 i is a zero of f 1x2 x2 4x 5. 98. (3.3) Currently, tickets to productions of the Shakespeare Community Theater cost $10.00, with an average attendance of 250 people. Due to market research, the theater director believes that for each $0.50 reduction in price, 25 more people will attend. What ticket price will maximize the theater’s revenue? What will the average attendance projected to become at that price?

7.7


A. Use additional algebraic

C. D. E.

100. (6.1) The largest Ferris wheel in the world, located in Yokohama, Japan, has a radius of 50 m. To the nearest hundredth of a meter, how far does a seat on the rim travel as the wheel turns through ␪ 292.5°?

General Trig Equations and Applications

LEARNING OBJECTIVES

B.

99. (7.5) Evaluate without using a calculator: 1 a. tan c sin1a b d b. sin 3tan1 112 4 2

techniques to solve trig equations Solve trig equations using multiple angle, sum and difference, and sum-toproduct identities Solve trig equations using graphing technology Solve trig equations of the form A sin(Bx C ) D k Use a combination of skills to model and solve a variety of applications

At this point you’re likely beginning to understand the true value of trigonometry to the scientific world. Essentially, any phenomenon that is cyclic or periodic is beyond the reach of polynomial (and other) functions, and may require trig for an accurate understanding. And while there is an abundance of trig applications in oceanography, astronomy, meteorology, geology, zoology, and engineering, their value is not limited to the hard sciences. There are also rich applications in business and economics, and a growing number of modern artists are creating works based on attributes of the trig functions. In this section, we try to place some of these applications within your reach, with the Exercise Set offering an appealing variety from many of these fields.

A. Trig Equations and Algebraic Methods We begin this section with a follow-up to Section 7.6, by introducing trig equations that require slightly more sophisticated methods to work out a solution.

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EXAMPLE 1

䊳

Solving a Trig Equation by Squaring Both Sides

Solution

䊳

Our first instinct might be to rewrite the equation in terms of sine and cosine, but that simply leads to a similar equation that still has two different functions 3 13 cos x ⫺ sin x ⫽ 1 4. Instead, we square both sides and see if the Pythagorean identity 1 ⫹ tan2x ⫽ sec2x will be of use. Prior to squaring, we separate the functions on opposite sides to avoid the mixed term 2 tan x sec x. given equation sec x ⫹ tan x ⫽ 13 2 2 subtract tan x and square 1sec x2 ⫽ 1 13 ⫺ tan x2 sec2x ⫽ 3 ⫺ 2 13 tan x ⫹ tan2x result

Find all solutions in 30, 2␲2: sec x ⫹ tan x ⫽ 13.

Since sec2x ⫽ 1 ⫹ tan2x, we substitute directly and obtain an equation in tangent alone. substitute 1 ⫹ tan2x for sec2x 1 ⫹ tan2x ⫽ 3 ⫺ 2 13 tan x ⫹ tan2x simplify ⫺2 ⫽ ⫺213 tan x 1 ⫽ tan x solve for tan x 13 tan x 7 0 in QI and QIII 7␲ ␲ [QI] and [QIII]. Since squaring an equation 6 6 sometimes introduces extraneous roots, both should be checked in the original equation. ␲ The check shows only x ⫽ is a solution. 6 The proposed solutions are x ⫽


䊳

Here is one additional example that uses a factoring strategy commonly employed when an equation has more than three terms. EXAMPLE 2

䊳

Solving a Trig Equation by Factoring

Solution

䊳

The four terms in the equation share no common factors, so we attempt to factor by grouping. We could factor 2 cos ␪ from the first two terms but instead elect to group the sin2␪ terms and begin there.

Find all solutions in 30°, 360°2: 8 sin2␪ cos ␪ ⫺ 2 cos ␪ ⫺ 4 sin2␪ ⫹ 1 ⫽ 0.

8 sin2␪ cos ␪ ⫺ 2 cos ␪ ⫺ 4 sin2␪ ⫹ 1 18 sin2␪ cos ␪ ⫺ 4 sin2 ␪2 ⫺ 12 cos ␪ ⫺ 12 4 sin2 ␪12 cos ␪ ⫺ 12 ⫺ 112 cos ␪ ⫺ 12 12 cos ␪ ⫺ 1214 sin2␪ ⫺ 12

⫽0 ⫽0 ⫽0 ⫽0

given equation rearrange and group terms remove common factors remove common binomial factors

Using the zero factor property, we write two equations and solve each independently. resulting equations 2 cos ␪ ⫺ 1 ⫽ 0 4 sin2 ␪ ⫺ 1 ⫽ 0 1 2 cos ␪ ⫽ 1 isolate variable term sin2␪ ⫽ 4 1 1 cos ␪ ⫽ solve sin ␪ ⫽ ⫾ 2 2 cos ␪ 7 0 in QI and QIV sin ␪ 7 0 in QI and QII ␪ ⫽ 60°, 300° sin ␪ 6 0 in QIII and QIV ␪ ⫽ 30°, 150°, 210°, 330° solutions

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Initially factoring 2 cos ␪ from the first two terms and proceeding from there would have produced the same result.

A. You’ve just seen how we can use additional algebraic techniques to solve trig equations


䊳

B. Solving Trig Equations Using Various Identities To solve equations effectively, a student should strive to develop all of the necessary “tools.” Certainly the underlying concepts and graphical connections are of primary importance, as are the related algebraic skills. But to solve trig equations effectively we must also have a ready command of commonly used identities. Observe how Example 3 combines a double-angle identity with factoring by grouping. EXAMPLE 3

䊳

Using Identities and Algebra to Solve a Trig Equation

Solution

䊳

Noting that one of the terms involves a double angle, we attempt to replace that term to make factoring a possibility. Using the double identity for sine, we have

Find all solutions in 3 0, 2␲2: 3 sin12x2 ⫹ 2 sin x ⫺ 3 cos x ⫽ 1. Round solutions to four decimal places as necessary.

312 sin x cos x2 ⫹ 2 sin x ⫺ 3 cos x ⫽ 1 16 sin x cos x ⫹ 2 sin x2 ⫺ 13 cos x ⫹ 12 ⫽ 0 2 sin x13 cos x ⫹ 12 ⫺ 113 cos x ⫹ 12 ⫽ 0 13 cos x ⫹ 1212 sin x ⫺ 12 ⫽ 0

substitute 2 sin x cos x for sin (2x) set equal zero and group terms factor using 3 cos x ⫹ 1 common binomial factor

Use the zero factor property to solve each equation independently. 3 cos x ⫹ 1 ⫽ 0 1 cos x ⫽ ⫺ 3 cos x 6 0 in QII and QIII x ⬇ 1.9106, 4.3726

B. You’ve just seen how we can solve trig equations using various identities

2 sin x ⫺ 1 ⫽ 0 resulting equations 1 isolate variable term sin x ⫽ 2 sin x 7 0 in QI and QII ␲ 5␲ solutions x⫽ , 6 6

Should you prefer the exact form, the solutions from the cosine equation could be 1 1 written as x ⫽ cos⫺1a⫺ b and x ⫽ 2␲ ⫺ cos⫺1a⫺ b. 3 3 Now try Exercises 17 through 26

䊳

C. Trig Equations and Graphing Technology A majority of the trig equations you’ll encounter in your studies can be solved using the ideas and methods presented both here and in the previous section. But there are some equations that cannot be solved using standard methods because they mix polynomial functions (linear, quadratic, and so on) that can be solved using algebraic methods, with what are called transcendental functions (trigonometric, logarithmic, and so on). By definition, transcendental functions are those that transcend the reach of standard algebraic methods. These kinds of equations serve to highlight the value of graphing and calculating technology to today’s problem solvers.

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EXAMPLE 4

䊳

Solving Trig Equations Using Technology Use a graphing calculator in radian mode to find all real roots of 3x 2 sin x ⫹ ⫺ 2 ⫽ 0. Round solutions to four decimal places. 5

Solution

䊳

When using graphing technology our initial concern is the size of the viewing window. After carefully entering the equation on the Y= screen, we note the term 2 sin x will never be larger than 2 or less than ⫺2 for any real number x. On the other 3x hand, the term becomes larger for larger values of x, which would seem to cause 5 3x 2 sin x ⫹ to “grow” as x gets larger. We conclude the standard window is a good 5 place to start, and the resulting graph is shown in Figure 7.51. Figure 7.51

Figure 7.52

10

⫺10

10

10

⫺10

0

20

⫺10

From this screen it appears there are three real roots, but to be sure none are hidden to the right, we extend the Xmax value to 20 (Figure 7.52). Using 2nd TRACE (CALC) 2:zero, we follow the prompts and enter a left bound of 0 (a number to the left of the zero) and a right bound of 2 (a number to the right of the zero—see Figure 7.52). The calculator then prompts you for a GUESS, which you can bypass by pressing . The smallest root is approximately x ⬇ 0.8435. Repeating this sequence we find the other roots are x ⬇ 3.0593 and x ⬇ 5.5541. ENTER


䊳

Some equations are very difficult to solve analytically, and even with the use of a graphing calculator, a strong combination of analytical skills with technical skills is 1 1 required to state the solution set. Consider the equation 5 sin a xb ⫹ 5 ⫽ cot a xb and 2 2 solutions in 3⫺2␲, 2␲2 . There appears to be no quick analytical solution, and the first 1 attempt at a graphical solution holds some hidden surprises. Enter Y1 ⫽ 5 sin a Xb ⫹ 5 2 1 and Y2 ⫽ on the Y= screen. Pressing ZOOM 7:ZTrig gives the screen in Figtan1 12X2 Figure 7.53 ure 7.53, where we note there are at least two 4 and possibly three solutions, depending on how the sine graph intersects the cotangent graph. We are also uncertain as to whether the graphs ␲ ␲ intersect again between ⫺ and . Increasing ⫺2␲ 2␲ 2 2 the maximum Y-value to Ymax ⫽ 8 shows they do indeed. But once again, are there now three or four solutions? In situations like this it may ⫺4

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be helpful to use the Zeroes Method for solving graphically. On the Y= screen, disable Y1 and Y2 and enter Y3 as Y1 ⫺ Y2. Pressing ZOOM 7:ZTrig at this point clearly shows that there are four solutions in this interval (Figure 7.54), which can easily be found using 2nd TRACE (CALC) 2:zero: x ⬇ ⫺5.7543, ⫺4.0094, ⫺3.1416, and 0.3390. See Exercises 33 and 34 for more practice with these ideas.

C. You’ve just seen how we can solve trig equations using graphing technology

Figure 7.54 4

⫺2␲

2␲

⫺4

D. Solving Equations of the Form A sin (Bx ⴞ C) ⴞ D ⴝ k You may remember equations of this form from Section 6.5. They actually occur quite frequently in the investigation of many natural phenomena and in the modeling of data from a periodic or seasonal context. Solving these equations requires a good combination of algebra skills with the fundamentals of trig. EXAMPLE 5

䊳

Solving Equations That Involve Transformations ␲ ␲ Given f 1x2 ⫽ 160 sin a x ⫹ b ⫹ 320 and x 僆 3 0, 2␲2 , for what real numbers x 3 3 is f(x) less than 240?

Algebraic Solution

䊳

We reason that to find values where f 1x2 6 240, we should begin by finding values where f 1x2 ⫽ 240. The result is ␲ ␲ 160 sin a x ⫹ b ⫹ 320 ⫽ 240 3 3 ␲ ␲ sin a x ⫹ b ⫽ ⫺0.5 3 3

equation

subtract 320 and divide by 160; isolate variable term

␲ ␲ ␲ At this point we elect to use a u-substitution for a x ⫹ b ⫽ 1x ⫹ 12 to obtain 3 3 3 a “clearer view.” ␲ sin u ⫽ ⫺0.5 substitute u for 1x ⫹ 12 3 sin u 6 0 in QIII and QIV 7␲ 11␲ u⫽ solutions in u u⫽ 6 6 ␲ 1x ⫹ 12 for u and solve. 3 ␲ 11␲ ␲ 1x ⫹ 12 ⫽ re-substitute 1x ⫹ 12 for u 3 3 6 11 3 x⫹1⫽ multiply both sides by ␲ 2 solutions x ⫽ 4.5

To complete the solution we re-substitute ␲ 7␲ 1x ⫹ 12 ⫽ 3 6 7 x⫹1⫽ 2 x ⫽ 2.5

We now know f 1x2 ⫽ 240 when x ⫽ 2.5 and x ⫽ 4.5 but when will f(x) be less than 2␲ 240? By analyzing the equation, we find the function has period of P ⫽ ␲ ⫽ 6 3

and is shifted to the left 1 units. This would indicate the graph peaks early in the interval 30, 2␲2 with a “valley” in the interior. We conclude f 1x2 6 240 in the interval (2.5, 4.5).

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Graphical Solution

䊳

␲ ␲ , enter Y1 ⫽ 160 sin a X ⫹ b ⫹ 320 and 3 3 Y2 ⫽ 240 on the Y= screen. To set an appropriate window, note the amplitude of Y1 is 160 and that the graph has been vertically shifted up 320 units. With the x-values ranging from 0 to 2␲, the 2nd TRACE (CALC) 5:intersect feature determines (2.5, 240) is a point of intersection of the two graphs (see Figure 7.55). In Figure 7.56, we find a second point of intersection is (4.5, 240). Observing that the graph of Y1 falls below 240 between these two points, we conclude that in the interval [0, 2␲), f 1x2 6 240 for x (2.5, 4.5). With the calculator in radian

MODE

Figure 7.55

Figure 7.56 600

600

0

2␲

0

0

0

D. You’ve just seen how we can solve trig equations of the form A sin(Bx C) D k

2␲


䊳

There is a mixed variety of equation types in Exercises 39 through 48.

E. Applications Using Trigonometric Equations Using characteristics of the trig functions, we can often generalize and extend many of the formulas that are familiar to you. For example, the formulas for the volume of a right circular cylinder and a right circular cone are well known, but what about the volume of a nonright figure (see Figure 7.57)? Here, trigonometry provides the answer, as the most general volume formula is V ⫽ V0 sin ␪, where V0 is a “standard” volume formula and ␪ is the complement of angle of deflection (see Exercises 51 and 52). As for other applications, consider the following from the environmental sciences. Natural scientists are very interested in the discharge rate of major rivers, as this gives an indication of rainfall over the inland area served by the river. In addition, the discharge rate has a large impact on the freshwater and saltwater zones found at the river’s estuary (where it empties into the sea).

Figure 7.57 ␣ h ␪

EXAMPLE 6

䊳

Solving an Equation Modeling the Discharge Rate of a River For May through November, the discharge rate of the Ganges River (Bangladesh) ␲ 2␲ b ⫹ 17,760 where t ⫽ 1 represents can be modeled by D1t2 ⫽ 16,580 sin a t ⫺ 3 3 May 1, and D(t) is the discharge rate in m3/sec. Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

a. What is the discharge rate in mid-October? b. For what months (within this interval) is the discharge rate over 26,050 m3/sec?

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Solution

䊳

727

a. To find the discharge rate in mid-October we simply evaluate the function at t ⫽ 6.5: ␲ 2␲ given function b ⫹ 17,760 D1t2 ⫽ 16,580 sin a t ⫺ 3 3 2␲ ␲ d ⫹ 17,760 substitute 6.5 for t D16.52 ⫽ 16,580 sin c 16.52 ⫺ 3 3 ⫽ 1180 compute result on a calculator In mid-October the discharge rate is 1180 m3/sec. b. We first find when the rate is equal to 26,050 m3/sec: D1t2 ⫽ 26,050. ␲ 2␲ substitute 26,050 for D(t) b ⫹ 17,760 26,050 ⫽ 16,580 sin a t ⫺ 3 3 ␲ 2␲ b 0.5 ⫽ sin a t ⫺ subtract 17,760; divide by 16,580 3 3 2␲ ␲ Using a u-substitution for a t ⫺ b we obtain the equation 3 3 0.5 ⫽ sin u sin u 7 0 in QI and QII ␲ 5␲ solutions in u u⫽ u⫽ 6 6 2␲ ␲ ␲ To complete the solution we re-substitute t ⫺ ⫽ 1t ⫺ 22 for u and solve. 3 3 3 ␲ ␲ ␲ 5␲ ␲ re-substitute 1t ⫺ 22 for u 1t ⫺ 22 ⫽ 1t ⫺ 22 ⫽ 3 3 6 3 6 3 t ⫺ 2 ⫽ 0.5 t ⫺ 2 ⫽ 2.5 multiply both sides by ␲ t ⫽ 2.5 t ⫽ 4.5 solutions The Ganges River will have a flow rate of over 26,050 m3/sec between mid-June (2.5) and mid-August (4.5). Now try Exercises 53 through 56

E. You’ve just seen how we can use a combination of skills to model and solve a variety of applications

To obtain a graphical solution to Example 6(b), ␲ 2␲ b ⫹ 17,760 on enter Y1 ⫽ 16,580 sin a X ⫺ 3 3 the Y= screen, then Y2 ⫽ 26,050. Set the window as shown in Figure 7.58 and locate the points of intersection. The graphs verify that in the interval [1, 8], D1t2 7 26,050 for t 僆 (2.5, 4.5). There is a variety of additional exercises in the Exercise Set. See Exercises 57 through 64.

䊳

Figure 7.58 40,000

0

8

0

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7.7 EXERCISES 䊳



1. The three Pythagorean identities are , and .

䊳

,

2. One strategy to solve equations with four terms and no common factors is by .

3. When an equation contains two functions from a Pythagorean identity, sometimes both sides will lead to a solution.

4. To combine two sine or cosine terms with different arguments, we can use the to formulas.

5. Regarding Example 4, discuss/explain the 3 relationship between the line y ⫽ x ⫺ 2 and the 5 graph shown in Figure 7.52.

6. Regarding Example 6, discuss/explain how to determine the months of the year the discharge rate is under 26,050 m3/sec, using the solution set given.


Solve each equation in [0, 2␲) using the method indicated. Round nonstandard values to four decimal places.

x x 21. 2 sin2 a b ⫺ 3 cos a b ⫽ 0 2 2

• Squaring both sides

x x 22. 2 cos2 a b ⫹ 3 sin a b ⫺ 3 ⫽ 0 3 3

16 2

8. cot x ⫺ csc x ⫽ 13

9. tan x ⫺ sec x ⫽ ⫺1

10. sin x ⫹ cos x ⫽ 12

7. sin x ⫹ cos x ⫽

4 11. cos x ⫹ sin x ⫽ 3

12. sec x ⫹ tan x ⫽ 2

13. cot x csc x ⫺ 2 cot x ⫺ csc x ⫹ 2 ⫽ 0 14. 4 sin x cos x ⫺ 213 sin x ⫺ 2 cos x ⫹ 13 ⫽ 0 15. 3 tan2x cos x ⫺ 3 cos x ⫹ 2 ⫽ 2 tan2x 16. 413 sin2x sec x ⫺ 13 sec x ⫹ 2 ⫽ 8 sin2x

25. sec4x ⫺ 2 sec2x tan2x ⫹ tan4x ⫽ tan2x

Find all roots in [0, 2␲) using a graphing calculator. State answers in radians rounded to four decimal places.

27. 5 cos x ⫺ x ⫽ 3

29. cos 12x2 ⫹ x ⫽ 3 2

31. x2 ⫹ sin12x2 ⫽ 1

• Using identities

1 ⫹ cot x ⫽2 cot2x

24. sin17x2cos14x2 ⫹ sin15x2 ⫺ cos17x2sin14x2 ⫹ cos x ⫽ 0 26. tan4x ⫺ 2 sec2x tan2x ⫹ sec4x ⫽ cot2x

• Factor by grouping

2

17.

23. cos13x2 ⫹ cos15x2cos12x2 ⫹ sin15x2sin12x2 ⫺ 1 ⫽ 0

4 1 ⫹ tan x ⫽ 3 tan2x 2

18.

19. 3 cos12x2 ⫹ 7 sin x ⫺ 5 ⫽ 0 20. 3 cos12x2 ⫺ cos x ⫹ 1 ⫽ 0

28. 3 sin x ⫹ x ⫽ 4

30. sin2 12x2 ⫹ 2x ⫽ 1

32. cos12x2 ⫺ x2 ⫽ ⫺5

33. 11 ⫹ sin x2 2 ⫹ cos 12x2 ⫽ 4 cos x11 ⫹ sin x2 x 34. 4 sin x ⫽ 2 cos2 a b 2

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State the period P of each function and find all solutions in [0, P). Round to four decimal places as needed.

41.

␲ ␲ 35. 250 sin a x ⫹ b ⫺ 125 ⫽ 0 6 3

13 1 ⫺ cos2x ⫽ 2 tan2x

42. 5 csc2x ⫺ 5 cot x ⫺ 5 ⫽ 0

␲ ␲ 36. ⫺75 12 sec a x ⫹ b ⫹ 150 ⫽ 0 4 6 37. 1235 cos a

729

␲ ␲ x ⫺ b ⫹ 772 ⫽ 1750 12 4

␲ ␲ 38. ⫺0.075 sin a x ⫹ b ⫺ 0.023 ⫽ ⫺0.068 2 3 Solve each equation in [0, 2␲) using any appropriate method. Round nonstandard values to four decimal places.

43. csc x ⫹ cot x ⫽ 1 44.

12 1 ⫺ sin2x ⫽ 2 2 cot x

45. sec x cos a 46. sin a

␲ ⫺ xb csc x ⫽ 13 2

47. sec2x tan a 48. 2 tan a

12 39. cos x ⫺ sin x ⫽ 2

␲ ⫺ xb ⫽ ⫺1 2

␲ ⫺ xb ⫽ 4 2

␲ 13 ⫺ xb sin2x ⫽ 2 2

40. 5 sec2x ⫺ 2 tan x ⫺ 8 ⫽ 0

䊳

WORKING WITH FORMULAS D x cos ␪ sin ␪ The trigonometric form of a linear equation is given by the formula shown, where D is the perpendicular distance from the origin to the line and ␪ is the angle between the perpendicular segment and the x-axis. For each pair of perpendicular lines given, (a) find the point (a, b) of their intersection; (b) compute the distance D ⫽ 2a2 ⫹ b2 and the b angle ␪ ⫽ tan⫺1a b, and give the equation of the line L1 in trigonometric form; and (c) a use the GRAPH or the TABLE feature of a graphing calculator to verify that both equations name the same line. 1 13 4 13 x⫹ I. L1: y ⫽ ⫺x ⫹ 5 II. L1: y ⫽ ⫺ x ⫹ 5 III. L1: y ⫽ ⫺ 2 3 3

49. The equation of a line in trigonometric form: y

L2: y ⫽ x

L2: y ⫽ 2x

Exercise 49 y

D ␪

L2: y ⫽ 13x

50. Rewriting y a cos x b sin x as a single function: y k sin(x ␪) Linear terms of sine and cosine can be rewritten as a single function using the formula shown, where a k ⫽ 2a2 ⫹ b2 and ␪ ⫽ sin⫺1a b. Rewrite the equations given using these relationships and verify they are k equivalent using the GRAPH or the TABLE feature of a graphing calculator: a. y ⫽ 2 cos x ⫹ 2 13 sin x b. y ⫽ 4 cos x ⫹ 3 sin x The ability to rewrite a trigonometric equation in simpler form has a tremendous number of applications in graphing, equation solving, working with identities, and solving applications.

x

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APPLICATIONS

51. Volume of a cylinder: The Exercise 51 volume of a cylinder is given by the formula V ⫽ ␲r2h sin ␪, where ␣ r is the radius and h is the height of the cylinder, and ␪ is the h indicated complement of the angle of deflection ␣. Note that when ␲ ␪ ␪ ⫽ , the formula becomes that 2 ␲ of a right circular cylinder (if ␪ ⫽ , then h is called 2 the slant height or lateral height of the cylinder). An old farm silo is built in the form of a right circular cylinder with a radius of 10 ft and a height of 25 ft. After an earthquake, the silo became tilted with an angle of deflection ␣ ⫽ 5°. (a) Find the volume of the silo before the earthquake. (b) Find the volume of the silo after the earthquake. (c) What angle ␪ is required to bring the original volume of the silo down 2%? 52. Volume of a cone: The Exercise 52 volume of a cone is given by 1 the formula V ⫽ ␲r2h sin ␪, 3 where r is the radius and h is the height of the cone, and ␪ h is the indicated complement ␣ of the angle of deflection ␣. ␲ Note that when ␪ ⫽ , the 2 ␪ formula becomes that of a ␲ right circular cone (if ␪ ⫽ , 2 then h is called the slant height or lateral height of the cone). As part of a sculpture exhibit, an artist is constructing three such structures each with a radius of 2 m and a slant height of 3 m. (a) Find the volume of the sculptures if the angle of deflection is ␣ ⫽ 15°. (b) What angle ␪ was used if the volume of each sculpture is 12 m3? 53. River discharge rate: For June through February, the discharge rate of the La Corcovada River (Venezuela) can be modeled by the function ␲ 9 D1t2 ⫽ 36 sin a t ⫺ b ⫹ 44, where t represents 4 4 the months of the year with t ⫽ 1 corresponding to June, and D(t) is the discharge rate in cubic meters per second. (a) What is the discharge rate in midSeptember 1t ⫽ 4.52 ? (b) For what months of the year is the discharge rate over 50 m3/sec? Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

54. River discharge rate: For February through June, the average monthly discharge of the Point Wolfe River (Canada) can be modeled by the function ␲ D1t2 ⫽ 4.6 sin a t ⫹ 3b ⫹ 7.4, where t represents 2 the months of the year with t ⫽ 1 corresponding to February, and D(t) is the discharge rate in cubic meters/second. (a) What is the discharge rate in mid-March 1t ⫽ 2.52 ? (b) For what months of the year is the discharge rate less than 7.5 m3/sec? Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

55. Seasonal sales: Hank’s Heating Oil is a very seasonal enterprise, with sales in the winter far exceeding sales in the summer. Monthly sales for the company can be modeled by ␲ ␲ S1x2 ⫽ 1600 cos a x ⫺ b ⫹ 5100, where S(x) is 6 12 the average sales in month x 1x ⫽ 1 S January2. (a) What is the average sales amount for July? (b) For what months of the year are sales less than $4000? 56. Seasonal income: As a roofing company employee, Mark’s income fluctuates with the seasons and the availability of work. For the past several years his average monthly income could be approximated by ␲ ␲ the function I1m2 ⫽ 2100 sin a m ⫺ b ⫹ 3520, 6 2 where I(m) represents income in month m 1m ⫽ 1 S January2. (a) What is Mark’s average monthly income in October? (b) For what months of the year is his average monthly income over $4500? 57. Seasonal ice thickness: The average thickness of the ice covering an arctic lake can be modeled by ␲ the function T1x2 ⫽ 9 cos a xb ⫹ 15, where 6 T(x) is the average thickness in month x 1x ⫽ 1 S January2. (a) How thick is the ice in mid-March? (b) For what months of the year is the ice at most 10.5 in. thick? 58. Seasonal temperatures: The function ␲ ␲ T1x2 ⫽ 19 sin a x ⫺ b ⫹ 53 6 2 models the average monthly temperature of the water in a mountain stream, where T(x) is the temperature 1°F2 of the water in month x 1x ⫽ 1 S January2 . (a) What is the temperature of the water in October? (b) What two months are most likely to give a temperature reading of 62°F? (c) For what months of the year is the temperature below 50°F?

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59. Coffee sales: Coffee sales fluctuate with the weather, with a great deal more coffee sold in the winter than in the summer. For Joe’s Diner, assume 2␲ ␲ x ⫹ b ⫹ 29 the function G1x2 ⫽ 21 cos a 365 2 models daily coffee sales (for non-leap years), where G(x) is the number of gallons sold and x represents the days of the year 1x ⫽ 1 S January 12. (a) How many gallons are projected to be sold on March 21? (b) For what days of the year are more than 40 gal of coffee sold? 60. Park attendance: Attendance at a popular state park varies with the weather, with a great deal more visitors coming in during the summer months. Assume daily attendance at the park can be modeled by the function 2␲ V1x2 ⫽ 437 cos a x ⫺ ␲b ⫹ 545 (for non-leap 365 years), where V(x) gives the number of visitors on day x 1x ⫽ 1 S January 12 . (a) Approximately how many people visited the park on November 1 111 ⫻ 30.5 ⫽ 335.52? (b) For what days of the year are there more than 900 visitors? 61. Exercise routine: As part of his yearly physical, Manu Tuiosamoa’s heart rate is closely monitored during a 12-min, cardiovascular exercise routine. His heart rate in beats per minute (bpm) is modeled ␲ by the function B1x2 ⫽ 58 cos a x ⫹ ␲b ⫹ 126 6 where x represents the duration of the workout in minutes. (a) What was his resting heart rate? (b) What was his heart rate 5 min into the workout? (c) At what times during the workout was his heart rate over 170 bpm?

䊳

731


62. Exercise routine: As part of her workout routine, Sara Lee programs her treadmill to begin at a slight initial grade (angle of incline), gradually increase to a maximum grade, then gradually decrease back to the original grade. For the duration of her workout, the grade is modeled by ␲ the function G1x2 ⫽ 3 cos a x ⫺ ␲b ⫹ 4, where 5 G(x) is the percent grade x minutes after the workout has begun. (a) What is the initial grade for her workout? (b) What is the grade at x ⫽ 4 min? (c) At G1x2 ⫽ 4.9%, how long has she been working out? (d) What is the duration of the treadmill workout? Geometry applications: Solve Exercises 63 and 64 graphically using a calculator. For Exercise 63, give ␪ in radians rounded to four decimal places. For Exercise 64, answer in degrees to the nearest tenth of a degree.

63. The area of a circular segment (the shaded portion shown) is given by the 1 formula A ⫽ r2 1␪ ⫺ sin ␪2, 2 where ␪ is in radians. If the circle has a radius of 10 cm, find the angle ␪ that gives an area of 12 cm2.

r

␪

Exercise 64

b 64. The perimeter of a trapezoid with parallel sides B and b, h altitude h, and base angles ␣ ␤ ␣ and ␤ is given by the formula B P ⫽ B ⫹ b ⫹ h1csc ␣ ⫹ csc ␤2. If b ⫽ 30 m, B ⫽ 40 m, h ⫽ 10 m, and ␣ ⫽ 45°, find the angle ␤ that gives a perimeter of 105 m.


65. As we saw in Chapter 6, cosine is the cofunction of sine and each can be expressed in terms of the other: ␲ ␲ cos a ⫺ ␪b ⫽ sin ␪ and sin a ⫺ ␪b ⫽ cos ␪. 2 2 This implies that either function can be used to model the phenomenon described in this section by adjusting the phase shift. By experimentation, (a) find a model using cosine that will produce results identical to the sine function in Exercise 58 and (b) find a model using sine that will produce results identical to the cosine function in Exercise 59. 66. Use multiple identities to find all real solutions for the equation given: sin15x2 ⫹ sin12x2cos x ⫹ cos12x2sin x ⫽ 0.

r

67. A rectangular Exercise 67 parallelepiped with square ends has 12 edges and six surfaces. If the sum of all edges is 176 cm and the total surface area is 1288 cm2, find (a) the length of the diagonal of ␪ the parallelepiped (shown in bold) and (b) the angle the diagonal makes with the base (two answers are possible).

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68. (6.7) Find the values of all six trig functions of an angle, given P1⫺51, 682 is on its terminal side.

Exercise 71

antenna extends an additional 280 ft into the air. Find the viewing angle ␪ for the antenna from a distance of 1000 ft (the angle formed from the base of the antenna to its top).

69. (4.3) Sketch the graph of f by locating its zeroes and using end-behavior: f 1x2 ⫽ x4 ⫺ 3x3 ⫹ 4x. 70. (5.3) Use a calculator and the change-of-base formula to find the value of log5279. 71. (6.6) The Willis Tower (formerly known as the Sears Tower) in Chicago, Illinois, remains one of the tallest structures in the world. The top of the roof reaches 1450 ft above the street below and the

280 ft

1450 ft

␪

1000 ft

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs A through H are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

⫺2␲

y

(b)

4

⫺4

2␲

y

4

⫺2␲

2␲

⫺2␲

2␲

⫺4

1. ____ f 1t2 ⫽ cos t, t 僆 30, ␲ 4 2. ____ tan t ⫽ ⫺1 ␲ ␲ 3. ____ f 1t2 ⫽ sin t, t 僆 c⫺ , d 2 2

y

(h)

4

⫺4

2␲

⫺4

y

(g)

4

⫺4

4

⫺2

y

(f)

4

⫺4

⫺2␲

4

y

(d)

2

⫺4

⫺4

(e)

y

(c)

4

4

⫺4

4

⫺4

4

⫺4

1 ␲ 9. ____ f a b ⫽ 2 6

10. ____ f 1t2 ⫽ cos⫺1t 11. ____ 2 sin t cos t

4. ____ y ⫽ arcsin t

12. ____ t ⫽ ⫺

5␲ ␲ 3␲ 7␲ ,⫺ , , 4 4 4 4

5. ____ sec t ⫽ 2

13. ____ t ⫽ ⫺

5␲ ␲ ␲ 5␲ ,⫺ , , 3 3 3 3

6. ____ sin(2t)

1 ␲ 14. ____ f a b ⫽ 2 3

7. ____ x ⫽ cos y

15. ____ x ⫽ sin y

␲ 13 b is on the graph 8. ____ a , 6 2

16. ____ y ⫽ cos2t ⫹ sin2t

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733


Fundamental Identities and Families of Identities

KEY CONCEPTS • The fundamental identities include the reciprocal, ratio, and Pythagorean identities. • A given identity can algebraically be rewritten to obtain other identities in an identity “family.” • Standard algebraic skills like distribution, factoring, combining terms, and special products play an important role in working with identities. C AD BC A gives an efficient method for combining rational terms. • The pattern B D BD • Using fundamental identities, a given trig function can be expressed in terms of any other trig function. • Once the value of a given trig function is known, the value of the other five can be uniquely determined using fundamental identities, if the quadrant of the terminal side is known. EXERCISES Verify using the method specified and fundamental identities. 1. multiplication 4. combine terms using 2 A C AD ⴞ BC sin x1csc x sin x2 cos x ⴞ ⴝ B D BD 2. factoring 2 2 sec x tan2x tan x csc x csc x sin x csc x csc x csc x sec2x 3. special product 1sec x tan x21sec x tan x2 sin x csc x Find the value of all six trigonometric functions using the information given. 12 25 5. cos ␪ ; ␪ in QIII 6. sec ␪ ; ␪ in QIV 37 23

SECTION 7.2

More on Verifying Identities

KEY CONCEPTS • The sine and tangent functions are odd functions, while cosine is even. • The steps used to verify an identity must be reversible. • If two expressions are equal, one may be substituted for the other and the result will be equivalent. • To verify an identity we mold, change, substitute, and rewrite one side until we “match” the other side. • Verifying identities often involves a combination of algebraic skills with the fundamental trig identities. A collection and summary of the Guidelines for Verifying Identities can be found on page 662. • To show an equation is not an identity, find any one value where the expressions are defined but the equation is false, or graph both functions on a calculator to see if the graphs are identical. EXERCISES Verify that each equation is an identity. csc2x11 cos2x2 cot2x 7. tan2x sin4x cos4x 9. tan x cot x sin x cos x

cot x csc x cot x1cos x csc x2 sec x tan x 1sin x cos x2 2 10. csc x sec x 2 sin x cos x 8.

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SECTION 7.3

The Sum and Difference Identities

KEY CONCEPTS The sum and difference identities can be used to • Find exact values for nonstandard angles that are a sum or difference of two standard angles. • Verify the cofunction identities and to rewrite a given function in terms of its cofunction. • Find coterminal angles in 3 0°, 360°2 for very large angles (the angle reduction formulas). • Evaluate the difference quotient for sin x, cos x, and tan x. • Rewrite a sum as a single expression: cos ␣ cos ␤ sin ␣ sin ␤ cos1␣ ␤2 . The sum and difference identities for sine and cosine can be remembered by noting • For cos1␣ ␤2 , the function repeats and the signs alternate: cos1␣ ␤2 cos ␣ cos ␤ sin ␣ sin ␤ • For sin1␣ ␤2 the signs repeat and the functions alternate: sin1␣ ␤2 sin ␣ cos ␤ cos ␣ sin ␤

EXERCISES Find exact values for the following expressions using sum and difference formulas. ␲ 11. a. cos 75° b. tana b 12. a. tan 15° 12 Evaluate exactly using sum and difference formulas. 13. a. cos 109° cos 71° sin 109° sin 71°

b. sin a

␲ b 12

b. sin 139° cos 19° cos 139° sin 19°

Rewrite as a single expression using sum and difference formulas. 14. a. cos 13x2 cos 12x2 sin 13x2 sin 12x2

x 3x x 3x b. sin a b cos a b cos a b sin a b 4 8 4 8

Evaluate exactly using sum and difference formulas, by reducing the angle to an angle in 3 0, 360°2 or 30, 2␲2. 57␲ 15. a. cos 1170° b. sin a b 4 Use a cofunction identity to write an equivalent expression for the one given. x 16. a. cos a b 8

b. sin ax

␲ b 12

17. Verify that both expressions yield the same result using sum and difference formulas: tan 15° tan145° 30°2 and tan 15° tan1135° 120°2 . 18. Use sum and difference formulas to verify the following identity. cos ax

␲ ␲ b cos ax b 13 cos x 6 6

SECTION 7.4

The Double-Angle, Half-Angle, and Product-to-Sum Identities

KEY CONCEPTS • When multiple angle identities (identities involving n␪) are used to find exact values, the terminal side of ␪ must be determined so the appropriate sign can be used. • The power reduction identities for cos2x and sin2x are closely related to the double-angle identities, and can be derived directly from cos12x2 2 cos2x 1 and cos12x2 1 2 sin2x. • The half-angle identities can be developed from the power reduction identities by using a change of variable and taking square roots. The sign is then chosen based on the quadrant of the half angle. • The product-to-sum and sum-to-product identities can be derived using the sum and difference formulas, and have important applications in many areas of science.

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EXERCISES Find exact values for sin12␪2, cos12␪2 , and tan12␪2 using the information given. 13 29 19. a. cos ␪ ; ␪ in QIV b. csc ␪ ; ␪ in QIII 85 20 Find exact values for sin ␪, cos ␪, and tan ␪ using the information given. 336 41 20. a. cos12␪2 b. sin12␪2 ; ␪ in QII ; ␪ in QII 841 625 Find exact values using the appropriate double-angle identity. 21. a. cos222.5° sin222.5°

b. 1 2 sin2a

␲ b 12

Find exact values for sin ␪ and cos ␪ using the appropriate half-angle identity. 5␲ 22. a. ␪ 67.5° b. ␪ 8 ␪ ␪ Find exact values for sin a b and cos a b using the given information. 2 2 24 65 23. a. cos ␪ ; 0° 6 ␪ 6 360°; ␪ in QIV b. csc ␪ ; 90° 6 ␪ 6 0; ␪ in QIV 25 33 24. Verify the equation is an identity. cos13␣2 cos ␣ 2 tan2␣ cos13␣2 cos ␣ sec2␣ 2

25. Solve using a sum-to-product formula. cos13x2 cos x 0

␪ ␪ 26. The area of an isosceles triangle (two equal sides) is given by the formula A x2sin a b cos a b, where the equal 2 2 sides have length x and the vertex angle measures ␪°. (a) Use this formula and the half-angle identities to find the area of an isosceles triangle with vertex angle ␪ 30° and equal sides of 12 cm. (b) Use substitution and a ␪ ␪ 1 double-angle identity to verify that x2sin a b cos a b x2sin ␪, then recompute the triangle’s area. Do the 2 2 2 results match?

SECTION 7.5

The Inverse Trig Functions and Their Applications

KEY CONCEPTS • In order to create one-to-one functions, the domains of y sin t, y cos t, and y tan t are restricted as follows: ␲ ␲ ␲ ␲ (a) y sin t, t 僆 c , d ; (b) y cos t, t 僆 3 0, ␲ 4 ; and (c) y tan t; t 僆 a , b. 2 2 2 2 • For y sin x, the inverse function is given implicitly as x sin y and explicitly as y sin1x or y arcsin x. • The expression y sin1x is read, “y is the angle or real number whose sine is x.” The other inverse functions are similarly read/understood. • For y cos x, the inverse function is given implicitly as x cos y and explicitly as y cos1x or y arccos x. • For y tan x, the inverse function is given implicitly as x tan y and explicitly as y tan1x or y arctan x. • The domains of y sec t, y csc t, and y cot t are likewise restricted to create one-to-one functions: ␲ ␲ ␲ ␲ (a) y sec t; t 僆 c 0, b ´ a , ␲ d ; (b) y csc t, t 僆 c , 0b ´ a0, d ; and (c) y cot t, t 僆 10, ␲2. 2 2 2 2 • In some applications, inverse functions occur in a composition with other trig functions, with the expression best evaluated by drawing a diagram using the ratio definition of the trig functions.

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• To evaluate y sec1t, we use y cos1a b; for y cot1t, use tan1a b; and so on. 1 1 t t • Trigonometric substitutions can be used to simplify certain algebraic expressions.

EXERCISES Evaluate without the aid of calculators or tables. State answers in both radians and degrees in exact form. 12 13 27. y sin1a 28. y csc12 29. y arccos a b b 2 2 Evaluate the following using a calculator. Answer in radians to the nearest ten-thousandth and in degrees to the nearest tenth. Some may be undefined. 7 30. y tan14.3165 31. y sin10.8892 32. f 1x2 arccos a b 8 Evaluate the following without the aid of a calculator, keeping the domain and range of each function in mind. Some may be undefined. 1 ␲ 33. c sin c sin1a b d 34. arcsec c sec a b d 35. cos1cos122 2 4 Evaluate the following using a calculator. Some may be undefined. 36. sin1 1sin 1.02452

37. arccos3 cos160°2 4

Evaluate each expression by drawing a right triangle and labeling the sides. 12 7 39. sin c cos1a b d 40. tan c arcsec a b d 37 3x

38. cot1 c cot a 41. cot c sin1a

11␲ bd 4

x 281 x2

bd

Use an inverse function to solve the following equations for ␪ in terms of x. 42. x 5 cos ␪

SECTION 7.6

43. 7 13 sec ␪ x

44. x 4 sin a␪

␲ b 6

Solving Basic Trig Equations

KEY CONCEPTS • When solving trig equations, we often consider either the principal root, roots in 3 0, 2␲2, or all real roots. • Keeping the graph of each function in mind helps to determine the desired solution set. • After isolating the trigonometric term containing the variable, we solve by applying the appropriate inverse function, realizing the result is only the principal root. • Once the principal root is found, roots in 30, 2␲2 or all real roots can be found using reference angles and the period of the function under consideration. • Trig identities can be used to obtain an equation that can be solved by factoring or other solution methods. EXERCISES Solve each equation without the aid of a calculator (all solutions are standard values). Clearly state (a) the principal root; (b) all solutions in the interval 3 0, 2␲2 ; and (c) all real roots. 45. 2 sin x 12 46. 3 sec x 6 47. 8 tan x 7 13 13 Solve using a calculator and the inverse trig functions (not by graphing). Clearly state (a) the principal root; (b) solutions in 3 0, 2␲2; and (c) all real roots. Answer in radians to the nearest ten-thousandth as needed. 2 1 48. 9 cos x 4 49. sin12␪2 50. 12 csc x 3 7 5 4

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Practice Test

General Trig Equations and Applications

SECTION 7.7

KEY CONCEPTS • In addition to the basic solution methods from Section 7.6, additional strategies include squaring both sides, factoring by grouping, and using the full range of identities to simplify an equation. • Many applications result in equations of the form Asin1Bx C2 D k. To solve, isolate the factor sin1Bx C2 (subtract D and divide by A), then apply the inverse function. • Once the principal root is found, roots in 3 0, 2␲2 or all real roots can be found using reference angles and the period of the function under consideration. EXERCISES Find solutions in 3 0, 2␲2 using the method indicated. Round nonstandard values to four decimal places. 51. squaring both sides 52. using identities 16 sin x cos x 3 cos12x2 7 sin x 5 0 2 53. factor by grouping 54. using any appropriate method 4 sin x cos x 213 sin x 2 cos x 13 0 csc x cot x 1 State the period P of each function and find all solutions in [0, P). Round to four decimal places as needed. ␲ ␲ ␲ ␲ 55. 750 sin a x b 120 0 56. 80 cos a x b 40 12 0 6 2 3 4 57. The revenue earned by Waipahu Joe’s Tanning Lotions fluctuates with the seasons, with a great deal more lotion ␲ ␲ sold in the summer than in the winter. The function R1x2 15 sin a x b 30 models the monthly sales of 6 2 lotion nationwide, where R(x) is the revenue in thousands of dollars and x represents the months of the year (x 1 S Jan). (a) How much revenue is projected for July? (b) For what months of the year does revenue exceed $37,000? 58. The area of a circular segment (the shaded portion shown in the diagram) is given by the 1 formula A r2 1␪ sin ␪2 ,where ␪ is in radians. If the circle has a radius of 10 cm, find 2 the angle ␪ that gives an area of 12 cm2.

r

␪

r

PRACTICE TEST Verify each identity using fundamental identities and the method specified. 1. special products 1csc x cot x21csc x cot x2 cos x sec x sin x cos x sin x cos x 1 cos x sin x 3

2. factoring

3

3. Find the value of all six trigonometric functions 48 given cos ␪ ; ␪ in QIV 73

4. Find the exact value of tan 15° using a sum or difference formula. 5. Rewrite as a single expression and evaluate: cos 81° cos 36° sin 81° sin 36° 6. Evaluate cos 1935° exactly using an angle reduction formula. 7. Use sum and difference formulas to verify sin ax

␲ ␲ b sin ax b 12 cos x. 4 4

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8. Find exact values for sin ␪, cos ␪, and tan ␪ given 161 cos12␪2 ; ␪ in QI. 289

18. Solve each equation using a calculator and inverse trig functions to find the principal root (not by graphing). Then state (a) the principal root, (b) all solutions in the interval 30, 2␲2 , and (c) all real roots. 1 2 I. sin12x2 II. 3 cos12x2 0.8 0 3 4

9. Use a double-angle identity to evaluate 2 cos275° 1. ␪ ␪ 10. Find exact values for sin a b and cos a b given 2 2 12 tan ␪ ; ␪ in QI. 35 11. The area of a triangle is given geometrically as 1 A base # height. The 2 trigonometric formula for

b ␣

a

c

1 the triangle’s area is A bc sin ␣, where ␣ is the 2 angle formed by the sides b and c. In a certain triangle, b 8, c 10, and ␣ 22.5°. Use the formula for A given here and a half-angle identity to find the area of the triangle in exact form. 12. The equation Ax2 Bxy Cy2 0 can be written in an alternative form that makes it easier to graph. This is done by eliminating the mixed xy-term using B the relation tan12␪2 to find ␪. We can then AC find values for sin ␪ and cos ␪, which are used in a conversion formula. Find sin ␪ and cos ␪ for 17x2 5 13xy 2y2 0, assuming 2␪ in QI. 13. Evaluate without the aid of calculators or tables. 1 1 b a. y tan1a b. y sin c sin1a b d 2 13 c. y arccos1cos 30°2 14. Evaluate the following. Use a calculator for part (a), give exact answers for part (b), and find the value of the expression in part (c) without using a calculator. Some may be undefined. a. y sin10.7528 b. y arctan1tan 78.5°2 7␲ c. y sec1 c seca b d 24 Evaluate the expressions by drawing a right triangle and labeling the sides. 15. cos c tan1a 16. cot c cos1a

56 bd 33 x

bd

225 x 17. Solve without the aid of a calculator (all solutions are standard values). Clearly state (a) the principal root, (b) all solutions in the interval 30, 2␲2, and (c) all real roots. I. 8 cos x 412 II. 13 sec x 2 4 2

19. Use a graphing calculator to solve the equations in the indicated interval. State answers in radians rounded to the nearest ten-thousandth. a. 3 cos12x 12 sin x; x 僆 3 ␲, ␲4 b. 2 1x 1 3 cos2 x; x 僆 3 0, 2␲2

20. Solve the following equations for x 僆 30, 2␲2 using a combination of identities and/or factoring. State solutions in radians using the exact form where possible. a. 2 sin x sin12x2 sin12x2 0 1 b. 1cos x sin x2 2 2 Solve each equation in [0, 2␲) by squaring both sides, factoring, using identities, or by using any appropriate method. Round nonstandard values to four decimal places. 21. 3 sin 12x2 cos x 0 22.

␲ 3 5 2 sin a2x b 3 6 2 6

23. The revenue for Otake’s Mower Repair is very seasonal, with business in the summer months far exceeding business in the winter months. Monthly revenue for the company can be modeled by the ␲ 4␲ b 12.5, where function R1x2 7.5 cos a x 6 3 R(x) is the average revenue (in thousands of dollars) for month x 1x 1 S Jan2. (a) What is the average revenue for September? (b) For what months of the year is revenue at least $12,500? 24. The lowest temperature on record for the even months of the year are given in the table for the city of Denver, Colorado. The equation y 35.223 sin10.576x 2.5892 6 is a fairly accurate model for this data. Use the equation to estimate the record low temperatures for the odd numbered months.

Month (Jan S 1)

Low Temp. (ⴗF )

2

30

4

2

6

30

8

41

10

3

12

25

Source: 2004 Statistical Abstract of the United States, Table 379.

25. Write the product as a sum using a product-to-sum identity: 2 cos11979␲t2cos1439␲t2.

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CALCULATOR EXPLORATION AND DISCOVERY Seeing the Beats as the Beats Go On When two sound waves of slightly different frequencies are combined, the resultant wave varies periodically in amplitude over time. These amplitude pulsations are called beats. In this Exploration and Discovery, we’ll look at ways to “see” the beats more clearly on a graphing calculator, by representing sound waves very simplistically as Y1 ⫽ cos1mX2 and Y2 ⫽ cos1nX2 and noting a relationship between m, n, and the number of beats in 3 0, 2␲ 4. Using a sum-to-product formula, we can represent the resultant wave as a single term. For Y1 ⫽ cos112X2 and Y2 ⫽ cos18X2 the result is 12X ⫹ 8X 12X ⫺ 8X b cos a b 2 2 ⫽ 2 cos110X2cos12X2

Y3 ⫽ cos112X2 ⫹ cos18X2 ⫽ 2 cos a

Figure 7.60 The window used and resulting graph are Figure 7.59 3 shown in Figures 7.59 and 7.60, and it appears that “silence” occurs four times in this interval— where the graph of the combined waves is tangent to (bounces off of) the x-axis. This indicates a 2␲ 0 total of four beats. Note the number of beats is equal to the difference m ⫺ n: 12 ⫺ 8 ⫽ 4. Further experimentation will show this is not a coincidence, and this enables us to construct two ⫺3 additional functions that will frame these pulsations and make them easier to see. Since Figure 7.62 the maximum amplitude of the resulting Figure 7.61 3 wave is 2, we use functions of the form k ⫾2 cos a xb to construct the frame, where k is 2 the number of beats in the interval 1m ⫺ n ⫽ k2. 0 2␲ For Y1 ⫽ cos112X2 and Y2 ⫽ cos18X2, we have k 12 ⫺ 8 ⫽ ⫽ 2 and the functions we use will be 2 2 Y4 ⫽ 2 cos12X2 and Y5 ⫽ ⫺2 cos12X2 as ⫺3 shown in Figure 7.61. The result is shown in Figure 7.62, where the frame clearly shows the four beats or more precisely, the four moments of silence. For each exercise, (a) express the sum Y1 ⫹ Y2 as a product, (b) graph YR on a graphing calculator for x 僆 3 0, 2␲4 k and identify the number of beats in this interval, and (c) determine what value of k in ⫾2 cos a xb would be used to 2 frame the resultant YR, then enter these as Y4 and Y5 to check the result.

Exercise 1: Y1 ⫽ cos114X2; Y2 ⫽ cos18X2




STRENGTHENING CORE SKILLS Trigonometric Equations and Inequalities The ability to draw a quick graph of the trigonometric functions is a tremendous help in understanding equations and inequalities. A basic sketch can help reveal the number of solutions in 30, 2␲2 and the quadrant of each solution. For nonstandard angles, the value given by the inverse function can then be used as a basis for stating the solution set for all real numbers. We’ll illustrate the process using a few simple examples, then generalize our observations to solve

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Figure 7.63 more realistic applications. Consider the function f 1x2 2 sin x 1, a sine wave with y amplitude 2, and a vertical translation of 1. To find intervals in 3 0, 2␲2 where f 1x2 7 2.5, we y 2.5 f (x) reason that f has a maximum of 2112 1 3 and a minimum of 2112 1 1, since 1 sin x 1. With no phase shift and a standard period of 2␲, we can easily draw a quick sketch of f by vertically translating x-intercepts and max/min points 1 unit up. After drawing the line y 2.5 (see Figure 7.63), it appears there are two intersections in the interval, one in QI and one in QII. More importantly, it is clear that f 1x2 7 2.5 between these two solutions. Substituting 2.5 for f 1x2 in f 1x2 2 sin x 1, we solve for sin x to obtain sin x 0.75, which we use to state the solution in exact form: f 1x2 7 2.5 for x 僆 1sin1 0.75, ␲ sin1 0.752. In approximate form the solution interval is x 僆 10.85, 2.292. If the function involves a horizontal shift, the graphical analysis will reveal which intervals should be chosen to satisfy the given inequality. The basic ideas remain the same regardless of the complexity of the equation, and we illustrate by studying the 2␲ ␲ d b 950 and the inequality R 1d2 7 1250. Remember—our current goal is not a function R 1d2 750 sin a 365 2 supremely accurate graph, just a sketch that will guide us to the solution using the inverse functions and the correct C quadrants. Perhaps that greatest challenge is recalling that when B 1, the horizontal shift is , but other than this a B fairly accurate sketch can quickly be obtained. 3

2␲

x

360

d

␲

3

2␲ ␲ d b 950, find intervals in 30, 365 4 where R 1d2 7 1250. 365 2 C Solution 䊳 This is a sine wave with a period of 365 (days), an amplitude of 750, shifted 91.25 units to the right B Figure 7.64 and 950 units up. The maximum value will be 1700 and the minimum value will be 200. For y convenience, scale the axes from 0 to 360 (as though the period were 360 days), and plot the x-intercepts and maximum/minimum values for a standard sine wave with amplitude 750 (by R(d) scaling the axes). Then shift these points about 90 units in the positive direction (to the right), and 950 units up, again using a scale that makes this convenient (see Figure 7.64). This sketch along with the graph of y 1250 is sufficient to show that solutions to R 1d2 1250 occur early in the second quarter and late in the third quarter, with solutions to R 1d2 7 1250 occurring between 2␲ ␲ d b 950, we substitute 1250 for R 1d2 and isolate the sine function, these solutions. For R 1d2 750 sina 365 2 2␲ ␲ ␲ 365 d b 0.4, which leads to exact form solutions of d asin10.4 b a b and obtaining sin a 365 2 2 2␲ ␲ 365 d a␲ sin10.4 b a b. In approximate form the solution interval is x 僆 3115, 250 4. 2 2␲

Illustration 1 䊳 Given R 1d2 750 sin a

2000

1000

180

500

Practice with these ideas by finding solutions to the following inequalities in the intervals specified. Exercise 1: f 1x2 3 sin x 2; f 1x2 7 3.7; x 僆 30, 2␲2 ␲ Exercise 2: g1x2 4 sin ax b 1; g1x2 2; x 僆 30, 2␲2 3 ␲ ␲ Exercise 3: h 1x2 125 sin a x b 175; h 1x2 150; x 僆 3 0, 122 6 2 2␲ ␲ Exercise 4: f 1x2 15,750 sin a x b 19,250; f 1x2 7 25,250; x 僆 30, 3602 360 4

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CUMULATIVE REVIEW CHAPTERS 1–7 1. Find f 1␪2 for all six trig functions, given P113, 842 is on the terminal side with ␪ in QII. 2. Find the lengths of the missing sides. 3. Verify that x 2 13 is a zero of g1x2 x2 4x 1.

3

10. Solve for x: 21x 32 4 1 55.

Exercise 2 60

5√3

4. Determine the domain of r1x2 29 x2. Answer in interval notation. 5. Standing 5 mi (26,400 ft) from the base of Mount Logan (Yukon) the angle of elevation to the summit is 36° 56¿ . How much taller is Mount McKinley (Alaska) which stands at 20,320 ft high? 6. Use the Guidelines for Graphing Polynomial Functions to sketch the graph of f 1x2 x3 3x2 4. 7. Use the Guidelines for Graphing Rational Functions x1 to sketch the graph of h1x2 2 x 4 8. The Petronas Towers in Malaysia are two of the tallest structures in the world. The top of the roof reaches 1483 ft above the street below and the stainless steel pinnacles extend an additional 241 ft into the air (see figure). Find the viewing angle ␪ for the pinnacles from a distance of 1000 ft (the angle formed from the base of the antennae to its top). 241 ft

1483 ft ␪

11. Solve for x: 3 ` x

1 ` 5 10 2

12. Earth has a radius of 3960 mi. Tokyo, Japan, is located at 35.4° N latitude, very near the 139° E latitude line. Adelaide, Australia, is at 34.6° S latitude, and also very near 139° E latitude. How many miles separate the two cities? 13. Since 1970, sulphur dioxide emissions in the United States have been decreasing at a nearly linear rate. In 1970, about 31 million tons were emitted into the atmosphere. In 2000, the amount had decreased to approximately 16 million tons. (a) Find a linear equation that models sulphur dioxide emissions. (b) Discuss the meaning of the slope ratio in this context. (c) Use the equation model to estimate the emissions in 1985 and 2010. Source: 2004 Statistical Abstract of the United States, Table 360.

14. List the three Pythagorean identities and three identities equivalent to cos12␪2. ␲ ␲ 15. For f 1x2 325 cos a x b 168, what values 6 2 of x in 3 0, 2␲2 satisfy f 1x2 7 330.5? 16. Write as a single logarithmic expression in simplest form: log1x2 92 log1x 12 log1x2 2x 32. 17. After doing some market research, the manager of a sporting goods store finds that when a four-pack of premium tennis balls are priced at $9 per pack, 20 packs per day are sold. For each decrease of $0.25, 1 additional pack per day will be sold. Find the price at which four-packs of tennis balls should be sold in order to maximize the store’s revenue on this item. 18. Write the equation of the function whose graph is given, in terms of a sine function. 19. Verify that the following is an 8␲ cos x 1 cos x identity: 2 sec x 1 tan x

1000 ft

9. A wheel with radius 45 cm is turning at 5 revolutions per second. Find the linear velocity of a point on the rim in kilometers per hour, rounded to the nearest hundredth.

Exercise 18 y 6

4␲

4␲

6

8␲ x

20. Given the zeroes of f 1x2 x4 18x2 32 are x 4 and x 12, estimate the following: a. the domain of the function b. intervals where f 1x2 7 0 and f 1x2 0

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21. Use the triangle shown to find the exact value of sin12␪2. ␪ √202

9

11

22. Use the triangle shown to find the exact value of sin1␣ ␤2. ␣ 51 ␤ 68

23. The amount of waste product released by a manufacturing company varies according to its production schedule, which is much heavier during the summer months and lighter in the winter. Waste product amount reaches a maximum of 32.5 tons in the month of July, and falls to a minimum of 21.7 tons in January 1t 12 . (a) Use this information to build a sinusoidal equation that models the amount of waste produced each month. (b) During what months of the year does output exceed 30 tons? 24. At what interest rate will $2500 grow to $3500 if it’s left on deposit for 6 yr and interest is compounded continuously? 25. Identify each geometric formula: a. y ␲r2h b. y LWH 1 c. y 2␲r d. y bh 2

Exercises 26 through 30 require the use of a graphing calculator. 26. Use the GRAPH and TABLE features of a calculator to support the validity of the identity 2 cos11.7x2 sec10.9x2 cos10.8x2 sec10.9x2 cos12.6x2 . Then convert all functions to sines and cosines, rewrite the identity, and verify this new identity using a sum-toproduct identity. 27. On a graphing calculator, the graph of 1.5x2 0.9x 2.4 r1x2 appears to be 0.5x3 0.8x2 x 1.6 continuous. However, Descartes’ rule of signs tells us the denominator must have at least one real zero. Since the graph does not appear to have any vertical asymptotes, (a) what does this tell you about r(x)? (b) Use the TABLE feature of a calculator with ¢Tbl ⴝ 0.1 to help determine the x-coordinate of this discontinuity. 28. Solve the following equation graphically. Round your answer to two decimal places. cos1x tan x 29. Use the intersection-of-graphs method to solve the inequality sin x cos x, for x 僆 30, 2␲2 . Answer using exact values. 30. The range of a certain projectile is modeled by the 625 sin ␪ cos ␪, where ␪ is the function r1␪2 4 angle at launch and r1␪2 is in feet. (a) Rewrite the function using a double angle formula, (b) determine the maximum range of the projectile, and (c) determine the launch angle that results in this maximum range.

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CONNECTIONS TO CALCULUS In calculus, as in college algebra, we often encounter expressions that are difficult to use in their given form, and so attempt to write the expression in an alternative form more suitable to the task at hand. Often, we see algebra and trigonometry working together to achieve this goal.

Simplifying Expressions Using a Trigonometric Substitution For instance, it is difficult to apply certain concepts from calculus to the equation x y⫽ , and we attempt to rewrite the expression using a trig substitution and a 29 ⫺ x2 Pythagorean identity. When doing so, we’re careful to ensure the substitution used represents a one-to-one function, and that it maintains the integrity of the domain. EXAMPLE 1

䊳

Simplifying Algebraic Expressions Using Trigonometry Simplify y ⫽

x

using the substitution x ⫽ 3 sin ␪ for ⫺

29 ⫺ x verify that the result is equivalent to the original function.

Solution

Check

䊳

䊳

3 sin ␪

y⫽

2

29 ⫺ 13 sin ␪2 2 3 sin ␪ 3 sin ␪ ⫽ ⫽ 2 29 ⫺ 9 sin ␪ 2911 ⫺ sin2␪2 3 sin ␪ ⫽ 29 cos2␪ 3 sin ␪ ⫽ 3 cos ␪ ⫽ tan ␪

␲ ␲ 6 ␪ 6 , then 2 2

substitute 3 sin ␪ for x 13 sin ␪2 2 ⫽ 9 sin2␪; factor substitute cos2␪ for 1 ⫺ sin2␪

29 cos2␪ ⫽ 3 cos ␪ since ⫺

␲ ␲ ⱕ␪ⱕ 2 2

result

Using the notation for inverse functions, we can rewrite y ⫽ tan ␪ as a function of x and use a calculator to compare it with the original function. For x ⫽ 3 sin ␪ we x x x obtain ⫽ sin ␪ or ␪ ⫽ sin⫺1a b. Substituting sin⫺1a b for ␪ in y ⫽ tan ␪ gives 3 3 3 x y ⫽ tan c sin⫺1a b d . With the calculator in radian 3 X X MODE , enter Y ⫽ and Y2 ⫽ tan csin⫺1a bd 1 2 3 29 ⫺ X on the

Y=

screen. Using TblStart ⫽ ⫺3 (due to

the domain), the resulting table seems to indicate that the functions are indeed equivalent (see the figure). Now try Exercises 1 through 6

䊳

Trigonometric Identities and Equations

7–91

While the tools of calculus are very powerful, it is the application of rudimentary concepts that makes them work. Earlier, we saw how basic algebra skills were needed to simplify expressions that resulted from applications of calculus. Here we illustrate the use of basic trigonometric skills combined with basic algebra skills to achieve the same end. 743

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EXAMPLE 2

䊳

Finding Maximum and Minimum Values Using the tools of calculus, it can be shown that the maximum and/or minimum 1 ⫹ cot ␪ values of f 1␪2 ⫽ will occur at the zero(s) of the function csc ␪ f 1␪2 ⫽

csc ␪1⫺csc2␪2 ⫺ 11 ⫹ cot␪2 1⫺csc ␪ cot ␪2 csc2␪

.

Simplify the right-hand side and use the result to find the location of any maximum or minimum values that occur in the interval 3 0, 2␲2.

Solution

䊳

Begin by simplifying the numerator. f 1␪2 ⫽

csc ␪1⫺csc2␪2 ⫺ 1⫺csc ␪ cot ␪ ⫺ csc ␪ cot2␪2

csc2␪ ⫺csc3␪ ⫹ csc ␪ cot ␪ ⫹ csc ␪ cot2␪ ⫽ csc2␪ csc ␪1cot2␪ ⫹ cot ␪ ⫺ csc2␪2 ⫽ csc2␪ 2 1csc ␪ ⫺ 12 ⫹ cot ␪ ⫺ csc2␪ cot ␪ ⫺ 1 ⫽ ⫽ csc ␪ csc ␪

distribute

simplify

factor, commute terms simplify, substitute csc2␪ ⫺ 1 for cot2␪; result

␲ This shows that f 1␪2 ⫽ 0 when cot ␪ ⫽ 1, or when ␪ ⫽ ⫹ ␲k, k 僆⺪. In the 4 5␲ ␲ interval 30, 2␲2, this gives ␪ ⫽ and . The function f has a maximum value 4 4 ␲ 5␲ of 12 at , with a minimum value of ⫺ 12 at . 4 4 Now try Exercises 7 through 10

䊳

Connections to Calculus Exercises For the functions given, (a) use the substitution indicated to find an equivalent function of ␪, (b) rewrite the resulting function in terms of x using an inverse trig function, and (c) use the TABLE feature of a graphing calculator to verify the two functions are equivalent for x ⴝ 0. 1. y ⫽

2169 ⫹ x2 ␲ ␲ ; x ⫽ 13 tan ␪, ␪ 僆 a⫺ , b x 2 2

2. y ⫽

2144 ⫺ x2 ␲ ␲ ; x ⫽ 12 sin ␪, ␪ 僆 a⫺ , b x 2 2

Rewrite the following expressions using the substitution indicated. 3.

x2

; x ⫽ 4 sin ␪ 216 ⫺ x2

5.

x

; x ⫽ 3 tan ␪ 29 ⫹ x2

4.

281 ⫺ x2 ; x ⫽ 9 sin ␪ x

6.

8

1x2 ⫺ 42 2 3

; x ⫽ 2 sec ␪

Using the tools of calculus, it can be shown that for each function f(x) given, the zeroes of f (x) give the location of any maximum and/or minimum values. Find the location of these values in the interval [0, 2␲), using trig identities as needed to solve f (x) ⴝ 0. Verify solutions using a graphing calculator. 7. f 1x2 ⫽ f 1x2 ⫽

1 ⫹ cos x ; sec x sec x1⫺sin x2 ⫺ 11 ⫹ cos x2sec x tan x 2

sec x

8. f 1x2 ⫽ sin x tan x; f 1x2 ⫽ sin x sec2x ⫹ tan x cos x

9. f 1x2 ⫽ 2 sin x cos x;

f 1x2 ⫽ 2 sin x1⫺sin x2 ⫹ 2 cos x cos x

10. f 1x2 ⫽

cos x ; 2 ⫹ sin x 12 ⫹ sin x2 1⫺sin x2 ⫺ cos x cos x f 1x2 ⫽ 12 ⫹ sin x2 2

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8.2 The Law of Cosines; the Area of a Triangle 759

When an airline pilot charts a course, it’s not as simple as pointing the airplane in the right direction. Wind currents must be taken into consideration, and compensated for by additional thrust or a change of heading that will help equalize the force of the wind and keep the plane flying in the desired direction. The effect of these forces working together can be modeled using a carefully drawn vector diagram, and with the aid of trigonometry, a pilot can easily determine any adjustments in navigation needed.

8.3 Vectors and Vector Diagrams 771

䊳

Applications of Trigonometry CHAPTER OUTLINE 8.1 Oblique Triangles and the Law of Sines 746

8.4 Vector Applications and the Dot Product 787


8.5 Complex Numbers in Trigonometric Form 802 8.6 De Moivre’s Theorem and the Theorem on nth Roots 813

As with other forms of problem solving, drawing an accurate sketch or diagram of the relationships involved has a large impact on our ability to understand vector applications and other applications of trigonometry. This Connections is certainly no less true in a calculus course and is the subject of the Chapter 8 Connections to Calculus. 745 to Calculus

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Oblique Triangles and the Law of Sines


A. Develop the law of sines and use it to solve ASA and AAS triangles B. Solve SSA triangles (the ambiguous case) using the law of sines C. Use the law of sines to solve applications

WORTHY OF NOTE As with right triangles, solving an oblique triangle involves determining the lengths of all three sides and the measures of all three angles.

Many applications of trigonometry involve oblique triangles, or triangles that do not have a 90° angle. For example, suppose a trolley carries passengers from ground level up to a mountain chateau, as shown in Figure 8.1. Assuming the cable could be held taut, what is its approximate length? Can we also determine the slant height of the mountain? To answer questions like these, we’ll develop techniques that enable us to solve acute and obtuse triangles using fundamental trigonometric relationships.

Figure 8.1

23 2000 m

67

A. The Law of Sines and Unique Solutions Consider the oblique triangle ABC pictured in Figure 8.2 Figure 8.2. Since it is not a right triangle, it seems B the trigonometric ratios studied earlier cannot be applied. But if we draw the altitude h (from vertex B), two right triangles are formed that share a common c a h side. By applying the sine ratio to angles A and C, we can develop a relationship that will help us solve the triangle. A C h b For ⬔A we have sin A or h c sin A. For c h ⬔C we have sin C or h a sin C. Since both products are equal to h, the transitive a property gives c sin A a sin C and we have c sin A a sin C c sin A a sin C ac ac sin A sin C a c

since h h divide by ac

simplify

Using the same triangle and the altitude drawn from C (Figure 8.3), we note a simh h ilar relationship involving angles A and B: sin A or h b sin A, and sin B or a b sin A sin B h a sin B. As before, we can then write . If ⬔A is obtuse, the altitude a b h actually falls outside the triangle, as shown in Figure 8.4. In this case, consider that sin1180° ␣2 sin ␣ from the difference formula for sines (Exercise 55, Section 7.3). h In the figure we note sin1180° ␣2 sin ␣, yielding h c sin ␣ and the c Figure 8.3

Figure 8.4 B

B

a

c

h

h

A

746

b

a c

180 ␣ C

␣ A

b

C

8–2

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Section 8.1 Oblique Triangles and the Law of Sines

preceding relationship can now be stated using any pair of angles and corresponding sides. The result is called the law of sines, which is usually stated by combining the three possible proportions. The Law of Sines For any triangle ABC, the ratio of the sine of an angle to the side opposite that angle is constant: sin A sin C sin B a c b As a proportional relationship, the law requires that we have three parts in order to solve for the fourth. This suggests the following possibilities:

WORTHY OF NOTE The case where three angles are known (AAA) is not considered since we then have a family of similar triangles, with infinitely many solutions.

1. 2. 3. 4. 5.

two angles and an included side (ASA) two angles and a side opposite one of these angles (AAS) two sides and an angle opposite one of these sides (SSA) two sides and an included angle (SAS) three sides (SSS)

Each of these possibilities is diagrammed in Figures 8.5 through 8.9. Figure 8.5 A

WORTHY OF NOTE

Figure 8.7

Figure 8.6 A Angle

ASA

When working with triangles, keeping these basic properties in mind will prevent errors and assist in their solution:

SSA

A

AAS

Side

1. The angles must sum to 180°. 2. The combined length of any two sides must exceed the length of the third side. 3. Longer sides will be opposite larger angles. 4. The sine of an angle cannot be greater than 1. 5. For y 僆 10, 12 , the equation y sin ␪ has two solutions in 10°, 180°2 that are supplements.

Angle B

Side

Angle C

Angle B

Side

C

B

Side

Angle C

Figure 8.9

Figure 8.8

SSS

SAS Angle Side

Side

Side

Side Side

Since applying the law of sines requires we have a given side opposite a known angle, it cannot be used in the case of SAS or SSS triangles. These require the law of cosines, which we will develop in Section 8.2. In the case of ASA and AAS triangles, a unique triangle is formed since the measure of the third angle is fixed by the two given angles (they must sum to 180°) and the remaining sides must be of fixed length. EXAMPLE 1

䊳

Solving a Triangle Using the Law of Sines Figure 8.10

Solve the triangle shown in Figure 8.10 and state your answer using a table. Use a calculator to verify your solution.

Solution

䊳

This is not a right triangle, so the standard ratios cannot be used. Since ⬔B and ⬔C are given, we know ⬔A 180° 1110° 32°2 38°. With ⬔A and side a, we have

B 110

39.0 cm

c A

32 b

C

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CHAPTER 8 Applications of Trigonometry

sin A sin B a b sin 110° sin 38° 39 b b sin 38° 39 sin 110° 39 sin 110° b sin 38° b ⬇ 59.5

WORTHY OF NOTE Although not a definitive check, always review the solution table to ensure the shortest side is opposite the smallest angle, the longest side is opposite the largest angle, and so on. If this is not the case, you should go back and check your work.

Check

A. You’ve just seen how we can develop the law of sines and use it to solve ASA and AAS triangles

䊳

Repeating this procedure using

law of sines applied to ⬔A and ⬔B

substitute given values multiply by 39b divide by sin 38° result

sin A sin C shows side c ⬇ 33.6 cm. In table a c

form we have Angles

Sides (cm)

A 38°

a ⴝ 39.0

B ⴝ 110ⴗ

b ⬇ 59.5

C ⴝ 32ⴗ

c ⬇ 33.6

Figure 8.11

Begin by comparing the table entries to the measures of the original triangle. In this case, we have correctly recorded that B 110°, C 32°, and a 39.0. Using the law of sines and a calculator in degree MODE will help to verify the remaining calculations (see Figure 8.11). Note the results are very close (agreeing to four decimal places), with the discrepancies attributed to the rounded values used.


䊳

B. Solving SSA Triangles—The Ambiguous Case Figure 8.12 15

12

25

15

12 25

To understand the concept of unique and nonunique solutions regarding the law of sines, consider an instructor who asks a large group of students to draw a triangle with sides of 15 and 12 units, and a nonincluded 25° angle. Unavoidably, three different solutions will be offered (see Figure 8.12). For the SSA case, there is some doubt as to the number of solutions possible, or whether a solution even exists. To further understand why, consider a triangle with side c 30 cm, ⬔A 30°, and side a opposite the 30° angle (Figure 8.13—note the length of side b is yet to be determined). From our work with 30-60-90 triangles, we know if a 15 cm, it is exactly the length needed to form a right triangle (Figure 8.14). Figure 8.13

Figure 8.14 B

B 15

12

c 30 cm

25 A

c 30 cm

a

60

One solution

30 b

C

A

30 b 15√3

C

a 15 cm

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Precalculus—

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749


By varying the length of side a, we note three other possibilities. If side a 6 15 cm, no triangle is possible since a is too short to contact side b (Figure 8.15), while if 15 cm 6 side a 6 30 cm, two triangles are possible since side a can then intersect side b at two locations, C1 and C2 (Figure 8.16). Figure 8.15

Figure 8.16

B c 30 cm

c 30 cm

a 10 cm No

A

30 solution b

Figure 8.17 B

B

C

A

c 30 cm

Two a 20 cm solutions

C1

30 C2 b

a 35 cm

30 A

C

b

For future use, note that when two triangles are possible, angles C1 and C2 must be supplements since an isosceles triangle is formed. Finally, if side a 7 30 cm, it will intersect side b only once, forming the obtuse triangle shown in Figure 8.17, where we’ve assumed a 35 cm. Since the final solution is in doubt until we do further work, the SSA case is called the ambiguous case of the law of sines. EXAMPLE 2

䊳

Analyzing the Ambiguous Case of the Law of Sines Given triangle ABC (Figure 8.18) with ⬔A 45° and side c 100 12 mm, a. What length for side a will produce a Figure 8.18 right triangle where ⬔C 90°? B b. How many triangles can be formed if side a 97 mm? c 100√2 a ?? mm c. If side a 120 mm, how many triangles can be formed? 45 d. If side a 145 mm, how many triangles A C can be formed?

Solution

䊳

a. Recognizing the sides of a 45-45-90 triangle are in proportion according to 1x:1x: 12x, side a must be 100 mm for a right triangle to be formed. b. If a 97 mm, it will be too short to contact side b and no triangle is possible. c. As shown in Figure 8.19, if a 120 mm, it will contact side b in two distinct places and two triangles are possible. d. If a 145 mm, it will contact side b only once, since it is longer than side c and will “miss” intersecting side b a second time as it pivots around B (see Figure 8.20). One triangle is possible. Figure 8.19

Figure 8.20 B

B

a 120 mm

c 100√2 45 A

C2

C1

a 145 mm

c 100√2 Misses side b A

C1

45 b


䊳

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The TABLE feature of a calculator can help increase our understanding of the ambiguous case. Consider triangle ABC from Example 2. Using the law of sines, we have the following sequence: sin 45° sin C ⫽ a 100 12 12 1100 122 2 sin C ⫽ a sin C ⫽

100 a

C ⫽ sin⫺1a

law of sines

sin 45° ⫽

12 , multiply by 100 12 2

simplify

100 b a

apply sin⫺1

We can enter this expression for ⬔C as Y1 on Figure 8.21 the Y= screen and then use the TABLE feature to evaluate the expression for different lengths of side a. If two triangles are possible, C1 and C2 must be complements (see comment on top of page 749) and the expression shown for Y2 in Figure 8.21 gives the second value for ⬔C. To investigate possibilities for side a, set TblStart ⫽ 97 with ⌬Tbl ⫽ 1. Pressing 2nd GRAPH (TABLE) produces the results shown in Figure 8.22. Note that when a ⫽ 100, C ⫽ 90⬚ and a Figure 8.22 right triangle is formed, confirming the solution of 100 Example 2(a). When a ⬍ 100 as in part (b), is greaa ter than 1 and the table generates an ERROR message, verifying that no triangle can be formed. To investigate part (d), scroll through the table until the screen shown in Figure 8.23 appears and note that when a ⫽ 145, the second possibility for ⬔C would be near 136.4⬚. Since ⬔A ⫹ ⬔C ⬇ 45° ⫹ 136.4° ⫽ 181.4° is already greater than 180⬚ (without even taking ⬔B into Figure 8.23 account), this second value of ⬔C must be discarded. Only one triangle is possible. For an even better understanding of the SSA (ambiguous) case, scaled drawings can initially be used along with a metric ruler and protractor. Begin with a horizontal line segment of undetermined length to represent the third (unknown) side, and use the protractor to draw the given angle on either the left or right side of this segment (we chose the left). Then use the metric ruler to draw an adjacent side of appropriate length, choosing a scale that enables a complete diagram. For instance, if the given sides are 3 ft and 5 ft, use 3 cm and 5 cm instead 11 cm ⫽ 1 ft2. If the sides are 80 mi and 120 mi, use 8 cm and 12 cm 11 cm ⫽ 10 mi2, and so on. Once the adjacent side is drawn, start at the free endpoint and draw a vertical segment to represent the remaining side. A careful sketch will often indicate whether none, one, or two triangles are possible (see the Reinforcing Basic Concepts feature on page 786).

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EXAMPLE 3

䊳

Solving the Ambiguous Case of the Law of Sines Solve the triangle with side b 100 ft, side c 60 ft, and ⬔C 28.0°.

Solution

䊳

Two sides and an angle opposite are given Figure 8.24 (SSA), and we draw a diagram to help A determine the possibilities. Draw the horizontal segment of some length and use a protractor to b 100 ft c 60 ft mark ⬔C 28°. Then draw a segment 10 cm long (to represent b 100 ft2 as the adjacent 28 side of the angle, with a vertical segment 6 cm C B1 long from the free end of b (to represent c 60 ft2. It seems apparent that side c will intersect the horizontal side in two places (see Figure 8.24), and two triangles are possible. We apply the law of sines to solve the first triangle, whose features we’ll note with a subscript of 1. sin B1 sin C c b sin B1 sin 28° 100 60 5 sin B1 sin 28° 3 B1 ⬇ 51.5°

WORTHY OF NOTE In Example 3, we found ⬔B2 using the property that states the angles must be supplements (see the isosceles triangle comment on top of page 749). We could also view ⬔B1 as a QI reference angle, which also gives a QII solution of 1180 51.52ⴗ 128.5ⴗ.

law of sines

substitute

solve for sin B1 apply arcsine

Since ⬔B1 ⬔B2 180°, we know ⬔B2 ⬇ 128.5°. These values give 100.5° and 23.5° as the measures of ⬔A1 and ⬔A2, respectively. By once again applying the law of sines to each triangle, we find side a1 ⬇ 125.7 ft and a2 ⬇ 51.0 ft. See Figure 8.25. Angles

Sides (ft)

Angles

Sides (ft)

A1 ⬇ 100.5°

a1 ⬇ 125.7

A2 ⬇ 23.5°

a2 ⬇ 51.0

B1 ⬇ 51.5°

b ⴝ 100

B2 ⬇ 128.5°

b ⴝ 100

C ⴝ 28ⴗ

c ⴝ 60

C ⴝ 28ⴗ

c ⴝ 60

Figure 8.25 First solution

A b 100 ft C

28.0

B2

b 100 ft

c 60 ft B1

C

28.0

Second solution

A1

100.5

51.5

a1 ≈ 125.7 ft

b 100 ft

c 60 ft B1

C

A2

23.5

c 60 ft

28.0 128.5

a2 ≈ 51.0 ft B2


䊳

Admittedly, the scaled drawing approach has some drawbacks—it takes time to draw the diagrams and is of little use if the situation is a close call. It does, however, offer a deeper understanding of the subtleties involved in solving the SSA case. Instead of a scaled drawing, we can use a simple sketch as a guide, while keeping in mind the properties mentioned in the Worthy of Note on page 747.

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8–8


EXAMPLE 4

䊳

Solving the Ambiguous Case of the Law of Sines Solve the triangle with side a 220 ft, side b 200 ft, and ⬔A 40°.

Solution

䊳

The information given is again SSA, and we apply the law of sines with this in mind. sin A sin B a b sin B sin 40° 220 200 200 sin 40° sin B 220 200 sin 40° B1 sin1a b 220

C law of sines

b 200 ft

a 220 ft

substitute

solve for sin B

40

A

B1

apply arcsine

B1 ⬇ 35.8°

calculator result

This is the solution from Quadrant I. The QII solution is about 1180 35.82° 144.2°. At this point our solution tables have this form: Angles

Sides (ft)

Angles

Sides (ft)

A ⴝ 40ⴗ

a ⴝ 220

A ⴝ 40ⴗ

a ⴝ 220

B1 ⬇ 35.8°

b ⴝ 200

B2 ⬇ 144.2°

b ⴝ 200

C1

c1

C2

c2

It seems reasonable to once again find the remaining angles and finish by reapplying the law of sines, but observe that the sum of the two angles from the second solution already exceeds 180°: 40° 144.2° 184.2°! This means no second solution is possible (side a is too long). We find that C1 ⬇ 104.2°, and applying the law of sines gives a value of c1 ⬇ 331.7 ft.

B. You’ve just seen how we can solve SSA triangles (the ambiguous case) using the law of sines


䊳

C. Applications of the Law of Sines As “ambiguous” as it is, the ambiguous case has a number of applications in engineering, astronomy, physics, and other areas. Here is an example from astronomy. EXAMPLE 5

䊳

Solving an Application of the Ambiguous Case— Planetary Distance The planet Venus can be seen from Earth with the naked eye, but as the diagram indicates, the position of Venus is uncertain (we are unable to tell if Venus is in the near position or the far position). Given Earth is 93 million miles from the Sun and Venus is 67 million miles from the Sun, determine the closest and farthest possible distances that separate the planets in this alignment. Assume a viewing angle of ␪ ⬇ 18° and that the orbits of both planets are roughly circular.

Solution

䊳

A close look at the information and diagram shows a SSA case. Begin by applying the law of sines where E S Earth, V S Venus, and S S Sun. sin E sin V e v sin V sin 18° 67 93 93 sin 18° sin V 67 V ⬇ 25.4°

Venus

law of sines

67 substitute given values

Venus ␪

solve for sin V apply arcsine

Earth

67 93

Sun

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753

This is the angle V1 formed when Venus is farthest away. The angle V2 at the closer distance is 180° 25.4° 154.6°. At this point, our solution tables have this form: Sides (106 mi)

Angles E ⴝ 18ⴗ

e ⴝ 67

V1 ⬇ 25.4°

v ⴝ 93

S1

s1

Sides (106 mi)

Angles E ⴝ 18ⴗ

e ⴝ 67

V2 154.6°

v ⴝ 93

S2

s2

For S1 and S2 we have S1 ⬇ 180° 118° 25.4°2 136.6° (larger angle) and S2 ⬇ 180° 118° 154.6°2 7.4° (smaller angle). Reapplying the law of sines for s1 shows the farther distance between the planets is about 149 million miles. Solving for s2 shows that the closer distance is approximately 28 million miles. Now try Exercises 47 and 48 EXAMPLE 6

䊳

䊳

Solving an Application of the Ambiguous Case —Radar Detection As shown in Figure 8.26, a radar ship is 30 mi off shore when a large fleet of ships leaves port at an angle of 43°. Figure 8.26 a. If the maximum range of the ship’s radar is 20 mi, Fleet will the departing fleet be detected? b. If the maximum range of the ship’s radar is 25 mi, Radar how far from port is the fleet when it is first detected? 20 mi

Solution

䊳

a. This is again the SSA (ambiguous) case. Applying the law of sines gives sin ␪ sin 43° 20 30 30 sin 43° sin ␪ 20 sin ␪ ⬇ 1.02299754

Radar ship

43 30 mi Port

law of sines solve for sin ␪ result

No triangle is possible and the departing fleet will not be detected. b. If the radar has a range of 25 mi, the radar beam will intersect the course of the fleet in two places. sin ␪ sin 43° 25 30 30 sin 43° sin ␪ 25 ␪ ⬇ 54.9°

Figure 8.27

Radar 25 mi Radar ship

␪

125.1

43 ␣ 30 mi Port

solve for sin ␪ apply arcsine

This is the acute angle ␪ related to the farthest point from port at which the fleet could be detected (see Figure 8.27). For the second triangle, we have 180° 54.9° 125.1° (the obtuse angle) giving a measure of 180° 1125.1° 43°2 11.9° for angle ␣. For d as the side opposite ␣ we have sin 11.9° sin 43° 25 d 25 sin 11.9° d sin 43° ⬇ 7.6

C. You’ve just seen how we can use the law of sines to solve applications

law of sines

law of sines solve for d simplify

This shows the fleet is first detected about 7.6 mi from port. Now try Exercises 49 and 50

䊳

There are a number of additional, interesting applications in the Exercise Set (see Exercises 51 through 66).

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8.1 EXERCISES 䊳



䊳

1. For the law of sines, if two sides and an angle opposite one side are given, this is referred to as the case, since the solution is in doubt until further work.

2. Two inviolate properties of a triangle that can be used to help solve the ambiguous case are: (a) the angles must sum to and (b) no sine ratio can exceed .

3. For positive k 1, the equation sin ␪ k has two solutions, one in Quadrant and the other in Quadrant .

4. After a triangle is solved, you should always check to ensure that the side is opposite the angle.

5. In your own words, explain why the AAS case results in a unique solution while the SSA case does not. Give supporting diagrams.

6. Explain why no triangle is possible in each case: a. A 34°, B 73°, C 52°, a 14¿, b 22¿, c 18¿ b. A 42°, B 57°, C 81°, a 7–, b 9–, c 22–


Solve each of the following equations for the unknown part (if possible). Round sides to the nearest hundredth and degrees to the nearest tenth.

7.

sin 32° sin 18.5° a 15

sin 63° sin C 9. 21.9 18.6 11.

8.

12.

14. side b 385 m ⬔B 47° ⬔A 108°

15. side b 10 13 in. ⬔A 30° ⬔B 60° 16.

19 in. 102

18. 89 yd

sin 38° sin B 125 190

Solve each triangle using the law of sines. If the law of sines cannot be used, state why. Draw and label a triangle or label the triangle given before you begin. Use a calculator to verify your results.

13. side a 75 cm ⬔A 38° ⬔B 47°

33

sin 30° sin 52° b 12

sin 105° sin B 10. 3.14 6.28

sin C sin 19° 48.5 43.2

17.

121 29

19.

⬔A 45° ⬔B 45° side c 15 12 mi

21.

⬔B 103.4° side a 42.7 km ⬔C 19.6°

20.

22. 13

126.2 mi

98 7.2 m 22 27

⬔A 20.4° side c 12.9 mi ⬔B 63.4°

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23.

755

34.

112 0.8 cm

a 432 cm

c 56 38 b 382 cm

A

24. 35.

47 37

C B

27.5 cm

Answer each question and justify your response using a diagram, but do not solve.

25. Given ÂBC with ⬔A 30° and side c 20 cm, (a) what length for side a will produce a right triangle? (b) How many triangles can be formed if side a 8 cm? (c) If side a 12 cm, how many triangles can be formed? (d) If side a 25 cm, how many triangles can be formed?

38

C

b

36.

B 398 mm

c a

465 mm

59

C

b

37.

25

mi

B

10 2.6 c

2.9 25

a

10

26. Given ÂBC with ⬔A 60° and side c 613 m, (a) what length for side a will produce a right triangle? (b) How many triangles can be formed if side a 8 m? (c) If side a 10 m, how many triangles can be formed? (d) If side a 15 m, how many triangles can be formed?

6.7 km

c

10.9 km a

mi

Solve using the law of sines and a scaled drawing. If two triangles exist, solve both completely.

31. side c 58 mi ⬔C 59° side b 67 mi

32. side b 24.9 km ⬔B 45° side a 32.8 km

Use the law of sines to determine if no triangle, one triangle, or two triangles can be formed from the diagrams given (diagrams may not be to scale), then solve. If two solutions exist, solve both completely. Note the arrowhead marks the side of undetermined length.

33.

A

67 ft b

59

C

B 6 .8 10 13 km a

c

51 b

For Exercises 39 to 44, assume the law of sines is being applied to solve a triangle. Solve for the unknown angle (if possible), then determine if a second angle 10ⴗ ⬍ ␪ ⬍ 180ⴗ2 exists that also satisfies the proportion.

39.

sin A sin 48° 12 27

40.

sin B sin 60° 32 9

41.

sin 57° sin C 35.6 40.2

42.

sin 65° sin B 5.2 4.9

43.

sin A sin 15° 280 52

44.

sin 29° sin B 121 321

a

58 ft c A

km

30. side c 10 13 in. ⬔A 60° side a 15 in.

10 13

29. side c 25.8 mi ⬔A 30° side a 12.9 mi

38.

28. side a 36.5 yd ⬔B 67° side b 12.9 yd

C

5.9

27. side b 385 m ⬔B 67° side a 490 m

62 b

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45. Triple angle formula for sine: sin13␪2 ⴝ 3 sin ␪ ⴚ 4 sin3 ␪ Most students are familiar with the double-angle formula for sine: sin12␪2 2 sin ␪ cos ␪. The triple-angle formula for sine is given here. Use the formula to find an exact value for sin 135°, then verify the result using a reference angle. B

b 46. Radius of a circumscribed circle: R ⴝ 2 sin B

51 c

Given ÂBC is circumscribed by a circle of radius R, the radius of the circle can be found using the formula shown, where side b is opposite angle B. Find the radius of the circle shown.

䊳

A

R a

b 15 cm

C

APPLICATIONS

47. Planetary distances: In a solar system that parallels our own, the planet Sorus can be seen from a Class M planet Exercise 47 with the naked eye, but Sorus as the diagram indicates, the position 51 of Sorus is uncertain. Sorus 51 sun Assume the orbits of both planets are ␪ 82 roughly circular and Class M that the viewing angle ␪ is about 20°. If the Class M planet is 82 million miles from its sun and Sorus is 51 million miles from this sun, determine the closest and farthest possible distances that separate the planets in this alignment. 48. Planetary distances: In a solar system that parallels our own, the planet Cirrus can be seen from a Class M planet Exercise 48 with the naked eye, Cirrus but as the diagram indicates, the position 70 of Cirrus is uncertain. 70 sun Assume the orbits of Cirrus ␪ both planets are 105 roughly circular and Class M that the viewing angle ␪ is about 15°. If the Class M planet is 105 million miles from its sun and Cirrus is 70 million miles from this sun, determine the closest and farthest possible distances that separate the planets in this alignment. 49. Radar detection: A radar ship is 15.0 mi off shore from a major port when a large fleet of ships leaves the port at the 35.0° angle shown. (a) If the

maximum range of the ship’s radar is 8.0 mi, will the departing fleet be detected? (b) If the maximum range of the ship’s radar is 12 mi, how far from port is the fleet when it is first detected?

Radar 8 mi Radar ship

35.0 15 mi Port

50. Motion detection: To notify environmentalists of the presence of big game, motion detectors are installed 200 yd from a Exercise 50 watering hole. A pride of lions has just visited the hole and is leaving the Range area at the 29.0° angle 90 yd 29.0 shown. (a) If the 200 yd Motion Water maximum range of the detector motion detector is 90 yd, will the pride be detected? (b) If the maximum range of the motion detector is 120 yd, how far from the watering hole is the pride when first detected? 51. Distance between cities: The cities of Van Gogh, Rembrandt, Pissarro, and Seurat are situated as shown in the Exercise 51 diagram. Assume R that triangle RSP is 80 km isosceles and use 55 km the law of sines to find the distance 40 V S P between Van Gogh and Seurat, and between Van Gogh and Pissarro.

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. in

C

B

A

4

55.

. in

In Exercises 55 and 56, three rods are attached via pivot joints so the rods can be manipulated to form a triangle. How many triangles can be formed if angle B must measure 26ⴗ? If one triangle, solve it. If two, solve both. Diagrams are not drawn to scale.

A 11

54. Distance to target: Quarterback As part of an All-Star competition, a quarterback stands as shown in the diagram 50 yd and attempts to hit a moving target with a football. (a) If the quarterback has a 53 maximum effective range of about 35 yd, can the target be hit? (b) What is the shortest range the quarterback can have and still hit the target? (c) If the quarterback’s range is 45 yd and the target is moving at 5 yd/sec, how many seconds is the target within range?

C

B

. in

53. Distance to target: Exercise 53 To practice for a competition, an archer stands as Archer shown in the diagram and attempts to hit a moving target. (a) If 246 ft the archer has a maximum effective range of about 180 ft, 55 can the target be hit? (b) What is the shortest range the archer can have and still hit the target? (c) If the archer’s range is 215 ft and the target is moving at 10 ft/sec, how many seconds is the target within range?

56.

??

Exercise 52 52. Distance between cities: The cities of M V O 33 Mozart, Rossini, Offenbach, and Verdi 100 km 75 km are situated as shown in the diagram. Assume that triangle R ROV is isosceles and use the law of sines to find the distance between Mozart and Verdi, and between Mozart and Offenbach.

757

In the diagrams given, the measure of angle C and the lengths of sides a and c are fixed. Side c can be rotated at pivot point B. Solve any triangles that can be formed. (Hint: Begin by using the grid to find lengths a and c, then find angle C.)

57.

y a

B

C

c A x

58.

y

B

a

C

c A x

Length of a rafter: Determine the length of both roof rafters in the diagrams given.

59.

B Rafter c

Rafter a

53

32 C

A

42 ft

60.

B Rafter a

Rafter c

29

45 A

50 ft

C

8

cm

cm

12

m

c ??

61. Distance to a fire: In Yellowstone Park, a fire is spotted by park rangers stationed in two towers that are known to be 5 mi apart. Using the line between them as a baseline, tower A reports the fire is at an angle of 39°, while tower B reports an angle of 58°. How far is the fire from the closer tower?

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Exercise 62 62. Width of a canyon: To find the distance across Waimea Canyon (on the island of Kauai), a A 1000 m B surveyor marks a 28 110 1000-m baseline along the southern rim. Using a C transit, she sights on a large rock formation on the north rim, and finds the angles indicated. How wide is the canyon from point B to point C?

Exercise 65 65. Height of a mountain: Approaching from the west, a group of hikers notes the angle of elevation to the summit of a 48 35 steep mountain is 1250 m 35° at a distance of 1250 meters. Arriving at the base of the mountain, they estimate this side of the mountain has an average slope of 48°. (a) Find the slant height of the mountain’s west side. (b) Find the slant height of the east side of the mountain, if the east side has an average slope of 65°. (c) How tall is the mountain?

63. Circumscribed Exercises 63 and 64 triangles: A triangle is circumscribed within 5 cm the upper semicircle drawn in the figure. Use 27 63 52 38 the law of sines to solve the triangle given the 6.8 cm measures shown. What is the diameter of the circle? What do you notice about the triangle?

66. Distance on a map: Coffeyville and Liberal, Kansas, lie along the state’s southern border and are roughly 298 mi apart. Olathe, Kansas, is very near Exercise 66 the state’s eastern border at an angle of KANSAS Olathe 23° with Liberal and 72° with Liberal Coffeyville 72 Coffeyville 23 298 mi (using the southern border as one side of the angle). (a) Compute the distance between these cities. (b) What is the shortest (straight line) distance from Olathe to the southern border of Kansas?

64. Circumscribed triangles: A triangle is circumscribed within the lower semicircle shown. Use the law of sines to solve the triangle given the measures shown. How long is the longer chord? What do you notice about the triangle? 䊳


67. Solve the triangle shown in three ways— 10.2 cm first by using the law of sines, 30 second using right triangle trigonometry, and third using the standard 30-60-90 triangle. Was one method “easier” than the others? Use these connections to express the irrational number 13 as a quotient of two trigonometric functions of an angle. Can you find a similar expression for 12? 68. Use the law of sines and any needed identities to solve the triangle shown.

69. Similar to the law of sines, there is a law of tangents. The law says for any triangle ABC, ab ab

20 m

B 45 cm 19

x A

C

.

31

A

50 m

1 tan c 1A B2 d 2

Use the law of tangents to solve the triangle shown.

B 2x

1 tan c 1A B2 d 2

C

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759

Section 8.2 The Law of Cosines; the Area of a Triangle

70. A UFO is sighted on a direct line between the towns of Batesville and Cave City, sitting stationary in the sky. The towns are 13 mi apart as the crow flies. A student in Batesville calls a friend in Cave City and both take measurements of the angle of elevation: 35° from Batesville and 42° from Cave City. Suddenly the UFO zips across the sky at a level altitude heading directly for Cave City, then stops and hovers long enough for an additional measurement from Batesville: 24°. If the UFO was in motion for 1.2 sec, at what average speed (in mph) did it travel? 䊳


71. (7.7) Find all solutions to the equation 2 sin x cos12x2

73. (4.2) Write a real polynomial equation with smallest degree possible, having the solutions x 2, x 1, and x 1 2i.

72. (7.2) Prove the given identity: tan2x sin2x tan2x sin2x

8.2

74. (1.4) Given the points 15, 32 and (4, 2), find (a) the equation of the line containing these points and (b) the distance between these points.

The Law of Cosines; the Area of a Triangle


A. Apply the law of cosines when two sides and an included angle are known (SAS) B. Apply the law of cosines when three sides are known (SSS) C. Solve applications using the law of cosines D. Use trigonometry to find the area of a triangle

The distance formula d 21x2 x1 2 2 1y2 y1 2 2 is traditionally developed by placing two arbitrary points on a rectangular coordinate system and using the Pythagorean theorem. The relationship known as the law of cosines is developed in much the same way, but this time by using three arbitrary points (the vertices of a triangle). After giving the location of one vertex in trigonometric form, we obtain a formula that enables us to solve SSS and SAS triangles, which cannot be solved using the law of sines alone.

A. The Law of Cosines and SAS Triangles In situations where all three sides are known (but no angles), the law of sines cannot be applied. The same is true when two sides and the angle between them are known, since we must have an angle opposite one of the sides. In these two cases, side-side-side (SSS) and side-angle-side (SAS), we use the law of cosines (Figure 8.28). Figure 8.28 Law of sines cannot be applied. B c 7 ft A

WORTHY OF NOTE Keep in mind that the sum of any two sides of a triangle must be greater than the remaining side. For example, if a 7, B 20, and C 12, no triangle is possible (see the figure).

12 cm

7 cm 20 cm

B

SSS

a 16 ft

b 18 ft

C

c 7 ft

95 SAS

A

To solve these cases, it’s evident we need additional insight on the unknown angles. Consider a general triangle ABC on the rectangular coordinate system conveniently placed with vertex A at the origin, side c along the x-axis, and the vertex C at some point (x, y) in QI (Figure 8.29). y x Note cos ␪ giving x b cos ␪, and sin ␪ b b

a 16 ft C

b

Figure 8.29 (b cos u, b sin u)

y

C

b

y

a A (0, 0)

u

B xc

c x

x

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or y b sin ␪. This means we can write the point (x, y) as (b cos ␪, b sin ␪) as shown, and use the Pythagorean theorem with side x c to find the length of side a of the exterior, right triangle. It follows that a2 1x c2 2 y2

1b cos ␪ c2 1b sin ␪2 2

Pythagorean theorem 2

substitute b cos ␪ for x and b sin ␪ for y

b cos ␪ 2bc cos ␪ c b sin ␪

square binomial, square term

b cos ␪ b sin ␪ c 2bc cos ␪

rearrange terms

2 2

2 2

2

2

2

2

2

2

b 1cos ␪ sin ␪2 c 2bc cos ␪ 2

2

2

2

b2 c2 2bc cos ␪

factor out b 2 substitute 1 for cos 2 ␪ sin2 ␪

We now have a formula relating all three sides and an included angle. Since the naming of the angles is purely arbitrary, the formula can be used in any of the three forms shown. For the derivation of the formula where ⬔B is acute, see Exercise 60. The Law of Cosines For any triangle ABC and corresponding sides a, b, and c,

"

a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C

"

760

Note the relationship between the squared side and indicated angle. In words, the law of cosines says that the square of any side is equal to the sums of the squares of the other two sides, minus twice their product times the cosine of the included angle. It is interesting to note that if the included angle is 90°, the formula reduces to the Pythagorean theorem since cos 90° 0. EXAMPLE 1

䊳

Verifying the Law of Cosines

Figure 8.30

For the triangle shown in Figure 8.30, verify: a. c2 a2 b2 2ab cos C b. b2 a2 c2 2ac cos B

Solution

Figure 8.31

䊳

B c 20

Note the included angle C is a right angle. A a. c2 a2 b2 2ab cos C 2 2 2 20 10 110 13 2 21102 11013 2 cos 90° 400 100 300 0 400 ✓ b. b2 a2 c2 2ac cos B 110 132 2 102 202 211021202cos 60° 1 300 100 400 400 a b 2 500 200 300 ✓

30 30 b 10 √3

60 a 10 C

A calculator check of part (b) is shown in Figure 8.31. Now try Exercises 7 through 14

䊳

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䊳

CAUTION

When evaluating the law of cosines, a common error is to combine the coefficient of cos ␪ with the squared terms (the terms shown in blue): a2 b2 c2 2bc cos A. Be sure to use the correct order of operations when simplifying the expression.

Once additional information about the triangle is known, the law of sines can be used to complete the solution. EXAMPLE 2

䊳

Solving a Triangle Using the Law of Cosines — SAS Figure 8.32

Solve the triangle shown in Figure 8.32. Write the solution in table form. Verify your answer with a calculator.

Solution

䊳

After using the law of cosines, we often use the law of sines to complete a solution. With a little foresight, we can avoid the ambiguous case—since the ambiguous case occurs only if ␪ could be obtuse (the largest angle of the triangle). After calculating the third side of a SAS triangle using the law of cosines, use the law of sines to find the smallest angle, since it cannot be obtuse.

Figure 8.33

c 7.0 ft

The given information is SAS. Apply the law of cosines with respect to side b and ⬔B: b b2 ⬇ b⬇ 2

WORTHY OF NOTE

B

a c 2ac cos B 1162 2 172 2 21162 172cos 95° 324.522886 18.0 2

2

95

A

a 16.0 ft C

b

law of cosines with respect to b substitute known values simplify 1324.522886 ⬇ 18.0

We now have side b opposite ⬔B, and complete the solution using the law of sines, selecting the smaller angle to avoid the ambiguous case (we could apply the law of cosines again, if we chose). sin C sin B c b sin 95° sin C ⬇ 7 18 sin 95° sin C ⬇ 7 # 18 7 sin 95° b C ⬇ sin1a 18 ⬇ 22.8°

law of sines applied to ⬔C and ⬔B

substitute given values

solve for sin C apply sin1 result

For the remaining angle, ⬔A:

Angles

Sides (ft)

A ⬇ 62.2° a ⴝ 16.0

180° 195° 22.8°2 62.2°

B ⴝ 95.0ⴗ b ⬇ 18.0

C ⬇ 22.8° c ⴝ 7.0 The finished solution is shown in the table (given information is in bold), and a verification (employing the law of sines) is shown in Figure 8.33. As in Section 8.1, the differences beyond the third decimal place are attributed to round-off error. A. You’ve just seen how we can apply the law of cosines when two sides and an included angle are known (SAS)


䊳

B. The Law of Cosines and SSS Triangles When three sides of a triangle are given, we use the law of cosines to find any one of the three angles. As a good practice, we first find the largest angle, or the angle opposite the largest side. This will ensure that the remaining two angles are acute, avoiding the ambiguous case if the law of sines is used to complete the solution.

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EXAMPLE 3

䊳

Solving a Triangle Using the Law of Cosines — SSS Solve the triangle shown. Write the solution in table form, with angles rounded to tenths of a degree.

Solution

䊳

C a 15 m

b 25 m

The information is given as SSS. Since side c is the longest side, we apply the law of cosines with respect to side c and ⬔C:

A

c2 a2 b2 2ab cos C 282 1152 2 1252 2 211521252cos C 784 850 750 cos C 66 750 cos C 0.088 cos C cos10.088 C 85.0° ⬇ C

B

c 28 m

law of cosines with respect to c substitute known values simplify isolate variable term divide solve for C result

We now have ⬔C opposite side c and finish up using the law of sines. sin A sin C a c sin 85° sin A ⬇ 15 28 sin 85° sin A ⬇ 15 # 28 ⬇ 0.5336757311 A ⬇ sin10.5336757311 ⬇ 32.3°

law of sines applied to ⬔A and ⬔C

substitute given values

solve for sin A simplify solve for A result

The remaining angle must be acute and we compute it directly. ⬔B: 180° 185° 32.3°2 62.7°

The finished solution is shown in the table, with the information originally given shown in bold.

B. You’ve just seen how we can apply the law of cosines when three sides are known (SSS)

Angles

Sides (m)

A ⬇ 32.3°

a ⴝ 15

B ⬇ 62.7°

b ⴝ 25

C ⬇ 85°

c ⴝ 28


䊳

C. Applications Using the Law of Cosines As with the law of sines, the law of cosines has a large number of applications from very diverse fields including geometry, navigation, surveying, and astronomy. EXAMPLE 4

䊳

Solving an Application of the Law of Cosines — Geological Surveys A vulcanologist needs to measure the distance across the base of an active volcano. Distance AB is measured at 1.5 km, while distance AC is 3.2 km. Using a theodolite (a sighting instrument used by surveyors), angle BAC is found to be 95.7°. What is the distance across the base?

B

C

1.5 km

3.2 km 95.7 A

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Solution

䊳

763

The information is given as SAS. To find the distance BC across the base of the volcano, we apply the law of cosines with respect to side a and ⬔A. a2 b2 c2 2bc cos A 11.52 2 13.22 2 211.52 13.22cos 95.7° ⬇ 13.44347 a ⬇ 3.7

law of cosines with respect to a substitute known values simplify solve for a

The volcano is approximately 3.7 km wide at its base. Now try Exercises 37 through 40

C. You’ve just seen how we can solve applications using the law of cosines

䊳

Recall from Section 6.6 that the bearings used by surveyors are stated as a number of degrees East or West from a due North or South orientation. Headings provide the same information, but are more commonly used in aerial and nautical navigation. They are stated simply as an amount of rotation from due north, in the clockwise direction 10° ␪ 6 360°2. For instance, the bearing N 25 W and a heading of 335 would indicate the same direction or course. See Exercises 41 through 44. A variety of additional applications of the law of cosines can be found in Exercises 45 through 52.

D. Trigonometry and the Area of a Triangle While you’re likely familiar with the most common formula for a triangle’s area, A 12bh, there are actually over 20 formulas for computing this area. Many involve basic trigonometric ideas, and we’ll use some Figure 8.34 of these ideas to develop three additional forB mulas here. 1 For A 2bh, recall that b represents the a length of a designated base, and h represents the c h length of the altitude drawn to that base (see Figure 8.34). If the height h is unknown, but sides a and b with angle C between them are C b A h known, h can be found using sin C , giving Figure 8.35 a h a sin C. Figure 8.35 indicates the same B result is obtained if C is obtuse, since sin1180° C2 sin C. Substituting a sin C for h in the formula A 12bh gives h c a A 12ba sin C, or A 12ab sin C in more common form. Since naming the angles in a triangle is arbitrary, the formulas A 12bc sin A C b A 180 C and A 12ac sin B can likewise be obtained. Area Given Two Sides and an Included Angle (SAS) 1 1. A ab sin C 2

1 2. A bc sin A 2

1 3. A ac sin B 2

In words, the formulas say the area of a triangle is equal to one-half the product of two sides times the sine of the angle between them.

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EXAMPLE 5

䊳

Finding the Area of a Nonright Triangle A

Find the area of ÂBC, if a 16.2 cm, b 25.6 cm, and C 28.3°.

Solution

䊳

25.6 cm

Since sides a and b and angle C are given, we apply the first formula. 1 ab sin C 2 1 116.22125.62 sin 28.3° 2 ⬇ 98.3 cm2

A

area formula

28.3 16.2 cm

C

B

substitute 16.2 for a, 25.6 for b, and 28.3° for C result

The area of this triangle is approximately 98.3 cm2. Now try Exercises 53 and 54

䊳

Using these formulas, a second formula type requiring two angles and one side 1 2A . (AAS or ASA) can be developed. Solving for b in A bc sin A gives b 2 c sin A 2A 1 Likewise, solving for a in A ac sin B yields a . Substituting these for b 2 c sin B 1 and a in A ab sin C gives 2 1# # # a b sin C 2 2A # 2A # sin C 2A c sin B c sin A c2 sin A # sin B 2A # sin C c2 sin A sin B A 2 sin C A

given formula substitute

2A 2A for a, for b; multiply by 2 c sin B c sin A

multiply by c sin A c sin B; divide by 2A solve for A

As with the previous formula, versions relying on side a or side b can also be found. Area Given Two Angles and Any Side (AAS/ASA) 1. A

EXAMPLE 6

䊳

c2 # sin A # sin B 2 sin C

2. A

a2 # sin B # sin C 2 sin A

3. A

b2 # sin A # sin C 2 sin B

Finding the Area of a Nonright Triangle Find the area of ÂBC if a 34.5 ft, B 87.9°, and C 29.3°.

Solution

䊳

A

Since side a is given, we apply the second version of the formula. First we find the measure of angle A, then make the appropriate substitutions: A 180° 187.9° 29.3°2 62.8° a2sin B sin C A 2 sin A 134.52 2sin 87.9°sin 29.3° 2 sin 62.8° ⬇ 327.2 ft2

C

29.3 34.5 ft

87.9 B

area formula — side a

substitute 34.5 for a, 87.9° for B, 29.3° for C, and 62.8° for A simplify

The area of this triangle is approximately 327.2 ft2. Now try Exercises 55 and 56

䊳

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765

Our final formula for a triangle’s area is a useful addition to the other two, as it requires only the lengths of the three sides. The development of the formula requires only a Pythagorean identity and solving for the angle C in the law of cosines, as follows. a2 b2 2ab cos C c2

law of cosines

a b c 2ab cos C

add 2ab cos C, subtract c 2

a b c cos C 2ab

divide by 2ab

2

2

2

2

2

2

Beginning with our first area formula, we then have 1 ab sin C 2 1 ab 21 cos2C 2 1 a2 b2 c2 2 ab 1 a b 2 B 2ab

A

previous area formula

sin2 C cos2 C 1 S sin C 21 cos2C

substitute

a2 b 2 c 2 for cos C 2ab

We can now find the area of any triangle given its three sides. While the formula certainly serves this purpose, it is not so easy to use. By working algebraically and using the perimeter of the triangle, we can derive a more elegant version. 1 a2 b2 c2 2 b d A2 a2b2 c 1 a 4 2ab a2 b2 c2 a2 b2 c2 1 bd c1 a bd a2b2 c 1 a 4 2ab 2ab 1 2 2 2ab a2 b2 c2 2ab a2 b2 c2 ab c dc d 4 2ab 2ab 2 2 2 1a2 2ab b2 2 c2 1 1a 2ab b 2 c c dc d 4 2 2 1 3 1a b2 2 c2 4 3 c2 1a b2 2 4 16 1 1a b c2 1a b c2 1c a b2 1c a b2 16

square both sides factor as a difference of squares 1

2ab ; combine terms 2ab

rewrite/regroup numerator; cancel a 2b 2

factor (binomial squares)

factor (difference of squares)

For the perimeter a b c p, we note the following relationships (subtract 2c, 2b, and 2a respectively, from both sides): a b c p 2c

c a b p 2b

c a b p 2a

and making the appropriate substitutions gives A2

1 p1p 2c2 1p 2b2 1p 2a2 16

substitute

While this would provide a usable formula for the area in terms of the perimeter, we p abc 1 4 1 . Since a b, can refine it further using the semiperimeter s 2 2 16 2 we can write the expression as p p 2c p 2b p 2a ba ba b A2 a 2 2 2 2 p p p p a cba bba ab 2 2 2 2 s1s c21s b2 1s a2

rewrite expression

simplify substitute s for

p 2

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Taking the square root of each side produces what is known as Heron’s formula. A 1s1s a21s b2 1s c2

Heron’s formula

Heron’s Formula Given ÂBC with sides a, b, and c and semiperimeter s the triangle is

EXAMPLE 7

䊳

abc , the area of 2

A 1s1s a2 1s b21s c2

Solving an Application of Heron’s Formula — Construction Planning A New York City developer wants to build condominiums on the triangular lot formed by Greenwich, Watts, and Canal Streets. How many square meters does the developer have to work with if the frontage along each street is approximately 34.1 m, 43.5 m, and 62.4 m, respectively?

Solution

䊳

The perimeter of the lot is p 34.1 43.5 62.4 140 m, so s 70 m. By direct substitution we obtain A 1s1s a21s b2 1s c2 170170 34.12170 43.52170 62.42 170135.92126.5217.62 1506,118.2 ⬇ 711.4

Heron’s formula substitute known values simplify multiply result

The developer has about 711.4 m2 of land to work with. A calculator verification is shown in the figure, where the keystrokes STO ALPHA LN were used to store the semiperimeter in memory location S. This greatly simplifies the use of Heron’s formula, which is then used to verify the preceding calculations. D. You’ve just seen how we can use trigonometry to find the area of a triangle


䊳

For a derivation of Heron’s formula that does not depend on trigonometry, see Appendix B.

8.2 EXERCISES 䊳



1. When the information given is SSS or SAS, the law of is used to solve the triangle.

2. Fill in the blank so that the law of cosines is complete: c2 a2 b2 cos C.

3. If the law of cosines is applied to a right triangle, the result is the same as the theorem, since cos 90° 0.

4. Write out which version of the law of cosines you would use to begin solving the A triangle shown:

B

52 m 17 37 m

C

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5. Solve the triangle in Exercise 2 using only the law of cosines, then by using the law of cosines followed by the law of sines. Which method was more efficient?

6. Begin with a2 b2 c2 2bc cos A and write cos A in terms of a, b, and c (solve for cos A). Why must b2 c2 a2 6 2bc hold in order for a solution to exist?


Determine whether the law of cosines can be used to begin the solution process for each triangle.

C

6.8 AU

C

49

26.

C

C

A 8 yd

15 yd

A

For each triangle, verify all three forms of the law of cosines.

14. B 50 km A 30 80 816 mi 105 577 mi 26.6 km 52.4 km 70 30 45 A C 1114.7 mi B

C

27. side c 10 13 in. side b 6 13 in. side a 1513 in. 28. side a 282 ft side b 129 ft side c 300 ft 29. side a 32.8 km side b 24.9 km side c 12.4 km 30. B

Solve each of the following equations for the unknown part.

432 cm 382 cm

15. 42 52 62 2152 162cos B

16. 12.92 15.22 9.82 2115.2219.82cos C

20. 202 182 98 2118221982cos B 2

208 cm

mi

i

21. side a 75 cm ⬔C 38° side b 32 cm 23. side c 25.8 mi ⬔B 30° side a 12.9 mi

22. side b 385 m ⬔C 67° side a 490 m

C

B

3

25

4.1

0 1

mi

C

Solve each triangle using the law of cosines. Use a calculator to verify your results.

C

32.

A 2.9 10 25 m

10 25

19. 102 122 152 21122 1152cos A

31.

2.3

18. b2 3.92 9.52 213.9219.52cos 30°

A

33. side a 12 13 yd side b 12.9 yd side c 9.2 yd 34. side a 36.5 AU side b 12.9 AU side c 22 AU

km 10 9 . 4 13

B

2.9 1013 km

17. a2 92 72 2192 172cos 52°

27.5 ft

32.5 ft

B 12.5 yd

C

B 141

B

12.

2

29 465 mm

A

88

C

538 mm

km

10.

92

29 49 cm

2

6.7 km A

50 cm

13.

98

10 1

A

B

A

10.9 km

B

B

11.

25. B

B

A

15 mi 12 mi

24. 15 in.

85

A

30 km

C

C

12 in.

15 km

C

A

28 km

9.

8.

B

8

7.

Solve using the law of cosines (if possible). Label each triangle appropriately before you begin. Use a calculator to verify your results.

6.

䊳

767


A

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35. Alternative form for the law of cosines: B

b ⴙc ⴚa 2bc 52 m By solving the law of 39 m cosines for the cosine of the angle, the formula A C 37 m can be written as shown. Derive this formula (solve for cos ␪), beginning from a2 b2 c2 2bc cos A, then use this form to begin the solution of the triangle given. 2

2

2

cos A ⴝ

䊳

36. The perimeter of a trapezoid: P ⴝ a ⴙ b ⴙ h (csc ␣ ⴙ csc ␤) The perimeter of a a trapezoid can be found using the h formula shown, b a where a and b b represent the lengths of the parallel sides, h is the height of the trapezoid, and ␣ and ␤ are the base angles. Find the perimeter of Trapezoid Park (to the nearest foot) if a 5000 ft, b 7500 ft, and h 2000 ft, with base angles ␣ 42° and ␤ 78°.

APPLICATIONS

37. Distance between cities: The satellite Mercury II measures its distance from Portland and from Green Bay using radio waves as shown. Using an on-board sighting device, the satellite determines that ⬔M is 99°. How many miles is it from Portland to Green Bay? Mercury II M 99 1435 mi

692 mi

P

G

Portland

Green Bay

m

38. Distance between cities: Voyager VII measures its distance from Los Angeles and from San Francisco using radio waves as shown. Using an onboard sighting device, the satellite determines ⬔V is 95°. How many kilometers separate Los Angeles and San Francisco?

39. Runway length: Surveyors are measuring a large, marshy area outside of the city as part of a feasibility study for the 1.8 mi construction of a new 2.6 mi airport. Using a 51 theodolite and the markers shown gives the information indicated. If the main runway must be at least 11,000 ft long, and environmental concerns are satisfied, can the airport be constructed at this site (recall that 1 mi 5280 ft)? 40. Tunnel length: An engineering firm decides to bid on a proposed tunnel through Harvest Mountain. In order to find the tunnel’s length, the measurements shown are taken. (a) How long will the tunnel be? (b) Due to previous tunneling experience, the firm estimates a cost of $5000 per foot for boring through this type of rock and constructing the tunnel according to required specifications. If management insists on a 25% profit, what will be their minimum bid to the nearest hundred?

Voyager VII V 95 488 km

311 km

685 yd

610 yd 79

S San Francisco

L v

Los Angeles

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41. Trip planning: A business executive is going to fly the corporate jet from Providence to College Cove. She calculates the distances shown using a map, with Mannerly Main for reference since it is due west of Providence. What is the measure of angle P? What heading should she set for this trip?

45. Geoboard geometry: A rubber band is placed on a geoboard (a board with all pegs 1 cm apart) as shown. Approximate the perimeter of the triangle formed by the rubber band and the angle formed at each vertex. (Hint: Use a standard triangle to find ⬔A and length AB.)

Exercise 41

Exercise 45 North West

College Cove C

B

East South

m 354 mi

p 198 mi

M Mannerly Main

C A

c 423 mi

P Providence

42. Trip planning: A troop of Scouts is planning a hike from Montgomery to Pattonville. They calculate the distances shown using a map, using Bradleyton for reference since it is due east of Montgomery. What is the measure of angle M? What heading should they set for this trip?

46. Geoboard geometry: A rubber band is placed on a geoboard as shown. Approximate the perimeter of the triangle formed by the rubber band and the angle formed at each vertex. (Hint: Use a Pythagorean triple, then find angle A.) Exercise 46

B

Exercise 42 p 21 mi

M

B

C A

b 18 mi

m 10 mi P

43. Aerial distance: Two planes leave Los Angeles International Airport at the same time. One travels due west (at heading 270°) with a cruising speed of 450 mph, going to Tokyo, Japan, with a group that seeks tranquility at the foot of Mount Fuji. The other travels at heading 225° with a cruising speed of 425 mph, going to Brisbane, Australia, with a group seeking adventure in the Great Outback. Approximate the distance between the planes after 5 hr of flight. 44. Nautical distance: Two ships leave Honolulu Harbor at the same time. One travels 15 knots (nautical miles per hour) at heading 150°, and is going to the Marquesas Islands (Crosby, Stills, and Nash). The other travels 12 knots at heading 200°, and is going to the Samoan Islands (Samoa, le galu a tu). How far apart are the two ships after 10 hr?

In Exercises 47 and 48, three rods are attached via pivot joints so the rods can be manipulated to form a triangle. Find the three angles of the triangle formed.

47.

C B A 12 cm

9 cm

20 cm

48. A

B C 6 in.

11 in. 16 in.

49. Pentagon perimeter: Find the approximate perimeter of a regular pentagon that is circumscribed by a circle with radius r 10 cm. 50. Hexagon perimeter: Find the perimeter of a regular hexagon that is circumscribed by a circle with radius r 15 cm.

Exercise 49

m

10 c

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Solve the following triangles. Give exact values for the sides and round angles to the nearest tenth. (Hint: Use Pythagorean triples.)

51.

52.

y 5

C

A

5

y A

B

B

5

x

C

5

56. Area of the Nile River Delta: The Nile River Delta is one of the world’s largest. The delta begins slightly upriver from the Egyptian capitol (Cairo) and stretches along the Mediterranean from Alexandria in the west to Port Said in the east (over 240 km). Approximate the area of this rich agricultural region using the two triangles shown.

x

Exercise 53 53. Billboard design: Creative Designs iNc. has designed a flashy, new billboard for "Spice one of its clients. Using a rectangular highway up" billboard measuring 20 ft a by 30 ft, the primary advertising area is a triangle formed using the diagonal of the billboard as one side, and one-half the base as another (see figure). Use the dimensions given to find the angle formed at the corner, then compute the area of the triangle using two sides and this included angle.

Alexandria

20° 42°

䊳

Nile Delta

20° 45°

Port Said

Cairo

57. Area of the Yukon Territory: The Yukon Territory in northwest Canada is roughly triangular in shape with sides of 1289 km, 1063 km, and 922 km. What is the approximate area covered by this territory?

54. Area caught by Exercise 54 surveillance camera: A stationary surveillance camera is set up to 110 ft monitor activity in the 225 ft 38° parking lot of a shopping mall. If the camera has a 38° field of vision, how many square feet of the parking lot can it tape using the dimensions given? 55. Pricing for undeveloped Exercise 55 lots: Undeveloped land in a popular resort area is selling for $3,000,000/acre. Given the dimensions of the lot 42° 65° shown, (a) find what 299 ft percent of a full acre is being purchased (to the nearest whole percent), and (b) compute the cost of the lot. Recall that 1 acre 43,560 ft2.

218 km

YUKON

58. Alternate method for computing area: Referring to Exercise 53, since the dimensions of the billboard are known, all three sides of the triangle can actually be determined. Find the length of the sides rounded to the nearest whole, then use Heron’s formula to find the area of the triangle. How close was your answer to that in Exercise 53?

EXTENDING THE CONCEPT Exercise 59

59. No matter how hard I try, I cannot solve the triangle shown. Why?

Sputnik 10

60. In Figure 8.29 (page 759), note that if the x-coordinate of vertex B is greater than the x-coordinate of vertex C, ⬔B becomes acute, and ⬔C obtuse. How does this change the relationship between x and c? Verify the law of cosines remains unchanged.

B

502 mi

387 mi

A

902 mi

C

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61. For the triangle shown, verify that c ⫽ b cos A ⫹ a cos B, then use two different forms of the law of cosines to show this relationship holds for any triangle ABC. 62. Most students are familiar with this double-angle formula for cosine: cos12␪2 ⫽ cos2␪ ⫺ sin2␪. The triple angle formula for cosine is cos13␪2 ⫽ 4 cos3␪ ⫺ 3 cos ␪. Use the formula to find an exact value for cos 135°. Show that you get the same result as when using a reference angle. 䊳

771

Section 8.3 Vectors and Vector Diagrams

Exercise 61 B 53.9 mi A

117⬚

25⬚

37 mi 38⬚

78 mi

C


63. (5.4) Write the expression as a single term in simplest form: 2 log24 ⫹ 2 log23 ⫺ 2 log26. 64. (6.2) State exact forms for each of the following: ␲ 7␲ ␲ sina b, cosa b, and tana b. 6 6 3

8.3


A. Represent a vector

C. D.

E.

quantity geometrically Represent a vector quantity graphically Perform defined operations on vectors Represent a vector quantity algebraically and find unit vectors Use vector diagrams to solve applications

Figure 8.37 B 300 mph A

300 mph C

66. (4.2) Use synthetic division to show f 1⫺22 7 0 for f 1x2 ⫽ ⫺x4 ⫺ x3 ⫹ 7x2 ⫹ x ⫺ 6.

Vectors and Vector Diagrams

LEARNING OBJECTIVES

B.

65. (6.7) Use fundamental identities to find the values of all six trig functions that satisfy the conditions. 5 sin x ⫽ ⫺ and cos x 7 0 13

D

The study of vectors is closely connected to the study of force, motion, velocity, and other related phenomena. Vectors enable us to quantify certain characteristics of these phenomena and to physically represent their magnitude and direction with a simple model (quantify: to assign a relative numeric value for purposes of study and comparison). While very uncomplicated, this model turns out to be a powerful mathematical tool.

A. The Notation and Geometry of Vectors Measurements involving time, area, volume, energy, and temperature are called scalar measurements or scalar quantities because each can be adequately described by their magnitude alone and the appropriate unit or “scale.” The related real number is simply called a scalar. Concepts that require more than a single quantity to describe their attributes are called vector quantities. Examples might inFigure 8.36 clude force, velocity, and displacement, which require knowA B ing a magnitude and direction to describe them completely. To begin our study, consider two identical airplanes flying Line AB at 300 mph, on parallel courses and in the same direction. A B Although we don’t know how far apart they are, what direction they’re flying, or if one is “ahead” of the other, we can still Segment AB model, “300 mph on parallel courses,” using directed line A B segments (Figure 8.36). Drawing these segments parallel with Directed segment AB the arrowheads pointing the same way models the direction of flight, while drawing segments the same length indicates the velocities are equal (Figure 8.37). The directed segment used to represent a vector quantity is simply called a vector. In this case the length of the vector models the magnitude of the velocity, while the arrowhead indicates the direction of travel. The origin of the segment is called the initial point, with the arrowhead pointing to the terminal point. These are labeled using capital letters as shown in Figure 8.37 and we call this a geometric representation of the vectors.

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Figure 8.38

Vectors can be !named using the initial and terminal points that define them (initial ! point first) as in AB and CD , or using a bold, lowercase letter with the favorites being v (first letter of the word vector) and u. Other lowercase, bold letters can be used, and subscripted vector names (v1, v2, v3, . . .) are also common.! Two vectors! are equal if they have the! same! magnitude and direction. For u ⫽ AB and v ⫽ CD , we can say u ⫽ v or AB ⫽ CD since both airplanes are flying at the same speed and in the same direction (Figure 8.37). Based on these conventions, it seems reasonable to represent an airplane flying at 600 mph with a vector that is twice as long as u or v, and one flying at 150 mph with a vector that is half as long. If all planes are flying in the same direction Figure 8.39 on parallel courses, we can represent them geometrically as shown in Figure 8.38, and state that v1 ⫽ 2v, v2 ⫽ 12v, and v1 ⫽ 4v2. The multiplication of a vector by a constant is called scalar multiplication, since the product changes only the scale or size of the vector v 300 and not its direction. Finally, consider the airplane represented by vector w, flying mph at 200 mph on a parallel course but in the opposite direction w 2 200 ⫽ (Figure 8.39). In this case, the directed segment will be 200 300 3 mph as long as v and point in the opposite or “negative” direction. In perspective we can now state: w ⫽ ⫺23v, w ⫽ ⫺13v1, w ⫽ ⫺43v2, or any equivalent form of these equations.

v 300 mph v2 v1

150 mph

600 mph

EXAMPLE 1

䊳

Using Geometric Vectors to Model Forces Acting on a Point Two tugboats are attempting to free a barge that is stuck on a sand bar. One is pulling with a force of 2000 newtons (N) in a certain direction, the other is pulling with a force of 1500 N in a direction that is perpendicular to the first. Represent the situation geometrically using vectors.

Solution

䊳

A. You’ve just seen how we can represent a vector quantity geometrically

We could once again draw a vector of arbitrary length and let it represent the 2000-N force applied by the first tugboat. For better perspective, we can actually use a ruler and choose a convenient length, say 6 cm. We then represent the pulling force of the second tug 3 with a vector that is 1500 2000 ⫽ 4 as long (4.5 cm), drawn at a 90° angle with relation to the first. Note that many correct solutions are possible, depending on the direction of the first vector drawn.

N


䊳

B. Vectors and the Rectangular Coordinate System Figure 8.40 y (3, 4) u

v 3 x 4

w 5

Representing vectors geometrically (with a directed line segment) is fine for simple comparisons, but many applications involve numerous vectors acting on a single point or changes in a vector quantity over time. For these situations, a graphical representation in the coordinate plane helps to analyze this interaction. The only question is where to place the vector on the grid, and the answer is . . . it really doesn’t matter. Consider the three vectors shown in Figure 8.40. From the initial point of each, counting four units in the vertical direction, then three units in the horizontal direction, puts us at the terminal point. This shows the vectors are all 5 units long (since a 3-4-5 triangle is formed) and are ¢y 4 ⫽ 2. In other words, they are equivalent vectors. all parallel (since slopes are equal: ¢x 3 Since a vector’s location is unimportant, we can replace any given vector with one whose initial point is (0, 0). This unique and equivalent vector is called the position vector.

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773

Position Vectors For a vector v with initial point (x1, y1) and terminal point (x2, y2), the position vector for v is v ⫽ Hx2 ⫺ x1, y2 ⫺ y1I,

an equivalent vector with initial point (0, 0) and terminal point 1x2 ⫺ x1, y2, ⫺y1 2. For instance, the initial and terminal points of vector w in Figure 8.40 are 12, ⫺42 and (5, 0), respectively, with 15 ⫺ 2, 0 ⫺ 1⫺422 ⫽ 13, 42. Since (3, 4) is also the terminal point of v (whose initial point is at the origin), v is the position vector for u and w. This observation also indicates that every geometric vector in the xy-plane corresponds to a unique ordered pair of real numbers (a, b), with a as the horizontal component and b as the vertical component of the vector. As indicated, we denote the vector in component form as Ha, bI, using the new notation to prevent confusing vector Ha, bI with the ordered pair (a, b). Finally, while each of the vectors in Figure 8.40 has a component form of H3, 4I, the horizontal and vertical components can be read directly only from v ⫽ H3, 4I, giving it a distinct advantage.

WORTHY OF NOTE For vector u, the initial and terminal points are 1⫺5, ⫺12 and 1⫺2, 32, respectively, yielding the position vector H⫺2 ⫺ 1⫺52, 3 ⫺ 1⫺12I ⫽ H3, 4I as noted.

EXAMPLE 2

䊳

Solution

䊳

Figure 8.41

⫺5

w H12, ⫺5I y

x

12

r

Verifying the Components of a Position Vector

Vector w ⫽ H12, ⫺5I has initial point 1⫺4, 32. a. Find the coordinates of the terminal point. b. Verify the position vector for w is H12, ⫺5I and find its length. c. Use the DRAW feature of a calculator to graph w and its position vector. a. Since w has a horizontal component of 12 and a vertical component of ⫺5, we add 12 to the x-coordinate and ⫺5 to the y-coordinate of the initial point. This gives a terminal point of 1⫺4 ⫹ 12, 3 ⫹ 1⫺52 2 ⫽ 18, ⫺22. b. To verify, we use the initial and terminal points to compute Hx2 ⫺ x1, y2 ⫺ y1I, giving a position vector of H8 ⫺ 1⫺42, ⫺2 ⫺ 3I ⫽ H12, ⫺5I. To find its length we can use either the Pythagorean theorem or simply note that a 5-12-13 Pythagorean triple is formed (see Figure 8.41). Vector w has a length of 13 units. c. Most graphing calculators have a feature that can actually draw a vector in the plane, in the form of a line segment that begins and ends at specified points. It’s a good practice to first delete any equations in the Y= screen, use the keystrokes 2nd PRGM 1 to clear any old drawings that may exist, and set an appropriate window before Figure 8.42 entering the vector. From the home screen, the keystrokes 2nd PRGM (DRAW) ENTER

2:Line(ⴚ4, 3, 8, ⴚ2) will draw the vector w from (⫺4, 3) (the first two entries of the argument), to the point (8, ⫺2). Returning to the home screen and drawing the position vector for w will produce the screen shown in Figure 8.42. Note these two vectors appear to have the same magnitude and direction, as expected. ENTER

5.4

4.8

14

7


䊳

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For the remainder of this section, vector v ⫽ Ha, bI will refer to the unique position vector for all those equivalent to v. Upon considering the graph of Ha, bI (shown in QI for convenience in Figure 8.43), several things are immediately evident. The length or magnitude of the vector, which is denoted 冟v冟, can be determined using the Pythagorean theorem: 0 v 0 ⫽ 2a2 ⫹ b2. In addition, basic trigonometry shows the horizontal a component can be found using cos ⫽ or a ⫽ 冟v冟cos , with the vertical component 冟v冟 b being sin ⫽ or b ⫽ 冟v冟sin . Finally, we note the angle can be determined using 冟v冟 b b tan ⫽ , or r ⫽ tan⫺1 ` ` and the quadrant of v. a a

Figure 8.43 y Ha, bI |v| b r a

x

Vector Components in Trig Form

For a position vector v ⫽ Ha, bI and angle , we have . . . horizontal component: a ⫽ 冟v冟cos

y

vertical component: b ⫽ 冟v冟sin , where and

r ⫽ tan⫺1 `

Ha, bI

b ` a

v

b r

冟v冟 ⫽ 2a2 ⫹ b2

x

a

The ability to model characteristics of a vector using these equations is a huge benefit to solving applications, since we must often work out solutions using only the partial information given. EXAMPLE 3

Solution

䊳

䊳

Finding the Magnitude and Direction Angle of a Vector For v1 ⫽ H⫺2.5, ⫺6I and v2 ⫽ H313, 3I, a. Graph each vector and name the quadrant where it is located. b. Find their magnitudes. c. Find the angle for each vector (round to tenths of a degree as needed). a. The graphs of v1 and v2 are shown in Figure 8.44. Using the signs of each coordinate, we note that v1 is in QIII, and v2 is in QI. b.

䊲

䊲

Algebraic Solution

冟v1冟 ⫽ 21⫺2.52 ⫹ 1⫺62 ⫽ 16.25 ⫹ 36 ⫽ 142.25 ⫽ 6.5 2

冟v2冟 ⫽ 213 132 2 ⫹ 132 2 ⫽ 127 ⫹ 9 ⫽ 136 ⫽6

2

Figure 8.44 y

H3√3, 3I v2

1

2 x

v1

H2.5, 6I

Graphical Solution

The 2nd APPS (ANGLE) 5:R 䊳Pr( command (introduced in Section 6.7) can be used to find the magnitude of any position vector, as shown in Figure 8.45. Figure 8.45

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c.

䊲

䊲

Algebraic Solution

⫺6 ` ⫺2.5 ⫽ tan⫺1 12.42 ⬇ 67.4° In QIII, 1 ⬇ 247.4°. For v1: r ⫽ tan⫺1 `

For v2: r ⫽ tan⫺1 `

3 ` 3 13 13 b ⫽ 30° ⫽ tan⫺1a 3 In QI, 2 ⫽ 30°.

Graphical Solution

Similarly, the 2nd APPS (ANGLE) 6:R䊳P␪( command (also from Section 6.7) can be used to find the angle for any position vector. Note in Figure 8.46, this feature returns a negative angle for v1. Adding 360⬚ in the second entry reveals this angle is coterminal with our algebraic result. Figure 8.46


EXAMPLE 4

Solution

䊳

䊳

䊳

Finding the Horizontal and Vertical Components of a Vector The vector v ⫽ Ha, bI is in QIII, has a magnitude of 冟v冟 ⫽ 21, and forms an angle of 25° with the negative x-axis (Figure 8.47). Find the horizontal and vertical components of the vector, rounded to tenths. Begin by graphing the vector and setting up the equations for its components. For r ⫽ 25°, ⫽ 205°.

Figure 8.47 205 x 25 |v| 21 Ha, bI y

For the horizontal component: a ⫽ 冟v冟cos ⫽ 21 cos 205° ⬇ ⫺19.0

For the vertical component: b ⫽ 冟v冟sin ⫽ 21 sin 205° ⬇ ⫺8.9

With v in QIII, its component form is approximately H⫺19, ⫺8.9I. See Figure 8.48. As a check, we use the R 䊳Pr and R 䊳P␪ features of a calculator as shown in Figure 8.49 ✓. Figure 8.48

Figure 8.49

19

x

25 |v| 21

8.9

Ha, bI B. You’ve just seen how we can represent a vector quantity graphically

y


䊳

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C. Operations on Vectors and Vector Properties The arithmetic operations defined for vectors have a close-knit graphical representation. Consider a local park having a large pond with pathways around both sides, so that a park visitor can enjoy the view from either side. Suppose v ⫽ H8, 2I is the position vector representing a person who decides to turn to the right at the pond, while u ⫽ H2, 6I represents a person who decides to first turn left. At (8, 2) the first person changes direction and walks to (10, 8) on the other side of the pond, while the second person arrives at (2, 6) and turns to head for (10, 8) as well. This is shown graphically in Figure 8.50 and demonstrates that (1) a parallelogram is formed (opposite sides equal and parallel), (2) the path taken is unimportant relative to the destination, and (3) the coordinates of the destination represent the sum of corresponding coordinates from the terminal points of u and v: 12, 62 ⫹ 18, 22 ⫽ 12 ⫹ 8, 6 ⫹ 22 ⫽ 110, 82. In other words, the result of adding u and v gives the new position vector u ⫹ v ⫽ w, called the resultant or the resultant vector. Note the resultant vector is a diagonal of the parallelogram formed. Geometrically or graphically, the addition of vectors can be viewed as a “tail-to-tip” combination of one with another, by shifting one vector (without changing its direction) so that its tail (initial point) is at the tip (terminal point) of the other vector. This is illustrated in Figures 8.51 through 8.53.

Figure 8.50 y (10 0 8) (22

v

) x

Figure 8.51 Given vectors u and v

Figure 8.52 Shift vector v x

u

v

Figure 8.53 Shift vector u x

u

v u

uv

uv

v Tail of v to the tip of u

WORTHY OF NOTE The geometry of vector subtraction is a key part of resolving a vector into perpendicular components that are nonquadrantal. Applications of this concept are wide ranging, and include thrust and drag forces, tension and stress limits in a cable, and others.

x

u

v H10, 9I

y

Tail of u to the tip of v

y

The subtraction of vectors can be understood as either u ⫺ v or u ⫹ 1⫺v2. Since the location of a vector is unimportant relative to the information it carries, vector subtraction can be interpreted as the tip-to-tip diagonal of the parallelogram from vector addition. In Figures 8.51 to 8.53, assume u ⫽ H1, ⫺5I and v ⫽ H9, ⫺4I. Then u ⫺ v ⫽ H1, ⫺5I ⫺ H9, ⫺4I ⫽ H1 ⫺ 9, ⫺5 ⫹ 4I giving the position vector H⫺8, ⫺1I. By repositioning this vector with its tail at the tip of v, we note the new vector points directly at u, forming the diagonal (see Figure 8.54). Scalar multiplication of vectors also has a graphical representation that corresponds to the geometric description given earlier.

H10, 9I y

Figure 8.54

x

v u uv (repositioned) v

y

Operations on Vectors

Given vectors u ⫽ Ha, bI, v ⫽ Hc, dI, and a scalar k,

1. u ⫹ v ⫽ Ha ⫹ c, b ⫹ dI 2. u ⫺ v ⫽ Ha ⫺ c, b ⫺ dI 3. ku ⫽ Hka, kbI for k 僆⺢

If k 7 0, the new vector points in the same direction as u. If k 6 0, the new vector points in the opposite direction as u.

u

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EXAMPLE 5

Solution

䊳

Representing Operations on Vectors Graphically

Given u H3, 2I and v H4, 6I compute each of the following and represent the result graphically: 1 1 a. 2u b. v c. 2u v 2 2 Note the relationship between part (c) and parts (a) and (b).

䊳

a. 2u 2 H3, 2I

b.

H6, 4I

1 1 v H4, 6I 2 2 H2, 3I

y

1 c. 2u v H6, 4I H2, 3I 2 H8, 1I

y

y

H6, 4I

H6, 4I 2u

2u x

u H3, 2I

x

qv

H8, 1I

qv

H2, 3I

x

H2, 3I

v H4, 6I

Now try Exercises 31 through 48 Figure 8.55

䊳

Most graphing calculators have built-in features that can be used to perform arithmetic operations on vectors. Essentially, we use left and right braces ” {” and ”}” to denote the vector, instead of ” H” and ” I”, respectively. In Figure 8.55, the computation from Example 5(c) is verified using this feature (recall the braces are accessed using ( ) the 2nd ( { ) and 2nd ( } ) keys). The properties that guide operations on vectors closely resemble the familiar properties of real numbers. Note we define the zero vector 0 H0, 0I as one having no magnitude or direction. Properties of Vectors For vector quantities u, v, and w and real numbers c and k, 1. 3. 5. 7. 9.

1u u uvvu 1u v2 w u 1v w2 u0u k1u v2 ku kv

2. 4. 6. 8. 10.

0u 0 k0 u v u 1v2 1ck2u c1ku2 k1cu2 u 1u2 0 1c k2u cu ku

Proof of Property 3

For u Ha, bI and v Hc, dI, we have

C. You’ve just seen how we can perform defined operations on vectors

u v Ha, bI Hc, dI Ha c, b dI Hc a, d bI Hc, dI Ha, bI vu

sum of u and v vector addition commutative property vector addition result

Proofs of the other properties are similarly derived (see Exercises 89 through 97).

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D. Algebraic Vectors, Unit Vectors, and i, j Form

While the bold, small case v and the Ha, bI notation for vectors has served us well, we now introduce an alternative form that is somewhat better suited to the algebra of vectors, and is used extensively in some of the physical sciences. Consider the vector H1, 0I, a vector 1 unit in length extending along the x-axis. It is called the horizontal unit vector and given the special designation i (not to be confused with the imaginary unit i ⫽ 1⫺1). Likewise, the vector H0, 1I is called the vertical unit vector and given the designation j (see Figure 8.56). Using scalar multiplication, the unit vector along the negative x-axis is ⫺i and along the negative y-axis is ⫺j. SimiFigure 8.56 larly, the vector 4i represents a position vector 4 units long along the y x-axis, and ⫺5j represents a position vector 5 units long along the (0, 1) negative y-axis. Using these conventions, any nonquadrantal vector Ha, bI can be written as a linear combination of i and j, with a and b j (1, 0) expressed as multiples of i and j, respectively: ai ⫹ bj. These ideas x can easily be generalized and applied to any vector. i WORTHY OF NOTE

Algebraic Vectors and i, j Form

Earlier we stated, “Two vectors are equal if they have the same magnitude and direction.” Note that this means two vectors are equal if their components are equal.

For the unit vectors i ⫽ H1, 0I and j ⫽ H0, 1I, any arbitrary vector v ⫽ Ha, bI can be written as a linear combination of i and j: v ⫽ ai ⫹ bj Graphically, v is being expressed as the resultant of a vector sum.

EXAMPLE 6

䊳

Finding the Horizontal and Vertical Components of Algebraic Vectors Vector u is in QII, has a magnitude of 15, and makes an angle of 20° with the negative x-axis. a. Graph the vector. b. Find the horizontal and vertical components (round to one decimal place) then write u in component form. c. Write u in terms of i and j.

Solution Figure 8.58 y Ha, bI

|u| 15

x

5.1j

20

14.1i

䊳

a. The vector is graphed in Figure 8.57. Figure 8.57 b. Horizontal Component Vertical Component a ⫽ 冟v冟cos b ⫽ 冟v冟sin Ha, bI |u| 15 ⫽ 15 cos 160° ⫽ 15 sin 160° ⬇ ⫺14.1 ⬇ 5.1 r 20 With the vector in QII, u ⬇ H⫺14.1, 5.1I in component form. c. In terms of i and j we have u ⬇ ⫺14.1i ⫹ 5.1j. See Figure 8.58.

y

160 x

Now try Exercises 49 through 62 Figure 8.59 y

H6, 8I v

10

8

6 x

䊳

Some applications require that we find a nonhorizontal, nonvertical vector one unit in length, having the same direction as a given vector v. To understand how this is done, consider vector v ⫽ H6, 8I. Using the Pythagorean theorem we find 冟v冟 ⫽ 10, and can form a 6-8-10 triangle using the horizontal and vertical components (Figure 8.59). Knowing that similar triangles have sides that are proportional, we can find a unit vector in the same direction as v by dividing all three sides by 10, giving a triangle with sides 35, 45, and 1. The new vector “u” (along the hypotenuse) indeed points in the same direction since we 4 2 9 16 3 2 ⫹ ⫽ 1. have merely shortened v, and is a unit vector since a b ⫹ a b S 5 5 25 25

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In retrospect, we have divided the components of vector v by its magnitude 冟v冟 (or multiplied components by the reciprocal of 冟v冟) to obtain the desired unit vector: H6, 8I 6 8 3 4 v ⫽ h , i ⫽ h , i . In general we have the following: ⫽ 冟v冟 10 10 10 5 5 Unit Vectors

For any nonzero vector v ⫽ Ha, bI ⫽ ai ⫹ bj, the vector a b v ⫽ i ⫹ j 冟v冟 2a2 ⫹ b2 2a2 ⫹ b2 is a unit vector in the same direction as v. u⫽

You are asked to verify this relationship in Exercise 100. In summary, for vector v ⫽ 6i ⫹ 8j, we find 冟v冟 ⫽ 262 ⫹ 82 ⫽ 10, so the unit vector pointing in the same v 3 4 direction is ⫽ i ⫹ j. See Exercises 63 through 74. 冟v冟 5 5 EXAMPLE 7

䊳

Using Unit Vectors to Find Coincident Vectors Vectors u and v form the 37° angle illustrated in the figure. Find the vector w (in red), which points in the same direction as v (is coincident with v) and forms the base of the right triangle shown.

Solution

䊳

WORTHY OF NOTE In this context, w is called the projection of u on v, an idea applied more extensively in Section 8.4.

D. You’ve just seen how we can represent a vector quantity algebraically and find unit vectors

Using the Pythagorean theorem we find 冟u冟 ⬇ 7.3 and 冟v冟 ⫽ 10. Using the cosine of 37° the magnitude of w is then 冟w冟 ⬇ 7.3 cos 37° or about 5.8. To ensure that w will point in the same direction as v, we simply multiply the 5.8 magnitude by the unit H8, 6I v ⫽ 15.82H0.8, 0.6I, vector for v: 冟w冟 ⬇ 15.82 冟v冟 10 and we find that w ⬇ H4.6, 3.5I. As a check we use the Pythagorean theorem: 24.62 ⫹ 3.52 ⫽ 133.41 ⬇ 5.8.

y

H2, 7I

u

H8, 6I v

37 w


x

䊳

E. Vector Diagrams and Vector Applications Applications of vectors are virtually unlimited, with many of these in the applied sciences. Here we’ll look at two applications that are an extension of our work in this section. In Section 8.4 we’ll see how vectors can be applied in a number of other creative and useful ways. In Example 1, two tugboats were pulling on a barge to dislodge it from a sand bar, with the pulling force of each represented by a vector. Using our knowledge of vector components, vector addition, and resultant forces (a force exerted along the resultant), we can now determine the direction and magnitude of the resultant force if we know the angle formed by one of the vector forces and the barge.

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EXAMPLE 8

䊳

Solving an Application of Vectors — Force Vectors Acting on a Barge Figure 8.60

Two tugboats are attempting to free a barge that is stuck on a sand bar, and are exerting the forces shown in Figure 8.60. Find the magnitude and direction of the resultant force.

Solution

䊳

Begin by orienting the diagram on a coordinate grid (see Figure 8.61). Since the angle between the vectors is 90°, we know the acute angle formed by the left-hand tugboat and the x-axis is 55°. With this information, we can write each vector in “i, j” form and add the vectors to find the resultant.

2000 N 1500 N

35°

For vector v1 (in QII): Horizontal Component

Vertical Component

a ⫽ 冟v1冟cos ⫽ 2000 cos 125° ⬇ ⫺1147

b ⫽ 冟v1冟sin ⫽ 2000 sin 125° ⬇ 1638

v1 ⬇ ⫺1147i ⫹ 1638j.

Figure 8.61

For vector v2 (in QI):

y

Horizontal Component

Vertical Component

a ⫽ 冟v2冟cos ⫽ 1500 cos 35° ⬇ 1229

b ⫽ 冟v2冟sin ⫽ 1500 sin 35° ⬇ 860

v1 2000 N

v2 ⬇ 1229i ⫹ 860j.

125⬚v

2

55⬚

This gives a resultant of v1 ⫹ v2 ⬇ 1⫺1147i ⫹ 1638j2 ⫹ 11229i ⫹ 860j2 ⫽ 82i ⫹ 2498j, with magnitude 冟v1 ⫹ v2冟 ⬇ 2822 ⫹ 24982 ⬇ 2499 N. To find the direction 2498 of the force, we have r ⫽ tan⫺1 ` ` , or about 88°. 82

35⬚

1500 N x


䊳

It’s worth noting that a single tugboat pulling at 88° with a force of 2499 N would have the same effect as the two tugs in the original diagram. In other words, the resultant vector 82i ⫹ 2498j truly represents the “result” of the two forces. Knowing that the location of a vector is unimportant enables us to model and solve a great number of seemingly unrelated applications. Although the final example concerns aviation, headings, and crosswinds, the solution process has a striking similarity to the “tugboat” example just discussed. In navigation, headings involve a single angle, which is understood to be the amount of rotation from due north in the clockwise direction. Several headings are illustrated in Figures 8.62 through 8.65. Figure 8.62

Figure 8.63

Figure 8.64

Figure 8.65

North West

East South

Heading 30

Heading 330

30 115

210 Heading 210

Heading 115

330

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In order to keep an airplane on course, the captain must consider the direction and speed of any wind currents, since the plane’s true course (relative to the ground) will be affected. Both the plane and the wind can be represented by vectors, with the plane’s true course being the resultant vector. EXAMPLE 9

䊳

Solving an Application of Vectors — Airplane Navigation An airplane is flying at 240 mph, heading 75°, when it suddenly encounters a strong, 60 mph wind blowing from the southwest, heading 10°. What is the actual course and speed of the plane (relative to the ground) as it flies through this wind?

Solution

䊳

Figure 8.66 y

Begin by drawing a vector p to represent the speed and direction of the airplane (Figure 8.66). Since the heading is 75°, the angle between the vector and the x-axis must be 15°. For convenience (and because location is unimportant) we draw it as a 60 ⫽ 14 as long, and position vector. Note the vector w representing the wind will be 240 can also be drawn as a position vector—with an acute 80° angle. To find the resultant, we first find the components of each vector, then add. For vector w (in QI): Horizontal Component

w

a ⫽ 冟w冟cos b ⫽ 冟w冟sin ⫽ 60 cos 80° ⫽ 60 sin 80° ⬇ 10.4 ⬇ 59.1 w ⬇ 10.4i ⫹ 59.1j. For vector p (in QI):

p

80 15

x

Figure 8.67

Vertical Component

The resultant is w ⫹ p ⬇ 110.4i ⫹ 59.1j2⫹1231.8i ⫹ 62.1j2 ⫽ 242.2i ⫹ 121.2j, with magnitude 冟w ⫹ p冟 ⬇ 21242.22 2 ⫹ 1121.22 2 ⬇ 270.8 mph (see Figure 8.67). To find the heading of the plane relative to the ground we use r ⫽ tan⫺1 `

wp p

80

Horizontal Component

a ⫽ 冟p冟cos b ⫽ 冟p冟sin ⫽ 240 cos 15° ⫽ 240 sin 15° ⫽ 231.8 ⬇ 62.1 p ⬇ 231.8i ⫹ 62.1j.

y

w

Vertical Component

15 x

121.2 `, 242.2 which shows r ⬇ 26.6°. The plane is flying on a course heading of approximately 90° ⫺ 26.6° ⫽ 63.4° at a speed of about 270.8 mph relative to the ground. Note the airplane has actually “increased speed” due to the wind. Now try Exercises 83 through 86

WORTHY OF NOTE Be aware that using the rounded values of intermediate calculations may cause slight variations in the final result. In Example 9, if we calculate w ⫹ p ⫽ 160 cos 80° ⫹ 240 cos 15°2 i ⫹ 160 sin 80° ⫹ 240 sin 15°2 j, then find 冟w ⫹ p冟, the result is actually closer to 270.9 mph.

Applications like those in Examples 8 and 9 can also be solved using what is called the parallelogram method, which takes its name from the tail-to-tip vector addition noted earlier (See Figure 8.68). The resultant will be a diagonal of the parallelogram, whose magnitude can be found using the law of cosines. For Example 9, we note the parallelogram has two acute angles of 180 ⫺ 152° ⫽ 65°, and since the adjacent angles must sum to 180°, the obtuse angles must be 115°. Using the law of cosines,

䊳

Figure 8.68 y

115

wp w

p

80 15

x

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E. You’ve just seen how we can use vector diagrams to solve applications

冟w ⫹ p冟2 ⫽ ⫽ ⫽ 冟w ⫹ p冟 ⬇

0 p 0 2 ⫹ 0 w 0 2 ⫺ 2 0 p 0 0w 0 cos 115° 2402 ⫹ 602 ⫺ 212402 1602 cos 115° 73,371.40594 270.9

law of cosines substitute 240 for 0 p0 , 60 for 0 w 0 compute result take square roots

Note this answer is slightly more accurate, since there was no rounding until the final stage.

8.3 EXERCISES 䊳



1. Measurements that can be described using a single number are called quantities.

4. Two vectors are equal if they have the same and .

2.

5. Discuss/Explain the geometric interpretation of vector addition. Give several examples and illustrations.

quantities require more than a single number to describe their attributes. Examples are force, velocity, and displacement.

3. To represent a vector quantity geometrically we use a segment.

䊳

6. Describe the process of finding a resultant vector given the magnitude and direction of two arbitrary vectors u and v. Follow up with an example.


Draw the comparative geometric vectors indicated.

7. Three oceanic research vessels are traveling on parallel courses in the same direction, mapping the ocean floor. One ship is traveling at 12 knots (nautical miles per hour), one at 9 knots, and the third at 6 knots. 8. As part of family reunion activities, the Williams Clan is at a bowling alley and using three lanes. Being amateurs they all roll the ball straight on, aiming for the 1 pin. Grand Dad in Lane 1 rolls his ball at 50 ft/sec. Papa in Lane 2 lets it rip at 60 ft/sec, while Junior in Lane 3 can muster only 30 ft/sec. 9. Vector v1 is a geometric vector representing a boat traveling at 20 knots. Vectors v2, v3, and v4 are geometric vectors representing boats traveling at 10 knots, 15 knots, and 25 knots, respectively. Draw these vectors given that v2 and v3 are traveling the same direction and parallel to v1, while v4 is traveling in the opposite direction and parallel to v1.

10. Vector F1 is a geometric vector representing a force of 50 N. Vectors F2, F3, and F4 are geometric vectors representing forces of 25 N, 35 N, and 65 N, respectively. Draw these vectors given that F2 and F3 are applied in the same direction and parallel to F1, while F4 is applied in the opposite direction and parallel to F1. Represent each situation described using geometric vectors.

11. Two tractors are pulling at a stump in an effort to clear land for more crops. The Massey-Ferguson is pulling with a force of 250 N, while the John Deere is pulling with a force of 210 N. The chains attached to the stump and each tractor form a 25° angle. 12. In an effort to get their mule up and plowing again, Jackson and Rupert are pulling on ropes attached to the mule’s harness. Jackson pulls with 200 lb of force, while Rupert, who is really upset, pulls with 220 lb of force. The angle between their ropes is 16°.

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8–39 Draw the vector v indicated, then graph the equivalent position vector. Verify your work by using the DRAW feature of a calculator to graph both vectors.

13. initial point 1⫺3, 22; terminal point (4, 5)

14. initial point 1⫺4, ⫺42; terminal point (2, 3) 15. initial point 15, ⫺32; terminal point 1⫺1, 22 16. initial point (1, 4); terminal point 1⫺2, 22

33. u ⫽ H7, ⫺2I; v ⫽ H1, 6I

34. u ⫽ H⫺5, ⫺3I; v ⫽ H6, ⫺4I 35. u ⫽ H⫺4, 2I; v ⫽ H1, 4I 36. u ⫽ H7, 3I; v ⫽ H⫺7, 3I

Use the graphs of vectors a, b, c, d, e, f, g, and h given to determine if the following statements are true or false.

For each vector v ⴝ Ha, bI and initial point (x, y) given, find the coordinates of the terminal point and the length of the vector.

y

17. v ⫽ H7, 2I; initial point 1⫺2, ⫺32

b

18. v ⫽ H⫺6, 1I; initial point 15, ⫺22

19. v ⫽ H⫺3, ⫺5I; initial point (2, 6)

20. v ⫽ H8, ⫺2I; initial point 1⫺3, ⫺52 For each position vector given, (a) graph the vector and name the quadrant, (b) compute its magnitude, and (c) find the acute angle ␪ formed by the vector and the nearest x-axis. Use the R䊳Pr and R䊳P␪ features of a calculator to verify your results for parts (b) and (c).

21. H8, 3I

23. H⫺2, ⫺5I

22. H⫺7, 6I

a

d g h c

e f

x

37. a ⫹ c ⫽ b

38. f ⫺ e ⫽ g

39. c ⫹ f ⫽ h

40. b ⫹ h ⫽ c

41. d ⫺ e ⫽ h

42. d ⫹ f ⫽ 0

For the vectors u and v shown, compute u ⴙ v and u ⴚ v and represent each result graphically.

43.

44.

y

x

H4, 1I

u

H1, 4I H3, 6I

H7, 2I

v

u

x

45.

46.

y v

25. 冟v冟 ⫽ 12; r ⫽ 25°; QII

H8, 3I

u

y

x

H1, 3I

H4, 4I

26. 冟u冟 ⫽ 25; r ⫽ 32°; QIII

H5, 2I u

27. 冟w冟 ⫽ 140.5; r ⫽ 41°; QIV 28. 冟p冟 ⫽ 15; r ⫽ 65°; QI

v

x

47.

29. 冟q冟 ⫽ 10; r ⫽ 15°; QIII

48.

y v

u

30. 冟r冟 ⫽ 4.75; r ⫽ 62°; QII

H5, 3I

b. u ⫺ v d. u ⫺ 2v

32. u ⫽ H⫺3, ⫺4I; v ⫽ H0, 5I

y

x

H4, 3I

H2, 3I

v u H5, 1I

For each pair of vectors u and v given, compute (a) through (d) and illustrate the indicated operations graphically.

31. u ⫽ H2, 3I; v ⫽ H⫺3, 6I

y v

24. H8, ⫺6I

For Exercises 25 through 30, the magnitude of a vector is given, along with the quadrant of the terminal point and the angle it makes with the nearest x-axis. Find the horizontal and vertical components of each vector and write the result in component form.

a. u ⫹ v c. 2u ⫹ 1.5v

783


x

Graph each vector and write it as a linear combination of i and j. Then compute its magnitude.

49. u ⫽ H8, 15I

51. p ⫽ H⫺3.2, ⫺5.7I

50. v ⫽ H⫺5, 12I

52. q ⫽ H7.5, ⫺3.4I

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For each vector here, ␪r represents the acute angle formed by the vector and the x-axis. (a) Graph each vector, (b) find the horizontal and vertical components and write the vector in component form, and (c) write the vector in i, j form. Round to the nearest tenth.

53. v in QIII, 冟v冟 ⫽ 12, r ⫽ 16° 54. u in QII, 冟u冟 ⫽ 10.5, r ⫽ 25° 55. w in QI, 冟w冟 ⫽ 9.5, r ⫽ 74.5° 56. v in QIV, 冟v冟 ⫽ 20, r ⫽ 32.6° For vectors v1 and v2 given, compute the vector sums (a) through (d) and find the magnitude and direction of each resultant.

a. v1 ⫹ v2 ⫽ p c. 2v1 ⫹ 1.5v2 ⫽ r

b. v1 ⫺ v2 ⫽ q d. v1 ⫺ 2v2 ⫽ s

Find a unit vector pointing in the same direction as the vector given. Verify that a unit vector was found.

63. u ⫽ H7, 24I

64. v ⫽ H⫺15, 36I

67. 20i ⫺ 21j

68. ⫺4i ⫺ 7.5j

65. p ⫽ H⫺20, 21I 69. 3.5i ⫹ 12j

70. ⫺9.6i ⫹ 18j

73. 6i ⫹ 11j

74. ⫺2.5i ⫹ 7.2j

71. v1 ⫽ H13, 3I

72. v2 ⫽ H⫺4, 7I

Vectors p and q form the angle indicated in each diagram. Find the vector r that points in the same direction as q and forms the base of the right triangle shown.

75.

76.

y

57. v1 ⫽ 2i ⫺ 3j; v2 ⫽ ⫺4i ⫹ 5j

y

H2, 7I

58. v1 ⫽ 7.8i ⫹ 4.2j; v2 ⫽ 5j

H10, 4I

p

59. v1 ⫽ 512i ⫹ 7j; v2 ⫽ ⫺3 12i ⫺ 5j

H7, 4I p

52

60. v1 ⫽ 6.8i ⫺ 9j; v2 ⫽ ⫺4i ⫹ 9j 61. v1 ⫽ 12i ⫹ 4j; v2 ⫽ ⫺4i

66. q ⫽ H12, ⫺35I

q

23 r

r

H9, 1I q

x

77.

62. v1 ⫽ 213i ⫺ 6j; v2 ⫽ ⫺4 13i ⫹ 2j

x

78.

y r

36

q

y

x H2, 7I

H8, 3I

p

p

q

H10, 5I

H4, 6I

48 r x

䊳


The magnitude of a vector in three dimensions: 冟v冟 ⴝ 2a2 ⴙ b2 ⴙ c2

79. The magnitude of a vector in three dimensional space is given by the formula shown, where the components of the position vector v are Ha, b, cI. Find the magnitude of v if v ⫽ H5, 9, 10I. 䊳

80. Find a cardboard box of any size and carefully measure its length, width, and height. Then use the given formula to find the magnitude of the box’s diagonal. Verify your calculation by direct measurement.

APPLICATIONS

81. Tow forces: A large van has careened off of the road into a ditch, and two tow trucks are attempting to winch it out. The cable from the first winch exerts a force of 900 lb, while the cable from the second exerts a force of W1 700 lb. Determine the angle for the first tow truck that will bring the 32 van directly out of the W2 ditch and along the line indicated.

82. Tow forces: Two tugboats are pulling a large ship into dry dock. The first is pulling with a force of 1250 N and the second with a force of 1750 N. Determine the angle for the second tugboat that will keep the ship moving straight forward and into the dock. T1

40 T2

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83. Projectile components: An arrow is shot into the air at an angle of 37° with an initial velocity of 100 ft/sec. Compute the horizontal and vertical components of the representative vector. 84. Projectile components: A football is punted (kicked) into the air at an angle of 42° with an initial velocity of 20 m/sec. Compute the horizontal and vertical components of the representative vector.

785

85. Headings and cross-winds: An airplane is flying at 250 mph on a heading of 75°. There is a strong 35 mph wind blowing from the southwest on a heading of 10°. What is the true course and speed of the plane (relative to the ground)? 86. Headings and currents: A cruise ship is traveling at 16 knots on a heading of 300°. There is a strong water current flowing at 6 knots from the northwest on a heading of 120°. What is the true course and speed of the cruise ship?

The lights used in a dentist’s office are multijointed so they can be configured in multiple ways to accommodate various needs. As a simple model, consider such a light that has the three joints, as illustrated. The first segment has a length of 45 cm, the second is 40 cm in length, and the third is 35 cm.

87. If the joints of the light are positioned so a straight line is formed and the angle made with the horizontal is 15°, determine the approximate coordinates of the joint nearest the light.

88. If the first segment is rotated 75° above horizontal, the second segment ⫺30° (below the horizontal), and the third segment is parallel to the horizontal, determine the approximate coordinates of the joint nearest the light.

30 15

䊳

75


For the arbitrary vectors u ⴝ Ha, bI, v ⴝ Hc, dI, and w ⴝ He, f I and the scalars c and k, prove the following vector properties using the properties of real numbers.

89. 1u ⫽ u

91. u ⫺ v ⫽ u ⫹ 1⫺v2

90. 0u ⫽ 0 ⫽ k0

99. Show that the sum of the vectors given, which form the sides of a closed polygon, is the zero vector. Assume all vectors have integer coordinates and each tick mark is 1 unit. y

92. 1u ⫹ v2 ⫹ w ⫽ u ⫹ 1v ⫹ w2 93. 1ck2u ⫽ c1ku2 ⫽ k1cu2 94. u ⫹ 0 ⫽ u

96. k1u ⫹ v2 ⫽ ku ⫹ kv

95. u ⫹ 1⫺u2 ⫽ 0

97. 1c ⫹ k2u ⫽ cu ⫹ ku

98. Consider an airplane flying at 200 mph at a heading of 45°. Compute the groundspeed of the plane under the following conditions. A strong, 40-mph wind is blowing (a) in the same direction; (b) in the direction of due north 10°2 ; (c) in the direction heading 315°; (d) in the direction heading 270°; and (e) in the direction heading 225°. What did you notice about the groundspeed for (a) and (b)? Explain why the plane’s speed is greater than 200 mph for (a) and (b), but less than 200 mph for the others.

s r t

p

v

u

x

100. Verify that for v ⫽ ai ⫹ bj and v 冟v冟 ⫽ 2a2 ⫹ b2, ⫽ 1. 冟v冟 (Hint: Create the vector u ⫽

v and find its magnitude.) 冟v冟

101. Referring to Exercises 87 and 88, suppose the dentist needed the pivot joint at the light (the furthest joint from the wall) to be at (80, 20) for a certain patient or procedure. Find at least one set of “joint angles” that will make this possible.

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102. (7.1) Derive the other two common versions of the Pythagorean identities, given cos2x ⫹ sin2x ⫽ 1. 103. (2.4/5.3) Evaluate each expression for x ⫽ 3 (if possible): 5 a. y ⫽ ln12x ⫺ 72 b. y ⫽ x⫺3 1 x⫺5 c. y ⫽ A3

104. (7.5) Evaluate the expression csc c tan⫺1a

55 b d by 48

drawing a representative triangle. 105. (4.3) Graph the function g1x2 ⫽ x3 ⫺ 7x and find its zeroes.

MID-CHAPTER CHECK 1. Beginning with

sin A sin B , solve for sin B. ⫽ a b

2. Given b2 ⫽ a2 ⫹ c2 ⫺ 2ac cos B, solve for cos B. Solve the triangles shown below using any appropriate method. Use a calculator to verify your results.

3.

B 207 m C 250 m

4.

31

A

B 17 cm 21 cm A

C

the sign casts a 75 ft shadow. Find the height of the sign if the angle of elevation (measured from a horizontal line) from the tip of the shadow to the top of the sign is 65°. 8. Modeled after an Egyptian obelisk, the Washington Monument (Washington, D.C.) is one of the tallest masonry buildings in the world. Find the height of the monument given the measurements shown (see the figure).

58 70

9. The circles shown here have 44 m radii of 4 cm, 9 cm, and 12 cm, and are tangent to each other. Find the angles formed by the line segments joining their centers.

25 cm

Solve the triangles described below using the law of sines. If more than one triangle exists, solve both.

5. A ⫽ 44°, a ⫽ 2.1 km, c ⫽ 2.8 km 6. C ⫽ 27°, a ⫽ 70 yd, c ⫽ 100 yd 7. A large highway sign is erected on a steep hillside that is inclined 45° from the horizontal. At 9:00 A.M.

75 ft

10. On her delivery route, Judy drives 23 mi to Columbus, then 17 mi to Drake, then 17 mi back home to Balboa. Use 21 the diagram given to find C the distance from Drake to Balboa.

D B 23 mi

45

REINFORCING BASIC CONCEPTS Scaled Drawings and the Laws of Sine and Cosine In mathematics, there are few things as satisfying as the tactile verification of a concept or computation. In this Reinforcing Basic Concepts, we’ll use scaled drawings to verify the relationships stated by the law of sines and the law of cosines. First, gather a blank sheet of paper, a ruler marked in centimeters/millimeters, and a protractor.

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Section 8.4 Vector Applications and the Dot Product

When working with scale models, always measure and mark as carefully as possible. The greater the care, the better the results. For the first illustration (see Figure 8.69), we’ll draw a 20-cm horizontal line segment near the bottom of the paper, then use the left endpoint to mark off a 35° angle. Draw the second side a length of 18 cm. Our first goal is to compute the length of the side needed to complete the triangle, then verify our computation by measurement. Since the current “triangle” is SAS, we use the law of cosines. Label the 35° as ⬔A, the top vertex as ⬔B, and the right endpoint as ⬔C. a2 ⫽ b2 ⫹ c2 ⫺ 2bc cos A ⫽ 1202 2 ⫹ 1182 2 ⫺ 212021182cos 35° ⬇ 724 ⫺ 589.8 ⫽ 134.2 a ⬇ 11.6

Figure 8.69 18 cm ?

90

35 180

0

20 cm

law of cosines with respect to a substitute known values simplify (round to tenths) combine terms solve for a

The computed length of side a is 11.6 cm, and if you took great care in drawing your diagram, you’ll find the missing side is indeed very close to this length. Exercise 1: Finish solving the triangle above using the law of sines. Once you’ve computed ⬔B and ⬔C, measure these angles from the diagram using your protractor. How close was the computed measure to the actual measure? Figure 8.70 For the second illustration (see Figure 8.70), draw any arbitrary triangle on a separate blank sheet, noting that the larger the triangle, the easier it is to measure the angles. After you’ve drawn it, measure the length of each side to the nearest millimeter (our triangle 13.3 cm 15.3 cm turned out to be 21.2 cm ⫻ 13.3 cm ⫻ 15.3 cm2. Now use the law of cosines to find one angle, then the law of sines to solve the triangle. The computations for our triangle gave angles of 95.4°, 45.9°, and 38.7°. What angles did your computations give? Finally, use 21.2 cm your protractor to measure the angles of the triangle you drew. With careful drawings, the measured results are often remarkably accurate! Exercise 2: Using sides of 18 cm and 15 cm, draw a 35° angle, a 50° angle, and a 70° angle, then complete each triangle by connecting the endpoints. Use the law of cosines to compute the length of this third side, then actually measure each one. Was the actual length close to the computed length?

8.4

Vector Applications and the Dot Product


A. Use vectors to investigate

In Section 8.3 we introduced the concept of a vector, with its geometric, graphical, and algebraic representations. We also looked at operations on vectors and employed vector diagrams to solve basic applications. In this section we introduce additional ideas that enable us to solve a variety of new applications, while laying a strong foundation for future studies.

forces in equilibrium

B. Find the components of C. D.

E.

F.

one vector along another Solve applications involving work Compute dot products and the angle between two vectors Find the projection of one vector along another and resolve a vector into orthogonal components Use vectors to develop an equation for nonvertical projectile motion and solve related applications

A. Vectors and Equilibrium

Figure 8.71 y

Much like the intuitive meaning of the word, vector forces are in equilibrium when they “counterbalance” each other. The simplest example is two vector forces of equal magnitude acting on the same point but in opposite directions. Similar to a tug-of-war with both sides equally matched, no one wins. If vector F1 has a magnitude of 500 lb in the positive direction, F1 ⫽ H500, 0I would need vector F2 ⫽ H⫺500, 0I to counter it. If the forces are nonquadrantal, we intuitively sense the components must still sum to zero, and that F3 ⫽ H600, ⫺200I would need F4 ⫽ H⫺600, 200I for equilibrium to occur (see Figure 8.71). In other

H600, 200I F4 x

F3

H600, 200I

F3 F4 H600, 200I H600, 200I H0, 0I 0

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words, two vectors are in equilibrium when their sum is the zero vector 0. If the forces have unequal magnitudes or do not pull in opposite directions, recall a resultant vector F ⫽ Fa ⫹ Fb can be found that represents the combined force. Equilibrium will then occur by adding the vector ⫺1(F) and this vector is sometimes called the equilibriant. These ideas can be extended to include any number of vector forces acting on the same point. In general, we have the following: Vectors and Equilibrium Given vectors F1, F2, p , Fn acting on a point P, 1. The resultant vector is F ⫽ F1 ⫹ F2 ⫹ ⭈⭈⭈ ⫹ Fn. 2. Equilibrium for these forces requires the vector ⫺1F, where F ⫹ 1⫺1F2 ⫽ 0.

EXAMPLE 1

䊳

Finding the Equilibriant for Vector Forces

y

Two force vectors F1 and F2 act on the point P as shown. Find a force F3 so equilibrium will occur, and sketch it on the grid.

Solution

A. You’ve just seen how we can use vectors to investigate forces in equilibrium

䊳

H4.0, 5.9I

5

F1 4.5

Begin by finding the horizontal and vertical components of each vector. For F1 we have 5 H4.5 cos 116°, 4.5 sin 116°I ⬇ H⫺2.0, 4.0I, and for F2 we have H6.3 cos 18°, 6.3 sin 18°I ⬇ H6.0, 1.9I. The resultant vector is F ⫽ F1 ⫹ F2 ⫽ H4.0, 5.9I, meaning H4.0, 5.9I equilibrium will occur by applying the force ⫺1F ⫽ H⫺4.0, ⫺5.9I (see figure).

F2

6.3 18

64 P

5

x

5


䊳

B. The Component of u along v: compvu As in Example 1, many simple applications involve position vectors where the angle and horizontal/vertical components are known or can easily be found. In these situations, the components are often quadrantal, that is, they lie along the x- and y-axes and meet at a right angle. Many other applications require us to find components of a vector that are nonquadrantal, with one of the components parallel to, or lying along a second vector. Given vectors u and v, as shown in Figure 8.72, we symbolize the component of u that lies along v as compvu, noting its value is simply 冟u冟cos compvu adj ⫽ . As the diagrams further indicate, compvu ⫽ 冟u冟cos since cos ⫽ 冟u冟 hyp Figure 8.72 u

u

v

u

p vu

com

u v

v

u

p vu

u comp v 0

com 2

2

u

regardless of how the vectors are oriented. Note that even when the components of a vector are not parallel to the x- or y-axes, they are still orthogonal (meet at a 90° angle).

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It is important to note that compvu is a scalar quantity (not a vector), giving only the magnitude of this component (the vector projection of u along v is studied later in this section). From these developments we make the following observations regarding the angle at which vectors u and v meet: Vectors and the Component of u Along v Given vectors u and v, which meet at an angle , 1. compvu ⫽ 冟u冟cos . 2. If 0° 6 6 90°, compvu 7 0; if 90° 6 6 180°, compvu 6 0. 3. If ⫽ 0°, u and v have the same direction and compvu ⫽ 冟u冟. 4. If ⫽ 90°, u and v are orthogonal and compvu ⫽ 0. 5. If ⫽ 180°, u and v have opposite directions and compvu ⫽ ⫺冟u冟.

EXAMPLE 2

䊳

Solution

䊳

Finding the Component of Vector G Along Vector v Given the vectors G and v with 冟G冟 ⫽ 850 lb as shown in the figure, find compvG.

p vG com 65

Using 冟G冟cos ⫽ compvG we have 850 cos 65° ⬇ 359 lb. The component of G along v is about 359 pounds.

v

850 lb

G


䊳

One interesting application of equilibrium and compvu involves the force of gravity acting on an object placed on a ramp or an inclined plane. The greater the incline, the greater the tendency of the object to slide down the plane (for this study, we assume there is no friction between the object and the plane). While the force of gravity continues to pull straight downward (represented by the vector G in Figure 8.73), G is now the resultant of a force acting parallel to the plane along vector v (causing the object to slide) and a force acting perpendicular to the plane along vector p (causing the object to press against the plane). If we knew the component of G along v (indicated by the shorter, bold segment), we would know the force required to keep the object stationary as the two forces must be opposites. For instance, Figure 8.74 shows an 850-lb object sitting on a ramp with a 25⬚ incline. Note that since G forms a right angle with the base of the incline, angle must be complementary to 25⬚ and measures 65⬚. Since the location of a vector doesn’t change its characteristics, vector p can be repositioned (from directly under the object) to a location that forms the right triangle shown. The force required to keep the object from sliding down the incline is now seen to be compvG, and you may recognize the result as the same diagram from Example 2, where we calculated that the force required was about 359 lb. Figure 8.73

Figure 8.74 compvG

850

lb

850 lb v

p G

v

25 p G

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EXAMPLE 3

䊳

Finding Components of Force for an Object on a Ramp A winch is being used to haul a 2000-lb block of granite up a ramp that is inclined at 15° (Figure 8.75). a. If the winch has a maximum tow rating of 500 lb, will it be successful? b. Use the TABLE feature of a calculator to v determine the heaviest block this winch can successfully pull up the incline.

Solution

䊳

B. You’ve just seen how we can find the components of one vector along another

a. We again need the component of G along the inclined plane: compvG ⫽ 2000 cos 75° ⬇ 518 lb. Since the capacity of the winch is exceeded, the attempt will likely not be successful. b. Begin by entering X cos 75⬚ as Y1 on the Y= screen. In part (a) we found that 2000 lb exceeded the capacity of this winch, so we set up a TABLE to begin at 2000 with ⌬Tbl ⴝ ⴚ1, essentially decreasing the weight in 1-lb increments until we find a weight less than or equal to the winch’s 500-lb capacity. After scrolling down to the screen shown in Figure 8.76, we find the winch can successfully haul a 1931-lb block up the ramp.

Figure 8.75 0 lb

200 15⬚

2000 lb

G

Figure 8.76


䊳

C. Vector Applications Involving Work In common, everyday usage, work is understood to involve the exertion of energy or force to move an object a certain distance. For example, digging a ditch is hard work and involves moving dirt (exerting a force) from the trench to the bankside (over a certain distance). In an office, moving a filing cabinet likewise involves work. If the filing cabinet is heavier, or the distance it needs to be moved is greater, more work is required to move it (Figures 8.77 and 8.78). To determine how much work was done by each person, we need to quantify the concept. Consider a constant force F, applied to move an object a distance D in the same direction as the force. In this case, work is defined as the product of the force applied and the distance the object is moved: Work ⫽ (Force)(Distance) or W ⫽ 冟F冟D. If the force is given in pounds and the distance in feet, the amount of work is measured in a unit called foot-pounds (ft-lb). If the force is in Newtons and the distance in meters, the amount of work is measured in Newton-meters (N-m).

Figure 8.77

D

Figure 8.78

D

EXAMPLE 4

䊳

Solving Applications of Vectors — Work and Force Parallel to the Direction of Movement While rearranging the office, Carrie must apply a force of 55.8 N to relocate a filing cabinet 4.5 m, while Bernard applies a 77.5 N force to move a second cabinet 3.2 m. Who did the most work?

Solution

䊳

For Carrie: W ⫽ 冟F冟D ⫽ 155.8214.52 ⫽ 251.1 N-m

For Bernard: W ⫽ 冟F冟D ⫽ 177.52 13.22 ⫽ 248 N-m

Carrie did 251.1 ⫺ 248 ⫽ 3.1 N-m more work than Bernard. Now try Exercises 31 and 32

䊳

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In many applications of work, the force F is not applied parallel to the direction of movement, as illustrated in Figures 8.79 and 8.80. Figure 8.79

Figure 8.80

F

F

30°

D

D

25°

In calculating the amount of work done, the general concept of (force)(distance) is preserved, but only the component of force in the direction of movement is used. In terms of the component forces discussed earlier, if F is a constant force applied at angle to the direction of movement, the amount of work done is the component of force along D times the distance the object is moved. Force Vectors and Work W WORTHY OF NOTE

In the formula W ⫽ 1冟F冟cos 21D2 , observe that if ⫽ 0°, we have the old formula for work when the force is applied in the direction of movement W ⫽ FD. If ⫽ 0°, cos ⫽ 1 and the “effective force” on the object becomes 冟F冟cos .

EXAMPLE 5

䊳

Given a force F applied in the direction of movement at the acute angle to an object, and D the distance it is moved, W ⫽ 1 冟F冟cos 21D2

F

D

兩F兩cos

Solving an Application of Vectors — Work and Force Applied at Angle ␪ to the Direction of Movement To help move heavy pieces of furniture across the floor, movers sometime employ a body harness similar to that used for a plow horse. A mover applies a constant 200-lb force to drag a piano 100 ft down a long hallway and into another room. If the straps make a 40° angle with the direction of movement, find the amount of work performed.

Solution

䊳

40° D

The component of force in the direction of movement is 200 cos 40° or about 153 lb. The amount of work done is W ⬇ 15311002 ⫽ 15,300 ft-lb. Now try Exercises 35 through 40

䊳

These ideas can be generalized to include work problems where the component of force in the direction of motion is along a nonhorizontal vector v. Consider Example 6. EXAMPLE 6

Solution

C. You’ve just seen how we can solve applications involving work

䊳

䊳

Solving an Application of Vectors — Forces Along a Nonhorizontal Vector The force vector F ⫽ H5, 12I moves an object along the vector v ⫽ H15.44, 2I as shown. Find the amount of work required to move the object along the entire length of v. Assume force is in pounds and distance in feet.

y

H5, 12I F

To begin, we first determine the angle between the vectors. H15.44, 2I 2 v ⫺1 12 ⫺1 60 ` ⬇ 60°. In this case we have ⫽ tan ` ` ⫺ tan ` x 5 15.44 For 冟F冟 ⫽ 13 (5-12-13 triangle), the component of force in the direction of motion is compvF ⫽ 13 cos 60° ⫽ 6.5. With 冟v冟 ⫽ 2115.442 2 ⫹ 122 2 ⬇ 15.57, the work required is W ⫽ 1compvF21 冟v冟 2 or 16.52115.572 ⬇ 101.2 ft-lb. Now try Exercises 41 through 44

䊳

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D. Dot Products and the Angle Between Two Vectors When the component of force in the direction of motion lies along a nonhorizontal vector (as in Example 6), the work performed can actually be computed more efficiently using an operation called the dot product. For any two vectors u and v, the dot product u # v is equivalent to 1compvu21 冟v冟 2, yet is much easier to compute (for the proof of u # v 1compvu21 冟v冟 2, see Appendix B). The operation is defined as follows: The Dot Product u # v

Given vectors u Ha, bI and v Hc, dI, u # v Ha, bI # Hc, dI ac bd. In words, it is the real number found by taking the sum of corresponding component products. Figure 8.81

Earlier, we used the braces ” {” and ” }” to perform addition, subtraction, and scalar multiplication of vectors on a calculator. This technique can be extended to compute the dot product of two vectors, as it is simply defined as “the sum of the component products.” For u H2, 5I and v H3, 7I, the operation {2, 5} * {3, 7} is used for the component products, returning {6, 35} as shown in Figure 8.81. For the sum (although very simple in this case), we use the keystrokes 2nd STAT (LIST) (MATH) to

access the 5:sum( option, then {6, 35}) . The result shows H2, 5I # H3, 7I 41. To be more efficient, the calculator can combine these two operations as seen in Example 7. ENTER

EXAMPLE 7

䊳

Using the Dot Product to Determine Force Along a Nonhorizontal Vector Verify the answer to Example 6 using the dot product u # v.

Solution

䊳

For u H5, 12I and v H15.44, 2I, we have u # v H5, 12I # H15.44, 2I giving 5115.442 12122 101.2. The result is 101.2, as in Example 6. A calculator check is shown in the figure.


䊳

Note that dot products can also be used in the simpler case where the direction of motion is along a horizontal distance (Examples 4 and 5). While the dot product offers a powerful and efficient way to compute the work performed, it has many other applications; for example, to find the angle between two vectors. Consider that for any two u v # (solve for vectors u and v, u # v 1 冟u冟cos ␪21 冟v冟 2 , leading directly to cos ␪ 冟u冟冟v冟 cos ␪). In summary, The Angle ␪ Between Two Vectors Figure 8.82 y (x, y)

Given the nonzero vectors u and v: u # v cos ␪ 冟u冟冟v冟

and

␪ cos1a

u v # b 冟u冟冟v冟

u 1 ␪

x

v compvu x

v 1

In the special case where u and v are unit vectors, this simplifies to cos ␪ u # v since 冟u冟冟v冟 1. This relationship is shown in Figure 8.82. The dot product u # v gives 1compvu21 冟v冟 2 , but 冟v冟 1 and the component of u along v is simply the adjacent side of a right triangle whose hypotenuse is 1. Hence, u # v cos ␪.

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EXAMPLE 8

䊳

793

Determining the Angle Between Two Vectors Find the angle between the vectors given. a. u ⫽ H⫺3, 4I; v ⫽ H5, 12I b. v1 ⫽ 2i ⫺ 3j; v2 ⫽ 6i ⫹ 4j

Solution

䊳

v1 v2 u v # # b. cos ⫽ 冟u冟冟v冟冟v1冟冟v2冟 ⫺3 4 5 12 ⫺3 6 4 2 ⫽ h , i # h , i , i # h , i ⫽ h 5 5 13 13 113 113 152 152 48 ⫺15 ⫺12 12 ⫹ ⫽ ⫹ ⫽ 65 65 1676 1676 33 0 ⫽ ⫽0 ⫽ 65 26 33 ⫽ cos⫺1a b ⫽ cos⫺10 65 ⬇ 59.5° ⫽ 90°

a. cos ⫽


䊳

Note we have implicitly shown that if u # v ⫽ 0, then u is orthogonal to v (see Exercises 55 through 60). As with other vector operations, recognizing certain properties of the dot product will enable us to work with them more efficiently. Properties of the Dot Product Given vectors u, v, and w and a constant k, 1. u # v ⫽ v # u 3. w # 1u ⫹ v2 ⫽ w # u ⫹ w # v 5. 0 # u ⫽ u # 0 ⫽ 0

2. u # u ⫽ 冟u冟2 4. k1u # v2 ⫽ ku # v ⫽ u # kv u#v u # v 6. ⫽ 冟u冟冟v冟冟u冟冟v冟

Property 6 offers an alternative to unit vectors when finding cos —the dot product of the vectors can be# computed first, and the result divided by the product of their u v magnitudes: cos ⫽ . Proofs of the first two properties are given here. Proofs of 冟u冟冟v冟 the others have a similar development (see Exercises 79 through 82). For any two nonzero vectors u ⫽ Ha, bI and v ⫽ Hc, dI: Property 1: u # v ⫽ Ha, bI # Hc, dI ⫽ ac ⫹ bd ⫽ ca ⫹ db

⫽ Hc, dI # Ha, bI ⫽v#u

Property 2: u # u ⫽ Ha, bI # Ha, bI ⫽ a2 ⫹ b2 ⫽ 冟u冟2

(since 冟u冟 ⫽ 2a2 ⫹ b2)

Using compvu ⫽ 冟u冟cos and u # v ⫽ 1compvu2 冟v冟, we can also state the following relationships, which give us some flexibility on how we approach applications of the dot product.

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For any two vectors u ⫽ Ha, bI and v ⫽ Hc, dI:

D. You’ve just seen how we can compute dot products and the angle between two vectors

(1) u # v ⫽ ac ⫹ bd (2) u # v ⫽ 1 冟u冟cos 2 1 冟v冟 2 (3) u # v ⫽ 1compvu21 冟v冟 2 u#v (4) ⫽ cos 冟u冟冟v冟 u#v ⫽ compvu (5) 冟v冟

standard computation of the dot product alternative computation of the dot product replace 冟u冟cos in (2) with compvu divide (2) by scalars 冟u冟 and 冟v冟 divide (3) by 冟v冟

See Exercises 61 through 66 for practice with relationship (5).

E. Vector Projections and Orthogonal Components In work problems and other simple applications, it is enough to find and apply compvu (Figure 8.83). However, applications involving thrust and drag forces, tension and stress limits in a cable, electronic circuits, and cartoon animations often require that we also find the vector form of compvu. This is called the projection of u along v or projvu, and is a vector in the same direction of v with magnitude compvu (Figures 8.84 and 8.85). Figure 8.83

Figure 8.84

u

u

Figure 8.85 u v

v

u comp v r) a l a c s (

v

u proj v r) o t c e v (

u proj v r) o t c e (v

v has a length of one and points in the same direction as 冟v冟 v v, so projvu can be computed as 1compvu2 a b (see Example 7, Section 8.3). Using 0v 0 equation (5) from this page and the properties shown earlier, an alternative formula for projvu can be found that is usually easier to simplify: By its design, the unit vector

projvu ⫽ 1compvu2 a

v b 0v 0

u#v v ba b 0v 0 0v 0 u#v ⫽a 2bv 0v 0 ⫽a

definition of a projection

substitute

u#v for compvu 0v0

rewrite factors

Vector Projections Given vectors u and v, the projection of u along v is the vector projvu ⫽ a

u#v bv 0v 0 2

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EXAMPLE 9A

䊳

Finding the Projection of One Vector Along Another

Solution

䊳

To begin, find u # v and 冟v冟.

y H6, 6I

u

projvu

u#v bv 0v 0 2 ⫺36 b H6, 6I ⫽a 72 ⫽ H⫺3, ⫺3I

projvu ⫽ a

v

H7, 1I

Given u ⫽ H⫺7, 1I and v ⫽ H6, 6I, find projvu. u # v ⫽ H⫺7, 1I # H6, 6I ⫽ ⫺42 ⫹ 6 ⫽ ⫺36

Figure 8.86

x

795

冟v冟 ⫽ 262 ⫹ 62 ⫽ 172 ⫽ 612 projection of u along v substitute ⫺36 for u # v, 172 for 冟v冟, and H6, 6I for v

Figure 8.87

result, see Figure 8.86

For the calculator check shown in Figure 8.87, u # v is shown in the first entry. The second entry u#v calculates the scalar multiplier 2 , and the final 冟v冟 entry confirms projvu ⫽ H⫺3, ⫺3I.

WORTHY OF NOTE Note that u2 ⫽ u ⫺ u1 is the shorter diagonal of the parallelogram formed by the vectors u and u1 ⫽ projvu. This can also be seen in the graph supplied for Example 9B.

A useful consequence of computing projvu is we can then resolve the vector u into orthogonal components that need not be quadrantal. One component will be parallel to v and the other perpendicular to v (the dashed line in Figure 8.86). In general terms, this means we can write u as the vector sum u1 ⫹ u2, where u1 ⫽ projvu and u2 ⫽ u ⫺ u1 (note u1 7 v). Resolving a Vector into Orthogonal Components Given vectors u, v, and projvu, u can be resolved into the orthogonal components u1 and u2, where u ⫽ u1 ⫹ u2, u1 ⫽ projvu, and u2 ⫽ u ⫺ u1.

EXAMPLE 9B

䊳

Solution

䊳

y H6, 6I H⫺7, 1I u2

Given u ⫽ H⫺7, 1I and v ⫽ H6, 6I from Example 9A, resolve u into orthogonal components u1 and u2, where u1 7 v and u2⬜v. Also verify u1⬜u2. In Example 9A, we found projvu ⫽ u1 ⫽ H⫺3, ⫺3I. Because u ⫽ u1 ⫹ u2 we have u2 ⫽ u ⫺ u1 ⫽ H⫺7, 1I ⫺ H⫺3, ⫺3I ⫽ H⫺4, 4I

v u

u1 ⴝ projvu

Resolving a Vector into Orthogonal Components

x

E. You’ve just seen how we can find the projection of one vector along another and resolve a vector into orthogonal components

subtract u1 from both sides

substitute H⫺7, 1I for u and H⫺3, ⫺3I for u1 result

To verify u1⬜u2, we need only show u1 ⭈ u2 ⫽ 0: u1 # u2 ⫽ H⫺3, ⫺3I # H⫺4, 4I ⫽ 1⫺321⫺42 ⫹ 1⫺32142 ⫽0✓


䊳

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F. Vectors and the Height of a Projectile Our final application of vectors involves projectile motion. A projectile is any object that is thrown or projected upward, with no source of propulsion to sustain its motion. In this case, the only force acting on the projectile is gravity (air resistance is neglected), so the maximum height and the range of the projectile depend solely on its initial velocity and the angle at which it is projected. In a college algebra course, the equation y ⫽ v0t ⫺ 16t2 is developed to model the height in feet (at time t) of a projectile thrown vertically upward with initial velocity of v0 feet per second. Here, we’ll modify the equation slightly to take into account that the object is now moving horizontally as well as vertically. As you can see in Figure 8.88, the vector v representing the initial velocity, as well as the velocity vector at other times, can easily be decomposed into horizontal and vertical components. This will enable us to find a more general relationship for the position of the projectile. For now, we’ll let vy represent the component of velocity in the vertical (y) direction, and vx represent the component of velocity in the horizontal (x) direction. Since gravity acts only in the vertical (and negative) direction, the horizontal component of the velocity remains constant at 冟vx冟 ⫽ 冟v冟cos . Using D ⫽ RT, the x-coordinate of the projectile at time t is x ⫽ 1 冟v冟cos 2t. For the vertical component vy we use the projectile equation developed earlier, substituting 冟v冟sin for v0, since the angle of projection is no longer 90°. This gives the y-coordinate at time t as y ⫽ v0t ⫺ 16t2 ⫽ 1 冟v冟sin 2t ⫺ 16t2. Figure 8.88 y

vy

vy

3 0 vy 0

v 2

vx

4

vx v 1

vx

vy x

vx

Projectile Motion Given an object is projected upward from the origin with initial velocity 冟v冟 at angle . The x-coordinate of its position at time t is x ⫽ 1 冟v冟cos 2t. The y-coordinate of its position at time t is y ⫽ 1 冟v冟sin 2t ⫺ 16t2.

EXAMPLE 10

䊳

Solving an Application of Vectors — Projectile Motion An arrow is shot upward with an initial velocity of 150 ft/sec at an angle of 50°. a. Find the position of the arrow after 3 sec. b. How many seconds does it take to reach a height of 190 ft?

Solution

䊳

a.

䊲

Algebraic Solution

Using the preceding equations yields these coordinates for its position at t ⫽ 3: x⫽ ⫽ ⬇

1 冟v冟cos 2t 1150 cos 50°2132 289

y ⫽ 1 冟v冟sin 2t ⫺ 16t2 ⫽ 1150 sin 50°2132 ⫺ 16132 2 ⬇ 201

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797

Section 8.4 Vector Applications and the Dot Product 䊲

Graphical Solution After entering (150 cos 50⬚)X as Y1 and (150 sin 50⬚)X ⫺ 16X2 as Y2 on the Y= screen, we can graph the functions and use the TRACE feature to find the arrow’s position at t ⫽ 3 (see Figures 8.89 and 8.90). Figure 8.89

Figure 8.90

750

750

0

8

0

8

⫺100

⫺100

The arrow has traveled a horizontal distance of about 289 ft and is about 201 ft high. b.

䊲

Algebraic Solution

To find the time required to reach 190 ft in height, set the equation for the y coordinate equal to 190, which yields a quadratic equation in t: y ⫽ 1 冟v冟sin 2t ⫺ 16t2 equation for y 2 190 ⫽ 1150 sin 50°2t ⫺ 16t substitute 150 for 冟v冟 and 50° for 0 ⬇ ⫺16t2 ⫹ 115t ⫺ 190 150 sin 50 ⬇ 115 Using the quadratic formula we find that t ⬇ 2.6 sec and t ⬇ 4.6 sec are solutions. 䊲

Graphical Solution Begin by deactivating Y1, entering to Y3 ⫽ 190, and changing the the dimensions shown for Figure 8.91. The intersection-of-graphs method confirms our algebraic solution. The result for the first intersection point is shown. The two solutions make sense, since the arrow reaches a given height once on the way up and again on the way down (as long as it hasn’t reached its maximum height).

Figure 8.91 250

WINDOW

F. You’ve just seen how we can use vectors to develop an equation for nonvertical projectile motion and solve related applications

0

8

0


䊳

Note the GRAPH of Y2 in Figure 8.91 does not represent the path of the arrow. For this important difference and more on projectile motion, see the Calculator Exploration and Discovery feature at the end of this chapter.

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8.4 EXERCISES 䊳



䊳

1. Vector forces are in when they counterbalance each other. Such vectors have a sum of .

4. The component of u along v is a quantity. The projection of u along v is a .

2. The component of a vector u along another vector v is written notationally as , and is computed as .

5. Explain/Discuss exactly what information the dot product of two vectors gives us. Illustrate with a few examples.

3. Two vectors that meet at a right angle are said to be .

6. Compare and contrast the projectile equations y ⫽ v0t ⫺ 16t2 and y ⫽ 1v0sin 2t ⫺ 16t2. Discuss similarities/differences using illustrative examples.


The force vectors given are acting on a common point P. Find an additional force vector so that equilibrium takes place.

7. F1 ⫽ H⫺8, ⫺3I; F2 ⫽ H2, ⫺5I 8. F1 ⫽ H⫺2, 7I; F2 ⫽ H5, 3I

9. F1 ⫽ H⫺2, ⫺7I; F2 ⫽ H2, ⫺7I; F3 ⫽ H5, 4I

10. F1 ⫽ H⫺3, 10I; F2 ⫽ H⫺10, 3I; F3 ⫽ H⫺9, ⫺2I 11. F1 ⫽ 5i ⫺ 2j; F2 ⫽ i ⫹ 10j

12. F1 ⫽ ⫺7i ⫹ 6j; F2 ⫽ ⫺8i ⫺ 3j 13. F1 ⫽ 2.5i ⫹ 4.7j; F2 ⫽ ⫺0.3i ⫹ 6.9j; F3 ⫽ ⫺12j 14. F1 ⫽ 3 12i ⫺ 2 13j; F2 ⫽ ⫺2i ⫹ 7j; F3 ⫽ 5i ⫹ 2 13j 15.

F1

16.

y

10 104

F2 6 25 20 x 9 F

3

F1

y

20 F2

17 19

35

F1

x

20. Three cowhands have roped y F1 a wild stallion and are attempting to hold him F 200 steady. The first and second 2 170 75 19 cowhands are pulling with ?F3 the magnitude and at the angles indicated in the diagram. If the stallion is held fast by the three cowhands, find the magnitude and angle of the rope from the third cowhand.

x

Find the component of u along v (compute compvu) for the vectors u and v given.

115 18

19. A new “Survivor” game y F2 involves a three-team tugof-war. Teams 1 and 2 are 2210 2500 pulling with the 50 40 magnitudes and at the ?F3 angles indicated in the diagram. If the teams are currently in a stalemate, find the magnitude and angle of the rope held by team 3.

x

21.

22. u

u

F3

17. The force vectors F1 and F2 are simultaneously acting on a point P. Find a third vector F3 so that equilibrium takes place if F1 ⫽ H19, 10I and F2 ⫽ H5, 17I. 18. The force vectors F1, F2, and F3 are simultaneously acting on a point P. Find a fourth vector F4 so that equilibrium takes place if F1 ⫽ H⫺12, 2I, F2 ⫽ H⫺6, 17I, and F3 ⫽ H3, 15I.

3.5 tons 50 kg 42

v

128

v

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23.

24.

u

65 v 1525 lb

221 lb v 70

u

25.

26.

u

30

2 tons

v

v

3010 kg

115 u

Determine the requested missing component.

27. A 500-lb crate is sitting on a ramp that is inclined at 35°. Find the force needed to hold the object stationary. 28. A 1200-lb skiff is being pulled from a lake, using a boat ramp inclined at 20°. Find the minimum force needed to dock the skiff. 䊳

500

35 G

1200

29. A 325-kg carton is sitting on a ramp, held stationary by 225 kg of tension in a restraining rope. Find the ramp’s angle of incline.

325

kg

225 kg

325 kg

G

30. A heavy dump 1.75 tons truck is being ? ton s winched up a ramp with an 18° incline. 18 Approximate G the weight of the truck if the winch is working at its maximum capacity of 1.75 tons and the truck is barely moving.

lb

500 lb

lb

1200 lb 20 G

Determine the amount of work done.

31. While rearranging the patio furniture, Rick has to push the weighted base of the umbrella stand 15 m. If he uses a constant force of 75 N, how much work did he do? 32. Vinny’s car just broke down in the middle of the road. Luckily, a buddy is with him and offers to steer if Vinny will get out and push. If he pushes with a constant force of 185 N to move the car 30 m, how much work did he do?


The range of a projectile: R ⴝ

v2sin ␪ cos ␪ 16

33. The range of a projected object (total horizontal distance traveled) is given by the formula shown, where v is the initial velocity and is the angle at which it is projected. If an arrow leaves the bow traveling 175 ft/sec at an angle of 45°, what horizontal distance will it travel? 䊳

799


34. A collegiate javelin thrower releases the javelin at a 40° angle, with an initial velocity of about 95 ft/sec. If the NCAA record is 280 ft, will this throw break the record? What is the smallest angle of release that will break this record? If the javelin were released at the optimum 45°, by how many feet would the record be broken?

APPLICATIONS

35. Plowing a field: An old-time farmer is plowing his field with a mule. How much work does the mule do in plowing one length of a field 300 ft long, if it pulls the plow with a constant force of 250 lb and the straps make a 30° angle with the horizontal?

36. Pulling a sled: To enjoy a beautiful snowy day, a mother is pulling her three 32° children on a sled along a level street. How much work (play) is done if the street is 100 ft long and she pulls with a constant force of 55 lb with the tow-rope making an angle of 32° with the street?

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37. Tough-man contest: As part of a “tough-man” contest, participants are required to pull a bus along a level street for 100 ft. If one contestant did 45,000 ft-lb of work to accomplish the task and the straps used made an angle of 5° with the street, find the tension in the strap during the pull. 38. Moving supplies: An arctic explorer is hauling supplies from the supply hut to her tent, a distance of 150 ft, in a sled she is dragging behind her. If 9000 ft-lb of work was done and the straps used made an angle of 25° with the snow-covered ground, find the tension in the strap during the task. 39. Wheelbarrow rides: To break up the monotony of a long, hot, boring Saturday, a father decides to (carefully) give his kids a ride in a wheelbarrow. He applies a force of 30 N to move the “load” 100 m, then stops to rest. Find the amount of work done if the wheelbarrow makes an angle of 20° with level ground while in motion. 40. Mowing the lawn: A homeowner applies a force of 40 N to push her lawnmower back and forth across the back yard. Find the amount of work done if the yard is 50 m long, requires 24 passes to get the lawn mowed, and the mower arm makes an angle of 39° with the level ground.

47. Use the dot product to verify the solution to Exercise 43. 48. Use the dot product to verify the solution to Exercise 44. For each pair of vectors given, (a) use a calculator to compute the dot product p # q and (b) find the angle between the vectors to the nearest tenth of a degree.

49. p ⫽ H5, 2I; q ⫽ H3, 7I

50. p ⫽ H⫺3, 6I; q ⫽ H2, ⫺5I

51. p ⫽ ⫺2i ⫹ 3j; q ⫽ ⫺6i ⫺ 4j 52. p ⫽ ⫺4i ⫹ 3j; q ⫽ ⫺6i ⫺ 8j 53. p ⫽ 712i ⫺ 3j; q ⫽ 212i ⫹ 9j 54. p ⫽ 12i ⫺ 3j; q ⫽ 312i ⫹ 5j Determine if the pair of vectors given are orthogonal.

55. u ⫽ H7, ⫺2I; v ⫽ H4, 14I

56. u ⫽ H⫺3.5, 2.1I; v ⫽ H⫺6, ⫺10I 57. u ⫽ H⫺6, ⫺3I; v ⫽ H⫺8, 15I 58. u ⫽ H⫺5, 4I; v ⫽ H⫺9, ⫺11I 59. u ⫽ ⫺2i ⫺ 6j; v ⫽ 9i ⫺ 3j

60. u ⫽ 3 12i ⫺ 2j; v ⫽ 2 12i ⫹ 6j Find compvu for the vectors u and v given.

61. u ⫽ H3, 5I; v ⫽ H7, 1I

62. u ⫽ H3, 5I; v ⫽ H⫺7, 1I

63. u ⫽ ⫺7i ⫹ 4j; v ⫽ ⫺10j Force vectors: For the force vector F and vector v given, find the amount of work required to move an object along the entire length of v. Assume force is in pounds and distance in feet.

41. F ⫽ H15, 10I; v ⫽ H50, 5I

42. F ⫽ H⫺5, 12I; v ⫽ H⫺25, 10I 43. F ⫽ H8, 2I; v ⫽ H15, ⫺1I

44. F ⫽ H15, ⫺3I; v ⫽ H24, ⫺20I 45. Use the dot product to verify the solution to Exercise 41. 46. Use the dot product to verify the solution to Exercise 42.

64. u ⫽ 8i; v ⫽ 10i ⫹ 3j 65. u ⫽ 7 12i ⫺ 3j; v ⫽ 6i ⫹ 5 13j 66. u ⫽ ⫺3 12i ⫹ 6j; v ⫽ 2i ⫹ 5 15j For each pair of vectors given, (a) find the projection of u along v (compute projvu) and (b) resolve u into vectors u1 and u2, where u1 || v and u2 ⬜v.

67. u ⫽ H2, 6I; v ⫽ H8, 3I

68. u ⫽ H⫺3, 8I; v ⫽ H⫺12, 3I

69. u ⫽ H⫺2, ⫺8I; v ⫽ H⫺6, 1I

70. u ⫽ H⫺4.2, 3I; v ⫽ H⫺5, ⫺8.3I 71. u ⫽ 10i ⫹ 5j; v ⫽ 12i ⫹ 2j 72. u ⫽ ⫺3i ⫺ 9j; v ⫽ 5i ⫺ 3j

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Projectile motion: A projectile is launched from a catapult with the initial velocity v0 and angle indicated. Find (a) the position of the object after 3 sec and (b) the time(s) at which the object is at a height of 250 ft. Use a calculator to verify your results.

77. At the circus, a “human cannon ball” is shot from a large cannon with an initial velocity of 90 ft/sec at an angle of 65° from the horizontal. How high is the acrobat after 1.2 sec? How long until the acrobat is again at this same height?

73. v0 ⫽ 250 ft/sec; ⫽ 60°

78. A center fielder runs down a long hit by an opposing batter and whirls to throw the ball to the infield to keep the hitter to a double. If the initial velocity of the throw is 130 ft/sec and the ball is released at an angle of 30° with level ground, how high is the ball after 1.5 sec? How long until the ball again reaches this same height?

74. v0 ⫽ 300 ft/sec; ⫽ 55° 75. v0 ⫽ 200 ft/sec; ⫽ 45° 76. v0 ⫽ 500 ft/sec; ⫽ 70°

䊳

801


For the arbitrary vectors u ⴝ Ha, bI, v ⴝ Hc, dI, and w ⴝ He, f I and the scalar k, prove the following vector properties using the properties of real numbers.

79. w # 1u ⫹ v2 ⫽ w # u ⫹ w # v

80. k1u # v2 ⫽ ku # v ⫽ u # kv

81. 0 # u ⫽ u # 0 ⫽ 0

82.

# u v # ⫽u v 冟u冟冟v冟冟u冟冟v冟

m2 ⫺ m1 u#v for finding the angle between two vectors, the equation tan ⫽ can be 冟u冟冟v冟 1 ⫹ m2m1 used, where m1 and m2 represent the slopes of the vectors. Find the angle between the vectors 1i ⫹ 5j and 5i ⫹ 2j using each equation and comment on which you found more efficient. Then see if you can find a geometric connection between the two equations.

83. As alternative to cos ⫽

84. Use the equations for the horizontal and vertical components of the projected object’s position to obtain the 16 equation of trajectory y ⫽ 1tan 2x ⫺ 2 2 x2. This is a quadratic equation in x. What can you say about its v cos graph? Include comments about the concavity, x-intercepts, maximum height, and so on.

䊳


85. (5.5) Solve for t: 2.9e⫺0.25t ⫹ 7.6 ⫽ 438

88. (8.3) A plane is flying 200 mph at heading 30⬚, with a 40 mph wind blowing from due west. Find the true course and speed of the plane.

86. (6.5) Graph the function using a reference rectangle and the rule of fourths: y ⫽ 3 cosa2 ⫺ b 4 87. (8.2) Solve the triangle shown, then compute A its perimeter and area.

Plane

200 mph 30

C 250 m

Wind 40 mph

32 172 m B

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Precalculus—

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Complex Numbers in Trigonometric Form


A. Graph a complex number B. Write a complex number C. D. E.

F.

in trigonometric form Convert from trigonometric form to rectangular form Interpret products and quotients geometrically Compute products and quotients in trigonometric form Solve applications involving complex numbers (optional)

WORTHY OF NOTE Surprisingly, the study of complex numbers matured much earlier than the study of vectors, and using directed line segments to represent complex numbers actually preceded their application to vector quantities.

EXAMPLE 1

䊳

Once the set of complex numbers became recognized and defined, the related basic operations matured very quickly. With little modification—sums, differences, products, quotients, and powers all lent themselves fairly well to the algebraic techniques used for real numbers. But roots of complex numbers did not yield so easily and additional tools and techniques were needed. Writing complex numbers in trigonometric form enables us to find complex roots (Section 8.6) and in some cases, makes computing products, quotients, and powers more efficient.

A. Graphing Complex Numbers In previous sections, we defined a vector quantity as one that required more than a single component to describe its attributes. The complex number z a bi certainly fits this description, since both a real number “component” and an imaginary “component” are needed to define it. In many respects, we can treat complex numbers in the same way we treated vectors and in fact, there is much we can learn from this connection. Since both axes in the xy-plane have real number values, it’s not possible to graph a complex number in ⺢ (the real plane). However, in the same way we used the x-axis for the horizontal component of a vector and the y-axis for the vertical, we can let the Figure 8.92 x-axis represent the real valued part of a complex number and the y-axis the imaginary part. The Imaginary yi axis result is called the complex plane ⺓. Every point (a + bi) b (a, b) in ⺓ can be associated with a complex number a bi, and any complex number a bi can z be associated with a point (a, b) in ⺓ (Figure 8.92). x The point (a, b) can also be regarded as the terminal a Real point of a position vector representing the complex axis number, generally named using the letter z. Graphing Complex Numbers

yi

Graph the complex numbers below on the same complex plane. a. z1 2 6i b. z2 5 4i c. z3 5 d. z4 4i

Solution

䊳

5 4i

4i z4

z2 z3

The graph of each complex number is shown in the figure.

x

5

z1 2 6i


A. You’ve just seen how we can graph a complex number

802

In Example 1, you likely noticed that from a vector perspective, z2 is the “resultant vector” for the sum z3 z4. To investigate further, consider z1 2 3i, z2 5 2i, and the sum z1 z2 z shown in Figure 8.93. The figure helps to confirm that the sum of complex numbers can be illustrated geometrically using the parallelogram (tail-to-tip) method employed for vectors in Section 8.3.

䊳

Figure 8.93 yi 3 5i

z2 2 3i

z

5 2i z2

z1 x

8–58

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Section 8.5 Complex Numbers in Trigonometric Form

B. Complex Numbers in Trigonometric Form The complex number z a bi is said to be in rectangular form since it can be graphed using the rectangular coordinates of the complex plane. Complex numbers can also be written in trigonometric form. Similar to how x represents the distance between the real number x and zero, z represents the distance between a bi and the origin in the complex plane. It is computed as z 2a2 b2. Figure 8.94 With any nonzero z, we can also associate yi an angle ␪, which is the angle in standard position whose terminal side coincides z a bi with the graph of z. If we let r represent r a b z, Figure 8.94 shows cos ␪ and r x r b sin ␪ , yielding r cos ␪ a and a r r sin ␪ b. The appropriate substitutions into a bi give the trigonometric form: z a bi r cos ␪ r sin ␪ # i Factoring out r and writing the imaginary unit as the lead factor of sin ␪ gives the b relationship in its more common form, z r1cos ␪ i sin ␪2, where tan ␪ . a The Trigonometric Form of a Complex Number

WORTHY OF NOTE

For the complex number z a bi and angle ␪ shown, z r 1cos ␪ i sin ␪2 is b the trigonometric form of z, where r 2a2 b2, and tan ␪ ; a 0. a yi • r z represents the magnitude of z (also called the modulus). z a bi • ␪ is often referred to as the argument r of z.

While it is true the trigonometric form can more generally be written as z r 3 cos1␪ 2␲k2 i sin1␪ 2␲k2 4 for k 僆⺪, the result is identical for any integer k and we will select ␪ so that 0 ␪ 6 2␲ or 0° ␪ 6 360°, depending on whether we are working in radians or degrees.

b

r

x

a

b b Be sure to note that for tan ␪ , tan1 ` ` is equal to ␪r (the reference angle a a for ␪) and the value of ␪ will ultimately depend on the quadrant of z.

EXAMPLE 2

䊳

Converting a Complex Number from Rectangular to Trigonometric Form State the quadrant of the complex number, then write each in trigonometric form. a. z1 2 2i b. z2 6 2i

Solution

䊳

Knowing that modulus r and angle ␪ are needed for the trigonometric form, we first determine these values. Once again, to find the correct value of ␪, it’s important to note the quadrant of the complex number. When ␪ is not a standard angle, we will often answer in exact form as seen in Example 2(b).

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a. z1 2 2i; QIII

yi 2 r

3

2 x

r

2 2i 3

b. z2 6 2i; QI

r 2122 2 122 2 18 2 12 2 ␪r tan1 ` ` 2 tan1 112 45°

r 2162 2 122 2 140 2 110 2 ␪r tan1 ` ` 6 1 tan1a b 3

yi 5 6 2i r r 6 x

4

5

With z1 in QIII, ␪ 225°, and we have z1 2 121cos 225° i sin 225°2.

1 z2 is in QI, so ␪ tan1a b and 3 1 z2 2 110 acos c tan1a b d 3 1 i sin c tan1a b d b. 3

See the figures. Now try Exercises 11 through 26

䊳

With a graphing calculator in “a bi” MODE , we can confirm the intermediate steps of Example 2(a). To confirm that the modulus of 2 2i is 2 12, and to find the modulus of other complex numbers, we press MATH then and select option 5:abs( from the CPX (complex) submenu. We then enter 2 2i Figure 8.95 with a right parenthesis and press . The result is shown in Figure 8.95 and confirms the modulus of z1 is 2 12. Similarly, option 4:angle( of the MATH :CPX submenu finds and displays the angle (argument) of a complex number. In the third entry of Figure 8.95, we find the argument of z1 is 135, coterminal with the ␪ 225 which we computed earlier (see Worthy of Note on page 803). Since the angle ␪ is repeated for both cosine and sine, we often use an abbreviated notation for the trigonometric form, called “cis” (sis) notation: z r 1cos ␪ i sin ␪2 r cis ␪. The results of Example 2(a) and 2(b) would then be 1 written z 2 12 cis 225° and z 2 210 cis c tan1a b d , respectively. 3 b As in Example 2(b), when ␪r tan1 ` ` is not a standard angle we can either a answer in exact form as shown, or use a four-decimal-place approximation: 2110 cis10.32182. ENTER

B. You’ve just seen how we can write a complex number in trigonometric form

C. Converting from Trigonometric Form to Rectangular Form Converting from trigonometric form back to rectangular form is simply a matter of evaluating r cis ␪. This can be done regardless of whether ␪ is a standard angle or in the b form tan1a b, since in the latter case we can construct a right triangle with side b a opposite ␪ and side a adjacent to ␪, and find the needed values as in Section 7.5.

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EXAMPLE 3

Solution

䊳

䊳

Converting a Complex Number from Trigonometric to Rectangular Form Graph the following complex numbers, then write them in rectangular form. ␲ 5 a. z 12 cis a b b. z 13 cis c tan1a b d 6 12 ␲ Figure 8.96 a. We have r 12 and ␪ , which yields the 6 yi graph in Figure 8.96. In the nonabbreviated z ␲ ␲ form we have z 12 ccos a b i sin a b d . 6√3 6i 6 6 12 Evaluating within the brackets gives 1 13 k i d 6 13 6i. z 12 c 2 2 x 5 b, we 12 construct the graph shown in Figure 8.97 with 5 opposite ␪, 12 adjacent to ␪, and a vector magnitude of 13. From the diagram we note that 5 12 cos ␪ and sin ␪ , meaning 13 13

b. For r 13 and ␪ tan1a

Figure 8.97 yi z 5

z 131cos ␪ i sin ␪2 13 c C. You’ve just seen how we can convert from trigonometric form to rectangular form

12 5i 13 5 ␪ tan112

12 5 id 13 13

12 x

12 5i. Now try Exercises 27 through 34

䊳

D. Interpreting Products and Quotients Geometrically The multiplication and division of complex numbers has some geometric connections that can help us understand their computation in trigonometric form. Note the relationship between the modulus and argument of the following product, with the moduli (plural of modulus) and arguments from each factor. EXAMPLE 4

Solution

䊳

Noting Graphical Connections for the Product of Two Complex Numbers

䊳

For z1 3 3i and z2 0 2i, a. Graph the complex numbers and compute their moduli and arguments. b. Compute and graph the product z1z2 and find its modulus and argument. Discuss any connections you see between the factors and the resulting product. Figure 8.98 a. The graphs of z and z are shown in Figure 8.98. 1

2

For the modulus and argument we have: z1 3 3i; QI r 2132 2 132 2 118 3 12 ␪ tan11 1 ␪ 45°

z2 0 2i; 1quadrantal2 r 2 directly ␪ 90° directly

yi

z3

z1

135 z2 45

x

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b. The product z1z2 is 13 ⫹ 3i2 12i2 ⫽ ⫺6 ⫹ 6i, which is in QII. The modulus is 21⫺62 2 ⫹ 162 2 ⫽ 172 ⫽ 6 12, with ␪r ⫽ tan⫺1 0 ⫺1 0 ⫽ 45° and an argument ␪ ⫽ 135° (QII). Note the product of the two moduli is equal to the modulus of the final product: 2 # 3 12 ⫽ 6 12. Also note that the sum of the arguments for z1 and z2 is equal to the argument of the product: 45° ⫹ 90° ⫽ 135°! See Figure 8.99.

Figure 8.99

Now try Exercises 35 and 36 D. You’ve just seen how we can interpret products and quotients geometrically

䊳

A similar geometric connection exists for the division of complex numbers. This connection is explored in Exercises 37 and 38 of the Exercise Set.

E. Products and Quotients in Trigonometric Form The connections in Example 4 are not a coincidence, and can be proven to hold for all complex numbers. Consider any two nonzero complex numbers z1 ⫽ r1 1cos ␣ ⫹ i sin ␣2 and z2 ⫽ r2 1cos ␤ ⫹ i sin ␤2. For the product z1z2 we have z1z2 ⫽ r1 1cos ␣ ⫹ i sin ␣2 r2 1cos ␤ ⫹ i sin ␤2

⫽ r1r2 3 1cos ␣ ⫹ i sin ␣21cos ␤ ⫹ i sin ␤2 4

product in trig form rearrange factors

⫽ r1r2 3cos ␣ cos ␤ ⫹ i sin ␤ cos ␣ ⫹ i sin ␣ cos ␤ ⫹ i2sin ␣ sin ␤4 ⫽ r1r2 3 1cos ␣ cos ␤ ⫺ sin ␣ sin ␤2 ⫹ i1sin

⫽ r1r2 3cos1␣ ⫹ ␤2 ⫹ i sin1␣ ⫹ ␤2 4

F-O-I-L commute ␤ cos ␣ ⫹ sin ␣ cos ␤2 4 terms; i 2 ⫽ ⫺1 use sum/difference identities for sine/cosine

In words, the proof says that to multiply complex numbers in trigonometric form, we multiply the moduli and add the arguments. For division, we divide the moduli and subtract the arguments. The proof for division resembles that for multiplication and is asked for in Exercise 71. Products and Quotients of Complex Numbers in Trigonometric Form For the complex numbers z1 ⫽ r1 1cos ␣ ⫹ i sin ␣2 and • •

EXAMPLE 5

䊳

z2 ⫽ r2 1cos ␤ ⫹ i sin ␤2, z1z2 ⫽ r1r2 3cos1␣ ⫹ ␤2 ⫹ i sin1␣ ⫹ ␤2 4 and z1 r1 ⫽ 3cos1␣ ⫺ ␤2 ⫹ i sin1␣ ⫺ ␤2 4 , z2 ⫽ 0. z2 r2

Multiplying Complex Numbers in Trigonometric Form For z1 ⫽ ⫺3 ⫹ i 13 and z2 ⫽ 13 ⫹ 1i, a. Write z1 and z2 in trigonometric form and compute z1z2. z1 b. Compute the quotient in trigonometric form. z2 c. Verify the product using the rectangular form.

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Solution

䊳

807

a. For z1 in QII we find r 2 13 and ␪ 150°, for z2 in QI, r 2 and ␪ 30°. In trigonometric form, z1 2 131cos 150° i sin 150°2 and z2 21cos 30° i sin 30°2: z1z2 2 131cos 150° i sin 150°2 # 21cos 30° i sin 30°2 multiply moduli, 2 13 # 2 3cos1150° 30°2 i sin1150° 30°2 4 add arguments 4 131cos 180° i sin 180°2 4 13 11 0i2 4 13 b.

2 131cos 150° i sin 150°2 z1 z2 21cos 30° i sin 30°2 133cos1150° 30°2 i sin1150° 30°2 4 131cos 120° i sin 120°2 1 13 ib 13 a 2 2 3 13 i 2 2

c. z1z2 13 i 132 1 13 1i2 3 13 3i 3i i2 13 4 13

divide moduli, subtract arguments

verified [matches part (a)]


䊳

Converting to trigonometric form for multiplication and division seems too clumsy for practical use, as we can often compute these results more efficiently in rectangular form. However, this approach leads to powers and roots of complex numbers, an indispensable part of advanced equation solving, and these are not easily found in rectangular form. In any case, note that the power and simplicity of computing products/quotients in trigonometric form is highly magnified when the complex numbers are given in trig form: E. You’ve just seen how we can compute products and quotients in trigonometric form

112 cis 50°2 13 cis 20°2 36 cis 70°

12 cis 50° 4 cis 30° 3 cis 20°

See Exercises 43 through 50.

F. (Optional) Applications of Complex Numbers Somewhat surprisingly, complex numFigure 8.100 bers have several applications in the real Magnetic flux world. Many of these involve a study of electricity, and in particular AC (alternating current) circuits. In simplistic terms, when an armature (molded wire) is rotated in a uniform S N magnetic field, a voltage V is generated that depends on the strength of the field. As the armature is rotated, the voltage varies between a maximum and a minimum value, with the amount of voltage modeled by V1␪2 Vmaxsin1B␪2, with ␪ in degrees. Here, Vmax represents the maximum voltage attained, and the input variable ␪ represents the angle the armature makes with the magnetic flux, indicated in Figure 8.100 by the dashed arrows between the magnets.

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When the armature is perpendicular to the flux, we say ␪ ⫽ 0°. At ␪ ⫽ 0° and ␪ ⫽ 180°, no voltage is produced, while at ␪ ⫽ 90° and ␪ ⫽ 270°, the voltage reaches its maximum and minimum values, respectively (hence the name alternating current). Many electric dryers and other large appliances are labeled as 220 volt (V) appliances, but use an alternating current that varies from 311 V to ⫺311 V (see Worthy of Note). This means when ␪ ⫽ 52°, V152°2 ⫽ 311 sin152°2 ⫽ 245 V is being generated. In practical applications, we use time t as the independent variable, rather than the angle of the armature. These large appliances usually operate with a frequency of 60 cycles 2␲ 1 1 , we obtain per second, or 1 cycle every of a second aP ⫽ b. Using B ⫽ 60 60 P B ⫽ 120␲ and our equation model becomes V1t2 ⫽ 311 sin1120␲t2 with t in radians. This variation in voltage is an excellent example of a simple harmonic model.

WORTHY OF NOTE You may have wondered why we’re using an amplitude of 311 for a 220-V appliance. Due to the nature of the sine wave, the average value of an alternating current is always zero and gives no useful information about the voltage generated. Instead, the root-mean-square (rms) of the voltage is given on most appliances. While the maximum voltage is 311 V, the rms voltage is 311 ⬇ 220 V. See Exercise 72. 12

EXAMPLE 6

䊳

Analyzing Alternating Current Using Trigonometry Use the equation V1t2 ⫽ 311 sin1120␲t2 to: a. Create a table of values illustrating the voltage produced every thousandth of a 1 second for the first half-cycle at ⫽ ⬇ 0.008b. 120 b. Use a graphing calculator to find the times t in this half-cycle when 160 V is being produced.

Solution

䊳

a. Starting at t ⫽ 0 and using increments of 0.001 s produces the table shown. Time t

Voltage

0

0

0.001

114.5

0.002

212.9

0.003

281.4

0.004

310.4

0.005

295.8

0.006

239.6

0.007

149.8

0.008

39.0

b. From the table we note V1t2 ⫽ 160 when t 僆 10.001, 0.0022 and t 僆 10.006, 0.0072 . Using the intersection-of-graphs method places these values at t ⬇ 0.0014 and t ⬇ 0.0069 (see graph). 400

0

0.0165

⫺400


Figure 8.102 Voltage and current are in phase (phase angle ⫽ 0⬚)

sin

90⬚

180⬚

270⬚

䊳

The chief components of AC circuits are voltage (V) Figure 8.101 and current (I). Due to the nature of how the current is genR XL XC erated, V and I can be modeled by sine functions. Other B C D characteristics of electricity include pure resistance (R), A inductive reactance (XL), and capacitive reactance (XC) (see Figure 8.101). Each of these is measured in a unit called ohms 1⍀2, while current I is measured in amperes (A), and voltages are measured in volts (V). These components of electricity are related by fixed and inherent traits, which include the following: (1) voltage across a resistor is always in phase with the current, meaning the phase shift or phase angle between them is 0° (Figure 8.102); (2) voltage across an inductor leads the current by 90° (Figure 8.103); (3) voltage across a capacitor lags the current by 90° (Figure 8.104); and (4) voltage is equal to the product of the current times the resistance or reactance: V ⫽ IR, V ⫽ IXL, and V ⫽ IXC.

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Figure 8.103 sin

90

Figure 8.105 yi XL

R x

XC

WORTHY OF NOTE While mathematicians generally use the symbol i to represent 11, the “i” is used in other fields to represent an electric current so the symbol j 11 is used instead. In conformance with this convention, we will temporarily use j for 11 as well.

EXAMPLE 7

䊳

Figure 8.104

Voltage leads current by 90 (phase angle 90)

180

270

Voltage lags current by 90 (phase angle 90)

sin

90

180

270

Different combinations of R, XL, and XC in a combined (series) circuit alter the phase angle and the resulting voltage. Since voltage across a resistance is always in phase with the current (trait 1), we can model the resistance as a vector along the positive real axis (since the phase angle is 0°). For traits (2) and (3), XL is modeled on the positive imaginary axis since voltage leads current by 90°, and XC on the negative imaginary axis since voltage lags current by 90° (see Figure 8.105). These natural characteristics make the complex plane a perfect fit for describing the characteristics of the circuit. Consider a series circuit (Figure 8.101), where R 12 , XL 9 , and XC 4 . For a current of I 2 amps through this circuit, the voltage across each individual element would be VR 1221122 24 V (A to B), VL 122192 18 V (B to C), and VC 122142 8 V (C to D). However, the resulting voltage across this circuit cannot be an arithmetic sum, since R is real, while XL and XC are represented by imaginary numbers. The joint effect of resistance (R) and reactance (XL, XC) in a circuit is called the impedance, denoted by the letter Z, and is a measure of the total resistance to the flow of electrons. It is computed Z R XL j XC j (see Worthy of Note), due to the phase angle relationship of the voltage in each element (XL and XC point in opposite directions, hence the subtraction). The expression for Z is more commonly written R 1XL XC 2 j, where we more clearly note Z is a complex number whose magnitude and angle with the x-axis can be found as before: Z 2R2 1XL XC 2 2 and XL XC ␪r tan1a b. The angle ␪ represents the phase angle between the voltage R and current brought about by this combination of elements. The resulting voltage of the circuit is then calculated as the product of the current with the magnitude of the impedance, or VRLC I Z (Z is also measured in ohms, 2.

Finding the Impedence and Phase Angle of the Current in a Circuit For the circuit diagrammed in the figure, (a) find the magnitude of Z, the phase angle between current and voltage, and write the result in trigonometric form; and (b) find the total voltage for a current of I 2 amps across this circuit.

Solution R A

12 Ω

XL B

9Ω

䊳

XC C

4Ω

D

a. Using the values given, we find Z R 1XL XC 2j 12 19 42 j 12 5j 1QI2:

This gives a magnitude of

F. You have just seen how we can solve applications involving complex numbers

Z 21122 2 152 2 1169 13 , with a phase angle of 5 ␪ tan1a b 22.6° (voltage leads the current by about 22.6°). 12 In trigonometric form Z 13 cis 22.6°. b. With I 2 amps, the total voltage across this circuit is VRLC I Z 21132 26 V. Now try Exercises 55 through 70

䊳

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8.5 EXERCISES 䊳



䊳

1. For a complex number written in the form z r 1cos ␪ i sin ␪2, r is called the and ␪ is called the .

4. To divide complex numbers in trigonometric form, we the moduli and the arguments.

␲ ␲ 2. The complex number z 2 c cosa b i sin a b d 4 4 can be written as the abbreviated “cis” notation as .

5. Write z 1 i 13 in trigonometric form and explain why the argument is ␪ 240° instead of 60° as indicated by your calculator’s evaluation of tan1 13.

3. To multiply complex numbers in trigonometric form, we the moduli and the arguments.

6. Discuss the similarities between finding the components of a vector and writing a complex number in trigonometric form.


Graph the complex numbers z1, z2, and z3 given, then express one as the sum of the other two.

7. z1 7 2i z2 8 6i z3 1 4i 9. z1 2 5i z2 1 7i z3 3 2i

8. z1 2 7i z2 3 4i z3 1 3i 10. z1 2 6i z2 7 2i z3 5 4i

State the quadrant of each complex number, then write it in trigonometric form. For Exercises 11 through 14, answer in degrees. For 15 through 18, answer in radians. Use the abs( and angle( features of a calculator to verify your results.

11. 2 2i

12. 7 7i

13. 513 5i

14. 2 2i13

15. 312 3i12

16. 517 5i 17

17. 413 4i

18. 6 6i 13

Write each complex number in trigonometric form. For Exercises 19 through 22, answer in degrees using both an exact form and an approximate form, rounding to tenths. For 23 through 26, answer in radians using both an exact form and an approximate form, rounding to four decimal places.

19. 8 6i

20. 9 12i

21. 5 12i

22. 8 15i

23. 6 17.5i

24. 30 5.5i

25. 6 10i

26. 12 4i

Graph each complex number using its trigonometric form, then convert each to rectangular form.

␲ 27. 2 cis a b 4

␲ 28. 12 cis a b 6

␲ 29. 413 cis a b 3

30. 5 13 cis a

31. 17 cis c tan1a

15 bd 8

7␲ b 6

3 32. 10 cis c tan1a b d 4

33. 6 cis c ␲ tan1a

5 bd 111

34. 4 cis c ␲ tan1a

17 bd 3

For the complex numbers z1 and z2 given, find their moduli r1 and r2 and arguments ␪1 and ␪2. Then compute their product in rectangular form. For modulus r and argument ␪ of the product, verify that r1r2 ⴝ r and ␪1 ⴙ ␪2 ⴝ ␪.

35. z1 2 2i;

z2 3 3i

36. z1 1 i 13; z2 3 i 13

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8–67 For the complex numbers z1 and z2 given, find their moduli r1 and r2 and arguments ␪1 and ␪2. Then compute their quotient in rectangular form. For modulus r and argument ␪ of the quotient, verify that r1 ⴝ r and ␪1 ⴚ ␪2 ⴝ ␪. r2

37. z1 13 i; z2 1 i13 38. z1 13 i; z2 3 0i z1 using the z2 trigonometric form. Answer in exact rectangular form where possible, otherwise round all values to two decimal places. Compute the product z1z2 and quotient

39. z1 413 4i 3 313 i z2 2 2 41. z1 213 0i 21 7 13 z2 i 2 2 䊳

5 13 5 40. z1 i 2 2 z2 0 6i

42. z1 0 6i 12 3 12 316 z2 i 2 2

43. z1 9 c cosa

␲ ␲ b i sina b d 15 15

z2 1.8 c cosa 44. z1 2 c cosa

2␲ 2␲ b i sina b d 3 3

3␲ 3␲ b i sina b d 5 5

␲ ␲ z2 8.4 c cosa b i sina b d 5 5 45. z1 101cos 60° i sin 60°2 z2 41cos 30° i sin 30°2 46. z1 71cos 120° i sin 120°2 z2 21cos 300° i sin 300°2 47. z1 5 12 cis 210° z2 2 12 cis 30°

48. z1 5 13 cis 240° z2 13 cis 90°

49. z1 6 cis 82° z2 1.5 cis 27°

50. z1 1.6 cis 59° z2 8 cis 275°


51. Equilateral triangles in the complex plane: u2 ⴙ v2 ⴙ w2 ⴝ uv ⴙ uw ⴙ vw If the line segments connecting the complex numbers u, v, and w form the vertices of an equilateral triangle, the formula shown above holds true. Verify that u 2 i 13, v 10 i13, and w 6 5i 13 form the vertices of an equilateral triangle using the distance formula, then verify the formula given. 䊳

811


52. The cube of a complex number: 1A ⴙ B2 3 ⴝ A3 ⴙ 3A2B ⴙ 3AB2 ⴙ B3 The cube of any binomial can be found using the formula here, where A and B are the terms of the binomial. Use the formula to compute the cube of 1 2i (note A 1 and B 2i2.

APPLICATIONS

53. Electric current: In the United States, electric power is supplied to homes and offices via a “120 V circuit,” using an alternating current that varies from 170 V to 170 V, at a frequency of 60 cycles/sec. (a) Write the voltage equation for U.S. households, (b) create a table of values illustrating the voltage produced every thousandth of a second for the first half-cycle, and (c) find the first time t in this half-cycle when exactly 140 V is being produced. 54. Electric current: While the electricity supplied in Europe is still not quite uniform, most countries employ 230-V circuits, using an alternating current that varies from 325 V to 325 V. However, the frequency is only 50 cycles per second. (a) Write the voltage equation for these European countries, (b) create a table of values illustrating the voltage produced every thousandth of a second for the first half-cycle, and (c) find the first time t in this half-cycle when exactly 215 V is being produced. AC circuits: For the circuits indicated in Exercises 55 through 60, (a) find the magnitude of Z, the phase angle between current and voltage, and write the result in trigonometric form; and (b) find the total voltage across this circuit. Recall Z R 1XL XC 2 j and Z 2R2 1XL XC 2 2. Exercises 55 through 58 55. R 15 , X 12 , and X 4 , with I 3 A L

C

R

56. R 24 , XL 12 , and XC 5 , with I 2.5 A 57. R 7 , XL 6 , and XC 11 , with I 1.8 A 58. R 9.2 , XL 5.6 , and XC 8.3 , with I 2.0 A

A

XL

XC B

C

D

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59. R ⫽ 12 ⍀ and XL ⫽ 5 ⍀, with I ⫽ 1.7 A 60. R ⫽ 35 ⍀ and XL ⫽ 12 ⍀, with I ⫽ 4 A

Exercises 59 and 60 R A

XL B

C

AC circuits — voltage: The current I and the impedance Z for certain AC circuits are given. Write I and Z in trigonometric form and find the voltage in each circuit. Recall V ⫽ IZ.

61. I ⫽ 13 ⫹ 1j A and Z ⫽ 5 ⫹ 5j ⍀

62. I ⫽ 13 ⫺ 1j A and Z ⫽ 2 ⫹ 2j ⍀

63. I ⫽ 3 ⫺ 2j A and Z ⫽ 2 ⫹ 3.75j ⍀

64. I ⫽ 4 ⫹ 3j A and Z ⫽ 2 ⫺ 4j ⍀

AC circuits — current: If the voltage and impedance are known, the current I in the circuit is calculated as the quotient V I ⫽ . Write V and Z in trigonometric form to find the current in each circuit. Z

65. V ⫽ 2 ⫹ 2 13j and Z ⫽ 4 ⫺ 4j ⍀

66. V ⫽ 4 13 ⫺ 4j and Z ⫽ 1 ⫺ 1j ⍀

67. V ⫽ 3 ⫺ 4j and Z ⫽ 4 ⫹ 7.5j ⍀

68. V ⫽ 2.8 ⫹ 9.6j and Z ⫽ 1.4 ⫺ 4.8j ⍀

Z1Z2 , where Z1 and Z2 represent Z1 ⫹ Z2 the impedance in each branch. Find the total impedance for the values given. Compute the product in the numerator using trigonometric form, and the sum in the denominator in rectangular form. Parallel circuits: For AC circuits wired in parallel, the total impedance is given by Z ⫽

69. Z1 ⫽ 1 ⫹ 2j and Z2 ⫽ 3 ⫺ 2j 䊳


71. Verify that for the complex numbers z1 ⫽ r1 1cos ␣ ⫹ i sin ␣2 and z2 ⫽ r2 1cos ␤ ⫹ i sin ␤2, r1 z1 ⫽ 3 cos1␣ ⫺ ␤2 ⫹ i sin1␣ ⫺ ␤2 4 . z2 r2 72. Using the Internet, a trade manual, or some other resource, find the voltage and frequency at which electricity is supplied to most of Japan (oddly enough — two different frequencies are in common use). As in Example 6, the voltage given will likely be the root-mean-square (rms) voltage. Use the information to find the true voltage and the equation model for voltage in most of Japan.

䊳

70. Z1 ⫽ 3 ⫺ j and Z2 ⫽ 2 ⫹ j

73. Recall that two lines are perpendicular if their slopes have a product of ⫺1. For the directed line segment representing the complex number z1 ⫽ 7 ⫹ 24i, find complex numbers z2 and z3 whose directed line segments are perpendicular to z1 and have a magnitude one-fifth as large. 74. The magnitude of the impedance is

冟 Z 冟 ⫽ 2R2 ⫹ 1XL ⫺ XC 2 2. If R, XL, and XC are all nonzero, what conditions would make the magnitude of Z as small as possible?


75. (7.7) Solve for x 僆 3 0, 2␲2: ␲ 350 ⫽ 750 sina2x ⫺ b ⫺ 25 4 76. (4.5) Name all asymptotes of the function 1 ⫹ x3 h1x2 ⫽ x2 77. (2.5) Graph the piecewise-defined function given: 2 f 1x2 ⫽ • x2 x

x 6 ⫺2 ⫺2 ⱕ x 6 1 xⱖ1

Exercise 78 78. (8.1) A ship is spotted by two observation posts that are 4 mi apart. Using the line between them for reference, the first 41⬚ 63⬚ post reports the 4 mi ship is at an angle of 41°, while the second reports an angle of 63°, as shown. How far is the ship from the closest post?

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De Moivre’s Theorem and the Theorem on nth Roots


A. Use De Moivre’s theorem to raise complex numbers to any power B. Use De Moivre’s theorem to check solutions to polynomial equations C. Use the n th roots theorem to find the n th roots of a complex number

The material in this section represents some of the most significant developments in the history of mathematics. After hundreds of years of struggle, mathematical scientists had not only come to recognize the existence of complex numbers, but were able to make operations on them commonplace and routine. This allowed for the unification of many ideas related to the study of polynomial equations, and answered questions that had puzzled scientists from many different fields for centuries. In this section, we will look at two fairly simple theorems that actually represent over 1000 years in the evolution of mathematical thought.

A. De Moivre’s Theorem Having found acceptable means for applying the four basic operations to complex numbers, our attention naturally shifts to the computation of powers and roots. Without them, we’d remain wholly unable to offer complete solutions to polynomial equations and find solutions for many applications. The computation of powers, squares, and cubes offer little challenge, as they can be computed easily using the formula for binomial squares 3 1A B2 2 A2 2AB B2 4 or by applying the binomial theorem. For larger powers, the binomial theorem becomes too time-consuming and a more efficient method is desired. The key here is to use the trigonometric form of the complex number. In Section 8.5, we noted the product of two complex numbers involved multiplying their moduli and adding their arguments: For z1 r1 1cos 1 i sin 1 2 and z2 r2 1cos 2 i sin 2 2 we have z1z2 r1r2 3cos11 2 2 i sin11 2 2 4

For the square of a complex number, r1 r2 and 1 2. Using itself yields z2 r2 3cos1 2 i sin1 2 4 r2 3cos122 i sin122 4

Multiplying this result by z r1cos i sin 2 to compute z3 gives

r2 3cos122 i sin122 4 r 1cos i sin 2 r3 3cos12 2 i sin12 2 4 r3 3cos132 i sin132 4.

WORTHY OF NOTE Sometimes the argument of cosine and sine becomes very large after applying De Moivre’s theorem. In these cases, we use the fact that 360°k and 2k represent coterminal angles for integers k, and use the coterminal angle where 0° 6 360° or 0 6 2.

The result can be extended further and generalized into De Moivre’s theorem. De Moivre’s Theorem For any positive integer n, and z r1cos i sin 2,

zn rn 3cos1n2 i sin1n2 4

For a proof of the theorem where n is an integer and n 1, see Appendix B. EXAMPLE 1

䊳

Using De Moivre’s Theorem to Compute the Power of a Complex Number Use De Moivre’s theorem to compute z9, given z 12 12i.

Solution

8–69

䊳

1 2 12 1 2 . With z in QIII, tan 1 yields a b a b 2 2 2 B 12 5 5 5 . The trigonometric form is z c cosa b i sina b d and applying 4 2 4 4 the theorem with n 9 gives Here we have r

813

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12 9 5 5 b c cos a9 # b i sin a9 # bd 2 4 4 45 45 12 c cos a b i sin a bd 32 4 4 5 5 12 c cos a b i sin a b d 32 4 4 12 12 12 a ib 32 2 2 1 1 i 32 32

z9 a

A calculator in a bi

MODE

De Moivre’s theorem

simplify

coterminal angles

evaluate functions

result

can be used to check our work (see figure). Now try Exercises 7 through 14

䊳

As with products and quotients, if the complex number is given in trigonometric form, computing any power of the number is both elegant and efficient. For instance, if z 2 cis 40°, then z4 16 cis 160°. See Exercises 15 through 18. For cases where is not a standard angle, De Moivre’s theorem requires an intriguing application of the skills developed in Chapter 7, including the use of multiple b angle identities and working from a right triangle drawn relative to r tan1 ` ` . a See Exercises 57 and 58.

A. You’ve just seen how we can use De Moivre’s theorem to raise complex numbers to any power

B. Checking Solutions to Polynomial Equations One application of De Moivre’s theorem is checking the complex roots of a polynomial, as in Example 2. EXAMPLE 2

䊳

Using De Moivre’s Theorem to Check Solutions to a Polynomial Equation Use De Moivre’s theorem to show that z 2 2i is a solution to z4 3z3 38z2 128z 144 0.

Solution

䊳

We will apply the theorem to the third and fourth degree terms, and compute the square directly. Since z is in QIII, the trigonometric form is z 212 cis 225°. In the following illustration, note that 900° and 180° are coterminal, as are 675° and 315°. 12 2i2 4 12 122 4cis14 # 225°2 12 122 4cis 900° 64 cis 180° 6411 0i2 64

12 2i2 3 12 122 3cis13 # 225°2 12 122 3cis 675° 12122 3cis 315° 12 12 1612a ib 2 2 16 16i

12 2i2 2 4 8i 12i2 2 4 8i 4i2 4 8i 4 0 8i 8i

Substituting back into the original equation gives z4 3z3 38z2 128z 144 0 1642 3116 16i2 3818i2 12812 2i2 144 0 64 48 48i 304i 256 256i 144 0 164 48 256 1442 148 304 2562i 0 00✓ Now try Exercises 19 through 26

䊳

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B. You’ve just seen how we can use De Moivre’s theorem to check solutions to polynomial equations

Section 8.6 De Moivre’s Theorem and the Theorem on nth Roots

815

Regarding Example 2, we know from a study of algebra that complex roots must occur in conjugate pairs, meaning 2 2i is also a root. This equation actually has two real and two complex roots, with z 9 and z 2 being the two real roots.

C. The nth Roots Theorem Having looked at De Moivre’s theorem, which raises a complex number to any power, we now consider the nth roots theorem, which will compute the nth roots of a complex number. If we allow that De Moivre’s theorem also holds for rational values 1 , instead of only the integers n illustrated previously, the formula for computing an n nth root would be a direct result: 1 1 1 1 zn r n c cos a b i sin a b d n n n 1r c cos a b i sin a b d n n

replace n with

1 in De Moivre’s theorem n

simplify

However, this formula would find only the principal nth root! In other words, periodic solutions would be ignored. As in Section 8.5, it’s worth noting the most general form of a complex number is z r 3cos1 360°k2 i sin1 360°k2 4 , for k 僆⺪. When De Moivre’s theorem is applied to this form for integers n, we obtain zn rn 3cos1n 360°kn2 i sin1n 360°kn2 4 , which returns a result identical to 1 rn 3cos1n2 i sin1n2 4 . However, for the rational exponent , the general form takes n additional solutions into account and will return all n, nth roots. 1 1 1 1 z n r n e cos c 1 360°k2 d i sin c 1 360°k2 d f n n

360°k 360°k n 1r c cos a b i sin a bd n n n n

De Moivre’s theorem for rational exponents

simplify

The nth Roots Theorem For z r1cos i sin 2, a positive integer n, and r 僆⺢, z has exactly n distinct nth roots determined by 360°k 360°k n n 1z 1r c cos a b i sin a bd n n n n where k 0, 1, 2, p , n 1.

For ease of computation, it helps to note that once the argument for the principal 2 360°k 360° root is found using k 0, simply adds aor b to the previous argun n n n ment for k 1, 2, 3, p , n 1. In Example 3 you’re asked to find the three cube roots of 1, also called the cube roots of unity, and graph the results. The nth roots of unity play a significant role in the solution of many polynomial equations. For an in-depth study of this connection, visit www.mhhe.com/coburn and go to Section 8.8: Trigonometry, Complex Numbers, and Cubic Equations.

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EXAMPLE 3

䊳

Finding nth Roots Use the nth roots theorem to solve the equation x3 1 0. Write the results in rectangular form and graph.

Solution

䊳

From x3 1 0, we have x3 1 and must find the three cube roots of unity. As before, we begin in trigonometric form: 1 0i 11cos 0° i sin 0°2 since tan1 102 0° (or by observation). With n 3, r 1, and 0°, we have 0° 360°k 3 3 √3 yi 1 0° 120°k. r 11 1, and 冢q, ]冣 3 3 The principal root 1k 02 is z1 z0 11cos 0° i sin 0°2 1. Adding 120° to each previous argument, we find the other 120 (1, 0) 120 roots are z x

0

z1 11cos 120° i sin 120°2 z2 11cos 240° i sin 240°2.

120 z2

1 13 i, In rectangular form these are 2 2 13 1 i, as shown in the figure. and 2 2

√3

冢q, ]冣


EXAMPLE 4

䊳

䊳

Finding nth Roots Use the nth roots theorem to find the five fifth roots of z 1613 16i.

Solution

䊳

In trigonometric form, 16 13 16i 321cos 30° i sin 30°2 since tan1a

16 b 30°. With n 5, r 32, and 30°, we have 16 13 30° 360°k 5 5 1r 132 2, and 6° 72°k. The principal root is 5 5 z0 21cos 6° i sin 6°2.

Adding 72° to each previous argument, we find the other four roots are

WORTHY OF NOTE Using a calculator in degree MODE to find the fifth root of z 1613 16i in Example 4 will only result in the principal root, as shown here (note the MODE was set to round to four decimal places). The nth roots theorem is necessary to find the other four roots!

z1 21cos 78° i sin 78°2 z3 21cos 222° i sin 222°2

z2 21cos 150° i sin 150°2 z4 21cos 294° i sin 294°2 Now try Exercises 41 through 44

Of the five roots in Example 4, only z2 21cos 150° i sin 150°2 uses a standard angle. Applying De Moivre’s theorem with n 5 gives 12 cis 150°2 5 32 cis 750° 32 cis 30° or 16 13 16i.✓ See Exercise 52. As a consequence of the arguments in a 360° , the solution being uniformly separated by n graphs of complex roots are equally spaced n about a circle of radius 1r. The five fifth roots from Example 4 are shown in Figure 8.106 (note each argument differs by 72°2. This

䊳

Figure 8.106 yi

z1 2 cis 78

z2 2 cis 150 2

z0 2 cis 6 x

z3 2 cis 222 z4 2 cis 294

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817

characteristic of nth roots also simplifies a search for all nth roots when some form of technology is used. For instance, once a template has been set, a graphing calculator can easily cycle through the values of k and display the results in a table. EXAMPLE 5

䊳

Using a Calculator to Find nth Roots Use the nth roots theorem and the TABLE feature of a graphing calculator to find all six sixth roots of z 4 4i 13.

Solution

䊳

For z 4 4i 13, we have r 2142 2 14132 2 216 48 8. With z in QIII, r 60 yields 240. Using the nth roots theorem with n 6 gives 360°k 360°k n n 1 z 1 r c cos a b i sin a bd n n n n 240° 240° 360°k 360°k 6 6 1 z 1 8 c cos a b i sin a bd 6 6 6 6 6 1 83cos140° 60°k2 i sin140° 60°k2 4

n th roots theorem substitute 6 for n, 8 for r, 240 for simplify

While the TABLE feature cannot display complex numbers (even in a bi MODE ), we can separate these roots into their real and imaginary parts. Consider the following. 1 z 1 83cos140° 60°k2 i sin140° 60°k2 4 6 6 1 8 cos140° 60°k2 1 8 sin140° 60°k2i 6

6

sixth roots of z 6 distribute 1 8

On the Y= screen, simply enter the real part as Y1 18 cos140 60X2 and the 6 coefficient of the imaginary part as Y2 18 sin140 60X2 as shown in Figure 8.107. After setting TBlStart 0 and Tbl 1, we can quickly produce all sixth roots of z 4 4i 23. Reading from the table shown in Figure 8.108, we find the six sixth roots of z (rounded to 3 decimal places) are 6

WORTHY OF NOTE In Example 5, when k 6, the argument of the root is 400. This is coterminal with 40, the argument when k 0. As seen in Figure 8.108, these coterminal angles produce the same complex root.

z0 1.083 0.909i z3 1.083 0.909i

z1 0.246 1.393i z4 0.246 1.393i

z2 1.329 0.484i z5 1.329 0.484i

As a check, each root zk from k 0 to k 5 can be raised to the sixth power on the home screen, using the expression Y1(k) iY2(k). The check for k 0 is shown in Figure 8.109 (note that 4 13 ⬇ 6.92820323). Figure 8.107

Figure 8.108

Figure 8.109


䊳

For additional insight into roots of complex numbers, we reason that the nth roots of a complex number must also be complex. To find the four fourth roots of z 8 8i 13 161cos 60° i sin 60°2, we seek a number of the form r1cos i sin 2 such that 3r1cos i sin 2 4 4 161cos 60° i sin 60°2. Applying De Moivre’s theorem to the left-hand side and equating equivalent parts we obtain r4 3cos142 i sin142 4 161cos 60° i sin 60°2, which leads to r4 16 and 4 60°

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From this it is obvious that r 2, but as with similar equations solved in Chapter 7, the equation 4 60° has multiple solutions. To find them, we first add 360°k to 60°, then solve for . 4 60° 360°k 60° 360°k 4 15° 90°k

add 360°k divide by 4 result

For convenience, we start with k 0, 1, 2, and so on, which leads to For k 0:

15° 90°102 15°

For k 1:

15° 90°112 105°

For k 2:

15° 90°122 195°

For k 3:

15° 90°132 285°

At this point it should strike us that we have four roots — exactly the number required. Indeed, using k 4 gives 15° 90°142 375°, which is coterminal with the 15° obtained when k 0. Hence, the four fourth roots are z0 21cos 15° i sin 15°2 z2 21cos 195° i sin 195°2

C. You’ve just seen how we can use the nth roots theorem to find the nth roots of a complex number

z1 21cos 105° i sin 105°2 z3 21cos 285° i sin 285°2 .

The check for these solutions is asked for in Exercise 51. As a final note, it must have struck the mathematicians who pioneered these discoveries with some amazement that complex numbers and the trigonometric functions should be so closely related. The amazement must have been all the more profound upon discovering an additional connection between complex numbers and exponential functions. For more on these connections, visit www.mhhe.com/coburn and review Section 8.7: Complex Numbers in Exponential Form.

8.6 EXERCISES 䊳



1. For z r1cos i sin 2, z5 is computed as according to theorem.

2. If z 6i, then z raised to an be real and z raised to an since

3. One application of De Moivre’s theorem is to check solutions to a polynomial equation.

4. The nth roots of a complex number are equally n spaced on a circle of radius 1 r, since their arguments all differ by degrees or radians.

5. From Example 4, go ahead and compute the value of z5, z6, and z7. What do you notice? Discuss how this reaffirms that there are exactly n, nth roots.

power will power will be .

6. Use a calculator to find 11 3i2 4. Then use it again to find the fourth root of the result. What do you notice? Explain the discrepancy and then resolve it using the nth roots theorem to find all four roots.

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Use De Moivre’s theorem to compute the following. Clearly state the values of r, n, and ␪ before you begin. Use a calculator to check your results.

7. 13 3i2 4

9. 11 i132

3

1 13 5 11. a ib 2 2 12 12 6 13. a ib 2 2 15. 14 cis 330°2 3 17. a

5 12 cis 135°b 2

8. 12 2i2 6

10. 1 13 i2

3

13 1 6 12. a ib 2 2 12 12 5 14. a ib 2 2 16. 14 cis 300°2 3 18. a

8 12 cis 135°b 2

Use De Moivre’s theorem to verify the solution given for each polynomial equation.

19. z4 3z3 6z2 12z 40 0; z 2i 20. z4 z3 7z2 9z 18 0; z 3i 21. z4 6z3 19z2 6z 18 0; z 3 3i 22. 2z4 3z3 4z2 2z 12 0; z 1 i 23. z5 z4 4z3 4z2 16z 16 0; z 13 i 24. z5 z4 16z3 16z2 256z 256 0; z 2 13 2i 25. z4 4z3 7z2 6z 10 0; z 1 2i 26. z4 2z3 7z2 28z 52 0; z 3 2i Find the nth roots indicated by writing and solving the related equation.

31. three cube roots of 27i 32. five fifth roots of 32i Find all solutions of each equation using the nth roots theorem.

33. x5 32 0

34. x5 243 0

35. x3 27i 0

36. x3 64i 0

37. x5 12 i 12 0 38. x5 1 i 13 0 39. Solve the equation x3 1 0 by factoring it as the difference of cubes and applying the quadratic formula. Compare results to those obtained in Example 3. 40. Use the nth roots theorem to find the four fourth roots of unity, then find all solutions to x4 1 0 by factoring it as a difference of squares. What do you notice? Use the nth roots theorem to find the nth roots. Clearly state r, n, and ␪ (from the trigonometric form of z) as you begin. Answer in exact form when possible, otherwise use a four decimal place approximation.

41. four fourth roots of 8 8i 13 42. five fifth roots of 16 16i13 43. four fourth roots of 7 7i 44. three cube roots of 9 9i Use the nth roots theorem and the TABLE feature of a graphing calculator to find all roots indicated. Clearly state r, ␪, n, and k for each root. Round to three decimal places when necessary.

27. five fifth roots of unity

45. three third roots of 423 4i

28. six sixth roots of unity

46. four fourth roots of 8 8i 23

29. five fifth roots of 243

47. four fourth roots of 81

30. three cube roots of 8

48. three third roots of 27i

䊳

819


4p3 ⴙ 27q2 108 For cubic equations of the form z3 ⴙ pz ⴙ q ⴝ 0, where p and q are real numbers, one solution has the form q q z ⴝ 3 ⴚ ⴙ 1D ⴙ 3 ⴚ ⴚ 1D, where D is called the discriminant. Compute the value of D for the cubic equations A 2 A 2 q q given, then use the nth roots theorem to find the three cube roots of ⴚ ⴙ 1D and ⴚ ⴚ 1D in trigonometric form 2 2 (also see Exercises 59 and 60). The discriminant of a cubic equation: D ⴝ

49. z3 6z 4 0

50. z3 12z 8 0

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APPLICATIONS

51. Powers and roots: Just after Example 5, the four fourth roots of z 8 8i13 were given as z0 21cos 15° i sin 15°2 z1 21cos 105° i sin 105°2 z2 21cos 195° i sin 195°2 z3 21cos 285° i sin 285°2. Verify these are the four fourth roots of z 8 8i13 using a calculator and De Moivre’s theorem.

52. Powers and roots: In Example 4 we found the five fifth roots of z 16 13 16i were z0 z1 z2 z3 z4

21cos 6° i sin 6°2 21cos 78° i sin 78°2 21cos 150° i sin 150°2 21cos 222° i sin 222°2 21cos 294° i sin 294°2

Verify these are the five fifth roots of 1613 16i using a calculator and De Moivre’s theorem.

Electrical circuits: For an AC circuit with three branches wired in parallel, the total impedance is given by Z1Z2Z3 ZT , where Z1, Z2, and Z3 represent the impedance in each branch of the circuit. If the impedance Z1Z2 Z1Z3 Z2Z3 in each branch is identical, Z1 Z2 Z3 Z, and the numerator becomes Z3 and the denominator becomes 3Z2, (a) use De Moivre’s theorem to calculate the numerator and denominator for each value of Z given, (b) find the total impedance by Z3 Z computing the quotient 2 , and (c) verify your result is identical to . 3 3Z

53. Z 3 4j in all three branches 䊳

54. Z 5 13 5j in all three branches


In Chapter 7, you were asked to verify that sin13␪2 ⴝ 3 sin ␪ ⴚ 4 sin3␪ and cos14␪2 ⴝ 8 cos4␪ ⴚ 8 cos2␪ ⴙ 1 were 17 identities (Section 7.4, Exercises 21 and 22). For z ⴝ 3 ⴙ i 17, verify 冟z冟 ⴝ 4 and ␪ ⴝ tanⴚ1a b, then draw a right 3 triangle with 17 opposite ␪ and 3 adjacent to ␪. Discuss how this right triangle and the identities given can be used in conjunction with De Moivre’s theorem to find the exact value of the powers given (also see Exercises 57 and 58).

55. 13 i172 3

56. 13 i 172 4

For cases where ␪ is not a standard angle, working toward an exact answer using De Moivre’s theorem requires the b use of multiple angle identities and drawing the right triangle related to ␪r ⴝ tanⴚ1 ` ` . For Exercises 57 and 58, use a De Moivre’s theorem to compute the complex powers by (a) constructing the related right triangle for ␪r, (b) evaluating sin14␪2 using two applications of double-angle identities, and (c) evaluating cos14␪2 using a Pythagorean identity and the computed value of sin14␪2.

57. z 11 2i2 4

58. 12 i 152 4

The solutions to the cubic equations in Exercises 49 and 50 (repeated in Exercises 59 and 60) can be found by adding q q the cube roots of ⴚ ⴙ 1D and ⴚ ⴚ 1D that have arguments summing to 360ⴗ. 2 2

59. Find the roots of z3 6z 4 0

60. Find the roots of z3 12z 8 0

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Making Connections

MAINTAINING YOUR SKILLS y

63. (1.4) Find an equation of the line whose graph is given.

61. (7.2) Prove the following is a identity: tan2x 1 ⫺ cos x ⫽ cos x sec x ⫹ 1

5

5 x

⫺1

62. (1.3) Given f 1x2 ⫽ 2x2 ⫺ 3x, determine:

64. (6.6) Solve the triangle given. Round lengths to hundredths of a meter.

