Thomas' Calculus (12th Edition)

THOMAS' CALCULUS Twelfth Edition Based on the original work by George B. Thomas, Jr. Massachusetts Institute of Techn...

Author: George B. Thomas | Maurice D. Weir | Joel Hass

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THOMAS'

CALCULUS Twelfth Edition

Based on the original work by

George B. Thomas, Jr. Massachusetts Institute of Technology as revised by

Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis

Addison-Wesley Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

Edltor-iD-Cblef: Deirdre Lynch Semor AcqailltiODl Editor: William Hoffinan Semor Project Editor: Rachel S. Reeve Associate Editor: Caroline Celano Associate Project Editor: Leah Goldberg Semor Managing Editor: Karen Wernhohn Semor Prodnctton Snpenlsor: Sheila Spinoey Senior Design Supervisor: Andrea Nix Digital Assets Manager: Mariaone Groth Media Producer: Lin Mahoney Software Development: Mary Dumwa1d and Bob Carroll EIecutive Marketing Manager: Jeff Weidenaar MarketingAsmlant: Kendra Bassi Senior Author Support/fec:hnology Specialist: Joe Vetere Senior Prepreill Supervilor: Caroline Fell Manufacturing Manager: Evelyn Beaton Production Coordinator: Kathy Diamond Composition: Nesbitt Graphics, Inc. lli••tratioDl: Karen Heyt, lllustraThch Cover Design: Rokusck Design Cover image: Forest Edge, Hokuto, Hokkaido, Japsn 2004 © Michael Kenna About the cover: The cover image of a tree line on a snow-swept landscape, by the photographer Michael Kenna, was taken in Hokkaido, Japan. The artist was not thinking of calculus when he composed the image, but rather, of a visual haiku consisting of a few elements that would spaIl< the viewer's imagination. Similarly, the minima1 design of this text allows the central ideas of calculus developed in this book to unfold to igoite the learner's imagination.

For permission to use copyrighted material, grateful acknowledgment is made to the copyright holders on page C-I, which is hereby made part of this copyright page. Many of the desigoations used by manufacturers and sellers to distinguish their products are claimed as tradenulrks. Where those designations appear in this book, and Addison-Wesley was aware of a trademark claim, the designations have been printed in initial caps or all caps.

Ubrary of Congress Cataloging-in-Publication Data Weir, Maurice D. Thomas' Calculus I Maurice D. Weir, Joel Hass, George B. Thomas.-12th ed. p.em ISBN 978-0-321-58799-2 1. CalculUl!-Textbooks. I. Hass, Joel. II. Thoroas, George B. (George Brinton), 1914--2006. III. Thomas, George B. (George Brinton), 1914--2006. Calculus. Iv. Title V. Title: Calculus. QA303.2.W452009b 515-dc22

2009023069

Copyright () 2010, 2005, 2001 Pearson Education, Inc. All rights reserved. No part of this publication may be reproduced, stated in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United

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Addison-Wesley is an imprint of

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PEARSON

ISBN-10: 0-321-58799-5

www.pearsoned.com

ISBN-13: 978-0-321-58799-2

CONTENTS

1

Preface

ix

I Functions

1 1.1 1.2 1.3 1.4

Functions and Their Graphs I CombiJring Functions; Shifting and Scaling Graphs Trigonometric Functions 22 Graphing with Calcolators and Computers 30 QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

34

35 37

AoDmONAL AND ADvANCED EXERCISES

2

Limits and Continuity 2.1 2.2 2.3 2.4 2.5 2.6

39

Rates of Change and Tangents to Curves 39 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 57 One-Sided Limits 66 Continuity 73 Limits Involving Infinity; Asymptotes of Graphs QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

98

Differentiation 3.1 3.2 3.3 3.4 3.5 3.6

84

96

97

AoDmONAL AND ADvANCED EXERCISES

3

14

102 Tangents and the Derivative at a Point 102 The Derivative as a Function 106 Differentiation Rules 115 The Derivative as a Rate of Change 124 Derivatives of Trigonometric Functions 135 The Chain Rule 142

iii

iv

Contents

3.7 3.8 3.9

4

Applications of Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7

5

Extreme Values of Functions 184 The Mean Value Theorem 192 Monotonic Functions and the First Derivative Test Concavity and Curve Sketching 203 Applied Optimization 214 Newton's Method 225 Antiderivatives 230 QuESTIONS TO GUIDE YOUR REVIEW 239 PRACTICE EXERCISES 240 ADDITIONAL AND ADvANCED EXERCISES 243

184

198

I Integration

246 5.1 5.2 5.3 5.4 5.5 5.6

6

Implicit Differentiation 149 Related Rates 155 Linearization and Differentials 164 QuESTIONS TO GUIDE YOUR REVIEW 175 PRACTICE EXERCISES 176 ADDITIONAL AND ADvANCED EXERCISES 180

Area and Estimating with Finite Sums 246 Sigma Notation and Limits of Finite Sums 256 The Definite Integral 262 The Fundamental Theorem of Calculus 274 Indefinite Integrals and the Substitution Method 284 Substitution and Area Between Curves 291 QuESTIONS TO GUIDE YOUR REVIEW 300 PRACTICE EXERCISES 301 ADDITIONAL AND ADvANCED EXERCISES 304

Applications of Definite Integrals 6.1 6.2 6.3 6.4 6.5 6.6

Volumes Using Cross-Sections 308 Volumes Using Cylindrical Shells 319 Arc Length 326 Areas of Surfaces of Revolution 332 Work and Fluid Forces 337 Moments and Centers of Mass 346 QuESTIONS TO GUIDE YOUR REVIEW 357 PRACTICE EXERCISES 357 ADDITIONAL AND ADvANCED EXERCISES 359

308

Contents

7

Transcendental Functions 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

8

9

Inverse Functions and Their Derivatives 361 Natural Logarithms 369 Exponential Functions 377 Exponential Change and Separable Differential Equations Indeterminate Forms and VHopitai's Rule 396 Inverse Trigonometric Functions 404 Hyperbolic Functions 416 Relative Rates of Growth 424 QuEsTIONS TO GUIDE YOUR REVIEW 429 PRACTICE EXERCISES 430 ADDmONAL AND ADvANCED EXERCISES 433

Integration by Parts 436 Trigonometric Integrals 444 Trigonometric Substitotions 449 Integration of Rational Functions by Partial Fractions Integral Tables and Computer Algebra Systems 463 Numerical Integration 468 Improper Integrals 478 QuEsTIONS TO GUIDE YOUR REVIEW 489 PRACTICE EXERCISES 489 ADDmONAL AND ADvANCED EXERCISES 491

453

496

Solutions, Slope Fields, and Euler's Method 496 First-Order Linear Equations 504 Applications 510 Graphical Solutions of Autonomous Equations 516 Systems of Equations and Phase Planes 523 QuEsTIONS TO GUIDE YOUR REVIEW 529 PRACTICE EXERCISES 529 ADDmONAL AND ADVANCED EXERCISES 530

Infinite Sequences and Series 10.1 10.2 10.3 lOA 10.5

387

435

First-Order Differential Equations 9.1 9.2 9.3 9.4 9.5

10

361

Techniques of Integration 8.1 8.2 8.3 8.4 8.5 8.6 8.7

V

Sequences 532 Infinite Series 544 The Integral Test 553 Comparison Tests 558 The Ratio and Root Thsts

532

563

vi

Contents

10.6 10.7 10.8 10.9 10.10

11

Parametric Equations and Polar Coordinates 11.1 11.2 11.3 11.4 11.5 11.6 11.7

12

660

Three-Dimensional Coordinate Systems 660 Vectors 665 The Dot Product 674 The Cross Product 682 Lines and Planes in Space 688 Cylinders and Quadric Surfaces 696 QuESTIONS TO GUIDE YOUR REVIEW 701 PRACTICE EXERCISES 702 ADDITIONAL AND ADvANCED EXERCISES 704

Vector-Valued Functions and Motion in Space 13.1 13.2 13.3 13.4 13.5 13.6

610

Parametrizations of Plane Curves 610 Calculus with Parametric Curves 618 Polar Coordinates 627 Graphing in Polar Coordinates 631 Areas and Lengths in Polar Coordinates 635 Conic Sections 639 Conics in Polar Coordinates 648 QuESTIONS TO GUIDE YOUR REVIEW 654 PRACTICE EXERCISES 655 ADDITIONAL AND ADvANCED EXERCISES 657

Vectors and the Geometry of Space 12.1 12.2 12.3 12.4 12.5 12.6

13

Alternating Series, Absolute and Conditional Convergence 568 Power Series 575 Taylor and Maclaurin Series 584 Convergence of Taylor Series 589 The Binomial Series and Applications of Taylor Series 596 QuESTIONS TO GUIDE YOUR REVIEW 605 PRACTICE EXERCISES 605 ADDITIONAL AND ADvANCED EXERCISES 607

Curves in Space and Their Tangents 707 Integrals of Vector Functions; Projectile Motion 715 Arc Length in Space 724 Curvatore and Normal Vectors ofa Curve 728 Tangential and Normal Components of Acceleration 734 Velocity and Acceleration in Polar Coordinates 739 QuESTIONS TO GUIDE YOUR REVIEW 742 PRACTICE EXERCISES 743 ADDITIONAL AND ADvANCED EXERCISES 745

707

Contents

14

Partial Derivatives 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10

15

16

747 Functions of Several Variables 747 Limits and Continuity in Higher Dimensions 755 Partial Derivatives 764 The Chain Rule 775 Directional Derivatives and Gradient Vectors 784 Tangent Planes and Differentials 791 Extreme Values and Saddle Points 802 Lagrange Multipliers 811 Taylor's Formula for Two Variables 820 Partial Derivatives with Constrained Variables 824 QUESTIONS TO GUIDE YOUR REVIEW 829 PRACTICE ExERCISES 829 ADomONAL AND ADvANCED EXERCISES 833

Multiple Integrals 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8

836 Double and Iterated Integrals over Rectangles 836 Double Integrals over General Regions 841 Area by Double Integration 850 Double Integrals in Polar Form 853 Triple Integrals in Rectangular Coordinates 859 Moments and Centers of Mass 868 Triple Integrals in Cylindrical and Spherical Coordinates Substitutions in Multiple Integrals 887 QUESTIONS TO GUIDE YOUR REVIEW 896 PRACTICE ExERCISES 896 ADomONAL AND ADvANCED EXERCISES 898

875

Integration in Vector Fields 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8

vii

Line Integrals 901 Vector Fields and Line Integrals: Work, Circulation, and Flux 907 Path Independence, Conservative Fields, and Potential Functions 920 Green's Theorem in the Plane 931 Surfaces and Area 943 Surface Integrals 953 Stokes' Theorem 962 The Divergence Theorem and a Unified Theory 972 QUESTIONS TO GUIDE YOUR REVIEW 983 PRACTICE ExERCISES 983 ADomONAL AND ADVANCED EXERCISES 986

901

viii

Contents

17

Second-Order Differential Equations 17.1 17.2 17.3 17.4 17.5

online

Second-Order Linear Equations Nonhomogeneous Linear Equations Applications Euler Equations Power Series Solutions

Appendices

AP-1 A.I A.2 A.3 AA A.5 A.6 A.7 A.S A.9

Real Numbers and the Real Line AP-I Mathematical Induction AP-6 Lines, Circles, and Parabolas AP-1O Proofs of Limit Theorems AP-IS Commonly Occurring Limits AP-2I Theory of the Real Numbers AP-23 Complex Numbers AP-25 The Distributive Law for Vector Cross Products AP-35 The Mixed Derivative Theorem and the Increment Theorem

AP-36

Answers to Odd-Numbered Exercises

A-1

Index

1-1

A Brief Table of Integrals

T-1

Credits

C-1

PREFACE We have significantly revised this edition of Thomas' Calculus to meet the changing needs of today's instructors and students. The result is a book with more examples, more midlevel exercises, more figures, better conceptual flow, and increased clarity and precision. As with previous editions, this new edition provides a modern introduction to calculus that supports conceptual understanding but retains the essential elements of a traditional course. These enhancements are closely tied to an expanded version for this text of MyMathLab® (discussed further on), providing additional support for students and flexibility for instructors. Many of our students were exposed to the terminology and computational aspects of calculus during high school. Despite this familiarity, students' algebra and trigonometry skills often hinder their success in the college calculus sequence. With this text, we have sought to balance the students' prior experience with calculus with the algebraic skill development they may still need, all without undermining or derailing their confidence. We have taken care to provide enough review material, fully stepped-out solutions, and exercises to support complete understanding for students of all levels. We encourage students to think beyond memorizing formulas and to generalize concepts as they are introduced. Our hope is that after taking calculus, students will be confident in their problem-solving and reasoning abilities. Mastering a beautiful subject with practical applications to the world is its own reward, but the real gift is the ability to think and generalize. We intend this book to provide support and encouragement for both.

Changes for the TweLfth Edition CONTENT In preparing this edition we have maintained the basic structure of the Table of Contents from the eleventh edition. Yet we have paid attention to requests by current users and reviewers to postpone the introduction of parametric equations until we present polar coordinates, and to treat I'Hopital 's Rule after the transcendental functions have been studied. We have made numerous revisions to most of the chapters, detailed as follows.

• Functions We condensed this chapter even more to focus on reviewing function concepts. Prerequisite material covering real numbers, intervals, increments, straight lines, distances, circles, and parabolas is presented in Appendices 1-3. • Limits To improve the flow of this chapter, we combined the ideas of limits involving infinity and their associations with asymptotes to the graphs of functions, placing them together in the final chapter section. • Differentiation While we use rates of change and tangents to curves as motivation for studying the limit concept, we now merge the derivative concept into a single chapter. We reorganized and increased the number of related rates examples, and we added new examples and exercises on graphing rational functions.

ix

X

Preface

• Antiderivatives and Integration We maintain the orgaJrization of the eleventh edition in placing antiderivatives as the fInal topic of the chapter covering applications of derivatives. Our focus is on ''recovering a function from its derivative" as the solution to the simplest type offirst-order differential equation. Integrals, as "limits of Riemann sums," motivated primarily by the problem of rmding the areas of general regions with curved boundaries, are a new topic fanning the substance of Chapter 5. After carefully developing the integral concept, we turn our attention to its evaluation and connection to antiderivatives captured in the Fundamental Theorem of Calculus. The ensuing applications then define the various geometric ideas of area, volume, lengths of paths, and centroids all as limits of Riemann sums giving definite integrals, which can be evaluated by finding an antiderivative of the integrand. We return later to the topic of solving more complicated first-order differential equations, after we derme and establish the transcendental functions and their properties. • Differential Equations Some universities prefer that this subject be treated in a course separate from calculus. Although we do cover solutions to separable differential equations when treating exponential growth and decay applications in the chapter on transcendental functions, we organize the bulk of our material into two chapters (which may be omitted for the calculus sequence). We give an introductory treatment of first-order differential equations in Chapter 9, including a new section on systems and phase planes, with applications to the competitive-hunter and predator-prey models. We present an introduction to second-order differential equations in Chapter 17, which is included in MyMathLab as well as the Thomas' Calculus Web site, www.pearsonhighered.comlthomas. • Series We retain the organizational structure and content of the eleventh edition for the topics of sequences and series. We have added several new figures and exercises to the various sections, and we revised some of the proofs related to convergence of power series in order to improve the accessibility of the material for students. The request stated by one of our users as, "anything you can do to make this material easier for students will be welcomed by our faculty;' drove our thinking for revisions to this chapter. • Parametric Equations Several users requested that we move this topic into Chapter II, where we also cover polar coordinates and conic sections. We have done this, realizing that many departments choose to cover these topics at the beginning of Calculus m, in preparation for their coverage of vectors and multivariable calculus. • Vector-Valued Functions We streamlined the topics in this chapter to place more emphasis on the conceptual ideas supporting the later material on partial derivatives, the gradient vector, and line integrals. We condensed the discussions of the Frenet frame and Kepler's three laws of planetary motion. • Multivariahle Calculus We have further enhanced the art in these chapters, and we have added many new figures, examples, and exercises. We reorganized the opening material on double integrals, and combined the applications of double and triple integrals to masses and moments into a single section covering both two- and threedimensional cases. This reorganization allows for better flow of the key mathematical concepts, together with their properties and computational aspects. As with the eleventh edition, we continue to make the connections of multivariable ideas with their aingle-variable analogues studied earlier in the book. • Vector Fields We devoted considerable effort to improving the clarity and mathematical precision of our treatment of vector integral calculus, including many additional examples, figures, and exercises. Important theorems and results are stated more clearly and completely, together with enhanced explanations of their hypotheses and mathematical consequences. The area of a surface is now organized into a single section, and surfaces defined implicitly or explicitly are treated as special cases of the more general parametric representation. Surface integrals and their applications then follow as a separate section. Stokes' Theorem and the Divergence Theorem are still presented as generalizations of Green's Theorem to three dimensions.

Preface

xi

EXERCISES AND EXAMPLES We know that the exercises and examples are critical components in learning calculus. Because of this importance, we have updated, improved, and increased the number of exercises in nearly every section of the book. There are over 700 new exercises in this edition. We continue our organization and grouping of exercises by topic as in earlier editions, progressing from computational problems to applied and theoretical problems. Exercises requiring the use of computer software systems (such as Maple® or Mathematica®) are placed at the end of each exercise section, labeled Computer Explorations. Most of the applied exercises have a subheading to indicate the kind of application addressed in the problem. Many sections include new examples to clarify or deepen the meaning of the topic being discussed, and to help students understand its mathematical consequences or applications to science and engineering. At the same time, we have removed examples that were a repetition of material already presented. ART Because of their importance to learning calculus, we have continued to improve existing figures in Thomas' Calculus and we have created a significant number of new ones. We continue to use color consistently and pedagogically to enhance the conceptual idea that is being illustrated. We have also taken a fresh look at all of the figure captions, paying considerable attention to clarity and precision in short statements.

y

No matter what

y=1

positive nwnber E is, the graph enters this band at x = -~ and stays.

FIGURE 2.50, page 85 The geometric explanation of a finite limit as x - ± 00.

FIGURE 16.9, page 908 A surface in a space occupied by a moving fluid.

MYMATHLAB AND MATH XL The increasing use of and demand for online homework systems has driven the changes to MyMathLab and MathXL® for Thomas' Calculus. The MyMathLab course now includes significantly more exercises of all types. New Java™ applets add to the already significant collection to help students visualize the concepts and generalize the material.

Continuing Features RIGOR The level of rigor is consistent with that of earlier editions. We continue to distinguish between formal and informal discussions, and to point out their differences. We think starting with a more intuitive, less formal approach helps students understand a new or difficult concept so they can then appreciate its full mathematical precision and outcomes. We pay attention to defining ideas carefully and to proving theorems appropriate for

xii

Preface

calculus students, while mentioning deeper or subtler issues they would study in a more advanced course. Our organization, and distinctions between informal and formal discussions, gives the instructor a degree of flexibility in the amount and depth of coverage of the various topics. For example, while we do not prove the Intermediate Value Theorem or the Extreme Value Theorem for continuous functions on a :5 x :5 b, we do state these theorems precisely, illustrate their meanings in numerous examples, and use them to prove other important results. Furthermore, for those instructors who desire greater depth of coverage, we discuss in Appendix 6 the reliance of the validity of these theorems on the completeness of the real numbers.

WRmNG EXERCISES Writing exercises placed throughout the text ask students to explore and explain a variety of calculus concepts and applications. In addition, the end of each chapter contains a list of questions for students to review and summarize what they have learned. Many of these exercises make good writing assignments. END-Of-CHAPTER REVIEWS AND PROJECTS In addition to problems appearing after each section, each chapter colminates with review questions, practice exercises covering the entire chapter, and a series of Additional and Advanced Exercises serving to include more challenging or synthesizing problems. Most chapters also include descriptions of several Technology Application Projects that can be worked by individual students, or groups of students, over a longer period of time. Thesc projects require the use of a computer, running Mathematica or Maple, and additional material that is available over the Internet at www.pearsonhigherecLcomlthomas and in MyMathLab. WRITING AND APPUCATIONS As always, this text continues to be easy to read, conversational, and mathematically rich. Each new topic is motivated by clear, easy-to-understand examples and is then reinforced by its application to real-world problems of immediate interest to students. A hallmark of this book has been the application of calculus to science and engineering. These applied problems have been updated, improved, and extended continually over the last several editions. TECHNOLOGY In a course using the text, technology can be incorporated according to the taste of the instructor. Each section contains exerciscs requiring the use of technology; these are marked with a D if suitable for calculator or computer use or are labeled Computer Explorations if a computer algebra system (CAS, such as Maple or Mathematica) is required.

Text Versions THOMAS' CALCULUS, Twelfth Edition Complete (Chapters 1-16), ISBN 0-321-58799-51978-0-321-58799-2 Single Variable Calculus (Chapters 1-11), ISBN 0-321-63742-91978-0-321-63742-0 Multivariable Calculus (Chapters 11-16), ISBN 0.321-64369-01978-0-321-64369-8

THOMAS' CALCULUS: EARLY TRANSCENDENTALS, Twelfth Edition Complete (Chapters 1-16), ISBN 0-321-58876-21978-0-321-58876-0 Single Variable Calculus (Chapters 1-11),0-321-62883-71978-0-321-62883-1 Multivariable Calculus (Chapters 11-16), ISBN 0.321-64369-01978-0-321-64369-8 The early transcendentals version of Thomas' Calculus introduces and integrates transcendental functions (such as inverse trigonometric, exponential, and logsritlnnic functions) into the exposition, examples, and exercises of the early chapters alongside the algebraic functions. The Multivariable book for Thomas' Calculus: Early 1'rnnscendentals is the same text as Thomas' Calculus, Multivariable.

Preface

xiii

Instructor's Editions Thomas , Calculus, ISBN 0-321-60075-41978-0-321-60075-2 Thomas 'Calculus: Early Transcentientals, ISBN 0-321-62718-01978-0-321-62718-6 In addition to including all of the answers present in the stodent editions, the Instructor ~ Editions include even-numbered answers for Chapters l--{j.

University Calculus (Early Transcendentals) University Calculus: Alternative Edition (Late Transcendentals) University Calculus: Elements with Early Transcendentals The University Calculus texts are based on Thomas' Calculus and featore a streamlined presentation of the contents of the calculus course. For more information about these titles, visit www.pearsonhighered.com.

Print Supplements INSTRUCTOR'S SOLUTIONS MANUAL Single Variable Calculus (Chapters 1-11), ISBN 0-321-60807-01978-0-321-60807-9 Multivariable Calculus (Chapters 11-16), ISBN 0-321-60072-X 1978-0-321-60072-1 The Instructor ~ Solutions Manual by William Ardis, Collin County Community College, contains complete worked-out solutions to all of the exercises in the text.

STUDENrs SOLUTIONS MANUAL Single Variable Calculus (Chapters 1-11), ISBN 0-321-60070-31978-0-321-60070-7 Multivariable Calculus (Chapters 11-16), ISBN 0-321-60071-11978-0-321-60071-4 The Student~ SolutiollS Manual by William Ardis, Collin County Community College, is designed for the stodent and contains carefully worked-out solutions to all the oddnumbered exercises in the text.

JUST-IN-TIME ALGEBRA AND TRIGONOMETRY FOR CALCULUS, Fourth Edition ISBN 0-321-67104-X 1978-0-321-67104-2 Sharp algebm and trigonometry skills are critical to mastering calculus, and Just-in-TIme Algebra and 1Hgonometry for Calculus by Guntram Mueller and Ronald I. Brent is designed to bolster these skills while stodents stody calculus. As stodents make their way through calculus, this text is with them every step of the way, showing them the necessary algebm or trigonometry topics and pointing out potential problem spots. The easy-to-use table of contents has algebm and trigonometry topics armnged in the order in which stodents will need them as they stody calculus.

CALCULUS REVIEW CARDS The Calculus Review Cards (one for Single Variable and another for Multivariable) are a stodent resource containing important formulas, functions, defmitions, and theorems that correspond precisely to Thomas' Calculus. These cards can work as a reference for completing homework assignments or as an aid in studying, and are available bundled with a new text. Contact your Pearson sales representative for more information.

Media and Online Supplements TECHNOLOGY RESOURCE MANUALS Maple Manual by James Stapleton, North Carolina State University Mathematica Manual by Marie Vanisko, Carroll College IT-Graphing Calculator Manual by Elaine McDonald-Newman, Sonoma State University These manuals cover Maple 13, Mathematica 7, and the TI-83 PlUS/TI-84 Plus and TI-89, respectively. Each manual provides detailed gnidance for integrating a specific software package or gmphing calculator throughout the course, including syntax and commands.

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These manuals are available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhigheredlirc, and MyMatbLab.

WEB SITE _w.pearsonhighered.com/thomas The Thomas' Calculus Web site contains the cbapter on Second-Order Differential Equations, including odd-numbered answers, and provides the expanded historical biographies and essays referenced in the text. Also available is a collection of Maple and Mathematica modules, as well as the Technology Application Projects, which can be used as projects by individual students or groups of stodents.

MyMathLab Online Course (access code required) MyMathLab is a text-specific, easily customizable ouline course that integrates interactive multimedia inatruction with textbook content. MyMathLab gives you the tools you need to deliver all or a portion of your course ouline, whether your stodents are in a lab setting or working from home. o Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help. o "Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with. o Personalized Study Plan, generated when students complete a test or quiz, indicates which topics have been mastered and links to tutorial exercises for topics stodents have not mastered. o Multimedia learning aids, such as video lectures, Java applets, animations, and a complete multimedia textbook, help stodents independently improve their understanding and performance. o

Assessment Manager lets you create ouline homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MyMatbLab exercise bank and instructor-created custom exercises.

o Gradebook, designed specifically for mathematics and statistics, automatically tracks students' results and gives you control over how to calculate f'mal grades. You can also add offiine (paper-and-pencil) grades to the gradebook. o MathXL Exercise Builder allows you to create static and algorithmic exercises for your ouline assignments. You can use the library of sample exercises as an easy starting point. o Pearson Tutor Center (www.pearsontutorservices.com) access is automatically included with MyMatbLab. The Tutor Center is staffed by qualified math instructors who provide textbook-specific tutoring for students via toll-free phone, fax, email, and interactive Web sessions. MyMathLab is powered by CourseCompass™, Pearson Education's ouline teaching and learning environment, and by MathXL, our ouline homework, tutorial, and assessment system. MyMatbLab is available to qualified adopters. For more information, visit www.mymathlab.com or contact your Pearson sales representative.

Video Lectures with Optional Captioning The Video Lectures with Optional Captioning feature an engaging team of mathematics instructors who present comprehensive coverage of topics in the text. The lectorers' presentations include examples and exercises from the text and support an approach that emphasizes visualization and problem solving. Available ouly through MyMathLab and MathXL.

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XV

MathXL Online Course (access code required) MathXL is an online homework, tutorial, and assessment system that accompanies Pearson's textbooks in mathematics or statistics. • Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help. • "Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with. • PersonaIized Study Plan, generated when students complete a test or qillz, indicates which topics have been mastered and links to tutorial exercises for topics students have not mastered. • Multimedia learning aids, such as video lectures, Java applets, and animations, help students independently improve their understanding and performance. • Gradebook, designed specifically for mathematics and statistics, automatically tracks students' results and gives you control over how to calculate fmal grades. • MathXL Exercise Builder allows you to create static and algorithmic exercises for your online assignments. You can use the library of sample exercises as an easy starting point. • Assessment Manager lets you create online homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MathXL exercise bank, or instructur-created custom exercises. MathXL is available to qualified adopters. For more information, visit our Web site at www.mathxl.com. or contact your Pearson sales representative.

TestGen® TestGen (www.pearsonhighered.comltestgen)enablesinstructurstobuild.edit.print. and administer tests using a computerized hank of questions developed to cover all the objectives of the text. TestGen is algorithmically based, allowing instructurs to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modifY test hank questions or add new questions. Tests can be printed or administered online. The software and testbank are available for dowuload from Pearson Education's online catalog.

PowerPoint® Lecture Slides These classroom presentation slides are geared specifically to the sequence and philosophy of Thomas' Calculus series. Key graphics from the book are included to help bring the concepts alive in the classroom.These files are available to qualified instructors tbrough the Pearson Jnstructur Resource Center, www.pearsonhigheredlirc, and MyMathLab.

Acknowledgments We would like to express our thanks to the people who made many valuable contributions to this edition as it developed through its various stages:

Accuracy Checkers Blaise DeSesa Paul Lorczak Kathleen Pellissier Lauri Semarne Sarab Streett Holly Zullo

xvi

Preface

Reviewers for the Twelfth Edition Meighan Dillon, Southern Polytechnic State University Anne Dougherty, University of Colorado Said Fariabi, San Antonio College Klaus Fischer, George Mason University Tim Flood, PiUsburg State University Rick Ford, California State University-Chico Robert Gardner, East Tennessee State University Christopher Heil, Georgia Institute ofTechnology Joshua Brandon Holden, Rose-Hulman Institute ofTechnology Alexander Hulpke, Colorado State University Jacqueline Jensen, Sam Houston State University Jennifer M. Johnson, Princeton University Hideaki Kaneko, Old Dominion University Przemo Kranz, University ofMississippi Xin Li, University of Central Florida Maura Mast, University ofMassachusetts-Boston Val Mohanakumar, Hillsborough Community College-Dale Mabry Campus Aaron Montgomery, Central Washington University Cynthia Piez, University ofIdaho Brooke Quinlan, Hillsborough Community College-Dale Mabry Campus Rebecca A. Segal, Virginia Commonwealth University Andrew V. Sills, Georgia Southern University Alex Smith, University ofWlSconsin-Eau Claire Mark A. Smith, Miami University Donald Solomon, University ofWISconsin-Milwaukee Blake Thornton, Washington University in St. Louis David Walnut, George Mason University Adrian Wilson, University ofMontevallo Bobby Winters, PiUsburg State University Dennis Wortman, University ofMassachusetts-Boston

1 FUNCTIONS OVERVIEW Functions are fundamental to the study of calculus. In this chapter we review what functions are and how they are pictured as graphs, how they are combined and transformed, and ways they can be classified. We review the trigonometric functions, and we discuss misrepresentations that can occur when using calculators and computers to obtain a function's graph. The real number system, Cartesian coordinates, straight lines, parabolas, and circles are reviewed in the Appendices. We treat inverse, exponential, and logarithmic functions in Chapter 7.

1.1

Functions and Their Graphs Functions are a tool for describing the real world in mathematical terms. A function can be represented by an equation, a graph, a numerical table, or a verbal description; we will use all four representations throughout this book. This section reviews these function ideas.

Functions; Domain and Range The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the length of time the investment is held. The area of a circle depends on the radius of the circle. The distance an object travels at constant speed along a straight-line path depends on the elapsed time. In each case, the value of one variable quantity, say y, depends on the value of another variable quantity, which we might call x. We say that "y is a function of x" and write this symbolically as y

= f(x)

("y equals f of x").

In this notation, the symbol f represents the function, the letter x is the independent variable representing the input value of f, and y is the dependent variable or output value of

fatx.

DEFINITION

A function f from a set D to a set Y is a rule that assigns a unique E Y to each element XED.

(single) element f(x)

The set D of all possible input values is called the domain of the function. The set of all values of f(x) as x varies throughout D is called the range of the function. The range may not include every element in the set Y. The domain and range of a function can be any sets of objects, but often in calculus they are sets of real numbers interpreted as points of a coordinate line. (In Chapters 13-16, we will encounter functions for which the elements of the sets are points in the coordinate plane or in space.)

1

2

Chapter 1: Functions

J

x --:-_.~

Input (donurin)

FIGURE 1.1

:"""":~~•• J(x)

Output

(nmge)

A diagram showing a

function as a kind of machine.

~

~J(a) D

= domain set

Vx

J(x)

Y = set containing tberange

A function from a setD to a set Yassigns a unique element of Y to each element in D. FIGURE 1.2

Often a function is given by a formula that describes how to calculate the output value from the input variable. For instance, the equation A = 1Tr2 is a rule that calculates the area A of a circle from its radius r (so r, interpreted as a length, can only be positive in this formula). When we derme a function y = f(x) with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real x-values for which the formula gives real y-values, the so-called natura! domain. If we want to restrict the domain in some way, we must say so. The domain of y = x 2 is the entire set of real numbers. To restrict the domain of the function to, say, positive values of x, we would write "y = x 2 ,x > 0." Changing the domain to which we apply a formula usually changes the range as well. The range of y = x 2 is [0, 00). The range of y = x 2, X ;;,: 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation (see Appendix 1), the rangeis {x 2 Ix;;,: 2} or{Yly;;': 4} or [4, 00). When the range of a function is a set of real numbers, the function is said to be rea!valued. The domains and ranges of many real-valued functions of a rea! variable are intervals or combinations of interva1s. The interva1s may be open, closed, or half open, and may be finite or infinite. The range of a function is not a!ways eaay to find. A function f is like a machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (Figure 1.1). The function keys on a calculator give an example of a function as a machine. For instance, the key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the key. A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element of the domain D with a unique or single element in the set Y. In Figure 1.2, the arrows indicate thatf(a) is associated with a, f(x) is associated with x, and so on. Notice that a function can have the same value at two different input elements in the domain (as occurs with f(a) in Figure 1.2), but each input element x is assigned a single output value f(x).

Vx

EXAMPLE 1 Let's verify the natural domains and associated ranges of some simple functions. The domains in each case are the values of x for which the formula makes sense. Function

Domain (x)

Range (y)

y = x2

(-00,00) (-00,0) U (0, 00) [0, 00) (-00,4] [-I, I]

[0, 00)

Y = I/x y=

Vx

y=~ y =

v'1=7

(-00,0) U (0, 00) [0, 00) [0, 00)

[0, I]

Solution The formula y = x 2 gives a rea! y-value for any real number x, so the domain is (- 00, 00). The range of y = x 2 is [0, 00) because the square of any real number is nonnegative and every nonnegative number y is the square of its own square root, y = (vY)2fory;;,: O. The formula y = l/x gives a real y-value for every x except x = O. For consistency in the rules of arithmetic, we cannot divide any number by zero. The range of y = I/x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = I/(I/y). That is, for y oF 0 the number x = I/y is the input assigned to the output valuey. The formula y = Vx gives a real y-value only if x ;;,: O. The range of y = Vx is [0, 00) because every nonnegative number is some number's square root (namely, it is the square root of its own square). In y = '\I'4-=x, the quantity 4 - x cannot be negative. That is, 4 - x ;;,: 0, or x :5 4. The formula gives real y-values for all x :5 4. The range of '\I'4-=x is [0, 00), the set of all nonnegative numbers.

1.1

Functions and Their Graphs

3

The formula y = ~ gives a real y-value for every x in the closed interval from - 1 to 1. Outside this domain. 1 - x 2 is negative and its square root is not a real number. The values of 1 - x 2 vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of ~ is [0. 1]. •

Graphs of Functions If j is a function with domain D, its graph consists of the points in the Cartesian plane whose coordinates are the input-output pairs for j. In set notation. the graph is {(x,j(x)) I XED}.

The graph of the function j(x) = x + 2 is the set of points with coordinates (x,y) for which y = x + 2. Its graph is the straight line sketched in Figure 1.3. The graph of a function j is a useful picture of its behavior. If (x. y) is a point on the graph, then y = j(x) is the height of the graph above the pointx. The height may be positive or negative, depending on the sign of j(x) (Figure 104).

y

X

y=xl

-2 -1 0 1

4 1 0 1

3

2

9 4

2

4

FIGURE 1.3 The graph of f(x) = x + 2 is the set of points (x,y) for whichy has the value x + 2.

FIGURE 1.4

f(x)

EXAMPLE 2

Graph the function y

=

If(x.y) lies on the graph of

f. then the value y = f(x) is the height of the graph above the poiut x (or below x if

is negative).

x 2 over the interval [-2,2].

Solution Make a table ofxy-pairs that satisfy the equation y = x 2• Plot the points (x,y) whose coordinates appear in the table, and draw a smooth curve (labeled with its equation) through the plotted points (see Figure 1.5). •

y

How do we know that the graph of y = x 2 doesn't look like one of these curves? (2.4)

y

--~--~~~~--~~X

-2

-1

FIGURE 1.5

Example 2.

0

2

Graph of the function in

y

4


To find out, we could plot more points. But how would we then connect them? The basic question still remains: How do we know for sure what the graph looks like between the points we plot? Calculus answers this question, as we will see in Chapter 4. Meanwhile we will have to settle for plotting points and connecting them as best we can.

Representing a Function Numerically We have seen how a function may be represented algebraically by a formula (the area function) and visually by a graph (Example 2). Another way to represent a function is numerically, through a table of values. Numerical representations are often used by engineers and scientists. From an appropriate table of values, a graph of the function can be obtained using the method illustrated in Example 2, possibly with the aid of a computer. The graph consisting of only the points in the table is called a scatterplot.

EXAMPLE 3 Musical notes are pressure waves in the air. The data in Table l.l give recorded pressure displacement versus time in seconds of a musical note produced by a toning fork. The table provides a representation of the pressure function over time. If we first make a scatterplot and then connect approximately the data points (t, p) from the table, we obtain the graph shown in Figure 1.6. p (pressure)

TABLE 1.1 Tuning fork data Time

Pressure

Time

0.00091 0.00108 0.00125 0.00144 0.00162 0.00180 0.00198 0.00216 0.00234 0.00253 0.00271 0.00289

-0.080

0.00362 0.00379 0.00398 0.00416 0.00435 0.00453 0.00471 0.00489 0.00507 0.00525 0.00543 0.00562

0.00307 0.00325 0.00344

0.200 0.480 0.693 0.816 0.844 0.771 0.603 0.368 0.099 -0.141 -0.309 -0.348 -0.248 -0.041

0.00579 0.00598

Pressure 0.217 0.480 0.681 0.810 0.827 0.749 0.581 0.346 0.077 -0.164 -0.320 -0.354 -0.248 -0.035

1.0 0.8 0.6 0.4 0.2

t (sec) -0.2 -0.4 -0.6

FIGURE 1.6 A smooth curve through the plotted points gives a graph of the pressure functioo represented by Table l.l (Example 3).

• The Vertical Line Test for a Function Not every curve in the coordinate plane can be the graph of a function. A function 1 can have ouly one value I(x) for each x in its domain, so 110 vertical line can intersect the graph of a function more than once. If a is in the domain of the function I, then the vertical line x = a will intersect the graph of 1 at the single point (a, I(a». A circle cannot be the graph of a function since some vertical lines intersect the circle twice. The circle in Figure 1.7a, however, does contain the graphs of two functions of x: the upper semicircle defmed by the function I(x) = ~ and the lower semicircle defined by the function g(x) = - ~ (Figures 1.7b and 1.7c).

1.1 Y


Y

5

Y

-I --~----~----~-->x

-I

0

--~----~----~~x

o

(C)Y~-~

FIGURE 1.7 (a) The circle is not the graph ofa function; itfails the vertical line test. (b) The upper semicircle is the graph ofa function f(x) = ~. (c) The lower semicircle is the graph ofa functiong(x) = -~.

y

Piecewise-Defined Functions Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value funetion FIGURE 1.8 The absolute value function has domain ( - 00, 00) and range [0, 00). y

Ixl =

x;;"

0

x

-----+1---~f---~x

(a)

(c)

(b)

FIGURE 1.20 Graphs of three algebraic functions.

Trigonometric Functions The six basic trigonometric functions are reviewed in Section 1.3. The graphs of the sine and cosine functions are shown in Figure 1.21. y

w~· _

w

0

w

~

-1

(a) f(x)

FIGURE 1.21

~

sin x

(b) f(x)

~

cosx

Graphs of the sine and cosine functions.

Exponential Functions Functions of the fonn f(x) = aX, where the base a > 0 is a positive constant and a oF 1, are called exponential fuoctions. All exponential functions have domain ( - 00, 00) and range (0, 00), so an exponential function never assumes the value O. We study exponential functions in Section 7.3. The graphs of some exponential functions are shown in Figure 1.22. y

-1

-0.5

y

12

12

10

10

8

8

6

6

4

4

2

2

0

0.5

1

x

(a)

FIGURE 1.22

Graphs of expooential functinos.

-1

-0.5

0 (b)

0.5

1.1


11

Logarithmic Functions These are the functions t(x) = log" x, where the base a '" 1 is a positive constant. They are the inverse jUnctions of the exponential functions, and the calculus of these functions is studied in Chapter 7. Figure 1.23 shows the graphs of four logarithmic functions with various bases. In each case the domain is (0, 00) and the range is (-00, 00). Y

y

--~O~~ ~~~~=Y=~~IO~~XX -1 ,

FIGURE 1.23 functions.

Y = lOglOx

Graphs offour logaritlunic

FIGURE 1.24 Graph ofa cateoary or hanging cable. (The Latin word calena means "chain.")

Transcendental Functions These are functions that are not algebraic. They include the trigonometric, inverse trigonometric, exponential, and logarithmic functions, and many other functions as well. A particular example of a transcendental function is a catenary. Its graph has the shape of a cable, like a telephone line or electric cable, strung from one support to another and hanging freely under its own weight (Figure 1.24). The function defming the graph is discussed in Section 7.7.

Exercises 1.1 Functions

8...

Y

b.

y

In Exercises 1--{j, find 1he domain and range of each function.

=

+ x' v'Sx +

=

4 3_ I

1. f(x) = I 3. F(x) S. f(t)

2. f(x) = 1 10

4. g(x)

=

6. G(t)

=

;Ix

v'x' - 3x 2 t' _ 16 ~------------~X

In Exercises 7 and 8, which of1he graphs are graphs of functions oh, and which are not? Give reasons for your answers.

7.a.y

b.y

o

~-------------+x

o

Finding Formulas for Functions 9. Express 1he area and perimeter of an equilateral triangle as a function of1he triangle's side leng1hx. 10. Express 1he side 1eng1h ofa square as a function of1he 1eng1h d of 1he square's diagonal. Then express 1he area as a function of 1he diagonalleng1h.

~------------~X

o

~-------------+x

o

11. Express 1he edge leng1h ofa cube as a function of1he cube's diagonal 1eng1h d. Then express 1he surface area and volume of 1he cube as a function of1he diagonalleng1h.

12


12. A point P in the first quadrant lies 00 the graph of the functioo I(x) ~ Express the coordinates of P as functions of the slope of the line joining P to the origin.

Vx.

31. a.

13. Consider the point (x, y) lying 00 the graph of the line a + 4y ~ 5. Let L he the distsnce from the point (x, y) to the origin (0, 0). WriteL as a functioo ofx. 14. Consider the point (x, y) lying on the graph of y = ~. Let L he the distsnce hetweeo the points (x,y) and (4, 0). WriteL as a function ofy.

Y

~x

17. g(x) =

a

16. I(x) = I -

'\J'Ix1

19. F(t) = t/ltl

2

20. G(t) =

!fit I

x' -

4 ~

h

(-2,-1)

Y

32. a.

b. Ix

-A

The Greatest and Least Integer Functions 33. For what values oh is

9

x' + -,--. x +4

+ Iyl = I

A

1:

b. rxl=O?

a.lxJ=O?

Iyl=x

Ixl

Y

1

n+-~!'--~x OTT

b. y'=x' 24. Graph the following equations and explain why they are not graphs of functioos of x. L

b.

o

23. Graph the following equations and explain why they are not graphs of functioos of x. L

- --=---t-i-- - x

x'

x~.

21. Find the domain ofy = 22. Find the range of y

a -

18. g(x) =

Y

2

3

I

Functions and Graphs Find the domain and graph the functions in Exercises 15-20.

15. I(x) = 5 -

b.

34. What real nwnhers x satisfY the equation l x J = r xl? 35. Does r -xl = -lxJ for all real x? Give reasons foryour answer. 36. Graph the function

I(x) =

{f:~:

+ yl = I

x~o

x

< O.

Why is I(x) called the integer part ofx?

Piece>rise-Defined Functions Graph the functions in Exercises 25-28.

25. I(x) = {x, 2-x, I - x 26. g(x) = { 2 _ x:

4 27. F(x) ~ { , x

28. G(x)

x'a'

+

~ {I/X, x,

Increasing and Decreasing Functions Graph the functioos in Exercises 37-46. What synnnetries, if any, do the graphs have? SpecifY the intervals over which the function is increasing and the intervals where it is decreasing.

0 :5 x :5 I 1 <x::S;2 Osxsl

1<x:52 x:S 1

,

37. y

= -x'

38.y=_I,

x> I x 1 + x·

+ (4/x)

to-

b. Confirm your findings in part (a) algebraically.

D 68.

a. Graph the functions j(x) = 3/(x - 1) and g(x) = 2/(x together to identify the values of x for which

+ 1)

_3_0

0

1, the graph is scaled:

y

= c f(x)

Y

=

cI f(x)

Stretches the graph of f vertically by a factor of c. Compresses the graph of f vertically by a factor of c.

f(cx) Compresses the graph of f horizontally by a factor of c. y = f(x/c) Stretches the graph of f horizontally by a factor of c. For c = -1, the graph is reflected: y = - f(x) Reflects the graph of f across the x-axis. Reflects the graph of f across the y-axis. y = f( -x)

y

=

EXAMPLE 4

Here we scale and reflect the graph of y =

Yx.

Yx

Yx

(a) Vertical: Multiplying the right-hand side of y = by 3 to get y = 3 stretches the graph vertically by a factor of 3, whereas multiplying by 1/3 compresses the graph by a factor of3 (Figure 1.32).

(b) Horizontal: The graph of y = ~ is a horizontal compression of the graph of y = by a factor of 3, and y = v;j3 is a horizontal stretching by a factor of 3 (Figure 1.33). Note that Y = ~ = so a horizontal compression may correspond to a vertical stretching by a different scaling factor. Likewise, a horizontal stretching may correspond to a vertical compression by a different scaling factor.

Yx

v'3 Yx

Yx

is a reflection of y = (c) Reflection: The graph of y = y = is a reflection across the y-axis (Figure 1.34).

h

y

Yx across the x-axis, and •

y 4

stretch

t

~~~=-;=.x -1 0 1 2 3 4 FIGURE 1.32 Vertically s1retching and compressing the graph y = Vi hy a factor of3 (Example 4a).

3 2

y~

1 ~~:::::::::~::::::"-Y F

Yx

--~~--~-r~~~~--~x

-3

-2

-1

v;j3

FIGURE 1.33 Horizontally s1retching and compressing the graph y = Vi hy a factor of 3 (Example 4b).

FIGURE 1.34 Reflections of the graph y = Vi across the coordinate axes

(Example 4c).

18


EXAMPLE 5

Given the functioof(x) =

X4 -

4x' + 10 (Figure 1.35a), fmdfannulas to

(a) compress the graph horizootally by a factor of 2 followed by a reflectioo across the y-axis (Figure 1.35b). (b) compress the graph vertically by a factor of2 followed by a reflection across the x-axis (Figure 1.35c). y

y

20

10 -1

x

0 -10

x

1

-1

+ 2x3 - 5

Y = _!x 4 2

x

0 -10

-20

-20 (a)

(b)

(c)

FIGURE 1.35 (a) The original graph off (b) The horizontal compression ofy = j(x) in part (a) by a factor of2, followed by a reflection across they-axis. (c) The vertical compression ofy = j(x) in part (a) by a factor of2, followed by a reflection across the x-axis (Example 5).

Solution (a) We multiply x by 2 to get the horizootal compressioo, and by -I to give reflectioo across the y-axis. The formula is obtained by substituting -2x for x in the right-hand side of the equatioo for f: y = f( -2x) = (-2x)4 - 4( -2x)' =

16>:4

+ 10

+ 32x' + 10.

(b) The fannula is

Y =

-!f(x)

=

-!X 4 + 2x' - 5.

•

Ellipses Although they are not the graphs of functions, circles can be stretched horizontally or vertically in the same way as the graphs of functions. The standard equation for a circle of radius r centered at the origin is x2 + y2 = r2. Substituting ex for x in the standard equation for a circle (Figure 1.36a) gives c2x2

y

+ y2

= r2.

(1)

y

y

r

--~----~-----4--~x -r r

-r

o

r c

-r

-r

(a) circle

FIGURE 1.36

--~--------------~---------------+--~x

(b) ellipse, 0 < c

1

1.2 Combining Functions; Shifting and Scaling Graphs

If 0 < c < I, the graph of Equation (I) horizontally stretches the circle; if c > I the circle is compressed horizontally. In either case, the graph of Equation (I) is an ellipse (Figure 1.36). Notice in Figure 1.36 that they-intercepts of all three graphs are always-r and r. In Figure 1.36b, the line segment joining the points (±rlc, 0) is called the major axis of the ellipse; the minor axis is the line segment joining (0, ±r). The axes of the ellipse are reversed in Figure 1.36c: The major axis is the line segment joining the points (0, ±r), and the minor axis is the line segment joining the points (±rlc, 0). In both cases, the major axis is the longer line segment. Ifwe divide both sides of Equation (I) by r', we obtain

y

b

----+-----~~----~--~x

-a

19

a

-b (2)

where a = ric andb = r. Ifa > b, the major axis is horizontal; if a < b, the major axis is vertical. The center of the ellipse given by Equation (2) is the origin (Figure 1.37). Substituting x - h for x, and y - k for y, in Equation (2) results in

FIGURE 1.37 Graph of tire ellipse 2

y2

+2 =

l,a > b,wherethemajor a b axis is horizontal. X

2

(x - h)' (y - kf -'----a"'--'-- + b'

=

(3)

I.

Equation (3) is the standard equation of an ellipse with center at (h, k). The geometric def"mition and properties of ellipses are reviewed in Section 11.6.

Exercises 1.2 Algebraic Combinations In Exercises I and 2, fmd tire domains and ranges of f, g, f

f·g· 1. I(x) = x, 2. I(x) =

+ g, and

-v;=J Vx+J, g(x) = -v;=J g(x) =

4. f(x)

=

g(x) = x 2

I, g(x)

=

I

+ I

+

Vx

+ 5 andg(x)

10. f(x)

=

I

• ,---c-;

vx

= x' - 3, fmd the following.

a. l(g(O))

b. g(/(O))

c. I(g(x))

II. g(/(x))

•• 1(1(-5))

r.

x+2

3 -x' g(x)

go 1(I(x))

b. g(g(x))

+

x'

= x' +

I' h(x)

=

d.y=x-6

•• y=2~

r.

y=

W-=3

13. Copy and coroplere the following table.

I), fmd the following.

g(x)

I(x)

(/0 g)(x)

a. x - 7

Vx

?

3x

?

a. f(g(1/2)) c. f(g(x)) e. 1(1(2))

b. g(/(1/2))

II. g(/(x))

c.

~

~

r.

b. g(g(x))

x x - I

?

go N(x))

x d. - - I x-

••

?

1+1 x

x

f.

I x

?

x

g(g(2))

In Exercises 7-10, write a formula for fog

8. I(x)

= x + I, = 3x + 4,

= 3x, g(x) = 2x

g(x)

h(x)

v2-x

b. y = x'/2

c. y = x 9

b. x + 2

7. I(x)

.~

= x - 3, g(x) = Vx, h(x) = x', and j(x) = 2x. Express each of the functions in Exercises 11 and 12 as a composite involving one or more of j,g, h, andj. 11••• y = Vx - 3 b. y = 2Vx c. y = xl/4 d. y = 4x •• Y = y"(x------:c )'" r. y = (2x - 6)' 3

g(g(2))

6. If f(x) = x - I andg(x) = I/(x

I

+ I, g(x) = x + 4' h(x) = x

12••• y=2x-3

Composites of Functions S. If I(x) = x

=

Let f(x)

In Exercises 3 and 4, fmd the domains and ranges of I, g, I/g, and g/f.

3. I(x) = 2,

9. f(x)

=4

- I, h(x)

0

h.

- x

= x2

?

20

Chapter 1: Functions 22. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs.

14. Copy and complete the following table.

g(x)

(j

f(x)

I x - I

L--

0

g)(x)

?

Ixl

b.

?

x - I x

c.

?

Vx

d.

Vx

?

x x

+I Ixl Ixl

15. Evaluate each expression using the giveo table of values

-2

-I

0

I

2

f(x)

I

0

-2

I

2

g(x)

2

I

0

-I

0

x

23. Match the equations listed in parts (a)-(d) to the graphs in the accompanying figure .

.. j(g(-I))

g{j(O» •. g(j( -2))

b.

d. g(g{2»

.. y

c. j(j( -I)) f.

j(g(I))

-x g(x) = { x ~ I,

.. j(g(O» d. j(j(2»

g{j(3» •. g{j(O»

b.

=

Vx+l,

g(x)

18. f(x) = x 2 , g(x) = I -

=

f.

j(g(I/2)) 0

f and

i-

Vx

x-

0

d. Y = (x

c. g{g{-I»

19. Let f(x) = ~2' Find a function y = g(x) so that (j

+2 + 3)2 - 2

b. Y = (x - 2)2

-2sx 0, we get the inequality sin2 8 + (I - cos 8)2 :5 0'.

:5

6'.

By taking square roots, this is equivalent to saying that and

11- cosfJl

and

-161

so

-161

:5 sinfJ:5

161

:5

I - cosll :5

161.

-"./6

"./4

These inequalities will be useful in the next chapter.

Exercises 1.3 Radians and Degrees 1. On a circle of radius 10m, how long is an arc that subtends a central angle of (8) 4'1f/5 radiaos? (b) 110'? 2. A central aogle in a circle of radius 8 is subtended by ao arc of length 101r. Find the aogle's radian and degree measures. 3. You waot to ma1re ao 80' aog1e by marking ao arc on the perimeter of. 12-in.-diarneter disk and drawing lines from the ends of the arc to the disk's ceoter. To the nearest tenth of an incb, how long should the arc be?

4. If you roll a I-m-diaroeter wheel forward 30 em over level grouod, through what aogle will the wheel turn? Aoswer in radiaos (to the nearest tenth) aod degrees (to the nearest degree). Evaluating Trigonometric Functions S. Copy aod complete the following table of function values. If the function is undefmed at a given angle, enter "UND." Do not use a calculator or tables.

fJ

-3"./2

-"./3

sinO

cos 8 tan 8 cot 8 sec 0 esc 8 In Exercises 7-12, one of sin x, cos x, and tan x is given. Find the other two if x lies in the specified interval.

. 7• smx

= 5'

3

9. cosx

= 3'

[f, 'If] XE[-f,o]

XE

I

8. tanx = 2, XE 5

10. cosx = -1 3'

12. sinx = (}

-".

-2"./3

o

"./2

S1r/6

-t,

[o,f] XE

XE

[f,'lT]

['If, 3;]

3"./4

sinO

cosO tan 8 cot 8 sec 0 esc 8 6. Copy and complete the following table of function values. If the function is undefmed at a given angle, enter "UND." Do not use a calculator or tables.

Graphing Trigonometric Functions Graph the functions in Exercises 13-22. What is the period of each function? 13. sin 2x

14. sin (x/2)

15. cos 71'X

16. cosT

'IfX

'If

. X 17• - 8m T

19. cos(X -

18. - cos 2'1TX

f)

29

1.3 Trigonometric Functions

21. Sin(X -

-;f) + I

22.

COS(X + 2:) -

Solving Trigonometric Equations For Exercises 51-54, solve for the angle 0, where 0 ,;; 0 ,;; 2'IT.

2

Graph the functioos in Exercises 23-26 in the Is-plane (t·axis horizoo· tal, s-axis vertical). What is the period of each function? What sym· metries do the graphs have?

23. s

= cot2t

24. s

25. s

= sec(~t)

26. s

= -tao 'ITt

t

52. sin' 0 = cos' 0

53. sin20 - cosO = 0

54. cos20

+ cosO

=

0

Theory and Examples 55. The tangent snm formula The staodard formula for the taogoot of the smn of two aogles is

= csc(f)

D 27.

D

51. sin' 0 =

a. Graph y = cos x and y = sec x together for - 3'IT/2 ,;; x ,;; 3'IT/2. Cornmeot on the behavior of sec x in relatioo to the signs and values of cos x. b. Grapby = sin x andy = c8Cxtogetherfor-w ~ x ~ 2'17". Comment on the behavior of esc x in relation to the signs and values of sin x. 28. Graph y = taox aod y = cotx together for -7 ,;; x ,;; 7. Comment on the behavior of cot x in relation to the signs and values of taox. 29. Graph y = sinx aod y = l sin x J together. What are the domain and raoge of l sin x J? 30. Graph y = sinx aod y = rsin x 1together. What are the domain

tao(A

+ B)

taoA+taoB

=

I - taoA taoB'

Derive the formula.

56. (ContinUIJtion ofExercise 55.) Derive a formula for tao (A - B). 57. Apply the law of cosines to the triaogle in the accoropaoying figure to derive the formula for cos (A - B). y

1

r

and raoge of sin x 1?

Using the Addition Formulas Use the addition formulas to derive the ideotities in Exercises 31-36.

I-) 33. Sin(X + f) 31. cos(x -

35. cos(A - B)

= sinx

cosx

=

32. cos 34.

Sin(X - -I-) =

-cosx

= cosA cosB + sinA sinB (Exercise 57 provides a =

sinA cosB - cosA sinB

38. What happeos if you take B = 2'IT in the addition formulas? Do the results agree with something you already know? In Exercises 39-42, express the giveo quaotity in terms of sin

cosx.

+ x)

41. Sin(3; -

40. sin(2'IT - x)

x)

42. cose;

+

x)

. ('IT 'IT) . 7'IT 43. Evaluatesffi12assm 4+3. 44. Evaluate cos

58. a. Apply the formula for cos(A - B) to the ideotity sinO cos(I -

37. What happeos if you take B = A in the trigoooroetric ideotity cos(A - B) = cosA cosB + sinA sinB? Does the result agree with something you already know?

39. cos('IT

o

(x + 1-) = -sinx

different derivation.)

36. sin(A - B)

----4---~~~~-l----+_--~x

x aod

0)

to obtain the addition formula for sin(A

b. Derive the formula for cos (A + B) by substituting - B for B in the formula for cos (A - B) from Exercise 35. 59. A triaogle has sides a the leogtil of side c.

= 2 aod b = 3 aod angle C = 60°. Find

60. A triaogle has sides a

= 2 aod b = 3 aod aogle C = 40°. Find

the leogtil ofside c.

The law ofsines says that if a, b, aod c are the sides opposite the angles A, B, and C in a triangle, then

61. The law of sine.

sinA sinB sin C abe .

A

45. Evaluate cos

~.

+ B).

Use the accoropaoying figures aod the identity sin('IT - 0) sin 0 , if required, to derive the law.

If; as cos (-;f + 2:).

=

=

A

46. Evaluate sin ; ; .

Using the Double-Angle Formulas Find the functioo values in Exercises 47-50. 47. cos2 '!!

8

. 2

'IT

49. sm 12

5'IT 48• cos2 12 231T

SO. sin

8

B " - - -"' .-

-'-----' C

B £..----:.,----.J

62. A triaogle has sides a = 2 aod b = 3 and aogle C = 60° (as in Exercise 59). Find the sine of aogle B using the law of sines.

30


63. A triangle has side c = 2 and angles A Find the length a of the side opposite A.

D 64.

= 1f/4 and B = 1f/3.

The appnDimation sin x '" x It is often useful to know that, whenx is measured in radians, sin x ~ x for n'lDDerically small values of x. In Sectioo 3.9, we will see why the approximation holds. The approximatioo error is less than 1 in 5000 if Ixl < 0.1. L With your grapher in radian mode, graph y = sinx and y = x together in a viewiog wiodow about the otigin. What do you see happening as x nears the origin? b. With your grapher in degree mode, graph y = sinx and y = x together about the origin again. How is the picture different froru the ooe obtained with radian mode?

General Sine Curves For f(x)

=

ASine: (x - C») + D,

identify A, B, C, and D for the sine functions in Exercises 65- O. (Logarithmic and exponential functions are presented in Chapter 7.) To obtain the full picture showing both branches, we can graph the function

f(x)

=

1~1·lxll/'.

xl/'

This function equals except at x = 0 (where f is undefined, although 0 1/3 = 0). The graph of f is shown in Figure 1.55b. •

Exercises 1.4 Choosing a Viewing Window

o In Exercises 1-4, use a graphing calculator or coroputer to detennine which of the given viewing windows displays the most appropriate graph of the specified function. 1. f(xl = x' - 7x 2 + 6x L

[-I, I]by[-I, I]

b. [-2, 2] by [-5, 5]

c. [-10, 10] by [-10, 10] d. [-5,5] by [-25, 15] = x' - 4x 2 - 4x + 16

19. f(xl

[-I, I] by [-5,5] c. [-5,5]by[-1O,20] 3. f(xl = 5 + 12x - x' L [-I, I]by[-I, I] c. [-4,4] by [-20,20] 4. f(xl = '1'5 + 4x - x 2 L

[-2,2] by [-2,2]

c. [-3, 7] by [0, 10]

b. [-3, 3] by [-10, 10]

22. f(xl

= -2-

2 23. f(xl = 6x - 15x + 6 4x 2 - lOx

24. f(xl = - - 2

window for the given 3

6. f(xl =

+3 +2

Chapter

8. f(xl

2

x x T - "2 -

16. Y

II

3

25. y

= sin 250x

26. y

= 3 cos 60x

27. y

= cos(;O)

28. y

=

29. y

=x +

_ 2 I 30. y - x + 50 cos IOOx

/OSin3Ox

110 sin (;0)

34. Graph two periods of the function f(x l = 3 cot

15. y = Ix 2

-

-

x-

33. Graph four periods of the function f(xl = - tan 2x.

9. f(xl = x~ 11. y = 2x - 3x 2/' 13. Y = 5x 2/' - 2x x 17. y = x

x' - 5x'

x2

d. [-10, 10] by [-10, 10]

4x' - x' 10. f(xl = x 2(6 - x'l 12. y = x 1/'(x 2 - 8l 14. Y = x 2/'(5 - xl

=

8 x - 9

b. [-2,6]by[-1,4]

functioo and use it to display its graph.

7. f(xl

x +I

32. Graph the upper branch of the hyperbola y2 - 16x 2

o In Exercises 5-30, rmd an appropriate viewing + 15 + 10

1

31. Graph the lower half of the circle dermed by the equation x 2 +2x=4+4y- y 2.

d. [-4,5]by[-15,25]

Finding a Viewing Window

5. f(xl = x' - 4x'

-

21. f(x l = -x2o"_--x"_~6

d. [-20, 20] by [-100, 100] b. [-5, 5] by [-10, 10]

x2

= -2-

2. f(xl L

2 x +2 x2 + I x - I

20. f(xl

=

18. y

=

2x

+I

=

Ix 2

=I

-

- x

xl I

+3

35. Graph the function f(xl

=

sin 2x

36. Graph the function f(xl

=

sin' x.

= I.

I + I.

+ cos 3x.

Graphing in Dot Mode

o Another

way to avoid incorrect conooctions when using a graphing device is through the use of a "dot mode:' which plots only the points. !fyour graphing utility allows that mode, use it to plot the functions in Exercises 37-40. I 38. y = sin~ 37. y = x - 3

39. y

=

xlxJ

40. y

x3

-

X

-

1 I

= -2--

Questions to Guide Your Review

1. What is a function? What is its doroain? Its range? What is an arrow diagraro for a functioo? Give examples. 2. What is the graph of a real-valued functioo of a real varisble? What is the vertical line test?

3. What is a piecewi..,.dermed functioo? Give examples.

4. What are the importsnt types of functions frequently encountered in calculus? Give an example of each type.

Chapter 1 Practice Exercises

35

5. What is meant by an increasing function? A decreasing function? Give an example of each.

11. What is the stsndard equation of an ellipse with center (h, k)? What is its major axis? Its minor axis? Give examples.

6. What is an even function? An odd function? What symmetry properties do the graphs of such functions have? What advantage can we take of this? Give an example of a function that is neither even nor odd.

12. What is radian measure? How do you convert from radians to degrees? Degrees to radians?

7. If f and g are real-valued functions, how are the domains of f + g, f - g, fg, and f / g related to the domains of f and g? Give examples.

14. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions?

8. When is it possible to compose one function with another? Give examples of composites and their values at various points. Does the order in which functions are composed ever matter? 9. How do you change the equation y = f(x) to shift its graph vertically up or down by Ikl units? Horizontally to the left or right? Give examples.

10. How do you change the equation y = f(x) to compress or stretch the graph by a factor c > I? Reflect the graph across a coordinate axis? Give examples.

Chapter

a

13. Graph the six basic trigonometric functions. What symmetries do the graphs have?

15. Stsrting with the identity sin2 8 + cos2 8 = I and the formulas for cos (A + B) and sin (A + B), show how a variety of other trigonometric identities may be derived.

16. How does the formula for the general sine function f(x) = A sin «2'IT/B)(x - e)) + D relate to the shifting, stretching, compressing, and reflection of its graph? Give examples. Graph the general sine curve and identifY the constsnts A, B, e,andD. 17. Name three issues that arise when functions are graphed using a calculator or computer with graphing software. Give examples.

Practice Exercises

Functions and Graphs 1. Express the area and circumference of a circle as functions of the circle's radius. Then express the area as a function of the circumference.

2. Express the radius of a sphere as a function of the sphere's surface area. Then express the surface area as a function of the volume. 3. A point P in the frrst quadrant lies on the parabola y = x 2 • Express the coordinates of P as functions of the angle of inclination of the line joining P to the origin. 4. A hot-air balloon rising straight up from a level field is traclred by a range fmder located 500 ft from the point ofliftoff. Express the balloon's height as a function of the angle the line from the range fmder to the balloon makes with the ground.

Iu Exercises 19. y =

11. y

= x2 + I = I - cosx

13• y

=~ 3

x - 2x 15. Y = x + cosx

12. y

= x' - x 3 = sec x tsnx

14. Y

=

10. y

16. y

fmd the (a) domain and (b) range.

2

20. y = -2

= Vl6 - x

23. y

= 2e~

25. y

=

27. y

= Iu (x

2

- 3

2sin(3x

+ 'IT)

- 3)

- I

+I

+~ +I

22. y =

32 ->

24. y

=

tsn(2x - 'IT)

26. y

=

x 2/'

28. y = -I

+ ~2 - x

29. State whether each function is increasing, decreasing, or neither. a. Volume of a sphere as a function of its radius b. The greatest integer function

c. Height above Earth's sea level as a function of atmospheric pressure (assumed nonzero) d. Kinetic energy as a function ofa particle~ velocity

Iu Exercises ~16, deterntine v.lrether the function is even, odd, or neith"" 9. y

Ixl -

21. y

Iu Exercises 5--1!, determine whether the graph of the function is sym-

metric about the y-axis, the origin, or neither. 5. y = x'i' 6. y = x 2/' 7. y = x 2 - 2x - I 8. y = e-"

1~28,

30. Find the largest interval on which the given function is increasing.

a. f(x)=lx-21+ I c. g(x) = (3x - 1)'/3

x

x - sinx

= xcosx

b. f(x)

= (x +

I)'

d. R(x)=~

Piecewise-Defined Functions Iu Exercises 31 and 32, fmd the (a) domain and (b) range.

17. Suppose that f and g are both odd functions defined on the entire real1ine. Which of the following (where defmed) are even? odd?

31

a. fg b. f3 c. f(sinx) d. g(sec x) •• Igl 18. If f(a - x) = f(a + x), show that g(x) = f(x + a) is an even function.

32.y=

_ .y-

{h, Yx,

-4 S x'" 0 0<xs4

-x - 2 {

x,

'

-x + 2,

-2::S; x::S; -1 -1 < x::s; 1

1 <x::S;2

36


In Exercises 33 and 34, write a piecewise formula for the function.

e. S1retch vertically by • factor of 5

33.

f. Compress horizontally by a factor of 5

Y

34.

Y (2,5)

50. Describe how each graph is obtained from the graph ofy = f(x).

~J---~--..x

o

4

(f" g)(-l).

d. (g" g)(x).

I x'

~

g(x) =

36. f(x) = 2 - x,

g(x) =

37. f(x) = 2 - x 2 ,

g(x) =

Yx,

=

d. y = f(2x + I)

e. Y=f(t)-4

f. Y = -3f(x)

51. y=

-~1 +~

52. y=

I-t

53. y

~2 +

54. Y

(-5x)1j3

=

I

V'x+l

g(x)

{=~,

=

-

=

f "g

and g "

f

and

Vx+2

~

-4::;;x:5-1 -1<x:51 l<x:52

2,

x - 2,

40.

f (x)

+ I, x _ I,

X

= {

-2 '" x < 0 0"'x"'2

Composition with absolute wines In Exercises 41-48, graph h and h together. Then describe how applying the absolute wlue function in h affects the graph of h

j,(x)

55. Y

= cos2x

57. Y

=

Ixl

42. x 2

Ixl

Ix'i Ix

(x - f). + (x + .;(J sin

In Exercises 61-{i4, ABC is a right triangle with the right angle at C. The sides opposite angles A, B, and C are a, b, and c, respectively. 61. a. Findaandbifc = 2,B = 1f/3. b. Findaandcifb = 2,B = 1f/3. 62. a. Express a in terms of A and c. b. Express a in terms of A and b.

b. Express c in terms of A and a.

2

b. Express sin A in terms of b and c.

+ xl 2

14 - x 1

xI

I Ixl

47.

Yx

vfxj

48. sinx

sinlxl

Shifting and Scaling Graphs 49. Suppose the graph of g is given. Write equations for the graphs that are obtsined from the graph of g by shifting, scaling, or

reflecting, as indicated.

tuni~

1fX

COST

64. a. Express sin A in terms of a and c.

46.

Up

58. Y =

sin1fX

60. Sketch the graph y = 1

2

44.x 2 +x

L

. x = sm 2

63. a. Express a in terms of Band b.

43. x' 45.4-x2

56• y

59. Sketch the graph y = 2 cos

hex)

41. x

=

Trigonometry In Exercises 55-58, sketch the graph of the given function. What is the pcriod of the function?

For Exercises 39 and 40, sketch the graphs of f and f " f.

39. f(x)

+ 4I

x+2

In Exercises 37 and 38, B

10. Show that the area of triangle ABC is given by Vs(s - a)(s - b)(s - c)wheres = (a + b seroiperimeter of the triangle.

+ O(x),

15

16. •• Find the slope of the line from the origin to the midpoint P of sideAB in the triangle in the accoropanying figure (a, b > 0).

+ c)/2isthe

11. Show that if 1 is both even and odd, then I(x) = 0 for every x in the domain of I. 12. a. Even-odd decompositions Let 1 be a function whose domain is symmetric about the origin, that is, -x belongs to the domain whenever x does. Show that 1 is the sum of an even function and an odd function:

I(x) = E(x)

10 Kilometers

Y

~~----------~A~~-,~O)~X

b. When is OP perpendicular to AB?

38


17. Consider the quarter-circle of radius I and right triangles ABE and ACD given in the accompanying figure. Use standard area formulas to conclude that I . Zsm6cos6

18. Let f(:x:) ~ ax + b and g(:x:) ~ ex + d. What conditioo must be satisfied by the coostants a, b, c, d in order that (f 0 g)(:x:) ~ (g 0 f)(:x:) for every value of:x:1

I sin6

6

< Z < Z cos 6'

y (0,1)

~~8~____~~ D~__~x A

Chapter

t!I

E

(1,0)

Technology Application Projects

All Overview ofMathemotica An overview of Mathematica sufficient to complete the Mathematica modules appearing on the Web site.

Mathematica/Maple Module: Modeling Change: Springs, Drivillg Safety, JlJJdioactivity, 1l-e.., Fish, and Mammals Coostruct and interpret mathematical models, analyze and improve them, and make predictioos using them.

2 LIMITS AND CONTINUITY OVERVIEW Mathematicians of the seventeenth century were keenly interested in the study of motion for objects on or near the earth and the motion of planets and stars. This study involved both the speed of the object and its direction of motion at any instant, and they knew the direction was tangent to the path of motion. The concept of a limit is fundamental to finding the velocity of a moving object and the tangent to a curve. In this chapter we develop the limit, first intuitively and then formally. We use limits to describe the way a function varies. Some functions vary continuously; small changes in x produce only small changes in f(x). Other functions can have values that jump, vary erratically, or tend to increase or decrease without bound. The notion of limit gives a precise way to distinguish between these behaviors.

2.1

Rates of Change and Tangents to Curves Calculus is a tool to help us understand how functional relationships change, such as the position or speed of a moving object as a function of time, or the changing slope of a curve being traversed by a point moving along it. In this section we introduce the ideas of average and instantaneous rates of change, and show that they are closely related to the slope of a curve at a point P on the curve. We give precise developments of these important concepts in the next chapter, but for now we use an informal approach so you will see how they lead naturally to the main idea of the chapter, the limit. You will see that limits playa major role in calculus and the study of change.

Average and Instantaneous Speed HISTORICAL BIOGRAPHY*

Galileo Galilei (1564- 1642)

In the late sixteenth century, Galileo discovered that a solid object dropped from rest (not moving) near the surface of the earth and allowed to fall freely will fall a distance proportional to the square of the time it has been falling. This type of motion is called free faU. It assumes negligible air resistance to slow the object down, and that gravity is the only force acting on the falling body. If y denotes the distance fallen in feet after t seconds, then Galileo's law is

y = 16t 2 ,

where 16 is the (approximate) constant of proportionality. (If y is measured in meters, the constant is 4.9.) A moving body's average speed during an interval oftime is found by dividing the distance covered by the time elapsed. The unit of measure is length per unit time: kilometers per hour, feet (or meters) per second, or whatever is appropriate to the problem at hand.

*To learn more about the historical figures mentioned in the text and the development of many major elements and topics of calculus, visit www.aw.comlthomas.

39

40

Chapter 2: Limits and Continuity

EXAMPLE 1

A rock breaks loose from the top of a tall cliff. What is its average speed

x

2+h NOTTOSCALB

FIGURE 2.4 Finding 1he slope of1he parabola y ~ x 2 at 1he point P(2, 4) as 1he

limit of secant slopes (Example 3).

2.1

Rates of Change and Tangents to Curves

43

The tangent to the parabola at P is the line through P with slope 4:

y = 4

+ 4(x

- 2)

Point-slope equation

•

Y = 4x - 4.

Instantaneous Rates of Change and Tangent Lines The rates at which the rock in Example 2 was falling at the instants t = 1 and t = 2 are called instantaneous rates of change. Instantaneous rates and slopes of tangent lines are intimately connected, as we will now see in the following examples.

EXAMPLE 4 Figure 2.5 shows how a populationp of fruit flies (Drosophila) grew in a 50-day experiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time t, and the points joined by a smooth curve (colored blue in Figure 2.5). Find the average growth rate from day 23 to day 45. Solution There were 150 flies on day 23 and 340 flies on day 45. Thus the number of flies increased by 340 - 150 = 190 in 45 - 23 = 22 days. The average rate of change of the population from day 23 to day 45 was /lp 340 - 150 190 . Average rate of change: T/ = 45 23 = 22 '" 8.6 flIes/day. p

350 ~

V

300

/

" 250

'S

.l!

200

PI ,2 3,150)

.~ 150 Z

~

100 /

50

/

o

10

V

/ ~ 4, -

V

~,~

/

/ Q(45 340)

-tJ.-p-=r 190

8. !lie day

22

'/ 20

30 Tim. (days)

40

50

FIGURE 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of chaage over 22 days is the slope !!'p/!!.t of the secant line (Example 4).

This average is the slope of the secant through the points P and Q on the graph in Figure 2.5. • The average rate of cbange from day 23 to day 45 calculated in Example 4 does not tell us how fast the population was cbanging on day 23 itself. For that we need to examine time intervals closer to the day in question.

EXAMPLE 5

How fast was the number of flies in the population of Example 4 growing

on day 23?

Solution To answer this question, we examine the average rates of change over increasingly short time intervals starting at day 23. In geometric terms, we find these rates by calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure 2.6).

44

Chapter 2: Limits and Continuity p

= fl.p / fl.t

Q

Slope of PQ (flies / day)

(45,340)

340 - 150 '" 8 6 45 - 23 .

'a

(40,330)

330 - 150 106 40-23'" .

iz

(35,310)

310 - 150 '" 13 3 35 - 23 .

(30,265)

265 - 150 '" 164 30 - 23 .

FIGURE 2.6

B(35, 350) ;,1 ~ /'

350

J: h ~/

300

~

250

/

Q(45,340)

AV

200

P(23,150»

ISO

fY

I('

100

/

50 /

0

..:::. ~

~ .

/ 10' ~ 30 \ 20 A(14,0) Time (days)

40

50

The positions and slopes offour secants through the pointP on the fruit fly grapb (Example 5).

The values in the table show that the secant slopes rise from 8.6 to 16.4 as the I-coordinate of Q decreases from 45 to 30, and we would expect the slopes to rise slightly higher as t continued on toward 23. Geometrically, the secants rotate about P and seem to approach the red tangent line in the figure. Since the line appears to pass through the points (14, 0) and (35, 350), it has slope

~;0_-1~

=

16.7 flies/day (approximately).

On day 23 the population was increasing at a rate of about 16.7 flies/day.

•

The instantaneous rates in Example 2 were found to be the values of the average speeds, or average rates of change, as the time interval ofiength h approached O. That is, the instantaneous rate is the value the average rate approaches as the length h of the interval over which the change occurs approaches zero. The average rate of change corresponds to the slope of a secant line; the instantaneous rate corresponds to the slope of the tangent line as the independent variable approaches a fixed value. In Example 2, the independent variable I approached the values I = I and I = 2. In Example 3, the independent variable x approached the value x = 2. So we see that instantaneous rates and slopes of tangent lines are closely connected. We investigate this connection thoroughly in the next chapter, but to do so we need the concept of a limil.

Exercises 2.1 Average Rate. of Change In Exercises I-j), fmd the average rate of chaoge of the function over the given interval or intervaIs. 1. !(x) ~ x 3

+

b.[-I,I]

2. g(x) ~ x 2

a. [-1,1] 3. h(t) ~ cot t a. ["./4,3"./4] 4. g(t) ~ 2 + cost •• [0,,,.]

R(O) ~

\I'49+l;

6. P(O) ~ 03 - 411'

[0,2]

+ 50;

[1,2]

Slope of a Curve at a Point

I

•• [2,3]

s.

b. [-2,0]

In Exercises 7-14, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (h) an equation of the tangent line atP.

7. y ~ x 2

-

3, P(2, I)

8. Y ~ 5 - x 2 , b. ["./6, "./2]

9. Y

~ x2 -

10. Y ~ x 2 11. Y ~ x 3,

-

P(I,4)

2x - 3,

P(2, -3)

4x, P(I, -3)

P(2, 8)

2.1

12. Y = 2 13. Y

=

x' -

x',

+ 4,

t. Use yoor graph to estimate the rate at which the profits were

P(2, 0)

changing in 2002.

D 18.

Instantaneous Rates of Change 15. Speed of a tar The acoompanying figure shows the time1D-distance graph for a sports car accelerating from a standstill. p

650 600

J;j-

D 19.

~

Q,

o

V

10

15

t. What does your table indicate is the rate of chaage of g(x)

20

with respect to x at x = I?

Elapsed time (sec)

a. Estimate the slopes of secants PQ" PQ2, PQ" and PQ., ammging them in order in a table like the one in Figure 2.6. What are the appropriate units for these slopes? b. Then estimate the car's speed at time I = 20 sec.

16. The accompanying figure shows the plot of distance falleo versus time for ao object that fell froro the lunar landing module a distance 80 m to the surface of the moon. a. Estimate the slopes of the secaots PQ" PQ2,PQ" andPQ., ammging them in a table like the one in Figure 2.6. b. About how fast was the object going when it hit the surfa..? y

80 I

Q / Q

o

-

10'

p

Q,

~'1 f-

D 20.

Let/(l) = 1/1 for I '" O.

a. Find the average rate of chaage of I with respect to I over the intervals (i) froro I = 2 to I = 3, aod (ii) froro I = 2 to I = T. b. Make a table of values of the average rate ofchaoge of I with respect to lover the interval [2, 11, for some values of T approachiog 2, say T = 2.1,2.01,2.001,2.0001,2.00001, aod 2.000001. t. What does your table indicate is the rate of chaage of I with

respect to I at I = 2? d. Calculate the limit as T approaches 2 of the average rate of chaoge of I with respect to lover the interval froro 2 to T. You will have to do soroe algebra before you cao substitute T = 2. 21. The accoropaoying graph shows the tota1 distance s traveled by a bicyclist after I hours.

./

s

5

10

The profits of a sma11 company for each of the first five years of its operation are given in the following table: Year

d. Calculate the limit as h approaches zero of the average rate of chaage ofg(x) with respect to x over the interval [I, I + h].

V

Elapsed time (sec)

D 17.

Vx folX "" O.

aod 0.000001.

/'

......

Letg(x) =

a. Find the average rate of chaage of g(x) with respect to x over the intervals [1,2], [I, 1.5] and [I, 1+ h]. b. Make a table of values of the average rate ofchaoge ofg with respect to x over the interval [I, I + h] for some values ofh approaching zero, say h = 0.1, O.o!, 0.001, 0.0001, 0.00001,

/ 5

Make a table of values for the function F(x) = (x + 2)/(x - 2) at the points x = 1.2, x = l1/lO,x = 101/100, x = 1001/1000, x = 10001/10000, aodx = 1.

a. Find the average rate ofchaage ofF(x) over the intervals [I,x] for each x '" I in your table. b. Exteoding the table if necessary, 1ry to determine the rate of change of F(x) at x = 1.

s

100

45

b. What is the average rate of increase of the profits between 2002 aod 2004?

P(I, I)

12x, P(I, -11)

14. Y = x' - 3x2

Rates of Change and Tangents to Curves

Profit in $1000.

!

~

40

~

20

J

10

"

/'

30 ./

,,./ /

/ 0

2

3

4

Elapsed time (br)

2000 2001 2002 2003 2004

6 27 62

III 174

a. Plot points representing the profit as a function of year, aod join them by as smooth a curve as you can.

a. Estimate the bicyclist's average speed over the time intervals [0, I], [1,2.5], aod [2.5, 3.5]. b. Estimate the bicyclist's instantaneous speed at the times t = 1= 2,aodl = 3.

!,

t. Estimate the bicyclist's maximum speed aod the specific time

at which it occurs.

46


22. The accompanying graph shows the total amount of gasoline A in the gas tank of an automobile after being driven for I days.

a. Estimate the average rate of gasoline consumption over the time intervals [0, 31, [0, 51, and [7, 101. b. Estimate the instantaneous rate of gaaoline consomption at the times I ~ 1,1 ~ 4,andl ~ 8.

A

:;

,,

16

r:

:(

c. Estimate the maximum rate of gasoline consumption and the specific time at which it occurs. ~

[,

4

,l!

.....

-

012345678910 Elapsed time (days)

Limit of a Function and Limit Laws

2.2

In Section 2.1 we saw that limits arise when finding the instsntsneous rate of change of a function or the tsngent to a curve. Here we begin with an infonnal definition of limit and show how we can calculate the values of limits. A precise definition is presented in the next section.

Limits of Function Values

HIsTORICAL ESSAY

Frequently when studying a function y ~ I(x) we rmd ourselves interested in the function's behavior near a particular point xo, but not at Xo. This might be the case, for instsnce, if Xo is an irrational number, like 1T or whose values can only be approximated by "close" rational numbers at which we actoally evaluate the function instead. Another situation occurs when trying to evaluate a function at Xo leads to division by zero, which is undefined. We encountered this last circumstsnce when seeking the instsntsneous rate of change in y by considering the quotient function tly/h for h closer and closer to zero. Here's a specific example where we explore numerically how a function behaves near a particular point at which we carmot directly evaluate the function.

Lintits

V2,

y

2

__~--~----~------~X

o

EXAMPLE 1

How does the function

I(x)

y

behave near x 2 y=x+l

o FIGURE 2.7 The graph off is identical with the line y ~ x + 1 except at x = 1, where f is not

defmed (Example 1).

~

~x

2 -

1

x-I

I?

Solution The given formula defines I for all real numbers x except x ~ 1 (we carmot divide by zero). For any x oF 1, we can simplify the formula by factoring the numerator and canceling common factors: () Ix=

(x - I)(x + 1)

x-I

~x+I

for

x oF 1.

The graph of I is the line y ~ x + 1 with the point (I, 2) removed. This removed point is shown as a "hole" in Figure 2.7. Even though 1(1) is not dermed, it is clear that we can make the value of I(x) as close as we want to 2 by choosing x close enough to 1 (Table 2.2).

•

2.2 Limit of a Function and Limit Laws

47

TABLE 2.2 The closer x gets to 1, the closer f(x) = (x' - 1)/(x - 1)

seems to get to 2 Values of x below and above 1

x, -1 !(x) = --1x- = x + 1,

0.9 l.l 0.99 1.01 0.999 1.001 0.999999 1.000001

1.9 2.1 1.99 2.01 1.999 2.001 1.999999 2.000001

x;ld

Let's generalize the idea illustrated in Example I. Suppose !(x) is defined on ao open interval about xo, except possibly at Xo itself. If !(x) is arbitrarily close to L (as close to L as we like) for all x sufficiently close to xo, we say that f approaches the limit L as x approaches xo, aod write lim !(x) = L,

x-x,

which is read ''the limit of f(x) as x approaches Xo is L." For instance, in Example I we would say that !(x) approaches the limit 2 as x approaches I, aod write lim f(x) = 2,

x-I

or

2 - I lim-x-- = 2. x-I

x-I

Essentially, the def'mition says that the values of f(x) are close to the number L whenever x is close to Xo (on either side of xo). This def'mition is ''informal'' because phrases like arbitrarily close aod suffICiently close are imprecise; their meaoing depends on the context. (fo a machinist maoufacturing a piston, close may meao within a few thousandths of an inch. To ao astronomer studying distant galaxies, close may mean within a few thousand light-years.) Nevertheless, the def'mition is clear enough to enable us to recognize aod evaluate limits of specific fimctions. We will need the precise def'mition of Section 2.3, however, when we set out to prove theorems about limits. Here are several more examples exploring the idea oflimits.

EXAM PLE 2 This example illustrates that the limit value of a function does not depend on how the function is def'med at the point being approached. Consider the three functions in Figure 2.8. The function! has limit 2 as x --+ I even though f is not defined at x = I. y

y

y

---cr.--c+--;-.x (a) f(x)

~ ~~

i

X2 -

1

- - , x'l-l x-I

(b) g(x) ~ {

I,

x

=

(c) h(x)

~

x

+1

1

FIGURE 2.8 The limits of f(x), g(x), aod h(x) all equal 2 as x approaches 1. However, ooIy h(x) has the same functioo value as its limit at x ~ 1 (Example 2).

48

Chapter 2: Limits and Continuity The function g has limit 2 as x .... 1 even though 2 i' g( 1). The function h is the only one of the three functions in Figure 2.8 whose limit as x .... 1 equals its value at x = 1. For h, we have limx~l h(x) = h(I). This equality oflimit and function value is significant, and we return to it in Section 2.5. •

y

EXAMPLE 3 (a) If f is the identity function f(x) = x, then for any value ofxo (Figure 2.9a), (a) Identity function

lim f(x) = lim x = Xo.

x-xo

y

X--+Xo

(h) If f is the constant function f(x) = k (function with the constant value k), then for any value ofxo (Figure 2.9b), lim f(x) = lim k = k. X-XcI

°

----------~----~----~x

xo

limx=3

x~3

(b) Constant function

FIGURE 2.9 The functioos in Example 3 have limits at all points xo.

X--+Xo

For instances of each of these rules we have and

lim (4) = lim(4) = 4.

x--+-7

x--+2

•

We prove these rules in Example 3 in Section 2.3.

Some ways that limits can fail to exist are illustrated in Figure 2.10 and described in the next example. y

y~{o. 1,

y

xO

{o..

sm

x sO

1

x'

x>O

2.2


49

Solution (8) It jumps: The unit step function U(x) has no limit as x --> 0 because its values jump at x = O. For negative values of x arbitrarily close to zero, V(x) = O. For positive values of x arbitrarily close to zero, U(x) = 1. There is no single value L approached by U(x) as x --> 0 (Figure 2. lOa).

(h) It grows too "large" to have a limit: g(x) has no limit as x --> 0 because the values ofg grow arbitrarily large in absolute value as x --> 0 aod do not stay close to any f'lXed real number (Figure 2.1 Ob). (0) It oscillates too much to have a limit: f(x) has no limit as x --> 0 because the functioo's values oscillate between + 1 aod -1 in every open interval cootaining O. The values • do not stay close to aoy one number as x --> 0 (Figure 2.10c).

The Limit Laws When discussing limits, sometimes we use the notatioo x --> Xo if we want to emphasize the point Xo that is being approached in the limit process (usually to enhance the clarity of a particular discussioo or example). Other times, such as in the statements of the following theorem, we use the simpler notation x --> c or x --> a which avoids the subscript in Xo. In every case, the symbols Xo, c, aod a refer to a single point 00 the x-axis that may or may not beloog to the domain of the functioo involved. To calculate limits of functioos that are arithmetic combinatioos offunctioos having known limits, we cao use several easy rules.

THEOREM l-Umit Laws lim f(x) = L

If L, M, c, aod k are real numbers aod aod

x~c

1. Sum Rule: 2. Difference Rule:

3. Constant Multiple Rule: 4. Product Rule: 5. Quotient Rule: 6. Puwer Rule:

7. Root Rule:

lim g(x) = M,

then

x~c

lim(j(x)

x-c

+ g(x»

=

L

+M

lim(j(x) - g(x» = L - M

x-c

lim(k' f(x)) = k· L

x-c

lim(j(x)' g(x» = L' M

x-c

lim f(x) = £ g(x) M'

x~c

M "i' 0

lim[f(x))" = L", n a positive integer

x-c

lim '{Y'fW = x~c

(If n is even, we assume that limf(x) = L x~c

>

"ifi =

L 1/", n a positive integer

0.)

In words, the Sum Rule says that the limit of a sum is the sum of the limits. Similarly, the next rules say that the limit of a difference is the difference of the limits; the limit of a coostant times a functioo is the constant times the limit of the functioo; the limit of a product is the product of the limits; the limit of a quotient is the quotient of the limits (provided that the limit of the denOOIinator is not 0); the limit ofa positive integer power (or root) ofa functioo is the integer power (or root) of the limit (provided that the root of the limit is a real number). It is reasonable that the properties in Theorem 1 are true (although these intuitive arguments do not constitute proofs). Ifx is sufficiently close to c, then f(x) is close to L aod g(x) is close to M, from our informal def'mitioo of a limit. It is then reasonable that f(x) + g(x) is close to L + M; f(x) - g(x) is close to L - M; kf(x) is close to kL; f(x)g(x) is close to LM; aod f(x)/g(x) is close to LIM if M is not zero. We prove the Sum Rule in Section 2.3, based on a precise def'mition oflimit. Rules 2-5 are proved in

50

Chapter 2: Limits and Continuity Appendix 4. Rule 6 is obtained by applying Rule 4 repeatedly. Rule 7 is proved in more advanced texts. The sum, difference, and product rules can be extended to any number of functions, not just two.

EXAMPLE 5 Use the observations lim....... , k = k and limx~, x = c (Example 3) and the properties oflimits to f"md the following limits. x

-1

I -

0

FIGURE 2.13 Aoy functioo u(x) whose graph lies io the regioo betweeo y ~ I + (x'/2) andy = I - (x'/4) has limit I as x --> 0 (Example 10).

4x

:5 u(x) :5 I

Solution

Since lim(1 - (x2/4» = I

and

+

(x 2/2» = I,

•

The Sandwich Theorem helps us establish several important limit rules:

(a) lim sin 0 = 0 -1r

lim(I

x~o

the Sandwich Theorem implies thatlimx~o u(x) = I (Figore 2.13).

EXAM PLE 11 ~------~~--------~ 8

for all x oF 0,

fmd lim..... o u(x) , no matter how complicated u is.

x~o

y

x2

+2

(h) lim cos 0 = I

6-0

8-0

(c) For any function f, lim If(x) I = 0 implies lim f(x) = x-c

x-c

o.

Solution (a)

y

(a) In Section 1.3 we established that -101:5 sinO :5101 for all 0 (see Figure 2.14a). Since lim'hO ( -I 0 I) = lim'hO 10 I = 0, we have lim sinO = O.

8~O

(h) From Section 1.3,0:5 I - cosO :5101 for all 0 (see Figore 2.14b), and we have Inn,,~o (I - cos 0) = 0 or

---_~2~~-1~~O~~1~-2~~8 lim cos 0 = I.

(b)

FIGURE 2.14 The Sandwich Theorem

coofl1'!I1S the limits io Example II.

8~O

(c) Since -If(x)l:5 f(x):5 If(x) I and-lf(x)1 and If(x) I have limit 0 as x-->c,it follows that limx~, f(x) = O. •

54


Another important property oflimits is given by the next theorem. A proof is given in the next section.

THEOREM 5 possibly at x then

TIf(x) :5 g(x) for ailxin some openintervai containingc, except x approaches c,

= c itself, and the limits of f and g both exist as

lim fix) :5 lim g(x).

x--"'c

x--"'c

The assertion resulting from replacing the less than or equal to (:5) inequality by the strict less than «) inequality in Theorem 5 is false. Figure 2.l4a shows that for 0 .. 0, -101 < sin 0 < 10 I, but in the limit as 0 --> 0, equality holds.

Exercises 2.2 Limits from Graphs 1. For the function g(x) graphed here, fmd the following limits or explain why they do not exist.

a. lim g(x) x-I

b. lim g(x) x-2

c. lim g(x)

d.

%-3

y

lim g(x)

x-2.S

y

4. Which of the following statements about the function y graphed here are 1rue, and which are false?

a. b. c. d.

2. For the function f(/) graphed here, fmd the following limits or explain why they do not exist.

a. lim f(/)

b. lim f(/)

1--2

1--1

c. lim f(/)

d.

1-0

lim f(/)

lim f(x) does not exist.

x~2

lim f(x) = 2 lim f(x) does not exist. x~l lim f(x) exists at every pointxo in (-I, I). x~2

x~

t---{).S

...

e. lim f(x) exists at every pointxo in (I, 3).

s

x~

...

IL

ro -2

-I

= f(x)

y

I

0

1

)

t

-I

I

3. Which of the following statements about the function y = f(x) graphed here are 1rue, and which are false? a. lim f(x) exists. x~o

b. lim f(x) = 0 x~o

c. lim f(x) = I x~o

d. lim f(x) = I x~l

e. lim f(x) = 0 x~l

f. lim f(x) exists at every point Xo in (-1, 1). x~

...

g. lim f(x) does not exist. x~l

Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. 5lim~

.

x~o

Ixl

6. lim _1-1 x-I x-

7. Suppose that a function f(x) is defined for all real values ofx except x = Xo. Can anything be said about the existence of lim~" f(x)? Give reasons for your answer. 8. Suppose that a function f(x) is defmed for all x in [-I, I]. Can anything be said about the existence of limx~o f(x)? Give rea-

sons for your answer.

2.2

9. Iflim,.....1 f(x) ~ 5, must f be defmed at x ~ I? If it is, must f( I) ~ 5? Can we conclude anything about the values of f at x ~ I? Explain. 10. If f(1) ~ 5, must lim'~1 f(x) exist? If it does, thea must lim,.....1 f(x) ~ 5? Can we conclude anything about lim,.....1 f(x)? Explain.


Using Limit Rules

51. Suppose limr-o f(x) ~ 1 and limz~o g(x) ~ -5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. lim 2f(x) - g(x) ~ ,~o (j(x)

+ 7)2/3

%--7

+ 7)2/3

(a)

14. lim

x~o

(b)

+ 7))2/3

2 lim f(x) - lim g(x)

z,,' + 4x + 8)

(x 3 -

x-o

( lim (J(x)

+ 5x - 2)

%-2

1-6

lim (j(x)

x-o

12. lim (-x'

13. lim 8(t - 5)(t - 7)

- g(x))

lim 2f(x) - lim g(x)

Find the limits in Exercises 11-22.

+ 5)

1!'!l, (2f(x) x~o

Calculating Limits

11. lim (z"

55

x-o

%-+-2

x-a

(c)

16. lim 38(Zs - 1) s-2/3

17. lim 3(z" - I)'

18.lim

19. lim (5 - y)4/3

20. lim (2z - 8)1/3

%--1

y--3

21. lim

h~O

(2)(1) - (-5) (1 + 7)'/3

y+2

y-2 y2

+

5y

+

6

z~o

3

v'3h+1 + I

22. lim

v5h+4 -

4-0

2

52. Let lim,.....1 h(x) ~ 5, limz~lp(X) ~ 1, and lim'~1 r(x) ~ 2. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.

h

Limits of quotients Find the limits in Exercises 23-42.

lim

x~1

23.lim x - 5 x-s x 2 - 25

24. lim

x'+3x-1O lim 25. %--5 X + 5

' x2 - 7x + 10 26. lun 2 %-2 x 2 28. lim t + 3t + 2 t--l t 2 - t - 2

29. lim -z" - 4 %-+-2 x 3 + 2,x2 ! - I

y-o

39. lim

x-I

Vx+3 - 2 W+12 - 4

x-2

X -

41. lim 2 x--3

2

w-=s +

X

3

y'5lim h(x) ,~1

,

38. lim

W+8 -

%-+-1

40. lim

X

+1

x

+2

53.

Supposelim,~,,(x) ~

+ 3g(x))

54. Suppose lim.~"(x) •• lim (g(x) + 3) %-4

~

4-x

Find the limits in Exercises

c. lim (g(x»2

44. lim sin2 x

45. lim sec x

46. lim tan x

47. lim 1 + x + sinx x-o 3 cos x

48. lim (x' - 1)(2 - cosx)

,~o

x-o

Vx+4 cos (x + 'IT)

,~o

x-o

x-o

SO. lim Y 7

x-o

+ sec' X

~ -

3. Find

%-4

r

d •

g(x)

x~ f(x} - 1

55. Suppose lim'~b f(x) ~ 7 and lim'~b g (x) ~ - 3. Find b. lim f(x)' g(x) •• lim (j(x) + g(x)) x-b

x-b

c. lim 4g(x)

43. lim (2 sin x - I)

r f(x) • ,~ f(x) - g(x)

0 and lim.~4 g (x) b. lim xf(x)

%-4

x~45-~

-2. Find

x-c

d

,~,

3

~

b. lim 2f(x)g(x)

x-c

c. lim (j(x)

x-I

5 2

5andlim,.....,g(x)

•• lim f(x)g(x)

x~-2~-3

42.lim

(c)

x-I

yI(5)(5)

X

4 -

x-I

(1)(4 - 2)

43-50.

%--'11"

x-I

x+l

16

(b)

( lim p(x»)( lim (4 - r(x»))

+ _1_

v-2 v 4 -

Limits with trigonometric functions

49. lim

,~1

x-I

36. lim x x '~42-Vx ,~1

vlim 5h(x)

( lim p(x»)( lim 4 - lim r(x»)

34. lim if - 8

37.lim

,~1

3y4 - 16y2 _1_ x-I

(a)

lim (P(x)(4 - r(x)))

p(x)(4 - r(x))

+ 4x + 3

x-o

x-

x-I

z~1

5y 3 + 8y' 30. lim --'-;c-----co

32. lim

31. lim -x--l

limY5h(x)

VShW

x+3

%-+-3 Xl

27.lim t'+t-2 I-I t2 - 1

7 4

d. lim f(x)/g(x)

x-b

x-b

56. Suppose that limr--2 p(x) lim,~_,,(x) ~ -3. Find

~

4,lim,....._2 r(x)

•• lim (P(x) + r(x) + ,(x» %--2

b.

lim p(x)' r(x)' ,(x)

%-+-2

c. lim (-4p(x) x--2

+ 5r(x))l8(x)

~

0, and

56


Umits of Average Rates of Change

h. Support your conclusion in part 0 such that for all x 0< Ix - xol < Il

I/(x) -

LI
0 that places the open interval (xo - Il, Xo + Il) centered atxo inside the interval (a, b). The inequality I/(x) - L I < E will hold for all x oF Xo in this Il-interval.

NOT TO SCALE

FIGURE 2.22 The function and intervals in Example 4.

EXAMPLE 5

Prove that limx~2 /(x) = 4 if

/(x) Solutton

{x2,

=

I,

Our task is to show that given E > 0 there exists all> 0 such that for all x 0< Ix - 21 < Il

y

I/(x) - 41

0 such that for all x

Ig(x) - MI

112

E/2.

< E/2.

Letll = mio {Il" 1l2},the smaller of Il, andll,. If 0 < Ix - cl < Ilthen Ix so If(x) - LI < E/2, and Ix - cl < 1l2' so Ig(x) - MI < E/2. Therefore

If(x) + g(x) - (L + Mll < ~ + ~ Thisshowsthatlim.-c(j(x)

+ g(x»

=

L

cl
0 such

y

f(x)~ ~ -'0

~

-1

L~2 E

that for all x

-'0 ~! 2

2.01

= 0.5

X~Xo

~~

f(x)

-x

0< Ix - xol < 8 31. !(x) = 3 - 2x,

= 0.01

E

Xo = 3,

32. !(x) = -3x - 2, 2.5 I I --i-I I I I I I I I I I I I I

-9

x2 - 4 33. !(x) = - - 2 ' 2 34. !(x) = x :

~ !2

0

X

0

36. !(x)

"1 1.99

2.01

~

x~4

16. !(x)

~

L

L

2x - 2,

17. !(x) ~ ~, 18. !(x) ~

Yx,

~

L ~ 1/2,

L

I/x,

X'o

L ~ I,

xo ~

1/4,

1/4,

Xo

~

v'3,

Xo ~

23. !(x) ~ x 2,

L ~ 4,

Xo ~ -2,

L

I/x,

25. !(x) ~ x 2 - 5, 26. !(x)

~

27. !(x)

~ nIX,

28. !(x)

~ nIX,

+ b,

30. !(x) = nIX E ~ 0.05

+ b,

-I,

L

~

L

=

m

>

0,

m

>

0,

120/x,

29. !(x) ~ nIX Xo = 1/2,

~

• = c

m

Xo 11,

~ 0.1

E

~

-I,

X'o

=

E

L

=

2m,

X'o = 2,

L

=

3m,

X'o

~

(m/2)

0,

L

~ I

48.

E

24,

=

~

x~o

0.1 I

I

3,

E

~

L = m

+ b,

+ b,

0.03

e=c>O

>0 m > 0,

4

if !(x)

~

2 {x , I,

X #-2 x =-2

' I 441 • llJlr. "2

I

x-v3 x

1~ !(x) ~ 0

49. lim

~

=

# I x = I X

46.lim x-I

~

I

-3

x2 - 1 =2 x-I x

E

E

0.4

38. lim (3x - 7)

5

x~l

0.02

~ 0.1

E ~

4,

L ~ 3,

~

~

41. lim !(x) = I

43.

E

Xo ~ 23,

22. !(x) ~ x 2,

24. !(x)

E ~

Xo ~ 10,

L ~ 4, ~

om

-2,

=

Xo ~ 0,

L ~ 3,

20. !(x) ~ ~,

E ~

X'o = 4,

-6,

=

19. !(x) ~ ~,

21. !(x)

5,

=

• = 0.05 • = 0.5

E ~

x~.

Eacb of Exercises 15-30 gives a function !(x) aod numbers L, XO, aod E > O. In each case, f'md an open interval about Xo on which the inequality I!(x) - L I < E holds. Then give a value for 8 > 0 such that forallxsatisfyingO < Ix - xol < 8 the inequality I!(x) - LI 1. y

The number L is the limit of f(x) as x approaches Xo if, given any E > 0, there exists a value ofx for which If(x) - LI
x

+B

(see Figure 2.28)

X--i'Xo+

if for every number E > 0 there exists a corresponding number Il > 0 such that for all x

FIGURE 2.28 Intervals associated with the defmition of right-hand limit.

If(x) - LI
0 be given. Here Xo = 0 and L = 0, so we want to find all> 0 such that for all x

Solution

~------f--* x-----v--~x

O<x --11- > cos II. Since lim(Ho· cos II = I (Example lib, Section 2.2), the Sandwich Theorem gives lim sin II = I. 8

8--+0+

Recall that sin II and II are both odd jUnctions (Section 1.1). Therefore, /(11) = (sin 11)/11 is an even jUnction, with a graph symmetric about the y-axis (see Figure 2.32). TIris symmetry implies that the left-hand limit at 0 exists and has the same value as the right-hand limit: lim sin II = I = lim sin II 8 8--+0+ 8 '

8--+0-

•

so limo~o (sin 11)/11 = I by Theorem 6.

EXAMPLE 5

Show that (a) lim cos h - I = 0 11-0 h

and

(b) lim sin 2x = ~.

x-a 5x

5

2.4

One·Sided Limits

71

Solution O,suchthatif x lies in I, then < •. What !intit is being verified and what is its value?

Vx-=s

48. Given. > O,f"md an interval I = (4 - S, 4), S > 0, such that if x lies in I, then ~ < •. What !intit is being verified and what is its value?

. h3h sm

,~o

33. lim sin (I - cos t) 1-0 1 - cost

0 cot 48

.~o sin' 0 cot' 20

0' cot 30

Formal Definitions of One-Sided Limits

Find the lintits in Exercises 21-42.

x

73

sons for your answer.

(J

b. lim(t - ltJ)

x-o

42. lim

tan 0

Continuity

.~o

Use the def"ntitions of right-hand and left-hand lintits to prove the lintit statements in Exercises 49 and 50. 49. lim -IXI = -I x-o-

X

50. lim ( x-2+ x

;1

=

I

51. Greatest integer function Find (a) lim,_.oo' l x J and (b) lim,_.oo- l x J ; then use limit definitions to verilY your f"mdings. (0) Based on your conclusions in parts (a) and (b), can you say anything about lim,........ l x J ? Give reasons for your answer.

< 00 x> .

52. One-sided1intits Letj(x) = {:';in(I/X), x

vx,

Find (a) lim,.....o' j(x) and (b) lim,_o- j(x); then use lintit definitions to verilY your f"mdings. (0) Based on your conclusions in parts (a) and (b), can you say anything about lim,~o j(x)? Give

reasons for your answer.

Continuity When we plot function values generated in a laboratory or collected in the field, we often connect the plotted points with an unbroken curve to show what the function's values are likely to have been at the times we did not measure (Figure 2.34). In doing so, we are assuming that we are working with a continuaus jUnction, so its outputs vary continuously with the inputs and do not jump from one value to another without taking on the values in between. The limit of a continuous function as x approaches c can be found simply by calculating the value of the function at c. (We found this to he true for polynomials in Theorem 2.) Intuitively, any function y = t(x) whose graph can be sketched over its domain in one continuous motion without lifting the pencil is an example of a continuous function. In this section we investigate more precisely what it means for a function to be continuous.

74


y

80

p

I Q

~9 f-

We also study the properties of continuous functions, and see that many of the function types presented in Section 1.1 are continuous.

Continuity at a Point

Q /

To understand continuity, it helps to consider a function like that in Figure 2.35, whose limits we investigated in Example 2 in the last section.

Q 1/

I-'"

..... o

5

10

Elapsed time (sec)

FIGURE 2.34 Connecting plotted points by an unbroken curve from experimental

data Q, , Q2, Q" ... for a falling object

EXAMPLE 1 Find the points at which the function 1 in Figure 2.35 is continuous and the points at which 1 is not continuous. The function 1 is continuous at every point in its domain [0,4] except at 2, and x = 4. At these points, there are breaks in the graph. Note the relationship between the limit of 1 and the value of 1 at each point of the function's domain.

Solution x = I, x

=

Points at which f is continuous:

y

2

--Xf(X) • 1

2

3

=

0,

lim I(x) = 1(0). x-o+

Atx

=

3,

AtO

x

Continuity Test A function f(x) is continuous at an interior point x = c of its domain if and only if it meets the following three conditions.

o

FIGURE 2.38 A function that has a jump discontinuity at the origin (Example 3).

Y

1. 2. 3.

3

Y~

2

LxJ

f(c) exists

(c lies in the doroain of f).

limx~,

f(x) exists

(f has a limit as x --> c).

limx~,

f(x) = f( c)

(the limit equals the function value).

For one-sided continuity and continuity at an endpoint, the limits in parts 2 and 3 of the test should be replaced by the appropriate one-sided limits.

-

4

1

75

is right-continuous at a and continuous at a right endpoint b of its domain if it is leftcontinuous at b. A function is continuous at an interior point c of its doroain if and only if it is both right-continuous and left-continuous at c (Figure 2.36).

y

-2

Continuity

EXAMPLE 4 Thefunctiony = lxJ introduced in Section l.l is graphed in Figure 2.39. It is discontinuous at every integer because the left-hand and right-hand limits are not equal

........0

........0

asx~n:

limJx J = n - I

........0

and

x~n

x

1

-I

---0

2

3

4

-2

FIGURE 2.39 The greatest integer function is continuous at every noninteger point It is right-continuous, but not left-continuous, at every integer point (Example 4).

Since l1l J = 11, the greatest integer function is right-continuous at every integer 11 (but not left-continuous). The greatest integer function is continuous at every real numher other than the integers. For example, lim lxJ = I = l1.5J.

x--+1.5

In general, if 11

-

I

O.

76

Chapter 2: Limits and Continuity y

y

y

y

1

x

Y ~ f(x)

0

(b)

(a)

x

(d)

(c)

y

J

Y~f(x)~l2

0

x

x x

(f)

(e)

FIGURE 2.40

The functioo in (a) is cootinuous at x

~

0; tire functioos in (b) through (f)

are not

Continuous Functions A function is continuous on an interval if and only if it is continuous at every point of the interval. For example, the semicircle function graphed in Figure 2.37 is continuous on the interval [ - 2, 2], which is its domain. A continuous function is one that is continuous at every point of its domain. A continuous function need not be continuous on every interval. Y

EXAMPLE 5 (a) The function y = l/x (Figure 2.41) is a continuous function because it is continuous at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = O.

--------~t-------~x

\ "

The functioo y ~ l/x is continuous at every value of x except x = O. It has a point of discontinuity at x = 0 (Example 5). FIGURE 2.41

(h) The identity function I(x) = x and constant functions are continuous everywhere by Example 3, Section 2.3. • Algebraic combinations of continuous functions are continuous wherever they are defined.

THEOREM B-Properties of Continuous Functions If the functions I and g are continuous at x = c, then the following combinations are continuous at x = c.

Differences:

I+g I-g

Constant multiples:

k· I, for any number k

Products:

I'g Ilg,

1. Sums:

2. 3. 4. 5. 6. 7.

Quotients: Powers:

r,

Roots:

"ifj

providedg(c) oF 0 n a positive integer provided it is deImed on an open interval

, containing c, where n is a positive integer

2.5

Continuity

77

Most of the results in Theorem 8 follow from the limit rules in Theorem I, Section 2.2. For instance, to prove the sum property we have

lim(j + g)(x)

x-c

= lim(j(x) x-c

= lim f(x) x-c

= f(c) =

This shows that f

+ g(x)) + lim g(x),

Sum Rule, Theorem 1

x-c

+ g(c)

Continuity of j. g at c

(j + g)(c).

+ g is continuous.

EXAMPLE 6 (a) Every polynomial P(x) = anx n + a._1Xn-1 lim P(x) = P(c) by Theorem 2, Section 2.2.

+ ... + ao

is continuous because

x~c

(b) If P(x) and Q(x) are polynomials, then the rational function P(x)/Q(x) is continuous wherever it is dermed(Q(c) oF 0) by Theorem 3, Section 2.2. _

EXAMPLE 7 The function f(x) = Ixl is continuous at every value of x. If x > 0, we have f(x) = x, a polynomial. Ifx < 0, we have f(x) = -x, another polynomial. Finally, at the origin, limx~o Ixl = 0 = 101. The functions y = sin x and y = cos x are continuous at x = 0 by Example 11 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 68). It follows from Theorem 8 that all six trigonometric functions are then continuous wherever they are dermed. For example, y = tanx is continuous on ... U (-7r/2, 7r/2) U (7r/2, 37r/2) U ....

Composites All composites of continuous functions are continuous. The idea is that if f(x) is continuous atx = c andg(x) is continuous atx = f(c), theng 0 f is continuous atx = c (Figure 2.42). In this case, the limit as x --> c is g(f(c».

Continuous at c g

f Continuous

•

c

FIGURE 2.42

ate

_

Continuous -:;.o C-_ _ _ _ atf(e) _ _ _ _=. . ~f(c) g(f(c»

Composites of continuous functions are continuous.

THEOREM 9-Composite of Continuous Functions If f is continuous at c and g is continuous at f(c), then the composite g 0 f is continuous at c.

Intuitively, Theorem 9 is reasonable because if x is close to c, then f(x) is close to f(c), and since g is continuous at f(c), it follows thatg(f(x» is close to g(f(c». The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of the proof of Theorem 9, see Exercise 6 in Appendix 4.

78


EXAMPLE 8 Show that the following functions are continuous everywhere on their respective domains. =

Yx 2

(e) y =

Ix2 _

(a) y

x 2/3

2x - 5

-

+ x4

(b) y = I

21

x - 2

(d) Y =

1:2S:~1

Solution

y

0.4 3

(a) The square root function is continuous on [0, 00) because it is a root of the continuous identity function fix) = x (part 7, Theorem 8). The given function is then the composite of the polynomial fix) = x 2 - 2x - 5 with the square root function g(t) = Vi, and is continuous on its domain.

(b) The numerator is the cube root of the identity function squared; the denominator is an everywhere-positive polynomial. Therefore, the quotient is continuous. (e) The quotient (x - 2)/(x 2 - 2) is continuous for all x # ± v2, and the function is the composition of this quotient with the continuous absolute value function (Example 7).

FIGURE 2.43 The graph suggests 1hat y = I(x sinx)/(x 2 + 2) I is continuous

(Example 8d).

(d) Because the sine function is everywhere-continuous (Exercise 68), the numerator term x sin x is the product of continuous functions, and the denominator term x 2 + 2 is an everywhere-positive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.43). • Theorem 9 is actually a consequence of a more general result which we now state and prove.

THEOREM lO-Limits of Continuous Functions b and limx~d(x) = b, then lim~, g(f(x» =

Proof that

Let E

>

g(b)

If g is continuous at the point

= g(lim~, fix»~.

0 be given. Since g is continuous at b, there exists a nornber 8 , Ig(y) - g(b)1

±OO. We further extend the concept of limit to infinite limits, which are not limits as before, but rather a new use of the term limit. Infmite limits provide useful symbols and langnage for describing the behavior of functions whose values become arlJitrarily large in magnitude. We use these limit ideas to analyze the graphs of functions having horizontal or vertical asymptotes.

Finite Limits as x --+ ±oo y

The symbol for infinity (oo) does not represent a real number. We use 00 to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For example, the function f{x) = I/x is defined for all x oF 0 (Figure 2.49). When x is positive and becomes increasingly large, 1/x becomes increasingly small. When x is negative and its magnitude becomes increasingly large, I/x again becomes small. We summarize these observations by saying that f{x) = I/x has limit 0 as x --> 00 or x --> -00, or that 0 is a limit o/f{x) = l/x at i'lfinity a1ld negative i'lfi1lity. Here are precise definitions.

4

3

y=i

2

-

1

2

3

4

x

DEFINmONs 1. We say that f(x) has the limit L as x approaches infinity and write

lim f{x) = L

x_oo

>

if, for every number E FIGURE 2.49 The graph of y = Ilx approaches 0 as x ~ 00 or x ~ - 00.

0, there exists a corresponding number M such tbat fur all x

x>M

=)

If{x)-LI<E.

2. We say that f{x) has the limit L as x approaches minus infinity and write lim f{x) = L

x--=

> 0, there exists a corresponding number N such that for all x 1+, we have (x - I) ..... 0+ and I/(x - 1) ..... oo.Asx ..... I-, we have (x - I) ..... 0- and I/(x - I) .....

y

-00.

} x 0

x-.o+

No matter how high B is, the graph

EXAMPLE 10

•

Discuss the behavior of

goes higher.

~ x

I f(x) = 2" x x

FIGURE 2.59 The graph of f(x) io Example 10 approaches inf"mity as x ..... O.

x-->

as

o.

As x approaches zero from either side, the values of l/x 2 are positive and become arbitrarily large (Figure 2.59). This means that

Solution

lim f(x) = lim

x-o

1, =

x---+Ox

00.

•

90

Chapter 2: Limits and Continuity The function y = I/x shows no consistent behavior as x --> o. We have I/x --> 00 if x-->O+, but I/x--> -00 ifx--> 0-. All we can say about lim.-o (I/x) is that it does not exist. The function y = l/x 2 is different. Its values approach infinity as x approaches zero from either side, so we can say that limx _ o (l/x 2 ) = 00.

EXAMPLE 11 These examples illustrate that rational functions can behave in various ways near zeros of the denominator. I· (x - 2)2 lim x - 2 · (x - 2f IlID 2 (a) x-2 lID ( 2 = 0 x - 4 = x-2 x - 2)(X + 2) = x-2 X + (b) lim x - 2 = lim .~2 X 2 -

(c) y

4

.~2

x - 2 = lim _1_ = ! (x - 2)(x + 2) .~2 x + 2 4

lim x - 3 = lim

.-2+ X 2 -

4

.-2+ (X -

(d) lim x - 3 = lim • -2- X 2 -

4

.-T

.-2

(f) lim

B ~-~--r~~--

---;;If--~-;-----;;____c:\---~x

Xo

+8

4

x - 3 = 00 (x - 2)(x + 2)

.-2 .-2

(e) lim x - 3 = lim • -2 X 2 -

x - 3 = _ 00 2)(x + 2)

x - 3 does not exist. (x - 2)(x + 2)

The values are negative for x > 2, x near 2. The values are positive forx < 2, x near 2 . See parts (c) and (d).

.-2

2-x = lim -(x-2)=lim -I =-00 (x - 2)3 (x - 2)3 (x - 2)2

In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f), where cancellation still leaves a zero factor in the denominator. •

Predse Definitions of Infinite Limits FIGURE 2.60 For XQ - 8 < x < xo + 8, the graph of f(x) lies above the lioe y = B. y

Instead of requiriog f(x) to lie arbitrarily close to a finite number L for all x sufficieotly close to xo, the defInitions of infinite limits require f(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seeo before. Figures 2.60 and 2.61 accompany these definitions.

DEFINmONS 1. We say that f(x) approaches infinity as x approaches -"0, and write lim f(x) = 00, X-Xo

if for every positive real number B there exists a correspoodiog Ii that for all x

>

0 such

f(x»B. o < Ix - xol < Ii 2. We say that f(x) approaches minus inf"mity as x approaches xo, and write lim f(x) = -00, X-Xo

FIGURE 2.61 For Xo - 8 < x < XQ the graph of f(x) lies below the lioe y = -B.

+ 8,

if for every negative real number - B there exists a correspondiog Ii that for all x f(x) < -B. o < Ix - xol < Ii

>

0 such

The precise defmitions of ooe-sided infinite limits at Xo are similar and are stated in the exercises.

2.6

EXAMPLE 12 Solution

Prove that lim

x-a

Given B

>

Limits Involving Infinity; Asymptotes of Graphs

~

91

= 00

X

>

0, we want to find 8

0 such that

1 implies 2>B. x

0; - 4

y

For x numerically large, 2x

~ 4 is near O.

For x near 2, this term is very large.

If we want to know how f behaves, this is the way to find out. It behaves like y = (x/2) + I when x is numerically large and the contribution ofl/(2); - 4) to the tota\ value of f is insignificant. It behaves like 1/(2); - 4) when x is so close to 2 that

-_~2~--_~I~~Ott-~~--~2~X

-5

1/(2); - 4) makes the dominant contribution. We say that (x/2) + I dominates when x is numerically large, and we say that 1/(2); - 4) dominates when x is near 2. Dominant terms like these help us predict a function's behavior.

Let f(x) = 3x 4 - 2>;3 + 3x 2 - 5x + 6 and g(x) = 3x4. Show that although f and g are quite different for numerically small values of x, they are virtually identical for Ix I very large, in the sense that their ratios approach I as x ..... 00 or

EXAMPLE 16

V'2

d. x-+-l

b. x-+2+

d.x-->2

e. What, if anything, can be said about the limit as x --> O?

72. f(2)

+2

73. 74. 75.

t~/3) as

~

b. t-+ 0-

:/3

x-+ 1+

.%-0+

~

2, and

0, lim f(x) ~ 0, lim f(x) ~ x-±oo

%-1-

~ -00,

I, f( -I)

~

~ -00,

and lim f(x) .%--1-

0, lim f(x) x_OO

~

and lim f(x)

lim f(x) ~

.1'--1+ ~ -00

0, lim f(x) .1'-0+

~

00,

~ 00,

I

%_-00

lim f(x)

~

0, lim f(x) ~

lim g(x)

~

0, lim g(x)

lim h(x)

~

-I, lim h(x)

x_±OO x_±OO

.%-2-

.%-3-

00,

and lim f(x) ~ x-2+

~ -00, ~

and lim g(x) ~ x-3+

I, lim h(x)

.%_00

00

.%-0-

~

00

-I, and

lim k(x) x_±OO

~

I, lim k(x) .%-1-

~ 00,

and lim k(x) x-I+

~ -00

for your answer.

C:/5 + 7) as

t-+ 0+

0, lim f(x)

79. How many horizootal asymptotes can the graph of a given rational func1ioo have? Give reasons for ynur answer.

d. x-+ 1-

Finding Limits of Differences when x--> ±oo Find the limits in Exercises 80-86. x~oo

81.

83.

b. x-+O-

x-+ 1+

d. x-+ 1-

(Vx+9 - Vx+4) lim (\I'x 2 + 25 - w-=J) lim (w-+3 + x) x_-oo lim (2x + yrc -_--c 4X"2-+--c3x 2)

80. lim

82.

~/3 - (x - I I)4/') as x

a. x-+o+ C.

~

~

77. Suppose that f(x) and g(x) are polynootials in x and that limx_oo (f(x)/g(x)) ~ 2. Can you conclude anything about limx_~ (f(x)/ g (x))? Give reasons for your answer.

t-+ 0+

62. lim(

-I, and

78. Suppose that f(x) and g(x) are polynootials in x. Can the graph of f(x)/g(x) have an asymptote if g(x) is never zero? Give reasons

x + (x - 2 I)2/') as a. x-+o+ b. x-+OC.

~

X~ ...

Find the limits in Exercises 5'HJ2.

61. lim(

-2, lim f(x)

lim h(x) ~ I

4x

e. What, if anything, can be said about the limit as x --> O?

a.

I

-2

.%_-00

as 76.

60. lim

+

In Exercises 73-76, fmd a func1ioo that satisfies the given cooditions and sketch its graph. (The answers here are not unique. Any func1ioo that satisfies the conditions is acceptable. Feel free to use formulas defmed in pieces if that will help.)

as

a. x-+o+ c. x-+2-

a.

~

x-o-

c.x-+l+

59. lim(2 -

2x

X~-OO

x_±OO

lim f(x)

a.x-+-2+

-

~

2, f( -I)

0, lim f(x)

lim f(x)

1 56. hm 2x + 4 as

x3

~

x-I+

-

x2-3x+2

~

0, f(l)

...

71. f(O) b. x-+O-

58. limx2 - 3x

~

lim f(x)

a. x-+o+

r

68. Y ~ x

X~OO

70. f(O)

• un x 3 -2x 2

x-3

lim f(x) ~ I

c. x-+ -1+

57

+3 +2

-3

y~-

Inventing Graphs and Functions In Exercises 61}-72, sketch the graph of a func1ioo y ~ f(x) that satisfies the given oouditioos. No form:u1as are required--just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer sec1ioo.)

69. f(O)

a.x-+l+

. x2

66.

65'Y~2x+4

54.lim+-as x - I

55. lim(x; -

I 64'Y~x+1

I

lim

95

Limits Involving Infinity; Asymptotes of Graphs

X~OO

x_-oo

84. lim (v'9x 2 - x - 3x) x~OO

96


85. lim (VX2

+ 11: -

86. lim (Vx2

+x

x~oo

X~OO

VX 2 - 2x)

Use the formal defmitioos from Exercise 93 to prove the lintit statemoots in Exercises 94-98.

- ~)

94. lim xl =

Using the Formal Definitions Use the formal defmitioos of limits as x---> ±oo to establish the limits in Exercises 87 and 88.

95. lim

00

%-0+

96. lim _1_ =

2

x_2-X -

-00

-00

97. lim _1_ =

2

x_2+X -

98. lim _1_2 =

00

00

%-1-1 - x

87. If j bas the coostant value j(x) = k, then lim j(x) = k.

1=

x_o-x

X~OO

88. If j bas the coostant value j(x) = k, then lim j(x) = k. X~-OO

Use formal defmitions to prove the lintit statements in Exercises 8c}'-92. -I 89. lim - 2 = -00 90. lim _III = 00

x-o

x-o x

91. lim

x~3

-2 (x - 3)2

92. lim

=-00

X~-,

Oblique Asymptotes Graph the ratioua1 functioos in Exercises 9c}'-104. Include the graphs and equations of the asymptotes. x2 x 2 +1 99. Y = - - I 100. Y = - - I

x-

x2

X

(x

I

+ 5)2

= 00

93. Here is the defmition ofinfinite right-hand limit.

x-

-4

-x=T

102. Y

x2 - 1 103. y=-x-

104• Y

101. Y

=

=

x 2 -1 2x + 4

=x3+1 X

2

Additional Graphing Exercises the curves in Exercises \05-\08. Explain the relationship between the curve's formula and what you see.

D Graph We say that j(x) approaches infmity as x approaches Xo from the right, and write lim j(x) =

105.Y=~2 4 - x

00,

x-XQ+

if.

for every positive real number B. there exists a corresponding nomber 8 > 0 such that for all x

:%'0<x<xo+8

j(x) > B.

107. Y = x 2/3

+ ~f3 x

l06.Y=~ 4 - x2 108. Y

= sin

(+-) x + I

D Graph the functions in Exercises 109 and 110. Thou aoswer the following questioos.

a. How does the graph behave as x ---> o+? h. How does the graph behave as x ---> ± oo?

Modify the defmition to cover the following cases. L

lim_j(x)

= 00

c. How does the graph behave near x = I and x = -I?

%-%0

lim j(x)

= -00

Give reasons for your answers.

c. limj(x)

= -00

109. Y

h.

%-xo+ %-Xo

Chapter

~

=

r

&(x - ~

3(

X )2/3

1l0'Y=I x - I


1. What is the average rate of chaoge of the function g(t) over the interval from t = a to t = b? How is it related to a secant line?

6. What theorems are available for calculating lintits? Give examples of bow the theorems are used

2. What lintit must be calculated to fmd the rate of change of a function g(t) att = to?

7. How are one-sided limits related to limits? How can this relationship sometimes be used to calculate a limit or prove it docs not exist? Give examples.

3. Give ao informal or inmitive defmition of the lintit lim j(x) = L. X~Xo

Why is the definition "informal''? Give examples. 4. Docs the existence aod value of the limit of a function j(x) as x approaches Xo ever depend on what happens at x = xo? Explain aod give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples.

8. What is the value oflim._o «sinO)/O)? Docs it matter whcther 0 is measured in degrees or radiaos? Explain. 9. What exactly does lim~Xo j(x) = L mean? Give an example in which you fmd a 8 > 0 for a given j, L, Xo, aod. > 0 in the precise defmition oflintit.

10. Give precise defmitions of the following staterneots.

a. lim,....2- j(x) = 5 c. lim,....2 j(x) = 00

h. lim~2' j(x) = 5

d. lim,....2 j(x) = -

00

97

Chapter 2 Practice Exercises 11. What conditions must be satisfied by a function if it is to be continuous at an interior point of its domain? At an endpoint?

12. How can looking at the graph of a function help you tell where the fiwnctionis continuous? 13. What does it mean for a function to be right-continuous at a point? Left-continuous? How are continuity and one-sided continuity related?

14. What does it mean for a fiwnction to be continuous on an interval? Give examples to illustrate the fact that a function that is not continuous on its entire domain may still be continuous on selected intervals within the domain.

15. What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example.

Chapter ~

16. What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graphingandsolvingtheequationf(x) ~ O?

17. Under what circumstances can you extend a function f(x) to be

= c? Give an example.

continuous at a point x

18. What exact1y do limr _

oo

f(x) = L and limr-~ f(x)

~

L mean?

Give examples.

19. What are limr_±<X> k (k a constant) and lim~±<x> (I/x)? How do you extend these results to other functions? Give examples. 20. How do you f"md the limit of a rational function as x ---+ ± oo? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.

Practice Exerdses

Lim;ts and ContinuUy 1. Graph the function

In Exercises 5 and 6, f"md the value that lim~og(x) must have if the given lintit statements hold.

I,

f(x) =

I

-~:

-x, I,

x:5

<x
c- L . Why was the left-hand limit needed?

(a)

r= 6cm

4. Controlling the flow from a draining tank Torricelli's law says that if you drain a tank like the one in the figure shown, the rate y at which water runs out is a constant times the square root of the water's depth x. The constant depends on the size and shape of the exit valve.

/'

t,

"-

/

T 1// h

----

\

Liquid volume V= 367Th

~ (b)

A l-L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6 cm

Suppose that y = Yx/2 for a certain tank. You are trying to maintain a fairly constant exit rate by adding water to the tank with a hose from time to time. How deep must you keep the water if you want to maintain the exit rate a. within 0.2 ft3/min of the rate Yo = I ft3/min? b. within 0.1 ft3/min of the rate Yo = I ft3/min? 5. Thermal expansion in precise equipment As you may know, most metals expand when heated and contract when cooled. The dimensions of a piece oflaboratory equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laboratory where the equipment is to be used. A typical aluminum bar that is 10 cm wide at 70°F will be y = 10

+ (t

- 70) X 10- 4

centimeters wide at a nearby temperature t. Suppose that you are using a bar like this in a gravity wave detector, where its width must stay within 0.0005 cm of the ideal 10 cm. How close to to = 70°F must you maintain the temperature to ensure that this tolerance is not exceeded? 6. Stripes on a measuring cup The interior of a typical l-L measuring cup is a right circular cylinder of radius 6 cm (see accompanying figure). The volume of water we put in the cup is therefore a function of the level h to which the cup is filled, the formula being

Precise Definition of Limit In Exercises 7- 10, use the formal definition of limit to prove that the function is continuous at xo .

7. f(x) = x 2

-

7,

Xo

9. h(x) =~,

= 1 Xo

8. g(x) = 1/(2x),

Xo

= 2 10. F(x) =~,

= 1/4 Xo

=5

11. Uniqueness of limits Show that a function cannot have two different limits at the same point. That is, if limx-->xo f(x) = L J and limx-->xo f(x) = L2, then LJ = L2 . 12. Prove the limit Constant Multiple Rule: lim kf(x) = k lim f(x)

x---+c

x---+c

13. One-sided limits

for any constant k.

If limx-->o+ f(x) = A and limx-->o- f(x) = B,

find a. limx-->o+ f(x 3

-

x)

b. limx-->o- f(x 3

-

x)

c. limx-->o+ f(x 2

-

x4)

d. limx-->o- f(x 2

-

x 4)

14. Limits and continuity

Which of the following statements are true, and which are false? If true, say why; if false, give a counterexample (that is, an example confirming the falsehood). a. If limx-->a f(x) exists but limx-->a g (x) does not exist, then limx-->a(J(x) + g(x)) does not exist.

b. If neither limx-->a f(x) nor limx-->a g (x) exists, then limx-->a (J(x) + g (x)) does not exist. c. If f is continuous at x, then so is 1f

I.

d. If 1f 1is continuous at a, then so is j.

100


In Exercises 15 and 16, use the formal defmitioo oflimit to prove that the function has a continuous extension to the given value of x.

15. f(x) =

x'-I

X+T'

x = -I

16. g(x) =

x'-2x-3 2x 6 '

x = 3

17. A function continuous at only one point Let

= f (x ) L

{x, ifxis ratiooa1

0, if x is irratiooa1.

Show that f is continuous at x

= O.

b. Use the fact that everynonempty open interval of real num-

hers cootains both ratiooa1 and irratiooa1 numbers to show that f is not continuous at any nonzero value of x. 18. The Dirichlet ruler function If x is a rational number, then x can be written in a unique way as a quotient of integers min where n > 0 and m and n have no commoo factors greater than I. (We say that such a fractioo is in luwest terms. For example, 6/4 written in lowest terms is 3/2.) Letf(x) be defmed for alb in the interval [0, I] by f(x)

= {I/n, ~x.= .m/~ is aratiooal number in lowest terms if x IS Irrational.

0,

Forinstauce,f(O) = f(1) = l,f(I/2) = 1/2,f(I/3) = f(2/3) = 1/3, f(1/4) = f(3/4) = 1/4, aod so on. L

exists, a lower bound for fooD aod say that f is bouoded froro below by M. We say that f is bounded 00 D if it is bouoded froro both above aod below. a. Show that f is bouoded ooD ifaod only if there exists a numberBsuchthat If(x) I '" B for allxinD.

b. Suppose that f is bouoded from above by N. Show that if limr->",f(x) = L, thenL '" N. c. Suppose that f is bouoded from below by M. Show that if limr->",f(x) = L, thenL '" M. 24. Mu {a, b} and min {a, b} a. Show that the expression

max {a, b}

c. Sketch the graph of f. Why do you think f is called the ''ruler function"? 19. Antipodal points Is there any reason to believe that there is always a pair of antipodal (diaroetrically opposite) points 00 Earth's equator where the temperatures are the same? Explain.

20. If lim (j(x) + g(x)) = 3 and lim (j(x) - g(x)) = -I, find x-c

x-c

lim f(x)g(x). x~,

21. Root. of a quadratic equation that is almost linear The equation ax 2 + 2x - 1 = 0, where a is a constant, has two roots if a > -1 and a '¢ O. one positive and one negative: T+ ( a )

_ -I -

+

'V'1+a a

'

r_(a)

=

-I -

'V'1+a a .

What happens to r+(a) as a --> 01 As a --> -1+1 b. What happens to r_(a) as a --> 01 As a --> -1+1 L

c. Support your cooclusions by graphing r+(a) aodr_(a) as functions of a. Describe what you see. d. For added support, graph f(x) = ax' + 2x - I simultaneously for a = 1,0.5,0.2, 0.1, aod 0.05. 22. Root of an equation Show that the equation x + 2 cos x = 0 has at least one solution. 23. Bounded functions A real-valued func1ioo f is bounded from above on a set D if there exists a number N such that f(x) '" N for all x in D. We call N, wheo it exists, ao upper bound for f 00 D aod say that f is bouoded froro above by N. In a similar marmer, we say that f is bounded from below on D if there exists a number M such that f(x) '" M for all x in D. We call M, wheo it

a +b -2-

la - bl

+

2

equals a if a '" b and equals b if b '" a. In other words, max {a, b} gives the larger of the two numhers a aodb. b. Find a similar expression for min {a, b}, the smaller of a aodb. Generalized

Limn. Involving 5'1: 8

The formula lime_o(sin 8)/8 = I cao be generalized. If lim,_, f(x) = 0 aod f(x) is never zero in an open interval containing the point x = c, except possibly c itself, theo lim sinf(x) = I f(x) .

Show that f is discootinuous at every ratiooa1 number in [0, I].

b. Show thatf is continuous at every irratiooa1 number in [0, I]. (Hint If. is a given positive number, show that there are only fmitelymaoyratiooa1 numbers rin [0, I] such that f(r) '" E.)

=

x~,

Here are several examples.

.

a. lim. smx

,

=

x-a x 2 .

1

2

.

2

2

b. lim smx = lim smx lim "-- = 1·0 = 0

x-a

x-o x 2 x-o

X

X

. sin (x' - x - 2) . sin (x' - x - 2) c. lim Ion • x--l X + 1 x--l (x 2 - X - 2) . (x' - x - 2) . (x + I)(x - 2) lim =1, lim =-3 x--l x + 1 x--l x + 1 . sin (I - '\!'x) . sin (I - '\!'x) I - '\!'x d. Ion = Ion = x-I x-I x-I 1 _ Vx x-I

I· lim (I - '\!'x)(1 + '\!'x) x~l (x - 1)(1 + '\!'x)

lim I-x x~l (x - 1)(1 +

Find the limits in Exercises 25-30. sin (1 - cosx) 26. lim 25. lim x x~o

x-O+

27. lim

sin (sinx) x

28. lim

29. lim

sin (x' - 4) 2

30. lim

x-o

%-2

x-o

sinx sinYx

sin (x' x

+ x)

sin ('\!'x x-9

X

'\!'x)

x

- 3) 9

Obl;que Asymptotes Find all possible oblique aaymptotes in Exercises 31-34. 31. Y

=

33. y=

+ 2x • ['

2x 3/2

vx + I

W+1

3

32. Y

= x + x sin (I/x)

34. y

=

\Ix' + 2x

I 2

Chapter 2 Technology Application Projects

Chapter

fJ

101


Mathematica,tMap1e Modules: 1IIke It to the Limit Part I Part II (Z.ro Rais.d to tb. P.....r Z.ro: What Does it M.an?) Part III (On.-Sid.d Limit.) Visualize and interpret the limit concept through graphical and numerical explorations.

Part IV (What a Diff.renc. a Pow.r Makes) See how sensitive limits can be with various powers of x. Going to Injmity Part I (EIploring Function Behavior as x ---+ oc or x ---+ - 00) lbis module provides four examples to explore the behavior of a function as x ---+ 00 or x ---+ - 00 . Part II (Rate. of Growtb) Observe graphs that appear to be continuous, yet the function is not continuous. Several issues of continuity are explored to obtain results that you may find surprising.

3 DIFFERENTIATION OVERVIEW In the beginning of Chapter 2 we discussed how to determine the slope of a curve at a point and how to measure the rate at which a function changes. Now that we have studied limits, we can define these ideas precisely and see that both are interpretations of the derivative of a function at a point. We then extend this concept from a single point to the derivative junction, and we develop rules for finding this derivative function easily, without having to calculate any limits directly. These rules are used to find derivatives of most of the common functions reviewed in Chapter 1, as well as various combinations of them. The derivative is one of the key ideas in calculus, and we use it to solve a wide range of problems involving tangents and rates of change.

3.1

Tangents and the Derivative at a Point In this section we define the slope and tangent to a curve at a point, and the derivative of a function at a point. Later in the chapter we interpret the derivative as the instantaneous rate of change of a function, and apply this interpretation to the study of certain types of motion.

y

Finding a Tangent to the Graph of a Function To find a tangent to an arbitrary curve y = f(x) at a point P(xo, f(xo)) , we use the procedure introduced in Section 2.1. We calculate the slope of the secant through P and a nearby point Q(xo + h, f(xo + h)). We then investigate the limit of the slope as h ~ 0 (Figure 3.1). If the limit exists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope.

-O~----~Xo--------~~----+X

DEFINITIONS

The slope ofthe curve y

= f(x) at the point P(xo, f(xo)) is the

number FIGURE 3.1 The slope of the tangent · p. 1· f(xo + h) - f(xo) 1me at IS un h

.

m = hm

h~O

f(xo

+ h) - f(xo) h

(provided the limit exists).

h..... O

The tangent line to the curve at P is the line through P with this slope.

In Section 2.1, Example 3, we applied these definitions to find the slope of the parabola f(x) = x 2 at the point P(2, 4) and the tangent line to the parabola at P. Let's look at another example.

102

3.1 Tangents and the Derivative at a Point

103

EXAMPLE 1

y

(a) Find the slope of the curve y = I/x at any point x = a '" O. What is the slope at the pointx = -I?

(b) Where does the slope equal -1/4? (c) What happens to the tangent to the curve at the point (a, I/a) as a changes? Solution (a) Here f(x) = I/x. The slope at (a, I/a) is slope is-l atx =-1

1

lim f(a

+ hh) - f(a)

h~O

FIGURE 3.2 The tangent slopes, steep near the origin, become more gradual as

the point of tangency moves away (Example I). y

=

I

a

1

+ hh - a = -h

·

h~ ha(a

+ h)

lim

1 a - (a + h)

h~O

h a(a + h)

lim

=

-I

h~O a(a

+ h)

I

a2 '

Notice how we had to keep writing "Iimh~o" before each fraction until the stage where we could evaluate the limit by substituting h = O. The number a may be positive or negative, but not O. When a = -1, the slope is -1/(-1)2 = -1 (Figure 3.2).

(b) The slope of y = I/x at the point where x = a is -1/a 2• It will be -1/4 provided that

y=l s1opels-"4 . 1

= lim h~O

(2. ~)

~'----t-~x s1 . 1 opels-

4

FIGURE 3.3 The two tangent lines to y = I/x having slope -1/4 (Example I).

This equation is equivalent to a 2 = 4, so a = 2 or a = -2. The curve has slope -1/4 at the two points (2, 1/2) and (-2, -1/2) (Figure 3.3). (c) The slope -1/a 2 is always negative if a '" O. As a --+ 0+, the slope approaches -00 and the tangent becomes increasingly steep (Figure 3.2). We see this situation again as a --+ 0- . As a moves away from the origin in either direction, the slope approaches 0 and the tangent levels off to become horizontal. •

Rates of Change: Derivative at a Point The expression

f(xo + h) - f(xo) h is called the difference quotient of I at Xo with increment h. If the difference quotient has a limit as h approaches zero, that limit is given a special name and notation.

DEFINmON

The derivative ofa function I at a point Xo, denoted f'(xo), is

f' (xo)

= lim h~O

f(xo + h) - f(xo) h

provided this limit exists.

If we interpret the difference quotient as the slope of a secant line, then the derivative gives the slope of the curve y = f(x) at the point P(xo, f(xo». Exercise 31 shows

104

Chapter 3: Differentiation that the derivative of the linear function f(x) = mx of the line, so

f'(xo)

=

+ b at any point Xo is simply the slope

m,

which is consistent with our definition of slope. If we interpret the difference quotient as an average rate of change (Section 2.1), the derivative gives the function's instantaneous rate of change with respect to x at the point x = Xo. We study this interpretation in Section 3.4.

EXAMPLE 2 In Examples 1 and 2 in Section 2.1, we studied the speed ofa rock falling freely from rest near the surface of the earth. We knew that the rock fell y = 161 2 feet during the first 1 sec, and we used a sequence of average rates over increasingly short intervals tu estimate the rock's speed at the instant 1 = 1. What was the rock's exacl speed at this time? Solution We let f(l) = 1612. The average speed of the rock over the interval between 1 = 1 and 1 = 1 + h seconds, for h > 0, was found tu be

f(1 + h) - f(l)

16(1

+ h)2 -

h

16(1)2

h

The rock's speed at the instant 1 = 1 is then

lim 16(h

h~O

+ 2)

= 16(0

+ 2)

= 32 ft/sec.

Our original estimate of 32 ft/ sec in Section 2.1 was right.

•

Summary We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, and the derivative of a function at a point. All of these ideas refer to the same limit.

The following are all interpretations for the limit of the difference quotient, lim f(xo h-O

+ h) - f(xo) h

1. The slope of the graph of y = f(x) at x = Xo

2. The slope of the tangent tu the curve y = f(x) at x = Xo 3. The rate of change of f(x) with respect to x at x = Xo

4. The derivative f'(xo) at a point

In the next sections, we allow the point Xo to vary across the domain of the function f.

3.1

Tangents and the Derivative at a Point

105

Exercises 3.1 Slopes and Tangent Unes In Exercises 1-4, use the grid and a straight edge to make a rough esti· mate of the slope of the curve (in y-units per x·unit) at the points PI

Tangent Unes with Spedfied Slopes At what points do the graphs of the functions in Exercises 23 aod 24 have horizontal tangents?

andP,.

23. f(x) ~ x'

1.

2.

y

y

+ 4x - I

24. g(x) ~ x' - 3x

25. Find equations of all lines having slope -I that are tangent to the curve y ~ I/(x - I). 26. Find an equatioo of the straight line having slope 1/4 that is tangent to the curve y ~ Yx.

1

Rates of Change 27. Object dropped from. tower Ao oi!ject is dropped from the top of a 100·m-high tower. Its height above ground after t sec is 100 - 4.9t' m. How fast is it falling 2 sec after it is dropped? 28. Speed of a rocket At t sec after liftoff, the height of a rocket is 3t' ft. How fast is the rocket climbing 10 sec after liftoff?

3.

4.

y

29. Circle'. changing area What is the rate of chaage of the area of a circle (A ~ 'lTr') with respect to the radius when the radius is r ~ 3?

y

30. Ball'. changing volume What is the rate of change of the volume ofa ball (V ~ (4/3)'lTr') with respect to the radius when the radius is r = 2? 31. Show that the line y point (xo, nlXo + b).

81

~ nIX

+b

is its own tangent line at any

32. Find the slope of the tangent to the curve y ~ I/Yx at the point where x = 4.

o

Testing for Tangents

33. Does the graph of In Exercises 5-10, fmd an equatioo for the tangent to the curve at the given point. Then sketch the curve and tangent together.

5. Y ~ 4 - x', 7. Y

~

9. Y ~

2Yx,

(-1,3)

6. Y ~ (x - I)' 8. Y

(1,2)

x', (-2, -8)

10. Y

~

1"x

+

(I, I)

I,

have a tangent at the origin? Give reasons for your answer.

(-I, I)

~ :',

34. Does the graph of

(-2,-t)

In Exercises 11-18, fmd the slope of the function's graph at the given point. Then find an equatioo for the line tangent to the graph there. 11. f(x) ~ x'

+ I, (2,5)

x 13. g(x) ~ x _ 2' 15. h(t) ~ t',

(3,3)

(2,8)

17. f(x) ~ Yx,

(4,2)

12. f(x) ~ x - 2>;',

18. f(x) ~

Vx+1,

+ 3t,

(1,4) (8,3)

In Exercises 1!l-22, find the slope of the curve at the point indicated. 19. Y

=

5x 2 ,

I

21. Y ~ x _ I'

20. Y = 1 - x 2,

-1

X =

x

~

3

x - I

22'Y~x+I'

g(x)

~

{""in (I/x), 0,

x # 0 x =0

have a tangent at the origin? Give reasons for your answer.

(1,-1)

(2,2)

14. g(x) ~

8 2:'

x 16. h(t) ~ t'

f(x) ~ {X'Sin(I/X), x # 0 0, x= 0

X =

2

x~O

Vertical Tangents We say that a continuous curve y ~ f(x) has a vertical tangent at the point where x ~ Xo if lim,_o (/(xo + h) - f(xo))/h ~ 00 or -00. For example, y ~ x 1/' has a vertical tangent at x ~ 0 (see accompanying figure): lim f(O 4-0

+ h) - f(O) h ~ lim _1_ ~ h-O h2j3

00

.

106

Chapter 3: Differentiation y

36. Does the graph of

{a,

U(x) =

x
0 2vx

lim

=

z-x

y

Vi-Vi (Vi - Vi) ( Vi + Vi)

. Iun

=

I

z-xVi

I

2Vi'

+ Vi

(b) The slope of the curve at x = 4 is

1'(4) = _1_ =

2V4

FIGURE 3.5 The curve y = Vi and its tangent at (4, 2). The tangent's slope is found by evaluating the derivative at x = 4

1. 4

The tangent is the line through the point (4, 2) with slope 1/4 (Figure 3.5): I

y=2+

y =

4 (x-4)

I 4x +

•

I.

(Example 2).

Notations There are many ways to denote the derivative of a function y = f(x), where the independent variable is x and the dependent variable is y. Some common alternative notations for the derivative are dy df d I'(x) = y' = dx = dx = dx f(x) = D(j)(x) = Dxf(x).

The symbols d/ dx and D indicate the operation of differentiation. We read dy/ dx as ''the derivative ofy with respect to x;' and df/dx and (d/dx)f(x) as "the derivative of f with respect to x." The "prime" notations y' and f' come from notations that Newton used for derivatives. The d/dx notations are similar to those used by Leibniz. The symbol dy/dx should not be regarded as a ratio (until we introduce the idea of "differentials" in Section 3.9). To indicate the value of a derivative at a specified number x = a, we use the notation

dyl dx

= -d/l

I'(a) = -

;c=a

dx

= -d f(x) dx

;c=a

I

.

;c=a

For instance, in Example 2

1'(4) =

A. Vii dx

= x=4

_I_I 2Vi

= _1_ = 1 x=4

2V4

4.

Graphing the Derivative We can often make a reasonable plot of the derivative of y = f(x) by estimating the slopes on the graph of f. That is, we plot the points (x, I' (x» in the xy-plane and connect them with a smooth curve, which represents y = I' (x).

109

3.2 The Derivative as a Function

EXAMPLE 3

Graph the derivative of the function y = I(x) in Figure 3.6a.

We sketch the tangents to the graph of I at frequent intervals and use their slopes to estimate the values of f' (x) at these points. We plot the corresponding (x, f' (x)) pairs and connect them with a smooth curve as sketched in Figure 3.6b. • Solution

What can we learn from the graph of y =

f' (x)? At a glance we can see

1. where the rate of change of I is positive, negative, or zero; 2. the rough size of the growth rate at any x and its size in relation to the size of I(x); 3. where the rate of change itself is increasing or decreasing. y

Slope 0 A

y ~f(X)/

......

10 "

-

Slo

,.-1

B,\ C

5

: ;/

\.

Slope 0

- 8y-units

/

- 4 x-units lp l~

5

0

8 = 2 y-um'Wx-umt . /Slope - 4

Slope _1

I

x

(a)

Slope

4

Y ~f'(xy

3

2 1

A"

D'

-1

)'

~~ 1O

-2

V!,:.;ca10

Ihl

lim --h

h--+O-

Ihl ~ -h whenh

= lim-I =-1. h--+O-

EXAMPLE 5

In Example 2 we found that for x

>

--_

- -1- -2

(c)

27.

(d)

28.

y

y

b. Repeat part (a) assuming that the graph starts at (-2,0) instead of( -2,3).

33. Growth in the economy The graph in the accompanying figore shows the average aonual percentage change y ~ /(t) in the u.s. gross natioua1 product (GNP) for the years 1983-1988. Graph dy/dt (where defmed). 7%

------~~~----~x

6 5

------~~~----~x

o

4

--fr\

I-

l \

3 2

............. /"

-

1

29.

30.

y

o

y

1983 1984 1985 1986 1987 1988 34. Fruit rues (Continuation of Example 4, Section 2.1.) Populations starting out in closed environments grow slowly at fust, when there are relatively few members, then more rapidly as the

---\--~~4---+-~x

number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment

31. a. The graph in the accompanying figore is made of line segments joined end to eml. At which points of the interval [-4, 6] is f' not defined? Give reasons for your answer.

a, Use the graphical tecbrtique of Example 3 to graph the derivative of the fruit fly population. The graph of the population is reproduced here. p

y

~

,.....

350 300 250

V

/

'a 200

Loo

150

Z

(1. -2)

(4.-2)

50

o

/ / ./

10

20

30

40

50

TIme (days)

b. Graph the derivative off. The graph should show a step function.

b. During what dsys does the population seem to be increasing fastest? Slowest?

114

Chapter 3: Differentiation

35. Temperature The giveo graph shows the temperature T in OF at Davis, CA, on April 18, 2008, between 6 A.M. and 6 P.M.

39.

40.

Y

Y

Y ~ J(x)

T

P(I. I)

80

.....

G:' 70

l- I-'

'"

------~--7-----~x

/

y=x

'\

--~Onr--7-------~X

In Exercises 41 and 42, determine if the piecewise defined function is differentiable at the origin.

o

3

6 A.M.

9 A.M.

6

9

12

12 NOON 3 P.M. Tune (brs)

41. f(x)

6 P.M.

a. Estimate the rate of temperature change at the times i) 7 A.M.

ii) 9 A.M.

iii) 2 P.M.

~

42. g(x) ~

{2x - I, x + 2x + 7, 2

{

X2/3,

X ~

xli',

x

crease most rapidly? What is the rate for each of those times? c. Use the graphical technique of Example 3 to graph the derivative of temperature Tversus time t. 36. Weight Loss Jared Fogle, also known as the "Subway Sandwich Guy;' weighed 425 Ib in 1997 before losing more than 240 Ib in 12 months (http://en.wikipedia.org/wiki/Jared_Fogle). A chart showing his possible dramatic weight loss is given in the accompanying figure.

:'.

122


I How to Read the Symbols for

If y" is differentiable, its derivative, y'" = dy" /dx = d'y/dx', is the third derivative ofy with respect to x. The names continue as you imagine, with

Derivatives

y' ''y prime" y. 'y double prime" d'y

dx' y~

yv.)

d"y

dx" D"

d d"y y(") = dxy("-l) = dx" = D"y

"d squared y dx squared"

denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y = f{x) at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.

''y triple prime" ''y super n"

"d to the n of y by dx to the n" "Dtothen"

EXAMPLE 8

The

flISt

four derivatives of y

x' - 3x' + 2 are

=

First derivative:

y' = 3x' - 6x

Second derivative:

y"

= 6x -

Third derivative:

y.'

=

Fourth derivative:

y(4) =

6

6 O.

The function has derivatives of all orders, the fifth and later derivatives all being zero.

_

Exercises 3.3 Derivative Calculations

25. v

In Exercises 1-12, fmd the fITSt and second derivatives.

1. y = -x' + 3

2. Y = x' + x + 8

3•• = 51' - 31'

4. w

4x' 5. y=T-x

x3 x2 x 6·y=T+T+4

7. w = 3z-2

8. •

-}

= 3z'

- 7z 3

+ 21z2

11.r=-I--~

3.'

=

X4

29. Y

=

10. Y

=4

31. Y

= (x

12. r

=

- 2x - x-,

12 - ~ 8 8'

+ ~4 8

In Exercises 13-16, find y' (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate.

37. w

=

Find the derivatives of the functions in Exercises 17-28.

39. p =

2x + 5 17. Y = 3x - 2 x2

19. g(x) = x

21. v

= (1

23. /(.)

-

18.

4

+ 0.5

- /)(1

= Vs Vs +

- 4x)

4 - 3x z=-,-3x + x

20. /(1) =

22. w

I I

24 =5x+1 • u 2v'i

= (2x

- 7,-'(x

34

.

(8 - 1)(8'

,

+

8

e;/Z):' - lOx - 5x-'

27. Y

I + x - 4v'i x

=

+ 5)

u'(O)

= -3,

v(O) = -1,

=0

v'(O) = 2.

Find the values of the following derivatives at x = O.

d d. dx (7v - 2u)

3.3 42. Suppose u and v are differentiable functions of x and that

Differentiation Rules

52. Find all points (x,y) on the graph of f(x)

x' with tangent lines

passing through the point (3, 8).

u(i) = 2, u'(I) = 0, v(l) = 5, v'(I) = -I. Find the values of the following derivatives at x = I.

d

=

123

f(x)

= xl

d d. dx (7v - 2u)

a. dx (uv)

Slope. and Tangents 43. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x' - 4x + I at the point (2, I). b. Smalle.t slope Wbat is the smallest slope on the curve? At what point on the curve does the curve have this slope?

c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8. 44. a. Horizontal tangents Find equations for the horizontal tan3x - 2. Also !md equations for gents to the curve y = the lines that are perpendicular to these tsngents at the points of tangency.

x' -

b. Smalle.t slope Wbat is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve~ tangent at this point. 45. Find the tangents to Newton's serpentine (graphed here) at the origin and the point (I, 2).

53. a. Find an equation for the line that is tangent to the curve y = xatthepoint(-I,O).

x' -

D b.

Graph the curve and tsngent line together. The tsngent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordiuates.

D c.

Conrmn your estimates of the coordinates of the secood intersection point by solving the equations for the curve and tsngent simultaneously (Solver key).

54. a. Find an equation for the line that is tsngent to the curve y = x 3 - 6x 2 + Sx at the origin.

D b.

Graph the curve and tsngent together. The tangent intersects

the curve at another point. Use Zoom and Trace to estimate

y

the point~ coordiuates.

D c.

Conf'mn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

Theory and Example. 46. Find the tangent to the Witch ofAgnesi (graphed here) at the point

(2, I).

For Exercises 55 and 56 evaluate each limit by !lrSt coovertiog each to a derivative at a particular x-value.

55. lim

x-I

y y=_8_ x2 + 4

0

2 3

I

50

56.

"----=--x-I

lim x'/9 - I x--l

X

+

1

57. Find the value of a that makes the following function differentiable for all x-values.

x

47. Quadratic tangent to identity funetlon The curve y = + bx + c passes through the point (I, 2) and is tsngent to the line y = x at the origin. Find a, b, and c.

49. Find all points (x, y) on the graph of f(x) = 3x' - 4x with tangent lines parallel to the line y = 8x + 5. SO. Find all points (x,y) on the graphofg(x) = tx' - ~x' tangent lines parallel to the line 8x - 2y = I.

+

I with

51. Find all points (x, y) on the graph of y = x/(x - 2) with tangent lines perpendicular to the line y = 2x + 3.

x:5

-I -1

59. The general polynomial of degree n has the fonn

P(x} where a.

¢'

=

a,.x n + a,._tXn-t

+ ... + a2x 2 + atX + ao

O. Find P' (x).

60. The body's re••tion to medicine The reaction of the body to • dose of medicine can sometimes be represented by an equation of the form

124


where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dRj dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see how to find the amount of medicine to which the body is most sensitive. 61. Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say?

64. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d (x -m) dx

p=~_an2 V - nb

=

V2

'

in which a, b, n, and R are constants. Find dPjdV. (See accompanying figure.)

a. The Reciprocal Rule says that at any point where the function vex) is differentiable and different from zero,

(1)v

l

65. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume Vby a formula of the form

62. The Reciprocal Rule

dx

-mx - m-

where m is a positive integer.

What does this say about the Derivative Constant Multiple Rule?

.E...

=

_l...dv v2dx'

Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.

63. Generalizing the Product Rule The Derivative Product Rule gives the formula d -(uv) dx

=

dv udx

+ vdudx

66. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is

for the derivative of the product uv of two differentiable functions ofx.

A(q) =

a. What is the analogous formula for the derivative of the product uvw of three differentiable functions of x? b. What is the formula for the derivative of the product u 1 u2 u3 U4 offour differentiable functions of x? c. What is the formula for the derivative of a product Ul U2U3 ..• Un of a finite number n of differentiable functions of x?

3.4

km

hq

q + cm + 2'

where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dAjdq and d 2Ajdq2.

The Derivative as a Rate of Change In Section 2.1 we introduced average and instantaneous rates of change. In this section we study further applications in which derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.

Instantaneous Rates of Change If we interpret the difference quotient (J(x + h) - !(x)) / h as the average rate of change in ! over the interval from x to x + h, we can interpret its limit as h ~ 0 as the rate at which! is changing at the point x.

3.4 The Derivative as a Rate of Change

DEFINmON

125

The instantaneous rate of change of / with respect to x at Xo is

the derivative /

'(x ) = lim /(xo

o

+ h)

- /(xo)

h

h-O

'

provided the limit exists.

Thus, instantaneous rates are limits of average rates.

It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change.

EXAM PLE 1

The area A of a circle is related to its diameter by the equation

A = TT D2 4 How fast does the area change with respect to the diameter when the diameter is 10m? Solution

The rate of change of the area with respect to the diameter is

~

= ; • 2D =

TTf.

When D = 10 m, the area is changing with respect to the diameter at the rate of (TT/2)10 = 5TT m2/m "" 15.71 m 2/m. •

Motion Along a Line: Displacement,. Velocity, Speed, Acceleration, and Jerk Suppose that an object is moving along a coordinate line (an s-axis), usually horizontal or vertical, so that we know its position s on that line as a function of time t:

s Position at time t ...

I'

•

s~f(t)

and at time t

i1s- -->l'1

/(t).

The displacement of the object over the time interval from tto t

+ At

.

=

Ils = /(t

) S

s+l!.s~f(t+l!.t)

FIGURE 3.12 The positions ofa body moving along a coordinate line at time t aod shortly later at time t + b.t. Here the coordinate line is horizontal.

+ Ilt (Figure 3.12) is

+ Ilt) - /(t) ,

and the average velocity of the object over that time interval is

Vav

=

displacement Ils /(t travel time = Ilt =

+ tlt) - /(t) Ilt

To find the body's velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + Ilt as Ilt shrinks to zero. This limit is the derivative of / with respect to t.

DEFINmON Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body's position at time tis s = /(t) , then the body's velocity at time t is

_ tis _

.

() vt-dt-Inn &-0

/(t + tlt) - /(t) Il t

126

Chapter 3: Differentiation Besides telling how fast an object is moving along the horizontal line in Figure 3.12, its velocity tells the direction of motion. When the object is moving forward (8 increasing), the velocity is positive; when the object is moving backward (. decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity (Figure 3.13). s

)

,

o

s increasing:

s decreasing:

positive slope so

negative slope so moving downward

moving upward

FIGURE 3.13 For motion. ~ /(,) along a straight line (tire vertical axis), v = dsldt is positive when s increases and negative whens decreases. The blue curves represent position along lhe lioe over time; lhey do not portray lhe palh of motion, which lies along lhe .-axis.

If we drive to a friend's house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show - 30 on the way back, even though our distance from home is decreasing. The speedometer always shows .peed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.

DEFINITION

Speed is the absolute value of velocity. Speed

~

IV(I)I

~ I~I

EXAMPLE 2 Figure 3.14 shows the graph of the velocity v = /'(1) ofa particle moving along a horizontal line (as opposed to showing a position function. = /(1) such as in Figure 3.13). In the graph of the velocity function, it's not the slope of the curve that tells us if the particle is moving forward or backward along the line (which is not shown in the figure), but rather the sign of the velocity. Looking at Figure 3.14, we see that the particle moves forward for the first 3 sec (when the velocity is positive), moves backward for the next 2 sec (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at 1 ~ 3 sec (when the velocity is zero) and reverses direction as the velocity starts to become negative. The particle is now moving backward and gaining in speed until 1 ~ 4 sec, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time 1 ~ 4, the particle starts to slow down again until it finally stops at time 1 = 5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at 1 ~ 6 sec, speeding up during the f"mal second of the forward motion indicated in the velocity graph. •


127

v

, MOVES FORWARD

,

I

(v> 0)

flQRWARD I

I

AGAIN

I

:

(v> 0)

:

Velocity v ~ 1'(')

,

I

I

I

I

"

I ,,

~ Speeds -+: up ,I

Speeds~Steady~ Slows ~

up

:(v

= const) :

down

,

I

, ,I

I

"

rI

:

I

Stands

till

S

,"

~I

I (v = 0) I

I I

I I

-0;;i"'-------;c----!2c----"::i...--~----,+.5:---6~--'7~ ' ,(sec)

,

4--- Speeds ~ Slows ~ I Up , down I I I I

I I MOVES BACKWARD

:

(v

< 0)

I I I

:

FIGURE 3.14 The velocity graph ofa particle moving aloog a horizoota1line, discussed in Example 2. HIsTORICAL BIOGRAPHY

Bernard Bolzano (1781-1848)

The rate at which a body's velocity changes is the body's acceleration. The acceleration measures how quickly the body picks up or loses speed. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt.

DEFINmONS Acceleration is the derivative of velocity with respect to time. Ifa body's position at time 1 is s = /(1), then the body's acceleration at time 1 is a(l) =

~~ = ~:~.

Jerk is the derivative of acceleration with respect to time:

j(l) =

~~ = ~:~.

Near the surface of the Earth all bodies fall with the same constant acceleration. Galileo's experiments with free fall (see Section 2.1) lead to the equation

s = 19t2 2

'

where s is the distance fallen and g is the acceleration due to Earth's gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are sigoificant. The value of g in the equation s = (1/2)gI2 depends on the units used to measure 1and s. With 1in seconds (the usual unit), the value ofg determined by measurement at sea level is approximately 32 ft/sec 2 (feet per second squared) in English units, and g = 9.8 m/sec2

128


t~O

t

=1

(meters per second squared) in metric units. (These gravitational constants depend on the distance from Earth's center of mass, and are slightly lower on top ofM!. Everest, for example.) The jerk associated with the constant acceleration of gravity (g = 32 ft/sec 2) is zero:

s (meters)

t (seconds) •

o

, ,

5

'-'

10 15

j =

20

30

EXAMPLE 3 Figure 3.15 shows the free fall of a heavy ball bearing released from rest at time I = Osee.

35

40

=3

~~:

(g) = O.

An object does not exhibit jerkiness during free fall.

25

t

!

(a) How many meters does the ball fall in the first 2 sec?

45

(b) What is its velocity, speed, and acceleration when t = 2?

FIGURE 3.15 A ball bearing falling from rest (Example 3).

Solution (a) The metric free-fall equation is s = 4.91 2. During the first 2 sec, the ball falls s(2) = 4.9(2)2 = 19.6 m. (b) At any time I, velocity is the derivative of position:

s

v(l) = s'(I) =

!

(4.91 2) = 9.81.

At I = 2, the velocity is v(2) = 19.6 m/sec

in the downward (increasing s) direetion. The speed at I = 2 is speed = Iv(2)1 = 19.6m/sec. The acceleration at any time I is

a(l) = v'(I) = s"(I) = 9.8 m/sec2.

t

All = 2, the acceleration is 9.8 m/ sec2.

s~O s.m~--

•

EXAMPLE 4

A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec (about 109 mph) (Figure 3.16a). It reaches a height of s = 1601 - 161 2 ft after t sec.

(a)

s,V

s ~ 160t - 16,z

(a) How bigh does the rock go? (b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?

(c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock bit the ground again? -160

Solution (b)

FIGURE 3.16 (a) The rock in Example 4. (b) The graphs of s and v as functions of time; s is largest when v = ds/dt ~ O. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock's velocity, graphed here as a straight line.

(a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is O. To find the maximum height, all we need to do is to find when v = 0 and evaluate s at this time. At any time I during the rock's motion, its velocity is V

tis d 2 = dt = dl (1601 - 161 ) = 160 - 321ft/sec.


129

The velocity is zero when

160 - 32t

=

0

or

t = 5 sec.

The rock's height at t = 5 sec is

s.... = s(5) = 160(5) - 16(5)' = 800 - 400 = 400 ft. See Figure 3.16h.

(b) To find the rock's velocity at 256 ft on the way up and again on the way down, we rust rmd the two values of t for which s(t) = 160t - 16t 2 = 256. To solve this equation, we write

16t2 16(t

+ 256 lOt + 16)

160t

2

-

= =

(t - 2)(t - 8) =

0 0 0

t = 2sec,t = 8 sec. The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock's velocities at these times are

v(2) = 160 - 32(2) = 160 - 64 = 96 ft/sec.

v(8) = 160 - 32(8) = 160 - 256 = -96 ft/sec. At both instants, the rock's speed is 96 ft/sec. Since v(2) > 0, the rock is moving upward (s is increasing) at t = 2 sec; it is moving downward (s is decreasing) at t = 8 because v(8) < O. (c) At any time during its flight following the explosion, the rock's acceleration is a constant

a -_dv_d( dt - dt 160 - 32t)_ - -32 ft/2 sec. The acceleration is always downward. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = O. The equation 160t - 16t 2 = 0 factors to give 16t(1O - t) = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward It retamed to the ground 10 sec later. _

Derivatives in Economics Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost ofproduction c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc/dx. Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton:

c(x

+ h) - c(x) h

average cost of each of the additional h tons of steel produced.

130

Chapter 3: Differentiation The limit of this ratio as h --+ 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.17):

Cost y (dollars)

dc

ax ~~--------~----~~x

o

x x+h Production (tons/week)

.

= hm

c(x

+ h)

h

h~O

- c(x). . = margmal cost of production.

Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit: ac

FIGURE 3.17 Weekly steel production: c(x) is the cost of producing x tons per week. TIre cost of producing an additional h tons is c(x + h) - c(x). y

c(x

+

!>.X

I) - c(x) 1

which is approximated by the value of dc/ax at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc/ax, which is the rise in the tangent line if!>.x = I (Figure 3.18). The approximation works best for large values of x. Economists often represent a total cost function by a cubic polynomial

c(x) = ax 3 + f3x 2

+ 1X + {J

where {J represents ['!Xed costs such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.

EXAMPLE 5

Suppose that it costs

c(x) = x 3

-

6x 2

+

15x

~~----L-----~~------~x

o

x

x+ 1

FIGURE 3.18 Tlremarginalcostdc/dds approximately the extra cost ac of producing = I more unit.

ax

dollars to produce x radiators when 8 to 30 radiators are produced and that

1'(x) = x 3 - 3x 2

+

12x

gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling II radiators a day?

Solution c'(IO):

The cost of producing one more radiator a day when 10 are produced is about

c'(x) = c'(IO)

:fx (x3 -

6x 2 + 15x) = 3x 2

-

12x

+

IS

= 3(100) - 12(10) + IS = 195.

The additional cost will be about $195. The marginal revenue is

r'(x) =

:fx (x 3 -

3x 2 + 12x) = 3x 2 - 6x

+

12.

The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about

r'(IO)

= 3(100) - 6(10)

if you increase sales to II radiators a day.

+

12 = $252

•


131

EXAM PLE 6 To get some feel for the iangoage of marginal rates, consider marginal tax mtes. If your marginal income tax mte is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes Twith respect to income is dT/dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal mte will increase. _

Sensitivity to Change When a small change in x produces a large change in the value of a function f(x), we say that the function is relatively sensitive to changes in x. The derivative f' (x) is a measure of this sensitivity.

EXAMPLE 7

Genetic Data and Sensitivity to Change

The Austrian monk Gregor Johann Mendel (1822-1884), working with garden peas and other piants, provided the first scientific explanation ofbybri x

o

2 2 Y = 4 + cotx - 2cscx 1!

In Exercises 27-32, fmddp/dq.

1 27. p = 5 + cotq sinq 29. p = 31. p =

28. p = (1

+ cosq

+

Y= 1 + Y2cscx+ cot x

cseq) cosq

tanq

cosq

+ tanq 3q + tanq

30. p = 1

qsinq

32. p =

-2--

q - 1

qsecq

Trigonometric Limits Find the limits in Exercises 47-54.

47. lim sin %-2

33. Find y" if L

Y

= cscX'.

b. y = secx.

48.

b. y = geosx.

49.

34. Find y(4) = d 4y/dx 4 if L

Y

= -2sinx.

Tangent Lines In Exercises 35-38, graph the curves over the given interva1s, together with their tangents at the given values of x. Label each curve aod tangent with its equation.

35. y x

=

sin x,

= -'IT,

-371/2:5 x :s; 2w

0, 3w/2

36. y = tanx,

-'1f/2 -'1f/2

'1f/2

<x
",1 = (i + cot(I/2)t2 = (1-'/4 sin 1)4/3 31 - 4)-'

48.y= ( 51+2

tions at the given value of x. •• 2j(x), x = 2

x =3 x= 2

+ g(x), x = 3 f(x)/g(x), x = 2

b. f(x)

e. f(x)' g(x),

d.

e. f(g(x)),

r. v'iW,

2

g. l/g (x),

X

=3

x=2 h. Vf(x) + g2(x),

X

= 2

148


74. Suppose that the functioos f and g and their dctivatives with respect to x have the following values at x ~ 0 and x ~ 1.

83. Running machinery too rast

Suppose that a piston is mnving straight up and down and that its position at time t sec is

s

x

f(x)

o

I 3

I

g(x)

/'(x)

g'(x)

I

5

-4

-1/3

1/3 -8/3

Find the derivatives with respect to x of the following combina-

tions at the given value of x. 5f(x) - g(x), x ~ I f(x) c. g(x) + I' x ~ I

e. g(f(x)), x g. f(x

~

+ g(x)),

75. Find d>/dtwhen 8

d. f(g(x)),

r.

0

(x 11

x

~

0

+ f(x))-2,

X

with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What efree! does doubliog the frequency have on the piston's velocity, acceleration, and jerk? (Once you fmd out, you will know why some machinery breaks when you ruo it too fast.)

~ I

x ~ 0

y

~ 3'1f/2 ifs ~ cos8andd8/dt ~ 5.

76. Find eIy/dt when x ~ I ify ~ x 2

+ 7x

- 5andd>/dt ~ 1/3.

What happens if you can write a :function as a composite in different

77. Find dy/dx ify ~ x hy using the Chain Rule withy as a composite of

b. y ~ I

+7

and

u ~ 5x - 35

+ (I/u)

and

u ~ I/(x - I).

y ~ u'

b. y~

Vu

and and

u~x'.

«x -

80. Find the tangeot to y ~

y'x 2

L

I)/(x -

X

+

1))2atx ~ O.

+ 7 at x ~

2.

Find the tangent to the curve y ~ 2 tan (=/4) at x ~ I.

b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval -2 < x < 2? Give reasons for your answer.

82. Slopes on sine curves L

25

ing when it is increasing at its fastest? y

60 H-+-+-+--b.>"'"""'~ d-+-+-+-+-+-H fi:'

1

'-- 40 H

~

0

-

1 . . . .'

Find equations for the tangents to the curves y ~ sin 2x and ~ -sin(x/2) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer.

y

b. Can anything be said about the tangents to the curves y ~ sinmxandy ~ -sin(x/m) at the origin (m a coostant 0# O)? Give reasons for your answer. c. For a giveo m, what are the largest values the slopes of the curvesy ~ sinmxandy ~ -sin(x/m) can ever have? Give

reasons for your answer. d. The function y ~ sin x completes one period on the interval [0, 2'1f], the function y ~ sin 2x coropletes two periods, the functiony ~ sin(x/2) completes half a period, and so on. Is there any relation between the number of periods y ~ sin mx coropletes on [0, 2'1f] and the slope of the curve y ~ sinmx at the origin? Give reasons for your answer.

_

H-f--71-+-+-+-+'"~-+--+--+-+-+-1 -

- - - '~ - - ----~

-v-

-~

u ~ '\IX

79. Findthetangeottoy ~

81.

1+

and is graphed in the accompanying figure.

20

78. Find eIy/dx if y ~ x'/2 hy using the Chain Rule with y as a composite of L

101)

.. On what day is the temperature increasing the fastest?

ways? Do you get the same dctivative each time? The Chain Rule says you should. Try it with the functions in Exercises 77 and 78.

y ~ (u/5)

~ 37 sin [;;; (x -

b. About how many degrees per day is the temperature increas-

Theory and Examples

L

A cos (2'1fbt) ,

84. Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbaoks, Alaaks, doring a typical 365-day year. The equation that approsirnates the temperature on day x is

x ~ 0

b. f(x)g'(x),

L

~

85. Particle motion ordinate line is s =

-

-

-

- 1'\ - - --~ --I>x

V+

position of a particle moving along a co-

1 41, with s in meters and t in seconds. Find the particle~ velocity and acceleration at t ~ 6 sec.

86. Constant acceleration Suppose that the velocity of a fa1liog body is v ~ kYs m/sec (k a constant) at the instant the body has fallen s m from its starliog point. Show that the body's accelera-

tion is constant. 87. Falling meteorite The velocity of a heavy meteorite entering Earth's atmosphere is inversely proportional to Ys when it is s kIn from Earth's center. Show that the meteorite's acceleration is inversely proportional to 8 2 . 88. Particle acceleration A particle moves along the x-axis with velocity dx/dt ~ f(x). Show that the particle's acceleration is

f(x)/'(x). 89. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation

T

~ 2'1f~'

where g is the constant acceleration of gravity at the pendulum's

location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is

3.7

made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,

Implicit Differentiation

•• Graph dg/dt (where defmed) over [-"" ".]. b. Findd//dt. c. Graph d// dt. Where does the approxinlation of dg/ dt by d//dt seem to be best? Least good? Approximations by 1rigonome1ric polynomia1s are important in the theories of

tiL du = kL.

heat and oscillation, but we must not expect too much of them, as we see in the next exercise.

Assuming Ibis to be the case, show that the rate at which the period changes with respect to temperature is kT/2. 90. Chain Rule

,

,

Ixl. Then the

Suppose that /(x) = x' and g(x) =

~

0

g)(x) =

lxi' =

x'

are both differentiable at x

and

(g

0

/)(x) =

Ix'i =

/"

x'

D 91.

The derivative of sin 2x Graph the function y = 2 cos 2x for -2 '" x s 3.5. Then, on the same screen, graph

y=

sin 2(x

+ h) -

sin 2x

h

for h = 1.0, 0.5, and 0.2. Expctiment with other values of h, including negative values. What do you see happening as h --> O? Explain Ibis behavior.

D 92.

The derivative of cos (x') Graph y = -2x sin (x') -2 '" x s 3. Then, on the same screen, graph cos «x

y=

o

-"

= 0 even though g itself is not differ-

entiable at x = O. Does Ibis contradict the Chain Rule? Explain.

g(t)

'~f(t)

/

composites (j

149

"

94. (Continuation of Exercise 93.) In Exercise 93, the 1rigonome1ric polynomial /(t) that approximated the sawtooth function g(t) on [-".,,,.] had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polyoomial to approximate a function in a reasonable way without its derivative approxinlating the function's derivative at all well. As a case in point, the "polynomial"

s = h(t) = 1.2732 sin 2t

+

0.4244 sin 6t

+ 0.25465 sin lOt

+ 0.18189 sin 14t + 0.14147 sin 18t

for

graphed in the accompanying figure approximates the step function s = k(t) shown there. Yet the derivative of h is nothing like the derivative of k.

+ h)') - oos (x') h

,

for h = 1.0,0.7, and 0.3. Expctiment with other values of h. What do you see happening as h --> O? Explain this behavior.

s

COMPUTER EXPLORATIONS

I

Trigonometric Polynomials

-"

93. As the accompanying figure shows, the 1rigonometric "polynomial"

k(t)

I. ~ /

1

-2"

~

"2

0

h(t)

"

-1

s = /(t) = 0.78540 - 0.63662 cos 21 - 0.07074cos6t - 0.02546 cos lOt - 0.01299 cos 14t

•• Graph dk/dt (where defmed) over [-'IT, 'IT].

gives a good approximation of the sawtooth function s = g(t) on the interval [-"" ".]. How well does the derivative of / approximate the derivative of g at the points where dg/ dt is defmed? To find out, carry out the following steps.

3.7

b. Find dh/ dt. c. Graph dh/ dt to see how badly the graph fits the graph of dk/dt. Comment on what you see.

Implicit Differentiation Most of the functions we have dealt with so far have been described by an equation of the form y = f(x) that expressesy explicitly in tenDs of the variable x. We have learned rules for differentiating functions defined in this way. Another situation occurs when we encounter equations like

x 3 + y3 - 9xy

= 0,

y' - x = 0,

or

x' + y' - 25

= O.

150

Chapter 3: Differentiation (See Figures 3.26, 3.27, and 3.28.) These equations defme an implicit relation between the variables x andy. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation F(x, y) = 0 in the fonn y = f(x) to differentiate it in the usual way, we may still be able to fmd dyj ax by implicit differentiation. This section describes the tecbnique.

Y

Implicitly Defined Functions We begin with examples involving familiar equations that we can solve for y as a function ofx to calculate dyjax in the usual way. Then we differentiate the equations inlplicitly, and find the derivative to compare the two methods. Following the examples, we summarize the steps involved in the new method. In the examples and exercises, it is always assumed that the given equation determines y inJplicitly as a differentiable function of x so that dyj ax exists.

EXAMPLE 1 FIGURE 3.26 The curve + y' - 9xy ~ 0 is not the graph of

x'

anyone function of x. The curve can,

however, be divided into separate arcs that are the graphs of functions ofx. This particular curve, called afolium, dates to Descartes in 1638.

Find dyjax ify2 = x.

The equation y2 = x defmes two differentiable functions of x that we can actoally find, namely Yl = VX and Y2 = -VX (Figure 3.27). We know how to calculate the derivative of each of these for x > 0:

Solution

dYI ax

I 2VX

dY2 ax

and

1 2VX'

But suppose that we knew only that the equation y2 = x defmed y as one or more differentiable functions of x for x > 0 without knowing exactly what these functions were. Could we still find dy j ax? The answer is yes. To fmd dyj ax, we simply differentiate both sides of the equation y2 = x with respect to x, treating y = f(x) as a differentiable function of x:

y2

I I I I

= -1

2Y2

x

The Chain Rule gives

ax

-Yx

dy ax

1

=--

2Yx

FIGURE 3.27 The equationy2 - x ~ 0, or y2 = X as it is usually written, dermes two differentiable functions of x on the interval x > O. Exarople 1 shows how to find the derivatives of these functions without solving the equation y2 = x for y.

! (y2)

=

d 2 dy dx [f(x)] ~ 2f(x)j'(x) ~ 2y dx'

2y dy = I

Q(x, -\IX)

Y2~

/

Slope

=

I 2y'

This one fonnula gives the derivatives we calculated for both explicit solutions Yl = VX andY2 = -VX:

dYI ax

EXAMPLE 2

I 2Yl

I 2VX

and

d)'2 ax

I 2Y2

I

2( -vx)

I 2VX'

•

Find the slope of the circle x 2 + y2 = 25 at the point (3, -4).

The circle is not the graph of a single function of x. Rather it is the combined graphs of two differentiable functions, Yl = V25 - x 2 and Y2 = - V25 - x 2 (Figure 3.28). The point (3, -4) lies on the graph of)'2, so we can fmd the slope by calculating the derivative directly, using the Power Chain Rule:

Solution

dY21

ax x~3 =

-

I

-2x 2V25 - x 2 x~3

-6 2v'25=9

3 4'

3.7 Implicit Differentiation y

We can solve this problem more easily by differentiating 1he given equation of the circle implicitly with respect to x:

A.. (x2) + A.. (y2) dx

dx

=

dy dx Y2~-V25-X2

The slope at (3, -4) is-.x.1

Y

(3, -4).

A.. (25) dx

dy 2x+ 2y dx=O

---_5~------~Or-------~5~~x

FIGURE 3.28 The circle combines 1he graphs of two functions. The graph of Y2 is 1he lower semicircle and passes through

151

= (3, ....)

x

y.

-~ = 1. -4 4

Notice that unlike the slope formula for dy2/ dx, which applies ouly to points below the x-axis, the formula dy/dx = -x/yapplies everywhere the circle has a slope. Notice also that the derivative involves both variables x and y, not just the independent variable x. • To calculate the derivatives of other implicitly defmed functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defming equation.

Implicit Differentiation 1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy/ dx on one side of the equation and solve for dy/ dx.

Y

EXAMPLE 3 Solution

Find dy/dx if y2 = x 2

(Figure 3.29).

We differentiate the equation implicitly. y2 =

x 2 + sinxy

:. (y2) = : . (x2)

FIGURE 3.29 The graph of y2 ~ x 2 + sin'9' in Example 3.

+ sinxy

+ :. (sinxy)

2y dy dx = 2x

+

d( ) (cosxy) dx xy

2y: = 2x

+

(cosxy)&

2Y: - (cosxy) (x : ) = 2x

+

(cosxy)y

dy (2y - xcosxy) dx = 2x dy dx

. .. treating Y as a function of and using the Chain Rule.

x

Treat xy as a product

Collect terms wilh dy/dx.

+ ycosxy

+ Y cosxy 2y - x cosxy

2x

+ x :)

Differentiate both sides with respect to x ...

Solve for dy/dx.

Notice that the formula for dy/dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x andy, not just the independent variable x. •

152


Derivatives of Higher Order Implicit differentiation can alao be used to find bigher derivatives.

EXAMPLE 4

Find d'y/dx 2 if 2x' - 3y2 = 8.

To start, we differentiate both sides of the equation with respect to x in order to findy' = dy/dx.

Solution

JI (2x' dx

- 3y2)

=

JI (8)

6Y.Y'

=

0

fu;2 -

dx

x

Treaty as a function ob. 2

y' =y,

when y # 0

Solve for y'.

We now apply the Quotient Rule to f'md y' .

y"

=:.. (~) =2xy / 2y ' =~ _;~.y'

Finally, we substitute y' = x 2/y to express y" in terms ofx andy.

2 _ 2x x (X2) _ 2x x' y" ----Y y2 Y Y y"

Curve of lens smface

•

wheny # 0

Lenses, Tangents, and Normal Lines In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular tu the surface of the lens at the point of entry (angles A and B in Figure 3.30). This line is called the normal tu the surface at the point of entry. In a profile view of a lens like the one in Figure 3.30, the normal is the line perpendicular to the tangent of the profile curve at the point of entry. FIGURE 3.30 TIre profile ofa leos, showing the beading (refraction) of a ray oflight as it passes through the leos surface.

y

EXAMPLE 5 Show that the point (2, 4) lies on the curve x' find the tangent and normal tu the curve there (Figure 3.31).

+ y' - 9xy

= O. Then

Solution

The point (2, 4) lies on the curve because its coordinates satisfy the equation given for the curve: 2' + 4' - 9(2)(4) = 8 + 64 - 72 = O. To find the slope of the curve at (2, 4), we first use implicit differentiation to f'md a fonnula for dy/ dx:

x'+y'-9xy=0

JI (x') + JI (y') dx

dx

-

JI (9xy)

=

JI (0)

Y

+ Y dx) dx

=

0

Differentiate both sides with respect to x.

+ 3x 2 - 9y

=

0

TreatX)' as a product andy as a function of x.

dy - 9 (xddx 3x2 + 3y2 dx dy (3y2 - 9x) dx

dx

dx

3(y2 - 3X): = 9y - 3x 2 FIGURE 3.31 Example 5 shows how to fmd equations for the taogeot and normal to the folium of Descartes at (2, 4).

dy dx

3y-x 2 y2 - 3x'

Solve for dy/d

X

(3, -4)

xV ~ 9,

(-1,3) 32. y2 - 2x - 4y - I ~ 0,

(-2, I)

33. 6x 2 + 3xy + 2y2 + 17y - 6 ~ 0, (-1,0) 34. x 2 - V3xy + 2y2 ~ 5, ( V3, 2 ) 35. 2xy

+ 'If siny

~ 2'1f,

36. xsin2y ~ ycos2x, 37. y

(I, 'If/2) ('If/4,'If/2)

~

2 sin ('IfX - y), (1,0) 2 38. x cos2y - siny ~ 0, (0,'If)

39. Parallel tangents Find the two points where the curve x 2 + xy + y2 ~ 7 crosses the x-axis, and show that the tangents to the curve at tlrese points are parallel. What is the common slope of these tangents? 40. Normals parallel to a lin. Find the normals to the curve xy + 2x - y ~ 0 that are parallel to the line 2x + y ~ O. 41. The eight curve Find the slopes of the curve y' ~ y2 - x 2 at

(See Figure 3.26.) a. Find the slope of the folium of Descartes x'

44. The folium of Deseartes

c. Find the coordinates of the point A in Figure 3.26, where the folium has a vertical tsngent.

Theory and Examples 45. Intersecting normal The line that is normal to tire curve x 2 + 2xy - 3y2 ~ 0 at (I, I) intersects the curve at what other point? 46. Power rule for rational ""ponents Let p aod q be integers with q > O. If y ~ xPi., differentiate the equivalent equation y' ~ xP implicitly aod show that, for y #' 0,

~ "'i.)-1 . x

+ y'

at the points (4, 2) and (2, 4).

3.8

Related Rates

155

x' D 10 Exercises 51 aod 52, fmd both dy/dx (treatingy as a differentiable

48. Is there anything special about the tangents to the curves y' ~ and lx' + 3y' ~ 5 at the points (I, ± I)? Give reasons for your

answer. y

function of x) and dx/dy (treating x as a differentiable function ofy). How do dy/dx aod dx/dy seern to be related? Explain the relationship geometrically in terms of the graphs. 51. xy' +

x'y

~ 6

52. x' + y' ~ sin'y

COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises

5~.

a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.

b. Using implicit differentiatioo, find a formula for the derivative dy/dx aod evaluate it at tlte given pnint P.

49. Verify that the following pairs of curves meet orthogoualiy.

a.

x' + y' ~ 4, x' ~ 3y'

b. x

3

50. The graph of y' ~ is called a semicubica! parabola aod is shown in the accoropaoying figure. Determine the coostant b so x + b meets this graph orthogoually. that the line y ~

-l

53.

x' -

xy

+ y'

~ 7,

55. y

,

_2+x

+ Y - I _ x' P(O, I)

56. y' + cos xy ~ 57.

x',

59.

60.

P(1,O)

x+tan(~) ~2, P(I,-;r) ~ I, 2y' + (xy)l/' ~ x' + 2,

58. xy' + tan (x + y)

3.8

P(2, I)

54. x' + y'x + yx' + y' ~ 4, P(I, I)

~ I - y', x ~ ly'

x'

c. Use tlte slope found in part (b) to fmd an equation for tlte tangent line to tlte curve at P. Then plot tlte implicit curve and tangent line togetlter no a single graph.

p(-;r, 0) P(1, 1)

xv'1+2.Y + y ~ x', P(I,O)

Related Rates In this section we look at problems that ask for the mte at which some variable changes when it is known how the mte of some other related variable (or perhaps several variables) changes. The problem of finding a mte of change from other known rates of change is called a related rates problem.

Related Rates Equations Suppose we are pumping air into a spherical balloon. Both the volume and radius of the balloon are increasing over time. If V is the volume and r is the radius of the balloon at an instant of time, then

156


Using the Chain Rule, we differentiate both sides with respect to t to rmd an equatiDJI relating the rates of change of V and r,

So if we know the radius r of the balloon and the rate dV/dt at which the volume is increasing at a given instant of time, then we can solve this last equation for dr/dt to fmd how fast the radius is increasing at that instant. Note that it is easier to directly measure the rate of increase of the volume (the rate at which air is being pumped into the balloon) than it is to measure the increase in the radius. The related rates equation allows us to calculate dr/d, from dV/dt. Very often the key to relating the variables in a related rates problem is drawing a picture that shows the geometric relations between them. as illustrated in the following example.

EXAMPLE 1 Water runs into a conical tank at the rate of 9 W/min. The tank stands point down and has a height of lOft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep? Solution

Figure 3.32 shows a partially filled conical tank. The variables in the problem are V = volume (W) of the water in the tank at time t (min)

x - radius (ft) of the surface of the water at time t y = depth (ft) of the water in the tank at time t. We assume that V, x, and y are differentiable functions of t. The constants arc the dimensions of the tank. We arc asked for dy/dt when FIGURE 3.32 The geometry of the conical tank and the rate at which water rills the tank dctmnine how fast the water level rises (Example 1).

y

~

6 ft

and

" d'v + ~-~-k';!---~ d:x;' k d:x;k d'v

be differentiable for all values of x?

+"'+u dx '"

b. Discuss the geometry of the resulting graph of g.

19. Odd differentiable function. Is there aoytIting special about the derivative of an odd differentiable function of x1 Give reasoos for your answer.

The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using

20. Even differentiable function.

m! m! (m)k + (m) \k + I ~ k!(m - k)! + (k + I)!(m - k -

Is there aoytIting special about the derivative of an eveo differentiable function of x1 Give reasons for your answer.

f and g are defmed throughout an open interval containing the point xo, that f is differentiable at:x:o, that f(xo) ~ 0, and that g is continuous at Xo. Show that the product fg is differentiable at Xo. This process shows, for example, that although Ixl is not differentiable at x ~ 0, the product x Ixl" differentiable at x ~ O.

21. Suppose that the functions

22. (Continuation of Exerc"e 21.) Use the result of Exercise 21 to show that the followiog functions are differentiable at x ~ O. L

Ixl sin x

d. h(x)

b. x'/' sinx

c."\Yx(l - cosx)

~ {x' sin (I/x), x

0# 0 x= 0

0,

23. Is the derivative of

h(x) ~ {x' sin (I/x), x 0# 0 0, x= 0 continuous atx ~ 01 How about the derivative ofk(x) Give reasons for your answers.

~

:x:h(x)1

24. Suppose that a function f satisfies the followiog cooditioos for all real values ofx andy:

i) f(x

+ y)

ii) f(x) ~ I

~

f(x)·f(y).

+ xg(x) , where linJ,....og(x)

~

I.

Show that the derivative I'(x) exists at every value oh and that I'(x) ~ f(x). 25. The generalized product rule Use mathematical induction to prove that if y = "1"2'" u" is a finite product of differentiable functions, then y is differentiable on their common domain and

du,

dy

ax

=

du,

dun

liX U2 "'u" + Ul(jX"'UII + ... + u t u2 "'ulI - l liX'

I)!'

27. The period of a dock pendulum

The period T of a clock pendulum (time for one full swing and back) is given by the formula T' ~ ~L/g, where Tis measured in seconds, g ~ 32.2 ft/sec', and L, the length of the pendulum, is measured in feet. Find approximately a. the length of a clock pendulum whose period is T ~ I sec.

b. the change dTin Tifthependulum in part (a) is lengthened

0.01 ft.

c. the amount the clock gains or loses in a day as a result of the period's changing by the amount dT found in part (b). 28. The melting ice cube Assume that an ice cube retains its cubical shape as it melts. Ifw. call its edge length s, its volume is V = 3 3 and its surface area is 63 2 . We assume that V and s are differentiable functions of time t. We assume also that the cube's volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms, dV ~ -k(6s') dt '

k> O.

The minus sign indicates that the volume is decreasing. We assume that the proportiona1ity factor k is coostant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperatore, and the incidence or absence of sunlight, to nanre only a fow.) Assume a particular set of conditions in which the cube lost 1/4 of its volume during the first hour, and that the volume is Vo when t ~ O. How long will it take the ice cube to melt?


Chapter ID

183


Mathematica,tMap1e Modules: Convergence ofSecant Slopes to the Derivative Function

You will visualize the secant line between successive points on a curve and observe what happens as the distance between them becomes small. The function, sample points, and secant lines are plotred on a single graph, while a second graph compares the slopes of the secant lines with the derivative function. DeriVlllives, Slopes, Tangent Lines, and Making Movies Part. I-ill. You will visualize the derivative at a point, the linearization of a function, and the derivative of a function. You learn how to plot the function and selected tangeots on the same graph. Part IV (plotting Many Tangent.) Part V (Making Movies). Parts N and V of the module can he used to aoimate tangeot lines as one moves along the graph ofa function. Convergence ofSecant Slopes to the Derivative Function

You will visualize right-hand and left-hand derivatives. MotWnAlollg tJ Stmight Line: Position .... Velocity .... AccelertJdDn

Observe dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration functions. Figures in the text can he animated.

4 ApPLICATIONS OF DERIVATIVES OVERVIEW In this chapter we use derivatives to find extreme values of functions, to determine and analyze the shapes of graphs, and to find numerically where a function equals zero. We also introduce the idea of recovering a function from its derivative. The key to many of these applications is the Mean Value Theorem, which paves the way to integral calculus in Chapter 5.

Extreme Values of Functions

4.1

This section shows how to locate and identify extreme (maximum or minimum) values of a function from its derivative. Once we can do this, we can solve a variety of problems in which we find the optimal (best) way to do something in a given situation (see Section 4.5). Finding maximum and minimum values is one of the most important applications of the derivative.

DEFINITIONS Let f be a function with domain D. Then f has an absolute maximum value on D at a point c if

f(x)

:5

f(c)

for all x inD

and an absolute minimum value on D at c if y

f(x) y

=

f(c)

for all x in D .

sin x

------+----+----~-----+x

FIGURE 4.1 Absolute extrema for the sine and cosine functions on [-7T/2, 7T/2]. These values can depend on the domain of a function.

184

2::

Maximum and minimum values are called extreme values of the function f. Absolute maxima or minima are also referred to as global maxima or minima. For example, on the closed interval [-7T/2, 7T/2] the function f(x) = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function g(x) = sinx takes on a maximum value of 1 and a minimum value of -1 (Figure 4.1). Functions with the same defining rule or formula can have different extrema (maximum or minimum values), depending on the domain. We see this in the following example.

4.1


185

EXAMPLE 1 The absolute extrema ofthe following functions on their domains can be seen in Figure 4.2. Notice that a function might not have a maxinnnn or minimmn if the domain is unbounded or fails to contain an endpoint

Function rule

DomainD

Absolute extrema on D

(a) y = x 2

(-00,00)

No absolute maximmn. Absolute minimmn of 0 at x =

[0,2]

Absolute maximmn of 4 at x = 2. Absolute minimmn of 0 at x = o.

(c) y = x 2

(0,2]

Absolute maximmn of 4 at x = 2. No absolute minimum.

(d) y = x 2

(0,2)

No absolute extrema.

- ---"......-----:cx 2 (a) abs min only

FIGURE 4.2

HISWRICAL BIOGRAPHY

Daniel Bernoulli (170()-1789)

o.

(b) y = x 2

- ---"......-----:c- x 2

x - ---" 0, and a local maximum at the endpoint x = b if f(x) :5 f(b) for all x in some half·open interval (b - Ii, b J, Ii > O. The inequalities are reversed for local minimum values. In Figure 4.5, the function f has local maxima at c and d and local min· ima at a, e, and b. Local extrema are also called relative extrema. Some functions can have inrmitely many local extrema, even over a finite interval. One example is the func· tion f(x) = sin (l/x) on the interval (0, IJ. (We graphed this function in Figure 2.40.)

x

4.1


187

Absolute maximum.

Local minimum No smaller value : off nearby. Absolute minimum. No smaller value of f anywhere. AlBo a I locaIminimum. : _ _ _ _ _ _ _ _ _ _ _ _ _ _ _LI

Local minimum. I No smaller value of :fnearby. : I

_ _ _ _ _ _~_ _ _ _ _ _LI

a

: : I

:

---------"; ' --~':-------_>x

e

d

b

FIGURE 4.5 How to ideotifY types of maxima and minima for a functioo with domain a ~ x ~ b.

An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list ofall local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.

Finding Extrema Local maximum value

The next theorem explains why we usually need to investigate only a few values to find a function's extrema.

If f has a local maximum or minimum value at an interior point c of its domain, and if f' is defined at c, then

THEOREM 2-The First Derivative Theorem for Local Extreme values

f'{c) = O. Secant slopes

2:

(never ,negative)

0

Secant slopes ::s 0 (never po~itive)

I I I I

--~I~------~------~-->,X

X

X

FIGURE 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of noopositive numbers and nonnegative numbers, is zero.

Proof To prove that f' (c) is zero at a local extremum, we show first that f' (c) cannot be positive and second that f' (c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what f' (c) must be. To begin, suppose that f has a local maximum value at x = c (Figure 4.6) so that f{x) - f{c) ,;; 0 for all values of x near enough to c. Since c is an interior point of f's domain, f' (c) is derfied by the (w(}.sided limit

r

x~

f{x) - f{c) xc·

This means that the right-hand and left-hand limits both exist at x = c and equal f'{c). When we examine these limits separately, we find that

f '{ C )

=

f '{ C )

=

l'

un

x-c+

f{x) - f{c) xc

0

-'

Because (x - c) > 0 and !(x) '" !(c)

(1)

Similarly,

l'

11D_

x-c

f{x) - f{c) X

C

-.

Because (x - c) < 0 and !(x) " !(c)

(2)

Together, Equations (1) and (2) imply f'{c) = O. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use f{x) 2: f{c) , which reverses the inequalities in Equations (1) and (2). •

188

Chapter 4: Applications of Derivatives Theorem 2 says that a function's first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function 1 can possibly have an extreme value (local or global) are

1. 2.

interior points where f' = 0,

3.

endpoints of the domain of I.

interior points where f' is undefined,

The following definition helps us to summarize. (a)

y

----~----~----~--->x

(b)

FIGURE 4.7 Critical points wi1hout extrem.evalues. (a)y' = 3x 2 isOatx = 0, but y = x 3 has no extremum there. (b) y' = (1/3)x-2/3 is undef"med atx = 0, but y = X 1/3 bas no extremum 1here.

DEFINITION An interior point of the domain of a function 1 where f' is zero or undef"med is a critical point of I.

Thus the only domain points where a function can assume extreme values are critical points and endpoints. However, be careful not to misinterpret what is being said here. A function may have a critical point at x = c without having a local extreme value there. For instance, both of the functions y = x' and y = x 1/' have critical points at the origin and a zero value there, but each function is positive to the right of the origin and negative to the left. So neither function has a local extreme value at the origin. Instead, each function has a point of inflection there (see Figure 4.7). We define and explore inflection points in Section 4.4. Most problems that ask for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can simply list these points and calculate the corresponding function values to fmd what the largest and smallest values are, and where they are located. Of course, if the interval is not closed or not fmite (such as a < x < b or a < x < 00), we have seen that absolute extrema need not exist. If an absolute maximum or minimum value does exist, it must occur at a critical point or at an included right- or left-hand endpoint of the interval.

How to Find the Absolute Extrema of a Continuous Function f on a Finite Closed Interval 1. Evaluate 1 at all critical points and endpoints. 2. Take the largest and smallest of these values.

EXAMPLE 2 [-2, I].

Find the absolute maximum and minimum values of I(x) = x 2 on

Solution

The function is differentiable over its entire domain, so the only critical point is where f' (x) = 2x = 0, namely x = O. We need to check the function's values at x = 0 and at the endpoints x = -2 and x = I: Critical point value: Endpoint values:

1(0) = 0

I( -2) = 4

1(1) = 1 The function has an absolute maximum value of 4 at x = -2 and an absolute minimum value of 0 atx = O. •

4.1

EXAMPLE 3 [-2,1].

y 7

(1.7)

189


Find the absolute maximum and minimum values of g(l) = 81 - 14 on

Solutton

The function is differentiable on its entire domain, so the only critical points occur where g'(I) = O. Solving this equation gives

y= 8t- t 4

8-41'=0

1=\0/2>1,

or

a point not in the given domain. The function's absolute extrema therefore occur at the endpoints, g( -2) = -32 (absolute minimum), and g(l) = 7 (absolute maximum). See • Figure 4.8.

EXAMPLE 4

Find the absolute maximum and minimum values of I(x) = x2/' on the

interval [-2,3]. FIGURE 4.8 The extreme values of g(t) = 8t - t 4 on [-2, I] (Example 3).

y

Solutton

We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The tITSt derivative

I'(x) y =x213 , -2sxs 3 Absolute maximum; also a local maximum

Loca1

-I

0 ~ 1

2

3

Critical point value:

1(0)

Endpoint values:

I( -2) = (_2)2/3 =

Absolute minimum; also a local minimum

1(3)

FIGURE 4.9 The extreme values of f(x) = x 213 0n [-2, 3] occuratx = oand x = 3 (Example 4).

~x-I/3 = _2_ 3 3\Yx

has no zeros but is undefined at the interior point x = O. The values of I at this one critical point and at the endpoints are

~--~~~~--~--~--~X

-2

=

=

0

= (3)'/3 =

V'4

'f9.

We can see from this list that the function's absolute maximum value is 'f9 .., 2.08, and it occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the • interior point x = 0 where the graph has a cusp (Figure 4.9).

Exercises 4.1 finding Extrema from Graphs

3.

y

4.

y

In Exercises l-{i, determine from the graph whether the function has any absolute ex1reme values on [a, b]. Then explain how your answer is consistent with Theorem 1.

1.

y

2.

y ~O~--aL---~c----~b--~X

~---L--~----~-+x

o

a

c

b

o

a

c

b

190 S.

Chapter 4: Applications of Derivatives 6.

Y

In Exercises 15-20, sketch 1Ire graph of each function and determine whether the functioo has any absolute extreme values on its domain.

y

(~'

Explain how your answer is consistent with Theorem 1. 16. Y

~---L--~----~-+x

o

a

c

lxi, -I <x1

In Exercises 67 and 68, give reasons for your aoswers. 67. Le!f(x) = (x - 2)'/3.

a. Does f' (2) exist? b. Show that the ooly local extreme value of f occurs at x = 2. c. Does the result in part (b) cootradict the Extreme Value Theorem? d. Repeatparts(a)and(b)forf(x) = (x - a)'/3,rep1acing2bya. 68. Le!f(x)

=

lx' - 9xl.

a. Does f' (0) exist? b. Does f' (3) exist? c. Does f' (- 3) exist? d. Determine all extreroa of f. Theory and Examples 69. A minimum with no derivative The functioo f(x) = Ixl has an absolute minimum value at x = 0 even though f is not differentiable at x

= O.

Is this consistent with Theorem 2? Give rea-

sons for your answer. 70. E..n functions If an even functioo f(x) has a local maximum value at x = c, can anything be said about the value of f at x = -c? Give reasons for your answer.

71. Odd functions If an odd functioo g(x) has a local minimum value at x = c, can anything be said about the value of g at x = -c? Give reasons for your answer.

191

72. We know how to f"md the extreroe values of a continuous function f(x) by m..stigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasoos for your answers. 73. The functioo

V(x) = x(1O - a)(16 - a),

0< x < 5,

models the volume of a box. a. Find the extreme values of V. b. Interpret any values found in part (a) in terms of the volume

of the box. 74. Cubic functions f(x)

Coosider the cubic functioo

= ax'

+ bx' + ex + d.

a. Show that f can have 0, I, or 2 critical points. Give examples and graphs to support your arguroent. b. How many local extreme values can f have?

75. Muimum height of a ..rticaUy moviog body The height of a body moving vertically is given by s

=

I -"2gt2

+ vot + So,

g> 0,

with s in m_s and t in seconds. Find the body's maximum height 76. Peak alternating current Suppose that at any given time t (in seconds) the curreot i (in amperes) in an alternatiog current circuit is i = 2 cos t + 2 sin t. What is the peak current for this circuit (largest magoitode)?

D Graph the functions in Exercises 77-80. Then f"md the extreme values of the function 00 the interval and say where they occur. 77. f(x) = Ix 78. g(x) = Ix 79. h(x)

=

Ix +

80. k(x) = Ix +

21 + II 21 II +

Ix+ Ix Ix Ix -

31, 51, 31, 31,

-5 "'x '" 5 -2'" x'" 7 -00 -00

<x < <x
0 on (1,00).

~

a. Show that I(x) '" I for all x. b. Mustf'(I) ~ O?Explain.

72. Let I(x) ~ px2 + qx + r be a quadratic function defined on a closed interval [a, b j. Show that there is exactly one point c in (a, b) at which I satisfies the conclusion of the Mean Value

Theorem.

Monotonic Functions and the First Derivative Test In sketching the graph of a dllferentiable function it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function to identifY whether local extreme values are present.

4.3

Monotonic Functions and the First Derivative Test

199

Increasing Functions and Decreasing Functions All another corollary to the Mean Value Theorem, we show that functions with positive derivatives are increasing functions and functions with negative derivatives are decreasing functions. A function that is increasing or decreasing on an interval is said to be monotonic on the interval.

COROLLARY (a, h).

3

If I'(x)

Suppose that

I

is continuous on [a,

> 0 at each point x E (a, h), then I

Ifl'(x) < 0 at each point X

E

hI and differentiable on

is increasing on [a,

(a, h), then/is decreasing on [a,

Proof Let Xl andx2 be any two points in [a, applied to I on [Xh x21 says that

I(X2) - l(xI)

=

hI with Xl I'(C)(X2 -

hI. hI.

< X2. The Mean Value Theorem

XI)

for some c between XI and X2. The sign of the right-hand side of this equation is the same as the sign of I'(c) because X2 - Xl is positive. Therefore, I(X2) > I(Xl) if I' is positive • on (a, h) and/(x2) < I(xtl if I' is negative on (a, h). Corollary 3 is valid for infinite as well as finite intervals. To find the intervals where a function I is increasing or decreasing, we first find all of the critical points of I. If a < h are two critical points for I, and if the derivative I' is continuous but never zero on the interval (a, h), then by the Intermediate Value Theorem applied to 1', the derivative must be everywhere positive on (a, h), or everywhere negative there. One way we can determine the sign of I' on (a, h) is simply by evaluating the derivative at a single point c in (a, h). Ifl'(c) > 0, thenl'(x) > 0 for all X in (a, h) so/isincreasing on [a, hI by Corollary 3; if I'(c) < 0, then I is decreasing on [a, hI. The next example illustrates how we use this procedore. y y~x'-I2x - 5

EXAMPLE 1 Find the critical points of I(x) = x3 - 12x - 5 and identify the intervals on which I is increasing and on which I is decreasing.

20

Solution

The function I is everywhere continuous and differentiable. The f"trst derivative

I'(X)

=

3X2 - 12

= 3(x

FIGURE 4.20 The function /(x) ~ 12x - 5 is monotonic on three separate intervals (Example I).

=

+ 2)(x

3(X2 - 4) - 2)

is zero at X = -2 and X = 2. These critical points subdivide the domain of I to create nonoverlapping open intervals (-00, -2), (-2,2), and (2, 00) on which I' is either positive or negative. We determine the sign of I' by evaluating I' at a convenient point in each subinterval. The behavior of I is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the greph of I is given in Fignre 4.20.

X3 -

Intervnl /' evnluated Sign of/, Behaviorofl

-00 < X 0 immediately to the left and f' < 0 immediately to the right. Thus, the function is increasing on the left of the maximum value and decreasing on its right. In summary, at a local extreme point, the sign of f' (x) changes. Absolute max I' undefined I I I I I

No extremum I'~ 0

11' Oforx < land/,(x) < Oforx> I; b. /'(x)

< Oforx < land/,(x) > Oforx> I;

+ 4x,

0:5 x

Oforx "" I;

b. Which of the extreme values, if any, are absolute?

x T -

0:5 x :5 2-rr

56. j(x) =

+3

main, and say where they occur.

47. h(x) =

Graph the functioo and its derivative together. Connnent on the behavior of j in relation to the signs and values of /'.

18x

= 2: 0-,

0+.

67. Discuss the extreme-value behavior of the function j(x) = x sin (I/x), x "" O. How many critical points does this function have? Where are they located 00 the x-axis? Does j have an

4.4 Concavity and Curve Sketching

69. Determine the values of constants a and b so that f(x) ax 2 + bx has an absolute maximum at the point (I, 2).

absolute minimum? An absolute maximum? (See Exercise 49 in Section 2.3.)

68. Find tire intervals on which the function f(x) ~ ax 2 + bx + c, a -=F 0, is increasing and decreasing. Describe the reasoning behind your answer.

4.4

203 ~

70. Determine the values of constants a, h, c, and d so that f(x) ~ ax 3 + bx2 + ex + d has a local maximum at tire point (0,0) and a local minimum at the point (I, -I).

Concavity and Curve Sketching We have seen how the first derivative tells us where a function is increasing, where it is decreasing, and whether a local maximum or local minimum occurs at a critical point. In this section we see that the second derivative gives us infonnation about how the graph of a differentiable function bends or turns. With this knowledge about the first and second derivatives, coupled with oUI previous understanding of asymptotic behavior and symmetry stodied in Sections 2.6 and 1.1, we can now draw an accurate graph of a function. By organizing all of these ideas into a coherent procedure, we give a method for sketching graphs and revealing visually the key featores of functions. Identifying and knowing the locations of these featores is of major importance in mathematics and its applications to science and engineering, especially in the graphical analysis and interpretation of data.

Concavity FIGURE 4.23 The graph of f(x) ~ x 3 is concave down on ( - 00, 0) and concave up on (0,00) (Example la).

As you can see in Figare 4.23, the curve y = x 3 rises as x increases, but the portions defmed on the intervals (-00,0) and (0, 00) turn in different ways. As we approach the origin from the left along the curve, the curve turns to OUI right and falls below its tangents. The slopes of the tangents are decreasing on the interval ( - 00, 0) . As we move away from the origin along the curve to the right, the curve turns to OUI left and rises above its tangents. The slopes of the tangents are increasing on the interval (0, 00). This turning or bending behavior defmes the concavity of the curve.

DEFINmON

The graph of a differentiable function y ~ j(x) is

(a) concave up on an open interval I if j' is increasing on I;

(b) concave down on an open interval I if j' is decreasing on!.

If y

~

j(x) has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to the first derivative function. We conclude that j' increases if f" > 0 on I, and decreases iff" < O.

The Second Derivative Test for Concavity Let y = j(x) be twice-differentiable on an interval!.

1. If f"

0 on I, the graph of j over I is concave up.

2.

0 on I, the graph of j over I is concave down.

> If f"
O. (b) The curve y = x 2 (Figure 4.24) is concave up on (-00, 00) because its second deriv_ ative y" = 2 is always positive. (8) The curve

EXAMPLE 2

y

=

Detennine the concavity of y = 3

+

sin x on [0, 21T].

Solution The first derivative of y = 3 + sin x is y' = cos x, and the second derivative is y" = -sinx.Thegraphofy = 3 + sin x is concave down on (O,1T),wherey" = -sinx is negative. It is concave up on (1T, 21T), where y" = -sinx is positive (Figure 4.25). _ FIGURE 4.24

The graph of f(x) = x 2

is concave up on every interval (Example Ib).

The curve y = 3 + sin x in Example 2 changes concavity at the point (1T, 3). Since the first derivative y' = cos x exists for all x, we see that the curve has a taogent line of slope -I at the point (1T, 3). This point is called a point of inflection of the curve. Notice from Figure 4.25 that the graph crosses its taogent line at this point and that the second derivative y" = -sin x has value 0 when x = 1T. In general, we have the following definition.

y

y=3+sinx

2

Points of Inflection

: I

:~

DEFINITION A point where the graph of a function has a taogent line and where the concavity changes is a point of inflection.

~t.-~~-L~~~--~~ x

0 ," / .. -1- - - - - - . y" = -sinx

2

Using the sign of y' to determine the concavity ofy (Example 2). FIGURE 4.25

We observed that the second derivative of f(x) = 3 + sin x is equal to zero at the inflection point (1T, 3). Generally, if the second derivative exists at a point of inflection (c, f(c», then f"(c) = O. This follows immediately from the Intermediate Value Theorem whenever f" is continuous over an interval containing x = c because the second derivative changes sign moving across this interval. Even if the continuity assumption is dropped, it is still true that f"(c) = 0, provided the second derivative exists (although a more advanced agrument is required in this noncontinuous case). Since a taogent line must exist at the point of inflection, either the first derivative f' (c) exists (is finite) or a vertical taogent exists at the point. At a vertical taogent neither the flISt nor second derivative exists. In summary, we conclude the following result.

At a point of inflection (c, f(c», either f"(c) = 0 or f"(c) fails to exist. y

The next example illustrates a function having a point of inflection where the flISt derivative exists, hut the second derivative fails to exist. EXAMPLE 3 The graph of f(x) = x'/3 has a horizontal taogent at the origin because f'(x) = (5/3)x2/3 = 0 when x = O. However, the second derivative

f"(x) = FIGURE 4.26 The graph of f(x) = x'/3 has a horizontal taogent at the origin where the coocavity chaoges, although f" does not exist at x = 0 (Example 3).

~G'X2/3)

=

10 x - 1/ 3 9

failstoexistatx = O.Nevertheless,f"(x) < ofor x < Oandf"(x) > ofor x > O,sothe second derivative changes sign at x = 0 and there is a point of inflection at the origin. The graph is shown in Figure 4.26. _

4.4 Concavity and Curve Sketching y

205

Here is an example sbowing that an inflection point need not occur even though both derivatives exist and f" = O.

EXAMPLE 4 The curve y = x' has no inflection point at x = 0 (Figure 4.27). Even though the second derivative y" = 12x 2 is zero there, it does not change sign. •

FIGURE 4.27 The grapb of y = x' bas no inflection point at the origin, even though y" = 0 there (Example 4).

As our fmal illustration, we show a situation in which a point of inflection occurs at a vertical tangent to the curve where neither the first nor the second derivative exists. The graph of y = x 1/ 3 has a point of inflection at the origin because the second derivative is positive for x < 0 and negative for x > 0:

EXAMPLE 5

y Point of

inflection

-----~ ~~------~x

FIGURE 4.28

A point of

inflection where y' and y" fail to exist (Example 5).

However, both y' = x -2/3/3 and y" fail to exist at x = 0, and there is a vertical tangent there. See Figure 4.28. • To study the motion of an o~ect moving along a line as a function of time, we often are interested in knowing when the object's acceleration, given by the second derivative, is positive or negative. The points of inflection on the gmph of the object's position function reveal where the acceleration changes sign.

EXAMPLE 6 A particle is moving along a horizontal coordinate line (positive to the right) with position function .(1)

=

21 3

-

1412

+ 221 -

5,

12: O.

Find the velocity and acceleration, and describe the motion of the particle.

Solutton

The velocity is

V(I)

= .'(t) = 6t 2

- 28t + 22

= 2(t -

1)(3t - 11),

and the acceleration is

a(t) = v'(t) = ."(t) = 12t - 28 = 4(3t - 7). When the function .(t) is increasing, the particle is moving to the right; when .(t) is de· creasing, the particle is moving to the left. Notice that the first derivative (v = .') is zero at the critical points 1 = 1 and 1 = 11/3.

Interval Sign ofv =.' Behavior of. Particle motion

0 0 so the graph is increasing; on the interval (-2, 0) the derivative is negative and the graph is decreasing. Similarly, the graph is decreasing on the interval (0, 2) and increasing on

(2,00). 5. 4

Y~ ~X2+ 4 2x

val (0,00).

(2,2) 2

y= ! 2

6.

__L -__L-~~~~__-L->x

-4

-2

2

The graph ofy =

+4 ----zx-

=

+00

lim (~+.2.) x-o2 x

-00' so the y·axis is a vertical asymptote. Also, as x ..... 00 or as x ..... - 00, the graph of

I' x2

From the rewritten formula for f(x), we see that

lim (~+.2.) x-o+ 2 x

4

~2 (-2, -2)

FIGURE 4.31

There are no points of inflection because /,,(x) < 0 whenever x < 0, /,,(x) > 0 whenever x > 0, and /" exists everywhere and is never zero throughout the domain of f. The graph is concave down on the interval (- 00, 0) and concave up on the inter·

and

=

f(x) approaches the line y = x/2. Thus y = x/2 is an oblique asymptote.

7.

•

The graph of f is sketched in Figure 4.3\.

(Example 9).

Graphical Behavior of Functions from Derivatives As we saw in Examples 7-9, we can learn much about a twice·differentiable function = f(x) by examining its firat derivative. We can find where the function's graph rises and falls and where any local extrema are located. We can differentiate y' to learn how the graph bends as it passes over the intervals of rise and fall. We can deter· mine the shape of the function's graph. Information we cannot get from the derivative is how to place the graph in the xy·plane. But, as we discovered in Section 4.2, the only additional information we need to position the graph is the value of f at one point. Information about the asymptotes is found using limits (Section 2.6). The fol· lowing figure summarizes how the derivative and second derivative affect the shape of a graph.

y

r:;d" 7

> 0 ~ rises from

Differentiable ~ smooth, connected; graph may rise and fall

y'

,/ or

( or \

y" > 0

"

~ concave up throughout; no waves; graph may rise or fall

C"\ or ~ y' changes sign => graph has local maximum or local minimum

left to righ~ may be wavy

y" < 0

~ concave down throughout; no waves; graph may rise or fall

f\

and y"O at a point; graph has local minimum

y'~

4.4 Concavity and Curve Sketching

211

Exercises 4.4 Analyzing Functions from Graphs Identify the inflectioo points and local maxima and minima of the functioos graphed in Exercises I-S. IdentifY the intervals on which the functions are concave up and concave down.

1.

2.

17. y ~ x' - z,,' = x'(x' - 2) 18. y ~ -x' + 6>;' - 4 ~ x'(6 - x') - 4 19. y ~ 4x 3 - x' = x 3(4 - x)

+ z,,3

=

x 3(x

21. y ~ x' - 5x'

=

x'(x - 5)

20. y ~ x'

--/- *-- --+--.x

+ 1)3

16. y ~ I - (x

_5)'

22. y

~ x(~

23. y

~

24. Y

= x - sinx,

25. y ~

x + sinx, 0 s x s 2.-

V3x -

0

27. Y

4.

y ~ ~(x2 _ 1f'3

=

'vJV.

31. y ~

x

~

0

2w

x '" 2.-

S

-1T'

'1T

"2 < x < 2

sinX' cos x,

28. y ~ cosx 29. y ~ x'/5

~

2 COU,

4 26. y ~ '3x - taox,

3.

+ 2)

O:s x :s

+ V3 sin x,

0

1T'

sx

30. y = x'/5

x

W+1

32. y

33. y ~ z" - 3x'/3 35. y

~ x'/3(t -

'3 Y

3

6. y=tanx-4x-1!:<x;' (x - 2)3 + 1

13. y ~ _z,,3 15. y ~

+3 3

57. y'

= sec

2

x,

10.y~6-z,,-x'

58. y' = tanx,

12. y ~ x(6 - z,,)' 14. y ~ I - 9x - 6>;' - x 3

59. y' =

cot~,

,

+ 3)

_,!!<x

0

y' = 0, y" > 0 y"

>

0

4

4

y' > 0, y"

~

0

4<x6

y' y' y'

0, sketch a curve y = f(x) that has f(1) = 0 aod f'(x) = I/x. Cao aoything be said about the coocavity of such a curve? Give reasons for your answer.

s

I

104. Cao anything be said about the graph ofa functiooy = f(x) that has a continuous second derivative that is never zero? Give reasons for your answer.

].

is 0

5

15

10

Ttme (sec)

lOS. If b, c, aod d are coostaots, for what value of b will the corve y = x 3 + bx 2 + ex + d have a point of inflection at x = I?

Give reasons for your answer. 106. Parabolas

s

9S.

213

] fil-

is

0

15

5 Tnne(sec)

99. Marginal cost The accompanying graph shows the hypothetical cost c = f(x) of manufacturing x items. At approximately what productioo level does the marginal cost chaoge from decreasing to increasing?

.. Find the coordinates of the vertex of the parabola y = ax 2 + bx + c, a¢' O. b. When is the parabola concave up? Concave down? Give reasons for your answers. 107. Quadratic curves What can you say about the ioflectioo points of a quadratic curve y = ax 2 + bx + c, a =F O? Give reasons for your answer. lOS. Cubic curve. What cao you say about the ioflectiou points of a cubic curve y = ax 3 + bx2 + ex + d, a =F O? Give reasons for your answer. 109. Suppose that the secood derivative of the functioo y = f(x) is y"

= (x +

I)(x - 2).

For what x-values does the graph of f have ao ioflectioo point? 110. Suppose that the secoodderivative of the functiooy = f(x) is

c

+ 3). For what x-values does the graph of f have ao ioflectioo point? y" = x 2(x - 2)'(x