1 f 1⫺12, f a b, f 1a2 , and f 1a ⫹ h2 . 3

B 52⬚

213 m

C

A

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. yi

(a)

C

(b)

5

yi

(c)

(d)

u

2

v b

a v ⫺5

⫺2

5 x

2

␪

x

B ⫺2

⫺5

(e)

yi

(f)

(g)

yi

(h)

5

2

c

b

v

v ␪

u

⫺2

2

x

⫺5

a

⫺2

1. ____ projuv

5 x

⫺5

9. ____ cos C ⫽

b2 ⫺ a2 ⫺ c2 ⫺2ac

2. ____ 4i 4j

10. ____ 2 cis 60⬚, 2 cis 180⬚, 2 cis 300⬚

3. ____ c ⫽ 2a2 ⫹ b2 ⫺ 2ab cos C

11. ____ ␪ ⫽ cos⫺1a

4. ____ x3 ⫺ 8 ⫽ 0

12. ____ ␪ ⫽ 45⬚

5. ____ | v| ⫽ 422, ␪ ⫽ 225⬚

13. ____ x3 ⫹ 8 ⫽ 0

6. ____ A ⫽ 2s1s ⫺ a2 1s ⫺ b21s ⫺ c2

u v # b 冟u冟冟v冟

14. ____ u ⴙ v

7. ____ 2, ⫺1 ⫹ i13, ⫺1 ⫺ i 23

15. ____

8. ____ u v

16. ____

sin B sin C ⫽ c b

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Oblique Triangles and the Law of Sines

KEY CONCEPTS sin A

sin C

sin B

. • In any triangle, the ratio of the sine of an angle to its opposite side is constant: a c b • The law of sines requires a known angle, a side opposite this angle and an additional side or angle, hence cannot

• • • •

be applied for SSS or SAS triangles. For AAS and ASA triangles, the law of sines yields a unique solution. When given two sides of a triangle and an angle opposite one of these sides (SSA), the number of solutions is in doubt, giving rise to the designation, “the ambiguous case.” SSA triangles may have no solution, one solution, or two solutions, depending on the length of the side opposite the given angle. When solving triangles, always remember: • The sum of all angles must be 180°: ⬔A ⬔B ⬔C 180°. • The sum of any two sides must exceed the length of the remaining side. • Longer sides are opposite larger angles. • k sin1 has no solution for 冟k冟 7 1. • k sin1 has two solutions in 3 0°, 360°2 for 0 6 冟k冟 6 1.

EXERCISES Solve the following triangles. 1.

A

2. B

293 cm B

A 28 142 52 yd C

123

21

C

3. A tree is growing vertically on a hillside. Find the height of the tree if it makes an angle of 110° with the hillside and the angle of elevation from the base of the hill to the top of the tree is 25° at a distance of 70 ft. 25

4. Find two values of that will make the equation true:

110

sin sin 50° . 14 31

5. Solve using the law of sines. If two solutions exist, find both (figure not drawn to scale). C 105 cm

67 cm

35 A

6. Jasmine is flying her tethered, gas-powered airplane at a local park, where a group of bystanders is watching from a distance of 60 ft, as shown. If the tether has a radius of 35 ft and one of the bystanders walks away at an angle of 40°, will he get hit by the plane? What is the smallest angle of exit he could take (to the nearest whole) without being struck by Jasmine’s plane?

Radius 35 ft Jasmine

40 60 ft Bystanders

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SECTION 8.2


823

The Law of Cosines; the Area of a Triangle

KEY CONCEPTS • The law of cosines is used to solve SSS and SAS triangles. • The law of cosines states that in any triangle, the square of any side is equal to the sums of the squares of the other two sides, minus twice their product times the cosine of the included angle: a2 b2 c2 2bc cos A

• When using the law of cosines to solve a SSS triangle, always begin with the largest angle or the angle opposite the largest side. • The area of a nonright triangle can be found using the following formulas. The choice of formula depends on the information given. • two sides a and b • two angles A and B • three sides a, b, and c abc with included angle C with included side c with s 1 c2 sin A sin B 2 A ab sin C A 2 2 sin C A 2s1s a21s b21s c2

EXERCISES 98 167 m 325 m 7. Solve for B: 92 122 152 211221152 cos B 8. Use the law of cosines to find the missing side of the triangle shown. 9. While preparing for the day’s orienteering meet, Rick finds that the distances 1250 yd 720 yd between the first three markers he wants to pick up are 1250 yd, 1820 yd, and 720 yd. Find the measure of each angle in the triangle formed so that Rick is sure to find all three markers. 1820 yd 10. The Great Pyramid of Giza, also known as Khufu’s pyramid, is the sole remaining member of the Seven Wonders of the Ancient World. It was built as a tomb for the Egyptian pharaoh Khufu from the fourth dynasty. This square pyramid is made up of four isosceles triangles, each with a base of 230.0 m and a slant height along the edges of about 218.7 m. Approximate the total surface area of Khufu’s pyramid (excluding the base).

SECTION 8.3

Vectors and Vector Diagrams

KEY CONCEPTS • Quantities/concepts that can be described using a single number are called scalar quantities. Examples are time, perimeter, area, volume, energy, temperature, weight, and so on. • Quantities/concepts that require more than a single number to describe their attributes are called vector quantities. Examples are force, velocity, displacement, pressure, and so on. • Vectors can be represented using directed line segments to indicate magnitude and direction. The origin of the segment is called the initial point, with the arrowhead pointing to the terminal point. When used solely for comparative analysis, they are called geometric vectors. • Two vectors are equal if they have the same magnitude and direction. • Vectors can be represented graphically in the xy-plane by naming the initial and terminal points of the vector or by giving the magnitude and angle of the related position vector [initial point at (0, 0)]. • For a vector with initial point (x1, y1) and terminal point (x2, y2), the related position vector can be written in the component form Ha, bI, where a x2 x1 and b y2 y1. • For a vector written in the component form Ha, bI, a is called the horizontal component and b is called the vertical component of the vector. For vector v Ha, bI, the magnitude of v is 冟v冟 2a2 b2. • • Vector components can also be written in trigonometric form. See page 774. • For u Ha, bI, v Hc, dI, and any scalar k, we have the following operations defined: u v Ha c, b dI

u v Ha c, b dI

ku Hka, kbI for k 僆⺢

If k 7 0, the new vector has the same direction as u; k 6 0, it has the opposite direction.

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• Vectors can be written in algebraic form using i, j notation, where i is an x-axis unit vector and j is a y-axis unit

vector. The vector Ha, bI is written as a linear combination of i and j: Ha, bI ai bj. v • For any nonzero vector v, vector u 冟冟 is a unit vector in the same direction as v. v • In aviation and shipping, the heading of a ship or plane is understood to be the amount of rotation in the clockwise direction from due north.

EXERCISES 11. Graph the vector v H9, 5I, then compute its magnitude and direction angle. 12. Write the vector u H8, 3I in i, j form and compute its magnitude and direction angle. 13. Approximate the horizontal and vertical components of the vector u, where 冟u冟 18 and 52°. 14. Compute 2u v, then find the magnitude and direction of the resultant: u H3, 5I and v H2, 8I. 15. Find a unit vector that points in the same direction as u 7i 12j. 16. Without computing, if u H9, 2I and v H2, 8I, will the resultant sum lie in Quadrant I or II? Why? Exercise 18 17. It’s once again time for the Great River Race, a 12 -mi swim across the Panache River. If Karl fails to take the river’s 1-mph current into account and he swims the race at 3 mph, how far from the finish marker does he end up when he makes it to the other side? 18. Two Coast Guard vessels are towing a large yacht into port. The first is pulling with a 18 ␪ force of 928 N and the second with a force of 850 N. Determine the angle for the second Coast Guard vessel that will keep the ship moving safely in a straight line.

SECTION 8.4

Vector Applications and the Dot Product

KEY CONCEPTS • Vector forces are in equilibrium when the sum of their components is the zero vector. • When the components of vector u are nonquadrantal, with one of its components lying along vector v, we call this component the “component of u along v” or compvu. For vectors u and v, compvu 冟u冟cos , where is the angle between u and v. • • Work done is computed as the product of the constant force F applied, times the distance D the force is applied: W 冟F冟 # D. • If force is not applied parallel to the direction of movement, only the component of the force in the direction of movement is used in the computation of work. If u is a force vector not parallel to the direction of vector v, the equation becomes W compvu # 冟v冟. • For vectors u Ha, bI and v Hc, dI, the dot product u # v is defined as the scalar ac bd. • The dot product u # v is equivalent to compuv # 冟v冟 and to 冟u冟冟v冟cos . u#v u v • The angle between two vectors can be computed using cos 冟冟 # 冟冟冟冟冟冟 . u v uv u#v • Given vectors u and v, the projection of u along v is the vector projvu defined by projvu a 2 b v. 冟v冟 • Given vectors u and v, projvu, u can be resolved into the orthogonal components u1 and u2 where u u1 u2, u1 projvu, and u2 u u1. • The horizontal distance x a projectile travels in t seconds is x 1 冟v冟cos 2t, where 冟v冟 is the magnitude of the initial velocity, and is the angle of projection. • The vertical height y of a projectile after t seconds is y 1 冟v冟sin 2t 16t2.

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EXERCISES 19. For the force vectors F1 and F2 given, find the resultant and an additional force vector so that equilibrium takes place: F1 H20, 70I; F2 H45, 53I. 20. Find compvu for u 12i 16j and v 19i 13j. 21. Find the component d that ensures vectors u and v are orthogonal: u H2, 9I and v H18, dI. 22. Compute p # q and find the angle between them: p H5, 2I; q H4, 7I. 23. Given force vector F H50, 15I and v H85, 6I, find the work required to move an object along the entire length of v. Assume force is in pounds and distance in feet. 24. A 650-lb crate is sitting on a ramp that is inclined at 40°. Find the force needed to Exercise 24 hold the crate stationary. lb 25. An arctic explorer is hauling supplies from the supply hut to her tent, a distance of 650 120 ft, in a sled she is dragging behind her. If the straps used make an angle of 25° 650 lb with the snow-covered ground and she pulls with a constant force of 75 lb, find the amount of work done. 40 26. A projectile is launched from a slingshot with an initial velocity of v0 280 ft/sec G at an angle of 50°. Find (a) the position of the object after 1.5 sec and (b) the time required to reach a height of 150 ft.

SECTION 8.5

Complex Numbers in Trigonometric Form

KEY CONCEPTS Imaginary yi • A complex number a bi 1a, b2 can be written in trigonometric (a, b) axis form by noting (from its graph) that a r cos and b r sin : (r cos ␪, r sin ␪) b a bi r 1cos i sin 2. • The angle is called the argument of z and r is called the modulus of z. r • The argument of a complex number z is not unique, since any rotation of x ␪ 2k (k an integer) will yield a coterminal angle. a Real • To convert from trigonometric to rectangular form, evaluate cos and sin axis and multiply by the modulus. • To multiply complex numbers in trig form, multiply the moduli and add the arguments. To divide complex numbers in trig form, divide the moduli and subtract the arguments. • Complex numbers have numerous real-world applications, particularly in a study of AC electrical circuits. • The impedance of an AC circuit is given as Z R j1XL XC 2, where R is a pure resistance, XC is the capacitive reactance, XL is the inductive reactance, and j 11.

• Z is a complex number with magnitude 0 Z 0 2R2 1XL XC 2 2 and phase angle tan1a ( represents the angle between the voltage and current). V • In an AC circuit, voltage V IZ; current I . Z

EXERCISES 27. Write in trigonometric form: z 1 13i

28. Write in rectangular form: z 3 12 c cisa b d 4

XL XC b R

29. Graph in the complex plane: z 51cos 30° i sin 30°2

z1 30. For z1 8 cis a b and z2 2 cis a b, compute z1z2 and . z2 4 6 31. Find the current I in a circuit where V 4 13 4j and Z 1 13j . 32. In the VRLC series circuit shown, R 10 , XL 8 , and XC 5 . Find the magnitude of Z and the phase angle between current and voltage. Express the result in trigonometric form.

R A

10 Ω

XL B

8Ω

XC C

5Ω

D

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SECTION 8.6

De Moivre’s Theorem and the Theorem on nth Roots

KEY CONCEPTS • For complex number z r1cos i sin 2, zn rn 3cos1n2 i sin1n2 4 (De Moivre’s theorem). • De Moivre’s theorem can be used to check complex solutions of polynomial equations. 2k 2k n n • For complex number z r1cos i sin 2 , 2z 2r c cosa n n b i sina n n b d , for k 1, 2, 3, p , n 1 (nth roots theorem). n • The nth roots of a complex number are equally spaced around a circle of radius 1r in the complex plane. EXERCISES 33. Use De Moivre’s theorem to compute the value of 11 i 132 5. 34. Use De Moivre’s theorem to verify that z 1 i is a solution of z4 z3 2z2 2z 4 0. 35. Use the nth roots theorem to find the three cube roots of 125i. 36. Solve the equation using the nth roots theorem: x3 216 0. 37. Given that z 2 2i is a fourth root of 64, state the other three roots. 38. Solve using the quadratic formula and the nth roots theorem: z4 6z2 25 0. 39. Use De Moivre’s theorem to verify the three roots of 125i found in Exercise 35.

PRACTICE TEST 1. Within the Kilimanjaro Game Reserve, a fire is spotted by park rangers stationed in two towers 39 68 known to be 10 mi apart. 10 mi Using the line between them as a baseline, tower A A B reports the fire is at an angle of 39°, while tower B reports an angle of 68°. How far is the fire from the closer tower? 2. At the circus, Mac and Joe are watching a high-wire act from first-row seats on opposite sides of the center ring. Find the height of the performing acrobat at the instant Mac measures an angle of elevation of 68° while Joe measures an angle of 72°. Assume Mac and Joe are 100 ft apart.

Exercise 2 Acrobat

72 68 Mac

Joe

3. Three rods are C B attached via two A joints and shaped 6 in. 15 in. into a triangle. ?? in. How many triangles can be formed if the angle at the joint B must measure 20°? If two triangles can be formed, solve both.

4. Jackie and Sam are rounding up cattle in the Range brush country, and are 3 mi communicating via 32 walkie-talkie. Jackie is at Water hole 6 mi Dead the water hole and Sam is Oak at Dead Oak, which are 6 mi apart. Sam finds some strays and heads them home at the 32° indicated. (a) If the maximum range of Jackie’s unit is 3 mi, will she be able to communicate with Sam as he heads home? (b) If the maximum range were 4 mi, how far from Dead Oak is Sam when he is first contacted by Jackie? 5. As part of an All-Star Exercise 5 competition, a group of soccer players (forwards) stand where shown in the diagram and attempt to hit a moving target with a twohanded overhead pass. If a Overhead 35 yd player has a maximum pass effective range of approximately (a) 25 yd, 53 can the target be hit? (b) about 28 yd, how many “effective” throws can be made? (c) 35 yd and the target is moving at 5 yd/sec, how many seconds is the target within range?

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6. The summit of Triangle 3.5 mi Peak can only be reached from one side, using a trail 24 straight up the side that is 5 mi approximately 3.5 mi long. If the mountain is 5 mi wide at its base and the trail makes a 24° angle with the horizontal, (a) what is the approximate length of the opposing side? (b) How tall is the peak (in feet)? 7. The Bermuda Triangle is 1025 mi B generally thought to be M the triangle formed by 977 mi Miami, Florida, San 1020 mi Juan, Puerto Rico, and Bermuda itself. If the P distances between these locations are the 1025 mi, 1020 mi, and 977 mi indicated, find the measure of each angle and the area of the Bermuda Triangle. 8. A helicopter is flying at 90 mph on a heading of 40°. A 20-mph wind is blowing from the NE on a heading of 190°. What is the true course and speed of the helicopter relative to the ground? Draw a diagram as part of your solution. 9. Two mules walking along a river bank are pulling a heavy barge up river. The 30 ␪ first is pulling with a force of 250 N and the second with a force of 210 N. Determine the angle for the second mule that will ensure the barge stays midriver and does not collide with the shore. Exercise 10 10. Along a production line, various tools are attached to 22 the ceiling with a multijointed arm so that 42 workers can draw one down, position it for use, then move 58 Joint it up out of the way for the next tool (see the diagram). If the first segment is 100 cm, the second is 75 cm, and the third is 50 cm, determine the approximate coordinates of the last joint. 11. Three ranch hands have roped F1 F2 a runaway steer and are 150 N attempting to hold him steady. 110 N The first and second ranch 67 42 hands are pulling with the ?F3 magnitude and at the angles indicated in the diagram. If the steer is held fast by the efforts of all three, find the magnitude of the tension and angle of the rope from the third cowhand.

827

Practice Test

12. For u H9, 5I and v H2, 6I, (a) compute the angle between u and v; (b) find the projection of u along v (find projvu); and (c) resolve u into vectors u1 and u2, where u1 7 v and u2⬜v. 13. A lacrosse player flips a long pass to a teammate way downfield who is near the opponent’s goal. If the initial velocity of the pass is 110 ft/sec and the ball is released at an angle of 50° with level ground, how high is the ball after 2 sec? How long after launch until the ball reaches this height again? 14. Compute the quotient

z1 , given z2

z1 6 15 cis a b and z2 3 15 cis a b. 8 12 15. Compute the product z z1z2 in trigonometric form, then verify 冟z1冟冟z2冟冟z冟 and 1 2 : z1 6 6i; z2 4 4i 13 16. Use De Moivre’s theorem to compute the value of 1 13 i2 4.

17. Use De Moivre’s theorem to verify 2 2i 13 is a solution to z5 3z3 64z2 192 0. 18. Use the nth roots theorem to solve x3 125i 0. 19. Solve using u-substitution, the quadratic formula, and the nth roots theorem: z4 6z2 58 0. 20. Due to its huge biodiversity, preserving Southeast Asia’s Coral Triangle has become a top priority for conservationists. Stretching from the northern Philippines (P), south along the coast of Borneo (B) to the Lesser Sunda Islands (L), then eastward to the Solomon Islands (S), this area is home to over 75% of all coral species known. Use Heron’s formula to help find the total area of this natural wonderland, given the dimensions shown. P

1275 mi Halmahera

B 2010 mi

2390 mi 23

L

1690 mi

1590 mi

S

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CALCULATOR EXPLORATION AND DISCOVERY Investigating Projectile Motion There are two important aspects of projectile motion that were not discussed earlier, the range of the projectile and the optimum angle that will maximize this range. Both can be explored using the equations for the horizontal and vertical components of the projectile’s position: horizontal S 1 冟v冟cos 2t and vertical S 1 冟v冟sin 2 16t2. In Example 10 of Section 8.4, an arrow was shot from a bow with initial velocity 冟v冟 150 ft/sec at an angle of 50°. Enter the equations above on the Y= screen as Y1 and Y2, using these values (Figure 8.110). To find the range of the arrow, we must find where the height (Y2) is zero (the arrow has hit the ground). From the GRAPH of Y1 and Y2, the zero in question can be found by applying the CALC:zero feature to Y2 with a left bound of 7 and right bound of 8. As Figure 8.111 shows, the arrow hits the ground after approximately 7.2 seconds. Pressing the up arrow key jumps the cursor to Y1 at this same X value, and we see in Figure 8.112 that the horizontal range of the arrow is approximately 692.4 feet. The question now is how to compute these values exactly. We begin with the equation for the arrow’s vertical position y 1 冟v冟sin 2t 16t2. Since the object returns to Earth when y 0, we substitute 0 for y and factor out t: 0 t1 冟v冟sin 16t2. Solving for t gives t 0 冟v冟sin . Since the component of velocity in the horizontal direction is or t 16 冟v冟cos , the basic distance relationship D r # t gives the horizontal range of 冟v冟sin 冟v冟2sin cos R 冟v冟cos # or . Checking the values given for the arrow 16 16 (冟v冟 150 ft/sec and 50°) verifies the range is R ⬇ 692.4. But what about the maximum possible range for the arrow? Using 冟v冟 150 for R results in an 1502sin cos , which we can enter as Y3 and equation in theta only: R12 16 investigate for various . After carefully entering R12 as Y3 and resetting TBLSET to TblStart 30 and ¢Tbl 5, the TABLE in Figure 8.113 shows a maximum range of about 703 ft at 45°. Resetting TBLSET to TblStart 40 and ¢Tbl 1 verifies this fact. For each of the following exercises, find (a) the height of the projectile after 1.75 sec, (b) the maximum height of the projectile, (c) the range of the projectile, and (d) the number of seconds the projectile is airborne.

Figure 8.110

Figure 8.111 750

0

8

100

Figure 8.112 750

0

8

100

Figure 8.113

Exercise 1: A javelin is thrown with an initial velocity of 85 ft/sec at an angle of 42°. Exercise 2: A cannonball is shot with an initial velocity of 1120 ft/sec at an angle of 30°. Exercise 3: A baseball is hit with an initial velocity of 120 ft/sec at an angle of 50°. Will it clear thecenter field fence, 10 ft high and 375 ft away? Exercise 4: A field goal (American football) is kicked with an initial velocity of 65 ft/sec at an angle of 35°. Will it clear the crossbar, 10 ft high and 40 yd away?

STRENGTHENING CORE SKILLS Vectors and Static Equilibrium In Sections 8.3 and 8.4, the concepts of vector forces, resultant forces, and equilibrium were studied extensively. A nice extension of these concepts involves what is called static equilibrium. Assuming that only coplanar forces are acting on an object, the object is said to be in static equilibrium if the sum of all vector forces acting on it is 0. This implies that the object is stationary, since the forces all counterbalance each other. The methods involved are simple and direct, with a wonderful connection to the systems of equations you’ve likely seen previously. Consider the following example.

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Illustration 1 䊳 As part of their training, prospective FBI agents must move handover-hand across a rope strung between two towers. An agent-in-training weighing 180 lb is two-thirds of the way across, causing the rope to deflect from the horizontal at the angles shown. What is the tension in each part of the rope at this point?

u 9

14 w 180 lb

Solution 䊳 We have three concurrent forces acting on the point where the agent grasps the rope. Begin by drawing a vector diagram and computing the components of each force, using the i, j notation. Note that w 180j. u 冟u冟cos19°2i 冟u冟sin19°2j ⬇ 0.9877冟u冟i 0.1564冟u冟 j v 冟v冟cos114°2i 冟v冟sin114°2j ⬇ 0.9703冟v冟i 0.2419冟v冟 j

v

y

9

v 14

u 180

x

w

For equilibrium, all vector forces must sum to the zero vector: u v w 0, which results in the following equation: 0.9877 0 u 0 i 0.1564 0 u 0 j 0.9703 0 v 0 i 0.2419 0 v 0 j180j 0i 0j. Factoring out i and j from the left-hand side yields 10.9877 |u | 0.9703| v| 2 i 10.1564 |u | 0.2419 | v | 1802 j 0i 0j. Since any two vectors are equal only when corresponding components are equal, we obtain a system in the two variables 冟u冟 and 冟v冟: e

0.9877冟u冟 0.9703冟v冟 2 0 . 0.1564冟u冟 0.2419冟v冟 180 0

Solving the system using matrix equations and a calculator (or any desired method), gives 冟u冟 ⬇ 447 lb and 冟v冟 ⬇ 455 lb. At first it may seem surprising that the vector forces (tension) in each part of the rope are so much greater than the 180-lb the agent weighs. But with a 180-lb object hanging from the middle of the rope, the tension required to keep the rope taut (with small angles of deflection) must be very great. This should become more obvious to you after you work Exercise 2. Exercise 1: A 500-lb crate is suspended by two ropes attached to the ceiling rafters. Find the tension in each rope.

25

20 500 lb

Exercise 2: Two people team up to carry a 150-lb weight by passing a rope through an eyelet in the object. Find the tension in each rope.

45

Exercise 3: Referring to Illustration 1, if the rope has a tension limit of 600-lb (before it snaps), can it suspend a 200-lb agent at the same location shown?

45 150 lb

CUMULATIVE REVIEW CHAPTERS 1–8 1. Solve using a standard triangle. a 20, b _____, c ____ 30°, ____, ____ A

B ␤ 20 m 30

B

2. Solve using trigonometric ratios. a ⬇ ____, b ⬇ ____, c 82

C

82 ft

63

____, 63°, ____ A ␣ C 3. A torus is a donut-shaped solid figure. Its surface area is given by the formula A 2 1R2 r2 2 , where R is the outer radius of the donut, and r is the inner radius. Solve the formula for R in terms of r and A.

4. For a complex number a bi, (a) verify the sum of a complex number and its conjugate is a real number, and (b) verify the product of a complex number and its conjugate is a real number. 5. State the values of all six trig functions given 3 tan with cos 7 0. 4 6. Sketch the graph of y 4 cos a x b using 6 3 transformations of y cos x. 7. Solve using the quadratic formula: 5x2 8x 2 0.

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8. Solve by completing the square: 3x2 72x 427 0. 9. Given cos 53° ⬇ 0.6 and cos 72° ⬇ 0.3, approximate the value of cos 19° and cos 125° without using a calculator. 10. Find all real values of x that satisfy the equation 13 2 sin12x2 2 13. State the answer in degrees. 11. a. Given that 2008 ft 1 acre 43,560 ft2, find the cost of a lot with the 25.9 dimensions shown (to the 1475 ft nearest dollar) if land in this area is going for $4500 per acre. North b. After an accident at (5, 12) 12 sea, a search-and10 rescue team decides 8 to focus their efforts 6 on the area shown due (12, 5) 4 to prevailing winds 2 and currents. Find the 0 2 4 6 8 10 12 14 East (0, 0) distances between each vertex (use Pythagorean triples and a special triangle) and the number of square miles in the search area.

12. State the domain of each function: a. f 1x2 12x 3

b. g1x2 logb 1x 32

16. A commercial fishery stocks a lake with 250 fish. Based on previous experience, the population of fish is expected to grow according to the model 12,000 P1t2 , where t is the time in months. 1 25e0.2t From on this model, (a) how many months are required for the population to grow to 7500 fish? (b) If the fishery expects to harvest three-fourths of the fish population in 2 yr, approximately how many fish will be taken? 17. A 900-lb crate is sitting on a ramp which is inclined at 28°. Find the force needed to hold the object stationary.

900

900 lb 28

18. A jet plane is flying at G 750 mph on a heading of 30°. There is a strong 50-mph wind blowing from due south (heading 0°). What is the true course and speed of the plane (relative to the ground)? 19. Use the Guidelines for Graphing to sketch the graph of function f given, then use it to solve f 1x2 6 0. f 1x2 x3 4x2 x 6

20. Use the Guidelines for Graphing to sketch the graph of function g given, then use it to name the intervals where g1x2T and g1x2c. x2 4 x2 1 21. Find 11 i 132 8 using De Moivre’s theorem. g1x2

x3 x2 5 d. v1x2 2x2 x 6 c. h1x2

22. Solve ln1x 22 ln1x 32 ln14x2 .

13. Write the following formulas from memory: a. slope formula b. midpoint formula c. quadratic formula d. distance formula e. interest formula (compounded continuously) Solve each triangle using the law of sines or the law of cosines, whichever is appropriate.

23. If I saved $200 each month in an annuity program that paid 8% annual interest compounded monthly, how long would it take to save $10,000? 24. Mount Tortolas lies on the Argentine-Chilean border. When viewed from a distance of 5 mi, the angle of elevation to the top of the peak is 38°. How tall is Mount Tortolas? State the answer in feet. 25. The graph given is of the form y A sin1Bx C2. Find the values of A, B, and C. y

14. A 2

27

1

19 in. ␲

C

2

112

15.

B

37 A

B 52 cm

1 2

C 31 cm

lb

␲ 2

␲

3␲ 2

2␲

x

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Exercises 26 through 30 require the use of a graphing calculator. 26. For v H2.0807, 4.5465I, use the RPr and RP␪ features of a calculator to find the magnitude and angle (in radians) for v. Round your answer to four decimal places. 27. Solve the inequality on the interval [0, 2). Round your values to two decimal places when necessary. cos1 1sin x2 tan x 28. Determine where the graphs of f(x) x1.234 and g(x) x0.234 intersect.

29. Given u H 1, 1 I and v H 1, 1I, use a calculator “ “to determine u v (recall that the braces {” and }” must be used). What does your result say about these two vectors?


831

30. The average fuel consumption of passenger cars produced by Alpha Motors can be modeled by the 3 function A1t2 51 10t 15, where A is in miles per gallon (mpg) and t 0 corresponds to the year 2005. In 2010, Beta Motors began producing passenger cars and the function B1t2 5 110t 10 models their average fuel consumption, with B in mpg and t 0 once again corresponding to the year 2005. When will the two companies’ passenger cars have the same average fuel consumption? What is that average gas mileage? Round your answers to two decimal places.

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CONNECTIONS TO CALCULUS Chapter 8 unites our study of trigonometry with many of the skills you likely developed in previous coursework. Certainly algebraic skills continue to play a significant role, but here we’ve chosen to review and highlight certain problem-solving skills as well. From the vector diagrams seen in Chapter 8 to the triangle diagrams used in numerous applications, a careful sketch can highlight relationships and suggest a solution process, and this will certainly serve us well throughout the calculus sequence.

Trigonometry and Problem Solving Perhaps to a higher degree in a calculus course, the quality of the diagrams we draw have a greater impact on our ability to understand and solve applications. Draw the diagrams very carefully and to scale, if possible. Label the various parts of the diagram, make a list of all given information, assign variable names to unknown quantities, and build any needed relationships—in short, use the total accumulation of your problemsolving skills. EXAMPLE 1

䊳

Determining Sight Angles A 20-ft-tall road sign is to be placed along a busy highway by attaching it to a support pole, with the bottom edge of the sign 13 ft above the ground. a. Find a simplified expression for the “sight angle” (the angle at a person’s eyes subtended by the sign) in terms of x, when the driver is x ft away from the base of the pole (assume that eye level for an average driver is 3 ft above the ground). b. Find the sight angle at the moment a driver is 50 ft away.

Solution

䊳

a. Draw a car that is x ft away from a road sign with the dimensions described. Note that the triangle from eye 20 ft level to the top of the sign, and the triangle from eye level to the bottom of the sign are right triangles, while the sight angle is not. Naming the acute angle of the larger right triangle (at the 10 ft ⫺ eyes of the driver) ␣, and the acute angle of the smaller right triangle ␤ (see the figure), we see that the x 3 ft sight angle is then ␣ ⫺ ␤. Since we 3 ft need to involve the distance x in our equation model, it appears the tangent function should be used, giving 10 30 tan ␣ ⫽ and tan ␤ ⫽ . Knowing the identity for tan1␣ ⫺ ␤2 expressed x x tan ␣ ⫺ tan ␤ in terms of tan ␣ and tan ␤ individually, we have tan1␣ ⫺ ␤2 ⫽ . 1 ⫹ tan ␣ tan ␤ Our algebraic expression is then 30 10 ⫺ x x tan1␣ ⫺ ␤2 ⫽ 30 10 1 ⫹ a ba b x x 20 x ⫽ 300 1⫹ 2 x

832

make substitutions

simplify

8–88

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20 x ⫽ 2 x ⫹ 300 x2 20x ⫽ 2 x ⫹ 300

833

combine terms

result

The expression for the sight angle is ␣ ⫺ ␤ ⫽ tan⫺1a

20x b x ⫹ 300 2

b. When x ⫽ 50 ft, ␣ ⫺ ␤ ⫽ tan⫺1a

1000 b, or about 19.65°. For comparison, the 2800 sight angle at 25 ft is about 28.39°, and at 100 ft the angle is about 10.99°. Now try Exercises 1 through 4

䊳

Vectors in Three Dimensions Many elements of a study of vectors in z two dimensions transfer directly to the study of vectors in three dimensions. To help visualize a three-dimensional vector, H3, 4, 2I consider a large 3 ft ⫻ 4 ft ⫻ 2 ft cardboard box placed snugly in the corner of a y room. The floor of the room forms the positive xy-plane (the axes are where the walls meet the floor, the x-axis to the left), with x the third dimension (the z-axis) formed where the two walls meet (see the figure). The origin is then 10, 0, 02 , where the three axes meet, and the position vector H3, 4, 2I forms an interior diagonal of the box. For three-dimensional vectors u ⫽ Ha, b, cI and v ⫽ Hd, e, f I, the formulas for the magnitude of v and the dot product u # v are an extension of the two-dimensional cases, and are offered here without proof: 冟u冟 ⫽ 2a2 ⫹ b2 ⫹ c2, and u # v ⫽ ad ⫹ be ⫹ cf. EXAMPLE 2

䊳

Solution

䊳

Finding the Angle between Two Vectors

For vector u ⫽ H3, 4, 2I forming the interior diagonal of the box and v ⫽ H3, 4, 0I forming the diagonal of the base, a. Find 冟u冟, then verify the result using the Pythagorean theorem and 冟v冟. u#v b. Use the formula cos ␪ ⫽ to find the angle ␪ between u and v, then verify 冟u冟冟v冟 the result using right-triangle trig. a. First find 冟u冟 ⫽ 232 ⫹ 42 ⫹ 22 ⫽ 129. To verify, we use 冟v冟 ⫽ 232 ⫹ 42 ⫹ 02 ⫽ 5 and the height of the box in the Pythagorean theorem (since the triangle formed by (0, 0, 0), (3, 4, 0), and (3, 4, 2) is a right triangle): 冟v冟2 ⫹ 22 ⫽ 25 ⫹ 4 ⫽ 29 ⫽ 冟u冟2 ✓. u#v 9 ⫹ 16 ⫹ 0 5 5 , cos ␪ ⫽ ⫽ b ⬇ 21.8°. b. For cos ␪ ⫽ , so ␪ ⫽ cos⫺1a 冟u冟冟v冟 5 129 129 129 2 To verify, we use 冟v冟 ⫽ 5 and the height of the box, giving ␪ ⫽ tan⫺1a b ⬇ 21.8°✓. 5 Now try Exercises 5 through 10

䊳

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Occasionally we need to determine the angle a three-dimensional vector forms with the x-, y-, or z-axis. To do so, we apply the method and formulas shown in Example 2, by utilizing a secondary vector lying along the axis in question. For example, the angle that u ⫽ H3, 4, 2I forms with the z-axis is exactly the angle ␪ between the vector u 2 b ⬇ 68.2°. itself and k ⫽ H0, 0, 1I. Verify that in this case, ␪ ⫽ cos⫺1a 229

Connections to Calculus Exercises For the following applications, draw a diagram and develop an appropriate equation model. Note that the emphasis on these exercises is shared equally between the quality of diagram and developing the correct equation model.

1. A jet plane is flying at a constant altitude of 9000 ft, straight toward a searchlight on the ground that is tracking it. (a) Find an expression for the angle the searchlight makes with the ground in terms of the horizontal distance d from the light to a point on the ground directly below the plane. (b) Find the angle ␪ at the moment d ⫽ 12,850 ft. 2. A light is hung from the rafters at a height h m above the floor, providing a circle of illumination. The illuminance E (in lumens/square meter or lm/m2) at any point P on the floor varies directly as the cosine of the angle ␪ at which the rays leave the light source toward P, and inversely as the square of the height h of the light source. Using k as the constant of variation, (a) write an expression for the illuminance in terms of k, h, and ␪. (b) If the distance from the light source to point P is 13 m, write an expression for the illuminance in terms of k and h alone. (c) For k ⫽ 15,600, what is the height of the light source if E ⫽ 100 lm/m2? (d) For k and h as in part (c), at what angle ␪ is the illuminance 83 lm/m2?

3. In a large factory, two automated carts run on concentric circular tracks. The inner track has a radius of 24 m, while the outer track has a radius of 32 m. (a) Write an expression for the straight-line distance x between the two carts in terms of the angle ␪ formed by the line segments from the center to each cart. (b) Determine the straight-line distance between the carts at the moment ␪ ⫽ 150°. (c) At what angle ␪ are the carts 45 m apart? 4. A rectangle is inscribed in a semicircle of radius r. A straight line from the center of the circle to the point where a vertex of the rectangle meets the circle, forms an angle ␪. (a) Find an expression for the area of the rectangle in terms of r and a sine function. (Hint: Write the length and width of the rectangle in terms of sine and cosine.) (b) Find the area of the circle if r ⫽ 5 cm and ␪ ⫽ 26°. (c) At what angle ␪ does the rectangle have an area of 22 cm2?

Solve each application of three-dimensional vectors using a careful sketch drawn from the information given. The emphasis on these exercises continues to be shared equally between the quality of diagram and developing the correct equation model.

5. A rectangular playground is 64 ft long and 48 ft wide. There is a light pole for evening play in one corner of the playground, with the light placed 18 ft high. (a) Find 冟v冟, the length of the vector formed from the top of the light pole to the opposite corner of the playground. (b) If u is the vector formed by the opposite end of the playground and the base of the pole, find the angle between these two vectors u#v using cos ␪ ⫽ and verify the result using right 冟u冟冟v冟 triangle trig.

6. A gallery is about to open an exhibit in a room that is 22 ft long, 24 ft wide, and 16 ft high. The edge of the door to the exhibit is on the 22-ft side, 4 ft from one wall. For special effects lighting as guests enter the door, let u be the vector representing the distance from the edge of the door (at the floor) to the farthest bottom corner of the room, and let v be the vector from the same point to the farthest upper corner. (a) Find 冟v冟, then (b) find the angle between u#v u and v using cos ␪ ⫽ and verify the result 冟u冟冟v冟 using right triangle trig.

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7. At a small jungle airport, bush pilots are practicing their approaches to drop zones for food aid. There is an observation post 525 ft from the middle of one end of an 1800-ft straight dirt runway. At the moment the pilot of the first drop plane is over the opposite end of the runway, her altitude is 1000 ft. Let d be the vector representing the distance from the observation post to the opposite end of the runway, and let D represent the vector for the straight-line distance from the post to the plane at this moment. (a) Find 冟D冟, then (b) find the angle d#D between d and D using cos ␪ ⫽ and verify 冟d冟冟D冟 the result using right triangle trig. 8. An ornithologist is in the field studying what ornithologists study (in this case, a formation of geese), and is 180 m south and 385 m west of where she parked her car. Let u be the vector representing the distance between the ornithologist and her car, and let v be the vector representing the straight-line distance from the ornithologist to the formation. If the lead goose flies directly over her car at an altitude of 319 m, (a) find 冟v冟 then, (b) find u#v the angle between u and v using cos ␪ ⫽ and 冟u冟冟v冟 verify the result using right triangle trig.


835

9. A person is sitting in the center seat, in the center row, on the inclined seating of a new movie theater. The movie screen is 25 ft tall with the bottom of the screen 8 ft above a level floor. The person’s eye level is 15 ft above the floor and the person’s seat is a horizontal distance of x ft from the movie screen. (a) Write an expression for the person’s sight angle to the screen in terms of x. (b) Find the sight angle if the person is 46 ft from the screen. 10. The main road in front of a school runs north/south along the east side. A sidewalk runs east/west along the south side of the school, leading to a crosswalk. A secondary road intersects the west side of the main road near the crosswalk, forming a 30⬚ angle with the sidewalk. For the safety of the school children using the sidewalk, engineers are interested in the sight angle that drivers have as they drive on the secondary road approaching this intersection. (a) Find an expression for the sight angle of a driver in terms of x, the distance between their car and a point on the sidewalk 100 ft from the intersection. (b) Find the sight angle at the moment x ⫽ 125 ft.


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Systems of Equations and Inequalities CHAPTER OUTLINE 9.1 Linear Systems in Two Variables with Applications 838 9.2 Linear Systems in Three Variables with Applications 853 9.3 Systems of Inequalities and Linear Programming 865 9.4 Partial Fraction Decomposition 879 9.5 Solving Linear Systems Using Matrices and Row Operations 893

9.6 The Algebra of Matrices 905

At the turn of the century, there was an explosion in the number of handheld electronic devices available to consumers. One device in particular, the popular MP3 player, experienced a phenomenal growth in demand. With high demand and a large market, competition between manufacturers and suppliers is often fierce, with each fighting to earn and hold a share of the market. One significant factor in who gains the largest share is the price charged for the player, with suppliers willing to supply more at a greater price, and consumers willing to buy more at a lesser price. Determining where the price will stabilize is an important component in the economics of “supply and demand.” 䊳

This application occurs as Exercise 78 in Section 9.1.

9.7 Solving Linear Systems Using Matrix Equations 917 9.8 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More 933

Recall that for exponential functions, the rate of growth is proportional to the current population as shown in (1) below. For logistic functions, the rate of growth is proportional to the current population times the difference Connections between the carrying capacity c and the current population (2). For further analysis, this relationship can be to Calculus rewritten as in (3), with a technique known as partial fraction decomposition then applied. This technique is presented in Section 9.4, and further explored in the Connections to Calculus feature for Chapter 9. ¢P ¢P ¢P (1) (2) (3) k¢t kP kP1c P2 837 ¢t ¢t P1c P2

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Linear Systems in Two Variables with Applications

LEARNING OBJECTIVES

In earlier chapters, we used linear equations in two variables to model a number of real-world situations. Graphing these equations gave us a visual image of how the variables were related, and helped us better understand this relationship. In many applications, two different measures of the independent variable must be considered simultaneously, leading to a system of two linear equations in two unknowns. Here, a graphical presentation once again supports a better understanding, as we explore systems and their many applications.


A. Verify ordered pair solutions

B. Solve linear systems by C. D. E.

F.

graphing Solve linear systems by substitution Solve linear systems by elimination Recognize inconsistent systems and dependent systems Use a system of equations to model and solve applications

A. Solutions to a System of Equations A system of equations is a set of two or more equations for which a common solution is sought. Systems are widely used to model and solve applications when the information given enables the relationship between variables to be stated in different ways. For example, consider an amusement park that brought in $3100 in revenue by charging $9.00 for adults and $5.00 for children, while selling 500 tickets. Using a for adult and c for children, we could write one equation modeling the number of tickets sold: a c 500, and a second modeling the amount of revenue brought in: 9a 5c 3100. To show that we’re considering both equations simultaneously, a large “left brace” is used and the result is called a system of two equations in two variables:

c

Children

700

e

9a 5c 3100

500

(150, 350)

300

0

100

300

500

700

Adults

EXAMPLE 1

䊳

amount of revenue

Verifying Solutions to a System Verify that (150, 350) is a solution to e

Solution

number of tickets

We note that both equations are linear and will have different slope values, so their graphs must intersect at some point. Since every point on a line satisfies the equation of that line, this point of intersection must satisfy both equations simultaneously and is the solution to the system. The figure that accompanies Example 1 shows the point of intersecion for this system is (150, 350).

a c 500 a

100

a c 500 9a 5c 3100

䊳

a c 500 . 9a 5c 3100

Substitute the 150 for a and 350 for c in each equation. a c 500 11502 13502 500 500 500 ✓

first equation

9a 5c 3100 second equation 911502 513502 3100 3100 3100 ✓

Since (150, 350) satisfies both equations, it is the solution to the system and we find the park sold 150 adult tickets and 350 tickets for children. Now try Exercises 7 through 18 䊳 To check the solution to Example 1 on a graphing calculator, recall that we can store values in any of the ALPHA characters using the STO key. The home screen shown in Figure 9.1 shows that we’ve stored a value of 150 in alpha location A, and 350 in location C. The solution to a system can then be checked using these alpha keys and the operations indicated, as shown in Figure 9. 2.

838

9–2

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Section 9.1 Linear Systems in Two Variables with Applications

Figure 9.1

Figure 9.2

A. You’ve just seen how we can verify ordered pair solutions

B. Solving Systems Graphically To solve a system of equations means we apply various methods in an attempt to find ordered pair solutions. As Example 1 suggests, one method for finding solutions is to graph the system. Any method for graphing the lines can be employed, but to keep important concepts fresh, the slope-intercept method is used here. EXAMPLE 2

䊳

Solving a System Graphically Solve the system by graphing: e

Solution

䊳

4x 3y 9 . 2x y 5

First write each equation in slope-intercept form (solve for y): y

4 y x3 3 • S• 2x y 5 y 2x 5 4x 3y 9

5

4 For the first line, ¢y ¢x 3 with y-intercept 10, 32. 2 The second equation yields ¢y ¢x 1 with 10, 52 as the y-intercept. Both are then graphed on the grid as shown. The point of intersection appears to be (3, 1), and checking this point in both equations gives

4x 3y 9 4132 3112 9 99✓

y 2x 5

(3, 1) 5

5

x

y 43 x 3

(0, 3) 5

2x y 5 substitute 3 2132 112 5 for x and 1 for y 5 5 ✓

(0, 5)

This verifies that (3, 1) is the solution to the system. Now try Exercises 19 through 22 䊳 Graphical solutions to a system of equations can be found using a graphing calculator and the intersection-of-graphs method seen earlier. After solving for y and entering the equations on the Y= screen (Figure 9.3), use the keystrokes 2nd TRACE (CALC) 5:Intersect to determine the point of intersection, if it exists (Figure 9.4). Figure 9.4

Figure 9.3

5

5

B. You’ve just seen how we can solve linear systems by graphing

5

5

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CHAPTER 9 Systems of Equations and Inequalities

C. Solving Systems by Substitution While a graphical approach best illustrates why the solution must be an ordered pair, it does have one obvious drawback — noninteger solutions are difficult to spot. The 4x y 4 ordered pair 1 25, 12 , but this would be difficult to 5 2 is the solution to e yx2 “pinpoint” as a precise location on a hand-drawn graph. To overcome this limitation, we next consider a method known as substitution. The method involves converting a system of two equations in two variables into a single equation in one variable by 4x y 4 , the second equation says “y is two using an appropriate substitution. For e yx2 more than x.” We reason that all points on this line are related this way, including the point where this line intersects the other. For this reason, we can substitute x 2 for y in the first equation, obtaining a single equation in x. EXAMPLE 3

䊳

Solving a System Using Substitution Solve using substitution: e

Solution

䊳

4x y 4 . yx2

Since y x 2, we can replace y with x 2 in the first equation. 4x y 4 4x 1x 22 4 5x 2 4 2 x 5

first equation substitute x 2 for y simplify result

The x-coordinate is 52. To find the y-coordinate, we substitute 25 for x into either of the original equations, a process known as back-substitution. Substituting in the second equation gives yx2 2 2 5 12 5

second equation substitute

2 for x 5

2 10 2 10 12 , 1 5 5 5 5

The solution to the system is 1 25, 12 5 2 , which can be verified using the intersection-ofgraphs method, and noting 25 0.4 and 12 5 2.4. See Figure 9.5. Recall that the keystrokes MATH 1:䉴FRAC can be used to convert decimal numbers to fractions.

Figure 9.5 Y1 4X 4, Y2 X 2 5

ENTER

5

5

5

Now try Exercises 23 through 32 䊳 If neither equation allows an immediate substitution, we first solve for one of the variables, either x or y, and then substitute. The method is summarized here, and can actually be used with either like variables or like variable expressions. See Exercises 33 to 36.

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Solving Systems Using Substitution

C. You’ve just seen how we can solve linear systems by substitution

1. Solve one of the equations for x in terms of y or y in terms of x. 2. Substitute for the appropriate variable in the other equation and solve for the variable that remains. 3. Substitute the value from step 2 into either of the original equations and solve for the other unknown. 4. Write the answer as an ordered pair and check the solution in both original equations.

D. Solving Systems Using Elimination 2x 5y 13 , where solving for any one of the 2x 3y 7 variables will result in fractional values. The substitution method can still be used, but often the elimination method is more efficient. The method takes its name from what happens when you add certain equations in a system (by adding the like terms from each). If the coefficients of either x or y are additive inverses—they sum to zero and are eliminated. For the system shown, “adding the equations” produces 2y 6, giving y 3, then x 1 using back-substitution (verify). If neither of the like-variable terms sum to zero, we can multiply one or both equations by a nonzero constant to “match up” the coefficients, so an elimination will take place. In doing so, we create an equivalent system of equations, meaning one that has 7x 4y 16 the same solution as the original system. For e , multiplying the 3x 2y 6 7x 4y 16 second equation by 2 produces e , and after adding the equations, 6x 4y 12 we see that x 4. Note the three systems produced are equivalent, and all have (4, 3) as a solution (y 3 was found using back-substitution). Now consider the system e

1. e

7x 4y 16 3x 2y 6

2. e

7x 4y 16 6x 4y 12

3. e

7x 4y 16 x4

In summary, Operations that Produce an Equivalent System 1. Changing the order of the equations. 2. Replacing an equation by a nonzero constant multiple of that equation. 3. Replacing an equation with the sum of two equations from the system. Before beginning a solution using elimination, check to make sure the equations are written in the standard form Ax By C, so that like terms will appear above/below each other. Throughout this chapter, we will use R1 to represent the equation in row 1 of the system, R2 to represent the equation in row 2, and so on. These designations are used to help describe and document the steps being used to solve a system, as in Example 4 where 2R1 R2 indicates the first equation has been multiplied by two, with the result added to the second equation. EXAMPLE 4

䊳

Solving a System by Elimination Solve using elimination: e

2x 3y 7 6y 5x 4

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Solution

Y1

䊳

The second equation is not in standard form, so we rewrite the system as 2x 3y 7 e . If we “added the equations” now, we would get 7x 3y 11, 5x 6y 4 with neither variable eliminated. However, if we multiply both sides of the first equation by 2, the y-coefficients will become additive inverses. The sum then results in an equation with x as the only unknown.

7 2X, 4 5X Y2 3 6

2R1 4x 6y 14 R2 5x 6y 4 sum 9x 0y 18 9x 18 x 2

{

5

5

5

add

solve for x

Substituting 2 for x back into either of the original equations yields y 1. The ordered pair solution is (2, 1). A graphical check is shown in the figure.

5

Now try Exercises 37 through 42 䊳 The elimination method is summarized here. If either equation has fraction or decimal coefficients, we can “clear” them using an appropriate constant multiplier. Solving Systems Using Elimination 1. Write each equation in standard form: Ax By C. 2. Multiply one or both equations by a constant that will create coefficients of x (or y) that are additive inverses. 3. Combine the two equations using vertical addition and solve for the variable that remains. 4. Substitute the value from step 3 into either of the original equations and solve for the other unknown. 5. Write the answer as an ordered pair and check the solution in both original equations.

WORTHY OF NOTE As the elimination method involves adding two equations, it is sometimes referred to as the addition method for solving systems.

EXAMPLE 5

䊳

Solving a System Using Elimination 5 x 34y 14 . Solve using elimination: e 18 2 2x 3y 1

Solution

䊳

Multiplying the first equation by 8 (8R1) and the second equation by 6 (6R2) will clear the fractions from each. 5x 6y 2 8R1 81 1 58 2x 81 1 34 2y 81 1 14 2 e6 1 Se 6 2 3x 4y 6 6R2 1 1 2 2x 1 1 3 2y 6112

The x-terms can now be eliminated if we use 3R1 15R22 . 3R1 15x 18y 6 5R2 15x 20y 30 sum 0x 2y 24 y 12

{

add

solve for y

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Substituting y 12 in either of the original equations yields x 14, and the solution is 114, 122. We can check this solution by graphing, as we did for the solution to Example 4, or by substituting the x- and y-values into the original equations, as we did for Example 1. The check using substitution is shown in the figure. Note that 14 STO X has scrolled out of view.

D. You’ve just seen how we can solve linear systems by elimination


䊳

CAUTION

Be sure to multiply all terms (on both sides) of the equation when using a constant multiplier. Also, note that for Example 5, we could have eliminated the y-terms using 2R1 with 3R2.

E. Inconsistent and Dependent Systems A system having at least one solution is called a consistent system. As seen in Example 2, if the lines have different slopes, they intersect at a single point and the system has exactly one solution. Here, the lines are independent of each other and the system is called an independent system. If the lines have equal slopes and the same y-intercept, they are identical or coincident lines. Since one is right atop the other, they intersect at all points, and the system has an infinite number of solutions. Here, one line depends on the other and the system is called a dependent system. Using substitution or elimination on a dependent system results in the elimination of all variable terms and leaves a statement that is always true, such as 0 0 or some other simple identity. EXAMPLE 6

䊳

Solving a Dependent System Solve using elimination: e

Solution

䊳

3x 4y 12 . 6x 24 8y

Writing the system in standard form gives 3x 4y 12 e . By applying 2R1, we 6x 8y 24 can eliminate the variable x: 2R1 6x 8y 24 e 6x 8y 24 R2 sum 0x 0y 0 0 0

y 5

3x 4y 12

(0, 3) (4, 0)

5

6x 24 8y

5

5

add variables are eliminated true statement

Although we didn’t expect it, both variables were eliminated and the final statement is true 10 02. This indicates the system is dependent, which the graph verifies (the lines are coincident). Writing both equations in slope-intercept form shows they represent the same line. e

3x 4y 12 6x 8y 24

e

4y 3x 12 8y 6x 24

3 y x3 4 μ 3 y x3 4

x

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The solutions of a dependent system are often written in set notation as the set of ordered pairs (x, y), where y is a specified function of x. Here the solution would be 5 1x, y2 0 y 34x 36. Using an ordered pair with an arbitrary variable, called a 3p 3b. parameter, is also common: ap, 4

WORTHY OF NOTE When writing the solution to a dependent system using a parameter, the solution can be written in many different ways. For instance, if we let p 4b for the first coordinate of the solution to Example 6, we have 314b2 3 3b 3 as the 4 second coordinate, and the solution becomes (4b, 3b 3) for any constant b.


Figure 9.7 5

5

5

5

Figure 9.8

If we had attempted to solve the system in Example 6 Figure 9.6 by graphing (Figure 9.6), we could be mislead into thinking something is wrong—because only one line is visible (Figure 9.7). In this case, using the TABLE feature of the calculator would help verify that the system is dependent. Since the ordered pair solutions are identical (try scrolling through positive and negative values), the equations must be dependent (Figure 9.8). Finally, if the lines have equal slopes and different y-intercepts, they are parallel and the system will have no solution. A system without a solution is called an inconsistent system. An “inconsistent system” produces an “inconsistent answer,” such as 12 0 or some other false statement when substitution or elimination is applied. In other words, all variable terms are once again eliminated, but the remaining statement is false. A summary of the three possibilities is shown in Figure 9.9 for arbitrary slope m and y-intercept (0, b).

Figure 9.9 Dependent m1 m2, b1 b2

Independent m1 m2 y

E. You’ve just seen how we can recognize inconsistent systems and dependent systems

Inconsistent m1 m2, b1 b2

y

x

One point in common

y

x

All points in common

x

No points in common

F. Systems and Modeling In previous chapters, we solved numerous real-world applications by writing all given relationships in terms of a single variable. Many situations are easier to model using a system of equations with each relationship modeled independently using two variables. We begin here with a mixture application. Although they appear in many different forms (coin problems, metal alloys, investments, merchandising, and so on), mixture problems all have a similar theme. Generally one equation is related to quantity (how much of each item is being combined) and one equation is related to value (what is the value of each item being combined). EXAMPLE 7

䊳

Solving a Mixture Application A jeweler is commissioned to create a piece of artwork that will weigh 14 oz and consist of 75% gold. She has on hand two alloys that are 60% and 80% gold, respectively. How much of each should she use?

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Solution

䊳

WORTHY OF NOTE As an estimation tool, note that if equal amounts of the 60% and 80% alloys were used (7 oz each), the result would be a 70% alloy (halfway in between). Since a 75% alloy is needed, more of the 80% gold will be used.

Y1 14 X, Y2

Let x represent ounces of the 60% alloy and y represent ounces of the 80% alloy. The first equation must be x y 14, since the piece of art must weigh exactly 14 oz (this is the quantity equation). The x ounces are 60% gold, the y ounces are 80% gold, and the 14 oz will be 75% gold. This gives the value equation: x y 14 0.6x 0.8y 0.751142. The system is e (after clearing decimals). 6x 8y 105 Solving for y in the first equation gives y 14 x. Substituting 14 x for y in the second equation gives

105 6X 8

6x 8y 105 6x 8114 x2 105 6x 112 8x 105 2x 112 105 7 x 2

15

5

845

15

second equation substitute 14 x for y distribute simplify solve for x

Substituting 72 for x in the first equation gives y 21 2 . She should use 3.5 oz of the 60% alloy and 10.5 oz of the 80% alloy. A graphical check is shown in the figure.

5

Now try Exercises 63 through 70 䊳 A second example involves an application of uniform motion (distance rate # time), and explores concepts of great importance to the navigation of ships and airplanes. As a simple illustration, if you’ve ever walked at your normal rate r on the “moving walkways” at an airport, you likely noticed an increase in your total speed. This is because the resulting speed combines your walking rate r with the speed w of the walkway: total speed r w. If you walk in the opposite direction of the walkway, your total speed is much slower, as now total speed r w. This same phenomenon is observed when an airplane is flying with or against the wind, or a ship is sailing with or against the current. EXAMPLE 8

䊳

Solving an Application of Systems—Uniform Motion An airplane flying due south from St. Louis, Missouri, to Baton Rouge, Louisiana, uses a strong, steady tailwind to complete the trip in only 2.5 hr. On the return trip, the same wind slows the flight and it takes 3 hr to get back. If the flight distance between these cities is 912 km, what is the cruising speed of the airplane (speed with no wind)? How fast is the wind blowing?

Solution

䊳

Let r represent the rate of the plane and w the rate of the wind. Since D RT, the flight to Baton Rouge can be modeled by 912 1r w2 12.52 , and the return flight by 912 1r w2132 . This produces the system e

䊲

Algebraic Solution

Dividing R1 by 2.5 and R2 by 3 produces the following sequence: R1 364.8 r w 2.5 912 2.5r 2.5w Se e R2 912 3r 3w 304.0 r w 3

䊲

912 2.5r 2.5w . 912 3r 3w

Graphical Solution

Using x for w and y for r, we solve each equation for y and obtain: Y1

912 2.5X 2.5

Y2

912 3X 3

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Using R1 R2 gives 668.8 2r, showing 334.4 r. The speed of the plane is 334.4 kph. Substituting 334.4 for r in the second equation, we have: 912 3r 3w 912 31334.42 3w 912 1003.2 3w 91.2 3w 30.4 w

We then set an appropriate window and graph these equations to find the point of intersection. 400

equation substitute multiply

0

50

subtract 1003.2 divide by 3

The speed of the wind is 30.4 kph.

200

The speed of the wind (x) is 30.4 kph, and the speed of the plane (y) is 334.4 kph. Now try Exercises 71 through 74 䊳

Systems of equations also play a significant role in cost-based pricing in the business world. The costs involved in running a business can broadly be understood as either a fixed cost k or a variable cost v. Fixed costs might include the monthly rent paid for facilities, which remains the same regardless of how many items are produced and sold. Variable costs would include the cost of materials needed to produce the item, which depends on the number of items made. The total cost can then be modeled by C1x2 vx k for x number of items. Once a selling price p has been determined, the revenue equation is simply R1x2 px (price times number of items sold). We can now set up and solve a system of equations that will determine how many items must be sold to break even, performing what is called a break-even analysis where C(x) R(x). EXAMPLE 9

䊳

Solving an Application of Systems: Break-Even Analysis In home businesses that produce items to sell on Ebay®, fixed costs are easily determined by rent and utilities, and variable costs by the price of materials needed to produce the item. Karen’s home business makes large decorative candles for all occasions. The cost of materials is $3.50 per candle, and her rent and utilities average $900 per month. If her candles sell for $9.50, how many candles must be sold each month to break even?

Solution

䊳

Let x represent the number of candles sold. Her total cost is C1x2 3.5x 900 (variable cost plus fixed cost), and projected revenue is R1x2 9.5x. This gives the C1x2 3.5x 900 system e . To break even, Cost Revenue which gives R 1x2 9.5x 9.5x 3.5x 900 6x 900 x 150

The analysis shows that Karen must sell 150 candles each month to break even. Now try Exercises 75 through 78 䊳

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WORTHY OF NOTE There are limitations to this model, and this interplay can be affected by the number of available consumers, production limits, “shelf-life” issues, and so on, but at any given moment in the life cycle of a product, consumer demand responds to price in this way. Producers also respond to price in a very predictable way.

EXAMPLE 10

䊳

In a “free-market” economy, also referred to as a “supply-and-demand” economy, there are naturally occurring forces that invariably come into play if no outside forces act on the producers (suppliers) and consumers (demanders). Generally speaking, the higher the price of an item, the lower the demand. A good advertising campaign can increase the demand, but the increasing demand brings an increase in price, which moderates the demand — and so it goes until a balance is reached. These free-market forces ebb and flow until market equilibrium occurs, at the specific price where the supply and demand are equal. In Exercises 75 to 78, the equation models were artificially constructed to yield a “nice” solution. In actual practice, the equations and coefficients are not so “well behaved” and are based on the collection and interpretation of real data. While market analysts have sophisticated programs and numerous models to help develop these equations, here we’ll use our experience with regression to develop the supply and demand curves. Using Technology to Find Market Equilibrium A manufacturer of MP3 players has hired a consulting firm to do market research on their “next-generation” player. Over a 10-week period, the firm collected the data shown for the MP3 player market (data includes MP3 players sold and expected to sell). a. Use a graphing calculator to simultaneously display the demand and supply scatterplots. b. Calculate a line of best fit for each and graph them with the scatterplots (identify each curve). c. Find the equilibrium point.

Solution

Figure 9.10

Figure 9.12

䊳

Price Supply (dollars) Demand (Inventory) 107.10

6900

12,200

85.50

7900

9900

64.80

13,200

8000

52.20

13,500

7900

108.00

6700

14,000

91.80

7600

12,000

77.40

9200

9400

46.80

13,800

6100

a. Begin by clearing all lists. This can be done 74.70 10,600 8800 manually, or by pressing 2nd + (MEM) 68.40 12,800 8600 and selecting option 4:ClrAllLists (the command appears on the home screen). Figure 9.11 Pressing will execute the command, 16,000 and the word DONE will appear. Carefully input price in L1, demand in L2, and supply in L3 (see Figure 9.10). With the window settings given in 40 115 Figure 9.11, pressing GRAPH will display the price/demand and price/supply scatterplots shown. If this is not the case, use 2nd Y= (STAT PLOT) to be sure 3000 that “On” is highlighted in Plot1 and Plot2, and that Plot1 uses L1 and L2, while Plot2 uses L1 and L3 (Figure 9.12). Note we’ve chosen a different mark to indicate the data points in Plot2. b. Calculate the linear regression equation for L1 and L2 (demand), and paste it in Y1: LinReg (ax ⴙ b) L1, L2, Y1 . Next, calculate the linear regression for (recall that L1 and L3 (supply) and paste it in Y2: LinReg (ax ⴙ b) L1, L3, Y2 Y1 and Y2 are accessed using the VARS key). The resulting equations and graphs are shown in Figures 9.13 and 9.14. c. Once again we use 2nd TRACE (CALC) 5:intersect to find the equilibrium point, which is approximately (80, 9931). Supply and demand for this MP3 player model are approximately equal at a price of about $80, with 9931 MP3 players bought and sold. ENTER

ENTER

ENTER

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16,000

40

115

3000

Now try Exercises 79 through 82 䊳 F. You’ve just seen how we can use a system of equations to model and solve applications

Other interesting applications can be found in the Exercise Set. See Exercises 83 through 88.

9.1 EXERCISES 䊳



䊳

1. Systems that have no solution are called systems.

2. Systems having at least one solution are called systems.

3. If the lines in a system intersect at a single point, the system is said to be and .

4. If the lines in a system are coincident, the system is referred to as and .

5. The given systems are equivalent. How do we obtain the second system from the first? 2 1 5 x y 4x 3y 10 2 3 • 3 e 2x 4y 10 0.2x 0.4y 1

6. For e

2x 5y 8 , 3x 4y 5 which solution method would be more efficient, substitution or elimination? Discuss/Explain why.


Show the lines in each system would intersect in a single point by writing the equations in slope-intercept form.

7x 4y 24 7. e 4x 3y 15

0.3x 0.4y 2 8. e 0.5x 0.2y 4

An ordered pair is a solution to an equation if it makes the equation true. Given the graph shown here, determine which equation(s) have the indicated point as a solution. If the point satisfies more than one equation, write the system for which it is a solution.

9. A

10. B

y 5

3x 2y 6

yx2 A

F

B

E 5

5 x

C x 3y 3 5

D

11. C

12. D

13. E

14. F

Substitute the x- and y-values indicated by the ordered pair to determine if it is a solution to the system. Also check using the ALPHA keys on the home screen of a graphing calculator.

15. e

3x y 11 13, 22 5x y 13;

16. e

3x 7y 4 16, 22 7x 8y 21;

17. e

8x 24y 17 7 5 a , b 12x 30y 2; 8 12

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18. e

4x 15y 7 1 1 a , b 8x 21y 11; 2 3

Solve using elimination. In some cases, the system must first be written in standard form. Verify solutions using a graphing calculator.

Solve each system by graphing manually. Check results by graphing the system on a graphing calculator, and locating any points of intersection.

3x 2y 12 19. e x y9

5x 2y 2 20. e 3x y 10

5x 2y 5 21. e x 3y 16

3x y 2 22. e 5x 3y 2

Solve each system using substitution. Write solutions as an ordered pair, and verify solutions using a graphing calculator.

23. e

x 5y 9 x 2y 6

24. e

4x 5y 7 2x 5 y

25. e

y 23x 7 3x 2y 19

26. e

2x y 6 y 34x 1

Identify the equation and variable that makes the substitution method easiest to use, then solve the system. Verify solutions using a graphing calculator.

3x 2y 19 x 4y 3

27. e

3x 4y 24 5x y 17

29. e

0.7x 2y 5 0.8x y 7.4 30. e 0.6x 1.5y 9.3 x 1.4y 11.4

31. e

5x 6y 2 x 2y 6

28. e

32. e

2x 5y 5 8x y 6

The substitution method can be used for like variables or for like expressions. Solve the following systems, using the expression common to both equations (do not solve for x or y alone).

33. e

2x 4y 6 x 12 4y

34. e

8x 3y 24 8x 5y 36

35. e

5x 11y 21 11y 5 8x

36. e

6x 5y 16 5y 6x 4

䊳

849

37. e

2x 4y 10 3x 4y 5

38. e

x 5y 8 x 2y 6

39. e

4x 3y 1 3y 5x 19

40. e

5y 3x 5 3x 2y 19

41. e

2x 3y 17 4x 5y 12

42. e

2y 5x 2 4x 17 6y

43. e

0.5x 0.4y 0.2 0.2x 0.3y 0.8 44. e 0.3y 1.3 0.2x 0.3x 0.4y 1.3

45. e

0.32m 0.12n 1.44 0.24m 0.08n 1.04

46. e

0.06g 0.35h 0.67 0.12g 0.25h 0.44

47. e

16u 14v 4 1 2 2 u 3 v 11

x 13y 2 1 2x 5y 3 3

48. e 43

Solve using any method and identify the system as consistent, inconsistent, or dependent. Verify solutions using a graphing calculator.

49. e

4x 34y 14 9x 58y 13

2 xy2 50. e 3 2y 56x 9

51. e

0.2y 0.3x 4 1.2x 0.4y 5 52. e 0.6x 0.4y 1 0.5y 1.5x 2

53. e

6x 22 y 3x 12y 11

55. e

10x 35y 5 2x 3y 4 56. e x 2.5y y 0.25x

57. e

7a b 25 2a 5b 14

58. e

2m 3n 1 5m 6n 4

59. e

4a 2 3b 6b 2a 7

60. e

3p 2q 4 9p 4q 3

54. e

15 5y 9x 3x 53y 5


61. Uniform motion with current: e

1R ⴙ C2T1 ⴝ D1 1R ⴚ C2T2 ⴝ D2

The formula shown can be used to solve uniform motion problems involving a current, where D represents distance traveled, R is the rate of the object with no current, C is the speed of the current, and T is the time. Chan-Li rows 9 mi up river (against the current) in 3 hr. It only took him 1 hr to row 5 mi downstream (with the current). How fast was the current? How fast can he row in still water?

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62. Fahrenheit and Celsius temperatures: e

y ⴝ 95x ⴙ 32 y ⴝ 59 1x ⴚ 322

ⴗF ⴗC

Many people are familiar with temperature measurement in degrees Celsius and degrees Fahrenheit, but few realize that the equations are linear and there is one temperature at which the two scales agree. Solve the system using the method of your choice and find this temperature. 䊳

APPLICATIONS

Solve each application by modeling the situation with a linear system. Be sure to clearly indicate what each variable represents. Check answers using a graphing calculator and the method of your choice.

Mixture 63. Theater productions: At a recent production of A Comedy of Errors, the Community Theater brought in a total of $30,495 in revenue. If adult tickets were $9 and children’s tickets were $6.50, how many tickets of each type were sold if 3800 tickets in all were sold? 64. Milkfat requirements: A dietician needs to mix 10 gal of milk that is 212 % milkfat for the day’s rounds. He has some milk that is 4% milkfat and some that is 112 % milkfat. How much of each should be used? 65. Filling the family cars: Cherokee just filled both of the family vehicles at a service station. The total cost for 20 gal of regular unleaded and 17 gal of premium unleaded was $144.89. The premium gas was $0.10 more per gallon than the regular gas. Find the price per gallon for each type of gasoline. 66. Household cleaners: As a cleaning agent, a solution that is 24% vinegar is often used. How much pure (100%) vinegar and 5% vinegar must be mixed to obtain 50 oz of a 24% solution? 67. Alumni contributions: A wealthy alumnus donated $10,000 to his alma mater. The college used the funds to make a loan to a science major at 7% interest and a loan to a nursing student at 6% interest. That year the college earned $635 in interest. How much was loaned to each student? 68. Investing in bonds: A total of $12,000 is invested in two municipal bonds, one paying 10.5% and the other 12% simple interest. Last year the annual interest earned on the two investments was $1335. How much was invested at each rate? 69. Saving money: Bryan has been doing odd jobs around the house, trying to earn enough money to buy a new Dirt-Surfer©. He saves all quarters and dimes in his piggy bank, while he places all nickels and pennies in a drawer to spend. So far, he has 225 coins

in the piggy bank, worth a total of $45.00. How many of the coins are quarters? How many are dimes? 70. Coin investments: In 1990, Molly attended a coin auction and purchased some rare “Flowing Hair” fifty-cent pieces, and a number of very rare twocent pieces from the Civil War Era. If she bought 47 coins with a face value of $10.06, how many of each denomination did she buy? Uniform Motion 71. Canoeing on a stream: On a recent camping trip, it took Molly and Sharon 2 hr to row 4 mi upstream from the drop in point to the campsite. After a leisurely weekend of camping, fishing, and relaxation, they rowed back downstream to the drop in point in just 30 min. Use this information to find (a) the speed of the current and (b) the speed Sharon and Molly would be rowing in still water. 72. Taking a luxury cruise: A luxury ship is taking a Caribbean cruise from Caracas, Venezuela, to just off the coast of Belize City on the Yucatan Peninsula, a distance of 1435 mi. En route they encounter the Caribbean Current, which flows to the northwest, parallel to the coastline. From Caracas to the Belize coast, the trip took 70 hr. After a few days of fun in the sun, the ship leaves for Caracas, with the return trip taking 82 hr. Use this information to find (a) the speed of the Caribbean Current and (b) the cruising speed of the ship. 73. Airport walkways: As part of an algebra field trip, Jason takes his class to the airport to use their moving walkways for a demonstration. The class measures the longest walkway, which turns out to be 256 ft long. Using a stop watch, Jason shows it takes him just 32 sec to complete the walk going in the same direction as the walkway. Walking in a direction opposite the walkway, it takes him 320 sec—10 times as long! The next day in class, Jason hands out a two-question quiz: (1) What was the speed of the walkway in feet per second? (2) What is my (Jason’s) normal walking speed? Create the answer key for this quiz.

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74. Racing pigeons: The American Racing Pigeon Union often sponsors opportunities for owners to fly their birds in friendly competitions. During a recent competition, Steve’s birds were liberated in Topeka, Kansas, and headed almost due north to their loft in Sioux Falls, South Dakota, a distance of 308 mi. During the flight, they encountered a steady wind from the north and the trip took 4.4 hr. The next month, Steve took his birds to a competition in Grand Forks, North Dakota, with the birds heading almost due south to home, also a distance of 308 mi. This time the birds were aided by the same wind from the north, and the trip took only 3.5 hr. Use this information to (a) find the racing speed of Steve’s birds and (b) find the speed of the wind. 75. Lawn service: Dave and his sons run a lawn service, which includes mowing, edging, trimming, and aerating lawns. His fixed cost includes insurance, his salary, and monthly payments on equipment, and amounts to $4000/mo. The variable costs include gas, oil, hourly wages for his employees, and miscellaneous expenses, which run about $75 per lawn. The average charge for fullservice lawn care is $115 per visit. Do a breakeven analysis to (a) determine how many lawns Dave must service each month to break even and (b) the revenue required to break even. 76. Production of mini-microwave ovens: Due to high market demand, a manufacturer decides to introduce a new line of mini-microwave ovens for personal and office use. By using existing factory space and retraining some employees, fixed costs are estimated at $8400/mo. The components to assemble and test each microwave are expected to run $45 per unit. If market research shows consumers are willing to pay at least $69 for this product, find (a) how many units must be made and sold each month to break even and (b) the revenue required to break even. 77. Farm commodities: One area where the law of supply and demand is clearly at work is farm commodities. Both growers and consumers watch this relationship closely, and use data collected by government agencies to track the relationship and make adjustments, as when a farmer decides to convert a large portion of her farmland from corn to soybeans to improve profits. Suppose that for x billion bushels of soybeans, supply is modeled by y 1.5x 3, where y is the current market price (in dollars per bushel). The related demand equation might be y 2.20x 12. (a) How many billion bushels will be supplied at a market price of $5.40? What will the demand be at this

price? Is supply less than demand? (b) How many billion bushels will be supplied at a market price of $7.05? What will the demand be at this price? Is demand less than supply? (c) To the nearest cent, at what price does the market reach equilibrium? How many bushels are being supplied/demanded? 78. Digital media: Market research has indicated that by 2015, sales of MP3 players and similar products will mushroom into a $70 billion dollar market. With a market this large, competition is often fierce—with suppliers fighting to earn and hold market shares. For x million MP3 players sold, supply is modeled by y 10.5x 25, where y is the current market price (in dollars). The related demand equation might be y 5.20x 140. (a) How many million MP3 players will be supplied at a market price of $88? What will the demand be at this price? Is supply less than demand? (b) How many million MP3 players will be supplied at a market price of $114? What will the demand be at this price? Is demand less than supply? (c) To the nearest cent, at what price does the market reach equilibrium? How many units are being supplied/demanded? 79. Pricing wakeboards: A water sports company that manufactures high-end wakeboards has hired an outside consulting firm to do some market research on their best wakeboard. This consulting firm collected the following supply and demand data for this and comparable wakeboards over a 10-week period. Find the equilibrium point. Round your answer to the nearest integer and dollar. Average Price (in U.S. dollars)

Quantity Demanded

Available Inventory

424.85

175

232

445.25

166

247

389.55

291

215

349.98

391

201

402.22

218

226

413.87

200

222

481.73

139

251

419.45

177

235

397.05

220

219

361.90

317

212

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80. Pricing pet care products: A metal shop that manufactures pens for pet rabbits has collected some data on sales and production over the past 8 weeks. The following table shows the supply and demand data for these pens. Find the equilibrium point (round to the nearest cent and whole cage). Average Price (in U.S. dollars)

Quantity Sold (Demand)

Production (Supply)

Average Price Quantity Demanded Available Inventory (in U.S. dollars) (in millions) (in millions) 9.40

0.84

1.23

8.51

1.17

0.95

8.78

1.05

1.11

10.82

0.68

1.29

6.77

1.47

0.77

9.33

0.91

1.21

1.25

0.88

22.99

12

7

8.34

21.49

14

6

10.37

0.76

1.27

1.09

1.02

23.99

11

7

8.62

26.99

9

11

8.44

1.21

0.92

1.18

0.97

1.01

1.17

25.99

8

10

8.58

27.99

8

13

8.96

24.49

10

9

26.49

9

11

81. Tracking supply and demand— oil products: The U.S. Bureau of Labor and Statistics tracks important data from many different markets. In May 2008, it collected the following supply-anddemand data for refined gasoline. Data were collected every Tuesday and Friday. Find the equilibrium point, rounding your answer to the nearest hundred thousand gallons and whole cent. Average Price Quantity Demanded Available Inventory (in U.S. dollars) (1 ⴛ 107 gal) (1 ⴛ 107 gal) 3.17

8.82

9.10

3.12

8.87

9.05

3.04

9.08

8.97

2.84

9.22

8.91

3.11

8.92

9.02

3.15

8.76

9.08

3.10

9.01

8.99

3.11

8.94

9.01

2.93

9.13

8.93

82. Tracking supply and demand—energy efficient lightbulbs: The U.S. Bureau of Labor and Statistics has collected the following supply and demand data for the energy-efficient fluorescent lightbulbs sold each month for the past year. Find the equilibrium point, rounding your answer to the nearest ten thousand lightbulbs and whole cent. What is the yearly demand at the equilibrium point?

Descriptive Translation 83. Important dates in U.S. history: If you sum the year that the Declaration of Independence was signed and the year that the Civil War ended, you get 3641. There are 89 yr that separate the two events. What year was the Declaration signed? What year did the Civil War end? 84. Architectural wonders: When it was first constructed in 1889, the Eiffel Tower in Paris, France, was the tallest structure in the world. In 1975, the CN Tower in Toronto, Canada, became the world’s tallest structure. The CN Tower is 153 ft less than twice the height of the Eiffel Tower, and the sum of their heights is 2799 ft. How tall is each tower? 85. Pacific islands land area: In the South Pacific, the island nations of Tahiti and Tonga have a combined land area of 692 mi2. Tahiti’s land area is 112 mi2 more than Tonga’s. What is the land area of each island group? 86. Card games: On a cold winter night, in the lobby of a beautiful hotel in Sante Fe, New Mexico, Marc and Klay just barely beat John and Steve in a close game of Trumps. If the sum of the team scores was 990 points, and there was a 12-point margin of victory, what was the final score?

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Section 9.2 Linear Systems in Three Variables with Applications

853

Given any two points, the equation of a line through these points can be found using a system of equations. While there are certainly more efficient methods, using a system here will lead to finding equations for polynomials of higher degree. The key is to note that each point will yield an equation of the form y ⴝ mx ⴙ b. For instance, the points (3, 6) and 1ⴚ2, ⴚ42 yield the system e

6 ⴝ 3m ⴙ b . ⴚ4 ⴝ ⴚ2m ⴙ b

87. Use a system of equations to find the equation of the line containing the points (2, 7) and 14, 52 . 88. Use a system of equations to find the equation of the line containing the points 19, 12 and 13, 72 . 䊳


89. Federal income tax reform has been a hot political topic for many years. Suppose tax plan A calls for a flat tax of 20% tax on all income (no deductions or loopholes). Tax plan B requires taxpayers to pay $5000 plus 10% of all income. For what income level do both plans require the same tax? 䊳

90. Suppose a certain amount of money was invested at 6% per year, and another amount at 8.5% per year, with a total return of $1250. If the amounts invested at each rate were switched, the yearly income would have been $1375. To the nearest whole dollar, how much was invested at each rate?


91. (2.2) Given the toolbox function f 1x2 冟x冟, sketch the graph of F1x2 冟x 3冟 2.

93. (5.5) Solve for x (rounded to the nearest thousandth): 33 77.5e0.0052x 8.37

92. (6.1) Find two positive and two negative angles that are coterminal with ␪ 112°.

94. (7.2) Verify that

sin x csc x cos2x is an csc x

identity.

9.2

Linear Systems in Three Variables with Applications


A. Visualize a solution in

The transition to systems of three equations in three variables requires a fair amount of “visual gymnastics” along with good organizational skills. Although the techniques used are identical and similar results are obtained, the third equation and variable give us more to track, and we must work more carefully toward the solution.

three dimensions

B. Check ordered triple solutions C. Solve linear systems in three variables D. Recognize inconsistent and dependent systems E. Use a system of three equations in three variables to solve applications

A. Visualizing Solutions in Three Dimensions The solution to an equation in one variable is the single number that satisfies the equation. For x 1 3, the solution is x 2 and its graph is a single point on the number line, a one-dimensional graph. The solution to an equation in two variables, such as x y 3, is an ordered pair (x, y) that satisfies the equation. When we graph this solution set, the result is a line on the xy-coordinate grid, a two-dimensional graph. The solutions to an equation in three variables, such as x y z 6, are the ordered triples (x, y, z) that satisfy the equation. When we graph this solution set, the result is a plane in space, a graph in three dimensions. Recall a plane is a flat surface having infinite length and width, but no depth. We can graph this plane using the intercept method and the result is shown in Figure 9.15. For graphs in three dimensions, the xy-plane is parallel to the ground (the y-axis points to the right) and z is the vertical axis. To find an additional point on this plane, we use any three numbers whose sum is 6, such as (2, 3, 1). Move 2 units along the x-axis, 3 units parallel to the y-axis, and 1 unit parallel to the z-axis, as shown in Figure 9.16.

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Figure 9.15

WORTHY OF NOTE We can visualize the location of a point in space by considering a large rectangular box 2 ft long 3 ft wide 1 ft tall, placed snugly in the corner of a room. The floor is the xy-plane, one wall is the xz-plane, and the other wall is the yz-plane. The z-axis is formed where the two walls meet and the corner of the room is the origin (0, 0, 0). To find the corner of the box located at (2, 3, 1), first locate the point (2, 3) in the xy-plane (the floor), then move up 1 ft.

Figure 9.16

z

z

(0, 0, 6)

(0, 0, 6)

(2, 3, 1) y

2 units along x

y

䊳

EXAMPLE 1

x

y

z

l lle

x

(0, 6, 0)

l lle

(6, 0, 0)

a ra sp nit 3u

(6, 0, 0)

a par nit 1u

(0, 6, 0)

Finding Solutions to an Equation in Three Variables Use a guess-and-check method to find four additional points on the plane determined by x y z 6.

䊳

Solution A. You’ve just seen how we can visualize a solution in three dimensions

We can begin by letting x 0, then use any combination of y and z that sum to 6. Two examples are (0, 2, 4) and (0, 5, 1). We could also select any two values for x and y, then determine a value for z that results in a sum of 6. Two examples are 12, 9, 12 and 18, 3, 12. Now try Exercises 7 through 10 䊳

B. Solutions to a System of Three Equations in Three Variables When solving a system of three equations in three variables, remember each equation represents a plane in space. These planes can intersect in various ways, creating different possibilities for a solution set (see Figures 9.17 to 9.20). The system could have a unique solution (a, b, c), if the planes intersect at a single point (Figure 9.17) (the point satisfies all three equations simultaneously). If the planes intersect in a line (Figure 9.18), the system is linearly dependent and there is an infinite number of solutions. Unlike the twodimensional case, the equation of a line in three dimensions is somewhat complex, and the coordinates of all points on this line are usually represented by a specialized ordered triple, which we use to state the solution set. If the planes intersect at all points, the system has coincident dependence (see Figure 9.19). This indicates the equations of the system differ by only a constant multiple—they are all “disguised forms” of the same equation. The solution set is any ordered triple (a, b, c) satisfying this equation. Finally, the system may have no solutions. This can happen a number of different ways, most notably if the planes either intersect or are parallel, as shown in Figure 9.20 (other possibilities are discussed in the exercises). In the case of “no solutions,” an ordered triple may satisfy none of the equations, only one of the equations, only two of the equations, but not all three equations. Figure 9.17

Unique solution

Figure 9.18

Linear dependence

Figure 9.19

Coincident dependence

Figure 9.20

No solutions

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EXAMPLE 2

䊳

Determining If an Ordered Triple Is a Solution

Determine if the ordered triple 11, 2, 32 is a solution to the systems shown. x 4y z 10 a. • 2x 5y 8z 4 x 2y 3z 4

Solution

䊳

855

3x 2y z 4 b. • 2x 3y 2z 2 x y 2z 9

Substitute 1 for x, 2 for y, and 3 for z in the first system.

x 4y z 10 112 4122 132 10 10 10 true a. • 2x 5y 8z 4 S • 2112 5122 8132 4 S • 16 4 false x 2y 3z 4 112 2122 3132 4 4 4 true No, the ordered triple 11, 2, 32 is not a solution to the first system. Now use the same substitutions in the second system. 3x 2y z 4 3112 2122 132 4 4 4 true b. • 2x 3y 2z 2 S • 2112 3122 2132 2 S • 2 2 true x y 2z 9 112 122 2132 9 9 9 true The ordered triple 11, 2, 32 is a solution to the second system only.


䊳

As with systems of two equations in two variables, solutions to larger systems can also be checked by storing values using the ALPHA keys, and entering each equation on the home screen of a graphing calculator. The screens shown in Figures 9.21 and 9.22 illustrate this process for Example 2(b). Figure 9.21

Figure 9.22

B. You’ve just seen how we can check ordered triple solutions

C. Solving Systems of Three Equations in Three Variables Using Elimination From Section 9.1, we know that two systems of equations are equivalent if they have the same solution set. The systems 2x y 2z 7 • x y z 1 2y z 3

and

2x y 2z 7 • y 4z 5 z1

are equivalent, as both have the unique solution 13, 1, 12. In addition, it is evident that the second system can be solved more easily, since R2 and R3 have fewer variables than the first system. In the simpler system, mentally substituting 1 for z into R2 immediately gives y 1, and these values can be back-substituted into the first equation to find that x 3. This observation guides us to a general approach for solving larger systems—we would like to eliminate variables in the second and third equations, until we obtain an equivalent system that can easily be solved by back-substitution. To begin, let’s review the three operations that “transform” a given system, and produce an equivalent system.

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Operations That Produce an Equivalent System 1. Changing the order of the equations. 2. Replacing an equation by a nonzero constant multiple of that equation. 3. Replacing an equation with the sum of two equations from the system. Building on the ideas from Section 9.1, we develop the following approach for solving a system of three equations in three variables. Solving a System of Three Equations in Three Variables 1. Write each equation in standard form: Ax By Cz D. 2. If the “x” term in any equation has a coefficient of 1, interchange equations (if necessary) so this equation becomes R1. 3. Use the x-term in R1 to eliminate the x-terms from R2 and R3. The original R1, with the new R2 and R3, form an equivalent system that contains a smaller “subsystem” of two equations in two variables. 4. Solve the subsystem for either x or y and keep the result as the new R3. The result is an equivalent system that can be solved using back-substitution. 2x y 2z 7 We’ll begin by solving the system • x y z 1 using the elimination 2y z 3 method and the procedure outlined. In Example 3, the notation 2R1 R2 S R2 indicates the equation in row 1 has been multiplied by 2 and added to the equation in row 2, with the result placed in the system as the new row 2. EXAMPLE 3

Solution

䊳

䊳

Solving a System of Three Equations in Three Variables 2x y 2z 7 Solve using elimination: • x y z 1 . 2y z 3 1. The system is in standard form. 2. If the x-term in any equation has a coefficient of 1, interchange equations so this equation becomes R1. 2x y 2z 7 x y z 1 R2 4 R1 S • x y z 1 • 2x y 2z 7 2y z 3 2y z 3 3. Use R1 to eliminate the x-term in R2 and R3. Since R3 has no x-term, the only elimination needed is the x-term from R2. Using 2R1 R2 will eliminate this term: 2R1 R2

2x 2y 2z

2

2x y 2z 7 0x 1y 4z 5 y 4z 5

sum simplify

The new R2 is y 4z 5. The original R1 and R3, along with the new R2 form an equivalent system that contains a smaller subsystem x y z 1 x y z 1 2R1 R2 S R2 S • • 2x y 2z 7 y 4z 5 R3 S R3 2y z 3 2y z 3

new equivalent system

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4. Solve the subsystem for either y or z, and keep the result as a new R3. We choose to eliminate y using 2R2 R3: 2R2 R3

The new R3 is z 1. •

x y z 1 y 4z 5 2y z 3

2y 8z 10 2y z 3 0y 7z 7 z 1

2R2 R3 S R3

S

•

sum simplify

x y z 1 y 4z 5 z1

new equivalent system

The new R3, along with the original R1 and R2 from step 3, form an equivalent system that can be solved using back-substitution. Substituting 1 for z in R2 yields y 1. Substituting 1 for z and 1 for y in R1 yields x 3. The solution is 13, 1, 12. Check this result using the ALPHA keys, as illustrated following Example 2. Now try Exercises 13 through 16

䊳

While not absolutely needed for the elimination process, there are two reasons for wanting the coefficient of x to be “1” in R1. First, it makes the elimination method more efficient since we can more easily see what to use as a multiplier. Second, it lays the foundation for developing other methods of solving larger systems. If no equation has an x-coefficient of 1, we simply use the y- or z-variable instead (see Example 7). Since solutions to larger systems generally are worked out in stages, we will sometimes track the transformations used by writing them between the original system and the equivalent system, rather than to the left as we did in Section 9.1. Here is an additional example illustrating the elimination process, but in abbreviated form. Verify the calculations indicated using a separate sheet. EXAMPLE 4

䊳

Solving a System of Three Equations in Three Variables 5y 2x z 8 Solve using elimination: • x 3z 2y 13 . z 3y x 5

Solution

䊳

2x 5y z 8 1. Write the equations in standard form: • x 2y 3z 13 x 3y z 5 2x 5y z 8 x 3y z 5 equivalent R3 4 R1 2. • x 2y 3z 13 S • x 2y 3z 13 system x 3y z 5 2x 5y z 8 3. Using R1 R2 will eliminate the x-term from R2, yielding 5y 2z 18. Using 2R1 R3 eliminates the x-term from R3, yielding 11y z 18. x 3y z 5 x 3y z 5 R1 R2 S R2 equivalent S • 5y 2z 18 system • x 2y 3z 13 2R1 R3 S R3 11y z 18 2x 5y z 8 4. Using 2R3 R2 will eliminate z from the subsystem, leaving 27y 54. x 3y z 5 x 3y z 5 2R3 R2 S R3 S • 5y 2z 18 • 5y 2z 18 equivalent system 11y z 18 27y 54

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Solving for y in R3 yields:

27y 54 , showing y 2. Substituting 2 for y in R2 gives, 27 27 5122 2z 18 10 2z 18 2z 8 z4

substitute 2 for y simplify subtract 10 divide by 2

Substituting 2 for y and 4 for z in R1 gives, x 3122 4 5 x25 x3 C. You’ve just seen how we can solve linear systems in three variables

substitute 2 for y, 4 for z simplify subtract 2

The solution is (3, 2, 4). Now try Exercises 17 through 20

䊳

D. Inconsistent and Dependent Systems As mentioned, it is possible for larger systems to have no solutions or an infinite number of solutions. As with our work in Section 9.1, an inconsistent system (no solutions) will produce inconsistent results, ending with a statement such as 0 3 or some other contradiction. EXAMPLE 5

䊳

Attempting to Solve an Inconsistent System 2x y 3z 3 Solve using elimination: • 3x 2y 4z 2 . 4x 2y 6z 7

Solution

䊳

1. This system has no equation where the coefficient of x is 1. 2. We can still use R1 to begin the solution process, but this time we’ll use the variable y since it does have coefficient 1. Using 2R1 R2 eliminates the y-term from R2, leaving 7x 2z 4. But using 2R1 R3 to eliminate the y-term from R3 results in a contradiction: 2R1 4x 2y 6z 6 R2 3x 2y 4z 2 7x 2z 4

2R1 R3

4x 2y 6z

6

4x 2y 6z 7 0x 0y 0z 1 0 1

contradiction

We conclude the system is inconsistent. The answer is the empty set , and we need work no further. Now try Exercises 21 and 22

䊳

Unlike our work with systems having only two variables, systems in three variables can have two forms of dependence— linear dependence (Figure 9.18) or coincident dependence (Figure 9.19). To help understand linear dependence, consider a system of 2x 3y z 5 two equations in three variables: e . Each of these equations x 3y 2z 1 represents a plane, and unless the planes are parallel, their intersection will be a line. As in Section 9.1, we can state solutions to a dependent system using set notation with two of the variables written in terms of the third, or as an ordered triple using a parameter. The relationships named can then be used to generate specific solutions to the system.

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Systems with two equations and two variables or three equations and three variables are called square systems, meaning there are exactly as many equations as there are variables. A system of linear equations cannot have a unique solution unless there are at least as many equations as there are variables in the system. EXAMPLE 6

䊳

Solving a Dependent System Solve using elimination: e

Solution

䊳

2x 5y 3z 5 . x 5y 2z 1

We immediately note that R1 R2 eliminates the y-term from R2, yielding 2x 5y 3z 5 x z 4 and the new system e . This means (x, y, z) will x z 4 satisfy both equations only when x z 4 (the x-coordinate must be 4 less than the z-coordinate). Since x is written in terms of z, we substitute z 4 for x in either equation to find how y is related to z. Using R2 in the original system we have: 1z 42 5y 2z 1, which yields y 15z 35 (verify). This means the y-coordinate of the solution must be 35 more than 15z. In set notation, the solution is 5 1x, y, z,2 | x z 4, y 15z 35, z 僆⺢6 . Randomly choosing z 3, 2, and 7, the solutions would be 17, 0, 32, 12, 1, 22, and 13, 2, 72 , respectively. Verify that these satisfy both equations. Using p as a parameter, the solution could be written 1 p 4, 15p 35, p2 in parameterized form. Now try Exercises 23 through 26

䊳

The system in Example 6 was nonsquare, and we knew ahead of time the system would be dependent. The system in Example 7 is square, but only by applying the elimination process can we determine the nature of its solution(s). EXAMPLE 7

䊳

Solving a Dependent System 3x 2y z 1 Solve using elimination: • 2x y z 5 . 10x 2y 8

Solution

䊳

This system has no equation where the coefficient of x is 1. We will still use R1, but eliminate z in R2 (there is no z-term in R3). Using R1 R2 eliminates the z-term from R2, yielding 5x y 4. 3x 2y z 1 R1 R2 S R2 S • 2x y z 5 R3 S R3 10x 2y 8

3x 2y z 1 • 5x y 4 10x 2y 8

We next solve the subsystem. Using 2R2 R3 eliminates the y-term in R3, but also all other terms: 2R2 R3

10x 2y 8 10x 2y 8 0x 0y 0 0 0

sum result

Since R3 is the same as 2R2, the system is linearly dependent and equivalent to 3x 2y z 1 e . We can solve for y in R2 to write y in terms of x: y 5x 4. 5x y 4 Substituting 5x 4 for y in R1 enables us to also write z in terms of x:

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3x 2y z 1 3x 215x 42 z 1 3x 10x 8 z 1 7x z 9 z 7x 9

R1 substitute 5x 4 for y distribute simplify solve for z

The solution set is 5 1x, y, z2 | x 僆⺢, y 5x 4, z 7x 96. Three of the infinite number of solutions are 10, 4, 92 for x 0, 12, 6, 52 for x 2, and 11, 9, 162 for x 1. Verify these triples satisfy all three equations. Again using the parameter p, the solution could be written as 1 p, 5p 4, 7p 92 . Now try Exercises 27 through 30

D. You’ve just seen how we can recognize inconsistent and dependent systems

䊳

For coincident dependence the equations in a system differ by only a constant multiple. After applying the elimination process—all variables are eliminated from the other equations, leaving statements that are always true (such as 2 2 or some other). For additional practice solving various kinds of systems, see Exercises 31 through 44.

E. Applications Applications of larger systems are simply an extension of our work with systems of two equations in two variables. Once again, the applications come in a variety of forms and from many fields. In the world of business and finance, systems can be used to diversify investments or spread out liabilities, a financial strategy hinted at in Example 8. EXAMPLE 8

䊳

Modeling the Finances of a Business A small business borrowed $225,000 from three different lenders to expand their product line. The interest rates were 5%, 6%, and 7%. Find how much was borrowed at each rate if the annual interest came to $13,000 and twice as much was borrowed at the 5% rate than was borrowed at the 7% rate.

Solution

䊳

Let x, y, and z represent the amounts borrowed at 5%, 6%, and 7%, respectively. This means our first equation is x y z 225 (in thousands). The second equation is determined by the total interest paid, which was $13,000: 0.05x 0.06y 0.07z 13. The third is found by carefully reading the problem. “twice as much was borrowed at the 5% rate than was borrowed at the 7% rate,” or x 2z. x y z 225 These equations form the system: • 0.05x 0.06y 0.07z 13. The x-term of x 2z the first equation has a coefficient of 1. Written in standard form we have: x y z 225 R1 • 5x 6y 7z 1300 R2 multiply R2 by 100 x 2z 0 R3 Using 5R1 R2 will eliminate the x term in R2, while R1 R3 will eliminate the x-term in R3. 5R1 R2

5x 5y 5z 1125 5x 6y 7z y 2z

1300 175

R1 R3

x y z 225 x

2z 0 y 3z 225

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The new R2 is y 2z 175, and the new R3 (after multiplying by 1) is x y z 225 y 3z 225, yielding the equivalent system • y 2z 175. y 3z 225 Solving the 2 2 subsystem using R2 R3 yields z 50. Back-substitution shows y 75 and x 100, yielding the solution (100, 75, 50). This means $50,000 was borrowed at the 7% rate, $75,000 was borrowed at 6%, and $100,000 at 5%. A “substitution check” on a graphing calculator verifies this solution is correct (see Figures 9.23 and 9.24). Figure 9.23

E. You’ve just seen how we can use a system of three equations in three variables to solve applications

Figure 9.24


䊳

9.2 EXERCISES 䊳



䊳

1. The solution to an equation in three variables is an ordered .

2. The graph of the solutions to an equation in three variables is a(n) .

3. Systems that have the same solution set are called .

4. If a 3 3 system is linearly dependent, the ordered triple solutions can be written in terms of a single variable called a(n) .

5. Find a value of z that makes the ordered triple 12, 5, z2 a solution to 2x y z 4. Discuss/Explain how this is accomplished.

6. Explain the difference between linear dependence and coincident dependence, and describe how the equations are related.


Find any four ordered triples that satisfy the equation given.

7. x 2y z 9 8. 3x y z 8 9. x y 2z 6 10. 2x y 3z 12

Use the ALPHA keys on a graphing calculator to determine if the given ordered triples are solutions to the system. If not a solution, identify which equation(s) are not satisfied.

x y 2z 1 11. • 4x y 3z 3 ; 3x 2y z 4 2x 3y z 9 12. • 5x 2y z 32; x y 2z 13

10, 3, 22 13, 4, 12 14, 5, 22 15, 4, 112

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Solve each system using elimination and back-substitution.

x y 2z 10 x y 2z 1 13. • x 14. • 4x y z1 3 z4 3x 6 x 3y 2z 16 x y 5z 1 15. • 2y 3z 1 16. • 4x y 1 8y 13z 7 3x 2y 8 x y 2z 10 x y 2z 1 17. • x y z 7 18. • 4x y 3z 3 2x y z 5 3x 2y z 4 2x 3y 2z 0 3x y z 6 19. • 3x 4y z 20 20. • 2x 2y z 5 x 2y z 16 2x y z 5 Solve using the elimination method. If a system is inconsistent or dependent, so state. For systems with linear dependence, write solutions in set notation and as an ordered triple in terms of a parameter.

3x y 2z 3 2x y 3z 8 21. • x 2y 3z 1 22. • 3x 4y z 4 4x 8y 12z 7 4x 2y 6z 5 4x y 3z 8 23. e x 2y 3z 2

4x y 2z 9 24. e 3x y 5z 5

6x 3y 7z 2 25. e 3x 4y z 6

2x 4y 5z 2 26. e 3x 2y 3z 7

Solve using elimination. If the system is linearly dependent, state the general solution in terms of a parameter (different forms of the solution are possible). Then find four specific ordered triples that satisfy the system (answers will vary).

3x 4y 5z 5 27. •x 2y 3z 3 3x 2y z 1 5x 3y 2z 4 28. • 9x 5y 4z 12 3x y 2z 12 x 2y 3z 1 29. • 3x 5y 8z 7 x y 2z 5 2x 3y 5z 3 30. • 5x 7y 12z 8 x y 2z 2

Solve using the elimination method. If a system is inconsistent or dependent, so state. For systems with linear dependence, write the answer in terms of a parameter. For coincident dependence, state the solution in set notation.

4x 2y 8z 24 31. • x 0.5y 2z 6 2x y 4z 12 2x 5y 4z 6 32. • x 2.5y 2z 3 3x 7.5y 6z 9 x 2y 2z 6 4x 5y 6z 5 33. • 2x 6y 3z 13 34. • 2x 3y 3z 0 3x 4y z 11 x 2y 3z 5 x 5y 4z 3 2x 3y 5z 4 35. • 2x 9y 7z 2 36. • x y 2z 3 3x 14y 11z 5 x 3y 4z 1

37.

u

1 1 1 x y z2 6 3 2 3 1 1 x y z9 4 3 2 1 1 x y z2 2 2

38.

u

y x 2 3 2x y 3 x 2y 6

z 2 2 z8 3z 6 2

Some applications of systems lead to systems similar to those that follow. Solve using elimination.

39. •

2A B 3C 21 B C1 A B 4

40. •

A 3B 2C 11 2B C 9 B 2C 8

A 2C 7 41. • 2A 3B 8 3A 6B 8C 33 A 2B 5 42. • B 3C 7 2A B C 1 C 2 43. • 5A 2C 5 4B 9C 16 C3 44. • 2A 3C 10 3B 4C 11

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45. Dimensions of a rectangular solid: 2w ⴙ 2h ⴝ P1 P2 16 cm (top) P1 14 cm • 2l ⴙ 2w ⴝ P2 P3 18 cm (small side) 2l ⴙ 2h ⴝ P3 h (large side) w

l

Using the formula shown, the dimensions of a rectangular solid can be found if the perimeters of the three distinct faces are known. Find the dimensions of the solid shown. 䊳

863

46. Distance from a point (x, y, z) to the plane Ax ⴙ By ⴙ Cz ⴚ D Ax ⴙ By ⴙ Cz ⴝ D: ` ` 2A2 ⴙ B2 ⴙ C2 The perpendicular distance from a given point (x, y, z) to the plane defined by Ax By Cz D is given by the formula shown. Consider the plane given in Figure 9.15 1x y z 62. What is the distance from this plane to the point (3, 4, 5)?

APPLICATIONS

Solve the following applications by setting up and solving a system of three equations in three variables. Note that some equations may have only two of the three variables used to create the system. Check answers using back-substitution or the ALPHA keys on the home screen of a graphing calculator. Investment/Finance and Simple Interest Problems

47. Investing the winnings: After winning $280,000 in the lottery, Maurika decided to place the money in three different investments: a certificate of deposit paying 4%, a money market certificate paying 5%, and some Aa bonds paying 7%. After 1 yr she earned $15,400 in interest. Find how much was invested at each rate if $20,000 more was invested at 7% than at 5%. 48. Purchase at auction: At an auction, a wealthy collector paid $7,000,000 for three paintings: a Monet, a Picasso, and a van Gogh. The Monet cost $800,000 more than the Picasso. The price of the van Gogh was $200,000 more than twice the price of the Monet. What was the price of each painting? Descriptive Translation

49. Major wars: The United States has fought three major wars in modern times: World War II, the Korean War, and the Vietnam War. If you sum the years that each conflict ended, the result is 5871. The Vietnam War ended 20 years after the Korean War and 28 years after World War II. In what year did each end? 50. Animal gestation periods: The average gestation period (in days) of an elephant, rhinoceros, and camel sum to 1520 days. The gestation period of a rhino is 58 days longer than that of a camel. Twice the camel’s gestation period decreased by 162 gives the gestation period of an elephant. What is the gestation period of each?

51. Moments in U.S. history: If you sum the year the Declaration of Independence was signed, the year the 13th Amendment to the Constitution abolished slavery, and the year the Civil Rights Act was signed, the total would be 5605. Ninety-nine years separate the 13th Amendment and the Civil Rights Act. The Civil Rights Act was signed 188 years after the Declaration of Independence. What year was each signed? 52. Aviary wingspan: If you combine the wingspan of the California Condor, the Wandering Albatross (see photo), and the prehistoric Quetzalcoatlus, you get an astonishing 18.6 m (over 60 ft). If the wingspan of the Quetzalcoatlus is equal to five times that of the Wandering Albatross minus twice that of the California Condor, and six times the wingspan of the Condor is equal to five times the wingspan of the Albatross, what is the wingspan of each? Mixtures

53. Chemical mixtures: A chemist mixes three different solutions with concentrations of 20%, 30%, and 45% glucose to obtain 10 L of a 38% glucose solution. If the amount of 30% solution used is 1 L more than twice the amount of 20% solution used, find the amount of each solution used. 54. Value of gold coins: As part of a promotion, a local bank invites its customers to view a large sack full of $5, $10, and $20 gold pieces, promising to give the sack to the first person able to state the number of coins for each

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denomination. Customers are told there are exactly 250 coins, with a total face value of $1875. If there are also seven times as many $5 gold pieces as $20 gold pieces, how many of each denomination are there? Nutrition

55. Industrial food production: Acampana Soups is creating a new sausage and shrimp gumbo that contains three different types of fat: saturated, monounsaturated, and polyunsaturated. As a new member of the “Heart Healthy” menu, this soup must contain only 2.8 g of total fat per serving. The head chef Yev Kasem demands that the recipe provides exactly twice as much saturated fat as polyunsaturated fat. At the same time, the lead nutrition expert Florencia requires the amount of saturated fat in a serving to be 0.4 g less than the combined amount of unsaturated fats. How many grams of each type of fat will a serving of this soup contain? 56. Geriatric nutrition: The dietician at McKnight Place must create a balanced diet for Fred, consisting of a daily total of 1600 calories. For his successful rehabilitation, exact quantities of complex carbohydrates, fat, and protein must provide these calories. In this diet, carbohydrates provide 160 more calories than fat and protein together, while fat provides 1.25 times more calories than protein alone. How many calories should each nutrient provide on a daily basis? 䊳

In Exercises 87 and 88 from Section 9.1, we found the equation of a line through two points by creating a system of two equations of the form y ⴝ mx ⴙ b. Just as any two points determine a unique line, three noncollinear points determine a unique parabola (as long as no two have the same first coordinate). Similar to the method used there, each point gives an equation of the form y ⴝ ax2 ⴙ bx ⴙ c, and we create a 3 ⴛ 3 system that can be solved using elimination. For instance, the point (2, 41) gives the equation 41 ⴝ a(2)2 ⴙ 2b ⴙ c.

57. Height of a soccer ball: One second after being kicked (t 1), a soccer ball is 26 ft high. After 2 sec (t 2), the ball is 41 ft high, and after 6 sec the ball is 1 ft above the ground. Use the ordered pairs (time, height) to find a function h(t) modeling the height of the ball after t sec, then use the equation to find (a) the maximum height of the kick, and (b) the height of the ball after 5.4 sec. 58. Height of an arrow: An archer is out in a large field testing a new bow. Pulling back the bow to near its breaking point, the archer lets the arrow fly. Suppose that 1 sec after release (t 1) the arrow was 184 ft high, four sec later (t 5) it was 600 ft high, and after 12 sec, the arrow was 96 ft high. Use the ordered pairs (time, height) to find a function h(t) modeling the height of the arrow after t sec, then use the equation to (a) find the maximum height of the shot and (b) determine how long the arrow was airborne.


x 2y z 2 59. The system • x 2y kz 5 is dependent if 2x 4y 4z 10 , and inconsistent otherwise. k

䊳

9–28


60. One form of the equation of a circle is x2 y2 Dx Ey F 0. Use a system to find the equation of the circle through the points 12, 12, 14, 32, and 12, 52.


61. (8.3) Given u H1, 7I and v H3, 12I, compute u 4v and 3u v. 62. (6.2) Given cot A 1.6831, use a calculator to find the acute angle A to the nearest tenth of a degree. 63. (5.5) Solve the logarithmic equation: log1x 22 log x log 3

y

64. (2.1) Analyze the graph of g shown. Clearly state the domain and range, the zeroes of g, intervals where g1x2 7 0, x intervals where g1x2 6 0, local maximums or minimums, and intervals where the function is increasing or decreasing. Assume each tick mark is one unit and estimate endpoints to the nearest tenths.

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Systems of Inequalities and Linear Programming

LEARNING OBJECTIVES

In this section, we’ll build on many of the ideas from Section 9.1, with a focus on systems of linear inequalities. While systems of linear equations have an unlimited number of applications, there are many situations that can only be modeled using linear inequalities. For example, decisions in business and industry are often based on a large number of limitations or constraints, with many different ways these constraints can be satisfied.


A. Solve a linear inequality in two variables

B. Solve a system of linear inequalities C. Solve applications using a system of linear inequalities D. Solve applications using linear programming

A. Linear Inequalities in Two Variables A linear equation in two variables is any equation that can be written in the form Ax By C, where A and B are real numbers, not simultaneously equal to zero. A linear inequality in two variables is similarly defined, with the “ ” sign replaced by the “ 6 ,” “ 7 ,” “ ,” or “ ” symbol: Ax By 6 C Ax By C

Ax By 7 C Ax By C

Solving a linear inequality in two variables has many similarities with the one variable case. For one variable, we graph a boundary point on a number line, decide whether the endpoint is included or excluded, and shade the appropriate half line. For x 1 3, we have the solution x 2 with the endpoint (boundary point) included and the line shaded to the left (Figure 9.25): Figure 9.25

Figure 9.26

3

y

2

1

[ 0

1

2

3

Interval notation: x (, 2]

5

(0, 3)

xy3

Upper half plane

5

(4, 1)

Lower half plane

5

x

5

EXAMPLE 1

䊳

For linear inequalities in two variables, we graph a boundary line, decide whether the boundary line is included or excluded, and shade the appropriate half plane. For x y 3, the boundary line x y 3 is graphed in Figure 9.26. Note it divides the coordinate plane into two regions called half planes, and it forms the boundary between the two regions. If the boundary is included in the solution set, we graph it using a solid line. If the boundary is excluded, a dashed line is used. Recall that solutions to a linear equation are ordered pairs that make the equation true. We use a similar idea to find or verify solutions to linear inequalities. If any one point in a half plane makes the inequality true, all points in that half plane will satisfy the inequality. Checking Solutions to an Inequality in Two Variables Determine whether the given ordered pairs are solutions to x 2y 2: a. 14, 32 b. 12, 12 c. 14, 12

Solution

9–29

䊳

a. Substitute 4 for x and 3 for y: 142 2132 2 10 2 14, 32 is a solution. b. Substitute 2 for x and 1 for y: 122 2112 2 42 12, 12 is not a solution. c. Substitute 4 for x and 1 for y: 142 2112 2 22 14, 12 is a solution.

substitute 4 for x, 3 for y true substitute 2 for x, 1 for y false substitute 4 for x, 1 for y true

Now try Exercises 7 through 10 䊳 865

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WORTHY OF NOTE This relationship is often called the trichotomy axiom or the “three-part truth.” Given any two quantities, they are either equal to each other, or the first is less than the second, or the first is greater than the second.

EXAMPLE 2

䊳

Earlier we graphed linear equations by plotting a small number of ordered pairs or by solving for y and using the slope-intercept method. The line represented all ordered pairs that made the equation true, meaning the left-hand expression was equal to the right-hand expression. To graph linear inequalities, we reason that if the line represents all ordered pairs that make the expressions equal, then any point not on that line must make the expressions unequal —either greater than or less than. These ordered pair solutions must lie in one of the half planes formed by the line, which we shade to indicate the solution region. Note this implies the boundary line for any inequality is determined by the related equation, created by temporarily replacing the inequality symbol with an “” sign. Solving an Inequality in Two Variables Solve the inequality x 2y 2.

Solution

䊳

The related equation and boundary line is x 2y 2. Since the inequality is inclusive (less than or equal to), we graph a solid line. Using the intercepts, we graph the line through (0, 1) and 12, 02 shown in Figure 9.27. To determine the solution region and which side to shade, we select (0, 0) as a test point, which results in a true statement: 102 2102 2 ✓. Since (0, 0) is in the “lower” half plane, we shade this side of the boundary (see Figure 9.28). Figure 9.27 Figure 9.28 y

y

5

Upper half plane

5

(4, 3)

(4, 3)

(0, 1)

(0, 1)

(2, 0)

(2, 0)

5

(0, 0) Test point

5

x

5

5

(0, 0) Test point

Lower half plane 5

x

5

Now try Exercises 11 through 14 䊳 The same solution would be obtained if we first solved for y and graphed the boundary line using the slope-intercept method. However, using the slope-intercept method offers a distinct advantage — test points are no longer necessary since solutions to “less than” inequalities will always appear below the boundary line and solutions to “greater than” inequalities appear above the line. Written in slope-intercept form, the inequality from Example 2 is y 12 x 1. Note that (0, 0) still results in a true statement, but the “less than or equal to” symbol now indicates directly that solutions will be found in the lower half plane. These observations lead to our general approach for solving linear inequalities: Solving a Linear Inequality 1. Graph the boundary line by solving for y and using the slope-intercept form. • Use a solid line if the boundary is included in the solution set. • Use a dashed line if the boundary is excluded from the solution set. 2. For “greater than” inequalities shade the upper half plane. For “less than” inequalities shade the lower half plane.

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Section 9.3 Systems of Inequalities and Linear Programming

EXAMPLE 3

䊳

Solving linear Inequalities in Two Variables Using Technology Solve the inequality 3x 5y 10 using a graphing calculator.

Solution

䊳

We begin by solving the inequality for y, so we can enter the equation on the Y= screen. 3x 5y 10 5y 3x 10 3 x2 y 5

given inequality subtract 3x (isolate y-term) divide by 5

3 Entering Y1 X 2 on the Y= screen, and graphing the boundary line using 5 ZOOM 6:ZStandard produces the graph shown in Figure 9.29. The “less than” inequality indicates the region below the line should be shaded, so we return to the Y= screen and move the cursor to the far left. From the default “connected line” setting, pressing three times brings the “shade below the graph” marker “” into view (Figure 9.30), and pressing GRAPH gives the solution shown in Figure 9.31. As a check, note that (0, 0) is in the solution region, and is a solution to 3x 5y 10. ENTER

Figure 9.31

Figure 9.29

Figure 9.30

10

10

10

10

10

10

10

10

Now try Exercises 15 through 18 䊳 A. You’ve just seen how we can solve a linear inequality in two variables

B. Solving Systems of Linear Inequalities To solve a system of inequalities, we apply the procedure outlined above to all inequalities in the system, and note the ordered pairs that satisfy all inequalities simultaneously. In other words, we find the intersection of all solution regions (where they overlap), which then represents the solution for the system. In the case of vertical boundary lines, the designations “above” or “below” the line cannot be applied, and instead we simply note that for any vertical line x k, points with x-coordinates larger than k will occur to the right. EXAMPLE 4

䊳

Solving a System of Linear Inequalities Solve the system of inequalities: e

Solution

䊳

2x y 4 . xy 6 2

Solving for y, we obtain y 2x 4 and y 7 x 2. The line y 2x 4 will be a solid boundary line (included), while y x 2 will be dashed (not included). Both inequalities are “greater than” and so we shade the upper half plane for each. The regions overlap and form the solution region (the lavender region shown). This sequence of events is illustrated here:

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CHAPTER 9 Systems of Equations and Inequalities Shade above y x 2 (in pink) y

Shade above y 2x 4 (in blue) y 5

Overlapping region y

5

2x y 4

5

2x y 4

Solution region

2x y 4 xy2

5

5

x

5

5

x

xy2 5

5

x

Corner point 5

5

5

The solutions are all ordered pairs found in this region and its included boundaries. To verify the result, test the point (2, 3) from inside the region, and 15, 22 from outside the region [the point (2, 0) is not a solution since it does not satisfy x y 6 2]. Now try Exercises 19 through 36 䊳 For future reference, the point of intersection (2, 0) is called a corner point or vertex of the solution region. If the point of intersection is not easily found from the graph, we can find it by solving a linear system using the two lines. For Example 4, the system is e

2x y 4 xy2

and solving by elimination gives 3x 6, x 2, and (2, 0) as the point of intersection. A graphing calculator can also be used to solve a system of linear inequalities. One method (there are several) involves these three steps, which are performed on each equation: 1. Solve for y and enter the results on the Y= screen to create the boundary lines. 2. Graph each line and test the related half plane. 3. Shade the appropriate half plane. This process is illustrated in Example 5. Since many real-world applications of linear inequalities do not use negative numbers, we often include x 0 and y 0 as part of the system, and set Xmin ⴝ 0 and Ymin ⴝ 0 as part of the size. The value of Xmax and Ymax will depend on equations or context given. WINDOW

EXAMPLE 5

䊳

Solving a System of Inequalities Using Technology 3x 2y 6 14 x 2y 6 8 Use a graphing calculator to solve the system: μ x0 y0

Solution

䊳

Following the steps outlined above, we have 1. Enter the related equations. For 3x 2y 14, we have y 1.5x 7. For x 2y 8, we have y 0.5x 4. Enter these as Y1 and Y2 on the Y= screen. 2. Graph the boundary lines. Note the x- and y-intercepts of both lines are less than 10, so we can graph them using a “friendly window” where x 僆 [0, 9.4] and y 僆 [0, 6.2]. After setting the window, press GRAPH to graph the lines.

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3. Shade the appropriate half plane. Since both equations are in slope-intercept form and solving for y resulted in two “less than” inequalities, we shade below both lines, using the “” feature located to the far left of Y1 and Y2. Simply overlay the diagonal line and press repeatedly until the symbol appears (Figure 9.32). After pressing the GRAPH key, the calculator draws both lines and shades the appropriate regions (Figure 9.33). Note the calculator uses two different kinds of shading. This makes it easy to identify the solution region — it will be the “checker-board area” where the horizontal and vertical lines cross. ENTER


6.2

0

9.4

0

As a final check, we can navigate the position marker into the solution region and test a few points in both inequalities. Using the test point (2, 2) from within the region yields 3x 2y 6 14 3122 2122 14 10 14

B. You’ve just seen how we can solve a system of linear inequalities

first inequality substitute 2 for x, 2 for y true

x 2y 6 8 122 2122 8 68

second inequality substitute 2 for x, 2 for y true


C. Applications of Systems of Linear Inequalities Systems of inequalities give us a way to model the decision-making process when certain constraints must be satisfied. A constraint is a fact or consideration that somehow limits or governs possible solutions, like the number of acres a farmer plants—which may be limited by time, size of land, government regulations, and so on. 䊳

EXAMPLE 6

Solving Applications of Linear Inequalities As part of their retirement planning, James and Lily decide to invest up to $30,000 in two separate investment vehicles. The first is a bond issue paying 9% and the second is a money market certificate paying 5%. A financial adviser suggests they invest at least $10,000 in the certificate and not more than $15,000 in bonds. What various amounts can be invested in each?

䊳

Solution C

B 15

40

(0, 30) Solution region 30

QI

20

(15, 15) C 10 10

(0, 10) (15, 10) 10

20

30

40

B

Consider the ordered pairs (B, C) where B represents the money invested in bonds and C the money invested in the certificate. Since they plan to invest no more than $30,000, the investment constraint would be B C 30 (in thousands). Following the adviser’s recommendations, the constraints on each investment would be B 15 and C 10. Since they cannot invest less than zero dollars, the last two constraints are B 0 and C 0. B C 30 B 15 μ C 10 B0 C0

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The resulting system is shown in the figure, and indicates solutions will be in the first quadrant. There is a vertical boundary line at B 15 with shading to the left (less than) and a horizontal boundary line at C 10 with shading above (greater than). After graphing C 30 B, we see the solution region is a quadrilateral with vertices at (0, 10), (0, 30), (15, 10), and (15, 15), as shown. Now try Exercises 61 and 62 䊳 From Example 6, any ordered pair in this region or on its boundaries would represent an investment of the form (money in bonds, money in CDs) S (B, C), and would satisfy all constraints in the system. A natural follow-up question would be—What combination of (money in bonds, money in CDs) would offer the greatest return? This would depend on the interest being paid on each investment, and introduces us to a study of linear programming, which follows soon.

C. You’ve just seen how we can solve applications using a system of linear inequalities

D. Linear Programming To become as profitable as possible, corporations look for ways to maximize their revenue and minimize their costs, while keeping up with delivery schedules and product demand. To operate at peak efficiency, plant managers must find ways to maximize productivity, while minimizing related costs and considering employee welfare, union agreements, and other factors. Problems where the goal is to maximize or minimize the value of a given quantity under certain constraints or restrictions are called programming problems. The quantity we seek to maximize or minimize is called the objective function. For situations where linear programming is used, the objective function is given as a linear function in two variables and is denoted f (x, y). A function in two variables is evaluated in much the same way as a single variable function. To evaluate f 1x, y2 2x 3y at the point (4, 5), we substitute 4 for x and 5 for y: f 14, 52 2142 3152 23. EXAMPLE 7

䊳

Determining Maximum Values Determine which of the following ordered pairs maximizes the value of f 1x, y2 5x 4y: (0, 6), (5, 0), (0, 0), or (4, 2).

Solution

䊳

Organizing our work in table form gives Given Point (0, 6) (5, 0) (0, 0) (4, 2)

Evaluate f 1x, y2 ⴝ 5x ⴙ 4y

f 10, 62 5102 4162 24

f 15, 02 5152 4102 25

f 10, 02 5102 4102 0

f 14, 22 5142 4122 28

The function f 1x, y2 5x 4y is maximized at (4, 2). Now try Exercises 51 through 54 䊳 When the objective is stated as a linear function in two variables and the constraints are expressed as a system of linear inequalities, we have what is called a linear programming problem. The systems of inequalities solved earlier produced solution regions that were either bounded (as in Example 6) or unbounded (as in Example 4).

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Figure 9.34

Convex

We interpret the word bounded to mean we can enclose the solution region within a circle of appropriate size. If we cannot draw a circle around the region because it extends indefinitely in some direction, the region is said to be unbounded. In this study, we will consider only situations that produce bounded solution regions, meaning the regions will have three or more vertices. The regions we study will also be convex, meaning that for any two points in the enclosed region, the line segment between them is also in the region (Figure 9.34). Under these conditions, it can be shown that the maximum or minimum values must occur at one of the corner points of the solution region, also called the feasible region.

Not convex

EXAMPLE 8

䊳

Finding the Maximum of an Objective Function

Solution

䊳

y Begin by noting that the solutions must be in QI, 8 since x 0 and y 0. Graph the boundary lines 7 y x 4 and y 3x 6, shading the lower 6 half plane in each case since they are “less than” 5 4 inequalities. This produces the feasible region (1, 3) 3 shown in lavender. There are four corner points to Feasible 2 this region: (0, 0), (0, 4), (2, 0), and (1, 3). Three of region 1 these points are intercepts and can be found quickly. 5 4 3 2 1 1 2 3 The point (1, 3) was found by solving the 1 xy4 2 . Knowing that the objective system e 3x y 6 function will be maximized at one of the corner points, we test them in the objective function, using a table to organize our work.

Find the maximum value of the objective function f 1x, y2 2x y given the xy4 3x y 6 constraints shown: μ . x0 y0

Corner Point (0, 0) (0, 4) (2, 0) (1, 3)

4

5

x

Objective Function f 1x, y2 ⴝ 2x ⴙ y

f 10, 02 2102 102 0

f 10, 42 2102 142 4

f 12, 02 2122 102 4

f 11, 32 2112 132 5

The objective function f 1x, y2 2x y is maximized at (1, 3). Now try Exercises 55 through 58 䊳 Figure 9.35 y 8 7 6 5 4

(1, 3)

3 2 1 5 4 3 2 1 1

1

K1

2

3

4

5

K5 K3

x

To help understand why solutions must occur at a vertex, note the objective function f (x, y) is maximized using only (x, y) ordered pairs from the feasible region. If we let K represent this maximum value, the function from Example 8 becomes K 2x y or y 2x K, which is a line with slope 2 and y-intercept K. The table in Example 8 suggests that K should range from 0 to 5 and graphing y 2x K for K 1, K 3, and K 5 produces the family of parallel lines shown in Figure 9.35. Note that values of K larger than 5 will cause the line to miss the solution region, and the maximum value of 5 occurs where the line intersects the feasible region at the vertex (1, 3). These observations lead to the following principles, which we offer without a formal proof.

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Linear Programming Solutions 1. If the feasible region is convex and bounded, a maximum and a minimum value exist. 2. If a unique solution exists, it will occur at a vertex of the feasible region. 3. If more than one solution exists, at least one of them occurs at a vertex of the feasible region with others on a boundary line. 4. If the feasible region is unbounded, a linear programming problem may have no solutions. Solving linear programming problems depends in large part on two things: (1) identifying the objective and the decision variables (what each variable represents in context), and (2) using the decision variables to write the objective function and constraint inequalities. This brings us to our five-step approach for solving linear programming applications. Solving Linear Programming Applications 1. Identify the main objective and the decision variables (descriptive variables may help) and write the objective function in terms of these variables. 2. Organize all information in a table, with the decision variables and constraints heading up the columns, and their components leading each row. 3. Complete the table using the information given, and write the constraint inequalities using the decision variables, constraints, and the domain. 4. Graph the constraint inequalities, determine the feasible region, and identify all corner points. 5. Test these points in the objective function to determine the optimal solution(s).

EXAMPLE 9

䊳

Solving an Application of Linear Programming The owner of a snack food business wants to create two nut mixes for the holiday season. The regular mix will have 14 oz of peanuts and 4 oz of cashews, while the deluxe mix will have 12 oz of peanuts and 6 oz of cashews. The owner estimates he will make a profit of $3 on the regular mixes and $4 on the deluxe mixes. How many of each should be made in order to maximize profit, if only 840 oz of peanuts and 348 oz of cashews are available?

Solution

䊳

Our objective is to maximize profit, and the decision variables could be r to represent the regular mixes sold, and d for the number of deluxe mixes. This gives P1r, d2 $3r $4d as our objective function. The information is organized in Table 9.1, using the variables r, d, and the constraints to head each column. Since the mixes are composed of peanuts and cashews, these lead the rows in the table. Table 9.1 ⴙ P1r, d2 ⴝ $3r T Regular r

$4 d T Deluxe d

Constraints: Total Ounces Available

Peanuts

14

12

840

Cashews

4

6

348

After filling in the appropriate values, reading the table from left to right along the “peanut” row and the “cashew” row, gives the constraint inequalities 14r 12d 840

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and 4r 6d 348. Realizing we won’t be making negative numbers of mixes, the remaining constraints are r 0 and d 0. The complete system is 14r 12d 840 4r 6d 348 μ r0 d0 Note once again that the solutions must be in QI, since r 0 and d 0. Graphing the first two inequalities using slope-intercept form gives d 76r 70 and d 23r 58 producing the feasible region shown in lavender. The four corner points are (0, 0), (60, 0), (0, 58), and (24, 42). Three of these points are intercepts and can be read from a table of values or the graph itself. The point (24, 42) was found by solving 14r 12d 840 the system e . Knowing the solution 4r 6d 348 must occur at one of these points, we test them in the objective function (Table 9.2).

d 100 90 80 70 60 50 40 30 20

Feasible region

10 10 20 30 40 50 60 70 80 90 100

r

Table 9.2 Corner Point (0, 0) (60, 0)

Objective Function P(r, d ) ⴝ $3r ⴙ $4d P10, 02 $3102 $4102 $0 P160, 02 $31602 $4102 $180

(0, 58)

P10, 582 $3102 $41582 $232

(24, 42)

P124, 422 $31242 $41422 $240

Profit will be maximized if 24 boxes of the regular mix and 42 boxes of the deluxe mix are made and sold. Now try Exercises 63 through 68

䊳

Linear programming can also be used to minimize an objective function, as in Example 10. EXAMPLE 10

䊳

Minimizing Costs Using Linear Programming A beverage producer needs to minimize shipping costs from its two primary plants in Kansas City (KC) and St. Louis (STL). All wholesale orders within the state are shipped from one of these plants. An outlet in Macon orders 200 cases of soft drinks on the same day an order for 240 cases comes from Springfield. The plant in KC has 300 cases ready to ship and the plant in STL has 200 cases. The cost of shipping each case to Macon is $0.50 from KC, and $0.70 from STL. The cost of shipping each case to Springfield is $0.60 from KC, and $0.65 from STL. How many cases should be shipped from each warehouse to minimize costs?

Solution

䊳

Our objective is to minimize costs, which depends on the number of cases shipped from each plant. To begin we use the following assignments: A S cases shipped from KC to Macon B S cases shipped from KC to Springfield C S cases shipped from STL to Macon D S cases shipped from STL to Springfield

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From this information, the equation for total cost T is T ⫽ 0.5A ⫹ 0.6B ⫹ 0.7C ⫹ 0.65D, an equation in four variables. To make the cost equation more manageable, note since Macon ordered 200 cases, A ⫹ C ⫽ 200. Similarly, Springfield ordered 240 cases, so B ⫹ D ⫽ 240. After solving for C and D, respectively, these equations enable us to substitute for C and D, resulting in an equation with just two variables. For C ⫽ 200 ⫺ A and D ⫽ 240 ⫺ B we have T1A, B2 ⫽ 0.5A ⫹ 0.6B ⫹ 0.71200 ⫺ A2 ⫹ 0.651240 ⫺ B2 ⫽ 0.5A ⫹ 0.6B ⫹ 140 ⫺ 0.7A ⫹ 156 ⫺ 0.65B ⫽ 296 ⫺ 0.2A ⫺ 0.05B The constraints involving the KC plant are A ⫹ B ⱕ 300 with A ⱖ 0, B ⱖ 0. The constraints for the STL plant are C ⫹ D ⱕ 200 with C ⱖ 0, D ⱖ 0. Since we want a system in terms of A and B only, we again substitute C ⫽ 200 ⫺ A and D ⫽ 240 ⫺ B in all the STL inequalities: C ⫹ D ⱕ 200

1200 ⫺ A2 ⫹ 1240 ⫺ B2 ⱕ 200

Cⱖ0

Dⱖ0

200 ⫺ A ⱖ 0

240 ⫺ B ⱖ 0

200 ⱖ A

240 ⱖ B

STL inequalities substitute 200 ⫺ A for C, 240 ⫺ B for D

440 ⫺ A ⫺ B ⱕ 200 240 ⱕ A ⫹ B

simplify result

Combining the new STL constraints with those from KC produces the following system and solution. All points of intersection were read from the graph or located using the related system of equations. 400 B A ⫽ 200

u

A ⫹ B ⱕ 300 A ⫹ B ⱖ 240 A ⱕ 200 B ⱕ 240 Aⱖ0 Bⱖ0

300

B ⫽ 240

(60, 240) 200

100

Feasible region

(200, 100)

(200, 40) 100

200

A 300

400

To find the minimum cost, we check each vertex in the objective function.

Vertices

Objective Function T(A, B) ⴝ 296 ⴚ 0.2 A ⴚ 0.05B

(0, 240)

P10, 2402 ⫽ 296 ⫺ 0.2102 ⫺ 0.0512402 ⫽ $284

(60, 240)

P160, 2402 ⫽ 296 ⫺ 0.21602 ⫺ 0.0512402 ⫽ $272

(200, 100)

P1200, 1002 ⫽ 296 ⫺ 0.212002 ⫺ 0.0511002 ⫽ $251

(200, 40)

P1200, 402 ⫽ 296 ⫺ 0.212002 ⫺ 0.051402 ⫽ $254

The minimum cost occurs when A ⫽ 200 and B ⫽ 100, meaning the producer should ship the following quantities:

D. You’ve just seen how we can solve applications using linear programming

A S cases shipped from KC to Macon ⫽ 200 B S cases shipped from KC to Springfield ⫽ 100 C S cases shipped from STL to Macon ⫽ 0 D S cases shipped from STL to Springfield ⫽ 140 Now try Exercises 69 and 70 䊳

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9.3 EXERCISES 䊳



䊳

1. Any line y mx b drawn in the coordinate plane divides the plane into two regions called .

2. For the line y mx b drawn in the coordinate plane, solutions to y 7 mx b are found in the region the line.

3. The overlapping region of two or more linear inequalities in a system is called the region.

4. If a linear programming problem has a unique solution (x, y), it must be a of the feasible region.

5. Suppose two boundary lines in a system of linear inequalities intersect, but the point of intersection is not a vertex of the feasible region. Describe how this is possible.

6. Describe the conditions necessary for a linear programming problem to have multiple solutions. (Hint: Consider the diagram in Figure 9.35, and the slope of the line from the objective function.)


Determine whether the ordered pairs given are solutions.

7. 2x y 7 3; (0, 0), 13, 52, 13, 42, 13, 92

8. 3x y 7 5; (0, 0), 14, 12 , 11, 52, 11, 22

9. 4x 2y 8; (0, 0), 13, 52 , 13, 22 , 11, 12

10. 3x 5y 15; (0, 0), 13, 52, 11, 62, 17, 32

Solve the linear inequalities by shading the appropriate half plane. Verify your answer using a graphing calculator.

11. x 2y 6 8

12. x 3y 7 6

13. 2x 3y 9

14. 4x 5y 15

Solve the following inequalities using a graphing calculator. Your answer should include a screen shot or facsimile, and comments regarding a test point.

15. 3x 2y 8

16. 2x 5y 10

17. 4x 5y 20

18. 6x 3y 18

Determine whether the ordered pairs given are solutions to the accompanying system.

5y x 10 19. e ; 5y 2x 5 12, 12, 15, 42, 16, 22, 18, 2.22

8y 7x 56 20. • 3y 4x 12 y 4; 11, 52, 14, 62, 18, 52, 15, 32 Solve each system of inequalities by graphing the solution region. Verify the solution using a test point.

21. e

x 2y 1 2x y 2

22. e

x 5y 6 5 x 2y 1

23. e

3x y 7 4 x 7 2y

24. e

3x 2y y 4x 3

25. e

2x y 6 4 2y 7 3x 6

26. e

x 2y 6 7 2x y 7 5

27. e

x 7 3y 2 x 3y 6

28. e

2x 5y 6 15 3x 2y 7 6

29. e

5x 4y 20 x1y

30. e

10x 4y 20 5x 2y 7 1

31. e

0.2x 7 0.3y 1 0.3x 0.5y 0.6

32. e

x 7 0.4y 2.2 x 0.9y 1.2

33. •

3 x 2 4y 6x 12

3x 4y 7 12 2 34. • y 6 x 3

y

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2 x 3 35. μ 1 x 2

3 y1 4 2y 3

2 1 x y5 2 5 36. μ 5 x 2y 5 6

x y 4 37. • 2x y 4 x 0, y 0

2x y 5 38. • x 3y 6 x 0, y 0

yx3 39. • x 2y 4 x 0, y 0

4y 6 3x 12 40. • y x 1 x 0, y 0

2x 3y 18 41. • 2x y 10 x 0, y 0

8x 5y 40 42. • x y 7 x 0, y 0

Use a graphing calculator to find the solution region for each system of linear inequalities. Your answer should include a screen shot or facsimile, and the location of any points of intersection.

y 2x 6 8 yx 6 6 43. μ x0 y0

x 2y 6 10 xy 6 7 44. μ x0 y0

2x y 7 8 x 2y 6 7 45. μ x0 y0

y 2x 10 2y x 11 46. μ x0 y0

Use the equations given to write the system of linear inequalities represented by each graph.

47.

48.

y 5 4 3 2 1 54321 1 2 3 4 5

䊳

9–40


y 5 4 3 2 1

yx1

1 2 3 4 5 x

xy3

54321 1 2 3 4 5

yx1

1 2 3 4 5 x

49.

50.

y 5 4 3 2 1 54321 1 2 3 4 5

yx1

1 2 3 4 5 x

xy3

y 5 4 3 2 1 54321 1 2 3 4 5

yx1

1 2 3 4 5 x

xy3

Determine which of the ordered pairs given produces the maximum value of f (x, y).

51. f 1x, y2 12x 10y; (0, 0), (0, 8.5), (7, 0), (5, 3) 52. f 1x, y2 50x 45y; (0, 0), (0, 21), (15, 0), (7.5, 12.5)

Determine which of the ordered pairs given produces the minimum value of f (x, y).

53. f 1x, y2 8x 15y; (0, 20), (35, 0), (5, 15), (12, 11)

54. f 1x, y2 75x 80y; (0, 9), (10, 0), (4, 5), (5, 4) For Exercises 55 and 56, find the maximum value of the objective function f (x, y) ⴝ 8x ⴙ 5y and where this value occurs, given the constraints shown.

x 2y 6 3x y 8 55. μ x0 y0

2x y 7 x 2y 5 56. μ x0 y0

For Exercises 57 and 58, find the minimum value of the objective function f (x, y) ⴝ 36x ⴙ 40y and where this value occurs, given the constraints shown.

3x 2y 18 3x 4y 24 57. μ x0 y0

2x y 10 x 4y 3 58. μ x2 y0

xy3


Area Formulas

59. The area of a triangle is usually given as A 12 BH, where B and H represent the base and height, respectively. The area of a rectangle can be stated as A BH. If the base of both a triangle and rectangle is equal to 20 in., what are the possible values for H if the triangle must have an area greater than 50 in2 and the rectangle must have an area less than 200 in2?

Volume Formulas

60. The volume of a cone is V 13␲r 2h, where r is the radius of the base and h is the height. The volume of a cylinder is V ␲r 2h. If the radius of both a cone and cylinder is equal to 10 cm, what are the possible values for h if the cone must have a volume greater than 200 cm3 and the volume of the cylinder must be less than 850 cm3?

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877

APPLICATIONS

Write a system of linear inequalities that models the information given, then solve. Verify the solution region using a graphing calculator.

61. Gifts to grandchildren: Grandpa Augustus is considering how to divide a $50,000 gift between his two grandchildren, Julius and Anthony. After weighing their respective positions in life and family responsibilities, he decides he must bequeath at least $20,000 to Julius, but no more than $25,000 to Anthony. Determine the possible ways that Grandpa can divide the $50,000. 62. Guns versus butter: Every year, governments around the world have to make the decision as to how much of their revenue must be spent on national defense and domestic improvements (guns versus butter). Suppose total revenue for these two needs was $120 billion, and a government decides they need to spend at least $42 billion on butter and no more than $80 billion on defense. Determine the possible amounts that can go toward each need. Solve the following applications of linear programming.

63. Land/crop allocation: A farmer has 500 acres of land to plant corn and soybeans. During the last few years, market prices have been stable and the farmer anticipates a profit of $900 per acre on the corn harvest and $800 per acre on the soybeans. The farmer must take into account the time it takes to plant and harvest each crop, which is 3 hr/acre for corn and 2 hr/acre for soybeans. If the farmer has at most 1300 hr to plant, care for, and harvest each crop, how many acres of each crop should be planted in order to maximize profits? 64. Coffee blends: The owner of a coffee shop has decided to introduce two new blends of coffee in order to attract new customers — a Deluxe Blend and a Savory Blend. Each pound of the deluxe blend contains 30% Colombian and 20% Arabian coffee, while each pound of the savory blend contains 35% Colombian and 15% Arabian coffee (the remainder of each is made up of cheap and plentiful domestic varieties). The profit on the deluxe blend will be $1.25 per pound, while the profit on the savory blend will be $1.40 per pound. How many pounds of each should the owner make in order to maximize profit, if only 455 lb of Colombian coffee and 250 lb of Arabian coffee are currently available?

65. Manufacturing screws: A machine shop manufactures two types of screws — sheet metal screws and wood screws, using three different machines. Machine Moe can make a sheet metal screw in 20 sec and a wood screw in 5 sec. Machine Larry can make a sheet metal screw in 5 sec and a wood screw in 20 sec. Machine Curly, the newest machine (nyuk, nyuk) can make a sheet metal screw in 15 sec and a wood screw in 15 sec. (Shemp couldn’t get a job because he failed the math portion of the employment exam.) Each machine can operate for only 3 hr each day before shutting down for maintenance. If sheet metal screws sell for 10 cents and wood screws sell for 12 cents, how many of each type should the machines be programmed to make in order to maximize revenue? (Hint: Standardize time units.) 66. Hauling hazardous waste: A waste disposal company is contracted to haul away some hazardous waste material. A full container of liquid waste weighs 800 lb and has a volume of 20 ft3. A full container of solid waste weighs 600 lb and has a volume of 30 ft3. The trucks used can carry at most 10 tons (20,000 lb) and have a carrying volume of 800 ft3. If the trucking company makes $300 for disposing of liquid waste and $400 for disposing of solid waste, what is the maximum revenue per truck that can be generated? 67. Maximizing profit — food service: P. Barrett & Justin, Inc., is starting up a fast-food restaurant specializing in peanut butter and jelly sandwiches. Some of the peanut butter varieties are smooth, crunchy, reduced fat, and reduced sugar. The jellies will include those expected and common, as well as some exotic varieties such as kiwi and mango. Independent research has determined the two most popular sandwiches will be the traditional P&J (smooth peanut butter and grape jelly), and the Double-T (three slices of bread). A traditional P&J uses 2 oz of peanut butter and 3 oz of jelly. The Double-T uses 4 oz of peanut butter and 5 oz of jelly. The traditional sandwich will be priced at $2.00, and a Double-T at $3.50. If the restaurant has 250 oz of smooth peanut butter and 345 oz of grape jelly on hand for opening day, how many of each should they make and sell to maximize revenue?

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68. Maximizing profit — construction materials: Mooney and Sons produces and sells two varieties of concrete mixes. The mixes are packaged in 50-lb bags. Type A is appropriate for finish work, and contains 20 lb of cement and 30 lb of sand. Type B is appropriate for foundation and footing work, and contains 10 lb of cement and 20 lb of sand. The remaining weight comes from gravel aggregate. The profit on type A is $1.20/bag, while the profit on type B is $0.90/bag. How many bags of each should the company make to maximize profit, if 2750 lb of cement and 4500 lb of sand are currently available? 69. Minimizing transportation costs: Robert’s Las Vegas Tours needs to drive 375 people and 19,450 lb of luggage from Salt Lake City, Utah, to Las Vegas, Nevada, and can charter buses from two companies. The buses from company X carry 45 passengers and 2750 lb of luggage at a cost of $1250 per trip. Company Y offers buses that carry 60 passengers

䊳

and 2800 lb of luggage at a cost of $1350 per trip. How many buses should be chartered from each company in order for Robert to minimize the cost? 70. Minimizing shipping costs: An oil company is trying to minimize shipping costs from its two primary refineries in Tulsa, Oklahoma, and Houston, Texas. All orders within the region are shipped from one of these two refineries. An order for 220,000 gal comes in from a location in Colorado, and another for 250,000 gal from a location in Mississippi. The Tulsa refinery has 320,000 gal ready to ship, while the Houston refinery has 240,000 gal. The cost of transporting each gallon to Colorado is $0.05 from Tulsa and $0.075 from Houston. The cost of transporting each gallon to Mississippi is $0.06 from Tulsa and $0.065 from Houston. How many gallons should be distributed from each refinery to minimize the cost of filling both orders?


71. Graph the feasible region formed by the system x0 y0 μ . (a) How would you describe this region? y3 x3 (b) Select random points within the region or on any boundary line and evaluate the objective function f 1x, y2 4.5x 7.2y. At what point (x, y) will this function be maximized? (c) How does this relate to optimal solutions to a linear programing problem?

䊳

9–42


72. Find the maximum value of the objective function f 1x, y2 22x 15y given the constraints 2x 5y 24 3x 4y 29 μ x 6y 26 . x0 y0


73. (6.7) Given the point 13, 42 is on the terminal side of angle ␪ with ␪ in standard position, find a. cos ␪ b. csc ␪ c. cot ␪

75. (2.6) The resistance to current flow in copper wire varies directly as its length and inversely as the square of its diameter. A wire 8 m long with a 0.004-m diameter has a resistance of 1500 . Find the resistance in a wire of like material that is 2.7 m long with a 0.005-m diameter.

74. (4.6) Solve the rational inequality. Write your x2 7 0 answer in interval notation. 2 x 9

76. (7.4) Use a half-angle identity to find an exact 7␲ value for cos a b. 12

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Partial Fraction Decomposition


A. Set up a decomposition template to help rewrite a rational expression as the sum of its partial fractions B. Decompose a rational expression using convenient values C. Decompose a rational expression by equating coefficients and using a system of equations D. Apply partial fraction decomposition to a telescoping sum

One application of linear systems often seen in higher mathematics involves rewriting a rational expression as a sum of its partial fractions. While most often used as a prelude to the application of other mathematical techniques, practical applications of the process range from a study of chemical reactions to the analysis of thermodynamic experiments.

A. Setting Up a Decomposition Template P1x2

, where P and Q are polynomials Q1x2 and Q1x2 ⫽ 0. The addition of rational expressions is widely taught in courses prior to college algebra, and involves combining two rational expressions into a single term using a common denominator. For the decomposition of rational expressions, we seek to reverse this process. To begin, we make the following observations: Recall that a rational expression is one of the form

5 7 ⫹ , noting both terms are proper fractions (the degree x⫹2 x⫺3 of the numerator is less than the degree of the denominator) and have distinct linear denominators.

1. Consider the sum

51x ⫹ 22 71x ⫺ 32 7 5 ⫹ ⫹ ⫽ x⫹2 x⫺3 1x ⫹ 22 1x ⫺ 32 1x ⫺ 321x ⫹ 22 71x ⫺ 32 ⫹ 51x ⫹ 22 ⫽ 1x ⫹ 22 1x ⫺ 32 12x ⫺ 11 ⫽ 1x ⫹ 22 1x ⫺ 32

common denominator

combine numerators

result

Assuming we didn’t have the original sum to look at, reversing the process would require us to begin with the template A B 12x ⫺ 11 ⫽ ⫹ x⫹2 x⫺3 1x ⫹ 22 1x ⫺ 32 and solve for the constants A and B. We know the numerators must be constant, otherwise the fraction(s) would be improper while the original expression is not. 3 5 , again noting both terms are proper ⫹ 2 2. Consider the sum x⫺1 x ⫺ 2x ⫹ 1 fractions. 3 5 5 3 ⫽ ⫹ 2 ⫹ x⫺1 x⫺1 1x ⫺ 12 1x ⫺ 12 x ⫺ 2x ⫹ 1 31x ⫺ 12 5 ⫹ ⫽ 1x ⫺ 12 1x ⫺ 12 1x ⫺ 12 1x ⫺ 12 13x ⫺ 32 ⫹ 5 ⫽ 1x ⫺ 121x ⫺ 12 3x ⫹ 2 ⫽ 1x ⫺ 12 2

factor denominators

common denominator

combine numerators

result

Note that while the new denominator is the repeated factor 1x ⫺ 12 2, both 1x ⫺ 12 and 1x ⫺ 12 2 were denominators in the original sum. Assuming we didn’t know the original sum, reversing the process would require us to begin with the template

9–43

3x ⫹ 2 A B ⫽ ⫹ 2 x⫺1 1x ⫺ 12 1x ⫺ 12 2

879

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and solve for the constants A and B. As before, the numerator of the first term must be constant. While the second term would still be a proper fraction if the numerator were linear (degree 1), the denominator is a repeated linear factor and using a single constant in the numerator of all such fractions will ensure we obtain unique values for A and B (see Exercise 58). Note that for any repeated linear factor 1x ⫹ a2 n in the original denominator, terms of the form A1 A2 A3 An⫺1 An ⫹ ⫹ ⫹p⫹ ⫹ must appear in 2 3 n⫺1 x⫹a 1x ⫹ a2 n 1x ⫹ a2 1x ⫹ a2 1x ⫹ a2 the decomposition template, although some of these numerators may turn out to be zero. EXAMPLE 1

䊳

Setting Up the Decomposition Template for Linear Factors Write the decomposition template for x⫺8 x⫹1 a. 2 b. 2 2x ⫹ 5x ⫹ 3 x ⫺ 6x ⫹ 9

Solution

䊳

x⫺8 . Since the denominator 12x ⫹ 32 1x ⫹ 12 consists of two distinct linear factors, the decomposition template is:

a. Factoring the denominator gives

A x⫺8 B ⫽ ⫹ 2x ⫹ 3 x⫹1 12x ⫹ 321x ⫹ 12

decomposition template

x⫹1 , where the denominator 1x ⫺ 32 2 consists of a repeated linear factor. Using our previous observations, the decomposition template would be:

b. After factoring the denominator we have

x⫹1 A B ⫽ ⫹ 2 x⫺3 1x ⫺ 32 1x ⫺ 32 2


Now try Exercises 7 through 16 䊳 When both distinct and repeated linear factors are present in the denominator, the decomposition template maintains the elements illustrated above. Each distinct linear factor appearing in the denominator, and all powers of a repeated linear factor, will have a constant numerator. EXAMPLE 2

䊳

Setting Up the Decomposition Template for Repeated Linear Factors Write the decomposition template for

Solution

䊳

x2 ⫺ 4x ⫺ 15 . x3 ⫺ 2x2 ⫹ x

x2 ⫺ 4x ⫺ 15 x2 ⫺ 4x ⫺ 15 or after factoring x1x2 ⫺ 2x ⫹ 12 x1x ⫺ 12 2 completely. With a distinct linear factor of x, and the repeated linear factor 1x ⫺ 12 2, the decomposition template becomes: Factoring the denominator gives

A C x2 ⫺ 4x ⫺ 15 B ⫽ ⫹ ⫹ 2 x x⫺1 x1x ⫺ 12 1x ⫺ 12 2


Now try Exercises 17 through 20 䊳 Returning to our observations:

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Section 9.4 Partial Fraction Decomposition

881

4 2x ⫹ 3 , noting the denominator of the first term is linear, ⫹ 2 x x ⫹1 while the denominator of the second is an irreducible quadratic.

3. Consider the sum

41x2 ⫹ 12 12x ⫹ 32x 4 2x ⫹ 3 ⫽ ⫹ 2 ⫹ 2 2 x x ⫹1 x1x ⫹ 12 1x ⫹ 12x 2 14x ⫹ 42 ⫹ 12x2 ⫹ 3x2 ⫽ x1x2 ⫹ 12 2 6x ⫹ 3x ⫹ 4 ⫽ x1x2 ⫹ 12

find common denominator

combine numerators

result

Here, reversing the process would require us to begin with the template A Bx ⫹ C 6x2 ⫹ 3x ⫹ 4 ⫽ ⫹ 2 2 x x1x ⫹ 12 x ⫹1 allowing that the numerator of the second term might be linear since it is quadratic but not a repeated linear factor, and noting the fraction would still be proper in cases where B ⫽ 0. x⫺2 1 ⫹ 2 , where we note the denominator of 4. Finally, consider the sum 2 x ⫹3 1x ⫹ 32 2 the first term is an irreducible quadratic, with the denominator of the second term being the same factor with multiplicity two. 11x2 ⫹ 32 x⫺2 x⫺2 1 ⫹ 2 ⫹ 2 ⫽ 2 2 2 2 x ⫹3 1x ⫹ 32 1x ⫹ 32 1x ⫹ 32 1x ⫹ 321x2 ⫹ 32 2 1x ⫹ 32 ⫹ 1x ⫺ 22 ⫽ 1x2 ⫹ 321x2 ⫹ 32 x2 ⫹ x ⫹ 1 ⫽ 1x2 ⫹ 32 2

WORTHY OF NOTE Note that the second term in the decomposition template would still be a proper fraction if the numerator were quadratic or cubic, but since the denominator is a repeated quadratic factor, using only a linear form ensures we obtain unique values for all coefficients.

EXAMPLE 3

䊳

common denominator

combine numerators

result after simplifying

Reversing the process would require us to begin with the template Cx ⫹ D Ax ⫹ B x2 ⫹ x ⫹ 1 ⫹ 2 ⫽ 2 2 2 1x ⫹ 32 x ⫹3 1x ⫹ 32 2 allowing that the numerator of either term might be nonconstant for the reasons in observation 3. Similar to our reasoning in observation 2, all powers of a repeated quadratic factor must be present in the decomposition template. Setting Up the Decomposition Template for Quadratic Factors Write the decomposition template for x2 x2 ⫹ 10x ⫹ 1 a. b. 1x ⫹ 121x2 ⫹ 3x ⫹ 42 1x2 ⫹ 22 3

Solution

䊳

a. With the denominator having one distinct linear factor and one irreducible quadratic factor, the decomposition template would be: A x2 ⫹ 10x ⫹ 1 Bx ⫹ C ⫽ ⫹ 2 2 x⫹1 1x ⫹ 121x ⫹ 3x ⫹ 42 x ⫹ 3x ⫹ 4


b. The denominator consists of a repeated quadratic factor. Using our previous observations, the decomposition template would be: Cx ⫹ D x2 Ax ⫹ B Ex ⫹ F ⫹ 2 ⫽ 2 ⫹ 2 2 3 2 1x ⫹ 22 x ⫹2 1x ⫹ 22 1x ⫹ 22 3



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When both distinct and repeated factors are present in the denominator, the decomposition template maintains the essential elements determined by observations 1 through 4. Using these observations, we can formulate a general approach to the decomposition template. The Decomposition Template For the rational expression

A. You’ve just seen how we can set up a decomposition template

P1x2 Q1x2

and constants A, B, C, D, ... ,

1. If the degree of P is greater than or equal to the degree of Q, find the quotient and remainder using polynomial division. Only the remainder portion need be decomposed into partial fractions. 2. Factor Q completely into linear factors and irreducible quadratic factors. 3. For the linear factors, each distinct linear factor and each power of a repeated linear factor must appear in the decomposition template and have a constant numerator. 4. For the irreducible quadratic factors, each distinct quadratic factor and each power of a repeated quadratic factor must appear in the decomposition template and have a linear numerator.

B. Decomposition Using Convenient Values Once the decomposition template is obtained, we multiply both sides of the equation by the factored form of the original denominator and simplify. The resulting equation is an identity (a true statement for all real numbers x), and in many cases, all that’s required is a choice of convenient values for x to identify the constants A, B, C, and so on. EXAMPLE 4

䊳

Decomposing a Rational Expression Using Convenient Values 4x ⫹ 11 into partial fractions. Use the TABLE x ⫹ 7x ⫹ 10 feature of a calculator to check your answer. Decompose the expression

Solution

䊳

2

4x ⫹ 11 , and we note there are two distinct 1x ⫹ 52 1x ⫹ 22 linear factors in the denominator. The decomposition template will be Factoring the denominator gives

A B 4x ⫹ 11 ⫽ ⫹ x⫹5 x⫹2 1x ⫹ 521x ⫹ 22


Multiplying both sides by 1x ⫹ 521x ⫹ 22 clears all denominators and yields 4x ⫹ 11 ⫽ A1x ⫹ 22 ⫹ B1x ⫹ 52

clear denominators

Since the equation must be true for all x, using x ⫽ ⫺5 will conveniently eliminate the term with B, and enable us to solve for A directly: 41⫺52 ⫹ 11 ⫽ A1⫺5 ⫹ 22 ⫹ B1⫺5 ⫹ 52 ⫺20 ⫹ 11 ⫽ ⫺3A ⫹ B102 ⫺9 ⫽ ⫺3A 3⫽A

substitute ⫺5 for x simplify term with B is eliminated solve for A

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883

To find B, we repeat this procedure, using a value that conveniently eliminates the term with A, namely x ⫽ ⫺2. 4x ⫹ 11 ⫽ A1x ⫹ 22 ⫹ B1x ⫹ 52 41⫺22 ⫹ 11 ⫽ A1⫺2 ⫹ 22 ⫹ B1⫺2 ⫹ 52 ⫺8 ⫹ 11 ⫽ A102 ⫹ 3B 3 ⫽ 3B 1⫽B

original equation substitute ⫺2 for x simplify term with A is eliminated solve for B

With A ⫽ 3 and B ⫽ 1, the complete decomposition is: 3 1 4x ⫹ 11 ⫽ ⫹ x⫹5 x⫹2 1x ⫹ 52 1x ⫹ 22

Check

䊳

In the

Y=

screen, enter the original expression as Y1 ⫽

4X ⫹ 11 and the X ⫹ 7X ⫹ 10 2

1 3 (see Figure 9.36). Next, set the TABLE to ⫹ X⫹5 X⫹2 start at –3 with a step size of 1 and press 2nd GRAPH to produce the table shown in Figure 9.37. Scrolling through the table, note the Y1 and Y2 outputs for any given X are the same, providing strong evidence that Y1 ⫽ Y2. For a rigorous check that 3 1 4x ⫹ 11 ⫽ ⫹ , simply add the rational expressions on the right. 2 x⫹5 x⫹2 x ⫹ 7x ⫹ 10 decomposition as Y2 ⫽

Figure 9.36

Figure 9.37

WORTHY OF NOTE Keep in mind the zeroes of any denominator in rational functions are excluded from the domain. The calculator reminds us of this in Example 4 by indicating that X ⫽⫺2 and X ⫽⫺5 result in an “ERROR,” for both Y1 and Y2.


EXAMPLE 5

䊳

Decomposing a Rational Expression Using Convenient Values Decompose the expression

Solution

䊳

9 into partial fractions. 1x ⫹ 52 1x ⫹ 7x ⫹ 102 2

9 9 or in 1x ⫹ 521x ⫹ 22 1x ⫹ 52 1x ⫹ 221x ⫹ 52 2 simplified form. With one distinct linear factor and a repeated linear factor, the A B C 9 ⫽ ⫹ ⫹ decomposition template will be . 2 x⫹2 x⫹5 1x ⫹ 221x ⫹ 52 1x ⫹ 52 2 Multiplying both sides by 1x ⫹ 22 1x ⫹ 52 2 clears all denominators and yields Factoring the denominator gives

9 ⫽ A1x ⫹ 52 2 ⫹ B1x ⫹ 22 1x ⫹ 52 ⫹ C1x ⫹ 22

Using x ⫽ ⫺5 will conveniently eliminate the terms with A and B, enabling us to solve for C directly: 9 ⫽ A1⫺5 ⫹ 52 2 ⫹ B1⫺5 ⫹ 22 1⫺5 ⫹ 52 ⫹ C1⫺5 ⫹ 22 9 ⫽ A102 ⫹ B1⫺32 102 ⫺ 3C 9 ⫽ ⫺3C ⫺3 ⫽ C

substitute ⫺5 for x simplify terms with A and B are eliminated solve for C

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Using x ⫽ ⫺2 will conveniently eliminate the terms with B and C, enabling us to solve for A: 9 ⫽ A1x ⫹ 52 2 ⫹ B1x ⫹ 221x ⫹ 52 ⫹ C1x ⫹ 22 9 ⫽ A1⫺2 ⫹ 52 2 ⫹ B1⫺2 ⫹ 22 1⫺2 ⫹ 52 ⫹ C1⫺2 ⫹ 22 9 ⫽ A132 2 ⫹ B102132 ⫹ C102 9 ⫽ 9A 1⫽A

original equation substitute ⫺2 for x simplify terms with B and C are eliminated solve for A

To find the value of B, we can substitute A ⫽ 1 and C ⫽ ⫺3 into the previous equation, and any value of x that does not eliminate B. For efficiency’s sake, we often elect to use x ⫽ 0 or x ⫽ 1 for this purpose (if possible). 9 ⫽ A1x ⫹ 52 2 ⫹ B1x ⫹ 221x ⫹ 52 ⫹ C1x ⫹ 22 9 ⫽ 110 ⫹ 52 2 ⫹ B10 ⫹ 22 10 ⫹ 52 ⫺ 310 ⫹ 22 9 ⫽ 25 ⫹ 10B ⫺ 6 ⫺10 ⫽ 10B ⫺1 ⫽ B

original equation substitute 1 for A, ⫺3 for C, 0 for x simplify

solve for B

With A ⫽ 1, B ⫽ ⫺1, and C ⫽ ⫺3, the complete decomposition is:

Figure 9.38

9 1 ⫺1 ⫺3 ⫹ ⫹ ⫽ , x⫹2 x⫹5 1x ⫹ 221x ⫹ 52 2 1x ⫹ 52 2 1 1 3 ⫺ ⫺ . Figure 9.38 shows a x⫹2 x⫹5 1x ⫹ 52 2 TABLE check, where the original expression and decomposition have been entered into Y1 and Y2, respectively. which can also be written as


EXAMPLE 6

䊳

Decomposing a Rational Expression Using Convenient Values Decompose the expression

Solution

䊳

3x ⫹ 11 into partial fractions. x3 ⫺ 3x2 ⫹ x ⫺ 3

After inspection, we note the denominator can be factored by grouping, resulting 3x ⫹ 11 in the expression . With one distinct linear factor and one 1x ⫺ 321x2 ⫹ 12 irreducible quadratic factor, the decomposition template will be A Bx ⫹ C 3x ⫹ 11 ⫽ . ⫹ 2 2 x⫺3 1x ⫺ 321x ⫹ 12 x ⫹1

Multiplying both sides by 1x ⫺ 32 1x2 ⫹ 12 yields

3x ⫹ 11 ⫽ A1x2 ⫹ 12 ⫹ 1Bx ⫹ C2 1x ⫺ 32

clear denominators

Using x ⫽ 3 will conveniently eliminate the term with B and C, giving 3132 ⫹ 11 ⫽ A132 ⫹ 12 ⫹ 1B 33 4 ⫹ C2 13 ⫺ 32 20 ⫽ 10A ⫹ 1B 33 4 ⫹ C2102 20 ⫽ 10A 2⫽A

substitute 3 for x simplify term with B and C is eliminated solve for A

Noting x ⫽ 0 will conveniently eliminate the term with B, we substitute 0 for x and 2 for A in order to solve for C: 3x ⫹ 11 ⫽ A1x2 ⫹ 12 ⫹ 1Bx ⫹ C2 1x ⫺ 32 3102 ⫹ 11 ⫽ 2102 ⫹ 12 ⫹ 1B 30 4 ⫹ C2 10 ⫺ 32

original equation substitute 2 for A and 0 for x

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11 ⫽ 2 ⫹ C1⫺32 9 ⫽ ⫺3C ⫺3 ⫽ C

885

simplify solve for C result

To find the value of B, we substitute 2 for A, ⫺3 for C, and any value of x that does not eliminate B. Here we chose x ⫽ 1. 3x ⫹ 11 ⫽ A1x2 ⫹ 12 ⫹ 1Bx ⫹ C2 1x ⫺ 32 3112 ⫹ 11 ⫽ 2112 ⫹ 12 ⫹ 1B 31 4 ⫹ 3⫺34 2 11 ⫺ 32 14 ⫽ 2122 ⫹ 1B ⫺ 321⫺22 14 ⫽ 4 ⫺ 2B ⫹ 6 4 ⫽ ⫺2B ⫺2 ⫽ B

original equation substitute 2 for A, ⫺3 for C, and 1 for x simplify distribute solve for B result

With A ⫽ 2, B ⫽ ⫺2, and C ⫽ ⫺3, the complete decomposition is: 2 3x ⫹ 11 ⫺2x ⫺ 3 ⫽ . ⫹ 2 2 x⫺3 1x ⫺ 32 1x ⫹ 12 x ⫹1 The result can actually be written with fewer negative signs, as 2 2x ⫹ 3 . Check the result by combining these fractions. ⫺ 2 x⫺3 x ⫹1

B. You’ve just seen how we can decompose a rational expression using convenient values


C. Decomposition Using a System of Equations As an alternative to using convenient values, a system of equations can be set up by multiplying out the right-hand side (after clearing fractions) and equating coefficients of the terms with like degrees. Here we’ll re-solve Example 6 using this method. EXAMPLE 7

䊳

Decomposing a Rational Expression Using a System of Equations 3x ⫹ 11 into partial fractions by equating x ⫺ 3x2 ⫹ x ⫺ 3 coefficients of like degree terms and using a system. Decompose the expression

Solution

䊳

3

From Example 6 we obtain A Bx ⫹ C 3x ⫹ 11 ⫽ ⫹ 2 2 x⫺3 1x ⫺ 321x ⫹ 12 x ⫹1


Multiplying both sides by 1x ⫺ 32 1x2 ⫹ 12 yields

3x ⫹ 11 ⫽ A1x2 ⫹ 12 ⫹ 1Bx ⫹ C21x ⫺ 32 ⫽ Ax2 ⫹ A ⫹ Bx2 ⫺ 3Bx ⫹ Cx ⫺ 3C ⫽ 1A ⫹ B2x2 ⫹ 1⫺3B ⫹ C2x ⫹ 1A ⫺ 3C2

clear denominators distribute/FOIL collect like terms

By comparing the like terms on the left and right, we find A ⫹ B ⫽ 0, A⫹B ⫽0 ⫺3B ⫹ C ⫽ 3, and A ⫺ 3C ⫽ 11, resulting in the system • ⫺3B ⫹ C ⫽ 3 . A ⫺3C ⫽ 11 ⫺3B ⫹ C ⫽ 3 , Using A ⫽ ⫺B (from R1) in R3 results in the 2 ⫻ 2 subsystem e ⫺B ⫺ 3C ⫽ 11 and 3R1 ⫹ R2 of this system yields ⫺10B ⫽ 20, giving B ⫽ ⫺2 as before. Back-substitution again verifies C ⫽ ⫺3 and A ⫽ 2. Now try Exercises 39 and 40 䊳

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In some cases, there are no “convenient values” to use and a system of equations is our best and only approach. More often than not, this happens when one or more of the denominators are irreducible quadratic factors. EXAMPLE 8

䊳

Decomposing a Rational Expression Using a System of Equations 5x2 ⫹ 2x ⫹ 12 into partial fractions by equating x4 ⫹ 6x2 ⫹ 9 coefficients of like degree terms and using a system. Decompose the expression

Solution

䊳

By inspection or using a u-substitution, we find the denominator factors into a perfect square: x4 ⫹ 6x2 ⫹ 9 ⫽ 1x2 ⫹ 32 2. The decomposition template is then Cx ⫹ D Ax ⫹ B 5x2 ⫹ 2x ⫹ 12 ⫹ 2 ⫽ 2 2 2 1x ⫹ 32 x ⫹3 1x ⫹ 32 2

After clearing fractions we obtain 5x2 ⫹ 2x ⫹ 12 ⫽ 1Ax ⫹ B21x2 ⫹ 32 ⫹ Cx ⫹ D. The only “candidate” for a convenient value is x ⫽ 0, but this still leaves the unknowns B and D. Instead, we multiply out the right-hand side and equate coefficients of like terms as before. 5x2 ⫹ 2x ⫹ 12 ⫽ 1Ax ⫹ B2 1x2 ⫹ 32 ⫹ Cx ⫹ D ⫽ Ax3 ⫹ 3Ax ⫹ Bx2 ⫹ 3B ⫹ Cx ⫹ D ⫽ Ax3 ⫹ Bx2 ⫹ 13A ⫹ C2x ⫹ 13B ⫹ D2

By equating coefficients we find A ⫽ 0, since the left-hand side has no cubic term, and B ⫽ 5 by direct comparison. This yields the system A⫽0 B⫽5 μ 3A ⫹ C ⫽ 2 3B ⫹ D ⫽ 12 With A ⫽ 0 the third equation shows C ⫽ 2, and by substituting 5 for B in the fourth equation we find that D ⫽ ⫺3. The final form of the decomposition is 2x ⫺ 3 5 5x2 ⫹ 2x ⫹ 12 ⫹ 2 ⫽ 2 2 2 1x ⫹ 32 x ⫹3 1x ⫹ 32 2

C. You’ve just seen how we can decompose a rational expression by equating coefficients and using a system of equations


D. Partial Fractions and Telescoping Sums From movies or the popular media (Pirates of the Caribbean, Dances With Wolves, etc.), you might be aware that the telescopes of old were retractable. Since they were constructed as a series of nested tubes, you could compress the length with the lengths of the interior tubes being “negated” (see figure). In a similar fashion, some very extensive sums, called telescoping sums, can be rewritten in a more “compressed” form, where the interior terms are likewise negated and the resulting sum easily computed. EXAMPLE 9

䊳

Using Decomposition to Evaluate a Telescoping Sum 3 x ⫹x a. Evaluate the function for x ⫽ 1, 2, 3, and 4, then find the sum of these four terms. b. Decompose the expression into partial fractions.

For t1x2 ⫽

2

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c. Evaluate the decomposed form for x ⫽ 1, 2, 3, and 4, leaving each result in unsimplified form. d. Rewrite the sum from part (a) using the decomposed form of each term, and see if you can detect a pattern that enables you to compute the original sum more efficiently. 3 3 ⫽ e. Note 2 and use the pattern noted in part (d) to find the 1x2 1x ⫹ 12 x ⫹x 3 3 3 3 following sum: . ⫹ # ⫹ # ⫹p⫹ # 1 2 2 3 3 4 24 # 25 Use a calculator to check your result.

Solution

䊳

a.

3 3 3 3 1 1 3 3 ⫹ ⫹ ⫹ ⫽ ⫹ ⫹ ⫹ 1⫹1 4⫹2 9⫹3 16 ⫹ 4 2 2 4 20 10 5 3 30 ⫹ ⫹ ⫹ ⫽ 20 20 20 20 48 12 ⫽ ⫽ 20 5

b. The denominator has two distinct linear factors and the decomposition template will be A B 3 ⫽ ⫹ x x⫹1 x1x ⫹ 12


Multiplying both sides by x1x ⫹ 12 clears all denominators and gives 3 ⫽ A1x ⫹ 12 ⫹ Bx

clear denominators

Since the equation must be true for all x, using x ⫽ ⫺1 will eliminate the term with A, and enable us to solve for B directly: 3 ⫽ A1⫺1 ⫹ 12 ⫹ B1⫺12 3 ⫽ ⫺B ⫺3 ⫽ B

substitute ⫺1 for x simplify, term with A is eliminated solve for B

Repeat this procedure using x ⫽ 1 to solve for A. 3 ⫽ A1x ⫹ 12 ⫹ Bx 3 ⫽ A11 ⫹ 12 ⫺ 3112 6 ⫽ 2A 3⫽A

from template substitute ⫺3 for B, 1 for x add 3 to both sides result

With A ⫽ 3 and B ⫽ ⫺3, the decomposition is 3 3 3 ⫽ ⫺ x x⫹1 x1x ⫹ 12 c. For t1x2 ⫽ 3 3 ⫺ 1 1⫹1 3 ⫽3⫺ 2

t112 ⫽

decomposed form

3 3 , we have ⫺ x x⫹1 3 3 ⫺ 2 2⫹1 3 3 ⫽ ⫺ 2 3

t122 ⫽

3 3 ⫺ 3 3⫹1 3 3 ⫽ ⫺ 3 4

t132 ⫽

3 3 ⫺ 4 4⫹1 3 3 ⫽ ⫺ 4 5

t142 ⫽

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d. Replacing each term in the given sum by the equivalent term in decomposed form yields 3 3 3 3 3 3 3 3 3 3 3 3 ⫹ ⫹ ⫹ ⫽a ⫺ b⫹a ⫺ b⫹a ⫺ b⫹a ⫺ b 1⫹1 4⫹2 9⫹3 16 ⫹ 4 1 2 2 3 3 4 4 5 and we note that all interior terms will cancel (add to zero) leaving only the 3 15 3 12 3 ⫺ ⫽ . first and last terms of the sum. The result is ⫺ ⫽ 1 5 5 5 5 e. From the pattern in part (d) we note that regardless of the number of terms in the sum, all interior terms will cancel (add to zero) leaving only the first and last terms. 3 1#2

⫹

3

3

⫹ ⫹ 2#3 3#4

p⫹

⫽a ⫺ b⫹a ⫺ b⫹a ⫺ b⫹ 24 # 25 1 2 2 3 3 4 3

3

3

3

3

3

3

p⫹a3 ⫺ 3b 24 25

3 75 3 3 ⫺ ⫽ ⫺ 1 25 25 25 72 ⫽ 25

⫽

Check

䊳

To check this result using a calculator, we begin by clearing out lists L1, L2, and L3 and entering the integers 1 through 24 into L1. These values will serve as the first factors of each denominator in the given sum. Next, move the cursor onto ) + 1 the L2 column header and use the keystrokes 2nd 1 (L1⫹1) to fill L2 with the second factors of each denominator. Now we can use L3 to generate each individual term in the sum, as shown in Figure 9.39. At this point, pressing will cause these terms appear and we QUIT to the home screen. From here, use the calculator to sum the L3 terms with the keystrokes 2nd STAT ) (LIST) 5:sum( 2nd 3 . As Figure 9.40 shows, this sum is equal 72 to and the result from part (e) checks. 25 ENTER

ENTER

ENTER

Figure 9.39

Figure 9.40


D. You’ve just seen how we can apply partial fraction decomposition to a telescoping sum

As a final note, if the degree of the numerator is greater than the degree of the denominator in the original expression, divide using long division and apply the methods above to the rational remainder. For instance, you can check that 2x ⫺ 7 3x3 ⫹ 6x2 ⫹ 5x ⫺ 7 ⫽ 3x ⫹ , and decomposing the remainder expression 2 x ⫹ 2x ⫹ 1 1x ⫹ 12 2 9 2 ⫺ . gives a final result of 3x ⫹ x⫹1 1x ⫹ 12 2

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889

9.4 EXERCISES 䊳



䊳

1. In order to rewrite a rational expression as the sum of its partial fractions, we must set up a decomposition .

2. Before beginning the process of partial fraction decomposition, the rational expression must be . If not, use polynomial division.

3. If the denominator of a rational expression contains a factor, each power of the factor must appear in the decomposition template and have a constant numerator. 8x ⫺ 3 5. Discuss/Explain the process of writing 2 as a x ⫺x sum of partial fractions.

4. If the denominator of a rational expression contains a distinct irreducible quadratic factor, it must appear in the decomposition template with a numerator. 6. Discuss/Explain the first steps of the process of x2 writing 2 as a sum of partial fractions. x ⫺ 7x ⫹ 12


The exercises below are designed solely to reinforce the various possibilities for decomposing a rational expression. All are proper fractions whose denominators are completely factored. Write the decomposition template, but do not solve.

7. 9. 11. 13. 15. 17. 19. 21. 23.

3x ⫹ 2 1x ⫹ 32 1x ⫺ 22 2x ⫹ 5 1x ⫺ 12 2

8. 10.

⫺4x ⫹ 1 1x ⫺ 22 1x ⫺ 52 x⫺7 1x ⫹ 32 2

3x2 ⫺ 2x ⫹ 5 ⫺2x2 ⫹ 3x ⫺ 4 12. 1x ⫺ 12 1x ⫹ 221x ⫺ 32 1x ⫹ 32 1x ⫹ 121x ⫺ 22

x2 ⫹ 5 x1x ⫺ 32 1x ⫹ 12 x2 ⫹ 2x ⫺ 4 1x ⫺ 52 3 x2 ⫹ x ⫺ 1 x2 1x ⫹ 22

14. 16. 18.

x3 ⫹ 2x ⫺ 5 x2 1x ⫺ 52 2

20.

x3 ⫹ 3x ⫺ 2 1x ⫹ 12 1x2 ⫹ 22 2

24.

x2 ⫺ 7 1x ⫹ 42 1x ⫺ 22x

x2 ⫹ 2x ⫹ 3 1x ⫹ 42 3

x2 ⫺ 3x ⫹ 5 1x ⫺ 32 1x ⫹ 22 2

3x2 ⫺ 5 x 12x ⫹ 12 2 2

2x2 ⫹ 3 7x2 ⫹ 3x ⫺ 1 22. 1x ⫺ 32 1x2 ⫹ 5x ⫹ 72 1x ⫹ 12 1x2 ⫹ 2x ⫹ 52 2x3 ⫹ 3x2 ⫺ 4x ⫹ 1 x1x2 ⫹ 32 2

Decompose each rational expression into partial fractions using convenient values. Use the TABLE feature of a calculator to verify your results.

25.

2x ⫺ 27 2x ⫹ x ⫺ 15

26.

⫺11x ⫹ 6 5x ⫺ 4x ⫺ 12

27.

8x2 ⫺ 3x ⫺ 7 x3 ⫺ x

28.

x2 ⫹ 24x ⫺ 12 x3 ⫺ 4x

29.

3x2 ⫹ 7x ⫺ 1 x3 ⫹ 2x2 ⫹ x

30.

⫺2x2 ⫺ 7x ⫹ 28 x3 ⫺ 4x2 ⫹ 4x

31.

3x3 ⫹ 3x2 ⫹ 3x ⫹ 5 x4 ⫹ 3x2 ⫹ 2

32.

2x3 ⫺ 2x2 ⫹ 6x x4 ⫹ 4x2 ⫹ 3

33.

6x2 ⫹ x ⫹ 13 x3 ⫹ 2x2 ⫹ 3x ⫹ 6

34.

3x2 ⫹ 4x ⫺ 1 x3 ⫺ 1

35.

x4 ⫺ 3x2 ⫺ 2x ⫹ 1 x5 ⫹ 2x3 ⫹ x

36.

⫺3x4 ⫺ 11x2 ⫹ x ⫺ 12 x5 ⫺ 4x3 ⫹ 4x

37.

3x3 ⫹ 2x2 ⫹ 7x ⫹ 3 x4 ⫹ x3 ⫹ 3x2

38.

2x3 ⫺ x ⫹ 6 x ⫺ 2x3 ⫹ 3x2

39.

3x2 ⫹ 10x ⫹ 4 8 ⫺ x3

40.

2x2 ⫺ 14x ⫺ 7 x3 ⫺ 2x2 ⫹ 5x ⫺ 10

2

2

4

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When the denominator of the rational expression contains repeated factors, using a system of equations sometimes offers a more efficient way to find the needed coefficients. Decompose each rational expression into partial fractions by equating coefficients and using a system of equations.

41.

5x ⫹ 13 2 x ⫹ 6x ⫹ 9

42.

14 ⫺ 3x 2 x ⫺ 8x ⫹ 16

43.

2x3 ⫹ x2 ⫹ 5x ⫹ 1 x4 ⫹ 2x2 ⫹ 1

44.

x3 ⫹ 5x2 ⫹ 6x ⫹ 37 x4 ⫹ 14x2 ⫹ 49

45.

2x2 ⫺ 4x ⫹ 5 x3 ⫺ 3x2 ⫹ 3x ⫺ 1

46.

5x2 ⫹ 20x ⫹ 21 x3 ⫹ 6x2 ⫹ 12x ⫹ 8

䊳


Logistic equations and population size: Logistic equations are often used to model the growth of various populations. Initially growth is very near to exponential, but then due to certain limiting factors (food, limited resources, space, etc.), growth slows and reaches a limit called the carrying capacity c. In the process of solving the logistic equation for P(t), we often need to decompose the expression shown into partial fractions, where P(0) represents the initial population. Decompose the expression for the following values given.

47. P102 ⫽ 100, c ⫽ 10 䊳

48. P102 ⫽ 80, c ⫽ 20

49. P102 ⫽ 10, c ⫽ 100

1 P aP102 ⫺

P 102 c

Pb

50. P102 ⫽ 20, c ⫽ 80

APPLICATIONS

Telescoping sums: For each function given, use the method illustrated in Example 9 to find a pattern that enables you to compute the sum shown quickly.

1 ; x ⫹x 1 1 1 1 ⫹ # ⫹ # ⫹p⫹ 1#2 2 3 3 4 49 # 50

51. t1x2 ⫽

2

2 ; x ⫹x 2 2 2 2 ⫹ # ⫹ # ⫹p⫹ # 1 2 2 3 3 4 19 # 20

52. t1x2 ⫽

䊳

2

1 ; 12x ⫺ 1212x ⫹ 12 1 1 1 1 ⫹ # ⫹ # ⫹p⫹ # 1 3 3 5 5 7 123 # 125

53. t1x2 ⫽

2x ⫹ 1 ; 1x ⫹ 12 1x2 ⫹ 2x ⫹ 22 1 3 5 21 ⫹ # ⫹ # ⫹p⫹ 1#2 2 5 5 10 101 # 122

54. t1x2 ⫽

2


Decompose the following expressions into partial fractions.

55. 56.

ln x ⫹ 2 1ln x ⫺ 22 1ln x ⫺ 12 2

1 ⫹ ex 1ex ⫺ 32 1e2x ⫺ 2ex ⫹ 12

57. As written, the rational expression x⫹2 given cannot be decomposed using 1x ⫺ 12 11 ⫺ x2 the standard template (two distinct linear factors). Try to apply this template and see what happens. What can be done to decompose the expression?

58. Try to decompose the rational expression 3x ⫹ 1 using the decomposition template x2 A Bx ⫹ C (note the second rational expression ⫹ x x2 in this template is a proper fraction). (a) How does this affect the decomposition process? Is the decomposition unique? (b) Next, use the standard template as outlined in this section. What do you notice?

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59. (4.1) Use polynomial division to rewrite r1x2 2x3 ⫺ 3x2 ⫹ 13x ⫺ 5 in the form q1x2 ⫹ . 2 d1x2 x ⫺x⫹6 60. (4.1) Use the remainder theorem to find f (3) for f 1x2 ⫽ 2x3 ⫺ x2 ⫹ 10. 61. (7.2) Verify that

62. (7.5) Evaluate the following expressions: 23 bd a. cos c cos ⫺1a⫺ 2 ␲ b. cos ⫺1 c cosa⫺ b d 6

cos3␪ ⫽ cot ␪ ⫺ cos ␪ sin ␪ is an sin ␪

identity.

MID-CHAPTER CHECK 1. Solve by graphing the system on a graphing calculator. State whether the system is consistent, inconsistent, or dependent. e

x ⫺ 3y ⫽ ⫺2 2x ⫹ y ⫽ 3

2. Solve the system using elimination. State whether the system is consistent, inconsistent, or dependent: x ⫺ 3y ⫽ ⫺4 e 2x ⫹ y ⫽ 13 3. Solve using a system of linear equations and any method you choose. How many ounces of a 40% acid should be mixed with 10 oz of a 64% acid to obtain a 48% acid solution? 4. Determine whether the ordered triple is a solution to the system. If not, identify which equation(s) are not satisfied. 5x ⫹ 2y ⫺ 4z ⫽ 22 • 2x ⫺ 3y ⫹ z ⫽ ⫺1 3x ⫺ 6y ⫹ z ⫽ 2

12, 0, ⫺32

5. The system given is a dependent system. Without solving, state why. x ⫹ 2y ⫺ 3z ⫽ 3 • 2x ⫹ 4y ⫺ 6z ⫽ 6 x ⫺ 2y ⫹ 5z ⫽ ⫺1 6. Solve the system of equations: •

x ⫹ 2y ⫺ 3z ⫽ ⫺4 2y ⫹ z ⫽ 7 5y ⫺ 2z ⫽ 4

7. Solve using elimination: 2x ⫹ 3y ⫺ 4z ⫽ ⫺4 • x ⫺ 2y ⫹ z ⫽ 0 ⫺3x ⫺ 2y ⫹ 2z ⫽ ⫺1 8. Decompose the expression partial fractions.

⫺x2 ⫺ 4x ⫹ 15 into 1x ⫹ 121x ⫺ 22 2

9. Child prodigies: If you add Mozart’s age when he wrote his first symphony, with the age of American chess player Paul Morphy when he began dominating the international chess scene, and the age of Blaise Pascal when he formulated his well-known Essai pour les coniques (Essay on Conics), the sum is 37. At the time of each event, Paul Morphy’s age was 3 yr less than twice Mozart’s, and Pascal was 3 yr older than Morphy. Set up a system of equations and find the age of each. 10. Candle manufacturing: David and Karen make table candles and holiday candles and sell them out of their home. Dave works 10 min and Karen works 30 min on each table candle, while Karen works 40 min and David works 20 min on each holiday candle. Dave can work at most 3 hr and 20 min (200 min) per day on the home business, while Karen can work at most 7 hr. If table candles sell for $4 and holiday candles sell for $6, how many of each should made to maximize their revenue?

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REINFORCING BASIC CONCEPTS Window Size and Graphing Technology Since most substantial applications involve noninteger values, technology can play an important role in applying mathematical models. However, with its use comes a heavy responsibility to use it carefully. A very real effort must be made to determine the best approach and to secure a reasonable estimate. This is the only way to guard against (the inevitable) keystroke errors, or ensure a window size that properly displays the results.

Rationale On October 1, 1999, the newspaper USA TODAY ran an article titled, “Bad Math added up to Doomed Mars Craft.” The article told of how a $125,000,000.00 spacecraft was lost, apparently because the team of scientists that plotted the course for the craft used U.S. units of measurement, while the team of scientists guiding the craft were using metric units. NASA’s space chief was later quoted, “The problem here was not the error, it was the failure of . . . the checks and balances in our process to detect the error.” No matter how powerful the technology, always try to begin your problem-solving efforts with an estimate. Begin by exploring the context of the problem, asking questions about the range of possibilities: How fast can a human run? How much does a new car cost? What is a reasonable price for a ticket? What is the total available to invest? There is no calculating involved in these estimates, they simply rely on “horse sense” and human experience. In many applied problems, the input and output values must be positive—which means the solution will appear in the first quadrant, narrowing the possibilities considerably. This information will be used to set the viewing window of your graphing calculator, in preparation for solving the problem using a system and graphing technology. Illustration 1 䊳 Erin just filled both her boat and Blazer with gas, at a total cost of $211.14. She purchased 35.7 gallons of premium for her boat and 15.3 gal of regular for her Blazer. Premium gasoline cost $0.10 per gallon more than regular. What was the cost per gallon of each grade of gasoline? Solution 䊳 Asking how much you paid for gas the last time you filled up should serve as a fair estimate. Certainly (in 2011) a cost of $6.00 or more per gallon in the United States is too high, and a cost of $1.50 per gallon or less would be too low. Also, we can estimate a solution by assuming that both kinds of gasoline cost the same. This would mean 51 gal were purchased for about $211, and a quick division would place the estimate at near 211 51 ⬇ $4.14 per gallon. A good viewing window would be restricted to the first quadrant 1since cost 7 02 with maximum values of Xmax ⫽ 6 and Ymax ⫽ 6. Exercise 1: Solve Illustration 1 using graphing technology. Exercise 2: Re-solve Exercises 63 and 64 from Section 9.1 using graphing technology. Verify results are identical.

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Solving Linear Systems Using Matrices and Row Operations


A. State the size of a matrix

Just as synthetic division streamlines the process of polynomial division, matrices and row operations streamline the process of solving systems using elimination. With the equations of the system in standard form, the location and order of the variable terms and constant terms are predetermined, and we simply apply the elimination process on the coefficients and constants alone.

and identify its entries

B. Form the augmented

C.

D.

E. F.

A. Introduction to Matrices

matrix of a system of equations Solve a system of equations using row operations Solve a system of equations using technology Recognize inconsistent and dependent systems Solve applications using matrix methods

EXAMPLE 1A

In general terms, a matrix is simply a rectangular arrangement of numbers, called the entries of the matrix. Matrices (plural of matrix) are denoted by enclosing the entries between a left and right bracket, and named using a capital letter, such as 2 1 3 1 3 2 A c d and B £ 4 6 2 § . They occur in many different sizes 5 1 1 1 0 1 as defined by the number of rows and columns each has, with the number of rows always given first. Matrix A is said to be a 2 3 (two by three) matrix, since it has two rows and three columns. Matrix B is a 3 3 (three by three) matrix.

䊳

Identifying the Size and Entries of a Matrix Determine the size of each matrix and identify the entry located in the second row and first column. 3 2 0.2 0.5 0.7 3.3 a. C £ 1 b. D £ 0.4 5 § 0.3 1 2 § 4 3 2.1 0.1 0.6 4.1

Solution

䊳

a. Matrix C is 3 2. The row 2, column 1 entry is 1. b. Matrix D is 3 4. The row 2, column 1 entry is 0.4. If a matrix has the same number of rows and columns, it’s called a square matrix. Matrix B above is a square matrix, while matrix A is not. For square matrices, the values on a diagonal line from the upper left to the lower right are called the diagonal entries and are said to be on the diagonal of the matrix. When solving systems using matrices, much of our focus is on these diagonal entries.

EXAMPLE 1B

䊳

Identifying the Diagonal Entries of a Square Matrix Name the diagonal entries of each matrix. 0.2 1 4 d a. E c b. F £ 0.4 2 3 2.1

Solution A. You’ve just seen how we can state the size of a matrix and identify its entries

䊳

0.5 0.3 0.1

0.7 1 § 0.6

a. The diagonal entries of matrix E are 1 and 3. b. For matrix F, the diagonal entries are 0.2, 0.3, and 0.6. Now try Exercises 7 through 9

䊳

B. The Augmented Matrix of a System of Equations 9–57

A system of equations can be written in matrix form by augmenting or joining the coefficient matrix, formed by the variable coefficients, with the matrix of constants. 893

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2x 3y z 1 2 3 1 0 1 § with The coefficient matrix for the system • x z 2 is £ 1 1 3 4 x 3y 4z 5 column 1 for the coefficients of x, column 2 for the coefficients of y, and so on. The 1 matrix of constants is £ 2 § . These two are joined to form the augmented matrix, 5 2 3 1 1 0 1 2 §. with a dotted line often used to separate the two as shown here: £ 1 1 3 4 5 It’s important to note the use of a zero placeholder for the y-variable in the second row of the matrix, signifying there is no y-variable in the corresponding equation. EXAMPLE 2

䊳

Forming Augmented Matrices Form the augmented matrix for each system, and name the diagonal entries of each coefficient matrix. 1 y 5 2x 2x y 11 2 5 a. e b. • x 3y 6z 11 x 3y 2 z6

Solution

䊳

2x y 11 ¡ c2 a. e 1 x 3y 2

1 3

11 d 2

Diagonal entries: 2 and 3. b. • x

1 y 5 1 0 5 2 5 2 5 2 ¡ £ y z 11 1 11 § 3 6 3 6 z6 0 0 1 6 1 2 Diagonal entries: 2, 3, and 1. Notice that in the third row of part (b), the fact that z 6 is indicated by zeroes in the x- and y-columns, 1 in the z-column, and a 6 in the column of constants. 1 2x


䊳

This process can easily be reversed to write a system of equations from a given augmented matrix. EXAMPLE 3

䊳

Writing the System Corresponding to an Augmented Matrix Write the system of equations corresponding to each matrix, then solve the system using back-substitution. 1 4 1 10 3 5 14 d 7 § a. c b. £ 0 3 10 0 1 4 0 0 1 1

Solution

䊳

a. c

3 0

5 1

14 ¡ 3x 5y 14 d e 4 1y 4

With y 4, we have 3x 5142 14 3x 20 14 3x 6 x2 The solution is (2, 4)

substitute 4 for y multiply add 20 divide by 3

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895

1 4 1 10 1x 4y 1z 10 ¡ b. £ 0 3 10 7 § • 3y 10z 7 0 0 1 1 1z 1 With z 1, we first substitute 1 for z in the second equation. substitute 1 for z 3y 10112 7 multiply 3y 10 7 3y 3 subtract 10 divide by 3 y1 We then substitute 1 for z and 1 for y in the first equation.

B. You’ve just seen how we can form the augmented matrix of a system of equations

x 4112 1112 10 substitute 1 for y and 1 for z x 3 10 simplify x 13 subtract 3 The solution is (13, 1, 1). Now try Exercises 13 through 18

䊳

C. Solving a System Using Matrices When a system of equations is written in augmented matrix form, we can solve the system by applying the same operations to each row of the matrix, that would be applied to the equations in the system. In this context, the operations are referred to as elementary row operations. Elementary Row Operations 1. Any two rows in a matrix can be interchanged. 2. The elements of any row can be multiplied by a nonzero constant. 3. Any two rows can be added together, and the sum used to replace one of the rows. In this section, we’ll use these operations to triangularize the augmented matrix, employing a solution method known as Gaussian elimination. A square matrix is said to be in triangular form when all of the entries below the diagonal are zero. For example, 1 4 1 10 1 4 1 10 the matrix £ 0 3 10 7 § is in triangular form: £ 0 3 10 7 §. 0 0 1 1 0 0 1 1 x 4y z 10 In system form we have • 3y 10z 7 , meaning a matrix written in triangular z 1 form can be used to solve the system using back-substitution. We’ll illustrate by 1x 4y 1z 4 solving • 2x 5y 8z 15, using elimination to the left, and row operations on the 1x 3y 3z 1 augmented matrix to the right. As before, R1 represents the first equation in the system and the first row of the matrix, R2 represents equation 2 and row 2, and so on. The calculations involved are shown for the first stage only and are designed to offer a careful comparison. In actual practice, the format shown in Example 4 is used.

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Elimination (System of Equations)

Row Operations (Augmented Matrix)

1x 4y 1z 4 • 2x 5y 8z 15 1x 3y 3z 1

1 £2 1

1 8 3

4 5 3

4 15 § 1

To eliminate the x-term in R2, we use 2R1 R2 S R2. Identical operations are performed on the matrix, which begins the process of triangularizing the matrix. System Form 2R1 R2 New R2

Matrix Form

2x 8y 2z 8 2x 5y 8z 15 3y 10z 7

2R1 R2 New R2

For R3, the operations are 1R1 R3 S R3. 1x 4y 1z 4 1R1 1R1 1x 3y 3z 1 R3 R3 1y 2z 3 New R3 New R3

2

2

8

2 5 8 0 3 10

15 7

1

8

4

1

4

1 3 3 1 0 1 2 3

As always, we should look for opportunities to simplify any equation in the system (and any row in the matrix). Note that 1R3 will make the coefficients and related matrix entries positive. Here is the new system and matrix. New System •

WORTHY OF NOTE The procedure outlined for solving systems using matrices is virtually identical to that for solving systems by elimination. Using a 3 3 system for illustration, the “zeroes below the first diagonal entry” indicates we’ve eliminated the x-term from R2 and R3, the “zeroes below the second entry” indicates we’ve eliminated the y-term from the subsystem, and the division “to obtain a ‘1’ in the final entry” indicates we have just solved for z.

1x

4y 1z 4 3y 10z 7 1y 2z 3

New Matrix 1 £0 0

4 3 1

1 10 2

4 7§ 3

On the left, we would finish by solving the 2 2 subsystem using R2 3R3 S R3. In matrix form, we eliminate the corresponding entry (third row, second column) to triangularize the matrix. 1 £0 0

4 3 1

1 10 2

4 1 7 § R2 3R3 S R3 £ 0 ¬¬¬¬¡ 3 0

4 3 0

1 10 16

4 7§ 16

Dividing R3 by 16 gives z 1 in the system, and entries of 0 0 1 1 in the augmented matrix. Completing the solution by back-substitution in the system gives the ordered triple (1, 1, 1). For practice using these row operations, see Exercises 19 through 27. The general solution process is summarized here. Solving Systems by Triangularizing the Augmented Matrix 1. 2. 3. 4. 5.

Write the system as an augmented matrix. Use row operations to obtain zeroes below the first diagonal entry. Use row operations to obtain zeroes below the second diagonal entry. Continue until the matrix is triangularized (entries below diagonal are zero). Divide to obtain a “1” in the last diagonal entry (if it is nonzero), then convert to equation form and solve using back-substitution.

Note: At each stage, look for opportunities to simplify row entries using multiplication or division. Also, to begin the process any equation with an x-coefficient of 1 can be made R1 by interchanging the equations.

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EXAMPLE 4

䊳

Solving Systems Using the Augmented Matrix 2x y 2z 7 Solve by triangularizing the augmented matrix: • x y z 1 2y z 3

Solution

䊳

2x y 2z 7 • x y z 1 2y z 3 1 £2 0

1 1 2

1 2 1

1 7 § 3

matrix form S

2 £1 0

1 1 2

2 1 1

7 1 § 3

2R1 R2 S R2

1 £0 0

1 1 2

1 4 1

1 5 § 3

R1 4 R2

1 £2 0

1 1 2

1 2 1

1 7 § 3

1R2 S R2

1 £0 0

1 1 2

1 4 1

1 5 § 3

1 1 1 1 1 1 1 1 1 1 1 1 R3 S R3 £0 1 4 5 § £0 1 4 5 § 2R2 R3 S R3 £ 0 1 4 5 § 7 0 0 1 1 0 2 1 3 0 0 7 7 Converting the augmented matrix back into equation form yields z 1. Back-substituting 1 for z in the second equation gives y 4112 5, with y 4 5 showing y 1. Back-substituting 1 for z and 1 for y in the first equation gives x 1 1 1, or x 2 1. Subtracting 2 from both sides shows x 3. The solution is 13, 1, 12 . Now try Exercises 28 through 32 䊳 As mentioned, the process used in Example 4 is called Gaussian elimination (Carl Friedrich Gauss, 1777–1855), with the last matrix written in row-echelon form. It’s also possible to solve a system entirely using only the augmented matrix, by continuing to use row operations until the diagonal entries are 1’s, with 0’s for all other entries 1 0 0 a in the coefficient matrix: £ 0 1 0 b § . The process is then called Gauss-Jordan 0 0 1 c elimination (Wilhelm Jordan, 1842–1899), with the final matrix written in reduced row-echelon form (see Appendix D). Note that with Gauss-Jordan elimination, our initial focus is less on getting 1’s along the diagonal, and more on obtaining zeroes for all entries other than the diagonal entries. This will enable us to work with integer values in the solution process, as shown in Example 5.

EXAMPLE 5

䊳

Solving a System Using Gauss-Jordan Elimination 2x 5z 15 2y 1 z. Solve using Gauss-Jordan elimination • 2x 3y 4y z 7 standard form

Solution

䊳

2x 2y 5z 15 • 2x 3y 1z 1 0x 4y 1z 7 2 2 5 £ 0 5 6 0 4 1

matrix form S

2 2 5 £ 2 3 1 0 4 1

2R2 5R1 S R1 10 15 16 § £0 7 4R2 5R3 S R3 0

0 5 0

13 6 29

15 1 § 7 43 16 § 29

R1 R2 S R2

R3 S R3 29

2 2 5 £ 0 5 6 0 4 1

15 16 § 7

10 £0 0

43 16 § 1

0 5 0

13 6 1

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10 £0 0

0 5 0

13 43 ⫺6 ⫺16 § 1 1

⫺13R3 ⫹ R1 S R1 6R3 ⫹ R2 S R2

10 £0 0

0 5 0

0 0 1

R1 S R1 10 R2 S R2 5

30 ⫺10 § 1

1 £0 0

The final matrix shows the solution is x ⫽ 3, y ⫽ ⫺2, and z ⫽ 1, or 13, ⫺2, 12 . C. You’ve just seen how we can solve a system of equations using row operations

0 1 0

0 0 1

3 ⫺2 § 1


䊳

D. Solving Systems of Equations Using Technology Graphing calculators offer a very efficient way to solve systems of equations using matrices. Once the system has been written in matrix form, it can easily be entered into a calculator and solved by having the calculator instantly perform the row operations needed to produce the diagonal of 1’s (with zeroes for all other coefficient matrix entries). On many calculators, pressing 2nd x (MATRIX) gives a screen similar to the one shown in Figure 9.41, where we begin by selecting the EDIT option (push the right arrow twice). Pressing places you on a screen where you can EDIT matrix A, changing the size as needed, then input the entries of the matrix. To use the 3 ⫻ 4 matrix from Example 4, we press 3 and then 4 , giving the screen shown in Figure 9.42. The dash marks to the right indicate that there is a fourth column that cannot be seen, but that comes into view as you enter the elements of the matrix. Begin entering the first row of the matrix, which is 52, 1, ⫺2, ⫺76. Press after each entry and the cursor automatically goes to the next position in the matrix (note the calculator automatically shifts left and right to allow all four columns to be entered). After entering the second row 51, 1, 1, ⫺16 and the third row 50, ⫺2, ⫺1, ⫺36 the completed matrix should look like the one shown in Figure 9.43 (the matrix is currently shifted to the right, showing the fourth column). -1

ENTER

ENTER

ENTER

ENTER

Figure 9.41

Figure 9.42

Figure 9.43

Most graphing calculators have an rref function, which is short for reduced row x echelon form. To access this function, press 2nd (MATRIX) and select the MATH option, then scroll upward (or downward) until you get to the B:rref option. Pressing places this function on the home screen, where we must tell it to perform the rref operation on matrix [A]. Pressing the 2nd x once again returns us to the matrix options, enabling us to select matrix NAMES Figure 9.44 (notice that NAMES is automatically highlighted). With the cursor over 1:[A] 3ⴛ4 we press and matrix [A] is placed on the home screen as the object of the rref function. After pressing ) and the calculator quickly computes the reduced row echelon form and displays it on the screen, as in Figure 9.44. The solution is easily read as x ⫽ ⫺3, y ⫽ 1, and z ⫽ 1, just as we found in Example 4. -1

ENTER

-1

ENTER

ENTER

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EXAMPLE 6

䊳

Solving a System of Equations Using Technology Solve the following system using a graphing calculator. 2x 3y z 4 • x 3y 2z 1 3x y 4z 5

Solution

䊳

Entering the system as matrix [A] produces the screen shown in Figure 9.45. Using the rref([A]) command as shown above produces the screen in Figure 9.46. The solution is the ordered triple (2, 1, 3). Check by substituting 2 for x, 1 for y, and 3 for z in the original equations. Figure 9.45

Figure 9.46

D. You’ve just seen how we can solve a system of equations using technology


䊳

E. Inconsistent and Dependent Systems Due to the strong link between a linear system and its augmented matrix, inconsistent and dependent systems can be recognized just as in Sections 9.1 and 9.2. An inconsistent system will yield an inconsistent or contradictory statement such as 0 12, meaning all entries in a row of the matrix of coefficients are zero, but the constant is not. A linearly dependent system will yield an identity statement such as 0 0, meaning all entries in one row of the matrix are zero. If the system has coincident dependence, all rows will be zero except one. EXAMPLE 7

䊳

Solving a Dependent System x y 5z 3 2z 1 Solve the system using Gauss-Jordan elimination: • x 2x y z 0

Solution

䊳

x y 5z 3 2z 1 • x 2x y z 0 1 1 £ 1 0 2 1

5 2 1

3 1 § 0

standard form S

R1 R2 S R2 2R1 R3 S R3

x y 5z 3 • x 0y 2z 1 2x y z 0 1 £0 0

1 5 1 3 3 9

3 2 § 6

matrix form S

1R2 R1 S R1 3R2 R3 S R3

1 1 £ 1 0 2 1 1 £0 0

0 2 1 3 0 0

5 2 1

3 1 § 0

1 2§ 0

Since all entries in the last row are zeroes and it’s the only row of zeroes, we conclude the system 2z 1 x is linearly dependent and equivalent to e (a calculator solution is shown in the y 3z 2 figure). As in Section 9.2, we demonstrate this dependence by writing the (x, y, z) solution in terms of

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a parameter. Solving for y in R2 gives y in terms of z: y 3z 2. Solving for x in R1 gives x in terms of z: x 2z 1. As written, the solutions all depend on z: x 2z 1, y 3z 2, and z z. Selecting p as the parameter (or some other “neutral” variable), we write the solution as 12p 1, 3p 2, p2. Two of the infinite number of solutions would be (1, 2, 0) for p 0, and 11, 1, 12 for p 1. Test these triples in the original equations. Now try Exercises 43 through 52

䊳

Since there was only one row of zeroes in the system from Example 7, we knew the system was linearly dependent. As mentioned previously, for coincident dependence the system will have only one nonzero row, with all other rows consisting entirely of zeroes. Note that if we change the constant in the third equation of the original system (the entry for row 3, column 4 was changed to a 5), an inconsistent system results as one equation is no longer dependent on the others (see Figures 9.47 and 9.48). Figure 9.47

Figure 9.48

When the number of variables and equations in a system increases, solutions using technology can be particularly effective. In addition, some applications generate systems where the number of equations and number of variables are unequal, which the technology still handles efficiently.

EXAMPLE 8

䊳

Solving a System of Equations Using Technology Solve the following system using a graphing calculator. x 2x

3y z 4 y 3z 7 μ 2y 4z 0 x 4y 2z 11

Solution

䊳

E. You’ve just seen how we can recognize inconsistent and dependent systems

After entering the system as matrix [A], using the rref([A]) command produces the screen shown in the figure. Note that in this case, while the solution contains a row of zeroes, the system is not linearly dependent as there are more equations than variables, and unique values have been found for x, y, and z. The solution is (1, 2, 1). Now try Exercises 53 through 58

䊳

F. Solving Applications Using Matrices As in other areas, solving applications using systems relies heavily on the ability to mathematically model information given verbally or in context. As you work through the exercises, read each problem carefully. Look for relationships that yield a system of two equations in two variables, three equations in three variables, and so on.

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EXAMPLE 9

䊳

Determining the Original Value of Collector’s Items A museum purchases a famous painting, a ruby tiara, and a rare coin for its collection, spending a total of $30,000. One year later, the painting has tripled in value, while the tiara and the coin have doubled in value. The items now have a total value of $75,000. Find the purchase price of each if the original price of the painting was $1000 more than twice the coin.

Solution

Let P represent the price of the painting, T the tiara, and C the coin.

䊳

Total spent was $30,000: One year later: Value of painting versus coin:

S P T C 30,000 S 3P 2T 2C 75,000 S P 2C 1000

P T C 30,000 1P 1T 1C 30,000 • 3P 2T 2C 75,000 standard form S • 3P 2T 2C 75,000 P 2C 1000 1P 0T 2C 1000 1 £3 1

1 2 0

1 £0 0

1 1 1

1 2 2

30000 75000 § 1000

1 1 3

30000 15000 § 29000

3R1 R2 S R2 1R1 R3 S R3

1R2 R3 S R3

1 £0 0

1 1 1

1 £0 0

1 1 0

1 1 3 1 1 2

30000 15000 § 29000

30000 15000 § 14000

matrix form S

1 £3 1

1R2 S R2 1R3 S R3 R3 S R3 2

1 2 0

1 2 2

30000 75000 § 1000

1 £0 0

1 1 1

1 1 3

30000 15000 § 29000

1 £0 0

1 1 0

1 1 1

30000 15000 § 7000

From R3 of the triangularized form, C $7000 directly. Since R2 represents T C 15,000, we find the tiara was purchased for T $8000. Substituting these values into the first equation shows the painting was purchased for $15,000. The solution is (15,000, 8000, 7000). The solution found using a graphing calculator is also shown.

F. You’ve just seen how we can solve applications using matrix methods


9.5 EXERCISES 䊳



1. A matrix with the same number of rows and columns is called a(n) matrix. 2 4 3 d is a by 1 2 1 matrix. The entry in the second row and third column is .

3. Matrix A c

5. The notation 2R1 R2 S R2 indicates that an equivalent matrix is formed by performing what operations/replacements?

2. When the coefficient matrix is used with the matrix of constants, the result is a(n) matrix. 4. Given matrix B shown here, the diagonal entries are , , and .

1 4 3 B £ 1 5 2§ 3 2 1 6. Describe how to tell an inconsistent system apart from a dependent system when solving using matrix methods (row reduction).

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Determine the size of each matrix and identify the third row and second column entry. If the matrix given is a square matrix, identify the diagonal entries.

1 7. £ 2.1 3 1 1 9. ≥ 5 2

0 1 § 5.8 0 3 1 3

1 8. £ 1 5

0 3 1

4 7 § 2

4 7 ¥ 2 9

Form the augmented matrix, then name the diagonal entries of the coefficient matrix.

2x 3y 2z 7 10. • x y 2z 5 3x 2y z 11 x 2y z 1 z3 11. • x 2x y z 3 2x 3y z 5 12. • 2y z 7 x y 2z 5 Write the system of equations for each matrix. Then use back-substitution to find its solution.

13. c

1 0

4 1

14. c

1 0

5 1

5 1 2

d 15 d 2

1 15. £ 0 0

2 1 0

1 2 1

0 2§ 3

1 16. £ 0 0

0 1 0

7 5 1

5 15 § 26

1 17. £ 0 0

3 1 0

4 32 1

29

1 18. £ 0 0

2 1 0

1

3

1 6

1

21 2

3

§

2 3 § 22 7

Perform the indicated row operation(s) in the order given and write the new matrix.

3 2

1 2

19. c

5

20. c

7 4

1 d 4

1 4 R2

3 d 12

4 8

2R1 S R1, 5R1 R2 S R2

S R2, R1 4 R2

2 21. £ 5 1

1 8 3

3 22. £ 1 4

2 1 1

0 2 3

0 6§ 2

3 23. £ 6 4

1 1 2

1 1 3

8 10 § 22

2 24. £ 3 4

1 1 3

0 3 3

1 1 2

4 5 § 2

3 0 § 3

R1 4 R3, 5R1 R2 S R2 R1 4 R2, 4R1 R3 S R3 2R1 R2 S R2, 4R1 3R3 S R3 3R1 2R2 S R2, 2R1 R3 S R3

What row operations would produce zeroes beneath the first entry in the diagonal?

1 25. £ 2 3

3 4 1

0 1 2

2 1§ 9

1 26. £ 3 5

1 0 3

4 1 2

3 5 § 3

1 27. £ 5 4

2 1 3

0 2 3

10 6 § 2

Solve each system by triangularizing the augmented matrix and using back-substitution. Simplify by clearing fractions or decimals before beginning.

28. e

2y 5x 4 5x 2 4y

29. e

0.15g 0.35h 0.5 0.12g 0.25h 0.1

1u 14v 1 30. e 15 1 10 u 2 v 7 2x 3y 2z 7 32. • x y 2z 5 3x 2y z 11

x 2y 2z 7 31. • 2x 2y z 5 3x y z 6

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Solve using Gauss-Jordan elimination.

x 2y z 1 33. • x z3 2x y z 3

2x 3y z 5 34. • 2y z 7 x y 2z 5

x y 2z 2 x y 2z 1 35. • x y z 1 36. • 4x y 3z 3 2x y z 4 3x 2y z 4 Solve each system of equations using a graphing calculator. Verify each solution on the home screen using the ALPHA keys, as in Section 9.2.

x 2y 3z 9 37. • 2x y 5z 20 3x 5y z 14 3x 5y z 6 38. • 4x 7y 2z 24 x 10y 3z 40 0.2x 0.5y z 13 39. • x 0.7y 0.4z 9 0.5x y 0.8z 4.6 0.1x 2y 0.6z 36.4 40. • 3x 0.8y 0.2z 35 0.5x y 0.4z 25 34y z 34 41. • x 14y 32z 58 1 1 5 4 x y 2 z 2 1 2x

12y z 14 42. • x 32y 78z 35 1 2 1 4x y 4z 5 5 4x

Solve each system by triangularizing the augmented matrix and using back-substitution. If the system is linearly dependent, give the solution in terms of a parameter. If the system has coincident dependence, answer in set notation as in Section 9.2.

4x 8y 8z 24 43. • 2x 6y 3z 13 3x 4y z 11 3x y z 2 44. • x 2y 3z 1 2x 3y 5z 3 x 3y 5z 20 45. • 2x 3y 4z 16 x 2y 3z 12

903

x 2y 3z 6 x y 2z 4 3x 6y 9z 18 3x 4y 2z 2 47. • 32x 2y z 1 6x 8y 4z 4 46. •

2x y 3z 1 48. • 4x 2y 6z 2 10x 5y 15z 5 49. •

2x y 3z 1 2y 6z 2 x 12y 32z 5

x 2y 3z 2 50. • 3x 4y z 6 4x 2y 2z 7 x 2y z 4 51. • 3x 4y z 4 6x 8y 2z 8 2x 4y 3z 4 52. • 5x 6y 7z 12 x 2y z 4 Solve each system of equations using a graphing calculator. Verify each solution on the home screen using the ALPHA keys, as in Section 9.2.

2x 3y z 0 x 2y z 5 53. μ 3x 2z 4 x 3y z 4 2x 5y 4 3x 4y z 0 54. μ x y 3z 8 5x 9y z 4 x 3y z 1 x 2z 7 55. μ 2y 3z 6 2x 3y z 8 2x 3y 4 2y 5z 19 56. μ x 4z 13 x 3y 4z 17

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x 2y 3z w 13 2x 3y z 2w 3 57. μ x y 2z 3w 4 3x 2y z w 5

䊳

9–68


2x 3y 4z w 10 x 2y 3w 6 58. μ 3x 5z 2w 1 x y 2z 5w 21


1 Area of a triangle in the plane: A (x1 y2 x2 y1 x2 y3 x3 y2 x3 y1 x1 y3) 2 The area of a triangle in the plane is given by the formula shown, where the vertices of the triangle are located at the points (x1, y1), (x2, y2), and (x3, y3), and the sign is chosen to ensure a positive value.

59. Find the area of a triangle whose vertices are 11, 32, (5, 2), and (1, 8). 䊳

60. Find the area of a triangle whose vertices are 16, 22, 15, 42 , and 11, 72.

APPLICATIONS

Model each problem using a system of linear equations. Then solve using the augmented matrix. Descriptive Translation

61. The distance (via air travel) from Los Angeles (LA), California, to Saint Louis (STL), Missouri, to Cincinnati (CIN), Ohio, to New York City (NYC), New York, is approximately 2480 mi. Find the length of each flight if the distance from LA to STL is 50 mi more than five times the distance between STL and CIN, and 110 mi less than three times the distance from CIN to NYC.

New York City St. Louis

Cincinnati

Los Angeles

62. In the 2006 NBA Championship Series, Dwayne Wade of the Miami Heat carried his team to the title after the first two games were lost to the Dallas Mavericks. If 187 points were scored in the title game and the Heat won by 3 points, what was the final score? 63. Moe is lecturing Larry and Curly once again (Moe, Larry, and Curly of The Three Stooges fame) claiming he is twice as smart as Larry and three times as smart as Curly. If he is correct and the sum of their IQs is 165, what is the IQ of each stooge? 64. A collector of rare books buys a handwritten, autographed copy of Edgar Allan Poe’s Annabel Lee, an original advance copy of L. Frank Baum’s The Wonderful Wizard of Oz, and a first print copy of The Caine Mutiny by Herman Wouk, paying a total of $100,000. Find the cost of each one, given that the cost of Annabel Lee and twice the cost of The Caine Mutiny sum to the price paid for The Wonderful Wizard of Oz, and The Caine Mutiny cost twice as much as Annabel Lee. Geometry

65. A right triangle has a hypotenuse of 39 m. If the perimeter is 90 m, and the longer leg is 6 m longer than twice the shorter leg, find the dimensions of the triangle. 66. In triangle ABC, the sum of angles A and C is equal to three times angle B. Angle C is 10 degrees more than twice angle B. Find the measure of each angle.

Investment and Finance

67. Suppose $10,000 is invested in three different investment vehicles paying 5%, 7%, and 9% annual interest. Find the amount invested at each rate if the interest earned after 1 yr is $760 and the amount invested at 9% is equal to the sum of the amounts invested at 5% and 7%. 68. The trustee of a union’s pension fund has invested funds in three ways: a savings fund paying 4% annual interest, a money market fund paying 7%, and government bonds paying 8%. Find the amount invested in each if the interest earned after one year is $0.178 million and the amount in government bonds is $0.3 million more than twice the amount in money market funds. The total amount invested is $2.5 million dollars.

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Section 9.6 The Algebra of Matrices

905


69. Given the drawing shown, use a system of equations and the matrix method to find the measure of the angles labeled as x and y. Recall that vertical angles are equal and that the sum of the angles in a triangle is 180°.

70. The system given here has a solution of 11, 2, 32. Find the value of a and b. 1 £ 2b 2a

71

a 2a 7

b 5 3b

1 13 § 8

x

y (x 59) 䊳


71. (8.5) a. Convert z1 1 3i to trigonometric form. 2␲ b. Convert z2 5 cisa b to rectangular 3 form. 72. (6.2) State the exact value of the following trig functions: ␲ 5␲ a. sina b b. cosa b 6 4 c. tana

2␲ b 3

d. csca

3␲ b 2

year will the number of installations be greater than 30,000? 74. (5.6) If a set amount of money p is deposited regularly (daily, weekly, monthly, etc.) n times per year at a fixed interest rate r, the amount of money A accumulated in t years is given by the formula shown. If a parent deposits $250 per month for 18 yr at 4.6% beginning when her first child was born, how much has been accumulated to help pay for college expenses?

73. (5.6) Since 2005, cable installations for an Internet company have been modeled by the function C1t2 15 ln1t 12 , where C(t) represents cable installations in thousands, t yr after 2005. In what

9.6

r nt p c a1 b 1 d n A r n

The Algebra of Matrices

LEARNING OBJECTIVES

B. Add and subtract

Matrices serve a much wider purpose than just a convenient method for solving systems. To understand their broader application, we need to know more about matrix theory, the various ways matrices can be combined, and some of their more practical uses. The common operations of addition, subtraction, multiplication, and division are all defined for matrices, as are other operations. Practical applications of matrix theory can be found in the social sciences, inventory management, genetics, operations research, engineering, and many other fields.

matrices C. Compute the product of two matrices

A. Equality of Matrices


A. Determine if two matrices are equal

To effectively study matrix algebra, we first give matrices a more general definition. For the general matrix A, all entries will be denoted using the lowercase letter “a,” with their position in the matrix designated by the dual subscript aij. The letter “i” gives the row and the letter “j ” gives the column of the entry’s location. The general m n matrix A is written

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col 1 row 1 S a11 row 2 S l a21 row 3 S a31 row i S row m S

p

col 2 a12 a22 a32

col 3 a13 a23 a33

p p p

col j a1j a2j a3j

p p p

o ai1

o ai2

o ai3

p

o aij

p

am1

o am2

o am3

p

o amj

p

z o

col n a1n a2n } a3n

o p ain

aij is a general matrix element

o { amn

The size of a matrix is also referred to as its order, and we say the order of general matrix A is m ⫻ n. Note that diagonal entries have the same row and column number, aij, where i ⫽ j. Also, where the general entry of matrix A is aij, the general entry of matrix B is bij, of matrix C is cij, and so on. EXAMPLE 1

䊳

Identifying the Order and Entries of a Matrix State the order of each matrix and name the entries corresponding to a22, a31; b22, b31; and c22, c31. 3 ⫺2 0.2 ⫺0.5 0.7 1 4 d 5 § 0.3 1 § a. A ⫽ c b. B ⫽ £ 1 c. C ⫽ £ ⫺1 ⫺2 ⫺3 ⫺4 3 2.1 ⫺0.1 0.6

Solution

䊳

a. matrix A: order 2 ⫻ 2. Entry a22 ⫽ ⫺3 (the row 2, column 2 entry is ⫺3). There is no a31 entry (A is only 2 ⫻ 2). b. matrix B: order 3 ⫻ 2. Entry b22 ⫽ 5, entry b31 ⫽ ⫺4. c. matrix C: order 3 ⫻ 3. Entry c22 ⫽ 0.3, entry c31 ⫽ 2.1. Now try Exercises 7 through 12

䊳

Equality of Matrices Two matrices are equal if they have the same order and their corresponding entries are equal. In symbols, this means that A ⫽ B if aij ⫽ bij for all i and j. EXAMPLE 2

䊳

Determining If Two Matrices Are Equal Determine whether the following statements are true, false, or conditional. If false, explain why. If conditional, find values that will make the statement true. 3 ⫺2 1 4 ⫺3 ⫺2 3 ⫺2 1 a. c b. £ 1 d ⫽ c d 5 § ⫽ c d ⫺2 ⫺3 4 1 5 ⫺4 3 ⫺4 3 1 4 a ⫺ 2 2b c. c d ⫽ c d ⫺2 ⫺3 c ⫺3

Solution

A. You’ve just seen how we can determine if two matrices are equal

䊳

1 4 ⫺3 ⫺2 d is false. The matrices have the same order and d ⫽ c ⫺2 ⫺3 4 1 entries, but corresponding entries are not equal. 3 ⫺2 3 ⫺2 1 b. £ 1 5 § ⫽ c d is false. Their orders are not equal. 5 ⫺4 3 ⫺4 3 1 4 a ⫺ 2 2b c. c d ⫽ c d is conditional. The statement is true when ⫺2 ⫺3 c ⫺3 a ⫺ 2 ⫽ 1 1a ⫽ 32, 2b ⫽ 4 1b ⫽ 22, c ⫽ ⫺2, and is false otherwise. a. c


䊳

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907

B. Addition and Subtraction of Matrices A sum or difference of matrices is found by combining the corresponding entries. This limits the operations to matrices of like orders, so that every entry in one matrix has a “corresponding entry” in the other. This also means the result is a new matrix of like order, whose entries are the corresponding sums or differences. If you attempt to add matrices of unlike size on a graphing calculator, an error message is displayed: ERR: DIM MISMATCH. Note that since aij represents a general entry of matrix A, [aij] represents the entire matrix. Addition and Subtraction of Matrices Given matrices A, B, and C having like orders. The sum A ⫹ B ⫽ C, where 3aij ⫹ bij 4 ⫽ 3cij 4. EXAMPLE 3

䊳

The difference A ⫺ B ⫽ C, where 3aij ⫺ bij 4 ⫽ 3cij 4.

Adding and Subtracting Matrices Compute the sum or difference of the matrices indicated. 2 6 A ⫽ £1 0 § 1 ⫺3 a. A ⫹ C

Solution

䊳

2 a. A ⫹ C ⫽ £ 1 1

B⫽ c

⫺3 ⫺5

b. A ⫹ B 6 3 0 § ⫹ £ 1 ⫺3 ⫺4

2 4

3 C⫽ £ 1 ⫺4

⫺1 d 3

⫺2 5 § 3

c. C ⫺ A ⫺2 5 § 3

sum of A and C

2⫹3 6 ⫹ 1⫺22 5 4 ⫽ £ 1⫹1 0 ⫹ 5 §⫽ £ 2 5§ 1 ⫹ 1⫺42 ⫺3 ⫹ 3 ⫺3 0 2 6 ⫺3 2 ⫺1 b. A ⫹ B ⫽ £ 1 0 § ⫹ c d ⫺5 4 3 1 ⫺3 3 ⫺2 2 6 c. C ⫺ A ⫽ £ 1 5 § ⫺ £1 0 § ⫺4 3 1 ⫺3 3⫺2 ⫺2 ⫺ 6 1 ⫺8 ⫽ £ 1⫺1 5⫺0 § ⫽ £ 0 5 § ⫺4 ⫺ 1 3 ⫺ 1⫺32 ⫺5 6

add corresponding entries

Addition and subtraction are not defined for matrices of unlike order.

difference of C and A

subtract corresponding entries


䊳

Operations on matrices can be very laborious for larger matrices and for matrices with noninteger or large entries. For these, we can turn to available technology for assistance. This shifts our focus from a meticulous computation of entries, to carefully entering each matrix into the calculator, double-checking each entry, and appraising results to see if they’re reasonable.

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EXAMPLE 4

䊳

Using Technology for Matrix Operations Use a calculator to compute the difference A ⫺ B for the matrices given. 2 11

A ⫽ £ 0.9 0

Solution

䊳

⫺0.5 3 4

6

6 5

⫺4 § 5 ⫺12

B⫽ £

⫺7 10

1 6 11 25

0

⫺4

⫺5 9

0.75 ⫺5 § ⫺5 12

The entries for matrix A are shown in Figure 9.49. After entering matrix B, exit to the home screen [ 2nd MODE (QUIT)], call up matrix A, press the ⴚ (subtract) key, then call up matrix B and press . The calculator quickly finds the difference and displays the results shown in Figure 9.50. The last line on the screen shows the result can be stored for future use in a new matrix C by pressing the STO key, calling up matrix C, and pressing . ENTER

ENTER

Figure 9.49

Figure 9.50


䊳

Figure 9.51

In Figure 9.50 the dots to the right on the calculator screen indicate there are additional digits or matrix columns that can’t fit on the display, as often happens with larger matrices or decimal numbers. Sometimes, converting entries to fraction form will provide a display that’s easier to read. Here, this is done by calling up the matrix C, and using the MATH 1: 䊳 Frac option. After pressing , all entries are converted to fractions in simplest form (where possible), as in Figure 9.51. The third column can be viewed by pressing the right arrow. Since the addition of two matrices is defined as the sum of corresponding entries, we find the properties of matrix addition closely resemble those of real number addition. Similar to standard algebraic properties, ⫺A represents the product ⫺1 # A and any subtraction can be rewritten as an algebraic sum: A ⫺ B ⫽ A ⫹ 1⫺B2. As noted in the properties box, for any matrix A, the sum A ⫹ 1⫺A2 will yield the zero matrix Z, a matrix of like size whose entries are all zeroes. Also note that matrix ⫺A is the additive inverse for A, while Z is the additive identity. ENTER

Properties of Matrix Addition Given matrices A, B, C, and Z are m ⫻ n matrices, with Z the zero matrix. Then, B. You’ve just seen how we can add and subtract matrices

I. A ⫹ B ⫽ B ⫹ A II. 1A ⫹ B2 ⫹ C ⫽ A ⫹ 1B ⫹ C2 III. A ⫹ Z ⫽ Z ⫹ A ⫽ A IV. A ⫹ 1⫺A2 ⫽ 1⫺A2 ⫹ A ⫽ Z

matrix addition is commutative matrix addition is associative Z is the additive identity ⫺A is the additive inverse of A

C. Matrices and Multiplication The algebraic terms 2a and ab have counterparts in matrix algebra. The product 2A represents a constant times a matrix and is called scalar multiplication. The product AB represents the product of two matrices.

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Scalar Multiplication Scalar multiplication is defined by taking the product of the constant with each entry in the matrix, forming a new matrix of like size. In symbols, for any real number k and matrix A, kA ⫽ 3kaij 4. EXAMPLE 5

䊳

Computing Operations on Matrices 4

3 3 1 § and B ⫽ £ 0 Given A ⫽ £ 0 ⫺3 ⫺4 1 1 a. 2 B b. ⫺4A ⫺ 2 B 1 2

Solution

䊳

⫺2 6 § , compute the following: 0.4

3 ⫺2 1 1 a. B ⫽ a b £ 0 6 § 2 2 ⫺4 0.4 3 1 12 2132 1 12 2 1⫺22 ⫺1 2 1 1 ⫽ £ 1 2 2102 1 2 2 162 § ⫽ £ 0 3 § ⫺2 0.2 1 12 21⫺42 1 12 210.42 1 1 b. ⫺4A ⫺ B ⫽ ⫺4A ⫹ a⫺ b B rewrite using algebraic addition 2 2 1⫺42142 1⫺42132 1⫺12 21⫺22 1⫺12 2132 ⫽ £ 1⫺42 1 12 2 1⫺42112 § ⫹ £ 1⫺12 2102 1⫺12 2162 § 1 1⫺42102 1⫺421⫺32 1⫺2 21⫺42 1⫺12 210.42 ⫺16 ⫺12 1 ⫺32 ⫽ £ ⫺2 ⫺4 § ⫹ £ 0 ⫺3 § simplify 0 12 2 ⫺0.2 ⫺16 ⫹ 1⫺32 2 ⫺12 ⫹ 1 ⫺11 ⫺35 2 ⫽ £ ⫺2 ⫹ 0 ⫺4 ⫹ 1⫺32 § ⫽ £ ⫺2 ⫺7 § result 0⫹2 12 ⫹ 1⫺0.22 2 11.8 Now try Exercises 25 through 28

䊳

Matrix Multiplication Consider a cable company offering three different levels of Internet service: Bronze— fast, Silver—very fast, and Gold—lightning fast. Table 9.3 shows the number and types of programs sold to households and businesses for the week. Each program has an incentive package consisting of a rebate and a certain number of free weeks, as shown in Table 9.4. Table 9.3 Matrix A Bronze

Silver

Table 9.4 Matrix B Gold

Rebate

Free Weeks

Homes

40

20

25

Bronze

$15

2

Businesses

10

15

45

Silver

$25

4

Gold

$35

6

To compute the amount of rebate money the cable company paid to households for the week, we would take the first row (R1) in Table 9.3 and multiply by the

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corresponding entries (bronze with bronze, silver with silver, and so on) in the first column (C1) of Table 9.4 and add these products. In matrix form, we have 15 # 3 40 20 25 4 £ 25 § ⫽ 40 # 15 ⫹ 20 # 25 ⫹ 25 # 35 ⫽ $1975. Using R1 of Table 9.3 35 with C2 from Table 9.4 gives the number of free weeks awarded to homes: 2 # 3 40 20 254 £ 4 § ⫽ 40 # 2 ⫹ 20 # 4 ⫹ 25 # 6 ⫽ 310. Using the second row (R2) of 6 Table 9.3 with the two columns from Table 9.4 will give the amount of rebate money and the number of free weeks, respectively, awarded to business customers. When all computations are complete, the result is a product matrix P with order 2 ⫻ 2. This is because the product of R1 from matrix A, with C1 from matrix B, gives the entry in 15 2 40 20 25 # 1975 310 d £ 25 4 § ⫽ c d. position P11 of the product matrix: c 10 15 45 2100 350 35 6 # Likewise, the product R1 C2 will give entry P12 (310), the product of R2 with C1 will give P21 (2100), and so on. This “row ⫻ column” multiplication can be generalized, and leads to the following. Given m ⫻ n matrix A and s ⫻ t matrix B, A 1m ⫻ n2

c

B 1s ⫻ t 2

A 1m ⫻ n2

c

c

matrix multiplication is possible only when n⫽s

B 1s ⫻ t 2

c

result will be an m ⫻ t matrix

In more formal terms, we have the following definition of matrix multiplication. Matrix Multiplication

Given the m ⫻ n matrix A ⫽ 3aij 4 and the s ⫻ t matrix B ⫽ 3bij 4. If n ⫽ s, then matrix multiplication is possible and the product AB is an m ⫻ t matrix P ⫽ 3 pij 4, where pij is product of the ith row of A with the jth column of B. In less formal terms, matrix multiplication involves multiplying the row entries of the first matrix with the corresponding column entries of the second, and adding them together. In Example 6, two of the matrix products [parts (a) and (b)] are shown in full detail, with the first entry of the product matrix color-coded.

EXAMPLE 6

䊳

Multiplying Matrices Given the matrices A through E shown here, compute the following products: a. AB

⫺2 A⫽ c 3

b. CD 1 d 4

4 B⫽ c 6

3 d 1

c. DC ⫺2 1 C⫽ £ 1 0 4 1

d. AE 3 2 § ⫺1

2 D ⫽ £4 0

e. EA 5 1 ⫺1 1 § 3 ⫺2

⫺2 E⫽ £ 3 1

⫺1 0 § 2

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Solution

䊳

⫺2 1 4 3 d ⫽ c dc 36 4 6 1 1⫺22142 ⫹ 112162 Computation: c 132 142 ⫹ 142 162

a. AB ⫽ c

⫺2 3

A B (2 ⫻ 2) (2 ⫻ 2)

⫺5 d 13 1⫺22 132 ⫹ 112 112 d 132132 ⫹ 142112

c

⫺2 1 d£ 3 4 1

⫺2 e. EA ⫽ £ 3 1

⫺1 ⫺2 0 §c 3 2

c

c

result will be a 2 ⫻ 2 matrix

C D (3 ⫻ 3) (3 ⫻ 3)

c

c

multiplication is possible since 3 ⫽ 3

1⫺22 152 ⫹ 112 1⫺12 ⫹ (3)132 112152 ⫹ 1021⫺12 ⫹ 122132 142 152 ⫹ 1121⫺12 ⫹ 1⫺12132

c result will be a 3 ⫻ 3 matrix

1⫺22 112 ⫹ 112112 ⫹ (3)1⫺22 112112 ⫹ 102112 ⫹ 1221⫺22 § 142112 ⫹ 112112 ⫹ 1⫺121⫺22 D (3 ⫻ 3)

c

C (3 ⫻ 3)

D (3 ⫻ 3)

c

c


A (2 ⫻ 2)

⫺1 0 § 2 1 1 d ⫽ £ ⫺6 4 4

c

C D (3 ⫻ 3) (3 ⫻ 3)

2 5 1 ⫺2 1 3 5 3 15 c. DC ⫽ £ 4 ⫺1 1 § £ 1 0 2 § ⫽ £ ⫺5 5 9 § 0 3 ⫺2 4 1 ⫺1 ⫺5 ⫺2 8

⫺2 d. AE ⫽ c 3

c


0 ⫺2 ⫺7 ⫺2 1 3 2 5 1 b. CD ⫽ £ 1 0 2 § £ 4 ⫺1 1 § ⫽ £ 2 11 ⫺3 § 12 16 7 4 1 ⫺1 0 3 ⫺2 1⫺22122 ⫹ 112142 ⫹ (3)(0) Computation: £ 112 122 ⫹ 102142 ⫹ 122(0) 142122 ⫹ 112 142 ⫹ 1⫺12(0)

A B (2 ⫻ 2) (2 ⫻ 2)

c

C (3 ⫻ 3)

c


E (3 ⫻ 2)

c

multiplication is not possible since 2 ⫽ 3

⫺6 3 § 9

E (3 ⫻ 2)

c

A (2 ⫻ 2)

c

E (3 ⫻ 2)

A (2 ⫻ 2)

c


c



䊳

Example 6 shows that in general, matrix multiplication is not commutative. Parts (b) and (c) show CD ⫽ DC since we get different results, and parts (d) and (e) show AE ⫽ EA, since AE is not defined while EA is. As with the addition and subtraction of matrices, matrix multiplication becomes cumbersome and time consuming for larger matrices, and we will often turn to the technology available in such cases. EXAMPLE 7

䊳

Using Technology for Matrix Operations Use a calculator to compute the product AB.

Solution

䊳

2 ⫺1 A⫽ ≥ 6 3

Carefully enter matrices A and B into the calculator, then press 2nd MODE (QUIT) to get to the home screen. Use [A][B] , and the calculator finds the product shown in the figure. Just for “fun,” we’ll

⫺3 5 0 2

0 1 2 4 ¥ B ⫽ £ 0.5 2 ⫺2 ⫺1

A (4 ⫻ 3)

⫺0.7 3.2 3 4

B (3 ⫻ 3)

A (4 ⫻ 3)

c

c

1 ⫺3 § 4 B (3 ⫻ 3)

ENTER

c


c


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double check the first entry of the product matrix: (2) 1 12 2 ⫹ (⫺3)(0.5) ⫹ (0)(⫺2) ⫽ ⫺0.5. ✓ 2 ⫺1 AB ⫽ ≥ 6 3

⫺3 5 0 2

0 1 2 4 ¥ £ 0.5 2 ⫺2 ⫺1

⫺0.7 3.2 3 4

1 ⫺3 § 4 Now try Exercises 41 through 52

䊳

Properties of Matrix Multiplication Earlier, Example 6 demonstrated that matrix multiplication is not commutative. Here is a group of properties that do hold for matrices. You are asked to check these properties in the Exercise Set using various matrices. See Exercises 53 through 56. Properties of Matrix Multiplication Given matrices A, B, and C for which the products are defined:

I. A1BC2 ⫽ 1AB2C II. A1B ⫹ C2 ⫽ AB ⫹ AC III. 1B ⫹ C2A ⫽ BA ⫹ CA IV. k1A ⫹ B2 ⫽ kA ⫹ kB

matrix multiplication is associative matrix multiplication is distributive from the left matrix multiplication is distributive from the right a constant k can be distributed over addition

We close this section with an application of matrix multiplication. There are many other interesting applications in the Exercise Set. EXAMPLE 8

䊳

Using Matrix Multiplication to Track Volunteer Enlistments In a certain country, the number of males and females that will join the military depends on their age. This information is stored in matrix A (Table 9.5). The likelihood a volunteer will join a particular branch of the military also depends on their age, with this information stored in matrix B (Table 9.6). (a) Compute the product P ⫽ AB and discuss/interpret what is indicated by the entries P11, P13, and P24 of the product matrix. (b) How many males are expected to join the Navy this year? Table 9.6 Matrix B

Table 9.5 Matrix A A

Solution

䊳

B

Age Group

Likelihood of Joining

Sex

18–19

20–21

22–23

Age Group

Army

Navy

Air Force

Marines

Female

1000

1500

500

18–19

0.42

0.28

0.17

0.13

Male

2500

3000

2000

20–21

0.38

0.26

0.27

0.09

22–23

0.33

0.25

0.35

0.07

a. Matrix A has order 2 ⫻ 3 and matrix B has order 3 ⫻ 4. The product matrix P can be found and is a 2 ⫻ 4 matrix. Carefully enter the matrices in your calculator. Figure 9.52 shows the entries of matrix B. Using 3A 4 3 B4 , the calculator finds the product matrix shown in Figure 9.53. Pressing the right arrow shows the complete product matrix is ENTER

P⫽ c

1155 2850

795 1980

750 1935

300 d. 735

The entry P11 is the product of R1 from A and C1 from B, and indicates that for the year, 1155 females are projected to join the Army. In like manner, entry P13 shows that 750 females are projected to join the Air Force. Entry P24 indicates that 735 males are projected to join the Marines.

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Figure 9.53

Figure 9.52

C. You’ve just seen how we can compute the product of two matrices

b. The product R2 (males) # C2 (Navy) gives P22 ⫽ 1980, meaning 1980 males are expected to join the Navy. Now try Exercise 59 through 66

䊳

9.6 EXERCISES 䊳



䊳

1. Two matrices are equal if they are like size and the corresponding entries are equal. In symbols, A ⫽ B ⫽ if .

2. The sum of two matrices (of like size) is found by adding the corresponding entries. In symbols, A⫹B⫽ .

3. The product of a constant times a matrix is called multiplication.

4. The size of a matrix is also referred to as its 1 2 3 d is The order of A ⫽ c . 4 5 6

5. Give two reasons why matrix multiplication is generally not commutative. Include several examples using matrices of various sizes.

6. Discuss the conditions under which matrix multiplication is defined. Include several examples using matrices of various sizes.

.


State the order of each matrix and name the entries in positions a12 and a23 if they exist. Then name the position aij of the “5” in each.

1 7. c 5 2 9. c 0 ⫺2 11. £ 0 5

⫺3 5 1 8 ⫺1

13. c

19 8. £ ⫺11 § 5

⫺3 d ⫺7 0.5 d 6 ⫺7 1§ 4

2 10. £ ⫺0.1 0.3 89 12. £ 13 2

Determine if the following statements are true, false, or conditional. If false, explain why. If conditional, find values of a, b, c, p, q, and r that will make the statement true.

3 2 14. ≥ ⫺1 2

0.4 5§ ⫺3 55 8 1

34 5 1

21 3§ 0

⫺2 15. £ 2b 0 16. £

14 132

11 116

⫺7 5 ⫺2 5 3 ⫺5 ⫺9

2p ⫹ 1 1 q⫹5

18 1 d ⫽ c 164 4

2 412

13 1.5 10 ¥ ⫽ c 1 ⫺0.5 3

⫺1.4 ⫺0.4

a c 4 § ⫽ £6 3c 0 ⫺5 12 9

3 ⫺5 ⫺3b

9 7 0 § ⫽ £ 1 ⫺2r ⫺2

2 12 d 8 1.3 d 0.3

⫺4 ⫺a § ⫺6 ⫺5 3r 3p

2⫺q 0 § ⫺8

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For matrices A through J as given, perform the indicated operation(s), if possible. If an operation cannot be completed, state why. Use a calculator only for those exercises designated by an icon.

A⫽ c

2 5

1 E ⫽ £0 4

0.5 5 § 3 ⫺2 ⫺1 3

⫺1 G⫽ £ 0 ⫺4 1 2 1 4 1 8

2 B⫽ £ 1 § ⫺3

3 d 8

2 C ⫽ £ 0.2 ⫺1

I⫽

9–78


0 2 § ⫺6

2 1 ⫺3 3 8 3 2 3 4

⫺

1 D ⫽ £0 0

0 ⫺2 § 6 1 4 5 ⫺ 8 5 ⫺ 2 ⫺

F⫽ c H⫽ c

0 1 0

6 12 8 ⫺5

9 d ⫺6

⫺3 d 2

7 32 J⫽ ≥ 5 ⫺ 16

A⫽ c

⫺5 3

5 8 ¥ 3 16

0 G⫽

⫺1 2 ⫺1 4

46. BH

47. DG

48. GD

2

21. H ⫹ J

22. A ⫺ J

23. G ⫹ I

24. I ⫺ G

25. 3H ⫺ 2A

26. 2E ⫹ 3G

31. AH

32. HA

33. CB

34. FH

35. HF

36. EB

37. H

2

39. FE

38. F2 40. EF

1 4 ⫺3 1 19 H⫽ ≥ 8 1 1 19 16

3 4 3 8 11 16

45. HB

51. FG

0 0§ 1

0 12 ⫺8 32 2 § F⫽ c d 4 8 16 ⫺6

44. GE

20. G ⫹ D

30. DE

0 1 0

43. EG

19. F ⫹ H

29. ED

1 D ⫽ £0 0

0 d 1

42. HA

49. C

2 28. F ⫺ F 3

1 0

41. AH

18. E ⫹ G

1 E ⫺ 3D 2

B⫽ c

13 3 § 2 13

⫺2 ⫺1 3

1 E ⫽ £0 4

17. A ⫹ H

27.

4 d 9

13 C⫽ £ 2 13

0 0§ 1

⫺3 0

For matrices A through H as given, use a calculator to perform the indicated operation(s), if possible. If an operation cannot be completed, state why.

4 57 ¥ 5 57

50. E2 52. AF

For Exercises 53 through 56, use a calculator and matrices A, B, and C to verify each statement.

⫺1 A⫽ £ 2 4

3 7 0

45 C ⫽ £ ⫺6 21

⫺1 10 ⫺28

5 ⫺1 § 6

0.3 B ⫽ £ ⫺2.5 1

⫺0.4 2 ⫺0.5

1.2 0.9 § 0.2

3 ⫺15 § 36

53. Matrix multiplication is not generally commutative: (a) AB ⫽ BA, (b) AC ⫽ CA, and (c) BC ⫽ CB. 54. Matrix multiplication is distributive from the left: A1B ⫹ C2 ⫽ AB ⫹ AC. 55. Matrix multiplication is distributive from the right: 1B ⫹ C2A ⫽ BA ⫹ CA. 56. Matrix multiplication is associative: 1AB2C ⫽ A1BC2.

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c


915


2 W

2 d 0

#

c

L Perimeter d ⴝ c d W Area

The perimeter and area of a rectangle can be simultaneously calculated using the matrix formula shown, where L represents the length and W represents the width of the rectangle. Use the matrix formula and your calculator to find the perimeter and area of the rectangles shown, then check the results using P ⫽ 2L ⫹ 2W and A ⫽ LW. 57.

6.374 cm

4.35 cm

䊳

58.

5.02 cm

3.75 cm

APPLICATIONS

59. Custom T’s designs and sells specialty T-shirts and sweatshirts, with factories in Verdi and Minsk. The company offers this apparel in three quality levels: standard, deluxe, and premium. Last fall the Verdi plant produced 3820 standard, 2460 deluxe, and 1540 premium T-shirts, along with 1960 standard, 1240 deluxe, and 920 premium sweatshirts. The Minsk plant produced 4220 standard, 2960 deluxe, and 1640 premium T-shirts, along with 2960 standard, 3240 deluxe, and 820 premium sweatshirts in the same time period. a. Write a 3 ⫻ 2 “production matrix” for each plant 3 V S Verdi, M S Minsk], with a T-shirt column, a sweatshirt column, and three rows showing how many of the different types of apparel were manufactured. b. Use the matrices from part (a) to determine how many more or fewer articles of clothing were produced by Minsk than Verdi. c. Use scalar multiplication to find how many shirts of each type will be made at Verdi and Minsk next fall, if each is expecting a 4% increase in business. d. Write a matrix that shows Custom T’s total production next fall (from both plants), for each type of apparel. 60. Terry’s Tire Store sells automobile and truck tires through three retail outlets. Sales at the Cahokia store for the months of January, February, and March broke down as follows: 350, 420, and 530 auto tires and 220, 180, and 140 truck tires. The Shady Oak branch sold 430, 560, and 690 auto tires and 280, 320, and 220 truck tires during the same 3 months. Sales figures for the downtown store were 864, 980, and 1236 auto tires and 535, 542, and 332 truck tires. a. Write a 2 ⫻ 3 “sales matrix” for each store 3C S Cahokia, S S Shady Oak, D S Downtown], with January, February, and

March columns, and two rows showing the sales of auto and truck tires respectively. b. Use the matrices from part (a) to determine how many more or fewer tires of each type the downtown store sold (each month) over the other two stores combined. c. Market trends indicate that for the same three months in the following year, the Cahokia store will likely experience a 10% increase in sales, the Shady Oak store a 3% decrease, with sales at the downtown store remaining level (no change). Write a matrix that shows the combined monthly sales from all three stores next year, for each type of tire. 61. Home improvements: Dream-Makers Home Improvements specializes in replacement windows, replacement doors, and new siding. During the peak season, the number of contracts that came from various parts of the city (North, South, East, and West) are shown in matrix C. The average profit per contract is shown in matrix P. Compute the product PC and discuss what each entry of the product matrix represents. Windows Doors Siding Windows 31500

N S E W 9 6 5 4 £7 5 7 6§ ⫽ C 2 3 5 2 Doors Siding 500 2500 4 ⫽ P

62. Classical music: Station 90.7 — The Home of Classical Music — is having their annual fund drive. Being a loyal listener, Mitchell decides that for the next 3 days he will donate money according to his favorite composers, by the number of times their music comes on the air: $3 for every piece by Mozart (M), $2.50 for every piece by Beethoven (B), and $2 for every piece by Vivaldi (V).

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This information is displayed in matrix D. The number of pieces he heard from each composer is displayed in matrix C. Compute the product DC and discuss what each entry of the product matrix represents. Mon. Tue. Wed. M 4 3 5 B £3 2 4§ ⫽ C V 2 3 3 M B V 3 3 2.5 2 4 ⫽ D 63. Pizza and salad: The science department and math department of a local college are at a pre-semester retreat, and decide to have pizza, salads, and soft drinks for lunch. The quantity of food ordered by each department is shown in matrix Q. The cost of the food item at each restaurant is shown in matrix C using the published prices from three popular restaurants: Pizza Home (PH), Papa Jeff’s (PJ), and Dynamos (D). a. What is the total cost to the math department if the food is ordered from Pizza Home? b. What is the total cost to the science department if the food is ordered from Papa Jeff’s? c. Compute the product QC and discuss the meaning of each entry in the product matrix. Pizza Science 8 c Math 10 PH Pizza 8 Salad £ 1.5 Drink 0.90

Salad Drink 12 20 d ⫽Q 8 18 PJ D 7.5 10 1.75 2 § ⫽C 1 0.75

64. Manufacturing pool tables: Cue Ball Incorporated makes three types of pool tables, for homes, commercial use, and professional use. The amount of time required to pack, load, and install each is summarized in matrix T, with all times in hours. The cost of these components in dollars per hour, is summarized in matrix C for two of its warehouses, one on the west coast and the other in the midwest. a. What is the cost to package, load, and install a commercial pool table from the coastal warehouse? b. What is the cost to package, load, and install a commercial pool table from the warehouse in the midwest? c. Compute the product TC and discuss the meaning of each entry in the product matrix.

Pack Load Install Home 1 0.2 1.5 Comm £ 1.5 0.5 2.2 § ⫽ T Prof 1.75 0.75 2.5 Coast Midwest Pack 10 8 Load £ 12 10.5 § ⫽ C Install 13.5 12.5 65. Joining a club: Each school year, among the students planning to join a club, the likelihood a student joins a particular club depends on their class standing. This information is stored in matrix C. The number of males and females from each class that are projected to join a club each year is stored in matrix J. Compute the product JC and use the result to answer the following: a. Approximately how many females joined the chess club? b. Approximately how many males joined the writing club? c. What does the entry p13 of the product matrix tells us? Fresh Female 25 c Male 22 Spanish Fresh 0.6 Soph £ 0.5 Junior 0.4

Soph Junior 18 21 d ⫽J 19 18 Chess 0.1 0.2 0.2

Writing 0.3 0.3 § ⫽ C 0.4

66. Designer shirts: The SweatShirt Shoppe sells three types of designs on its products: stenciled (S), embossed (E), and applique (A). The quantity of each size sold is shown in matrix Q. The retail price of each sweatshirt depends on its size and whether it was finished by hand or machine. Retail prices are shown in matrix C. Assuming all stock is sold, compute the product QC and use the result to answer the following. a. How much revenue was generated by the large sweatshirts? b. How much revenue was generated by the extra-large sweatshirts? c. What does the entry p11 of the product matrix QC tell us? S med 30 large £ 60 x-large 50

E 30 50 40

A 15 20 § ⫽ Q 30

Hand S 40 E £ 60 A 90

Machine 25 40 § ⫽ C 60

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Section 9.7 Solving Linear Systems Using Matrix Equations


67. For the matrix A shown, use your calculator to compute A2, A3, A4, and A5. Do you notice a pattern? Try to write a “matrix formula” for An, where n is a positive integer, then use your formula to find A6. Check results using a calculator. 1 0 1 A ⫽ £1 1 1§ 1 0 1 䊳

917

68. The matrix M ⫽ c

2 1 d has some very ⫺3 ⫺2 interesting properties. Compute the powers M 2, M 3, M 4, and M 5, then discuss what you find. Try to find/create another 2 ⫻ 2 matrix that has similar properties.


69. (9.2) Solve the system using elimination. x ⫹ 2y ⫺ z ⫽ 3 • ⫺2x ⫺ y ⫹ 3z ⫽ ⫺5 5x ⫹ 3y ⫺ 2z ⫽ 2

72. (4.1) Find the quotient using synthetic division, then check using multiplication.

70. (7.5) Evaluate cos1cos⫺10.32112 .

9.7

71. (8.6) Solve z4 ⫺ 81i ⫽ 0 using the nth roots theorem. Leave your answer in trigonometric form.

x3 ⫺ 9x ⫹ 10 x⫺2

Solving Linear Systems Using Matrix Equations


A. Recognize the identity matrix for multiplication B. Find the inverse of a square matrix C. Solve systems using matrix equations D. Use determinants to find whether a matrix is invertible

While using matrices and row operations offers a degree of efficiency in solving systems, we are still required to solve for each variable individually. Using matrix multiplication we can actually rewrite a given system as a single matrix equation, in which the solutions are computed simultaneously. As with other kinds of equations, the use of identities and inverses are involved, which we now develop in the context of matrices.

A. Multiplication and Identity Matrices From the properties of real numbers, 1 is the identity for multiplication since n # 1 ⫽ 1 # n ⫽ n. A similar identity exists for matrix multiplication. Consider the 2 ⫻ 2 1 4 matrix A ⫽ c d . While matrix multiplication is not generally commutative, ⫺2 3 if we can find a matrix B where AB ⫽ BA ⫽ A, then B is a prime candidate for the identity matrix, which is denoted I. For the products AB and BA to be possible and have the same order as A, we note B must also be a 2 ⫻ 2 matrix. Using the arbitrary matrix a b B⫽ c d , we have the following. c d

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EXAMPLE 1A

䊳

Solving AB A to Find the Identity Matrix For c

1 4 a b 1 4 d c d c d , use matrix multiplication, the equality of 2 3 c d 2 3 matrices, and systems of equations to find the value of a, b, c, and d.

Solution

䊳

a 4c b 4d 1 4 d c d. 2a 3c 2b 3d 2 3 Since corresponding entries must be equal (shown by matching colors), we can find a 4c 1 b 4d 4 a, b, c, and d by solving the systems e and e . For 2a 3c 2 2b 3d 3 the first system, 2R1 R2 shows a 1 and c 0. Using 2R1 R2 for the second 1 0 d is a candidate for the identity matrix. shows b 0 and d 1. It appears c 0 1 The product on the left gives c

Before we name B as the identity matrix, we must show that AB BA A. EXAMPLE 1B

䊳

Verifying AB BA A Given A c

Solution

䊳

1 2

1 4 d and B c 0 3

0 d , determine if AB A and BA A. 1

AB c

1 4 1 0 d c d 2 3 0 1 1112 4102 1102 4112 d c 2112 3102 2102 3112 1 4 c d A✓ 2 3

BA c

1 0 1 4 d c d 0 1 2 3 1112 0122 1142 0132 c d 0112 1122 0142 1132 1 4 c d A✓ 2 3

Since AB A BA, B is the identity matrix I. Now try Exercises 7 through 10 By replacing the entries of A c

䊳

1 4 d with those of the general matrix 2 3 a11 a12 1 0 d , we can show that I c c d is the identity for all 2 2 matrices. In a21 a22 0 1 considering the identity for larger matrices, we find that only square matrices have inverses, since AI IA is the primary requirement (the multiplication must be possible in both directions). This is commonly referred to as multiplication from the right and multiplication from the left. Using the same procedure as before we can show 1 0 0 £ 0 1 0 § is the identity for 3 3 matrices (denoted I3). The n n identity matrix 0 0 1 In consists of 1’s down the main diagonal and 0’s for all other entries. Also, the identity In for a square matrix is unique.

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As in Section 9.6, a graphing calculator can be used to investigate operations on matrices and matrix properties. For the 3 3 matrix 2 5 1 1 0 0 A £ 4 1 1 § and I3 £ 0 1 0 § , a 0 3 2 0 0 1 calculator will confirm that AI3 A I3 A. Carefully enter A into your calculator as matrix A, and I3 as matrix B. Figure 9.54 shows AB A and BA A. See Exercises 11 through 14.

A. You’ve just seen how we can recognize the identity matrix for multiplication

919

Figure 9.54

B. The Inverse of a Matrix Again from the properties of real numbers, we know the multiplicative inverse for a is 1 a1 1a 02, since the products a # a1 and a1 # a yield the identity 1. To show a 6 5 d and that a similar inverse exists for matrices, consider the square matrix A c 2 2 a b d . If we can find a matrix B, where AB BA I, then B an arbitrary matrix B c c d is a prime candidate for the inverse matrix of A, which is denoted A1. In an attempt to find such a matrix B, we proceed as in Examples 1A and 1B. EXAMPLE 2A

䊳

Solving AB I to find A1 For c

6 5 a b 1 0 d c d c d , use matrix multiplication, the equality of matrices, 2 2 c d 0 1 and systems of equations to find the entries of B.

Solution

䊳

6a 5c 6b 5d 1 0 d c d . Since corresponding 2a 2c 2b 2d 0 1 entries must be equal (shown by matching colors), we find the values of a, b, c, and 6a 5c 1 6b 5d 0 . Using 3R2 R1 d by solving the systems e and e 2a 2c 0 2b 2d 1 for the first system shows a 1 and c 1, while 3R2 R1 for the second a b 1 2.5 d c d is the system shows b 2.5 and d 3. Matrix B c c d 1 3 prime candidate for A1. The product on the left gives c

It may have occurred to you that the systems used in Example 2A have exactly 6 5 1 6 5 0 the same coefficients. In matrix form, these systems are c d and c d. 2 2 0 2 2 1 Instead of solving these two systems separately, we can solve them simultaneously by sim6 5 1 0 ply using the coefficient matrix augmented with the identity matrix: c d. 2 2 0 1 Row operations are then used to transform the given matrix into the identity: 1 0 e f c d , in a sense “preserving” the inverse operations needed. The inverse 0 1 g h e f matrix is then c d. g h

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EXAMPLE 2B

䊳

Finding a 2 2 Inverse Using the Augmented Matrix Use the augmented matrix method to find A1 for A c calculator to verify A1A AA1 I.

Solution

䊳

First we augment the given matrix with the identity: c

6 2 row operations to transform A into the identity matrix. c c

6 0

6 c 0

6 2

5 2

5 1 0 1

1 1 6 1

5 2

0 d . Then we use 1

1 0

3R2 R1 S R2

c

6 0

5 1

1 1

0 d 3

0 d 3

5R2 R1 S R1

c

6 0

0 1

6 1

15 d 3

15 d 3

R1 S R1 6 1R2 S R2

c

1 0

0 1

1 1

2.5 d 3

1 0

0 d 1

5 d . Then use a 2

6 2

As in Example 2A, a prime candidate for A1 is c

1 1

2.5 d . Using a calculator 3

with A 3 A4 and A1 3B 4 confirms that A1A AA1 I (see figure). Now try Exercises 15 through 22

䊳

These observations guide us to the following definition of an inverse matrix. The Inverse of a Matrix Given an n n matrix A, if there exists an n n matrix A1 such that AA1 A1A In, then A1 is the inverse of matrix A. We will soon discover that while only square matrices have inverses, not every square matrix has an inverse. If an inverse exists, the matrix is said to be invertible. For 2 2 matrices that are invertible, a simple formula exists for computing the inverse. The formula is derived in the Strengthening Core Skills feature at the end of Chapter 9. The Inverse of a 2 2 Matrix If A c

a b 1 d d , then A1 c c d ad bc c

b d provided ad bc 0 a

To “test” the formula, again consider the matrix A c c 2, and d 2: A1

2 1 c 162122 152122 2

1 1 2 5 c d c 2 2 6 1

6 2

5 d , where a 6, b 5, 2

5 d 6 2.5 d✓ 3

See Exercises 63 through 66 for more practice with this formula.

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Almost without exception, real-world applications involve much larger matrices, with entries that are not integer-valued. Although the augmented matrix method from Example 2 can be extended to find the inverse of larger matrices, the process becomes very tedious and too time consuming to be useful. The process for a 3 3 matrix is discussed in the Strengthening Core Skills feature at the end of Chapter 9. But for practical reasons, we will rely on a calculator to produce these larger inverse matrices. This is done by (1) carefully entering a square matrix A into the calculator, (2) returning to the home screen, and (3) calling up matrix A and pressing the x key and to find A1. In the context of matrices, calculators are programmed to compute an inverse matrix using this key, rather than to somehow find a reciprocal. See Exercises 23 through 26. -1

B. You’ve just seen how we can find the inverse of a square matrix

ENTER

C. Solving Systems Using Matrix Equations One reason matrix multiplication has its row column definition is to assist in writing a linear system of equations as a single matrix equation. The equation consists of the matrix of constants B on the right, and a product of the coefficient matrix A with the x 4y z 10 matrix of variables X on the left: AX B. For • 2x 5y 3z 7, the matrix 8x y 2z 11 1 4 1 x 10 equation is £ 2 5 3 § £ y § £ 7 § . Note that computing the product on the left 8 1 2 z 11 will yield the original system. Once written as a matrix equation, the system can be solved using an inverse matrix and the following sequence. If A represents the matrix of coefficients, X the matrix of variables, B the matrix of constants, and I the appropriate identity, the sequence is 112

122 132

142 152

AX B

1

A

matrix equation

1AX2 A B 1

multiply from the left by the inverse of A

1A A2X A B 1

1

associative property

1

IX A B X A1B

A 1 A I IX X

Lines 1 through 5 illustrate the steps that make the method work. In actual practice, after carefully entering the matrices, only step 5 is used when solving matrix equations using technology. Once matrix A is entered, the calculator will automatically find and use A1 as we enter A1B. EXAMPLE 3

䊳

Using Technology to Solve a Matrix Equation Use a calculator and a matrix equation to solve the system x 4y z 10 • 2x 5y 3z 7. 8x y 2z 11

Solution

䊳

1 As before, the matrix equation is £ 2 8

4 5 1

1 x 10 3 § £ y § £ 7 § . 2 z 11

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Carefully enter (and double-check) the matrix of coefficients as matrix A in your calculator, and the matrix of constants as matrix B (see Figure 9.55). The product A 1B shows the solution is x 2, y 3, z 4 (see Figure 9.56). Verify by substitution. Figure 9.55 Figure 9.56


The matrix equation method does have a few shortcomings. Consider the system 4 10 x 8 d c d c d . After entering whose corresponding matrix equation is c 2 5 y 13 the matrix of coefficients A and matrix of constants B, attempting to compute A1B results in the error message shown in Figure 9.57. The calculator is unable to return a solution due to something called a “singular matrix.” To investigate further, we attempt 4 10 d using the formula for a 2 2 matrix. With a 4, to find A1 for c 2 5 b 10, c 2, and d 5, we have A1

C. You’ve just seen how we can solve systems using matrix equations

䊳

1 d c ad bc c

5 b 1 c d a 142 152 1102 122 2

1 5 10 d c 4 0 2

10 d 4

Since division by zero is undefined, we conclude that matrix A has no inverse. A matrix having no inverse is said to be singular or noninvertible. Solving systems using matrix equations is only possible when the matrix of coefficients is nonsingular.

D. Determinants and Singular Matrices As a practical matter, it becomes important to know ahead of time whether a particular matrix has an inverse. To help with this, we introduce one additional operation on a square matrix, that of calculating its determinant. For a 1 1 matrix the determinant a11 a12 d , the determinant of A, written is the entry itself. For a 2 2 matrix A c a21 a22 as det(A) or denoted with vertical bars as 冟A冟, is computed as a difference of diagonal products beginning with the upper-left entry:

det1A2 `

a11 a21

2nd diagonal product a12 ` a11a22 a21a12 a22 1st diagonal product

The Determinant of a 2 2 Matrix Given any 2 2 matrix A c

a11 a12 d, a21 a22

det1A2 冟A冟 a11a22 a21a12

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EXAMPLE 4

䊳

Calculating Determinants Compute the determinant of each matrix. 3 2 5 2 1 a. B c b. C c d d 1 6 1 3 4

Solution

䊳

c. D c

4 2

10 d 5

a. det1B2 `

3 2 ` 132162 112122 20 1 6 b. Determinants are only defined for square matrices (see figure). 4 10 c. det1D2 ` ` 142 152 122 1102 20 20 0 2 5


䊳

4 10 d is zero, and this is the 2 5 same matrix we earlier found had no inverse. This observation can be extended to larger matrices and offers the connection we seek between a given matrix, its inverse, and matrix equations. Notice from Example 4(c), the determinant of c

Singular Matrices If A is a square matrix and det1A2 0, the inverse matrix does not exist and A is said to be singular or noninvertible.

WORTHY OF NOTE For the determinant of a general n n matrix using cofactors, see Appendix D.

In summary, inverses exist only for square matrices, but not every square matrix has an inverse. If the determinant of a square matrix is zero, an inverse does not exist and the method of matrix equations cannot be used to solve the system. To use the determinant test for a 3 3 system, we need to compute a 3 3 determinant. At first glance, our experience with 2 2 determinants appears to be of little help. However, every entry in a 3 3 matrix is associated with a smaller 2 2 matrix, formed by deleting the row and column of that entry and using the entries that remain. These 2 2’s are called the associated minor matrices or simply the minors. Using a general matrix of coefficients, we’ll identify the minors associated with the entries in the first row. a11 a12 a13 £ a21 a22 a23 § a31 a32 a33

a11 £ a21 a31

a12 a22 a32

a13 a23 § a33

Entry: a11 associated minor a22 a23 c d a32 a33

Entry: a12 associated minor a21 a23 c d a31 a33

a11 £ a21 a31

a12 a22 a32

a13 a23 § a33

Entry: a13 associated minor a21 a22 d c a31 a32

To illustrate, consider the system shown, and (1) form the matrix of coefficients, (2) identify the minor matrices associated with the entries in the first row, and (3) compute the determinant of each minor.