~~~~~~~~ X

20 40 60 80100120 Thousands of ooits produced

100. The accompanying graph shows the moothly reveoue of the Widget Corporatioo for the last 12 year>. During approximately what time intervals was the marginal revenue increasing? Decreasing?

101. Suppose the derivative of the functioo y = f(x) is

= (x

- I)'(x - 2).

At what points, if aoy, does the graph of f have a local minimum, local maximum, or point of ioflectioo? (Hint: Draw the sigo pattern for y' .) 102. Suppose the derivative of the functioo y

y'

112. Find the values of coostaots a, b, aod c so that the graph of y = (x 2 + a)/(bx + c) has a local minimum at x = 3 aod a I()cal maximum at (-I, -2). COMPUTER EXPLORATIONS 10 Exercises 113-116, fmd the ioflectioo points (if aoy) 00 the graph of

y

y'

111. Find the values of coostaots a, b, aod c so that the graph of y = ax 3 + bx2 + ex has a local maximum at x = 3, local minimumatx = -1,aodioflectioopointat(l, 11).

= (x

=

f(x) is

- I)'(x - 2)(x - 4).

At what points, if aoy, does the graph of f have a local minimum, local maximum, or point of inflection?

the function aod the coordinates of the points on the graph where the functioo has a local maximum or local minimum value. Thcu graph the functiou in a regioo large coough to show all these points simuitaoously. Add to your picture the graphs of the function's fl!St and secood derivatives. How are the values at which these graphs intersect the x-axis related to the graph of the functiou? 10 what other ways are the grapha of the derivatives related to the graph of the functioo? 113. y = x' - 5x' - 240 114. Y = x 3 - 12x 2 115. Y

= ~x' + 16>;2 -

116. Y

= "4 - "3 -

X4

x3

25

4x 2 + 12x + 20

117. Graph f(x) = 2x' - 4x 2 + I and its frrst two derivatives lgether. Comment on the behavior off in relation to the sigos aod values of f' aod f" . 11S. Graph f(x) = x cos x aod its second derivative together for o ,;; x ,;; 2".. Comment 00 the behavior of the graph of fin relatioo to the sigos aod values of f" .

214

Chapter 4: AppLications of Derivatives

Applied Optimization

4.5

What are the dimensions of a rectangle with IlXCd perimeter having maximum area? What are the dimensions for the least expensive cylindrical can of a given volume? How many items should be produced for the most profitable production run? Each of these questions asks for the best, or optimal, value of a given function. In this section we use derivatives to solve a variety of optimization problems in business, mathematics, physics, and economics.

Solving Applied Optimization Problems 1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized? 2. Draw a picture. Label any part that may be important to the problem. 3. hltroduce variables. List every relation in the picture and in the problem as an equation or algebraic expression, and identify the unknown variable. 4. Write an equation for the unknown quantity. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. This may require considerable manipulation. S. Test the critical points and endpoints in the domain ofthe Il1Iknown. Use what you know about the shape of the function's graph. Use the lust and second derivatives to identify and classify the function's critical points.

(.)

EXAMPLE 1 An open-top box is to be made by cutting small congruent squares from the comers of a 12-in.-by-12-in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible? (b)

FIGURE 4.32 An open box made by cutting the corners from a square sheet of

Solution We start with a picture (Figure 4.32). In the figure, the corner squares are x in. on a side. The volume of the box is a function of this variable:

tin What size comers maximize the box's volmne (Example I)?

, y - x(12 - 'lx'jl,

O:Sx:S6

Since the sides of the sheet of tin are only 12 in. long, x :S 6 and the domain of Vis the interval 0 :S X :S 6. A graph of V (Figure 4.33) suggests a minimum value of 0 at x = 0 and x = 6 and amaximum near x = 2. To learn more, we examine the first derivative of V with respect

to x: min

o

V=hlw

:

~

144 - 9&

+ 12%' ~

12(12 - 8x

+ x') ~

12(2 - x)(6 - x).

/,

•

FIGURE 4.33 The volume of the box in Figure 4.32 graphed 18 a fimction of x.

Of the two zeros, x - 2 and x - 6, only x - 2 lies in the interior of the function's domain

and makes the critical-point list. The values of Vat this one critical point and two endpoints are Critical-point value: ilDdpointvalucs:

1'(2) - 128 1'(0)

~

0,

1'(6)

~

o.

The lll8Ximum volume is 128 in3 • The cutout squares should be 2 in. on a side.

•

4.5

AppLied Optimization

215

EXAMPLE 2 You have been asked to design a one-liter can shaped like a right circular cylinder (Figure 4.34). What dimensions will use the least material?

+----2r--~) 1

T h

1

SoLution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is I liter = 1000 crn3

Surface area ofcan:

A = 27Tr2

+ 27Trh '---v---'

FIGURE 4.34 This one-liter can uses the least material when h = 2r (Example 2).

circular cylindrical ends wall

How can we interpret the phrase "least material"? For a first approximation we can ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible while satisfying the constraint 7Tr 2h = 1000. To express the surface area as a function of one variable, we solve for one of the variables in 7Tr 2h = 1000 and substitute that expression into the surface area formula. Solving for h is easier:

Thus,

Our goal is to find a value of r that such a value exists.

> 0 that minimizes the value of A. Figure 4.35 suggests

A

Tall and thin can Short and wide can A = 271"r2 Tall and thin

+

2000, r r

/> 0

--r-------~--------------~r

o

Short and wide

FIGURE 4.35

~

The graph of A = 27fr2

+ 2000/r is concave up.

Notice from the graph that for small r (a tall, thin cylindrical container), the term 2000/r dominates (see Section 2.6) and A is large. For large r (a short, wide cylindrical container), the term 27Tr2 dominates and A again is large.

216

Chapter 4: Applications of Derivatives Since A is differentiable on r > 0, an interval with no endpoints, it can have a minimum value only where its fITst derivative is zero.

dA dr

2000 r2

- - = 4'ITr - - -

0= 4'ITr - 2000 -r2

SetdA/dr ~

4",,3 = 2000 r =

~5~0

o.

Multiply by r'.

'" 5.42

Solve forr.

What happens at r = "¢'500/'IT? The second derivative 2

d A = 4'11' + 4000 dr 2 r3 is positive throughout the domain of A. The graph is therefore everywhere concave up and the value of A at r = "¢'500/'IT is an absolute minimum. The corresponding value of h (after a little algebm) is

h = 1000 = 7f'r2

2~500

= 2r.

'1T

The one-liter can that uses the least material has height equal to twice the mdius, here with r '" 5.42 em andh '" 10.84 em. •

Examples from Mathematics and Physics EXAMPLE 3

A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?

y x'+y'~4

(x. V4 - X2) 2 ~L-~----~~------~~ x

-2 -x

0

x 2

FIGURE 4.36 The rectangle inscribed in the semicircle in Example 3.

Solution Let (x, ~) be the coordinates of the comer of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.36). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower right-hand comer: Length: 2x,

Height:

V4 -

x 2,

Area:2x~.

Notice that the values of x are to be found in the interval 0 :5 x :5 2, where the selected comer of the rectangle lies. Our goal is to find the absolute maximum value of the function

A(x)=2x~ on the domain [0, 2]. The derivative

dA dx

~+2~ 4 - x2

is not defined when x = 2 and is equal to zero when

-2x

2

~

+2~=0

-2x 2 + 2(4 - x 2 )

= 0 2 8-4x =0

x 2 = 20rx =

±v'2.

4.5

V2

V2,


217

V2

Of the two zeros, x = and x = only x = lies in the interior of A's domain and makes the critical-point list. The values of A at the endpoints and at this one critical point are

A( V2)

Critical-point value:

=

2V2v'4=2 =

A(O) = 0,

Endpoint values:

4

A(2) = O.

The area has a maximum value of 4 when the rectangle is 2x = 2 uuits long.

V2

V4 -

x2 =

V2 units high and •

EXAMPLE 4 The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat's principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Describe the path that a my of light will follow in going from a point A in a medium where the speed of light is c, to a point B in a second medium where its speed is C2.

HiSTORICAL BIOGRAPHY

Willebrord Snell van Rayen

(158()"'1626)

y

Solution Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xy-plane and that the line separating the two media is the x-axis (Figure 4.37). In a uuiform medium, where the speed of light remains constant, "shortest time" means "shortest path;' and the my oflight will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance traveled equals rate times time, so

Medium I

-ooot---,----"''Ir- -d\ - - - ' Medium 2

d-x B

Time = distance rate

FIGURE 4.37 Alightrayrefracted (deflected from its path) as it passes from

From Figure 4.37, the time required for light to travel from A to P is

one medium to a denser medium (Example 4).

t}

AP

= Ct

=

Va 2 + x 2 Ct

.

From P to B, the time is

PB

t2=C2=

Vb 2 + (d -

x)'

C2

The time from A toB is the sum of these: 1 = I,

+ 12

=

Va 2 + x 2 C,

+

Vb 2 + (d C2

x)'

This equation expresses 1 as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of 1 on this closed interval. We find the derivative

d-x

dtIdx negative

dtIdx zero

( - - - -: /,. + + + + + + + (

o

0

and observe that it is continuous. In terms of the angles 6, and 62 in Figure 4.37,

dtldx positive >x

dl dx

sin 6,

sin 62

d

FIGURE 4.38 The sign pattern of dt/dx in Example 4.

The function 1 has a negative derivative at x = 0 and a positive derivative at x = d. Since dl/dx is continuous over the interval [0, d], by the Intermediate Value Theorem for continuous functions (Section 2.5), there is a point Xo E [0, d] where dl/dx = 0 (Figure 4.38).

218

Chapter 4: Applications of Derivatives

There is only one such point because dt/ dx is an increasing function of x (Exercise 62). At this unique point we then have sin II,

sin 112 C2·

Ct

This equation is SneU's Law or the Law of Refraction, and is an important principle in the theory of optics. It describes the path the ray of light follows. •

Examples from Economics Suppose that r(x) = the revenue from selling x items c (x) = the cost of producing the x items p(x) = r(x) - c(x) = the profit from producing and selling x items. Although x is usually an integer in many applications, we can learn about the behavior of these functions by defining them for all nonzero real numbers and by assuming they are differentiable functions. Economists use the terms marginal revenue, marginal cost, and marginal profit to name the derivatives r'(x), c'(x), and p'(x) of the revenue, cost, and profit functions. Let's consider the relationship of the profit p to these derivatives. If r(x) and c(x) are differentiable for x in some interval of production possibilities, and if p(x) = r(x) - c(x) has a maximum vaIue there, it occurs at a critical point ofp(x) or at an endpoint of the interval. If it occurs at a critical point, then p'(x) = r'(x) c'(x) = 0 and we see that r'(x) = c'(x). In economic terms, this last equation means that

At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.39).

y

I

i Maximum profit, c'{x) = r'(x) I I I

I I

"

I

Local maximum. for loss (minimum profit), c'(x) = ,'(x)

~~ _ _L -______~ ____J-~

o

____________->x

Items produced

FIGURE 4.39 The graph of a typical cost functioo starts coocave down aod larer turns coocave up. It crosses the revenue curve at the break-even point B. To the left of B, the compaoy operates at a loss. To the right, the compaoy operates at a profit, with the maximmn profit occurriug where c' (x) = r' (x). Farther to the right, cost exceeds revenue (perllaps because of a combinatioo of rising labor aod material costs aod market saturatioo) aod productioo levels become unprofitable again.

4.5 Applied Optimization

219

EXAMPLE 5 Supposethatr(x) = 9xandc(x) = x 3 - 6>;2 + 15x, where x represents millions ofMP3 players produced. Is there a production level that maximizes profit? If so, what is it? y

Solution

Notice that r'(x) = 9 and c'(x) = 3x2

-

IZl: + 15.

2

3x - IZl: + 15 = 9 3x 2 -1Zl:+6=0

Setc'(x) ~ r'(x).

The two solutions of the quadratic equation are

x,

= 12 -6

V72

= 2 -

Vz "" 0.586

X2

= 12 +6

V72

= 2 +

Vz "" 3.414.

: Maximum : for profit I I I

Local maximum for loss ~-L~---J----J-~~-----c>x

o 2-v2

2

2+v2

NOT TO SCALB

FIGURE 4.40 The cost and revenue curves for Example 5.

and

The possible production levels for maximum profit are x "" 0.586 million MP3 players or x"" 3.414 million. The second derivative of p(x) = r(x) - c(x) is p"(x) = -c"(x) since r"(x) is everywhere zero. Thus, p"(x) = 6(2 - x), which is negative atx = 2 + and positive at x = 2 By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graphs of r(x) and c(x) are shown in Figure 4.40. •

Vz

Vz.

Exercises 4.5 Mathematical Applications Whenever you are maximizing or minimizing a function of a single variable, we mge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your aoswer. 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in2 , and what are its dimensions? 2. Show that among all rectangles with an 8-m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. &.

Express the }"oordinate of P in terms of x. (Hint: Write an equation for the line AB.)

b. Express the area of the rectangle in terms or.. c. What is the largest area the rectangle can have, and what are its dimensions? y

4. A rectangle has its base on the x-axis and its upper two vertices on the parabola y ~ 12 - x 2 • Whst is the largest area the rectangle can have, and what are its dimensions? 5. You are plauning to make an open rectangular box from an 8-in.by-15-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. Whst are the dimensions of the box of largest volume you can make this way, and wbat is its volume?

6. You are planning to close off a comer of the fIrst quadrant with a line segment 20 units long running from (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when

a = b. 7. The best fencing plan A rectangular plot offarmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its

dimensions? 8. The .hortest fence A 216 m 2 rectangu1ar pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. Whst dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 9. Designing. tank Your iron works bas contracted to design and build a 500 ft', square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to f"md dimensions for the base and height that will make the tank weigh as little as possible.

220


a. What dimensions do you tell the shop to use? b. Briefly describe how you took weight into account.

oJ

~

u

10. Catching rainwater A 1125 ft3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.

'" l3

!3z

I 1

I+-x-+j

I+-x+j T

IT

I

10"

Base

Lid

T

x .i

T

I+-x-+j

x .i

I+-x+j IS"

l-

-I

a. If the total costis

c = 5(x 2

a. Write a formula V(x) for the volume of the box.

+ 4xy) + lOX)!,

b. Find the domain of V for the problem situation and graph V over this domain.

what values of x and y will minimize it?

c. Use a graphical method to find the maximum volume and the value of x that gives it.

b. Give a possible scenario for the cost function in part (a). 11. Designing a poster You are designing a rectangular poster to contain 50 in2 of printing with a 4-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.

d. Confirm your result in part (c) analytically.

D 17.

Designing a suitcase A 24-in.-by-36-in. sheet of cardboard is folded in half to form a 24-in.-by-18-in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. e. Find a value of x that yields a volume of 1120 in 3 . f. Write a paragraph describing the issues that arise in part (b).

13. Two sides of a triangle have lengths a and b, and the angle between them is 8 . What value of 8 will maximize the triangle's area? (Hint: A = O/2)ab sin 8.) 14. Designing a can What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 1000 cm3 ? Compare the result here with the result in Example 2. 15. Designing a can You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be

A = 8r 2 + 21Trh rather than the A = 21Tr2 + 21Trh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now?

D 16.

Designing a box with a lid A piece of cardboard measures 10 in. by 15 in. Two equal squares are removed from the corners of a IO-in. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with lid.

I

I I I I I I I I I I

24"

1

x

x

24"

x

x

:

I,

I

)1

36"

~ 18" ------;.I

1

The sheet is then unfolded.

I 1 24"

I,

Base

36"

)1

18. A rectangle is to be inscribed under the arch of the curve y = 4 cos (0.5x) from x = -1T to x = 1T . What are the dimensions of the rectangle with largest area, and what is the largest area?

4.5


221

19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?

cylindrical sidewall. Determine the dimensions to be used if the volume is fIxed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.

20. a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?

24. The trough in the figure is to be made to the dimensions shown. Only the angle 8 can be varied. What value of 8 will maximize the trough's volume?

Girth = distance

o b. Graph the volume of a 108-in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). 21. (Continuation ofExercise 20.)

25. Paper folding A rectangular sheet of 8.5-in.-by-II-in. paper is placed on a flat surface. One of the comers is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. a. Show that L 2 = 2x 3/(2x - 8.5) . b. What value of x minimizes L 2 ?

a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by wand the girth is 2h + 2w . What dimensions will give the box its largest volume now?

c. What is the minimum value of L?

Girth

Q (originally at A)

h

o b.

Graph the volume as a function of h and compare what you see with your answer in part (a).

22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fIxed. Find the proportions of the window that will admit the most light Neglect the thickness of the frame.

26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values ofx andy give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?

x y I

_-------1--------, /

23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the

(a)

(b)

,

222

Chapter 4: AppLications of Derivatives

27. CoaJtructlDg conel A right triangle whose hypotcuuac ia m long is mvolwd about one of it! less to gcnmate a right circular OODC. Find the radiwJ, height, and vulumc of the OODC of ~t volume that can be made this way.

v'3

• 28. Find the point on the line ~

Building

,

+~=

1 that is closest to the origin

29. Find a positive mnnbcr for which the sum of it and its reciprocal ia the !lll8lJ.e!1: (least) possible. 30. Find a postitive mnnbcr for which the sum of it! reciprocal and four timca it! square is the smallest poYible. 31. A wire b m long is cut into two pieces. One piece is bent into an cquila.tcmJ. triangle and the other is bent into a circle. If the !IUlD. of the areas cnclo!Cd by each part ia a minjrnnm, what is the length of each part? 32.

~

and I in seconds. L

3

34. Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle ofradiWl 3. (See accompanying figure.)

35. What value of a makes f(x} = x'1.

•

I

~

b. a point of inflection at x - I 1

+ a;t2 + bx have

a local maximum at x - - 1 and a local mjnjrnmn at x - 3 1

b. a local minimum. at x = 4 and a point of inflection at x = 11

Physical Applications 37. Ver11c.J. motloa The height above ground of an object moving vertically ia givm by

b. What is the farthest apart that the particles ever get?

c. When in the interval 0

:S t :S 2Tr is the distance between the particles changing the fastest?

41. Projectile motioa The runge R of a projectile flICd from the origin am: horizontal ground is the distance from the origin to the point of impact. If the projectile is flICd with an initial velocity VI) at an angle a with the horizonlal, then in Chapter 13 we find that

vJ.

+ 96t + 112,

o 43. Streqtb. of.

beam. The strength S of a rectangular wooden beam is proportional to it! width timca the square of its depth. (See the accompanying figm:.) L

Find the dimensions of the strongest beam that can be cut from a 12-in..-diamctcr cylindrical log.

b. GraphSas a function of the beam's width w, assuming the proportionality constant to be 1 - 1. Reconcile what you see with your answer in part (a).

c. On the same screen, graph S as a function of the beam's depth d, again taking 1 - 1. Compare the graphs with one mother and with your answer inpart (a). What would be the cifect of chansins to some other value of 11 Try it.

with a in feet and t in seconds. Find L

2Tr do the particles

wbcrc g is the downward acceleration due ttl gravity. Find the angle a for which the range R is the laIjest possible.

a local minimum. at x = 21

a - -161 2

~

m<et1

R = -gsm2a,

+ (a/x) have

36. What values of a and b make f(x} - x 3

At what time(s) in the in1m"Val 0

41. The intensity of illumination at my point from a light source is proportional to the square of the reciprocal of the diIltance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. Haw far from the stronger light is the total illumination least?

33. Determine the dimensions of the rectanglc of ialBCllt area that can be :imcribed in the right triangle shawn in the accompanying; figwe.

L

40. Motioa OD • Hoe The positions of two parti.cJ.es on the s-axis are 61 - sin I and.!"'}. - sin(1 + Tr/3), with 61 and 62 in meterJi

Excrciac 31 if one picoc is bent into a square and the

other into a circle.

L

39. Shortat beam. The 8-ft wall shawn here stands 27 ft from. the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall

the object's velocity when t - 0

b. its maximmn height and when it occurs

c. itsvclocity when" - o. 38. Quickest nnrte lane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point Dearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the leaat amount of time1

12"

d

w

4.5 Applied Optimization

o 44. StHfneu of. beam

The stiffiless S of a rectangular beam is proportional to its width times the cube of its depth. L

Find the dimensions of the stiffest beam. that can be cut from a 12-in.-diameter cylindrical log.

b. Graph S as a function of 1he beam's wid1h w, assuming 1he proportionality constant to be k - 1. Reconcile what you see with your answer in part (a).

e. On 1he same screen, graph S as a fimction of the beam's dep1h d, again taking k - 1. Compare the graphs wi1h one another and wi1h your answer in part (a). What would be the effect of changing to some other value of k? Try it. 45. FrictiODlell urt A small frictionless cart, att:aclled to the wall by a spring. is pulled 10 em. from its rest position and released at time t - 0 to roll back and for1h for 4 sec. Its position at time tis 9 = 1000s"ll'1. L What is the cart's maximum. speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart's speed then?

~ o

10

223

c. The visibility that day was 5 nauti.cal. miles. Did the ships ever sight each other?

o d. Graphs

and dY/dt together as fimctions oftfor -1 :s t :s 3, using different colors if posSIble. Compare the graphs and reconcile what you see with your answers in parts (b) and (c).

e. The graph of ds/dt looks as ifit might have a horizont:al. asymptote in the rust quadrant. This in turn suggests that ds/dt approaches a limiting value as t - 00. What is this value? What is itll relation to the ships' individual. speeds? 48. Fermat's prlDdple iD optks Light from. a source A is reflected by a plane mirror to a receiver at point B, as shawn in the accompanying fJgUlC. Shaw that for 1he light to obey Fermat's principle, the angle of incidence must cqual the angle of reflection, bo1h measured from. the line normal to 1he reflecting surface. (This result can also be derived without calculus. There is a purely goometric argument, which you may prefer.) N""",,

Angle of

Angle of

incidence

rdlcction

"

"""

B

II,

"

46. 1Wo masses hanging side by side from springs have positions 9l = 2 sin t and 92 = sin 2t, respectively. L At what times in the interval 0 < t dD the masses pass each other? (Hint: sin2t = 2 sin t cos t.) b. When in the interval 0 :s t :s 2'11" is the vertical distance between the masses the greatest? What is this distance? (Hint: cas2t = 2co:flt - 1.)

49. Tin peat When metallic tin is kept below I3.2°C, it slowly becomes brittle and crumbles to a gray powder. 1m objects eventually cnnnble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that control! the rate of reaction without undergoing any permanent change in itself. An autOC4tolytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at Hrst if tile amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abtmdant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate v - dx/dt of the reaction is proportional. both to the amount of the original substance present and to the amount of product. That is, v may be considered to be a fimction of x alone, and v = .b;(a - x) = kin; - kx 2,

, 47. Distanee between two ddpi At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 lcnot:! (nautical miles per hour; a nautical mile is 2000 yd) and continued to dD so all day. Ship B was sailing east at 8 knots and continued to dD so all day. L Start counting time wi1h t = 0 at noon and express the distance 9 between the ships as a fimction oft. b. How rapidly was the distance between the ships changing at noon? One hour later?

wh...

x - the amount of product

a

= the amount of substance at the beginning

k = a positive constant. At what value of x does the rate v have a maximum? What is the maxinnnn value of v? SO. Airplane landing path An airplane is flying at altitude H wilen it begins Us descent to an aiIport runway that is at horizootaJ. ground distance L from the airplane, as shown in the fJgL1IC. Assume that 1he

224


landing path of the airplane is the graph of a cubic polynmial fimctioo y ~ ax' + bx2 + ex + d, where y ( - L) ~ H and

y(O) L

~

O.

55. Showthatifr(x) ~ 6xande(x) ~

Whatisdy/dxatx ~ O?

b. What is dy/dx atx

~

~

Oandy(-L)

~

x' - 6x

2

+

15xareyourrev-

enue and cost functions, then the best you can do is break. even

-L?

c. Use the values forth>/dx atx y(O)

54. Production level Prove that the productioo level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost

~

oand x

~

-Ltogetherwith

Htoshowthat

(have revenue equal cost). 56. Production level Suppose that e(x) ~ x' - 2Ox 2 + 20,00Ox is the cost of manufacturing x iterus. Find a productioo level that

will minimize the average cost of making x items. 57. You are to construct an open rectangular box with a square base and a volume of48 ft'. Ifmaterial for the bottom costs $6/ft' and materisl for the sides costs S4/ft', what dimensions will result in

y

Landing path

the least expensive box? What is the minimum cost? 58. The 800-room Mega Motel chain is filled to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?

H = Cruising altitude

Airport

~I'-------------L------------~

Business and Economics 51. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by

n~ x ~ e

+ b(IOO

Biology 59. Sensitivity to medicine (Continuation of Exercise 60, Section 3.3.) Find the amount of medicine to which the body is most sensitive by f'mding the value of M that maximizes the derivative dR/ dM, where

- x),

where a and b are positive constants. What selling price will bring a maximum profit?

and C is a constant 60. How we cough

52. You operate a tour service that offers the following rates:

.. When we cough, the trachea (windpipe) contracts to

$200 per persoo if 50 people (the minimum number to book the tour) go on the tour.

increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the

For each additiooal persoo, up to a maximum of 80 people total, the rate per petliOO is reduced by $2.

velocity and whether it really cootraets that much when we cough.

Under reasonable assumptions about the elasticity of the

It costs $6000 (a fixed cost) plus $32 per petlion to cooduct the tour. How many people does it take to maximize your profit?

tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity v can be modeled by the

53. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is

A(q) ~

km

equation

v ~ c(ro - r)r 2 em/sec,

hq

placing an order (the same, no matter how often you order), c is

L

Your job, as the inventory manager for your store, is to fmd the quantity that will minimize A(q). What is it? (The fonnula you get for the answer is called the Wilson lot size formula.)

b. Shipping costs sometimes depend 00 order size. When they do, it is more realistic to replace k by k + bq, the sum of k and a constant multiple of q. What is the most economical quantity to order now?

70,

length of the trachea. Show that v is greatest when r ~ (2/3)ro; that is, when the trachea is about 33% cootraeted. The remarkable fact is that X-ray photographs conf'mn that the trachea contracts about this much during a cough.

where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of

(a constant that takes into account things such as space, utilities, insurance, and security).

0) intersect at one point x = r. Use Newton's method to estimate the value of r to four decimal places.

in two dimensions

------~~~--~~--~--~x

o

D 27.

"

2

SOOObu~"(2, -!)

Curve. that are nearly flat at the root Some curves are so flat that, in practice, Newton's method stops too far from the root to give a useful estimate. Try Newton's method on f(x) = (x - 1)'0 with a starting value of .:to = 2 to see how close your machine comes to the root x = 1. See the accompanying graph.

230

Chapter 4: Applications of Derivatives y

28. The accompanying figure shows a circle of radius r with a chord ofleogth 2 aod ao arc S ofleogth 3. Use Newton's method to solve for r aod 8 (radians) to four decimal places. Assume 0 < 6 < TT.

y ~ (x - I)'"

Slope ~-4O

Slope

I

(2, I)

~

40

Nearly flat

o

4.7

I

2

Antiderivatives We have studied how to find the derivative of a function. However, many problems require that we recover a function from its known derivative (from its knownmte of change). For instance, we may know the velocity function of an object falling from an initial height and need to know its height at any time. More generally, we want to fmd a function F from its derivative f. If such a function F exists, it is called an antiderivative of f. We will see in the next chapter that antiderivatives are the link connecting the two major elements of calculus: derivatives and defmite integrals.

Finding Antiderivatives DEFINITION F'(x)

=

A function F is an antiderivative of

f

on an interval I if

f(x) forallxinI.

The process of recovering a function F(x) from its derivative f(x) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function f, G to represent an antiderivative of g, and 80 forth.

EXAMPLE 1 (a) f(x) = 2x

Find an antiderivative for each of the following functions. (b) g(x) = cosx

(0) h(x) =

2x + cosx

Solution

We need to think backward here: What function do we know has a derivative equal to the given function?

(a) F(x) = x 2

(b) G(x) = sinx

(0)

H(x)

=

x 2 + sinx

Each answer can be checked by differentiating. The derivative of F(x) = x 2 is 2x. The derivative of G(x) = sinx is cosx and the derivative of H(x) = x 2 + sinx is 2x + cosx. •

4.7 Antiderivatives

231

The function F(x) = x 2 is not the only function whose derivative is 2x. The function x + I has the same derivative. So does x 2 + C for any constant C. Are there others? Corollary 2 of the Mean Value Theorem in Section 4.2 gives the answer: Any two antiderivatives of a function differ by a constant. So the functions x 2 + C, where C is an arbitrary constant, form all the antiderivatives of f(x) = 2x. More generally, we have the following result. 2

THEOREM 6

If Fis an antiderivative off on an interval I, then the most general antiderivative off on I is F(x)

+C

where C is an arbitrnry constant.

Thus the most general antiderivative of f on I is a family of functions F(x) + C whose graphs are vertical translations of one another. We can select a particular antiderivative from this family by assigning a specific value to C. Here is an example showing how such an assignment might be made.

EXAMPLE 2

y

C~2

Solution

Find an antiderivative of f(x) = 31;2 that satisfies F(I) = -1.

Since the derivative of x' is 3x 2 , the general antiderivative

C~l

F(x) = x'

C~O C~-l

..--------C=-2

+

C

gives all the antiderivatives of f(x). The condition F( I) = - I detennines a specific value for C. Substituting x = I into F(x) = x 3 + C gives

F(I) = (1)3 + C = I + SinceF(I) = -I, solving I

+

C = -I for C gives C = -2. So

F(x) = x 3

FIGURE 4.48 The curves y ~ x' + C fill the coordinate plane without overlapping. In Example 2, we identify the curve y = x 3 - 2 as the one that passes through the given point (I, -I).

c.

-

2

is the antiderivative satisfying F( I) = - I. Notice that this assignment for C selects the particular curve from the family of curves y = x 3 + C that passes through the point (I, -I) in the plane (Fignre 4.48). • By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrnry constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number of important functions. The rules in Table 4.2 are easily verified by differentiating the general antiderivative formula to obtain the function to its left. For example, the derivative of (tan lex)/! + Cis sec2 la, whatever the value of the constants Cork '" 0, and this establishes Formula 4 for the most general antiderivative of sec2 lex.

EXAMPLE 3 (8) f(x) = XS

Find the general antiderivative of each of the following functions. (b) g(x) =

I Vx

(0) h(x) = sin 2x

(d) i(x) = cos ~

232


TABLE 4.2 Antiderivative formulas, k a nonzero constant Function

General antiderivative

1.

x'

_1_ x ·+ 1 n+l

+C n '

2.

sinkx

1 --cos kx k

+C

3.

coskx

tsinkx

+C

4.

sec' kx

ttankx

+C

oF -I

Function


5.

csc2 kx

1 --cotkx k

6.

sec kxtan kx

k seckx + C

7.

esc kx cot kx

1 --esc kx k

1

6

F(x)

+C

In each case, we can use one of the formulas listed in Table 4.2.

Solution (a)

+C

=

6x +

Formula 1 withn=5

C

(b) g(x) =x-1/2 ,so Xl/2

G(x) = 1/2 (0)

H(x)

-cos 2x 2

=

(d) f(x) =

+C

sin (x/2) 1/2

•

=

2

V

r

x

Formula 1 with. ~ -1/2

+C

Formula 2

+C

+

withk~2

. x C = 2 sm 2

+

Formula 3 withk ~ 1/2

C

•

Other derivative rules also lead to corresponding antiderivative rules. We can add and subtract antiderivatives and multiply them by constants.

TABLE 4.3 Antiderivative linearity rules

1. 2. 3.

Constant Multiple Rule: Negative Rule: Sum or Difference Rule:

Function


kf(x)

kF(x)

-fix) fix) ± g(x)

+ C, ka constant -F(x) + C F(x) ± G(x) + C

The formulas in Table 4.3 are easily proved by differentiating the antiderivatives and verifying that the result agrees with the original function. Formula 2 is the special case k = - 1 in Formula 1.

EXAMPLE 4

Find the general antiderivative of fix) =

~ + sin2x.

Solution We have that fix) = 3g(x) + h(x) for the functions g and h in Example 3. Since G(x) = 2Vx is an antiderivative of g(x) from Example 3b, it follows from the

4.7

Antiderivatives

233

Constant Multiple Rule for antiderivatives that 3G(x) = 3· 2Vx = 6Vx is an antiderivativeof3g(x) = 3/Vx. Likewise, from Example 3c we know that H(x) = (-1/2) cos2x is an antiderivative of h(x) = sin 2x. From the Sum Rule for antiderivatives, we then get that

F(x)

=

3G{x)

+ H(x) + C

~cos2x + C

= 6Vx -

is the general antiderivative fannula for f(x), where C is an arbitrary constant.

•

Initial Value Problems and Differential Equations Antiderivatives play several important roles in mathematics and its applications. Methods and techniques for rmding them are a major part of calculus, and we take up that study in Chapter 8. Finding an antiderivative for a function f(x) is the same problem as finding a functiony(x) that satisfies the equation

dy

dx

=

f(x).

This is called a differential equation, since it is an equation involving an uuknown function y that is being differentiated To solve it, we need a function y(x) that satisfies the equation. This function is found by taking the antiderivative of f(x). We fix the arbitrary constant arising in the antidifferentiation process by specifying an initial condition

y(xo)

=

Yo.

This condition means the function y(x) has the value Yo when x = Xo. The combination of a differential equation and an initial condition is called an initial value problem. Such problems play important roles in all branches of science. The most general antiderivative F(x) + C (such as x 3 + C in Example 2) of the function f(x) gives the general solution y = F(x) + C of the differential equation dy/ dx = f(x). The general solution gives all the solutions of the equation (there are infinitely many, one fur each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by rmding the partieulsr solution that satisfies the initial condition y(xo) = Yo. In Example 2, the function y = x 3 - 2 is the particular solution of the differential equationdy/dx = 3x2 satisfYing the initial condition y(l) = -l.

Antiderivatives and Motion We have seen that the derivative of the position function of an object gives its velocity, and the derivative of its velocity function gives its acceleration. Ifwe know an object's acceleration, then by finding an antiderivative we can recover the velocity, and from an antiderivative of the velocity we can recover its position function. This procedure was used as an application of Corollary 2 in Section 4.2. Now that we have a terminology and conceptual framework in terms of antiderivatives, we revisit the problem from the point of view of differential equations.

EXAMPLE 5 A hot-air balloon ascending at the rate of 12 ft/sec is at a height 80 ft above the ground when a package is dropped How long does it take the package tu reach the ground? Solution Let v(l) denote the velocity of the package at time I, and let s(l) denote its height above the ground The acceleration of gravity near the surface of the earth is 32 ft/ sec'. Assuming no other forces act on the dropped package, we have

dv dt

= -32

.

Negative because gravity acts in the direction of decreasing s

234

Chapter 4: AppLications of Derivatives This leads to the following initial value problem (Figure 4.49):

0, I'(e

< 8

. I'(e + h) hm '---.....,.-h--->O h

=

+ h)

h

0, but no point beyond b. Determine its velocity at t = 0.

Vi - (1/Vi).

28. A particle moves with acceleration a = Assuming that the velocity v = 4/3 and the position s = -4/15 when t=O,find a. the velocity v in terms of t. b. the position s in terms of t.

Chapter 4 Technology Application Projects 29. Suppose f(x) ~ ax' + 2bx + c with a > O. By coosidering the minimum, prove that f(x) '" 0 for all real x if, and ooly if, b 2 - ac :s; O. 30. SchWlllZ'. inequolity

and deduce Schwarz's inequality: (al b l

+ a,b, + ... + a"b.)' '" (ai' + a,' + ... + a,,')(bl' + b,' + ... + b.').

b. Show that equality holds in Schwarz's inequality only ifthcre exists a real number x that makes a,x equal -hi for every value of i from I to n.

•• In Exercise 29, let

Chapter 1:3

245


Mathematica,tMap1e Modules: MOMIIAlolIg tJ Stmight Line: POSitioll .... Velocity .... Accelertllioll You will observe the shape of a graph through dramatic animated visualizatioos of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated Newtoll's Method: EstimtJte 'fr to How MtJlly PltJces? Plot a function, observe a root, pick a starting point near the root, and use Newton's Iteration Procedure to approximate the root to a desired accuracy. The numbers 'fr, e, and v2 are approsimated.

5 INTEGRATION OVERVIEW A great achievement of classical geometry was obtaining formulas for the areas and volumes of triangles, spheres, and cones. In this chapter we develop a method to calculate the areas and volumes of very general shapes. This method, called integration, is a tool for calculating much more than areas and volumes. The integral is of fundamental importance in statistics, the sciences, and engineering. We use it to calculate quantities ranging from probabilities and averages to energy consumption and the forces against a dam's floodgates. We study a variety of these applications in the next chapter, but in this chapter we focus on the integral concept and its use in computing areas of various regions with curved boundaries.

5.1

Area and Estimating with Finite Sums

y

~------~~-----7--~X

FIGURE 5.1

The area of the region R cannot be found by a simple formula.

246

The definite integral is the key tool in calculus for defining and calculating quantities important to mathematics and science, such as areas, volumes, lengths of curved paths, probabilities, and the weights of various objects, just to mention a few. The idea behind the integral is that we can effectively compute such quantities by breaking them into small pieces and then summing the contributions from each piece. We then consider what happens when more and more, smaller and smaller pieces are taken in the summation process. Finally, if the number of terms contributing to the sum approaches infinity and we take the limit of these sums in the way described in Section 5.3, the result is a definite integral. We prove in Section 5.4 that integrals are connected to antiderivatives, a connection that is one of the most important relationships in calculus. The basis for formulating definite integrals is the construction of appropriate finite sums. Although we need to define precisely what we mean by the area of a general region in the plane, or the average value of a function over a closed interval, we do have intuitive ideas of what these notions mean. So in this section we begin our approach to integration by approximating these quantities with finite sums. We also consider what happens when we take more and more terms in the summation process. In subsequent sections we look at taking the limit of these sums as the number of terms goes to infinity, which then leads to precise definitions of the quantities being approximated here.

Area Suppose we want to find the area of the shaded region R that lies above the x-axis, below the graph of y = 1 - x 2 , and between the vertical lines x = 0 and x = 1 (Figure 5.1). Unfortunately, there is no simple geometric formula for calculating the areas of general shapes having curved boundaries like the region R. How, then, can we find the area of R? While we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the region R. Each rectangle has width 1/2 and they have heights 1 and 3/4, moving from left to right. The height of each rectangle is the maximum value of the function j,

5.1

Area and Estimating with Finite Sums

y

247

y

y=1-x 2 1+,(;;; 0'", 1)::-------,

1

0.5

(0, 1)

0.5 R

~------~~----~--~X

o

o

0.5

0.25

0.5

(a)

0.75

(b)

FIGURE 5.2 (a) We get an upper estimate of the area of R by using two rectang1es containing R. (b) Four rectaogles give a better upper estimate. Both estimates oversboot the true value for the area by the amount shaded in light red.

obtained by evaluating f at the left endpoint of the subinterval of [0, I] forming the base of the rectangle. The total area of the two rectangles approximates the area A of the region R, A .., I·

2I + 43 . 2I = 87 =

0.875.

This estimate is larger than the true area A since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of each rectangle as the maximum (uppermost) value of f(x) for a point x in the base interval of the rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1/4, which taken together contain the region R. These four rectangles give the approximation A .., 1·

1

15

1

3

1

7

1

4 + 16 . 4 + 4 . 4 + 16 . 4 =

25 32 = 0.78125,

which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained imide the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1/4 as before, but the rectangles are y

y

1

1

0.5

0.5

~--~~--~--~~--~~X

o

0.25

0.5

(a)

0.75

(U)

~--~~-L~~~~~--~ x

o

0.25 0.5 0.75 1 0.125 0.375 0.625 0.875 (b)

FIGURE 5.3 (a) Rectang1es contained in R give an estimate for the area that uodershoots the true value by the amoout shaded in light blue. (b) The midpoint rule uses rectsngles whose height is the value of y = f(x) at the midpoints of their bases. The estimate appears closer to the true value of the area because the light red overshoot areas roughly balaoce the light blue undershoot areas.

248

Chapter 5: Integration shorter and lie entirely beneath the graph of j. The function f(x) = I - x 2 is decreasing on [0, 1], so the height of each of these rectangles is given by the value of f at the right endpoint of the subinterval fonning its base. The foorth rectangle has zero height and therefore contributes no area. Summing these rectangles with heights equal to the mini· mum value of f(x) for a point x in each base subinterva1 gives a lower sum approximation to the area, A ..,

y

15

1

3

1

1

1

17 32 = 0.53125.

This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums:

1

0.53125

o (a)

l

13 I)

3

t

250

Chapter 5: Integration With / evaluated at t = 0, I, and 2, we have D '" /(t1) !!.t

+ /(t2)

+ /(t3) !!.t + [160 - 9.8(1)](1) +

!!.t

= [160 - 9.8(0)](1)

[160 - 9.8(2)](1)

= 450.6.

(b) Three subintervals a/length I, with / evaluated at right endpoints giving a lower sum:

o

tl

t2

13

1

2

3

•

•

e) t

~. ' - 1

k- l

n

Riemann Sums In Exercises 33-36, graph each function I(x) over the given interval.

Partition the ioterval ioto four suhiotervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum ~~- d(c.) .1x.. given that c. is the (a) left-hand endpoiot, (b) righthand endpoint, (e) midpoint of the kth subinterval. (Make a separate sketch for each set of rectangles.)

33. I(x)

=

x2

35. I(x)

=

siox,

-

I,

[0,2]

34. I(x)

=

-x2 ,

[-'1f, '1f]

36. I(x)

=

siox

37. Fiod the norm of the partition P

=

38. Fiod the norm of the partition P

=

[0, I]

+ I,

[-'1f, '1f]

{O, 1.2, 1.5,2.3,2.6, 3}. {-2, -1.6, -0.5, 0, 0.8, I}.

k=l

" - I) d. ~(b.

1)

'-1

Evaluate the sums io Exercises 19-32. 10

+ b.)

Fl

k=l

"

k=l

" e. ~(b. - 2a.)

d. ~(a. - b.)

'-1

c. ~(a.

I>. ~-

k=l

e. ~(a.

•

• b.

" ~3a.

L

7 k' -~.t=1 4

j;=1

>=-1

16

=1

30. a. ~k

32. a.

~k

500

,.

+ 1)

CY

28. I>. ~7

Fl

Express the sums io Exercises 11-16 io sigma notation. The form ofyour answer will depend on your choice of the lower limit of summation.

k=l

C),

29. a. ~3

+ If

I>. ~ (k

L ~(k - 1)2

26. ~k(2k

+ 5)

5 k' ~ 225 + ~k

10. Which formula is not equivalent to the other two? 4

7

5

9. Which formula is not equivalent to the other two?

10

10

I>. ~k2

c. ~k'

'-1 13

>=-1 13

I>. ~k2

c. ~k'

... 1 5 '1fk 22. ~T5

'-1

Limits nf Riemann Sums For the functions in Exercises 39-46, fmd a formula for the Riemann sum obtained by dividiog the ioterval [a, b] ioto n equal subintervals and using the right-hand endpoint for each c•. Then tske a limit of these sums as n ----i' 00 to calculate the area under the curve over [a, b].

39. I(x) = I - x 2 over the interval [0, I]. 40. I(x)

=

2x over the ioterval[O, 3].

41. I(x) = x 2

42. I(x)

=

+

I over the interval [0, 3].

2

3x overtheioterval[O, I].

43. I(x) = x

+ x 2 0ver the interval [0, I].

44. I(x) = 3x + 2x 2 over the ioterval [0, I].

'-1

•

24. ~(k2 - 5)

'-1

45. I(x)

=

2x' over the ioterval [0, I] .

46. I(x) = x 2

-

x' over the interval [-1,0].

The Definite Integral In Section 5.2 we investigated the limit of a finite sum for a function defined over a closed interval [a, b] usiog n subiotervals of equal width (or length), (b - a)/n. In this section we consider the limit of more general Riemann sums as the norm of the partitions of [a, b] approaches zero. For general Riemann sums the subintervals of the partitions need not have equal widths. The limiting process then leads to the definition of the definite integrol of a function over a closed interval [a, b].

Definition of the Definite Integral The definition of the defioite integral is based on the idea that for certain functions, as the norm of the partitions of [a, b] approaches zero, the values of the corresponding Riemann

5.3 The Definite Integral

263

sums approach a limiting value J. What we mean by this limit is that a Riemann sum will be close to tbe number J provided that tbe nann of its partition is sufficiently small (so that all of its subintervals have tbin enough widtha). We introduce tbe symbol E as a small positive number that specifies how close to J tbe Riemann sum must be, and tbe symbolll as a second small positive number that specifies how small tbe nann of a partition must be in order for convergence to happen. We now defme this limit precisely.

DEFINmON Let f(x) be a function defmed on a closed interval [a, hI. We say that a number J is tbe defmite integral of f over [a, bl and tbat J is tbe limit oftbe Riemann sums ~~=t!(c.) Ax. iftbe following condition is satisfied: Given any number E > 0 tbere is a corresponding number Il > 0 such tbat for every partition P = {xo, x" ... , xn } of [a, hI witb Ilpll < Il and any choice of in [Xk-" xii, we have

c.

The defmition involves a limiting process in which tbe nann of tbe partition goes to zero. In tbe cases where tbe subintervals all have equal widtb ~x = (h - a)/n, we can fann

each Riemann sum as

•

• (h ; a ),

S. = ~ f(ck) ~Xk = ~ f(ck)

ax, = ax = (b - alln for all k

c.

where is chosen in tbe subinterval Ax•. If tbe limit of tbese Riemann sums as n ---> exists and is equal to J, tbenJ is tbe defmite integral of f over [a, hI, so J = lim

~f(ck)(h; a)

n --+ 00 (:1

=

lim

n--+ 00

~f(ck) Ax.

{:1

00

ax = (b - all.

Leibniz introduced a notation for tbe defmite integral that captures its construction as a limit of Riemann sums. He envisioned tbe fmite sums ~~=1 f(Ck) Ax. becoming an infinite sum of function values f(x) multiplied by "infmitesimal" subinterval widtbs dx. The sum symbol ~ is replaced in tbe limit by tbe integral symbol j, whose origin is in tbe letter "S." The function values f(ck) are replaced by a continuous selection of function values f(x). The subinterval widtbs Ax. become tbe differential dx. It is as if we are summing all products of tbe fann f(x) • dx as x goes from a to h. While tbis notation captures tbe process of constructing an integral, it is Riemann's defmition that gives a precise meaning to tbe defmite integral. The symbol for tbe number J in tbe defmition of tbe definite integral is

tf(X)dx,

which is read as ''the integral from a to h off of x dee x" or sometimes as ''the integral from a to h off of x witb respect to x." The component parts in tbe integral symbol also have names: The function is the integrand. Upper limit of integration/

-""f;(x) ,h~r-"-... . ~ Lower limit of mtegration

When you :find the value of the integral, you have Integral off from a to b ~ evaluated the integral. ,

y

264

Chapter 5: Integration When the condition in the dermition is satisfied, we say the Riemann swns of I on [a, b] converge to the definite integral J = I{x} ax and that I is integrable over [a, b]. We have many choices for a partition P with norm going to zero, and many choices of points CO for each partition. The definite integral exists when we always get the same limit J, no matter what choices are made. When the limit exists we write it as the definite integral

t

lim f/{Ck}

11111~O

.... ,

~k =

J =

l

a

b, {X} ax.

When each partition has n equal subintervals, each of width ~ = {b - a}Jn, we will also write n

lim ~/(ck} ~ = J =

11_ 00

1=1

lb a

I{x} ax.

The limit of any Riemann swn is always taken as the norm of the partitions approaches zero and the nwnber of subintervals goes to inImity. The value of the dermite integral of a function over any particular interval depends on the function, not on the letter we choose to represent its independent variable. If we decide to use t or u instead of x, we simply write the integral as instead of

or

[/{X}ax.

No matter how we write the integral, it is still the same nwnber that is dermed as a limit of Riemann swns. Since it does not matter what letter we use, the variable of integration is called a dummy variable.

Integrable and Nonintegrable Functions Not every function dermed over the closed interval [a, b] is integrable there, even if the function is bounded. That is, the Riemann swns for some functions may not converge to the same limiting value, or to any value at all. A full development of exactly which functions defined over [a, b] are integrable requires advanced mathematical analysis, but fortonately most functions that commonly occur in applications are integrable. In particular, every continuous function over [a, b] is integrable over this interval, and so is every function having no more than a finite nwnber of iwnp discontinuities on [a, b]. (The latter are called piecewise-continuous /Unctions, and they are defined in Additional Exercises 11-18 at the end of this chapter.) The following theorem, which is proved in more advanced courses, establishes these results.

THEOREM l-Integralrility of Continuous Functions If a function I is continuous over the interval [a, b], or if I has at most finitely many iwnp discontinuities there, then the dermite integral I{x} ax exists and I is integrable over [a, b].

t

The idea behind Theorem 1 for continuous functions is given in Exercises 86 and 87. Briefly, when I is continuous we can choose each Ck so that I{ Ck} gives the maximwn value of I on the subinterval [Xk-), Xk], resulting in an upper swn. Likewise, we can choose CO to give the minimwn value of Ion [Xk-), xii to obtain a lower swn. The upper and lower swns can be shown to converge to the same limiting value as the norm of the partition P tends to zero. Moreover, every Riemann swn is trapped between the values of the upper and lower swns, so every Riemann swn converges to the same limit as well. Therefore, the nwnber J in the definition of the definite integral exists, and the continuous function I is integmble over [a, b]. For integrability to fail, a function needs to be sufficiently discontinuous that the region between its graph and the x-axis cannot be approximated well by increasingly thin rectangles. The next example shows a function that is not integrable over a closed interval.


EXAMPLE 1

265

The function

{I,0,

f(x) =

if x is rational if x is irrational

has no Riemann iotegral over [0, I]. Underlying this is the fact that between any two numbers there is both a rational number and an irrational number. Thus the function jwnps up and down too erratically over [0, I] to allow the region beneath its graph and above the x-axis to be approximated by rectangles, no matter how thio they are. We show, io fact, that upper sum approximations and lower sum approximations converge to different limitiog values. If we pick a partition P of [0, I] and choose to be the poiot giving the maximum value for f on [x._ 10 x.tJ then the correspondiog Riemann sum is

c.

•

•

U = ~f(C.) Ax. = ~(1) ~x. = I,

sioee each subioterval [Xk-- 10 x.] contaios a rational number where f(c.) = 1. Note that the lengtha of the iotervals io the partition sum to I, ~f~l~X' = 1. So each such Riemann sum equals I, and a limit of Riemann sums using these choices equals 1. On the other hand, if we pick to be the poiot giving the minimum value for f on [x._1o x.tJ, then the Riemann sum is

c.

•

L = ~ f(c.)

Ax.

•

~(O) Ax.

=

=

0,

sioee each subioterval [x.-1ox.tJ contaios an irrational number c. where f(C.) = 0. The limit of Riemann sums usiog these choices equals zero. Sioce the limit depends on the choices of c., the function f is not iotegrable. _ Theorem I says nothiog about how to calculate defioite integrals. A method of calculation will be developed io Section 5.4, through a connection to the proeess of taking antiderivatives.

Properties of Definite Integrals In defming tf(x) dx as a limit of sums ~f~!/(c.) Ax.. we moved from left to right across the ioterval [a, b]. What would happen if we iostead move right to left, starting with Xo = b and endiog at x. = a? Each Ax. io the Riemann sum would change its sign, with x. - Xk-I now negative iostead of positive. With the same choices of io each subioterval, the sign of any Riemann sum would change, as would the sign of the limit, the iotegral f(x) dx. Sioce we have not previously given a meaniog to iotegrating backward, we are led to defme

c.

[f(X)dx

= -

lb'

tf(X)dx.

Although we have ouly defmed the integral over an ioterval [a, b] when a < b, it is convenient to have adefmition for the iotegral over [a, b] when a = b, that is, for the iotegral over an ioterval of zero width. Sioce a = b gives ~x = 0, whenever f(a) exists we defioe

[f(X)dx

=

0.

Theorem 2 states basic properties of iotegrais, given as rules that they satisfy, iocludiog the two just discussed. These rules become very useful io the process of computing iotegrais. We will refer to them repeatedly to simplify our calculations. Rules 2 through 7 have geometric ioterpretations, shown io Figure 5.11. The graphs io these figures are of positive functions, but the rules apply to general iotegrable functions.

THEOREM 2 Whenf andg are iotegrable over the ioterval [a, b], the definite iotegral satisfies the rules io Table 5.4.

266

Chapter 5: Integration

TABLE 5.4

RuLes satisfied by definite integrals bf - l (x) ax

1.

Ordero/Integration: l"f(x) ax

2.

Zero Width Interval:

l

3.

Constant Multiple:

l\f(x)ax

4.

Sum and Difference:

l bU(x) ± g(x}} ax

5.

Additivity:

lbf(x) ax + l'f(x) ax

6.

Max-Min Inequality:

If f has maximum value max f and minimum value minf on [a, b], then

=

A Def"mition

a

A Def'mition when!(a) exists

f(x) ax - 0

"-

=

bf kl (x)ax =

Anycons_k

bf (x) ax ± l

l

b g(x) ax

l'f(x) ax

=

bf minf·(b - a) " ' l (x)ax '" maxf·(b - a). 7.

y

Domination:

f(x)

20

g(x) on [a, b] => l

f(x)

20

Oon[a,b] => l

y

-O~~ a--------~ b~x

(a) Zero Width Interval:

(b) Constant Multiple: (k

a

[f(X)dx

~0

l ' kf(x) dx

y

bf (x)ax

20

l'

~k

~

2)

f(x) dx

o

l

0

(Special Case)

a

y

~ !(x)

y

~

y

~ !(x)

x

b

l'(f(x)

+ g(x))dx

~ l'f(x)dx + l'g(x)dx

y

min! y

[f(X)dx

+ [f(X)dx

~ [f(X)dx

~~----------~~x

o

a

b

(e) Max-Min Inequality:

mmf'(b - a)

"'1'

f(x)dx

'" maxf'(b - a) FIGURE 5.11

Geometric interpretations of Rules 2-7 in Table 5.4.

+ g(x)

g(x)

max!

(d) Additivity for definite integrals:

(x) ax

(c) Sum: (areas add)

y

-O~--~a----~b~-------c~~x

bg

20

y

~------~----~x

o

bf (x) ax

~

g(x)

~~--------~~x

oa

b

(f) Domination:

f(x)

=>

20

l'

g(x) on [a, b]

f(x) dx

20

l'

g(x) dx


267

While Rules I and 2 are def'mitions, Rules 3 to 7 of Table 5.4 must be proved. The following is a proof of Rule 6. Similar proofs can be given to verify the other properties in Table 5.4.

Proof of Rule 6 Rule 6 says that the integral of I over [a, b] is never smaller than the minimum value of I times the length of the interval and never larger than the maximum value of I times the length of the interval. The reason is that for every partition of [a, b] and for every cboice of the points Ck, min/·(b - a) = mini'

• L dXk k=1 Constant Multiple Rule

min! S !(c,)

Constant Multiple Rule

maxi' (b - a).

=

In short, all Riemann sums for I on [a, b] satisfy the inequality

•

min I' (b - a) ,,; ~/(ck) dxk"; max I' (b - a).

•

Hence their limit, the integral, does too.

EXAMPLE 2

To illustrate some of the rules, we suppose that

[1(X)dx

j1

4/

= 5,

(X) dx

-2,

=

th(x) dx

and

J-I

= 7.

Then

1. [I(X) dx = 2.

3.

[[2 /

(X)

f'1(x) dx

1-1

EXAMPLE 3

j4/

(x) dx

+ 3h(x)] dx

= -( -2) =

=

2[I(X) dx

=

2(5)

+ 3[h(X) dx

+ 3(7)

j41/

Rulel

=

Rules 3 and 4

31

=

5

+ (-2)

Show that the value of fa!.'/!

+

cosxdx is less than or equal to

=

+

2

tl(x) dx

1-1

(x) dx

=

3

•

Rule 5

V2.

The Max-Min Inequality for def'mite integrals (Rule 6) says that min I' (b - a) is a lower bound for the value of I(x) dx and that max I' (b - a) is an upper bound.

Solution

The maximum value of

VI

f:

+

11VI

cosx on [0, I] is

+

cosxdx";

v'1+l = V2, so

V2'(1 -

0) =

V2.

•

268


Area Under the Graph of a Nonnegative Function We now return to the problem that started this chapter, that of defining what we mean by the area of a region having a curved boundary. In Section 5.1 we approximated the area under the graph of a nonnegative continuous function using several types of finite sums of areas of rectangles capturing the region-upper sums, lower sums, and sums using the midpoints of each subinterva1-all being cases of Riemann sums constructed in special ways. Theorem I guarantees that all of these Riemann sums converge to a single def'mite integral as the norm of the partitions approaches zero and the number of subintervals goes to inf'mity. As a result, we can now define the area under the graph of a nonnegative integrable function to be the value of that definite integral.

DEFINITION If Y = f{x) is nonnegative and integrable over a closed f{x> over [II, b] is the interval [a, b], then the area under the curve y integml of f from a to b,

=

For the first time we have a rigorous def'mition for the area of a region whose boundary is the graph of any continuous function. We now apply this to a simple example, the area under a straight line, where we can verifY that our new definition agrees with our previous notion of area.

EXAMPLE 4

Compute

J: ax x

and find the area A under

y=

x over the interva1

[O,b],b>O. Solution

~~--------~~-+x

o

b

FIGURE 5.12 The region in Example 4 is a triangle.

The region of interest is a triangle (Figure 5.12). We compute the area in two ways.

To compute the def'mite integral as the limit of Riemann sums, we calculate lim lPl~o ~~,f{ck) dxk for partitions whose norms go to zero. Theorem 1 tells us that it does not matter how we choose the partitions or the points Ck as long as the norms approach zero. All choices give the exact same limit So we consider the partition P that subdivides the interval [0, b] into n subinterva1s of equal width dx = {b - O)ln = bin, and we choose Ck to be the right endpoint in each subinterval. The partition is p = { 0,

t 2:, 3:,,,. ,n: }andck

= ,,:. So

Constant Multiple Rule

=

b2 n2

n{n + 1) •

2

Sum of First n Integers

5.3 y

269

All n --> 00 and Ilpll --> 0, this last expression on the right has the limit b'/2. Therefore,

r

b

Jo xdx b a ~~--L-----+-~x

o

The Definite Integral

a I

h(x) dx

~ 4.

-2f(x) dx

b.

1,'

[f(x)

+ h(x)] dx

c. 1,'[2f(x) - 3h(x)] dx

d. J.' f (X) dx

e. l'f(x) dx

f. J.'[h(X) - f(x)] dx

11. Suppose that f.,'f(x) dx ~ 5. Find

l'

Y3f(z)

dz

a. l'f(U) du

b.

c. J,'f (l)dl

d.l'[-f(X)]dx

t, g(l) dl ~ v'2. Find

l:

a. 1,-'g(l) dl

b.

c·l°[

d.l° ~~

-3

-g(x)] dx

-3

13. Suppose thatf is integrable and that

104 f(z) dz ~ 7. Find

a.

1,4f (z) dz

g (u) du

v2

dr

fa' f(z) dz ~ 3 and

b. /.' f(l) dl

14. Suppose that h is integrable and that f~, h(r) dr ~ 0 and

l, h(r) dr ~ 6. Find

a.l'h(r)dr

+

x'dx

_1,' h(U)dU

0

28. f(x)

on •• [-1,0], b. [-1,1]

Evaluattng Definite Integrals Use the results of Equations (I) and (3) to evaluate the integrals in Exercises 29-40.

32.

i,

xdx

30.

sv'2 rdr

33.

J{'/2 v2

35. io

I'dl

l

v'3a xdx

1','

xdx

J.

34.

io{"-'8' ds

w

05

{V'7

io x'dx 36. 1,w/2IP dO 39.

'W

31.

OdO

37·1"xdx

{~ x'dx

io

io(" x'dx

40.

Use the rules in Table 5.4 and Equations (1)-(3) to evaluate the integrals in Exercises 41-50.

41. 1,'7 dx

42.1,'sxdx

43.1,'(21 - 3) dl

44.

45.[(I+I)dz

46. 1,0(2z - 3)

47·1'3U'dU

48. (' 24u' du

49.

1,\3x' + x -

{v2

5) dx

io

(I -

v'2) dl

dz

i'/2 50.1°(3x' + x -

5) dx

Finding Allla by Definite Integrals 10 Exercises 51-54, use a definite integral to rmd the area of the region between the giveo curve and the x-axis on the interval [0, b].

51. Y ~ b.

>

on a. [-2,2], b. [0,2]

38..

12. Suppose that

J~v'16 -

26.1'3Id', O 00 and ax ~ (b - a)/n --> O. 86. Suppose that f is continuous and nonnegative over [a, b], as in the accompanying figure. By inserting points XloX2,··· ,.:t,t-loX} •••• , Xn-l

as shown, divide [a, b] into n subintervals oflengths ax, ~ x, - a, dx2 = Xl - Xl, • •• , dA" = b - XII_I, which need not be equal. •• !fm, ~ min {f(x) for x in the kth subinterval}, explain the connection between the lower sum L = ml

axl + m2 dx2 + ... + mil 4x1l

and the shaded regions in the first part of the figure. b. !f M, ~ max {f(x) for x in the kth subinterval}. explain the

connection between the upper sum U~

M,ax, + M2ax2 + ... + M.ax.

and the shaded regions in the second part of the figure. c. Explain the connection between U - L and the shaded regions along the curve in the third part of the figure.

0

II

I,

~ b-a

l-j

,I

87. We say f is uniformly continuous on [a, b] if given any E > 0, thereisa8 > osuch thatifx"x2 are in [a, b] and lx, - x21 < 8, then If(x,) - f(X2) I < E. It can be shown that a continuous function on [a, b] is uniformly continuous. Use this and the figure for Exercise 86 to show that iff is continuous and E > 0 is given, it is possible to make U - L '" E' (b - a) by making the largest of the ax,'s sufficiently small. 88. If you average 30 mijh on a 150·mi trip and then return over the same 150 mi at the rate of 50 mi/h, what is your average speed for the trip? Give reasons for your answer. COMPUTER EXPLORATIONS

!fyour CAS can draw rectaugles associated with RieD2anD sums, use it to drsw rectangles associated with Riemann sums that converge to the integrals in Exercises 8g..-94. Use n ~ 4, 10,20, and 50 subintervals of equal length in each case. 89.

[(I -

x)dx

~k

274 90.


1'<x'

+ I) tb;

91.

J~ cos x tb; =

92.

Jo

93.

[ixl

=

b. Partition 1he interval into n = 100, 200, and 1000 subinter-

1

vals of equallengtb, and evaluate 1he function at 1he ntidpoint of each subinterval.

0

c. Compute the average value of the function values generated in part (b).

[~/4

d. Solve 1he equationf(x)

= (average value) for x using 1he average value calculated in part (c) for 1he n = 1000 partitioning. 95. f(x) = sinx on [0, 'IT] 96. f(x) = sin' x on [0, 'IT]

sec' x tb; = I tb; =

94.1' ~tb;

I

(The integra1'svalue is about 0.693.)

10 Exercises 95-98, use a CAS to perform 1he following steps:

Plot 1he functions over 1he given interval.

L

97. f(x) = x sin

~

98.f(x)=xsin'~

[~,,,.]

on

on

[~,'IT]

5.4 I The Fundamental Theorem of Calculus In this section we present the Fundamental Theorem of Calculus, which is the central theorem of iutegral calculus. It connects iutegration and differentiation, enabliug us to compute iutegrals usiug an antiderivative of the iutegrand function rather 1han by taking limits of Riemann sums as we did iu Section 5.3. Leibniz and Newton exploited this relationship and started ma1hematical developments that fueled the scientific revolution for the next 200 years. Along the way, we present an integral version of the Mean Value Theorem, which is another important theorem of iutegral calculus and is used to prove the Fundamental Theorem.

HISTORICAL BIOGRAPHY

Sir Isaac Newton (1642-1727) y

f(e), average height ~~------L---~~x

o

a

c ~b- a

FIGURE 5.16

b

------->I

The value f(e)

in 1he

Mean Value Theorem. is, in a sense, the

average (or mean) height off on [a, b]. wtwn f '" 0, 1he area of1he rectaogle is 1he area under 1he graph of f from a

Mean Value Theorem for Definite Integrals In the previous section we defiued the average value of a continuous function over a closed iuterva1 [a, hI as the def'mite iutegral I{x) dx divided by the length or width h - a of the iuterva1. The Mean Value Theorem for Definite Integrals asserts that this average value is always taken on at least once by the function 1 iu the iuterva1. The graph iu Figure 5.16 shows a positive continuous function y = I{x) defiued over the iuterval [a, hI. Geometrically, the Mean Value Theorem says that there is a number c iu [a, hI such that the rectangle with height equal to the average value I{c) of the function and base width h - a has exactly the same area as the region beneath the graph of1 from a to h.

1:

to h,

f(e)(b - a) =

l'

f(x) tb;.

THEOREM 3-The Mean Value Theorem for Definite Integrals ous on [a, hI, then at some poiut c iu [a,

hI,

If 1 is continu-

1 b

I{c) = h _I a a I{x) dx.

Proof Ifwe divide both sides of the Max-Miu Inequality (Table 5.4, Rule 6) by (h - a), we obtain

5.4 The Fundamental Theorem of Calculus y

, 1 2

Average value 112

----------- notassumed

-nt-----t-----~-------+x

o

2

FIGURE 5.17

A discontinuous function

275

Since I is continuous, the Intermediate Value Theorem for Continuous Functions (Section 2.5) says that I must assume every value between min I and max f. It must therefore assume the value (1/(b - a»I:/(x) dx at some pointe in [a, b]. • The continuity of I is important here. It is possible that a discontinuous function never equals its average value (Figure 5.17).

EXAMPLE 1

Sbow thatifl is continuous on [a, b],a '" b,andif

1

need not assume its average value.

b/(x) dx

=

0,

then I(x) = 0 at least once in [a, b]. Solution

The average value of I on [a, b] is

1 b

av(f) = b _I a

a

I(x) dx = b _I a . 0 = O.

By the Mean Value Theorem, I assumes this value at some point e

E

[a, b].

•

Fundamental Theorem, Part 1 area

y

~

If I(t) is an integrable function over a fiuite interval I, then the integral from any rlXed number a E I to another number x E I defines a new function F whose value at x is

F(x)

F(x) = [I(t) dt.

o

a

x

b

FIGURE 5.18 The funetionF(x) dermed by Equation (I) gives the area under the graph of f from a to x when

f is nonnegative and x > a.

(1)

For example, if I is nonnegative and x lies to the right of a, then F(x) is the area under the graph from a tox (Figure 5.18). The variable x is the upper limit of integration of an integra!, but F is just like any other real-valued function of a real variable. For each value of the input x, there is a well-dermed numerical output, in this case the dermite integra! of I from a to x. Equation (1) gives a way to derme new functions (as we will see in Section 7.2), but its importance now is the connection it makes between integrals and derivatives. If I is any continuous function, then the Fundamental Theorem asserts that F is a differentiable function of x whose derivative is I itself. At every value of x, it asserts that d dx F(x) = I(x).

y

To gain some insight into why this result holds, we look at the geometry behind it. If I ;;,: 0 on [a, b], then the computation of F'(x) from the dermition of the derivative means taking the limit as h --> 0 of the difference quotient F(x b

FIGURE 5.19

In Equation (I), F(x) is the area to the left of x. Also, F(x + h) is the area to the left of

+ h. The difference quotient [F(x + h) - F(x)l/h is then

x

approximately equal to f(x), the height of the rectangle shown here.

+ h) - F(x) h

For h > 0, the numerator is obtaioed by subtracting two areas, so it is the area under the graph of I from x to x + h (Figure 5.19). If h is small, this area is approximately equal to the area of the rectaogle ofheight/(x) and width h, which can be seen from Figure 5.19. That is, F(x

+ h) - F(x) '" h/(x).

Dividing both sides of this approximation by h and letting h --> 0, it is reasonable to expect that F'(x) = lim F(x h~O

+ h) - F(x)

=

I(x).

h

This result is true even if the function I is not positive, and it forms the first part of the Fundamental Theorem of Calculus.

276


THEOREM 4-The Fundamental Theorem of Calculus, Part 1 If I is continuous on [a, b], thenF(x) = I(t) dt is continuous on [a, b] and differentiable on (a, b) and its derivative is I(x):

J:

X

(2)

F'(x) = ax d ia I(t) dt = I(x).

Before proving Theorem 4, we look at several examples to gain a better understanding of what it says. In each example, notice that the independent variable appears in a limit of integration, possibly in a formula.

EXAMPLE 2 (a) y =

Solution

i

Use the Fundamental Theorem to tmd dy/ax if

X

(t 3

+ 1) dt

(h) y = lS3tSintdt

(e) y = ] ' " costdt

We calculate the derivatives with respect to the independent variable x.

(a) -dy = -d ax ax

i a

X

(t 3

+

I)dt = x 3

+

Eq.(2)withf(') ~,'

1

dy = ax diS d(f' ) (h) ax x 3tsintdt = ax - Js 3tsintdt

+

I

Table 5.4, Rule I

rx

= -

d axJs 3tsintdt

= -

3x sin X

Eq. (2) with f(') ~ 3tsin'

(e) The upper limit of integration is not x but functions,

x 2• This

makes y a composite of the two

and

y = ] " costdt

We must therefore apply the Chain Rule when finding dy/ ax.

dy dy du ax=du'ax =

(~]" costdt). du 1

= cosu .

du ax

du ax

= cos(x 2 ). 2x =

2xcosx 2

•

Proof of Theorem 4 We prove the Fundamental Theorem, Part I, by applying the definition of the derivative directly to the function F(x), when x and x + h are in (a, b). This means writing out the difference quotient F(x + h) - F(x) h

(3)

5.4 The Fundamental Theorem of Calculus

277

and showing that its limit as h --+ 0 is the number I(x) for each x in (a, b). Thus,

F'(x)

Fi,,-(x_+~h)'----_Fi-"(x-'-) _

= lim h~O =

h

l~d [[+O/(t) dt l

= lim -h h--+O

l

[I(t) dt]

X

+hI(t)

x

dt

Table 5.4, Rule 5

According to the Mean Value Theorem for Definite Integrals, the value before taking the limit in the last expression is one of the values taken on by 1 in the interval between x and x + h. That is, for some number c in this interval,

11'H

Ii,

I(t) dt

=

I(c).

(4)

All h --+ 0, x + h approaches x, forcing c to approach x also (because c is trapped between x and x + h). Since 1 is continuous atx, I(c) approaches I(x): lim I(c) = I(x).

(5)

h-O

In conclusion, we have

F'(x)

= lim -hi h-O

= lim

h~O

=

l

x

X +h

I(t) dt

I(c)

I(x).

Eq. (4) Eq. (5)

If x = a or b, then the limit of Equation (3) is interpreted as a one-sided limit with h --+ 0+ or h --+ 0-, respectively. Then Theorem I in Section 3.2 shows that F is continuous for every point in [a, b]. This concludes the proof. •

Fundamental Theorem, Part 2 (The Evaluation Theorem) We now come to the second part of the Fundamental Theorem of Calculus. This part describes how to evaluate definite integrals without having to calculate limits of Riemann sums. Instead we find and evaluate an antiderivative at the upper and lower limits of integration.

If1 is continuous at every point in [a, b] and F is any antiderivative of 1 on [a, b], then

THEOREM 4 (Continued)-The Fundamental Theorem of Calcuius, Part 2

l b,

(x) dx

=

F(b) - F(a).

Proof Part I of the Fundamental Theorem tells us that an antiderivative of 1 exists, namely

G(x)

=

[I(t) dt.

Thus, if F is allY antiderivative of I, then F(x) = G(x) + C for some constant C for a < x < b (by Corollary 2 of the Mean Value Theorem for Derivatives, Section 4.2). Since both F and G are continuous on [a, b], we see that F(x) = G(x) + C also holds when x = a and x = b by taking one-sided limits (as x --+ a + and x --+ r).

278

Chapter 5: Integration Evaluating F(b) - F(a), we have

F(b) - F(a) = [G(b)

+ C] -

[G(a)

+ C]

=

G(b) - G(a)

=

J.b/(t) dt - J."/(t) dt

=

J.b/(t) dt - 0

=

J.b/(t) dt.

•

The Evaluation Theorem is important because it says that to calculate the definite integral of / over an interval [a, b] we need do only two things: 1.

Find an antiderivative F of /, and

2.

Calculate the number F(b) - F(a), which is equal to

1: /(x) dx.

This process is much easier than using a Rieroann sum computation. The power of the theorem follows from the realization that the definite integral, which is defmed by a complicated process involving all of the values of the function / over [a, b], can be found by knowing the values of any antiderivative F at only the two endpoints a and b. The usual notation for the difference F(b) - F(a) is

F(x)

1:

or

depending on whether F has one or more terms.

EXAMPLE 3 We calculate several definite integrals using the Evaluation Theorem, rather than by taking limits of Riemann sums. (a)

l

u

cosx dx = sin x

I

d

.

dx smx

=

cosx

d dxsecx

=

sec x tan x

= sin". - sinO = 0 - 0 = 0

(b)

r

J-w/4

sec x tan x dx = secx]O -w/4 = secO- sec ( - ; ) = 1-

(e)

l' (~Vi - x~)

dx =

[x

3/2

+

[(4)3/2 =

[8

Vi

iI + ~]- [(1)3/2 +

+ 1] - [5]

=

t]

4.

•

Exercise 66 offers another proof of the Evaluation Theorem, bringing together the ideas of Riemann sums, the Mean Value Theorem, and the defmition of the definite integral.

The Integral of a Rate We can interpret Part 2 of the Fundaments! Theorem in another way. If F is any antiderivative of /, then F' = f. The equation in the theorem can then be rewritten as

J.bF'(X) dx = F(b) - F(a).

5.4 The Fundamental Theorem of Calculus

279

Now F'(x) represents the rate of change of the function F(x) with respect to x, so the integral of F' is just the net change in F as x changes from a to b. Fonnally, we have the following result.

THEOREM 5-The Net Change Theorem

The net change in a function F(x) over an interval a :5 x :5 b is the integral of its rate of change:

F(b) - F(a)

EXAMPLE 4

=

l

b

F'(x) dx.

(6)

Here are several interpretations of the Net Change Theorem.

(a) If c(x) is the cost of producing x units of a certain commodity, then c'(x) is the marginal cost (Section 3.4). From Theorem 5, x

l..

,C'(X) dx

=

C(X2) - C(XI),

which is the cost of increasing production from XI units to X2 units.

(b) If an object with position function .(t) moves along a coordinate line, its velocity is v(t) = .'(t). Theorem 5 says that

l"

"v(t) dt = .(t2) - .(tl),

so the integral of velocity is the displacement over the time interval tl :5 t :5 t2' On the other hand, the integral of the speed Iv(t) I is the total distaoce traveled over the • time interval. This is consistent with our discussion in Section 5.1.

If we rearrange Equation (6) as

F(b)

=

F(a) + lbF'(X) dx,

we see that the Net Change Theorem also says that the f'mal value ofa functionF(x) over an interval [a, b] equals its initial value F(a) plus its net change over the interval. So ifv(t) represents the velocity function of an object moving along a coordinate line, this means that the object's final position .(t2) over a time interval tl :5 t :5 t2 is its initial position S(tl) plus its net change in position along the line (see Example 4b).

EXAMPLE 5 Consider again our analysis of a heavy rock blown straight up from the ground by a dynamite blast (Example 3, Section 5.1). The velocity of the rock at any time t during its motion was given as v(t) = 160 - 32t ft./sec. (a) Find the displacement of the rock during the time period 0 :5 t :5 8.

(b) Find the total distance traveled during this time period. Solution (a) From Example 4b, the displacement is the integral

10rv(t) dt = 10r(160 =

32t) dt = [160t - 16t 2]8 0

(160)(8) - (16)(64) = 256.

This means that the height of the rock is 256 ft. above the ground 8 sec after the explosion, which agrees with our conclusion in Example 3, Section 5.1.

280


(b) As we noted in Table 5.3, the velocity function v(t) is positive over the time interval [0,5] and negative over the interval [5, 8]. Therefore, from Example 4b, the total distance traveled is the integral

[lv(t)ldt

=

[lv(t)ldt + 18Iv(t)ldt

1 8

= 1'(160 =

32t) dt -

(160 - 32t) dt

[160t - 16t2]~ - [160t - 16t 2]:

= [(160)(5) - (16)(25)] - [(160)(8) - (16)(64) - «160)(5) - (16)(25))] = 400 - (-144) = 544.