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2x 3y z 1 • x 4y 2z 3 1 3x y 2 122 £ 1 3

3 1 4 2 § 1 0

2 (1) Matrix of coefficients £ 1 3

3 1 4 2§ 1 0

2 £1 3

3 4 1

1 2§ 0

2 £1 3

1 2§ 0

3 4 1

Entry a11: 2 associated minor 4 2 c d 1 0



(3) Determinant of minor

Determinant of minor

Determinant of minor

142 102 112122 2

112 102 132122 6 112 112 132142 13

For computing a 3 3 determinant, we illustrate a technique called expansion by minors. The Determinant of a 3 3 Matrix — Expansion by Minors For the matrix M shown, det(M) is the unique number computed matrix M as follows: a11 a12 a13 1. Select any row or column and form the product of each £ a21 a22 a23 § entry with its minor matrix. The illustration here uses the a31 a32 a33 entries in row 1: det1M2 a11 `

a22 a23 a21 a23 a21 a22 ` a12 ` ` a13 ` ` a32 a33 a31 a33 a31 a32

2. The signs used between terms of the expansion depends on the row or column chosen, according to the sign chart shown.

Sign Chart £

§

The determinant of a matrix is unique and any row or column can be used. For this reason, it’s helpful to select the row or column having the most zero, positive, and/or smaller entries. EXAMPLE 5

䊳

Calculating a 3 3 Determinant 2 1 Compute the determinant of M £ 1 1 2 1

Solution

䊳

3 0 §. 4

Since the second row has the “smallest” entries as well as a zero entry, we compute the determinant using this row. According to the sign chart, the signs of the terms will be negative–positive–negative, giving 1 3 2 3 2 1 ` 112 ` ` 102 ` ` 1 4 2 4 2 1 114 32 11218 62 10212 22 7 (2) 0 9 The value of det1M2 is 9.

det1M2 112 `


䊳

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Try computing the determinant of M two more times, using a different row or column each time. Since the determinant is unique, you should obtain the same result. There are actually other alternatives for computing a 3 3 determinant. The first is called determinants by column rotation, and takes advantage of patterns generated from the expansion of minors. This method is applied to the matrix shown, which uses alphabetical entries for simplicity. a det £ d g

b e h

c a1ei fh2 b1di fg2 c1dh eg2 f § aei afh bdi bfg cdh ceg i aei bfg cdh afh bdi ceg

expansion using R1 distribute rewrite result

Although history is unsure of who should be credited, notice that if you repeat the first two columns to the right of the given matrix (“rotation of columns”), identical products are obtained using the six diagonals formed—three in the downward direction using addition, three in the upward direction using subtraction. a £d g

b e h

gec c f§ i aei

hfa idb a b d e g h bfg cdh

Adding the products in blue (regardless of sign) and subtracting the products in red (regardless of sign) gives the determinant. This method is more efficient than expansion by minors, but can only be used for 3 3 matrices! EXAMPLE 6

䊳

Calculating det(A) Using Column Rotation 1 Use the column rotation method to find the determinant of A £ 2 3

Solution

䊳

5 3 8 0 § . 11 1

Rotate columns 1 and 2 to the right, and compute the diagonal products. 1 £ 2 3

72 0 10 1 5 5 3 8 0 § 2 8 3 11 11 1 8 0 66

Adding the products in blue (regardless of sign) and subtracting the products in red (regardless of sign) shows det1A2 4: 8 0 66 72 0 1102 4.


䊳

The final method is presented in the Extending the Concept feature of the Exercise Set, and shows that if certain conditions are met, the determinant of a matrix can be found using its triangularized form. As with the operations studied in Section 9.6, the process of computing a determinant becomes very cumbersome for larger matrices, or those with rational or radical entries. Most graphing calculators are programmed to handle these computations easily. x After accessing the matrix menu ( 2nd ), calculating a determinant is the first option under the MATH submenu (Figure 9.58). The calculator results for det([A]) and det([B]) as defined are shown in Figures 9.59 and 9.60. See Exercises 57 and 58. -1

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Figure 9.58

EXAMPLE 7

䊳

Figure 9.59

Figure 9.60

Solving a System after Verifying A is Invertible Given the system shown here, (1) form the matrix equation AX B; (2) compute the determinant of the coefficient matrix and determine if you can proceed; and (3) if so, solve the system using a matrix equation. 2x 1y 3z 11 • 1x 1y 1 2x 1y 4z 8

Solution

䊳

1. Form the matrix equation AX B: 2 £ 1 2

1 3 x 11 1 0 § £ y § £ 1 § 1 4 z 8

2. Since det(A) is nonzero (from Example 5 and Figure 9.61), we can proceed. 3. For X A1B, input A1B on your calculator and press (Figure 9.62). Notice that matrix name [B] has scrolled out of view. ENTER

Figure 9.61

Figure 9.62

The solution is the ordered triple 13, 2, 12 . Now try Exercises 59 through 62

䊳

We close this section with an application involving a 4 4 system. There is a large variety of additional applications in the Exercise Set. EXAMPLE 8

䊳

Solving an Application Using Technology and Matrix Equations A local theater sells four sizes of soft drinks: 32 oz @ $2.25; 24 oz @ $1.90; 16 oz @ $1.50; and 12 oz @ $1.20/each. As part of a “free guest pass” promotion, the manager asks employees to try and determine the number of each size sold, given the following information: (1) the total revenue from soft drinks was $719.80; (2) there were 9096 oz of soft drink sold; (3) there was a total of 394 soft drinks sold; and (4) the number of 24-oz and 12-oz drinks sold was 12 more than the number of 32-oz and 16-oz drinks sold. Write a system of equations that models this information, then solve the system using a matrix equation.

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Solution

䊳

927

If we let x, l, m, and s represent the number of 32-oz, 24-oz, 16-oz, and 12-oz soft drinks sold, the following system is produced: 2.25x 1.90l 1.50m 1.20s 719.8 revenue: 32x 24l 16m 12s 9096 ounces sold: μ x l m s 394 quantity sold: l s x m 12 amounts sold: When written as a matrix equation the system becomes: 2.25 32 ≥ 1 1

1.9 24 1 1

1.5 1.2 x 719.8 16 12 l 9096 ¥ ≥ ¥ ≥ ¥ 1 1 m 394 1 1 s 12

To solve, carefully enter the matrix of coefficients as matrix A (see Figure 9.63), and the matrix of constants as matrix B, then compute A1B X [since det1A2 0]. This gives a solution of 1x, l, m, s2 1112, 151, 79, 522 1Figure 9.642. Figure 9.63

D. You’ve just seen how we can use determinants to find whether a matrix is invertible

Figure 9.64


䊳

9.7 EXERCISES 䊳



1. The n n identity matrix In consists of 1’s down the and for all other entries.

2. The product of a square matrix A and its inverse A1 yields the matrix.

3. Given square matrices A and B of like size, B is the inverse of A if . Notationally we write B .

4. If the determinant of a matrix is zero, the matrix is said to be or , meaning no inverse exists.

5. Explain why inverses exist only for square matrices, then discuss why some square matrices do not have an inverse. Illustrate each point with an example.

6. What is the connection between the determinant of a 2 2 matrix and the formula for finding its inverse? Use the connection to create a 2 2 matrix that is invertible, and another that is not.

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Use matrix multiplication, equality of matrices, and the a b 1 0 arbitrary matrix given to show that c d c d. c d 0 1

2 7. A c 3

5 a b 2 5 dc d c d 7 c d 3 7

8. A c

9 5

7 a dc 4 c

b 9 d c d 5

7 d 4

9. A c

0.4 0.3

0.6 a dc 0.2 c

b 0.4 d c d 0.3

0.6 d 0.2

1 4 1d 8

1

10. A c 21 3

For I2 c

1 0

c

a c

1 4 1d 8

1 b d c 21 d 3

1 0 d , I3 £ 0 1 0

0 1 0

3 4

8 d 10

0 0 § , and 1

4 1 13. £ 9 5 0 2

12. c 6 3§ 1

0.5 0.7

9 2 14. ≥ 4 0

0.2 d 0.3

1 3 1 0 5 3 ¥ 6 1 0 2 4 1

Find the inverse of each 2 2 matrix using matrix multiplication, equality of matrices, and a system of equations.

15. c

5 2

4 d 2

16. c

1 0

5 d 4

Find the inverse of each matrix by augmenting of the the identity matrix and using row operations.

17. c

1 4

3 d 10

18. c

4 0

5 d 2

1

5 8 1d 2

B c4 0

2 0.4 d 1 0.8

Demonstrate that B A1, by showing AB BA I. Do not use a calculator.

19. A c

1 2

5 d 9

20. A c

2 6 d 4 11

B c

9 2

5 d 1

B c

5.5 3 d 2 1

22. A c

2 5 d 3 4 4

B c 73 7

5 7 2d 7

Use a calculator to find A1 B, then confirm the inverse by showing AB BA I.

2 3 1 23. A £ 5 2 4 § 2 0 1 0.5 24. A £ 0 1

1 0 0 0 0 1 0 0 I4 ≥ ¥ , show AI IA A for the 0 0 1 0 0 0 0 1 matrices of like size. Use a calculator for Exercise 14.

11. c

21. A c

0.2 0.3 0.4

0.1 0.6 § 0.3

7 5 3 25. A £ 1 9 0 § 2 2 5 12 1 0 26. A ≥ 12 12 0

6 4 12 12

3 0 8 12 ¥ 0 0 0 12

Write each system in the form of a matrix equation. Do not solve.

27. e

2x 3y 9 5x 7y 8

28. e

0.5x 0.6y 0.6 0.7x 0.4y 0.375

x 2y z 1 29. • x z 3 2x y z 3 2x 3y 2z 4 30. • 14x 25y 34z 1 3 2x 1.3y 3z 5 2w x 4y 5z 3 2w 5x y 3z 4 31. μ 3w x 6y z 1 w 4x 5y z 9 1.5w 2.1x 0.4y z 1 0.2w 2.6x y 5.8 32. μ 3.2x z 2.7 1.6w 4x 5y 2.6z 1.8

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Precalculus—

9–93 Write each system as a matrix equation and solve (if possible) using inverse matrices and your calculator. If the coefficient matrix is singular, write no solution.

33. e

0.05x 3.2y 15.8 0.02x 2.4y 12.08

34. e

0.3x 1.1y 3.5 0.5x 2.9y 10.1

35.

12a 13b 216 16a b 4 12

37. e

5 3 3 2 a 5 b 10 3 7 5 16 a 2 b 16

38. e

3 12a 213b 12 512a 313b 1

1 52. D £ 2.5 3

4x 5y 6z 33 35y 54z 9 0.5x 2.4y 4z 32

2w 3x 4y 5z 3 0.2w 2.6x y 0.4z 2.4 43. μ 3w 3.2x 2.8y z 6.1 1.6w 4x 5y 2.6z 9.8 2w 5x 3y 4z 7 1.6w 4.2y 1.8z 5.4 44. μ 3w 6.7x 9y 4z 8.5 0.7x 0.9z 0.9 Compute the determinant of each matrix and state whether an inverse matrix exists. Do not use a calculator.

1.2 0.3

0.8 d 0.2

0.8 2 § 2.5

2 53. £ 4 1

3 1 1 5 § 0 2

1 55. £ 3 4

1 2 2 4 § 3 1

54.

3 £ 1 3

56.

5 6 2 £ 2 1 2 § 3 4 1

-1

• 18x

47. c

4 2 § 2

2 4 2 0 § 1 5

Use a calculator to compute the determinant of each matrix. If the determinant is zero, write singular matrix. If the determinant is nonzero, find A1 and store the result as matrix B ( STO 2nd x 2: [B] ). Then verify the inverse by showing AB BA I.

x 2y 2z 9 41. • 2x 1.5y 1.8z 12 2 1 3 3 x 2y 5 z 4

7 d 5

2 5 0

1 2§ 0

Compute the determinant of each matrix using the column rotation method.

1.7x 2.3y 2z 41.5 40. • 1.4x 0.9y 1.6z 10 0.8x 1.8y 0.5z 16.5

4 3

0 2 2 2 1 1 § 50. B £ 0 1 1 4 4 4

2 3 51. C £ 0 6 1 1.5

0.2x 1.6y 2z 1.9 39. • 0.4x y 0.6z 1 0.8x 3.2y 0.4z 0.2

45. c

Compute the determinant of each matrix without using a calculator. If the determinant is zero, write singular matrix.

1 49. A £ 0 2

1 u 14v 1 e 16 2 2 u 3 v 2

36. e

42.

929


46. c

0.6 0.4

0.3 d 0.5

48. c

2 3

6 d 9

1 2 57. A ≥ 8 0

0 5 15 8

3 0 6 4

1 2 0 1 58. M ≥ 1 0 2 1

ENTER

4 1 ¥ 5 1 1 3 2 1

1 2 ¥ 3 4

For each system shown, form the matrix equation AX B; compute the determinant of the coefficient matrix and determine if you can proceed; and if possible, solve the system using the matrix equation.

x 2y 2z 7 2x 3y 2z 7 59. • 2x 2y z 5 60. • x y 2z 5 3x y z 6 3x 2y z 11 x 3y 4z 1 5x 2y z 1 61. • 4x y 5z 7 62. • 3x 4y 9z 2 3x 2y z 3 4x 3y 5z 6

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The inverse of a 2 2 matrix: A c

a b 1 #c d d S A 1 c d ad bc c

b d a

The inverse of a 2 2 matrix can be found using the formula shown, as long as ad bc 0. Use the formula to find inverses for the matrices here (if possible), then verify by showing A # A1 A # A1 I.

63. A c 䊳

3 2

5 d 1

64. B c

2 3 d 5 4

65. C c

0.3 0.6

0.4 d 0.8

66. c

0.2 0.4

0.3 d 0.6

APPLICATIONS

Solve each application using a matrix equation. Descriptive Translation

67. Convenience store sales: The local Moto-Mart sells four different sizes of Slushies — behemoth, 60 oz @ $2.59; gargantuan, 48 oz @ $2.29; mammoth, 36 oz @ $1.99; and jumbo, 24 oz @ $1.59. As part of a promotion, the owner offers free gas to any customer who can tell how many of each size were sold last week, given the following information: (1) The total revenue for the Slushies was $402.29; (2) 7884 ounces were sold; (3) a total of 191 Slushies were sold; and (4) the number of behemoth Slushies sold was one more than the number of jumbo. How many of each size were sold? 68. Cartoon characters: In America, four of the most beloved cartoon characters are Foghorn Leghorn, Elmer Fudd, Bugs Bunny, and Tweety Bird. Suppose that Bugs Bunny is four times as tall as Tweety Bird. Elmer Fudd is as tall as the combined height of Bugs Bunny and Tweety Bird. Foghorn Leghorn is 20 cm taller than the combined height of Elmer Fudd and Tweety Bird. The combined height of all four characters is 500 cm. How tall is each one? 69. Rolling Stones music: One of the most prolific and popular rock-and-roll bands of all time is the Rolling Stones. Four of their many great hits include: Jumpin’ Jack Flash, Tumbling Dice, You Can’t Always Get What You Want, and Wild Horses. The total playing time of all four songs is 20.75 min. The combined playing time of Jumpin’ Jack Flash and Tumbling Dice equals that of You Can’t Always Get What You Want. Wild Horses is 2 min longer than Jumpin’ Jack Flash, and You Can’t Always Get What You Want is twice as long as Tumbling Dice. Find the playing time of each song. 70. Mozart’s arias: Mozart wrote some of vocal music’s most memorable arias in his operas,

including Tamino’s Aria, Papageno’s Aria, the Champagne Aria, and the Catalogue Aria. The total playing time of all four arias is 14.3 min. Papageno’s Aria is 3 min shorter than the Catalogue Aria. The Champagne Aria is 2.7 min shorter than Tamino’s Aria. The combined time of Tamino’s Aria and Papageno’s Aria is five times that of the Champagne Aria. Find the playing time of all four arias. Manufacturing

71. Resource allocation: Time Pieces Inc. manufactures four different types of grandfather clocks. Each clock requires these four stages: (1) assembly, (2) installing the clockworks, (3) inspection and testing, and (4) packaging for delivery. The time required for each stage is shown in the table, for each of the four clock types. At the end of a busy week, the owner determines that personnel on the assembly line worked for 262 hr, the installation crews for 160 hr, the testing department for 29 hr, and the packaging department for 68 hr. How many clocks of each type were made? Dept.

Clock A

Clock B

Clock C

Clock D

Assemble

2.2

2.5

2.75

3

Install

1.2

1.4

1.8

2

Test

0.2

0.25

0.3

0.5

Pack

0.5

0.55

0.75

1.0

72. Resource allocation: Figurines Inc. makes and sells four sizes of metal figurines, mostly historical figures and celebrities. Each figurine goes through four stages of development: (1) casting, (2) trimming, (3) polishing, and (4) painting. The time required for each stage is shown in the table, for each of the four sizes. At the end of a busy week, the manager finds that the casting department put in 62 hr, and

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the trimming department worked for 93.5 hr, with the polishing and painting departments logging 138 hr and 358 hr, respectively. How many figurines of each type were made? Dept.

Small

Medium

Large

X-Large

Casting

0.5

0.6

0.75

1

Trimming

0.8

0.9

1.1

1.5

Polishing

1.2

1.4

1.7

2

Painting

2.5

3.5

4.5

6

64°C 73. Thermal conductivity: In lab experiments designed to measure the heat conductivity p2 p1 70°C 80°C of a square metal plate of uniform density, the p3 p4 edges are held at four different (constant) 96°C temperatures. The mean-value principle from physics tells us that the temperature at a given point pi on the plate is equal to the average temperature of nearby points. Use this information to form a system of four equations in four variables, and determine the temperature at interior points p1, p2, p3, and p4 on the plate shown. (Hint: Use the temperature of the four points closest to each.)

74. Thermal conductivity: Repeat Exercise 73 if (a) the temperatures at the top and bottom of the plate were increased by 10°, with the temperatures at the left and right edges decreased by 10° (what do you notice?); (b) the temperature at the top and the temperature to the left were decreased by 10°, with the temperatures at the bottom and right held at their original temperature. Curve Fitting

75. Quadratic fit: Use a matrix equation to find a quadratic function of the form y ⫽ ax2 ⫹ bx ⫹ c such that 1⫺4, ⫺52 , 10, ⫺52 , and (2, 7) are on the graph of the function. 76. Quadratic fit: Use a matrix equation to find a quadratic function of the form y ⫽ ax2 ⫹ bx ⫹ c such that 1⫺4, ⫺02 , (1, 5), and, 12, ⫺62 are on the graph of the function. 77. Cubic fit: Use a matrix equation to find a cubic function of the form y ⫽ ax3 ⫹ bx2 ⫹ cx ⫹ d such that 1⫺4, ⫺62, 1⫺1, 02, 11, ⫺162, and (3, 8) are on the graph of the function. 78. Cubic fit: Use a matrix equation to find a cubic function of the form y ⫽ ax3 ⫹ bx2 ⫹ cx ⫹ d such that 1⫺2, 52, 10, 12, 12, ⫺32, and (3, 25) are on the graph of the function.

931

Investing

79. Wise investing: Morgan received an $800 gift from her grandfather, and showing wisdom beyond her years, decided to place the money in a certificate of deposit (CD) and a money market fund (MM). At the time, CDs were earning 3.5% and MMs were earning 2.5%. At the end of 1 yr, she cashed both in and received a total of $824.50. How much was deposited in each? 80. Baseball cards: Gary has a passion for baseball, which includes a collection of rare baseball cards. His most prized cards feature Willie Mays (1953 Topps) and Mickey Mantle (1959 Topps). The Willie Mays card has appreciated 28% and the Mickey Mantle card 25% since he purchased them, and together they are now worth $17,100. If he paid a total of $13,507.50 at auction for both cards, what was the original price of each?

Willie Mays

(1953 Topps) $7187.50

81. Retirement planning: Using payroll deduction, Jeanette was able to put aside $4800 per month last year for her impending retirement. Last year, her company retirement fund paid 4.2% and her mutual funds returned 5.75%, but her stock fund actually decreased 2.5% in value. If her net gain for the year was $104.50 and $300 more was placed in stocks than in mutual funds, how much was placed in each investment vehicle? 82. Charitable giving: The hyperbolic funnels seen at many shopping malls are primarily used by nonprofit organizations to raise funds for worthy causes. A coin is launched down a ramp into the funnel and seemingly makes endless circuits before finally disappearing down a “black hole” (the collection bin). During one such collection, the bin was found to hold $112.89, and 1450 coins consisting of pennies, nickels, dimes, and quarters. How many of each denominator were there, if the number of quarters and dimes was equal to the number of nickels, and the number of pennies was twice the number of quarters.

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Nutrition

83. Animal diets: A zoo dietician needs to create a specialized diet that regulates an animal’s intake of fat, carbohydrates, and protein during a meal. The table given shows three different foods and the amount of these nutrients (in grams) that each ounce of food provides. How many ounces of each should the dietician recommend to supply 20 g of fat, 30 g of carbohydrates, and 44 g of protein? 84. Training diet: A physical trainer is designing a workout diet for one of her clients, and wants to supply him with 24 g of fat, 244 g of carbohydrates, and 40 g of protein for the noontime meal. The table given shows three different foods and the amount of these nutrients (in grams) that each ounce of food provides. How many ounces of each should the trainer recommend?

䊳

Nutrient

Food I

Food II

Food III

Fat

2

4

3

Carb.

4

2

5

Protein

5

6

7

Food I

Food II

Food III

Fat

2

5

0

Carb.

10

15

18

Protein

2

10

0.75

Nutrient


85. Some matrix applications require that you solve a matrix equation of the form AX B C, where A, B, and C are matrices with the appropriate number of rows and columns and A1 exists. Investigate the solution process 2 3 4 12 x d, B c d, C c d , and X c d , then solve AX B C for X for such equations using A c 5 4 9 4 y 1 symbolically (using A , I, and so on). 86. Another alternative for finding determinants uses the triangularized form of a matrix and is offered without proof: If nonsingular matrix A is written in triangularized form using standard row operations but without exchanging any rows and without using the operation kRi to replace any row (k a constant), then det(A) is equal to the product of resulting diagonal entries. Compute the determinant of each matrix using this method. Be careful not to interchange rows and do not replace any row by a multiple of that row in the process. 1 a. £ 4 2

䊳

2 3 5 6 § 5 3

2 b. £ 2 4

5 1 3 4 § 6 5

2 4 c. £ 5 7 3 8

1 2 § 1

3 d. £ 0 2

1 4 2 6 § 1 3


87. (6.3) Find the amplitude and period of y 125 cos13t2.

89. (2.3) Solve the absolute value inequality: 3冟2x 5冟 7 19.

88. (2.2/5.3) Match each equation to its related graph. Justify your answers.

90. (7.6) Find all solutions of 7 tan2x 21 in 30, 2␲2.

a.

y log2 1x 22 y

5 4 3 2 1 1 1 2 3 4 5 6

x2

1 2 3 4 5 6 7 8 9 10 x

y log2 x 2 b.

y 5 4 3 2 1 1 1 2 3 4 5 6

1 2 3 4 5 6 7 8 9 10 x

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Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More


A. Solve a system using

B.

C.

D. E.

determinants and Cramer’s rule Use determinants in applications involving geometry in the coordinate plane Decompose a rational expression into partial fractions using matrices and technology Use matrices to solve static systems Use matrices for encryption/decryption

In addition to solving systems, matrices can be used to accomplish such diverse things as finding the volume of a three-dimensional solid or establishing certain geometrical relationships in the coordinate plane. Numerous uses are also found in higher mathematics, such as checking whether solutions to a differential equation are linearly independent.

A. Solving Systems Using Determinants and Cramer’s Rule In addition to identifying singular matrices, determinants can actually be used to develop a formula for the solution of a system. Consider the following solution to a general 2 2 system, which parallels the solution to a specific 2 2 system. With a view toward a solution involving determinants, the coefficients of x are written as a11 and a21 in the general system, and the coefficients of y are a12 and a22. Specific System e

General System

2x 5y 9 3x 4y 10

e

eliminate the x-term 3R1 2R2

a11x a12y c1 a21x a22y c2 eliminate the x-term a21R1 a11R2

sums to zero

sums to zero

3 # 2x 3 # 5y 3 # 9 a21a11x a21a12y a21c1 e e 2 # 3x 2 # 4y 2 # 10 a11a21x a11a22y a11c2 # # # # 2 4y 3 5y 2 10 3 9 a11a22y a21a12y a11c2 a21c1 Notice the x-terms sum to zero in both systems. We are deliberately leaving the solution on the left unsimplified to show the pattern developing on the right. Next we solve for y. Factor Out y

12 # 4 3 # 52y 2 # 10 3 # 9 2 # 10 3 # 9 y # 2 43#5

Factor Out y

divide

1a11a22 a21a12 2y a11c2 a21c1 a11c2 a21c1 divide y a11a22 a21a12

On the left we find y 7 7 1 and back-substitution shows x 2. But more important, on the right we obtain a formula for the y-value of any 2 2 system: a11c2 a21c1 y . If we had chosen to solve for x, the solution would be a11a22 a21a12 a22c1 a12c2 . Note these formulas are defined only if a11a22 a21a12 0. x a11a22 a21a12 You may have already noticed, but this denominator is the determinant of the matrix of a11 a12 coefficients c d from the previous section! Since the numerator is also a difference a21 a22 of two products, we investigate the possibility that it too can be expressed as a determinant. Working backward, we’re able to reconstruct the numerator for x in determinant c1 a12 d , where it is apparent this matrix was formed by replacing the coefform as c c2 a22 ficients of the x-variables with the constant terms. (removed)

a11 à 21 9–97

a12 a22 `

remove coefficients of x

`

a12 a22 `

c1 a12 `c a ` 2 22 replace with constants

933

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It is also apparent the numerator for y can be also written in determinant form as a11 c1 ` , or the determinant formed by replacing the coefficients of the y-variables ` a21 c2 with the constant terms: a11 à 21

a12 a22 `

remove coefficients of y

a11 à 21

(removed)

a11 à

`

21

c1 c2 `

replace with constants

If we use the notation Dy for this determinant, Dx for the determinant where x coefficients were replaced by the constants, and D as the determinant for the matrix of coefficients—the solution of the system can be written as shown next, with the result known as Cramer’s rule. Cramer’s Rule for 2 ⴛ 2 Systems Given a 2 2 system of linear equations e

a11x a12y c1 a21x a22y c2

the solution is the ordered pair (x, y), where c1 a12 ` c2 a22 Dx x a11 a12 D ` ` a21 a22 `

and

`

a11 c1 ` a21 c2 y a11 a12 D ` ` a21 a22 Dy

provided D 0.

EXAMPLE 1

䊳

Solving a System Using Cramer’s Rule Use Cramer’s rule to solve the system e

Solution

䊳

2x 5y 9 . 3x 4y 10

Begin by finding the value of D, Dx, and Dy. 2 5 9 5 ` Dx ` ` 3 4 10 4 122142 132152 192142 1102152 7 14

D `

Dy `

2 9 ` 3 10 1221102 132192 7

Dy Dx 14 7 2 and y 1. The solution is D 7 D 7 12, 12. Check by substituting these values into the original equations. This gives x


䊳

Regardless of the method used to solve a system, always be aware that a consistent, y 2x 3 inconsistent, or dependent system is possible. The system e yields 4x 6 2y 2x y 3 2 1 e in standard form, with D ` ` 122 122 142112 0. 4x 2y 6 4 2 Since D 0, Cramer’s rule cannot be applied, and the system is either inconsistent or dependent. To find out which, we write the equations in function form (solve

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935

Section 9.8 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More

for y). The result is e

y 2x 3 , showing the system consists of two parallel lines y 2x 3

and has no solutions.

Cramer’s Rule for 3 ⴛ 3 Systems Cramer’s rule can be extended to a 3 3 system of linear equations, using the same pattern as for 2 2 systems. Given the general 3 3 system a11x a12y a13z c1 • a21x a22y a23z c2 a31x a32y a33z c3 Dy Dz Dx , y , and z , where Dx, Dy, and Dz are again formed D D D by replacing the coefficients of the indicated variable with the constants, and D is the determinant of the matrix of coefficients 1D 02 . the solutions are x

Cramer’s Rule Applied to 3 ⴛ 3 Systems Given a 3 3 system of linear equations a11x a12y a13z c1 • a21x a22y a23z c2 a31x a32y a33z c3 The solution is an ordered triple (x, y, z), where

x

c1 † c2 c3

a12 a22 a32

a13 a23 † a33

a11 † a21 a31

a12 a22 a32

a13 a23 † a33

y

a11 c1 a13 † a21 c2 a23 † a31 c3 a33 a11 a12 a13 † a21 a22 a23 † a31 a32 a33

z

a11 † a21 a31

a12 a22 a32

c1 c2 † c3

a11 † a21 a31

a12 a22 a32

a13 a23 † a33

,

provided D 0.

EXAMPLE 2

䊳

Solving a 3 ⴛ 3 System Using Cramer’s Rule x 2y 3z 1 Solve using Cramer’s rule: • 2x y 5z 1 3x 3y 4z 2

Solution

䊳

Begin by computing the determinant of the matrix of coefficients, to ensure that Cramer’s rule can be applied. Using the third row, we have 1 D † 2 3

2 1 3

3 2 3 1 3 1 5 † 3 ` ` 3` ` 4` 1 5 2 5 2 4 3172 3112 4132 6

2 ` 1

Since D 0 we continue, electing to compute the remaining determinants using a calculator (see figures). 1 Dx † 1 2

2 3 1 5 † 12 3 4

1 Dy † 2 3

1 1 2

3 5 † 0 4

1 Dz † 2 3

2 1 3

1 1 † 6 2

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As the final screen shows, the solution is x

Dy Dx 12 0 2, y 0, and D 6 D 6

6 1, or 12, 0, 12 in triple form. Check this D 6 solution in the original equations. Dz

z


A. You’ve just seen how we can solve a system using determinants and Cramer’s rule

䊳

B. Determinants, Geometry, and the Coordinate Plane As mentioned in the introduction, the use of determinants extends far beyond solving systems of equations. Here, we’ll demonstrate how determinants can be used to find the area of a triangle whose vertices are given as three points in the coordinate plane. The Area of a Triangle in the xy-Plane Given a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3), Area `

EXAMPLE 3

䊳

det1T2 2

x1 ` where T £ x2 x3

y1 y2 y3

1 1§ 1

Finding the Area of a Triangle Using Determinants Find the area of a triangle with vertices at (3, 1), (2, 3), and (1, 7) (see Figure 9.65).

Solution

䊳

Figure 9.65 y 7

2

(1, 7)

6 5 4

(2, 3)

3 2

(3, 1)

1 5 4 3 2 1 1 2 3

1

2

3

Begin by forming matrix T and computing det(T) (see Figure 9.66): Figure 9.66 3 1 1 x1 y1 1 T [A] det1T2 † x y 1 † † 2 3 1 †

4

5

x

2

x3 y3 1 1 7 1 313 72112 12 1114 32 12 3 1172 26 det1T 2 26 ` ` ` Compute the area: A ` 2 2 13 The area of this triangle is 13 units2.


䊳

As an extension of this formula, what if the three points were collinear? After a moment, it may occur to you that the formula would give an area of 0 units2, since no triangle could be formed. This gives rise to a test for collinear points.

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937

Test for Collinear Points Three points (x1, y1), (x2, y2), and (x3, y3) are collinear if

B. You’ve just seen how we can use determinants in applications involving geometry in the coordinate plane

x1 det1A2 † x2 x3

y1 y2 y3

1 1 † 0. 1

See Exercises 33 through 38.

C. More on Partial Fraction Decomposition Occasionally, the decomposition template for a rational expression becomes rather extensive. Instead of attempting a solution using convenient values or row operations as in Section 9.4, solutions can more readily be found using a matrix equation and technology. EXAMPLE 4

䊳

Decomposing a Rational Expression Using Matrices and Technology Use matrix equations and a graphing calculator to decompose the given expression 12x3 62x2 102x 56 . into partial fractions: x4 6x3 12x2 8x

Solution

䊳

The degree of the numerator is less than that of the denominator, so we begin by factoring the denominator. After removing the common factor of x and applying the rational roots theorem with synthetic division 1c 22, the completely 12x3 62x2 102x 56 factored form is . The decomposed form will be x1x 22 3 A C D 12x3 62x2 102x 56 B . 3 1 2 x x1x 22 1x 22 1x 22 1x 22 3 Clearing denominators and simplifying yields 12x3 62x2 102x 56 A1x 22 3 Bx1x 22 2 Cx1x 22 Dx

clear denominators

After expanding the powers on the right, grouping like terms, and factoring, we obtain 12x3 62x2 102x 56 1A B2x3 16A 4B C2x2 112A 4B 2C D2x 8A

Figure 9.67

By equating the coefficients of like terms, the following system and matrix equation are obtained: A B 12 1 6A 4B C 62 6 μ S≥ 12A 4B 2C D 102 12 8A 56 8

1 4 4 0

After carefully entering the matrices F (coefficients) and G (constants) (see Figure 9.67), we obtain the solution A 7, B 5, C 0, and D 2 as shown in Figure 9.68. The decomposed form is

C. You’ve just seen how we can decompose a rational expression into partial fractions using matrices and technology

0 1 2 0

0 A 12 0 B 62 ¥≥ ¥ ≥ ¥. 1 C 102 0 D 56 Figure 9.68

12x3 62x2 102x 56 7 2 5 . x 1x 22 1 1x 22 3 x1x 22 3


䊳

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D. Solving Static Systems with Varying Constraints When the considerations of a business or industry involve more than two variables, solutions using matrix methods have a distinct advantage over other methods. Companies often have to perform calculations using basic systems weekly, daily, or even hourly, to keep up with trends, market changes, changes in cost of raw materials, and so on. In many situations, the basic requirements remain the same, but the frequently changing inputs require a recalculation each time they change. EXAMPLE 5

䊳

Determining Supply Inventories Using Matrices BNN Soft Drinks receives new orders daily for its most popular drink, Saratoga Cola. It can deliver the carbonated beverage in a twelve-pack of 12-ounce (oz) cans, a six-pack of 20-oz bottles, or in a 2-L bottle. The ingredients required to produce a twelve-pack include 1 gallon (gal) of carbonated water, 1.25 pounds (lb) of sugar, 2 cups (c) of flavoring, and 0.5 grams (g) of caffeine. For the six-pack, 0.8 gal of carbonated water, 1 lb of sugar, 1.6 c of flavoring, and 0.4 g of caffeine are needed. The 2-L bottle contains 0.47 gal of carbonated water, 0.59 lb of sugar, 0.94 c of flavoring, and 0.24 g of caffeine. How much of each ingredient must be on hand for Monday’s order of 300 twelve-packs, 200 six-packs, and 500 2-L bottles? What quantities must be on hand for Tuesday’s order: 410 twelve-packs, 320 six-packs, and 275 2-L bottles?

Solution

䊳

Begin by setting up a general system of equations, letting x represent the number of twelve-packs, y the number of six-packs, and z the number of 2-L bottles: 1x 0.8y 0.47z gallons of carbonated water 1.25x 1y 0.59z pounds of sugar μ 2x 1.6y 0.94z cups of flavoring 0.5x 0.4y 0.24z grams of caffeine As a matrix equation we have 1 0.8 1.25 1 ≥ 2 1.6 0.5 0.4

w 0.47 x s 0.59 ¥ £y§ ≥ ¥ f 0.94 z c 0.24

Enter the 4 3 matrix as matrix A, and the size of the order as matrix as B. Using a calculator, we find 1 1.25 AB ≥ 2 0.5

0.8 1 1.6 0.4

0.47 695 300 870 0.59 ¥ £ 200 § ≥ ¥, 1390 0.94 500 0.24 350

and BNN Soft Drinks will need 695 gal of 410 carbonated water, 870 lb of sugar, 1390 c of flavoring, and 350 g of caffeine for Monday’s order. After entering C £ 320 § 275 for Tuesday’s orders, computing the product AC shows 795.25 gal of carbonated water, 994.75 lb of sugar, 1590.5 c of flavoring, and 399 g of caffeine are needed for Tuesday. Now try Exercises 43 through 46

䊳

Example 5 showed how the creation of a static matrix can help track and control inventory requirements. In Example 6, we use a static matrix to solve a system that will identify the amount of data traffic used by a company during various hours of the day.

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EXAMPLE 6

䊳

939

Identifying the Source of Data Traffic Using Matrices Mariño Imports is a medium-size company that is considering upgrading from a 1.544 megabytes per sec (Mbps) T1 Internet line to a fractional T3 line with a bandwidth of 7.72 Mbps. They currently use their bandwidth for phone traffic, office data, and Internet commerce. The IT (Information Technology) director devises a plan to monitor how much data traffic each resource uses on an hourly basis. Because of the physical arrangement of the hardware, she cannot monitor each resource individually. The table shows the information she collected for the first 3 hr. Determine how many gigabytes (GB) each resource used individually during these 3 hr.

Solution

䊳

Phone, Data, and Commerce

Phone and Data

Data and Commerce

9–10:00 A.M.

5.4 GB

4.0 GB

4.2 GB

10–11:00 A.M.

5.3 GB

3.8 GB

4.2 GB

11–12:00 P.M.

5.1 GB

3.5 GB

3.6 GB

Using p to represent the phone traffic, d for office data, and c p d c 5.4 for Internet commerce, we create the system shown, which •p d 4.0 models data use for the 9:00 o’clock hour. Since we actually d c 4.2 need to solve two more systems whose only difference is the constant terms (for the 10 and 11 o’clock hours), using a matrix equation to solve the system (Section 9.7) will be most convenient. Begin by writing the related matrix equation for this system: 1 AX B: £ 1 0

1 1 1

1 p 5.4 0 § £ d § £ 4.0 § 1 c 4.2

Using X A1B to solve the system (see figure), we find there was 1.2 GB of phone traffic, 2.8 GB of office data, and 1.4 GB of Internet commerce during this hour. Note that the IT director may make this calculation 10 or more times a day (once for every hour of business). While we could solve for the 10 and 11 o’clock hours 5.3 5.1 using C £ 3.8 § and D £ 3.5 § , then calculate A1C and A1D, these 4.2 3.6 calculations can all be performed simultaneously by combining the matrices B, C, and D into one 3 3 matrix and multiplying by A1. Due to the properties of matrix multiplication, each column of the product will represent the data information for a given hour, as shown here: 3A

1

D. You’ve just seen how we can use matrix equations to solve static systems

5.4 4 £ 4.0 4.2

5.3 3.8 4.2

5.1 1.2 3.5 § £ 2.8 3.6 1.4

1.1 2.7 1.5

1.5 2.0 § 1.6

In the second hour, there was 1.1 GB of phone traffic, 2.7 GB of office data, and 1.5 GB of Internet commerce. In the third hour, 1.5 GB of phone traffic, 2.0 GB of office data, and 1.6 GB of Internet commerce bandwidth was used. Now try Exercises 47 through 50

䊳

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E. Using Matrices to Encrypt Messages In The Gold-Bug, by Edgar Allan Poe, the narrator deciphers the secret message on a treasure map by accounting for the frequencies of specific letters and words. The coding used was a simple substitution cipher, which today can be broken with relative ease. In modern times where information wields great power, a simple cipher like the one used there will not suffice. When you pay your tuition, register for classes, make on-line purchases, and so on, you are publicly transmitting very private data, which needs to be protected. While there are many complex encryption methods available (including the now famous symmetric-key and public-key techniques), we will use a matrix-based technique. The nature of matrix multiplication makes it very difficult to determine the exact matrices that yield a given product, and we’ll use this fact to our advantage. Beginning with a fixed, invertible matrix A, we will develop a matrix B such that the product AB is possible, and our secret message is encrypted in AB. At the receiving end, they will need to know A1 to decipher the message, since A1 1AB2 1A1A2B B, which is the original message. Note that in case an intruder were to find matrix A (perhaps purchasing the information from a disgruntled employee), we must be able to change it easily. This means we should develop a method for generating a matrix A, with integer entries, where A is invertible and A1 also has integer entries. EXAMPLE 7

䊳

Finding an Invertible Matrix A Where Both A and Aⴚ1 Have Integer Entries Find an invertible 3 3 matrix A as just described, and its inverse A1.

Solution

䊳

Begin with any 3 3 matrix that has only 1s or 1s on its main diagonal, and 0s below the diagonal. The upper triangle can consist of any integer values you choose, as in 1 £ 0 0

WORTHY OF NOTE Performing row operations is explained in more detail in the graphing calculator manual accompanying this text.

5 1 0

1 8§. 1

Now, use any of the elementary row operations to make the matrix more complex. For instance, we’ll use a calculator to create a new matrix by (1) using R1 R2 S R2 to create matrix [C], then (2) using R1 R3 S R3 to create matrix [D], then (3) using R3 R2 S R2 to create matrix [E], and finally (4) 2 R1 R3 S R3 to obtain our final matrix [A]. To begin, enter the initial matrix as matrix [B]. For (1) R1 R2 S R2, go to the (MATRIX) MATH submenu, select option D:rowⴙ( and press to bring this option to the home screen. This feature requires us to name the matrix we’re using, and to indicate what rows to add, so we enter D:rowⴙ([B], 1, 2). The screen shown in Figure 9.69 indicates we’ve placed the result in matrix [C]. For R1 R3 S R3, recall D:rowⴙ([B], 1, 2) using 2nd and change it to D:rowⴙ([C], 1, 3) STO [D] (Figure 9.70). Repeat this process for (3) R3 R2 S R2 to create matrix [E]: D:rowⴙ([D], 3, 2) STO [E] (Figure 9.71). Finally, we compute (4) 2R1 R3 S R3 using the (new) option F:*rowⴙ(ⴚ2, [E], 1, 3) STO [A]

Figure 9.69

Figure 9.70

ENTER

ENTER

Figure 9.71

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Figure 9.72

Figure 9.73

and the process is complete (Figure 9.72). The entries of matrix A are all integers, and A1 exists and also has integer entries (Figure 9.73). This will always be the case for matrices created in this way. Now try Exercises 51 through 56

EXAMPLE 8

䊳

䊳

Using Matrices to Encrypt Messages Set up a substitution cipher to encode the message MATH IS SWEET, and then use the matrix A from Example 7 to encrypt it.

Solution

䊳

For the cipher, we will associate a unique number to every letter in the alphabet. This can be done randomly or using a systematic approach. Here we choose to associate 0 with a blank space, and assign 1 to A, 1 to B, 2 to C, 2 to D, and so on. Blank

A

B

C

D

E

F

G

H

I

J

K

L

M

0

1

1

2

2

3

3

4

4

5

5

6

6

7

N

O

P

Q

R

S

T

U

V

W

X

Y

Z

7

8

8

9

9

10 10 11 11 12 12 13

13

Now encode the secret message as shown: M

A

7

1

T

H

10 4

0

I

S

5

10

0

S

W

E

E

T

10

12

3

3

10

We next enter the coded message into a new matrix B, by entering it letter by letter into the columns of B. Note that since the encrypting matrix A is 3 3, B must have 3 rows for multiplication to be possible. The result is M B £A T

H * I

S * S

W E E

T 7 *§ £ 1 * 10

4 0 5

10 0 10

12 3 3

10 0§ 0

Since the message is too short to fill matrix B, we use blank spaces to complete the final column. Computing the product AB encrypts the message, and only someone with access to A1 will be able to read it: 1 AB £ 2 1 8 £ 73 18

5 11 5 1 43 6

1 7 7§ £ 1 2 10 20 50 30

0 30 3

4 0 5

10 0 10

12 3 3

10 0§ 0

10 20 § 10

The encrypted message is 8, 73, 18, 1, 43, 6, 20, 50, 30, 0, 30, 3, 10, 20, 10. Now try Exercises 57 through 62 䊳

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EXAMPLE 9

䊳

Deciphering Encrypted Messages Using an Inverse Matrix Decipher the encrypted message from Example 8 using A1 from Example 7.

Solution

䊳

The received message is 8, 73, 18, 1, 43, 6, 20, 50, 30, 0, 30, 3, 10, 20, 10, and is the result of the product AB. To find matrix B, we apply A1 since A1 1AB2 1A1 A2B B. Writing the received message in matrix form we have 1 43 6

8 AB £ 73 18

20 50 30

0 30 3

10 20 § 10

Next multiply AB by A1 on the left, to determine matrix B: 57 A1 1AB2 £ 11 1 7 £ 1 10

E. You’ve just seen how we can use matrices for encryption/decryption

5 1 0

46 8 9 § £ 73 1 18

4 0 5

10 0 10

12 3 3

1 43 6

20 50 30

0 30 3

10 20 § 10

10 0§ B 0

Writing matrix B in sentence form gives 7, 1, 10, 4, 0, 5, 10, 0, 10, 12, 3, 3, 10, 0, 0, and using the substitution cipher to replace numbers with letters, reveals the message MATH IS SWEET. 䊳


9.8 EXERCISES 䊳



1. The determinant ` as:

a11 a21

a12 ` is evaluated a22

.

3. Given the matrix of coefficients D, the matrix Dx is formed by replacing the coefficients of x with the terms. 5. Discuss/Explain the process of writing

8x 3 as a x2 x

sum of partial fractions. 䊳

2.

rule uses a ratio of determinants to solve for the unknowns in a system.

4. The three points (x1, y1), (x2, y2), and (x3, y3) are x1 y1

1

x3 y3

1

collinear if 0T 0 † x2 y2 1 † has a value of

.

6. Discuss/Explain why Cramer’s rule cannot be applied if D 0. Use an example to illustrate.


Write the determinants D, Dx, and Dy for the systems given. Do not solve.

7. e

2x 5y 7 3x 4y 1

8. e

x 5y 12 3x 2y 8

Solve each system of equations using Cramer’s rule, if possible. Do not use a calculator.

9. e

4x y 11 3x 5y 60

10. e

x 2y 11 y 2x 13

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y x ⫹ ⫽1 8 4 11. μ y x ⫽ ⫹6 5 2 13. e

3 7 2 x⫺ y⫽ 3 8 5 12. μ 3 11 5 x⫹ y⫽ 6 4 10

Use Cramer’s rule to solve each system of equations. Verify computations using a graphing calculator.

x ⫹ 2y ⫹ 5z ⫽ 10 x ⫹ 3y ⫹ 5z ⫽ 6 17. • 3x ⫹ 4y ⫺ z ⫽ 10 18. • 2x ⫺ 4y ⫹ 6z ⫽ 14 x ⫺ y ⫺ z ⫽ ⫺2 9x ⫺ 6y ⫹ 3z ⫽ 3

0.6x ⫺ 0.3y ⫽ 8 ⫺2.5x ⫹ 6y ⫽ ⫺1.5 14. e 0.8x ⫺ 0.4y ⫽ ⫺3 0.5x ⫺ 1.2y ⫽ 3.6

y ⫹ 2z ⫽ 1 19. • 4x ⫺ 5y ⫹ 8z ⫽ ⫺8 8x ⫺ 9z ⫽ 9

The two systems given in Exercises 15 and 16 are identical except for the third equation. For the first system given, (a) write the determinants D, Dx, Dy, and Dz then (b) determine if a solution using Cramer’s rule is possible by computing |D| without the use of a calculator (do not solve the system). Then (c) compute | D| for the second system and try to determine how the equations in the second system are related.

x ⫹ 2y ⫹ 5z ⫽ 10 20. • 3x ⫺ z ⫽ 8 ⫺y ⫺ z ⫽ ⫺3 w ⫹ 2x ⫺ 3y ⫽ ⫺8 x ⫺ 3y ⫹ 5z ⫽ ⫺22 21. μ 4w ⫺ 5x ⫽ 5 ⫺y ⫹ 3z ⫽ ⫺11

4x ⫺ y ⫹ 2z ⫽ ⫺5 2x ⫹ 3z ⫽ ⫺2 15. •⫺3x ⫹ 2y ⫺ z ⫽ 8 16. • ⫺x ⫹ 5y ⫹ z ⫽ 12 x ⫺ 5y ⫹ 3z ⫽ ⫺3 3x ⫺ 2y ⫹ z ⫽ ⫺8 4x ⫺ y ⫹ 2z ⫽ ⫺5 •⫺3x ⫹ 2y ⫺ z ⫽ 8 x⫹ y⫹ z⫽3 䊳

943

w ⫺ 2x ⫹ 3y ⫺ z ⫽ 11 3w ⫺ 2y ⫹ 6z ⫽ ⫺13 22. μ 2x ⫹ 4y ⫺ 5z ⫽ 16 3x ⫺ 4z ⫽ 5

2x ⫹ 3z ⫽ ⫺2 • ⫺x ⫹ 5y ⫹ z ⫽ 12 3x ⫹ 5y ⫹ 4z ⫽ 10


L r2 ␲ † . The determinant shown can be used to find the area of a Norman ⴚ W 2 W window (rectangle ⴙ half-circle) with length L, width W, and radius r ⴝ . Find the area of the following windows. 2 Area of a Norman window: A ⴝ †

23.

24. 16 in.

32 cm 20 in.

䊳

58 cm

APPLICATIONS

Area of a triangle: Find the area of the triangle with the vertices given. Assume units are in centimeters.

25. (2, 1), (3, 7), and (5, 3)

26. 1⫺2, 32, 1⫺3, ⫺42, and 1⫺6, 12 Area of a parallelogram: Find the area of the parallelogram with vertices given (Hint: Use two triangles.) Assume units are in feet.

27. 1⫺4, 22, 1⫺6, ⫺12, 13, ⫺12, and 15, 22

28. 1⫺5, ⫺62 , (5, 0), (5, 4), and 1⫺5, ⫺22

Volume of a pyramid: The volume of a triangular pyramid is given by the formula V ⫽ 13Bh, where B represents the area of the triangular base and h is the height of the pyramid. Find the volume of a triangular pyramid whose height is given and whose base has the coordinates shown. Assume units are in meters.

29. h ⫽ 6 m; vertices (3, 5), 1⫺4, 22, and 1⫺1, 62

30. h ⫽ 7.5 m; vertices 1⫺2, 32, 1⫺3, ⫺42, and 1⫺6, 12

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Volume of a prism: The volume of a right prism is given by the formula V Bh, where B represents the area of the base and h is the height of the prism. Find the volume of a triangular prism whose height is given and whose base has the coordinates shown. Assume units are in inches.

31. (3, 0), (9, 0), (6, 4), h 8 32. (2, 3), (9, 5), (0, 6), h 5 Determine if the following sets of points are collinear.

33. (1, 5), 12, 12 , and (4, 11)

much of each raw material they need to have on hand to fill these orders. 44. In March, Slammin’ Drums’ orders consisted of 19 bass drums, 19 toms, and 25 snares. Use your calculator and a matrix equation to determine how much of each raw material they need to have on hand to fill their orders. (See Exercise 43.) 45. The following table represents Slammin’s orders for the months of April through July. Use your calculator and a matrix equation to determine how much of each raw material they need to have on hand to fill these orders. (See Exercise 43.) (Hint: Using a clever 4 3 and 3 1 matrix can reduce this problem to a single step.)

34. (1, 1), 13, 52, and 12, 92

35. 12.5, 5.22, 11.2, 5.62 , and 12.2, 8.52 36. 10.5, 2.552, 12.8, 1.632, and (3, 3.95)

For each linear equation given, substitute the first two points to verify they are solutions. Then use the test for collinear points to determine if the third point is also a solution.

37. 2x 3y 7; 12, 12, 11.3, 3.22, 13.1, 4.42 38. 5x 2y 4; 12, 32, 13.5, 6.752, 12.7, 8.752

Decompose each rational expressing using a matrix equation and a graphing calculator.

39.

x4 x2 2x 1 x5 2x3 x

3x4 13x2 x 12 40. x5 4x3 4x x3 17x2 76x 98 41. 2 1x 6x 92 1x2 2x 32 42.

16x3 66x2 98x 54 12x2 3x2 14x2 12x 92

43. Slammin’ Drums manufactures several different types of drums. Its most popular drums are the 22 bass drum, the 12 tom, and the 14 snare drum. The 22 bass drum requires 7 ft2 of skin, 8.5 ft2 of wood veneer, 8 tension rods, and 11.5 ft of hoop. The 12 tom requires 2 ft2 of skin, 3 ft2 of wood veneer, 6 tension rods, and 6.5 ft of hoop. The 14 snare requires 2.5 ft2 of skin, 1.5 ft2 of wood veneer, 10 tension rods, and 7 ft of hoop. In February, Slammin’ Drums received orders for 15 bass drums, 21 toms, and 27 snares. Use your calculator and a matrix equation to determine how

April

May

June

July

Bass drum

23

21

17

14

Tom

20

18

15

17

Snare drum

29

35

27

25

46. The following table represents Slammin’s orders for the months of August through November. Use your calculator and a matrix equation to determine how much of each raw material they need to have on hand to fill their orders (see Exercise 43). (Hint: Using a clever 4 3 and 3 1 matrix can reduce this problem to a single step.) August

September

October

November

Bass drum

17

22

16

12

Tom

15

14

13

11

Snare drum

32

28

27

21

47. Roll-X Watches makes some of the finest wristwatches in the world. Their most popular model is the Clam. It comes in three versions: Silver, Gold, and Platinum. Management thinks there might be a thief in the production line, so they decide to closely monitor the precious metal consumption. A Silver Clam contains 1.2 oz of silver and 0.2 oz of gold. A Gold Clam contains 0.5 oz of silver, 0.8 oz of gold, and 0.1 oz of platinum. A Platinum Clam contains 0.2 oz of silver, 0.5 oz of gold, and 0.7 oz of platinum. During the first week of monitoring, the production team used 10.9 oz of silver, 9.2 oz of gold, and 2.3 oz of platinum. Use your calculator and a matrix equation to determine the number of each type of watch that should have been produced. 48. The following table contains the precious metal consumption of the Roll-X Watch production line

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during the next five weeks (see Exercise 47). Use your graphing calculator to determine the number of each type of watch that should have been produced each week. For which week does the data seem to indicate a possible theft of precious metal? Ounces Silver

Week 1

Week 2

Week 3

Week 4

Week 5

13.1

9

12.9

11.9

11.2

Gold

11

7.7

8.6

8.4

9.5

Platinum

2.5

1.5

0.9

2.8

1.7

49. There are three classes of grain, of which three bundles from the first class, two from the second, and one from the third make 39 measures. Two of the first, three of the second, and one of the third make 34 measures. And one of the first, two of the second, and three of the third make 26 measures. How many measures of grain are contained in one bundle of each class? (This is the historic problem from the Chiu chang suan shu.) 50. During a given week, the measures of grain that make up the bundles in Exercise 49 can vary slightly. Three local Chinese bakeries always buy the same numbers of bundles, as outlined in Exercise 49. That is to say, bakery 1 buys three bundles of the first class, two of the second, and one of the third. Bakery 2 buys two of the first, three of the second, and one of the third. And finally, bakery 3 buys one of the first, two of the second, and three of the third. The following table outlines how many measures of grain each bakery received each day. How many measures of grain were contained in one bundle of each class, on each day? Mon

Tues

Wed

Thurs

Fri

Bakery 1 (measures)

39

38

38

37.75

39.75

Bakery 2 (measures)

34

33

33.5

32.5

35

Bakery 3 (measures)

26

26

27

26.25

27.25

For Exercises 51–56, use the criteria indicated to find 3 ⴛ 3 matrices A and Aⴚ1, where the entries of both are all integers.

51. The lower triangle is all zeroes. 52. The upper triangle is all zeroes. 53. a2,1 5 54. a3,2 2 55. a3,1 1 and a2,3 2 56. a2,1 3 and a1,3 1

945

57. Use the matrix A you created in Exercise 51 and the substitution cipher from Example 8 to encrypt your full name. 58. Use the matrix A you created in Exercise 52 and the substitution cipher from Example 8 to encrypt your school’s name. 59. Design your own substitution cipher. Then use it and the matrix A you created in Exercise 53 to encrypt the title of your favorite movie. 60. Design your own substitution cipher. Then use it and the matrix A you created in Exercise 54 to encrypt the title of your favorite snack food. 61. Design your own substitution cipher. Then use it and the matrix A you created in Exercise 55 to encrypt the White House switchboard phone number, 202-456-1414. 62. Design your own substitution cipher. Then use it and the matrix A you created in Exercise 56 to encrypt the Casa Rosada switchboard phone number 54-11-4344-3600. The Casa Rosada, or Pink House, consists of the offices of the president of Argentina. 63. Use the matrix A1 from Exercise 51, and the appropriate substitution cipher to decrypt the message from Exercise 57. 64. Use the matrix A1 from Exercise 52, and the appropriate substitution cipher to decrypt the message from Exercise 58. 65. Use the matrix A1 from Exercise 53, and the appropriate substitution cipher to decrypt the message from Exercise 59. 66. Use the matrix A1 from Exercise 54, and the appropriate substitution cipher to decrypt the message from Exercise 60. 67. Use the matrix A1 from Exercise 55, and the appropriate substitution cipher to decrypt the message from Exercise 61. 68. Use the matrix A1 from Exercise 56, and the appropriate substitution cipher to decrypt the message from Exercise 62. Use a linear system to model each application. Then solve using Cramer’s rule.

69. Return on investments: If $15,000 is invested at a certain interest rate and $25,000 is invested at another interest rate, the total return was $2900. If the investments were reversed the return would be $2700. What was the interest rate paid on each investment?

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70. CD and DVD clearance sale: To generate interest in a music store clearance sale, the manger sets out a large box full of $2 CDs and $7 DVDs, with an advertised price of $800 for the lot. When asked how many of each are in the box, the manager will only say the box holds a total of 200 disks. How many CDs and DVDs are in the box? 71. Cost of fruit: Many years ago, two pounds of apples, 2 lb of kiwi, and 10 lb of pears cost $3.26. Three pounds of apples, 2 lb of kiwi, and 7 lb of pears cost $2.98. Two pounds of apples, 3 lb of kiwi, and 6 lb of pears cost $2.89. Find the cost of a pound of each fruit. 72. Vending machine receipts: The vending machines at an amusement park are stocked with various candies that sell for 5¢, 10¢, and 25¢. At week’s end, the park collected $54.30 from one of the machines. How many of each type of candy were sold, if 484 total sales were made and the machine vended twice as many 5¢ candies as 25¢ candies? 䊳

73. Coffee blends: To make its morning coffees, a coffee shop uses three kinds of beans costing $1.90/lb, $2.25/lb, and $3.50/lb, respectively. By the end of the week, the shop went through 24 lb of coffee beans, having a total value of $58. Find how many pounds of each type of bean were used, given that the number of pounds used of the cheapest beans was four more than the most expensive beans. 74. Manufacturing ball bearings: A ball bearing producer makes three sizes of bearings, halfinch-diameter size weighing 30 grams (g), a threequarter-inch-diameter size weighing 105 g, and a 1-in.-diameter size weighing 250 g. A large storage container holds ball bearings that have been rejected due to small defects. If the net weight of the container’s contents is 95.6 kilograms (95,600 g), and automated tallies show 920 bearings have been rejected, how many of each size is in the reject bin, given that statistical studies show there are twice as many rejects of the smallest bearing, as compared to the largest?


75. Solve the given system four different ways: (1) elimination, (2) row reduction, (3) Cramer’s rule, and (4) using a matrix equation. Which method seems to be the least error-prone? Which method seems most efficient (takes the least time)? Discuss the advantages and drawbacks of each method. x 3y 5z 6 • 2x 4y 6z 14 9x 6y 3z 3 76. Find the area of the pentagon whose vertices are: 15, 52, 15, 52, (8, 6), 18, 62, and (0, 12.5). 77. The polynomial form for the equation of a circle is x2 y2 Dx Ey F 0. Find the equation of the circle that contains the points 11, 72, (2, 8), and 15, 12. 78. For square matrix A, calculating A2 is a simple matter of matrix multiplication. But what of those applications that require higher powers? a b d , we define the For any matrix A c c d characteristic polynomial of A to be a b p12 ` ` 1 a21 d2 cb. c d 䊳

9–110


The Cayley-Hamilton theorem states that if the original matrix A is substituted for , the result must be zero, or in other words, every square matrix satisfies its own characteristic equation. This fact is often used to compute higher powers of a square matrix, since all higher powers can then be expressed in terms of A. Specifically, for 1 2 1 2 A c d , p12 ` ` 3 4 3 4

1 121 42 132122 2 5 2, with the theorem stating that A2 5A 2I2 0 (note I2 is used to ensure we remain in the context of 2 2 matrices). For A2 5A 2I2, we have the follow sequence for any power of A: A3 A2A A4 A3A 15A 2I2 2A 127A 10I2 2A 5A2 2A 27A2 10A 515A 2I2 2 2A 2715A 2I2 2 10A 27A 10I2 145A 54I2 2 1 d , find A2 in terms of A and I2, 4 1 then use the Cayley-Hamilton theorem to find A3, A4, and A5. Verify A3 using scalar multiplication and matrix addition. For A c


79. (4.3) Graph the polynomial using information about end-behavior, y-intercept, x-intercept(s), and midinterval points: f 1x2 x3 2x2 7x 6.

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Making Connections

81. (8.1) Solve the triangle: side a 8.7 in. side b 11.2 in. ⬔A 49.0°

80. (2.2) Which is the graph (left or right) of g1x2 冟x 1冟 3? Justify your answer. y

y

5 4 3 2 1

5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

54321 1 2 3 4 5

82. (6.4) Graph y 3 tan12␲t2 over the interval 312, 12 4. Note the period, asymptotes, zeroes and value of A.

1 2 3 4 5 x

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

(b)

5

5

y

(c)

5

5 x

5

5

5

5 x

y

(f)

5

5

5

5 x

5

1. ____ `

(g)

5

2. ____ e

1 2

x y1 x 2y 5

y x1 3. ____ μ y 0 5 1 y x 2 2

9. ____ three equations in three variables 10. ____ (3, 2) does not satisfy any inequality y x1 11. ____ • x 2y 5 x 0, y 0 12. ____ `

1 1

1 1

1 ` 3

13. ____ (1, 2) is the solution

5. ____ inconsistent

14. ____ (3, 2) is a solution

6. ____ linear dependence

15. ____ e

8. ____ independent and consistent

5 x

5

4. ____ coincident dependence

7. ____ vertices: (1, 2), (5, 0), (0, 0), (0, 1)

5

5

5 x

1 ` 5

y

(h)

5

1 1

5 x

5

5

y

(e)

y

(d)

5

y x1 x 2y 5

1 1 16. ____ A † 1 2 5

0 2 0

1 1† 1

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9–112


Linear Systems in Two Variables with Applications

KEY CONCEPTS • A solution to a linear system in two variables is any ordered pair (x, y) that makes all equations in the system true. • Since every point on the graph of a line satisfies the equation of that line, a point where two lines intersect must satisfy both equations and is a solution of the system. • A system with at least one solution is called a consistent system. • If the lines have different slopes, there is a unique solution to the system (they intersect at a single point). The system is called a consistent and independent system. • If the lines have equal slopes and the same y-intercept, they form identical or coincident lines. Since one line is right atop the other, they intersect at all points with an infinite number of solutions. The system is called a consistent and dependent system. • If the lines have equal slopes but different y-intercepts, they will never intersect. The system has no solution and is called an inconsistent system. EXERCISES Solve each system by graphing manually. Verify your answer by graphing on a graphing calculator. If the system is inconsistent or dependent, so state. 3x 2y 4 0.2x 0.5y 1.4 2x y 2 1. e 2. e 3. e x 3y 8 x 0.3y 1.4 x 2y 4 Solve using substitution. Indicate whether each system is consistent, inconsistent, or dependent. Write unique solutions as an ordered pair. Check your answer using a graphing calculator. y5x xy4 x 2y 3 4. e 5. e 6. e 2x 2y 13 0.4x 0.3y 1.7 x 4y 1 Solve using elimination. Indicate whether each system is consistent, inconsistent, or dependent. Write unique solutions as an ordered pair. Check your answer using a graphing calculator. 2x 4y 10 2x 3y 6 7. e 8. e 3x 4y 5 2.4x 3.6y 6 9. When it was first constructed in 1968, the John Hancock building in Chicago, Ilinois, was the tallest structure in the world. In 1974, the Willis Tower in Chicago (formerly known as the Sears Tower) became the world’s tallest structure. The Willis Tower is 323 ft taller than the John Hancock Building, and the sum of their heights is 2577 ft. How tall is each structure? 10. The manufacturer of a revolutionary automobile spark plug finds that demand for the plug can be modeled by the function D(p) 0.8p 110, where D(p) represents the number of plugs bought (demanded, in tens of thousands) at price p in cents. The supply of these spark plugs is modeled by S(p) 0.24p 14.8, where S(p) represents the number of plugs manufactured/supplied (in tens of thousands) at price p. Find the price for market equilibrium using a graphing calculator.

SECTION 9.2

Linear Systems in Three Variables with Applications

KEY CONCEPTS • The graph of a linear equation in three variables is a plane. • A linear system in three variables has the following possible solution sets: • If the planes intersect at a point, the system could have one unique solution (x, y, z). • If the planes intersect at a line, the system has linear dependence and the solution (x, y, z) can be written as linear combinations of a single variable (a parameter).

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949

• If the planes are coincident, the equations in the system differ by a constant multiple, meaning they are all “disguised forms” of the same equation. The solutions have coincident dependence, and the solution set can be represented by any one of the equations. • In all other cases, the system has no solutions and is an inconsistent system.

EXERCISES Solve using elimination. If a system is inconsistent or dependent, so state. For systems with linear dependence, give the answer as an ordered triple using a parameter. Verify solutions on the home screen using the ALPHA keys. x y 2z 1 x y 2z 2 3x y 2z 3 11. • 4x y 3z 3 12. • x y z 1 13. • x 2y 3z 1 3x 2y z 4 2x y z 4 4x 8y 12z 7 Solve using a system of three equations in three variables. 14. A large coin jar is full of nickels, dimes, and quarters. There are 217 coins in all. The number of nickels is 12 more than the number of quarters. The value of the dimes is $4.90 more than the value of the nickels. How many coins of each type are in the bank? 15. If the point (x, y) is on the graph of a parabola, it must satisfy the equation y ax2 bx c. Use a system of three equations in three variables to find the equation of the parabola containing the points 10, 92, (2, 7), and (6, 15).

SECTION 9.3

Systems of Inequalities and Linear Programming

KEY CONCEPTS • To solve a system of inequalities, we find the intersecting or overlapping regions of the solution regions from the individual inequalities. The common area is called the solution region for the system. • The process known as linear programming seeks to maximize or minimize the value of a given quantity under certain constraints or restrictions. • The quantity we attempt to maximize or minimize is called the objective function. • The solution(s) to a linear programming problem occur at one of the corner points of the feasible (solution) region. • The process of solving a linear programming application contains these six steps: • Identify the main objective and the decision variables. • Write the objective function in terms of these variables. • Organize all information in a table, using the decision variables and constraints. • Fill in the table with the information given and write the constraint inequalities. • Graph the constraint inequalities and determine the feasible region. • Identify all corner points of the feasible region and test these points in the objective function. EXERCISES Graph the solution region for each system of linear inequalities by first solving equation for y. Verify each solution using a test point. Solve Exercise 18 using a graphing calculator. x y 7 2 x 4y 5 x 2y 1 16. e 17. e 18. e x y 6 4 x 2y 0 2x y 2 19. Carefully graph the feasible region for the system of inequalities shown, then maximize the 2x y 10 objective function: f 1x, y2 30x 45y • 2x 3y 18 x 0, y 0 20. After retiring, Oliver and Lisa Douglas buy and work a small farm (near Hooterville) that consists mostly of milk cows and egg-laying chickens. Although the price of a commodity is rarely stable, suppose that milk sales bring in an average of $85 per cow and egg sales an average of $50 per chicken over a period of time. During this time period, the new ranchers estimate that care and feeding of the animals took about 3 hr per cow and 2 hr per

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chicken, while maintaining the related equipment took 2 hr per cow and 1 hr per chicken. How many animals of each type should be maintained in order to maximize profits, if at most 1000 hr can be spent on care and feeding, and at most 525 hr on equipment maintenance?

SECTION 9.4

Partial Fraction Decomposition

KEY CONCEPTS • Partial fraction decomposition is an attempt to reverse the process of adding rational expressions. • The process begins by factoring the denominator and setting up a “decomposition template.” A1 , A1 a constant. • For each linear factor 1x c2 of the denominator, the template will have a term of the form xc • For each irreducible quadratic factor 1ax2 bx c2 of the denominator, the template will have a term of the form A1x B1 for constants A1 and B1. ax2 bx c Ai for i 1 to n. • If the factor 1x c2 is repeated n times the template will have n terms of the form 1x c2 i • If the irreducible quadratic factor is repeated n times, the template will have n terms of the form Ai x Bi , for i 1 to n. 2 1ax bx c2 i • After the template has been set 1original expression decomposition template2, multiply both sides by the LCD and (1) use convenient values or (2) simplify the right-hand side and equate coefficients to find the constants Ai and Bi needed. Note that some of the coefficients found may be zero. • If the degree of the numerator is greater than the degree of the denominator, divide using long division and decompose the remainder portion. EXERCISES Decompose each of the following expressions into partial fractions. 6x 14 16x 1 21. 2 22. 2 x 2x 15 2x 5x 3 2x2 3x 19 2x2 13x 24 24. 3 25. x 5x2 3x 15 x3 27 2x2 8x 33 x2 15x 22 27. 3 28. x x2 8x 12 x3 3x2 9x 5

SECTION 9.5

x2 2x 9 x3 4x2 x 4 6x2 2x 7 26. x3 1 2x2 x 1 29. 4 x 6x2 9 23.

Solving Linear Systems Using Matrices and Row Operations

KEY CONCEPTS • A matrix is a rectangular arrangement of numbers. An m n matrix has m rows and n columns. • The matrix derived from a system of linear equations is called the augmented matrix. It is created by augmenting the coefficient matrix (formed by the variable coefficients) with the matrix of constants. • One matrix method for solving systems involves the augmented matrix and row-reduction. • If possible, interchange equations so that the coefficient of x is a “1” in R1. • Write the system in augmented matrix form (coefficient matrix with matrix of constants). • Use row operations to obtain zeroes below the first entry of the diagonal. • Use row operations to obtain zeroes below the second entry of the diagonal. • Continue until the matrix is triangularized (all entries below the diagonal are zero). • Convert the augmented matrix back into equation form and solve for z. • Graphing calculators can quickly and accurately perform row-reduction.

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EXERCISES Solve by triangularizing the augmented matrix. Use a calculator for Exercise 32. 30. e

x 2y 6 4x 3y 4

SECTION 9.6

x 2y 2z 7 31. • 2x 2y z 5 3x y z 6

2w x 2y 3z 19 w 2x y 4z 15 32. μ x 2y z 1 3w 2x 5z 60

The Algebra of Matrices

KEY CONCEPTS • The entries of a matrix are denoted aij, where i gives the row and j gives the column of its location. • The m n size of a matrix is also referred to as its order. • Two matrices A and B are equal if corresponding entries are equal: A B if aij bij. • The sum or difference of two matrices is found by combining corresponding entries: A B 3aij bij 4. • The identity matrix for addition is an m n matrix whose entries are all zeroes. • The product of a constant times a matrix is called scalar multiplication, and is found by taking the product of the scalar with each entry, forming a new matrix of like size. For matrix A: kA kaij. • Matrix multiplication is performed as row entry column entry according to the following procedure: For an m n matrix A 3 aij 4 and a p q matrix B 3bij 4, matrix multiplication is possible if n p, and the result will be an m q matrix P. In symbols A # B 3pij 4, where pij is product of the ith row of A with the jth column of B. • When technology is used to perform operations on matrices, the focus shifts from a meticulous computation of new entries, to carefully entering each matrix into the calculator, double checking that each entry is correct, and appraising the results to see if they are reasonable. EXERCISES Compute the operations indicated below (if possible), using the following matrices. 1 3 4 3 1 7 6 4 4 B c C £ 5 2 0§ A c 1 d d 7 1 2 8 8 6 3 2 33. A B 34. B A 35. C B 36. 8A 38. C D 39. D C 40. BC 41. 4D

SECTION 9.7

2 3 D £ 0.5 1 4 0.1 37. BA

0 1 § 5

42. CD

Solving Linear Systems Using Matrix Equations

KEY CONCEPTS • The identity matrix I for multiplication has 1’s on the main diagonal and 0’s for all other entries. For any n n matrix A, the identity matrix is also an n n matrix In, where AIn InA A. • For an n n (square) matrix A, the inverse matrix for multiplication is a matrix B such that AB BA I. Only square matrices have inverses. For matrix A the inverse is denoted A1. • Any n n system of equations can be written as a matrix equation and solved (if solutions exist) using an inverse 2x 3y z 5 x 5 2 3 1 # matrix. For • 3x 2y 4z 13, the matrix equation is £ 3 2 4 § £ y § £ 13 § . x 5y 2z 3 z 3 1 5 2 • Every square matrix has a real number associated with it called its determinant. For a 2 2 matrix a11 a12 A ` ` , det(A) a11a22 a21a12. a21 a22 • If the determinant of a matrix is zero, the matrix is said to be singular or noninvertible. • For matrix equations, if the coefficient matrix is noninvertible, the system has no unique solution.

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EXERCISES Complete Exercises 43 through 45 using the following matrices: 1 0 0.2 0.2 2 1 10 d A c d B c C c d D c 0 1 0.6 0.4 3 1 15 43. Exactly one of the matrices given is singular. Compute each determinant to identify it. 44. Show that AB BA B. What can you conclude about matrix A? 45. Show that BC CB I. What can you conclude about matrix C? Use a graphing calculator to complete Exercises 46 through 49, using these matrices: 1 2 3 1 1 1 1 0 0 E £ 2 1 5 § F £ 0 1 0 § G £0 1 0§ 1 1 2 2 1 1 0 0 1

6 d 9

1 H £ 0 2

0 1 1

1 0 § 1

46. Exactly one of the matrices above is singular. Determine which one. 47. Show that GF FG F. What can you conclude about the matrix G? 48. Show that FH HF I. What can you conclude about the matrix H? 49. Verify that EH HE and EF FE and comment. Solve manually using a matrix equation. 2x 5y 14 50. e 3y 4x 14 Solve using a matrix equation and your calculator. 0.5x 2.2y 3z 8 51. • 0.6x y 2z 7.2 x 1.5y 0.2z 2.6

SECTION 9.8

Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More

KEY CONCEPTS • Cramer’s rule uses a ratio of determinants to solve systems of equations (if they exist). • To compute the value of 3 3 and larger determinants, a calculator is generally used. • Determinants can be used to find the area of a triangle in the plane if the vertices are known, and as a test to see if three points are collinear. • A system of equations can be used to write a rational expression as a sum of its partial fractions. EXERCISES Solve using Cramer’s rule. Use a graphing calculator for Exercise 54. 2x y z 1 2x y 2 5x 6y 8 52. e 53. • x 2y z 5 54. • x y 5z 12 10x 2y 9 3x y 2z 8 3x 2y z 8 55. Find the area of a triangle whose vertices have the coordinates (6, 1), 11, 62 , and 16, 22. 7x2 5x 17 . 56. Use a matrix equation and a graphing calculator to find the partial fraction decomposition for 3 x 2x2 3x 6

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Practice Test

57. Midwest Petroleum (MP) produces three types of combustibles using common refined gasoline and vegetable products. The first is E10 (also known as gasohol), the second is E85, and the third is biodiesel. One gallon of E10 requires 0.90 gal of gasoline, 2 lb of corn, 1 oz of yeast, and 0.5 gal of water. One gallon of E85 requires 0.15 gal of gasoline, 17 lb of corn, 8.5 oz of yeast, and 4.25 gal of water. One gallon of biodiesel requires 20 lb of corn and 3 gal of water. One week’s production at MP consisted of 100,000 gal of E10, 15,000 gal of E85, and 7000 gal of biodiesel. Use your calculator and a matrix equation to determine how much of each raw material they used to fill their orders. 58. The following table represents Midwest Petroleum’s production for the next 3 weeks. Use your calculator and a matrix equation to determine the total amount of raw material they used to fill their orders. See Exercise 57. 59. In addition to being one of the world’s first celebrity chefs, Marie-Antoine Carême (1784 –1833) was a master pastry chef. Had he used matrices to encode his favorite culinary math secret, perhaps his message would have been:

Week 2

Week 3

Week 4

E10

110,000

95,000

105,000

E85

17,000

18,000

20,000

6000

8000

10,000

Biodiesel

11, 10, 25, 25, 6, 59, 33, 13, 76, 34, 43, 69, 3, 0, 6 1 2 3 If Carême used the cipher on page 941 (Example 8) and the matrix A £ 1 3 1 § for the 2 4 7 encoding process, decode the message to find the unconventional culinary secret that is actually familiar to students of mathematics.

PRACTICE TEST Solve each system and state whether the system is consistent, inconsistent, or dependent.Verify solutions using a graphing calculator. 1. Solve graphically:

0.5 C £ 0.4 0.1

3x 2y 12 e x 4y 10 2. Solve using substitution:

5x 8y 1 3x 7y 5

0.5 D £ 0.1 0.3

0.1 0.1 0.4

0.2 0 § 0.8

8. Solve using Cramer’s rule: 2x 3y z 3 • x 2y z 4 x y 2z 1

4. Solve using elimination: x 2y z 4 • 2x 3y 5z 27 5x y 4z 27

9. Solve using a matrix equation and your calculator: e

5. Given matrices A and B, compute: 2 a. A B b. B c. AB 5 1 d. A e. 冟A冟 3 5

0.2 0 § 0.1

4x 5y 6z 5 • 2x 3y 3z 0 x 2y 3z 5

3. Solve using elimination:

A c

0 0.5 0.4

7. Solve using matrices and row reduction:

3x y 2 e 7x 4y 6 e

6. Given matrices C and D, use a calculator to find: a. C D b. 0.6D c. DC 1 冟冟 d. D e. D

2 d 4

B c

3 3

2x 5y 11 4x 7y 4

10. Solve using a matrix equation and your calculator:

3 d 5

x 2y 2z 7 • 2x 2y z 5 3x y z 6

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Use a system of equations to model each exercise, then solve using the method of your choice. 11. The perimeter of a “legal-size” paper is 114.3 cm. The length of the paper is 7.62 cm less than twice the width. Find the dimensions of a legal-size sheet of paper. 12. The island nations of Tahiti and Tonga have a combined land area of 692 mi2. Tahiti’s land area is 112 mi2 more than Tonga’s. What is the land area of each island group? 13. After inheriting $30,000 from a rich aunt, David decides to place the money in three different investments: a savings account paying 5%, a bond account paying 7%, and a stock account paying 9%. After 1 yr he earned $2080 in interest. Find how much was invested at each rate if $8000 less was invested at 9% than at 7%. 14. The company from Price Demand Supply Exercise 10 of the (cents) (10,000s) (10,000s) Summary and 135 2.9 19.7 Concept Review, 90 37.2 8.0 hired a market 62 59.6 0.45 research firm to collect data about 112 19.7 12.7 the next-generation 75 46.0 3.69 spark plug that will 121 12.4 15.9 increase fuel efficiency by over 20%. The data they collected are shown in the table. Enter the data into a graphing calculator, calculate the regression line for each set of data, then (a) display the lines and scatterplot on a single screen, and (b) use the lines to find the estimated price and quantity at which market equilibrium is achieved.

Using a radar gun placed in a protective cage beneath a falling baseball, the following data points are taken: (1, 128), (2, 80), (2.5, 44). Assume the data are of the form (time in seconds, distance from radar gun). 15. Use the data to set up a system of three equations in three variables to find the quadratic equation that models the ball’s height at time t. 16. From what height is the baseball being released? In how many seconds will it strike the cage protecting the radar gun? 17. Solve the system of inequalities by graphing. e

xy2 x 2y 8

18. Maximize the objective function: P 50x 12y x 2y 8 • 8x 5y 40 x 0, y 0 Solve the linear programming problem. 19. A company manufactures two types of T-shirts, a plain T-shirt and a deluxe monogrammed T-shirt. To produce a plain shirt requires 1 hr of working time on machine A and 2 hr on machine B. To produce a deluxe shirt requires 1 hr on machine A and 3 hr on machine B. Machine A is available for at most 50 hr/week, while machine B is available for at most 120 hr/week. If a plain shirt can be sold at a profit of $4.25 each and a deluxe shirt can be sold at a profit of $5.00 each, how many of each should be manufactured to maximize the profit? 20. Decompose the expression into partial fractions: 4x2 4x 3 . x3 27

CALCULATOR EXPLORATION AND DISCOVERY Cramer’s Rule In Section 9.8, we saw that one interesting application of matrices is Cramer’s rule. You may have noticed that when technology is used with Cramer’s rule, the chances of making an error are fairly high, as we need to input the entries for numerous matrices. However, as we mentioned in the chapter introduction, one of the advantages of matrices is that they are easily programmable, and we can actually write a very simple program that will make Cramer’s rule a more efficient method. To begin, press the PRGM key, and then the right arrow twice and to create a name for our program. At the prompt, we’ll enter CRAMER2. As we write the program, note that the needed commands (ClrHome, Disp, Pause, Prompt, Stop) are all located in the submenus of the PRGM key, and the = sign is found under the TEST menu, accessed using the 2nd MATH keys (the arrows “S” are used to indicate the STO key. The following program takes the coefficients and constants of a 2 2 linear system, and returns the ordered pair solution in the form of x h and y k (for constants h and k). Even with minimal programming experience, reading through the program will help you identify that Cramer’s rule is being used. ENTER

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ClrHome Disp ''22 SYSTEMS'' Pause Disp ''AXBY C'' Disp ''DXEY F'' Disp '''' Disp ''ENTER THE VALUES'' Disp ''FOR A,B,C,D,E,F'' Disp '''' Prompt A,B,C,D,E,F

955

(CE–BF)/(AE–BD)SX (AF–DC)/(AE–BD)SY ClrHome Disp ''THE SOLUTION IS'' Disp '''' Disp ''X'' Disp X Disp ''Y'' Disp Y Stop

Exercise 1: Use the program to check the answers to Exercises 1, 2, and 3 of the Practice Test. Exercise 2: Create 2 2 systems of your own that are (a) consistent, (b) inconsistent, and (c) dependent. Then verify results using the program. Exercise 3: Use the box on page 935 of Section 9.8 to write a similar program for 3 3 systems. Call the program CRAMER3, and repeat parts (a), (b), and (c) from Exercise 2.

STRENGTHENING CORE SKILLS Augmented Matrices and Matrix Inverses The formula for finding the inverse of a 2 2 matrix has its roots in the more general method of computing the inverse of an n n matrix. This involves augmenting a square matrix M with its corresponding identity In on the right (forming an n 2n matrix), and using row operations to transform M into the identity. In some sense, as the original matrix is transformed, the “identity part” keeps track of the operations we used to convert M and we can use the results to “get back home,” so to speak. We’ll illustrate with the 2 2 matrix from Section 9.7, Example 2B, where we found that 1 2.5 6 5 6 5 c d was the inverse matrix for c d . We begin by augmenting c d with the 2 2 identity matrix. 1 3 2 2 2 2 c c

6 0

6 2

1 0

0 6 d 3R2 R1 S R2 c 1 0

5 1 1 1

0 6 d 1R2 S R2 c 3 0

5 1 1 1

0 6 d 5R2 R1 S R1 c 3 0

0 1

15 R1 1 d S R1 c 3 6 0

5 2

6 1

1 0 d 1 3

5 1 0 1

1 1

2.5 d 3

As you can see, the identity is automatically transformed into the inverse matrix when this method is applied. a b d results in the formula given earlier. As you might Performing similar row operations on the general matrix c c d imagine, attempting this on a general 3 3 matrix is problematic at best, and instead we simply apply the augmented matrix method to find A1 for the 3 3 matrix shown in blue. 2 1 £ 1 3 3 1 14 R2 7R1 S R1 → £ 0 5R2 7R3 S R3 0

0 1 0 0 2 0 1 0 § 2 0 0 1 0 4 6 7 4 1 0 8 16

14 0 0 4R3 R2 S R2 → £ 0 7 0 4R3 R1 S R1 0 0 1

R1 2R2 S R2→ 3R1 2R3 S R3 2 2 10

14 7 7 7 2 1.25

0 0§ 14 7 7 § 1.75

2 £0 0

1 0 7 4 5 4

R3 S R3 8

14 £ 0 0

R1 S R1 14 R2 S R2 7

1 £0 0

1 0 0 1 2 0§ 3 0 2

0 4 6 7 4 1 0 1 2 0 1 0

0 0 1

2 2 1.25

1 0.5 1 1 2 1.25

0 0 § 1.75

0.5 1 §. 1.75

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9–120


1

To verify, we show AA

0 1 0.5 2 § £ 1 1 2 2 1.25

2 1 I: £ 1 3 3 1

0.5 1 1 § £0 1.75 0

0 1 0

0 0 § ✓ (A1A I also checks). 1

Exercise 1: Use the preceding inverse and a matrix equation to solve the system 2x y 2 • x 3y 2z 15 . 3x y 2z 9

CUMULATIVE REVIEW CHAPTERS 1–9

3. A torus is a donut-shaped solid figure. Its surface area is given by the formula A ␲2 1R2 r2 2, where R is the outer radius of the donut, and r is the inner radius. Solve the formula for R in terms of r and A.

11. Given A 13 4 , y B is a point on the unit circle, find the value of sin ␪, cos ␪, and tan ␪ if y 7 0. 12. State the domain of each function: a. f 1x2 12x 3 b. g 1x2 logb 1x 32 x3 c. h 1x2 2 x 5 13. Write the following formulas from memory: a. slope formula b. midpoint formula c. quadratic formula d. distance formula e. interest formula (compounded continuously)

4. State the value of all six trig functions given (21, 28) is a point on the terminal side of ␪.

14. Find the equation of the line perpendicular to the line 4x 5y 20, that contains the point (0, 1).

1. Solve each equation by factoring. a. 9x2 12x 4 b. x2 7x 0 c. 3x3 15x2 6x 30 d. x3 4x 3x2 2. Solve for x. 1 3 3x 2 x3 x2 x x6

␲ ␲ 5. Sketch the graph of y 4 cos a x b using 6 3 transformations of y cos x. 6. A jai alai player in an open-air court becomes frustrated and flings the pelota out and beyond the court. If the initial velocity of the ball is 136 mph and it is released at height of 5 ft, (a) how high is the ball after 3 sec? (b) What is the maximum height of the ball? (c) How long until the ball hits the ground (hopefully in an unpopulated area)? Recall the projectile equation is h1t2 16t2 v0t k. 7. For a complex number a bi, (a) verify the sum of a complex number and its conjugate is a real number. (b) Verify the product of a complex number and its conjuate is a real number. 8. Solve using the quadratic formula: 5x2 8x 2 0. 9. Solve by completing the square: 3x2 72x 427 0. 10. Given cos 53° ⬇ 0.6 and cos 72° ⬇ 0.3, approximate the value of cos 19° and cos 125° without using a calculator.

15. For the force vectors F1 and F2 given, find a third force vector F3 that will bring these vectors into equilibrium: F1 H5, 12I, F2 H8, 6I 16. A commercial fishery stocks a lake with 250 fish. Based on previous experience, the population of fish is expected to grow according to the model 12,000 P 1t2 , where t is the time in months. 1 25e0.2t From on this model, (a) how many months are required for the population to grow to 7500 fish? (b) If the fishery expects to harvest three-fourths of the fish population in 2 yr, approximately how many fish will be taken? 17. Evaluate each expression by drawing a right triangle and labeling the sides. x bd a. sec c sin1a 2121 x2 b. sin c csc1a

29 x2 bd x

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18. An luxury ship is traveling at 15 mph on a heading of 10°. There is a strong, 12 mph ocean current flowing from the southeast, at a heading of 330°. What is the true course and speed of the ship? 19. Use the Guidelines for Graphing to sketch the graph of function f given, then use it to solve f 1x2 6 0: f 1x2 x3 4x2 x 6 20. Use the Guidelines for Graphing to sketch the graph of function g given, then use it to name the intervals x2 4 where g 1x2T and g 1x2c: g1x2 2 x 1 8 21. Find 11 i132 using De Moivre’s theorem.

22. Solve ln1x 22 ln1x 32 ln14x2.

23. If I deposited $200 each month into an annuity program that paid 8% annual interest compounded monthly, how long would it take to save $10,000? 24. Mount Tortolas lies on the Argentine-Chilean border. When viewed from a distance of 5 mi, the angle of elevation to the top of the peak is 38°. How tall is Mount Tortolas? State the answer in feet. 25. The graph given is of the form y A sin1Bx C2. Find the values of A, B, and C. y 2 1

␲

2

1 2

␲ 2

␲

3␲ 2

2␲

x

957

Exercises 26 through 30 require the use of a graphing calculator. 26. At Leroy’s Butcher Shop, last week’s specials were boneless chuck steaks, chicken breasts, and chorizo sausages. Pamela spent a total of $17.81 on 3.21 lb of chuck, 2.45 lb of chicken, and 1.15 lb of chorizo. Shannon purchased 1.57 lb of chuck, 4.21 lb of chicken, and 0.89 lb of chorizo with $15.24, while the 2.41 lb of chuck, 1.39 lb of chicken, and 1.46 lb of chorizo Dusty took home cost a total of $14.05. To the nearest cent, how much did each type of meat cost per pound? 27. Apply long division to find the quotient and remainder ␲2x2 ␲x . Determine where the graph x2 ␲ of the function crosses the horizontal asymptote. for v1x2

28. Find all real solutions of the equation: tan t cos t cos t. 29. Find the local minimum value of y x3 3x2. 30. In New Zealand’s Abel Tasman National Park, the depth of the water in one particular tidal pool can be ␲ modeled by the function d1t2 1.9 sin a tb 2.2, 6 where d(t) is the depth in feet and t is the time of the day (using a 24-hr clock). Determine the times of the day when there is at most 2 ft of water in this pool. Round your answers to the nearest minute.

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Precalculus—

CONNECTIONS TO CALCULUS As mentioned in the chapter opener, in calculus as well as algebra we often look for expressions that fit a special form, then apply a general formula to all expressions of that form. Here this general idea is applied to partial fraction decomposition, a technique we studied in Section 9.4 and an important part of integral calculus.

More on Partial Fraction Decomposition 10 as a sum of its partial fractions, we would use the x ⫺ 25 A 10 B ⫽ ⫹ template and proceed as in Section 9.4. Using the x⫹5 x⫺5 1x ⫹ 521x ⫺ 52 ideas developed there, we’ll attempt to develop a decomposition formula that works 2k for all rational expressions of the form 2 , a form that certainly applies to the x ⫺ k2 expression at hand. To rewrite the expression

EXAMPLE 1

䊳

Solution

䊳

2

Decomposing Special Forms

2k Use the techniques from Section 9.4 to rewrite the expression 2 as the sum x ⫺ k2 of its partial fractions. After factoring the denominator, we obtain the following template: A B 2k ⫽ ⫹ x⫹k x⫺k 1x ⫹ k21x ⫺ k2 2k ⫽ A1x ⫺ k2 ⫹ B1x ⫹ k2 ⫽ Ax ⫺ Ak ⫹ Bx ⫹ Bk ⫽ 1A ⫹ B2x ⫹ 1B ⫺ A2k

decomposition template clear denominators distribute collect like terms

A⫹B⫽0 , yielding the solutions B ⫽ 1 and A ⫽ ⫺1. As it B⫺A⫽2 2k turns out, any rational expression of the form 2 can be written in decomposed x ⫺ k2 ⫺1 1 form as ⫹ . x⫹k x⫺k This gives the system e


EXAMPLE 2

䊳

Solution

䊳

Decomposing Special Forms

10 Use the “formula” from Example 1 to decompose the expression 2 as the x ⫺ 25 sum of its partial fractions, then verify the result. 2k 10 10 with 2 , we note 2k ⫽ 10 and k ⫽ 5, so 2 must 2 x ⫺k x ⫺ 25 x ⫺ 25 ⫺1 1 be equal to . To check, we combine the fractions using the LCD ⫹ x⫹5 x⫺5 ⫺11x ⫺ 52 ⫹ 11x ⫹ 52 ⫺x ⫹ 5 ⫹ x ⫹ 5 10 or 2 ✓. 1x ⫹ 521x ⫺ 52, giving ⫽ 1x ⫹ 52 1x ⫺ 52 1x ⫹ 521x ⫺ 52 x ⫺ 25 By comparing

2

Now try Exercises 3 through 8 958

䊳

䊳

9–122

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959


The Geometry of Vectors and Determinants Although vectors (Section 8.3) and determinants (Section 9.8) are studied in two separate chapters in Precalculus, in a calculus course you’ll see there is an intriguing connection between the two. In Section 8.3 we noted the tip-to-tail addition of vectors u and v form a parallelogram, and used the parallelogram to illustrate a vector sum. In a calculus course, we say the vectors span the parallelogram, and in certain applications we need to know the area of this

Figure 9.74 Ha, bI Hc, dI

u#v for the 冟u冟冟v冟 angle ␪ between them, it can be shown that the area of the parallelogram is the absolute a b value of the determinant ` ` (see Figure 9.74). For a complete proof, see Appendix B. c d figure. Using the vectors u ⫽ Ha, bI and v ⫽ Hc, dI and relation cos ␪ ⫽

EXAMPLE 3

䊳

Solution

䊳

Using Determinants to Find Area

For the vectors u ⫽ H12, 5I and v ⫽ H8, 15I, a. Compute the vector sum w ⫽ u ⫹ v using the parallelogram (tail-to-tip) method. b. Compute 冟w冟 and find the angle ␪ between u and w. 1 c. Use the formula A ⫽ 冟u冟冟w冟sin ␪ from Section 8.2 to find the area of the 2 triangle formed. a b 12 5 d. Use the determinant ` ` ⫽ ` ` to find the area of the c d 8 15 parallelogram spanned by these vectors, then verify the area is twice as large as the triangle found in part (c). a. The vector sum is H12, 5I ⫹ H8, 15I ⫽ H20, 20I ⫽ w.

b. 冟w冟 ⫽ 2202 ⫹ 202 ⫽ 1800 ⫽ 20 12. For u # w ⫽ 121202 ⫹ 51202 ⫽ 340, 340 17 and 冟u冟 ⫽ 13 (verify this), we have cos ␪ ⫽ and ␪ ⫽ cos⫺1a b. 1132 120 122 1312 This indicates ␪ ⬇ 22.38°. 1 c. A ⫽ 冟u冟冟w冟sin ␪ area formula 2 1 ⬇ 1132 120 122sin 22.38° substitute 2 ⬇ 70 units2 result

d. For u ⫽ H12, 5I and v ⫽ H8, 15I,

a b 12 5 the determinant ` ` ⫽ ` ` ⫽ c d 8 15 11221152 ⫺ 182152 ⫽ 140 units2✓, and the area of the parallelogram is indeed twice the area of the triangle.

H20, 20I

20 18

H8, 15I

16 14 12 10 8 6

H12, 5I

4 2 2

4

6

8 10 12 14 16 18 20


䊳

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9–124


Connections to Calculus Exercises Use the techniques from Section 9.4 to complete Exercises 1 and 2.

1. A common form of partial fraction decomposition involves a constant over a factorable quadratic expression. (a) Rewrite the expression k as the sum of partial fractions to 1x ⫹ a21x ⫹ b2 A⫹B⫽0 . (b) Use show the system obtained is e Ab ⫹ Ba ⫽ k the fact that A ⫹ B ⫽ 0 to show the system can be A⫹B⫽0 rewritten as e . (c) Solve for A in the A1b ⫺ a2 ⫽ k second equation and use the result to quickly write 16 as the sum of its partial fractions. 1x ⫹ 321x ⫺ 52

2. Another common form involves a linear expression x ⫹ k over a factorable quadratic expression. x⫹k (a) Rewrite the expression as the 1x ⫹ a21x ⫹ b2 sum of partial fractions to show the system obtained A⫹B⫽1 is e . (b) Use the fact that A ⫹ B ⫽ 1 Ab ⫹ Ba ⫽ k to show that system can be written as A⫹B⫽1 e . (c) Solve for A in the A1b ⫺ a2 ⫹ a ⫽ k second equation, and use the result to write x ⫹ 10 as the sum of its partial fractions. 1x ⫹ 321x ⫹ 42

Use the “formulas” developed in Example 1 and Exercises 1 and 2 to decompose each expression into a sum of partial fractions.

3.

6 x ⫺9

4.

18 x ⫺ 81

5.

⫺22 x ⫹ 3x ⫺ 28

6.

15 x ⫺ 3x ⫺ 4

7.

x ⫹ 11 x ⫹ 13x ⫹ 40

8.

x ⫺ 17 x ⫺ 9x ⫹ 14

2

2

2

2

2

2

Find the area of the parallelogram spanned by the vectors given, then verify your answer using the method shown in Example 3.

9. u ⫽ H2, 8I and v ⫽ H15, 3I

11. 6i ⫹ 3j and 6i ⫺ 3j

10. u ⫽ H⫺4, 1I and v ⫽ H4, 4I 12. 9i and 7j

a b Similar to how the absolute value of ` ` gives the area of a parallelogram spanned by Ha, bI and Hc, dI, the absolute c d a b c value of the determinant † d e f † gives the volume of a parallelepiped spanned by Ha, b, c,I, Hd, e, f I, and Hg, h, iI in g h i three dimensions.

13. Find the volume of the parallelepiped spanned by H5, 0, 0I, H0, 6, 0I, and H0, 0, 8I using the determinant formula, then verify your answer using a more familiar formula.

14. Find the volume of the parallelepiped spanned by u ⫽ H2, 5, ⫺8I, H3, 0, 4I, and H0, 10, ⫺7I.

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CHAPTER CONNECTIONS

Analytical Geometry and the Conic Sections CHAPTER OUTLINE 10.1 A Brief Introduction to Analytical Geometry 962

Cathedral windows are just one place where we can appreciate the beauty and intricacy of polar graphs. The characteristics and features of these graphs are explored in this chapter, and introduce a historic alternative to graphing relations by simply computing points (x, y). While graphing relations using a rectangular coordinate system is much more widely known, both polar graphs and rectangular graphs seemed to have matured simultaneously, in the early 1600s. 䊳

The graphs shown appear as Exercises 73 and 81 in Section 10.6.

10.2 The Circle and the Ellipse 969 10.3 The Hyperbola 984 10.4 The Analytic Parabola 997 10.5 Nonlinear Systems of Equations and Inequalities 1007 10.6 Polar Coordinates, Equations, and Graphs 1018 10.7 More on the Conic Sections: Rotation of Axes and Polar Form 1035

10.8 Parametric Equations and Graphs 1051

When using polar curves in the design of windows or other forms of architechture, finding where two graphs intersect aids in the calculation of the area between curves and in determining the arclength of certain portions of the graph. These are important considerations in a study of calculus, and techniques for working Connections 961 to Calculus with systems of polar equations are explored in the Connections to Calculus for Chapter 10.

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10.1

A Brief Introduction to Analytical Geometry

LEARNING OBJECTIVES

Generally speaking, analytical geometry is a study of geometry using the tools of algebra and a coordinate system. These tools include the midpoint and distance formulas; the algebra of parallel, perpendicular, and intersecting lines; and other tools that help establish geometric concepts. In this section, we’ll use these tools to verify certain relationships, then use these relationships to introduce a family of curves known as the conic sections.


A. Verify theorems from basic geometry involving the distance between two points B. Verify that points (x, y) are an equal distance from a given point and a given line C. Use the defining characteristics of a conic section to find its equation

A. Verifying Relationships from Plane Geometry For the most part, the algebraic tools used in this study were introduced in previous chapters. As the midpoint and distance formulas play a central role, they are restated here for convenience. Algebraic Tools Used in Analytical Geometry

Given two points P1 1x1, y1 2 and P2 1x2, y2 2 in the xy-plane. Midpoint Formula

Distance Formula

The midpoint of line segment P1P2 is x1 x2 y1 y2 , b 1x, y2 a 2 2

The distance from P1 to P2 is d 21x2 x1 2 2 1y2 y1 2 2

These formulas can be used to verify the conclusion of many theorems from Euclidean geometry, while providing important links to an understanding of the conic sections. EXAMPLE 1

䊳

Verifying a Theorem from Basic Geometry y

A theorem from basic geometry states: The midpoint of the hypotenuse of a right triangle is an equal distance from all three vertices. Verify this statement for the right triangle formed by 14, 22, 14, 22 , and (4, 4).

Solution

䊳

5

(0, 1)

R (4, 4) M

After the plotting points and drawing a 5 5 Q P triangle, we note the hypotenuse has endpoints (4, 2) (4, 2) 14, 22 and (4, 4), with midpoint 4 142 4 122 a , b 10, 12. 5 2 2 Using the distance formula to find the distance from (0, 1) to (4, 4) gives

x

d 214 02 2 14 12 2 2142 2 132 2 125 5 From the definition of midpoint, (0, 1) is also 5 units from 14, 22 . Checking the distance from (0, 1) to the vertex 14, 22 gives

d 214 02 2 12 12 2 242 132 2 125 5 The midpoint of the hypotenuse is an equal distance from all three vertices (see the figure).

962

Now try Exercises 7 through 12 䊳 10–2

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963

Section 10.1 A Brief Introduction to Analytical Geometry

Figure 10.1 Recall from Section 1.1 that a circle is the set y of all points that are an equal distance (called the radius) from a given point (called the center). If all 5 three vertices of a triangle lie on the circumference of a circle, we say the circle circumscribes the triangle. Based on our earlier work, it appears we (0, 1) could also state the theorem in Example 1 as: For 5 any circle in the xy-plane whose center (h, k) is the midpoint of the hypotenuse L of a right triangle, the circle defined by 1x h2 2 1y k2 2 1 L2 2 2 (4, 2) circumscribes the triangle. The circle and trian5 gle from Example 1 illustrate this theorem in Figure 10.1, where the equation of the circle is 2 1x 02 2 1y 12 2 1 10 2 2 . See Exercises 13 through 20.

A. You’ve just seen how we can verify theorems from basic geometry involving the distance between two points

Figure 10.2

(4, 4)

x

5

(4, 2)

B. The Distance Between a Point and a Line In a study of analytical geometry, we are also interested in the distance d between a point and a line. This is always defined as the perpendicular distance, or the length of a line segment perpendicular to the given line, with the given point and the point of intersection as endpoints (see Figure 10.2).

d

䊳

EXAMPLE 2

Locating Points That Are an Equal Distance from a Given Point and Line In Figure 10.3, the origin (0, 0) is seen to be an equal distance from the point (0, 2) and the line y 2. Show that the following points are also an equal distance from (0, 2) and y 2: a. 12, 12 2 b. (4, 2) c. (8, 8)

䊳

Solution

Since the given line is horizontal, the perpendicular distance from the line to each point can be found by vertically counting the units. It remains to show that this is also the distance from the given point to (0, 2) (see Figure 10.4). a. The distance from 12, 12 2 to y 2 is 2.5 units. The distance from 12, 12 2 to (0, 2) is Figure 10.3 d 210 22 2 12 0.52 2

Figure 10.4 y d1 d2

5

(0, 2)

(8, 8)

d1 (4, 2)

d2

(2, 0.5) 5 (0, 0)

5 3

x

y 2

B. You’ve just seen how we can verify that points (x, y) are an equal distance from a given point and a given line

2122 2 11.52 2 16.25 2.5 ✓ b. The distance from (4, 2) to y 2 is 4 units. The distance from (4, 2) to (0, 2) is d 210 42 2 12 22 2

y

5

(0, 2)

2142 2 102 2 5 (0, 0) 116 4✓ 3 c. The distance d2 from (8, 8) to y 2 is 10 units. The distance d1 from (8, 8) to (0, 2) is d1

x

5

y 2

210 82 2 12 82 2 2182 2 162 2 1100 10 ✓


䊳

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10–4

CHAPTER 10 Analytical Geometry and the Conic Sections

C. Characteristics of the Conic Sections

Figure 10.5 Axis

Nappe Generator Vertex

Nappe

Examples 1 and 2 bring us one step closer to the wider application of these ideas in a study of the conic sections. But before the connection is clearly made, we’ll introduce some background on this family of curves. In common use, a cone might bring to mind the conical paper cups found at a water cooler. The point of the cone is called the vertex and the sheet of paper forming the sides is called a nappe. In mathematical terms, a cone has two nappes, formed by rotating a nonvertical line (called the generator), about a vertical line (called the axis), at their point of intersection—the vertex (see Figure 10.5). The conic sections are so named because all curves in the family can be formed by a section of the cone, or more precisely the intersection of a plane and a cone. Figure 10.6 shows that if the plane does not go through the vertex, the intersection will produce a circle, ellipse, parabola, or hyperbola. Figure 10.6

Circle

Circle

Ellipse

Parabola

Hyperbola

Ellipse

Parabola

Hyperbola

If the plane does go through the vertex, the result is a single point, a single line (if the plane contains the generator), or a pair of intersecting lines (if the plane contains the axis). The connection we seek to make is that each conic section can be defined in terms of the distance between points in the plane, as in Example 1, or the distance between a given point and a line, as in Example 2. In Example 1, we noted the points 14, 22, 14, 22, and (4, 4) were all on a circle of radius 5 with center (0, 1), in line with the analytic definition of a circle: A circle is the set of all points that are an equal distance (called the radius) from a given point (called the center). Figure 10.7 In Example 2, you may have noy ticed that the points seemed to form the right branch of a parabola (see Figure 10.7), and in fact, this example il(8, 8) lustrates the analytic definition of a parabola: A parabola is the set of all d1 points that are an equal distance from 5 a given point (called the focus), and a Focus d2 given line (called the directrix). (4, 2) The focus and directrix are not ac(0, 2) tually part of the graph, they are sim5 (0, 0) x ply used to locate points on the graph. 5 For this reason all foci (plural of focus) y 2 directrix will be represented by a “” symbol d1 d2 rather than a point.

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EXAMPLE 3

䊳

965

Finding an Equation for All Points That Form a Certain Parabola With Example 2 as a pattern, use the analytic definition to find a formula (equation) for the set of all points that form the parabola.

Solution

Y1

䊳

1 2 X , Y2 2 8

Use the ordered pair (x, y) to represent an arbitrary point on the parabola. Since any point on the line y 2 has coordinates 1x, 22 , we set the distance from 1x, 22 to (x, y) equal to the distance from (0, 2) to (x, y). The result is

9

12

12

6

21x x2 2 3y 122 4 2 21x 02 2 1y 22 2 distances are equal simplify 21y 22 2 2x2 1y 22 2 power property 1y 22 2 x2 1y 22 2 2 2 2 expand binomials y 4y 4 x y 4y 4 simplify 8y x2 1 result y x2 8 All points satisfying these conditions are on the parabola defined by y 18x2. See the figure. Now try Exercises 27 and 28

䊳

At this point, it seems reasonable to ask what happens when the distance from the focus to (x, y) is less than the distance from the directrix to (x, y). For example, what if the distance is only two-thirds as long? As you might guess, the result is one of the other conic sections, in this case an ellipse. If the distance from the focus to a point (x, y) is greater than the distance from the directrix to (x, y), one branch of a hyperbola is formed. While we will defer a development of their general equations until later in the chapter, the following diagrams serve to illustrate this relationship for the ellipse, and show why we refer to the conic sections as a family of curves. In Figure 10.8, the line segment from the focus to each point on the graph (shown in blue), is exactly two-thirds the length of the line segment from the directrix to the same point (shown in red). Note the graph of these points forms the right half of an ellipse. In Figure 10.9, the lines and points forming the first half are removed to more clearly show the remaining points that form the complete graph. Figure 10.8

EXAMPLE 4

䊳

Figure 10.9

Finding an Equation for All Points That Form a Certain Ellipse Suppose we arbitrarily select the point (1, 0) as a focus and the (vertical) line x 4 as the directrix. Use these to find an equation for the set of all points where the distance from the focus to a point (x, y) is 12 the distance from the directrix to (x, y).

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Solution

䊳

Since any point on the line x 4 has coordinates (4, y), we have: 1 Distance from 11, 02 to 1x, y2 3 distance from 14, y2 to 1x, y2 4 in words 2 1 21x 12 2 3y 102 4 2 21x 42 2 1y y2 2 resulting equation 2 1 21x 12 2 y2 21x 42 2 simplify 2 1 1x 12 2 y2 1x 42 2 power property 4 1 x2 2x 1 y2 1x2 8x 162 expand binomials 4 1 x2 2x 1 y2 x2 2x 4 distribute 4 3 2 1 3 x y2 3 simplify: 1x 2 x 2 x 2 4 4 4 3x2 4y2 12 polynomial form All points satisfying these conditions are on the ellipse defined by 3x2 4y2 12. Now try Exercises 29 and 30

Figure 10.10 f1

f2

C. You’ve just seen how we can use the defining characteristics of a conic section to find its equation

䊳

Actually, any given ellipse has two foci (see Figure 10.10) and the equation from Example 4 could also have been developed using the left focus (with the directrix also on the left). This symmetrical relationship leads us to an alternative definition for the ellipse, which we will explore further in Section 10.2: For foci f1 and f2, an ellipse is the set of all points Figure 10.11 (x, y) where the sum of the distances from f1 to (x, y) and f2 to (x, y) is constant. d1 d2 See Figure 10.11 and Exercises 31 and 32. Both the f1 focus/directrix definition and the two foci definition have d3 f2 d4 merit, and simply tend to call out different characteris- (x, y) tics and applications of the ellipse. The hyperbola also d1 d2 d3 d4 has a focus/directrix definition and a two foci definition. See Exercises 33 and 34.

10.1 EXERCISES 䊳



1. Analytical geometry is a study of the tools of .

using

2. The distance formula is d the midpoint formula is M

; .

3. The distance between a point and a line always refers to the distance.

4. The conic sections are formed by the intersection of a and a .

5. If a plane intersects a cone at its vertex, the result is a , a line, or a pair of lines.

6. A circle is defined relative to an equal distance between two . A parabola is defined relative to an equal distance between a and a .

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The three points given form a right triangle. Find the midpoint of the hypotenuse and verify that the midpoint is an equal distance from all three vertices.

7. P1 15, 22 P2 11, 22 P3 15, 62 9. P1 12, 12 P2 16, 52 P3 12, 72

11. P1 110, 212 P2 16, 92 P3 13, 32

8. P1 13, 22 P2 13, 142 P3 18, 22

15. Find an equation of the circle that circumscribes the triangle in Exercise 9. 16. Find an equation of the circle that circumscribes the triangle in Exercise 10. 17. Find an equation of the circle that circumscribes the triangle in Exercise 11.

10. P1 10, 52 P2 16, 42 P3 16, 12

18. Find an equation of the circle that circumscribes the triangle in Exercise 12. 19. Of the following six points, identify which four are an equal distance from the point A(2, 3). B(7, 15) D(9, 14) C110, 82 E13, 92

12. P1 16, 62 P2 112, 182 P3 120, 422

F15, 4 3 1102

G12 2 130, 102

20. Of the following six points, identify which four are an equal distance from the point P11, 42 . Q19, 102 R(5, 12) S17, 112 T14, 4 5132

13. Find an equation of the circle that circumscribes the triangle in Exercise 7. 14. Find an equation of the circle that circumscribes the triangle in Exercise 8. 䊳

U11 416, 62

V17, 4 1512


The Perpendicular Distance from a Point to a Line: d ⴝ

円Ax1 ⴙ By1 ⴙ C円

Exercise 21 . The perpendicular

2A2 ⴙ B2 distance from a point (x1, y1) to a given line can be found using the formula shown, where Ax ⴙ By ⴙ C ⴝ 0 is the equation of the line in standard form (A, B, and C are integers).

21. Use the formula to verify that P16, 22 and Q(6, 4) are an equal distance from the line y 12x 3.

䊳

967


22. Find the value(s) for y that ensure (1, y) is this same distance from y 12x 3.

(x1, y1) d

Ax By C 0

APPLICATIONS

23. Of the following four points, identify which three are an equal distance from the point A(0, 1) and the line y 1. E14 12, 82 B16, 92 C14, 42 D12 12, 62 24. Of the following four points, identify which three are an equal distance from the point P(2, 4) and the line y 4. Q110, 92 R12 4 12, 32 S110, 42 T12 415, 52 25. Consider the fixed point 10, 42 and the fixed line y 4. Verify that the distance from each point given to 10, 42 , is equal to the distance from the point to the line y 4. 25 C14 12, 22 A14, 12 B a10, b 4 D1815, 202

26. Consider the fixed point 10, 22 and the fixed line y 2. Verify that the distance from each point given to 10, 22 , is equal to the distance from the point to the line y 2. 9 R1415, 102 P112, 182 Qa6, b 2 S14 16, 122 27. The points from Exercise 25 are on the graph of a parabola. Find an equation of the parabola. 28. The points from Exercise 26 are on the graph of a parabola. Find an equation of the parabola.

29. Using 10, 22 as the focus and the horizontal line y 8 as the directrix, find an equation for the set of all points (x, y) where the distance from the focus to (x, y) is one-half the distance from the directrix to (x, y).

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33. From the focus/directrix definition of a hyperbola: If the distance from the focus to a point (x, y) is greater than the distance from the directrix to (x, y), one branch of a hyperbola is formed. Using (2, 0) as the focus and the vertical line x 12 as the directrix, find an equation for the set of all points (x, y) where the distance from the focus to (x, y), is twice the distance from the directrix to (x, y).

30. Using (4, 0) as the focus and the vertical line x 9 as the directrix, find an equation for the set of all points (x, y) where the distance from the focus to (x, y) is two-thirds the distance from the directrix to (x, y). 31. From Exercise 29, verify the points 13, 22 and 1 112, 02 are on the ellipse defined by 4x2 3y2 48. Then verify that d1 d2 d3 d4.

Exercise 31 y 5

f2 (0, 2)

(3, 2) d1 5

32. From Exercise 30, verify the points 14, 10 3 2 and 13, 1152 are on the ellipse defined by 5x2 9y2 180. Then verify that d1 d2 d3 d4.

( 12, 0)

d3

d2 (0, 2)

5 x

f1

d4

5

34. From the two foci definition of a hyperbola: For foci f1 and f2, a hyperbola is the set of all points (x, y) where the difference of the distances from f1 to (x, y) and f2 to (x, y) is constant. Verify the points (2, 3) and 13, 2 162 are on the graph of the hyperbola from Exercise 33. Then verify d1 d2 d3 d4.

Exercise 32

Exercise 34

y

y

5

5

冢4, j冣 d1 (4, 0) 6 f1

䊳

d1

f2 5

d4 5

d4

(3, 2 6)

(2, 3) d2 (2, 0) f1 5 x

d3 5


35. Properties of a circle: A theorem from elementary geometry states: If a radius is perpendicular to a chord, it bisects the chord. Verify this is true for the circle, radii, and chords shown.

䊳

(2, 0)

(4, 0) 6 x

d3 (3, 15)

d2

f2

y 5

(3, 4) Q (4, 2) T

C

P 5

5 x

S (2, 4) U 5

R

(4, 3)

36. Verify that points C12, 32 and D12 12, 162 are points on the ellipse with foci at A12, 02 and B(2, 0), by verifying d1AC2 d1BC2 d1AD2 d1BD2. The expression that results has the form 1U V 1U V, which prior to the common use of technology, had to be simplified using the formula 1U V 1U V 2a 1b, where a 2U and b 41U 2 V 2 2 . Use this relationship to simplify the equation above.


37. (7.4) Verify the following is an identity: cos 12x2 sin2x cot2 1x2 1 cos2x 38. (7.7) Find all solutions in [0, 2␲) ␲ 225 600 825 sin ax b 6

39. (5.5) Solve for x in both exact and approximate form: 10 1 9e0.5x b. 345 5e0.4x 75 a. 5

x2 9 . x2 4 Clearly label all intercepts and asymptotes.

40. (4.4) Sketch a complete graph of h1x2

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A. Use the characteristics of a circle and its graph to understand the equation of an ellipse B. Use the equation of an ellipse to graph central and noncentral ellipses C. Locate the foci of an ellipse and use the foci and other features to write the equation D. Solve applications involving the foci

EXAMPLE 1

In Section 10.1, we introduced the equation of an ellipse using analytical geometry and the focus-directrix definition. Here we’ll take a different approach, and use the equation of a circle to demonstrate that a circle is simply a special ellipse. In doing so, we’ll establish a relationship between the foci and vertices of the ellipse, that enables us to apply these characteristics in context.

A. The Equation and Graph of a Circle Recall that the equation of a circle with radius r and center at (h, k) is 1x h2 2 1y k2 2 r2.

As in Section 1.1, the standard form can be used to construct the equation of the circle given the center and radius as in Example 1, or to graph the circle as in Example 2.

䊳

Determining the Equation of a Circle Given Its Center and Radius Find an equation of the circle with radius 5 and center at (2, 1), then graph the relation on a calculator.

Solution

䊳

With a center of (2, 1), we have h 2, k 1, and r 5. Making the corresponding substitutions into the standard form we obtain 1x h2 2 1y k2 2 r2 1x 22 2 3y 112 4 2 52 1x 22 2 1y 12 2 25

standard form substitute 2 for h, 1 for k, and 5 for r simplify

The equation of this circle is 1x 22 1y 12 2 25. Recall from Section 1.1 that circles (and other relations) can be graphed by solving for y, then graphing the upper and lower halves of the circle. 2

1x 22 2 1y 12 2 25 1y 12 2 25 1x 22 2 y 1 225 1x 22 2 y 225 1x 22 2 1

original equation isolate term containing y take square roots subtract 1

Y1 225 1X 22 1, Y2 225 1X 22 2 1 2

6.2

The graph is shown in the figure using a square window. Note the point (5, 3) satisfies the original equation and is a point on the graph and that 15, 52, 11, 52 , and 11, 32 must also be on the graph due to symmetry.

9.4

9.4

6.2


䊳

If the equation is given in polynomial form, recall that we first complete the square in x and y to identify the center and radius. 10–9

969

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EXAMPLE 2

䊳

Completing the Square to Graph a Circle Find the center and radius of the circle whose equation is given, then sketch its graph: x2 y2 6x 4y 3 0.

Solution

䊳

Begin by completing the square in both x and y.

1x2 6x __ 2 1y2 4y __ 2 3 1x2 6x 92 1y2 4y 42 3 9 4 adds 9 to left side

adds 4 to left side

1x 32 2 1y 22 2

group x- and y-terms; add 3

complete the square add 9 4 to right side 16 factor and simplify

The center is at (3, 2), with radius r 116 4. y 3

(3, 2)

Circle Center at (3, 2)

2

8

r4 (3, 2)

(1, 2)

7

Radius: r 4

x

Diameter: 2r 8

(7, 2)

Endpoints of horizontal diameter (1, 2) and (7, 2) Endpoints of vertical diameter (3, 2) and (3, 6)

(3, 6)


䊳

The equation of a circle in standard form provides a useful link to some of the other conic sections, and is obtained by setting the equation equal to 1. In the case of a circle, this means we simply divide by r2. 1x h2 2 1y k2 2 r2

1x h2 r

A. You’ve just seen how we can use the characteristics of a circle and its graph to understand the equation of an ellipse

2

2

1y k2 r2

standard form

2

1

divide by r 2

In this form, the value of r in each denominator gives the horizontal and vertical distances, respectively, from the center to the graph. This is not so important in the case of a circle, since this distance is the same in any direction. But for other conics, these horizontal and vertical distances are not the same, making the new form a valuable tool for graphing. To distinguish the horizontal from the vertical distance, r2 is replaced by a2 in the “x-term” (horizontal distance), and by b2 in the “y-term” (vertical distance). This distinction leads us directly into our study of the ellipse.

B. The Equation of an Ellipse It then seems reasonable to ask, “What happens to the graph when a b?” To answer, 1x 32 2 1y 22 2 1 (after consider the equation from Example 2. We have 42 42 1y 22 2 1x 32 2 1, where a 4 dividing by 16), which we now compare to 42 32 and b 3. The center of the graph is still at (3, 2), since h 3 and k 2 remain unchanged. Substituting y 2 to find additional points, eliminates the y-term and gives two values for x:

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Section 10.2 The Circle and the Ellipse

1x 32 2

WORTHY OF NOTE If you point a flashlight at the floor keeping it perpendicular to the ground, a circle is formed with the bulb pointing directly at the center and every point along the outer edge of the beam an equal distance from this center. If you hold the flashlight at an angle, the circle is elongated and becomes an ellipse, with the bulb pointing directly at one focus.

42

12 22 2

32 1x 32 2

1

01 42 1x 32 2 16

x 3 4

x34 x 7 and x 1

substitute 2 for y

simplify multiply by 42 16 property of square roots add 3

This shows the horizontal distance from the center to the graph is still a 4, and the points (1, 22 and (7, 2) are on the graph (see Figure 10.12). Similarly, for x 3 we have 1y 22 2 9, giving y 5 and y 1, and showing the vertical distance from the center to the graph is now b 3, with points (3, 1) and (3, 5) on the graph. Using this information to sketch the curve reveals the “circle” is elongated and has become a horizontal ellipse. For this ellipse, the line segment through the center, parallel to the x-axis, and with endpoints on the ellipse is called the major axis, with the endpoints of the major axis called vertices. The segment perpendicular to and bisecting the major axis (with its endpoints on the ellipse) is called the minor axis, as shown in Figure 10.13. Figure 10.12 y 3

Figure 10.13

(3, 1)

Major axis b3

2

(1, 2)

8

a4

(3, 2) 5

a

x

(7, 2)

b Vertex

Vertex

Ellipse (3, 5)

The case where a>b

Minor axis

• If a2 7 b2, the major axis is horizontal (parallel to the x-axis) with length 2a, and the minor axis is vertical with length 2b (see Example 3). • If a2 6 b2 the major axis is vertical (parallel to the y-axis) with length 2b, and the minor axis is horizontal with length 2a (see Example 4). Generalizing this observation we obtain the equation of an ellipse in standard form. The Equation of an Ellipse in Standard Form 1x h2 2

1y k2 2

1. a2 b2 If a b the equation represents the graph of an ellipse with center at (h, k). • 冟a冟 gives the horizontal distance from center to graph. • 冟b冟 gives the vertical distance from center to graph. Given

Finally, note the line segment from center to vertex is called the semimajor axis, with the perpendicular line segment from center to graph called the semiminor axis. EXAMPLE 3

䊳

Graphing a Horizontal Ellipse Sketch the graph of the ellipse defined by 1x 22 2 1y 12 2 1. 25 9

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Solution

䊳

Noting a b, we have an ellipse with center 1h, k2 12, 12. The horizontal distance from the center to the graph is a 5, and the vertical distance from the center to the graph is b 3. After plotting the corresponding points and connecting them with a smooth curve, we obtain the graph shown.

y Ellipse (2, 2)

(3, 1)

b3 a5 (2, 1)

x

(7, 1)

(2, 4)


As with the circle, the equation of an ellipse can be given in polynomial form, and here our knowledge of circles is helpful. For the equation 25x2 4y2 100, we know the graph cannot be a circle since the coefficients are unequal, and the center of the graph must be at the origin since h k 0. To actually draw the graph, we convert the equation to standard form. Note that a circle whose center is at (0, 0) is called a central circle, and an ellipse with center at (0, 0) is called a central ellipse.

WORTHY OF NOTE In general, for the equation Ax2 By2 F (A, B, F 7 0), the equation represents a circle if A B, and an ellipse if A B.

EXAMPLE 4

䊳

䊳

Graphing a Vertical Ellipse For 25x2 4y2 100, a. Write the equation in standard form and identify the center and the values of a and b. b. Identify the major and minor axes and name the vertices. c. Sketch the graph. d. Graph the relation on a graphing calculator using a “friendly” window, then use the TRACE feature to find four additional points on the graph whose coordinates are rational.

Solution

䊳

The coefficients of x2 and y2 are unequal, and 25, 4, and 100 have like signs. The equation represents an ellipse with center at (0, 0). To obtain standard form: a. 25x2 4y2 100 given equation 4y2 25x2 1 divide by 100 100 100 y2 x2 1 standard form 4 25 y2 x2 1 write denominators in squared form; a 2, b 5 22 52 b. The result shows a 2 and b 5, indicating the major axis will be vertical and the minor axis will be horizontal. With the center at the origin, the x-intercepts will Figure 10.14 be 12, 02 and (2, 0), y with the vertices (and Vertical ellipse (0, 5) y-intercepts) at Center at (0, 0) 10, 52 and (0, 5). b5 Endpoints of major axis (vertices) c. Plotting these (0, 5) and (0, 5) intercepts and (2, 0) (2, 0) Endpoints of minor axis x sketching the ellipse (2, 0) and (2, 0) a 2 results in the graph Length of major axis 2b: 2(5) 10 shown in Length of minor axis 2a: 2(2) 4 Figure 10.14. (0, 5)

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d. As with the circle, we begin by solving for y. 25x2 4y2 100 4y2 100 25x2 100 25x2 y2 4 100 25x2 y B 4 Y1

original equation isolate term containing y divide by 4

take square roots

100 25X2 100 25X2 , Y2 B B 4 4 Figure 10.15 6.2

The graph is shown in Figure 10.15, where we note that (1.6, 3) is a point on the graph. Due to the symmetry of the ellipse, 11.6, 32 , 11.6, 32 , and 11.6, 32 are also on the graph.

9.4

9.4

6.2

Now try Exercises 25 through 36 WORTHY OF NOTE After writing the equation in standard form, it is possible to end up with a constant that is zero or negative. In the first case, the graph is a single point. In the second case, no graph is possible since roots of the equation will be complex numbers. These are called degenerate cases. See Exercise 84.

EXAMPLE 5

䊳

If the center of the ellipse is not at the origin, the polynomial form has additional linear terms and we must first complete the square in x and y, then write the equation in standard form to sketch the graph (see the Reinforcing Basic Concepts feature for more on completing the square). Figure 10.16 illustrates how the central ellipse and the shifted ellipse are related.

䊳

Figure 10.16 y Ellipse with center at (h, k) k

(h, k)

Central ellipse (0, b) (a, 0) (a, 0) (0, 0) (0, b)

All points shift h units horizontally, k units vertically, opposite the sign (x h)2 (y k)2 1 a2 b2 a2 b2

x h x2 y2 2 1 a2 b a2 b2

Completing the Square to Graph an Ellipse Sketch the graph of 25x2 4y2 150x 16y 141 0, then state the domain and range of the relation.

Solution

䊳

The coefficients of x2 and y2 are unequal and have like signs, and we assume the equation represents an ellipse but wait until we have the factored form to be certain (it could be a degenerate ellipse). 25x2 4y2 150x 16y 141 0 25x2 150x 4y2 16y 141 2 251x 6x __ 2 41y2 4y __ 2 141 251x2 6x 92 41y2 4y 42 141 225 16 c

c

adds 25192 225

c

c adds 4142 16

add 225 16 to right

given equation (polynomial form) group like terms; subtract 141 factor out leading coefficient from each group complete the square

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251x 32 2 41y 22 2 41y 22 2 251x 32 2 100 100 1y 22 2 1x 32 2 4 25 1x 32 2 1y 22 2 22 52 The result is a vertical ellipse with center at 13, 22, with a 2 and b 5. The vertices are a vertical distance of 5 units from center, and the endpoints of the minor axis are a horizontal distance of 2 units from center. Note this is the same ellipse as in Example 4, but shifted 3 units left and 2 up. The domain of this relation is x 僆 3 5, 14 , and the range is y 僆 3 3, 7 4 .

100 100 100

factor divide both sides by 100

1

simplify (standard form)

1

write denominators in squared form

(3, 7)

y

Vertical ellipse Center at (3, 2)

(5, 2)

(3, 2)

Endpoints of major axis (vertices) (3, 3) and (3, 7) (1, 2)

Endpoints of minor axis (5, 2) and (1, 2) x Length of major axis 2b: 2(5) 10 Length of minor axis 2a: 2(2) 4

(3, 3)

Now try Exercises 37 through 44 B. You’ve just seen how we can use the equation of an ellipse to graph central and noncentral ellipses

䊳

C. The Foci of an Ellipse In Section 10.1, we noted that an ellipse could also be defined in terms of two special points called the foci. The Museum of Science and Industry in Chicago, Illinois (http://www.msichicago.org), has a permanent exhibit called the Whispering Gallery. The construction of the room is based on some of the reflective properties of an ellipse. If two people stand at designated points in the room and one of them whispers very softly, the other person can hear the whisper quite clearly—even though they are over 40 ft apart! The point where each person stands is a focus of an ellipse. This reflective property also applies to light and radiation, giving the ellipse some powerful applications in science, medicine, acoustics, and other areas. To understand and appreciate these applications, we introduce the analytic definition of an ellipse.

WORTHY OF NOTE You can easily draw an ellipse that satisfies the definition. Press two pushpins (these form the foci of the ellipse) halfway down into a piece of heavy cardboard about 6 in. apart. Take an 8-in. piece of string and loop each end around the pins. Use a pencil to draw the string taut and keep it taut as you move the pencil in a circular motion—and the result is an ellipse! A different length of string or a different distance between the foci will produce a different ellipse.

Definition of an Ellipse Given two fixed points f1 and f2 in a plane, an ellipse is the set of all points (x, y) where the distance from f1 to (x, y) added to the distance from f2 to (x, y) remains constant.

y P(x, y) d1

d1 d2 k The fixed points f1 and f2 are called the foci of the ellipse, and the points P(x, y) are on the graph of the ellipse.

f1

d2 f2

x

d1 d2 k

6 in. 3 in.

5 in.

To find the equation of an ellipse in terms of a and b we combine the definition just given with the distance formula. Consider the ellipse shown in Figure 10.17 (for

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Figure 10.17 y (0, b) P(x, y) (a, 0) x

(a, 0) (c, 0)

(c, 0)

calculating ease we use a central ellipse). Note the vertices have coordinates 1a, 02 and (a, 0), and the endpoints of the minor axis have coordinates 10, b2 and (0, b) as before. It is customary to assign foci the coordinates f1 S 1c, 02 and f2 S 1c, 02. We can calculate the distance between (c, 0) and any point P(x, y) on the ellipse using the distance formula: 21x c2 2 1y 02 2

Likewise the distance between 1c, 02 and any point (x, y) is 21x c2 2 1y 02 2

(0, b)

According to the definition, the sum must be constant: 21x c2 2 y2 21x c2 2 y2 k EXAMPLE 6

䊳

Finding the Value of k from the Definition of an Ellipse Use the definition of an ellipse and the diagram given to determine the constant k used for this ellipse (also see the following Worthy of Note). Note that a 5, b 3, and c 4. y (0, 3)

P(3, 2.4)

(5, 0) (4, 0)

(4, 0)

(5, 0) x

(0, 3)

Solution

䊳

21x c2 2 1y 02 2 21x c2 2 1y 02 2 k

213 42 12.4 02 213 42 12.4 02 k 2

WORTHY OF NOTE Note that if the foci are coincident (both at the origin) the “ellipse” will k actually be a circle with radius ; 2 2x2 y2 2x2 y2 k leads to k2 x2 y2 . In Example 6 we 4 10 5, and if found k 10, giving 2 we used the “string” to draw the circle, the pencil would be 5 units from the center, creating a circle of radius 5.

2

2

2

2112 2.4 27 2.4 k 16.76 154.76 k 2.6 7.4 k 10 k The constant value for this ellipse is 10 units. 2

2

2

2

given substitute add simplify radicals compute square roots result

Now try Exercises 45 through 48 In Example 6, the sum of the distances could also be found by moving the point (x, y) to the location of a vertex (a, 0), then using the symmetry of the ellipse. The sum is identical to the length of the major axis, since the overlapping part of the string from (c, 0) to (a, 0) is the same length as from (a, 0) to (c, 0) (see Figure 10.18). This shows the constant k is equal to 2a regardless of the distance between foci. As we noted, the result is

䊳

Figure 10.18 y d1 d2 2a d1

d2

(a, 0) (c, 0)

21x c2 2 y2 21x c2 2 y2 2a

(c, 0)

(a, 0) x

These two segments are equal substitute 2a for k

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10–16


The details for simplifying this expression are given in Appendix E, and the result is very close to the standard form seen previously: y2 x2 1 a2 a2 c2 y2 y2 x2 x2 1 1, we might with a2 b2 a2 a2 c2 suspect that b2 a2 c2, and this is indeed the case. Note from Example 6 the relationship yields By comparing the standard form

b2 a2 c2 32 52 42 9 25 16 Additionally, when we consider that (0, b) is Figure 10.19 a point on the ellipse, the distance from (0, b) to y (c, 0) must be equal to a due to symmetry (the (0, b) “constant distance” used to form the ellipse is always 2a). We then see in Figure 10.19, that a a b2 c2 a2 (Pythagorean Theorem), yielding b (a, 0) (a, 0) 2 2 2 b a c as above. x (c, 0) (c, 0) With this development, we now have the ability to locate the foci of any ellipse—an important step toward using the ellipse in practical (0, b) applications. Because we’re often asked to find the location of the foci, it’s best to rewrite the relationship in terms of c2, using absolute value bars to allow for a major axis that is vertical: c2 冟a2 b2冟. EXAMPLE 7

䊳

Completing the Square to Graph an Ellipse and Locate the Foci For the ellipse defined by 25x2 9y2 100x 54y 44 0, find the coordinates of the center, vertices, foci, and endpoints of the minor axis. Then sketch the graph.

Solution

䊳

25x2 9y2 100x 54y 44 0 25x2 100x 9y2 54y 44 2 251x 4x __ 2 91y2 6y __ 2 44 251x2 4x 42 91y2 6y 92 44 100 81 c

c

adds 25142 100

c

c adds 9192 81

251x 22 2 91y 32 2 91y 32 2 251x 22 2 225 225 1x 22 2 1y 32 2 9 25 2 1y 32 2 1x 22 32 52

given group terms; add 44 factor out lead coefficients

add 100 81 to right-hand side

225 225 225

factored form divide by 225

1

simplify (standard form)

1


The result shows a vertical ellipse with a 3 and b 5. The center of the ellipse is at (2, 3). The vertices are a vertical distance of b 5 units from center at (2, 8) and (2, 2). The endpoints of the minor axis are a horizontal distance of a 3 units from center at (1, 3) and (5, 3). To locate the foci, we use the foci formula

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for an ellipse: c2 冟a2 b2冟, giving c2 冟32 52冟 16. This shows the foci “” are located a vertical distance of 4 units from center at (2, 7) and (2, 1). y (2, 8)

Vertical ellipse Center at (2, 3)

(2, 7)

(1, 3)

(2, 3)

(2, 1) (2, 2)

Endpoints of major axis (vertices) (2, 8) and (2, 2) (5, 3)

x

Endpoints of minor axis (1, 3) and (5, 3) Location of foci (2, 7) and (2, 1) Length of major axis: 2b 2(5) 10 Length of minor axis: 2a 2(3) 6


䊳

For an ellipse, a focal chord is a line segment perpendicular to the major axis, through a focus and with endpoints on the ellipse. In the Exercise Set, you are asked 2m2 to verify that the focal chord of an ellipse has length L , where m is the length n of the semiminor axis and n is the length of the semimajor axis. This means the m2 distance from the foci to the graph (along a focal chord) is , a fact that can often be used n to help graph an ellipse. For Example 7, m 3 and n 5, so the horizontal distance Figure 10.20 from focus to graph (in either direction) is 9.2 9 32 . From the upper focus (2, 7), we can 5 5 now graph the additional points 12 1.8, 72 11.4 10.2, 72 and (2 1.8, 7) 13.8, 72, and 7.4 from the lower focus 12, 12 we obtain 10.2, 12 and 13.8, 12 without having to evaluate the original equation. Graphical ver3.2 ification is provided in Figure 10.20. Also see Exercises 83 and 85. For future reference, remember the foci of an ellipse always occur on the major axis, with a 7 c and a2 7 c2 for a horizontal ellipse, and b 7 c and b2 7 c2 for a vertical ellipse. This makes it easier to remember the foci formula for ellipses: c2 冟a2 b2冟. If any two of the values for a, b, and c are known, the relationship between them can be used to construct the equation of the ellipse.

EXAMPLE 8

䊳

Finding the Equation of an Ellipse Find the equation of the ellipse (in standard form) that has foci at (0, 2) and (0, 2), with a minor axis 6 units in length. Then graph the ellipse a. By hand. b. On a graphing calculator. m2 c. Find the distance from foci to graph along a focal chord ausing b, and use n the result to verify that the endpoints of both focal chords are all on the graph.

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Solution

䊳

LOOKING AHEAD

Since the foci are on the y-axis and an equal distance from (0, 0), we know this is a vertical and central ellipse with c 2 and c2 4. The minor axis has a length of 2a 6 units, meaning a 3 and a2 9. To find b2, use the foci equation and solve. Figure 10.21 foci equation (ellipse) c2 冟a2 b2冟 4 冟9 b2冟 4 9 b2 4 9 b2 b2 13 b2 5

For the hyperbola, we’ll find that c 7 a, and the formula for the foci of a hyperbola will be c2 a2 b2.

y

substitute

(0, √13)

solve the absolute value equation result

2

Since we know b must be greater than a2 (the major axis is always longer), b2 5 can be discarded. The

(0, 2) (3, 0)

2

y x2 1. 2 3 1 2132 2 a. The graph is shown in Figure 10.21. b. For a calculator generated graph, begin by solving for y.

(3, 0) x

(0, 2)

standard form is

(0, √13)

y2 x2 original equation 1 9 13 13x2 9y2 117 clear denominators isolate y-term 9y2 117 13x2 2 117 13x y2 divide by 9 9.4 9 117 13x2 y take square roots B 9 117 13X2 117 13X2 Y1 , Y2 B 9 B 9 The graph is shown in Figure 10.22. c. From the discussion prior to Example 8, the horizontal distance from foci to graph m2 9 must be . Using the TRACE feature and entering X 9/ 213 verifies n 213

Figure 10.22 6.2

9.4

6.2

Figure 10.23 6.2

9.4

9.4

6.2

that a

9

, 2b is a point on the graph (Figure 10.23), and that a

9

, 2b, 213 213 9 9 a , 2b, and a , 2b must also be on the graph due to symmetry. 213 213 Now try Exercises 55 through 62

C. You’ve just seen how we can locate the foci of an ellipse and use the foci and other features to write the equation

EXAMPLE 9

䊳

D. Applications Involving Foci Applications involving the foci of a conic section can take various forms. In many cases, only partial information about the conic section is available and the ideas from Example 8 must be used to “fill in the gaps.” In other applications, we must rewrite a known or given equation to find information related to the values of a, b, and c. 䊳

Solving Applications Using the Characteristics of an Ellipse In Washington, D.C., there is a park called the Ellipse located between the White House and the Washington Monument. The park is surrounded by a path that forms an ellipse with the length of the major axis being about 1502 ft and the minor axis having a length of 1280 ft. Suppose the park manager wants to install water fountains at the location of the foci. Find the distance between the fountains rounded to the nearest foot.

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Solution

䊳

979

Since the major axis has length 2a 1502, we know a 751 and a2 564,001. The minor axis has length 2b 1280, meaning b 640 and b2 409,600. To find c, use the foci equation: c2 a2 b2 564,001 409,600 154,401 c ⬇ 393 and c ⬇ 393

since we know a 7 b substitute subtract square root property

The distance between the water fountains would be 213932 786 ft.

D. You’ve just seen how we can solve applications involving the foci


䊳

10.2 EXERCISES 䊳



1. For an ellipse, the relationship between a, b, and c is given by the foci equation , since c 6 a or c 6 b.

2. The greatest distance across an ellipse is called the and the endpoints are called .

3. For a vertical ellipse, the length of the minor axis is and the length of the major axis is .

4. To write the equation 2x2 y2 6x 7 in standard form, the in x.

5. Explain/Discuss how the relations a 7 b, a b and a 6 b affect the graph of a conic section with

6. Suppose foci are located at (3, 2) and (5, 2). Discuss/Explain the conditions necessary for the graph to be an ellipse.

equation 䊳

1x h2 2 a2

1y k2 2 b2

1.


Find an equation of the circle satisfying the conditions given, then graph the result on a graphing calculator and locate two additional points on the graph.

15. x2 y2 4x 10y 4 0 16. x2 y2 4x 6y 3 0


17. x2 y2 6x 5 0


18. x2 y2 8y 5 0


Sketch the graph of each ellipse.

19.

1x 12 2

20.

1x 32 2

21.

1x 22 2

22.

1x 52 2

10. center (0, 4), radius 15 11. diameter has endpoints (4, 9) and (2, 1) 12. diameter has endpoints (2, 32 , and (3, 9) Write each equation in standard form to identify the center and radius of the circle, then sketch its graph.

13. x2 y2 12x 10y 52 0 14. x2 y2 8x 6y 11 0

9 4

25 1

1y 22 2

1y 12 2

1y 32 2

1y 22 2

16 25 4

16

1 1 1 1

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23.

1x 12 2

24.

1x 12

16

1y 22 2

1y 32

2

36

47.

1

9

48.

(0, b) y (0, 8)

(0, b) y

(4.8, 6) (0, 28)

2

1

9

(6, 0) x

(6, 0)

(96, 0) x

(96, 0) (0, 28 )

For each exercise, (a) write the equation in standard form, then identify the center and the values of a and b, (b) state the coordinates of the vertices and the coordinates of the endpoints of the minor axis, (c) sketch the graph, and (d) for 25–28 (only) graph the relations on a graphing calculator and identify four additional points on the graph whose coordinates are rational.

(0, 8)

(76.8, 60)

(0, b)

(0, b)

Find the coordinates of the (a) center, (b) vertices, (c) foci, and (d) endpoints of the minor axis. Then (e) sketch the graph.

49. 4x2 25y2 16x 50y 59 0

25. x2 4y2 16

26. 9x2 y2 36

50. 9x2 16y2 54x 64y 1 0

27. 16x2 9y2 144

28. 25x2 9y2 225

51. 25x2 16y2 200x 96y 144 0

29. 2x2 5y2 10

30. 3x2 7y2 21

52. 49x2 4y2 196x 40y 100 0 53. 6x2 24x 9y2 36y 6 0

Identify each equation as that of an ellipse or circle, then sketch its graph.

54. 5x2 50x 2y2 12y 93 0

31. 1x 12 41y 22 16 2

2

32. 91x 22 2 1y 32 2 36

Find the equation of the ellipse (in standard form) that satisfies the following conditions. Then (a) graph the ellipse by hand, (b) confirm your graph by graphing the ellipse on a graphing calculator, and (c) find the length of the focal chords and verify the endpoints of the chords are on the graph.

33. 21x 22 2 21y 42 2 18 34. 1x 62 2 y2 49

35. 41x 12 2 91y 42 2 36 36. 251x 32 2 41y 22 2 100 Complete the square in both x and y to write each equation in standard form. Then draw a complete graph of the relation and identify all important features, including the domain and range.

37. 4x2 y2 6y 5 0

55. vertices at (6, 0) and (6, 0); foci at (4, 0) and (4, 0) 56. vertices at (8, 0) and (8, 0); foci at (5, 0) and (5, 0) 57. foci at (3, 6) and (3, 2); length of minor axis: 6 units 58. foci at (4, 3) and (8, 3); length of minor axis: 8 units

38. x2 3y2 8x 7 0 39. x2 4y2 8y 4x 8 0

Use the characteristics of an ellipse and the graph given to write the related equation and find the location of the foci.

40. 3x2 y2 8y 12x 8 0 41. 5x2 2y2 20y 30x 75 0

59.

60.

y

y

42. 4x 9y 16x 18y 11 0 2

2

43. 2x2 5y2 12x 20y 12 0 44. 6x2 3y2 24x 18y 3 0

x

x

Use the definition of an ellipse to find the constant k for each ellipse (figures are not drawn to scale).

45.

46.

y (0, 8) (6, 6.4) (a, 0) (6, 0)

(0, 8)

61.

y

62.

y

y

(0, 12)

(6, 0)

(9, 9.6)

(a, 0) (a, 0) x

(9, 0)

(0, 12)

(9, 0)

(a, 0) x

x

x

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63. Area of an Ellipse: A ⴝ ␲ab The area of an ellipse is given by the formula shown, where a is the distance from the center to the graph in the horizontal direction and b is the distance from center to graph in the vertical direction. Find the area of the ellipse defined by 16x2 9y2 144. 䊳

981

a2 ⴙ b2 B 2 The perimeter of an ellipse can be approximated by the formula shown, where a represents the length of the semimajor axis and b represents the length of the semiminor axis. Find the perimeter of the y2 x2 1. ellipse defined by the equation 49 4

64. The Perimeter of an Ellipse: P ⴝ 2␲

APPLICATIONS

65. Decorative fireplaces: A bricklayer intends to build an elliptical fireplace 3 ft high and 8 ft wide, with two glass doors that open at the middle. The hinges to these doors are to be screwed onto a spine that is perpendicular to the hearth and goes through the foci of the ellipse. How far from center will the spines be located? How tall will each spine be? 8 ft

68. Medical procedures: The medical procedure called lithotripsy is a noninvasive medical procedure that is used to break up kidney and bladder stones in the body. A machine called a lithotripter uses its three-dimensional semielliptical shape and the foci properties of an ellipse to concentrate shock waves generated at one focus, on a kidney stone located at the other focus (see diagram—not drawn to scale). If the lithotripter has a length (semimajor axis) of 16 cm and a radius (semiminor axis) of 10 cm, how far from the vertex should a kidney stone be located for the best result? Round to the nearest hundredth.

3 ft

Exercise 68 Vertex Spines

Focus Lithotripter

66. Decorative gardens: A retired math teacher decides to present her husband with a beautiful elliptical garden to help celebrate their 50th anniversary. The ellipse is to be 8 m long and 5 m across, with decorative fountains located at the foci. How far from the center of the ellipse should the fountains be located (round to the nearest 100th of a meter)? How far apart are the fountains? 67. Attracting attention to art: As part of an art show, a gallery owner asks a student from the local university to design a unique exhibit that will highlight one of the more significant pieces in the collection, an ancient sculpture. The student decides to create an elliptical showroom with reflective walls, with a rotating laser light on a stand at one focus, and the sculpture placed at the other focus on a stand of equal height. The laser light then points continually at the sculpture as it rotates. If the elliptical room is 24 ft long and 16 ft wide, how far from the center of the ellipse should the stands be located (round to the nearest 10th of a foot)? How far apart are the stands?

Exercise 69 69. Elliptical arches: In some situations, bridges are built using uniform elliptical 8 ft archways as shown in the 60 ft figure given. Find the equation of the ellipse forming each arch if it has a total width of 30 ft and a maximum center height (above level ground) of 8 ft. What is the height of a point 9 ft to the right of the center of each arch?

70. Elliptical arches: An elliptical arch bridge is built across a one-lane highway. The arch is 20 ft across and has a maximum center height of 12 ft. Will a farm truck hauling a load 10 ft wide with a clearance height of 11 ft be able to go under the bridge without damage? (Hint: See Exercise 69.)

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71. Plumbing: By allowing the free flow of air, a properly vented home enables water to run freely throughout its plumbing system, while helping to prevent sewage gases from entering the home. Find the equation of the elliptical hole cut in a roof in order to allow a 3-in. vent pipe to exit, if the roof 4 has a slope of . 3 72. Light projection: Standing a short distance from a wall, Kymani’s flashlight projects a circle of radius 30 cm. When holding the flashlight at an angle, a vertical ellipse 50 cm long is formed, with the focus 10 cm from the vertex (see Worthy of Note, page 971). Find the equation of the circle and ellipse, and the area of the wall that each illuminates. As a planet orbits around the Sun, it traces out an ellipse. If the center of the ellipse were placed at (0, 0) on a coordinate grid, the Sun would be actually offcentered (located at the focus of the ellipse). Use this information and the graphs provided to complete Exercises 73 through 78.

y

Sun x

70.5 million miles

Mercury

72 million miles

Exercise 74 y Pluto

Sun x

3650 million miles

3540 million miles

74. Orbit of Pluto: The approximate orbit of the Kuiper object formerly known as Pluto is shown in the figure given. Find an equation that models this orbit.

75. Planetary orbits: Except for small variations, a planet’s orbit around the Sun is elliptical with the Sun at one focus. The aphelion (maximum distance from the Sun) of the planet Mars is approximately 156 million miles, while the perihelion (minimum distance from the Sun) of Mars is about 128 million miles. Use this information to find the lengths of the semimajor and semiminor axes, rounded to the nearest million. If Mars has an orbital velocity of 54,000 miles per hour (1.296 million miles per day), how many days does it take Mars to orbit the Sun? (Hint: Use the formula from Exercise 64.) 76. Planetary orbits: The aphelion (maximum distance from the Sun) of the planet Saturn is approximately 940 million miles, while the perihelion (minimum distance from the Sun) of Saturn is about 840 million miles. Use this information to find the lengths of the semimajor and semiminor axes, rounded to the nearest million. If Saturn has an orbital velocity of 21,650 miles per hour (about 0.52 million miles per day), how many days does it take Saturn to orbit the Sun? How many years? 77. Orbital velocity of Earth: The planet Earth has a perihelion (minimum distance from the Sun) of about 91 million mi, an aphelion (maximum distance from the Sun) of close to 95 million mi, and completes one orbit in about 365 days. Use this information and the formula from Exercise 64 to find Earth’s orbital speed around the Sun in miles per hour. 78. Orbital velocity of Jupiter: The planet Jupiter has a perihelion of 460 million mi, an aphelion of 508 million mi, and completes one orbit in about 4329 days. Use this information and the formula from Exercise 64 to find Jupiter’s orbital speed around the Sun in miles per hour.

Exercise 73

73. Orbit of Mercury: The approximate orbit of the planet Mercury is shown in the figure given. Find an equation that models this orbit.

10–22

79. Area of a race track: Suppose the Toronado 500 is a car race that is run on an elliptical track. The track is bounded by two ellipses with equations of 4x2 9y2 900 and 9x2 25y2 900, where x and y are in hundreds of yards. Use the formula given in Exercise 63 to find the area of the race track. Exercise 80 80. Area of a border: The tablecloth for a large oval table is elliptical in shape. It is designed with two concentric ellipses (one within the other) as shown in the figure. The equation of the outer ellipse is 9x2 25y2 225, and the equation of the inner ellipse is 4x2 16y2 64 with x and y in feet. Use the formula given in Exercise 63 to find the area of the border of the tablecloth.

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81. Whispering galleries: Due to their unique properties, ellipses are used in the construction of whispering galleries like those in St. Paul’s Cathedral (London) and Statuary Hall in the U.S. Capitol. Regarding the latter, it is known that John Quincy Adams (1767–1848), while a member of the House of Representatives, situated his desk at a focal point of the elliptical ceiling, easily eavesdropping on the private conversations of other House members located near the other focal point. Suppose a whispering gallery was built using the equation y2 x2 1, with the dimensions in feet. (a) How tall is the ceiling at its highest point? (b) How wide is 2809 2025 the gallery vertex to vertex? (c) How far from the base of the doors at either end, should a young couple stand so that one can clearly hear the other whispering, “I love you.”? 82. Elliptical billiards: While an elliptical billiard table has little practical value, it offers an excellent illustration of elliptical properties. A ball placed at one focus and hit with the cue stick from any angle, will hit the cushion and immediately rebound to the other focus and continue through each focus until coming to rest. Suppose one such table was constructed using the y2 x2 1 as a model, with the dimensions in feet. equation 9 4 (a) How far apart are the vertices? (b) How far apart are the foci? As a side note, Lewis Carroll (1832–1898) did invent a game of circular billiards, complete with rules.

䊳


83. For 6x2 36x 3y2 24y 74 28, does the equation appear to be that of a circle, ellipse, or parabola? Write the equation in factored form. What do you notice? What can you say about the graph of this equation?

䊳

84. Algebraically verify that for the ellipse y2 x2 1 with b 7 a, the length of the focal a2 b2 2a2 chord is still . b


85. (5.4) Evaluate the expression using the change-ofbase formula: log320. 86. (2.6) The resistance R to current flow in an electrical wire varies directly as the length L of the wire and inversely as the square of its diameter d. (a) Write the equation of variation; (b) find the constant of variation if a wire 2 m long with diameter d 0.005 m has a resistance of 240 ohms (); and (c) find the resistance in a similar wire 3 m long and 0.006 m in diameter.

87. (8.6) Use De Moivre’s theorem to compute the value of z 11 i 132 6. 88. (8.3) Find the true direction and groundspeed of the airplane shown, given the direction and speed of the wind (indicated in blue).

Exercise 88 N

250 mph heading 20

30 mph heading 90

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The Hyperbola


A. Use the equation of a hyperbola to graph central and noncentral hyperbolas B. Distinguish between the equations of circles, ellipses, and hyperbolas C. Locate the foci of a hyperbola and use the foci and other features to write its equation D. Solve applications involving foci

EXAMPLE 1

䊳

As seen in Section 10.1 (see Figure 10.24), a hyperbola is a conic section formed by a plane that cuts both nappes of a right circular cone. A hyperbola Axis has two symmetric parts called branches, which open in opposite directions. Although the branches appear to resemble parabolas, we will soon discover they are actually a very different curve.

Figure 10.24

Hyperbola

A. The Equation of a Hyperbola In Section 10.2, we noted that for the equation Ax2 By2 F, if A B, the equation is that of a circle, if A B, the equation represents an ellipse. Both cases contain a sum of second-degree terms. Perhaps driven by curiosity, we might wonder what happens if the equation has a difference of second-degree terms. Consider the equation 9x2 16y2 144. It appears the graph will be centered at (0, 0) since no shifts are applied (h and k are both zero). Using the intercept method to graph this equation reveals an entirely new curve, called a hyperbola. Graphing a Central Hyperbola Graph the equation 9x2 16y2 144 using intercepts and additional points as needed.

Solution

9x2 16y2 144 9102 2 16y2 144 16y2 144 y2 9

䊳

given substitute 0 for x simplify divide by 16

Since y2 can never be negative, we conclude that the graph has no y-intercepts. Substituting y 0 to find the x-intercepts gives 9x2 16y2 144 9x2 16102 2 144 9x2 144 x2 16 x 116 and x 116 x 4 and x 4 (4, 0) and 14, 02

given substitute 0 for y simplify divide by 9 square root property simplify

x-intercepts

Knowing the graph has no y-intercepts, we select inputs greater than 4 and less than 4 to help sketch the graph. Using x 5 and x 5 yields 9x2 16y2 144 9152 2 16y2 144 91252 16y2 144 225 16y2 144 16y2 81 81 y2 16 9 9 y y 4 4 y 2.25 y 2.25 15, 2.252 15, 2.252 984

given substitute for x 5 152 2 25 2

simplify subtract 225 divide by 16

square root property decimal form ordered pairs

9x2 16y2 144 9152 2 16y2 144 91252 16y2 144 225 16y2 144 16y2 81 81 y2 16 9 9 y y 4 4 y 2.25 y 2.25 15, 2.252 15, 2.252 10–24

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Section 10.3 The Hyperbola

Plotting these points and connecting them with a smooth curve, while knowing there are no y-intercepts, produces the graph in the figure. The point at the origin (in blue) is not a part of the graph, and is given only to indicate the “center” of the hyperbola. The points 14, 02 and (4, 0) are called vertices, and the center of the hyperbola is always the point halfway between them.

y Hyperbola (5, 2.25) (5, 2.25) (4, 0)

(4, 0) (0, 0)

x

(5, 2.25) (5, 2.25)


䊳

As with the circle and ellipse, the hyperbola fails the vertical line test and we must graph the relation on a calculator by writing its equation in two parts, each of which is a function. For the hyperbola in Example 1, this gives 9x2 16y2 144 original equation 2 2 isolate y-term 16y 144 9x multiply by 1 16y2 9x2 144 9x2 144 y2 divide by 16 16 9x2 144 y take square roots B 16 9.4 2 2 9X 144 9X 144 , Y2 Y1 B 16 B 16

Figure 10.25 6.2

9.4

The graph is shown in Figure 10.25. 6.2 Since the hyperbola in Example 1 crosses a horizontal line of symmetry, it is referred to as a horizontal hyperbola. If the center is at the origin, we have a central hyperbola. The line passing through the center and both vertices is called the transverse axis (vertices are always on the transverse axis), and the line passing through the center and perpendicular to this axis is called the conjugate axis (see Figure 10.26). In Example 1, the coefficient of x2 was positive and we were subtracting 16y2: 2 9x 16y2 144. The result was a horizontal hyperbola. If the y2-term is positive and we subtract the term containing x2, the result is a vertical hyperbola (Figure 10.27). Figure 10.26

Figure 10.27

y Conjugate axis

Center Vertex

Transverse axis

Transverse axis Vertex

Horizontal hyperbola

x

y

Vertex Vertex Center

Vertical hyperbola

Conjugate axis x

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EXAMPLE 2

䊳

Identifying the Axes, Vertices, and Center of a Hyperbola from Its Graph For the hyperbola shown, state the location of the vertices and the equation of the transverse axis. Then identify the location of the center and the equation of the conjugate axis.

Solution

䊳

By inspection we locate the vertices at (0, 0) and (0, 4). The equation of the transverse axis is x 0. The center is halfway between the vertices at (0, 2), meaning the equation of the conjugate axis is y 2.

y

5

5

5

x

4


䊳

Standard Form As with the ellipse, the polynomial form of the equation is helpful for identifying hyperbolas, but not very helpful when it comes to graphing a hyperbola (since we still must go through the laborious process of finding additional points). For graphing, standard form is once again preferred. Consider the hyperbola 9x2 16y2 144 from Example 1. To write the equation in standard form, we divide by 144 and obtain y2 x2 1. By comparing the standard form to the graph, we note a 4 represents 42 32 the distance from center to vertices, similar to the way we used a previously. But since the graph has no y-intercepts, what could b 3 represent? The answer lies in the fact that branches of a hyperbola are asymptotic, meaning they will approach and become very close to imaginary lines that can be used to sketch the graph. For b a central hyperbola, the slopes of the asymptotic lines are given by the ratios and a b b b , with the related equations being y x and y x. The graph from Example 1 a a a is repeated in Figure 10.28, with the asymptotes drawn. For a clearer understanding of how the equations for the asymptotes were determined, see Exercise 87. A second method of drawing the asymptotes involves drawing a central rectangle with dimensions 2a by 2b, as shown in Figure 10.29. The asymptotes will be the extended diagonals of this rectangle. This brings us to the equation of a hyperbola in standard form. Figure 10.28

Figure 10.29

y Slope m

y

34

rise b3 (4, 0)

Slope m

3 4

run a4 (4, 0) (0, 0)

(4, 0)

2b

x

2a

Slope method

Central rectangle method

x

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The Equation of a Hyperbola in Standard Form The equation 1x h2 a2

2

1y k2 b2

The equation 1y k2 2

2

1

b2

represents a horizontal hyperbola with center (h, k) • transverse axis y k • conjugate axis x h • 冟a冟 gives the distance from center to vertices.

1x h2 2 a2

1

represents a vertical hyperbola with center (h, k) • transverse axis x h • conjugate axis y k • 冟b冟 gives the distance from center to vertices.

b • Asymptotes can be drawn by starting at (h, k) and using slopes m . a

EXAMPLE 3

䊳

Graphing a Hyperbola Using Its Equation in Standard Form Sketch the graph of 161x 22 2 91y 12 2 144, and label the center, vertices, and asymptotes.

Solution

䊳

Begin by noting a difference of the second-degree terms, with the x2-term occurring first. This means we’ll be graphing a horizontal hyperbola whose center is at (2, 1). Continue by writing the equation in standard form. 161x 22 2 91y 12 2 91y 12 2 161x 22 2 144 144 1y 12 2 1x 22 2 9 16 1x 22 2 1y 12 2 32 42

144

144 144

given equation divide by 144

1

simplify

1


Since a 3 the vertices are a horizontal distance of 3 units from the center (2, 1), giving 12 3, 12 S 15, 12 and 12 3, 12 S 11, 12 . After plotting the center and b 4 vertices, we can begin at the center and count off slopes of m , or a 3 draw a rectangle centered at (2, 1) with dimensions 2132 6 (horizontal dimension) by 2142 8 (vertical dimension) to sketch the asymptotes. The complete graph is shown here. Horizontal hyperbola

y

Center at (2, 1)

md

Vertices at (1, 1) and (5, 1) Transverse axis: y 1 Conjugate axis: x 2

(2, 1) (1, 1)

(5, 1) x

m d

冢

Width of rectangle horizontal dimension and distance between vertices 2a 2(3) 6

冣

Length of rectangle (vertical dimension) 2b 2(4) 8


䊳

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If the hyperbola in Example 3 were a central hyperbola, the equations of the 4 4 asymptotes would be y x and y x. But the center of this graph has been 3 3 shifted 2 units right and 1 unit up. Using our knowledge of shifts and translations, the equations for the asymptotes of the shifted hyperbola must be 4 4 5 1. 1y 12 1x 22 , or y x in simplified form, and 3 3 3 4 11 4 2. 1y 12 1x 22 or y x . 3 3 3 Using Y1

161X 22 2 144

161X 22 2 144

1, and Y2 1 B 9 B 9 (obtained by solving for y in the original equation), a calculator generated graph of the hyperbola and its asymptotes is shown here (Figures 10.30 and 10.31). Figure 10.31 Figure 10.30

9.3

7.4

11.4

9.3

Polynomial Form If the equation is given as a polynomial in expanded form, complete the square in x and y, then write the equation in standard form.

EXAMPLE 4

䊳

Graphing a Hyperbola by Completing the Square Graph the equation 9y2 x2 54y 4x 68 0 by completing the square. Label the center and vertices and sketch the asymptotes. Then graph the hyperbola on a graphing calculator and use the TRACE feature with a “friendly” window to locate four additional points whose coordinates are rational.

Solution

䊳

Since the y2-term occurs first, we assume the equation represents a vertical hyperbola, but wait for the factored form to be sure (see Exercise 91). 9y2 x2 54y 4x 68 0 9y2 54y x2 4x 68 2 91y 6y ___ 2 11x2 4x ___ 2 68 91y2 6y 92 11x2 4x 42 68 81 142 c

c

adds 9 19 2 81

c

c

adds 1 14 2 4

91y 32 2 11x 22 2 9 1x 22 2 1y 32 2 1 1 9 1y 32 2 1x 22 2 1 12 32

add 81 14 2 to right

given collect like-variable terms; subtract 68 factor out 9 from y-terms and 1 from x-terms complete the square

factor S vertical hyperbola divide by 9 (standard form)


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The center of the hyperbola is 12, 32 with a 3, b 1, and a transverse axis of x 2. The vertices are at 12, 3 12 and 12, 3 12 S 12, 22 and 12, 42 . After plotting the center and vertices, we draw a rectangle centered at 12, 32 with a horizontal “width” of 2132 6 and a vertical “length” of 2112 2 to sketch the asymptotes. The completed graph is given in Figure 10.32. Figure 10.32 Vertical hyperbola

y

Center at (2, 3) Vertices at (2, 2) and (2, 4)

m 3 1

Transverse axis: x 2 Conjugate axis: y 3

x 1 m3 (2, 3) center

(2, 2)

Width of rectangle (horizontal dimension) 2a 2(3) 6

(2, 4)

冢

Length of rectangle vertical dimension and distance between vertices 2b 2(1) 2

冣

To graph the hyperbola on a calculator, we again solve for y.

91y 32 2 1x 22 2 9

factored form

91y 32 9 1x 22 isolate term containing y 2 9 1x 22 1y 32 2 divide by 9 9 9 1x 22 2 y3 take square roots B 9 9 1x 22 2 y subtract 3 3 B 9 9 1X 22 2 9 1X 22 2 Y1 3 3, Y2 Figure 10.33 B B 9 9 2

2

Using the “friendly” window shown, we can use the arrow keys to TRACE though x-values and find the points 15.2, 0.42 and 19.2, 0.42 on the upper branch, with 15.2, 5.62 and 19.2, 5.62 on the lower branch. Note how these points show that a hyperbola is symmetric to its center, as well as the horizontal line and vertical line through its center. The graph is shown in Figure 10.33.

3.2

7.4

11.4

9.2

Now try Exercises 45 through 48 A. You’ve just seen how we can use the equation of a hyperbola to graph central and noncentral hyperbolas

䊳

B. Distinguishing Between the Equations of Circles, Ellipses, and Hyperbolas So far we’ve explored numerous graphs of circles, ellipses, and hyperbolas. In Example 5 we’ll attempt to identify a given conic section from its equation alone (without graphing the equation). As you’ve seen, the corresponding equations have unique characteristics that can help distinguish one from the other.

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EXAMPLE 5

䊳

Identifying a Conic Section from Its Equation Identify each equation as that of a circle, ellipse, or hyperbola. Justify your choice and name the center, but do not draw the graphs. a. y2 36 9x2 b. 4x2 16 4y2 c. x2 225 25y2 d. 25x2 100 4y2 2 2 e. 31x 22 41y 32 12 f. 41x 52 2 36 91y 42 2

Solution

B. You’ve just seen how we can distinguish between the equations of circles, ellipses, and hyperbolas

䊳

a. Writing the equation as y2 9x2 36 shows h 0 and k 0. Since the equation contains a difference of second-degree terms, it is the equation of a (vertical) hyperbola (A and B have opposite signs). The center is at (0, 0). b. Rewriting the equation as 4x2 4y2 16 and dividing by 4 gives x2 y2 4. The equation represents a circle of radius 2 1A B2 , with the center at (0, 0). c. Writing the equation as x2 25y2 225 we note a sum of second-degree terms with unequal coefficients. The equation is that of an ellipse 1A B2 , with the center at (0, 0). d. Rewriting the equation as 25x2 4y2 100 we note the equation contains a difference of second-degree terms. The equation represents a central (horizontal) hyperbola (A and B have opposite signs), whose center is at (0, 0). e. The equation is in factored form and contains a sum of second-degree terms with unequal coefficients. This is the equation of an ellipse 1A B2 with the center at 12, 32 . f. Rewriting the equation as 41x 52 2 91y 42 2 36 we note a difference of second-degree terms. The equation represents a horizontal hyperbola (A and B have opposite signs) with center 15, 42. Now try Exercises 49 through 60

䊳

C. The Foci of a Hyperbola Like the ellipse, the foci of a hyperbola play an important part in their application. A long distance radio navigation system (called LORAN for short), can be used to determine the location of ships and airplanes and is based on the characteristics of a hyperbola (see Exercises 85 and 86). Hyperbolic mirrors are also used in some telescopes, and have the property that a beam of light directed at one focus will be reflected to the second focus. To understand and appreciate these applications, we use the analytic definition of a hyperbola: Definition of a Hyperbola Given two fixed points f1 and f2 in a plane, a hyperbola is the set of all points (x, y) such that the distance d1 from f1 to (x, y) and the distance d2 from f2 to (x, y), satisfy the equation 冟d1 d2冟 k. In other words, the difference of these two distances is a positive constant. The fixed points f1 and f2 are called the foci of the hyperbola, and all such points (x, y) are on the graph of the hyperbola.

y d1

(x, y) d2

f1

f2

|d1 d2| k k>0

x

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Figure 10.34 y

(x, y)

(c, 0) x

(c, 0) (a, 0)

(a, 0)

EXAMPLE 6

䊳

991

As with the analytic definition of the ellipse, it can be shown that the constant k is again equal to 2a (for horizontal hyperbolas). To find the equation of a hyperbola in terms of a and b, we use an approach similar to that of the ellipse (see Appendix E), y2 x2 and the result is identical to that seen earlier: 2 2 1 where b2 c2 a2 a b (see Figure 10.34). We now have the ability to find the foci of any hyperbola —and can use this information in many significant applications. Since the location of the foci play such an important role, it is best to remember the relationship as c2 a2 b2 (called the foci formula for hyperbolas), noting that for a hyperbola, c 7 a and c2 7 a2 (also c 7 b and c2 7 b2). Be sure to note that for ellipses, the foci formula is c2 冟a2 b2冟 since a 7 c (horizontal ellipses) or b 7 c (vertical ellipses). Graphing a Hyperbola and Identifying Its Foci by Completing the Square For the hyperbola defined by 7x2 9y2 14x 72y 200 0, find the coordinates of the center, vertices, foci, and the dimensions of the central rectangle. Then sketch the graph, including the asymptotes.

Solution

䊳

given 7x2 9y2 14x 72y 200 0 2 2 group terms; add 200 7x 14x 9y 72y 200 factor out leading coefficients 71x2 2x ____ 2 91y2 8y ____ 2 200 2 2 71x 2x 12 91y 8y 162 200 7 11442 complete the square c

c

c

S add 7 11442 to right-hand side

c

adds 9 116 2 144

adds 7112 7

71x 12 2 91y 42 2 63 1y 42 2 1x 12 2 1 9 7 1x 12 2 1y 42 2 1 32 1 172 2

factored form divide by 63 and simplify


This is a horizontal hyperbola with a 3 1a2 92 and b 17 1b2 72. The center is at (1, 4), with vertices 12, 42 and (4, 4). Using the foci formula c2 a2 b2 yields c2 9 7 16, showing the foci are 13, 42 and (5, 4) (4 units from center). The central rectangle is 2132 6 by 2 17 ⬇ 5.29. Drawing the rectangle and sketching the asymptotes results in the graph shown. Horizontal hyperbola

y 10

Center at (1, 4) Vertices at (2, 4) and (4, 4) (1, 4)

(3, 4) 10

(2, 4)

10

(4, 4)

Transverse axis: y 4 Conjugate axis: x 1 Location of foci: (3, 4) and (5, 4)

(5, 4) 10

x

冢

Width of rectangle horizontal dimension and distance between vertices 2a 2(3) 6

冣

Length of rectangle (vertical dimension) 2b 2(√7) ≈ 5.29


䊳

A focal chord for a horizontal hyperbola is a vertical line segment through a focus with endpoints on the hyperbola. Similar to a focal chord of an ellipse, we can use its length to find additional points on the graph of the hyperbola. The total length 2b2 is once again L (for a horizontal hyperbola), meaning the distance from the foci a

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Figure 10.35 b2 . 10.2 a 2 For Example 6, a 3 and b 7, so the vertical distance from focus to graph (in either 7 direction) is 2.3. From the left focus 9.4 9.4 3 13, 42 , we can now graph the additional 13, 4 2.32 13, 6.32 , and points 13, 4 2.32 13, 1.62 . From the right 2.2 focus (5, 4), we obtain (5, 6.3) and (5, 1.6). Graphical verification is provided in Figure 10.35. Also see Exercise 80. As with the ellipse, if any two of the values for a, b, and c are known, the relationship between them can be used to construct the equation of the hyperbola. See Exercises 71 through 78.

to the graph (along the focal chord) is

C. You’ve just seen how we can locate the foci of a hyperbola and use the foci and other features to write its equation

D. Applications Involving Foci Applications involving the foci of a conic section can take many forms. As before, only partial information about the hyperbola may be available, and we’ll determine a solution by manipulating a given equation or constructing an equation from given facts. EXAMPLE 7

䊳

Applying the Properties of a Hyperbola—The Path of a Comet Comets with a high velocity cannot be captured by the Sun’s gravity, and are slung around the Sun in a hyperbolic path with the Sun at one focus. If the path illustrated by the graph shown is modeled by the equation 2116x2 400y2 846,400, how close did the comet get to the Sun? Assume units are in millions of miles and round to the nearest million.

Solution

䊳

y

(0, 0) x

We are essentially asked to find the distance between a vertex and focus. Begin by writing the equation in standard form: 2116x2 400y2 846,400 given y2 x2 1 divide by 846,400 400 2116 y2 write denominators in x2 1 squared form 2 2 20 46 This is a horizontal hyperbola with a 20 1a2 4002 and b 46 1b2 21162. Use the foci formula to find c2 and c. c2 a2 b2 c2 400 2116 c2 2516 c ⬇ 50 and c ⬇ 50 Since a 20 and 冟c冟 ⬇ 50, the comet came within about 50 20 30 million miles of the Sun. Now try Exercises 81 through 84

䊳

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EXAMPLE 8

䊳

Applying the Properties of a Hyperbola — The Location of a Storm Two amateur meteorologists, living 4 km apart (4000 m), see a storm approaching. The one farthest from the storm hears a loud clap of thunder 9 sec after the one nearest. Assuming the speed of sound is 340 m/sec, determine an equation that models possible locations for the storm at this time.

Solution

D. You’ve just seen how we can solve applications involving foci

䊳

Let M1 represent the meteorologist nearest the storm and M2 the farthest. Since M2 heard the thunder 9 sec after M1, M2 must be 9 # 340 3060 m farther away from the storm S. In other words, from our definition of a hyperbola, we have 冟d1 d2冟 3060. The set of all points that satisfy this equation will be on the graph of a hyperbola, and we’ll use this fact to develop an equation model for possible locations of the storm. Let’s place the information on a coordinate grid. For convenience, we’ll use the straight line distance between M1 and M2 as the x-axis, with the origin an equal distance from each. With the constant difference equal to 3060, we have 2a 3060, a 1530 from M2 the definition of a hyperbola, giving 3 2 y2 x2 2 1. With c 2000 m 15302 b (the distance from the origin to M1 or M2), we find the value of b using

y

S

2 1

M1 1

1

2

x in 1000s

3

1 2

the equation c2 a2 b2: 20002 15302 b2 or b2 120002 2 115302 2 1,659,100 ⬇ 12882. The equation that models possible y2 x2 ⬇ 1. locations of the storm is 15302 12882 Now try Exercises 85 and 86

10.3 EXERCISES 䊳



1. The line that passes through the vertices of a hyperbola is called the axis.

2. The center of a hyperbola is located between the vertices.

3. The conjugate axis is axis and contains the of the hyperbola.

4. The center of the hyperbola defined by

to the

1x 22 2 42

1y 32 2 52

1 is at

.

䊳

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5. Compare/Contrast the two methods used to find the asymptotes of a hyperbola. Include an example illustrating both methods.

䊳

10–34


6. Explore/Explain why A1x h2 2 B1y k2 2 F, 1A, B 7 02 results in a hyperbola regardless of whether A B or A B. Illustrate with an example.


Graph each hyperbola. Label the center, vertices, and any additional points used.

7.

y2 x2 1 9 4

8.

y2 x2 9. 1 4 9 11. 13. 15. 17.

12.

y2 x2 1 36 16

14.

y2 x2 1 9 1

16.

y2 x2 1 12 4

18.

2

19. 21.

2

y x 1 9 9 2

27.

1y 12 2

1x 22 2 y2 x2 1 28. 1 4 25 4 9 1x 32 2 1y 22 2 1 29. 36 49 30.

1x 22 2

1y 12 2

31.

1y 12 2

1x 52 2

y2 x2 1 1 4

32.

1y 32 2

1x 22 2

y2 x2 1 9 18

33. 1x 22 2 41y 12 2 16

y2 x2 1 16 9

y2 x2 10. 1 25 16

y2 x2 1 49 16

20.

2

y x 1 36 25

22.

Sketch a complete graph of each equation, including the asymptotes. Be sure to identify the center and vertices.

y2 x2 1 25 9 y2 x2 1 81 16

9 7

16

4 9 5

1 1 1

2

2

34. 91x 12 2 1y 32 2 81

2

2

36. 91y 42 2 51x 32 2 45

y x 1 4 4

35. 21y 32 2 51x 12 2 50

y x 1 16 4

37. 121x 42 2 51y 32 2 60 38. 81x 42 2 31y 32 2 24

For the graphs given, state the location of the vertices and the equation of the transverse axis. Then identify the location of the center and the equation of the conjugate axis. Note the scale used on each axis.

23.

24.

y 10

10

10

25.

10 x

26.

y

41. 9y2 4x2 36

42. 25y2 4x2 100

43. 12x2 9y2 72

44. 36x2 20y2 180

Graph each hyperbola by writing the equation in standard form. Label the center and vertices, and sketch the asymptotes. Then use a graphing calculator to graph each relation and locate four additional points whose coordinates are rational.

45. 4x2 y2 40x 4y 60 0

10

10

40. 16x2 25y2 400

y 10

10

10 x

39. 16x2 9y2 144

46. x2 4y2 12x 16y 16 0

y 10

47. 4y2 x2 24y 4x 28 0 48. 9x2 4y2 18x 24y 9 0

10

10 x

10

10

10 x

10

Classify each equation as that of a circle, ellipse, or hyperbola. Justify your response (assume all are nondegenerate).

49. 4x2 4y2 24 50. 9y2 4x2 36

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51. x2 y2 2x 4y 4

67. 4y2 16x2 24y 28 0

52. x2 y2 6y 7

68. 4y2 81x2 162x 405 0

53. 2x2 4y2 8

69. 9x2 3y2 54x 12y 33 0

54. 36x2 25y2 900

70. 10x2 60x 5y2 20y 20 0

55. x2 5 2y2

Find the equation of the hyperbola (in standard form) that satisfies the following conditions:

56. x y2 3x2 9 57. 2x2 2y2 x 20

71. vertices at (6, 0) and (6, 0); foci at (8, 0) and (8, 0)

58. 2y 3 6x 8 2

2

72. vertices at (4, 0) and (4, 0); foci at (6, 0) and (6, 0)

59. 16x2 5y2 3x 4y 538 60. 9x2 9y2 9x 12y 4 0 Use the definition of a hyperbola to find the distance between the vertices and the dimensions of the rectangle centered at (h, k). Figures are not drawn to scale. Note that Exercises 63 and 64 are vertical hyperbolas.

61.

62.

y

(5, 2.25) (a, 0) (5, 0)

(5, 0)

(15, 0)

64.

y

(0, b)

75.

(0, 13) (6, 7.5)

(9, 6.25)

x

y 5 4 3 2 1 54321 1 2 3 4 5

y

(0, 10) (0, b)

(a, 0) (15, 0) x

(a, 0)

(a, 0)

63.

74. foci at (5, 2) and (7, 2); length of conjugate axis: 8 units Use the characteristics of a hyperbola and the graph given to write the related equation and state the location of the foci (75 and 76) or the dimensions of the central rectangle (77 and 78).

y

(15, 6.75)

x

73. foci at 12, 3122 and 12, 3122; length of conjugate axis: 6 units

76.

1 2 3 4 5 x

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

(0, b) (0, b)

(0, 10) (0, 13)

Write each equation in standard form to find and list the coordinates of the (a) center, (b) vertices, (c) foci, and (d) dimensions of the central rectangle. Then (e) sketch the graph, including the asymptotes.

x

77.

y 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

78.

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

65. 4x2 9y2 24x 72y 144 0 66. 4x2 36y2 40x 144y 188 0 䊳

WORKING WITH FORMULAS 36 ⴚ 4x2 B ⴚ9 The “upper half ” of a certain hyperbola is given by the equation shown. (a) Simplify the radicand, (b) state the domain of the expression, and (c) enter the expression as Y1 on a graphing calculator and graph. What is the equation for the “lower half” of this hyperbola?

79. Equation of a semi-hyperbola: y ⴝ

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2m2 n The focal chords of a hyperbola are line segments parallel to the conjugate axis with 10 8 endpoints on the hyperbola, and containing points f1 and f2 (see grid). The length of the 6 chord is given by the formula shown, where n is the distance from center to vertex and m is 4 f 2 the distance from center to one side of the central rectangle. Use the formula to find the length of the focal chord for the hyperbola indicated, then compare the calculated value with 1086422 4 6 the length estimated from the given graph:

80. Focal chord of a hyperbola: L ⴝ

1

1x 22 2 4

䊳

1y 12 2 5

y

f2 2 4 6 8 10 x

8 10

1.

APPLICATIONS

81. Stunt pilots: At an air show, a stunt plane dives along a hyperbolic path whose vertex is directly over the grandstands. If the plane’s flight path can be modeled by the hyperbola 25y2 1600x2 40,000, what is the minimum altitude of the plane as it passes over the stands? Assume x and y are in yards. 82. Flying clubs: To test their skill as pilots, the members of a flight club attempt to drop sandbags on a target placed in an open field, by diving along a hyperbolic path whose vertex is directly over the target area. If the flight path of the plane flown by the club’s president is modeled by 9y2 16x2 14,400, what is the minimum altitude of her plane as it passes over the target? Assume x and y are in feet. 83. Charged particles: It has been shown that when like particles with a common charge are hurled at each other, they deflect and travel along paths that are hyperbolic. Suppose the paths of two such particles is modeled by the hyperbola x2 9y2 36. What is the minimum distance between the particles as they approach each other? Assume x and y are in microns. 84. Nuclear cooling towers: The natural draft cooling towers for nuclear power stations are called hyperboloids of one sheet. The perpendicular cross sections of these hyperboloids form two branches of a hyperbola. Suppose the central cross section of one such tower is modeled by the hyperbola 1600x2 4001y 502 2 640,000. What is the minimum distance between the sides of the tower? Assume x and y are in feet.

85. Locating a ship using radar: Under certain conditions, the properties of a hyperbola can be used to help locate the position of a ship. Suppose two radio stations are located 100 km apart along a straight shoreline. A ship is sailing parallel to the shore and is 60 km out to sea. The ship sends out a distress call that is picked up by the closer station in 0.4 milliseconds (msec — one-thousandth of a second), while it takes 0.5 msec to reach the station that is farther away. Radio waves travel at a speed of approximately 300 km/msec. Use this information to find the equation of a hyperbola that will help you find the location of the ship, then find the coordinates of the ship. (Hint: Draw the hyperbola on a coordinate system with the radio stations on the x-axis at the foci, then use the definition of a hyperbola.) 86. Locating a plane using radar: Two radio stations are located 80 km apart along a straight shoreline, when a “Mayday” call (a plea for immediate help) is received from a plane that is about to ditch in the ocean (attempt a water landing). The plane was flying at low altitude, parallel to the shoreline, and 20 km out when it ran into trouble. The plane’s distress call is picked up by the closer station in 0.1 msec, while it takes 0.3 msec to reach the other. Use this information to construct the equation of a hyperbola that will help you find the location of the ditched plane, then find the coordinates of the plane. Also see Exercise 85.

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Section 10.4 The Analytic Parabola

997


87. For a greater understanding as to why the branches y2 x2 of a hyperbola are asymptotic, solve 2 2 1 a b for y, then consider what happens as x S q (note that x2 k ⬇ x2 for large x).

88. Which has a greater area: (a) The central rectangle of the hyperbola given by 1x 52 2 1y 42 2 57, (b) the circle given by 1x 52 2 1y 42 2 57, or (c) the ellipse given by 91x 52 2 101y 42 2 570?

89. It is possible for the plane to intersect only the vertex of the cone or to be tangent to the sides. These are called degenerate cases of a conic section. Many times we’re unable to tell if the equation represents a degenerate case until it’s written in standard form. Write the following equations in standard form and comment. a. 4x2 32x y2 4y 60 0 b. x2 4x 5y2 40y 84 0

䊳


90. (8.3) In weight-lifting competitions, Ursula Unger has shown she can lift up to 350 lb. Use a vector analysis to determine whether she will be able to pull the crate up the frictionless ramp shown.

700

lb

25

91. (6.1) The wheels on a motorized scooter are rotating at 403 rpm. If the wheels have a 2.5 in. radius, how fast is the scooter traveling in miles per hour?

10.4

92. (4.2) The number z 1 i 12 is a solution to two out of the three equations given. Which two? a. x4 4 0 b. x3 6x2 11x 12 0 c. x2 2x 3 0 93. (9.3) A government-approved company is licensed to haul toxic waste. Each container of solid waste weighs 800 lb and has a volume of 100 ft3. Each container of liquid waste weighs 1000 lb and is 60 ft3 in volume. The revenue from hauling solid waste is $300 per container, while the revenue from liquid waste is $350 per container. The truck used by this company has a weight capacity of 39.8 tons and a volume capacity of 6960 ft3. What combination of solid and liquid waste containers will produce the maximum revenue?

The Analytic Parabola


A. Graph parabolas with a horizontal axis of symmetry B. Identify and use the focus-directrix form of the equation of a parabola C. Solve applications of the analytic parabola

In previous coursework, you likely learned that the Figure 10.36 graph of a quadratic function was a parabola. Parabolas Parabola are actually the fourth and final member of the family of conic sections, and as we saw in Section 10.1, the Axis graph can be obtained by observing the intersection of Element a plane and a cone. If the plane is parallel to the generator of the cone (shown as a dark line in Figure 10.36), the intersection of the plane with one nappe forms a parabola. In this section we develop the general equation of a parabola from its analytic definition, opening a new realm of applications that extends far beyond those involving only zeroes and extreme values.

A. Parabolas with a Horizontal Axis An introductory study of parabolas generally involves those with a vertical axis, defined by the equation y ax2 bx c. Unlike the previous conic sections, this equation has only one second-degree (squared) term in x and defines a function. As a

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Figure 10.37 1. Opens upward

y

4. Axis of symmetry

review, the primary characteristics are listed here and illustrated in Figure 10.37. See Exercises 7 through 12. Vertical Parabolas

2. y-intercept

3. x-intercepts

For a second-degree equation of the form y ax2 bx c, the graph is a vertical parabola with these characteristics: 1. opens upward if a 7 0, downward if a 6 0. 2. y-intercept: (0, c) (substitute 0 for x) 3. x-intercept(s): substitute 0 for y and solve. b 4. axis of symmetry: x 2a b 4ac b2 5. vertex: 1h, k2 a , b 2a 4a

x

5. Vertex

Horizontal Parabolas Similar to our study of horizontal and vertical hyperbolas, the graph of a parabola can open to the right or left, as well as up or down. After interchanging the variables x and y in the standard equation, we obtain the parabola x ay2 by c, noting the resulting graph will be a reflection about the line y x. Here, the axis of symmetry is a horizontal line and factoring or the quadratic formula is used to find the y-intercepts (if they exist). Note that although the graph is still a parabola—it is not the graph of a function. Horizontal Parabolas For a second-degree equation of the form x ay2 by c, the graph is a horizontal parabola with these characteristics: 1. opens right if a 7 0, left if a 6 0. 2. x-intercept: (c, 0) (substitute 0 for y) 3. y-intercept(s): substitute 0 for x and solve. b 4. axis of symmetry: y 2a 4ac b2 b 5. vertex: 1h, k2 a , b 4a 2a EXAMPLE 1

䊳

Graphing a Horizontal Parabola Graph the relation whose equation is x y2 3y 4, then state the domain and range of the relation. y

Solution

䊳

10 Since the equation has a single squared term in y, the graph will be a horizontal parabola. With a 7 0 1a 12, the parabola opens to the right. (0, 1) The x-intercept is 14, 02. Factoring shows the (4, 0) y-intercepts are y 4 and y 1. The axis of 10 10 symmetry is y 3 2 1.5, and substituting (6.25, 1.5) y 1.5 this value into the original equation gives (0, 4) x 6.25. The coordinates of the vertex are 16.25, 1.52. Using horizontal and vertical 10 boundary lines we find the domain for this relation is x 僆 36.25, q 2 and the range is y 僆 1q, q 2. The graph is shown.


x

䊳

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As with the vertical parabola, the equation of a horizontal parabola can be written as a transformation: x a1y k2 2 h by completing the square. Note that in this case, the vertical shift is k units opposite the sign, with a horizontal shift of h units in the same direction as the sign. EXAMPLE 2

䊳

Graphing a Horizontal Parabola by Completing the Square Graph by completing the square: x 2y2 8y 9, then state the domain and range.

Solution

䊳

y

Using the original equation, we note the graph will be a horizontal parabola opening to the left 1a 22 and have an x-intercept of 19, 02. Completing the square gives x 21y2 4y 42 9 8, so the equation of this parabola in shifted form is

10

(9, 0) 10

y 2

x 21y 22 1 2

The vertex is at 11, 22 and y 2 is the axis of symmetry. This means there are no y-intercepts, a fact that comes to light when we attempt to solve the equation after substituting 0 for x: 21y 22 2 1 0

(0, 1) 10

x

(1, 2)

(9, 4)

substitute 0 for x

1 1y 22 2 2

no real roots

Using symmetry, the point 19, 42 is also on the graph. After plotting these points we obtain the graph shown. From the graph, the domain is x 僆 1q, 1 4 and the range is y 僆⺢. Now try Exercises 19 through 36

Figure 10.38

As with the other relations graphed in this chapter, a horizontal parabola also fails the vertical line test and we must graph the relation in two pieces. Using the equation from Example 2 we have x 21y 22 2 1 shifted form

4.2

13.4

5.4

8.2

A. You’ve just seen how we can graph parabolas with a horizontal axis of symmetry

䊳

x 1 21y 22 2

isolate y-term

x1 1y 22 2 2

divide by 2

x1 y2 B 2

take square roots

x1 y B 2

solve for y

2

Y1 2

X1 X1 , Y2 2 B 2 B 2

The graph is given in Figure 10.38, and shows that (3, 1) is also a point on the graph.

B. The Focus-Directrix Form of the Equation of a Parabola As with the ellipse and hyperbola, many significant applications of the parabola rely on its analytical definition rather than its algebraic form. From the construction of radio telescopes to the manufacture of flashlights, the location of the focus of a parabola is

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critical. To understand these and other applications, we use the analytic definition of a parabola first introduced in Section 10.1. Definition of a Parabola

Given a fixed point f and fixed line D in the plane, a parabola is the set of all points (x, y) such that the distance from f to (x, y) is equal to the distance from line D to (x, y). The fixed point f is the focus of the parabola, and the fixed line is the directrix.

d1

(x, y)

f d2 Vertex D

d1 d2

The general equation of a parabola can be obtained by combining this definition with the distance formula. With no loss of generality, we can assume the parabola shown in the definition box is oriented in the plane with the vertex at (0, 0) and the focus at (0, p). As the diagram in Figure 10.39 indicates, this gives the directrix an equation of y p with all points on D having coordinates of 1x, p2. Using d1 d2 the distance formula yields

WORTHY OF NOTE For the analytic parabola, we use p to designate the focus since c is so commonly used as the constant term in y ax2 bx c.

Figure 10.39 y

1x 02 1y p2 1x x2 1y p2 2

2

2

2

2

d2

y p (0, 0) D

2

x y 2py p 0 y 2py p 2

P(x, y)

F

21x 02 2 1y p2 2 21x x2 2 1y p2 2 2

d1

(0, p)

2

x 2py 2py 2

(0, p)

x (x, p)

from the definition square both sides simplify; expand binomials subtract p 2 and y 2

x2 4py

isolate x 2

The resulting equation is called the focus-directrix form of a vertical parabola with center at (0, 0). If we had begun by orienting the parabola so it opened to the right, we would have obtained the equation of a horizontal parabola with center (0, 0): y2 4px. The Equation of a Parabola in Focus-Directrix Form Vertical Parabola

Horizontal Parabola

x 4py

y2 4px

2

focus (0, p), directrix: y p If p 7 0, opens upward. If p 6 0, opens downward.

focus at (p, 0), directrix: x p If p 7 0, opens to the right. If p 6 0, opens to the left.

For a parabola, note there is only one second-degree term.

EXAMPLE 3

䊳

Locating the Focus and Directrix of a Parabola Find the vertex, focus, and directrix for the parabola defined by x2 12y. Then sketch the graph, including the focus and directrix.

Solution

䊳

Since the x-term is squared and no shifts have been applied, the graph will be a vertical parabola with a vertex of (0, 0). Use a direct comparison between the given

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equation and the focus-directrix form to determine the value of p: x2 12y T x2 4py

y 10

x0

4p 12 p 3

y3 (0, 0)

10

f

focus-directrix form

This shows:

D

(6, 3)

given equation

10 x (6, 3)

(0, 3)

10

Since p 3 1p 6 02, the parabola opens downward, with the focus at 10, 32 and directrix y 3. To complete the graph we need a few additional points. Since 36 is divisible by 12, we can use inputs of x 6 and x 6 3 162 2 62 36 4, giving the points 16, 32 and 16, 32. Note the axis of symmetry is x 0. The graph is shown. Now try Exercises 37 through 48

Figure 10.40 y 2p f d1 Vertex y p

P

(x, y) d2 p x p D

EXAMPLE 4

䊳

䊳

As an alternative to calculating additional points to sketch the graph, we can use what is called the focal chord of the parabola. Similar to the ellipse and hyperbola, the focal chord is the line segment that contains the focus, is parallel to the directrix, and has its endpoints on the graph. Using the definition of a parabola and the diagram in Figure 10.40, we note the vertical distance from (x, y) to the directrix y p is 2p. Since d1 d2 a line segment parallel to the directrix from the focus to the graph will also have a length of 冟2p冟, and the focal chord of any parabola has a total length of 冟4p冟. Note that in Example 3, the points we happened to choose were actually the endpoints of the focal chord. Finally, if the vertex of a vertical parabola is shifted to (h, k), the equation will have the form 1x h2 2 4p1y k2. As with the other conic sections, both the horizontal and vertical shifts are “opposite the sign.” Locating the Focus and Directrix of a Parabola Find the vertex, focus, and directrix for the parabola whose equation is given, then sketch the graph, including the focus, focal chord, and directrix: x2 6x 12y 15 0.

Solution

䊳

Since only the x-term is squared, the graph will be a vertical parabola. To find the end-behavior, vertex, focus, and directrix, we complete the square in x and use a direct comparison between the shifted form and the focus-directrix form: x2 6x 12y 15 0 x2 6x ___ 12y 15 x2 6x 9 12y 24 1x 32 2 121y 22

y 10

x3 y5 (3, 2)

(3, 1)

(9, 1)

10

10

(3, 1)

10

x

given equation complete the square in x add 9 factor

Notice the parabola has been shifted 3 units right and 2 up, so all features of the parabola will likewise be shifted. Since we have 4p 12 (the coefficient of the linear term), we know p 3 1p 6 02 and the parabola opens downward. If the parabola were in standard position, the vertex would be at (0, 0), the focus at (0, 3) and the directrix a horizontal line at y 3. But since the parabola is shifted 3 right and 2 up, we add 3 to all x-values and 2 to all y-values to locate the features of the shifted parabola. The vertex is at 10 3, 0 22 13, 22. The focus is 10 3, 3 22 13, 12 and the directrix is y 3 2 5. Finally, the horizontal distance from the focus to the graph is 冟2p冟 6 units (since 冟4p冟 12), giving us the additional points 13, 12 and 19, 12 as endpoints of the focal chord. The graph is shown. Now try Exercises 49 through 60

䊳

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In many cases, we need to construct the equation of the parabola when only partial information in known, as illustrated in Example 5. EXAMPLE 5

䊳

Constructing the Equation of a Parabola Find the equation of the parabola with vertex (4, 4) and focus (1, 4). Then graph the parabola using the equation and focal chord.

Solution

䊳

B. You’ve just seen how we can identify and use the focus-directrix form of the equation of a parabola

As the vertex and focus are on a horizontal line, we have a horizontal parabola with general equation 1y k2 2 4p1x h2 . The distance p from vertex to focus is 3 units, and with the focus to the left of the vertex, the parabola opens left so p 3. Using the focal chord, the vertical distance from (1, 4) to the graph is 冟2p冟冟2132 冟 6, giving points (1, 10) and (1, 2). The vertex is shifted 4 units right and 4 units up from (0, 0), showing h 4 and k 4, and the equation of the parabola must be 1y 422 121x 42 , with directrix x 7. The graph is shown.

y 14

10

10

x

6


䊳

C. Application of the Analytic Parabola Here is just one of the many ways the analytic definition of a parabola can be applied. There are several others in the Exercise Set. Many applications use the parabolic property that light or sound coming in parallel to the axis of a parabola will be reflected to the focus. EXAMPLE 6

䊳

Locating the Focus of a Parabolic Receiver The diagram shows the cross section of a radio antenna dish. Engineers have located a point on the cross section that is 0.75 m above and 6 m to the right of the vertex. At what coordinates should the engineers build the focus of the antenna?

Solution

䊳

Focus

(6, 0.75) (0, 0)

By inspection we see this is a vertical parabola with center at (0, 0). This means its equation must be of the form x2 4py. Because we know (6, 0.75) is a point on this graph, we can substitute (6, 0.75) in this equation and solve for p: x2 4py 162 2 4p10.752 36 3p p 12

equation for vertical parabola, vertex at (0, 0) substitute 6 for x and 0.75 for y simplify result

With p 12, we see that the focus must be located at (0, 12), or 12 m directly above the vertex. Now try Exercises 79 through 86 C. You’ve just seen how we can solve applications of the analytic parabola

䊳

Note that in many cases, the focus of a parabolic dish may be above the rim of the dish.

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10.4 EXERCISES 䊳



1. The equation x ay2 by c is that of a(n) parabola, opening to the if a 7 0 and to the left if .

2. If point P is on the graph of a parabola with directrix D, the distance from P to line D is equal to the distance between P and the .

3. Given y2 4px, the focus is at equation of the directrix is

4. Given x2 16y, the value of p is the coordinates of the focus are

and the .

5. Discuss/Explain how to find the vertex, directrix, and focus from the equation 1x h2 2 4p1y k2.

䊳

and .

6. If a horizontal parabola has a vertex of 12, 3) with a 7 0, what can you say about the y-intercepts? Will the graph always have an x-intercept? Explain.


Find the x- and y-intercepts (if they exist) and the vertex of the parabola. Then sketch the graph by using symmetry and a few additional points or completing the square and shifting a parent function. Scale the axes as needed to comfortably fit the graph and state the domain and range.

7. y x2 2x 3

8. y x2 6x 5

9. y 2x2 8x 10 10. y 3x2 12x 15 11. y 2x2 5x 7

12. y 2x2 7x 3

Find the x- and y-intercepts (if they exist) and the vertex of the graph. Then sketch the graph using symmetry and a few additional points (scale the axes as needed). Finally, state the domain and range of the relation.

29. x 3 8y 2y2

30. x 2 12y 3y2

33. x 1y 32 2 2

34. x 1y 12 2 4

31. y 1x 22 2 3

35. x 21y 32 2 1

32. y 1x 22 2 4

36. x 21y 32 2 5

Find the vertex, focus, and directrix for the parabolas defined by the equations given, then use this information to sketch a complete graph (illustrate and name these features). For Exercises 49 to 60, also include the focal chord.

37. x2 8y

38. x2 16y

39. x2 24y

40. x2 20y

13. x y2 2y 3

14. x y2 4y 12

41. x2 6y

42. x2 18y

15. x y2 6y 7

16. x y2 8y 12

43. y2 4x

44. y2 12x

17. x y2 8y 16

18. x y2 6y 9

45. y2 18x

46. y2 20x

47. y2 10x

48. y2 14x

Sketch by completing the square and using symmetry and shifts of a basic function. Be sure to find the x- and y-intercepts (if they exist) and the vertex of the graph, then state the domain and range of the relation.

49. x2 8x 8y 16 0 50. x2 10x 12y 25 0

19. x y2 6y

20. x y2 8y

51. x2 14x 24y 1 0

21. x y2 4

22. x y2 9

52. x2 10x 12y 1 0

23. x y2 2y 1

24. x y2 4y 4

53. 3x2 24x 12y 12 0

25. x y2 y 6

26. x y2 4y 5

54. 2x2 8x 16y 24 0

27. x y2 10y 4

28. x y2 12y 5

55. y2 12y 20x 36 0

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56. y2 6y 16x 9 0 57. y2 6y 4x 1 0 58. y2 2y 8x 9 0 59. 2y2 20y 8x 2 0

For the graphs in Exercises 73–76, only two of the following four features are displayed: vertex, focus, directrix, and endpoints of the focal chord. Find the remaining two features and the equation of the parabola.

73. 4

60. 3y2 18y 12x 3 0

y

74.

y

6

(1, 4)

4

2

(4, 2) 2

For Exercises 61–72, find the equation of the parabola in standard form that satisfies the conditions given.

61. focus: (0, 2) directrix: y 2


63. focus: (4, 0) directrix: x 4




67. vertex: (2, 2) focus: (1, 2)

68. vertex: (4, 1) focus: (1, 1)

69. vertex: (4, 7) focus: (4, 4)

70. vertex: (3, 4) focus: (3, 1)



2

4

(2, 2) 6

(1, 4)

4

2

x

6

2

4

2

2

x

2

x 3 y

75. 6

y5

4

(2, 2)

2 4

2

4

6

8

10 x

6

8

10 x

2

76.

y 2

(4, 0) 4

2

2

4

2 4

y 6

䊳

WORKING WITH FORMULAS y

77. The area of a right parabolic segment: A ⴝ 23 ab A right parabolic segment is that part of a parabola formed by a line perpendicular to its axis, which cuts the parabola. The area of this segment is given by the formula shown, where b is the length of the chord cutting the parabola and a is the perpendicular distance from the vertex to this chord. What is the area of the parabolic segment shown in the figure? b2 4a ⴙ 2b2 ⴙ 16a2 1 lna b 78. The arc length of a right parabolic segment: 2b2 ⴙ 16a2 ⴙ 2 8a b Although a fairly simple concept, finding the length of the parabolic arc traversed by a projectile requires a good deal of computation. To find the length of the arc ABC shown, we use the formula given where a is the maximum height attained by the projectile, b is the horizontal distance it traveled, and “ln” represents the natural log function. Suppose a baseball thrown from centerfield reaches a maximum height of A 20 ft and traverses an arc length of 340 ft. Will the ball reach the catcher 310 ft away without bouncing? 䊳

10 8 (3, 4) 6 4 2 108642 2 4 6 8 10

APPLICATIONS

79. Parabolic car headlights: The cross section of a typical car headlight can be modeled by an equation similar to 25x 16y2, where x and y are in inches and x 僆 3 0, 44 . Use this information to graph the relation for the indicated domain.

80. Parabolic flashlights: The cross section of a typical flashlight reflector can be modeled by an equation similar to 4x y2, where x and y are in centimeters and x 僆 30, 2.25 4 . Use this information to graph the relation for the indicated domain.

2 4 6 8 10 x

B a C b

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81. Parabolic sound receivers: Sound technicians at professional sports events often use parabolic receivers as they move along the sidelines. If a two-dimensional cross section of the receiver is modeled by the equation y2 54x, and is 36 in. in diameter, how deep is the parabolic receiver? What is the location of the focus? [Hint: Graph the parabola on the coordinate grid (scale the axes).]


Exercise 81 y

x

82. Parabolic sound receivers: Private investigators will often use a smaller and less expensive parabolic receiver (see Exercise 81) to gather information for their clients. If a two-dimensional cross section of the receiver is modeled by the equation y2 24x, and the receiver is 12 in. in diameter, how deep is the parabolic dish? What is the location of the focus? 83. Parabolic radio wave receivers: The program known as S.E.T.I. (Search for Extra-Terrestrial Intelligence) involves a group of scientists using radio y telescopes to look for radio signals from possible intelligent species in x outer space. The radio telescopes are actually parabolic dishes that vary in size from a few feet to hundreds of feet in diameter. If a particular radio telescope is 100 ft in diameter and has a cross section modeled by the equation x2 167y, how deep is the parabolic dish? What is the location of the focus? [Hint: Graph the parabola on the coordinate grid (scale the axes).] 䊳

1005

84. Solar furnace: Another form of technology that uses a parabolic dish is called a solar furnace. In general, the rays of the Sun are reflected by the dish and concentrated at the focus, producing extremely high temperatures. Suppose the dish of one of these parabolic reflectors has a 30-ft diameter and a cross section modeled by the equation x2 50y. How deep is the parabolic dish? What is the location of the focus?

85. Commercial flashlights: The reflector of a large, commercial flashlight has the shape of a parabolic dish, with a diameter of 10 cm and a depth of 5 cm. What equation will the engineers and technicians use for the manufacture of the dish? How far from the vertex (the lowest point of the dish) will the bulb be placed? (Hint: Analyze the information using a coordinate system.) 86. Industrial spotlights: The reflector of an industrial spotlight has the shape of a parabolic dish with a diameter of 120 cm. What is the depth of the dish if the correct placement of the bulb is 11.25 cm above the vertex (the lowest point of the dish)? What equation will the engineers and technicians use for the manufacture of the dish? (Hint: Analyze the information using a coordinate system.)


87. In a study of quadratic graphs from the equation y ax2 bx c, no mention is made of a parabola’s focus and directrix. Generally, when a 1, the focus of a parabola is very near its vertex. Complete the square of the function y 2x2 8x and write the result in the form 1x h2 2 4p1y k2 . What is the value of p? What are the coordinates of the vertex?

88. Like the ellipse and hyperbola, the focal chord of a parabola (also called the latus rectum) can be used to help sketch its graph. From our earlier work, we know the endpoints of the focal chord are 2p units from the focus. Write the equation 12y 15 x2 6x in the form 4p1y k2 1x h2 2, and use the endpoints of the focal chord to help graph the parabola.

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89. (4.2/4.3) Use the function f 1x2 x5 2x4 17x3 34x2 18x 36 to comment and give illustrations of the tools available for working with polynomials: (a) synthetic division, (b) rational roots theorem, (c) the remainder and factor theorems, (d) the test for x 1 and x 1, (e) the upper/lower bounds property, (f) Descartes’ rule of signs, and (g) roots of multiplicity (bounces, cuts, alternating intervals). 90. (3.1/4.2) Find all roots (real and complex) to the equation x6 64 0. (Hint: Begin by factoring the expression as the difference of two perfect squares.)

91. (7.6) Find all real solutions to sec ␪ 1.1547 (round to the nearest degree). 92. (6.3) The graph shown 3 displays the variation in daylight from an average of 12 hours per 0 60 120 180 240 300 360 day (i.e., the maximum is 15 hours and the minimum is 9). Use the 3 Day of year graph to approximate the number of days in a year there are 10.5 or less hours of daylight. Answers may vary. Hours

䊳

10–46


MID-CHAPTER CHECK c.

Sketch the graph of each conic section. 1. 1x 42 1y 32 9 2

y 10 8 6 4 2

2

2. x2 y2 10x 4y 4 0 1x 22 2 1y 32 2 1 3. 16 1

108642 2 4 6 8 10

4. 9x2 4y2 18x 24y 9 0

6. 9x 4y 18x 24y 63 0 2

7. Find the equation of each relation and state its domain and range. a. b. y

(3, 5)

(5, 1)

54321 1 2 3 (3, 3)4 5

y 10 8 6 4 (1, 2) 2

5 4 3 2 (1, 1) 1 1 2 3 4 5 x

108642 2 4 6 8 10

(3, 4)

8. Find the equation of a parabola with focus (1, 2) and directrix x 2.

1x 32 2 1y 42 2 1 5. 9 4 2

2 4 6 8 10 x

(3, 6) (7, 2)

9. Find the equation of the ellipse (in standard form) if the vertices are (4, 0) and (4, 0) and the distance between the foci is 4 13 units. 10. The radio signal emanating from a tall radio tower spreads evenly in all directions with a range of 50 mi. If the tower is located at coordinates (20, 30) and my home is at coordinates (10, 78), will I be able to pick up this station on my home radio? Assume coordinates are in miles.

2 4 6 8 10 x

(3, 2)

REINFORCING BASIC CONCEPTS More on Completing the Square From our work so far in Chapter 10, we realize the process of completing the square has much greater use than simply as a tool for working with quadratic equations. It is a valuable tool in the application of the conic sections, as well as other areas. The purpose of this Reinforcing Basic Concepts is to strengthen the ability and confidence needed to apply the process correctly. This is important because in some cases the values of a and b are rational or irrational numbers. No matter what the context,

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Section 10.5 Nonlinear Systems of Equations and Inequalities

1007

1. The process begins with a coefficient of 1. For 20x2 120x 27y2 54y 192 0, we recognize the equation of an ellipse, since the coefficients of the squared terms are positive and unequal. To study or graph this ellipse, we’ll use the standard form to identify the values of a, b, and c. Grouping the like-variable terms gives 120x2 120x

2 127y2 54y

2 192 0

and to complete the square, we factor out the lead coefficient of each group (to get a coefficient of 1): 201x2 6x

2 271y2 2y

2 192 0

Subtracting 192 from both sides brings us to the fundamental step for completing the square. 2 1 # linear cofficientb will complete a trinomial square. For this example we obtain 2 2 1 1# 2 a 6b 9 for x, and a # 2b 1 for y, with these numbers inserted in the appropriate group: 2 2

2. The quantity a

201x2 6x 92 271y2 2y 12 192

complete the square

Due to the distributive property, we have in effect added 20 # 9 180 and 27 # 1 27 (for a total of 207) to the left side of the equation: 201x2 6x 92 271y2 2y 12 192 adds 20 # 9 180 adds 27 # 1 27 to left side

to left side

This brings us to the final step. 3. Keep the equation in balance. Since the left side was increased by 207, we also increase the right side by 207. 201x2 6x 92 271y2 2y 12 192 207 adds 20 # 9 180 adds 27 # 1 27 add 180 27 207 to left side

to left side

to right side

The quantities in parentheses factor, giving 201x 32 271y 12 15. We then divide by 15 and simplify, 41x 32 2 91y 12 2 obtaining the standard form 1. Note the coefficient of each binomial square is not 1, 3 5 even after setting the equation equal to 1. In the Strengthening Core Skills feature of this chapter, we’ll look at how to write equations of this type in standard form to obtain the values of a and b. For now, practice completing the square using these exercises. 2

2

Exercise 1: 100x2 400x 18y2 108y 554 0 Exercise 2: 28x2 56x 48y2 192y 195 0

10.5

Nonlinear Systems of Equations and Inequalities


A. Visualize possible solutions B. Solve nonlinear systems using substitution C. Solve nonlinear systems using elimination D. Solve systems of nonlinear inequalities E. Solve applications of nonlinear systems

Equations where the variables have exponents other than 1 or that are transcendental (like logarithmic and exponential equations), are all nonlinear equations. A nonlinear system of equations has at least one nonlinear equation, and these systems occur in a great variety.

A. Possible Solutions for a Nonlinear System When solving nonlinear systems, it is often helpful to visualize the graphs of each equation in the system. This can help determine the number of possible intersections and further assist the solution process.

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EXAMPLE 1

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Visualizing the Number of Possible Intersections for the Graphs in a System Identify each equation in the system as that of a line, parabola, circle, ellipse, or hyperbola. Then determine the number of solutions possible by 4x2 9y2 36 considering the different ways the graphs might intersect: e . 2x 3y 6 Finally, solve the system by graphing.

Solution

䊳

The first equation contains a sum of second-degree terms with unequal coefficients, and we recognize this as a central ellipse. The second equation is obviously linear. This means the system may have no solution, one solution, or two solutions, as shown in Figure 10.41. The graph of the system is shown in Figure 10.42 and the two points of intersection appear to be (3, 0) and (0, 2). After checking these in the original equations we find that both are solutions to the system. Figure 10.41

Figure 10.42 y

y No solutions

(0, 2)

One solution (3, 0)

(3, 0) x

x

Two solutions (0, 2)

A. You’ve just seen how we can visualize possible solutions


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B. Solving Nonlinear Systems by Substitution Since manually graphing nonlinear systems at best offers an estimate for the solution (points of intersection may not have rational values), we more often turn to algebraic methods or the use of a graphing calculator. Recall the substitution method involves solving one of the equations for a variable or expression that can be substituted in the other equation, to eliminate one of the variables. As with our study of linear systems, be aware that some nonlinear systems have no solutions. EXAMPLE 2A

䊳

Solving a Nonlinear System Using Substitution and Graphing Technology Solve the system e

y x2 2x 3 using 2x y 7

a. Substitution. b. A graphing calculator.

Solution

䊳

a. Substitution: The first equation is that of a parabola. The second equation is linear. Since the first equation is already written with y in terms of x, we can substitute x2 2x 3 for y in the second equation to solve. 2x y 7 2x 1x2 2x 32 7 2x x2 2x 3 7 x2 4x 3 7 x2 4x 4 0 1x 22 2 0 We find that x 2 is a repeated root.

second equation substitute x 2 2x 3 for y distribute combine like terms set equal to zero factor

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Since the second equation is simpler than the first, we substitute 2 for x in this equation and find y 3. The ordered pair 12, 32 checks in both equations and the system has only one (repeated) solution at 12, 32 . b. Graphing calculator: The first equation is already in function form (y is written in terms of x), and we can immediately enter y x2 2x 3 as Y1 on the Y= screen. After solving 2x y 7 for y, we obtain y 2x 7, which we enter as Y2. Using the standard window ( ZOOM 6:ZStandard), we obtain the graphs shown in Figure 10.43. Using the keystrokes 2nd TRACE (CALC) 5:intersect, we obtain the solution shown in Figure 10.44. We suspect that the calculator is trying to display x 2 and y 3 as the point of intersection, and for confirmation we use the TABLE feature (Figure 10.45), which shows this is indeed the case. Figure 10.43

Figure 10.44

10

10

10

10

10

10

10

10

EXAMPLE 2B

Figure 10.45

䊳

Solving a System Using Substitution and Graphing Technology Solve the following system e

y x2 4x 4 using 1x 22 2 y

a. Substitution. b. A graphing calculator.

Solution

䊳

a. Substitution: Expanding the binomial square in the second equation gives y 1x2 4x 42 , then y x2 4x 4 after simplification. y x2 4x 4 The system then becomes: e . Substituting y x2 4x 4 x2 4x 4 for y in the first equation gives the following: first equation y x2 4x 4 substitute x 2 4x 4 for y x 4x 4 x2 4x 4 2 add x 2 and 4; subtract 4x 0 2x 8 2 isolate squared term 8 2x 2 divide by 2 4 x x 14 or x 14 square root property result x 2i or x 2i From this result we see that the system has no real solutions, but we continue working to provide the complete complex-number solution. Substituting 2i for x in the first equation gives 2

y x2 4x 4 12i2 2 412i2 4 4112 8i 4 8i

first equation substitute 2i for x simplify result

One solution is (2i, 8i). Substituting 2i for x shows the second solution is (2i, 8i). Both solutions can be verified using back-substitution.

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b. Graphing calculator: Both equations can be entered directly on the Y= screen (both have y in terms of x) as Y1 and Y2, respectively. Using the standard window, we can easily see that the graphs do not intersect (see figure), indicating there are no real-number solutions.

10

10

10

10

B. You’ve just seen how we can solve nonlinear systems using substitution


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C. Solving Nonlinear Systems by Elimination When both equations in the system have second-degree terms with like variables, it is generally easier to use the elimination method, rather than substitution. EXAMPLE 3

䊳

Solving a Nonlinear System Using Elimination Solve the system using elimination: e

Solution

䊳

2y2 5x2 13 . 3x2 4y2 39

The first equation contains a difference of second-degree terms and is the equation of a central hyperbola. The second has a sum of second-degree terms with unequal coefficients, and represents a central ellipse. By mentally visualizing the possibilities, there could be zero, one, two, three, or four points of intersection (see Example 1). However, with both centered at (0, 0), we find there can only be zero, two, or four solutions. After writing the system with the x- and y-terms in the same order, we find that multiplying the first equation by 2 will help eliminate the y-term. e

10x2 4y2 26 3x2 4y2 39 13x 0 2

x 1

13 x2 1 or x 1

rewrite first equation and multiply by 2 original second equation add divide by 13 square root property

Substituting x 1 and x 1 into the second equation we obtain: 3112 2 4y2 39 3 4y2 39 4y2 36 y2 9 y 3

y

3112 2 4y2 39 3 4y2 39 4y2 36 y2 9 y 3

5

(1, 3)

(1, 3)

5

Since 1 and 1 each generated two outputs, there are a total of four ordered pair solutions: 11, 32, (1, 3), 11, 32, and 11, 32. The graph is shown and supports our results.

2y2 5x2 13

5

x

3x2 4y2 39 (1, 3)

(1, 3) 5


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Nonlinear systems like the one in Example 3 can also be solved by graphing the system on a graphing calculator and looking for points of intersection. Solving each equation for y yields the following results.

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2y2 5x2 13 2y2 13 5x2 y2

13 5x2 2

y Y1

13 5x2 2 B

3x2 4y2 39

original equation

4y2 39 3x2

isolate y-term divide by coefficient

Y3

39 3x2 4

y

take square roots

13 5X2 13 5X2 , Y2 B 2 B 2

y2

39 3x2 B 4

39 3X2 39 3X2 , Y4 B 4 B 4

The equations and graphs are shown in Figures 10.46 and 10.47, and verify that the point (1, 3) is a solution to the system. Using symmetry, the other solutions are (1, 3), (1, 3), and (1, 3). See Exercises 27 through 30. Figure 10.47

Figure 10.46

6.2

9.4

9.4

6.2

Nonlinear systems may involve other relations as well, including power, polynomial, logarithmic, or exponential functions. These are solved using the same methods. EXAMPLE 4

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Solving a System of Logarithmic Equations Solve the system using the method of your choice: e

y log1x 72 2 . y log1x 42 1

Verify your solution using a graphing calculator.

Solution

䊳

Since both equations have y written in terms of x, substitution appears to be the most convenient choice. The result is a logarithmic equation, which we can solve using the techniques from Chapter 5. log1x 42 1 log1x 72 2 log1x 42 log1x 72 1 log1x 42 1x 72 1 1x 42 1x 72 101 x2 11x 18 0 1x 92 1x 22 0 x 9 0 or x 2 0 x 9 or x 2

3.1

4.7

4.7

3.1

substitute log(x 4) 1 for y in first equation add log(x 7); subtract 1 product property of logarithms exponential form eliminate parentheses and set equal to zero factor zero factor theorem possible solutions

By inspection, we see that x 9 is extraneous, since log 19 42 and log19 72 are not real numbers. Substituting 2 for x in the second equation we find one form of the (exact) solution is 12, log 2 12. If we substitute 2 for x in the first equation the exact solution is 12, log 5 22. Using a calculator we can verify the ordered pairs are equivalent and approximately equal to 12, 1.32. Using the ZOOM 4:ZDecimal feature of graphing calculator produces the screen shown in the figure, which confirms the solution (2, log 2 1) obtained by substitution. Now try Exercises 31 through 42

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For practice solving more complex systems using a graphing calculator, see Exercises 43 through 48.

C. You’ve just seen how we can solve nonlinear systems using elimination

D. Solving Systems of Nonlinear Inequalities

Figure 10.48 y x2 4y2 25

(5, 0)

(5, 0)

EXAMPLE 5

x

䊳

As with our previous work with inequalities in two variables, nonlinear inequalities can be solved by graphing the boundary given by the related equation, and checking the regions that result using a test point. For example, the inequality x2 4y2 6 25 is solved by first graphing x2 4y2 25, which is a central ellipse with vertices at 15, 02 and (5, 0). We then decide whether the boundary is included or excluded (in this case it is not included), and use a test point from either “outside” or “inside” the region formed. The test point (0, 0) results in a true statement since 102 2 4102 2 6 25, so the inside of the ellipse is shaded to indicate the solution region (Figure 10.48). For a system of nonlinear inequalities, we identify regions where the solution sets for each inequality overlap, paying special attention to points of intersection. Solving a System of Nonlinear Inequalities Solve the system e

x2 4y2 6 25 by graphing. Verify results using a graphing x 4y 5

calculator.

Solution

䊳

y x 4y 5 (3, 2) (5, 0)

x2 4y2 25

x

We recognize the first inequality from Figure 10.48, an ellipse with vertices at 15, 02 and (5, 0), and a solution region in the interior. The second inequality is linear and after solving for x in the related equation, we use substitution to find points of intersection (if they exist). From x 4y 5, we obtain x 4y 5 and substitute 4y 5 for x in x2 4y2 25: x2 4y2 25 14y 52 2 4y2 25 16y2 40y 25 4y2 25 20y2 40y 25 25 y2 2y 0 y1y 22 0 y 0 or y 2

given substitute 4y 5 for x expand simplify subtract 25; divide by 20 factor result

Back-substitution shows the graphs intersect at 15, 02 and (3, 2). Graphing the line through these points and using (0, 0) as a test point shows the upper half plane is the solution region for the linear inequality 3 102 4102 5 is false]. The overlapping (solution) region for both inequalities is the elliptical sector shown in purple. Note the points of intersection are graphed using “open dots” (see figure) since points on the graph of the ellipse are excluded from the solution set. Our manual sketch indicates the boundaries of the solution region are formed by the line and the upper half of the ellipse. For x2 4y2 25, the equation for the 25 x2 Figure 10.49 upper half of the ellipse is y , 4 B 4 which we enter as Y1. Solving the second x5 equation for y we obtain y and 4 6 6 enter this as Y2. The resulting graph is shown in Figure 10.49 using the window displayed. As the test point indicated, the region below the half-ellipse and above the line forms the 4 solution region. As in Section 9.3, we use the

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shading capability of the graphing calculator by selecting the “” and “” symbols as shown in Figure 10.50. The resulting graph is shown in Figure 10.51, and appears to support the graphical solution completed by hand. Figure 10.51

Figure 10.50

4

6

6

4

D. You’ve just seen how we can solve systems of nonlinear inequalities


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E. Solving Applications of Nonlinear Systems In the business world, a fast growing company can often reduce the average price of its products using what are called the economies of scale. These would include the ability to buy necessary materials in larger quantities, integrating new technology into the production process, and other means. However, there are also countering forces called the diseconomies of scale, which may include the need to hire additional employees, rent more production space, and the like. Companies often use a break-even analysis to determine the production level at which these forces stabilize. EXAMPLE 6

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Solving an Application of Nonlinear Systems—Economies of Scale Suppose the cost to produce a new and inexpensive shoe made from molded plastic is modeled by the function C1x2 x2 5x 18, where C(x) represents the cost to produce x thousand of these shoes. Similarly, revenue from the sales of these shoes is modeled by R1x2 x2 10x 4. Use a break-even analysis to find the quantity of sales that will cause the company to break even.

Solution

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Essentially we are asked to solve the system formed by the two equations: C1x2 x2 5x 18 . Since we want to know the point where the company e R1x2 x2 10x 4 breaks even, we set C1x2 R1x2 and solve. C1x2 R1x2 x2 5x 18 x2 10x 4 2x2 15x 22 0 12x 1121x 22 0 11 or x 2 x 2

break even: cost revenue substitute for C 1x2 and R 1x2 set equal to zero factored form result

With x in thousands, it appears the company will break even if either 2000 shoes or 5500 shoes are made and sold. Now try Exercises 59 and 60

䊳

There are actually a large number of significant applications that involve nonlinear systems. Solutions to many of these are difficult to solve manually, and it’s here that a

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skillful use of technology displays its ultimate value. One such application involves the manufacture of cardboard boxes (a billion dollar industry) used to ship goods and commodities all over the globe. Designing a box with the desired volume and needed dimensions often involves the use of a nonlinear system. EXAMPLE 7

䊳

Manufacturing Cardboard Boxes In order to transport seedlings from the nursery to the market, a manufacturing engineer designs an open box with a square bottom and four sides of equal height. For the most efficient fit, the volume of the box must be 900 in3 and use 465 in2 of cardboard stock. Find the dimensions of the box.

Solution

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Figure 10.52

x in.

y in. x in.

After carefully drawing a diagram (see Figure 10.52), we let x represent the length and width of the square bottom, and y represent the height. The volume of the box is then represented by x2y 900 in3 (V LWH), and the surface area by x2 4xy 465 x2y 900 1SA S2 4SH2 . The resulting system is e 2 . Solving for y in x 4xy 645 900 the first equation, we obtain y 2 and can now substitute this result into the x second equation to obtain an equation in x alone: x2 4xy 465 900 x2 4x a 2 b 465 x 3600 x2 465 x x3 3600 465x x3 465x 3600 0

Figure 10.53 7500

30

E. You’ve just seen how we can solve applications of nonlinear systems

substitute

900 x2

for y

simplify multiply by x (clear denominator) set equal to zero

The zeroes of this cubic equation will give the dimensions of the base. A comprehensive graph of this function (using window size [30, 30] for x and [2000, 7500] for y) shows there are three zeroes, but one of them is negative and is discounted in this context (Figure 10.53). Using the 2nd TRACE (CALC) 2:zero option, we find the positive zeroes are x ⬇ 9.7 (shown) and x 15, and there turns out to be two solutions. Substituting 9.7 for x in the first equation, 30

2000

second equation

x2y 900 19.72 2y 900 194.092y 900 y ⬇ 9.6

first equation substitute 9.7 for x square 9.7 divide

yields y ⬇ 9.6 in. for the height of the box, which does not seem practical for transporting seedlings. Substituting 15 for x gives y 4 in. as the height, which seems more reasonable. The dimensions of the box will most likely be 15 in. 15 in. 4 in. Now try Exercises 61 and 62

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For additional applications of nonlinear systems, see Exercises 63 through 70.

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10.5 EXERCISES 䊳



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1. Draw sketches showing the different ways each pair of relations can intersect and give one, two, three, and/or four points of intersection. If a given number of intersections is not possible, so state. a. circle and line b. parabola and line c. circle and parabola d. circle and hyperbola e. hyperbola and ellipse f. circle and ellipse

2. By inspection only, identify the systems having no solutions and justify your choices. y2 x2 16 y x2 4 e a. e 2 b. x y2 9 x2 4y2 4 yx1 c. e 2 3x 4y2 12

3. The solution to a system of nonlinear inequalities is a(n) of the plane where the for each individual inequality overlap.

4. When both equations in the system have at least one -degree term, it is generally easier to use the method to find a solution.

5. Suppose a nonlinear system contained a central hyperbola and an exponential function. Are three solutions possible? Are four solutions possible? Explain/Discuss.

6. Solve the system twice, once using elimination, then again using substitution. Compare/Contrast each process and comment on which is more 4x2 y2 25 . efficient in this case: e 2 2x y 5


Identify each equation in the system as that of a line, parabola, circle, ellipse, or hyperbola, and solve the system by graphing.

7. e

x y6 xy4

9. e

y x 1 4x2 y2 100

11. e

2

2

x y 9 x2 y2 41 2

2

8. e

2x y 4 4x2 y2 16

10. e

x y 25 x2 y 13

12. e

4x y 36 y2 9x2 289

2

2

2

2

Solve using substitution. Find both real and complex solutions (if they exist). In Exercises 17 and 18, substitute for x2 (not y).

13. e

x2 y2 25 yx1

15. e

14. e

x 7y 50 x2 y2 100

x 4y 25 x 2y 7

16. e

x 2y 8 xy6

17. e

x2 y 13 9x2 y2 81

18. e

y x2 10 4x2 y2 100

19. e

y x2 6x y 23 1x 32 2

20. e

x2 y2 16 y x 6

2

2

2

2

Solve using elimination. Find both real and complex solutions (if they exists).

21. e

x2 y2 25 2x2 3y2 5

22. e

y2 x2 12 x2 y2 20

23. e

x2 y2 6 yx4

24. e

25. e

5x2 2y2 75 2x2 3y2 125

y x2 8x y 34 1x 42 2

26. e

3x2 7y2 20 4x2 9y2 45

Solve the following systems using a graphing calculator. Round approximate solutions to three decimal places.

1x 22 2 y2 20 27. • x2 y8 4 28. e 29. e 30. e

4x2 1y 122 2 441 x2 1y 122 2 1764

1x 102 2 1y 102 2 144 1x 42 2 y2 144

31x 242 2 y2 196 4x 4y 31

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Solve using the method of your choice. Find both real and complex solutions (if they exist). Answer in exact form.

31. e

y 5 log x y 6 log1x 32

32. e

y log1x 42 1 y 2 log1x 72 log1x 1.12 y 3 y 4 log1x2 2

33. e

y ln1x2 2 1 y 1 ln1x 122

34. e

35. e

y 9 e2x 3 y 7ex

36. e

y 2e2x 5 y 1 6ex

45. e

y 2x 3 y 2x2 9

1 2 1x 32 2 47. • 1x 32 2 y2 10 y

46. e

y 2 log1x 82 y x3 4x 2

y2 x2 5 1 48. • y 2 x1

Solve each system of inequalities. Verify results using a graphing calculator.

y 4x3 2 37. e y 2x 3x 0

y 3x 2x 0 38. e y 9x2

49. e

y x2 1 xy3

50. e

x2 y2 25 x 2y 5

x3 y 2x 39. e y 5x 6

y x3 2 40. e y 4 3x

51. e

x2 y2 7 16 x2 y2 64

52. e

y 4 x2 x2 y2 34

53. e

y x2 16 y2 x2 6 9

54. e

x2 y2 16 x 2y 7 10

55. e

y2 x2 25 冟x冟 1 7 y

56. e

y2 x2 4 xy 6 4

41. e

x2 6x y 4 y 2x 8

2

42. e

y x 2 y 4x x2

Solve each system using a graphing calculator. Round solutions to hundredths (as needed).

43. e

䊳

x2 y2 34 y2 1x 32 2 25

44. e

5x2 5y2 40 y 2x x2 6

WORKING WITH FORMULAS d 2 1ⴚa b a B

57. Tunnel clearance: h ⴝ b

The maximum rectangular clearance allowed by an elliptical tunnel can be found using the formula y2 x2 shown, where e 2 2 1 models the tunnel’s a b elliptical cross section and h is the height of the

䊳

tunnel at a distance d from the center. If a 50 and b 30, find the maximum clearance at distances of d 20, 30, and 40 ft from center. 58. Manufacturing cylindrical vents: e

A ⴝ 2␲rh V ⴝ ␲r2h

In the manufacture of cylindrical vents, a rectangular piece of sheet metal is rolled, riveted, and sealed to form the vent. The radius and height required to form a vent with a specified volume, using a piece of sheet metal with a given area, can be found by solving the system shown. Use the system to find the radius and height if the volume required is 4071 cm3 and the area of the rectangular piece is 2714 cm2.

APPLICATIONS

Solve the following applications of economies of scale.

59. World’s most inexpensive car: Early in 2008, The Tata Company (India) unveiled the Tata Nano, the world’s most inexpensive car. With its low price and 54 miles per gallon, the car may prove to be very popular. Assume the cost to produce these cars is modeled by the function

C1x2 2.5x2 120x 3500, where C(x) represents the cost to produce x thousand cars. Suppose the revenue from the sales of these cars is modeled by R1x2 2x2 180x 500. Use a break-even analysis to find the quantity of sales (to the nearest hundred) that will cause the company to break even.

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60. Document reproduction: In a world of technology, document reproduction has become a billion dollar business. With very stiff competition, the price of a single black and white copy has varied greatly is recent years. Suppose the cost to produce these copies is modeled by the function C1x2 0.1x2 1.2x 7, where C(x) represents the cost to produce x hundred thousand copies. If the revenue from the sales of these copies is modeled by R1x2 0.1x2 1.8x 2, use a break-even analysis to find the quantity of copies sold (to the nearest thousand) that will cause the copy company to break even. Build the system of nonlinear equations needed to solve the following manufacturing applications, then solve.

61. Dimensions of a pool: A homeowner wants to build a new swimming pool in her backyard. Due to size limitations, she decides to build a square pool with a flat bottom. If the volume of the pool must be 2000 ft3 and the tile surface for the sides and bottom will be 800 ft2, what will be the final dimensions of the pool? 62. Box manufacturing: A cardboard box manufacturer receives an order for a large number of boxes for wrapping gifts during the Christmas season. The boxes are to have a square top and bottom, with a volume of 6750 cm3 and a surface area of 2700 cm2. Find the dimensions of the box fitting these requirements. Solve by setting up and solving a system of nonlinear equations.

63. Dimensions of a flag: A large American flag has an area of 85 m2 and a perimeter of 37 m. Find the dimensions of the flag.

Exercise 64

64. Dimensions of a sail: The sail on a boat is a right triangle with a perimeter of 36 ft and a hypotenuse of 15 ft. Find the height and width of the sail. 䊳

1017

65. Dimensions of a tract: The area of a rectangular tract of land is 45 km2. The length of a diagonal is 1106 km. Find the dimensions of the tract. 66. Dimensions of a deck: A rectangular deck has an area of 192 ft2 and the length of the diagonal is 20 ft. Find the dimensions of the deck. 67. Dimensions of a trailer: The surface area of a closed rectangular trailer with square ends is 928 ft2. If the sum of all edges of the trailer is 164 ft, find its dimensions. 68. Dimensions of a cylindrical tank: The surface area of a closed cylindrical tank is 192␲ m2. Find the dimensions of the tank if the volume is 320␲ m3 and the radius is as small as possible. 69. Supply and demand: Suppose the monthly market demand D (in ten-thousands of gallons) for a new synthetic oil is related to the price P in dollars by the equation 10P2 6D 144. For the market price P, assume the amount D that manufacturers are willing to supply is modeled by 8P2 8P 4D 12. (a) What is the minimum price at which manufacturers are willing to begin supplying the oil? (b) Use this information to create a system of nonlinear equations, then solve the system to find the market equilibrium price (per gallon) and the quantity of oil supplied and sold at this price. 70. Supply and demand: The weekly demand D for organically grown carrots (in thousands of pounds) is related to the price per pound P by the equation 8P2 4D 84. At this market price, the amount that growers are willing to supply is modeled by the equation 8P2 6P 2D 48. (a) What is the minimum price at which growers are willing to supply the organically grown carrots? (b) Use this information to create a system of nonlinear equations, then solve the system to find the market equilibrium price (per pound) and the quantity of carrots supplied and sold at this price.


71. The area of a vertical parabolic segment is given by A 23BH, where B is the length of the horizontal base of the segment and H is the height from the base to the vertex. Investigate how this formula can be used to find the area of the solution region for the general system

of inequalities shown. e H

B

y x2 bx c y c bx x2

(Hint: Begin by investigating with b 6 and c 8, then use other values and try to generalize what you find.)

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72. For what values of r will the volume of a sphere be numerically equal to its surface area? For what values of r will the volume of a cylinder be numerically equal to its lateral surface area? Can a similar relationship be found for the volume and lateral surface area of a cone? Why or why not?

䊳

10–58


73. A rectangular fish tank has a bottom and four sides made out of glass. Use a system of equations to help find the dimensions of the tank if the height is 18 in., surface area is 4806 in2, the tank must hold 108 gal 11 gal 231 in3 2 , and all three dimensions are integers.


74. (8.4) Determine the angle ␪ between the vectors u 12i 35j and v 20i 21j. 75. (8.1) Estimate the length L to the base of the water tank, if the angle of elevation to the base of the tower is 10°, the angle of elevation to the base of the tank is 35°, and the tower makes a 112° angle with the hillside. W L 100 ft 10

10.6


A. Plot points given in

C.

D. E.

a. y 2冟x 3冟 1

3 b. y 1 x23

c. y 1x 42 2 3

d. y 2x1 3

77. (1.4) In 2008, a small business purchased a copier for $4500. In 2011, the value of the copier had decreased to $3300. Assuming the depreciation is ¢value linear: (a) find the rate-of-change m and ¢time discuss its meaning in this context; (b) find the depreciation equation; and (c) use the equation to predict the copier’s value in 2015. (d) If the copier is traded in for a new model when its value is less than $700, how long will the company use this copier?

Polar Coordinates, Equations, and Graphs

LEARNING OBJECTIVES

B.

112

76. (2.2/5.2) Sketch each transformation:

polar form Express a point in polar form Convert between polar form and rectangular form Sketch basic polar graphs using an r-value analysis Use symmetry and families of curves to write a polar equation given a polar graph or information about the graph

One of the most enduring goals of mathematics is to express relations with the greatest tan ␪ cot ␪ sin ␪ cos ␪, we would possible simplicity and ease of use. For tan2␪ cot2␪ definitely prefer working with sin ␪ cos ␪, although the expressions are equivalent. Similarly, we would prefer computing 13 i 132 6 in trigonometric form rather than algebraic form—and would quickly find the result is 1728. In just this way, many equations and graphs are easier to work with in polar form rather than rectangular form. In rectangular form, a circle of radius 2 centered at (0, 2) has the equation x2 1y 22 2 4. In polar form, the equation of the same circle is simply r 4 sin ␪. As you’ll see, polar coordinates offer an alternative method for plotting points and graphing relations.

A. Plotting Points Using Polar Coordinates Suppose a Coast Guard station receives a distress call from a stranded boat. The boater could attempt to give the location in rectangular form, but this might require imposing an arbitrary coordinate grid on an uneven shoreline, using uncertain points of reference. However, if the radio message said, “We’re stranded 4 miles out, at heading 60°,” the Coast Guard could immediately locate the boat and send help. In polar coordinates, the location “4 miles out, at heading 60°” would be written 1r, ␪2 14, 30°2, with r representing the distance from the station and ␪ 7 0 measured from a horizontal axis in the counterclockwise direction (see Figure 10.54). If we placed the scenario on a rectangular grid (assuming a straight shoreline), the coordinates of the boat would be

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4 mi 2 mi 60 30 Coast Guard (pole)

1019

Section 10.6 Polar Coordinates, Equations, and Graphs

shoreline

2兹3 mi

12 1 3, 22 using basic trigonometry. As you see, the polar coordinate system uses angles and distances to locate a point in the plane. In this example, the Coast Guard station would be considered the pole or origin, with the x-axis as the polar axis or axis of reference (Figure 10.55). A distinctive feature of polar coordinates is that we allow r to be negative, in which case P1r, ␪2 is the point 冟r冟 units from the pole in a direction opposite 1180°2 to that of ␪ (Figure 10.56). For convenience, polar graph paper is often used when working with polar coordinates. It consists of a series of concentric circles that share the same center and have integer radii. The standard angles are marked off in multiples of ␲ 15° depending on whether you’re working in radians or degrees (Figure 10.57). To 12 plot the point P1r, ␪2, go a distance of 冟r冟 at 0° then move ␪° counterclockwise along a circle of radius r. If r 7 0, plot a point at that location (you’re finished). If r 6 0, the point is plotted on a circle of the same radius, but 180° in the opposite direction.

Figure 10.55

Figure 10.57

Figure 10.56 P(r, ␪) r0

120 135 150

r r Pole

␪

Pole

␪

165

Polar axis

Polar axis

105

5 4 3 2 1

75

45 30 15

195

P(r, ␪) r0

60

345 330 315 300

210 225 240

255 285 Polar graph paper

EXAMPLE 1

䊳

Plotting Points in Polar Coordinates Plot each point P1r, ␪2 given: A14, 45°2; B15, 135°2; C13, 30°2; D a2,

Solution

䊳

2␲ ␲ ␲ b; E a5, b; and F a3, b. 3 3 6

For A14, 45°2 go 4 units at 0°, then rotate 45° counterclockwise and plot point A. For B15, 135°2, move 冟5冟 5 units at 0°, rotate 135°, then actually plot point B 180° in the opposite direction, as shown. Point C13, 30°2 is plotted by moving 冟3冟 3 units at 0°, rotating 30°, then plotting point C 180° in the opposite 2␲ ␲ direction (since r 7 0). See Figure 10.58. The points D a2, b, E a5, b, and 3 3 ␲ F a3, b are plotted on the gird in Figure 10.59. 6 Figure 10.58 Figure 10.59 120 135 150 C (3, 30)

5 4 3 2 1

60 3␲ 4

45 A

30

(4, 45)

2␲ 3

5␲ 6

(2, ) 2␲ 3

5 4 3 D 2 1

␲ 3

␲ 4 ␲ 6

F 210 225 240

(5, 135)

B

330 315 300

␲ 6

(3, )

7␲ 6

E 5␲ 4

4␲ 3

(

5,

␲ 3

) 5␲ 3

11␲ 6 7␲ 4


䊳

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␲ While plotting the points B15, 135°2 and F a3, b, you likely noticed that the 6 coordinates of a point in polar coordinates are not unique. For B15, 135°2 it appears ␲ more natural to name the location 15, 315°2, while for F a3, b, the expression 6 11␲ b is just as reasonable. In fact, for any point P1r, ␪2 in polar coordinates, a3, 6 P1r, ␪ 2␲2 and P1r, ␪ ␲2 name the same location. See Exercises 23 through 36.

A. You’ve just seen how we can plot points given in polar form

B. Expressing a Point in Polar Coordinates Conversions between rectangular and polar coordiFigure 10.60 y nates is a simple application of skills from previous sections, and closely resembles the conversion P(x, y) from the rectangular form to the trigonometric r ␪ y form of a complex number. To make the connection, we first assume r 7 0 with ␪ in Quadrant II ␪r x x (see Figure 10.60). In rectangular form, the coordinates of the point are simply (x, y), with the lengths of x and y forming the sides of a right triangle. The distance r from the origin to point P resembles the modulus of a complex number and is computed in the same way: y r 2x2 y2. As long as x 0, we have ␪r tan1 ` ` , noting ␪r is a reference x angle if the terminal side is not in Quadrant I. If needed, refer to Section 6.2 for a review of reference arcs and reference angles. Converting from Rectangular to Polar Coordinates Any point P1x, y2 in rectangular coordinates can be represented as P1r, ␪2 in polar coordinates, y where r 2x2 y2 and ␪r tan1 ` `, x 0. x

y

P(x, y) r

y

␪ ␪r

x

x

EXAMPLE 2

䊳

Converting a Point from Rectangular Form to Polar Form Convert from rectangular to polar form, with r 7 0 and 0° ␪ 6 360º (round values to one decimal place as needed). a. P15, 122 b. P13 12, 3122

Solution

䊳

a. Point P15, 122 is in Quadrant II. r 2152 2 122 1169 13 P15, 122 S P113, 112.6°2

␪r tan1a ␪r ⬇ 67.4° ␪ ⬇ 112.6°

12 b 5

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b. Point P13 1 2, 3122 is in Quadrant IV. r 313122 2 13122 2 136 6 P13 12, 3122 S P16, 315°2

␪r tan1a

3 12 b 3 12

␪r 45° ␪ 315°


In Sections 6.7 and 8.3, we used the calculator features 2nd APPS (ANGLE) 5:RSPr( and 6:RSP␪( without identifying what the notation actually means. Now we recognize that RSPr( takes a point expressed in Rectangular form and returns its Polar coordinate r, while RSP␪( returns its coordinate ␪. In Figure 10.61, we use these commands to confirm the results of Example 2(b), noting that 45° is coterminal with 315°.

B. You’ve just seen how we can express a point in polar form

䊳

C. Converting Between Polar Coordinates and Rectangular Coordinates The conversion from polar form to rectangular form is likewise straightforward. From Figure 10.62 y x we again note cos ␪ and sin ␪ , giving r r x r cos ␪ and y r sin ␪. The conversion simply consists of making these substitutions and simplifying.

Figure 10.62 y

P(x, y) r

y

␪ ␪r

x

x

Converting from Polar to Rectangular Coordinates Any point P1r, ␪2 in polar coordinates can be represented as P(x, y) in rectangular coordinates, P(r cos ␪, r sin ␪) where x r cos ␪ and y r sin ␪. r

y y ␪

␪r

x

x

Once again, these conversion formulas have been built in to many calculators. To calculate the Rectangular x coordinate of a point in Polar form, we select option 7:PSRx( from the 2nd APPS (ANGLE) menu, input the Polar coordinates of the point, and press EXAMPLE 3

䊳

ENTER

. Similarly, option 8:PSRy( calculates and returns the y coordinate.

Converting a Point from Polar Form to Rectangular Form Convert from polar to rectangular form (round values to one decimal place as needed). 5␲ a. P a12, b. P16, 240°2 b 3

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Solution

䊳

a. Point P a12,

5␲ b is in Quadrant IV. 3

Algebraic Solution

Graphical Solution

x r cos ␪

y r sin ␪ The screen shown is from a calculator in radian MODE . 5␲ 5␲ 12 cosa b 12 sin a b 3 3 1 13 12 a b b 12 a 2 2 6 6 13 5␲ b S P16, 6132 ⬇ P16, 10.42 Pa12, 3 b. Point P16, 240°2 is in Quadrant III.

Algebraic Solution

Graphical Solution

x 6 cos 240° y 6 sin 240° The screen shown is from a calculator in degree MODE . 1 23 6 a b b 6a 2 2 3 313 P16, 240°2 S P13, 3132 ⬇ P13, 5.22


䊳

Using the relationships x r cos ␪ and y r sin ␪, we can convert an equation given in rectangular form to an equivalent equation in polar form. EXAMPLE 4

Solution

䊳

䊳

Converting an Equation from Rectangular Form to Polar Form Convert the following equations to polar form. 1 a. x2 y2 16 b. y x a. The graph of this equation is a circle of radius 4 centered at the origin. x 2 y 2 16 given equation 1r cos ␪2 2 1r sin ␪2 2 16 substitute r cos ␪ for x and r sin ␪ for y r2cos2␪ r2sin2␪ 16 expand squares r2 1cos2␪ sin2␪2 16 factor out r 2 on the left r2 112 16 Pythagorean identity, cos2 ␪ sin2 ␪ 1 r2 16 Simplify square root of both sides r4 In polar form, the equation of this circle is expressed simply as r 4.

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1023

b. For this rational function, we proceed as before, simplifying with algebra and trigonometric identities. 1 y given equation x 1 r sin ␪ substitute r cos ␪ for x and r sin ␪ for y r cos ␪ 1r sin ␪21r cos ␪2 1 multiply both sides by r cos ␪ reorder terms r2sin ␪ cos ␪ 1 2 multiply both sides by 2 2r sin ␪ cos ␪ 2 r2sin 2␪ 2 double angle identity: 2 sin ␪ cos ␪ sin 2␪ Multiplying both sides by 2 enabled us to use a double-angle identity and express the rational function in polar form as r2sin 2␪ 2. Note the similarity to rectangular coordinates, where there are many different forms in which an equation may be written.

WORTHY OF NOTE Note in part b of Example 4, the equation was actually simpler in rectangular coordinates.


䊳

Converting an equation given in polar form to rectangular form relies primarily on y the relationships x r cos ␪, y r sin ␪, tan ␪ , and x2 y2 r2. With experix ence, you will become familiar with various “tricks of the trade,” which include 1. Multiplying both sides of an equation by r to yield terms of r cos ␪ or r sin ␪. 2. Taking the tangent of both sides. 3. Squaring both sides of an equation to yield r2. EXAMPLE 5

䊳

Converting an Equation from Polar Form to Rectangular Form Convert the following equations to rectangular form. ␲ a. ␪ b. r 4 sin ␪ 4

Solution

䊳

a. With ␪ isolated on the left side, we must introduce a trig function. ␲ ␪ given equation 4 ␲ tan ␪ tan take tangent of both sides 4 y y ␲ 1 substitute for tan ␪; tan 1 x 4 x multiply both sides by x yx In rectangular form, we recognize the equation y x as a line with slope 1 passing through the origin. b. Hoping to use the relationship y r sin ␪, we proceed as follows: r 4 sin ␪ r2 4r sin ␪ x2 y2 4y

given equation multiply both sides by r substitute x 2 y 2 for r 2 and y for r sin ␪

Using algebra, we can write the equation x2 y2 4y as x2 1y 22 2 4, which is the standard form of a circle with center (0, 2) and radius 2. Now try Exercises 61 through 68

䊳

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C. You’ve just seen how we can convert between polar form and rectangular form

While squaring both sides of the equation in Example 5(b) would also have yielded r2 on the left, the right would have become 16 sin2␪, which does not easily convert to rectangular form. Often, you will need to try various approaches before finding one that works.

D. Basic Polar Graphs and r-Value Analysis To really understand polar graphs, an intuitive Figure 10.63 5␲ 7␲ sense of how they’re developed is needed. Polar 2␲ 12 12 ␲ 5 3 3 ␲ equations are generally stated in terms of r and 3␲ 4 4 4 trigonometric functions of ␪, with ␪ being the input ␲ 5␲ 3 6 6 value and r being the output value. First, it helps to ␲ 2 view the length r as the long second hand of a 11␲ 12 12 1 clock, but extending an equal distance in both directions from center (Figure 10.63). This “sec- 13␲ 23␲ ond hand” ticks around the face of the clock in the 12 12 7␲ 11␲ counterclockwise direction, with the angular 6 6 5␲ 7␲ ␲ 4 4 4␲ radians 15°. As measure of each tick being 5␲ 3 17␲ 19␲ 3 12 12 12 each angle “ticks by,” we locate a point somewhere along the radius, depending on whether r is positive or negative, and plot it on the face of the clock before going on to the next tick. For the purposes of this study, we will allow that all polar graphs are continuous and smooth curves, without presenting a formal proof. EXAMPLE 6

䊳

Graphing Basic Polar Equations Graph the polar equations. ␲ 4 a. For r 4, we’re plotting all points of the form 14, ␪2 where r has a constant value and ␪ varies. As the second hand “ticks around the polar grid,” we plot all points a distance of 4 units from the 5␲ 7␲ 2␲ 12 pole. As you might imagine, the graph is 12 ␲ r, ␲4 5 3 3 ␲ 3␲ a circle with radius 4. See Example 4(a). 4 4 4 ␲ 5␲ ␲ 3 6 6 b. For ␪ , all points have the form ␲ 2 11␲ 4 12 12 1 ␲ ␲ ar, b with constant and r varying. In 4 4 13␲ 23␲ 12 ␲ 12 7␲ this case, the “second hand” is frozen at , 11␲ etc. 6 4 6 5␲ 7␲ and we plot any selection of r-values, 4 4 4␲ 5␲ 3 17␲ 19␲ 3 (4, ␪) producing the straight line shown in the 12 12 figure. See Example 5(a). a. r 4

Solution

䊳

b. ␪

( )


䊳

To develop an “intuitive sense” that allows for the efficient graphing of more sophisticated equations, we use a technique called r-value analysis. This technique basically takes advantage of the predictable patterns in r sin ␪ and r cos ␪ taken from their graphs, including the zeros and maximum/minimum values. We begin with the r-value analysis for r sin ␪, using the graph shown in Figure 10.64. Note the analysis occurs in the four colored parts corresponding to Quadrants I, II, III, and IV, and that the maximum value of 冟sin ␪冟 1.

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Figure 10.64 r 1

(1)

(2) r sin ␪ ␲ 2

␲

␪

2␲

3␲ 2

(3)

(4)

1

␲ , sin ␪ is positive and 冟sin ␪冟 increases from 0 to 1. 2 1 for r sin ␪, r is increasing ␲ 2. As ␪ moves from to ␲, sin ␪ is positive and 冟sin ␪冟 decreases from 1 to 0. 2 1 for r sin ␪, r is decreasing 3␲ , sin ␪ is negative and 冟sin ␪冟 increases from 0 to 1. 3. As ␪ moves from ␲ to 2 1 for r sin ␪, 冟r冟 is increasing 3␲ 4. As ␪ moves from to 2␲, sin ␪ is negative and 冟sin ␪冟 decreases from 1 to 0. 2 1 for r sin ␪, 冟r冟 is decreasing 1. As ␪ moves from 0 to

WORTHY OF NOTE It is important to remember that if r 6 0, the related point on the graph is 冟r冟 units from center, 180° in the opposite direction: 1r, ␪2 S 1r, ␪ 180°2.

In summary, note that the value of 冟r冟 goes through four cycles, two where it is increasing from 0 to 1 (in red), and two where it is decreasing from 1 to 0 (in blue). EXAMPLE 7

䊳

Graphing Polar Equations Using an r-Value Analysis Sketch the graph of r 4 sin ␪ using an r-value analysis.

Solution WORTHY OF NOTE While the same graph is obtained by simply plotting points, using an r-value analysis is often more efficient, particularly with more complex equations.

␪

r ⴝ 4 sin ␪

0 ␲ 6 ␲ 4 ␲ 3 ␲ 2 2␲ 3 3␲ 4 5␲ 6 ␲

0 2 212 ⬇ 2.8

䊳

Begin by noting that r 0 at ␪ 0, and will increase from 0 to 4 as the clock ␲ ␲ ␲ “ticks” from 0 to , since sin ␪ is increasing from 0 to 1. (1) For ␪ , , and 2 6 4 ␲ ␲ , r 2, r ⬇ 2.8, and r ⬇ 3.5, respectively (at ␪ , r 4). See Figure 10.65. 3 2 ␲ 冟冟 (2) As ␪ continues “ticking” from to ␲, r decreases from 4 to 0, since sin ␪ 2 2␲ 3␲ 5␲ , , and , r ⬇ 3.5, r ⬇ 2.8, and r 2, is decreasing from 1 to 0. For ␪ 3 4 6 3␲ , 冟r冟 increases respectively (at ␪ ␲, r 0). See Figure 10.66. (3) From ␲ to 2 from 0 to 4, but since r 6 0, this portion of the graph is reflected back into

213 ⬇ 3.5

Figure 10.65

4 213 ⬇ 3.5

3␲ 4

2␲ 3

5␲ 6

212 ⬇ 2.8

5 4 3 2 1

Figure 10.66

␲ 3

␲ 4

3␲ 4 ␲ 6

2␲ 3

5␲ 6

(2) and (4)

(1)

5 4 3 2 1

␲ 3

␲ 4 ␲ 6

(1) and (3)

2 0 7␲ 6 5␲ 4 4␲ 3

5␲ 3

11␲ 6 7␲ 4

7␲ 6 5␲ 4

11␲ 6 4␲ 3

5␲ 3

7␲ 4

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3␲ ␲ . (4) From to 2 2 ␲ 2␲, 冟r冟 decreases from 4 to 0, overlapping the portion drawn from to ␲. We 2 conclude the graph is a closed figure limited to Quadrants I and II as shown in Figure 10.66. This is a circle with radius 2, centered at (0, 2) [see Example 5(b)]. In summary: Quadrant I, overlapping the portion already drawn from 0 to

r ⴝ 4 sin ␪ ␲ 2

␪

0 to

冟r冟

0 to 4

␲ to ␲ 2

␲ to

4 to 0

0 to 4

3␲ 2

3␲ to 2␲ 2 4 to 0

Now try Exercises 73 and 74 Figure 10.67

䊳

Most graphing calculators are programmed to evaluate and graph four kinds of relations—functions, sequences, polar equations, and parametric equations. This is the reason the “default” variable key X,T,␪,n is embossed with four different input variables. The X is used for functions, T for parametric equations (Section 10.8), ␪ for polar equations, and n for relations defined by a sequence (Section 11.1). To change the operating mode of the calculator, press the MODE key, highlight the desired operating mode, and press (Figure 10.67). In this case, our main interest is polar equations. Note the Y= screen obtained while in Pol (polar mode) is similar to that obtained in Func (function mode), except the dependent variable is now ri instead of Yi (Figure 10.68). In addition, a ␪ now appears when the variable key X,T,␪,n is pressed, rather than an X as when in function mode. Verify by entering the equation from Example 7, r1 4 sin ␪, as shown. Since polar graphs are by nature different from polynomial graphs, the screen also takes on a different appearance (Figure 10.69). We can still set a desired window size or use the ZOOM options as before, and we can also state the desired interval for ␪, and indicate the interval between plotted points using ␪step. The standard settings ( ZOOM 6) are shown in Figure 10.69, where we note that ␲ ␪step ⬇ 0.13. Knowing that the graph of r1 is a circle of radius 4 centered at (0, 2), 24 we set the window appropriately to obtain the GRAPH shown in Figure 10.70. ENTER

WINDOW

Figure 10.68

Figure 10.70

Figure 10.69

4.6

4.7

4.7

1.6

Although it takes some effort, r-value analysis offers an efficient way to graph polar equations, and gives a better understanding of graphing in polar coordinates. In addition, it often enables you to sketch the graph with a minimum number of calculations and plotted points. As you continue using the technique, it will help to have Figure 10.64 in plain view for quick reference, as well as the corresponding analysis of y cos ␪ for polar graphs involving cosine (see Exercise 114).

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EXAMPLE 8

䊳

1027

Graphing Polar Equations Using an r-Value Analysis Sketch the graph of r 2 2 sin ␪ using an r-value analysis. Use a calculator to verify your graph.

Solution

䊳

Figure 10.71 3␲ 4 5␲ 6 11␲ 12

(2)

5 4 3 2 1

(3)

13␲ 12 7␲ 6

7␲ 2␲ 12 3

5␲ 4

4␲ 3 17␲ 12

5␲ 12

␲ 3

(1)

␲ 4

Since the minimum value of sin ␪ is 1, we note that r will always be greater than or equal to zero. At ␪ 0, r has a value of 2 (sin 0 0), and will increase from 2 ␲ to 4 as the clock “ticks” from 0 to (sin ␪ is positive and 冟sin ␪冟 is increasing). 2 ␲ From to ␲, r decreases from 4 to 2 (sin ␪ is positive and 冟sin ␪冟 is decreasing). 2 3␲ From ␲ to , r decreases from 2 to 0 (sin ␪ is negative and 冟sin ␪冟 is increasing); 2 3␲ and from to 2␲, r increases from 0 to 2 (sin ␪ is negative and 冟sin ␪冟 is decreasing). 2 ␲ We conclude the graph is a closed figure containing the points (2, 0), a4, b, 12, ␲2 , 2 3␲ ␲ 5␲ b. Noting that ␪ and ␪ and a0, will produce integer values, we 2 6 6 ␲ 5␲ b. evaluate r 2 2 sin ␪ and obtain the additional points a3, b and a3, 6 6 Using these points and the r-value analysis produces the graph shown in Figure 10.71, called a cardioid (from the limaçon family of curves). In Figure 10.72, a calculator verifies our graph. In summary we have: Figure 10.72 ␪

␲ 6 ␲ 12

(2) (4)

5␲ 19␲ 3 12

7␲ 4

23␲ 12 11␲ 6

␲ 2

2 to 4

␲ to ␲ 2

4 to 2

3␲ 2

2 to 0

(1) 0 to

(3) ␲ to (4)

4.6

r ⴝ 2 ⴙ 2 sin ␪

3␲ to 2␲ 2

4.7

4.7

1.6

0 to 2

Now try Exercises 75 through 78 D. You’ve just seen how we can sketch basic polar graphs using an r-value analysis

WORTHY OF NOTE In mathematics we refer to the tests for polar symmetry as sufficient but not necessary conditions. The tests are sufficient to show symmetry (if the test is satisfied, the graph must be symmetric), but the tests are not necessary to show symmetry (the graph may be symmetric even if the test is not satisfied).

䊳

E. Symmetry and Families of Polar Graphs Even with a careful r-value analysis, some polar graphs require a good deal of effort to produce. In many cases, symmetry can be a big help, as can recognizing certain families of equations and their related graphs. As with other forms of graphing, gathering this information beforehand will enable you to graph relations with a smaller number of plotted points. Figures 10.73 to 10.76 offer some examples of symmetry for polar graphs. The tests for symmetry in polar coordinates bear a strong resemblance to those for rectangular coordinates, but there is a major difference. Since there are many different ways to name a point in polar coordinates, a polar graph may actually exhibit a form of symmetry without satisfying the related test. In other words, the tests are sufficient to establish symmetry, but not necessary.

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Figure 10.73 Vertical-axis symmetry: r 2 2 sin ␪

Figure 10.74 Polar-axis symmetry: r 5 cos ␪

Figure 10.75 Polar symmetry: r 5 sin12␪2

Figure 10.76 Polar symmetry: r 2 25 cos12␪2

5 4 3 2 1

5 4 3 2 1

5 4 3 2 1

5 4 3 2 1

The formal tests for symmetry are explored in Exercises 116 to 118. For our purposes, we’ll rely on a somewhat narrower view, one that is actually a synthesis of our observations here and our previous experience with the sine and cosine. Symmetry for Graphs of Certain Polar Equations

Given the polar equation r f 1␪2 , 1. If f 1␪2 represents an expression in terms of sine(s), the graph will be ␲ symmetric to ␪ : 1r, ␪2 and 1r, ␲ ␪2 are on the graph. 2 2. If f 1␪2 represents an expression in terms of cosine(s), the graph will be symmetric to ␪ 0: 1r, ␪2 and (r, ␪) are on the graph. While the fundamental ideas from ExFigure 10.77 amples 7 and 8 go a long way toward graphy ing other polar equations, our discussion 1 (1) (2) (5) (6) would not be complete without a review of y sin(2␪) the period of sine and cosine. Many polar equations have factors of sin1n␪2 or cos1n␪2 ␲ 3␲ ␲ 2␲ ␪ 2 2 in them, and it helps to recall the period (3) (4) (7) (8) 2␲ formula P . Comparing r 4 sin ␪ from 1 n Example 7 with r 4 sin12␪2, we note the period of sine changes from P 2␲ to 2␲ ␲, meaning there will be twice as many cycles and 冟r冟 will now go through P 2 eight cycles—four where 冟sin12␪2 冟 is increasing from 0 to 1 (in red), and four where it is decreasing from 1 to 0 (in blue). See Figure 10.77. EXAMPLE 9

䊳

Sketching Polar Graphs Using Symmetry and r-Values Sketch the graph of r 4 sin12␪2 using symmetry and an r-value analysis. Use a calculator to verify your graph.

Solution

䊳

Since r is expressed in terms of sine, the graph will be symmetric to ␪

␲ . We 2

n␲ , where n is any integer, and the graph will go through 2 the pole at these points. This also tells us the graph will be a closed figure. From note that r 0 at ␪

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␲ 3␲ 5␲ , , , and 4 4 4 7␲ ␲ 3␲ 5␲ 7␲ b, a4, b, and a4, b. , so the graph will include the points a4, b, a4, 4 4 4 4 4 Only the analysis of the first four cycles is given next, since the remainder of the graph can be drawn using symmetry. the graph of sin12␪2 in Figure 10.80, we see 冟sin12␪2 冟 1 at ␪

Table 10.1 ␪ 0 ␲ 6 ␲ 4 ␲ 3 ␲ 2 2␲ 3 3␲ 4 5␲ 6 ␲

r ⴝ 4 sin 12␪2

Cycle

0

0 to

␲ 4

冟r冟 increases from 0 to 4

QI 1r 7 02

(2)

␲ ␲ to 4 2

冟r冟 decreases from 4 to 0

QI 1r 7 02

0

(3)

␲ 3␲ to 2 4

冟r冟 increases from 0 to 4

QIV 1r 6 02

2 13 ⬇ 3.5

(4)

3␲ to ␲ 4

冟r冟 decreases from 4 to 0

QIV 1r 6 02

4 2 13 ⬇ 3.5

4

Plotting the points and applying the r-value analysis with the symmetry involved produces the graph in the figure, called a four-leaf rose. At any time during this process, additional points such as those shown in Table 10.1 can be calculated to “round-out” the graph shown in Figure 10.78.

2 13 ⬇ 3.5 0

7␲ 2␲ 12 3

5␲ 6 11␲ 12

5 4 3 2 1

5␲ 12

r ⴝ 4 sin12␪2 ␲ 3

(2)

13␲ 12 7␲ 6 5␲ 4

Location of Graph

(1)

2 13 ⬇ 3.5

Figure 10.78 3␲ 4

r-Value Analysis

(3)

4␲ 3 17␲ 12

␲ 4

␪ ␲ 6

(1)

␲ 12

(4)

23␲ 12 11␲ 6

7␲ 5␲ 4 19␲ 3 12

␲ 4 ␲ ␲ to 4 2 ␲ 3␲ to 2 4 3␲ to ␲ 4

0 to

Figure 10.79 4

|r| 0 to 4 4 to 0

6.1

6.1

0 to 4 4 to 0

4

The calculator GRAPH of r1 4 sin12␪2 with ␪min 0, ␪max 2␲, and ␲ ␪step is shown in Figure 10.79. 24 Now try Exercises 79 through 86

䊳

Graphing Polar Equations To assist the process of graphing polar equations: 1. Carefully note any symmetries you can use. 2. Have graphs of y sin1n␪2 and y cos1n␪2 in view for quick reference. 3. Use these graphs to analyze the value of r as the “clock ticks” around the polar grid: (a) determine the max/min r-values and write them in polar form, and (b) determine the polar-axis intercepts and write them in polar form. 4. Plot the points, then use the r-value analysis and any symmetries to complete the graph.