Again, this calculation agrees with our conclusion in Example 3, Section 5.1. That is, the total distance of 544 ft traveled by the rock during the time period 0 :5 t :5 8 is (i) the maximum height of 400 ft it reached over the time interval [0, 5] plus (ii) the additional distance of 144 ft the rock fell over the time interval [5, 8]. •

The Relationship between Integration and Differentiation The conclusions of the Fundamental Theorem tell us several things. Equation (2) can be

rewritten as

di

dx

a

X

j(t) dt

=

j(x),

which says that if you frrst integrate the function j and then differentiate the result, you get the function j back again. Likewise, replacing b by x and x by t in Equation (6) gives

i

X

F'(t) dt

=

F(x) - F(a),

so that if you first differentiate the function F and then integrate the result, you get the function F back (adjusted by an integration constant). In a sense, the processes of integration and differentiation are "inverses" of each other. The Fundamental Theorem also says that every continuous function j has an antiderivative F. It shows the importance of finding antiderivatives in order to evaluate definite integrals easily. Furthermore, it says that the differential equation dy/dx = j(x) has a solution (namely, any of the functions y = F(x) + C) for every continuous function j.

Total Area The Riemann sum contains terms such as j( Ck) ,lxk that give the area of a rectangle when j(ck) is positive. When j(ck) is negative, then the product j( Ck) ,lxk is the negative of the rectangle's area. When we add up such terms for a negative function we get the negative of the area between the curve and the x-axis. If we then take the absolute value, we obtain the correct positive area.

EXAMPLE 6 g(x) = 4 -

Figure 5.20 shows the graph of j(x) = x 2 - 4 and its mirror image across the x-axis. For each function, compute

x 2 reflected

J~ds 58 + 4

J J J J

3yV7 - 3y2
cos (3z

+ 4) dz

tan2x sec' x dx 7X

tan

2 se.,-_2X2 dx

jr4 (7 - ~~)' dr

x 1/2 sin (x 3/2 + I) dx

esc - 2 - cot - 2 - dv

sin(2t+ I) dt eos2 (2t I)

38.

J J~ J J~x

40.

Jl~x2 3 2

32. 34. 36.

t 3(1 + t 4 )3 dt

~

dx

42.

secztanz dz

~

cos(V'i

+ 3) dt

cos Yo dO Yosin2 Yo ;, I dx

x

J~x

I dx

x

X 3 '

- I

dx

291

5.6 Substitution and Area Between Curves

43. 45. 47.

49.

! ! ! x'Vx'+! ! ~

x(x - 1)'0 tb; (x

44.

+ 1)'(1 - x)' tb;

(x 2

46. 48.

tb;

4)' tb;

SO.

!X~tb;

! ! 3x'W+! ! ~

(x + 5)(x - 5)'/3 tb;

(x

tb;

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplifY the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions in Exercises 51 and 52. 2 2 18 tan x sec x tb; 51. (2 + tan' x)2

!

b. u = tan'x,followedbyv = 2

~; =

59.

~;~ =

2 3cos

(~- 0).

-4 sin (21 -

!

VI

+v

63. It looks as if we can integrate 2 sin x cosx with respect to x in three different ways:

a. !2SinXCOSXdx = !2UdU

+ sin2 (x - I) sin (x - I) cos (x - I) tb;

b.

!

+u

2 sin x cos x ax

sin Yo dO VOcos'v1i

57.

=

~

4x (x 2

+

= 8sin2 (I

5.6

8r'/', +

~).

=

sinx

u

=

cosx

C2

= -cos2 x +

C2

2sinxC08X=sin2x

+ C,.

Cao all three integrations be correct'/ Give reasons for your answer. 64. (Continuation ofExample 9.) •• Show by evaluating the integral in the expression

/,'/60 V_sin 120'ITldl

(1/60) - 0 0

Solve the initial value problems in Exercises 55--60.

56. :

-2u du

= _ co~2x

1

~

!

= -u 2 +

Initial Value Problems 55.

=

c. !2SinXCOSXdx=! sin2xdx

!

u

= u2 + Cl = 8in2X + Cl

2

Evaluate the integrals in Exercises 53 and 54. (2r - I) cos V3(2r - I)' + 6 53. dr V3(2r - I)' + 6

= 121 (31 2 - I)',

.(0) = 0

62. The acceleration of a particle moving back and forth on a line is a = d 2./dI 2 = ~ cos 'lT1 m/sec' for all I. If • = 0 and v = 8 mlsec when I = 0, fmd. when I = I sec.

a. u = x - I, followed by v = sinu,thenbyw = I + if

54.!

.'(0) = 100,

61. The velocity of a particle moving back and forth on a line is v = tis/dl = 6 sin 21 m/sec for all I. If. = 0 when 1= 0, fmd the value of. when I = 'IT/2 sec.

+u

b. u = sin (x - I), followed by v = I C. U = I + sin2 (x - I)

f),

f

d 2y 60. tb;2 = 4 sec2 2xtan 2x, y'(O) = 4, y(O) = -I

c.u=2+tan3 x

52.

r(O) =

Theory and Examples

4)' tb;

a. u = tanx, followed byv = u', then byw = 2

58.

.(i) = 3 y(O)

=

that the average value of V = V_sin 120 'lT1 over a full cycle is zero. b. The circuit that runs your eleel1'ic stove is rated 240 volts rms. What is the peak value of the allowable voltage?

0

c. Showthat

.(0) = 8

/,o

'/60

(V, )' (V.,..),sin2 120'ITldl= ;;; .

Substitution and Area Between Curves There are two methods for evaluating a defmite integral by substitution. One method is to fmd an antiderivative using substitution and then to evaluate the defmite integral by applying the Evaluation Theorem. The other method extends the process of substitution directly to definite integrals by changing the limits of integration. We apply the new formula introduced here to the problem of computing the area between two curves.

The Substitution Formula The following formula shows how the limits of integration change when the variable of integration is changed by substitution.

292


THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, bland f is continuous on the range of g(x) = u, then

l

b

f(g(x))' g'(x) dx =

19(b)

a

f(u) duo

g(a)

Proof Let F denote any antiderivative of f. Then,

d d e dB. When e = ",/4, u = cot (",/4) = 1. When e = ",/2, u = cot (",/2) = o.

Let u

("/2 cotlicsc'lidli

=

lOU' (-du)

J"/4

1

=

=

-lOUdU

• Definite Integrals of Symmetric Functions

y

The Substitotion Formula io Theorem 7 simplifies the calculation of definite iotegrais of even and odd functions (Section 1.1) over a symmetric ioterval [-a, a] (Figure 5.24).

THEOREM 8 -a

o

(a)

a

Let f be continuous on the aymmetric ioterval [ -a, a].

Iff is even, then 1>(X) dx

(a)

(b) Iff is odd, then

y

1:

f(x) dx

=

af Zl (X) dx.

= O.

Proof of Part (a)

i (b)

FIGURE 5.24

=

=

(a)feven, J~f(x); . a. W+1

4x

Jo

0

11. a.

Jor

x4 + 9

l6

14. a.

Jo v'4

1°

w/2( 2 + tan ~

b.

dw

15. J.1 v'I'+2I (5t' + 2) dt

sec'Zdl

dz

+ 3 sinz

~/2

~/2 (3 + 2 cos wJ2

o

"

,

cos31)sin3ldl

cos Z dz + 3 sinz

sin w

~--~--------~~X

d>;

w/6

2

{2",

y = (1 - cou) sinx

dv

~/'

(I - cos31)sin3ldl

12. a. E/2(2 + tanf)sec fd' 13. a.

y

d>;

10. a. {' k d > ;

Jo

26. y

o

16 •

smw (3+2cosw)'

]4 2Vy(1 + Vy)' ely

1

28.

27. y

y = ~(COSX)(sin(1T + 1Tsinx»

y

298


29.

36.

30.

Y

y

y

y=!sec 2 t 2

1

Y~ 1

0

"2

2

"

o

x

37.

38.

39.

40.

-4 y

31. (-2,8)

(2.8)

8

Y NOTTOSCAIE

32.

Y

y=~

3

o

(3, -5)

33.

Find the areas of the regions enclosed by the lines and curves in Exercises 41-50. 41. Y = x 2

x2

42. Y = 1. -

34.

35. Y

Y

and y = 2

2

-

and y = -3

43.y=x' and y=8x 44.y=x2 -1. and y=x 45. y=x2 and y= -x 2 +4x 46. y=7-1. 2 and y=x2+4 47. y = x' - 4x 2 + 4 and y = x 2 48. Y

=

49. Y

=

xv'a 2 -

v'R

x 2,

a

>

0,

and

5y

=

x

and y

+6

=

0

(How many intersection points

are there?) 50. y y = _44

-2

= Ix 2

-

41

and y

= (x 2/2) + 4

Find the areas of the regions enclosed by the lines and curves in Exercises 51-58. 51. x = 2y 2,

X

= 0,

and y = 3

5.6

52. x ~ y'

and x ~ y

53. y' - 4x ~ 4

and

54. x - y' ~ 0

and x

55. x ~ y' - y

and

56. x - y'/3 ~ 0 57. x ~ y' - I

+2

Sub,titution and Area Between (urves

299

y

4x - y ~ 16

+ 2y'

~ 3

x ~ 2y' - 2y - 6

and x

+ y'

~ 2

and x ~ lylv'i=Y'

58. x ~ y' - y'

and x ~ 2y

~--~~----~--~X

o

2

Find the areas of the regions enclosed by the curves in Exercises 59-{j2.

59. 4x'+y~4 60. x'-y~O 61. x 62. x

and x4_y~ I

+ 4y' ~ 4 + y' ~ 3

79. The figure here shows triaogle AOC inscribed in the region cut from the parabola y ~ by the lioe y ~ Fiod the limit of the ratio of the area of the ttiangle to the area of the parabolic region as a approaches zero.

x'

3x'-y~4

and and

x

and 4x

+ y' + y'

~ I,

for

x "" 0

~ 0

a'.

y

Find the areas of the regions eoclosed by the lines and curves in Exercises 63-70.

63. Y

= 2 sin x and y = sin2x, 0

64. y ~ 8 cosx

and y ~ sec' x,

65. y ~ cos ('lTx/2) 66. y

~

sin ('lTx/2)

~

x

-'IT/3:5 x :5 'IT/3

and y ~ I - x'

and y

~

x

67. y ~ sec' x, y ~ tan' x, x ~ -'IT/4, 68. x ~ tan'y

and

x ~ -tan'y,

69. x ~ 3 siny v'cosy 70. y ~ sec' ('lTx/3)

~ 'IT

x ~ 'IT/4

-'IT/4 s y '" 'IT/4

and x ~ 0,

and y ~ xli',

and

0:5 Y :5 'IT/2 -I '" x :5 I

71. Find the area of the propeller-shaped region enclosed by the curve x - y' ~ o and the line x - y ~ O. 72. Find the area of the propeller-shaped region enclosed by the curves x - yl/3 ~ o and x - yl/' ~ O.

80. Suppose the area of the region betweeo the graph of a positive continuous function f and the x-axis from x = a to x = b is 4 square uoits. Find the area between the curves y ~ /(x) and y ~ 2/(x) from x ~ atox ~ b.

81. Show that the area of the shaded region equals 1/6 for all values ofz.

73. Find the area of the region in the f1I1!t quadrant bounded by the liney ~ x,the line x ~ 2,thecurvey ~ I/x',andthex-axis. 74. Find the area of the ''triangular'' region in the first quadrant bounded on the left by the y-axis and on the right by the curves y = sinx andy = cosx.

x'

75. The region bounded below by the parabola y ~ and above by the line y ~ 4 is to be partitioned into two subsections of equal area by cutting across it with the horizontal line y = c. ~ c across it that looks about right. In terms of c, what are the coordinates of the points where the line and parabola intersect? Add them to your figure.

a. Sketch the region and draw a line y

82. True, sometimes true, or never true? The area of the region between the graphs of the continuous functions y ~ /(x) and y ~ g(x)andtheverticsllinesx ~ a and x ~ b(a < b) is

b. Find c by integrating with respect to y. (This puts c in the limits ofintegration.)

c. Find c by integrating with respect to x. (This puts c into the integrand as well.) 76. Find the area of the region between the curve y ~ 3 line y ~ -I by integrating with reopect to (a) x, (b) y.

x' and the

77. Find the area of the region in the fJrst quadraot bounded on the left by the y-axis, below by the line y ~ x/4, above left by the curvey ~ I + v'i,andaboverightbythecurvey ~ 2/v'i. 78. Find the area of the region in the fJrst quadraot bounded on the left by they-axis, below by the curve x ~ 2Vy, above left by the curve x ~ (y - I)', and above right by the line x ~ 3 - y.

l\/(x) - g(x)] dx. Give reasons for your answer. Theory and Examples 83. Suppose thatF(x) is an antiderivative of /(x) ~ (sinx)/x, x Express {' sin2x dx x

11 io terms ofF.

> O.

300


84. Show that iff is continuous, then

l

lf

(X) dx =

l

lf(1 -

y

x) dx.

85. Suppose that

Find tf(X)dx if (a) f is odd, (h) f is even. 86. L Show that if f is odd on [-a, aJ, 1hen

89. Use a substitution to verify Equation (I). 90. For each of 1he following functions, graph f(x) over [a, bJ and f(x + c) over [a - c, b - cJ to convince yourself that Equation (1) is reasonable.

i:f(X)dx = O. b. Test1heresultinpar!(a)wi1hf(x) = sinxanda = '1f/2.

&.

87. Iff is a continuous function, fmd 1he value oftha integral ('

1=

Jo

b. f(x)

f(x) dx f(x) + f(a - x)

by making 1he substitution u integral to l.

=

c. f(x)

a - x and adding 1he resultiog

88. By using a substitution, prove 1hat for all positive numbers x andy,

t~'f(x + c) dx.

a. Plot the curves toge1her to see what they look like and how many points of intersection they have. b. Use 1he numerical equation solver in your CAS to fmd a111he points of intersection. c. lotegrate If(x) - g(x) I over consecutive pairs of intersection values. d. Sum toge1her 1he integrals found in par! (c).

(1)

The equation holds wbeoever f is integrable and defmed for 1he necessary values of x. For example, in 1he accompanying figure show that

r- (x + 2)' dx = Jo(lx'dx J-2 l

because the areas of the shaded regions are congruent.

Chapter

= x 2 , a = 0, b = I, c = I = sinx, a = 0, b = '1f, C = '1f/2 = v;=4, a = 4, b = 8, c = 5

COMPUTER EXPLORATIONS 10 Exercises 91-94, you will find tha area between curves in tha plane when you caonot fmd 1heir points of intersection using siruple algebra. Use a CAS to perform 1he following steps:

The Sbift Property for Definite Integral. A basic property of definite integrals is thair invatiance under translation, as expressed by 1he equation [f(x) dx =

f(x)

3

2

91. f(x) =

x x T - 2 -

92. f(x)

=

2x' -

93. f(x)

=

x

94. f(x)

= x2

3x'

2x

+

1 + 3'

10,

+ sin (2x), cos x, g(x)

g(x)

g(x) =

g(x) = x - I

=

=

8 - 12x

x'

x' - x


1. How can you sometimes estimate quantities like distance traveled, area, and average value wi1h fmite sums? Why ntight you want to do so? 2. What is sigma notation? What advantage does it offer? Give examples. 3. What is a Riemann sum? Why ntight you want to consider such a S'lD1l?

4. What is 1he norm of a partition of a closed interval? 5. What is 1he defmite integral of a function f over a closed interval [a, bJ? When can you be sure it exists?

6. What is 1he relation between defmite integrals and area? Describe some other interpretations of def'mite integrals. 7. What is the average value of an integrable function over a closed interval? Must the function assume its average value? Explain. 8. Describe the rules for working wi1h defntite integrals (Table 5.4). Give examples. 9. What is 1he Fundamental Theorem of Calculus? Why is it so iruportant? Illustrate each part of the 1heorem with an example. 10. What is the Net Change Theorero? What does it say about the integral of velocity? The integral of margina1 cost?

301

Chapter 5 Practice Exercises 11. Discuss how the processes of inregration and differentiation can be considered as "inverses" of each other.

14. How can you sometimes evaluare indefutire integrals by substitution? Give examples.

12. How does the Fundamental Theorem provide a solution to the initial value problem dyldx = f(x), y(xo) = Yo, when f is continuous?

15. How does the method of substitution worl< for defutite inregrals? Give examples.

16. How do you defme and calculare the area of the region between

13. How is inregration by substitution relared to the Chain Rule?

Chapter

~

the graphs of two continuous functions? Give an example.

Practice Exerdses

Finite Sums and Estimates 1. The accompanying figure shows the graph of the velocity (fr/sec) of a model rocket for the first 8 sec after launch. The rocket accel· erared straight up for the fIrSt 2 sec and then coasred to reach its maximum height at t = 8 sec.

(5 )

10

c. ~(ak + bk - I)

10 - - b d. ~ k k- I 2

k- l

'0

20

4. Suppose that ~ak = 0 and ~bk = 7. Find the values of 20

'0

a. ~3ak

b. ~(ak

k- l

k- I

150 1 ¢l

+ bk)

'0

d. ~(ak - 2)

~IOO

.~

~

Definite Integrals

50

o

2

6

4

8

Time after launch (sec)

a. Assuuting that the rocket was launched from ground level, about how high did it go? (This is the rocket in Section 3.3, Exercise 17, but you do not need to do Exercise 17 to do the exercise here.)

b. Sketch a graph of the rocket's height above ground as a function oftime for 0 :5 t :5 8. 2. a. The accompanying figure shows the velocity (mj sec) of a body moving along the s·axis duting the time in1llrva1 from t = 0 to t = 10 sec. About how far did the body travel duting those 10 sec?

b. Sketch a graph of s as a function of t for 0 ,;; t ,;; 10 assum· ings(O) = O.

v ....

5

I

\

6

8

10

Time (sec) 10

10

3. Suppose that ~ak = -2 and ~bk = 25. Find the value of i-I

7. 8.

•

lim ~e, x,

-'If/3:5 x :5 'If/3

29. Find the extren3e values of f(x) ~ x' - 3x 2 and rmd the area of the region enclosed by the graph off and the x-axis. 30. Find the area of the regioo cut frOO3 the rlrst quadrant by the curve x 1/2 + yl/2 ~ a1/2. 31. Find the total area of the region enclosed by the curve x ~ y2/3 aod the lines x ~ y and y ~ -I.

• ]0,

50.

{I (88'

Jo {27 Jl X- a) to generate a solid shaped like a doughnut 2 - y2 dy = and called a torus. Find its volume. (Hint: J~a

57. Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about 3 L if you make it 9 cm deep and give the sphere a radius of 16 cm. To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? (1 L = 1000 cm3.)

Ya

y(cm)

'Tfa 2/2, since it is the area ofa semicircle of radius a.) /---- --...

52. Volume of a bowl A bowl has a shape that can be generated by revolving the graph of y = x 2/2 between y = 0 and y = 5 about they-axis.

/

I

I

x 2 + y2 = 16 2 = 256

' "

I

a. Find the volume of the bowl.

o

b. Related rates If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep?

x (cm)

__

53. Volume of a bowl

~~~ drep

a. A hemispherical bowl of radius a contains water to a depth h. Find the volume of water in the bowl. b. Related rates Water runs into a sunken concrete hemispherical bowl of radius 5 m at the rate of 0.2 m 3/ sec. How fast is the water level in the bowl rising when the water is 4mdeep? 54. Explain how you could estimate the volume of a solid of revolution by measuring the shadow cast on a table parallel to its axis of revolution by a light shining directly above it. 55. Volume of a hemisphere Derive the formula V = (2/3)'TfR 3 for the volume of a hemisphere of radius R by comparing its cross-sections with the cross-sections of a solid right circular cylinder of radius R and height R from which a solid right circular cone of base radius R and height R has been removed, as suggested by the accompanying figure. YR2_ h 2

h

~t

•

R

l' h

_ t_

R

58. Max-min The arch y = sinx, 0 ::::; x ::::; 'Tf, is revolved about the line y = c, 0 ::::; c ::::; 1, to generate the solid in the accompanying figure. a. Find the value of c that minimizes the volume of the solid. What is the minimum volume? b. What value of c in [0, 1] maximizes the volume of the solid?

D

c. Graph the solid's volume as a function of c, first for o : : ; c ::::; 1 and then on a larger domain. What happens to the volume of the solid as c moves away from [0, I]? Does this make sense physically? Give reasons for your answers. y

6.2 Volumes Using Cylindrical Shells 59. Consider the region R bounded by the graphs of y = f(x) > 0, x = a > 0, x = b > a, and y = 0 (see acoomanying figure). If the volume of the solid formed by revolving R about the x-axis is 4"., and the volume of the solid formed by revolving R about the line y = -I is 8"., fmd the area of R.

319

60. Consider the region R given in Exercise 59. If the volume of the solid formed by revolving R arouod the x-axis is 6"., and the volume of the solid formed byrevolvingR around the line y = -2 is 10"., fmd the area of R.

y

~ :

R

:

I I --;;-t-~-----:-~x

o

6.2

a

b

Volumes Using Cylindrical Shells

1:

In Section 6.1 we defmed the volume ofa solid as the defmite integral V = A(x) dx, where A(x) is an integrable cross-sectional area of the solid from x = a to x = b. The area A(x) was obtained by slicing through the solid with a plane perpendicular to the x-axis. However, this method of slicing is sometimes awkward to apply, as we will illustrate in our flISt example. To overcome this difficulty, we use the same integral definition for volume, but obtain the area by slicing through the solid in a different way.

Slicing with Cylinders Suppose we slice through the solid using circular cylinders of increasing radii, like cookie cutters. We slice straight down through the solid so that the axis of each cylinder is parallel to the y-axis. The vertical axis of each cylinder is the same line, but the radii of the cylinders increase with each slice. In this way the solid is sliced up into thin cylindrical shells of constant thickness that grow outward from their common axis, like circular tree rings. Umolling a cylindrical shell shows that its volume is approximately that of a rectangular slab with area A(x) and thickness ax. This slab interpretation allows us to apply the same integral defmition for volume as before. The following example provides some insight before we derive the general method. The region enclosed by the x-axis and the parabola y = j(x) = 3x - x 2 is revolved about the vertical line x = - 1 to generate a solid (Figure 6.16). Find the volume of the solid.

EXAMPLE 1

Solutton Using the washer method from Section 6.1 would be awkward here because we would need to express the x-values of the left and right sides of the parabola in Figure 6.l6a in terms ofy. (These x-values are the inner and outer radii for a typical washer, requiring us to solve y = 3x - x 2 for x, which leads to complicated formulas.) Instead of rotating a horizontal strip of thickness ,ly, we rotate a vertical strip of thickness ,lx. This rotation produces a cylindrical shell of height Yk above a point Xk within the base of the vertical strip and of thickness ,lx. An example ofa cylindrical shell is shown as the orange-shaded region in Figure 6.17. We can think of the cylindrical shell shown in the figure as approximating a slice of the solid obtained by cutting straight down through it, parallel to the axis of revolution, all the way around close to the inside hole. We then cut another cylindrical slice around the enlarged hole, then another, and so on, obtaining n cylinders. The radii of the cylinders gradually increase, and the heights of

320

Chapter 6: Applications of Definite Integrals

G t I-

\

__ 1. __ L __ 'L

-_-2~-_~ I--~---L--~~~-+x

I

I

1 I

1 I

r,

-- ~ --f01

-

_1_- _1 __

-1

Axis of revolution x =-1

Axis of revolution x = -1

-2

(b)

(a)

r

FIGURE 6.16 (a) The graph of the region in Example I, before revolution. (b) The solid formed when the region in part (a) is revolved about the axis of revolution x = - I.

~f 1

r

__

Yk

L_~_ J--~

T

° r~3

~"'"---l-I-'r x =-1

FIGURE 6.17 A cylindrical shell of height Yk obtained by rotating a vertical strip of thickness .:lxk about the line x = -I. The outer radius of the cylinder occurs at Xb where the height of the parabola is Yk = 3Xk - Xk2 (Example I).

x

the cylinders follow the contour of the parabola: shorter to taller, then back to shorter (Figure 6.16a). Each slice is sitting over a subinterval of the x-axis oflength (width) f:.xk. Its radius is approximately (1 + Xk), and its height is approximately 3Xk - x/. If we unroll the cylinder at Xk and flatten it out, it becomes (approximately) a rectangular slab with thickness f:.xk (Figure 6.18). The outer circumference of the kth cylinder is 21T' radius = 21T(1 + Xk), and this is the length of the rolled-out rectangular slab. Its volume is approximated by that of a rectangular solid, f:. Vk

= circumference

X height X thickness

= 21T( 1 + Xk) • (3Xk - Xk2) • f:.xk. Summing together the volumes f:. Vk of the individual cylindrical shells over the interval [0,3] gives the Riemann sum n

n

k=l

k=l

~f:.Vk = ~21T(Xk + 1)(3Xk - Xk2) f:.xk.

Outer circumference

L Radius ~

=

= 2'Fr • radius =

2 'Fr(1

+ Xk)

1 +xk

--- =--

I

-F

Llxk = thickness

FIGURE 6.18 Cutting and unrolling a cylindrical shell gives a nearly rectangular solid (Example 1).

6.2

Taking the limit as the thickness D..xk - 0 and n -


00

321

gives the volume integral

n

V = lim ~ 27T(Xk n~OO k=l

+ O(3Xk - Xk2) D..xk

= 1327T(X + O(3x - x2) dx = 1327T(3X2 + 3x - x3 - x2) dx

1 3

(2x 2 + 3x - x3) dx

=

27T

=

2 3 + -x2 3 27T [-x - -x 4

1]3

3

457T 2 .

240

•

We now generalize the procedure used in Example 1.

The Shell Method

I

The volume of a cylindrical shell of height h with inner radius r and outer radius R is

7TR 2h - 7Tr 2 h

=

27T (R ; r )OO k=!

Ib

fllk -

0 and n -

00

gives the volume of the

27T(shell radius)(shell height) dx.

a

=

I

b

27T(X - L)f(x) dx.

We refer to the variable of integration, here x, as the thickness variable. We use the first integral, rather than the second containing a formula for the integrand, to emphasize the process of the shell method. This will allow for rotations about a horizontal line L as well.

Shell Formula for Revolution About a Vertical Line The volume of the solid generated by revolving the region between the x-axis and the graph of a continuous function y = f(x) 2: 0, L :5 a :5 x :5 b, about a vertical line x = L is

-l

V-

a

b

(shell ) ( shell ) dx 27T radius height .

EXAMPLE 2 The region bounded by the curve y = Vx, the x-axis, and the line x is revolved about the y-axis to generate a solid. Find the volume of the solid.

=

4

SoLution Sketch the region and draw a line segment across it parallel to the axis of revolution (Figure 6.20a). Label the segment's height (shell height) and distance from the axis of revolution (shell radius). (We drew the shell in Figure 6.20b, but you need not do that.)

y=

Vx (4,2)

y=

2

Vx

Shell height

o ~+-------+-~----~----~X

integration

x

Interval of integration (a)

(b)

FIGURE 6.20 (a) The region, shell dimensions, and interval of integration in Example 2. (b) The shell swept out by the vertical segment in part (a) with a width ~x.

6.2


323

The shell thickness variable is x, so the limits of integration for the shell formula are a = 0 and b = 4 (Figure 6.20). The volume is then

v= =

l sh~ll 14 Yx) 1 b

a

27T (

radIUs

) (

27T(X) (

s~ell )

heIght

dx

dx

]24 =~. 128 = 27T 4 x 3/ 2 dx = 27T [- 2 X 5/

o

5

0

5

•

So far, we have used vertical axes of revolution. For horizontal axes, we replace the x's withy's.

EXAMPLE 3 The region bounded by the curve y = Yx, the x-axis, and the line x = 4 is revolved about the x-axis to generate a solid. Find the volume of the solid by the shell method. Solution This is the solid whose volume was found by the disk method in Example 4 of Section 6.1 . Now we find its volume by the shell method. First, sketch the region and draw a line segment across it parallel to the axis of revolution (Figure 6.21a). Label the segment's length (shell height) and distance from the axis of revolution (shell radius). (We drew the shell in Figure 6.21b, but you need not do that.) In this case, the shell thickness variable is y, so the limits of integration for the shell formula method are a = 0 and b = 2 (along the y-axis in Figure 6.21). The volume of the solid is

v= =

l 1 1 b

a

sh~ll (s~ell

27T( ) ) d radIUs heIght Y

227T(Y)(4 - y2) dy

= 27T 2(4Y - y3) dy

y 2

y

o 0 .... OJ ''''c :>

~

B~

~ .5

12

y

4 _ y2 Shell height

[;:1x 3 ;J) 0

~

8.y~2x,

v3

x~r

x, y = -x/2, x ~ 2 y=x/2, x~1 9. y=x2 , y=2-x, x=O, forx~O 10. y = 2 - xl, Y = xl, X = 0 7. y

Y

Fv2

Revolution About the y-Axis Use the shell method to find the volumes of the solids generated by revolving the regions bouoded by the curves and lines in Exercises 7-12 about the y-axis.

4.

3.

0

2

v3

11. y ~ 2x 12. y ~

I, y ~ Yx, x ~ 0

3/{2Yx), y

=

0, x ~ I, x

=

4

6.2

13. Letj(x) = {(Sinx)fx,

0

1,

< x :5

25. y=x+2, y=x' L The line x = 2

'Tr

x = 0

a. Showthatxj(x) = sin x, 0:5

y

{

sinX

o<x

1,

x =

x'

S'11'

d. The lioe y = 4

26. y = x', y = 4 - 3x' L

The line x = I

c. The x-axis

In Exertises 27 and 28, use 1he shell metbod to fiod 1he volumes of 1he solids generated by revolving 1he shaded regions about 1he indicated axes.

27.

0

L

The x-axis

b. The line y = I

•• The line y = 8/5

o

= {o(tan, x)'/x,

0

1

< x :5 'Tr/4

x=o a. Showthatxg(x) = (tan x)', Os

X

b. The lioe y = -2/5

y

~~--------~~-+x

14. Letg(x)

325

b. Thelioex =-1

c. The x-axis

'Tr.

X S

b. Find the volwne of 1he solid generated by revolving the shaded regioo about the y-axis in 1he accompanying figure.

y=


x

= 12(y' -

y'J

:5 'Tr/4.

b. Find the volwne of the solid generated by revolving the

shaded region about 1he y-axis in the accompanying figure. y

tan' x

y

'1T

= { ---'O<x:Sx 4 x= 0

0,

~F---------~----------~x

28. a. The x-axis

b. The lioe y = 2

•• The line y = 5

d. The lioe y = -5/8

y ~I'---:o----+x

o

(2.2)

Revolution About the x-AxIs Use the shell me1hod to find the volwnes of1he solids generated by reo volving the regions bounded by 1he curves and lines in Exercises 15-22 about the x-axis.

15. x =

yY,

16. x = y2,

x = -y, y = 2 X = -y, y

=

17. x = 2y - y', x = 0

18. x = 2y - y', x = Y

19. y = lxi, y = I

20. y = x, y = 2x, y = 2

21. y=

Yx,

y=O, y=x-2

22.y=Yx, y=O, y=2-x Revolution About Horizontal and Vertical Lines In Exercises 23-26, use 1he shell me1hod to f'md 1he volwnes of the solids generated by revolving the regions bounded by 1he given curves

about the given lines. 23.y=3x, y=O, x=2 a. The y-axis •• The line x = -I e. Theliney = 7 24. y = x', y = 8, x = 0 a. The y-axis •• The line x = -2 e. Theliney = 8

~t-----~----~~x

2, Y ~ 0

o

Choosing the Washer Method or Shell Method For soroe regions, both the washer and shell me1hods work well for 1he solid generated by revolving the region about 1he coordinate axes, but this is not always the case. When a region is revolved about the y-axis, for example, and washers are used, we must integrate wi1h respect toy. It may not be possible, however, to express the integrand in terms ofy. In such a case, the shell method allows us to integrate with respect to x instead. Exercises 29 and 30 provide some insight

29. Compute 1he volwne of1he solid generated by revolving 1he regioo bounded by y = x and y = x' about eath coordinste axis using •• 1he shell me1hod.

b. The line x = 4

d. The x-axis f. The line y = -2

2

b. 1he washer method.

30. Compute the volwne of the solid generated by revolving 1he triangu1ar region bounded by the lines 2y = x + 4, y = x, and x = 0 about

•• 1he x-axis using 1he washer me1hod. b. The line x = 3

b. 1he y-axis using 1he shell method.

d. The x-axis f. Theliney=-I

•• 1he line x = 4 using 1he shell method. d. 1he line y = 8 using 1he washer me1hod.

326


Io Exercises 31-36, find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so.

use to f"md the volume of the solid? How many integrals would be required in each case? Explain. y

31. The triangle with vertices (I, I), (I, 2), and (2, 2) about L

the x-axis

c. the line x

b. the y-axis ~

10/3

d. the line y

1 ~

I

32. The region bounded by y ~ Yx, y ~ 2, x ~ 0 about L

the x-axis

c. the line x

~~----~~--~~X

-2

0

b. the y-axis ~

4

d. the line y

~

2

33. The region in the f"ITst quadrant bounded by the curve x ~ y - y' and the y-axis about L

the x-axis

b. the line y

~

I

40. The region shown here is to be revolved about the y-axis to generate a solid. Which of the methods (disk, washer, shell) could you use to f"md the volume of the solid? How many integrals would be required in each case? Give reasons for your answers.

34. The region in the first quadrant bounded by x ~ y - y',x ~ I, and y ~ I about L

the x-axis

c. the line x

y

b. the y-axis ~

I

d. the line y

~

I

35. The region bounded by y ~ Yx and y ~ x 2/8 about L

the x-axis the y-axis

x2

o

and y ~ x about

b. the line x

~

I

37. The region in the f"ITSt quadrant that is bounded above by the curvey ~ I/x'/"ootheleftbythelinex ~ 1/16,andbelowby the line y

--;cf"""':....--+.-----> x

b. the y-axis

36. The region bounded by y ~ 2x L

= 1 is revolved about the x-axis to generate a solid.

-1

41. A bead is formed from a sphere of radius 5 by drilling through a diameter of the sphere with a dri11 bit of radius 3.

Find the volume of the solid by L

the washer method.

.. Find the volume of the bead. b. the shell method.

38. The region in the f"ITSt quadrant that is bounded above by the curve y ~ I/Yx, on the left by the line x ~ 1/4, and below by the line y = 1 is revolved about the y-axis to generate a solid. Find the volume of the solid by L

the wasber method.

b. the shell method.

Choosing Disks, Washers, or Shells 39. The region shown here is to be revolved about the x-axis to generate a solid. Which of the methods (disk, washer, shell) could you

6.3

(I, 1)

b. Find the volume of the removed portion of the sphere.

42. A Bundt cake, well known for having a ringed shape, is formed by revolving around the y-axis the region bouoded by the graph of y ~ sin (x 2 - I) and the x-axis over the interval I :5 x :5 vT+1T. Find the volume of the cake. 43. Derive the formula for the volume of a right circular cone of height h and radius r using an appropriate solid of revolution.

44. Derive the equation for the volume of a sphere of radius r using the shell method.

Arc Length We know what is meant by the length of a straight line segment, but without calculus, we have no precise dermition of the length of a general winding curve. If the curve is the graph of a continuous function dermed over an interval, then we can rmd the length of the curve using a procedure similar to that we used for derming the area between the curve and the x-axis. This procedure results in a division of the curve from point A to point B into II1lIl1Y pieces andjoining successive points of division by straight line segments. We then sum the lengths of all these line segments and derme the length of the curve to be the limiting value of this sum as the number of segments goes to inf'mity.

Length of a Curve y

~ f(x)

Suppose the curve whose length we want to find is the graph of the function y ~ f(x) from ~ a to x ~ b. In order to derive an integral formula for the length of the curve, we assume that fhas a continuous derivative at every point of [a, bl. Such a function is called smooth, and its greph is a smooth curve because it does not have any breaks, corners, or cusps.

x

6.3

y

327

Arc Length

ll.-l

--~-,--,-----,-----r#------~--~~--~~------~x

FIGURE 6.22 The length of the polygonal path PoPlP2... p. approximates the length of the curve y = f(x) from point A to pointB. y

I

AYk

I

I

L1

r ------AXl

P,

I

I I I

FIGURE 6.23 The arc P'-IP, of the curve y = f(x) is approximated by the straight line segment shown here, which has length L, = v'(ax.,)2 + (ay.,)2.

We partition the interval [a, b] into n subintervals with a = Xo < Xl < X2 < ... < X. = b. IfYk = j(Xk), then the corresponding point Pk(Xk, Yk) lies on the curve. Next we connect successive points Pk-l and Pk with straight line segments that, taken together, form a polygonal path whose length approximates the length of the curve (Figure 6.22). If Axk = Xk - Xk-l and ~Yk = Yk - Yk- t. then a representative line segment in the path bas length (see Figure 6.23)

so the length of the curve is approximated by the sum (1)

We expect the approximation to improve as the partition of [a, b] becomes rmer. Now, by the Mean Value Theorem, there is a point with Xk-l < Ck < X., such that

c'"

~Yk = f'(et)

Axk.

With this substitution for ~Y'" the sums in Equation (I) take the form

•

~ Lk

=

1=1

•

~ V(Axk)2 + (f'(Ck)~Xk)2 1=1

=

•

~ VI + [('(Ck»)' ~Xk'

(2)

1=1

Because VI + [('(x»)' is continuous on [a, b], the limit of the Riemann sum on the righthand side of Equation (2) exists as the norm of the partition goes to zero, giving

..

lim ~ Lk = lim ~ VI

n--i'OO 1=1

n_OO k=l

+

[('(et»)' Axk =

l

b

a

VI + [('(x»)' dx.

We derme the value of this limiting integral to be the length of the curve.

DEFINmON If j' is continuous on [a, b], then the length (arc length) of the curvey = j(X) from the point A = (a,j(a»tothepointB = (b,j(b» is the value of the integral

L

=

t

VI + [{'(x)l'dx

=

t ~I (:y +

dx.

(3)

328


y

EXAMPLE 1

Find the length of the curve (Figure 6.24)

Y Solution

=

3/ 2 4v'Zx 3

I ,O";x";l.

-

We use Equation (3) with a = 0, b = I, and

y = 4v'Zx 3/ 2 3

I

_

x=1,y~O.89

dy = 4v'Z • lx'/2 = 2V2xl/2

FIGURE 6.24 The length of the curve is slightly larger than the length of the line segmeot joining points A and B (Example 1).

dx

(:Y

3

2

(2V2xl/2y =

=

&.

The length of the curve over x = 0 to x = I is

L= =

[)l+ (:Ydx= [~dx t· t(I +

&)3/2

J:

Eq. (3) with

a=O,b=l Letu = 1 + ~, integrate, and replace u by I + Ilx.

Ii '" 2.17.

=

Notice that the length of the curve is slightly larger than the length of the straight-lioe segment joining the points A = (0, -I) and B = (I, 4 v'Z/3 - I) on the curve (see Figure 6.24):

,112 + (1.089)2 '" 2.14

2.17>

EXAMPLE 2

Solution

•

Find the length of the graph of

f(x) y

Decimal approximations

=

x3

I

12 + x'

I ,,; x ,,; 4.

A graph of the function is shown in Figure 6.25. To use Equation (3), we find

f'(x)

=

x2

I

4 - x2

so

I + [f'(x)f I + (x24 _ l)2 I + (x'16 _ !2 + l) x x' x' I I (x2 I )2 =T6+z+ '= 4+ =

2

=

x

X

2

'

The length of the graph over [I, 4] is FIGURE 6.25 The curve in Example 2, where A = (1, 13/12) andB = (4,67/12).

L

=

=

l VI

+ [f'(x)]2dx

[~~ f

I

-

=

=

l (~

+ :2)dx

-

(~i - i) - C~ I) i~ 6. =

=

•

Dealing with Discontinuities in dy/dx At a point on a curve where dy/dx fails to exist, dx/dy may exist. In this case, we may be able to find the curve's length by expressing x as a function of y and applying the following analogue of Equation (3):

6.3

329

Arc Length

Formula for the Length of x = g(y), c ,;; y ,;; d Ifg'iscontinuouson[c,d],thelengthofthecurvex = g(y) from A = (g(c),c) to B = (g(d), d) is

L

EXAMPLE 3 Solution

y

=

t ~I

+

(~y dy =

t

(4)

Find the length of the curve y = (X/2)2/3 from x = 0 to x = 2.

The derivative

is not defined at x = 0, so we cannot find the curve's length with Equation (3). We therefore rewrite the equation to express x in terms of y:

y= y3/2

=

(~t Raise both sides to the power 3/2.

'! 2

x = 2y 3/2.

FIGURE 6.26 The graph of y = (X/2)2/3 from x = Otox = 2 is also the graph ofx = 2y 3/2 from y = 0 to y = I (Example 3).

VI + [g'(y)]2dy.

Solve forx.

From this we see that the curve whose length we want is also the graph of x = 2y 3/2 from y = 0 to y = I (Figure 6.26). The derivative

~ = 2(~}'/2 =

3y1/2

is continuous on [0, I]. We may therefore use Equation (4) to find the curve's length:

L = =

=

t ~I I 2 -'-(1 9 3

+

(~y dy = 1'v'i+9Ydy

+ 9y)3/2 ]'

0

{7 (IOv'iO - I) ,., 2.27.

Eq. (4) with c=O,d=l. Letu= 1 +9y, du/9 = dy, integrate, and substitute back.

•

The Differential Formula for Arc Length If y = f(x) and if I' is continoous on [a, b], then by the Fundamental Theorem of Calculus we can defme a new function

1 x

s(x) =

VI + [I'(t)f dt.

(5)

From Equation (3) and Figure 6.22, we see that this function s(x) is continoous and measures the length along the curve y = f(x) from the initial point Po(a, f(a» to the point Q(x, f(x» for each x E [a, b]. The function s is called the arc length function for y = f(x). From the Fundamental Theorem, the function s is differentiable on (a, b) and

:

=

VI + [1'(x)]2

=

~I +

(:r

330


Then the differential of arc length is

y

(6) A useful way to remember Equation (6) is to write

ds = -4--------------~x

o

(aJ y

Vax' + dy',

(7)

which can be integrated between appropriate limits to give the total length of a curve. From this point of view, all the arc length fonnulas are simply dllferent expressions for the equation L = ds. Figure 6.27a gives the exact interpretation of ds corresponding to Equation (7). Figure 6.27b is not strictly accurate, but is to be thought of as a simplified approximation of Figure 6.27a. That is, ds RJ fl •.

J

EXAMPLE 4 A

=

Find the arc length function for the curve in Example 2 taking (1, 13/12) as the starting point (see Figure 6.25).

Solution

In the solution to Example 2, we found that

~--------------~X

1

o

+

+ ~,

[I'(x»)' = (x:

r

(bJ

Therefore the arc length function is given by FIGURE 6.27 Diagrams for remembering the equation tis = v'fix' + dy'.

.(x) =

[VI + 3 1

[I'(I»)'dl=

[(~+I~)dl

1]' =12-:X+12' 1

= [ 12 -I

x

I

3

11

To compute the arc length along the curve from A instance, we simply calculate 43

I

=

(1, 13/12) to B

=

(4,67/12), for

11

.(4) = 12 - 4 + 12 = 6.

•

TIris is the same result we obtained in Example 2.

Exercises 6.3 finding Lengths of Curves Find the lengths of the curves in Exercises 1-10. If you have a grapher, you may want to graph these curves to see what they look like. 1. y = (I/3)(x' + 2)3/2 from x = 0 tox = 3

y=x'/' from x=Otox=4 x = (y'/3) + 1/(4y) from y = I toy = 3 x = (y'/2/3) - yl/2 from y = I toy = 9 x = (y'/4) + 1/(8y') from y = I toy = 2 6. x = (y'/6) + 1/(2y) from y = 2 toy = 3 7. y = (3/4)x'/3 - (3/8)x'/3 + 5, I,;; x ,;; 8 8. y = (x'/3) + x' + x + 1/(4x + 4), 0,;; x ,;; 2

2. 3. 4. 5.

9. x = 10. y =

D

J.Y v'sec't -

I dt,

1\/:31'-=-! dt, -,

-1f/4,;; y '" 1f/4 -2';; x ,;; -I

Finding Integrals for Lengths of Curves In Exercises 11-18, do the following.

.. Set up an integral for the length of the curve. b. Graph the curve to see what it looks like.

c. Use your grapher's or computer's integral evaluator to find the curve's length numerically.

= x', -I '" x '" 2 Y = tanx, -",/3 '" x '" x = siny, O::s; y ::s; 'IT

24. Circumference of a circle Set up an integral to find the circumference of a circle of radius r centered at the origin. You will learn how to evaluate the integral in Section 8.3.

11. Y

12. 13.

0

14. x =~, -1/2 '" Y '" 1/2 15. y' + 2y = 2x + I from (-I, -I) to (7, 3)

sinx - x COS x,

16. y

=

17. Y

=

18. x

= J.Y V

I.'

tan I dl,

O:s; x :s;

25. If9x' = y(y - 3)', show that

, _ (y

ds-

+ I)' 4y

dy. ,

'11'

26. If 4x' - y' = 64, show that

0 '" x '" ",/6

sec' I - I dl,

331

Arc Length

6.3

ds' = 4, (5x' - 16) dx'.

-",/3 '" Y '" ",/4

y 27. Is t1rere a smooth (continuously differentiable) curve y

=

j(x)

whose length over the interval 0 '" x '" a is a1ways Yza? Give

reasons for your answer.

Theory and Example. 19••• Find a curve through the point (I, I) whose length integral (Equation 3) is

L =

[~1+ lxdx.

b. How many such curves are there? Give reasons for your answer.

20. •• Find a curve through the point (0, I) whose length integral (Equation 4) is

L=

28. Using tangent fins to derive the length formula for curves Assume that j is smooth on [a, h1and partition the interval [a, h1 in the usual way. In each subinterval [X.-I,x.tl, construct the langenlfin at the point (X.-b j(x._I)) , as shown in the accompanying figure . •• Show that the length of the kth tangent rm over the interval

[XH, x.tl equals V(&t.)'

+ (!'(XH) &t.)'.

b. Showthat

lim ±(lengthofkthtangentftn) =

[~I+ :.dy.

n-OO.t-l

l'

VI

a

+ (j'(x))'dx,

which is the lengthL of the curve y = j(x) from a to h.

b. How many such curves are there? Give reasons for your answer.

21. Find the length of the curve

y =

I.'

Vcos21dl

I

~ Tangent fin : with slope

(xk_l. !(xk_l»

from x = 0 to x = 1T/4. 22. The length of an astroid The graph of the equation x'i' + y'/3 = I is one of a family of curves called astroids (not "asteroids',) because of their starlike appeanmce (see the accompanying figure). Find the length of this particular astroid by rmding the length of half the first-quadrant portion, y = (I - X'/3)3/2, v'2/4 '" x '" I, and multiplying by 8.

-----f!""'--------:-Ilx-.----~, H'.-l) I I I

'.

29. Approximate the arc length of one-quarter of the unit circle (which is~) by computing the length of the polygonal approximation with n = 4 segments (see accompanying figure).

y y

1 x2i3

+ y'1J3 = 1

~~------~------~~ ,

-1

1

~~~~~~~~,

o

-1

23. Length of a line .egment Use the arc length formula (Equation 3) to rmd the length of the line segment y = 3 - 2x, 0 '" x '" 2. Check your answer by rmding the length of the segment as the hypotenuse of a right 1riangle.

0.25 0.5 0.75 1

30. Distance between two points Assume that the tv. points (Xl, Y1) and (x"Yz) lie on the graph of the slraight line y = nIX + h. Use the arc length formula (Equation 3) to rmd the distance between the two points. 31. Find the arc length function furthe graph ofj(x) = 2x 3/2using(O,O) as the stilting point What is the length ofthe curve from (0, 0) to (I, 2)?

332


32. Find the arc length function for the curve in Exercise 8, using (0, 1/4) as the starting point. What is the length of the curve from (0, 1/4) to (1,59/24)?

c. Evaluate the length of the curve using an integral. Compare your approximations for n = 2, 4, 8 with the actual length given by the integral. How does the actual length compare with the approximations as n increases? Explain your answer.


33. f(x) =~,

In Exercises 33- 38, use a CAS to perform the following steps for the given graph of the function over the closed interval.

34. f(x)

a. Plot the curve together with the polygonal path approximations for n = 2, 4, 8 partition points over the interval. (See Figure 6.22.) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments.

= x l/3

+

35. f(x) = sin (1TX 2 ) , 36. f(x) =

x2

-I ~ x ~ I

x 2/ 3,

cos x,

0 ~ x ~ 2 0 ~ x ~

v2

0 ~ x ~ 1T

37. f(x) = x2 - I ' 4x + I

-l<xe1ed by the ",,'s cen1roid during the revolution. If p is the distance from the axis of revolution to the centroid, then

s=

21TpL.

(11)

355

6.6 Moments and Centers of Mass

The proof we give assumes that we can model the axis of revolution as the x-axis and the arc as the graph of a continuously differentiable function of x.

y

Proof We draw the axis of revolution as the x-axis with the arc extending from x = a to x = b in the f"ITSt quadrant (Figure 6.59). The area of the surface generated by the arc is S=

o

l

x~b

lx~b

27T)1 tis = 2".

x- a

Y tis.

(12)

x- a

The y-coordinate of the arc's centroid is FIGURE 6.59

_ E:bytls E:bytls

Figure for proving

Pappus's Theorem for surface area. The arc lenglh differential tis is given by Equation (6) in Section 6.3.

y =

l

x- b

tis

L

=

L

= J ds is the arc's

length and

y ~ y.

x=a

Hence

E:bytls

=

YL.

Substituting yL for the last integral in Equation (12) gives S = 2"'YL. With p equal to y, = 2".pL. •

we have S

EXAMPLE 8

Use Pappus's area theorem tof"md the surface area of the torus in Example 6.

Solution From Figure 6.57, the surface of the torus is generated by revolving a circle of radius a about the z-axis, and b ;" a is the distance from the centroid to the axis ofrevolution. The arc length of the smooth curve generating this surface of revolution is the circumference of the circle, so L = 2".a. Substituting these values into Equation (II), we f"md the surface area of the torus to be

• Exercises 6.6 10. a. The region cut from the frrst quadrant by 1he circle x 2

Thin Plates with Constant Density In Exercises 1-12, find the cenrer of mass ofa thin plate of coostant deosity B covering the given region.

1. The region bounded by the parabola y = x 2 and the line y = 4 2. The region bounded by the parabola y = 25 - x 2 and the x·axis 3. The region bounded by the parabola y = x - x 2 and the line y = -x 4. The region enclosed by the parabolas y = x 2

3 and y = - 2x 2

+ y2

= 9

b. The region bounded by 1he x·axis and the semicircle

y=

vI9=7

Compare your answer in part (b) wi1h 1he answer in part (a). 11. The "ttiangular" region in 1he ftrst quadrant between the circle x 2 + y2 = 9 and 1he lines x = 3 and y = 3. (Hint: Use geometry to fmd the area.)

s.

12. The region bounded above by 1he curve y = I/x', below by the curve y = -I/x', and on 1he left and right by 1he lines x = I and x = a > 1. Also, f"md lima_ oo j.

7. The region bounded by the x·axis and the curve y = cos x,

Thin Plates with varying Density 13. Find the center of mass of a thin plate covering the region between the x·axis and 1he curve y = 2/X2, I '" x s 2, if the

-

The region bounded by the y-axis and the curve x = y - y', O"'y'" I 6. The region bounded by the parabola x = y2 - Y and the line y = x

-"./2 '"

x"''''/2

8. The region betweeo 1he curve y = sec' x, -"./4 s x s "./4 and the x-axis 9. The region bounded by the parabolas y = 2x 2

y=2x-x 2

-

4x and

plate's deosity at 1he point (x,y) is B(x) = x 2.

14. Find the cenrer of mass of a thin plate coveriog the region bounded below by 1he parabola y = x 2 and above by 1he line y = x if1he plate's deosity at the point (x,y) is B(x) = l2x.

356


±4/Yx

15. The region bounded by the curves y = and the lines x = I and x = 4 is revolved about the y-axis to generate • solid. L

25. Variable density Suppose that the density of the wire in Example 4 is 8 = hin6 (kconstant). Find the center of mass.

16. Variable density Suppose that the density of the wire in Exam-

Find the volume of the solid.

b. Find the center of mass of. thin plate covering the region if the plate's density at the point (x, y) is 8(x) = I/x. c. Sketch the plate and show the center of mass in your sketch.

16. The region between the curve y = 2/x and the x-axis fium x to x = 4 is revolved about the x-axis to generate a solid

=I

a. Find the volume of the solid.

ple4is8 = I

Yx.

c. Sketch the plate aod show the center of mass in your sketch.

Centroids of Triangles 17. The centroid of a triangle lies at the intersection of the triangle's medians You may recall that the point inside a triaogie that lies ace-third of the way from esch side toward the opposite vertex is the point where the triaog\e's three medians interse O.

18. (-1,0), (I , 0), (0, 3) 19. (0,0), (I, 0), (0, I) 10. (0,0), (a, 0), (0, a) 21. (0,0), (a, 0), (0, b) 11. (0,0), (a , 0), (a/2 , b)

Thin Wires 23. Constant deoslty Find the moment about the x-axis of a wire of constant density that lies along the curve y = from x = 0 tox = 2.

Yx

14. Constant density Find the momeot about the x-axis of a wire of constant deosity that lies along the curve y

tox = I.

= x'

from x

=0

The Theorems of Pappus 33. The square region with vertices (0, 2), (2, 0), (4, 2), and (2, 4) is revolved about the x-axis to generate a solid. Fted the volume and surface area of the solid. 34. Use a theorem of Pappus to fmd the volume generated by revolving about the line x = 5 the triangular region bounded by the coordinate axe. and the line 2x + Y = 6 (see Exercise 17). 35. Find the volume of the torus generated by revolving the circle (x - 2)' + y' = I about the y-axis.

Chapter 6 Practice Exercises 36. Use the theorems of Pappus to fmd the lateral surface area and the volume of a right-circular cone. 37. Use Pappus's Theorem for sorface area and the fact that the surface area of a sphere of radius a is 4'1Ta 2 to fmd the centroid of the semicircle y ~ Va 2 - x 2 • 38. As found in Exercise 37, the centroid of the semicircle y ~ Va 2 - x 2 lies at the pcint (0, 2a/lr). Find the area of the sorface swept out by revolving the semicircle about the line y

~

a.

39. The area of the region R enclosed by the semiellipse y ~ (b/a)Va 2 - x 2 and the x-axis is (I/2)'lTab, and the volume of the ellipsoid generated by revolving R about the x-axis is (4/3)'lTab 2 . Find the centroid of R. Notice that the location is independent of a. 40. As found in Example 7, the centroid of the region enclosed by the x-axis and the semicircle y = Va 2 - x2 lies at the point (0, 4a/3'IT). Find the volume of the solid generated by revolving this region about the line y ~ -a.

357

42. As found in Exercise 37, the centroid of the semicircle y ~ Va 2 - x 2 lies at the pcint (0, 2a/'IT). Find the area of the sorface generated by revolving the semicircle about the line y = x-a. In Exercises 43 and 44, use a theorem of Pappus to fmd the centroid

of the given triangle. Use the fact that the volume of a cone of radius r and height h is V ~ ~ 'lTr2h. 43.

y (0, b)

~~--------~~x (0,0) (a, 0)

44.

y

(a, c)

41. The region of Exercise 40 is revolved about the line y = x - a to generate a solid. Find the volume of the solid.

Chapter

!


1. How de you define and calculate the volumes of solids by the method of slicing? Give an example. 2. How are the disk and waaber methods for calculating volumes derived from the method of slicing? Give examples of volume calculations by these methods. 3. Describe the method of cylindrical sbells. Give an example. 4. How do you fmd the length of the graph of a smooth function over a closed interva1? Give an example. What about functions that do not have continuous flISt derivatives? 5. How do you defme and calculate the area of the sorface swept out by revolving the graphofa smooth functiony ~ f(x), a :5 x:5 b, about the x-axis? Give an example.

Chapter

[j

6. How de you defme and calculate the work done by a variable force directed along a pcrtion of the x-axis? How do you calculate the work it takes to pomp a liquid from a tank? Give examples. 7. How de you calculate the force exerted by a liquid against a portion of a flat vertical wall? Give an example. 8. What is a center of mass? a centroid? 9. How do you locate the center of mass ofa thin flat plate ofmatrial? Give an example. 10. How de you locate the center of mass of a thin plate bounded by _ curves y ~ f(x) andy ~ g(x) over a :5 x :5 b?

Practice Exercises

Volumes Find the volumes of the solids in Exercises 1-16.

the solid perpendicular to the x-axis are equilateral triangles whose bases stretch from the line to the curve.

1. The solid lies between planes perpendicular to the x-axis at x ~ 0 and x ~ 1. The cross-sections perpendicu1ar to the x-axis between these planes are circular disks whose diameters run from the parabola y ~ x 2 to the parabola y ~ Yx.

3. The solid lies between planes perpendicular to the X- f(XI) when X2 > Xl, is one-to-one and has an inverse. Decreasing functions also have an inverse. Functions that are neither increasing nor decreasing may still be one-to-one and have an inverse, as with the function f(x) = l/x for X oF 0 and f(O) = 0, defined on ( - 00, (0) and passing the horizontal line test.

Finding Inverses The graphs of a function and its inverse are closely related. To read the value of a function from its graph, we start at a point X on the x-axis, go vertically to the graph, and then move horizontally to the y-axis to read the value of y. The inverse function can be read from the grapb by reversing this process. Start with a point y on the y-axis, go horizontally to the graph of y = f(x), and then move vertically to the x-axis to read the value of x = r'(y) (Figure 7.2). y

y

--~--~~----~-------.x

o

X DOMAINOF

o

f

X RANGE OF

1-1

r'

(b) The graph of is the graph off, but with x and y interchanged. To find the x that gave y. we start at y and go over to the curve and down to the x-axis. The domain off-1 is the range off, The range off-1 is the domain off.

(8) To find the value off at x, we start at X, go up to the curve, and then over to the y-axis.

y

y ~r'(x)

__1-__~O~/~__~~______~y

-------o~--~~------~x

/

/ / / / /

/ /

(c) To draw the graph ofr' in the

(d) Then we interchange the letters x andy. We now have a normal-looking graph ofj-l as a function of x.

more usual way. we reflect the

system. across the line y = x.

r'

FIGURE 7.2 Detennining tlie graph ofy ~ (x) from tlie graph ofy = j(x). The graph of is obtaioed by reflecting the graph of j about the line y ~ x.

r'

r

We want to set up the graph of I so that its input values lie along the x-axis, as is usually done for functions, rather than on the y-axis. To achieve this we interchange the x and y axes by reflecting across the 45° line y = x. After this reflection we have a new graph that represents The value of r'(x) can now be read from the graph in the usual way, by starting with a point x on the x-axis, going vertically to the graph, and then horizontally

r'.

364

Chapter 7: Transcendental Functions to the y-axis to get the value of r'(x). Figure 7.2 indicates the relationship between the The graphs are interchanged by reflection through the line y = x. graphs of I and Although this geometric treatment is not a proof, it does make reasonable the fact that the inverse of a one·to·one continuous function defmed on an interval is also continuous. The process of passing from I to 1-' can be summarized as a twc>-step procedure.

r' .

y

1.

Solve the equation y = I(x) for x. This gives a formula x = r'(y) where x is ex· pressed as a function ofy.

2.

Interchange x and y, obtaining a formula y = r'(x) where r' is expressed in the conventional format with x as the independent variable and y as the dependent variable.

y~2x-2

EXAMPLE 3

Find the inverse of y =

~ x + I, expressed as a function ofx.

Solution 1.

2. FIGURE 7.3 Graphingf(x) = (1/2)x + I and r'(x) = 2x - 2 together shows the graphs' symmetry with respect to the line y = x (Example 3).

Solve for x in terms ofy:

Interchange x and y:

y =

y =

2y

=

x

=

1 2x +

x

I

+2

2y - 2. 2x - 2.

The inverse of the function I(x) = (1/2)x + I is the function r'(x) = 2x - 2. (See Figure 7.3.) To check, we verify that both composites give the identity function:

r'(f(x»

=

2(~X + I) - 2 =

I(r'(x»

=

~(2x

- 2) + I

=

+2

x

- 2 = x

x - I + I

x.

=

•

y

EXAMPLE 4

Find the inverse of the function y = x 2 , X

;;"

0, expressed as a function

ofx.

Solution

We flISt solve for x in terms of y:

y = x2

Ixl ~~------------~X

FIGURE 7.4 The functions y = Vi xl, X ~ 0, are inverses of one another (Example 4).

because x

2:

0

We then interchange x andy, obtaining

,0

and y =

= x

y=

Yx.

The inverse of the functiony O,isthefunctiony = Yx (Figure 7.4). Notice that the function y = x 2 , X ;;" 0, with domain restricted to the nonnegative real numbers, is one·to·one (Figure 7.4) and has an inverse. On the other hand, the func· tion y = x 2, with no domain restrictions, is not one·to·one (Figure 7.lb) and therefore has = x 2,x;;"

•

no inverse.

Derivatives of Inverses of Differentiable Functions If we calculate the derivatives of I(x) = (1/2)x + 1 and

its inverse r'(x) = 2x - 2

from Example 3, we see that

~/(x)=~(~x+I)=~ ~r'(x)

=

~(2x

- 2)

=

2.

365

7.1 Inverse Functions and Their Derivatives

The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. (See Figure 7.3.) This is not a special case. Reflecting any nonhorizontal or nonverticalline across the line y = x always inverts the line's slope. If the original line has slope m oF 0, the reflected line has slope I/m.

The slopes are reciprocal: (r') '(b) ~ _I_or (f-') '(b) ~ 1'(0)

IV

I '(b»

FIGURE 7.5 The graphs of inverse functions hsve reciprocal slopes at corresponding points.

r

The reciprocal relationship between the slopes of f and 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = f(x) at the point (a,f(a»isj'(a) andj'(a) oF O,thentheslopeofy = (x) at the point (f(a), a) is the reciprocal 1/ j'(a) (Figure 7.5). Ifwe set b = f(a), then

r'

(r ' ),(b)

=

j'~a)

= j'(f \b))'

r'

If y = f(x) has a horizontal tangent line at (a, f(a» then the inverse function has a vertical tangent line at (f(a), a), and this wmite slope implies that is not differentiable at f(a). Theorem I gives the conditions under which is differentiable in its domain (which is the same as the range of f).

r'

r'

THEOREM I-The Derivative Rule for Inverses If f has an interval I as domain and j' (x) exists and is never zero on I, then 1 is differentiable at every point in its domain (the range of f). The value of(r y at a point b in the domain of ' isthereciprocalofthevalueofj' at the point a = r'(b):

r

r'

(f - 1)'(b) _

I - j'(f '(b))

(i)

or

dr' I ax

x- b -

_l_

dfl ax

x - r'(b)

Theorem 1 makes two assertions. The trrst of these has to do with the conditions under which f- 1 is differentiable; the second assertion is a formula for the derivative of

366

Chapter 7: Transcendental Functions

I -, when it exists. While we omit the proof of the first assertion, the second one is proved in the following way:

I(r'(x))

=

x

Inverse function relationship

fx I(r'(x))

=

I

Differentiating both sides

j'(r'(x))' ~ r'(x) dx d

dx

= 1

Chain Rule

1-'( ) _

I

Solving for the derivative

x - j'(j '(x»"

Thefunction/(x) = x 2 ,x;" oand its inverse r'(x) = Yxhavederiva· tivesj'(x) = 2xand(r'),(x) = 1/(2Yx). Let's verify that Theorem I gives the same fannula for the derivative of r'(x):

EXAMPLE 5

(r'),(x)

j'(j \x))

=

1'(x) ~ 2xwithxreplaced by r1(x)

I

--ccfL----!--~____!---"-----~ x

o

1

2

3

4

FIGURE 7.6 The derivative of r'(x) = Vi at the point (4, 2) is the reciprocal of the derivative of f(x) = x 2 at (2, 4) (Example 5).

Theorem I gives a derivative that agrees with the known derivative of the square root function. Let's examine Theorem 1 at a specific point. We pick x = 2 (the number a) and 1(2) = 4 (the value b). Theorem I says that the derivative of I at 2, j'(2) = 4, and the derivative of at 1(2), )'(4), are reciprocals. It states that

r'

(r'

I _ I (/ -')'(4) -- j'(j I '(4)) -_ j'(2) - 2x

I.-2-_4'I

•

See Figure 7.6.

y

6

y=x 3 -

(2, 6)

2 Slope 3.1;2 ~ 3(2)2 ~ 12

We will use the procedure illustrated in Example 5 to calculate fannulas for the derivatives of many inverse functions throughout this chapter. Equation (I) sometimes enables us to find specific values of dr'/dx without knowing a formula for r'.

Let I(x) = x 3 finding a fannula for r'(x).

EXAMPLE 6

Reciprocal slope:

lz

L _ J-- - - -- 0, and suppose that j has a differentiable inverse, Revolve about the y-axis the regioo bounded by the graph of j and the lines x = a aod y = j(b) to geoerate a solid. Then the values of the integrals giveo by the washer and shell methods for the volume have ideotica1 values:

r' .

i

f

(') 1T«r'(y»' - a') t(y =

f(a)

l'

21TX(f(b) - j(x)) tIx.

To prove this equality, defme

i

e. Plot the functions j and g, the ideotity, the two tangoot lines, and the line segmoot joining the points (xo, j(xo» and (f(xo), xo). Discuss the symmetries you see across the main diagonal.

- 3x+2 62 ,y-2x-11'

-2< _ x O.

EXAMPLE 1

(2)

We use Equation (2) to fmd derivatives.

dId I I (8) dx In 2x = 2x dx (2x) = 2x (2) = x' (b) Equation (2) with u = x 2

iL ln (x 2 dx

+ 3)

=

x

>0

_1_.iL(x 2 x 2 +3 dx

+ 3)

+ 3 gives =

_1_. 2x =~. x 2 +3 x 2 +3

•

Notice the remarkable occurrence in Example la. The function y = In 2x has the same derivative as the function y = Inx. This is true of y = In bx for any constant b, provided that bx > 0:

dI d -Inbx = -'-(bx) dx bxdx If x < 0 and b < 0, then bx b = -I we get

> d

=

I -(b) bx

I

=-.

0 and Equation (3) still applies. In particular, if x

dx In (-x)

=

1

X

for x

0 and x > 0, the natural logarithm satisfies the following rules:

b

1. Product Rule:

Inbx = Inb + Inx

2. Quotient Rule:

b Inx = Inb - Inx

3. Reciprocal Rule:

I Inx = -Inx

Rule 2 with b

4. Power Rule:

Inx T = rlnx

For r rational

=

1

For now we consider ouly rational exponents in Rule 4. In Section 7.3 we will see that the rule bolds for all real exponents as well.

EXAMPLE 2 (a) 1n4 + In sin x = In (4 sin x)

Product

x + I (b) 1n2x _ 3 = In(x + I) - 1n(2x - 3)

Quotient

(e) Int = -inS

Reciprocal

= -In23 = -3ln2

•

Power

We now give the proof of Theorem 2. The properties are proved by applying Corollary 2 of the Mean Value Theorem to each of them. Proof that In bx = In b + In x have the same derivative:

The argument starts by observing that In bx and In x

d bid dx In (bx) = bx = x = dx In x. According to Corollary 2 of the Mean Value Theorem, the functions must differ by a constant, which means that

Inbx = Inx + C for some constant C. Since this last equation holds for all positive values of x, it must hold for x = I. Hence, In(b'l) = Inl + C Inb=O+C

lnl~O

C = Inb.

By substituting we conclude that

Inbx = Inb + Inx.

•

7.2

Proof that Inx'

=

Natural Logarithms

3 73

rlnx (assuming r rational) We use the same-derivative argument

again. For all positive values of x,

l-"-

Eq. (2) with u

=

1 r-l -rx

General Power Rule for derivatives, r rational

=

r·k =

-"- In x T = T (x T ) dx x dx x'

=

x7

fx (rlnx).

Since In x' and r In x have the same derivative, Inx T = rlnx

+C

for some constant C. Taking x to be I identifies C as zero, and we're done. (Exercise 46 in Section 3.7 indicates a proof of the General Power Rule for derivatives when r is rational.) You are asked to prove Rule 2 in Exercise 86. Rule 3 is a special case of Rule 2, obtained by setting b = I and noting that In I = O. This covers all cases of Theorem 2. • y

We have not yet proved Rule 4 for r irrational; however, the rule does hold for all r, rational or irrational. We will show this in the next section after we def'me exponential functions and irrational exponents.

1 1 ~

-+______

______

~~

__

2

O~-------+1--------~2~~x

Y

57. y

= ~I ~

I

58.

59. y =

v'i+3 sinO

61. y

1(1

=

+

1)(1

+ 2)

y=~I(I~ I)

60. y = (tanO)v'29+1 62

• y

=

I(t

I 1)(1

+

0+5 63. y = OcosO

64.y= OsinO

65 =xW+! • y (x + 1)2/3

66. y

3

(x - 2) x 2 +1

+ 2)

~

68. y

(x

= =

+ 1)10

(2x 3

+ I)'

x(x + I)(x - 2) (x 2

+

1)(2x

+ 3)

Theory and Applications 69. Locate and identifY the absolute extreme values of

26. y = In(sccO + tanO) _~

1

~

67.y=

2S. y = O(sin (In 0) + cos (In 0))

27. y=ln

48.

~/4

-In2

L

2

3sec't 6 + 3tanl dl

50. 51.

4.

J

1

40. 42. 44

•

J

8rdr 4r2 - 5

/,o

~/3

l' 2

~

xlnx

1

16

46.

2

4' 0 sm dO 1-4cosO

dx

2x~

•• In (cos x) on [-1T/4,1T/3], b. cos (In x) on [1/2, 2].

70. a. Provethatf(x) = x -Inxisincressingforx b. Using part (a), show that Inx

> 1.

< xifx > 1.

71. Find the area between the curves y = Inx and y = In 2x from x=ltox=5.

72. Find the area between the curve y = tanx and the x-axis from x = -1T/4tox = 1T/3. 73. The region in the first quadrant bounded by the coordinate "'"'s, the line y = 3, and the curve x = 2/vy+! is revolved about

the y-axis to generate a solid. Find the volume of the solid. 74. The region between the curve y = ~ and the x-axis from x = 1T/6 to x = 1T/2 is revolved about the x-axis to generate a solid. Find the volume of the solid. 75. The region between the curve y = l/x2 and the x-axis from x = 1/2 to x = 2 is revolved about the y-axis to generate a solid. Find the volume of the solid.

76. In Section 6.2, Exercise 6, we revolved about the y-axis the region between the curve y = 9x/W+9 and the x-axis from x = 0 to x = 3 to generate a solid of volume 361T. What volume do you get if you revolve the region about the x-axis instead? (See Section 6.2, Exercise 6, for a graph.) 77. Find the lengths of the following curves.

a. y = (x 2/8) - Inx, 4 '" x '" 8 b. x = (y/4), - 21n (y/4), 4 '" y '" 12

7.3 Exponential Functions 78. Find a curve through the point (I, 0) whose length from x = I to x

[~I + :,dx.

L =

D 79.

D 85.

2is

=

I to x

=

=

a. Derive the linearization 10 (1

2. Give the coordinates to

b. Sketcb the region and sbow the centroid in your sketch.

+ lox is one-ro-one.

D 87• ••

84.

d'y

dx' =

I

I

+ x'

D 88.

y(1)

=

3

sec' x, y(O)

=

0

aod y'(O)

+ sinx) for a

= 2,

I

Does the graph of y = Yx - lox, x> 0, have an inflection point? Try to answer this question (a) by graphing, (b) by using calculus.

Having developed the theory of the function In x, we introduce its inverse, the exponential function exp x = eX. We study its properties and compute its derivative and integral. We prove the power rule for derivatives involving general real exponents. Finally, we introduce general exponential functions, aX, and generallogaritbmic functions, log.,x.

y y = In-ix

or 7

Graph y = sinxaod the curves y = In (a 4, 8, 20, and 50 together for 0 ,;; x ,;; 23.

Exponential Functions

7.3 8

=

O.

b. Why do the curves flatten as a increases? (Hint: Find an a- 0

Chain Rule for e", Eq. (2)

=

In short, whenever x

For x < 0, if y

x"·!! x

Definition and derivative of 1n x

> 0,

=

x n, y' , and X,,-l all exist, then

lnlyl

=

lnlxl'

n lnlx!-

=

Using implicit differentiation (which assumes the existence of the derivative y') and Equation (4) in Section 7.2, we have

Y' y

'!

X'

Solving for the derivative, y'

=

y

nX-

=

x'

nx

=

nx n- t .

382

Chapter 7: Transcendental Functions It can be shown directly from the defInition of the derivative that the derivative equals 0 wben x = 0 and n ;;,: I. This completes the proof of the general version of the Power Rule for all values of x. •

EXAMPLE 4

> O.

Differentiate /(x) = xx, x

Solution We cannot apply the power rule here because the exponent is the variable x mther than being a constant value n (mtional or irrational). However, from the def"mition of the general exponential function we note !bat /(x) = XX = exlnx , and differentiation gives /,(x) =

~ (e xInX ) dx

=

e xlnx ~ (x In x) dx

=

e Xlnx ( Inx

=

xX(lnx + I).

Eq. (2) with u

~

.In.

+ x.:})

•

.>0

The Number e Expressed as a Limit We have def"med the number e as the number for which In e = I, or equivalently, the value exp (I). We see that e is an important constant for the logarithmic and exponential functions, but what is its numerical value? The next theorem shows one way to calculate e as a limit. The number e can be calculated as the

THEOREM 4-The Number. as a Um;t

limit e = lim (I • -0

Proof If /(x)

=

+ x)l/x .

Inx, then /,(x) = I/x, so /,(1) = 1. But, by the def"mition of derivative,

/,(1) = lim /(1 h-O

=

lim In (I

x-o =

lim In (I

x-o

+ h) - /(1)

=

lim /(1

h

x-o

+ x)

- In I = lim

x-o

X

+ x)l/x

=

In [lim(1

x-o

+ x)

- /(1)

x

lin (I + x)

Inl~O

X

+ x)l/x]

In is continuous;

use Theorem lOin Chapter 2.

Because /,(1) = I, we have In [lim(1

.-0

+ X)I/x]

=

I

Therefore, exponentiating both sides we get lim (I + x)l/x .-0

=

e.

•

Approximating the limit in Theorem 4 by taking x very small gives approximations to e. Its value is e '" 2.718281828459045 to 15 decimal places as noted before.

The Derivative of aU To f"md tbis derivative, we start with the defIning equation aX = exlna . Then we have

383

7.3 Exponential Functions

We now see why eX is the exponential function preferred in calculus. If a = e, then In a = I and the derivative of aX simplifies to

With the Chain Rule, we get the following form for the derivative of the general exponential function.

If a > 0 and u is a differentiable function of x, then a" is a differentiable function ofx and

%x a"

= a"lna :,.

(3)

The integral equivalent of this last result gives the general antiderivative

y

=

lOx y

=

3x

y~ F ,....-----'--

-1

J

(4)

From Equation (3) with u = x, we see that the derivative of aX is positive if In a > 0, or a > I, and negative if In a < 0, or 0 < a < I. Thus, aX is an increasing function of x if a > I and a decreasing function of x if 0 < a < I. In each case, aX is one-to-one. The second derivative d2 d -(aX) = -(axIna) = (InaJ2a x dx 2 dx

1

FIGURE 7.11 Exponential functions decrease if 0 < a < 1 and increase if a> 1.Asx---+00,wehaveax---+Oif O 1.As x---+ -00, we have aX ---+ 00 if 0 < a < 1 and aX ---+ 0 if a > 1.

a" + C.

a"du = Ina

=2x

is positive for all x, so the graph of aX is concave up on every interval of the real line. Figure 7.11 displays the graphs of several exponential functions.

EXAMPLE 5 (a)

~3x

(b)

~rx

dx dx

We fmd derivatives and integrals using Equations (3) and (4).

= 3x ln3

=

rX(ln3)~(-x) dx

=

-rxln3

d· . d . (e) dx 30nx = 3,mx(ln3) dx (sinx) = 3,mx(ln3) cosx

(d) (e)

J J

2 x dx =

~x2 + C

2 .... cosxdx =

Eq.(3)wi1ha

~ 3,u ~-x

... ,u

~

2"du =

2 sinx

= In2

+

C

~"2

+

C

u

~

x

~

,inx

Eq.(4)wi1ha

J

3,u

Eq.(3)wi1ha

~ 2,u ~ x

~ sinx,du ~

COSXtU,

u replaced by sin x

andEq. (4)

•

Logarithms with Base Q If a is any positive number other than I, the function aX is one-to-one and has a nonzero derivative at every point. It therefore has a differentiable inverse. We call the inverse the logarithm ofx with base a and denote it by log.x.

384


DEFINITION

For any positive number a

'¢'

I,

log"x is the inverse function of aX.

~=-----;;of-,f-c:.~~~~-~x

The graph of y = log"x can be obtained by reflecting the graph of y = aX across the 45° line y = x (Figure 7.12). When a = e, we have logex = inverse of eX = Inx. (The function 10glO x is sometimes written simply as log x and is called the common logarithm of x.) Since log. x and aX are inverses of one another, composing them in either order gives the identity function.

FIGURE 7.12 Thegraphof2x andits inverse, log2 x.

Inverse Equations for aX and log.x

a log.,x = x

(x> 0)

log" (aX) = x

(all x)

The function log" x is actually just a nmnerical multiple of In x. To see this, we let Y = log"x and then take the natural logarithm of both sides of the equivalent equation

a'

=

x to obtain Y In a

= In x. Solving for y gives

Inx log"x = Ina'

TABLE 7.2

Rules for base a

Logarithms For any nmnbers x

> 0 and

y> 0, 1.

2.

3.

Product Rule: log" xy = log" x

+ log" y

The algebraic rules satisfied by log. x are the same as the ones for In x. These rules, given in Table 7.2, can be proved using Equation (5) by dividing the corresponding rules for the natural logarithm function by In a. For example, Inxy = Inx

+ Iny

Inxy Inx Ina = Ina

+

log" xy = log" x

Iny Ina

+

log" y.

Rule 1 for natura110garitbms ... . .. djvided by In a ...

. .. gives Rule 1 for base a logarithms.

Quotient Rule: log" ~ = log" x - log" Y

Derivatives and Integrals Involving log. x

Reciprocal Rule:

To f"md derivatives or integrals involving base a logarithms, we convert them to natural logarithms. If u is a positive differentiable function of x, then

I log" y = -log" Y 4.

(5)

Power Rule: log" x' = Y log" x

d d (lnu) I d I I du tU (Iog"u) = tU Ina = Ina tU (lnu) = lna'll tU'

d I I du -(Iog"u) = -.-tU In a u tU

7.3

Exponential Functions

385

EXAMPLE 6 d

(a) dx log1O(3x

+

lid

1) = In 10 • 3x

+

r°'6)X dx = ~2J In/ dx

(b)

=

~2

J

U

I dx (3x

u

3

+

1) = (In 1O)(3x

1)

~ :;

log,x

du

+

~ lnx,

du

~ ~tU

I u2 I (lnxj2 (lnxj2 =ln2T+ C =ln2-2-+ C = 2ln2 +C

•

Exercises 7.3 Solving Exponential Equations

35.

In Exercises 1-4, solve for I.

1. a.

e--()·3t

=

~

2. a. e.....011

b.e kt

27

1000

b. e

tt

I

=-

c.

e(lnO.2)t =

~

c.

e(Jn2)'

2

/0

~

0.4

t

37.

39. 41.

Finding Derivatives In Exercises 5--24, rmd the derivative of y with respect to x, I, or 8, as appropriate. 6. Y ~ e2>/3 5. y ~ 8. y = e(4v'i+x1) 7. y = e5 - 7:r:

e-"

10. y ~ (1

9.y=xe x -e x 11. Y ~ (x 2

-

2x + 2)e'

13. y ~ e'(sin 8 + cos 8) 15. Y ~ cos (e-6')

+ 2x)e-2> - fu + 2)e"

16. y ~

~ In (3te-l)

18. Y

~ In (2e-' sin I)

19. Y

~ In (----"'-) 1 + eO

20. y

~ inC :~)

21. y

= e(coIHlnt)

22. y ~ e""(InI 2 + I)

23. y

~ /,'.. sin e' dl

In Exercises 25--28, find 25. Iny

~

eY sinx

27. e2> ~ sin (x

+ 3y)

24. Y

~

l.

Inldl

26. In xy = e X +Y 28. !any

~

e'

+ Inx

Finding Integrals

31.

f l f

(e"

+ 5e~) fix

30.

8e«+I) fix

J-1n2

34.

f

38'] e;;

21 e-" dt

40.

f V/r f f e'l'

42.

7fix

44.

f f e~o(~+t) l f

dr

I'e(") dl e-II" ---;>fix

J.~j2 (1 + e""') csc2 8 d8

e sec frt sec 7ft tan wI dt

esc (."

+ I) cot (." + I) dl

2evcosevdv

48.

In ('11'/6)

49.

f f

fix

~/4

ln(wj2)

47.

44

/,v.;;,:

2 2 2xeX cos(eX )dx

0

e' I + e' dr

50.

f

fix I + e'

Initial Value Problems Solve the initial value problems in Exercises 51-54.

51. :;;

~ e'sin(e' -

52. :;;

~ e-lsec2 (."e-'),

d 2y

53. fix2 = 2e-',

d 2y

-2

dl

2), y(ln2)

0

y(ln4) = 2/."

y(O) = I

= I - e 2t ,

~

aod y'(O) ~ 0

y(l) = -I

aod y'(I) ~ 0

(2e' - 3e -2» fix

32. ('

1n2

33.

f

in,

e' fix

46.

54.

Evaluate the integrals in Exercises 2!1-50.

29.

45.

e

eV;

'b

e4Vi.

dy/ fix.

lln4

r /4 (1 + e""') see> 8 d8

lI'e-29 cos 58

17. y

36. 10

43. 10

12. y ~ (9x 2

14. y ~ In (38e-')

[m 16

{In' e,j2 fix

e~ fix

2e (2)-1) fix

Differentiation In Exercises 55-112, rmd the derivative of Y with respect to the given independent variable. 55.y~2'

56.y=3~

57. Y ~ 5V;

58. Y =

2(")

386

Chapter 7: Transcendental Functions Logarithmic Differentiation

59. y = x"

60. y = t l -

61. y = (cos8)v2

62. y = (ln8)"

63. y = 7sec 81n 7 65. y = 2sin3t

64. y = 3tan 8 In 3

67. y = IOg258

68. y = IOg3(l

66. y =

69. y = IOg4x + IOg4x2 71. y = x 3 10g lO x 73. y = log3 (

(~ ~

U

l n3)

e

In Exercises 111- 118, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. 111. y = (x + I)X 112. y = x 2 + x2x

5 - cOS 21

+ 81n 3)

113. y = (VI)t

114. y = t Vi 116. y = x sinx 118. y = (lnx)lnx

70. y

=

IOg25 eX - IOg5vX

115. y = (sinx)X

72. y

=

log3 r· log9 r

117. y = sinxx

r-;----,--;-~

(3}~ 2 r 5

74. y = IOg5

Theory and Applications 119. Find the absolute maximum and minimum values of f(x) = eX - 2x on [0, I].

75. y = 8 sin (lOg7 8)

76

77. y

858 78. y = 2 - IOg5 8

10glO eX

=

= I

(sin8cos8) Og7 e 8 28

• y

79. y = 310g ,t

80. y = 3 logg (lOg2 t)

81. y = IOg2 (8t ln2 )

82. y = t IOg3 (e(Sin t)(ln 3))

120. Where does the periodic function f(x) = 2e sin (x/2) take on its extreme values and what are these values? y

~,

Integration Evaluate the integrals in Exercises 83- 92.

83.

J

SX dx

84.

J ~x 3

121. Let f(x) = xe- x.

a. Find all absolute extreme values for f.

3x dx

b. Find all inflection points for f. eX 122. Letf(x) = --2x' I +e a. Find all absolute extreme values for f.

(I3')

r,,/4

90.

Jo

92.

J

tan 1

b. Find all inflection points for f. sec2 t dt

where it is assumed.

x2x'

D 124.

---,dx I + 2x

Evaluate the integrals in Exercises 93- 106.

J 95.1 J i

3x v'3 dx

93.

94.

3

(0 + I)x v1dx

10glOx -x-dx

97.

103.

Jo

+ 2) + 2 dx 2 10glO (x + I) x + I dx

105.

x

r

-x-dx

126. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = ex / 2 , below by the curve y = e - x/2 , and on the right by the line x = 2 In 2 .

21n 10 log 10 X x dx

127. Find a curve through the origin in the xy-plane whose length from x = 0 to x = I is

i i 1 1 J I

r210g2 (x

Jo

J

10

102.

log 10 (lOx) x dx

L =

11)1 +

:teXdx .

1/10

104.

3 2 IOg2 (x - I) x _ I dx

2

106.

dx x log 10 X

Graph f(x) = (x - 3)2e x and its first derivative together. Comment on the behavior of f in relation to the signs and values of f' . Identify significant points on the graphs with calculus, as necessary.

125. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = e 2x , below by the curve y = eX, and on the right by the line x = In 3 .

410g2X e

100.

I

101.

x v2- 1dx

I

41n210g2X x dx

99.

J

96·iexOn2) - ldx

98.

123. Find the absolute maximum value of f(x) = x 2 1n (I/x) and say

128. Find the area of the surface generated by revolving the curve x = (e Y + e- Y )/2,0 ::s y ::s In 2, about the y-axis. eY + e-Y

dx x(logg x)2

X=---

2

Evaluate the integrals in Exercises 107- 110.

107. 109.

i i

lnx

I

I

l /X

I [dt,

x

I [dt,

x > 0

> I

108.

iex I I [dt

I 110. Ina

i

I

X

I [dt,

x> 0

x

387

7.4 Exponential Change and Separable Differential Equations In Exercises 129--132, find the length of each curve. 129. y = t(e X

+ e~)fromx

130. y = In (eX - I) -In(e'

137. Find the area of the region between the curve y = Zr/(l and the interval -2 ~ x ::s; 2 of the x-axis.

= Otox = I

+

I) from x = 1n2tox = 1n3

+ c.

graphing.

o 140. Could

Xln2 possibly be the same as 21n' for two functions and explain what you see.

b. Find the average value oflnx over [I, e].

134. Find the average value of f(x) = I/x on [I, 2]. 135. Thelinearizalion or.x at x = 0

a. Derive the linear approximation eX

~

1

+ x at x

= O.

x

eX

x :5 x :5 colors, if availahle. On what intervals does the approximation appear to overestimate eX? Underestimate eX?

136. The geometric, logarithmic, and arithmetic mean inequality

a. Show that the graph of e' is concave up over every interval of x-values. b. Show, by reference to the accompanying figure, that if 0< a < bthen

e(ln.+lnb)/2'(lnb -Ina)
O? Graph the

141. Thelinearizalion ofl'

o b. Estimate to five decimal places the magnitude of the error involved in replacing by I + on the interval [0, 0.2]. o c. Graph and I + together for - 2 2. Use different eX

2

= 2' bas three sclutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by

132. y = In (esc x) from x = 'IT/6 to x = 'IT/4 133. a. ShowthatJlnxdx =xlnx -x

138. Find the area of the region between the curve y = 2 1- x and the interval -1 :5 x :5 1 of the x-axis.

o 139. The equation x

131. y = In (cos x) from x = Otox = 'IT/4

+ x 2)

a. Find the linearization of f(x)

= 2' at x = O. Then round its coefficients to two decin2a1 places.

o b. Graph the linearization and function together for -3:5 3and-l 1. x:5

:5 x:5

142. Thelinearizalion oflog, x

a. Find the linearization of f(x) = log, x at x its coefficients to two decimal places.

=

3. Then round

o b. Graph the linearization and function together in the

window

O:5x:5 8and2:5x :54.

o 143. Which is bigger, 'IT' or e ? W

Calculators have taken scme of the mystery out of this once-challenging question. (Go ahead and check; you will see that it is a surprisingly close call.) You can answer the question without a calculator, though.

&.

Find an equation for the line through the origin tangent to the graph of y = In x.

E

B,

__~A~:Ina____~__~~__~~__~,x Ina+lnb

2

b. Give an argument hased on the grapha of y = In x and the tangent line to explain why In x < x/e for all positive x #' e.

c_ Show that In (x')

NorTOSCALB

c. Use the inequality in part (b) to conclude that

.rc

b-a a+b vab < Inb - Ina < --2--'

This inequality says that the geome1ric mean of two positive numbers is less than their logarithmic mean, which in turn is less than their arithmetic mean.

7.4

[-3,6] by [-3, 3]

d. Conclude that xt'

< x for all positive x #' e. < eX for all positive x =F e.

e. So which is bigger, ",e or e""?

o 144. A decimal representalion of.

Find e to as many decimal places as your calculator allows by sclving the equation Inx = I using Newton's method in Section 4.6.

Exponential Change and Separable Differential Equations Exponential functions increase or decrease very rapidly with changes in the independent variable. They describe growth or decay in many natural and industrial situations. The variety of models based on these functions partly accounts for their importance. We nOW investigate the basic proportionality assumption that leads to such exponential change.

388


Exponential Change In modeling many real-world situations, a quantity y increases or decreases at a rate proportional to its size at a given time I. Examples of such quantities include the amount of a decaying radioactive material, the size of a population, and the temperature difference between a hot object and its surrounding medium. Such quantities are said to undergo exponential change. If the amount present at time I = 0 is called Yo, then we can find y as a function of t by solving the following initial value problem:

dy dl

Differential equation:

y = Yo

Initial condition:

(la)

= ky

when

t

= O.

(lb)

Ify is positive and increasing, then It is positive, and we use Equation (Ia) to say that the rate of growth is proportional to what has already been accumulated. If y is positive and decreasing, then k is negative, and we use Equation (Ia) to say that the rate of decay is proportional to the amount still left. We see right away that the constant function y = 0 is a solution of Equation (Ia) if Yo = O. To find the nonzero solutions, we divide Equation (Ia) by y: Idy

y"O

y'dl=k

J~:dl=Jkdl In

Iyl Iyl Iyl

=

Integrate with respect to t;

kl + C

= ekt+c

J(1ju) du ~ In lui + C. Exponentiate.

= eC'e kt

= ±eCekt y = Ae kt .

y

If Iyl ~

T, tbeny

~

±T.

A is a shorter name for ±e c ,

By allowing A to take on the value 0 in addition to all possible values ±ec, we can include the solution y = 0 in the formula. We fmd the value of A for the initial value problem by solving for A when y = Yo and

1=0:

Yo

=

Ae k ' O = A.

The solution of the initial value problem is therefore y = yoe kt .

(2)

Quantities changing in this way are said to undergo exponential growth if It > 0, and exponential decay if k < O. The number k is called the rate constant of the change. The derivation of Equation (2) shows also that the only functions that are their own derivatives are constant multiples of the exponential function. Before presenting several examples of exponential change, let's consider the process we used to derive it.

Separable Differential Equations Exponential change is modeled by a differential equation of the form dy/dx = ky for some nonzero constant k. More generally, suppose we have a differential equation of the form

dy dx = f(x,y),

(3)

7.4 Exponential Change and Separable Differential Equations

389

where f is a function of both the independent and dependent variables. A solution of the equation is a differentiable function y = y(x) defined on an interval of x-values (perhaps infmite) such that d

dx y(x) = f(x,y(x»

on that interval. 1bat is, wheny(x) and its derivative y'(x) are substituted into the differential equation, the resulting equation is true for all x in the solution interval. The general solution is a solutiony(x) that contains all possible solutions and it always contains an arlJitrary constant. Equation (3) is separable if f can be expressed as a product of a function of x and a function of y. The differential equation then has the fonn dy dx = g{x)H(y).

g is a function of x; H is a function of y.

When we rewrite this equation in the form dy dx

g(x) h(y) ,

I H(y) ~ hey)

its differential fonn allows us to collect all y terms with dy and all x terms with dx: h(y) dy = g(x) dx.

Now we simply integrate both sides of this equation:

J

h(y) dy =

J

(4)

g(x) dx.

After completing the integrations we obtain the solution y defmed implicitly as a function

ofx. The justification that we can simply integrate both sides in Equation (4) is based on the Substitution Rule (Section 5.5):

J

h(y) dy =

=

=

EXAMPLE 1

J J J

h(y(x)) : dx g(x) h(y(x)) h(y(x)) dx

ely

g(x)

;2

~ = yer' + 2vY er'

I

dy

20. fix=xy+3x-2y-6

27. Working underwater The intensity L(x) oflightx feet beneatlt the surface of the ocean satisfies the differential equation

tiL = eX - Y

+ eX +

-0.6y

26. The inversion of sngar The processing of raw sugar has a step called ''inversion'' that changes the sugar's molecular s1ructw'e. Once the process has begun, the rate of chaoge of the amount of raw sugar is proportional to the ao30Unt of raw sugar remaining. If 1000 kg ofraw sugar reduces to 800 kg ofraw sugar during the 'IrSt 10 hours, how much raw sugar will remain after aoother 14 hours?

= e Y+';"

I

dy-

Differential equation: Initial condition:

to express p in tenns of h. Determine tlte values of po aod k from tlte given altitude-pressure data.

x>O 8. x"y' = xy - y2,

24. Atmospheric pressure The earth's atmospheric pressme p is often modeled by asson3ing that the rate dp/dh at which p chaoges witlt the altitode h above sea level is proportional to p.

candidate

2

6. y' = e""" - 2xy 7. xy'

Solution

Initial condition

equation

theconditionthaty = 0.99yowhent = 1000 to find the value of k in the equation y = yoe it . Then use this value of k to answer the following questions.

b. In about how many years will humao teeth be 90% of their present size?

3

vT+7dt, y'

23. Human evolution continues The aoalysis of tooth shrinkage by C. Loring Brace and colleagues at the University of Michlgao's Musenm of Anthropology indicates that humao tooth size is continuing to decrease aod that tlte evolutionary process did not come to a halt some 30,000 years ago as many scientists contend In northern Emopeans, for example, tooth size reduction now has a rate of I % per 1000 years.

e-Y

+1

Applications and Examples

D The answers to most of the following exercises are in terms of logarithms aod exponentials. A calculator can be helpful, enabling you to express the aoswers in decimal form.

fix

=

-AL.

As a diver, you know from experience that diving to 18 ft in tlte Caribbean Sea cuts tlte intensity in half. You caonot work witltout artificial light when the intensity falls below on.,.tenth of the surface value. About how deep can you expect to work witltout artificiallight?

7.4 Exponential Change and Separable Differential Equations 28. Voltage in • discharging capacitor Suppose that electricity is draiuiug from a capacitor at a rate that is proportional to the voltage V across its tenninals and that, if t is measured in seconds,

Solve this equation for V, using Vo to denote the value of V when t = O. How long will it take the voltage to drop to 10% of its original value? 29. Chol.ra b.ct.ria Suppose that the bactetia in a colony can grow unchecked, by the law of expouential cbange. The colony starts with I bacterium aod doubles every ba1f-hour. How many bacteria will the colony contain at the end of24 hours? (Under fa-

vorable laboratory conditions, the number of cholera bacteria can double every 30 min. In ao infected petlion, many bacteris are destroyed, but this example helps explain why a person whn feels well in the morning may be dangerously ill by evening.) 30. Growth of bacteria A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are Ml,OOO. How many bacteria were present initially?

dis....

31. Th. incidence of. (Continuation of Example 3.) Suppose that in any given year the number of cases can be reduced by 25% instead of 20%.

a. How long will it take to reduce the number of cases to 1000? b. How long will it take to eradicate the disease, that is, reduce

the number of cases to less than I?

u.s.

32. Th. population The US. Census Bureau keeps a running clock totaling the US. popu1ation. On March 26, 2008, the tota1 was increasing at the rate of I petlion every 13 sec. The population figure for 2:31 P.M. EST on that day was 303,714,725.

a. Assuming exponential growth at a constaot rate, fmd the rate constaot for the population's growth (people per 365-day year).

b. At this rate, what will the US. population be at 2:31 P.M. EST on March 26, 2015? 33. Oil depl.tion Suppose the amount of oil pumped from one ofthe canyon wells in Whittier, California, decreases at the contiouous rate of 10% per year. When will the well's output fall to on.,.fifth

ofits present value? 34. Continuous price discounting To encourage buyers to place 100-unit ordetli, your firm's sales department applies a continuous discouot that makes the unit price a functionp(x) of the number of units x ordered. The discount decreases the price at the rate of $0.0 I per unit ordered. The price per unit for a lOO-unit order is p( 100) = $20.09.

a. Find p(x) by solving the following initial value problem: Differential equation: Initial condition:

dp I dx = -lOOP p( 100)

=

20.09.

b. Find the unit price p(1 0) for a 10-unit order and the uoit price P(90) for a 90-unit order. c. The sales department has asked you to find out if it is discouoting so much that the firm's revenue, r(x) = X' p(x) , will actually be less for a 100-unit order than, say, for a

395

90-unit order. Reassure them by showing that r has its maximum value atx = 100. d. Graph the reveoue function r(x) = xp(x) for 0 :5 x :5 200. 35. Plutonium-239 The balf-life of the plutooium isotope is 24,360 years. If 10 g of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for 80% of the is()tope to decay? 36. Polonium-210 The balf-life of polonium is 139 days, but your sample will not be useful to you after 95% of the radioactive nuclei preseot on the day the sample arrives bas disintegrated. For about how many days after the sample arrives will you be able to

use the polonium? 37. Th. mean life of. radio.ctive nucleus Physicists using the radioactivity equation y = yoe -kt call the number I/k the mean life

of a mdioactive nucleus. The mean life of a radon nucleus is about 1/0.18 = 5.6 days. The meao life ofa carbon-14 nucleus is more than 8000 years. Show that 95% of the radioactive nuclei origiually preseot in a sample will disintegrate within three meao lifetimes, i.e., by time t = 3/k. Thus, the meonlife ofanucleus gives a quick way to eslimste how long the radioactivity of a sample will last. 38. Californiuno-252 What costs $27 million per gram and can be used to treat brain cancer, analyze coal for its sulfur content, and detect explosives in luggage? The answer is californium-252, a radioactive isotope so rare that only 8 g of it bave been made in the western world since its discovery by Gleno Seaborg in 1950. The ba1f-life of the isotope is 2.645 years--Iong enough for a us.,. ful service life and short enough to bave a high radioactivity per unit mass. One microgram of the isotope releases 170 million

neutrons per minute. a. What is the value of k in the decay equation for this isotope?

b. What is the isotope's meao life? (See Exercise 37.) c. How long will it take 95% of a sample's radioactive nuclei to disintegrate? 39. Cooting .oup Suppose that a cup of soup cooled from 90'C to 60°C after 10 min in a room whose temperature was 20D e. Use Newton's law of cooling to aoswer the following questions. a. How much longer would it take the soup to cool to 35'C?

b. Instead ofbeing left to staod in the room, the cop of90'C soup is put in a freezer whose temperature is -15°C. How long will it take the soup to cool from 90'C to 35'C? 40. A beam of unknown temperatore An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at 65'F. After 10 min, the beam warmed to 35'F and after another 10 min it was 50'F. Use Newton's law of cooling to estimate the beam's initial temperature. 41. Surrounding medium of unknown temperatore A pan of warm water (46'C) was put in a refiigerator. Ten minutes later, the water's temperature was 39'C; 10 min after that, it was 33'C. Use Newton's law of cooling to estimate how cold the refiigerator was. 42. SiIv.r cooling in air The temperature of an ingot of silver is 60'C above room temperature right now. Twenty minutes ago, it

was 70D e above room temperature. How far above room temperature will the silver be

•• 15 min from now?

b. 2 hours from now?

c. When will the silver be lODe above room temperature?

396


43. The age of Crater Lake The charcoal from a tree killed in the volcanic eruption that formed Crater Lake in Oregon cootained 44.5% of the carbon-14 found in living matter. About how old is Crater Lake? 44. The sensitivity of carbon-14 dating to measurement To see the effect of a relatively sma1l error in the estimate of the amount of carbon-14 in a sample being dated, consider this hypothetical situation: L A fossilized bone found in centrallllinois in the year A.D. 2000 cootains 17% ofits original carbon-14 cootent. Estimate the year the animal died.

45. Carbon-14 The oldest known frozen humao mummy, discovered in the Schoalslal glacier of the Italian Alps in 1991 and called Olzi, was found wearing straw shoes and a leather coal with goat fur, and holdiog a copper ax and stone dagger. It was estimated that Otzi died 5000 years before he was discovered in the melting glacier. How much of the originsl carboo-14 remained in Otzi al the time of his discovery? 46. Art forgery A painting attributed to Vermeer (1632-1675), which should cootain no more than 96.2% of its originsl carboo14, contains 99.5% instead. Aboulhow old is the forgery?

b. Repeat part (a) assuming 18% instead of 17%.

c. Repeat part (a) assuming 16% instead of 17%.

7.5 HISTORICAL

Indeterminate Forms and L'Hopital's Rule

BIOGRAPHY

Guillaume Fran~ois Antoine de I'H6pital (\661-1704) Johano Bernoulli (1667-1748)

John (Joharm) Bernoulli discovered a rule using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or + 00 . The rule is known today as I'HopitaI'. Rule, after Guillaume de I'HOpitai. He was a French nobleman who wrote the first introductory differential calculus text, where the rule 1mt appeared in print. Limits involving transcendental functions often require some use of the rule for their calculation.

Indeterminate Form 0/0 If we want to know how the function

F(x) = x - sinx x3 behaves near x = 0 (where it is undefined), we can examine the limit of F(x) as x --+ O. We carmol apply the Quotient Rule for limits (Theorem I of Chapter 2) because the limit of the denominator is O. Moreover, in this case, both the numerator and denominator approach 0, and % is unde1med. Such limits may or may not exist in general, but the limit does exist for the function F(x) under discussion by applying I'HOpital's Rule, as we will see in Example I d. If the continuous functions f(x) andg(x) are both zero at x = a, then lim f(x) x~a

g(x)

cannot be found by substitoting x = a. The substitotion produces 0/0, a meaningless expression, which we carmot evaluate. We use % as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as 00/00, 00 00, 0, 00 - 00, 0°, and 1 , which cannot be evaluated in a consistent way; these are called indeterminate forms as well. Sometimes, but not always, limits that lead to indetermiuate forms may be found by cancellation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find limx~o (sinx)/x. But we have had success with the limit

'( ) f a

=

Ii

m

x~a

f(x) - f(a) x a '

7.5

Indeterminate Forms and rH6pitars Rule

397

from which we calculate derivatives and which produces the indetenninant fann % when we substitute x = a. I:Hopitai's Rule enables us tu draw on our success with derivatives to evaluate limits that otherwise lead to indetenninate forms.

THEOREM 5- L'H6Jrital's Rule Suppose 1hatj(a) = g(a) = 0, thatj and g are differentiable on an open interval I containing a, and that g' (x) # 0 on !if x # a. Then lim j(x) g(x)

x~a

=

lim f'(x) x~a

g'(x) ,

assuming that the limit on the right side of this equation exists. We give a proof of Theorem 5 at the end of this section.

I

Caution

To apply I'H6pital's RnIe to Ilg, divide 1he derivative of I by 1he derivative of g. Do not fall into 1he trap of taking 1he detivative of IIg. The quotient to use is ng', not (f!g)'.

EXAMPLE 1 The following limits involve % indeterminate forms, so we apply I'HOpitai's Rule. In some cases, it must be applied repeatedly. (8) lim 3x - sinx = lim 3 - cosx = 3 - cosxl = Z x-o X x-o 1 1 x=o

1 (b) lim

x-o

(c) lim

yIJ+x - I

=

x

yIJ+x -

lim

x-o

zyIJ+x 1

I 2

1 - x/2

o

2

x-o

o

X

=

=

xr'/ x-o 2x . -(1/4)(1 + xr lim lim (I/Z)(I +

2

-

32 /

x-a

2

I/Z

Still ~; differentiate again.

I 8

=--

o

(d) lim x - sinx

x-o

Not~; limit is found

o

x3

=

lim 1 - cosx

x-o =

Still Q

o

3x 2

lim sinx

x-o 6x

Not~;limitisfound Here is a summary of the procedure we followed in Example 1. Using UHopital's Rule Tofmd

lim j(x) x-a g(x)

by I'HOpital's Rule, continue to differentiate j and g, so long as we still get the form % at x = a. But as soon as one or the other of these derivatives is different from zero at x = a we stop differentiating. I:HOpitai's Rule does not apply

when either the numemtor or denominator has a fmite nonzero limit.

•

398


EXAMPLE 2

Be careful to apply I'Hopital's Rule correctly:

o o

lim I - cosx

x-o =

X+X2

. sin x ;~ I + 2x

0

=

T = O.

Not ~; limit is found.

Up to now the calculation is correct, but if we continue to differentiate in an attempt to apply I'HOpital's Rule once more, we get

!

lim cosx =

x-o 2

2'

wbich is not the correct limit. I:HOpital's Rule can ouly be applied to limits that give indeterminate forms, and 0/ I is not an indeterminate form. • I:HOpital's Rule applies to one-sided limits as well.

EXAMPLE 3

In this example the one-sided limits are different.

o o

I Recall that thing.

00

Positive for x > 0

and + 00 mean the same

o

. sin x (b) Iu n -2-

o

x-o- x =

lim cosx x-o- 2x

Indeterminate Forms

=

-00

•

Negative for x < 0

ot:J / ot:J, ot:J •

0,

ot:J -

ot:J

Sometimes when we try to evaluate a limit as x --+ a by substitoting x = a we get an indeterminant form like 00/00, 00·0, or 00 - 00, instead of 0/0. We first consider the form

00/00. In more advanced treatments of calculus it is proved that I'HOpital's Rule applies to the indeterminate form 00/00 as well as to 0/0. Iff(x) --+ ±oo andg(x) --+ ±oo asx --+ a, then lim fix) = lim f'(x) x~.

g(x)

x~.

g'(x)

provided the limit on the right exists. In the notation x --+ a, a may be either finite or infinite. Moreover, x --+ a may be replaced by the one-sided limits x --+ a + or x --+ a -.

EXAMPLE 4 (a)

lim

x~"/2 I

Find the limits of these 00/00 forms: secx

(b) lim

+ tanx

x-OO

lnx

2Vx

• eX (c) lun 2'

x-oo x

Solution O+

00 • 0

=

lim x-O+

=

EXAMPLE 6

lim

x-o+

I/x -1/2x 3/ 2

(-2Vx)

I/x.

converted to

00/00

I'H8pital's Rule

•

= 0

Find the limit of this

00 -

00

lim ( _ I x-a sinx

Solution

~

form:

_1)

x·

Ifx--+ 0+, then sinx--+ 0+ and

I - -~oo I -.smx x

00

Similarly, if x --+ 0-, then sin x --+ 0- and

I - .I - - ---+ smx x

00 -

(-00) = -00

+

00.

Neither form reveals what happens in the limit. To find out, we flfst combine the fractions: I sin x

I

----

x

x - sinx x sin x

Common denominator is x sin x.

Then we apply I'HOpital's Rule to the result: lim

(_._1__ 1) =

x-a smx

x

lim x - .sinx

x-a x smx

= lim

I - cosx

x-a sinx + xcosx sin x = Q= 0 x_o2cosx-xsinx 2 .

= lim

o o Sti1I()

o

•

400


Indeterminate Powers Limits that lead to the indeterminate fonns 100, 0°, and 00° can sometimes be bandied by first taking the logarithm of the function. We use I'Hllpital's Rule to find the limit of the logarithm expression and then exponentiate the result to find the original function limit. This procedure is justified by the continuity of the exponential function and Theorem lOin Section 2.5, and it is fonnulated as follows. (The fonnula is also valid for one-sided limits.)

Iflimx~a

In j(x) = L, then lim j(x) = lim elnf(x) = eL. x-+a

x-a

Here a may be either finite or infinite.

EXAMPLE 7

Apply I'Hllpital's Rule to show thatlimx~o+ (I

+ x) I/x

= e.

The limit leads to the indeterminate fonn 100. We let j(x) = (I find limx~o+ In j(x). Since

Solution

In j(x) = In (I

+ x)I/X

+ x)I/X

and

i In (1 + x),

=

I'Hllpital's Rule now applies to give

+ x)

lim In j(x) = lim In (I

x-o+

x-o+

X

0 0

I I +x Iim - x-o+ 1

=+=1. Therefore, lim (I x-+o+

EXAMPLE 8

+ x)l/x =

lim j(x)

x-+o+

=

lim elnf~)

x-o+

= el =

e.

•

Find limx~oo xl/x.

Solution

The limit leads to the indeterminate fonn 00°. We let j(x) = xl/x and f"md lim_oo In j(x). Since

~x,

Inj(x) = Inxl/x =

I 'Hllpital's Rule gives lim In j(x) = lim Inx x_OO x

x_OO

=

=

Therefore lim xl/x x-+ 00

=

lim j(x) x-+ 00

=

00

I/x lim-

x-+oo

¥

=

lim elnf(x) x- 00

00

1

o. =

eO

= I.

•

Proof of L'Hiipital's Rule The proof of I'H6pital's Rule is based on Caucby's Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Caucby's Theorem first and then show how it leads to I'H6pital's Rule.

7.5

Indeterminate Forms and rH6pitars Rule

401

HisTORICAL BIOGRAPHY

Augustin-Louis Cauchy (178!1-1857)

THEOREM 6-Cauchy's Mean Value Theorem Suppose functions f and g are continuous on [a, bl and differentiable throughout (a, b) and also suppose g'(x) oF 0 throughout (a, b). Then there exists a number c in (a, b) at which

f'(c) g'(c)

f(b) - f(a) g(b) - g(a)"

Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that g(a) oF g(b). For ifg(b) did equalg(a), then the Mean Value Theorem would give g'(c)

=

g(b) - g(a) b-a

=

0

for some c between a and b, which cannot happen because g'(x) oF 0 in (a, b). We next apply the Mean Value Theorem to the function

F(x)

=

f(b) - f(a) f(x) - f(a) - g(b) _ g(a) [g(x) - g(a)].

This function is continuous and differentiable where f and g are, and F(b) = F(a) = O. Therefore, there is a number c between a and b for which F' (c) = O. When expressed in terms of f and g, this equation becomes

F'(c)

=

f'(c) -

:~:~ =~~~ [g'(c)1 =

0

so that

f'(c) g'(c)

y j '(c) slope ~,«c) -

--.!

A (g(a).j(a)) ~--------------------~X

o

FIGURE 7.13 There is at least one point P on the curve C for which the slope of the tangent to the curve at P is the same as the slope of the secaot lioe joioing the poiots A(g(a), f(a)) aodB(g(b), f(b)).

f(b) - f(a) g(b) - g(a)"

•

Notice that the Mean Value Theorem in Section 4.2 is Theorem 6 with g(x) = x. Cauchy's Mean Value Theorem has a geometric interpretation for a general winding curve C in the plane joining the two points A = (g(a),/(a)) and B = (g(b), f(b)). In Chapter 11 you willleam how the curve C can be formulated so that there is at least one point P on the curve for which the tangent to the curve at P is parallel to the secant line joining the points A andB. The slope of that tangent line toms out to be the quotient f' /g' evaluated at the number c in the interval (a, b), as asserted in Theorem 6. Because the slope of the secant line joining A andB is

f(b) - f(a) g(b) - g(a)' the equation in Cauchy's Mean Value Theorem says that the slope of the tangent line equals the slope of the secant line. This geometric interpretation is shown in Figure 7.13. Notice from the figure that it is possible for more than one point on the curve C to have a tangent line that is parallel to the secant line joining A and B.

Proof of rHClpital's Rule We flfst establish the limit equation for the case x --> a +. The method needs almost no change to apply to x --> a -, and the combination of these two cases establishes the result. Suppose that x lies to the right of a. Then g'(x) oF 0, and we can apply Cauchy's Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that f'(c) g'(c)

f(x) - f(a) g(x) - g(a)"

402

Chapter 7: Transcendental Functions But f(a) = g(a) = 0, so

j'(c) g'(c)

f(x) g(x)"

As x approaches a, c approaches a because it always lies between a and x. Therefore, lim f(x) = lim j'(c) = lim j'(x)

.r-a+ g(x)

c-a+ g'{c)

.r-a+ g'(x) ,

which establishes I'HOpital's Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy's Mean Value Theorem to the closed interval [x, aj, x < a. •

Exercises 7.5 Finding Limits in Two Ways In Exercises l--{j, use I'HOpital's Ru1e to evaluare the limit. Then evaluare the limil using a method studied in Chapter 2.

. 3x - I 30. lun~1 x-o~

31. lim In(x + I)

33. lim

3. lim 5x' - 3x x_OO 7x 2 + 1

%-0+

35. lim

5. lim I - cosx x-o x 2

8.

t2

1--3

-

+

15 t - 12

11. lim 5x' - 2x x_OO 7x 3 + 3

,

.

13. lim sm 1 1-0

15.

t

8x '

lim=~.

x-o cosX'

lim 20 - 'If · '~w/' cos (2'1f 0)

17

19

lim

I - sinO • 6-1T/2 1 + cos 26

,

10. lim

' 3"'" - I lun 27. ,~o 0

41

r

+ I))

(lnx)'

x~+ In (sinx) lim (_1___1_) x-I Inx

• x-I+

I' - I

x-a

t- 3

71'X

In (esc x) x~w/' (x - ('If/2))'

(3X + I__smX' .1_) X

42. lim (cscx - calx x-o+

e' -

<ex - I)' .

lim 0 - sinO cosO

49

• "~O

lanO - 0

r

SO. x~

sin3x-3x+x 2 sin x sin 2x

Indeterminate Powers and Products Find the limits in Exercise 51--{j6. 51. lim x 1/(1-x) x-I+

53. lim (lnx)l/x x~oo

lim

+ cosx)

t

1 sin 1

x-(n/2)-

"~O

40. lim x-o+

+ ' 45. lim _et_1_

lim

28. lim

%-0+

.

20.lim

1_01 -

,a>O

x-o x smx

r 30 + 'If . '~'!!;/' sin (6 + ('If/3))

26.

y

38. lim (lnx - In sin x)

48. lun

xl

x-? x_tInx - SIn

~-a

43. lim cos 0 - I 8_0efJ - 8-1 1_00

18

(-SInt

(x - ~2) sec x

•

36. lim

y~O

16. lim sinX' - x

24. lim

lim

39

14. lim sin 5/ 1-0 21

/(1 - cos I) 23. lim .

x-(7r/2)-

25 5

12. lim x - 8x' x_OO 12x2 + 5x

22.

25.

+

1-14t 3 -

21 lim x • x~o In (sec x) 1-0

x' -

lim

x--5 x

9. lim I' - 41

y

X~OO

log, x

34. lim In (eX - I) x-o+ Inx

+ 25 - 5

37. lim (In 2x - In(x

Use I'H6pital's rule 10 fmd the limits in Exercises 7-50. 7.lim x - 2 x_2x2 - 4

Y5y

r

• x!'.?o log, (x + 3)

In(x' + 2x) In x

y~O

Applying l'H6pital's Rule

32

lo~x

x_OO

1. lim x+2 x_-2x2 - 4

-

cost

(~2 - x) Ian x

(1/2)" - I 6

57. lim - tanx

X

•• Estimate the value of

lim sin x olnx

%-0+

xl

oc - oc Form

X~O+

68. lim

x_OO ~

+ "- + sin bX) = O?

lim (tan 2x x-a x3

Theory and Applications I:H6pital's Rule does not help with the limits in Exercises 67-74. Try it-you just keep on eycling. Find the limits some other way. 67. lim

403

+

I)Yx + 2

1 hy graphing. Then confmn your estimate with I'H6pital's Rule. x-I

X

84. This exercise explores the difference between the limit

a. lim x-3 = lim 1...=1. x-3 x 2 - 3 x-3 2x 6 b. lim x-3 =Q=o %-3 x 2 - 3 6

and the limit

76. Which one is correct, and which one is wrong? Give reasons for

. (1+"I)X =e.

your answers.

lim

X~OO

2x - 2 a. lim x '-2x = l i m x_ox2 - sinx x-O 2x - cosx = lim x_o2 b.

2

+ sin x

•• Use I'H6pital's Rule to show that

lim x' - 2x = lim 2x - 2 x-O 2x cosx _ sinx

-2

.%_ox2

0 - 1 = 2

others wrong? Give reasons for your answers. a. lim xlnx = 0·(-00) = 0

x-o+ b. lim xlnx = 0'(-00) = -00 x-o+

lim Inx -00 c. x-'r!1l+ X In X = x~o+ (I/x) = """00 =

lim

2+0

77. Only one of these calculations is correct Which one? Why are the

d.

. (1+"I)X =e.

= _2_ = I

-

1

X~OO

D b. Graph

f(x)

=

(I +:S

lim (I

.... 00

a. f(x) = x,

g(x) = x',

(a,b) = (-2,0)

b. f(x) = x,

g(x)

(a, b) arbitrary

=

c. f(x) = x'/3 - 4x,

x',

g(x) = x',

(a, b) = (0,3)

79. Continuous OItension Find a value of c thst makes the function f(x) =

{

9X - 3sin3x, x # 0 5x'

c,

x= 0

continuous at x = O. Explain why your value of c works.

(I +~)'

85. Show thst

lim x In x = lim In/x X~O+ (I x)

78. Find all values of c thst satisfY the conclusion of Cauchy's Mean Value Theoreru for the given functions and interval.

=

together for x '" O. How does the behavior off compare with thst of g? Estimate the value oflim~oo f(x). c. Confmn your estimate oflimx->oo f(x) by calculating it with I'H6pital's Rule.

X~O+

= lim (I/x) = lim (-x) = 0 x-o+ (-1/x 2) x-o+

and g(x)

86. Given that x a. Xl/x

>

1")'

+k

=

e'.

0, fmd the maximum value, if any, of

b. xl/i'

c. xlix' (n a positive integer) d. Show that 1imx-oo xl/x· = 1 for every positive integer n. 87. Use limits to find horizontal asymptotes for each function.

•. y=xtan(~)

b. y

= { 0,

3x + e2x "'2x'--'+-'e~3x"

x # 0

e-11i'

88. Findf'(O) for f(x)

=

'

x

= O.

404

D 89.


D 90.

The continuous extension of(sinxf to [0, 'IT[

°

The function (sin x)... • (Continuation ofExercise 89.)

a. Graphf(x) ~ (sinxf on the interval :5 x:5 'IT. What value would you assign to f to make it continuous at x = O?

a. Graphf(x) ~ (sin x)"'" on the interva1-7 s x '" 7. How do you account for the gaps in the graph? How wide are the gaps?

b. VerifY your conclusion in part (a) by rmding lim,....o· f(x)

b. Now graph f on the interval :5 x '" 'IT. The function is not dermed at x ~ 'IT/2, hut the graph has no break at this point. What is going on? What value does the graph appear to give for f atx ~ 'IT/2? (Hint: Use I'H6pital~ Rule to rmd limf as x ..... ('IT/2)- and x ..... ('IT/2)+.)

°

with I'H6pital~ Rule. c. Returning to the graph, estimate the maximwn value off on [0, 'IT]. About where is max f taken on? d. Sharpen your estimate in part (c) by graphing f' in the same window to see where its graph crosses the x-axis. To simplifY your work, you might want to delete the esponential factor from the expression for f' and graph just the factor that has a zero.

Inverse Trigonometric Functions

7.6

Inverse trigonometric functions arise when we want to calculate angles from side measurements in triangles. They also provide useful antiderivatives and appear frequently in the solutions of differential equations. This section shows how these functions are derIDed, graphed, and evaluated, how their derivatives are computed, and why they appear as important antiderivatives.

y x

= siny

2) VDomain: 'IT'

.

Y = sin-Ix

-1';x,,1

Range: -wl2 :S y :S 'lT12

_~

c. Continuing with the graphs in part (b), rmd max f and min f as accurately as you can and estimate the values of x at which they are taken on.

1

x

2

FIGURE 7.14 The graph ofy ~ sin-1 x.

Defining the Inverses The six basic trigonometric functions are not one-ro-one (their values repeat periodically). However, we can restrict their domains to intervals on which they are one-ro-one. The sine function increases from -I at x = -1f/2 to + I at x = 1f/2. By restricting its domain to the interval [-1f/2, 1f/2], we make it on.,.to-one, so that it has an inverse sin- 1 x (Figure 7.14). Similar domain restrictions can be applied to all six trigonometric functions.

Domain restrictions that make the trigonometric functions one-to-one Y

Y

sin x

= sinx Domain: [-'IT/2, 'IT/2] Range: [-1,1]

y

= cosx Domain: [0, 'IT] Range: [-1, 1]

y

Y

2

i( ~ cotx Domain: (0, 'IT) Range: (-00, 00)

= secx Domain: [0, 'IT/2) U ('IT/2, 'IT] Range: (-00, -1] U [I, 00)

y

~ tao x Domain: (-'IT/2, 'IT/2) Range: (-00, 00)

y

sec x

-o;;t---,,;;;---;;,,~x

y

tan x

----;;,,;--c;cot---c,,;;-~x

-~

2

= cscx Domain: [-'IT/2, 0) U (0, 'IT/2] Range: (-00, -1] U [1, 00)

y

405

7.6 Inverse Trigonometric Functions

I

Since these restricted functions are now one-to-one, they have inverses, which we denote by

The "Arc" in Arcsine and Arccosine The accompanying figure gives a

geometric interpretation ofy = sin-I x andy = COS-IX for radian angles in the frrst quadrant. For a unit circle, the = r8 becomes s = (J, so cen1ral angles and the arcs they subtend have the same measure. If x = sin y, then, in addition to being the angle whose sine is.:t, y is also the length of arc on the unit circle that subtends an angle whose sine is x. So we call y ''the arc

equation s

whose sine is .:t,"

y = sin-Ix

or

y = cos- t x

or

y

= tan-I

x

or

y

= coC I X

or

sec- 1 x

or

Y = arccotx y = arcsecx

y = csc- I x

or

y

y =

Y = arcsinx Y = arccosx y = arctanx

= arccscx

These equations are read "y equals the arcsine of x" or "y equals arcsin x' and so on. Caution The - 1 in the expressions for the inverse means ''inverse.'' It does not mean reciprocal. For example, the reciprocal of sin x is (sinx)-I = l/sinx = cscx.

y

x2

+ y2 = 1

Arc whose sine is x

~-+--

Angle whose sine is x --~------~~---L~--t---~x

o

x

1

The graphs of the six inverse trigonometric functions are shown in Figure 7.15. We can obtain these graphs by reflecting the graphs of the restricted trigonometric functions through the line y = x, as in Section 7.1. We now take a closer look at these functions and their derivatives. Domain: -1 =S x =S 1 Range' _1!:s yS 'I!

Domain: -1:S X:S 1 Range: 0 :S Y :S '1r

y

y

.

2

Angle whose

2

Domain: -00 < x < Range:

OQ

-I I; y(2)

_ I_ _ 2 I+x' ~'

~

y(O)

~

'If

2

AppUcations and Theory 107. You are sitting in a classroom next to the wall looking at the blackboard at 1he froot of 1he room. TIre blackboard is 12 ft loog aod starts 3 ft from 1he wall you are sitting oext to.

a. Show 1hat your viewing angle is a = coC1 -.-£ - coCt !. IS 3 if you are x ft from 1he froot wall.

b. Find x so 1hat a is as large as possible.

T11 0

12'

~

1 3'

"-

I' L'Hop;tal's Rule

x-a

93. lim r~""

x

x

xtan-I~

. tan-I x 2 95. lun--x-O x sin- 1 X

92. lim x-l+

w-=-l sec 1 x

x

'I

erate a solid. Find 1he volume of 1he solid. y

94. lim 2 tan-I 3x'

7x 2

x-a 96. lim eX tan-I eX x_OO e2%

98.

Wall

108. The region between 1he curve y ~ sec-I x and the x-axis from x ~ I to x ~ 2 (shown here) is revolved about 1hey-axis to geo-

Find 1he limits in Exercises 91-9S. . -IS

91. lim 8m

You

a

+x

. -1 2

x x-o+(sin lx)2 lim

sm

3"

~t---~--~~x

o

2

7.6 Inverse Trigonometric Functions 109. The slant height of the cone shown here is 3 m. How large should the indicated angle be to maximize the cone's volume?

415

+ tan- I (l/x) is constant.

114. Show that the sum tan- I x 115. Use the identity

csc- Iu = ~ - sec- Iu 2 to derive the formula for the derivative of csc- I u in Table 7.3 from the formula for the derivative of sec- I u . 116. Derive the formula dy

dx

1

+ x2

for the derivative of y = tan- I x by differentiating both sides of the equivalent equation tan y = x.

110. Find the angle a .

117. Use the Derivative Rule, Theorem 1, to derive

d - I dx sec x

1

-lx-I-V---=x=2=-=I'

=

> 1.

Ixl

118. Use the identity 111. Here is an informal proof that tan- II Explain what is going on.

+ tan- I 2 + tan- I 3

coc l =

U

=

1T.

~ - tan- I u 2

to derive the formula for the derivative of coC I u in Table 7.3 from the formula for the derivative of tan- I u. 119. What is special about the functions

f( ) - . - I x-I x - sm x + l'

x;:: 0,

and

g(x) = 2 tan- I\IX?

Explain. 120. What is special about the functions

f(x)

=

sin- I

1

W+l

and

g(x)

=

tan- I}?

Explain. 121. Find the volume of the solid of revolution shown here. 112. Two derivations ofthe identity sec- 1 (-x) =

sec- 1 x a. (Geometric) Here is a pictorial proof that sec- I (-x) = 1T - sec- I x . See if you can tell what is going on. 7T -

y

:=17--

17

-V3 -3-

y

\/ ~~\--l / 1

__ -1 ______ _ 1 7T

y= _ _ 1_

~ V3

7------+--4--x

"2

1

_ _- L _ _- L _ _ _ _~--J---J-_+x

-x

-1

o

x

b. (Algebraic) Derive the identity sec- I (-x) = 1T - sec- I x by combining the following two equations from the text: cos- I x I sec- Ix = cos- (l/x)

cos- I (-x) =

1T -

122. Arc length Find the circumference of a circle of radius r using Eq. (3) in Section 6.3. Find the volumes of the solids in Exercises 123 and 124.

Eq. (3) Eq. (5)

113. The identity sin- 1 x + cos- 1 X = 7T /2 Figure 7.21 establishes the identity for 0 < x < 1. To establish it for the rest of [ - 1, 1], verify by direct calculation that it holds for x = 1, 0, and - 1 . Then, for values of x in (- 1, 0), let x = - a, a > 0, and apply Eqs. (1) and (3) to the sum sin- I( -a) + cos- I( -a).

123. The solid lies between planes perpendicular to the x-axis at x = - I and x = I . The cross-sections perpendicular to the x-axis are a. circles whose diameters stretch from the curve y =

-I/~tothecurvey

=

I/~ .

b. vertical s uares whose base edges run from the curve y = -1/ I + x 2 to the curvey = I/~ .

416


124. The solid lies between ~lanes perpendicular to the x-axis at x ~ - Yz/2 and x ~ Yz/2. The cross-sections perpendicular

to the x-axis are

D 125. Find the values of the following. c. coCl 2

D 126. Find the values of the following. •• sec- I( -3)

b. csc- I 1.7

C.

coCI (-2)

D 10 Exercises 127-129, fmd the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. 127••• Y ~ tan-I (tanx) b. y ~ tan (tan- I x) 128••• Y ~ sin-I (sin x)

7.7

D Use your graphing utility for Exercises 13()-134. sec.

b. squares whose diagonals stretch from the x-axis to the curve y ~ 2/~I-x'. b. csc- I (-1.5)

b. Y ~ cos (cos-I x)

130. Graph y ~ sec (sec-I x) ~ sec (cos-I(1/x)). Explain what you

a. circles whose diameters stretch from the x-axis to the curve y ~ 2/~I-x'.

•• sec-I 1.5

129... Y ~ cos-I (cos x)

131. Newton'. .erpentine Graph y ~ 4x/(x' + I), knowo as Newton's serpentine. Then graph y ~ 2 sin (2 tan-I x) in the same graphing window. What do you see? Explain. 132. Graph the rational function y ~ (2 - x')/x'. Then graph y ~ cos (2 sec-I x) in tire same graphing window. What do you see? Explain. 133. Graph f(x) ~ sin-I x together with its first two derivatives. Comment on the behavior of f and the shape of its graph in relation to the signs and values of f' and

r.

134. Graph f(x) ~ tan-I x together with its frrst two derivatives. Comment on the behavior of f and the shape of its graph in relation to the signs and values of f' and

r.

b. Y ~ sin (sin- I x)

Hyperbolic Functions The hyperbolic functions are formed by taking combinations of the two exponential functions eX and e -x. The hyperbolic functions simplify many mathematical expressions and occur frequently in mathematical applications. In this section we give a brief introduction to these functions, their graphs, and their derivatives.

Definitions and Identities The hyperbolic sine and hyperbolic cosine functions are defined by the equations

.

,ex~-c'e,--x_

sinhx ~ -

2

coshx ~

and

eX

+ e-x 2

.

We pronounce sinh x as "cinch x," rhyming with ''pinch x," and cosh x as "kosh x:' rhyming with "gosh x." From this basic pair, we defme the hyperbolic tangent, cotangent, secant, and cosecant functions. The defining equations and graphs of these functions are shown in Table 7.5. We will see that the hyperbolic functions bear many similarities to the trigonometric functions after which they are named. Hyperbolic functions satisfy the identities in Table 7.6. Except for differences in sign, these resemble identities we know for the trigonometric functions. The identities are proved directly from the defmitions, as we show here for the second one: 2 sinh x cosh x = 2 ( eX -2 e2:x -

e-X) (eX +2 e-X)

e-2:x

2 = sinh2x.

The other identities are obtained similarly, by substitoting in the defmitions of the hyperbolic functions and using algebra. Like many standard functions, hyperbolic functions and their inverses are easily evaluated with calculators, which often have special keys for that purpose.

7.7

HyperboLic Functions

417

The six basic hyperboLic functions

TABLE 7.5

y

Yy

=

coshx

eX 3

Y

= 2 2 Il ~ L "

Y

=

sinhx

-3-2-1

1 2 3

(a)

(b)

Hyperbolic sine: .

sinhx

(c)

Hyperbolic tangent:

Hyperbolic cosine:

eX - e-x

= "'---020"'--

coshx

+ e-x

eX

=

tanh x

2

= _sinh __ x = cosh x

.. e~x~-...'e'C~~ eX + e-x

Hyperbolic cotangent:

cosh x

cothx

y

Identities for hyperbolic functions

TABLE 7.6

cosh2 X - sinh2 X = I sinh 2x = 2 sinh x cosh x cosh 2x = cosh2 X + sinh2 X cosh2 X = cosh 2x 2

+

y=sechx (e)

(d)

Hyperbolic secant:

sechx = _1_ = 2 cosh x eX + e-x

Hyperbolic cosecant:

cschx = _._1_ = sinhx

Derivatives of hyperboLic functions

TABLE 7.7

:fx (sinh u) =

cosh u::

:fx (cosh u) =

sinh u::

JL (coth u) dx

2

eX - e-x

For any rea! number u, we know the point with coordinates (cos u, sin u) lies on the unit circle x 2 + y2 = 1. So the trigonometric functions are sometimes called the circular functions. Because of the first ideotity sinh2 U =

COSh2 U -

sech2

eX + e-x eX _ e-x

I

sinh2 X = cosh 2x - I 2 2 tanh X = I - sech2 X coth2 X = I + csch2 X

:fx (tanh u) =

= sinhx =

with u substituted for x in Table 7.6, the point having coordinates (cosh u, sinh u) lies on the right-hand branch of the hyperbola x 2 - y2 = 1. This is where the hyperbolic functions get their names (see Exercise 86).

Derivatives and Integrals of Hyperbolic Functions The six hyperbolic functions, being mtiona! combinations of the differentiable functions

u::

= -csch2 U

1,

du dx

d du dx (sech u) = -sech u tanh u dx

eX and e""", have derivatives at every point at which they are defined (Table 7.7). Again,

there are similarities with trigonometric functions. The derivative formulas are derived from the derivative of e U :

:fx (sinhu) = :fx (e

U -;

e U du/dx

e~)

+ e-u du/dx 2

dcsch () dx u = -csch u coth u •dx = coshu

du dx

Def'mition of sinh u

Derivative of e" Defmition of cosh u

418


IntegraL formulas for hyperboLic functions

TABLE 7.8

J J J J J J

coshu

+C

coshudu = sinhu

+C

sinhudu

=

sech'udu

= tanhu + C

csch'udu

= -cothu + C

sechutanhudu = -sechu

TIris gives the first derivative formula. From the definitioo, we can calculate the derivative of the hyperbolic cosecant functioo, as follows:

~(CSChU) = ~ (s~U)

Def'mition of csch u

coshu du sinh'u dx

Quotient Rule

I coshu du sinh u sinh u dx

Rearrange terms.

----

------

= -csch u coth u

+C

cschucothudu = -cschu + C

du dx

Def'mitions of csch u and coth u

The other formulas in Table 7.7 are obtained similarly. The derivative formulas lead to the integral formulas in Table 7.8.

EXAMPLE 1 (a)

:1 (tanh v'l+7)

=

(b) J

coth5x dx

t Jo

sinh5x

sinh'xdx = = =

(d)

u = sinh5x, du = 5 cosh5xrh

=JCOsh5Xdx=lJdU =

(c)

v'l+7. ~ (v'l+7) 1 sech' v'l+7 v'l+7

= sech'

tlo

lui

5

+C

{ ' cosh2x -

Jo

2Jo

{In' 4ex sinh x dx =

Jo

=

tlo I

sinh 5xl

=

+C

I dx

Table 7.6

2

1 {' (cosh2x

su:

u

2 -

~

_ 1) dx =

1 [Sinh2x 2

- xl ' 0

Evaluate with a calculator.

,., 0.40672

{In' 4ex eX - e .... dx = 2

Jo

[e 2x -

2

{In' (2e 2x

Jo

_

2) dx

2xl~' = (e'In' - 2102) - (I - 0)

= 4 - 2102 -

I,., 1.6137

•

Inverse Hyperbolic Functions The inverses of the six basic hyperbolic functioos are very useful in integratioo (see Chapter 8). Since d(sinhx)/dx = coshx > 0, the hyperbolic sine is an increasing function of x. We deoote its inverse by

y = sinh- 1 x. For every value of x in the interval - 00 < x < 00, the value of y = sinh-1 x is the number whose hyperbolic sine is x. The graphs of y = sinh x and y = sinh-1 x are shown in Figure 7.25a. The functioo y = coshx is not ooe-to-ooe because its graph in Table 7.5 does not pass the horizootal line test. The restricted function y = cosh x, x ;" 0, however, is oneto-ooe and therefore has an inverse, deooted by

y = cosh- 1 x.

7.7

419

Hyperbolic Functions

For every value of x ;", I, Y = cosh-I x is the number in the interval 0 ,,;; y < 00 whose hyperbolic cosine is x. The graphs of y = cosh x, x ;", 0, and y = cosh-I x are shown in Figure 7.25b. Y

y

y=siDhx y=x

Y = coshx, X2!=O y=x

y

/

/

/

// Y

=

sinh-I X

3

(x = sinhy)

/

o

=

/

/

/

/

/

y

/

(x ~ coshy,y" 0) x

/ /

/ /

cosh-1 X

/ / / /

y '" 0)

2

/

y

y=x

y = sech-1 x (x = seehy.

/

/

=

sechx

X2:0

~~+-~~~--> x

o

123 4 5 6 7 8

2

3

(c)

(b)

(a)

FIGURE 7.25 The graphs of the inverse hyperbolic sine, cosine, and secant ofx. Notice the symmetries about Ibe line y = x.

Like Y = coshx, the function y = sechx = I/coshx fails to be one-to-one, but its restriction to nonnegative values of x does have an inverse, denoted by y = sech- I x. For every value of x in the interval (0, II, y = sech- I x is the nonnegative number whose hyperbolic secant is x. The graphs of y = sech x, x ;", 0, and y = sech-I x are shown in Figure 7.25c. The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains and therefore have inverses, denoted by y = tanh-I x, y = coth- I x, y = csch- I X. These functions are graphed in Figure 7.26. y

y ,

,

x=cothy i y = coth-Ix :

x= ltanhy

Y = :tanh-1x

, , , ,

, , , ,

x

-I , , , , , , ,

(a)

TABLE 7.9

Identities for inverse hyperbolic functions

csch- I X = sinh-II x

FIGURE 7.26

1

y

il

x = cschy

y

,

0

(b)

x

=

esch-Ix

~

X

(c)

The graphs oflbe inverse byperbolic tangent, cotangent, and cosecant ofx.

Useful Identities We use the identities in Table 7.9 to calculate the values ofsech- I x, csch- I x, and coth- I x on calculators that give only cosh-I x, sinh-I x, and tanh-I x. These identities are direct consequences of the defmitions. For example, if 0 < x ,,;; I, then

420

Chapter 7: Transcendental Functions We also know that sech (sech- 1 x) = x, so because the hyperbolic secant is one-to-one on (0, I], we have

Derivatives of Inverse Hyperbolic Functions An important use of inverse hyperbolic functions lies in antiderivatives that reverse the derivative formulas in Table 7.10. TABLE 7.10

d(sinh- 1 u)

dx d(cosh- 1 u)

dx d(tanh- 1 u)

Derivatives of inverse hyperbolic functions I du ~dx I du w-=-ldx'

u

I du 1-u2dx '

lui
1 by applying Theorem 1 with f(x) = coshx andr ' (x) = cosh- 1 x. Theorem I can be applied because the derivative of cosh x is positive for 0 < x.

(r' )'(x) = !'u I, (x)) 1 sinh (cosh

Theorem 1

1 x)

1 Ycosh2(cosh 1 x) - I

1

1'(u)

~

cosb2 U sinhu

=

sinh u -

sinh2 U

=

I,

v'cosh2 u - 1

421


7.7 The Chain Rule gives the final result:


Sonya Kova1evsky (!8S()-1891)

d ( sh- 1 ) _ dx co u -

du

1

•

~dx'

With appropriate substitutions, the derivative formulas in Table 7.10 lead tu the integration formulas in Table 7.11. Each of the formulas in Table 7.11 can be verified by differentiating the expression on the right-hand side. Integrals leading to inverse hyperbolic functions

TABLE 7.11

1.

1 1V2

du 2

Va + u

2.

3.

5.

a>O

du = sh-1 (1l)+c a ' u - a 2 co

1 1V 1 V2 2

-

a

2

u =

a

u2

'

< a2

u2 > a2

(*) +

u

du =_1 a sech-l(Il)+C a ' a2 - u 2

O

Integration Fonnulas

1

VetifY the integration formulas in Exercises 37-40. 37.

38. 39. 40.

J J J J J L

sechx dt = tan-I (sinh x)

+C

b.

sechx dt = sin- l (tanh x)

+C

xsech-lxdt xcoth-lxdt tanh-lxdt

Use the formulas in the box here to express the ntanhers in Exercises 61-{i6 in tenns ofnaturallogatithms. 61. sinh- l (-5/12) 62. cosh- l (5/3)

~ ~ sech-lx - k~ + C ~ x' ~

I coth-lx

+~+C

~ X tanh-I x + kln(1

- x')

+C

63. tanh-I (-1/2)

64. coth- l (5/4)

65. sech-l (3/5) Evaluate the integrals in Exercises 67-74 in tenns of

a. inverse hyperbolic functions. b. natural logarithms.

7.7

67.

69

•

J.

'v3

v4 +

o

l'

68.

----""-

dx

1 --:-7'==='" l/S

73. J.~

xVI -

80. Volume

6 dx

J. VI + 9x' 12 70. J.1 ~ o 1- x

x'

5/41-x 2 3/13

71.

113

dx

.~

16>;'

cos x dx

•

l' 1

xVI

The region enclosed by the curve y = sech x, the

V3

81. An: length

Find the lengih of the graph of y = (1/2) cosh 2x

fromx = Otox = In

Vs.

82. Use the definitions of the hyperbolic funetions to fmd each of the following lintits.

1x~

74

(a) lim tanh

dx

lim tanhx

x~oo

+ (lnx)'

(c) lim sinhx

(d)

(e) lim sechx

(f) lim cotbx

x~oo

Appllcattons and Examples 75. Show that if a function f is defmed on an interval symme1ric ahout the origin (so that f is defmed at -x whenever it is defined at x), then

_ f(x) + f( -x) ( ) fx 2

+

f(x) - f( -x) 2

423

x-axis, and the lines x = ± In is revolved about the x-axis to generate a solid. Find the volume of the solid.

0

72·1'-,=dx===

o VI + sin'x


(1)

+ f( -x))/2 is even and that (f(x) - f( -x))/2 is odd.

Then show that (f(x)

W+i)

76. Derive the fornrula sinh- 1 x = In (x + for all real x. Explain in your derivation why the plus sign is used with the square root instead of the minus sign. 77. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body's velocity I sec into the fall satisfies the differential equation

x~oo

lim sinhx

x_-oo x~oo

(g) lim coth x

x-o+

(i)

lim csch x

x_-oo

83. Hanging cable. ln3agine a cahle, like a telephone line or TV cable, strung from one support to another and hanging freely. The cahle's weight per unit lengih is a constant w and the horizontal tension at its lcmest point is a vector of length H. Ifwe choose a c0ordinate system for the plane of the cahle in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cahle lies at the point y = H/w on the y-asis (see accompanying fignre), then it can be shown that the cahle lies along the graph of the hypetbolic cosine

H

w

y = wcoshJjx.

dv , mdt=mg-kv, where k is a constant that depends on the body's aerodyuamic properties and the density of the air. CWe assume that the fall is short enough so that the variation in the air's density will not affect the outconre significantly.)

a. Showthat

satisfies the differential equation and the initial condition that v = o when I = O. b. Find the body's limiting velocity, Jim,....oo v. c. For a 160-lb skydiver (mg = 160), with time in seconds and distance in feet, a typical value for k is 0.005. What is the diver~ limiting velocity'l

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning "chain." a. LetP(x,y) denote an arbitrary point on the cable. The next accompanying fignre displays the tension at P as a vector of lengih (magnitude) T, as well as the tension H at the lowest pnintA. Show that the cable's slope atP is

dy . w tan = dx = sinhJjx.

78. Accelerations whose magnitndes IIl"O proportionlll to displacement Suppose that the position of a body moving along a coordinate line at time t is

y

y= l1cosh~x w H

a. s = acoskt + bsinkt. b. s = a coshkl + b sinhkl. Show in hoth cases that the acceleration d's/ dl' is proportional to s bot that in the frrst case it is directed toward the origin, whereas in the second case it is directed away from the origin. 79. Volume A region in the flIS! quadrant is hounded shove by the curve y = cosh x, below by the curve y = sinh x , and on the left and right by the y-axis and the line x = 2, respectively. Find the volume of the solid generated by revolving the region ahout the x-axis.

P(X,y)~.....

TcostfJ H

o

424


b. Using the result from part (a) aod the fact that the horizontal tension at P must equal H (the cable is not moving), show that

T = "y. Hence, the magnitude of the tension atP(x,y) is exactly equal to the weight of y units of cable. 84. (Continuation of Exercise 83.) TIre length of arc AP in the Exercise 83 figure is s = (1/a) sinh ax, where a = wlH. Show that the coordinates ofP may be expressed in terms of s as

x=

~Sinh-l as,

y =

a. Show that the areaA(u) of sector AOP is A(u) = tCOshuSinhu

~S2 + :2"

85. Area Show that the area of the region in the fIrSt quadrant enclosed by the curve y = (1/a) cosh ax, the coordinate axes, and the line x = b is the sarae as the area of a rectangle of height 1/a and length s, where s is the length of the curve from x = 0 to x = b. Draw a figure illustrating this result. 86. The hyperbolic in hyperbolic functions Just as x = cos u and y = sin u are identified with points (x, y) on the unit circle, the functions x = cosh u and y = sinh u are identified with points (x, y) on the right-hand branch of the unit hyperbola, X2 _

Another artalogy between hyperbolic and circular functions is that the variable u in the coordinates (cosh u, sinh u) for the points of the right-hand branch of the hyperbola x 2 - y2 = I is twice the area of the sector AOP pictured in the accompanying figure. To see why this is so, carry out the followiog steps.

-1°O""~ fix.

b. Differentiate both sides of the equation in part (a) with respect to u to show that A'(u) =

t.

c. Solve this last equation for A(u). What is the value of A(O)? What is the value of the constant of integration C in your solution? With C determined, what does your solution say about the relationship of u to A(u)? y

y 2=1. y

y x2

+

r =1

P(cos U, sin u)

"...--,-.

~

------t-~~------~x

Since cosh2 U - sinh2 U = I, the point (cosh u, sinh u) lies on the right-hand branch of the hyperbola x 2 - y2 = I for every value of u (Exercise 86).

7.8

One of the artalogies between hyperbolic and circular functions is revealed by these two diagrams (Exercise 86).

Relative Rates of Growth It is often important in mathematics, computer science, and engineering to compare the rates at which functions of x grow as x becomes large. Exponential functions are important in these comparisons because of their very fast growth, and logarithmic functions because of their very slow growth. In this section we introduce the liUle-oh and big-oh notation used to describe the results of these comparisons. We restrict our attention to functions whose values eventoa1ly become and remain positive as x ..... 00 .

y

160 140 120 100

Growth Rates of Functions

80 60 40 20

-d~~~~~~--x o 1 2 3 4 5 6 7 FIGURE 7.27 andx2 .

The graphs of eX, 2x,

You may have noticed that exponential functions like 2" and eX seem to grow more rapidly as x gets large than do polynomials and rational functions. These exponentials certaiuly grow more rapidly than x itself, and you can see 2" outgrowing x 2 as x increases in Figure 7.27. In fact, as x ..... 00, the functions 2X and eX grow faster than any power of x, even x',ooo,ooo (Exercise 19). In contrast, logarithmic functions like y = log2x and y = lnx grow more slowly as x ..... 00 than any positive power of x (Exercise 21). To get a feeling fur how rapidly the values of y = eX grow with increasing x, think of graphing the function on a large blackboard, with the axes scaled in centimeters. At x = I cm,

7.8

425

the graph is e ' '" 3 em above the x-axis. At x = 6 em, the graph is e 6 '" 403 em '" 4 m high (it is about to go through the ceiliog if it hasn't done so already). At x = 10 em, the graph is e 'O '" 22,026 em '" 220 m high, higher than most buildings. At x = 24 cm, the graph is more than halfway to the moon, and at x = 43 em from the origin, the graph is high enough to reach past the sun's closest stellar neighbor, the red dwarf star Proxima Centauri. By contrast, with axes scaled in centimeters, you have to go nearly 5 light-years out on the x-axis to find a point where the graph of y = lnx is even y = 43 em high. See Figure 7.28. These important comparisons of exponential, polynomial, and logarithmic functions can be made precise by defining what it means for a function I(x) to grow faster than another function g(x) as x --+ 00 .

y

y

Relative Rates of Growth

= eX

70 60 50 40 30 20

10 0

Y = lnx 10

20

30

40

50

60

FIGURE 7.28 Scale drawings of the graphs of e' and \0 x.

x

Rates of Growth as x --+

DEFINmON

QC)

Let I(x) andg(x) be positive for x sufficiently large.

1.

1 grows faster than g as x --+ 00 if lim I(x) =

00

x_oo g(x)

or, equivalently, if

.

g(x)

lim I( X ) = O.

x--+OO

2.

We also say that g grows slower than 1 as x --+ 00 . 1 and g grow at the same rate as x --+ 00 if lim I(x) = L x_oo

g(x)

where L is finite and positive.

According to these defmitions, y = 2x does not grow faster than y = x. The two functions grow at the same rate because lim

2x

x--+OO X

=1im2=2 x--+ OO

'

which is a finite, positive limit. The reason for this departure from more common usage is that we want "I grows faster than Ft' to mean that for large x-values g is negligible when compared with f.

EXAMPLE 1

Let's compare the growth rates of several common functions.

(a) eX grows faster than x 2 as x --+

e'

Iim -2 x

,.--+00

lim eX 2x

=

x-co

00/00 (b) 3' grows faster than

00

because

lim eX = 2

00.

x_OO

Using I'H6pital's Rule twice

00/00 2x

as x --+

00

because

. 3x

. (3)X -2 =

Inn 2x = lim

x--i>OO

x--i>OO

00.

426

Chapter 7: Transcendental Functions (e) x 2 grows faster than In x as x --->

00

because

2

lim x = lim k = lim 2>;2 = x---+OO Inx x--+OO l/x x--+OO (d) In x grows slower than x '/n as X --->

x_oo

j'H6pital's Rule

for any positive integer n because

00

lim Inx = lim x_oo xl/n

00

I/x (I/n)x(l/n)-'

j'H6pital's Rule

= lim --"-= O. x_ co x l / 1I

11

is constant.

(e) As Part (b) suggests, exponential functions with different bases never grow at the same rate as X---> 00. If a > b > 0, then a' grows faster than b'. Since (a/b) > I, lim

x_OO

ba: =

(-ba )' =

lim

x_OO

00.

>

(f) In contrast to exponential functions, logarithmic functions with different bases a and b > I always grow at the same rate as x ---> 00 :

I

. log" x . In x/In a Inb hm--=hm =10llbX x_OO Inx/lnb Ina'

x_OO

•

The limiting ratio is always finite and never zero.

x --->

If 1 grows at the same rate as g as x ---> 00, and g grows at the same rate as h as 00, then 1 grows at the same rate as h as x ---> 00. The reason is that lim _gl = L,

x_OO

and

together imply

If L, andL 2 are fmite and nonzero, then so is L,L2 • Show that ~ and (2Vx

EXAMPLE 2

-

1)2 grow at the same rate asx--->

00.

Solution We show that the functions grow at the same rate by showing that they both grow at the same rate as the function g(x) = x:

lim

x_OO

lim

~= lim~l+ x52 =1 ' x x_OO

(2Vx -

x-oo

1)2 = lim

x

x-oo

(2Vx Vi

1)2 = lim x-oo

(2 __1_)2 = Vi

4.

•

Order and Oh-Notation The "little-oh" and ''big-oh'' notation was invented by number theorists a hundred years ago and is now commonplace in mathematical analysis and computer science.

DEFINITION

A function

1

is of smaller order than g as x --->

lim g(/(X» = O. We indicate this by writing f

x-OO

x

00

= o(g) ("I is little-oh of g'').

if

7.8

Notice that saying f

=

o(g) as x --->

00

Relative Rates of Growth

427

is another way to say that I grows slower than gas

x~oo.

EXAMPLE 3 (a) lox

=

Here we use little-oh notation.

o(x) as x --->

(b) x 2 = 0(x 3

00

because

+ I) as x ---> 00 because

2

lim _x_= 0

x_ oo x 3

+ 1

•

DEFINmON Let I(x) and g(x) be positive for x sufficiently large. Then I is of at most the order of g as x --+ 00 if there is a positive integer M for which I(x) < M g(x) ,

for x sufficiently large. We indicate this by writing /

EXAMPLE 4

= O(g) ("I is big-oh of g").

Here we use big-oh notation.

(a) x

+ sinx

(b) eX

+ x2

=

=

O(x) as x ---> O( eX) as x --+

00

00

because because

x + sinx . x ,;; 2 for x sufficiently large. eX

+ x2 e

x

~lasx--+oo

X. ---> 0 as x ---> 00 . • e If you look at the defmitions again, you will see that I = o(g) implies 1= O(g) forfunctions that are positive for x sufficiently large. Also, if I and g grow at the same rete, then I = O(g) and g = OU) (Exercise ll). (c) x = O( eX) as x --->

00

because

Sequential vs. Binary Search Computer scientists often measure the efficiency of an algorithm by counting the number of steps a computer must take to execute the algorithm. There can be significant differences in how efficiently algorithms perform, even if they are designed to accomplish the saroe task. These differences are often described in big-oh notation. Here is an exarople. Websters International Dictionary lists about 26,000 words that begin with the letter a. One way to look up a word, or to learn if it is not there, is to read through the list one word at a time until you either fmd the word or determine that it is not there. This method, called sequential search, makes no particular use of the words' alphabetical armngement. You are sure to get an answer, but it might take 26,000 steps. Another way to find the word or to learn it is not there is to go straight to the middle of the list (give or take a few words). If you do not find the word, then go to the middle of the half that contains it and forget about the half that does not. (You know which half contains it because you know the list is ordered alphabetically.) This method, called a binary search, eliminates roughly 13,000 words in a single step. If you do not find the word on the second try, then jump to the middle of the half that contains it. Continue this way until you have either found the word or divided the list in half so many times there are no words left. How many times do you have to divide the list to find the word or learn that it is not there? At most 15, because (26,000/2 15 ) < 1. That certainly beats a possible 26,000 steps.

428

Chapter 7: Transcendental Functions For a list of length n, a sequential search algorithm takes on the order of n steps to find a word or determine that it is not in the list. A binary search, as the second algorithm is called, takes on the order of 10(U n steps. The reason is that if 2m - 1 < n ,;; 2m , then m - I < 10(U n ,;; m, and the number of bisections required to narrow the list to one word will be at most m = 10(U n the integer ceiling for 10(U n. Big-oh notation provides a compact way to say all this. The number of steps in a sequential search of an ordered list is O(n); the number of steps in a binary search is O(iog2 n). In OUI example, there is a big difference between the two (26,000 ys. 15), and the difference can only increase with n because n grows faster than log2 n as n --> 00.

r

1,

Exercises 7.8 Comparisons with the Exponential e' 1. Which of the following functions grow faster thao eX as x ..... 00 ? Which grow at the same rate as eX? Which grow slower?

3

b. x 3

c. Yx e. (3/2)'

d.4x

LX -

g. e

X

f.

/2

+ 8in2 x

ex/2

h. 10glOX

2. Which of the following functions grow faster thao eX as x ..... 00 ? Which grow at the same rate as eX? Which grow slower?

+ 30x +

L

lOx'

c.

vT+7

I

b. x Inx - x

6. Which of the following functions grow faster than In x as x .... oo? Which grow at the same rate as In x? Which grow slower?

a. logz (x 2 )

b. 10glO lOx

c. I/Yx e. x - 2Inx

d. l/x 2 f. e-x

g. In (In x)

b. In (2x

Ordering Functions by Growth Rates 7. Order the following functions from slowest growing to fastest growing as x ---+

d. (S/2)'

+ 5)

a.

00.

OX

c. (lnx)'

Comparisons with the Power i' 3. Which of the following functions grow faster than x 2 as x ..... 00 ? Which grow at the same rate as x 2 ? Which grow slower?

8. Order the following functions from slowest growing to fastest growing as x ---+ 00. a. 2% b. x 2 d.

c. (ln2)'

OX

Big-oh and Litt1e-oh; Order 9. True, or false? As x ---+

e. xInx g.

x 3e-x

X

f. 2 h.

&x 2 2

4. Which of the following functions grow faster than x as x ..... 00 ? Which grow at the same rate as x 2 ? Which grow slower? L

x2

+ Yx

b. IOx 2

d. 10glO (x 2 ) f. (1/10)'

g. (1.1)'

h. x 2

+ IOOx

a. x = o(x) c. x = O(x + 5) e. 0% = 0(02%)

b.x=o(x+S)

g. Inx = 0(1n 2x)

b.

10. True, or false? As x ---+

x! = o(}) c. } - ;tI = o(})

a.

e. eX

5. Which of the following functions grow faster thao In x as x ---+ oo? Which grow at the same rate as In x? Which grow slower? L

log,x

b. In 2x

c. InYx

d. Yx

e. x

f. Slnx

g.

l/x

h. eX

d. x = 0(2x) f. x

+

Inx = O(x)

W+5 =

O(x)

00,

3

2

Comparisons with the Logarithm In x

00,

+x

=

O(e X)

g. In (In x) = O(lnx)

d. 2

+ cosx

= 0(2)

f. x Inx = 0(x 2)

b. In (x) = 0(1n (x 2 + I))

11. Show that ifpositive functions f(x) and g(x) grow at the same rate as x"" 00, thenf= O(g) andg = O(f). 12. When is a polynomial f(x) of smaller order thao a polynomial g(x) as x ---+ oo? Give reasons for your answer. 13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x ---+ oo? Give reasons for your answer.

Chapter 7 Questions to Guide Your Review 14. What do the conclusions we drew in Section 2.4 about the limits of rational functions tell us about the relative growth of polynomi· a1sasx~ oo?

429

Xl/l,ooo,ooo > Inx. You might start by observing that when x > 1 the equation Inx = X 1/1,000,000 is equivalent to the equation In (In x) ~ (lnx)/I,OOO,OOO.

o c. Even x takes a long time to overtake In x. Experiment with a calculator to fmd the value of x at which the graphs of x l/lO

o

Other Comparisons 15. Investigate

lim In (x + I) Inx

%_00

1/10

and

and Inx cross, or, equivalently, at which Inx ~ 10 In (lnx) . Bracket the crossing point between powers of 10 and then close in by successive halving.

. In (x + 999) lim In X .

.%_00

o d. (Continuation ofpart (c).) The value

ofx at which Inx ~ 10 In (lnx) is too far out for some graphers and root fmders to identify. Try it on the equipment available to you and see what happens. 22. The function In x grows slower than any polynomial Show that In x grows slower as x ---+ 00 than any nonconstant polynomial.

Then use I'HOpital's Rule to explain what you fmd. 16. (Continuation ofExercise 15.) Show that the value of

lim In (x + a) Inx

];_00

is the same no matter what value you assign to the constant a. What does this say about the relative rates at which the functions f(x) ~ In (x + a)andg(x) ~ Inxgrow? 17. Show that v'lOx + 1 and Vx+l grow at the same rate as x ---> 00 by showing that they both grow at the same rate as Vi as

Algorithms and Searches 23. •• Suppose you have three different algorithms for solving the same problem and each algorithm takes a number of steps that is of the order of one of the functions listed here:

n lo~ n, n 3/2,

X~OO.

18. Show that ~ and v'x' - x' grow at the same rate as x ---+ 00 by showing that they both grow at the same rate as x 2 as x ---+ 00.

19. Show that eX grows faster as x ---+ 00 than x" for any positive integer n, even x 1,000,000 . (Hint: What is the nth derivative of x"?) 20. The function e' outgrows any polynomial Show that e' grows faster as x ---> 00 than any polynomial

n(lo~ n? .

Which of the algorithms is the most efficient in the long ruo? Give reasons for your answer.

o b. Graph the functions in part (a) together to get a sense ofhow rapidly each one grows. 24. Repeat Exercise 23 for the functions n,

Vn log, n,

(log, n? .

o 25. Suppose you are looking for an item in an ordered list one million items long. How many steps might it take to fmd that item with a sequential search? A binary search?

21. a. Show that Inx grows slower as x ---> 00 than xii' for any positive integer n, even x 1/1,000,000 •

You are looking for an item in an ordered list 450,000 items long o b. Although the values ofxl/l,ooo,ooo eventually overtake the val- o 26. (the length of Webster. Third New International Dictionary).

ues of In x, you have to go way out on the x-axis before this happens. Find a value of x greater than I for which

Chapter

,

How many steps might it take to fmd the item with a sequential search? A binary search?


1. What functions have inverses? How do you know if two functions f and g are inverses of one another? Give examples of functions that are (are not) inverses of one another. 2. How are the domains, raoges, and graphs of functions and their inverses related? Give an example.

3. How can you sometimes express the inverse of a function of x as a function ofx? 4. Under what circumstances can you be sure that the inverse of a function f is differentiable? How are the derivatives of f and l related?

r

5. What is the natural logarithm function? What are its domain, raoge, and derivative? What arithmetic properties does it have? Comment on its graph. 6. What is logarithmic differentiation? Give an example.

7. What integrals lead to logarithms? Give examples. What are the integrals of tan x and cot x? 8. How is the exponential function e' defined? What are its domain, range, and derivative? What laws of exponents does it obey? Comment on its graph. 9. How are the functions a' and log.x defmed? Are there any rs1rictions on a? How is the graph oflog. x related to the graph of In x? What troth is there in the statement that there is really only one exponential function and one logarithmic function? 10. How do you solve separable first-order differential equations? 11. What is the law of exponential change? How can it be derived from an initial value problem? What are some of the applications of the law?

12. Describe I'HOpital's Rule. How do you know when to use the rule and when to stop? Give an example.

430


13. How can you sometimes handle limits that lead to indeterminate forms 00/00, 00' 0, and 00 - 001 Give examples.

14. How can you sometimes handle limits that lead to indeterminate forms rx>, 00, and 00 001 Give examples. 15. How are the inverse trigonometric functions deImed? How can you sometimes use right triangles to find values of these functions? Give examples. 16. What are the derivatives of the inverse trigonometric functions? How do the domains of the derivatives compare with the domains of the functions? 17. What integrals lead to inverse trigonometric functions? How do substitution and completing the square broaden the application of these integrals? 18. What are the six basic hyperbolic functions? Comment on their domains, ranges, and graphs. What are some of the identities relating them?

Chapter 'f)

1. y = LOe~/'

2. y

= VieW.

I. 3. Y -_I"" "4xe - 16 e.

4. y

=

=

In (sin2 0)

7. y = 10!!2 (x /2) 9. y = 8-1 11. y = 5x,·6

+ 2)"+2

15. y = sin-I~, 16. y = sin-I

x 221 e- x

6. y = In (see> 0)

2

13. y = (x

X~OO?

23. What roles do the functions e% and In x play in growth comparisons?

24. Describe big-oh and little-oh notation. Give examples. 25. Which is more efficient--- tanffix

41. J.'_2_I_ dl o 12 - 25

I

26. y

eY csc (eY

32. j e' cos (3e' - 2) dl

37.

45.

_ 2(x + I) 25. y - .,----;:v cos 2x + 1)(1 27. y (I _ 2)(1 + 3)

j

35. j see> (x)e"'" fix

21 z

= (lnx)I/(In%)

30. y

Evaluating Integrals Evaluate the integrals in Exercises 31-78.

14. y = 2(lnx)"f2

I 17. y

w-=-l,

2u2"

W+1

33. j e% sec2 (e% - 7) fix

19. y = I tan-I I - tlnl

=

22. How do you compare the growth rates of positive functions as

20. y = (I + 12)

21. What integrals lead naturally to inverse hyperbolic functions?

8. Y = log, (It - 7) 10. y = 9" 12. y = y'l,,-Yz

18. y=zcos-Iz- ~

21. y

20. How are the inverse hyperbolic functions deImed? Comment on their domains, ranges, and graphs. How can you fmd values of sech-I X, csch- I x, and coth-I x using a calculator's keys for cosh- 1 X, sinh-1 x, and tanh-I x?

Practice Exercises

Finding Derivatives In Exercises 1-24, Imd the derivative ofy with respect to the approptiate vatiable.

5. y

19. What are the derivatives of the six basic hyperbolic functions? What are the corresponding integral formulas? What simi1arities do you see here with the six basic trigonometric functions?

47.

j j j }csc>

tan (In v) v dv

(lnx)-' -x-fix (I + Inr) dr

42.

[~fix

1

~/6

cos I

-",/2 1 - sm t

dl

44.j~ vlnv 46.

j

ln(x - 5) x 5 fix

48.

j

cos(I-lnv) v dv

49. jX3%' fix

SO. j 2"'" see> x fix

51. [ffix

52.

[2

;x fix

431

Chapter 7 Practice Exercises

53.

54.1 (~

l' (~+ ~)dx

55.1-

56.1

1

e --(>0+1) dx

/.ln5 e'(3e' + 1)-'/2 dr

' l 1

+ 7lnxt l/'

dx

(In (v + I»)' v+1 dv

1 1

58.

60

"

65.

-6- d6

1

'/4

-'/4

6dx

4x 2

66.

68

J

dy yV4y2 - I

74

dx V-2x-x 2 1

77 •

V

2

t-O+

4 - 25x2

-liS

~ 24dy

J I J

--=-r~==

yVy2 - 16

•

-2tVs

•

dl

(I + I)VI 2 + 21 - 8

78 •

dy :-::,~==

Iy IV5y2 - 3

dx V-x 2 +4x-1

76 11

+ 4v + 5

-1

J

3 dv 4v 2 + 4v + 4 dl

(31 + I)V912 + 61

Solving Equations In Exercises 751--84, solve for y. 79. 3" = 2"+1

80. 4" = 3"+2

81. ge'" = x 2

82. 3"

83. In (y - I)

106. lim e-l/Y lny

= x + Iny

y-o+

107. lim (eX + I x_OO

=

d. f(x)

=

e. f(x)

= csc-

= 1n5x

tan~

x-o x + smX'

· sinmx 90• Inn . x-o smnx

89. Inn - ( 2)

x-o tan x

l

x,

tan-I X

=

g(x) = I/x g(x)

eX

=

.. f(x)

= 3~,

b. f(x) = 1n2x,

g(x)

= 2~

g(x)

=

Inx 2

c. f(x) = lOx' + 2x 2, g(x) = eX d. f(x) = tan- l (1jx), g(x) = I/x

= sechx,

g(x) = l/x2 g(x) =

e~

.. ~2 + ~ = O(~) 2

b.

c. X = o(x + Inx) e. tan-I x = 0(1)

d. In (lnx) = o(lnx) f. coshx = O(e X )

x

x4

x

~ + ~ = O(~) x 2 x4 x4

~= O(~+~) x4 x2 x4

c. Inx = o(x + I) e. sec-l x = 0(1)

d. 1n2x = O(lnx) f. sinhx

rl

x~o

G'

g(x)

x,

= O(e

X

)

113. Thefimctioof(x) = eX + x,beingdifferentishleaodone-to-one, has a differentiable inverse (x). Find the value of /dx at the point f(ln 2).

x-n/2-

+ I - Vx2 - xl

,)

lim _x_ _ _ x_ 2-1 x 2 +1

'x_oo

....

Theory and Applications

lim sec 7x cos 3x

93. lim (esc x - cotx)

96

(1+ ?)X

112. Tme, or false? Give reasons for your answers.

2 8in X

X

x-o+

110. Does f grow faster, slower, or at the same rate as g as x ..... oo? Give reasons for your answers.

..

X~OO

108. lim

f. f(x) = sinh,

X

95. lim (VX2 +

)lnX

111. True, or false? Give reasons for your answers.

3 Inx

84. In (lOlny)

88. lim

91.

eX - 1

Comparing Growth Rates of Functions 109. Does f grow faster, slower, or at the same rate as g as x ..... oo? Give reasons for your answers. .. f(x) = 10l!2 x, g(x) = log, x I b. f(x) = x, g(x) = x + X g(x) = xe~ c. f(x) = x/100,

f. f(x)

' x2+3x-4 85. Inn I

•

+ 21)

I - In (1 2 t

e. f(x) = sin-l (l/x),

Apply;ng L:H6pUal's Rule Use I'HOpital's Rule to find the limits in Exercises 85-108. x-I

100. lim Z-.mx - I x-o eX-l

6dx

-V.tVs

72.

J 75.1- --,-~2~d~v_c: J -2

98. lim 3' - I 6-0 8

x

101. lim 5 - 5 cou x_oe x x 1 103. lim

dx

'8ln310g,6 d6 6

{'

70.

2/3 dy 71. /, :-::-r~= Yz/3IY IV9y2 - I •

1)1/2 d6

• Jv'33 + 12

• _24+31

73

1 1V 1/5

-:;:~~

'19 -

l"-Ix~

x-o

99. lim 2'"" - I x_oex-l

dw

/.1n9 e'(e' _

64'1

6712~2 69 •

e2w

97. lim lOX - I

62. ],4(1 + 1n1)llnldl

81084 6

63.

0

-In2

59.1' ~ (1 61.

-

3x

1

-2

57.

~) dx x2

8

drl

114. Find the inverse of the fimctioof(x) = I + (I/x),x # O.Then showthatr l (f(x» = f(r l (x» = x and that

drll dx

__1_ ft.) -

f'(x)"

432


In Exercises 115 and 116, fmd the absolure maximum and minimum values of each functioo 00 the given interval.

[i., fJ

115. y = xlo2x - x, 116. y = Hh:(2 - Inx),

(0,

127. y.y' = secy' sec' x

.'l = 2(lnx)/x and the

Find the area between the curve y x-axis from x = 1 to x = e.

117. Area

118. a. Show that the area hetweoo the curve y = I/x and the x-axis from x = 10 to x = 20 is the same as the area between the curve and the x-axis from x = 1 to x = 2. b. Show that the area between the curve y = I/x and the x-axis from lea to kh is the same as the area between the curve and the x-axis from x = a tox = b (0 < a < b, k > 0).

119. A particle is traveling upward and to the right along the curve y = Inx. Its x-coordinare is increasing at the rate (fix/dt) = Vi mlsec. At what rare is the y-coordinare changiog at the point (.',2)? 120. A girl is sliding dowo a slide shaped like the curve y = 9. -xl3 . Her y-coordinare is changiog at the rate dy/ dt = ( -1/4)~ Nsec. At approximarely what rate is her x-coordinate changiog when she reaches the bottom of the slide at x = 9 ft? (Take. 3 to be 20 and rouod your ans",", to the nearest sec.)

ftl

121. The rectangle shown here has one side

In Exercises 125-128 solve the differential equatioo. dy _, 3y(x + I)' 125. fix = v'Ycos' vy 126. y' = y I

00

the positive y-axis,

one side on the positive x-axis, and its upper right-hand vertex on the curve y = e -r. What dimensions give the rectangle its largest area, and what is that area? y

128. y cos' x dy

+ sin x fix

= 0

In Exercises 129-\32 solve the initial value problem.

.-'-Y-',

129. :

=

dy 130. fix

=-

131. xdy -

ylny

I

+x

"

y(O) = -2

y(0)

=•

(y + v'Y) fix =

132. y-' dyfix =

,

0, y(l) = I

,f--, y(O) = • +I

I

133. What is the age ofa sample of charcoal in which 90% of the carbon-14 origioally present has decayed? 134. Cooling a pie A deep-dish apple pie, whose inrerna\ temperature was 2200f when removed from the oven, was set out on a breezy 40°F porch to cool. Fifteen minures later, the pie's internal tcmperatore was 180°F. How long did it take the pie to cool from there to 70°F? 135. Locating a solar station You are under cootract to build a solar statioo at grouod level on tlte east-west line between the two buildings shown here. How far from the taller building should you place the statioo to maximize the number of hours it will be in the sun 00 a day when the sun passes directly overhead? Begio by observing that

8

= 'If

-

-1 X

cot

C1

60 - co

50 - x

~.

Then fmd the value of x that maximizes 8. ~~------~--~ x

o

60m 122. The rectangle shown here has one side

00

the positive y-axis,

one side on the positive x-axis, and its upper right-hand vertex on the curve y = (lnx)/x'. What dimeosions give the rectangle its largest area, and what is that area? 0.2 0.1

t ,_ ..

o

D

~

~I

II

'x

123. Graph the following functioos and use what you see to locate and estimare the extreme values, identify the coordinares of the inflectioo points, and identifY the intervals 00 which the grapha

are concave up and concave down. Then confrrm your estimates by working with the functions' derivatives.

a. y

=

(lnx)/Vi

D 124. Graph f(x)

b. y = .-..'

c. Y

=

(l

+ x).~

= x In x. Does the function appear to have an absolure minimum value? Cnnfmn your answer with calculus.

00 00 00 00

00 00

9

30m

~-+----~--~~~x

o

x

SOm

136. A rouod underwater transmissioo cable consists of a core of copper wires surrounded by noocooducting insulatioo. 1f x denores the ratio of the radius of the core to the thickness of the insulation, it is known that the speed of the transmission signal is given by the equation v = x'in (I/x). If the radius of the core is I em, what insulatioo thickness h will allow the greatest transmissioo speed? Insulation

Chapter 7 Additional and Advanced Exercises

Chapter

f)

Additional and Advanced Exercises

Limits

16. Let g be a function that is differentiable throughout an open interval containing the origin. Suppose g has the following properties:

Find the limits in Exercises l--{j.

[h dx b-l-jo ~

2. lim} {' tan-I I dl

1. lim

%_00

3. lim (cos Yx)l/x %-0+

4. lim (x

Jo

• I) g(x

x

+ ex)2/x

+n+2 +... +~) 2n

.~o

_1_

iii) lim g(h)

1n (e1/1l + e2/11 + ... + e(II-1)/1I + en/II)

volving the region about the x-axis. Find the following limits.

b. lim V(I)/A(t)

c. lim V(I)/A(I)

t-oo

1_00

1-0+

8. Varying a logarithm's bas.

o

a. Findlimlog.2asa .... 0+, 1-, 1+, and 00. b. Graph y ~ log. 2 as a function of a over the interval 0< a :5 4.

Theory and Examples 9. Find the areas between the curves y ~ 2(loll2x)/x and y ~ 2(log.,x)/x and the x-axis from x ~ I to x ~ e. What is the ratio of the larger area to the smaller?

0 10. Graph f(x) ~ tan-I x + tan-l(l/x) for -5

:5 x :5 5. Then use calculus to explain what you see. How would you expect f to behave beyond the interval [-5,5]? Give reasons for your answer.

11. For what x

>

0 does x(x') ~ (xx)'? Give reasons for your

answer. 0

12. Graphf(x) ~ (sinx)"nxover[O, 31T]. Explain what you see. 13. Findj'(2) if f(x)

~

e«X) andg(x)

~ (X_t_, dl.

l,

~

0

~ I

h

a. Showthatg(O)

7. l.e!A(t) be the area of the region in the 'ITst quadrant enclosed by the coordinate axes, the curve y = e -x, and the vertical line x ~ I, t > O. Let V(I) be the volume of the solid generated by rea. lim A(t)

g(x) + g(y) ~ I _ g(x)g(y) forallrealnurnbers x,y, and

+ y in the domain of g

h-O 1I_00

+ y)

il) lim g(h)

%_00

5. lim (_1_ 11_00 n+ 1

6. lim

433

1+ t

14. a. Find df/dx if

~

O.

b. Showthatg'(x) ~ I

+ [g(x)]'.

c. Find g(x) by solving the differential equation in part (b).

17. Center of mas. Find the center of mass ofa thin plate of constant density covering the region in the first and fourth quadnmts enclosed by the curves y ~ 1/(1 + x') and y ~ -1/(1 + x') and by the lines x ~ 0 and x ~ 1. 18. Solid of revolution The region between the curve y ~ 1/(2Yx) and the x-axis from x ~ 1/4 to x ~ 4 is revolved about the x-axis to generate a solid. •• Find the volume of the solid.

b. Find the centroid of the region. 19. Th. best branching angl•• for blood v ••••l. and pipe. When a smaller pipe branches offfrom a larger one in a flow systero, we may want it to run off at an angle that is best from SOD2O energysaving point of view. We might require, for instance, that energy loss due to friction be minimized along the sectionAOB showo in the accompanying figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe Ul'" stream from B, aod 0 is the point where the braochiog occurs. A law due to Poiseuille states that the loss of energy due to friction in nontorbulent flow is proportional to the length of the path aod inversely proportional to the fourth power of the radius. Thus, the loss along AO is (ledl)/R' and along OB is (Ied,)/r' , where /r; is a constant, dl is the length of AO, d, is the length of OB, R is the radius of the larger pipe, and r is the radius of the smaller pipe. The angle 8 is to be chosen to minimize the sum of these two losses:

f(x)~ ['2~ldt.

dl

d,

R

r

L~/r;4+/r;4·

b. Find f(O). c. What can you conclude about the graph off? Give reasons

for your answer. 15. Even-odd decompositions

a. Suppose that g is an even function of x and h is an odd function ofx. Show that if g(x) + h(x) ~ 0 for all x then g(x) ~ for aIIxandh(x) ~ for allx.

o

A

o

b. If f(x) ~ IE (x) + f 0 (x) is the sum of an even function h(x) and an odd function fo(x) , then show that lE(x) ~ f(x)

+ f( -x) 2

and

fo(x) ~ f(x) - f( -x)

2

c. What is the sigoificance of the result in part (b)?

In our model, we assume that AC Thus we have the relations

~

a and BC

d,sinO

~

b

~

b are fixed.

434


D 20.

so that d,=bcscO,

dl

=a

- d,cosO

=a

- beatO.

We can express the total loss L as a function of 0:

Urban gardening A vegetable gardeo 50 ft wide is to be growo between two buildiogs, which are 500 ft apart along an east-west line. If the buildiogs are 200 ft and 350 ft ta1l, where should the garden be placed in order to receive the maximum llUll1ber ofbours of sunlight exposure? (Hint: Determine the value of x in the accompa-

nying fIgure that maximizes sunlight exposure for the garden.)

L= k(a - bcotO + bCSCO). R4

L

74

00 00 00 00 00

Show that the critical value of 0 for which dL/ dO equals zero is

200 ft tall

b. If the ratio of the pipe radii isr/R = 5/6, estimate to the nearest degree the optimal branching angle given in part (a).

The mathematical analysis described here is also used to explain the angles at which arteries branch in an animal's body.

West

00 00 00 x

50

350 ft tall

East

8 TECHNIQUES OF INTEGRATION OVERVIEW The Fundamental Theorem tells us how to evaluate a definite integral once we have an antiderivative for the integrand function. Table 8.1 summarizes the forms of antiderivatives for many of the functions we have studied so far, and the substitution method helps us use the table to evaluate more complicated functions involving these basic ones. In this chapter we study a number of other important techniques for finding anti derivatives (or indefinite integrals) for many combinations of functions whose antiderivatives cannot be found using the methods presented before.

TABLE 8.1

1.

3.

4. 5. 6.

7. 8. 9. 10. 11.

Basic integration formulas

I

kdx = kx

+

I~ = In Ixl

I

eX dx = eX

lax

I I I I I I

dx =

+

C

+

12.

(n # -1)

13. 14.

C

C

15.

LIna + C

sinxdx = -cosx

cosxdx = sinx 2

(any number k)

sec xdx = tanx

(a>O,a#l)

+

+ +

csc2 xdx = -cotx

C

17.

C

18.

C

+

16.

19. C

20.

I I I I I I I v'ar:- x (~) I ~ ~tan-1 (~) I:!. I I v' tanxdx = In Isecxl

+

C

cotxdx = In Isinxl

+

C

secxdx

= In Isecx + tanxl + C

cscxdx

= -In Icscx + cotxl + C

sinh x dx

= coshx + C

coshxdx

= sinhx + C 2 =

a2

x

secxtanxdx = secx

+

cscxcotxdx = -cscx

C

+

x2 =

sin-

1

+

+

dx =1 a sec-1 a x 2 - a2

C

C

+

C

(a> 0) C

(x> a

>

0)

435

436

Chapter 8: Techniques of Integration

8.1

Integration by Parts Integration by parts is a technique for simplifying integrals of the fann

J f(x)g(x) dx. It is useful when f can be differentiated repeatedly and g can be integrated repeatedly without difficulty. The integrals

and

JxCOSXdx

are such integrals because f(x) = x or f(x) = x 2 can be differentiated repeatedly to become zero, and g(x) = cos x or g(x) = eX can be integrated repeatedly without difficulty. Integration by parts also applies to integrals like and

J Inxdx

J

eX cosxdx.

In the rlISt case,j(x) = In x is easy to differentiate and g(x) = I easily integrates to x. In the second case, each part of the integrand appears again after repeated differentiation or integration.

Product Rule in Integral Form If f and g are differentiable functions of x, the Product Rule says that

d dx [f(x)g(x)] = f'(x)g(x)

+ f(x)g'(x).

In terms of indermite integrals, this equation becomes

J

fx [f(x)g(x)] dx

= J

[f'(x)g(x)

+

f(x)g'(x)] dx

or

J

fx [f(x)g(x)] dx

= J

f'(x)g(x) dx

+ J f(x)g'(x) dx.

Rearranging the terms of this last equation, we get

J f(x)g'(x) dx = J

fx [f(x)g(x)] dx -

J f'(x)g(x) dx,

leading to the integration by parts formula

J f(x)g'(x) dx = f(x)g(x) - J f'(x)g(x) dx

{l}

Sometimes it is easier to remember the formula if we write it in differential form. Let u = f(x} and v = g(x}. Then du = f'(x) dx and dv = g'(x) dx. Using the Substitution Rule, the integration by parts fannula becomes

8.1

437

Integration by Parts

Integration by Part. Formula

j udv

=

uv - j vdu

(2)

J

J

This formula expresses one integral, u dv, in terms of a second integral, v du. With a proper choice of u and v, the second integral may be easier to evaluate than the first. In using the formula, various choices may be available for u and dv. The next examples illustrate the technique. To avoid mistakes, we always list our choices for u and dv, then we add to the list our calculated new terms du and v, and finally we apply the formula in Equation (2).

EXAMPLE 1

Find

j We use the formula j

Solution

u dv

xcosxd!:.

=

uv - j v du with

dv = c08xdx, v = sinx.

u =x, du = d!:,

Simplest antiderivative of cos x

Then

j xcosxd!:

=

x sin x - j sinxd!:

=

x sin x

+ cosx +

C.

•

There are four choices available for u and dv in Example 1:

1. Letu = 1 anddv = xcosxd!:. Letu = xcosx and dv = d!:.

2. 4.

3.

Letu = xanddv = cosxd!:. Letu = cosxanddv = xd!:.

Choice 2 was used in Example I. The other three choices lead to integrals we don't know how to integrate. For instance, Choice 3 leads to the integral

j{xcosx - x 2 sinx)d!:.

J

The goal of integration by parts is to go from an integral u dv that we don't see how to evaluate to an integral v du that we can evaluate. Generally, you choose dv first to be as much of the integrand, including d!:, as you can readily integrate; u is the leftover part. When finding v from dv, any antiderivative will work and we usually pick the simplest one; no aIbitrary constant of integration is needed in v because it would simply cancel out of the right-hand side of Equation (2).

J

EXAMPLE 2

Find

j Inxd!:. Solution

J udv = u du

Since

uv -

= lnx =

1

xd!:,

J In x d!: can J vduwith

be written as

Simplifies when differentiated

J In x • 1 d!:, dv

= d!:

v =x.

we use the formula Easy to integrate

Simplest antiderivative

438

Chapter 8: Techniques of Integration Then from Equation (2),

1

lnxdx

=

1

xlnx -

x·fdx

1

xlnx -

=

dx

=

xlnx - x + C.

•

Sometimes we have to use integration by parts more than once.

EXAMPLE 3

Evaluate

1

2

X

x e dx.

Solution

With u = x 2 , dv

= eX dx,

1

du = 2xdx, and v =

x 2e Xdx

x 2e x -

=

21

eX,

we have

xeXdx.

The new integral is less complicated than the original because the exponent on x is reduced by one. To evaluate the integral on the right, we integrate by parts again with u = x, dv = eX dx. Then du = tix, v = eX, and

1

xeXdx

=

xe X -

1

eXdx

=

xe X - eX +

c.

Using this last evaluation, we then obtain

1

x 2e Xdx

=

x 2e x -

= x 2e x -

21

xeXdx

2xe X

+

2e X

•

+ c.

J

The technique of Example 3 works for any integral x"e X dx in which n is a positive integer, because differentiating x" will eventually lead to zero and integrating eX is easy. Integrals like the one in the next example occur in electrical engineering. Their evaluation requires two integrations by parts, followed by solving for the unknown integral.

EXAMPLE 4

Evaluate

1

eXcosxdx.

Solution

Letu

=

eX and dv

1

=

cosxdx. Then du

eXcosxdx

=

eX sin x -

= eXdx,v

sinx,and

=

1

eXsinxdx.

The second integral is like the flISt except that it has sin x in place of cos x. To evaluate it, we use integration by parts with

dv Then

1

eX cosx dx

= sinxdx,

v =

du=exdx.

-C08X,

1 1

=

eX sin x - (-ex cosx -

=

eX sin x + eXcosx -

(-cosx)(e Xdx»)

eXcosxdx.


8.1

439

The unknown integral now appears on both sides of the equation. Adding the integral to both sides and adding the constant of integration give

2J

e'cosxdx

=

e'sinx + e'cosx + CI.

,.

Dividing by 2 and renaming the constant of integration give

J

e , cosx dx -- e

EXAMPLE 5

+' 2 e cosx +

SfiX

C.

•

Obtain a fonnula that expresses the integral

J

cos' xdx

in terms of an integral of a lower power of cos x. We may think of cos' x as cos·- I x • cos x. Then we let

Solution

u

=

cosn-

1X

dv

and

c08xdx,

=

so that

du = (n - I) cos·- 2 x(-sinx dx)

v = sinx.

and

Integration by parts then gives

J

cos'xdx = cos·-Ixsinx

+

(n

-1)J

sin2 xcos·- 2 xdx

J

= cos·- I x sin x

+ (n - 1)

= cos·-Ixsinx

+ (n

(1 - cos2 x) cos·- 2 X dx

-1)J

cos·- 2 xdx - (n -

1)J

cos·xdx.

Ifwe add (n

-1)J

cos'xdx

to both sides of this equation, we obtain

J

n

cos' xdx = cos·- I x sin x

+

J

(n - 1)

We then divide through by n, and the final result is

J

.-1' cos• x dx -- cos nx smx

IJ

+~ n

cos·- 2 xdx.

cosn-2 x dx .

•

The fonnula found in Example 5 is called a reduction formula because it replaces an integral containing some power of a function with an integral of the same fonn having the power reduced. When n is a positive integer, we may apply the fonnula repeatedly until the remaining integral is easy to evaluate. For example, the result in Example 5 tells us that

J

3

2

cos xdx =

cos xsinx 3 2

2J

+3

= tcos xsinx

cosxdx

+ tsinx +

c.

440


Evaluating Definite Integrals by Parts The integration by parts fonnula in Equation (I) can be combined with Part 2 of the Fundamental Theorem in order to evaluate definite integrals by parts. Assuming that both f' andg' are continuous over the interval [a, b], Part 2 of the Fundamental Theorem gives

Integration by Parts Formula for Definite Integrals

tf{x)g'{x) dx

=

f{x)g(x)

l: - tf'{x)g(x) dx

(3)

In applying Equation (3), we normally use the u and v notation from Equation (2) because it is easier to remember. Here is an example. y

EXAMPLE 6 Find the area of the region bounded by the curve y = xe-x and the x-axis fromx = Otox = 4. Solution

0.5

The region is shaded in Figure 8.1. Its area is

1'xe-X dx. -1

1

2

3

4

x

Letu

=

x,dv

e-xdx,v

=

-e-x,anddu = tix. Then,

1'xe-x dx

=

-xe -x l~

=

[-4e-4 - (O)] +

=

-4e-4 - e-x]~

=

-4e-4 - e-4 - (-eO)

=

FIGURE 8.1 The region in Example 6.

-14 (

-e -X) dx

14

e-X dx

= I -

5e-4 '" 0.91.

•

Tabular Integration

J

We have seen that integrals of the fonn f{x)g(x) dx, in which f can be differentiated repeatedly to become zero and g can be integrated repeatedly without difficulty, are natural candidates for integration by parts. However, if many repetitions are required, the calculations can be cumbersome; or, you choose substitutions for a repeated integration by parts that just ends up giving back the original integral you were trying to find. In situations like these, there is a way to organize the calculations that prevents these pitfalls and makes the work much easier. It is called tabular integration and is illustrated in the following examples.

EXAMPLE 7

Evaluate

J

2 x

x e dx.

Solution

With f{x) = x 2 and g{x) = eX, we list:

f(x) and its derivatives

g(x) and its integrals


8.1

441

We combine tbe products of tbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain

J

x'e' dx

x'e' - 2xe'

=

+ 2e' + C.

•

Compare tbis witb tbe result in Example 3.

EXAMPLE 8

Evaluate

J

x 3 sinxdx.

Witb f(x) = x 3 and g(x) = sin x, we list:

Solution

f(x) and its derivatives

g(x) and its integrals

x3

(+)

sin x

3x'

(-)

-cosx

(+) (-)

.... -sinx

6x

----------

6

----------

0

----------

~

cos x

~

sin x

Again we combine tbe products of tbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain

J

x 3 sinxdx

= -x 3 cosx

+ 3x'sinx + 6xcosx

- 6 sin x

+ C.

•

The Additional Exercises at tbe eod of!bis chapter show how tabular integration can be used wheo neitber function f nor g can be differentiated repeatedly to become zero.

Exercises 8.1 Integration by Parts Evaluate the integrals in Exercises 1-24 using integmtion by parts.

15. / x'exdx

(x' - 5x)e Xdx

1. / xsinIdx

2. /oeOS1rOdO

3. / t'eostdt

4. / x sinxdx

19. / x'exdx

6. /,e x31nx dx

21. /

5.

/,2

xlnxdx

7. /xeXdx 9.

/x e-

11. /

2

X

2

8. dx

tan-1ydy

13. / xsec2 xdx

17. /

10.

/xe

3x

12. /

23. / e lx cos 3x dx

dx

/(X 2- 2x + l)e sin-1 ydy

14. /4xsec 2 2xdx

e'sinOdO

2x

dx

16. / p

e" dp

4

18. /

(r' + r + l)e'dr

20. /

t'e 4t dt

22. /

e-Yeosydy

24. /

e-2x sin2xdx

Using Substitution Evaluate the integrals in Exercises 25-30 by using a substitutioo prior to integration by parts. 25. /

e v'hl9 tis

26.

J,lx~dx

442 27.

29.


J.

~/3

xtan2 xdx

0

j

28.

30.

sin (lnx) dx

j j

+ x 2)dx

In(x

d. What pattern do you see? What is the area b-.eo the curve

and the x-axis for 2 (~)

z(lnz)2 dz

n an arbitrary positive integer? Give reasons for your answer.

Evaluating Integrals Evaluate the integrals in Exercises 31-50. Some integrals do not require integration by parts.

31. jxsecX2 dx

32.

33. jXQnx)2dx

34.

45.

j ~: j j v?"+l j j e~ j Vx

47.

J.~/2 8 2 sin 28 d8

35. 37. 39. 41. 43.

49.

46.

48.

J.~/2 x 3 cos 2x dx

50.

J.

36.

x 3 e x'dx

38. dx

40.

sin 3x cos 2x dx

42.

sin eX fix

44.

dx

cos

tsec-Itdt

('

12;v3

(lnx)3

x

dx

x 2 sin x 3 dx sin 2x cos 4x dx

dx

e Vi dx

1;0.

0

~ 'IT.

55. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes and the curve y = cos x, 0 :5 x :s; 71'/2, about b. the line x = 1T/2. 56. Finding volume Find the volume of the solid generated by revolving the region bounded by the x-axis aod the curve y = xsinx,O:5 x:s; '11',about

a. the y-axis.

y=xsinx

5

o -5 52. Finding area Find the area of the region enclosed by the curve y = x cos x aod the x-axis (see the accompaoying figure) for

a. 1T/2 s x s 31T/2. 51T/2.

c. 51T/2 s x s 71T/2.

andx

=

e.

a. Find the area of the region. b. Find the volume of the solid formed by revolving this region

about the x-axis.

y

10

7T'.

(See Exercise 51 for a graph.)

nonnegative integer? Give reasons for your answer.

S

a. about the y-axis. b. about the line x = I.

57. Consider the region boonded by the graphs of y = In x, Y = 0,

d. What pattern do you see here? What is the area b-.eo the curve and the x-axis for mr s x s (n + 1)1T, n an arbitnlry

'" x

54. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes, the curve y = e -r, and the line x = 1

b. the line x =

1T'

b. 31T/2

53. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes, the curve y = eX. and the line x = In 2 about the line x = In2.

a. the y-axis. 2xsin-1 (x 2 )dx

:s x :s 21T. c. 21T :s x :s 31T.

b.

-10

-x-dx

51. Finding area Find the area of the region eoc1osed by the curve y = x sin x aod the x-axis (see the accompaoying figure) for ~

10

1_2 dx xQnx)

Theory and Examples

a. 0

y

jco~dx

j __ j j x' e" j j j -:x j Vx

dx

x3

«~) 2 'fr,

< '1T_X_

c. Find the volume of the solid formed by revolving this region about the line x = -2. d. Find the centroid of the region. 58. Consider the region bounded by the graphs of y aodx = 1.

= tan-I x, y = 0,

a. Find the area of the region. b. Find the volume of the solid formed by revolving this region about the y-axis. 59. Average value A retarding force, symbolized by the dashpot in the accompaoying fIgure, slows the motion of the weighted spring so that the mass's position at time t is

t '" O.

443

8.1 Integration by Parts Find the average value ofy aver the intervaJ. 0

:!;';

,

t

:!;';

2'11".

The idea is to take the most complicated part of the integral, in this case l(x), and simplify it fIrst For the integral ofln x, we get

r

J

lnxdx

~

J

, - lnx,

ye'dy

X _ 1I 7

dx - Il'rIy

+C x + C.

= ye' - II'

= xlnx -

o

,

For the integral of COlI-I X we get

f

-I

C08-l xdx - xoos-Ix

y

cosydy

=

coa-1x

= xoos-Ix - siny + C

Use the formula

60. Avenge value In a mass-spring-dashpot system l:ilcc the one in Exercise 59, the mass's position at time t is

y = ~-t(sint - cost),

t:2!:

Find the average value ofy aver the interval 0

O.

Reduction Formulas

61.

61.

f x~

cosxdt = x"sinx -

1

1

x· sinxQ = -x· cosx +

to cvaluate the integrals in Exercises 67-70. Express your answers in terms ob.

69.

f

70·/IO~Xdx

sec-I x ttt

r

1

1

x·- I COSXQ

If

(4)

fly) dy

Another way to integrate rl(x) (when course) is to use integration by parts with u rewrite the integral. of 1 as

X,,-I sinxQ

If

J

r'(x) dx - xr'(x) -

:s: t :s: 2'11".

In Exercises 61-64, use integration by parts to establish the reduction

-.....

J

rl(x)dt

=

xr1(x)

-I

r 1 is integrable, of r 1(x) and dv = dx to

x(!r1(x»)dt.

(5)

Exm:ises 71 and 72 compare the:results ofusing Equations (4) and (5). 64.

f

(lnx)"Q = x(lnx)· -

nf

71. Equations (4) and (5) give diffen:ul: form.ulas for the integral. of

(lnX),,-I Q

cos-1 x:

65. Show that

L

f

b.1 66. Use inqration by paI19 to obtain the formula

Integration by parts leads to a rule for integrating inverses that usually gives good results:

b.

J

~ yf(y) -

ofor -~2 < (J < '!! 2

+ tan0l + C

From Fig. 8.4

C.

and

451

Notice how we expressed Isec 0 + tan 0 I in tenDs of x: We drew a reference triangle for _ the original substitution x = 2 tan 0 (Figore 8.4) and read the ratios from the triangle.

EXAMPLE 2

Solution

Evaluate

We set

_1r5'

452

Chapter 8: Techniques of Integration to put the radicand in the form x'

- a'. We then substitute

2 x = Ssec 0,

is

x' - (~)' =

2 dx = Ssec 0 lanO dO,

sec' 0 -

~lan'O

=

25

25

~x' - (~)' ~ ItanOI ~lanO. =

J FIGURE 8.6 Ifx = (2/5)sec 6, 0< 6 < 'IT/2,then6 = sec- I (5. o for 0< 0 < ",/2

=

With these substitutions, we have dx v'25x' - 4 =

3, y(5) = 1n3

51. (x 1

+ 4)

52. (xl

+ 1)2 :

:

= 3, y(2) = 0 -

W+I,

y(O) - 1

J x 3 ~_dx using

a. integration by p!lrts. b. a v-substitution.

e. a trigonometric substitution. 58. Path of. water dder Suppose that a boat is positioned at the mgin with a water skier tethered to the boat at the point (30. 0) on a rope 30 ft long. As the boat travels along the positiw y-axis. the skier is pulled behind fue boat along an unknown path y = f(x), as shown in the acccwnpanying figure. -\1'900 Xl L Showthatf'(x) = X (!lint: Assume that the skier is always pointed directly at the boat and the rope is on a line tangent to the path y = f(x).)

AppLications and Examples 53. Area Find the area of the :region in the first quadrant that is enclosed by the coordinate axes and the curve y = v'9 - x 2/3. 54. Area

I

b. Solve the equation in part (a) fur f(;&), using f(30) = y y - f{x) pGb. of mel"

Find the area enclosed by the ellipse

x2

o.

y2

2+2=1.

a b 55. ConsideIthcregi.on bmmdedbythe graphs ofy = sin-I :C,Y = 0, andx = 1/2.

a. Find the area of the region. b. Find the centroid of the regicm..

8.4

~~L-____~~~_ x

0];

(30,0)

Integration of Rational Functions by Partial Fractions This section shows how to express a rational function (a quotient of polynomials) as a sum of simpler fractions, called partial fractions, which are easily integrated. For instance, the ,..tiona! function (S.% - 3)/(x' - 2x - 3) can be rewri_ as --,-5~x"c-;-,3"--:o ~ _2_ Xl 2x 3 x+l

+ _3_

x-3'

You can verify this equation algebraically by placing the fractions on the right side over a common denominator (x + l)(x - 3). The skill acquired in writing rational functions as such a sum. is useful in other settings as well (for instance, when usmg certain transfonn methods to solve di:ffcrcntial equations). To integrate the rational function

454


(5x - 3)/(x 2

- 2x - 3) on the left side of our previous expression, we simply swn the integrals of the fractions on the right side:

J (x

5x - 3

dx = J_2_ dx x + I

+ I)(x - 3)

= 2ln

+ J_3_ dx x - 3

Ix + II +

3ln

Ix -

31 +

c.

The method for rewriting rational functions as a swn of simpler fractions is called the method of partial fractions. In the case of the preceding example, it consists of finding constants A and B such that

5x-3 =_A_+~ x 2 -2x-3 x+1 x-3·

(1)

(Pretend for a moment that we do not know that A = 2 and B = 3 will work.) We call the fractions A/(x + I) andB/(x - 3) partial fractions because their denominators are only part of the original denominator x 2 - 2x - 3. We call A and B undetermined coefficients until proper values for them have been found. To find A and B, we flfst clear Equation (I) of fractions and regroup in powers of x, obtaining

5x - 3 = A(x - 3) + B(x + I) = (A + B)x - 3A + B. This will be an identity in x if and only if the coefficients of like powers of x on the two sides are equal: A

+B

= 5,

-3A

+B

=

-3.

Solving these equations simultaneously gives A = 2 and B = 3.