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Similar to polynomial graphs, polar graphs come in numerous shapes and varieties, yet many of them share common characteristics and can be organized into certain families. Some of the more common families are illustrated in Appendix F, and give the general equation and related graph for common family members. Also included are characteristics of certain graphs that will enable you to develop the polar equation given its graph or information about its graph. For further investigations using a graphing calculator, see Exercises 87 through 92. EXAMPLE 10

䊳

Graphing a Limaçon Using Stated Conditions Find the equation of the polar curve satisfying the given conditions, then sketch the graph: limaçon, symmetric to ␪ 90°, with a 2 and b 3.

Solution

䊳

Figure 10.80 (2, 180)

(2, 0) (1, 90)

(5, 270)

The general equation of a limaçon symmetric to ␪ 90° is r a b sin ␪, so our desired equation is r 2 3 sin ␪. Since 冟a冟 6 冟b冟, the limaçon has an inner loop of length 3 2 1 and a maximum distance from the origin of 2 3 5. The polaraxis intercepts are 12, 0°2 and 12, 180°2 . With b 6 0, the graph is reflected across the polar axis (facing “downward”). The complete graph is shown in Figure 10.80. ␪

r ⴝ 2 ⴚ 3 sin ␪

0° 45° 90° 135° 180° 225° 270° 315° 360°

2 ⬇0.1 1 ⬇0.1 2 ⬇4.1 5 ⬇4.1 2

Figure 10.81 2

6.1

6.1

6

A calculator in degree MODE with ␪min 0°, ␪max 360°, ␪step 7.5°, and window settings shown verifies our graph (Figure 10.81). Now try Exercises 95 through 110

EXAMPLE 11

䊳

䊳

Modeling the Flight Path of a Scavenger Bird Scavenger birds sometimes fly over dead or dying animals (called carrion) in a “figure-eight” formation, closely resembling the graph of a lemniscate. Suppose the flight path of one of these birds was plotted and found to contain the polar coordinates (81, 0°) and (0, 45°). Find the equation of the lemniscate. If the bird lands at the point (r, 136°), how far is it from the carrion? Assume r is in yards. Use a calculator to check your result.

Solution

䊳

Since (81, 0°) is a point on the graph (Figure 10.82), the lemniscate is symmetric to the polar axis and the general equation is r2 a2cos12␪2 . The point (81, 0°) indicates 812 a2, hence the equation is r2 6561 cos12␪2 . At ␪ 136° we have r2 6561 cos 272°, and the bird has landed r ⬇ 15 yd away. ␪

r ⴝ 16561 cos12␪2

0° 30° 60° 90° 120° 150° 180°

81 ⬇ 57.3 error error error ⬇ 57.3 81

Figure 10.82

(15, 136) (81, 0)

Lemniscate

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Check

䊳

WORTHY OF NOTE

1031

To begin, we take the square root of each side of the general equation r2 6561 cos12␪2 and enter r1 16561 cos12␪2 in our calculator. Next, set the calculator to G-T MODE to simultaneously display the graph and table (Figure 10.83). With settings of ␪min 0°, ␪max 360°, ␪step 1°, Xmin 100, Xmax 100, Ymin 100, and Ymax 100, our calculator produces the graph of the lemniscate shown in Figure 10.84. The keystrokes 2nd TRACE (CALC) 1:value enables us to enter 136°, see the rectangular coordinates of the bird, and verify the distance r ⬇ 15 yd using the table (Figure 10.84). WINDOW

Because of the polar symmetry of lemniscates, the points described by r 16561 cos12␪2 will be plotted by r1 as the “second hand ticks around the graph.”

Figure 10.83

Figure 10.84


E. You’ve just seen how we can use symmetry and families of curves to write a polar equation given a polar graph or information about the graph

䊳

You’ve likely been wondering how the different families of polar graphs were named. The roses are easy to figure as each graph has a flower-like appearance. The limaçon (pronounced li-ma-sawn) family takes its name from the Latin words limax or lamacis, meaning “snail.” With some imagination, these graphs do have the appearance of a snail shell. The cardioids are a subset of the limaçon family and are so named due to their obvious resemblance to the human heart. In fact, the name stems from the Greek kardia meaning heart, and many derivative words are still in common use (a cardiologist is one who specializes in a study of the heart). Finally, there is the lemniscate family, a name derived from the Latin lemniscus, which describes a certain kind of ribbon. Once again, a little creativity enables us to make the connection between ribbons, bows, and the shape of this graph.

10.6 EXERCISES 䊳



1. The point (r, ␪) is said to be written in coordinates. 3. The point (4, 135°) is located in Q (4, 135°) is located in Q .

2. In polar coordinates, the origin is called the and the horizontal axis is called the axis. , while

5. Write out the procedure for plotting points in polar coordinates, as though you were explaining the process to a friend.

4. If a polar equation is given in terms of cosine, the graph will be symmetric to . 6. Discuss the graph of r 6 cos ␪ in terms of an r-value analysis, using y cos ␪ and a color-coded graph.

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DEVELOPING YOUR SKILLS Match each (r, ␪) given to one of the points A, B, C, or D shown.

Plot the following points using polar graph paper.

7. a4,

␲ b 2

8. a3,

␲ 10. a4.5, b 3

3␲ b 2

11. a5,

9. a2,

5␲ b 4

Exercise 27–36

5␲ 7␲ b 12. a4, b 6 4

3␲ 4

Express the points shown using polar coordinates with ␪ in radians, 0 ⱕ ␪ ⬍ 2␲ and r ⬎ 0.

16.

y

13␲ 12 7␲ 6

x

18.

y

y

(4, 4) x

(4, 4)

19.

20.

y

␲ 4

B

␲ 6

4␲ 3 17␲ 12

19␲ 12

5␲ 3

7␲ 4

␲ 12

23␲ 12 11␲ 6

5␲ b 6

28. a4,

5␲ b 4

29. a4,

␲ b 6

30. a4,

3␲ b 4

33. a4,

5␲ b 4

13␲ b 6

35. a4,

y

␲ 3

27. a4,

31. a4,

x

5␲ 12

D

(4, 0) x

5 4 3 2 1

C 5␲ 4

y

(0, 4)

17.

A

5␲ 6 11␲ 12

2␲ ␲ 13. a3, b 14. a4, b 3 4

15.

7␲ 2␲ 12 3

21␲ b 4

␲ 32. a4, b 4 34. a4,

19␲ b 6

36. a4,

35␲ b 6

(4, 4√3) x

Convert from rectangular coordinates to polar coordinates. A diagram may help. Use a calculator to verify your results.

x

(4√3, 4)

21.

39. (4, 4)

22. y

(4, 4)

37. (8, 0) 41. (5 12, 5 12)

y

43. (5, 12) x

x

(4√3, 4)

List three alternative ways the given points can be expressed in polar coordinates using r ⬎ 0, r ⬍ 0, and ␪ 僆 3ⴚ2␲, 2␲).

23. a3 12, 25. a2,

3␲ b 4

11␲ b 6

24. a4 13, 26. a3,

5␲ b 3

7␲ b 6

38. (0, 7)

40. 14 13, 42

42. (6, 613) 44. (3.5,12)

Convert from polar coordinates to rectangular coordinates. A diagram may help. Use a calculator to verify your results.

45. (8, 45°)

46. (6, 60°)

47. a4,

48. a5,

3␲ b 4

49. a2,

7␲ b 6

51. 15, 135°2

5␲ b 6

50. a10,

4␲ b 3

52. 14, 30°2

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Convert the following equations to polar form.

54. y 13x

53. x 25 y 2

55. x

2

3 y

56. 6xy 1

57. y 3x2 x

58. y2 5x

59. x y x 2x y y 2

2

4

2 2

4

Convert the following equations to rectangular form.

61. r 6 cos ␪

62. r 2 sin ␪

63. r 2 sec ␪

64. r 3 csc ␪

65. r 1r cos ␪ 1

66. r sin ␪ r 1cos 2␪

1 1 sin ␪

68. r

6 2 4 sin ␪

Sketch each polar graph using an r-value analysis (a table may help), symmetry, and any convenient points. Use a calculator to verify your results from Exercises 75 through 86.

69. r 5 71. ␪ 䊳

␲ 6

75. r 3 3 sin ␪

76. r 2 2 cos ␪

77. r 2 4 sin ␪

78. r 1 2 cos ␪

79. r 5 cos12␪2

80. r 3 sin14␪2

81. r 4 sin 2␪

82. r 6 cos15␪2

83. r 9 sin12␪2

84. r2 16 cos12␪2

␪ 85. r 4 sina b 2

␪ 86. r 6 cosa b 2

Use a graphing calculator in polar mode to produce the following polar graphs.

87. r 4 21 sin2 ␪, a hippopede 88. r 3 csc ␪, a conchoid 89. r 2 cos ␪ cot ␪, a cissoid 90. r cot ␪, a kappa curve 91. r 8 sin ␪ cos2␪, a bifoliate 92. r 8 cos ␪14 sin2␪ 22 , a folium

70. r 6 72. ␪

3␲ 4


93. The midpoint formula in polar coordinates: r cos ␣ ⴙ R cos ␤ r sin ␣ ⴙ R sin ␤ Mⴝa , b 2 2 The midpoint of a line segment connecting the points (r, ␣) and (R, ␤) in polar coordinates can be found using the formula shown. Find the midpoint of the line segment between 1r, ␣2 16, 45°2 and 1R, ␤2 18, 30°2 , then convert these points to rectangular coordinates and find the midpoint using the “standard” formula. Do the results match?

䊳

74. r 2 sin ␪

2

60. x2 y2 2xy

67. r

73. r 4 cos ␪

1033

94. The distance formula in polar coordinates: d ⴝ 2R2 ⴙ r2 ⴚ 2Rr cos1␣ ⴚ ␤2 Using the law of cosines, it can be shown that the distance between the points (R, ␣) and (r, ␤) in polar coordinates is given by the formula indicated. Use the formula to find the distance between 1R, ␣2 16, 45°2 and 1r, ␤2 18, 30°2 , then convert these to rectangular coordinates and compute the distance between them using the “standard” formula. Do the results match?

APPLICATIONS

Polar graphs: Find the equation of a polar graph satisfying the given conditions, then sketch the graph.

95. limaçon, symmetric to polar axis, a 4 and b 4 97. rose, five petals, one petal symmetric to the polar axis, a 4

96. rose, four petals, two petals symmetric to the polar axis, a 6 ␲ 98. limaçon, symmetric to ␪ , a 2 and b 4 2

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100. lemniscate, a ⫽ 8 through a8,

99. lemniscate, a ⫽ 4 through (␲, 4) ␲ ␲ 101. circle, symmetric to ␪ ⫽ , center at a2, b, 2 2 ␲ containing a2, b 6

␲ b 4

102. circle, symmetric to polar axis, through 16, ␲2

Matching: Match each graph to its equation a through h, which follow. Justify your answers.

103.

104.

105.

106.

6

6

6

107.

6

108.

109.

110. 6

6

6 6

a. r ⫽ 6 cos ␪ e. r2 ⫽ 36 sin12␪2

b. r ⫽ 3 ⫺ 3 sin ␪ f. r ⫽ 2 ⫹ 4 sin ␪

c. r ⫽ 6 cos14␪2 g. r ⫽ 6 sin ␪

d. r2 ⫽ 36 cos12␪2 h. r ⫽ 6 sin15␪2

111. Figure eights: Waiting for help to arrive on foot, a light plane is circling over some stranded hikers using a “figure eight” formation, closely resembling the graph of a lemniscate. Suppose the flight path of the plane was plotted (using the hikers as the origin) and found to contain the polar coordinates (7200, 45°) and (0, 90°) with r in meters. Find the equation of the lemniscate. 112. Animal territories: Territorial animals often prowl the borders of their territory, marking the boundaries with various bodily excretions. Suppose the territory of one such animal was limaçon shaped, with the pole representing the den of the animal. Find the polar equation defining the animal’s territory if markings are left at (750, 0°), (1000, 90°), and (750, 180°). Assume r is in meters. 113. Prop manufacturing: The propellers for a toy boat are manufactured by stamping out a rose with n petals and then bending each blade. If the manufacturer wants propellers with five blades and a radius of 15 mm, what two polar equations will satisfy these specifications? 114. Polar curves and cosine: Do a complete r-value analysis for graphing polar curves involving cosine. Include a color-coded graph showing the relationship between r and ␪, similar to the analysis for sines that preceded Example 8. 䊳


1 115. The polar graph r ⫽ a␪ is called the Spiral of Archimedes. Consider the spiral r ⫽ ␪. As this graph spirals 2 around the origin, what is the distance between each positive, polar intercept? In QI, what is the distance between ␲ consecutive branches of the spiral each time it intersects ␪ ⫽ ? What is the distance between consecutive 4 ␲ branches of the spiral at ␪ ⫽ ? What can you conclude? 2

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Section 10.7 More on the Conic Sections: Rotation of Axes and Polar Form

1035

As mentioned in the exposition, tests for symmetry of polar graphs are sufficient to show symmetry (if the test is satisfied, the graph must be symmetric), but the tests are not necessary to show symmetry (the graph may be symmetric even if the test is not satisfied). For r ⴝ f (␪), the formal tests for the symmetry are: (1) the graph will be ␲ symmetric to the polar axis if f (␪) ⫽ f (ⴚ␪); (2) the graph will be symmetric to the line ␪ ⴝ if f (␲ ⫺ ␪) ⫽ f (␪); 2 and (3) the graph will be symmetric to the pole if f (␪) ⫽ ⫺f (␪).

116. Sketch the graph of r 4 sin12␪2. Show the equation fails the first test, yet the graph is still symmetric to the polar axis. 117. Why is the graph of every lemniscate symmetric to the pole? 118. Verify that the graph of every limaçon of the form r a b cos ␪ is symmetric to the polar axis. 䊳

119. The graphs of r a sin1n␪2 and r a cos1n␪2 are from the rose family of polar graphs. If n is odd, there are n petals in the rose, and if n is even, there are 2n petals. An interesting extension of this fact is that the n petals enclose exactly 25% of the area of the circumscribed circle, and the 2n petals enclose exactly 50%. Find the area within the boundaries of the rose defined by r 6 sin15␪2.


120. (7.4) Verify the following is an identity: cos2 x sin2x 1 sin12x2 tan x. 121. (7.7) Solve for t 僆 3 0, 2␲2: 20 5 30 sina2t

123. (2.5) Graph the piecewise-defined function shown and state its domain and range. x 2 5 x 6 1 f 1x2 • x 1 x 6 2 4 2 6 x5

␲ b. 6

122. (2.3) Solve the absolute value inequality. Answer in interval notation: 3冟2x 5冟 7 7 19

10.7

More on the Conic Sections: Rotation of Axes and Polar Form


A. Graph conic sections that have nonvertical and nonhorizontal axes (rotated conics) B. Identify conics using the discriminant of the polynomial form—the invariant B 2 4AC C. Write the equation of a conic section in polar form D. Solve applications involving the conic sections in polar form

Our study of conic sections would not be complete without considering conic sections whose graphs are not symmetric to a vertical or horizontal axis. The axis of symmetry still exists, but is rotated by some angle. We’ll first study these rotated conics using the equation in its polynomial form, then investigate some interesting applications of the polar form.

A. Rotated Conics and the Rotation of Axes It’s always easier to understand a new idea in terms of a known idea, so we begin our 1 study with a review of the reciprocal function y . From the equation we note: x 1. The denominator is zero when x 0, and the y-axis is a vertical asymptote (the vertical line x 0). 2. Since the degree of the numerator is less than the degree of the denominator, the x-axis is a horizontal asymptote (the horizontal line y 0). 3. Since x 6 0 implies y 6 0 and x 7 0 implies y 7 0, the graph will have two branches—one in the first quadrant and one in the third.

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Figure 10.85

Note the polynomial form of this equation is xy 1. The resulting graph is shown in Figure 10.85, and is actually the graph of a hyperbola with a transverse axis of y x. Using the 45°-45°-90° triangle indicated, we find the distance from the origin to each vertex is 12. If we rotated the hyperbola 45° clockwise, we would obtain a more “standard” graph with a horizontal transverse axis and vertices at 1a, 02 S 112, 02. The asymptotes would be b y 1x, and since y x is the general form we a know b 12. This information can be used to find the equation of the rotated hyperbola.

y

yx

(1, 1) 1 1

(1, 1)

EXAMPLE 1

䊳

Finding the Equation of a Rotated Conic from Its Graph

Solution

䊳

Using the standard form

x

The hyperbola xy 1 is rotated clockwise 45°, with new vertices at 1 12, 02, asymptotes at y 1x and b 12. Find the equation and graph the hyperbola. y2 x2 1 and a2 b2 substituting 12 for a and b, the equation of the rotated hyperbola is y2 x2 1 or x2 y2 2 in polynomial 2 2 form. The resulting graph is the central hyperbola shown.

y

(√2, 0)

(√2, 0) x

Now try Exercises 7 and 8 Figure 10.86 y x r cos ␣ y r sin ␣

(x, y)

r

␣ x

䊳

It’s important to note the equation of the rotated hyperbola is devoid of the mixed “xy” term. In nondegenerate cases, the equation Ax2 Cy2 Dx Ey F 0 is the polynomial form of a conic with axes that are vertical/horizontal. However, the most general form of the equation is Ax2 Bxy Cy2 Dx Ey F 0, and includes this Bxy term. As noted in Example 1, the inclusion of this term will rotate the graph through some angle ␤. Based on these observations, we reason that one approach to graphing these conics is to find the angle of rotation ␤ with respect to the xy-axes. We can then use ␤ to rewrite the equation so that it corresponds to a new set of XY-axes, which are parallel to the axes of the conic. The mixed xy-term will be absent from the new equation and we can graph the Figure 10.87 conic on the new axes using the same ideas as y Y before (identifying a, b, foci, and so on). To find ␤, X r cos(␣ ␤) recall that a point (x, y) in the xy-plane can be writY r sin(␣ ␤) (X, Y) ten x r cos ␣, y r sin ␣, as in Figure 10.86. The diagram in Figure 10.87 shows the axes of a new XY-plane, rotated counterclockwise by angle ␤. In r this new plane, the coordinates of the point (x, y) ␣ X become X r cos1␣ ␤2 and Y r sin1␣ ␤2 as ␣␤ shown. Using the difference identities for sine and ␤ cosine and substituting x r cos ␣ and y r sin ␣ x leads to:

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X r cos1␣ ␤2

Y r sin1␣ ␤2

r1cos ␣ cos ␤ sin ␣ sin ␤2

r1sin ␣ cos ␤ cos ␣ sin ␤2

r cos ␣ cos ␤ r sin ␣ sin ␤

r sin ␣ cos ␤ r cos ␣ sin ␤

x cos ␤ y sin ␤

y cos ␤ x sin ␤

The last two equations can be written as a system, which we will use to solve for x and y in terms of X and Y. e e

X x cos ␤ y sin ␤ Y y cos ␤ x sin ␤

original system

X cos ␤ x cos2␤ y sin ␤ cos ␤ Y sin ␤ y sin ␤ cos ␤ x sin2␤

X cos ␤ Y sin ␤ x cos2␤ x sin2␤ X cos ␤ Y sin ␤ x

multiply second equation by cos ␤ multiply second equation by sin ␤ first equation – second equation factor out x 1cos2␤ sin2␤ 12

Re-solving the system for y results in y X sin ␤ Y cos ␤, yielding what are called the rotation of axes formulas (see Exercise 79).

WORTHY OF NOTE If you are familiar with matrices, it may be easier to remember the rotation formulas in their matrix form, since the pattern of functions is the same, with only a difference in sign: x cos␤ sin␤ X c d c dc d y sin␤ cos␤ Y

Rotation of Axes Formulas If the x- and y-axes of the xy-plane are rotated counterclockwise by the (acute) angle ␤ to form the X- and Y-axes of an XY-plane, the coordinates of the points (x, y) and (X, Y ) are related by the formulas

X cos ␤ sin␤ x c d c dc d Y sin␤ cos␤ y See Exercises 86 and 87.

x X cos ␤ Y sin ␤ y X sin ␤ Y cos ␤

X x cos ␤ y sin ␤ Y x sin ␤ y cos ␤

EXAMPLE 2

䊳

Naming the Location of a Point After Rotating the Axes

Solution

䊳

Using the formulas with x 1, y 13, and ␤ 60°, we obtain

Given the point 11, 132 in the xy-plane, find the coordinates of this point in the XY-plane given the angle ␤ between the xy-axes and the XY-axes is 60°. X x cos ␤ y sin ␤ 1 cos 60° 13 sin 60° 1 13 a b 13 a b 2 2 2

Y x sin ␤ y cos ␤ 1 sin 60° 13 cos 60° 13 13 2 2 0

The coordinates of P(X, Y) would be (2, 0). 䊳

Now try Exercises 9 through 16 Figure 10.88 y (2,

0)

X

-pl

an e:

Y

xy-plane: (1, √3)

XY

The diagram in Figure 10.88 provides a more intuitive look at the rotation from Example 2. As you can see, a 30°-60°-90° triangle is formed with a hypotenuse of 2, giving coordinates (2, 0) in the XY-plane.

√3 60 1

x

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EXAMPLE 3

䊳

Writing the Equation of a Rotated Conic The equation X2 4Y2 16 describes a central ellipse in the XY-plane. Given the XY-axes are formed by a 45° counterclockwise rotation of the xy-axes, what is the corresponding equation in the xy-plane?

Solution

䊳

We proceed as before, using the rotation formulas X x cos ␤ y sin ␤ and 12 Y y cos ␤ x sin ␤. With ␤ 45° we have cos ␤ sin ␤ , yielding 2 X2 4Y2 1x cos ␤ y sin ␤2 2 41y cos ␤ x sin ␤2 2 12 2 12 2 12 12 x yb 4a y xb a 2 2 2 2 1 1 1 1 a x2 xy y2 b 4a x2 xy y2 b 2 2 2 2 1 1 2 x xy y2 2x2 4xy 2y2 2 2 5 2 5 x 3xy y2 2 2

16 16

use rotation formulas

16

substitute

16

square binomials

16

distribute

16

result

12 for sin ␤ and cos ␤ 2


䊳

In Section 10.2, we used our calculator to graph central ellipses like X2 4Y2 16 from Example 3 by solving for Y and, in this case, entering the two halves as X2 X2 Figure 10.89 Y1 4 and Y2 4 (see B 4 B 4 4 Figure 10.89). Similarly, a calculator can graph the rotated ellipse described by the equation 5 2 5 x 3xy y2 16. Begin by rewriting 2 2 6.1 6.1 5 2 5 2 the equation as y 3xy x 16 0 2 2 with the goal of solving for y. Since the equation is quadratic (in terms of y), we 4 5 5 2 substitute a , b 3x, and c x 16 Figure 10.90 2 2 4 into the quadratic formula and find 2 3x 2160 16x y (verify this). Now 5 3X 2160 16X2 6.1 6.1 we can easily enter Y1 5 3X 2160 16X2 and Y2 into our 5 calculator and graph the rotated ellipse (see 4 Figure 10.90). Note the XY-axes have also been graphed as Y3 X and Y4 X. Note the equation of the conic in the standard xy-plane contains the “mixed” Bxy-term. In practice, we seek to reverse this procedure by starting in the xy-plane, and finding the angle ␤ needed to eliminate the Bxy-term. Using the rotation formulas and the appropriate angle ␤, the equation Ax2 Bxy Cy2 Dx Ey F 0 becomes aX2 cY2 dX eY f 0, where the xy-term is absent. To find the angle ␤, note that without loss of generality, we can assume D E 0 since only the

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second-degree terms are used to identify a conic. Starting with the simplified equation Ax2 Bxy Cy2 F 0 and using the rotation formulas we obtain Ax2

Bx

#

y

Cy2

F0

A1X cos ␤ Y sin ␤2 2 B1X cos ␤ Y sin ␤2 1X sin ␤ Y cos ␤2 C1X sin ␤ Y cos ␤2 2 F 0 Expanding this expression and collecting like terms (see Exercise 80), gives the following expressions for coefficients a, b, and c of the corresponding equation aX2 bXY cY2 f 0: a S A cos2␤ B sin ␤ cos ␤ C sin2␤

a is the coefficient of X 2

b S 2A sin ␤ cos ␤ B1cos2␤ sin2␤2 2C sin ␤ cos ␤

b is the coefficient of XY

c S A sin ␤ B sin ␤ cos ␤ C cos ␤ fSF 2

2

c is the coefficient of Y 2 f F (the constant remains unchanged)

To accomplish our purpose, we require the coefficient b to be zero. While this expression looks daunting, the double-angle identities for sine and cosine simplify it very nicely: b S A12 sin ␤ cos ␤2 B1cos2␤ sin2␤2 C12 sin ␤ cos ␤2 0 A sin12␤2 B cos12␤2 C sin(2␤) 0

1C A2sin12␤2 B cos12␤2

B CA B tan12␤2 ;AC AC

tan12␤2

(1) (2) (3) (4)

(5)

Note from line (3) that A C would imply cos12␤2 0, giving 2␤ 90° or 90°, with ␤ 45° or 45° (for the sake of convenience, we select the angle in QI). B This fact can many times be used to great advantage. If A C, tan12␤2 and AC we choose 2␤ between 0 and 180° so that ␤ will be in the first quadrant 3 0 6 ␤ 6 90° 4. The Equation of a Conic After Rotating the Axes For a conic defined by Ax2 Bxy Cy2 Dx Ey F 0 and its graph in the B xy-plane, an angle ␤ can be determined using tan12␤2 and used in the AC rotation formulas to find a polynomial aX2 cY2 dX eY f 0 in an XY-plane, where the conic is either vertical or horizontal.

EXAMPLE 4

䊳

Rotating the Axes to Eliminate the Bxy-Term For x2 2 13xy 3y2 13x y 16 0, eliminate the xy-term using a rotation of axes and identify the conic associated with the resulting equation. Then sketch the graph of the rotated conic in the XY-plane.

Solution

䊳

B 2 13 13. , giving tan12␤2 AC 13 13 This shows 2␤ tan1 13, yielding 2␤ 60° so ␤ 30°. Using cos 30° 2 1 and sin 30° along with the rotation formulas we obtain the following 2 XY-equation, with corresponding terms shown side-by-side for clarity: Since A C, we find ␤ using tan12␤2

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Given Term in xy-Plane x2 2 13xy 3y2

13x y 16

Corresponding Term in XY-Plane → a

1 2 3 1 13 13 X Yb X2 XY Y2 2 2 4 2 4

13 1 1 13 3 2 3 2 → 213a 2 X 2 Yba 2 X 2 Yb 2 X 13XY 2 Y 13 2 3 2 9 2 13 1 → 3a 2 X 2 Yb 4 X 3 2 XY 4 Y → 13a

1 3 13 13 X Yb X Y 2 2 2 2

13 1 13 1 → a 2 X 2 Yb 2 X 2 Y → 16

Adding the like terms on the far right, the X2-terms (in red), the Y-terms (in bold), and the mixed XY-terms (in blue) sum to zero, leaving the equation 2X 4Y2 16 0, which is the parabola defined by Y2 12 1X 82. This parabola is symmetric to the X-axis and opens to the right, with a vertex at (8, 0), Y-intercepts at (0, 2) and (0, 2), focus at 163 8 , 02 and directrix through 165 8 , 02 . The graph is shown in the figure.

y Y X (0, 2) 30 x

(0, 2) (8, 0)


A. You’ve just seen how we can graph conic sections that have nonvertical and nonhorizontal axes (rotated conics)

䊳

In Example 4, the angle ␤ was a standard angle and easily found. In general, this is not the case and finding exact values of cos ␤ and sin ␤ for use in the rotation formulas sin12␤2 , the corresponding (triangle) diagram, and the requires using tan12␤2 cos12␤2 1 cos12␤2 1 cos12␤2 identities cos ␤ and sin ␤ . See Exercises 31, 32, B 2 2 A 84, and 85 for further study.

B. Identifying Conics Using the Discriminant In addition to rotating the axes, the inclusion of the “xy-term” makes it impossible to identify the conic section using the tests seen earlier. For example, having A C no longer guarantees a circle, and A 0 or C 0 does not guarantee a parabola. Rather than continuing to look at what the mixed term and the resulting rotation changes, we now look at what the rotation does not change, called invariants of the transformation. These invariants can be used to double-check the algebra involved and to identify the conic using the discriminant. These are given here without proof.

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Invariants of a Rotation and Classification Using the Discriminant By rotating the coordinate axes through a predetermined angle ␤, the equation Ax2 Bxy Cy2 Dx Ey F 0 can be transformed into aX2 cY2 dX eY f 0 in which the xy-term is absent. This rotation has the following invariants: (1) F f

(2) A C a c

(3) B2 4AC b2 4ac.

The discriminant of a conic equation in polynomial form is B2 4AC. Except in degenerate cases, the graph of the equation can be classified as follows: If B2 4AC 0, the graph will be a parabola. If B2 4AC 6 0, the graph will be a circle or an ellipse. If B2 4AC 7 0, the graph will be a hyperbola.

EXAMPLE 5A

䊳

Verifying the Invariants of a Rotation of Axes Verify the invariants just given using the equations from Example 4. Also verify the discriminant test.

Solution

䊳

From the equation x2 2 13xy 3y2 13x y 16 0, we have A 1, B 2 13, C 3, D 13, E 1, and F 16. After applying the rotation the equation became 2X 4Y2 16 0, with a 0, b 0, c 4, d 2, e 0, and f 16. Checking each invariant gives: (1) 16 16 ✓ (2) 1 3 0 4 ✓ (3) 12 132 2 4112132 102 2 4102 142 ✓

With B2 4AC 0, the discriminant test indicates the conic is a parabola ✓.

EXAMPLE 5B

䊳

Identifying the Equation of a Conic Using the Discriminant Use the discriminant to identify each equation as that of a circle, ellipse, parabola, or hyperbola, but do not graph the equation. a. 3x2 4xy 3y2 6x 12y 2 0 b. 4x2 9xy 4y2 8x 24y 9 0 c. 6x2 7xy y2 5 0 d. x2 6xy 9y2 6x 0

Solution

䊳

B. You’ve just seen how we can identify conics using the discriminant of the polynomial form—the invariant B2 ⴚ 4AC

a. A 3; B 4; C 3 B2 4AC 142 2 4132 132 20 circle or ellipse

b. A 4; B 9; C 4 B2 4AC 192 2 4142142 17 hyperbola

c. A 6; B 7; C 1 B2 4AC 172 2 4162 112 25 hyperbola

d. A 1; B 6; C 9 B2 4AC 162 2 4112192 0 parabola Now try Exercises 33 through 36

䊳

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C. Conic Equations in Polar Form Figure 10.91 y P(r, ␪)

D

r

x

␪ F(0, 0)

A

B

You might recall that earlier in this chapter we defined ellipses and hyperbolas in terms of a distance between two points, but a parabola in terms of a distance between a point and a line (the focus and directrix). Actually, all conic sections can be defined using a focus/directrix development and written in polar form. This serves to unify and greatly simplify their study. We begin by revisiting the focus/directrix development of a parabola, using a directrix l and placing the focus at the origin. With the polar axis as the axis of symmetry and the point P1r, ␪2 in polar coordinates, we obtain the graph shown in Figure 10.91. Given D and A are points on l (with A on the polar axis), we note the following: (1) DP FP (2) DP AB (3) FB r cos ␪

d

(4) AB AF FB

Directrix

definition of a parabola equal line segments FB cos ␪ r sum of line segments

Using the preceding equations and representing the distance AF by the constant d, we obtain this sequence:

ᏸ

AB d r cos ␪

substitute d for AF and r cos ␪ for FB

FP d r cos ␪ r d r cos ␪

substitute FP for AB since FP DP AB substitute r for FP

d , 1 cos ␪ which is the equation of a parabola in polar form with its focus at the origin, vertex at 3␲ d ␲ a , ␲b, and y-intercepts at ad, b and ad, b. Note the constant “1” in the denomi2 2 2 nator is a key characteristic of polar equations, and helps define the standard form. Solving the last equation for r we have r r cos ␪ d, then r

EXAMPLE 6A

䊳

Identifying a Conic from Its Polar Equation 6 Verify the equation r represents a parabola, then describe and 3 3 cos ␪ sketch the graph. Use a calculator to verify your sketch.

Solution

䊳

Write the equation in standard form by dividing the numerator and denominator by 3, obtaining 2 r . From this we see d 2 and the 1 cos ␪ equation represents a parabola symmetric to the polar axis, with vertex at (1, ␲) and y-intercepts ␲ 3␲ b, as shown in Figure 10.92. at a2, b and a2, 2 2 A calculator in radian and polar MODE with ␲ ␪min 0, ␪max 2␲, ␪step and 24 window settings shown produces the graph in Figure 10.93 and verifies our sketch.

Figure 10.93 5

7.6

7.6

5

Figure 10.92 2␲ 3 3␲ 4

␲ 3

ᏸ

5␲ 6

4 3 (2, ␲2 )

␲ 4 ␲ 6

(1,␲)

(2, 3␲ 2 ) 7␲ 6 5␲ 4

11␲ 6 4␲ 3

5␲ 3

7␲ 4

The polar equation for a parabola depended on DP and FP being equal in length, FP 1. But what if this ratio is not equal to 1? Similar to our introduction with ratio DP

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Figure 10.94 1 FP and 2 DP investigate the graph that results. Cross-multiplying DP2 2FP2 DP1 2FP1 gives 2FP DP, which states that the distance D P2 P1 F from D to P is twice the distance from F to P. Note that we are able to locate two points P1 and P2 on the Directrix polar axis that satisfy this relation, rather than only ᏸ one as in the case of the parabola. Figure 10.94 illustrates the location of these points. Using the focal chord for convenience, two additional points P3 and P4 can be located that also satisfy the stated condition (see Figure 10.95). In fact, we can locate an 1 FP infinite number of these points using , and the resulting graph appears to be an 2 DP ellipse (and is definitely not a parabola). These illustrations provide the basis for FP stating a general focus/directrix definition of the conic sections. The ratio is often DP represented by the letter e, and represents the eccentricity of the conic. Using FP r FP r e, which and DP d r cos ␪ from our initial development, d r cos ␪ DP enables us to state the general equation of a conic in polar form. Solving for r leads to de , where the type of conic depends solely on e. Depending the equation r 1 e cos ␪ on the orientation of the conic, the general form may involve sine instead of cosine, and have a sum of terms in the denominator rather than a difference. Note once again that if e 1, the relation simplifies into the parabolic equation seen earlier.

to conics in Section 10.1, we assume

Figure 10.95 P3

D1 F P1

D1P3 2FP3 D2P4 2FP4

D2 ᏸ

P4

P2

The Standard Equation of a Conic in Polar Form Given a conic section with eccentricity e, one foci at the pole of the r␪-plane, and directrix l located d units from this focus. Then the polar equations r

de 1 e cos ␪

and

r

de 1 e sin ␪

represent one of the conic sections as determined by the value of e. • If e 1, the graph is a parabola. • If 0 6 e 6 1, the graph is an ellipse. • If e 7 1, the graph is a hyperbola. For the ellipse and hyperbola, the major axis and transverse axis (respectively) are both perpendicular to the directrix and contain the vertices and foci. From our development of ellipses in Section 10.2, eccentricity can then be expressed in terms of a and c, c as the ratio e . a As in our previous study of polar equations, if the equation involves cosine the ␲ graph will be symmetric to the polar axis. If the graph involves sine, the line ␪ is 2 the axis of symmetry. In addition, if the denominator contains a difference of terms (as in Example 6A), the graph will be above or to the right of the directrix (depending on whether the equation involves sine or cosine). If the denominator contains a sum of terms, the graph will be below or to the left of the directrix.

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EXAMPLE 6B

䊳

Using the Standard Equation to Graph a Conic in Polar Form 10 represents a parabola, ellipse, or 5 ⫺ 3 sin ␪ hyperbola. Then describe and sketch the graph. Check your results with a calculator. Figure 10.96 To write the equation in standard form, we divide 2␲ ␲ both numerator and denominator by 5, obtaining 3 (5, ␲2 ) 3 ␲ 3␲ 2 4 4 4 . From the standard the equation r ⫽ 5␲ ␲ 3 6 6 3 1 ⫺ sin ␪ 2 5 1 3 (2, 0) (2, ␲) form we note e ⫽ so the equation represents 5 an ellipse. With a difference of terms and the sine (1.25, 3␲ 2 ) 11␲ function involved, the graph is symmetric to 7␲ 6 ␲ 6 ᏸ ␪ ⫽ and is above the directrix. Given so much 5␲ 2 7␲ 4 4 information by the equation, we require very few 4␲ 5␲ 3 3 points to sketch the graph and settle for those ␲ 3␲ Figure 10.97 , yielding generated by ␪ ⫽ 0, , ␲, and 2 2 6 ␲ 5 3␲ the points (2, 0), a5, b, 12, ␲2, and a , b. 2 4 2 The graph is shown in Figure 10.96. A 6.8 calculator in radian MODE with ␪min ⫽ 0, ⫺6.8 ␲ ␪max ⫽ 2␲, ␪step ⫽ and window 24 settings shown verifies our graph (Figure 10.97). Determine if the equation r ⫽

Solution

䊳

C. You’ve just seen how we can write the equation of a conic section in polar form

⫺3


D. Applications of Conics in Polar Form For centuries it has been known that the orbits of the planets around the Sun are elliptical, with the Sun at one focus. In addition, comets may approach our Sun in elliptical, hyperbolic, or parabolic paths with the Sun again at the foci. This makes planetary studies a very natural application of the conic sections in polar form. To aid this study, it helps to know that in an elliptical orbit, the maximum distance of a planet from the Sun is called its aphelion, and the shortest distance is the perihelion (Figure 10.98). This means the length of the major axis is “aphelion ⫹ perihelion,” enabling us to find the value of c if the aphelion and c perihelion are known (Figure 10.99). Using e ⫽ , we can a then find the eccentricity of the planet’s orbit.

EXAMPLE 7

䊳

Figure 10.98

Sun Perihelion

Aphelion

Figure 10.99 a Perihelion c

Determining the Eccentricity of a Planet’s Orbit In its elliptical orbit around the Sun, Mars has an aphelion of 154.9 million miles and a perihelion of 128.4 million miles. What is the eccentricity of its orbit?

䊳

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Solution

䊳

1045

The length of the major axis would be 2a 1154.9 128.42 million mi, yielding a semimajor axis of a 141.65 million miles. Since a c perihelion (Figure 10.99), we have 141.65 c 128.4 so c 13.25. The eccentricity c 13.25 of the orbit is e or about 0.0935. a 141.65 Now try Exercises 59 and 60

䊳

We can also find the perihelion and aphelion directly in terms of a (semimajor axis) and e (eccentricity) if these quantities are known. Using a c perihelion, we c obtain: perihelion a c. For e , we have ea c and by direct substitution we a obtain: perihelion a ea a11 e2. For Example 8, recall that “AU” designates an astronomical unit, and represents the mean distance from the Earth to the Sun, approximately 92.96 million miles. EXAMPLE 8

䊳

Determining the Perihelion of a Planet’s Orbit The orbit of the planet Jupiter has a semimajor axis of 5.2 AU (1 AU ⬇ 92.96 million miles) and an eccentricity of 0.0489. What is the closest distance from Jupiter to the Sun?

Solution

䊳

With perihelion a11 e2, we have 5.211 0.04892 ⬇ 4.946. At its closest approach, Jupiter is 4.946 AU from the Sun (about 460 million miles). Now try Exercises 61 and 62

䊳

To find the polar equation of a planetary orbit, Figure 10.100 it’s helpful to write the general polar equation in terms of the semimajor axis a, which is often known d or easily found, rather than in terms of the distance d from directrix to focus, which is often unknown. D P 1 F C P2 Consider the diagram in Figure 10.100, which a shows an elliptical orbit with the Sun at one focus, vertices P1 and P2 (perihelion and aphelion), and ᏸ the center C of the ellipse. Assume the point P used to define the conic sections is FP1 at position P1, giving e. From the discussion preceding Example 8 we DP1 have FP1 a11 e2. Substituting a11 e2 for FP1 and solving for DP1 gives a11 e2 DP1 . Using d DP1 FP1, we obtain the following sequence: e d DP1 FP1 a11 e2 a11 e2 e

substitute

a11 e2 e

for DP1 and a11 e2 for FP1

a11 e2 ae11 e2 e e

common denominator

a11 e211 e2 e

combine terms, factor out a11 e2

a11 e2 2 e de a11 e2 2

11 e211 e2 1 e 2 multiply by e

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Substituting a11 e2 2 for de in the standard equation r

de gives the 1 e cos ␪ 2 a11 e 2 . equation of the orbit entirely in terms of a and e: r 1 e cos ␪ EXAMPLE 9

䊳

Writing the Polar Equation of an Ellipse from Given Information At its aphelion, the Kuiper object Pluto is the most distant from the Sun at 4538 million miles. It has a perihelion of 2756 million miles. Use this information to find the polar equation Perihelion that models the orbit of Pluto, then find the length of the focal chord for this ellipse.

Solution

䊳

D. you’ve just seen how we can solve applications involving the conic sections in polar form

a c Center

Aphelion

With all figures in millions of miles, the major axis is 2a 4538 2756 7294, so the semimajor axis has length a 3647. With a c perihelion, we obtain 891 3647 c 2756 or c 891. The eccentricity of the orbit is e ⬇ 0.244. 3647 2 13647211 3 0.2444 2 The polar equation for the orbit of Pluto is r ⬇ or 1 3 0.2444 cos ␪ ␲ 3430 r⬇ . Substituting ␪ (since the left-most focus is at the pole), 1 0.244 cos ␪ 2 we obtain r 3430, so the length of the focal chord is 2134302 6860 million miles. Now try Exercises 65 through 70

䊳

10.7 EXERCISES 䊳



1. The set of points (x, y) in the xy-plane are related to points (X, Y) in the XY-plane by the formulas. To find the angle ␤ between the original axes and the rotated axes, we use tan12␤2 .

2. For a point P on the graph of a conic with focus F FP and D a point on the directrix, the ratio gives DP the of the graph. For the eccentricity e, if e 1 the graph is a , if e 7 1 the graph is a , and if 0 6 e 6 1 the graph will be an ellipse.

3. Features or relationships that do not change when certain transformations are applied are called of the transformation.

4. The

5. Discuss the advantages of graphing a rotated conic using the rotation of axes over graphing by simply plotting points.

6. Discuss the primary advantages of using a11 e2 2 de r rather than r to 1 e cos ␪ 1 e cos ␪ develop an equation of planetary orbit.

form of the equation of a conic de is r if the graph is symmetric to the 1 e cos ␪ de axis, and r if symmetric 1 e sin ␪ to the line .

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The graph of a conic rotated in the xy-plane is given. Use the graph (not the rotation of axes formulas) to find the equation of the conic in the XY-plane.

7.

y Y

X

For the given conics in the xy-plane, (a) use a rotation of axes to find the corresponding equation in the XY-plane (clearly state the angle of rotation ␤), and (b) sketch its graph. Be sure to indicate the characteristic features of each conic in the XY-plane.

21. x2 4xy y2 2 0 (2, 2)

22. x2 2xy y2 12 0 45

23. 5x2 6xy 5y2 16

x

24. 5x2 26xy 5y2 72

(2, 2)

25. x2 10 13xy 11y2 64 26. 37x2 42 13xy 79y2 400 0 8.

Y

27. 3x2 2 13xy y2 8x 813y 0

y X

28. 6x2 4 13xy 2y2 2x 2 13y 0

(3√3, 3)

29. 13x2 6 13xy 7y2 100 0

(√3, 3)

30. x2 4xy y2 12 0

30 x

(√3, 3)

(3√3, 3)

Given the point (x, y) in the xy-plane, find the coordinates of this point in the XY-plane given the angle ␤ between the xy-axes and the XY-axes is 45ⴗ.

9. 1612, 62

11. (0, 5)

10. 14, 3 122 12. (8, 0)

Given the point (X, Y) in the XY-plane, find the coordinates of this point in the xy-plane given the angle ␤ between the xy-axes and the XY-axes is 30ⴗ.

13. 12, 2132

14. 1 13, 32

15. (3, 4)

16. (12, 5)

The angle ␤ between the xy-axes and the XY-axes is given. Find the corresponding equation of the given conic sections in the xy-plane.

17. X2 Y2 9; 60°

18. X2 Y 4; 60°

The conic sections whose equations are given in the xy-plane are rotated by the indicated angle. What is the corresponding equation in the XY-plane?

19. 3x2 2xy 3y2 9; 45° 20. x2 13xy 2y2 8; 60°

Identify the graph of each equation using the discriminant, then find the value of cos12␤2 using sin12␤2 and the related triangle diagram. tan12␤2 ⴝ cos12␤2 Finally, find sin ␤ and cos ␤ using the half-angle 1 ⴙ cos12␤2 identities cos ␤ ⴝ and B 2 1 ⴚ cos12␤2 sin ␤ ⴝ . B 2

31. 12x2 24xy 5y2 40x 30y 25 32. 25x2 840xy 16y2 400 0 For the following equations, (a) use the discriminant to identify the equation as that of a circle, ellipse, parabola, or hyperbola; (b) find the angle of rotation ␤ and use it to find the corresponding equation in the XY-plane; and (c) verify all invariants of the transformation.

33. x2 2xy y2 5 0 34. 2x2 3xy 2y2 0 35. 3x2 13xy 4y2 4x 1 36. 3x2 8 13xy 5y2 12y 2

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10–88


Match each graph to its corresponding equation. Justify your answers (two equations have no match).

37.

2␲ 3

3␲ 4

ᏸ

6 5 4 ␲ 3 (3, 2 ) 2 1

5␲ 6

(1.5, ␲)

␲ 3

41.

2␲ 3

3␲ 4 5␲ 6

␲ 4

(3, ␲2 )

␲ 6

(3,

7␲ 6

(3,

7␲ 6

3␲ 2)

4␲ 3

7␲ 4

5␲ 3 ␲ 2

38. 2␲ ᏸ 3 3␲ 4

5 4 3 2 1

5␲ 6

(1.3, ␲)

␲ 6

(3, ␲)

11␲ 6

4␲ 3

42.

5␲ 3 ␲ 3

2␲ 3

3␲ 4

␲ 4

3␲ 2)

6 5 4 3 2 (1, ␲2 ) 1 (2, 0)

5␲ 6

␲ 3

␲ 3

5␲ 4

11␲ 6

5

5␲ 4

6 5 4 3 2 1

␲ 4 ␲ 6

7␲ 4

␲ 4 ␲ 6

ᏸ

(4, 0) 7␲ 6

7␲ 6 4␲ 3

39. 3␲ 4 5␲ 6

(4,

a. r

␲ 3

2␲ 3 3␲ 2 )

6 5 4 3 2 1 (0.8, ␲2 )

␲ 4

c. r ␲ 6

e. r g. r

7␲ 6

11␲ 6 5␲ 4 4␲ 3

40. 3␲ 4

ᏸ

4␲ 3

7␲ 4

5␲ 3

5␲ 6

2␲ 3

4 3 ␲ 2 (0.84, 2 ) 1 (1.4, ␲)

5␲ 3 ␲ 3

7␲ 4

␲ 4 ␲ 6

(1.4, 0)

7␲ 6

11␲ 6 5␲ 4

4␲ 3

5␲ 3

5

5␲ 4

11␲ 6 5␲ 4

11␲ 6

7␲ 4

5 3 4 2

5␲ 3

10 5 sin ␪ 5.4 2 sin ␪ 12 6 sin ␪ 4 3 sin ␪

7␲ 4

b. r d. r f. r h. r

4 3 2 3

8 2 cos ␪ 4.2 2 sin ␪ 6 2 cos ␪ 9 6 cos ␪

For the conic equations given, determine if the equation represents a parabola, ellipse, or hyperbola. Then describe and sketch the graphs using polar graph paper. Verify your sketch with a graphing calculator.

43. r

4 2 2 sin ␪

44. r

10 5 5 sin ␪

45. r

12 6 3 sin ␪

46. r

6 4 3 cos ␪

47. r

6 2 4 cos ␪

48. r

2 2 3 sin ␪

49. r

5 5 4 cos ␪

50. r

2 4 5 sin ␪

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Write the equation of a conic that satisfies the conditions given. Assume each has one focus at the pole. Note for Exercises 51, 52, and 56, there are four possible equations.

52. hyperbola, e 1.25, directrix to focus: d 6

䊳

54. ellipse, e 0.35, vertex at (4, 0) ␲ b 2

56. parabola, directrix to focus: d 5.4

WORKING WITH FORMULAS 58. Polar form of an ellipse with center at the pole:

57. Equation of a line in polar form: C rⴝ A cos ␪ ⴙ B sin ␪

r2 ⴝ

For the line Ax By C in the xy-plane with A C slope m and y-intercept a0, b, the B B corresponding equation in the r␪-plane is given by the formula shown. (a) Given the line 2x 3y 12 in the xy-plane, find the corresponding polar equation and (b) verify r1␲/22 A . that B r102 䊳

53. parabola, vertex at 12, ␲2

55. hyperbola, e 1.5, vertex at a3,

51. ellipse, e 0.8, directrix to focus: d 4

1049

a 2b 2 a 2sin 2␪ ⴙ b 2cos 2␪

If an ellipse in the r␪-plane has its center at the pole (with major axis parallel to the x-axis), its equation is given by the formula here, where 2a and 2b are the lengths of the major and minor axes, respectively. (a) Given an ellipse with center at the pole has a major axis of length 8 and a minor axis of length 4, find the equation of the ellipse in polar form and (b) graph the result on a calculator and verify that 2a 8 and 2b 4.

APPLICATIONS

Planetary motion: The perihelion, aphelion, and orbital period of the planets Jupiter, Saturn, Uranus, and Neptune are shown in the table. Use the information to answer or complete the following exercises. The formula L 2␲ 20.51a2 b2 2 can be used to estimate the length of the orbital path. Recall for an ellipse, c2 a2 b2.

Planet

Perihelion (106 mi)

Aphelion (106 mi)

Period (yr)

Jupiter

460

507

11.9

Saturn

840

941

29.5

Uranus

1703

1866

84

Neptune

2762

2824

164.8

59. Find the eccentricities of the planets Jupiter and Saturn. 60. Find the eccentricities of the planets Uranus and Neptune. 61. The orbit of Pluto (a Kuiper object) has a semimajor axis of 3647 million miles and an eccentricity of e 0.2443. Find the perihelion of Pluto. 62. The orbit of Ceres (a large asteroid) has a semimajor axis 257 million miles and an eccentricity of e 0.097. Find the perihelion of Ceres.

63. Which of the four planets in the table given has the greatest orbital eccentricity? 64. Which of these four planets has the greatest orbital velocity? 65. Find the polar equation modeling the orbit of Jupiter. 66. Find the polar equation modeling the orbit of Saturn. 67. Find the polar equation modeling the orbit of Uranus. 68. Find the polar equation modeling the orbit of Neptune. 69. Suppose all four major planets arrived at the focal ␲ chord of their orbit a␪ b simultaneously. Use 2 the equations in Exercises 65 to 68 to determine the distance between each of the planets at this moment. 70. The polar equation for the orbit of Pluto (a Kuiper object) was developed in Example 9. From an earlier exercise, the polar equation for the orbit of 2793 Neptune is r ⬇ . Using the TABLE 1 0.011 cos ␪ of your graphing calculator, determine if Pluto is always the farthest planet from the Sun. If not, how much further from the Sun is Neptune than Pluto at their perihelion?

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Mirror manufacturing: A modern manufacturer of oval (elliptical) mirrors for consumer use has programmed the equipment to automatically cut the glass for each mirror (major axis horizontal). The most popular mirrors are those that fit within a golden rectangle (ratio of L to W is approximately 1 to 0.618). Find the polar equation the manufacturer should use to program the equipment for mirror orders of the following lengths. Recall that c2 a2 b2 c and e and assume one focus is at the pole. a

71. L 4 ft

72. L 3.5 ft

73. L 1.5 m

74. L 0.5 m

75. Referring to Exercises 71 to 74, find the total cost of each mirror (to the consumer) if they sell for $75 per square foot ($807 per square meter). The area of an ellipse is given by A ␲ab. 76. Referring to Exercises 71 to 74, find the total cost of an elliptical frame for each mirror (to the consumer) if the frame sells for $12.50 per linear foot ($41.01 per meter). The circumference of an ellipse is approximated by C ␲ 221a2 b2 2. 77. Home location: Candice is an enthusiastic golfer and an avid swimmer. After being transferred to a new city, she decides to buy a house that is an equal distance from the local golf course and the river running through the city. If the distance between the river and the 䊳

10–90


golf course at the closest point is 3 mi, find the polar equation of the parabola that will trace through the possible locations for her new home. Assume the golf course is at the focus of the parabola.

Exercise 77 R

d

H Home

River d ␪ G B Golf course

78. Home location: Referring to Exercise 77, assume Candice finds the perfect dream house in a ␲ subdivision located at a6, b. Does this home fit 3 the criteria (is it an equal distance from the river and golf course)? 79. Solve the system below for y to verify the rotation formula for y given on page 1037. e

X x cos ␤ y sin ␤ Y y cos ␤ x sin ␤

80. Rotation of a conic section: Expand the following, collect like terms, and simplify. Show the result is the equation aX2 bXY cY2 f 0, where the coefficients a, b, c, and f are as given on page 1039. A1X cos ␤ Y sin ␤2 2 B1X cos ␤ Y sin ␤2 1X sin ␤ Y cos ␤2 C1X sin ␤ Y cos ␤2 2 F0


81. Using the rotation of axes formulas in the general equation Ax2 Bxy Cy2 F 0 1D E 02, we were able to obtain the equation aX2 bXY cY2 f 0 (see page 1037), where a S A cos2␤ B sin ␤ cos ␤ C sin2␤ b S 2A sin ␤ cos ␤ B1cos2␤ sin2␤2 2C sin ␤ cos ␤ c S A sin2␤ B sin ␤ cos ␤ C cos2␤ and f S F a. Use these to verify b2 4ac B2 4AC.

b. Use these to verify a c A C.

82. A short-period comet is one that orbits the Sun in 200 yr or less. Two of the best known are Halley’s Comet and Encke’s Comet. Using any of the resources available to you, find the perihelion and aphelion of each comet and use the information to find the lengths of the semimajor and semiminor axes. Also find the period of each comet. If the length of an elliptical (orbital) path is approximated by L 2␲20.51a2 b2 2, find the approximate average speed of each comet in miles per hour. Finally, determine the polar equation of each orbit.

c. Explain why the invariant f F must always hold.

83. In the r␪-plane, the equation of a circle having radius R, center at 1R, ␤2, and going through the pole is given by r 2R cos1␪ ␤2. Consider the circle defined by x2 y2 6 12x 6 12y 0 in the xy-plane. Verify this circle goes through the origin, then find the equation of the circle in polar form.

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Section 10.8 Parametric Equations and Graphs

1051

For the given conics in the xy-plane, use a rotation of axes to find the corresponding equation in the XY-plane. See Exercises 31 and 32.

84. 12x2 24xy 5y2 40x 30y 25

85. 25x2 840xy 16y2 400 0

86. A right triangle in the xy-plane had vertices at (0, 0), (8, 0), and (8, 6). Use the matrix equation X cos ␤ sin ␤ # x c d c d c d to find the vertices y Y sin ␤ cos ␤ in the XY-plane after the triangle is rotated 60°.

87. A square in the XY-plane has vertices at (0, 0), 12 13, 22, 12 13 2, 2 2 132 and 12, 2132. Use the matrix equation x cos ␤ sin ␤ # X c d c d c d to find the vertices Y y sin ␤ cos ␤ in the xy-plane after the triangle is rotated 30°.

䊳


88. (9.2) Solve the system using elimination. x 2y z 3 • 2x 6y z 4 5x 4y 2z 3 89. (5.5) Solve for x (to the nearest tenth): 21.7 77.5e0.0052x 44.95 90. (6.5) Use the graph shown to write an equation of the form y A sec1Bx C2. Clearly state the values of A, B, and C.

91. (8.3) A ship is moving at 12 mph on a heading of 325°, with a 5 mph current flowing at a 100° heading. Find the true course and speed of the ship. N

12 mph ship

y 5

100 325 ␲

␲

␲ 2

2

␲

3␲ 2

x

5 mph current

5

10.8

Parametric Equations and Graphs


A. Sketch the graph of a parametric equation B. Write parametric equations in rectangular form C. Graph curves from the cycloid family D. Solve applications involving parametric equations

A large portion of the mathematics curriculum is devoted to functions, due to their overall importance and widespread applicability. But there are a host of applications for which nonfunctions are a more natural fit. In this section, we show that many nonfunctions can be expressed as parametric equations, where each is actually a function. These equations can be appreciated for the diversity and versatility they bring to the mathematical spectrum.

A. Sketching a Curve Defined Parametrically Suppose you were given the set of points in the table on the next page, and asked to come up with an equation model for the data. To begin, you might plot the points to see if any patterns or clues emerge, but in this case the result seems to be a curve we’ve never seen before (see Figure 10.101).

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Figure 10.101 1

y

⫺1

1

Figure 10.102

⫺1

y

Lissajous figure

1

0

13 2

13 2

0

y

1

13 2

1 2

0

⫺

13 2

⫺

13 2

0

1 2

⫺

13 2

⫺1

⫺

x

⫺1

1

x

x

⫺1

You also might consider running a regression on the data, but it’s not possible since the graph is obviously not a function. However, a closer look at the data reveals the y-values could be modeled independently of the x-values by a cosine function, y ⫽ cos t for t 僆 3 0, ␲ 4. This observation leads to a closer look at the x-values, which we find could be modeled by a sine function over the same interval, namely, x ⫽ sin12t2 for t 僆 3 0, ␲ 4. These two functions combine to name all points on this curve, and both use the independent variable t called a parameter. The functions x ⫽ sin12t2 and y ⫽ cos t are called the parametric equations for this curve. The complete curve, shown in Figure 10.102, is called a Lissajous figure, or a closed graph (coincident beginning and ending points) that crosses itself to form two or more loops. Note that since the maximum value of x and y is 1 (the amplitude of each function), the entire figure will fit within a 2 ⫻ 2 rectangle centered at the origin. This observation can often be used to help sketch parametric graphs with trigonometric parameters. In general, parametric equations can take many forms, including polynomial, exponential, trigonometric, and other forms. Parametric Equations

Given the set of points P(x, y) such that x ⫽ f 1t2 and y ⫽ g1t2, where f and g are both defined on an interval of the domain, the equations x ⫽ f 1t2 and y ⫽ g1t2 are called parametric equations, with parameter t.

EXAMPLE 1

䊳

Graphing a Parametric Curve Where f and g Are Algebraic Graph the curve defined by the parametric equations x ⫽ t2 ⫺ 3 and y ⫽ 2t ⫹ 1.

Solution

䊳

Begin by creating a table of values using t 僆 3⫺3, 3 4. After plotting ordered pairs (x, y), the result appears to be a parabola, opening to the right. y

x ⴝ t2 ⴚ 3

t

x

y ⴝ 2t ⴙ 1

⫺3

6

⫺5

⫺2

1

⫺3

⫺1

⫺2

⫺1

0

⫺3

1

1

⫺2

3

2

1

5

3

6

7

Now try Exercises 7 through 12, part a 䊳

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Figure 10.103 Most graphers have a parametric MODE that enables you to enter the equations for x and y separately, and graph the resulting points as a single curve. In this mode, the variable key X,T,␪,n produces a T when pressed, and the Y= screen prompts you for both XiT and YiT. Verify this by entering X1T ⫽ T2 ⫺ 3 and Y1T ⫽ 2T ⫹ 1 from Example 1 (Figure 10.103). The screen for parametric equations is similar to the screen for polar equations, and as before, we can use the preset ZOOM options or manually set our Figure 10.104 desired window size along with Tmin, Tmax, 7.6 and Tstep. To verify the Example 1 result and produce the graph shown in Figure 10.104, size the window as shown and set Tmin ⫽ ⫺3, Tmax ⫽ 3, and Tstep ⫽ 0.1. To see more of the ⫺10 10 parabola, we must change the window and increase the range of T-values the calculator uses. If the parametric equations contain trig functions, we’ll often use standard angles as in⫺5.6 puts to simplify calculations and the period of the function(s) to help sketch the resulting graph. Also note that successive values of t give rise to a directional evolution of the graph, meaning the curve is traced out in a direction dictated by the points that correspond to the next value of t. The arrows drawn along the graph illustrate this direction, also known as the orientation of the graph. WINDOW

EXAMPLE 2

䊳

Graphing a Parametric Curve Where f and g Are Trig Functions Graph the curve defined by the parametric equations x ⫽ 4 cos t and y ⫽ 2 sin t. Use a calculator to verify your graph.

Solution WORTHY OF NOTE Since trig functions are used so frequently in parametric curves, the ZOOM 6:Zstandard feature automatically sets 30, 2␲ 4 as the range of T-values.

䊳

Using standard angle inputs and knowing the maximum value of any x- and y-coordinate will be 4 and 2, respectively, we begin computing and graphing a few points. After going from 0 to ␲, we note the graph appears to be a horizontal ellipse. This is verified using standard values from ␲ to 2␲. Plotting the points and connecting them with a smooth curve produces the ellipse shown in Figure 10.105. Note the ellipse has a counterclockwise orientation. Figure 10.105 y

x

t

x ⴝ 4 cos t

y ⴝ 2 sin t

0

4

0

␲ 6

213

1

␲ 3

2

13

␲ 2

0

2

2␲ 3

⫺2

13

5␲ 6

⫺213

1

␲

⫺4

0

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With your calculator in radian MODE , enter X1T ⫽ 4 cos T and Y1T ⫽ 2 sin T in the Y= screen. With Tmin ⫽ 0, Tmax ⫽ 2␲, ␲ and Tstep ⫽ , the ZOOM 4:ZDecimal 24 feature will produce the graph shown in Figure 10.106, verifying our sketch.

Figure 10.106 3.1

⫺4.7

4.7

⫺3.1

A. You’ve just seen how we can sketch the graph of a parametric equation

Now try Exercises 13 through 18, part a

䊳

B. Writing Parametric Equations in Rectangular Form When graphing parametric equations, there are sometimes alternatives to simply plotting points. One alternative is to try and eliminate the parameter, writing the parametric equations in standard, rectangular form. To accomplish this we use some connection that enables us to “rejoin” the parameterized equations, such as variable t itself, a trigonometric identity, or some other connection.

EXAMPLE 3

䊳

Eliminating the Parameter to Obtain the Rectangular Form Eliminate the parameter from the equations in Example 1: x ⫽ t2 ⫺ 3 and y ⫽ 2t ⫹ 1.

Solution

䊳

Solving for t in the second equation gives t ⫽ the first. The result is x ⫽ a

y⫺1 , which we then substitute into 2

y⫺1 2 1 b ⫺ 3 ⫽ 1y ⫺ 12 2 ⫺ 3. Notice this is indeed a 2 4

horizontal parabola, opening to the right, with vertex at 1⫺3, 12.

Now try Exercises 7 through 12, part b

EXAMPLE 4

䊳

䊳

Eliminating the Parameter to Obtain the Rectangular Form Eliminate the parameter from the equations in Example 2: x ⫽ 4 cos t and y ⫽ 2 sin t.

Solution

䊳

Instead of trying to solve for t, we note the parametrized equations involve sine and cosine functions with the same argument (t), and opt to use the identity cos2t ⫹ sin2t ⫽ 1. Squaring both equations and solving for cos2t and sin2t yields y2 y2 x2 x2 ⫽ cos2t and ⫽ sin2t. This shows cos2t ⫹ sin2t ⫽ ⫹ ⫽ 1, and as we 16 4 16 4 suspected — the result is a horizontal ellipse with vertices at 1⫾4, 02 and endpoints of the minor axis at 10, ⫾22 . Now try Exercises 13 through 16, part b

䊳

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1055

It’s important to realize that a given curve can be represented parametrically in infinitely many ways. This flexibility sometimes enables us to simplify the given form, or to write a given polynomial form in an equivalent nonpolynomial form. The easiest way to write the function y ⫽ f 1x2 in parametric form is x ⫽ t; y ⫽ f 1t2, which is valid as long as t is in the domain of f (t). EXAMPLE 5

䊳

Writing an Equation in Terms of Various Parameters Write the equation y ⫽ 41x ⫺ 32 2 ⫹ 1 in three different parametric forms.

Solution

B. You’ve just seen how we can write parametric equations in rectangular form

䊳

1. If we let x ⫽ t, we have y ⫽ 41t ⫺ 32 2 ⫹ 1. 2. Letting x ⫽ t ⫹ 3 simplifies the related equation for y, and we begin to see some of the advantages of using a parameter: x ⫽ t ⫹ 3; y ⫽ 4t2 ⫹ 1. 1 3. As a third alternative, we can let x ⫽ tan t ⫹ 3, which gives 2 2 1 1 ␲ ␲ x ⫽ tan t ⫹ 3; y ⫽ 4 a tan tb ⫹ 1 ⫽ tan2t ⫹ 1 or y ⫽ sec2t for t 僆 a⫺ , b. 2 2 2 2 Now try Exercises 19 through 26

䊳

C. Graphing Curves from the Cycloid Family The cycloids are an important family of curves, and are used extensively to solve what are called brachistochrone applications. The name comes from the Greek brakhus, meaning short, and khronos, meaning time, and deal with finding the path along which a weight will fall in the shortest time possible. Cycloids are an excellent example of why parametric equations are important, as it’s very difficult to write them in rectangular form. Consider a point fixed to the circumference of a wheel as it rolls from left to right. If we trace the path of the point as the wheel rolls, the resulting curve is a cycloid. Figure 10.107 shows the location of the point every one-quarter turn. Figure 10.107

By superimposing a coordinate grid on the diagram in Figure 10.107, we can construct parametric equations that will produce the graph. This is done by developing equations for the location of a point P(x, y) on the circumference of a circle with center (h, k), as the circle rotates through angle t. After a rotation of t rad, the x-coordinate of P(x, y) is x ⫽ h ⫺ a (Figure 10.108), and the y-coordinate is y ⫽ k ⫺ b. Using a right triangle with the radius as the hypotenuse, we a b find sin t ⫽ and cos t ⫽ , giving r r

Figure 10.108 y

(h, k) r k

P(x, y)

t

b

a

h

x

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a ⫽ r sin t and b ⫽ r cos t. Substituting into x ⫽ h ⫺ a and y ⫽ k ⫺ b yields x ⫽ h ⫺ r sin t and y ⫽ k ⫺ r cos t. Since the circle has radius r, we know k ⫽ r (the “height” of the center is constantly k ⫽ r). The arc length subtended by t is the same as the distance h (see Figure 10.108), meaning h ⫽ rt (t in radians) Substituting rt for h and r for k in the equations x ⫽ h ⫺ r sin t and y ⫽ k ⫺ r cos t, gives the equation of the cycloid in parametric form: x ⫽ rt ⫺ r sin t and y ⫽ r ⫺ r cos t, sometimes written x ⫽ r 1t ⫺ sin t2 and y ⫽ r 11 ⫺ cos t2. After entering the equation of the cycloid formed by a circle of radius r ⫽ 3 (Figure 10.109), set the viewing window with Ymin ⫽ ⫺1 and Ymax at slightly more than 6 (since r ⫽ 3). Since the cycloid completes one cycle every 2␲r, we set Xmax at 2␲rn, where n is the number of cycles we’d like to see. In this case, we set it for four cycles 12␲2132142 ⫽ 24␲ (Figure 10.110). With r ⫽ 3 we can set Xscl at 312␲2 ⫽ 6␲ ⬇ 18.8 to tick each cycle, or Xscl ⫽ 3␲ ⬇ 9.4 to tick each half cycle (Figure 10.110). We also set Tmin ⫽ 0, Tmax ⫽ 8␲ ⬇ 25.1 for the four cycles, and ␲ Tstep ⫽ ⬇ 0.52. The resulting graph is shown in Figure 10.111, which doesn’t look 6 much like a cycloid because the current settings do not produce a square viewing window. Using ZOOM 5:ZSquare (and changing Yscl) produces the graph shown in Figure 10.112, which looks much more like the cycloid we expected.

Figure 10.109

Figure 10.110

Figure 10.111

Figure 10.112

7

28.9

⫺␲

24␲

⫺␲

24␲

⫺22.9

⫺1

䊳

EXAMPLE 6

Using Technology to Graph a Cycloid Use a graphing calculator to graph the curve defined by the equations x ⫽ 3 cos3t and y ⫽ 3 sin3t, called a hypocycloid with four cusps.

䊳

Solution Figure 10.113 r y

A hypocycloid is a curve traced out by the path of a point on the circumference of a circle as it rolls inside a larger circle of radius r (see Figure 10.113). Here r ⫽ 3 and we set Xmax and Ymax accordingly. Knowing ahead of time the hypocycloid will have four cusps, we set Tmax ⫽ 412␲2 ⬇ 25.13 to show all four. The window settings used and the resulting graph are shown in Figures 10.114 and 10.115. Figure 10.115

x

⫺r

Figure 10.114

5

r

⫺r

C. You’ve just seen how we can graph curves from the cycloid family

⫺5

5

⫺5


䊳

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1057


D. Common Applications of Parametric Equations In Example 1 the parameter was simply the real number t, which enabled us to model the x- and y-values of an ordered pair (x, y) independently. In Examples 2 and 6, the parameter t represented an angle. Here we introduce yet another kind of parameter, that of time t. A projectile is any object thrown, dropped, or projected in some way with no continuing source of propulsion. The parabolic path traced out by the projectile (assuming negligible air resistance) was fully developed in Section 8.4. It is stated again here in parametric terms. For the projectile’s location P(x, y) and any time t in seconds, the x-coordinate (horizontal distance from point of projection) is given by x ⫽ v0t cos ␪, where v0 is the initial velocity in feet per second and t is the time in seconds. The y-coordinate (vertical height) is y ⫽ v0t sin ␪ ⫺ 16t2. EXAMPLE 7

䊳

Using Parametric Equations in Projectile Applications As part of a circus act, Karl the Human Cannonball is shot out of a specially designed cannon at an angle of 40° with an initial velocity of 120 ft/sec. Use a graphing calculator to graph the resulting parametric curve. Then use the graph to determine how high the ringmaster must place a circular ring for Karl to be shot through at the maximum height of his trajectory, and how far away the net must be placed to catch Karl.

Solution

䊳

The information given leads to the equations x ⫽ 120t cos 40° and y ⫽ 120t sin 40° ⫺ 16t2. Enter these equations on the Y= screen of your calculator, remembering to reset the MODE to degrees (circus clowns may not know or understand radians). To set the window size, we can use trial and error, or estimate using ␪ ⫽ 45° (instead of 40°) and an estimate for t (the time that Karl 12 b ⫽ 36012 for will stay aloft). With t ⫽ 6 we get estimates of x ⫽ 120162 a 2 the horizontal distance. To find a range for y, use t ⫽ 3 since the maximum height of the parabolic path will occur halfway through the flight. This gives an 12 b ⫺ 16192 ⫽ 180 12 ⫺ 144 for y. The results are shown in estimate of 120132 a 2 Figures 10.116 and 10.117. Using the TRACE feature or 2nd GRAPH (TABLE) feature, we find the center of the net used to catch Karl should be set at a distance of about 440 ft from the cannon, and the ring should be located 220 ft from the cannon at a height of about 93 ft. Figure 10.117 Figure 10.116

100

0

509

0


䊳

It is well known that planets orbit the Sun in elliptical paths. While we’re able to model their orbits in both rectangular and polar form, neither of these forms can give a true picture of the direction they travel. This gives parametric forms a great advantage, in that they can model the shape of the orbit, while also indicating the direction of travel. We illustrate in Example 8 using a “planet” with a very simple orbit.

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EXAMPLE 8

䊳

Modeling Elliptical Orbits Parametrically The elliptical orbit of a certain planet is defined parametrically as x ⫽ 4 sin t and y ⫽ ⫺3 cos t. Graph the orbit and verify that for increasing values of t, the planet orbits in a counterclockwise direction.

Solution

䊳

y2 x2 ⫹ ⫽ 1, 16 9 or the equation of an ellipse with center at (0, 0), major axis of length 8, and minor axis of length 6. The path of the planet is traced out by the ordered pairs (x, y) generated by the parametric equations, shown in the table for t 僆 30, ␲4 . Starting at ␲ t ⫽ 0, P(x, y) begins at 10, ⫺32 with x and y both increasing until t ⫽ . Then from 2 ␲ t ⫽ to t ⫽ ␲, y continues to increase as x decreases, indicating a counterclockwise 2 orbit in this case. The orbit is illustrated in the figure. Eliminating the parameter as in Example 4, we obtain the equation

y

(0, ⫺3)

t

x ⴝ 4 sin t

y ⴝ ⴚ3 cos t

0

0

⫺3

2

(4, 0)

␲ 6

⫺2.6

x (3.46, ⫺1.5)

␲ 3

3.46

⫺1.5

␲ 2

4

0

2␲ 3

3.46

1.5

5␲ 6

2

2.6

␲

0

3

(2, ⫺2.6)


䊳

Most graphing calculators have features that make it easy (and fun) to explore parametric equations by using a circular cursor to trace the path of the plotted points, as they are generated by the equations. This can be used to illustrate the path of a projectile, the distance of a runner, or the orbit of a planet. Operations can also be applied to the parameter T to give the effect of “speed” (the points from one set of equations are plotted faster than the points of a second set). To help illustrate their use, consider the simple, elliptical orbit of a planet in Example 8. Physics tells us the closer a planet is to the Sun, the faster its orbit. In fact, the orbital speed of Mercury is about twice that of Mars and about 10 times as fast as the Kuiper object Pluto (29.8, 15, and 2.9 mi/sec, respectively). With this information, we can explore a number of interesting questions (see Exercises 52 and 53). On the Y= screen of a calculator in radian MODE , let the orbits of Planet 1 and Planet 2 be modeled parametriFigure 10.118 cally by the equations shown in Figure 10.118. Since the orbit of Planet 1 is “smaller” (closer to the Sun), we have T-values growing at a rate that is four times as fast as for Planet 2. Notice to the far left of X1T, there is a symbol that looks like an old key “⫺0.” By moving the cursor to the far left of the equation, you can change how the graph will look by repeatedly pressing . With this symbol in view, the calculator will trace ENTER

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10

⫺10

10

⫺10

D. You’ve just seen how we can solve applications involving parametric equations

out the curve with a circular cursor, which in this case represents the planets as they orbit (be sure you are in simultaneous MODE ). Setting the window as in Figure 10.119 and pressing GRAPH produces Figure 10.120, which displays their elliptical paths as they race around the Sun. Notice the inner planet has already completed one orbit while the outer planet has just completed one-fourth of an orbit. Finally, you may recall from your previous work with linear 3 ⫻ 3 systems, that a dependent system occurs when one of the three equations is a linear combination of the other two. The result is a system with more variables than equations, with solutions expressed in terms of a parameter, or in parametric form. These solutions can be explored on a graphing calculator using ordered triples of the form (t, f(t), g(t)), where X1T ⫽ f 1t2 and Y1T ⫽ g1t2 (see Exercises 54 through 57).

10.8 EXERCISES 䊳



䊳

1. When the coordinates of a point (x, y) are generated independently using x ⫽ f 1t2 and y ⫽ g1t2 , t is called a(n) .

2. The equations x ⫽ f 1t2 and y ⫽ g1t2 used to generate the ordered pairs (x, y) are called equations.

3. Parametric equations can both graph a curve and indicate the traveled by a point on the curve.

4. To write parametric equations in rectangular form, we must the parameter to write a single equation.

5. Discuss the connection between solutions to dependent systems and the parametric equations studied in this section.

6. In your own words, explain and illustrate the process used to develop the equation of a cycloid. Illustrate with a specific example.


For Exercises 7 through 18, (a) graph the curves defined by the parametric equations using the specified interval and identify the graph (if possible) and (b) eliminate the parameter (Exercises 7 to 16 only) and write the corresponding rectangular form.

7. x ⫽ t ⫹ 2; t 僆 3⫺3, 3 4 y ⫽ t2 ⫺ 1

8. x ⫽ t ⫺ 3; t 僆 3 ⫺5, 54 y ⫽ 2 ⫺ 0.5t2

9. x ⫽ 12 ⫺ t2 2; t 僆 30, 5 4 y ⫽ 1t ⫺ 32 2

10. x ⫽ t3 ⫺ 3; t 僆 3⫺2, 2.5 4 y ⫽ t2 ⫹ 1

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5 11. x ⫽ , t ⫽ 0; t 僆 3⫺3.5, 3.5 4 t y ⫽ t2 t3 12. x ⫽ ; t 僆 3 ⫺5, 5 4 10 y ⫽ 冟t冟

15. x ⫽ 4 sin12t2; t 僆 3 0, 2␲2 y ⫽ 6 cos t ␲ 3␲ 16. x ⫽ 4 cos12t2; t 僆 c , d 2 2 y ⫽ 6 sin t ⫺3 ; t 僆 10, ␲2 tan t y ⫽ 5 sin12t2

17. x ⫽

␲ , t 僆 3 0, ␲ 4 2

Write each function in three different parametric forms by altering the parameter. For Exercises 19–22 use at least one trigonometric form, restricting the domain as needed.

19. y ⫽ 3x ⫺ 2

21. y ⫽ 1x ⫹ 32 2 ⫹ 1

23. y ⫽ tan2 1x ⫺ 22 ⫹ 1

2 , y ⫽ 8 sin t cos t, serpentine curve tan t 8 sin3t , cissoid of Diocles cos t

31. x ⫽ 21cos t ⫹ t sin t2, y ⫽ 21sin t ⫺ t cos t2, involute of a circle

14. x ⫽ 2 sin t; t 僆 3 0, 2␲2 y ⫽ ⫺3 cos t

y ⫽ 3 cos t

29. x ⫽

30. x ⫽ 8 sin2t, y ⫽

13. x ⫽ 4 cos t; t 僆 3 0, 2␲2 y ⫽ 3 sin t

18. x ⫽ tan2t; t ⫽

28. x ⫽ 8 cos t ⫹ 4 cos12t2, y ⫽ 8 sin t ⫺ 4 sin12t2, hypocycloid (3-cusp)

20. y ⫽ 0.5x ⫹ 6 22. y ⫽ 21x ⫺ 52 2 ⫺ 1 24. y ⫽ sin12x ⫺ 12

25. Use a graphing calculator or computer to verify that the parametric equations from Example 5 all produce the same graph. 26. Use a graphing calculator or computer to verify that your parametric equations from Exercise 21 all produce the same graph. The curves defined by the following parametric equations are from the cycloid family. (a) Use a graphing calculator or computer to draw the graph and (b) use the graph to approximate all x- and y-intercepts, and maximum and minimum values to one decimal place.

27. x ⫽ 8 cos t ⫹ 2 cos14t2, y ⫽ 8 sin t ⫺ 2 sin14t2, hypocycloid (5-cusp)

32. x ⫽ ⫺5 cos3t, y ⫽ ⫺

10 3 sin t, evolute of an ellipse 3

33. x ⫽ 3t ⫺ sin t, y ⫽ 3 ⫺ cos t, curtate cycloid 34. x ⫽ t ⫺ 3 sin t, y ⫽ 1 ⫺ 3 cos t, prolate cycloid

35. x ⫽ 2 33 cos t ⫺ cos13t2 4 , y ⫽ 2 33 sin t ⫺ sin13t2 4 , nephroid Use a graphing calculator or computer to draw the following parametrically defined graphs, called Lissajous figures (Exercise 37 is a scaled version of Figure 10.102). Then find the dimensions of the rectangle necessary to frame the figure and state the number of times the graph crosses itself.

36. x ⫽ 6 sin13t2 y ⫽ 8 cos t

37. x ⫽ 6 sin12t2 y ⫽ 8 cos t

38. x ⫽ 8 sin14t2 y ⫽ 10 cos t

39. x ⫽ 5 sin17t2 y ⫽ 7 cos14t2

40. x ⫽ 8 sin14t2 y ⫽ 10 cos13t2

41. x ⫽ 10 sin11.5t2 y ⫽ 10 cos12.5t2

42. Use a graphing calculator to experiment with parametric equations of the form x ⫽ A sin1mt2 and y ⫽ B cos1nt2. Try different values of A, B, m, and n, then discuss their effect on the Lissajous figures. 43. Use a graphing calculator to experiment with a parametric equations of the form x ⫽ and tan t y ⫽ b sin t cos t. Try different values of a and b, then discuss their effect on the resulting graph, called a serpentine curve. Also see Exercise 29.

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44. The Folium of Descartes: 3kt 3kt2 x1t2 ⴝ ; y1t2 ⴝ 1 ⴙ t3 1 ⴙ t3 The Folium of Descartes is a parametric curve developed by Descartes in order to test the ability of Fermat to find its maximum and minimum values. a. Graph the curve on a graphing calculator with k ⫽ 1 using a decimal window ( ZOOM 4), with Tmin ⫽ ⫺6, Tmax ⫽ 6, and Tstep ⫽ 0.1. Locate the coordinates of the tip of the folium (the loop). b. This graph actually has a discontinuity (a break in the graph). At what value of t does this occur? c. Experiment with different values of k and generalize its effect on the basic graph.

䊳


45. The Witch of Agnesi: x1t2 ⴝ 2kt; y1t2 ⴝ

2k 1 ⴙ t2

The Witch of Agnesi is a parametric curve named by Maria Agnesi in 1748. Some believe various translations confused the Italian word for witch with a similar word that means curve, or turning. In any case, the name stuck. The curve can also be stated in trigonometric form: x1t2 ⫽ 2k cot t and y ⫽ 2k sin2t. a. Graph the curve with k ⫽ 1 on a calculator or computer on a decimal window ( ZOOM 4) using both of the forms shown with Tmin ⫽ ⫺6, Tmax ⫽ 6, and Tstep ⫽ 0.1. Try to determine the maximum value. b. Explain why the x-axis is a horizontal asymptote. c. Experiment with different values of k and generalize its effect on the basic graph.

APPLICATIONS

Model each application using parametric equations, then solve using the GRAPH and TRACE features of a graphing calculator.

46. Archery competition: At an archery contest, a large circular target 5 ft in diameter is laid flat on the ground with the bull’s-eye exactly 180 yd (540 ft) away from the archers. Marion draws her bow and shoots an arrow at an angle of 25° above horizontal with an initial velocity of 150 ft/sec (assume the archers are standing in a depression and the arrow is shot from ground level). (a) What was the maximum height of the arrow? (b) Does the arrow hit the target? (c) What is the distance between Marion’s arrow and the bull’s-eye after the arrow hits?

47. Football competition: As part of their contribution to charity, a group of college quarterbacks participate in a contest. The object is to throw a football through a hoop whose center is 30 ft high and 25 yd (75 ft) away, trying to hit a stationary (circular) target laid on the ground with the center 56 yd (168 ft) away. The hoop and target both have a diameter of 4 ft. On his turn, Lance throws the football at an angle of 36° with an initial velocity of 75 ft/sec. (a) Does the football make it through the hoop? (b) Does the ball hit the target? (c) What is the approximate distance between the football and the center of the target when the ball hits the ground? 48. Walk-off home run: It’s the bottom of the ninth, two outs, the count is full, and the bases are loaded with the opposing team ahead 5 to 2. The home team has Heavy Harley, their best hitter at the plate; the opposition has Raymond the Rocket on the mound. Here’s the pitch . . . it’s a hit . . . a long fly ball to left-center field! If the ball left the bat at an angle of 30° with an initial velocity of 112 ft/sec, will it clear the home run fence, 9 ft high and 320 ft away?

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49. Last-second win: It’s fourth-and-long, late in the fourth quarter of the homecoming football game, with the home team trailing 29 to 27. The coach elects to kick a field goal, even though the goal posts are 50 yd (150 ft) away from the spot of the kick. If the ball leaves the kicker’s foot at an angle of 29° with an initial velocity of 80 ft/sec, and the kick is “true,” will the home team win (does the ball clear the 10-ft high cross bar)?

50 yards (150 feet)

50. Particle motion: The motion of a particle is modeled by the parametric equations x ⫽ 5t ⫺ 2t2 e . Between t ⫽ 0 and t ⫽ 1, is the y ⫽ 3t ⫺ 2 particle moving to the right or to the left? Is the particle moving upward or downward? 51. Electron motion: The motion of an electron as it orbits the nucleus is modeled by the parametric x ⫽ 6 cos t equations e with t in radians. Between y ⫽ 2 sin t t ⫽ 2 and t ⫽ 3, is the electron moving to the right or to the left? Is the electron moving upward or downward? 52. Verify that the inner planet shown in Figure 10.120 completes four orbits for every single orbit of the outer planet. 53. Suppose that due to some cosmic interference, the orbit of the faster planet from Figure 10.120 begins to decay at a rate of T0.84 (replace T with T0.84 in both equations for the inner planet). By observation, about how many orbits did the inner planet make for the first revolution of the outer planet? What is the ratio of orbits for the next complete orbit of the outer planet? Systems applications: Solve the following systems using elimination. If the system is dependent, write the general solution in parametric form and use a calculator to generate several solutions.

2x ⫺ y ⫹ 3z ⫽ ⫺3 54. • 3x ⫹ 2y ⫺ z ⫽ 4 8x ⫹ 3y ⫹ z ⫽ 5

x ⫺ 5y ⫹ z ⫽ 3 55. • 5x ⫹ y ⫺ 7z ⫽ ⫺9 2x ⫹ 3y ⫺ 4z ⫽ ⫺6

⫺5x ⫺ 3z ⫽ ⫺1 56. • x ⫹ 2y ⫺ 2z ⫽ ⫺3 ⫺2x ⫹ 6y ⫺ 9z ⫽ ⫺10 x ⫹ y ⫺ 5z ⫽ ⫺4 57. • 2y ⫺ 3z ⫽ ⫺1 x ⫺ 3y ⫹ z ⫽ ⫺3 58. Regressions and parameters: x y Draw a scatterplot of the data 0 0 given in the table. Note that 12 0.25 connecting the points with a smooth curve will not result in 2 2 a function, so a standard 12 6.75 regression cannot be run on 0 16 the data. Now consider the ⫺ 12 31.25 x-values alone — what do you ⫺2 54 notice? Find a sinusoidal ⫺12 85.75 model for the x-values, using T ⫽ 0, 1, 2, 3, . . . , 8. Use the 0 128 same inputs to run some form of regression on the y-values, then use the results to form the “best-fit” parametric equations for this data (use L1 for T, L2 for the x-values, and L3 for the y-values). With your calculator in parametric MODE , enter the equations as X 1T and Y1T, then graph these along with the scatterplot (L2, L3) to see the finished result. Use the TABLE feature of your calculator to comment on the accuracy of the model. 59. Regressions and parameters: x y Draw a scatterplot of the data 1 0 given in the table, and connect ⫺1.75 the points with a smooth curve. 1.2247 The result is a function, but no ⫺3 1.5 standard regression seems to ⫺3.75 1.8371 give an accurate model. The ⫺4 2.25 x-values alone are actually ⫺3.75 2.7557 generated by an exponential ⫺3 3.375 function. Run a regression 4.1335 ⫺1.75 on these values using T ⫽ 0, 1, 2, 3, . . . , 8 as inputs 5.0625 0 to find the exponential model. Then use the same inputs to run some form of regression on the y-values and use the results to form the “best-fit” parametric equations for this data (use L1 for T, L2 for the x-values, and L3 for the y-values). With your calculator in parametric MODE , enter the equations as X 1T and Y1T, then graph these along with the scatterplot (L2, L3) to see the finished result. Use the TABLE feature of your calculator to comment on the accuracy of the model.

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1063

EXTENDING THE CONCEPT x ⫽ 6 sin14t2 . Initially, y ⫽ 8 cos t use only the 2nd GRAPH (TABLE) feature of your calculator (not the graph) to name the intervals for which the particle is moving (a) to the left and upward and (b) to the left and downward. Answer to the nearest tenth (set ¢Tbl ⫽ 0.1). Is it possible for this particle to collide with another particle in this medium whose movement is x ⫽ 3 cos t ⫹ 7 modeled by e ? Discuss why or why not. y ⫽ 2 sin t ⫹ 2

60. The motion of a particle in a certain medium is modeled by the parametric equations e

1 61. Write the function y ⫽ 1x ⫹ 32 2 ⫺ 1 in parametric form using the substitution x ⫽ 2 cos t ⫺ 3 and the 2 appropriate double-angle identity. Is the result equivalent to the original function? Why or why not?

䊳


62. (1.5) The price of a popular video game is reduced by 20% and is selling for $39.96. By what percentage must the sale price be increased to return the item to its original price? 63. (6.6) When the tip of the antenna atop the Eiffel Tower is viewed at a distance of 265 ft from its base, the angle of elevation is 76°. Is the Eiffel Tower taller or shorter than the Chrysler Building (New York City) at 1046 ft? 64. (4.3) Graph f 1x2 ⫽ x3 ⫹ 2x2 ⫺ 5x ⫺ 6 using information about end-behavior, y-intercept, x-intercept(s), and midinterval points.

65. (7.6) The maximum height a projectile will attain depends on the angle it is projected and its initial velocity. This phenomena is modeled by the v2 sin2␪ , where v is the initial function H ⫽ 64 velocity (in feet/sec) of the projectile and ␪ is the angle of projection. Find the angle of projection if the projectile attained a maximum height of 151 ft, and the initial velocity was 120 ft/sec.

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MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs y

(a)

y

(b)

5

⫺4

⫺5

6 x

y

⫺5

⫺6

⫺4

1 1x ⫺ 12 2 4

3. ____ foci at 1⫺1, 1 ⫾ 152

⫺6

10. ____ 41y ⫺ 12 2 ⫺ 1x ⫹ 12 2 ⫽ 16 11. ____ center at 10, ⫺22 12. ____ focus at (1, 1)

1 5. ____ x ⫽ ⫺ 1y ⫹ 12 2 ⫹ 3 2 6. ____ domain: x 僆 3⫺3, 5 4 , range: y 僆 3 ⫺3, 54

13. ____ 41x ⫹ 22 2 ⫹ 1y ⫹ 12 2 ⫽ 16

8. ____ x2 ⫹ 1y ⫺ 22 2 ⫽ 9

⫺5

9. ____ vertices at 1⫺3, 12 and (5, 1)

4. ____ transverse axis y ⫽ 1

7. ____ 41x ⫹ 12 2 ⫺ 1y ⫺ 12 2 ⫽ 16

5 x

4 x

1. ____ 1x ⫺ 12 2 ⫹ 1y ⫺ 12 2 ⫽ 16 2. ____ y ⫽

5

5 x

5 x

⫺5

y

(h)

4

⫺5

⫺5

⫺5

y

(g)

6

5 x

5 x

⫺4

y

(f)

5

5

⫺5

5 x

⫺5

y

(d)

6

⫺5

⫺5

(e)

y

(c)

5

14. ____ 1x ⫺ 12 2 ⫹ 41y ⫺ 12 2 ⫽ 16 15. ____ axis of symmetry: y ⫽ ⫺1 16. ____ domain: x 僆 3 ⫺4, 04 , range: y 僆 3⫺5, 3 4


A Brief Introduction to Analytical Geometry

KEY CONCEPTS • The midpoint and distance formulas play important roles in the study of analytical geometry: x2 ⫹ x1 y2 ⫹ y1 midpoint: 1x, y2 ⫽ a , b distance: d ⫽ 21x2 ⫺ x1 2 2 ⫹ 1y2 ⫺ y1 2 2 2 2

• The perpendicular distance from a point to the line is the length of the line segment perpendicular to the given line with the given point and the point of intersection as endpoints. • Using these tools, we can verify or construct relationships between points, lines, and curves in the plane; verify properties of geometric figures; prove theorems from Euclidean geometry; and construct relationships that define the conic sections.

EXERCISES 1. Verify the closed figure with vertices (⫺3, ⫺4), (⫺5, 4), (3, 6), and (5, ⫺2) is a square. 2. Find the equation of the circle that circumscribes the square in Exercise 1.

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3. A theorem from Euclidean geometry states: If any two points are equidistant from the endpoints of a line segment, they are on the perpendicular bisector of the segment. Determine if the line through (3, 6) and (6, 9) is the perpendicular bisector of the segment through (5, 2) and (5, 4). 4. Four points are given. Verify that the distance from each point to the line y 1 is the same as the distance from the given point to the fixed point (0, 1): (6, 9), (2, 1), (4, 4), and (8, 16).

SECTION 10.2

The Circle and the Ellipse

KEY CONCEPTS • The equation of a circle centered at (h, k) with radius r is 1x h2 2 1y k2 2 r2.

• Dividing both sides by r2, we obtain the standard form vertical distance from center to graph is r. • The equation of an ellipse in standard form is

1x h2 2 2

1x h2 2 r2

1y k2 2 2

1y k2 2 r2

1, showing the horizontal and y

1. The center of

a b the ellipse is (h, k), with horizontal distance a and vertical distance b from center to graph.

(x, y) d1

d2 (a, 0)

(a, 0) (c, 0)

(c, 0)

x

• Given two fixed points f1 and f2 in a plane (called the foci), an ellipse is the set of all

d1 d2 k points (x, y) such that the distance from the first focus to (x, y), plus the distance from the second focus to (x, y), remains constant. • For an ellipse, the distance from center to vertex is greater than the distance c from center to one focus. • To find the foci of a horizontal ellipse, use: a2 b2 c2 (since a 7 c), or c2 |a2 b2|.

EXERCISES Sketch the graph of each equation in Exercises 5 through 9. 5. x2 y2 16 6. x2 4y2 36 7. 9x2 y2 18x 27 0 2 2 1x 32 1y 22 8. x2 y2 6x 4y 12 0 9. 1 16 9 10. Find the equation of the ellipse with minor axis of length 6 and foci at 14, 02 and (4, 0). Then graph the equation on a graphing calculator using a “friendly” window and use the TRACE feature to locate four additional points on the graph with coordinates that are rational. 11. Find the equation of the ellipse with vertices at (a) 113, 02 and (13, 0), foci at 112, 02 and (12, 0); (b) foci at 10, 162 and (0, 16), major axis: 40 units.

12. Write the equation in standard form and sketch the graph, noting all of the characteristic features of the ellipse. 4x2 25y2 16x 50y 59 0

SECTION 10.3

The Hyperbola

KEY CONCEPTS 1x h2 2 1y k2 2 1. The center of the hyperbola • The equation of a horizontal hyperbola in standard form is a2 b2 is (h, k) with horizontal distance a from center to vertices, and vertical distance b from center to the midpoint of y the sides of the central rectangle. Given two fixed points f and f in a plane (called the foci), a hyperbola is the set of all (x, y) • 1 2 d1 points (x, y) such that the distance from one focus to point (x, y), less the distance from the d2 (a, 0) (a, 0) other focus to (x, y), remains a positive constant: |d1 d2| k. (c, 0) (c, 0) x • For a hyperbola, the distance from center to one vertex is less than the distance from center to the focus c. • To find the foci of a hyperbola: c2 a2 b2 (since c 7 a). |d1 d2| k

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EXERCISES Sketch the graph of each equation in Exercises 13 through 17, indicating the center, vertices, and asymptotes. 1x 22 2 1y 32 2 1x 22 2 1y 12 2 2 2 1 1 13. 4y 25x 100 14. 15. 16 9 9 4 16. 9y2 x2 18y 72 0 17. x2 4y2 12x 8y 16 0 4 18. Find the equation of the hyperbola with vertices at (3, 0) and (3, 0), and asymptotes of y x. Then graph 3 the equation on a graphing calculator using a “friendly” window and use the TRACE feature to locate two additional points with rational coordinates. 19. Find the equation of the hyperbola with (a) vertices at (15, 0), foci at (17, 0), and (b) foci at (0, 5) with vertical dimension of central rectangle 8 units. 20. Write the equation in standard form and sketch the graph, noting all of the characteristic features of the hyperbola. 4x2 9y2 40x 36y 28 0

SECTION 10.4

The Analytic Parabola

KEY CONCEPTS • Horizontal parabolas have equations of the form x ay2 by c; a 0.

• A horizontal parabola will open to the right if a 7 0, and to the left if a 6 0. The axis of symmetry is y

b y or by completing the square and 2a writing the equation in shifted form: 1x h2 a1y k2 2. d1 (x, y) Given a fixed point f (called the focus) and fixed line D (called the directrix) in the plane, a f d1 d2 d2 parabola is the set of all points (x, y) such that the distance from f to (x, y) is equal to the Vertex x D distance from (x, y) to line D. The equation x2 4py describes a vertical parabola, opening upward if p 7 0, and opening downward if p 6 0. The equation y2 4px describes a horizontal parabola, opening to the right if p 7 0, and opening to the left if p 6 0. p is the distance from the vertex to the focus (or from the vertex to the directrix). The focal chord of a parabola is a line segment that contains the focus and is parallel the directrix, with its endpoints on the graph. It has a total length of 冟4p冟, meaning the distance from the focus to a point on the graph (as described) is 冟2p冟. It is commonly used to assist in drawing a graph of the parabola. with the vertex (h, k) found by evaluating at y

• • • • •

b , 2a

EXERCISES For Exercises 21 and 22, find the vertex and x- and y-intercepts if they exist. Then sketch the graph using symmetry and a few points or by completing the square and shifting a parent function. 21. x y2 4 22. x y2 y 6 For Exercises 23 and 24, find the vertex, focus, and directrix for each parabola. Then sketch the graph using this information and the focal chord. Also graph the directrix. 23. x2 20y

SECTION 10.5

24. x2 8x 8y 16 0

Nonlinear Systems of Equations and Inequalities

KEY CONCEPTS • Nonlinear systems of equations can be solved using substitution or elimination. • First identify the graphs of the equations in the system to help determine the possible number of real solutions. • For systems of nonlinear inequalities, graph the related equation for each inequality given, then use a test point to decide what region to shade as the solution.

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• The solution for a system of inequalities is the overlapping region (if it exists) created by the areas shaded for the individual inequalities. • If the boundary is included, graph it using a solid line; if the boundary is not included (for strict inequalities) use a dashed line.

EXERCISES Solve Exercises 25–28 using substitution or elimination. For Exercises 29 and 30, graph the solution set. Identify the graph of each relation before you begin. Verify your answers using a graphing calculator if possible. x2 y2 25 x y2 1 x2 y 1 25. e 26. e 27. e 2 y x 1 x 4y 5 x y2 7 x2 y2 7 9 x2 y2 10 y x2 2 28. e 29. e 2 30. e 2 2 2 y 3x 0 x y 16 x y 3

SECTION 10.6

Polar Coordinates, Equations, and Graphs

KEY CONCEPTS P(r, ␪) • In polar coordinates, the location of a point in the plane is denoted 1r, ␪2, where r is the r0 distance to the point from the origin or pole, and ␪ is the angle between a stipulated polar axis and a ray containing P. r • In the polar coordinate system, the location 1r, ␪2 of a point is not unique for two reasons: (1) the angles ␪ and ␪ 2␲n are coterminal (n an integer), and (2) r may be negative. ␪ Pole Polar axis • The point P1r, ␪2 can be converted to P(x, y) in rectangular coordinates where x r cos ␪ and y r sin ␪ The point P(x, y) in rectangular coordinates can be converted to P1r, ␪2 in polar coordinates, where r 2x2 y2 • y and ␪r tan1 ` ` . x • To sketch a polar graph, we view the length r as being along the second hand of a clock, ticking in a ␲ counterclockwise direction. Each “tick” is rad or 15°. For each tick we locate a point on the radius and plot it 12 on the face of the clock before going on. • For graphing, we also apply an “r-value” analysis, which looks where r is increasing, decreasing, zero, maximized, and/or minimized. ␲ • If the polar equation is given in terms of sines, the graph will be symmetric to ␪ . 2 • If the polar equation is given in terms of cosines, the graph will be symmetric to the polar axis. • The graphs of several common polar equations are given in Appendix F. EXERCISES Sketch using an r-value analysis, symmetry, and any convenient points. Use a graphing calculator to verify your results. 31. r 5 sin ␪ 32. r 4 4 cos ␪ 33. r 2 4 cos ␪ 34. r 8 sin12␪2

SECTION 10.7

More on the Conic Sections: Rotation of Axes and Polar Form

KEY CONCEPTS • Using a rotation, the conic equation Ax2 Bxy Cy2 Dx Ey F 0 in the xy-plane can be transformed into aX2 cY2 dX eY f 0 in the XY-plane, in which the mixed xy-term is absent. B ; 0 6 2␤ 6 180°. • The required angle of rotation ␤ is found using tan12␤2 AC • The change in coordinates from the xy-plane to the XY-plane is accomplished using the rotation formulas: x X cos ␤ Y sin ␤ y X sin ␤ Y cos ␤

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• In the process of this conversion, certain quantities, called invariants, remain unchanged and can be used to

check that the conversion was correctly performed. These invariants are (1) F f, (2) A C a c, and (3) B2 4AC b2 4ac. • The invariant B2 4AC b2 4ac is called the discriminant and can be used to classify the type of graph the equation will give, except in degenerate cases: • If B2 4AC 0, the equation is that of a parabola. • If B2 4AC 6 0, the equation is that of a circle or an ellipse. • If B2 4AC 7 0, the equation is that of a hyperbola. • All conics (not only the parabola) can be stated in terms of a focus/directrix definition. This is done using the concept of eccentricity, symbolized by the letter e. FP e • If F is a fixed point and l a fixed line in the plane with the point D on l, the set of all points P such that DP (e a constant) is the graph of a conic section. If e 1, the graph is a parabola. If 0 6 e 6 1, the graph is an ellipse. If e 7 1, the graph is a hyperbola. Given a conic section with eccentricity e, one focus at the pole of the r␪-plane, and directrix l located d units from • de de this focus, then the polar equations r and r represent one of the conic sections as 1 e cos ␪ 1 e sin ␪ determined by the value of e.

EXERCISES For the given conics in the xy-plane, use a rotation of axes to find the corresponding equation in the XY-plane, then sketch its graph. 35. 2x2 4xy 2y2 812y 24 0 36. x2 6 13xy 7y2 160 0 For the conic equations given, determine if the equation represents a parabola, ellipse, or hyperbola. Then determine the eccentricity e and sketch the graphs using polar graph paper. Verify solutions by graphing them on a calculator. 4 9 8 37. r 38. r 39. r 3 2 cos ␪ 4 6 cos ␪ 3 3 sin ␪ 40. Mars has a perihelion of 128.4 million miles and an aphelion of 154.9 million miles. Use this information to find a polar equation that models the elliptical orbit, then find the length of the focal chord.

SECTION 10.8

Parametric Equations and Graphs

KEY CONCEPTS • If we consider the set of points P(x, y) such that the x-values are generated by f (t) and the y-values are generated by g(t) (assuming f and g are both defined on an interval of the domain), the equations x f 1t2 and y g1t2 are called parametric equations, with parameter t. • Parametric equations can be converted to rectangular form by eliminating the parameter. This can sometimes be done by solving for t in one equation and substituting in the other, or by using trigonometric forms. • A function can be written in parametric form many different ways, by altering the parameter or using trigonometric identities. • The cycloids are an important family of curves, with equations x r1t sin t2 and y r11 cos t2. • The solutions to dependent systems of equations are often expression in parametric form, with the points (t, f(t), g(t)) given by the parametric equations generating solutions to the system. EXERCISES Graph the curves defined by the parametric equations over the specified interval and identify the graph. Then eliminate the parameter and write the corresponding rectangular form. 41. x t 4: t 僆 3 3, 3 4 : 42. x 12 t2 2: t 僆 30, 5 4: 43. x 3 sin t: t 僆 30, 2␲2: 2 2 y 1t 32 y 4 cos t y 2t 3 2 44. Write the function in three different forms by altering the parameter: y 21x 52 1 45. Use a graphing calculator to graph the Lissajous figure indicated, then state the size of the rectangle needed to frame it: x 4 sin15t2; y 8 cos t

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Practice Test

PRACTICE TEST For Exercises 16 and 17, identify and graph each conic section from the parametric equations given. Then remove the parameter and convert to rectangular form.

By inspection only (no graphing), match each equation to its correct description. 1. x2 y2 6x 4y 9 0 3. x2 4y2 4x 12y 20 0 4. y x2 4x 20 0 a. Parabola b. Hyperbola

c. Circle

18. Use a graphing calculator to graph the cycloid, then identify the maximum and minimum values, and the period. x 4T 4 sin T, y 4 4 cos T

d. Ellipse

19. Solve each nonlinear system using any method. 4x2 y2 16 2y2 x2 4 a. e b. e 2 yx2 x y2 8

Identify and then graph each of the following conic sections. State the center, vertices, foci, asymptotes, and other important points when applicable. 5. x2 y2 4x 10y 20 0

20. Halley’s comet has a perihelion of 54.5 million miles and an aphelion of 3253 million miles. Use this information to find a polar equation that models its elliptical orbit. How does its eccentricity compare with that of the planets in our solar system?

6. 251x 22 41y 12 100 2

2

12 10 8. r 5 4 cos ␪ 5 5 cos ␪ 1y 32 2 1x 22 2 1 9. 9 16 7. r

21. The soccer match is tied, with time running out. In a desperate attempt to win, the opposing coach pulls his goalie and substitutes a forward. Suddenly, Marques gets a break-away and has an open shot at the empty net, 165 ft away. If the kick is on-line and leaves his foot at an angle of 28° with an initial velocity of 80 ft/sec, is the ball likely to go in the net and score the winning goal?

10. 41x 12 2 251y 22 2 100 Use the equation 80x2 120xy 45y2 100y 44 0 to complete Exercises 11 and 12. 11. Use the discriminant B2 4AC to identify the graph, and tan12␤2

B to find cos ␤ and sin ␤. AC

22. The orbit of Mars around the Sun is elliptical, with the Sun at one focus. When the orbit is expressed as a central ellipse on the coordinate grid, its equation is y2 x2 1. Use this information to 1141.652 2 1141.032 2 find the aphelion of Mars and the perihelion of Mars in millions of miles.

12. Find the equation in the xy-plane and use a rotation of axes to draw a neat sketch of the graph in the XY-plane. Graph each polar equation. Verify your results with a graphing calculator. 13. r 3 3 cos ␪

17. x 1t 32 2 1 yt2

16. x 4 sin t y 5 cos t

2. 4y2 x2 4x 8y 20 0

14. r 4 8 cos ␪

15. r 6 sin12␪2 Determine the equation of each relation and state its domain and range. For the parabola and the ellipse, also give the location of the foci. 23.

24.

y

25.

y

(2, 6)

y

(3, 4)

(1, 0)

(3, 0) x

(4, 1)

(5, 1)

(6, 1)

(1, 1)

x

(0, 3)

(1, 4)

x

(5, 2) (2, 4)

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CALCULATOR EXPLORATION AND DISCOVERY Elongation and Eccentricity Technically speaking, a circle is an ellipse with both foci at the center. As the distance between foci increases, the ellipse becomes more elongated. We saw other instances of elongation in stretches and compressions of parabolic graphs, and in hyperbolic graphs where the asymptotic slopes varied depending on the values a and b. The measure c used to quantify this elongation is called the eccentricity e, and is determined by the ratio e . For this Exploration and a Discovery, we’ll use the repeat graph feature of a graphing calculator to explore the eccentricity of the graph of a conic. The “repeat graph” feature enables you to graph a family of curves by enclosing changes in a parameter in braces “{ }.” For instance, entering 52, 1, 0, 1, 26 X 3 as Y1 on the Y= screen will automatically graph these five lines: y 2x 3, y x 3, y 3, y x 3, and y 2x 3. We’ll use this feature to graph a family of ellipses, observing the result and calculating the eccentricity for each y2 x2 curve in the family. The standard form is 2 2 1, which we’ll solve for y and enter as Y1 and Y2. After a b x2 simplification the result is y b 1 2 , but for this investigation we’ll use the constant b 2 and vary the B a x2 . Note from Figure 10.121 that parameter a using the values a 2, 4, 6, and 8. The result is y 2 1 B 54, 16, 36, 646 we’ve set Y2 Y1 to graph the lower half of the ellipse. Using the “friendly” window shown (Figure 10.122) gives the result shown in Figure 10.123, where we see the ellipse is increasingly elongated in the horizontal direction (note when a 2 the result is a circle since a b). Figure 10.123 Figure 10.121

Figure 10.122

6.2

9.4

9.4

6.2

Using a 2, 4, 6, and 8 with b 2 in the foci formula c 2a2 b2 gives c 0, 2 13, 4 12, and 2 115, 2 115 0 2 13 4 12 respectively, with these eccentricities: e , , and , . While difficult to see in radical form, we find 2 4 6 8 that the eccentricity of an ellipse always satisfies the inequality 0 6 e 6 1 (excluding the circle ellipse case). To two decimal places, the values are e 0, 0.87, 0.94, and 0.97, respectively. c As a final note, it’s interesting how the e definition of eccentricity relates to our everyday use of the word a “eccentric.” A normal or “noneccentric” person is thought to be well-rounded, and sure enough e 0 produces a well-rounded figure—a circle. A person who is highly eccentric is thought to be far from the norm, deviating greatly from the center, and greater values of e produce very elongated ellipses. Exercise 1: Perform a similar exploration using a family of hyperbolas. What do you notice about the eccentricity? Exercise 2: Perform a similar exploration using a family of parabolas. What do you notice about the eccentricity?

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STRENGTHENING CORE SKILLS Ellipses and Hyperbolas with Rational/Irrational Values of a and b Using the process known as completing the square, we were able to convert from the polynomial form of a conic section to the standard form. However, for some equations, values of a and b are somewhat difficult to identify, since the coefficients are not factors. Consider the equation 20x2 120x 27y2 54y 192 0, the equation of an ellipse. 20x2 120x 27y2 54y 192 0 201x 6x ___ 2 271y2 2y ___ 2 192 201x2 6x 92 271y2 2y 12 192 27 180 201x 32 2 271y 12 2 15 41x 32 2 91y 12 2 1 3 5 2

original equation subtract 192 complete the square in x and y factor and simplify standard form

Unfortunately, we cannot easily identify the values of a and b, since the coefficients of each binomial square are not “1.” In these cases, we can write the equation in standard form by using a simple property of fractions—the numerator and denominator of any fraction can be divided by the same quantity to obtain an equivalent fraction. 1y 12 2 1x 32 2 1. Although the result may look odd, it can nevertheless be applied here, giving a result of 3 5 4 9 We can now identify a and b by writing these denominators in squared form, which gives the following expression: 1x 32 2 1y 12 2 15 13 1. The values of a and b are now easily seen as a ⬇ 0.866 and b ⬇ 0.745. 2 2 2 3 13 15 b b a a 2 3 Use this idea to complete the following exercises. Exercise 1: Write the equation in standard form, then identify the values of a and b and use them to graph the ellipse. 41x 32 2 49

251y 12 2 36

1

Exercise 2: Write the equation in standard form, then identify the values of a and b and use them to graph the hyperbola. 91x 32 2 80

41y 12 2 81

1

Exercise 3: Identify the values of a and b by writing the equation 100x2 400x 18y2 108y 230 0 in standard form. Exercise 4: Identify the values of a and b by writing the equation 28x2 56x 48y2 192y 195 0 in standard form.

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CUMULATIVE REVIEW CHAPTERS 1–10 1. Find the equation of the parabola with vertex at (2, 3) and directrix x 0. Solve each equation.

Graph each relation. Include vertices, x- and y-intercepts, asymptotes, and other features. 14. f 1x2 冟x 2冟 3 15. y 1x 3 1

2. x2 6x 13 0 3. 4 # 2x1 18

16. g1x2 1x 321x 121x 42

4. 3x2 7

17. h1x2

x2 x2 9 19. f 1x2 log2 1x 12

5. log381 x 6. log x log1x 32 1

20. x2 y2 10x 4y 20 0

7. 6 tan x 213

21. 41x 12 2 361y 22 2 144

␲ ␲ 8. 25 sin a x b 3 15.5 3 6 9.

22. y 2 cos ax

sin 27° sin x 18 35

␲ b1 4

23. r 4 cos12␪2

10. Use De Moivre’s theorem to find the three cube roots of 8i. Write the roots in a bi form. 11. The price of beef in Argentina varies directly with demand and inversely with supply. In the small town of Chascomus, the tender-cut lomito was selling for 18 pesos/kg last week. There were 1000 kg available, and 850 kg were bought. Next week there is a 3-day weekend, so the demand is expected to be closer to 1400 kg, but the butchers will only be able to supply 1200 kg. What will a kilogram of tendercut lomito cost next week? 12. Find the inverse of f 1x2 3 sin12x 12 .

13. A surveyor needs to estimate the width of a large rock formation in Canyonlands National Park. From her current position she is 540 yd from one edge of the formation and 850 yd from the other edge. If the included angle is 110°, how wide is the formation?

540 yd

850 yd 110

18. y 2x 3

24. x 2 sin t y tan t

25. Use the dot product to find the angle between the vectors u H4, 5I and v H3, 7I. Solve each system of equations with a graphing calculator. Use a matrix equation for Exercise 26, and the intersection-of-graphs method for Exercise 27. 4x 3y 13 26. • 9y 5z 19 x 4z 4 27. e

x2 y2 25 64x2 12y2 768

28. Use a graphing calculator to solve 1x 2 2 13x 4 29. Use a graphing calculator to decompose 3x3 2x2 x 3 into partial fractions. x4 x2 30. In the summer, Hollywood releases its big budget, big star, big money movies. Suppose the weekly summer revenue generated by ticket sales was modeled by the function R1w2 w4 25w3 200w2 560w 234, where R(w) represents the revenue generated in week w and 1 w 12. Use a calculator to determine the amount of revenue generated in week 5.

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CONNECTIONS TO CALCULUS As with other relations and functions we’ve studied in precalculus, there is a high level of interest in finding rates of change in polar functions, and in solving systems of polar equations. From their use in architecture to their application in studies of planetary motion, both skills play a fundamental role in our continuing study.

Polar Graphs and Instantaneous Rates of Change Using the tools of calculus, it can be shown that the slope of the tangent line for many different polar graphs can be generated from a specific template. It can also be shown that under certain conditions, the zeroes of the numerator will give the location of horizontal tangent lines, while the zeroes of the denominator will given the location of vertical tangent lines. EXAMPLE 1

䊳

Finding Slopes of Tangent Lines For the cardioid r 1 sin ␪, it can be shown that

cos ␪ sin ␪ 11 sin ␪2cos ␪

cos ␪ cos ␪ 11 sin ␪2sin ␪ gives the slope of the tangent line at ␪. a. Simplify the expression using double-angle identities. 2␲ b. Use the result to find the slope of the tangent line at ␪ . 3 Then, determine the value(s) of ␪ for which a tangent line is c. Horizontal. d. Vertical.

Solution

䊳

a. After using the distributive property in the numerator and denominator, we have cos ␪ sin ␪ 11 sin ␪2cos ␪

cos ␪ sin ␪ cos ␪ sin ␪ cos ␪ distribute cos ␪ cos ␪ 11 sin ␪2sin ␪ cos2␪ sin ␪ sin2␪ 2 sin ␪ cos ␪ cos ␪ combine terms cos2␪ sin2␪ sin ␪ sin12␪2 cos ␪ double angle identities cos12␪2 sin ␪ 2␲ 4␲ 1 13 sina b cosa b 3 3 2␲ 2 2 1. b. For ␪ , the expression gives 3 1 13 4␲ 2␲ cosa b sina b 2 2 3 3 For the cardioid r 1 sin ␪, the slope of Figure 10.124 2␲ the tangent line at ␪ is 1 (Figure 10.124). 3 5 c. For horizontal tangents, set the numerator 4 equal to zero and solve. 3 sin12␪2 cos ␪ 0 2 sin ␪ cos ␪ cos ␪ 0 cos ␪ 12 sin ␪ 12 0

zeroes of numerator double angle identity factor cos ␪

cos ␪ 0 or sin ␪

10–113

1 2

2 1 54321 1 2 3 4 5

1 2 3 4 5

1073

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11␲ ␲ 7␲ Horizontal tangent lines will occur at ␪ , , and . Note that a horizontal 2 6 6 3␲ tangent does not occur at ␪ , as this value also makes the denominator of 2 the expression equal to zero. d. For vertical tangents, set the denominator equal to zero and solve. cos12␪2 sin ␪ 0 cos ␪ sin2␪ sin ␪ 0 1 sin2␪ sin2␪ sin ␪ 0 1 sin ␪ 2 sin2␪ 0 11 2 sin ␪211 sin ␪2 0 1 sin ␪ or sin ␪ 1 2 2

Figure 10.125

zeroes of denominator

3

double angle identity

2


1

combine terms

3 2 1

factor

Vertical tangent lines will occur at ␪

1

2

3

1 2

5␲ ␲ and 6 6

3

(see enlargement in Figure 10.125). A vertical tangent does not occur at ␪ as this value makes both the numerator and denominator of the expression equal to zero.

3␲ , 2


䊳

Systems of Polar Equations In the “Connections” feature from Chapter 3, we investigated the area bounded by a given curve, the x-axis, and given vertical lines. In many applications, we need to find the area bounded between two curves, which requires that we find where the curves intersect. This can be accomplished by solving a system consisting of the two given r f 1␪2 functions. For polar equations, the system takes the form e . r g1␪2 EXAMPLE 2

䊳

Finding Intersections of Polar Curves To find the area of the region inside the circle r 4 sin ␪ and outside the limacon r 1 2 sin ␪, we must find where the curves intersect. Set up and solve the required system (see figure).

Solution

䊳

5 4 3 2 1

For r 4 sin ␪ and r 1 2 sin ␪, the system is 54321 1 2 3 4 5 r 4 sin ␪ 1 e . Substituting 4 sin ␪ for r in the 2 r 1 2 sin ␪ 3 second equation gives 4 sin ␪ 1 2 sin ␪, 2 sin ␪ 1 1 sin ␪ 2 1 ␪ sin1a b 2

substitute 4 sin ␪ for r subtract 2 sin ␪ divide by 2

apply inverse sine

4 5

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␲ 5␲ and ␪ . It appears we’re finished, 6 6 but an inspection of the equations (or the graph) shows there must be a third point of intersection at the pole (0, 0). The reason we missed it is that in polar coordinates each point can be represented many different ways. Solutions using a system of polar equations finds only the intersections represented by the identical ordered pair, and a separate analysis is needed for the case where r 0. Solving the equation 7␲ 0 4 sin ␪ gives ␪ ␲k, while 0 1 2 sin ␪ gives ␪ 2␲k and 6 11␲ ␪ 2␲k. Although the ordered pairs are different, both represent the pole 6 (0, 0) and the graphs must intersect there. This shows the graphs intersect at ␪


䊳

Connections to Calculus Exercises For each polar function given, the expression for the slope of the tangent line is also given. (a) Simplify the expression. (b) Find the slope of the tangent line at the values of ␪ shown. Then determine the value(s) of ␪ in [0, 2␲) for which a tangent line is (c) horizontal, and (d) vertical.

1. r 1 cos ␪ (cardioid), slope of the tangent line: cos ␪ sin ␪1sin ␪2 cos ␪ cos ␪ sin ␪ 2 cos ␪1sin ␪2

;␪

␲ 11␲ , 2 6

3. r cos ␪ sin ␪ (circle), slope of the tangent line: cos ␪ cos ␪ sin ␪1sin ␪2 2 sin ␪ cos ␪

2. r 6 cos ␪ (circle), slope of the tangent line: ␲ 2␲ 6 sin ␪ sin ␪ 6 cos ␪ cos ␪ ;␪ , 6 sin ␪ cos ␪ 6 cos ␪ sin ␪ 6 3 4. r sin ␪ cos ␪ (circle), slope of the tangent line: 2 sin ␪1cos ␪2 3cos ␪ cos ␪ sin ␪1sin ␪2 4

2 cos ␪1sin ␪2 3sin ␪1sin ␪2 cos ␪ cos ␪ 4

cos ␪ cos ␪ sin ␪1sin ␪2 2 cos ␪1sin ␪2

␪

␪

␲ 3␲ , 4 4

␲ 3␲ , 2 2

Solve each system. For Exercises 7 and 8, identify the graph of each function before you begin. Verify results using a graphing calculator.

r

3 2 5. • r 1 cos2␪

6. e

r 2 sin2␪ r 1 sin ␪

r sin12␪2 r cos ␪

8. e

r 2 sin12␪2 r 2 sin ␪

7. e

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9. Find the points of intersection for the circle r 4 cos ␪ and the curve r 4 cos ␪ r 4 sin12␪2 , by solving the system e . State your answer in r 8 sin ␪ cos ␪ both (r, ␪) form and (x, y) form. Note from the graph there are three points of intersection, and that a polar grid has been superimposed on a rectangular grid.

10. Find the points of intersection of the two four-leaf roses formed by r 4 sin12␪2 r 4 sin12␪2 and r 4 cos12␪2 , by solving the system e r 4 cos12␪2 (there are nine in all). Some of these points can be found by solving 4 cos12␪2 4 sin12␪2 , others can be found using symmetry [since r can be positive or negative, we consider the possibility r1 r2, or 4 cos12␪2 4 sin12␪2 ]. Be sure to include an analysis of the case where r 0. Answer in (r, ␪) form.

Exercise 9 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5

Exercise 10 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5

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Additional Topics in Algebra CHAPTER OUTLINE 11.1 Sequences and Series 1078 11.2 Arithmetic Sequences 1089 11.3 Geometric Sequences 1098 11.4 Mathematical Induction 1112 11.5 Counting Techniques 1120

For a corporation of any size, decisions made by upper management often depend on a large number of factors, with the desired outcome attainable in many different ways. For instance, consider a legal firm that specializes in family law, with a support staff of 15 employees—6 paralegals and 9 legal assistants. Due to recent changes in the law, the firm wants to send some combination of five support staff to a conference dedicated to the new changes. In Chapter 11, we’ll see how counting techniques and probability can be used to determine the various ways such a group can be randomly formed, even if certain constraints are imposed. 䊳


11.6 Introduction to Probability 1132 11.7 The Binomial Theorem 1145

While the notation and properties of summation were introduced as tools to work with finite sums, in Section 11.3 we will note that under certain conditions, consideration of an infinite sum is possible. The process will involve an ever larger sum of ever smaller pieces, a process further illustrated in the Connections to Calculus Connections to Calculus feature following Chapter 11. 1077

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Sequences and Series

LEARNING OBJECTIVES

A sequence can be thought of as a pattern of numbers listed in a prescribed order. A series is the sum of the numbers in a sequence. Sequences and series come in countless varieties, and we’ll introduce some general forms here. In following sections we’ll focus on two special types: arithmetic and geometric sequences. These are used in a number of different fields, with a wide variety of significant applications.


A. Write out the terms of a

B.

C. D.

E.

sequence given the general or n th term Work with recursive sequences and sequences involving a factorial Find the partial sum of a series Use summation notation to write and evaluate series Use sequences to solve applications

WORTHY OF NOTE Sequences can actually start with any natural number. For instance, 2 the sequence an ⫽ must start n⫺1 at n ⫽ 2 to avoid division by zero. In addition, we will sometimes use a0 to indicate a preliminary or inaugural element, as in a0 ⫽ $10,000 for the amount of money initially held, prior to investing it.

EXAMPLE 1A

䊳

A. Finding the Terms of a Sequence Given the General Term Suppose a person had $10,000 to invest, and decided to place the money in government bonds that guarantee an annual return of 7%. From our work in Chapter 5, we know the amount of money in the account after x years can be modeled by the function f 1x2 ⫽ 10,00011.072 x. If you reinvest your earnings each year, the amount in the account would be (rounded to the nearest dollar): Year: Value:

f (2)

f (3)

f (4)

T

T

T

T

f (5) p T

$10,700

$11,449

$12,250

$13,108

$14,026 p

Note the relationship (year, value) is a function that pairs 1 with $10,700, 2 with $11,449, 3 with $12,250, and so on. This is an example of a sequence. To distinguish sequences from other algebraic functions, we commonly name the functions a instead of f, use the variable n instead of x, and employ a subscript notation. The function f 1x2 ⫽ 10,00011.072 x would then be written an ⫽ 10,00011.072 n. Using this notation a1 ⫽ 10,700, a2 ⫽ 11,449, and so on. The values a1, a2, a3, a4, p are called the terms of the sequence. If the account were closed after a certain number of years (for example, after the fifth year) we have a finite sequence. If we let the investment grow indefinitely, the result is called an infinite sequence. The expression an that defines the sequence is called the general or nth term and the terms immediately preceding it are called the 1n ⫺ 12st term, the 1n ⫺ 22nd term, and so on. Sequences A finite sequence is a function an whose domain is the set of natural numbers from 1 to n. The terms of the sequence are labeled a1, a2, a3, p , ak, ak⫹1, p , an⫺1, an where ak represents an arbitrary “interior” term and an also represents the last term of the sequence. An infinite sequence is a function an whose domain is the set of all natural numbers. Computing Specified Terms of a Sequence For an ⫽

1078

f (1)

n⫹1 , find a1, a3, a6, and a7. n2

1⫹1 ⫽2 12 6⫹1 7 a6 ⫽ ⫽ 2 36 6

3⫹1 4 ⫽ 2 9 3 7⫹1 8 a7 ⫽ ⫽ 2 49 7

Solution

䊳

a1 ⫽

a3 ⫽

EXAMPLE 1B

䊳

Computing the First k Terms of a Sequence

Find the first four terms of the sequence an ⫽ 1⫺12 n2n. Write the terms of the sequence as a list. 11–2

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Section 11.1 Sequences and Series

Solution

䊳

WORTHY OF NOTE When the terms of a sequence alternate in sign as in Example 1B, we call it an alternating sequence.

a1 ⫽ 1⫺12 121 ⫽ ⫺2 a3 ⫽ 1⫺12 323 ⫽ ⫺8

1079

a2 ⫽ 1⫺12 222 ⫽ 4 a4 ⫽ 1⫺12 424 ⫽ 16

The sequence can be written ⫺2, 4, ⫺8, 16, p , or more generally as ⫺2, 4, ⫺8, 16, p , 1⫺12 n2n, p to show how each term was generated. Now try Exercises 7 through 22

Much of the beauty and power of studying sequences comes from patterns detected within the sequence, or the ability to find a particular term quickly. Here, the calculator becomes an invaluable tool, aiding computations to be sure p but also enabling explorations not generally possible with paper and pencil alone. Most graphing calculators offer a “seq(” feature (often in a STAT or LIST menu), Figure 11.1 where the left parenthesis indicates we need to enter the following four items of information: the nth term formula defining the sequence, the variable in use, the starting value, and the ending value. The sequence generated can be seen on the home screen, or stored in a list for future use. In Figure 11.1, this feature was used to generate the first four terms of the sequence defined in Example 1B using the keystrokes 2nd STAT (LIST) (OPS) 5:seq(. In addition to the seq( feature, most calculators have a sequence MODE , which is especially useful when working with several sequences simultaneously. In sequence MODE (Figure 11.2A), the Y= screen presents a very Figure 11.2B different look (Figure 11.2B), first of all showing that the minimum value of n is 1, then naming the functions u(n), v(n), and w(n) instead of Y1, Y2, and Y3. The entries following each, for example “u(nMin) ⫽,” are used in a study of the recursive sequences, which are covered later. Note that function names for the sequences (u, v, and w) are the 2nd function for the numbers 7, 8, and 9, respectively.

Figure 11.2A

EXAMPLE 2

䊳

䊳

Finding the Terms of a Sequence Using Technology Use a calculator in sequence MODE to a. Find the sixth term (as a fraction) for the sequence defined in Example 1A. b. Generate the first five terms for the sequence defined in Example 1B and store the results in a list.

Solution Figure 11.3

䊳

a. On the

screen, define the sequence u(n) n⫹1 as u1n2 ⫽ (Figure 11.3), leaving the n2 second line blank. Then go to the home screen, access the “seq(” feature as before, and supply the information required. Since we only want the sixth term (a6), we enter seq(u(n), n, 6, 6) 䉴Frac, and after pressing , we obtain 7 the expected result (Figure 11.4). 36 Y=

ENTER

Figure 11.4

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CHAPTER 11 Additional Topics in Algebra

Figure 11.5

b. For the first five terms, we change u(n) to u1n2 ⫽ 1⫺12 n # 2n then go to the home screen and use the “5:seq(” command once again, this time for terms one through five. Knowing that we want the results stored in a list, we follow the command with STO 2nd 1 (L1) (Figure 11.5). The list is shown in Figure 11.6.

A. You’ve just seen how we can write out the terms of a sequence given the general or nth term

Figure 11.6


䊳

B. Recursive Sequences and Factorial Notation Sometimes the formula defining a sequence uses the preceding term or terms to generate those that follow. These are called recursive sequences and are particularly useful in writing computer programs. Because of how they are defined, recursive sequences must give an inaugural term or seed element(s), to begin the recursion process. Perhaps the most famous recursive sequence is associated with the work of Leonardo of Pisa (A.D. 1180–1250), better known to history as Fibonacci. In the Fibonacci sequence, each successive term is the sum of the previous two, beginning with 1, 1, p .

EXAMPLE 3

䊳

Computing the Terms of a Recursive Sequence Write out the first eight terms of the Fibonacci sequence, which is defined by c1 ⫽ 1, c2 ⫽ 1, and cn ⫽ cn⫺1 ⫹ cn⫺2.

Solution

WORTHY OF NOTE One application of the Fibonacci sequence involves the Fibonacci spiral, found in the growth of many ferns and the spiral shell of many mollusks.

䊳

The first two terms are given, so we begin with n ⫽ 3. c3 ⫽ c3⫺1 ⫹ c3⫺2 c4 ⫽ c4⫺1 ⫹ c4⫺2 c5 ⫽ c5⫺1 ⫹ c5⫺2 ⫽ c2 ⫹ c1 ⫽ c3 ⫹ c2 ⫽ c4 ⫹ c3 ⫽1⫹1 ⫽2⫹1 ⫽3⫹2 ⫽2 ⫽3 ⫽5 At this point we can simply use the fact that each successive term is simply the sum of the preceding two, and find that c6 ⫽ 3 ⫹ 5 ⫽ 8, c7 ⫽ 5 ⫹ 8 ⫽ 13, and c8 ⫽ 8 ⫹ 13 ⫽ 21. The first eight terms are 1, 1, 2, 3, 5, 8, 13, and 21. Now try Exercises 33 through 38

Figure 11.8

䊳

Since a recursive sequence is defined using a preceding term or terms, the first term(s) must be given in order to determine those that follow. To generate the sequence using technology, we enter these initial terms as “u1nMin2 ⫽” on the Y= screen, enclosed in braces. For the Fibonacci sequence from Example 3, we would enter Figure 11.7 u1nMin2 ⫽ 51, 16” as shown in Figure 11.7, with the result shown in Figure 11.8 (use the right arrow to view any remaining terms). Some sequences may involve the computation of a factorial, which is the product of a given natural number with all those that precede it. The expression 5! is read, “five factorial,” and is evaluated as: 5! ⫽ 5 # 4 # 3 # 2 # 1 ⫽ 120. Factorials For any natural number n,

n! ⫽ n # 1n ⫺ 12 # 1n ⫺ 22 # p # 3 # 2 # 1

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1081

Rewriting a factorial in equivalent forms often makes it easier to simplify certain expressions. For example, we can rewrite 5! as 5 # 4! or 5! ⫽ 5 # 4 # 3!, and so on. Consider Example 4. EXAMPLE 4

䊳

Simplifying Expressions Using Factorial Notation Simplify by writing the numerator in an equivalent form. 9! 11! 6! a. b. c. 7! 8!2! 3!5!

Solution

䊳

a.

9! 9 # 8 # 7! ⫽ 7! 7! ⫽9#8 ⫽ 72

b.

11 # 10 # 9 # 8! 11! ⫽ 8!2! 8!2! 990 ⫽ 2 ⫽ 495

c.

6! 6 # 5! ⫽ 3!5! 3!5! 6 ⫽ 6 ⫽1


䊳

Most calculators have a factorial option or key. On many calculator models it is located on a submenu of the MATH key: MATH PRB 4: !. EXAMPLE 5

䊳

Computing a Specified Term from a Sequence Defined Using Factorials Find and simplify the third term of each sequence. a. an ⫽

Algebraic Solution

䊳

Technology Solution

䊳

n! 2n

3! 23 6 3 ⫽ ⫽ 8 4

a. a3 ⫽

For this exercise we enter 1⫺1 2n 12n ⫺ 1 2!

b. cn ⫽

1⫺12 n 12n ⫺ 12!

b. c3 ⫽

1⫺12 32132 ⫺ 1 4 !

n!

3

3! 1⫺12 3 5 # 4 # 3!4 1⫺1215!2 ⫽ ⫽ 3! 3! ⫽ ⫺20

n! as u(n), and 2n

Figure 11.9, 11.10

as v(n). Once these functions n! have been defined on the Y= screen, we can evaluate them on the home screen as with other functions (Figure 11.9), or use the TABLE feature (Figure 11.10).

B. You’ve just seen how we can work with recursive sequences and sequences involving a factorial

Figure 11.11


䊳

C. Series and Partial Sums Sometimes the terms of a sequence are dictated by context rather than a formula. Consider the stacking of large pipes in a storage yard. If there are 10 pipes in the bottom row, then 9 pipes, then 8 (see Figure 11.11), how many pipes are in the stack if there is a single pipe at the top? The sequence generated is 10, 9, 8, p , 3, 2, 1 and to answer the question we would have to compute the sum of all terms in the sequence. When the terms of a finite sequence are added, the result is called a finite series.

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Finite Series Given the sequence a1, a2, a3, a4, p , an, the sum of the terms is called a finite series or partial sum and is denoted Sn: Sn ⫽ a1 ⫹ a2 ⫹ a3 ⫹ p ⫹ an⫺1 ⫹ an

EXAMPLE 6

䊳

Computing a Partial Sum Given an ⫽ 2n, find the value of

Solution

䊳

a. S4

b. S7.

Since we eventually need the sum of the first seven terms, begin by writing out these terms: 2, 4, 6, 8, 10, 12, and 14. a. S4 ⫽ a1 ⫹ a2 ⫹ a3 ⫹ a4 b. S7 ⫽ a1 ⫹ a2 ⫹ a3 ⫹ a4 ⫹ a5 ⫹ a6 ⫹ a7 ⫽2⫹4⫹6⫹8 ⫽ 2 ⫹ 4 ⫹ 6 ⫹ 8 ⫹ 10 ⫹ 12 ⫹ 14 ⫽ 20 ⫽ 56 Now try Exercises 51 through 56

Figure 11.12

䊳

Figure 11.13 There are several ways of computing a partial sum using technology, with the most common being (1) storing the sequence in a list then computing the sum of the list elements, or (2) computing the sum of a sequence directly on the home screen using the appropriate commands. On many calculators, the “sum(” option is in a MATH subSTAT menu, accessed using 2nd (LIST) (MATH) 5:sum(. For the sum in Example 6(b), we have u1n2 ⫽ 2n, with option (1) demonstrated in Figure 11.12 and option (2) in Figure 11.13.

C. You’ve just seen how we can find the partial sum of a series

D. Summation Notation When the general term of a sequence is known, the Greek letter sigma © can be used to write the related series as a formula. For instance, to indicate the sum of the first four terms 4

of an ⫽ 3n ⫹ 2, we write

兺 13i ⫹ 22 with this notation indicating we are to compute

i⫽1

the sum of all terms generated as i cycles from 1 through 4. This result is called summation or sigma notation and the letter i is called the index of summation. The letters j, k, l, and m are also used as index numbers, and the summation need not start at 1. EXAMPLE 7

䊳

Computing a Partial Sum Compute each sum: 4

6 1 c. 1⫺12 kk2 i⫽1 j⫽1 j k⫽3 d. Check each sum using a graphing calculator.

a.

兺

5

13i ⫹ 22

b.

兺

兺

4

Solution

䊳

a.

兺 13i ⫹ 22 ⫽ 13 # 1 ⫹ 22 ⫹ 13 # 2 ⫹ 22 ⫹ 13 # 3 ⫹ 22 ⫹ 13 # 4 ⫹ 22

i⫽1 5

b.

⫽ 5 ⫹ 8 ⫹ 11 ⫹ 14 ⫽ 38

1

1

1

1

1

1

兺j ⫽1⫹2⫹3⫹4⫹5

j⫽1

⫽

60 30 20 15 12 137 ⫹ ⫹ ⫹ ⫹ ⫽ 60 60 60 60 60 60

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Section 11.1 Sequences and Series 6

c.

兺 1⫺12 k

k 2

1083

⫽ 1⫺12 3 # 32 ⫹ 1⫺12 4 # 42 ⫹ 1⫺12 5 # 52 ⫹ 1⫺12 6 # 62

⫽ ⫺9 ⫹ 16 ⫹ 1⫺252 ⫹ 36 ⫽ 18 d. Begin by entering the functions in parts (a), (b), and (c) as u(n), v(n), and w(n) respectively on the Y= screen (Figure 11.14). Note that while different indices are used in this example, all are entered into the calculator using the variable n. The results are shown in Figures 11.15 and 11.16. k⫽3

Figure 11.14

Figure 11.16

Figure 11.15


䊳

If a definite pattern is noted in a given series expansion, this process can be reversed, with the expanded form being expressed in summation notation using the nth term. EXAMPLE 8

䊳

Writing a Sum in Sigma Notation Write each of the following sums in summation (sigma) notation. a. 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 b. 6 ⫹ 9 ⫹ 12 ⫹ 15 ⫹ p

Solution

䊳

a. The series has five terms and each term is an odd number, or 1 less than a 5

multiple of 2. The general term is an ⫽ 2n ⫺ 1, and the series is

兺 12n ⫺ 12.

n⫽1

WORTHY OF NOTE

b. The raised ellipsis “ p ” indicates the sum continues infinitely. Since the terms are multiples of 3, we identify the general term as an ⫽ 3n, while noting the series starts at n ⫽ 2 (instead of n ⫽ 1). Since the sum continues indefinitely, we use the infinity symbol q as the “ending” value in sigma notation. The q

By varying the function given and/or where the sum begins, more than one acceptable form is possible.

series is

兺 3n.

n⫽2

For Example 8(b) 13 ⫹ 3k2 k⫽1 also works. q

兺


䊳

Since the commutative and associative laws hold for the addition of real numbers, summations have the following properties: Properties of Summation Given any real number c and natural number n, n

(I)

兺 c ⫽ cn

i⫽1

If you add a constant c “n” times the result is cn. n

(II)

兺

n

cai ⫽ c

i⫽1

兺a

i

i⫽1

A constant can be factored out of a sum.

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n

(III)

n

n

兺 1a ⫾ b 2 ⫽ 兺 a ⫾ 兺 b i

i

i

i⫽1

i

i⫽1

i⫽1

A summation can be distributed to two (or more) sequences. m

(IV)

兺

n

i⫽1

n

兺

ai ⫹

ai ⫽

i⫽m⫹1

兺a; 1 ⱕ m 6 n i

i⫽1

A summation is cumulative and can be written as a sum of smaller parts. The verification of property II depends solely on the distributive property. n

Proof:

兺 ca ⫽ ca i

i⫽1

1

⫹ ca2 ⫹ ca3 ⫹ p ⫹ can

expand sum

⫽ c1a1 ⫹ a2 ⫹ a3 ⫹ p ⫹ an 2

factor out c

⫽c

write series in summation form

n

兺a

i

i⫽1

The verifications of properties III and IV simply use the commutative and associative properties. You are asked to prove property III in Exercise 94. EXAMPLE 9

䊳

Computing a Sum Using Summation Properties 4

Recompute the sum

兺 13i ⫹ 22 from Example 7(a) using summation properties.

i⫽1 4

Solution

䊳

兺

i⫽1

13i ⫹ 22 ⫽

4

兺

4

兺2

property III

兺i ⫹ 兺2

property II

3i ⫹

i⫽1 4

⫽3

i⫽1

i⫽1 4 i⫽1

⫽ 31102 ⫹ 2142 ⫽ 38

D. You’ve just seen how we can use summation notation to write and evaluate series

1 ⫹ 2 ⫹ 3 ⫹ 4 ⫽ 10; property I result


䊳

E. Applications of Sequences To solve applications of sequences, (1) identify where the sequence begins (the initial term), (2) write out the first few terms to help identify the nth term, and (3) decide on an appropriate approach or strategy. EXAMPLE 10

䊳

Solving an Application — Accumulation of Stock Hydra already owned 1420 shares of stock when her company began offering employees the opportunity to purchase 175 discounted shares per year. If she made no purchases other than these discounted shares each year, how many shares will she have 9 yr later? If this continued for the 25 yr she will work for the company, how many shares will she have at retirement?

Solution

䊳

To begin, it helps to simply write out the first few terms of the sequence. Since she already had 1420 shares before the company made this offer, we let a0 ⫽ 1420 be the inaugural element, showing a1 ⫽ 1595 (after 1 yr, she owns 1420 ⫹ 175 ⫽ 1595 shares). The first few terms are 1595, 1770, 1945, 2120, and so on. This supports a general term of an ⫽ 1595 ⫹ 1751n ⫺ 12.

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After 9 years

After 25 years

a9 ⫽ 1595 ⫹ 175182 ⫽ 2995

a25 ⫽ 1595 ⫹ 1751242 ⫽ 5795

1085

After 9 yr she would have 2995 shares. Upon retirement she would own 5795 shares of company stock. Now try Exercises 85 through 90

䊳

Surprisingly, sequences and series have a number of other interesting properties, applications, and mathematical connections. For instance, some of the most celebrated numbers in mathematics can be approximated using a series, as demonstrated in Example 11. EXAMPLE 11

䊳

Use a calculator to find the partial sums S4, S8, and S12 for the sequences given. If any sum seems to be approaching a fixed number, name that number. a. an ⫽

Solution

䊳

1 n!

b. an ⫽

1 3n

1 as u(n) on the Y= screen. Using the “sum(” and “seq(” n! commands as before produces the results shown in Figures 11.17 through 11.19, where it appears the sum becomes very close to e ⫺ 1 for larger values of n.

a. Begin by entering

Figure 11.17

Figure 11.18

Figure 11.19

1 as v(n) on the Y= screen, we once again compute the sums 3n indicated as shown in Figures 11.20 through 11.22. It appears the sum becomes 1 very close to for larger values of n. 2

b. After entering

Figure 11.20

E. You’ve just seen how we can use sequences to solve applications

Figure 11.21

Figure 11.22


䊳

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11.1 EXERCISES 䊳



1. A sequence is a(n) specific .

䊳

of numbers listed in a

2. A series is the given sequence.

of the numbers from a

3. A sequence that uses the preceding term(s) to generate those that follow is called a sequence.

4. The notation n! represents the natural number n, with all those

5. Describe the characteristics of a recursive sequence and give one example.

6. Describe the characteristics of an alternating sequence and give one example.

of the n.


Find the first four terms, then find the 8th and 12th term for each nth term given.

1 n 29. an ⫽ a1 ⫹ b ; a10 n

1 n 30. an ⫽ an ⫹ b ; a9 n

7. an ⫽ 2n ⫺ 1

8. an ⫽ 2n ⫹ 3

9. an ⫽ 3n ⫺ 3

10. an ⫽ 2n3 ⫺ 12

31. an ⫽

1 ; a4 n12n ⫹ 12

12. an ⫽

32. an ⫽

1 ; a5 12n ⫺ 12 12n ⫹ 12

2

11. an ⫽ 1⫺12 nn 13. an ⫽

n n⫹1

1 n 15. an ⫽ a b 2 17. an ⫽ 19. an ⫽

1 n

1 n 14. an ⫽ a1 ⫹ b n 2 n 16. an ⫽ a b 3 18. an ⫽

1⫺12 n

n1n ⫹ 12

21. an ⫽ 1⫺12 n2n

1⫺12 n n

20. an ⫽

1 n2 1⫺12 n⫹1 2n ⫺ 1 2

22. an ⫽ 1⫺12 n2⫺n

Use a calculator to: (a) find the indicated term for each sequence, and (b) generate the first five terms of each sequence and store the results in a list. Use fractions if possible; round to tenths when necessary.

23. an ⫽ n2 ⫺ 2; a9

24. an ⫽ 1n ⫺ 22 2; a9

25. an ⫽

26. an ⫽

1⫺12 n⫹1 ; a5 n

1 n⫺1 27. an ⫽ 2a b ; a7 2

1⫺12 n⫹1 2n ⫺ 1

; a5

1 n⫺1 28. an ⫽ 3a b ; a7 3

Find the first five terms of each recursive sequence.

33. e

a1 ⫽ 2 an ⫽ 5an⫺1 ⫺ 3

34. e

a1 ⫽ 3 an ⫽ 2an⫺1 ⫺ 3

35. e

a1 ⫽ ⫺1 an ⫽ 1an⫺1 2 2 ⫹ 3

36. e

a1 ⫽ ⫺2 an ⫽ an⫺1 ⫺ 16

38. e

c1 ⫽ 1, c2 ⫽ 2 cn ⫽ cn⫺1 ⫹ 1cn⫺2 2 2

c1 ⫽ 64, c2 ⫽ 32 37. • cn⫺2 ⫺ cn⫺1 cn ⫽ 2

Simplify each factorial expression.

39.

8! 5!

40.

12! 10!

41.

9! 7!2!

42.

6! 3!3!

43.

8! 2!6!

44.

10! 3!7!

Write out the first four terms in each sequence.

45. an ⫽ 47. an ⫽

n! 1n ⫹ 12!

46. an ⫽

13n2!

48. an ⫽

1n ⫹ 12!

n! 1n ⫹ 32!

1n ⫹ 32! 12n2!

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nn n!

49. an ⫽

50. an ⫽

71. a. 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ p 1 1 1 1 1 b. 1 ⫹ ⫹ ⫹ ⫹ ⫹ ⫹p 2 4 8 16 32 72. a. 0.1 ⫹ 0.01 ⫹ 0.001 ⫹ 0.0001 ⫹ p

2n n!

Find the indicated partial sum for each sequence.

51. an ⫽ n; S5

52. an ⫽ n2; S7

53. an ⫽ 2n ⫺ 1; S8

54. an ⫽ 3n ⫺ 1; S6

1 55. an ⫽ ; S5 n

56. an ⫽

b. 1 ⫹

n ; S4 n⫹1

57.

兺 13i ⫺ 52

5

58.

i⫽1 5

60.

兺 12i ⫺ 32

兺 1k

⫹ 12

61.

4

4

兺兺

64.

j⫽3 2

j

兺 1⫺12 2

k k

k⫽1 7

兺i 8

67.

62.

2

i⫽2

j

5

兺 1⫺12 k

k⫽1

兺 12k ⫺ 32 2

73. an ⫽ n ⫹ 3; S5 74. an ⫽

n2 ⫹ 1 ; S4 n⫹1

75. an ⫽

n2 ; third partial sum 3

k⫽1

k

k⫽1

7

59.

7

2

i2 63. i⫽1 2 66.

5

i⫽1

65.

兺 2j

j⫽3

1⫺12 k

6

兺 k1k ⫺ 22

68.

k⫽3

1⫺12 k⫹1

兺k

k⫽2

2

⫺1

Write each sum using sigma notation. Answers are not necessarily unique.

76. an ⫽ 2n ⫺ 1; sixth partial sum 77. an ⫽

Compute each sum by applying properties of summation. 5

69. a. 4 ⫹ 8 ⫹ 12 ⫹ 16 ⫹ 20 b. 5 ⫹ 10 ⫹ 15 ⫹ 20 ⫹ 25 70. a. ⫺1 ⫹ 4 ⫺ 9 ⫹ 16 ⫺ 25 ⫹ 36 b. 1 ⫺ 8 ⫹ 27 ⫺ 64 ⫹ 125 ⫺ 216

81.

兺

i⫽1

80.

13k2 ⫹ k2

82.

4

兺

6

14i ⫺ 52

k⫽1

兺 13 ⫹ 2i2

i⫽1 4

兺 12k

3

⫹ 52

k⫽1


83. Sum of an ⴝ 3n ⴚ 2: Sn ⴝ

n13n ⴚ 12 2

The sum of the first n terms of the sequence defined by an ⫽ 3n ⫺ 2 ⫽ 1, 4, 7, 10, p , 13n ⫺ 22, p is given by the formula shown. Find S5 using the formula, then verify by direct calculation. 䊳

n ; sum for n ⫽ 3 to 7 2n

78. an ⫽ n2; sum for n ⫽ 2 to 6

79.

䊳

1 1 1 1 1 ⫹ ⫹ ⫹ ⫹ ⫹p 2 6 24 120 720

For the given general term an, write the indicated sum using sigma notation.

Expand and evaluate each series. Verify results using a graphing calculator. 4

1087

n13n ⴙ 12 2 The sum of the first n terms of the sequence defined by an ⫽ 3n ⫺ 1 ⫽ 2, 5, 8, 11, p , 13n ⫺ 12, p is given by the formula shown. Find S8 using the formula, then verify by direct calculation. Observing the formulas from Exercises 83 and 84, can you now state the sum formula for an ⫽ 3n ⫺ 0?

84. Sum of an ⴝ 3n ⴚ 1: Sn ⴝ

APPLICATIONS

Use the information given in each exercise to determine the nth term an for the sequence described. Then use the nth term to list the specified number of terms.

85. Wage increases: Latisha gets $7.25 an hour for filling candy machines for Archtown Vending. Each year she receives a $0.50 hourly raise. List Latisha’s hourly wage for the first 5 yr. How much will she make in the fifth year if she works 8 hr per day for 240 working days?

86. Average birth weight: The average birth weight of a certain animal species is 900 g, with the baby gaining 125 g each day for the first 10 days. List the infant’s weight for the first 10 days. How much does the infant weigh on the 10th day?

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87. Blue-book value: Steve’s car has a blue-book value of $6000. Each year it loses 20% of its value (its value each year is 80% of the year before). List the value of Steve’s car for the next 5 yr. (Hint: a0 ⫽ 6000.) 88. Effects of inflation: Suppose inflation (an increase in value) will average 4% for the next 5 yr. List the growing cost (year by year) of a DVD that costs $15 right now. (Hint: a0 ⫽ 15.) 89. Stocking a lake: A local fishery stocks a large lake with 1500 bass and then adds an additional 100 mature bass per month until the lake nears maximum capacity. If the bass population grows at a rate of 5% per month through natural reproduction, the number of bass in the pond after n months is given by the recursive sequence b0 ⫽ 1500, bn ⫽ 1.05bn⫺1 ⫹ 100. How many bass will be in the lake after 6 months? 䊳

11–12


90. Species preservation: The Interior Department introduces 50 wolves (male and female) into a large wildlife area in an effort to preserve the species. Each year about 12 additional adult wolves are added from capture and relocation programs. If the wolf population grows at a rate of 10% per year through natural reproduction, the number of wolves in the area after n years is given by the recursive sequence w0 ⫽ 50, wn ⫽ 1.10wn⫺1 ⫹ 12. How many wolves are in the wildlife area after 6 years? Use your calculator to find the partial sums for n ⴝ 4, n ⴝ 8, and n ⴝ 12 for the summations given, and attempt to name the number the summation approximates: n

91.

2k ⫹ 3k 6k k⫽1 n

1 k k⫽1 2

兺

92.

兺


93. Verify that a constant can be factored out of a sum. That is, verify that the following statement is true: n

兺

n

caj ⫽ c

j⫽1

兺a

94. Verify that a summation may be distributed to two (or more) sequences. That is, verify that the following statement is true: n

j

兺

j⫽1

i⫽1

1ai ⫾ bi 2 ⫽

n

兺

n

ai ⫾

i⫽1

兺b . i

i⫽1

Regarding Exercises 91 and 92, sometimes a series will approach a fixed number very slowly, and many more terms must be added before this value is recognized. Use your graphing calculator to compute the sums S10, S25, and S50 for the following sequences to see if you can recognize the number. Add more terms if necessary.

95. an ⫽

䊳

1 n1n ⫹ 12 1n ⫹ 22

96. an ⫽

1 12n ⫺ 12 12n ⫹ 12


97. (7.7) Solve csc x sina

␲ ⫺ xb ⫽ ⫺1 2

98. (3.6) Set up the difference quotient for f 1x2 ⫽ 1x, then rationalize the numerator.

99. (8.2) Given a triangle where a ⫽ 0.4 m, b ⫽ 0.3 m, and c ⫽ 0.5 m, find the three corresponding angles. 100. (9.7) Solve the system using a matrix equation. 25x ⫹ y ⫺ 2z ⫽ ⫺14 • 2x ⫺ y ⫹ z ⫽ 40 ⫺7x ⫹ 3y ⫺ z ⫽ ⫺13

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11.2

Arithmetic Sequences

LEARNING OBJECTIVES

Similar to the way polynomials fall into certain groups or families (linear, quadratic, cubic, etc.), sequences and series with common characteristics are likewise grouped. In this section, we focus on sequences where each successive term is generated by adding a constant value, as in the sequence 1, 8, 15, 22, 29, p , where 7 is added to a given term in order to produce the next term.


A. Identify an arithmetic sequence and its common difference B. Find the n th term of an arithmetic sequence C. Find the n th partial sum of an arithmetic sequence D. Solve applications involving arithmetic sequences

A. Identifying an Arithmetic Sequence and Finding the Common Difference An arithmetic sequence is one where each successive term is found by adding a fixed constant to the preceding term. For instance 3, 7, 11, 15, p is an arithmetic sequence, since adding 4 to any given term produces the next term. This also means if you take the difference of any two consecutive terms, the result will be 4 and in fact, 4 is called the common difference d for this sequence. Using the notation developed earlier, we can write d ⫽ ak⫹1 ⫺ ak, where ak represents any term of the sequence and ak⫹1 represents the term that follows ak. Arithmetic Sequences Given a sequence a1, a2, a3, p , ak, ak⫹1, p , an, where k, n 僆⺞ and k 6 n, if there exists a common difference d such that ak⫹1 ⫺ ak ⫽ d for all k, then the sequence is an arithmetic sequence. The difference of successive terms can be rewritten as ak⫹1 ⫽ ak ⫹ d (for k ⱖ 12 to highlight that each following term is found by adding d to the previous term.

EXAMPLE 1

䊳

Identifying an Arithmetic Sequence Determine if the given sequence is arithmetic. If yes, name the common difference. If not, try to determine the pattern that forms the sequence. 77 29 a. 2, 5, 8, 11, p b. 12, 56, 13 12 , 60 , 20 , p

Solution

䊳

a. Begin by looking for a common difference d ⫽ ak⫹1 ⫺ ak. Checking each pair of consecutive terms we have 5⫺2⫽3

8⫺5⫽3

11 ⫺ 8 ⫽ 3 and so on.

This is an arithmetic sequence with common difference d ⫽ 3. b. Checking each pair of consecutive terms yields 5 1 5 3 ⫺ ⫽ ⫺ 6 2 6 6 2 1 ⫽ ⫽ 6 3

13 5 13 10 ⫺ ⫽ ⫺ 12 6 12 12 3 1 ⫽ ⫽ 12 4

13 77 65 77 ⫺ ⫽ ⫺ 60 12 60 60 12 1 ⫽ ⫽ 60 5

Since the difference is not constant, this is not an arithmetic sequence. It appears the sequence is formed by adding 1k to each previous term, for natural numbers k. Now try Exercises 7 through 18

11–13

䊳

1089

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EXAMPLE 2

䊳

Writing the First k Terms of an Arithmetic Sequence Write the first five terms of the arithmetic sequence, given the first term a1 and the common difference d. a. a1 ⫽ 12 and d ⫽ ⫺4 b. a1 ⫽ 12 and d ⫽ 13

Solution

䊳

A. You’ve just seen how we can identify an arithmetic sequence and its common difference

a. a1 ⫽ 12 and d ⫽ ⫺4. Starting at a1 ⫽ 12, add ⫺4 to each new term to generate the sequence: 12, 8, 4, 0, ⫺4. b. a1 ⫽ 12 and d ⫽ 13. Starting at a1 ⫽ 12 and adding 13 to each new term will generate the sequence: 12, 56, 76, 32, 11 6 . Note that since the common denominator is 6, terms of the sequence can quickly be found by adding 13 ⫽ 26 to the previous term and reducing if possible. Now try Exercises 19 through 30

䊳

B. Finding the nth Term of an Arithmetic Sequence If the values a1 and d from an arithmetic sequence are known, we could generate the terms of the sequence by adding multiples of d to the first term, instead of adding d to each new term. For example, we can generate the sequence 3, 8, 13, 18, 23 by adding multiples of 5 to the first term a1 ⫽ 3: 3 ⫽ 3 ⫹ 1025

a1 ⫽ a1 ⫹ 0d

13 ⫽ 3 ⫹ 1225

a3 ⫽ a1 ⫹ 2d

8 ⫽ 3 ⫹ 1125

a2 ⫽ a1 ⫹ 1d

18 ⫽ 3 ⫹ 1325

a4 ⫽ a1 ⫹ 3d

23 ⫽ 3 ⫹ 1425

current term

initial term

S

S

a5 ⫽ a1 ⫹ 4d coefficient of common difference

It’s helpful to note the coefficient of d is 1 less than the subscript of the current term (as shown): 5 ⫺ 1 ⫽ 4. This observation leads us to a formula for the nth term. The n th Term of an Arithmetic Sequence The nth term of an arithmetic sequence is given by

an ⫽ a1 ⫹ 1n ⫺ 12d

where d is the common difference.

EXAMPLE 3

䊳

Finding a Specified Term in an Arithmetic Sequence Find the 24th term of the sequence 0.1, 0.4, 0.7, 1, p .

Solution

䊳

Instead of creating all terms up to the 24th, we determine the constant d and use the nth term formula. By inspection we note a1 ⫽ 0.1 and d ⫽ 0.3. an ⫽ a1 ⫹ 1n ⫺ 12d ⫽ 0.1 ⫹ 1n ⫺ 120.3 ⫽ 0.1 ⫹ 0.3n ⫺ 0.3 ⫽ 0.3n ⫺ 0.2

n th term formula substitute 0.1 for a1 and 0.3 for d eliminate parentheses simplify

To find the 24th term we substitute 24 for n: a24 ⫽ 0.31242 ⫺ 0.2 ⫽ 7.0

substitute 24 for n result


䊳

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Section 11.2 Arithmetic Sequences

EXAMPLE 4

䊳

Determining the Number of Terms in an Arithmetic Sequence Find the number of terms in the arithmetic sequence 2, ⫺5, ⫺12, ⫺19, p , ⫺411.

Solution

䊳

By inspection we see that a1 ⫽ 2 and d ⫽ ⫺7. As before, an ⫽ a1 ⫹ 1n ⫺ 12d ⫽ 2 ⫹ 1n ⫺ 12 1⫺72 ⫽ 2 ⫺ 7n ⫹ 7 ⫽ ⫺7n ⫹ 9

n th term formula substitute 2 for a1 and ⫺7 for d distribute ⫺7 simplify

Although we don’t know the number of terms in the sequence, we do know the last or nth term is ⫺411. Substituting ⫺411 for an gives ⫺411 ⫽ ⫺7n ⫹ 9 60 ⫽ n

substitute ⫺411 for an solve for n

There are 60 terms in this sequence. Now try Exercises 43 through 50

䊳

Note that in both Examples 3 and 4, the nth term had the form of a linear equation 1y ⫽ mx ⫹ b2 after simplifying: an ⫽ 0.3n ⫺ 0.2 and an ⫽ ⫺7n ⫹ 9. This is a characteristic of arithmetic sequences, with the common difference d corresponding to the slope m. This means the graph of an arithmetic sequence will always be a set of discrete points that lie on a straight line. After entering u1n2 ⫽ 0.3n ⫺ 0.2 (from Example 3), Figure 11.23 shows the table of values for an ⫽ 0.3n ⫺ 0.2, with the graph in Figure 11.24. Figure 11.24

Figure 11.23

5

10

0

⫺1.5

Figure 11.25 In sequence MODE , we still set the size of the viewing window as before, but we can also stipulate the range of values of n to be used (nMin and nMax), which term we want to plot as a beginning (PlotStart), and whether we want all following terms graphed (PlotStep ⫽ 1), every second term graphed (PlotStep ⫽ 2), and so on. The graph in Figure 11.24 was generated using the values shown in Figure 11.25 (Ymin ⫽ ⫺1.5, Ymax ⫽ 5, and Yscl ⫽ 1 cannot be seen). To see the graph of a sequence more distinctly, we can enter the natural numbers 1 through 10 in L1, then define L2 as u(L1) (Figures 11.26 and 11.27) and plot these points (Figure 11.28).

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Figure 11.27

5

10

0

⫺1.5

One additional advantage of this approach is that we can easily verify the common difference is d using the “ ¢ List(” feature, which automatically computes the difference between each successive term in a specified list. This option is located in the same submenu as the “seq(” option, accessed using 2nd STAT (LIST) (OPS) 7: ¢ List(L2) . See Figures 11.29 and 11.30. ENTER

Figure 11.29

EXAMPLE 5

䊳

Figure 11.30

Graphing Arithmetic Sequences Enter the natural numbers 1 through 6 in L1, and the terms of the sequence ⫺0.45, ⫺0.1, 0.25, 0.6, 0.95, 1.3 in L2. Then verify the sequence in L2 is arithmetic by a. Graphing the related points to see if they appear linear. b. Using the ¢ List( feature. c. Finding the nth term that defines the sequence and graph the sequence.

Solution

䊳

a. The plotted points are shown in Figure 11.31 and appear to be linear. b. The ¢ List( feature shows there is a common difference of 0.35 (Figure 11.32). c. With a1 ⫽ ⫺0.45 and d ⫽ 0.35, the nth term must be an ⫽ a1 ⫹ 1n ⫺ 12d ⫽ ⫺0.45 ⫹ 1n ⫺ 12 10.352 ⫽ 0.35n ⫺ 0.8

Figure 11.31 2

⫺1

simplify

The nth term for this sequence is an ⫽ 0.35n ⫺ 0.8 . The graph is shown in Figure 11.33. Figure 11.33 Figure 11.32 2

7

0

n th term formula substitute for a1 and d

7

0

⫺1


䊳

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If the term a1 is unknown but a term ak is given, the nth term can be written an ⫽ ak ⫹ 1n ⫺ k2d

(the subscript of the term ak and coefficient of d sum to n). EXAMPLE 6

䊳

Finding the First Term of an Arithmetic Sequence Given an arithmetic sequence where a6 ⫽ 0.55 and a13 ⫽ 0.9, find the common difference d and the value of a1.

Solution

䊳

At first it seems that not enough information is given, but recall we can express a13 as the sum of any earlier term and the appropriate multiple of d. Since a6 is known, we write a13 ⫽ a6 ⫹ 7d (note 13 ⫽ 6 ⫹ 7 as required). a1 is unknown a13 ⫽ a6 ⫹ 7d 0.9 ⫽ 0.55 ⫹ 7d substitute 0.9 for a13 and 0.55 for a6 0.35 ⫽ 7d subtract 0.55 d ⫽ 0.05 solve for d Having found d, we can now solve for a1. a13 ⫽ a1 ⫹ 12d 0.9 ⫽ a1 ⫹ 1210.052 0.9 ⫽ a1 ⫹ 0.6 a1 ⫽ 0.3

B. You’ve just seen how we can find the nth term of an arithmetic sequence

n th term formula for n ⫽ 13 substitute 0.9 for a13 and 0.05 for d simplify solve for a1

The first term is a1 ⫽ 0.3 and the common difference is d ⫽ 0.05. Now try Exercises 55 through 60

䊳

C. Finding the nth Partial Sum of an Arithmetic Sequence Using sequences and series to solve applications often requires computing the sum of a given number of terms. To develop and understand the approach used, consider the sum of the first 10 natural numbers. Using S10 to represent this sum, we have S10 ⫽ 1 ⫹ 2 ⫹ 3 ⫹ 4 ⫹ 5 ⫹ 6 ⫹ 7 ⫹ 8 ⫹ 9 ⫹ 10. We could just use brute force, but if we rewrite the sum a second time but in reverse order, then add it to the first, we find that each column adds to 11. S10 ⫽ 1 ⫹ 2 ⫹ 3 ⫹ 4 ⫹ 5 ⫹ 6 ⫹ 7 ⫹ 8 ⫹ 9 ⫹ 10 S10 ⫽ 10 ⫹ 9 ⫹ 8 ⫹ 7 ⫹ 6 ⫹ 5 ⫹ 4 ⫹ 3 ⫹ 2 ⫹ 1 2S10 ⫽ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 Since there are 10 columns, the total is 11 ⫻ 10 ⫽ 110 but this is twice the actual sum and we find that 2S10 ⫽ 110, so S10 ⫽ 55. Now consider the sequence a1, a2, a3, a4, p , an with common difference d. Use Sn to represent the sum of the first n terms and write the original series, then the series in reverse order underneath. Since one row increases at the same rate the other decreases, the sum of each column remains constant, and for simplicity’s sake we choose a1 ⫹ an to represent this sum. Sn ⫽ a1 ⫹ a2 ⫹ a3 ⫹ p ⫹ an⫺2 ⫹ an⫺1 ⫹ an add p columns Sn ⫽ an ⫹ an⫺1 ⫹ an⫺2 ⫹ ⫹ a3 ⫹ a2 ⫹ a1 p 2Sn ⫽ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ⫹ ⫹ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ↓ vertically To understand why each column adds to a1 ⫹ an, consider the sum in the second column: a2 ⫹ an⫺1. From a2 ⫽ a1 ⫹ d and an⫺1 ⫽ an ⫺ d, we obtain a2 ⫹ an⫺1 ⫽ 1a1 ⫹ d2 ⫹ 1an ⫺ d2 by adding the equations, which gives a result of a1 ⫹ an. Since there are n columns, we end up with 2Sn ⫽ n1a1 ⫹ an 2, and solving for Sn gives the formula for the first n terms of an arithmetic sequence.

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The nth Partial Sum of an Arithmetic Sequence Given an arithmetic sequence with first term a1, the nth partial sum is given by Sn ⫽

n 1a1 ⫹ an 2. 2

In words: The sum of an arithmetic sequence is one-half the number of terms times the sum of the first and last term.

EXAMPLE 7

䊳

Computing the Sum of an Arithmetic Sequence Use the summation formula to find the sum of the first 75 positive odd integers: 75

兺 12n ⫺ 12 . Verify the result using a graphing calculator.

n⫽1

Solution

䊳

The initial terms of the sequence are 1, 3, 5, p and we note a1 ⫽ 1, d ⫽ 2, and n ⫽ 75. To use the sum formula, we need the value of a75: 21752 ⫺ 1 ⫽ 149. n 1a1 ⫹ an 2 2 75 ⫽ 1a1 ⫹ a75 2 2 75 ⫽ 11 ⫹ 1492 2 ⫽ 5625

Sn ⫽ S75

sum formula

substitute 75 for n

substitute 1 for a1, 149 for a75 result

The sum of the first 75 positive odd integers is 5625. To verify, we enter u1n2 ⫽ 2n ⫺ 1 on the Y= screen, and find the sum of the first 75 terms of the sequence on the home screen as before. See figure.


䊳

By substituting the nth term formula directly into the formula for partial sums, we’re able to find a partial sum without actually having to find the nth term: n 1a1 ⫹ an 2 2 n ⫽ 1a1 ⫹ 3a1 ⫹ 1n ⫺ 12d 4 2 2 n ⫽ 3 2a1 ⫹ 1n ⫺ 12d 4 2

Sn ⫽

C. You’ve just seen how we can find the nth partial sum of an arithmetic sequence

sum formula substitute a1 ⫹ 1n ⫺ 12d for an

alternative formula for the nth partial sum

See Exercises 67 through 72 for more on this alternative formula.

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D. Applications In the evolution of certain plants and shelled animals, sequences and series seem to have been one of nature’s favorite tools. The sprials found on many ferns and other plants are excellent examples of sequences in nature, as are the size of the chambers in nautilus shells (see Figures 11.34 and 11.35). Sequences and series also provide a good mathematical model for a variety of other situations as well.

Figure 11.34

spiral fern

EXAMPLE 8

Figure 11.35

nautilus

䊳

Solving an Application of Arithmetic Sequences: Seating Capacity Cox Auditorium is an amphitheater that has 40 seats in the first row, 42 seats in the second row, 44 in the third, and so on. If there are 75 rows in the auditorium, what is the auditorium’s seating capacity?

Solution

䊳

D. You’ve just seen how we can solve applications involving arithmetic sequences

The number of seats in each row gives the terms of an arithmetic sequence with a1 ⫽ 40, d ⫽ 2, and n ⫽ 75. To find the seating capacity, we need to find the total number of seats, which is the sum of this arithmetic sequence. Since the value of n a75 is unknown, we opt for the alternative formula Sn ⫽ 3 2a1 ⫹ 1n ⫺ 12d4 . 2 n sum formula Sn ⫽ 3 2a1 ⫹ 1n ⫺ 12d 4 2 75 S75 ⫽ 3 21402 ⫹ 175 ⫺ 12122 4 substitute 40 for a1, 2 for d, and 75 for n 2 75 simplify ⫽ 12282 2 ⫽ 8550 result The seating capacity of Cox Auditorium is 8550. Now try Exercises 75 through 80

䊳

11.2 EXERCISES 䊳



1. Consecutive terms in an arithmetic sequence differ by a constant called the .

2. The sum of the first n terms of an arithmetic sequence is called the nth .

3. The formula for the nth partial sum of an

4. The nth term formula for an arithmetic sequence is an ⫽ term , where a1 is the and d is the .

arithmetic sequence is Sn ⫽ is the term.

, where an

5. Discuss how the terms of an arithmetic sequence can be written in various ways using the relationship an ⫽ ak ⫹ 1n ⫺ k2d.

6. Describe how the formula for the nth partial sum was derived, and illustrate its application using a sequence from the exercise set.

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Determine if the sequence given is arithmetic. If yes, name the common difference. If not, try to determine the pattern that forms the sequence.

7. ⫺5, ⫺2, 1, 4, 7, 10, p

41. a1 ⫽ ⫺0.025, d ⫽ 0.05; find a50 42. a1 ⫽ 3.125, d ⫽ ⫺0.25; find a20 Find the number of terms in each sequence.

8. 1, ⫺2, ⫺5, ⫺8, ⫺11, ⫺14, p

43. a1 ⫽ 2, an ⫽ ⫺22, d ⫽ ⫺3

9. 0.5, 3, 5.5, 8, 10.5, p

44. a1 ⫽ 4, an ⫽ 42, d ⫽ 2

10. 1.2, 3.5, 5.8, 8.1, 10.4, p

45. a1 ⫽ 0.4, an ⫽ 10.9, d ⫽ 0.25

11. 2, 3, 5, 7, 11, 13, 17, p

46. a1 ⫽ ⫺0.3, an ⫽ ⫺36, d ⫽ ⫺2.1

12. 1, 4, 8, 13, 19, 26, 34, p

47. ⫺3, ⫺0.5, 2, 4.5, 7, p , 47

13.

1 1 1 1 5 24 , 12 , 8 , 6 , 24 ,

14.

1 1 1 1 1 12 , 15 , 20 , 30 , 60 ,

48. ⫺3.4, ⫺1.1, 1.2, 3.5, p , 38

p

49.

p

15. 1, 4, 9, 16, 25, 36, p 16. ⫺125, ⫺64, ⫺27, ⫺8, ⫺1, p 17. ␲,

5␲ 2␲ ␲ ␲ ␲ , , , , ,p 6 3 2 3 6

18. ␲,

7␲ 3␲ 5␲ ␲ , , , ,p 8 4 8 2

1 1 1 5 1 12 , 8 , 6 , 24 , 4 ,

p , 89

50.

1 1 1 1 12 , 15 , 20 , 30 ,

p , ⫺14

For Exercises 51 through 54, enter the natural numbers 1 through 6 in L1 on a graphing calculator, and the terms of the given sequence in L2. Then determine if the sequence is arithmetic by (a) graphing the related points to see if they appear linear, and (b) using the ¢ List( feature. If an arithmetic sequence, (c) find the nth term and graph the sequence.

51. 1.5, 2.25, 3, 3.75, 4.5, 5.25, p Write the first four terms of the arithmetic sequence with the given first term and common difference.

19. a1 ⫽ 2, d ⫽ 3

20. a1 ⫽ 8, d ⫽ 3

21. a1 ⫽ 7, d ⫽ ⫺2

22. a1 ⫽ 60, d ⫽ ⫺12

23. a1 ⫽ 0.3, d ⫽ 0.03

24. a1 ⫽ 0.5, d ⫽ 0.25

25. a1 ⫽

1 26. a1 ⫽ 15, d ⫽ 10

3 2,

d⫽

1 2

27. a1 ⫽ 34, d ⫽ ⫺18

28. a1 ⫽ 16, d ⫽ ⫺13

29. a1 ⫽ ⫺2, d ⫽ ⫺3

30. a1 ⫽ ⫺4, d ⫽ ⫺4

Identify the first term and the common difference, then write the expression for the general term an and use it to find the 6th, 10th, and 12th terms of the sequence.

31. 2, 7, 12, 17, p

32. 7, 4, 1, ⫺2, ⫺5, p

33. 5.10, 5.25, 5.40, p

34. 9.75, 9.40, 9.05, p

35. 32, 94, 3, 15 4,p

3 36. 57, 14 , ⫺27, ⫺11 14 , p

52.

53. 9, 8, 6, 3, ⫺1, ⫺6, p

55. a3 ⫽ 7, a7 ⫽ 19

40. a1 ⫽

d⫽

1 ⫺10 ;

find a9

56. a5 ⫽ ⫺17, a11 ⫽ ⫺2

58. a6 ⫽ ⫺12.9, a30 ⫽ 1.5 27 59. a10 ⫽ 13 18 , a24 ⫽ 2

60. a4 ⫽ 54, a8 ⫽ 94

Evaluate each sum. For Exercises 65 and 66, use the summation properties from Section 11.1. Verify all results on a graphing calculator. 30

兺

n⫽1 37

63.

38. a1 ⫽ 9, d ⫽⫺2; find a17 12 25 ,

1 1 1 1 1 1 , , , , , ,p 1 2 3 4 5 6

57. a2 ⫽ 1.025, a26 ⫽ 10.025

Find the indicated term using the information given.

1 39. a1 ⫽ 32, d ⫽ ⫺12 ; find a7

54.

Find the common difference d and the value of a1 using the information given.

61. 37. a1 ⫽ 5, d ⫽ 4; find a15

47 19 29 10 11 1 , , , , , ,p 18 9 18 9 18 9

13n ⫺ 42

兺 a 4 n ⫹ 2b 3

29

62.

20

64.

n⫽1 15

65.

兺

n⫽4

13 ⫺ 5n2

兺 14n ⫺ 12

n⫽1

兺 a 2 n ⫺ 3b 5

n⫽1 20

66.

兺 17 ⫺ 2n2

n⫽7

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1097

Use the alternative formula for the nth partial sum to compute the sums indicated.

67. The sum S15 for the sequence ⫺12 ⫹ 1⫺9.52 ⫹ 1⫺72 ⫹ 1⫺4.52 ⫹ p

68. The sum S20 for the sequence 92 ⫹ 72 ⫹ 52 ⫹ 32 ⫹ p 69. The sum S30 for the sequence 0.003 ⫹ 0.173 ⫹ 0.343 ⫹ 0.513 ⫹ p

䊳

71. The sum S20 for the sequence 12 ⫹ 2 12 ⫹ 3 12 ⫹ 4 12 ⫹ p 72. The sum S10 for the sequence 12 13 ⫹ 10 13 ⫹ 8 13 ⫹ 613 ⫹ p


73. Sum of the first n natural numbers: Sn ⴝ

n1n ⴙ 12 2

The sum of the first n natural numbers can be found using the formula shown, where n represents the number of terms in the sum. Verify the formula by adding the first six natural numbers by hand, and then evaluating S6. Then find the sum of the first 75 natural numbers.

䊳

70. The sum S50 for the sequence 1⫺22 ⫹ 1⫺72 ⫹ 1⫺122 ⫹ 1⫺172 ⫹ p

74. Sum of the squares of the first n natural n1n ⴙ 1212n ⴙ 12 numbers: Sn ⴝ 6 If the first n natural numbers are squared, the sum of these squares can be found using the formula shown, where n represents the number of terms in the sum. Verify the formula by computing the sum of the squares of the first six natural numbers by hand, and then evaluating S6. Then find the sum of the squares of the first 20 natural numbers: 112 ⫹ 22 ⫹ 32 ⫹ p ⫹ 202 2.

APPLICATIONS

75. Temperature fluctuation: At 5 P.M. in Coldwater, the temperature was a chilly 36°F. If the temperature decreased by 3°F every half-hour for the next 7 hr, at what time did the temperature hit 0°F? 76. Arc of a baby swing: When Mackenzie’s baby swing is started, the first swing (one way) is a 30-in. arc. As the swing slows down, each successive arc is 23 in. less than the previous one. Find (a) the length of the tenth swing and (b) how far Mackenzie has traveled during the 10 swings. 77. Computer animations: The animation on a new computer game initially allows the hero of the game to jump a (screen) distance of 10 in. over booby traps and obstacles. Each successive jump is limited to 34 in. less than the previous one. Find (a) the length of the seventh jump and (b) the total distance covered after seven jumps. 78. Seating capacity: The Fox Theater creates a “theater in the round” when it shows any of Shakespeare’s plays. The first row has 80 seats, the second row has 88,

the third row has 96, and so on. How many seats are in the 10th row? If there is room for 25 rows, how many chairs will be needed to set up the theater? 79. Sales goals: At the time that I was newly hired, 100 sales per month was what I required. Each following month — the last plus 20 more, as I work for the goal of top sales award. When 2500 sales are thusly made, it’s Tahiti, Hawaii, and piña coladas in the shade. How many sales were made by this person in the seventh month? What were the total sales after the 12th month? Was the goal of 2500 total sales met? 80. Bequests to charity: At the time our mother left this Earth, she gave $9000 to her children of birth. This we kept and each year added $3000 more, as a lasting memorial from the children she bore. When $42,000 is thusly attained, all goes to charity that her memory be maintained. What was the balance in the sixth year? In what year was the goal of $42,000 met?

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81. From a study of numerical analysis, a function is known to be linear if its “first differences” (differences between successive outputs) are constant. Likewise, a function is known to be quadratic if its “first differences” form an arithmetic sequence. Use this information to determine if the following sets of output come from a linear or quadratic function: a. 19, 11.8, 4.6, ⫺2.6, ⫺9.8, ⫺17, ⫺24.2, p b. ⫺10.31, ⫺10.94, ⫺11.99, ⫺13.46, ⫺15.35, p 䊳

82. From elementary geometry it is known that the interior angles of a triangle sum to 180°, the interior angles of a quadrilateral sum to 360°, the interior angles of a pentagon sum to 540°, and so on. Use the pattern created by the relationship between the number of sides and the number of angles to develop a formula for the sum of the interior angles of an n-sided polygon. The interior angles of a decagon (10 sides) sum to how many degrees?

MAINTAINING YOUR SKILLS 85. (1.4) In 2000, the deer population was 972. By 2005 it had grown to 1217. Assuming the growth is linear, find the function that models this data and use it to estimate the deer population in 2008.

83. (6.5) Identify the amplitude (A), period (P), horizontal shift (HS), vertical shift (VS) and endpoints of the translated interval corresponding ␲ ␲ to [0, 2␲) for f 1t2 ⫽ 7 sin a t ⫺ b ⫹ 10. 3 6

86. (7.1) Verify

84. (3.3) Graph by completing the square. Label all important features: y ⫽ x2 ⫺ 2x ⫺ 3.

11.3


A. Identify a geometric

C. D. E.

identity.

Geometric Sequences

LEARNING OBJECTIVES

B.

tan x sin x sin x ⫺ 1 is an ⫺ ⫽ csc x cos x cot x

sequence and its common ratio Find the n th term of a geometric sequence Find the n th partial sum of a geometric sequence Find the sum of an infinite geometric series Solve application problems involving geometric sequences and series

Recall that arithmetic sequences are those where each term is found by adding a constant value to the preceding term. In this section, we consider geometric sequences, where each term is found by multiplying the preceding term by a constant value. Geometric sequences have many interesting applications, as do geometric series.

A. Geometric Sequences A geometric sequence is one where each successive term is found by multiplying the preceding term by a fixed constant. Consider growth of a bacteria population, where a single cell splits in two every hour over a 24-hr period. Beginning with a single bacterium 1a0 ⫽ 12, after 1 hr there are 2, after 2 hr there are 4, and so on. Writing the number of bacteria as a sequence we have: hours: bacteria:

a1 T 2

a2 T 4

a3 T 8

a4 T 16

a5 T 32

p p

The sequence 2, 4, 8, 16, 32, p is a geometric sequence since each term is found by multiplying the previous term by the constant factor 2. This also means that the ratio of any two consecutive terms must be 2 and in fact, 2 is called the common ratio r ak⫹1 , where for this sequence. Using the notation from Section 11.1 we can write r ⫽ ak ak represents any term of the sequence and ak⫹1 represents the term that follows ak.

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Section 11.3 Geometric Sequences

1099

Geometric Sequences Given a sequence a1, a2, a3, p , ak, ak⫹1, p , an, where k, n 僆⺞ and k 6 n, ak⫹1 if there exists a common ratio r such that ⫽ r for all k, ak then the sequence is a geometric sequence. The ratio of successive terms can be rewritten as ak⫹1 ⫽ akr (for k ⱖ 12 to highlight that each term is found by multiplying the preceding term by r. EXAMPLE 1

Solution

䊳

䊳

Testing a Sequence for a Common Ratio Determine if the given sequence is geometric. If not geometric, try to determine the pattern that forms the sequence. 120 a. 1, 0.5, 0.25, 0.125, p b. 17, 27, 67, 24 7, 7 ,p ak⫹1 Apply the definition to check for a common ratio r ⫽ . ak a. For 1, 0.5, 0.25, 0.125, p , the ratio of consecutive terms gives 0.5 0.25 0.125 ⫽ 0.5, ⫽ 0.5, ⫽ 0.5, and so on. 1 0.5 0.25 This is a geometric sequence with common ratio r ⫽ 0.5. 120 b. For 17, 27, 67, 24 7 , 7 , p , we have: 2 1 2 7 6 2 6 7 6 24 # 7 24 and so on. ⫼ ⫽ # ⫼ ⫽ # ⫼ ⫽ 7 7 7 1 7 7 7 2 7 7 7 6 ⫽2 ⫽3 ⫽4 Since the ratio is not constant, this is not a geometric sequence. The sequence n! appears to be formed by dividing n! by 7: an ⫽ . 7 Now try Exercises 7 through 24

EXAMPLE 2

䊳

䊳

Writing the Terms of a Geometric Sequence Write the first five terms of the geometric sequence, given the first term a1 ⫽ ⫺16 and the common ratio r ⫽ 0.25.

Solution

䊳

Given a1 ⫽ ⫺16 and r ⫽ 0.25. Starting at a1 ⫽ ⫺16, multiply each term by 0.25 to generate the sequence. a2 ⫽ ⫺16 # 0.25 ⫽ ⫺4 a4 ⫽ ⫺1 # 0.25 ⫽ ⫺0.25

A. You’ve just seen how we can identify a geometric sequence and its common ratio

a3 ⫽ ⫺4 # 0.25 ⫽ ⫺1 a5 ⫽ ⫺0.25 # 0.25 ⫽ ⫺0.0625

The first five terms of this sequence are ⫺16, ⫺4, ⫺1, ⫺0.25, and ⫺0.0625. Now try Exercises 25 through 32

䊳

B. Find the nth Term of a Geometric Sequence If the values a1 and r from a geometric sequence are known, we could generate the terms of the sequence by applying additional factors of r to the first term, instead of multiplying each new term by r. If a1 ⫽ 3 and r ⫽ 2, we simply begin at a1, and

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continue applying additional factors of r for each successive term. a1 ⫽ a1r0

6 ⫽ 3 # 21

a2 ⫽ a1r1

12 ⫽ 3 # 22

a3 ⫽ a1r2

24 ⫽ 3 # 23

a4 ⫽ a1r3

48 ⫽ 3 # 24

a5 ⫽ a1r4

current term

initial term

S

3 ⫽ 3 # 20

S

1100

exponent on common ratio

From this pattern, we note the exponent on r is always 1 less than the subscript of the current term: 5 ⫺ 1 ⫽ 4, which leads us to the formula for the nth term of a geometric sequence. The n th Term of a Geometric Sequence The nth term of a geometric sequence is given by an ⫽ a1r n⫺1 where r is the common ratio.

EXAMPLE 3

䊳

Finding a Specific Term in a Sequence Identify the common ratio r, and use it to write the expression for the nth term. Then find the 10th term of the sequence: 3, ⫺6, 12, ⫺24, p .

Solution

䊳

By inspection we note that a1 ⫽ 3 and r ⫽ ⫺2. This gives an ⫽ a1rn⫺1 ⫽ 31⫺22 n⫺1

n th term formula substitute 3 for a1 and ⫺2 for r

To find the 10th term we substitute n ⫽ 10: a10 ⫽ 31⫺22 10⫺1 ⫽ 31⫺22 9 ⫽ ⫺1536

substitute 10 for n simplify


EXAMPLE 4

䊳

䊳

Determining the Number of Terms in a Geometric Sequence 1 . Find the number of terms in the geometric sequence 4, 2, 1, p , 64

Solution

䊳

Observing that a1 ⫽ 4 and r ⫽ 12, we have an ⫽ a1 rn⫺1 1 n⫺1 ⫽ 4a b 2

n th term formula substitute 4 for a1 and

1 for r 2

Although we don’t know the number of terms in the sequence, we do know the last 1 1 . Substituting an ⫽ 64 or nth term is 64 gives 1 1 n⫺1 ⫽ 4a b 64 2 1 1 n⫺1 ⫽a b 256 2

substitute

1 for an 64

1 divide by 4 amultiply by b 4

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1101

From our work in Chapter 5, we attempt to write both sides as exponentials with a like base, or apply logarithms. Since 256 ⫽ 28, we equate bases. 1 8 1 n⫺1 a b ⫽a b 2 2 S8 ⫽ n ⫺ 1 9⫽n

write

1 1 8 as a b 256 2

like bases imply exponents must be equal solve for n

This shows there are nine terms in the sequence. Now try Exercises 47 through 58

WORTHY OF NOTE The sequence from Example 3 is an alternating sequence, and the exponential characteristics of its graph can be seen by taking a look at the graph of an ⫽ |31⫺22 n⫺1 | .

䊳

Note that in both Examples 3 and 4, the nth term had the form of an exponential 1 n⫺1 equation 1y ⫽ a # bn 2 after simplifying: an ⫽ 31⫺22 n⫺1 and an ⫽ 4a b . This is in 2 fact a characteristic of geometric sequences, with the common ratio r corresponding to the base b. This means the graph of a geometric sequence will always be a set of discrete points that lie on an exponential graph (see Worthy of Note). After entering 1 n⫺1 (from Example 4), Figure 11.36 shows the table of values for this u1n2 ⫽ 4a b 2 sequence, with the graph in Figure 11.37. As before, we can see the graph of the sequence more distinctly by entering the natural numbers 1 through 8 in L1, then defining L2 as u(L1). The resulting graph is shown in Figure 11.38.

Figure 11.36

Figure 11.37

Figure 11.38

5

5

10

0

⫺1.5

10

0

⫺1.5

One additional advantage of using a list is that we can easily verify whether or not a common ratio r exists. We do this by duplicating L2 in L3 (define L3 ⫽ L2), then deleting the first entry of L3 and the last entry of L2 (to keep the same number of terms ak⫹1 in each list). See Figure 11.39. This enables us to find the ratio for the entire list ak by defining L4 as the ratio L3/L2 (which automatically computes the ratio of successive terms). See Figures 11.40 and 11.41. Figure 11.39

Figure 11.40

Figure 11.41

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EXAMPLE 5

䊳

Graphing Geometric Sequences Enter the natural numbers 1 through 6 in L1, and the terms of the sequence 0.25, 1.25, 6.25, 31.25, 156.25, 781.25 in L2. Then determine if the sequence is geometric by a. Graphing the related points to see if they appear to form along an exponential graph. b. Finding the successive ratios between terms. c. If the sequence is geometric, find the nth term and graph the sequence.

Solution

䊳

a. The plotted points are shown in Figure 11.42 and appear to lie along an exponential graph. Figure 11.42 Figure 11.43

1000

8

0

⫺250

b. Using the approach described prior to Example 5, we find there is a common ratio of r ⫽ 5 (Figure 11.43). c. With a1 ⫽ 0.25 and r ⫽ 5, the nth term must be an ⫽ a1rn⫺1 ⫽ 0.25152 n⫺1

Figure 11.44 1000

n th term formula 8

0

substitute for a1 and r

The nth term for this sequence is an ⫽ 0.25152 n⫺1. The graph is shown in Figure 11.44.

⫺250

Now try Exercises 59 through 62 If the term a1 is unknown but a term ak is given, the nth term can be written an ⫽ akrn⫺k, (the subscript on the term ak and the exponent on r sum to n). EXAMPLE 6

䊳

Finding the First Term of a Geometric Sequence Given a geometric sequence where a4 ⫽ 0.075 and a7 ⫽ 0.009375, find the common ratio r and the value of a1.

Solution

䊳

Since a1 is not known, we express a7 as the product of a known term and the appropriate number of common ratios: a7 ⫽ a4r3 17 ⫽ 4 ⫹ 3, as required). a7 ⫽ a4 # r3 0.009375 ⫽ 0.075r3 0.125 ⫽ r3 r ⫽ 0.5

a1 is unknown substitute 0.009375 for a7 and 0.075 for a4 divide by 0.075 solve for r

䊳

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1103


Having found r, we can now solve for a1 a7 ⫽ a1r6 0.009375 ⫽ a1 10.52 6 0.009375 ⫽ a1 10.0156252 a1 ⫽ 0.6

n th term formula substitute 0.009375 for a7 and 0.5 for r simplify solve for a1

The first term is a1 ⫽ 0.6 and the common ratio is r ⫽ 0.5.

B. You’ve just seen how we can find the nth term of a geometric sequence


䊳

C. Find the nth Partial Sum of a Geometric Sequence As with arithmetic series, applications of geometric series often involve computing sums of consecutive terms. We can adapt the method for finding the sum of an arithmetic sequence to develop a formula for adding the first n terms of a geometric sequence. For the nth term an ⫽ a1rn⫺1, we have Sn ⫽ a1 ⫹ a1r ⫹ a1r2 ⫹ a1r3 ⫹ p ⫹ n⫺1 a1r . If we multiply Sn by ⫺r then add the original series, the “interior terms” sum to zero. ⫺ rSn ⫽ ⫺a1r ⫹ 1⫺a1r2 2 ⫹ 1⫺a1r 3 2 ⫹ p ⫹ 1⫺a1rn⫺1 2 ⫹ 1⫺a1rn 2 ------S --S --S S a rn⫺1 ⫹ Sn ⫽ a1 ⫹ a1r ⫹S a1r 2 ⫹ p ⫹ a1rn⫺2 ⫹ 1 Sn ⫺ rSn ⫽

a1 ⫹

0

⫹

0

⫹ 0 ⫹

0

⫹

0

⫹ 1⫺a1rn 2

We then have Sn ⫺ rSn ⫽ a1 ⫺ a1rn, and can now solve for Sn: Sn 11 ⫺ r2 ⫽ a1 ⫺ a1rn

factor out Sn

a1 ⫺ a1r solve for Sn (divide by 1 ⫺ r ) 1⫺r The result is a formula for the nth partial sum of a geometric sequence. n

Sn ⫽

The nth Partial Sum of a Geometric Sequence Given a geometric sequence with first term a1 and common ratio r, the nth partial sum (the sum of the first n terms) is a1 11 ⫺ rn 2 a1 ⫺ a1rn ⫽ ,r⫽1 Sn ⫽ 1⫺r 1⫺r In words: The sum of the first n terms of a geometric sequence is the difference of the first and 1n ⫹ 12st term, divided by 1 minus the common ratio.

EXAMPLE 7

Solution

䊳

Computing a Partial Sum

䊳

3i (the first nine powers Use the preceding summation formula to find the sum: i⫽1 of 3). Verify the result on a graphing calculator. The initial terms of this series are 3 ⫹ 9 ⫹ 27 ⫹ p , and we note a1 ⫽ 3, r ⫽ 3, and n ⫽ 9. We could find the first nine terms and add, but using the partial sum formula is much faster and gives Sn ⫽ S9 ⫽

a1 11 ⫺ rn 2 1⫺r 311 ⫺ 39 2 1⫺3

9

兺

sum formula

substitute 3 for a1, 9 for n, and 3 for r

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⫽

31⫺19,6822

⫺2 ⫽ 29,523

simplify result

To verify, we enter u1n2 ⫽ 3 on the Y= screen, and find the sum of the first nine terms of the sequence on the home screen as before. See figure. n

C. You’ve just seen how we can find the nth partial sum of a geometric sequence


䊳

D. The Sum of an Infinite Geometric Series To this point we’ve considered only partial sums of a geometric series. While it is impossible to add an infinite number of these terms, some of these “infinite sums” appear to have what is called a limiting value. The sum appears to get ever closer to this value but never exceeds it—much like the asymptotic behavior of some graphs. We will define the sum of this infinite geometric series to be this limiting value, if it exists. Consider the illustration in Figure 11.45, where a standard sheet of typing paper is cut in half. One of the halves is again cut in half and the process is continued indefinitely, as shown. 1 1 p , 32, with a1 ⫽ 12 and r ⫽ 12. Notice the “halves” create an infinite sequence 21, 14, 18, 16 1 1 1 1 1 The corresponding infinite series is 2 ⫹ 4 ⫹ 8 ⫹ 16 ⫹ 32 ⫹ p ⫹ 21n ⫹ p . Figure 11.45 1 2

1 8 1 2

1 4

1 4 1 8

1 16

1 16

1 32 1 32

1 64

1 64

and so on

Figure 11.46 1 64

1 16

1 32

1 4 1 8

1 2

WORTHY OF NOTE The formula for the sum of an infinite geometric series can also be derived by noting that Sq ⫽ a1 ⫹ a1r ⫹ a1r2 ⫹ a1r3 ⫹ p can be rewritten as Sq ⫽ a1 ⫹ r1a1 ⫹ a1r ⫹ a1r2 ⫹ a1r3 ⫹ p 2 ⫽ a1 ⫹ rSq. Sq ⫺ rSq ⫽ a1 Sq 11 ⫺ r2 ⫽ a1 a1 . Sq ⫽ 1⫺r

If we arrange one of the halves from each stage as shown in Figure 11.46, we would be rebuilding the original sheet of paper. As we add more and more of these halves together, we get closer and closer to the size of the original sheet. We gain an intuitive sense that this series must add to 1, because the pieces of the original sheet of paper must add to 1 whole sheet. To explore this idea further, consider what happens to 1 12 2 n as n becomes large. 1 4 n ⫽ 4: a b ⫽ 0.0625 2

1 8 n ⫽ 8: a b ⬇ 0.004 2

1 12 n ⫽ 12: a b ⬇ 0.0002 2

Further exploration with a calculator seems to support the idea that as n S q, 1 12 2 n S 0, although a definitive proof is left for a future course. In fact, it can be shown that for any 冟r冟 6 1, rn becomes very close to zero as n becomes large. a1 ⫺ a1rn a1 a1rn n ⫽ ⫺ , note that if In symbols: as n S q, r S 0. For Sn ⫽ 1⫺r 1⫺r 1⫺r 冟r冟 6 1 and “we sum an infinite number of terms,” the second term becomes zero, leaving a1 . only the first term. In other words, the limiting value (represented by Sq) is Sq ⫽ 1⫺r Infinite Geometric Series

Given a geometric sequence with first term a1 and 0 r 0 6 1, the sum of the related infinite series is given by a1 ;r⫽1 Sq ⫽ 1⫺r If 冟r冟 7 1, no finite sum exists.

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EXAMPLE 8

䊳

1105

Computing an Infinite Sum Find the limiting value of each infinite geometric series (if it exists). a. 1 ⫹ 2 ⫹ 4 ⫹ 8 ⫹ p b. 3 ⫹ 2 ⫹ 43 ⫹ 89 ⫹ p c. 0.185 ⫹ 0.000185 ⫹ 0.000000185 ⫹ p

Solution

䊳

Begin by determining if the infinite series is geometric with 冟r冟 6 1. If so, a1 use Sq ⫽ . 1⫺r a. Since r ⫽ 2 (by inspection), a finite sum does not exist. b. Using the ratio of consecutive terms we find r ⫽ 23 and the infinite sum exists. With a1 ⫽ 3, we have 3 3 Sq ⫽ ⫽ 1 ⫽9 2 1⫺3 3 c. This series is equivalent to the repeating decimal 0.185185185 p ⫽ 0.185. The common ratio is r ⫽ 0.000185 0.185 ⫽ 0.001 and the infinite sum exists: Sq ⫽

0.185 5 ⫽ 1 ⫺ 0.001 27 Now try Exercises 93 through 108

䊳

While it is impossible for us to sum an infinite number of terms, we can often see a pattern that strongly suggests a limiting value exists. As in Section 11.2, we do this by increasing the number terms we sum, until such a limiting value seems apparent. But be very careful to note that such a calculator approach is far from an actual proof, and a more definitive means must be used to actually prove a limiting value exists. For the 2 n⫺1 sequence in Example 8(b), we enter 3a b as u(n), and compute the sums S10, S20, 3 S30, and S40 (Figures 11.47 and 11.48), where it indeed appears the limiting value is 9. Figure 11.47

Figure 11.48

D. You’ve just seen how we can find the sum of an infinite geometric series

E. Applications Involving Geometric Sequences and Series Here are a few of the ways these ideas can be put to use. EXAMPLE 9

䊳

Solving an Application of Geometric Sequences: Pendulums A pendulum is any object attached to a fixed point and allowed to swing freely under the influence of gravity. Suppose each swing is 0.9 the length of the previous one. Gradually the swings become shorter and shorter and at some point the pendulum will appear to have stopped (although theoretically it never does). a. How far does the pendulum travel on its eighth swing, if the first was 2 m? b. What is the total distance traveled by the pendulum for these eight swings? c. How many swings until the length of each swing falls below 0.5 m? d. What total distance does the pendulum travel before coming to rest? Verify your response to parts (a) through (c) using a graphing calculator.

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Solution

䊳

a. The lengths of each swing form the terms of a geometric sequence with a1 ⫽ 2 and r ⫽ 0.9. The first few terms are 2, 1.8, 1.62, 1.458, and so on. For the 8th term we have: an ⫽ a1rn⫺1 a8 ⫽ 210.92 8⫺1 ⬇ 0.957

n th term formula substitute 8 for n, 2 for a1, and 0.9 for r

The pendulum travels about 0.957 m on its 8th swing. See Figure 11.49. b. For the total distance traveled after eight swings, we compute the value of S8. Sn ⫽ S8 ⫽

a1 11 ⫺ rn 2

1⫺r 211 ⫺ 0.98 2

1 ⫺ 0.9 ⬇ 11.4

Figure 11.49

Figure 11.50

n th partial sum formula substitute 2 for a1, 0.9 for r, and 8 for n

The pendulum has traveled about 11.4 m by the end of the 8th swing. See Figure 11.50. c. To find the number of swings until the length of each swing is less than 0.5 m, we solve for n in the equation 0.5 ⫽ 210.92 n⫺1. This yields 0.25 ⫽ 10.92 n⫺1 ln 0.25 ⫽ 1n ⫺ 12ln 0.9 ln 0.25 ⫹1⫽n ln 0.9 14.16 ⬇ n

Figure 11.51

divide by 2 take the natural log, apply power property solve for n (exact form) solve for n (approximate form)

After the 14th swing, each successive swing will be less than 0.5 m. See Figure 11.51. d. For the total distance traveled before coming Figure 11.52 to rest, we consider the related infinite geometric series, with a1 ⫽ 2 and r ⫽ 0.9. a1 infinite sum formula Sq ⫽ 1⫺r 2 substitute 2 for a1 and 0.9 for r Sq ⫽ 1 ⫺ 0.9 result ⫽ 20 The pendulum would travel 20 m before coming to rest. Note that summing a larger number of terms on a calculator takes an increasing amount of time. The values for S15 and S150 are shown in Figure 11.52. Now try Exercises 111 and 112

䊳

As mentioned in Section 11.1, sometimes the sequence or series for a particular application will use the preliminary or inaugural term a0. For example, we might consider an initial amount of money that is deposited before any interest is earned, or the efficiency of a new machine after purchase—prior to any wear and tear.

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EXAMPLE 10

䊳

1107

Equipment Efficiency—Furniture Manufacturing The manufacturing of mass-produced furniture requires robotic machines to drill numerous holes for the bolts used in the assembly process. When new, the drill bits are capable of drilling through hardwood at a rate of 6 cm/sec. As the bit becomes worn, it loses 4% of its drilling speed per day. a. How many cm/sec can the bit drill through after a 5-day workweek? b. When the drilling speed falls below 3.6 cm/sec, the bit must be replaced. After how many days must the bit be replaced?

Solution

䊳

The efficiency of a new drill bit (prior to use) is given as a0 ⫽ 6 cm/sec. Since the bit loses 4% ⫽ 0.04 of its efficiency per day, it maintains 96% ⫽ 0.96 of its efficiency, showing that after 1 day of use a1 ⫽ 0.96162 ⫽ 5.76. This means the nth term formula will be an ⫽ 5.7610.962 n⫺1. a. At the end of day 5 we have a5 ⫽ 5.7610.962 5⫺1 ⫽ 5.7610.962 4 ⬇ 4.9 After 5 days, the bit can drill through the hardwood at about 4.9 cm/sec. b. To find the number of days until the efficiency falls below 3.6 cm/sec, we replace an with 3.6 and solve for n.

E. You’ve just seen how we can solve application problems involving geometric sequences and series

3.6 ⫽ 5.7610.962 n⫺1 0.625 ⫽ 0.96n⫺1 ln 0.625 ⫽ ln 0.96n⫺1 ln 0.625 ⫽ 1n ⫺ 12 ln 0.96 ln 0.625 ⫽n⫺1 ln 0.96 ln 0.625 ⫹1⫽n ln 0.96 12.5 ⬇ n

substitute 3.6 for an divide take the natural log of both sides power property divide

solve for n (exact form) solution (approximate form)

The drill bit must be replaced after approximately 12.5 days of use. Now try Exercises 113 through 126

䊳

11.3 EXERCISES 䊳



1. In a geometric sequence, each successive term is found by the preceding term by a fixed value r.

2. In a geometric sequence, the common ratio r can be found by computing the of any two consecutive terms.

3. The nth term of a geometric sequence is given by an ⫽ , for any n ⱖ 1.

4. For the general sequence a1, a2, a3, p , ak, p , the fifth partial sum is given by S5 ⫽ .

5. Describe/Discuss how the formula for the nth partial sum is related to the formula for the sum of an infinite geometric series.

6. Describe the difference(s) between an arithmetic and a geometric sequence. How can a student prevent confusion between the formulas?

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11–32



Determine if the sequence given is geometric. If yes, name the common ratio. If not, try to determine the pattern that forms the sequence.

7. 4, 8, 16, 32, p

14. 12, 0.12, 0.0012, 0.000012, p 15. ⫺1, 3, ⫺12, 60, ⫺360, p 16. ⫺ 23, 2, ⫺8, 40, ⫺240, p

47. a1 ⫽ 9, an ⫽ 729, r ⫽ 3 48. a1 ⫽ 1, an ⫽ ⫺128, r ⫽ ⫺2 1 , r ⫽ 12 49. a1 ⫽ 16, an ⫽ 64 1 , r ⫽ 12 50. a1 ⫽ 4, an ⫽ 512

p

51. a1 ⫽ ⫺1, an ⫽ ⫺1296, r ⫽ 16

18. ⫺36, 24, ⫺16, 32 3,p

52. a1 ⫽ 2, an ⫽ 1458, r ⫽ ⫺ 13

1 19. 12, 14, 18, 16 ,p

53. 2, ⫺6, 18, ⫺54, p , ⫺4374

8 16 20. 23, 49, 27 , 81, p

54. 3, ⫺6, 12, ⫺24, p , ⫺6144

12 48 192 21. 3, , 2 , 3 , p x x x

55. 64, 3212, 32, 1612, p , 1 56. 243, 81 13, 81, 27 13, p , 1

10 20 40 22. 5, , 2 , 3 , p a a a

57. 38, ⫺34, 32, ⫺3, p , 96 5 5 , 9, ⫺53, ⫺5, p , ⫺135 58. ⫺27

23. 240, 120, 40, 10, 2, p 24. ⫺120, ⫺60, ⫺20, ⫺5, ⫺1, p Write the first four terms of the sequence, given a1 and r.

25. a1 ⫽ 5, r ⫽ 2 27. a1 ⫽ ⫺6, r ⫽

26. a1 ⫽ 2, r ⫽ ⫺4 ⫺12

28. a1 ⫽ 23, r ⫽ 15

29. a1 ⫽ 4, r ⫽ 13

30. a1 ⫽ 15, r ⫽ 15

31. a1 ⫽ 0.1, r ⫽ 0.1

32. a1 ⫽ 0.024, r ⫽ 0.01

Write the expression for the nth term, then find the indicated term for each sequence.

For Exercises 59 through 62, enter the natural numbers 1 through 6 in L1 on a graphing calculator, and the terms of the given sequence in L2. Then determine if the sequence is geometric by (a) graphing the related points to see if they appear to lie on an exponential curve, and (b) computing the successive ratios of all terms. If a geometric sequence, find the nth term and graph the sequence.

59. 131.25, 26.25, 5.25, 1.05, 0.21, 0.042, p 60. 2, 2 25, 10, 10 25, 50, 50 15, p 61. 20, 16, 12, 8, 4, 0, p

33. a1 ⫽ ⫺24, r ⫽ 12; find a7

36. a1 ⫽

1, 12, 2, p

Find the number of terms in each sequence.

13. 3, 0.3, 0.03, 0.003, p

3 20 ,

42. 625, 125, 25, 5, 1, p

46. 0.5, ⫺0.35, 0.245, ⫺0.1715, p

12. ⫺13, ⫺9, ⫺5, ⫺1, 3, p

35. a1 ⫽

12 2 ,

40. ⫺78, 74, ⫺72, 7, ⫺14, p

45. 0.2, 0.08, 0.032, 0.0128, p

11. 2, 5, 10, 17, 26, p

1 ⫺20 ,

⫺19, 13, ⫺1, 3, p

44. 36 13, 36, 12 13, 12, 4 13, p

10. 128, ⫺32, 8, ⫺2, p

34. a1 ⫽ 48, r ⫽

1 27 ,

43. 12,

9. 3, ⫺6, 12, ⫺24, 48, p

17. 25, 10, 4,

39.

41. 729, 243, 81, 27, 9, p

8. 2, 6, 18, 54, 162, p

8 5,

Identify a1 and r, then write the expression for the nth term an ⴝ a1rnⴚ1 and use it to find a6, a10, and a12.

⫺13;

find a6

r ⫽ ⫺5; find a4

r ⫽ 4; find a5

62.

1 1 1 2 5 , , , , , 1, p 6 3 2 3 6

Find the common ratio r and the value of a1 using the information given (assume r 7 0).

37. a1 ⫽ 2, r ⫽ 12; find a7

63. a3 ⫽ 324, a7 ⫽ 64

64. a5 ⫽ 6, a9 ⫽ 486

38. a1 ⫽ 13, r ⫽ 13; find a8

65. a4 ⫽

2 66. a2 ⫽ 16 81 , a5 ⫽ 3

4 9,

a8 ⫽

9 4

67. a4 ⫽ 32 3 , a8 ⫽ 54

68. a3 ⫽ 16 25 , a7 ⫽ 25

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Find the partial sum indicated.

9 89. a3 ⫽ 49, a7 ⫽ 64 ; find S6

69. a1 ⫽ 8, r ⫽ ⫺2; find S12

2 90. a2 ⫽ 16 81 , a5 ⫽ 3 ; find S8

70. a1 ⫽ 2, r ⫽ ⫺3; find S8

91. a3 ⫽ 2 12, a6 ⫽ 8; find S7

71. a1 ⫽ 96, r ⫽ 13; find S5

92. a2 ⫽ 3, a5 ⫽ 9 13; find S7

72. a1 ⫽ 12, r ⫽ 73. a1 ⫽ 8, r ⫽

1 2;

3 2;

find S8

Determine whether the infinite geometric series has a finite sum. If so, find the limiting value.

find S7

74. a1 ⫽ ⫺1, r ⫽ ⫺32; find S10 75. 2 ⫹ 6 ⫹ 18 ⫹ p ; find S6

93. 9 ⫹ 3 ⫹ 1 ⫹ p

76. 2 ⫹ 8 ⫹ 32 ⫹ p ; find S7 77. 16 ⫺ 8 ⫹ 4 ⫺ p ; find S8

95. 3 ⫹ 6 ⫹ 12 ⫹ 24 ⫹ p 96. 4 ⫹ 8 ⫹ 16 ⫹ 32 ⫹ p

78. 4 ⫺ 12 ⫹ 36 ⫺ p ; find S8 1 79. 43 ⫹ 29 ⫹ 27 ⫹ p ; find S9

97. 25 ⫹ 10 ⫹ 4 ⫹ 85 ⫹ p 98. 10 ⫹ 2 ⫹ 2 ⫹ 2 ⫹ p

80.

1 18

94. 36 ⫹ 24 ⫹ 16 ⫹ p

5

⫺ 16 ⫹ 12 ⫺ p ; find S7

5

兺4

10

j

82.

兺2

k

j⫽1

k⫽1

8

7

2 k⫺1 83. 5a b 3 k⫽1

兺 10

1 i⫺1 85. 9 a⫺ b 2 i⫽4

兺

25

99. 6 ⫹ 3 ⫹ ⫹ ⫹ p 3 2

Find the partial sum indicated, and verify the result using a graphing calculator. For Exercises 85 and 86, use the summation properties from Section 11.1.

81.

1109

1 j⫺1 84. 3a b 5 j⫽1

兺

1 i⫺1 86. 5 a⫺ b 4 i⫽3

100. ⫺49 ⫹ 1⫺72 ⫹ 1⫺17 2 ⫹ p 101. 6 ⫺ 3 ⫹ 3 ⫺ 3 ⫹ p 2

4

102. 10 ⫺ 5 ⫹ ⫺ 54 ⫹ p 5 2

103. 0.3 ⫹ 0.03 ⫹ 0.003 ⫹ p 104. 0.63 ⫹ 0.0063 ⫹ 0.000063 ⫹ p 105.

3 2 k a b k⫽1 4 3

兺

106.

1 i 5a b 2 i⫽1

107.

5 j 9 a⫺ b 4 j⫽1

108.

4 k 12 a b 3 k⫽1

q

8

兺

3 4

q

q

兺

兺 q

兺

Find the indicated partial sum using the information given. Write all results in simplest form. 1 87. a2 ⫽ ⫺5, a5 ⫽ 25 ; find S5

88. a3 ⫽ 1, a6 ⫽ ⫺27; find S6 䊳


109. Sum of the cubes of the first n natural numbers: n2 1n ⴙ 12 2 Sn ⴝ 4 3 Compute 1 ⫹ 23 ⫹ 33 ⫹ p ⫹ 83 using the formula given. Then confirm the result by direct calculation.

110. Student loan payment: An ⴝ P11 ⴙ r2 n If P dollars is borrowed at an annual interest rate r with interest compounded annually, the amount of money to be paid back after n years is given by the indicated formula. Find the total amount of money that the student must repay to clear the loan, if $8000 is borrowed at 4.5% interest and the loan is paid back in 10 yr.

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11–34

APPLICATIONS

Write the nth term formula for each application, then solve.

111. Pendulum movement: On each swing, a pendulum travels only 80% as far as it did on the previous swing. If the first swing is 24 ft, how far does the pendulum travel on the 7th swing? What total distance is traveled before the pendulum comes to rest? 112. Tire swings: Ernesto is swinging to and fro on his backyard tire swing. Using his legs and body, he pumps each swing until reaching a maximum height, then suddenly relaxes until the swing comes to a stop. With each swing, Ernesto travels 75% as far as he did on the previous swing. If the first arc (or swing) is 30 ft, find the distance Ernesto travels on the 5th arc. What total distance will he travel before coming to rest?

Identify the inaugural term and write the nth term formula for each application, then solve.

113. Depreciation— automobiles: A certain new SUV depreciates in value about 20% per year (meaning it holds 80% of its value each year). If the SUV is purchased for $46,000, how much is it worth 4 yr later? How many years until its value is less than $5000? 114. Depreciation—business equipment: A new photocopier under heavy use will depreciate about 25% per year (meaning it holds 75% of its value each year). If the copier is purchased for $7000, how much is it worth 4 yr later? How many years until its value is less than $1246? 115. Equipment aging— industrial oil pumps: Tests have shown that the pumping power of a heavyduty oil pump decreases by 3% per month. If the pump can move 160 gallons per minute (gpm) new, how many gpm can the pump move 8 months later? If the pumping rate falls below 118 gpm, the pump must be replaced. How many months until this pump is replaced?

116. Equipment aging—lumber production: At the local mill, a certain type of saw blade can saw approximately 2 log-feet/sec when it is new. As time goes on, the blade becomes worn, and loses 6% of its cutting speed each week. How many log-feet/sec can the saw blade cut after 6 weeks? If the cutting speed falls below 1.2 log-feet/sec, the blade must be replaced. During what week of operation will this blade be replaced? 117. Population growth—United States: At the beginning of the year 2000, the population of the United States was approximately 277 million. If the population is growing at a rate of 2.3% per year, what was the population in 2010, 10 yr later? 118. Population growth—space colony: The population of the Zeta Colony on Mars is 1000 people. Determine the population of the Colony 20 yr from now, if the population is growing at a constant rate of 5% per year. 119. Creating a vacuum: To create a vacuum, a hand pump is used to remove the air from an air-tight cube with a volume of 462 in3. With each stroke of the pump, two-fifths of the air that remains in the cube is removed. How much air remains inside after the 5th stroke? How many strokes are required to remove all but 12.9 in3 of the air? 120. Atmospheric pressure: In 1654, scientist Otto Von Guericke performed his famous demonstration of atmospheric pressure and the strength of a vacuum in front of Emperor Ferdinand III of Hungary. After joining two hemispheres with mating rims, he used a vacuum pump to remove all of the air from the sphere formed. He then attached a team of 15 horses to each hemisphere and despite their efforts, they could not pull the hemispheres apart. If the sphere held a volume of 4200 in3 of air and one-tenth of the remaining air was removed with each stroke of the pump, how much air was still in the sphere after the 11th stroke? How many strokes were required to remove 85% of the air? 121. Treating swimming pools: In preparation for the summer swim season, chlorine is added to swimming pools to control algae and bacteria. However, careful measurements must be taken as levels above 5 ppm (parts per million) can be highly irritating to the eyes and throat, while levels below 1 ppm will be ineffective (3.0 to 3.5 ppm is ideal). In addition, the water must be treated daily since within a 24-hr period, about 25% of the chlorine will dissipate into the air. If the chlorine

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level in a swimming pool is 8 ppm after its initial treatment, how many days should the County Pool Supervisor wait before opening it up to the public? If left untreated, how many days until the chlorine level drops below 1 ppm? 122. Venting landfill gases: The gases created from the decomposition of waste in landfills must be carefully managed, as their release can cause terrible odors, harm the landfill structure, damage vegetation, or even cause an explosion. Suppose the accumulated volume of gas is 50,000 ft3, and civil engineers are able to vent 2.5% of this gas into the atmosphere daily. What volume of gas remains after 21 days? How many days until the volume of gas drops below 10,000 ft3? 123. Population growth—bacteria: A biologist finds that the population of a certain type of bacteria doubles each half-hour. If an initial culture has 50 bacteria, what is the population after 5 hr? How long will it take for the number of bacteria to reach 204,800? 䊳

1111

124. Population growth—boom towns: Suppose the population of a “boom town” in the old west doubled every 2 months after gold was discovered. If the initial population was 219, what was the population 8 months later? How many months until the population exceeded 28,000? 125. Elastic rebound—super balls: Megan discovers that a rubber ball dropped from a height of 2 m rebounds four-fifths of the distance it has previously fallen. How high does it rebound on the 7th bounce? How far does the ball travel before coming to rest? 126. Elastic rebound—computer animation: The screen saver on my computer is programmed to send a colored ball vertically down the middle of the screen so that it rebounds 95% of the distance it last traversed. If the ball always begins at the top and the screen is 36 cm tall, how high does the ball bounce on its 8th rebound? How far does the ball travel before coming to rest (and a new screen saver starts)?


127. A standard piece of typing paper is approximately 0.001 in. thick. Suppose you were able to fold this piece of paper in half 26 times. How thick would the result be? As tall as a hare, as tall as a hen, as tall as a horse, as tall as a house, or over 1 mi high? Find the actual height by computing the 27th term of a geometric sequence. Discuss what you find. 128. As part of a science experiment, identical rubber balls are dropped from a given height onto these surfaces: slate, cement, and asphalt. When dropped onto slate, the ball rebounds 80% of the height from which it last fell. Onto cement the figure is 75% and onto asphalt the figure is 70%. The ball is dropped from 130 m onto the slate, 175 m onto the cement, and 200 m onto the asphalt. Which ball has traveled the shortest total distance at the time of the fourth bounce? Which ball will travel farthest before coming to rest?

䊳


129. Consider the following situation. A person is hired at a salary of $40,000 per year, with a guaranteed raise of $1750 per year. At the same time, inflation is running about 4% per year. How many years until this person’s salary is overtaken and eaten up by the actual cost of living? 130. Find an alternative formula for the sum n

Sn ⫽

兺 log k, that does not use the sigma notation.

k⫽1

131. Verify the following statements: a. If a1, a2, a3, p , an is a geometric sequence with r and a1 greater than zero, then log a1, log a2, log a3, p , log an is an arithmetic sequence. b. If a1, a2, a3, p , an is an arithmetic sequence, then 10a1, 10a2, p , 10an, is a geometric sequence.


132. (3.2) Find the zeroes of f using the quadratic formula: f 1x2 ⫽ x2 ⫹ 5x ⫹ 9. 133. (8.3) Find a unit vector in the same direction as 3i ⫺ 7j. 134. (4.5) Graph the rational function: x2 h1x2 ⫽ x⫺1

135. (6.1) The cars on the Millenium Ferris Wheel are 100 ft from the center axle. If the top speed of the wheel is 1.5 revolutions per minute, find the linear velocity of a passenger in a car. Round your answer to the nearest whole number. Also, give the velocity in miles per hour.

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11.4

Mathematical Induction


A. Use subscript notation to evaluate and compose functions B. Apply the principle of mathematical induction to sum formulas involving natural numbers C. Apply the principle of mathematical induction to general statements involving natural numbers

EXAMPLE 1

䊳

Solution

䊳

A. You’ve just seen how we can use subscript notation to evaluate and compose functions

Since middle school (or even before) we have accepted that, “The product of two negative numbers is a positive number.” But have you ever been asked to prove it? It’s not as easy as it seems. We may think of several patterns that yield the result, analogies that indicate its truth, or even number line illustrations that lead us to believe the statement. But most of us have never seen a proof (see www.mhhe.com/coburn). In this section, we introduce one of mathematics’ most powerful tools for proving a statement, called proof by induction.

A. Subscript Notation and Function Notation One of the challenges in understanding a proof by induction is working with the notation. Earlier in the chapter, we introduced subscript notation as an alternative to function notation, since it is more commonly used when the functions are defined by a sequence. But regardless of the notation used, the functions can still be simplified, evaluated, composed, and even graphed. Consider the function f 1x2 ⫽ 3x2 ⫺ 1 and the sequence defined by an ⫽ 3n2 ⫺ 1. Both can be evaluated and graphed, with the only difference being that f(x) is continuous with domain x 僆⺢, while an is discrete (made up of distinct points) with domain n 僆⺞. Using Subscript Notation for a Composition

For f 1x2 ⫽ 3x2 ⫺ 1 and an ⫽ 3n2 ⫺ 1, find f 1k ⫹ 12 and ak⫹1. f 1k ⫹ 12 ⫽ 31k ⫹ 12 2 ⫺ 1 ⫽ 31k2 ⫹ 2k ⫹ 12 ⫺ 1 ⫽ 3k2 ⫹ 6k ⫹ 2

ak⫹1 ⫽ 31k ⫹ 12 2 ⫺ 1 ⫽ 31k2 ⫹ 2k ⫹ 12 ⫺ 1 ⫽ 3k2 ⫹ 6k ⫹ 2 Now try Exercises 7 through 18

䊳

No matter which notation is used, every occurrence of the input variable is replaced by the new value or expression indicated by the composition.

B. Mathematical Induction Applied to Sums Consider the sum of odd numbers 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ 13 ⫹ p . The sum of the first four terms is 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫽ 16, or S4 ⫽ 16. If we now add a5 (the next term in line), would we get the same answer as if we had simply computed S5? Common sense would say, “Yes!” since S5 ⫽ 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫽ 25 and S4 ⫹ a5 ⫽ 16 ⫹ 9 ⫽ 25✓. In diagram form, we have add next term a5 ⫽ 9 to S4

>

1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ 13 ⫹ 15 ⫹ p c c S4 S5

sum of 4 terms sum of 5 terms

Our goal is to develop this same degree of clarity in the notational scheme of things. For a given series, if we find the kth partial sum Sk (shown next) and then add the next term ak⫹1, would we get the same answer if we had simply computed Sk⫹1? In other words, is Sk ⫹ ak⫹1 ⫽ Sk⫹1 true?

1112

11–36

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Section 11.4 Mathematical Induction

1113

add next term ak1

>

a1 a2 a3 p ak1 ak ak1 p an–1 an c c sum of k terms Sk sum of k 1 terms

Sk1

Now, let’s return to the sum 1 3 5 7 p 12n 12. This is an arithmetic series with a1 1, d 2, and nth term an 2n 1. Using the sum formula for an arithmetic sequence, an alternative formula for this sum can be established. Sn

n1a1 an 2

2 n11 2n ⴚ 12 2

No matter how distant the city or how many relay stations are involved, if the generating plant is working and the kth station relays to the (k 1)st station, the city will get its power.

substitute 1 for a1 and 2nⴚ1 for an

n12n2

n2

WORTHY OF NOTE

summation formula for an arithmetic sequence

2

simplify result

This shows that the sum of the first n positive odd integers is given by Sn n2. As a check we compute S5 1 3 5 7 9 25 and compare to S5 52 25✓. We also note S6 62 36, and S5 a6 25 11 36, showing S6 S5 a6. For more on this relationship, see Exercises 19 through 24. While it may seem simplistic now, showing S5 a6 S6 and Sk ak1 Sk1 (in general) is a critical component of a proof by induction. Unfortunately, general summation formulas for many sequences cannot be established from known formulas. In addition, just because a formula works for the first few values of n, we cannot assume that it will hold true for all values of n (there are infinitely many). As an illustration, the formula an n2 n 41 yields a prime number for every natural number n from 1 to 40, but fails to yield a prime for n 41. This helps demonstrate the need for a more conclusive proof, particularly when a relationship appears to be true, and can be “verified” in a finite number of cases, but whether it is true in all cases remains in doubt. Proof by induction is based on a relatively simple idea. To help understand how it works, consider n relay stations that are used to transport electricity from a generating plant to a distant city. If we know the generating plant is operating, and if we assume that the kth relay station (any station (k 1)st kth Generating plant in the series) is making the transrelay relay fer to the 1k 12st station (the next station in the series), then we’re sure the city will have electricity. This idea can be applied mathematically as follows. Consider the statement, “The sum of the first n positive even integers is n2 n.” In other words, 2 4 6 8 p 2n n2 n. We can certainly verify the statement for the first few even numbers: 112 2 1 2 The first even number is 2 and p 122 2 2 6 The sum of the first two even numbers is 2 4 6 and p The sum of the first three even numbers is 2 4 6 12 and p 132 2 3 12 The sum of the first four even numbers is 2 4 6 8 20 and p 142 2 4 20

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While we could continue this process for a very long time (or even use a computer), no finite number of checks can prove a statement is universally true. To prove the statement true for all positive integers, we use a reasoning similar to that applied in the relay stations example. If we are sure the formula works for n 1 (the generating station is operating), and if the truth of n k implies that n k 1 is true [the kth relay station is transferring electricity to the 1k 12st station], then the statement is true for all n (the city will get its electricity). The case where n 1 is called the base case of an inductive proof, and the assumption that the formula is true for n k is called the induction hypothesis. When the induction hypothesis is applied to a sum formula, we attempt to show that Sk ak1 Sk1. Since k and k 1 are arbitrary, the statement must be true for all n. Mathematical Induction Applied to Sums Let Sn be a sum formula involving positive integers. If 1. S1 is true, and 2. the truth of Sk implies that Sk1 is true, then Sn must be true for all positive integers n.

WORTHY OF NOTE To satisfy our finite minds, it might help to show that Sn is true for the first few cases, prior to extending the ideas to the infinite case.

EXAMPLE 2

䊳

Both parts 1 and 2 must be verified for the proof to be complete. Since the process requires the terms Sk, ak1, and Sk1, we will usually compute these first.

Proving a Statement Using Mathematical Induction Use induction to prove that the sum of the first n perfect squares is given by n1n 12 12n 12 . 1 4 9 16 25 p n2 6

Solution

䊳

Given an n2 and Sn

n1n 12 12n 12 6

, the needed components are p

ak1 1k 12 2 For an n2: ak k2 and n1n 12 12n 12 k1k 12 12k 12 1k 12 1k 2212k 32 : Sk For Sn and Sk1 6 6 6 1. Show Sn is true for n 1. Sn S1

n1n 12 12n 12 1122 132

sum formula

6

base case: n 1

6 1✓

result checks, the first term is 1

2. Assume Sk is true, 1 4 9 16 p k2

k1k 1212k 12 6

and use it to show the truth of Sk1 follows. That is,

1k 121k 2212k 32

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎬ ⎭

1 4 9 16 p k2 1k 12 2

induction hypothesis: Sk is true

6

Sk

ak1

Sk1

Working with the left-hand side, we have 1 4 9 16 p k2 1k 12 2 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

k1k 12 12k 12

1k 12 6 k1k 12 12k 12 61k 12 2 6

2

induction hypothesis: substitute

k1k 1212k 12

for 1 4 9 16 25 p k2 common denominator

6

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B. You’ve just seen how we can apply the principle of mathematical induction to sum formulas involving natural numbers

1k 12 3 k12k 12 61k 12 4

1115

factor out k 1

6 1k 12 3 2k2 7k 64

multiply and combine terms

6 1k 121k 2212k 32

factor the trinomial, result is Sk1

6

Since the truth of Sk1 follows from Sk, the formula is true for all n. Now try Exercises 27 through 38

䊳

C. The General Principle of Mathematical Induction Proof by induction can be used to verify many other kinds of relationships involving a natural number n. In this regard, the basic principles remain the same but are stated more broadly. Rather than using Sn to represent a sum, we will use Pn to represent any proposed statement or relationship we might wish to verify. This broadens the scope of the proof and makes it more widely applicable, while maintaining its connection to the sum formulas verified earlier. The General Principle of Mathematical Induction Let Pn be a statement involving natural numbers. If 1. P1 is true, and 2. the truth of Pk implies that Pk1 is also true then Pn must be true for all natural numbers n.

EXAMPLE 3

䊳

Proving a Statement Using the General Principle of Mathematical Induction Use the general principle of mathematical induction to show the statement Pn is true for all natural numbers n. Pn: 2n n 1

Solution

䊳

The statement Pn is defined as 2n n 1. This means that Pk is represented by 2k k 1 and Pk1 by 2k1 k 2. 1. Show Pn is true for n 1: Pn: P1:

2n n 1 21 1 1 2 2✓

given statement base case: n 1 true

Although not a part of the formal proof, a table of values can help to illustrate the relationship we’re trying to establish. It appears that the statement is true. n

1

2

3

4

5

2n

2

4

8

16

32

n1

2

3

4

5

6

2. Assume that Pk is true. Pk: 2k k 1

induction hypothesis

and use it to show the truth of Pk1. That is, Pk1: 2k1 1k 12 1 k2

Begin by working with the left-hand side of the inequality, 2k1.

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2k1 212k 2

21k ⴙ 12 2k 2

properties of exponents

induction hypothesis: substitute k 1 for 2k (symbol changes since 2k is greater than or equal to k 1) distribute

Since k is a positive integer, 2k1 2k 2 k 2, showing 2k1 k 2.

WORTHY OF NOTE Note there is no reference to an, ak, or ak+1 in the statement of the general principle of mathematical induction.

EXAMPLE 4

Since the truth of Pk1 follows from Pk, the formula is true for all n. Now try Exercises 39 through 42

䊳

䊳

Proving Divisibility Using Mathematical Induction Let Pn be the statement, “4n 1 is divisible by 3 for all positive integers n.” Use mathematical induction to prove that Pn is true.

Solution

䊳

If a number is evenly divisible by three, it can be written as the product of 3 and some positive integer we will call p. 1. Show Pn is true for n 1: Pn: 4n 1 3p P1: 4112 1 3p 3 3p ✓

given statement, p 僆⺪ substitute 1 for n statement is true for n 1

2. Assume that Pk is true. Pk: 4k 1 3p 4k 3p 1

induction hypothesis isolate 4k

and use it to show the truth of Pk1. That is, Pk1: 4k1 1 3q for q 僆⺪ is also true. Beginning with the left-hand side we have: 4k1 1 4 # 4k 1 4 # 13p ⴙ 12 1 12p 3 314p 12 3q

properties of exponents induction hypothesis: substitute 3p 1 for 4k distribute and simplify factor

The last step shows 4 1 is divisible by 3. Since the original statement is true for n 1, and the truth of Pk implies the truth of Pk1, the statement, “4n 1 is divisible by 3” is true for all positive integers n. k1


C. You’ve just seen how we can apply the principle of mathematical induction to general statements involving natural numbers

䊳

We close this section with some final notes. Although the base step of a proof by induction seems trivial, both the base step and the induction hypothesis are necessary 1 1 parts of the proof. For example, the statement n 6 is false for n 1, but true for 3 3n all other positive integers. Finally, for a fixed natural number p, some statements are false for all n 6 p, but true for all n p. By modifying the base case to begin at p, we can use the induction hypothesis to prove the statement is true for all n greater than p. For example, n 6 13n2 is false for n 6 4, but true for all n 4.

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11.4 EXERCISES 䊳



1. No statement

䊳

number of verifications can prove a true.

3. Assuming that a statement/formula is true for n k is called the .

4. The graph of a sequence is is made up of distinct points.

5. Explain the equation Sk ak1 Sk1. Begin by saying, “Since the kth term is arbitrary p ” (continue from here).

6. Discuss the similarities and differences between mathematical induction applied to sums and the general principle of mathematical induction.

7. an 10n 6 9. an n 11. an 2n1

8. an 6n 4 10. an 7n

12. an 213n1 2

For the given sum formula Sn, find S4, S5, Sk, and Skⴙ1.

13. Sn n15n 12 15. Sn

n1n 12

2 17. Sn 2n 1

14. Sn n13n 12 16. Sn

7n1n 12

2 18. Sn 3n 1

Verify that S4 ⴙ a5 ⴝ S5 for each exercise. Note that each Sn is identical to those in Exercises 13 through 18.

19. an 10n 6; Sn n15n 12 20. an 6n 4; Sn n13n 12 21. an n; Sn

n1n 12

22. an 7n; Sn 23. an 2

2 7n1n 12 2

; Sn 2 1

n1

24. an 213

n1

n

2; Sn 3n 1


25. Sum of the first n cubes (alternative form): (1 ⴙ 2 ⴙ 3 ⴙ 4 ⴙ p ⴙ n)2 Earlier we noted the formula for the sum of the n2 1n 12 2 . An alternative is first n cubes was 4 given by the formula shown. a. Verify the formula for n 1, 5, and 9. 䊳

, meaning it


For the given nth term an, find a4, a5, ak, and akⴙ1.

䊳

2. Showing a statement is true for n 1 is called the of an inductive proof.

b. Verify the formula using 123pn

n1n 12 2

.

26. Powers of the imaginary unit: in ⴙ 4 ⴝ in, where i ⴝ 1 ⴚ1 Use a proof by induction to prove that powers of the imaginary unit are cyclic. That is, that they cycle through the numbers i, 1, i, and 1 for consecutive powers.

APPLICATIONS

Use mathematical induction to prove the indicated sum formula is true for all natural numbers n.

27. 2 4 6 8 10 p 2n; an 2n, Sn n1n 12 28. 3 7 11 15 19 p 14n 12; an 4n 1, Sn n12n 12

29. 5 10 15 20 25 p 5n; 5n1n 12 an 5n, Sn 2 30. 1 4 7 10 13 p 13n 22; n13n 12 an 3n 2, Sn 2

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31. 5 9 13 17 p 14n 12; an 4n 1, Sn n12n 32 32. 4 12 20 28 36 p 18n 42; an 8n 4, Sn 4n2 33. 3 9 27 81 243 p 3n; 313n 12 an 3n, Sn 2 34. 5 25 125 625 p 5n; 515n 12 an 5n, Sn 4 35. 2 4 8 16 32 64 p 2n; an 2n, Sn 2n1 2 36. 1 8 27 64 125 216 p n3; n2 1n 12 2 an n3, Sn 4

37.

1 1 1 1 p ; 1132 3152 5172 12n 1212n 12 1 n an , Sn 2n 1 12n 12 12n 12

1 1 1 1 p ; 1122 2132 3142 n1n 12 1 n , Sn an n 1 n1n 12 Use the principle of mathematical induction to prove that each statement is true for all natural numbers n. 38.

39. 3n 2n 1

40. 2n n 1

41. 3 # 4n1 4n 1

42. 4 # 5n1 5n 1

43. n2 7n is divisible by 2 44. n3 n 3 is divisible by 3 45. n3 3n2 2n is divisible by 3 46. 5n 1 is divisible by 4 47. 6n 1 is divisible by 5

䊳


48. You may have noticed that the sum formula for the first n integers was quadratic, and the formula for the first n integer squares was cubic. Is the formula for the first n integer cubes, if it exists, a quartic (degree four) function? Use your calculator to run a quartic regression on the first five perfect cubes (enter 1 through 5 in L1 and the cumulative sums in L2). What did you find? How is this exercise related to Exercise 36? xn 1 11 x x2 x3 p xn1 2. x1 50. Use mathematical induction to prove that for 14 24 34 p n4, where an n4,

49. Use mathematical induction to prove that

Sn 䊳

n1n 12 12n 12 13n2 3n 12 30

.


51. (7.2) Verify the identity 1sin ␪ cos ␪2 2 1sin ␪ cos ␪2 2 2. 52. (2.5) State the domain and range of the piecewise function shown here. y 5 4 3 (1, 1) 2 1 54321 1 2 3 4 5

53. (1.1) State the equation of the circle whose graph is shown here. y 10 (1, 7) 8 6 (4, 3) 4 2 108642 2 4 6 8 10

1 2 3

(3, 2)

2 4 6 8 10 x

5 x

54. (8.4) Given p H13, 1I and q H1, 1I, find a. the dot product p # q b. the angle between the vectors

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MID-CHAPTER CHECK

1. an ⫽ 7n ⫺ 4

Find the number of terms in each series and then find the sum. Verify results on a graphing calculator. 12. 2 ⫹ 5 ⫹ 8 ⫹ 11 ⫹ p ⫹ 74

2. an ⫽ n2 ⫹ 3

13.

In Exercises 1 through 3, the nth term is given. Write the first three terms of each sequence and find a9.

3. an ⫽ 1⫺12 n 12n ⫺ 12

⫹ 32 ⫹ 52 ⫹ 72 ⫹ p ⫹ 31 2

14. For an arithmetic series, a3 ⫽ ⫺8 and a7 ⫽ 4. Find S10.

4

4. Evaluate the sum

1 2

兺3

n⫹1

15. For a geometric series, a3 ⫽ ⫺81 and a6 ⫽ 3. Find S10.

n⫽1

5. Rewrite using sigma notation. 1 ⫹ 4 ⫹ 7 ⫹ 10 ⫹ 13 ⫹ 16

16. Identify a1 and the common ratio r. Then find an expression for the general term an. a. 2, 6, 18, 54, p

Match each formula to its correct description. n1a1 ⫹ an 2 6. Sn ⫽ 7. an ⫽ a1rn⫺1 2 a1 8. Sq ⫽ 9. an ⫽ a1 ⫹ 1n ⫺ 12d 1⫺r a1 11 ⫺ rn 2 10. Sn ⫽ 1⫺r a. sum of an infinite geometric series

1 b. 12, 14, 18, 16 ,p

17. Find the number of terms in the series then compute the sum. 541 ⫹ 181 ⫹ 16 ⫹ p ⫹ 812 18. Find the infinite sum (if it exists). ⫺49 ⫹ 1⫺72 ⫹ 1⫺12 ⫹ 1⫺17 2 ⫹ p 19. Barrels of toxic waste are stacked at a storage facility in pyramid form, with 60 barrels in the first row, 59 in the second row, and so on, until there are 10 barrels in the top row. How many barrels are in the storage facility? Verify results using a graphing calculator.

b. nth term formula for an arithmetic series c. sum of a finite geometric series d. summation formula for an arithmetic series

20. As part of a conditioning regimen, a drill sergeant orders her platoon to do 25 continuous standing broad jumps. The best of these recruits was able to jump 96% of the distance from the previous jump, with a first jump distance of 8 ft. Use a sequence/ series to determine the distance the recruit jumped on the 15th try, and the total distance traveled by the recruit after all 25 jumps. Verify results using a graphing calculator.

e. nth term formula for a geometric series 11. Identify a1 and the common difference d. Then find an expression for the general term an. a. 2, 5, 8, 11, p b. 32, 94, 3, 15 4,p

REINFORCING BASIC CONCEPTS Applications of Summation The properties of summation play a large role in the development of key ideas in a first semester calculus course, and the following summation formulas are an integral part of these ideas. The first three formulas were verified in Section 11.4, while proof of the fourth was part of Exercise 48 on page 1118. n

(1)

兺

n

c ⫽ cn

i⫽1

(2)

兺

i⫽

i⫽1

n1n ⫹ 12 2

n

(3)

兺

i2 ⫽

n1n ⫹ 1212n ⫹ 12

i⫽1

6

n

(4)

兺

i3 ⫽

i⫽1

n2 1n ⫹ 12 2 4

To see the various ways they can be applied consider the following. Illustration 1 䊳 Over several years, the owner of Morgan’s LawnCare has noticed that the company’s monthly profits (in thousands) can be approximated by the sequence an ⫽ 0.0625n3 ⫺ 1.25n2 ⫹ 6n, with the points plotted in Figure 11.53 (the continuous graph is shown for effect only). Find the company’s approximate annual profit.

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Solution 䊳 The most obvious approach would be to simply compute terms a1 through a12 (January through December) and find their sum: sum(seq(Y1, X, 1, 12), which gives a result of 35.75 or $35,750. As an alternative, we could add the amount of profit earned by the company in the first 8 months, then add the amount the company lost (or broke even) during the last 4 months. In other words, we could apply summation property IV: 12

兺a

n

8

⫽

Figure 11.53 12

0

12

兺a

12

n

⫹

兺a

n

[(see Figure 11.54), which gives the same result:

42 ⫹ 1⫺6.252 ⫽ 35.75 or $35,750]. As a third option, we could use summation properties along with the appropriate summation formulas, and compute the result manually. Note the function is now written in terms of “i.” Distribute summation and factor out constants (properties II and III): i⫽1

i⫽1

12

兺

i⫽1

i⫽9

10.0625i3 ⫺ 1.25i2 ⫹ 6i2 ⫽ 0.0625

12

兺

12

i3 ⫺ 1.25

i⫽1

兺

⫺5

Figure 11.54

12

i2 ⫹ 6

i⫽1

兺i

i⫽1

Replace each summation with the appropriate summation formula, then substitute 12 for n: ⫽ 0.0625 c ⫽ 0.0625 c

n2 1n ⫹ 12 2 4 2 1122 1132 2

d ⫺ 1.25 c

d ⫺ 1.25 c

n1n ⫹ 12 12n ⫹ 12 6 1122 1132 1252

4 6 ⫽ 0.0625160842 ⫺ 1.2516502 ⫹ 61782 ⫽ 35.75

d ⫹ 6c

d ⫹ 6c

n1n ⫹ 12

1122 1132 2

2

d

d

As we expected, the result shows profit was $35,750. While some approaches seem “easier” than others, all have great value, are applied in different ways at different times, and are necessary to adequately develop key concepts in future classes. Exercise 1: Repeat Illustration 1 if the profit sequence is an ⫽ 0.125x3 ⫺ 2.5x2 ⫹ 12x.

11.5

Counting Techniques


A. Count possibilities using B.

C.

D.

E.

lists and tree diagrams Count possibilities using the fundamental principle of counting Quick-count distinguishable permutations Quick-count nondistinguishable permutations Quick-count using combinations

How long would it take to estimate the number of fans sitting shoulder-to-shoulder at a sold-out basketball game? Well, it depends. You could actually begin counting 1, 2, 3, 4, 5, p , which would take a very long time, or you could try to simplify the process by counting the number of fans in the first row and multiplying by the number of rows. Techniques for “quick-counting” the objects in a set or various subsets of a large set play an important role in a study of probability.

A. Counting by Listing and Tree Diagrams Consider the simple spinner shown in Figure 11.55, which is divided into three equal parts. What are the different possible outcomes for two spins, spin 1 followed by spin 2? We might begin by organizing the possibilities using a tree diagram. As the name implies, each choice or possibility appears as the branch of a tree, with the total possibilities being equal to the number of (unique)

Figure 11.55 B A

C

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Section 11.5 Counting Techniques

Figure 11.56

paths from the beginning point to the end of a branch. Figure 11.56 shows how the spinner exercise would appear (possibilities for two spins). Moving from top to bottom we can trace nine possible paths: AA, AB, AC, BA, BB, BC, CA, CB, and CC.

Begin

A A

EXAMPLE 1

䊳

B

B C

A

B

C C

A

B

C

Listing Possibilities Using a Tree Diagram A basketball player is fouled and awarded three free throws. Let H represent the possibility of a hit (basket is made), and M the possibility of a miss. Determine the possible outcomes for the three shots using a tree diagram.

Solution

䊳

Each shot has two possibilities, hit (H) or miss (M), so the tree will branch in two directions at each level. As illustrated in the figure, there are a total of eight possibilities: HHH, HHM, HMH, HMM, MHH, MHM, MMH, and MMM. Begin

H

M

H H

M M

H

H M

H

M M

H

M


WORTHY OF NOTE Sample spaces may vary depending on how we define the experiment, and for simplicity’s sake we consider only those experiments having outcomes that are equally likely.

䊳

To assist our discussion, an experiment is any task that can be done repeatedly and has a well-defined set of possible outcomes. Each repetition of the experiment is called a trial. A sample outcome is any potential outcome of a trial, and a sample space is a set of all possible outcomes. In our first illustration, the experiment was spinning a spinner, there were three sample outcomes (A, B, or C), the experiment had two trials (spin 1 and spin 2), and there were nine elements in the sample space. Note that after the first trial, each of the three sample outcomes will again have three possibilities (A, B, and C). For two trials we have 32 9 possibilities, while three trials would yield a sample space with 33 27 possibilities. In general, for N equally likely outcomes we have A “Quick-Counting” Formula for a Sample Space If an experiment has N sample outcomes that are equally likely and the experiment is repeated t times, the number of elements in the sample space is N t.

EXAMPLE 2

䊳

Counting the Outcomes in a Sample Space Many combination locks have the digits 0 through 39 arranged along a circular dial. Opening the lock requires stopping at a sequence of three numbers within this range, going counterclockwise to the first number, clockwise to the second, and counterclockwise to the third. How many three-number combinations are possible?

5

35

10

30 25

15 20

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Solution

䊳

A. You’ve just seen how we can count possibilities using lists and tree diagrams

There are 40 sample outcomes 1N 402 in this experiment, and three trials 1t 32. The number of possible combinations is identical to the number of elements in the sample space. The quick-counting formula gives 403 64,000 possible combinations. Now try Exercises 11 and 12

䊳

B. Fundamental Principle of Counting The number of possible outcomes may differ depending on how the event is defined. For example, some security systems, license plates, and telephone numbers exclude certain numbers. For example, phone numbers cannot begin with 0 or 1 because these are reserved for operator assistance, long distance, and international calls. Constructing a three-digit area code is like filling in three blanks with three digit digit digit

digits. Since the area code must start with a number between 2 and 9, there are eight choices for the first blank. Since there are 10 choices for the second digit and 10 choices for the third, there are 8 # 10 # 10 800 possibilities in the sample space. EXAMPLE 3

䊳

Counting Possibilities for a Four-Digit Security Code A digital security system requires that you enter a four-digit PIN (personal identification number), using only the digits 1 through 9. How many codes are possible if a. Repetition of digits is allowed? b. Repetition is not allowed? c. The first digit must be even and repetitions are not allowed?

Solution

䊳

a. Consider filling in the four blanks

digit digit digit digit

with the number of

ways the digit can be chosen. If repetition is allowed, the experiment is similar to that of Example 2 and there are N t 94 6561 possible PINs. b. If repetition is not allowed, there are only eight possible choices for the second digit of the PIN, then seven for the third, and six for the fourth. The number of possible PIN numbers decreases to 9 # 8 # 7 # 6 3024. c. There are four choices for the first digit (2, 4, 6, 8). Once this choice has been made there are eight choices for the second digit, seven for the third, and six for the last: 4 # 8 # 7 # 6 1344 possible codes. Now try Exercises 13 through 16

䊳

Given any experiment involving a sequence of tasks, if the first task can be completed in p possible ways, the second task has q possibilities, and the third task has r possibilities, a tree diagram will show that the number of possibilities in the sample space for task1–task2–task3 is p # q # r. This situation is simply a variation of the previous quick-counting formula. Even though the examples we’ve considered to this point have varied a great deal, this idea was fundamental to counting all possibilities in a sample space and is, in fact, known as the fundamental principle of counting (FPC). Fundamental Principle of Counting (Applied to Three Tasks) Given any experiment with three defined tasks, if there are p possibilities for the first task, q possibilities for the second, and r possibilities for the third, the total number of ways the experiment can be completed is p # q # r. This fundamental principle can be extended to include any number of tasks, and can be applied in many different ways. See Exercises 17 through 20.

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䊳

EXAMPLE 4

1123

Counting Possibilities for Seating Arrangements Adrienne, Bob, Carol, Dax, Earlene, and Fabian bought tickets to see The Marriage of Figaro. Assuming they sat together in a row of six seats, how many different seating arrangements are possible if a. Bob and Carol are sweethearts and must sit together? b. Bob and Carol are enemies and must not sit together?

䊳

Solution Figure 11.57 Bob 1

Carol 2

3

4

5

6

1

Bob 2

Carol 3

4

5

6

1

2

Bob 3

Carol 4

5

6

1

2

3

Bob 4

Carol 5

6

1

2

3

4

Bob 5

Carol 6

B. You’ve just seen how we can count possibilities using the fundamental principle of counting

a. Since a restriction has been placed on the seating arrangement, it will help to divide the experiment into a sequence of tasks: task 1: they sit together; task 2: either Bob is on the left or Bob is on the right; and task 3: the other four are seated. Bob and Carol can sit together in five different ways, as shown in Figure 11.57, so there are five possibilities for task 1. There are two ways they can be side-by-side: Bob on the left and Carol on the right, as shown, or Carol on the left and Bob on the right. The remaining four people can be seated randomly, so task 3 has 4! ⫽ 24 possibilities. Under these conditions they can be seated 5 # 2 # 4! ⫽ 240 ways. b. This is similar to part (a), but now we have to count the number of ways they can be separated by at least one seat: task 1: Bob and Carol are in nonadjacent seats; task 2: either Bob is on the left or Bob is on the right; and task 3: the other four are seated. For tasks 1 and 2, be careful to note there is no multiplication involved, just a simple counting. If Bob sits in seat 1 (to the left of Carol), there are four nonadjacent seats on the right. If Bob sits in seat 2, there are three nonadjacent seats on the right. With Bob in seat 3, there are two nonadjacent seats to his right. Similar reasoning for the remaining seats shows there are 10 # 2 ⫽ 20 possibilities for Bob and Carol not sitting together (by symmetry, Bob could also sit to the right of Carol). Multiplying by the number of ways the other four can be seated (task 3) gives 20 # 4! ⫽ 480 possible seating arrangements. We could also reason that since there are 6! ⫽ 720 random seating arrangements and 240 of them consist of Bob and Carol sitting together, the remaining 720 ⫺ 240 ⫽ 480 must consist of Bob and Carol not sitting together. More will be said about this type of reasoning in Section 11.6. Now try Exercises 21 through 28

䊳

C. Distinguishable Permutations In the game of Scrabble® (Milton Bradley), players attempt to form words by rearranging letters. Suppose a player has the letters P, S, T, and O at the end of the game. These letters could be rearranged or permuted to form the words POTS, SPOT, TOPS, OPTS, POST, or STOP. These arrangements are called permutations of the four letters. A permutation is any new arrangement, listing, or sequence of objects obtained by changing an existing order. A distinguishable permutation is a permutation that produces a result different from the original. For example, a distinguishable permutation of the digits in the number 1989 is 8199. Example 4 considered six people, six seats, and the various ways they could be seated. But what if there were fewer seats than people? By the FPC, with six people and four seats there could be 6 # 5 # 4 # 3 ⫽ 360 different arrangements, with six people and three seats there are 6 # 5 # 4 ⫽ 120 different arrangements, and so on. These rearrangements are called distinguishable permutations. You may have noticed that for six people and six seats, we will use all six factors of 6!, while for six people and four seats we used the first four, six people and three seats required only the first three, and so on. Generally, for n people and r seats, the first r factors of n! will be used. The notation and formula for distinguishable permutations of n objects taken r at a time is n! . By defining 0! ⫽ 1, the formula includes the case where all n objects nPr ⫽ 1n ⫺ r2! n! n! n! ⫽ ⫽ ⫽ n!. are selected, which of course results in nPn ⫽ 0! 1 1n ⫺ n2!

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Distinguishable Permutations: Unique Elements

If r objects are selected from a set containing n unique elements 1r n2 and placed in an ordered arrangement, the number of distinguishable permutations is nPr

EXAMPLE 5

䊳

n! 1n r2!

or

nPr

n1n 121n 22 # # # 1n r 12

Computing a Permutation Compute each value of nPr using the methods described previously. a. 7P4 b. 10P3

Solution

䊳

n! , noting the Begin by evaluating each expression using the formula nPr 1n r2! third line (in bold) gives the first r factors of n!. 7! 10! a. 7P4 b. 10P3 17 42! 110 32! 7 # 6 # 5 # 4 # 3! 10 # 9 # 8 # 7! 3! 7! 7#6#5#4 10 # 9 # 8 840 720 Now try Exercises 29 through 36

䊳

Figure 11.58 When the number of objects is very large, the formula for permutations can become somewhat unwieldy and the computed result is often a very large number. When needed, most graphing calculators have the ability to compute permutations, with this option accessed using MATH (PRB) 2:nPr. Figure 11.58 verifies the computation for Example 5(b), and also shows that if there were 15 people and 7 chairs, the number of possible seating arrangements exceeds 32 million! Note that the value of n is entered first, followed by the nPr command and the value of r.

EXAMPLE 6

䊳

Counting the Possibilities for Finishing a Race As part of a sorority’s initiation process, the nine new inductees must participate in a 1-mi race. Assuming there are no ties, how many first- through fifth-place finishes are possible if it is well known that Mediocre Mary will finish fifth and Lightning Louise will finish first?

Solution

䊳

To help understand the situation, we can diagram the possibilities for finishing first through fifth. Since Louise will finish first, this slot can be filled in only one way, by Louise herself. The same goes for Mary and her fifth-place finish: Mary

Louise 1st

C. You’ve just seen how we can quick-count distinguishable permutations

2nd

3rd

4th

5th

The remaining three slots can be filled in 7P3 7 # 6 # 5 different ways, indicating that under these conditions, there are 1 # 7 # 6 # 5 # 1 210 different ways to finish. Now try Exercises 37 through 42

䊳

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D. Nondistinguishable Permutations As the name implies, certain permutations are nondistinguishable, meaning you cannot tell one apart from another. Such is the case when the original set contains elements or sample outcomes that are identical. Consider a family with four children, Lyddell, Morgan, Michael, and Mitchell, who are at the photo studio for a family picture. Michael and Mitchell are identical twins and cannot be told apart. In how many ways can they be lined up for the picture? Since this is an ordered arrangement of four children taken from a group of four, there are 4P4 24 ways to line them up. A few of them are Lyddell Morgan Michael Mitchell Lyddell Michael Morgan Mitchell Michael Lyddell Morgan Mitchell

Lyddell Morgan Mitchell Michael Lyddell Mitchell Morgan Michael Mitchell Lyddell Morgan Michael

But of these six arrangements, half will appear to be the same picture, since the difference between Michael and Mitchell cannot be distinguished. In fact, of the 24 total permutations, every picture where Michael and Mitchell have switched places will be nondistinguishable. To find the distinguishable permutations, we need to take the total permutations (4P4) and divide by 2!, the number of ways the twins can be 24 4P4 12 distinguishable pictures. permuted: 2 122! These ideas can be generalized and stated in the following way. Nondistinguishable Permutations: Nonunique Elements In a set containing n elements where one element is repeated p times, another is repeated q times, and another is repeated r times 1p q r n2, the number of nondistinguishable permutations is n! nPn p!q!r! p!q!r! The idea can be extended to include any number of repeated elements. EXAMPLE 7

䊳

Counting Nondistinguishable Permutations A Scrabble player starts the game with the seven letters S, A, O, O, T, T, and T in her rack. How many distinguishable arrangements can be formed as she attempts to play a word?

Solution

䊳

D. You’ve just seen how we can quick-count nondistinguishable permutations

Essentially the exercise asks for the number of distinguishable permutations of the seven letters, given T is repeated three times and O is repeated twice (for S and 7P7 420 distinguishable permutations. A, 1! 1). There are 3!2! Now try Exercises 43 through 54

䊳

E. Combinations WORTHY OF NOTE In Example 7, if a Scrabble player is able to play all seven letters in one turn, he or she “bingos” and is awarded 50 extra points. The player in Example 7 did just that. Can you determine what word was played?

Similar to nondistinguishable permutations, there are other times the total number of permutations must be reduced to quick-count the elements of a desired subset. Consider a vending machine that offers a variety of 40¢ candies. If you have a quarter (Q), dime (D), and nickel (N), the machine wouldn’t care about the order the coins are deposited. Even though QDN, QND, DQN, DNQ, NQD, and NDQ give the 3P3 6 possible permutations, the machine considers them as equal and will vend your snack. Using sets, this is similar to saying the set A 5X, Y, Z6 has only one subset with three elements, since {X, Z, Y}, {Y, X, Z}, {Y, Z, X}, and so on, all represent the same set. Similarly, there are six two-letter permutations of X, Y, and Z 1 3P2 62: XY, XZ, YX,

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YZ, ZX, and ZY, but only three two-letter subsets: {X, Y}, {X, Z} and {Y, Z}. When permutations having the same elements are considered identical, the result is the number of possible combinations and is denoted nCr. Since the r objects can be selected in r! nPr , ways, we divide nPr by r! to “quick-count” the number of possibilities: nCr r! which can be thought of as the first r factors of n!, divided by r!. By substituting n! for nPr in this formula, we find an alternative method for computing nCr is 1n r2! n! . Take special note that when r objects are selected from a set with n elements r!1n r2! and the order they’re listed is unimportant (because you end up with the same subset), the result is a combination, not a permutation. Combinations The number of combinations of n objects taken r at a time is given by nCr

EXAMPLE 8

䊳

nPr

r!

or

nCr

n! r!1n r2!

Computing Combinations Using a Formula Compute each value of nCr given. a. 7C4 b. 8C3 c. 5C2

Solution

䊳

7#6#5#4 4! 35

a. 7C4

8#7#6 3! 56

5#4 2! 10

b. 8C3

c. 5C2

Now try Exercises 55 through 64 As with permutations, when the number of objects is very large, the formula for combinations can also become somewhat cumbersome. Most graphing calculators have the ability to compute combinations, with this option accessed on the same submenu as nPr: MATH (PRB) 3:nCr. Figure 11.59 verifies the computation from Example 8(b), and also shows that in a Political Science class with 30 students, 5 can be picked at random to attend a seminar in the nation’s capitol 142,506 ways. EXAMPLE 9

䊳

䊳

Figure 11.59

Applications of Combinations-Lottery Results A small city is getting ready to draw five Ping-Pong balls of the nine they have numbered 1 through 9 to determine the winner(s) for its annual raffle. If a ticket holder has the same five numbers, they win. In how many ways can the winning numbers be drawn?

Solution

䊳

Since the winning numbers can be drawn in any order, we have a combination of 9 things taken 5 at a time. The five numbers can be drawn in 9C5

9#8#7#6#5 126 ways. 5! Now try Exercises 65 and 66

䊳

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Somewhat surprisingly, there are many situations where the order things are listed is not important. Such situations include • The formation of committees, since the order people volunteer is unimportant • Card games with a standard deck, since the order cards are dealt is unimportant • Playing BINGO, since the order the winning numbers are called is unimportant When the order in which people or objects are selected from a group is unimportant, the number of possibilities is a combination, not a permutation. Another way to tell the difference between permutations and combinations is the following memory device: Permutations have Priority or Precedence; in other words, the Position of each element matters. By contrast, a Combination is like a Committee of Colleagues or Collection of Commoners; all members have equal rank. For permutations, a-b-c is different from b-a-c. For combinations, a-b-c is the same as b-a-c. EXAMPLE 10

䊳

Applications of Quick-Counting — Committees and Governance The Sociology Department of Lakeside Community College has 12 dedicated faculty members. (a) In how many ways can a three-member textbook selection committee be formed? (b) If the department is in need of a Department Chair, Curriculum Chair, and Technology Chair, in how many ways can the positions be filled?

Solution

䊳

a. Since textbook selection depends on a Committee of Colleagues, the order members are chosen is not important. This is a Combination of 12 people taken 3 at a time, and there are 12C3 220 ways the committee can be formed. b. Since those selected will have Position or Priority, this is a Permutation of 12 people taken 3 at a time, giving 12P3 1320 ways the positions can be filled. Now try Exercises 67 through 78

E. You’ve just seen how we can quick-count using combinations

䊳

The Exercise Set contains a wide variety of additional applications. See Exercises 81 through 107.

11.5 EXERCISES 䊳



1. A(n) and has a(n)

is any task that can be repeated set of possible outcomes.

2. When unique elements of a set are rearranged, the result is called a(n) permutation.

3. If an experiment has N equally likely outcomes and is repeated t times, the number of elements in the sample space is given by .

4. If some elements of a group are identical, certain rearrangements are identical and the result is a(n) permutation.

5. A three-digit number is formed from digits 1 to 9. Explain how forming the number with repetition differs from forming it without repetition.

6. Discuss/Explain the difference between a permutation and a combination. Try to think of new ways to help remember the distinction.

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DEVELOPING YOUR SKILLS 7. For the spinner shown here, (a) draw a tree diagram illustrating all possible outcomes for two spins and (b) create an ordered list showing all possible outcomes for two spins.

Z

W

Y

X

Heads

8. For the fair coin shown here, (a) draw a tree diagram illustrating all possible outcomes for four flips and (b) create an ordered list showing the possible outcomes for four flips.

Tails

9. A fair coin is flipped five times. If you extend the tree diagram from Exercise 8, how many possibilities are there? 10. A spinner has the two equally likely outcomes A or B and is spun four times. How is this experiment related to the one in Exercise 8? How many possibilities are there? 11. An inexpensive lock uses the numbers 0 to 24 for a three-number combination. How many different combinations are possible? 12. Grades at a local college consist of A, B, C, D, F, and W. If four classes are taken, how many different report cards are possible? License plates. In a certain (English-speaking) country, license plates for automobiles consist of two letters followed by one of four symbols (■, ◆, ❍, or ●), followed by three digits. How many license plates are possible if

13. Repetition is allowed? 14. Repetition is not allowed? 15. A remote access door opener requires a five-digit (1–9) sequence. How many sequences are possible if (a) repetition is allowed? (b) repetition is not allowed? 16. An instructor is qualified to teach Math 020, 030, 140, and 160. How many different four-course schedules are possible if (a) repetition is allowed? (b) repetition is not allowed? Use the fundamental principle of counting and other quick-counting techniques to respond.

17. Menu items: At Joe’s Diner, the manager is offering a dinner special that consists of one choice of entree (chicken, beef, soy meat, or pork), two vegetable

servings (corn, carrots, green beans, peas, broccoli, or okra), and one choice of pasta, rice, or potatoes. How many different meals are possible? 18. Getting dressed: A frugal businessman has five shirts, seven ties, four pairs of dress pants, and three pairs of dress shoes. Assuming that all possible arrangements are appealing, how many different shirt-tie-pants-shoes outfits are possible? 19. Number combinations: How many four-digit numbers can be formed using the even digits 0, 2, 4, 6, 8, if (a) no repetitions are allowed; (b) repetitions are allowed; (c) repetitions are not allowed and the number must be less than 6000 and divisible by 10. 20. Number combinations: If I was born in March, April, or May, after the 19th but before the 30th, and after 1949 but before 1981, how many different MM–DD–YYYY dates are possible for my birthday? Seating arrangements: William, Xayden, York, and Zelda decide to sit together at the movies. How many ways can they be seated if

21. They sit in random order? 22. York must sit next to Zelda? 23. York and Zelda must be on the outside? 24. William must have the aisle seat? Course schedule: A college student is trying to set her schedule for the next semester and is planning to take five classes: English, art, math, fitness, and science. How many different schedules are possible if

25. The classes can be taken in any order. 26. She wants her science class to immediately follow her math class. 27. She wants her English class to be first and her fitness class to be last. 28. She can’t decide on the best order and simply takes the classes in alphabetical order. Find the value of nPr in two ways: (a) compute r factors n! of n! and (b) use the formula nPr . 1n r2!