General Description of the Method Success in writing a rational function f(x)/g(x) as a swn of partial fractions depends on two things:

• The degree of f(x) must be less than the degree of g(x). That is, the fraction must be proper. If it isn't, divide f(x) by g(x) and work with the remainder term. See Example 3 of this section. • We must how the factors of g(x). In theory, any polynomial with real coefficients can be written as a product of real linear factors and real quadratic factors. In practice, the factors may be bard to find. Here is how we fmd the partial fractions ofa proper fraction f(x)/g(x) when the factors ofg are known. A quadratic polynomial (or factor) is irreducible if it cannot be written as the product of two linear factors with real coefficients. That is, the polynomial has no real roots.

Method of Partial Fractions (f(x)/g(x) Proper) 1. Let x - r be a linear factor of g(x). Suppose that (x - r)m is the highest power ofx - r that dividesg(x). Then, to this factor, assign the swn of the m partial fractions:

Aj

(x - r)

+

A2 + ... + Am (x - r)2 (x - r)m·

Do this for each distinct linear factor of g(x).

continued

8.4 Integration of Rational Functions by Partial Fractions

455

2. Let x 2 + px + q be an irreducible quadratic factor of g(x) so that x 2 + px + q bas no real roots. Suppose that (x 2 + px + q)n is the highest power of this factor that divides g(x). Then, to this factor, assign the sum of the n partial fractions:

BIX

+

Cl

c--.~-~c-

~+~+~

+

B2X

+

C2

~+~+~

B.x + C. + ... + -;-c;c-"--~'-=

~+~+~.

Do this for each distinct quadratic factor of g(x). 3. Set the original fraction f(x)/ g(x) equal to the sum of all these partial fractions. Clear the resulting equation of fractions and arrange the terms in decreasing powers of x. 4. Equate the coefficients of corresponding powers of x and solve the resulting equations for the undetennined coefficients.

EXAMPLE 1

Use partial fractions to evaluate

J

x2+4x+1 (x - I)(x + I)(x + 3)

Solution

ax .

The partial fraction decomposition bas the fonn

2 x +4x+1 =_A_+~+~ (x - I)(x + I)(x + 3) x - I x + I x +3. To find the values of the undetermined coefficients A, B, and C, we clear fractions and get

x2

+ 4x + I

= A(x

+ I)(x + 3) + B(x - I)(x + 3) + C(x - I)(x + I) + 4x + 3) + B(x 2 + 2x - 3) + C(x 2 - I)

= A(x 2 = (A

+B +

C)x 2

+

(4A

+ 2B)x +

(3A - 3B - C).

The polynomials on both sides of the above equation are identical, so we equate coefficients of like powers of x, obtaining Coefficient of x 2: Coefficient of Xl: Coefficient ofxo:

A+ B+C=I 4A+2B =4 3A-3B-C=1

There are several ways of solving such a system oflinear equations for the unknowns A, B, and C, including elimination of variables or the use of a calculator or computer. Whatever method is used, the solution is A = 3/4, B = 1/2, and C = -1/4. Hence we bave

J

X2 + 4x + I (x - I)(x + I)(x

+

3)

ax -

=

J[l_l_ 4x - I

+ 1_1_ _ l _I _]ax 2x + I 4x +3

"In Ix - II + 21n Ix + II - "In Ix + 31 + 3

I

I

K,

where K is the arbitrary constant of integration (to avoid confusion with the undetennined coefficient we labeled as C). •

EXAMPLE 2


J

6x+7

(x

+ 2)'

ax .

456

Chapter 8: Techniques of Integration Solution First we express the integrand as a sum of partial fractions with undetennined coefficients.

6x+ 7 (x + 2)'

=

_A_+ B x +2 (x + 2)'

+7

=

A(x

=

Ax

6x

+ 2) + B + (2A + B)

Multiply both sides by (x

+ 2)'.

Equating coefficients of corresponding powers of x gives

A=6

2A

and

+B

=

12

+B

7,

=

A

or

= 6

and

B = -5.

Therefore,

j

EXAMPLE 3

C! 2-

(~++2;2 dx =

j

=

6

=

6ln

j '! x

2)') dx

(x :

2 - 5

j

(x +

2r dx 2

Ix + 21 + 5(x + 2r t + C.

•

Use partial fractions to evaluate 3

2

-X-3 dx. j 2x x-4x-2x-3 2

Solution First we divide the denominator into the numerator to get a polynomial plus a proper fraction. 2x x 2 - 2x - 3)2x 3 - 4x 2 - X - 3 2x 3

-

4x 2

-

6x 5x - 3

Then we write the improper fraction as a polynomial plus a proper fraction.

2x 3 -4x 2 -x-3 x 2 -2x-3

~--.~=--~~

= 2x

5x-3 + --,-'~~---c2 x -2x-3

We found the partial fraction decomposition of the fraction on the right in the opening

example, so 3

2

2x -4x -X-3 dx =j2xdx+j 5x-3 dx x 2 -2x-3 j x 2 -2x-3 = j2xdx+ j x i l dx + =

EXAMPLE 4

x2

jX~3dx

+ 2ln Ix + 11 + 3ln Ix - 31 + c.

•


j

-2x + 4 (x 2 + 1)(x _ 1)' dx.

Solution The denominator has an irreducible quadratic factor as well as a repeated linear factor, so we write

(x 2

-2x + 4 + 1)(x - 1)2

=

Ax

x2

+B +~ + D + 1 x-I (x - 1)'·

(2)


457

Clearing the equation of fractions gives -Zl; + 4 = (Ax + B)(x - 1)2 + C(X - 1)(x2 + I) + D(x 2 + 1) = (A + C)x 3 + (-2A + B - C + D)x2 + (A - 2B + C)X + (B - C + D).

Equatiog coefficients oflike tenns gives Coefficients of x 3 :

O=A+C

°

2

-2A + B - C -2 = A - 2B + C 4=B-C+D

Coefficients of x : Coefficients of x I : Coefficients ofxo:

=

+D

We solve these equations simultaneously to find the values ofA,B, C, andD: -4 = -2A,

A = 2

Subtract fourth equation from second.

C = -A =-2

B = (A

From the flfSt equation

+ C + 2)/2

=

I

D=4-B+C=1.

From the third equation and C

~

-A

From the fourth equation

We substitute these values into Equation (2), obtaining -Zl; + 4 = Zl; + I _ _ 2_ + I (x 2 + I)(x - 1)2 x2 + I x - I (x - 1)2 . Finally, using the expansion above we can integrate:

!

-Zl;+4 dx_!(Zl;+I _ _ 2_+ I (x 2 + I)(x - 1)2 ~2 + I x - I (x - 1)2

)dx

_!C'~_+_I_ _ _2_+

-

=

EXAMPLE 5

~2 + I

x - I

I (x - 1)2

10 (x 2 + 1) + tan-I x - 210 Ix - II - x

~

)dx I + C. •


! Solution

x2 + I

x(x 2

a:: 1)2 .

The fonn of the partial fraction decomposition is I =4+Bx+C+ Dx+E x(x 2 + 1)2 x x2 + I (x 2 + 1)2

Multiplying by x(x 2 + 1)2, we have I = A(x 2 + 1)2 + (Bx + C)x(x 2 + I) + (Dx + E)x = A(x4 + Zl;2 + I) + B(x 4 + x 2 ) + C(x 3 + x) + Dx 2 + Ex = (A + B)x4 + Cx 3 + (2A + B + D)x 2 + (C + E)x + A

If we equate coefficients, we get the system A+B=O,

C = 0,

2A

+B +D

=

0,

C+E=O,

A=l.

458

Chapter 8: Techniques of Integration Solving this system gives A = I,B = -I, C = O,D = -I, andE = O. Thus,

J

Iix

x(x 2

-

+

I)' -

=

J[1 ----=-'---Jx - J J x

+

x2 + I

Iix

+

(x 2

-x

+

xlix x2 + I -

I)'

] Iix

xlix (x 2 + I)' u =

du

+ I, 2xdx

Xl

=

I I =Inlxl-zlnlul +2u+ K

= In Ixl = In

t

ln (x 2

+ I) +

+

I

Ixl ~

2(x2

+ I)

2(X,'+ I)

+K

+ K.


The Heaviside uCover-upu Method for Linear Factors

Oliver Heaviside

When the degree of the polynomial I(x) is less than the degree of g(x) and

•

(185()-1925)

g(x) = (x - r,)(x - r2)'" (x - rn) is a product of n distinct linear factors, each raised to the lITst power, there is a quick way to expand I(x)/g(x) by partial fractions.

EXAMPLE 6

Find A, B, and C in the partial fraction expansion

x + I =_A_+~+~ (x - I)(x - 2)(x - 3) x - I x - 2 x - 3. 2

Solution

(3)

If we multiply both sides of Equation (3) by (x - I) to get x2 + I B(x - 1) C(x - I) -;---"-::-cO---'----;:-c = A + + ----'--~ (x - 2)(x - 3) x - 2 x - 3

and set x = I, the resulting equation gives the value of A:

(I)' + I (1 - 2)(1 - 3)

= A

+ 0 + 0,

A=l. Thus, the value of A is the number we would have obtained if we had covered the factor (x - I) in the denominator of the original fraction

x2 + I (x - I)(x - 2)(x - 3) and evaluated the rest at x = I:

(1)' + I

A =

(:= x= -=I~ ) "1

[1 =

n

Cover

(~I--~2:-:)(-1-----:-:-3)

2 = I (-1)( -2)

(4)


459

Similarly, we find tbe value of B in Equation (3) by covering tbe factor (x - 2) in Expression (4) and evaluating tbe rest at x = 2:

(2)' + I B = -(2---1-)~ I (~ x =- =2)= '1'---(2---3)

5 (1)(-1) = -5.

f

Covor

Finally, C is found by covering tbe (x - 3) in Expression (4) and evaluating tbe rest at

x

=

3: 10 (3)2 + I C = -:-(3---1-:-)(:-'3-'---2~ )1 =:'(x= _==C )"'11 = (2)( I) = 5. 3

•

f

Cover

Heaviside Method

1. Write the quotient with g(x)factored: f(x) g(x)

f(x) (x - r,)(x - r2)'" (x - rn) .

2. Cover the factors (x - ri) ofg(x) one at a time, each time replacing all tbe uncoveredx's by tbe number rio This gives a number Ai for each root ri:

A,

=

A2 =

f(r,) (r, - r2)'" (r, - rn) f(r2) (r2 - r,)(r2 - 1'3) ... (r2 - rn)

3. Write the partialjraction expansion off(x)/g(x) as f(x) g(x)

EXAMPLE 7

Use tbe Heaviside Metbod to evaluate

J

x3

Solution g(x) = x 3

x+4

+ 3x 2 - lOx

ax .

The degree of f(x) = x + 4 is less tban tbe degree oftbe cubic polynomial 3x 2 - lOx, and, witbg(x) factored,

+

x+4 x3

+ 3x 2 -

lOx

x+4 x(x - 2)(x + 5) .

460

Chapter 8: Techniques of Integration The roots ofg(x) are'l = 0,'2 = 2, and'3 = -5. We find Al

=

0 +4 = 4 = _~ 2)(0 + 5) (-2)(5) 5

GJ (0 f

Cover

A _ 2 -

2+4 2 1 (x - 2) (2 + 5)

6 (2)(7)

1

3 7

f

Cover

-5 + 4

A3 = -

-I (-5)( -7)

__-----"'--"--~===J

+ 5)

(-5)(-5 - 2) 1 (x

1

I 35 .

n

Cover

Therefore,

x +4 = _~ x(x - 2)(x + 5) 5x

+

I

3 7(x - 2)

35(x

+ 5)'

and

J

x+4

x(x _ 2)(x

+

2 5)dx = -slnlxl

3

I

+ 7 1n Ix - 21 - 35 In Ix + 51 + C.

•

Other Ways to Determine the Coefficients Another way to determine the constants that appear in partial fractions is to differentiate, as in the next example. Still another is to assign selected numerical values to x.

EXAMPLE 8

FindA,B, and Cin the equation

x - I (x

+

_A_

=

1)3

X

+

+

+

B

(x

I

+

C

(x

1)2

+

1)3

by clearing fractions, differentiating the result, and substitoting x = -I.

Solution

We fIrst clear fractions: x - I = A(x

+ I)' + B(x + I) + C.

Substituting x = - I shows C = -2. We then differentiate both sides with respect to x, obtaining

+

I = 2A(x

1)

+ B.

Substituting x = - I shows B = I. We differentiate again to get 0 = 2A, which shows A = O. Hence,

x - I (x

+

2

I

1)3

(x

+

1)3'

•

In some problems, assigning small values to x, such as x = 0, ±I, ±2, to get equations in A, B, and C provides a fast alternative to other methods.

EXAMPLE 9

FindA,B, and Cin the expression

x +I =_A_+~+~ (x - I)(x - 2)(x - 3) x - I x - 2 x - 3 2

by assigning numerical values to x.

8.4 Integration of Rational Functions by Partial Fractions Solution

461

Clear fractions to get x2

+I

3)

= A(x - 2)(x -

+ B(x

- I)(x - 3)

+ C(x

- I)(x - 2).

Then let x = 1,2,3 successively to f'mdA, B, and C:

(1)2 + 1 = A( -1)( -2) + B(O) + C(O)

x = 1:

x

2 = 2A A=1 (2)2 + 1 = A(O) 5 =-B

= 2:

+ B(I)( -1) + C(O)

B =-5 x

(3)2

= 3:

+

1 = A(O)

10 C

= =

+ B(O) + C(2)(l)

2C 5.

Cooclusioo: 2

x + 1 =_1_ _ _ 5_+_5_ (x - l)(x - 2)(x - 3) x-I x - 2 x - 3.

•

Exercises 8.4 Expanding Quotients into Partial Fractions

20 /

Expand the quotients in Exercises 1-8 by partial fractions.

1 5x-13 · (x - 3)(x - 2)

3.

x+4 , + I)

(x

5 z + I 'z'(z-I)

(' + 8 - 5t + 6

7. , (

2.

Sx - 7

, x -3x+2

4 2x + 2 'x'-2x+1

8.

(' + 9 t' + 9t'

11./ 13.

x+4 dx x'+Sx-6

/.8 ,

ydy

'Y -2y-3

15 / ·

dt ('+t'-2t

Irreducible Quadratic Factors

10./~ x + 2x

12./

2x+1 dx x' - 7x + 12

[' y

14.

1, '

dx . , (x + I)(x' + I)

23. /

10 Exercises 9-16, express the integrand as a sum of partial fractions and evaluate the integrals.

x'dx (x - I)(x' + 2x + I)

10 Exercises 21-32, express the integrand as a sum of partial fractions and evaluate the integrals.

21

6 z ·z3_ z 2_6z

Nonrepeated Linear Factors

9./~ I - x

.

y'+2Y +I (y' + I)' dy

25/ 23+2 tis . (8' + 1)(8 - I)'

+ 2 dx

27. /

x' - x

29.

x' -,--dx x - I

/

x' - I

+4

l'/2 y' + y dy x + 3 dx 2x' - 8x

16. /

Improper Fractions Repeated Linear Factors 10 Exercises 17-20, express the integrand as a sum of partial fractions and evaluate the integrals. 17

r

1 --o;-__x,-'... dx___

• Jo x2+2x+ 1

18 •

l' . .

~ x,-'__ dx___

-1

x2

-

2x + 1

10 Exercises 33-38, perfor03long division on the integrand, write the proper fraction as a sum. of partial fractions, and then evaluate the integral.

33./ 2x'-2x'+l dx x2 - x

34.

/

x' -,--dx x - I

462


35. /9X'-3X+ Idx x 3 - xl

37.

y' /

+ y2 - I , dY y +Y

56. The y-axis

16x' dx 4x -4x+1

36. /

2

_ 2 y - (x + 1)(2

y

2'

11--.

38 Y dY • / y' - y2 + y - I

x)

.....

Evaluating Integrals Evaluate the integrals in Exercises 39-50.

39 / •

etdt e" + 3e'

41. /

cosydY sin2 y + siny - 6

+2

40.

/

e

4t

+

2e

e2'

2t

e

-

+I

"""'O+---------""'"I--+ x

t

dt

(x - 2)'tan- 1 (2x) - 12x' - 3x (4x2 + I)(x _ 2)' dx

43. /

D 57.

Find, to two decimal places, the x-oordiuate of the centroid of the region in the first quadrant bounded the x-axis, the curve y = tan-Ix, and the line x = V3.

D 58.

Find the x-coordiuate of the centroid of this region to two decimal places.

sinO dO 0082 8 + cosB - 2

42 / •

by

(x + I)'tan- I (3x) + 9x' + x 44. /

+

(9x2

I)(x

+

y (3. 1.83)

dx

1)2

46./ I Vi (xl/3 - I)

dx

(5,0.98)

(Hint: Letx = u'.)

47.

(Hint: Let x

49./ I+ x(x'

+

x x

o

9

D 59.

I = u2.)

50/ I

dx

I)

~~~~~--------~--~x

48./ ~dx +

/~dx

•

x' (x'

+ 4)

dx

by::.)

(Hint: Multiply

Initial Value Problems

3

Social diffusion Sociologists s0fi3ctimes use the phrase "social diffusion" to describe the way information spreads througb a population. The information might be a rumor, a cultural fad, or news about a tecbnica1 innovation. In a sufficiently large population, the nun3ber of people x who have the information is treated as a differentiable function of time t, and the rate of diffusion, dx/ dt, is asSun3ed to be proportional to the number of people who have the information times the number of people who do not. This leads to the equation

Solve the initial value problems in Exercises 51-54 for x as a function oft.

51. (t

2

-

3t + 2): = I

I):

52. (3t' + 4t 2 + 53. (12 54. (t

+ 2t):

+

2x

=

I ) : = x2

=

dx dt

(t > 2), x(3) = 0

2V3,

x(1) =

5

=

kx(N - x)

'

where N is the number of people in the population. Suppose t is in days, k ~ 1/250, and two people start a mor at time t ~ 0 in a population of N = 1000 people.

-1fV3/4

TU-

a. Find x as a function of t.

+ 2 (t,x > 0), x(1)

+ I (t > -I),

=

I

b. When will half the population have heard the rumor? (This is when the rumor will be spreading the fastest.)

D 60.

x(O) = 0

Applications and Examples In Exercises 55 and 56, fmd the volwne of the solid generated volving the shaded region about the indicated axis.

by re-

Seoond-order chemical reaction. Many chemical reactions are the result of the interaction of two molecules that uodergo a change to produce a new product. The rate of the reaction typically depends on the concentrations of the two kinds of moleoules. If a is the amount of substance A and b is the amount of substance B at time t ~ 0, and ifx is the an30unt ofproduct at time t, then the rate of formation of x may be given the differential equation

by

55. The x-axis y~

dx dt

3

V3x-x 2

Y (0.5. 2.68)

(2.5, 2.68)

2

k(a - x)(b - x),

or

I dx = k (a - x)(b - x) dt '

~~~----------~~x

o

=

0.5

2.5

where k is a constant for the reaction. Integrate both sides of this equation to obtain a relation between x aod t (a) if a ~ b, aod (h) if a #' b. Assume in each case that x = 0 when t = O.

8.5 Integral Tables and Computer Algebra Systems

8.5

463

Integral Tables and Computer Algebra Systems In this section we discuss how to use tables and computer algebra systems to evaluate integrals.

Integral Tables A Brief Table of Integrals is provided at the back of the book, after the index. (More extensive tables appear in compilations such as CRC Mathematical Tables, which contain thousands of integrals.) The integration fonnulas are stated in terms of constants a, b, c, m, n, and so on. These constants can usually assume any real value and need not be integers. Occasional limitations on their values are stated with the formulas. Formula 5 requires n oF - I , for example, and Formula 11 requires n oF -2. The formulas also assume that the constants do not take on values that require dividing by zero or taking even roots of negative numbers. For example, Formula 8 assumes that a oF 0, and Formulas 13a and 13b cannot be used unless b is positive.

EXAMPLE 1

Find

J

+ 5r l dx.

x(2x

We use Formula 23 at the back of the book (not 22, which requires n oF -I):

Solution

J

x(ax

+ b)-ldx

:2

=

~-

=

I - ~ In 12x + 51 + C.

In lax

+ bl + C.

With a = 2andb = 5,wehave

J

x(2x

EXAMPLE 2

+ 5r l dx

•

Find

Jxvb' Solution

We use Formula 29a:

J

dx xVax-b

=

~tan-l~ax Vb

- b b

+

C.

With a = 2 andb = 4, we have

JxV2x dx

EXAMPLE 3

=

4

~tan-l ~2x -

v'4

+

JV

n . -1 x n+ 1 . -1 a x sm axdx--+ Ism ax--+ n n I

J

C = tan-I ~x - 2 2

+ C.

Find

We begin by using Formula 106:

Solution

4

4

xn+l dx 1 _ a 2x 2

'

n oF -\.

•

464

Chapter 8: Techniques of Integration With" = 1 and a = 1, we have

/

1/

2 . -I x. -I dx _ x xsm x -2 sm Z

2 x dx

~.

Next we use Fonnula 49 to f"md the integral on the right:

/

x2 dx Va2-x2

=

2 a sin-I 2

(~)

_ lxVa 2 _ x 2 + C. 2

Witha = 1,

x2dx /

~

1. -I X - -1'''------'1 =-sm X V I - X -2 2 2

+ C.

The combined result is /

X

sin-I xdx = x; sin-I X

=

-

tx~ +

t (tsin-I x -

C)

x2 1). -I + 1 '''------'12 + C' . ( 2-4 sm x 4 XV1 -x-

•

Reduction Formulas The time required for repeated integrations by parts can sometimes be shortened by applying reduction fonnulas like

tannxdx

=

/ / (lnx)n dx sin"xcosmxdx=

_1_ tann-I x - /tan n- 2xdx " - 1

(1)

x(lnx)n - " / (lnx)n-I dx

(2)

=

m+1 1/ . n-I sm mX~O: X+~~n sin,,-2xcosmxdx

/

(" "" -m). (3)

By applying such a fonnula repeatedly, we can eventually express the original integral in terms of a JXlVI"f low enough to be evaluated directly. The next example illuatrates this procedure.

EXAMPLE 4

Find

/ tanSxdx. Solution

We apply Equation (1) with 11 = 5 to get

/ tan'xdx

=

itan4x - / tan3xdx.

We then apply Equation (1) again, with" = 3, to evaluate the remaining integral:

/ tan3xdx

2

= ttan x - /

tanxdx

2

= ttan x

+ In Icosxl +

C.

The combined result is

/ tan'xdx

=

itan4x - ttan2 x -In Icosxl + C'.

•

As their fonn suggests, reduction fonnulas are derived using integration by parts. (See Example 5 in Section 8.1.)


465

Integration with a CAS A powerful capability of compoter algebm systems is their ability to integrate symbolically. This is performed with the integrate command specified by the particular system (for example, int in Maple, Integrate in Mathematica).

EXAMPLE 5

Suppose that you want to evaluate the indef'mite integml of the function

f(x) = x 2Ya 2 + x 2. Using Maple, you first define or name the function:

> f:=

x A2 * sqrt (a A2

+ xA2);

Then you use the integmte command on f, identifYing the variable ofintegmtion:

> int(j,x); Maple retorns the answer

t

x(a 2 + x 2)3/2 - ta 2xYa 2 + x 2 -

ta 4Jn (x + Ya 2 + X2).

If you want to see if the answer can be simplified, enter

>

simplify(%);

Maple retorns

ta 2xYa 2 + x 2 +

t x3Ya2 + x 2 - ta 4Jn (x + Ya2 + X2).

If you want the def'mite integmlfor 0 :5 X :5 1T/2, you can use the format

>

int(j, x = 0.. Pi/2);

Maple will retorn the expression

~1T(4a2 + ,.2)(3/2) - ~a21TY4a2 + 1T2 + la 4 Jn(2) 64

32

-

ta 4Jn(1T +

8

Y4a 2 +,.2) +

1~a4Jn(a2).

You can also f'md the definite integral for a particular value of the constant a:

> a:= 1;

> int(j,x

=

0.. 1);

Maple retorns the numerical answer

iYz + tJn(Yz - I). EXAMPLE 6

Solution

•

Use a CAS to find

With Maple, we have the entry

>

int«sin A2)(x)

* (cosA3)(x),x);

with the immediate return -tsin(x) cos (X)4

+ 1~

cos(xj2sin(x)

+

2 15 sin(x).

•

Compoter algebm systems vary in how they process integmtions. We used Maple in Examples 5 and 6. Mathematica would have returned somewhat different results:

466


1.

In Example 5, given

In [1J:= Integrate [x'2' Sqrt [a'2

+ x'2],x]

Mathematica returns Out[lJ= Va 2 + x 2

(a~x + x;)

-

t

a4LOg[X

+ Va 2 + X2]

without having to simplify an intermediate result. The answer is close to Formula 22 in the integral tables. 2.

The Mathematica answer to the integral

In [2J:= Integrate [Sin [x]Az' Cos [x],3, x] in Example 6 is Sin [x] I. I . Out[2J= - - - -Sm[3x] - -Sm[5x] 8 48 80 differing from the Maple answer. Both answers are correct. Although a CAS is very powerful and can aid us in solving difficult problems, each CAS has its own limitations. There are even situations where a CAS may further complicate a problem (in the sense of producing an answer that is extremely difficult to use or interpret). Note, too, that neither Maple nor Mathematica retorns an arbitrary constant +C. On the other hand, a little mathematical thinking on your part may reduce the problem to one that is quite easy to handle. We provide an example in Exercise 67.

Nonelementary Integrals The development of computers and calculators that find antiderivatives by symbolic manipulation has led to a renewed interest in determining which antiderivatives can be expressed as finite combinations of elementary functions (the functions we have been stodying) and which cannot. Integrals of functions that do not have elementary antiderivatives are called nonelementary integrals. They require infinite series (Chapter 10) or numerical methods for their evaluation, which give only an approximation. Examples of nonelementary integrals include the error function (which measures the probability of random errors) 2 erf(x) = ...;:;;.

10f'

e-I'dt

and integrals such as

f

2

sinx d!;

and

that arise in engineering and physics. These and a number of others, such as

f

ex

-xdx,

f ~xd!;, f

in (in x) d!;,

f

sinx d!; x '

0< k < 1, look so easy they tempt us to try them just to see how they tom out. It can be proved, however, that there is no way to express these integrals as fmite combinations of elementary functions. The same applies to integrals that can be changed into these by substitution. The integrands all have antiderivatives, as a consequence of the Fundamental Theorem of Calculus, Part 1, because they are continuous. However, none of the antiderivatives are elementary. None of the integrals you are asked to evaluate in the present chapter fall into this category, but you may encounter nonelementary integrals in your other work.


467

Exercises 8.S Using Integral Tables Use the table of integrals at the back of the book to evaluate the integrals in Exercises 1-26.

1/ · xVx-=-3

2/ · xVx+4

dx

dx

xdx

4 / •

(2x

5./X~dx

6. /

x(7x + 5)'/2 dx

7·/7

8 / dx • x 2v'4x - 9

3 /

·

9. /

11 /

·

13.

\Ix=2

dx

/~dx

xv'4x -x2 dx

10.

dx xv7+?-

12 /

dx x~

·

/~ dx

14.

xdx + 3)'/2

/~dx

15. /

e 2t cos 31 dt

16. /

e-3t sin 4t dt

17. /

X COS-IX tb:

18. /

xtao-l xdx

19. / 21. /

X

2tao- l xdx

20.

sin 3x cos 2x dx

I

23. /

8 sin 4t sin

25. /

cos~cos£ dO

dt

2

38. /

40. /

x v'x 2 - 4x

+5

dx

Using Reduction Fonnulas Use reduction fonnulas to evaluate the integrals in Exercises 41-50. sin' 2x dx

42. /

8 cos' 27ft dt

43. / sin2 20 cos' 20 dO

44. /

2 sin2 t sec' t dt

41. /

45. /

4 tao' 2x dx

46.

47. /2sec'7fXdx 49. /

cscs x dx

24• /

. Sffi

26. /

cOS~COS70dO

t . t dt

3 S1n"6

51. /

e'sec' (e' - l)dt

/

29. / 31. / 33. /

34. / 36. /

(x 2 + 1)2 dx

28.

(x2

+ 3)' dx

sin- l Vidx Vi

~

0
0, the improper integral can be interpreted as the (finite) area be• neath the curve and above the x-axis (Figure 8.15).

The Integral

l

""dx Ii X

1

The function y = l/x is the boundary between the convergent and divergent improper integrals with integrands of the form y = l/x p • Ai; the next example shows, the improper integral converges if p > 1 and diverges if p :5 1.

EXAMPLE 3

For what values ofp does the integral tegral does converge, what is its value?

Solution

Jt' dx/x

P

converge? When the in-

If p "" 1,

I

1

bdx -

xP

=

x -p+ 1

-p

+1

]b 1

= _1_ b-p+l -

1- p(

1 = -1- (1 - - - 1) ) 1 - P b P- 1

•

Thus,

=

b~oo [1 - p (b

I

lim _1_ _ 1__ 1 P- 1

= ) ]

p> 1

P - l' { co,

p

< 1

because lim _1_ = {O,

h-OO bP - 1

00,

p>1 p 0+:

Therefore the area under the curve from 0 to 1 is finite and is defined to be

DEFINmON Integrals of functions that become infmite at a point within the interval of integration are improper integrals of Type II. 1. If fix) is continuous on (a, bl and discontinuous at a, then b fix) dx = lim+ib fix) dx.

i i a

C-Q

c

2. If fix) is continuous on [a, b) and discontinuous at b, then

a

b fix) dx = lim_ic fix) dx. c-b

a

3. If fix) is discontinuous at e, where a [a, c) U (e, bl, then [

fix) dx = [f{X) dx

+

O. Is the sine-integral function everywhere increasing or decreasing? Do you think Si (x) = 0 for x > O? Check your answers by graphing the function Si (x) for 0 ,.;; x ,.;; 25.

=

The function

- -I - e _1(=)2 2 u

u\,f2;

is called the normal probability density jUnction with mean f.L and standard deviation u. The number f.L tells where the distribution is centered, and u measures the "scatter" around the mean. From the theory of probability, it is known that

b

idx.

1:

y

f(x) dx

=

1.

In what follows, let f.L = 0 and u = I.

D

o

a. Draw the graph of j. Find the intervals on which f is increasing, the intervals on which f is decreasing, and any local extreme values and where they occur. b. Evaluate

However, the integral

forn = 1,2,and3. c. Give a convincing argument that

1:

for the volume of the solid converges.

f(x) dx = 1.

a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But ifwe fill the hom with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction. 75. Sine-integral function

1

e -x/2 for x

> I, and for b > I,

00

e-x/ 2 dx --+ 0

as

b --+ 00.)

78. Show that if f(x) is integrable on every interval of real numbers and a and b are real numbers with a < b, then

laoo f(x) dx both converge if and only if f~oo f(x) dx and hoo f(x) dx both converge.

x

Si(x) =

< f(x)
"" --.:

21. Centmid of a region Find the centroid of the region in the flISt quadrant that is bounded below by the »-axis, above by the curve y - In x, and on the right by the line;t - e.

8. lim ;til c~t dt l:--->o+ % t

sintdt

Evaluate the limits in Exercises 9 and 10 by identifying them with definite integrals and evaluating the integrals.

.-,

10oHm~

....... "" .1;-0

1

Vn 2 - t2

Applfcatlons 11. Finding are length Find the length of the curve Y

-1%

Vcos2tdt,

21. Centmid of a region Find the centroid of the region in the plane enclosed by the curves y - ±( I - ;t2tl/2 and the lines ;t - Oand;t - 1. 23. Length of. curve Find the length of the curve y - In;t from ;t - lto;t - e. 24. Finding mrfaee area Find the area of the surface generated by revolving the curve in Exercise 23 about the y-axis.

25. The.urfaee poenrted by an astroid The gnlph of the equation ;t2f3 + yl/3 _ I ill an tJ.ftroid (see accompanying ftgure). Find the area of the surface generated by revolving the curve about the x-Ws.

O:!;';;t:!;'; 'lr/4.

11. Finding are lengtb Find the length of the graph of the fimction y - In(1 _;t2), O:!;'; x:!;'; 1/2.

y

13. Finding volume The region in the ftrst ~t that is enclosed by the ;t-axis and the curve y = 1x"~ is revolved about the y-axis to generate a solid. Find the volume of the solid. 1... FIndIng volume The region in the first quadrant that is enclosed by the".ui>. the ""'" Y ~ '/(x~)o the .... ~ 1 and ;t = 4 is revolved about the »-axis to gwerate a solid. Find the volume of the solid.

""

x

15. FIndIng volume The region in the flIst quadrant enclosed by the coordinate axes, the curve y = er • and the line ;t = 1 is revolved about the y-axis to generate a solid. Find the volume of the solid. 16. Finding volume The region in the flISt quadrant that ill bounded above by the curve y - er - 1, below by the x-axis, and on the

-I

Z6. Lengthofaclll'ft Find the leug1h of the curve y= [%y'V,-ldt.

right by the line ;t - In 2 is revolved about the line ;t - In 2 to generate a solid. Find the volume of the solid.

27. For what value or values of a does

17. Finding volume Let R be the ''triangular"' region in the flIst quadrant that is bounded above by the line y = 1, below by the curve y = In;t, and on the left by the line ;t = 1. Find the

1-

volume of the solid generated by revolving R about L the ;t-axis. b. the line y - 1.

18. FInding volume (Continuation of Exercise 17.) Find the volume of the solid generated by revolving the region R about L the y-axis. b. the line x = 1. 19. FInding volume The region between the;t-axis and the curve Y

~

f(x)

~ {Oo

xlnx.

;t=0

0<x:s2

is revolved about the »-axis to generate the solid shown here. L Show that f is continuous at;t = O. b. Find the volume of the solid.

[G::

Isxsl6.

ix)dx

converge? Evaluate the corresponding integral(s). 28. For each ;t > O. let G(;t) for each x > O.

10

00

e --,.t dt. Prove that ;tG(;t) - 1

29. lnf"mite area IIDd (mite wlume What values of p have the folIawing property: The area of the region between the curve y - ;t---P, 1 :!;'; ;t < 00. and the ;t-axis is infmite but the volume of the solid generated by revolving theregi.on aboutthe;t-axis is finite. 30. Infbdte area and fbdte volume What values ofp have the :follawing property: The area of the region in the flISt quadrant enclosed by the curve y - ;t---p , the y-axis, the line x - I, and the interval [0, I] on the x-axis ill infinite but the volume of the solid genemted by revolving the region about one of the coon1ina1J; axes is finite.

The Gamma Function and Stlrlfng's Formula Euler's gamma fimction f(;t) ("gamma ofr'; f is a Greek capital g) uses an integral to extend the factorial function from the nonnegative integers to ot:heI" real. values. The formula is

_1

00

f(;t)

r1e-tdt,

;t> O.

For each positive;t. the number f{x) is the integral oftr-1e ~ with respect to t from. 0 to 00. Figure 8.21 shDws the graph of f near the origin. You will sec how to calculate f(l/2) if you do Additional Exercise 23 in Chapter 14.

Chapter 8 Additional and Advanced Exercises y

493

As you will see if you do Exercise 104 in Section 10.1, Equation (4) leads to the approximation

~"'~.

(5)

o b. Compare your calculator's value for n! with the value given by Stirling's approximation for n = 10, 20, 30, ... , as far as your calculator can go.

o c. A refmoment of Equation (2) gives

---r--±--+~+-~--~~~->X

o

2

3

-1

f(x) =

-2 -3

(~Y l#e1/(l2x)(1

or f(x) '"

FIGURE 8.21 Euler's gamma function f(x) is a continuous function ofx whose value at each positive integer n + 1 is n!. The deftning integral fonnula for f is valid only for x > 0, but we can extend f to negative noninteger values of x with the fonnula f(x) = (f(x + 1»/x, which is the subject of Exercise 31. 31. H II is a nonnegative integer, f(1I

+ 1)

(6) Compare the values given for 1O! by your calculator, Stirling's approximation, and Equation (6).

Tabular Integratton The technique of tabular integration also applies to integrals of the f(x )g(x) fix when neither function can be differentiated reform peatedly to become zero. For example, to evaluate

J

= II!

b. Then apply integration by parts to the integral for f(x show that f(x + I) = xf(x). This gives

+

f(4) = 3f(3) = 6

~andits derivatives

= nf(n) = n!

(1)

Use mathematical induction to verify Equation (1) for every nonnegative integer n.

32. Stirling'. formula

Scottish mathematician James Stirling (1692-1770) showed that

,~('iYfIrf(X)=

integral.

e'"'~

cos x

2e2x~inX (+) ~ -cosx

4e2¥

<E-

Stophere:Rowissameas [lI'St row except for multiplicative constants (4 on the left, -Ion tire right).

We stop differentiating and integrating as soon as we reach a row that

is the same as the fIrst row except for multiplicative constants. We in-

.(x) --+ 0 as x --+ oc.

(2)

f21r (ex)' \j--x

J

=

+(e2x siox) - (2e'"'(-cosx»

+ (Stirling'. formula).

(3)

a. Stirling'. approximation for II! Use Equation (3) and the fact that n! = nf(n) to show that

(~)" VZ;;;

terpret the table as saying

e2x cosx fix

Dropping .(x) leads to the approximation

n! '"

cos x and its

I,

so, for large x,

f(x) '"

e2;rcosxdt

we begin as before with a table listing successive derivatives of e2x and integrals of cos x:

f(3) = 2f(2) = 2

c.

J

I) to

f(2) = If(1) = I

+ 1)

(~Y l # e W2x),

which tells us that

a. Show that f(l) = 1.

[(n

+ .(x»

(Stirling'. approximation).

(4)

J

(4 e 2x)( -cos x) fix.

We take signed products from the diagonal arrows and a signed iot.,. gra1 for the last horizontal arrow. Transposing the integral on the righthand side over to the left-hand side now gives

5

J

e2xcos x dx

=

e2x sinx

+ 2e2xcosx

494


or

.

2z

(9)

smx=--2·

/

. " dx -- e " smX' +5 2" e cosX' e cosx

I

+ C'

+z

Finally, x ~ 2tan-1 z,so

after dividing by 5 and adding the constant of integration. Use tabullIr integration to evaluate the integrals in Exercises 33-40. 33. /

e2t"cos3xdt

34. /

e"'sin4xdx

35. /

sin 3x sinxdt

36. /

cos 5x sin 4x dx

37. /

e~sinbxdx

38. /

e~ cos bx dx

(10)

Examples _/ I dx~/I+.'~ • I+cosx 2 1+.'

=/dz=Z+C

40. / x2 1n (ax) dx

39. / In (ax) dx

= tan The Substitution z The substitotion

~

(f) + C

tan (x/2)

z = tan~ 2

(7)

I

b. /

2

+ sinx

dx = /

reduces the problem of integrating a rational expression in sin x and cos x to a problem of integrating a rational function of z. This in turn can be integrated by partial fractions. From the accornpaoying figure

2

I + z2 2dz + 2z + 2z2 1 + z2

= / .' +

~+ I = /

(z +

(I/~)2 + 3/4

=/,;~,; =

~tan-l (~)

+ C

= ~tan-12z + I + C yI3 yI3 2 _II + 2 tan (x/2) =yI3tan yI3 +C

Use the substitotions in Equations (7}-{1O) to evaluate the integrals in Exercises 41-48. Integrals like these arise in calculating the average aogular velocity of the output shaft of a uoiversal joint wben the input and output shafts are not aligned.

we can read the relation

tan'! 2

=

1

sin x

41. /

+ cosx·

To see the effect of the substitution, we calculate 43.

cosx

~ 2 cos

2

I cosX'

+

1-

(f) - I ~

sec2

~X/2) -

I - dx. smX'

W!2

I

W!2 45. /. 0

2 _1~_2_ _ 1 tan2 (x/2) I + z2

47. / .

z2

= --2' I +z

dx

. smX'

44.

dO 2+cosO

46

1

/.o

+

dl

smt - cost

(8)

~I~~ . ....dx'"'"c-_ _ + SInX' + cosX'

42. /

l

W !2

'TT/3

1

2w

• ",/2

48. /

/3

dx 1 - cosX' cos 0 dO sinO cos 0 + sinO

cosldl 1 - cost

Use the substilution z = tan (0/2) to evaluate the integrals in Exercises 49 and 50.

and

. . x x sin (x/2) 2 (x) smx ~ 2 sm Icosl ~ 2 cos (x/2) • cos I

~ 2tan"-.

I 2 sec (x/2) 2

2 tan (x/2) I + tan2 (x/2)

49. /

secOdO

50. /

cscOdO


Chapter

3


Mathematica/Maple Modules: Riemllllll, Tropezoidol, IIIId SiIIIpsonApproxilllations Part I: Visualize the error involved in using Riemann sums to approximate the area under a curve.

l'lIrt II: Build a table of values and compute the relative magnitude of the error as a function of the step size 4. -I

Ld;+R'~O dt l ,

0 < 6 < 1r/2

which is Equation (5) with V

Solve the initial value problems in Exercises 15-20.

+ 2y

16. 1dy dl

~ 3,

c. Show that the value of the current when 1 ~ L/R is I/e. (The significaoce of this time is esplained in the next exercise.)

+ 2y -_ 3 I, 1 > 0, y (2)

~

I

6> 0, Y(1r/2) ~ I

dy _ 3 18. 6 d6 - 2y - 6 sec 6 tan 6,

6> 0,

dy

y(1r/3)

~

2

x'

19. (x

+ I) dx - 2(x 2 + x)y ~ x e+ I' x > -I, y(O) ~ 5

dy 20. dx

+ xy

~

O.

fall to half its original value?

y(O) ~ I

dy . 17. 6 d6 + Y ~ sm6,

~

a. Solve the equation to express i as a function of t. b. How loog after the switch is thrown will it take the current to

Solving In;tial Value Problems dy 15. dl

509

26. Current in an open RL circuit If the switch is thrown opeo after the current in an RL circuit has built up to its steady-state value I ~ VIR, the decaying current (see accompaoying figure) obeys the equatioo

0 < 6 < 1r/2

tan6,

First-Order Linear Equations

3!-

x, y(O) ~ -6

R

21. Solve the expooeotial growth/decay initial value probl= for y as • functioo of 1 by thinking of the differential equation as a rITStorder linear equation with P(x) ~ -kandQ(x) ~ 0:

dy dl

~

ky

(keonstant),

y(O)

~

Yo

22. Solve the following initial value probl= for u as a functioo of t.

~~ + ~u ~

0 (kaodm positive coostants), u(O)

~

Uo

27. TIme constants Engineers call the number L/R the time constanl of the RL circuit in Figure 9.9. The sigoificance of the time coostant is that the current will reach 95% of its final value within 3 time coostants of the time the switch is closed (Figure 9.9). Thus, the time coostant gives a built-in measure of how rapidly an individual circuit will reach equilibrium. •• Find the value of; in Equation (7) that correspoods to 1 ~ 3L/R and show that it is about 95% of the steady-state value I ~ V/ R.

b. Approximately what percentage of the steady-state current will be flowing in the circuit 2 time coostants after the switch is closed (i.e., wheo 1 ~ 2L/R)?

a. as a flIst-order linear equation. b. as a separable equatioo.

28. Derivation of Equation (7) in Example 4 a. Show that the solution of the equation

Theory and Examples 23. Is either of the following equatioos correct? Give reasoos for your

answers. b.

x/

tdx

~ xlnlxl + ex

24. Is either of the following equatioos correct? Give reasoos for your

answers. a.

co~x/ cosX' dx =

1/

b. cosX'

cosX' ax

=

is

; ~ l' + Ce -(R/Ll' tanx

+C

tanx

+

C

cosX'

25. Current in a closed RL circuit How maoy seeonds after the switch in an RL circuit is closed will it take the current ; to reach half of its steady-state value? Notice that the time depeods 00 R and L and not 00 how much voltage is applied.

R

b. Theo use the initial cooditioo ;(0) ~ 0 to determine the value of C. This will complete the derivation ofEquatioo (7). c. Showthati ~ VIR is a solution of Equation (6) and that i = Ce -{RILlt satisfies the equation

di dl

R.

+ I'

~

o.

510

Chapter 9: First-Order Differential Equations we have n ~ 2,

James Bernoulli

so thatu ~ yl-2 ~ y-l anddu/dx ~ -y-2dy/dx.Thendy/dx ~ -y 2 du/dx ~ -u-2 du/dx.

(1654-1705)

Substitution into the original equation gives

HiSTORICAL BIOGRAPHY

A Bernoulli differential equation is of the form

dy dx

Observe that, if n

+ P(x)y

~

Q(x)y".

or, equivalently,

0 or I, the Bernoulli equation is lioear. For other values of n, the substitution u = y I-n transforms ~

du dx

+ u = -e-x.

the Bernoulli equation into the linear equation

: + (I -

n)P(x)u

~ (1 -

This last equation is linear in the (unknown) dependent variable u.

n)Q(x).

Solve the Bernoulli equations in Exercises 2~32.

For example, in the equation dy - y dx

9.3

~ e~y2

29. y'

-y~

_y2

30. y' - Y ~xy2

31. xy'

+y

y-2

32.

~

x'y' + 2xy

~ y'

AppLications We now look at four applications of flfSt-order differential equations. The flfSt application analyzes an object moving along a straight line while subject to a fOrce opposing its motion. The second is a model of population growth. The third application considers a curve or curves intersecting each curve in a second family of curves orthogonally (that is, at right angles). The fmal application analyzes chemical concentrations entering and leaving a container. The various models involve separable or liuear flfSt-<mler equations.

Motion with Resistance Proportional to Velocity In SOIl3e cases it is reasonable to assume that the resistance encountered by a moving object,

such as a car coasting to a stop, is proportional to the object's velocity. The faster the object moves, the more its forward progress is resisted by the air through which it passes. Pictore the object as a mass m moving along a coordinate liue with position function s and velocity vat time t. From Newton's second law of motion, the resisting force opposing the motion is Force = rnass X acceleration

~

m ~~ .

If the resisting force is proportional to velocity, we have dv

m dt = -kv

or

dv k dt ~ - m v

(k

> 0).

This is a separable differential equation representing exponential change. The solution to the equation with initial condition v ~ Vo at t = 0 is (Section 7.4) (1)

What can we learn from Equation (I)? For one thing, we can see that if m is something large, like the rnass of a 20,000-ton ore boat in Lake Erie, it will take a long time for the velocity to approach zero (because t must be large in the exponent of the equation in order to make kt/m large enough for v to be small). We can learn even more ifwe integmte Equation (1) to find the position s as a function of time t. Suppose that a body is coasting to a stop and the only force acting on it is a resistance proportional to its speed. How far will it coast? To find out, we start with Equation (1) and solve the initial value problem ds _

dt - voe

-(klm)t

,

s(O)

~

O.

9.3

Applications

511

Integrating with respect to I gives v~m e-(klm)t

S = -

Substitoting S

~

0 when I

~

+ C.

0 gives

vom o ~--+ k

C

vom k

C~-

and

The body's position at time I is therefore

_ vom -("m)t ()st -Te "

vom _ vom (

+ T - T I-e

-(klm)t)

.

(2)

To find how far the body will coast, we fmd the limitofs(l) as 1--> 00. Since -(kim) < 0, we know that e -(klm)t --+ 0 as I --+ 00 , so that

lim S(I) ~ lim vom (I t_OO k

e-(klm),)

t_OO

~ v~m (I _ 0) ~ v~m .

Thus, Distance coasted

=

vom

T

(3)

The number vom/k is only an upper bound (albeit a useful one). It is true to life in one respect, at least: if m is large, the body will coast a long way.

l In the English system, where weight is measmed in pounds, mass is measured in slugs. Thus, Pounds

~

slugs X 32,

assuming the gravitational constant is 32 ft./sec'.

EXAMPLE 1

For a 192-lb ice skater, the k in Equation (I) is about 1/3 slug/sec and

m = 192/32 = 6 slugs. How long will it take the skater to coast from 11 ftjsec (7.5 mph)

to I ft/sec? How far will the skater coast before coming to a complete stop? Solution We answer the flISt question by solving Equation (I) for I: lIe-tl18 ~ e-t118 ~

Eq. (I) with k ~ 1/3, m = 6, Vo = 11, v = 1

I

1/11

-1/18

~

In (1/11)

~

I

~

18ln 11 '" 43 sec.

-In 11

We answer the second question with Equation (3):

. D.stance coasted

vom

=

T

=

198 ft.

~

11·6 1/3

•

Inaccuracy of the Exponential Population Growth Model In Section 7.4 we modeled popolation growth with the Law of Exponential Change: dP iii =

P(O)

!P,

~

Po

whereP is the population at time t, k > 0 is a constant growth mte, and Po is the size of the population at time I ~ O. In Section 7.4 we found the solution P ~ Poe kt to this model. To assess the model, notice that the exponential growth differential equation says that dPt

dl

= k

(4)

512

Chapter 9: First-Order Differential Equations

TABLE 9.3 WorLd popuLation (midyear)

f

7000

World population (1980-2008)

( 5000

~O~------~I~O---Av

I

30

)

t

FIGURE 9.10 Notice that the value of the solutionP ~ 4454e°.0 17t is 7169 when t ~ 28, which is nearly 7% more than the actual population in 2008.

Orthogooal trajectory

Year

Population (millions)

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989

4454 4530 4610 4690 4770 4851 4933 5018 5105 5190

76/4454 80/4530 80/4610 80/4690 81/4770 82/4851 85/4933 87/5018 85/5105

'" '" '" '" '" '" '" '" '"

0.0171 0.0177 0.0174 0.0171 0.0170 0.0169 0.0172 0.0173 0.0167

Source: u.s. Bureau of the Census (Sept., 2007): WWW.ceDSUS .gov/ipc/www/idb.

is constant. This rate is called the relative growth rate. Now, Table 9.3 gives the world population at midyear for the years 1980 to 1989. Taking dt = I and tiP '" l1P, we see from the table that the relative growth rate in Equation (4) is approximately the constant 0.017. Thus, based on the tabled data with t = 0 representing 1980, t = I representing 1981, and so forth, the world population could be modeled by the initial value problem,

tiP Iii =

FIGURE 9.11 An nrthogonal trajectory intersects the family of curves at right angles, or nrthogonally.

l1P/P

0.017P,

P(O) = 4454.

The solution to this initial value problem gives the population function P = 4454eo.017t.1n year 2008 (so t = 28), the solution predicts the world population in midyear to be about 7169 million, or 7.2 billion (Figure 9.10), which is more than the actoa1 population of 6707 million from the U.S. Bureau of the Census. A more realistic model would consider environmental and other factors affectiog the growth rate, which has been steadily decliniog to about 0.012 since 1987. We consider one such model in Section 9.4.

Orthogonal Trajectories y

An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family at right angles, or orthogonally (Figure 9.11). For instance, each straight line through the origin is an orthogonal trajectory of the family of circles x2 + y2 = a 2 , centered at the origin (Figure 9.12). Such mutoa1ly orthogonal systems of curves are ofparticular importance in physical problems related to electrical potential, where the curves in one family correspond to strength of an electric field and those in the other family correspond to constant electric potential. They also occur in hydrodynamics and heat-flow problems.

EXAMPLE 2 Find the orthogonal trajectories of the family of curves xy a .. 0 is an arhitrary constant.

FIGURE 9.12 Every straight line through the origin is orthogonal to the family of circles centered at the origin.

= a, where

Solution The curves xy = a form a family of hyperbolas having the coordinate axes as asymptotes. First we find the slopes of each curve in this family, or their dy/ ax values. Differentiatiog xy = a implicitly gives

dy

xax+y=O

or

dy

Y

ax

X'

9.3

Applications

513

Thus the slope of the tangent line at any point (x, y) on one of the hyperbolas xy = a is y' = -y/x. On an orthogonal trajectory the slope of the tangent line at this same point must be the negative reciprocal, or x/yo Therefore, the orthogonal trajectories must satisfy the differential equation

dy dx

x y'

This differential equation is separable and we solve it as in Section 7.4:

ydy = xdx

Separate variables. Integrate both sides.

FIGURE 9.13 Each curve is orthogonal to every curve it meets in the other family

(Example 2).

(5)

where b = 2C is an arbitrary constant. The orthogonal trajectories are the family ofhyperbolas given by Equation (5) and sketched in Figure 9.13. •

Mixture Problems Suppose a chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas) with, possibly, a specified amount of the chemical dissolved as well. The nrixture is kept uniform by stirring and flows out of the container at a known mte. In this process, it is often important to know the concentration of the chemical in the container at any given time. The differential equation describing the process is based on the formula (mte at WhiCh) (mte at WhiCh) Rate of change of amount = chemical chemical in container arrives departs.

(6)

If y(t) is the amount of chemical in the container at time t and V(t) is the total volume of liquid in the container at time t, then the departure mte of the chemical at time t is y(t) Departuremte = V(t) • (outflow rate)

=

concentration in ) ( container at time t • (outflow rate) .

(7)

Accordingly, Equation (6) becomes

dy . . y(t) dt = (chenncal'samvalmte) - V(t) . (outflow rate).

(8)

If, say, y is measured in pounds, V in gallons, and t in minutes, the uuits in Equation (8) are pounds minutes

pounds pounds gallons minutes - gallons • minutes'

EXAMPLE 3 In an oil refinery, a storege tank contains 2000 gal of gasoline that initially has 100 lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 lb of additive per gallon is pumped into the tank at a rate of 40 gal/min.

514


The wt:ll-mixed solution is pumped out at a rate of 45 gal/min. How much of the additive is in the taJ:Jk 20 min after the pumping process begins (Figure 9.14)? 40 gaIImin containing 21b1pl

• FIGURE 9.14 The storage tank in Example 3 mixes input liquid with stored liquid to produce an output liquid.

Lety be the amolDlt (in pounds) of additive in the taJ:Jk at time I. We know that y = 100 when I = O. The number of gallons of gasoline and additive in solution in the taJ:Jk at any time I is

Solution

~ 2000 gal + (40: -

V(I)

~

45 : ) (tmin)

(2000 - 51) gal.

Thcrefon:,

y(1)

Rate out

=

- (200: ~

Eq.(7)

Y(I)' outflow rate

Ootflawrate ia45 gaJjmin and v - 2000 - St.

51) 45

45y Ib 2000 51 min .

Also, Hate in

~ (2~)(4O:) ~801b. mm

The diffcrcn:tial equation modeling the

mnture process is

dy 45y dt ~ 80 - 2000 51

Eq. (8)

in pmmds per minute. To solve this dift"erent:i.al equation, wt: rll"St write it in standard linear form: dy dt

45

51 Y ~ 80.

+ 2000

Thus, P(I) - 45/(2000 - 51) and Q(I) - 80. Th. integnUing factnr is v(l)

=

eJPdt

=

eJ~dt

= e-9In(2000-St)

~ (2000 -

5Ir'.

2000-5t> 0

9.3

Applications

515

Multiplying both sides of the standard equation by v(l) and integrating both sides gives 9 (2000 - 51r •

(2000 - 51r9

(t + 200~5_ Y)

t+

51

= 80(2000 - 51r

9

45(2000 - 51)-10 Y = 80(2000 - 51r9

1

:1 [(2000 - 51)-9y = 80(2000 - 51r9

(2000 - 51)-9y =

J

80(2000 - 51)-9 dl

-9 (2000 - 51r' (2000 - 51) Y = 80· (-8)(-5)

+ c.

The general solution is

y

= 2(2000 - 51)

+

C(2000 - 51)9.

Because y = 100 when I = 0, we can detennine the value of C: 100 = 2(2000 - 0)

C=

+

C(2000 - 0)9

3900 (2000)9·

The particular solution of the initial value problem is

The amount of additive 20 min after the pumping begins is y(20) = 2[2000 - 5(20)1 - ( 3900)9 [2000 - 5(20)1 9 .., 1342Ib. 2000

•

Exercises 9.3 Motion Along a Line 1. Coa.ting bicycle A 66·kg cyclist on a 7·kg bicycle starts coast· ing on level ground at 9 m/sec. The k in Equation (I) is about 3.9 kgfsec.

a. About how far will the cyclist coast before reaching a com· plete stop?

b. How long will it take the cyclist~ speed to drop to I m/sec? 2. Coa.ting battleship Suppose that an Iowa class battleship has rnass around 51,000 metric tons (51,000,000 kg) and a k value in

Equation (I) of about 59,000 kgfsec. Assume that the ship loses power when it is moving at a speed of9 mfsec. a. About how far will the ship coast before it is dead in the water?

b. About how long will it take the ship~ speed to drop to I mfsec? 3. The data in Table 9.4 were collected with a motion detector and a CBLTM by Valerie Sharritts, a mathematics teacher at St. Francis DeSales High School in Columbus, Ohio. The table shows the distance s (meters) coasted on in·line skates in t sec by her daughter Ashley when she was 10 years old. Find a model for Ashley's

516


position given by the data in Table 9.4 in the form of Equation (2). Her initial velocity was Vo ~ 2.75 mlsec, her mass m ~ 39.92 kg (she weighed 88Ib), and her total coasting distaoce was 4.91 m.

Orthogonal Trajectories In Exercises S-10, fmd the orthogonal trajectories of the family of curves. Sketch several members of each family. 6.y~cx2

5.y~mx

7. k:x;2

TABLE 9.4 Ashley Sharritts skating data

t (sec)

s(m)

t (sec)

s(m)

t (sec)

s(m)

0 0.16 0.32 0.48 0.64 0.80 0.96 1.12 1.28 1.44 1.60 1.76 1.92 2.08

0 0.31 0.57 0.80 1.05 1.28 1.50 1.72 1.93 2.09 2.30 2.53 2.73 2.89

2.24 2.40 2.56 2.72 2.88 3.04 3.20 3.36 3.52 3.68 3.84 4.00 4.16 4.32

3.05 3.22 3.38 3.52 3.67 3.82 3.96 4.08 4.18 4.31 4.41 4.52 4.63 4.69

4.48 4.64 4.80 4.96 5.12 5.28 5.44 5.60 5.76 5.92 6.08 6.24 6.40 6.56

4.77 4.82 4.84 4.86 4.88 4.89 4.90 4.90 4.91 4.90 4.91 4.90 4.91 4.91

+ y2

~

I

11. Show that the

8.2x2+y2~c2

10. y

9. y = ce --x curves 2x 2

+ 3y2

=

eb

~ S and y2 ~ x 3 are orthogonal.

12. Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families.

••

xdx+ydy~O

b.xdy-2ydx~0

Mixture Problems 13. Salt miItnre A tack initially contains \00 gal of brine in which SO Ib of salt are dissolved. A brine containing 2 Ib/gal of salt runs into the tack at the rate of S gal/min. The mixture is kept uniform by stirring and flows out of the tack at the rate of4 gal/min. •• At what rate (pounda per minute) does salt enter the tack at time I?

b. What is the volume of brine in the tank at time I? •• At what rate (pounda per minute) does salt leave the tack at

time t? d. Write down and solve the initial value problem describing ilie mixing process. e. Find the concentration of salt in ilie tank 25 min afler the

process starts. 4. Coasting to ••Iop Thhle 9.S shows the distaoce s (meters) coasU:d on in-line skates in terms of time t (seconds) by Kelly Schmilzer. Find a model for her position in the form of Equation (2). Her initial velocity was Vo ~ 0.80 mlsec, her mass m ~ 49.90 kg (110 Ib), and her total coasting distaoce was 1.32 m.

TABLE 9.5

•• At what time willilie tack be full?

b. At ilie time the tack is full, how many pounds of concentrate will it contain?

Kelly Schmitzer skating data

t (sec)

s(m)

t (sec)

s(m)

t (sec)

s(m)

0 0.1 0.3 0.5 0.7 0.9 1.1 1.3

0 0.07 0.22 0.36 0.49 0.60 0.71 0.81

1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9

0.89 0.97 1.05 1.11 1.17 1.22 1.25 1.28

3.1 3.3 3.5 3.7 3.9 4.1 4.3 4.5

1.30 1.31 1.32 1.32 1.32 1.32 1.32 1.32

9.4

14. MiItnre problem A 200-gal tack is half full of distilled water. At time t ~ 0, a solution containing O.Slb/gal of concentrate enters ilie tack at the rate of 5 gal/min, and the well-stirred mixture is wiilidrawn at ilie rate 00 gal/min.

15. Fertilizer mhture A tack contains 100 gal offresh water. A solution containing I Ib/gal of soluble \awn fertilizer ruos into the tack at ilie rate of 1 gal/min, and ilie mixture is pumped out ofilie tack at the rate 00 gal/min Find ilie maxinuun amount of fertilizer in ilie tack and the time required to reach the maxinuun. 16. Carbon monmide pollution An executive conference room of a corporation contains 4S00 It' of air initially free of carbon monoxide. Starting at time t ~ 0, cigarette smoke containing 4% carbon monoxide is blown into the racon at the rate of 0.3 It'/min. A ceiling fan keeps the air in the room well circulated and the air leaves ilie room at the same rate of 0.3 It'/min. Find the time when the concentration of carbon monoxide in the room reaches 0.01 %.

Graphical Solutions of Autonomous Equations In Chapter 4 we learned that the sign of the first derivative tells where the graph of a function is increasing and where it is decreasing. The sign of the second derivative tells the concavity of the graph. We can build on our knowledge of how derivatives determine the shape of a graph to solve differential equations graphically. We will see that the ability to

9.4 Graphical Solutions of Autonomous Equations

517

discern physical behavior from gmphs is a powerful tool in understanding real-world systems. The starting ideas for a gmphical solution are the notions of phase line and equilibrium value. We arrive at these notions by investigating, from a point of view quite different from that studied in Chapter 4, what happens when !be derivative of a differentiable function is zero.

Equilibrium Values and Phase Lines When we differentiate implicitly the equation

I

Sin (5y - 15)

= x

+ I,

we obtain

I( 5 )dY

S

5y - 15

dx = I.

Solving for y' = dy/dx we find y' = 5y - 15 = 5(y - 3). In this case the derivative y' is a function ofy only (the dependent variable) and is zero when y = 3. A differential equation for which dy/ dx is a function ofy only is called an autonomous differential equation. Let's investigate what happens when the derivative in an autonomous equation equals zero. We assume any derivatives are continuous.

DEnNmON If dy/dx = g(y) is an autonomous differential equation, then the values of y for which dy/ dx = 0 are called equilihrium values or rest points.

Thus, equilibrium values are those at which no change occurs in the dependent variable, so y is at rest. The emphasis is on the value of y where dy/ dx = 0, not the value of x, as we studied in Chapter 4. For example, the equilibrium values for the autonomous differential equation dy dx = (y

+

I)(y - 2)

arey = -I andy = 2. To construct a graphical solution to an autonomous differential equation, we fIrst make a phase line for the equation, a plot on the y-axis that shows the equation's equilibrium values along with the intervals where dy/dx and d'y/dx 2 are positive and negative. Then we know where the solutions are increasing and decreasing, and the concavity of the solution curves. These are the essential features we found in Section 4.4, so we can determine the shapes of the solution curves without having to find formulas for them.

EXAMPLE 1

Draw a phase line for the equation dy dx = (y

+ I)(y - 2)

and use it to sketch solutions to the equation.

Solution 1.

Draw a number line for y and mark the equilibrium values y = - I and y = 2, where dy/dx = O. ------1@)------------i@~--->, Y

-1

2

518

Chapter 9: First-Order Differential Equations 2.

IdentifY and label the intervals where y' > 0 and y' < 0, This step resembles what we did in Section 4.3, only now we are marking the y-axis instead of the x-axis, I I I I

y'>o

I I I I

y' 2, we have y' > 0, so a solution with a y-value greater than 2 will increase from there without bound, In short, solution curves below the horizontal line y = - I in the xy-plane rise toward y = - L Solution curves between the lines y = -I and y = 2 fall away from y = 2 toward Y = - L Solution curves above y = 2 rise away from y = 2 and keep going up,

3.

Calculate y' and mark the intervals where y' > 0 and y" entiate y' with respect to x, using implicit differentiation, y' = (y

+ I)(y - 2)

= y2 - Y -

2

O y">O

dx

differentiated implicitly with respect to x

2Y.)1' - y'

= (2y -

I)y'

= (2y -

I)(y

+

I)(y - 2),

From this formula, we see that y" changes sign at y = - I, y = 1/2, and y = 2, We add the sign information to the phase line. y'>O y" o. Figure 9.17 adds this information to the phase line. The completed phase line shows that if the temperatore of the object is above the equilibrium value of 15°C, the graph of H(t) will be decreasing and concave upward. If the temperature is below 15°C (the temperature of the surrounding medium), the graph of H(t) will be increasing and concave downward. We use this information to sketch typical solution curves (Figure 9.18).

520

Chapter 9: First-Order Differential Equations From the upper solution curve in Figure 9.18, we see that as the object cools down, the rate at which it cools slows down because 1iH/ dt approaches zero. This observation is implicit in Newton's law of cooling and contained in the differential equation, but the flattening of the graph as time advances gives an immediate visual representation of the phenomenon.

A Falling Body Encountering Resistance Newton observed that the rate of change in momentum encountered by a moving equal to the net force applied to it. In mathematical terms,

F

=

d

o~ect

is

(2)

dt (mv),

where F is the net force acting on the object, and m and v are the object's mass and velocity. Ifm varies with time, as it will if the o~ect is a rocket burning fuel, the right-hand side of Equation (2) expands to

using the Derivative Product Rule. In many situations, however, m is constant, dm/dt = 0, and Equation (2) takes the simpler form or

F=

'Ina,

(3)

s

known as Newton second law ofmotion (see Section 9.3). In free fall, the constant acceleration due to gravity is denoted by g and the one force acting downward on the falling body is

y

Fp FIGURE 9.19 An object falling under 1he influence of gravity wi1h a resistive force assumed to be proportional to 1he velocity.

=

mg,

the force due to gravity. If, however, we think of a real body falling through the air---1!ay, a penny from a great height or a parachutist from an even greater height-we know that at some point air resistance is a factor in the speed of the fall. A more realistic model of free fall would include air resistance, shown as a force F, in the schematic diagram in Figure 9.19. For low speeds well below the speed of sound, physical experiments have shown that F, is approximately proportional to the body's velocity. The net force on the falling body is therefore

F = Fp - F" giving

dv mdt=mg-ku dv k dt =g- m V '

(4)

We can use a phase line to analyze the velocity functions that solve this differential equation. The equilibrium point, obtained by setting the right-hand side of Equation (4) equal to

-

dv>O dt ____

I I I

dv v

---{@~~

mg

zero, is

mg v=T'

k

FIGURE 9.20 Initial pbase line for 1he falling body encountering resistance.

If the body is initially moving faster than this, dv/dt is negative and the body slows down. If the body is moving at a velocity below mg/k, then dv/ dt > 0 and the body speeds up. These observations are captured in the initial phase line diagram in Figure 9.20.

9.4 Graphical Solutions of Autonomous Equations

dv>O dt 2 d v dt 2

0 dt

I I I I

• , @

dP 0 if 0 < P < M and dP/ dt < 0 if P > M. These observations are recorded on the phase line in Figure 9.23. We determine the concavity of the population curves by differentiating both sides of Equation (6) with respect to t:

(7)

If P = M/2, then d 2p/dt 2 = O. If P < M/2, then (M - 2P) and dP/dt are positive and d 2p/dt 2 > O. If M/2 < P < M, then (M - 2P) < 0, dP/dt > 0, and d 2p/dt 2 < O.

522


I I I

I

If P > M, then (M - 2P) and dP/dt are both negative and d 2p/dt 2 > O. We add this information to the phase line (Figure 9.24). The lines P = M/2 and P = M divide the llfst quadrant of the tP-plane into horizontal bands in which we know the signs of both dP/dt and d 2p/dt 2• In each band, we know how the solution curves rise and fall, and how they bend as time passes. The equilibrium lines P = 0 and P = M are both population curves. Population curves crossing the line P = M/2 have an inflection point there, giving them a sigmoid shape (curved in two directions like a letter S). Figure 9.25 displays typical population curves. Notice that each population curve approaches the limiting population M as t --+ 00.

dP >0 dt I

2

' 4X>o I dt 2

@l o

J'

2 : d P I

dt2

I,

8. y' = y3 _ y2

a

Models of Population Growth The autonomous diffi:rential equations in Exercises 9-12 represeot models for populatioo growth. For each exercise, use a phase line analysis to sketch solution curves for p(t), selecting different startiog values P(O). Which equilibria are stahle, aod which are uostable?

dP 9. dt = I - 2P dP 11. dt

=

2P(P - 3)

dP 10. dt

= P( I

and that the current populatioo Po is fairly close to the carrying capacity Mo. You ntight imagine a populatioo of fish living in a freshwater lake in a wilderuess area. Suddeoly a catastrophe such as the Mount St Helens volcauic eruption contaminates the lake and destroys a siguificant part of the food and oxygen 00 which the fish depend. The result is a new enviromnent with a carrying capacity Ml coosiderably less than Mo and, in fact, less than the current population Po. Startiug at some time before the catastrophe, slretch a ''before-and-after'' curve that shows how the fish populatioo responds to the change in environment 14. Controlling a population The fish and game depar1ment in a certain state is planning to issue huntiog permits to cootrol the deer population (one deer per permit). It is knowo that if the deer populatioo falls below a certain level m, the deer will become extioct. It is also knowo that if the deer population rises above the carrying capacity M, the population will decrease back to M throngh disease and maIoutrition.

a. Discuss the reasonableness of the following model for the growth rate of the deer population as a function of time:

dP dt

= rP(M -

P)(P - m),

- 2P)

12 dP = 3P(i _ P)(p • dt

_1) 2

13. Catastropbic change in logistic growth Suppose that a healthy population of some species is growing in a limited environment

where P is the population of the deer and r is a positive constant of proportionality. Ioclude a phase line.

b. Explain how this model differs from the logistic model dP/ dt = rP(M - Pl. Is it better or worse than the logistic model?

9.5

> Mforallt,thenlim,_ooP(t) What happens if P < m for all t?

c. ShowthatifP d.

~

M.

e. Discuss the solutions to the differential equation. What are the equilibrium points of the model? Explain the dependence of the steady-stste value of P on the initial values of P. About how many pcrruits should he issued?

Applications and Examples 15. Skydiving If a body of mass m falling from rest under the ac· tion of gravity encouoters an air resistsoce proportional to the square of velocity, then the body's velocity t seconds into the fall

satisfies the equation dv 2 m-=mg-kv dt '

Systems of Equations and Phase Planes

523

a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of X' and X".

c. Sketch representative solution curves. d. Predict the value of X for which the information is spreading most rapidly. How many people eventually receive the infor-

mation? 19. Current in an RL-circuit The accoropanying diagram represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. From Section 9.2, we have

k>O

di

.

Ldt+RI~V,

where k is a constsot that depends on the body's aerodynamic properties and the density of the air. CWe assume that the fall is too short to be affected by changes in the air's density.)

where i is the current in amperes and t is the time in seconds.

v

a. Draw a phase line for the equation.

+ d~ -

b. Sketch a typical velocity curve.

~--b

c. For al60·lb skydiver (mg ~ 160) and with time in seconds and distsoce in feet, a typical value of k is 0.005. What is the diver's terminal velocity?

Switch

Vv

16. Resistance proportional to A body ofmassm is projected vertically downward with initial velocity Vo. Assume that the reo sisting force is proportional to the square root of the velocity and [mel the terruioal velocity from a graphical analysis. 17. Sailing A sailboat is running along a slraight course with the wind providing a constsot forward force of 50 lb. The only other force acting on the boat is resistance as the boat moves through the water. The resisting force is numerically equal to five times the boat's speed, and the initial velocity is 1 ft/sec. What is the maximum velocity in feet per second of the boat under this wind? 18. The spread of information Sociologists recognize a phenomenon called social diffUsion, which is the spreading of a piece of informa· tion, technological innovation, or cultoral fad among a population. The members of the population can he divided into tv. classes: those who have the information and those who do not In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the infor· mation times the number yet to receive it IfX denotes the nwnhcr of individuals who have the information in a population of N people, then a mathematical model for social diffusion is given by

L

Use a phase line analysis to sketch the solution curve assuming that the switch in the RL-circuit is closed at time t ~ O. What happens to the current as t ---> oo? This value is called the steady-state solution. 20. A pearl in shampoo Suppose that a pearl is sinking in a thick fluid, like shampoo, subject to a frictional force opposing its fall and proportional to its velocity. Suppose that there is also a resistive buoyant force exerted by the shampoo. According to Archimedes'principle, the buoyant force equala the weight of the fluid displaced by the pearl. Using m for the mass of the pearl and P for the mass of the shampoo displaced by the pearl as it desceods, coroplete the following steps. a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Figore 9.19. b. Using v(t) for the pearl's velocity as a func1ion of time t, write a differential equation modeling the velocity of the pearl as a falling body. c. Construct a phase line displaying the signs of v' and v".

dX

dt ~ kX(N - X),

d. Sketch typical solution curves.

where t represents time in days and k is a positive constant.

9.5

R

e. What is the terminal velocity of the pearl?

Systems of Equations and Phase Planes In some situations we are led to consider not one, but several first-order differential equations. Such a collection is called a system of differential equations. In this section we present an approach to understanding systems through a graphical procedure known as a phase-plane analysis. We present this analysis in the context of modeling the populations of trout and bass living in a common pond.

524


Phase Planes A general system of two first-order clifferential equations may tske the form

dx dt = F(x,y), dy dt = G(x,y). Such a system of equations is called autonomous because dx/ dt and dy/ dt do not depend on the independent variable time t, but only on the dependent variables x andy. A solution of such a system consists of a pair of functions x(t) and yet) that satisfies both of the differential equations simultaneously for every t over some time interval (finite or infinite). We cannot look at just one of these equations in isolation to find solutions x(t) or y(t) since each derivative depends on both x andy. To gain insight into the solutions, we look at both dependent variables together by plotting the points (x(t), yet»~ in the xy-plane starting at some specified point. Therefore the solution functions define a solution curve through the specified point, called a trajectnry of the system. The xy-plane itself, in which these trajectories reside, is referred to as the phase plane. Thus we consider both solutions together and study the behavior of all the solution trajectories in the phase plane. It can be proved that two trajectories can never cross or touch each other. (Solution trajectories are examples of parametric curves, which are studied in detail in Chapter 11.)

A Competitive-Hunter Model Imagine two species of fish, say trout and bass, competing for the same limited resources (such as food and oxygen) in a certain pond. We let x(t) represent the number of trout and yet) the number of bass living in the pond at time t. In reality x(t) and y(t) are always integer valued, but we will approximate them with real-valued differentiable functions. This allows us to apply the methods of clifferential equations. Several factors affect the rates of change of these populations. As time passes, each species breeds, so we assume its population increases proportionally to its size. Taken by itself, this would lead to exponential growth in each of the two populations. However, there is a countervailing effect from the fact that the two species are in competition. A large number of hass tends to cause a decrease in the number of trout, and vice-versa. Our model takes the size of this effect to be proportional to the frequency with which the two species interact, which in turn is proportional to xy, the product of the two populations. These considerations lead to the following model for the growth of the trout and hass in the pond:

dx dt = (a - by)x,

(la)

dy dt = (m - nx)y.

(lb)

Here x(t) represents the trout population, yet) the bass population, and a, b, m, n are positive constants. A solution of this system then consists of a pair of functions x(t) and yet) that gives the population of each fish species at time t. Each equation in (I) contains both of the unknown functions x and y, so we are unable to solve them individually. Instead, we will use a graphical analysis to study the solution trajectories of this competitive-hunter model. We now examine the nature of the phase plane in the trout-bass population model. We will be interested in the I st quadrant of the xy-plane, where x ;" 0 and y ;" 0, since populations cannot be negative. First, we determine where the bass and trout populations are both constant. Noting that the (x(t),y(t» values remain unchanged when dx/dt = 0 and dy/dt = 0, Equations (Ia and Ib) then become

(a - by)x = 0, (m - nx)y =

o.

This pair of simultaneous equations has two solutions: (x, y) = (0, 0) and (x, y) = (m/n, a/b). At these (x,y) values, called equilibrium or rest points, the two populations

9.5

525

Systems of Equations and Phase Planes

remain at constant values over all time. The point (0, 0) represeots a pond containing no members of either fish species; the point (m/n, a/b) corresponds to a pond with an unchanging number of each fish species. Next, we note that ify = alb, then Equation (Ia) implies dx/dt = 0, so the trout popillation x(t) is constant. Similarly, if x = min, theo Equation (Ib) implies dy/dt = 0, and the bass popu1ation y(t) is constant. This information is recorded in Figure 9.26. y Bass

y Bass

y Bass I

!!

-I, y(O) x

> 0,

+ 3y2) tho + Y dx = 0 (Hint: d(xy) = y dx + x tho) x tho + (3y - x-2 cosx) dx = 0, x> 0

20. x tho

21 . .>:y'

+ (y -

+ (x

22. y dx

cosx)dx = 0,

- 2)y

+ (3x

= h3e~,

- .>:y

y(l)

=

=

I

I

y(O) = -I

amx

13. (l + eX) tho + (ye x + e~) dx = 0 14. e~ tho + (e~y - 4x) dx = 0

15. (x

9. Why is the exponential model unrealistic for predicting long-term population grow1h? How does lbe logistic model correct for the deficiency in the exponential model for population grow1h? What is the logistic differential equation? What is the form of its solution? Desctibe the graph of the logistic solution.

In Exercises 17-22 solve lbe initial value problem.

6. y' =

7. x(x - I) tho - ydx = 0 9. 2y' - y

8. How do you construct lbe phase line for an autonomous differential equation? How does lbe phase line help you produce a graph which qualitatively depicts a solution to the differential equation?

Practice Exercises

In Exercises 1-16 solve the differential equation.

3. secxtho

7. What is an autonomous differential equation? What are its equilibtium values? How do they differ from ctitical points? What is a stable equilibtiwn value? Unstable?

+ 2) dy

y(~) y(l)

= 0,

=

= 0

0

y(2) = -I, y < 0

530


Euler's Method

Massm mgR2

In Exercises 23 and 24. use Euler's method to solve the initial value problem 00 the given interval starting at Xo with dx ~ 0.1.

0 23.

y' ~ y

+

cosx,

y(O) ~ 0;

0,;; x ,;; 2;

Xo

F~---

,2

~

~ 0

0 24. y' ~ (2 - y)(2x + 3), y(-3) ~ I; -3:sx:s -1; Xo =-3 In Exercises 25 and 26, use Euler's method with dx ~ 0.05 to estimate y(c) where y is the solution to the given initial value problem.

0 25. c ~ 3;

dy x-2y dx ~ X+T' y(O) ~ I

0 26. c ~ 4;

dy x 2 -2y+1 dx ~ x ,y(l) ~ I

In Exercises 27 and 28, use Euler's method to solve the initial value problem graphically, starting at Xo ~ 0 with L

0 27. dxdy

dx

~

0.1.

~

;

eX Y

+2'

b. dx

y(O)

~

~

-0.1.

2gR.

-

v'2iii

30. y' ~ I/x

29.y'=x

32. y' ~ I/y

xy

Autonomous Differential Equations and Phase Lines In Exercises 33 and 34: IdentifY the equilibrium values. Which are stahle and which are unstable?

b. Construct a phase line. IdentifY the signs of y' and y'. c. Sketch a representative selection of solution curves.

dy

dy

33'dx~y2-1

34.dx~y_y2

Applications 35. Escape velocity

+ VOl

The gravitationsl attraction F exerted by an

airless moon on a body of mass m at a distance s:from the moon's center is given by the equation F = -mg R 2s -2, where g is the acceleration of gravity at the moon's surface and R is the moon's radius (see accompanying figaro). The force F is negative because it acts in the direction of decreasing s.

Chapter

v'2iii,

The velocity Vo = is the moon's escape velocity. A body projected upward with this velocity or a greater one will escape from the moon's gravitational pull.

Slope Fields

L

2gR2 = -.-

Thus, the velocity remains positive as loog as Vo '"

In Esercises 2!1-32, sketch part of the equatioo's slope field. Then add to your sketch the solutioo curve that pssses through the point P( I, -I). Use Euler~ method with Xo ~ I and dx ~ 0.2 to estimate y(2). Round your answers to four decimal places. Find the exact value of y(2) for comparison. ~

v2

-2

dy x2 + Y 0 28. dx ~ - eY + x' y(O) ~ 0

31. y'

a. If the body is projected vertically upward from the moon's surface with an initial velocity Vo at time t = O. use Newton's second law, F ~ rna, to show that the body's velocity at position s is given by the equation

b. Show that if Vo ~

v'2iii, then 3vo

.~RI+2Rt (

)2/3 .

36. Coasting to a stop

Table 9.6 shows the distance. (meters) coasted 00 in-line skates in t sec by 10huathon Krueger. Find a model for his position in the form of Equatioo (2) ofSectioo 9.3. His initial velocity was Vo ~ 0.86 m/sec, his mass m ~ 30.84 kg (be weighed 68 Ib), and his total coasliog distance 0.97 m.

TABLE 9.6 Johnathon Krueger skating data

t (sec)

s(m)

t (sec)

s(m)

t (sec)

s(m)

0 0.13 0.27 0.40 0.53 0.67 0.80

0 0.08 0.19 0.28 0.36 0.45 0.53

0.93 1.06 1.20 1.33 1.46 1.60 1.73

0.61 0.68 0.74 0.79 0.83 0.87 0.90

1.86 2.00 2.13 2.26 2.39 2.53 2.66

0.93 0.94 0.95 0.96 0.96 0.97 0.97


Theory and Applications 1. Transport through a cell membrane Under some conditions, the result of the movement ofa dissolved substance across a cell's membrane is described by the equation

dy A dt ~ kV(c - y). In this equation, y is the concentration of the substance inside the cell and dy/dtis the rate at whichy changes over time. The letter>

Chapter 9 Technology Application Projects k,A, V, and c stand for constants, kbeing the permeability coefficient (a property of the membrane), A the surface area of the membrane. V the cell's volume, and c the concentration of the

•• Vetify that y(x) problem

substance outside the cell. The equation says that the rate at which the cooceotratioo changes within the cell is proportional to the difference between it and the outside concentration.

b. For the integratiog factor v(x) = ef P«)"', show that

y'

In this equation, m is the mass of the system at time t, v is its velocity, and v + u is the velocity of the mass that is entering (or leaving) the aystern at the rate dm/ dt. Suppose that a rocket of initial mass mo starts from rest, but is driven upward by ftring some of its mass directly backward at the constant rate of dm/dt = -b units per second and at constant speed relative to the rocket u = -c. The ooly external force acting on the rocket is F = -mg due to gravity. Under these assumptions, show that the height of the rocket above the ground at the end of t secoods (t smal\ compared to molb) is

_ [ mo - bt In mo - bt]_ 1 2 y-ct+ b mo 2gt· 3••• Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Foodameotal Theorem of Calculus, Part I to show that aoy function y satisfYing the equation

v(x)y

=

J

v(x)Q(x) tb:

+C

for v(x) = ef p(.) '" is a solutioo to the tirst-rder Hoear equstion

dy tb: + P(x)y = Q(x).

J:'

b. If C = yov(xo) v(t)Q(t) dt, then show that any solutioo y in part (a) satisfies the initial conditioo y(xo) = Yo.

4. (Continuation ofExercise 3.) Assume the hypotheses of Exercise 3, and assume that Yl (x) and Y2(X) are both solutions to the firstorder Hoear equation satisfYing the initial condition y(xo) = Yo.

Chapter

y(xo) = O.

o.

c. Fnnn part (a), we have Yl (xo) - Y2(XO) = O. Since v(x) > 0 fora < x < b, usc part (b) to estshlish that Yl(X) - Y:z(x) = 0 on the interwl (a, b). ThusYl(X) = Y2(X) for all a < x < b.

tim

+ (v + u)Tt·

= F

+ P(x)y = 0,

Conclude that v(x )[Yl (x) - Y2(X)] = constant.

2. Height of • rocket If an exteroa\ force F acts upon a aystem whose mass varies with time, Newton's law of motion is

d(mv)

Yl (x) - Y2(X) satisfies the initial value

d tb: (V(X)[yl(X) - Y2(X)]) =

•• Solve the equation for y(t), using Yo to deoote y(O). b. Find the steady-state cooceotratioo, lim,...ooy(t).

----;It

=

531

Homogeneous Equations A rITst-order differeotiaI equation of the form

is called homogeneous. It can be transformed into an equation whose variables are separable by defIning the new variable v = y/x. Then, y = vxand dy dv tb:=v+xtb:. Substitution into the original differential equation and collecting terms with \ike vatiables then gives the separable equstioo tb:

dv

x + v _ F(v) =

O.

Afler solving this separable equation, the solution of the original equation is obtsined when we replace v by y/x .

Solve the branogeoeous equstioos in Exercises 5-10. First put the equation in the funn of a homogeneous equstioo. 5. (x 2 + y2)tb: + xydy = 0 6. x 2dy + (y2 - xy)tb: = 0 7. (xeYi'

+ y)tb: - xdy

= 0

8. (x +y)dy+ (x -y)tb: 9. y'

10.

Y

=i +

=

0

y-x cos - x -

(x sin ~ - ycos~) tb: + xcos ~ dy


Mathematica/Maple Modules: Drllg Dosages: Are They Effective? Are They S"/e? Formulate and solve an initial value model for the absorption of a drug in the bloodstream. First-Ortkr Differenlifll Eqllfltions and Slope Fields Plot slope fields and solution curves for various initial conditions to selected rrrst-order differeotiaI equations.

= 0

10 INFINITE SEQUENCES AND SERIES OVERVIEW Everyone knows how to add two numbers together, or even several. But how do you add infinitely many numbers together? In this chapter we answer this question, which is part of the theory of infinite sequences and series. An important application of this theory is a method for representing a known differentiable function f(x) as an infinite sum of powers of x, so it looks like a ''polynomial with infinitely many terms." Moreover, the method extends our knowledge of how to evaluate, differentiate, and integrate polynomials, so we can work with even more general functions than those encountered so far. These new functions are often solutions to important problems in science and engineering.

10.1 HrSlORICAL ESSAY

Sequences and Series

Sequences Sequences are fundamental to the study of infinite series and many applications of mathematics. We have already seen an example of a sequence when we studied Newton's Method in Section 4.6. There we produced a sequence of approximations Xn that became closer and closer to the root of a differentiable function. Now we will explore general sequences of numbers and the conditions under which they converge.

Representing Sequences A sequence is a list of numbers

in a given order. Each of aJ, a2, a3 and so on represents a number. These are the terms of the sequence. For example, the sequence 2,4,6,8, 10, 12, ... , 2n, ... has first term al = 2, second term a2 = 4, and nth term an = 2n. The integer n is called the index of an, and indicates where an occurs in the list. Order is important. The sequence 2, 4, 6, 8 ... is not the same as the sequence 4, 2, 6, 8 .... We can think of the sequence

as a function that sends 1 to al , 2 to a2, 3 to a3, and in general sends the positive integer n to the nth term an. More precisely, an infinite sequence of numbers is a function whose domain is the set of positive integers. The function associated with the sequence 2, 4, 6, 8, 10, 12, ... , 2n, ... sends 1 to al = 2, 2 to a2 = 4, and so on. The general behavior of this sequence is described by the formula an = 2n.

532

10.1 Sequences

533

We can equally well make the domain the integers larger than a given number no, and we allow sequences of!bis type also. For example, the sequence 12, 14, 16, 18,20,22 ... is described by the fonnula a" = 10 + 2n. It can also be described by the simpler fonnula bn = 2n, where the index n starts at 6 and increases. To allow such simpler fonnulas, we let the first index of the sequence be any integer. In the sequence above, {an} starts with a, while {b.} starts withb 6 • Sequences can be described by writing rules that specify their terms, such as • r an=vn,

bn = (_I)n+11 n'

dn = (_I)n+1,

n - I cn=-n-'

or by listing !enns:

{an}

=

{Vi, Vi, v'3, ... , Vn, ... }

{b n }

=

{1,-!,t,-i, .. ·,(-I r1 k, ... }

{cn}

=

1234 n-I } { 0'2'3'4'5'''''-n-''''

{dn }

= {I, -I, I, -I, I, -I, ... ,

(-I).+" ... }.

We also sometimes write

{a,,}

=

{Vn }:,.

Figure 10.1 shows two ways to represent sequences graphically. The fIrst marks the rrrs! few points from a" a2, a3,"" an,'" on the real axis. The second method shows the graph of the function defIning the sequence. The function is dermed only on integer inputs, and the graph consists of some points in the xy-plane located at (I, ad, (2, a2), ... , (n, a,,), ....

-+--~~~I--~I--~ 'n

2

3

4

5

FIGURE 10.1 Sequences can be represented as peints on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis all is its value.

Convergence and Divergence Sometimes the numbers in a sequence approach a single value as the index n increases. This happens in the sequence

534

Chapter 10: Infinite Sequences and Series whose terms approach 0 as n gets large, and in the sequence

{o,~, t, ~'~'"'' I

-

k, .. ·}

whose terms approach I. On the other hand, sequences like

{VI, Yz, V3, ... , V;;, ... } L-E

o

L L+E

•• ••• • (• • -I- )

~------------------- L+E

L

have tenDs that get larger than any number as n increases, and sequences like

------------(n,a,,)- ~ - . ---

L-E

{I, -I, I, -I, I, -I, ... , O.

DEFINmONS The sequence {a,} converges to the number L iffor every positive number E there corresponds an integer N such that for all n,

~~~~----~~~----~n

o

2 3

N

n

la, - LI
N

In the representation ofa sequence as points in the plane, all ~ L if y = L is a horizontal asymptote of the sequence of points (in, a,)}. In this figure,

FIGURE 10.2

E.

Ifno such number L exists, we say that {an} diverges. If {an} converges to L, we write lim,~oo an = L, or simply a, --+ L, and call L the limit of the sequence (Figure 10.2).

all the a,/s after aN lie within E of L. The definition is very similar to the definition of the limit ofa function fix) asx tends to 00 (limx~oof<x) in Section 2.6). We will exploit this connection to calculate limits of HISTORICAL BIOGRAPHY

sequences.

Nicole Oresme (ca. l32()-1382)

EXAMPLE 1 "

FIGURE 10.5 As n --+ 00, lin --+ 0 and 21/· --+ 2° (Example 6). The terms of {lin} are shown on the x-axis; the terms of {21/.} are shown as the y-values 00 the graph of f(x) ~ 2".

The next theorem formalizes the connection between Iim,,~oo an and limx~oo f(x). It enahles US to use I'Hllpital's Rule to f'md the limits of some sequences.

THEOREM 4 Suppose that f(x) is a function defined for all x ;;,: no and that {an} is a sequence of real numbers such that an ~ f(n) for n ;;,: no. Then ~

lim f(x)

lim an

L

x~OO

~

L.

n~OO

Proof Suppose that lim_oo f(x) her M such that for all x,

~

L. Then for each positive number E there is a numIf(x) - LI
M

E.

Let N be an integer greater than M and greater than or equal to no. Then

n >N

EXAMPLE 7

=>

a.

~

f(n)

and

lao -

LI

~

If(n) - LI
odefine a sequence that, when it converges, gives a solution to the equation sin x - x 2 = o. •

Bounded Monotonic Sequences Two concepts that play a key role in determining the convergence of a sequence are those of a bounded sequence and a monotonic sequence.

DEFINmONS A sequence {a.} is bounded Crom above if there exists a number M such that a. ,,; M for all n. The number M is an upper bound for {a.}. If M is an upper bound for {a.} but no number less than M is an upper bound for {a.}, then M is the least upper bound for {a,,}. A sequence {a,,} is bounded from below if there exists a number m such that a. ;" m for all n. The number m is a lower bound for {a,,}. If m is a lower bound for {a,,} but no number greater than m is a lower bound for {a.}, then m is the greatest lower bound for {a,,}. If {a.} is bounded from above and below, the {a.} is bounded. If {a.} is not bounded, then we say that {a.} is an unbounded sequence. EXAMPLE 11 (a) The sequence 1,2,3, ... , n, ... has no upper bound since it eventually surpasses every number M. However, it is bounded below by every real number less than or equal to I. The number m = 1 is the greatest lower bound of the sequence.

123

n

.

(b) Thesequence Z'3'4"'" n + 1"" lSbounded above by every real number greater than or equal to 1. The upper bound M = 1 is the least upper bound (Exercise 125). The sequence is also bounded below by every number less than or equal to its greatest lower bound

t,

which is •

540

Chapter 10: Infinite Sequences and Series

I

Convergent .equences are bounded

If a sequence {a,,} converges to the number L, then by detmition there is a number N such that Ia" - L I < 1 if n > N. That is,

L-l N and n > N

~

I .... -

a" 1

-

133. Sequence. generated by N_on'. method Newton's method, applied to a differentiable functioo f(x), begins with a starting value Xo and coostructs from it a sequence of numbers {x,} that under favorable circumstances converges to a zero of f. The

recursion formula for the sequence is Xn+l

I

Which of the sequences in Exercises 115-124 converge, and which di· 115.a,~I-1i

a n +l =

I)!

114. a" ~ 2 - Ii - 2'

I

1,

al =

125. The sequence {II/(II + I)} bas a lea.t upper bound of 1 Show that if M is a number less thao I, then the terms of {n/(n + I)} eventually exceed M. That is, if M < I there is ao inreger N such that n/(n + I) > M whenever n > N. Since n/(n + 1) < I for every n, this proves that I is a least upper bouodfor {n/(n + I)}.

132. Prove that a sequence {a,} converges to 0 ifaod only if the quence of absolure values {I an I} cooverges to O.

In Exercises 111-114, detennine if the sequence is monotooic aod if it ~

124.

131. For a sequence {a,,} the tenDs of even index are denored by au and the terms of odd index by au+ I. Prove that if au ---+ L and a2k+1 -+ L, then a" -+ L.

107. Prove that lim.~oo 'C",; ~ 1.

111. a,

4"+ 1 + 3" 4'

n+ I 122. an = - n -

129. Uniqueness of limit. Prove that limits of sequences are uoique. That is, show that if LI aod L, are numbers such that On -+ LI and all -+ L2, then Ll = L,..

if c is any positive constant.

109. Prove Theorero 2.

~ I +~

123. a" ~

+ I)}.

max {x.. cos (n

127. Is it true that a sequence {a,} of positive numbers must cooverge if it is bouoded from above? Give reasons for your answer.

vi,;! ~ ~ for large values of n.

D b.

I)

=X

f(xn) ll -

!'(x

)'

lI

.. Show that the recursioo formula for f(x) ~ x 2 caobewrittelIaSxn+l ~ (xn + a/x,)/2.

D b.

-

a, a

> 0,

Starting with Xo ~ I aod a ~ 3, calculate successive tenDs of the sequence uotil the display begins to repeat. What number is being apprnsimared? Explain.

544


D 134. Arecursi.. defioi1ionof"./2

Ifyou start with Xl = I anddefme the subsequent terms of {x,} by the rule x, = X,_I + COSX,_I, you geoerate a sequeoce that converges rapidly to 1f/2. (a) Try it. (b) Use the accompaoying figure to explain why the conver-

a. Calculate and theo plot the frrst 25 terms of the sequeoce. Does the sequeoce appear to be bouoded from above or be-

gence is so rapid.

b. If the sequeoce converges, find an integer N such that Ia" - L I :5 0.01 for n '" N. How far in the sequeoce do you have to get for the terms to lie within 0.0001 of L?

low? Does it appear to converge or diverge? If it does cooverge, what is the limit L?

y

135. a, =

COS~_l

o

I

138.

an+l = an

at = =

1,

(I + 0~5)"

= an + 5"

QII+I

+ (-2)11 140. a"

sinn

sinn

COMPUTER EXPLORATIONS Use a CAS to perform the followiog steps for the sequeoces in Exercises 135-146.

10.2

136. a" =

137. at = I,

139. an

~¥-~~--J------>x

%

=

n sin

k

Inn

141. a, = --n-

142. a" = "...

143. a, = (0.9999)'

144. a" = (123456)1/,

145. an

=

n 41

8' I

146. a" = 19"

n.

Infinite Series An infinite series is the sum of an infmite sequence of numbers

al + a2 + a, + ... + a, + ... The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infmitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at the result of summing the first n terms of the sequence and stopping. The sum of the first n terms Sn =

al

+

a2

+ a3 + ... + all

is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n geta larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in Section 10.1. For example, to assign meaning to an expression like

1+

1

1

1

2 + 4 + "8 +

1

16 + ...

we add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.

Partial sum

Value

Suggestive expression for partial sum

First:

Sl =

1

1

2 - 1

Second:

s2=1+ 21

Third:

1 1 s,=I+-+2 4

3 2 7 4

1 2-2 1 2-4

nth:

1 1 1 s, =1+-+-+···+-2 4 2,-1

2' - 1 2n - 1

2 -n1-1 2 -

10.2

Infinite Series

545

Indeed there is a pattern. The partial sums form a sequence whose nth tenn is SrI

=

1 2 - 2,,-1.

This sequence of partial sums converges to 2 because 1im,,~00(I/2·-1) = O. We say "the sum of the infinite series 1

+ !2 + !4 + ... + _1_ + ... 2.- 1

is 2."

Is the sum of any fmite numberoftenns in this series equal to 2?No. Can we actually add an infinite number oftenns one by one? No. But we can still define their sum by defming it to be the limit of the sequence of partial sums as n --> 00, in this case 2 (Figure 10.8). Our knowledge of sequences and limits enables us to break away from the confines of finite sums.

1

114

~

o

112

FIGURE 10.8 As the lengths 1, ,/2, '/.,

1/8

2

'I" ... are added one by one, the sum

approaches 2.


Blaise Pascal (1623-1662)

DEFINmONS

Given a sequence of numbers {a.}, an expression of the form

al+a2+ a3+ .. ·+ a.+ .. · is an inf"mite series. The number a. is the 11th term of the series. The sequence {s.} defmed by

is the sequence of partial sums of the series, the number s. being the 11th partial sum. If the sequence of partial sums converges to a limit L, we say that the series converges and that its sum is L. In this case, we also write 00 al + az + ... + a. + ... = ~ a. = L . •- 1

If the sequence of partial sums of the series does not converge, we say that the series diverges.

When we begin to study a given series al + az + ... + an + .. " we might not know whether it converges or diverges. In either case, it is convenient to use sigma notation to write the series as or

A useful shor1hand when smnmation from 1 to 00 is

understood

546


Geometric Series Geometric series are series of the form 00

a + aT + ar2 + ... + ar n- t + ...

=

L ar n- 1 11=1

in which a and r are fixed real numbers and a oF ~:o ar". The ratio r can be positive, as in

o. The

I I

(1)"-1 + ...

1-t+~-···+

(-tf' + ....

1+-+-+···+ 2

4

2

series can also be written as

r~I/2,a~1

'

or negative, as in r~ -1/3,a~

1

If r = I, the nth partial sum of the geometric series is

s"

= a + a(l) + a(I)2 + ... + a(I)"-1 = na,

and the series diverges because lim"~oo s" = ± 00, depending on the sign of a. If r = -I, the series diverges because the nth partial sums alternate between a and O. If Ir I oF 1, we can determine the convergence or divergence of the series in the following way:

rSn

=

+ aT + ar2 + ... + arl'l-l aT + ar2 + ... + arn- 1 + ar1!

Sn - rSn

=

a - ar"

Sn =

a

s.(1 - r) = a(1 - r") 8n =

Multiply s. by r. Subtract rSn from Sn. Most of the terms on the right cancel.

Factor.

a(1 - r") 1- r '

(r oF 1).

We can solve for 311 if r =F 1.

If Irl < I, then rn --> 0 as n --> 00 (as in Section 10.1) and s" --> a/(I - r). Iflrl then Ir"I--> 00 and the series diverges.

Iflrl < 1, the geometric series a to a/(I - r):

+ ar + ar2 + ... + ar"-l + ...

> I,

converges

00

L ar n - 1 =

11=1

_a_, 1- r

Irl

< I.

If Ir I ;;,: I, the series diverges.

We have determined when a geometric series converges or diverges, and to what value. Often we can determine that a series converges without knowing the value to which it converges, as we will see in the next several sections. The formula a/ (I - r) for the sum of a geometric series applies only when the summation index begins with n = I in the expression ~:"l ar"-l (or with the index n = 0 if we write the series as ~:o ar").

EXAMPLE 1

The geometric series with a

I

I

I

= 1/9 and r = 1/3 is

I (I )"-'

9 + 27 + Sf + ... = ~ 9 3 EXAMPLE 2

00

1/9 I - (1/3)

The series 00

(-I)n5 4"

5

5

5

4

16

64

~--=5--+---+···

n~O

I 6·

•

10.2

is a geometric series with a

a

547

= 5 and r = -1/4. It converges to 5

a

+ (l/4)

1- r

ar - -

Infinite Series

=

•

4.

EXAMPLE 3 You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find the total distance the ball travels up and down (Figure 10.9).

--_( -_ _ _

ar2 __ _

Solution

The total distance is

ar 3 - - S

2ar = a + 2ar + 2ar 2 + 2ar 3 + ... = a + - = all+r -l-r -r This sum is 2ar/ (I - r) .

If a = 6 m and r = 2/3, for instance, the distance is (a)

1

S

EXAMPLE 4

~ o o

Solution

o o a

+ (2/3)

(5/3)

= 6 1 _ (2/3) = 6 1/3

•

= 30 m.

Express the repeating decimal 5.232323 ... as the ratio of two integers.

From the definition of a decimal number, we get a geometric series

5.232323 ... = 5

+ 12030 + ~ + ~ + .. .

o

(l00)

(l00)

1

( 1 100

23 (

Q

= 5 + 100 1 + 100 + o

)2 + ... )

a = 1, r = 1/ 100

1/ (1 - 0.01)

00. n=1

(b)

00,,+1. ,,+1 L -,,diverges because - , , - --> I. n=1

(c)

L (- 1).+1 diverges because lim.~oo( - 1).+1 does not exist.

00

.-1 00

-n

(d) ~ 2n

+

EXAMPLE 8

.

.

-n

5 diverges because lim.~oo 2n

+5

1

= -

2 oF o.

•

The series

IIIIII II I 1+-+-+-+-+-+-+···+-+-+···+-+··· 224444 2'2' 2' ~

2 terms

4 terms

2" terms

diverges because the terms can be grouped into infinitely many clusters each of which adds to I, so the partial sums increase without bound. However, the terms of the series form a sequence that converges to O. Example 1 of Section 10.3 shows that the harmonic series also behaves in this manner. _

10.2 Infinite Series

549

Combining Series Whenever we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series.

If ~an = A and ~bn = B are convergent series, then

THEOREM 8

~(an + bn) = ~a. + ~bn = A + B

1. Sum Rule:

2. Difference Rule: 3. Constant Multiple Rule:

(any number k).

Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section lO.1. To prove the Sum Rule for series, let ~=~+_+

...

+~

~=b,+~+···+~.

Then the partial sums of ~(an + bn) are Sn = (a, + b,) + (a2 + b2) + ... + (an + bn) = (a, + ... + an) + (b, + ... + bn) =

An

+ Bn·

Sinee An --+ A and Bn --+ B, we have Sn --+ A + B by the Sum Rule for sequences. The proof of the Difference Rule is similar. To prove the Constant Multiple Rule for series, observe that the partial sums of ~kan form the sequence Sn = ka, + ka2 + ... + kan = kia, + a2 + ... + an) = kAn.

•

which converges to kA by the Constant Multiple Rule for sequences. All corollaries of Theorem 8, we have the following results. We omit the proofs.

1. Every nonzero constant multiple of a divergent series diverges. 2. If ~an converges and ~bn diverges, then ~(an + bn) and ~(a. diverge.

- bn) both

Remember that ~(an + bn) can converge when ~a. and ~bn both diverge. = I + I + I + "'and~bn = (-I) + (-I) + (-1) +"'diverge, whereas ~(an + bn) = 0 + 0 + 0 + ... converges to O. Caution

Forexample,~an

EXAMPLE 9 00

(8)

L n- l

Find the sums of the following series.

3n-' _ 1 6n ,

00

=

L n- l 00

=

L

11=1

(

1 1 ) 2n-' - 6n-'

1

00

L

I

2n-' - 11=1 6n-' I 1

1 - (1/2)

=2- Q =± 5

5

1 - (1/6)

Difference Rule Geometric series with a ~ 1 andr ~ 1/2,1/6

550


(b)

Cons_ Multiple Rule

=

4C - ~1/2»)

Geometric series with a

I, r

=

=

1/2

•

=8

Adding or Deleting Terms We can add a finite number of terms to a series or delete a rmite number of terms without altering the series' convergence or divergence, although in the case of convergence this will usually change the sum. If ~:;"~1 a, converges, then ~:;;'ka. converges for any k> land 00

Lan n=1

00

+

= at

a2

+ ... +

Conversely, if ~:;"-ka. converges for any It

>

ak-t

+

Lan" 1I=k

I, then ~:::'1 a, converges. Thus,

and


Richard Dedekind (1831-1916)

Reindexing As long as we preserve the order of its terms, we can reindex any series without altering its convergence. To raise the starting value of the index h units, replace the 11 in the fonnula fora, by 11 - h: 00

00

L an = 11-L1+11: an-h = 11 - 1

at

+

a2

+ a3 + ....

To lower the starting value of the index h units, replace the 11 in the fonnula for a. by 11 00

+ h:

00

:L an = 11=1-11: :L an+h = 11=1

at

+

a2

+

Q3

+ ....

We saw this reindexing in starting a geometric series with the index 11 = 0 instead of the index 11 = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions.

EXAMPLE 10

We can write the geometric series

1

00

~ 2,-1

1

=

I

1

+ 2 + 4 + ...

as 00

I

~2n'

00

1

~ 2,-5' 11=5

00

or even

1

n~2n+4'

The partial sums remain the same no matter what indexing we choose.

•

10.2 Infinite Series

551

Exercises 10.2 Finding nth Partial Sums

Using the nth-Tenn Test

Iu Exercises l--{j, fmd a formula for the nth partial sum of each series and use it to fmd the series' sum if the series converges.

Iu Exercises 27-34, use the nth-Term Test for divergence to show that

1. 2+~+~+l+ ... +~+ ... 3 9 27 3,-1 2 ~ + _9_ + _9_ + ... + _9_ + ... • 100 100' 100' 100' 3.1_

1 + 1 _ 1 +"'+(_1),-1_1_+ ... 2

4

8

2,-1

4.1- 2 +4 - 8 + ... + (-1),-12,-1 + ...

the series is divergent, or state that the test is inconclusive. 00 n ~ n(n + I) 27. ~ n + 10 28. ,-I (n + 2)(n + 3) 00 00 I 30. ~+29. ~~4 11""0 n 11- 1 n + 3 00 , 00 I 31. ~ cos" 32. ~+-+ 11"'0 e n

,-I

34. ~ cosmT

I I I I 5. 2. 3 + 3· 4 + 4· 5 + ... + (n + I)(n + 2) + ... 5 5 5 5 6. 1'2 + 2'3 + 3'4 + ... + n(n + I) + ...

Series with Geometric Terms Iu Exercises 7-14, write out the fITst few tenDs of each series to show how the series starts. Then fmd the sum of the series. 00 (-I)'

7. ~-4'

, -0

00 7

(5 + I) (I + -----sn (-I)') ~

m (~y +

+

(~)' + (~)' + .. .

1& (~2y + (~2)' + (~2)' + (~2)' + (~2)6 + ... Repeating Decimals Express each of the numbers in Exercises 19-26 as the ratio of two integers. 19. 0.23 = 0.23 23 23 ... 20. 0.234 = 0.234 234 234 ... 21. 0.7 = 0.7777 .. . 22. O.d = O.dddd ... , where d is a digit

= 0.06666 .. .

24. 1.414

(I"-~I I) n

00 37. ~ (In v;;-+!

,-I

(t) + (t)' + (t)' + (t)' + (t)' + ...

23. 0.06

11- 1

00 40. ~

211

16. 1+ (-3) + (-3)2 + (-3)' + (-3)' + .. . 17.

00 35. ~

-

36.~(3

3) (n + I)'

'-1~'

In Vn)

311

In Exercises 15--18, determine if the geometric series converges or diverges. If a series converges. fmd its sum.

15. I +

Iu Exercises 35-40, fmd a formula for the nth partial sum of the series and use it to detennine if the series converges or diverges. If a series converges, ftnd its sum.

11=1

00 11. ~ 211 00

Telescoping Series

38. ~ (tan (n) - tan (n - I))

9. ,-I ~4'

13.

11=0

1.414414414 ...

=

25. 1.24123

=

26. 3.142857

1.24123 123 123 .. .

= 3.142857 142857 .. .

(Vn+4 - Vn+3)

Find the sum of each series in Exercises 41-48. 41

~.

4

. ::1 (4n - 3)(4n + I)

43

~ 40n • ,~1 (2n - 1)'(2n + I)'

45.

00(1 I) ~ Vn - v;;-+!

47

~.

'.;:1

(

42

~

. ::1 (2n

6

- 1)(2n + I)

44.~2n+1 ,~1 n'(n

46.

00(1 ~ 21/,

+ I)' I) - 21/(,+ 1)

I _ I ) In(n+2) In(n+l)

48. ~(tan-l (n) - tao-I (n + I))

,-I

Convergence or Divergence Which series in Exercises 49--{j8 converge, and which diverge? Give reasons for your answers. If a series converges, f"md its sum.

00(1)' 49. ~ • r.:: 11=0 v2 51.

~ (-1)'+1

11=1

00 53. ~ cosmT 11=0

3, 2

00 SO. ~(V2r n=O

52. ~ ( -1)'+1n n=l

00 54. ~ cos:'11' n=O 5

552


00

,-I

59.

61.

~~

11=-1

83. Show by example that ~(aJb,) may diverge even though and ~b, converge and no b, equals O.

56. ~ In 3'

00

57.

I

00

e-'" ,-0

55. ~

00

58.

10"

00

2"_ 1

00

1

84. Find convergent geometric series A ~ ~a, and B ~ ~b, that illustrate the fact that ~a"b. may converge without being equal toAB.

I

~"' Ixl> 11=0 X

I

I)'

~3"

00 ( 60.~ I-Ii

~ I~O'

00

•

00

21'.

85. Show by example that ~(a,/b,) may converge to something other than A/Beven when A ~ ~a., B ~ ~b. #' 0, and no b, equals O.

62. ~"1Ft n!

2"+ 3" 63. ~11=1 4" 00

64.

86. If ~a, converges and a, > 0 for all n, can anything be said about ~(l/a,,)? Give reasons for your answer.

+ 4"

~311+411

87. What happens if you add a fmite number of terms to a divergent series or delete a ftnite number of tenns :from a divergent series? Give reasons for your answer.

65.~1n(n :1) 11=-1

Geometric Series with a Variable x In each of the geometric series in Exercises 69-72, write out the fl!St few terms of the series to fmd a and r, and fmd the sum of the series. Then express the inequality Ir I < I in terms of x and fmd the values of x for which the inequality holds and the series converges.

69. ~(-I)'x'

70. ~(-I)'x'"

.-0

(-I)' (

~-2- 3 + 00

88. If ~a" converges and ~b. diverges, can anything be said about their term-by-term sum ~(a, + b,)? Give reasons for your answer. 89. Make up a geometric series ~ar"-l that converges to the number 5if a. a

=2

b. a ~ 13/2.

90. Find the value of b for which

1 + eb + e2JJ + e3b + ... = 9. 91. For what values of r dues the infinite series

,-0

72.

~a"

I

sinx

)' converge? Find the sum of the series when it converges.

10 Exercises 73-78, find the values of x for which the given geometric senes converges. Also, fmd the sum of the series (as a function ofx) for those values of x. 00

00

73. ~2'x'

74. ~(-!)y'"

,-0

,-0

00

92. Show that the error (L - so) obtained by replacing a convergent geometric series with one of its partial sums s" is ar"/( 1 - r). 93. The accompanying figure shows the first five of a sequence of squares. The outermost square has an area of 4 m2 • Each of the other squares is obtained by joining the midpoints of the sides of the squares before it Find the sum of the areas of all the squares.

I)"(x - 3)' 00 ( - 76. ~

75. ~(-I)'(x + I)' 11=0

11=0

00

00

.-

2

78. ~(lnx)'

77. ~ sin"x

Theory and Examples 79. The series in Exercise 5 can also be written as

~

I

,~I (n + I)(n + 2)

d ~ an ,~_I (n

Write it as a sum beginning with (a) n (0) n ~ 5.

~

I

+ 3)(n + 4) . -2, (b) n

~

94. Helga von Kocb's snowflake CIIl"W Helga von Koch's snowflake is • curve of infmite length that encloses a region of fmite area. To see why thls is so, suppose the curve is generated by starting with an equilateral triangle whose sides have length 1.

0,

80. The series in Exercise 6 can also be written as 00

5

~ n(n + I)

00

and

~ (n +

5 !)(n

Write it as a sum beginning with (a) n (0) n ~ 20.

+ 2) . ~

-I, (b) n

~

3,

81. Make up an infinite series of nonzero terms whose sum. is L I b . -3 c. O. 82. (Continuation o/Exercise 81.) Can you make an write series of nonzero terms that converges to any number you want? Explain.

.. Find the length L, of the nth curve C, and show that limn. . . oo L" = 00. b. Find the area An of the region enclosed by C, and show that lim,~ooA. ~ (8/5)A , .

LOO O C2

c,

10.3 The Integral Test

10.3

553

The Integral Test Given a series, we want to know whether it converges or not. In this section and the next two, we study series with nonnegative terms. Such a series cooverges if its sequence of partial sums is bounded. If we establish that a given series does converge, we generally do not have a formula available for its sum, so we investigate methods to approximate the sum instead.

Nondecreasing Partial Sums Suppose that ~~, O.

the series converges by the Integral Test. We emphasize that the sum of the p-series is not I/(p - I). The series converges, but we don't know the value it converges to. Ifp < I,thenl-p > oand

roo ~dx = x

11

P

_1_ lim (bl-p - I) = 1 - Pb_OO

00.

The series diverges by the Integral Test. If p = I, we have the (divergent) hannonic series I I I 1+-+-+···+-+··· 2 3 n We have convergence for p

>

I but divergence for all other values ofp.

•

The p-series with p = I is the harmonic series (Example I). The p-Series Test shows that the hannonic series is just barely divergent; if we increase p to 1.000000001, for in-

stance, the series converges! The slowness with which the partial sums of the hannonic series approach infinity is impressive. For instance, it takes more than 178 million terms of the hannonic series to move the partial sums beyond 20. (See also Exercise 43b.)

2+ I)) is not ap-series, but it converges by the

The series ~::'~I (1/(n Integral Test. The function j(x) = 1/(x 2 x ~ 1,and

EXAMPLE 4

+

I) is positive, continuous, and decreasing for

{"" _1_ dx = lim [arctan x2 + I b~oo

11

=

x]b 1

lim [arctan b - arctan I]

b~oo

1T

1T

1T

2

4

4'

---

Again we emphasize that 1T/4 is not the sum of the series. The series converges, but we do not know the value of its sum. •

Error Estimation If a series ~a. is shown to be convergent by the Integral Test, we may want to estimate the size of the remainder R. between the total sum S of the series and its nth partial sum s•. That is, we wish to estimate Rn =

S-

9"

= an+l +

tln+2

+

an+3

+

556

Chapter 10: Infinite Sequences and Series To get a lower bound for the remainder, we compare the sum of the areas of the rectangles with the area under the curve y = f(x) for x ;;,: n (see Figure 1O.lla). We see that

1

00

R. = a.+1

+

a.+2

+

a,,+3

+ ... ;;,:

f(x) fix .

• +1

Similarly, from Figure 10.llb, we fmd an upper bound with

1

00

R.

= a,,+1

+

a.+2

+

a,,+3

+ ... ,,;;

f(x) fix.

These comparisons prove the following result giving bounds on the size of the remainder.

Bounds for the Remainder in the Integral Test Suppose {at} is a sequence of positive terms with at = f(!), wbere f is a continuous positive decreasing function of x for all x ;;,: n, and that l:a" converges to S. Then the remainder R. = S - s. satisfies the inequalities

1

00

1

00

f(x) fix ,,;; R. ,,;;

11+1

f(x) fix.

(1)

11

Ifwe add the partial sum s. to each side of the inequalities in (I), we get

1

00

s.

+

1

00

f(x) fix ,,;; S ,,;; s.

+

n+l

f(x) fix

(2)

n

since s. + R. = s. The inequalities in (2) are useful for estimating the error in approximating the sum of a convergent series. The error can be no larger than the length of the interval containing S, as given by (2).

EXAMPLE 5 n = 10.

Estimate the sum of the series l:(I/n 2 ) using the inequalities in (2) and

Solution We have that

Using this result with the inequalities in (2), we get SIO

Taking S10 = I equalities give

I<S< + IT - -

SIO

+1 10·

+ (1/4) + (1/9) + (1/16) + ... + (1/100)

'" 1.54977, these last in-

1.64068 ,,;; S,,;; 1.64997. If we approximate the sum S by the midpoint of this interval, we find that 00

I

~ 2: '" 1.6453.

n=1 n

The error in this approximation is less !ban half the length of the interval, so the error is • less !ban 0.005.

10.3 The Integral Test

557

Exercises 10.3 Applying the Integral Test Use the Integral Test to determine if the series in Exercises 1-10 con· verge or diverge. Be sure to check that the conditions of the Integral Test are satisfied 00 I 00 I

1. ~2 11 - 1 n

2. ~0:2 11"'1 n

I 3. ~-211=-1 n + 4

4. i-In +4

00

41

~ 00 • /I _ I

43.

B.

.-2

.-1 00

00

7.~+1Ft n + 4

In

(n 2)

10.

i

2

1I"'2n

D b.

n - 4 -2n+l

Give reasons for your answers. (When you check an answer, remem-

00

00

11. ~_I 1Ft 10"

00

i.

~ 00

29.

19.

11=1

Inn

00

21.

2'

~3'

22.

00

24'~2~1 11- 1 n

-2 ----=t=l n I

Vn(Vn +

27. I)

~

(lin)

.~, (lnn)Yln2 n - I

00 I 33. "5"-nsinJi

::1 00

+

II- I

(1 I( v;;-+l)) diverge.

Use the accompanying graph to show that the partia1 sum Sso =

~;'!,1

(I/( Vn+J)) satisfies

2'

25. ~----=t=l

• -1

00

n

1 Vx+l 51

I

dt

Conclude that 11.5

I (ln3)'

30.

~

32.

i n(1 + 1n n) .-1 I

< Sso < {SO. ~

Jo

2

1·

The number 1, whose value is 0.5772 ... ,is calledEuler~

(p a positive coostant)

>

I

Thus. the sequence

I

a. Show that the improper integral

I

0< In (n + I) - Inn '" I + 2 + ... + " - Inn '" 1.

55. Logarithmic p-series constant.

58. Use the Integral Test to show that the series

1.

b. What implicatioos does the fact in part (a) have for the convergence of the series

f-I-? ,-2 n(1n n)P . Give reasons for your answer. 56. (Continuation ofExercise 55.) Use the result in Exercise 55 to determine which of the following series converge and which diverge. Support your answer in each case. 00 1 00 1 a. ~n(lnn) b. ~n(lnn)1.01

10.4

I

+ I) '" I + 2+ "'+,,'" I + Inn

or

52. How many terms of the convergent series ~~,(I/n(lnn)') should be used to estimate its value with error at most 0.01? 53. The Cauchy condeusation test The Cauchy coodensatioo test says: l.ct {a,} be • nooincreasing sequence (a, '" a,+1 for all n) of positive terms that converges to O. Then ~an converges if and ooIy if ~2'a2" converges. For example, ~(l/n) diverges because ~2" (1/2') = ~I diverges. Show why the test works.

l/xintheproofofThe0ren39,showthat

=

converges.

59. a. For the series ~(I/n'), use the inequalities in Equatioo (2) with n = 10 to Imd an interval cootaining the sum S. b. Ai; in Example 5, use the midpoint of the interval found in part (a) to approximate the sum of the series. What is the maximum error for your approximation? 60. Repeat Exercise 59 using the series ~(I/n').

Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their tenns to those of a series whose convergence is known.

10.4 Comparison Tests

THEOREM lO-The Comparison Test Let ~an. ~cn. and nonnegative terms. Suppose that for some integer N

~d"

559

be series with

n >N.

for all

0 and b,

>

a. is a

0 for

N (N an integer).

a

1. If lim b, = c n

11_00

>

a

0, then

2. If lim b, = 0 and 11- 00

~a, and ~b, hoth converge or hoth diverge.

~b, converges, then ~a, converges.

II

a

3. If lim b, = 11_00

00

11

and ~b, diverges, then

~a. diverges.

Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 55a and b. Since c/2 > 0, there exisla an integer N such that for all n

-

Limit defInition with E = c/2,L = c,and a" replaced by a"fb,

n> N ~ I~ cl < I. Thus, for n

>

N,

If ~b, converges, then ~(3c/2)b, converges and ~a. converges by the Direct Comparison Test. If ~b, diverges, then ~(c/2)b, diverges and ~a. diverges by the Direct Comparison Test. _

EXAMPLE 2

Which of the following series converge, and which diverge?

(a) 1 + ~ + l

4

9

+ ~ + ... 16 25

(b) 1 + 1 + 1 +

1

3

7

~ + ... = 15

00

=

00

~ 2n + I = ~

,~l

(n + 1)'

14

2n + I n2 + 2n + 1

5: _1 2' - 1

,~l

(c) 1 + 2m 2 + 1 + 3m 3 + 1 + 4m 4 + ... 9

,~l

21

=

5: 1 + +n 5 n m

,~2

n2

10.4 Comparison Tests

561

We apply the Limit Comparison Test to each series. (8) Let a" = (2n + I )/(n 2 + 2n + I). For large n, we expect a" to behave like 2n/n2 = 2/n since the leading terms dominate for large n, so we let b. = I/n. Since 00 00 I ~b. = ~ n diverges

Solution

n- l

n- l

and lim a" = lim 2n2 + n = 2, bn 11--+00 n2 + 2n + 1

11--+00

~a. diverges by Part 1 of the Limit Comparison Test. We could just as well have taken b. = 2/n, but I/n is simpler.

(b) Let a. = 1/(2' - I). For large n, we expect a. to behave like 1/2", so we let b. = 1/2'. Since 00 00 1 ~bn = ~ 21l converges

and lim a. = lim ~ b ll 11_00 211 1

11_00

=

lim

.~OO

1 1 - (1/2')

= 1, ~a.

converges by Part I of the Limit Comparison Test. (0) Leta" = (1 + nlnn)/(n 2 + 5). For large n, we expecta. to behave like (nlnn)/n 2 = (lnn)/n,whichisgreaterthanl/nforn;;" 3,soweletb. = l/n.Since 00 00 I ~b. = ~ n diverges ,,=2

11=2

and

= 00, ~a.

diverges by Part 3 of the Limit Comparison Test.

EXAMPLE 3

00

Does ~

11=1

Inn 3/2

n

-

converge?

Because In n grows more slowly than n' for any positive constant c (Section 10.1, Exercise 105), we can compare the series to a convergent p-series. To get the p-series, we see that Inn nl /4 1 Solution

-3 2< =32 n

n

n'/4 3 for n sufficiently large. Then taking a. = (In n)/n /2 and b. = I/n'/" we have /

/

lim a. = lim Inn btl 11--+00 n 1/ 4

11_00

=

=

lim

l/n

c-~----;o,-;

.~OO (1/4)n 3/4

lim ;/4 II-con

=

Since ~b. = ~(I/n'/4)isap-serieswithp of the Limit Comparison Test.

l'H6pital's Rule

O.

> I,itconverges,so ~a.convergesbyPart2 _

562


Exercises 10.4 Comparison Test

00

10 Exercises 1--1!, use the Comparison Test to determine if each series

37.~.

converges or diverges. I

00

1. ~ ~,.--o-::-::11=-1 n + 30 3.

~

I

'~'Vn-I 00

,

7.

~~n/4

+4

n

11=1

4.~~+2

41. ~-2' 11 - 1 n

1Ft

n - n

42.

~In(n: I)

I

00

8.~Vn+1

(Hint: First show that (lin!) '" (1/n(n - I» for n '" 2.) 00 I 00 • I ~ (n - I)! 45. ~ sm n 46. ~!ann 44. 11=-1 £- (n + 2)1. 00 tan-I n 00 -1 00 th 47. "5"_ -,-.,48 ")'_ sec n 49. ~_ co 2 n

,~'W+3

,-,

~ 00

50.

-

tanhn

~~n, + I

• 11=1

n +2 (Hint: Limit Comparison with ~:::, (1/Vn)) ~ n(n+1) 00 2' 11. ~ (n' + I)(n _ I) 12. ~ 3 + 4'

51.

n

1

~

nl.3 I

00

~-2-

53~

,-,

• ~

n

11=1

(Hint: Limit Comparisoo with ~::"~, (lin'» 10.

2"+311

43. ~., ~ n.

n-2 n2 + 3

n3

1

00 I 6. ~-3' 11=1 n

series converges or diverges.

~

+

3'

40·~3'+4'

oo2n _n

10 Exercises ~ 16, use the Limit Comparison Test to detennioe if each

11=-1

00

+ 3n 5n

n

3"-1

::1

~ ~ + I .-.L

39.

Umit Comparison Test

9.

(Xl

38. Y

+1

~ n-I 2. £ - - . - /I -I n + 2 11=2

5.~co:n 11=1 n /2

,I

~3n

00

~ .,r

~~

52.

lI=lnvn

I

+ 2 + 3 + ... + n

S4~ •

11=1

n

dr

11=1 n

1 + 22

I

+ 32 + ... + n 2

11=1

S' 13 ~.• .-=1 Vn4n 00 I 15. ~·-In ~ n 00

14.

~ 00

Theory and Examples 55. Prove (a) Part 2 and (h) Part 3 of the Limit Comparison Thst. S6. If ~:l all is a convergent series of nonnegative numbers, can anything be said about ~';,(aJn)? Explain.

(2nSn ++ 43)'

57. Suppose that a" > 0 and b, > 0 for n '" N (N an integer). If 1in3.~oo(a./b,) ~ 00 and ~a, converges, can anything be said

shout ~ b, ? Give reasons for yoor answer.

58. Prove that if ~all is a convergent series of nonnegative tenns,

(Hint: Limit Comparisoo with ~::"~, (lIn»

then ~a,2converges.

16. ~ In (I + 1,) n

59. Suppose that an

11=-1

(Hint: Limit Comparisoo with ~::"~, (lIn'»

20.

~

I

,~,

2Vn +

~

I +

-%

~osn

18. 21.

n

11=1

~

3 ,~, n + Vn

19.

~~

22.

n=t

23~

~ (3n:

S

26.

~~

29~.

31~.

32.

:.:1

n

I

":.:1 1 + 1nn

34.

~ 11=-1

n

,Vn

+

,~,

2

~n 1Ft

+ I n 2Vn

27.

j,1n(~n)

30.

~.----,p:-

00

• ~

35. 1

~ Sin:n

Sn'-3n , _, n'(n - 2)(n' + S)

(Inn)' 28. ~. - , 00

= 00.

,~OO

Prove that ~an diverges.

60. Suppose that a" > 0 and 1in3 n'a" ~ O. Prove that ~a, converges. 61. Show that ~::"~2 «In n)qlnP) converges for p>1.

-00

lor P is in-

566


Proof (8) P < 1. Choose an e > 0 so small that p + e < I. Since %;; ---> p, the terms %;; eventually get closer than e to p. In other words, there exists an index M 2: N such that

%;; I. _

. the senes · · WI t h terms an = CODSl'der agam

EXAMPLE 2

{n/2" /.

I 2,

nodd

n even.

Does ~a. converge?

Solution We apply the Root Test, finding that

%: = {"'I;/n/2, a.

nodd n even.

1/2,

Therefore,

t ,; %;; ,; fn· Since "'I;/n ---> 1 (Section 10.1, Theorem 5), we have Iim"~",, %;; = 1/2 by the Sandwich _ Theorem. The limit is less than I, so the series converges by the Root Test.

EXAMPLE 3 ""

(a)

2

~ ~.

(b)

.-1

Solution

Which of the followiog series converge, and which diverge?

~ 2: n

"" (

~

(c)

n=-1

1

1

+n

).

We apply the Root Thst to each series.

"" n 2

(a) ~ 2' converges because _1

"" 2' .

(b) ~ --, diverges because ,,=1 n

""(I)'

(c) ~ -+I ,,=1 n

1f.n W ("'I;/ny 12 2' = _.r.:;. = --2- ---> 2 < I. 2

v~

~2' --, n

2

r)3

= ( •• Vn

converges because

2

---> --,

1

> I.

~(1)' 1In ---> 0 < I. -+I = -+ n

-

10.5 The Ratio and Root Tests

567

Exercises 10.5 Using the Ratio Test In Exercises 1-8, use the Ratio Test to determine if each series con-

verges or diverges. 00 2' 1. /I - I n. ~ (n - I)! 3. ~ 2 ,~I (n + I) S. 7.

2.

~

in~2 00

4.

4

6.

~

37.

3

11",1

~:II 00

+ 2)!

8

2"+1 n3',-1

f. 3Inn i (2n + 3)n5'In (n + I) 11

2

+

Using the Root Test In Exercises 9-16, use the Root Thst to detennine if each series con· verges or diverges. 00 7 00 4' 10. (3 )" 9. 11 - 1 (2n + 5 )' 11"'1 n

L

L

3n - 5

11 - 1

13 ~ • ,;';:1 (3 15.

i

,-I

+

(I -

8 (1In))2n

14.

i i

(In(e sin'

2

+

k) r'

(~)

k)'"

(Hint: lim (I + xln)' ,~OO

= e")

I

00

16.

12.

"5'.

~n

1+,

Determining Convergence or Divergence In Exercises 17-44, use any method to detennine if tire series con·

verges or diverges. Give reasons for your answer. 00 V2 00 17.

"5' n ,

:.:1

00

00

19. Ln!e-' 11=1

21.

L InO' ,-I 00

23.

~

2

+ (-I)' 1.25'

00 ( 25'LI-

,-I

27.

i

~ Inri,

22.

i (n; 2)' ,-I

3)'

n

(-2)' , 24. ,-I 3 00

L-

OO(

I)'

26'LI-11"'1

3n

(Inn)' L-,n 00

In;

28.

11 - 1 n

(I I)

II - I

(I I )'

29. ~ li- n 2

30. ~ Ii - n 2

~. ~n

~_ n~n

00

31. 33.

32.

,;":1 ~ ~

/I - I

(n

00

+ I)(n + 2) , n.

,;":1

+ I)!

3'n.,

00

,

~n 00

L L

40

L n ,~2 (In n )(,/2)

42.

3' L--'--;; n2

•

00

11- 1

(n!? (2n)'.

00

43.

~

n2'(n

11 _

38. "5' n;

I)!

,,=-2

L

,,"" I

~

44.~ ,-I

(2n + 3)(2' + 3) 3'+2

Recunively Def"med Terms Which of the series ~;l an defmed by the formulas in Exercises 45-54 converge, and which diverge? Give reasons for your answers. all +l =

al =

1,

a,,+1 =

a, =

I

47.

3'

""+I =

48.

al

49.

al =

2,

SO.

a, =

5,

51.

a,

= 3,

1 + sinn n

all

1 + tan-I n

46.

n

a"

3n - I

2n + 5 "" n

an+1 = n

+ 1 a"

n

+ Inn

= I,

I

2'

=

a,,+1 =

a, t, ""+1 54. a, t, ""+I 53.

,

20.

I.

00

,

~ (2n:

52. al

18. Yn 2e-

~ 2

oo

36.

(Inn )' n 00 'In 41 n. n . ,_I n(n + 2)!

39.

• ,~I

3211

11. i (4n + 3)'

+ 3)! 3!n!3"

(n

00

11=2

n2 (n n!

~ 00

L, 00

00

35.

n

+ 10

an

~

=

=

=

= (",,),+1

Convergence or Divergence Which of the series in Exercises 55--62 converge, and which diverge? Give reasons for your answers. 00 2' , , 00 (3n)! 55 L~ 56. ~ n!(n + I)!(n + 2)! • ,~I (2n)! 00 (n!)" 57. II- I n ')2

L( 00

59. Y

,

n( ')

:.:1 2 00

IJ

1.3 .... '(2n - 1) 4"2IJn !

61.

~

62.

~ [2 -4- ...• (2n)](3' + I)

00 00

1-3- ... '(2n -I)

568


Theory and Examples 63. Neither the Ratio Test nor the Root Thst helps with p-series. Try them an

n/2',

65. Let a, = { 1/2', Does

I ~11=1 nP 00

~an

if n is a prime number otherwise.

converge? Give reasons for your answer.

66. Show that ~:_ I 2(.i')/n! diverges. Recall from the Laws ofExponents that 2('~ = (2')'.

and show that both tests fail to provide information shout convergence. 64. Show that neither the Ratio Test nor the Root Test provides information about the convergence of

(p constant).

10.6

Alternating Series, Absolute and Conditional Convergence A series in which the terms are alternately positive and negative is an alternating series. Here are three examples:

1

1

1

1

2 + 3 - 4 + 5 - ... +

I -

(_1),+1

n

+ ...

(1)

1 1 1 (-1)'4 -2 + 1 - 2 + 4 - 8 + ... + ----z'" + ... 1- 2

(2)

+ 3 - 4 + 5 - 6 + ... + (-I)·+l n + ...

(3)

We see from these examples that the nth term of an alternating series is of the fonn

an

= (_1).+l u•

or

a. = (-I)·u.

where u. = Ia.1 is a positive number. Series (1), called the alternating harmonic series, converges, as we will see in a moment Series (2), a geometric series with ratio r = -1/2, converges to -2/[1 + (1/2)] = -4/3. Series (3) diverges because the nth term does not approach zero. We prove the convergence of the alternating barmonic series by applying the Alternating Series Test. The Test is for convergence of an alternating series and cannot be used to conclude that such a series diverges.

THEOREM 14-The Alternating Series Test (Leibniz's Test)

The series

00

~(_1)'+IU. = UI - U2

+ U3

-

U4

+ ...

n=1

converges if all three of the following conditions are satisfied: 1. The u. 's are all positive. 2. The positive un's are (eventuslly) nonincreasing: u. ;;" U.+I for all n ;;" N, for some integer N.

3. u...... O. Proof Assume N = I. If n is an even integer, say n = 2m, then the sum of the {lISt n tenns is S2m =

(UI - U2)

+

(U3 -

= UI - (U2 - U3) -

U4) (U4 -

+ ... +

(U2m-1 -

us) - ... -

U2m)

(U2m-2 -

U2m-I) -

U2m.

10.6 Alternating Series, Absolute and Conditional Convergence

569

The first equality shows that S2m is the sum of m nonnegative terms since each tenn in parentheses is positive or zero. Hence S2m+2 ;;,: S2m, and the sequence {S2m} is nondecreasing. The second equality shows that S2m ,;; U1. Since {S2m} is nondecreasing and bounded from above, it has a limit, say (4)

If n is an odd integer, say n = 2m S2m+l

=

82m

+

I, then the sum of the first n terms is

+ U2m+l. Sinceun~O,

and,asm----i>00,

+

S2m+l = 82m

U2m+l --+ L

+0=

Iim,,~oo Sn ~

Combining the results of Equations (4) and (5) gives Exercise 131).

EXAMPLE 1

(S)

L.

L (Section 10.1, _

The alternating harmonic series 00

L(-l)n+11 =

n~l

I

n

_1 + 1_1 + ... 2

3

4

clearly satisfies the three requirements of Theorem 14 with N verges.

~

I; it therefore con-

-

Rather !ban directly verifying the defInition Un ;;,: Un+1o a second way to show that the sequence {un} is nonincreasing is to define a differentiable function j(x) satisfying j{n) = Un. That is, the values of j match the values of the sequence at every positive integer n. If j' (x) ,;; 0 for all x greater than or equal to some positive integer N, then j(x) is nonincreasing for x ;;,: N. It follows that j(n) ;;,: j(" + I), or Un ;;,: Un+ 10 for" ;;,: N.

EXAMPLE 2 Consider the sequence where Un ~ IOn/(,,2 + 16). Detme IOx/(x 2 + 16). Then from the Derivative Quotient Rule, j '() X

j(x) ~

2

~

1O{16 - x ) 1, the series converges absolutely. If 0 tionally. Conditional convergence: Absolute convergence:

p>O

< P ,;; 1, the series converges condi-

1 __1_+_1___1_+ ...

v'2

v'3

v'4

1 __1_+_1___1_+ ... 23(2

33/2

4 3/ 2

-

572


Rearranging Series We can always rearrange the terms of a finite sum. The same result is true for an inimite series that is absolutely convergent (see Exercise 68 for an outline of the proof).

THEOREM 17-The Reammgement Theorem for Absolutely Convergent Series If ~:::, a. converges absolutely, and bj, b2, ... , b., ... is any arrangement of the sequence {a.} , then ~b. converges absolutely and 00

00

Lb. = n=1 La.·

11=1

If we rearrange the terms of a conditionally convergent series, we get different results. In fact, it can be proved that for any real number r, a given conditionally convergent series can be rearranged so its sum is equal to r. (We omit the proof of this fact.) Here's an example of summing the terms of a conditionally convergent series with different orderings, with each ordering giving a different value for the sum.

EXAMPLE 6 We know that the alternating harmonic series ~:::, (-I).+I/n converges to some number L. Moreover, by Theorem 15, L lies between the successive partial sums 82 = 1/2 and 83 = 5/6, so L '¢' O. If we multiply the series by 2 we obtain

2L=2~

=2(1_1+1_1+1_1+1_1+1_~+~_ ...) .-1(_1).+1 n 2 3 4 5 6 7 8 9 10 11 212121212 =2-1 +3-2+5-3+7-4+9-5+0- .. ·.

Now we change the order of this last sum by grouping each pair of tenns with the same odd denominator, but leaving the negative terms with the even denominators as they are placed (so the denominators are the positive integers in their natural order). This rearrangement gives

(2 - 1) -

! + (t -t) - t + (t -t) - i + (% - t) - i + ... (1 -! + t - t + t - i + t - i + ~ - 1~ + 1\ _ .. -) 00

L 11=1

(-1).+1 n

= L.

So by rearranging the terms of the conditionally convergent series ~:::,2( _I).+I/ n, the series becomes ~:;O~I ( -1)'+ '/ n, which is the alternating harmonic series itself. If the two • series are the same, it would imply that 2L = L, which is clearly false since L '¢' O. Example 6 shows that we cannot rearrange the terms of a conditionally convergent series and expect the new series to be the same as the original one. When we are using a conditionally convergent series, the terms must be added together in the order they are given to obtain a correct result. On the other hand, Theorem 17 guarantees that the terms of an absolutely convergent series can be summed in any order without affecting the result.

Summary of Tests We have developed a variety of tests to determine convergence or divergence for an infinite series of constants. There are other tests we have not presented which are sometimes given in more advanced courses. Here is a summary of the tests we have considered.

10.6 Alternating Series, Absolute and Conditional Convergence

573

1. The 11th-Term Test: Unless a, --> 0, the series diverges. 2_ Geometric series: ~ar' converges if Ir I < I; otherwise it diverges. 3. p-series: ~ I/n" converges if p > I; otherwise it diverges.

4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test or the Limit Comparison Test. ~la,1 converge? If yes, so does ~a, since absolute convergence implies convergence.

5. Series with some negative terms: Does 6. Alternating series:

~a. converges if the series satisfies the conditions of the

Alternating Series Test.

Exercises 10.6 Determining Convergence or Divergence

00

In Exercises 1-14, detennine if the alternating series converges or

27. L( -1)'n'(2/3)'

diverges. Some of the series do not satisfY the conditions of the Alternating Series Test.

29.

,~I

::1

4. L ( - I ) ' - )' ( 11=2 In n 00 '5 6. L(-I),+I

;-±n + 4

11=1

10"

(Xl

8.

,&1 (-I)' (n + I)!

10. i(-l)n+I_I-

Inn

11"'2

12. i(-l)nln(1

14.

~(-I)'+I

+

k)

3v;;-:t:l" Vn+1

~

00

33.

00

37.

00

19.

L (- 1),+1

1Ft 00

21.

++ n

1

I

(-I)' ,~I I + Vn

n; n

5

+n +n

00 (-2),+1 24·L +S' 11"'1 n

25. i(_I),+11

~n

26. L( - I)n+I("'\Ylo)

23. i(-I)'+l 3 11 - 1

11=-1

n

00

,-I

11-

(_1),-1

~,c'-='---1 n + 2n + 1

00

36. ,~_. CO~n1T ~

(-I)'(n + I)' (2n)n

00

38.

~

(-I),+I(n!)' (2n)!

00 (nl)'3' 40. ~(-I)' (2n.+ I)!

00

00

11=1

11=1

41. L(-I)'(v;;-:t:l" - Vn)42. L(-I)'(~ - n) 00

43. L(-I)'(Vn

+ Vn - Vn)

11=1

44.

L _r 11""1

(-I)' _ r-:--c

00

45. L(-I)' sechn

,-I

vn+ vn+l

00

46. L(-I)'cschn

,-I I

I

I

I

I

I

"4 - 6 + "8 - 10 + IT - 14 + ...

48. I +

1111111 2S + 36 - 49 - 64 + ...

"4 - "9 - 16 +

•

22. L(-I)n sm,n 11=1

~

L

2

11=1 00

~(-I)' n + 3

I

00

20. L( - I)n+l

00

34.

00 ,(2n)! 39. ~(-I) 2"n!n

47.

18. i

~

35. ~_ cos mr ~ nVn

n-I

17. i(-I),_I,~I Vn

32. Y(-st'

(-100)' I n.

00

Which of the series in Exercises 15-48 converge absolutely, which converge, and which diverge? Give reasons for your answers. 00 00 (0 I)' 15. L(_I),+I(O.I)' 16. L(-I)'+I---j,

n

~ I

00

Absolute and Conditional Convergence

~-1

L /I _ I

Inn

n - Inn

00

31. Y(-I)'

4

00

3. i(-I)'+I-.!.". 1Ft n3

11- 1

00

2. i(-I)'+I+ 11=1 n /2

Vn

30. i(-I)'

1

n

nlnn

11=2

i( -I)' ~-I+ n

II- I

1. i(_I),+I_I-

28. i(_I),+I_I-

,-I

Errnr Estimation In Exercises 4g...S2, estimate the magoitude of the error involved in using the sum of the first four tenns to approximate the sum of the entire series.

574

Chapter 10: Infinite Sequences and Series is the same as the sum of the ftrSt n terms of the series

N. you will see in Section 10.7, the sum is In(l.Ol).

52. I

~t~

i.(

I

-I)'t',

0< t

Idl.

Proof The proof uses the Comparison Test, with the given series compared to a converging geometric series. Suppose the series ~:.o a.c' converges. Then lim,~co a.c' = 0 by the nth-Term Test. Hence, there is an integer N such that Ia.c'l < 1 for all n > N, so that forn>N. Now take anyx such thatlxl

Idl. The corollary to Theorem 18 asserts the

existence of a radius of convergence R '" O.

Since Ixlc I < 1, it follows that the geometric series ~~o Ixl c converges. By the Comparison Test (Theorem 10), the series ~~o la,llx'l converges, so the original power series ~~o a,x' converges absolutely for -I c I < x < Ie I as claimed by the theorem. (See Figure 10.16.) Now suppose that the series ~::'~o a,x' diverges at x ~ d. If x is a number with Ixl > Idl and the series converges at x, then the first half of the theorem shows that the series also converges at d, contrary to our assumption. So the series diverges for all x with Ixl > Idl· • To simplify the notation, Theorem 18 deals with the convergence of series of the form - a)' we can replace x - a by x' and apply the results to the series ~a.(x')'. ~a,x'. For series of the form ~a,(x

The Radius of Convergence of a Power Series The theorem we have just proved and the examples we have studied lead to the conclusion that a power series ~c.(x - a)' behaves in one of three possible ways. It might converge only at x ~ a, or converge everywhere, or converge on some interval of radius R centered at x ~ a. We prove this as a Corollary to Theorem 18.

COROLLARY TO THEOREM 18 The convergence of the series ~c,(x - a)' is described by one of the following three cases: 1. There is a positive number R such that the series diverges for x with Ix - a I > R but converges absolutely for x with Ix - a I < R. The series may or may not converge at either of the endpoints x ~ a - R and x ~ a + R. 2. The series converges absolutely for every x (R ~ 00). 3. The series converges at x ~ a and diverges elsewhere (R = 0).

10.7

Power Series

579

Proof We first consider the case where a = 0, so that we have a power series ~:o c"x' centered at O. If the series converges everywhere we are in Case 2. If it converges only at x = 0 then we are in Case 3. Otherwise there is a nonzero number d such that ~:;"~o c.d· diverges. Let S be the set of values of x for which ~:o c"x' converges. The set S does not include any x with Ixl > since Theorem 18 implies the series diverges at all such values. So the set S is bounded. By the Completeness Property of the Real Numbers (Appendix 7) S has a least upper bound R. (This is the smallest number with the property that all elements of S are less than or equal to R.) Since we are not in Case 3, the series converges at some number b oF 0 and, by Theorem 18, also on the open interval (-Ibl, Ibl). ThereforeR > O. Iflxl < R then there is a number c in S with Ixl < c < R, since otherwise R would not be the least upper bound for S. The series converges at c since c E S, so by Theorem 18 the series converges absolutely at x. Now suppose Ixl > R. If the series converges at x, then Theorem 18 implies it converges absolutely on the open interval (-lxi, Ixl), so that S contains this interval. Since R is an upper bound for S, it follows that Ix I :5 R, which is a contradiction. So if Ix I > R then the series diverges. This proves the theorem for power series centered at a = O. For a power series centered at an arbitrary point x = a, set x' = x - a and repeat the argument above replacing x with x' . Since x' = 0 when x = a, convergence of the series ~:olc'(x') I' on a radiusR open interval centered at x' = 0 corresponds to convergence oftheseries~:olc.(x - a)I' onaradiusR open interval centeredatx = a. _

14

R is called the radius of convergence of the power series, and the interval of radius R centered at x = a is called the interval of convergence. The interval of convergence may be open, closed, or half-open, depending on the particular series. At points x with Ix - a I < R, the series converges absolutely. If the series converges for all values of x, we say its radius of convergence is infinite. If it converges only at x = a, we say its radius of convergence is zero.

How to Test a Power Series for Convergence

1. Use the Ratio Test (or Root Test) to find the interval where the series converges absolutely. Ordinarily, this is an open interval Ix - al

I

Test each endpoint ofthe (rmite)

2.

interval of convergence.

R (it does not even converge conditiona1ly) because the nth term does not approach zero for those values of x.

Operations on Power Series On the intersection of their intervals of convergence, two power series can be added and subtracted term by term just like series of constants (Theorem 8). They can be multiplied just as we multiply polynomials, but we often limit the computation of the product to the 1rrst few terms, which are the most important. The following result gives a formula for the coefficients in the product, but we omit the proof.

580


THEOREM

A(x)

=

19-The Series Multiplication Theorem for Power Series

~::'~o a.x· and B(x) = ~:;;,o b.x· converge absolutely for Ixl

c. = aob.

If

< R, and

•

= ~akb.-.,

+ a,b.-, + a2b.-2 + ... + an-,b, + anbo

k~O

then ~::'~o c.x· converges absolutely to A(x)B(x) for Ixl

< R:

Finding the general coefficient c. in the product of two power series can be very tedious and the term may be unwieldy. The following computation provides an illustration of a product where we find the first few terms by multiplying the terms of the second series by each term of the fITst series:

00 ) • (00 .+1 ) ~(-l)'_x( ~x' ~o ~o n + I Multiply second series ...

(x -

f + X; - .. -) +

(x2 -

by!

~ + X; - .. -) +

t

(x3 -

by.

+

X; - .. -) + ...

by.' and gather the IllS! four powers.

We can also substitute a function j(x) for x in a convergent power series.

If ~:;;'o a.x· converges absolutely for Ix I < R, then ~:o an (j(x))" converges absolutely for any continuous function j on Ij(x) I < R.

THEOREM

20

Since I/(i - x) = ~:;;'ox' converges absolutely for Ixl < I, it follows from Theorem 20 that 1/(1 - 4x 2 ) = ~::'_o(4x2)'convergesabsolutelyforI4x21 < 1 orlxl < 1/2. A theorem from advanced calculus says that a power series can be differentiated term by term at each interior point of its interval of convergence. THEOREM 21-The Term-by-Term Differentiation Theorem

radius of convergence R

>

If ~cn(x - a)' has

0, it defines a function

00

j(x)

= ~cn(x

- a)'

on the interval

a - R

0 as k---> 00, whatever the value of x, so R2k+ 1(x) ---> 0 and the Maclaurin series for sin x converges to sin x for every x. Thus, 00 (_I)k 2k+1 3 5 7 x . '" xxx smx = ~ (2k + I)! = x - 3! + 5! - 7!

+ ...

(4)

•

592


EXAMPLE 3

Show that the Taylor series for cos x at x = 0 converges to cos x for every

value ofx. Solution We add the remainder term to the Taylor polynomial for cos x (Section 10.8, Example 3) to obtain Taylor's formula for cos x with n = 2k: x2

x 2k

x4

cosx = I - 2! + 4! - ... + (-I)k (2k)! + R2I.{x). Because the derivatives of the cosine have absolute value less thao or equal to I, the Remainder Estimation Theorem with M = 1 gives I R 2k(x) I

For every value of x, R2k(x) --+ 0 as k--+ value of x. Thus, 00

cosx = ~

,;;

Ixl2k+ 1

I· (2k + I)! .

00. Therefore, the

series conveIgeS to cos x for every

I)kx2k x 2 x' x. (2k)! = 1 - 2! + 4! - 6! + ...

(-

(5)

• Using Taylor Series Since every Taylor series is a power series, the operations of adding, subtracting, and multiplying Taylor series are all valid on the intersection of their intervals of convergence.

EXAMPLE 4 Using known series, f"md the first few terms of the Taylor series for the given function using power series operations. 1 (8) 3(2x + x cos x)

(b) eX cos x

Solution 1 21(X2X4 x2k) (8) 3(2x + x cos x) = 3 x + 3 x 1- 2! + 4! - ... + (-I)k (2k)! + ...

2

3

4

x + -x+ ' " x ( l + x +2!- + 3! 4!

= 1

x3

x4

3

6

+x----+ .. ·

)

-

•

By Theorem 20, we can use the Taylor series of the function f to find the Taylor series of f(u(x» where u(x) is any continuous function. The Taylor series resulting from this substitution will converge for all x such that u(x) lies within the interval of convergence of the Taylor

10.9 Convergence of Taylor Series

593

series of f. For instance, we can find the Taylor series for cos 2x by substituting 2x for x in the Taylor series for cos x: (-1)1(2x)21 (2k)! =

00

cos2x =

=

~

22X 2

24x4

(2x)'

(2x)4

(2x).

Eq. (5) with 2xfor x

I - 2! + 4! - 6 ! + ... 2.x.

1 -2!- +4!- - -6!+ .. ·

For what values ofx can we replace sinxby x - (x 3/3!) with an error of magnitude no greater than 3 X 1O-4?

EXAMPLE 5

Solution Here we can take advantage of the fact that the Taylor series for sinx is an alternating series for every nonzero value of x. According to the Alternating Series Estimation Theorero (Section 10.6), the error in truncating

. = x - -x 3 ! smx 3! I I

5

+ -x - ... 5!

after (x 3/3!) is no greater than

I~; I

lxi' 120 .

Therefore the error will be less than or equal to 3 X 10-4 if

lxi' < 120

3 X 10-4

or

Ixl
b is nearly the same.

f"(a) j(n)(a) f(a) + f'(a)(x - a) + ~(x - a)' + ... + ----;;r-(x - a)n

and its flISt n derivatives match the function f and its first n derivatives at x = a. We do not disturb that matching if we add another term of the form K(x - a )n+ 1 , where K is any constant, because such a term and its first n derivatives are all equal to zero at x = a. The new function

cf>.{x)

Pn(x) + K(x - a)n+l

=

and its frrst n derivatives still agree with f and its first n derivatives at x = a. We now choose the particular value of K that makes the curve y = cf>.{x) agree with the original curve y = f(x) at x = b. In symbols,

f(b)

=

Pn(b) + K(b - ar , '

or

K

=

--:f(--,-b)_---cP~n(b,_'_) (b - a)n+l

(7)

With K defined by Equation (7), the function

F(x)

=

f(x) - cf>.{x)

measures the difference between the original function f and the approximating function cf>n for each x in [a, b]. We now use Rolle's Theorem (Section 4.2). First, because F(a) = F(b) = 0 and both FandF' are continuous on [a, b], we know that

F'(Cl)

=

0

for some Cl in (a, b).

Next,becauseF'(a) = F'(Cl) = OandbothF'andF" are continuous on [a, c,],weknow that

F"(C2)

=

0

for some C2 in (a,

Cl).

Rolle's Theorem, applied successively to F", F"', ... , p(n-l) implies the existence of

C3

such that F"'(C3) = 0,

in (a, C2)

such thatF(4)(C4) = 0,

C4 in (a, C3) Cn

in (a, cn -,)

Finally, because F(n) is continuous on [a, cn] and differentiable on (a, cn), and = F(n)(cn) = 0, Rolle's Theorem implies that there is a number Cn+l in (a, cn) such that

F(n)(a)

(8)

Ifwe differentiate F(x) = f(x) - P.{x) - K(x - a)n+l a total ofn

F(n+l)(X)

=

+

I times, we get

j 0 has a Taylor series that converges to the function at every point of (-R, R). Show this by showing that the Taylor selies generated by f(x) = ~::'-o a.x· is the series ~:o a"x" itself. An immediate consequence of this is that series like •

2

6

8

5!

7!

x4 -+ x- - -x + ...

3!

approximation with an error less than 10-2 ? b. What is the maximum error we could expect ifwe replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions :2 21

4

' Y - tao-I Y 33. lun ,

y-o

dl,

(a) [0,0.5]

(b) [0, I]

35. lim x 2(e- 1/ x'

-

x~oo

1n(1 + x ) 37. lim """"'"''-----"x-a 1 cosX' 2

sinB - B + (11'/6) 32. 9-0 lim "' tr '

34. lun

y-o

y

I)

3'x'

+ 41 - 61 +'" X S + x 7 - x 9 + XII - •••

51. -I +2>;-3x 2 +4x'-5x 4 + ...

Theory and Examples 53. Replace x by -x in the Taylor series for In (I + x) to ob1ain a series for In (I - x). Then subtract this from the Thylor series for In (I + x) to showthatforlxl < I,

54. How many lem3s of the Taylor series for In (I + x) should you

2

t

34x4

+ ...

add to be sure of calculatiog In (1.1) with ao error of magnitude

less than 10-8? Give reasons for your answer.

I - cos 1 - (1 2/2)

1-0

2'

2 3 22.t-4 23x 5 2 4x 6 50. x -2>; +21-"""31+41-'"

(b) [0, I]

Indeterminate Forms

x-a

3"7!

In I + _ x =2 ( x+ x' + x' + . . .) . I x 3 5

Use series to evaluate the limits in Exercises 29--40. eX - (I + x) 29. lim

3"5!

"3 - 3"3 + 3"5 - 3"7 x 3 + X4 + x 5 + x 6 + ...

=

I

_1_ _ _1_ + ... 3'2' 4'24

1r - -r- + -1f'5- - - 1T'7 - + ...

. 3

48.1-

10 Exercises 25--28, fmd a polynomial that will approximate F(x) throughout the given interval with ao error of magnitude less than

/.X sin 12 dl, [0, I] 26. F(x) /,xI2e -t'dl, [0, I] 27. F(x) = /.X tao-I dl, (a) [0,0.5]

. 2

47.

0.1

s~x dx

0

1. __1_ +

44

4S

19--22 with an error of magnitude less than 10-8 .

19.

+ x') sinx 2

41. I + I + 2! + 3! + 4! + ...

D 10 Exercises 15--18, use series to estimate the integrals' values with an

/.°

X'

Using Table 10.1 10 Exercises 41-52, use Table 10.1 to fmd the sum of each series.

+ x 2)'

12. (1

. 3

sm x 1 cos 2x

603

_tao_-_'~y_-_sin ____ y

, y cosy

36. lim (x + I) sin _+1 I x_OO x

55. According to the Alternatiog Series Estimation The0ren3, how many lem3s of the Thylor series for tao-I I ""uld you have to add to be sure of fmding 'If/4 with ao error of magnitude less than

10-31 Give reasons for your answer. 56. Show that the Thylor series for !(x) Ixl> I.

D 57.

= tao-I x

diverges for

Estimating Pi About how many tenns of the Taylor series for tao-I x v.uld you have to use to evaluate each term on the righthand side of the equation

_ 48

'If -

-1

tao

I + 18

32

-1

tao

I 20 -1 I 57 tao 239

with ao error of magnitude less than IO.... ? 10 contrast, the convergence of ~:~I(I/n2) to ~/6 is so slow that even 50 terms will not yield two-place accoracy.

604


58. Integrate the rlrst three nonzero terms of the Taylor series for tan t from 0 to x to obtain the fIrst three nonzero terms of the Taylor series for In sec x.

59.

L

by integrating the series

Use the binomial series and the fact that

in the first case from x to tox. to generate the fIrst four nonzero terms of the Taylor series

for sin-1 x. What is the radius of convergence? b. Series for cos-1 x Use your result in part (a) to rmd the rrrst five nonzero terms of the Taylor series for 008- 1 x. 60.

L

Seri.. for sinh-1 x Taylor series for

Find the rrrst four nonzero terms of the

sinb-1 x

=

r

dt

Jo \IT+7

+ x)'

from the series for

+ x).

62. Use the Taylor series for 1/(1 - x') to obtain a series for 22/(1 - x')'.

D 63.

00

Euler's Identity 67. Use Equation (4) to write the followiog powers of e in the form a + hi. a. e-l'll' 68. Use Equation (4) to show that

69. Establish the equations in Exercise 68 by combining the formal Taylor series for e ifJ and e-ifJ • 70. Show that

a. coshiO = cosO,

estimation error. 61. Obtain the Taylor series for 1/(1

and in the second case from -

.

D b. Use the rrrst three terms of the series in part (a) to estimate sinb-1 0.25. Give an upper bound for the magnitude of the

-1/(1

00

Estimating Pi The English mathernaticiao Wallis discovered the fom2Ula ~

2,4,4,6,6,8, .. .

4

3,3,5,5,7,7, ... .

b. sinhi8

= isin8.

71. By multiplying the Taylor series for eX and sin x, rmd the terms through x 5 of the Taylor series for eX sin x. This series is the imaginary part of the series for eXoeix

=

e(l+i~.

Use this fact to check your anawer. For what values of x should the series for eX sin x converge? 72. Wheo a aod b are real, we derme e(a+ib).< with the equation e(a+ib)x

=

etIXoeibx

=

etIX(cosbx

+ isinbx).

Find ~ to two decin2al places with this formula. Differentiate the right-hand side of this equation to show that

64. Use the followiog steps to prove Equation (I). L

:fx

Differentiate the series f(x)

=

I

+

~ (:)x'

e(a+ib).

n

n~OO

(

1-

ofor all n,

a. Show that ~: 1 a,? converges.

nx

(n + 1)(2); + I)n

b. Does ~:;"~1 an/(I - an) converge? Explain.

28. (Continuation of Exercise 27.) If ~::;'1 a" converges, and if I> a" > Oforalln,abowthat~:lln(1 - an)eonverges.

Does the value of

.

n2 •

(2n - I)' Ul ~ I, Un+l ~ (2n)(2n + I) Un·

converges absolutely.

hm

j(n)

n

Apply Raabe's Test to determine whether the series converges.

00

L

+ f: +

where Ij(n)l < K for n '" N, then ~::;'1 Un converges ifC > I and diverges if C '" I. Show that the results of Raabe's Test agree with what you know about the series ~:;"- 1 (I/n') and ~:;"- 1 (I/n).

I - - , dx. l+x

18. Find all values of x for which

D 19.

I

26. (Continuation ofExercise 25. ) Suppose that the terms of ~:;"- 1 Un are defmed recursively by the formulas

17. Evaluate

~

~

(Hint: First show that Iln(1 - an) I '" a"f(1 - a,,).)

eos(a/n»)n a constant, n •

appear to depend on the value of a? If so, how?

I

b. Does the value of . ( cos (a/n»)n hm 1bn ' 11_00

a and b constant, b

Prove Nicole Oresme's Theorem that

29. Nicole Oresme's Theorem

¢'

0,

+ 1. 2

2

+ 1 . 3 + ... + --"-+ ... ~ n 1 4

2

-

4.

(Hint: Differentiate both sides of the equatioo 1/(1 - x) 1 + ~:_ IX".)

~

30. a. Show that

appear to depend on the value of b? If so, how? c. Use calculus to confirm your fmdings in parts (a) and (b). 20. Show that if ~:;"~1 a" converges, then

00

~I n-

n(n

+ I) Xn

2>;'

~(x - I )'

for Ix I > I by differentiating the identity 00

LXn+

,

1

IFI

converges.

=_x_ 1 - x

twice, multiplying the result by x, and then replacing x by I/x.

21. Find a value for the eonstant b that will make the radius of con-

vergence of the power series

b. Use part (a) to fmd the real solmioo greater than I of the equation

x~fn(n;l). 11=1

equal to 5.

x

31. Quality control

22. How do yoo know that the functioos sinx, lnx, and e' are not polyoomia\s? Give reasons for your answer.

23. Find the value of a for which the limit lim sin (ax) - sinx - x

x-a

x3

is fmite and evaluate the limit. 24. Find values of a and b for which . cos(ax)-b hm , 2x

x-o

~-l.

25. Raahe'. (or Gau••'.) Te.t The fo\lowiog test, which we state without proof, is an extension of the Ratio Test.

a. Differentiate the series _1_= 1 I-x

+X+X2+ ... +X"+'"

toobtainaseriesforl/(l-x)'.

b. In ooe throw of two dice, the probability of gettiog a roll of7 is p ~ 1/6. If you throw the dice repeatedly, the probability that a 7 will appear for the fll'St time at the nth throw is qn-lp , where q ~ I - p ~ 5/6. The expected number of throws until a 7 frrst appears is ~:=1 nq"-Ip . Find the sum of this series. c. As an engineer applying statistical control to an industrial operation, yoo inspect items taken at randem from the assembly line. You classifY each sampled item as either "good" or

Chapter 10 Technology Application Projects ''bad.'' If the probability of an item's beiog good is p and of an item's beiog bad is q ~ I - p, the probability that the fust bad item found is the nth one inspected is p,,-lq. The average number iospected up to and iocluding the fust bad itero found is ~:lnpll-lq.Evaluatethis sum, assuming 0 < P < 1. 32. EIpected value Suppose that a random variable X may assome the values 1, 2, 3, ... , with probabilities Pt,P2,P3, ... , where Pi is the probability that X equals k (k ~ I, 2, 3, ... ). Suppose also that Pk '" 0 and that ~:'Pk ~ 1. The OIpected value of X, denoted by E(X), is the number ~:,kp1' provided the series converges. In each of the following cases, show that ~~'Pk ~ I and fmdE(X) ifit exists. (Hint: See Exercise 31.)

C. P1~

D 33.

c. Ifk ~ 0.01 h- 1 and to ~ IOh, fmd the smallestn such that

R. > (1/2)R. (Source: Prescribing Safe and Efficnve Dosage, B. Horelick and S. Kooot, COMAP, Inc., Lexington, MA.)

34. Time b_een drug do... (ContinUfJtion of Exercise 33.) If a drug is known to be ineffective below a concentration CL and harmful above some higher concentration CH , one needs to fmd values of Co and to that will produce a concentration that is sare (not above CH ) but effective (not below CL ). See the accompanyiog figure. We therefore want to fmd values for Co and to for which

R

51--1

~

CL

and

Co

+R

b.Pk~1

a.Pk=2--j;

6

I

I

"§ :0 C H

.S

Safe and effective dosage The concentration io the blood resultiog from a siogle dose of a drug nonnaily decreases with time as the drug is eliotioated from the body. Doses may therefore need to be repeated periodically to keep the conccn1ration from dreppiog below some particular level. One model for the effect of repeated doses gives the residual concentration just before the (n + I)stdoseas

Coe-kto

+ Coe-2kto + ... + Coe-nkto ,

where Co ~ the change io conceotratioo achievable by a single dose (mgjmL), k ~ the elimination constant (h-1) , and to ~ time between doses (h). See the accompanying figure.

~

CH.

C

I

k(k+ 1) ~k- k+ I

RII =

609

.g•

~ ~

Highest safe !evel

t

Co

CL

.j,

I

I

Lowest effective level

~ to~ 0

Time

Thus Co ~ CH - CL . Wheo !bese values are substituted io !be equatioo for R ohtsioed io part (a) of Exercise 33, the resultiog equatioo simplifies to

C

To reach an effective level rapidly, one might administer a "Ioadiog" dose that would produce a conccn1ration of CH mg/mL. This could be followed every to hours by a dose that raises the conceotratioo by Co ~ CH - CL mg/mL. a. VerifY the preceding equation for to.

Time (h)

a. Write R" in closed form as a single fraction, and fmd R =

litnn_oo R".

b. CaicuiateR, and RIO for Co ~ I mg/mL, k ~ 0.1 h- 1 , and to ~ 10 h. How good an estimate of R is RIO?

Chapter

2J!)

b. If k ~ 0.05 h- 1 and the highest safe conccn1ration is e times the lowest effective concen1ration, fmd the leogth of time b _ doses that will assure safe and effective concentrations. c. GivenCH~ 2mg/mL,CL ~ O.5mg/mL,andk~ 0.02h- 1 , determine a scheroe for administeriog the drug. d. Suppose that k ~ 0.2 h-1 and that the sma11est effective coocen1ratioo is 0.03 mgjmL. A siogle dose that produces a coocen1ratioo of 0.1 mg/mL is administered. About how long will the drug reroaio effective?


Mathematica/Maple Modules: Bouncing BaH The model predicts the height of a bounciog ball, and the time until it stops bounciog. ~Ior Po/ynomiDIApproximll/ion.

ofII Function

A graphical animation shows the convergence of the Taylor polyoomials to functioos baviog derivatives of all orders over an ioterval io their domaios.

11 PARAMETRIC EQUATIONS AND POLAR COORDINATES OVERVIEW In this chapter we study new ways to define curves in the plane. Instead of thinking of a curve as the graph of a function or equation, we consider a more general way of thinking of a curve as the path of a moving particle whose position is changing over time. Then each of the x- and y-coordinates of the particle's position becomes a function of a third variable t. We can also change the way in which points in the plane themselves are described by using polar coordinates rather than the rectangular or Cartesian system. Both of these new tools are useful for describing motion, like that of planets and satellites, or projectiles moving in the plane or space. In addition, we review the geometric definitions and standard equations of parabolas, ellipses, and hyperbolas. These curves are called conic sections, or conics, and model the paths traveled by projectiles, planets, or any other object moving under the sole influence of a gravitational or electromagnetic force.

11.1

Parametrizations of PLane Curves In previous chapters, we have studied curves as the graphs of functions or equations involving the two variables x and y. We are now going to introduce another way to describe a curve by expressing both coordinates as functions of a third variable t.

Parametric Equations Figure 11.1 shows the path of a moving particle in the xy-plane. Notice that the path fails the vertical line test, so it cannot be described as the graph of a function of the variable x. However, we can sometimes describe the path by a pair of equations, x = f(t) and y = g(t), where f and g are continuous functions. When studying motion, t usually denotes time. Equations like these describe more general curves than those like y = f(x) and provide not only the graph of the path traced out but also the location of the particle (x, y) = (f(t), g(t» at any time t.

DEFINITION

Ifx andy are given as functions

x = f(t),

FIGURE 11.1 The curve or path traced by a particle moving in the xy-plane is not always the graph of a function or single equation.

610

y = g(t)

over an interval loft-values, then the set of points (x, y) = (f(t), g(t» defined by these equations is a parametric curve. The equations are parametric equations for the curve.

The variable t is a parameter for the curve, and its domain I is the parameter interval. If I is a closed interval, a :5 t :5 b, the point (f(a), g(a») is the initial point ofthe curve and (f(b), g(b) is the terminal point. When we give parametric equations and a parameter

11.1

611

Parametrizations ofplan. Curves

interval for a curve, we say that we have parametrized the curve. The equations and interval together constitute a parametrization of the curve. A given curve can be represented by different sets of parametric equations. (See Exercises 19 and 20.)

EXAMPLE 1

Sketch the curve defined by the parametric equations

y = I

+ I,

-00 < 1 O.

t'

Solution We make a brief table of values in Table 11.2, plot the points, and draw a smooth curve through them, as we did in Example I. Next we eliminate the parameter I from the equations. The procedure is more complicated than in Example 2. Taking the difference between x and y as given by the parametric equations, we fmd that x - y =

(I + +) - (I - +) t. =

If we add the two parametric equations, we get

4.8 9.9

x

+y

=

(I + +) + (I - +)

=

21.

We can then eliminate the parameter I by multiplying these last equations together: y

10

(x - y)(x

+ y)

(t) 0, for which x is always positive. That is, the parametric equations do not yield any points on the left branch of the hyperbola given by Equation (I), points where the x-coordinate would be negative. For small positive values of I, the path lies in the -10 fourth quadrant and rises into the first quadrant as I increases, crossing the x-axis when t = 0.1 I = I (see Figure 11.6). The parameter domain is (0, 00) and there is no starting point and _ FIGURE 11.6 The curve for x = t + (l/t), no terminal point for the path. y = t - (I/t), t > 0 in Example 7. (The Examples 4,5, and 6 illustrate that a given curve, or portion ofi!, can be represented by part shown is for 0.1 s t '" 10.) different parametrizations. In the case of Example 7, we can also represent the right-hand branch of the hyperbola by the parametrization ~~~----~------~--~x

x=

\1'4+7,

y

=

t,

-00

x

)

Find all the polar coordinates of the point P(2, 1r/6).

Solution We sketch the initial ray of the coordinate system, draw the ray from the origin that makes an angle of 1r/6 radians with the initial ray, and mark the point (2, TT/6) (Figure 11.21). We then find the angles for the other coordinate pairs of P in which r = 2 andr = -2. For r = 2, the complete list of angles is 7T

6'

7T

7T

7T

6 ± 21r, 6 ± 41r, 6 ± 61r, ....

628

Chapter 11: Parametric Equations and Polar Coordinates

(2. *) ~ (_2._5;) ~ (-2. 7;)

For r = -2, the angles are

_ 5'1f ± 4'Il" 6 '

5'1f -T± 6'1f, ....

The corresponding coordinate pairs of P are

(2, ~ + 2n'lf).

n

=

0, ±l, ±2, ...

n

=

and

FIGURE 11.21 The point P(2. 'IT/6) has inImitely many polar coordinate pairs (Example 1).

5'1f + ( -2'-6

2n'lf) '

0, ±l, ±2, ....

When n = 0, the formulas give (2,'If/6) and (-2, -5'1f/6). When n = I, they give (2, 13'1f/6) and (-2, 7'1f/6) , and so on. _

Polar Equations and Graphs If we hold r fixed at a constant value r = a oF 0, the point P(r, 6) will lie Ia Iunits from the origin O. As 0 varies over any interval oflength 2'1f, P then traces a circle of radius Ia I centered at 0 (Figure 11.22). If we hold 6 fixed at a constant value 6 = 60 and let r vary between - 00 and 00, the point P(r, 0) traces the line through 0 that makes an angle of measure 00 with the initial ray. FIGURE 11.22 The polar equation for a circle is r = a.

Equation

Graph

r=a 6 = 60

Circle of radius Ia Icentered at 0 Line through 0 making an angle 60 with the initial ray

y

(a)

1::sr:S2.0:S(J:S~

EXAMPLE 2 ----c+---1'-'-~x

o

1

2

0, the parabola partitions the rectsngular region bounded by these lines and the coordinate axes into two arualler regions, A and B. a. If the two arualler regions are revolved about the y-axis, show that they generate solids whose volumes have the ratio 4: 1. b. What is the ratio of the volumes generated by revolving the regions about the x-axis? y p

B ~~----------~~X

o

73. Area Find the dimensions of the rectsng1e of largest area that can be insctibed in the ellipse x 2 + 4y2 ~ 4 with its sides parallel to the coordinate axes. What is the area of the rectangle? 74. Volume Find the volUD2e of the solid generated by revolving the region enclosed by the ellipse 9x' + 4yl ~ 36 about the 0

Polar Coordinates 17.

L

Find an equation in polar coordinates for the curve

x = e 2 tcost. y = e2t sint;

-00

- 0 from (I), tanI/J

= tan(t/>

tant/> - tanO

- 0)

= I + tant/>tanO'

Furthermore,

p

dy

dy/dO

tant/> = 4 The exterior of the sphere x 2 + y2 + z2 = 4. 2 (d) x + y2 + z2 = 4, z :5 0 The lower hemisphere cut from the sphere x 2 +

y2

+ z2

= 4 by the xy-plane (the plane

z = 0) .

•

Just as polar coordinates give another way to locate points in the xy-plane (Section 11.3), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for three-dimensional space. We examine two of these coordinate systems in Section 15.7.

Exercises 12.1 Geometric Interpretations of Equations

Geometric Interpretations of Inequalities and Equations

In Exercises 1-16, give a geometric description of the set of points in

In Exercises 17-24, describe the sets of points in space whose coordinates satisfY the given inequalities or combinatioos of equatioos aod inequalities.

space whose coordinates satisfY the given pairs of equations. 1. x = 2, Y = 3 2. x = -1, z = 0

= 0, 2 x + y2

3. y

z = 0

S.

=

2

4,

z = 0

+z 2 =4, y=O 9. x 2 + y2 + z2 = I, x = 0

7.

X

11. 12. 13. 14. 15.

16.

+ y2 = 4, z = 8. y2 + z2 = I, x =

6. xl

+ y2 + z2 = 25, Y = -4 x 2 + y2 + (z + 3)' = 25, z = 0 x 2 + (y - 1)2 + z2 = 4, Y = 0 x 2 + y2 = 4, z = Y x 2 + y2 + z2 = 4, Y = x Y = x 2, z = 0 z = y2, X = 1

10. x 2

17. a. x

4. x = I, Y = 0 -2 0

~

0, y

0,

~

Z

18.•. 0 '" x '" 1 c. 0 ~ x ~ 1, 0

~

y

= 0

~

b. x

1,

~

+ y2 + z2

:s 1

20. a. x 2

+ y2 :s 1,

Z

+ y2:5

norestrictiononz

a.

C. x 2

1,

= 0

~

b. 0 '" x '" I, 0 ~z~ 1

x2

19.

0, y

b.

Xl

b. x 2

0,

Z

= 0

0 '" Y s 1

+ y2 + z2 >

1

+ y2 :s

=3

1,

Z

21. a. 1 :sx 2 +y2+z2 :S4 b. x 2

+ y2 + z2 :s

22. a. x = y,

23. a. y ~ x 2,

z = 0 Z

24. a. z = 1 - y, b. z =

y3,

X

~ 0

I, z ~ 0 b. x

=

y,

b. x

S

y2,

no restriction onx

=2

no restriction on z

0

S Z

s 2

662

Chapter 12: Vectors and the Geometry of Space

Distance and Spheres in Space

Z

The formula for the distance between two points in the xy-plane extends to points in space.

The Distance Between P1(XhYh Zl) and P 2(X2,Y2, Z2) is

+

IPI P21 = V(X2 - XI)2

x

FIGURE 12.5 We find the distance between PI and P 2 by applying the Pythagorean theorem to the right triangles PIAB and PIBP2.

(Y2 - YI)2

+

(Z2 - ZI)2

Proof We construct a rectangular box with faces parallel to the coordinate planes and the points PI and P2 at opposite comers of the box (Figure 12.5). If A(X2, YI, zd and B(X2, Y2, Zl) are the vertices of the box indicated in the figure, then the three box edges P I A, AB, and BP2 have lengths

Because triangles PIBP2 and PIAB are both right-angled, two applications of the Pythagorean theorem give IPIP212 = IPIBI 2 + IBP21 2

and

(see Figure 12.5). So IPIP212 = IPIBI 2 + IBP21 2 IPIA 12

+

IABI2

+

Substitute IP]BI 2 = IP]AI 2 + IABI2 .

IBP212

IX2 - xl1 2 + IY2 - YI1 2 + IZ2 - zI1 2 =

(X2 - XI)2

+

(Y2 - Ylf

+

(Z2 - zlf

Therefore

• EXAMPLE 3

The distance between P I (2, 1,5) and P2( -2,3,0) is IPIP21 = V(-2 - 2f

Z

Po(Xo, Yo, zo)

\

P(x,Y,z)

al

I

=

V16

=

V45

+4+ I"::j

+

(3 - 1)2

+

(0 - 5)2

25

•

6.708.

We can use the distance formula to write equations for spheres in space (Figure 12.6). A point P(x, y, z) lies on the sphere of radius a centered at Po(xo,Yo, zo) precisely when IPoPI = a or

I

-- I f"

,,/ 1--

The Standard Equation for the Sphere of Radius a and Center (xo,Yo, zo) (x - xof

+ (y -

YO)2

+

(z - zO)2

= a2

Y x

FIGURE 12.6 The sphere of radius a centered at the point (xo,Yo, zo).

EXAMPLE 4

Find the center and radius of the sphere x2

+ y2 + z2 +

3x - 4z

+ 1 = O.

SoLution We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the X-, Y-, and z-terms as necessary and write each

664


In Exercises 25-34, describe the given set with a single equatioo or

with a pair of equations. 25. The plaoe perpendicular to the L

x-axis at (3, 0, 0)

b. y-axis

c. z-axis

27. The plaoe througb the point (3, -I, I) parallel to the xy-plane

b. yz-plane

c. xz-plane

28. The circle of radius 2 centered at (0, 0, 0) and lying in the L

xy-plane

b. yz-plane

c. xz-plane

29. The circle of radius 2 centered at (0, 2, 0) and lying in the L

xy-plane

b. yz-plane

parallel to the xy-plane

b. yz-plane

Sphel'l!s Find the centers and radii of the spheres in Exercises 47-50.

+2)' + +(z - 2)' = 8 48. I)' +&+tY +(z +3)' 25 49. (x - Vz)' +{y - Vz)2 +(z +Vz)' 50. x+ &+tY +(z-tY ~ 47. (x

y2

(x -

=

2

b. y-axis

2

=

Find equations for the spheres whose centers aud radii are given in Exercises 51-54.

c. xz-plaoe

Center

Radius

31. The line through the point (I, 3, -I) parallel to the

LX-axis

=

c. planey = 2

30. The circle of radius I centered at (-3,4, I) and lying in a plane L

P2(2, -2, -2)

46. P I (5, 3, -2), P2(0, 0, 0)

26. The plaoe througb the point (3, -1,2) perpendicular to the

L

P 2(2, 3, 4)

45. PI(O, 0, 0), b. y-axis at (0, -I, 0)

c. z-axis at (0, 0, -2) LX-axis

44. P I (3, 4, 5),

c. z-axis

32. The set of points in space equidistant from the origin and the point (0, 2, 0)

53.

33. The circle in whicb the plane through the point (I, 1,3) perpendicular to 1Ire z-axis meets the sphere of radius 5 centered at the origin 34. The set of points in space that lie 2 units from 1Ire point (0, 0, I) and, at 1Ire saroe time, 2 units from the point (0, 0, -I) Inequalities to Describe Sets of Points Write inequalities to describe the sets in Exercises 35-40.

35. The slab bounded by the planes z = 0 and z = I (planes

Vi4

51. (1,2,3) 52. (0, -1,5)

2

(-I,t,-t)

4 9

54. (0, -7,0)

7

Find the centers and radii of the spheres in Exercises 55-58. 55. x 2 y2 z2 4x - 4z = 0

+++ 56. ++ + 57. + + +++ 9 58. 3x + + + 9 x2

lx

y2

2 2

z2 -

8z = 0

6y

2y2

2z2

X

3y2

3z2

2y -

Y

z =

2z =

included) 36. The solid cube in the frrst octant bounded by 1Ire coordioate plaoes and 1Ire planes x = 2, y = 2, and z = 2

37. The half-space consisting of the points 00 and below the xy-plane 38. The upper hemisphere of the sphere of radius I centered at the origin

39. The (a) interior and (b) exterior of the sphere of radius I centered at the point (I, I, I) 40. The closed regioo bounded by the spheres of radius I and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open regioo bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out)

Distance In Exercises 41-46, fmd the distsnce between points PI and P2. 41. Pl(l, I, I),

P 2(3, 3, 0)

42. Pl( -I, 1,5), P2(2, 5, 0) 43. PI(I, 4, 5),

P2(4, -2,7)

Theory and Examples 59. Find a formula for the distance from the point P(x, y, z) to the

a. x-axis

c. z-axis

b. y-axis

60. Find a formula for the distance from the point P(x, y, z) to the

a. xy-plane

c. xz-plane

b. yz-plane

61. Find the perimeter of the triangle with vertices A( -1,2, I), B(I, -I, 3), and C(3, 4, 5). 62. Show that the point p(3, I, 2) is equidistant from the points A(2, -1,3) andB(4, 3, I). 63. Find an equation for the set of all points equidistant from the planesy = 3 andy = -\. 64. Find an equation for the set of all points equidistant from the point (0, 0, 2) and the xy-plane. 65. Find the point on the sphere x 2 nearest •• thexy-plane.

+(y - + + 3)'

(z

5)' = 4

b. thepoint(O, 7, -5).

66. Find the point equidistant from the points (0, 0, 0), (0,4, 0), (3, 0, 0), and (2, 2, -3).

12.2

12.2

Vectors

665

I~ ~_ ct_ or_ s __________________________________ Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit of measure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body's displacement, we have to say in what direction it moved as well as how far. To describe a body's velocity, we have to know where the body is headed as well as how fast it is going. In this section we show how to represent thiogs that have both magnitude and direction in the plane or in space. Terminal

FIGURE 12.7

The directed line segment

AB is called a vector.

Component Form A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Fignre 12.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen uuit. For example, a force vector points in the direction in which the force acts and its length is a measure of the force's strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 12.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 13.) y

y B _______ D

A

C p ----~~~-----~x

o

FIGURE 12.9 The four arrows io the plane (directed lioe segments) shown here have the same length and direction. They therefore represeot the same vector, aodwewriteAB = cD = OP = EF.

~~~~----~y

o

F

E

o

~--------~X

x

(a) two dimensions

(b) three dllnensions

FIGURE 12.8 The velocity vector of a particle moving aloog a path o. The cross-sections in planes perpendicular to the z-axis above and below the xy-plane are hyperbolas. The cross-sections in planes perpendicular to the other axes are parabolas.

Near the origin, the surface is shaped like a saddle or mountain pass. To a person traveling along the surface in the yz-plane the origin looks like a minimum. To a person traveling the xz-plane the origin looks like a maximum. Such a point is called a saddle point of a surface. We will say more about saddle points in Section 14.7. _ Table 12.1 shows graphs of the six basic types of quadric surfaces. Each surface shown is symmetric with respect to the z-axis, but other coordinate axes can serve as well (with appropriate changes to the equation).

12.6

699

Cylinder.; and Quadric Surfaces

TABLE 12.1 Graphs of Quadric Surfaces

,

, Tho PEIobolu _ ~l:3

ill Ibo ",,"pImo

*~~+,, -T'---, ,, ,,

" ......

, 'Ibe p,nbolu _

'~/

~:I

____

;"'1110 1f'1Ilmo

,

,

• ELLIPSOID

ELLIPTICAL PARABOLOID

•

,/

)f--r-"

HYPERBOLOID OF ONE SHEET

ELLIPTICAL CONE

,

. .,.

'Ibe e1Hpee -

+- _1

.'f - cV'i

,

,

. HYPERBOLOID OF 1WO SHEETS

HYPERBOLIC PARABOLOID

% C'

.>0

700

Chapter 12: Vector.; and the Geometry of Space

Exercises 12.6 Matchtng Equations with Surfaces In Exercises 1-12, match the equation with the surface it defines.

k.

,

L

Also, identify each surface by type (paraboloid, ellipsoid, etc.) The swfaces arc labeled (a)-{I).

1. x:1 +y:1 + 4z2 = 10 3. 9y2 +z:1 = 16 5. x _ y 2_ z 2 7.x2 +2z 2 _ S

2.z:1+4y 2_4x:1=4 4.y:1+ z 2=X 2 6.x _ _ 2_ 2 z y 8.z 2 +x:1- y 2 _ 1 10. z = _4.1: 2 _ y2

9. X=z2_ y :1 11. x:1 + 4z:1 = y2

12. 9x:1

+ 4y:1 + 2z2

Drawtng

Sketch the surfaces in Exercises 13-44.

= 36

,

b.

L

,

CYUNDERS 13. X2 +y2=4 15. x2+~ _ 16

14.z=y2-1 16. 4x 2 + y2 _ 36

ELUPSOIDS 17. 9;t2 + y2 + z2 = 9 19. 4x 2 + 9y2 + ~ _ 36

18. 4x:1 20. 9;t2

+ 4y:1 + z:1 = 16 + 4y2 + 36z2 _ 36

PARABOLOIDS AND CONES

,

,.

,

d.

:n.z=S-:x2 -y:1

21. z=x:1+4y:1 23. x=4-4y:1-~ 25. X:1+ y 2=z:1

24.y=I-:x 2 -z2 26. 4x:1 + ~ = 9y2

HYPERBOLOIDS 28.y:1+ z 2_ x 2=1 30. (y'/4) - (.'/4) -

27.x:1+ y 2- z :1=1 29. z2_ X:1_ y 2= 1

y

z'

~

1

HYPERBOUC PARABOLOIDS

••

,

f•

31. y2-:x:1 _ z

,

ASSORTED

y

..

,

h•

x2

38.x2+~ _ y

40. 16y2 + 9z:1 _ 4x 2 42.y:1_ x 2_ z 2=1 44.x 2 +y:1=z

+z2

_ 1 41. z = _(X2 + y2) 43.4y 2+ z 2_4x:1=4

39.

y

34. 4x:1 + 4y:1 = z2 36. 1&2 + 4y2 _ 1

33. z= 1 +y2_X:1 35. Y _ _ (x:1 + z2) 37. x 2 +y2_z2 _ 4

Theory and Examples 45. L Express the area.4. of the cross-section cut from. the ellipsoid

, ,

x:1+L+~=1 4 9

L

,

J.

,

by the plane z - c as a function of c. (The area of an ellipse with semiaxesa andbis7rab.) b. Use slices perpendicular to the z-axis to fmd the volume of

the ellipsoid in part (a). c. Now find the volume of the ellipsoid

x2

y2

z:1

-+-+- 1. a:1 b2 c2 y

Does your formula give the volume of a sphere of radius a if a - b - c?

Chapter 12

701

h. Express your answer in part (a) in terms of h and the areas Ao and Ah of the regions cut by the hyperboloid from the planes z = 0 andz = h.

46. The barrel shown here is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the z-axis. The crosssections perpendicular to the z-axis are circular. The barrel is 2h units high. its midsection radius is R, and its end radii are both r. Find a formula for the barrel's volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder of radius R and height 2h. Does your formula give the cylinder's volume? Second, suppose r = 0 and h = R so the barrel is a sphere. Does your formula give the sphere's volume?

z


c. Show that the volume in part (a) is also given by the formula V=

"6h (Ao + 4Am + Ah ),

where Am is the area of the region cut by the hyperboloid from the plane z = h/2.

Viewing Surfaces

D Plot the surfaces in Exercises 49-52 over the indicated domains.

If

you can, rotate the surface into different viewing positions.

-2 ~ x ~ 2,

49. z = y2,

-2 ~ x ~ 2,

50. z = 1 - y2, y

51. z = x 2 52. z = x 2

47. Show that the volume ofthe segment cut from the paraboloid x2 y2 z -2+ - = a b2 c

by the plane z = h equals half the segment's base times its altitude. 48. a. Find the volume of the solid bounded by the hyperboloid x2 y2 z2 -2+ -2- - = 1 a b c2

and the planes z = 0 and z = h, h

Chapter

>

~

c. -2 d. -2

~

x

~

x

-2 ~ y ~ 2

x ~ 3,

-3 ~ y ~ 3

~

3,

-3

~

y

~

3

~x~

1, 2,

-2

~

y

~

3

~

-2

~

y

~

2

~

2,

-1

~y~

1

x

COMPUTER EXPLORATIONS Use a CAS to plot the surfaces in Exercises 53-58. Identify the type of quadric surface from your graph. x2

53.

y2

9 + 36

55. 5x 2

o.

+ y2, -3 ~ + 2y2 over

a. -3 h. -1

-0.5 ~ y ~ 2

=

9 -

1 - 25

z2 - 3y2

x2

57.

z2 =

y2

1

=

54.

56. z2

16 + 2

x2

z2

9 - 9 y2

16

58. y -

=

1-

=

1-

x2

y2

16

9 +z

V4 -

z2 = 0


1. When do directed line segments in the plane represent the same vector? 2. How are vectors added and subtracted geometrically? Algebraically? 3. How do you find a vector's magnitude and direction? 4. If a vector is multiplied by a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative? 5. Define the dot product (scalar product) of two vectors. Which algebraic laws are satisfied by dot products? Give examples. When is the dot product of two vectors equal to zero? 6. What geometric interpretation does the dot product have? Give examples. 7. What is the vector projection of a vector u onto a vector v? Give an example of a useful application of a vector projection. 8. Define the cross product (vector product) of two vectors. Which algebraic laws are satisfied by cross products, and which are not? Give examples. When is the cross product of two vectors equal to zero? 9. What geometric or physical interpretations do cross products have? Give examples.

10. What is the determinant formula for calculating the cross product of two vectors relative to the Cartesian i, j, k-coordinate system? Use it in an example. 11. How do you find equations for lines, line segments, and planes in space? Give examples. Can you express a line in space by a single equation? A plane? 12. How do you find the distance from a point to a line in space? From a point to a plane? Give examples. 13. What are box products? What significance do they have? How are they evaluated? Give an example. 14. How do you find equations for spheres in space? Give examples. 15. How do you find the intersection of two lines in space? A line and a plane? Two planes? Give examples. 16. What is a cylinder? Give examples of equations that define cylinders in Cartesian coordinates. 17. What are quadric surfaces? Give examples of different kinds of ellipsoids, paraboloids, cones, and hyperboloids (equations and sketches).

702


Chapter

Practice Exercises

Vedor Calculations in Two Dimensions In Exercises 1-4, let u = (-3,4) and v = (2, -5). Find (a) the componeot form of the vector aod (b) its magnitude.

In Exercises 25 and 26, fmd (a) the area of the parallelogram determined by vectors u and v and (b) the volume of the parallelepiped determined by the vectors u, v, and w.

1.3u-4v

2.u+v

2S.u=i+j-k,

3. -2u

4. 5v

26.u=i+j,

In Exercises 5-8, fmd the component form of the vector.

5. The veetor obtained by rotating (0, I) througb an angle of21f/3

radians 6. The unit vector that makes an angle of 1f/6 radian with the positivex-axis 7. The vector 2 units loog in the directioo 4i - j

v=2i+j+k,

v=j,

w=i+j+k

Lines, Planes, and Distances 27. Suppose that n is DOnnai to a plane and that v is parallel to the plane. Describe how you would fmd a veetor n that is both perpendicular to v and parallel to the plane. 28. Find a veetor in the plane parallel to the line ax + by = c. In Exercises 29 and 30, fmd the distance from the point to the line.

8. The veetor 5 units loog in the directioo opposite to the directioo of(3/5)i + (4/5)j

29. (2,2,0); x

=

-t, Y

=

t,

Z =

Express the vectors in Exercises 9-12 in terms of their lengtha aod directioos.

12. Velocity vector v = (etcost - etsint)i + (etsint + etcost)j wheol = In 2.

Vedor Calculations in Three Dimensions Express the veetors in Exercises 13 and 14 in terms of their leogtha and directions.

+ 6k

14. i

Z

= I

31. Parametrize the line that passes througb the point (I, 2, 3) parallel to the vector v = - 3i + 7k. 32. Parametrize the line segment joining the points P(I, 2, 0) and

10. -i - j

11. Velocityveetorv = (-2sinl)i + (2cosI)jwheol = 1f/2.

13. 2i - 3j

+t

-1

30. (0,4, I); x = 2 + I, y = 2 + I,

9. v2i + v2j

w=-i-2j+3k

+ 2j - k

15. Find a vector 2 units long in the direction ofv = 4i - j + 4k. 16. Find a vector 5 units loog in the direction opposite to the directioo ofv = (3/5) i + (4/5)k.

Q(I,3, -I). In Esercises 33 and 34, fmd the distance from the point to the plane.

33. (6,0, -6),

x - y = 4

34. (3,0, 10),

2x

+

3y

+Z

= 2

35. Find an equation for the plane that passes through the point (3, -2, 1) normal to the veetor n = 2i + j + k. 36. Find an equatioo for the plane that passes through the point (-1,6,0) perpendicular to the line x = -I + I,y = 6 - 21, Z = 31. In Exercises 37 and 38, fmd an equatioo for the plane througb points

P,Q,andR. In Esercises 17 and 18, fmd lvi, lui, V'u, U'v, v X u, u X v,

37. P(i, -1,2),

Iv X u I, the angle between v and u, the scalar compooent of u in the direction afv, and the vector projection ofo onto v.

38. P(i, 0, 0),

17. v = i + j

18. v = i + j + 2k

u=2i+j-2k

u=-i-k

In Exercises 19 and 20, fmd proj. u.

19. v=2i+j-k u=i+j-5k

Q(2, 1,3),

R( -1,2, - 1)

Q(O, 1,0), R(O, 0, I)

39. Find the points in which the line x = I + 21, y = -I - I, z = 3t meets the three coordinate planes. 40. Find the point in which the line througb the origin perpendicular to the plane 2x - Y - Z = 4 meets the plane 3x - 5y + 2z = 6. 41. Find the acute angle between the planes x = 7 and x + y = -3.

v2z

20. u=i-2j

v=i+j+k

+

42. Find the acute angle betweeo the planes x + y = I and

y+z= 1. In Exercises 21 and 22, draw coordinate axes and then sketch u, v, and

u X v as vectors at the origin. 21. u=i.,

v=i+j

22. u=i-j,

v=i+j

23. !flvl = 2, Iwl = 3,andthe angie between v andw is 1f/3,fmd Iv - 2wl· 24. For what value or values of a will the vectors u = 2i + 4j - 5k and v = -4i - 8j + ak be parallel?

43. Find pararne1ric equations for the line in which the planes x + 2y + Z = I and x - y + 2z = -8 intersect. 44. Show that the line in which the planes

x+2y-2z=5

and

5x-2y-z=0

intersect is parallel to the line

x = -3 + 21, y = 31, z = I + 41.

Chapter 12 45. The planes 3x L

+ 6z = I and 2x + 2y - z = 3 intcncct in a Iinc.

Shaw that the planes are orthogonal.

L (21 - 3J + 3k)' b. x - 3-t,

b. Find equatiOlli for the line of intersection.

046. Find an equation for the plane that pasSCll through the point (1,2, 3)paraIleltou - 2i + 3j + kandv - i - j + 2k. 47. Is v = 2i - 4j + k related in any !pCcial way to the plane 2x + Y = 5? Give reasons for your answer. 48. The equation.' P-;P = 0 represents ~plane through Po normal to D. What set docs the inequality D' PoP> 0 tcpICscn.t?

e. (x

«x + 2)1 + (y -

y - -Ilt,

+ 2) + l1(y

Practice Exercises

703

I)J + zk) - 0

z - 2-3t

- 1) - 3z

d. (2i - 3j + 3k) X «x + 2)i + (y - I)j + zk) = 0 e. (2i - j + 3k) X (-3i + k)·«x + 2)i + (y - I)j + zk) ~O

6l. The puallelogram !hawn here has vertices at A(2, -1,4), B(I, 0, -I), C(I, 2, 3), andD. Find

,

49. Find the distance from the point P(I, 4, 0) to the plane through A(O. o. 0). B(2. o. -1). "'" C(2. -1.0).

D),

50. Find the distance from the point (2, 2, 3) to the plane 2x + 3y + 5z - O. 51. Find a vector parallel to the plane 2x - y - z - 4 and orthogonaltot+J+k. A(2,-l,4)

51. Find a unit vector orthogonal to A in the plane of B and C if A - 2i - j + k,B - i+ 2j + k,andC - i + j - 2k.

C(l.2, 3)

53. Find a vector of magnitude 2 parallel to the line of intcncction of theplane8x + 2y +z - I = and x -y + 2z+ 7 = O.

o

54. Find the point in which the line through the origin perpendicular totheplane2x - y - z = 4mcctsthep1mc3x - 5y +22' = 6.

y B(l, 0. -1)

55. Find the point in which the line through p(3, 2, I) normal ttl the plane 2x - y + 2z = -2 meets the plane. 56. What angle doc! the line of intcncction of the planca 2x + Y - z = 0 and x + Y + 2z = 0 make with the positive

x-axis?

L the coordinates of D, b. the cosine of the interior angle atB,

e. the vector projection of BA. onto iiC,

57. The line

d. the area of the parallelogram, L:

x=3+2t,

y=2t,

z=t

in1eDcctll the plane x + 3y - z = -4 in a point P. Find the coonIin.atcs of P and find equations for the line in the plane through P perpendicular ttl L.

58. Shaw that for every real number k the plane

x - 2y+ z + 3 + k(2x - y- z + I) = 0

contains the line of intmscction of the planes

c. an equatioo for the plane of the parallelogram. f. the areas of the orthogonal projections of the panillelogram

on the three coord.i.na.tJ; planes. 63. Distmte betweea lineI Find the distance bctwccn the line L, through the points .4(1,0,-1) and B(-I,I,O) and the line L2 through the points C(3, I, -I) andD(4, 5, -2). The distance is ttl be meuured along the line perpendicular In the two lines. FltBt fmd • vector D perpendicular to both lines. Thm project AC onto D. 64. (Continuatio1l o/Exerci.!e 63.) Find the distance between the line through A(4, 0, 2) and B(2, 4, 1) and the line through C(t, 3, 2)

""'D(2. 2. 4). x - 2y + z + 3 = 0 and 2x - y - z + I = O. Quadric SumCt!!s 59. Find an equation for the plane through A(-2,O, -3) and B(I, -2, I) that lies parallel to the line through

Identify and sketch the surfaces in Exercises 65-76.

C( -2. -13/'. 26/') ""'D(16/'. -13/'. 0). 60. Is the line x = 1 + 2t, y = -2 + 31, z = -5t related in any way to the plane -4x - 6y + 10z - 91 Give reasoos for your

67.4x 2 +4y 2+ Z l=4 69. z = _(x 2 + y2)

""""'.