29. 10P3

30.

12P2

32. 5P3

33. 8P7

31. 9P4 34. 8P1

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Determine the number of three-letter permutations of the letters given, then use an organized list to write them all out. How many of them are actually words or common names?

35. T, R, and A

36. P, M, and A

37. The regional manager for an office supply store needs to replace the manager and assistant manager at the downtown store. In how many ways can this be done if she selects the personnel from a group of 10 qualified applicants? 38. The local chapter of Mu Alpha Theta will soon be electing a president, vice-president, and treasurer. In how many ways can the positions be filled if the chapter has 15 members? 39. The local school board is going to select a principal, vice-principal, and assistant viceprincipal from a pool of eight qualified candidates. In how many ways can this be done? 40. From a pool of 32 applicants, a board of directors must select a president, vice-president, labor relations liaison, and a director of personnel for the company’s day-to-day operations. Assuming all applicants are qualified and willing to take on any of these positions, how many ways can this be done? 41. A hugely popular chess tournament now has six finalists. Assuming there are no ties, (a) in how many ways can the finalists place in the final round? (b) In how many ways can they finish first, second, and third? (c) In how many ways can they finish if it’s sure that Roberta Fischer is going to win the tournament and that Geraldine Kasparov will come in sixth? 42. A field of 10 horses has just left the paddock area and is heading for the gate. Assuming there are no ties in the big race, (a) in how many ways can the horses place in the race? (b) In how many ways can they finish in the win, place, or show positions? (c) In how many ways can they finish if it’s sure that John Henry III is going to win, Seattle Slew III will come in second (place), and either Dumb Luck II or Calamity Jane I will come in tenth? Assuming all multiple births are identical and the children cannot be told apart, how many distinguishable photographs can be taken of a family of six, if they stand in a single row and there is

43. one set of twins 44. one set of triplets 45. one set of twins and one set of triplets

1129

46. one set of quadruplets 47. How many distinguishable numbers can be made by rearranging the digits of 105,001? 48. How many distinguishable numbers can be made by rearranging the digits in the palindrome 1,234,321? How many distinguishable permutations can be formed from the letters of the given word?

49. logic

50. leave

51. lotto

52. levee

A Scrabble player (see Example 7) has the six letters shown remaining in her rack. How many distinguishable, six-letter permutations can be formed? (If all six letters are played, what was the word?)

53. A, A, A, N, N, B 54. D, D, D, N, A, E Find the value of nCr: (a) using nCr

nPr

r!

(r factors of n!

n! . r!1n r2!

over r!) and (b) using nCr

55. 9C4

56.

10C3

58. 6C3

59. 6C6

57. 8C5 60. 6C0

Use a calculator to verify that each pair of combinations is equal.

61. 9C4, 9C5

62.

10C3, 10C7

63. 8C5, 8C3

64. 7C2, 7C5

65. A platoon leader needs to send four soldiers to do some reconnaissance work. There are 12 soldiers in the platoon and each soldier is assigned a number between 1 and 12. The numbers 1 through 12 are placed in a helmet and drawn randomly. If a soldier’s number is drawn, then that soldier goes on the mission. In how many ways can the reconnaissance team be chosen? 66. Seven colored balls (red, indigo, violet, yellow, green, blue, and orange) are placed in a bag and three are then withdrawn. In how many ways can the three colored balls be drawn? 67. When the company’s switchboard operators went on strike, the company president asked for three volunteers from among the managerial ranks to temporarily take their place. In how many ways can the three volunteers “step forward,” if there are 14 managers and assistant managers in all?

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68. Becky has identified 12 books she wants to read this year and decides to take four with her to read while on vacation. She chooses Pastwatch by Orson Scott Card for sure, then decides to randomly choose any three of the remaining books. In how many ways can she select the four books she’ll end up taking? 69. A new garage band has built up their repertoire to 10 excellent songs that really rock. Next month they’ll be playing in a Battle of the Bands contest, with the winner getting some guaranteed gigs at the city’s most popular hot spots. In how many ways can the band select 5 of their 10 songs to play at the contest? 70. Pierre de Guirré is an award-winning chef and has just developed 12 delectable, new main-course recipes for his restaurant. In how many ways can he select three of the recipes to be entered in an international culinary competition? For each exercise, determine whether a permutation, a combination, counting principles, or a determination of the number of subsets is the most appropriate tool for obtaining a solution, then solve. Some exercises can be completed using more than one method.

72. If you flip a fair coin five times, how many different outcomes are possible? 73. Eight sprinters are competing for the gold, silver, and bronze medals. In how many ways can the medals be awarded? 74. Motorcycle license plates are made using two letters followed by three numbers. How many plates can be made if repetition of letters (only) is allowed? 75. A committee of five students is chosen from a class of 20 to attend a seminar. How many different ways can this be done? 76. If onions, cheese, pickles, and tomatoes are available to dress a hamburger, how many different hamburgers can be made? 77. A caterer offers eight kinds of fruit to make various fruit trays. How many different trays can be made using four different fruits? 78. Eighteen females try out for the basketball team, but the coach can only place 15 on her roster. How many different teams can be formed?

71. In how many ways can eight second-grade children line up for lunch?

䊳


79. Stirling’s Formula: n! ⬇ 12␲ # 1nnⴙ0.5 2 # eⴚn Values of n! grow very quickly as n gets larger (13! is already in the billions). For some applications, scientists find it useful to use the approximation for n! shown, called Stirling’s Formula. a. Compute the value of 7! on your calculator, b. Compute the value of 10! on your calculator, then use Stirling’s Formula with n 7. By then use Stirling’s Formula with n 10. By what percent does the approximate value what percent does the approximate value differ from the true value? differ from the true value? 80. Factorial formulas: For n, k 僆⺧, where n 7 k, a. Verify the formula for n 7 and k 5. 䊳

n! ⴝ n1n ⴚ 12 1n ⴚ 22 p 1n ⴚ k ⴙ 12 1n ⴚ k2! b. Verify the formula for n 9 and k 6.

APPLICATIONS

81. Yahtzee: In the game of “Yahtzee”® (Milton Bradley) five dice are rolled simultaneously on the first turn in an attempt to obtain various arrangements (worth various point values). How many different arrangements are possible?

82. Twister: In the game of “Twister”® (Milton Bradley) a simple spinner is divided into four quadrants designated Left Foot (LF), Right Hand (RH), Right Foot (RF), and Left Hand (LH), with four different color possibilities in each quadrant (red, green, yellow, blue). Determine the number of possible outcomes for three spins.

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83. Clue: In the game of “Clue”® (Parker Brothers) a crime is committed in one of nine rooms, with one of six implements, by one of six people. In how many different ways can the crime be committed? Phone numbers in North America have 10 digits: a threedigit area code, a three-digit exchange number, and the four final digits that make each phone number unique. Neither area codes nor exchange numbers can start with 0 or 1. Prior to 1994 the second digit of the area code had to be a 0 or 1. Sixteen area codes are reserved for special services (such as 911 and 411). In 1994, the last area code was used up and the rules were changed to allow the digits 2 through 9 as the middle digit in area codes.

1131

Seating arrangements: In how many different ways can eight people (six students and two teachers) sit in a row of eight seats if

96. the teachers must sit on the ends 97. the teachers must sit together Television station programming: A television station needs to fill eight half-hour slots for its Tuesday evening schedule with eight programs. In how many ways can this be done if

98. there are no constraints 99. Seinfeld must have the 8:00 P.M. slot

84. How many different area codes were possible prior to 1994?

100. Seinfeld must have the 8:00 P.M. slot and The Drew Carey Show must be shown at 6:00 P.M.

85. How many different exchange numbers were possible prior to 1994?

101. Friends can be aired at 7:00 or 9:00 P.M. and Everybody Loves Raymond can be aired at 6:00 or 8:00 P.M.

86. How many different phone numbers were possible prior to 1994? 87. How many different phone numbers were possible after 1994? Aircraft N-Numbers: In the United States, private aircraft are identified by an “N-Number,” which is generally the letter “N” followed by five characters and includes these restrictions: (1) the N-Number can consist of five digits, four digits followed by one letter, or three digits followed by two letters; (2) the first digit cannot be a zero; (3) to avoid confusion with the numbers zero and one, the letters O and I cannot be used; and (4) repetition of digits and letters is allowed. How many unique N-Numbers can be formed

88. that have four digits and one letter? 89. that have three digits and two letters?

Scholarship awards: Fifteen students at Roosevelt Community College have applied for six available scholarship awards. How many ways can the awards be given if

102. there are six different awards given to six different students 103. there are six identical awards given to six different students Committee composition: The local city council has 10 members and is trying to decide if they want to be governed by a committee of three people or by a president, vice-president, and secretary.

104. If they are to be governed by committee, how many unique committees can be formed? 105. How many different president, vice-president, and secretary possibilities are there?

90. that have five digits? 91. that have three digits, two letters with no repetitions of any kind allowed? Seating arrangements: Eight people would like to be seated. Assuming some will have to stand, in how many ways can the seats be filled if the number of seats available is

92. eight

93. five

94. three

95. one

106. Team rosters: A soccer team has three goalies, eight defensive players, and eight forwards on its roster. How many different starting line-ups can be formed (one goalie, three defensive players, and three forwards)? 107. e-mail addresses: A business wants to standardize the e-mail addresses of its employees. To make them easier to remember and use, they consist of two letters and two digits (followed by @esmtb.com), with zero being excluded from use as the first digit and no repetition of letters or digits allowed. Will this provide enough unique addresses for their 53,000 employees worldwide?

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Tic-Tac-Toe: In the game Tic-Tac-Toe, players alternately write an “X” or an “O” in one of nine squares on a 3 ⫻ 3 grid. If either player gets three in a row horizontally, vertically, or diagonally, that player wins. If all nine squares are played with neither person winning, the game is a draw. Assuming “X” always goes first, 108. How many different “ending boards” are possible if the game ends after five plays? 䊳

109. How many different “ending boards” are possible if the game ends after six plays?


110. (9.3) Solve the given system of linear inequalities by graphing. Shade the feasible region. 2x ⫹ y 6 6 x ⫹ 2y 6 6 μ xⱖ0 yⱖ0

112. (7.3) Rewrite cos12␣2cos13␣2 ⫺ sin12␣2sin13␣2 as a single expression. 113. (10.3) Graph the hyperbola that is defined by 1x ⫺ 22 2 4

⫺

1y ⫹ 32 2 9

⫽ 1.

12 , determine the other five 13 trig functions of the acute angle ␪.

111. (6.2) Given sin ␪ ⫽

11.6

Introduction to Probability

LEARNING OBJECTIVES

There are few areas of mathematics that give us a better view of the world than probability and statistics. Unlike statistics, which seeks to analyze and interpret data, probability (for our purposes) attempts to use observations and data to make statements concerning the likelihood of future events. Such predictions of what might happen have found widespread application in such diverse fields as politics, manufacturing, gambling, opinion polls, product life, and many others. In this section, we develop the basic elements of probability.


A. Define an event on a sample space

B. Compute elementary probabilities C. Use certain properties of probability D. Compute probabilities using quick-counting techniques E. Compute probabilities involving nonexclusive events

EXAMPLE 1

A. Defining an Event In Section 11.5 we defined the following terms: experiment and sample outcome. Flipping a coin twice in succession is an experiment, and two sample outcomes are HH and HT. An event E is any designated set of sample outcomes, and is a subset of the sample space. One event might be E1: (two heads occur), another possibility is E2: (at least one tail occurs).

䊳

Stating a Sample Space and Defining an Event Consider the experiment of rolling one standard, six-sided die (plural is dice). State the sample space S and define any two events relative to S.

Solution A. You’ve just seen how we can define an event on a sample space

䊳

S is the set of all possible outcomes, so S ⫽ 51, 2, 3, 4, 5, 66. Two possible events are E1: (a 5 is rolled) and E2: (an even number is rolled). Now try Exercises 7 through 10

䊳

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Section 11.6 Introduction to Probability

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B. Elementary Probability When rolling a die, we know the result can be any of the six equally likely outcomes in the sample space, so the chance of E1:(a five is rolled) is 16. Since three of the elements in S are even numbers, the chance of E2:(an even number is rolled) is 36 12. This suggests the following definition. The Probability of an Event E Given S is a sample space of equally likely events and E is an event relative to S, the probability of E, written P(E), is computed as n1E2 P1E2 n1S2 where n(E) represents the number of elements in E, and n(S) represents the number of elements in S.

WORTHY OF NOTE Our study of probability will involve only those sample spaces with events that are equally likely.

A standard deck of playing cards consists of 52 cards divided in four groups or suits. There are 13 hearts (♥), 13 diamonds 1䉬2, 13 spades (♠), and 13 clubs (♣). As you can see in Figure 11.60, each of the 13 cards in a suit is labeled A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, and K. Also notice that 26 of the cards are red (hearts and diamonds), 26 are black (spades and clubs), and 12 of the cards are “face cards” (J, Q, K of each suit).

EXAMPLE 2

䊳

Figure 11.60

Stating a Sample Space and the Probability of a Single Outcome A single card is drawn from a well-shuffled deck. Define S and state the probability of any single outcome. Then define E as a King is drawn and find P(E).

Solution

䊳

Sample space: S 5the 52 cards6 . There are 52 equally likely outcomes, 1 so the probability of any one outcome is 52 . Since S has four Kings, n1E2 4 or about 0.077. P1E2 52 n1S2


EXAMPLE 3

䊳

䊳

Stating a Sample Space and the Probability of a Single Outcome A family of five has two girls and three boys named Sophie, Maria, Albert, Isaac, and Pythagoras. Their ages are 21, 19, 15, 13, and 9, respectively. One is to be selected randomly. Find the probability a teenager is chosen.

Solution B. You’ve just seen how we can compute elementary probabilities

䊳

The sample space is S 59, 13, 15, 19, 216. Three of the five are teenagers, meaning the probability is 35, 0.6, or 60%.


䊳

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C. Properties of Probability A study of probability necessarily includes recognizing some basic and fundamental properties. For example, when a fair die is rolled, what is P(E) if E is defined as a 1, 2, 3, 4, 5, or 6 is rolled? The event E will occur 100% of the time, since 1, 2, 3, 4, 5, 6 are the only possibilities. In symbols we write P(outcome is in the sample space) or simply P1S2 ⫽ 1 (100%). What percent of the time will a result not in the sample space occur? Since the die has only the six sides numbered 1 through 6, the probability of rolling something else is zero. In symbols, P1outcome is not in sample space2 ⫽ 0 or simply P1~S2 ⫽ 0.

WORTHY OF NOTE In probability studies, the tilde “~” acts as a negation symbol. For any event E defined on the sample space, ~E means the event does not occur.

Properties of Probability Given sample space S and any event E defined relative to S. 1. P1S2 ⫽ 1

EXAMPLE 4

䊳

2. P1~S2 ⫽ 0

3. 0 ⱕ P1E2 ⱕ 1

Determining the Probability of an Event A game is played using a spinner like the one shown. Determine the probability of the following events: E1: A nine is spun.

Solution

䊳

2 8

P1E2 2 ⫽

5 7

The sample space consists of eight equally likely outcomes. 0 ⫽0 8

4

1

E2: An integer greater than 0 and less than 9 is spun. P1E1 2 ⫽

3

6

8 ⫽ 1. 8

Technically, E1: A nine is spun is not an “event,” since it is not in the sample space and cannot occur, while E2 contains the entire sample space and must occur. Now try Exercises 17 and 18

䊳

Because we know P1S2 ⫽ 1 and all sample outcomes are equally likely, the probabilities of all single events defined on the sample space must sum to 1. For the experiment of rolling a fair die, the sample space has six outcomes that are equally likely. Note that P112 ⫽ P122 ⫽ P132 ⫽ P142 ⫽ P152 ⫽ P162 ⫽ 16, and 16 ⫹ 16 ⫹ 16 ⫹ 16 ⫹ 16 ⫹ 16 ⫽ 1. Probability and Sample Outcomes Given a sample space S with n equally likely sample outcomes s1, s2, s3, p , sn. n

兺 P1s 2 ⫽ P1s 2 ⫹ P1s 2 ⫹ P1s 2 ⫹ p ⫹ P1s 2 ⫽ 1 i

1

2

3

n

i⫽1

The complement of an event E is the set of sample outcomes in S not contained in E. Symbolically, ~E is the complement of E. Probability and Complementary Events Given sample space S and any event E defined relative to S, the complement of E, written ~E, is the set of all outcomes not in E and: 1. P1E2 ⫽ 1 ⫺ P1~E2

2. P1E2 ⫹ P1~E2 ⫽ 1

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EXAMPLE 5

䊳

1135

Stating a Probability Using Complements Use complementary events to answer the following questions: a. A single card is drawn from a well-shuffled deck. What is the probability that it is not a diamond? b. A single letter is picked at random from the letters in the word “divisibility.” What is the probability it is not an “i”?

Solution

䊳

WORTHY OF NOTE Probabilities can be written in fraction form, decimal form, or as a percent. For P(~D) from Example 5(a), the probability could be written 34 , 0.75, or 75%.

EXAMPLE 6

a. Since there are 13 diamonds in a standard 52-card deck: 39 P1~D2 ⫽ 1 ⫺ P1D2 ⫽ 1 ⫺ 13 52 ⫽ 52 ⫽ 0.75. b. Of the 12 letters in d-i-v-i-s-i-b-i-l-i-t-y, 5 are “i’s.” This means 5 7 P1~i2 ⫽ 1 ⫺ P1i2, or 1 ⫺ 12 ⫽ 12 . The probability of choosing a letter other than i is 0.583. Now try Exercises 19 through 22

䊳

䊳

Stating a Probability Using Complements Inter-Island Waterways has just opened hydrofoil service between several islands. The hydrofoil is powered by two engines, one forward and one aft, and will operate if either of its two engines is functioning. Due to testing and past experience, the company knows the probability of the aft engine failing is P1aft engine fails2 ⫽ 0.05, the probability of the forward engine failing is P1forward engine fails2 ⫽ 0.03, and the probability that both fail is P1both engines simultaneously fail2 ⫽ 0.012. What is the probability the hydrofoil completes its next trip?

Solution

䊳

Although the answer may seem complicated, note that P(trip is completed) and P(both engines simultaneously fail) are complements. P1trip is completed2 ⫽ 1 ⫺ P1both engines simultaneously fail2 ⫽ 1 ⫺ 0.012 ⫽ 0.988 There is close to a 99% probability the trip will be completed. Now try Exercises 23 and 24

䊳

The chart in Figure 11.61 shows all 36 possible outcomes (the sample space) from the experiment of rolling two fair dice. Figure 11.61

EXAMPLE 7

䊳

Stating a Probability Using Complements Two fair dice are rolled. What is the probability the sum of both dice is greater than or equal to 5, P1sum ⱖ 52?

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Solution

䊳

C. You’ve just seen how we can use certain properties of probability

See Figure 11.61. For P 1sum ⱖ 52 it may be easier to use complements as there are far fewer possibilities: P1sum ⱖ 52 ⫽ 1 ⫺ P1sum 6 52 , which gives 1 5 6 ⫽ 1 ⫺ ⫽ ⫽ 0.83. 1⫺ 36 6 6 Now try Exercises 25 and 26

䊳

D. Probability and Quick-Counting Quick-counting techniques were introduced earlier to help count the number of elements in a large or more complex sample space, and the number of sample outcomes in an event. EXAMPLE 8A

䊳

Stating a Probability Using Combinations Five cards are drawn from a shuffled 52-card deck. Calculate the probability of E1:(all five cards are face cards) or E2:(all five cards are hearts).

Solution

䊳

The sample space for both events consists of all five-card groups that can be formed from the 52 cards or 52C5. For E1 we are to select five face cards from the 12 that are available (three from each suit), or 12C5. The probability of five face n1E2 792 12C5 ⫽ cards is ⬇ 0.0003. For E2 we are to select five , which gives 2,598,960 n1S2 52C5 n1E2 13C5 ⫽ hearts from the 13 available, or 13C5. The probability of five hearts is , n1S2 52C5 1287 ⬇ 0.0005. which is 2,598,960

䊳

Stating a Probability Using Combinations and the Fundamental Principle of Counting

WORTHY OF NOTE It seems reasonable that the probability of 5 hearts is slightly higher, as 13 of the 52 cards are hearts, while only 12 are face cards.

EXAMPLE 8B

Of the 42 seniors at Jacoby High School, 23 are female and 19 are male. A group of five students is to be selected at random to attend a conference in Reno, Nevada. What is the probability the group will have exactly three females?

Solution

䊳

The sample space consists of all five-person groups that can be formed from the 42 seniors or 42C5. The event consists of selecting 3 females from the 23 available (23C3) and 2 males from the 19 available (19C2). Using the fundamental principle of counting n1E2 ⫽ 23C3 # 19C2 and the probability the group has 3 females is # n1E2 302,841 23C3 19C2 ⫽ , which gives ⬇ 0.356. There is approximately a 850,668 n1S2 42C5 35.6% probability the group will have exactly 3 females. Now try Exercises 27 through 34 䊳 While the fundamentals of probability are usually introduced using dice, cards, and very basic applications, this should not take away from its true power and utility. We use probability to help us to analyze things quantitatively (with numbers) so that important decisions can be made. In manufacturing, the probability a product becomes defective during its warranty period is crucial to the company’s financial stability. Health organizations rely heavily on probability to plan against the anticipated spread

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of a symptom through a given population. In these and many other situations, the sample spaces are very large and the defined events likewise large, making the use of technology an integral part of probability studies. The computed result of Example 8A is shown in Figure 11.62 (in scientific notation—shift the decimal four places to the left). The computed result of Example 8B appears in Figure 11.63. Figure 11.62

Figure 11.63

D. You’ve just seen how we can compute probabilities using quick-counting techniques

E. Probability and Nonexclusive Events Figure 11.64 Sometimes the way events are defined S causes them to share sample outcomes. Using a standard deck of playE1 E2 J♠ A♣ ing cards once again, if we define the 2♣ Q♠ 3♣ J♣ events E1:(a club is drawn) and E2:(a J♦ K♠ 4♣ 5♣ face card is drawn), they share the outQ♦ Q♣ 6♣ comes J♣, Q♣, and K♣ as shown in K♦ K♣ J♥ 8♣ 7♣ Figure 11.64. This overlapping region Q♥ 9♣ K♥ is the intersection of the events, or 10♣ E1 傽 E2. If we compute n1E1 ´ E2 2 as n1E1 2 ⫹ n1E2 2 as before, this intersecting region gets counted twice! In cases where the events are nonexclusive (not mutually exclusive), we maintain the correct count by subtracting one of the two intersections, obtaining n1E1 ´ E2 2 ⫽ n1E1 2 ⫹ n1E2 2 ⫺ n1E1 傽 E2 2. This leads to the following calculation for the probability of nonexclusive events:

WORTHY OF NOTE This can be verified by simply counting the elements involved: n1E1 2 ⫽ 13 and n1E2 2 ⫽ 12 so n1E1 2 ⫹ n1E2 2 ⫽ 25. However, there are only 22 possibilities in E1 ´ E2—the J♣, Q♣, and K♣ got counted twice.

P1E1 ´ E2 2 ⫽ ⫽

n1E1 2 ⫹ n1E2 2 ⫺ n1E1 傽 E2 2

n1E1 2 n1S2

n1S2

⫹

n1E1 2 n1S2

⫺

n1E1 傽 E2 2 n1S2

⫽ P1E1 2 ⫹ P1E2 2 ⫺ P1E1 傽 E2 2

definition of probability

property of rational expressions definition of probability

Probability and Nonexclusive Events Given sample space S and nonexclusive events E1 and E2 defined relative to S, the probability of E1 or E2 is given by P1E1 ´ E2 2 ⫽ P1E1 2 ⫹ P1E2 2 ⫺ P1E1 傽 E2 2

EXAMPLE 9A

䊳

Stating the Probability of Nonexclusive Events What is the probability that a club or a face card is drawn from a standard deck of 52 well-shuffled cards?

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Solution

䊳

As before, define the events E1:(a club is drawn) and E2:(a face card is drawn). 12 Since there are 13 clubs and 12 face cards, P1E1 2 ⫽ 13 52 and P1E2 2 ⫽ 52 . But three 3 . This leads to of the face cards are clubs, so P1E1 傽 E2 2 ⫽ 52 P1E1 ´ E2 2 ⫽ P1E1 2 13 ⫹ ⫽ 52 22 ⫽ ⬇ 52

⫹ P1E2 2 ⫺ P1E1 傽 E2 2 12 3 ⫺ 52 52

0.423

nonexclusive events substitute

combine terms

There is about a 42% probability that a club or face card is drawn.

EXAMPLE 9B

䊳

Stating the Probability of Nonexclusive Events A survey of 100 voters was taken to gather information on critical issues and the demographic information collected is shown in the table. One out of the 100 voters is to be drawn at random to be interviewed on the 5 P.M. News. What is the probability the person is a woman (W) or a Republican (R)?

Solution

䊳

Women

Men

Totals

Republican

17

20

37

Democrat

22

17

39

Independent

8

7

15

Green Party

4

1

5

Tax Reform

2

2

4

53

47

100

Totals

Since there are 53 women and 37 Republicans, P1W2 ⫽ 0.53 and P1R2 ⫽ 0.37. The table shows 17 people are both female and Republican so P1W 傽 R2 ⫽ 0.17. P1W ´ R2 ⫽ P1W2 ⫹ P1R2 ⫺ P1W 傽 R2 ⫽ 0.53 ⫹ 0.37 ⫺ 0.17 ⫽ 0.73

nonexclusive events substitute combine

There is a 73% probability the person is a woman or a Republican. Now try Exercises 35 through 48 䊳 Two events that have no common outcomes are called mutually exclusive events (one excludes the other and vice versa). For example, in rolling one die, E1:(a 2 is rolled) and E2:(an odd number is rolled) are mutually exclusive, since 2 is not an odd number. For the probability of E3:(a 2 is rolled or an odd number is rolled), we note that n1E1 傽 E2 2 ⫽ 0 and the previous formula simply reduces to P1E1 2 ⫹ P1E2 2 . See Exercises 49 and 50. There is a large variety of additional applications in the Exercise Set. See Exercises 53 through 68. When probability calculations require a repeated use of permutations or combinations, tables can make the work more efficient and help to explore the patterns they generate. For instance, to choose r children from a group of six children 1n ⫽ 62, we can set TBLSET to AUTO, then press Y= and enter 6 nCr X as Y1 (Figure 11.65). Access the TABLE ( 2nd GRAPH ) and note that the calculator has automatically computed the value of 6C0, 6C1, 6C2, p , 6C6 (Figure 11.66) and the pattern of outputs is symmetric. For calculations like those required in Example 8B 1 23C3 # 19C2 2, we can enter Y1 ⫽ 23 nCr X, Y2 ⫽ 19 nCr 15 ⫺ X2, and Y3 ⫽ Y1 # Y2 to further explore the number of ways groups with different numbers of females can be chosen. Also see Exercise 70.

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Figure 11.65

1139

Figure 11.66

E. You’ve just seen how we can compute probabilities involving nonexclusive events

11.6 EXERCISES 䊳



1. Given a sample space S and an event E defined relative to S, P1E2 ⫽

䊳

n1S2

.

2. In elementary probability, we consider all events in the sample space to be likely.

3. Given a sample space S and an event E defined ⱕ P1E2 ⱕ relative to S: , P1S2 ⫽ , and P1~S2 ⫽ .

4. The of an event E is the set of sample outcomes in S which are not contained in E.

5. Discuss/Explain the difference between mutually exclusive events and nonexclusive events. Give an example of each.

6. A single die is rolled. With no calculations, explain why the probability of rolling an even number is greater than rolling a number greater than four.


State the sample space S and the probability of a single outcome. Then define any two events E relative to S (many answers possible). Exercise 8

7. Two fair coins (heads and tails) are flipped. 8. The simple spinner shown is spun.

1 4

2 3

9. The head coaches for six little league teams (the Patriots, Cougars, Angels, Sharks, Eagles, and Stars) have gathered to discuss new changes in the rule book. One of them is randomly chosen to ask the first question. 10. Experts on the planets Mercury, Venus, Mars, Jupiter, Saturn, Uranus, Neptune, and the Kuiper object formerly known as the planet Pluto have gathered at a space exploration conference. One group of experts is selected at random to speak first.

Find P(E) for the events defined.

11. Nine index cards numbered 1 through 9 are shuffled and placed in an envelope, then one of the cards is randomly drawn. Define event E as the number drawn is even. 12. Eight flash cards used for studying basic geometric shapes are shuffled and one of the cards is drawn at random. The eight cards include information on circles, squares, rectangles, kites, trapezoids, parallelograms, pentagons, and triangles. Define event E as a quadrilateral is drawn. 13. One card is drawn at random from a standard deck of 52 cards. What is the probability of a. drawing a Jack b. drawing a spade c. drawing a black card d. drawing a red three

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14. Pinochle is a card game played with a deck of 48 cards consisting of 2 Aces, 2 Kings, 2 Queens, 2 Jacks, 2 Tens, and 2 Nines in each of the four standard suits [hearts (♥), diamonds (䉬), spades (♠), and clubs (♣)]. If one card is drawn at random from this deck, what is the probability of a. drawing an Ace b. drawing a club c. drawing a red card d. drawing a face card (Jack, Queen, King)

Use the complementary events to complete Exercises 19 through 22.

15. A group of finalists on a game show consists of three males and five females. Hank has a score of 520 points, with Harry and Hester having 490 and 475 points, respectively. Madeline has 532 points, with Mackenzie, Morgan, Maggie, and Melanie having 495, 480, 472, and 470 points, respectively. One of the contestants is randomly selected to start the final round. Define E1 as Hester is chosen, E2 as a female is chosen, and E3 as a contestant with fewer than 500 points is chosen. Find the probability of each event.

22. A corporation will be moving its offices to Los Angeles, Miami, Atlanta, Dallas, or Phoenix. If the site is randomly selected, what is the probability Dallas is not chosen?

16. Soccer coach Maddox needs to fill the last spot on his starting roster for the opening day of the season and has to choose between three forwards and five defenders. The forwards have jersey numbers 5, 12, and 17, while the defenders have jersey numbers 7, 10, 11, 14, and 18. Define E1 as a forward is chosen, E2 as a defender is chosen, and E3 as a player whose jersey number is greater than 10 is chosen. Find the probability of each event. 17. A game is played using a spinner like the one shown. For each spin, a. What is the probability the 1 2 arrow lands in a shaded region? 4 3 b. What is the probability your spin is less than 5? c. What is the probability you spin a 2? d. What is the probability the arrow points to prime number? 18. A game is played using a spinner like the one shown here. For each spin, a. What is the probability the 2 1 arrow lands in a lightly 3 6 shaded region? 5 4 b. What is the probability your spin is greater than 2? c. What is the probability the arrow lands in a shaded region? d. What is the probability you spin a 5?

19. One card is drawn from a standard deck of 52. What is the probability it is not a club? 20. Four standard dice are rolled. What is the probability the sum is less than 24? 21. A single digit is randomly selected from among the digits of 10!. What is the probability the digit is not a 2?

23. A large manufacturing plant can remain at full production as long as one of its two generators is functioning. Due to past experience and the age difference between the systems, the plant manager estimates the probability of the main generator failing is 0.05, the probability of the secondary generator failing is 0.01, and the probability of both failing is 0.009. What is the probability the plant remains in full production today? 24. A fire station gets an emergency call from a shopping mall in the mid-afternoon. From a study of traffic patterns, Chief Nozawa knows the probability the most direct route is clogged with traffic is 0.07, while the probability of the secondary route being clogged is 0.05. The probability both are clogged is 0.02. What is the probability they can respond to the call unimpeded using one of these routes? 25. Two fair dice are rolled (see Figure 11.61). What is the probability of a. a sum less than four b. a sum less than eleven c. the sum is not nine d. a roll is not a “double” (both dice the same) “Double-six” dominos is a game played with the 28 numbered tiles shown in the diagram.

26. The 28 dominos are placed in a bag, shuffled, and then one domino is randomly drawn. What is the probability the total number of dots on the domino a. is three or less b. is greater than three c. does not have a blank half d. is not a “double” (both sides the same)

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Find P(E) given the values for n(E) and n(S) shown.

27. n1E2 ⫽ 6C3 # 4C2; n1S2 ⫽ 10C5 28. n1E2 ⫽

12C9

# 8C7; n1S2 ⫽ 20C16

29. n1E2 ⫽ 9C6 # 5C3; n1S2 ⫽ 14C9 30. n1E2 ⫽ 7C6 # 3C2; n1S2 ⫽ 10C8 31. Five cards are drawn from a well-shuffled, standard deck of 52 cards. Which has the greater probability: (a) all five cards are red or (b) all five cards are numbered cards? How much greater? 32. Five cards are drawn from a well-shuffled pinochle deck of 48 cards (see Exercise 14). Which has the greater probability, (a) all five cards are face cards (King, Queen, or Jack) or (b) all five cards are black? How much greater?

1141

41. Two fair dice are rolled. What is the probability the sum of the dice is a. a multiple of 3 and an odd number b. a sum greater than 5 and a 3 on one die c. an even number and a number greater than 9 d. an odd number and a number less than 10 42. Eight Ball is a game played on a pool table with 15 balls numbered 1 through 15 and a cue ball that is solid white. Of the 15 numbered balls, 8 are a solid (nonwhite) color and numbered 1 through 8, and seven are striped balls numbered 9 through 15. The fifteen numbered pool balls (no cueball) are placed in a large bowl and mixed, then one is drawn out. What is the probability of drawing

33. A dietetics class has 24 students. Of these, 9 are vegetarians and 15 are not. The instructor receives enough funding to send six students to a conference. If the students are selected randomly, what is the probability the group will have a. exactly two vegetarians b. exactly four nonvegetarians c. at least three vegetarians 34. A large law firm has a support staff of 15 employees: six paralegals and nine legal assistants. Due to recent changes in the law, the firm wants to send five of them to a forum on the new changes. If the selection is done randomly, what is the probability the group will have a. exactly three paralegals b. exactly two legal assistants c. at least two paralegals Find the probability indicated using the information given.

35. Given P1E1 2 ⫽ 0.7, P1E2 2 ⫽ 0.5, and P1E1 ¨ E2 2 ⫽ 0.3, compute P1E1 ´ E2 2. 36. Given P1E1 2 ⫽ 0.6, P1E2 2 ⫽ 0.3, and P1E1 ¨ E2 2 ⫽ 0.2, compute P1E1 ´ E2 2.

37. Given P1E1 2 ⫽ 38, P1E2 2 ⫽ 34, and P1E1 ´ E2 2 ⫽ 56; compute P1E1 ¨ E2 2.

38. Given P1E1 2 ⫽ 12, P1E2 2 ⫽ 35, and P1E1 ´ E2 2 ⫽ 17 20 ; compute P1E1 ¨ E2 2. 39. Given P1E1 ´ E2 2 ⫽ 0.72, P1E2 2 ⫽ 0.56, and P1E1 ¨ E2 2 ⫽ 0.43; compute P(E1).

40. Given P1E1 ´ E2 2 ⫽ 0.85, P1E1 2 ⫽ 0.4, and P1E1 ¨ E2 2 ⫽ 0.21; compute P(E2).

a. b. c. d. e. f. g. h.

the eight ball a number greater than fifteen an even number a multiple of three a solid color and an even number a striped ball and an odd number an even number and a number divisible by 3 an odd number and a number divisible by 4

43. A survey of 50 veterans was taken to gather information on their service career and what life is like out of the military. A breakdown of those surveyed is shown in the table. One out of the 50 will be selected at random for an interview and a biographical sketch. What is the probability the person chosen is Women

Men

Totals

6

9

15

Corporal

10

8

18

Sergeant

4

5

9

Private

Lieutenant

2

1

3

Captain

2

3

5

24

26

50

Totals

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a. b. c. d. e.

a woman and a sergeant a man and a private a private and a sergeant a woman and an officer a person in the military

44. Referring to Exercise 43, what is the probability the person chosen is a. a woman or a sergeant b. a man or a private c. a woman or a man d. a woman or an officer e. a captain or a lieutenant A computer is asked to randomly generate a three-digit number. What is the probability the

45. ten’s digit is odd or the one’s digit is even 46. first digit is prime and the number is a multiple of 10 A computer is asked to randomly generate a four-digit number. What is the probability the number is

49. Two fair dice are rolled. What is the probability of a. boxcars (sum of 12) or snake eyes (sum of 2) b. a sum of 7 or a sum of 11 c. an even-numbered sum or a prime sum d. an odd-numbered sum or a sum that is a multiple of 4 e. a sum of 15 or a multiple of 12 f. a sum that is a prime number 50. Suppose all 16 balls from a game of pool (see Exercise 42) are placed in a large leather bag and mixed, then one is drawn out. Consider the cue ball as “0.” What is the probability of drawing a. a striped ball b. a solid-colored ball c. a polka-dotted ball d. the cue ball e. the cue ball or the eight ball f. a striped ball or a number less than five g. a solid color or a number greater than 12 h. an odd number or a number divisible by 4

47. at least 4000 or a multiple of 5 48. less than 7000 and an odd number 䊳


51. Games involving a fair spinner (with numbers 1 through 4): P1n2 ⴝ 1 14 2 n Games that involve moving pieces around a board using a fair spinner are fairly common. If a fair spinner has the numbers 1 through 4, the probability that any one number is spun n times in succession is given by the formula shown, where n represents the number of spins. What is the probability (a) the first player spins a two? (b) all four players spin a two? (c) Discuss the graph of P(n) and explain the connection between the graph and the probability of consistently spinning a two. 䊳

52. Games involving a fair coin (heads and tails): P1n2 ⴝ 1 12 2 n When a fair coin is flipped, the probability that heads (or tails) is flipped n times in a row is given by the formula shown, where n represents the number of flips. What is the probability (a) the first flip is heads? (b) the first four flips are heads? (c) Discuss the graph of P(n) and explain the connection between the graph and the probability of consistently flipping heads.

APPLICATIONS

53. To improve customer service, a company tracks the number of minutes a caller is “on hold” and waiting for a customer service representative. The table shows the probability that a caller will wait m minutes. Based on the table, what is the probability a caller waits a. at least 2 minutes b. less than 2 minutes c. 4 minutes or less d. over 4 minutes

e. less than 2 or more than 4 minutes f. 3 or more minutes Wait Time (minutes m)

Probability

0

0.07

0 6 m 6 1

0.28

1ⱕm 6 2

0.32

2ⱕm 6 3

0.25

3ⱕm 6 4

0.08

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54. To study the impact of technology on American families, a researcher first determines the probability that a family has n computers at home. Based on the table, what is the probability a home a. has at least one computer b. has two or more computers c. has fewer than four computers d. has five computers e. has one, two, or three computers f. does not have two computers Number of Computers

Probability

0

9%

1

51%

2

28%

3

9%

4

3%

55. Jolene is an experienced markswoman and is able to hit a 10 in. by 20 in. target 100% of the time at a range of 100 yd. Assuming the 10 in. probability she hits a target is related to its area, what is the 20 in. probability she hits the shaded portions shown? a. isosceles triangle b. right triangle

c. isosceles right triangle

56. a. square

b. circle

c. isosceles trapezoid with b ⫽ B2

1143

57. A circular dartboard has a 2 total radius of 8 in., with 4 circular bands that are 2 in. 6 wide, as shown. You are 8 skilled enough to hit this board 100% of the time so you always score at least two points each time you throw a dart. Assuming the probabilities are related to area, on the next dart that you throw what is the probability you a. score at least a 4? b. score at least a 6? c. hit the bull’s-eye? d. score exactly 4 points? 58. Three red balls, six blue balls, and four white balls are placed in a bag. What is the probability the first ball you draw out is a. red b. blue c. not white d. purple e. red or white f. red and white 59. Three red balls, six blue balls, and four white balls are placed in a bag, then two are drawn out and placed in a rack. What is the probability the balls drawn are a. first red, second blue b. first blue, second red c. both white d. first blue, second not red e. first white, second not blue f. first not red, second not blue 60. Consider the 210 discrete points found in the first and second quadrants where ⫺10 ⱕ x ⱕ 10, 1 ⱕ y ⱕ 10, and x and y are integers. The coordinates of each point is written on a slip of paper and placed in a box. One of the slips is then randomly drawn. What is the probability the point (x, y) drawn a. is on the graph of y ⫽ 冟x冟 b. is on the graph of y ⫽ 2冟x冟 c. is on the graph of y ⫽ 0.5冟x冟 d. has coordinates 1x, y 7 ⫺22 e. has coordinates 1x ⱕ 5, y 7 ⫺22 f. is between the branches of y ⫽ x2

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61. Your instructor surprises you with a True/False quiz for which you are totally unprepared and must guess randomly. What is the probability you pass the quiz with an 80% or better if there are a. three questions b. four questions c. five questions 62. A robot is sent out to disarm a timed explosive device by randomly changing some switches from a neutral position to a positive flow or negative flow position. The problem is, the switches are independent and unmarked, and it is unknown which direction is positive and which direction is negative. The bomb is harmless if a majority of the switches yield a positive flow. All switches must be thrown. What is the probability the device is disarmed if there are a. three switches b. four switches c. five switches 63. A survey of 100 retirees was taken to gather information concerning how they viewed the Vietnam War back in the early 1970s. A breakdown of those surveyed is shown in the table. One out of the hundred will be selected at random for a personal, taped interview. What is the probability the person chosen had a a. career of any kind and opposed the war b. medical career and supported the war c. military career and opposed the war d. legal or business career and opposed the war e. academic or medical career and supported the war Career

Support

Military

9

3

12

Medical

8

16

24

Legal

15

12

27

Business

18

6

24

3

10

13

53

47

100

Academics Totals

Opposed

Total

64. Referring to Exercise 63, what is the probability the person chosen a. had a career of any kind or opposed the war b. had a medical career or supported the war

c. supported the war or had a military career d. had a medical or a legal career e. supported or opposed the war 65. The Board of Directors for a large hospital has 15 members. There are six doctors of nephrology (kidneys), five doctors of gastroenterology (stomach and intestines), and four doctors of endocrinology (hormones and glands). Eight of them will be selected to visit the nation’s premier hospitals on a 3-week, expenses-paid tour. What is the probability the group of eight selected consists of exactly a. four nephrologists and four gastroenterologists b. three endocrinologists and five nephrologists 66. A support group for hodophobics (an irrational fear of travel) has 32 members. There are 15 aviophobics (fear of air travel), eight siderodrophobics (fear of train travel), and nine thalassophobics (fear of ocean travel) in the group. Twelve of them will be randomly selected to participate in a new therapy. What is the probability the group of 12 selected consists of exactly a. two aviophobics, six siderodrophobics, and four thalassophobics b. five thalassophobics, four aviophobics, and three siderodrophobics 67. A trained chimpanzee is given a box containing eight wooden cubes with the letters p, a, r, a, l, l, e, l printed on them (one letter per block). Assuming the chimp can’t read or spell, what is the probability he draws the eight blocks in order and actually forms the word “parallel”? 68. A number is called a “perfect number” if the sum of its proper factors is equal to the number itself. Six is the first perfect number since the sum of its proper factors is six: 1 ⫹ 2 ⫹ 3 ⫽ 6. Twenty-eight is the second since: 1 ⫹ 2 ⫹ 4 ⫹ 7 ⫹ 14 ⫽ 28. A young child is given a box containing eight wooden blocks with the following numbers (one per block) printed on them: four 3’s, two 5’s, one 0, and one 6. What is the probability she draws the eight blocks in order and forms the fifth perfect number: 33,550,336?

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Section 11.7 The Binomial Theorem


69. The function f 1x2 ⫽ 1 12 2 x gives the probability that x number of flips will all result in heads (or tails). Compute the probability that 20 flips results in 20 heads in a row, then use the Internet or some other resource to find the probability of winning a state lottery. Which is more likely to happen (which has the greater probability)? Were you surprised? 70. Recall that a function is a relation in which each element of the domain is paired with only one element from the range. Is the relation defined by C1x2 ⫽ nCx (n is a constant) a function? To 䊳

1145

investigate, plot the points generated by C1x2 ⫽ 8Cx for x ⫽ 0 to x ⫽ 8 using a “friendly” window 1x 僆 30, 9.4 4 , y 僆 3 0, 93 4 2 and answer the following questions: a. Is the resulting graph continuous or discrete (made up of distinct points)? b. Does the resulting graph pass the vertical line test? c. Discuss the features of the relation and its graph, including the domain, range, maximum or minimum values, and symmetries observed.


71. (6.2) Given csc ␪ ⫽ 3 and cos ␪ 6 0, find the values of the remaining five trig functions of ␪. 72. (5.4) Complete the following logarithmic properties: logb b ⫽ __

logb 1 ⫽ __

logb b ⫽ __

blogbn ⫽ __

n

11.7


A. Use Pascal’s triangle to

C. D. E.

74. (11.3) A rubber ball is dropped from a height of 25 ft onto a hard surface. With each bounce, it rebounds 60% of the height from which it last fell. Use sequences/series to find (a) the height of the sixth bounce, (b) the total distance traveled up to the sixth bounce, and (c) the distance the ball will travel before coming to rest.

The Binomial Theorem

LEARNING OBJECTIVES

B.

73. (7.4) Find exact values for sin12␪2, cos12␪2, and 21 tan12␪2 given cos ␪ ⫽ ⫺ and ␪ is in Quadrant II. 29

find 1a ⫹ b2 n Find binomial coefficients n using a b notation k Use the binomial theorem to find 1a ⫹ b2 n Find a specific term of a binomial expansion Solve applications of binomial powers

Strictly speaking, a binomial is a polynomial with two terms. This limits us to terms with real number coefficients and whole number powers on variables. In this section, we will loosely regard a binomial as the sum or difference of any two terms. Hence 1 13 1 3x2 ⫺ y4, 1x ⫹ 4, x ⫹ , and ⫺ ⫹ i are all “binomials.” Our goal is to develop x 2 2 an ability to raise a binomial to any natural number power, with the results having important applications in genetics, probability, polynomial theory, and other areas. The tool used for this purpose is called the binomial theorem.

A. Binomial Powers and Pascal’s Triangle Much of our mathematical understanding comes from a study of patterns. One area where the study of patterns has been particularly fruitful is Pascal’s triangle (Figure 11.67), named after the French scientist Blaise Pascal (although the triangle was well known before his time). It begins with a “1” at the vertex of the triangle, with 1’s extending diagonally downward to the left and right as shown. The entries on the interior of the triangle are found by adding the two entries directly above and to the left and right of each new position.

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Figure 11.67 1

First row

1 1 1 1 1

?

Third row

1

2 3

?

Second row

1

6

Fourth row

1

3

1

?

? ? and so on

1

?

There are a variety of patterns hidden within the triangle. In this section, we’ll use the horizontal rows of the triangle to help us raise a binomial to various powers. To begin, recall that 1a ⫹ b2 0 ⫽ 1 and 1a ⫹ b2 1 ⫽ 1a ⫹ 1b (unit coefficients are included for emphasis). In our earlier work, we saw that a binomial square (a binomial raised to the second power) always followed the pattern 1a ⫹ b2 2 ⫽ 1a2 ⫹ 2ab ⫹ 1b2. Observe the overall pattern that is developing as we include 1a ⫹ b2 3: 1a ⫹ b2 0 1a ⫹ b2 1 1a ⫹ b2 2 1a ⫹ b2 3

1 1a ⫹ 1b 1a2 ⫹ 2ab ⫹ 1b2 1a3 ⫹ 3a2b ⫹ 3ab2 ⫹ 1b3

row 1 row 2 row 3 row 4

Apparently the coefficients of 1a ⫹ b2 will occur in row n ⫹ 1 of Pascal’s triangle. Also observe that in each term of the expansion, the exponent of the first term a decreases by 1 as the exponent on the second term b increases by 1, keeping the degree of each term constant (recall the degree of a term with more than one variable is the sum of the exponents). n

1a3b0 ⫹ 3a2b1 ⫹ 3a1b2 ⫹ 1a0b3 3⫹0 degree 3

2⫹1 degree 3

1⫹2 degree 3

0⫹3 degree 3

These observations help us to quickly expand a binomial power. EXAMPLE 1

䊳

Expanding a Binomial Using Pascal’s Triangle

Solution

䊳

Working step-by-step we have

Use Pascal’s triangle and the patterns noted to expand 1x ⫹ 12 2 4. 1. The coefficients will be in the fifth row of Pascal’s triangle. 1 4 6 4 1 2. The exponents on x begin at 4 and decrease, while the exponents on 12 begin at 0 and increase. 1 0 1 1 1 2 1 3 1 4 1x4a b ⫹ 4x3a b ⫹ 6x2 a b ⫹ 4x1 a b ⫹ 1x0 a b 2 2 2 2 2 3. Simplify each term. 3 1 1 The result is x4 ⫹ 2x3 ⫹ x2 ⫹ x ⫹ . 2 2 16 Now try Exercises 7 through 10

䊳

If the exercise involves a difference rather than a sum, we simply rewrite the expression using algebraic addition and proceed as before.

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EXAMPLE 2

䊳

Raising a Complex Number to a Power Using Pascal’s Triangle

Solution

䊳

Begin by rewriting 13 ⫺ 2i2 5 as 33 ⫹ 1⫺2i2 4 5.

1147

Use Pascal’s triangle and the patterns noted to compute 13 ⫺ 2i2 5.

1. The coefficients will be in the sixth row of Pascal’s triangle. 5 10 10 5 1 1 2. The exponents on 3 begin at 5 and decrease, while the exponents on 1⫺2i2 begin at 0 and increase. 1135 2 1⫺2i2 0 ⫹ 5134 21⫺2i2 1 ⫹ 10133 21⫺2i2 2 ⫹ 10132 21⫺2i2 3 ⫹ 5131 21⫺2i2 4 ⫹ 1130 21⫺2i2 5 3. Simplify each term. 243 ⫺ 810i ⫺ 1080 ⫹ 720i ⫹ 240 ⫺ 32i The result is ⫺597 ⫺ 122i. Now try Exercises 11 and 12

䊳

Expanding Binomial Powers 1a ⴙ b2 n

1. The coefficients will be in row n ⫹ 1 of Pascal’s triangle. 2. The exponents on the first term begin at n and decrease, while the exponents on the second term begin at 0 and increase. 3. For any binomial difference 1a ⫺ b2 n, rewrite the base as 3a ⫹ 1⫺b2 4 n using algebraic addition and proceed as before, then simplify each term.

A. You’ve just seen how we can use Pascal’s triangle to find 1a ⴙ b2 n

B. Binomial Coefficients and Factorials

Pascal’s triangle can easily be used to find the coefficients of 1a ⫹ b2 n, as long as the exponent is relatively small. If we needed to expand 1a ⫹ b2 25, writing out the first 26 rows of the triangle would be rather tedious. To overcome this limitation, we introduce a formula that enables us to find the coefficients of any binomial expansion. The Binomial Coefficients

n For natural numbers n and r where n ⱖ r, the expression a b, read “n choose r,” r is called the binomial coefficient and evaluated as: n n! a b⫽ r r!1n ⫺ r2!

Notice the formula for determining binomial coefficients is identical to that for and it turns out the coefficients are actually found using a combination, with the new notation used primarily as a convenience. In Example 1, we found the coefficients of 1a ⫹ b2 4 using the fifth or 1n ⫹ 12st row of Pascal’s triangle. In Example 3, these coefficients are found using the formula for binomial coefficients. nCr,

EXAMPLE 3

䊳

Computing Binomial Coefficients n n! Evaluate a b ⫽ as indicated: r r!1n ⫺ r2! 4 a. a b 1

4 b. a b 2

4 c. a b 3

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Solution

䊳

4 # 3! 4 4! ⫽ a. a b ⫽ ⫽4 1 1!3! 1!14 ⫺ 12! 4 # 3 # 2! 4 4! 4#3 ⫽ ⫽ ⫽6 b. a b ⫽ 2 2!2! 2 2!14 ⫺ 22! 4 4! 4 # 3! ⫽ ⫽4 c. a b ⫽ 3 3!1! 3!14 ⫺ 32! Now try Exercises 13 through 20

䊳

4 4 4 Note a b ⫽ 4, a b ⫽ 6, and a b ⫽ 4 give the interior entries in the fifth row of 1 2 3 Pascal’s triangle: 1 4 6 4 1. For consistency and symmetry, we define 0! ⫽ 1, which enables the formula to generate all entries of the triangle, including the “1’s.” 4! 4 4! 4 4! ⫽ apply formula a b ⫽ apply formula a b⫽ 4 0 4! # 0! 0!14 ⫺ 02! 4!14 ⫺ 42! 4! 4! ⫽ 0! ⫽ 1 ⫽ 0! ⫽ 1 ⫽1 ⫽1 # 1 4! 4! # 1 n The formula for a b with 0 ⱕ r ⱕ n now gives all coefficients in the 1n ⫹ 12st r row. For n ⫽ 5, we have 5 a b 0 1 EXAMPLE 4

䊳

5 a b 1 5

5 a b 2 10

5 a b 3 10

5 a b 4 5

5 a b 5 1

Computing Binomial Coefficients Compute the binomial coefficients: 9 a. a b 0

Solution

䊳

9 b. a b 1

9 9! a. a b ⫽ 0 0!19 ⫺ 02! 9! ⫽ ⫽1 9! 6 6! c. a b ⫽ 5 5!16 ⫺ 52! 6! ⫽ ⫽6 5!

6 c. a b 5

6 d. a b 6 9 9! b. a b ⫽ 1 1!19 ⫺ 12! 9! ⫽ ⫽9 8! 6 6! d. a b ⫽ 6 6!16 ⫺ 62! 6! ⫽ ⫽1 6! Now try Exercises 21 through 24

B. You’ve just seen how we can find binomial coefficients n using a b notation k

䊳

n As mentioned, the formulas for a b and nCr yield like results for given values of n r and r. For future use, it will help to commit the general results from Example 4 to n n n n memory: a b ⫽ 1, a b ⫽ n, a b ⫽ n, and a b ⫽ 1. 0 1 n⫺1 n

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1149

C. The Binomial Theorem n Using a b notation and the observations made regarding binomial powers, we can now r state the binomial theorem. Binomial Theorem

For any binomial 1a ⫹ b2 and natural number n, n n n 1a ⫹ b2 n ⫽ a b anb0 ⫹ a b an⫺1b1 ⫹ a b an⫺2b2 ⫹ p 0 1 2 n n ⫹a b a1bn⫺1 ⫹ a b a0bn n⫺1 n The theorem can also be stated in summation form as 1a ⫹ b2 n ⫽

n

兺 arba n

n⫺r r

b

r⫽0

The expansion actually looks overly impressive in this form, and it helps to summarize the process in words, as we did earlier. The exponents on the first term a begin at n and decrease, while the exponents on the second term b begin at 0 and increase, n keeping the degree of each term constant. The a b notation simply gives the coefficients r n of each term. As a final note, observe that the r in a b gives the exponent on b. r EXAMPLE 5

䊳

Solution

䊳

Expanding a Binomial Using the Binomial Theorem Expand 1a ⫹ b2 6 using the binomial theorem.

6 6 6 6 6 6 6 1a ⫹ b2 6 ⫽ a b a6b0 ⫹ a b a5b1 ⫹ a b a4b2 ⫹ a b a3b3 ⫹ a b a2b4 ⫹ a b a1b5 ⫹ a b a0b6 0 1 2 3 4 5 6 6! 6 6! 5 1 6! 4 2 6! 3 3 6! 2 4 6! 1 5 6! 6 ⫽ a ⫹ ab ⫹ ab ⫹ ab ⫹ ab ⫹ ab ⫹ b 0!6! 1!5! 2!4! 3!3! 4!2! 5!1! 6!0! ⫽ 1a6 ⫹ 6a5b ⫹ 15a4b2 ⫹ 20a3b3 ⫹ 15a2b4 ⫹ 6ab5 ⫹ 1b6 Now try Exercises 25 through 32

EXAMPLE 6

䊳

Using the Binomial Theorem to Find the Initial Terms of an Expansion

Solution

䊳

Use the binomial theorem with a ⫽ 2x, b ⫽ y2, and n ⫽ 10.

䊳

Find the first three terms of 12x ⫹ y2 2 10.

12x ⫹ y2 2 10 ⫽ a

C. You’ve just seen how we can use the binomial theorem to find 1a ⴙ b2 n

10 10 10 b12x2 10 1y2 2 0 ⫹ a b12x2 9 1y2 2 1 ⫹ a b12x2 8 1y2 2 2 ⫹ p 0 1 2 10! 256x8y4 ⫹ p ⫽ 1121024x10 ⫹ 1102512x9y2 ⫹ 2!8! ⫽ 1024x10 ⫹ 5120x9y2 ⫹ 1452256x8y4 ⫹ p ⫽ 1024x10 ⫹ 5120x9y2 ⫹ 11,520x8y4 ⫹ p

first three terms a

10 10 b ⫽ 1, a b ⫽ 10 0 1 10! ⫽ 45 2!8! result


䊳

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D. Finding a Specific Term of the Binomial Expansion In some applications of the binomial theorem, our main interest is a specific term of the expansion, rather than the expansion as a whole. To find a specified term, it helps to consider that the expansion of 1a ⫹ b2 n has n ⫹ 1 terms: 1a ⫹ b2 0 has one term, n 1a ⫹ b2 1 has two terms, 1a ⫹ b2 2 has three terms, and so on. Because the notation a b r always begins at r ⫽ 0 for the first term, the value of r will be 1 less than the term we are seeking. In other words, for the seventh term of 1a ⫹ b2 9, we use r ⫽ 6. The k th Term of a Binomial Expansion

For the binomial expansion 1a ⫹ b2 n, the kth term is given by n a b an⫺rbr, where r ⫽ k ⫺ 1. r

EXAMPLE 7

䊳

Solution

䊳

Finding a Specific Term of a Binomial Expansion Find the eighth term in the expansion of 1x ⫹ 2y2 12.

By comparing 1x ⫹ 2y2 12 to 1a ⫹ b2 n we have a ⫽ x, b ⫽ 2y, and n ⫽ 12. Since we want the eighth term, k ⫽ 8 and r ⫽ 7. The eighth term of the expansion is a

D. You’ve just seen how we can find a specific term of a binomial expansion

12 5 12! b x 12y2 7 ⫽ 128x5y7 7 7!5! ⫽ 179221128x5y7 2 ⫽ 101,376x5y7

27 ⫽ 128 1 12 7 2 ⫽ 792 result


䊳

E. Applications One application of the binomial theorem involves a binomial experiment and binomial probability. For binomial probabilities, the following must be true: (1) The experiment must have only two possible outcomes, typically called success and failure, and (2) if the experiment has n trials, the probability of success must be constant for all n n trials. If the probability of success for each trial is p, the formula a b11 ⴚ p2 nⴚkpk k gives the probability that exactly k trials will be successful. Binomial Probability Given a binomial experiment with n trials, where the probability for success in each trial is p. The probability that exactly k trials are successful is given by n a b11 ⫺ p2 n⫺k pk. k

EXAMPLE 8

䊳

Applying the Binomial Theorem — Binomial Probability Paula Rodrigues has a free-throw shooting average of 85%. On the last play of the game, with her team behind by three points, she is fouled at the three-point line, and is awarded two additional free throws via technical fouls on the opposing coach (a total of five free-throws). What is the probability she makes at least three (meaning they at least tie the game)?

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Solution

䊳

1151

Here we have p ⫽ 0.85, 1 ⫺ p ⫽ 0.15, and n ⫽ 5. The key idea is to recognize the phrase at least three means “3 or 4 or 5.” So P(at least 3) ⫽ P13 ´ 4 ´ 52. “or” implies a union P1at least 32 ⫽ P13 ´ 4 ´ 52 ⫽ P132 ⫹ P142 ⫹ P152 sum of probabilities (mutually exclusive events) 5 5 5 ⫽ a b 10.152 2 10.852 3 ⫹ a b 10.152 1 10.852 4 ⫹ a b 10.152 0 10.852 5 4 5 3 ⬇ 0.1382 ⫹ 0.3915 ⫹ 0.4437 ⫽ 0.9734

Paula’s team has an excellent chance 1⬇97.3% 2 of at least tying the game.


䊳

As you can see, calculations involving binomial probabilities can become quite extensive. Here again, a conceptual understanding of what the numbers mean can be combined with the use of technology to solve significant applications of the idea. Most graphing calculators provide a binomial probability distribution function, abbreviated “binompdf(” and accessed using 2nd VARS (DISTR) 0:binompdf(. The function requires three inputs: the number of trials n, the probability of success p for each trial, and the value of k. As with the evaluation of other functions, k can be a single value or a list of values enclosed in braces: “{ }.” The resulting calculation for Example 8 is shown in Figure 11.68, and verifies each of the individual probabilities, although we must use the right arrow to see them all. To find the sum of these probabilities, we simply precede the “binompdf(” command with the “sum(” feature used previously. The final result is shown in Figure 11.69, and verifies our earlier calculation. See Exercises 47 and 48. Figure 11.68

Figure 11.69

E. You’ve just seen how we can solve applications of binomial powers

11.7 EXERCISES 䊳



1. In any binomial expansion, there is always more term than the power applied.

3. To expand a binomial difference such as 1a ⫺ 2b2 5, we rewrite the binomial as and proceed as before. 5. Discuss why the expansion of 1a ⫹ b2 n has n ⫹ 1 terms.

2. In all terms in the expanded form of 1a ⫹ b2 n, the exponents on a and b must sum to . 4. In a binomial experiment with n trials, the probability there are exactly k successes is given by the formula . 6. For any defined binomial experiment, discuss the relationships between the phrases, “exactly k success,” and “at least k successes.”

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Use Pascal’s triangle and the patterns explored to write each expansion.

7. 1x ⫹ y2 5

10. 1x2 ⫹ 13 2 3

8. 1a ⫹ b2 6

11. 11 ⫺ 2i2 5

Evaluate each of the following

7 13. a b 4 9 16. a b 5 40 19. a b 3 5 22. a b 0

䊳

8 14. a b 2 20 17. a b 17 45 20. a b 3 15 23. a b 15

9. 12x ⫹ 32 4

12. 12 ⫺ 5i2 4 5 15. a b 3 30 18. a b 26 6 21. a b 0 10 24. a b 10

Use the binomial theorem to expand each expression. Write the general form first, then simplify.

25. 1c ⫹ d2 5

26. 1v ⫹ w2 4

31. 11 ⫺ 2i2 3

32. 12 ⫹ i 132 5

28. 1x ⫺ y2 7

29. 12x ⫺ 32 4

27. 1a ⫺ b2 6

30. 1a ⫺ 2b2 5

Use the binomial theorem to write the first three terms.

33. 1x ⫹ 2y2 9

36. 1 12a ⫺ b2 2 10

34. 13p ⫹ q2 8

35. 1v2 ⫺ 12w2 12

Find the indicated term for each binomial expansion.

37. 1x ⫹ y2 7; 4th term

38. 1m ⫹ n2 6; 5th term

41. 12x ⫹ y2 12; 11th term

42. 13n ⫹ m2 9; 6th term

39. 1p ⫺ 22 8; 7th term

40. 1a ⫺ 32 14; 10th term


n 1 k 1 nⴚk 43. Binomial probability: P1k2 ⴝ a b a b a b k 2 2 The theoretical probability of getting exactly k heads in n flips of a fair coin is given by the formula above. What is the probability that you would get exactly 5 heads in 10 flips of the coin?

䊳

11–76


n 1 k 4 nⴚk 44. Binomial probability: P1k2 ⴝ a b a b a b k 5 5 A multiple choice test has five options per question. The probability of guessing correctly k times out of n questions is found using the formula shown. What is the probability a person scores a 70% by guessing randomly (7 out of 10 questions correct)?

APPLICATIONS

45. Batting averages: Tony Gwynn (San Diego Padres) had a lifetime batting average of 0.347, ranking him as one of the greatest hitters of all time. Suppose he came to bat five times in any given game. a. What is the probability that he will get exactly three hits? b. What is the probability that he will get at least three hits?

47. Late rental returns: The manager of Victor’s DVD Rentals knows that 6% of all DVDs rented are returned late. Of the eight videos rented in the last hour, what is the probability that a. exactly five are returned on time b. exactly six are returned on time c. at least six are returned on time d. none of them will be returned late

46. Pollution testing: Erin suspects that a nearby iron smelter is contaminating the drinking water over a large area. A statistical study reveals that 83% of the wells in this area are likely contaminated. If the figure is accurate, find the probability that if another 10 wells are tested a. exactly 8 are contaminated b. at least 8 are contaminated

48. Opinion polls: From past experience, a research firm knows that 20% of telephone respondents will agree to answer an opinion poll. If 20 people are contacted by phone, what is the probability that a. exactly 18 refuse to be polled b. exactly 19 refuse to be polled c. at least 18 refuse to be polled d. none of them agree to be polled

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Making Connections


49. If you sum the entries in each row of Pascal’s triangle, a pattern emerges. Find a formula that generalizes the result for any row of the triangle, and use it to find the sum of the entries in the 12th row of the triangle. 䊳

1153

50. The derived polynomial of f (x) is f 1x ⫹ h2 or the original polynomial evaluated at x ⫹ h. Use Pascal’s triangle or the binomial theorem to find the derived polynomial for f 1x2 ⫽ x3 ⫹ 3x2 ⫹ 5x ⫺ 11. Simplify the result completely.


51. (2.5) Graph the function shown and find f (3): f 1x2 ⫽ e

x⫹2 xⱕ2 1x ⫺ 42 2 x 7 2

52. (6.2) Given the point (⫺0.6, y) is a point on the unit circle in the third quadrant, find y.

53. (4.3/4.6) Graph the function g1x2 ⫽ x3 ⫺ x2 ⫺ 6x. Clearly indicate all intercepts and intervals where g1x2 7 0. 54. (7.5) Evaluate arcsin c sina

5␲ bd. 6

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight situations are described in (a) through (h) below. Match the characteristics, formulas, operations, or results indicated in 1 through 16 to one of the eight situations. (a)

(b) ⫺2 ⫹ 0.5 ⫹ 3 ⫹ 5.5 ⫹ 8 ⫹ 10.5 ⫹ p ⫹ 33

7

3i⫺1 i⫽1 18

兺

10

(c)

(d)

5.5

0

⫺6

(e)

16a4 ⫺ 32a3b ⫹ 24a2b2 ⫺ 8ab3 ⫹ b4

(f) ⫺29, ⫺23, ⫺17, ⫺11, p

(g)

1, 1, 2, 3, 5, 8, 13, …

(h) a4 ⫹ 8a3b ⫹ 24a2b2 ⫹ 32ab3 ⫹ 16b4

1. ____ alternating sequence

9. ____ a22 ⫽ ⫺11.8

2. ____ Fibonacci sequence

10. ____ geometric series

3. ____ 232.5

11. ____ r ⫽ 3

4. ____ d ⫽ ⫺0.7

12. ____ Sq ⫽

5. ____ an ⫽ 3.6 ⫺ 0.7n

13. ____ 12a ⫺ b2 4

6. ____ S39 ⫽ 3315

16 3

14. ____ an ⫽ 6n ⫺ 35

7. ____ arithmetic series

15. ____ recursively defined

8. ____ 1a ⫹ 2b2 4

16. ____

1093 18

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Sequences and Series

KEY CONCEPTS • A finite sequence is a function an whose domain is the set of natural numbers from 1 to n. • The terms of the sequence are labeled a1, a2, a3, p , ak⫺1, ak, ak⫹1, p , an⫺2, an⫺1, an. • The expression an, which defines the sequence (generates the terms in order), is called the nth term. • An infinite sequence is a function whose domain is the set of natural numbers. • When each term of a sequence is larger than the preceding term, it is called an increasing sequence. • When each term of a sequence is smaller than the preceding term, it is called a decreasing sequence. • When successive terms of a sequence alternate in sign, it is called an alternating sequence. • When the terms of a sequence are generated using previous term(s), it is called a recursive sequence. • Sequences are sometimes defined using factorials, which are the product of a given natural number with all natural numbers that precede it: n! ⫽ n # 1n ⫺ 12 # 1n ⫺ 22 # p # 3 # 2 # 1. • Given the sequence a1, a2, a3, a4, p , an the sum is called a finite series and is denoted Sn. • Sn ⫽ a1 ⫹ a2 ⫹ a3 ⫹ a4 ⫹ p ⫹ an. The sum of the first n terms is called a partial sum. k

• In sigma notation, the expression

兺a ⫽ a i

1

⫹ a2 ⫹ p ⫹ ak represents a finite series,

i⫽1

and the letter “i ” is called the index of summation.

EXERCISES Write the first four terms that are defined and the value of a10. 1. an ⫽ 5n ⫺ 4

2. an ⫽

n⫹1 n2 ⫹ 1

Find the general term an for each sequence, and the value of a6. 3. 1, 16, 81, 256, p 4. ⫺17, ⫺14, ⫺11, ⫺8, p Find the eighth partial sum (S8). 5. 12, 14, 18, p

6. ⫺21, ⫺19, ⫺17, p

Evaluate each sum. 7

7.

兺

5

n2

8.

n⫽1

兺 13n ⫺ 22

n⫽1

Write the first five terms that are defined. n! 9. an ⫽ 1n ⫺ 22!

10. e

a1 ⫽ 12 an⫹1 ⫽ 2an ⫺ 14

Write as a single summation and evaluate. 7

11.

7

兺i

2

i⫽1

⫹

兺 13i ⫺ 22

i⫽1

12. A large wildlife preserve brings in 40 rare hawks (male and female) in an effort to repopulate the species. Each year they are able to add an average of 10 additional hawks in cooperation with other wildlife areas. If the population of hawks grows at a rate of 12% through natural reproduction, the number of hawks in the preserve after x yr is given by the recursive sequence h0 ⫽ 40, hn ⫽ 1.12hn⫺1 ⫹ 10. (a) How many hawks are in the wildlife preserve after 5 yr? (b) How many years before the number of hawks exceeds 200?

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SECTION 11.2


1155


KEY CONCEPTS • In an arithmetic sequence, successive terms are found by adding a fixed constant to the preceding term. • In a sequence, if there exists a number d, called the common difference, such that ak⫹1 ⫺ ak ⫽ d, then the sequence is arithmetic. Alternatively, ak⫹1 ⫽ ak ⫹ d for k ⱖ 1. • The nth term n of an arithmetic sequence is given by an ⫽ a1 ⫹ 1n ⫺ 12d, where a1 is the first term and d is the common difference. If • the initial term is unknown or is not a1 the nth term can be written an ⫽ ak ⫹ 1n ⫺ k2d, where the subscript of the term ak and the coefficient of d sum to n. • For an arithmetic sequence with first term a1, the nth partial sum (the sum of the first n terms) is given by n1a1 ⫹ an 2 Sn ⫽ . 2 EXERCISES Find the general term (an) for each arithmetic sequence. Then find the indicated term. 13. 2, 5, 8, 11, p ; find a40 14. 3, 1, ⫺1, ⫺3, p ; find a35 Find the sum of each series. 15. ⫺1 ⫹ 3 ⫹ 7 ⫹ 11 ⫹ p ⫹ 75 17. 3 ⫹ 6 ⫹ 9 ⫹ 12 ⫹ p ; S20

16. 1 ⫹ 4 ⫹ 7 ⫹ 10 ⫹ p ⫹ 88 18. 1 ⫹ 3 ⫹ 1 ⫹ 1 ⫹ p ; S15 4

2

4

25

19.

兺 13n ⫺ 42

n⫽1

20. From a point just behind the cockpit, the width of a modern fighter plane’s swept-back wings is 1.25 m. The width of the wings, measured in equal increments, increases according to the pattern 1.25, 2.15, 3.05, 3.95, p . Find the width of the wings on the eighth measurement.

SECTION 11.3

Geometric Sequences

KEY CONCEPTS • In a geometric sequence, successive terms are found by multiplying the preceding term by a nonzero constant. ak⫹1 • In other words, if there exists a number r, called the common ratio, such that a ⫽ r, then the sequence is k ⫽ a r for k ⱖ 1. geometric. Alternatively, we can write a k⫹1

k

• The nth term an of a geometric sequence is given by an ⫽ a1rn⫺1, where a1 is the first term and an represents the general term of a finite sequence. • If the initial term is unknown or is not a1, the nth term can be written an ⫽ akrn⫺k, where the subscript of the term ak and the exponent on r sum to n. a1 11 ⫺ rn 2 . • The nth partial sum of a geometric sequence is Sn ⫽ 1⫺r a1 . • If 冟r 冟 6 1, the sum of an infinite geometric series is Sq ⫽ 1⫺r

EXERCISES Find the indicated term for each geometric sequence. 21. a1 ⫽ 5, r ⫽ 3; find a7 22. a1 ⫽ 4, r ⫽ 12; find a7

23. a1 ⫽ 17, r ⫽ 17; find a8

Find the indicated sum, if it exists. 24. 16 ⫺ 8 ⫹ 4 ⫺ p ; find S7 27. 4 ⫹ 8 ⫹ 16 ⫹ 32 ⫹ p

1 ⫹ 25 ⫹ 15 ⫹ 10 ⫹ p ; find S12 3 3 29. 6 ⫺ 3 ⫹ 2 ⫺ 4 ⫹ p

25. 2 ⫹ 6 ⫹ 18 ⫹ p ; find S8 28. 5 ⫹ 0.5 ⫹ 0.05 ⫹ 0.005 ⫹ p

26.

4 5

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30.

11–80


2 k 5a b 3 k⫽1

兺

4 k 12a b 3 k⫽1 q

31.

兺

1 k 5a b 2 k⫽1 q

32.

兺

33. Sumpter reservoir contains 121,500 ft3 of water and is being drained in the following way. Each day one-third of the water is drained (and not replaced). Use a sequence/series to compute how much water remains in the pond after 7 days. 34. Credit-hours taught at Cody Community College have been increasing at 7% per year since it opened in 2001 and taught 1225 credit-hours. For the new faculty, the college needs to predict the number of credit-hours that will be taught in 2015. Use a sequence/series to compute the credit-hours for 2015 and to find the total number of credit hours taught through the 2015 school year.

SECTION 11.4

Mathematical Induction

KEY CONCEPTS • Functions written in subscript notation can be evaluated, graphed, and composed with other functions. • A sum formula involving only natural numbers n as inputs can be proven valid using a proof by induction. Given that Sn represents a sum formula involving natural numbers, if (1) S1 is true and (2) Sk ⫹ ak⫹1 ⫽ Sk⫹1, then Sn must be true for all natural numbers. • Proof by induction can also be used to validate other relationships, using a more general statement of the principle. Let Pn be a statement involving the natural numbers n. If (1) P1 is true (Pn for n ⫽ 12 and (2) the truth of Pk implies that Pk⫹1 is also true, then Pn must be true for all natural numbers n. EXERCISES Use the principle of mathematical induction to prove the indicated sum formula is true for all natural numbers n. 35. 1 ⫹ 2 ⫹ 3 ⫹ 4 ⫹ 5 ⫹ p ⫹ n; 36. 1 ⫹ 4 ⫹ 9 ⫹ 16 ⫹ 25 ⫹ 36 ⫹ p ⫹ n2; n1n ⫹ 12 n1n ⫹ 1212n ⫹ 12 an ⫽ n and Sn ⫽ . an ⫽ n2 and Sn ⫽ . 2 6 Use the principle of mathematical induction to prove that each statement is true for all natural numbers n. 37. 4n ⱖ 3n ⫹ 1 38. 6 # 7n⫺1 ⱕ 7n ⫺ 1 39. 3n ⫺ 1 is divisible by 2

SECTION 11.5

Counting Techniques

KEY CONCEPTS • An experiment is any task that can be repeated and has a well-defined set of possible outcomes. • Each repetition of an experiment is called a trial. • Any potential outcome of an experiment is called a sample outcome. • The set of all sample outcomes is called the sample space. • An experiment with N (equally likely) sample outcomes that is repeated t times, has a sample space with N t elements. • If a sample outcome can be used more than once, the counting is said to be with repetition. If a sample outcome can be used only once, the counting is said to be without repetition. • The fundamental principle of counting states: If there are p possibilities for a first task, q possibilities for the second, and r possibilities for the third, the total number of ways the experiment can be completed is pqr. This fundamental principle can be extended to include any number of tasks. • If the elements of a sample space have precedence or priority (order or rank is important), the number of elements is counted using a permutation, denoted nPr and read, “the distinguishable permutations of n objects taken r at a time.” n! . • To expand nPr, we can write out the first r factors of n! or use the formula nPr ⫽ 1n ⫺ r2!

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1157

• If any of the sample outcomes are identical, certain permutations will be nondistinguishable. In a set containing n

elements where one element is repeated p times, another is repeated q times, and another r times 1p ⫹ q ⫹ r ⫽ n2, n! nPn the number of distinguishable permutations is given by ⫽ . p!q!r! p!q!r! • If the elements of a set have no rank, order, or precedence (as in a committee of colleagues) permutations with the n! . same elements are considered identical. The result is the number of combinations, nCr ⫽ r!1n ⫺ r2!

EXERCISES 40. Three slips of paper with the letters A, B, and C are placed in a box and randomly drawn one at a time. Show all possible ways they can be drawn using a tree diagram. 41. The combination for a certain bicycle lock consists of three digits. How many combinations are possible if (a) repetition of digits is not allowed and (b) repetition of digits is allowed. 42. Jethro has three work shirts, four pairs of work pants, and two pairs of work shoes. How many different ways can he dress himself (shirt, pants, shoes) for a day’s work? 43. From a field of 12 contestants in a pet show, three cats are chosen at random to be photographed for a publicity poster. In how many different ways can the cats be chosen? 44. Compute the following values by hand, showing all work: c. 7C4 a. 7! b. 7P4 45. Six horses are competing in a race at the McClintock Ranch. Assuming there are no ties, (a) how many different ways can the horses finish the race? (b) How many different ways can the horses finish first, second, and third place? (c) How many finishes are possible if it is well known that Nellie-the-Nag will finish last and Sea Biscuit will finish first? 46. How many distinguishable permutations can be formed from the letters in the word “tomorrow”? 47. Quality Construction Company has 12 equally talented employees. (a) How many ways can a three-member crew be formed to complete a small job? (b) If the company is in need of a Foreman, Assistant Foreman, and Crew Chief, in how many ways can the positions be filled?

SECTION 11.6

Introduction to Probability

KEY CONCEPTS • An event E is any designated set of sample outcomes. • Given S is a sample space of equally likely sample outcomes and E is an event relative to S, the probability of E, n1E2 , where n(E) represents the number of elements in E, and n(S) written P(E), is computed as P1E2 ⫽ n1S2 represents the number of elements in S. • The complement of an event E is the set of sample outcomes in S, but not in E and is denoted ⬃E. • Given sample space S and any event E defined relative to S: 112 P1⬃S2 ⫽ 0, 122 0 ⱕ P1E2 ⱕ 1, 132 P1S2 ⫽ 1, 142 P1E2 ⫽ 1 ⫺ P1⬃E2, and 152 P1E2 ⫹ P1⬃E2 ⫽ 1. • Two events that have no outcomes in common are said to be mutually exclusive. • If two events are not mutually exclusive, P1E1 or E2 2 S P1E1 ´ E2 2 ⫽ P1E1 2 ⫹ P1E2 2 ⫺ P1E1 ¨ E2 2. • If two events are mutually exclusive, P1E1 or E2 2 S P1E1 ´ E2 2 ⫽ P1E1 2 ⫹ P1E2 2 . EXERCISES 48. One card is drawn from a standard deck. What is the probability the card is a ten or a face card? 49. One card is drawn from a standard deck. What is the probability the card is a Queen or a face card? 50. One die is rolled. What is the probability the result is not a three? 51. Given P1E1 2 ⫽ 38, P1E2 2 ⫽ 34, and P1E1 ´ E2 2 ⫽ 56, compute P1E1 ¨ E2 2.

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52. Find P(E) given that n1E2 ⫽ 7C4

# 5C3 and n1S2 ⫽ 12C7.

53. To determine if more physicians should be hired, a medical clinic tracks the number of days between a patient’s request for an appointment and the actual appointment date. The table given shows the probability that a patient must wait d days. Based on the table, what is the probability a patient must wait a. at least 20 days c. 40 days or less e. less than 40 and more than 10 days

SECTION 11.7

b. less than 20 days d. over 40 days f. 30 or more days

Wait (days d)

Probability

0

0.002

0 6 d 6 10

0.07

10 ⱕ d 6 20

0.32

20 ⱕ d 6 30

0.43

30 ⱕ d 6 40

0.178

The Binomial Theorem

KEY CONCEPTS • To expand 1a ⫹ b2 n for n of “moderate size,” we can use Pascal’s triangle and observed patterns. n • For any natural numbers n and r, where n ⱖ r, the expression a b (read “n choose r”) is called the binomial r n n! . coefficient and evaluated as a b ⫽ r r!1n ⫺ r2! If n is large, it is more efficient to expand using the binomial coefficients and binomial theorem. • • The following binomial coefficients are useful/common and should be committed to memory: n n n n a b⫽n a b⫽n a b⫽1 a b⫽1 1 n⫺1 n 0 1 1 n n! ⫽ ⫽ ⫽ 1. • We define 0! ⫽ 1; for example a b ⫽ 0! 1 n n!1n ⫺ n2! n n n n n b a1bn⫺1 ⫹ a b a0bn. • The binomial theorem: 1a ⫹ b2 n ⫽ a b anb0 ⫹ a b an⫺1b1 ⫹ a b an⫺2b2 ⫹ p ⫹ a 0 1 2 n⫺1 n n • The kth term of 1a ⫹ b2 n can be found using the formula a b an⫺rbr, where r ⫽ k ⫺ 1. r EXERCISES 54. Evaluate each of the following: 7 8 a. a b b. a b 5 3

55. Use Pascal’s triangle to expand the expressions: a. 1x ⫺ y2 4

Use the binomial theorem to: 56. Write the first four terms of

b. 11 ⫹ 2i2 5

57. Find the indicated term of each expansion.

a. 1a ⫹ 132 b. 15a ⫹ 2b2 a. 1x ⫹ 2y2 7; fourth b. 12a ⫺ b2 14; 10th 58. Mark Leland is a professional bowler who is able to roll a strike (knocking down all 10 pins on the first ball) 91% of the time. (a) What is the probability he rolls at least four strikes in the first five frames? (b) What is the probability he rolls five strikes (and scares the competition)? 8

7

PRACTICE TEST 1. The general term of a sequence is given. Find the first four terms and the 8th term. 1n ⫹ 22! 2n a. an ⫽ b. an ⫽ n⫹3 n! c. an ⫽ e

a1 ⫽ 3 an⫹1 ⫽ 21an 2 2 ⫺ 1

2. Expand each series and evaluate. 6

a.

兺

k⫽2 5

c.

12k2 ⫺ 32

j⫽1

兺 1⫺12 a j ⫹ 1 b j

j⫽2 q

j

兺 1⫺22a 4 b 3

6

b. d.

兺 7a 2 b

k⫽1

1

k

j

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3. Identify the first term and the common difference or common ratio. Then find the general term an. a. 7, 4, 1, ⫺2, p b. ⫺8, ⫺6, ⫺4, ⫺2, p c. 4, ⫺8, 16, ⫺32, p

d. 10, 4, 85, 16 25 , p

4. Find the indicated value for each sequence. a. a1 ⫽ 4, d ⫽ 5; find a40 b. a1 ⫽ 2, an ⫽ ⫺22, d ⫽ ⫺3; find n c. a1 ⫽ 24, r ⫽ 12; find a6 d. a1 ⫽ ⫺2, an ⫽ 486, r ⫽ ⫺3; find n 5. Find the sum of each series. a. 7 ⫹ 10 ⫹ 13 ⫹ p ⫹ 100 37

b.

1159

Practice Test

兺

k⫽1

13k ⫹ 22

c. For 4 ⫺ 12 ⫹ 36 ⫺ 108 ⫹ p , find S7 d. 6 ⫹ 3 ⫹ 3 ⫹ 3 ⫹ p 2

4

6. Each swing of a pendulum (in one direction) is 95% of the previous one. If the first swing is 12 ft, (a) find the length of the seventh swing and (b) determine the distance traveled by the pendulum for the first seven swings. 7. A rare coin that cost $3000 appreciates in value 7% per year. Find the value after 12 yr. 8. A car that costs $50,000 decreases in value by 15% per year. Find the value of the car after 5 yr. 9. Use mathematical induction to verify that for 5n2 ⫺ n an ⫽ 5n ⫺ 3, the sum formula Sn ⫽ is true 2 for all natural numbers n.

Juliet (Shakespeare), four identical copies of Faustus (Marlowe), and four identical copies of The Faerie Queen (Spenser). If these books are to be arranged on a shelf, how many distinguishable permutations are possible? 16. A company specializes in marketing various cornucopia (traditionally a curved horn overflowing with fruit, vegetables, gourds, and ears of grain) for Thanksgiving table settings. The company has seven fruit, six vegetable, five gourd, and four grain varieties available. If two from each group (without repetition) are used to fill the horn, how many different cornucopia are possible? 17. Use Pascal’s triangle to expand/simplify: a. 1x ⫺ 2y2 4 b. 11 ⫹ i2 4 18. Use the binomial theorem to write the first three terms of (a) 1x ⫹ 122 10 and (b) 1a ⫺ 2b3 2 8. 19. Michael and Mitchell are attempting to make a nonstop, 100-mi trip on a tandem bicycle. The probability that Michael cannot continue pedaling for the entire trip is 0.02. The probability that Mitchell cannot continue pedaling for the entire trip is 0.018. The probability that neither one can pedal the entire trip is 0.011. What is the probability that they complete the trip? 20. The spinner shown is spun once. What is the probability of spinning a. a striped wedge b. a shaded wedge

11 10 9 8

c. a clear wedge

12 1 2 3 4 7 6

10. Use the principle of mathematical induction to verify that Pn: 2 # 3n⫺1 ⱕ 3n ⫺ 1 is true for all natural numbers n.

d. an even number

11. Three colored balls (aqua, brown, and creme) are to be drawn without replacement from a bag. List all possible ways they can be drawn using (a) a tree diagram and (b) an organized list.

g. a shaded wedge or a number greater than 12

12. Suppose that license plates for motorcycles must consist of three numbers followed by two letters. How many license plates are possible if zero and “Z” cannot be used and no repetition is allowed? 13. If one icon is randomly chosen from the following set, find the probability a mailbox is not chosen: {,,,,,}. 14. Compute the following values by hand, showing all work: (a) 6! (b) 6P3 (c) 6C3 15. An English major has built a collection of rare books that includes two identical copies of The Canterbury Tales (Chaucer), three identical copies of Romeo and

5

e. a two or an odd number f. a number greater than nine h. a shaded wedge and a number greater than 12 21. To improve customer service, a cable company tracks the number of days a customer must wait until their cable service is Wait (days d ) Probability installed. The table 0 0.02 shows the probability that a customer must 0 6 d 6 1 0.30 wait d days. Based on 1ⱕd 6 2 0.60 the table, what is the 2ⱕd 6 3 0.05 probability a customer 3ⱕd 6 4 0.03 waits a. at least 2 days b. less than 2 days c. 4 days or less

d. over 4 days

e. less than 2 or at least 3 days f. three or more days

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22. An experienced archer can hit the rectangular target shown 100% of the time at a range of 75 m. Assuming the probability the target is hit is 64 cm related to its area, what is the probability the archer hits within the a. triangle b. circle

in the right column. One out of the hundred will be selected at random for a personal interview. What is the probability the person chosen is a a. woman or a craftsman

48 cm

b. man or a contractor c. man and a technician d. journeyman or an apprentice 24. Cheddar is a 12-year-old male box turtle. Provolone is an 8-year-old female box turtle. The probability that Cheddar will live another 8 yr is 0.85. The probability that Provolone will live another 8 yr is 0.95. Find the probability that a. both turtles live for another 8 yr

c. circle but outside the triangle d. lower half-circle e. rectangle but outside the circle f. lower half-rectangle, outside the circle 23. A survey of 100 union workers was taken to register concerns to be raised at the next bargaining session. A breakdown of those surveyed is shown in the table Expertise Level

Women

Men

Total

Apprentice

16

18

34

Technician

15

13

28

Craftsman

9

9

18

Journeyman

7

6

13

Contractor

3

4

7

50

50

100

Totals

b. neither turtle lives for another 8 yr c. at least one of them will live another 8 yr 25. The quality control department at a lightbulb factory has determined that the company is losing money because their manufacturing process produces a defective bulb 12% of the time. If a random sample of 10 bulbs is tested, (a) what is the probability that none are defective? (b) What is the probability that no more than 3 bulbs are defective?

CALCULATOR EXPLORATION AND DISCOVERY Infinite Series, Finite Results Although there were many earlier flirtations with infinite processes, it may have been the paradoxes of Zeno of Elea (⬃450 B.C.) that crystallized certain questions that simultaneously frustrated and fascinated early mathematicians. The first paradox, called the dichotomy paradox, can be summarized by the following question: How can one ever finish a race, seeing that one-half the distance must first be traversed, then one-half the remaining distance, then one-half the distance that then remains, and so on an infinite number of times? Although we easily accept that races can be finished, the subtleties involved in this question stymied mathematicians for centuries and were not satisfactorily resolved until the 1 ⫹ p 6 1. This is a geometric series eighteenth century. In modern notation, Zeno’s first paradox says 12 ⫹ 14 ⫹ 18 ⫹ 16 1 1 with a1 ⫽ 2 and r ⫽ 2. 1 1 1 and r ⫽ , the nth term is an ⫽ n . Use the “sum(” and 2 2 2 “seq(” features of your calculator to compute S5, S10, and S15 (see Section 11.1). Does Figure 11.70 the sum appear to be approaching some “limiting value?” If so, what is this value? Now compute S20, S25, and S30. Does there still appear to be a limit to the sum? What happens when you have the calculator compute S35? Illustration 1 䊳 For the geometric sequence with a1 ⫽

Solution 䊳 the calculator and enter sum(seq (0.5^X, X, 1, 5)) on the home screen. Pressing gives S5 ⫽ 0.96875 (Figure 11.70). Press 2nd to recall the expression and overwrite the 5, changing it to a 10. Pressing shows S10 ⫽ 0.9990234375. Repeating these steps gives S15 ⫽ 0.9999694824, and it seems that “1” may be a limiting value. Our conjecture receives further support as S20, S25, and S30 are closer and closer to 1, but do not exceed it. CLEAR

ENTER

ENTER

ENTER

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1161

Note that the sum of additional terms will create a longer string of 9’s. That the sum of an infinite number of these terms is 1 can be understood by converting the repeating decimal 0.9 to its fractional form (as shown). For x ⫽ 0.9, 10x ⫽ 9.9 and it follows that 10x ⫽ 9.9 ⫺x ⫽ ⫺0.9 9x ⫽ 9 x⫽ 1 a1 . However, there are 1⫺r many nongeometric, infinite series that also have a limiting value. In some cases these require many, many more terms before the limiting value can be observed. For a geometric sequence, the result of an infinite sum can be verified using Sq ⫽

Use a calculator to write the first five terms and to find S5, S10, and S15. Decide if the sum appears to be approaching some limiting value, then compute S20 and S25. Do these sums support your conjecture? Exercise 1: a1 ⫽ 13 and r ⫽ 13

Exercise 2: a1 ⫽ 0.2 and r ⫽ 0.2

Exercise 3: an ⫽

1 1n ⫺ 12!

Additional Insight: Zeno’s first paradox can also be “resolved” by observing that the “half-steps” needed to complete the race require increasingly shorter (infinitesimally short) amounts of time. Eventually the race is complete.

STRENGTHENING CORE SKILLS Probability, Quick-Counting, and Card Games The card game known as Five Card Stud is often played for fun and relaxation, using toothpicks, beans, or pocket change as players attempt to develop a winning “hand” from the five cards dealt. The various “hands” are given here with the higher value hands listed first (e.g., a full house is a better/higher hand than a flush). Five Card Hand

Description

Probability of Being Dealt

royal flush

five cards of the same suit in sequence from 10 to Ace

0.000 001 540

straight flush

any five cards of the same suit in sequence (exclude royal)

0.000 013 900

four of a kind

four cards of the same rank, any fifth card

full house

three cards of the same rank, with one pair

flush

five cards of the same suit, no sequence required

straight

five cards in sequence, regardless of suit

three of a kind

three cards of the same rank, any two other cards

two pairs

two cards of the one rank, two of another rank, one other card

0.047 500

one pair

two cards of the same rank, any three others

0.422 600

0.001 970

For this study, we will consider the hands that are based on suit (the flushes) and the sample space to be five cards dealt from a deck of 52, or 52C5.

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A flush consists of five cards in the same suit, a straight consists of five cards in sequence. Let’s consider that an Ace can be used as either a high card (as in 10, J, Q, K, A) or a low card (as in A, 2, 3, 4, 5). Since the dominant characteristic of a flush is its suit, we first consider choosing one suit of the four, then the number of ways that the straight can be formed (if needed). Illustration 1 䊳 What is the probability of being dealt a royal flush? Solution 䊳 For a royal flush, all cards must be of one suit. Since there are four suits, it can be chosen in 4C1 ways. A royal flush must have the cards A, K, Q, J, and 10 and once the suit has been decided, it can happen in only this (one) # 4C1 1C1 ⬇ 0.000 001 540. way or 1C1. This means P 1royal flush2 ⫽ 52C5 Illustration 2 䊳 What is the probability of being dealt a straight flush? Solution 䊳 Once again all cards must be of one suit, which can be chosen in 4C1 ways. A straight flush is any five cards in sequence and once the suit has been decided, this can happen in 10 ways (Ace on down, King on down, Queen on down, and so on). By the FCP, there are 4C1 # 10C1 ⫽ 40 ways this can happen, but four of these will be royal flushes that are of a higher value and must be subtracted from this total. So in the intended context we have # 4C1 10C1 ⫺ 4 ⬇ 0.000 013 900 P 1straight flush2 ⫽ 52C5 Using these examples, determine the probability of being dealt Exercise 1: a simple flush (no royal or straight flushes) Exercise 2: three cards of the same suit and any two other (nonsuit) cards Exercise 3: four cards of the same suit and any one other (nonsuit) card Exercise 4: a flush having no face cards

CUMULATIVE REVIEW CHAPTERS 1–11 1. Robot Moe is assembling memory cards for computers. At 9:00 A.M., 52 cards had been assembled. At 11:00 A.M., a total of 98 had been made. Assuming the production rate is linear Table for a. Find the slope of this line and Exercise 3 explain what it means in this x y context. b. Determine how many cards Moe 0 can assemble in an eight-hour day. ␲ c. Find a linear equation model for 6 this data. ␲ d. Determine the approximate time 4 that Moe began work this ␲ morning. 3

2. When using a calculator to find 13 , yet sin 120°, you get 2 13 b ⫽120°. Explain why. sin⫺1a 2 3. Complete this table of special values for y ⫽ cos x without using a calculator.

␲ 2 2␲ 3 5␲ 6 ␲

4. Sketch the graph of y ⫽ 1x ⫹ 4 ⫺ 3 using transformations of a toolbox function. Label the x- and y-intercepts and state what transformations were used. 5. Solve using the quadratic formula: 3x2 ⫹ 5x ⫺ 7 ⫽ 0. State your answer in exact and approximate form. 6. The orbit of Venus around the Sun is nearly circular, with a radius of 67 million miles. The planet completes one revolution in about 225 days. Calculate the planet’s (a) angular velocity in radians per hour and (b) the planet’s orbital velocity in miles per hour. 7. For the graph of g(x) shown, state where a. g1x2 ⫽ 0 b. g1x2 6 0 5 c. g1x2 7 0 d. g1x2c e. g1x2T f. local max g. local min h. g1x2 ⫽ 2 ⫺5 i. g(4) j. g1⫺12 k. as x S ⫺1 ⫹ , g1x2 S ⫺5 l. as x S q, g1x2 S m. the domain of g(x)

y

g(x)

5 x

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15. Write each expression in exponential form: a. 3 ⫽ logx 11252 b. ln12x ⫺ 12 ⫽ 5

8. Match each equation to its corresponding graph. a. y ⫽ sin1␲x2 b. y ⫽ sin1␲x ⫺ ␲2 ␲ c. y ⫽ sin12␲x ⫺ ␲2 d. y ⫽ sina␲x ⫺ b 2 ␲ e. y ⫽ sin12␲x2 f. y ⫽ sina␲x ⫹ b 2 (1) y (2) y 1

1

2

x

⫺1

⫺1

(3)

1

(4)

1

2

x

y 1

⫺1

1

2

x

⫺1

⫺1

1

2

x

(6)

1

2

⫺1

x

⫺1

1

2

19. Solve using a calculator and inverse matrices. 0.7x ⫹ 1.2y ⫺ 3.2z ⫽ ⫺32.5 • 1.5x ⫺ 2.7y ⫹ 0.8z ⫽ ⫺7.5 2.8x ⫹ 1.9y ⫺ 2.1z ⫽ 1.5 20. Find an equation of the hyperbola with foci at 1⫺6, 02 and (6, 0) and vertices at 1⫺4, 02 and (4, 0).

x

⫺1

9. Graph the piecewise function and state the domain and range. ⫺2 y ⫽ •x x2

18. Solve using matrices and row reduction: 2a ⫹ 3b ⫺ 6c ⫽ 15 • 4a ⫺ 6b ⫹ 5c ⫽ 35 3a ⫹ 2b ⫺ 5c ⫽ 24

22. Use properties of sequences to determine a20 and S20. a. 262,144, 65,536, 16,384, 4096, ... 7 27 19 11 b. , , , , ... 8 40 40 40

y 1

1

b. log13x ⫺ 22 ⫹ 1 ⫽ 4

21. Identify the center, vertices, and foci of the conic section defined by x2 ⫹ 4y2 ⫺ 24y ⫹ 6x ⫹ 29 ⫽ 0.

⫺1

y

⫺1

17. Solve for x. a. e2x⫺1 ⫽ 217

⫺1

y

(5)

16. What interest rate is required to ensure that $2000 will double in 10 yr if interest is compounded continuously?

1

⫺1

1163

⫺3 ⱕ x ⱕ ⫺1 ⫺1 6 x 6 2 2ⱕxⱕ3

10. For u ⫽ 3i ⫹ 4j and v ⫽ ⫺12i ⫹ 8j, find the resultant vector w ⫽ u ⫹ v, then use the dot product to compute the angle between u and w. 11. Compute the difference quotient for each function given. 1 a. f 1x2 ⫽ 2x2 ⫺ 3x b. h1x2 ⫽ x⫺2 12. Graph the polynomial function given. Clearly indicate all intercepts. f 1x2 ⫽ x3 ⫹ x2 ⫺ 4x ⫺ 4

2x2 ⫺ 8 . Clearly x2 ⫺ 1 indicate all asymptotes and intercepts.

13. Graph the rational function h1x2 ⫽

14. Write each expression in logarithmic form: 1 ⫽ 3⫺4 a. x ⫽ 10 y b. 81

23. Use the difference identity for cosine to ␲ a. verify that cosa ⫺ ␪b ⫽ sin ␪ and 2 b. find the value of cos 15° in exact form. 24. Caleb’s grandparents live in a small town that lies 125 mi away at a heading of 110°. Having just received his pilot’s license, he sets out on a heading of 110° in a rented plane, traveling at 125 mph (total flight time 1 hr). Unfortunately, he forgets to account for the wind, which is coming from the northeast at 20 mph on a heading of 190°. (a) If Caleb starts out at coordinates (0, 0), what are the coordinates of his grandparent’s town? (b) What are the vector coordinates of the plane 1 hr later? (c) How many miles is he actually away from his grandparent’s town? 25. Empty 55-gal drums are stacked at a storage facility in the form of a pyramid with 52 barrels in the bottom row, 51 barrels in the next row, and so on, until there are 10 barrels in the top row. Use properties of sequences to determine how many barrels are in this stack.

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Exercises 26 through 30 require the use of a graphing calculator. 26. Solve the system. For this system x, y, and z are the variables, with ␲ and e the well-known constants. Round your answer to two decimal places. ␲x ⫹ 12y ⫹ ez ⫽ 5 • 12x ⫹ ey ⫹ ␲z ⫽ 5 ex ⫹ ␲y ⫹ 12z ⫽ 5 27. Find the solution region for the system of linear inequalities. Your answer should include a screen shot or facsimile, and the location of any points of intersection, rounded to four decimal places. 112x ⫹ 39y 7 438 57x ⫺ 64y 6 101 μ xⱖ0 yⱖ0

28. A triangle has vertices at (112.3, 98.5), (67.7, ⫺39), and (⫺27␲, 21.5). Use the determinant formula to determine its area, rounded to the nearest tenth. 29. A recursive sequence is defined by a1 ⫽ 0.3 and ak⫹1 ⫽ a2k ⫺ 0.4ak. Find S40, rounded to four decimal places. 30. Solve the equation. Round your answer to two decimal places. log x ⫺ 2 ln x ⫹ 3 log3 x ⫽ 6.

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CONNECTIONS TO CALCULUS The properties of summation play an important role in the development of many key ideas in a first semester calculus course. Here, we’ll see how the summation properties (Section 11.1) are combined with the summation formulas (Section 11.4) and the concept of the area under a curve (Connections to Calculus Chapter 3), to produce some very interesting results. For convenience, the summation formulas are restated here: n

(1)

兺

n

c ⫽ cn

(2)

i⫽1 n

(3)

兺

i2 ⫽

i⫽1

兺

i⫽

i⫽1

2n3 ⫹ 3n2 ⫹ n 6

n2 ⫹ n 2

n

(4)

兺

i3 ⫽

i⫽1

n4 ⫹ 2n3 ⫹ n2 4

Applications of Summation For this study, consider the area under the line f 1x2 ⫽ x ⫹ 5 in the interval [0, 4] (between the y-axis and the vertical line x ⫽ 4, Figure 11.71). To find the total area, we could add the area of the rectangle and triangle shown, or use f 102 ⫽ 5 and f 142 ⫽ 9 as the bases of a trapezoid and h simply apply the formula A ⫽ 1b1 ⫹ b2 2 2 (giving A ⫽ 28 units2). Instead of being so direct, we’ll use this simple geometric shape to develop some powerful ideas, to be used when the desired area is much more complex. Using what we’ll call the rectangle method, we first approximate the total area by adding the area of the four rectangles shown in Figure 11.72. Note this will give an area greater than the true value, but yields a reasonable estimate. Each rectangle will have a width of 1, since there are four rectangles in an interval four units wide 1 44 ⫽ 12, and the length of each rectangle will be f 1x2 ⫽ x ⫹ 5 for x ⫽ 1, 2, 3, and 4. Using the formula A ⫽ LW, we can write this as 4

兺LW ⫽ 兺 f 1i2 112, where the function is written

Figure 11.71 y 10 9 8 7 6 5 4 3 2 1 1

2

3

4

5

6

x

5

6

x

5

6

x

Figure 11.72 y 10 9 8 7 6 5 4 3 2 1 1

2

3

4

in terms of i 3 f 1i2 ⫽ i ⫹ 5 4 for the sake of the summation notation. i⫽1

4

兺 f 1i2 112 ⫽ f 112 112 ⫹ f 122 112 ⫹ f 132 112 ⫹ f 142 112

i⫽1

Figure 11.73

⫽ 6 ⫹ 7 ⫹ 8 ⫹ 9 ⫽ 30 units2

To refine this estimate, consider the eight rectangles shown in Figure 11.73. The original fourunit interval is now divided into eight parts, so each part is 48 ⫽ 12 unit wide. The length of each rectangle is still given by f 1x2 ⫽ x ⫹ 5, but we now evaluate the function in increments of 12:

y 10 9 8 7 6 5 4 3 2 1 1

11–89

2

3

4

1165

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11–90

Connections to Calculus 8 rectangles

giving

one-half unit wide 8

兺

LW ⫽

兺 f a 2 iba 2 b 1

1

i⫽1

for i ⫽ 1, 2, 3, . . . , 8

evaluate f each 1/2 unit

Since the length of each rectangle is multiplied by W ⫽ 12, we can factor out 12 and 1 1 8 f a ib for convenience. write the expression as 2 i⫽1 2

兺

EXAMPLE 1

䊳

Computing a Sum of Rectangular Areas Find the sum

Solution

䊳

1 1 8 f a ib by writing out each term. 2 i⫽1 2

兺

1 8 1 1 1 3 5 7 f a ib ⫽ c f a b ⫹ f 112 ⫹ f a b ⫹ f 122 ⫹ f a b ⫹ f 132 ⫹ f a b ⫹ f 142 d 2 i⫽1 2 2 2 2 2 2 1 11 13 15 17 ⫹7⫹ ⫹8⫹ ⫹ 9b ⫽ a ⫹6⫹ 2 2 2 2 2 11 13 7 15 17 9 ⫽ ⫹3⫹ ⫹ ⫹ ⫹4⫹ ⫹ ⫽ 29 units2 4 4 2 4 4 2

兺


䊳

This approximation is still too large, but closer Figure 11.74 to the true value of 28 units2. At this point, we y could get an even better estimate using 16 rectan- 10 gles (see Figure 11.74), with each width being 9 4 1 8 16 ⫽ 4 unit wide, and the length still computed by 7 1 f 1x2 ⫽ x ⫹ 5 but using increments of 4 . This 6 16 5 1 1 1 16 1 f a ib a b ⫽ f a ib. 4 would yield LW ⫽ 4 4 4 i⫽1 4 i⫽1 3 1 16 1 2 f a ib 1 As you might imagine, finding the sum 4 i⫽1 4 5 6 x 1 2 3 4 by writing out each term would be tedious and time consuming. Instead, the properties of summation and the summation formulas can be used to develop an expression where we can find the sum for n ⫽ 16 directly. 1 1 The key idea is to note that since f a ib ⫽ i ⫹ 5, we can write the summation as 4 4 1 16 1 a i ⫹ 5b and work as in Example 2. 4 i⫽1 4

兺

兺

兺

兺

兺

EXAMPLE 2

䊳

Computing a Sum of Rectangular Areas 1 n 1 a i ⫹ 5b, use the summation properties and summation formulas 4 i⫽1 4 1 16 1 to develop an expression where a i ⫹ 5b can be found by substituting n ⫽ 16 4 i⫽1 4 directly. For the sum

兺

兺

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Solution

䊳

n 1 n 1 1 n 1 a i ⫹ 5b ⫽ a i⫹ 5b 4 i⫽1 4 4 i⫽1 4 i⫽1 n 1 1 n ⫽ a i⫹ 5b 4 4 i⫽1 i⫽1 1 1 n2 ⫹ n b ⫹ 5n d ⫽ c a 4 4 2 1 1 162 ⫹ 16 b ⫹ 51162 d ⫽ c a 4 4 2 1 272 b ⫹ 20 ⫽ a 16 2 ⫽ 8.5 ⫹ 20 ⫽ 28.5

兺

兺

兺

兺

summation properties (distribute)

兺

factor

1167

1 from first summation 4

use summation formulas

substitute 16 for n

distribute and simplify result

2

The approximate area is 28.5 units . Now try Exercises 3 and 4

䊳

The significance of this result cannot be overstated. As we’ll see in Chapter 12, similar calculations can be used if n rectangles are assumed, with the area easily calculated for any value of n. Further, you might imagine the amazement of the early students of mathematics, when they realized that these ideas could still be applied for areas where no elementary formula existed, as for the area under a nonlinear graph.

Connections to Calculus Exercises

1. For f 1x2 ⫽ ⫺x ⫹ 6 and x 僆 3 0, 4 4, (a) graph the function, then approximate the area under the graph in this interval using four rectangles of width 1 and length f (x) for x ⫽ 1, 2, 3, and 4. Begin by developing the summation formula for

兺LW, then

expand the sum. (b) Repeat part (a) using eight rectangles of width one-half and length f (x). Draw both the graph and the rectangles in each case, and verify the increased number of rectangles gives a better estimate.

2. For f 1x2 ⫽ x ⫹ 4 and x 僆 3 0, 6 4, (a) graph the function, then approximate the area under the graph in this interval using six rectangles of width 1 and length f (x) for x ⫽ 1, 2, 3, 4, 5, and 6. Begin by developing the summation formula for

兺LW, then expand the sum. (b) Repeat part (a) using 12 rectangles of width one-half and length f(x). Draw both the graph and the rectangles in each case, and verify the increased number of rectangles gives a better estimate.

For Exercises 3 and 4, show all work requested using a methodical step-by-step process, similar to that illustrated in Example 2. 3. For f 1x2 ⫽ ⫺x ⫹ 6 and x 僆 30, 4 4 from Exercise 1, (a) discuss how the area under the graph of f in this interval can be approximated using the rectangle method and n ⫽ 32 rectangles without expanding the sum. (b) Find the area for n ⫽ 32 by applying summation properties and formulas.

4. For f 1x2 ⫽ x ⫹ 4 and x 僆 3 0, 64 from Exercise 2, (a) discuss how the area under the graph of f in this interval can be approximated using the rectangle method and n ⫽ 36 rectangles without expanding the sum. (b) Find the area for n ⫽ 36 by applying summation properties and formulas.


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CHAPTER CONNECTIONS

Bridges to Calculus: An Introduction to Limits CHAPTER OUTLINE 12.1 An Introduction to Limits Using Tables and Graphs

1170

12.2 The Properties of Limits 1180 12.3 Continuity and More on Limits 1191 12.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

1203

This chapter introduces one of the most central and powerful ideas in calculus, the concept of a limit. While the study of limits isn’t a recent phenomenon (Archimedes inscribed a polygon within a circle and considered an ever increasing number of sides to estimate the circumference), a complete development of the idea had to wait for the talents of Augustin-Louis Cauchy (c. 1821) and a maturation of the notation needed to work with the concept effectively. Using a rational function to illustrate, essentially the concept involves recognizing the distinction between x2 ⫹ 2x ⫺ 3 the value of g1x2 ⫽ when x ⫽ 1 x⫺1 (the function is not defined at 1), and the limit of g(x) as x becomes very close to 1. 䊳


1169

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12.1

An Introduction to Limits Using Tables and Graphs

LEARNING OBJECTIVES

The concept of a limit is the foundation of all calculus. If the concept is faulty, the conclusions of calculus can be neither supported nor sustained, and the whole becomes a great leap of faith. The inventors of calculus (Isaac Newton and Gottfried Leibniz) tried mightily to bridge the gap between the finite and the infinite, but it was only after a long evolution that the ideas would mature to a point where infinity could be tamed and calculus placed on an unassailable footing.


A. Distinguish between a limit and an approximation B. Estimate limits using tables C. Evaluate one-sided limits D. Determine when a limit does not exist

A. Distinguishing between Limits and Approximations To begin our study, consider the area of a regular polygon inscribed in a circle of radius r (inscribed: all vertices of the polygon are on the circumference). The radii drawn from the center to each vertex divide the polygons into isosceles triangles, enabling us to approximate the area of the circle by summing the areas of these triangles (Figure 12.1). Figure 12.1

A3

A4

A5

A6

A8

If we let An represent the area of an inscribed polygon with n sides, we note that as the number of sides increases, the area of the polygon gives a better and better approximation of the area of the circle. In fact, it seems intuitive that as n becomes infinitely large, the area of the polygon becomes infinitely close to the area of the circle. Using the notation seen in previous chapters we would say, as n S q, An S ␲r2. In a study of limits, limit notation is used, and the same statement is written lim An ⫽ ␲r2

nSq

In words, “the limit of An as n becomes infinitely large is ␲r2.” For practice using the new notation, see Exercises 7 through 12. In a study of limits it becomes extremely important to accept that no finite number of sides for our polygon will give the exact area of the circle (even with 100 sides we still have only a very good approximation). But that’s not what the concept and notation are saying. They only say that ␲r2 is the limiting value, and that we can become as close as we like to this limit, by choosing n sufficiently large. EXAMPLE 1

䊳

WORTHY OF NOTE From the area formula 1 A ⫽ ab sin C, note that for each 2 triangle within the circle, the angle 2␲ at the center C is , and a ⫽ b ⫽ r. n This shows each triangle has area 2␲ 1 A ⫽ r2 sina b and there are n of n 2 these triangles.

1170

Finding the Area of an Inscribed Polygon If an n-sided regular polygon is inscribed in a circle of radius r, its area is given by 2␲ nr2 An ⫽ sina b (see Worthy of Note). Assume the circle has a radius of n 2 r ⫽ 10 cm. Use the TABLE feature of a graphing calculator to determine the number of sides needed (to the nearest 100) for the area of the polygon to approximate the area of the circle rounded to two decimal places (A ⬇ 314.16 units2 rounded).

12–2

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Section 12.1 An Introduction to Limits Using Tables and Graphs

Solution

䊳

A. You’ve just seen how we can distinguish between a limit and an approximation

1171

With r ⫽ 10, the formula becomes 2␲ An ⫽ 50n sina b, where n represents n the number of sides. Enter this equation as Y1 in your calculator (in function and radian MODE ), and set up a table to start at 100 with ¢Tbl ⫽ 100. In the figure, we note that when n ⫽ 700, the area of the polygon approximates the area of the circle rounded to two decimal places. Now try Exercises 13 through 16

䊳

B. Estimating Limits Using Tables and Graphs

The limit in the discussion prior to Example 1 1 lim An 2 is called a limit at infinity, or a nSq limit as n becomes infinitely large. These are discussed in more detail in Section 12.3. Here we turn our attention to limits at a constant c (the limit as x S c), an idea we explored in our study of asymptotic behavior in Section 4.5. Consider the function sin x f 1x2 ⫽ (x in radians), noting that x ⫽ 0 is not in the domain of f. However, the x fact that we could not evaluate lim An for n ⫽ q in Example 1, did not stop us from nSq sin x at x ⫽ 0, but we evaluating An as n S q . Similarly, we cannot evaluate f 1x2 ⫽ x can evaluate f for values of x slightly less than zero, or slightly greater than zero, to sin x investigate the limit of as x S 0. For additional practice with the limit notation, x see Exercises 17 through 22.

EXAMPLE 2

䊳

Determining Whether a Limiting Value Exists Use the TABLE feature of a graphing calculator to evaluate f 1x2 ⫽

sin x as x S 0. x If the function seems to approach a limiting value, write the relationship in words and using the limit notation.

Solution Figure 12.2

Figure 12.3

䊳

For this exercise it seems appropriate to begin a short distance from zero, and sin X approach it in small increments. Enter Y1 ⫽ in your calculator and set up the X table to ASK for the independent variable values. Noting that 0 can be “approached” from either side, we use values less than zero (from the left-hand side) and values greater than zero (from the right-hand side). Beginning at x ⫽ ⫺0.5 and approaching 0 from the left, our calculator produces the screen shown in Figure 12.2. Starting at x ⫽ 0.5 and approaching from the right yields Figure 12.3. Note the calculator rounds each entry in the Y1 column to five decimal places. Because of this, the final Y1 value shown in each figure is 1. After using the arrow keys to highlight this value, the calculator displays a more accurate approximation (rounded to 12 decimal places) along the bottom of the screen. These values in both sin x figures seem to indicate that as x becomes very close to 0, becomes very close x sin x ⫽ 1. to 1, or seems to have a limiting value of 1. In limit notation we write lim xS0 x Now try Exercises 23 through 28

䊳

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CHAPTER 12 Bridges to Calculus: An Introduction to Limits

sin x x as close as we like to 1 by using values of x that are sufficiently close to 0. This suggests the following definition of a limit, a definition, however, that will be refined and made more precise as we gain a greater understanding of the limit concept. Similar to our work with the area of a polygon and circle, note that we can make

The Definition of a Limit For a fixed real number L and constant c,

lim f 1x2 ⫽ L

xSc

means that values of f(x) can be made arbitrarily close (infinitely close) to L, by taking values of x sufficiently close to c, x ⫽ c. The graph of (1) f 1x2 ⫽

sin x is shown in Figure 12.4. As we did in Section 2.5, x we could create a piecewise function that defines f (x) at x ⫽ 0, and two possibilities are listed here, followed by their graphs (Figures 12.5 and 12.6, respectively). sin x x⫽0 122 f 1x2 ⫽ • x 2 x⫽0 Figure 12.4

sin x 132 f 1x2 ⫽ • x 1

Figure 12.5

y

x⫽0 x⫽0

Figure 12.6

y

2.5

y

2.5

2

2

1.5

1.5

2.5

(0, 2)

2 1.5

1

1

1

0.5

0.5

0.5

⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺0.5

2

4

6

8 10

x

⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺0.5

2

4

6

8 10

x

⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺0.5

⫺1

⫺1

⫺1

⫺1.5

⫺1.5

⫺1.5

⫺2

⫺2

⫺2

⫺2.5

⫺2.5

⫺2.5

(0, 1)

2

4

6

8 10

x

In (2), f (x) defines the function at (0, 2) but leaves a hole in the graph at (0, 1), while in (3), f (x) is a continuous function as we can now draw the entire graph without lifting our pencil (see Exercises 29 and 30). Note that in all three cases, lim f 1x2 ⫽ 1 xS0 remains true, as the definition of a limit is still satisfied in every respect. In other words, the limit of a function as x S c can exist even if (2) the function is defined at c but f 1c2 ⫽ L sin x , x⫽0 sin x at c as in f 1x2 ⫽ , as in f 1x2 ⫽ • x , x 2 x⫽0 sin x , x⫽0 or (3) the function is defined at c and f 1c2 ⫽ L as in f 1x2 ⫽ • x . x⫽0 1

(1) the function is not defined

EXAMPLE 3

䊳

Determining Whether a Limiting Value Exists ⫺x , as x⫺2 x S 2. If the function seems to approach a limiting value, write the relationship in words and using the limit notation. Use the TABLE feature of a graphing calculator to evaluate g1x2 ⫽

1 2 2x

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Section 12.1 An Introduction to Limits Using Tables and Graphs 䊳

Solution

Similar to Example 2, enter this equation as Y1 in your calculator, start a short distance from 2 on the left- and right-hand sides, and approach 2 in small increments. Starting from the left at 1.6 produces the table shown in Figure 12.7. Starting from the right at 2.4 yields Figure 12.8. These screen shots seem to indicate that as x becomes very close to 2, g(x) becomes very close to 1, or seems to have a limiting value of 1. In limit notation, 1 2 2x ⫺ x ⫽ 1. The graph of y ⫽ g1x2 is shown in Figure 12.9. lim xS2 x ⫺ 2

Figure 12.7

Figure 12.8

Figure 12.9 y 5 4 3 2

g(x)

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3 ⫺4 ⫺5

B. You’ve just seen how we can estimate limits using tables


䊳

C. One-Sided Limits In Examples 2 and 3 our discussion focused on what are called two-sided limits, as we approached the limiting value by selecting inputs to the left and to the right of c x2 ⫺ 2x (x ⫽ c). But what of functions like y ⫽ 12 ⫺ x, y ⫽ and many others, 20 1x ⫺ 2 that may only be defined to the left or right of c ⫽ 2? Can these functions approach a limiting value as x approaches c from only one side? EXAMPLE 4

䊳

Determining Whether a Limit Exists for a One-Sided Approach Use the TABLE feature of a graphing calculator to help determine if x2 ⫺ 2x f 1x2 ⫽ has a limiting value as x S 2 from the right. 20 1x ⫺ 2

Solution

䊳

With the radical 1x ⫺ 2 in the denominator, the domain of f is 12, q2 . For Y1 ⫽ f 1x2 , we input values close to 2 but greater than 2, since 2 can only be “approached” from the right-hand side. The result is the TABLE shown, and seems to indicate that as x becomes very close to 2 from the right, Y1 becomes very close to 0. Further, it appears we can make Y1 as close to zero as we like, by choosing values of x sufficiently close to 2 but greater than 2. This suggests the limit of f (x) as x approaches 2 from the right is 0. Now try Exercises 37 through 42

䊳

For functions like those in Example 4, we can define a one-sided limit, using the general definition as a template and introducing a modified notation to represent the idea more clearly.

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Right-Hand Limits For a fixed real number L and constant c,

lim f 1x2 ⫽ L

xSc⫹

means that values of f(x) can be made arbitrarily close (infinitely close) to L, by taking values of x sufficiently close to c and to the right of c 1x 7 c2 . Left-Hand Limits For a fixed real number L and constant c,

lim f 1x2 ⫽ L

xSc⫺

means that values of f(x) can be made arbitrarily close (infinitely close) to L, by taking values of x sufficiently close to c and to the left of c 1x 6 c2 . Note how the superscript on c indicates the direction from which x is approaching c (c can be any real number). For x S c⫺, x is approaching c from the left (the “negative side”), for x S c ⫹ , x is approaching c from the right (the “positive side”). For practice with this notation, see Exercises 43 through 46. EXAMPLE 5

䊳

Determining Left- and Right-Handed Limits 冟x ⫺ 2冟 For h1x2 ⫽ , use the TABLE feature of a gaphing calculator to evaluate x⫺2 the expressions a. lim⫺ h1x2 b. lim⫹ h1x2 xS2

xS2

If a limiting value appears to exist from either side, write the relationship in words and using the limit notation.

Solution

䊳

Begin by noting the domain of h is x 僆⺢, x ⫽ 2. a. For lim⫺ h1x2 , we use a table to track the value of h as x approaches 2 from xS2

the left. After entering h(x) as Y1 (recall the abs( command can be found under the MATH NUM menu), the TABLE produced is shown in Figure 12.10, 冟x ⫺ 2冟 ⫽ ⫺1. In words, the limit of h(x) as x approaches 2 and suggests lim⫺ xS2 x⫺2 from the left is ⫺1. Figure 12.11

Figure 12.10 Figure 12.12 y 5 4 3 2

h(x)

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3 ⫺4 ⫺5

C. You’ve just seen how we can evaluate one-sided limits

b. Similarly, Figure 12.11 suggests lim⫹ xS2

冟x ⫺ 2冟 ⫽ 1, or the limit of h(x) as x x⫺2

approaches 2 from the right is 1. The graph of h is shown in Figure 12.12 and seems to support these conclusions. Now try Exercises 47 through 54

䊳

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1175

D. Limits That Fail to Exist Unlike the function in Example 4 (where x could approach 2 only from the right), the functions in Examples 3 and 5 allowed approaches from both the left and right sides. Note that the statement, “taking values of x sufficiently close to c” from the general definition of a limit implies a two-sided approach to c, and we must approach c from both sides wherever possible. Since this is not possible in Example 4, we use the notation for x2 ⫺ 2x ⫽ 0. For the function g(x) in Example 3, a one-sided limits to state: lim⫹ xS2 20 1x ⫺ 2 “two-sided” approach was possible and we saw the value approached from each side was equal to 1. Using the general definition we were able to write lim g1x2 ⫽ 1 (the xS2

limit is L ⫽ 1). For the function h(x) in Example 5, x could also approach c from both sides, but the left-hand limit was not equal to the right, and there is no single number L that h(x) approaches as x S 2. This leads us to the following statement: The Existence of a Limit For a fixed real number L and constant c, lim f 1x2 ⫽ L if and only if

xSc

lim f 1x2 ⫽ lim⫹ f 1x2 ⫽ L.

xSc ⫺

xSc

In words, “the limit of f(x) as x approaches c is L, if and only if the left-hand limit as x approaches c is equal to the right-hand limit as x approaches c, and both are also equal to L.” EXAMPLE 6

Solution

䊳

䊳

Determining Whether a Limit Exists Use the TABLE feature of a graphing calculator to evaluate the expression x2 ⫺ 3x lim . xS3 2x2 ⫺ 6x ⫹ 9 X2 ⫺ 3X Similar to the previous examples, we enter Y1 ⫽ into our 2X2 ⫺ 6X ⫹ 9 calculator. As a matter of choice we first evaluate the left-hand limit, and x2 ⫺ 3x Figure 12.13 suggests lim⫺ ⫽ ⫺3. For the right-hand limit, xS3 2x2 ⫺ 6x ⫹ 9 x2 ⫺ 3x Figure 12.14 suggests lim⫹ ⫽ 3. xS3 2x2 ⫺ 6x ⫹ 9 Figure 12.14 Figure 12.13

Figure 12.15 y 7 6 5

Since the left-hand limit ⫽ right-hand limit, lim

4

xS3

3 2

The graph of y ⫽

1 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2

3

4

5

6

7

x

x2 ⫺ 3x

x2 ⫺ 3x 2x2 ⫺ 6x ⫹ 9

does not exist.

is shown in Figure 12.15. As a matter of order 2x2 ⫺ 6x ⫹ 9 we will adopt the following notation for the case where the left-hand limit is not equal to the right-hand limit. Using dne as an abbreviation for “does not exist,” dne B . we will write lim f 1x2 ⫽ A LH⫽RH xSc


䊳

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In Example 7 we note a second way that a limit can fail to exist. EXAMPLE 7

䊳

Determining Whether a Limit Exists Given f 1x2 ⫽

Solution

䊳

9x , use a TABLE to evaluate lim f 1x2 . xS2 1x ⫺ 22 2

For the left-hand limit, we obtain the TABLE shown and note as x S 2 ⫺ , the value of f(x) increases without bound [for x ⫽ 1.999, f(x) is almost 18,000,000]. A similar check shows that as x S 2 ⫹ , the values of f(x) likewise become infinitely large. Since f(x) becomes infinitely large and positive as x S 2, the limit does not exist and . we write lim f 1x2 ⫽ A dne q B xS2


䊳

From our work with rational functions in Section 4.5, we know the graph of 9x has a vertical asymptote at x ⫽ 2, verifying it is impossible for the f 1x2 ⫽ 1x ⫺ 22 2 function to have a limit as x S 2. Finally, we consider a third way that a limit can fail to exist. EXAMPLE 8

䊳

Determining Whether a Limit Exists 1 For h1x2 ⫽ 2 ` sina b ` , try to find a fixed number L such that lim f 1x2 ⫽ L. x xS0

Solution

䊳

WORTHY OF NOTE For students and instructors with a sense of humor, Example 8 is sometimes called the Christmas Case, because there’s No-el.

Using an approach from the right gives the result shown in Figure 12.16. This time there is no apparent pattern in the output values, and a graph of the function (Figure 12.17) reveals why. As x S 0, the function begins to oscillate wildly between 0 and 2, with no approach to any fixed number L. A similar thing happens for a left-handed approach. As per our definition, the limit does not exist because L does not exist. In this case, we write lim h1x2 ⫽ A dne B . xS2

L

Figure 12.16

Figure 12.17 y 4 3

h(x)

2 1

⫺0.4

⫺0.2

0.2

0.4

x

⫺1

D. You’ve just seen how we can determine when a limit does not exist


䊳

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12.1 EXERCISES 䊳



1. The limit of f (x) as x S q is called a limit at .

2. lim g1x2 ⫽ L means that values of g(x) can be

3. To evaluate the limit of a function as x S c, we must find the limit using values of x less than c, and the limit using values of x than c.

4. If a rational function v(x) has a vertical asymptote at x ⫽ c, the limit as x S c does not exist and we write lim v1x2 ⫽ .

xSc

made

5. Discuss/Explain why lim g1x2 ⫽ 5 for xS3

x ⫺x⫺6 g1x2 ⫽ , but g132 ⫽ 5. Can we redefine x⫺3 g(x) to create a piecewise-defined function where lim g1x2 ⫽ g132 ? 2

to L, by taking values of sufficiently close to .

xSc

6. Discuss/Explain why lim f 1x2 does not exist for f 1x2 ⫽

x2 ⫺ x

xS1

, even though the left-hand limit 21x ⫺ 12 2 and the right-hand limit do exist.

xS3

䊳


Write each of the following statements using limit notation.

needed (to the nearest 100) for the area of the polygon to approximate the area of the circle correct to two decimal places when rounded. Compare your result with that of Example 1.

4 7. As n approaches q , Vn approaches ␲r3. 3 8. As n approaches q , An approaches a 3 b 2 x ⫹ x ⫹ cx ⫹ d. 3 2 9. As t approaches ⫺q , e f(t) approaches 0. 10. As t approaches ⫺q , sin[g(t)] approaches 1. 1 11. As x increases without bound, cosa b approaches 1. x 5x ⫺ 3 2x2 ⫺ x ⫺ 1 2

12. As x decreases without bound, approaches 52 .

Use the TABLE feature of a graphing calculator for the following exercises.

13. If an n-sided regular polygon circumscribes a circle of radius r (see the figure), its area is ␲ given by An ⫽ nr2 tan a b. For n a circle with radius r ⫽ 10 cm, determine the number of sides

A8 r

14. If an n-sided regular polygon is inscribed in a circle of radius r, its perimeter is given by ␲ P ⫽ 2nr sin a b. For a circle with radius n r ⫽ 50 mm, determine the number of sides needed (to the nearest 25) for the perimeter of the polygon to approximate the circumference of the circle correct to two decimal places when rounded. 15. If an n-sided regular polygon circumscribes a circle of radius r, its perimeter is given by ␲ P ⫽ 2nr tan a b. For a circle with radius n r ⫽ 50 mm, determine the number of sides needed (to the nearest 25) for the perimeter of the polygon to approximate the circumference of the circle correct to two decimal places when rounded. Compare your results with those in Exercise 14. 16. One of the most famous and useful numbers in all of mathematics is the one defined as 1 x lim a1 ⫹ b ⫽ e. Determine the smallest x to x xSq (the nearest 250) that can be used to approximate the value of e to three decimal places.

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Write each of the following statements using limit notation.

17. As t approaches 5, st approaches 5r. 18. As t approaches ⫺3, dt approaches 29 ⫹ r2. 19. As x approaches a, tan⫺1 3 g1x2 4 approaches ␲3 . 20. As x approaches b, csc2[ f(x)] approaches 43.

x⫹3 21. As x S ⫺3, 2 approaches ⫺16 . x ⫺9 22. As x S ␲,

sin x approaches 2. x cosa b 2

Use the TABLE feature of a graphing calculator to evaluate each function as x approaches the value indicated. If the function seems to approach a limiting value, write the relationship in words and using the limit notation.

23. p1x2 ⫽ cos x ⫹ sina

3x b; x S ␲ 2

24. q1x2 ⫽ tan x ⫹ 3 sin 2x; x S ␲ cosa b x

25. v1x2 ⫽

sin1␲x2

26. w1x2 ⫽

␲ 4

;xS2

tan1x ⫺ 22 x⫺2

;xS2

27. s1x2 ⫽

2 cos x ⫺ 2 ;xS0 x

28. t1x2 ⫽

␲ sin x ⫺ 1 ;xS 2 sin12x2

The functions in Exercises 29 and 30 have a hole (discontinuity) in their graphs at x ⴝ 2. Write a related piecewise-defined function that creates a continuous graph for each.

2x2 ⫺ 7x ⫹ 6 29. r1x2 ⫽ sin1x ⫺ 22 30. s1x2 ⫽

12–10


sin1x2 ⫺ 4x ⫹ 42 x⫺2

32. v1x2 ⫽

x2 ⫺ 3x ⫺ 10 ; x S ⫺2 2x ⫹ 4

33. g1x2 ⫽

x2 ⫹ 2x ⫺ 3 ;xS1 x⫺1

34. w1x2 ⫽

x1x ⫺ 22 ⫺ 8 41x ⫺ 42

35. f 1x2 ⫽ 3x2 ⫺ x ⫺ 2; x S 1 36. g1x2 ⫽ 10x ⫺ x2; x S ⫺2

See if the TABLE feature of a graphing calculator suggests a limit exists for the functions and approaches indicated.

37. f 1x2 ⫽ 2x2 ⫺ 3x as x S ⫺3 from the right. 1 38. g1x2 ⫽ 6x2 ⫺ x as x S ⫺ from the right. 2 39. f 1x2 ⫽

x3 ⫹ 8 1 2x 4

⫹1

31. f 1x2 ⫽

x2 ⫺ 2x ;xS2 x2 ⫺ 4

as x S ⫺2 from the left.

40. g1x2 ⫽

x ⫺1 as x S ⫺1 from the left. x⫹1

41. f 1x2 ⫽

sin 1x as x S 0 from the right. 1x

42. g1x2 ⫽

x as x S 0 from the right. ln x

Write each statement using limit notation.

43. As x approaches 3 from the left, Ix approaches 3 cos2 1R1 ⫹ R2 2 .

44. As x approaches ⫺1 from the right, Sx approaches log1␸ ⫹ 12 .

45. As x approaches m from the left, f approaches L. 46. As x approaches n from the right, g approaches L. Evaluate the following limits using the TABLE feature of a graphing calculator.

47. Given f 1x2 ⫽ a. lim⫺ f 1x2 xS␲

Use the TABLE feature of a graphing calculator to evaluate each function as x approaches the value indicated. If the function seems to approach a limiting value, write the relationship in words and using the limit notation.

;xS4

sin x , find 冟sin x冟

b. lim⫹ f 1x2 xS␲

cos x 48. Given g1x2 ⫽ ⫺ , find 冟cos x冟 g1x2 g1x2 a. lim b. lim ␲⫺ ␲⫹ xS 2

49. Given f 1x2 ⫽

xS 2

x2 ⫺ 10x ⫹ 24

, find 2x2 ⫺ 12x ⫹ 36 a. lim⫺ f 1x2 b. lim⫹ f 1x2 xS6

xS6

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x2 ⫹ 2x ⫺ 3

50. Given g1x2 ⫽ a.

␲ 4 ␲ 59. Given f 1x2 ⫽ μ tan x x ⫽ , find 4 ␲ cos x x 7 4 sin x

, find

2x2 ⫹ 6x ⫹ 9 b. lim ⫹ g1x2 lim ⫺ g1x2

xS⫺3

xS⫺3

51. Given f 1x2 ⫽

1 2 ⫺ 3x2, find x a. lim⫺ f 1x2 b. lim⫹ f 1x2

xS4

xSln 7.3

53. For f 1x2 ⫽ e a. lim⫺ f 1x2

x ⫺4 sin1x2 ⫺ 42 2

xS2

lim

xSln 7.3 ⫹

x ⱕ ⫺1 x 7 ⫺1

xS⫺5

xS⫺5

⫺1x ⫹ 12 x ⱕ 10 , find log x x 7 10 lim ⫺ g1x2 b. lim ⫹ g1x2

56. Given g1x2 ⫽ e

xS10

c. lim g1x2 xS10

57. Given f 1x2 ⫽ a. lim⫺ f 1x2 xS␲

c. lim f 1x2 xS␲

58. Given g1x2 ⫽ a. lim⫺ g1x2 xS0

c. lim g1x2 xS0

xS 4

xS 4

5␲ 4 5␲ 60. Given g1x2 ⫽ μ cot x x ⫽ , find 4 5␲ csc x x 7 4 g1x2 g1x2 a. lim b. lim 5␲ 5␲ sec x

, find

xS

xS

2x2 ⫺ 7 x 6 ⫺5 55. Given f 1x2 ⫽ e , find 3 ⫺ 2x x ⱖ ⫺5 a. lim ⫺ f 1x2 b. lim ⫹ f 1x2 c. lim f 1x2

b. lim f 1x2 ␲⫹

⫺

⫺

x 6

xS

4

⫹

4

c. lim5␲ g1x2

b.

xS⫺1

xS⫺5

x 6

c. lim␲ f 1x2

g1x2

xS2

lim ⫺ g1x2

xS10

xS 4

x 6 2 , find xⱖ2 b. lim⫹ f 1x2

␲ 3 tan c 1x ⫹ 22 d 4 54. For g1x2 ⫽ • 2 2x ⫹ 8

a.

a. lim␲ f 1x2

xS4

52. Given g1x2 ⫽ ex ⫺ 5, find lim ⫺ g1x2 a. b.

a.

1179


11 ⫺ cos 12x2

, find 12 sin x b. lim⫹ f 1x2 xS␲

sin3 ␲1x ⫺ 12 4 , find 冟x冟 b. lim⫹ g1x2 xS0

4

61. Given f 1x2 ⫽ x2 ⫹

3 , find lim f1x2 . xS⫺2 1x ⫹ 22 2

62. Given g1x2 ⫽ 2x2 ⫺ 3x ⫹ 4x⫺2, find lim g1x2 xS0

␲ ⫺4 cosa xb 4 63. Given f 1x2 ⫽ , find lim f 1x2 . x⫺2 xS2 64. Given g1x2 ⫽ 冟csc1x2 ⫺ 17x ⫹ 522 冟, find lim g1x2 . xS13

65. Given f 1x2 ⫽

2x , find lim f 1x2 . xS⫺6 冟x3 ⫹ 216冟

66. Given g1x2 ⫽

6x2 ⫺ x ⫺ 1 , find lim1 g1x2 . 2x ⫺ 1 xS 2

67. Given f 1x2 ⫽ sina

x⫹1 b, find lim f 1x2 . x⫺1 xS1

68. Given g1x2 ⫽ cos a

x⫺1 b, find lim g1x2 . x⫹1 xS1

69. Given f 1x2 ⫽ tan1cos x2 , find lim␲ f 1x2 . xS 2

70. Given f 1x2 ⫽ cos1tan x2 , find lim␲ f 1x2 . xS 2

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71. (4.2) Use the rational zeroes theorem to write the polynomial in completely factored form: 3x4 ⫺ 19x3 ⫹ 15x2 ⫹ 27x ⫺ 10.

73. (8.2) Use Heron’s formula to find the area of a triangle with sides a ⫽ 5 in., b ⫽ 8 in., c ⫽ 9 in., rounded to two decimal places.

72. (3.1) Find the sum, difference, product, and quotient of 1 ⫹ 3i and 1 ⫺ 3i.

74. (6.5) Write a sinusoidal equation given the maximum is 100, the minimum is ⫺20 and the period is 365.

12.2

The Properties of Limits


A. Establish limit properties using a table

B. Evaluate limits using the limit properties C. Distinguish between a declarative statement and a proof

After the invention of calculus, but before the concept of a limit had been fully developed, the ideas of Newton (the method of fluxions) and Leibniz (the method of differentials) were often met with skepticism and sarcasm even within the academic world. In a tract called The Analyst (1734), mathematician and philosopher George Berkeley once said, “he who can digest a second or third fluxion ... need not, methinks, be squeamish about any point in divinity.” In this section, we introduce the properties of limits, which helped to place the study of calculus on a sure footing. The plausibility of these properties will be illustrated using a table or graph, with their formal proof left for a future course.

A. Establishing the Limit Properties While our work in Section 12.1 seems plausible, using a table of values to draw conclusions can sometimes be misleading (see objective C that follows), and it seems we need a more definitive method for evaluating limits. As it turns out, limits possess certain properties that, once established, enable us to evaluate even complicated limits with complete confidence. The properties we’re interested in involve the various ways that limits can be combined using basic operations. For instance, from our study of operations on functions, we know 1f ⫹ g21x2 ⫽ f 1x2 ⫹ g1x2 . Is it similarly true that lim 3f 1x2 ⫹ g1x2 4 ⫽ lim f 1x2 ⫹ lim g1x2 ? While a formal proof of xSc

xSc

xSc

the properties that follow will be offered in a first calculus course, for the time being we will accept them as true after demonstrating their plausibility using tables and our intuition (despite their shortcomings).

EXAMPLE 1

䊳

Combining Limits to Establish Limit Properties Using a table of values and the techniques from Section 12.1, it can be shown that for f 1x2 ⫽ x and g1x2 ⫽ ⫺x2 ⫹ 7, lim f 1x2 ⫽ 3 and lim g1x2 ⫽ ⫺2. Create a new function

h1x2 ⫽ 1f ⫹ g21x2 and use the TABLE feature of a graphing calculator to investigate whether lim 3f 1x2 ⫹ g1x2 4 ⫽ lim f 1x2 ⫹ lim g1x2 . xS3

xSc

xSc

xS3

xSc

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Section 12.2 The Properties of Limits

Solution

䊳

For f and g as given, we have h1x2 ⫽ x ⫹ 1⫺x2 ⫹ 72 . This means we need to investigate lim h1x2 ⫽ lim 3x ⫹ 1⫺x2 ⫹ 72 4 . With Y1 ⫽ X ⫹ 1⫺X2 ⫹ 72, the xS3

xS3

tables shown in Figures 12.18 and 12.19 seem to indicate lim h1x2 ⫽ 1. xS3

Figure 12.18

Figure 12.19

It appears that such a property exists, as the tables suggest

lim 3 f 1x2 ⫹ g1x2 4 ⫽ lim f 1x2 ⫹ lim g1x2

xS3

xS3

1 ⫽ 3 ⫹ 1⫺22 1⫽1✓

xS3


䊳

Other investigations might involve whether limits have a “multiplicative” property. In lim g1x2 g1x2 xSc ⫽ other words, is lim 3f 1x2g1x2 4 ⫽ lim f 1x2 lim g1x2 ? Or, is lim a true xSc xSc xSc xSc f 1x2 lim f 1x2 xSc statement? We explore these questions in Example 2. EXAMPLE 2

䊳

Combining Limits to Establish Limit Properties Using a table of values and the techniques from Section 12.1 suggests that for f 1x2 ⫽ x2 and g1x2 ⫽ ⫺3x, lim f 1x2 ⫽ 4 and lim g1x2 ⫽ 6. Note this yields xS⫺2

lim f 1x2

xS⫺2

xS⫺2

lim g1x2

#

lim g1x2 ⫽ 4162 ⫽ 24 and

xS⫺2

xS⫺2

lim f 1x2

⫽

6 3 ⫽ . Use this information to 4 2

xS⫺2

a. Create a new function m1x2 ⫽ f 1x2g1x2 to investigate whether lim m1x2 ⫽ lim f 1x2 lim g1x2 . xS⫺2

xS⫺2

b. Create a new function v1x2 ⫽

g1x2

to investigate

xS⫺2

䊳

a. For the product of f and g we have m1x2 ⫽ 1x2 2 1⫺3x2 ⫽ ⫺3x3. Now using a single table to view the approach from the left and from the right (indicated by the arrows), Table 12.1 suggests lim m1x2 ⫽ lim 1⫺3x3 2 ⫽ 24. xS⫺2

xS⫺2

x

m(x)

⫺1.9

20.577

⫺1.99

23.642

⫺1.999

23.964

xS⫺2

f 1x2 lim g1x2 xS⫺2 whether lim v1x2 ⫽ . xS⫺2 lim f 1x2

Solution

Table 12.1

⫺2 ⫺2.001

24.036

⫺2.01

24.362

⫺2.1

27.783

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WORTHY OF NOTE While the limits illustrated in Tables 12.1 and 12.2 used functions that were defined at x ⫽ ⫺2, remember that the limit does not depend on the functions being defined there.

b. For the quotient of g and f we have ⫺3x 3 v1x2 ⫽ 2 ⫽ ⫺ 1x ⫽ 02 . Table 12.2 indicates x x ⫺3 3 lim v1x2 ⫽ lim ⫽ . xS⫺2 xS⫺2 x 2 The tables seem to indicate that such properties exist, as they suggest lim 3f 1x2g1x2 4 ⫽ lim f 1x2 # lim g1x2 ⫽ 24, xS⫺2

g1x2

xS⫺2

xS⫺2

lim g1x2

3 ⫽ ⫽ . and lim xS⫺2 f 1x2 2 lim f 1x2 xS⫺2

Table 12.2 x

v(x)

⫺1.9

1.5789

⫺1.99

1.5075

⫺1.999

1.5008

⫺2 ⫺2.001

1.4993

⫺2.01

1.4925

⫺2.1

1.4286

xS⫺2


䊳

The preceding observations seem to imply that the limit of a product can be found by taking the limit of each factor, and the limit of a quotient is equal to the quotient of the limits (if the limit of the denominator is not 0). In fact, each of these observations (and others) can be generalized and formally established using the definition of a limit. Based on these observations, we might expect that limit properties exist for the other basic operations, and this is indeed the case. The end result is the following seven properties. Properties of Limits

Given that the limits lim f 1x2 and lim g1x2 exist, xSc

xSc

(I) lim 3f 1x2 ⫹ g1x2 4 ⫽ lim f 1x2 ⫹ lim g1x2 xSc

xSc

xSc

the limit of a sum is the sum of the limits (II) lim 3f 1x2 ⫺ g1x2 4 ⫽ lim f 1x2 ⫺ lim g1x2 xSc

xSc

xSc

the limit of a difference is the difference of the limits (III) lim 3f 1x2g1x2 4 ⫽ lim f 1x2 lim g1x2 xSc

xSc

xSc

the limit of a product is the product of the limits (IV) lim 3kg1x2 4 ⫽ k lim g1x2 , k a constant xSc

WORTHY OF NOTE A formal proof of these properties, requiring a more sophisticated approach to limits, is offered in a calculus course.

xSc

the limit of a constant times a function is the constant times the limit of the function lim f 1x2 f 1x2 xSc ⫽ (V) lim , provided lim g1x2 ⫽ 0 xSc g1x2 xSc lim g1x2 xSc

the limit of a quotient is the quotient of the limits By repeatedly applying property III with f 1x2 ⫽ g1x2 , we obtain the property related to the limit of a power. (VI) lim 3f 1x2 4 n ⫽ 3 lim f 1x2 4 n, for n a natural number xSc

xSc

the limit of a power is the power of the limit A similar property holds for the limit of an nth root.

(VII) lim 2f 1x2 ⫽ 2 lim f 1x2 , for n 僆⺞, n 7 1 [if n is even, then f 1x2 7 0] n

A. You’ve just seen how we can establish limit properties using a table

xSc

n

xSc

the limit of an nth root is the nth root of the limit

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B. Finding Limits Using the Limit Properties Before we begin applying these properties to evaluate limits, we will accept the validity of the following basic limits, most of which are supported by the previous examples. These are used extensively in our application of the limit properties. Basic Limits 1. lim k ⫽ k

2. lim x ⫽ c

3. lim xn ⫽ cn, n a natural number

4. lim 1x ⫽ 1c, n a natural number,

xSc

xSc

n

xSc

n

xSc

n 7 1 (if n is even, then c 7 0)

EXAMPLE 3

䊳

Using Limit Properties to Determine a Limit Find the following limits using the limit properties. State the property that justifies each step. x4 ⫹ 3x2 ⫺ 5 a. lim 1x3 ⫹ 3x ⫺ 42 b. lim xS⫺3 xS2 7 ⫺ 4x

Solution

䊳

a. lim 1x3 ⫹ 3x ⫺ 42 ⫽ lim x3 ⫹ lim 3x ⫺ lim 4 xS⫺3

xS⫺3

xS⫺3

⫽ lim x3 ⫹ 3 lim x ⫺ lim 4

property IV

⫽ 1⫺32 ⫹ 31⫺32 ⫺ 4

basic limits 3, 2, and 1

xS⫺3 3

xS⫺3

xS⫺3

⫽ ⫺40 x4 ⫹ 3x2 ⫺ 5 b. lim ⫽ xS2 7 ⫺ 4x ⫽

result

lim 1x4 ⫹ 3x2 ⫺ 52

xS2

lim 17 ⫺ 4x2

lim x ⫹ lim 3x2 ⫺ lim 5

xS2

xS2

properties I and II

xS2 2

lim x ⫹ 3 lim x ⫺ lim 5

xS2

xS2

xS2

lim 7 ⫺ 4 lim x

xS2

⫽

xS2

lim 7 ⫺ lim 4x

2 ⫹ 312 2 ⫺ 5 4

⫽

property V

xS2 4

xS2 4

⫽

properties I and II

xS⫺3

property IV

xS2

2


7 ⫺ 4122

23 ⫽ ⫺23 ⫺1

result

Note the general strategy is to “break-up” the expressions using limit properties, then use the basic limits to evaluate the result. Now try Exercises 37 through 48

EXAMPLE 4

䊳

Using the Limit Properties to Determine a Limit Evaluate the following limits using the limit properties. 3 a. lim 2x3 ⫺ 7x ⫹ 28

xS⫺4

9 2x2 ⫹ 4 xS0 x2 ⫹ 5

b. lim

䊳

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Solution

䊳

3 3 3 a. lim 2 x ⫺ 7x ⫹ 28 ⫽ 2 lim 1x3 ⫺ 7x ⫹ 282

xS⫺4

property VII

xS⫺4

3 ⫽2 lim x3 ⫺ lim 7x ⫹ lim 28

properties I and II

3 ⫽2 lim x3 ⫺ 7 lim x ⫹ lim 28

property IV

3 ⫽2 1⫺42 3 ⫺ 71⫺42 ⫹ 28


xS⫺4

xS⫺4

xS⫺4

xS⫺4

xS⫺4

xS⫺4

⫽ 2⫺8 ⫽ ⫺2 3

9 2x2 ⫹ 4 ⫽ b. lim xS0 x2 ⫹ 5 ⫽

lim 19 2x2 ⫹ 42

xS0

lim 1x2 ⫹ 52

result

property V

xS0

9 lim 2x2 ⫹ 4 xS0

lim 1x2 ⫹ 52

property IV

xS0

⫽

9 2 lim 1x2 ⫹ 42 xS0

lim 1x2 ⫹ 52

property VII

xS0

⫽

9 2 lim x2 ⫹ lim 4 xS0 2

xS0

lim x ⫹ lim 5

xS0

properties I and II

xS0

9 202 ⫹ 4 02 ⫹ 5 9122 18 ⫽ ⫽ 5 5

⫽


result


EXAMPLE 5

䊳

Using Limit Properties to Determine Limits Find the following limits using the limit properties. If a limit does not exist, use the TABLE feature of a graphing calculator to state why. x2 x2 ⫹ 2x x⫺9 a. lim 2 b. lim c. lim 2 xS⫺1 x ⫹ 2x ⫹ 1 xS⫺2 2x ⫹ 4x ⫹ 4 xS9 2x ⫺ 3

Solution

䊳

lim x2 x2 xS⫺1 ⫽ a. lim 2 xS⫺1 x ⫹ 2x ⫹ 1 lim 1x2 ⫹ 2x ⫹ 12

property V

xS⫺1

lim x2 ⫽

xS⫺1

lim x2 ⫹ lim 2x ⫹ lim 1

xS⫺1

⫽

xS⫺1

1⫺12 2

1⫺12 2 ⫹ 21⫺12 ⫹ 1 1 ⫽ 0

property I

xS⫺1

property IV and basic limits 3, 2, and 1 result

In this case, applying the limit properties results in an undefined expression, and we resort to a table of values to see what’s happening as x S ⫺1.

䊳

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Figure 12.20

1185

Figure 12.21

Figures 12.20 and 12.21 show that as x S ⫺1, function values grow infinitely x2 ⫽ A dne B . large and positive, and we conclude lim 2 q xS⫺1 x ⫹ 2x ⫹ 1 b. lim xS⫺2

x2 ⫹ 2x 2x2 ⫹ 4x ⫹ 4

⫽

lim 1x2 ⫹ 2x2

xS⫺2

property V

lim 2x2 ⫹ 4x ⫹ 4

xS⫺2

⫽

lim x2 ⫹ lim 2x

xS⫺2

xS⫺2

2 lim 1x ⫹ 4x ⫹ 42

properties I and VII

2

xS⫺2

⫽

lim x2 ⫹ lim 2x

xS⫺2

xS⫺2

lim x B A xS⫺2

2

xS⫺2

⫹ 2 lim x xS⫺2

2 A lim x B ⫹ 4 lim x ⫹ lim 4

properties VI and IV

2

xS⫺2 2

⫽

property I

2 lim x ⫹ lim 4x ⫹ lim 4 xS⫺2

⫽

xS⫺2

2

xS⫺2

1⫺22 ⫹ 21⫺22

xS⫺2

21⫺22 ⫹ 41⫺22 ⫹ 4 0 ⫽ 0 2

⫽

0

basic limits 2 and 1

20

result

Here, application of the limit properties results in an indeterminate expression, and we again resort to a table of values to check the behavior of the function as x S ⫺2. Figure 12.23

Figure 12.22

Figures 12.22 and 12.23 imply that lim ⫺ xS⫺2

x2 ⫹ 2x

x2 ⫹ 2x 2x2 ⫹ 4x ⫹ 4

⫽ 2 (the left-hand

⫽ ⫺2 (the right-hand limit is ⫺2). 2x2 ⫹ 4x ⫹ 4 Since these do not agree, the limit does not exist and we write x2 ⫹ 2x lim ⫽ A dne B . LH⫽RH xS⫺2 2x2 ⫹ 4x ⫹ 4 limit is 2), while lim ⫹ xS⫺2

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c. lim xS9

x⫺9 2x ⫺ 3

⫽

lim 1x ⫺ 92

xS9

lim 1 2x ⫺ 32

property V

xS9

lim x ⫺ lim 9

⫽

xS9

xS9

lim 2x ⫺ lim 3

xS9

9⫺9

⫽

property II

xS9

⫽

9⫺9 3⫺3


29 ⫺ 3 0 ⫽ result 0 Once again the result is an indeterminate expression and we use a table of values to check the behavior of the function as x S 9. To our surprise, the tables in Figures 12.24 and 12.25 indicate that the limit is 6, and we conclude x⫺9 ⫽ 6. lim xS9 2x ⫺ 3 Figure 12.24 Figure 12.25


B. You’ve just seen how we can evaluate limits using the limit properties

䊳

k As a summary note, Example 5(a) demonstrates why expressions of the form are said 0 to be undefined, as any limit resulting in this form will approach positive or negative 0 infinity. Similarly, Examples 5(b) and 5(c) show why expressions of the form are 0 said to be indeterminate, as the existence or final value of the original limit cannot be determined without further investigation.

C. Distinguishing between a Declarative Statement and a Proof Consider the function p1x2 ⫽ x2 ⫺ x ⫹ 41. Evaluating p for the natural numbers 1 through 7 reveals that all outputs are prime numbers (Table 12.3). Made curious by this discovery, we investigate further and find that values of x from 8 through 14 also generate prime numbers (Table 12.4), as do the inputs 15 through 40. At this point it might Table 12.4

Table 12.3 x

x

p(x)

p(x)

1

41

8

97

2

43

9

113

3

47

10

131

4

53

11

151

5

61

12

173

6

71

13

197

7

83

14

223

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1187

seem reasonable to declare that p generates a prime number for all natural numbers n. This exercise points out the dangers of making a declarative statement without formal proof, because when x ⫽ 41, the function returns a value of 1681, which is not a prime number (41 # 41 ⫽ 1681). Also, in Section 11.4 we noted 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫽ 16 (first 4 odd numbers)

1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫽ 36 (first 6 odd numbers)

1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ 13 ⫹ 15 ⫹ 17 ⫹ 19 ⫹ 21 ⫹ 23 ⫽ 144 (first 12 odd numbers)

and it appeared the sum of the first n odd numbers was equal to n2. While this declaration turns out to be true, no number of finite checks can prove the statement is true for all n, and a formal proof had to be rendered using mathematical induction. In exactly x2 ⫺ 9 ⫽ 6, using values the same way, we can view a table of values to “declare” lim xS3 x ⫺ 3 closer and closer to 3, then even closer still, but prior to the development of the limit properties how could we prove the limit is 6? It was this type of question that prompted additional sarcasm from George Berkeley. “And what are these fluxions? The velocities of evanescent increments? And what are these same increments? They are neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not then call them the ghosts of departed quantities?” Example 6 demonstrates that in the absence of a formal proof, it appears Mr. Berkeley’s skepticism was justified and that a table of values is insufficient “proof.” EXAMPLE 6

䊳

Evaluating Limits Using a Table Evaluate lim ax2 ⫹ xS0

Solution

䊳

cos x ⫹ 1b using the TABLE feature of a graphing calculator. 1000

For the left- and right-hand limits our graphing calculator provides Figures 12.26 and 12.27 respectively, and it appears we can declare that lim f 1x2 ⫽ 1. xS0

Figure 12.26

C. You’ve just seen how we can distinguish between a declarative statement and a proof

Figure 12.27

However, if we continue the approach using values of x that are even closer to c ⫽ 0, we find f 10.0012 1.0010001, f 10.00012 1.00100001, and it turns out the limit is actually 1.001. Our declarative statement then has no standing and our work falls far short of a proof. Now try Exercises 69 through 72

䊳

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12.2 EXERCISES 䊳



1. The limit of a sum is the

of the

3. The limit of an nth

is the

. root of

the limit: lim 2f 1x2 ⫽ 2 lim f 1x2 , provided n

2. The limit of a product is the 4. The limit of a

is the quotient of the f 1x2 lim f 1x2 xSc , lim , provided ⫽ xSc g1x2 lim g1x2

n

xSc

xSc

if n is even.

xSc

5. Discuss/Explain how applying the limit properties can result in an undefined or indeterminate expression. What should be done in these cases? 䊳

of the limits.

6. Discuss/Explain the relationship between limit properties VI and VII, and basic limits 3 and 4. How is basic limit 2 involved here?


The following functions will be used for Exercises 7 through 36.

a1x2 ⫽ 3 ⫺ 2x c1x2 ⫽ 1 ⫺ x2 f 1x2 ⫽ 1x ⫹ 7 h1x2 ⫽ sin1x ⫹ 32 k1x2 ⫽ x2 ⫺ 16

b1x2 ⫽ x ⫺ 2 d1x2 ⫽ 5x ␲ g1x2 ⫽ cosa b 2x j 1x2 ⫽ log3 15x2 2 4

l1x2 ⫽ 2x2 ⫺ 10x ⫹ 25

For Exercises 7 through 16, use the TABLE feature of a graphing calculator to evaluate each limit.

7. lim a1x2

8. lim b1x2

xS⫺3

xS2

20. lim 3b1x2 ⫹ d1x2 ⫹ j1x2 4 xS2

21. lim 3h1x2 ⫺ c1x2 4 xS⫺3

xS⫺3

24. lim 3j1x2 ⫺ d1x2 ⫹ l1x2 4 xS2

25. lim 3k1x2 ⫹ k1x2 4 xS⫺3

xS2

11. lim f 1x2

12. lim g1x2

xS⫺3

xS2

13. lim h1x2

14. lim j1x2

xS⫺3

xS2

15. lim k1x2

16. lim l1x2

xS⫺3

xS2

For Exercises 17 through 26, use the TABLE feature of a graphing calculator to evaluate each limit. Compare each result with that of the corresponding Exercises in 7 through 16.

17. lim 3 a1x2 ⫹ c1x2 4 xS⫺3

18. lim 3 d1x2 ⫹ g1x2 4

19. lim 3 f 1x2 ⫹ h1x2 ⫹ k1x2 4 xS⫺3

xS2

26. lim 3d1x2 ⫹ d1x2 4 xS2

For Exercises 27 through 36, use a table of values to evaluate each limit. Compare each result with that of the corresponding Exercises in 7 through 16.

27. lim 3c1x2f 1x2 4

10. lim d1x2

xS⫺3

xS2

23. lim 3f 1x2 ⫹ k1x2 ⫺ a1x2 4

xS⫺3

9. lim c1x2

22. lim 3b1x2 ⫺ g1x2 4

29. lim xS⫺3

k1x2 f 1x2

28. lim 3j1x2d1x2 4 xS2

30. lim xS2

d1x2 g1x2

31. lim 3a1x2h1x2k1x2 4

32. lim 3b1x2g1x2l1x2 4

33. lim 3c1x2c1x2 4

34. lim 3d1x2d1x2 4

xS⫺3

xS⫺3

35. lim xS⫺3

a1x2c1x2 f 1x2

xS2

xS2

36. lim xS2

d1x2 g1x2j1x2

For Exercises 37 through 56, evaluate the limits using the limit properties.

37. lim 1x3 ⫺ 52

38. lim 1 1x ⫺ 162

39. lim 12x ⫺ x ⫺ 72

40. lim 1x4 ⫺ 2x ⫹ 52

xS⫺4

2

xS⫺4

xS16

xS2

.

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41. lim 1x2 ⫺ 5 1x ⫹ 32

For Exercises 69 and 70, (a) complete the table given for each function and state the apparent limit as x S 0, then (b) continue the approach using values even closer to 0 and comment.

xS4

1 3 42. lim a x2 ⫺ 1 x ⫹ 3b xS⫺8 2 2x2 ⫺ 5 43. lim xS2 x ⫹ 3

x2 ⫺ 2x ⫹ 3 44. lim xS⫺3 x⫺3

1 ⫺ x2 x⫹1 45. lim xS1 1 ⫹ 2x2 ⫺ x x

1 ⫹ 3x2 x⫺2 46. lim xS3 2 1 x ⫺6⫹ x⫺1

47. lim 1x2 ⫺ 32 3 xS3

48. lim 13x2 ⫺ 2x2 2 xS2

49. lim 5 22x ⫹ 7

50. lim ⫺22x ⫺ 6

51. lim 1 1x ⫹ 7 ⫺ 7x2

52. lim 12x ⫹ 14 ⫺ x2

3

xS10

xS⫺3

53. lim xS2

x3 ⫺ 2x ⫺ 10 3 22 5x2 ⫹ 2x ⫹ 3

2 ⫺ 3x x⫹1 55. lim xS3 1x ⫺ 2 ⫹ 1

3

xS⫺5

54. lim xS2

11 ⫺ 3x2 2x2 ⫹ 3x ⫺ 1

2 x a b ⫺ 3x 2x ⫹ 1 56. lim 2 xS⫺1 2x ⫹ 3 ⫹ 1

3x ⫺ 11x ⫺ 4 2 xS⫺2 x ⫺ 2x ⫺ 8 2

57. lim

x ⫺ 5x ⫺ 14 59. lim xS7 x⫺7 2

x2 ⫹ 5x ⫹ 6 2 xS⫺2 x ⫹ 4x ⫹ 4

61. lim

2x ⫺ x ⫺ 3 x2 ⫺ 1 2

58. lim xS1

x2 ⫹ 7x ⫹ 12 60. lim xS⫺4 2x ⫹ 8 62. lim xS1

x2 ⫺ 4x ⫹ 3 x2 ⫺ 2x ⫹ 1

2x ⫹ 14x ⫹ 49 x2 ⫹ 8x ⫹ 7

xS⫺7

64. lim

2x2 ⫹ 2x ⫹ 1 1x ⫹ 32 2 ⫺ 9 65. lim x xS0 x3 ⫺ x2

2x ⫺ 2x ⫹ 1 x3x ⫺ 1 68. lim xS1 x ⫺ 1 xS1

2

x

f (x)

x

f (x)

0.5

⫺0.5

0.4

⫺0.4

0.3

⫺0.3

0.2

⫺0.2

0.1

⫺0.1

0.01

⫺0.01

0.001

⫺0.001 2

1000x 3 ⫺ 1 70. f 1x2 ⫽ 1000 x

f (x)

x

0.5

⫺0.5

0.4

⫺0.4

0.3

⫺0.3

0.2

⫺0.2

0.1

⫺0.1

0.01

⫺0.01

0.001

⫺0.001

71. a. Complete the table shown. yⴝxⴙ

x

1x ⴚ 2 10 x

2.7 2.8

3.1

2x2 ⫺ x ⫺ 3

xS⫺1

sin x 1000x

2.9

2

63. lim

69. f 1x2 ⫽ x2 ⫹

xS⫺2

For Exercises 57 through 68, evaluate the limits using limit properties. If a limit does not exist, state why.

67. lim

1189


3.2

1x ⫺ 12 3 ⫹ 1 66. lim x xS0

3.3

b. Complete the table shown. x 2.99 2.999 2.9999 3.0001 3.001 3.01

yⴝxⴙ

1x ⴚ 2 10 x

f (x)

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c. Based on the table in part (a), estimate 1x ⫺ 2 b. lim ax ⫹ xS3 10x

b. Complete the table shown. x

⫺1.9999 ⫺1.99999 ⫺2.00001 ⫺2.0001

72. a. Complete the table shown. ⴝ 2x ⴙ

⫺2.001

xⴙ3 401x ⴚ 32 2

c. Based on the table in part (a), estimate

⫺1.7 ⫺1.8

.

⫺1.9 ⫺2.1

d. Based on the table in part (b), what appears to be x⫹3 the actual value of lim c 2x ⫹ d? xS⫺2 401x ⫺ 32 2 Verify the limit using limit properties.

⫺2.2 ⫺2.3

䊳

xⴙ3 401x ⫺ 32 2

⫺1.999

d. Based on the table in part (b), what appears to be 1x ⫺ 2 b? Given the actual value of lim ax ⫹ xS3 10x lim 10x ⫽ 103, verify the limit using limit xS3 properties.

x

y ⴝ 2x ⴙ


73. (3.2) If p1x2 ⫽ 2x2 ⫺ x ⫺ 3, in what interval(s) is p1x2 ⱕ 0?

75. (7.2) Verify the identity: sec y ⫺ cos y ⫽ tan y sin y

74. (5.5) Solve the logarithmic equation log1x ⫹ 22 ⫹ log x ⫽ log 3.

76. (7.7) Use any appropriate method to solve in [0, 2␲): sin 2x ⫽ cos x.

MID-CHAPTER CHECK 1. Express the following statement using limit notation: 6x ⫺ 3 12x3 ⫺ x ⫺ 1 2

As x decreases without bound, approaches 0.

Use the TABLE feature of a graphing calculator to complete Exercises 2 through 7. 2. Evaluate each limit. sin h a. lim hS0 h

cos h ⫺ 1 hS0 h

b. lim

3. The formula for interest compounded n times per r nt year is A ⫽ P a1 ⫹ b , where P represents the n

principal deposit, r is the annual interest rate, t is the number of years the funds are on deposit, and n is the number of compoundings per year. If we let n S q , this formula can be transformed into the formula for interest compounded continuously: A ⫽ Pert. Verify that $1000 invested at 3% compounded continuously for 1 yr grows to $1030.45. Then find the minimum number of compoundings per year needed to yield the same amount with the same $1000 principal. 4. The function f(x) shown has a pointwise discontinuity. Create a related piecewise-defined function F(x) that is continuous for x 僆⺢. f 1x2 ⫽

6x2 ⫺ 19x ⫺ 7 2x ⫺ 7

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Section 12.3 Continuity and More on Limits

⫺x 5. Given g1x2 ⫽ • 2x ⫹ 5 cos1x ⫹ 52 following: a. lim ⫺ g1x2 b. xS⫺5

x 6 ⫺5

b. Based on your table, it appears that ␲ lim cosa b is approaching what value? x xS0 ␲ Graph y ⫽ cosa b on a graphing calculator, x and “Zoom In” a few times. Is your prior observation still valid?

, find the

x ⱖ ⫺5 lim g1x2

xS⫺5 ⫹

c. lim g1x2 xS5

6. Given h1x2 ⫽ cota

x2 ⫺ 4 b, find lim h1x2 . x⫹2 xS2

For Exercises 8 through 10, evaluate each limit using the limit properties. If a limit does not exist, state why.

7. a. Complete the table of values shown. x

␲ y ⴝ cos a b x

0.1 0.01 0.001

8. lim xS2

4x2 ⫹ x ⫺ 5 x5 ⫺ 3x

9. lim

2x ⫹ 10

3 22 11 ⫺ x2 2x2 ⫺ 4x ⫹ 4 10. lim xS2 x⫺2

xS⫺3

⫺0.001 ⫺0.01 ⫺0.1

12.3

Continuity and More on Limits


A. Determine if a function is continuous at c and find limits by direct substitution B. Evaluate limits using algebra and the properties of limits C. Evaluate limits at infinity D. Use the definition of a limit to evaluate limits graphically

Historically, the careful development of the limit concept also affected the concept of continuity and the two seem to have matured simultaneously. In this section, we’ll see how limits help define the continuity of a function more precisely, and once established, how continuity makes finding certain limits an easier task. We’ll also see how the algebraic skills developed in previous courses (adding rational expressions, simplifying complex fractions, multiplying by conjugates, etc.) are used to help determine some very important limits.

A. Continuity and Finding Limits by Direct Substitution In previous chapters, we stated that a function was continuous if you could draw the entire graph without lifting your pencil. We also noted there were several ways that continuity could be interrupted, such as the holes and gaps seen in a study of piecewise-defined functions (Section 2.5), and the vertical asymptotes in some rational and trigonometric graphs. These “interruptions” are called discontinuities and can occur with other types of functions as well. The graph of f(x) given in Figure 12.28 shows a vertical asymptote at x ⫽ ⫺1, which interrupts the continuity of the graph and indicates that f is not defined at ⫺1.

Figure 12.28 y 4 3 2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

x ⫽ ⫺1

1

2

3

4

5

x

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This is called an asymptotic discontinuity. The graph also shows that continuity is interrupted at x ⫽ 1 (there is a “hole” in the graph), even though f is still defined there: f 112 ⫽ 3. This is called a removable discontinuity, because the discontinuity can be removed/repaired by redefining the function as f 1x2 ⫽ 1 when x ⫽ 1. The discontinuity at x ⫽ 4 is also removable, even though the function is not defined at 4. A break in the graph also occurs at x ⫽ 3, but this time it leaves a gap in the output values, and there is no way to redefine f to “fill the gap.” This is called a jump discontinuity or a nonremovable discontinuity. In fact, of all the points called out in Figure 12.28, the function is only continuous at x ⫽ ⫺2. By comparing the graph at x ⫽ ⫺2 with the graph of f at other points, we find the following conditions are required for continuity. Continuity A function f is continuous at c if 1. f is defined at c

2. lim⫺ f 1x2 ⫽ lim⫹ f 1x2 xSc

xSc

3. lim f 1x2 ⫽ f 1c2 xSc

In Figure 12.28, the discontinuities at x ⫽ ⫺1 and x ⫽ 4 violate condition 1, the discontinuity at x ⫽ 3 violates condition 2, and the discontinuity at x ⫽ 1 violates condition 3. EXAMPLE 1

䊳

y

The graph of a function f is given. Use the graph to comment on the continuity of the function for all integer values from ⫺2 to 3. If the function is discontinuous at any value, state which of the three conditions are violated.

WORTHY OF NOTE The continuity of a function really boils down to condition 3, since if this condition is satisfied, conditions 1 and 2 are satisfied.

Solution

Analyzing the Continuity of a Function Graphically

䊳

x⫽3

5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

x ⫽ ⫺2: discontinuous, condition 2 is violated

x ⫽ ⫺1: continuous

x ⫽ 0: discontinuous, condition 3 is violated

x ⫽ 1: continuous



1

2

3

4

5

x

⫺2 ⫺3


䊳

The fact that condition 3, lim f 1x2 ⫽ f 1c2 , is one of the requirements for continuity xSc offers a great advantage when evaluating the limit of many common functions. For example, using properties of limits it can be shown that polynomials are continuous for all real numbers, and rational, root, and trigonometric functions are continuous wherever they are defined. This means as long as c is in the domain of these functions, we can find the limit by direct substitution since lim f 1x2 ⫽ f 1c2 can be assumed. xSc

Finding Limits Using Direct Substitution If f is a polynomial, rational, root, or trigonometric function and c is in the domain of f, then lim f 1x2 ⫽ f 1c2

xSc

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EXAMPLE 2

䊳

1193

Finding Limits Using Direct Substitution Evaluate the following limits using direct substitution, if possible. a. lim 1x3 ⫹ 3x ⫺ 42 xS⫺3

Solution

䊳

b. lim xS2

x4 ⫹ 3x2 ⫺ 5 7 ⫺ 4x

x2 ⫹ 3x ⫺ 10 xS2 x⫺2

c. lim

a. Since x ⫹ 3x ⫺ 4 is a polynomial, the limit can be found by direct substitution: 3

lim 1x3 ⫹ 3x ⫺ 42 ⫽ 1⫺32 3 ⫹ 31⫺32 ⫺ 4

xS⫺3

⫽ ⫺27 ⫹ 1⫺92 ⫺ 4 ⫽ ⫺40

substitute ⫺3 for x simplify result

See Example 3 of Section 12.2 where this limit was found using limit properties. x4 ⫹ 3x2 ⫺ 5 b. Since lim is a rational function and 2 is in the domain, the limit xS2 7 ⫺ 4x can be found by direct substitution: 24 ⫹ 3122 2 ⫺ 5 x4 ⫹ 3x2 ⫺ 5 ⫽ lim xS2 7 ⫺ 4x 7 ⫺ 4122 16 ⫹ 12 ⫺ 5 ⫽ 7⫺8 ⫽ ⫺23

A. You’ve just seen how we can determine if a function is continuous at c and find limits by direct substitution

substitute 2 for x

simplify result

See Example 3 of Section 12.2 where this limit was found using limit properties. x2 ⫹ 3x ⫺ 10 c. Since lim is a rational function and 2 is not in the domain, the xS2 x⫺2 limit cannot be found by direct substitution. This doesn’t mean the limit doesn’t exist, only that direct substitution cannot be used. Applying the limit x2 ⫹ 3x ⫺ 10 properties shows that lim ⫽ 7 [also see Example 3(a)]. xS2 x⫺2 Now try Exercises 17 through 26

䊳

B. Evaluating Limits Using Algebra and Limit Properties

Recall that for lim f 1x2 ⫽ L, the difference between x and c is made very small by xSc

taking values of x very close to c but not equal to c. In many cases, knowing that x⫺c ⫽ 1 (even if x ⫺ c is very small), can help evaluate limits where a direct substix⫺c tution is not initially possible. This is done by using algebra to rewrite the function prior to taking the limit.

EXAMPLE 3

䊳

Finding Limits Using Algebra and Limit Properties Evaluate the following limits. x2 ⫹ 3x ⫺ 10 a. lim xS2 x⫺2

b. lim xS5

1x ⫹ 4 ⫺ 3 x⫺5

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Solution

䊳

a. Although the function given is rational, x ⫽ 2 is not in the domain and the limit cannot be found using direct substitution. But noting the numerator is factorable, we can work as follows. lim xS2

1x ⫹ 521x ⫺ 22 x2 ⫹ 3x ⫺ 10 ⫽ lim x⫺2 xS2 x⫺2 1x ⫹ 521x ⫺ 22 ⫽ lim xS2 x⫺2 ⫽ lim 1x ⫹ 52

factor the numerator

factors reduce result is a polynomial

xS2

⫽2⫹5⫽7

evaluate by direct substitution

As in Example 2(c), a table of values will support this result. b. Since x ⫽ 5 is not in the domain, we again attempt to rewrite the expression in a form that would make direct substitution possible. Note that multiplying the numerator and denominator by 1x ⫹ 4 ⫹ 3 (the conjugate of 1x ⫹ 4 ⫺ 3) will eliminate the radical in the numerator since 1A ⫺ B21A ⫹ B2 ⫽ A2 ⫺ B2. This yields lim xS5

1 1x ⫹ 4 ⫺ 32 1 1x ⫹ 4 ⫹ 32 1x ⫹ 4 ⫺ 3 ⫽ lim x⫺5 xS5 1x ⫺ 521 1x ⫹ 4 ⫹ 32 3 1x ⫹ 42 ⫺ 9 4 ⫽ lim xS5 1x ⫺ 521 1x ⫹ 4 ⫹ 32 1x ⫺ 52 ⫽ lim xS5 1x ⫺ 521 1x ⫹ 4 ⫹ 32 1 ⫽ lim xS5 1x ⫹ 4 ⫹ 3 lim 1 xS5 ⫽ lim 1 1x ⫹ 4 ⫹ 32

multiply by

1x ⫹ 4 ⫹ 3 1x ⫹ 4 ⫹ 3

1A ⫺ B2 1A ⫹ B2 ⫽ A2 ⫺ B2 simplify numerator

factors reduce

property V

xS5

⫽

1 15 ⫹ 4 ⫹ 3


⫽

1 1 ⫽ 6 19 ⫹ 3

result

The result can be supported using a table of values. Now try Exercises 27 through 36

䊳

In Example 4, we need to find the limit as h S 0 rather than x S c, but the concepts are h identical a ⫽ 1 as long as h is not 0b. If h is not in the domain, the limit cannot be h found by direct substitution, and we again attempt to write the expression in an alternative form.

EXAMPLE 4

䊳

Finding Limits Using Algebra and Limit Properties Evaluate the following limits. a. lim hS0

3 1x ⫹ h2 2 ⫹ 31x ⫹ h2 4 ⫺ 1x2 ⫹ 3x2 h

1 1 ⫺ x x⫹h b. lim hS0 h

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Section 12.3 Continuity and More on Limits 䊳

Solution

1195

a. Here we begin by simplifying the numerator. lim

3 1x ⫹ h2 2 ⫹ 31x ⫹ h2 4 ⫺ 1x2 ⫹ 3x2 h

hS0

x ⫹ 2xh ⫹ h ⫹ 3x ⫹ 3h ⫺ x2 ⫺ 3x hS0 h 2

2

⫽ lim

2xh ⫹ h2 ⫹ 3h hS0 h h12x ⫹ h ⫹ 32 ⫽ lim hS0 h

⫽ lim

square binomial, distribute 3, distribute ⫺1

combine like terms

factor out h

⫽ lim 12x ⫹ h ⫹ 32

factors reduce since h ⫽ 0

⫽ 2x ⫹ 3


hS0

b. Here we begin by simplifying the complex fraction, hoping to rewrite the expression in a form that allows a direct substitution. 1x ⫹ h2 1 1 x ⫺ ⫺ x x⫹h x1x ⫹ h2 x1x ⫹ h2 lim ⫽ lim hS0 h hS0 h x ⫺ 1x ⫹ h2 ⫽ lim

hS0

x1x ⫹ h2 h

⫺h x1x ⫹ h2 ⫽ lim hS0 h ⫺h #1 x1x ⫹ h2 h ⫺1 ⫽ lim hS0 x1x ⫹ h2 ⫺1 ⫽ 2 x ⫽ lim

LCD for the numerator is x 1x ⫹ h2

combine terms in the numerator

simplify

invert and multiply

hS0

B. You’ve just seen how we can evaluate limits using algebra and the properties of limits

factors reduce since h ⫽ 0



䊳

C. Evaluating Limits at Infinity Figure 12.29 y 10 8 6 4 2 ⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

h(x) 2

4

6

8 10

x

The limit of a function as x S c necessarily has x moving infinitely close to a number c from the left or right. In contrast, limits at infinity are concerned with the value of a function as x moves infinitely away from an arbitrary number c to the left or right, meaning as x S ⫺q or x S q . 4 The graph of h1x2 ⫽ 2 is shown in Figure 12.29, and we note as x S q , x ⫹1 values of h(x) become closer and closer to 0. Similar to our previous work, we could say that values of h(x) can be made arbitrarily close to 0, by taking values of x sufficiently large and positive. This suggests the following definition.

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Limits at Positive Infinity For a function f defined on an interval (c, q ), lim f 1x2 ⫽ L xSq

means that values of f(x) can be made arbitrarily close to L, by taking values of x sufficiently large and positive. In the case illustrated in Figure 12.29, we have lim xSq

EXAMPLE 5

䊳

4 ⫽ 0. x2 ⫹ 1

Evaluating Limits at Infinity 3x2 ⫺ x ⫺ 12 , use the TABLE feature of a graphing calculator to 2x2 ⫺ 8 evaluate lim f 1x2 .

For f 1x2 ⫽

xSq

Solution

䊳

Begin by entering f (x) as Y1 on the Y= screen of a graphing calcualtor. From our work with rational functions in Chapter 4 (ratio of leading terms), we 3 suspect lim f 1x2 ⫽ . This is supported by the xSq 2 TABLE shown. Now try Exercises 45 through 52

䊳

The limit of a function as x S ⫺q is likewise defined.

WORTHY OF NOTE Since q is not a number, the notation x S q should no longer be read, “as x approaches infinity,” and instead we say, “as x becomes infinitely large,” or “as x increases without bound.”

Limits at Negative Infinity For a function f defined on an interval (⫺q , c)

lim f 1x2 ⫽ L

xS⫺q

means that values of f(x) can be made arbitrarily close to L, by taking values of x sufficiently large and negative.

EXAMPLE 6

䊳

Finding Limits at Infinity 1 For f 1x2 ⫽ , evaluate lim f 1x2 and lim f 1x2 . x xS⫺q xSq

Solution

䊳

You might recognize this function from our work in Chapter 2, where we observed that as x becomes very 1 large, becomes very small and close to zero (see x figure). The result was a horizontal asymptote at y ⫽ 0 (the x-axis). In fact, we can make f (x) as close to 0 as we like, by taking x to be a sufficiently large positive number, or a sufficiently large negative 1 ⫽ 0 and number. In limit notation, we have lim xS⫺q x 1 lim ⫽ 0. xSq x

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3 ⫺4 ⫺5


䊳

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1197

From Example 6 we learn two important things. First, horizontal asymptotes can be defined in terms of limits at infinity. Horizontal Asymptotes The line y ⫽ L is a horizontal asymptote if lim f 1x2 ⫽ L

or

xS⫺q

lim f 1x2 ⫽ L

xSq

Second, by repeatedly applying limit property VI (the limit properties also hold for limits at infinity), we can state the following result. Limits of Reciprocal Powers For any positive integer k, lim xS⫺q

1 ⫽0 xk

and

lim xSq

1 ⫽0 xk

These limits are a great asset to finding limits at infinity. We often need to rewrite an expression and apply these limits of reciprocal powers in order to find the originally desired limit. For rational functions, this involves dividing numerator and denominator by the highest power of x occurring in the denominator. EXAMPLE 7

䊳

Finding Limits at Infinity for a Rational Function 5x2 ⫺ 6x ⫹ 3 . 2 xSq 4x ⫺ 12x ⫹ 9 Begin by dividing the numerator and denominator by x2, the highest power of x in the denominator. 5x2 6x 3 ⫺ 2 ⫹ 2 2 2 5x ⫺ 6x ⫹ 3 x x x ⫽ lim lim divide numerator and denominator by x2 2 2 xSq 4x ⫺ 12x ⫹ 9 xSq 4x 12x 9 ⫺ 2 ⫹ 2 x2 x x 6 3 5⫺ ⫹ 2 x x simplify ⫽ lim xSq 12 9 4⫺ ⫹ 2 x x 6 3 lim a5 ⫺ ⫹ 2 b x xSq x property V ⫽ 12 9 lim a4 ⫺ ⫹ 2b x xSq x 6 3 lim 5 ⫺ lim ⫹ lim 2 xSq xSq x xSq x properties I and II ⫽ 12 9 lim 4 ⫺ lim ⫹ lim 2 xSq xSq x xSq x 5⫺0⫹0 5 k ⫽ ⫽ lim n ⫽ 0 xSq x 4⫺0⫹0 4 Evaluate lim

Solution

䊳

y 10 8 6 4 2 ⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2

4

6

8 10

x

5x2 ⫺ 6x ⫹ 3 5 A similar calculation for x S ⫺q shows lim is also . The 2 xS⫺q 4x ⫺ 12x ⫹ 9 4 graph is shown in the figure. Now try Exercises 61 through 64

䊳

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If the function contains a radical term in the numerator or denominator, we attempt to rewrite the expression as a single radical term. We can then apply limit property VII and treat the radicand as a rational function. When doing so, you must particularly note whether you’re taking the limit as x S ⫺q or as x S q , since this may affect the sign of the final result. EXAMPLE 8

䊳

Finding Limits at Infinity for a Radical Function Evaluate each limit: 8x a. lim xSq 24x2 ⫹ 1

Solution

䊳

b.

8x

lim xS⫺q

24x2 ⫹ 1

a. Since x is becoming an infinitely large positive number, we know 8x can be written as ⫹218x2 2 ⫽ ⫹ 264x2 (signs added for emphasis). The original expression can then be rewritten as follows. lim xSq

8x 24x2 ⫹ 1

⫽ lim

264x2

xSq

24x2 ⫹ 1

64x2 2 xSq B 4x ⫹ 1

⫽ lim ⫽

lim a

B xSq

64x2 b 4x2 ⫹ 1

8x ⫽ 264x2

quotient property of radicals

property VII

We are now taking the limit of a rational expression, so we next divide the numerator and denominator by x2. 64x2 x2 ⫽ lim ± 2 ≤ xSq 4x 1 ⫹ 2 b x2 x ⫽

lim R

⫽

⫽

64

1 x2 lim 64 xSq

xSq

divide by x2

simplify

4⫹

1 lim a4 ⫹ 2 b R xSq x lim 64 xSq

1 lim 4 ⫹ lim 2 xSq x R xSq 64 ⫽ A4 ⫹ 0 ⫽ 116 ⫽ 4

property V

property I

lim xSq

result

k ⫽0 xn

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b. For part (b), x is becoming an infinitely large 8x negative number, so the sign of will 24x2 ⫹ 1 be negative for all such values. For this reason 8x must be written as ⫺218x2 2 ⫽ ⫺ 264x2. This is the only change to be made, and after

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3

applying algebra and the limit properties we ⫺4 8x ⫺5 find lim ⫽ ⫺4. The graph is xS⫺q 24x2 ⫹ 1 shown in the figure and shows that this function has two distinct horizontal asymptotes.

C. You’ve just seen how we can evaluate limits at infinity


䊳

D. Evaluating Limits Graphically As a complement to the equations and tables we’ve used to explore the limit concept, we now look at how these concepts appear graphically. This will offer a heightened understanding of important ideas, and an excellent summary of the definitions, properties, and relationships involved. The graphs of two general functions f and g are given (Figures 12.30 and 12.31, respectively) for use in the following examples. Figure 12.31

Figure 12.30 y

y

5

f (x)

5

g(x)

4

3

2

2

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

EXAMPLE 9

䊳

4

3

1 1

2

3

4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

⫺2

⫺2

⫺3

⫺3

⫺4

⫺4

⫺5

⫺5

1

2

3

4

5

x

Evaluating and Combining Function Values Using Graphs Use the graphs of f and g to perform the operations indicated. If any computation is not possible, state why. a. f 1⫺12 ⫺ g142 b. g122 ⫹ f 142 c. f 1⫺32 ⫺ f 122 d. g122 g1⫺32

Solution

䊳

a. f 1⫺12 ⫺ g142 ⫽ 3 ⫺ 2 b. g122 ⫹ f 142 ⫽ 1 ⫹ 2 ⫽1 ⫽3 c. f 1⫺32 ⫺ f 122 ⫽ X d. g122 g1⫺32 ⫽ 112102 ⫽0 not possible since f (2) is not defined


䊳

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EXAMPLE 10

䊳

Evaluating One-Sided Limits Graphically Use the graphs of f and g to evaluate the one-sided limits necessary to complete each calculation. If any computation is not possible, state why. a. lim⫹ f 1x2 ⫹ lim⫺ g1x2

b. lim⫹ f 1x2 ⫺

c.

d.

xS1

Solution

䊳

a. c.

xS4

xS3

lim g1x2 lim ⫹ f 1x2

xS⫺2 ⫹

xS⫺3

lim f 1x2 ⫹ lim⫺ g1x2 ⫽ 0 ⫹ 2

xS1 ⫹

lim ⫹ g1x2

xS⫺2

xS3

⫽2 lim ⫹ f 1x2 ⫽ X

xS⫺3

xS5

lim g1x2 ⫽ 1 ⫺ 1⫺32

xS⫺5 ⫹

⫽4 lim ⫹ f 1x2 ⫹ lim⫹ g1x2 ⫽ X

d.

xS⫺1

not possible since lim f 1x2 ⫽ A dne B

xS5

not possible since 5 is not in the domain of g

⫺q

xS⫺3 ⫹

lim f 1x2 ⫹ lim⫹ g1x2

xS⫺1 ⫹

b. lim⫹ f 1x2 ⫺

xS4

lim g1x2

xS⫺5 ⫹


EXAMPLE 11

䊳

䊳

Evaluating Limits Graphically Use the graphs of f and g to evaluate the limits necessary to complete each calculation. If any computation is not possible, state why. a. lim 3f 1x2 ⫹ g1x2 4

b. lim 33f 1x2 ⫺ g1x2 4

xS⫺2

c. lim xS1

Solution

䊳

xS⫺1

d. lim 3 1f 1x2 2 2 ⫹ 1g1x2 4

g1x2 f 1x2

xS2

a. lim 3f 1x2 ⫹ g1x2 4 ⫽ lim f 1x2 ⫹ lim g1x2 xS⫺2

xS⫺2

⫽1⫹3⫽4 b. lim 33f 1x2 ⫺ g1x2 4 ⫽ lim 3f 1x2 ⫺ lim g1x2 xS⫺1

xS⫺1

xS1

g1x2 f 1x2

property II

xS⫺1

property IV

⫽ 3122 ⫺ 3 ⫽ 3

result

xS⫺1

lim g1x2 ⫽

result

⫽ 3 lim f 1x2 ⫺ lim g1x2 xS⫺1

c. lim

property I

xS⫺2

property V (if lim f 1x2 ⫽ 0)

xS1

lim f 1x2

xS1

xS1

not possible since lim f1x2 ⫽ 0

limit of the denominator is 0

xS1

d. lim 3 1f 1x22 2 ⫹ 1g1x2 4 ⫽ lim 3f 1x2 4 2 ⫹ lim 1g1x2 xS2

xS2

⫽ 3 lim f 1x2 4 2 ⫹ 1 lim g1x2 xS2

D. You’ve just seen how we can use the definition of a limit to evaluate limits graphically

property I

xS2

⫽ 122 2 ⫹ 14 ⫽ 6

properties VI and VII

xS2

result


䊳

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1201


12.3 EXERCISES 䊳



1. There are three types of discontinuities we may encounter. These are (1) , (2) and (3) discontinuities.

,

3. If f (x) is continuous at c, we can evaluate lim f 1x2 xSc by . 5. Discuss/Explain the relationship between the horizontal asymptote of a rational function f, and lim f 1x2 .

2. In order for a function f (x) to be continuous at c, it must be at c. Furthermore, lim⫺ f 1x2 xSc

with lim f 1x2 ⫽

must equal

.

xSc

4. The notation x S q is read, “as x becomes ,” or “as x increases without

.”

6. Discuss/Explain how the three different types of discontinuities appear on the graph of a function.

xSq

䊳


For Exercises 7 through 16, use the given graphs to comment on the continuity of the function at the given x-values. If the function is discontinuous at any value, state which of the three conditions for continuity are violated. y 5 4 3 2 1 ⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

y 5 4 3 2 1

f (x)

1 2 3 4 5 x

7. f(x) at x ⫽ ⫺1 9. f(x) at x ⫽ 1

⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

g(x)

x⫹1 2 xS⫺1 x ⫺ 1

24. lim

25. lim 2x2 ⫺ 6x

26. lim 13x ⫺ 5

xS⫺5

10. f(x) at x ⫽ 2

11. f(x) at x ⫽ 0

12. g(x) at x ⫽ ⫺1

13. g(x) at x ⫽ ⫺3

14. g(x) at x ⫽ 1

15. g(x) at x ⫽ ⫺2

16. g(x) at x ⫽ 0

Evaluate the following limits using direct substitution, if possible. If not possible, state why.

x⫹1 2 xS⫺1 x ⫺ 1 x⫺4 lim xS4 1x ⫺ 2 2x2 ⫺ 3x ⫺ 9 lim xS3 x⫺3 1x ⫹ 7 ⫺ 2 lim xS⫺3 x⫹3 3 x ⫹ 8x2 ⫹ 16x lim 2 xS⫺4 x ⫹ 7x ⫹ 12

28. lim

29.

30.

31. 33. 35.

18. lim 13x2 ⫺ 5x ⫺ 22

19. lim 13x ⫺ 19

20. lim 2x ⫹ 7x

38. lim

xS5

21. lim xS2

x 5x ⫺ 2

hS0

xS⫺2 xS⫺6

2

22. lim xS8

2x ⫺ 5 x2 ⫺ 5x

32. 34. 36.

Evaluate the following limits. Write your answer in simplest form.

17. lim 12x2 ⫺ 5x ⫹ 32

2

4 ⫺ x2 2 xS⫺2 x ⫹ 5x ⫹ 6 x ⫺ 16 lim xS16 4 ⫺ 1x 3x2 ⫹ 7x ⫹ 2 lim xS⫺2 x⫹2 12x ⫹ 1 ⫺ 5 lim xS12 x ⫺ 12 3 2x ⫺ 12x2 ⫹ 18x lim xS3 x2 ⫺ 7x ⫹ 12

27. lim

37. lim

xS⫺3

xS6

Evaluate the following limits by rewriting the given expression as needed.

1 2 3 4 5 x

8. f(x) at x ⫽ 3

4 ⫺ x2 2 xS⫺2 x ⫹ 5x ⫹ 6

23. lim

hS0

3 21x ⫹ h2 2 ⫺ 1x ⫹ h2 4 ⫺ 12x2 ⫺ x2 h

3 31x ⫹ h2 ⫺ 1x ⫹ h2 2 4 ⫺ 13x ⫺ x2 2 h

3 3 ⫺ x⫹h⫹2 x⫹2 39. lim hS0 h

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12–34


For Exercises 61 through 64, evaluate the limits by dividing the numerator and denominator by the highest power of x occurring in the denominator.

2 2 ⫺ x⫹h⫺1 x⫺1 40. lim hS0 h 1x ⫹ h ⫹ 2 ⫺ 1x ⫹ 2 41. lim hS0 h

61. lim

2x ⫹ h ⫺ 1 ⫺ 1x ⫺ 1 42. lim hS0 h

62. lim

43. lim hS0

44. lim

xSq

xSq

1x ⫹ h ⫹ 22 3 ⫺ 1x ⫹ 22 3

63.

1x ⫹ h ⫺ 42 3 ⫺ 1x ⫺ 42 3

64.

Use the TABLE feature of a graphing calculator to evaluate the following limits as x increases without bound.

8x ⫺ 3x2 46. lim 2 xSq 7x ⫺ x ⫹ 1

65. lim

x2 ⫹ 6x ⫹ 9 10 ⫺ 3x2 lim 52. 3 xSq xSq 10 ⫺ 3x 2x3 Use the TABLE feature of a graphing calculator to evaluate the following limits as x decreases without bound.

xSq

66.

xS⫺q

x2 ⫹ 1 55. lim xS⫺q 2x ⫺ 11

lim xS⫺q

6x2 ⫺ x ⫹ 2 2x2 ⫹ 1

7x3 56. lim 2 xS⫺q 5x ⫹ 3x

10 ⫺ 3x2 x2 ⫹ 6x ⫹ 9 lim 58. 3 xS⫺q 10 ⫺ 3x xS⫺q 2x3 1 59. Given lim ⫽ 0, find the smallest positive value xSq x 1 of x such that ⱕ 0.001. x 57.

lim

lim

1 ⫽ 0, find the largest negative value 60. Given lim xS⫺q x 1 of x such that ⱖ ⫺0.01. x

236x2 ⫺ 11 3x

212x2 ⫺ 6x ⫹ 1 xSq 7x

67. lim 68.

212x2 ⫺ 6x ⫹ 1 xS⫺q 7x lim

3 2 216x3 ⫹ 36x2 ⫺ 6x ⫹ 1 69. lim xSq 2x

51. lim

5x3 ⫹ 2 54. 10x3 ⫺ 2x ⫹ 1

8x3 ⫺ 27 x4 ⫺ 1

236x2 ⫺ 11 3x

xS⫺q

3x ⫺ 5x3 3 2 xSq x ⫹ 2x ⫺ 3x ⫹ 7 7x3 x2 ⫹ 1 lim 49. lim 50. 2 xSq 5x ⫹ 3x xSq 2x ⫺ 11

lim

lim

2x2 ⫺ 1 x ⫹ 2x ⫹ 12 3

For Exercises 65 through 70, evaluate each limit.

48. lim

53.

lim

xS⫺q

h

5x3 ⫹ 2 45. lim 3 xSq 10x ⫺ 2x ⫹ 1 2 6x ⫺ x ⫹ 2 47. lim xSq 2x2 ⫹ 1

5x2 ⫹ 11x ⫺ 3 12x2 ⫹ 7x ⫹ 5

xS⫺q

h

hS0

3x2 ⫺ 2x ⫹ 1 8x2 ⫹ 5

70.

3 2 216x3 ⫹ 36x2 ⫺ 6x ⫹ 1 xS⫺q 2x

lim

The graphs of two functions f and g are given here, for use in Exercises 71 through 88. y

f (x)

5 4 3 2 1

⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

y 5 4 3 2 1

1 2 3 4 5 x

⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

g(x) 1 2 3 4 5 x

For Exercises 71 through 76, use the graphs of f and g to perform the operations indicated. If any computation is not possible, state why.

71. f 102 ⫹ g1⫺12

72. f 152 ⫺ g1⫺32

75. g102 ⫹ f 1⫺42

76. g152 ⫹ f 1⫺12

73. g142 ⫺ f 122

74. f 1⫺22 ⫹ g122

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Section 12.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

For Exercises 77 through 82, use the graphs of f and g to evaluate the one-sided limits necessary to complete each calculation. If any computation is not possible, state why.

77.

lim f 1x2 ⫺ lim⫹ g1x2

xS⫺4 ⫺

lim ⫺ f 1x2

78. lim⫹ g1x2 ⫹ xS4

79.

xS0

85. lim c

xS⫺2

lim ⫹ f 1x2 ⫺ lim⫺ g1x2

xS⫺4 xS4

81. lim⫺ f 1x2 ⫹

lim f 1x2

86. lim e

xS⫺2 ⫹

xS5

xS⫺1

lim ⫺ g1x2

xS⫺2

82. lim⫹ g1x2 ⫹ lim⫹ f 1x2 xS2

䊳

3f 1x2

xS⫺3

xS0

80. lim⫺ g1x2 ⫹

For Exercises 83 through 88, use the graphs of f and g to evaluate the limits necessary to complete each calculation. If any computation is not possible, state why.

83. lim 3f 1x2 ⫺ g1x2 4

xS0

87.

g1x2

84. lim 3g1x2 ⫺ f 1x2 4 xS2

d

3 3g1x2 4 2 f 1x2

f

3 lim 3f 1x2 ⫺ 2 g1x2 4 88. lim 32g1x2f 1x2 4

xS⫺q

xS6

1203

xS3


89. (3.2) Solve each equation: a. 3x2 ⫹ 4x ⫺ 12 ⫽ 0 b. 13x ⫹ 1 ⫺ 12x ⫽ 1 2 3 1 ⫽ ⫹ 2 c. x⫹2 x⫹3 x ⫹ 5x ⫹ 6 ␲ 90. (7.5) Find cos⫺1 c cosa⫺ b d . 6 91. (6.6) Solve the right triangle shown. Round answers to the nearest tenth.

92. (1.4) In 2008, a small business purchased a copier for $4500. By 2011, the value of the copier had decreased to $3300. Assuming the depreciation is ¢ value linear, (a) find the rate-of-change m ⫽ and ¢ time discuss its meaning in this context. (b) Find the depreciation equation and (c) use the equation to predict the copier’s value in 2015. (d) If the copier is traded in for a new model when its value is less than $700, how long will the company use this copier?

A

C

12.4

35⬚ 13.7 cm

B

Applications of Limits: Instantaneous Rates of Change and the Area under a Curve


A. Evaluate the limit of a difference quotient to find instantaneous rates of change B. Evaluate the limit of a sum to find the area under a curve

At this point, the concept of rates of change is a familiar theme. From their first introy2 ⫺ y1 ¢y b to investigations of nonlinear functions ⫽ duction in a linear context a x2 ⫺ x1 ¢x using the difference quotient, the idea has proven powerful and useful. In Example 11 from Section 3.6, we used the difference quotient to approximate the velocity of a falling wrench with great accuracy. While this study certainly answers Galileo’s historic questions regarding the speed of a falling body, business, science, industry, education, and government present us with equally compelling questions that can likewise be answered using the difference quotient.

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A. The Limit of a Difference Quotient In Example 4 of Section 3.4, the height of a soccer ball kicked straight up into the air was modeled by the function d1t2 ⫽ ⫺16t2 ⫹ 64t. Using this function and the d1t2 2 ⫺ d1t1 2 ¢d ⫽ d , we found that between t ⫽ 0.5 and rate-of-change formula c ¢t t2 ⫺ t1 t ⫽ 1.0 sec, the average velocity of the ball was 40 ft/sec. Graphically, this was represented by the slope of the secant line through (0.5, 28) and (1.0, 48). See Figure 12.32 and the related calculation. For the time interval [1.0, 1.5], the average velocity had slowed to 24 ft/sec (the slope of this secant line is m ⫽ 24). See Figure 12.33 and the related calculation. Figure 12.32

Figure 12.33

d(t) ⌬d

70

20 m⫽ 0.5 ⫽ 40

60

⌬t

⫽

d(t2) ⫺ d(t1) t2 ⫺ t1

50

d(1) ⫺ d(0.5) ⫽ 1 ⫺ 0.5

40

⫽

48 ⫺ 28 0.5 ⫽ 40

48 ⫺ 28 ⫽ 20

30

70

d(t) m ⫽ 40

⌬d

12 0.5 ⫽ 24

m⫽

60

⌬t

d(1.5) ⫺ d(1) 1.5 ⫺ 1

60 ⫺ 48 0.5 ⫽ 24

⫽

60 ⫺ 48 ⫽ 12

50

⫽

1.5 ⫺ 1 ⫽ 0.5 40 30

1 ⫺ 0.5 ⫽ 0.5 20

20

10

10

0

1

2

3

4

t

0

1

2

3

4

t

Using smaller and smaller increments of time Figure 12.34 d(t) enables a better and better estimate of the velocity near t ⫽ 1, but requires a new calculation for each interval. Instead, we consider applying the difference quotient, as it 60 allows for a fixed point t and a second point 1t ⫹ h2 that 50 d1t ⫹ h2 ⫺ d1t2 . becomes arbitrarily close to t as h S 0: h 40 Graphically, this means the two points used for the secant line are becoming arbitrarily close, and in fact, 30 can be made as close as we like by applying the concept 20 of a limit. In Figure 12.34, several secant lines are shown, each 10 with a left endpoint at t ⫽ 1. Selecting right endpoints that become closer and closer to 1 11 ⫹ h as h S 02 , t 0 1 2 3 4 we obtain the sequence of secant lines shown in black 1h ⫽ 1.5, m ⫽ 82 , then purple 1h ⫽ 1, m ⫽ 162 , then red 1h ⫽ 0.5, m ⫽ 242 (for h ⫽ 0.01, the difference quotient returns a value of 31.84). As 1 ⫹ h gets infinitely close to 1, the secant line reaches a “limiting position” (the yellow line) as the two points forming the line become infinitely close. The end result is a tangent line to the graph of d at t ⫽ 1 (tangent: touching the curve at only one point in the interval near t ⫽ 1). As the slope of each secant line gives the average rate of change between two points, in this case the velocity of the ball, the slope of the tangent line gives the instantaneous velocity at precisely t ⫽ 1. As the

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graph indicates, the slope of the tangent line is the limit of the slopes of the secant lines and we write, mtan ⫽ lim msec ⫽ lim hS0

d1t ⫹ h2 ⫺ d1t2

hS0

h

For a general function f (x), we have the following result. The Slope of a Tangent Line Given a function f(x) that is smooth and continuous over the interval containing x and x ⫹ h, the slope of a line drawn tangent to the graph of f at x is given by the function f ¿1x2 , where f 1x ⫹ h2 ⫺ f 1x2 f ¿ 1x2 ⫽ lim hS0 h The notation f ¿ 1x2 is read “f prime of x,” and is commonly used to denote the slope of the line drawn tangent to the curve at x. EXAMPLE 1

䊳

Computing the Limit of a Difference Quotient to Find d ¿(t) For the function d1t2 ⫽ ⫺16t2 ⫹ 64t (see page 1204) a. Find the limit of the difference quotient for d, to obtain a function d¿ 1t2 that represents the slope of the line drawn tangent to d at t, and the instantaneous velocity at time t. b. Use d¿ 1t2 to find the velocity at t ⫽ 1, t ⫽ 2, and t ⫽ 3.5.

Solution

䊳

a. For d1t2 ⫽ ⫺16t2 ⫹ 64t, d1t ⫹ h2 ⫽ ⫺161t ⫹ h2 2 ⫹ 641t ⫹ h2 . After squaring the binomial and applying the distributive property, we obtain d1t ⫹ h2 ⫽ ⫺161t2 ⫹ 2th ⫹ h2 2 ⫹ 64t ⫹ 64h ⫽ ⫺16t2 ⫺ 32th ⫺ 16h2 ⫹ 64t ⫹ 64h

To find d¿ 1t2 , we then have lim hS0

d1t ⫹ h2 ⫺ d1t2 h

⫽ lim

⫺16t2 ⫺ 32th ⫺ 16h2 ⫹ 64t ⫹ 64h ⫺ 1⫺16t2 ⫹ 64t2

apply difference quotient

h ⫺16t ⫺ 32th ⫺ 16h ⫹ 64t ⫹ 64h ⫹ 16t2 ⫺ 64t ⫽ lim distribute ⫺1 hS0 h ⫺32th ⫺ 16h2 ⫹ 64h ⫽ lim combine like terms hS0 h h1⫺32t ⫺ 16h ⫹ 642 ⫽ lim factor out h hS0 h h ⫽ lim 1⫺32t ⫺ 16h ⫹ 642 ⫽ 1, h ⫽ 0 hS0

2

2

hS0

d¿ 1t2 ⫽ ⫺32t ⫹ 64

h

result

b. For the velocity of the ball at t ⫽ 1, t ⫽ 2, and t ⫽ 3.5, we evaluate d¿ 1t2 . t ⫽ 1:

d¿ 112 ⫽ ⫺32112 ⫹ 64 ⫽ ⫺32 ⫹ 64 ⫽ 32

substitute 1 for t simplify

At the moment t ⫽ 1 sec, the velocity is 32 ft/sec. t ⫽ 2:

d¿ 122 ⫽ ⫺32122 ⫹ 64 ⫽ ⫺64 ⫹ 64 ⫽ 0

substitute 2 for t simplify

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12–38


At the moment t ⫽ 2 sec, the velocity is 0 ft/sec. This seems reasonable, as we expect the ball will slow as it approaches its maximum height, where it momentarily stops (velocity is 0), then begins its return to the ground. t ⫽ 3.5: d¿ 13.52 ⫽ ⫺3213.52 ⫹ 64 ⫽ ⫺112 ⫹ 64 ⫽ ⫺48

substitute 3.5 for t simplify

At the moment t ⫽ 3.5 sec, the velocity is ⫺48 ft/sec. This indicates the ball is falling back to Earth (velocity is negative) at a rate of 48 ft/sec. Now try Exercises 7 through 12 䊳 Most graphing calculators have a built-in feature that can quickly compute the slope of a line tangent to the graph of f at a specific x. Begin by placing the nDeriv( command on the home screen with the keyFigure 12.35 strokes MATH 8 . After entering the function, the independent variable x, and the specific x-value (all separated by commas), pressing calculates and displays the slope of the tangent line at that x-value. The results from Example 1(b) are confirmed in Figure 12.35 using this feature. Note the 2nd (ENTRY) feature (used to recall and edit the previous entry) is very useful for computing the slopes of various tangent lines of the same function. In Example 11 of Section 3.6 (page 285), we modeled the distance a wrench had fallen from a skyscraper with the function d1t2 ⫽ 16t2. Using a very small time interval and the difference quotient, we obtained a close approximation for the velocity of the wrench in various intervals. Using the concept of a limit we can now find the instantaneous velocity at any time t, again denoted d¿ 1t2 . ENTER

ENTER

EXAMPLE 2

䊳

Finding an Instantaneous Rate of Change For the function d1t2 ⫽ 16t2, a. Find the limit of the difference quotient for d to obtain a function d¿1t2 that represents the instantaneous velocity of the falling wrench. b. Use d¿ 1t2 to find the velocity at t ⫽ 2, t ⫽ 7, and t ⫽ 9. Verify your results with a calculator.

Solution

䊳

a. In the referenced example we obtained the expression 32t ⫹ 16h after applying the difference quotient. Using the ideas developed here, we know the velocity of the wrench will be d¿ 1t2 ⫽ lim 132t ⫹ 16h2 hS0

⫽ 32t by direct substitution. b. For the instantaneous velocity at t ⫽ 2, t ⫽ 7, and t ⫽ 9, we evaluate d¿ 1t2 . t ⫽ 2: d¿ 122 ⫽ 32122 ⫽ 64

substitute 2 for t

At t ⫽ 2 sec, the velocity of the wrench is 64 ft/sec. t ⫽ 7: d¿ 172 ⫽ 32172 ⫽ 224

substitute 7 for t

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1207

At t ⫽ 7 sec, the velocity of the wrench is 224 ft/sec. t ⫽ 9: d¿ 192 ⫽ 32192 ⫽ 288

substitute 9 for t

At the instant t ⫽ 9 sec, the velocity of the wrench has increased to 288 ft/sec. In fact, in the absence of air resistance the velocity of the wrench will continue to increase until it hits the ground (with a wham!). The nDeriv( feature of our calculator verifies these velocities (see figure). Now try Exercises 13 through 16

䊳

Mathematical models of real-world phenomena come in many different forms, and in a calculus course these ideas are applied to a very wide variety of functions. For practice with the algebra required to make these ideas work, see Exercises 17 through 20. The following applications demonstrate how these ideas are applied to root and rational function models. EXAMPLE 3

䊳

Computing a Rate of Change for the Distance to the Horizon Many new high-rise buildings have external elevators, allowing an unobstructed view of the city as it extends toward the horizon. The distance that Melody can see as the elevator takes her higher is modeled by the function d1x2 ⫽ 1.151x, where d(x) represents her sight distance in nautical miles, at a height of x feet. a. Find the limit of the difference quotient for d1x2 , to obtain a function d¿ 1x2 that represents the instantaneous rate of change in the sight distance at x. b. Find the instantaneous rate of change at heights of 225, 400, and 625 ft.

Solution

䊳

lim hS0

a. For d1x2 ⫽ 1.15 1x, d1x ⫹ h2 ⫽ 1.15 1x ⫹ h. To find d¿ 1x2 , we have

d1x ⫹ h2 ⫺ d1x2 h

1.15 1x ⫹ h ⫺ 1.15 1x hS0 h 1x ⫹ h ⫺ 1x ⫽ 1.15 lim hS0 h ⫽ lim

apply difference quotient

factor out the constant

Noting that multiplying the numerator by its conjugate will “free up” the h in the first radicand, we multiply numerator and denominator by 1x ⫹ h ⫹ 1x. ⫽ 1.15 lim

1 1x ⫹ h ⫺ 1x2 1 1x ⫹ h ⫹ 1x2

h 1 1x ⫹ h ⫹ 1x2 x⫹h⫺x ⫽ 1.15 lim 1A ⫺ B21A ⫹ B2 ⫽ A2 ⫺ B2 hS0 h1 1x ⫹ h ⫹ 1x2 h combine like terms ⫽ 1.15 lim hS0 h1 1x ⫹ h ⫹ 1x2 1 h ⫽ 1.15 lim ⫽ 1, h ⫽ 0 h hS0 1x ⫹ h ⫹ 1x 1 apply limit ⫽ 1.15a b 1x ⫹ 1x 1.15 result d¿ 1x2 ⫽ 2 1x hS0

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b. To find the instantaneous rate of change at x ⫽ 225, 400, and 625 ft, we evaluate d¿ 1x2 . d¿ 12252 ⫽

1.15 1.15 ⫽ 21152 2 1225 ⬇ 0.038

substitute 225 for x

At x ⫽ 225 ft, the sight distance is increasing at a rate of about 0.038 mi/ft (about 3.8 nautical miles per 100 ft). d¿ 14002 ⫽

1.15 1.15 ⫽ 21202 2 1400 ⬇ 0.029


At x ⫽ 400 ft, the sight distance is increasing at a rate of about 0.029 mi/ft (about 2.9 nautical miles per 100 ft). d¿ 16252 ⫽

1.15 1.15 ⫽ 21252 2 1625 ⫽ 0.023


At x ⫽ 625 ft, the sight distance is increasing at a rate of 0.023 mi/ft (2.3 nautical miles per 100 ft). These rates of change are verified in the figure. Now try Exercises 21 through 24 䊳 Note that for square root functions, the algebra involves rationalizing the numerator to “free up” the constant h in hopes that the limit can eventually be found by direct substitution. For rational functions, the difference quotient yields a difference of rational expressions in the numerator, which we then combine for the same reason. EXAMPLE 4

䊳

Computing Instantaneous Rates of Change for a Cost Function A county government finds the cost of removing pollutants from a stream can be 2x modeled by the function C1x2 ⫽ , where C(x) represents the cost in millions 1⫺x of dollars, to remove x # 100 percent of the pollutants [C1x2 7 0, x is a decimal less than 1]. a. Find the limit of the difference quotient for C, to obtain a function C¿1x2 that represents the instantaneous rate of change in the cost of removing pollutants. b. Find the rate the cost is increasing at the moment 60%, 70%, and 80% of the pollutants are removed. c. Use the information from part (b) to estimate the rate the cost will be increasing at the moment 90% of the pollutants are removed, then compute the actual rate. Were you surprised?

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Solution

䊳

a. For C1x2 ⫽

21x ⫹ h2 2x . To find C¿ 1x2 we then have , C1x ⫹ h2 ⫽ 1⫺x 1 ⫺ 1x ⫹ h2 21x ⫹ h2

lim hS0

C1x ⫹ h2 ⫺ C1x2 h

1209

⫺

2x 1⫺x

1 ⫺ 1x ⫹ h2 apply difference quotient hS0 h x x⫹h ⫺ 1⫺x⫺h 1⫺x factor out 2, distribute ⫺1 ⫽ 2 lim hS0 h 1x ⫹ h211 ⫺ x2 ⫺ x11 ⫺ x ⫺ h2

⫽ lim

⫽ 2 lim

hS0

⫽ 2 lim

hS0

⫽ 2 lim

hS0

⫽ 2 lim

hS0

⫽ 2 lim

hS0

C¿ 1x2 ⫽

11 ⫺ x ⫺ h211 ⫺ x2 C AD ⫺ BC A ⫺ ⫽ B D BD h 2 2 x ⫺ x ⫹ h ⫺ xh ⫺ x ⫹ x ⫹ xh F-O-I-L, distribute 11 ⫺ x ⫺ h2 11 ⫺ x2 h h 11 ⫺ x ⫺ h211 ⫺ x2 simplify numerator h h invert and multiply 11 ⫺ x ⫺ h211 ⫺ x2h 1 h ⫽ 1, h ⫽ 0 h 11 ⫺ x ⫺ h211 ⫺ x2

2 11 ⫺ x2 2

apply limit, multiply by 2

b. Evaluating C¿ 1x2 for x ⫽ 0.6, 0.7, 0.8 (60%, 70%, and 80%) yields C¿ 10.62 ⫽

2 2 ⫽ 2 0.16 11 ⫺ 0.62 ⫽ 12.5

substitute 0.6 for x and simplify

At x ⫽ 0.6, the cost is increasing at a rate of $12,500,000 for each additional percentage point of pollution removed. C¿ 10.72 ⫽

2 2 ⫽ 0.09 11 ⫺ 0.72 2 ⫽ 22.2


At x ⫽ 0.7, the cost is increasing at a rate of about $22,222,222 for each additional percentage point. C¿ 10.82 ⫽

2 2 ⫽ 2 0.04 11 ⫺ 0.82 ⫽ 50

A. You’ve just seen how we can evaluate the limit of a difference quotient to find instantaneous rates of change


At x ⫽ 0.8, the cost is increasing at a rate of $50,000,000 for each additional percentage point. c. Evaluating C(x) for x ⫽ 0.9 shows the cost rises dramatically to a rate of $200,000,000 per percentage point. Removing anywhere near 100% of the pollutants would become prohibitively expensive. Now try Exercises 25 through 32 䊳

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B. Limits and the Area under a Curve Figure 12.36 y 10

In the Connections to Calculus from Chapter 11, the area under the graph of f 1x2 ⫽ x ⫹ 5 for x in the interval [0, 4] was approximated using the rectangle method with n ⫽ 4 and n ⫽ 8 rectangles (see Example 1). While we agreed that using 16 rectangles would give a better estimate (see Figure 12.36), expanding the related summation 16 1 1 LW ⫽ f a ib a b seemed too tedious and instead we opted to use the properties 4 4 i⫽1 of summation to write the expression in a form where the sum could be found by substituting n ⫽ 16 directly. Here we extend this idea to a situation where n rectangles are used, as we can then take the limit of this sum as n S q . The key to this extension is 1 a result from Section 12.3, where we noted for any positive integer k, lim k ⫽ 0 and nS⫺q n 1 lim k ⫽ 0 (note the change of variable from x to n, which is more commonly used nSq n for summations). The original 4-unit interval would then be divided into n parts, so each 4 rectangle is units wide. The length of each rectangle is still given by f 1x2 ⫽ x ⫹ 5, n n 4 4 4 but we now evaluate f in increments of , giving LW ⫽ f a ib a b. Since n n n i⫽1

兺

9 8 7 6 5 4 3 2 1 1

2

3

4

5

6

x

兺

兺

兺

n 4 4 4 4 f a ib ⫽ i ⫹ 5, we can write the sum as a i ⫹ 5b and proceed as before. n n n n i⫽1

兺

EXAMPLE 5

䊳

Finding the Limit of a Sum n

Evaluate the following limit: lim

nSq

Solution

䊳

兺 a ni ⫹ 5b n . 4

4

i⫽1

Begin by applying summation properties and formulas to write the expression in a 4 form where the limit properties can be applied. Since is a constant, we begin by n factoring it out of the summation (summation property II). 4 n

n

4 4 n 4 a i ⫹ 5b ⫽ a i⫹ n i⫽1 n i⫽1 n

兺

兺

⫽

4 4 a n n

n

n

兺 5b

兺 i ⫹ 兺 5b

i⫽1


i⫽1 n

factor

i⫽1

4 4 n ⫹n b ⫹ 5n d c a n n 2 16 n2 ⫹ n ⫽ 2a b ⫹ 20 2 n 16 n2 ⫹ n b ⫹ 20 ⫽ a 2 n2 n2 n ⫽ 8 a 2 ⫹ 2 b ⫹ 20 n n 1 8 ⫽ 8 a1 ⫹ b ⫹ 20 ⫽ 28 ⫹ n n

4 from first summation n

2

⫽

We can now take the limit of this result and find 8 lim a28 ⫹ b ⫽ 28. n

nSq

apply summation formulas

distribute

4 n

rewrite denominators (commute)

decompose rational expression

result

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1211

This shows n

lim

nSq

兺 a n i ⫹ 5b n ⫽ 28, 4

4

i⫽1

and that the area under the graph of f 1x2 ⫽ x ⫹ 5 between x ⫽ 0 and x ⫽ 4 is exactly 28 units2! Now try Exercises 33 through 40 䊳 This result may seem unimpressive, as we could have quickly found the area using simple geometry. But as mentioned earlier, this method is easily applied to areas under a general function f(x). In the Connections to Calculus feature from Chapter 3, we noted how the area under a graph could represent the distance covered by a runner. Our work here now takes on added meaning, as the area under a general curve has many other real-world implications in a study of calculus. EXAMPLE 6

䊳

Using Limits to Find the Area under a Curve

f(x)

10 During a training session for the Boston Marathon, 9 Georgiana plans to begin at a leisurely walk of 2 mi/hr 8 and gradually increase her speed over a 5-hr period, 7 6 finishing up at a clip of over 8 mi/hr. Her speed might 5 1 2 then be modeled by the function f 1x2 ⫽ x ⫹ 2, where 4 4 3 f(x) represents the velocity in miles per hour, and x is the 2 1 time in hours. The area under the curve in the interval [0, 5] then represents the distance Georgiana has run in ⫺2 ⫺1 1 2 3 4 5 6 x these 5 hr. a. Show that using the rectangle method results in the expression 5 n 1 5 2 c a ib ⫹ 2 d . n i⫽1 4 n b. Determine if Georgiana has run a “marathon distance” (about 26.2 mi) in these 5 hr, by applying summation properties and formulas, and taking the limit of this sum as n S q .

兺

Solution

䊳

a. The area described is shown in the figure. The given interval is 5 units wide, 5 and after dividing it into n parts, each rectangle will have a width of . The n 5 1 length of each one is given by f 1x2 ⫽ x2 ⫹ 2, evaluated at each of the n 4 subintervals. The total area is then

兺

n

LW ⫽

5 5 5 5 n 5 n 1 5 2 f a ib a b ⫽ f a ib ⫽ c a ib ⫹ 2 d . n n n i⫽1 n n i⫽1 4 n i⫽1

兺

兺

兺

b. Proceed as in Example 5. 5 n

n

n 1 5 2 5 n 1 5 2 c a ib ⫹ 2 d ⫽ c a ib ⫹ 2d n i⫽1 4 n i⫽1 4 n i⫽1 n 5 n 1 25 2 ⫽ c a 2 bi ⫹ 2d n i⫽1 4 n i⫽1

兺

⫽ ⫽

兺

兺

兺

兺

5 n 25 2 i ⫹ c n i⫽1 4n2

兺

5 2 25 a ib ⫽ 2 i 2 n n

n

兺 2d

multiply

i⫽1

5 25 n 2 b i ⫹ ca n 4n2 i⫽1

兺


n

兺 2d

i⫽1

factor

25 n2

from first summation

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⫽

5 25 2n3 ⫹ 3n2 ⫹ n a b ⫹ 2n d c n 4n2 6

125 2n3 ⫹ 3n2 ⫹ n a b ⫹ 10 6 4n3 125 2n3 ⫹ 3n2 ⫹ n ⫽ a b ⫹ 10 24 n3 ⫽

⫽

3 1 125 a2 ⫹ ⫹ 2 b ⫹ 10 n 24 n

apply summation formulas

distribute

5 n

rewrite denominators (commute) decompose rational expression

We can now take the limit of this result, and after applying the needed limit properties we obtain 125 3 1 lim a2 ⫹ ⫹ 2b ⫹ lim 10 n 24 nSq nSq n 125 b122 ⫹ 10 ⫽ 20.416 ⫽a 24

⫽

apply properties of limits

1 The area under the graph of f 1x2 ⫽ x2 ⫹ 2 between x ⫽ 0 and x ⫽ 5 is 4 20.416 units2, which is numerically equivalent to the distance run. Georgiana is still about 5.8 mi short of a marathon. Now try Exercises 41 through 44 䊳

B. You’ve just seen how we can evaluate the limit of a sum to find the area under a curve

Note: For Learning Objective B, the functions and intervals offered in the Exercise Set were deliberately chosen for their simplicity. In particular, all are polynomials where f 1x2 7 0 in the interval selected. In a calculus course, the same ideas are applied to a greater variety of functions, not all intervals begin at 0, and the function may vary from positive to negative within the interval. Some of these ideas are explored in the Calculator Exploration and Discovery for this chapter, but regardless, the general ideas demonstrated here remain consistent.

12.4 EXERCISES 䊳



1. The slope of a line drawn tangent to the graph of any function f(x) can be found by taking the limit of the quotient for f, as h approaches zero.

2. The instantaneous rate of change of a function f (x) can be found by evaluating the expression

3. In order to obtain a better approximation for the area under a curve, we can use narrower (and thus more) in our computation.

4. As we let the number of approximating rectangles increase without bound, the resulting limit gives a(n) value for the area under the curve.

5. Discuss/Explain the relationship between the difference quotient and the formula for finding the slope of a line given two points.

6. Do some research on geometric shapes other than rectangles that can be used to approximate the area under a curve. Return to the graph on page 1210, and comment on why approximating rectangles consistently give an overestimate of the area.

.

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1213


Use the following information to answer Exercises 7 through 12. Two model rockets are launched at a gathering of the National Association of Rocketry (NAR: www.nar.org). Frank’s Apollo II motor burns out at a height of 500 m, at which point the rocket has a velocity of 88.2 meters per second (m/sec). His rocket’s height in meters, t sec after engine burnout, is given by f 1t2 ⫽ 500 ⫹ 88.2t ⫺ 4.9t2. Gwen’s Icarus Alpha motor burns out at a height of 600 m, at which point the rocket has a velocity of 78.4 m/sec. Her rocket’s height in meters, t sec after burnout, is given by g1t2 ⫽ 600 ⫹ 78.4t ⫺ 4.9t2.

7. Find the limit of the difference quotient for f, to obtain a function f ¿ 1t2 that represents the instantaneous velocity at time t. 8. Find the limit of the difference quotient for g, to obtain a function g¿ 1t2 that represents the instantaneous velocity at time t. 9. Use the result from Exercise 7 to find the instantaneous velocity of Frank’s rocket at a. t ⫽ 5 b. t ⫽ 9 c. t ⫽ 13 Verify your results with a calculator. 10. Use the result from Exercise 8 to find the instantaneous velocity of Gwen’s rocket at a. t ⫽ 2 b. t ⫽ 8 c. t ⫽ 10 Verify your results with a calculator. 11. Use the result from Exercise 7 to find the maximum height of Frank’s rocket. This occurs when v ⫽ 0. 12. Use the result from Exercise 8 to find the maximum height of Gwen’s rocket. This occurs when v ⫽ 0. 13. To replicate Galileo’s famous test as to whether the velocity of a falling body depends on its weight, a science class is dropping bowling balls of different weights from an 11-story building onto the lawn below. The ball’s height in meters, t sec after it is released, is modeled by d1t2 ⫽ ⫺4.9t2 ⫹ 44.1. Find the limit of the difference quotient for d, to obtain a function d¿1t2 that represents the instantaneous velocity of the bowling ball at time t. 14. A rock climber’s carabiner falls off her harness 256 ft above the floor of the Grand Canyon. It’s height in feet, t sec after it falls, can be modeled by d1t2 ⫽ ⫺16t2 ⫹ 256. Find the limit of the difference quotient for d, to obtain a function d¿ 1t2 that represents the instantaneous velocity of the “biner” at time t.

15. Use the results of Exercise 13 to find the instantaneous velocity of the bowling ball at a. t ⫽ 1 b. t ⫽ 2 c. t ⫽ 3 (time of impact) 16. Use the results of Exercise 14 to find the instantaneous velocity of the “biner” at a. t ⫽ 1 b. t ⫽ 3 c. t ⫽ 4 (time of impact) Find the limit of the difference quotient for each function f(x) given, to obtain a function f ¿(x) that represents the instantaneous rate of change at x for each function.

1 17. f 1x2 ⫽ x ⫹ 5 2

18. f 1x2 ⫽ x2 ⫺ 3x

19. f 1x2 ⫽ x3

20. f 1x2 ⫽

1 x

21. The population of a small town can be modeled by the function p1t2 ⫽ 1.2 1t ⫹ 40, where p is measured in thousands and t is the number of years after 2008. Find the limit of the difference quotient for p to obtain a function p¿ 1t2 that represents the instantaneous rate of change of population at time t. 22. The number of bacteria in a person’s body, after they begin a regimen of antibiotics, can be modeled by the function b1t2 ⫽ 6 ⫺ 1t, where b is measured in tens of thousands and t is the number of hours after the first dose. Find the limit of the difference quotient for b to obtain a function b¿ 1t2 that represents the instantaneous rate of change of number of bacteria at time t. 23. Use the results of Exercise 21 to find the instantaneous rate of change of population of the town in a. 2009 b. 2012 c. 2024 24. Use the results of Exercise 22 to find the instantaneous rate of change of number of bacteria after a. 1 hr b. 16 hr c. 25 hr For Exercises 25 through 28, find the limit of the difference quotient of the given function to obtain a function that represents the slope of a line drawn tangent to the curve at x.

25. f 1x2 ⫽

2 x⫺1

26. g1x2 ⫽

3 x⫹2

27. h1x2 ⫽

5x x⫹5

28. j1x2 ⫽

x 21x ⫹ 12

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Use the results from the corresponding Exercises 25 through 28 for the following exercises.

29. Find the slope of the line drawn tangent to the graph of f (x) at x ⫽ 3. 30. Find the slope of the line drawn tangent to the graph of g(x) at x ⫽ ⫺1. 31. Find the slope of the line drawn tangent to the graph of h(x) at x ⫽ 0.

For Exercises 33 through 36, graph each function over the interval [0, 7]. Then use geometry to find the area of the region below the graph, and above the x-axis in the interval [0, 6].

1 33. q1x2 ⫽ x 2

1 34. r1x2 ⫽ x 3

1 35. t1x2 ⫽ x ⫹ 1 2

1 36. s1x2 ⫽ x ⫹ 2 3

For Exercises 37 through 40, evaluate the following limits and compare your result to the corresponding exercise in 33 through 36. n

37. lim

兺

a

nSq i⫽1 n

39. lim

n

1 6 6 # ib 2 n n

38. lim

兺 a 3 # n ib n 1 6

6

nSq i⫽1

兺 a 2 # n i ⫹ 1b n 1 6

6

nSq i⫽1 n

40. lim

兺 a 3 # n i ⫹ 2b n 1 6

y after x days on the job. The 12 11 area under this curve in the 10 interval x 僆 30, 4 4 then 9 8 represents the total number 7 6 of parts produced in the first 5 4 4 days. Using the rectangle 3 method results in the 2 1 expression n 2 ⫺5⫺4⫺3⫺2⫺1 1 2 3 4 5 x 1 4 4 c a ib ⫹ 3 d . Find n i⫽1 2 n the total number of parts produced by applying the summation properties/formulas and taking the limit as n S q .

兺

32. Find the slope of the line drawn tangent to the 1 graph of j(x) at x ⫽ ⫺ . 2

6

nSq i⫽1

41. For a new machine shop employee, the rate of production for a specialized part is modeled by 1 the function p1x2 ⫽ x2 ⫹ 3 (production 2 increases quickly with experience), where p(x) represents the number of parts completed per day

䊳

12–46


42. Water has been leaking undetected from a reservoir formed by an earthen dam. Due to the accelerating erosion, the rate of water loss can be modeled by the function 1 g1x2 ⫽ ⫺ x2 ⫹ 8, 4

Exercise 42 y 12 11 10 9 8 7 6 5 4 3 2 1 ⫺6⫺5⫺4⫺3⫺2⫺1

1 2 3 4 5 6 x

x 7 0, where g(x) represents hundreds of gallons of water lost per hour. The area under the curve in the interval [0, 5] then represents the total water loss in 5 hr. Using the rectangle method results in 5 n 1 5 2 c ⫺ a ib ⫹ 8 d . Find the the expression n i⫽1 4 n total amount of water lost by applying summation properties and formulas and taking the limit as n S q.

兺

Find the area under the curve for each function and interval given, using the rectangle method and n subintervals of equal width.

1 43. f 1x2 ⫽ ⫺ x2 ⫹ 4x, x 僆 30, 6 4 2 1 44. f 1x2 ⫽ ⫺ x3 ⫹ 6x, x 僆 30, 3 4 2


45. (5.5) Solve: ⫺350 ⫽ 211 e⫺0.025x ⫺ 450

46. (8.3) Given u ⫽ H⫺2, ⫺3I and v ⫽ H⫺3, ⫺5I, find 3u ⫺ v.

47. (4.2) Find all solutions by factoring: x3 ⫺ 5x2 ⫹ 3x ⫺ 15 ⫽ 0. 48. (6.2) Evaluate all six trigonometric functions at 2␲ t⫽⫺ 3

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1215

Making Connections

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs f(x)

(a)

f(x)

(b)

5

5

4

4

3

3

2

2

1

1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

x

⫺2

⫺2

⫺3

⫺3

⫺4

⫺4

⫺5

⫺5

f(x)

(d)

1

(c)

2

3

4

5

x

(e)

f(x)

(f)

5

5

4

4

3

3

2

2

1

1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

x

⫺2

⫺2

⫺3

⫺3

⫺4

⫺4

⫺5

⫺5

f(x)

(g)

f(x)

(h)

5

5

4

4

3

3

2

2

1

1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

x

⫺2

⫺2

⫺3

⫺3

⫺4

⫺4

⫺5

⫺5

1. ____ msec ⫽ 1

1

2

3

4

5

x

9. ____ 122 132 ⫹ 122 142

2. ____ lim⫺ f 1x2 ⫽ f(2), lim⫹ f 1x2 ⫽ f 122

10. ____

3. ____ lim⫺ f 1x2 ⫽ 2

11. ____ lim f 1x2 ⫽ ⫺3

xS2

xS2

xS2

lim f 1x2 ⫽ 3

xS ⫺ q

xSq

4. ____ lim f 1x2 ⫽ 3

12. ____ f ¿ 122 ⫽ ⫺1

13. ____ lim f 1x2 ⫽ 1

xSq

5. ____ lim⫹ f 1x2 ⫽ A dne ⫺q B

xS2

xS2

6. ____ lim⫺ f 1x2 ⫽ A dne ⫺q B

14. ____ lim

xS2

hS0

7. ____ msec ⫽ ⫺1 n

2

1 4i 4 c⫺ a b ⫹ 4da b 8. ____ lim n nSq i⫽1 4 n

兺

f 12 ⫹ h2 ⫺ f 122 h

⫽1

15. ____ lim⫹ f 1x2 ⫽ 3 xS2

16. ____ removable discontinuity

1

2

3

4

5

x

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12–48



An Introduction to Limits Using Tables and Graphs

KEY CONCEPTS • The concept of a limit involves the incremental approach of a variable x to a fixed constant c (the cause), with a focus on the behavior the function f during this approach (the effect). lim f 1x2 ⫽ L means values of f (x) can be made arbitrarily close to L by taking values of • The notation xSc x sufficiently close to c, but not equal to c. L is called the limit of f as x approaches c. • Limits can be investigated using a table of values or the graph of a function. Choosing values of x that are closer and closer to c, we observe the resulting values of f (x) to see if it appears a limiting value exists. lim f 1x2 may exist even if f (x) is not defined at c, if f (x) is defined at c but f 1c2 ⫽ L, or if f (x) is defined • xSc at c and f 1c2 ⫽ L. lim f 1x2 ⫽ L means values of f (x) can be made • We can also define a left-hand limit and a right-hand limit: xSc arbitrarily close to L by taking values of x sufficiently close to c and to the left of c. The right-hand limit is similarly defined. • A limit can fail to exist in one of three ways. dne B , (1) the left-hand limit is not equal to the right-hand limit: lim f 1x2 ⫽ A LH⫽RH (2) as x S c, f (x) increases or decreases without bound: lim f 1x2 ⫽ A dne , or ⫾qB xSc

(3) as x S c, there is no fixed number L that f(x) approaches: lim f 1x2 ⫽ A dne . L B xSc

xSc

EXERCISES 1. Estimate the following limits (if they exist) using the tables shown. a. lim⫺ f 1x2 b. lim⫹ f 1x2 c. lim f 1x2 xS2

xS2

xS2

x

f(x)

x

f(x)

1.5

⫺2.5

2.5

5.5

1.6

⫺2.6

2.4

5.4

1.7

⫺2.7

2.3

5.3

1.8

⫺2.8

2.2

5.2

1.9

⫺2.9

2.1

5.1

1.99

⫺2.99

2.01

5.01

1.999

⫺2.999

2.001

5.001

Use the TABLE feature of a graphing calculator to evaluate each limit. 2. lim xS5

x2 ⫺ 2x ⫺ 15 x2 ⫺ 6x ⫹ 5

SECTION 12.2

3. lim3 xS 2

sin12x ⫺ 32 2x ⫺ 3

x⫺4 x ⱕ ⫺2 4. lim f 1x2, f 1x2 ⫽ • 冟x冟 xS⫺2 x2 ⫺ 1 x 7 ⫺2

The Properties of Limits

KEY CONCEPTS • When evaluating limits using a table, great care must be exercised as results can sometimes be misleading. Using a graph and some common sense will sometimes help. • Limits possess certain properties that seem plausible using a table of values. These include the sum, difference, product, and quotient properties, as well as the power and root properties. For a review of these properties, see page 1182. • Once the limit properties have been established, they can be used to evaluate the limit of many different kinds of functions.

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1217

• These four basic limits play an important role in the evaluation of general limits. (1) lim k k xSc

(2) lim x c

(3) lim xk ck

xSc

k k (4) lim 1x 1c, if c 7 0 when k is even

xSc

xSc

• If the limit properties lead to an undefined or indeterminate expression, a limit may still exist and other means should be used to investigate.

EXERCISES Determine the following limits using limit properties. 5. lim 12x3 5x 12

6. lim

xS3

SECTION 12.3

xS7

111 x

7. lim

2x 3x 1 3

2

xS1

1x 1x 12x 92 5

8. lim

2

xS1

22 x 1 1x

Continuity and More on Limits

KEY CONCEPTS • Generally, a function is continuous if you can draw the entire graph without lifting your pencil. • Specifically, a function f is continuous at c if these three conditions are met: (1) c is in the domain of f (2) lim f 1x2 lim f 1x2 (3) lim f 1x2 f 1c2 . xSc

xSc

xSc

• The continuity of a function can be interupted by (1) asymptotic, (2) jump (non-removable), or (3) removable discontinuities. lim f 1x2 can be found by direct substitution. • If a function is continuous at c, the limit xSc • Polynomial functions are continuous for x 僆⺢, while rational, root, and trigonometric functions are continuous everywhere they are defined. • Using algebraic methods, a limit that cannot be evaluated by direct substitution can sometimes be rewritten in a form where a direct substitution becomes possible. f 1x2 L, provided f (x) can be • Limits can also occur as x becomes infinitely large, positively or negatively: xSlim q made arbitrarily close to L, by taking values of x sufficiently large.

EXERCISES 9. Name the x-value(s) where the graph has a. an asymptotic discontinuity b. a jump discontinuity c. a removable discontinuity

f (x)

54321 1 2 3 4 5

Use the graph of f to evaluate the following limits. 10. lim f 1x2 11. lim f 1x2 12. lim f 1x2 lim f 1x2

xSq

13.

xS3

SECTION 12.4

14. lim f 1x2 xS3 xS3

y 5 4 3 2 1

15. lim f 1x2 xS3 xS1

1 2 3 4 5 x

Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

KEY CONCEPTS • The rate of change of f (x) in the interval [x1, x2] can be approximated using the rate-of-change formula:

f 1x2 2 f 1x1 2 ¢f . x2 x1 ¢x • The rate of change of f (x) in the arbitrary interval [x, x h] can be approximated using the difference quotient: f 1x h2 f 1x2 h

.

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• Graphically, the slope of the secant line gives a better and better approximation for the instantaneous rate of change as h S 0, and we define the instantaneous rate of change as the limiting value of this slope: mtan lim msec lim hS0

f 1x h2 f 1x2 h

hS0

.

• The instantaneous rate of change of f at point x is the limit of the difference quotient (if it exists): f ¿ 1x2 lim

hS0

f 1x h2 f 1x2 h

.

• For f 1x2 7 0 in the first quadrant, the area under the graph of f in the interval [0, b] can be found using the

b rectangle method and n rectangles of equal width. The width of each rectangle will be , the height of each is n b found by evaluating f in increments of : n n n b b LW lim f a ib . A lim nSq nSq n n i1 i1

兺

兺

EXERCISES Find the limit of the difference quotient as h S 0 to determine the instantaneous rate of change for the functions given. 1 16. f 1x2 x2 5x 2 17. g1x2 12x 1 18. v1x2 x3 19. Take the limit of a difference quotient to find the slope of the line drawn tangent to f 1x2 x2 3x at x 4. 20. Use the rectangle method to find the area in QI, under the graph of f 1x2 x2 6x 9 in the interval [0, 3].

PRACTICE TEST 1. Write the expression in words, as though you were reading it out loud: lim f 1x2 10 xS5

2. Fill in the blanks: lim f 1x2 L, means that values of xSc

can be made arbitrarily close to by taking values of x close to

, .

State true or false and justify your answer. x2 1 , lim f 1x2 cannot exist since x 1 x 1 xS1 is not in the domain.

3. For f 1x2

x 1, x 1 , lim f 1x2 must exist x 1, x 6 1 xS1 since x 1 is in the domain.

4. For f 1x2 e

5. For g1x2 1x 1 2, state which of the following one-sided limits exist. Justify your answer and then find the limit. a. lim 1 1x 1 22 b. lim 1 1x 1 22 xS1 xS1

6. Match each expression to its conclusion. 1 a. lim cosa b I. A dne q B xS0 x 5x b. lim 2 II. A dne L B xS2 x 4x 4 sin x dne B c. lim III. A LHRH xS0 x 2x2 2x d. lim IV. 1 xS1 2x2 2x 1 Determine the following using the graphs of f (x) and g(x) shown. 7. a.

lim f 1x2

xS1

8. a. lim g1x2 xS4

9. a. f 102 g102 b. g152 f 142 10. a. [g(1)]

2

b.

b. lim f 1x2 xS3

b. lim g1x2 xS3

f 142

g132 11. a. lim 3f 1x2 g1x2 4 b. lim f 1x2g1x2 xS4

xS1

y

f (x)

5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

y

g(x)

5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

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12. a.


lim 1g1x2

xS5

b.

lim

xS1

3g1x2 4 2

13. Evaluate the following limits using the TABLE feature of a calculator. If the limit does not exist, state why. cos x 1 x3 1 a. lim b. lim x xS0 xS1 21x 12 2 c. lim xS0

sin x 2x tan x

d. lim

sin13x2 3x

xS0

14. Each of the following limits does not exist. State why. 冟x2 1冟 1 a. lim b. lim tana b x xS1 x 1 xS0 2 x 1 c. lim d. lim 1x 4 xS2 x2 xS3 Evaluate each limit. x3 1 xS1 x 1

15. lim 17. lim

xS25

1x 5 x 25

16.

lim xSq

18. lim xS0

3x2 5x 4 2x2 x 3

cos12x2 cos2 x

1219

19. A controlled explosion at a strip mine causes a huge shower of dirt and rock. The height of the debris that was projected vertically upward by the blast can be modeled by the function d1t2 16t2 224t, where d(t) is the height in feet after t sec. (a) Compute the limit of the difference quotient for d to find a function d¿ 1t2 that gives the instantaneous velocity of the debris at time t. (b) Find the velocity at t 2, 6, 7, and 11, and explain the significance of each result. 20. In preparation for another sold-out performance by the Danish pianist and comedian Victor Borge (Phonetic Punctuation), stage hands move a large grand piano 15 ft across the main stage and then carefully into position. The force applied can be modeled by the function f 1x2 225 x2 (the force is much greater as they start than when they finish). Since Work Force # Distance, the area under the curve from 0 to x represents the amount of work done in foot-pounds. Use the rectangle method to (a) find an expression representing the amount of work done in moving the piano the first 6 ft, and (b) find the amount of work done in the interval [0, 6].

CALCULATOR EXPLORATION AND DISCOVERY Technology and the Area Under a Curve In Section 12.4, we used approximating rectangles and limit properties to determine the area under a curve. In a calculus course, we will discover a much more efficient method to determine this same area, but for now we can turn to our calculators to verify our work. Most graphing calculators have at least two features that enable us to quickly compute the area under a curve, between two x-values. Figure 12.37 In Example 5 from Section 12.4, we used the limit of the sum of these rectangles to determine that the area under the graph of f 1x2 x 5 between x 0 and x 4 is 28 units2. From the home screen of many calculators, the keystrokes MATH 9 call up the fnInt( feature, which can be used to calculate these types of areas. To use this feature, we must enter (separated by commas) the function, the variable x, the lower limit of the x-values, and the upper limit of the x-values. The screenshot shown in Figure 12.37 verifies our result from Example 5. While the fnInt( command is convenient to use, it ignores the underlying idea of area and reduces these problems to a dry computation. Our calculators have Figure 12.38 a very nice feature that provides the same information, while highlighting the 11 area connection. Begin by entering the function in the Y= screen, in this case Y1 X 5. Setting the x-values of the to range from 9.4 to 9.4 and the y-values to range from 1.4 to 11 will create a “friendly” GRAPH of this function, as shown in Figure 12.38. Now for the fun part! The 2nd TRACE 9.4 9.4 (CALC) 7:兰f(x)dx feature not only calculates the specified area under the curve, but also uses an animated shading to display the region. After we provide WINDOW

1.4

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Figure 12.39

Figure 12.40

11

⫺9.4

11

⫺9.4

9.4

⫺1.4

9.4

⫺1.4

4 the lower and upper x limits with the keystrokes 0 (for this example), this feature of our calculator once again confirms the area [which is 冮f 1x2dx in calculus notation] is 28 units2. See Figure 12.39. The graph of the region in Figure 12.39 once again reminds us that we could have found the area of this trapezoid using simple geometry. The usefulness of the approximating rectangles and our calculators becomes apparent when we look to more complex shapes. For instance, in Example 6 of Section 12.4 we used the rectangle technique to 1 determine the area under the graph of f 1x2 ⫽ x2 ⫹ 2 in the interval [0, 5] was 20.416 units2. After we enter 4 1 2 Y1 ⫽ X ⫹ 2 and graph it in the shown in Figure 12.40, our calculator verifies this result. 4 ENTER

ENTER

WINDOW

Exercise 1: Use both the fnInt( and 2nd TRACE (CALC) 7:兰f(x)dx features to determine the area under the graph of 1 f 1x2 ⫽ ⫺ x2 ⫹ 4x over the interval [0, 6]. See Section 12.4, Exercise 43. 2 Exercise 2: Use both the fnInt( and 2nd TRACE (CALC) 7:兰f(x)dx features to determine the area under the graph of 1 f 1x2 ⫽ ⫺ x3 ⫹ 6x over the interval [0, 3]. See Section 12.4, Exercise 44. 2 Exercise 3: Use the 2nd TRACE (CALC) 7:兰f(x)dx feature to determine the area “under” the graph of 1 f (x) ⫽ x2 ⫺ 4x over the interval [0, 6]. Compare your result to Exercise 1 and comment on what you notice. 2 Exercise 4: Use the 2nd TRACE (CALC) 7:兰f(x)dx feature to determine the area “under” the graph of 1 f 1x2 ⫽ ⫺ x3 ⫹ 6x over the interval 3 ⫺3, 34 . Comment on what you notice. 2

CUMULATIVE REVIEW CHAPTERS 1–12 1. Graph the function f 1x2 ⫽ 2x2 ⫺ 12x ⫹ 15 by completing the square. 2. Solve the system. x ⫹ 3y ⫺ 2z ⫽ 6 • 2x ⫹ y ⫹ z ⫽ 2 ⫺3x ⫹ 4y ⫺ 2z ⫽ 3 3. Write as a sum/difference of logarithmic terms. ln c

2 3

x 2x2 ⫹ 3 d 12x ⫹ 12 4

4. Standing 134 ft from the St. Louis Arch, the angle of elevation to the top of the arch is 78°. How tall is the arch?

5. Verify that a

8 15 , ⫺ b is a point on the unit circle, 17 17 then use it to find the value of sin ␪, cos ␪, and tan ␪.

6. Find the sum of the infinite series 1 1 1 1 ⫹ ⫹ ⫹ ⫹p 3 9 27 81 7. Use the Guidelines for Graphing Rational Functions x to graph v1x2 ⫽ 2 . x ⫺9 8. Write f 1x2 ⫽ x3 ⫺ 2x2 ⫺ 5x ⫹ 6 in completely factored form, then graph the function.

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9. Evaluate the following limits: 0x 1 0 0x 1 0 a. lim 2 and lim 2 xS1 x 1 xS1 x 1 0x 1 0 b. lim 2 xS1 x 1 10. For z 2 2i, use De Moivre’s theorem to find z6 12 2i2 6.

11. Solve the equation 2 13 tan 1␣ 12 3. Find all solutions in [0, 2␲). 12. I am a heavy pencil user. My new pencils are 19 cm long, but after 30 days of heavy use they now average only 9 cm in length. Assuming their length decreases linearly, (a) find a function L(x) that models the length of my pencils after x days of use. (b) Use the model to find the length of a pencil I’ve been using for 15 days. (c) If one of my pencils is 11 cm long, how many days has it been in use? 13. An airline pilot is having to fight a stiff crosswind to stay on course. If the plane is flying at 250 mph on a heading of 20°, and the wind is from the west at 50 mph, what would be the course and heading of the plane if no course corrections are made? 14. Graph the piecewise-defined function given, then state whether its graph is continuous. f 1x2 e 15. Solve for x: 3e

2x1

1x 22 2 x 0 x2 x 7 0

28.08.

16. Graph the function y 2 cosax

␲ b. 4

17. Evaluate the limit: 6x

lim

xSq 24x2

5 18. Verify the following is an identity: sin4x cos4x 2 sin2x 1 19. Decompose the rational expression

x2 x 4 into x2 x 2

a sum of partial fractions. 20. Find all real solutions for 3510 7000 sin a4␪

␲ b 10. 6

21. Find the equation of the ellipse with vertices at (4, 1) and (6, 1), and foci at (3, 1) and (5, 1). 22. Graph the hyperbola by completing the square in x and y, then writing the equation in standard form: x2 4y2 24y 6x 43 0.

Determine the following using the graph of f (x) shown: 23. a. b.

lim f 1x2

y

xS3

f (x)

lim f 1x2

xS3

c. lim f 1x2

5 4 3 2 1

54321 1 2 3 4 5

xS2

d. lim f 1x2 xS5

24. a. f 132 f 102

1 2 3 4 5 x

b. f 122 f 142

c. f 112 f 132 f 102

f 122 25. Hi’ilani has to reset her five-digit PIN number. How many PINs are possible if no repetitions are allowed and two different common vowels must be followed by three different nonzero digits? d.

Exercises 26 through 30 require the use of a graphing calculator. 26. Solve the system using a graphing calculator. Round your answers to two decimal places. e

x y2 10 3x2 2y2 30

27. Solve the following system using inverse matrices and your calculator. w x y z 2 2w x 3y z 9 μ w x y z 4 3w 2x 2y 3z 19 28. In Cuivre River State Park, there are two dominant squirrel species: gray-tailed and red-tailed. The function g1t2 ln冟t5 3t4 2t3 5t2 t 3冟 models the population of gray-tailed squirrels from 1990 to 2010, while r1t2 ln冟2t3 5t2 3t 1冟 3 models the red-tailed population during this same 20-yr period. These two functions give the respective number of squirrels (in hundreds) living in this park during year t, where t 0 corresponds to 1990. Using integervalued years and the TABLE feature of your calculator, determine when these two populations were closest in size. What was the difference in populations that year? 29. Determine the intervals over which the function f 1x2 4x3 4x2 32x 17 is increasing. 30. Find all solutions (if they exist) using a graphing calculator. x x2 ex1 Bx 1 2


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Appendix A A Review of Basic Concepts and Skills APPENDIX OUTLINE A.1 Algebraic Expressions and the Properties of Real Numbers A-1 A.2 Exponents, Scientific Notation, and a Review of Polynomials A-10 A.3 Solving Linear Equations and Inequalities A-24 A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring A-38 A.5 Rational Expressions and Equations A-52 A.6 Radicals, Rational Exponents, and Radical Equations A-64

A.1

Algebraic Expressions and the Properties of Real Numbers

LEARNING OBJECTIVES In Section A.1 you will review how to:

A. Identify terms,

B. C. D.

E.

coefficients, and expressions Create mathematical models Evaluate algebraic expressions Identify and use properties of real numbers Simplify algebraic expressions

To effectively use mathematics as a problem-solving tool, we must develop the ability to translate written or verbal information into a mathematical model. After obtaining a model, many applications require that you work effectively with algebraic terms and expressions. The basic ideas involved are reviewed here.

A. Terms, Coefficients, and Algebraic Expressions An algebraic term is a collection of factors that may include numbers, variables, or expressions within parentheses. Here are some examples: (a) 3

(b) ⫺6P

(c) 5xy

(d) ⫺8n2

(e) n

(f ) 21x ⫹ 32

If a term consists of a single nonvariable number, it is called a constant term. In (a), 3 is a constant term. Any term that contains a variable is called a variable term. We call the constant factor of a term the numerical coefficient or simply the coefficient. The coefficients for (a), (b), (c), and (d) are 3, ⫺6, 5, and ⫺8, respectively. In (e), the coefficient of n is 1, since 1 # n ⫽ 1n ⫽ n. The term in (f ) has two factors as written, 2 and 1x ⫹ 32. The coefficient is 2. An algebraic expression can be a single term or a sum or difference of terms. To avoid confusion when identifying the coefficient of each term, the expression can be rewritten using algebraic addition if desired: A ⫺ B ⫽ A ⫹ 1⫺B2. For instance, 4 ⫺ 3x ⫽ 4 ⫹ 1⫺3x2 shows the coefficient of x is ⫺3. To identify the coefficient of a rational term, it sometimes helps to decompose the term, rewriting it using a unit fraction as in n ⫺5 2 ⫽ 15 1n ⫺ 22 and 2x ⫽ 12x. A-1


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A-2

APPENDIX A A Review of Basic Concepts and Skills

EXAMPLE 1

䊳

Identifying Terms and Coefficients State the number of terms in each expression as given, then identify the coefficient of each term. x⫹3 ⫺ 2x a. 2x ⫺ 5y b. c. ⫺1x ⫺ 122 d. ⫺2x2 ⫺ x ⫹ 5 7

Solution

䊳

We can begin by rewriting each subtraction using algebraic addition. Rewritten: Number of terms: Coefficient(s):

A. You’ve just seen how we can identify terms, coefficients, and expressions

a. 2x ⫹ 1⫺5y2 b. 17 1x ⫹ 32 ⫹ 1⫺2x2 c. ⫺11x ⫺ 122 d. ⫺2x2 ⫹ 1⫺1x2 ⫹ 5 two 2 and ⫺5

two 1 7

and ⫺2

one

three

⫺1

⫺2, ⫺1, and 5


䊳

B. Translating Written or Verbal Information into a Mathematical Model The key to solving many applied problems is finding an algebraic expression that accurately models relationships described in context. First, we assign a variable to represent an unknown quantity, then build related expressions using words from the English language that suggest mathematical operations. Variables that remind us of what they represent are often used in the modeling process, such as D ⫽ RT for Distance equals Rate times Time. These are often called descriptive variables. Capital letters are also used due to their widespread appearance in other fields. EXAMPLE 2

䊳

Translating English Phrases into Algebraic Expressions Assign a variable to the unknown number, then translate each phrase into an algebraic expression. a. twice a number, increased by five b. eleven less than eight times the width c. ten less than triple the payment d. two hundred fifty dollars more than double the amount

Solution

䊳

a. Let n represent the number. Then 2n represents twice the number, and 2n ⫹ 5 represents twice the number, increased by five. b. Let W represent the width. Then 8W represents eight times the width, and 8W ⫺ 11 represents 11 less than eight times the width. c. Let p represent the payment. Then 3p represents triple the payment, and 3p ⫺ 10 represents 10 less than triple the payment. d. Let A represent the amount in dollars. Then 2A represents double the amount, and 2A ⫹ 250 represents 250 dollars more than double the amount. Now try Exercises 15 through 28

䊳

Identifying and translating such phrases when they occur in context is an important problem-solving skill. Note how this is done in Example 3. EXAMPLE 3

䊳

Creating a Mathematical Model The cost for a rental car is $35 plus 15 cents per mile. Express the cost of renting a car in terms of the number of miles driven.

Solution

䊳

B. You’ve just seen how we can create mathematical models

Let m represent the number of miles driven. Then 0.15m represents the cost for each mile and C ⫽ 35 ⫹ 0.15m represents the total cost for renting the car. Now try Exercises 29 through 40

䊳


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Section A.1 Algebraic Expressions and the Properties of Real Numbers

A-3

C. Evaluating Algebraic Expressions We often need to evaluate expressions to investigate patterns and note relationships. Evaluating a Mathematical Expression 1. Replace each variable with open parentheses ( ). 2. Substitute the values given for each variable. 3. Simplify using the order of operations. In this process, it’s best to use a vertical format, with the original expression written first, the substitutions shown next, followed by the simplified forms and the final result. The numbers substituted or “plugged into” the expression are often called the input values, with the result called the output value. EXAMPLE 4

䊳

Evaluating an Algebraic Expression Evaluate the expression x3 ⫺ 2x2 ⫹ 5 for x ⫽ ⫺3.

Solution

䊳

WORTHY OF NOTE In Example 4, note the importance of the first step in the evaluation process: replace each variable with open parentheses. Skipping this step could easily lead to confusion as we try to evaluate the squared term, since ⫺32 ⫽ ⫺9, while 1⫺32 2 ⫽ 9. Also see Exercises 55 and 56.

For x ⫽ ⫺3:

x3 ⫺ 2x2 ⫹ 5 ⫽ 1 2 3 ⫺ 21 2 2 ⫹ 5 ⫽ 1⫺32 3 ⫺ 21⫺32 2 ⫹ 5 ⫽ ⫺27 ⫺ 2192 ⫹ 5 ⫽ ⫺27 ⫺ 18 ⫹ 5 ⫽ ⫺40

replace variables with open parentheses substitute ⫺3 for x

simplify: 1⫺32 3 ⫽ ⫺27, 1⫺32 2 ⫽ 9 simplify: 2192 ⫽ 18 result

When the input is ⫺3, the output is ⫺40. Now try Exercises 41 through 60

䊳

If the same expression is evaluated repeatedly, results are often collected and analyzed in a table of values, as shown in Example 5. As a practical matter, the substitutions and simplifications are often done mentally or on scratch paper, with the table showing only the input and output values. EXAMPLE 5

䊳

Evaluating an Algebraic Expression Evaluate x2 ⫺ 2x ⫺ 3 to complete the table shown. Which input value(s) of x cause the expression to have an output of 0?

Solution

䊳

Input x ⫺2 ⫺1

Output x2 ⴚ 2x ⴚ 3

1⫺22 2 ⫺ 21⫺22 ⫺ 3 ⫽ 5 0

0

⫺3

1

⫺4

2

⫺3

3

0

4

5

The expression has an output of 0 when x ⫽ ⫺1 and x ⫽ 3. Now try Exercises 61 through 66

䊳


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A-4


Graphing calculators provide an efficient means of evaluating many expressions. After entering the expression on the Y= screen (Figure A.1), we can set up the table using the keystrokes 2nd (TBLSET). For this exercise, we’ll put the table in the “Indpnt: Auto Ask” mode, which will have the calculator “automatically” generate the input and output values. In this mode, we can tell the calculator where to start the inputs (we chose TblStart ⫽ ⫺2), and have the calculator produce the input values using any increment desired (we choose ¢Tbl ⫽ 1), as shown in Figure A.2. We access the completed table using 2nd GRAPH (TABLE), and the result for Example 5 is shown in Figure A.3. For exercises that combine the skills from Examples 3 through 5, see Exercises 91 to 98.

Figure A.1

Figure A.2

WINDOW

C. You’ve just seen how we can evaluate algebraic expressions

Figure A.3

D. Properties of Real Numbers While the phrase, “an unknown number times five,” is accurately modeled by the expression n5 for some number n, in algebra we prefer to have numerical coefficients precede variable factors. When we reorder the factors as 5n, we are using the commutative property of multiplication. A reordering of terms involves the commutative property of addition. The Commutative Properties Given that a and b represent real numbers: ADDITION: a ⫹ b ⫽ b ⫹ a

MULTIPLICATION: a # b ⫽ b # a

Terms can be combined in any order without changing the sum.

Factors can be multiplied in any order without changing the product.

Each property can be extended to include any number of terms or factors. While the commutative property implies a reordering or movement of terms (to commute implies back-and-forth movement), the associative property implies a regrouping or reassociation of terms. For example, the sum 1 34 ⫹ 35 2 ⫹ 25 is easier to compute if we regroup the addends as 34 ⫹ 1 35 ⫹ 25 2 . This illustrates the associative property of addition. Multiplication is also associative. The Associative Properties Given that a, b, and c represent real numbers: ADDITION:

EXAMPLE 6

䊳

MULTIPLICATION:

1a ⫹ b2 ⫹ c ⫽ a ⫹ 1b ⫹ c2

1a # b2 # c ⫽ a # 1b # c2

Terms can be regrouped.

Factors can be regrouped.

Simplifying Expressions Using Properties of Real Numbers Use the commutative and associative properties to simplify each calculation.

Solution

䊳

b. 3⫺2.5 # 1⫺1.22 4 # 10

a.

3 8

⫺ 19 ⫹ 58

a.

3 8

⫺ 19 ⫹ 58 ⫽ ⫺19 ⫹ 38 ⫹ 58 ⫽ ⫺19 ⫹ 1 38 ⫹ 58 2 ⫽ ⫺19 ⫹ 1 ⫽ ⫺18

commutative property (order changes) associative property (grouping changes) simplify result


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Precalculus—


WORTHY OF NOTE Is subtraction commutative? Consider a situation involving money. If you had $100, you could easily buy an item costing $20: $100 ⫺ $20 leaves you with $80. But if you had $20, could you buy an item costing $100? Obviously $100 ⫺ $20 is not the same as $20 ⫺ $100. Subtraction is not commutative. Likewise, 100 ⫼ 20 is not the same as 20 ⫼ 100, and division is not commutative.

b. 3⫺2.5 # 1⫺1.22 4 # 10 ⫽ ⫺2.5 # 3 1⫺1.22 # 10 4 ⫽ ⫺2.5 # 1⫺122 ⫽ 30

A-5

associative property (grouping changes) simplify result


䊳

For any real number x, x ⫹ 0 ⫽ x and 0 is called the additive identity since the original number was returned or “identified.” Similarly, 1 is called the multiplicative identity since 1 # x ⫽ x. The identity properties are used extensively in the process of solving equations. The Additive and Multiplicative Identities Given that x is a real number, x⫹0⫽x Zero is the identity for addition.

1#x⫽x One is the identity for multiplication.

For any real number x, there is a real number ⫺x such that x ⫹ 1⫺x2 ⫽ 0. The number ⫺x is called the additive inverse of x, since their sum results in the additive identity. Similarly, the multiplicative inverse of any nonzero number x is 1x , since x # 1x ⫽ 1 p q (the multiplicative identity). This property can also be stated as q # p ⫽ 1 1p, q ⫽ 02 for q p p any rational number q. Note that q and p are reciprocals. The Additive and Multiplicative Inverses

Given that p, q, and x represent real numbers 1p, q ⫽ 02: x ⫹ 1⫺x2 ⫽ 0

p q # ⫽1 q p p q

x and ⫺x are additive inverses. EXAMPLE 7

䊳

Determining Additive and Multiplicative Inverses Replace the box to create a true statement: # ⫺3 x ⫽ 1 # x a. b. x ⫹ 4.7 ⫹ 5

Solution

䊳

q

and p are multiplicative inverses.

a.

⫽

5 ⫺3 5 # , since ⫽1 ⫺3 ⫺3 5

⫽x ⫽ ⫺4.7, since 4.7 ⫹ 1⫺4.72 ⫽ 0

b.


䊳

Note that if no coefficient is indicated, it is assumed to be 1, as in x ⫽ 1x, 1x2 ⫹ 3x2 ⫽ 11x2 ⫹ 3x2 , and ⫺1x3 ⫺ 5x2 2 ⫽ ⫺11x3 ⫺ 5x2 2. The distributive property of multiplication over addition is widely used in a study of algebra, because it enables us to rewrite a product as an equivalent sum and vice versa. The Distributive Property of Multiplication over Addition Given that a, b, and c represent real numbers: a1b ⫹ c2 ⫽ ab ⫹ ac A factor outside a sum can be distributed to each addend in the sum.

ab ⫹ ac ⫽ a1b ⫹ c2 A factor common to each addend in a sum can be “undistributed” and written outside a group.


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Precalculus—

A-6


EXAMPLE 8

䊳

Simplifying Expressions Using the Distributive Property Apply the distributive property as appropriate. Simplify if possible. a. 71p ⫹ 5.22

Solution

䊳

WORTHY OF NOTE From Example 8b we learn that a negative sign outside a group changes the sign of all terms within the group: ⫺12.5 ⫺ x2 ⫽ ⫺2.5 ⫹ x.

b. ⫺12.5 ⫺ x2

c. 7x3 ⫺ x3

d.

1 5 n⫹ n 2 2

a. 71p ⫹ 5.22 ⫽ 7p ⫹ 715.22 ⫽ 7p ⫹ 36.4

b. ⫺12.5 ⫺ x2 ⫽ ⫺112.5 ⫺ x2 ⫽ ⫺112.52 ⫺ 1⫺121x2 ⫽ ⫺2.5 ⫹ x

c. 7x3 ⫺ x3 ⫽ 7x3 ⫺ 1x3 ⫽ 17 ⫺ 12x3 ⫽ 6x3

d.

5 1 5 1 n ⫹ n ⫽ a ⫹ bn 2 2 2 2 6 ⫽ a bn 2 ⫽ 3n

D. You’ve just seen how we can identify and use properties of real numbers


䊳

E. Simplifying Algebraic Expressions Two terms are like terms only if they have the same variable factors (the coefficient is not used to identify like terms). For instance, 3x2 and ⫺17x2 are like terms, while 5x3 and 5x2 are not. We simplify expressions by combining like terms using the distributive property, along with the commutative and associative properties. Many times the distributive property is used to eliminate grouping symbols and combine like terms within the same expression. EXAMPLE 9

䊳

Simplifying an Algebraic Expression

Solution

䊳

712p2 ⫹ 12 ⫺ 11p2 ⫹ 32

Simplify the expression completely: 712p2 ⫹ 12 ⫺ 1p2 ⫹ 32 . ⫽ 14p ⫹ 7 ⫺ 1p ⫺ 3 ⫽ 114p2 ⫺ 1p2 2 ⫹ 17 ⫺ 32 ⫽ 114 ⫺ 12p2 ⫹ 4 ⫽ 13p2 ⫹ 4 2

2

original expression; note coefficient of ⫺1 distributive property commutative and associative properties (collect like terms) distributive property result


䊳

The steps for simplifying an algebraic expression are summarized here: To Simplify an Expression 1. Eliminate parentheses by applying the distributive property. 2. Use the commutative and associative properties to group like terms. 3. Use the distributive property to combine like terms. E. You’ve just seen how we can simplify algebraic expressions

As you practice with these ideas, many of the steps will become more automatic. At some point, the distributive property, the commutative and associative properties, as well as the use of algebraic addition will all be performed mentally.


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Precalculus—


A-7

A.1 EXERCISES 䊳



䊳

1. A term consisting of a single number is called a(n) term.

4. When 3 # 14 # 23 is written as 3 # 23 # 14, the property has been used.

2. A term containing a variable is called a(n) term.

5. Discuss/Explain why the additive inverse of ⫺5 is 5, while the multiplicative inverse of ⫺5 is ⫺15.

3. The constant factor in a variable term is called the .

6. Discuss/Explain how we can rewrite the sum 3x ⫹ 6y as a product, and the product 21x ⫹ 72 as a sum.


Identify the number of terms in each expression and the coefficient of each term.

7. 3x ⫺ 5y 9. 2x ⫹

x⫹3 4

8. ⫺2a ⫺ 3b 10.

n⫺5 ⫹ 7n 3

11. ⫺2x2 ⫹ x ⫺ 5

12. 3n2 ⫹ n ⫺ 7

13. ⫺1x ⫹ 52

14. ⫺1n ⫺ 32

Translate each phrase into an algebraic expression.

15. seven fewer than a number 16. a number decreased by six 17. the sum of a number and four 18. a number increased by nine

Create a mathematical model using descriptive variables.

29. The length of the rectangle is three meters less than twice the width. 30. The height of the triangle is six centimeters less than three times the base. 31. The speed of the car was fifteen miles per hour more than the speed of the bus. 32. It took Romulus three minutes more time than Remus to finish the race. 33. Hovering altitude: The helicopter was hovering 150 ft above the top of the building. Express the altitude of the helicopter in terms of the building’s height.

19. the difference between a number and five is squared 20. the sum of a number and two is cubed 21. thirteen less than twice a number 22. five less than double a number 23. a number squared plus the number doubled 24. a number cubed less the number tripled 25. five fewer than two-thirds of a number 26. fourteen more than one-half of a number 27. three times the sum of a number and five, decreased by seven 28. five times the difference of a number and two, increased by six

34. Stacks on a cruise liner: The smoke stacks of the luxury liner cleared the bridge by 25 ft as it passed beneath it. Express the height of the stacks in terms of the bridge’s height. 35. Dimensions of a city park: The length of a rectangular city park is 20 m more than twice its width. Express the length of the park in terms of the width.


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Precalculus—

A-8


36. Dimensions of a parking lot: In order to meet the city code while using the available space, a contractor planned to construct a parking lot with a length that was 50 ft less than three times its width. Express the length of the lot in terms of the width. 37. Cost of milk: In 2010, a gallon of milk cost two and one-half times what it did in 1990. Express the cost of a gallon of milk in 2010 in terms of the 1990 cost. 38. Cost of gas: In 2010, a gallon of gasoline cost two and one-half times what it did in 1990. Express the cost of a gallon of gas in 2010 in terms of the 1990 cost. 39. Pest control: In her pest control business, Judy charges $50 per call plus $12.50 per gallon of insecticide for the control of spiders and certain insects. Express the total charge in terms of the number of gallons of insecticide used. 40. Computer repairs: As his reputation and referral business grew, Keith began to charge $75 per service call plus an hourly rate of $50 for the repair and maintenance of home computers. Express the cost of a service call in terms of the number of hours spent on the call. Evaluate each algebraic expression given x ⴝ 2 and y ⴝ ⴚ3.

41. 4x ⫺ 2y

42. 5x ⫺ 3y

43. ⫺2x2 ⫹ 3y2

44. ⫺5x2 ⫹ 4y2

45. 2y2 ⫹ 5y ⫺ 3

46. 3x2 ⫹ 2x ⫺ 5

47. ⫺213y ⫹ 12

48. ⫺312y ⫹ 52

49. 3x2y

50. 6xy2

51. 1⫺3x2 2 ⫺ 4xy ⫺ y2 52. 1⫺2x2 2 ⫺ 5xy ⫺ y2 53. 12x ⫺ 13y

55. 13x ⫺ 2y2 57.

54. 32x ⫺ 12y

2

⫺12y ⫹ 5 ⫺3x ⫹ 1

59. 1⫺12y # 4

56. 12x ⫺ 3y2 58.

2

12x ⫹ 1⫺32 ⫺3y ⫹ 1

60. 7 # 1⫺27y

Evaluate each expression for integers from ⴚ3 to 3 inclusive. Verify results using a graphing calculator. What input(s) give an output of zero?

61. x2 ⫺ 3x ⫺ 4

62. x2 ⫺ 2x ⫺ 3

63. ⫺311 ⫺ x2 ⫺ 6

64. 513 ⫺ x2 ⫺ 10

65. x3 ⫺ 6x ⫹ 4

66. x3 ⫹ 5x ⫹ 18

Rewrite each expression using the given property and simplify if possible.

67. Commutative property of addition a. ⫺5 ⫹ 7 b. ⫺2 ⫹ n c. ⫺4.2 ⫹ a ⫹ 13.6 d. 7 ⫹ x ⫺ 7 68. Associative property of multiplication a. 2 # 13 # 62 b. 3 # 14 # b2 # # c. ⫺1.5 16 a2 d. ⫺6 # 1⫺56 # x2 Replace the box so that a true statement results.

69. a. x ⫹ 1⫺3.22 ⫹ b. n ⫺ 56 ⫹ 70. a. b.

⫽x

⫽n

# 23x ⫽ 1x #

n ⫽ 1n ⫺3

Apply the distributive property and simplify if possible.

71. ⫺51x ⫺ 2.62 72. ⫺121v ⫺ 3.22 73. 23 1⫺15p ⫹ 92

2 74. 56 1⫺15 q ⫹ 242

75. 3a ⫹ 1⫺5a2

76. 13m ⫹ 1⫺5m2 77. 23x ⫹ 34x 78.

5 12 y

⫺ 38y

Simplify by removing all grouping symbols (as needed) and combining like terms.

79. 31a2 ⫹ 3a2 ⫺ 15a2 ⫹ 7a2 80. 21b2 ⫹ 5b2 ⫺ 16b2 ⫹ 9b2 81. x2 ⫺ 13x ⫺ 5x2 2

82. n2 ⫺ 15n ⫺ 4n2 2

83. 13a ⫹ 2b ⫺ 5c2 ⫺ 1a ⫺ b ⫺ 7c2

84. 1x ⫺ 4y ⫹ 8z2 ⫺ 18x ⫺ 5y ⫺ 2z2 85. 35 15n ⫺ 42 ⫹ 58 1n ⫹ 162 86. 23 12x ⫺ 92 ⫹ 34 1x ⫹ 122

87. 13a2 ⫺ 5a ⫹ 72 ⫹ 212a2 ⫺ 4a ⫺ 62

88. 213m2 ⫹ 2m ⫺ 72 ⫺ 1m2 ⫺ 5m ⫹ 42


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Precalculus—

Section A.1 Algebraic Expressions and the Properties of Real Numbers 䊳


89. Electrical resistance: R ⴝ

kL d2

The electrical resistance in a wire depends on the length and diameter of the wire. This resistance can be modeled by the formula shown, where R is the resistance in ohms, L is the length in feet, and d is the diameter of the wire in inches. Find the resistance if k ⫽ 0.000025, d ⫽ 0.015 in., and L ⫽ 90 ft

䊳

A-9

k V If temperature remains constant, the pressure of a gas held in a closed container is related to the volume of gas by the formula shown, where P is the pressure in pounds per square inch, V is the volume of gas in cubic inches, and k is a constant that depends on given conditions. Find the pressure exerted by the gas if k ⫽ 440,310 and V ⫽ 22,580 in3.

90. Volume and pressure: P ⴝ

APPLICATIONS

Translate each key phrase into an algebraic expression, then evaluate as indicated.

91. Cruising speed: A turbo-prop airliner has a cruising speed that is one-half the cruising speed of a 767 jet aircraft. (a) Express the speed of the turbo-prop in terms of the speed of the jet, and (b) determine the speed of the airliner if the cruising speed of the jet is 550 mph. 92. Softball toss: Macklyn can throw a softball twothirds as far as her father. (a) Express the distance that Macklyn can throw a softball in terms of the distance her father can throw. (b) If her father can throw the ball 210 ft, how far can Macklyn throw the ball? 93. Dimensions of a lawn: The length of a rectangular lawn is 3 ft more than twice its width. (a) Express the length of the lawn in terms of the width. (b) If the width is 52 ft, what is the length? 94. Pitch of a roof: To obtain the proper pitch, the crossbeam for a roof truss must be 2 ft less than three-halves the rafter. (a) Express the length of the crossbeam in terms of the rafter. (b) If the rafter is 18 ft, how long is the crossbeam? 䊳

95. Postage costs: In 2009, a first class stamp cost 29¢ more than it did in 1978. Express the cost of a 2009 stamp in terms of the 1978 cost. If a stamp cost 15¢ in 1978, what was the cost in 2009? 96. Minimum wage: In 2009, the federal minimum wage was $4.95 per hour more than it was in 1976. Express the 2009 wage in terms of the 1976 wage. If the hourly wage in 1976 was $2.30, what was it in 2009? 97. Repair costs: The TV repair shop charges a flat fee of $43.50 to come to your house and $25 per hour for labor. Express the cost of repairing a TV in terms of the time it takes to repair it. If the repair took 1.5 hr, what was the total cost? 98. Repair costs: At the local car dealership, shop charges are $79.50 to diagnose the problem and $85 per shop hour for labor. Express the cost of a repair in terms of the labor involved. If a repair takes 3.5 hr, how much will it cost?


99. If C must be a positive odd integer and D must be a negative even integer, then C2 ⫹ D2 must be a: a. positive odd integer. b. positive even integer. c. negative odd integer. d. negative even integer. e. cannot be determined.

100. Historically, several attempts have been made to create metric time using factors of 10, but our current system won out. If 1 day was 10 metric hours, 1 metric hour was 10 metric minutes, and 1 metric minute was 10 metric seconds, what time would it really be if a metric clock read 4:3:5? Assume that each new day starts at midnight.


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Precalculus—

A.2

Exponents, Scientific Notation, and a Review of Polynomials


A. Apply properties of

A. The Properties of Exponents

exponents

B. Perform operations in C. D. E.

An exponent is a superscript number or letter occurring to the upper right of a base number, and indicates how many times the base occurs as a factor. For b # b # b b3, we say b3 is written in exponential form. In some cases, we may refer to b3 as an exponential term. Exponential Notation For any positive integer n, bn b # b # b # . . . # b

b # b # b # . . . # b bn

and

⎞ ⎜ ⎜ ⎜ ⎬ ⎜ ⎜ ⎠

scientific notation Identify and classify polynomial expressions Add and subtract polynomials Compute the product of two polynomials Compute special products: binomial conjugates and binomial squares

⎞ ⎜ ⎜ ⎜ ⎬ ⎜ ⎜ ⎠

F.

In this section, we review basic exponential properties and operations on polynomials. Although there are five to eight exponential properties (depending on how you count them), all can be traced back to the basic definition involving repeated multiplication.

n times

n times

The Product and Power Properties There are two properties that follow immediately from this definition. When b3 is multiplied by b2, we have an uninterrupted string of five factors: b3 # b2 1b # b # b2 # 1b # b2, which can then be written as b5. This is an example of the product property of exponents. Product Property of Exponents

WORTHY OF NOTE In this statement of the product property and the exponential properties that follow, it is assumed that for any expression of the form 0m, m 7 0 (hence 0m 0).

For any base b and positive integers m and n: bm # bn bmn In words, the property says, to multiply exponential terms with the same base, keep the common base and add the exponents. A special application of the product property uses repeated factors of the same exponential term, as in (x2)3. Using the product property, we have 1x2 2 1x2 21x2 2 x6. Notice the same result can be found more quickly by # multiplying the inner exponent by the outer exponent: 1x2 2 3 x2 3 x6. We generalize this idea to state the power property of exponents. In words the property says, to raise an exponential term to a power, keep the same base and multiply the exponents. Power Property of Exponents For any base b and positive integers m and n:

1bm 2 n bm n #

EXAMPLE 1

䊳

Multiplying Terms Using Exponential Properties Compute each product. a. 4x3 # 12x2 b. 1p3 2 2 # 1p4 2 5

Solution

䊳

# 12x2 14 # 12 21x3 # x2 2

a. 4x3

122 1x 2 2 x5 # # 1p4 2 5 p3 2 # p4 5 p6 # p20 p620 p26 32

b. 1p 2

3 2

#

commutative and associative properties simplify; product property result power property simplify product property result

Now try Exercises 7 through 12 䊳 A-10


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Precalculus—

Section A.2 Exponents, Scientific Notation, and a Review of Polynomials

A-11

The power property can easily be extended to include more than one factor within the parentheses. This application of the power property is sometimes called the product to a power property and can be extended to include any number of factors. We can also raise a quotient of exponential terms to a power. The result is called the quotient to a power property. In words the properties say, to raise a product or quotient of exponential terms to a power, multiply every exponent inside the parentheses by the exponent outside the parentheses. Product to a Power Property For any bases a and b, and positive integers m, n, and p: 1ambn 2 p amp # bnp

Quotient to a Power Property For any bases a and b 0, and positive integers m, n, and p: a EXAMPLE 2

䊳

am p amp b np bn b

Simplifying Terms Using the Power Properties Simplify using the power property (if possible): 5a3 2 a. 13a2 2 b. 3a2 c. a b 2b

Solution

䊳

WORTHY OF NOTE Regarding Examples 2a and 2b, note the difference between the expressions 13a2 2 13 # a2 2 and 3a2 3 # a2. In the first, the exponent acts on both the negative 3 and the a; in the second, the exponent acts on only the a and there is no “product to a power.”

a. 13a2 2 132 2 # 1a1 2 2 9a2 152 2 1a3 2 2 5a3 2 b c. a 2b 22b2 25a6 4b2

b. 3a2 is in simplified form

Now try Exercises 13 through 24 䊳 Applications of exponents sometimes involve linking one exponential term with another using a substitution. The result is then simplified using exponential properties.

EXAMPLE 3

䊳

Applying the Power Property after a Substitution The formula for the volume of a cube is V S3, where S is the length of one edge. If the length of each edge is 2x2: a. Find a formula for volume in terms of x. b. Find the volume if x 2.

Solution

䊳

a. V S

3

S

substitute 2x 2 for S

12x2 2 3 8x6

2x2 2x2

2x2

b. For V 8x , V 8122 6 substitute 2 for x # 8 64 or 512 122 6 64 The volume of the cube would be 512 units3. 6



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Precalculus—

A-12


The Quotient Property of Exponents x 1 for x 0, we note a x x5 x # x # x # x # x x3, pattern that helps to simplify a quotient of exponential terms. For 2 # x x x the exponent of the final result appears to be the difference between the exponent in the numerator and the exponent in the denominator. This seems reasonable since the subtraction would indicate a removal of the factors that reduce to 1. Regardless of how many factors are used, we can generalize the idea and state the quotient property of exponents. In words the property says, to divide two exponential terms with the same base, keep the common base and subtract the exponent of the denominator from the exponent of the numerator. By combining exponential notation and the property

Quotient Property of Exponents For any base b 0 and positive integers m and n: bm bmn bn

Zero and Negative Numbers as Exponents If the exponent of the denominator is greater than the exponent in the numerator, the x2 quotient property yields a negative exponent: 5 x25 x3. To help understand x what a negative exponent means, let’s look at the expanded form of the expression: x2 x # x1 1 3 . A negative exponent can literally be interpreted as “write 5 # # # # x x x x x x x the factors as a reciprocal.” A good way to remember this is

!

!

23

three factors of 2 written as a reciprocal

1 1 23 3 1 8 2

Since the result would be similar regardless of the base used, we can generalize this idea and state the property of negative exponents. Property of Negative Exponents For any base b 0 and integer n: bn 1 n 1 b WORTHY OF NOTE The use of zero as an exponent should not strike you as strange or odd; it’s simply a way of saying that no factors of the base remain, since all terms have been reduced to 1. 8 23 For 3 , we have 1, or 8 2 1

1

1

2#2#2 1, or 233 20 1. 2#2#2

1 bn bn 1

a n b n a b a b ;a0 a b

x3 x3 by division, and 1 x33 x0 using the x3 x3 quotient property, we conclude that x0 1 as long as x 0. We can also generalize this observation and state the meaning of zero as an exponent. In words the property says, any nonzero quantity raised to an exponent of zero is equal to 1. Finally, when we consider that

Zero Exponent Property For any base b 0: b0 1


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Precalculus—

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EXAMPLE 4

Solution

䊳

䊳

Simplifying Expressions Using Exponential Properties Simplify using exponential properties. Answer using positive exponents only. 2a3 2 a. a 2 b b. 13hk2 2 3 16h2k3 2 2 b 12m2n3 2 5 c. 13x2 0 3x0 32 d. 14mn2 2 3 2a3 2 b2 2 a. a 2 b a 3 b property of negative exponents b 2a 1b2 2 2 2 3 2 power properties 2 1a 2

b4 4a6

result

b. 13hk2 2 3 16h2k3 2 2 33h3 1k2 2 3 # 62 1h2 2 2 1k3 2 2 3 3 6

3hk

3

2

3

#6

2 4 6

#6 hk # h34 # k66

27h k 36

1 32

1 9

Notice in Example 4(c), we have 13x2 0 13 # x2 0 1, while 3x0 3 # x0 3112. This is another example of operations and grouping symbols working together: 13x2 0 1 because any quantity to the zero power is 1. However, for 3x0 there are no grouping symbols, so the exponent 0 acts only on the x and not the 3: 3x0 3 # x0 3112 3.

12m2n3 2 5 14mn2 2 3

122 5 1m2 2 5 1n3 2 5 43m3 1n2 2 3

result

power property

32m10n15 64m3n6

simplify

1m7n9 2

quotient property

m7n9 2

1 b 36

zero exponent property; property of negative exponents

simplify: (3x )0 1, 3x 0 3 1 3

1 37 4 9 9 d.

1 62

result 1k 0 12

c. 13x2 0 3x0 32 1 3112

WORTHY OF NOTE

product property a62

3h7 4

4

simplify

simplify

7 0

power property

result



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Precalculus—

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Summary of Exponential Properties For real numbers a and b, and integers m, n, p (excluding 0 raised to a nonpositive power) Product property: bm # bn bmn # Power property: 1bm 2 n bm n Product to a power: 1ambn 2 p amp # bnp am p amp Quotient to a power: a n b np 1b 02 b b m b Quotient property: bmn 1b 02 bn Zero exponents: b0 11b 02 1 1 a n b n bn Negative exponents: n , n bn, a b a b 1a, b 02 a 1 b b b

A. You’ve just seen how we can apply properties of exponents

B. Exponents and Scientific Notation In many technical and scientific applications, we encounter numbers that are either extremely large or very, very small. For example, the mass of the Moon is over 73 quintillion kilograms (73 followed by 18 zeroes), while the constant for universal gravitation contains 10 zeroes before the first nonzero digit. When computing with numbers of this magnitude, scientific notation has a distinct advantage over the common decimal notation (base-10 place values). WORTHY OF NOTE

Scientific Notation

Recall that multiplying by 10’s (or multiplying by 10k, k 7 02 shifts the decimal point to the right k places, making the number larger. Dividing by 10’s (or multiplying by 10k, k 7 0) shifts the decimal point to the left k places, making the number smaller.

A non-zero number written in scientific notation has the form N 10k

where 1 0N 0 6 10 and k is an integer. To convert a number from decimal notation into scientific notation, we begin by placing the decimal point to the immediate right of the first nonzero digit (creating a number less than 10 but greater than or equal to 1) and multiplying by 10k. Then we determine the power of 10 (the value of k) needed to ensure that the two forms are equivalent. When writing large or small numbers in scientific notation, we sometimes round the value of N to two or three decimal places.

EXAMPLE 5

䊳

Converting from Decimal Notation to Scientific Notation The mass of the Moon is about 73,000,000,000,000,000,000 kg. Write this number in scientific notation.

Solution

䊳

Place decimal to the right of first nonzero digit (7) and multiply by 10k. 73,000,000,000,000,000,000 7.3 10k To return the decimal to its original position would require 19 shifts to the right, so k must be positive 19. 73,000,000,000,000,000,000 7.3 1019 The mass of the Moon is 7.3 1019 kg. Now try Exercises 67 and 68 䊳


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Precalculus—


A-15

Converting a number from scientific notation to decimal notation is simply an application of multiplication or division with powers of 10. EXAMPLE 6

䊳

Converting from Scientific Notation to Decimal Notation The constant of gravitation is 6.67 1011. Write this number in common decimal form.

Solution

䊳

Since the exponent is negative 11, shift the decimal 11 places to the left, using placeholder zeroes as needed to return the decimal to its original position: 6.67 1011 0.000 000 000 066 7 Now try Exercises 69 and 70 䊳 Computations that involve scientific notation typically use real number properties and the properties of exponents.

EXAMPLE 7

䊳

Storage Space on a Hard Drive A typical 320-gigabyte portable hard drive can hold about 340,000,000,000 bytes of information. A 2-hr DVD movie can take up as much as 8,000,000,000 bytes of storage space. Find the number of movies (to the nearest whole movie) that can be stored on this hard drive.

Solution

䊳

Using the ideas from Example 5, the hard drive holds 3.4 1011 bytes, while the DVD requires 8.0 109 bytes. Divide to find the number of DVDs the hard drive will hold. 3.4 1011 3.4 1011 8.0 8.0 109 109 0.425 102 42.5

rewrite the expression divide; subtract exponents result

The drive will hold approximately 42 DVD movies. A calculator check is shown in the figure. B. You’ve just seen how we can perform operations in scientific notation


C. Identifying and Classifying Polynomial Expressions A monomial is a term using only whole number exponents on variables, with no variables in the denominator. One important characteristic of a monomial is its degree. For a monomial in one variable, the degree is the same as the exponent on the variable. The degree of a monomial in two or more variables is the sum of exponents occurring on variable factors. A polynomial is a monomial or any sum or difference of monomial terms. For instance, 12x2 5x 6 is a polynomial, while 3n2 2n 7 is not (the exponent 2 is not a whole number). Identifying polynomials is an important skill because they represent a very different kind of real-world model than nonpolynomials. In addition, there are different families of polynomials, with each family having different characteristics. We classify polynomials according to their degree and number of terms. The degree of a polynomial in one variable is the largest exponent occurring on the variable. The degree of a polynomial in more than one variable is the largest sum of exponents in any one term. A polynomial with two terms is called a binomial (bi means two) and a polynomial with three terms is called a trinomial (tri means three). There are special names for polynomials with four or more terms, but for these, we simply use the general name polynomial (poly means many).


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EXAMPLE 8

䊳

Classifying and Describing Polynomials For each expression: a. Classify as a monomial, binomial, trinomial, or polynomial. b. State the degree of the polynomial. c. Name the coefficient of each term.

Solution

䊳

Expression

Classification

Degree

5x y 2xy

binomial

three

5, 2

x2 0.81

binomial

two

1, 0.81

z3 3z2 9z 27

polynomial (four terms)

three

1, 3, 9, 27

binomial

one

trinomial

two

2

3 4 x 2

5

2x x 3

Coefficients

3 4 ,

5

2, 1, 3

Now try Exercises 73 through 78 䊳 A polynomial expression is in standard form when the terms of the polynomial are written in descending order of degree, beginning with the highest-degree term. The coefficient of the highest-degree term is called the leading coefficient. EXAMPLE 9

䊳

Writing Polynomials in Standard Form Write each polynomial in standard form, then identify the leading coefficient.

Solution

䊳

Polynomial

Standard Form

9x

x 9

5z 7z2 3z3 27

3z3 7z2 5z 27

2

2

1 3 4 2x 2

3 2x x

2

3 4 x 2

2

2x x 3

C. You’ve just seen how we can identify and classify polynomial expressions

Leading Coefficient 1 3 3 4

2


D. Adding and Subtracting Polynomials Adding polynomials simply involves using the distributive, commutative, and associative properties to combine like terms (at this point, the properties are usually applied mentally). As with real numbers, the subtraction of polynomials involves adding the opposite of the second polynomial using algebraic addition. This can be viewed as distributing 1 to the second polynomial and combining like terms. EXAMPLE 10

䊳

Adding and Subtracting Polynomials Perform the indicated operations:

10.7n3 4n2 82 10.5n3 n2 6n2 13n2 7n 102.

Solution

䊳

0.7n3 4n2 8 0.5n3 n2 6n 3n2 7n 10 0.7n3 0.5n3 4n2 1n2 3n2 6n 7n 8 10 1.2n3 13n 18

eliminate parentheses (distributive property) use properties to collect like terms combine like terms



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Sometimes it’s easier to add or subtract polynomials using a vertical format and aligning like terms. Note the use of a placeholder zero in Example 11. EXAMPLE 11

䊳

Subtracting Polynomials Using a Vertical Format Compute the difference of x3 5x 9 and x3 3x2 2x 8 using a vertical format.

Solution

䊳

x3 0x2 5x 9 x3 0x2 5x 9 1x3 3x2 2x 82 ¡ x3 3x2 2x 8 3x2 7x 17 The difference is 3x2 7x 17.

D. You’ve just seen how we can add and subtract polynomials


E. The Product of Two Polynomials Monomial Times Monomial The simplest case of polynomial multiplication is the product of monomials shown in Example 1a. These were computed using exponential properties and the properties of real numbers.

Monomial Times Polynomial To compute the product of a monomial and a polynomial, we use the distributive property. EXAMPLE 12

䊳

Solution

䊳

Multiplying a Monomial by a Polynomial Find the product: 2a2 1a2 2a 12.

2a2 1a2 2a 12 2a2 1a2 2 12a2 212a1 2 12a2 2112 2a4 4a3 2a2

distribute simplify


Binomial Times Polynomial For products involving binomials, we still use a version of the distributive property— this time to distribute one polynomial to each term of the other polynomial factor. Note the distribution can be performed either from the left or from the right. EXAMPLE 13

䊳

Multiplying a Binomial by a Polynomial Multiply as indicated: a. 12z 12 1z 22

Solution

䊳

b. 12v 32 14v2 6v 92

a. 12z 12 1z 22 2z1z 22 11z 22 distribute to every term in the first binomial eliminate parentheses (distribute again) 2z2 4z 1z 2 simplify 2z2 3z 2 2 2 b. 12v 32 14v 6v 92 2v14v 6v 92 314v2 6v 92 distribute 8v3 12v2 18v 12v2 18v 27 simplify 8v3 27 combine like terms



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The F-O-I-L Method By observing the product of two binomials in Example 13(a), we note a pattern that can make the process more efficient. The product of two binomials can quickly be computed using the First, Outer, Inner, Last (FOIL) method, an acronym giving the respective position of each term in a product of binomials in relation to the other terms. We illustrate here using the product 12x 12 13x 22.

WORTHY OF NOTE

Consider the product 1x 321x 22 in the context of area. If we view x 3 as the length of a rectangle (an unknown length plus 3 units), and x 2 as its width (the same unknown length plus 2 units), a diagram of the total area would look like the following, with the result x2 5x 6 clearly visible.

Combine like terms 6x2 x 2

Inner Outer

The first term of the result will always be the product of the first terms from each binomial, and the last term of the result is the product of their last terms. We also note that here, the middle term is found by adding the outermost product with the innermost product. As you practice with the F-O-I-L process, much of the work can be done mentally and you can often compute the entire product without writing anything down except the answer. 䊳

Multiplying Binomials Using F-O-I-L Compute each product mentally: a. 15n 121n 22 b. 12b 32 15b 62 a. 15n 12 1n 22:

5n2 9n 2

product of first two terms

S

䊳

S

Solution

10n (1n) 9n

S

(x 3)(x 2) x2 5x 6

EXAMPLE 14

S

6

S

2x

S

2

S

3x

S

x2

S

x

First Outer Inner Last

12x 12 13x 22 S

3

6x2 4x 3x 2

Last First S

x

The F-O-I-L Method for Multiplying Binomials

sum of outer and inner products

product of last two terms

12b 15b 3b

product of first two terms

S

S

S

b. 12b 3215b 62: 10b2 3b 18 sum of outer and inner products

E. You’ve just seen how we can compute the product of two polynomials

product of last two terms


F. Special Polynomial Products Certain polynomial products are considered “special” for two reasons: (1) the product follows a predictable pattern, and (2) the result can be used to simplify expressions, graph functions, solve equations, and/or develop other skills.

Binomial Conjugates Expressions like x 7 and x 7 are called binomial conjugates. For any given binomial, its conjugate is found by using the same two terms with the opposite sign


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A-19

between them. Example 15 shows that when we multiply a binomial and its conjugate, the “outers” and “inners” sum to zero and the result is a difference of two squares. EXAMPLE 15

䊳

Multiplying Binomial Conjugates Compute each product mentally: a. 1x 721x 72 b. 12x 5y2 12x 5y2

2 2 c. ax b ax b 5 5

7x 7x 0x

Solution

䊳

a. 1x 72 1x 72 x2 49

difference of squares 1x2 2 172 2

10xy (10xy) 0xy

b. 12x 5y212x 5y2 4x2 25y2

difference of squares: 12x2 2 15y2 2

52 x 25 x 0

2 2 4 c. ax b ax b x2 5 5 25

2 2 difference of squares: x2 a b 5

Now try Exercises 117 through 124 䊳 In summary, we have the following. The Product of a Binomial and Its Conjugate Given any expression that can be written in the form A B, the conjugate of the expression is A B and their product is a difference of two squares: 1A B2 1A B2 A2 B2

Binomial Squares

Expressions like 1x 72 2 are called binomial squares and are useful for solving many equations and sketching a number of basic graphs. Note 1x 72 2 1x 721x 72 x2 14x 49 using the F-O-I-L process. The expression x2 14x 49 is called a perfect square trinomial because it is the result of expanding a binomial square. If we write a binomial square in the more general form 1A B2 2 1A B21A B2 and compute the product, we notice a pattern that helps us write the expanded form more quickly. LOOKING AHEAD Although a binomial square can always be found using repeated factors and F-O-I-L, learning to expand them using the pattern is a valuable skill. Binomial squares occur often in a study of algebra and it helps to find the expanded form quickly.

1A B2 2 1A B2 1A B2

repeated multiplication

A AB AB B

F-O-I-L

A 2AB B

simplify (perfect square trinomial)

2 2

2

2

The first and last terms of the trinomial are squares of the terms A and B. Also, the middle term of the trinomial is twice the product of these two terms: AB AB 2AB. The F-O-I-L process shows us why. Since the outer and inner products are identical, we always end up with two. A similar result holds for 1A B2 2 and the process can be summarized for both cases using the symbol.


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The Square of a Binomial

Given any expression that can be written in the form 1A B2 2, 1. 1A B2 2 A2 2AB B2 2. 1A B2 2 A2 2AB B2 䊳

CAUTION

EXAMPLE 16

䊳

Solution

䊳

Note the square of a binomial always results in a trinomial (three terms). In particular, 1A B2 2 A2 B2.

Find each binomial square without using F-O-I-L: a. 1a 92 2

F. You’ve just seen how we can compute special products: binomial conjugates and binomial squares

b. 13x 52 2

a. 1a 92 2 a2 21a # 92 92 a2 18a 81 b. 13x 52 2 13x2 2 213x # 52 52 9x2 30x 25 c. 13 1x2 2 9 213 # 1x2 1 1x2 2 9 61x x

c. 13 1x2 2

1A B2 2 A 2 2AB B 2

simplify 1A B2 2 A 2 2AB B 2 simplify 1A B2 2 A 2 2AB B 2 simplify

Now try Exercises 125 through 136 䊳 With practice, you will be able to go directly from the binomial square to the resulting trinomial.

A.2 EXERCISES 䊳



1. The equation 1x2 2 3 x6 is an example of the property of exponents. 2. The equation 1x3 2 2 property of

1 is an example of the x6 exponents.

3. The sum of the “outers” and “inners” for 12x 52 2 is , while the sum of the outers and inners for 12x 52 12x 52 is . 䊳

4. The expression 2x2 3x 10 can be classified as a of degree , with a leading coefficient of . 5. Discuss/Explain why one of the following expressions can be simplified further, while the other cannot: (a) 7n4 3n2; (b) 7n4 # 3n2. 6. Discuss/Explain why the degree of 2x2y3 is greater than the degree of 2x 2 y 3. Include additional examples for contrast and comparison.


Determine each product using the product and/or power properties.

7.

2 2# n 21n5 3

10. 11.5vy2 2 18v4y2

8. 24g5

# 3 g9 8

11. 1a2 2 4 # 1a3 2 2 # b2 # b5

9. 16p2q2 12p3q3 2 12. d 2 # d 4 # 1c5 2 2 # 1c3 2 2


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Simplify using the product to a power property.

13. 16pq2 2 3

14. 13p2q2 2

15. 13.2hk2 2 3 17. a

16. 12.5h5k2 2

p 2 b 2q

18. a

19. 10.7c 2 110c d 2 4 2

21.

1 34x3y2 2

3 2 2

b 3 b 3a

5m2n3 2 b 2r4

45. a

5p2q3r4

2 2 3

22. 1 45x3 2 2

24. 1 23m2n2 2 # 1 12mn2 2

47.

25. Volume of a cube: The 3x2 formula for the volume of a cube is V S3, where S is the length of one edge. If the length of each edge is 3x2, 3x2 a. Find a formula for volume in terms of the variable x. b. Find the volume of the cube if x 2.

49.

23. 138x2 2 116xy2 2

26. Area of a circle: The formula for the area of a circle is A ␲r2, where r is the length of the radius. If the radius is given as 5x3, a. Find a formula for area in terms of the variable x. b. Find the area of the circle if x 2.

3x2

5x3

Simplify using the quotient property or the property of negative exponents. Write answers using positive exponents only.

27.

6w 2w2

28.

8z 16z5

29.

12a3b5 4a2b4

30.

5m3n5 10mn2

5

7

31. 1 23 2 3 33.

2 h3

32. 1 56 2 1 34.

3 m2

35. 122 3

36. 142 2

37.

38.

3 1 1 2 2

2 1 2 3 2

Simplify each expression using the quotient to a power property.

39. a

2p4

b

40. a

5v4 2 b 7w3

41. a

0.2x2 3 b 0.3y3

42. a

0.5a3 2 b 0.4b2

q3

2

2pq2r

44. a 2

b 4

46. a

4p3 3x2y

3

b

9p3q2r3 12p5qr

3

b 2

Use properties of exponents to simplify the following. Write the answer using positive exponents only.

20. 12.5a 2 13a b 2 3 2

43. a

A-21

51.

9p6q4 12p4q6 20h2 12h5 1a2 2 3 a4 # a5

a3 # b 4 b c2 612x3 2 2

48. 50. 52.

53. a

54.

55.

56.

57.

10x2

14a3bc0 713a2b2c2 3

59. 40 50 61. 21 51 63. 30 31 32 65. 5x0 15x2 0

58.

5m5n2 10m5n 5k3 20k2 153 2 4

59 1p4q8 2 2 p5q2

18n3 813n2 2 3 312x3y4z2 2 18x2yz0

60. 132 0 172 0 62. 41 81

64. 22 21 20 66. 2n0 12n2 0

Convert the following numbers to scientific notation.

67. In mid-2009, the U.S. Census Bureau estimated the world population at nearly 6,770,000,000 people. 68. The mass of a proton is generally given as 0.000 000 000 000 000 000 000 000 001 670 kg. Convert the following numbers to decimal notation.

69. The smallest microprocessors in common use measure 6.5 109 m across. 70. In 2009, the estimated net worth of Bill Gates, the founder of Microsoft, was 5.8 1010 dollars. Compute using scientific notation. Show all work.

71. The average distance between the Earth and the planet Jupiter is 465,000,000 mi. How many hours would it take a satellite to reach the planet if it traveled an average speed of 17,500 mi per hour? How many days? Round to the nearest whole. 72. In fiscal terms, a nation’s debt-per-capita is the ratio of its total debt to its total population. In the year 2009, the total U.S. debt was estimated at $11,300,000,000,000, while the population was estimated at 305,000,000. What was the U.S. debtper-capita ratio for 2009? Round to the nearest whole dollar.


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Identify each expression as a polynomial or nonpolynomial (if a nonpolynomial, state why); classify each as a monomial, binomial, trinomial, or none of these; and state the degree of the polynomial.

73. 35w3 2w2 112w2 14 74. 2x3 23x2 12x 1.2

4 75. 5n2 4n 117 76. 3 2.7r2 r 1 r 2 3 77. p 5 78. q3 2q2 5q Write each polynomial in standard form and name the leading coefficient.

79. 7w 8.2 w3 3w2 80. 2k2 12 k 81. c3 6 2c2 3c

82. 3v3 14 2v2 112v2 83. 12 23x2

84. 8 2n 7n 2

96. 1s 3215s 42

97. 1x 32 1x2 3x 92

98. 1z 521z2 5z 252

99. 1b2 3b 2821b 22

100. 12h2 3h 821h 12 101. 17v 42 13v 52 103. 13 m2 13 m2

102. 16w 1212w 52 104. 15 n2 15 n2

105. 1p 2.52 1p 3.62 106. 1q 4.921q 1.22 107. 1x 12 21x 14 2

109. 1m 34 21m 34 2

108. 1z 13 2 1z 56 2

110. 1n 25 21n 25 2

111. 13x 2y2 12x 5y2 112. 16a b21a 3b2

113. 14c d2 13c 5d2 114. 15x 3y212x 3y2 115. 12x2 521x2 32 116. 13y2 2212y2 12

For each binomial, determine its conjugate and find the product of the binomial with its conjugate.

117. 4m 3

118. 6n 5

85. 13p3 4p2 2p 72 1p2 2p 52

119. 7x 10

120. c 3

121. 6 5k

122. 11 3r

87. 15.75b2 2.6b 1.92 12.1b2 3.2b2

123. x 16

124. p 12

Find the indicated sum or difference.

86. 15q2 3q 42 13q2 3q 42

88. 10.4n2 5n 0.52 10.3n2 2n 0.752

Find each binomial square.

90. 1 59n2 4n 12 2 1 23n2 2n 34 2

127. 14g 32 2

89. 1 34x2 5x 22 1 12x2 3x 42

91. Subtract q5 2q4 q2 2q from q6 2q5 q4 2q3 using a vertical format. 92. Find x4 2x3 x2 2x decreased by x4 3x3 4x2 3x using a vertical format. Compute each product.

93. 3x1x2 x 62

94. 2v2 1v2 2v 152 95. 13r 521r 22 䊳

125. 1x 42 2

126. 1a 32 2

129. 14p 3q2 2

130. 15c 6d2 2

131. 14 1x2 2

128. 15x 32 2

132. 1 1x 72 2

Compute each product.

133. 1x 32 1y 22

134. 1a 32 1b 52

135. 1k 52 1k 62 1k 22

136. 1a 62 1a 12 1a 52


137. Medication in the bloodstream: M ⴝ 0.5t 4 ⴙ 3t 3 ⴚ 97t 2 ⴙ 348t If 400 mg of a pain medication are taken orally, the number of milligrams in the bloodstream is modeled by the formula shown, where M is the number of milligrams and t is the time in hours, 0 t 6 5. Construct a table of values for t 1 through 5, then answer the following. a. How many milligrams are in the bloodstream after 2 hr? After 3 hr? b. Based on part a, would you expect the number of milligrams in the bloodstream after 4 hr to be less or more? Why? c. Approximately how many hours until the medication wears off (the number of milligrams in the bloodstream is 0)?


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Aa 138. Amount of a mortgage payment: M ⴝ

A-23

r r n b a1 ⴙ b 12 12

r n b ⴚ1 12 The monthly mortgage payment required to pay off (or amortize) a loan is given by the formula shown, where M is the monthly payment, A is the original amount of the loan, r is the annual interest rate, and n is the term of the loan in months. Find the monthly payment (to the nearest cent) required to purchase a $198,000 home, if the interest rate is 6.5% and the home is financed over 30 yr.

䊳

a1 ⴙ

APPLICATIONS

139. Attraction between particles: In electrical theory, the force of attraction between two particles P and kPQ Q with opposite charges is modeled by F 2 , d where d is the distance between them and k is a constant that depends on certain conditions. This is known as Coulomb’s law. Rewrite the formula using a negative exponent. 140. Intensity of light: The intensity of illumination from a light source depends on the distance from k the source according to I 2 , where I is the d intensity measured in footcandles, d is the distance from the source in feet, and k is a constant that depends on the conditions. Rewrite the formula using a negative exponent. 141. Rewriting an expression: In advanced mathematics, negative exponents are widely used because they are easier to work with than rational expressions. 3 2 5 Rewrite the expression 3 2 1 4 using x x x negative exponents. 142. Swimming pool hours: A swimming pool opens at 8 A.M. and closes at 6 P.M. In summertime, the

䊳

number of people in the pool at any time can be approximated by the formula S1t2 t 2 10t, where S is the number of swimmers and t is the number of hours the pool has been open (8 A.M.: t 0, 9 A.M.: t 1, 10 A.M.: t 2, etc.). a. How many swimmers are in the pool at 6 P.M.? Why? b. Between what times would you expect the largest number of swimmers? c. Approximately how many swimmers are in the pool at 3 P.M.? d. Create a table of values for t 1, 2, 3, 4, . . . and check your answer to part b. 143. Maximizing revenue: A sporting goods store finds that if they price their video games at $20, they make 200 sales per day. For each decrease of $1, 20 additional video games are sold. This means the store’s revenue can be modeled by the formula R 120 1x2 1200 20x2, where x is the number of $1 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue. 144. Maximizing revenue: Due to past experience, a jeweler knows that if they price jade rings at $60, they will sell 120 each day. For each decrease of $2, five additional sales will be made. This means the jeweler’s revenue can be modeled by the formula R 160 2x2 1120 5x2, where x is the number of $2 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue.


145. If 13x2 kx 12 1kx2 5x 72 12x2 4x k2 x2 3x 2, what is the value of k?

1 2 1 b 5, then the expression 4x2 2 2x 4x is equal to what number?

146. If a2x


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A.3

Solving Linear Equations and Inequalities

LEARNING OBJECTIVES

In a study of algebra, you will encounter many families of equations, or groups of equations that share common characteristics. Of interest to us here is the family of linear equations in one variable, a study that lays the foundation for understanding more advanced families. This section will also lay the foundation for solving a formula for a specified variable, a practice widely used in science, business, industry, and research.

In Section A.3 you will review how to:

A. Solve linear equations

B.

C. D. E.

F.

using properties of equality Recognize equations that are identities or contradictions Solve linear inequalities Solve compound inequalities Solve basic applications of linear equations and inequalities Solve applications of basic geometry

CAUTION

A. Solving Linear Equations Using Properties of Equality An equation is a statement that two expressions are equal. From the expressions 31x 12 x and x 7, we can form the equation 31x 12 x x 7,

Table A.1 x

31x ⴚ 12 ⴙ x

ⴚx ⴙ 7

2

11

9

1

7

8

which is a linear equation in one variable (the 0 exponent on any variable is a 1). To solve an equa1 tion, we attempt to find a specific input or x-value 2 that will make the equation true, meaning the left3 hand expression will be equal to the right. Using 4 Table A.1, we find that 31x 12 x x 7 is a true equation when x is replaced by 2, and is a false equation otherwise. Replacement values that make the equation true are called solutions or roots of the equation. 䊳

3

7

1

6

5

5

9

4

13

3

From Appendix A.1, an algebraic expression is a sum or difference of algebraic terms. Algebraic expressions can be simplified, evaluated or written in an equivalent form, but cannot be “solved,” since we’re not seeking a specific value of the unknown.

Solving equations using a table is too time consuming to be practical. Instead we attempt to write a sequence of equivalent equations, each one simpler than the one before, until we reach a point where the solution is obvious. Equivalent equations are those that have the same solution set, and can be obtained by using the distributive property to simplify the expressions on each side of the equation. The additive and multiplicative properties of equality are then used to obtain an equation of the form x constant. The Additive Property of Equality

The Multiplicative Property of Equality

If A, B, and C represent algebraic expressions and A B,

If A, B, and C represent algebraic expressions and A B,

then A C B C

then AC BC and

B A , 1C 02 C C

In words, the additive property says that like quantities, numbers, or terms can be added to both sides of an equation. A similar statement can be made for the multiplicative property. These properties are combined into a general guide for solving linear equations, which you’ve likely encountered in your previous studies. Note that not all steps in the guide are required to solve every equation.

A-24


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Section A.3 Solving Linear Equations and Inequalities

A-25

Guide to Solving Linear Equations in One Variable • Eliminate parentheses using the distributive property, then combine any like terms. • Use the additive property of equality to write the equation with all variable terms on one side, and all constants on the other. Simplify each side. • Use the multiplicative property of equality to obtain an equation of the form x constant. • For applications, answer in a complete sentence and include any units of measure indicated. For our first example, we’ll use the equation 31x 12 x x 7 from our initial discussion.

EXAMPLE 1

䊳

Solving a Linear Equation Using Properties of Equality Solve for x: 31x 12 x x 7.

Solution

䊳

31x 12 x x 7 3x 3 x x 7 4x 3 x 7 5x 3 7 5x 10 x2

original equation distributive property combine like terms add x to both sides (additive property of equality) add 3 to both sides (additive property of equality) multiply both sides by 15 or divide both sides by 5 (multiplicative property of equality)

As we noted in Table R.1, the solution is x 2. Now try Exercises 7 through 12

䊳

To check a solution by substitution means we substitute the solution back into the original equation (this is sometimes called back-substitution), and verify the lefthand side is equal to the right. For Example 1 we have: 31x 12 x x 7

original equation

312 12 2 2 7

substitute 2 for x

3112 2 5 5 5✓

simplify solution checks

If any coefficients in an equation are fractional, multiply both sides by the least common denominator (LCD) to clear the fractions. Since any decimal number can be written in fraction form, the same idea can be applied to decimal coefficients.

EXAMPLE 2

䊳

Solution

䊳

A. You’ve just seen how we can solve linear equations using properties of equality

Solving a Linear Equation with Fractional Coefficients Solve for n: 14 1n 82 2 12 1n 62. 1 4 1n 82 1 4n 2

2 12 1n 62 2 12n 3 1 1 4n 2n 3 1 41 4n2 41 12n 32 n 2n 12 n 12 n 12

original equation distributive property combine like terms multiply both sides by LCD 4 distributive property subtract 2n multiply by 1

Verify the solution is n 12 using back-substitution. Now try Exercises 13 through 30

䊳


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B. Identities and Contradictions Example 1 illustrates what is called a conditional equation, since the equation is true for x 2, but false for all other values of x. The equation in Example 2 is also conditional. An identity is an equation that is always true, no matter what value is substituted for the variable. For instance, 21x 32 2x 6 is an identity with a solution set of all real numbers, written as 5x|x 僆⺢6, or x 僆 1q, q 2 in interval notation. Contradictions are equations that are never true, no matter what real number is substituted for the variable. The equations x 3 x 1 and 3 1 are contradictions. To state the solution set for a contradiction, we use the symbol “” (the null set) or “{ }” (the empty set). Recognizing these special equations will prevent some surprise and indecision in later chapters.

EXAMPLE 3

䊳

Solving Equations (Special Cases) Solve each equation and state the solution set. a. 21x 42 10x 8 413x 12 b. 8x 16 10x2 24 613x 52

Solution

䊳

a. 21x 42 10x 8 413x 12 2x 8 10x 8 12x 4 12x 8 12x 12 8 12

original equation distributive property combine like terms subtract 12x; contradiction

Since 8 is never equal to 12, the original equation is a contradiction. The solution set is empty: { }

b. 8x 16 10x2 24 613x 52 8x 6 10x 24 18x 30 18x 6 18x 6 6 6

original equation distributive property combine like terms subtract 18x; identity

The result shows that the original equation is an identity, with an infinite number of solutions: 5x | x 僆⺢6 . You may recall this notation is read, “the set of all numbers x, such that x is a real number.” Now try Exercises 31 through 36

B. You’ve just seen how we can recognize equations that are identities or contradictions

䊳

In Example 3(a), our attempt to solve for x ended with all variables being eliminated, leaving an equation that is always false —a contradiction (8 is never equal to 12). There is nothing wrong with the solution process, the result is simply telling us the original equation has no solution. In Example 3(b), all variables were again eliminated but the end result was always true —an identity (6 is always equal to 6). Once again we’ve done nothing wrong mathematically, the result is just telling us that the original equation will be true no matter what value of x we use for an input.

C. Solving Linear Inequalities A linear inequality resembles a linear equality in many respects: Linear Inequality

Related Linear Equation

(1)

x 6 3

x3

(2)

3 p 2 12 8

3 p 2 12 8

A linear inequality in one variable is one that can be written in the form ax b 6 c, where a, b, and c 僆⺢ and a 0. This definition and the following properties also apply when other inequality symbols are used. Solutions to simple inequalities are easy to spot. For instance, x 2 is a solution to x 6 3 since 2 6 3. For more involved inequalities we use the additive property of inequality and the multiplicative property of


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inequality. Similar to solving equations, we solve inequalities by isolating the variable on one side to obtain a solution form such as variable 6 number. The Additive Property of Inequality If A, B, and C represent algebraic expressions and A 6 B, then A C 6 B C Like quantities (numbers or terms) can be added to both sides of an inequality. While there is little difference between the additive property of equality and the additive property of inequality, there is an important difference between the multiplicative property of equality and the multiplicative property of inequality. To illustrate, we begin with 2 6 5. Multiplying both sides by positive three yields 6 6 15, a true inequality. But notice what happens when we multiply both sides by negative three: 2 6 5

original inequality

2132 6 5132 6 6 15

multiply by negative three false

The result is a false inequality, because 6 is to the right of 15 on the number line. Multiplying (or dividing) an inequality by a negative quantity reverses the order relationship between two quantities (we say it changes the sense of the inequality). We must compensate for this by reversing the inequality symbol. 6 7 15

change direction of symbol to maintain a true statement

For this reason, the multiplicative property of inequality is stated in two parts. The Multiplicative Property of Inequality

EXAMPLE 4

䊳

If A, B, and C represent algebraic expressions and A 6 B, then AC 6 BC

If A, B, and C represent algebraic expressions and A 6 B, then AC 7 BC

if C is a positive quantity (inequality symbol remains the same).

if C is a negative quantity (inequality symbol must be reversed).

Solving an Inequality Solve the inequality, then graph the solution set and write it in interval notation: 2 1 5 3 x 2 6.

Solution

WORTHY OF NOTE As an alternative to multiplying or dividing by a negative value, the additive property of inequality can be used to ensure the variable term will be positive. From Example 4, the inequality 4x 2 can be written as 2 4x by adding 4x to both sides and subtracting 2 from both sides. This gives the solution 12 x, which is equivalent to x

12 .

䊳

1 5 2 x 3 2 6 2 1 5 6 a x b 162 3 2 6 4x 3 5 4x 2 4x 2 4 4 1 x 2

original inequality

clear fractions (multiply by LCD) simplify subtract 3 divide by 4, reverse inequality sign result

1 2

• Graph:

3 2 1

[ 0

1

2

3

• Interval notation: x 僆 312, q 2

4


䊳


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To check a linear inequality, you often have an infinite number of choices—any number from the solution set/interval. If a test value from the solution interval results in a true inequality, all numbers in the interval are solutions. For Example 4, using x 0 results in the true statement 12 56 ✓. Some inequalities have all real numbers as the solution set: 5x| x 僆⺢6, while other inequalities have no solutions, with the answer given as the empty set: { }.

EXAMPLE 5

䊳

Solving Inequalities Solve the inequality and write the solution in set notation: a. 7 13x 52 21x 42 5x b. 31x 42 5 6 21x 32 x

Solution

䊳

a. 7 13x 52 21x 42 5x 7 3x 5 2x 8 5x 3x 2 3x 8 2 8

original inequality distributive property combine like terms add 3x

Since the resulting statement is always true, the original inequality is true for all real numbers. The solution is all real numbers ⺢. b. 31x 42 5 3x 12 5 3x 7 7

6 6 6 6

21x 32 x 2x 6 x 3x 6 6

original inequality distribute combine like terms subtract 3x

Since the resulting statement is always false, the original inequality is false for all real numbers. The solution is { }. C. You’ve just seen how we can solve linear inequalities


䊳

D. Solving Compound Inequalities In some applications of inequalities, we must consider more than one solution interval. These are called compound inequalities, and these require us to take a close look at the operations of union “ ´ ” and intersection “ ¨”. The intersection of two sets A and B, written A ¨ B, is the set of all elements common to both sets. The union of two sets A and B, written A ´ B, is the set of all elements that are in either set. When stating the union of two sets, repetitions are unnecessary.

EXAMPLE 6

䊳

Finding the Union and Intersection of Two Sets

Solution

䊳

A ¨ B is the set of all elements in both A and B: A 傽 B 51, 2, 36. A ´ B is the set of all elements in either A or B: A ´ B 52, 1, 0, 1, 2, 3, 4, 56.

WORTHY OF NOTE For the long term, it may help to rephrase the distinction as follows. The intersection is a selection of elements that are common to two sets, while the union is a collection of the elements from two sets (with no repetitions).

For set A 52, 1, 0, 1, 2, 36 and set B 51, 2, 3, 4, 56, determine A ¨ B and A ´ B.


䊳

Notice the intersection of two sets is described using the word “and,” while the union of two sets is described using the word “or.” When compound inequalities are formed using these words, the solution is modeled after the ideas from Example 6. If “and” is used, the solutions must satisfy both inequalities. If “or” is used, the solutions can satisfy either inequality.


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EXAMPLE 7

䊳

Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x 1 6 4 or 4x 3 6 6.

Solution

䊳

Begin with the statement as given: 3x 1 6 4 3x 6 3 x 7 1

4x 3 6 6 4x 6 9 9 x 6 4

or or or

original statement isolate variable term solve for x, reverse first inequality symbol

The solution x 7 1 or x 6 94 is better understood by graphing each interval separately, then selecting both intervals (the union). WORTHY OF NOTE

x 1:

The graphs from Example 7 clearly show the solution consists of two disjoint (disconnected) intervals. This is reflected in the “or” statement: x 6 94 or x 7 1, and in the interval notation. Also, note the solution x 6 94 or x 7 1 is not equivalent to 94 7 x 7 1, as there is no single number that is both greater than 1 and less than 94 at the same time.

x 9 : 4

)

8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

9 4

)

8 7 6 5 4 3 2 1

x 9 or x 1: 4

0

1

2

3

4

5

6

9 4

)

)

8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

9 Interval notation: x 僆 aq, b ´ 11, q 2. 4 Now try Exercises 59 and 60

EXAMPLE 8

䊳

䊳

Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x 5 7 13 and 3x 5 6 1.

Solution

䊳

Begin with the statement as given: 3x 5 7 13 3x 7 18 x 7 6

3x 5 6 1 3x 6 6 x 6 2

and and and

original statement subtract five divide by 3

The solution x 7 6 and x 6 2 can best be understood by graphing each interval separately, then noting where they intersect. WORTHY OF NOTE x 6: x 2: x 6 and x 2:

)

8 7 6 5 4 3 2 1

)

8 7 6 5 4 3 2 1

)

The inequality a 6 b (a is less than b) can equivalently be written as b 7 a (b is greater than a). In Example 8, the solution is read, “x 7 6 and x 6 2,” but if we rewrite the first inequality as 6 6 x (with the “arrowhead” still pointing at 62, we have 6 6 x and x 6 2 and can clearly see that x must be in the single interval between 6 and 2.

0

1

2

3

4

5

6

0

1

2

3

4

5

6

0

1

2

3

4

5

6

)

8 7 6 5 4 3 2 1

Interval notation: x 僆 16, 22. Now try Exercises 61 through 72

䊳


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The solution from Example 8 consists of the single interval 16, 22, indicating the original inequality could actually be joined and written as 6 6 x 6 2, called a joint inequality. We solve joint inequalities in much the same way as linear inequalities, but must remember they have three parts (left, middle, and right). This means operations must be applied to all three parts in each step of the solution process, to obtain a solution form such as smaller number 6 x 6 larger number. The same ideas apply when other inequality symbols are used.

EXAMPLE 9

䊳

Solving a Joint Inequality Solve the joint inequality, then graph the solution set and write it in interval 2x 5 6. notation: 1 7 3

Solution

䊳

2x 5 6 3 3 6 2x 5 18 8 6 2x 13 13 4 6 x 2 1 7

original inequality multiply all parts by 3; reverse the inequality symbols subtract 5 from all parts divide all parts by 2 13 2

• Graph:

)

5 4 3 2 1

[ 0

• Interval notation: x 僆 14,

D. You’ve just seen how we can solve compound inequalities

1

13 2 4

2

3

4

5

6

7

8


䊳

E. Solving Basic Applications of Linear Equations and Inequalities Applications of linear equations and inequalities come in many forms. In most cases, you are asked to translate written relationships or information given verbally into an equation using words or phrases that indicate mathematical operations or relationships. Here, we’ll practice this skill using ideas that were introduced in Appendix A.1, where we translated English phrases into mathematical expressions. Very soon these skills will be applied in much more significant ways.

EXAMPLE 10

䊳

Translating Written Information into an Equation Translate the following relationships into equations, then solve: In an effort to lower the outstanding balance on her credit card, Laura paid $10 less than triple her normal payment. If she sent the credit card company $350.75, how much was her normal payment? (See Section A.1, Examples 2 and 3.)

Solution

䊳

Let p represent her normal payment. Then “triple her normal payment” would be 3p, and “ten less than triple” would be 3p 10. Since “she sent the company $350.75,” we have


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3p 10 350.75 3p 360.75 p 120.25

equation form add 10 divide by 3

Laura’s normal payment is $120.25 per month. A calculator check is shown in the figure.


䊳

Inequalities are widely used to help gather information, and to make comparisons that will lead to informed decisions.

EXAMPLE 11

䊳

Using an Inequality to Compute Desired Test Scores Justin earned scores of 78, 72, and 86 on the first three out of four exams. What score must he earn on the fourth exam to have an average of at least 80?

Solution

䊳

The current scores are 78, 72, and 86. An average of at least 80 means A 80. In organized form: Test 1

Test 2

Test 3

Test 4

Computed Average

Minimum

78

72

86

x

78 72 86 x 4

80

Let x represent Justin’s score on the fourth exam, then represents his average score. 78 72 86 x 80 4 78 72 86 x 320 236 x 320 x 84 E. You’ve just seen how we can solve applications of linear equations and inequalities

78 72 86 x 4

average must be greater than or equal to 80 multiply by 4 simplify solve for x (subtract 236)

Justin must score at least an 84 on the last exam to earn an 80 average. Now try Exercises 91 through 100

䊳

F. Solving Applications of Basic Geometry As your translation skills grow, your ability to solve a wider range of more significant applications will grow as well. In many cases, the applications will involve some basic geometry and the most often used figures and formulas appear here. For a more complete review of geometry, see Appendix II Geometry Review with Unit Conversions, which is posted online at www.mhhe.com/coburn.

Perimeter and Area Perimeter is a measure of the distance around a two dimensional figure. As this is a linear measure, results are stated in linear units as in centimeters (cm), feet (ft), kilometers (km), miles (mi), and so on. If no unit is specified, simply write the result as x units. Area is a measure of the surface of a two dimensional figure, with results stated in square units as in x units2. Some of the most common formulas involving perimeter and area are given in Table A.2A.


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Table A.2A Perimeter Formula (linear units or units)

Definition and Diagram

Area Formula (square units or units2)

a three-sided polygon s1

Triangle

s2

P s1 s2 s3

h

s3

A

1 bh 2

b

a quadrilateral with four right angles and opposite sides parallel Rectangle W

P 2L 2W

A LW

P 4S

A S2

L a rectangle with four equal sides Square

S

a quadrilateral with one pair of parallel sides (called bases b1 and b2) b1

s2

Trapezoid s1

s3 s4

Circle

sum of all sides P s1 s2 s3 s4

h

A

h 1b1 b2 2 2

b2

the set of all points lying in a plane that are an equal distance (called the radius r) from a given point (called the center C).

r C

C 2␲r or C ␲d

A ␲r2

If an exercise or application uses a formula, begin by stating the formula first. Using the formula as a template for the values substituted will help to prevent many careless errors.

EXAMPLE 12A

䊳

Computing the Area of a Trapezoidal Window A basement window is shaped like an isosceles trapezoid (base angles equal, nonparallel sides equal in length), with a height of 10 in. and bases of 1.5 ft and 2 ft. What is the area of the glass in the window?

1.5 ft

10 in.

2 ft


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Solution

䊳

Before applying the area formula, all measures must use the same unit. In inches, we have 1.5 ft 18 in. and 2 ft 24 in. h 1b1 b2 2 2 10 in. 118 in. 24 in.2 A 2 A 15 in.2 142 in.2 A 210 in2 A

given formula substitute 10 for h, 18 for b1, and 24 for b2 simplify result

The area of the glass in the window is 210 in2. Now try Exercises 101 and 102

䊳

Volume Volume is a measure of the amount of space occupied by a three dimensional object and is measured in cubic units. Some of the more common formulas are given in Table A.2B. Table A.2B Volume Formula (cubic units or units3)

Definition and Diagram Rectangular solid

Cube

a six-sided, solid figure with opposite faces congruent and adjacent faces meeting at right angles

H

V LWH

W

L

a rectangular solid with six congruent, square faces

V S3 S

Sphere

the set of all points in space, an equal distance (called the radius) from a given point (called the center)

Right circular cylinder

union of all line segments connecting two congruent circles in parallel planes, meeting each at a right angle

Right circular cone

Right pyramid

union of all line segments connecting a given point (vertex) to a given circle (base) and whose altitude meets the center of the base at a right angle union of all line segments connecting a given point (vertex) to a given square (base) and whose altitude meets the center of the base at a right angle

V

r

C

h

4 3 ␲r 3

V ␲r2h

r 1 V ␲r2h 3

h r

1 V s2h 3 h s


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EXAMPLE 12B

䊳

Computing the Volume of a Composite Figure Sand at a cement factory is being dumped from a conveyor belt into a pile shaped like a right circular cone atop a right circular cylinder (see figure). How many cubic feet of sand are there at the moment the cone is 6 ft high with a diameter of 10 ft?

Solution

䊳

F. You’ve just seen how we can solve applications of basic geometry

Total Volume volume of cylinder volume of cone 1 V ␲r2h1 ␲r2h2 3 1 ␲152 2 132 ␲152 2 162 3 75␲ 50␲ 125␲

6 ft 3 ft 10 ft

verbal model formula model (note h1 h2 2 substitute 5 for r, 3 for h1, and 6 for h2 simplify result (exact form)

There are about 392.7 ft3 of sand in the pile. Now try Exercises 103 and 104

A.3 EXERCISES 䊳



䊳

of sets A and B is written A 傽 B. of sets A and B is written A ´ B.

1. A(n) is an equation that is always true, regardless of the value while a(n) is an equation that is always false, regardless of the value.

4. The The

2. For inequalities, the three ways of writing a solution set are notation, a number line graph, and notation.

5. Discuss/Explain the similarities and differences between the properties of equality for equations and those for inequalities.

3. The mathematical sentence 3x 5 6 7 is a(n) inequality, while 2 6 3x 5 6 7 is a(n) inequality.

6. Discuss/Explain the use of the words “and” and “or” in the statement of compound inequalities. Include a few examples to illustrate.


Solve each equation. Check your answer by substitution.

7. 4x 31x 22 18 x 8. 15 2x 41x 12 9

9. 21 12v 172 7 3v

10. 12 5w 9 16w 72

11. 8 13b 52 5 21b 12 12. 2a 41a 12 3 12a 12

Solve each equation.

13. 15 1b 102 7 13 1b 92

14. 61 1n 122 14 1n 82 2 15. 23 1m 62 1 2

16. 45 1n 102 8 9

17. 12x 5 13x 7 19.

x3 x 7 5 3

18. 4 23y 12y 5 20.

z4 z 2 6 2

䊳


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21. 15 6

3p 8

22. 15

2q 21 9

Solve each inequality and write the solution in set notation.

23. 0.2124 7.5a2 6.1 4.1

47. 7 21x 32 4x 61x 32

24. 0.4117 4.25b2 3.15 4.16

48. 3 61x 52 217 3x2 1

25. 6.2v 12.1v 52 1.1 3.7v

26. 7.9 2.6w 1.5w 19.1 2.1w2 n 2 n 2 5 3 2 m m 28. 3 5 4 p p 29. 3p 5 2p 6 4 6 q q 30. 1 3q 2 4q 6 8 27.

Identify the following equations as an identity, a contradiction, or a conditional equation, then state the solution.

31. 314z 52 15z 20 3z 32. 5x 9 2 512 x2 1 33. 8 813n 52 5 611 n2 34. 2a 41a 12 1 312a 12 35. 414x 52 6 218x 72 36. 15x 32 2x 11 41x 22 Write the solution set illustrated on each graph in set notation and interval notation.

37. 38. 39. 40.

[

3 2 1

0

1

0

[

1

2

50. 8 16 5m2 7 9m 13 4m2 51. 61p 12 2p 212p 32

52. 91w 12 3w 215 3w2 1 Determine the intersection and union of sets A, B, C, and D as indicated, given A ⴝ 5ⴚ3, ⴚ2, ⴚ1, 0, 1, 2, 36, B ⴝ 52, 4, 6, 86, C ⴝ 5ⴚ 4, ⴚ2, 0, 2, 46, and D ⴝ 54, 5, 6, 76 .

53. A 傽 B and A ´ B 54. A 傽 C and A ´ C 55. A 傽 D and A ´ D 56. B 傽 C and B ´ C

57. B 傽 D and B ´ D 58. C 傽 D and C ´ D Express the compound inequalities graphically and in interval notation.

59. x 6 2 or x 7 1 60. x 6 5 or x 7 5 61. x 6 5 and x 2 62. x 4 and x 6 3 63. x 3 and x 1

Solve the compound inequalities and graph the solution set.

65. 41x 12 20 or x 6 7 9 66. 31x 22 7 15 or x 3 1

69. 35x 12 7

0

1

2

0

1

2

) 4

Solve the inequality and write the solution in set notation. Then graph the solution and write it in interval notation.

41. 5a 11 2a 5 42. 8n 5 7 2n 12 43. 21n 32 4 5n 1 44. 51x 22 3 6 3x 11 3x x 45. 6 4 8 4

3 10

and 4x 7 1

70. 23x 56 0 and 3x 6 2

3

3

64. x 5 and x 7

68. 3x 5 17 and 5x 0

3

[

[

3 2 1

49. 413x 52 18 6 215x 12 2x

67. 2x 7 3 and 2x 0

3

)

3 2 1

3 2 1

2

2y y 46. 6 2 5 10

A-35

71.

x 3x 6 3 or x 1 7 5 8 4

72.

x 2x 6 2 or x 3 7 2 5 10

73. 3 2x 5 6 7 74. 2 6 3x 4 19 75. 0.5 0.3 x 1.7 76. 8.2 6 1.4 x 6 0.9 77. 7 6 34x 1 11 78. 21 23x 9 6 7


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79. Euler’s Polyhedron Formula: V ⴙ F ⴚ E ⴝ 2 Discovered by Leonhard Euler in 1752, this simple but powerful formula states that in any regular polyhedron, the number of vertices V and faces F is always two more than the number of edges E. (a) Verify the formula for a simple cube. (b) Verify the formula for the octahedron shown in the figure. (c) If a dodecahedron has 12 faces and 30 edges, how many vertices does it have? 80. Area of a Regular Polygon: A ⴝ

1 ap 2

The area of any regular polygon can be found a using the formula shown, where a is the apothem of the polygon (perpendicular distance from center to any edge), and p is the perimeter. (a) Verify the formula using a square with sides of length 6 cm. (b) If the hexagon shown has an area of 259.8 cm2 with sides 10 cm in length, what is the length a of the apothem? 䊳

81. Body mass index: B ⴝ

704W H2

The U.S. government publishes a body mass index formula to help people consider the risk of heart disease. An index “B” of 27 or more means that a person is at risk. Here W represents weight in pounds and H represents height in inches. If your height is 5¿8– what range of weights will help ensure you remain safe from the risk of heart disease? Source: www.surgeongeneral.gov/topics.

82. Lift capacity: 75S ⴙ 125B ⱕ 750 The capacity in pounds of the lift used by a roofing company to place roofing shingles and buckets of roofing nails on rooftops is modeled by the formula shown, where S represents packs of shingles and B represents buckets of nails. Use the formula to find (a) the largest number of shingle packs that can be lifted, (b) the largest number of nail buckets that can be lifted, and (c) the largest number of shingle packs that can be lifted along with three nail buckets.

APPLICATIONS

Write an equation to model the given information and solve.

83. Celebrity Travel: To avoid paparazzi and overzealous fans, the arrival gates of planes carrying celebrities are often kept secret until the last possible moment. While awaiting the arrival of Angelina Jolie, a large crowd of fans and photographers had gathered at Terminal A, Gate 18. However, the number of fans waiting at Gate 32 was twice that number increased by 5. If there were 73 fans at Gate 32, how many were waiting at Gate 18? (See Section A.1, Example 2a.) 84. Famous Architecture: The Hall of Mirrors is the central gallery of the Palace of Versailles and is one of the most famous rooms in the world. The length of this hall is 11 m less than 8 times the width. If the hall is 73 m long, what is its width? (See Section A.1, Example 2b.) 85. Dietary Goals: At the picnic, Mike abandoned his diet and consumed 13 calories more than twice the number of calories he normally allots for lunch. If he consumed 1467 calories, how many calories are normally allotted for lunch?

86. Marathon Training: While training for the Chicago marathon, Christina’s longest run of the week was 5 mi less than double the shortest. If the longest run was 11.2 mi, how long was the shortest? 87. Actor’s Ages: At the time of this writing, actor Will Smith (Enemy of the State, Seven Pounds, others), was 1 yr older than two-thirds the age of Samuel Jackson (The Negotiator, Die Hard III, others). If Will Smith was 41 at this time, how old was Samuel Jackson? 88. Football versus Fútbol: The area of a regulation field for American football is about 410 square meters (m2) less than three-fifths of an Olympicsized soccer field. If an American football field covers 5350 m2, what is the area of an Olympic soccer field?


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89. Forensic Studies: In forensic studies, skeletal remains are analyzed to determine the height, gender, race, age, and other characteristics of the decedent. For instance, the height of a male individual is approximated as 34 in. more than three and one-third times the length of the radial bone. If a live individual is 74 in. tall, how long is his radial bone?

A-37

90. Famous Waterways: The Suez Canal and the Panama Canal are two of the most important waterways in the world, saving ships thousands of miles as they journey from port to destination. The length of the Suez Canal is 39 kilometers (km) less than three times the length of the Panama Canal. If the Egyptian canal is 192 km long, how long is the Central American canal?

Write an inequality to model the given information and solve.

91. Exam scores: Jacques is going to college on an academic scholarship that requires him to maintain at least a 75% average in all of his classes. So far he has scored 82%, 76%, 65%, and 71% on four exams. What scores are possible on his last exam that will enable him to keep his scholarship? 92. Timed trials: In the first three trials of the 100-m butterfly, Johann had times of 50.2, 49.8, and 50.9 sec. How fast must he swim the final timed trial to have an average time of at most 50 sec? 93. Checking account balance: If the average daily balance in a certain checking account drops below $1000, the bank charges the customer a $7.50 service fee. The table Weekday Balance gives the daily balance Monday $1125 for one customer. What Tuesday $850 must the daily balance be Wednesday $625 for Friday to avoid a service charge? Thursday $400 94. Average weight: In the Lineman Weight National Football Left tackle 318 lb League, many consider an offensive line to be Left guard 322 lb “small” if the average Center 326 lb weight of the five down Right guard 315 lb linemen is less than Right tackle ? 325 lb. Using the table, what must the weight of the right tackle be so that the line will not be considered small? 95. Area of a rectangle: Given the rectangle shown, what is the range of values for the width, in order to keep the area less than 150m2? 20 m

w

96. Area of a triangle: Using the triangle shown, find the height that will guarantee an area equal to or greater than 48 in2.

h

12 in.

97. Heating and cooling subsidies: As long as the outside temperature is over 45°F and less than 85°F 145 6 F 6 852, the city does not issue heating or cooling subsidies for low-income families. What is the corresponding range of Celsius temperatures C? Recall that F 95C 32. 98. U.S. and European shoe sizes: To convert a European male shoe size “E” to an American male shoe size “A,” the formula A 0.76E 23 can be used. Lillian has five sons in the U.S. military, with shoe sizes ranging from size 9 to size 14 19 A 142. What is the corresponding range of European sizes? Round to the nearest half-size. 99. Power tool rentals: Sunshine Equipment Co. rents its power tools for a $20 fee, plus $4.50/hr. Kealoha’s Rentals offers the same tools for an $11 fee plus $6.00/hr. How many hours h must a tool be rented to make the cost at Sunshine a better deal? 100. Moving van rentals: Stringer Truck Rentals will rent a moving van for $15.75/day plus $0.35 per mile. Bertz Van Rentals will rent the same van for $25/day plus $0.30 per mile. How many miles m must the van be driven to make the cost at Bertz a better deal?


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101. Cost of drywall: After the studs are up, the 3 ft wall shown in the figure 15 ft must be covered in 10 ft 7 ft drywall. (a) How many square feet of drywall 19 ft are needed? (b) If drywall is sold only in 4-ft by 8-ft sheets, approximately how many sheets are required for this job?

5 in. 103. Trophy bases: The base of a new trophy 7 in. has the form of a cylinder sitting atop a rectangular solid. 2 in. If the base is to be 10 in. cast in a special 10 in. aluminum, determine the volume of aluminum to be used.

102. Paving a walkway: Current plans 104. Grain storage: The dimensions of call for building a circular fountain a grain silo are shown in the figure. 6m 6 m in diameter with a circular If the maximum storage capacity of walkway around it that is 1.5 m the silo is 95% of the total volume wide. (a) What is the approximate of the silo, how many cubic meters area of the walkway? (b) If the 1.5 m of corn can be stored? concrete for the walkway is to be 6 cm deep, what volume of cement must be used 11 cm 0.01 m2 ? 䊳

16 m

6m


105. Solve for x: 314x2 5x 22 7x 614 x 2x2 2 19

106. Solve for n: 553 3 4 215 9n2 4 6 15 655 23 n 1019 n2 4 6 107. Use your local library, the Internet, or another resource to find the highest and lowest point on each of the seven continents. Express the range of altitudes for each continent as a joint inequality. Which continent has the greatest range? 108. The sum of two consecutive even integers is greater than or equal to 12 and less than or equal to 22. List all possible values for the two integers.

A.4

Place the correct inequality symbol in the blank to make the statement true.

109. If m 7 0 and n 6 0, then mn

0.

110. If m 7 n and p 7 0, then mp

np.

111. If m 6 n and p 7 0, then mp

np.

112. If m n and p 6 0, then mp

np.

113. If m 7 n, then m 114. If 0 6 m 6 n, then

n. 1 m

1 n.

115. If m 7 0 and n 6 0, then m2 116. If m 6 0, then m

3

n. 0.

Factoring Polynomials and Solving Polynomial Equations by Factoring

LEARNING OBJECTIVES In Section A.4 you will review:

A. Factoring out the greatest common factor

It is often said that knowing which tool to use is just as important as knowing how to use the tool. In this section, we review the tools needed to factor an expression, an important part of solving polynomial equations. This section will also help us decide which factoring tool is appropriate when many different factorable expressions are presented.

B. Common binomial factors and factoring by grouping C. Factoring quadratic polynomials D. Factoring special forms and quadratic forms E. Solving Polynomial Equations by Factoring

A. The Greatest Common Factor To factor an expression means to rewrite the expression as an equivalent product. The distributive property is an example of factoring in action. To factor 2x2 6x, we might first rewrite each term using the common factor 2x: 2x2 6x 2x # x 2x # 3, then apply the distributive property to obtain 2x1x 32. We commonly say that we have factored out 2x. The greatest common factor (or GCF) is the largest factor common to all terms in the polynomial.


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Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

EXAMPLE 1

䊳

Factoring Polynomials Factor each polynomial: a. 12x2 ⫹ 18xy ⫺ 30y

Solution

䊳

A-39

b. x5 ⫹ x2

a. 6 is common to all three terms: 12x2 ⫹ 18xy ⫺ 30y ⫽ 612x2 ⫹ 3xy ⫺ 5y2

mentally: 6 # 2x2 ⫹ 6 # 3xy ⫺ 6 # 5y

b. x2 is common to both terms:

A. You’ve just seen how we can factor out the greatest common factor

x5 ⫹ x2 ⫽ x2 1x3 ⫹ 12

mentally: x 2 # x 3 ⫹ x 2 # 1


䊳

B. Common Binomial Factors and Factoring by Grouping If the terms of a polynomial have a common binomial factor, it can also be factored out using the distributive property. EXAMPLE 2

䊳

Factoring Out a Common Binomial Factor Factor: a. 1x ⫹ 32x2 ⫹ 1x ⫹ 325

Solution

䊳

b. x2 1x ⫺ 22 ⫺ 31x ⫺ 22

a. 1x ⫹ 32x2 ⫹ 1x ⫹ 325 ⫽ 1x ⫹ 32 1x2 ⫹ 52

b. x2 1x ⫺ 22 ⫺ 31x ⫺ 22 ⫽ 1x ⫺ 221x2 ⫺ 32 Now try Exercises 9 and 10

䊳

One application of removing a binomial factor involves factoring by grouping. At first glance, the expression x3 ⫹ 2x2 ⫹ 3x ⫹ 6 appears unfactorable. But by grouping the terms (applying the associative property), we can remove a monomial factor from each subgroup, which then reveals a common binomial factor. x3 ⫹ 2x2 ⫹ 3x ⫹ 6 ⫽ x2 1x ⫹ 22 ⫹ 31x ⫹ 22 ⫽ 1x ⫹ 22 1x2 ⫹ 32

This grouping of terms must take into account any sign changes and common factors, as seen in Example 3. Also, it will be helpful to note that a general four-term polynomial A ⫹ B ⫹ C ⫹ D is factorable by grouping only if AD ⫽ BC. EXAMPLE 3

䊳

Factoring by Grouping Factor 3t3 ⫹ 15t2 ⫺ 6t ⫺ 30.

Solution

䊳

Notice that all four terms have a common factor of 3. Begin by factoring it out. 3t3 ⫹ 15t2 ⫺ 6t ⫺ 30 ⫽ 31t3 ⫹ 5t2 ⫺ 2t ⫺ 102 ⫽ 31t3 ⫹ 5t2 ⫺ 2t ⫺ 102 ⫽ 3 3 t2 1t ⫹ 52 ⫺ 21t ⫹ 52 4 ⫽ 31t ⫹ 52 1t2 ⫺ 22

original polynomial factor out 3 group remaining terms factor common monomial factor common binomial


䊳


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B. You’ve just seen how we can factor common binomial factors and factor by grouping

When asked to factor an expression, first look for common factors. The resulting expression will be easier to work with and help ensure the final answer is written in completely factored form. If a four-term polynomial cannot be factored as written, try rearranging the terms to find a combination that enables factoring by grouping.

C. Factoring Quadratic Polynomials A quadratic polynomial is one that can be written in the form ax2 ⫹ bx ⫹ c, where a, b, c 僆⺢ and a ⫽ 0. One common form of factoring involves quadratic trinomials such as x2 ⫹ 7x ⫹ 10 and 2x2 ⫺ 13x ⫹ 15. While we know 1x ⫹ 521x ⫹ 22 ⫽ x2 ⫹ 7x ⫹ 10 and 12x ⫺ 321x ⫺ 52 ⫽ 2x2 ⫺ 13x ⫹ 15 using F-O-I-L, how can we factor these trinomials without seeing the original expression in advance? First, it helps to place the trinomials in two families—those with a leading coefficient of 1 and those with a leading coefficient other than 1.

ax 2 ⴙ bx ⴙ c, where a ⴝ 1

When a ⫽ 1, the only factor pair for x2 (other than 1 # x2 2 is x # x and the first term in each binomial will be x: (x )(x ). The following observation helps guide us to the complete factorization. Consider the product 1x ⫹ b2 1x ⫹ a2: 1x ⫹ b21x ⫹ a2 ⫽ x2 ⫹ ax ⫹ bx ⫹ ab

⫽ x ⫹ 1a ⫹ b2x ⫹ ab 2

F-O-I-L distributive property

Note the last term is the product ab (the lasts), while the coefficient of the middle term is a ⫹ b (the sum of the outers and inners). Since the last term of x2 ⫺ 8x ⫹ 7 is 7 and the coefficient of the middle term is ⫺8, we are seeking two numbers with a product of positive 7 and a sum of negative 8. The numbers are ⫺7 and ⫺1, so the factored form is 1x ⫺ 721x ⫺ 12. It is also helpful to note that if the constant term is positive, the binomials will have like signs, since only the product of like signs is positive. If the constant term is negative, the binomials will have unlike signs, since only the product of unlike signs is negative. This means we can use the sign of the linear term (the term with degree 1) to guide our choice of factors. Factoring Trinomials with a Leading Coefficient of 1 If the constant term is positive, the binomials will have like signs: 1x ⫹ 2 1x ⫹ 2 or 1x ⫺ 2 1x ⫺ 2 ,

to match the sign of the linear (middle) term. If the constant term is negative, the binomials will have unlike signs: 1x ⫹ 2 1x ⫺ 2,

with the larger factor placed in the binomial whose sign matches the linear (middle) term.

EXAMPLE 4

䊳

Factoring Trinomials Factor these expressions: a. ⫺x2 ⫹ 11x ⫺ 24

Solution

䊳

b. x2 ⫺ 10 ⫺ 3x

a. First rewrite the trinomial in standard form as ⫺11x2 ⫺ 11x ⫹ 242. For x2 ⫺ 11x ⫹ 24, the constant term is positive so the binomials will have like signs. Since the linear term is negative, ⫺11x2 ⫺ 11x ⫹ 242 ⫽ ⫺11x ⫺ 21x ⫺ 2 ⫽ ⫺11x ⫺ 821x ⫺ 32

like signs, both negative 1⫺82 1⫺32 ⫽ 24; ⫺8 ⫹ 1⫺32 ⫽ ⫺11


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b. First rewrite the trinomial in standard form as x2 ⫺ 3x ⫺ 10. The constant term is negative so the binomials will have unlike signs. Since the linear term is negative, x2 ⫺ 3x ⫺ 10 ⫽ 1x ⫹ 2 1x ⫺ 2 ⫽ 1x ⫹ 22 1x ⫺ 52

unlike signs, one positive and one negative 5 7 2, 5 is placed in the second binomial; 122 1⫺52 ⫽ ⫺10; 2 ⫹ 1⫺52 ⫽ ⫺3


䊳

Sometimes we encounter prime polynomials, or polynomials that cannot be factored. For x2 ⫹ 9x ⫹ 15, the factor pairs of 15 are 1 # 15 and 3 # 5, with neither pair having a sum of ⫹9. We conclude that x2 ⫹ 9x ⫹ 15 is prime. ax 2 ⴙ bx ⴙ c, where a ⴝ 1 If the leading coefficient is not one, the possible combinations of outers and inners are more numerous. Furthermore, the sum of the outer and inner products will change depending on the position of the possible factors. Note that 12x ⫹ 321x ⫹ 92 ⫽ 2x2 ⫹ 21x ⫹ 27 and 12x ⫹ 921x ⫹ 32 ⫽ 2x2 ⫹ 15x ⫹ 27 result in a different middle term, even though identical numbers were used. To factor 2x2 ⫺ 13x ⫹ 15, note the constant term is positive so the binomials must have like signs. The negative linear term indicates these signs will be negative. We then list possible factors for the first and last terms of each binomial, then sum the outer and inner products. Possible First and Last Terms for 2x2 and 15 1. 12x ⫺ 121x ⫺ 152 2. 12x ⫺ 1521x ⫺ 12 3. 12x ⫺ 321x ⫺ 52 4. 12x ⫺ 521x ⫺ 32 WORTHY OF NOTE The number of trials needed to factor a polynomial can also be reduced by noting that the two terms in any binomial cannot share a common factor (all common factors are removed in a preliminary step).

EXAMPLE 5

䊳

Sum of Outers and Inners ⫺30x ⫺ 1x ⫽ ⫺31x ⫺2x ⫺ 15x ⫽ ⫺17x ⫺10x ⫺ 3x ⫽ ⫺13x

d

⫺6x ⫺ 5x ⫽ ⫺11x

As you can see, only possibility 3 yields a linear term of ⫺13x, and the correct factorization is then 12x ⫺ 321x ⫺ 52. With practice, this trial-and-error process can be completed very quickly. If the constant term is negative, the number of possibilities can be reduced by finding a factor pair with a sum or difference equal to the absolute value of the linear coefficient, as we can then arrange the sign in each binomial to obtain the needed result as shown in Example 5. Factoring a Trinomial Using Trial and Error Factor 6z2 ⫺ 11z ⫺ 35.

Solution

䊳

Note the constant term is negative (binomials will have unlike signs) and 冟⫺11冟 ⫽ 11. The factors of 35 are 1 # 35 and 5 # 7. Two possible first terms are: (6z )(z ) and (3z )(2z ), and we begin with 5 and 7 as factors of 35.

(6z

Outer and Inner Products

)(z

)

Sum

Difference

1. (6z

5)(z

7)

42z ⫹ 5z 47z

42z ⫺ 5z 37z

2. (6z

7)(z

5)

30z ⫹ 7z 37z

30z ⫺ 7z 23z

(3z

Outer and Inner Products

)(2z

)

Sum

Difference

3. (3z

5)(2z

7)

21z ⫹ 10z 31z

21z ⫺ 10z 11z

4. (3z

7)(2z

5)

15z ⫹ 14z 29z

15z ⫺ 14z 1z


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Since possibility 3 yields a linear term of 11z, we need not consider other factors of 35 and write the factored form as 6z2 ⫺ 11z ⫺ 35 ⫽ 13z 5212z 72. The signs can then be arranged to obtain a middle term of ⫺11z: 13z ⫹ 5212z ⫺ 72, ⫺21z ⫹ 10z ⫽ ⫺11z ✓. C. You’ve just seen how we can factor quadratic polynomials


䊳

D. Factoring Special Forms and Quadratic Forms Next we consider methods to factor each of the special products we encountered in Appendix A.2.

The Difference of Two Squares WORTHY OF NOTE In an attempt to factor a sum of two perfect squares, say v2 ⫹ 49, let’s list all possible binomial factors. These are (1) 1v ⫹ 721v ⫹ 72, (2) 1v ⫺ 721v ⫺ 72, and (3) 1v ⫹ 721v ⫺ 72. Note that (1) and (2) are the binomial squares 1v ⫹ 72 2 and 1v ⫺ 72 2, with each product resulting in a “middle” term, whereas (3) is a binomial times its conjugate, resulting in a difference of squares: v2 ⫺ 49. With all possibilities exhausted, we conclude that the sum of two squares is prime!

EXAMPLE 6

䊳

Multiplying and factoring are inverse processes. Since 1x ⫺ 721x ⫹ 72 ⫽ x2 ⫺ 49, we know that x2 ⫺ 49 ⫽ 1x ⫺ 721x ⫹ 72. In words, the difference of two squares will factor into a binomial and its conjugate. To find the terms of the factored form, rewrite each term in the original expression as a square: 1 2 2. Factoring the Difference of Two Perfect Squares Given any expression that can be written in the form A2 ⫺ B2, A2 ⫺ B2 ⫽ 1A ⫹ B21A ⫺ B2

Note that the sum of two perfect squares A2 ⫹ B2 cannot be factored using real numbers (the expression is prime). As a reminder, always check for a common factor first and be sure to write all results in completely factored form. See Example 6(c). Factoring the Difference of Two Perfect Squares Factor each expression completely. a. 4w2 ⫺ 81 b. v2 ⫹ 49 c. ⫺3n2 ⫹ 48

Solution

䊳

a. 4w2 ⫺ 81 ⫽ 12w2 2 ⫺ 92 ⫽ 12w ⫹ 92 12w ⫺ 92 b. v2 ⫹ 49 is prime. c. ⫺3n2 ⫹ 48 ⫽ ⫺31n2 ⫺ 162 ⫽ ⫺33 n2 ⫺ 142 2 4 ⫽ ⫺31n ⫹ 42 1n ⫺ 42 1 d. z4 ⫺ 81 ⫽ 1z2 2 2 ⫺ 1 19 2 2 ⫽ 1z2 ⫹ 19 2 1z2 ⫺ 19 2 ⫽ 1z2 ⫹ 19 2 3 z2 ⫺ 1 13 2 2 4 ⫽ 1z2 ⫹ 19 2 1z ⫹ 13 2 1z ⫺ 13 2 2 e. x ⫺ 7 ⫽ 1x2 2 ⫺ 1 172 2 ⫽ 1x ⫹ 1721x ⫺ 172

1 d. z4 ⫺ 81

e. x2 ⫺ 7

write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2 factor out ⫺3 write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2 write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2

write as a difference of squares (z 2 ⫹ 19 is prime) result write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2


Perfect Square Trinomials

䊳

Since 1x ⫹ 72 2 ⫽ x2 ⫹ 14x ⫹ 49, we know that x2 ⫹ 14x ⫹ 49 ⫽ 1x ⫹ 72 2. In words, a perfect square trinomial will factor into a binomial square. To use this idea effectively, we must learn to identify perfect square trinomials. Note that the first and last terms of x2 ⫹ 14x ⫹ 49 are the squares of x and 7, and the middle term is twice the product of these two terms: 217x2 ⫽ 14x. These are the characteristics of a perfect square trinomial.


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Factoring Perfect Square Trinomials Given any expression that can be written in the form A2 ⫾ 2AB ⫹ B2, 1. A2 ⫹ 2AB ⫹ B2 ⫽ 1A ⫹ B2 2 2. A2 ⫺ 2AB ⫹ B2 ⫽ 1A ⫺ B2 2

EXAMPLE 7

䊳

Factoring a Perfect Square Trinomial Factor 12m3 ⫺ 12m2 ⫹ 3m.

Solution

12m3 ⫺ 12m2 ⫹ 3m ⫽ 3m14m2 ⫺ 4m ⫹ 12

䊳

check for common factors: GCF ⫽ 3m factor out 3m

For the remaining trinomial 4m2 ⫺ 4m ⫹ 1 p 1. Are the first and last terms perfect squares?

4m2 ⫽ 12m2 2 and 1 ⫽ 112 2 ✓ Yes.

2. Is the linear term twice the product of 2m and 1? 2 # 2m # 1 ⫽ 4m ✓ Yes.

Factor as a binomial square: 4m2 ⫺ 4m ⫹ 1 ⫽ 12m ⫺ 12 2 This shows 12m3 ⫺ 12m2 ⫹ 3m ⫽ 3m12m ⫺ 12 2. Now try Exercises 19 and 20

CAUTION

䊳

䊳

As shown in Example 7, be sure to include the GCF in your final answer. It is a common error to “leave the GCF behind.”

In actual practice, these calculations can be performed mentally, making the process much more efficient.

Sum or Difference of Two Perfect Cubes Recall that the difference of two perfect squares is factorable, but the sum of two perfect squares is prime. In contrast, both the sum and difference of two perfect cubes are factorable. For either A3 ⫹ B3 or A3 ⫺ B3 we have the following: 1. Each will factor into the product of a binomial and a trinomial: 2. The terms of the binomial are the quantities being cubed: 3. The terms of the trinomial are the square of A, the product AB, and the square of B, respectively: 4. The binomial takes the same sign as the original expression 5. The middle term of the trinomial takes the opposite sign of the original expression (the last term is always positive):

(

)(

binomial

) trinomial

(A

B)(

)

(A

B)(A2

AB

B 2)

(A ⫾ B)(A2

AB

B 2)

(A ⫾ B)(A2 ⫿ AB ⫹ B 2)

Factoring the Sum or Difference of Two Perfect Cubes: A3 ⫾ B3 1. A3 ⫹ B3 ⫽ 1A ⫹ B2 1A2 ⫺ AB ⫹ B2 2 2. A3 ⫺ B3 ⫽ 1A ⫺ B21A2 ⫹ AB ⫹ B2 2


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EXAMPLE 8

䊳

Factoring the Sum and Difference of Two Perfect Cubes Factor completely: a. x3 ⫹ 125

Solution

䊳

a.

b.

b. ⫺5m3n ⫹ 40n4

x3 ⫹ 125 ⫽ x3 ⫹ 53 Use A3 ⫹ B3 ⫽ 1A ⫹ B21A2 ⫺ AB ⫹ B2 2 x3 ⫹ 53 ⫽ 1x ⫹ 52 1x2 ⫺ 5x ⫹ 252

write terms as perfect cubes factoring template A S x and B S 5

⫺5m3n ⫹ 40n4 ⫽ ⫺5n1m3 ⫺ 8n3 2 ⫽ ⫺5n 3m3 ⫺ 12n2 3 4 Use A3 ⫺ B3 ⫽ 1A ⫺ B21A2 ⫹ AB ⫹ B2 2 m3 ⫺ 12n2 3 ⫽ 1m ⫺ 2n2 3 m2 ⫹ m12n2 ⫹ 12n2 2 4 ⫽ 1m ⫺ 2n21m2 ⫹ 2mn ⫹ 4n2 2 3 4 1 ⫺5m n ⫹ 40n ⫽ ⫺5n1m ⫺ 2n21m2 ⫹ 2mn ⫹ 4n2 2.

check for common factors 1GCF ⫽ ⫺5n2 write terms as perfect cubes factoring template A S m and B S 2n simplify factored form

The results for parts (a) and (b) can be checked using multiplication. Now try Exercises 21 and 22

䊳

Quadratic Forms and u-Substitution For any quadratic expression ax2 ⫹ bx ⫹ c in standard form, the degree of the leading term is twice the degree of the middle term. Generally, a trinomial is in quadratic form if it can be written as a1 __ 2 2 ⫹ b1 __ 2 ⫹ c, where the parentheses “hold” the same factors. The equation x4 ⫺ 13x2 ⫹ 36 ⫽ 0 is in quadratic form since 1x2 2 2 ⫺ 131x2 2 ⫹ 36 ⫽ 0. In many cases, we can factor these expressions using a placeholder substitution that transforms them into a more recognizable form. In a study of algebra, the letter “u” often plays this role. If we let u represent x2, the expression 1x2 2 2 ⫺ 131x2 2 ⫹ 36 becomes u2 ⫺ 13u ⫹ 36, which can be factored into 1u ⫺ 92 1u ⫺ 42. After “unsubstituting” (replace u with x2), we have 1x2 ⫺ 92 1x2 ⫺ 42 ⫽ 1x ⫹ 321x ⫺ 321x ⫹ 22 1x ⫺ 22. EXAMPLE 9

䊳

Factoring a Quadratic Form

Solution

䊳

Expanding the binomials would produce a fourth-degree polynomial that would be very difficult to factor. Instead we note the expression is in quadratic form. Letting u represent x2 ⫺ 2x (the variable part of the “middle” term), 1x2 ⫺ 2x2 2 ⫺ 21x2 ⫺ 2x2 ⫺ 3 becomes u2 ⫺ 2u ⫺ 3.

Write in completely factored form: 1x2 ⫺ 2x2 2 ⫺ 21x2 ⫺ 2x2 ⫺ 3.

u2 ⫺ 2u ⫺ 3 ⫽ 1u ⫺ 32 1u ⫹ 12

factor

To finish up, write the expression in terms of x, substituting x2 ⫺ 2x for u. ⫽ 1x2 ⫺ 2x ⫺ 32 1x2 ⫺ 2x ⫹ 12

substitute x2 ⫺ 2x for u

The resulting trinomials can be further factored. ⫽ 1x ⫺ 32 1x ⫹ 12 1x ⫺ 12 2

x2 ⫺ 2x ⫹ 1 ⫽ 1x ⫺ 12 2


D. You’ve just seen how we can factor special forms and quadratic forms

䊳

It is well known that information is retained longer and used more effectively when it’s placed in an organized form. The “factoring flowchart” provided in Figure A.4 offers a streamlined and systematic approach to factoring and the concepts involved. However, with some practice the process tends to “flow” more naturally than following a chart, with many of the decisions becoming automatic.


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A-45

Factoring Polynomials

Standard Form: decreasing order of degree

Greatest Common Factor (positive leading coefficient)

Number of Terms

Three

Two

Difference of squares

Difference of cubes

Sum of cubes

Trinomials (a ⫽ 1)

• Can any result be factored further?

Four

Trinomials (a ⫽ 1)

Factor by grouping

Advanced methods (Section 4.2)

• Polynomials that cannot be factored are said to be prime.

Figure A.4

For additional practice with these ideas, see Exercises 25 through 52.

E. Polynomial Equations and the Zero Product Property The ability to solve linear and quadratic equations is the foundation on which a large percentage of our future studies are built. Both are closely linked to the solution of other equation types, as well as to the graphs of these equations. In standard form, linear and quadratic equations have a known number of terms, so we commonly represent their coefficients using the early letters of the alphabet, as in ax2 ⫹ bx ⫹ c ⫽ 0. However, these equations belong to the larger family of polynomial equations. To write a general polynomial, where the number of terms is unknown, we often represent the coefficients using subscripts on a single variable, such as a1, a2, a3, and so on. Polynomial Equations A polynomial equation of degree n is one of the form anxn ⫹ an⫺1xn⫺1 ⫹ p ⫹ a1x1 ⫹ a0 ⫽ 0 where an, an⫺1, p , a1, a0 are real numbers and an ⫽ 0. As a prelude to solving polynomial equations of higher degree, we’ll first look at quadratic equations and the zero product property. As before, a quadratic equation is one that can be written in the form ax2 ⫹ bx ⫹ c ⫽ 0, where a, b, and c are real numbers and a ⫽ 0. As written, the equation is in standard form, meaning the terms are in decreasing order of degree and the equation is set equal to zero.


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Quadratic Equations A quadratic equation can be written in the form ax2 ⫹ bx ⫹ c ⫽ 0, with a, b, c 僆⺢, and a ⫽ 0. Notice that a is the leading coefficient, b is the coefficient of the linear (first degree) term, and c is a constant. All quadratic equations have degree two, but can have one, two, or three terms. The equation n2 ⫺ 81 ⫽ 0 is a quadratic equation with two terms, where a ⫽ 1, b ⫽ 0, and c ⫽ ⫺81. EXAMPLE 10

䊳

Determining Whether an Equation Is Quadratic State whether the given equation is quadratic. If yes, identify coefficients a, b, and c. ⫺3 a. 2x2 ⫺ 18 ⫽ 0 b. z ⫺ 12 ⫺ 3z2 ⫽ 0 c. x⫹5⫽0 4 d. z3 ⫺ 2z2 ⫹ 7z ⫽ 8 e. 0.8x2 ⫽ 0

Solution WORTHY OF NOTE The word quadratic comes from the Latin word quadratum, meaning square. The word historically refers to the “four sidedness” of a square, but mathematically to the area of a square. Hence its application to polynomials of the form ax2 ⫹ bx ⫹ c, where the variable of the leading term is squared.

䊳

Standard Form

Quadratic

a.

2x ⫺ 18 ⫽ 0

yes, deg 2

a⫽2

b.

⫺3z ⫹ z ⫺ 12 ⫽ 0

yes, deg 2

a ⫽ ⫺3

c.

⫺3 x⫹5⫽0 4

no, deg 1

(linear equation)

d.

z3 ⫺ 2z2 ⫹ 7z ⫺ 8 ⫽ 0

no, deg 3

(cubic equation)

e.

0.8x ⫽ 0

yes, deg 2

2

2

2

Coefficients b⫽0

a ⫽ 0.8

c ⫽ ⫺18

b⫽1

b⫽0

c ⫽ ⫺12

c⫽0


䊳

With quadratic and other polynomial equations, we generally cannot isolate the variable on one side using only properties of equality, because the variable is raised to different powers. Instead we attempt to solve the equation by factoring and applying the zero product property. Zero Product Property If A and B represent real numbers or real-valued expressions and A # B ⫽ 0, then A ⫽ 0 or B ⫽ 0. In words, the property says, If the product of any two (or more) factors is equal to zero, then at least one of the factors must be equal to zero. We can use this property to solve higher degree equations after rewriting them in terms of equations with lesser degree. As with linear equations, values that make the original equation true are called solutions or roots of the equation.


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EXAMPLE 11

䊳

A-47

Solving Equations Using the Zero Product Property Solve by writing the equations in factored form and applying the zero product property. a. 3x2 ⫽ 5x b. ⫺5x ⫹ 2x2 ⫽ 3 c. 4x2 ⫽ 12x ⫺ 9

Solution

3x2 ⫽ 5x 3x ⫺ 5x ⫽ 0 x13x ⫺ 52 ⫽ 0 x ⫽ 0 or 3x ⫺ 5 ⫽ 0 5 x ⫽ 0 or x⫽ 3 b. ⫺5x ⫹ 2x2 ⫽ 3 2x2 ⫺ 5x ⫺ 3 ⫽ 0 12x ⫹ 121x ⫺ 32 ⫽ 0 or x ⫺ 3 ⫽ 0 2x ⫹ 1 ⫽ 0 1 x⫽⫺ or x⫽3 2 c. 4x2 ⫽ 12x ⫺ 9 2 4x ⫺ 12x ⫹ 9 ⫽ 0 12x ⫺ 3212x ⫺ 32 ⫽ 0 2x ⫺ 3 ⫽ 0 or 2x ⫺ 3 ⫽ 0 3 3 x⫽ or x⫽ 2 2

䊳

a.

2

given equation standard form factor set factors equal to zero (zero product property) result given equation standard form factor set factors equal to zero (zero product property) result given equation standard form factor set factors equal to zero (zero product property) result

3 This equation has only the solution x ⫽ , which we call a repeated root. 2 Now try Exercises 65 through 88

䊳

CAUTION

EXAMPLE 12

䊳

䊳

Consider the equation x2 ⫺ 2x ⫺ 3 ⫽ 12. While the left-hand side is factorable, the result is 1x ⫺ 321x ⫹ 12 ⫽ 12 and finding a solution becomes a “guessing game” because the equation is not set equal to zero. If you misapply the zero factor property and say that x ⫺ 3 ⫽ 12 or x ⫹ 1 ⫽ 12, the “solutions” are x ⫽ 15 or x ⫽ 11, which are both incorrect! After subtracting 12 from both sides, x2 ⫺ 2x ⫺ 3 ⫽ 12 becomes x2 ⫺ 2x ⫺ 15 ⫽ 0 giving 1x ⫺ 521x ⫹ 32 ⫽ 0 with solutions x ⫽ 5 or x ⫽ ⫺3.

Solving Polynomials by Factoring Solve by factoring: 4x3 ⫺ 40x ⫽ 6x2.

Solution

䊳

4x3 ⫺ 40x ⫽ 6x2 4x ⫺ 6x2 ⫺ 40x ⫽ 0 2x 12x2 ⫺ 3x ⫺ 202 ⫽ 0 2x12x ⫹ 521x ⫺ 42 ⫽ 0 2x ⫽ 0 or 2x ⫹ 5 ⫽ 0 or x ⫺ 4 ⫽ 0 x ⫽ 0 or x ⫽ ⫺5 or x⫽4 2 3

given equation standard form common factor is 2x factored form zero product property result — solve for x

Substituting these values into the original equation verifies they are solutions. Now try Exercises 89 through 92

䊳

Example 12 reminds us that in the process of factoring polynomials, there may be a common monomial factor. This factor is also set equal to zero in the solution process (if the monomial is a constant, no solution is generated).


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EXAMPLE 13

䊳

Solving Higher Degree Equations Solve each equation by factoring. a. x3 ⫺ 4x ⫹ 20 ⫽ 5x2

Solution

䊳

a.

b. x4 ⫺ 10x2 ⫹ 9 ⫽ 0

x3 ⫺ 4x ⫹ 20 ⫽ 5x2 x ⫺ 5x2 ⫺ 4x ⫹ 20 ⫽ 0

original equation

3

x 1x ⫺ 52 ⫺ 41x ⫺ 52 ⫽ 0 1x ⫺ 521x2 ⫺ 42 ⫽ 0 1x ⫺ 52 1x ⫹ 221x ⫺ 22 ⫽ 0 or x ⫺ 2 ⫽ 0 x ⫺ 5 ⫽ 0 or x ⫹ 2 ⫽ 0 x ⫽ ⫺2 or x⫽2 x ⫽ 5 or 2

standard form; factor by grouping remove common factors from each group factor common binomial factored form zero product property solve

The solutions are x ⫽ 5, x ⫽ ⫺2, and x ⫽ 2. b. The equation appears to be in quadratic form and we begin by substituting u for x2 and u2 for x4. original equation x4 ⫺ 10x2 ⫹ 9 ⫽ 0 substitute u for x 2 and u 2 for x 4 u2 ⫺ 10u ⫹ 9 ⫽ 0 factored form 1u ⫺ 921u ⫺ 12 ⫽ 0 zero product property u ⫺ 9 ⫽ 0 or u ⫺ 1 ⫽ 0 2 2 substitute x 2 for u x ⫺ 9 ⫽ 0 or x ⫺ 1 ⫽ 0 1x ⫹ 321x ⫺ 32 ⫽ 0 or 1x ⫹ 12 1x ⫺ 12 ⫽ 0 factor x ⫽ ⫺1 or x ⫽ 1 zero product property x ⫽ ⫺3 or x ⫽ 3 or The solutions are x ⫽ ⫺3, x ⫽ 3, x ⫽ ⫺1, and x ⫽ 1. Now try Exercises 93 through 100 E. You’ve just seen how we can solve polynomial equations by factoring

䊳

In Examples 12 and 13, we were able to solve higher degree polynomial equations by “breaking them down” into linear and quadratic forms. This basic idea can be applied to other kinds of equations as well.

A.4 EXERCISES 䊳



1. To factor an expression means to rewrite the expression as an equivalent .

2. If a polynomial will not factor, it is said to be a(n) polynomial.

3. The difference of two perfect squares always factors into the product of a(n) and its .

4. The expression x2 ⫹ 6x ⫹ 9 is said to be a(n) trinomial, since its factored form is a perfect (binomial) square.

5. Discuss/Explain why 4x2 ⫺ 36 ⫽ 12x ⫺ 62 12x ⫹ 62 is not written in completely factored form, then rewrite it so it is factored completely.

6. Discuss/Explain why a3 ⫹ b3 is factorable, but a2 ⫹ b2 is not. Demonstrate by writing x3 ⫹ 64 in factored form, and by exhausting all possibilities for x2 ⫹ 64 to show it is prime.


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A-49


Factor each expression using the method indicated. Greatest Common Factor

7. a. ⫺17x2 ⫹ 51 c. ⫺3a4 ⫹ 9a2 ⫺ 6a3

b. 21b3 ⫺ 14b2 ⫹ 56b

8. a. ⫺13n2 ⫺ 52 b. 9p2 ⫹ 27p3 ⫺ 18p4 c. ⫺6g5 ⫹ 12g4 ⫺ 9g3 Common Binomial Factor

9. a. 2a1a ⫹ 22 ⫹ 31a ⫹ 22 b. 1b2 ⫹ 323b ⫹ 1b2 ⫹ 322 c. 4m1n ⫹ 72 ⫺ 111n ⫹ 72 10. a. 5x1x ⫺ 32 ⫺ 21x ⫺ 32 b. 1v ⫺ 522v ⫹ 1v ⫺ 523 c. 3p1q2 ⫹ 52 ⫹ 71q2 ⫹ 52

b. z2 ⫺ 18z ⫹ 81 d. 16q2 ⫹ 40q ⫹ 25

Sum/Difference of Perfect Cubes

21. a. 8p3 ⫺ 27 c. g3 ⫺ 0.027

b. m3 ⫹ 18 d. ⫺2t 4 ⫹ 54t

22. a. 27q3 ⫺ 125 c. b3 ⫺ 0.125

8 b. n3 ⫹ 27 d. 3r4 ⫺ 24r

b. x4 ⫹ 13x2 ⫹ 36

25. Completely factor each of the following (recall that “1” is its own perfect square and perfect cube). a. n2 ⫺ 1 b. n3 ⫺ 1 3 c. n ⫹ 1 d. 28x3 ⫺ 7x

2

Trinomial Factoring where 円a円 ⴝ 1

13. a. ⫺p2 ⫹ 5p ⫹ 14 c. n2 ⫹ 20 ⫺ 9n

b. q2 ⫺ 4q ⫹ 12

14. a. ⫺m2 ⫹ 13m ⫺ 42 c. v2 ⫹ 10v ⫹ 15

b. x2 ⫹ 12 ⫹ 13x

Trinomial Factoring where a ⴝ 1

15. a. 3p2 ⫺ 13p ⫺ 10 c. 10u2 ⫺ 19u ⫺ 15

b. 4q2 ⫹ 7q ⫺ 15

16. a. 6v ⫹ v ⫺ 35 c. 15z2 ⫺ 22z ⫺ 48

b. 20x ⫹ 53x ⫹ 18 2

Difference of Perfect Squares

17. a. 4s2 ⫺ 25 c. 50x2 ⫺ 72 e. b2 ⫺ 5

b. 9x2 ⫺ 49 d. 121h2 ⫺ 144

18. a. 9v ⫺ c. v4 ⫺ 1 e. x2 ⫺ 17

b. 25w ⫺ d. 16z4 ⫺ 81

1 25

20. a. x2 ⫹ 12x ⫹ 36 c. 25p2 ⫺ 60p ⫹ 36

24. a. x6 ⫺ 26x3 ⫺ 27 b. 31n ⫹ 52 2 ⫹ 21n ⫹ 52 ⫺ 21 c. 21z ⫹ 32 2 ⫹ 31z ⫹ 32 ⫺ 54

12. a. 6h ⫺ 9h ⫺ 2h ⫹ 3 b. 4k3 ⫹ 6k2 ⫺ 2k ⫺ 3 c. 3x2 ⫺ xy ⫺ 6x ⫹ 2y

2

b. b2 ⫹ 10b ⫹ 25 d. 9n2 ⫺ 42n ⫹ 49

23. a. x4 ⫺ 10x2 ⫹ 9 c. x6 ⫺ 7x3 ⫺ 8

11. a. 9q3 ⫹ 6q2 ⫹ 15q ⫹ 10 b. h5 ⫺ 12h4 ⫺ 3h ⫹ 36 c. k5 ⫺ 7k3 ⫺ 5k2 ⫹ 35

2

19. a. a2 ⫺ 6a ⫹ 9 c. 4m2 ⫺ 20m ⫹ 25

u-Substitution

Grouping

3

Perfect Square Trinomials

2

1 49

26. Carefully factor each of the following trinomials, if possible. Note differences and similarities. a. x2 ⫺ x ⫹ 6 b. x2 ⫹ x ⫺ 6 2 c. x ⫹ x ⫹ 6 d. x2 ⫺ x ⫺ 6 e. x2 ⫺ 5x ⫹ 6 f. x2 ⫹ 5x ⫺ 6 Factor each expression completely, if possible. Rewrite the expression in standard form (factor out “⫺1” if needed) and factor out the GCF if one exists. If you believe the expression will not factor, write “prime.”

27. a2 ⫹ 7a ⫹ 10

28. b2 ⫹ 9b ⫹ 20

29. 2x2 ⫺ 24x ⫹ 40

30. 10z2 ⫺ 140z ⫹ 450

31. 64 ⫺ 9m2

32. 25 ⫺ 16n2

33. ⫺9r ⫹ r2 ⫹ 18

34. 28 ⫹ s2 ⫺ 11s

35. 2h2 ⫹ 7h ⫹ 6

36. 3k2 ⫹ 10k ⫹ 8

37. 9k2 ⫺ 24k ⫹ 16

38. 4p2 ⫺ 20p ⫹ 25

39. ⫺6x3 ⫹ 39x2 ⫺ 63x

40. ⫺28z3 ⫹ 16z2 ⫹ 80z

41. 12m2 ⫺ 40m ⫹ 4m3

42. ⫺30n ⫺ 4n2 ⫹ 2n3

43. a2 ⫺ 7a ⫺ 60

44. b2 ⫺ 9b ⫺ 36

45. 8x3 ⫺ 125

46. 27r3 ⫹ 64


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47. m2 ⫹ 9m ⫺ 24

48. n2 ⫺ 14n ⫺ 36

49. x3 ⫺ 5x2 ⫺ 9x ⫹ 45

50. x3 ⫹ 3x2 ⫺ 4x ⫺ 12

51. Match each expression with the description that fits best. a. prime polynomial b. standard trinomial a ⫽ 1 c. perfect square trinomial d. difference of cubes e. binomial square f. sum of cubes g. binomial conjugates h. difference of squares i. standard trinomial a ⫽ 1 3 A. x ⫹ 27 B. 1x ⫹ 32 2 C. x2 ⫺ 10x ⫹ 25 D. x2 ⫺ 144 2 E. x ⫺ 3x ⫺ 10 F. 8s3 ⫺ 125t3 G. 2x2 ⫺ x ⫺ 3 H. x2 ⫹ 9 I. 1x ⫺ 72 and 1x ⫹ 72 52. Match each polynomial to its factored form. Two of them are prime. a. 4x2 ⫺ 9 b. 4x2 ⫺ 28x ⫹ 49 c. x3 ⫺ 125 d. 8x3 ⫹ 27 e. x2 ⫺ 3x ⫺ 10 f. x2 ⫹ 3x ⫹ 10 g. 2x2 ⫺ x ⫺ 3 h. 2x2 ⫹ x ⫺ 3 i. x2 ⫹ 25 A. 1x ⫺ 521x2 ⫹ 5x ⫹ 252 B. 12x ⫺ 321x ⫹ 12 C. 12x ⫹ 3212x ⫺ 32 D. 12x ⫺ 72 2 E. prime trinomial F. prime binomial G. 12x ⫹ 321x ⫺ 12 2 H. 12x ⫹ 3214x ⫺ 6x ⫹ 92 I. 1x ⫺ 521x ⫹ 22 Determine whether each equation is quadratic. If so, identify the coefficients a, b, and c. If not, discuss why.

53. 2x ⫺ 15 ⫺ x2 ⫽ 0 55.

2 x⫺7⫽0 3

54. 21 ⫹ x2 ⫺ 4x ⫽ 0 56. 12 ⫺ 4x ⫽ 9

63. 1x ⫺ 12 2 ⫹ 1x ⫺ 12 ⫹ 4 ⫽ 9

64. 1x ⫹ 52 2 ⫺ 1x ⫹ 52 ⫹ 4 ⫽ 17 Solve using the zero factor property. Be sure each equation is in standard form and factor out any common factors before attempting to solve. Check all answers in the original equation.

65. x2 ⫺ 15 ⫽ 2x

66. z2 ⫺ 10z ⫽ ⫺21

67. m2 ⫽ 8m ⫺ 16

68. ⫺10n ⫽ n2 ⫹ 25

69. 5p2 ⫺ 10p ⫽ 0

70. 6q2 ⫺ 18q ⫽ 0

71. ⫺14h2 ⫽ 7h

72. 9w ⫽ ⫺6w2

73. a2 ⫺ 17 ⫽ ⫺8

74. b2 ⫹ 8 ⫽ 12

75. g2 ⫹ 18g ⫹ 70 ⫽ ⫺11 76. h2 ⫹ 14h ⫺ 2 ⫽ ⫺51 77. m3 ⫹ 5m2 ⫺ 9m ⫺ 45 ⫽ 0 78. n3 ⫺ 3n2 ⫺ 4n ⫹ 12 ⫽ 0 79. 1c ⫺ 122c ⫺ 15 ⫽ 30

80. 1d ⫺ 102d ⫹ 10 ⫽ ⫺6 81. 9 ⫹ 1r ⫺ 52r ⫽ 33 82. 7 ⫹ 1s ⫺ 42s ⫽ 28

83. 1t ⫹ 421t ⫹ 72 ⫽ 54

84. 1g ⫹ 1721g ⫺ 22 ⫽ 20 85. 2x2 ⫺ 4x ⫺ 30 ⫽ 0

86. ⫺3z2 ⫹ 12z ⫹ 36 ⫽ 0 87. 2w2 ⫺ 5w ⫽ 3 88. ⫺3v2 ⫽ ⫺v ⫺ 2 89. 22x ⫽ x3 ⫺ 9x2 90. x3 ⫽ 13x2 ⫺ 42x 91. 3x3 ⫽ ⫺7x2 ⫹ 6x 92. 7x2 ⫹ 15x ⫽ 2x3 93. p3 ⫹ 7p2 ⫺ 63 ⫽ 9p 94. q3 ⫺ 4q ⫹ 24 ⫽ 6q2 95. x3 ⫺ 25x ⫽ 2x2 ⫺ 50 96. 3c2 ⫹ c ⫽ c3 ⫹ 3

1 57. x2 ⫽ 6x 4

58. 0.5x ⫽ 0.25x

59. 2x2 ⫹ 7 ⫽ 0

60. 5 ⫽ ⫺4x2

2

61. ⫺3x2 ⫹ 9x ⫺ 5 ⫹ 2x3 ⫽ 0

62. z2 ⫺ 6z ⫹ 9 ⫺ z3 ⫽ 0

97. x4 ⫺ 29x2 ⫹ 100 ⫽ 0 98. z4 ⫺ 20z2 ⫹ 64 ⫽ 0

99. 1b2 ⫺ 3b2 2 ⫺ 141b2 ⫺ 3b2 ⫹ 40 ⫽ 0

100. 1d2 ⫺ d 2 2 ⫺ 81d2 ⫺ d2 ⫹ 12 ⫽ 0


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Section A.4 Factoring Polynomials and Solving Polynomial Equations by Factoring 䊳

A-51


101. Surface area of a cylinder: 2␲r2 ⴙ 2␲rh

102. Volume of a cylindrical shell: ␲R2h ⴚ ␲r2h

The surface area of a cylinder is given by the formula shown, where h is the height of the cylinder and r is the radius. Factor out the GCF and use the result to find the surface area of a cylinder where r ⫽ 35 cm and h ⫽ 65 cm. Answer in exact form and in approximate form rounded to the nearest whole number.

r The volume of a cylindrical shell (a larger cylinder with a smaller cylinder removed) can be found using the formula shown, where R is the radius of the larger cylinder and r is the radius of the smaller. Factor the expression R completely and use the result to find the volume of a shell where R ⫽ 9 cm, r ⫽ 3 cm, and h ⫽ 10 cm. Answer in exact form and in approximate form rounded to the nearest whole number.

䊳

APPLICATIONS

In many cases, factoring an expression can make it easier to evaluate as in the following applications.

103. Conical shells: The volume of a conical shell (like the shell of an ice cream cone) is given by the 1 1 formula V ⫽ ␲R2h ⫺ ␲r2h, where R is the outer 3 3 radius and r is the inner radius of the cone. Write the formula in completely factored form, then find the volume of a shell when R ⫽ 5.1 cm, r ⫽ 4.9 cm, and h ⫽ 9 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 104. Spherical shells: The volume of a spherical shell (like the outer r shell of a cherry cordial) is given R 4 4 3 3 by the formula V ⫽ 3␲R ⫺ 3␲r , where R is the outer radius and r is the inner radius of the shell. Write the right-hand side in completely factored form, then find the volume of a shell where R ⫽ 1.8 cm and r ⫽ 1.5 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 105. Volume of a box: The volume of a rectangular box x inches in height is given by the relationship V ⫽ x3 ⫹ 8x2 ⫹ 15x. Factor the right-hand side to determine: (a) The number of inches that the width exceeds the height, (b) the number of inches the length exceeds the height, and (c) the volume given the height is 2 ft. 106. Shipping textbooks: A publisher ships paperback books stacked x copies high in a box. The total number of books shipped per box is given by the relationship B ⫽ x3 ⫺ 13x2 ⫹ 42x. Factor the

right-hand side to determine (a) how many more or fewer books fit the width of the box (than the height), (b) how many more or fewer books fit the length of the box (than the height), and (c) the number of books shipped per box if they are stacked 10 high in the box. 107. Space-Time relationships: Due to the work of Albert Einstein and other physicists who labored on space-time relationships, it is known that the faster an object moves the shorter it appears to become. This phenomenon is modeled by the v 2 Lorentz transformation L ⫽ L0 1 ⫺ a b , c B where L0 is the length of the object at rest, L is the relative length when the object is moving at velocity v, and c is the speed of light. Factor the radicand and use the result to determine the relative length of a 12-in. ruler if it is shot past a stationary observer at 0.75 times the speed of light 1v ⫽ 0.75c2 . 108. Tubular fluid flow: As a fluid flows through a tube, it is flowing faster at the center of the tube than at the sides, where the tube exerts a backward drag. Poiseuille’s law gives the velocity of the flow G 2 1R ⫺ r2 2 , at any point of the cross section: v ⫽ 4␩ where R is the inner radius of the tube, r is the distance from the center of the tube to a point in the flow, G represents what is called the pressure gradient, and ␩ is a constant that depends on the viscosity of the fluid. Factor the right-hand side and find v given R ⫽ 0.5 cm, r ⫽ 0.3 cm, G ⫽ 15, and ␩ ⫽ 0.25.


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Solve by factoring.

109. Envelope sizes: Large mailing envelopes often come in standard sizes, with 5- by 7-in. and 9- by 12-in. envelopes being the most common. The next larger size envelope has an area of 143 in2, with a length that is 2 in. longer than the width. What are the dimensions of the larger envelope?

a length that is 6 in. longer than the width. What are the dimensions of the Ledger size paper?

Letter

Legal Ledger

110. Paper sizes: Letter size paper is 8.5 in. by 11 in. Legal size paper is 812 in. by 14 in. The next larger (common) size of paper has an area of 187 in2, with

䊳


111. Factor out a constant that leaves integer coefficients for each term: a. 12x4 ⫹ 18x3 ⫺ 34x2 ⫹ 4 b. 23b5 ⫺ 16b3 ⫹ 49b2 ⫺ 1 112. If x ⫽ 2 is substituted into 2x3 ⫹ hx ⫹ 8, the result is zero. What is the value of h? 113. Factor the expression: 192x3 ⫺ 164x2 ⫺ 270x. 114. As an alternative to evaluating polynomials by direct substitution, nested factoring can be used. The method has the advantage of using only products and sums — no powers. For P ⫽ x3 ⫹ 3x2 ⫹ 1x ⫹ 5, we begin by grouping all variable terms and factoring x: P ⫽ 3 x3 ⫹ 3x2 ⫹ 1x 4 ⫹ 5 ⫽

A.5

In Section A.5 you will review how to:

A. Write a rational

C. D. E.

Factor each expression completely.

115. x4 ⫺ 81

116. 16n4 ⫺ 1

117. p6 ⫺ 1

118. m6 ⫺ 64

119. q4 ⫺ 28q2 ⫹ 75

120. a4 ⫺ 18a2 ⫹ 32

Rational Expressions and Equations

LEARNING OBJECTIVES

B.

x 3 x2 ⫹ 3x ⫹ 14 ⫹ 5. Then we group the inner terms with x and factor again: P ⫽ x 3 x2 ⫹ 3x ⫹ 14 ⫹ 5 ⫽ x 3 x1x ⫹ 32 ⫹ 14 ⫹ 5. The expression can now be evaluated using any input and the order of operations. If x ⫽ 2, we quickly find that P ⫽ 27. Use this method to evaluate H ⫽ x3 ⫹ 2x2 ⫹ 5x ⫺ 9 for x ⫽ ⫺3.

expression in simplest form Multiply and divide rational expressions Add and subtract rational expressions Simplify compound fractions Solve rational equations

A rational number is one that can be written as the quotient of two integers. Similarly, a rational expression is one that can be written as the quotient of two polynomials. We can apply the skills developed in a study of fractions (how to reduce, add, subtract, multiply, and divide) to rational expressions, sometimes called algebraic fractions.

A. Writing a Rational Expression in Simplest Form A rational expression is in simplest form when the numerator and denominator have no common factors (other than 1). After factoring the numerator and denominator, we apply the fundamental property of rational expressions. Fundamental Property of Rational Expressions If P, Q, and R are polynomials, with Q, R ⫽ 0, P P P#R P#R and 122 ⫽ # ⫽ 112 # Q R Q Q Q R


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Section A.5 Rational Expressions and Equations

A-53

In words, the property says (1) a rational expression can be simplified by canceling common factors in the numerator and denominator, and (2) an equivalent expression can be formed by multiplying numerator and denominator by the same nonzero polynomial. EXAMPLE 1

Simplifying a Rational Expression Write the expression in simplest form:

Solution

1x ⫺ 121x ⫹ 12 x2 ⫺ 1 ⫽ 2 1x ⫺ 121x ⫺ 22 x ⫺ 3x ⫹ 2

1x ⫺ 121x ⫹ 12

x2 ⫺ 1 . x ⫺ 3x ⫹ 2 2

factor numerator and denominator

1

⫽

1x ⫺ 121x ⫺ 22 x⫹1 ⫽ x⫺2

common factors reduce to 1

simplest form


WORTHY OF NOTE If we view a and b as two points on the number line, we note that they are the same distance apart, regardless of the order they are subtracted. This tells us the numerator and denominator will have the same absolute value but be opposite in sign, giving a value of ⫺1 (check using a few test values).

CAUTION

When simplifying rational expressions, we sometimes encounter expressions of a⫺b a⫺b . If we factor ⫺1 from the numerator, we see that ⫽ the form b⫺a b⫺a ⫺11b ⫺ a2 ⫽ ⫺1. b⫺a

When reducing rational numbers or expressions, only common factors can be reduced. It is incorrect to reduce (or divide out) individual terms: x⫹1 1 ⫽ (except for x ⫽ 0) x⫹2 2

Note that after simplifying an expression, we are actually saying the resulting (simpler) expression is equivalent to the original expression for all values where both are defined. The first expression is not defined when x ⫽ 1 or x ⫽ 2, the second when x ⫽ 2 (since the denominators would be zero). The calculator screens shown in Figure A.5 help to illustrate this fact, and it appears that we would very much prefer to be working with the simpler expression!

⫺6 ⫹ 423 ⫽ ⫺3 ⫹ 423, and 2

Figure A.5


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EXAMPLE 2

Simplifying a Rational Expression Write the expression in simplest form:

Solution

213 ⫺ x2 6 ⫺ 2x ⫽ 2 1x ⫺ 321x ⫹ 32 x ⫺9 1221⫺12 ⫽ x⫹3 ⫺2 ⫽ x⫹3

6 ⫺ 2x . x2 ⫺ 9

factor numerator and denominator

reduce:

3⫺x ⫽ ⫺1 x⫺3

simplest form

A. You’ve just seen how we can write a rational expression in simplest form


B. Multiplication and Division of Rational Expressions Operations on rational expressions use the factoring skills reviewed earlier, along with much of what we know about rational numbers. Multiplying Rational Expressions Given that P, Q, R, and S are polynomials with Q, S ⫽ 0, P R # ⫽ PR Q S QS 1. Factor all numerators and denominators completely. 2. Reduce common factors. 3. Multiply numerator ⫻ numerator and denominator ⫻ denominator.

EXAMPLE 3

Multiplying Rational Expressions Compute the product:

Solution

2a ⫹ 2 3a2 ⫺ a ⫺ 2 # . 3a ⫺ 3a2 9a2 ⫺ 4

21a ⫹ 12 13a ⫹ 221a ⫺ 12 2a ⫹ 2 # 3a2 ⫺ a ⫺ 2 # ⫽ 2 2 3a11 ⫺ a2 13a ⫺ 2213a ⫹ 22 3a ⫺ 3a 9a ⫺ 4 ⫽ ⫽

21a ⫹ 12

#

1⫺12

13a ⫹ 22 1a ⫺ 1 2 1

3a11 ⫺ a2 13a ⫺ 2213a ⫹ 22 ⫺21a ⫹ 12 3a13a ⫺ 22

factor

reduce:

a⫺1 ⫽ ⫺1 1⫺a

simplest form


To divide fractions, we multiply the first expression by the reciprocal of the second (we sometimes say, “invert the divisor and multiply”). The quotient of two rational expressions is computed in the same way.


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A-55

Dividing Rational Expressions Given that P, Q, R, and S are polynomials with Q, R, S ⫽ 0, R P S PS P ⫼ ⫽ # ⫽ Q S Q R QR Invert the divisor and multiply.

EXAMPLE 4

Dividing Rational Expressions Compute the quotient

Solution

4m3 ⫺ 12m2 ⫹ 9m 10m2 ⫺ 15m ⫼ . m2 ⫺ 49 m2 ⫹ 4m ⫺ 21

10m2 ⫺ 15m 4m3 ⫺ 12m2 ⫹ 9m ⫼ m2 ⫺ 49 m2 ⫹ 4m ⫺ 21 4m3 ⫺ 12m2 ⫹ 9m # m2 ⫹ 4m ⫺ 21 ⫽ m2 ⫺ 49 10m2 ⫺ 15m 2 m14m ⫺ 12m ⫹ 92 1m ⫹ 72 1m ⫺ 32 # ⫽ 1m ⫹ 721m ⫺ 72 5m12m ⫺ 32

m 12m ⫺ 3212m ⫺ 32 1m ⫹ 721m ⫺ 32 1

⫽ ⫽

1

invert and multiply

factor

1

1m ⫹ 721m ⫺ 72 12m ⫺ 32 1m ⫺ 32

#

5m12m ⫺ 32

factor and reduce

lowest terms

51m ⫺ 72

Note that we sometimes refer to simplest form as lowest terms. Now try Exercises 21 through 42

CAUTION

1w ⫹ 721w ⫺ 72 w ⫺ 2 # , it is a common mistake to think that all factors 1w ⫺ 721w ⫺ 22 w ⫹ 7 “cancel,” leaving an answer of zero. Actually, all factors reduce to 1, and the result is a value of 1 for all inputs where the product is defined. For products like

1w ⫹ 72 1w ⫺ 72 w ⫺ 2 # ⫽1 1w ⫺ 721w ⫺ 22 w ⫹ 7 1

B. You’ve just seen how we can multiply and divide rational expressions

1

1

C. Addition and Subtraction of Rational Expressions Recall that the addition and subtraction of fractions requires finding the lowest common denominator (LCD) and building equivalent fractions. The sum or difference of the numerators is then placed over this denominator. The procedure for the addition and subtraction of rational expressions is very much the same. Note that the LCD can also be described as the least common multiple (LCM) of all denominators. Addition and Subtraction of Rational Expressions 1. 2. 3. 4.

Find the LCD of all rational expressions. Build equivalent expressions using the LCD. Add or subtract numerators as indicated. Write the result in lowest terms.


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EXAMPLE 5

Adding and Subtracting Rational Expressions Compute as indicated: 7 3 a. ⫹ 10x 25x2

Solution

b.

10x 5 ⫺ x⫺3 x ⫺9 2

a. The LCM for 10x and 25x2 is 50x2. 3 3 2 7 7 # 5x # ⫹ ⫹ ⫽ 2 10x 10x 5x 25x 25x2 2 35x 6 ⫽ ⫹ 2 50x 50x2 35x ⫹ 6 ⫽ 50x2

find the LCD write equivalent expressions

simplify

add the numerators and write the result over the LCD

The result is in simplest form. b. The LCM for x2 ⫺ 9 and x ⫺ 3 is 1x ⫺ 32 1x ⫹ 32. 5 5 10x 10x #x⫹3 ⫺ ⫺ ⫽ 2 x⫺3 x⫺3 x⫹3 1x ⫺ 321x ⫹ 32 x ⫺9 10x ⫺ 51x ⫹ 32 ⫽ 1x ⫺ 321x ⫹ 32 10x ⫺ 5x ⫺ 15 ⫽ 1x ⫺ 321x ⫹ 32 5x ⫺ 15 ⫽ 1x ⫺ 32 1x ⫹ 32

find the LCD write equivalent expressions

subtract numerators, write the result over the LCD distribute

combine like terms

1

⫽

51x ⫺ 32

1x ⫺ 321x ⫹ 32

⫽

5 x⫹3

factor and reduce


EXAMPLE 6

Adding and Subtracting Rational Expressions Perform the operations indicated: n⫺3 c b2 5 a. b. ⫺ 2 ⫺ 2 a n⫹2 n ⫺4 4a

Solution

a. The LCM for n ⫹ 2 and n2 ⫺ 4 is 1n ⫹ 22 1n ⫺ 22. 5 n⫺3 n⫺3 5 #n⫺2⫺ ⫽ ⫺ 2 n⫹2 1n ⫹ 22 n ⫺ 2 1n ⫹ 22 1n ⫺ 22 n ⫺4 51n ⫺ 22 ⫺ 1n ⫺ 32 ⫽ 1n ⫹ 221n ⫺ 22 5n ⫺ 10 ⫺ n ⫹ 3 ⫽ 1n ⫹ 221n ⫺ 22 4n ⫺ 7 ⫽ 1n ⫹ 221n ⫺ 22

write equivalent expressions subtract numerators, write the result over the LCD distribute

result


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b. The LCM for a and 4a2 is 4a2:

b2 c c 4a b2 ⫺ ⫺ # ⫽ a a 4a 4a2 4a2 b2 4ac ⫽ 2⫺ 2 4a 4a b2 ⫺ 4ac ⫽ 4a2

write equivalent expressions simplify subtract numerators, write the result over the LCD


CAUTION

A-57

When the second term in a subtraction has a binomial numerator as in Example 6a, be sure the subtraction is applied to both terms. It is a common error to write 51n ⫺ 22 n⫺3 5n ⫺ 10 ⫺ n ⫺ 3 X in which the subtraction is applied ⫺ ⫽ 1n ⫹ 221n ⫺ 22 1n ⫹ 221n ⫺ 22 1n ⫹ 221n ⫺ 22 to the first term only. This is incorrect!

C. You’ve just seen how we can add and subtract rational expressions

D. Simplifying Compound Fractions Rational expressions whose numerator or denominator contain a fraction are called 3 2 ⫺ 3m 2 compound fractions. The expression is a compound fraction with a 3 1 ⫺ 4m 3m2 2 3 3 1 numerator of ⫺ and a denominator of ⫺ . The two methods commonly 3m 2 4m 3m2 used to simplify compound fractions are summarized in the following boxes. Simplifying Compound Fractions (Method I) 1. Add/subtract fractions in the numerator, writing them as a single expression. 2. Add/subtract fractions in the denominator, also writing them as a single expression. 3. Multiply the numerator by the reciprocal of the denominator and simplify if possible. Simplifying Compound Fractions (Method II) 1. Find the LCD of all fractions in the numerator and denominator. 2. Multiply the numerator and denominator by this LCD and simplify. 3. Simplify further if possible. Method II is illustrated in Example 7.

EXAMPLE 7

Simplifying a Compound Fraction Simplify the compound fraction: 2 3 ⫺ 3m 2 3 1 ⫺ 4m 3m2


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Solution

The LCD for all fractions is 12m2. 2 2 3 3 12m2 a ⫺ ⫺ ba b 3m 2 3m 2 1 ⫽ 1 1 3 3 12m2 ⫺ ⫺ b a ba 4m 4m 1 3m2 3m2 2 12m2 3 12m2 a ba b ⫺ a ba b 3m 1 2 1 ⫽ 12m2 1 3 12m2 b ⫺ a 2 ba b a ba 4m 1 1 3m 8m ⫺ 18m2 ⫽ 9m ⫺ 4

multiply numerator and denominator by 12m 2 ⫽

12m 2 1

distribute

simplify

⫺1

⫽

2m14 ⫺ 9m2 9m ⫺ 4

⫽ ⫺2m

D. You’ve just seen how we can simplify compound fractions

factor and write in lowest terms


E. Solving Rational Equations In Appendix A.3 we solved linear equations using basic properties of equality. If any equation contained fractional terms, we “cleared the fractions” using the least common denominator (LCD). We can also use this idea to solve rational equations, or equations that contain rational expressions. Solving Rational Equations 1. Identify and exclude any values that cause a zero denominator. 2. Multiply both sides by the LCD and simplify (this will eliminate all denominators). 3. Solve the resulting equation. 4. Check all solutions in the original equation.

EXAMPLE 8

Solving a Rational Equation Solve for m:

Solution

4 2 1 . ⫽ 2 ⫺ m m⫺1 m ⫺m

Since m2 ⫺ m ⫽ m1m ⫺ 12, the LCD is m1m ⫺ 12, where m ⫽ 0 and m ⫽ 1. 4 2 1 d multiply by LCD b ⫽ m1m ⫺ 12 c m1m ⫺ 12a ⫺ m m⫺1 m1m ⫺ 12 m1m ⫺ 12 2 m1m ⫺ 12 m1m ⫺ 12 1 4 a b⫺ a b⫽ a b distribute and simplify m 1 1 m⫺1 1 m1m ⫺ 12 denominators are eliminated 21m ⫺ 12 ⫺ m ⫽ 4 2m ⫺ 2 ⫺ m ⫽ 4 distribute m⫽6 solve for m


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A-59

Checking by substitution we have: 4 2 1 ⫽ 2 original equation ⫺ m m⫺1 m ⫺m 1 4 2 ⫺ ⫽ substitute 6 for m 2 162 162 ⫺ 1 162 ⫺ 162 1 4 1 ⫺ ⫽ simplify 3 5 30 3 2 5 ⫺ ⫽ common denominator 15 15 15 2 2 ⫽ ✓ result 15 15 A calculator check is shown in the figure. Now try Exercises 75 through 80 Multiplying both sides of an equation by a variable sometimes introduces a solution that satisfies the resulting equation, but not the original equation—the one we’re trying to solve. Such “solutions” are called extraneous roots and illustrate the need to check all apparent solutions in the original equation. In the case of rational equations, we are particularly aware that any value that causes a zero denominator is outside the domain and cannot be a solution. EXAMPLE 9

Solving a Rational Equation Solve: x ⫹

Solution

4x 12 ⫽1⫹ . x⫺3 x⫺3

The LCD is x ⫺ 3, where x ⫽ 3. multiply both 4x 12 b ⫽ 1x ⫺ 32a1 ⫹ b 1x ⫺ 32ax ⫹ sides by LCD x⫺3 x⫺3 12 x⫺3 4x x⫺3 distribute and ba ba b ⫽ 1x ⫺ 32 112 ⫹ a b simplify 1x ⫺ 32x ⫹ a 1 x⫺3 1 x⫺3 x2 ⫺ 3x ⫹ 12 ⫽ x ⫺ 3 ⫹ 4x denominators are eliminated set equation equal to zero x2 ⫺ 8x ⫹ 15 ⫽ 0 factor 1x ⫺ 321x ⫺ 52 ⫽ 0 zero factor property x ⫽ 3 or x ⫽ 5 Checking shows x ⫽ 3 is an extraneous root, and x ⫽ 5 is the only valid solution.

E. You’ve just seen how we can solve rational equations



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A.5 EXERCISES



1. In simplest form, 1a ⫺ b2/1a ⫺ b2 is equal to , while 1a ⫺ b2/1b ⫺ a2 is equal to 3. As with numeric fractions, algebraic fractions require a for addition and subtraction.

.

2. A rational expression is in when the numerator and denominator have no common factors, other than . 4. Since x2 ⫹ 9 is prime, the expression 1x2 ⫹ 92/ 1x ⫹ 32 is already written in .

State T or F and discuss/explain your response.

5.

x x⫹1 1 ⫺ ⫽ x⫹3 x⫹3 x⫹3

6.

1x ⫹ 321x ⫺ 22

1x ⫺ 221x ⫹ 32

⫽0


Reduce to lowest terms.

7. a. 8. a.

a⫺7 ⫺3a ⫹ 21

b.

x⫺4 ⫺7x ⫹ 28

b.

x2 ⫺ 5x ⫺ 14 9. a. 2 x ⫹ 6x ⫺ 7

2x ⫹ 6 4x2 ⫺ 8x 3x ⫺ 18 6x2 ⫺ 12x

a2 ⫹ 3a ⫺ 28 b. a2 ⫺ 49

16. a. c.

x3 ⫹ 4x2 ⫺ 5x x3 ⫺ x 12y2 ⫺ 13y ⫹ 3 27y3 ⫺ 1

b. d.

5p2 ⫺ 14p ⫺ 3 5p2 ⫹ 11p ⫹ 2 ax2 ⫺ 5x2 ⫺ 3a ⫹ 15 ax ⫺ 5x ⫹ 5a ⫺ 25

Compute as indicated. Write final results in lowest terms.

17.

a2 ⫺ 4a ⫹ 4 # a2 ⫺ 2a ⫺ 3 a2 ⫺ 9 a2 ⫺ 4

10. a.

r2 ⫹ 3r ⫺ 10 r2 ⫹ r ⫺ 6

b.

m2 ⫹ 3m ⫺ 4 m2 ⫺ 4m

18.

11. a.

x⫺7 7⫺x

b.

5⫺x x⫺5

b b2 ⫹ 5b ⫺ 24 # 2 2 b ⫺ 6b ⫹ 9 b ⫺ 64

19.

12. a.

v2 ⫺ 3v ⫺ 28 49 ⫺ v2

b.

u2 ⫺ 10u ⫹ 25 25 ⫺ u2

x2 ⫺ 7x ⫺ 18 # 2x2 ⫹ 7x ⫹ 3 x2 ⫺ 6x ⫺ 27 2x2 ⫹ 5x ⫹ 2

20.

13. a.

⫺12a3b5 4a2b⫺4

b.

7x ⫹ 21 63

6v2 ⫹ 23v ⫹ 21 # 4v2 ⫺ 25 3v ⫹ 7 4v2 ⫺ 4v ⫺ 15

21.

m3n ⫺ m3 d. 4 m ⫺ m4n

22.

a3 ⫺ 4a2 a2 ⫹ 3a ⫺ 28 ⫼ 3 2 a ⫹ 5a ⫺ 14 a ⫺8

y2 ⫺ 9 c. 3⫺y 14. a. c. 15. a. c.

p3 ⫺ 64 p3 ⫺ p2

⫼

p2 ⫹ 4p ⫹ 16 p2 ⫺ 5p ⫹ 4

5m⫺3n5 ⫺10mn2

b.

⫺5v ⫹ 20 25

23.

n2 ⫺ 4 2⫺n

d.

w4 ⫺ w4v w3v ⫺ w3

3⫺x 3x ⫺ 9 ⫼ 4x ⫹ 12 5x ⫹ 15

24.

2n3 ⫹ n2 ⫺ 3n n3 ⫺ n2

b.

6x2 ⫹ x ⫺ 15 4x2 ⫺ 9

2⫺b 5b ⫺ 10 ⫼ 7b ⫺ 28 5b ⫺ 20

25.

x3 ⫹ 8 x2 ⫺ 2x ⫹ 4

d.

mn2 ⫹ n2 ⫺ 4m ⫺ 4 mn ⫹ n ⫹ 2m ⫹ 2

a2 ⫹ a 3a ⫺ 9 # a2 ⫺ 3a 2a ⫹ 2


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26.

2 p2 ⫺ 36 # 2 4p 2p 2p ⫹ 12p

Compute as indicated. Write answers in lowest terms [recall that a ⴚ b ⴝ ⴚ1(b ⴚ a)].

8 # 1a2 ⫺ 2a ⫺ 352 27. 2 a ⫺ 25

43.

3 5 ⫹ 2 2x 8x

44.

15 7 ⫺ 2 16y 2y

m2 ⫺ 5m m2 ⫺ m ⫺ 20

45.

7 1 ⫺ 4x2y3 8xy4

46.

5 3 ⫹ 6a3b 9ab3

28. 1m2 ⫺ 162 29.

#

xy ⫺ 3x ⫹ 2y ⫺ 6 x2 ⫺ 3x ⫺ 10

xy ⫺ 3x xy ⫺ 5y

⫼

47.

4p p ⫺ 36 2

⫺

2 p⫺6

48.

3q q ⫺ 49 2

⫺

3 2q ⫺ 14

30.

ab ⫺ 2a 2a ⫺ ab ⫹ 7b ⫺ 14 ⫼ 2 ab ⫺ 7a b ⫺ 14b ⫹ 49

49.

4 m ⫹ 4⫺m m ⫺ 16

50.

p 2 ⫹ 2 p⫺2 4⫺p

31.

m2 ⫺ 16 m2 ⫹ 2m ⫺ 8 ⫼ m2 ⫺ 2m m2

51.

2 ⫺5 m⫺7

52.

4 ⫺9 x⫺1

32.

2x2 ⫺ 18 18 ⫺ 6x ⫼ 3 2 x ⫺ 25 x ⫺ 2x2 ⫺ 25x ⫹ 50

53.

33.

y⫹3

2

#y

3y2 ⫹ 9y

⫹ 7y ⫹ 12 y2 ⫺ 16

⫼

y2 ⫹ 4y

2

y⫹1 y ⫹ y ⫺ 30 2

⫺

2 y⫹6

54.

3 4n ⫺ 4n ⫺ 20 n ⫺ 5n

y2 ⫺ 4y

2

34.

x2 ⫺ 1 x ⫹ 1 x2 ⫹ 4x ⫺ 5 # ⫼ x2 ⫺ 5x ⫺ 14 x2 ⫺ 4 x ⫹ 5

55.

1 a ⫹ 2 a⫹4 a ⫺ a ⫺ 20

35.

x2 ⫺ x ⫹ 0.21 x2 ⫺ 0.49 ⫼ x2 ⫹ 0.5x ⫺ 0.14 x2 ⫺ 0.09

56.

x⫺5 2x ⫺ 1 ⫺ 2 x ⫹ 3x ⫺ 4 x ⫹ 3x ⫺ 4

36.

x2 ⫺ 0.8x ⫹ 0.15 x2 ⫺ 0.25 ⫼ x2 ⫹ 0.1x ⫺ 0.2 x2 ⫺ 0.16

57.

4 4 n2 ⫹ n ⫹ 3 9 37. ⫼ 13 1 2 2 n ⫺ n⫹ n2 ⫺ 15 15 25 n2 ⫺

q2 ⫺ 38. q2 ⫺

40. 41.

9 25

3 1 q⫺ 10 10

17 3 q⫹ 20 20 1 q2 ⫺ 16

q2 ⫹ ⫼

2

p3 ⫹ p2 ⫺ 49p ⫺ 49 p2 ⫹ 6p ⫺ 7

42. a

4x ⫺ 25 2x ⫺ x ⫺ 15 4x ⫹ 25x ⫺ 21 ⫼ 2 b# x ⫺ 11x ⫹ 30 x ⫺ 9x ⫹ 18 12x2 ⫺ 5x ⫺ 3 2

2

y ⫹ 2y ⫹ 1

m⫺5 2 ⫹ 2 m ⫺9 m ⫹ 6m ⫹ 9

60.

m⫹6 m⫹2 ⫺ 2 2 m ⫺ 25 m ⫺ 10m ⫹ 25

2

y⫹2 5y ⫹ 11y ⫹ 2 2

⫹

5 y ⫹y⫺6 2

m m⫺4 ⫹ 2 3m ⫺ 11m ⫹ 6 2m ⫺ m ⫺ 15 2

Write each term as a rational expression. Then compute the sum or difference indicated.

p3 ⫺ 1

4n2 ⫺ 1 6n2 ⫹ 5n ⫹ 1 # 12n2 ⫺ 17n ⫹ 6 # 12n2 ⫺ 5n ⫺ 3 2n2 ⫹ n 6n2 ⫺ 7n ⫹ 2 2

2y ⫺ 5 2

59.

62.

p2 ⫹ p ⫹ 1

y ⫹ 2y ⫹ 1

⫺

⫺2 7 ⫺ 2 3a ⫹ 12 a ⫹ 4a

2

⫼

3y ⫺ 4 2

58.

61.

6a ⫺ 24 3a ⫺ 24a ⫺ 12a ⫹ 96 ⫼ 3 2 a ⫺ 11a ⫹ 24 3a ⫺ 81 3

39.

4 9

2

2

63. a. p⫺2 ⫺ 5p⫺1

64. a. 3a⫺1 ⫹ 12a2 ⫺1

b. x⫺2 ⫹ 2x⫺3

b. 2y⫺1 ⫺ 13y2 ⫺1


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Simplify each compound fraction. Use either method.

5 1 ⫺ a 4 65. 25 1 ⫺ 2 16 a

1 8 ⫺ 3 27 x 66. 2 1 ⫺ x 3

1 p⫺2 67. 1 1⫹ p⫺2

3 y⫺6 68. 9 y⫹ y⫺6

3 2 ⫹ 3⫺x x⫺3 69. 4 5 ⫹ x x⫺3

2 1 ⫺ y⫺5 5⫺y 70. 3 2 ⫺ y y⫺5

2 y ⫺ y ⫺ 20 71. 3 4 ⫺ y⫹4 y⫺5

2 x ⫺ 3x ⫺ 10 72. 6 4 ⫺ x⫹2 x⫺5

p⫹

2

1⫹

74. a.

1 ⫹ 3m⫺1 1 ⫺ 3m⫺1

b.

4 ⫺ 9a⫺2 3a⫺2

b.

75.

2 5 1 ⫽ 2 ⫹ x x⫹1 x ⫹x

76.

1 5 3 ⫽ ⫺ 2 m m⫹3 m ⫹ 3m

77.

3 21 ⫽ a⫹2 a⫺1

78.

4 7 ⫽ 2y ⫺ 3 3y ⫺ 5

79.

1 1 1 ⫺ ⫽ 2 3y 4y y

81. x ⫹

80.

1 1 3 ⫺ ⫽ 2 5x 2x x

2x 14 ⫽1⫹ x⫺7 x⫺7

82.

2x 10 ⫹x⫽1⫹ x⫺5 x⫺5

83.

5 20 6 ⫽ ⫹ 2 n⫹3 n⫺2 n ⫹n⫺6

84.

1 ⫹ 2x⫺2 1 ⫺ 2x⫺2

2 1 7 ⫽⫺ ⫺ 2 p⫹2 p ⫹ 3 p ⫹ 5p ⫹ 6

85.

3 ⫹ 2n⫺1 5n⫺2

3 2a2 ⫹ 5 a ⫽ ⫺ 2 2a ⫹ 1 a⫺3 2a ⫺ 5a ⫺ 3

86.

⫺18 3n 4n ⫹ ⫽ 2n ⫺ 1 3n ⫹ 1 6n ⫺ n ⫺ 1

2

Rewrite each expression as a compound fraction. Then simplify using either method.

73. a.

Solve each equation. Identify any extraneous roots.

2


87. Cost to seize illegal drugs: C ⴝ

450P 100 ⴚ P

The cost C, in millions of 450P P dollars, for a government to find 100 ⴚ P and seize P% 10 ⱕ P 6 1002 of 40 a certain illegal drug is modeled 60 by the rational equation shown. 80 Complete the table (round to the nearest dollar) and answer the 90 following questions. 93 a. What is the cost of seizing 95 40% of the drugs? Estimate 98 the cost at 85%. 100 b. Why does cost increase dramatically the closer you get to 100%? c. Will 100% of the drugs ever be seized?

88. Chemicals in the bloodstream: C ⴝ

200H2 H3 ⴙ 40

Rational equations are often used 200H2 to model chemical concentrations H H3 ⴙ 40 in the bloodstream. The percent 0 concentration C of a certain drug 1 H hours after injection into muscle tissue can be modeled by 2 the equation shown (H ⱖ 0). 3 Complete the table (round to the 4 nearest tenth of a percent) and 5 answer the following questions. 6 a. What is the percent 7 concentration of the drug 3 hr after injection? b. Why is the concentration virtually equal at H ⫽ 4 and H ⫽ 5? c. Why does the concentration begin to decrease? d. How long will it take for the concentration to become less than 10%?


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APPLICATIONS

89. Stock prices: When a hot new stock hits the market, its price will often rise dramatically and then taper 5017d 2 ⫹ 102 off over time. The equation P ⫽ d3 ⫹ 50 models the price of stock XYZ d days after it has “hit the market.” (a) Create a table of values showing the price of the stock for the first 10 days (rounded to the nearest dollar) and comment on what you notice. (b) Find the opening price of the stock. (c) Does the stock ever return to its original price? 90. Population growth: The Department of Wildlife introduces 60 elk into a new game reserve. It is projected that the size of the herd will grow 1016 ⫹ 3t2 , where according to the equation N ⫽ 1 ⫹ 0.05t N is the number of elk and t is the time in years. (a) Approximate the population of elk after 14 yr. (b) If recent counts find 225 elk, approximately how many years have passed? 91. Typing speed: The number of words per minute that a beginner can type is approximated by the 60t ⫺ 120 , where N is the number equation N ⫽ t

A-63

of words per minute after t weeks, 3 6 t 6 12. Use a table to determine how many weeks it takes for a student to be typing an average of forty-five words per minute. 92. Memory retention: A group of students is asked to memorize 50 Russian words that are unfamiliar to them. The number N of these words that the average student remembers D days later is modeled 5D ⫹ 35 1D ⱖ 12. How many by the equation N ⫽ D words are remembered after (a) 1 day? (b) 5 days? (c) 12 days? (d) 35 days? (e) 100 days? According to this model, is there a certain number of words that the average student never forgets? How many? 93. Pollution removal: For a steel mill, the cost C (in millions of dollars) to remove toxins from the 22P , where resulting sludge is given by C ⫽ 100 ⫺ P P is the percent of the toxins removed. What percent can be removed if the mill spends $88,000,000 on the cleanup? Round to tenths of a percent.


94. One of these expressions is not equal to the others. Identify which and explain why. 20n a. b. 20 # n ⫼ 10 # n 10n 20 n 1 # c. 20n # d. 10n 10 n 95. The average of A and B is x. The average of C, D, and E is y. The average of A, B, C, D, and E is: 3x ⫹ 2y 2x ⫹ 3y a. b. 5 5 21x ⫹ y2 31x ⫹ y2 c. d. 5 5

3 2 and , what is the 5 4 reciprocal of the sum of their reciprocals? Given c a that and are any two numbers—what is the b d reciprocal of the sum of their reciprocals?

96. Given the rational numbers


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A-64


A.6

Radicals, Rational Exponents, and Radical Equations


A. Simplify radical

B.

C.

D. E.

F.

expressions of the form n 1a n Rewrite and simplify radical expressions using rational exponents Use properties of radicals to simplify radical expressions Add and subtract radical expressions Multiply and divide radical expressions; write a radical expression in simplest form Solve equations and use formulas involving radicals

Square roots and cube roots come from a much larger family called radical expressions. Expressions containing radicals can be found in virtually every field of mathematical study, and are an invaluable tool for modeling many real-world phenomena. n

A. Simplifying Radical Expressions of the Form 1an In previous coursework, you likely noted that 1a ⫽ b only if b2 ⫽ a. This definition cannot be applied to expressions like 1⫺16, since there is no number b such that b2 ⫽ ⫺16. In other words, the expression 1a represents a real number only if a ⱖ 0 (for a full review of the real numbers and other sets of numbers, see Appendix I at www.mhhe.com/coburn). Of particular interest to us now is an inverse operation for a2. In other words, what operation can be applied to a2 to return a? Consider the following.

EXAMPLE 1

䊳

Evaluating a Radical Expression

Evaluate 2a2 for the values given: a. a ⫽ 3 b. a ⫽ 5 c. a ⫽ ⫺6

Solution

䊳

a. 232 ⫽ 19 ⫽3

b. 252 ⫽ 125 ⫽5

c. 21⫺62 2 ⫽ 136 ⫽6


䊳

The pattern seemed to indicate that 2a2 ⫽ a and that our search for an inverse operation was complete—until Example 1(c), where we found that 21⫺62 2 ⫽ ⫺6. Using the absolute value concept, we can “repair” this apparent discrepancy and state a general rule for simplifying these expressions: 2a2 ⫽ 冟a冟. For expressions like 249x2 and 2y6, the radicands can be rewritten as perfect squares and simplified in the same manner: 249x2 ⫽ 217x2 2 ⫽ 7冟x冟 and 2y6 ⫽ 21y3 2 2 ⫽ 冟y3冟. The Square Root of a2: 2a2 For any real number a, 2a2 円 a円.

EXAMPLE 2

䊳

Simplifying Square Root Expressions Simplify each expression. a. 2169x2

Solution

䊳

b. 2x2 ⫺ 10x ⫹ 25

a. 2169x2 ⫽ 冟13x冟 ⫽ 13冟x冟 b. 2x ⫺ 10x ⫹ 25 ⫽ 21x ⫺ 52 ⫽ 冟x ⫺ 5冟 2

since x could be negative 2

since x 5 could be negative


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Section A.6 Radicals, Rational Exponents, and Radical Equations

䊳

CAUTION

A-65

In Appendix A.2, we noted that 1A ⫹ B2 2 ⫽ A2 ⫹ B2, indicating that you cannot square the individual terms in a sum (the square of a binomial results in a perfect square trinomial). In a similar way, 2A2 ⫹ B2 ⫽ A ⫹ B, and you cannot take the square root of individual terms. There is a big difference between the expressions 2A2 ⫹ B2 and 21A ⫹ B2 2 ⫽ 冟A ⫹ B冟. Try evaluating each when A ⫽ 3 and B ⫽ 4. 3 3 3 To investigate expressions like 2x3, note the radicand in both 18 and 1⫺64 can be written as a perfect cube. From our earlier definition of cube roots we know 3 3 3 3 18 ⫽ 2 122 3 ⫽ 2, 1⫺64 ⫽ 2 1⫺42 3 ⫽ ⫺4, and that every real number has only one real cube root. For this reason, absolute value notation is not used or needed when taking cube roots.

3 3 a The Cube Root of a3: 2

For any real number a, 3 3 2 a a.

EXAMPLE 3

䊳

Simplifying Cube Root Expressions Simplify each expression. 3 3 a. 2 b. 2 ⫺27x3 ⫺64n6

Solution

䊳

3 3 a. 2⫺27x3 ⫽ 2 1⫺3x2 3 ⫽ ⫺3x

3 3 b. 2⫺64n6 ⫽ 2 1⫺4n2 2 3 ⫽ ⫺4n2


䊳

We can extend these ideas to fourth roots, fifth roots, and so on. For example, the 5 fifth root of a is b only if b5 ⫽ a. In symbols, 1a ⫽ b implies b5 ⫽ a. Since an odd number of negative factors is always negative: 1⫺22 5 ⫽ ⫺32, and an even number of negative factors is always positive: 1⫺22 4 ⫽ 16, we must take the index into account n when evaluating expressions like 1an. If n is even and the radicand is unknown, absolute value notation must be used. n

WORTHY OF NOTE 2 Just as 1 ⫺16 is not a real number, 4 6 1 ⫺16 and 1 ⫺16 do not represent real numbers. An even number of repeated factors is always positive!

EXAMPLE 4

䊳

The nth Root of an: 2an For any real number a, n 1. 1an ⫽ 冟a冟 when n is even. Simplifying Radical Expressions Simplify each expression. 4 4 a. 1 b. 1 81 ⫺81 4 5 4 e. 216m f. 232p5

Solution

A. You’ve just seen how we can simplify radical n expressions of the form 1a n

䊳

n

2. 1an ⫽ a when n is odd.

4 1 81 ⫽ 3 5 132 ⫽ 2 4 4 2 16m4 ⫽ 2 12m2 4 ⫽ 冟2m冟 or 2冟m冟 6 g. 2 1m ⫹ 52 6 ⫽ 冟m ⫹ 5冟

a. c. e.

5 c. 1 32 6 g. 2 1m ⫹ 52 6

5 d. 1 ⫺32 7 h. 2 1x ⫺ 22 7

4 b. 1⫺81 is not a real number 5 d. 1 ⫺32 ⫽ ⫺2 5 5 f. 232p5 ⫽ 2 12p2 5 ⫽ 2p 7 h. 2 1x ⫺ 22 7 ⫽ x ⫺ 2


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B. Radical Expressions and Rational Exponents As an alternative to radical notation, a rational (fractional) exponent can be used, 3 along with the power property of exponents. For 1a3 ⫽ a, notice that an exponent of one-third1 can 3 replace the cube root notation and produce the same result: 3 3 2 a ⫽ 1a3 2 3 ⫽ a3 ⫽ a. In the same way, an exponent of one-half can replace the 1 2 square root notation: 2a2 ⫽ 1a2 2 2 ⫽ a2 ⫽ 冟a冟. In general, we have the following: Rational Exponents If a is a real number and n is an integer greater than 1, 1 n n then 1a ⫽ 2a1 ⫽ an n provided 1a represents a real number.

EXAMPLE 5

䊳

Simplifying Radical Expressions Using Rational Exponents Simplify by rewriting each radicand as a perfect nth power and converting to rational exponent notation. 3 3 4 4 3 8w a. 2⫺125 b. ⫺216x20 c. 2⫺81 d. B 27

Solution

䊳

a. 2⫺125 ⫽ 2 1⫺52 3 1 ⫽ 3 1⫺52 3 4 3 3 ⫽ 1⫺52 3 ⫽ ⫺5

b. ⫺216x20 ⫽ ⫺ 2 12x5 2 4 1 ⫽ ⫺ 3 12x5 2 4 4 4 ⫽ ⫺ 0 2x5 0 ⫽ ⫺2冟x冟5

4 c. 2 ⫺81 ⫽ 1⫺812 4 is not a real number

d.

3

3

1

4

4

3 8w3 3 2w ⫽ a b B 27 B 3 1 2w 2w 3 3 ⫽ ca b d ⫽ 3 3 3

Now try Exercises 15 and 16 n

WORTHY OF NOTE Any rational number can be decomposed into the product of a unit fraction and an m 1 integer: ⫽ # m. n n

n

䊳

1

Figure A.6 When a rational exponent is used, as in 1a ⫽ 2a1 ⫽ an, the denominator of the exponent represents the index number, while the m numerator of the exponent represents the original power on a. This an n m is true even when the exponent on a is something other than one! In (兹a ) 4 other words, the radical expression 2163 can be rewritten as 1 3 1 3 1163 2 4 ⫽ 1161 2 4 or 164. This is further illustrated in Figure A.6 where we see the rational exponent has the form, “power over root.” To evaluate this expression without the aid of a calculator, we use the commutative property to rewrite 3 1 1 3 1 3 1161 2 4 as 1164 2 1 and begin with the fourth root of 16: 1164 2 1 ⫽ 23 ⫽ 8. m In general, if m and n have no common factors (other than 1) the expression a n can be interpreted in the following two ways.

Rational Exponents If mn is a rational number expressed in lowest terms with n ⱖ 2, then 112 a n ⫽ 1 2a2 m m

n

n

or

122 a n ⫽ 2am m

n

(compute 1a, then take the mth power), (compute am, then take the nth root), n provided 1a represents a real number.


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Expressions with rational exponents are generally easier to evaluate if we compute the root first, then apply the exponent. Computing the root first also helps us determine whether or not an expression represents a real number.

EXAMPLE 6

䊳

Simplifying Expressions with Rational Exponents Simplify each expression, if possible. b. 1⫺492 2

3

3

a. ⫺492

Solution

3

WORTHY OF NOTE

While the expression 1⫺82 3 ⫽ 1⫺8 represents the real number ⫺2, the 2 6 expression 1⫺82 6 ⫽ 1 1⫺82 2 is not a 2 1 real number, even though ⫽ . 3 6 Note that the second exponent is not in lowest terms. 1

3

2

b. 1⫺492 2 ⫽ 3 1⫺492 2 4 3, ⫽ 1 1⫺492 3 not a real number

1

3

c. 1⫺82 3 ⫽ 3 1⫺82 34 2 3 ⫽ 1 1⫺82 2 2

d. ⫺8⫺3

2

a. ⫺492 ⫽ ⫺1492 2 3 ⫽ ⫺1 1492 3 ⫽ ⫺172 3 or ⫺343

䊳

c. 1⫺82 3

1

d.

⫽ 1⫺22 2 or 4

1

⫺8⫺3 ⫽ ⫺183 2 ⫺2 3 ⫽ ⫺1 1 82 ⫺2 2

1

⫽ ⫺2⫺2 or ⫺

B. You’ve just seen how we can rewrite and simplify radical expressions using rational exponents

1 4


䊳

C. Using Properties of Radicals to Simplify Radical Expressions The properties used to simplify radical expressions are closely connected to the properties of exponents. For instance, the1 product to a power1 property holds even when n 1 1 1 1 is a rational number. This means 1xy2 2 ⫽ x2y2 and 14 # 252 2 ⫽ 42 # 252. When the second statement is expressed in radical form, we have 14 # 25 ⫽ 14 # 225, with both forms having a value of 10. This suggests the product property of radicals, which can be extended to include cube roots, fourth roots, and so on. Product Property of Radicals n

n

If 1A and 1B represent real-valued expressions, then n

n

n

1AB ⫽ 1A # 1B

CAUTION

䊳

and

n

n

n

1A # 1B ⫽ 1AB.

Note that this property applies only to a product of two terms, not to a sum or difference. In other words, while 29x2 ⫽ 冟3x冟, 29 ⫹ x2 ⫽ | 3 ⫹ x |!

One application of the product property is to simplify radical expressions. In genn eral, the expression 1a is in simplified form if a has no factors (other than 1) that are perfect nth roots.

EXAMPLE 7

䊳

Simplifying Radical Expressions Write each expression in simplest form using the product property. a. 118

Solution

䊳

3 b. 5 2 125x4

a. 118 ⫽ 19 # 2 ⫽ 1912 ⫽ 3 12

c.

⫺4 ⫹ 220 2

3 3 d. 1.22 16n4 2 4n5

3 3 b. 5 2 125x4 ⫽ 5 # 2 125 # x4 3 3 3 # 3 1 These steps can ⫽ 5 # 2125 # 2 x 2x e 3 be done mentally # # # ⫽ 5 5 x 1x 3 ⫽ 25x 1 x


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WORTHY OF NOTE

c.

For expressions like those in Example 7(c), students must resist the “temptation” to reduce individual terms as in ⫺4 ⫹ 120 ⫽ ⫺2 ⫹ 120. 2 Remember, only factors can be reduced.

⫺4 ⫹ 120 ⫺4 ⫹ 14 # 5 ⫽ 2 2 ⫺4 ⫹ 2 15 ⫽ 2 1 21⫺2 ⫹ 152 ⫽ 2 ⫽ ⫺2 ⫹ 15

3 3 3 d. 1.2 2 16n4 2 4n5 ⫽ 1.2 2 64 # n9 3 3 9 ⫽ 1.2 264 2 n 3 3 ⫽ 1.2 2 64 2 1n3 2 3 ⫽ 1.2142n3 ⫽ 4.8n3

䊳


When radicals are combined using the product property, the result may contain a perfect nth root, which should be simplified. Note that the index numbers must be the same in order to use this property. The quotient property of radicals can also be established using exponential 100 1100 ⫽ ⫽ 2 suggests the following: properties. The fact that A 25 125 Quotient Property of Radicals n

n

If 1A and 1B represent real-valued expressions with B ⫽ 0, then n n 1A 1A n A n A ⫽ n and n ⫽ . AB 1B 1B A B Many times the product and quotient properties must work together to simplify a radical expression, as shown in Example 8A.

EXAMPLE 8A

䊳

Simplifying Radical Expressions Simplify each expression: 218a5 a. 22a

Solution

䊳

a.

218a5 22a

⫽

b.

18a5 B 2a

b.

⫽ 29a4 ⫽ 3a2

81 A 125x3 3

3 81 1 81 ⫽ 3 A 125x3 2125x3 3 1 27 # 3 ⫽ 5x 3 3 13 ⫽ 5x 3

Radical expressions can also be simplified using rational exponents.

EXAMPLE 8B

䊳

Using Rational Exponents to Simplify Radical Expressions Simplify using rational exponents: 3 4 a. 236p4q5 b. v 2 v

Solution

C. You’ve just seen how we can use properties of radicals to simplify radical expressions

䊳

a. 236p4q5 ⫽ 136p4q5 2 2 1 4 5 ⫽ 362p2q2 4 1 ⫽ 6p2q12 ⫹ 2 2 1 ⫽ 6p2q2q2 ⫽ 6p2q2 1q 1

3 c. 1 m1m 4

3 b. v 2v4 ⫽ v1 # v3 3 4 ⫽ v3 # v3 7 ⫽ v3 6 1 ⫽ v3v3 3 ⫽ v2 1 v

1

1

3 c. 1m1m ⫽ m3m2 1 1 ⫽ m3 ⫹ 2 5 ⫽ m6 6 ⫽ 2 m5


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D. Addition and Subtraction of Radical Expressions 3 Since 3x and 5x are like terms, we know 3x ⫹ 5x ⫽ 8x. If x ⫽ 17, the sum becomes 3 3 3 3 17 ⫹ 5 17 ⫽ 8 17, illustrating how like radical expressions can be combined. Like radicals are those that have the same index and radicand. In some cases, we can identify like radicals only after radical terms have been simplified.

EXAMPLE 9

䊳

Adding and Subtracting Radical Expressions Simplify and combine (if possible). 3 3 a. 145 ⫹ 2 120 b. 2 16x5 ⫺ x 2 54x2

Solution

䊳

a. 145 ⫹ 2 120 ⫽ 3 15 ⫹ 212 152 ⫽ 3 15 ⫹ 4 15 ⫽ 7 15

simplify radicals: 145 ⫽ 19 # 5; 120 ⫽ 14 # 5 like radicals result

3 b. 216x ⫺ x 254x ⫽ 28 # 2 # x # x ⫺ x 2 27 # 2 # x2 3 3 ⫽ 2x 2 2x2 ⫺ 3x 2 2x2 3 2 ⫽ ⫺x 22x 3

5

3

2

3

3

D. You’ve just seen how we can add and subtract radical expressions

2

simplify radicals result


䊳

E. Multiplication and Division of Radical Expressions; Radical Expressions in Simplest Form Multiplying radical expressions is simply an extension of our earlier work. The multiplication can take various forms, from the distributive property to any of the special products reviewed in Appendix A.2. For instance, 1A ⫾ B2 2 ⫽ A2 ⫾ 2AB ⫹ B2, even if A or B is a radical term.

EXAMPLE 10

䊳

Multiplying Radical Expressions Compute each product and simplify. a. 5 131 16 ⫺ 4 132 b. 12 12 ⫹ 6 132 13 110 ⫹ 1152 c. 1x ⫹ 172 1x ⫺ 172 d. 13 ⫺ 122 2

Solution

䊳

a. 5 131 16 ⫺ 4 132 ⫽ 5 118 ⫺ 201 132 2 ⫽ 5132 1 2 ⫺ 1202132 ⫽ 15 12 ⫺ 60

distribute simplify: 1 18 ⫽ 31 2, 1 132 2 ⫽ 3 result

b. 1212 ⫹ 613213110 ⫹ 1152 ⫽ 6 120 ⫹ 2 130 ⫹ 18130 ⫹ 6145 F-O-I-L ⫽ 12 15 ⫹ 20 130 ⫹ 18 15 extract roots and simplify ⫽ 30 15 ⫹ 20 130 result c. 1x ⫹ 172 1x ⫺ 172 ⫽ x2 ⫺ 1 172 2 ⫽ x2 ⫺ 7

LOOKING AHEAD Notice that the answer for Example 10(c) contains no radical terms, since the outer and inner products sum to zero. This result will be used to simplify certain radical expressions in this section and later in Chapter 2.

1A ⫹ B2 1A ⫺ B2 ⫽ A 2 ⫺ B 2 result

d. 13 ⫺ 122 2 ⫽ 132 2 ⫺ 2132 1 122 ⫹ 1 122 2 ⫽ 9 ⫺ 6 12 ⫹ 2 ⫽ 11 ⫺ 6 12

1A ⫺ B 2 2 ⫽ A 2 ⫺ 2AB ⫹ B 2 simplify each term result


䊳

One application of products and powers of radical expressions is to evaluate certain quadratic expressions, as illustrated in Example 11.


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EXAMPLE 11

䊳

Evaluating a Quadratic Expression Show that when x2 ⫺ 4x ⫹ 1 is evaluated at x ⫽ 2 ⫹ 13, the result is zero.

Solution

䊳

x2 ⫺ 4x ⫹ 1 original expression 12 ⫹ 132 2 ⫺ 412 ⫹ 132 ⫹ 1 substitute 2 ⫹ 13 for x 4 ⫹ 4 13 ⫹ 3 ⫺ 8 ⫺ 4 13 ⫹ 1 square binomial; distribute 14 ⫹ 3 ⫺ 8 ⫹ 12 ⫹ 1413 ⫺ 4 132 commutative and associative properties 0✓ A calculator check is shown in the figure.


䊳

When we applied the quotient property in Example 8A, we obtained a denominator free of radicals. Sometimes the denominator is not automatically free of radicals, and the need to write radical expressions in simplest form comes into play. This process is called rationalizing the denominator. Radical Expressions in Simplest Form A radical expression is in simplest form if: 1. The radicand has no perfect nth root factors. 2. The radicand contains no fractions. 3. No radicals occur in a denominator. As with other types of simplification, the desired form can be achieved in various ways. If the denominator is a single radical term, we multiply the numerator and denominator by the factors required to eliminate the radical in the denominator [see Examples 12(a) and 12(b)]. If the radicand is a rational expression, it is generally easier to build an equivalent fraction within the radical having perfect nth root factors in the denominator [see Example 12(c)].

EXAMPLE 12

䊳

Simplifying Radical Expressions Simplify by rationalizing the denominator. Assume a, x ⫽ 0. 2 ⫺7 a. b. 3 1x 5 23

Solution

䊳

a.

b.

2 # 13 2 ⫽ 5 13 5 13 13 2 13 2 13 ⫽ ⫽ 2 15 51 132 ⫺7 3 1 x

⫺71 1x211x2 3

⫽ ⫽

simplify—denominator is now rational

3

3 3 3 1 x1 1 x211 x2 3 2 ⫺7 2x

3 3 2 x 3 2 ⫺72 x ⫽ x

multiply numerator and denominator by 13

3 multiply using two additional factors of 1x

product property

3 3 2 x ⫽x


䊳


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In some applications, the denominator may be a sum or difference containing a radical term. In this case, the methods from Example 12 are ineffective, and instead we multiply by a conjugate since 1A ⫹ B21A ⫺ B2 ⫽ A2 ⫺ B2. If either A or B is a square root, the result will be a denominator free of radicals.

EXAMPLE 13

䊳

Simplifying Radical Expressions Using a Conjugate Simplify the expression by rationalizing the denominator. 2 ⫹ 13 Write the answer in exact form and approximate form 16 ⫺ 12 rounded to three decimal places.

Solution

䊳

2 ⫹ 13 2 ⫹ 13 # 16 ⫹ 12 ⫽ 16 ⫺ 12 16 ⫺ 12 16 ⫹ 12 2 16 ⫹ 2 12 ⫹ 118 ⫹ 16 ⫽ 1 162 2 ⫺ 1 122 2 3 16 ⫹ 2 12 ⫹ 3 12 6⫺2 3 16 ⫹ 5 12 ⫽ 4 ⬇ 3.605 ⫽

E. You’ve just seen how we can multiply and divide radical expressions and write a radical expression in simplest form

multiply by the conjugate of the denominator FOIL difference of squares simplify

exact form approximate form


䊳

F. Equations and Formulas Involving Radicals A radical equation is any equation that contains terms with a variable in the radicand. To solve a radical equation, we attempt to isolate a radical term on one side, then apply the appropriate nth power to free up the radicand and solve for the unknown. This is an application of the power property of equality. The Power Property of Equality n

n

If 1u and v are real-valued expressions and 1u ⫽ v, then 1 1u2 n ⫽ vn u ⫽ vn for n an integer, n ⱖ 2. n

Raising both sides of an equation to an even power can also introduce a false solution (extraneous root). Note that by inspection, the equation x ⫺ 2 ⫽ 1x has only the solution x ⫽ 4. But the equation 1x ⫺ 22 2 ⫽ x (obtained by squaring both sides) has both x ⫽ 4 and x ⫽ 1 as solutions, yet x ⫽ 1 does not satisfy the original equation. This means we should check all solutions of an equation where an even power is applied.


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EXAMPLE 14

䊳

Solving Radical Equations Solve each radical equation: a. 13x ⫺ 2 ⫹ 12 ⫽ x ⫹ 10

Solution

䊳

3 b. 2 1 x⫺5⫹4⫽0

a. 13x ⫺ 2 ⫹ 12 ⫽ x ⫹ 10 13x ⫺ 2 ⫽ x ⫺ 2 1 13x ⫺ 22 2 ⫽ 1x ⫺ 22 2 3x ⫺ 2 ⫽ x2 ⫺ 4x ⫹ 4 0 ⫽ x2 ⫺ 7x ⫹ 6 0 ⫽ 1x ⫺ 62 1x ⫺ 12 x ⫺ 6 ⫽ 0 or x ⫺ 1 ⫽ 0 x ⫽ 6 or x ⫽ 1

Check

䊳

x ⫽ 6:

Check

䊳

x ⫽ 1:

original equation isolate radical term (subtract 12) apply power property, power is even simplify, square binomial set equal to zero factor apply zero product property result, check for extraneous roots

13162 ⫺ 2 ⫹ 12 ⫽ 162 ⫹ 10 116 ⫹ 12 ⫽ 16 16 ⫽ 16 ✓

13112 ⫺ 2 ⫹ 12 ⫽ 112 ⫹ 10 11 ⫹ 12 ⫽ 11 13 ⫽ 11x

The only solution is x ⫽ 6; x ⫽ 1 is extraneous. A calculator check is shown in the figures. 3 b. 2 1 x⫺5⫹4⫽0 3 1 x ⫺ 5 ⫽ ⫺2 3 1 1x ⫺ 52 3 ⫽ 1⫺22 3 x ⫺ 5 ⫽ ⫺8 x ⫽ ⫺3

original equation isolate radical term (subtract 4, divide by 2) apply power property, power is odd 3 simplify: 1x ⫺ 52 3 ⫽ x ⫺ 5

solve

Substituting ⫺3 for x in the original equation verifies it is a solution.


䊳

Sometimes squaring both sides of an equation still results in an equation with a radical term, but often there is one fewer than before. In this case, we simply repeat the process, as indicated by the flowchart in Figure A.7.

EXAMPLE 15

䊳

Solving Radical Equations Solve the equation: 1x ⫹ 15 ⫺ 1x ⫹ 3 ⫽ 2.

Solution

䊳

1x ⫹ 15 ⫺ 1x ⫹ 3 ⫽ 2 1x ⫹ 15 ⫽ 1x ⫹ 3 ⫹ 2 1 1x ⫹ 152 2 ⫽ 1 1x ⫹ 3 ⫹ 22 2 x ⫹ 15 ⫽ 1x ⫹ 32 ⫹ 4 1x ⫹ 3 ⫹ 4 x ⫹ 15 ⫽ x ⫹ 4 1x ⫹ 3 ⫹ 7 8 ⫽ 4 1x ⫹ 3 2 ⫽ 1x ⫹ 3 4⫽x⫹3 1⫽x

original equation isolate one radical power property 1A ⫹ B2 2; A ⫽ 1x ⫹ 3, B ⫽ 2 simplify isolate radical divide by four power property possible solution


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Figure A.7

Check

1x ⫹ 15 ⫺ 1x ⫹ 3 ⫽ 2 1112 ⫹ 15 ⫺ 1112 ⫹ 3 ⫽ 2 116 ⫺ 14 ⫽ 2 4⫺2⫽2 2 ⫽ 2✓

䊳

Radical Equations

Isolate radical term

A-73

original equation substitute 1 for x simplify solution checks


䊳

Since rational exponents are so closely related to radicals, the solution process for each is very similar. The goal is still to “undo” the radical (rational exponent) and solve for the unknown.

Apply power property

Does the result contain a radical?

Power Property of Equality YES

For real-valued expressions u and v, with positive integers m, n, and mn in lowest terms: If m is odd m and u n ⫽ v,

NO

If m is even m and u n ⫽ v 1v 7 02,

then 1u n 2 m ⫽ vm m n

Solve using properties of equality

then 1u n 2 m ⫽ ⫾v m

n

m n

n

n

n

u ⫽ vm

u ⫽ ⫾vm

The power property of equality basically says that if certain conditions are satisfied, both sides of an equation can be raised to any needed power.

Check results in original equation

EXAMPLE 16

䊳

Solving Equations with Rational Exponents Solve each equation: 3 a. 31x ⫹ 12 4 ⫺ 9 ⫽ 15

Solution

䊳

3

a. 31x ⫹ 12 4 ⫺ 9 ⫽ 15 1x ⫹ 12 ⫽ 8 3 4

3 1x ⫹ 12 4 ⫽ 8 x ⫹ 1 ⫽ 16 x ⫽ 15 3 4 4 3

Check

䊳

4 3

3 4

3115 ⫹ 12 ⫺ 9 ⫽ 15

3116 2 ⫺ 9 ⫽ 15 3122 3 ⫺ 9 ⫽ 15 3182 ⫺ 9 ⫽ 15 15 ⫽ 15 ✓ 1 4

b.

3

1x ⫺ 32 ⫽ 4 2 3

3 1x ⫺ 32 4 ⫽ ⫾ 4 x⫺3⫽ ⫾8 x⫽3⫾8 2 3 3 2

3 2

b. 1x ⫺ 32 3 ⫽ 4 2

original equation; mn ⫽ 34 isolate variable term (add 9, divide by 3) apply power property, note m is odd simplify 383 ⫽ 183 2 4 ⫽ 16 4 4

1

result substitute 15 for x in the original equation simplify, rewrite exponent 4

116 ⫽ 2 23 ⫽ 8 solution checks original equation; mn ⫽ 23 apply power property, note m is even simplify 342 ⫽ 142 2 3 ⫽ 8 4 3

1

result

The solutions are 3 ⫹ 8 ⫽ 11 and 3 ⫺ 8 ⫽ ⫺5. Verify by checking both in the original equation. Now try Exercises 55 through 58

䊳


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䊳

CAUTION

As you continue solving equations with radicals and rational exponents, be careful not to arbitrarily place the “⫾” sign in front of terms given in radical form. The expression 118 indicates the positive square root of 18, where 118 ⫽ 312. The equation x2 ⫽ 18 becomes x ⫽ ⫾ 118 after applying the power property, with solutions x ⫽ ⫾312 1x ⫽ ⫺312, x ⫽ 3122, since the square of either number produces 18.

In Appendix A.4, we used a technique called u-substitution to factor expressions in quadratic form. The following equations are also in quadratic form since the de2 1 gree of the leading term is twice the degree of the middle term: x3 ⫺ 3x3 ⫺ 10 ⫽ 0 and 1x2 ⫹ x2 2 ⫺ 81x2 ⫹ x2 ⫹ 12 ⫽ 0. The first equation and its solution appear in Example 17.

EXAMPLE 17

䊳

Solving Equations in Quadratic Form 2

1

Solve using a u-substitution: x3 ⫺ 3x3 ⫺ 10 ⫽ 0.

Solution

䊳

This equation is in quadratic form since it can be rewritten as: 1 1 1x3 2 2 ⫺ 31x3 2 1 ⫺ 10 ⫽ 0, where the2 degree of leading term is twice that of second 1 term. If we let u ⫽ x3, then u2 ⫽ x3 and the equation becomes u2 ⫺ 3u1 ⫺ 10 ⫽ 0, which is factorable. 1u ⫺ 521u ⫹ 22 ⫽ 0 or u⫽5 u ⫽ ⫺2 1 1 or x3 ⫽ 5 x3 ⫽ ⫺2 1 1 or 1x3 2 3 ⫽ 1⫺22 3 1x3 2 3 ⫽ 53 or x ⫽ 125 x ⫽ ⫺8

factor solution in terms of u 1

resubstitute x 3 for u cube both sides: 13 132 ⫽ 1 solve for x

Both solutions check. Now try Exercises 59 and 60 Figure A.8 Hypotenuse

Leg 90⬚

Leg

䊳

A right triangle is one that has a 90° angle (see Figure A.8). The longest side (opposite the right angle) is called the hypotenuse, while the other two sides are simply called “legs.” The Pythagorean theorem is a formula that says if you add the square of each leg, the result will be equal to the square of the hypotenuse. Furthermore, we note the converse of this theorem is also true. Pythagorean Theorem 1. For any right triangle with legs a and b and hypotenuse c, a2 ⫹ b2 ⫽ c2 2. For any triangle with sides a, b, and c, if a2 ⫹ b2 ⫽ c2, then the triangle is a right triangle. A geometric interpretation of the theorem is given in Figure A.9, which shows

Figure A.9

32 ⫹ 42 ⫽ 52.

Area 16 in2

4

5

ea Ar in2 25

3

Area 9 in2

25

13 12

24

⫹ ⫽ 25 ⫹ 144 ⫽ 169 52

122

c

7

5

132

b

⫹ ⫽ 49 ⫹ 576 ⫽ 625 72

242

252

⫹ b2 ⫽ c2 general case

a2

a


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EXAMPLE 18

䊳

Applying the Pythagorean Theorem An extension ladder is placed 9 ft from the base of a building in an effort to reach a third-story window that is 27 ft high. What is the minimum length of the ladder required? Answer in exact form using radicals, and in approximate form by rounding to one decimal place.

Solution

䊳

We can assume the building makes a 90° angle with the ground, and use the Pythagorean theorem to find the required length. Let c represent this length. c2 ⫽ a2 ⫹ b2 c2 ⫽ 192 2 ⫹ 1272 2 c2 ⫽ 81 ⫹ 729 c2 ⫽ 810 c ⫽ 1810 c ⫽ 9 110 c ⬇ 28.5 ft

Pythagorean theorem substitute 9 for a and 27 for b 92 ⫽ 81, 272 ⫽ 729

c

add

27 ft

definition of square root; c 7 0 exact form: 1810 ⫽ 181 # 10 ⫽ 9 110 approximate form

The ladder must be at least 28.5 ft tall.

9 ft

F. You’ve just seen how we can solve equations and use formulas involving radicals


A.6 EXERCISES 䊳


Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary. n

1. 1an ⫽ 冟a冟 if n 7 0 is a(n)

integer.

3. By decomposing the rational exponent, we can 3 ? rewrite 16 4 as 116 ? 2 ?. 5. Discuss/Explain what it means when we say an expression like 1A has been written in simplest form. 䊳

2. The conjugate of 2 ⫺ 13 is

.

4. 1x2 2 3 ⫽ x2 3 ⫽ x1 is an example of the property of exponents. 3 2

#

3 2

6. Discuss/Explain why it would be easier x12 to simplify the expression given using 1 rational exponents rather than radicals. x3


Evaluate the expression 2x2 for the values given.

7. a. x ⫽ 9

b. x ⫽ ⫺10

8. a. x ⫽ 7

b. x ⫽ ⫺8

Simplify each expression, assuming that variables can represent any real number.

9. a. 249p2 c. 281m4

b. 21x ⫺ 32 2 d. 2x2 ⫺ 6x ⫹ 9

10. a. 225n2 c. 2v10 3 11. a. 1 64 3 c. 2 216z12 3 12. a. 1 ⫺8 3 c. 2 27q9

b. 21y ⫹ 22 2 d. 24a2 ⫹ 12a ⫹ 9 3 b. 2 ⫺216x3

d.

v3 B ⫺8 3

3 b. 2 ⫺125p3

d.

w3 B ⫺64 3

䊳


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6 13. a. 1 64 5 c. 2243x10 5 e. 2 1k ⫺ 32 5

4 14. a. 1 81 5 c. 21024z15 5 e. 2 1q ⫺ 92 5

3 15. a. 1 ⫺125

6 b. 1 ⫺64 5 d. 2⫺243x5 6 f. 2 1h ⫹ 22 6

4 b. 1 ⫺81 5 d. 2⫺1024z20 6 f. 2 1p ⫹ 42 6

4 b. ⫺ 281n12

10

c. 2⫺36

49v B 36 4 b. ⫺ 216m24 d.

16. a. 1 ⫺216 3

c. 2⫺121

25x6 d. B 4

24. a. 28x6 2 3 c. 227a2b6 9 12 ⫺ 248 e. 8 25. a. 2.5 218a 22a3 c.

26. a. 5.1 22p 232p5 c.

3

17. a. 8

16 2 b. a b 25

2 3

4 ⫺2 c. a b 25 3

d. a

27. a.

⫺27p6 8q3

2 3

b

28. a.

4 b. a b 9

3

16 ⫺4 c. a b 81 3

2

⫺125v9 3 b d. a 27w6 b. a⫺

3

19. a. ⫺1442 c. 1⫺272 ⫺3 2

d. ⫺a b. a⫺

3

20. a. ⫺1002 ⫺23

21. a. 12n p 2

2 ⫺25 5

22. a. a

1

4x2

c.

227y7 23y 20 B 4x4

5 29. a. 2 32x10y15

⫺20 ⫹ 232 4 2 b. ⫺ 23b212b2 3 f.

3 3 d. 29v2u23u5v2

4 b. ⫺ 25q220q3 5 d. 25cd2 125cd 3

b.

4 b 25

c. 31 b 4 e. 2b1 b 4

3 ⫺43

27x b 64

3

4 30. a. 2 81a12b16 4

c. 32a 3 4 e. 1 c1 c

3 2

49 b 36

x9 ⫺3 d. ⫺a b 8

3

3 2 108n4

3 2 4n 81 d. 12 3 9 A 8z

b.

3 2 72b5

23b2 3

125 d. ⫺9 3 A 27x6 4 5 b. x 2 x

3 2

Use properties of exponents to simplify. Answer in exponential form without negative exponents.

3

28m5

d. 254m6n8

d.

3 1 6

26

5 6 b. a 2 a

d.

3 1 3 4 2 3

4

c. 1⫺1252

24x8

ab2 25ab4 B 3 B 27

22m 45 c. B 16x2

3 2

18. a. 92

x3y 4x5y B 3 B 12y

3 b. 3 2128a4b2

2

b

b. a

3

8y4 3

64y2

1 3

b

b. 12x y 2 ⫺14

3 4

4

Simplify each expression. Assume all variables represent nonnegative real numbers.

23. a. 218m2 3 3 c. 264m3n5 8 ⫺6 ⫹ 228 e. 2

3 b. ⫺2 2 ⫺125p3q7

d. 232p3q6 f.

27 ⫺ 272 6

Simplify and add (if possible).

31. a. b. c. d.

12 272 ⫺ 9 298 8 248 ⫺ 3 2108 7 218m ⫺ 250m 2 228p ⫺ 3 263p

32. a. b. c. d.

⫺3280 ⫹ 2 2125 5 212 ⫹ 2 227 3 212x ⫺ 5 275x 3 240q ⫹ 9 210q

3 3 33. a. 3x 1 54x ⫺ 5 216x4 b. 14 ⫹ 13x ⫺ 112x ⫹ 145 c. 272x3 ⫹ 150 ⫺ 17x ⫹ 127 3 3 34. a. 5 254m3 ⫺ 2m216m3 b. 110b ⫹ 1200b ⫺ 120 ⫹ 140

c. 275r3 ⫹ 132 ⫺ 127r ⫹ 138


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Compute each product and simplify the result.

35. a. 17 122 b. 131 15 ⫹ 172 c. 1n ⫹ 1521n ⫺ 152 d. 16 ⫺ 132 2

46. a.

7 17 ⫹ 3

b.

12 1x ⫹ 13

47. a.

110 ⫺ 3 13 ⫹ 12

b.

7 ⫹ 16 3 ⫺ 312

37. a. 13 ⫹ 2 17213 ⫺ 2 172 b. 1 15 ⫺ 1142 1 12 ⫹ 1132 c. 1212 ⫹ 6 162 13110 ⫹ 172

48. a.

1 ⫹ 12 16 ⫹ 114

b.

1 ⫹ 16 5 ⫹ 213

2

36. a. 10.3 152 b. 151 16 ⫺ 122 c. 14 ⫹ 13214 ⫺ 132 d. 12 ⫹ 152 2 2

38. a. 15 ⫹ 4 110211 ⫺ 2 1102 b. 1 13 ⫹ 122 1 110 ⫹ 1112 c. 1315 ⫹ 4 122 1 115 ⫹ 162

Use a substitution to verify the solutions to the quadratic equation given. Verify results using a calculator.

39. x2 ⫺ 4x ⫹ 1 ⫽ 0 a. x ⫽ 2 ⫹ 13

b. x ⫽ 2 ⫺ 13

40. x ⫺ 10x ⫹ 18 ⫽ 0 a. x ⫽ 5 ⫺ 17

b. x ⫽ 5 ⫹ 17

41. x2 ⫹ 2x ⫺ 9 ⫽ 0 a. x ⫽ ⫺1 ⫹ 110

b. x ⫽ ⫺1 ⫺ 110

42. x ⫺ 14x ⫹ 29 ⫽ 0 a. x ⫽ 7 ⫺ 2 15

b. x ⫽ 7 ⫹ 2 15

2

2

Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities.

3 112 27 c. B 50b 5 e. 3 1a

43. a.

20 B 27x3 1 d. 3 A 4p b.

⫺4 125 44. a. b. B 120 12n3 5 3 c. d. 3 B 12x A 2m2 ⫺8 e. 3 3 15 Simplify the following expressions by rationalizing the denominators. Where possible, state results in exact form and approximate form, rounded to hundredths. 45. a.

8 3 ⫹ 111

b.

6 1x ⫺ 12

A-77

Solve each equation and check your solutions by substitution. Identify any extraneous roots.

49. a. ⫺313x ⫺ 5 ⫽ ⫺9 b. x ⫽ 13x ⫹ 1 ⫹ 3 50. a. ⫺2 14x ⫺ 1 ⫽ ⫺10 b. ⫺5 ⫽ 15x ⫺ 1 ⫺ x 3 51. a. 2 ⫽ 13m ⫺ 1 3 b. 2 1 7 ⫺ 3x ⫺ 3 ⫽ ⫺7 3 1 2m ⫹ 3 ⫹2⫽3 c. ⫺5 3 3 d. 1 2x ⫺ 9 ⫽ 1 3x ⫹ 7 3 52. a. ⫺3 ⫽ 1 5p ⫹ 2 3 b. 3 1 3 ⫺ 4x ⫺ 7 ⫽ ⫺4 3 1 6x ⫺ 7 ⫺ 5 ⫽ ⫺6 c. 4 3 3 d. 3 1 x ⫹ 3 ⫽ 2 1 2x ⫹ 17

53. a. b. c. d.

1x ⫺ 9 ⫹ 1x ⫽ 9 x ⫽ 3 ⫹ 223 ⫺ x 1x ⫺ 2 ⫺ 12x ⫽ ⫺2 112x ⫹ 9 ⫺ 124x ⫽ ⫺3

54. a. b. c. d.

1x ⫹ 7 ⫺ 1x ⫽ 1 12x ⫹ 31 ⫹ x ⫽ 2 13x ⫽ 1x ⫺ 3 ⫹ 3 13x ⫹ 4 ⫺ 17x ⫽ ⫺2

Write the equation in simplified form, then solve. Check all answers by substitution. 3

55. a. x5 ⫹ 17 ⫽ 9 3 b. ⫺2x4 ⫹ 47 ⫽ ⫺7 5

56. a. 0.3x2 ⫺ 39 ⫽ 42 5 b. 0.5x3 ⫹ 92 ⫽ ⫺43 2

57. a. 21x ⫹ 52 3 ⫺ 11 ⫽ 7 4 b. ⫺31x ⫺ 22 5 ⫹ 29 ⫽ ⫺19 1

3 58. a. 3x3 ⫺ 10 ⫽ 1x 2 5 b. 2 1x 2 ⫺ 4 ⫽ x5 2

1

59. x3 ⫺ 2x3 ⫺ 15 ⫽ 0 3

60. x3 ⫺ 9x2 ⫽ ⫺8


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WORKING WITH FORMULAS 1

61. Fish length to weight relationship: L 1.131W2 3 The length to weight relationship of a female Pacific halibut can be approximated by the formula shown, where W is the weight in pounds and L is the length in feet. A fisherman lands a halibut that weighs 400 lb. Approximate the length of the fish (round to two decimal places). 䊳

1s 4 The time it takes an object to fall a certain distance is given by the formula shown, where t is the time in seconds and s is the distance the object has fallen. Approximate the time it takes an object to hit the ground, if it is dropped from the top of a building that is 80 ft in height (round to hundredths).

62. Timing a falling object: t

APPLICATIONS

63. Length of a cable: A radio tower is secured by cables that are anchored in the ground 8 m from its base. If the cables are attached to the tower 24 m above the ground, what is the length of each cable? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

24 m

c

8m 64. Height of a kite: Benjamin Franklin is flying his kite in a storm once again. John Adams has walked to a position directly under the kite and is 75 ft from Ben. If the kite is 50 ft above John Adams’ head, how much string S has Ben let out? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

S

50 ft

75 ft

The time T (in days) required for a planet to make one revolution around3 the sun is modeled by the function T 0.407R2, where R is the maximum radius of the planet’s orbit (in millions of miles). This is known as Kepler’s third law of planetary motion. Use the equation given to approximate the number of days required for one complete orbit of each planet, given its maximum orbital radius.

65. a. Earth: 93 million mi b. Mars: 142 million mi c. Mercury: 36 million mi 66. a. Venus: 67 million mi b. Jupiter: 480 million mi c. Saturn: 890 million mi 67. Accident investigation: After an accident, police officers will try to determine the approximate velocity V that a car was traveling using the formula V ⫽ 2 26L, where L is the length of the skid marks in feet and V is the velocity in miles per hour. (a) If the skid marks were 54 ft long, how fast was the car traveling? (b) Approximate the speed of the car if the skid marks were 90 ft long. 68. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V ⫽ Ak that depends on the size and efficiency of the generator. Rationalize the radical expression and use the new version to find the velocity of the wind if k ⫽ 0.004 and the generator is putting out 13.5 units of power.


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69. Surface area: The lateral surface area (surface area excluding the base) h S of a cone is given by the formula 2 2 S ⫽ ␲r 2r ⫹ h , where r is the r radius of the base and h is the height of the cone. Find the lateral surface area of a cone that has a radius of 6 m and a height of 10 m. Answer in simplest form. 70. Surface area: The lateral surface a area S of a frustum (a truncated cone) is given by the formula h S ⫽ ␲1a ⫹ b2 2h2 ⫹ 1b ⫺ a2 2, b where a is the radius of the upper base, b is the radius of the lower base, and h is the height. Find the surface area of a frustum where a ⫽ 6 m, b ⫽ 8 m, and h ⫽ 10 m. Answer in simplest form. 71. Planetary motion: The time T (in days) for a planet to make one revolution (elliptical orbit) 3 around the sun is modeled by T ⫽ 0.407R2, where R is the maximum radius of the planet’s orbit in

millions of miles (Kepler’s third law of planetary motion). Use the equation to approximate the maximum radius of each orbit, given the number of days it takes for one revolution. (See Exercises 65 and 66.) a. Mercury: 88 days b. Venus: 225 days c. Earth: 365 days d. Mars: 687 days e. Jupiter: 4333 days f. Saturn: 10,759 days 72. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V ⫽ Ak that depends on the size and efficiency of the generator. Given k ⫽ 0.004, approximately how many units of power are being delivered if the wind is blowing at 27 miles per hour? (See Exercise 68.)


The expression x2 7 is not factorable using integer values. But the expression can be written in the form x2 1 272 2, enabling us to factor it as a “binomial” and its conjugate: 1x 272 1x 272. Use this idea to factor the following expressions.

73. a. x2 ⫺ 5 b. n2 ⫺ 19 74. a. 4v2 ⫺ 11 b. 9w2 ⫺ 17 75. The following terms form a pattern that continues until the sixth term is found: 23x ⫹ 29x ⫹ 227x ⫹ p (a) Compute the sum of all six terms; (b) develop a system (investigate the pattern further) that will enable you to find the sum of 12 such terms without actually writing out the terms.

76. Find a quick way to simplify the expression without the aid of a calculator.

a a a a a3 b 5 6

3 2

4 5

3 4

2 5

aaaa

䊳

A-79

10 3

1 1 1 1 9 77. If 1x2 ⫹ x⫺2 2 2 ⫽ , find the value of x2 ⫹ x⫺2. 2

78. Rewrite by rationalizing the numerator: 1x ⫹ h ⫺ 1x h Determine the values of x for which each expression represents a real number. 79.

1x ⫺ 1 x2 ⫺ 4

80.

x2 ⫺ 4 1x ⫺ 1


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OVERVIEW OF APPENDIX A Prerequisite Definitions, Properties, Formulas, and Relationships Notation and Relations concept

• Set notation:

notation {members}

僆 • Is an element of ⭋ or { } • Empty set Is a proper subset of ( • 5x | x p6

• Defining a set

description braces enclose the members of a set

indicates membership in a set a set having no elements indicates the elements of one set are entirely contained in another the set of all x, such that x . . .

Sets of Numbers • Natural: ⺞ ⫽ 51, 2, 3, 4, p6

• Integers: ⺪ ⫽ 5p , ⫺3, ⫺2, ⫺1, 0, 1, 2, 3, p6 • Irrational: ⺘ ⫽ {numbers with a nonterminating,

example set of even whole numbers A ⫽ 50, 2, 4, 6, 8, p6 14 僆 A odd numbers in A S ⫽ 50, 6, 12, 18, 24, p6 S ( A S ⫽ 5x |x ⫽ 6n for n 僆⺧6

• Whole: ⺧ ⫽ 50, 1, 2, 3, p6

• Rational: ⺡ ⫽ e q , where p, q 僆⺪; q ⫽ 0 f • Real: ⺢ ⫽ {all rational and irrational numbers} p

nonrepeating decimal form}

Absolute Value of a Number

Distance between numbers a and b on the number line

• The distance between a number n

冟a ⫺ b冟 or 冟b ⫺ a冟 and zero (always positive) n if n ⱖ 0 0n 0 ⫽ e ⫺n if n 6 0 For a complete review of these ideas, go to www.mhhe.com/coburn.

A.1 Properties of Real Numbers: For real numbers a, b, and c, Commutative Property

• Addition: a ⫹ b ⫽ b ⫹ a • Multiplication: a # b ⫽ b # a Identities

Associative Property

• Addition: 1a ⫹ b2 ⫹ c ⫽ a ⫹ 1b ⫹ c2 • Multiplication: 1a # b2 # c ⫽ a # 1b # c2 Inverses

• Additive: a ⫹ 1⫺a2 ⫽ 0

• Additive: 0 ⫹ a ⫽ a • Multiplicative: 1 # a ⫽ a

p q • Multiplicative: q # p ⫽ 1; p, q ⫽ 0

A.2 Properties of Exponents: For real numbers a and b, and integers m, n, and p • • • •

(excluding 0 raised to a nonpositive power), Product property: b • Power property: 1bm 2 n ⫽ bmn am p amp Product to a power: 1ambn 2 p ⫽ amp # bnp Quotient to a power: b ⫽ np 1b ⫽ 02 a • bm bn b m⫺n 1b ⫽ 02 Quotient property: n ⫽ b 0 Zero exponents: ⫽ 1 1b ⫽ 02 b b • ⫺n n 1 a b • Scientific notation: N ⫻ 10k; 1 ⱕ 冟N冟 6 10, k 僆⺪ Negative exponents: b⫺n ⫽ n ; a b ⫽ a b a b b 1a, b ⫽ 02 m

# bn ⫽ bm⫹n


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A.2 Polynomials • A polynomial is a sum or difference of monomial terms • Polynomials are classified as a monomial, binomial, trinomial, or polynomial, depending on the number of terms • The degree of a polynomial in one variable is the same as the largest exponent occuring on the variable in any term • A polynomial expression is in standard form when written with the terms in descending order of degree A.3 Solving Linear Equations and Inequalities • Properties of Equality: additive property If A and B are algebraic expressions, where A ⫽ B, then A ⫹ C ⫽ B ⫹ C.

• • • • • • • •

multiplicative property If A and B are algebraic expressions, where A ⫽ B, then A # C ⫽ B # C B A [C can be positive or negative] and ⫽ ;C ⫽ 0 C C A linear equation in one variable is one that can be written in the form ax ⫹ b ⫽ c, where the exponent on the variable is a 1. To solve a linear equation, we attempt to isolate the term containing the variable using the additive property, then solve for the variable using the multiplicative property. An equation can be an identity (always true), a contradiction (never true) or conditional (true or false depending on the input value[s]). If an equation contains fractions, multiplying both sides by the least common denominator of all fractions will “clear the denominators” and reduce the amount of work required to solve. Inequalities are solved using properties similar to those used for solving equations. The one exception is when multiplying or dividing by a negative quantity, as the inequality symbol must then be reversed to maintain the truth of the resulting statement. Solutions to an inequality can be given using a simple inequality, graphed on a number line, stated in set notation, or stated using interval notation. Given two sets A and B: A intersect B 1A ¨ B2 is the set of elements shared by both A and B (elements common to both sets). A union B 1A ´ B2 is the set of elements in either A or B (elements are combined to form a larger set). Compound inequalities are formed using the conjunction “and” or the conjunction “or.” The result can be either a joint inequality as in ⫺3 6 x ⱕ 5, or a disjoint inequality, x 6 ⫺2 or x 7 7.

A.4 Special Factorizations • A2 ⫺ B2 ⫽ 1A ⫹ B21A ⫺ B2 • A2 ⫾ 2AB ⫹ B2 ⫽ 1A ⫾ B2 2 3 3 2 2 • A ⫺ B ⫽ 1A ⫺ B21A ⫹ AB ⫹ B 2 • A3 ⫹ B3 ⫽ 1A ⫹ B21A2 ⫺ AB ⫹ B2 2 • Certain equations of higher degree can be solved using factoring skills and the zero product property. A.5 Rational Expressions: For polynomials P, Q, R, and S with no denominator of zero, P#R P P#R P Equivalence: ⫽ ⫽ • Q#R Q Q Q#R P R P#R PR R P S PS P Multiplication: # ⫽ ⫽ • Division: ⫼ ⫽ # ⫽ # Q S Q S QS Q S Q R QR Q P⫹Q Q P⫺Q P P Addition: ⫹ ⫽ • Subtraction: ⫺ ⫽ R R R R R R Addition/subtraction with unlike denominators: 1. Find the LCD of all rational expressions. 2. Build equivalent expressions using LCD. 3. Add/subtract numerators as indicated. 4. Write the result in lowest terms. To solve rational equations, first clear denominators using the LCD, noting values that must be excluded. Multiplying an equation by a variable quantity sometimes introduces extraneous solutions. Check all results in the original equation.

• Lowest terms: • • •

• •


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A.6 Properties of Radicals • 1a is a real number only for a ⱖ 0 n • 1a ⫽ b, only if bn ⫽ a

• 1a ⫽ b, only if b2 ⫽ a n • If n is even, 1a represents a real number only

if a ⱖ 0 n • For any real number a, 1an ⫽ a when n is odd m If a is a real number and n is an integer greater • If n is a rational number written in lowest terms 1 n n m m than 1, then 1 a ⫽ an provided 1 a represents a n n with n ⱖ 2, then a n ⫽ 11a2 m and a n ⫽ 1am real number n provided 2a represents a real number. n n n n If 1A and 1B represent real numbers and B ⫽ 0, If 1A and 1B represent real numbers, • n n n n 1AB ⫽ 1 A # 1B 1A n A ⫽ n BB 1B A radical expression is in simplest form when: 1. the radicand has no factors that are perfect nth roots, 2. the radicand contains no fractions, and 3. no radicals occur in a denominator. To solve radical equations, isolate the radical on one side, then apply the appropriate “nth power” to free up the radicand. Repeat the process if needed. See flowchart on page 74. For equations with a rational exponent mn, isolate the variable term and raise both sides to the mn power. If m is even, there will be two real solutions.

n • For any real number a, 1an ⫽ 冟a冟 when n is even

•

•

•

• •

A.6 Pythagorean Theorem • For any right triangle with legs a and b and hypotenuse c: a2 ⫹ b2 ⫽ c2.

• For any triangle with sides a, b, and c, if

a2 ⫹ b2 ⫽ c2, then the triangle is a right triangle.

PRACTICE TEST 1. State true or false. If false, state why. 7 a. 13 ⫹ 42 2 ⫽ 25 b. ⫽ 0 0 c. x ⫺ 3 ⫽ ⫺3 ⫹ x d. ⫺21x ⫺ 32 ⫽ ⫺2x ⫺ 3 2. State the value of each expression. 3 a. 1 121 b. 1 ⫺125 c. 1⫺36 d. 1400 3. Evaluate each expression: 7 1 5 1 a. ⫺ a⫺ b b. ⫺ ⫺ 8 4 3 6 c. ⫺0.7 ⫹ 1.2 d. 1.3 ⫹ 1⫺5.92 4. Evaluate each expression: 1 a. 1⫺42a⫺2 b b. 1⫺0.621⫺1.52 3 ⫺2.8 c. d. 4.2 ⫼ 1⫺0.62 ⫺0.7 # 0.08 12 10 5. Evaluate using a calculator: 2000 a1 ⫹ b 12 6. State the value of each expression, if possible. a. 0 ⫼ 6 b. 6 ⫼ 0

7. State the number of terms in each expression and identify the coefficient of each. c⫹2 ⫹c a. ⫺2v2 ⫹ 6v ⫹ 5 b. 3 8. Evaluate each expression given x ⫽ ⫺0.5 and y ⫽ ⫺2. Round to hundredths as needed. y a. 2x ⫺ 3y2 b. 12 ⫺ x14 ⫺ x2 2 ⫹ x 9. Translate each phrase into an algebraic expression. a. Nine less than twice a number is subtracted from the number cubed. b. Three times the square of half a number is subtracted from twice the number. 10. Create a mathematical model using descriptive variables. a. The radius of the planet Jupiter is approximately 119 mi less than 11 times the radius of the Earth. Express the radius of Jupiter in terms of the Earth’s radius. b. Last year, Video Venue Inc. earned $1.2 million more than four times what it earned this year. Express last year’s earnings of Video Venue Inc. in terms of this year’s earnings.


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Practice Test

11. Simplify by combining like terms. a. 8v2 ⫹ 4v ⫺ 7 ⫹ v2 ⫺ v b. ⫺413b ⫺ 22 ⫹ 5b c. 4x ⫺ 1x ⫺ 2x2 2 ⫹ x13 ⫺ x2 12. Factor each expression completely. a. 9x2 ⫺ 16 b. 4v3 ⫺ 12v2 ⫹ 9v 3 2 c. x ⫹ 5x ⫺ 9x ⫺ 45 13. Simplify using the properties of exponents. 5 a. ⫺3 b. 1⫺2a3 2 2 1a2b4 2 3 b 5p2q3r4 2 m2 3 c. a b d. a b 2n ⫺2pq2r4 14. Simplify using the properties of exponents. ⫺12a3b5 a. 3a2b4 b. 13.2 ⫻ 10⫺17 2 ⫻ 12.0 ⫻ 1015 2 a⫺3 # b ⫺4 c. a ⫺2 b d. ⫺7x0 ⫹ 1⫺7x2 0 c 15. Compute each product. a. 13x2 ⫹ 5y2 13x2 ⫺ 5y2 b. 12a ⫹ 3b2 2 16. Add or subtract as indicated. a. 1⫺5a3 ⫹ 4a2 ⫺ 32 ⫹ 17a4 ⫹ 4a2 ⫺ 3a ⫺ 152 b. 12x2 ⫹ 4x ⫺ 92 ⫺ 17x4 ⫺ 2x2 ⫺ x ⫺ 92 Simplify or compute as indicated. x⫺5 4 ⫺ n2 b. 2 5⫺x n ⫺ 4n ⫹ 4 3 3x2 ⫺ 13x ⫺ 10 x ⫺ 27 c. 2 d. x ⫹ 3x ⫹ 9 9x2 ⫺ 4 x2 ⫹ x ⫺ 20 x2 ⫺ 25 e. 2 ⫼ 2 3x ⫺ 11x ⫺ 4 x ⫺ 8x ⫹ 16 2 m⫹3 f. 2 ⫺ 51m ⫹ 42 m ⫹ m ⫺ 12

17. a.

18. a. 21x ⫹ 112 2 ⫺3

25 2 c. a b 16 e. 7 140 ⫺ 190 2 g. B 5x

3 ⫺8 A 27v3 ⫺4 ⫹ 132 d. 8 f. 1x ⫹ 152 1x ⫺ 152 8 h. 16 ⫺ 12

b.

19. Maximizing revenue: Due to past experience, the manager of a video store knows that if a popular video game is priced at $30, the store will sell 40 each day. For each decrease of $0.50, one additional sale will be made. The formula for the store’s revenue is then R ⫽ 130 ⫺ 0.5x2140 ⫹ x2, where x represents the number of times the price is decreased. Multiply the binomials and use a table of values to determine (a) the number of 50¢ decreases that will give the most revenue and (b) the maximum amount of revenue. 20. Diagonal of a rectangular prism: Use the Pythagorean theorem to determine the length of the diagonal of the rectangular prism shown in the figure. (Hint: First find the diagonal of the base.)

42 cm

32 cm

24 cm

21. Solve each linear equation. a. ⫺2b ⫹ 7 ⫽ ⫺5 b. 312n ⫺ 62 ⫹ 1 ⫽ 7 1 2 3 c. 4m ⫺ 5 ⫽ 11m ⫹ 2 d. x ⫹ ⫽ 2 3 4 e. ⫺813p ⫹ 52 ⫺ 9 ⫽ 613 ⫺ 4p2 g 5g 1 f. ⫺ ⫽ 3 ⫺ ⫺ 6 2 12 22. If one-fourth of the sum of a number and twelve is added to three, the result is sixteen. Find an equation model for this statement, then use it to find the number. 23. Solve each polynomial equation by factoring. a. x3 ⫺ 7x2 ⫽ 4x ⫺ 28 b. ⫺7r3 ⫹ 21r2 ⫹ 28r ⫽ 0 c. g4 ⫺ 10g2 ⫹ 9 ⫽ 0 24. Solve each rational equation. 7 1 3 ⫹ ⫽ a. 5x 10 4x 1 7 3h ⫽ ⫺ 2 b. h⫹3 h h ⫹ 3h 3 4n ⫹ 20 n ⫺ ⫽ 2 c. n⫹2 n⫺4 n ⫺ 2n ⫺ 8 25. Solve each radical equation. 2x2 ⫹ 7 ⫹3⫽5 a. 2 b. 3 1x ⫹ 4 ⫽ x ⫹ 4 c. 13x ⫹ 4 ⫽ 2 ⫺ 1x ⫹ 2


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Appendix B Proof Positive — A Selection of Proofs from Precalculus The Remainder Theorem If a polynomial p(x) is divided by (x c) using synthetic division, the remainder is equal to p(c). Proof of the Remainder Theorem From our previous work, any number c used in synthetic division will occur as the factor (x c) when written as 1quotient21divisor2 remainder: p1x2 1x c2q1x2 r. Here, q(x) represents the quotient polynomial and r is a constant. Evaluating p(c) gives p1x2 1x c2q1x2 r

p1c2 1c c2q1c2 r 0 # q1c2 r r✓

The Factor Theorem Given a polynomial p(x), (1) if p(c) 0, then x c is a factor of p(x), and (2) if x c is a factor of p(x), then p(c) 0. Proof of the Factor Theorem

1. Consider a polynomial p written in the form p1x2 1x c2q1x2 r. From the remainder theorem we know p1c2 r, and substituting p(c) for r in the equation shown gives: p1x2 1x c2q1x2 p1c2

and x c is a factor of p(x), if p1c2 0

p1x2 1x c2q1x2 ✓

2. The steps from part 1 can be reversed, since any factor (x c) of p(x), can be written in the form p1x2 1x c2q1x2 . Evaluating at x c produces a result of zero: p1c2 1c c2q1x2 0✓

Complex Conjugates Theorem Given p(x) is a polynomial with real number coefficients, complex solutions must occur in conjugate pairs. If a bi, b 0 is a solution, then a bi must also be a solution. To prove this for polynomials of degree n 7 2, we let z1 a bi and z2 c di be complex numbers, and let z1 a bi, and z2 c di represent their conjugates, and observe the following properties:

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APPENDIX B Proof Positive — A Selection of Proofs from Precalculus

A-85

1. The conjugate of a sum is equal to the sum of the conjugates. sum of conjugates: z1 z2 sum: z1 z2 1a bi2 1c di2

1a c2 1b d2i

1a bi2 1c di2

1a c2 1b d2i ✓

→ conjugate of sum →

2. The conjugate of a product is equal to the product of the conjugates. product: z1 # z2

product of conjugates: z1 # z2

ac adi bci bdi2

ac adi cbi bdi2

1a bi2 # 1c di2

1ac bd2 1ad bc2i

1a bi2 # 1c di2

1ac bd2 1ad bc2i ✓

→ conjugate of product →

Since polynomials involve only sums and products, and the complex conjugate of any real number is the number itself, we have the following: Proof of the Complex Conjugates Theorem Given polynomial p1x2 anxn an1xn1 p a1x1 a0, where an, an1, p , a1, a0 are real numbers and z a bi is a zero of p, we must show that z a bi is also a zero. anzn an1zn1 p a1z1 a0 p1z2 evaluate p (x ) at z p 1z2 0 given

anzn an1zn1 p a1z1 a0 0 anzn an1zn1 p a1z1 a0 0 anzn an1zn1 p a1z1 a0 0 1 n n1 an 1z 2 an1 1z 2 p a1 1z 2 a0 0

an 1z 2 an1 1z n

n1

conjugate both sides property 1 property 2

1 2 p a1 1z 2 a0 0

conjugate of a real number is the number

p1z2 0

✓

result

An immediate and useful result of this theorem is that any polynomial of odd degree must have at least one real root.

Linear Factorization Theorem If p(x) is a complex polynomial of degree n 1, then p has exactly n linear factors and can be written in the form p(x) an(x c1)(x c2) # . . . # (x cn), where an 0 and c1, c2, . . . , cn are complex numbers. Some factors may have multiplicities greater than 1 (c1, c2, . . . , cn are not necessarily distinct). Proof of the Linear Factorization Theorem Given p(x) anxn an1xn1 . . . a1x a0 is a complex polynomial, the Fundamental Theorem of Algebra establishes that p(x) has a least one complex zero, call it c1. The factor theorem stipulates (x c1) must be a factor of P, giving p1x2 1x c1 2q1 1x2

where q1(x) is a complex polynomial of degree n 1. Since q1(x) is a complex polynomial in its own right, it too must also have a complex zero, call it c2. Then (x c2) must be a factor of q1(x), giving p1x2 1x c1 21x c2 2q2 1x2

where q2(x) is a complex polynomial of degree n 2. Repeating this rationale n times will cause p(x) to be rewritten in the form p1x2 1x c1 21x c2 2

#

...

#

1x cn 2qn 1x2

where qn(x) has a degree of n n 0, a nonzero constant typically called an. The result is p1x2 an 1x c1 21x c2 2 # . . . # 1x cn 2 , and the proof is complete.


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The Product Property of Logarithms Given M, N, and b 1 are positive real numbers, logb(MN) logbM logbN. Proof of the Product Property For P logb M and Q logb N, we have bP M and bQ N in exponential form. It follows that logb 1MN2 logb 1bPbQ 2 logb 1b

PQ

PQ

2

logb M logb N

substitute b P for M and b Q for N properties of exponents log property 3 substitute logb M for P and logb N for Q

The Quotient Property of Logarithms Given M, N, and b 1 are positive real numbers, M logb a b logb M logb N. N Proof of the Quotient Property For P logb M and Q logb N, we have bP M and bQ N in exponential form. It follows that logb a

M bP b logb a Q b N b

logb 1bPQ 2

substitute b P for M and b Q for N properties of exponents

PQ

log property 3

logb M logb N

substitute logb M for P and logb N for Q

The Power Property of Logarithms Given M, N, and b 1 are positive real numbers and any real number x, logb Mx x logb M. Proof of the Power Property For P logb M, we have bP M in exponential form. It follows that logb 1M2 x logb 1bP 2 x logb 1bPx 2 Px

1logb M2x x logb M

substitute b P for M properties of exponents log property 3 substitute logb M for P rewrite factors

Proof of the Determinant Formula for the Area of a Parallelogram 1 Since the area of a triangle is A ab sin ␪, the area of the corresponding parallelogram 2 is twice as large: A ab sin ␪. In terms of the vectors u Ha, bI and v Hc, dI, we have A 冟u冟冟v冟 sin ␪, and it follows that


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Area ⫽ 冟u冟冟v冟 21 ⫺ cos2␪, ⫽ 冟u冟冟v冟

u ␪

v

⫽ 冟u冟冟v冟

B B

1⫺

1u # v2


2

substitute for cos ␪

冟u冟2冟v冟2

冟u冟2冟v冟2 ⫺ 1u # v2 2

common denominator

冟u冟2冟v冟2

⫽ 2冟u冟2冟v冟2 ⫺ 1u # v2 2

simplify

⫽ 21a2 ⫹ b2 2 1c2 ⫹ d2 2 ⫺ 1ac ⫹ bd2 2

substitute

⫽ 2a c ⫹ a d ⫹ b c ⫹ b d ⫺ 1a c ⫹ 2acbd ⫹ b d 2 2 2

2 2

A-87

2 2

2 2

2 2

2 2

expand

⫽ 2a2d2 ⫺ 2acbd ⫹ b2c2

simplify

⫽ 21ad ⫺ bc2

factor

2

⫽ 冟ad ⫺ bc冟

result

Proof of Heron’s Formula Using Algebra Note that 2a2 ⫺ d2 ⫽ h ⫽ 2c2 ⫺ e2. It follows that a

2a2 ⫺ d2 ⫽ 2c2 ⫺ e2

c

h

d

a2 ⫺ d2 ⫽ c2 ⫺ e2

e

a2 ⫺ 1b ⫺ e2 2 ⫽ c2 ⫺ e2

b

a2 ⫺ b2 ⫹ 2be ⫺ e2 ⫽ c2 ⫺ e2 a2 ⫺ b2 ⫺ c2 ⫽ ⫺2be b2 ⫹ c2 ⫺ a2 ⫽e 2b This shows:

1 A ⫽ bh 2 1 ⫽ b 2c2 ⫺ e2 2 1 b2 ⫹ c2 ⫺ a2 2 b ⫽ b c2 ⫺ a 2 B 2b 1 b4 ⫹ 2b2c2 ⫺ 2a2b2 ⫺ 2a2c2 ⫹ a4 ⫹ c4 ⫽ b c2 ⫺ a b 2 B 4b2 1 4b2c2 b4 ⫹ 2b2c2 ⫺ 2a2b2 ⫺ 2a2c2 ⫹ a4 ⫹ c4 ⫽ b ⫺ a b 2 B 4b2 4b2 1 4b2c2 ⫺ b4 ⫺ 2b2c2 ⫹ 2a2b2 ⫹ 2a2c2 ⫺ a4 ⫺ c4 ⫽ b 2 B 4b2 1 ⫽ 22a2b2 ⫹ 2a2c2 ⫹ 2b2c2 ⫺ a4 ⫺ b4 ⫺ c4 4 1 ⫽ 2 3 1a ⫹ b2 2 ⫺ c2 4 3 c2 ⫺ 1a ⫺ b2 2 4 4 1 ⫽ 21a ⫹ b ⫹ c21a ⫹ b ⫺ c2 1c ⫹ a ⫺ b21c ⫺ a ⫹ b2 4

From this point, the conclusion of the proof is the same as the trigonometric development found on page 765.


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Proof that u ⴢ v ⴝ compv u ⴛ |v| Consider the vectors in the figure shown, which form a triangle. Applying the Law of Cosines to this triangle yields:

z u–v

v ␪ x

0 u v 0 2 0 u 0 2 0 v 0 2 2 0 u 0 0 v 0 cos ␪

Using properties of the dot product (page 680), we can rewrite the left-hand side as follows:

u y

0u v 0 2 1u v2 # 1u v2

u#uu#vv#uv#v 0u 0 2 2 u # v 0v 0 2

Substituting the last expression for 0 u v 0 2 from the Law of Cosines gives 0 u 0 2 2 u # v 0v 0 2 0 u 0 2 0 v 0 2 2 0 u 0 0v 0 cos ␪ 2 u # v 2 0 u 0 0 v 0 cos ␪ u # v 0u 0 0 v 0 cos ␪

0 u 0 cos ␪ 0v 0 .

Substituting compv u for 0u 0 cos ␪ completes the proof: u # v compv u 0 v 0

Proof of DeMoivre’s Theorem: (cos x ⴙ i sin x)n ⴝ cos(nx) ⴙ i sin(nx) For n 7 0, we proceed using mathematical induction. 1. Show the statement is true for n 1 (base case):

1cos x i sin x2 1 cos11x2 i sin11x2 cos x i sin x cos x i sin x ✓

2. Assume the statement is true for n k (induction hypothesis): 1cos x i sin x2 k cos1kx2 i sin1kx2

3. Show the statement is true for n k 1:

1cos x i sin x2 k1 1cos x i sin x2 k 1cos x i sin x2 1 3cos1kx2 i sin1kx2 4 1cos x i sin x2 induction hypothesis cos1kx2cos x sin1kx2sin x i 3cos1kx2sin x sin1kx2cos x 4 F-O-I-L cos 3 1k 12x 4 i sin 3 1k 12x 4 ✓ sum/difference identities By the principle of mathematical induction, the statement is true for all positive integers. For n 6 0 (the theorem is obviously true for n 0), consider a positive integer m, where n m. 1cos x i sin x2 n 1cos x i sin x2 m 1 1cos x i sin x2 m 1 cos1mx2 i sin1mx2 cos1mx2 i sin1mx2 cos1mx2 i sin1mx2 cos1nx2 i sin1nx2

negative exponent property

DeMoivre’s theorem for n 7 0 multiply numerator and denom by cos1mx2 i sin1mx2 and simplify even/odd identities n m


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Appendix C More on Synthetic Division As the name implies, synthetic division simulates the long division process, but in a condensed and more efficient form. It’s based on a few simple observations of long division, as noted in the division (x3 2x2 13x 172 1x 52 shown in Figure AI.1. Figure AI.1 x2 3x 2 x 5冄 x3 2x2 13x 17

1x 5x 2 3

Figure AI.2

x 5冄 1

2

3 13

2 17

5

↓

3x 13x 2

1 2

13x 15x2 2

3 15

↓

2x 17

2

12x 102 7

10 7

remainder

remainder

A careful observation reveals a great deal of repetition, as any term in red is a duplicate of the term above it. In addition, since the dividend and divisor must be written in decreasing order of degree, the variable part of each term is unnecessary as we can let the position of each coefficient indicate the degree of the term. In other words, we’ll agree that 1

2

13

17 represents the polynomial 1x3 2x2 13x 17.

Finally, we know in advance that we’ll be subtracting each partial product, so we can “distribute the negative,” shown at each stage. Removing the repeated terms and variable factors, then distributing the negative to the remaining terms produces Figure AI.2. The entire process can now be condensed by vertically compressing the rows of the division so that a minimum of space is used (Figure AI.3).

1

x 5冄 1

Figure AI.3 3

Figure AI.4 2

quotient

1 2

3 13

2 17

dividend

10

products

2

13

17

dividend

5

15

10

products

↓

5

15

3

2

7

sums

1

3

2

x 5冄 1

7

remainder

quotient

Further, if we include the lead coefficient in the bottom row (Figure AI.4), the coefficients in the top row (in blue) are duplicated and no longer necessary, since the quotient and remainder now appear in the last row. Finally, note all entries in the product row (in red) are five times the sum from the prior column. There is a direct connection between this multiplication by 5 and the divisor x 5, and in fact, it is the zero of the divisor that is used in synthetic division (x 5 from x 5 0). A simple change in format makes this method of division easier to use, and highlights the location of the divisor and remainder (the blue brackets in Figure AI.5). Note the process begins by “dropping the lead coefficient into place” (shown in bold). The full process of synthetic division is shown in Figure AI.6 for the same exercise. A-89


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APPENDIX C More on Synthetic Division

Figure AI. 5 2 13

divisor (use 5, not 5) →

5 1 ↓ drop lead coefficient into place → 1

17

← coefficients of dividend ← quotient and remainder appear in this row

We then multiply this coefficient by the “divisor,” place the result in the next column and add. In a sense, we “multiply in the diagonal direction,” and “add in the vertical direction.” Continue the process until the division is complete.

5 1 multiply by 5

↓ 1

The result is x2 3x 2

2 5 3

↓

A-90

Figure AI. 6 13 17 15 10 2 7

← coefficients of dividend ← quotient and remainder appear in this row

7 , read from the last row. x5


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Appendix D Reduced Row-Echelon Form and More on Matrices Reduced Row-Echelon Form A matrix is in reduced row-echelon form if it satisfies the following conditions: 1. All null rows (zeroes for all entries) occur at the bottom of the matrix. 2. The first non-zero entry of any row must be a 1. 3. For any two consecutive, nonzero rows, the leading 1 in the higher row is to the left of the 1 in the lower row. 4. Every column with a leading 1 has zeroes for all other entries in the column. Matrices A through D are in reduced row-echelon form. 0 A £0 0

1 0 0

0 0 0

0 1 0

0 0 1

1 B £0 0

5 3§ 2

0 1 0

0 0 0

5 3§ 0

1 C £0 0

0 1 0

0 3 0

5 2 § 0

1 D £0 0

0 1 0

5 2 0

0 0§ 1

Where Gaussian elimination places a matrix in row-echelon form (satisfying the first three conditions), Gauss-Jordan elimination places a matrix in reduced row-echelon form. To obtain this form, continue applying row operations to the matrix until the fourth condition above is also satisfied. For a 3 3 system having a unique solution, the diagonal entries of the coefficient matrix will be 1’s, with 0’s for all other entries. To illustrate, we’ll extend Example 3 from Section 9.2 until reduced row-echelon form is obtained. EXAMPLE 4B

䊳

Solving Systems Using the Augmented Matrix

2x y 2z 7 Solve using Gauss-Jordan elimination: • x y z 1 2y z 3 2x y 2z 7 • x y z 1 2y z 3

R 1 4 R2

x y z 1 • 2x y 2z 7 2y z 3

1 1 1 1 £2 1 2 7 § 0 2 1 3

2R1 R2 S R2

1 1 1 1 £ 0 1 4 5 § 0 2 1 3

1 1 1 1 £0 1 4 5§ 0 2 1 3

2R2 R3 S R3

1 £0 0

1 1 0

R2 R1 S R1

1 £0 0

0 3 6 1 4 5§ 0 1 1

1 £0 0

1 1 0

1 1 4 5§ 1 1

1 1 4 5§ 7 7

matrix form →

1 1 1 1 £2 1 2 7 § 0 2 1 3

1R2

1 1 1 1 £0 1 4 5§ 0 2 1 3

R3 S R3 7

3R3 R1 S R1 4R3 R2 S R2

1 £0 0

1 1 0

1 1 4 5§ 1 1

1 £0 0

0 1 0

0 3 0 1§ 1 1

The final matrix is in reduced row-echelon form with solution (3, 1, 1) just as in Section 6.1. A-91


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APPENDIX D Reduced Row-Echelon Form and More on Matric

The Determinant of a General Matrix To compute the determinant of a general square matrix, we introduce the idea of a cofactor. For an n n matrix A, Aij 112 ij 冟 Mij 冟 is the cofactor of matrix element aij, where 冟Mij冟 represents the determinant of the corresponding minor matrix. Note that i j is the sum of the row and column of the entry, and if this sum is even, 112 ij 1, while if the sum is odd, 112 ij 1 (this is how the sign table for a 3 3 determinant was generated). To compute the determinant of an n n matrix, multiply each element in any row or column by its cofactor and add. The result is a tierlike process in which the determinant of a larger matrix requires computing the determinant of smaller matrices. In the case of a 4 4 matrix, each of the minor matrices will be size 3 3, whose determinant then requires the computation of other 2 2 determinants. In the following illustration, two of the entries in the first row are zero for convenience. For 2 1 A ≥ 3 0 2 we have: det1A2 2 # 112 11 † 1 3

0 2 1 3 0 4 2

3 0 4 2

0 2 ¥, 1 1

2 1 1 † 132 # 112 13 † 3 1 0

2 1 3

2 1† 1

Computing the first 3 3 determinant gives 16, the second 3 3 determinant is 14. This gives: det1A2 21162 31142 74


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Appendix E The Equation of a Conic The Equation of an Ellipse In Section 10.2, the equation 21x c2 2 y2 21x c2 2 y2 2a was developed using the distance formula and the definition of an ellipse. To find the standard form of the equation, we treat this result as a radical equation, isolating one of the radicals and squaring both sides. 21x c2 2 y2 2a 21x c2 2 y2

1x c2 y 4a 4a 21x c2 y 1x c2 y 2

2

2

2

2

2

2

isolate one radical square both sides

We continue by simplifying the equation, isolating the remaining radical, and squaring again. x2 2cx c2 y2 4a2 4a 21x c2 2 y2 x2 2cx c2 y2 4cx 4a2 4a 21x c2 2 y2 a 21x c2 y a cx a2 3 1x c2 2 y2 4 a4 2a2cx c2x2 2 2 2 a x 2a cx a2c2 a2y2 a4 2a2cx c2x2 a2x2 c2x2 a2y2 a4 a2c2 x2 1a2 c2 2 a2y2 a2 1a2 c2 2 y2 x2 1 a2 a2 c2 2

2

2

simplify isolate radical; divide by 4 square both sides expand and distribute a 2 on left add 2a 2cx and rewrite equation factor divide by a 2 1a 2 c 2 2

Since a 7 c, we know a2 7 c2 and a2 c2 7 0. For convenience, we let b2 a2 c2 and it also follows that a2 7 b2 and a 7 b (since c 7 0). Substituting b2 for a2 c2 we obtain the standard form of the equation of an ellipse (major axis y2 x2 horizontal, since we stipulated a 7 b): 2 2 1. Note once again the x-intercepts a b are (a, 0), while the y-intercepts are (0, b).

The Equation of a Hyperbola In Section 10.3, the equation 21x c2 2 y2 21x c2 2 y2 2a was developed using the distance formula and the definition of a hyperbola. To find the standard form of this equation, we apply the same procedures as before. isolate one radical 21x c2 2 y2 2a 21x c2 2 y2 2 2 2 2 2 1x c2 y 4a 4a 21x c2 y 1x c2 2 y2 square both sides 2 expand x 2cx c2 y2 4a2 4a 21x c2 2 y2 x2 2cx c2 y2 binomials simplify 4cx 4a2 4a 21x c2 2 y2 cx a2 a 21x c2 2 y2 isolate radical; divide by 4 2 2 2 4 2 2 2 square both sides c x 2a cx a a 3 1x c2 y 4 2 2 2 4 2 2 2 2 2 2 2 c x 2a cx a a x 2a cx a c a y expand and distribute a 2 on the right add 2a 2cx and rewrite equation c2x2 a2x2 a2y2 a2c2 a4 2 2 2 2 2 2 2 2 factor x 1c a 2 a y a 1c a 2 y2 x2 divide by a 2 1c 2 a 2 2 2 1 A-93 a2 c a2


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Page A-94

Precalculus—

A-94

APPENDIX E The Equation of a Conic

From the definition of a hyperbola we have 0 6 a 6 c, showing c2 7 a2 and c2 a2 7 0. For convenience, we let b2 c2 a2 and substitute to obtain the stany2 x2 dard form of the equation of a hyperbola (transverse axis horizontal): 2 2 1. a b Note the x-intercepts are (0, a) and there are no y-intercepts.

The Asymptotes of a Central Hyperbola From our work in Section 10.3, a central hyperbola with a horizontal axis will have b asymptotes at y x. To understand why, recall that for asymptotic behavior we a investigate what happens to the relation for large values of x, meaning as 冟x冟 S q . y2 x2 Starting with 2 2 1, we have a b b2x2 a2y2 a2b2

clear denominators

ay bx ab 2 2

2 2

2 2

a2y2 b2x2 a1

a b x2 b2 a2 y2 2 x2 a1 2 b a x b a2 y x 1 2 a B x

As 冟x冟 S q,

isolate term with y

2

factor out b 2x 2 from right side

divide by a 2

square root both sides

b a2 S 0, and we find that for large values of x, y ⬇ x. a x2


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Page A-95

Precalculus—

Appendix F Families of Polar Curves Circles and Spiral Curves r

a

Circle ra 2

a

Spiral r k␪

Circle Circle r a cos ␪ r a sin ␪ a represents the diameter of the circle

Roses: r ⴝ a sin(n␪) (illustrated here) and r ⴝ a cos(n␪)

Four-petal rose r a sin(2␪)

a

a

a

a

Five-petal rose Eight-petal rose Three-petal rose r a sin(5␪) r a sin(4␪) r a sin(3␪) If n is odd S there are n petals, if n is even S there are 2n petals. |a| represents the maximum distance from the origin (the radius of a circumscribed circle)

Limaçons: r ⴝ a ⴙ b sin ␪ (illustrated here) and r ⴝ a ⴙ b cos ␪ |a| |b| |b| |a|

|a| |b| |a| |b|

a

a

|a| |b|

|a| |b| |a| |b|

a

a

Cardioid Apple Limaçon Eye-ball (limaçon where |a| |b|) (limaçon where |a| |b|) (limaçon where |a| 2|b|) (inner loop if |a| |b|) r a b sin ␪ r a b sin ␪ r a b sin ␪ r a b sin ␪ |a| |b| represents the maximum distance from the origin (along the axis of symmetry)

Lemniscates: r 2 ⴝ a2sin(2␪) and r 2 ⴝ a2cos(2␪) a a

Lemniscate Lemniscate r2 a2 cos(2␪) r2 a2 sin(2␪) a represents the maximum distance from the origin (the radius of a circumscribed circle)

A-95


cob19537_saa_SA1-SA7.qxd

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Page SA-1

Student Answer Appendix 25. 10, 22, 15, 02

CHAPTER 1

10

Exercises 1.1, pp. 14–18 1. first, second 3. radius, center 5. Answers will vary. in 7. Year D 51, 2, 3, 4, 56 college GPA R 52.75, 3.00, 3.25, 3.50, 3.756 1

2.75

2

3.00

3

3.25

4

3.50

5

3.75

9. D 51, 3, 5, 7, 96; R 52, 4, 6, 8, 106 11. D 54, 1, 2, 36; R 50, 5, 4, 2, 36 13. 15. 10 8 6 4 2 108642 2 4 6 8 10

y

10 8 6 4 2

2 4 6 8 10 x

x

y

6

108642 2 4 6 8 10

15 12 9 6 3

2 4 6 8 10 x

y

2

0

54321 3 6 9 12 15

1 2 3 4 5 x

x

y

3

0

2, 2

0

1

1

3, 3

2

3

3

1

3

5, 5

0

1

6

3

6

8, 8

2

3

8

13 3

9, 9

3

8

4

10 8 6 4 2 108642 2 4 6 8 10

x

21. D: x 1 R: y 僆⺢

y

10 8 6 4 2

2 4 6 8 10 x

y

108642 2 4 6 8 10

8

15

23. D: x 僆⺢ R: y 僆⺢

y

5 4 3 2 1 108642 1 2 3 4 5

2 4 6 8 10 x

x

y

29. 13, 12

10

10

31. 10.7, 0.32

39. 2234 41. 10 47. right triangle 49. x2 y2 9 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

x

y

3, 3

9

2

4

3

10

3

4

5

2, 2

2

1

0

5

4

13, 13

1

0

2

121

3

4

1.25

4

3

1

2

10

10

3

19. D: 5 x 5 R: 0 y 5

10

y

3

7

10

27. 10, 02, 10, 42

17. D: x 僆⺢ R: y 1

y

x 5

10

1, 1

0

1

0.5, 0.5

4

3 1 5

0

7

2

108642 2 4 6 8 10

y

2 4 6 8 10 x

y

108642 2 4 6 8 10

2 4 6 8 10 x

y

2 4 6 8 10 x

61. 1x 72 2 1y 12 2 100 20 16 12 8 4 2016 1284 4 8 12 16 20

37. 11, 02

51. 1x 52 2 y2 3 10 8 6 4 2 108642 2 4 6 8 10

y

4 8 12 16 20 x

y

2 4 6 8 10 x

55. 1x 72 2 1y 42 2 7 10 8 6 4 2

57. 1x 12 2 1y 22 2 9 10 8 6 4 2

35. 10, 12

43. right triangle 45. not a right triangle

53. 1x 42 2 1y 32 2 4 10 8 6 4 2

1 1 , 24 2 33. 1 20

108642 2 4 6 8 10

y

2 4 6 8 10 x

59. 1x 42 2 1y 52 2 12 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

63. 1x 32 2 1y 42 2 41 15 12 9 6 3 1512963 3 6 9 12 15

y

3 6 9 12 15 x

SA-1


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Page SA-2

Student Answer Appendix

65. 1x 52 2 1y 42 2 9 10 8 6 4 2 108642 2 4 6 8 10

y

67. (2, 3), r 2, x 僆 30, 4 4, y 僆 3 1, 5 4 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

300

2 4 6 8 10 x

71. 14, 02, r 9, x 僆 313, 5 4 , y 僆 3 9, 9 4

200

10 8 6 4 2 108642 2 4 6 8 10 15 12 9 6 3

73. 1x 52 2 1y 62 2 57, (5, 6), r 157

75. 1x 52 2 1y 22 2 25, 15, 22, r 5

77. x 1y 32 14, 10, 32, r 114

15 12 9 6 3 1512963 3 6 9 12 15

10 8 6 4 2 108642 2 4 6 8 10

2

79. 1x 22 2 1y 52 2 11, 12, 52, r 111

83. 1x 32 1y 52 32, 13, 52, r 4 12 2

(3, 183) 0 1 2 3 4 5 6 7 8 9 1011 x

1512963 3 6 9 12 15

81. 1x 72 2 y2 37, 17, 02, r 137

b. $473

400

69. 11, 22, r 2 13, x 僆 3 1 213, 1 2 134, y 僆 3 2 2 13, 2 2 134

2

93. a. (3, 183), (5, 241), (7, 299), (9, 357), (11, 415); yes y c. 2014 d. (11, 415)

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

15 12 9 6 3 1512963 3 6 9 12 15

2

15 12 9 6 3 1512963 3 6 9 12 15

85. a. y x2 6x 87. b. y x2 1y 32 2 36 89. f. 1x 12 2 1y 22 2 49 91. j. 6x y x2 9

y

2 4 6 8 10 x

95. a. 1x 52 1y 122 2 625 b. no 97. Red: 1x 22 2 1y 22 2 4; Blue: 1x 22 2 y2 16; Area blue 12␲ units2 y 99. 15 2

12 9 6 3

y

1512963 3 6 9 12 15

3 6 9 12 15 x

y

3 6 9 12 15 x

101. 103. 105. c. r2 109.

Exercises 1.2, pp. 28–32 7.

2 4 6 8 10 x

y

2 4 6 8 10 x

3. negative, downward 5.

x

y

6

6

3

4

0

2

3

0

y

10 8 6 (6, 6) 4 (3, 4) 2 (0, 2) (3, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

11. 0.5 32 132 4 13. 0.5 92 4 0.5 0.5 ✓ 19 3 1 4 2 12 2 4 19 3 4 4 4

y

19 4

15. 2 4 6 8 10 x

19 4 ✓

10 8 6 (0, 1) 4 2 (5, 0)

21. 3 6 9 12 15 x

10 8 6 4 (6, 0)2

17.

y

(5, 2)

108642 2 4 6 8 10

y

y

27.

10 8 6 4 (5, 0) 2

y

108642 2 2 4 6 8 10 x (0, 2) 4 6 (5, 4) 8 10

yes no m1 m2 m1 # m2 1 y 9. 10 x y

23.

29.

2

1

0

4

2

7

4

10

(4, 10)

8 (2, 7) 6 4 (0, 4) (2, 1) 2

108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 (0, 6) 4 (1, 3) 2 (2, 0) 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 (2, 8) 6 4 (T, 0) 2 (0, 3) 108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 2 4 6 8 10 x 4 (0, 4) (3, 6) 6 8 10

y

3 6 9 12 15 x

No, distance between centers is less than sum of radii.

y 4.8; y 3.6, Answers will vary. Answers will vary. a. center: 16, 22; r 0 (degenerate case) b. center: (1, 4); r 5 1; degenerate case 107. x 74 n 1 is a solution, n 2 is extraneous

1. 0, 0 y

3 6 9 12 15 x

y

108642 2 4 6 8 10

(2, 0)

25.

(4, 6) (2, 3) (0, 0) 2 4 6 8 10 x

y 80

(1, 75)

1 , 0 40 (0, 25) 2

(

8

)

4 40

4

8

x

80

31.

y

y

10 8 6 4 (4, R) 2

108642 2 2 4 6 8 10 x 4 6 (0, R) 8 10

2 4 6 8 10 x

10 8 6 4 2 (0, 0) (4, 2) 108642 2 2 4 6 8 10 x 4 6 (2, 1) 8 10

10 8 6 4 2

19.

y

10 8 6 4 (0, 4) 2 (3, 0) 108642 2 2 4 6 8 10 x 4 6 (6, 4) 8 10


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Page SA-3

SA-3

Student Answer Appendix 33. m 1; 12, 42 and 11, 32 y

8

8

y

4

4

8

8

x

4

4

4

4

8

8

y

8

8

(3, 6) 4 8

8

x

4

4 8

x

(4, 2) 4

8

x

4

(1, 8)

8

41. a. m 125, cost increased $125,000 per 1000 sq ft b. $375,000 43. a. m 22.5, distance increases 22.5 mph b. about 186 mi 45. a. m 23 6 , a person weighs 23 lb more for each additional 6 in. in height b. ⬇3.8 47. In inches: (0, 6) and (576, 18): m 1 48 . The sewer line is 1 in. deeper for each 48 in. in length. y y 49. 51. 10 10 8

8 6 4 2

(3, 5) 6 (3, 0)

4 2

108642 2 (3, 4) 4 6 8 10

108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

(2, 6)

60

Exercises 1.3, pp. 43–47 1. first 3. range 5. Answers will vary. 7. function 9. Not a function. The Shaq is paired with two heights. 11. Not a function; 4 is paired with 2 and 5. 13. function 15. function 17. Not a function; 2 is paired with 3 and 4. 19. function 21. function 23. Not a function; 0 is paired with 4 and 4. 25. function 27. Not a function; 4 is paired with 1 and 1. 29. function y y 31. 33. 5 7 4 3 2 1 54321 1 2 3 4 5

function

6 5 4 3 2 1

1 2 3 4 5 x

7654321 1 2 3

41.

4 3 2 1

0

43.

4 3 2 1

0

45.

43 21 0 1 2 3 4 5 6 7

0

1

2

3

2

3

4

5

6

m 僆 1q, 5 4

2

3

4

5

6

x 僆 1q, 12 ´ 11, q2

function

[ 1

)) 1

x 僆 12, 52

)

)

5x| x 26; 3 2, q 2 49. 5x|2 x 16; 32, 1 4 function, x 僆 3 4, 5 4 , y 僆 3 2, 3 4 function, x 僆 3 4, q2 , y 僆 34, q2 function, x 僆 3 4, 4 4 , y 僆 35, 1 4 function, x 僆 1q, q 2 , y 僆 1q, q 2 not a function, x 僆 33, 5 4 , y 僆 33, 3 4 not a function, x 僆 1q, 3 4 , y 僆 1q, q2 x 僆 1q, 52 ´ 15, q 2 65. a 僆 3 5 3 , q2 x 僆 1q, 52 ´ 15, 52 ´ 15, q 2 v 僆 1q, 3 122 ´ 13 12, 3 122 ´ 13 12, q 2 x 僆 1q, q 2 73. n 僆 1q, q 2 75. x 僆 1q, q2 x 僆 1q, 22 ´ 12, 52 ´ 15, q2 x 僆 32, 52 2 ´ 1 52 , q 2 81. x 僆 14, q 2 83. x 僆 13, q2

1 7 87. f 162 0, f 1 32 2 15 4 , f 12c2 c 3, f 1c 12 c 2 2 89. f 162 132, f 1 32 2 34 , f 12c2 12c2 8c, f 1c 12 3c2 2c 1 1 3 9 91. h132 1, h1 2 3 2 2 , h13a2 , h1a 22 a a2 93. h132 5, h1 2 3 2 5, h13a2 5 if a 6 0 or 5 if a 7 0, h1a 22 5 if a 7 2 or 5 if a 6 2 3 95. g142 8␲, g a b 3␲, g12c2 4␲c, g1c 32 2␲1c 32 2 9 3 97. g142 16␲, ga b ␲, g12c2 4␲c2, g1c 32 1c2 6c 92␲ 2 4 3 99. p152 213, pa b 26, p13a2 26a 3, p1a 12 22a 1 2 14 3 27a2 5 7 3a2 6a 2 101. p152 , p a b , p 13a2 , p1a 12 2 2 5 2 9 9a a 2a 1 103. a. D: 51, 0, 1, 2, 3, 4, 56 b. R: 52,1, 0, 1, 2, 3, 46 c. 1 d. 1 105. a. D: 3 5, 5 4 b. y 僆 33, 4 4 c. 2 d. 4 and 0 107. a. D: 3 3, q 2 b. y 僆 1q, 44 c. 2 d. 2 and 2 1 109. a. 186.5 lb b. 37 lb 111. A 182 22 1 25 units2 2 113. a. N1g2 2.5g b. g 僆 30, 5 4 ; N 僆 3 0, 12.54 115. a. 30, q ) b. about 2356 units3 c. 800 units3 117. a. c1t2 42.50t 50 b. $156.25 c. 5 hr d. t 僆 3 0, 10.6 4 ; c 僆 30, 5004 119. a. Yes. Each x is paired with exactly one y. b. 10 P.M. c. 0.9 m d. 7 P.M. and 1 A.M. 121. negative outputs become positive y x

10 8 6 4 2

108642 2 4 6 8 10

y

y

10 x 8 6 y x 4 2

yx 2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

3 123. a. x 僆 1q, 22 ´ 12, q2 ; x 2y 1 y ; y 僆 1q, 12 ´ 11, q2

b. x 僆⺢ x 1y 3; y 僆 33, q 2 1 2 3 x

p 僆 1q, 32

3 2 1

85. x 僆 1 73 , q 2

(2, 4) (2, 0)

53. L1: x 2; L2: y 4; point of intersection (2, 4) Justices 55. a. For any two points chosen m 0, indicating 109 8 there has been no increase or decrease in the number 7 6 Nonwhite, nonmale 5 of supreme court justices. b. For any two points 4 1 3 chosen m 10 , which indicates that over the last 5 2 decades, one nonwhite or nonfemale justice has been 1 10 20 30 40 50 Time t added to the court every 10 yr. 57. parallel 59. neither 61. parallel 63. right triangle 65. not a right triangle 67. right triangle 69. a. 78.2 yr b. 2015 71. v 1250t 8500 a. $3500 b. 5 yr 73. h 3t 300 a. 273 in. b. 20 months 30 75. Yes they will meet, the two roads are not parallel: 38 12 9.5 . 77. a. $7360 b. 2015 79. a. 21% b. 2016 81. a 6 83. a. 142 b. 83 c. 9 d. 27 2 85. a. 1015 b. 4 87. a. 1x 321x 221x 22 b. 1x 2421x 12 c. 1x 521x2 5x 252

4

)

39.

47. 51. 53. 55. 57. 59. 61. 63. 67. 69. 71. 77. 79.

y

4 4

8

(4, 5)

39. m 4 7 ; 110, 102 and 111, 22

8

4

(10, 3)

4

37. m 15 4 ; 15, 232 and 17, 222 (3, 7)

37. 250 6 T 6 450; T 僆 1250, 4502

8

(4, 6) (3, 5)

4

35. w 45; w 僆 3 45, q 2

35. m 43 ; 17, 12 and 11, 92

127. a. 1x 321x 521x 52 c. 12x 52 14x2 10x 252

125. a. 1916

b. 12x 321x 82

b. 1


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Student Answer Appendix 5 5 33. y 5 4 x 5, f 1x2 4 x 5, m 4 , y-intercept (0, 5) 1 1 1 35. y 3 x, f 1x2 3 x, m 3, y-intercept (0, 0) 3 3 37. y 3 4 x 3, f 1x2 4 x 3, m 4 ,y-intercept (0, 3)

Mid-Chapter Check, p. 48 y

1.

10 8 6 4 (6, 4) 2 (3, 0) 108642 2 2 4 6 8 10 x 4 (0, 4) 6 8 10

39. y 23 x 1

45. y 4x 10 51. y 35 x 4

2000

108642 2 4 6 8 10

75

D: x 僆 1q, q2, R: y 僆 1q, 72 D: x 僆 3 1, 7 4 , R: y 僆 3 6, 9 4 D: x 僆 1q, 94 , R: y 僆 3 3, q 2 D: x 僆 1q, q2, R: y 僆 1q, 9 4

3. 2.5

7. y 4 5 x 2

5. Answers will vary. 11. y 5 3 x 5

9. y 2x 7

x

y

x

y

x

y

6

5

3

5

2

18 5

2

3

2

10 3 5 3

0

2

0

7

0

5

1

6 5 2 5

1

9

1

20 3

3

13

3

10

13. y 2x 3: 2, 3

5 5 x 7: 3 , 7 15. y 3

35 17. y 35 6 x 4: 6 , 4 y 19. 21. 10 8 6 (0, 5) 4 2 (3, 1) 108642 2 2 4 6 8 10 x 4 (6, 3) 6 8 10

59.

y

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

(0, 5)

61.

y

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

2 4 6 8 10 x

8 6 4 2

108642 2 4 6 8 10

8642 2 4 6 8 10

2 4 6 8 10 x

y 4 38 1x 32

y 5 21x 22 y

10 8 6 4 2

2 4 6 8 10 12 x

2 4 6 8 10 x

y 3.1 0.51x 1.82

5

3

108642 2 4 6 8 10

2 4 6 8 10 x

y

8 6 4 2

108642 2 4 6 8 10

Exercises 1.4, pp. 59–63

, (0, 3)

10 8 6 4 2

49. y 75 2 x 150 55.

29 63. y 25 x 4 65. y 5 67. y 12 3 x 7 5 x 5 69. y 5 71. perpendicular 73. neither 75. parallel 5 77. a. y 3 b. y 43 x 20 79. a. y 49 x 31 4 x 2 3 9 9 3 1 b. y 4 x 4 81. a. y 2 x 2 b. y 2x 2 y y 83. 85. 10 10

87.

Reinforcing Basic Concepts, pp. 48–49

7 4

10 8 6 4 2 108642 2 4 6 8 10

200 5. x 3; no; input 3 is paired with more than one output. 6. y 2; yes 7. a. 0 b. x 僆 3 3, 5 4 c. 1 and 1 d. y 僆 3 4, 5 4 8. from x 1 to x 2; steeper line S greater slope 3 9. ¢F ¢p 4 ; For each increase of 4000 pheasants, the fox population increases by 300; 1100 foxes. 10. a. x 僆 53, 2, 1, 0, 1, 2, 3, 46, y 僆 53,2,1, 0, 1, 2, 3, 46 b. x 僆 3 3, 4 4 , y 僆 33, 4 4 c. x 僆 1q, q2 , y 僆 1q, q2

1.

10 8 6 4 (0, 4) 2

57.

0

47. y 53.

43. y 3x 2

3 2 x 1 y 23 x 5

y

2. m 18 3. positive, loss is decreasing (profit is increasing); m 32 , 7 1.5 yes; 1 , each year Data.com’s loss decreases by 1.5 million. 4. a. E1x2 7.5x 950 b. $1100, $1175, $1250 c. x 僆 30, 75, 5 4 and y 僆 3200, 2000, 200 4 d. 47 snowboards

1. 2. 3. 4.

41. y 3x 3

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

(0, 2) (2, 5) (4, 8)

23.

y

10 8 6 4 (0, 3) (3, 4) 2 (3, 2) 108642 2 4 6 8 10

2 4 6 8 10 x

25. a. 3 b. y 3 c. The coeff. of x is the slope and the 4 4 x 3 constant is the y-intercept. 27. a. 25 b. y 25 x 2 c. The coeff. of x is the slope and the constant is the y-intercept. 29. a. 45 b. y 45 x 3 c. The coeff. of x is the slope and the constant is the y-intercept. 2 2 31. y 2 3 x 2, f 1x2 3 x 2, m 3 , y-intercept (0, 2)

89. y 2 65 1x 42 ; For each 5000 additional sales, income rises $6000. 91. y 100 20 1 1x 0.52 ; For every hour of television, a 1 student’s final grade falls 20%. 93. y 10 35 2 1x 2 2 ; Every 2 in. of rainfall increases the number of cattle raised per acre by 35. 95. C 97. A c 99. B 101. D 103. m a a. m 3 b , y-intercept b 4 , y-intercept 2 5 (0, 2) b. m 5 , y-intercept (0, 3) c. m 6 , y-intercept (0, 2) d. m 53 , y-intercept (0, 3) 105. a. As the temperature increases 5°C, the velocity of sound waves increases 3 m/s. At a temperature of 0°C, the velocity is 331 m/s. b. 343 m/s c. 50°C 107. a. V 20 3 t 150 b. Every 3 yr the value of the coin increases by $20; the initial value was $150. c. $223.33 d. 15 years, in 2013 e. 3 yr 109. a. N 7t 9 b. Every 1 yr the number of homes with Internet access increases by 7 million. c. 1993 d. 86 million e. 13 yr f. 2010 111. a. P 58,000t 740,000 b. Each year, the prison population increases by 58,000. c. 1,900,000 113. Answers will vary.

115. 1. d.

2. a.

3. c.

4. b. 5. f.

6. h.

117. a. 9

119. 113.10 yd2

Exercises 1.5, pp. 75–78 1. intersection-of-graphs, Y1, Y2, x-coordinate, intersection 3. literal, two

5. Answers will vary.

b. 9|x|


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Page SA-5


15. 17. 19. 21.

y 10 8 6 4 2

(3, 2)

108642 2 4 6 8 10

1 4 6 8 10 x

a. linear b. positive c. strong d. m ⬇ 4.2 a. nonlinear b. positive c. NA d. NA a. nonlinear b. negative c. NA d. NA a. b. positive c. f 1x2 2.4x 62.3, y f 152 74.3174,3002, f 1212 112.7 1112,7002 100

x3 9. x 3 11. x 僆 1q, q2 13. x 9 15. no solutions 17. x 3, answers match 19. x 2 21. x 0 23. no solutions 25. x 僆 1q, q2 27. x 僆 13, q 2 , verified graphically in Exercise 7 29. x 僆 1q, 04 31. x 僆 15, q 2 P C 33. x 僆 1q, q2 35. no solutions 37. C 39. r 1M 2␲ 2Sn T1P2V2 3V 41. T2 43. h 45. n P1V1 a1 an 4␲r2 21S B2 A C 16 20 47. P 49. y 51. y x x S B B 9 3 4 53. y x 5 55. a 3; b 2; c 19; x 7 5 16 57. a 6; b 1; c 33; x 3 59. a 7; b 13; c 27; x 2 61. h 17 cm 63. 510 ft 65. 56 in 67. 3084 ft 69. 48; 50 71. 5; 7 73. 11:30 A.M. 75. 36 min 77. 4 quarts; 50% O.J. 79. 16 lb; $1.80 lb 81. 12 lb 83. 16 lb 85. Answers will vary. 87. 69 89. x 7 91. a. 12x 3212x 32 b. 1x 321x2 3x 92

Officers (1000s)

7.

90 80 70 60 3 5 7 9 11 13 x Year (1990 → 0)

23. Using (5, 7.6) and (20, 23.3): y ⬇ 1.05x 2.37; GDP in 2010 will be near 44,370 25. a.

40

Exercises 1.6, pp. 87–93 1. scatterplot 7. a.

3. linear

SA-5

0

8

5. Answers will vary. b. positive

y

0

1600

b. positive, larger radius 1 larger area c. perfect correlation d. m 2␲ 27. a. b. linear c. positive y d. y 0.96x 1.55, 63.95 in.

1400 1200 1000 x

9. a.

4

8

12 16 20 24 28

75.5

Year (1980 → 0)

Wingspan (in.)

0

b. linear c. positive

y

150 Congresswomen

125 100

63.5 57.5 51.5 51 57 63 69 75 x Height (in.)

75 50 25

29. a.

x

0

5

10 15 20 25 30 35 Year (1970 → 0)

11. a.

b. linear c. positive d. y 9.55x 70.42; about 271,000. The number of applications, since the line has a greater slope.

y 260

b. positive

c. strong

y 50 40

220 180 140 100

x 3

30 20

5 7 9 11 13 Year (1990 → 0)

31. a. 0

10 20 30 40 50 x Year (1970 → 0)

13. a. (A) (D) (C) (B) y b. 60 y 60 55 50 45 40 35 30 25

55 50 45 40 35 30 25

y 60 55 50 45 40 35 30 25

y 60 55 50 45 40 35 30 25

65

86 Percent of men

10

Percent of women

GDP per capita (1000s)

69.5

Patents (1000s)

Cost (cents)

1800

55 45 35 25

82 78 74 70

0

10 20 30 40 50x

Year (1950 → 0)

0

10 20 30 40 50 x Year (1950 → 0)

b. women: linear c. positive; b. men: linear c. negative d. yes, |slope| is greater

0 1 2 3 4 5 6 7 8 9 x

0 1 2 3 4 5 6 7 8 9 x

0 1 2 3 4 5 6 7 8 9 x

0 1 2 3 4 5 6 7 8 9 x

c. positive, d. m ⬇ 3.8;

c. positive, d. m ⬇ 4.2;

c. negative, d. m ⬇ 2.4;

c. negative, d. m ⬇ 4.6


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33. a.

10. a. parallel

2000

11. a.

10 8 6 4 2

b.

y

12. a. 0 a. linear b. y 108.18x 330.20 c. $1736.54 billion; about $2601.98 billion 35. a. r ⬇ 0.9783 b. r ⬇ 0.9783 c. they are almost identical; context, pattern of scatterplot, anticipated growth, etc. 37. No. Except for the endpoints of the domain, one x is mapped to two y’s. AP 39. r Pt

108642 2 4 6 8 10

b.

y

10 8 6 (0, 2) 4 2 (3, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

3. a 5. b

7. c

9. f

11. d

13. f

13. a. vertical b. horizontal c. neither

5

2.

3 2

5

1

3

0

0

8

5 4 3 2 1 54321 1 2 3 4 5

5 2

21. a.

1 2 3 4 5 x

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2

54321 1 2 3 4 5

10, 52, 13, 02

10 7.

y

1 2 3 4 5 x

y 5 4 3 2 1 54321 1 2 3 4 5

8. 1x 1.52 2 1y 22 2 6.25 y 9. a. b. 10

10 8 6 9 4 2

8 6 4 2

5 108642 2 2 4 6 8 10 x 4 9 6 8 10

5 9 ,

108642 2 4 6 8 10

10

5 4 3 2 1

114, 72

108642 2 4 6 8 10

1 3,

5x

b. 1

4 20. a. y 4 3 x 4, m 3 , y-intercept (0, 4) 5 5 b. y 3 x 5, m 3 , y-intercept 10, 52

10

6.

(5, 0)

14. yes 15. m 23 , y-intercept (0, 2); when the rodent population increases by 3000, the hawk population increases by 200. 16. a. x 僆 3 5 b. x 僆 1q, 22 ´ 12, 32 ´ 13, q 2 4 , q2

22. a.

5. 1 52 , 32

x5

(0, ) 5

10

4. 65 mi

y

y 4

x 僆 3 5, 5 4 y 僆 3 0, 54

3.

(w, 0)

2 17. 14; 26 18. It is a function. 9 ; 18a 9a 19. I. a. D 51, 0, 1, 2, 3, 4, 56, R 52, 1, 0, 1, 2, 3, 46 c. 2 II. a. x 僆 1q, q 2, y 僆 1q, q 2 b. 1 c. 3 III. a. x 僆 33, q 2, y 僆 3 4, q 2 b. 1 c. 3 or 3

y

x y 5 0 4 3 2 221 ⬇ 4.58 0 5 2 221 ⬇ 4.58 4 3 5 0

y

2y x 5

15. a

1. x 僆 57, 4, 0, 3, 56 y 僆 52, 0, 1, 3, 86 4

(2, 2)

5

Summary and Concept Review, pp. 94–98 7

10 8 6 4 2

2 4 6 8 10 x

108642 2 2 4 6 8 10 x 4 (0, 2) 6 8 10

Making Connections, p. 93 1. d

y

10 8 6 4 2 (0, 1)

(1, 1)

108642 2 2 4 6 8 10 x 4 (0, 2) 6 8 10

13

0

b. perpendicular

10, 32

1 2 3 4 5 x

b.

y

108642 2 4 6 8 10

2 4 6 8 10 x

b.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

11 23. y 5, x 2; y 5 24. y 3 25. f 1x2 43 x 4 x 4 2 2 26. m 5 , y-intercept (0, 2), y 5 x 2. When the rabbit population increases by 500, the wolf population increases by 200. 27. a. 1y 902 15 b. (14, 0), (0, 105) c. f 1x2 15 2 1x 22 2 x 105 d. f 1202 45, x 12 28. x 6 29. x 僆 1q, q2 30. x 僆 1q, 12 31. h ␲rV 2 32. L P 2 2W 33. x c a b 2 9 2 2 34. y 3 x 2 35. 8 gal 36. 12 8 ␲ ft ⬇ 15.5 ft 37. 23 hr 40 min 100 38. a.

0

y

10 8 6 4 2

130

3 2 4 6 8 10 x

50 b. linear

c. positive


2/2/11

2:52 PM

Page SA-7

SA-7

Student Answer Appendix 39. a. f 1x2 0.35x 56.10 b.

b. y 2 12 1x 32 ,

3. a. 12 , increasing y 12 x 12 c. 10, 12 2, 11, 02

100

0

130

10 8 6 4 2 108642 2 4 6 8 10

b. y 4 34 1x 52 ,

4. a. 34 , increasing y 34 x 14 1 c. 10, 1 4 2, 1 3 , 02

10 8 6 4 2 108642 2 4 6 8 10

50 c. strong 40. f 11202 98.1, over 98%

Practice Test, pp. 99–100 1. 2. 5. 6. 7.

P a. x 27 b. x 2 c. C 1 d. W P 2 2L k 30 gal 3. S 177 4. about 5.1 sec a and c are nonfunctions, they do not pass the vertical line test neither y y 8. 12,32; r 4 5 3 4 3 2

y 11

54321 1 2 3 4 5

2 1

x 4 1 2 3 4 5 x

321 1 2 3 4 5 6 7

1 2 3 4 5 6 7 x

20 2 2 9. V 20 10. y 6 3 t 3 , 663 mph 5 x 5 11. a. (7.5, 1.5), b. ⬇ 61.27 mi 12. L1: x 3 L2: y 4 13. a. x 僆 54, 2, 0, 2, 4, 66 y 僆 52, 1, 0, 1, 2, 36 b. x 僆 32, 6 4 y 僆 3 1, 44

14. a. 300

b. 30

c. W1h2 25 2h

5. a. 3 4 , decreasing 7 y 3 4 x 2 c. 10, 72 2, 1 14 3 , 02

b. y 5 3 4 1x 22 ,

6. a. 1 2 , decreasing y 1 2 x 6 c. 10, 62 , 112, 02

b. y 7 1 2 1x 22 ,

108642 2 4 6 8 10

15 12 9 6 3 1512963 3 6 9 12 15

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

3 6 9 12 15 x

Connections to Calculus, p. 104

d. Wages are $12.50 per hr.

7 a2 6a 7 e. h 僆 3 0, 404 ; w 僆 3 0, 500 4 15. a. b. 2 2 a 6a 9 ¢sales 13.5 16. a. ¢time 1 b. sales are increasing at a rate of 13.5 million phones per year c. 2008: about 15 million sales, 2010: about 42 million sales, 2011: about 55.5 million sales 17. a. x 9 b. no solution 18. a. x 僆 33, q2 b. x 僆 1q, q 2 19. a. b. linear c. positive 30

1. y 4x 9

3. y 9

1 5 9. y x 2 2

1 3 11. y x 2 2

5. y 44

7. y 36x 1

3 7 13. y x 4 4

CHAPTER 2 Exercises 2.1, pp. 114–119 1. linear, bounce 3. increasing 5. Answers will vary. y 7. 9. even 11. even 5

5

0

10 8 6 4 2

y

5 x

5

50 13.

15. odd

y

17. not odd

10

3

20. a. f 1x2 0.91x 10.78

b. f 1502 ⬇ 35

c. strong: r ⬇ 0.974

10

10 x

Strengthening Core Skills, pp. 100–101 1. a. 13 , increasing b. y 5 13 1x 02 , y 13 x 5 c. (0, 5), 115, 02

2. a. 7 3 , decreasing y 7 3 x 9 c. (0, 9), 1 22 7 , 02

b. y 9 7 3 1x 02 ,

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

2 4 6 8 10 x

10

19. neither 21. odd 23. neither 25. x 僆 3 1, 14 ´ 33, q2 27. x 僆 1q, 12 ´ 11, 12 ´ 11, q 2 29. x 僆 3 2, q2 31. x 僆 1q, 24 33. V1x2c: x 僆 13, 12 ´ 14, 62 ; V1x2T: x 僆 1q, 32 ´ 11, 42 ; constant: none 35. f 1x2c: x 僆 11, 42 ; f 1x2T: x 僆 12, 12 ´ 14, q2 ; constant: x 僆 1q, 22 37. a. p1x2c: x 僆 1q, q2 ; p1x2T: none b. down, up 39. a. f 1x2c: x 僆 13, 02 ´ 13, q 2 ; f 1x2T: x 僆 1q, 32 ´ 10, 32 b. up, up 41. a. x 僆 1q, q2; y 僆 1q, 5 4 b. x 1, 3 c. H1x2 0: x 僆 31, 3 4 ; H1x2 0: x 僆 1q, 14 ´ 3 3, q2 d. H1x2c: x 僆 1q, 22 ; H1x2T: x 僆 12, q 2 e. local max: y 5 at (2, 5)


SA-8

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43. a. x 僆 1q, q2; y 僆 1q, q2 b. x 1, 5 c. g1x2 0: x 僆 3 1, q 2 ; g1x2 0: x 僆 1q, 1 4 ´ 53.56 d. g1x2c: x 僆 1q, 12 ´ 15, q2 ; g1x2T: x 僆 11, 52 e. local max: y 6 at (1, 6); local min: y 0 at (5, 0) 45. a. x 僆 34, q 2; y 僆 1q, 34 b. x 4, 2 c. Y1 0: x 僆 3 4, 2 4 Y1 0: x 僆 32, q 4 d. Y1c: x 僆 14, 22 ; Y1T: x 僆 12, q2 e. local max: y 3 at (2, 3); endpoint min y 0 at 14, 02 47. a. x 僆 1q, q2, y 僆 1q, q 2 b. x 4 c. p1x2 0: x 僆 3 4, q2; p1x2 0: x 僆 1q, 4 4 d. p1x2c: x 僆 1q, 32 ´ 13, q 2; p1x2T: never decreasing e. local max: none; local min: none 49. max: y ⬇ 1.58 at x ⬇ 0.78; min: y ⬇ 0.47 at x ⬇ 2.55 51. max: y ⬇ 1.54 at x ⬇ 6.21, y ⬇ 3.28 at x ⬇ 2.55; min: y ⬇ 3.28 at x ⬇ 2.55, y ⬇ 1.54 at x ⬇ 6.21 53. max: y ⬇ 3.08 at x 83 ; min: y 0 at x 4 (endpoint) 55. a. x 僆 1q, 3 4 ´ 3 3, q 2 ; y 僆 30, q2 b. 13, 02 , (3, 0) c. f 1x2c: x 僆 13, q2 ; f 1x2T: x 僆 1q, 32 d. even 9y2 36 e. x 57. a. x 僆 3 0, 2604 , y 僆 3 0, 80 4 b. 80 ft B 2 c. 120 ft d. yes e. (0, 120) f. (120, 260) 59. a. x 僆 1q, q2; y 僆 31, q2 b. (1, 0), (1, 0) c. f 1x2 0: x 僆 1q, 1 4 ´ 31, q2; f 1x2 6 0: x 僆 11, 12 d. f 1x2c: x 僆 10, q2, f 1x2T: x 僆 1q, 02 e. min: (0, 1) 61. a. D: t 僆 [1983, 2009], R: I 僆 [5, 14]; b. I1t2c: t 僆 (1983, 1984) ´ (1986, 1987) ´ (1993, 1994) ´ (1998, 2000) ´ (2005, 2006); I1t2T: t 僆 (1984, 1986) ´ (1989, 1993) ´ (1994, 1998) ´ (2000, 2003) ´ (2006, 2009); I(t) constant: (1987, 1989) ´ (2003, 2005); c. global max: I 14 in 1984, global min: I 5 in 2009 (also an endpoint min); d. greatest increase: (1993, 1994), greatest decrease: (1985, 1986) y 63. zeroes: (8, 0), (4, 0), (0, 0), (4, 0); 8 min: 110, 62,(2, 1), (4, 0); max: (6, 2), (2, 2)

25. a. cubic; b. up/down, (1, 0), (1, 0), (0, 1); c. D: x 僆⺢, R: y 僆⺢ 27. a. cubic; b. down/up, (0, 1), (1, 0), (0, 1); c. D: x 僆⺢, R: y 僆⺢ 29. a. cube root; b. down/up, (1, 1), (2, 0), (0, 2); c. D: x 僆⺢, R: y 僆⺢ 31. square root function; y-int (0, 2); x-int (3, 0); initial point (4, 2); up on right; D: x 僆 34, q2, R: y 僆 32, q 2 33. cubic function; y-int (0, 2); x-int (2, 0); inflection point (1, 1); up, down; D: x 僆⺢, R: y 僆⺢ 35. 10

5

10

5 the graph of g is f shifted up 2 units; the graph of h is f shifted down 3 units 37.

10

10

10

10

4

8

4

4

8

the graph of q is p shifted down 5 units; the graph of r is p shifted up 2 units

x

4 8

39.

41.

y

y 8

8

65. no; no; answers will vary. 67. a. Thorpe; Rosolino b. two times, at 190 sec and 219 sec c. about 29 sec 1219 190 29 sec2 d. Thorpe e. about 2 sec 1223 221 2 sec2 f. 221 sec 3 min 41 sec 69. h1k2 h1k2 1 1 3 1k2 3 4 2 1k3 2 2 1 1 1k3 2 2 1k3 2 2 1 1 1k3 2 2 1k3 2 2 ✓ 71. a. 4 12 x2 b. 4 9 x2 73. V 5184␲ cm3, SA 1152␲ cm2

8

4

43.

Exercises 2.2, pp. 130–135 1. stretch, compression 3. (5, 9), upward 5. Answers will vary. 7. a. quadratic; b. up/up, (2, 4), x 2, (4, 0), (0, 0), (0, 0); c. D: x 僆⺢, R: y 僆 3 4, q 2 9. a. quadratic; b. up/up, (1, 4), x 1, (1, 0), (3, 0), (0, 3); c. D: x 僆⺢, R: y 僆 3 4, q2 11. a. quadratic; b. up/up, (2, 9), x 2, (1, 0), (5, 0), (0, 5); c. D: x 僆⺢, R: y 僆 39, q 2 13. a. square root; b. up to the right, (4, 22 , (3, 0), (0, 2); c. D: x 僆 34, q2 , R: y 僆 3 2, q 2 15. a. square root; b. down to the left, (4, 3), (3, 0), (0, 3); c. D: x 僆 1q, 4 4 , R: y 僆 1q, 3 4 17. a. square root; b. up to the left, (4, 0), (4, 0), (0, 4); c. D: x 僆 1q, 4 4 , R: y 僆 3 0, q 2 19. a. absolute value; b. up/up, (1, 4), x 1, (3, 0), (1, 0), (0, 2); c. D: x 僆⺢, R: y 僆 3 4, q2 21. a. absolute value; b. down/down, (1, 6), x 1, (4, 0), (2, 0), (0, 4); c. D: x 僆⺢, R: y 僆 1q, 64 23. a. absolute value; b. down/down, (0, 6), x 0, (2, 0), (2, 0), (0, 6); c. D: x 僆⺢, R: y 僆 1q, 6 4

4

4 4

8

8

x

4

4

4

4

8

8

8

x

10

10

10

10 the graph of q is p shifted left 5 units 45.

10

10

10

10 the graph of Y2 is Y1 shifted right 4 units


2/2/11

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SA-9

Student Answer Appendix 47.

49.

y

8

4

4

4

8

8

x

4

4

4

8

87. left 2, reflected across x-axis, compressed vertically, down 1

y

8

4 8

51.

y

8

4

8

8

x

4

4

4

4

8

8

8

x

89. left 1, reflected across x 1, reflected across x-axis, stretched vertically, up 3

y

53.

y

4 8

8

4

4

8

4

8

(3, 1)

987654321 1 2 3 4 5

x

8

x

4

4

2 1

(2, 1)4

4

y 8

(1, 3) 3

8 4

y

5 4

8

91. right 3, compressed vertically, up 1

8

1 x

4

4

8

x

4

8

x

4

8

x

4 8

8

93. a. 55.

b.

c.

y

10

8

10

8

8

4

4

4

4

8

8

x

4

4

8

8

8

95. a.

b. y

4

8

8

x

4

8

8

8

10

y 8

4

4

8

4 8

63. g 65. i 67. e 75. left 2, down 1

8

4

4

4

4

8

8

x

(2, 1) 4

4

8

8

8

79. left 3, down 1

(3, 2)

8

600 400

2

200 4

109. a. vertical stretch by a factor of 2 b. 12.5 ft 4

8

x

4 8

85. left 1, reflected across x-axis, stretched vertically, down 3

d(t) 14 12 10 8 6 4 2 1

8

8

4

4

2

(1, 2)

4

8

x

8

4

4

8

4

4 (3, 2)

8

8

x

8

4

4

t

y

f(x)

8 4 8 4

8

4 (1, 3) 8

3

111. x 僆 10, 42 ; yes, x 僆 14, q 2 ; yes

y

4

800

20 40 60 80 100 x

4

y

107. a. compressed vertically P(v) 1000 b. 216 W

1

4

x

r

3

3

8

8

2

4

y

8

4

x

99. p1x2 1.52x 3 101. f 1x2 45 0x 4 0 b. about 65 units3, V⬇65.4 units3; yes; 3 3 c. r 2 4␲ V

105. a. compressed vertically T(x) b. 2.25 sec 5

4

x

g(x) 4

4 8

8

20 1

83. left 3, reflected across x-axis, down 2

y

4

40

8

81. left 1, down 2

4

4

60

4

4

x

4

100

(3, 1) 8

8

V(r) 艐 4.2r3

y

8

8

120

2 4 6 8 10 x

69. j 71. l 73. c 77. left 3, reflected across x-axis, down 2

y

4

80

108642 2 4 6 8 10

x

4

4

97. f 1x2 1x 22 2 103. a.

8 6 4 2

4

4

d.

8

8

8

x

4

8

4

8

4

y

8

4

4

4

c.

10 the graph of Y2 is Y1 vertically stretched; the graph of Y3 is Y1 vertically compressed y y 59. 61. 10

8

4

4

10

y

8

4 8

4 4

8

10

8

x

4

y

57.

8

4

d. 10 the graph of q is p vertically stretched; the graph of r is p vertically compressed

y

8

4

10

y

8

8

x

8

12

16

20

24

28 v


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113. Any points in Quadrants III and IV will reflect across the x-axis and move to Quadrants I and II. 5 4 3 2 1 5 4321 1 2 3 4 5

f(x) f(x) x2 4 1 2 3 4 5 x

F(x)

5 4 3 2 1

F(x) 兩x2 4兩

5 4321 1 2 3 4 5

1 2 3 4 5 x

7. a. as x S q, y S 2 as x S q, y S 2 b. as x S 1 , y S q as x S 1 , y S q

9. a. as x S q, y S 1 as x S q, y S 1 b. y 1 c. as x S 2 , y S q as x S 2 , y S q 11. down 1, x 僆 1q, 02 ´ 10, q 2 , y 僆 1q, 12 ´ 11, q2 y

115. P 14x2 16x 142 units f 1x2T : x 僆 1q, 42

117. f 1x2c: x 僆 14, q 2 ;

x 0

8 4

(1, 0) 8

4

4

4 y 1

8

8

Exercises 2.3, pp. 143–146 1. reverse 3. 7, 7 5. no solution; answers will vary. 7. 54, 66 9. 52, 126 11. 53.35, 0.856 13. 5 87 , 26 1 1 15. 52 , 2 6 17. { } 19. 510, 66 21. {3.5, 11.5} 23. 51.6, 1.66 25. 53, 12 6 27. 15, 32 29. { } 7 8 14 3 31. c , d 33. 5 6 35. a1, b 37. c , 0 d 3 3 5 4 39. 1q, 102 ´ 14, q 2 41. 1q, 3 4 ´ 3 3, q 2 3 7 7 43. aq, d ´ c , qb 45. aq, d ´ 3 1, q 2 3 3 7 7 2 ´ 11, q 2 49. 1q, 02 ´ 15, q 2 47. 1q, 15 51. { } 53. 1q, q2 55. a. x 2 and x 6 b. 1q, 2 4 ´ 36, q 2 c. [2, 6] 57. a. x 3 and x 0.2 b. 1q, 3 4 ´ 30.2, q 2 c. 13, 0.22 59. a. 45 d 51 in. b. d L x d L 61. in feet: [32,500, 37,600]; yes 63. in feet: d 6 210 or d 7 578 65. a. 冟s 37.58冟 3.35 b. [34.23, 40.93] 67. a. 冟s 125冟 23 b. [102, 148] 69. a. 冟d 42.7冟 6 0.03 b. 冟d 73.78冟 6 1.01 c. 冟d 57.150冟 6 0.127 d. 冟d 2171.05冟 6 12.05 e. golf: t ⬇ 0.0014 71. a. x 4 b. 3 43 , 4 4 c. x 0 d. 1q, 35 4 e. { } 73. a. 52, 86 b. 1q, 32 ´ 111, q 2 c. 3 6, 2 4 d. 52, 106 75. ␳ V2W 2 CA

77. x 僆 1q, 52 2

13. left 2, x 僆 1q, 22 ´ 12, q 2 , y 僆 1q, 02 ´ 10, q2 y

8

x 2 8

4

y 0

4

4

x

8

15. right 2, reflected across x-axis, x 僆 1q, 22 ´ 12, q 2 , y 僆 1q, 02 ´ 10, q 2 y

8

冢0, q冣 8

4

x 2

4

4 4

8

y 0

x

8

17. left 2, down 1, x 僆 1q, 22 ´ 12, q 2 , y 僆 1q, 12 ´ 11, q 2 y

8 4

x 2 4

(1, 0) 4

4 8

1. neither 2. max: y ⬇ 11.12 at x ⬇ 0.50, y ⬇ 8.55 at x ⬇ 1.47; min: y ⬇ 7.80 at x ⬇ 0.75 3. increasing on 1q, 0.502 ´ 10.75, 1.472, decreasing on 10.50, 0.752 ´ 11.47, q2 4. g1x2 2x 4 2 5. a. cubic b. up on the left, down on the right; inflection point: (2, 2); x-int: (4, 0); y-int: (0, 5) c. 1q, q2 ; 1q, q 2 d. k 1 q(x) is a reflection of 6. 10 p(x) across the x-axis, and r(x) is the same as q(x), but compressed by a factor of 12 10 10

8

4

8

Mid-Chapter Check, pp. 146–147

冢0, q冣

8

y 1 冢0, q冣

x

19. right 1, x 僆 1q, 12 ´ 11, q 2 , y 僆 10, q 2

y 8

(0, 1) 4 y 0

8

b. { }

c. k 僆 1q, q 2

Reinforcing Basic Concepts, p. 147 1. x 3 or x 7 2. x 僆 3 5, 3 4

3. x 僆 1q, 1 4 ´ 3 4, q2

y 8 4

y 0

8

1. as x S q, y S 2 3. vertical, y 2

4

4

x

8

x 2 y 8 4

冢√q, 0冣 y 2 8 4

冢√q, 0冣 4

8

x

4 8

x 0

25. left 2, up 1, x 僆 1q, 22 ´ 12, q 2, y 僆 11, q2

y 8 4

y 1 8

4

冢0, @冣 4

8


8

冢0, ~冣

4

4


x

x 1

21. left 2, reflected across x-axis, x 僆 1q, 22 ´ 12, q 2 , y 僆 1q, 02

b. 566 19 23 b. y 僆 aq, b ´ a , q b 2 2

10. w 僆 3 8, 26 4 ; no, yes

8

8

10 8. a. q 僆 18, 02

9. a. d 僆 1q, 0 4 ´ 3 4, q 2

4 4

23. down 2, x 僆 1q, 02 ´ 10, q 2 , y 僆 12, q2 7. a. 54, 146

4

x 2

8

x


2/2/11

2:56 PM

Page SA-11

SA-11

Student Answer Appendix 1

27. reciprocal quadratic, S1x2

2

1x 12 1 2 29. reciprocal function, Q1x2 x1 1 5 33. S 2 31. reciprocal quadratic, v1x2 1x 22 2 35. S q

2

4 2 43. g1x2 increases faster a 7 b a. x 1, 0, and 1, 5 3 b. 11, 02 ´ (0, 1), c. 1q, 12 ´ 11, q 2

47. 49. 55. 59. 61. 63.

100

2500

40,000

37. 1, q

39. g1x2 increases faster 13 7 22 a. x 0 and 1, b. 1q, 02 ´ (0, 1), c. 11, q 2 41. f 1x2 increases faster 14 7 22 a. x 1, 0, and 1, b. 11, 02 ´ (0, 1), c. 1q, 12 ´ 11, q 2

45.

77. a.

1 1 g 1x2 increases faster a 7 b a. x 0 and 1, b. (0, 1), c. 11, q2 3 6 5 2 g1x2 increases faster a 7 b a. x 0 and 1, b. (0, 1), c. 11, q2 4 3 3 0, q2 51. 1q, q 2 53. 1q, q 2 defined: b, c, d; undefined: a 57. defined: a, b, d; undefined: c F is the graph of f shifted left 1 unit and down 2; verified P is the graph of p shifted right 2 units and reflected across the x-axis; verified d2F a. F becomes very small b. y x12 c. m2 km 1

65. a. It decreases; 75, 25, 15 c. as p S 0, D S q D(p)

b. It approaches 0.

10 b. S 1.687a

0.386

d. about 37,200 mi2

c. about 33 species

79. The area is always 1 unit ; The area is always 1x units2 2

81. y 2 3 x 5,

83. c 2mE

y 8 4 8

4

(3, 3) 4

8

x

4 8

Exercises 2.5, pp. 172–176 1. continuous 3. smooth 5. Each piece must be continuous on the corresponding interval, and the function values at the endpoints of each interval must be equal. Answers will vary. x2 6x 10 7. a. f 1x2 e 3 5 2x 2

0x5 b. y 僆 31, 11 4 5 6 x9

9. 2, 2, 12 , 0, 2.999, 5 11. 5, 5, 0, 4, 5, 11 13. D: x 僆 3 6, q 2 ; R: y 僆 34, q2 p122 4 p102 2

90 70 50 30 10 10 30 50 70 90 p

67. a. It decreases; 100, 25, 11.1. c. as d S 0, l S q 2200 I(d)

b. toward the light source

1800 1400 1000 600 200

d 5 15 25 35 45 55

69. a. $20,000, $80,000, $320,000; cost increases dramatically b. Cost ($1000)

15. D: x 僆 3 2, q 2 ; R: y 僆 34, q2

y 8 4

900

8

4

4

x

8

4

700 500

8

300 100

Percent 10 30 50 70 90

17. D: x 僆 1q, 92 ; R: y 僆 32, q2

c. as p S 100, C S q 71. a. 253 ft/sec (about 172 mph) b. approx. 791 ft 73. a. size 11 b. approx. 5 ft, 5 in. 75. a.

2

4 8

b. P 32.251 w0.246 c. about 63 days d. about 6.9 kg

75

y 8

4

4 8

19. D: x 僆 1q, q 2 ; R: y 僆 30, q2

y 8 4

30 8

4

4 4 8

10

x

8

4

8

x


SA-12

2/2/11

2:57 PM

Page SA-12

Student Answer Appendix 43. c1m2 e

y

5 4 3 2 1 54321 1 2 3 4 5

3.3m 7m 111

0 m 30 m 7 30

350

Cost of call in cents

21. D: x 僆 1q, q 2 ; R: y 僆 1q, 32 ´ 13, q2

1 2 3 4 5 x

300 250 200 150 100

(30, 99)

50

$2.11

10 20 30 40 50 60

Duration of call in min

4 8

4

4

8

4 8

25. discontinuity at x 1, redefine f 1x2 3 at x 1; c 3

y 8 4 8

4

4

8

x

4

x

0 2 45. C1a2 5 7 5

a 6 2 10 8 2 a 6 13 6 13 a 6 20 4 20 a 6 65 2 0 10 20 30 40 50 60 70 a 65 Age in years $38 47. a. C1w 12 17 <w 1 = 88 b. 0 6 w 13 c. 88¢ d. 173¢ e. 173¢ f. 173¢ g. 190¢ y 5 x 3 4 49. yes; h1x2 • 2x 1 3 6 x 6 2 2 5 x2 Cost of admission

y 8

⎞ ⎪ ⎬ ⎪ ⎠

23. discontinuity at x 3, redefine f 1x2 6 at x 3; c 6

4

8

1 x1 27. f 1x2 e 2 3x 6

2

2 4

4 x 6 2 x2

51. f(x) has a removable discontinuity at x 2; g(x) has a discontinuity at x 2 53. x 7, x 4 55. y 43 x 2

x2 2x 3 x 1 29. p1x2 e x1 x 7 1 31. Graph is discontinuous at x 0; f 1x2 1 for x 7 0; f 1x2 1 for x 6 0.

y


8 4 8

4

4

1. constant 3. directly, height, square 7. d kr 9. F ka 11. y 0.025x

x

8

4 8

t2 6t 0 t 5 33. a. S1t2 e b. S1t2 僆 3 0, 9 4 5 t 7 5 35. a. Year (0 S 1950) Percent 5 7.33 15 14.13 25 14.93 35 22.65 45 41.55 55 60.45 65 79.35

x

y

500

12.5

650

16.25

750

18.75


Number of stairs

13. w 9.18h; $321.30; the hourly wage; k $9.18/hr 192 15. a. k 192 b. 360 47 , S 47 h 320 c. 330 stairs d. S 331; yes 280 192 240

s(h) 冢 47 冣h

200 160 120 80 40 10 20 30 40 50 60 70 80 90100

Height h (in meters)

b. Each piece gives a slightly different value due to rounding of coefficients in each model. At t 30, we use the “first” piece: P1302 13.08. 0.09h 0 h 1000 200 C(h) 180 37. C1h2 e 160 0.18h 90 h 7 1000 140

C112002 $126 39. C1t2 e

4

2

120 100 80 60 40 20

0.75t 0 t 25 1.5t 18.75 t 7 25 C 1452 $48.75

17. A kS2 19. P kc2 21. k 0.112; p 0.112q2

h 200

60 55 50 45 40 35 30 25 20 15 10 5

600

1000

1400

p

45

226.8

55

338.8

70

548.8

23. a. Area varies directly as a side squared. b. A ks2 c. 75,000,000

C(t)

t 10

20

30

40

50

4000

Spending (in billions of dollars)

0 1.35t2 31.9t 152 0 t 12 41. S1t2 e 2.5t2 80.6t 950 12 6 t 22

$498 billion, $653 billion, $931 billion

q

450

(12, 340)

350 250 150 50 2

6 10 14 18 22

t (years since 1980)

7,500,000

x


2/2/11

2:57 PM

Page SA-13

Student Answer Appendix d.

SA-13

47. T 48 49. M 16 E; ⬇41.7 kg V ; 32 volunteers 51. D 21.6 1S; ⬇144.9 ft 53. C 8.5LD; $76.50 pp 55. C ⬇ 14.4 104 2 d1 22 ; about 222 calls

57. a. about 23.39 cm3 b. about 191% 59. a. M kwh2 1 L1 2 b. 180 lb 61. 6.67 10 7

e. k 6; A 6s2; 55,303,776 m2 25. a. Distance varies directly as time squared. c. 500

9y2

65. a. x 僆 1q, 42 ´ 14, 42 ´ 14, q2 4x2 b. x 僆 1q, 42 ´ 14, q 2 63.

b. D kt2

Making Connections, p. 188 1. g

3. a

5. d

7. a 9. b

11. h

13. c 15. f

Summary and Concept Review, pp. 189–193

1. D: x 僆 1q, q 2 , R: y 僆 3 5, q 2 , f 1x2c: x 僆 12, q 2 , f 1x2T: x 僆 1q, 22 , f 1x2 7 0: x 僆 1q, 12 ´ 15, q 2 , f 1x2 6 0: x 僆 11, 52 2. D: x 僆 33, q 2 , R: y 僆 1q, 04 , f 1x2c: none, f 1x2T: x 僆 13, q 2 , f 1x2 7 0: none, f 1x2 6 0: x 僆 13, q 2 3. D: x 僆 1q, q 2, R: y 僆 1q, q 2 , f 1x2c: x 僆 1q, 32 ´ 11, q 2 , f 1x2T: x 僆 13, 12 , f 1x2 7 0: x 僆 15, 12 ´ 14, q 2 , f 1x2 6 0: x 僆 1q, 52 ´ 31, 42 4. a. odd b. even c. neither d. odd y 5. zeroes: 16, 02 , (0, 0), 8 (10, 6) (6, 0) (9, 0) (3, 4) 4 min: 13, 82, 8 4 4 8 x 17.5, 22 (7.5, 2) 4 max: 16, 02, 13, 42

8

0

50 d.

8

e. k 16; d 16t2; about 3.5 sec; 121 ft 27. F dk2 29. S Lk 31. Y

12,321 Z

Z2

Y

37

9

74

2.25 1

111 33. w

3,072,000,000 r2

; 48 kg

35. l krt 37. A kh1B b2

39. V ktr2 41. C 6.75R S2

R

S

C

120

6

22.5

200

12.5

350

15

8.64

6. max: y 0.73 at x 0.48; min: y 0.73 at x 0.48 7. squaring function a. up on left/up on the right; b. x-intercept: 14, 02, 10, 02 ; y-intercept: (0, 0) c. vertex 12, 42 d. x 僆 1q, q2, y 僆 34, q2 8. square root function a. down on the right; b. x-intercept: (0, 0); y-intercept: (0, 0) c. initial point 11, 22 ; d. x 僆 31, q 2 , y 僆 1q, 2 4 9. cubing function a. down on left/up on the right b. x-intercept(s): 12, 02 ; y-intercept: (0, 2) c. inflection point: (1, 1) d. x 僆 1q, q2 , y 僆 1q, q 2 10. absolute value function a. down on left/down on the right b. x-intercepts: (1, 0), (3, 0); y-intercept: (0, 1) c. vertex: (1, 2); d. x 僆 1q, q 2 , y 僆 1q, 2 4 11. cube root function a. up on left, down on right b. x-intercept: (1, 0); y-intercept: (0, 1) c. inflection point: (1, 0) d. x 僆 1q, q2 , y 僆 1q, q 2 12. quadratic

13. absolute value y

10.5

y

8

8

4

43. E 0.5mv2; 612.50 J 45. a. cube root family b. answers will vary c. 0.054 or 5.4% d. A 1R 12 3 Amount A

Rate R

1.0

0.000

1.05

0.016

1.10

0.032

1.15

0.048

1.20

0.063

1.25

0.077

4

(3, 0) 8

4

4

8

x

8

4

4

(2, 5) 4

4

8

8

14. cubic

8

x

15. square root y

y 8

8

4

4

(9, 4) (5, 2) (6, 3) 8

4

4

4 8

(0, 1)

8

x

8

4

4 4 8

8

x


SA-14

2/2/11

2:58 PM

Page SA-14

Student Answer Appendix 20x x2 39. • 30x 20 2 6 x 4 40x 60 x 7 4 For 5 hr the total cost is $140.

y 8 4

(2, 0) 8

4

4

8

160

Cost

16. cube root

120 80 40

x

0

4

(2, 3)

5 4 3 2 1

54321 1 2 3 4 5

3 40. k 17.5; y 17.5 1x

b.

y

(1.5, 5)

(3, 3) (4, 1) (0.5, 1)

5 4 3 2 1

冢4, w冣 (1, 1)

54321 1 2 3 4 5

1 2 3 4 5 x

c.

y

5 4 3 2 1

8

8

4

4 4

4

8

8

x

54321 1 (1.5, 0.5)2 3 4 5

1 2 3 4 5 x

1 2 3 4 5 x

4

4

8

(0, 0.5)

105

196

7

2.88

0.343

12.25

38.75

1.25

17.856

729

157.5

24

0.6

48

w

700

1. a. D: x 僆 34, q 2 ; R: y 僆 33, q 2 b. f 112 ⬇ 2.2 c. f 1x2 6 0: x 僆 14, 32 ; f 1x2 7 0: x 僆 13, q 2 d. f 1x2c: x 僆 14, q2 ; f 1x2T: none e. f 1x2 31x 4 3 y y 2. 3. 5 5

5 4321 1 2 3 4 5

C(p)

600 500 400 300 200 100

p 20

40

60

80 100

100

5

7

4 3 2 1 5 4321 1 2 3 4 5

1 2 3 4 5 x

8

2 b. 570 km 5 x 4 4 6 x 3 b. R: y 僆 32, q 2 35. a. f 1x2 • x 1 3 1x 3 1 x 7 3 y 36. D: x 僆 1q, q2, 8 R: y 僆 1q, 82 ´ 18, q 2 , 4 h(x) discontinuity at x 3; 8 4 4 8 x define h1x2 8 at x 3 34. a. 88.4 hr,

(3, 6)

4 8

37. 4, 4, 4.5, 4.99, 3 13 9, 313.5 9 38. D: x 僆 1q, q 2 R: y 僆 3 4, q 2

1 2 3 4 5 x

4. max: y 8 at x 2; min: y 7 at x ⬇ 5.87 and y 7 at x ⬇ 1.87 5. I. a. square root b. x 僆 34, q2, y 僆 33, q2 c. 12, 02, 10, 12 d. up on right e. x 僆 12, q 2 f. x 僆 34, 22 II. a. cubic b. x 僆 1q, q2 , y 僆 1q, q 2 c. 12, 02, 10, 12 d. down on left, up on right e. x 僆 12, q 2 f. x 僆 1q, 22 III. a. absolute value b. x 僆 1q, q2 , y 僆 1q, 44 c. 11, 02, 13, 02, 10, 22 d. down on left, down on right e. x 僆 11, 32 f. x 僆 1q, 12 ´ 13, q 2 IV. a. quadratic b. x 僆 1q, q 2 ; y 僆 35.5, q2 c. 10, 02, 15, 02, 10, 02 d. up on left, up on right e. x 僆 1q, 02 ´ 15, q 2 f. x 僆 10, 52 y y 6. 7. 8

(0, 5) 4

Domain of f (x) is 1q, q 2 ; Domain of g(x) and h(x) is 30, q2 .

z

Practice Test, pp. 193–195

4 3 2 1

b.

3

v

42. t 160 43. 4.5 sec

x

8

32. a. ⬇$32,143; $75,000; $175,000; $675,000; cost increases dramatically c. as p S 100, C S q

41. k 0.72; z 0.72v w2

216

(0, 3.25)4

8

33.

8

y

冢1, w冣

(1, 0) 4

6

x

y

18. 54, 106 19. 57, 36 20. 55, 86 21. 54, 16 22. 1q, 62 ´ 12, q 2 23. [4, 32] 24. { } 25. { } 26. 1q, q 2 27. 3 2, 64 28. 1q, 2 4 ´ 3 10 29. a. 0 r 2.5 0 1.7 3 , q2 b. highest: 4.2 in., lowest: 0.8 in. 30. 31. y y

8

4

Hours

8

17. a.

2

(3, 2)

(2, 3) 8

4

4

8

8

x

4

4

4

4

4

8

8

22 8. 1q, 20 3 2 ´ 1 3 , q2 y 11. 5 x 3

9. 34, 02 12. 13.5 sec

8

10. x 0.75

冢0, s冣 y 0

5

5

13. a. 1q, q2 15. a.

5 x

b. 30, q 2

c. 3 0, q 2

14. VA: x 2; HA: y 1

25

y

8 4

8

4

4

8

x

x

0.1

1.1

4 8

3 b. S(t) 17.27 t 2.50

c. 3.05 mm

d. 0.95 sec


2/3/11

5:55 PM

Page SA-15

SA-15

Student Answer Appendix 16. a 1, b 4 17. 1617 KWH/year y 18. a. 4, 4, 6.25 b.

Connections to Calculus Exercises, p. 202 3 6 3 sec, t sec 5. If the difference between x and 5 2 2 is very small, then the difference between h(x) and 6 is very small: 3 if ` x ` 6 ␦, then 冟h1x2 6冟 6 ⑀. 7. If the difference between x and 2 2 is very small, then the difference between w(x) and 14 is very small: if 冟x 2冟 6 ␦, then 冟w1x2 14冟 6 ⑀.

4 8

(2, 4)

4

4

8

(2, 4) 4

x

8

19. M kd3 1 P12 2, approx. 2.2788 108

20. 520 lb

CHAPTER 3

Strengthening Core Skills, p. 196 1. k

1 3 2

or

6


1 26

3

Cumulative Review Chapters 1–2, pp. 197–198 1 7 1. f 122 23, f a b 2 4 7. a.

1 3

b.

9.

3. 29.45 cm

5. x 1

3 5

y 12 x 72

y

11.

5

y

8 4 ⫺8

⫺4

4

8

⫺5

x

5 x

⫺4 ⫺8

⫺5

13. a. D: x 僆 1q, 84 , R: y 僆 3 4, q2 b. 5, 3, 3, 1, 2 c. 12, 02 d. f 1x2 6 0: x 僆 12, 22 ; f 1x2 7 0: x 僆 1q, 22 ´ 32, 8 4 e. min: (0, 4), max: (8, 7) f. f 1x2c: x 僆 10, 82 ; f 1x2T: x 僆 1q, 02 y

8 4 8

4

4

8

x

4 8

15. a.

3. t

1. 4.24 in.

8

x7 1x 521x 22

b.

b2 4ac 4a2

17. x2 2

19. center 13, 62, r 3 21. W 31 cm, L 47 cm

5 23. a. x 4 b. x 5, 13, 13 3 ,2 25. P 15 197 units ⬇ 24.8 units. No, it is not a right triangle. 52 1 1972 2 102 27. 3.1

4.7

4.7

1. 3 2i 3. 2, 3 12 5. (b) is correct. 7. a. 12i b. 7i c. 3 13 d. 6 12 9. a. 3i 12 b. 5i 12 c. 15i d. 6i 2 13 3 12 11. a. i 119 b. i 131 c. i d. i 5 8 13. a. 1 i b. 2 i 13 15. a. 4 2i b. 2 i 12 17. a. 5 0i; a 5, b 0 b. 0 3i; a 0, b 3 12 12 19. a. 0 18i; a 0, b 18 b. 0 i; a 0, b 2 2 21. a. 4 5i 12; a 4, b 5 12 b. 5 3i 13; a 5, b 313 7 7 12 i; a 4 8 25. a. 19 i b. 2

7 7 12 1 110 1 110 ,b i; a , b b. 4 8 2 2 2 2 4i c. 9 10i 13 1 27. a. 3 2i b. 8 c. 2 8i 29. a. 2.7 0.2i b. 15 i 12 1 c. 2 i 31. a. 15 0i b. 16 0i 33. a. 21 35i 8 b. 42 18i 35. a. 12 5i b. 1 5i 37. a. 41 b. 74 39. a. 11 b. 17 41. a. 5 12i b. 7 24i 36 43. a. 21 20i b. 7 6i 12 45. a. 4 5i; 41 b. 3 i 12; 11 47. a. 7i; 49 b. 12 23 i; 25 49. no 51. yes 36 53. yes 55. no 57. yes 59. verified 61. a. 1 b. 1 c. i 2 4 21 14 15 10 d. i 63. a. 0 i b. 0 i 65. a. i b. i 7 5 13 13 13 13 2 3 67. a. 1 i b. 1 i 69. a. 113 b. 5 c. 111 4 3 71. A B 10 AB 40 73. 17 5i2 75. 125 5i2 V 23. a.

77. 1 74 i2

79. a. 1a bi21a bi2 a2 abi abi 1bi2 2 a2 b2 112 a2 b2 ✓ b. 1x 6i2 1x 6i2 c. 1m i 132 1m i 132 d. 1n 2i 1321n 2i 132 e. 12x 7i2 12x 7i2 83. 5.6 hr (5 hr 36 min) 85. John

81. 8 6i

Exercises 3.2, pp. 230–235 1. exact, approximate 3. discriminant, 2

3.1 29.

7. x 4 or 3

10

9. x 3 2 or 1


11. x 1 or 2

13. x ⬇ 4.19 or 1.19

15. x ⬇ 1.61 or 3.11

17. x ⬇ 1.14 or 2.64

19. { }

21. m 4

23. y 2 17; y ⬇ 5.29 25. no real solutions 10

10

27. x

121 4 ;

x ⬇ 1.15 29. n 9; n 3

31. w 5 13; w ⬇ 3.27 or w ⬇ 6.73 33. no real solutions 35. m 2 3 12 37. 9; 1x 32 2 7 ; m ⬇ 2.61 or m ⬇ 1.39

10

39. 94 ; 1n 32 2 2 41. 19 ; 1p 13 2 2 43. x 1; x 5 45. p 3 16; p ⬇ 5.45 or p ⬇ 0.55

47. p 3 15; p ⬇ 0.76 or p ⬇ 5.24 49. m

3 2

113 2 ;

m ⬇ 0.30 or m ⬇ 3.30


SA-16 51. 53. 55. 57. 59.

2/2/11

3:04 PM

Page SA-16


n 52 3 15 2 ; n ⬇ 5.85 or n ⬇ 0.85 a. x 12 or x 4 b. x 0.5 or x 4 a. n 3 or n 3 b. n 3 or n 1.5 2 a. p 38 141 b. p ⬇ 1.18 or p ⬇ 0.43 8 a. m 72 133 b. m ⬇ 6.37 or m ⬇ 0.63 2

15. right 74 , stretched vertically, down 25 8 y

(q, 0) 8

73. 77. 79. 81. 83. 85.

8

115. x 僆 3 17, 174 53 4

133 2 ,

32

133 2 4

Exercises 3.3, pp. 244–248 1. 25 3. 0, f(x) 5. Answers will vary. 2 7. left 2, down 9 9. right 1, reflected across x-axis, up 4 y

y

8

8

4

(5, 0) 8

(0, 3) 4 (1, 0)

(1, 0) 4

4 4

8

x

8

4

y

y 16

4

(0, 7) 8 (0.7, 0)

(0.6, 0) 4 4

(1, 8) 8

x

13. right 2, reflected across x-axis, stretched vertically, up 15

8

4

8

8

11. left 1, stretched vertically, down 8

8

(3, 0)

4

(2, 9) 8

(2.6, 0)

(1, 4)

4

(0, 5)

(0, 5)

8

x

8

4

(2, 15) (4.7, 0) 4

8 16

21. left 1, down 7 y

4

(4.6, 0) 8

(3.6, 0) 8

x

8

x

(1.6, 0)

4

4 4

(e, *)

(1, 7)

8

23. right 2, reflected across x-axis, up 6

8

(0, 2) 4

冢3, e冣

(4.4, 0) 4

8

8

x

4

4 4

8

8

27. right 52 , reflected across x-axis, stretched vertically, up 11 2

y

(0, 7)

4

(0, 3)

(4.2, 0)

4

x

8

冢e, y冣

(0.8, 0) 8

8

29. right 32 , stretched vertically, down 6

y

4

(0, 7)

4

(0.4, 0) 4

8

x

y

y

4

8

(0, 6)

25. left 3, compressed vertically up 52

8 (2, 6)

8

x

8

4 8

8

8

4 4

´ 3 1, q2 121. x 僆 1q, q2 123. { } x 僆 1q, x 僆 1q, 52 ´ 15, q 2 127. { } 129. x 僆 1q, q2 x 僆 1q, q 2 133. 13, 12 135. 1q, 3 2 4 ´ 3 2, q 2 1q, 1.32 ´ 11.3, q 2 139. {2.9} 141. { } 143. 1q, q2 x 僆 1q, 5 4 ´ 3 5, q2 147. x 僆 1q, 0 4 ´ 35, q 2 { } 151. a 153. b v 1v2 64h 155. t 157. t 3 1138 sec, t ⬇ 8.87 sec 2 32 2 159. a. P x 120x 2000 b. 10,000 161. t 2.5 sec, 6.5 sec 163. 36 ft, 78 ft 165. 30,000 ovens 167. x ⬇ 13.5, or the year 2008 169. a. 7x2 6x 16 0 b. 6x2 5x 14 0 c. 5x2 x 6 0 3 171. z 2i; z 5i 173. z i; z 2i 4 175. z 1 i; z 13 i 177. a. P 2L 2W, A LW b. P 2␲r, A ␲r2 c. P c h b1 b2, A 12 h1b1 b2 2 d. P a b c, A 12 bh 179. 700 $30 tickets; 200 $20 tickets 119. 125. 131. 137. 145. 149.

4 4

8

8

(s, 0)

4

4 (0, 2)

113. x 僆 11, 72 2

117. x 僆 332

8

(74, 258)

(0.4, 0)

107. x 54 3 17 109. x 僆 10, 42 4 i

111. x 僆 1q, 5 4 ´ 31, q2

(0, 6) (3, 0)

x

y

0.80i

89. p 23 1394 6 ; p ⬇ 3.97 or p ⬇ 2.64 91. two rational; factorable 93. two nonreal 95. two rational; factorable 97. two nonreal 99. two irrational 101. one repeated; factorable 103. x 32 12 i 13 2 i

8

19. right 52 , down 17 4

87. a 3 3 12 2 ; a ⬇ 0.88 or a ⬇ 5.12

105. x 12

4

y

8 4

(3, 0)

4

15 2 ; n ⬇ 2.12 or n ⬇ 0.12 2 w 3 or w 1 2 71. n 32 m 32 16 2 i; m ⬇ 1.5 1.12i 123 4 1 w 5 or w 2 75. a 6 6 i; a ⬇ 0.16 p 35 2 16 5 ; p ⬇ 1.58 or p ⬇ 0.38 1 121 w 10 10 ; w ⬇ 0.56 or w ⬇ 0.36 a 34 131 4 i; a ⬇ 0.75 1.39i 3 12 p 1 2 i; p ⬇ 1 2.12i 12 w 1 3 3 ; w ⬇ 0.14 or w ⬇ 0.80

(0, 3)

4

4

65. n 1 69.

12

(C, 121 ) 12

8

61. x 6 or x 3 63. m 52 67.

17. left 67 , reflected across x-axis, stretched vertically, up 121 12

4

8

8

x

4

(2.7, 0) 4

4

(0.3, 0) 4

8

8

8

x

冢w, 6冣

31. left 3, compressed vertically, down 19 2 y 8 4

(7.4, 0) 8

(1.4, 0) 4

冢3, p冣

4 8 4 (0, 5)

x

8

33. y 11x 22 2 1 35. y 11x 22 2 4 3 37. y 1x 22 2 3 39. i. x 3 15 ii. x 4 13 2 1 iii. x 4 214 iv. x 2 12 v. t 2.7, t 1.3 vi. t 1.4, t 2.6 41. a. 10, 66,0002; when no cars are produced, there is a loss of $66,000. b. (20, 0), (330, 0); no profit will be made if fewer than 20 or more than 330 cars are produced. c. 175 d. $240,250 43. a. $2 b. $44 c. $8800 d. $23; $44,100 45. 6000; $3200 47. a. h1t2 16t2 240t 544 b. 544 ft; that is when the fuel is exhausted. c. 1344 ft d. 1344 ft e. It is coming back down. f. 1444 ft g. 17 sec 49. a. 14.4 ft b. 41 ft c. 48.02 ft d. 90 ft 51. a. 25 ft b. approx. 3.43 sec c. 67.25 ft 53. a. 2500 ft2, 50 ft 50 ft b. 5000 ft2, 50 ft 100 ft 55. a. approx. 29.5" wide by 18.7" long b. approx. 930 in2 57. a. $1.25 each, $781.25 b. about $0.85 each 59. Answers will vary. 7 1 2 7 61. y ax b 63. m 43 , y-intercept (0, 3) 18 2 2 3 65. g1x2 1 x13


2/3/11

10:54 AM

Page SA-17


Mid-Chapter Check, p. 249 1. sum 4, product 13; both yield real numbers

2. i

11 2i2 2 211 2i2 5 0 1 4i 4i2 2 4i 5 0 1 142 2 5 0 0 0✓ Yes. 7 1 4. x 1 and 3 5. x 2 253 6. x 4 417 7. 1q, 1 4 ´ 3 0.37, q2 8. a. about 29.6 in. b. about 7.17 ft c. yes; 3 7.17 7 21.5 3.

9.

x 2

5

10. f 1x2 49 1x 32 2 2

y

SA-17

¢F 9.8 35. ¢y 37. ¢y 39. ¢m ¢x 39 ¢x 1 ¢A 41. ¢r ⬇ 37.70 43. a. 14,570 ft b. 14,174 ft c. 198 ft/sec d. h1102 13,340 ft; h1122 12,624 ft; average rate of change 358 ft/sec, almost twice as fast 45. a. R 43.07t2 976.53t 126.8 b. 4598 c. t ⬇ 8.26 days (early in the ninth day) d. about 5408 participants 47. a. T 5.92n2 83.13n 349.86 b. about 82 sec c. 16 tourists d. about 58 sec with seven tourists 50 49. a. ¢weight ¢time 1 , positive, 50 g are gained each week b. 25th to 29th: ¢w 50 ¢w 250 ¢t 1 ; 32nd to 36th: ¢t 1 ; the weight gain is five time greater in the later weeks. 51. a. 7 b. 7 y c. They are the same. 10 8 (2, 8) d. Slopes are equal. 6 4

(2, 2) (4, 0)

(1, 1) 2

(0, 0)

7

54321 2 4 6 (2, 8) 8 10

3 x

5

Reinforcing Basic Concepts, pp. 249–250 7 5 b 112 ✓ 2 2 a 7# 7 c 112 ✓ 2 2 a 2 312 4 b 2 312 Exercise 2: ✓ 2 2 2 a 2 3 12 2 3 12 14 7 c # ✓ 2 2 4 2 a b Exercise 3: 15 2 13i2 15 2 13i2 10 ✓ a c 15 213i215 2 13i2 25 12 37 ✓ a b b b Exercise 4: x1 x2 since radical terms sum to 0, and 2a b ; a 2a a 2 2 2 2 2 c b 2b 4ac b b 4ac 4ac c x1x2 since a b a b 2 2 a 2a 2a a 4a 4a2 4a Exercise 1:

Exercises 3.4, pp. 257–261 1. quadratic, negative 3. average, slope, secant line 5. Answers will vary. 7. a. y 1.45x2 2.59x 4.55 b. 4.1475 c. x 6.446 9. a. y 0.851x2 3.153x 64.428 b. 65.037 c. x 7.623 11. a. y 3.485x2 26.424x 60.505 b. 10.422 c. x 1.746 or 5.836 13. a. y 0.113x2 5.796x 61.583 b. 12.882 c. x 7.368 or 44.019 15. 0 17. 1.4 19. 0.2 21. 1.6 23. 0.1 25. 1.2 27. a. 48 ft/sec b. 32 ft/sec c. 16 ft/sec d. 32 ft/sec 29. a. 0 m/sec b. 0 m/sec c. 4.9 m/sec d. 4.9 m/sec 31. Answers will vary. 500 450 400 350 300 250 200 150 100 50 0

(5, 400) (4, 384) (7, 336) (3, 336) (2, 258)

33. Answers will vary. 80

(2.5, 57.725)

70

(3.5, 62.625) (4.5, 57.725)

60 50

(2, 51.6)

40 30

(5, 51.6) (1, 32)

(6, 32)

20 10 0

1

2

3

53. a. 176 ft b. 320 ft c. 144 ft/sec d. 144 ft/sec; The arrow is going down. 55. a. 17.89 ft/sec; 25.30 ft/sec b. 30.98 ft/sec; 35.78 ft/sec c. Between 5 and 10. d. 1.482 ft/sec, 0.96 ft/sec 57. a. about 3569 in3 b. about 3800 in3 c. ¢r ¢t 0.74 in./sec when 3 t 僆 30, 1 4 and 0.04 in./sec when t 僆 3 18, 194 ; ¢V ¢r ⬇ 64.745 in /in. when t 僆 30, 1 4 and 22.602 when t 僆 3 18, 19 4 . Answers will vary. 59. 1 12 , q 2 |x| x 6 1 61. a. 1q, 12 ´ 12, q 2 b. { } 63. g1x2 e 1 x1 2 1x 32


1. 1f g21x2 , A 傽 B 3. intersection, g(x) 5. Answers will vary. 7. a. x 僆⺢ b. f 122 g122 13 9. a. h1x2 x2 6x 3 b. h122 13 c. they are identical 11. a. x 僆 3 3, q 2 b. h1x2 1x 3 2x3 54 c. h142 75, 2 is not in the domain of h. 13. a. x 僆 35, 3 4 b. r1x2 1x 5 13 x c. r122 17 1, 4 is not in the domain of r. 15. a. x 僆 34, q 2 b. h1x2 1x 412x 32 c. h142 0, h1212 225 17. a. x 僆 31, 7 4 b. r 1x2 1x2 6x 7 c. 15 is not in the domain of r, r 132 4 19. a. x 僆 1q, 42 ´ 14, q 2 b. h1x2 x 4, x 4 21. a. x 僆 1q, 42 ´ 14, q 2 b. h1x2 x2 2, x 4 23. a. x 僆 1q, 12 ´ 11, q 2 b. h1x2 x2 6x, x 1 x1 25. a. x 僆 1q, 52 ´ 15, q 2 b. h1x2 ,x5 x5 2x 3 27. a. x 僆 1q, 22 b. r 1x2 12 x 15 c. 6 is not in the domain of r, r 162 2 x5 29. a. x 僆 15, q 2 b. r1x2 1x 5 c. r162 1, 6 is not in the domain of r. x2 36 13 c. r 162 0, r162 0 , q b b. r 1x2 2 12x 13 2x 4 33. a. h1x2 b. x 僆 1q, 32 ´ 13, q 2 c. x 2, x 0 x3 35. sum: 3x 1, x 僆 1q, q 2 ; difference: x 5, x 僆 1q, q2 ; product: 2x2 x 6, x 僆 1q, q2 ; 2x 3 quotient: , x 僆 1q, 22 ´ 12, q 2 x2 2 37. sum: x 3x 5, x 僆 1q, q 2 ; difference: x2 3x 9, x 僆 1q, q 2 ; product: 3x3 2x2 21x 14, x 僆 1q, q2 ; 31. a. x 僆 a

1 2 3 4 5 6 7 8 9 10

4

5

6

7

8

(1, 1) 1 2 3 4 5 x

quotient:

x2 7 2 2 , x 僆 aq, b ´ a , q b 3x 2 3 3


SA-18

2/2/11

3:06 PM

Page SA-18


39. sum: x2 3x 4, x 僆 1q, q 2 ; difference: x2 x 2, x 僆 1q, q2 ; product: x3 x2 5x 3, x 僆 1q, q 2 ; quotient: x 3, x 僆 1q, 12 ´ 11, q2 41. sum: 3x 1 1x 3, x 僆 3 3, q2 ; difference: 3x 1 1x 3, x 僆 3 3, q2 ; product: 13x 12 1x 3, x 僆 3 3, q2 ; 3x 1 quotient: , x 僆 13, q2 1x 3 2 43. sum: 2x 1x 1, x 僆 3 1, q 2 ; difference: 2x2 1x 1, x 僆 3 1, q2 ; product: 2x2 1x 1, x 僆 3 1, q 2 ; 2x2 quotient: , x 僆 11, q 2 1x 1 7x 11 45. sum: , x 僆 1q, 22 ´ 12, 32 ´ 13, q2 ; 1x 321x 22 3x 19 difference: , x 僆 1q, 22 ´ 12, 32 ´ 13, q2 ; 1x 321x 22 10 product: , x 僆 1q, 22 ´ 12, 32 ´ 13, q2 ; 1x 321x 22 2x 4 quotient: , x 僆 1q, 22 ´ 12, 32 ´ 13, q 2 51x 32 47. a. 6000 b. 3000 c. 8000 d. C192 T192 ; 4000 49. a. $1 billion b. $5 billion c. 2003, 2007, 2010 d. t 僆 12000, 20032 ´ 12007, 20102 e. t 僆 12003, 20072 f. R152 C152 ; $4 billion 1 51. a. 4 b. 0 c. 2 d. 3 e. f. 6 g. 3 h. 1 i. 1 3 2 j. undefined 53. h1x2 3 x 4 55. h1x2 4x x2 57. a. 1q, q2 b. 1q, 22 ´ 12, q2 59. a. 31, q 2 b. 11, q 2 61. a. 3 2, 34 b. 3 2, 32 63. a. 3 4, q 2 b. 34, q 2 65. a. 1q, q2 b. 1q, 42 ´ 14, q2 67. a. 3 4, 34 b. 34, 32 69. a. V 400 400e0.08t b. f 1t2 400, g1t2 400e0.08t, V112 ⬇ 400 369 31 ft/sec, V122 ⬇ 400 341 59 ft/sec, V1202 ⬇ 400 81 319 ft/sec c. 400 ft/sec 71. a. P1x2 12,000x 108,000 b. nine boats must be sold 73. a. P1n2 11.45n 0.1n2 b. $123 c. $327 d. C11152 7 R11152 75. km 115 77. Anywhere between km 115 and km 199. Answers will vary. 79. a. 4 sec b. about 494 ft 81. a. 1995 to 1996; 1999 to 2004 b. 30; 1995 c. 20 seats; 1997 d. The total number of seats in the senate (50); the number of additional seats held by the majority 83. 67 a. x 23 , where Y1 Y2; 67 b. x 4, where Y1 0; 68 a. x 0, where Y1 Y2; 68 b. x 1.2 or 3, the zeroes of Y1 and Y2. 85.

x

148 9

87. a. 6x 5y 13 b. d 261

23. h132 2, h1 222 5 h1 12 2 5.5, h152 ERR; g(5) is not in the domain of f

25. a. 1 f ⴰ g2 1x2 : For g(x) to be defined, x 0. 2g1x2 5 For f 3g1x2 4 , g1x2 3 so x . g1x2 3 3 domain: all real numbers except x 3 and x

b. 1g ⴰ f 2 1x2 : For f (x) to be defined, x 3. 5 , f 1x2 0 so x 0. For g 3 f 1x2 4 f 1x2 domain: all real numbers except x 3 and x 0 10 5x 15 ; 1g ⴰ f 2 1x2 c. 1 f ⴰ g2 1x2 ; the domain of a 5 3x 2x composition cannot always be determined from the composed form 27. a. 1 f ⴰ g2 1x2 : For g(x) to be defined, x 5. 4 , g1x2 0 and g(x) is never zero For f 3g1x2 4 g1x2 domain: {x| x 5} b. 1g ⴰ f 2 1x2 : For f(x) to be defined, x 0. 4 1 , f 1x2 5 so x . For g 3 f 1x2 4 f 1x2 5 5 4 domain: all real numbers except x 0 and x 5 x c. 1 f ⴰ g2 1x2 4x 20; 1g ⴰ f 2 1x2 ; the domain of a 4 5x composition cannot always be determined from the composed form 29. a. 41 b. 41 31. g1x2 1x 2 1, f 1x2 x3 5 33. p1x2 21x 42 2 3, q1x2 12x 72 2 1 35. a. 2 b. 2 c. 2 d. 1 e. 1 f. 1 g. 3 h. ⬇0.5 37. a. f 1x2: x 1, g1x2: x 2 c. h1x2: x 僆 1q, 22 ´ 31, q 2 39. 2 41. 2x h 43. 2x 2 h 45. x1x2 h2 ¢g ¢g ¢g 2x 2 h b. 3.9 c. 3.01 47. a. ¢x ¢x ¢x y d. The rates of change have opposite sign, with the 7 6 secant line to the left being slightly more steep. 5 4 3 2 1 654321 1 2 3

¢g 49. a. d.

19. h132 6, h1 222 ⬇ 9.071 h1 12 2 2.75, h152 50 21. h132 1, h1 222 ⬇ 145.91 h1 12 2 64, h152 ERR; x 5 is not in the domain of g

¢x

1 2 3 4 x

¢g 12.61 c. ⬇ 0.49 ¢x ¢x Both lines have a positive slope, but the line at x 2 is much steeper.

3x2 3xh h2 y

20 16 12 8 4

Exercises 3.6, pp. 287–291 1. composition 3. domain, g(x) 5. Answers will vary. 7. 0, 0, 4a2 10a 14, a2 9a 9. a. h1x2 12x 2 b. H1x2 2 1x 3 5 c. D of h(x): x 僆 3 1, q 2 ; D of H(x): x 僆 3 3, q2 11. a. h1x2 13x 1 b. H1x2 3 1x 3 4 c. D of h(x): x 僆 3 13 , q2 ; D of H(x): x 僆 3 3, q 2 13. a. h1x2 x2 x 2 b. H1x2 x2 3x 2 c. D of h(x): x 僆 1q, q 2 ; D of H(x): x 僆 1q, q 2 15. a. h1x2 x2 7x 8 b. H1x2 x2 x 1 c. D of h(x): x 僆 1q, q2 ; D of H(x): x 僆 1q, q 2 17. a. h1x2 冟3x 1冟 5 b. H1x2 3冟x冟 16 c. D of h(x): x 僆 1q, q2 ; D of H(x): x 僆 1q, q2

5 3

54321 4 8 12 16 20

¢g

b.

1 2 3 4 5 x

2x h ; x2 1x h2 2 ¢j 30.50, 0.51 4: ⬇ 15.5; ¢x ¢j 31.50, 1.51 4: ⬇ 0.6; ¢x Answers will vary. ¢g 3x 2 3xh h2; 53. ¢x ¢g 3 2.01, 2.00 4 : ⬇ 12.1; ¢x ¢g ⬇ 0.5; 3 0.40, 0.414: ¢x Answers will vary. ¢j

51.

¢x


2/2/11

3:08 PM

Page SA-19

SA-19

Student Answer Appendix 55. a. f 3g1x2 4 1x 22 2 41x 22 3 x2 4x 4 4x 8 3 x2 1✓ b. verified 57. h1x2 x 2.5; 10.5 59. a. 4160 b. 45,344 c. M1x2 453.44x; yes 61. a. 6 ft b. 36␲ ft2 c. A1t2 9␲t 2; yes 63. a. L102 500 lions and H15002 400 hyenas b. H3 L1x2 4 400 0.0075x, 1H ⴰ L2116,0002 520 hyenas c. prior to an increase of 30,000 ¢d ¢d ⬇ 0.2 b. ⬇ 0.04 65. a. ¢h ¢h c.

48 42 36 30 24 18 12 6 6 12

y

As height increases you can see farther, but the sight distance increases at a slower rate. x

21.

8 4

(5, 0)

(6, 0)

8

108642 2 2 4 6 8 10 x 4 6 8 10 (3, 9)

23.

y

22.

y

10 8 6 4 2 (0, 0)

(3, 0) 4

24.

y

x

y 8

(0, 3) 4

4

(2, 1) 4

8

8

8

8

4

(4, 1)4

4

(2.7, 0)

(0.3, 0) 8

8

x

4

4

4

4

8

8

(0, 5)

25. a. 0 ft b. 108 ft 26. $3.75; 3000 27. a.

c. 2.25 sec

8

x

冢w, 6冣

d. 144 ft, t 3 sec

400

100 200 300 400 500

¢d ¢d ⬇ 15, June: ⬇ 3, 5 times faster ¢t ¢t ¢d 6 0b; 5000 units/month b. t 6.75, late June c. decreasing a ¢t 69. Answers will vary. 1 ¢y 1 ¢y ⬇ 3.91; For y 2 : ⬇ 15.53 71. a. For y : x ¢x x ¢x 1 ¢y ⬇ 1.54; b. Less—decrease is more gradual; For y : x ¢x 1 ¢y ⬇ 3.83 For y 2 : x ¢x y y y 73. a. b. c. 67. a. March:

8 4 8

4

8

f(x) 4

4 8

8

x

4

0

50

4

4

8

x

8

4

h(x) 4

4

4

4

8

8

8

8

x

0 75. y

, quadratic

b. A1x2 2.144x2 1.010x 25.847, about $48.2 billion c. about $460.2 billion d. year 16 S 2016 28. a. 2000

8

g(x)

15

80

32 x

Making Connections, p. 292 1. b

3. h

5. d

7. a

9. g

11. f

13. c

15. a

Summary and Concept Review, pp. 292–297 1. 6i12 2. 24i 13 3. 2 i 12 4. 3i 12 5. i 6. 21 20i 7. 2 i 8. 5 7i 9. 13 10. 20 12i 15i2 2 9 34 11. 15i2 2 9 34 2 25i 9 34 25i2 9 34 25 9 34✓ 25 9 34✓ 12. 12 i 152 2 412 i 152 9 0 4 4i 15 5i2 8 4i 15 9 0 5 152 0 ✓ 12 i 152 2 412 i 152 9 0 4 4i 15 5i2 8 4i 15 9 0 5 152 0✓ 13. a. x 5 or x 2 b. x 5 or x 5 c. x 53 or x 3 d. x 2 or x 2 or x 3 14. a. x 3 b. x 2 15 c. x i 15 d. x 5 15. a. x 3 or x 5 b. x 8 or x 2 c. x 1 110 2 ; x ⬇ 2.58 or x ⬇ 0.58 d. x 2 or x 13 16. a. x 2 i 15; x ⬇ 2 2.24i b. x 32 12 c. x 32 12 i 2 ; x ⬇ 2.21 or x ⬇ 0.79 17. a. 1q, 22 ´ 13, q2 b. 3 1, 1 4 c. 1q, q 2 18. a. 3 4, 1 4 b. 1q, 52 ´ 14, q 2 c. {2} 19. a. 1.3 sec b. 4.7 sec c. 6 sec 20. a. 0.8 sec b. 3.2 sec c. 5 sec

200

, quadratic

b. Fd 1x2 0.35x 0.3x, 1242 units c. about 82 mph 2.95 11.93 ¢N ¢N ⬇ 29. a. , 295 stores per year b. for 2000 to 2002: , ¢t 1 ¢t 1 1193 stores per year; 295 4 ⬇ 1180 ✓ c. for 2002 to 2004: ¢N 13.42 ¢N 13.20 ⬇ ⬇ , 1342 stores per year; for 2006 to 2008: , 1320 ¢t 1 ¢t 1 stores per year, very close 30. a. 20 ft3 b. approx. 18.251 ft3 c. approx. 1.749 ft3/sec d. approx. 0.149 ft3/sec e. t ⬇ 22.4 sec 31. a2 7a 2 32. 147 33. x 僆 1q, 23 2 ´ 1 23 , q 2 34. a. 4 b. 6 c. 1 d. 14 5 35. a. P1x2 84.95n 10.002n2 20n 30,0002 2

0.002n2 64.95n 30,000 b. $3700 c. $344,750 d. 456 36. 4x2 8x 3 37. 99 38. x; x 39. f 1x2 1x 1; g1x2 3x 2 1 40. f 1x2 x2 3x 10; g1x2 x3 41. A1t2 ␲12t 32 2 42. a. 0 b. 7 c. 7 d. 2 e. 4 43. 2x 1 h; 3.01


1. a. 10, 11 b. 1q, 12 ´ 11, q 2 c. 1q, 73 4 ´ 33, q 2 d. { } 2 2 2. x 5i 3. x 1 i 23 4. x 23 , x 6 5. a. t 5 (May) b. t 9 (Sept.) c. July; $3000 more 6. a. 43 7. a. 1 b. i 13 c. 1 8. 32 32 i 9. 34 10. 12 3i2 2 412 3i2 13 0 4 12i 9 8 12i 13 0 13 13 0 00✓

15 3 i

b. i


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12 2

11. a. x 5

b. x 54

13 3

12. a. x 3

5.

17 4 i

b. x 1 3i

13. a. f 1x2 1x 52 2 9

8

410 20 30 x 3 40 50

y

y

20 16 12 8 4

(6, 21) (3, 3)

b. g1x2 12 1x 42 2 8 18 16 (0, 16) 14 12 10 8 6 4 2

(5, 9) (8, 0) (4, 8)

42 4 2 4 6 8 10 12 14 16 x 8 (2, 0) 12 16 (0, 16) 20

108642

14. 12, 02, y 2x2 4x 15. 16. a. 1300

6.

y 50 40 30 20 (0, 21) 10 4

4 8

8

4

冢#, ≥冣 4

8

x

4 8

x#

3. a. 1x 12 1x2 x 12 b. 1x 321x 221x 22 1009 11 5. all reals 7. verified 9. y x ; 39 min, driving time 60 60 increases 11 min every 60 days 11. Month 9 13. a. f (x) b. g(x) y 15. 5 R1R2 R2

1. R R1

2 4 6 8 10 x

b. 49 ft

c. 14 sec

5

14

, quadratic

0

5 x

5

b. y ⬇ 6.68x2 3.48x 176.30 1300

0

x

冢r, 8冣

8

Cumulative Review Chapters 1–3, pp. 301–302

a. 40 ft, 48 ft

0

y (0, 8)

17. X 63 19. a. f 142 1, g122 4, 1 f ⴰ g2 122 1 b. g142 0, f 182 4, 1g ⴰ f 2182 0 g 4 c. 1 fg2 102 122142 8, 1 f 2102 2 2 d. 1 f g2112 3 5 2, 1g f 2192 2 2 0 21. 2x2 9.2x 14.5 0 and 1.2x2 5.52x 8.7 23. y 3x2 1.5x 7 25. x 僆 1q, 0.22 ´ 10.2, 0.22 ´ 10.2, q 2

14

0 c. about 396,000; about 4,264,000 17. a. 4750 books per year b. approx. 11,267 books per year c. 2400 books per year and 18,100 books per year 18. 3x 1; x 僆 3 13 , q2 19. a. No, new company and sales should be growing b. 15 for [4, 5]; 19 for [5, 6] c. ¢S ¢t 4t 3 2h. For small h, sales volume is approximately units 37,000 unit units in month 10, 69,000 in month 18, and 93,000 in month 24 1 mo 1 mo 1 mo 20. a. V1t2 43 ␲1 1t2 3 b. 36␲ in3

Strengthening Core Skills, pp. 300–301 1.

x 1

2.

y 8

(1 2兹2, 0)

x e (5, 9)

4

冢e, %冣 8

4

4

8

x

8

4

4 4

4 4 8

8

x3

y

8

x

1 2 3 4 5 6 7 8 9 10 t

(5, 8) x5

4

(5 2兹2, 0)

(5 2兹2, 0)

(3, 2)

9 8 7 6 5 4 3 2 1

A 24, distance 24 ft 8

(6, 11)

8

x

8

8

4.

y (0, 11)12

8

4 4

(2, 7) (0, 7) (1, 8) 8

3.

4

1. a. 3 b. as h S 0, 3 remains constant 3. a. 2x 3 h 1 b. as h S 0, 2122 3 h S 1 5. a. 1x h2x 1 2x h 1 b. as h S 0, 7. a. 2 S 12 h2 122 4 2x 1x h2 2 2122 h 1 b. as h S 0, S 2 2 8 2122 12 h2 9. 10 v(t) 3 11. 10 v(t) 兩t 4兩 7 9 8 7 6 5 4 3 2 1

y 8 (0, 9)

(1 2兹2, 0)

4

Connections to Calculus Exercises, pp. 305–306

4

4

8

x

4 8 12

(0, 17) 16

(10, 17)

1 2 3 4 5 6 7 8 9 10 t

A 40, distance 40 ft

CHAPTER 4 Exercises 4.1, pp. 316–320 1. synthetic, zero 3. P(c), remainder 5. Answers will vary. 7. x3 5x2 4x 23 1x 22 1x2 3x 102 3 9. 2x3 5x2 4x 17 1x 32 12x2 x 72 4 11. x3 8x2 11x 20 1x 52 1x2 3x 42 0 2x2 5x 3 0 13. a. 12x 12 x3 x3 b. 2x2 5x 3 1x 3212x 12 0


2/2/11

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SA-21

Student Answer Appendix x3 3x2 14x 8 0 1x2 5x 42 x2 x2 b. x3 3x2 14x 8 1x 221x2 5x 42 0 15. a.

x3 5x2 4x 23 3 1x2 3x 102 x2 x2 b. x3 5x2 4x 23 1x 221x2 3x 102 3 17. a.

2x3 5x2 11x 17 13 12x2 3x 12 x4 x4 b. 2x3 5x2 11x 17 1x 4212x2 3x 12 13 21. x3 5x2 7 1x 121x2 4x 42 11 23. x3 13x 12 1x 421x2 4x 32 0 25. 3x3 8x 12 1x 1213x2 3x 52 7 27. n3 27 1n 321n2 3n 92 0 29. x4 3x3 16x 8 1x 221x3 5x2 10x 42 0 19. a.

31.

2x3 7x2 x 26 x2 3 x4 5x2 4x 7

12x 72

7x 5

x2 3 4x 3

35. 41. 47. 51.

1x2 42 2 x2 1 x 1 a. 30 b. 12 37. a. 2 b. 22 39. a. 1 b. 3 a. 31 b. 0 43. a. 10 b. 0 45. a. yes b. yes a. no b. yes 49. a. yes b. yes 3 1 2 5 6 53. 2 1 0 7 6 3 3 6 2 4 6 1 1 2 0 1 2 3 0

55.

2 3

33.

9 18 4 8 6 16 8 9 24 12 0

57. P1x2 1x 221x 321x 52, P1x2 x3 4x2 11x 30 59. P1x2 1x 221x 2321x 232, P1x2 x3 2x2 3x 6 61. P1x2 1x 521x 22321x 2 232, P1x2 x3 5x2 12x 60 63. P1x2 1x 121x 221x 21021x 2102, P1x2 x4 x3 12x2 10x 20 65. P1x2 1x 221x 321x 42 67. p1x2 1x 32 2 1x 321x 12 69. f 1x2 21x 32 2 1x 221x 52 71. p1x2 1x 32 1x 32 2 73. p1x2 1x 22 3 75. p1x2 1x 321x 32 3 77. p1x2 1x 321x 32 2 1x 42 2 79. 4-in. squares; 16 in. 10 in. 4 in. 81. a. week 10, 22.5 thousand b. one week before closing, 36 thousand c. week 9 83. a. 198 ft3 b. 2 ft c. about 7 ft 85. k 10 87. k 3 89. The theorems also apply to complex zeroes of polynomials. 91. S3 36; S5 225 93. Yes, John wins. 95. G1t2 1400t 5000

Exercises 4.2, pp. 332–336 1. coefficients 3. a bi 5. b; 4 is not a factor of 6 7. P1x2 1x 221x 221x 3i21x 3i2 x 2, x 2, x 3i, x 3i 9. Q1x2 1x 221x 221x 2i21x 2i2 x 2, x 2, x 2i, x 2i 11. P1x2 1x 121x 121x 12 x 1, x 1, x 1 13. Q1x2 1x 521x 521x 52 x 5, x 5, x 5 15. 1x 52 3 1x 92 2; x 5, multiplicity 3; x 9, multiplicity 2 17. 1x 72 2 1x 22 2 1x 72; x 7, multiplicity 2; x 2, multiplicity 2; x 7, multiplicity 1 19. P1x2 x3 3x2 4x 12 21. P1x2 x4 x3 x2 x 2 23. P1x2 x4 6x3 13x2 24x 36 25. P1x2 x4 2x2 8x 5 27. P1x2 x4 4x3 27 29. a. yes b. yes 31. 3 4.7, 4.6 4 , 3 2.3, 2.2 4 , [0.9, 1] 3 5 1 15 3 5 33. 51, 15, 3, 5, 14 , 15 4 , 4 , 4 , 2 , 2 , 2 , 2 6 1 15 3 5 35. 51, 15, 3, 5, 2 , 2 , 2 , 2 6 1 7 2 7 1 7 28 4 37. 51, 28, 2, 14, 4, 7, 16 , 14 3 , 3 , 3 , 3 , 6 , 2 , 2 , 3 , 3 6 1 1 1 1 1 3 3 3 3 3 39. 51, 3, 32 , 2 , 16 , 4 , 8 , 32 , 2 , 16 , 4 , 8 6

41. 1x 421x 121x 32, x 4, 1, 3 43. 1x 321x 221x 52, x 3, 2, 5 45. 1x 321x 121x 42, x 3, 1, 4 47. 1x 221x 321x 52, x 2, 3, 5 49. 1x 421x 121x 22 1x 32, x 4, 1, 2, 3 51. 1x 721x 221x 12 1x 32, x 7, 2, 1, 3 53. 12x 3212x 12 1x 12; x 32 , 12 , 1 55. 12x 32 2 1x 12; x 32 , 1 57. 1x 221x 1212x 52; x 2, 1, 52 59. 1x 1212x 12 1x 152 1x 152; x 1, 12 , 15, 15 61. 1x 2213x 22 1x 2i21x 2i2; x 2, 23 , 2i, 2i 63. x 1, 2, 3, 3 65. x 2, 1, 2 67. x 2, 3 2 3 2 ,4 5 69. x 3, 1, 3 71. x 1, 2, 3, i17 73. x 2, 23 , 1, i 13 75. x 1, 2, 4, 2 77. x 3, 4, 12 79. x 1, 32 , i 13 81. x 12 , 1, 2, i 13 83. a. possible roots: 51, 8, 2, 46; b. neither 1 nor 1 is a root; c. 3 or 1 positive roots, 1 negative root; d. roots must lie between 2 and 2 85. a. possible roots: 51, 26; b. 1 is a root; c. 2 or 0 positive roots, 3 or 1 negative roots; d. roots must lie between 3 and 2 87. a. possible roots: 51, 12, 2, 6, 3, 46; b. x 1 and x 1 are roots; c. 4, 2, or 0 positive roots, 1 negative root; d. roots must lie between 1 and 4 89. a. possible roots: 51, 20, 2, 10, 4, 5, 12 , 52 6; b. x 1 is a root; c. 1 positive root, 1 negative root; d. roots must lie between 2 and 1 91. 1x 4212x 32 12x 32; x 4, 32 , 32 93. 12x 1213x 22 1x 122; x 12 , 23 , 12 13 95. 1x 221x 122 14x2 32; x 2, 12, 13 2 i, 2 i 97. a. 5 b. 13 c. 2 99. yes 101. yes 103. a. 4 cm 4 cm 4 cm b. 5 cm 5 cm 5 cm 105. length 10 in., width 5 in., height 3 in. 107. 1994, 1998, 2002, about 5 yr 109. a. 8.97 m, 11.29 m, 12.05 m, 12.94 m b. 9.7 m, 3.7 111. a. yes b. no c. about 14.88 113A. a. 1x 5i21x 5i2 b. 1x 3i2 1x 3i2 c. 1x i 17 2 1x i 172 113B. a. x 17, 17 b. x 2 13, 2 13 c. x 3 12, 3 12 115. a. C1z2 1z 4i21z 32 1z 22 b. C1z2 1z 9i2 1z 42 1z 12 c. C1z2 1z 3i2 1z 1 2i2 1z 1 2i2 d. C1z2 1z i21z 2 5i21z 2 5i2 e. C1z2 1z 6i2 1z 1 i 132 1z 1 i 132 f. C1z2 1z 4i21z 3 i 122 1z 3 i 122 g. C1z2 1z 2 i21z 3i2 1z i2 h. C1z2 1z 2 3i2 1z 5i21z 2i2 117. a. w 150 ft, l 300 b. A 15,000 ft2 119. r 1x2 21x 4 2

Exercises 4.3, pp. 349–354 1. zero, m 3. bounce, flatter 5. Answers will vary. 7. polynomial, degree 3 9. not a polynomial, sharp turns 11. polynomial, degree 2 13. up/down 15. down/down 17. down/up; 10, 22 19. down/down; 10, 62 21. up/down; 10, 62 23. a. even b. 3 odd, 1 even, 3 odd c. f 1x2 1x 32 1x 12 2 1x 32, deg 4 d. x 僆⺢, y 僆 3 9, q2 25. a. even b. 3 odd, 1 odd, 2 odd, 4 odd c. f 1x2 1x 32 1x 121x 22 1x 42, deg 4 d. x 僆⺢, y 僆 1q, 25 4 27. a. odd b. 1 even, 3 odd c. f 1x2 1x 12 2 1x 32, deg 3 d. x 僆⺢, y 僆⺢ 29. degree 6; up/up; 10, 122 31. degree 5; up/down; 10, 242 33. degree 6; up/up; 10, 1922 35. degree 5; up/down; 10, 22 37. b 39. e 41. c y y y 43. 45. 47. 10 10 10 8 6 4 2 108642 2 4 6 8 10

8 6 4 2 2 4 6 8 10 x

108642 2 4 6 8 10

8 6 4 2 2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x


SA-22 49.

108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

61.

10 8 6 4 2 108642 2 4 6 8 10

67.

10 8 6 4 2 54321 2 4 6 8 10

73.

3:13 PM

Page SA-22


10 8 6 4 2

55.

2/2/11

40 32 24 16 8 54321 8 16 24 32 40

51.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

57.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

63.

y

20 16 12 8 4 54321 4 8 12 16 20

2 4 6 8 10 x

69.

y

200 160 120 80 40 54321 40 80 120 160 200

1 2 3 4 5 x

75.

y

1 2 3 4 5 x

30 24 18 12 6 54321 6 12 18 24 30

53.

y

20 16 12 8 4 108642 4 8 12 16 20

2 4 6 8 10 x

59.

y

20 16 12 8 4 108642 4 8 12 16 20

2 4 6 8 10 x

65.

y

20 16 12 8 4 54321 4 8 12 16 20

1 2 3 4 5 x

71.

y

1 2 3 4 5 x

40 32 24 16 8 54321 8 16 24 32 40

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

y

20

y

1 2 3 4 5 x

120 60 60 120 180 240 300

b. 5

c. B1x2

421x 92, $80,000

700

0

, quartic;

1 2 3 4 5 x

h1x2 1x 421x 2321x 2321x i 2321x i 232 f 1x2 21x 52 21x 2221x 2221x 2321x 232 P1x2 16 1x 421x 121x 32, P1x2 16 1x3 13x 122 a. 132 2 112 2 122 2 142 2 122 2 21132 9 1 4 16 4 26 30 30 ✓ b. 1x 321x 121x 221x 42 x4 2x3 13x2 14x 24 ✓ 85. a. 280 vehicles above average, 216 vehicles below average, 154 vehicles above average b. 6:00 A.M. 1t 02, 10:00 A.M. 1t 42 , 3:00 P.M. 1t 92 , 6:00 P.M. 1t 122 c. max: about 300 vehicles above average at 7:30 A.M.; 300 v(t) 240 min: about 220 vehicles below average at 12 noon 180

87. a. 3 89. a.

12

0 1 2 3 4 5 x

77. 79. 81. 83.

1 4 x1x

b. t ⬇ 1.7 (7:42 A.M.), 227 vehicles; t ⬇ 9.9 (3:54 P.M.), 551 vehicles c. t ⬇ 7.93 (1:56 P.M.) and t ⬇ 11.27 (5:16 P.M.) 91. a. 150

2 4 6 8 10 12 14 16 t

b. morning: t ⬇ 1.72 (11:43 A.M.); evening: t ⬇ 9.11, (7:07 P.M.) c. t ⬇ 4.98 (2:59 P.M.), about 33 customers d. t ⬇ 7.92 (5:55 P.M.) and t ⬇ 10.02 (8:01 P.M.) 93. a. f 1x2 S q, f 1x2 S q b. g1x2 S q, g1x2 S q; x4 0 for all x 95. c 18 97. verified 99. verified, x 1 2i 101. yes

Mid-Chapter Check, p. 354

1. a. x3 8x2 7x 14 1x2 6x 521x 22 4 x3 8x2 7x 14 4 b. x2 6x 5 x2 x2 2. f 1x2 12x 321x 12 1x 12 1x 22 3. f 122 7 4. f 1x2 x3 2x 4 5. g122 8 and g132 5 have opposite signs 6. f 1x2 1x 221x 121x 22 1x 42 7. x 2, x 1, x 1 3i y y 8. 9. 10 15 8 6 4 2

12 9 6 3 108642 3 6 9 12 15

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

10. a. degree 4; three turning points b. 2 sec c. A1t2 1t 12 2 1t 32 1t 52 , A1t2 t 4 10t3 32t2 38t 15 A122 3; altitude is 300 ft above hard-deck, A142 9; altitude is 900 ft below hard-deck 12

Reinforcing Basic Concepts, p. 355 Exercise 1. 1.532 100

, quartic;

Exercise 2. 2.152, 1.765

Exercises 4.4, pp. 366–371 1. as x S q, y S 2 3. denominator, numerator 5. about x 98 7. x 3, x 僆 1q, 32 ´ 13, q 2 9. x 3, x 3, x 僆 1q, 32 ´ 13, 32 ´ 13, q2 5 5 11. x 5 2 , x 1, x 僆 1q, 2 2 ´ 12 , 12 ´ 11, q2


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SA-23

Student Answer Appendix 13. No V.A., x 僆 1q, q2 15. x 3, yes; x 2, yes 17. x 3, no 19. x 2, yes; x 2, no 21. y 0, crosses at 1 32 , 02 23. y 4, crosses at 121 25. y 3, does not cross 4 , 42 27. q1x2 0, r 1x2 8x directly; the graph will cross the horizontal asymptote at x 0.

65. a. 5 hr; about 0.28 b. 0.019, 0.005; As the number of hours increases, the rate of change decreases. c. h S q, C S 0; horizontal asymptote 67. a. $20,000, $80,000, $320,000; cost increases dramatically b. c. as p S 100 , C S q Cost ($1000) 900 700 500

5

300 100

Percent 10 30 50 70 90

69. a. 2; 10 7

7

50

b. 10; 20 c. On average, 6 words will be remembered for life.

W(t) (1, 46)

40 30 20 10

5

29. q1x2 2, r 1x2 8x 8; the graph will cross the horizontal asymptote at x 1.

10 20 30 40 50 t

71. a. b. 35%; 62.5%; 160 gal c. 160 gal; 200 gal d. 70%; 75%

y

10

0.9 0.7 0.5 0.3 0.1

10

10

70 140 210 280 350 x

73. a. $225; $175 b. 2000 heaters c. 4000 heaters d. The horizontal asymptote at y 125 means the average cost approaches $125 as monthly production gets very large. Due to limitations on production (maximum of 5000 heaters) the average cost will never fall below A150002 135. 75. a. 5 b. 18 c. The horizontal asymptote at y 95 means her average grade will approach 95 as the number of tests taken increases; no d. 6 77. a. 16.0, 28.7, 65.8, 277.8 b. 12.7, 37.1, 212.0 c. a. 22.4, 40.2, 92.1, 388.9 b. 17.8, 51.9, 296.8; answers will vary. 4 1 3 79. y 81. 16 x 3 , 4 ; 13x 162 14x 32 0 3 3

10

31. (0, 0) cross, (3, 0) cross 33. 14, 02 cross, (0, 4) 35. (0, 0) cross, (3, 0) bounce 37.

10 8 6 4 2 108642 2 4 6 8 10

43.

10 8 6 4 2 108642 2 4 6 8 10

49.

10 8 6 4 2 108642 2 4 6 8 10

55. f 1x2 59.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

45.

y

10 8 6 4 2

47.

y

53.

y

2 4 6 8 10 x

57. f 1x2

x2 4

v(x)

2 4 6 8 10 x

61.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

51.

41.

y

10 8 6 4 2

2 4 6 8 10 x

1x 221x 32

108642 2 4 6 8 10

8

39.

1x 421x 12

10 8 6 4 2

63. a.

y

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

Exercises 4.5, pp. 380–384 1. slant 3. two y

y

2 4 6 8 10 x

108642 2 4 6 8

g(x)

6 4 2

3x 2x2 2x 3 11. H1x2 μ 3 2

108642 2 (2, 4) 4 6 8 10

10 8 6 4 2

x 1 x 1

3 2 3 x 2

10 8 6 4 2 108642 2 4 6 8 10

x2

15 25 35 45

Distance (mi)

2 4 6 8 10 x

y

y

冢w, w冣 2 4 6 8 10 x

y 16

x2

12

(2, 12)

8 4 8

5

y

108642 2 2 4 6 8 10 x 4 (1, 4) 6 8 10

x

2 4 6 8 10 x

Population density approaches zero far from town. b. 10 mi, 20 mi c. 4.5 mi, 704 people per square mi

Population (100s)

x 2

x2 2x 3 9. G1x2 • x 1 4

x3 8 13. P1x2 • x 2 12

10 8 6 4 2

x 2

2 4 6 8 10 x

9 x2 12 10 8 6 4 2


x2 4 7. F1x2 • x 2 4

4 2

4

8

x


SA-24

2/2/11

17. R1x2 μ

x 1

x2 2x 3 2 2

10 8 6 4 (2, 0) 2

10 8 6 4 (兹3, 0) 2

yx (2, 0)

108642 2 4 6 8 10

2 4 6 8 10 x

25.

31.

33.

y

10 8 yx5 6 4 (0.8, 0) 2 (1, 0) 108642 2 4 6 8 10

10 8 6 4 (2, 0) 2

(4.8, 0)

37.

39.

y

10

6 4 (0, 1) 2 yx1 108642 2 4 6 8 10

yx

(4, 0)

(4, 0)

49.

108642 2 4 6 8 10

(0, 0.2)

41. 10

x1 8 6 4 (2, 0) 2

16 12 8 4

6 4 2 4 8 12 16

y x3

y

(0, d) yx1

108642 2 4 6 8 10

108642 2 4 6 8 10

47.

2 4 6 8 10 x

y

yx x1 (2, 0)

2 4 6 8 10 x

(0, 0)

y

8 7 6 (0, 4) 5 4 3 2 y x2 1 1

yx (1, 0) 2 4 6 8 10 x

54321 1 2

51. 119.1

y y 14 x

x3

10 8 6 4 (2, 0) 2

108642 2 2 4 6 8 10 x 4 6 (0, 0) 8 10

(3.2, 0)

10 8 6 4 (2, 0) 2

x3

(2, 0)

y

(1, 0) 2 4 6 8 10 x

35.

x1

45.

y

yx

(2, 0)

y yx1

(0, 2)

4

53. a. a 5, y 3a 15 b. 60.5

1 2 3 4 5 x

1. vertical, multiplicity 3. empty 5. Answers will vary. 7. x 僆 13, 52 9. x 僆 34, q 2 ´ 516 11. x 僆 1q, 2 4 ´ 526 ´ 34, q 2 13. x 僆 12 13, 2 132 15. x 僆 1q, 3 4 ´ 516 17. x 僆 13, 12 ´ 12, q 2 19. x 僆 1q, 32 ´ 11, 12 ´ 13, q 2 21. x 僆 1q, 22 ´ 12, 12 ´ 13, q 2 23. x 僆 31, 1 4 ´ 536 25. x 僆 1q, 22 ´ 12, 32 27. x 僆 1q, 2 4 ´ 11, 12 ´ 33, q 2 29. x 僆 33, 22 31. x 僆 1q, 22 ´ 12, 12 33. x 僆 1q, 22 ´ 3 2, 32 35. x 僆 1q, 52 ´ 10, 12 ´ 12, q2 37. x 僆 14, 2 4 ´ 11, 24 ´ 13, q2 39. x 僆 17, 32 ´ 12, q 2 41. x 僆 1q, 2 4 ´ 10, 22 43. x 僆 1q, 172 ´ 12, 12 ´ 17, q2 45. x 僆 13, 7 47. x 僆 12, q 2 49. x 僆 11, q 2 4 4 ´ 12, q 2 51. 1q, 32 ´ 13, q 2 53. x 僆 1q, 3 4 ´ 35, q 2 55. x 僆 33, 0 4 ´ 3 3, q 2 57. d 59. b 61. a. verified 1 b. D 41p 34 21p 32 2; p 3, q 2; p 3 4 ,q 4

c. 1q, 32 ´ 13, 3 d. verified 4 2 63. d1x2 k1x3 192x 10242 a. x 僆 15, 8 4 b. 320k units c. x 僆 30, 32 d. 2 ft 65. a. verified b. horizontal: r2 20, as r1 increases, r2 decreases to maintain R 40; vertical: r1 20, as r1 decreases, r2 increases to maintain R 40 c. r1 僆 120, 402 67. R1t2 0.01t 2 0.1t 30 a. 30°, 30°2 b. 120°, q2 c. 150°, q 2 69. a. n 4 b. n 9 c. 13 x2 71. a. yes, x2 0 b. yes, 2 0 x 1 x1x 22 73. x1x 22 1x 12 2 7 0; 7 0 1x 12 2 75. R1x2 6 0 for x 僆 12, 82 ´ 18, 142 3 2 1 54321 1 2 3 4 5 6 7

55. a. A1x2 c. 8, $116.25 d.

; S1x2

2 69. a. P 30 cm, b. CD 60 13 cm, c. 30 cm , 4320 2 2 d. A 750 169 cm , and A 169 cm

c. 10

4x2 53x 250 ; x 0, y 4x 53 x b. cost: $307, $372, $445; Avg. cost: $307, $186, $148.33

2

␲r3 2V ; r ⬇ 3.1 in., h ⬇ 3 in. r

65. S

67. y 34 x 4, m 34 , 10, 42

77.

6 x

12

d. r ⬇ 5.76 cm, h ⬇ 11.51 cm; S ⬇ 625.13 cm2

(3.2, 0) 2

b. y

Exercises 4.6, pp. 391–395 y

108642 2 4 6 8 10

108642 2 2 4 6 8 10 x 4 (0, 4) 6 8 10

2 4 6 8 10 x

10 8 6 4 2

2

(1, 0) 2 4 6 8 10 x

10 8 yx1 6 4 (2, 0) 2 (2, 0)

x 1 8

2 4 6 8 10 x

10 8 6 4 2

108642 2 2 4 6 8 10 x 4 (1, 0) 6 8 10

2 4 6 8 10 x

yx

108642 2 4 6 8 10

2␲r3 2V r


y

10 8 6 4 2

yx3

108642 2 4 6 8 10

2 4 6 8 10 x

c. S

29.

y

10 8 6 4 2

(1, 2) 2 4 6 8 10 x

23.

(兹3, 0)

27.

y

10 8 6 (3, 2) 4 2 108642 2 4 6 8 10

y

(2, 0)

108642 2 4 6 8 10

1 2 3 4 5 x

108642 2 2 4 6 8 10 x 4 6 8 y x 10

y yx2

10 8 6 4 (1, 0) 2

54321 2 (1, 4) 4 6 8 10

x 3 x1

2x3 48 x x c. S(x) is asymptotic to y 2x2. d. x ⬇ 2 ft 3.5 in.; y ⬇ 2 ft 3.5 in. 2x 55 59. a. A1x, y2 xy; R1x, y2 1x 2.52 1y 22 b. y ; x 2.5 2 2x 55x c. A(x) is asymptotic to y 2x 60 A1x2 x 2.5 V 2V d. x ⬇ 11.16 in.; y ⬇ 8.93 in. 61. a. h b. S 2␲r2 r ␲r2 57. a. S1x, y2 2x2 4xy; V1x, y2 x2y

x 3, x 1

21.

y

y

10 8 6 4 2

x 1

x3 3x2 x 3

43.

Page SA-24


x3 7x 6 15. Q1x2 • x 1 4

19.

3:15 PM

y

F1x2 e

f 1x2 6

x 4 x 4

79. x 僆 33, 9 4

1 2 3 4 5x

Making Connections, p. 396 1. e

3. b

5. a 7. h

9. d

11. a

13. g

15. c

300


0

20

0

1. q1x2 x2 6x 7; r 8 2. q1x2 x 1; r 3x 4 3. 7 2 13 6 9 14 14 7 7 14 2 1 1 2 0 Since r 0, 7 is a root and x 7 is a factor. 4. x3 4x 5 1x 22 1x2 2x2 5


2/2/11

3:17 PM

Page SA-25

SA-25

Student Answer Appendix 5. 1x 421x 121x 32 6. h1x2 1x 121x 421x2 2x 22 7. 12 4 8 3 1 2 5 1 4 10 2 0 Since r 0, 12 is a root and 1x 12 2 is a factor. 8. 3i 1 2 9 18 3i 9 6i 18 1 2 3i 6i 0 Since r 0, 3i is a zero 9. 7 1 9 13 10 7 14 7 1 2 1 3 h172 3 10. P1x2 x3 x2 5x 5 11. C1x2 x4 2x3 5x2 8x 4 12. a. C102 350 customers b. more at 2 P.M., 170 c. busier at 1 P.M., 760 7 710 13. 51, 10, 2, 5, 12 , 52 , 14 , 54 6 14. x 12 , 2, 52 15. P1x2 12x 321x 421x 12 16. only possibilities are 1, 3, none give a remainder of zero 17. [1, 2], [4, 5]; verified 18. one sign change for g1x2 S 1 positive zero; three sign changes for g1x2 S 3 or 1 negative zeroes; 1 positive, 3 negative, 0 complex, or 1 positive, 1 negative, 2 complex; 1 positive, 1 negative, 2 complex, verified 19. degree 5; up/down; 10, 42 20. degree 4; up/up; 10, 82 y y y 21. 22. 23. 10 20 15 8 6 4 2 54321 2 4 6 8 10

16 12 8 4

12 9 6 3

54321 4 8 12 16 20

1 2 3 4 5 x

54321 3 6 9 12 15

1 2 3 4 5 x

108642 2 4 6 8 10

1 2 3 4 5 x

29. V1x2

8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

x2 x 12

; V102 2 x2 x 6 30. a. y 15; as 冟x冟 S q A1x2 S 15. As production increases, average cost decreases and approaches 15. b. x 7 2000 31. removable discontinuity at 12, 52 ;

y

10 8 6 4 2

32. H1x2 •

x 3x 4 x1 5

33.

20 16 12 8 4 108642 4 8 12 16 20

y

2 4 6 8 10 x

34.

x 1

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

b. about 2450 favors c. about $2.90 ea. 36. factored form 1x 421x 121x 22 7 0 Neg

Pos

4

37.

0

2

1

Pos 2

Neg

0

1x 221x 12 x1x 22

Pos

2

Neg 2

Pos

outputs are positive or zero for x 僆 32, 22 ´ 3 5, q2

5

Pos

1

outputs are positive for x 僆 14, 12 ´ 12, q2

0

0

Neg 1

Pos

outputs are negative or zero for x 僆 3 2, 02 ´ 3 1, 22

2

1. a. f 1x2 1x 52 2 9 b. g1x2 12 1x 42 2 8 y

y

20 16 12 8 4

18 16 (0, 16) 14 12 10 8 6 4 2

(5, 9) (8, 0)

42 4 2 4 6 8 10 12 14 16 x 8 (2, 0) 12 16 (0, 16) 20

(4, 8)

108642

2 4 6 8 10 x

2. 12, 02, y 2x 4x 3. a. 40 ft, 48 ft b. 49 ft c. 14 sec 14x 3 2 4. x 5 2 5. x2 2x 9 x2 x 2x 1 0 15 10 24 6. 3 1 3 9 18 24 1 3 6 8 0 r0✓ 7. 1 8. P1x2 x3 2x2 9x 18 9. Q1x2 1x 22 2 1x 12 2 1x 12, 2 mult 2, 1 mult 2, 1 mult 1 10. a. 1, 18, 2, 9, 3, 6 b. 1 positive zero, 3 or 1 negative zeroes; 2 or 0 complex zeroes c. C1x2 1x 221x 121x 3i21x 3i2 11. a. 2002, 2004, 2008 b. 4 yr c. deficit of $7.5 million y y y 12. 13. 14. 20 10 500 2

16 12 8 4 2 4 6 8 10 x

300 500

y

Pos

1x 52 1x 22 x2 3x 10 0 x2 x2

Neg

38.

Neg

300

2 4 6 8 10 x

108642 2 4 6 8 10

y

2 4 6 8 10 x

100

10 8 6 4 2

x 1

108642 2 4 6 8 10

108642 100

108642 2 4 6 8 10

2

y 10 8 6 4 2


24. a. even b. x 2, odd; x 1, even; x 1, odd c. deg 6: P1x2 1x 221x 12 2 1x 12 3 25. a. 5x冟x 僆⺢; x 1, 46 b. HA: y 1; VA: x 1, x 4 c. V102 94 (y-intercept); x 3, 3 (x-intercepts) d. V112 43 26. No—even multiplicity; yes—odd multiplicity y y 27. 28. 10 10 8 6 4 2

35. a.

108642 4 8 12 16 20

8 6 4 2 2 4 6 8 10 x

15. a. removal of 100% of the contaminants dramatic increase c. 88% y y 16. a. b. 50 10 40 30 20 10

2 4 6 8 10 x

108642 10 20 30 40 50

108642 2 4 6 8 10

2 4 6 8 10 x

b. $500,000; $3,000,000;

8 6 4 2 2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

17. 800 18. a. x 僆 1q, 3 4 ´ 3 1, 44 19. a. 1.2 y

b. x 僆 1q, 42 ´ 10, 22

0.8 0.4 4 0

4

8 12 16 20x

b. h 1 55; no c. 28.6%, 29.6% d. ⬇11.7 hr e. 4 hr, 43.7% f. The amount of the chemical in the bloodstream becomes neglible. x2 x 6 20. V1x2 2 ; V102 2 x 2x 3 3


SA-26

2/2/11

3:19 PM

Page SA-26


Strengthening Core Skills, pp. 402–403

Exercise 1. x 僆 1q, 3 4 Exercise 2. x 僆 12, 12 ´ 12, q2 Exercise 3. x 僆 1q, 42 ´ 11, 32 Exercise 4. x 僆 32, q2 Exercise 5. x 僆 1q, 22 ´ 12, q 2 Exercise 6. x 僆 33, 1 4 ´ 3 3, q 2

Cumulative Review Chapters 1–4, pp. 403–404 1. R

3. a. 1x 121x2 x 12

R1R2 R1 R2

7. verified 9. y

5. all reals

increases 11 min every 60 days 15.

5

b. 1x 321x 22 1x 22

1009 11 x ; 39 min, driving time 60 60

11. month 9

13. f

1

1x2

x 3 2 3

17. X 63

y

21. not one-to-one; h1x2 6 3, corresponds to two x-values 23. one-to-one 25. not one-to-one; y 3 corresponds to more than one x-value 27. f 1 1x2 5 11, 22, 14, 12, 15, 02, 19, 22, 115, 526 29. v1 1x2 5 13, 42, 12, 32, 11, 02, 10, 52, 11, 122, 12, 212, 13, 3226 5 x3 x 35. f 1 1x2 31. f 1 1x2 x 5 33. p1 1x2 4 4 37. t1x x3 4 39. x 僆⺢, y 僆⺢; f 1 1x2 x3 2, x 僆⺢, y 僆⺢; verified 3 41. x 僆⺢, y 僆⺢; f 1 1x2 2x 1, x 僆⺢, y 僆⺢; verified

43. x 2, y 0; f 1 1x2 8x 2, x 0, y 2; verified

45. x 1, y 1; f 1 1x2 1

5

51. 53. 57.

5 x

5

19.

10 8 6 4 2 54321 2 4 6 8 10

61.

21. 87.91, 80.09, 1.99

y

67. 1 2 3 4 5 x

73. 75.

59 2 23. a. 349.36 131.38i b. i c. 27.63 14.59i 41 41 d. 0 i 25. a. Y6 b. Y7 c. Y4 d. Y3 e. Y5 f. Y2

77. 79.

3. x 僆 3 3, 3 4, as x S 3, y S 0, as x S 3, y S 0; (0, 0); (3, 0), (0, 0), (3, 0)

y

54321 1

f (x) 1 2 3 4 5 x

2 3 4 5

5 4 3 2 1

81.

5 4 3 2 1

54321 1

h(x)

54321 1 2 3 4 5

1 2 3 4 5 x

2 3 4 5

5. x 僆 1q, q 2, down/up; 10, 62; 13, 02, 11, 02, (2, 0)

83.

5 4 3 2 1

y 10 8 6 4 2 54321 2 4 6 8 10

p(x)

1 2 3 4 5 x

7. max f 1x2 2 at x 2 9. max h1x2 4.5 at x 1.5 12; min h1x2 4.5

2 119 2 119 , pa bb or about 3 3 12.11, 4.062. The skater has a maximum anxiety level of near 4, 2 119 2 119 , pa bb or about 2 min before starting his routine; a 3 3 about (0.79, 8.21). The skater has a minimum anxiety level of near 8, shortly after starting his routine. at x 1.512

54321 1 2 3 4 5

y

5 4 3 2 1

11. a

b. f

2 1

Connections to Calculus Exercises, pp. 407–408 1. x 僆 1q, 34 , as x S q, y S q; (0, 0); (0, 0), (3, 0)

x 1, y 1; verified

1x2 1x 5, x 0, y 5 8 1 3, x 7 0, y 7 3 a. x 7 3, y 7 0 b. v 1x2 Ax 1 a. x 4, y 2 b. p 1x2 1x 2 4, x 2, y 4 1 f ⴰ g2 1x2 x, 1g ⴰ f 21x2 x 55. 1 f ⴰ g21x2 x, 1g ⴰ f 21x2 x 1 f ⴰ g2 1x2 x, 1g ⴰ f 21x2 x 59. 1 f ⴰ g21x2 x, 1g ⴰ f 21x2 x x5 f 1 1x2 63. f 1 1x2 2x 5 65. f 1 1x2 2x 6 3 x3 1 3 3 f 1 1x2 2 x 3 69. f 1 1x2 71. f 1 1x2 2 2x 1 2 2 2 x2 2 , D: x 0, R: y D: x , R: y 0; f 1 1x2 3 3 3 x2 3, D: x 0, R: y 3 D: x 3, R: y 0; p1 1x2 4 1 D: x 0, R: y 3; v 1x2 2x 3, D: x 3, R: y 0 y D: x 僆 30, q 2 , R: y 僆 3 2, q 2 ; 5 4 x 僆 32, q2 , R: y 僆 3 0, q2 D: 3

47. a. x 5, y 0 49.

x x,

1

54321 1 2 3 4 5

1 2 3 4 5 x

D: x 僆 10, q2 , R: y 僆 1q, q2 ; D: x 僆 1q, q 2 , R: y 僆 10, q 2

y

1 2 3 4 5 x

D: x 僆 1q, 44 , R: y 僆 1q, 44 ; D: x 僆 1q, 44 , R: y 僆 1q, 4 4

y

1 2 3 4 5 x

85. a. f 1 1x2 x 2 1 b. 13, 52 , (0, 1), and (1, 3) are on the graph of f ; 15, 32 , (1, 0), and (3, 1) are on the graph of f 1 c. verified d. 5

7.6

7.6

CHAPTER 5 Exercises 5.1, pp. 418–421

1. second, one 3. 111, 22, 15, 02, 11, 22, 119, 42 5. False, answers will vary. 7. one-to-one 9. one-to-one 11. not one-to-one, fails horizontal line test: x 3, x 0.5 and x 2 are paired with y 0 13. not a function 15. one-to-one 17. not oneto-one, y 1 is paired with x 6 and x 8 19. one-to-one

87. a. h 1x2 (0, 0), 1 12 , 12 , and 1

5

x 1 x 1 23 , 22

b. (0, 0), 11, 12 2 , and 12, 23 2 are on the graph of h; are on the graph of h1 c. verified


2/2/11

3:21 PM

Page SA-27

SA-27

Student Answer Appendix d.

37.

3.1

10 8

39.

y

4 3 2 1

(1, 5.4) 6 4

(2.3, 0) 2

4.7

108642 2 (3, 1) 4 6 8 10

4.7

3.1


y

10

(0, 1) 2 4 6 8 10 x

y0

6 4 2 (0, 1) 108642 2 4 6 8 10

increasing

15. y 3x; up 2

17. y 3x; left 3

10 8 6 (1, 5) 4 (0, 3) 2

108642 2 4 6 8 10

y0

2 4 6 8 10 x

19. y 3x; reflect across y-axis (2, 9) 108

y0

y

6 4 2 (0, 1) 108642 2 4 6 8 10

108642 2 4 6 8 10

31. 2.718282

108642 2 4 6 8 10

y1

2 4 6 8 10 x

25. e

(2, 1) 2 4 6 8 10 x

35. 4.113250

27. a

y

8 6 4 2 (0, 2)

108642 2 4 6 8 10

2 4 6 8 10 x

33. 7.389056

y

21. y 1 13 2 x; up 1

y 10 8 (0, 9) 6 4 2

(1, 9)10 8 6 4 (3, 1) 2

(2, 10) 10

23. y 1 13 2 x; right 2

y0

2 4 6 8 10 x

decreasing

y

y2

y

(2, 9) 8

(2, 9)

108642 2 4 6 8 10

(0, 1) 1 2 3 4 x

108642 2 4 6 8 10

(2, 5.4)

2 4 6 8 10 x

40,000 30,000 20,000 10,000

1

2

3 4 days

5t

a. $100,000 b. 3 yr 71. a. ⬇ $86,806 b. 3 yr a. $40 million b. 7 yr 75. no, they will have to wait about 5 min 32% transparent 79. 17% transparent 81. ⬇$25,526 a. 8 g b. 48 min 85. 51 87. 75 89. 9.5 10 7; answers will vary 7 91. 5; ; 2a2 3a; 2a2 4ah 2h2 3a 3h 9 93. a. no solution b. 55, 66 69. 73. 77. 83.


1. bx, b, b, x 3. a, 1 5. False; for 冟b冟 6 1 and x2 7 x1, bx2 6 bx1, so 1 1 function is decreasing 7. 16, 2, 8, 11.036 9. 1, 64 , 4 , 64 11. 13.

y0

y

10 8 6 (0, 6.4) 4 2 (2, 0)

43. 3 45. 32 47. 13 49. 4 51. 3 53. 3 55. 2 57. 2 59. 2 61. 3 63. x ⬇ 2.8 65. x ⬇ 3.2 67. a. 1732, 3000, 5196, 9000 b. yes c. as t S q, P S q d. 50,000 P

89. a. 31.5 cm b. The result is 80 cm. It gives the distance of the projector from the screen. 91. a. 63.5°F b. f 1 1x2 2 7 1x 592; independent: temperature, dependent: altitude c. 22,000 ft 1x 93. a. 144 ft b. f 1 1x2 , independent: distance fallen, dependent: 4 time fallen c. 7 sec 3x 95. a. 28,260 ft3 b. f 1 1x2 3 , independent: volume, dependent: B␲ height c. 9 ft 97. Answers will vary. 99. a. P 2l 2w b. A ␲r2 c. V ␲r2h d. V 13 ␲r2h e. C 2␲r f. A 12 bh g. A 12 1b1 b2 2h h. V 43 ␲r3 i. a2 b2 c2 101. ⬇0.472, ⬇0.365; rate of change is greater in [1, 2] due to shape of the graph.

10 8 6 4 2

654321 1 2 3 4 5 6

2 4 6 8 10 x

41.

y

2 4 6 8 10 x

29. b

1. logb x, b, b, greater 1

3. 11, 02 , 0

5. 5; answers will vary 7. 23 8

1 3

9. 7 11. 9 1 13. 8 2 15. 21 2 17. 72 49 2 19. 10 100 21. e4 ⬇ 54.598 23. log464 3 25. log319 2 1 7

0

27. 0 ln 1

1 33. log100 2 1 1 3 3 35. log48 2 37. log48 2 39. 1 41. 2 43. 1 45. 47. 2 2 49. 2 51. 1.6990 53. 0.4700 55. 5.4161 57. 0.7841 59. shift up 3 61. shift right 2, up 3 10 8 6 4 2 108642 2 4 6 8 10

29. log13 27 3

31. log 1000 3

y

10 8 6 4 2

(1, 3)

108642 2 4 6 8 10

2 4 6 8 10 x

63. shift left 1 10 8 6 4 2

y

(3, 3) 2 4 6 8 10 x

65. shift left 1, reflect across x-axis y

y

10 8 6 4 2 (0, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

108642 2 2 4 6 8 10 x 4 6 (0, 0) 8 10

67. II 69. VI 71. V 73. x 僆 1q, 12 ´ 13, q 2 75. x 僆 1 32 , q 2 77. x 僆 13, 32 79. pH ⬇ 4.1; acid 81. a. ⬇4.7 b. ⬇7.9 83. about 398 times 85. about 3.2 times 87. a. ⬇2.4 b. ⬇1.2 89. a. 20 dB b. 120 dB 91. about 501 times 93. about 3162 times 95. 6194 m 97. a. about 5434 m b. 4000 m 99. a. 2225 items b. 2732 items c. $117,000 101. a. about 58.6 cfm b. about 1605 ft2 103. a. 95% b. 67% c. 39% 105. ⬇4.3; acid 107. Answers will vary. a. 0 dB b. 90 dB c. 15 dB d. 120 dB e. 100 f. 140 dB 109. a. 2 b. 3 c. 5 3 2 2 y 111. D: x 僆⺢, R: y 僆⺢ 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

113. x 僆 1q, 52; f 1x2 1x 521x 42 2 x3 3x2 24x 80


SA-28

2/2/11

3:23 PM

Page SA-28


35. lna

29. log1x2 12

x5 b 37. ln1x 22 x

31. log34

1 2

x b x1

41. log5 1x 22

39. log242

43. 1x 22 log 8 45. 12x 12 ln 5 47.

33. log a

log 22 49. 4 log5 3

51. 3 log a log b 53. ln x 14 ln y 55. 2 ln x ln y 1 2 3 log1x

ln 60 57. 22 log x 4 59. ; 2.104076884 ln 7 log 1.73205 ln 152 ; 3.121512475 63. ; 0.499999576 61. ln 5 log 3 log 0.125

;3 log 0.5 log1x2 ; f 152 ⬇ 1.4650; f 1152 ⬇ 2.4650; f 1452 ⬇ 3.4650; 67. f 1x2 log132 outputs increase by 1; f 133 # 52 ⬇ 4.4650 log1x2 ; h122 ⬇ 0.3155; h142 ⬇ 0.6309; h182 ⬇ 0.9464; 69. h1x2 log192 outputs are multiples of 0.3155; h124 2 ⬇ 410.31552 ⬇ 1.2619 71. verified 73. a. N AXm b. ⬇ 3500 people 75. no, pH ⬇ 7.32 77. no, pH ⬇ 5.9 and the soil must be treated further 79. 600601 81. zeroes at x 3 and x 2; HA: y 1, VA x 1, x 1 83. x 1 or x 9 65.

Mid-Chapter Check, p. 456 5

1. a. 23 log279 b. 54 log81243 2. a. 83 32 b. 12960.25 6 3. a. x 5 b. b 54 4. a. x 3 b. b 5 5. a. $71,191.41 b. 6 yr 6. F1x2 4 # 5x3 2 7. f 1 1x2 1x 12 2 3, D: x 僆 3 1, q2 ; R: y 僆 3 3, q 2 ; verified 8. a. 4 log381, verified 2 b. 4 ⬇ ln 54.598, verified 9. a. 273 9, verified 1.4 b. e ⬇ 4.0552, verified 10. ⬇7.9 times more intense

Reinforcing Basic Concepts, p. 457 Exercise 1. Answers will vary. b. ln1x2 42 c. logx x 3

Exercise 2. a. log1x2 3x2

Exercises 5.5, pp. 465–468 1. variable, constant 3. uniqueness, one, one 5. False; answers will vary. 7. x 32 9. x 6.4 11. x 20, 5 is extraneous 13. x 2, 52 is extraneous 15. x 0 17. x 52 19. x 23 e2 63 ; x ⬇ 6.1790 21. x 32 23. x 19 25. x 9 9 3 27. x 2; 9 is extraneous 29. x 3e 12 ; x ⬇ 59.7566 31. no solution 33. x 2 13; 2 13 is extraneous ln 128,967 ln 231 ; x ⬇ 2.7968 37. x ; x ⬇ 2.4371 35. x ln 7 3 ln 5 ln 2 ; x ⬇ 1.7095 41. x ⬇ 4.815, x ⬇ 102.084 39. x ln 3 ln 2 43. x ⬇ 2.013, x ⬇ 3.608 45. x ⬇ 46.210 C p 1 b ln a a , t ⬇ 55.45 49. about 3.2 cmHg 47. t k 51. a. 30 fish b. about 37 months 53. about 50.2 min

Exercises 5.6, pp. 477–482 1. Compound 3. Q0ert 5. Answers will vary. 7. $4896 9. 250% 11. $2152.47 13. 5.25 yr 15. 80% 17. 4 yr 19. 16 yr 21. $7561.33 23. about 5 yr 25. 7.5 yr 27. no 29. a. no b. 9.12% 31. 7.9 yr 33. 7.5 yr 35. a. no b. 9.4% 37. a. no b. approx 13,609 euros 39. No; $234,612.02 41. about 7 yr 43. 22 yr 45. a. no b. $298.31 Ap nt A A 47. a. t b. p 49. a. r n a 1b pr 1 rt Bp Q1t2 A ln a b ln a b p Q1t2 Q0 b. t 51. a. Q0 rt b. t r r e n ln a1 b n 53. $709.74 55. a. 5.78% b. 91.67 hr 57. 0.65 g 59. about 816 yr 61. about 12.4% 63. $17,027,502.21 65. 7.2% 3 67. a. f 1x2 x3, f 1x2 x, f 1x2 1x, f 1x2 1 x, f 1x2 1x 1 b. f 1x2 冟x冟, f 1x2 x2, f 1x2 2 c. f 1x2 x, f 1x2 x3, f 1x2 1x, x 1 1 3 f 1x2 1 x d. f 1x2 , f 1x2 2 x x 69. P1x2 x4 4x3 6x2 4x 15

Exercises 5.7, pp. 487–495 1. scatterplot, context, situation 3. beyond 5. (1) clear out old data, (2) enter new data, (3) display the data, (4) calculate the regression equation, (5) display and use the results; Answers will vary. 7. e 9. a 11. d 13. linear 15. exponential 17. logistic 19. exponential 21. As time increases, the amount of radioactive 0.9 material decreases but will never truly reach 0 or 0.7 become negative. Exponential with b 6 1 and k 7 0 0.5 x is the best choice. y ⬇ 11.04220.5626 0.3 Time (hours)

27. ln12x2 14x2

55. $15,641 57. a. 6 hr b. 18.0% 59. Mf 52.76 tons 61. a. 26 planes b. 9 days 63. x 1.609438 y 2x1 65. a. b. y 2 ln1x 32 x 2y1 x 2 ln1 y 32 x ln x 1y 12 ln 2 ln1y 32 2 x ln x y1 e2 y 3 ln 2 x ln x 1y y e2 3 ln 2 x 67. a. y ex ln 2 eln 2 2x; x y 2 1 ln y x ln 2, eln y ex ln 2 1 y ex ln 2 b. y bx, ln y x ln b, eln y ex ln b, y exr for r ln b 69. a. d b. e c. b d. f e. a f. c 71. a. x 僆 3 32 , q 2, y 僆 3 0, q 2 b. x 僆 1q, q 2, y 僆 3 3, q2 73. 13.5 tons

0.1

0

1

2

3

4

Grams

23. Sales will increase rapidly, then level off as the market is saturated with ads and advertising becomes less effective, possibly modeled by a logarithmic function. y ⬇ 120.4938 217.2705 ln1x2

900 700 Sales

1. e 3. extraneous 5. Answers will vary; Yes, 1.5663025 1.5663025 7. x ⬇ 29.964 9. x ⬇ 1.778 11. x ⬇ 2.200 13. x ⬇ 1.260 65 15. x ln 2, x ⬇ 4.7881 17. x log1782 5, x ⬇ 3.1079 4 8 ln 2.32 19. x , x ⬇ 1.1221 21. x e3 4, x ⬇ 10.3919 0.75 e0.4 5 23. x 5 101.25, x ⬇ 12.7828 25. x , x ⬇ 1.7541 2

500 300 100 5 15 25 35 Cost

25. a.

b. about 1750 c. y ⬇

y 2000 1750

Cumulative cases


1500 1250 1000 750 500 250 10

30

50

70

Days after outbreak

90 x

logistic

27. 4.95 29. 6.25 31. 5.75 33. 6.84 35. a. about 19 boards b. about 15 days

1719 1 10.2e0.11x

5

6


2/2/11

3:25 PM

Page SA-29

SA-29


55. logistic, y ⬇

c. 34.8 lb

222.133 1 32.280e0.336x

;

45 35 25 15

y Subscriptions (millions)

Weight ( pounds)

37. logarithmic, y ⬇ 27.4 13.5 ln x y a. 9.2 lb b. 29 days

150

Logistic

125 100 75 50 25 2

5 10

30

50

70

90

6

10

14

18

Year (1990 S 0)

x

Time (days)

57.

39. logarithmic, y ⬇ 78.8 10.3 ln x y a. 51,000 b. 1977 Offices (1000s)

175 Exponential

x

about 55 million, about 184 million, about 214 million; 2014

power regression, a. y ⬇ x0.665, 9.5 AU; b. 84.8 yr

7.5

c. 29,900

90 70 50

0

30

15

x

10 0

20

40

60

80 100

Year (1900 S 0)

c. 2013

0

450

59.

350 250

4000

Rodents

150 50 2

6

10

14

18

Year (1980 S 0)

x

2000

20

40

60

80

100

x

Predators

61. a.

linear, W ⬇ 1.24L 15.83, 32.5 lb, 35.3 in.

W 25

440

Weight (lb)

Chicken production (millions)

3000

1000

43. quadratic y ⬇ 0.576x2 8.879x 394 y a. 360 million b. about 513 million 480 c. from 1984 to 1990 400 360 320 2

6

10

14

18

Year (1980 S 0)

b. 329.2 million

c. 2010

61. b.

150 50 7

9

11

Year (1990 S 0)

20

24

28

32

Length (in.)

250

5

10

16

350

3

15

5

450

1

20

x

45. linear, y ⬇ 6.555x 165.308 y a. 224.3 million Debit cards (millions)

a. power regression, y ⬇ 58,555.89x1.056; b. about 295 rodents c. about 17 predators

y 5000

L

logarithmic, C1a2 ⬇ 37.9694 3.4229 ln 1a2, about 49.3 cm, about 34 mo

C(a) Circumference (cm)

Number of farms (1000s)

41. exponential, y ⬇ 346.7910.942 x y a. 155,100 b. 54,200

52 48 44 40 36 5

x

10

15

20

25

a

Age (months)

Percent of U.S. Population

47. linear, P1t2 ⬇ 0.51t 22.51, 2005: 40.4%, 2010: 43.0%, 2015: 45.5%

63. D: x 僆 1q, 22 ´ 12, 12 ´ 11, 52 ´ 15, q 2, x 65. max: 10.4, 1.82 min: 13.5, 3.52, 12.3, 1.42

y 50 45 40 35 30 25 20 15 10 5

1 2

f 1x2c: x 僆 13.5, 0.42 ´ 12.3, q 2

f 1x2T: x 僆 1q, 3.52 ´ 10.4, 2.32

5 10 15 20 25 30 35 40 x

Year (1970 → 0)

49. exponential, y ⬇ 103.83 11.05952 y a. 220 b. The 22nd note, or F# 210 c. frequency doubles, yes 190 Frequency

x

Making Connections, p. 495 1. a

3. e

5. c

7. e

9. b

11. g

13. c

15. d

170


150 130 110 2

4

6

8

1. no

10 12 x

y Monthly charge

44 36 28 20 12 4 4

12

20

28

Year (1980 S 0)

53. quadratic, y ⬇ 1.18x 10.99x 4.60; month 8 2

36 x

8 6 4 2

y 6 3

Profit

3

1

3. yes

4. f 1 1x2

x2 3

5. f 1 1x2 1 x 2

6. f 1 1x2 x2 1; x 0 7. f 1x2: D: x 僆 3 4, q 2, R: y 僆 3 0, q2; f 1 1x2: D: x 僆 30, q2, R: y 僆 34, q 2 8. f 1x2: D: x 僆 1q, q 2, R: y 僆 1q, q2; f1 1x2: D: 1q, q2, R: y 僆 1q, q 2 9. f 1x2: D: x 僆 1q, q 2, R: y 僆 10, q2; f 1 1x2: D: x 僆 10, q 2, R: y 僆 1q, q2 2 10. a. $3.05 b. f 1 1t2 t 0.15 , f 1 13.052 7 c. 12 days y y y 11. 12. 13. 10 10 10

Note number

51. exponential, y ⬇ 8.02 11.05642 x $41.59/mo, $54.72/mo

2. no

3

5

7

9 15 21 27

Month

9

11

x

108642 2 4 6 8 10

y3 2 4 6 8 10 x

8 6 4 2 108642 y 1 2 4 6 8 10

8 6 4 2 2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

y 2

1 14. 2 15. 2 16. 52 17. 12.1 yr 18. 32 9 19. 53 125 20. e3.7612 ⬇ 43 21. log525 2 22. ln 0.7788 ⬇ 0.25 23. log381 4 24. 5 25. 1 26. 12


SA-30 27.

3:27 PM

Page SA-30


10 8 6 4 2 108642 2 4 6 8 10

28.

y

33. a. x e

32

x 3

29.

y

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

30. x 僆 132 , q 2

34. a. x

2/2/11

b. x 10

2.38

ln 4 , x ⬇ 2.7726 0.5

c. x ln 9.8 b. x

108642 2 4 6 8 10

2 4 6 8 10 x

31. x 僆 1q, 02 ´ 16, q 2 log 19 0.2

10 8 6 4 2

y x1

2 4 6 8 10 x

32. a. 4.79 b. 107.3I0 1 d. x log 7 2

, x ⬇ 6.3938

103 , x ⬇ 333.3333 d. x e2.75, x ⬇ 0.0639 3 3 35. a. ln 42 b. log930 c. ln 1 xx d. log1x2 x2 12 c. x

36. a. 2 log59

b. 2 log74

37. a. ln x ln y b. 1 4

1 3

c. 12x 12ln 5 d. 13x 22ln 10

4 3

5 2

47. 18.5%

45. a. 17.77%

b. 23.98 days

108642 2 4 6 8 10

2 4 6 8 10 x

y x2

2 4 6 8 10 x

11. a. 4.19 b. 0.81 12. f is a parabola (hence not one-to-one), x 僆⺢, y 僆 33, q2 ; vertex is at 12, 32 , so restricted domain could be x 僆 3 2, q 2 to create a one-to-one function; f 1 1x2 1x 3 2, x 僆 3 3, q 2, y 僆 32, q 2 . 13. x 1 lnln 89 14. x 1, x 5 is extraneous 3 , x ⬇ 5.0857 15. ⬇5 yr 16. a. P kM␣ b. P ⬇ 15.3 lb 1M 202 17. 19.1 months 18. a. no b. $54.09 19. a. 10.2 lb b. 34 weeks 39.1156 20. logistic; y ; 0.89 sec 1 314.6617e5.9483x 40

2

0

5

46. 38.6 cmHg b. $268.93

Strengthening Core Skills, p. 503 Exercise 1. about 126 times hotter

Exercise 2. about 4.2 hb

Cumulative Review Chapters 1–5, p. 504 20

0

35

a. logarithmic, y ⬇ 12.772 1.595 ln x c. the year 2011

2 1. x 2 7i 3. 14 5i2 814 5i2 41 0 9 40i 32 40i 41 0 0 0 ✓ 5. f 1g1x22 x; g1 f 1x2 2 x; Since 1 f ⴰ g2 1x2 1g ⴰ f 21x2, they are inverse functions. 7. a. T1t2 455t 2645 11991 S year 12 ¢T 455 b. , triple births increase by 455 each year ¢t 1 c. T162 5375 sets of triplets, T1172 10,380 sets of triplets y 9. D: x 僆 310, q2, R: y 僆 39, q2 10 8 h1x2c: x 僆 12, 02 ´ 13, q 2 h1x2T: x 僆 10, 32 6 4 2

8

52.

10 8 6 4 2

y3

43. x 5; 2 is extraneous

48. Almost, she needs $42.15 more. 49. a. no

50. 55.0% 51.

108642 2 4 6 8 10

10.

y

3 2

42. x e2 1, x ⬇ 6.3891 44. x 4.25

10 8 6 4 2

ln p ln q

c. log x log y log x log y d. log 4 53 log p 43 log q 32 log p log q log 45 log 128 ln 124 38. a. ⬇ 2.215 b. ⬇ 4.417 c. ⬇ 6.954 log 6 log 3 ln 2 ln 0.42 7 d. ⬇ 0.539 39. x ln ln 2 , x ⬇ 2.8074 ln 5 5 40. x ln 41. 1 2 ln 3 , x ⬇ 0.9530 ln 3 1, 0 ⬇ x .4650 5 3

9.

b. 16.9 mi/gal

108642 2 4 6 8 10

2 4 6 8 10 x

2V 11. x 3, x 2 1multiplicity 22; x 4 13. 2␲a b 5x 3 1 y 15. a. f 1 1x2 b. c. f 1 f 1x22 x 10 2 8

10

6 4 2

0

a. logistic, g1t2

108642 2 4 6 8 10

8

0 8.0

b. ⬇ 6.8 microns 1 1.0e0.7t c. 14 1.962 7.84; 5.56 hr (about 5 hr 34 min)

Practice Test, p. 501 1. 34 81 2. log255 12 3. 52 logbx 3 logby logbz 5 m1n3 4. logb 5. x 10 6. x 7. 2.68 8. 1.24 3 1p

2 4 6 8 10 x

17. x 5, x 6 is an extraneous root 19. a. about 88 hp for sport wagon, about 81 hp for minivan b. ⬇3294 rpm c. minivan, 208 hp at 5800 rpm 21. x ⬇ 5.064 23. x ⬇ 0.649, x ⬇ 4.967 25. x ⬇ 2.013, x ⬇ 3.608

Connections to Calculus Exercises, p. 507 3 1 1. 5x ln x 3. 5 log x 3 log y log z 2 2 1 y 5. A1x2 x2 2 yt x A(x) x

0

7. verified (factor out 1 x1 9. t lna b 2 1x

1 2,

t

then combine like terms and simplify)


2/2/11

3:42 PM

Page SA-31


CHAPTER 6 Exercises 6.1, pp. 521–526 1. Complementary, 180, less, greater 3. r ␪; 12 r2␪; radians 5. Answers will vary. 7. a. 77.5° b. 30.8° 9. 53° 11. 42.5° 13. 67.555° 15. 285.0025° 17. 45.7625° 19. 20° 15¿ 00– 21. 67° 18¿ 25.2– 23. 275° 19¿ 48– 25. 5° 27¿ 9– 27. No, 19 ⫹ 16 6 40 29. 69° 31. 25° 33. 62.5 m 35. 4112 ft ⫹ 10 ft 68 ft 37. ⫺645°, ⫺285°, 435°, 795° 39. ⫺765°, ⫺405°, 315°, 675° 41. s ⫽ 980 m 43. ␪ ⫽ 0.75 rad 45. r 1760 yd 47. s 8.38 49. r 9.4 km 51. A ⫽ 115.6 km2 53. ␪ ⫽ 0.6 rad 55. r 3 m 57. ␪ ⫽ 1.5 rad; s ⫽ 7.5 cm; r ⫽ 5 cm; A ⫽ 18.75 cm2 59. ␪ ⫽ 4.3 rad; s ⫽ 43 m; r ⫽ 10 m; A ⫽ 215 m2 61. ␪ ⫽ 3 rad; A ⫽ 864 mm2; s ⫽ 72 mm; r ⫽ 24 mm ␲ 7␲ ⫺2␲ 63. 2␲ rad 65. rad 67. rad 69. rad 71. 0.4712 rad 4 6 3 73. 3.9776 rad 75. 60° 77. 30° 79. 120° 81. 720° 83. 165° 85. 186.4° 87. h 7.06 cm; m 3.76 cm; n 13.24 cm 89. approx. 960.7 mi apart 91. a. 50.3 m2 b. 80° c. 17 m 93. a. 1.5␲ rad/sec b. about 15 mi/hr 95. a. 40␲ rad/min b. ␲6 ft/sec 0.52 ft/sec c. about 11.5 sec 97. a. 3142 radians per minute b. 524 feet per minute c. 34 seconds 99. a. 1000 m b. 1000 m c. 100012 m 1414.2 m 45⬚

1000 m

√2 (1000) m

45⬚ 1000 m

101. 5012 or about 70.7 mi apart 103. a. 50.3°/day; 0.8788 rad/day b. 0.0366 rad/hr c. 6.67 mi/sec 105. 34.9 m/hr 107. Answers will vary. 109. Shift left 3 units, reflect across x-axis, stretch by a factor of 2, shift down 1 unit 111. y ⫽ 2x ⫺ 3

Exercises 6.2, pp. 537–542 y 3. x, y, ; sec t, csc t, cot t 5. Answers will vary. x ⫺ 111 15 5 ⫺12 111 5 b 11. a , b 13. a , b 7. 1⫺0.6, ⫺0.82 9. a , 13 13 6 6 4 4 15. 1⫺0.9769, ⫺0.21372 17. 1⫺0.9928, 0.11982 ⫺ 13 ⫺1 13 1 13 ⫺1 , b, a , b, a , b 19. a 2 2 2 2 2 2 ⫺111 ⫺5 ⫺ 111 5 111 5 , b, a , b, a , b 21. a 6 6 6 6 6 6 23. 1⫺0.3325, 0.94312, 1⫺0.3325, ⫺0.94312, 10.3325, ⫺0.94312 25. (0.9937, 0.1121), 1⫺0.9937, 0.11212, 1⫺0.9937, ⫺0.11212 ␲ ⫺ 13 ⫺1 ␲ ⫺ 12 ⫺ 12 , b 31. ; a , b 27. verified 29. ; a 4 2 2 6 2 2 ␲ ⫺ 12 12 ␲ 13 1 12 12 , b 35. ; a , b 37. a. 33. ; a b. 4 2 2 6 2 2 2 2 ⫺ 12 ⫺12 ⫺ 12 12 ⫺ 12 12 c. d. e. f. g. h. 2 2 2 2 2 2 13 ⫺ 13 ⫺ 13 39. a. ⫺1 b. 1 c. 0 d. 0 41. a. b. c. 2 2 2 13 13 13 ⫺ 13 13 d. e. f. g. h. 2 2 2 2 2

1. x, y, origin

SA-31

43. a. 0 b. 0 c. undefined d. undefined 45. sin t ⫽ 0.6, cos t ⫽ ⫺0.8, tan t ⫽ ⫺0.75, csc t ⫽ 1.6, sec t ⫽ ⫺1.25, cot t ⫽ ⫺1.3 5 12 13 13 47. sin t ⫽ ⫺12 13 , cos t ⫽ ⫺13 , tan t ⫽ 5 , csc t ⫽ ⫺12 , sec t ⫽ ⫺ 5 , 5 cot t ⫽ 12 5 121 121 ⫺2 ⫺ 121 49. sin t ⫽ , cos t ⫽ , tan t ⫽ , csc t ⫽ , 5 5 2 21 ⫺2 121 ⫺5 sec t ⫽ , cot t ⫽ 2 21 ⫺3 12 ⫺212 ⫺1 51. sin t ⫽ , cos t ⫽ , tan t ⫽ 2 12, csc t ⫽ , 3 3 4 12 53. QI, 0.7 55. QIV, 0.7 57. QI, 1 sec t ⫽ ⫺3, cot t ⫽ 4 2␲ 7␲ 2␲ ␲ 3␲ 5␲ , 59. QII, 1.1 61. 63. 65. 67. 69. 3 6 3 2 4 4 ␲ 3␲ 3␲ 5␲ 4 , 71. , 73. 75. 0, ␲ 77. a. 1 35 , 45 2 b. 1 ⫺3 5 , 52 2 2 4 4 79. 2.3416 81. 1.7832 83. 3.5416 5 12 5 2 25 144 169 12 2 , 13 , 12, 1 13 2 ⫹ 1 12 85. a. 1 13 13 2 ⫽ 169 ⫹ 169 ⫽ 169 ⫽ 1; sin t ⫽ 13 , 5 12 13 13 5 cos t ⫽ 13 , tan t ⫽ 5 , csc t ⫽ 12 , sec t ⫽ 5 , cot t ⫽ 12 7 24 7 2 49 576 625 24 7 2 , 25 , 12, 1 25 2 ⫹ 1 24 b. 1 25 25 2 ⫽ 625 ⫹ 625 ⫽ 625 ⫽ 1; sin t ⫽ 25 , cos t ⫽ 25 , 25 25 7 tan t ⫽ 24 7 , csc t ⫽ 24 , sec t ⫽ 7 , cot t ⫽ 24 35 12 2 35 2 144 1225 1369 35 c. 1 12 37 , 37 , 12, 1 37 2 ⫹ 1 37 2 ⫽ 1369 ⫹ 1369 ⫽ 1369 ⫽ 1; sin t ⫽ 37 , 35 37 37 12 cos t ⫽ 12 , tan t ⫽ , csc t ⫽ , sec t ⫽ , cot t ⫽ 37 12 35 12 35 9 40 9 2 81 1600 1681 40 2 , 41 , 12, 1 41 2 ⫹ 1 40 d. 1 41 41 2 ⫽ 1681 ⫹ 1681 ⫽ 1681 ⫽ 1; sin t ⫽ 41 , 9 40 41 41 9 cos t ⫽ 41 , tan t ⫽ 9 , csc t ⫽ 40 , sec t ⫽ 9 , cot t ⫽ 40 87. (⫺0.4161, 0.9093) 89. (0.2837, ⫺0.9589) 91. 193.2 cm, 51.8 cm 93. 50 cm right and 86.6 cm above center of circle 95. a. 5 rad b. 30 rad 97. a. 5 dm b. 6.28 dm 99. a. 2.5 AU b. 6.28 AU 101. yes 103. range of sin t and cos t is [⫺1, 1] 105. a. 2t 2.2 b. QI c. cos t 0.5 d. No 107. a. d ⫽ 10 b. midpoint: (1, ⫺1) 3 c. m ⫽ 109. a. x ⫽ ⫺6, 4 b. x ⫽ 24 4


1. increasing 3. 1⫺q, q 2 , 3 ⫺1, 1 4 7. y ⫽ cos t t ␲ ⫺1 7␲ 6 5␲ 4 4␲ 3 3␲ 2 5␲ 3 7␲ 4 11␲ 6 2␲

13 2 12 ⫺ 2 1 ⫺ 2 ⫺

0 1 2 12 2 13 2 1



SA-32

2/2/11

3:42 PM

Page SA-32


␲ ␲ ␲ ␲ , , b. sin 0.25882 c. same as second row 12 6 4 12 11. a. II b. V c. IV d. I e. III y y 13. 15. 9. a.

1

3␲

2

␲

1

␲

␲ 2

2

t

1

␲

␲ 2

2

2␲ t

3␲ 2

␲

1

17. 3 19. 0.75 21. A ⫽ 3, P ⫽ 2␲

23. A ⫽ 2, P ⫽ 2␲

y

63. a. 3 ft

␲ tb b. D 3.86 c. 72° 12 67. a. D ⫽ 15 cos1␲t2 b. at center c. Swimming leisurely. One complete cycle in 2 sec. 69. a. Graph a b. 76 days c. 96 days 71. a. 480 nm S blue b. 620 nm S orange 73. I ⫽ 30 sin150␲t2, I 21.2 amps 75. Since m ⫽ ⫺M, 0; avg. value ⫽ 3; shifted up 3 units; avg. value ⫽ 1 t y 0 ␲ 2 ␲ 3␲ 2 2␲

2 3

2␲

␲

t

3

2␲

t

2

25. A ⫽ 12 , P ⫽ 2␲

27. A ⫽ 1, P ⫽ ␲

y

2␲

␲

y

0.8

4

␲ 2

␲

2␲

t

⫺0.8

4␲

t

⫺4

33. A ⫽ 3, P ⫽ 12

1 t 1477

1 2954

79. distance ⫽

200 yd 115.5 yd 13

81. a. 3 ⫺ 4i

b. ⫺1 ⫹ 6i

c. 7 ⫺ 3i

3 7 d. ⫺ ⫺ i 2 2

Exercises 6.4, pp. 571–577 ␲ 3. odd, ⫺f 1t2 , ⫺0.268 5. a. Use reciprocals of standard B b. Use reciprocals of given values.

1. ␲, P ⫽

35. A ⫽ 4, P ⫽ 65

y

3

1

31. A ⫽ 4, P ⫽ 4␲

y

1

t

⫺1

29. A ⫽ 0.8, P ⫽ ␲

3

y

␲ 2

t

q

5

1

1

␲

3

77. g(t) has the shortest period;

y

q

␲ xb 40

65. a. D ⫽ ⫺4 cos a

y

␲

c. h ⫽ 1.5 cos a

b. 80 mi

values.

y

7.

9.

y

y

4 3

1

~

Q

t

⫺3

E

T

1

␲ 2

t

⫺4

1 37. A ⫽ 2, P ⫽ 128

t

13.

y

2

2␲

4␲

t

␲

2␲

t

1

11.

y

␲

1

y

3 2

1 256

1 128

␲

t

2␲

t 2

3

⫺2

39. Y2 0.86392, Xmin ⫽ 0, Xmax ⫽ 1/1336, Xscl ⫽ 1/13,360, Ymin ⫽ ⫺2, Ymax ⫽ 2, Yscl ⫽ 1 ␲ 41. A ⫽ 2, P ⫽ , g 43. 0 A 0 ⫽ 3, P ⫽ ␲, f 45. 0 A 0 ⫽ 34 , P ⫽ 5␲, b 2 2␲ 1 , d 49. y ⫽ ⫺34 cos 18t2 51. y ⫽ 6 cos a tb 47. A ⫽ 4, P ⫽ 72 3 3 sin110␲t2 55. y ⫽ ⫺22 3␲ 7␲ , 57. red: y ⫽ ⫺cos x; blue: y ⫽ sin x; x ⫽ 4 4 59. red: y ⫽ ⫺2 cos x; blue: y ⫽ 2 sin13x2; 3␲ 3␲ 7␲ 11␲ 7␲ 15␲ x⫽ , , , , , 8 4 8 8 4 8 2 61. a. 100␲ cm 314.2 cm2 b. 200 cm2; it is a square. c. The area of the polygon seems to be n A approaching the area of the circle. 10 293.89 20 309.02 30 311.87 100 313.95

53. y ⫽ 3 sin1 15 t2

1 , 1, 13, und. 17. 1.6, 0.8, 0.5, 1.4, 0.7, 1.2 13 19. a. ⫺1 b. 13 c. ⫺1 d. 13 15. 0,

7␲ 5␲ 3␲ c. d. 6 3 4 1 ⫺13␲ 35␲ 59␲ 23. und., 13, 1, , 0 25. , , 24 24 24 13 ␲ 27. ⫺1.6, 4.6, 7.8 29. ⫹ ␲k, k 僆 Z 10 ␲ 31. ⫹ ␲k, k 僆 Z 12 21. a.

7␲ 4

b.


2/2/11

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Page SA-33

SA-33


y 2 tan t

35.

y y tan t

5 4 3 2 1

3␲ ␲ 2␲ 2 ␲1 2 2 3 4 5

␲ 2

37.

␲

3␲ 2␲ 2

39.

1 ␲


y 3 cot t

5 4 3 2 1

3␲ ␲ 2␲ 2 ␲1 2 2 3 4 5

t

y

2

y cot x

␲ 2

␲

3␲ 2␲ 2

t

y

1

␲

␲ 4

4

␲ 2

␲ 8

t

␲ 4

1. a. 36.11°N, 115.08°W b. 2495.7 mi. 2. ␪ ⫽ 4.3; A ⫽ 860 cm2 ⫺ 12 13 3. a. b. 4. a. 1.0353 b. 8.9152 3 2 ⫺3 15 ⫺2 ⫺2 ⫺3 ⫺ 15 ; sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , sec ␪ ⫽ , 5. y ⫽ 3 3 2 3 5 15 2 15 tan ␪ ⫽ , cot ␪ ⫽ 6. 221.8°, 3.8713 5 2 7. asymptotes: x ⫽ ⫺5, ⫺3, ⫺1, 1, 3, 5; 8. A ⫽ 3, P ⫽ 4; y

t

y

3 3

6

41.

43.

y

4

2

1 2 3 4 5 6

4

t

2

y

2

5 ␲ 8

45.

␲ 4

3␲ 2

t

47.

y

3 1

3␲

t

1

1 4

1 2

1

1 51. P ⫽ 14 , d 53. y ⫽ 2 csc 15␲t2 55. y ⫽ 3 tana tb 2 2␲ ␲ 3␲ 57. y ⫽ 2 cota tb 59. , 61. about 137.8 ft 3 8 8 63. a. 20␲ cm 62.8 cm b. 80 cm; it is a square. c. n P

64.984 63.354 63.063 62.853 getting close to 20␲ 65. a. no; 35° b. 1.05 c. Angles will be greater than 68.2°; soft rubber on sandstone 67. a. 5.67 units b. 86.5° c. Yes. Range of tan ␪ is (⫺q, q ). d. The closer ␪ gets to 90°, the longer the line segment gets. ⫺1 ⫹ 25 69. sin 0.6662394325 ⫽ 0.618033989 ; cos x ⫽ tan x can be 2 2 rewritten as sin x ⫽ 1 ⫺ sin x, which can in turn be converted to sin2 1⫺x2 ⫽ 1 ⫹ sin 1⫺x2, which is the basis of the golden ratio. 71. 21,266,032 km2 73. t x y 0 ␲ 4 ␲ 2 ␲ 3␲ 4 3␲ 2 2␲

1

0

12 2

12 2

0

1

⫺1

0

⫺12 2

12 2

0

⫺1

1

0

b. 2␲ ⫺ 5.94 0.343 c. sin t, tan t 3␲ 8 10. a. A ⫽ 6, P ⫽ b. f 1t2 ⫽ ⫺6 cosa tb c. f 1␲2 ⫽ 3 4 3

23 2 2,

5␲ , negative since x 6 0 6 ␲ 4. QI, cos t ⫽ 12 , sin t ⫽ 13 2 ,t ⫽ 3

cos t ⫽ ⫺12 , sin t ⫽

t

49. P ⫽ 4␲, b

10 20 30 100

t

9. a. QIV

1. 1⫺12 , 1 2

t

4


y

2

2 4

2 3

3

3. QIV, negative since y 6 0

23 2

2. t ⫽

Exercises 6.5, pp. 590–595 1. 3. b. c.

y ⫽ A sin1Bt ⫹ C2 ⫹ D, y ⫽ A cos1Bt ⫹ C2 ⫹ D 0 ⱕ Bt ⫹ C 6 2␲ 5. Answers will vary. 7. a. A ⫽ 50, P ⫽ 24 ⫺25 c. [1.6, 10.4] 9. a. A ⫽ 200, P ⫽ 3 b. ⫺175 [1.75, 2.75]

11. y ⫽ 40 sina 100

␲ tb ⫹ 60 15

y

13. y ⫽ 8 sina 20

80

16

60

12

40

8

20

y

4 6

12

18

40

30 t

24

15. a. y ⫽ 5 sina b.

␲ tb ⫹ 12 180

60

␲ tb ⫹ 34 12

y

240 300 360 t

␲ 17. a. y ⫽ ⫺6.4 cosa tb ⫹ 12.4 6 b. 20 y

32

16

24

12

16

8

8

120 180

4 5

10

15

25 t

20

2

4

6

8

10

12 t

c. 134 days c. 1:30 A.M., 10:30 A.M. 19. a. P ⫽ 11 yr b. 1500 P c. max ⫽ 1200, min ⫽ 700 d. about 2 yr. 1200 900 600 300 2

1 ␲ tan a b is undefined because is und. 2 0

4

6

8

10

12 t

2␲ 2␲ 21. P1t2 ⫽ 250 cos c 1t ⫺ 2.752 d ⫹ 950; P1t2 ⫽ 250 sina tb ⫹ 950 11 11 23. A ⫽ 120; P ⫽ 24; HS: 6 units right; VS: (none); PI: 6 ⱕ t 6 30 25. A ⫽ 1; P ⫽ 12; HS: 2 units right; VS: (none); PI: 2 ⱕ t 6 14 27. A ⫽ 1; P ⫽ 8; HS: 23 unit left; VS: (none); PI: ⫺23 ⱕ t 6 22 3 29. A ⫽ 24.5; P ⫽ 20; HS: 2.5 units right; VS: 15.5 units up; PI: 2.5 ⱕ t 6 22.5


SA-34

2/2/11

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Page SA-34


31. A ⫽ 28; P ⫽ 12; HS: 52 units right; VS: 92 units up; PI: 52 ⱕ t 6 29 2 33. A ⫽ 2500; P ⫽ 8; HS: 13 unit left; VS: 3150 units up; PI: ⫺13 ⱕ t 6 23 3 ␲ ␲ ␲ 35. y ⫽ 250 sina tb ⫹ 350 37. y ⫽ 5 sina t ⫹ b ⫹ 13 12 50 2 ␲ ␲ t⫹ b⫹7 39. y ⫽ 4 sina 180 4 41. 100 y 43. 4 y 80

3

60

2

40

1

20

0 1

2

4

6

8

10 t

Exercises 6.6, pp. 604–610 ␲ 4

␲ 2

2

3␲ 4

␲t

1

t

3 4

45.

14 12 10 8 6 4 2 8 42 4 6 8 10

49.

47.

y

4 3 2 1 1

4

5 4 3 2 1 54321 1 2 3 4 5

51.

25 20 15 10 5 1 5 10 15 20 25

1 2 3 4 5 t

y

1 2 3 4 5 6

8 12 16 t

y

y

1

16

365

8

1. ␪ ⫽ tan⫺1x 3. opposite, hypotenuse 5. To find the measures of all three angles and all three sides. 13 13 12 5 7. sin ␪ ⫽ 12 13 , csc ␪ ⫽ 12 , sec ␪ ⫽ 5 , tan ␪ ⫽ 5 , cot ␪ ⫽ 12 85 13 84 85 9. cos ␪ ⫽ 13 , sec ␪ ⫽ , cot ␪ ⫽ , sin ␪ ⫽ , csc ␪ ⫽ 85 13 84 85 84 5 15 2 , sec ␪ ⫽ 5 15 , tan ␪ ⫽ 11 2 11. sin ␪ ⫽ 511 2 , csc ␪ ⫽ 11 , cos ␪ ⫽ 5 15 15 13. Angles 15. Sides Angles Sides A ⫽ 30° a ⫽ 98 cm A ⫽ 45° a ⫽ 9.9 mm B ⫽ 60° b ⫽ 98 13 cm B ⫽ 45° b ⫽ 9.9 mm C ⫽ 90° c ⫽ 196 cm C ⫽ 90° c ⫽ 9.9 12 mm 17.

2

t

2␲ 2␲ 1 1 2␲ ,B⫽ ;f⫽ ,P⫽ ;B⫽ ⫽ 2␲f. 53. P ⫽ B P P f 1/f 1 A sin1Bt2 ⫽ A sin 3 12␲f 2 t 4 55. a. P ⫽ 4 sec, f ⫽ cycle/sec 4 b. ⫺4.24 cm, moving away c. ⫺4.24 cm, moving toward d. about 1.76 cm; avg. vel. ⫽ 3.52 cm/sec; greater, still gaining speed 5␲ 57. d1t2 ⫽ 15 cosa tb 59. red S D2; blue S A2 4 61. D2: y ⫽ sin 3 146.84 12␲t2 4 ; P 0.0068 sec; G3: y ⫽ sin 3 39212␲t2 4 ; P 0.00255 sec ␲ 63. y ⫽ 5.2 tana xb; P ⫽ 12; asymptotes at x ⫽ 6 ⫹ 12k, k 僆⺪; 12 using (3, 5.2), A ⫽ 5.2; at x ⫽ 2, model gives y 3.002; at x ⫽ ⫺2, model gives y ⫺3.002; answers will vary. 65. Answers will vary; y ⫽ 11.95 tan ␪; P ⫽ 180°; asymptotes at ␪ ⫽ 90° ⫹ 180°k; A ⫽ 11.95 from 130°, 6.9 cm2; pen is 12 cm long. ␲ 67. a. L1t2 ⫽ 10 csc a tb b. 20 m 6 69. a. Caracas: 11.4 hr, Tokyo: 9.9 hr b. (i) Same # of hours on 79th day & 261st day (ii) Caracas: 81 days, Tokyo: 158 days

0

71. a. Adds 12 hr. The sinusoidal behavior is actually based on hours more/less than an average of 12 hr of light. b. Means 12 hr of light and dark on March 20, day 79 (Solstice). c. Additional hours of deviation from average. In the north, the planet is tilted closer toward the Sun or farther from Sun, depending on date. Variations will be greater. 73. Answers will vary. 75. QIII; 3.7 ⫺ ␲ 0.5584 i25 ⫺2 77. sum: ⫺2, difference: 2i 25, product: 6, quotient: ⫺ 3 3

21.

23. 35. 47. 55.

Angles A ⫽ 22° B ⫽ 68°

Sides a ⫽ 14 m b 34.65 m

C ⫽ 90° verified

c 37.37 m

Angles A ⫽ 32° B ⫽ 58° C ⫽ 90°

Sides a ⫽ 5.6 mi b 8.96 mi c 10.57 mi

verified

Angles Sides A ⫽ 65° a ⫽ 625 mm B ⫽ 25° b 291.44 mm C ⫽ 90° c 689.61 mm verified 0.4540 25. 0.8391 27. 1.3230 29. 0.9063 31. 27° 33. 40° 40.9° 37. 65° 39. 44.7° 41. 20.2° 43. 18.4° 45. 46.2° 61.6° 49. 21.98 mm 51. 3.04 mi 53. 177.48 furlongs They have like values. 57. They have like values. 1 13 13 13 1 2 13 , , , , , 13, 2, , 13 2 2 3 2 2 3 6 ⫹ 2 13 67. 7 ⫹ 413 69. ␪ 11.0°, ␤ 23.9°, ␥ 145.1° approx. 300.6 m 73. approx. 481.1 m 75. 87 ft 9 in. a. approx. 250.0 yd b. approx. 351.0 yd c. approx. 23.1 yd approx. 1815.2 ft; approx. 665.3 ft 81. approx. 118.1 mph approx. 386.0 ⍀ a. 875 m b. 1200 m c. 1485 m; 36.1°

59. 43° 65. 71. 77. 79. 83. 85.

19.

61. 21°

63.

875 m ␪ 1200 m 87. approx. 450 ft 89. a. approx. 20.2 cm for each side b. approx. 35.3° 91. S 85° W 93. a. approx. 3055.6 mi b. approx. 9012.8 mi c. approx. 7 hr, 13 min 95. a. local max: 1⫺5, 22, (2, 3); local min: 1⫺2, ⫺12, 1⫺7, ⫺22, 16, ⫺32 b. zeroes: x ⫽ ⫺6, ⫺3, ⫺1, 4 c. T1x2T: x 僆 1⫺5, ⫺22 ´ 12, 62; T1x2c: x 僆 1⫺7, ⫺52 ´ 1⫺2, 22 d. T1x2 7 0: x 僆 1⫺6, ⫺32 ´ 1⫺1, 42; T1x2 6 0: x 僆 1⫺7, ⫺62 ´ 1⫺3,⫺12 ´ 14, 62 97. d 53.74 in. D 65.82 in.


2/2/11

3:44 PM

Page SA-35


Exercises 6.7, pp. 618–621 1. origin, x-axis 3. positive, clockwise 5. Answers will vary. 7. slope ⫽ 13, equation: y ⫽ 13x, 1 13 , cos 60° ⫽ , tan 60° ⫽ 13 sin 60° ⫽ 2 2 y 9. QI/III; 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

3 3 ; (⫺4, ⫺3): sin ␪ ⫽ ⫺ 5 5 4 4 cos ␪ ⫽ ⫺ cos ␪ ⫽ 5 5 3 3 tan ␪ ⫽ tan ␪ ⫽ 4 4 y QII/QIV; 5

(4, 3): sin ␪ ⫽

11.

4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

13.

15.

17.

19.

21.

23.

25.

27.

13, ⫺ 132: sin ␪ ⫽ ⫺

1 2 13 13 cos ␪ ⫽ ⫺ cos ␪ ⫽ 2 2 1 1 tan ␪ ⫽ ⫺ tan ␪ ⫽ ⫺ 13 13 15 17 8 17 15 sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , sec ␪ ⫽ , tan ␪ ⫽ , 17 15 17 8 8 8 cot ␪ ⫽ 15 29 21 ⫺20 sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , 29 21 29 ⫺29 ⫺21 ⫺20 sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 20 20 21 ⫺ 12 12 sin ␪ ⫽ , csc ␪ ⫽ ⫺ 12, cos ␪ ⫽ , 2 2 sec ␪ ⫽ 12, tan ␪ ⫽ ⫺1, cot ␪ ⫽ ⫺1 1 13 , sin ␪ ⫽ , csc ␪ ⫽ 2, cos ␪ ⫽ 2 2 2 1 sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 13 13 13 4 117 1 sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , 4 117 117 1 sec ␪ ⫽ 117, tan ␪ ⫽ 4, cot ␪ ⫽ 4 ⫺2 ⫺ 113 ⫺3 sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , 2 113 113 ⫺ 113 2 3 sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 3 3 2 6 161 ⫺5 sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , 6 161 161 ⫺ 161 ⫺6 ⫺5 sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 5 5 6 ⫺2 15 121 1 , sin ␪ ⫽ , csc ␪ ⫽ ⫺ , cos ␪ ⫽ 121 2 15 121 ⫺1 sec ␪ ⫽ 121, tan ␪ ⫽ ⫺215, cot ␪ ⫽ 2 15

1 (⫺3, 13): sin ␪ ⫽ ; 2

SA-35

29. r ⫽ 10 31. x ⫽ 0, y ⫽ k; k 7 0; r ⫽ k; k 0 k sin 90° ⫽ , cos 90° ⫽ , tan 90° ⫽ , k k 0 sin 90° ⫽ 1, cos 90° ⫽ 0, tan 90° is undefined csc 90° ⫽ 1, sec 90° is undefined cot 90° ⫽ 0 33. 60° 35. 45° 37. 45° 39. 68° 41. 40° 43. 11.6° 1 13 1 47. QII 49. sin ␪ ⫽ ⫺ ; cos ␪ ⫽ ; tan ␪ ⫽ ⫺ 2 2 13

45. QII

⫺ 12 12 ; cos ␪ ⫽ ; tan ␪ ⫽ ⫺1 2 2 ⫺ 13 ⫺1 ; cos ␪ ⫽ ; tan ␪ ⫽ 13 53. sin ␪ ⫽ 2 2 1 ⫺ 13 1 ; tan ␪ ⫽ 55. sin ␪ ⫽ ⫺ ; cos ␪ ⫽ 2 2 13 4 ⫺3 ⫺5 , csc ␪ ⫽ , cos ␪ ⫽ , 57. x ⫽ 4, y ⫽ ⫺3, r ⫽ 5; QIV; sin ␪ ⫽ 5 3 5 ⫺4 5 ⫺3 sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 4 4 3 ⫺37 ⫺35 , csc ␪ ⫽ , 59. x ⫽ ⫺12, y ⫽ ⫺35, r ⫽ 37; QIII; sin ␪ ⫽ 37 35 12 ⫺12 ⫺37 35 cos ␪ ⫽ , sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 37 12 12 35 2 12 1 , 61. x ⫽ 2 12, y ⫽ 1, r ⫽ 3; QI; sin ␪ ⫽ , csc ␪ ⫽ 3, cos ␪ ⫽ 3 3 3 1 sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 212 2 12 2 12 ⫺8 ⫺7 , csc ␪ ⫽ , 63. x ⫽ ⫺ 115, y ⫽ ⫺7, r ⫽ 8; QIII; sin ␪ ⫽ 8 7 115 ⫺ 115 8 7 , sec ␪ ⫽ ⫺ cos ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 8 7 115 115 65. 52° ⫹ 360°k 67. 87.5° ⫹ 360°k 69. 225° ⫹ 360°k 13 ⫺1 1 13 1 , , ⫺ 13 75. ⫺ , , 71. ⫺107° ⫹ 360°k 73. 2 2 2 2 13 ⫺ 13 ⫺1 , cos ␪ ⫽ , tan ␪ ⫽ 13 77. sin ␪ ⫽ 2 2 ⫺ 13 1 , cos ␪ ⫽ ⫺ , tan ␪ ⫽ 13 79. sin ␪ ⫽ 2 2 ⫺1 ⫺13 1 , cos ␪ ⫽ , tan ␪ ⫽ 81. sin ␪ ⫽ 2 2 13 ⫺13 1 ⫺1 , cos ␪ ⫽ , tan ␪ ⫽ 83. sin ␪ ⫽ 2 2 13 85. QIV, neg., ⫺0.0175 87. QIV, neg., ⫺1.6643 89. QIV, neg., ⫺1.5890 91. QI, pos., 0.0872 93. a. approx. 144.78 units2 b. 53° c. The parallelogram is a ab rectangle whose area is A ⫽ ab. d. A ⫽ sin ␪ 2 95. ␪ ⫽ 60° ⫹ 360°k and ␪ ⫽ 300° ⫹ 360°k 97. ␪ ⫽ 240° ⫹ 360°k and ␪ ⫽ 300° ⫹ 360°k 99. ␪ ⫽ 61.1° ⫹ 360°k and ␪ ⫽ 118.9° ⫹ 360°k 101. ␪ ⫽ 113.0° ⫹ 180°k 103. 1890°; 90° ⫹ 360°k 105. head first; 900° 107. approx. 701.6° 109. Answers will vary. 111. a. 12,960° b. 125.66 in. c. 15,080 in. d. 85.68 mph ln 0.32 113. t ⫽ 22.79 115. y ⫽ ⫺54 x ⫹ 2 ⫺0.05 51. sin ␪ ⫽


SA-36

2/2/11

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Page SA-36

Student Answer Appendix M⫹m b 2 m⫺M 2m ⫺ M ⫺ m ⫽ ⫽ ⫺1 ⫽ M⫺m M⫺m M⫺m 2

m⫺a

Exercises 6.8, pp. 628–633 m⫺D ⫽ 33. A

1. sin(Bx ⫹ C), A ⫹ D 3. critical 5. Answers will vary. ␲ ␲ ␲ 7. y ⫽ 25 sin a xb ⫹ 50 9. y ⫽ 2.25 sin a x ⫹ b ⫹ 5.25 6 12 4 ␲ 2␲ b ⫹ 782 11. y ⫽ 503 sin a x ⫹ 6 3 13. a. y 49.26 sin10.213x ⫺ 1.1042 ⫹ 51.43 b. y 49 sin10.203x ⫺ 0.9632 ⫹ 51 c. at day 31 5.6 15. a. y 5.88 sin10.523x ⫺ 0.5212 ⫹ 16.00 b. y 6 sin10.524x ⫺ 0.5242 ⫹ 16 c. at month 9 0.12 ␲ 17. a. D ⫽ 2000 cos a tb b. north, 1258.6 mi 60 ␲ 4␲ 19. a. T1x2 ⫽ 19.6 sin a x ⫹ b ⫹ 84.6 b. about 94.4°F 6 3 c. beginning of May 1x 5.12 to end of August 1x 8.92 ␲ 13␲ b ⫹ 98.6 21. a. T1t2 ⫽ 0.4 sina t ⫹ 12 12 b. at 11 A.M. and 11 P.M. c. from t ⫽ 1 to t ⫽ 9, about 8 hr ␲ ␲ 23. P ⫽ 12, B ⫽ , C ⫽ ; using (4, 3) gives A ⫽ ⫺313, 12 2 ␲ ␲ so f 1x2 ⫽ ⫺313 tan a x ⫹ b a. f 12.52 6.77 12 2 b. f 1x2 ⫽ 16 for x 1.20 ␲ 25. a. using (18, 10) gives A 4.14; H1d2 ⫽ 4.14 tan a db 48 b. 12.2 cm c. 21.9 mi 27. a. T1m2 15.328 sin10.461m ⫺ 1.6102 ⫹ 85.244 b. 110 Month

0

13

35. a. 30, q 2

b. 1⫺q, q 2

c. 1⫺q, q 2

37. 15␲ 47.1 ft /min

Making Connections, p. 633 1. h

3. d

5. e

7. e 9. c 11. b

13. f

15. h

Summary and Concept Review, pp. 634–642 1. 147.613°

2. 32° 52¿ 12–

4. approx. 692.82 yd 8. ⫺

1 2

5. 120°

3. 10.125 ⫻ 13.5 ⫻ 16.875 7␲ 6. 7. approx. 4.97 units 6

9. s ⫽ 25.5 cm, A ⫽ 191.25 cm2

10. r 41.74 in., A 2003.48 in2 11. ␪ ⫽ 4.75 rad, s ⫽ 38 m 12. a. approx. 9.4248 rad/sec b. approx. 3.9 ft/sec c. about 15.4 sec 6 113 6 113 6 113 6 13. y ⫽ ⫺ , a⫺ , b, a⫺ , ⫺ b, and a , b 7 7 7 7 7 7 7 17 4 14. sin t ⫽ ⫺ , csc t ⫽ ⫺ , 4 17 3 4 17 3 cos t ⫽ , sec t ⫽ , tan t ⫽ ⫺ , cot t ⫽ ⫺ 4 3 3 17 ␲ 2␲ and 16. t 2.44 3 3 18. A ⫽ 3, P ⫽ 2␲ 15.

1

71

3

82

5

95

7

101

9

94

19. A ⫽ 1, P ⫽ ␲

y

4 3 2 1

Temp. (°F)

17. a. approx. 19.6667 rad b. 25 rad

y 1 0.5 ␲ 2

1 2 3 4

␲

␲

2␲ t

3␲ 2

2

␲ 2

0.5

t

1

␲ 2

20. A ⫽ 1.7, P ⫽ 2

21. A ⫽ 2, P ⫽

y

1 2

y 2

11

60

80

1

1

0

c. max difference is about 1°F in months 6 and 8 29. a. f 1x2 49.659 sin10.214x ⫺ 0.6892 ⫹ 48.328 b. about 26.8% 7␲ 2␲ b ⫹ 49.5; values for A, B, and D are very c. g1x2 ⫽ 49.5 sina x ⫺ 31 62 close; some variation in C. 31. a. Reno: R1t2 0.452 sin10.396t ⫹ 1.8312 ⫹ 0.750 b. The graphs intersect at t 2.6 and t 10.5. Reno gets more rainfall than Cheyenne for about 4 months of the year.

␲ 8

1

␲ 4

␲t 2

3␲ 8

1

4

1 8

1

1 4

t

2

2

22. A ⫽ 3, P ⫽

23. y ⫽ 0.75 sin 16t2

1 199

24. green; red

y

4

1

8

2 0

3

1 398

2

1 199

t

4

25. P ⫽ 2␲ 0

y

12

10 8 6 4 2 ␲

␲ 2

2 2 4 6 8 10

0.6

␲

29.

t

3␲ 2

30.

y 12 6 2␲ ␲

6

12

26. y ⫽ 4 csc13␲t2 7␲ 27. a. tan a b ⫽ ⫺1 4 ␲ 1 b. cot a b ⫽ 3 13 2␲ 2␲ 28. a. t ⫽ b. t ⫽ 3 3

␲

2␲

t

5 4 3 2 1 1 0.51 2 3 4 5

y

0.5

1

t


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c. The largest difference from 0 ⱕ x ⱕ 365 is about 372 at x ⫽ 85.

31. 1.55 ⫹ k␲ radians; k 僆 Z 32. 3.5860 33. 151.14 m 34. a. A ⫽ 240, P ⫽ 12, HS: 3 units right, VS: 520 units up b. 760 y 520

280 0

3

6

9

15 t

12

35. a. A ⫽ 3.2, P ⫽ 8, HS: 6 units left, VS: 6.4 units up y b. 8

52. a. y ⫽ 19.424 sin10.145x ⫺ 0.7482 ⫹ 79.581 100 ⫺ 67 b. A ⫽ ⫽ 16.5; 2 D ⫽ 67 ⫹ 16.5 ⫽ 83.5;

6 4 2

t 8

6

4

2

2

36. A ⫽ 125, P ⫽ 24, HS: 3 units right, VS: 175 units up, ␲ y ⫽ 125 cos c 1t ⫺ 32 d ⫹ 175 12 3␲ 37. A ⫽ 75, P ⫽ , HS: (none), 8 16 VS: 105 units up, y ⫽ 75 sin a tb ⫹ 105 3 ␲ 38. a. P1t2 ⫽ 0.91 sin a tb ⫹ 1.35 b. August: 1.81 in., Sept: 1.35 in. 6 39. a. A 0.80 b. A 64.3° 40. a. cot 32.6° b. cos 170° 29¿ 45– 2 41.

Angles

Sides

A ⫽ 49° B ⫽ 41° C ⫽ 90°

a ⫽ 89 in. b 77.37 in. c 117.93 in.

42.

Angles A 43.6° B 46.4° C ⫽ 90°

2␲ ⫽ 31, B 2␲ ⫽B 31 2␲ ␲ 3␲ C⫽ ⫺ 1312 ⫽ ⫺ 2 31 2 2␲ ␲ y ⫽ 16.5 sin a x ⫺ b ⫹ 83.5 31 2 c. The largest difference is about 7, at x ⫽ 26.

Sides a ⫽ 20 m b ⫽ 21 m c ⫽ 29 m

43. approx. 5.18 m 44. a. approx. 239.32 m b. approx. 240.68 m apart 45. approx. 54.5° and 35.5° 46. 207° ⫹ 360°k; answers will vary. 47. 28°, 19°, 30° ⫺12 35 37 ⫺37 48. a. sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , sec ␪ ⫽ , 37 35 37 12 ⫺35 ⫺12 tan ␪ ⫽ , cot ␪ ⫽ 12 35 ⫺3 ⫺ 113 113 2 b. sin ␪ ⫽ , csc ␪ ⫽ , cos ␪ ⫽ , sec ␪ ⫽ , 3 2 113 113 ⫺3 ⫺2 tan ␪ ⫽ , cot ␪ ⫽ 2 3 3 5 49. a. x ⫽ 4, y ⫽ ⫺3, r ⫽ 5; QIV; sin ␪ ⫽ ⫺ , csc ␪ ⫽ ⫺ , 5 3 4 5 ⫺3 ⫺4 cos ␪ ⫽ , sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 5 4 4 3 ⫺12 ⫺13 b. x ⫽ 5, y ⫽ ⫺12, r ⫽ 13; QIV; sin ␪ ⫽ , csc ␪ ⫽ , 13 12 5 13 ⫺12 ⫺5 cos ␪ ⫽ , sec ␪ ⫽ , tan ␪ ⫽ , cot ␪ ⫽ 13 5 5 12 50. a. ␪ ⫽ 135° ⫹ 180°k b. ␪ ⫽ 30° ⫹ 360°k or ␪ ⫽ 330° ⫹ 360°k c. ␪ 76.0° ⫹ 180°k d. ␪ ⫺27.0° ⫹ 360°k or ␪ ⫽ 207.0° ⫹ 360°k 51. a. y ⫽ 2187.723 sin 10.017x ⫹ 1.7512 ⫹ 2307.437 4450 ⫺ 90 b. A ⫽ ⫽ 2180; 2 D ⫽ 90 ⫹ 2180 ⫽ 2270; 359 2␲ 2␲ 3␲ 2␲ ⫽ 365, ⫽ B; C ⫽ ⫺ 11842 ⫽ ␲ B 365 2 365 730 359 2␲ y ⫽ 2180 sina x⫹ ␲b ⫹ 2270 365 730

53. a. y ⫽ 16.800 sin10.602x ⫺ 2.3412 ⫹ 70.968 b. x ⫽ 7 y ⫽ 87.01°F c. The model alternates between slightly overpredicting and underpredicting output values, and appears to be a fairly accurate model.

Practice Test, pp. 643–644 1. a. 45° b. 30° c. 3. a. 100.755° 5.

␲ 6

d.

␲ 3

b. 48° 12¿ 45–

2. 30° ⫹ 360°k; k 僆 Z 4. a. 430 mi

b. 21513 372 mi

t

sin t

cos t

tan t

csc t

sec t

cot t

0

0

1

0

undefined

1

undefined

2␲ 3

13 2

⫺

1 2

⫺ 13

2 13 3

⫺2

⫺ 13 3

7␲ 6

⫺

1 2

⫺

13 2

13 3

⫺2

⫺213 3

13

⫺

12 2

1

⫺ 12

⫺ 12

1

2

⫺ 13 3

213 3

13

5␲ 4

⫺

12 2

5␲ 3

⫺

13 2

13␲ 6

1 2

1 2

⫺ 13

13 2

13 3

⫺

213 3 2


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Student Answer Appendix ␲ 2␲ , HS: units right, VS: 19 units up 3 12 ␲, 3␲ ␲ y 4 31 32 ⱕt 6 PI: 12 4 24

5 ⫺ 121 ⫺ 121 ⫺5 6. sec ␪ ⫽ , sin ␪ ⫽ , tan ␪ ⫽ , csc ␪ ⫽ , 2 5 2 121 2 2 1 ⫺2 8 ⫺2 12 1 cot ␪ ⫽ b ⫽ ⫹ ⫽ 1; 7. a b ⫹ a 3 3 9 9 121

16. A ⫽ 12, P ⫽

16

⫺2 22 ⫺3 22 1 , csc ␪ ⫽ , cos ␪ ⫽ , 3 4 3 ⫺ 22 sec ␪ ⫽ 3, tan ␪ ⫽ ⫺222, cot ␪ ⫽ 4 23␲ rad/sec 0.1505 rad/sec 8. a. 225 ft 9.6 in. b. 480 c. 11.29 ft/sec 7.7 mph 9. Angles Sides

8

sin ␪ ⫽

A ⫽ 33° B ⫽ 57° C ⫽ 90°

12. a.

7␲ 6

b.

t

␲ 2

␲

3␲ 2

2␲ t

11␲ 6

c.

3␲ 4

b. D: t ⫽ 2␲k, k 僆 Z; R: y 僆 R, P ⫽ 2␲, y 4 2

␲ 2

2

␲

3␲ 2

2␲ t

4

␲ ␲ 19. y ⫽ 7.5 sina t ⫺ b ⫹ 12.5 6 2 20. a. t 4 b. t 2.3

10 t

1

Strengthening Core Skills, pp. 646–647

2

Exercise 1.

␲ b. D: t ⫽ 12k ⫹ 12 for k 僆 Z, 2 R: y 僆 1⫺q, ⫺1 4 ´ 3 1, ⫹ q2, P ⫽ 2␲ 3

␲

8

1

8

5␲ 3

4

4

y

6

2␲ 3

y

␲ tb ⫹ 34.1, in ten-thousands of gallons 12 b. 433,000 gal; 249,000 gal 14. a. D: t 僆 R, R: y 僆 3 ⫺2, 2 4 , P ⫽ 10, A ⫽ 2;

4

␲ 2

8

13. a. W1t2 ⫽ 18.4 sina

2

␲ 3

17. 1260° ␲ ␲ 18. a. D: t ⫽ 12k ⫹ 12, k 僆 Z; R: y 僆 R; P ⫽ ; 4 2

a 8.2 cm b 12.6 cm c ⫽ 15.0 cm

10. about 67 cm, 49.6° 11. 57.9 m

2

, 7␲ 12 7 ␲ 6

y

2

t

0

␲ 6

␲ 4

␲ 3

␲ 2

sin t ⫽ y

0

1 2

12 2

13 2

1

cos t ⫽ x

1

13 2

12 2

1 2

0

0

13 3

1

13

—

2␲ 3

3␲ 4

5␲ 6

␲

7␲ 6

5␲ 4

13 2

12 2

1 2

0

⫺1 2

⫺ 12 2

⫺1 2

⫺ 12 2

⫺ 13 2

⫺1

⫺ 13 2

⫺ 12 2

⫺ 13

⫺1

⫺ 13 3

0

13 3

1

1 ␲ 2

1

␲

3␲ 2

2␲ t

2 3

␲ ␲ c. D: t ⫽ 12k ⫹ 12 for k 僆 Z, R: y 僆 R, P ⫽ 6 3 y 8 4

4

␲ 3

2␲ 3

␲

4␲ 3

5␲ 3

2␲ t

8

15. a. y ⫽ 35.223 sin10.576x ⫺ 2.5892 ⫹ 6.120 b. Month Low (Jan S 1) Temp. (°F) 1

⫺26

3

⫺21

5

16

7

41

9

25

11

⫺14

tan t ⫽

y x

␲ 7␲ 4␲ 5␲ ␲ 7␲ Exercise 2. a. t ⫽ b. t ⫽ , c. t ⫽ , , 3 3 4 4 6 6 ␲ 7␲ d. t ⫽ , 4 4 Exercise 3. a. no solution b. t 1.2310, t 5.0522 c. t 2.8966, t 6.0382 d. t 1.9823, t 4.3009


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Cumulative Review Chapters 1–6, pp. 648–649

89

39

4

sin ␪ ⫽

hyp ⫽ 89; ␪ 64°; 90 ⫺ ␪ ⫽ 26° 4 3 ⫺ 17 ⫺ 17 5. cos t ⫽ , sin t ⫽ , tan t ⫽ , sec t ⫽ , 4 4 3 3 ⫺3 ⫺4 ⫺4 17 ⫺3 17 csc t ⫽ ⫽ , cot t ⫽ ⫽ 7 7 17 17 3 7. a. D: x 僆 c , qb, R: y 僆 30, q2 2 b. D: x 僆 1⫺q, ⫺72 ´ 1⫺7, 72 ´ 17, q2, R: y 僆 1⫺q, q2 9. a. max: (⫺2, 4), endpoint max: (4, 0) min: (2, ⫺4), endpoint min: (⫺4, 0) b. f 1x2 ⱖ 0: x 僆 3⫺4, 0 4 ´ 546 f 1x2 6 0: x 僆 10, 42 c. f 1x2c: x 僆 1⫺4, ⫺22 ´ 12, 42 f 1x2T: x 僆 1⫺2, 22 d. function is odd: f 1⫺x2 ⫽ ⫺f 1x2 11. 114.3 ft 13.

5

y

x 1 (0, 1)

5

5 x

y 2 5

15. x ⫽ ⫺9, y ⫽ 40, r ⫽ 41, QII; ⫺9 41 40 ⫺40 ⫺41 cos ␪ ⫽ , sin ␪ ⫽ , tan ␪ ⫽ , sec ␪ ⫽ , csc ␪ ⫽ , 41 41 9 9 40 ⫺9 cot ␪ ⫽ , ␪ 102.7° 40 17. S ⫽ 18 m; A ⫽ 135 m2 3 ␲ 1 19. y ⫽ sina4t ⫺ b ⫹ 2 2 2 3 y 21. 23. m ⫽ , y-intercept (0, ⫺2) 5 4 4 f(x) 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

f 1 (x)

25. about 6.85%

27.

x

y

⫺0.2

0.98

⫺0.4

0.92

⫺0.6

0.8

⫺0.8

0.6

29. x 14.61

Connections to Calculus Exercises, pp. 650–652 1. hyp ⫽ x ⫹ 4 4 2x2 ⫹ 8x , cos ␪ ⫽ x⫹4 x⫹4 x⫹4 x⫹4 csc ␪ ⫽ , sec ␪ ⫽ 4 2x2 ⫹ 8x sin ␪ ⫽

4 2x ⫹ 8x 2

x

␪

80

tan ␪ ⫽

3. x2 16

1. ⫺5 6 x 6 3 3. ␪

SA-39

, cot ␪ ⫽

2x2 ⫹ 8x 4

x

, cos ␪ ⫽

4

2x ⫹ 16 2x ⫹ 16 2x2 ⫹ 16 2x2 ⫹ 16 , sec ␪ ⫽ x 4 4 cot ␪ ⫽ x 2u2 ⫹ 169 2 13 5. cot ␪ ⫽ , sec ␪ ⫽ 7. r ⫽ u 13 sin ␪ 3 9. r ⫽ 11. x2 ⫹ y2 ⫺ 5y ⫽ 0; circle sin ␪ ⫹ 2 cos ␪ 13. 3x ⫺ 2y ⫽ 6; line 2

2

csc ␪ ⫽

CHAPTER 7 Exercises 7.1, pp. 658–661 1. sin ␪; sec ␪; cos ␪ 3. cos2 ␪, cot2 ␪ ⫹ 1 ⫽ csc2␪ 1 ⫺ sin2␪ 5. ; Answers will vary. sin ␪ sec ␪ y2 y2 x2 ⫹ y2 x2 r2 7. 1 ⫹ tan2␪ ⫽ 1 ⫹ 2 ⫽ 2 ⫹ 2 ⫽ ⫽ 2 ⫽ sec2␪ 2 x x x x x y y sin ␪ r 9. ⫽ ⫽ ⫽ tan ␪ cos ␪ x x r 11. Answers may vary; sec ␪ sin ␪ sec ␪ 1 sec ␪ 1 sin ␪ tan ␪ ⫽ ; ⫽ ; ⫽ ; ⫽ csc ␪ cos ␪ csc ␪ cot ␪ csc ␪ cot ␪ cos ␪ 13. Answers may vary: 1 ⫽ sec2␪ ⫺ tan2␪; tan2␪ ⫽ sec2␪ ⫺ 1; 1 ⫽ 1sec ␪ ⫹ tan ␪21sec ␪ ⫺ tan ␪2; tan ␪ ⫽ ⫾ 2sec2␪ ⫺ 1 cos ␪ 15. sin ␪ cot ␪ ⫽ sin ␪ ⫽ cos ␪ sin ␪ 1 cos2␪ 1 ⫽ ⫽ csc2␪ 17. sec2␪ cot2␪ ⫽ 2 2 cos ␪ sin ␪ sin2␪ 19. cos ␪1sec ␪ ⫺ cos ␪2 ⫽ cos ␪ sec ␪ ⫺ cos2␪ ⫽ 1 ⫺ cos2␪ ⫽ sin2␪ 21. sin ␪1csc ␪ ⫺ sin ␪2 ⫽ sin ␪ csc ␪ ⫺ sin2␪ ⫽ 1 ⫺ sin2 ␪ ⫽ cos2␪ 23. tan ␪1csc ␪ ⫹ cot ␪2 ⫽ tan ␪ csc ␪ ⫹ tan ␪ cot ␪ ⫽ sin ␪ 1 1 ⫹ 1 ⫽ sec ␪ ⫹ 1 ⫹1⫽ cos ␪ sin ␪ cos ␪ 2 2 2 25. tan ␪ csc ␪ ⫺ tan ␪ ⫽ tan2␪ 1csc2␪ ⫺ 12 ⫽ tan2␪ 1cot2␪2 ⫽ 1 sin ␪ 1cos ␪ ⫹ 12 sin ␪ cos ␪ ⫹ sin ␪ sin ␪ ⫽ ⫽ ⫽ tan ␪ 27. cos ␪ 11 ⫹ cos ␪2 cos ␪ cos ␪ ⫹ cos2␪ 11211 ⫹ sin ␪2 1 ⫹ sin ␪ 1 ⫽ ⫽ ⫽ sec ␪ 29. cos ␪ ⫹ cos ␪ sin ␪ 1cos ␪211 ⫹ sin ␪2 cos ␪ sin ␪1tan ␪ ⫹ 12 sin ␪ tan ␪ ⫹ sin ␪ ⫽ ⫽ 31. tan ␪11 ⫹ tan ␪2 tan ␪ ⫹ tan2␪ sin ␪ sin ␪ sin ␪ cos ␪ ⫽ cos ␪ ⫽ ⫽ tan ␪ sin ␪/cos ␪ sin ␪ 2 1sin ␪ ⫹ cos ␪2 2 sin ␪ ⫹ 2 sin ␪ cos ␪ ⫹ cos2␪ ⫽ ⫽ 33. cos ␪ cos ␪ 2 2 cos ␪ ⫹ sin ␪ ⫹ 2 sin ␪ cos ␪ 1 ⫹ 2 sin ␪ cos ␪ ⫽ ⫽ cos ␪ cos ␪ 2 sin ␪ cos ␪ 1 ⫹ ⫽ sec ␪ ⫹ 2 sin ␪ cos ␪ cos ␪ 35. 11 ⫹ sin ␪2 11 ⫺ sin ␪2 ⫽ 1 ⫺ sin2␪ ⫽ cos2␪ 1csc ␪ ⫺ cot ␪2 1csc ␪ ⫹ cot ␪2 csc2␪ ⫺ cot2␪ 1 ⫽ ⫽ ⫽ cot ␪ 37. tan ␪ tan ␪ tan ␪ 2 2 2 sin ␪ cos ␪ ⫹ sin ␪ cos ␪ 1 ⫹ ⫽ ⫽ ⫽ csc ␪ 39. sin ␪ 1 sin ␪ sin ␪


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sin ␪ cos ␪ ⫺ 1 tan ␪ sin ␪ tan ␪ cos ␪ ⫺ sin ␪ csc ␪ cos ␪ sin ␪ ⫺ 1 ⫺ ⫽ ⫽ ⫽ 41. csc ␪ cos ␪ csc ␪ cos ␪ 1 cot ␪ cos ␪ sin ␪ sec ␪ tan2␪ csc ␪ sec2␪ ⫺ sin ␪ csc ␪ sec2␪ ⫺ 1 43. ⫺ ⫽ ⫽ ⫽ ⫽ tan ␪ 1 sin ␪ sec ␪ sin ␪ sec ␪ tan ␪ sin ␪ cos ␪ sin ␪ ⫾ 21 ⫺ sin2␪ 1 45. 47. ⫾ ⫹ 1 49. B cot2␪ sin ␪ ⫾21 ⫺ sin2␪ 29 29 21 21 20 , tan ␪ ⫽ ⫺ , sec ␪ ⫽ ⫺ , csc ␪ ⫽ , cot ␪ ⫽ ⫺ 29 20 20 21 21 8 15 17 17 8 53. cos ␪ ⫽ ⫺ , sin ␪ ⫽ ⫺ , sec ␪ ⫽ ⫺ , csc ␪ ⫽ ⫺ , cot ␪ ⫽ 17 17 8 15 15 x 5 5 , sin ␪ ⫽ , tan ␪ ⫽ , 55. cos ␪ ⫽ x 2x2 ⫹ 25 2x2 ⫹ 25

15.

51. sin ␪ ⫽

2x2 ⫹ 25 2x2 ⫹ 25 , csc ␪ ⫽ x 5 1120 2 130 7 13 ⫽⫺ , tan ␪ ⫽ , sec ␪ ⫽ ⫺ , 57. cos ␪ ⫽ ⫺ 13 13 2 130 2 130 13 2130 csc ␪ ⫽ ⫺ , cot ␪ ⫽ 7 7 132 4 12 7 4 12 ⫽ , cos ␪ ⫽ ⫺ , tan ␪ ⫽ ⫺ , 59. sin ␪ ⫽ 9 9 9 7 9 7 csc ␪ ⫽ , cot ␪ ⫽ ⫺ 412 4 12

17.

sec ␪ ⫽

19.

61. V ⫽ 211 ⫺ cos2␪2 63. cos3␪ ⫽ cos ␪ cos2␪ ⫽ cos ␪11 ⫺ sin2␪2 65. tan ␪ ⫹ tan3␪ ⫽ tan ␪11 ⫹ tan2␪2 ⫽ tan ␪ sec2␪ 67. tan2␪ sec ␪ ⫺ 4 tan2␪ ⫽ tan2␪1sec ␪ ⫺ 42 ⫽ 1sec ␪ ⫺ 421tan2␪2 ⫽ 1sec ␪ ⫺ 421sec2 ⫺ 12 ⫽ 1sec ␪ ⫺ 421sec ␪ ⫺ 121sec ␪ ⫹ 12 69. cos2␪ sin ␪ ⫺ cos2␪ ⫽ cos2␪1sin ␪ ⫺ 12 ⫽ 11 ⫺ sin2␪21sin ␪ ⫺ 12 ⫽ 11 ⫹ sin ␪211 ⫺ sin ␪21sin ␪ ⫺ 12 ⫽ 11 ⫹ sin ␪211 ⫺ sin ␪21⫺1211 ⫺ sin ␪2 ⫽ ⫺111 ⫹ sin ␪211 ⫺ sin ␪2 2 m2 ⫺ m1 71. tan ␪ ⫽ 1 ⫹ m1m2 73. tan ␪ ⫽ 1 75. ⫺2 sin4␪ ⫹ 23 sin3␪ ⫹ 2 sin2␪ ⫺ 23 sin ␪ ⫽ sin ␪1⫺2 sin3␪ ⫹ 23 sin2␪ ⫹ 2 sin ␪ ⫺ 232 ⫽ sin ␪ 3 1⫺2 sin3␪ ⫹ 23 sin2␪2 ⫹ 12 sin ␪ ⫺ 232 4 ⫽ sin ␪ 3 ⫺sin2␪12 sin ␪ ⫺ 232 ⫹ 112 sin ␪ ⫺ 232 4 ⫽ sin ␪11 ⫺ sin2␪212 sin ␪ ⫺ 232 ⫽ sin ␪ cos2␪12 sin ␪ ⫺ 232 y 77. about 1148 ft 79. 2 y = 2 sin(2t)

21.

23.

1

t 1

π 2

π

3π 2

2π

2

25.

Exercises 7.2, pp. 666–669 1. identities, symmetry 3. complicated, simplify, build 5. Because we don’t know if the equation is true. 7. 11 ⫹ sin x2 3 1 ⫹ sin1⫺x2 4 ⫽ 11 ⫹ sin x211 ⫺ sin x2 ⫽ 1 ⫺ sin2x ⫽ cos2x 9. sin2 1⫺x2 ⫹ cos2x ⫽ 1⫺sin x2 2 ⫹ cos2x ⫽ sin2x ⫹ cos2x ⫽ 1 1 ⫺ sin1⫺x2 1 ⫹ sin x 1 ⫹ sin x 11. ⫽ ⫽ cos x ⫹ cos 1⫺x2 sin x cos x ⫹ cos x sin x cos x 11 ⫹ sin x2 1 ⫽ ⫽ sec x cos x

sin2x

⫽ sin2x ⫽ 1 ⫺ cos2x cos2x sin x cos x tan x ⫹ cot x ⫽ ⫹ cos x sin x sin2x ⫹ cos2x ⫽ cos x sin x 1 ⫽ cos x sin x 1 1 ⫽ cos x sin x ⫽ sec x csc x 1 csc x ⫺ sin x ⫽ ⫺ sin x sin x 1 ⫺ sin2x ⫽ sin x cos2x ⫽ sin x cos x ⫽ sin x cos x cos x ⫽ tan x 1 1sin x cos x2 cos x sec x ⫽ cot x ⫹ tan x sin x cos x ⫹ b1sin x cos x2 a sin x cos x sin x ⫽ cos2x ⫹ sin2x sin x ⫽ 1 ⫽ sin x 1 sin x ⫺ sin x sin x ⫺ csc x ⫽ csc x 1 sin x 1 b1sin x2 asin x ⫺ sin x ⫽ 1 sin x sin x 2 sin x ⫺ 1 ⫽ 1 ⫽ ⫺cos2x 1 1 ⫽ csc x ⫺ sin x 1 ⫺ sin x sin x 11sin x2 ⫽ 1 a ⫺ sin xb1sin x2 sin x sin x ⫽ 1 ⫺ sin2x sin x ⫽ cos2x sin x 1 ⫽ cos x cos x ⫽ tan x sec x cos ␪ cos ␪ # 1 ⫹ sin ␪ ⫽ 1 ⫺ sin ␪ 1 ⫺ sin ␪ 1 ⫹ sin ␪ cos ␪11 ⫹ sin ␪2 ⫽ 1 ⫺ sin2␪ cos ␪11 ⫹ sin ␪2 ⫽ cos2␪ 1 ⫹ sin ␪ ⫽ cos ␪ sin ␪ 1 ⫹ ⫽ cos ␪ cos ␪ ⫽ sec ␪ ⫹ tan ␪

13. cos2x tan2x ⫽ cos2x


2/2/11

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27.

29.

31.

33.

35.

cos x cos x # 1 ⫺ sin x ⫽ 1 ⫹ sin x 1 ⫹ sin x 1 ⫺ sin x cos x11 ⫺ sin x2 ⫽ 1 ⫺ sin2x cos x11 ⫺ sin x2 ⫽ cos2x 1 ⫺ sin x ⫽ cos x cos x csc2x ⫺ cos2x csc x ⫺ ⫽ cos x csc x cos x csc x csc2x ⫺ 11 ⫺ sin2x2 ⫽ 1 cos x sin x csc2x ⫺ 1 ⫹ sin2x ⫽ cot x cot2x ⫹ sin2x ⫽ cot x sin x11 ⫺ sin x2 ⫺ sin x11 ⫹ sin x2 sin x sin x ⫺ ⫽ 1 ⫹ sin x 1 ⫺ sin x 11 ⫹ sin x2 11 ⫺ sin x2 sin x ⫺ sin2x ⫺ sin x ⫺ sin2x ⫽ 1 ⫺ sin2x ⫺2 sin2x ⫽ cos2x ⫽ ⫺2 tan2x cot x11 ⫺ csc x2 ⫺ cot x11 ⫹ csc x2 cot x cot x ⫺ ⫽ 1 ⫹ csc x 1 ⫺ csc x 11 ⫹ csc x2 11 ⫺ csc x2 cot x ⫺ cot x csc x ⫺ cot x ⫺ cot x csc x ⫽ 1 ⫺ csc2x ⫺2 cot x csc x ⫽ ⫺cot2x 2 csc x ⫽ cot x 1 2 sin x ⫽ cos x sin x 2 ⫽ cos x ⫽ 2 sec x sec2x sec2x ⫽ 1 ⫹ cot2x csc2x 1 ⫽

⫽

cos2x 1

41.

43.

45.

47.

sin2x sin2x

cos2x ⫽ tan2x 37. sin2x1cot2x ⫺ csc2x2 ⫽ sin2x cot2x ⫺ sin2x csc2x cos2x ⫽ sin2x 2 ⫺ 1 sin x ⫽ cos2x ⫺ 1 ⫽ ⫺sin2x cos x 39. cos x cot x ⫹ sin x ⫽ cos x ⫹ sin x sin x 2 cos x ⫽ ⫹ sin x sin x cos2x ⫹ sin2x ⫽ sin x 1 ⫽ sin x ⫽ csc x

49.

51.

11 ⫹ sin x2 11 ⫹ sin x2 1 ⫹ sin x ⫽ 1 ⫺ sin x 11 ⫺ sin x2 11 ⫹ sin x2 1 ⫹ 2 sin x ⫹ sin2x ⫽ 1 ⫺ sin2x 1 ⫹ 2 sin x ⫹ sin2x ⫽ cos2x 1 sin x 1 sin2x ⫽ ⫹2 ⫹ 2 cos x cos x cos x cos2x 2 ⫽ sec x ⫹ 2 tan x sec x ⫹ tan2x ⫽ 1sec x ⫹ tan x2 2 ⫽ 1tan x ⫹ sec x2 2 1cos x ⫺ sin x2 1cos x ⫹ sin x2 cos x ⫺ sin x ⫽ 1 ⫺ tan x 11 ⫺ tan x2 1cos x ⫹ sin x2 1cos x ⫺ sin x21cos x ⫹ sin x2 ⫽ sin2x cos x ⫹ sin x ⫺ sin x ⫺ cos x 1cos x ⫺ sin x2 1cos x ⫹ sin x2 ⫽ sin2x cos x a1 ⫺ b cos2x 1cos x ⫺ sin x2 1cos x ⫹ sin x2 ⫽ cos x11 ⫺ tan2x2 1cos x ⫺ sin x2 1cos x ⫹ sin x2 ⫽ cos x11 ⫺ tan x2 11 ⫹ tan x2 1cos x ⫺ sin x21cos x ⫹ sin x2 ⫽ 1cos x ⫺ sin x211 ⫹ tan x2 cos x ⫹ sin x ⫽ 1 ⫹ tan x 2 2 1tan x ⫹ cot x2 1tan x ⫺ cot x2 tan x ⫺ cot x ⫽ tan x ⫺ cot x tan x ⫺ cot x ⫽ tan x ⫹ cot x sin x cos x ⫽ ⫹ cos x sin x sin2x ⫹ cos2x ⫽ cos x sin x 1 ⫽ cos x sin x 1 1 ⫽ cos x sin x ⫽ sec x csc x ⫽ csc x sec x cos x 1cos x sin x2 sin x cot x ⫽ cos x sin x 1cos x sin x2 cot x ⫹ tan x a ⫹ b sin x cos x 2 cos x ⫽ cos2x ⫹ sin2x cos2x ⫽ 1 ⫽ 1 ⫺ sin2x 1sec2x ⫺ tan2x2 1sec2x ⫹ tan2x2 sec4x ⫺ tan4x ⫽ 2 2 sec x ⫹ tan x 1sec2x ⫹ tan2x2 2 ⫽ sec x ⫺ tan2x ⫽1 1cos2x ⫺ sin2x2 1cos2x ⫹ sin2x2 cos4x ⫺ sin4x ⫽ cos2x cos2x 1cos2x ⫺ sin2x2 112 ⫽ cos2x cos2x sin2x ⫽ ⫺ 2 cos x cos2x 2 ⫽ 1 ⫺ tan x ⫽ 1 ⫺ 1sec2x ⫺ 12 ⫽ 1 ⫺ sec2x ⫹ 1 ⫽ 2 ⫺ sec2x

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53. 1sec x ⫹ tan x2 2 ⫽ sec2x ⫹ 2 sec x tan x ⫹ tan2x 1 2 sin x sin2x ⫽ ⫹ ⫹ 2 2 cos x cos x cos2x 1 ⫹ 2 sin x ⫹ sin2x ⫽ cos2x 11 ⫹ sin x2 2 ⫽ cos2x 1sin x ⫹ 12 2 ⫽ cos2x cos x sin x csc x cos2x sec x ⫹ sin2x sec x ⫹ csc x sin x cos x ⫹ ⫹ ⫽ 55. sin x cos x sec x sin x cos x sec x sec x1cos2x ⫹ sin2x2 ⫹ 112cos x ⫽ sin x112 sec x ⫹ cos x ⫽ sin x 4 4 1sin2x ⫹ cos2x21sin2x ⫺ cos2x2 sin x ⫺ cos x 57. ⫽ sin3x ⫹ cos3x 1sin x ⫹ cos x21sin2x ⫺ sin x cos x ⫹ cos2x2 1121sin x ⫹ cos x21sin x ⫺ cos x2 ⫽ 1sin x ⫹ cos x21sin2x ⫹ cos2x ⫺ sin x cos x2 sin x ⫺ cos x ⫽ 1 ⫺ sin x cos x 59. Answers will vary. 61. Answers will vary. 63. Answers will vary. 65. identity 67. not an identity 69. a. d2 ⫽ 120 ⫹ x cos ␪2 2 ⫹ 120 ⫺ x sin ␪2 2 ⫽ 400 ⫹ 40x cos ␪ ⫹ x2cos2␪ ⫹ 400 ⫺ 40 x sin ␪ ⫹ x2sin2␪ ⫽ 800 ⫹ 40x1cos ␪ ⫺ sin ␪2 ⫹ x2 1cos2␪ ⫹ sin2␪2 ⫽ 800 ⫹ 40x1cos ␪ ⫺ sin ␪2 ⫹ x2 b. 42.2 ft 71. a. h ⫽ 1cot x ⫹ tan x; h 3.76 cos x sin x b. cot x ⫹ tan x ⫽ ⫹ sin x cos x cos2x ⫹ sin2x ⫽ sin x cos x 1 ⫽ sin x cos x ⫽ csc x sec x; h ⫽ 1csc x sec x h 3.76; yes 73. D2 ⫽ 400 ⫹ 40x cos ␪ ⫹ x2 D 40.5 ft 75. sin ␣ ⫽ cos ␪ 1sin3x ⫺ cos3x21sin3x ⫹ cos3x2 sin6x ⫺ cos6x 77. ⫽ sin4x ⫺ cos4x 1sin2x ⫹ cos2x21sin2x ⫺ cos2x2 ⫽

b. cos1120° ⫺ 45°2 ⫽ cos 120° cos 45° ⫹ sin 120° sin 45° ⫽ ⫺ 22 ⫹ 26 26 ⫺ 22 ⫽ 4 4 13 16 ␲ 13. cos15␪2 15. 17. ⫺ 19. sin 33° 21. cot a b 2 65 12 ␲ 23. cos a ⫹ ␪b 25. sin(8x) 27. tan 13␪2 29. 1 31. 13 3 ⫺304 ⫺304 16 ⫹ 12 16 ⫹ 12 33. a. b. 35. 37. 425 297 4 4 1 13 3 ⫹ 13 ⫽⫺ 39. ⫺ 41. 3 13 3 ⫺ 13

16 ⫺ 12 4 b. sin1135° ⫺ 120°2 ⫽ sin 135° cos 120° ⫺ cos 135° sin 120° 12 1 ⫺ 12 13 ⫽a b a⫺ b ⫺ a ba b 2 2 2 2 ⫺ 12 16 ⫽ ⫹ 4 4 16 ⫺ 12 ⫽ 4 43. a. sin145° ⫺ 30°2 ⫽ sin 45° cos 30° ⫺ cos 45° sin 30° ⫽

45. 49. 51. 53. 55.

57.

59.

61.

63.

1sin x ⫺ cos x21sin2x ⫹ sin x cos x ⫹ cos2x2 1sin x ⫹ cos x21sin2x ⫺ sin x cos x ⫹ cos2x2 1121sin x ⫹ cos x21sin x ⫺ cos x2

⫽ 11 ⫹ sin x cos x211 ⫺ sin x cos x2 ⫽ 1 ⫺ sin2x cos2x y 79. 81. 752.3 yd 5

65.

4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

67.

Exercises 7.3, pp. 675–680 1. false, QII 3. repeat, opposite 12 ⫺ 16 12 ⫺ 16 7. 9. 4 4


11. a. cos145° ⫹ 30°2 ⫽ cos 45° cos 30° ⫺ sin 45° sin 30° ⫽

26 ⫺ 22 4

69.

319 319 ⫺12 ⫺ 16 480 47. a. b. c. ⫺ 4 481 481 360 3416 3416 1767 a. b. ⫺ c. 4505 4505 2937 12 ⫹ 5 13 12 13 ⫺ 5 12 ⫹ 5 13 a. b. c. 26 26 12 13 ⫺ 5 96 247 247 190° ⫺ ␣2 ⫹ ␪ ⫹ 190° ⫺ ␤2 ⫽ 180° a. b. c. 265 265 96 sin1␲ ⫺ ␣2 ⫽ sin ␲ cos ␣ ⫺ cos ␲ sin ␣ ⫽ 0 ⫺ 1⫺12sin ␣ ⫽ sin ␣ ␲ ␲ ␲ cosax ⫹ b ⫽ cos x cosa b ⫺ sin x sina b ⫽ 4 4 4 12 12 12 b ⫺ sin xa b⫽ 1cos x ⫺ sin x2 cos xa 2 2 2 ␲ tan x ⫹ tana b ␲ 4 tan x ⫹ 1 1 ⫹ tan x tanax ⫹ b ⫽ ⫽ ⫽ 4 ␲ 1 ⫺ tan x 1 ⫺ tan x 1 ⫺ tan x tana b 4 cos1␣ ⫹ ␤2 ⫹ cos1␣ ⫺ ␤2 ⫽ cos ␣ cos ␤ ⫺ sin ␣ sin ␤ ⫹ cos ␣ cos ␤ ⫹ sin ␣ sin ␤ ⫽ 2 cos ␣ cos ␤ cos12t2 ⫽ cos1t ⫹ t2 ⫽ cos t cos t ⫺ sin t sin t ⫽ cos2t ⫺ sin2t sin13t2 ⫽ sin12t ⫹ t2 ⫽ sin12t2 cos t ⫹ cos12t2 sin t ⫽ 2 sin t cos t cos t ⫹ 1cos2t ⫺ sin2t2 sin t ⫽ 2 sin t cos2t ⫹ sin t cos2t ⫺ sin3t ⫽ 3 sin t cos2t ⫺ sin3t ⫽ 3 sin t11 ⫺ sin2t2 ⫺ sin3t ⫽ 3 sin t ⫺ 3 sin3t ⫺ sin3t ⫽ ⫺4 sin3t ⫹ 3 sin t ␲ ␲ ␲ cosax ⫺ b ⫽ cos x cosa b ⫹ sin x sina b 4 4 4 12 12 ⫽ cos xa b ⫹ sin xa b 2 2 12 ⫽ 1cos x ⫹ sin x2 2 Wk 1 ⫺ 13 F⫽ c 1 ⫹ 13


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Student Answer Appendix cos s cos t ␻C sin1s ⫹ t2 cos s cos t ⫽ ␻C1sin s cos t ⫹ cos s sin t2 1 cos s cos t cos s cos t ⫽ 1 ␻C1sin s cos t ⫹ cos s sin t2 cos s cos t 1 ⫽ sin s cos t cos s sin t b ␻C a ⫹ cos s cos t cos s cos t 1 ⫽ ␻C1tan s ⫹ tan t2 sin ␪ cos190° ⫺ ␪2 A 73. ⫽ B cos ␪ sin190° ⫺ ␪2 sin ␪ 1cos 90° cos ␪ ⫹ sin 90° sin ␪2 A ⫽ B cos ␪ 1sin 90° cos ␪ ⫺ cos 90° sin ␪2 sin ␪10 ⫹ sin ␪2 ⫽ cos ␪ 1cos ␪ ⫺ 02 sin2␪ ⫽ cos2␪ ⫽ tan2␪ 71. R ⫽

75. verified using sum identity for sine f 1x ⫹ h2 ⫺ f 1x2

sin1x ⫹ h2 ⫺ sin x ⫽ h h sin x cos h ⫹ cos x sin h ⫺ sin x sin x cos h ⫺ sin x ⫹ cos x sin h ⫽ ⫽ h h sin x1cos h ⫺ 12 ⫹ cos x sin h cos h ⫺ 1 sin h ⫽ sin x a b ⫹ cos x a b ⫽ h h h

77.

1 ⫺12 79. 81. 2 2 83. D ⫽ d, so D2 ⫽ d2, and D2 ⫽ 1cos ␣ ⫺ cos ␤2 2 ⫹ 1sin ␣ ⫺ sin ␤2 2 ⫽ cos2␣ ⫺ 2 cos ␣ cos ␤ ⫹ cos2␤ ⫹ sin2␣ ⫺ 2 sin ␣ sin ␤ ⫹ sin2␤ ⫽ 2 ⫺ 2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ d2 ⫽ sin2 1␣ ⫺ ␤2 ⫹ 3 cos1␣ ⫺ ␤2 ⫺ 1 4 2 ⫽ sin2 1␣ ⫺ ␤2 ⫹ cos2 1␣ ⫺ ␤2 ⫺ 2 cos1␣ ⫺ ␤2 ⫹ 1 ⫽ 2 ⫺ 2 cos1␣ ⫺ ␤2 D2 ⫽ d2 so 2 ⫺ 2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ ⫽ 2 ⫺ 2 cos1␣ ⫺ ␤2 ⫺2 cos1␣ ⫺ ␤2 ⫺2 cos ␣ cos ␤ ⫺ 2 sin ␣ sin ␤ ⫽ ⫺2 ⫺2 cos ␣ cos ␤ ⫹ sin ␣ sin ␤ ⫽ cos1␣ ⫺ ␤2 ␲ 85. a. P ⫽ 16 b. P ⫽ 87. about 19.3 ft 2

Exercises 7.4, pp. 688–693 1. sum, ␣ ⫽ ␤

3. 2x, x 5. Answers will vary. 119 120 120 , cos12␪2 ⫽ , tan12␪2 ⫽ ⫺ 7. sin12␪2 ⫽ ⫺ 169 169 119 1519 720 720 , cos12␪2 ⫽ ⫺ , tan12␪2 ⫽ 9. sin12␪2 ⫽ ⫺ 1681 1681 1519 2184 6887 2184 , cos12␪2 ⫽ , tan12␪2 ⫽ 11. sin12␪2 ⫽ 7225 7225 6887 721 5280 ⫺5280 , cos12␪2 ⫽ , tan12␪2 ⫽ 13. sin12␪2 ⫽ ⫺ 5329 5329 721 7 ⫺24 24 15. sin12␪2 ⫽ ⫺ , cos12␪2 ⫽ , tan12␪2 ⫽ 25 25 7

4 3 4 17. sin ␪ ⫽ , cos ␪ ⫽ , tan ␪ ⫽ 5 5 3 20 21 21 19. sin ␪ ⫽ , cos ␪ ⫽ , tan ␪ ⫽ 29 29 20 21. sin13␪2 ⫽ sin12␪ ⫹ ␪2 ⫽ sin12␪2cos ␪ ⫹ cos 12␪2sin ␪ ⫽ 12 sin ␪ cos ␪2cos ␪ ⫹ 11 ⫺ 2 sin2␪2sin ␪ ⫽ 2 sin ␪ cos2␪ ⫹ sin ␪ ⫺ 2 sin3␪ ⫽ 2 sin ␪11 ⫺ sin2␪2 ⫹ sin ␪ ⫺ 2 sin3␪ ⫽ 2 sin ␪ ⫺ 2 sin3␪ ⫹ sin ␪ ⫺ 2 sin3␪ ⫽ 3 sin ␪ ⫺ 4 sin3␪ 1 1 12 1 23. 25. 27. 1 29. 4.5 sin(6x) 31. ⫺ cos14x2 4 2 8 8 9 3 3 33. ⫹ cos12x2 ⫹ cos14x2 8 2 8 5 7 3 1 35. ⫺ cos12x2 ⫹ cos14x2 ⫺ cos12x2cos14x2 8 8 8 8 22 ⫺ 12 22 ⫹ 12 , cos ␪ ⫽ , tan ␪ ⫽ 12 ⫺ 1 37. sin ␪ ⫽ 2 2 22 ⫺ 13 22 ⫹ 13 , cos ␪ ⫽ , tan ␪ ⫽ 2 ⫺ 13 39. sin ␪ ⫽ 2 2 22 ⫹ 12 22 ⫺ 12 , cos ␪ ⫽ , tan ␪ ⫽ 12 ⫹ 1 41. sin ␪ ⫽ 2 2 22 ⫹ 12 22 ⫺ 12 , cos ␪ ⫽ , tan ␪ ⫽ 12 ⫹ 1 43. sin ␪ ⫽ 2 2 45. 51. 55. 57. 59. 61. 63. 65. 69. 75.

22 ⫺ 12 ⫹ 12 22 ⫺ 12 ⫹ 13 47. 49. cos 15° 2 2 tan 2␪ 53. tan x ␪ ␪ ␪ 3 2 3 sin a b ⫽ , cos a b ⫽ , tan a b ⫽ 2 2 2 2 113 113 ␪ ␪ ␪ 3 1 sin a b ⫽ , cos a b ⫽ , tan a b ⫽ 3 2 2 2 110 110 ␪ ␪ ␪ 7 5 7 sin a b ⫽ , cos a b ⫽ , tan a b ⫽ 2 2 2 5 174 174 ␪ ␪ ␪ 1 15 1 sin a b ⫽ , cos a b ⫽ , tan a b ⫽ 2 2 2 15 1226 1226 ␪ ␪ ␪ 5 2 5 sin a b ⫽ , cos a b ⫽ ⫺ , tan a b ⫽ 2 2 2 ⫺2 129 129 1 3 cos112␪2 ⫺ cos14␪2 4 67. sin15t2 ⫺ sin12t2 2 1 ⫹ 13 12 cos11540␲t2 ⫹ cos12418␲t2 71. 73. 2 4 3 13 5 2 cosa hbcosa hb 77. 2 cos x sin a xb 2 2 8

79. 2 cos11072␲t2 cos1375␲t2 sin12x2 2 sin x cos x 16 12 ⫽ ⫽ tan12x2 83. ⫺ 85. 2 2 2 2 cos12x2 cos x ⫺ sin x 2 2 2 87. 1sin x ⫹ cos x2 ⫽ sin x ⫹ 2 sin x cos x ⫹ cos x ⫽ sin2x ⫹ cos2x ⫹ 2 sin x cos x ⫽ 1 ⫹ 2 sin x cos x ⫽ 1 ⫹ sin12x2 89. cos18␪2 ⫽ cos12 # 4␪2 ⫽ cos2 14␪2 ⫺ sin2 14␪2 cos 12␪2 cos2␪ ⫺ sin2␪ ⫽ 91. 2 sin ␪ sin2␪ cos2␪ sin2␪ ⫽ ⫺ sin2␪ sin2␪ ⫽ cot2␪ ⫺ 1 81.

SA-43


SA-44 93. tan 12␪2 ⫽

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Student Answer Appendix 2 tan ␪ 1 ⫺ tan2␪ 12 tan ␪2

1 tan ␪ ⫽ 1 11 ⫺ tan2␪2 tan ␪ 2 ⫽ 1 ⫺ tan ␪ tan ␪ 2 ⫽ cot ␪ ⫺ tan ␪ 2 95. 2 csc 12x2 ⫽ sin 12x2 2 ⫽ 2 sin x cos x 1 ⫽ sin x cos x sin2x ⫹ cos2x ⫽ sin x cos x sin2x cos2x ⫽ ⫹ sin x cos x sin x cos x sin x cos x ⫽ ⫹ cos x sin x ⫽ tan x ⫹ cot x x x x 97. cos2a b ⫺ sin2a b ⫽ cosa2 # b ⫽ cos x 2 2 2 99. 1 ⫺ 4 sin2␪ ⫹ 4 sin4␪ ⫽ 11 ⫺ 2 sin2␪2 2 ⫽ 3 cos 12␪2 4 2 ⫽ cos2 12␪2 ⫽ 1 ⫺ sin2 12␪2 sin1120␲t2 ⫹ sin180␲t2 2 sin1100␲t2cos120␲t2 101. ⫽ cos1120␲t2 ⫺ cos180␲t2 ⫺2 sin1100␲t2sin120␲t2 cos120␲t2 ⫽⫺ sin120␲t2 ⫽ ⫺cot120␲t2 103. sin2␣ ⫹ 11 ⫺ cos ␣2 2 ⫽ sin2␣ ⫹ 1 ⫺ 2 cos ␣ ⫹ cos2␣ ⫽ sin2␣ ⫹ cos2␣ ⫹ 1 ⫺ 2 cos ␣ ⫽ 1 ⫹ 1 ⫺ 2 cos ␣ ⫽ 2 ⫺ 2 cos ␣ ⫽ 2 11 ⫺ cos ␣2 1 ⫺ cos ␣ ⫽4a b 2 ␣ ⫽ 4 sin2a b 2 ␣ 2 ⫽ c 2 sina b d 2 105. sin12␣2 ⫽ sin1␣ ⫹ ␣2 ⫽ sin ␣ cos ␣ ⫹ cos ␣ sin ␣ ⫽ sin ␣ cos ␣ ⫹ sin ␣ cos ␣ ⫽ 2 sin ␣ cos ␣ tan12␣2 ⫽ tan1␣ ⫹ ␣2 tan ␣ ⫹ tan ␣ ⫽ 1 ⫺ tan ␣ tan ␣ 2 tan ␣ ⫽ 1 ⫺ tan2␣ 107. cos ␣ cos ␤ ⫹ sin ␣ sin ␤ ⫽ cos1␣ ⫺ ␤2 ⫺ 1cos ␣ cos ␤ ⫺ sin ␣ sin ␤2 ⫽ ⫺cos1␣ ⫹ ␤2 2 sin ␣ sin ␤ ⫽ cos1␣ ⫺ ␤2 ⫺ cos1␣ ⫹ ␤2 1 sin ␣ sin ␤ ⫽ 3 cos1␣ ⫺ ␤2 ⫺ cos1␣ ⫹ ␤2 4 2 2 2 109. a. M ⫽ , M 3.9 b. M ⫽ , M 2.6 22 ⫺ 13 22 ⫺ 12 c. ␪ ⫽ 60°

111. a. 1288 ⫺ 144 122 ft 84.3 ft b. 1288 ⫺ 144122 ft 84.3 ft 113. y(t) ⫽ 2 cos(2174␲t)cos(780␲t) 115. cos 32␲112092t 4 ⫹ cos 32␲19412t 4; the * key ␲t 117. d1t2 ⫽ ` 6 sin a b ` 60 1 ␲t ⫽ ` 6 sin a # b ` 2 30 ␲t 1 ⫺ cosa b 30 ⫽ ∞ 6 °⫾ ¢∞ Q 2 1 ⫺ cosa ⫽6

R

␲t b 30

2 1 ⫺ cosa

⫽

R

36

␲t b 30

2

␲t 18 c 1 ⫺ cosa b d 30 A 119. A: sin12␪ ⫺ 90°2 ⫹ 1 ⫽ sin12␪2cos 90° ⫺ cos12␪2sin 90° ⫹ 1 ⫽ 0 ⫺ cos12␪2 ⫹ 1 ⫽ 1 ⫺ cos12␪2 B: 2 sin2␪ ⫽ sin2␪ ⫹ sin2␪ ⫽ 1 ⫺ cos2␪ ⫹ sin2␪ ⫽ 1 ⫺ 1cos2␪ ⫺ sin2␪2 ⫽ 1 ⫺ cos12␪2 C: 1 ⫹ sin2␪ ⫺ cos2␪ ⫽ 1 ⫺ 1cos2␪ ⫺ sin2␪2 ⫽ 1 ⫺ cos12␪2 D: 1 ⫺ cos12␪2 ⫽ 1 ⫺ cos12␪2 121. a. 0.9659; 0.9659 22 ⫹ 13 2 16 ⫹ 12 2 b. a b ⱨa b 2 4 6 ⫹ 2 112 ⫹ 2 2 ⫹ 13 ⱨ 4 16 8 ⫹ 4 13 2 ⫹ 13 ⱨ 4 16 2 ⫹ 13 2 ⫹ 13 ⫽ 4 4 13 123. Answers will vary; One example is , 1, ⫺2 ⫺ 13 3 ⫽

125.

22 ⫹ 13 2 32 ⫹ 22 ⫹ 13 cos 7.5° ⫽ 2 cos 15° ⫽

cos 3.75° ⫽

42 ⫹ 32 ⫹ 22 ⫹ 13 0.9979 2

52 ⫹ 42 ⫹ 32 ⫹ 22 ⫹ 13 0.9995; 2 They are getting close to 1. cos 1.875° ⫽ 127.

60 2x

1x

30 √3x

129. y ⫽ 2 sin ax ⫹

␲ b⫹1 4


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Mid-Chapter Check, p. 693 1. sin x3 csc x ⫺ sin x4 ⫽ sin x csc x ⫺ sin2x 1 ⫽ sin x ⫺ sin2x sin x ⫽ 1 ⫺ sin2x ⫽ cos2x cos2x 2. cos2x ⫺ cot2x ⫽ cos2x ⫺ sin2x 1 ⫽ cos2xa1 ⫺ b sin2x 2 2 ⫽ cos x11 ⫺ csc x2 ⫽ cos2x1⫺cot2x2 ⫽ ⫺cos2x cot2x 2 sin x cos x 2 sin x csc x ⫺ cos x sec x ⫺ ⫽ 3. sec x csc x sec x csc x 2112 ⫺ 1 ⫽ sec x csc x 1 ⫽ sec x csc x ⫽ cos x sin x 4. 1 ⫹ sec2x ⫽ tan2x 1 ⫹ sec20 ⫽ tan20 1 ⫹ 12 ⫽ 02 1⫹1⫽0 2 ⫽ 0 False 1sin x ⫹ cos x21sin2x ⫺ sin x cos x ⫹ cos2x2 sin3x ⫹ cos3x 5. a. ⫽ sin x ⫹ cos x 1sin x ⫹ cos x2 ⫽ sin2x ⫹ cos2x ⫺ sin x cos x ⫽ 1 ⫺ sin x cos x 1 1⫹ 1 ⫹ cos x cos x 1 ⫹ cos x 1 ⫹ sec x b. ⫺ ⫽ ⫺ csc x cot x 1 cos x sin x sin x sin x sin x ⫽ asin x ⫹ b⫺a ⫹ sin xb cos x cos x sin x sin x ⫽ sin x ⫹ ⫺ ⫺ sin x cos x cos x ⫽0 sec2x ⫺ tan2x sec2x tan2x 6. a. ⫽ ⫺ 2 2 sec x sec x sec2x sin2x ⫽1⫺

7. 8. 9. 10.

cos2x 1

cos2x ⫽ 1 ⫺ sin2x ⫽ cos2x cot x ⫺ tan x cot x tan x b. ⫽ ⫺ csc x sec x csc x sec x csc x sec x sin x cos x sin x cos x ⫺ ⫽ 1 # 1 1 # 1 sin x cos x sin x cos x cos2x sin x sin2x cos x ⫽ ⫺ sin x cos x ⫽ cos2x ⫺ sin2x 3193 456 456 a. b. ⫺ c. 5785 5785 5767 7 ⫹ 24 13 24 ⫺ 713 7 ⫹ 24 13 sin A ⫽ , cos A ⫽ , tan A ⫽ 50 50 24 ⫺ 7 13 ␪ ␪ 4 1 sina b ⫽ , cosa b ⫽ 2 2 117 117 527 336 336 , cos12␣2 ⫽ , tan12␣2 ⫽ sin12␣2 ⫽ 625 625 527

Reinforcing Basic Concepts, pp. 693–694 1. cos2x ⫹ sin2x ⫽ 1 cos2x sin2x 1 ⫹ ⫽ 2 sin x sin2x sin2x cot2x ⫹ 1 ⫽ csc2x ✓ 2 cos x ⫹ sin2x ⫽ 1 cos2x sin2x 1 ⫹ ⫽ 2 cos x cos2x cos2x 1 ⫹ tan2x ⫽ sec2x ✓ 2. cos12␣2 ⫽ cos1␣ ⫹ ␣2 ⫽ cos ␣ cos ␣ ⫺ sin ␣ sin ␣ ⫽ cos2␣ ⫺ sin2␣ ⫽ cos2␣ ⫺ 11 ⫺ cos2␣2 ⫽ 2 cos2␣ ⫺ 1 ⫽ cos2␣ ⫺ sin2␣ ⫽ 11 ⫺ sin2␣2 ⫺ sin2␣ ⫽ 1 ⫺ 2 sin2␣


␲ ␲ 3. 3⫺1, 1 4 , c ⫺ , d 5. cos⫺1 1 15 2 2 2 ␲ 1 ␲ ␲ ␲ 7. 0; ; ⫺ ; ⫺ 9. 11. 13. 1.0956, 62.8° 2 6 2 4 2 ␲ 12 15. 0.3876, 22.2° 17. 19. 21. 45° 23. 0.8205 2 3 13 ␲ ; 120°; ␲ 27. 25. 0; 29. ␲ 31. 1.4352; 82.2° 2 3 ␲ 12 3␲ 33. 0.7297; 41.8° 35. 37. 0.5560 39. ⫺ 41. 4 2 4 ␲ ␲ ␲ 43. 0; ⫺ 1 3; 30°; 1 3; 45. ⫺ 47. 49. ⫺1.1170, ⫺64.0° 3 6 3 13 ␲ 51. 0.9441, 54.1° 53. ⫺ 55. 57. 12 59. 120° 6 3 ␲ 61. cannot evaluate tana b 2 ␲ 63. csc ⫽ 12 7 1, not in domain of sin⫺1x. 4 3 4 3 65. sin ␪ ⫽ , cos ␪ ⫽ , tan ␪ ⫽ 5 5 4 2x2 ⫺ 36 6 2x2 ⫺ 36 , cos ␪ ⫽ , tan ␪ ⫽ 67. sin ␪ ⫽ x x 6 y y 24 15 69. 71. 25 25 3 24 1. horizontal, line, one, one

3

␪ 7

73.

225 ⫺ 9x2 3x

x

5 ␪ √25 9x2

√5

␪ 2

3x

75.

12 B 12 ⫹ x2

√12 x2 ␪

x

x

√12

␲ ␲ or 30° 81. or 30° 83. 80.1° 6 6 85. 67.8° 87. 3⫺1, 1 4 89. 1⫺q, q2 91. a. FN 2.13 N; FN 1.56 N b. ␪ 63° for FN ⫽ 1 N, ␪ 24.9° for FN ⫽ 2 N 93. 30° 95. ␪ 72.3°; straight line distance; 157.5 yd 75 50 97. a. ␪ ⫽ tan⫺1a b ⫺ tan⫺1a b b. d 僆 139.2, 95.72 d d c. ␪ 11.5° at d 61.2 ft 94 70 99. a. ␪ ⫽ tan⫺1a b ⫺ tan⫺1a b b. ␪ 8.4° at d 81.1 ft d d 77. 0; 2; 30°; ⫺1; ␲

79.

101. a. ␪ 15.5°; ␪ 0.2710 rad b. 27 mi 103. sin1␣ ⫹ ␤2 ⫽ 1 77 105. a. P ⫽ t ⫹ 28 b. sales increasing by 15.4 packages/year c. 182 5


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3␲ ␲ ␲ 3␲ ␲ ; ⫹ 2␲k; ⫹ 2␲k 1. principal, 3 0, 2␲2 , real 3. ; ; 4 4 4 4 4 5. Answers will vary. 11.

␪

sin ␪

7. a. QIV cos ␪

b. 2 roots

0

1

0

␲ 6

1 2

13 2

13 3

␲ 3

13 2

1 2

13

␲ 2

1

0

und.

2␲ 3

13 2

⫺

5␲ 6

1 2

␲

0

7␲ 6 4␲ 3

⫺ ⫺

1 2

13 2

⫺

13 2

⫺1 ⫺

⫺ 13 ⫺

13 3

45.

13 3

1 2

13

␲ ␲ ␲ ␲ ␲ ␲ 15. ⫺ 17. 19. ⫺ 21. ␲ 23. 25. 4 4 6 3 3 6 ␲ 5␲ 2␲ 5␲ 3␲ 7␲ ␲ 5␲ 7␲ 11␲ , , , , 29. , 31. 33. 35. , 6 6 3 3 4 4 6 6 6 6 ␲ 3␲ ␲ 2␲ 4␲ 5␲ ␲ 3␲ 5␲ 7␲ , , , , 37. , 39. , 41. , 3 3 3 3 4 4 4 4 2 2 13.

27.

5␲ 6

7␲ 5␲ ⫹ 2␲k or ⫹ 2␲k 4 4 3␲ 3␲ 5␲ x⫽ ⫹ 2␲k or ⫹ 2␲k 55. x ⫽ ⫹ ␲k 4 4 4 3␲ ␲ 2␲ ␲ x ⫽ ⫹ ␲k or ⫹ ␲k 59. x ⫽ ⫹ k 61. x ⫽ 3␲ ⫹ 6␲k 3 3 8 2 ␲ ␲ ␲ ␲ 5␲ x ⫽ ⫹ ␲k 65. x ⫽ ⫹ k or ⫹ ␲k or ⫹ ␲k 2 6 3 12 12 a. x 1.2310 b. x 1.2310 ⫹ 2␲k, 5.0522 ⫹ 2␲k a. x 1.2094 b. x 1.2094 ⫹ 2␲k, 5.0738 ⫹ 2␲k a. ␪ 0.3649 b. ␪ 0.3649 ⫹ ␲k, 1.2059 ⫹ ␲k a. ␪ 0.8861 b. ␪ 0.8861 ⫹ ␲k, 2.2555 ⫹ ␲k 10␲ ␲ 5␲ 2␲ 4␲ 4␲ x ⫽ ⫹ ␲k or ⫹ ␲k 77. x ⫽ ⫹ k or ⫹ k 6 6 9 3 9 3 ␲ ␪ ⫽ k 81. ␪ 0.3398 ⫹ 2␲k or 2.8018 ⫹ 2␲k 2 22.1° and 67.9° 85. 0°; the ramp is horizontal 30.7°; smaller 89. ␣ ⫽ 35°, ␤ 25.5° k 1.36, ␣ 20.6° 93. a. 7 in. b. 1.05 in. and 5.24 in. ␲ ⫹ ␲k, explanations will vary 2 f 12 ⫹ i2 ⫽ 12 ⫹ i2 2 ⫺ 412 ⫹ i2 ⫹ 5 ⫽ 4 ⫹ 4i ⫹ i2 ⫺ 8 ⫺ 4i ⫹ 5 ⫽ 4 ⫹ 4i ⫺ 1 ⫺ 8 ⫺ 4i ⫹ 5 ⫽0 1 12 a. ⫺ b. ⫺ 2 13

49. x ⫽ 30°, 150°, 210°, or 330° 51. x ⫽

63. 67. 69. 71. 73. 75. 79. 83. 87. 91. 95. 97.

99.

␲ b. D ⫽ 112.5, ␪ ⫽ , y ⫽ 4

␲ 112.5 ⫺ x cosa b 4

␲ sina b 4 2 15 ⫺ x cos 1.1071 b. D ⫽ 2 15, ␪ 1.1071, y ⫽ sin 1.1071

c. verified II. a. (2, 4) c. verified

47. x ⫽ 120°, 240°, or x 82.8° or 277.2°

57.

␲ 5␲ 13␲ 17␲ , , , 12 12 12 12

47.

5 5 49. I. a. a , b 2 2

43. ␪ 70.5° or 289.5° 45. x ⫽ 30°, 90°, 150°, or 270°

53.

3␲ 7␲ , 4 4

0

13 2

⫺

b. 2 roots

tan ␪

0

1 2

9. a. QIV

1. cos2x ⫹ sin2x ⫽ 1, 1 ⫹ tan2x ⫽ sec2x, cot2x ⫹ 1 ⫽ csc2x 3. squaring 5. Answers will vary. ␲ 5␲ ␲ 5␲ ␲ 5␲ , , , 7. 9. 0 11. 0.4456, 1.1252 13. , 12 12 4 4 6 6 ␲ 3␲ 5␲ 7␲ ␲ 3␲ 5␲ 7␲ , , , 0.8411, 5.4421 17. , , , 15. , 4 4 4 4 4 4 4 4 2␲ ␲ 5␲ , 0.7297, 2.4119 21. 19. , 6 6 3 ␲ 2␲ 5␲ 2␲ ␲ 3␲ 5␲ 7␲ ⫹ k, ⫹ k; k ⫽ 0, 1, 2 25. , , , 23. 9 3 9 3 4 4 4 4 27. x 0.7290 29. x 2.6649 31. x 0.4566 33. x 1.1706, 4.1287 35. P ⫽ 12; x ⫽ 3; x ⫽ 11 ␲ 17␲ , 37. P ⫽ 24; x 0.4909, x 5.5091 39. 12 12 ␲ 3␲ a is extraneousb 41. 0.3747, 5.9085, 2.7669, 3.5163 43. 2 2

␲ III. a. (1, 13) b. D ⫽ 2, ␪ ⫽ , y ⫽ 3

␲ 2 ⫺ x cosa b 3

c. verified ␲ sina b 3 51. a. 2500␲ ft3 7853.98 ft3 b. 7824.09 ft3 c. ␪ 78.5° 53. a. 78.53 m3/sec b. during the months of August, September, October, and November 55. a. $3554.52 b. during the months of May, June, July, and August 57. a. 12.67 in. b. during the months of April, May, June, July, and August 59. a. 8.39 gal b. approx. day 214 to day 333 61. a. 68 bpm b. 176.2 bpm c. from about 4.6 min to 7.4 min 63. 1.1547 2␲ ␲ 65. a. y ⫽ 19 cosa␲ ⫺ xb ⫹ 53 b. y ⫽ ⫺21 sina xb ⫹ 29 6 365 67. a. L 25.5 cm.

b. ␪ 38.9° or 31.6°

69. 1⫺1, 02 , (0, 0), (2, 0) (multiplicity 2): up/up; y

3

3

3

x

3

71. ␪ 4.56°

Making Connections, p. 732 1. b

3. e

5. f

7. h

9. g

11. c

13. f

15. g

Summary and Concept Review, pp. 733–737 1. sin x1csc x ⫺ sin x2 ⫽ sin x csc x ⫺ sin x sin x 1 ⫽ sin x ⫺ sin2x sin x ⫽ 1 ⫺ sin2x ⫽ cos2x


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2.

tan2x csc x ⫹ csc x sec2x

⫽ ⫽

csc x1tan2x ⫹ 12 sec2x csc x sec2x

sec2x ⫽ csc x 1sec x ⫺ tan x21sec x ⫹ tan x2 sec2x ⫺ tan2x 3. ⫽ csc x csc x 1 ⫹ tan2x ⫺ tan2x ⫽ csc x 1 ⫽ csc x ⫽ sin x sec2x sec2x ⫺ sin x csc x 4. ⫺ sin x ⫽ csc x csc x sec2x ⫺ 1 ⫽ csc x tan2x ⫽ csc x 35 37 12 35 37 5. sin ␪ ⫽ ⫺ , csc ␪ ⫽ ⫺ , cot ␪ ⫽ , tan ␪ ⫽ , sec ␪ ⫽ ⫺ 37 35 35 12 12 4 16 25 23 4 16 6. sin ␪ ⫽ ⫺ , csc ␪ ⫽ ⫺ , , cot ␪ ⫽ ⫺ , tan ␪ ⫽ ⫺ 25 23 4 16 4 16 23 cos ␪ ⫽ 25 csc2x11 ⫺ cos2x2 csc2x sin2x 7. ⫽ tan2x tan2x 1 ⫽ tan2x ⫽ cot2x cot x csc x 1 8. ⫺ ⫽ cot x ⫺ cot x csc x sec x tan x sec x ⫽ cot x cos x ⫺ cot x csc x ⫽ cot x1cos x ⫺ csc x2 1sin2x ⫺ cos2x21sin2x ⫹ cos2x2 sin4x ⫺ cos4x 9. ⫽ sin x cos x sin x cos x 1sin2x ⫺ cos2x2112 ⫽ sin x cos x sin x sin x cos x cos x ⫽ ⫺ sin x cos x sin x cos x sin x cos x ⫽ ⫺ cos x sin x ⫽ tan x ⫺ cot x 1sin x ⫹ cos x2 2 sin2x ⫹ 2 sin x cos x ⫹ cos2x ⫽ 10. sin x cos x sin x cos x sin2x ⫹ cos2x 2 sin x cos x ⫽ ⫹ sin x cos x sin x cos x 1 ⫽ ⫹2 sin x cos x ⫽ csc x sec x ⫹ 2 16 ⫺ 12 11. a. cos 75° ⫽ 4 1 13 ⫺ 12 2 ␲ 13 ⫺ 1 ⫽ ⫽ 2 ⫺ 13 b. tana b ⫽ 12 2 1 ⫹ 13 1 13 ⫺ 12 2 13 ⫺ 1 ⫽ ⫽ 2 ⫺ 13 12. a. tan 15° ⫽ 2 1 ⫹ 13 13 ␲ 12 ⫺ 16 b. sina⫺ b ⫽ 13. a. cos 180° ⫽ ⫺1 b. sin 120° ⫽ 12 4 2 5x 14. a. cos x b. sina b 15. a. cos 1170° ⫽ cos 90° ⫽ 0 8 57␲ ␲ 12 ␲ x x b ⫽ sina b ⫽ b. sina 16. a. cosa b ⫽ sina ⫺ b 4 4 2 8 2 8

SA-47

␲ 7␲ b ⫽ cosa ⫺ xb 12 12 tan 45° ⫺ tan 30° 17. tan145° ⫺ 30°2 ⫽ 1 ⫹ tan 45° tan 30° 13 3 ⫺ 13 13 1⫺ 1⫺ 3 3 3 ⫽ ⫽ ⫽ 13 13 3 ⫹ 13 1⫹1# 1⫹ 3 3 3 131 13 ⫺ 12 3 ⫺ 13 13 ⫺ 1 3 3 ⫺ 13 ⫽ ⫽ ⫽ ⫽ 3 3 ⫹ 13 3 ⫹ 13 131 13 ⫹ 12 13 ⫹ 1 tan 135° ⫺ tan 120° tan1135° ⫺ 120°2 ⫽ 1 ⫹ tan 135° tan 120° ⫺1 ⫹ 13 13 ⫺ 1 13 ⫺ 1 ⫽ ⫽ ⫽ 1 ⫹ 1⫺12 1⫺ 132 1 ⫹ 13 13 ⫹ 1 ␲ ␲ 18. cos ax ⫹ b ⫹ cos ax ⫺ b 6 6 ␲ ␲ ␲ ␲ ⫽ cos x cos a b ⫺ sin x sin a b ⫹ cos x cos a b ⫹ sin x sin a b 6 6 6 6 ␲ 13 ⫽ 2 cos x cosa b ⫹ 0 ⫽ 2 cos xa b ⫽ 13 cos x 6 2 ⫺2184 19. a. sin12␪2 ⫽ 7225 ⫺6887 cos12␪2 ⫽ 7225 2184 tan12␪2 ⫽ 6887 840 b. sin12␪2 ⫽ 841 41 cos12␪2 ⫽ 841 840 tan12␪2 ⫽ 41 21 20 21 20. a. sin ␪ ⫽ , cos ␪ ⫽ ⫺ , tan ␪ ⫽ ⫺ , 29 29 20 ⫺7 7 24 ⫺24 ⫺7 b. sin ␪ ⫽ or sin ␪ ⫽ , cos ␪ ⫽ or cos ␪ ⫽ , tan ␪ ⫽ 25 25 25 25 24 ⫺24 or tan ␪ ⫽ 7 12 ␲ 13 21. a. cos 45° ⫽ b. cosa b ⫽ 2 6 2 22 ⫹ 12 22. a. sin 67.5° ⫽ 2 22 ⫺ 12 cos 67.5° ⫽ 2 5␲ 22 ⫹ 12 b. sina b ⫽ 8 2 5␲ 22 ⫺ 12 cosa b ⫽ ⫺ 8 2 ␪ 1 ␪ 23. a. sina b ⫽ , in QII 2 5 12 2 ␪ ⫺7 ␪ cosa b ⫽ , in QII 2 5 12 2 ␪ ⫺3 ␪ b. sina b ⫽ , in QIV 2 1130 2 ␪ 11 ␪ cosa b ⫽ , in QIV 2 1130 2 cos13␣2 ⫺ cos ␣ ⫺2 sin12␣2 sin ␣ 24. ⫽ cos13␣2 ⫹ cos ␣ 2 cos12␣2 cos ␣ ⫺2 sin2␣ 2 sin2␣ ⫽ ⫽ 2 2 2 cos ␣ ⫺ sin ␣ sin ␣ ⫺ cos2␣ 2 2 sin ␣ 2 tan2␣ ⫽ ⫽ 2 1 ⫺ 2 cos ␣ sec2␣ ⫺ 2 b. sinax ⫺


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25. cos13x2 ⫹ cos x ⫽ 0 S 2 cos12x2cos x ⫽ 0 ␲ ␲ cos12x2 ⫽ 0: x ⫽ ⫹ k; k 僆⺪ 4 2 ␲ cos x ⫽ 0: x ⫽ ⫹ ␲k; k 僆⺪ 2 30° 30° 1 ⫺ cos 30° 1 ⫹ cos 30° 2 26. a. A ⫽ 12 sina bcosa b ⫽ 144 2 2 A 2 A 2 13 13 1⫺ 1⫹ 2 ⫺ 13 2 ⫹ 13 2 2 ⫽ 144 ⫽ 144 R 2 R 2 B 4 B 4 14414 ⫺ 3 2 ⫽ ⫽ 36 cm 4 ␪ ␪ ␪ b. Let u ⫽ , then x2sina b cosa b ⫽ x2sin u cos u ⫽ 2 2 2 1 2 1 x 12sin u cos u2 ⫽ x2 sin 12u2 2 2 1 1 1 ⫽ x2sin ␪; A ⫽ 1122 2 sin130°2 ⫽ 72a b ⫽ 36 cm2; yes 2 2 2 5␲ ␲ ␲ 27. or 45° 28. or 30° 29. or 150° 30. 1.3431 or 77.0° 4 6 6 ␲ 1 31. 1.0956 or 62.8° 32. 0.5054 or 29.0° 33. 34. 2 4 3␲ 35. undefined 36. 1.0245 37. 60° 38. 4 39.

40.

y

37

y

35 7

␪ 12

x

sin ␪ ⫽ 41.

35 37

3x

tan ␪ ⫽ x 42. ␪ ⫽ cos⫺1a b 5

y

√49 9x2

␪ x

249 ⫺ 9r2 3x 43. ␪ ⫽ sec⫺1a

x b 7 13

Practice Test, pp. 737–738 1csc x ⫺ cot x2 1csc x ⫹ cot x2

csc2x ⫺ cot2x sec x sec x 1 ⫽ sec x ⫽ cos x 3 3 1sin x ⫺ cos x2 1sin2x ⫹ sin x cos x ⫹ cos2x2 sin x ⫺ cos x 2. ⫽ 1 ⫹ cos x sin x 1 ⫹ cos x sin x 1sin x ⫺ cos x2 11 ⫹ sin x cos x2 ⫽ 1 ⫹ cos x sin x ⫽ sin x ⫺ cos x 73 48 55 73 55 3. sin ␪ ⫽ ⫺ , sec ␪ ⫽ , cot ␪ ⫽ ⫺ , tan ␪ ⫽ ⫺ , csc ␪ ⫽ ⫺ 73 48 55 48 55 ⫺ 12 13 ⫺ 1 12 4. 5. cos 45° ⫽ 6. 2 2 13 ⫹ 1 ␲ ␲ 7. sin ax ⫹ b ⫺ sin ax ⫺ b 4 4 ␲ ␲ ␲ ␲ ⫽ sin x cos a b ⫹ cos x sin a b ⫺ sin x cos a b ⫹ cos x sin a b 4 4 4 4 ␲ ␲ ⫽ sin a b cos x ⫹ sin a b cos x 4 4 ␲ ⫽ 2 sin a b cos x 4 12 ⫽2 cos x 2 ⫽ 12 cos x 6 15 8 15 ⫺ 13 1 ; 8. sin ␪ ⫽ , cos ␪ ⫽ , tan ␪ ⫽ 9. 10. 17 17 8 2 137 137 16 ⫺ 12 16 ⫹ 12 0.2588; 0.9659 11. 20 22 ⫺ 12 12. 4 4 1 13. a. y ⫽ 30° b. y ⫽ c. y ⫽ 30° 14. a. y ⫽ 0.8523 rad or 2 7␲ 157␲ y ⫽ 48.8° b. y ⫽ 78.5° or rad c. y ⫽ rad 360 24 x y y 15. 16. 33 cot ␪ ⫽ cos ␪ ⫽ 5 65 1.

⫽

65

x

56

√2

5

x

x2

2

1

√8 ␪

x

9

cot ␪ ⫽

9 x

␲ 3␲ x ␲ ␲ 44. ␪ ⫽ sin⫺1a b ⫹ 45. a. b. , 4 6 4 4 4 2␲ 4␲ ␲ 3␲ 2␲ ⫹ 2␲k, k 僆⺪ 46. a. , c. x ⫽ ⫹ 2␲k or b. 4 4 3 3 3 2␲ 5␲ 2␲ 4␲ ␲ ⫹ 2␲k or ⫹ 2␲k, k 僆⺪ 47. a. ⫺ , c. b. 3 3 3 3 3 2␲ ⫹ ␲k, k 僆⺪ 48. a. 1.1102 b. 1.1102, 5.1729 c. 3 c. 1.1102 ⫹ 2␲k or 5.1729 ⫹ 2␲k, k 僆⺪ 49. a. 0.3376 b. 0.3376, 1.2332, 3.4792, 4.3748 c. 0.3376 ⫹ ␲k or 1.2332 ⫹ ␲k, k 僆⺪ 50. a. 0.3614 b. 0.3614, 2.7802 ␲ 5␲ c. 0.3614 ⫹ 2␲k or 2.7802 ⫹ 2␲k, k 僆⺪ 51. x ⫽ , 12 12 ␲ 5␲ 52. x 0.7297, 2.4119; x ⫽ , 6 6 ␲ ␲ 5␲ 11␲ , 53. x ⫽ , 54. x ⫽ 55. P ⫽ 12; x 2.6931, x 9.3069 6 6 6 2 9 56. P ⫽ 6; x ⫽ 0, x ⫽ 57. a. $43,000 b. April through August 2 58. ␪ 1.1547

5

␪

␪ 33

x

3␲ 4

b. x ⫽

x

x

3␲ 5␲ 3␲ 5␲ , ⫹ 2␲k or ⫹ 2␲k, k 僆⺪ c. x ⫽ 4 4 4 4 ␲ ␲ 11␲ ␲ 11␲ ⫹ 2␲k, k 僆⺪ II. a. b. x ⫽ , c. x ⫽ ⫹ 2␲k or 6 6 6 6 6 18. I. a. x 0.1922 b. x 0.1922, 1.3786, 3.3338, 4.5202 c. x 0.1922 ⫹ ␲k or 1.3786 ⫹ ␲k, k 僆⺪ II. a. x 0.9204 b. x 0.9204, 2.2212, 4.0620, 5.3628 c. x 0.9204 ⫹ ␲k or 2.2212 ⫹ ␲k, k 僆⺪ 19. a. x ⫺1.6875, ⫺0.3413, 1.1321, 2.8967 b. x 0.9671, 2.6110, 3.4538 7␲ 11␲ 19␲ 23␲ ␲ 3␲ 7␲ 11␲ 20. a. x ⫽ 0, , ␲, b. x ⫽ , , , , , 2 2 6 6 12 12 12 12 ␲ 3␲ 5␲ 11␲ 21. x ⫽ , ; x 3.3090, 6.1157 22. x ⫽ , 2 2 6 6 23. a. $6,000 b. January through July 24. 25. cos 12418␲t2 ⫹ cos 11540␲t2 Month Low (Jan S 1) Temp. (ⴗF ) 17. I. a.

1

⫺26

3

⫺21

5

16

7

41

9

24

11

⫺14


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29. B 60°, C 90°, b 12.9 13 mi 31. A 39°, B 82°, a 42.6 mi or A 23°, B 98°, a 26.4 mi 33. A 39°, B 82°, a 42.6 ft or A 23°, B 98°, a 26.4 ft 35. not possible 37. A 80.0°, B 38.0°, b 1.8 1025 mi or A 100.0°, B 18.0°, b 9.1 1024 mi 39. A1 19.3°, A2 160.7°, 48° 160.7° 7 180°; no second solution possible 41. C1 71.3°, C2 108.7°, 57° 108.7° 6 180°; two solutions possible 43. not possible, sin A 7 1

Strengthening Core Skills, pp. 739–740 Exercise 1. Exercise 2. Exercise 3. Exercise 4.

x僆 x僆 x僆 x僆

10.6025, 2.53912 3 0, 0.79454 ´ 3 4.4415, 2␲2 3 0, 2.61544 ´ 3 9.3847, 122 167.3927, 202.60732

Cumulative Review Chapters 1–7, pp. 741–742 85 13 85 1. sin ␪ 84 85 , csc ␪ 84 , cos ␪ 85 , sec ␪ 13 , 84 13 tan ␪ 13 , cot ␪ 84 3. g12 132 12 132 2 412 132 1 4 4 13 3 8 4 13 1 0 y 5. about 474 ft 7. 5

45. 12 47. 34.5 million miles or 119.7 million miles 2 49. a. No b. 3.9 mi 51. V 4 S 41.8 km, V 4 P 53. a. No b. about 201.5 ft c. 15 sec 55. Two triangles Angles Angles Sides A2 138.9° A1 41.1° a 12 cm B 26° B 26° b 8 cm C2 15.1° C1 112.9° c1 16.8 cm

4 3 2 1 54321 1 2 3 4 5

9. 50.89 km/hr

11. x 僆 c

1 2 3 4 5 x

57.

9 11 , d 2 2

1 13. a. y x 31 b. every 2 years, the amount of emissions 2 decreases by 1 million tons. c. 23.5 million tons; 11 million tons 15. x 僆 11, 52 17. $7 cos x1sec x 12 cos x 19. sec x 1 1sec x 121sec x 12 1 cos x sec2x 1 cos x 1 tan2x 99 ␲ 2␲ b 27.1 21. 23. a. y 5.4 sina x 101 6 3 b. from early May until late August 25. a. volume of a cylinder b. volume of a rectangular solid c. circumference of a circle d. area of a triangle 27. a. The function is not in simplified form b. x 1.6 ␲ 5␲ d 29. x 僆 c , 4 4

Angles A1 48.0° B1 107.6° C 24.4°

10.2(√3) cm

Exercises 8.2, pp. 766–771 1. cosines 3. Pythagorean 5. B 33.1°, C 129.9°, a 19.8 m; law of sines 7. yes 9. no 11. yes 13. verified 15. B 41.4° 17. a 7.24 19. A 41.6° 21. A 120.4°, B 21.6°, c 53.5 cm 23. A 23.8°, C 126.2°, b 16 mi 25. B

59.8

A

19 in.

31.

mi

A 103.5

43.4

13.7 in. 45 10.6 in.

21. ⬔A 57°, b 49.5 km, c 17.1 km 23.

12

0.8 cm 3.6 cm 56

d. 1

29 465 mm

C

27. not possible

2.9 10 25 m

i 33.1

25

4.1

0 1

B

mi

C

3.2 cm 112

c. 2

91.2

10 25

1. ambiguous 3. I, II 5. Answers will vary. 7. a 8.98 9. C 49.2° 11. C 21.4° 13. ⬔C 78°, b 109.5 cm, c 119.2 cm 15. ⬔C 90°, a 10 in., c 20 in. 17. 19. ⬔C 90°, a 15 mi, b 15 mi

2.3


b. 0

538 mm

260.9 mm

27. A 137.9°, B 15.6°, C 26.5° 29. A 119.3°, B 41.5°, C 19.2°

CHAPTER 8

25. a. 10 cm

Sides a9 b2 4.9 c5

69. A 19°, B 31°, C 130°, a 45 cm, b 71.2 cm, c 105.9 cm 71. x 0.3747 2␲k or 2.7669 2␲k, k 僆⺪ 73. x4 3x3 5x2 x 10 0

x b d c. verified 3. 4 tan ␪ sec ␪ 13 2␲ 4␲ ␲ 3␲ 5␲ 7␲ , , ␲ 9. x , , , 7. 0, 3 3 4 4 4 4

102

Angles A2 132.0° B2 23.6° C 24.4°

Sides a 12 cm b 8 cm c2 4.8 cm

30

1. a. y csc ␪, b. y csc c tan1a

33

Sides a9 b1 11.6 c5

80.8 km

59. a 33.7 ft, c 22.3 ft 61. 3.2 mi 63. angle 90°; sides 9.8 cm, 11 cm; diameter 11 cm; it is a right triangle. 65. a. about 3187 m b. about 2613 m c. about 2368 m sin 90° sin 60° 60 67. 20.4 cm ; 12 13 sin 30° sin 45° 10.2 cm

Connections to Calculus Exercises, p. 744

5. sin ␪

SA-49

33. A 139.7°, B 23.7°, C 16.6° 35. A 48.5° 37. about 1688 mi 39. It cannot be constructed (available length 10,703.6 ft). 41. P 27.7°; heading 297.7° 43. 1678.2 mi 45. P 22.4 cm, A 135°, B 23.2°, C 21.8° 47. A 20.6°, B 15.3°, C 144.1° 49. 58.78 cm


SA-50

2/2/11

5:43 PM

Page SA-50


51. a 13 A 133.2° b5 B 16.3° c 182 C 30.5° 53. 33.7; 150 ft2 55. a. 65% b. $1,950,000 57. about 483,529 km2 59. 387 502 889 6 902 61. (1) a2 b2 c2 2bc cos A; (2) b2 a2 c2 2ac cos B, use substitution for a2 and (2) becomes b2 1b2 c2 2bc cos A2 c2 2ac cos B. Then 0 2c2 2bc cos A 2ac cos B, 2bc cos A 2ac cos B 2c2, b cos A a cos B c 63. 2 5 13 12 13 65. sin x , csc x , cos x , sec x , 13 5 13 12 5 12 tan x , cot x 12 5

33. a. H8, 4I


35. a. H3, 6I

1. scalar 7.

3. directed, line 12 knots

5. Answers will vary. 9.

2

u

c. H15.5, 5I y

5

M (3, 2)

321 1 2 3 4 5

15. H6, 5I

8

4 4

(5, 14)

5

b. H5, 2I

y

y 5

5

r

u

v 5

3 x

r

(5, 2)

5 x

5

c. H6.5, 10I

d. H6, 6I

y

(6.5, 10)

y

10 3

H7, 3I

1.5v

u

1 2 3 4 5 6 7 x

r

4 (3) 7 523

5

5

3

2v

5

(6, 6)

x

37. True 39. False 41. True 43. u v H8, 6I 45. u v H9, 6I u v H6, 2I u v H7, 0I y

y

8 x (5, 3)

5

5

uv

uv 5

19. Terminal point: (1, 1), length: 234 21. a.

H8, 3I QI

4

4

8

5

QIII

2 4

t

2

2

4

2

8

4

5

x

4

47. u v H3, 6I u v H7, 0I

49. u 8i 15j u 17 y

y

16 5

H2, 5I

y

5

(1, 9) v

y

u

r 55 5

u

uv

b. 173 c. 20.6° b. 129 c. 68.2° 25. H10.9, 5.1I 27. H106, 92.2I 29. H9.7, 2.6I 31. a. H1, 9I b. H5, 3I

5

x

5

uv 5

2

51. p 3.2i 5.7j p 6.54

v

4

5 x

53. a.

c. H0.5, 15I

y

5

10

5

5 x

5

y

6

u 5

1.5v 10

r

2v

x

55. a.

10

y

r 10

2u 5

8 x

(8, 9)

5

w

5 x

r 74.5 5

5

b. v H11.5, 3.3I c. v 11.5i 3.3j

p

(0.5, 15)

y

v

d. H8, 9I

20

4

r 16

y

(5, 3) 5 x

8 x 4

r

u 5

x

uv

y 4

8 4

uv

5

x

5 5

23. a.

y

x

r

2u 5

8

8

2v

10

2u

(4, 5)

(1, 2)

4

r 20 x

10

3

y

x

u

1.5v

5

2 ( 3) 5

4

y

3

(3, 6)

17. Terminal point: (5, 1), length: 253

y 8 1 5 6

d. H5, 14I

(15.5, 5)

r

V4

5 4 3 2 1

v

r

5

u

13.

25 J 210 N

x

5

(6, 8)

v

V3

x

5

u

5

V2

250 N

2

v

9 knots

11.

y

(8, 4) r

V1

6 knots

b. H6, 8I

y

5

b. w H2.5, 9.2I c. w 2.5i 9.2j

5 x


2/3/11

10:56 AM

Page SA-51

SA-51

Student Answer Appendix p ⫽ ⫺2i ⫹ 2j; 冟p冟 ⫽ 2 12, ␪ ⫽ 135° q ⫽ 6i ⫺ 8j; 冟q冟 ⫽ 10, ␪ ⬇ 306.9° r ⫽ ⫺2i ⫹ 1.5j; 冟r冟 ⫽ 2.5, ␪ ⬇ 143.1° s ⫽ 10i ⫺ 13j; 冟s冟 ⬇ 16.4, ␪ ⬇ 307.6° p ⫽ 212i ⫹ 2j; 冟p冟 ⬇ 3.5, ␪ ⬇ 35.3° q ⫽ 812i ⫹ 12j; 冟q冟 ⬇ 16.5, ␪ ⬇ 46.7° r ⫽ 5.5 12i ⫹ 6.5j; 冟r冟 ⬇ 10.1, ␪ ⬇ 39.9° s ⫽ 11 12i ⫹ 17j; 冟s冟 ⬇ 23.0, ␪ ⬇ 47.5° p ⫽ 8i ⫹ 4j; 冟p冟 ⬇ 8.9, ␪ ⬇ 26.6° q ⫽ 16i ⫹ 4j; 冟q冟 ⬇ 16.5, ␪ ⬇ 14.0° r ⫽ 18i ⫹ 8j; 冟r冟 ⬇ 19.7, ␪ ⬇ 24.0° s ⫽ 20i ⫹ 4j; 冟s冟 ⬇ 20.4, ␪ ⬇ 11.3° ⫺20 21 7 24 63. h , i , verified 65. h , i , verified 25 25 29 29 7 20 21 24 67. i ⫺ j, verified 69. i ⫹ j, verified 29 29 25 25 6 13 3 11 71. h , i , verified 73. i⫹ j, verified 1178 1178 1157 1157 5 2 75. ⬇4.48 h , i ⬇ H4.16, 1.66I 129 129 ⫺3 8 77. ⬇5.83 h , i ⬇ H5.46, ⫺2.05I 79. ⬇14.4 81. ⬇24.3° 173 173 83. hor. comp. ⬇ 79.9 ft/sec; vert. comp. ⬇ 60.2 ft/sec 85. heading 68.2° at 266.7 mph 87. ⬇ 182.10 cm, 22.00 cm2 89. 1Ha, bI ⫽ H1a, 1bI ⫽ Ha, bI 91. Ha, bI ⫺ Hc, dI ⫽ Ha ⫺ c, b ⫺ dI ⫽ Ha ⫹ 1⫺c2, b ⫹ 1⫺d2I ⫽ Ha, bI ⫹ H⫺c, ⫺dI ⫽ Ha, bI ⫹ ⫺1Hc, dI ⫽ u ⫹ 1⫺1v2 93. 1ck2u ⫽ Hcka, ckbI ⫽ cHka, kbI ⫽ c1ku2 c1ku2 ⫽ Hcka, ckbI ⫽ Hkca, kcbI ⫽ kHca, cbI ⫽ k1cu2 95. u ⫹ 1⫺u2 ⫽ Ha, bI ⫹ H⫺a, ⫺bI ⫽ Ha ⫺ a, b ⫺ bI ⫽ H0, 0I 97. 1c ⫹ k2u ⫽ 1c ⫹ k2Ha, bI ⫽ H1c ⫹ k2a, 1c ⫹ k2bI ⫽ Hca ⫹ ka, cb ⫹ kbI ⫽ Hca, cbI ⫹ Hka, kbI ⫽ cu ⫹ ku 99. H1, 3I ⫹ H3, 3I ⫹ H4, ⫺1I ⫹ H2, ⫺4I ⫹ H⫺4, ⫺3I ⫹ H⫺6, 2I ⫽ H0, 0I 101. Answers will vary, one possibility: 0°, 81.4°, ⫺33.9° 103. a. not a real number b. not possible c. not a real number 105. x ⫽ 0, ⫾ 27 57. a. b. c. d. 59. a. b. c. d. 61. a. b. c. d.

10 8 6 4 2 ⫺5⫺4⫺3⫺2⫺1 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

y

1 2 3 4 5 x

1. Angles

Sides

A ⫽ 35° B ⬇ 81.5° C ⬇ 63.5°

a ⬇ 11.6 cm a ⫽ 20 cm c ⫽ 18 cm

2. For ⬔A ⫽ 35°, a ⬇ 10.3 For ⬔A ⫽ 50°, a ⬇ 14.2 For ⬔A ⫽ 70°, a ⬇ 19.1

Very close.


1. equilibrium, zero 3. orthogonal 5. Answers will vary 7. H6, 8I 9. H⫺5, 10I 11. ⫺6i ⫺ 8j 13. ⫺2.2i ⫹ 0.4j 15. H⫺11.48, ⫺9.16I 17. H⫺24, ⫺27I 19. 冟F3冟 ⬇ 3336.8; ␪ ⬇ 268.5° 21. 37.16 kg 23. 644.49 lb 25. 2606.74 kg 27. approx. 286.79 lb 29. approx. 43.8⬚ 31. 1125 N-m 33. approx. 957.0 ft 35. approx. 64,951.91 ft-lb 37. approx. 451.72 lb 39. approx. 2819.08 N-m 41. 800 ft-lb 43. 118 ft-lb 45. verified 47. verified 49. a. 29 b. 45⬚ 51. a. 0 b. 90° 53. a. 1 b. 89.4° 55. yes 57. no 59. yes 61. 3.68 63. ⫺4 65. 3.17 67. a. H3.73, 1.40I b. u1 ⫽ H3.73, 1.40I, u2 ⫽ H⫺1.73, 4.60I 69. a. H⫺0.65, 0.11I b. u1 ⫽ H⫺0.65, 0.11I, u2 ⫽ H⫺1.35, ⫺8.11I 71. a. 10.54i ⫹ 1.76j b. u1 ⫽ 10.54i ⫹ 1.76j, u2 ⫽ ⫺0.54i ⫹ 3.24j 73. a. projectile is about 375 ft away, and 505.52 ft high b. approx. 1.27 sec and 12.26 sec 75. a. projectile is about 424.26 ft away, and 280.26 ft high b. approx. 2.44 sec and 6.40 sec 77. about 74.84 ft; t ⬇ 3.9 ⫺ 1.2 ⫽ 2.7 sec 79. w # 1u ⫹ v2 ⫽ He, f I # Ha ⫹ c, b ⫹ dI ⫽ e1a ⫹ c2 ⫹ f 1b ⫹ d2 ⫽ ea ⫹ ec ⫹ fb ⫹ fd ⫽ 1ea ⫹ fb2 ⫹ 1ec ⫹ fd2 ⫽ He, f I # Ha, bI ⫹ He, f I # Hc, dI ⫽w#u⫹w#v 81. 0 # u ⫽ H0, 0I # Ha, bI ⫽ 01a2 ⫹ 01b2 ⫽ 0 u # 0 ⫽ Ha, bI # H0, 0I ⫽ a102 ⫹ b102 ⫽ 0 83. ␪ ⬇ 56.9°; answers will vary. 85. t ⬇ ⫺20 87. a ⬇ 138.4°, B ⬇ 106.8° C ⬇ 41.2°; P ⬇ 560.4 m, A ⬇ 11,394.3 m2

Exercises 8.5, pp. 810–812 1. modulus, argument 3. multiply, add 5. 21cos 240° ⫹ i sin 240°2, z is in QIII 7. z2 ⫽ z1 ⫹ z3 9. z2 ⫽ z1 ⫹ z3

Mid-Chapter Check, p. 786 b sin A a2 ⫹ c2 ⫺ b2 1. sin B ⫽ 2. cos B ⫽ a 2ac 3. a ⬇ 129 m, B ⬇ 93.3°, C ⬇ 55.7° 4. A ⬇ 42.3°, B ⬇ 81.5°, C ⬇ 56.2° 5. A ⫽ 44° 6. A ⬇ 18.5° a ⫽ 2.1 km B ⬇ 68.1° b ⬇ 2.8 km B ⬇ 134.5° C ⬇ 67.9° c ⫽ 2.8 km C ⫽ 27° or A ⫽ 44° a ⫽ 2.1 km B ⬇ 23.9° b ⬇ 1.2 km C ⬇ 112.1° c ⫽ 2.8 km 7. about 60.7 ft 8. 169 m 9. ␣ ⬇ 49.6° ␤ ⬇ 92.2° ␥ ⬇ 38.2° 10. 9.4 mi


yi 4 2

yi

z2

6

z3 z1

x z3

7 8 x

a ⫽ 70 yd b ⬇ 157.2 yd c ⫽ 100 yd

z1 z2

11. QIII; 2 121cos 225° ⫹ i sin 225°2 13. QIII; 101cos 210° ⫹ i sin 210°2 11␲ 3␲ 3␲ 11␲ 15. 6 c cosa b ⫹ i sina b d 17. 8 c cosa b ⫹ i sina bd 4 4 6 6 6 19. 10 cis c tan⫺1a b d ; 10 cis 36.9° 8 12 21. 13 cis c 180° ⫹ tan⫺1a b d ; 13 cis 247.4° 5 ⫺1 17.5 b d ; 18.5 cis 1.2405 23. 18.5 cis c tan a 6 5 25. 2 134 cis c ␲ ⫹ tan⫺1a⫺ b d ; 2 134 cis 2.1112 3


SA-52

2/2/11

5:48 PM

Page SA-52


␲ 4 ␲ z 2 cis a b 4 12 i 12

55. a. 17 cis 28.1° b. 51 V 57. a. 8.60 cis 324.5° b. 15.48 V 59. a. 13 cis 22.6° b. 22.1 V 61. I 2 cis 30°; Z 5 12 cis 45°; V 10 12 cis 75° 17 17113 63. I 113 cis 326.3°; Z cis 61.9°; V cis 28.2° 4 4 12 cis 105° 65. V 4 cis 60°; Z 4 12 cis 315°; I 2 10 cis 245° 67. V 5 cis 306.9°; Z 8.5 cis 61.9°; I 17 165 cis 29.7° 69. 71. verified 4 24 7 24 7 i, z3 i 73. z2 5 5 5 5 5␲ 13␲ 29␲ 37␲ , , , 75. 24 24 24 24 y 77.

yi

27. r 2, ␪

4 3

(√2, √2)

2 2

1 2

1

4

1

2

x

1

␲ 3 ␲ z 413 cis a b 3 213 6i

yi

29. r 4 13, ␪

(2√3, 6) 5

4√3

3

5

5 x 2

15 b 8 15 z 17 cis c tan1a b d 8 8 15 17a ib 8 15i 17 17

31. r 17, ␪ tan1a

5 b 33. r 6, ␪ ␲ tan1a 111 5 z 6 cis c ␲ tan1 d 111 111 5 6a ib 111 5i 6 6

20

yi (8, 15)

10

17 tan1 8 15

2

x

5

(√11, 5)

0

0 62.6

0.002

116.4

0.003

153.8

0.004

169.7

0.005

161.7

0.006

131.0

0.007

81.9

0.008

21.3

1 2 3 4

x

5

6 5 tan1√11


5

3 x 3

35. r1 212, r2 3 12, ␪1 135°, ␪2 45°; z z1z2 12 0i 1 r 12, ␪ 180°; r1r2 2 1213122 12 ✓ ␪1 ␪2 135° 45° 180° ✓ 37. r1 2, r2 2, ␪1 30°, ␪2 60°; r1 z1 13 2 1 z i 1 r 1, ␪ 30°; 1 ✓ z2 2 2 r2 2 ␪1 ␪2 30° 60° 30° ✓ z1 4 413 39. z1z2 24 0i, i z2 3 3 z1 13 1 i 41. z1z2 21 13 21i, z2 7 7 z1 43. z1z2 10.84 12.04i, 1.55 4.76i z2 z1 5 5 13 i 45. z1z2 0 40i, z2 4 4 z1 5 0i 47. z1z2 10 10i 13, z2 2 z1 49. z1z2 2.93 8.51i, 2.29 3.28i z2 51. verified; verified, u2 v2 w2 uv uw vw 11 4i132 197 20i 132 139 60i 132 117 12i 132 13 16i 132 145 56i 132, 59 84i 13 59 84i 13 53. a. V1t2 170 sin1120␲t2 b. c. t 0.00257 sec t V(t)

0.001

3 1

yi

1. r5 3cos15␪2 i sin15␪2 4 , De Moivre’s 3. complex 5. z5 2 cis 366° 2 cis 6°, z6 2 cis 438° 2 cis 78°, z7 2 cis 510° 2 cis 150°; Answers will vary. 7. r 3 12; n 4; ␪ 45°; 324 9. r 2; n 3; ␪ 120°; 8 1 13 11. r 1; n 5; ␪ 300°; i 2 2 13. r 1; n 6; ␪ 315°; i 15. r 4; n 3; ␪ 330°; 64i 1 1 12 17. r ; n 5; ␪ 135°; i 2 8 8 19. verified 21. verified 23. verified 25. verified 27. roots: 1, 0.3090 0.9511i, 0.8090 0.5878i 29. roots: 3, 0.9271 2.8532i, 2.4271 1.7634i 31. roots: 3i,

3 13 3 3 13 3 i, i 2 2 2 2

33. 2, 0.6180 1.9021i, 1.6180 1.1756i 313 3 3 13 3 35. i, i, 3i 2 2 2 2 37. 1.1346 0.1797i, 0.1797 1.1346i, 1.0235 0.5215i, 0.8123 0.8123i, 0.5215 1.0235i 1 13 39. x 1, i. These are the same results as in Example 3. 2 2 41. r 16; n 4; ␪ 120°; roots: 13 i, 1 i 13, 13 i, 1 i 13 43. r 712; n 4; ␪ 225°; roots: 0.9855 1.4749i, 1.4749 0.9855i, 0.9855 1.4749i, 1.4749 0.9855i 45. r 8, ␪ 210, n 3; k 0 → 0.684 1.879i; k 1 → 1.970 0.347i; k 2 → 1.286 1.532i 47. r 81, ␪ 180, n 4; k 0 → 2.121 2.121i; k 1 → 2.121 2.121i; k 2 → 2.121 2.121i; k 3 → 2.121 2.121i 1 1 1 49. D 4, z0 86 cis 45°, z1 86 cis 165°, z2 86 cis 285°, 1 1 1 z0 86 cis 75°, z1 86 cis 195°, z2 86 cis 315° 51. verified 53. a. numerator: 117 44j, denominator: 21 72j 4 b. 1 j c. verified 55. Answers will vary. 3


2/2/11

5:50 PM

Page SA-53

Student Answer Appendix 57. 7 24i 59. z 2.7321, z 0.7321, z 2. Note: Using sum and difference identities, all three solutions can actually be given in exact form: 1 13, 1 13, 2. tan2x sec2x 1 61. sec x 1 sec x 1 1sec x 121sec x 12 sec x 1 sec x 1 1 cos x cos x cos x 1 cos x cos x 4 12 63. y x 5 5

Making Connections, p. 821 1. e 3. b 5. h

7. c

9. g

11. d

13. f

15. a

Summary and Concept Review, pp. 822–826 1.

Angles A 36° B 21° C 123°

Sides a 205.35 cm b 125.20 cm c 293 cm

2. Angles Sides A 28° a 140.59 yd B 10° b 52 yd C 142° c 184.36 yd 3. approx. 41.84 ft 4. approx. 20.2° and 159.8° 5. Angles Sides A 35° a 67 cm B1 64.0° b 105 cm C1 81.0° c1 115.37 cm Angles A 35° B2 116.0° C2 29.0°

Sides a 67 cm b 105 cm c2 56.63 cm

6. no; 36° 7. approx. 36.9° 8. approx. 385.5 m 9. 133.2°, 30.1°, and 16.7° 10. 85,570.7 m2 y 11.

5 10.30 29.1 9 x

12. 8i 3j; u 8.54; ␪ 159.4° 13. horiz. comp. 11.08, vertical comp. 14.18 14. H4, 2I; 2u v 4.47, ␪ 206.6° 7 12 i j 16. QII; since the x-component will be negative 15. 1193 1193 and the y-component is positive. 17. 16 mi 18. approx. 19.7° 19. H25, 123I 20. approx. 0.87 21. 4 22. p # q 6; ␪ 97.9° 23. 4340 ft-lb 24. approx. 417.81 lb 25. approx. 8156.77 ft-lb 26. a. x 269.97 ft; y 285.74 ft b. approx. 0.74 sec 27. 21cos 240° i sin 240°2 28. 3 3i yi 29. \, e 5√3

3 2

30 4 5 x

SA-53

5␲ z1 ␲ b; 4 cis a b 31. 2 13 2j 12 z2 12 32. Z 10.44; ␪ 16.7°, 10.44 cis 16.7° 33. 16 16i13 5 13 5 5 13 5 34. verified 35. i, i, 5i 36. 6, 3 3i 13 2 2 2 2 37. 2 2i, 2 2i 38. 1 2i, 1 2i 39. verified 30. z1z2 16 cis a

Practice Test, pp. 826–827 1. 6.58 mi 2. 137.18 ft 3. Angles Sides (in.) A1 58.8° a 15 B 20° b6 C1 101.2° c1 17.21

Angles A2 121.2° B 20° C2 38.8°

Sides (in.) a 15 b6 c2 11.0

4. a. No b. 2.66 mi 5. a. No b. 1 c. 8.43 sec 6. a. 2.30 mi b. 7516.5 ft 7. A 438,795 mi2, P 61.7°, B 61.2°, M 57.1° 8. speed 73.36 mph, bearing 47.8° 9. ␪ 36.5° 10. 63.48 cm to the right and 130.05 cm down from the initial point on the ceiling 11. F3 212.94 N, ␪ 251.2° 12. a. ␪ 42.5° b. projvu H2.4, 7.2I c. u1 H2.4, 7.2I, u2 H6.6, 2.2I ␲ 13. 104.53 ft; 3.27 sec 14. 2 cis a b 15. 48 12 cis 75°; verified 24 5 13 5 5 13 5 16. 8 8 13i 17. verified 18. i, i, 5i 2 2 2 2 19. 2.3039 1.5192i, 2.3039 1.5192i 20. 2, 414,300 mi2

Strengthening Core Skills, pp. 828–829 Exercise 1. 664.46 lb, 640.86 lb Exercise 3. yes

Exercise 2. 106.07 lb, 106.07 lb

Cumulative Review Chapters 1–8, pp. 829–831 1 2A 1␲r2 2 ␲ 4 3 5 5 4 5. sin ␪ 3 5 ; cos ␪ 5 ; tan ␪ 4 ; csc ␪ 3 ; sec ␪ 4 ; cot ␪ 3 4 26 7. 9. cos 19° 0.94, cos 125° 0.58 5 11. a. about $66,825 b. 13, 13, 7 12; A 59.5 mi2 y2 y1 x2 x1 y2 y1 13. a. m b. a , b x2 x1 2 2 2 b 2b 4ac c. x d. d 21x2 x1 2 2 1y2 y1 2 2 2a e. A Pert 15. ⬔A 37°, a 33 cm, ⬔B 34.4°, b 31 cm, ⬔C 108.6°, c 52 cm 17. about 422.5 lb y 19. 1. 20 23, 40, 60°, 90°

3. R

8 4 54321 4

1 2 3 4 5 x

8

x 僆 1q, 12 ´ 12, 32 21. 128 128i 13 23. about 3.6 yr 25. A 2, B 1, C ␲ 3␲ 27. 30, 0.71 4 ´ a , 4.37 d ´ a , 2␲b 2 2 29. 0; The vectors are orthogonal.

␲ 4


SA-54

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Connections to Calculus Exercises, pp. 834–835

15.

yes

17.

yes

1. Plane 9000 ft ␪ d

Light

9000 a. ␪ tan a b b. ␪ 35° d 3. 1

32 m ␪ 24 m

a. d 2242 322 212421322cos ␪ b. d 54.13 m c. about 106.1° 64 ft 5. v

19.

18 ft

108642 2 4 6 8 10

␪ u u

48 ft

10 8 6 4 2

y

Y1

12 3X , Y2 X 9 2 10

2 4 6 8 10 x

(6, 3)

10 a. v 2482 642 182 82 ft b. u H48, 64, 0I, v H48, 64, 18I H48, 64, 0I ⴢ H48, 64, 18I cos ␪ 18021822 1 40 ␪ cos a b 12.68°; verified 41 1800 ft 7.

10

10 21.

D

1000 ft

␪ d 525 ft

10 8 6 4 2 108642 2 4 (1, 5) 6 8 10

5 5X 16 X Y1 , Y2 2 3 10

y

2 4 6 8 10 x

d

10

10

Post

a. D 25252 18002 10002 2125 ft b. d H525, 1800, 0I, D H525, 1800, 1000I H525, 1800, 0I ⴢ H525, 1800, 1000I cos ␪ 118752121252 1 15 ␪ cos a b 28.07°; verified 17 9.

10

18 ft

␤ ␣

25 ft x 7 ft

15 ft 8 ft

a. ␪ ␣ ␤

b. ␪ tan1a

7 18 tan1a b tan1a b x x

7 18 b tan1a b 46 46

30.02°

CHAPTER 9 Exercises 9.1, pp. 848–853 1. inconsistent 3. consistent, independent 5. Multiply the first equation by 6 and the second equation by 10. 7. y 74 x 6, y 4 3 x 5 9. y x 2 11. x 3y 3 13. y x 2, x 3y 3

23. 14, 12 25. 13, 52 27. second equation, y, 14, 32 29. second equation, x, 110, 12 31. second equation, x, 1 52 , 74 2 33. 12, 52 2 35. 12, 12 37. 13, 12 39. 12, 32 41. 1 11 43. 12, 32 45. 13, 42 47. 16, 122 2 , 22 49. (2, 8); consistent/independent 51. ; inconsistent 53. 5 1x, y2 | 6x y 226; consistent/dependent 55. (4, 1); consistent/independent 57. 13, 42; consistent/independent 4 59. 1 1 61. 1 mph; 4 mph 63. 2318 adult 2 , 3 2; consistent/independent tickets; 1482 child tickets 65. premium: $3.97, regular: $3.87 67. nursing student $6500; science major $3500 69. 150 quarters, 75 dimes 71. a. 3 mph b. 5 mph 73. a. 3.6 ft/sec b. 4.4 ft/sec 75. a. 100 lawns/mo b. $11,500/mo 77. a. 1.6 billion bu, 3 billion bu, yes b. 2.7 billion bu, 2.25 billion bu, yes c. $6.65, 2.43 billion bu 79. about 227 boards at $410 a piece 81. about 90,000,000 gal at $3.04/gal 83. 1776; 1865 85. Tahiti: 402 mi2, Tonga: 290 mi2 87. y 2x 3 89. $50,000 y 91. 93. x 120.716 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x


2/2/11

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Page SA-55

SA-55



27.

1. triple 3. equivalent, systems 5. z 5 7. Answers will vary. 9. Answers will vary. 11. yes, no; R2, R3 13. (5, 7, 4) 15. (2, 4, 3) 17. (4, 0, 3) 19. (5, 12, 13) 21. no solution, inconsistent 23. 51x, y, z2 | x 僆⺢, y 2 x, z 2 x6; 1p, 2 p, 2 p2 , other solutions possible 5 2 5 2 25. e1x, y, z2| x z , y z 2, z 僆⺢ f ; a p , p 2, pb, 3 3 3 3 other solutions possible 27. 1p, 2p, p 12 29. 1p 9, p 4, p2 1 31. e 1x, y, z2 ` x y 2z 6 f 33. 11, 3 2 , 22 2 35. 1p 17, p 4, p2 37. (12, 6, 4) 39. 11, 5, 62 1 1 41. 11, 2, 32 43. a , , 2b 45. (5 cm, 3 cm, 4 cm) 5 2 47. $80,000 at 4%; $90,000 at 5%; $110,000 at 7% 49. World War II, 1945; Korean, 1953; Vietnam, 1973 51. Declaration of Independence, 1776; 13th Amendment, 1865; Civil Rights Act, 1964 53. 1 L 20% solution; 3 L 30% solution; 6 L 45% solution 55. saturated: 1.2 g, monounsaturated: 1.0 g, polyunsaturated: 0.6 g 57. h(t) 5t2 30t 1 a. 46 ft b. 17.2 ft 59. 2 61. H11, 5I; H6, 43 63. x 1 2I

108642 2 4 6 8 10

33.

39.

15.

29.

y

108642 2 4 6 8 10

2 4 6 8 10 x

35.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

41.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

43.

10 8 6 4 2

108642 2 4 6 8 10

31.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

37.

y

2 4 6 8 10 x

5 4 3 2 1 54321 1 2 3 4 5

y

2 4 6 8 10 x

y

1 2 3 4 5 x

y

2 4 6 8 10 x

9.3

0

8 6 4 2 2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

1. half; planes 3. solution 5. The feasible region may be bordered by three or more oblique lines, with two of them intersecting outside and away from the feasible region. 7. No, No, No, No 9. No, Yes, Yes, No y y 11. 13. 10 10

108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10


8 6 4 2

10 8 6 4 2

9.4

0 (2, 4) 45.

9.3

2 4 6 8 10 x

(0, 0) is in the solution region, and 3(0) 2(0) 8 is true.

10

10

0

9.4

0

10

(3, 2)

yx 1 yx 1 49. • x y 6 3 51. (5, 3) 53. (12, 11) xy 7 3 y0 55. 26 at (2, 2) 57. 264 at (4, 3) 59. 5 6 H 6 10 61. 100

10 17.

(0, 0) is in the solution region, and 4(0) 5(0) 20 is true.

10

10

108642 2 4 6 8 10

2 4 6 8 10 x

23.

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

60 40 20 40

60

80 100

A (in thousands)

10

8 6 4 2

80

20

10

19. No, No, No, Yes y 21. 10

J (in thousands)

47. e

25.

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

J A 50,000 J 20,000 A 25,000 63. 300 acres of corn; 200 acres of soybeans 65. 240 sheet metal screws; 480 wood screws 67. 65 traditionals, 30 Double-T’s 69. 3 buses from company X; 4 buses from company Y y 71. a. the region is a square 10 8 b. maximum is 35.1 at (3, 3) 6 (3, 3) 4 c. optimal solutions occur at vertices 2 108642 2 4 6 8 10

73. a. 35 b.

2 4 6 8 10 x

5 4

c. 34

75. 324


SA-56

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Page SA-56


Exercises 9.4, pp. 889–891 1. template 7. 11. 15. 19. 23. 27. 31. 37. 41. 45. 49. 55.

3. repeated linear 5. Answers will vary

B A x3 x2

B A x1 1x 12 2 A B C A B C 13. x1 x2 x3 x x3 x1 A B C A B C 2 17. 2 3 x5 x x 2 1x 52 1x 52 x A B C D Bx C A 2 2 21. x x5 x3 x 1x 52 2 x 5x 7 A Bx C Dx E 3 4 2 2 25. x1 2x 5 x3 x 2 1x 22 2 2 1 4 5 7 1 29. x x1 x1 x x1 1x 12 2 2 3x 1 5 x1 5x 2 1 2 2 33. 35. 2 x2 x x2 1 x 2 x 3 1x 12 2 2 1 x1 3 4 2 2 39. x 2 x x x x3 4 2x x2 2 3x 5 2x 1 2 43. 2 x3 1x 32 2 x 1 1x 12 2 9.

2 3 x1 1x 12 3 1 10

P

1 100

1 10 10 P

1 100

47.

P

49 51. 50

53.

1 10

100 10P 62 125

4 4 3 ln x 2 ln x 1 1ln x 12 2

57. Factor out 1 from the denominator: 1 x2 x 2 3 x2 1x 1211 x2 1x 121121x 12 x1 1x 12 2 1x 12 2 59. 2x 1

1

x2 x 6

61. Verified

Mid-Chapter Check, p. 891 1. (1, 1); consistent 2. (5, 3); consistent 3. 20 oz 4. no; R2, R3 5. The second equation is a multiple of the first equation. 6. (1, 2, 3) 2 3 1 7. (1, 2, 3) 8. x1 x2 1x 22 2 9. Mozart 8 yr; Morphy 13 yr; Pascal 16 yr 10. 2 table candles, 9 holiday candles

Reinforcing Basic Concepts, p. 892 Exercise 1: Premium: $4.17/gal, Regular: $4.07/gal, 15.3R 35.7P 211.14 e P R 0.10 Exercise 2: Verified

Exercises 9.5, pp. 901–905 1. square 3. 2, 3, 1 5. Multiply R1 by 2 and add that result to R2. This sum will be the new R2. 7. 3 2, 5.8 9. 4 3, 1 1 11. £ 1 2

2 0 1

1 1 1

1 3 § ; diagonal entries 1, 0, 1 3 x 2y z 0 x 4y 5 1 13. e 15. S 13, 2 • y 2z 2 S 111, 4, 32 2 y 12 z3 x 3y 4z 29 1 6 2 17. • 19. c d y 32 z 21 2 S 14, 15, 32 0 28 6 z3

1 3 3 2 3 1 1 23 12 15 § 23. £ 0 3 3 21. £ 0 2 1 0 4 0 10 13 25. 2R1 R2 S R2 27. 5R1 R2 S R2 3R1 R3 S R3 4R1 R3 S R3

8 6 § 34

31. (1, 6, 9) 33. (1, 1, 2) 35. (1, 1, 1) 37. (2, 1, 3) 1 1 (10, 6, 8) 41. a , 1, b 43. 11, 3 2 , 22 2 4 linear dependence (p 4, 2p 8, p) coincident dependence {(x, y, z)|3x 4y 2z 2} inconsistent, no solution 51. linear dependence, 1p, 13 p, 4 53 p2 (2, 1, 1) 55. (3, 0, 2) 57. (1, 2, 3, 1) 59. 28 units2 LA to STL, 1600 mi; STL to CIN, 310 mi; CIN to NY, 570 mi Moe 90, Larry 45, Curly 30 65. 15 m, 36 m, 39 m $2000 at 5%; $3000 at 7%; $5000 at 9% 69. x 84°; y 25° 5 513 a. z1 110 cis 3 ␲ tan1 132 4 b. z2 i 2 2 C 30,000 in the year 2011 (t 6.39)

29. (20, 10) 39. 45. 47. 49. 53. 61. 63. 67. 71. 73.

Exercises 9.6, pp. 913–917 1. aij, bij 3. scalar 5. Answers will vary. 7. 2 2, a12 3, a21 5 9. 2 3, a12 3, a23 6, a22 5 11. 3 3, a12 1, a23 1, a31 5 10 0 d 13. true 15. conditional, c 2, a 4, b 3 17. c 0 10 19. different orders, sum not possible 13 12 14 8 263 20 15 19 32 8 1 5 d 21 21. c 85 23. £ 4 25. c 2 8 § 35 d 25 10 16 16 7 31 9 8 4 2 5 2

2 0 1 0 1 2 § 31. c d 0 1 3 6 12 24 90 d 33. matrix mult. not possible 35. c 6 15 57 79 30 42 18 60 1 0 d 39. c d 41. c d 37. c 50 19 12 42 36 0 1 27. £ 0 2

1 7 2 3 2

0 1 § 6

1 29. £ 0 4

1 3 0 1 0 0 4 4 4 3 57 1 3 1 £ d 43. £ 0 1 0 § 45. c 19 47. 2 8 8 § 1 5 19 57 1 11 1 0 0 1 4 16 16 4 28 4 1.75 2.5 d 51. c d 49. c 53. verified 55. verified 7.5 13 8 17 3 57. P 21.448 cm; A 27.7269 cm2 59. a. T S T S S 3820 1960 S 4220 2960 V D £ 2460 1240 § M D £ 2960 3240 § P 1540 920 P 1640 820 b. 3900 more by Minsk c. d. 8361.6 5116.8 3972.8 2038.4 £ 5636.8 4659.2 § V £ 2558.4 1289.6 § 3307.2 1809.6 1601.6 956.8 4388.8 3078.4 M £ 3078.4 3369.6 § 1705.6 852.8

61. 322,000 19,000 23,500 14,000 4 ; Total Profit N: $22,000, S: $19,000, E: $23,500, W: $14,000 63. a. $108.20 b. $101 First row, total cost for science from each 100 101 119 d restaurant; second row, total cost for math c. c 108.2 107 129.5 from each restaurant. 32.4 10.3 21.3 d a. 10 b. 20 65. c 29.9 9.6 19.5 c. p13 gives the approximate number of females expected to join the writing club


2/2/11

5:56 PM

Page SA-57

Student Answer Appendix 1 3 1 1 1 2 4 2 4 39. x x1 x1 1x 12 2 1x 12 2

0 2n1 2n1 n 67. £ 2 1 1 2n 1 § 69. 11, 1, 22 2n1 0 2n1 9␲ ␲ 5␲ 13␲ 71. z 3 cisa b, 3 cisa b, 3 cisa b, 3 cisa b 8 8 8 8

3. AB, BA, I, A1 5. Answers will vary.

7. verified 9. verified 11. verified 13. verified 1

9 15. c 1

23.

9 2 39 £ 13 4 39

2 9 5 d 18 1 13

0 2 13

17. c 10 39 1 3 § 19 39

5 2

1.5 d 0.5 9 80

1 25. £ 80

1 20

19. verified 21. verified 31 400 41 400 1 100

27 400 3 400 § 17 100

27. c

2 5

2 1 3 43. 214.5 ft2 of skin, 231.0 ft2 of wood x1 x3 1x 32 3 veneer, 516 tension rods, and 498 ft of hoop 45. 955 ft2 of skin, 1021.5 ft2 of wood veneer, 2180 tension rods, and 2129.5 ft of hoop 47. 5 Silver, 9 Gold, and 2 Platinum 49. one bundle of first class 9.25 measures of grain; one bundle of second class 4.25 measures of grain; one bundle of third class 2.75 measures of grain 51. Answers will vary. 53. Answers will vary. 55. Answers will vary. 57. Answers will vary. 59. Answers will vary. 61. Answers will vary. 63. Answers will vary. 65. Answers will vary. 67. Answers will vary. 69. $15,000 at 6%, $25,000 at 8% 71. apples: 29¢/lb; kiwi: 39¢/lb; pears: 19¢/lb 73. 10 lb of $1.90, 8 lb of $2.25, 6 lb of $3.50 75. Answers will vary. 77. x2 y2 4x 6y 12 0 41.

Exercises 9.7, pp. 927–932 1. diagonal, zeroes

SA-57

3 x 9 dc d c d 7 y 8

1 2 1 x 1 29. £ 1 0 1 § £y§ £3§ 2 1 1 z 3 2 1 4 5 w 3 2 5 1 3 x 4 31. ≥ ¥ ≥ ¥ ≥ ¥ 3 1 6 1 y 1 1 4 5 1 z 9 33. (4, 5) 35. (12, 12) 37. 1 15 , 13 2 39. 11.5, 0.5, 1.52

41. (3, 2, 5) 43. 11, 0.5, 1.5, 0.52 45. 1, yes 47. 0, no 51. singular matrix 53. 34 55. 7 57. singular matrix 1

13 59. det1A2 5; (1, 6, 9) 61. det1A2 0 63. A1 c 2 13

79.

10 8 6 4 2 54321 2 4 6 8 10

81. ⬔B 76.3°, ⬔C 54.7°, side c 9.4 in.

y

1 2 3 4 5 x

49. 1

Making Connections, p. 947 5 13 3 d, 13

verified

65. singular 67. 31 behemoth, 52 gargantuan, 78 mammoth, 30 jumbo 69. Jumpin’ Jack Flash: 3.75 min Tumbling Dice: 3.75 min You Can’t Always Get: 7.5 min Wild Horses: 5.75 min 71. 30 of clock A; 20 of clock B; 40 of clock C; 12 of clock D 73. p1 72.25°, p2 74.75°, p3 80.25°, p4 82.75° 75. y x2 4x 5 77. y x3 2x2 9x 10 79. $450 in the CD, $350 in the MM 81. $1500 in retirement fund, $1500 in mutual fund, $1800 in stock fund 83. 2 oz Food I, 1 oz Food II, 4 oz Food III 2␲ 85. Answers will vary. 87. A 125, period 3 9 1 89. x 僆 aq, d ´ c , qb 2 2

Exercises 9.8, pp. 942–947 1. a11a22 a21a12 3. constant 5. Answers will vary. 2 5 7 5 2 7 ` ; Dx ` 7. D ` ` ; Dy ` ` 9. 15, 92 3 4 1 4 3 1 26 25 11. a , b 13. not possible 3 3 4 1 2 5 1 2 2 1 † Dx † 8 2 1 † 15. a. D † 3 1 5 3 3 5 3 4 5 2 4 1 5 Dy † 3 8 1 † Dz † 3 2 8†, 1 3 3 1 5 3 b. | D| 22, solutions possible c. | D| 0, Cramer’s rule cannot be used: coefficients R1 R2 R3 3 5 1 17. (1, 2, 1) 19. a , , b 21. 10, 1, 2, 32 4 3 3 2 23. 320 32␲ 420.5 in 25. 8 cm2 27. 27 ft2 29. 19 m3 31. 96 in3 33. yes 35. no 37. yes; yes; yes

1. h

3. d

5. a

7. f

9. b

11. f

13. h

15. c

Summary and Concept Review, pp. 948–953 1.

(4, 4)

108642 2 4 6 8 10

3.

5

2.

y 3x 2y 4

10 8 6 4 x 3y 8 2

2 4 6 8 10 x

(4, 4)

10 8 6 4 2 54321 2 4 6 8 10

y

x 0.3y 1.4

1 2 3 4 5 x

(q, 3)

0.2x 0.5y 1.4

1 12 , 32

4. no solution; inconsistent

y

2x y 2 4 3 2 1 54321 1 2

x 2y 4 3

1 2 3 4 5 x

(U, T)

4 5

A 85, 6 5 B

5. 15, 12; consistent 6. (7, 2); consistent 7. 13, 12; consistent 1 8. A 11 9. Willis Tower is 1450 ft; Hancock Building is 4 , 6 B ; consistent 1127 ft. 10. $1.20 11. (0, 3, 2) 12. (1, 1, 1) 13. no solution, inconsistent 14. 72 nickels, 85 dimes, 60 quarters 15. y x2 10x 9 y y 16. 17. 10 10 8 6 4 2 108642 2 4 6 8 10

18.

8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

9.3

10

10

9.3


SA-58 19.

2/2/11

5:58 PM

Page SA-58

Student Answer Appendix Maximum of 270 occurs at both (0, 6) and (3, 4)

y 10 8 6 4 2

2 4 6 8 10 x

4 2 2 7 22. x3 x5 x3 2x 1 2 x2 x1 1 3 3 23. 24. 25. 2 2 2 x4 x5 x3 x 1 x 3 x 3x 9 5 x2 1 3 5 26. 27. 2 x1 x2 x3 x x1 1x 22 2 2 1 3 2 x5 28. 29. 2 2 x1 x5 1x 12 2 x 3 1x 32 2 30. 12, 42

7.25 33. c 0.875

21.

5.25 d 2.875

6.4 17 20.8

34. c

6.75 1.125

17 2 § 13

44. It’s an identity matrix.

Practice Test, pp. 953–954 y

10 3x 2y 12 8 x 4y 10 6 4 2 (2, 3)

108642 2 4 6 8 10

2 4 6 8 10 x

19. 30 plain; 20 deluxe

20.

3x 2 1 2 x3 x 3x 9

Strengthening Core Skills, pp. 955–956 Exercise 1: (1, 4, 1)

(2, 3)

2 4 2. a , b 5 5

3. 13, 22

4. 12, 1, 42

5 1.2 d b. c 9 1.2 1 d e. 2 1.5 0.1 0 0.6 0 § 0.8 0.9

0.31 c. £ 0.01 0.39

0.13 0.05 0.52

0.08 0.02 § 0.02

1.2 d 2

c. c

3 3

c. x 5, i 12

d. x 1, 0, 4

6 4 2 2 2

2

4

6

8 10 12 x

4 6

7. a. 1a bi2 1a bi2 2a b. 1a bi2 1a bi2 a2 1bi2 2 a2 b2 115 113 13 9. x 12 11. sin ␪ , tan ␪ , cos ␪ 3 4 4 y2 y1 x2 x1 y2 y1 b 13. a. m b. a , b c. x x2 x1 2 2

139 3 2b2 4ac 2a

d. d 21x2 x1 2 2 1y2 y1 2 2 e. A Pert 15. H3, 18I 2121 x2 x 17. a. b. 11 29 x2 y 19. x 僆 1q, 12 ´ 12, 32

2 4 6 8 10 x

6 8 2 d. c 2.5 0 6. a. £ 0.5 0.2

2 b. x 0, 7 3 1 3. R 2A 1␲r2 2 ␲ y 5. 1. a. x

5 2x 1 56. 57. 92,250 gal gasoline, 595,000 lb corn, 227,500 2 x2 x 3 oz yeast, and 134,750 gal water 58. 287,250 gal gasoline, 2,035,000 lb corn, 777,500 oz yeast, and 460,750 gal water 59. PIE ARE SQUARE

8

1 d 5

4 54321 4

1 2 3 4 5 x

8

0.3 b. £ 0.06 0.18 d.

40 17 40 £ 17 35 17

0.06 0.06 0.24 0 10 5

10 17 10 17 § 30 17

0.12 0 § 0.48 e. 0.034

97 18 1 b 8. 13, 2, 32 9. a , b 10. (1, 6, 9) 3 34 17 2 11. 21.59 cm by 35.56 cm 12. Tahiti 402 mi ; Tonga 290 mi2 13. $15,000 at 7%; $8000 at 5%; $7000 at 9% 7. a2, 1,

8 6 4 2

Cumulative Review Chapters 1–9, pp. 956–957 43. D

45. It’s the inverse of B. 46. E 47. It’s an identity matrix. 48. It’s the inverse of F. 49. matrix multiplication is not generally commutative 50. 18, 62 51. 12, 0, 32 19 25 91 37 36 31 52. a , b 53. 11, 1, 22 54. a , , b 55. units2 35 14 19 19 19 2

108642 2 4 6 8 10

20 b. equilibrium is achieved when approx 150,000 plugs are sold at a price near $1.18 15. h1t2 16t2 144 16. 144 ft, 3 sec y 17. 18. maximum 250 at (5, 0) 10

31. (1, 6, 9) 32. (2, 7, 1, 8)

6.75 d 35. not possible 1.125 1 0 4 2 6 1 0 36. c d 37. c d 38. £ 5.5 1 1 § 1 7 0 1 10 2.9 7 3 6 4 8 12 0 39. £ 4.5 3 1 § 40. not possible 41. £ 2 4 4 § 2 3.1 3 16 0.4 20

5. a. c

140

50

20. 50 cows, 425 chickens

1.

70

(3, 4)

6 4 2 2 4 6

15.5 42. £ 9 18.5

14. a.

21. 128 128i 13

23. about 3.6 yr ␲ 25. A 2, B 1, C 4 27. q1x2 ␲2, r1x2 ␲x ␲3, crosses at x ␲2 1 9.872 29. y 4 at x 2

Connections to Calculus Exercises, p. 960 k A B ; 1A B2x 1Ab Ba2 k 1x a21x b2 xa xb b. B A 1 Ab 1Aa2 k; A1b a2 k k 16 2 2 c. A ; b a 1x 32 1x 52 x3 x5 1. a.


2/2/11

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SA-59

Student Answer Appendix 13. 1x 62 2 1y 52 2 9 center: (6, 5), r 3

1 1 2 2 5. x3 x3 x7 x4 1 2 7. 9. 114 units2 11. 36 units2 x8 x5 13. 240 units3; V LWH 152162182 240 units3 3.

10

15. 1x 22 2 1y 52 2 25 center: 12, 52, r 5

y

10

r3 (6, 5) 10

10

10 x

r5

10 x

(2, 5)

CHAPTER 10

10

Exercises 10.1, pp. 966–968 1. geometry, algebra 3. perpendicular 5. point, intersecting 7. 12, 22; verified 9. 12, 22; verified 11. 1 132 , 92; verified 13. 1x 22 2 1y 22 2 52 15. 1x 22 2 1y 22 2 52 13 2 25 2 17. ax b 1y 92 2 a b 19. d 13; B, C, E, G 2 2 1 815 21. Verified, d 23. B, C, E 25. Verified 27. y x2 5 16 29. 4x2 3y2 48 31. Verified, verified 33. 3x2 y2 3 35. Verified 37.

cos2x sin2x sin2x sin2x

39. a. x 2 ln 9, x 4.39 b. x

cos2x sin2x

10

17. 1x 32 2 y2 14 center: (3, 0), r 114 10

19.

5 ln 54, x 9.97 2

10

r 3.7 10

10 x

10

10 x

(3, 0) 10

21.

10

23.

y

10

10 x

2


9.3

y

y

10

3. 2a, 2b

10

10

y

cot2x

Exercises 10.2, pp. 979–983 1. c2 | a2 b2 | 7. x2 y2 49

y

10

10 x

10

2

y x 1, (0, 0), a 4, b 2 16 4 b. (4, 0), (4, 0), (0, 2), (0, 2) d. 3.1

25. a.

c.

10

y

10

14.1

14.1

9.3

4.7

10

4.7

3.1 (2.4, 0.8), (2.4, 0.8), (2.4, 0.8), (2.4, 0.8) y2 x2 27. a. 1, (0, 0), a 3, b 4 9 16 b. (0, 4), (0, 4), 13, 02 , (3, 0) c. d. 3.1

(4.2, 5.6), (4.2, 5.6) 9. 1x 52 2 y2 3 3.1

10 x

10

y

10

10 x

9.4

0

4.7

10

4.7

3.1

16, 122, 16, 122 11. 1x 12 2 1y 52 2 25

3.1 (2.4, 0.8), (2.4, 0.8), (2.4, 0.8), (2.4, 0.8) y2 x2 29. a. 1, (0, 0), a 15, b 12 5 2 b. ( 15, 0), ( 15, 0), (0, 12), (0, 12) c.

12.4

9.4

y 5

9.4 5

5 x

5

0 (5, 8), (5, 2)


SA-60

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Page SA-60


31. ellipse

10

33. circle

y

10

y

10

55.

10

10 x

y2 x2 1 36 20

a.

10 x

r3 10

35. ellipse

10

37. x 2

y

10

(2, 3) (6, 1)

1

4

41.

1x 32 2 4

y

10

10

1y 52 2 10

1

y

10

10

D: x 僆 3 6, 24 , R: y 僆 3 1, 3 4

y 3

1y 22 2 10

2

8 x

(5, 5) (3, 5) (3, 5 √10)

D: x 僆 3 1, 54 , R: y 僆 3 5 110, 5 1104

7

b.

5

1

y

6

(3, 2 √10) (3, 2) 10

6.2 c. L 6.6 d. verified 1x 32 2 1y 22 2 57. 1 a. 9 25

10 x

(1, 5)

10

(3, 5 √10)

(2, 1) 10 x

25

(0, 5)

(2, 1)

(2, 1)

43.

9.4

(1, 3)

D: x 僆 3 1, 1 4, R: y 僆 35, 1 4

1y 12 2

10

1x 32 2

9.4

5 x

(1, 3) (0, 3) 5

10

6.2

y

(0, 1)

10

16

b.

1

4

10 x

10 x

10

1y 32 2

5

1x 22 2

10

10

5

39.

y 10

12

10 x (8, 2

(2, 2) 10

(3, 2 √10)

D: x 僆 32, 84 , R: y 僆 3 2 110, 2 1104 45. k 20 47. k 20 49. a. (2, 1) b. (3, 1) and (7, 1) c. (2 121, 1) and (2 121, 1) y d. (2, 3) and (2, 1) e. 10

10

10 x

10

51. a. (4, 3) b. (4, 2) and (4, 8) y and (8, 3) e.

c. (4, 0) and (4, 6)

d. (0, 3)

10

10

10 x

10

53. a. (2, 2) b. (5, 2) and (1, 2) c. (2 13, 2) and (2 13, 2) d. (2, 2 16 ) and y (2, 2 16 ) e. 10

7 c. L 3.6 d. verified y2 x2 59. 1, 1 17, 02 16 9 2 1x 32 1y 12 2 61. 1, 13, 1 2 132 4 16 63. A 12␲ units2 65. 27 2.65 ft, 2.25 ft y2 y2 x2 x2 67. 8.9 ft, 17.9 ft 69. 2 1; 6.4 ft 71. 1 2 6.25 2.25 15 8 2 y x2 73. 1 75. a 142 million miles, b 141 million 2 36 135.252 2 miles, orbit time 686 days 77. about 66,697 mph 79. 90,000 ␲ yd2 81. a. 45 ft b. 106 ft c. 25 ft 83. ellipse, since squared terms are positive and A B; 61x 32 2 31y 42 2 0, the constant term becomes zero; the graph is the single point 13, 42 log 20 85. 2.73 87. z 64 log 3

Exercises 10.3, pp. 993–997 1. transverse 3. perpendicular, transverse, center 5. Answers will vary. y y y 7. 9. 11. 10 10 10 (0, 0)

(0, 0)

10

(7, 0)

10 x

(3, 0) 10

(2, 0)

(3, 0)

10

10 x

10

10

(2, 0) 10 x

10

(7, 0) (0, 0)

10

10 x

10


2/2/11

5:28 PM

Page SA-61

SA-61


10

15.

y

10

17.

y

y

10

(0, 0)

(0, 3) (0, 0)

(6, 0) 10

10

10 x

(0, 2√3) (0, 0) 10

10 x

10 x (0, 2√3)

(0, 3) 10

19.

10

10

21.

y

10

1x 32 2

1y 42 2

1 a. (3, 4) b. (0, 4) and (6, 4) 9 4 c. 13 113, 42 and 13 113, 42 d. 2a 6, 2b 4 y e. 10

65.

(6, 0)

10

y

10

10 x

(0, 3) (0, 6) 10

(0, 0)

(0, 0) 10

10

10 x

10 x

10

1y 32 2

x2 1 a. (0, 3) b. (0, 7), (0, 1) 16 4 c. 10, 3 2 152, 10, 3 2 152 d. 2a 4, 2b 8 y e. y 2x 3, y 2x 3 14

(0, 6)

(0, 3)

67.

10

23. 14, 22, 12, 22, y 2, 11, 22, x 1 25. 14, 12, 14, 32, x 4, 14, 12, y 1 y y 27. 29. 31. 10

y

10

10

(9, 2)

5, 1 7

(0, 1)

10 10

10 x (0, 1) 10

33.

10

10

(0, 3)

10

35.

y

10

x

(3, 2) (3, 2)

5, 1 7

10

37.

y 6

10

(1, 2) (6, 1) 10

10

10

39.

10

41.

10

10

(0, 0) (3, 0)

43.

y

10

10

10

10

(√6, 0) 10

10 x

45.

9 10

10

1y 22 2 36

a. 13,22

b. 11, 22 and 15, 22

d. 2a 4, 2b 413

10

10 x

10

y2 y2 1x 22 2 x 1 73. 1 36 28 9 9 2 2 1y 12 2 y 1x 22 2 x 75. 1, 1 113, 02 77. 1, 4 by 215 4 9 4 5 79. a. y 23 2x2 9 b. x 僆 1q, 3 4 ´ 33, q 2 2 c. y 2 3 2x 9 6.2 71.

(√6, 0) 10 x

10

1

y

9.3

(8, 2)

10

1

y

(0, 2)

1x 52 2

1y 22 2

2

(0, 0) 10 x

4 5, 3

(0, 0)

(0, 2)

(3, 0)

6

10 x

14

y

1x 32 2

4 12 c. 11, 22 and 17, 22 y e. 10

y (4, 3)

10

(1, 8)

(2, 1)

(2, 1)

69.

4 5, 3

10 x (1, 3)

10 x

10 x

10 x

(5, 1)

10 x

9.4

(2, 2)

18.8

10

(5, 2)

9.4

9.4

6.2 15.3

81. 40 yd

(0, 10), (0, 6), (10, 10), (10, 6)

47.

1y 32 2 1

14

1x 22 2 4

85.

1

y

y2 x 1, about (24.1, 60) or 124.1, 602 225 2275

14.4

87. y

(2, 4) (2, 3) 14

(2, 2) 6

6 x

89. a. 20.8

83. 12 microns

2

16.8

b2 Ba

2

1x 42 2 1 4

x2 b2, as x S q, y S 1y 22 2 0

b2 B a2

b. 1x 22 2

b x2 x a 1y 42 2 1 5

0

91. 6mph 93. 42 solid, 46 liquid

Exercises 10.4, pp. 1003–1006 10.4 (2.8, 5.6), (2.8, 0.4), (6.8, 0.4), (6.8, 5.6) 49. circle; A B 51. circle; A B 53. hyperbola; A, B opposite signs 55. hyperbola; A, B opposite signs 57. circle; A B 59. ellipse; A B 61. 8, 2a 8, 2b 6 63. 12, 2a 16, 2b 12

1. horizontal, right, a 0 3. ( p, 0), x p 5. Answers will vary. 7. x 僆 1q, q 2, y 僆 34, q 2 9. x 僆 1q, q2, y 僆 1q, 18 4 10

y

20

y (2, 18)

(0, 10) (1, 0) 10

(0, 3) 10

(1, 0)

(3, 0) 10 x

3

(5, 0) 7 x

(1, 4) 20


SA-62

2/2/11

5:28 PM

Page SA-62


11. x 僆 1q, q2, y 僆 310.125, q 2

13. x 僆 3 4, q 2, y 僆 1q, q2

51.

8

(1, 0)

(5, 6)

4

4

8

(3, 0) 4 (0, 1)

x

(7, 2)

57.

16 8

(16, 0)

(7, 0)

8 16 4 (0, 1)

10 8

16

x

y

10

(9, 3) 8 4 (0, 6)

(0, 2)

(4, 2.5)

(4, 0)

(0, 0) 8

10

x

25

x 64

10 x

4

, 25 64 0

(0, 2)

8

1

10

23. x 僆 1q, 04 , y 僆 1q, q 2

81. 85. 87. 89. 91.

(0, 2) (0, 1) 10

10 x

10 x

(1, 0) (0, 3)

(6.25, 0.5) 10

27. x 僆 3 21, q2, y 僆 1q, q2

10

29. x 僆 1q, 11 4, y 僆 1q, q2

y

15 (0, 5 21)

15

0,

(0, 5 21) 25

4 22 2

y


0,

(11, 2) 4 22 2

10

4

15

4.

15 x

10

10

7. a. (0, 2)

10

10 x

y 2

(0, 0) 20 x

(19, 0)

10

41.

y

10

43.

y

10

y x1

(1, 0)

(0, 0) 10

y

10 (0, 6)

3 2

10

47.

y

10

(0, 0)

10

10 x

10 x

10

49.

y

9 9 ( 2,

10

5.

y

6.

y

y 4

12 8

8

4

12 8

4

1 2 3 4 x

5

0) 10 x

2, 0

Exercise 2: (4, 0)

(0, 0)

10

10 x

10

4 4

4 4 4

8

x

12

1y 12 2

15

y 2 10


y (4, 2)

(0, 0) 10

1x 32 2

Exercise 1:

5

x2

10

12

1 2 3 4 5 6 7 x

3

20 x

x 2

321 1 2 3 4 5

0, 2

(0, 0) 20

10

16 x

9. y6

45.

12

y2 x2 1 16 4 10. yes, distance d 49 mi

10

10

8

y

5 4 3 2 1

1; D: x 僆 3 5, 1 4 ; R: y 僆 3 3, 54 4 16 2 2 b. 1x 32 1y 22 16; D: x 僆 31, 7 4; R: y 僆 32, 64 c. y 1x 32 2 4; D: x 僆 1q, q2; R: y 僆 34, q 2 8. 1y 22 2 61x 12 2

y

y

(1, 3)

4 8

654321 1 2

10

37.

4 4

8 7 6 5 4 3 2 1

(11, 0)

35. x 僆 31, q2, y 僆 1q, q2

x

12

(2, 3) 10 x

12

8

y

(2, 3)

8

4

33. x 僆 3 2, q2, y 僆 1q, q2

y

3.

y 8 4

4

(0, 7)

39.

2.

y

15 x

15

31. x 僆 1q, q 2, y 僆 3 3, q 2

20

5 x

(3, 0)

15

10

4

4

(4, 0)

10

3

6 in.; (13.5, 0) 83. 14.97 ft, (0, 41.75) y2 5x or x2 5y, 1.25 cm 1x 22 2 12 1y 82; p 18 ; 12, 82 Answers will vary. ␪ 30° 360°k, 330° 360°k

1.

15

25 x

10

2

(4, 2.5)

y

10

10

1

25. x 僆 3 6.25, q2, y 僆 1q, q2

y

(6, 0)

(21, 5)

15

61. x2 8y 63. y2 16x 65. x2 20y 67. 1y 22 2 121x 22 69. 1x 42 2 121y 72 71. 1x 32 2 81y 22 73. y2 81x 12; vertex (1, 0); focus (1, 0) 75. 1x 22 2 121y 22 ; focus: (2, 12 ; endpoints (4, 1) and (8, 1) 77. 16 units2 y 79.

21. x 僆 3 4, q2, y 僆 1q, q2

y

10

10 x

4 (0, 4)

4

19. x 僆 3 9, q2, y 僆 1q, q2

4

10

10 x

(6, 5)

8

4

(2, 3)

10

16 8

x

x7

(5, 5)

8

8

8

20

y

15

x3

(1, 3)

y

(16, 3)

10

59.

y

10

17. x 僆 1q, 0 4, y 僆 1q, q2

y

4

10 x

(4, 3)

y 4

y 8

10

10

10 x

8

15. x 僆 1q, 16 4 , y 僆 1q, q2 (0, 7)

10

25 x

x

12 (1.25, 10.125)

8

20 (0, 6)

(4, 2)

25

(0, 3)

(4, 1)

6 6 12 6 (0, 7)

y

x 5

(7, 4)

4

12

12 6

55.

y

10

8

y

(3.5, 0)

53.

y

10

y

15 x

251x 22 2 2 281x 12 2 25

91y 32 2

1 4 2 481y 22 1 25

8

x


2/2/11

5:28 PM

Page SA-63

SA-63


Exercises 10.5, pp. 1015–1018 1. a. 3 or 4 not possible

63. 8.5 m 10 m 65. 5 km 9 km 67. 8 ft 8 ft 25 ft 69. $1.83; $3 90,000 gal 10P2 6D 144 e 2 8P 8P 4D 12 71. Answers will vary. 73. 18 in. by 18 in. by 77 in. 75. 62 ft 77. a. m 400 1 , the copier depreciates by $400 a year. b. y 400x 4500 c. $1700 d. 9.5 yr

b. 3 or 4 not possible

c.

d.

e.

Exercises 10.6, pp. 1031–1035

f.

1. polar 3. II, IV 2␲ ␲ 7. 3 3 5␲ 6

3. region, solutions 5. Answers will vary. 7. first: parabola; second: line 9. first: parabola; second: ellipse y

10 8 6 (1, 5) 4 2

(2, 2)

108642 2 4 6 8 10

(3, 8)

3 2 1

7␲ 6

11. first: hyperbola; second: circle

(5, 4)

y

10 8 6 4 2

2 4 6 8 10 x

(5, 4)

3 15. a4, b, (3, 2) 2 17. 1 110, 32, 1 110, 32, 15, 122, 15, 122 19. 14i, 16 24i2, 14i, 16 24i2 21. (4, 3), 14, 32, 14, 32, 14, 32 23. 12 i, 2 i2, 12 i, 2 i2 25. 15, 52, (5, 5), 15, 52, 15, 52 27. (4, 4), (6.187, 1.571) 29. 11.863, 11.8082, 17.863, 1.8082 31. (5, log 5 5) 33. (3, ln 9 1), (4, ln 16 1) 35. (0, 10), (ln 6, 45) 37. (3, 1), (2, 1024) 39. (3, 21), (1, 1), (2, 4) 41. (2, 4), (6, 4) 43. (3, 5), (3, 5) 45. 12.43, 2.812, (2, 1) 47. (0.72, 2.19), (2, 3), (4, 3), (5.28, 2.19) y y 49. 51. 10 10 13. (4, 3), (3, 4)

8 6 4 2 108642 2 4 6 8 10

20 16 12 8 4 2016 1284 4 8 12 16 20

108642 2 4 6 8 10

2 4 6 8 10 x

y

4 8 12 16 20 x

7␲ 6

7␲ 5␲ 4 3 2␲ 3

5 4 3 2 1

␲ 6

11␲ 6

7␲ 5␲ 4 3

P

5␲ 6

7␲ 6

␲ 3 ␲ 4

5 4 3 2 1

3␲ 4

␲ 6

11␲ 6

5␲ 4 4␲ 3

13.

␲ 3 ␲ 4

P 5␲ 4 4␲ 3

3␲ 4

7␲ 5␲ 4 3

2␲ 3

5 4 3 2 1

␲ 3 ␲ 4

P

␲ 6

11␲ 6

5␲ 4 4␲ 3

7␲ 5␲ 4 3

␲ 3␲ ␲ 2␲ b 17. a4 12, b 19. a8, b 21. a412, b 2 4 3 4 5␲ 7␲ ␲ 23. a3 12, b, a3 12, b, a3 12, b 4 4 4 5␲ 7␲ ␲ 25. a2, b, a2, b, a2, b 27. C 29. C 31. D 33. B 6 6 6 ␲ 35. D 37. (8, 180°) or 18, ␲2 39. 14 12, 45°2 or a4 12, b 4 ␲ 41. 110, 45°2 or a10, b 43. 113, 247.4°2 or 113, 4.31762 4 5 12 512 45. 14 12, 4 122 47. 12 12, 2 122 49. 1 13, 12 51. a , b 2 2 tan ␪ 1 53. r 5 55. r2sin 2␪ 6 57. tan ␪ 3r cos ␪ 1 or r 3 cos ␪ 59. r2 cos 2␪ 61. x2 y2 6x 63. x 2 65. x2 y2 x 1 1 1 67. y x2 2 2 ␲ ␲ 69. 3␲ 2␲3 71. 3␲ 2␲3 3 ␲ 3 ␲ 5␲ 6

7␲ 6

8 6 4 2

53. no solution

5␲ 6

15. a4,

(5, 4)

108642 2 (5, 4) 4 6 8 10

3␲ 4

␲ 6

11␲ 6

5␲ 4 4␲ 3

5␲ 6

2 4 6 8 10 x

␲ 4

5

P 4

11.

108642 2 4 6 8 10

2 4 6 8 10 x

7␲ 6

y

10 (3, 8) 8 6 4 2

3␲ 4

5. Answers will vary. 2␲ 9. 3

55.

5 4 3 2 1 54321 1 2 3 4 5

2 4 6 8 10 x

5 4 3 2 1

4

5␲ 6

3␲ 4

2␲ 3

␲ 6

5␲ 6

11␲ 6

5␲ 4 4␲ 3

73.

4

7␲ 6

7␲ 5␲ 4 3

5 4 3 2 1

␲ 3 ␲ 4

5 4 3 2 1

4

75. 5␲ 6

11␲ 6

7␲ 6

3␲ 4

␲ 6

11␲ 6

5␲ 4 4␲ 3

␲ 6

4

7␲ 5␲ 4 3

2␲ 3

5 4 3 2 1

␲ 3 ␲ 4

␲ 6

y 7␲ 6

1 2 3 4 5 x

57. h 27.5 ft; h 24 ft; h 18 ft 59. The company breaks even if either 18,400 or 48,200 cars are sold. x2y 2000 61. e 2 ; approx. (12.4, 13) or (20, 5); The pool will x 4xy 800 likely have the dimensions 20 ft by 20 ft by 5 ft.

5␲ 4 4␲ 3

77. 5␲ 6

7␲ 6

3␲ 4

2␲ 3

5␲ 4 4␲ 3

7␲ 5␲ 4 3

5 4 3 2 1

6

␲ 3 ␲ 4

5␲ 4 4␲ 3

79. ␲ 6

5␲ 6

11␲ 6

7␲ 6

7␲ 5␲ 4 3

3␲ 4

2␲ 3

5␲ 4 4␲ 3

7␲ 5␲ 4 3

5 4 3 2 1

␲ 3

␲ 4

11␲ 6

␲ 6

11␲ 6

(5, 90 )

7␲ 5␲ 4 3


SA-64 81. 5␲ 6

7␲ 6

3␲ 4

5␲ 6

3␲ 4

5:29 PM

Page SA-64


2␲ 3

83.

␲ 3 ␲ 4

5 4 3 2 1

␲ 6

5␲ 6

11␲ 6

5␲ 4 4␲ 3

85.

2/2/11

7␲ 6

7␲ 5␲ 4 3

2␲ 3

␲ 6

2␲ 3

␲ 3 ␲ 4

5 4 3 2 1

5␲ 6

3␲ 4

␲ 6

11␲ 6

5␲ 4 4␲ 3

87.

␲ 3 ␲ 4

5 4 3 2 1

3␲ 4

7␲ 5␲ 4 3

2␲ 3

␲ 3 ␲ 4

5 4 3 2 1

5 12 5 12 X, Y 13. 0 x, 4 y 2 2 y2 x2 3 13 3 15. 2 x; 2 13 y 17. xy 13 9 2 2 2 2 2 2 19. 4X 2Y 9 21. a. 3X2 Y2 2 23. a. 4X2 Y2 8 y y b. b. 11.

Y

45

11␲ 6 7␲ 5␲ 4 3

5␲ 4 4␲ 3

89. 5␲ 6

3␲ 4

2␲ 3

␲ 3 ␲ 4

5 4 3 2 1

7␲ 6

91. ␲ 6

11␲ 6 7␲ 5␲ 4 3

5␲ 4 4␲ 3

5␲ 6

3␲ 4

2␲ 3

␲ 3 ␲ 4

5 4 3 2 1

45

␲ 6

foci: 2√6, 0 3 asymptotes: Y √3X

25. a. Y2 4X2 16 y b.

7␲ 6

7␲ 5␲ 4 3

11␲ 6

5␲ 4 4␲ 3

X

413 3 12 4 3 12 93. a , b; 1312, 3 122; 14 13, 42; yes 2 2 312 4 13 3 12 4 Ma , b 2 2 95. r 4 4 cos ␪ 97. r 4 cos15␪2 5␲ 6

3␲ 4

2␲ 3

10 8 6 4 2

␲ 3 ␲ 4

␲ 6

5␲ 6

3␲ 4

5 4 72 3 2 1

144

X Y x

␲ 3 ␲ 4

vertex: (0, 0)

foci: (2√5, 0)

focus: (1, 0)

asymptotes: Y 2X

directrix: X 1

29. a. X 4Y 25 y b. 2

x

60

vertices: (4, 0)

7␲ 5␲ 4 3

2␲ 3

27. a. Y2 4X 0 y b.

Y 60

11␲ 6

5␲ 4 4␲ 3

x

foci: (0, √6) minor axis endpoints: (√2, 0)

Open dot 7␲ 6

X

vertices: (0, 2√2)

x

vertices: √6, 0 3 7␲ 6

Y

X

␲ 6

ᏸ

2

X

Y 60

vertices: (5, 0)

x

foci: 5√3, 0 2 minor axis 0, 5 endpoints: 2

␲ 6

4 7␲ 6

5␲ 4 4␲ 3

99. r2 16 cos12␪2 5␲ 6

3␲ 4

2␲ 3

7␲ 6

11␲ 6 7␲ 5␲ 4 3

5 4 3 2 1

␲ 3 ␲ 4

216 288

5␲ 4 4␲ 3

11␲ 6

7␲ 5␲ 4 3

4

7 4 3 ; cos ␤; sin ␤ 25 5 5

33. a. parabola b. ␤ 45°; 2Y2 5 c. verified

101. r 4 sin ␪ ␲ 6

31. 336 7 0; hyperbola; cos12␤2

␲ 2

␲ 6

9 5Y2 2X 2 13Y 1 35. a. circle or ellipse b. ␤ 60°; X2 2 2 (ellipse) c. verified 37. f 39. g 41. h

2 7␲ 6

5␲ 4 4␲ 3

43. parabola

11␲ 6

7␲ 5␲ 4 3

103. a; this is a circle through 16, 0°2 symmetric to the polar axis. ␲ ␲ 105. g; this is a circle through a6, b symmetric to ␪ . 2 2 ␲ 107. f; this is a limaçon symmetric to ␪ with an inner loop. Thus a 6 b. 2 ␲ 3␲ 109. b; this is a cardioid symmetric to ␪ through a6, b. 2 2 2 2 111. r 7200 sin12␪2 113. r 15 cos15␪2 or r 15 sin15␪2 115. ␲; ␲; ␲; Answers will vary. 117. Consider r a 1cos12␪2 and r a 1cos12␪2 ; both satisfy r2 a2cos12␪2. Thus, 1r, ␪2 and 1r, ␪2 will both be on the curve. The same is true with a 1sin12␪2 and a 1sin12␪2. 2␲ 5␲ 119. 9␲ units2 121. t 0, , ␲, 3 3 y 123. D: x 僆 3 5, 22 ´ 12, 5 4 5 4 R: y 僆 33, 22 ´ 546 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

Exercises 10.7, pp. 1046–1051 1. rotation of axes; Y2 X2 7. 1 8 8

B AC

3. invariants


9. 6 3 12 X, 6 312 Y

7␲ 2␲ 12 3␲ 3 4 5␲ 6 11␲ 12 13␲ 12 7␲ 6

5␲ 12 ␲ 3 ␲ 4 ␲ 6

(1, ␲2 )

(2, 0)

45. ellipse

␲ 12

23␲ 12 11␲ 7␲ 6 5␲ 4 19␲ 3 12

(4, 7␲ 6) 5␲ 4 4␲ 3 17␲ 12

47. hyperbola 7␲ 2␲ 12 3␲ 3 4 5␲ 6 11␲ 12

(1, 0)

13␲ 12 7␲ 6

5␲ 12 ␲ 3 ␲ 4 ␲ 6

7␲ 2␲ 12 3␲ 3 4 5␲ 6 11␲ (2 2 , 5␲ ) 3 6 12 13␲ 12 7␲ 6

5␲ 12 ␲ 3 ␲ 4 ␲ 6

(2, 0) (1

5␲ 4 4␲ 3 17␲ 12

␲ 12

23␲ 1 3␲ 3 , 2 ) 12

11␲ 7␲ 6 5␲ 4 19␲ 3 12

49. ellipse

␲ 12

(⫺3, ␲)

23␲ 12 11␲ (1.5, 5␲ 3 ) 7␲ 6 5␲ 4 4␲ 5␲ 4 19␲ 3 3 17␲ 12 12

7␲ 2␲ 12 3␲ 3 4 5␲ 6 11␲ (5, ) 12 13␲ 12 7␲ 6

5␲ 4 4␲ 3 17␲ 12

5␲ 12 ␲ 3 ␲ 4 ␲ 6

(1, ␲2 ) ( 59 , 0)

␲ 12

23␲ 12 11␲ 6 7␲ 4 5␲ 19␲ 3 12

3.2 3.2 3.2 ,r ,r , or 1 0.8 cos ␪ 1 0.8 cos ␪ 1 0.8 sin ␪ 3.2 4 7.5 r 53. r 55. r 1 0.8 sin ␪ 1 cos ␪ 1 1.5 sin ␪ r1␲/22 12 2 2 A and 57. a. r b. 2 cos ␪ 3 sin ␪ r102 3 B 3 59. Jupiter: e 0.0486, Saturn: e 0.0567 61. about 2756.0 million miles 63. Saturn: e 0.0567 1780.77 482.36 65. r 67. r 1 0.0486 cos ␪ 1 0.0457 cos ␪ 69. In millions of miles (approx): JS: 405.3, JU: 1298.4, JN: 2310.3, SU: 893.1, SN: 1905.0, UN: 1011.9 0.2864 0.7638 71. r 73. r 1 0.7862 cos ␪ 1 0.7862 cos ␪ 51. r


2/2/11

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Page SA-65

Student Answer Appendix 3 1 cos ␪ verified 81. a. verified b. verified c. Answers will vary. ␲ r 12 cos a␪ b or r 6 12 1cos ␪ sin ␪2 4 425X2 416Y2 400 0 87. (0, 0), (4, 0), (4, 4), (0, 4) x 29.0 91. 9.2 mph at heading 347.7°

75. $582.45; $445.94; $881.32; $97.92 79. 83. 85. 89.

77. y

SA-65

25. verified 27. a.

11

18

18

Exercises 10.8, pp. 1059–1063 1. parameter 3. direction 5. Answers will vary. 7. a. parabola with vertex at (2, 1) b. y x2 4x 3

11 y

3

3

1

9. a. parabola b. y x 2 1x 1

x

3

b. x-intercepts: t 0, x 10, y 0 and t ␲, x 6, y 0; y-intercepts: t 1.757, x 0, y 6.5 and t 4.527, x 0, y 6.5; minimum x-value is 8.1; maximum x-value is 10; minimum y-value is 9.5; maximum y-value is 9.5 29. a. 7

y

10 5

11. a. power function with p 2 25 b. y 2 , x 0 x

1

10

5 x

1

7 b. x-intercepts none, y-intercepts none; no minimum or maximum x-values; minimum y-value is 4 and maximum y-value is 4 31. a. 40

y (2, 6.25)

(2, 6.25) 6

(5, 1) 5

(5, 1) 2

2

5 x

6

60.6

13. a. ellipse y2 x2 1 b. 16 9

60.6

y (0, 3) (4, 0)

(4, 0)

40

x (0, 3)

15. a. Lissajous figure 1 x b. y 6 cos c sin1a b d 2 4

b. x-intercepts: t 0, x 2, y 0 and t 4.493, x 9.2, y 0; infinitely many others; y-intercepts: t 2.798, x 0, y 5.9 and t 6.121, x 0, y 12.4; infinitely many others; no minimum or maximum values for x or y 33. a. 5

y 6

4

4

x

6

17.

y

10

18␲

0

10

10 x

10

1 19. x t, y 3t 2; x t, y t 2; x tan t, y 3 tan t 2, 3 ␲ ␲ t 僆 a , b 2 2 21. x t, y 1t 32 2 1; x t 3, y t2 1; x tan t 3, ␲ ␲ y sec2t, t 僆 a , b 2 2 ␲ 23. x t, y tan2 1t 22 1, t ␲k 2, k 僆⺪; x t 2, 2 1 y sec2t, t ak b␲, k 僆⺪; x tan1t 2, y t2 1 2

5 b. no x-intercepts; y-intercept is t 0, x 0, y 2; no minimum or maximum x-values; minimum y-value is 2; maximum y-value is 4 35. a. 9

12

12

9


SA-66

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b. x-intercepts: t 0, x 4, y 0 and t ␲, x 4, y 0; ␲ 3␲ y-intercepts: t , x 0, y 8 and t , x 0, y 8; minimum 2 2 and maximum x-values are approx. 5.657; minimum and maximum y-values are 8 37. 9

61. x 2 cos t 3, y cos12t2 ; no, the range of cos(2t) is [1, 1] 63. Taller (The Eiffel Tower is about 1063 ft tall) 65. ␪ 55°


3. f

5. g

7. c 9. h

11. e

13. b

15. g


12

1. verified (segments are perpendicular and equal length) 2. x2 1y 12 2 34 3. yes 4. verified y y 5. 6. 7. 10 10

12

10

9 width 12 and height 16; including the endpoint t 2␲, the graph crosses itself two times from 0 to 2␲. 39. 8

10

10 x

8.

9.

y

5

10 x

10

2

2

10.

y

10

5 x

5

9

10

10

y x 1 25 9 6.2

8 width 10 and height 14; including the endpoint t 2␲, the graph crosses itself nine times from 0 to 2␲. 41. 12

12

9.4

9.4

6.2 Four possibilities: (3, 2.4), (3, 2.4), (3, 2.4), (3, 2.4)

12

11. a. 12.

12 width 20 and height 20; including the endpoint t 4␲, the graph crosses itself 23 times from 0 to 4␲. 43. The maximum value (as the graph swells to a peak) is at b 1x, y2 aa, b. The minimum value (as the graph dips to the valley) 2 b is at 1x, y2 aa, b. 2 3␲ 3␲ 45. a. The curve is approaching y 2 as t approaches , but cota b is 2 2 undefined, and the trig form seems to indicate a hole at 3␲ t , x 0, y 2. The algebraic form does not have this problem and 2 shows a maximum defined at t 0, x 0, y 2. b. As t S q, y1t2 S 0 c. The maximum value occurs at (0, 2k). 47. a. Yes b. Yes c. 0.82 ft 49. No, the kick is short. 51. The electron is moving left and downward. 6t 6 13t 21 53. 3 orbits; about 2.1 orbits 55. at, , b 17 17 17 17 57. Inconsistent, no solutions y 59. x 1.22475t 1 y 0.25t2 2t 5 x The parametric equations fit the data very well. 5

y2 x2 1 169 25

1x 22 2 25 5 4

1y 12 2 4

14.

20

1 13.

y 8 4

(2 21, 1)

2 1

321 1 2 3 4 5

y2 x2 1 144 400

y

(2 21, 1) 3 (3, 1)

b.

(2, 1)

8

15.

y

y

15

3 4

12

15

20 x

9

3 3

3

9

15 x

9 15

20

center: (2, 1) vertices: (5, 1), (1, 1)

center: (2, 3) vertices: (2, 1), (2, 7)

17.

y

15 9

12 (0, 1)

y (6, 1)

3

4 20 12 4 4

x

9

12

20

8

center: (0, 0) vertices: 0, 5)

4

16.

4 8

1 2 3 4 5 6 7 x

12

20 12 4 4

4 4

(7, 1)

4

12

20 x

12 20

center: (0, 1) vertices: (0, 2), (0, 4)

25 15 5 3

5

15

y

10 x

10

5

9

10

10 x

10

10

25

9 15

center: (6, 1) vertices: (2, 1), (10, 1)

35 x


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Page SA-67

SA-67


18.

35. Y2 2Y 2X 6 0 7 1Y 12 2 2aX b 2

y2 x2 1 9 16 6.2

y

36. 5X2 Y2 80 0 X2 Y2 1 16 80 y X

X

Y

Y

ᏸ 45

60

x

9.4

1

9.4

Y

vertex: 7 , 1 2 foci: (3, 1) Y-intercepts: (0, √7 1) and (0, √7 1)

37. ellipse, e 23 ; 6.2 Answers may vary. One possibility: 17.8, 9.62 . 2

y x2 1 225 64

19. a. 20.

2

1x 52 2 10

1y 22 2

9

b.

4

y

16

x2 1 9

1 21.

10

y

2

y 3x

4 3

(4, 0)

(5, 2)

10

(2, 2)

10 x

(8, 2)

6

(0, 2)

14 x 2

y 3 x

2 4 6 8 10

10

4 2 8

23.

10

y y5

(0, 2)

(6, 0)

(0, 0) 10

10

10 x

(6.25, 0.5)

(0, 3)

24.

10

5 4 3 2 1

10864 2 4 6 8 10

5432 1 2 3 4 5

2 4 6 8 10

(0, 5) (10, 5)

6

4 2

2 4 6 8 10 12 14 16

y

10

1 2 3 4 5

y

y 4

8 2 x

2

6 4

4

2

2 2

2

4

2 2

4

6

8

44. Answers will vary. 45. x 僆 34, 4 4: y 僆 3 8, 8 4

4

10 x

2

y 8

y

10

15

10 x

(10, 5)

10

39. parabola, e 1;

de with e 0.0935 and d 1501.1; focal cord: 1 e cos ␪ 280.82 million miles y2 x2 41. y 21x 42 2 3 42. y 11 1x2 2 43. 1 9 16

16 3

y

10

38. hyperbola, e 32 ; 10 8 6 4 2

10

22.

asymptotes: Y √5X

40. r

(0, 2)

y

foci: (4√6, 0)

10 8 6 4 2 108642 2 4 6 8 10

x

vertices: (4, 0)

4

(0, 2)

(4, 2)

y 2

(4, 0)

4

2

2

4

x

4

(8, 2)

8

15 x

10

25. circle, line, (4, 3), (3, 4) 26. parabola, line, (3, 2) 27. parabola, circle, 1 13, 22, 1 13, 22, 1i 12, 32, 1i 12, 32 i 110 10 i 110 10 , b, a , b 28. circle, parabola, (1, 3), (1, 3), a 3 3 3 3 29. Parabola, circle 30. Circle, parabola y

1. c 2. d 3. b 4. a 5. circle; center 12, 52; radius 3

6 5 4 3 2 1 654321 1 2 3 4 5 6

1 2 3 4 5 6 x

31.

3 (1, 5)

8 6 42 2 4 6 8

1 2 3 4 5

1 2 3 4 5 6 x

2 4 6 8

34. 6

6. ellipse; center 12, 12; vertices 12, 42, 12, 62; foci 12, 1 1212, 12, 1 1212

2

6

2

4

6

8 6 42 2 4 6 8

2 4 6 8

(2, 6)

y (2, 5.6)

(4, 1)

(2, 1)

(0, 1) x

(2, 3.6)

7. ellipse; center 1 40 9 , 02; vertices 80 1 10 9 , 02 , (10, 0); foci (0, 0), 1 9 , 02

8. parabola; vertex 11.2, 02; focus (0, 0); directrix at y 2.4

8 6 4 2

4

4

(5, 5)

(2, 5) (2, 8)

8 6 4 2

33.

6 4 2 2

x

(2, 2)

32. 5 4 3 2 1 5432 1 2 3 4 5

y

y

6 5 4 3 2 1 654321 1 2 3 4 5 6

Practice Test, p. 1069

(2, 4) 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10

x


SA-68

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9. hyperbola; center: 12, 32; vertices: (2, 0), 12, 62; foci: 12, 82, (2, 2); asymptotes: 3 y 34 x 92 , y 3 4 x 2

6 4 2

1x 32 2

y

8642 2 4 6 8 1012 x 2 4 6 8 10 12 14

10. hyperbola; center 11, 22; vertices 14, 22, 16, 22; foci 11 129, 22 14.39, 22 11 129, 22 16.39, 22: asymptotes: y 25 x 85 , y 25 x 12 5

1y 12 2 2 1 92 2

(3, 1)

10

9 5 1; a 4 1 3 3, b 2

(6, 1)

y

(0, 1)

10

1x 22

y 4 3 2 1 12 8 10 642 1 2 4 6 8 1012 x 2 3 4 5 6 7 8

11. parabola; ␤ 36.87°; cos ␤ 45 , sin ␤ 35 3 11 2 12. Y 25 16 X 4 X 20

2 1 4 15 3 2

2.

3.

2

2 1 22 5 2

a

22 5 ,

1y 32

2

2 1 23 2

1

4.

1x 12 27 1 5 14 2

1y 22

2

2

10 x

23 1 5 12 2

10

2

2

1

27 23 a 5 14 , b 5 12

b 23

Cumulative Review Chapters 1–10, p. 1072 1. 1y 32 2 81x 22

y Y X

5

x

3. x 6 5. x 4 7.

5␲ k␲, k 僆⺪ 6

9. x 61.98° 360°k; k 僆⺪ x 118.02° 360°k; k 僆⺪ 11. about 24.7 pesos/kg 13. The formation is 1152.4 yd wide. y 15. 8

13.

14. 4 2 6 4 2

6

15. 12 8 4

6

2

2

4

12 8 4

6

6

2 (3, 1) 2 2

4

6

4

y x 1 16 25

y 5 4 3 2 1

10 9 8 7 6 5 4 3 2 1

1x 22

1y 12

9 25 foci 12, 52, 12, 32 25.

1x 32 2 2 1 72 2

1y 12 2 2 1 65 2

(2, 0) 2

4

6 x

2

y

10

2

2␲

4␲

6␲

x

1; D: x 僆 35, 1 4 ; R: y 僆 3 4, 6 4 ;

(3, 2.2) (6.5, 1) 10

y

(0.5, 1) 10 x

(3, 0.2) 10

8 x

y 2 16 8

8

16

x

2 4

(1, 2)

6

23. 5 4 3 2 1

e is very close to 1. This makes its orbit a very

10

6

21. center 11, 22 ; foci 11 2 110, 22 17.32, 22, 11 2 110, 22 15.32, 22 ; asymptotes y 13 x 73 , y 13 x 53 vertices (5, 2), (7, 2)

b. (2, 2), 12, 22 , 12, 22 , 12, 22

1; a 72 , b 65

4

4

Strengthening Core Skills, p. 1071 1.

2 6 4 2

2

y

2 4

2

y x3

4

2

1 0.967 cos ␪ elongated ellipse, where the orbit of most planets is nearly circular. 21. The ball is 0.43 ft above the ground at x 165 ft, and will likely go into the goal. 22. Perihelion: 128.41 million miles, Aphelion: 154.27 million miles 23. y 1x 12 2 4; D: x 僆⺢; R: y 僆 3 4, q2 ; focus: (1, 3.75) 24. 1x 12 2 1y 12 2 25; D: x 僆 3 4, 6 4 ; R: y 僆 3 4, 64 2

6

(0, 0) 2

2

20. r

x 1

y

4

12, 02

x 3

6

19.

8 6

165411 0.9672 2

10 x

4

18. max: y 8; min: y 0; P 2␲

19. a.

8

1 2 3 4 5 x

21 1 2 3 4 5 6 7 8 9 10 x 2

513 2,

6

4

2 3 4 5

17. parabola; x 1y 52 2 1

4

2

17. horizontal asymptote: y 0 vertical asymptotes: x 3, x 3, x-intercept: (2, 0); y-intercept: 10, 29 2

6

54321 1

1313 ,

(7, 3)

2

2

2

16. ellipse;

4

4

6 4 2 2

4 8 12

4 8 12

4

6

54321 1 2 3 4 5 2 3 4 5

25. 61.9° 27. 13, 42 , (3, 4), 13, 42, 13, 42

29.

3 2x 1 1 2 2 x x x 1

Connections to Calculus Exercises, pp. 1075–1076 cos 12␪2 cos ␪

2␲ 4␲ ␲ 5␲ , , d. 0, 3 3 3 3 ␲ 5␲ 9␲ 13␲ , , 3. a. b. 1, 1 c. , sin12␪2 cos12␪2 8 8 8 8 ␲ 3␲ 5␲ 7␲ 3␲ 7␲ 11␲ 15␲ , , , , , d. 5. , ; verified 8 8 8 8 4 4 4 4 ␲ ␲ 5␲ 3␲ , 7. four-leaf rose, circle; , , ; verified 6 2 6 2 3␲ ␲ 5␲ ␲ b, a0, b; (x, y): (0, 0), 9. (r, ␪): a0, b, a2 13, b, a2 13, 2 6 6 2 13, 132 , 13, 132 1. a.

sin 12␪2 sin ␪ cos12␪2 sin12␪2

b. 1, 1

c.


2/3/11

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CHAPTER 11

53. a. appears nonlinear b. no common difference ⫺472 55. d ⫽ 3, a1 ⫽ 1 57. d ⫽ 0.375, a1 ⫽ 0.65 59. d ⫽ 115 126 , a1 ⫽ 63 61. 1275

Exercises 11.1, pp. 1086–1088 1. pattern, order 3. recursive 5. formula defining the sequence uses the preceding term(s); answers will vary. 7. 1, 3, 5, 7; a8 ⫽ 15; a12 ⫽ 23 9. 0, 9, 24, 45; a8 ⫽ 189; a12 ⫽ 429 11. ⫺1, 2, ⫺3, 4; a8 ⫽ 8; a12 ⫽ 12 1 2 3 4 8 12 1 1 1 1 1 1 13. , , , ; a8 ⫽ ; a12 ⫽ 15. , , , ; a8 ⫽ ; a12 ⫽ 2 3 4 5 9 13 2 4 8 16 256 4096 1 1 1 1 1 17. 1, , , ; a8 ⫽ ; a12 ⫽ 2 3 4 8 12 ⫺1 1 ⫺1 1 1 1 19. , , , ; a8 ⫽ ; a12 ⫽ 2 6 12 20 72 156 21. ⫺2, 4, ⫺8, 16; a8 ⫽ 256; a12 ⫽ 4096 1 ⫺1 1 23. a. 79 b. {⫺1, 2, 7, 14, 23} 25. a. 51 b. {1, ⫺1 2 , 3, 4 , 5} 1 625 7776 27. a. 32 b. {2, 1, 21 , 14 , 18 } 29. a. approx. 2.6 b. {2, 49 , 64 27 , 256 , 3125 } 1 1 1 1 1 31. a. 36 b. {13 , 10 , 21 , 36 , 55 } 33. 2, 7, 32, 157, 782 35. ⫺1, 4, 19, 364, 132,499 37. 64, 32, 16, 8, 4 39. 336 41. 36 1 1 1 43. 28 45. 21 , 13 , 14 , 15 47. 31 , 120 49. 1, 2, 92 , 32 , 15,120 , 3,991,680 3 137 51. 15 53. 64 55. 60 57. ⫺2 ⫹ 1 ⫹ 4 ⫹ 7 ⫽ 10 59. ⫺1 ⫹ 5 ⫹ 15 ⫹ 29 ⫹ 47 ⫽ 95 61. ⫺1 ⫹ 2 ⫹ 1⫺32 ⫹ 4 ⫹ 1⫺52 ⫹ 6 ⫹ 1⫺72 ⫽ ⫺4 63. 0.5 ⫹ 2 ⫹ 4.5 ⫹ 8 ⫽ 15 65. 6 ⫹ 8 ⫹ 10 ⫹ 12 ⫹ 14 ⫽ 50 1 1 1 1 1 1 27 67. ⫺ ⫹ ⫹ a⫺ b ⫹ ⫹ a⫺ b ⫹ ⫽⫺ 3 8 15 24 35 48 112 5 5 q q 1 69. a. 4n b. 5n 71. a. 12k ⫺ 12 b. 2 n⫽1 n⫽1 k⫽1 k⫽1 k 5 3 7 n2 n 73. 77. 79. 35 81. 100 1n ⫹ 32 75. n n⫽1 n⫽1 3 n⫽3 2 83. 35, verified 85. $7.25, $7.75, $8.25, $8.75, $9.25; $17,760 87. an ⫽ 600010.82 n⫺1; 6000, 4800, 3840, 3072, 2457.60, 1966.08 89. ⬇2690 91. approaches 1

兺

兺

兺

兺

兺

兺 ca ⫽ ca j

j⫽1

1

⫽ c1a1 ⫹ a2 ⫹ a3 ⫹ ⫽c

兺a

p

p

⫹ can⫺1 ⫹ can

⫹ an⫺1 ⫹ an 2

j

j⫽1

95.

1 or 0.25 4

97.

3␲ 7␲ , 4 4

65. ⫺534

兺

⫹ ca2 ⫹ ca3 ⫹

n

63. 601.25

兺

n

93.

SA-69

99. ⬔A ⬇ 53.1°, ⬔B ⬇ 36.9°, ⬔C ⫽ 90°

67. 82.5 69. 74.04 71. 21022 73. S6 ⫽ 21; S75 ⫽ 2850 75. at 11 P.M. 77. 5.5 in.; 54.25 in. 79. a7 ⫽ 220; a12 ⫽ 2520; yes 81. a. linear function b. quadratic 1 13 1 83. A ⫽ 7, P ⫽ 6, HS: unit right, VS: 10 units up, ⱕ t 6 . 2 2 2 85. f 1x2 ⫽ 49x ⫹ 972, 1364 deer

Exercises 11.3, pp. 1107–1111 Exercises 11.2, pp. 1095–1098 n1a1 ⫹ an 2

1. common, difference 3. , nth 5. Answers will vary. 2 7. arithmetic; d ⫽ 3 9. arithmetic; d ⫽ 2.5 11. not arithmetic; 1 all prime 13. arithmetic; d ⫽ 24 15. not arithmetic; an ⫽ n2 ⫺␲ 17. arithmetic; d ⫽ 6 19. 2, 5, 8, 11 21. 7, 5, 3, 1 23. 0.3, 0.33, 0.36, 0.39 25. 32 , 2, 52 , 3 27. 34 , 58 , 12 , 38 29. ⫺2, ⫺5, ⫺8, ⫺11 31. a1 ⫽ 2, d ⫽ 5, an ⫽ 5n ⫺ 3, a6 ⫽ 27, a10 ⫽ 47, a12 ⫽ 57 33. a1 ⫽ 5.10, d ⫽ 0.15, an ⫽ 0.15n ⫹ 4.95, a6 ⫽ 5.85, a10 ⫽ 6.45, a12 ⫽ 6.75 35. a1 ⫽ 32 , d ⫽ 34 , an ⫽ 34 n ⫹ 34 , 33 39 a6 ⫽ 21 4 , a10 ⫽ 4 , a12 ⫽ 4 37. 61 39. 1 41. 2.425 43. 9 45. 43 47. 21 49. 26 51. a. appears linear b. d ⫽ 0.75 c. an ⫽ 0.75n ⫹ 0.75 7

7

0

⫺2

1. multiplying 3. a1rn⫺1 5. Answers will vary. 7. r ⫽ 2 9. r ⫽ ⫺2 11. not geometric; an ⫽ n2 ⫹ 1 13. r ⫽ 0.1 15. not geometric; ratio of terms decreases by 1 17. r ⫽ 25 240 19. r ⫽ 12 21. r ⫽ 4x 23. not geometric; an ⫽ n! 3 25. 5, 10, 20, 40 27. ⫺6, 3, ⫺3 29. 4, 413, 12, 1213 2 ,4 31. 0.1, 0.01, 0.001, 0.0001 1 n⫺1 3 1 25 33. an ⫽ ⫺24a b ; a7 ⫽ ⫺ 35. an ⫽ ⫺ 1⫺52 n⫺1; a4 ⫽ 2 8 20 4 37. an ⫽ 21 222 n⫺1 ⫽ 1 222 n⫹1; a7 ⫽ 16 1 1 39. a1 ⫽ 27 , r ⫽ ⫺3, an ⫽ 27 1⫺32 n⫺1, a6 ⫽ ⫺9, a10 ⫽ ⫺729, a12 ⫽ ⫺6561 1 1 41. a1 ⫽ 729, r ⫽ 13 , an ⫽ 7291 13 2 n⫺1, a6 ⫽ 3, a10 ⫽ 27 , a12 ⫽ 243

43. a1 ⫽ 12 , r ⫽ 12, an ⫽ 12 1 122 n⫺1, a6 ⫽ 212, a10 ⫽ 8 12, a12 ⫽ 1612 45. a1 ⫽ 0.2, r ⫽ 0.4, an ⫽ 0.210.42 n⫺1 a6 ⫽ 0.002048, a10 ⫽ 0.0000524288, a12 ⫽ 0.000008388608 47. 5 49. 11 51. 9 53. 8 55. 13 57. 9


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59. a. appears exponential b. r 0.2 c. an 131.2510.22 n1 200

8

0

61. 63. 69. 77. 81.

50 a. appears nonexponential b. no common ratio 32 r 23 , a1 729 65. r 32 , a1 243 67. r 32 , a1 256 81 3872 2059 10,920 71. 27 143.41 73. 8 257.375 75. 728 85 79. 1.60 8 10.625 1364

131. a. For an arithmetic sequence, the difference d ak ak1 must be constant. For ak a1rk1 and ak1 a1rk2, we have d log1a1rk1 2 log1a1rk2 2 log1a1 2 log1rk1 2 3log1a1 2 log1rk2 2 4 log1a1 2 1k 12log1r2 log1a1 2 1k 22log1r2 log 1r2 3 1k 12 1k 22 4 log1r2 ✓ ak b. For a geometric sequence, the ratio r must be constant. ak1 For ak a1 1k 12d and ak1 a1 1k 22d we have r

10a1 1k12d

10a1 1k22d 10a1101k12d 10a1101k22d 101k12d

101k22d 101k12d 1k22d 10d 31k12 1k224 10d✓ 133.

7 3 i j 158 158

135. 942 ft per min, 10.7 mph

Exercises 11.4, pp. 1117–1118

83.

31,525 2187

85.

387 512

87. 521 89. 3367 91. 14 15 1 2 93. 27 95. no 97. 125 25 1296 7 3 1 3 99. 12 101. 4 103. 3 105. 2 107. No finite sum exists. n1 109. 1296 111. an 2410.82 , a7 6.3 ft, Sq 120 ft 113. a0 46,000; an 36,80010.82 n1, a4 $18,841.60, 10 yr 115. a0 160; an 155.210.972 n1, a8 125.4 gpm, 10 mo 117. a0 277; an 283.3711.0232 n1, a10 347.7 million 3 n1 119. a0 462; an 277.2 a b , a5 35.9 in3, 7 strokes 5 121. a0 8, an 610.752 n1, 2 days, 8 days 123. a0 50; an 100122 n1, a10 51,200 bacterium, 12 half-hours (6 hr) 4 n1 125. a0 2 m; an 1.6 a b , a7 0.42 m, 5 total distance a0 2Sq 18 m 127. about 67,109 in. This is almost 1.06 mi. 129. 40,000 1750x vs. 40,000(1.04)x; 6 yr

1. finite, universally 3. induction, hypothesis 5. Answers will vary. 7. an 10n 6 a4 10142 6 40 6 34; a5 10152 6 50 6 44; ak 10k 6; ak1 101k 12 6 10k 10 6 10k 4 9. an n a4 4; a 5 5; ak k; ak1 k 1 11. an 2n1 a4 241 23 8; a5 251 24 16; ak 2k1; ak1 2k11 2k 13. Sn n15n 12 S4 415142 12 4120 12 41192 76; S5 515152 12 5125 12 51242 120; Sk k15k 12; Sk1 1k 12151k 12 12 1k 1215k 5 12 1k 1215k 42 n1n 12 15. Sn 2 414 12 4152 S4 10; 2 2 515 12 5162 S5 15; 2 2 k1k 12 Sk ; 2 1k 121k 1 12 1k 121k 22 Sk1 2 2 17. Sn 2n 1 S4 24 1 16 1 15; S5 25 1 32 1 31; Sk 2k 1; Sk1 2k1 1 19. an 10n 6; Sn n15n 12 S4 415142 12 4120 12 41192 76; a5 10152 6 50 6 44; S5 515152 12 5125 12 51242 120; S4 a5 S5


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21.

23.

25.

27.

29.

76 44 120 120 120 Verified n1n 12 an n; Sn 2 4152 414 12 10; S4 2 2 a5 5; 5162 515 12 15; S5 2 2 S4 a5 S5 10 5 15 15 15 Verified an 2n1; Sn 2n 1 S4 24 1 16 1 15; a5 251 24 16; S5 25 1 32 1 31; S4 a5 S5 15 16 31 31 31 Verified a. an n3; Sn 11 2 3 4 p n2 2 S1 12 13 S5 11 2 3 4 52 2 152 225 1 8 27 64 125 225 S9 11 2 p 92 2 452 2025 1 8 p 729 2025 n1n 12 2 n2 1n 12 2 b. c d 2 4 1. Show Sn is true for n 1. S1 111 12 1122 2 Verified 2. Assume Sk is true: 2 4 6 8 10 p 2k k1k 12 and use it to show the truth of Sk1 follows. That is: 2 4 6 p 2k 21k 12 1k 121k 1 12 Sk ak1 Sk1 Working with the left hand side: 2 4 6 p 2k 21k 12 k1k 12 21k 12 1k 121k 22 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 1. Show Sn is true for n 1. 511211 12 5122 S1 5 2 2 Verified 2. Assume Sk is true: 5k1k 12 5 10 15 p 5k 2 and use it to show the truth of Sk1 follows. That is: 51k 121k 1 12 5 10 15 p 5k 51k 12 2 Sk ak1 Sk1 Working with the left hand side: 5 10 15 p 5k 51k 12 5k1k 12 51k 12 2 5k1k 12 101k 12 2

SA-71

1k 1215k 102 2 51k 121k 22

2 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 31. 1. Show Sn is true for n 1. S1 112112 32 5 Verified 2. Assume Sk is true: 5 9 13 17 p 14k 12 k12k 32 and use it to show the truth of Sk1 follows. That is: 5 9 13 17 p 14k 12 341k 12 1 4 1k 12121k 12 32 Sk ak1 Sk1 Working with the left hand side: 5 9 13 17 p 14k 12 341k 12 14 k12k 32 4k 5 2k2 3k 4k 5 2k2 7k 5 1k 1212k 52 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 33. 1. Show Sn is true for n 1. S1

3131 12 2

313 12 2

3122 2

3

Verified 2. Assume Sk is true: 3 9 27 p 3k

313k 12

2 and use it to show the truth of Sk1 follows. That is:

3 9 27 p 3k 3k1

313k1 12

2 Sk ak1 Sk1 Working with the left hand side: 3 9 27 p 3k 3k1 313k 12 3k1 2 313k 12 213k1 2 2 3k1 3 213k1 2 2 313k1 2 3 2 313k1 12 2 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 35. 1. Show Sn is true for n 1. S1 211 2 22 2 4 2 2 Verified 2. Assume Sk is true: 2 4 8 p 2k 2k1 2 and use it to show the truth of Sk1 follows. That is: 2 4 8 p 2k 2k1 2k2 2 Sk ak1 Sk1 Working with the left hand side: 2 4 8 p 2k 2k1 2k1 2 2k1 212k1 2 2 2k2 2 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n.


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37. 1. Show Sn is true for n 1. 1 1 1 S1 2112 1 21 3 Verified 2. Assume Sk is true: 1 1 1 1 k p 3 15 35 12k 1212k 12 2k 1 and use it to show the truth of Sk1 follows. That is: 1 1 1 1 p 3 15 35 12k 1212k 12

1 k1 121k 12 12121k 12 12 21k 12 1 Sk ak1 Sk1 Working with the left hand side: 1 1 1 1 1 p 3 15 35 12k 1212k 12 12k 12 12k 32 k 1 2k 1 12k 1212k 32 k12k 32 1 12k 1212k 32 2k2 3k 1 12k 1212k 32 12k 121k 12 12k 1212k 32 k1 2k 3 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 39. 1. Show Pn is true for n 1. P1: 31 2112 1 3 21 3 3 Verified 2. Assume Pk: 3k 2k 1 is true and use it to show the truth of Pk1 follows. That is: 3k1 21k 12 1. Working with the left hand side: 3k1 313k 2 312k 12 6k 3 Since k is a positive integer, 6k 3 2k 3 Showing Pk1: 3k1 2k 3 Since the truth of Pk1 follows from Pk, the statement is true for all n. 41. 1. Show Pn is true for n 1. P1: 3 # 411 41 1 3 # 40 4 1 3#13 33 Verified 2. Assume Pk: 3 # 4k1 4k 1 is true and use it to show the truth of Pk1 follows. That is: 3 # 4k11 4k1 1. Working with the left hand side: 3 # 4k 3 # 414k1 2 4 # 314k1 2 414k 12 4k1 4 Since k is a positive integer, 4k1 4 4k1 1 Showing Pk1: 3 # 4k 4k1 1 Since the truth of Pk1 follows from Pk, the statement is true for all n. 43. n2 7n is divisible by 2 1. Show Pn is true for n 1. Pn: n2 7n 2m P1:

112 2 7112 2m 1 7 2m 6 2m Verified 2. Assume Pk: k2 7k 2m for m 僆⺪ and use it to show the truth of Pk1 follows. That is: 1k 12 2 71k 12 2p for p 僆⺪. Working with the left hand side: 1k 12 2 71k 12 k2 2k 1 7k 7 k2 7k 2k 6 2m 2k 6 21m k 32 is divisible by 2. Since the truth of Pk1 follows from Pk, the statement is true for all n. 45. n3 3n2 2n is divisible by 3 1. Show Pn is true for n 1. Pn: n3 3n2 2n 3m P1: 112 3 3112 2 2112 3m 1 3 2 3m 6 3m 2m Verified 2. Assume Pk: k3 3k2 2k 3m for m 僆⺪ and use it to show the truth of Pk1 follows. That is: 1k 12 3 31k 12 2 21k 12 3p for p 僆⺪. Working with the left hand side: 1k 12 3 31k 12 2 21k 12 k3 3k2 3k 1 31k2 2k 12 2k 2 k3 3k2 2k 31k2 2k 12 3k 3 k3 3k2 2k 31k2 2k 12 31k 12 3m 31k2 2k 12 31k 12 is divisible by 3. Since the truth of Pk1 follows from Pk, the statement is true for all n. 47. 6n 1 is divisible by 5 1. Show Pn is true for n 1. Pn: 6n 1 5m P1: 61 1 5m 6 1 5m 5 5m 1m Verified 2. Assume Pk: 6k 1 5m or 6k 5m 1 for m 僆⺪ and use it to show the truth of Pk1 follows. That is: 6k1 1 5p for p 僆⺪. Working with the left hand side: 6k 1 616k 2 1 615m 12 1 30m 6 1 30m 5 516m 12 is divisible by 5. Since the truth of Pk1 follows from Pk, the statement is true for all n. 49. verified 51. verified 53. 1x 42 2 1y 32 2 25

Mid-Chapter Check, p. 1119 1. 3, 10, 17, a9 59

2. 4, 7, 12, a9 84

3. 1, 3, 5, a9 17

6

4. 360

5.

13k 22

6. d

7. e

8. a 9. b

10. c

k1

11. a. a1 2, d 3, an 3n 1 b. a1 d an 34 n 34 12. n 25, S25 950 13. n 16, S16 128 14. S10 5 3 2,

3 4,

14,762 16. a. a1 2, r 3, an 2132 n1 27 b. a1 12 , r 12 , an 1 12 2 n 17. n 8, S8 1640 18. 343 27 6 19. 1785 20. 4.5 ft; 127.9 ft 15. S10

Reinforcing Basic Concepts, pp. 1119–1120 Exercise 1: $71,500


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Exercises 11.5, pp. 1127–1132

51.

108642 2 4 6 8 10

Begin

W

W

X

X

Y

Z

W

X

Y

Y

Z

W

X

Z

Y

Z

W

X

8 6 4 2 2 4 6 8 10 x

Exercises 11.6, pp. 1139–1145 1. n1E2 3. 0, 1, 1, 0 5. Answers will vary. 7. S 5HH, HT, TH, TT6, 14 9. S {coach of Patriots, 1 Cougars, Angels, Sharks, Eagles, Stars}, 16 11. P1E2 49 13. a. 13 1 1 1 1 5 3 b. 4 c. 2 d. 26 15. P1E1 2 8 , P1E2 2 8 , P1E3 2 4 17. a. 43 b. 1 c. 14 d. 12 19. 34 21. 76 23. 0.991 1 11 60 25. a. 12 b. 12 c. 89 d. 56 27. 10 29. 143 31. b, about 12% 21 7 33. a. 0.3651 b. 0.3651 c. 0.3969 35. 0.9 37. 24 39. 0.59 2 1 7 1 4 2 9 41. a. 6 b. 36 c. 9 d. 9 43. a. 25 b. 50 c. 0 d. 25 e. 1 5 1 2 8 3 1 45. 34 47. 11 49. a. b. c. d. e. f. 15 18 9 9 4 36 12 1 51. 14 ; 256 ; answers will vary. 53. a. 0.33 b. 0.67 c. 1 d. 0 1 9 e. 0.67 f. 0.08 55. a. 21 b. 12 c. 81 57. a. 16 b. 14 c. 16 1 5 3 3 1 9 2 11 d. 16 59. a. 26 b. 26 c. 13 d. 26 e. 13 f. 26 61. a. 8 5 1 3 47 2 3 9 11 b. 16 c. 16 63. a. 100 b. 25 c. 100 d. 50 e. 100 65. a. 429 8 1 1 b. 2145 67. 3360 69. 1,048,576 ; answers will vary; 20 heads in a row. 1 212 12 3 12 71. sin ␪ , cos ␪ , tan ␪ , sec ␪ , 3 3 4 4 cot ␪ 2 12 840 41 840 , cos 12␪2 , tan 12␪2 73. sin 12␪2 841 841 41

Exercises 11.7, pp. 1151–1153

f 132 1

y

2 4 6 8 10 x

53. g1x2 7 0: x 僆 12, 02 ´ 13, q 2 Y

b. WW, WX, WY, WZ, XW, XX, XY, XZ, YW, YX, YY, YZ, ZW, ZX, ZY, ZZ 9. 32 11. 15,625 13. 2,704,000 15. a. 59,049 b. 15,120 17. 360 if double veggies are not allowed, 432 if double veggies are allowed. 19. a. 120 b. 625 c. 12 21. 24 23. 4 25. 120 27. 6 29. 720 31. 3024 33. 40,320 35. 6; 3 37. 90 39. 336 41. a. 720 b. 120 c. 24 43. 360 45. 60 47. 60 49. 120 51. 30 53. 60, BANANA 55. 126 57. 56 59. 1 61. verified 63. verified 65. 495 67. 364 69. 252 71. 8! 40,320 73. 8nPr 3 336 75. 20 nCr 5 15,504 77. 8 nCr 4 70 79. a. 1.2% b. 0.83% 81. 7776 83. 324 85. 800 87. 6,272,000,000 89. 518,400 91. 357,696 93. 6720 95. 8 97. 10,080 99. 5040 101. 2880 103. 5005 105. 720 107. 52,650, no 109. 6 · (6 nCr 3 2) 2 · 20 148 5 12 13 13 5 111. cos ␪ , tan ␪ , csc ␪ , sec ␪ , cot ␪ 13 5 12 5 12 y 113. 10

108642 2 4 6 8 10

10 8 6 4 2

3. N t 5. Answers will vary.

1. experiment, well-defined 7. a. 16 possible

1. one 3. 1a 12b22 5 5. Answers will vary. 7. x5 5x4y 10x3y2 10x2y3 5xy4 y5 9. 16x4 96x3 216x2 216x 81 11. 41 38i 13. 35 15. 10 17. 1140 19. 9880 21. 1 23. 1 25. c5 5c4d 10c3d2 10c2d3 5cd4 d5 27. a6 6a5b 15a4b2 20a3b3 15a2b4 6ab5 b6 29. 16x4 96x3 216x2 216x 81 20 2 31. 11 2i 33. x9 18x8y 144x7y2 35. v24 6v22w 33 2v w 37. 35x4y3 39. 1792p2 41. 264x2y10 43. 0.25 45. a. 17.8% b. 23.0% 47. a. 0.89% b. 7.0% c. 99.0% d. 61.0% 49. 2n1, 2048

Z

10 8 6 4 (2, 0) 2 108642 2 4 6 8 10

y

(0, 0) (3, 0) 2 4 6 8 10 x

Making Connections, p. 1153 1. c

3. b

5. d

7. b

9. d

11. a 13. e 15. g

Summary and Concept Review, pp. 1154–1158 5 11 1. 1, 6, 11, 16; a10 46 2. 1, 35 , 25 , 17 ; a10 101 3. an n4; a6 1296 4. an 17 1n 12132; a6 2 5. 255 6. 112 7. 140 8. 35 9. 2, 6, 12, 20, 30 10. 12 , 34 , 54 , 94 , 17 256 4 7

11.

1i

2

3i 22; 210

12. a. about 134 hawks b. 8 yr

i1

an 2 31n 12; 119 14. an 3 1221n 12; 65 740 16. 1335 17. 630 18. 11.25 19. 875 20. 7.55 m 819 3645 22. 32 23. 2401 24. 10.75 25. 6560 26. 512 63,050 does not exist 28. 50 29. 4 30. 31. does not exist 32. 5 9 6561 2 n1 33. a0 121,500, a1 81,000, an 81,000 a b , a7 7111 ft3 3

13. 15. 21. 27.

34. a0 1225, a1 1311, an 131111.072 n1, a15 3380, S15 32,944 111 12 35. (1) Show Sn is true for n 1: S1 1✓ 2 k1k 12 (2) Assume Sk is true: 1 2 3 p k 2 Use it to show the truth of Sk1: 1 2 3 p k 1k 12

1k 121k 1 12

left hand side: 1 2 3 p k 1k 12

k1k 12

1k 12

2

k1k 12 21k 12

2 1k 121k 22 2

36. (1) Show Sn is true for n 1: S1

2 111 12 3 2112 1 4 6

(2) Assume Sk is true: 1 4 9 p k2

1✓

k1k 1212k 12

6 Use it to show the truth of Sk1: 1k 121k 1 12 321k 12 1 4 1 4 9 p k2 1k 12 2 6 left hand side: 1 4 9 p k2 1k 12 2

k1k 1212k 12

61k 12 2

1k 12 3 12k2 k 6k 64

6 6 6 1k 1212k2 7k 62 1k 121k 2212k 32 6 6 37. (1) Show Pn is true for n 1: P1: 41 3112 1✓ (2) Assume Pk is true: 4k 3k 1 Use it to show the truth of Pk1: 4k1 31k 12 1 3k 4 left hand side: 4k1 414k 2 413k 12 12k 4 Since k is a positive integer, 12k 4 3k 4 showing 4k1 3k 4


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Student Answer Appendix 2 # 31k12 1 2 # 31k12 1 2 # 31k12 # 3 13k 12 # 3 k k1 # 13 12 3 3 3 3k1 1✓ 11. a. Begin

38. (1) Show Pn is true for n 1: P1: 6 # 711 71 1✓ (2) Assume Pk is true: 6 # 7k1 7k 1 Use it to show the truth of Pk1: 6 # 7k 7k1 1 left hand side: 6 # 7k 7 # 6 # 7k1 7 # 7k 1 7k1 7 Since k is a positive integer, 7k1 7 7k1 1. 39. (1) Show Pn is true for n 1: P1: 31 1 2 or 2112 ✓ (2) Assume Pk is true: 3k 1 2p or 3k 2p 1 for p 僆⺪ Use it to show the truth of Pk1: 3 k1 1 2q for q 僆⺪ left hand side: 3k1 1 3 # 3k 1 3 # 12p 12 1 3 # 2p 3 1 3 # 2p 2 213p 12 2q is divisible by 2 40. six ways Begin

B

A

B

C

A

A

B

C

A

C

A

B

C

B

C

A

B

A

B

Strengthening Core Skills, pp. 1161–1162 Exercise 1.

C

B

C

A

B

A

Exercise 2. 41. 720; 1000 42. 24 43. 220 44. a. 5040 b. 840 c. 35 45. a. 720 b. 120 c. 24 46. 3360 47. a. 220 b. 1320 4 3 7 175 48. 13 49. 13 50. 65 51. 24 52. 396 53. a. 0.608 b. 0.392 c. 1 d. 0 e. 0.928 f. 0.178 54. a. 21 b. 56 55. a. x4 4x3y 6x2y2 4xy3 y4 b. 41 38i 56. a. a8 8 13a7 84a6 168 13a5 b. 78,125a7 218,750a6b 262,500a5b2 175,000a4b3 57. a. 280x4y3 b. 64,064a5b9 58. a. about 93.3% b. about 62.4%

Practice Test, pp. 1158–1160 1. c. 3. b. d. 5. 7. 9.

# 13C5 40

4C1

52C5

4 # 13C3 # 39C2 52C5

4

0.001 970

0.326 170

# 13C4 # 39C1

0.042 917 52C5 4 # 10C5 Exercise 4. 0.000 388 52C5 Exercise 3.

Cumulative Review Chapters 1–11, pp. 1162–1164 1. a. 23 cards are assembled each hour. c. y 23x 155 d. 6:45 A.M. 3.

a. 12 , 45 , 1, 87 ; a8 16 b. 6, 12, 20, 30; a8 90 11 311 3, 212, 17, 16; a8 12 2. a. 165 b. 420 c. 2343 d. 7 512 a. a1 7, d 3, an 10 3n a1 8, d 2, an 2n 10 c. a1 4, r 2, an 4122 n1 a1 10, r 25 , an 101 25 2 n1 4. a. 199 b. 9 c. 34 d. 6 a. 1712 b. 2183 c. 2188 d. 12 6. a. 8.82 ft b. 72.4 ft $6756.57 8. $22,185.27 ak 5k 3, ak1 51k 12 3, 51k 12 2 1k 12 5k2 k Sk , Sk1 ; 2 2 5112 2 1 For n 1: S1 2✓ 2

5k2 k Assume: Sk is true, 2 2 51k 12 2 1k 12 5k k Prove: 51k 12 3 2 2 101k 12 6 1k 12 351k 12 14 5k2 k 2 2 2 2 1k 1215k 42 5k 9k 4 2 2 15k 421k 12 15k 421k 12 ✓ 2 2 11 1 10. For n 1: 2 # 3 3 1 2112 2✓ Assume: 2 # 3k1 3k 1 Prove: 2 # 31k12 1 3k1 1

C

b. ABC, ACB, BAC, BCA, CAB, CBA 12. 302,400 13. 31 14. a. 720 b. 120 c. 20 15. 900,900 16. 302,400 17. a. x4 8x3y 24x2y2 32xy3 16y4 b. 4 18. a. x10 1012 x9 90x8 b. a8 16a7b3 112a6b6 5 5 7 19. 0.989 20. a. 14 b. 12 c. 31 d. 12 e. 12 f. 14 g. 12 h. 0 21. a. 0.08 b. 0.92 c. 1 d. 0 e. 0.95 f. 0.03 22. a. 0.1875 59 b. 0.589 c. 0.4015 d. 0.2945 e. 0.4110 f. 0.2055 23. a. 100 53 13 47 b. 100 c. 100 d. 100 24. a. 0.8075 b. 0.0075 c. 0.9925 25. a. about 27.9% b. about 97.6%

C

C

B

A

x

y

0

1

␲ 6

13 2

␲ 4

12 2

␲ 3

1 2

␲ 2

0

2␲ 3

5␲ 6 ␲ 5. x 7. d. g. k.

b. 184 cards

1 2

13 2

1 5 1109 ; x 0.91; x 2.57 6

a. x 0 b. x 僆 11, 02 c. x 僆 1q, 12 ´ 10, q2 x 僆 1q, 12 ´ 11, 12 e. x 僆 11, q2 f. y 3 at (1, 3) none h. x 2.3, 0.4, 2 i. g142 0.25 j. does not exist q l. 0 m. x 僆 1q, 12 ´ 11, q 2


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D: x 僆 3 ⫺3, 3 4 R: y 僆 5⫺26 ´ 1⫺1, 22 ´ 34, 9 4

y

10 8 6 4 2

⫺1 1x ⫹ h ⫺ 221x ⫺ 22

(0, 8) y⫽2

⫺10⫺8⫺6⫺4⫺2 ⫺3 (⫺2, 0) ⫺6 ⫺9 ⫺12 ⫺15

2 4 6 8 10 x

10

0

, (3.4044, 1.4539)

0 29. 0.2794

Connections to Calculus Exercises, p. 1167 4

4

i⫽1

i⫽1

LW ⫽ f 1i2112 ⫽ 1 f 1i2

1. a. y

4

7 6 5 4 3 2 1

1

Exercises 12.1, pp. 1177–1180

1 ⫹ ln 217 15. a. x3 ⫽ 125 b. e5 ⫽ 2x ⫺ 1 17. a. x ⫽ 3.19 2 b. x ⫽ 334 19. (5, 10, 15) 21. (⫺3, 3); (⫺7, 3), (1, 3); (⫺3 ⫺ 2 13, 3), 16 ⫹ 12 (⫺3 ⫹ 2 13, 3) 23. a. verified b. 25. 1333 4 27. 20

f 1i2 ⫽ 13 f 112 ⫹ f 122 ⫹ f 132 ⫹ f 142 4

i⫽1

⫽5⫹4⫹3⫹2 ⫽ 14 units2

1 2 3 4 5 6 7 x

8

1 1 1 8 1 f a ib a b ⫽ f a ib 2 2 2 i⫽1 2 i⫽1

1. infinity 3. left-hand; right-hand; greater 5. Answers will vary. 4 1 7. lim Vn ⫽ ␲r3 9. lim e f 1t2 ⫽ 0 11. lim cos a b ⫽ 1 nSq 3 tS⫺q xSq x 13. 500 sides 15. 425 sides 17. lim st ⫽ 5r tS5

x⫹3 1 ⫽⫺ 6 x2 ⫺ 9 23. As x approaches ␲, p(x) approaches ⫺2: lim p1x2 ⫽ ⫺2 xS␲ 25. As x approaches 2, v(x) approaches 1 1 : lim v1x2 ⫽ 4 xS2 4 27. As x approaches 0, s(x) approaches 0: lim s1x2 ⫽ 0 xS0 2x2 ⫺ 7x ⫹ 6 x⫽2 29. R1x2 ⫽ • sin1x ⫺ 22 1 x⫽2 1 1 31. As x approaches 2, f (x) approaches : lim f 1x2 ⫽ 2 xS2 2 33. As x approaches 1, g(x) approaches 4: lim g1x2 ⫽ 4 ⫺1

19. lim tan xSa

␲ 3g1x2 4 ⫽ 3

21. lim

xS⫺3

xS1

35. As x approaches 1, f (x) approaches 0: lim f 1x2 ⫽ 0 xS1 37. 27 39. 24 41. 1 43. lim⫺ Ix ⫽ 3 cos2 1R1 ⫹ R2 2 45. lim⫺ f ⫽ L 47. a. 1 b. ⫺1 xS3

xSm

11 11 49. a. ⫺2 b. 2 51. a. ⫺ b. ⫺ 2 2 dne B 57. a. 1 55. a. 43 b. 13 c. A LH⫽RH

53. a. 0 b. ⫺1

b. 0 c.

dne B A LH⫽RH

12 12 12 b. c. 61. A dne 63. ␲ 65. A dne q B ⫺q B 2 2 2 2 dne 67. A B 69. 0 71. 1x ⫺ 52 1x ⫺ 221x ⫹ 1213x ⫺ 12 73. 19.90 in 59. a.

L

y 7 6 5 4 3 2 1

CHAPTER 12

(2, 0)

x ⫽ ⫺1 x ⫽ 1

32 1 1 c⫺ i⫹ 6d 8 8 i⫽1 i⫽1 2 1 1 32 ⫹ 32 ⫽ c⫺ a b ⫹ 61322d 8 8 2 1 1056 ⫽⫺ a b ⫹24 64 2 ⫽ ⫺8.25 ⫹ 24 ⫽ 15.75 For n ⫽ 32 the approximate area under the graph is 15.75 units2, even closer to the known area of 16 units2.

y

15 12 9 6 3

LW ⫽

⫽

1 2 3 4 5 x

11. a. 4x ⫹ 2h ⫺ 3 b.

b.

32 1 32 1 1 32 1 a⫺ i ⫹ 6b ⫽ c a⫺ ib ⫹ 6d 8 i⫽1 8 8 i⫽1 8 i⫽1 32

⫺5⫺4⫺3⫺2⫺1 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

13.

b.

1 8 1 1 3 5 1 f a ib ⫽ c f a b ⫹ f 112 ⫹ f a b ⫹ f 122 ⫹ f a b 2 i⫽1 2 2 2 2 2

7 ⫹ f 132 ⫹ f a b ⫹ f 142 d 2 9 1 11 7 5 ⫽ c ⫹ 5 ⫹ ⫹ 4 ⫹ ⫹ 3 ⫹ ⫹ 2d 2 2 2 2 2 1 ⫽ 3 30 4 ⫽ 15 units2 2 2 True area is 16 units , more rectangles S better estimate.

1 2 3 4 5 6 7 x

3. a. Since the interval [0, 4] is 4 units wide and we’re using 32 subintervals of equal length, the width of each interval (the width of each rectangle) 4 ⫽ 18 . The length of each rectangle is determined by a point of the will be 32 graph of f 1x2 ⫽ ⫺x ⫹ 6, so the length of the first rectangle is f 1 18 2, the second length is f 1 28 2, the third is f 1 38 2, and so on up to the 32nd rectangle. Since A ⫽ LW, we multiply each length f 1 18 i2 by width 18 and sum the areas of all such rectangles. Using i for a counter, this can be written as 32 32 1 1 1 A⫽ LW ⫽ f a iba b. Since all lengths are multiplied by 8 8 i⫽1 i⫽1 8 (the counter i does not affect the constant 18), we can factor out this term 1 1 1 32 a⫺ i ⫹ 6b. and evaluate f 1x2 ⫽ ⫺x ⫹ 6 at x ⫽ i. The result is 8 8 i⫽1 8

Exercises 12.2, pp. 1188–1190

1. sum; limits 3. root; n th; f 1x2 7 0 5. Answers will vary. 7. 9 9. ⫺8 11. 2 13. 0 15. 7 17. 1 19. 9 21. 8 23. 0 25. 14 7 27. ⫺16 29. 31. 0 33. 64 35. ⫺36 37. ⫺69 39. 29 41. 9 2 17 3 1 43. 45. ⫺ 47. 216 49. 15 51. 23 53. ⫺1 55. ⫺ 5 4 4 dne dne dne 57. A dne 59. 9 61. 63. 65. 6 67. A ⫺q B A LH⫽RH B A LH⫽RH B q B 69. a. limit appears to be 0 b. limit is actually ⫺0.001 x

f (x)

x

f (x)

0.5

0.2510

⫺0.5

0.2510

0.4

0.1610

⫺0.4

0.1610

0.3

0.0910

⫺0.3

0.0910

0.2

0.0410

⫺0.2

0.0410

0.1

0.0110

⫺0.1

0.0110

0.01

0.0011

⫺0.01

0.0011

0.001

0.001001

⫺0.001

0.001001


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71. a.

Exercises 12.4, pp. 1212–1214

b.

yⴝxⴙ

x

1x ⴚ 2 10 x

x

yⴝxⴙ

1x ⴚ 2 10 x

2.7

2.7017

2.99

2.991

2.8

2.8014

2.999

3

2.9

2.9012

2.9999

3.0009

3.1

3.1008

3.0001

3.0011

3.2

3.2007

3.001

3.002

3.3006

3.3

c. 3 d. 3.001, verified

3.01 73. 31, 32 4

3.011

75. verified

Mid-Chapter Check, pp. 1190–1191 6x2 3

0 2. a. 1 b. 0 3. 49 per year (almost weekly) 12x3 x 1 2 6x 19x 7 7 x 2x 7 2 dne B 4. F1x2 μ 5. a. 1 b. 1 c. A LHRH 7 23 x 2 2 6. A dne q B 7. a. b. Answers may vary. 1.

lim

xSq

1. difference 3. rectangles 5. Answers will vary. 7. f ¿ 1t2 88.2 9.8t 9. a. 39.2 m/sec b. 0 m/sec c. 39.2 m/sec 11. 896.9 m 13. d¿ 1t2 9.8t 15. a. 9.8 m/sec b. 19.6 m/sec c. 29.4 m/sec 1 0.6 17. f ¿1x2 19. f ¿1x2 3x2 21. p ¿ 1t2 2 1t 23. a. 600 people/yr b. 300 people/yr c. 150 people/yr 1 2 25 25. f ¿1x2 27. h¿1x2 29. 2 1x 12 2 1x 52 2 31. 1

33. 9 units2 n

41.

35. 15 units2

37. 9

39. 15

2

4 1 4 c a ib 3 d n i1 2 n n 4 n 1 4 2 c a ib 3 d summation properties (distribute) n i1 2 n i1 n n 4 1 16 2 c i 3 d simplify n 2 i1 n2 i1

4 16 c n 2n2

n

n

i

2

i1

3d

factor

i1 2

16 n2

from first summation

4 8 2n3 3n n c a b 3n d apply summation formula n n2 6 4 32 2n3 3n2 n 3a b 12 distribute 6 n n

32 2n3 3n2 n a b 12 rewrite denominators 6 n3 16 3 1 a2 2 b 12 decompose rational expression 3 n n 16 3 1 lim a2 2 b lim 12, and Applying the limit properties gives 3 nSq n nSq n

x

␲ y ⴝ cos a b x

0.1

1

0.01

1

0.001

1

0.001

1

16 68 b 122 12 units2. The new employee 3 3 has produced 22 complete parts.

0.01

1

43. A

0.1

1

1 8. 2

9. 1 4 3

10.

the area under the curve is a n

LW

i1 n

A LHRH B

dne

xS0

2 2 110 89. a. x b. x 0, x 8 c. x 2 3 91. A 55°, C 90°, b 9.6 cm, c 16.7 cm

6

6

area formula, rectangle method

6 6 f a ib n i1 n

6 n 1 6 2 6 c a ib 4a ib d n i1 2 n n

n 6 1 n 6 2 6 c a ib 4 a ib d n 2 i1 n i1 n

Exercises 12.3, pp. 1201–1203 1. asymptotic; removable; jump 3. direct substitution 5. Answers will vary. 7. not continuous, condition 1 is violated 9. continuous 11. not continuous, condition 2 is violated 13. continuous 15. not continuous, condition 3 is violated 17. 36 19. Direct substitution not possible, 5 is not in the domain. 1 21. 23. Direct substitution not possible, 1 is not in the domain. 2 1 1 25. 155 27. 29. 4 31. 9 33. 35. 0 37. 4x 1 2 4 3 1 1 39. 41. 43. 31x 22 2 45. 47. 3 2 2 1x 2 1x 22 2 1 49. A dne 51. 0 53. 55. A dne 57. 0 59. x 1000 q B q B 2 3 2 13 61. 63. 0 65. 2 67. 69. 3 71. 3 8 7 73. not possible since f(2) not defined 75. 3 77. 0 79. 1 81. not possible, lim g1x2 does not exist A dne q B xS2 dne B 85. 3 87. 3 83. not possible, lim g1x2 does not exist A LHRH

f a n ib a n b

i1 n

6 n

n

factor

n

distribute summation

n

6 1 36 2 6 c id i 4 n 2 i1 n2 i1 n

6 evaluate f at i n

simplify

n

6 24 36 i2 id c 2 n n i1 2n i1

6 and n n2 summation 6 18 2n3 3n2 n 24 n2 n c 2 a b a b d formulas n 6 n 2 n 6 108 2n3 3n2 n 144 n2 n 3 a b 2 a b distribute 6 2 n n n

factor

36

108 2n3 3n2 n 144 n2 n a b a b rewrite 6 2 n3 n2 denominators 3 1 1 decompose rational 18 a2 2 b 72 a1 b expression n n n 3 1 1 As n S q, , 2 , S 0, and the area is 1182 122 72 36 units2. n n n 45. x 29.87 47. x 5, i 13


2/9/11

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SA-77


25. 152142192 182 172 ⫽ 10,080 PINs 27. 1w, x, y, z2 ⫽ 11, ⫺2, 3, ⫺42 ⫺4 b ´ 12, q2 29. a⫺q, 3


3. e

5. g

7. f

9. b

11. c

13. a

15. e

Summary and Concept Review, pp. 1216–1218 1. a. ⫺3

b. 5

c.

dne B A LH⫽RH

2. 2

3. 1

4.

dne B A LH⫽RH

2 1 1 7. ⫺ 8. 9. a. x ⫽ ⫺3 b. x ⫽ ⫺2, ⫺1, 3, 4 3 32 2 c. x ⫽ 1, 2 10. 3 11. ⫺1 12. 2 13. A dne B 14. A dne B 5. ⫺38

6.

⫺q

16. f ¿ 1x2 ⫽ 2x ⫹ 5

15. ⫺2

18. v¿ 1x2 ⫽

⫺1

1x ⫹ 32 2

LH⫽RH

1 17. g¿ 1x2 ⫽ 12x ⫺ 1

19. at x ⫽ 4, mtan ⫽ ⫺5

20. 9 units2

Practice Test, pp. 1218–1219 1. The limit of f(x) as x approaches 5 is 10. 2. f(x); L; sufficiently; c 3. False, a limit can exist even if c is not in the domain. 4. False, a limit can fail to exist even if a function is defined at c. 5. As the domain of g is x ⱖ 1, the limit in b. exists and the limit in a. does not. lim⫹ 1 1x ⫺ 1 ⫹ 22 ⫽ 2. xS1

6. a. II b. I

c. IV

d. III 7. a. 2

9. a. 3

b. 12

10. a. 4

12. a. 2

b. 9

13. a. 0

14. a.

dne B A LH⫽RH

b. A dne B L

b. A dne B ⫺q

8. a. 1

10 8 6 4 2 ⫺2⫺1 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

兺

兺

y

1 2 3 4 5 6 7 8 x

x⫽3

2 1 3. ln x ⫹ ln1x2 ⫹ 32 ⫺ 4 ln12x ⫹ 12 3 2 15 8 15 5. sin ␪ ⫽ ⫺ , cos ␪ ⫽ , tan ␪ ⫽ ⫺ 17 17 8 y 7. 10 8 6 4 2 ⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

1. constant 3. coefficient 5. Answers will vary 7. two; 3 and ⫺5 9. two; 2 and 14 11. three; ⫺2, 1, and ⫺5 13. one; ⫺1 15. n ⫺ 7 17. n ⫹ 4 19. 1n ⫺ 52 2 21. 2n ⫺ 13 23. n2 ⫹ 2n 25. 23 n ⫺ 5 27. 31n ⫹ 52 ⫺ 7 29. Let w represent the width in meters. Then 2w represents twice the width and 2w ⫺ 3 represents three meters less than twice the width. 31. Let b represent the speed of the bus. Then b ⫹ 15 represents 15 mph more than the speed of the bus. 33. h ⫽ b ⫹ 150 35. L ⫽ 2W ⫹ 20 37. M ⫽ 2.5N 39. T ⫽ 12.50g ⫹ 50 41. 14 43. 19 45. 0 47. 16 49. ⫺36 51. 51 53. 2 55. 144 57. ⫺41 59. 24 5 61. x 63. x Output Output

2 4 6 8 10 x

1 1 ␲ 7␲ ⫹1 9. a. ⫺ , b. A dne B 11. ␣ ⫽ ⫹ 1, LH⫽RH 2 2 6 6 13. The plane would be on a heading of 16.7°, traveling 297.5 mph. ln 9.36 ⫹ 1 15. x ⫽ , x ⬇ 1.618 17. ⫺3 2 1y ⫺ 12 2 1x ⫺ 12 2 ⫺2 2 ⫹ ⫹ ⫽1 19. 1 ⫹ 21. x⫹2 x⫺1 25 9 23. a. 1 b. 0 c. ⫺3 d. 5 not in domain, limit does not exit

⫺3 ⫺18 ⫺2 ⫺15 ⫺1 ⫺12 0 ⫺9 1 ⫺6 2 ⫺3 3 0 3 gives an output of 0.

14 ⫺3 6 ⫺2 0 ⫺1 0 ⫺4 1 ⫺6 2 ⫺6 3 ⫺4 ⫺1 gives an output of 0.

⫺q

Cumulative Review Chapters 1–12, pp. 1220–1221 1.

Exercises A.1, pp. A-7–A-9

b. 0

b. undefined, g132 ⫽ 0 11. a. 4 b. ⫺4 1 b. A dne B c. d. 1 LH⫽RH 3 c. A dne B d. 3 is not in the domain

3 1 15. 3 16. 17. 18. 1 2 10 19. a. d¿ 1t2 ⫽ ⫺32t ⫹ 224 b. d¿ 122 ⫽ 160, the debris is rising at a velocity of 160 ft/sec; d¿ 162 ⫽ 32, the upward velocity of the debris has slowed to 32 ft/sec; d¿ 172 ⫽ 0, the debris has reached its maximum height (velocity is 0 ft/sec); d¿ 1112 ⫽ ⫺128, the velocity of the debris is now in the downward direction (v 6 0) at 128 ft/sec. n n 6 6 6 n 6 2 20. a. lim LW ⫽ lim f a ib ⫽ lim c225 ⫺ a ib d nSq i⫽1 nSq i⫽1 n n nSq n i⫽1 n b. 1278 ft-lb

兺

APPENDIX A

65.

x Output ⫺3 ⫺5 ⫺2 8 ⫺1 9 0 4 ⫺1 1 2 0 3 13 2 gives an output of 0. 67. a. 7 ⫹ 1⫺52 ⫽ 2 b. n ⫹ 1⫺22 c. a ⫹ 1⫺4.22 ⫹ 13.6 ⫽ a ⫹ 9.4 d. x ⫹ 7 ⫺ 7 ⫽ x 69. a. 3.2 b. 56 71. ⫺5x ⫹ 13 2 73. ⫺15 79. ⫺2a2 ⫹ 2a 81. 6x2 ⫺ 3x p ⫹ 6 75. ⫺2a 77. 17 12 x 29 38 83. 2a ⫹ 3b ⫹ 2c 85. 8 n ⫹ 5 87. 7a2 ⫺ 13a ⫺ 5 89. 10 ohms 91. a. t ⫽ 12 j b. t ⫽ 275 mph 93. a. L ⫽ 2W ⫹ 3 b. 107 ft 95. t ⫽ c ⫹ 29; 44¢ 97. C ⫽ 25t ⫹ 43.50; $81 99. a. positive odd integer

Exercises A.2, pp. A-20–A-23 1. power 3. 20x; 0 5. a. cannot be simplified, unlike terms b. can be simplified, like bases 7. 14n7 9. ⫺12p5q4 11. a14b7 13. 216p3q6

15. 32.768h3k6

17.

p2 4q2

19. 49c14d 4

21.

9 6 2 16 x y

b. 1728 units3 27. 3w3 29. ⫺3ab 4p8 27 ⫺1 8x6 25m4n6 31. 33. 2h3 35. 37. ⫺8 39. 6 41. 43. 9 8 8 q 27y 4r8 2 2 2 12 25p q 3p 5 1 a ⫺12 45. 47. 49. 51. 3 53. 4 8 55. 4 ⫺4q2 3h7 a bc 5x4 7 ⫺2b 7 57. 59. 2 61. 10 63. 13 65. ⫺4 67. 6.77 ⫻ 109 9 27a9c3 69. 0.000 000 006 5 71. 26,571 hrs; 1107 days 73. polynomial, none of these, degree 3 75. nonpolynomial because exponents are not whole numbers, NA, NA 77. polynomial, binomial, degree 3 79. ⫺w3 ⫺ 3w2 ⫹ 7w ⫹ 8.2; ⫺1 81. c3 ⫹ 2c2 ⫺ 3c ⫹ 6; 1 ⫺2 2 83. ⫺2 85. 3p3 ⫺ 3p2 ⫺ 12 87. 7.85b2 ⫺ 0.6b ⫺ 1.9 3 x ⫹ 12; 3 1 2 89. 4 x ⫺ 8x ⫹ 6 91. q6 ⫹ q5 ⫺ q4 ⫹ 2q3 ⫺ q2 ⫺ 2q 23. 94 x3y2

25. a. V ⫽ 27x6


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93. ⫺3x3 ⫹ 3x2 ⫹ 18x 95. 3r2 ⫺ 11r ⫹ 10 97. x3 ⫺ 27 99. b3 ⫺ b2 ⫺ 34b ⫺ 56 101. 21v2 ⫺ 47v ⫹ 20 103. 9 ⫺ m2 9 105. p2 ⫹ 1.1p ⫺ 9 107. x2 ⫹ 34 x ⫹ 18 109. m2 ⫺ 16 2 2 2 2 111. 6x ⫹ 11xy ⫺ 10y 113. 12c ⫹ 23cd ⫹ 5d 115. 2x4 ⫺ x2 ⫺ 15 117. 4m ⫹ 3; 16m2 ⫺ 9 119. 7x ⫹ 10; 49x2 ⫺ 100 121. 6 ⫺ 5k; 36 ⫺ 25k2 123. x ⫺ 16; x2 ⫺ 6 125. x2 ⫹ 8x ⫹ 16 127. 16g2 ⫹ 24g ⫹ 9 129. 16p2 ⫺ 24pq ⫹ 9q2 131. 16 ⫺ 81x ⫹ x 133. xy ⫹ 2x ⫺ 3y ⫺ 6 135. k3 ⫹ 3k2 ⫺ 28k ⫺ 60 137. a. 340 mg, 292.5 mg b. Less, amount is decreasing. c. after 5 hr 139. F ⫽ kPQd⫺2 141. 5x⫺3 ⫹ 3x⫺2 ⫹ 2x⫺1 ⫹ 4 143. $15 145. 6

79. 81. 85. 89. 93. 95.

a. 8 ⫹ 6 ⫺ 12 ⫽ 2 ✓ b. 6 ⫹ 8 ⫺ 12 ⫽ 2 ✓ c. 20 vartices W 6 177.34 lb 83. 2f ⫹ 5 ⫽ 73; 34 fans 2c ⫹ 13 ⫽ 1467; 727 cal 87. 41 ⫽ 23 S ⫹ 1; 60 yr 91. 82 ⫹ 76 ⫹ 565 ⫹ 71 ⫹ x ⱖ 75, at least 81% 74 ⫽ 10 3 r ⫹ 34; 12 in. 1125 ⫹ 850 ⫹ 625 ⫹ 400 ⫹ x ⱖ 1000, at least $2000 5 0 6 20 w 6 150, 0 6 w 6 7.5 m

97. 45 6 95 C ⫹ 32 6 85, 7.2° 6 C 6 29.4° 99. S ⫽ 4.5 h ⫹ 20, K ⫽ 6 h ⫹ 11, more than 6 hr 101. a. 216.5 ft2 b. 7 sheets will be needed 103. about 337.4 in3 1 105. x ⫽ 107. Answers may vary. 109. 6 111. 6 113. 6 2 115. 7

Exercises A.3, pp. A-34–A-38 1. identity; unknown; contradiction; unknown

3. simple; compound 6 5. Answers will vary. 7. x ⫽ 3 9. v ⫽ ⫺11 11. b ⫽ 5 27 13. b ⫽ ⫺15 15. m ⫽ ⫺ 17. x ⫽ 12 19. x ⫽ 12 21. p ⫽ ⫺56 4 20 12 23. a ⫽ ⫺3.6 25. v ⫽ ⫺0.5 27. n ⫽ 29. p ⫽ 21 5 11 31. contradiction; { } 33. conditional; n ⫽ ⫺ 10 35. identity; 5x 0 x 僆⺢6 37. 5x 0 x ⱖ ⫺26; 3 ⫺2, q2 39. 5x 0 ⫺2 ⱕ x ⱕ 16; 3 ⫺2, 1 4 41. 5a| a ⱖ 26;

[

⫺3 ⫺2 ⫺1

0

1

43. 5n |n ⱖ 16;

2

3

4

5

[ 0

45. 5x| x 6

1

2

⫺32 5 6;

3

)

⫺10 ⫺9 ⫺8 ⫺7 ⫺6 ⫺5 ⫺4⫺3 ⫺2 ⫺1 0

47. 53. 55. 57. 59.

; n 僆 3 1, q 2 ⫺32 ; x 僆 1⫺q, 5 2

{ } 49. 5x |x ⑀ ⺢6 51. 5p | p ⑀ ⺢6 526; 5⫺3, ⫺2, ⫺1, 0, 1, 2, 3, 4, 6, 86 5 6; 5⫺3, ⫺2, ⫺1, 0, 1, 2, 3, 4, 5, 6, 76 54, 66; 52, 4, 5, 6, 7, 86 x 僆 1⫺q,⫺22 ´ 11, q 2;

)

)

⫺4 ⫺3 ⫺2 ⫺1

0

61. x 僆 3⫺2, 52 ;

1

2

[

⫺3 ⫺2 ⫺1

0

3

) 1

2

3

4

5

6

3

4

63. no solution 65. x 僆 1⫺q, q2 ; ⫺4 ⫺3 ⫺2 ⫺1

67. x 僆 3⫺5, 0 4 ;

0

1

2

[

[

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

69. x 僆 1⫺13 , ⫺14 2; ⫺1

⫺0.5

⫺4 ⫺3 ⫺2 ⫺1

73. x 僆 3⫺4, 12;

0

0.5

0

1

2

[

0

75. x 僆 3⫺1.4, 0.8 4 ; ⫺1.4 ⫺1

77. x 僆 3⫺16, 82;

3

4

1

2

3

[ 1

[

2

) 0

4

1. product 3. binomial; conjugate 5. Answers will vary. 7. a. ⫺171x2 ⫺ 32 b. 7b13b2 ⫺ 2b ⫹ 82 c. ⫺3a2 1a2 ⫹ 2a ⫺ 32 9. a. 1a ⫹ 2212a ⫹ 32 b. 1b2 ⫹ 3213b ⫹ 22 c. 1n ⫹ 7214m ⫺ 112 11. a. 13q ⫹ 2213q2 ⫹ 52 b. 1h ⫺ 1221h4 ⫺ 32 c. 1k2 ⫺ 721k3 ⫺ 52 13. a. ⫺11p ⫺ 72 1p ⫹ 22 b. prime c. 1n ⫺ 421n ⫺ 52 15. a. 13p ⫹ 221p ⫺ 52 b. 14q ⫺ 521q ⫹ 32 c. 15u ⫹ 3212u ⫺ 52 17. a. 12s ⫹ 52 12s ⫺ 52 b. 13x ⫹ 72 13x ⫺ 72 c. 215x ⫹ 6215x ⫺ 62 d. 111h ⫹ 122 111h ⫺ 122 e. 1b ⫹ 2521b ⫺ 252 19. a. 1a ⫺ 32 2 b. 1b ⫹ 52 2 c. 12m ⫺ 52 2 d. 13n ⫺ 72 2 21. a. 12p ⫺ 3214p2 ⫹ 6p ⫹ 92 b. 1m ⫹ 12 2 1m2 ⫺ 12 m ⫹ 14 2 c. 1g ⫺ 0.321g2 ⫹ 0.3g ⫹ 0.092 d. ⫺2t1t ⫺ 321t2 ⫹ 3t ⫹ 92 23. a. 1x ⫹ 321x ⫺ 321x ⫹ 12 1x ⫺ 12 b. 1x2 ⫹ 921x2 ⫹ 42 c. 1x ⫺ 22 1x2 ⫹ 2x ⫹ 421x ⫹ 12 1x2 ⫺ x ⫹ 12 25. a. 1n ⫹ 121n ⫺ 12 b. 1n ⫺ 121n2 ⫹ n ⫹ 12 c. 1n ⫹ 121n2 ⫺ n ⫹ 12 d. 7x12x ⫹ 1212x ⫺ 12 27. 1a ⫹ 521a ⫹ 22 29. 21x ⫺ 221x ⫺ 102 31. ⫺113m ⫹ 8213m ⫺ 82 33. 1r ⫺ 32 1r ⫺ 62 35. 12h ⫹ 32 1h ⫹ 22 37. 13k ⫺ 42 2 39. ⫺3x12x ⫺ 721x ⫺ 32 41. 4m1m ⫹ 52 1m ⫺ 22 43. 1a ⫹ 521a ⫺ 122 45. 12x ⫺ 5214x2 ⫹ 10x ⫹ 252 47. prime 49. 1x ⫺ 521x ⫹ 321x ⫺ 32 51. a. H b. E c. C d. F e. B f. A g. I h. D i. G 53. a ⫽ ⫺1; b ⫽ 2; c ⫽ ⫺15 55. not quadratic 1a ⫽ 02 57. a ⫽ 14 ; b ⫽ ⫺6; c ⫽ 0 59. a ⫽ 2; b ⫽ 0; c ⫽ 7 61. not quadratic (degree 3) 63. a ⫽ 1; b ⫽ ⫺1; c ⫽ ⫺5 65. x ⫽ 5 or x ⫽ ⫺3 67. m ⫽ 4 69. p ⫽ 0 or p ⫽ 2 71. h ⫽ 0 or h ⫽ ⫺1 73. a ⫽ 3 or a ⫽ ⫺3 75. g ⫽ ⫺9 2 77. m ⫽ ⫺5 or m ⫽ ⫺3 or m ⫽ 3 79. c ⫽ ⫺3 or c ⫽ 15 81. r ⫽ 8 or r ⫽ ⫺3 83. t ⫽ ⫺13 or t ⫽ 2 85. x ⫽ 5 or x ⫽ ⫺3 87. w ⫽ ⫺12 or w ⫽ 3 89. x ⫽ ⫺2, x ⫽ 0, x ⫽ 11 91. x ⫽ ⫺3, x ⫽ 0, x ⫽ 23 93. p ⫽ ⫺7, p ⫽ ⫺3, p ⫽ 3 95. x ⫽ ⫺5, x ⫽ 2, x ⫽ 5 97. x ⫽ ⫺5, x ⫽ ⫺2, x ⫽ 2, x ⫽ 5 99. b ⫽ ⫺2, b ⫽ ⫺1, b ⫽ 4, b ⫽ 5 101. 2␲r 1r ⫹ h2,7000␲ cm2; 21,991 cm2 1 103. V ⫽ ␲h1R ⫹ r2 1R ⫺ r2; 6␲ cm3; 18.8 cm3 3 105. V ⫽ x1x ⫹ 52 1x ⫹ 32 a. 3 in. b. 5 in. c. V ⫽ 2412921272 ⫽ 18,792 in3 v v a1 ⫹ b a1 ⫺ b, L ⫽ 12 211 ⫹ 0.75211 ⫺ 0.752 B c c ⫽ 3 27 in. ⬇ 7.94 in. 109. 11 in. by 13 in. 1 111. a. 18 14x4 ⫹ x3 ⫺ 6x2 ⫹ 322 b. 18 112b5 ⫺ 3b3 ⫹ 8b2 ⫺ 182 113. 2x116x ⫺ 272 16x ⫹ 52 115. 1x ⫹ 32 1x ⫺ 32 1x2 ⫹ 92 117. 1p ⫹ 121p2 ⫺ p ⫹ 121p ⫺ 12 1p2 ⫹ p ⫹ 12

107. L ⫽ L0

119. 1q ⫹ 521q ⫺ 52 1q ⫹ 2321q ⫺ 232


0.8

0

⫺20 ⫺16⫺12 ⫺8 ⫺4

1

)

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

[

1

)(

71. x 僆 1⫺q, q2;

⫺2

; a 僆 3 2, q 2


8

12

1. 1; ⫺1 3. common denominator 5. F; numerator should be ⫺1 a⫺4 1 x⫹3 7. a. ⫺ b. 9. a. simplified b. 3 2x1x ⫺ 22 a⫺7 x⫹3 ⫺1 9 11. a. ⫺1 b. ⫺1 13. a. ⫺3ab b. c. ⫺11y ⫹ 32 d. 9 m


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SA-79

Student Answer Appendix 1a ⫺ 22 1a ⫹ 12 2n ⫹ 3 3x ⫹ 5 b. c. x ⫹ 2 d. n ⫺ 2 17. n 2x ⫹ 3 1a ⫹ 32 1a ⫹ 22 1p ⫺ 42 2 81a ⫺ 72 y ⫺15 3 1 21. 23. 25. 27. 29. 2 4 2 a⫺5 x p 1 n⫹ y⫹3 m x ⫹ 0.3 5 33. 35. 37. m⫺4 3y1y ⫹ 42 x ⫺ 0.2 2 n⫹ 3 31a2 ⫹ 3a ⫹ 92 14y ⫺ x 2n ⫹ 1 3 ⫹ 20x 41. 43. 45. 2 n 8x2 8x2y4 ⫺y ⫹ 11 2 ⫺3m ⫺ 16 ⫺5m ⫹ 37 49. 51. 53. p⫹6 1m ⫹ 421m ⫺ 42 m⫺7 1y ⫹ 621y ⫺ 52 2a ⫺ 5 1 m2 ⫺ 6m ⫹ 21 57. 59. 1a ⫹ 421a ⫺ 52 y⫹1 1m ⫹ 32 2 1m ⫺ 32

15. a. 19.

31.

39. 47. 55.

y2 ⫹ 26y ⫺ 1

1 5 1 ⫺ 5p 63. a. 2 ⫺ ; 15y ⫹ 121y ⫹ 321y ⫺ 22 p p p2 1 4a 2 x⫹2 x b. 2 ⫹ 3 ; 65. 67. p ⫺ 1 69. a ⫹ 20 313x ⫺ 42 x x x3 3 2 1⫹ 1⫹ 2 2 ⫺2 m m⫹3 x x ⫹2 ; ; 71. 73. a. b. 75. x ⫽ 1 y ⫹ 31 3 m⫺3 2 x2 ⫺ 2 1⫺ 1⫺ 2 m x 77. a ⫽ 32 79. y ⫽ 12 81. x ⫽ 3; x ⫽ 7 is extraneous 83. n ⫽ 7 85. a ⫽ ⫺1, a ⫽ ⫺8 87. a. $300 million; $2550 million 89. Price rises rapidly for first four b. It would require many resources. days, then begins a gradual c. No decrease. Yes, on the 35th day of trading. 450P P Day Price 100 ⫺ P 0 10 40 300 1 16.67 2 32.76 60 675 3 47.40 80 1800 4 53.51 5 52.86 90 4050 6 49.25 93 5979 7 44.91 95 8550 8 40.75 9 37.03 98 22050 10 33.81 100 ERROR 61.

91. t ⫽ 8 weeks 93. P ⬇ 80% 12x ⫹ 3y2 95. b. 5


1. even 3. 1164 2 3 5. Answers will vary. 7. a. 冟9冟 ⫽ 9 b. 冟⫺10冟 ⫽ 10 9. a. 7冟p冟 b. 冟x ⫺ 3冟 c. 9m2 d. 冟x ⫺ 3冟 11. a. 4 b. ⫺6x c. 6z4 v d. 13. a. 2 b. not a real number c. 3x2 d. ⫺3x e. k ⫺ 3 ⫺2 7冟v5冟 f. 冟h ⫹ 2冟 15. a. ⫺5 b. ⫺3冟n3冟 c. not real number d. 6 9p4 64 125 ⫺1728 17. a. 4 b. c. d. 19. a. 125 8 4q2 1 ⫺256 32n10 1 b. not a real number c. d. 21. a. b. 1 4 2 9 81x p 2y4 3 3 2 2 3 3 23. a. 3m12 b. 10pq 1q c. mn 2n d. 4pq 12p 2 x4 1y 9 e. ⫺3 ⫹ 17 f. ⫺ 12 25. a. 15a2 b. ⫺4b 1b c. 2 3 3 3 15 3 18 1 2 3 2 d. 3u v1v 27. a. 2m b. 3n c. d. 4x z3 1

6 5 3 2 6 e. b4 6 6 16 31. a. 9 12 b. 14 13 c. 16 12m d. ⫺5 17p 3 33. a. ⫺x 12x b. 2 ⫺ 13x ⫹ 315 c. 6x22x ⫹ 522 ⫺ 27x ⫹ 323 35. a. 98 b. 115 ⫹ 121 c. n2 ⫺ 5 d. 39 ⫺ 12 13 37. a. ⫺19 b. 110 ⫹ 165 ⫺ 217 ⫺ 1182 c. 12 15 ⫹ 2114 ⫹ 36115 ⫹ 6142

29. a. 2x2y3

4 12 b. x2 1 x c. 1 b d.

39. Verified 41. Verified 43. a. 3 2 2p2

1

⫽

23 2

b.

2215x 9x2

c.

3 26b 10b

3 2 61x ⫹ 6 12 52 a 45. a. ⫺12 ⫹ 4 111; 1.27 b. 2p a x⫺2 47. a. 130 ⫺ 215 ⫺ 313 ⫹ 312; 0.05 7 ⫹ 712 ⫹ 16 ⫹ 213 b. ; ⫺7.60 49. a. x ⫽ 14 3 ⫺3 b. x ⫽ 8, x ⫽ 1 is extraneous 51. a. m ⫽ 3 b. x ⫽ 5 c. m ⫽ ⫺64 d. x ⫽ ⫺16 53. a. x ⫽ 25 b. x ⫽ 7; x ⫽ ⫺2 is extraneous c. x ⫽ 2, x ⫽ 18 d. x ⫽ 6; x ⫽ 0 is extraneous 55. a. x ⫽ ⫺32 b. x ⫽ 81 57. a. x ⫽ ⫺32, x ⫽ 22 b. x ⫽ ⫺30, x ⫽ 34 59. x ⫽ ⫺27, x ⫽ 125 61. 8.33 ft 63. a. 8210 m b. about 25.3 m 65. a. 365.02 days b. 688.69 days c. 87.91 days 67. a. 36 mph b. 46.5 mph 69. 12␲234 ⬇ 219.82 m2 71. a. 36 million mi b. 67 million mi c. 93 million mi d. 142 million mi e. 484 million mi f. 887 million mi 73. a. 1x ⫹ 152 1x ⫺ 152 b. 1n ⫹ 11921n ⫺ 1192

d.

e.

75. a. 13 13x ⫹ 391x 79. x 僆 31, 22 ´ 12, q 2

b. Answers will vary.

77.

322 2

Practice Test, pp. A-82–A-83 1. d. 3. 5. b.

a. False; parentheses first b. False; undefined c. True False; ⫺2x ⫹ 6 2. a. 11 b. ⫺5 c. not a real number d. 20 a. 98 b. ⫺7 c. 0.5 d. ⫺4.6 4. a. 28 b. 0.9 c. 4 d. ⫺7 6 3 ⬇ 4439.28 6. a. 0 b. undefined 7. a. 3; ⫺2, 6, 5 2; 13 , 1 8. a. ⫺13 b. ⬇ 7.29 9. a. x3 ⫺ 12x ⫺ 92 n 2 b. 2n ⫺ 3a b 10. a. Let r represent Earth’s radius. Then 11r ⫺ 119 2 represents Jupiter’s radius. b. Let e represent this year’s earnings. Then 4e ⫹ 1.2 million represents last year’s earnings. 11. a. 9v2 ⫹ 3v ⫺ 7 b. ⫺7b ⫹ 8 c. x2 ⫹ 6x 12. a. 13x ⫹ 4213x ⫺ 42 b. v12v ⫺ 32 2 m6 25 2 2 c. 1x ⫹ 52 1x ⫹ 321x ⫺ 32 13. a. 5b3 b. 4a12b12 c. d. p q 3 4 8n 14. a. ⫺4ab

b. 6.4 ⫻ 10⫺2 ⫽ 0.064

c.

a12 b4c8

d. ⫺6

15. a. 9x ⫺ 25y b. 4a ⫹ 12ab ⫹ 9b 16. a. 7a4 ⫺ 5a3 ⫹ 8a2 ⫺ 3a ⫺ 18 b. ⫺7x4 ⫹ 4x2 ⫹ 5x 17. a. ⫺1 31m ⫹ 72 2⫹n x⫺5 x⫺5 b. c. x ⫺ 3 d. e. f. 2⫺n 3x ⫺ 2 3x ⫹ 1 51m ⫹ 421m ⫺ 32 1 ⫺2 64 12 18. a. 冟x ⫹ 11冟 b. c. d. ⫺ ⫹ e. 11 110 3v 125 2 2 110x f. x2 ⫺ 5 g. h. 21 16 ⫹ 122 19. ⫺0.5x2 ⫹ 10x ⫹ 1200; 5x a. 10 decreases of 0.50 or $5.00 b. Maximum revenue is $1250. 1 20. 58 cm 21. a. b ⫽ 6 b. n ⫽ 4 c. m ⫽ ⫺1 d. x ⫽ 6 e. 5 6 (contradiction) f. g ⫽ 10 22. 3 ⫹ 14 1n ⫹ 122 ⫽ 16, the number is 40 23. a. x ⫽ ⫺2, x ⫽ 2, x ⫽ 7 b. r ⫽ 0, r ⫽ ⫺1, r ⫽ 4 1 5 c. g ⫽ ⫺3, g ⫽ ⫺1, g ⫽ 1, g ⫽ 3 24. a. x ⫽ ⫺ b. h ⫽ ⫺ , h ⫽ 2 2 3 c. n ⫽ 13 (⫺2 is extraneous) 25. a. x ⫽ ⫺3, x ⫽ 3 b. x ⫽ ⫺4, x ⫽ 5 c. x ⫽ ⫺1 (7 is extraneous) 4

2

2

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Index Absolute maximum, 112 Absolute value of complex numbers, 213, 334 explanation of, 173, A–80 on graphing calculator, 137, 201 limits and, 201 multiplicative property of, 138 Absolute value equations distance and, 147 explanation of, 136, 191 on graphing calculator, 137 graphs to solve, 142 method to solve, 136–138 Absolute value functions characteristics of, 141, 163 explanation of, 120, 191 Absolute value inequalities applications of, 142, 143 distance and, 147 explanation of, 191 on graphing calculator, 142 graphs to solve, 142 limits and, 201 methods to solve, 139 solutions to “greater than,” 140–141 solutions to “less than,” 139–140 Accumulated value, 472, 473 Acute angles, 510, 596 Addition associative property of, A–4, A–80 commutative property of, A–4, A–80 of complex numbers, 206–207 distributive property of multiplication over, A–5 of matrices, 907–908 of ordinates, 273 of polynomials, A–16 – A–17 of radical expressions, A–69 of rational expressions, A–55 – A–57 Addition method. See Elimination method Additive identity, A–5, A–80 Additive inverse, A–5, A–80 Additive property of equality, A–24 Additive property of inequality, A–26 – A–27 Algebra to evaluate limits, 1193–1195 of functions, 262–268, 270, 295 fundamental theorem of, 320–323, 397 of matrices, 905–913, 951 of vectors, 776–778 to verify identities, 655–656 Algebraic equations. See Equations Algebraic expressions evaluation of, A–3 – A–4 explanation of, A–1, A–24 method to simplify, A–6 translating English phrases into, A–2 Algebraic fractions. See Rational expressions Algebraic methods to find inverse functions, 412–415 trigonometric equations and, 721–723 Algebraic terms, A–1, A–2 Algebraic vectors, 778 Algorithms, division, 309 Allometric studies, 157 Alpha (␣), 510 Alternating current (AC), 213, 807–809 Ambiguous case analysis of, 749 applications of, 752–753 explanation of, 748–749 on graphing calculator, 750 method to avoid, 761 method to solve, 751–752 scaled drawings and, 750–751

Amortization, 472–475, 480 Amplitude explanation of, 552 of sine and cosine functions, 551–552 variable, 645–646 Analytical geometry distance formula and, 962 explanation of, 962, 1064 midpoint formula and, 962 Analytic parabola, 997–1002, 1066 Angle-angle-side (AAS) triangles, 747, 764 Angle measurement degree, 510–512 degree-radian conversions and, 517–520 on graphing calculator, 515, 518–519 radian, 514–517 Angle reduction formulas, 679 Angles acute, 510 central, 515, 530 complementary, 510, 511 coterminal, 515, 617, 618 of elevation/depression, 601–604 explanation of, 510, 634 between intersecting lines, 620 negative, 514 obtuse, 510 positive, 514 quadrantal, 515, 613 reference, 531–532, 614–616, 635, 647 right, 510 standard, 646–647, 1040 straight, 510 supplementary, 510, 511 Angle-side-angle (ASA) triangles, 747, 764 Angular velocity, 520–521 Annuities, 472–475 Annuity formula, 473 Aphelion, 1044 Applications of absolute value, 142, 143 of algebra of functions, 267–268 of analytic parabola, 1002 of annuities and amortization, 472–475 of arithmetic sequence, 1095 of basic geometry, A–31 – A–34 of binomial theorem, 1150–1151 of combinations, 1126–1127 of completing the square, 18 of complex numbers, 213, 807–809 of composition of functions, 284–285 of conics in polar form, 1044–1046 of difference quotient, 283–286 of direct variation, 178–180 of exponential functions, 422, 428–429 of exponential growth and decay, 475–477 of foci, 978–979, 992–993 of functions, 105 of geometric series, 1107 of identities, 686–688 of inequalities, 390–391, A–31 of interest, 469–472, 499 of inverse functions, 417 of inverse trigonometric functions, 704–705 of inverse variation, 181 of joint variation, 182 of law of cosines, 762–763 of law of sines, 752–753 of limits, 1203–1212, 1217–1218 of linear equations, 27–28, 56–58, A–30 – A–31 of linear programming, 872–874 of logarithms, 438–442, 452–453 of matrices, 900–901, 938–939

of parametric equations, 1057–1059, 1063 of piecewise-defined functions, 170–171 of polynomials, 347–348 of power functions, 155–158 problem-solving guide for, 71 of Pythagorean theorem, 9, 598, 759, 760, 778, A–75 of quadratic equations, 228–229 of quadratic functions, 239–243, 251–253 of quadratic inequalities, 229–230 of rate of change, 255–256 of rational functions, 155–158, 365, 377–380 of regression, 86–87, 486–487, 626–627 of remainder theorem, 316, 331 of right triangles, 601–604 of sequences, 1084–1085, 1095 of sinking funds, 474 of slope, 20 of summation, 1119–1120 of system of linear inequalities, 869–870 of systems of linear equations, 844–848, 860–861 of systems of nonlinear equations, 1013–1014 of trigonometric equations, 721, 726–727 of trigonometric functions, 579–580, 585–586, 588–589, 617–618 of vectors, 779–782, 790–791 Approximations limits vs., 1170–1171 real zeroes, 355 Archimedes, 1169 Arc length, 516, 1004 Arc length formula, 516, 519–520 Area. See also Surface area of circle, A–21, A–32 of circular sector, 516–517, 731 under curve, 304–305, 1210–1212, 1219–1220 of ellipse, 981 explanation of, A–31 of inscribed polygon, 1170–1171 of Norman window, 942 of parallelogram, 620, A–86 – A–87 of polygon, 558, 660, 1170–1171 of rectangle, 876, 915, A–32 of right parabolic segment, 1004 of square, A–32 of trapezoid, A–32 of triangle, 381, 668, 763–766, 823, 876, 904, 936 of vertical parabolic segment, 1017 Area functions, 505–506 Arithmetic sequences applications of, 1095 explanation of, 32, 1089, 1155 finding nth partial sum of, 1093–1094 finding nth term of, 32, 1090–1093 graphs of, 1092 identifying and finding common difference in, 1089–1090 quadratic functions and, 1098 Associated minor matrices, 923, 924 Associations linear/nonlinear, 80–81, 98 positive/negative, 79–80, 98 Associative property of addition, A–4, A–80 Associative property of multiplication, A–4, A–80 Asymptotes of central hyperbola, A–94 on graphing calculator, 156, 361 to graph rational functions, 151–152 horizontal, 150, 359–361 of hyperbola, 986 nonlinear, 374, 375 oblique, 374–376 vertical, 151, 356–359, 363

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Index

Asymptotic behavior, 150, 151 Asymptotic discontinuity, 1192 Augmented matrices matrix inverses and, 955–956 of system of equations, 893–895 triangularizing, 895–897 Average distance, 9 Average rate of change applications of, 255–256 calculation of, 253–254 difference quotient and, 281–283 explanation of, 253, 295 projectile velocity and, 255 Average rate of change formula, 254–255 Average value of function, 552, 560 Axes conjugate, 985 rotation of, 1037–1039, 1041 transverse, 985 vertical, 853 Axis of symmetry, 121 Back-substitution, A–25 Base-b, 460 Base case of inductive proof, 1114 Base-e exponential functions, 425–426 Base function, 300 Bearings, 602 Beats, 739 Berkeley, George, 1180, 1187 Beta (␤), 510 Binomial coefficients, 1147–1148 Binomial conjugates, A–18 – A–19 Binomial cubes, 213 Binomial experiment, 1150 Binomial factors, A–39– A–40 Binomial probability, 1150–1152 Binomials. See also Polynomials expansion of, 146–1147, 1149, 1150 explanation of, 1145, A–15 multiplication of, 207, A–17 – A–19 Pascal’s triangle and, 1145–1147 Binomial squares, A–19– A–20 Binomial theorem applications of, 1150–1151 explanation of, 1145, 1149, 1150, 1158 Biorhythm theory, 629 Bisection, 355 Body mass index formula, A–36 Boundary, 865 Boundary lines horizontal, 37–38, 96 vertical, 36–37, 96 Boundary point, 36 Brachistochrone applications, 1055 Branches, of hyperbola, 984, 986 Break-even analysis, 846 Brewster’s law of reflection, 678 Calculators. See Graphing calculators Capacitive reactance, 808 Cardano, Girolomo, 213 Cardano’s formula, 213 Carrying capacity, 485 Cauchy, Augustin-Louis, 1169 Cayley-Hamilton theorem, 946 Ceiling functions, 171 Celsius temperature, 46, 850 Center of circle, 10, 12 of hyperbola, 985 Centimeters of mercury, 440 Central angle, 515, 530 Central circle, 10, 515, 527, 635, 972 Central ellipse, 972 Central hyperbola, 985, A–94 Central rectangle, 986 Change in x, 21 Change in y, 21 Change-of-base formula, 452, 498 Characteristic polynomials, 946 Circles area of, A–21, A–32 center of, 10, 12

central, 10, 515, 635, 972 circumference of, 90, 516–517 circumscribed, 17, 756, 963 equations of, 8–13, 94, 527, 969, 970, 1065 explanation of, 10, 963, A–95 on graphing calculator, 13–14, 528, 530, 969 graphs of, 9–13, 670, 969 involute of, 621 perimeter of, A–32 radius of, 10, 12, 756 standard form of equation of, 10, 11, 970 symmetry and, 528, 529, 531–532 unit, 527–534, 635–636 Circumference, 90, 516–517 Circumscribed circle, 17, 756, 963 Closed form solution, 219 Coefficients binomial, 1147–1148 explanation of, A–1 frictional, A–25 polynomial equations with complex, 234 tangent and cotangent functions and, 568–569 Cofactors, A–92 Cofunctions, 600–601 Coincident dependence, 854, 860 Coincident lines, 843 Collinear points, 936–937 Combinations applications of, 1126–1127 explanation of, 1125–1126 on graphing calculator, 1126 Combined variation, 182 Common binomial factors, A–39 Common difference d, of arithmetic sequence, 1089 Common logarithms explanation of, 434 method to find, 435–436 Common ratio r, 1098–1099 Commutative property of addition, A–4, A–80 Commutative property of multiplication, A–4, A–80 Complementary angles, 510, 511 Complements of event, 1134–1136 to write equivalent expressions, 600–601 Completely factored form, A–40 Completing the square applications of, 18 to graph ellipse, 976–977 to graph horizontal parabola, 999 to graph hyperbola, 988–989 to graph quadratic functions, 235–237 process of, 1006–1007 to solve quadratic equations, 219–221, 235 Complex conjugates explanation of, 208, 332 product of, 208 use of, 210 Complex conjugates theorem proof, A–84 – A–85 Complex numbers absolute value of, 213, 334 addition and subtraction of, 206–207 applications of, 213, 807–809 converted from trigonometric form to rectangular form, 804–805 cube of, 811 De Movie’s theorem to compute power of, 813–814 division of, 210, 806–807 explanation of, 205, 292–293 geometric connections for product and quotients of, 805–806 on graphing calculator, 207–210, 804, 806 graphs of, 802 historical background of, 204 multiplication of, 207–209, 806–807 nth roots of, 815–818 principal square root of, 334 standard form of, 205–206 in trigonometric form, 803–804, 806–807, 825 Complex plane, 802, 811 Complex polynomial functions, 320–321 Complex roots, substitution to check, 208 Complex zeroes, 401–402 Component form, 773 Composite functions, 278–279

Composition of functions applications of, 284–286 decomposition and, 278–279 explanation of, 274–286, 296 on graphing calculator, 277, 278 method to find, 276–278 notation for, 275 numerical and graphical view of, 279–281 transformations via, 288 Compound annual growth, 288 Compound fractions, 663, A–57–A–58 Compound inequalities explanation of, A–28 method to solve, A–28 – A–30 Compound interest, 470–471 Compound interest formula, 471 Compressions, vertical, 126–127, 568, 569 Conditional equations, A–26 Cones characteristics of, 964 volume of, 730, 876 Conic equations. See also specific conics ellipse, 965–966, 970–975, 1065, A–93 hyperbola, 984–987, 995, 1065, A–93 – A–94 in polar form, 1042–1044 Conic sections. See also Circles; Ellipses; Hyperbolas; Parabolas applications of, 1044–1046 characteristics of, 964–966 eccentricity of, 1043 equations of, 1038, 1042–1044 explanation of, 964, 1067–1068 graphs of, 964 identified by their equations, 990 in polar form, 1042–1046 rotated, 1035–1040 use of discriminant to identify, 1040–1041 Conjugate axis, 985 Conjugates binomial, A–18 – A–19 complex, 208, 210, 332 simplifying radical expressions using, A–71 verifying identities by multiplying, 664 Consecutive integers, 72 Consistent system, 844–845 Constant functions, 51 Constant of variation, 177 Constraint inequalities, 872 Constraints, 870 Continuous functions explanation of, 163, 1191, 1192 piecewise and, 167, 169 Continuous graphs, 5 Continuously compounded interest, 471–472 Contour interval, 525 Contradictions, A–26 Convenient values decomposition using, 882–885 explanation of, 882 Convex, 871 Coordinate grid, 3 Coordinate plane, 610–616, 641 Corner point, 868 Correlation, 81–83 Correlation coefficient, 92, 98 Cosecant function application of, 589 characteristics of, 563 explanation of, 532 graphs of, 561–562, 637 inverse, 702–704 Cosine law of, 759–763, A–88 origin of term for, 600 table of values for, 549 Cosine function amplitude of, 551–552 applications of, 585, 586 characteristics of, 550 explanation of, 532 graphs of, 549–554, 636–637 inverse, 698–699, 704 period and coefficient B of, 552–554 Cost-based pricing, 846


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Index Cotangent function characteristics of, 567 explanation of, 532 graphs of, 566–567, 637–638 hyperbolic, 506 inverse, 702, 703 table of values for, 566 Coterminal angles, 515, 617, 618, 813 Counting techniques combinations and, 1125–1127 distinguishable permutations and, 1123–1124 fundamental principle of, 1122–1123 key concepts on, 1156–1157 by listing using tree diagram, 1120–1122 nondistinguishable permutations and, 1125 Cramer’s rule, 934–936, 954–955 Critical points explanation of, 622, 642 sinusoidal models and, 622–626 Cube root function, 120 Cube roots, A–65. See also Radical expressions Cubes binomial, 213 of complex number, 811 perfect, A–65 sum or difference of two perfect, A–43 – A–44 volume of, A–21 Cubic equations, 393, 819 Cubic functions, 568 Cubic units, A–33 Current, 808, 849 Curves area under, 304–305, 1210–1212, 1219–1220 conic sections as family of, 965 from cycloid family, 1055–1056 families of polar, A–95 intersections of polar, 1074–1075 parametric, 1051–1054 Cycloids, 1055–1056 Cylinders surface area of, 76, 233, 271, 381, A–51 volume of, 90, 730, A–51 Cylindrical vents, 1016 Dampening factor, 632 Decay rate, 475 Decay rate constant, 475 Decibel (dB), 438 Decimal degrees, 511–512 Decimal notation, A–14 – A–15 Decision variables, 872 Declarative statements, 1186–1187 Decomposition of composite functions, 278–279 to evaluate telescoping sum, 886–888 explanation of, A–1 on graphing calculator, 884, 888 partial fraction, 837, 879–888, 950, 958 of rational expressions, 879 using convenient values, 882–885 using system of equations, 885–886 Decomposition template, 879–882 Decreasing functions, 110–111 Degenerate cases, 18, 973, 997 Degrees as angle measurement, 510–512 conversion between radians and, 517–520 decimal, 511–512 of polynomials, A–15 Degrees-minutes-seconds (DMS) method, 511 Delta (⌬), 21 Delta/epsilon form, 201–202 Demographics, 422 De Moivre’s theorem to check solutions to polynomial equations, 814–815 explanation of, 813–814 nth root theorem and, 815–818 proof of, A–88 Denominator lowest common, A–55 – A–59 rationalizing the, A–70 – A–71 Density, 307

Dependent systems of linear equations in three variables, 858–860 of linear equations in two variables, 843–844 matrices and, 899–900 Dependent variables, 2 Depreciation, 57, 58, 428 Depression, angle of, 601–604 Derivatives, 201 Descartes, René, 204 Descartes’ rule of signs, 328–329, 355, 397 Determinant formula for area of parallelogram, A–86 – A–87 Determinants of column rotation, 925 to find area of triangle, 936 of general matrix, A–92 geometry of, 959 singular matrices and, 922–927 to solve system of linear equations, 933–934, 952 Diagonal entries, of matrix, 893 Dichotomy paradox, 1160 Difference of two squares, A–19, A–42 Difference quotient applications of, 283–286 explanation of, 281, 296 limit of, 1204–1209 polynomial functions and, 303 radical functions and, 304 rates of change and, 259, 281–283, 303–304 Directed line segments, 771 Directrix explanation of, 964, 965 of hyperbola, 991 of parabola, 1000–1001 Direct substitution, 1192–1193 Direct variation applications of, 178–180 explanation of, 177–178, 193 Discontinuities asymptotic, 1192 explanation of, 1191 jump, 1192 nonremovable, 1192 removable, 169, 372–373, 399, 1192 Discontinuous functions, 168–169 Discriminant of cubic equations, 393, 819 explanation of, 223 identifying conics using, 1040–1041 of quadratic formula, 223–224 Diseconomies of scale, 1013 Distance absolute value and, 147 area under graph to determine, 305 perpendicular, 963, 967 between point and line, 963 from point to plane, 863 viewing, 256 Distance formula, 9–10, 668, 962, 1033, 1064 Distinguishable permutations, 1123–1124 Distributive property of multiplication over addition, A–5, A–38 Division of complex numbers, 210 factor theorem and, 313–315 on graphing calculator, 313 long, 308–309 of polynomials, 308–315, 327 of rational expressions, A–55 remainder theorem and, 312–313 synthetic, 309–312, 396–397, A–89 – A–90 Division algorithm, 309 Divisor, 309, 311 Domain of exponential functions, 422 of functions, 36–39, 42, 95, 191, 262–264 implied, 38–39 of logarithmic functions, 437–438 of piecewise-defined functions, 164–165 of power functions, 153–154 of rational functions, 149, 191 of relations, 2, 48–49 restricting, 413–414

I-3

solving quadratic inequality to determine, 227–228 of trigonometric functions, 535 Dominant term, 339 Dot products angle between two vectors and, 792–793 explanation of, 792, 824 properties of, 793–794 Double-angle formula, 682 Double-angle identities explanation of, 680–681, 694, 734 use of, 681, 682 Dynamic trigonometry, 610 e, 425, 426 Eccentricity application of, 1044–1045 of conics, 1043 on graphing calculator, 1070 Economies of scale, 1013 Electrical resistance, A–9 Elementary row operations, 895 Elements of set, 37 Elevation, angle of, 601–604 Elimination method explanation of, 841 Gaussian, 895 Gauss-Jordan, 897–898, A–91 to solve system of linear equations in three variables, 855–858 to solve system of linear equations in two variables, 841–843 to solve system of nonlinear equations, 1010–1012 Ellipses applications of, 978–979 area of, 981 central, 972 definition of, 974, 975 equation of, 965–966, 970–975, 1065, A–93 foci of, 974–978 on graphing calculator, 977, 978, 1038 graphs of, 971–974, 976–977 horizontal, 971–972 perimeter of, 981 polar equation of, 1046 polar form of, 1049 with rational/irrational values of a and b, 1071 standard form of equation of, 971–973, 976 vertical, 972–973 Empty set, A–80 Encryption, 940–942 End-behavior to analyze rational inequalities, 389 explanation of, 111, 338 of polynomial graphs, 338–341 of quadratic functions, 338 of rational functions, 149, 150 Endpoint maximum, 112 Endpoints, 36 Entries, of matrix, 893 Equality additive property of, A–24 of matrices, 905–906 multiplicative property of, A–24 power property of, A–71, A–73 square root property of, 217–219 Equation form, relations in, 3, 94 Equations. See also Linear equations; System of linear equations in three variables; System of linear equations in two variables; System of nonlinear equations; specific types of equations absolute value, 136–138, 142, 191 of circle, 8–13, 94, 527, 969, 970, 1065 conditional, A–26 of conic sections, 1038, 1042–1044 cubic, 393, 819 depreciation, 57, 58 of ellipse, 965–966, 970–975, 1065, A–93 equivalent, A–24 explanation of, 64 exponential, 426–429, 433–435, 447–448, 460–463, 499 families of, A–24 in function form, 51


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Index

Equations—Cont. of hyperbola, 984–987, 995, 1065, A–93 – A–94 intersection-of-graphs method to solve, 65–67, 74, 97 of line, 52–55, 729, 1049 literal, 69–70 logarithmic, 447, 457–460, 499, 1011 logistic, 463–464, 485–486, 502, 890 matrix, 921–922, 926–927, 951 of parabola, 965, 999–1002 parametric, 393, 1051–1059, 1068 of piecewise-defined function, 165, 169–170 polar, 1024, 1027, 1028 polynomial, 234, 814–815, A–45 properties of equality to solve, A–24 – A–25 quadratic, 214–215, 217–224, 249–250, 293, A–45 – A–46 in quadratic form, A–74 of quadratic functions, 238 radical, A–71 – A–75 rational, A–58 – A–59 of rational functions, 152, 364 roots of, A–24 sinusoidal, 582–583 solutions of, 199–200 standard form of, 841 that are not identities, 665–666 translating written information in, A–30 – A–31 trigonometric, 711–717, 721–727, 736, 737 variation, 177, 179 written from graphs, 554–555 Equilateral triangles, 524, 811 Equilibrium explanation of, 584 force required to maintain, 678 market, 847–848 static, 828–829 vectors and, 787–789 Equivalent equations, A–24 Equivalent system of equations, 841 Equivalent vectors, 772–775 Euclid, 510 Euclidean geometry, 962 Euler, Leonhard, 204, A–36 Euler’s polyhedron formula, A–36 Even functions, 106–107 Even multiplicity, 322 Events complementary, 1134–1136 explanation of, 1132 mutually exclusive, 1138 nonexclusive, 1137–1139 probability of, 1133, 1134 ex, 506 Exact solutions, 219 Existence theorem, 323 Experiment binomial, 1150 explanation of, 1121, 1132 Exponential decay, 475–477 Exponential equations explanation of, 499 logarithmic form and, 434, 435 methods to solve, 447–448, 460–463 uniqueness property to solve, 426–428 with unlike bases, 461 Exponential form, 152, 434, 449, A–10 Exponential functions applications of, 422, 428–429, 463 base-e, 425–426 characteristics of, 424 evaluation of, 423 explanation of, 422, 433, 496–497 on graphing calculator, 427–428 graphs of, 423–425 Exponential growth, 475–476 Exponential growth formula, 475 Exponential regression model, 483–485 Exponents explanation of, A–10 negative, A–12, A–14 power property of, A–10 – A–11, A–14 product property of, A–10 – A–11, A–14 product to power property of, A–11, A–14 properties of, 422, A–10 – A–14, A–80 quotient property of, A–12

quotient to power property of, A–11, A–14 rational, A–66 – A–67 scientific notation and, A–14 – A–15 zero, A–12, A–14 Extraneous roots, 457, A–59 Extrapolation, 486 Extreme values explanation of, 238 on graphing calculator, 239, 241–243 quadratic functions and, 238–243 Factorial formulas, 1130 Factorials, 1080–1081 Factors/factoring common binomial, A–39 difference of two squares, A–42 flowchart of, A–45 greatest common, A–38 – A–39 by grouping, A–39 – A–40 perfect square trinomials, A–42 – A–43 quadratic forms, A–42 – A–45 quadratic polynomials, A–40 A–42 to solve higher degree equations, A–48 to solve polynomial equations, A–47 to solve trigonometric equations, 722–723 sum or difference of two perfect cubes, A–43 – A–44 Factor theorem explanation of, 313, 397 finding zeroes using, 315 polynomials and, 314–314 proof of, A–84 Fahrenheit temperature, 46, 850 Families of equations, A–24 of functions, 120–121 of identities, 655 of polar curves, A–95 of polar graphs, 1027–1031 of polynomials, A–15 Feasible region, 871 Fibonacci (Leonardo of Pisa), 1080 Fibonacci sequence, 1080 Finite sequences, 1078 Five Card Stud, 1161–1162 Fixed costs, 846 Floor functions, 170–171 Fluid motion formula, 134 Focal chord of ellipse, 977 of hyperbola, 991, 996 of parabola, 1001 Foci applications of, 978–979, 992–993 of ellipse, 974–978 explanation of, 964, 974 of hyperbola, 990–992 of parabola, 1000–1001 Foci formula for ellipse, 977 for hyperbola, 991 Focus-directrix form, 965–966, 969, 999–1002 F-O-I-L method, 207, 325, 506, A–18 Folium of Descartes, 393, 1061 Foot-pounds (ft-lb), 790 Force between charged particles, 186 components of, 790, 791 to maintain equilibrium, 678 normal, 708 resultant, 779 Force vectors, 779–780, 791 Formulas absolute value, 173, 213, 334 alternative form for law of cosines, 768 amount of mortgage payment, 480, A–23 angle between intersecting lines, 620 angle reduction, 679 annuity, 473 arc length, 516, 519–520, 1004 area of circle, A–21 area of ellipse, 981 area of Norman window, 943 area of parallelogram, 620 area of rectangle, 876, 915 area of regular polygon, 660, A–36

area of regular polygon inscribed in circle, 558 area of right parabolic segment, 1004 area of triangle, 381, 668, 876, 904 average rate of change, 254–255 bicycle sales, 489 binomial coefficients, 1147 binomial cubes, 213 binomial probability, 1152 biorhythm theory, 629 body mass index, A–36 Brewster’s law of reflection, 6678 Celsius to Fahrenheit conversion, 46 change-of-base, 452, 498 chemicals in bloodstream, A–62 circumference of circle, 90, 516–517 compound annual growth, 288 compound interest, 471 conic sections, 116 continuously compounded interest, 471–472 cost of removing pollutants, 368 cost to seize illegal drugs, A–62 cube of complex number, 811 density, 307 difference quotient, 259, 281 dimensions of rectangular solid, 863 discriminant of reduced cubic, 393 distance, 9–10, 668, 863 double-angle, 682 electrical resistance, A–9 equation of line in polar form, 1049 equation of line in trigonometric form, 729 equation of semi-hyperbola, 995 equilateral triangles in complex plane, 811 estimating time of death, 466 Euler’s polyhedron, A–36 explanation of, 69 factorial, 1130 Fahrenheit and Celsius temperatures, 850 falling objects, 259, A–78 fine-tuning golf swing, 719 fish length to weight relationship, A–78 fluid motion, 134 focal chord of hyperbola, 996 Folium of Descartes, 393, 1061 force and equilibrium, 678 force between charged particles, 186 force normal to object on inclined plane, 708 games, 1142 general linear equation, 61 gravitational attraction, 160 growth of bacteria, 431 half-angle, 684 haversine function, 660 heat flow on cylindrical pipe, 708 height of equilateral triangle, 524 height of object calculated from distance, 574 height of projected image, 420 height of projectile, 233, 240–241 Heron’s, 766, A–87 human life expectancy, 31 hydrostatics, surface tension, and contact angles, 558 ideal weight for males, 46 illumination of surface, 606 interest earnings, 31 learning curve, 489 lift capacity, A–36 logistics equations and population size, 890 magnitude of vector in three dimensions, 784 Malus’s law, 691 manufacturing cylindrical vents, 1016 medication in bloodstream, A–22 midpoint, 9, 1033 number of daylight hours, 592 orbiting distance North and South of Equator, 629 perimeter of ellipse, 981 perimeter of rectangle, 915 perimeter of trapezoid, 768 period, 553, 560, 569 perpendicular distance from point to line, 967 pH level, 443 Pick’s theorem, 46 population density, 367 position of image reflected from spherical lens, 574 power reduction, 682–683 powers of imaginary unit, 1117


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Index probability, 431 Pythagorean theorem, A–82 Pythagorean triples, 540 quadratic, 222–224 radius, 17, 420 range of projectile, 719 relationship between coefficient B, frequency f, and period P, 592 relationships in right triangle, 524 rewriting y ⫽ a cos x ⫹ b as single function: y ⫽ k sin (x ⫹ ␪), 729 roots tests for quartic polynomials, 351 rotation of axes, 1037 sand dune function, 173 simple interest, 469 sine of angle between sides of triangle, 606 slope, 21–22, 24, 95, 100 spring oscillation, 145 square root of complex numbers, 334 Stirling’s, 1130 student loan payment, 1109 summation, 1094, 1113, 1119, 1165–1166 sum of cubes of n natural numbers, 1109 sum of first n cubes, 1109 sum of first n natural numbers, 1097 sum of infinite geometric series, 1104 sum of n terms of sequence, 1087 supersonic speeds, sound barrier, and Mach numbers, 691 surface area of cylinder, 76, 233, 271, 381, A–51 surface area of rectangular box with square ends, 245 time required to amortize mortgage, 480 time required to double investment, 443 trigonometric graphs, 116 triple angle formula for sine, 756 tunnel clearance, 1016 uniform motion, 849 velocity, 160, 271 vertex/intercept, 245 volume and pressure, A–9 volume of cone, 876 volume of cube, A–21 volume of cylinder, 90, 876, A–51 volume of open box, 318 volume of sphere, 134 Witch of Agnesi, 1061 Four-leaf rose, 1029 45-45-90 triangles, 513, 514, 595, 596 Fractional coefficients, A–25 Fractions compound, 663, A–57 – A–58 partial, 879, 886–888 (See also Partial fraction decomposition) Frequency, 584 Function notation, 39–41, 1112 Functions. See also Polynomial functions; Trigonometric functions; specific types of functions absolute value, 120, 141, 163, 191 of acute angles, 596 algebra of, 262–268, 270, 295 area, 505–506 base, 300 ceiling, 171 composite, 278–279 composition of, 274–280, 284–285 constant, 51 continuous, 163, 167, 169, 1191, 1192 cube root, 121 cubic, 568 discontinuous, 168–169 domain of, 36–39, 42, 95, 262–264 equations of, 152, 238, 265, 269–270 evaluation of, 40–41, 275 even, 106–107 explanation of, 33–34, 95, 189 exponential, 422–429, 496–497 in exponential form, 152, 434 family of, 120–121 floor, 170–171 graphical and numerical views of operations on, 265–267 on graphing calculator, 41, 127, 129, 373 graphs of, 106–108, 127–130, 151–155, 189 growth, 424

hyperbolic cotangent, 506 identity, 51 intervals and increasing or decreasing, 110–111 intervals and positive or negative, 108–110 inverse, 411–415 inverse trigonometric, 695–705, 735–736 linear, 51–52, 1098 logarithmic, 434, 436–438, 497 logistic, 837 maximum and minimum value of, 112–113, 405–407 monotonic, 423 nonlinear, 253–254 objective, 871, 873–874 odd, 107, 108, 546 one-to-one, 410–411, 424, 496 parent, 120, 128 periodic, 545 piecewise-defined, 163–171, 192 power, 152–158, 191, 196, 426 products and quotients of, 263–265 projectile, 246 quadratic, 151, 215–216, 235–243, 251–253, 321, 1098 radical, 304, 405, 1198–1199 range of, 37–38, 42, 95 rational, 148–152, 191, 356–365, 398–399, 1197 reciprocal, 148–151, 1035 relations as, 33–34 root, 152 sand dune, 173 sinusoidal, 632 smooth, 163 square root, 120 step, 170–171 sums and differences of, 262–263 symmetry and, 106–108 toolbox, 120–126, 177–180, 190 transcendental, 433, 723 transformations to graph, 127–130, 235 vertical line test for, 35–36, 95 wrapping, 577–578 zeroes of, 108, 109 Fundamental identities. See also Identities algebra to verify, 655–656 explanation of, 654–655, 662, 733 trigonometric functions and, 657–658 Fundamental principle of counting (FPC), 1122, 1136 Fundamental properties of logarithms explanation of, 446 to solve equations, 447–449 Fundamental property of rational expressions, A–52 – A–53 Fundamental theorem of algebra, 320–323, 397 Galileo Galilei, 152 Games, 1142 Gamma (␥), 510 Gauss, Carl Friedrich, 204, 320 Gaussian elimination, 895 Gauss-Jordan elimination, 897–898, A–91 General form of equation of circle, 12 of linear equations, 63 of quadratic functions, 321 General function, transformations of, 127–130 General linear equations, 61 General matrix, A–92 Geometric sequences applications of, 1105–1106 explanation of, 1098–1099, 1155 finding nth partial sum of, 1103–1104 finding nth term of, 1099–1103 graphs of, 1102 Geometric series applications of, 1107 sum of infinite, 1104–1105 Zeno’s first paradox as, 1160 Geometry analytical, 962 applications of, A–31 – A–34 of determinants, 959 perimeter and area formulas, A–31 – A–33 of vectors, 959 volume, A–33 – A–34

I-5

Global maximum, 112, 238 Global minimum, 238 Golden ratio, 576 Goodness of fit, 81 Graphical solutions intersect method for, 64–66 for linear inequalities, 68–69 for specified variable in literal equations, 69–70 x-intercept/zeroes method for, 66–68, 97 Graphing calculator features factorial option, 1081 fnInt, 1219–1220 intersect, 64–65 split screen viewing, 325 TABLE, 6–8, 373, 441, 750 window size, 892 Graphing calculators absolute value inequalities on, 142 absolute value on, 137, 201 angle measure on, 515, 518–519 area of polygon on, 1171 area under curve on, 1219–1220 asymptotes on, 156, 361, 374 beats on, 739 binomial probability distribution on, 1151 circle on, 13–14, 528, 530, 969 combinations on, 1126 complex numbers on, 207–210 composition of functions on, 277, 278 decomposition on, 884, 888 degrees on, 511–512 eccentricity on, 1070 ellipse on, 977, 978, 1038 equations on, 28 evaluating expressions on, 101–102, A–4 exponential functions on, 427–428 extreme values of function on, 239, 241–243 functions on, 41, 127, 129, 373 Heron’s formula and, 766 horizontal translations on, 124 hyperbola on, 985, 988, 989, 992 imaginary numbers on, 207 interest on, 472 intermediate value theorem on, 324–325 intersection-of-graphs method on, 65–66, 74, 429, 464 inverse functions on, 414, 416–417 inverse trigonometric functions on, 600, 696–699, 701, 704, 705 limits on, 1171–1176, 1180–1181, 1185–1187, 1196 linear inequalities on, 867 logarithms on, 435, 436, 438, 447–450, 459–465 market equilibrium on, 847–848 matrices on, 898–901, 908, 911–913, 919–923, 926–927, 938–941 maximums and minimums on, 112–113, 239, 241–243, 378, 380 nth roots on, 816, 817 parallel lines on, 54 parametric equations on, 1053, 1054, 1056–1059 partial sum on, 1082–1083, 1085 patterns on, 101–102 permutations on, 1124 polar coordinates on, 1026, 1027, 1029–1031 polynomial inequalities on, 386, 387 power functions on, 153 probability on, 1137–1139 projectile motion on, 687 quadratic equations on, 220–221 quadratic functions on, 215–216 radical expressions on, A–70, A–72 rational expressions on, A–53, A–59 rational functions on, 150, 389 rational inequalities on, 389, 390 rational zeroes on, 328, 330 regression on, 85–87, 92, 158, 298–299, 484–487, 626, 627 relations on, 6–8 scatterplots on, 348 sequences on, 1079–1081, 1091–1092, 1094, 1101–1102, 1104–1106, 1160–1161 slope of line tangent on, 1206 system of linear equations on, 838–840, 843–848, 898–899


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Index

Graphing calculators—Cont. system of linear inequalities on, 868–869 system of nonlinear equations on, 1008–1014 transformations on, 122–123 trigonometric equations on, 714, 717, 723–725 trigonometric functions on, 545, 549, 551–554, 561–567, 598–601, 613, 658, 665, 666, 714, 717 variation on, 180, 181, 195 x-intercept on, 215–216 zeroes method on, 67–68 Graphs. See also Trigonometric graphs of arithmetic sequence, 1092 of circle, 9–13, 969, 970 of complex numbers, 802–804 continuous, 5 of cosecant function, 561–562, 637 of cosine function, 549–554, 636–637 of cotangent function, 566–567, 637, 638 of cycloids, 1055–1056 of discontinuous functions, 168–169 of ellipse, 971–974, 976–977 end-behavior of, 111, 338–341 to evaluate limits, 1199–1200 of exponential functions, 423–425 of function and its inverse, 415–417 of function family, 120 of functions, 106–108, 127–130, 151–155, 189, 265–267 of geometric sequences, 1102 of hyperbola, 116, 984–991 of linear equations, 19–20, 50–56, 95 of logarithmic functions, 436–437 maximum and minimum values on, 112–113, 239, 241–243, 405–407 method to read and interpret, 41–42 one-dimensional, 853 orientation of, 1053 of parabola, 235–236, 414, 998–1002 parent, 121 of piecewise-defined functions, 165–170 polar, 1024–1027, 1067, 1073–1074 of polar equations, 1024, 1027, 1029 polynomial, 337–346, 398 of power functions, 153–155 of quadratic functions, 235–238, 300–301, 424 of rational functions, 151–152, 356–365, 372–376, 398–399 of reflections, 125 of relations, 3–6, 48–49 of secant function, 562–563, 637 of semicircle, 6 of sine function, 542–549, 551–554, 636–637 sinusoidal, 580 to solve inequalities, 109–110 stretches and compressions in, 126–127 symmetry and, 345 of system of equations, 839 of system of nonlinear equations, 1008, 1010 of tangent function, 563–566, 587, 637–638 of toolbox functions, 120–121 of transformations, 121, 127–130, 155 two-dimensional, 853 of variation, 178 vertical distance between, 270 writing equations from, 554–555 Gravity, 152, 160, 187 Greatest common factor (GCF), A–38 – A–39 Greatest integer functions, 170 Grid lines, 3 Grouping, factoring by, A–39 – A–40 Growth functions, 424 Growth rate, 475 Growth rate constant, 475 Gunter, Edmund, 600 Half-angle formulas, 684 Half-angle identities, 682–684, 694, 734 Half-life of radioactive substances, 476 Half planes, 865 Harmonic motion explanation of, 584 sound waves and, 585–586 springs and, 584–585

Haversine formula, 660 Haversine function, 660 Height of falling object, 259 formulas for, 259, 524, 574 of projectile, 233, 240–241, 796–797 HerdBurn scale (hb), 503 Heron’s formula explanation of, 766 proof of, A–87 Hexigons, 526 Horizontal asymptotes, 150, 359–361 Horizontal boundary lines, 37–38, 96 Horizontal change, 21 Horizontal component of vector, 773, 775, 788 Horizontal ellipse, 971–972 Horizontal hyperbola, 985, 987, 991 Horizontal lines explanation of, 23–24, 95 of orientation, 601 slope of, 24 Horizontal line test, 410 Horizontal reflections, 125–126 Horizontal shifts, 123–124, 581–583 Horizontal translations, 123–124 Horizontal unit vectors, 778 Human life expectancy, 31 Hyperbolas applications of, 992–993 branches of, 984, 986 central, 985, A–94 definition of, 990–991 equation of, 984–987, 995, 1036, 1065 focal chord of, 991, 996 foci of, 990–992 on graphing calculator, 985, 988, 989, 992 graphs of, 116, 984–991 horizontal, 985, 987, 991 with rational/irrational values of a and b, 1071 rotated, 1036 semi-hyperbola, 995 standard form of equation of, 986–988 vertical, 988 Hyperbolic cotangent, 506 Hypocycloid with four cusps, 1056 Hypotenuse, 595, A–74 Identities. See also specific identities additive, A–5, A–80 applications of, 686–688 cofunction, 673 connections and relationships in, 693–694 double-angle, 680–682, 694, 734 due to symmetry, 661–662 equations that are not, 665–666 explanation of, 654, A–26 families of, 655 fundamental, 654–658, 662, 733 half-angle, 682–684, 694, 734 multiplicative, A–5, A–80 power reduction, 682–683, 694 product-to-sum, 685–686, 694, 734 Pythagorean, 654, 662, 693, 694, 765 ratio, 654, 662, 663, 693, 694 reciprocal, 654, 662, 693, 694 to solve trigonometric equations, 723 sum and difference, 669–675, 734 sum-to-product, 686 verification of, 655–656, 662–664, 674–675, 733 Identity functions, 51 Identity matrices, 917–919 Imaginary numbers calculations with, 207 explanation of, 204 on graphing calculator, 207 historical background of, 213 identifying and simplifying, 204–205, 209 Imaginary unit, 204, 209, 1117 Impedance, 809 Implied domain, 38–39 Inclusion, of endpoint, 36 Inconsistent system of linear equations in three variables, 858–860 of linear equations in two variables, 844–845 matrices and, 899–900

Increasing functions, 110–111 Independent system, 844–845 Independent variables, 2 Index of refraction, 720 Index of summation, 1082 Induction. See Mathematical induction Induction hypothesis, 1114 Inductive reactance, 808 Inequalities. See also Linear inequalities absolute value, 139–143, 191 additive property of, A–26 – A–27 applications of, 390–391, A–31 compound, A–28– A–30 constraint, 872 on graphing calculator, 142, 386, 387, 389, 390 graphs to solve, 109–110 interval test method to solve, 226–227 joint, A–30 linear in two variables, 865–867 method to solve, A–26 – A–28 multiplicative property of, A–26 – A–27 polynomial, 385–388, 400 push principle to solve, 402–403 quadratic, 225–228, 294 rational, 387–390 system of linear, 867–870 system on nonlinear, 1012–1013 zeroes and, 401–402 Inequality symbol, 139, 141 Infinite geometric series, 1104–1105 Infinite sequences, 1078 Infinity, limits at, 1171, 1195–1199 Infinity symbol, 37, 1083, 1197 Initial points, 771, 772 Initial side, 514 Input values, 39, A–3 Instantaneous rates of change, 1073–1074, 1204–1209, 1217–1218 Instantaneous velocity, 1204–1207 Integers, consecutive, 72 Intercept-intercept form of linear equations, 61 Intercept method explanation of, 20, 50, 95 to graph hyperbola, 984–985 Interest applications of, 469–472, 499 compound, 470–471 continuously compounded, 471–472 on graphing calculator, 472 simple, 185, 469 Interest earnings, 31 Interest rate r, 469 Intermediate value theorem (IVT) explanation of, 323–325, 397 on graphing calculator, 324–325 Interpolation, 486 Intersection-of-graphs method on graphing calculator, 65–66, 74, 429, 464 to solve equations, 64–67, 97, 217 to solve linear inequalities, 68–69 Intersection of sets, A–28 Interval notation, 36 Intervals where function is increasing or decreasing, 110–111 where function is positive or negative, 108–110 Intervals of convergence, 201 Interval tests to solve polynomial and rational inequalities, 387–390 to solve quadratic inequalities, 226–227 Invariants, 1040 Inverse additive, A–5, A–80 graphs of function and its, 415–417 of matrices, 919–921, 930, 955–956 multiplicative, A–5, A–80 Inverse functions algebraic method to find, 412–415 applications of, 417 explanation of, 411–412, 416 on graphing calculator, 414, 416–417 graphs of, 415–417 method to find and verify, 415


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Index Inverse trigonometric functions applications of, 704–705 cosecant, 702–704 cosine, 698–699, 704 cotangent, 702, 703 facts about, 735–736 on graphing calculator, 600 principal roots and, 712 secant, 702–704 sine, 695–698, 704 summary of, 704 tangent, 699–700, 704 used to evaluate compositions, 700–702 Inverse variation, 180–181, 193 Involute of circle, 621 Irrational numbers, 425 Irrational zeroes, 355 Irreducible quadratic factors, 322 Joint inequality, A–30 Joint variation explanation of, 182, 193 on graphing calculator, 195 Jump discontinuity, 1192 Latitude, 519 Law of cosines alternative form for, 768 applications of, 762–763, A–88 explanation of, 759–760, 823 scaled drawings and, 786–787 side-angle-side triangles and, 759–761 side-side-side triangles and, 761–762 Law of sines ambiguous case of, 748–752, 761 applications of, 752–753 explanation of, 746–747, 822 scaled drawings and, 786–787 unique solutions and, 747–748 Leading coefficients, A–40 – 41 Learning curve, 489 Left-hand limits, 1174 Leibniz, Gottfried von, 1180 Lemniscates, A–95 Leonardo of Pisa (Fibonacci), 1080 Lift capacity formula, A–36 Limaçons, 1030, A–95 Limiting value, 1170 Limit notation, 1170, 1171 Limits absolute value inequalities and, 201–202 applications of, 1203–1212, 1217–1218 approximations vs., 1170–1171 area under curve and, 1210–1212 continuity and, 1191, 1192, 1217 definition of, 1172 of difference quotient, 1204–1209 existence of, 1171–1172, 1175–1176 on graphing calculator, 1171–1176, 1180–1181, 1185–1187, 1196 historical background of, 1169 at infinity, 1171, 1195–1199 instantaneous rates of change and, 1204–1209, 1217–1218 one-sided, 1173–1174, 1200 properties of, 1180–1182, 1216–1217 two-sided, 1171–1173 use of algebra and limit properties to find, 1193–1195 use of direct substitution to find, 1191–1193 use of graphs to evaluate, 1171–1173, 1199–1200, 1216 use of limit properties to find, 1183–1186 use of tables to evaluate, 1171–1173, 1187, 1216 Linear association, 80–81, 98 Linear combination, 778 Linear dependence, 854 Linear depreciation, 57, 58 Linear equations. See also Equations; System of linear equations applications of, 27–28, 56–58 explanation of, 19, 95 forms of, 100–101 in function form, 51 general, 61 on graphing calculator, 28

graphs of, 19–20, 50–56, 95 horizontal and vertical lines and, 23–25 identities and contradictions as, A–26 – A–28 intercept-intercept form of, 61 intercept method to graph, 20, 50 methods to solve, 50–51, A–81, A–A–24 – A–25 in one variable, A–24 of parallel and perpendicular lines, 25–27, 54–55 in point-slope form, 55–56 properties of equality to solve, A–24 – A–25 in slope-intercept form, 51–55 slope of line and rates of change and, 20–23 standard form of, 100 in two variables, 19–20 Linear factorization theorem explanation of, 321, 322, 397 proof of, A–85 Linear function models, 83–85, 98 Linear functions, 51–52, 1098. See also Functions Linear inequalities. See also Inequalities; System of linear inequalities explanation of, 865 on graphing calculator, 867, 868–869 intersection-of-graphs method to solve, 68–69, 74 method to solve, 866–867, A–26 – A–28, A–81 system of, 867–870 in two variables, 865–867 Linear programming explanation of, 870, 949 solutions to problems in, 870–874 Linear regression on graphing calculator, 298 line of best fit and, 85–87 Linear systems. See System of linear equations in three variables; System of linear equations in two variables Linear velocity, 520–521 Line of best fit explanation of, 85 linear regression and, 85–87 Lines characteristics of, 101 coincident, 843 distance between points and, 963 equations of, 52–55, 729, 1049 horizontal, 23–24, 101 horizontal boundary, 37–38, 96 parallel, 25–26, 53–55, 101 perpendicular, 26–27, 53–55, 101 relationships between, 101 secant, 54, 97, 576 slope-intercept form and graph of, 52–53, 96 tangent, 103, 576, 1073–1074, 1204, 1206 vertical, 23–24, 101 vertical boundary, 36–37, 96 Line segments directed, 771 midpoint of, 8–9 Lissajous figure, 1052 Literal equations, 69–70 Local maximum, 112 Logarithmic equations explanation of, 499 forms of, 462 on graphing calculator, 459–463 methods to solve, 447, 457–460 solutions to system of, 1011 Logarithmic form, 434, 435, 449 Logarithmic functions domain of, 437–438 explanation of, 434, 497 graphs of, 436–437 Logarithmic scales, 438, 439 Logarithms applications of, 438–442, 452–453, 463 change-of-base formula and, 452 common, 434 explanation of, 497 fundamental properties of, 446–449 on graphing calculator, 435, 436, 438, 442, 447–450, 459–465 natural, 434 product, quotient, and power properties of, 449–451, A–86

properties of, 456, 457, 498, 505 uniqueness property of, 458–459 Logistic equations explanation of, 463 on graphing calculator, 463, 464 investigation of, 502 method to solve, 463–464 regression models and, 485–486 use of, 890 Logistic functions, 837 Logistic growth, 464, 466, 485 Logistic growth model, 485 Long division, 308–309 Longitude, 519 Lower bound, 330 Lowest common denominator (LCD), A–55 – A–59 Magnitude of vector, 774–775, 784 of velocity, 771 Malus’s law, 691 Mapping notation, 2, 3 Market equilibrium, 847–848 Mathematical induction applied to sums, 1112–1115 explanation of, 1156 general principle of, 1115–1116 Mathematical models, translating written or verbal information into, A–2 Matrices addition and subtraction of, 907–908 algebra of, 905–913, 951 applications of, 900–901, 938–939, 952 associated minor, 923 augmented, 893–895, 955–956 coefficient, 893–894 of constants, 893–894 decomposing rational expressions using, 937 determinants of, 922–926, A–92 to encrypt messages, 940–942 entries of, 893 equality of, 905–906 explanation of, 893 on graphing calculator, 898–901, 908, 911–913, 919–923, 926–927, 938–941 identity, 917–919 inconsistent and dependent systems and, 899–901 inverse of, 919–921, 930 multiplication and, 908–913 reduced row-echelon form of, 897, 898, A–91 singular, 922–927 to solve systems, 895–898 square, 893, 895 Matrix equations, 921–922, 926–927, 951 Matrix inverses, 955–956 Maximum values explanation of, 112 of functions, 112, 238–243 on graphing calculator, 112–113, 239, 241–243, 645, 646 method to locate, 405–407, 744 Midinterval points, 344 Midpoint formula, 9, 962, 1033, 1064 Midpoint of line segment, 8–9, 94 Minimum values explanation of, 112 of functions, 112 on graphing calculator, 112–113, 378, 380, 645, 646 method to locate, 405–407, 744 Minors, 923–924. See also Associated minor matrices Minutes, 511 Mixture problems, 73–74, 844–845 Modeling/models exponential growth, 422 harmonic, 584–586 linear function, 83–85 piecewise-defined functions, 170 polynomial, 347–348 rational expressions and, A–64 rational functions, 365 regression, 250, 483–386 root and rational function, 1207–1209 sinusoidal, 559, 622–627

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Index

Modeling/models—Cont. step function, 171 system of linear equations, 844–848 trigonometric equations, 622–627, 642 Modulus, 213 Monomials, A–15, A–17. See also Polynomials Monotonic functions, 423 Motion projectile, 686–687, 796–797, 828 simple harmonic, 584–586 tidal, 645–646 uniform, 72–73, 845–846, 849 Multiplication. See also Product associative property of, A–4, A–80 commutative property of, A–4, A–80 of complex numbers, 207–209 identity matrices and, 917–919 of polynomials, 207 of radical expressions, A–69 – 70 of rational expressions, A–54, A–55, A–59 scalar, 772, 908–913 using exponential properties, A–10 Multiplicative identity, A–5, A–80 Multiplicative inverse, A–5, A–80 Multiplicative property of absolute value, 138 Multiplicative property of equality, A–24 Multiplicative property of inequality, A–26 – A–27 Multiplicity even, 322 odd, 322 rational functions and, 358–359 zeroes of, 225, 322, 341–344 Mutually exclusive events, 1138 Nappe, 964 Natural exponential functions, 426. See also Base-e exponential functions Natural logarithms explanation of, 434 method to find, 435–436 Negative angles, 514 Negative association, 79–80, 98 Negative functions, 108–109 Negative numbers, A–12 Negative slope, 22 Newton, Isaac, 1180 Newton-meters (N-m), 790 Newton’s law of cooling, 428–429 Newton’s law of universal gravitation, 187 Nondistinguishable permutations, 1125 Nonexclusive events, 1137–1139 Noninvertible matrix. See Singular matrix Nonlinear association, 80–81, 98 Nonlinear asymptotes, 374, 375 Nonlinear functions, 253–254 Nonlinear system of equations. See System of nonlinear equations Nonlinear system of inequalities, 1012–1013 Nonremovable discontinuity, 1192 Nonrigid transformations, 128 Normal, 103 Normal force, 708 Notation/symbols angle, 510 composition of functions, 275 decimal, A–14 – 15 degree, 510, 511, 518 delta (⌬), 21 factorial, 1080–1081 function, 39–41, 1112 inequality, 139, 141 infinity, 37, 1083, 1196 intersection, A–28 interval, 36 limit, 1170, 1171 mapping, 2, 3 overview of, A–80 ␲, 425, 695 probability, 1134 radical, A–66 rate of change, 21, 177 scientific, A–14 – A–15 sequence, 1078 set, 36, 37

subscript, 1112 summation, 1082–1084, 1165 union, A–28 vector, 771–772 nth roots on graphing calculator, 816, 817 method to find, 816–818, A–65 nth roots theorem, 815–816 nth term of arithmetic sequence, 32, 1090–1093 of geometric sequence, 1099–1103 of sequence, 1078, 1987 Number line, 3, 36 Numbers. See also Complex numbers imaginary, 204–205 irrational, 425 perfect, 1144 rational, A–52 – A–59 real, A–4 – A–6 sets of, A–80 Numerical coefficients, A–1. See also Coefficients Objective function, 871, 873–874 Objective variables, 69–70, 872 Oblique asymptotes, 374–376 Oblique triangles, 746–752, 822. See also Triangles Obtuse angles, 510 Odd functions, 107, 108, 546 Odd multiplicity, 322 One-dimensional graphs, 853 One-sided limits, 1173–1174, 1200 One-to-one functions explanation of, 410, 424, 496 identification of, 410–411 inverse of, 695 restricting domain to create, 413–414 Optimum angle, 828 Order, matrix, 906 Ordered pairs, 2, 5, 64, 94 Ordered triples, 853, 855 Orientation, of graphs, 1053 Origin, symmetry to, 107 Orthogonal components of vectors, 788, 794–795 Oscillation, spring, 145 Output values, 39, A–3 Parabolas. See also Quadratic functions analytic, 997–1002, 1066 applications of, 1002 definition of, 1000 equation of, 965, 999–1002 focal chord of, 1001 focus-directrix form of equation of, 999–1002 graphs of, 235, 236, 414, 998–1002 horizontal, 998–999 with horizontal axis, 997–1002 vertex of, 5 vertical, 5, 1000 Parallel lines equations for, 54–55 explanation of, 25–26, 101 on graphing calculator, 54 slope of, 25, 53–55 Parallelogram method, 781, 802 Parallelograms area of, 620 proof of determinant formula for area of, A–86 – A–87 Parameter, 844, 1052 Parametric curves, 1051–1054 Parametric equations applications of, 1057–1059, 1063 explanation of, 393, 1068 on graphing calculator, 1053, 1054, 1056–1059 in rectangular form, 1054–1055 Parent function, 120, 128 Parent graph, 121 Pareto, Vilfredo, 455 Pareto’s law, 455 Partial fraction decomposition. See also Decomposition convenient values used for, 882–885 explanation of, 837, 879–882, 950, 958 matrices and, 937 system of equations used for, 885–886

Partial fractions explanation of, 879 telescoping sums and, 886–888 Partial sum of arithmetic sequence, 1093–1094 explanation of, 1082 of geometric sequence, 1103–1104 on graphing calculator, 1082, 1083 method to compute, 1082–1083, 1085, 1094 Pascal’s triangle, 1145–1147 Patterns investigation of, 101–102 Pascal’s triangle and, 1145–1146 Perfect cubes, A–65 Perfect numbers, 1144 Perfect square trinomials explanation of, 219, A–19 factoring, A–42 – A–43 Perihelion, 1044, 1045 Perimeter of circle, A–32 of ellipse, 981 explanation of, A–31 formulas for, A–32 of polygon, 575 of rectangle, 915, A–32 of square, A–32 of trapezoid, 731, 768, A–32 Period formula explanation of, 560 for sine and cosine, 553 for tangent and cotangent, 569 Periodic functions, 545 Periods of cosecant and secant functions, 561 of sine and cosine functions, 546, 552–554 of tangent and cotangent functions, 569, 570 Permutations distinguishable, 1123–1124 on graphing calculator, 1124 nondistinguishable, 1125 Perpendicular distance, 963, 967 Perpendicular lines equations for, 54–55 explanation of, 26–27, 101 slope of, 26, 53–55 Phase angle, 582, 808, 809 pH level, 443, 452, 453 pH scale, 452 Pick’s theorem, 46 Piecewise-defined functions applications of, 170–171 continuous, 163, 167, 169 discontinuous, 168–169 domain of, 164–165 equation of, 165, 169–170 explanation of, 163, 192 on graphing calculator, 166 graphs of, 165–170 step functions as, 170–171 ␲, 425, 695 Placeholder substitution, A–44 Plane complex, 802, 811 half, 865 in space, 853 Point of inflection, 38 Points corner, 868 critical, 622–626, 642 distance between lines and, 963 quadrantal, 527 trigonometric ratios and, 610–614 on unit circle, 527–528 Point-slope form explanation of, 55, 96 linear equations in, 55–56, 100 Point-wise defined relations, 2, 3, 94 Polar axis, 1019 Polar-axis symmetry, 1028 Polar coordinates converting between rectangular coordinates and, 1021–1024 distance formula in, 1033


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Index explanation of, 1018, 1067 expressing points in, 1020–1021 on graphing calculator, 1026, 1027, 1029–1031 midpoint formula in, 1033 plotting points using, 1018–1020 Polar coordinate system, 1019 Polar equations explanation of, 1067 graphs of, 1024, 1027, 1029 of planetary orbit, 1045 symmetry for graphs of certain, 1028 systems of, 1074–1075 Polar form applications of conics in, 1044–1046 conic equations in, 1042–1044 converted to rectangular form, 652 of ellipse, 1049 of equation of line, 1049 use of, 1018 Polar graphs explanation of, 1067, A–95 instantaneous rates of change and, 1073–1074 r-value analysis and, 1024–1027 symmetry in, 1027–1031, 1035 Polar symmetry, 1028 Pole, 1019 Polygons area of, 558, 660, 1170–1171 perimeter of, 575 Polynomial equations with complex coefficients, 234 complex coefficients solve, 234 de Moivre’s theorem to check solutions to, 814–815 explanation of, A–45 factoring to solve, A–47 Polynomial form, 988 Polynomial functions applying difference quotient to, 303 complex, 320–321 dominant term of, 339 graphs of, 337–346, 398 zeroes of, 320–331, 397 Polynomial graphs end-behavior of, 338–341 guidelines for, 345–346 identification of, 337–338 symmetry in, 345 turning points and, 337, 338 zeroes of multiplicity and, 341–344 Polynomial inequalities on graphing calculator, 386, 387 interval tests to solve, 387–388 method to solve, 385–387, 400 push principle to solve, 402–403 Polynomial modeling, 347–348 Polynomials addition and subtraction of, A–16 – A–17 applications of, 347–348 characteristic, 946 degree of, A–15 division of, 308–315, 327 explanation of, 422, A–15, A–81 families of, A–15 identifying and classifying, A–15 – A–16 linear factorization theorem and, 322–323 prime, A–41 as product of linear factors, 321 products of, A–17 – A–20 quadratic, A–40 – A–42 quartic, 351 rational zeroes of, 326–330 in standard form, A–16 types of, A–15 use of remainder theorem to evaluate, 312–313 Population density, 367 Position vectors, 772, 773 Positive angles, 514 Positive association, 79–80, 98 Positive functions, 108–109 Positive slope, 22 Power functions domain of, 153–154 explanation of, 152, 191, 426 transformations of, 154–155 variation and, 196

Power property, product to, A–11 Power property of equality, A–71, A–73 Power property of exponents, A–10 – A–11 Power property of logarithms, 449–451, 498, A–86 Power reduction formula, 682–683 Power reduction identities, 682–683, 694 Pressure, A–9 Pressure wave, 585–586 Prime polynomials, A–41 Principal p, 469 Principal roots explanation of, 711 inverse functions and, 712 trigonometric equations and, 711–713 Principal square roots, 204, 334 Probability binomial, 1150–1152 elementary, 1132–1133 of events, 1133, 1134 explanation of, 1132, 1157 on graphing calculator, 1137–1139 of nonexclusive events, 1137–1139 properties of, 1134–1136 quick-counting and, 1136–1137 Problem solving. See also Applications; Applications index guide to, 71 trigonometry and, 832–833 Product property of exponents, A–10 – A–11, A–14 Product property of logarithms, 449–451, 498, A–86 Product property of radicals, A–67 Products. See also Multiplication of complex conjugates, 208 of complex numbers, 805–806 of functions, 263–264 of polynomials, A–17 – A–20 in trigonometric form, 806–807 Product to power property of exponents, A–11, A–14 Product-to-sum identities, 685–686, 694, 734 Projectile function, 246 Projectile motion, 86–687, 796–797, 828 Projectiles explanation of, 1057 height of, 233, 240–241, 796–797 range of, 719, 799, 828 Projectile velocity, 255 Proofs declarative statements vs., 1186–1187 by induction, 1112–1115, 1156 list of, A–84 – A–88 Properties of equality, to solve linear equations, A–24 – A–25 Property of negative exponents, A–12 Protractors, 510 Push principle, 402–403 Pythagorean identities explanation of, 654, 662, 693, 694, A–74 use of, 765, 774, A–75 Pythagorean theorem applications of, 9, 598, 759, 760, 778 explanation of, A–82 Pythagorean triples, 540 Quadrantal angles, 515, 613 Quadrantal points, 527 Quadrant analysis, 616, 657–658 Quadrants, 3 Quadratic equations checking solutions to, 249–250 completing the square to solve, 219–221, 235 explanation of, 214–215, 293, A–45 – A–46 on graphing calculator, 220–221 quadratic formula to solve, 222–224 square root property of equality to solve, 217–219 standard form of, 215, 222 Quadratic factors decomposition template for, 881–882 irreducible, 322 Quadratic form equations in, A–74 explanation of, A–44 factoring, A–42 – A–45 Quadratic formula applications of, 229–230 discriminant of, 223–224

I-9

explanation of, 222 to solve quadratic equations, 222–224 Quadratic functions. See also Parabolas completing the square to graph, 235–237 end-behavior of, 338 equation of, 238 explanation of, 215, 235, 315, 1098 extreme values and, 238–243 general form of, 321 on graphing calculator, 215–216 graphs of, 235–238, 301, 424 reciprocal, 151 vertex formula to graph, 237–238 zeroes of, 215–217 Quadratic graphs, 300–301 Quadratic inequalities applications of, 229–230 domain and, 227–228 explanation of, 225, 293 interval test method for, 226–227 method to solve, 225–226 Quadratic models, 250–253, 294 Quadratic polynomials explanation of, A–40 factoring, A–40 – A–42 Quadratic regression, 250–251, 299 Quartic polynomials, 351 Quick-counting techniques, 1136–1137 Quotient property of exponents, A–12, A–14 Quotient property of logarithms, 449–451, 498, A–86 Quotient property of radicals, A–67 Quotients of complex numbers, 805–806 difference, 259, 281–286, 303–304, 1204–1209 of functions, 263–265 in trigonometric form, 806–807 Quotient to power property of exponents, A–11, A–14 Radians arc length in, 516 conversion between degrees and, 517–520 explanation of, 516 Radical equations explanation of, A–71 methods to solve, A–72 – A–75 power property of equality and, A–71 Radical expressions. See also Cube roots; Square roots addition and subtraction of, A–69 explanation of, A–64, A–82 on graphing calculator, A–70, A–72 methods to simplify, A–64 – 65, A–67–68 multiplication and division of, A–69 – A–71 rational exponents and, A–66 – A–67 Radical functions applying difference quotient to, 304 graphs of, 405 limits at infinity for, 304, 405, 1198–1199 Radical properties, A–67 Radical symbol, A–66 Radioactive elements, 476, 477 Radius of circle, 10, 12, 756 of sphere, 420 Range of functions, 37–38, 42, 95 of projectile, 719, 799, 828 of relations, 2, 48–49 Rate of change applications of, 255–256 average, 253–256, 281–283, 295 difference quotient and, 259, 281–283, 303–304 limits and instantaneous, 1204–1209, 1217–1218 nonlinear functions and, 253–254 notation for, 21, 177 slope as, 21–23, 54, 56 slope-intercept form and, 53 Rate of growth, 1, 837 Ratio identities, 654, 662, 663, 693, 694 Rational equations, A–58 – A–59 Rational exponents explanation of, A–66 simplifying expressions with, A–67 solving equations with, A–73


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Index

Rational expressions addition and substraction of, A–55 – A–57 decomposition of, 879 explanation of, 879, A–52, A–81 fundamental property of, A–52 – A–54 on graphing calculator, A–53, A–59 multiplication and division of, A–54 – A–55, A–59 simplifying compound fractions and, A–57 – A–58 Rational functions application of, 155–158, 365, 377–380 end-behavior of, 149, 150 equation of, 152, 364 explanation of, 148, 191, 356, 398 on graphing calculator, 150, 358, 361, 365 graphs of, 151–152, 356–365, 372–376, 398–399 horizontal asymptotes of, 150, 359–361 limits at infinity for, 1197 multiplicities and, 358–359 with oblique or nonlinear asymptotes, 374–376 reciprocal function and, 148–149 removable discontinuities and, 169, 372–373 vertical asymptotes of, 356–359, 363 Rational inequalities analysis to solve, 388–389 end-behavior to analyze,400, 389 on graphing calculator, 389, 390 interval test method to solve, 387–390, 400 Rationalizing the denominator, A–70 – A–71 Rational zeroes theorem, 325–330, 355, 397 Ratios, trigonometric, 595–597, 610–614 Raw data, 79, 98 Ray, 510, 514 Real numbers properties of, A–4 – A–6, A–80 trigonometry of, 534–537, 577–578 Real roots, 714–717 Real zeroes, 355 Reciprocal function application of, 181 explanation of, 148, 149, 1035 transformations of, 151 trigonometric, 533 Reciprocal identities, 654, 662, 693, 694 Reciprocal quadratic functions, 151 Reciprocals, 26 Reciprocal square functions, 148–150 Rectangles area of, 876 central, 986 explanation of, A–32 perimeter and area formulas for, A–32 reference, 547, 548, 550–551, 554, 562 Rectangular coordinates converted to polar form, 1020–1021 converted to trigonometric form, 651–652 converting between polar coordinates and, 1021–1024 Rectangular coordinate system explanation of, 3 vectors and, 772–775 Rectangular form complex numbers in, 803–804 converted to trigonometric form, 651, 803–804 converting from polar to, 652 converting from trigonometric form to, 804–805 explanation of, 803 parametric equations in, 1054–1055 Rectangular method, 1165 Rectangular solid, 863 Recursive sequences, 1080–1081 Reduced row-echelon form, 897, 898, A–91 Reference angles explanation of, 531–532, 635, 646 trigonometric functions and, 614–616 Reference arc, 534 Reference intensity, 439 Reference rectangles explanation of, 547 to graph trigonometric functions, 548, 550–551, 554, 562 Reflections horizontal, 125–126 vertical, 124–125

Regression applications of, 86–87, 486–487 forms of, 482 on graphing calculator, 85–87, 92, 158, 298–299, 484–487, 626, 627 linear, 85–86 quadratic, 250–251, 299 sinusoidal, 626–627 Regression equations, 500 Regression line, 85 Regression models exponential, 483–484 logarithmic, 484–485 logistic equations and, 485–486 selection of, 250, 500 sinusoidal, 627 Regular polygon, 558, 660, A–36 Relations domain of, 2, 48–49 in equation form, 3, 94 explanation of, 2, 94 on graphing calculator, 6–8 graphs of, 3–6, 48–49 pointwise-defined, 2, 3, 94 range of, 2, 48–49 Remainder theorem applications of, 316, 331 to evaluate polynomials, 312–313 explanation of, 312, 396 proof of, A–84 Removable discontinuity explanation of, 372, 1192 rational functions and, 169, 372–373, 399 Repeated zeroes, 401–402 Residuals, 298, 482 Resistance, 808, 809, A–9 Resultant force, 779 Resultant vectors, 776, 802 Right angles, 510 Right-hand limits, 1174 Right parabolic segment, 1004 Right triangles applications of, 601–604 explanation of, 524, A–74 to find function values, 597 relationships in, 524, 650 solutions to, 597–600 solved given one angle and one side, 597–599 solved given two sides, 599–600 trigonometry of, 595–601, 640 Rigid transformations, 128 Rise, 21 Root functions explanation of, 152 transformations of, 154–155 Roots. See also Cube roots; Square roots cube, A–65 of equations, A–24 extraneous, 457, A–59 principal, 711 real, 714–717 substitution to check complex, 208, 209 trigonometric equations and, 711–715 Roots tests for quartic polynomials, 351 Roses, A–95 Rotated conics, 1035–1040 Rotation of axes, 1037–1039, 1041 Rotation of axes formulas, 1037 Rule of fourths, 547, 551, 579, 580 r-value analysis, 1024–1027 Sample outcomes, 1121, 1134 Sample space, 1121–1122, 1133 Sand dune function, 173 Scalar, 771 Scalar measurement, 771 Scalar multiplication, 772, 908–913 Scalar quantities, 771 Scaled drawings law of cosines and, 786–787 law of sines and, 750–751, 786–787 Scatterplots correlation and, 81–82 explanation of, 79, 98

on graphing calculator, 348 linear/nonlinear association and, 80–81 positive/negative association and, 79–80 Scientific notation, exponents and, A–14 – A–15 Secant function characteristics of, 563 explanation of, 532 graphs of, 562–563, 637 inverse, 702–704 Secant lines explanation of, 54, 97 slope of, 576 Seconds, 511 Seed elements, 1080 Semicircle, graph of, 6 Semi-hyperbola, 995 Semimajor axis, 971 Semiminor axis, 971 Sequences. See also Series; specific types of sequences applications of, 1084–1085, 1095, 1105–1106 arithmetic, 32, 1089–1095, 1155 explanation of, 1078, 1154 Fibonacci, 1080 finding terms of, 1078–1080 finite, 1078 geometric, 1098–1107, 1155 on graphing calculator, 1079–1081, 1091–1092, 1094, 1101–1102, 1104–1106, 1160–1161 infinite, 1078 nth term of, 1078, 1087 recursive, 1080–1081 summation notation and, 1082–1084 terms of, 1078 Series. See also Sequences explanation of, 1078, 1081, 1154 finite, 1081–1082 summation notation and, 1082–1084 Set notation, 36, 37, A–80 Sets intersection of, A–28 of numbers, A–80 Shifted form, of trigonometric functions, 582, 583 Shifts horizontal, 123–124, 581–583 vertical, 122–123, 579 Side-angle-side (SAS) triangles, 747, 759–761, 763–764 Side-side-angle (SSA) triangles, 747–752 Side-side-side (SSS) triangles, 747, 761–762 Sigma notation. See Summation notation Sign analysis, 616 Similar triangles, 512, 513 Simple interest, 185, 469 Simplest form radical expressions in, A–70, A–71 rational expressions in, A–52 – A–54 Sine law of, 746–753 sum and difference identities for, 672–675 table of values for, 542, 543 triple angle formula for, 756 Sine function amplitude of, 551–552 applications of, 579–580, 585, 586 characteristics of, 546 explanation of, 532 graphs of, 542–549, 551–554, 636–637 inverse, 695–698, 704 period and coefficient B of, 552–554 Singular matrix, 922–927 Sinking funds, 474 Sinusoidal equations, 582–583 Sinusoidal functions, 632, 642 Sinusoidal graphs, 580 Sinusoidal models, 559, 622–626 Sinusoidal pattern, 578 Sinusoidal regression, 626–627 Slope applications of, 20 explanation of, 20–21, 95 of horizontal and vertical lines, 24 of parallel lines, 25 of perpendicular lines, 26


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Index positive and negative, 22 as rate of change, 21–23, 54, 56 of secant lines, 54 of tangent lines, 103, 1073–1074, 1205 Slope formula explanation of, 21–22, 100 as rate of change, 22 use of, 24, 95 Slope-intercept form explanation of, 51–52, 96, 100 graph of line using, 52–53, 96 rate of change and, 53 system of equations and, 839 Smooth functions, 163 Snell’s law, 720 Solution region, 866, 870–871 Solutions exact or closed form, 219 explanation of, A–24 unique, 854 Sound energy, 585 Sound waves, 585–586, 687–688, 739 Spheres, volume of, 124 Spiral of Archimedes, 621, 1034 Spring oscillation, 145 Springs, 584–585 Square matrix, 893, 895 Square root function, 120 Square root property of equality, 217 to solve equations, 217–219 Square roots. See also Radical expressions of complex numbers, 334 explanation of, A–64 principal, 204, 334 simplification of, A–64 – A–65 Squares. See also Completing the square binomial, A–19– A–20 difference of two, A–19, A–42 explanation of, A–32 perimeter and area formulas for, A–32 Standard angles, 646, 647, 1040 Standard form of complex numbers, 205–206 of equation of circle, 10, 11, 970 of equation of ellipse, 971–973, 976 of equation of hyperbola, 986–988 of equations, 841 of linear equations, 100 of polynomial equations, A–45 of polynomial expressions, A–16 of quadratic equations, 215, 222 of trigonometric functions, 582, 583 Static equilibrium, 828–829 Static trigonometry, 610 Statistics, 1132 Step functions, 170–171 Stevens, Stanley, 501 Stevens’ law, 501 Stirling’s formula, 1130 Stretches, vertical, 126–127 Subscript notation, 1112 Subsets, A–80 Substitution applying power property after, A–11 back-substitution, A–25 to check complex roots, 208, 209 direct, 1192–1193 explanation of, 840 placeholder, A–44 to solve system of equations, 840–841 to solve system of nonlinear equations, 1008–1010 trigonometric, 743 u-substitution, 716, 727, A–74 Subtends, 515–516 Subtraction of complex numbers, 206–207 of matrices, 907–908 of polynomials, A–16 – A–17 of radical expressions, A–69 of rational expressions, A–55 – A–57 of vectors, 776, 777 Sum and difference identities background of, 680 for cosine, 669–672

explanation of, 669, 694, 734 for sine and tangent, 672–675 to verify other identities, 674–675 Summation applications of, 1119–1120, 1165–1167 index of, 1082 properties of, 1083–1084, 1119, 1165, 1212 Summation formulas, 1094, 1113, 1119, 1165 Summation notation, 1082–1084, 1165 Sum-to-product identities, 686 Supplementary angles, 510, 511 Surface area. See also Area of cylinder, 76, 233, 271, 381, A–51 of rectangular box, 245 Symbols. See Notation/symbols Symmetry axis of, 121 even functions and, 106–107 to find function values, 543–544 graphs and, 345 identities due to, 661–662 to locate points on unit circle, 528, 529 odd functions and, 107, 108 in polar graphs, 1027–1031, 1035 in polynomial functions, 345 with respect to origin, 107 with respect to y-axis, 106, 107 unit circle and, 528, 529, 531–532, 661 Synthetic division explanation of, 309–312, 396–397, A–89 – A–90 factor theorem and, 313–315, 397 polynomials and, 309–315, 327 remainder theorem and, 312–313, 396 System of linear equations Cramer’s rule to solve, 934–936 decomposition using, 885–886 determinants to solve, 933–934 explanation of, 838 graphs to solve, 839 matrices to solve, 895–901, 950 matrix equations to solve, 921–922, 951 System of linear equations in three variables applications to, 860–861 Cramer’s rule and, 935–936 elimination to solve, 855–858 explanation of, 853, 948–949 inconsistent and dependent, 858–860 solutions to, 854–855 visualizing solutions to, 853–854 System of linear equations in two variables Cramer’s rule and, 934–935 dependent, 844 elimination to solve, 841–843 equivalent, 841 explanation of, 838, 948 on graphing calculator, 838–840, 843–848 graphs to solve, 839 inconsistent and dependent, 843–844 independent, 844 modeling and, 844–848 solutions to, 838–839 substitution to solve, 840–841 in two unknowns, 838 System of linear inequalities applications of, 869–870 explanation of, 867, 949 on graphing calculator, 868–869 linear programming and, 870–874 method to solve, 867–869 solution regions of, 870–871 System of nonlinear equations applications of, 1013–1014 elimination to solve, 1010–1012 explanation of, 1007, 1066 on graphing calculator, 1008–1014 possible solutions for, 1007–1008 substitution to solve, 1008–1010 System of nonlinear inequalities, 1012–1013, 1066–1067 System of polar equations, 1074–1075 Tangent meaning of term, 576 sum and difference identities for, 672–675 table of values for, 565

I-11

Tangent functions applications of, 588 characteristics of, 567 explanation of, 532 graphs of, 563–566, 587, 637, 638 inverse, 699–700, 704 Tangent lines explanation of, 576, 1204 slope of, 103, 1073–1074, 1205, 1206 Telescoping sums, 886–888 Temperature, 46, 850 Terminal points, 771, 772 Terminal side, 514, 612, 613 Terms constant, A–1 explanation of, A–1, A–2 of sequence, 1078–1080 variable, A–1 Test for collinear points, 936–937 Theta (␪), 510 30-60-90 triangles, 513, 595, 596 3 ⫻ 3 systems. See System of linear equations in three variables Threshold audibility, 438 Tidal motion, 645–646 Toolbox functions direct variation and, 177–180, 193 explanation of, 120–121, 190 graphs of, 120–121 horizontal reflections and, 125–126 horizontal translations and, 123–124 transformations and, 121, 190 vertical reflections and, 124–125 vertical translations and, 122–123 Transcendental functions, 433, 723. See also Logarithmic functions; Trigonometric functions Transformations area under curve and, 304–305 explanation of, 121 of general functions, 127–130 to graph functions, 127–130, 235, 425, 436–437 on graphing calculator, 122–123 horizontal reflections and, 125–126 horizontal translations and, 123–124 nonrigid, 128 of parent function, 128 of parent graph, 121 of power functions, 154–155 of reciprocal functions, 151 rigid, 128 of root functions, 154–155 solving equations involving, 725–726 of trigonometric graphs, 547, 639 vertical reflection and, 124–125 vertical stretches or compressions and, 126–127 vertical translations and, 122–123 via composition, 288 Translations horizontal, 123–124, 581–584, 586–597 vertical, 122–123, 578–581, 586–587 Transverse axis, of hyperbola, 985 Trapezoids area of, A–32 – A–33 explanation of, A–32 perimeter of, 731, 768, A–32 Tree diagrams, 1120–1122 Trial-and-error method, A–41 – A–42 Trials, 1121 Triangles angle-angle-side, 747, 764 angle-side-angle, 747, 764 area of, 381, 668, 763–766, 823, 904, 936, A–32 equilateral, 524, 811 explanation of, 512, 634, A–32 45-45-90, 513, 514 law of sines to solve, 746–753 oblique, 746–752, 822 Pascal’s, 1145–1147 perimeter formula for, A–32 properties of, 512–513, 634 right, 524, 595–604, 640, 650, A–74 side-angle-side, 747, 759–761, 763–764 side-side-angle, 747–752


I-12

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Index

Triangles—Cont. side-side-side, 747, 761–762 similar, 512, 513 30-60-90, 513 unit circle and special, 529–532 Triangular form of matrix, 895 Triangularizing augmented matrix, 895–897 Trigonometric equation models applications of, 622, 721 critical points and sinusoidal models and, 622–626, 642 sinusoidal regression and, 626–627 Trigonometric equations algebraic methods and, 721–723 applications of, 721, 726–727 explanation of, 711 of form A sin (Bx ⫾ C ) ⫾ D ⫽ k, 725–726 on graphing calculator, 714, 717, 723–725 identities to solve, 723 methods to solve, 36, 712–717, 722–726, 737, 739–740 roots of, 711–712 Trigonometric form complex numbers in, 803–804, 826 converted to rectangular form, 804–805 converting from rectangular form to, 651 equation of line in, 729 products and quotients in, 806–807 Trigonometric functions. See also Inverse trigonometric functions; specific trigonometric functions of acute angles, 596 applications of, 559, 579–580, 585, 586, 617–618 domain of, 535 evaluation of, 533–534, 612, 614 explanation of, 532–534 fundamental identities and, 657–658 on graphing calculator, 545, 549, 551–554, 561–567, 598–601, 613, 658, 665, 666, 714, 717 graphs of, 542–554, 561–566 points on unit circle and, 534, 535 real numbers and, 534–537 reference angles and, 531–532, 614–616, 646 standard angles and, 646, 647 standard form of, 582, 583 value of, 536, 537 written in terms of another, 657–658 Trigonometric graphs applications of, 588–589, 639 of cosecant function, 561–562 of cosine function, 549–554 of cotangent function, 566–567 explanation of, 116, 542 of secant function, 562–563 of sine function, 542–549, 551–554 of tangent function, 563–566 transformations of, 547, 578–584, 639 Trigonometric identities, 743–744 Trigonometric ratios points and, 610–614 values of, 595–597 Trigonometric substitution, 743 Trigonometry area of triangle and, 763–766 coordinate plane and, 610–616, 641 dynamic, 610 explanation of, 532 problem solving and, 832–833 of real numbers, 527–537, 577–578 of right triangles, 595–604, 640 static, 610 Trinomials. See also Polynomials explanation of, 1145, A–15 factoring, A–40 – A–41 perfect square, 219, A–19, A–42 – A–43 Triple angle formula for sine, 756 Tunnel clearance, 1016 Two-dimensional graphs, 853 Two-sided limits, 1171–1173 2 ⫻ 2 systems. See System of linear equations in two variables

Uniform motion, 72–73, 845–846, 849 Union, A–28 Uniqueness property to solve exponential equations, 426–428 to solve logarithms, 458–459 Unique solutions, 854 Unit circle equation of, 527 explanation of, 527, 635–636 points on, 527–528 special triangles and, 529–532 symmetry and, 528, 529, 531–532, 661 trigonometric functions and points on, 532–534 Unit vectors, 779 Upper and lower bounds property, 328, 330 Upper bound, 330 u-substitution explanation of, 716 to factor expressions in quadratic form, A–44 to solve equations in quadratic form, A–74 to solve trigonometric equations, 716, 727 Variable amplitude, 645–646 Variable costs, 846 Variables decision, 872 dependent, 2 independent, 2 objective, 69–70, 872 Variable terms, A–1 Variation constant of, 177 direct, 177–180, 193 on graphing calculator, 180 graphs of, 178 inverse, 180–182, 193 joint or combined, 182, 193, 195 power functions and, 196 Variation equations, 177, 179 Vector projections, 789, 794–795 Vector quantities, 771 Vectors algebraic, 778 applications of, 779–782, 790–791, 824 arithmetic operations on, 776–777 component of u along v, 788–790 dot products and angle between, 792–794, 824 equilibrium and, 787–789 equivalent, 772 explanation of, 771, 823–824 force, 779–780, 791 geometry of, 959 on graphing calculator, 773–775, 777, 792, 795, 797 horizontal and vertical components of, 773, 775, 778, 788 horizontal unit, 778 magnitude of, 774–775, 784 notation and geometry of, 771–772 orthogonal components of, 788, 794–795 position, 772, 773 projectile height and, 796–797 projvu and, 794–795 properties of, 777 rectangular coordinate system and, 772–773 resultant, 776, 802 static equilibrium and, 828–829 in three dimensions, 833–834 unit, 779 vertical unit, 778 Velocity angular, 520–521 average rate of change applied to projectile, 255 of falling object, 271 instantaneous, 1204–1207 linear, 520–521 Velocity formulas, 160, 271 Vertex of cone, 964 explanation of, 510

of hyperbola, 985 of parabola, 5 of solution region, 868 Vertex formula, 237–238 Vertex/intercept formula, 245 Vertical asymptotes, 151, 356–359, 363 Vertical-axis symmetry, 1028 Vertical boundary lines, 36–37, 96 Vertical change, 20–21 Vertical component of vectors, 773, 775, 788 Vertical compressions, 126–127, 568, 569 Vertical ellipse, 972–973 Vertical format, A–3 Vertical hyperbola, 988 Vertical lines explanation of, 23–24, 95 slope of, 24 Vertical line test, 35–36, 95, 149 Vertical parabola, 5, 1000 Vertical parabolic segment, 1017 Vertical reflections, 124–125 Vertical shifts, 122–123, 579 Vertical stretches, 568 Vertical translations, 122–123 Vertical unit vectors, 778 Voltage, 808, 809 Volume of box, 318 of composite figure, A–34 of cone, 730, 876 of cube, A–21, A–33 of cylinder, 90, 730, A–51 explanation of, A–33 pressure and, A–9 of rectangular solid, A–33 of right circular cone, A–33 of right circular cylinder, A–33 of right pyramid, A–33 of sphere, 134, A–33 Witch of Agnesi, 1061 Work, 790–791 Wrapping function, 577–578 Written information translated into equations, A–30 – A–31 translated into mathematical model, A–2 x-intercepts explanation of, 20, 101 on graphing calculator, 215–216 method to find, 61 of quadratic functions, 215–216 as zeroes of function, 108–109 x-intercept/zeroes method. See Zeroes/x-intercept method xy-plane, 3, 936 y-axis, 106, 107 y-intercept explanation of, 20 method to find, 61 Zeno of Elea, 1160 Zeroes approximation of real, 355 complex, 401–402 factor theorem to find, 315 of functions, 108, 109 intermediate value theorem to find, 324–325 irrational, 355 of multiplicity, 225, 322, 341–344 of polynomial functions, 320–331, 397 of quadratic functions, 215–217 rational, 325–330 repeated, 401–402 Zeroes/x-intercept method, 66–67, 97, 218, 220 Zero exponent property, A–12 Zero exponents, A–12, A–14 Zero placeholder, 310–311 Zero product property, A–46 – A–47

cob19537_es.indd Page Sec1:2 29/01/11 9:51 PM s-60user

/Volume/208/MHDQ265/cob19537_disk1of1/0073519537/cob19537_pagefiles

▼

Special Products 1x a21x b2 x2 1a b2x ab

1a b21a b2 a2 b2

1a b2 2 a2 2ab b2

1a b2 2 a2 2ab b2

1a b2 3 a3 3a2b 3ab2 b3 ▼

1a b2 3 a3 3a2b 3ab2 b3

Special Factorizations a2 2ab b2 1a b2 2

a2 2ab b2 1a b2 2

x 2 1c d2 x cd 1x c21x d2

abx 2 1ad bc2 x cd 1ax c21bx d2

a3 b3 1a b21a2 ab b2 2

a3 b3 1a b21a2 ab b2 2

a2 b2 1a b21a b2

▼

a2 b2 is prime over the real numbers

Formulas from Plane Geometry: P S perimeter, C S circumference, A S area Rectangle

Square

w

P 2l 2w

Regular Polygon s

P 4s

l

Parallelogram A bh

A

b B

Triangle Sum of angles

h 1a b2 2

C

A

A

b

c

1 bh 2

h b

Circle

a

r

A r2

Pythagorean Theorem b

a2 b2 c2

Ellipse

Triangle

h

Right Triangle

A B C 180°

C 2r d

Right Parabolic Segment

A ab

C 221a b 2 2

a

Trapezoid

h

a

P ns a A P 2

A s2

A lw

s

a

4 A ab 3

a b

2

b

▼

Formulas from Solid Geometry: S S surface area, V S volume Rectangular Solid H

V LWH S 21LW LH WH2

L

W

Cube

Right Circular Cylinder

V s3

V r2h

S 6s2

S 2r 1r h2

s

h r

Right Circular Cone

Right Square Pyramid

Sphere

1 V r2h 3

1 V b2h 3

4 V r3 3

S r 1r s2

ISBN: 0-07-351953-7 Author: John W. Coburn & J.D.Herdlick Title: Precalculus

Front endsheets Color: 5 Pages: 1, 2

h r

S b2 b2b2 4h2

h b

S 4r2

r


Formulas from Analytical Geometry: P1 S (x1, y1), P2 S (x2, y2) Distance between P1 and P2 d 21x2 x1 2 1y2 y1 2 2

Slope of Line Containing P1 and P2 y2 y1 ¢y m x2 x1 ¢x

2

Equation of Line Containing P1 and P2

Point-Slope Form

Slope-Intercept Form (slope m, y-intercept b)

y

identity

constant

y

y mx b

y

yb

yx

(0, b) x

x y mx b

m 0, b 0

Parallel Lines

Perpendicular Lines

Slopes Are Equal: m1 m2

Slopes Have a Product of 1: m1m2 1

Intersecting Lines

Dependent (Coincident) Lines

y x

Slopes Are Unequal: m1 m2

Slopes and y-Intercepts Are Equal: m1 m2, b1 b2

x

absolute value

x

m 0, b 0

m 1, b 0

squaring

cubing

y

y

x

m 0, b 0

square root

y

y

y x2

y x3

x

y x x

x

Logarithms and Logarithmic Properties logb 1 0

logb bx x

blogb x x

logb 1MN 2 logb M logb N

M logb a b logb M logb N N

logc x

cube root

logb x logb c

ceiling function

logb MP P # logb M

n S compounding periods/year

r S interest rate per year

r R S interest rate per time period a b n

t S time in years

reciprocal square

A

y

AR 11 R2 nt 1

Sequences and Series: Geometric Sequences

a1, a2 a1 d, a3 a1 2d, ... , an a1 1n 12d

a1, a2 a1r, a3 a1r2, ... , an a1r n1 a1 a1rn Sn 1r a1 ; r 6 1 Sq 1r

n Sn 1a1 an 2 2 n Sn 2a1 1n 12d 2

Binomial Theorem n n n 1a b2 n a b anb0 a b an1b1 a b an2b2 0 1 2 n! n1n 121n 22 # # # 132122112;



###

a

0! 1

n n b a 1bn1 a b a0bn n1 n n n! a b k k!1n k2!

1 x

x

logistic y yc y

c 1 aebx

1

▼

a1 S 1st term, an S nth term, Sn S sum of n terms, d S common difference, r S common ratio


x

y y logb x (b 0)

y bx (b 0) x

Payments Required to Accumulate Amount A P

y

logarithmic

y 1 x2

A Pert

P 11 R2 nt 1 R

y

1 2

exponential

y

Interest Compounded Continuously

r A P a1 b n

Accumulated Value of an Annuity

x

x

P S initial deposit, P S periodic payment

reciprocal

y y x

3 y x

Applications of Exponentials and Logarithms A S amount accumulated

floor function

y y x

y

x

1

(0, 1 c a )

x

Transformations of Basic Graphs Given Function

Transformation of Given Function

y f 1x2

y a f 1x h2 k vertical reflections vertical stretches/compressions

▼

x

S

logb b 1

S

y logb x 3 b y x

nt

▼

linear

y

y mx b, where b y1 mx1

Interest Compounded n Times per Year

▼

linear

S

▼

The Toolbox and Other Functions

(0, b)

Equation of Line Containing P1 and P2 y y1 m1x x1 2

▼

▼

▼




Average Rate of Change of f(x) For linear function models, the average rate of change on the interval 3 x1, x2 4 is constant, and given by the slope formula: ¢y y2 y1 . The average rate of change for other function models is nonconstant. By writing the slope formula in function form x2 x1 ¢x using y1 f 1x1 2 and y2 f 1x2 2, we can compute the average rate of change of other functions on this interval: f(x2) ⴚ f(x1) ⌬y ⴝ ⌬x x2 ⴚ x1


Formulas from Analytical Geometry: P1 S (x1, y1), P2 S (x2, y2) Distance between P1 and P2 d 21x2 x1 2 1y2 y1 2 2

Slope of Line Containing P1 and P2 y2 y1 ¢y m x2 x1 ¢x

2

Equation of Line Containing P1 and P2

Point-Slope Form

Slope-Intercept Form (slope m, y-intercept b)

y

identity

constant

y

y mx b

y

yb

yx

(0, b) x

x y mx b

m 0, b 0

Parallel Lines

Perpendicular Lines

Slopes Are Equal: m1 m2

Slopes Have a Product of 1: m1m2 1

Intersecting Lines

Dependent (Coincident) Lines

y x

Slopes Are Unequal: m1 m2

Slopes and y-Intercepts Are Equal: m1 m2, b1 b2

x

absolute value

x

m 0, b 0

m 1, b 0

squaring

cubing

y

y

x

m 0, b 0

square root

y

y

y x2

y x3

x

y x x

x

Logarithms and Logarithmic Properties logb 1 0

logb bx x

blogb x x

logb 1MN 2 logb M logb N

M logb a b logb M logb N N

logc x

cube root

logb x logb c

ceiling function

logb MP P # logb M

n S compounding periods/year

r S interest rate per year

r R S interest rate per time period a b n

t S time in years

reciprocal square

A

y

AR 11 R2 nt 1

Sequences and Series: Geometric Sequences

a1, a2 a1 d, a3 a1 2d, ... , an a1 1n 12d

a1, a2 a1r, a3 a1r2, ... , an a1r n1 a1 a1rn Sn 1r a1 ; r 6 1 Sq 1r

n Sn 1a1 an 2 2 n Sn 2a1 1n 12d 2

Binomial Theorem n n n 1a b2 n a b anb0 a b an1b1 a b an2b2 0 1 2 n! n1n 121n 22 # # # 132122112;



###

a

0! 1

n n b a 1bn1 a b a0bn n1 n n n! a b k k!1n k2!

1 x

x

logistic y yc y

c 1 aebx

1

▼

a1 S 1st term, an S nth term, Sn S sum of n terms, d S common difference, r S common ratio


x

y y logb x (b 0)

y bx (b 0) x

Payments Required to Accumulate Amount A P

y

logarithmic

y 1 x2

A Pert

P 11 R2 nt 1 R

y

1 2

exponential

y

Interest Compounded Continuously

r A P a1 b n

Accumulated Value of an Annuity

x

x

P S initial deposit, P S periodic payment

reciprocal

y y x

3 y x

Applications of Exponentials and Logarithms A S amount accumulated

floor function

y y x

y

x

1

(0, 1 c a )

x

Transformations of Basic Graphs Given Function

Transformation of Given Function

y f 1x2

y a f 1x h2 k vertical reflections vertical stretches/compressions

▼

x

S

logb b 1

S

y logb x 3 b y x

nt

▼

linear

y

y mx b, where b y1 mx1

Interest Compounded n Times per Year

▼

linear

S

▼

The Toolbox and Other Functions

(0, b)

Equation of Line Containing P1 and P2 y y1 m1x x1 2

▼

▼

▼




Average Rate of Change of f(x) For linear function models, the average rate of change on the interval 3 x1, x2 4 is constant, and given by the slope formula: ¢y y2 y1 . The average rate of change for other function models is nonconstant. By writing the slope formula in function form x2 x1 ¢x using y1 f 1x1 2 and y2 f 1x2 2, we can compute the average rate of change of other functions on this interval: f(x2) ⴚ f(x1) ⌬y ⴝ ⌬x x2 ⴚ x1

ob19537_es.indd Page Sec1:4 29/01/11 9:51 PM s-60user

▼

Commonly used, small case Greek letters

▼


alpha zeta rho

beta theta sigma

delta mu psi

Trigonometric Form

ƒ z ƒ 2a b

z r 1cos i sin 2

distance from (0, 0) to (a, b)

where r ƒ z ƒ

2

2

Powers and DeMoivres Theorem

▼

Products and Quotients

z1z2 r1r2 3cos11 2 2 i sin11 2 2 4 z1 r1 3cos11 2 2 i sin 11 2 2 4 z2 r2

▼

defined as: u • v Ha, bI • Hc, dI ac bd.

v

▼ b a

x

k

(h, k b)

(h, k)

(h, k) (x h)2 (y k)2 r2

0.2618 12 13 0.5774 3

cos ␪

tan ␪

csc ␪

sec ␪

cot ␪

0

1

0

—

1

—

6 4 3 2

1 2

13 2

1 13

2

2 13

13

12 2

1

12

12

1

13 2

1 2

13

2 13

2

1 13

0

—

1

—

B

45 √2x

12 2

1

B

A 45

60 2x

1x

1x

30

A

C

(y k)2 b2

y (x, y)

y sin r

r

y

y tan , x 0 x

r csc , y 0 y

␪r ␪ x

x

x cot , y 0 y

Right Triangle Trigonometry adj hyp

sin

hyp sec adj

1

(0, b)

C

√3x

0


cos

(h a, k)

1x

For right ^ ABC with indicated sides adjacent and opposite to acute angle :

ellipse with center at (h, k), a b

(x h)2 a2

(h, k b)

central ellipse

sin ␪

r sec , x 0 x

(h a, k)

0.5236 6 13 0.8660 2

Special Triangles and Special Angles

x cos r

▼

r

0.7854 4 12 0.7071 2

For P1x, y2 a point on the terminal side of an angle in standard position:

␪

y

circle with center at (h, k)

▼

opp hyp

tan

hyp csc opp

hyp

opp adj

opp

␪

adj cot opp

adj

Trigonometric Functions of a Real Number

t (x, y)

(0, 0) x2

13 1.7321

90°

8a, b9

Conic Sections

(x, y)

12 1.4142

45°

u v . • ƒuƒ ƒvƒ

• Given the nonzero vectors u and v and angle between them, cos

r

e 2.7183

30°

y

• Given the vectors u Ha, bI and v Hc, dI, their dot product is denoted u • v and is

central circle

1.0472 3

60°

• For a position vector, v Ha, bI and angle as shown, a ƒ v ƒ cos and b ƒ v ƒ sin , b where r tan1 ` ` and ƒ v ƒ 2a2 b2. a v • For any nonzero vector v Ha, bI ai bj, the vector u is a unit vector in the ƒ v ƒ same direction as v.

k

1.5708 2

n

Vectors and the Dot Product

y

3.1416

0° 0

2k 2k i sin b for k 0, 1, 2, p , n 1 1 z 1 r acos n n n

n

Special Constants

␪

Roots and the nth Roots Theorem

z r 1cos n i sin n2 for positive integers n n

▼

epsilon pi omega

Complex Numbers z ⴝ a ⴙ bi Absolute Value

▼

gamma lambda phi

x

h

y2

(a, 0)

(c, 0)

x2 a2

(0, b) hyperbola with center at (h, k)

k (h, k) central hyperbola

(x h)2 a2

(y k)2 b2

1

(c, 0)

y2 b2

1

x2 4py vertical parabola focus (0, p) directrix y p y

▼

y

cos t x

sin t y

1 sec t ; x 0 x

1 csc t ; y 0 y

p0

x

y p

1 p0

a2 b2


r1

y tan t ; x 0 x x cot t ; y 0 y

1

Graphs of the Trigonometric Functions y csc t

y

y

x

x p

1

y2 4px horizontal parabola focus ( p, 0) directrix x p

1

y sec t

y

␲ 2

␲

3␲ 2

␲4 , 1 y cos t

1

y sin t 2␲

t

1

␲ 2

␲

3␲ 2

Domain: t 僆 1q, q 2 Range: sin t 僆 3 1, 1 4

2␲

t

␲

␲

␲ 2

2

4

2

2

y tan t

4

2

2

( p, 0)

x

If term containing y leads, the hyperbola is oriented vertically.


y2 b2

If a b, the ellipse is oriented vertically.

(0, p) h

(a, 0)

x

c2 |a2 b2|

(a, 0) (c, 0)

c2

For any real number t and point P1x, y2 on the unit circle associated with t:

(a, 0) h

y

x2 a2

(c, 0)

r2

Domain: t 僆 1q, q 2 Range: cos t 僆 3 1, 1 4

␲

y cot t

12k 12; k 僆 2 Range: tan t 僆 Domain: t

3␲ 2

t


▼

Commonly used, small case Greek letters

▼


alpha zeta rho

beta theta sigma

delta mu psi

Trigonometric Form

ƒ z ƒ 2a b

z r 1cos i sin 2

distance from (0, 0) to (a, b)

where r ƒ z ƒ

2

2

Powers and DeMoivres Theorem

▼

Products and Quotients

z1z2 r1r2 3 cos11 2 2 i sin11 2 2 4 z1 r1 3 cos11 2 2 i sin 11 2 2 4 z2 r2

▼

defined as: u • v Ha, bI • Hc, dI ac bd.

v

▼ b a

x

k

(h, k b)

(h, k)

(h, k) (x h)2 (y k)2 r2

0.2618 12 13 0.5774 3

cos ␪

tan ␪

csc ␪

sec ␪

cot ␪

0

1

0

—

1

—

6 4 3 2

1 2

13 2

1 13

2

2 13

13

12 2

1

12

12

1

13 2

1 2

13

2 13

2

1 13

0

—

1

—

B

45 √2x

12 2

1

B

A 45

60 2x

1x

1x

30

A

C

(y k)2 b2

y (x, y)

y sin r

r

y

y tan , x 0 x

r csc , y 0 y

␪r ␪ x

x

x cot , y 0 y

Right Triangle Trigonometry adj hyp

sin

hyp sec adj

1

(0, b)

C

√3x

0


cos

(h a, k)

1x

For right ^ ABC with indicated sides adjacent and opposite to acute angle :

ellipse with center at (h, k), a b

(x h)2 a2

(h, k b)

central ellipse

sin ␪

r sec , x 0 x

(h a, k)

0.5236 6 13 0.8660 2

Special Triangles and Special Angles

x cos r

▼

r

0.7854 4 12 0.7071 2

For P1x, y2 a point on the terminal side of an angle in standard position:

␪

y

circle with center at (h, k)

▼

opp hyp

tan

hyp csc opp

hyp

opp adj

opp

␪

adj cot opp

adj

Trigonometric Functions of a Real Number

t (x, y)

(0, 0) x2

13 1.7321

90°

8a, b9

Conic Sections

(x, y)

12 1.4142

45°

u v . • ƒuƒ ƒvƒ

• Given the nonzero vectors u and v and angle between them, cos

r

e 2.7183

30°

y

• Given the vectors u Ha, bI and v Hc, dI, their dot product is denoted u • v and is

central circle

1.0472 3

60°

• For a position vector, v Ha, bI and angle as shown, a ƒ v ƒ cos and b ƒ v ƒ sin , b where r tan1 ` ` and ƒ v ƒ 2a2 b2. a v • For any nonzero vector v Ha, bI ai bj, the vector u is a unit vector in the ƒ v ƒ same direction as v.

k

1.5708 2

n

Vectors and the Dot Product

y

3.1416

0° 0

2k 2k i sin b for k 0, 1, 2, p , n 1 1 z 1 r acos n n n

n

Special Constants

␪

Roots and the nth Roots Theorem

z r 1cos n i sin n2 for positive integers n n

▼

epsilon pi omega

Complex Numbers z ⴝ a ⴙ bi Absolute Value

▼

gamma lambda phi

x

h

y2

(a, 0)

(c, 0)

x2 a2

(0, b) hyperbola with center at (h, k)

k (h, k) central hyperbola

(x h)2 a2

(y k)2 b2

1

(c, 0)

y2 b2

1

x2 4py vertical parabola focus (0, p) directrix y p y

▼

y

cos t x

sin t y

1 sec t ; x 0 x

1 csc t ; y 0 y

p0

x

y p

1 p0

a2 b2


r1

y tan t ; x 0 x x cot t ; y 0 y

1

Graphs of the Trigonometric Functions y csc t

y

y

x

x p

1

y2 4px horizontal parabola focus ( p, 0) directrix x p

1

y sec t

y

␲ 2

␲

3␲ 2

␲4 , 1 y cos t

1

y sin t 2␲

t

1

␲ 2

␲

3␲ 2

Domain: t 僆 1q, q 2 Range: sin t 僆 3 1, 1 4

2␲

t

␲

␲

␲ 2

2

4

2

2

y tan t

4

2

2

( p, 0)

x

If term containing y leads, the hyperbola is oriented vertically.


y2 b2

If a b, the ellipse is oriented vertically.

(0, p) h

(a, 0)

x

c2 |a2 b2|

(a, 0) (c, 0)

c2

For any real number t and point P1x, y2 on the unit circle associated with t:

(a, 0) h

y

x2 a2

(c, 0)

r2

Domain: t 僆 1q, q 2 Range: cos t 僆 3 1, 1 4

␲

y cot t

12k 12; k 僆 2 Range: tan t 僆 Domain: t

3␲ 2

t


▼

Fundamental Identities Reciprocal Identities

▼

▼


Ratio Identities

sec

1 cos

tan

sin cos

csc

1 sin

cot

cos sin

cot

1 tan

Pythagorean Identities sin2 cos2 1

Cofunction Identities

▼

tan2 1 sec2

cos12 cos

1 cot2 csc2

tan12 tan

Sum and Difference Identities

b cos 2

cos a

b sin 2

cos1 2 cos cos sin sin

tan a

b cot 2

cot a

b tan 2

sin1 2 sin cos cos sin

sec a

b csc 2

csc a

b sec 2

tan1 2

Double-Angle Identities

▼

Half-Angle Identities

▼

Power Reduction Identities

1 cos sin a b 2 A 2

sin2

1 cos 2 2

cos122 cos2 sin2

1 cos cos a b 2 A 2

cos2

1 cos 2 2

1 cos tan a b 2 sin

tan2

1 cos 2 1 cos 2

sin 1 cos

Product-to-Sum Identities

▼

Sum-to-Product Identities b cos a b 2 2

sin cos

1 sin1 2 sin1 2 2

sin sin 2 sin a

cos sin

1 sin1 2 sin1 2 2

sin sin 2 cos a

b sin a b 2 2

cos cos

1 cos1 2 cos1 2 2

cos cos 2 cos a

b cos a b 2 2

sin sin

1 cos1 2 cos1 2 2

cos cos 2 sin a

Law of Sines

▼

C sin B sin A sin C a c b ▼

tan tan 1 tan tan

sin122 2 sin cos

1 2 sin2

▼

sin12 sin

sin a

2 cos2 1

▼

Identities due to Symmetry

Area of a Triangle 1 Area bc sin A 2


b A

B


Law of Cosines a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B

a c

b sin a b 2 2

c2 a2 b2 2ab cos C

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