61. Which of the following are equations for the plane through the points P(I, I, -I), Q(3, 0, 2), andR(-2, I, O)?

65. x 2 + y2 + z2 = 4

71. X2 +y2=z2 73. X2 +y2_z2=4 75. yl_x2_zl= 1

66. x 2 + (y - 1)2 + z2 = I 68. 3fu:l + 91 2 + 4z2 = 36

70. Y = _(Xl

+ zZ)

71. Xl+zZ=yl

74. 4yl+zZ-ob:l =4 76. z2 _x 2 _ y2 = I

704


Chapter


1. Submarine hunting Two surface ships on maneuvers are trying to determine a submarine's course and speed to prepare for an aircraft intercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is located at (0, 5, 0). All coordinates are given in thousands of feet. Ship A locates the submarine in the direction of the vector 2i + 3j - (I/3)k, and ship B locates it in the direction of the vector 18i - 6j - k. Four minutes ago, the submarine was located at (2, -1, -1/3). The aircraft is due in 20 min. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?

clockwise when we look toward the origin fromA. Find the velocity v of the point of the body that is at the position B(l, 3, 2).

z

IA~(:: l,~l,~l!-)

_---I.

B

I

:~: : ----.J!___ v: / I

z

---

x

(l, 3, 2)

I

/

I

1/

I~

I /

-----_

I

-------1//

~y

//

ShipB

Ship A

(4,0,0)

(0,5,0)

y

x

5. Consider the weight suspended by two wires in each diagram. Find the magnitudes and components of vectors F I and F 2, and angles a and {3.

a.

-\ Submarine

Nor TO SCALE

2. A helicopter rescue Two helicopters, HI and H 2, are traveling together. At time t = 0, they separate and follow different straight-line paths given by

HI:

x = 6 + 40t,

Y = -3 + lOt,

z = -3 + 2t

H2:

x = 6 + 11 Ot,

Y = -3 + 4t,

z = -3 + t.

Time t is measured in hours and all coordinates are measured in miles. Due to system malfunctions, H2 stops its flight at (446, 13, 1) and, in a negligible amount of time, lands at (446, 13,0). Two hours later, HI is advised of this fact and heads toward H2 at 150 mph. How long will it take HI to reach H2? 3. Torque The operator's manual for the Toro® 21 in. lawnmower says "tighten the spark plug to 15 ft-lb (20.4 N· m)." If you are installing the plug with a lO.5-in. socket wrench that places the center of your hand 9 in. from the axis of the spark plug, about how hard should you pull? Answer in pounds.

b.

(Hint: This triangle is a right triangle.)

6. Consider a weight of w N suspended by two wires in the diagram, where T I and T 2 are force vectors directed along the wires.

a. Find the vectors T I and T 2 and show that their magnitudes are w cos (3 ITII = sin (a + (3) 4. Rotating body The line through the ongm and the point A( 1, 1, 1) is the axis of rotation of a right body rotating with a constant angular speed of3/2 rad/sec. The rotation appears to be

and T

-

121-

wcosa sin (a+{3)

705

Chapter 12 Additional and Advanced Exercises b. For a fIxed {j determine the value of a which minimizes the

magnitude IT ,I. c. For a fIxed a determine the value of (j which minimizes the magnitude IT21· 7. Determinants and planes

X

X2 -

X

X3 -

x

Y' - Y Y2 - Y y, - Y

Zt -

z

Z2 -

Z

Z3 -

z

d

b. What set of points in space is described by the equation

Y

Z

Xl

Yl

Zl

X2

Y2

Z2

X3

Y3

Z3

8. Determinant. and line.

lax, + by,

-

cl

~ ~'--;=.c===.

0, there exists a corresponding number II

LI
'1T/4

t---1>'1T/4

+

+

tk, then

(lim sin t)j t-w/4

+

lim t)k ( l-w/4

=V2·+V2·+1T k J 2'

2

•

4'

We derme continuity for vector functions the same functions.

WH:j

we derme continuity for scalar

DEFINmON A vector function r{t) is continuous at a point t = to in its domain if lim,....~ r(t) = r(to). The function is continuous if it is continuous at every point in its domain. From Equation (3), we see that r(t) is continuous at t = to if and only if each component function is continuous there (Exercise 31).

EXAMPLE 3 (a) All the space curves shown in Figures 13.2 and 13.4 are continuous because their component functions are continuous at every value of t in ( - 00, 00).

(b) The function g(t) = (cost)i

+

(sint)j

+

ltJk

is discontinuous at every integer, where the greatest integer function discontinuous.

l t J is •

Derivatives and Motion Suppose that r(t) = I(t)i + g(t)j + h(t)k is the position vector of a particle moving along a curve in space and that I, g, and h are differentiable functions of t. Then the difference between the particle's positions at time t and time t + !J.t is ~r =

r(t

+ !J.t) - r(t)

710

Chapter 13: Vector-Valued Functions and Motion in Space

(Figure 13.5a). In terms of components, ar = r(t =

= ~----------------~y

(a)

+

M) - r(t)

+ at)i + g(t + M)j + h(t + M)k] - [/(t)i + g{t)j + h(t)k] [j(t + at) - I(t)]i + [g(t + M) - g{t)]j + [j(t

[h(t

+

M) - h(t)]k.

As at approaches zero. three things seem to happen simultaneously. First. Q approaches P along the curve. Second, the secant line PQ seems to approach a limiting position tangent to the curve atP. Third, the quotient arl M (Figure 13.5b) approaches the limit

lim arat [lim I(t + at)at -

I(t)]i

=

.1.,-0

[lim g{t + at)at - g(t)]J.

+

.l.t-O

+

[lim h(t

+ at) - h(t)]k

.1.,-0 =

.l.t-O

M

g [d/]i + [d ].J + [dh]k dt' dt dt

We are therefore led to the following definition. ~----------------~y

(b) x

FIGURE 13.5 As M .... O. thepointQ approaches the point P along the curve C. In the limit, the vector PQ/ at becomes the tangent vector r'(t).

DEFINmON The vector function r(t) = I(t)i + g{t)j + h(t)k has a derivative (is differentiable) at t if I, g, and h have derivatives at t. The derivative is the vector function '(t) r

= dr =

l' r(t dt,J!!!o

+ at)

- r(t)

at

dl. dt I

+ dg. + dt J

dh k dt'

A vector function r is differentiable if it is differentiable at every point of its domain. The curve traced by r is smootb if dr/dt is continuous and never 0, that is, if I, g, and h have continuous first derivatives that are not simultaneously O. The geometric significance of the definition of derivative is shown in ~ure 13.5. The points P and Q have position vectors r(t) and r(t + at), and the vector PQ is represented by r(t + at) - r(t). For at > 0, the scalar multiple (1/at)(r(t + at) - r(t» points in the same direction as the vector As at ..... 0, this vector approaches a vector that is tangent to the curve atP (Figure l3.5b). The vector r'(t), when different from 0, is defined to be the vector tangent to the curve at P. The tangent line to the curve at a point U(to), g{to), h(to» is defined to be the line through the point parallel to r'(to). We require dr/dt "" 0 for a smooth curve to make sure the curve has a continuously turoing tangent at each point. On a smooth curve, there are no sharp corners or cusps. A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth (Figure 13.6). Look once again at Figure 13.5. We drew the figure for at positive, so ar points forward, in the direction of the motion. The vector arl at, having the same direction as ar, points forward too. Had M been negative, ar would have pointed backward, against the direction of motion. The quotient arI at, however, being a negative scalar multiple of ar, would once again have pointed forward. No matter how ar points, arl at points forward and we expect the vector dr/dt = lim&--+o arl M, when different from O. to do the same. This means that the derivative drI dt, which is the rate of change of position with respect to time, always points in the direction of motion. For a smooth curve, drI dt is never zero; the particle does not stop or reverse direction.

PQ.

FIGURE 13.6 A piecewise smooth curve made up offive smooth curves connected end to end in • continuous fashion. The curve here is not smooth at the points joining the five smooth curves.

=

13.1 Curves in Space and Their Tangents

711

DEFINmONS If r is the position vector of a particle moving along a smooth curve in space, then dr V(I) = dl is the particle's velocity vector, tangent to the curve. At any time I, the direction of v is the direction of motion, the magnitude of v is the particle's speed, and the derivative 8 = dv/dl, when it exists, is the particle's acceleration vector. In

summary, dr v = dl'

1. Velocity is the derivative of position:

2. Speed is the magnitude of velocity:

Speed = Ivl.

3. Acceleration is the derivative of velocity: 4. The unit vectorv/lvl is the direction of motion at time I.

EXAMPLE 4 Find the velocity, speed, and acceleration of a particle whose motion in space is given by the position vector r( I) = 2 cos I i + 2 sin I j + 5 cos2 I k. Sketch the velocity vector V(77T/4). Solution

The velocity and acceleration vectors at time I are V(I) = r'(I) = -2sinli + 2coslj - 10 cos I sinlk

z

= -2 sin I i + 2 cos I j - 5 sin 21 k, 8(1) = r"(I) = -2cosli - 2sinlj - 10cos21k, and the speed is

y t

=

7," 4

IV(I) I = Y(-2sinl)2 + (2 cos 1)2 + (-5 sin 21)2 = Y4 + 25si02 21. When 1= 77T/4, we have

x

FIGURE 13.7 The curve and ilie velocity vector when t = 7'IT/4foriliemotion given in Example 4.

V(7:)=Vzi+ Vz i+ 5 k, A sketch of the curve of motion, and the velocity vector when 1= 77T/4, can be seen in Figure 13.7. • We can express the velocity of a moving particle as the product of its speed and direction: Velocity =

IVIC~I) =

(speed)(direction).

Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.

712


Differentiation Rules for Vector Functions Let u and v be clifferentiable vector functions of t, C a constant vector, c any scalar, and / any differentiable scalar function. d

1. Constant Function Rule:

dtC

2. Scalar Multiple Rules:

:t [cu(t)]

=

0

cu'(t)

=

:t [f(t)u(t)]

I When you use the Cross Product Rule,

remember to preserve the order of the factors. Ifu comes fIrst on the left side of the equation, it must also come first on the right or the signs will be wrong.

!,(t)u(t) + /(t)u'(t)

=

3. Sum Rule:

:t [u(t) + v(t)]

4. Difference Rule:

dt [u(t) - v(t)]

S. Dot Product Rule:

:t [u(t)· v(t)]

6. Cross Product Rule:

dt [u(t) X v(t)]

7. Chain Rule:

:t [u(f(t))]

d

=

d

=

=

u'(t) + v'(t)

=

u'(t) - v'(t)

u'(t)· v(t) + u(t)· v'(t) =

u'(t) X v(t) + u(t) X v'(t)

!'(t)u'(f(t))

We will prove the product rules and Chain Rule but leave the rules for constants, scalar multiples, sums, and differences as exercises. Proof of the Dot Product Rule

Suppose that

u = ul(t)i

+ U2(t)j + u3(t)k

and Then d

dt(u'v)

=

d

dt(U1Vl + U2 V2 + U3 V3)

= U1VI

+ ulv2 +

U3V3

+ UtVl + U2V2 + U3V3.

u' ·v

u·v'

•

Proof of the Cross Product Rule We model the proof after the proof of the Product Rule for scalar functions. According to the detmition of derivative,

E. (

)_ .

dtUXV-~

u(t + h) X v(t + h) - u(t) X v(t) h

To cbange this fraction into an equivalent one that contains the difference quotients for the derivatives ofu and v, we subtract and add u(t) X v(t + h) in the numerator. Then

:t

(u X v) = 1~~u(~t_+~h~)_X~v(~t_+~h~)_-~u(~~~X~v~(t_+~h)~+~u~(t~)_x_v~(t~+~h~)_-~u~(t~)_x_v~(~t)

~o

.

= lun

h~O

=

h [U(t

+ h)h - u(t) Xvt+h ( ) () +ut

X

v(t

+ h)h - V(t)]

I~ u(t + h) - u(t) X I~ v(t + h) + I~ u(t) X I~ v(t + h) - v(t) h-O

h

h-O

h-O

h-O

h

13.1

713

Curves in Space and Their Tangents

The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 32). As h approaches zero, vet + h) approaches v(t) because v, being differentiable at t, is continuous at t (Exercise 33). The two fractions approach the values of du/dt and dv/dt at t. In short,

d (

tit

IAs an a1~braic convenience, we

sometimes write the product of a scalar c and a vector v as vc instead. of cv. Thill permi11l us, for instance, to write the Chain Rule in a familiar form.:

~

)

du Xv oX dv =tit + dt·

•

Proof of the Chain Rule Suppose that o(s) = a(s)i + b(s)j + c(s)k is a differentiable vector function ofs and thats = J(t) is a differentiable scalar function oft. Then a, b, and c are differentiable functions of t, and the Chain Rule for differentiable real-valued functions gives

~[u(')l ~

du da ds (ji - didt'

whore,

oXv

:1 + 0 = (cosht)i - (sinht)j + tk

11. r(t) = (e'cost)i

12. r(t) 13. r(t) 14. r(t) 15. r(t)

16. r(t)

More on Curvature 17. Show that the parabola y = ax', a #' 0, has its largest curvature at its vertex and has no minimmn curvature. (Note: Since the curvature ofa curve remains the same if the curve is translated orrotated, this result is true for any parabola.)

734


18. Show that the ellipse x = a cos I, y = b sin I, a > b > 0, has its largest cmvature on its major axis and its smallest curvature on its minor axis. (As in Exercise 17, the same is true for any ellipse.) 19. Muimizing the curvature of. helli In Example 5, we found the curvature of the helix r(l) = (a cos I)i + (a sin I)j + blk (a, b '" 0) to be " = a/(a' + b'). What is the largest value" can have for a given value of b? Give reasons for your answer. 20. Total curvature We find the tutal curvature of the portion of a smooth curve that runs from s = So to s = 91 > So by integrating K from So to Sl • If the curve has some other parameter, say t, then the total curvature is

where to and t, correspond to So and L

s, .

Find the tota! curvatures of

The portion of the helix r(l) = (3 cos I)i 0:5 t:5 41T.

h. The parabola y = x 2 ,

-00

<X
0

tcost)i,

in the form a = aTT + lINN. (The path of the motion is the involute of the circle in Figure 13.27. See also Section 13.3, Exercise 19.)

Solution

We use the rlISt of Equations (2) to rmd aT: v =

~;

=

(-sint + sint + tcost)i + (cost - cost + tsint)i

= (tcost)i

+ (tsint)j

Ivl =

Vt 2 cos2 t + t 2 sin2 t

aT =

dt

d

d Ivl = dt(t)

=

=

W

It I = t

=

t> 0 Eq. (2)

1.

Knowing aT, we use Equation (3) to find aN:

FIGURE 13.27 The tangential and normal components of the acceleration of

a = (cost - tsint)i

laI 2 =t2 +1 aN = Vlal 2

themotionr(l) = (cos I + I sin I)i + (sin I - I cos I)j, for I > O. If a string

After some algebra

~~~

wound around a ftxed circle is unwound while held taut in the plaoe of the circle, its endP traces an involute of the circle (Example I).

+ (sint + tcost)i

=

V(t2

al

-

+ 1) - (I)

=

W

t.

=

We then use Equation (I) to rmd a: a = aTT

+ lINN

=

(I)T + (t)N = T + tN.

•

Torsion How does dB/lis behave in relation to T, N, and B? From the rule for differentiating a cross product, we have

dB = d(T X N) = dT X N

lis

lis

+ T X dN

lis

ds'

Since N is the direction of dT/Iis, (dT/ds) X N = 0 and

dB = 0

lis

+T

X dN = T X dN.

ds

ds

From this we see that dB/lis is orthogonal to T since a cross product is orthogonal to its factors. Since dB/lis is also orthogonal to B (the latter bas constant length), it follows that dB/lis is orthogonal to the plane of B and T. In other words, dB/ds is parallel to N, so dB/lis is a scalar multiple ofN.1n symbols,

dB

ds

= -TN.

The negative sign in this equation is traditional. The scalar 'I' is called the torsion along the curve. Notice that

dB

ds • N

= -TN' N = -'1'(1) =

We use this equation for our next definition.

-'I'.

13.5

I

Rectifying

plane

Binormal

DEFINmON

Let B

=

Tangential and Normal Components of Acceleration

737

T X N. The torsion function of a smooth curve is

Normal plane

T=

Principal normal

< A

Unit tangent

FIGURE 13.28 The names of the three planes determined by T. N, and B.

dB -dS·N.

(4)

Unlike the curvature K, which is never negative, the torsion T may be positive, negative, or zero. The three planes determined by T, N, and B are named and shown in Figure 13.28. The curvature K = IdT / ds Ican be thought of as the rete at which the normal plane turns as the point P moves along its path. Similarly, the torsion T = -(dB/ds)' N is the rete at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists. Look at Figure 13.29. If P is a !min climbing up a curved track, the rete at which the headlight turns from side to side per unit distance is the curvature of the tmck. The rete at which the engine tends to twist out of the plane formed by T and N is the torsion. In a more advanced course it can be shown that a space curve is a helix if and only if it has constant nonzero curvature and constant nonzero torsion.

The tmsion

I ~ ~ l(dT/d,)I· "

/

The curvature at P

s increases

~-

p

I

8=0

FIGURE 13.29 Every moving body travels with a TNB frame that characterizes the geometty of its path of motion.

Computational Formulas The most widely used formula for torsion, derived in more advanced texts, is

x x T=

y )i

x' y

Iv X al

z z 'z' 2

(ifv X a .. 0).

(5)

The dots in Equation (5) denote differentiation with respect to t, one derivative for each dot. Thus, x ("x dot") means dx/dt, x ("x double dof') means d 2x/dt 2 , and x ("x triple dot'') means d 3x/dt 3 • Similarly, y = dy/dt, and so on. There is also an easy-to-use formula for curvature, as given in the following summary table (see Exercise 21).

738


Computation Formulas for Curves in Space v

[Vj

Unit tangeot vector:

T

Principal unit normal vector:

dT/dt N = IdT/dtl

Bioormal vector:

B=TXN

=

Curvatore:

Torsion:

T =

Tangeotial and normal scalar componeots of acceleration:

a

dB

tis

d aT=dt

aN =

Y

z

x

ji "ji

z

x ·Z· 2 Iv x al

--'N =

= aTT

i:

+ aNN

lvl

Klvl2 = Vlal 2-

al

Exercises 13.5 Finding Tangential and Normal Components In Exercises 1 and 2, write a in the form • finding T and N. 2. r(l) = (I

+ 31)i + (I -

4. r(l) = S. r(l) = 6. r(l) =

+

14. r(l) =

(sin I)j

+

II 0

Physical Applications

+ 21j + 12k, I = I (I cos I)i + (I sin I)j + 12k, I = 0 12; + (I + (I/3)1')j + (I - (I/3)1')k, (e' cos I)i + (e' sin I)j + Y2e'l in terms of r and riJ by evaluating the dot produets v· i and v· j.

r

x

b. Express and r iI in terms of and j> by evaluating the dot products v • Ur and v· lie.

6. Express the curvatore of a twice-differentiable curve r = 1(0) in the polar coordinate plane in terms of 1 and its derivatives. 7. A slender rod through the origin of the polar coordinate plane r0tates (in the plane) about the origin at the rate of 3 rad/min. A beetle starting from the point (2, 0) crawls along the rod toward the origin at the rate of I in/min.

Positive z-axis .-------- points down.

1 z

2. Suppose the curve in Exercise I is replaced by the conical helix r = aO, z = bO shown in the accompanying figure.

a. Express the angular velocity dOjdt as a function ofO. h. Express the distance the particle travels along the helix as a function of o.

•• Find the beetle's acceleration and velocity in polar form when it is halfway to (I in. froru) the origin.

D b.

To the nearest tenth of an inch, what will be the length of the path the beetle has traveled by the time it reaches the origin? 8. Arc length in cylindrical coordinate. a. Show that when you express ds' = dx' + ely' + dz'in terms of cylindrical coordinates, you get ds' = dr' + r'do' + dz'. b. Interpret this result geometrically in terms of the edges and a

diagonal of a box. Sketch the box. c. Use the result in part (a) to fmd the length of the curve r = e 6,z = e 6,O::S O::s 8InS. Conical helix T

1

= a8,z =

be

Conez= ~r

Positive z-axis points down.

z

9. Unit vecton for position and motion in cylindrical coordinates When the position of a particle moving in space is given in cylindrical coordinates, the unit vectors we use to describe its position and motion are

u, = (cosO)i + (sinO)j,

... = -(sinOli +

(cosO)j,

and k (see accompanying figure). The particle's position vector is then r = rUr + zk, where r is the positive polar distance coordinate of the particle's position.

14 PARTIAL DERIVATIVES OVERVIEW Many functions depend on more than one independent variable. For instance, the volume of a right circular cylinder is a function V = 7Tr 2h of its radius and its height, so it is a function V(r, h) of two variables rand h. In this chapter we extend the basic ideas of single variable calculus to functions of several variables. Their derivatives are more varied and interesting because of the different ways the variables can interact. The applications of these derivatives are also more varied than for single-variable calculus, and in the next chapter we will see that the same is true for integrals involving several variables.

14.1

Functions of SeveraL VariabLes In this section we define functions of more than one independent variable and discuss

ways to graph them. Real-valued functions of several independent real variables are defined similarly to functions in the single-variable case. Points in the domain are ordered pairs (triples, quadruples, n-tuples) of real numbers, and values in the range are real numbers as we have worked with all along.

DEFINITIONS Suppose D is a set ofn-tuples of real numbers (x], X2, " " xn). A real-valued function f on D is a rule that assigns a unique (single) real number w =

f(Xl , X2, " . ,

xn)

to each element in D. The set D is the function's domain. The set of w-values taken on by f is the function's range. The symbol w is the dependent variable of f, and f is said to be a function of the n independent variables Xl to X n . We also call the x/ s the function's input variables and call w the function's output variable.

If f is a function of two independent variables, we usually call the independent variables X and y and the dependent variable z, and we picture the domain of f as a region in the xy-plane (Figure 14.1). Iff is a function of three independent variables, we call the independent variables x, y, and z and the dependent variable w, and we picture the domain as a region in space. In applications, we tend to use letters that remind us of what the variables stand for. To say that the volume of a right circular cylinder is a function of its radius and height, we might write V = f(r , h). To be more specific, we might replace the notation f(r, h) by the formula that calculates the value of V from the values of rand h, and write V = 7Tr 2h. In either case, rand h would be the independent variables and V the dependent variable of the function.

747

748

Chapter 14: Partial Derivatives

f f(a, b) --~~~~--~~~Z

o

FIGURE 14.1

f(x,y)

An arrow diagram for the functionz = j(x, y).

As usual, we evaluate functions defined by fannulas by substituting the values of the independent variables in the fannula and calculating the corresponding value of the 2 + y2 + z2 at the point dependent variable. For example, the value of f(x,y,z) = (3,0,4) is

Yx

f(3,0,4) =

Y(W + (0)2 + (4)2 = V25 =

5.

Domains and Ranges In defining a function of more than one variable, we follow the usual practice of excluding x 2,y cannot inputs that lead to complex numbers or division by zero. If f(x,y) = 2 be less than x . If f(x,y) = I/(xy),xy cannot be zero. The domain of a function is assumed to be the largest set for which the defining rule generates real numbers, unless the domain is otherwise specified explicitly. The range consists of the set of output values for the dependent variable.

Yy -

EXAMPLE 1 (a) These are functions of two variables. Note the restrictions that may apply to their domains in order to obtain a real value for the dependent variable z. Function z =

Yy -

x2

I z=xy z

=

sinxy

Domain

Range

y ~ x2

[0, 00)

xy '" 0

(-00,0) U (0, 00)

Entire plane

[-1,1]

(b) These are functions of three variables with restrictions on some of their domains.

Function w=

Yx 2 + y2 + z2

w=

~---'o---~

W

I

x2

+ y2 + z2

= xyInz

Domain

Range

Entire space

[0, 00)

(x,y,z) '" (0,0,0)

(0, 00)

Half-space z

>

0

(-00,00)

•

Functions of Two Variables Regions in the plane can have interior points and boundary points just like intervals on the real line. Closed intervals [a, b] include their boundary points, open intervals (a, b) don't include their boundary points, and intervals such as [a, b) are neither open nor closed.

14,1 Functions of Several Variables

749

DEFINmONS A point (xo, YO) in a region (set) R in Ibe .!J'-plane is an interior point of R if it is Ibe center of a disk of positive radius !bat lies entirely in R (Figure 14.2). A point (xo, YO) is a boundary point of R if every disk centered at (xo, YO) contains points !bat lie outside of R as well as points !bat lie in R. (The boundary point itself need not belong to R.) The interior points of a region, as a set, make up Ibe interior of Ibe region. The region's boundary points make up its boundary. A region is open ifit consists entirely of interior points. A region is closed if it contains all its boundary points (Figure 14.3).

(a) Interior point

y

y

y

, 1

R

«.' yol

o

/

o

(b) Boundary point

FIGURE 14,2 Interior points and boundary points of a plane regioo R, An interior point is necessarily a point of R. A

Ix

((x,y) 2 + y2< 1) Open unit disk. Every point an interior point

{(x,Y)lx2+y2~ 1)

Boundary of unit disk. (The unit circle.)

Ix

((x,y) 2 + y2,,;; 1) Closed unit disk. Contains all boundary points.

boundary point of R need not belong to R. FIGURE 14.3

Interior points and boundary points of 1he unit disk in 1he plane.

As wilb a half-open interval of real numbers [a, b), some regions in Ibe plane are neiIber open nor closed. If you start wilb Ibe open disk in Figure 14.3 and add to it some of but not all its boundary points, Ibe resulting set is neilber open nor closed. The boundary points !bat are Ibere keep Ibe set from being open. The absence of Ibe remaining boundary points keeps Ibe set from being closed.

DEFINmONS A region in Ibe plane is bounded if it lies inside a disk offIxed radius. A region is unbounded if it is not bounded.

y Interior points, wherey - x 2 > 0

/

Examples of bounded sets in Ibe plane include line segments, triangles, interiors of triangles, rectangles, circles, and disks. Examples of unbounded sets in Ibe plane include lines, coordinate axes, Ibe graphs of functions defined on infinite intervals, quadrants, half-planes, and Ibe plane itself.

EXAMPLE 2

Describe Ibe domain oflbe function f(x,y) =

Yy -

x 2•

Since f is defmed only where y - x 2 ;" 0, Ibe domain is Ibe closed, unbounded region shown in Figure 14.4. The parabola y = x 2 is Ibe boundary oflbe domain. The points above Ibe parabola make up Ibe domain's interior. •

Solutton

------~~~~--~----~x

-1

0

FIGURE 14,4 The domain of f(x, y) in Example 2 consists of1he shaded regioo and its bounding parabola.

Graphs, Level Curves, and Contours of Functions of Two Variables There are two standard ways to picture Ibe values of a function fix, y). One is to draw and label curves in Ibe domain on which f has a constant value. The olber is to sketch Ibe surface z = fix, y) in space.

750


DEFINITIONS The set of points in the plane where a function f(x, y) has a constant value f(x, y) = c is called a level curve of f. The set of all points (X, y, f(x, y) in space, for (X, y) in the domain of f, is called the graph of f. The graph of f is also called the surface z f(x,y) .

=

The surface z =f(x,y) = 100 - x 2 _ y2 is the graph off.

f(x, y) = 75

X'

X

f(x,y) = 51 (a typical level curve in the function's domain)

y

EXAMPLE 3 Graph f(x , y) = 100 - x 2 - y2 and plot the level curves f(x,y) f(x, y) = 51, and f(x, y) = 75 in the domain of f in the plane.

SoLution The domain of f is the entire xy-plane, and the range of f is the set of real numbers less than or equal to 100. The graph is the paraboloid z = 100 - x 2 - y2, the positive portion of which is shown in Figure 14.5. The level curve f(x, y) = 0 is the set of points in the xy-plane at which f(x,y)

The contour curve f(x , y) = 100 - x 2 - y2 = 75 is the circle x 2 + y2 = 25 in the plane z = 75.

f1~"

\

= 75

~

-P~

z = 100 - x 2 _ y2

> 100, then the values of f(x, y) are negative. For example, the circle x 2 + y2 = 144, which is the circle centered at the origin with radius 12, gives the constant value f(x, y) = -44 and is a level curve of f. • The curve in space in which the plane z = c cuts a surface z = f(x, y) is made up of the points that represent the function value f(x, y) = c. It is called the contour curve f(x, y) = c to distinguish it from the level curve f(x, y) = c in the domain of f. Figure 14.6 shows the contour curve f(x,y) = 75 on the surface z = 100 - x 2 - y2 defined by the function f(x, y) = 100 - X2 - Y 2. The contour curve lies directly above the circle x 2 + y2 = 25, which is the level curve f(x,y) = 75 in the function's domain. Not everyone makes this distinction, however, and you may wish to call both kinds of curves by a single name and rely on context to convey which one you have in mind. On most maps, for example, the curves that represent constant elevation (height above sea level) are called contours, not level curves (Figure 14.7).

Functions of Three Variables

/

x The level curvef(x, y) = 100 - x 2 - y2 = 75 is the circle x 2 + y2 = 25 in the xy-plane.

FIGURE 14.6 A plane z = c parallel to the xy-plane intersecting a surface z = f(x, y) produces a contour curve.

In the plane, the points where a function of two independent variables has a constant value f(x,y) = c make a curve in the function's domain. In space, the points where a function of three independent variables has a constant value f(x, y, z) = c make a surface in the function's domain. DEFINITION The set of points (x, y, z) in space where a function of three independent variables has a constant value f(x, y, z) = c is called a level surface of f.

Since the graphs of functions of three variables consist of points (x,y, z, f(x,y, z)) lying in a four-dimensional space, we cannot sketch them effectively in our three-dimensional frame of reference. We can see how the function behaves, however, by looking at its threedimensional level surfaces. EXAMPLE 4

Describe the level surfaces of the function

f(x,y, z) = Yx 2

+ y2 + z2 .

14.1

Functions of Several Variables

751

FIGURE 14.7 Contours on Mt. Washington in New Hampshire. (Reproduced by permission from the Appalachian Mountain Club.) FIGURE 14.8 The level surfaces of f(x,y,z) = Yx 2 + y2 + z2 are concentric spheres (Example 4).

(a) Interior point

(b) Boundary point

FIGURE 14.9 Interior points and boundary points of a region in space. As with regions in the plane, a boundary point need not belong to the space region R.

Solution The value of f is the distance from the origin to the point (x, y, z). Each level surface x 2 + Y 2 + Z 2 = c, c > 0, is a sphere of radius c centered at the origin. Figure 14.8 2 + y2 + z2 = 0 shows a cutaway view of three of these spheres. The level surface consists of the origin alone. We are not graphing the function here; we are looking at level surfaces in the function's domain. The level surfaces show how the function's values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from a point on one sphere to a point on another, the function's value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. The dependence of change on direction is important. We return to it in Section 14.5. •

Y

Yx

The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra dimension, we use solid balls of positive radius instead of disks.

DEFINITIONS A point (xo, Yo, zo) in a region R in space is an interior point of R if it is the center of a solid ball that lies entirely in R (Figure 14.9a). A point (xo, Yo, zo) is a boundary point of R if every solid ball centered at (xo, Yo , zo) contains points that lie outside of R as well as points that lie inside R (Figure 14.9b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary.

Examples of open sets in space include the interior of a sphere, the open half-space z > 0, the first octant (where x, y, and z are all positive), and space itself. Examples of closed sets in space include lines, planes, and the closed half-space z 2: O. A solid sphere

752

Chapter 14: PartiaL Derivatives with part of its boundary removed or a solid cube with a missing face, edge, or comer point is neither open nor closed. Functions of more than three independent variables are also important. For example, the temperature on a surface in space may depend not only on the location of the point p(x, y, z) on the swface but also on the time 1 when it is visited, 80 we would write T = /(x, y, z, t).

computer Graphing Three-dimensional graphing programs for computers and calculators make it possible to graph functions of two variables with only a few keystrokes. We can often get information more quickly from a graph than from a formula. w

EXAMPLE 5 The temperature w beneath the Earth's surface is a function of the depth x beneath the swface and the time t of the year. If we measure x in feet and t as the number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function w = 008(1.7 X 10-21 - O.2x)e-o·~.

FIGURE 14.10 Thisgrapbshowsthe seasonal variation of the temperatuIe below ground as a fraction of surface

tmnpenrtmc (Exampl' 5).

(The temperature at 0 ft is scaled to vary from +1 to -1, so that the variation atx feet can be interpreted as a fraction of the variation at the surface.) Figure 14.10 shows a graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 25 ft, there is almost no variation during the year. The graph also shows that the temperature 15 ft below the surface is about half a year out of phase with the surface temperature. When the temperature is lowt:st on the surface (late January, say), it is at its highest 15 ft below. Fifteen feet below the ground, the seasons are reversed. _ Figure 14.11 shows computer-generated graphs of a number of functions of two variables together with their level curves.

y

x

(a) z - sin x

FIGURE 14.11

+ 2 sin]

(b) l - (4x2

+ ]2.}o!-r-r

Computer-generated graphs and level curves of typical functions of two variables.

14.1 Functions of Several Variables

753

Exercises 14.1 Domain, Range, and Level Curves

21. f(x,y)

~

23. f(x,y)

~ \I'

In Exercises 1-4, find the specific function values.

+ xy'

1. f(x,y) ~ x'

b. f(-I, I)

a. f(O,O) c. f(2,3) 2. f(x,y) ~ sin (xy)

d. f(-3,-2)

16 -

25. f(x,y) ~ In (x' 27. f(x,y)

f( -3, ;;) d. f( -f, -7)

22. f(x,y) ~ y/x'

xy I

x' _ y'

+ y')

~ sin-1 (y -

x)

+ y' -

24. f(x,y)

~ \1'9 - x' - y'

26. f(x,y) ~ .""",'+Y') 28. f(x,y)

~ tan-I (~)

b.

29. f(x,y) ~ In (x'

a. f(3, -1,2)

b.

Exercises 31-36 show level curves for the functions graphed in (aHf) on the following page. Match each set of curves with the appropriate function.

c. f(O,-t,O)

d. f(2,2, 100)

a. f(2,');) c.

f(,."t)

3. f(x,y,z)

~

1) 30. f(x,y) ~ In (9 - x' - y')

Matching Surfaces with Level Curves

x-y -,--, y +z

f(l,t, -t)

31.

32. y

4. f(x,y,z) ~ \1'49 - x' - y' -

a. f(O,O,O)

z'

I(~t

b. f(2,-3,6)

c. f(-1,2,3)

d.

( 4 5 6)

f \1'2' \1'2' \1'2

~¥~'

In Exercises 5-12, fmd and sketch the domain for each function.

5. f(x,y) ~ \l'y - x - 2 6. f(x,y) ~ In (x'

+ y'

(x - I)(y

- 4)

+ 2)

7. f(x,y) ~ (y _ x)(y _ x') sin (xy)

8. f(x,y)

~,

9. f(x,y)

~ cos- 1 (y

x

+y

10. f(x,y) ~ In(xy

,

33.

34. y

- 25

- x')

+x - y -

1)

11. f(x,y) ~ \I'(x' - 4)(y' - 9)

I

12. f(x,y) ~ In (4 _ x' _ y') In Exercises 13-16, fmd and sketch the level curves f(x, y) ~ c on the same set of coordinate axes for the given values of c. We refer to these level curves as a contour map.

13. f(x,y)

~

x

+y

- I, c

~

-3, -2, -I, 0, 1,2,3

14. f(x,y) ~ x' + y', c ~ 0, 1,4,9, 16,25 15. f(x,y)

~

xy,

c

~

-9, -4, -I, 0, 1,4,9

16. f(x,y) ~ \1'25 - x' - y',

c ~ 0, 1,2,3,4

In Exercises 17-30, (a) fmd the function's domain, (b) fmd the function's range, (oj describe the function's level curves, (dJ fmd the boondary of the function's don3ain, (eJ determine if the don3ain is an open reo gion, a closed region, or neither, and (f) decide if the don3ain is boonded or unbounded

17. f(x,y)

~

y - x

19. f(x,y) ~ 4x'

+ 9y'

18. f(x,y) ~ ~ 20. f(x,y) ~ x' - y'

35.

36. y

1111

IIlIJi'

754

Chapter 14: PartiaL Derivatives

L

L

b.

,.

Functions of Two variables Display the values of the :functions in Exercises 37--48 in two ways: (a) by sketching the surfacez = f(",y) and (b) by drawing an assortment oflevel curves in the function's dmnain. Label each level curve with its function value.

Vx

37. f(x,y) = yl

38. f(x,y) =

39. f(x,y) _ xl + yl 41. f(x,y) _ xl - y

40. f(x,y) _ V"x,cc+c-y" 42. f(x,y) _ 4 - "l _ yl

43. f(x,y) _ 4"l ... f(x.y) - 1

47. f(",y) =

+ yl

44. f(x,y) - 6 - 2x - 3y

-Iyl

... f(x.y) - 1

yGx ,i'+'---y,OC+-04

48. f(x,y) =

-Ixl -Iyl

y,,2 + yl

4

FInding Level Curves d.

In Exercises 49-52, fmd an equation for and sketch the graph of the level curve of the function f(X, y) that passes through the given point.

,

4•• !(x.y) = 16 - x' - y'.

(20.0)

50. !(x.y) -

Vx'=l.

(1.0)

51. f(",y) =

y" + yl

3,

52. f(x,y) - x

(3,-1)

2y -"

+ Y + I' (-1, I)

y

Sketching Level Surfaces In Exercises 53-60, sketch a typical level surface for the :functioo..

••

+ yl + z2 54. f(x,y,z) _ 10(,,2 + yl + z2) 56. f(x,y,z) = z S5. f(x,y,z) =" + z 58. f(x,y,z) _ y2 + z2 57. f(x,y,z) _,,2 + yl 53. f(x,y,z) _ ,,2

59. f(x,y,z) - z - x2 _ y2 60. f(x,y,z) = (x 2/25) + (y2/16)

+ (z2/9)

FInding Level Surfaces In Exercises 61-64, fmd an equation for the level surface of the function through the given point.

61. f(",y,z) = ~ - loz, 62. f(x,y,z) _

1o(x 2

(3, -1, I)

+ y+z2), (-1,2,1)

14.2 63. g(x,y,z)

= Yx 2 + y2 + Z2, (I, -I, v'2)

64. g(x,y, z)

= ,_

Limits and Continuity in Higher Dimensions

71. f(x, y) = sin (x + 2 cos y), - 2". '" X '" 2"., -2". "'y '" 2"., P(".,,,.) 72. f(x,y) = e(z'Ly) sin (x 2 + y2), 0'" X '" 2".,

x-y+z + y-z ,(1,0, -2)

.LA.

-2". '" Y ",,,.,

In Exercises 65--{j8, rmd and sketch 1he domain of f. Then find an equation for 1he level curve or surface of1he function passing through 1he given point.

65. f(x,y)

=

it. ~Y. 00

66.g(x,y,z)=~

(x

1 1~ dO

vl-OZ

Y

68. g(x,y,z) =

%

I

+t

(0, I)

+ {' ~, (0, I, \13) Jo 4 - OZ

COMPUTER EXPLORATIONS Use a CAS to perfonn 1he following steps for each of 1he functioos in Exercises 6!1--72. L

Plot 1he surface over the given rectangle.

c. Plot 1he level curve of f through 1he given point.

69. f(x, y) = x sin ~

+ Y sin 2>:,

0 '" x '" 5".,

0 '" Y '" 5".,

14.2

- (cosy)Yx 2

0 '" x '" 5".,

+ z2

=

2

some parameter interval I, you can sometimes describe surfaces in

u,

0:5 u:S; 2,

78.x=ucosv, y=usinv, z=v, o :5 v :s; 2'1T

0::;; u:s; 2,

=

u cos v, y :s 21T

o :5 v =

=

u sin v, z

(2 + cosu) cos v, y

o :s; u :s; 2'17",

70. f(x,y) = (sinx)(cosy)e Vbl"/', o '" Y '" 5"., P(41T, 4".)

I

space wi1h a1riple of equations x = f(u, v),y = g(u, v),z = h(u, v) defined on some parameter rectangle a ::;; u :s; h, c :s; v :s; d. Many computer algebra systeros permit you to plot such surfaces in parametric mode. (parametrized surfaces are discussed in detail in Sectioo 16.5.) Use a CAS to plot 1he surfaces in Exercises 77-80. Also plot sevetallevel curves in 1he xy-plaoe.

79. x

P(3"., 3".)

(~)

74. x 2 + z2 = I

I

=

Parametrized Surfaces Just as you describe curves in 1he plaoe parametrically wi1h a pair of equations x = f(t),y = g(t) dermed on

77. x

b. Plot several level cutves in 1he rectangle.

+ z2)

+ y2 - 3z 2 =

76. sin

(ln4,ln9,2)

_ r.--:;,'

%

75. x

+ y)' I"

y

67. f(x, y) =

(1,2)

P("., -".)

Use a CAS to plot 1he implicitly dermed level surfaces in Exercises 73-76. 73. 4ln (x 2 + y2

n.z

11=0

755

O:s;

V

=

=

(2 + cosu) sin v, z

=

sinu,

:s; 2'17"

80. x = 2cosucosv, y = 2 cos u sin v, z o :5 u :s; 2'1T, 0:5 v :s; '11'

=

28inU,

Limits and Continuity in Higher Dimensions This section treats limits and continuity for multivariable functions. These ideas are analogous to limits and continuity for single-variable functions, but including more independent variables leads to additional complexity and important differences requiring some new ideas.

Limits for Functions of Two Variables If the values of f(x, y) lie arbitrnrily close to a f"lxed real number L for all points (x, y) suff"lciently close to a point (xo, yo), we say that f approaches the limit L as (x, y) approaches (xo, yo). This is similar to the informal definition for the limit of a function of a single variable. Notice, however, that if (xo, Yo) lies in the interior of j's domain, (x, y) can approach (xo, yo) from any direction. For the limit to exist, the same limiting value must be obtained whatever direction of approach is taken. We illustrate this issue in several examples following the detmition.

756


DEFINITION We say that a function /(x, y) approaches the limit L as (x, y) approaches (xo, yo), and write lim

(x, y)-(x. Yo)

/(x,y) = L

if, for every number E > 0, there exists a corresponding number 6 for all (x, y) in the domain of /,

O.

While we won't prove Theorem I here, we give an informal discussion ofwby it's true. If (x, y) is sufficiently close to (XO,yo), then f(x, y) is close to Land g(x, y) is close to M (from the informa! interpretation of limits). It is then reasonable that f(x, y) + g(x, y) is closetoL + M;f(x,y) - g(x,y) is close to L - M; kf(x,y) is close to kL; f(x,y)g(x,y) is close toLM; and f(x,y)/g(x,y) is close to L/Mif M # O. When we apply Theorem I to polynomials and rational functions, we obtsin the useful result that the limits of these functions as (x, y) ---> (xo, Yo) can be ca!culated by evaluating the functions at (xo, yo). The only requirement is that the rational functions be def"med at (xo, yo).

EXAMPLE 1 In this example, we can combiae the three simple results following the limit def"mition with the results in Theorem I to calculate the limits. We simply substitule the x and y values of the point being approached into the functional expression to f"md the limiting value. ( ) a

(h)

x - xy + 3

lim

(x,y)~(O,l) x2y + 5xy - y3 lim

(x,y)~(3,

EXAMPLE 2

-4)

Vx 2

+ y2

=

0 - (0)(1) + 3 = -3 (0)2(1) + 5(0)(1) - (1)3

V(3)2

+ (-4)2

=

v'25 =

-

xy

Find

x2 lim

(x,y)-(O,O)

.Vr .r· x - VY

5

•

758


Vx - vY

Solution Since the denominator approaches 0 as (x,y) --+ (0, 0), we cannot use the Q1!otient Rule from Theorem 1. If we multiply numerator and denominator by + Vy, however, we produce an equivalent fraction whose limit we can find:

Vx

(x 2 - .!JI)( Vx + vY) (x, y)~(O,O) Vx - vY (x,y)~(O,O) (Vx - vY)(Vx + vY) x2

lim

-

n,

lim

-J

x(x lim

Y)(Vx + vY) x

(x,y)-(O,O)

=

lim

(x,y)~(O,O)

Algebra

y

Cancel the nonzero fa 0 be given, but arbitrary. We want to rmd all> 0 such that 4xy2

1x2 + y2

-

01
0, there exists a 6 > 0 such thatfor all r and 8,

Give reasons for your answer. 56. Does knowing that

Irl+,+1) 14. I(x,y) ~ e~ sin (x + y)

+ y)

15. I(x,y)

~

17. I(x,y)

~ sin2 (x

In (x

- 3y)

16. I(x,y)

~

e"lny

18. I(x,y) ~ cos (3x - y2) 2

19. I(x,y)

~

21. I(x,y)

~

20. I(x,y) ~ log, x

x'

l'

(g cootinuous for all t)

g(t) dt

22. I(x,y) ~ ~(xy)'

.-0

(Ixyl
, 6) = p sin t/> cos 6 39. Work done by the heart

Using the Partial Derivative Definition

57. /(x,y)

sinh (xy - z2)

=

- x

_

+ 2x -

58. /(x,y) - 4

59. /(x,y)

(Section 3.9, Exercise 49)

=

60. /(x,y)

=

2

a/

A(c, h, k,m, q)

kin

=

q + em + T

Find all the second-order partia1 derivatives of the functions in Exercises 41-50.

+ Y + xy

x"y

=

44. h(x,y)

= xe Y

46. s(x,y)

= taJ.-1 (y/x)

+y + I

48. w = ye r - y

45. r(x,y)

In (x

+ y)

49. w = x sin (x"y)

x-y

= -2-X +y

+ 3y) xy2 + x"y' + x'y4

w,..

52. w = e'

fIrst? Try to answer without writing anything down.

a. /(x,y)

=

b. /(x,y)

= I/x =y +

c. /(x,y) e.

xsiny

+ eY

f. /(x,y)

+ y2

'

ay

and

a/ ay

a/ iJy

at(I,2) at (-2, I)

at (-2,3)

(x,y) oF (0,0) (x,y) = (0,0),

0,

a/

a/ ax

and

iJy

at

(0,0)

62. Let /(x,y) = x 2 + y'. Find the slope of the line tsngent to this surface at the point (-I, I) and lying in the a. plane x = -I b. plane y = I.

63. Three \'lUiabl.. Let w = /(x, y, z) be a function of three independent varisbles and write the formal defInition of the partial derivative a//az at (xo,Yo, zo). Use this defirtition to fmdaf/az at (I,2,3)for/(x,y,z) =x"yz2. 64. Three \'lUiabl.. Let w = /(x, y, z) be a function of three independent varisbles and write the formal defmition of the partial derivative a/lay at (XO,YO, zo). Use this defmition to fmda//ay at (-1,0,3) for /(x,y, z) = _2xy2 + yz2

defmes z as a function of the two independent variables x and y and the partial derivative exists.

66. Find the vaIue ofiJx/az at the point (I, -I, -3) if the equation

xz + ylnx - x 2 + 4 = 0 defmes x as a function of the two independent varisbles y and z and the partial derivative exists. Exercises 67 and 68 are about the triangle shown here.

(x/y)

+ x"y + 4y' - In (y2 + I) 2 /(x,y) = x + 5xy + sinx + 7e' =

+ y4)

ax

a/

xy+z'x-2yz=0

+ xlny + ylnx w = 54. w = x siny + y sinx + xy Which order of differeotiation will calculate f", faster: x first or y

51. w = 1n(2x

d. /(x,y)

and

a/

2

I,

and

Differentiating Implldtly 65. Find the vaIue ofaz/iJx at the point (I, I, I) if the equation

In Exercises 51-54, vetifY that w", =

55.

=

47. w=x 2 tan(xy)

Mixed Partial Derivatives

53.

= sinxy

+ cosy + ysinx

43. g(x,y)

50. w

42. /(x,y)

ax

{

ax

3y - xy,

+ 3y -

x2

a/

- 3x"y,

61. Let /(x, y) = 2x + 3y - 4. Find the slope of the line tsngent to this sorface at the point (2, -I) and lying in the a. plane x = 2 b. plane y = -I.

hq

Calculating Second-Order Partial Derivatives

41. /(x,y) = x

V2x

sin (x'

VIlv zg

(Section 4.5, Exercise 51)

+y

=I

38. g(r, 6, z) = r(1 - cos 6) - z

W(P, V, 6, v, g) = PV+ 40. Wilson lot size formula

773

In Exercises 57-{j(), use the limit defmition of partial detivative to compute the partial derivatives of the functions at the specifled points.

e-"'"

=

Partial Derivatives

y

= xlnxy

56. The fIfth-order partial derivative a'//iJx 2ay' is zero for each of the followiog functioos. To show this as quickly as possible, which variable would you differentiate with respect to first: x or y? Try to answer without writing anything down. a. /(x,y) = y'x 4e' + 2 b. /(x,y) = y2 + y(sinx - x 4) c. /(x,y) = x 2 + 5xy + sinx + 7e' d. /(x,y) = xe Y'/2

B~ C

b

A

67. Express A iroplicitly as a function of a, b, and c and calculate aA/aa and aA/ab. 68. Express a iroplicitly as a function of A, b, and B aod calculate aa/aA and aa/aB. 69. Two dependent variable. Express v, in terms of u and y if the equations x = v In u and y = u In v define u and v as functions of the independent variables x and y, and if v, exists. (Hint: Differeotiate both equations with respect to x and solve for v, by elintinating u,.)

774


70. Two dependent variable. Find ax/au and ay/au if the equations u = x 2 - y2 and v = x 2 - y defme x and y as functions of the indepeodeot variables u and v, and the partial derivatives exist (See the hint in Exercise 69.) Then let s = x 2 + y2 and fmd

as/au.

73. I(x,y, z) = x 2 + y2 - 2z2 74. I(x,y, z) = 2z 3 - 3(x 2 + y2)z 75. I(x, y) =

y3, 71. Let I(x,y) = { 2 -y,

y '" 0 0 y< .

Find I~ Iy. I"" and Iyx, and state the domain for each partial derivative. 72. Let I(x,y) =

Show that each function in Exercises 73-80 satisfies a Laplace equation.

Yx { X.2 '

x~0 X < O.

Find I ~ I yo I "Y' and I yx, and state the domain for each partial derivative. Theory and Examples

The three-dimensional Laplace equation

e -2y cos

2x

76. I(x,y) = lnv'x 2 + y2 77. I(x,y)

=

3x

+ 2y

78. I(x,y) = tan-I

- 4

Y

79. I(x,y, z) = (x 2 + y2 + z2)-I/2 80. !(x,y, z)

=

e 3r+4y cos 5z

The Wave Equation If we stand on an ocean shore and take a snapshot of the waves, the picture shows a regular pattern of peaks and valleys in an instant of time. We see periodic vertical motion in space, with respect to distance. If we stand in the water, we can feel the rise and fall of the water as the waves go by. We see periodic vertical motion in time. In pbysics, this beautiful symmetry is expressed by the

one-dimensional wave equation

a'w= e 2a'w 2 2 is satisfied by steady-state temperature distributions T = I(x, y, z) in space, by gravitational potentials, and by electrostatic potentials. The

two-dimensional Laplace equation

al

ax '

where w is the wave height, x is the distance variable, t is the time variable, and e is the velocity with which the waves are propagated.

a21 a21 ax ay

2+-2=0,

w

obtained by dropping the a2 f!az 2 term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) maybe treated as a thin slice of the solid (b) perpeodicular to the z-axis.

;)' + Inyz + Inxz, Po(!, I, I) In (x' + y' - I) + y + 6z, Po(1, 1,0)

Tangent LInes to Level Curves

31. Zero directional derivative In what directioo is the derivative of I(x, y) = >;)' + y' at P(3, 2) equal to zero? 32. Zero directional derivative In what directions is the derivative of I(x,y) = (x' - y')/(x' + y') atp(l, I) equal to zero? 33. Is there a direction n in which the rate of change of I(x, y) = 3>;)' + 4y' at P(I, 2) equals 14? Give reasoos for your

x' -

answer. 34. Changing temperatnre along a circle Is there a direction u in which the rate of change of the temperatore functioo T(x, y, z) = 2>;Y - yz (temperature in degrees Celsius, distance in feet) at P(I, -I, I) is -3°C/It? Give reasons for your answer.

In Exercises 25-28, sketch the curve I(x, y) = c together with VI and the tangent line at the given point Then write an equation for the tangent line.

35. The derivative of I(x,y) at Po(I,2) in the direction ofi + j is 2V2 and in the direction of -2j is -3 . What is the derivative of I in the directioo of -i - 2j? Give reasons for yoor answer.

x' + y' = 4, (v'2, v'2) 26.x'-y=l, (v'2,I)

36. The derivative of I(x, y, z) at a point P is greatest in the direction of v = i + j - k. In this direction, the value of the derivative is

25.

27. >;)' = -4,

2v'3.

(2, -2)

28.x'->;),+y'=7,

a. What is Vf at P? Give reasons for your answer.

(-1,2)

b. What is the derivative of I at P in the direction ofi

Theory and Examples 29. Let I(x,y) = x' - >;)' + y' - y. Find the directions u and the values of D. 1(1, -I) for which

b.

c. Du/(1, -I) e. Du/(1, -I)

d. D./(!,-I)

30. Let I(x,y) =

D./(-H)

= =

0 -3

~~ ~ ~~. Find the directions u

=

4

derivative of a differentiable function I(x,y, z) at a point Po in the direction of a unit vector u related to the scalar component of (V/)p, in the direction ofn? Give reasoos foryoor answer.

+~) is largest

b. D.

c. Du/(

+~) = 0

d.

38. Directional derivative. and partial derivative. Assuming that the necessary derivatives of I(x, y, z) are dermed, how are DI/, DJ/, and D.I related to I., I y , and I.? Give reasons for your

answer.

and the values of

39. Line. in the xy-plane Show that A(x - xo) + B(y - Yo) = 0 is an equation for the line in the xy-plane through the point (xo,Yo) norma1 to the vectorN = Ai + Bj.

for which

a. Du/(

e.

37. Directional derivatives and scalar components How is the

Do I(!, -I) is smallest

a. Du 1(1, -I) is largest

1(+ ~)

is smallest

D./(-t,~) =-2

40. The algebra rule. for gradient. gradients

Du/(+~) = I

14.6

+ j?

VI

= al i + al. + al k, ax

ayJ

az

Given a constant k and the

ag ag ag Vg=-i+-j+-k, ax ay az

establish the algebra rules for gradients.

Tangent Planes and Differentials In this section we derme the tangent plane at a point on a smooth surface in space. Then we show how to calculate an equation of the tangent plane from the partial derivatives of

the function defining the surface. This idea is similar to the dermition of the tangent line at a point on a curve in the coordinate plane for single-variable functions (Section 3.1). We then study the total differential and linearization of functions of several variables.

Tangent Planes and Normal Lines If r = g(t)i + h(t)j + k(t)k is a smooth curve on the level surface f(x, y, z) = c of a differentiable function f, then f(g(t), h(t), k(t)) = c. Differentiating both sides of this

792


equation with respect to t leads to

d dtf(g(t), h(t), k(t))

=

d dt (c)

+ af dh + af dk

=

0

af dg ax dt

FIGURE 14.32 The gradient VI is orthogonal to the velocity vector of every smooth curve in the surface through Po. The velocity vectors at Po therefore lie in a common plane, which we call the tangent plane at Po.

f ( a~ i

+ af. + af ~J

'

~

ay dt

az

Chain Rule

dt

k)' (d~g i + dh. + dk k) O. ~J ~ =

~

(1)

' --~

VI

dr/dt

At every point along the curve, Vf is orthogonal to the curve's velocity vector. Now let us restrict our attention to the curves that pass through Po (Figure 14.32). All the velocity vectors at Po are orthogonal to Vf at Po, so the curves' tangent lines all lie in the plane through Po normal to Vf. We now define this plane.

DEFINITIONS The tangent plane at the point Po(xo, Yo, zo) on the level surface f(x, y, z) = c of a differentiable function f is the plane through Po normal to Vflpo' The normal line of the surface at Po is the line through Po parallel to Vf

IPo'

From Section 12.5, the tangent plane and normal line have the following equations:

Tangent Plane to f(x,y, z)

fx(Po)(x - xo)

=

cat Po(xo,Yo, zo)

+ fy(Po)(y - Yo) + fz(Po)(z - zo) =

0

(2)

Normal Line to f(x,y, z) = cat Po(xo,Yo, zo)

x z

Po(l, 2, 4)

1

:/

EXAMPLE 1

The surface

xZ+r+z-

9 =O

I

=

Xo

+ fiPo)t,

z = Zo

+ fAPo)t

(3)

Find the tangent plane and normal line of the surface

f(x,y, z)

=

x 2 + y2

+z

- 9 = 0

A circular paraboloid

at the point Po(1, 2, 4).

I

Nonnalline

SoLution The surface is shown in Figure 14.33. The tangent plane is the plane through Po perpendicular to the gradient of f at Po . The gradient is

Vflpo = (2xi

+

2yj

+

k)(1 ,2,4)

= 2i + 4j + k.

The tangent plane is therefore the plane x

/

FIGURE 14.33 The tangent plane and normal line to this surface at Po (Example I).

2(x - 1)

+ 4(y

- 2)

+

(z - 4) = 0,

or

2x

+ 4y + z =

14.

The line normal to the surface at Po is

x = 1

+

2t,

y = 2

+ 4t,

z = 4

+

t.

•

To find an equation for the plane tangent to a smooth surface z = f(x, y) at a point = f(xo, Yo), we first observe that the equation z = f(x, y) is

Po(xo, Yo, zo) where Zo

793

14.6 Tangent Planes and Differentials

equivalent to f(x,y) - z = O. The surface z = f(x,y) is therefore the zero level surface of the function F(x, y, z) = f(x, y) - z. The partial derivatives of Fare

o

Ix -

F. = ox (f(x,y) - z) =

0 =

Ix

o

Fy = oy (f(x,y) - z) = /y - 0 = /y

o

Fz = oz(f(x,y) - z) = 0 - 1 = -1.

The formula F'(Po)(x - xo)

+ Fy(Po)(y - Yo) + F.{Po)(z - zo)

=

0

for the plane tangent to the level surface at Po therefore reduces to f.(xo,yo)(x - xo)

+ fy(xo,yo)(Y - Yo) - (z - zo)

=

O.

Plane Tangent to a Surface z = f(x,y) at (Xo,Yo, fVco,yo» The plane tangent to the surface z = f(x, y) of a differentiable function f at the poiot Po(xo,Yo, zo) = (xo,Yo, f(xo,yo)) is f.(xo,yo)(x - xo)

EXAMPLE 2

+

fY(xo,yo)(Y - Yo) - (z - zo) = O.

(4)

Fiod the plane tangent to the surface z = xcosy - ye' at (0,0,0).

Solution We calculate the partial derivatives of f(x, y) = x cos y - ye' and use Equation (4): f.(O,O) = (cosy - ye')(o.o) = 1 - 0·1 = 1

/y(O,O)

=

(-x sioy - e')(o.o)

0- 1

=

-1.

=

The tangent plane is therefore 1'(x - 0) - 1'(Y - 0) - (z - 0) = 0,

or x - y - z =

EXAMPLE 3

Eq.(4)

o.

•

The surfaces f(x,y,z) = x 2

+ y2

- 2 = 0

A cylinder

and g(x,y,z) = x

+z -

4 = 0

ApJane

meet in an ellipse E (Figure 14.34). Fiod parametric equations for the lioe tangent to E at the poiot Po(I, 1,3). Solution The tangent lioe is orthogonal to both Vf and Vg at Po, and therefore parallel tov = Vf X Vg. The components ofv and the coordinates of Po give us equations for the lioe. We bave Vfl(l.l~) = (2xi

Vgl (l.l~)

FIGURE 14.34 This cylinder and plane intersect in an ellipse E (Example 3).

=

(i

+ 2yj)(1.1.3)

+ k)(1.1.3)

v = (2i

=

i

=

2i

+ 2j

+k

+ 2j) X (i + k)

=

2 1

j 2 0

k 0 = 2i - 2j - 2k. 1

794

Chapter 14: Partial Derivatives The tangent line is

y=I-2t,

x=I+2t,

•

z=3-2t.

Estimating Change in a Specific Direction The directional derivative plays the role of an ordinary derivative when we want to estimate how much the value of a function / changes if we move a small distance tis from a point Po to another point nearby. If / were a function of a single variable, we would have

d/

=

f'(Po) tis.

Ordinary derivative x increment

For a function of two or more variables, we use the formula

d/

=

(V/lpo"u) tis,

Directional derivative x increment

where u is the direction of the motion away from Po.

Estimating the Change in / in a Direction u To estimate the change in the value of a differentiable function / when we move a small distance tis from a point Po in a particular direction u, use the formula

d/

=

(V/lpo" u)

tis

Directional Distance derivative increment

EXAMPLE 4

Estimate how much the value of

/(x,y,z) = ysinx + 2yz will change if the point P(x,y, z) moves 0.1 unit from Po(O, 1,0) straight toward Pt(2, 2, -2).

Solution We first fmd the derivative of / at Po in the direction of the vector P-;;Pt = 2i + i - 2k. The direction of this vector is

The gradient of / at Po is

V/I(O.1.0)

=

((ycosx)i

+

(sinx

+ 2zli + 2yk)(O.1.0)

= i

+ 2k.

Therefore,

V/lpo·u = (i

+

2k)· (t + ti - tk) t -t -to i

=

=

The change d/ in / that results from moving tis = 0.1 unit away from Po in the direction of u is approximately

d/

=

(V/lpo"u)(tis) =

(-t)tO.I) '"

-0.067unit.

•

How to Linearize a Function of Two Variables Functions of two variables can be complicated, and we sometimes need to approximate them with simpler ones that give the accuracy required for specific applications without being so difficult to work with. We do this in a way that is similar to the way we fmd linear replacements for functions of a single variable (Section 3.9).

14,6 Tangent Planes and Differentials

A point _ near (xQ. Yo)

(x, y)

Suppose the function we wish to approximate is z = I(x, y) near a point (xo, Yo) at which we know the values of I, Ix> and Iy and at which I is differentiable. Ifwe move from (xo, Yo) to any nearby point (x, y) by increments Ax = x - Xo and ~y = y - Yo (see Figure 14.35), then the defmition of differentiability from Section 14.3 gives the change

I(x,y) - I(xo,yo) A point where f is differentiable

795

=

fx(xo,Yo)Ax

+ ly(xo,Yo)~y + EtAx +

E2~y,

where E1, E2 --> 0 as Ax, ~y --> O. If the increments Ax and ~y are small, the products EJAx and E2~y will eventoally be smaller still and we have the approximation

/

I(x,y) "" I(xo,yo)

+ fx(xo,Yo)(x - xo) + /y(xo,Yo)(y - yo). L(x,y)

If! is differentiable at (xo, yo), then the value of! at aoy point (x, y) nearby is approximately !(xo,Yo) + fx(xo,Yo)/!x + !y(xO,yo)4y. FIGURE 14.35

In other words, as long as as the linear function L.

DEFINmONS

I

~x

and ~y are small, I will have approximately the same value

The linearization of a function I(x, y) at a point (xo, Yo) where

is differentiable is the function

L(x,y)

I(xo,yo) + fx(xo,Yo)(x - xo) + /y(xo,Yo)(y - yo).

=

(5)

The approximation

I(x,y) "" L(x,y) is the standard linear approximation of I at (xo, yo).

From Equation (4), we fmd that the plane z = L(x,y) is tangent to the surface z = I(x, y) at the point (xo, yo). Thus, the linearization of a function of two variables is a tangent-plane approximation in the same way that the linearization of a function of a single variable is a tangent-line approximation. (See Exercise 63.)

EXAMPLE 5

Find the linearization of

I(x,y)

=

x 2 - xy

+ ty2 + 3

at the point (3, 2).

Solutton

We first evaluate 1./.. and Iy at the point (xo, Yo) = (3,2):

1(3,2)

(x2 - xy

=

fx(3,2)

= :

/y(3,2)

=

x

+ 1 y2 +

(x2 - xy

aa (x2 - xy Y

2

3)

+ -21 y2 +

= (3,2)

3)(3,2)

+ -21 y2 + 3)

(3,2)

8 =

(2x - y)(3,2)

=

(-x

+ Y)(3,2)

=

4

=

-I,

giving

L(x, y)

=

I(xo, Yo) + fx(xo, Yo)(x - xo) + /y(xo, Yo)(Y - Yo)

=

8

+ (4)(x - 3) + (-I)(y - 2)

The linearization of I at (3, 2) is L(x,y) = 4x - y - 2.

=

4x - y - 2.

•

When approximating a differentiable function I(x, y) by its linearization L(x, y) at (xo, Yo), an important question is how accurate the approximation might be.

796


y

If we can f'md a common upper bound M for II", I, IIyy I, and II", I on a rectangle R centered at (xo,Yo) (Figure 14.36), then we can bound the error E throughout R by using a simple formula (derived in Section 14.9). The error is def'med by E(x,y) =

I(x,y) - L(x,y).

R ~--------------~X

o

FIGURE 14.36 The rectangular region R: Ix -.tol '" h, Iy - yol '" kinthe xy·plane.

The Error in the Standard Linear Approximation If I has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at (xo, YO) and if M is any upper bcund for the values of 1/",1, I/yyl, and 1/",1 on R, then the error E(x, y) incurred in replacing I(x, y) on R by its linearization

L(x, y)

=

I(xo, YO) + fx(xo, yo)(x - xo) + /y(xo, YO)(y - YO)

satisfies the inequality

IE(x,y) I'" t M(lx - xol + Iy - yoll'·

To makeIE(x,y) I small for a givenM, we just make Ix - xol andly - yol small.

EXAMPLE 6 Find an upper bound for the error in the approximation I(x, y) '" L(x, y) in Example 5 over the rectangle R:

Ix - 31 '" 0.1,

Iy - 21'" 0.1.

Express the upper bound as a percentage of 1(3, 2), the value of I at the center of the rectangle. Solution

We use the inequality

IE(x,y) I'" t M(lx - xol + Iy - yoll'· To f'md a suitable value for M, we calculate I ",,j"', and I yy, f'mding, after a routine differentiation, that all three derivatives are constant, with values 1/",1 = 121 = 2,

I/yyl = III = I.

1/",1 = I-II = I,

The largest of these is 2, so we may safely takeMto be 2. With (xo, Yo) = (3,2), we then know that, throughout R,

IE(x,y) I '" t(2)(lx - 31

+ Iy

- 21)2 = (Ix - 31

+ Iy

- 21)'.

Finally, since Ix - 31 '" 0.1 andly - 21 '" 0.1 onR, we have

IE(x,y) I '" (0.1 + 0.1)2

= 0.04.

As a percentage of 1(3,2) = 8, the error is no greater than

0.~4

X 100 = 0.5%.

•

Differentials Recall from Section 3.9 that for a function of a single variable, y = I(x), we defined the change in I as x changes from a to a +
801

55. You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer. 56. a. Around the point (1, 0), is f(x, y) = X 2(y + 1) more sensitive to changes in x or to changes in y? Give reasons for your answer. b. What ratio of dx to dy will make df equal zero at (1, O)?

y

57. Error carryover in coordinate changes y P(3 ± 0.01,4 ± 0.01)

52. Consider a closed container in the shape of a cylinder of radius 10 cm and height 15 cm with a hemisphere on each end, as shown in the accompanying figure. 15 cm JlOcm

- -l Oat(a,b). - Ix'? < oat (a, b).

iv) the test is inconclusive at (a, b) if 1",lyy - Ix'? = 0 at (a, b). In this case, we must fmd some other way to determine the behavior of I at (a, b).

(a)

The expression Ixxlyy - Ix'? is called the discriminant or Hessian of I. It is sometimes easier to remember it in determinant form,

Ixxlyy

-!xl =

I~: ~:I·

Theorem 11 says that if the discriminant is positive at the point (a, b), then the surface curves the same way in all directions: downward if Ixx < 0, giving rise to a local maximum, and upward if Ixx > 0, giving a local minimum. On the other hand, if the discriminant is negative at (a, b), then the surface curves up in some directions and down in others, so we have a saddle point. (b)

(a) The origin is a saddle point of the function j(x, y) = y2 - x 2. There are no local extreme values (Example 2). (b) Level curves for the function j io Example 2. FIGURE 14.44

EXAMPLE 3

Find the local extreme values of the function

I(x,y) = xy - x 2

y2 - 2x - 2y

-

+ 4.

Solution The function is defined and differentiable for all x and y and its domain has no boundary points. The function therefore has extreme values only at the points where Ix and Iy are simultaneously zero. This leads to Ix = y - 2x - 2 = 0,

Iy = x - 2y - 2 = 0,

or

x =y

=

-2.

Therefore, the point ( -2, -2) is the only point where I may take on an extreme value. To see if it does so, we calculate

Ixx = -2,

Iyy = -2,

Ixy = I.

The discriminant of I at (a, b) = (-2, -2) is

Ixxlyy - Ixy2 = (-2)(-2) - (1)2 = 4 - 1 = 3. The combination

Ixx tells us that

I

I( -2, -2)

8.

=

EXAMPLE 4 Solution

;4 + y' - 2>;2 - 2y2 + 3, -3/2'" x'" 3/2, - 3/2 ,;; Y ,;; 3/2 73. I(x,y) = 5x' + lax' - 3Ox' + 3Oxy2 - 12Ox', -4 :s x :s 3, -2:s y :s 2 (x,y) # (0,0) 74. I(x,y) = {~:In(x2 + y2), (x,y) = (0,0)' -2 :s x :5 2, -2:s y :5 2

Ir-La~g~~_n= ge__ M_ UL_ ti~ Pl_ ie_~________________________________

HIsTORICAL BIOGRAPHY

Joseph Louis Lagrange (1736-1813)

Sometimes we need to find the extreme values of a function whose domain is constrained to lie within some particular subset of the plane---a disk, for example, a closed triangular region, or along a curve. In this section, we explore a powerful method for rmding extreme values of constrained functions: the method of Lagrange multipliers.

Constrained Maxima and Minima We first consider a problem where a constrained minimum can be found by eliminating a variable.

EXAMPLE 1

FindthepointP(x,y,z)ontheplane2x

+ y -z -

5 =

othat is closest

to the origin.

Solution

The problem asks us to rmd the minimum value of the function

IoFl = Y(x - 0)2 + (y = Yx2 + y2 + z2

- 0)2

+

(z - 0)2

subject to the constraint that

2x+y-z-5=0.

oF I has a minimum value wherever the function

Since I

f(x,y, z) = x 2

+ y2 + z2

has a minimum value, we may solve the problem by finding the minimum value of f(x, y, z) subject to the constraint 2x + y - z - 5 = 0 (thus avoiding square roots). Ifwe regard x and y as the independent variables in this equation and write z as z=2x+y-5, our problem reduces to one of finding the points (x,y) at which the function

h(x, y) = f(x, y, 2x

+y

- 5) = x 2

+ y2 +

(2x

+y

- 5)'

812

Chapter 14: Partial Derivatives has its minimum value or values. Since the domain of h is the entire xy-plane, the First Derivative Test of Section 14.7 tells us that any minima that h might have must occur at points where hy = 2y + 2(2x + y - 5) = O.

hx = 2x + 2(2x + y - 5)(2) = 0, This leads to

lOx + 4y = 20,

4x

+ 4y

=

10,

and the solution

x

5

=

3'

We may apply a geometric argument together with the Second Derivative Thst to show that these values minimize h. The z·coordinate of the corresponding point on the plane

z=2x+y-5is 5 6' Therefore, the point we seek is Closest point:

•

The distance fromPto the origin is 5jV6 '" 2.04.

Attempts to solve a constrained maximum or minimum problem by substitotion, as we might call the method of Example 1, do not always go smoothly. This is one of the rea· sons for learning the new method of this section.

z

EXAMPLE 2

Find the points on the hyperbolic cylinder x 2 - z2 - I = 0 that are clos·

est to the origin. Solution 1 The cylinder is sbowo in Figure 14.49. We seek the points on the cylinder closest to the origin. These are the points whose coordinates minimize the value of the function

/(x,y,z) = x 2 + y2

+

z2

Square of the distance

subject to the constraint that x 2 - z2 - I = O. Ifwe regard x andy as independent vari· abIes in the constraint equation, then

and the values of /(x,y, z) = x 2

+ y2 + z2 on the cylinder are given by the function

h(x, y) = x 2 + y2 + (x 2 - I) = 2x 2 + y2 - 1. To f'md the points on the cylinder whose coordinates minimize /, we look for the points in the :ry·plane whose coordinates minimize h. The only extreme value of h occurs where FIGURE 14.49

x' - z' -

The hyperbolic cylinder

I = 0 in Example 2.

hx =4x=0

and

hy =2y=0,

that is, at the point (0, 0). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened was that the First Derivative Test found (as it sbonld have) the point in the domain of h where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire

14.8 Lagrange Multipliers The hyperbolic cylinder x 2 On this part,

x= v?+l

- Z2 =

1

On this part,

z x=-v?+l

xy-plane, the domain from which we can select the first two coordinates of the points (x, y, z) on the cylinder is restricted to the "shadow" ofthe cylinder on the xy-plane; it does not include the band between the lines x = -1 and x = 1 (Figure 14.50). We can avoid this problem if we treat y and z as independent variables (instead of x and y) and express x in terms of y and z as

+

x 2 = z2

k(y, z)

1.

+ y2 + z2 becomes

With this substitution, I(x,y, z) = x 2

FIGURE 14.50 The region in the ;ry-plane from which the first two coordinates of the points (x,y, z) on the hyperbolic cylinder x 2 - z2 = 1 are selected excludes the band -1 < x < 1 in the xy-plane (Example 2).

813

= (z2 + 1) + y2 + z2 = 1 + y2 + 2z2

and we look for the points where k takes on its smallest value. The domain of k in the yzplane now matches the domain from which we select the y- and z-coordinates of the points (x, y, z) on the cylinder. Hence, the points that minimize k in the plane will have corresponding points on the cylinder. The smallest values of k occur where

ky = 2y = 0 or where y

kz

and

=

4z = 0,

= z = O. This leads to x2

= z2 + 1 = 1,

x = ±1.

The corresponding points on the cylinder are (± 1,0,0). We can see from the inequality

k(y, z)

= 1 + y2 + 2z2

~ 1

that the points (± 1, 0, 0) give a minimum value for k. We can also see that the minimum distance from the origin to a point on the cylinder is 1 unit.

\

Solution 2 Another way to find the points on the cylinder closest to the origin is to imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the cylinder (Figure 14.51). At each point of contact, the cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting and

x

/

FIGURE 14.51 A sphere expanding like a soap bubble centered at the origin until it just touches the hyperbolic cylinder x 2 - z2 - 1 = 0 (Example 2).

g(x,y,z)

= x 2 - z2 - 1

equal to 0, then the gradients VI and Vg will be parallel where the surfaces touch. At any point of contact, we should therefore be able to find a scalar A ("lambda") such that

VI = AVg, or

2xi

+

2yj

+

2zk = A(2xi - 2zk).

Thus, the coordinates x, y, and z of any point of tangency will have to satisfy the three scalar equations

2x

= 2Ax,

2y

= 0,

2z

= -2Az.

For what values of A will a point (x, y, z) whose coordinates satisfy these scalar equations also lie on the surface x 2 - z2 - 1 = O? To answer this question, we use our knowledge that no point on the surface has a zero x-coordinate to conclude that x O. Hence, 2x = 2Ax only if

*'

2 = 2A,

or

A=1.

For A = 1, the equation 2z = -2Az becomes 2z = -2z. If this equation is to be satisfied as well, z must be zero. Since y = 0 also (from the equation 2y = 0), we conclude that the points we seek all have coordinates of the form

(x, 0, 0).

814

Chapter 14: Partial Derivatives What points on the surface x 2 points (x, 0, 0) for which

x2

-

(0)'

z2 =

-

=

I have coordinates oflhis fonn? The answer is the

x2

I,

=

I,

or

x

=

± I.

The points on the cylinder closest to the origin are the points (± 1,0,0).

•

The Method of Lagrange Multipliers In Solution 2 of Example 2, we used the method of Lagrange multipliers. The method says that the extreme values of a function I(x, y, z) whose variables are subject to a constraint g(x,y,z) = 0 are to be found on the surface g = 0 among the points where VI = AVg for some scalar A (called a Lagrange multiplier). To explore the method further and see why it works, we first make the following observation, which we state as a theorem.

THEOREM 12-The Orthogonal Gradient Theorem

Suppose that I(x, y, z) is differentiable in a region whose interior contains a smooth curve C:

r(t)

=

g(t)i

+ h(t)i + k(t)k.

If Po is a point on C where I has a local maximum or minimum relative to its values on C, then VI is orthogonal to C at Po.

Proof We show that VI is orthogonal to the curve's velocity vector at Po. The values of I on C are given by the composite I(g(t), h(t), k(t», whose derivative with respect to tis

dl dt At any point Po where curve, dl/dt = 0, so

I

=

al dg ax dt

al dh

al dk

+ ay dt + az dt

= VI' v.

has a local maximum or minimum relative to its values on the

V/'Y

=

o.

•

By dropping the z-tenns in Theorem 12, we obtain a similar result for functions of two variables.

COROLLARY OF THEOREM 12 At the points on a smooth curve r(t) = g(t)i + h(t)j where a differentiable function I(x, y) takes on its local maxima and minima relative to its values on the curve, VI' Y = 0, where Y = dr/dt.

Theorem 12 is the key to the method of Lagrange multipliers. Suppose that I(x, y, z) and g(x, y, z) are differentiable and that Po is a point on the surface g(x, y, z) = 0 where I has a local maximum or minimum value relative to its other values on the surface. We assume also that Vg '" 0 at points on the surface g(x,y,z) = o. Then I takes on a local maximum or minimum at Po relative to its values on every differentiable curve through Po on the surface g(x,y, z) = o. Therefore, Viis orthogonal to the velocity vector of every such differentiable curve through Po. So is Vg, moreover (because Vg is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefure, at Po, VI is some scalar multiple A ofVg.

14.8

815

Lagrange Multipliers

The Method of Lagrange Multipliers Suppose that I(x, y, z) and g(x, y, z) are clifferentiable and Vg '" 0 when g(x,y,z) = O. To find the local maximum and minimum values of I subject to the constraint g(x, y, z) = 0 (if these exist), f"md the values of x, y, z, and A that simultaneously satisfY the equations

VI

AVg

=

g(x,y, z)

and

o.

=

(1)

For functions of two independent variables, the condition is similar, but without the variable z.

Some care must be used in applying this method. An extreme value may not actually exist (Exercise 41). y

EXAMPLE 3 X2

Y2

8+

Find the greatest and smallest values that the function

y2

I(x,y)

2~1

=

xy

takes on the ellipse (Figure 14.52) x

0

2Y2 We want to find the extreme values of I(x, y)

Solution FIGURE 14.52 Example 3 shows how to f"md the largest and smallest values of the product"" on this ellipse.

g(x,y)

=

x'

y'

"8 + "2 - I

xy subject to the constraint

=

=

o.

To do so, we first find the values ofx,y, and A for which

VI

=

AVg

g(x,y)

and

=

O.

The gradient equation in Equations (1) gives yi

+ xj

=

Ay,

=

~ xi + Ayj,

from which we find y

xy=-2

xy=2

x

so that y

-~~i~;~~~~~~

=

0 or A = ±2. We now consider these two cases.

Case 1: If y = 0, then x = y = O. But (0, 0) is not on the ellipse. Hence, y '" O. Case2: If y '" 0, then A = ±2 and x = ±2y. Substituting this in the equation g(x, y) = 0 gives

x

(±2y)' -8xy=2/

FIGURE 14.53

eonstraiot g(x, y)

and

y'

+ "2

=

I,

4y' + 4y'

=

8

and

y

=

±l.

The function I(x,y) = xy therefore takes on its extreme values on the ellipse at the four points (±2, 1), (±2, -I). The extreme values are xy = 2 andxy = -2.

xy=-2

When subjected to the = x'/8

+ y'/2 - I

=

0, The Geometry of the Solution The level curves of the function I(x, y) = xy are the hyperbolas xy = c (Figure 14.53). The farther the hyperbolas lie from the origin, the larger the absolule value of I. We want to f"md the es1reme values of I(x, y), given that the point (x, y) also lies on the ellipse x' + 4y' = 8. Which hyperbolas intersecting the ellipse lie farthest from the origin? The hyperlJolas that just graze the ellipse, the ones that are tangent to it, are

the functioo f(x, y) = "" takes 00 extreme values at the four points (±2, ± I ~ These are the points 00 the ellipse when Vf (red) is a scalar multiple ofVg (blue) (Example 3).

816

Chapter 14: Partial Derivatives farthest. At these points, any vector nonna! to the hyperbola is nonna! to the ellipse, so VI = yi + xj is a multiple (A ~ ±2) ofVg ~ (x/4)i + yj. At the point (2, I), for example, VI

~

i

+

Vg ="2' 1. + J, .

2j,

aod

VI = 2Vg.

aod

VI

At the point ( - 2, I), VI = i - 2j,

EXAMPLE 4 3x

Vg =

-ti +

j,

~

-2Vg.

•

Find the maximum aod minimum values of the function I(x,y) = ~ 1.

+ 4y on the circle x 2 + y2

Solution

We model this as a Lagraoge multiplier pmblem with

g(x,y) ~ x 2 + y2 - I

+ 4y,

I(x,y) ~ 3x

aod look for the values ofx,y, aod A that satisfy the equations VI ~ AVg:

g(x,y) ~

+ 4j = 2xAi + 2yAj 0: x 2 + y2 - I = O. 3i

The gradient equation in Equations (I) implies that A "" 0 aod gives

x

3 = 2A'

Y

~

2

X'

These equations tell us, among other things, that x aod y have the same sign. With these values for x aod y, the equation g(x, y) ~ 0 gives

(iAY + (iY -

I

~ 0,

so

9

+ 16

= 4A2,

4A2 = 25,

5 A ~ ±"2'

aod

Thus,

y

3

3

x~2A=±S'

2

4

Y ~ X = ±S'

aod/(x,y) = 3x + 4yhas extreme values at (x,y) ~ ±(3/5,4/5). By calculating the value of 3x + 4y at the points ±(3/5, 4/5), we see that its maximum aod minimum values on the circle x 2 + y2 ~ I are 3x+4y=5

aod 3x

+ 4y =-5

FIGURE 14.54 The functionj(x,y) ~ 3x + 4y takes 00 its largest value 00 the unitcircleg(x,y) ~ x 2 + y2 - I ~ Oat the point (3/5, 4/5) and its smallest value at the point (-3/5, -4/5) (Example 4). At each of these points, Vj is a scalar multiple ofVg. The figure shows the gradients at the Il!St point but not the second.

The Geometry of the Solution The level curves of I(x, y) ~ 3x + 4y are the lines 3x + 4y ~ c (Figure 14.54). The farther the lines lie from the origin, the larger the absolute value of f. We want to find the extreme values of I(x,y) given that the point (x,y) also lies on the circle x 2 + y2 ~ I. Which lines intersecting the circle lie farthest from the origin? The lines taogent to the circle are farthest. At the points oftaogency, aoy vector nonna! to the line is normal to the circle, so the gradient VI ~ 3i + 4j is a multiple (A ~ ±5/2) of the gradient Vg ~ 2xi + 2yj. At the point (3/5, 4/5), for example, VI = 3i

+

4j,

Vg

~

6. + 8. S' SJ,

aod

5

VI = "2Vg.

•

14.8 Lagrange Multipliers

817

Lagrange Multipliers with Two Constraints Many problems require us to find the extreme values of a differentiable function I(x, y, z) whose variables are subject to two constraints. If the constraints are and

c

\

and gl and g2 are differentiable, with Vg 1 not parallel to Vg2, we find the constrained local maxima and minima of I by introducing two Lagrange multipliers A and JL (mu, pronounced "mew"). That is, we locate the points P(;x, y , z) where I takes on its constrained extreme values by finding the values of x, y , z, A, and JL that simultaneously satisfy the equations

gl = 0

FIGURE 14.55 The vectors Vg 1 and Vg2 lie in a plane perpendicular to the curve C because Vg 1 is normal to the surface gl = 0 and Vg2 is normal to the surface

g2 = O.

(2)

°

°

Equations (2) have a nice geometric interpretation. The surfaces gl = and g2 = (usually) intersect in a smooth curve, say C (Figure 14.55). Along this curve we seek the points where I has local maximum and minimum values relative to its other values on the curve. These are the points where VI is normal to C, as we saw in Theorem 12. But Vg 1 and Vg2 are also normal to C at these points because C lies in the surfaces gl = and g2 = 0. Therefore, VI lies in the plane determined by Vg 1 and Vg2, which means that VI = AVgl + JL Vg2 for some A and JL . Since the points we seek also lie in both surfaces, their coordinates must satisfy the equations gl(X, y, z) = and g2(X,y, z) = 0, which are the remaining requirements in Equations (2).

°

°

EXAMPLE 5 The plane x + y + z = 1 cuts the cylinder x 2 + y2 = 1 in an ellipse (Figure 14.56). Find the points on the ellipse that lie closest to and farthest from the origin. Solution

We find the extreme values of

I(x,y, z)

=

x 2 + y2

+ z2

(the square of the distance from (x, y , z) to the origin) subject to the constraints

gl(X, y , Z)

=

x 2 + y2 - 1

g2(X, y, z)

=

x

=

°

(3)

+ Y + z - 1 = 0.

(4)

The gradient equation in Equations (2) then gives

= AVg 1 + JLVg2 2xi + 2yj + 2zk = A(2xi + 2yj) + JLO + j + k) 2xi + 2yj + 2zk = (2Ax + JL)i + (2Ay + JL)j + JLk VI

or 2x = 2Ax FIGURE 14.56 On the ellipse where the plane and cylinder meet, we find the points closest to and farthest from the origin. (Example 5).

+ JL,

2y

=

2Ay

+ JL,

2z = JL .

(5)

The scalar equations in Equations (5) yield

+ 2z~(1 - A)x = z, 2Ay + 2z~(1 - A)y = z.

2x = 2Ax

2y

=

(6)

°

Equations (6) are satisfied simultaneously if either A = 1 and z = or A*-1 and x = y = z/(1 - A). If z = 0, then solving Equations (3) and (4) simultaneously to find the corresponding points on the ellipse gives the two points (1, 0, 0) and (0, 1,0). This makes sense when you look at Figure 14.56.

818

Chapter 14: Partial Derivatives If x = y, then Equations (3) and (4) give X2 +X 2

x+x+z- 1=0

-1=0 2>;2 = I

z=I-2>;

x=±v'Z 2

z= I 'f

v'Z.

The corresponding points on the ellipse are PI

=

('?, '?,

1-

v'Z)

and

Here we need to be careful, however. Although PI and P2 both give local maxima of f on the ellipse, P2 is farther from the origin than PI. The points on the ellipse closest to the origin are (I, 0, 0) and (0, 1,0). The point on the ellipse farthest from the origin is P2 • •

Exercises 14.8 Two Independent Variables with One Constraint 1. Extrema on an ellipse Find the points 00 the ellipse x 2 + 2y2 = I where f(x, y) = xy has its extreme values. 2. Extrema on a circle Find the extreme values of f(x, y) = xy subject to the constraint g(x, y) = x 2 + y2 - 10 = 0. 3. Maximum on a line Find the maximum value of f(x, y) = 49 - x 2 - y2 00 the line x + 3y = 10.

4. Extrema on a line Find the local extreme values of f(x, y) 00 the line x + y = 3.

= x7

S. Constrained minimum nearest the origin.

Find the points on the curve xy2 = 54

6. Constrained minimum

Find the points

00

the curve x7 = 2

nearest the origin. 7. Use the method of Lagrange multipliers to find L

Minimum on a hyperbola The minimum value of x subject to the coostraints xy = 16, x > 0, Y > 0

+ y,

b. Muimum on a line The maximum value ofxy, subject to the constraint x + y = 16.

Comment on the geometry of each solution. 8. EItrema on a curve Find the points on the curve x + xy + y2 = I in the xy-plane that are nearest to aod farthest from the origio. 2

9. Minimum .urface area with ("ned volume Find the dimensioos of the closed right circular cylindrical cao of smallest surface area whose volume is 16'7T cm.3• 10. Cylinder in a .phere Find the radius aod height of the open right cireolar cylinder oflargest surface area that cao be inscribed in a sphere of radius a. What is the largest surface area? 11. Rectaogle of greate.t area in an ellipse Use the method of Lagrange multipliers to f"md the dimensions of the rectsng1e of greatest area that can be inscribed in the ellipse x 2/16 + y2/9 = I with sides parallel to the courdinate ases.

ll. Rectaogle of longest perimeter in an ellipse Find the dimensioos of the rectangle oflargest perimeter that cao be inscribed in

the ellipse x 2/a 2 + y2/b 2 = I with sides parallel to the coordinate ases. What is the largest perimeter? 13. Extrema on a circle Find the maximum aod minimum values of x 2 + y2 subject to the coostraint x 2 - 2x + y2 - 4y = O. 14. Extrema on a circle Find the maximum and minimum values of 3x - y + 6 sui!ject to the constraint x 2 + y2 = 4. 15. Ant on a metal plate TIre temperature at a point (x, y) on a metal plate is T(x, y) = 4x 2 - 4xy + y2. An aot on the plate wa1ks arouod the circle of radius 5 centered at the origio. What are the highest aod lowest temperatures encoontered by the aot? 16. Cheape.t storage taok Yoor f"mn has been asked to design a storage tauk for liquid petroleum gas. The customer's specifications call for a cylindrical tank with hemispherical ends, aod the tank is to hold 8000 m' of gas. The customer also wants to use the smallest arooont of materisl possible in building the tank. What radius and height do you recommend for the cylindrical portion of the tank?

Thllle Independent Variables with One Cons!ra;nt 17. Minimnm distance to a point Find the point x + 2y + 3z = 13 closest to the point (I, I, I).

00

the plane

18. Maximnm distaoce to a point Find the point on the sphere x 2 + y2 + z2 = 4 farthest from the point (I, -I, I). 19. Minimnm di.laoce to the origin

from the surface x 2

Find the minimum distance

z2 = 1 to the origin.

- y2 -

20. Minimnm distance to the origin z = xy + 1 nearest the origin.

Find the point on the surface

21. Minimum di.laoce to the origin Find the points on the surface z2 = xy + 4 closest to the origin. 22. Minimnm distaoce to the origin face xyz = I closest to the origio.

Find the point(s) on the sur-

23. Extrema on a sphere Find the maximoro aod minimum values of

f(x,y,z) on the sphere

x2

=

+ y2 + z2

x - 2y = 30.

+ 5z

819

14.8 Lagrange Multipliers 24. Extrema on a sphere Find the points on the sphere x 2 + y2 + z2 ~ 25 where I(x,y, z) ~ x + 2y + 3z has its

maximum and minimum values.

34. Minimize the function I(x,y,z) ~ x 2 + y2 + z2 subject to the constraints x + 2y + 3z ~ 6 and x + 3y + 9z ~ 9.

25. Minjmjzing a sum of squares Find three real numbers whose sum. is 9 and the sum of whose squares is as small as possible.

35. Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes y + 2z ~ 12 and x + y ~ 6.

26. Maximizing a product Find the largest product the positive numbersx,y,andz can have if x + y + z2 = 16.

36. Muimum value on line of intersection Find the maximum

27. Rectangular bOJ[ of largest volume in a sphere Find the di· mensions of the closed rectangular box with maximum volume that can be inseribed in the unit sphere.

37. EItrema on a curve of intersection Find the extreme values of

28.

])OJ[ with vertex on a

plane Find the volume of the largest closed rectangular box in the fIrSt oclaot having three faces in the coordinareplanesandavertexontheplanex/a + y/b + z/c ~ I,where a> O,b > o,andc > O.

29. Hottest point on a space probe A space probe in the shape of the ellipsoid 4x 2 + y2 + 4z2 ~ 16 enters Earth's atmosphere and its surface begins to heat After I hour, the temperature at the point (x,y, z) on the probe's surface is

T(x,y,z) ~

ax 2 + 4yz -

16z + 600.

Find the hotrest point on the probe's surface.

30. Extreme temperatores on a sphere Suppose that the Celsius temperature at the point (x, y, z) on the sphere x 2 + y2 + z2 ~ I is T ~ 40Oxyz2 . Locare the highest and lowest temperatures on the sphere. 31. Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts x and y of two capital goods G1 and G2 is sometimes measured by a function U(x, y). For example, Gl and (h might be two chemicals a pharmaceutical coropany needs to have on hand and U(x, y) the gain from manufactoring a product whose synthesis requires different amounts of the chemicals depending on the process used. If Gl costs a dollars per kilogram, (h costs b dollars per kilogram, and the total amount allocared for the purchase of G 1 and G2 together is c dollars, then the company's managers want to maximize U(x, y) given that ax + by ~ c. Thus, they need to solve a typical Lagrange multiplier problem. Suppose that

U(x,y) ~ xy and that the equation ax

+

by

2x + Y

~

~

+ 2x

c simplifies to

30.

Find the maximum value of U and the corresponding values of x and y subject to this latter constraint

32. Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you waot to place it where the magnetic field of the plauet is weakest. The plauet is spherical, with a radius of 6 units. Based on a coordinare system whose origin is at the center of the plauet, the streugth of the magnetic field is given by M(x, y, z) ~ 6x - y2 + xz + 60. Where should you locare the radio relescope?

Extreme value. Subject to Two Constraints

33. Maximize the function I(x,y, z) ~ x 2 + 2y - z2 subject to the constraints 2x - y ~ 0 aod y + z ~ O.

value that I(x, y, z) ~ x 2 + 2y - z2 can have on the line of intersection of the planes 2x - Y ~ 0 and y + z ~ O.

I(x, y, z) ~ x'yz

+

1 on the intersection of the plane z ~ I with

the sphere x 2 + y2 + z2 = 10. 38. a. Muimum on line of intersection Find the maximum value of w ~ xyz on the line of intersection of the two planes x + y + z ~ 40 and x + y - z ~ O. b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of w.

39. EItrema on a circle ofintenection Find the extreme values of the function/(x,y, z) ~ xy + z2 on the circle in which the plane y - x = 0 intersects the sphere x 2 + y2 + z2 = 4.

40. Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane 2y the cone z2 = 4x 2 + 4y2.

+ 4z

~

5 and

Theory and Examples 41. The condition VI = AVgisnotsnllicient Although VI ~ AVg is a necessary condition for the occurrence of an extreme value of I(x, y) subject to the conditions g(x, y) ~ 0 and Vg #' 0, it does

not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to fmd a maximum value of I(x, y) ~ x + y subject to the constraint that xy ~ 16. The method will identifY the two points (4, 4) and (-4, -4) as candidares for the location of extreme values. Yet the sum (x + y) has no maximUD2 value on the hyperbola xy ~ 16. The farther you go from the origin on this hyperbola in the first quadrant, the larger the sUD2/(x,y) ~ x + ybecomes.

42. A lea.t squa...s plane The plaue z ~ Ax "fitted" to the following points (x" Yb z,):

(0,0,0),

(0, I, I),

+ By + C

is to be

(1,0, -I).

(I, I, I),

Find the values of A, B, and C that minimize

f-1• (Ax, + By, + C -

z,)"

the sum of the squares of the deviations. 43. a. Muimum on a sphere Show that the maximum value of a 2b 2c 2 on a sphere of radius r centered at the origin of a Cartesian abc-coordinare systero is (r 2/3)3 b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers a, b, and c,

a (abc ) 1/3 < -

+b +c. 3

'

that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean. 44. Sum of products Let at. a2, ... , an be n positive numbers. Find the maximum of l:?=1 a,x, subject to the constraint l:1=1 x? = 1.

820



46. Minimize J(x, y, z) = :x:yz subject to the constraints x 2 + y2 - 1 = 0 and x - z = O. 47. Maximize J(x, y, z) = x 2 + y2 + z2 sui!ject to the constraints 2y + 4z - 5 = 0 and4x 2 + 4y2 - z2 = O.

In Exercises 45--50, use a CAS to perform the following steps implementing tire method of Lagrange multipliers for fmding constrained extrema: L

Formthefunctionh = J - AlgI - A2g2,whereJ is the function to optimize subject to tire constraints gl = 0 and g2 = O.

48. Minimize J(x,y,z) = x 2 + y2 + z2 subject to the constraints x 2 - :x:y + y2 _ z2 - I = 0 andx 2 + y2 - I = O. 49. Minimize J(x,y,z, w) = x 2 + y2 + z2 +w 2 subject to the

h. Determine all the f!rst partial derivatives of h, including the partials with respect to Al andA2, and set them equal to O.

2:x:-y+z-w-I=O

constraints

and

x+y-z+

w - I = O.

c. Solve the system of equations found in part (b) for all the unknowns, including Al and A2.

50. Determine the distance from the line y = x + I to the parabola y2 = x. (Hint: Let (x, y) be a point on tire line and (w, z) a point on the parabola. You want to minimize (x - w)' + (y - z)'.)

eI. Evaluate J at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the

exercise. 45. Minimize J(x,y,z) = :x:y + yz sui!ject to x 2 + y2 - 2 = 0 andx 2 + z2 - 2 = O.

14.9

rl

the

constraints

_~~y~Lo_(_s_F_o_ rm_u_La__ ro_r_Tw _ o__ Va_n_'a_b_Le_s______________-----------In this section we use Taylor's formula to derive the Second Derivative Test for local extreme values (Section 14.7) and the error formula for linearizations of functions of two independent variables (Section 14.6). The use of Taylor's formula in these derivations leads to an extension of the formula that provides polynomial approximations of all orders for functions of two independent variables.

Derivation of the Second Derivative Test t

~

1 S(a

+ h,b + k)

Parametrized

Let I(x, y) have continuous partial derivatives in an open region R containing a point P( a, b) where Ix = Iy = 0 (Figure 14.57). Let h and k be increments small enough to put the

point S(a + h, b segment PS as

+ k) and the line segment joining it to P inside R. We parametrize the x=a+th,

If F(t) = I(a

O:5t:51.

y=b+tk,

+ th, b + tk), the Chain Rule gives

,

t= 0

F (t)

P(a, b)

=

ax

Ix dt

dy

+ Iy dt

=

hlx

+ k/y.

Part of open region R

FIGURE 14.57 We begin the derivation of the Second Derivative Test atp(a, b) by parametrizing a typical line segment from P to a point S nearby.

Since Ix and Iy are differentiable (they have continuous partial derivatives), F' is a differentiable function of t and

F"

=

aF' ax + aF'dy axdt aydt

= h 21xx

=

~(hf + kf)'h + ~(hf + kf)'k ax x y ay x y f", ~ fyx

+ 2hklxy + elyy .

Since F and F' are continuous on [0, I] and F' is differentiable on (0, 1), we can apply Taylor's formula with n = 2 and a = 0 to obtain

F(I)

=

F(O) + F'(O)(1 - 0) + F"(c)

F(I)

=

F(O) + F'(O)

t

+ F"(c)

(1

2

0)' (1)

14.9 Taylor's Formula for Two Variables

821

for some c between 0 and 1. Writing Equation (1) in terms of I gives I(a

+ h, b + k)

= I(a, b)

+ hlx(a, b) +

kly(a, b)

+ ~ (h 2lxx + 2hkl"1 + k 2lyy) I

.

(2)

(a+ch, b+cl)

Since Ix(a, b) = I,(a, b) = 0, this reduces to I(a

I(a

+ h,b + k)

- I(a,b) =

~(h2Ixx + 2hkl"1 + k 2lyy) I(a+ch, b+ck).

(3)

The presence of an extremum of I at (a, b) is determined by the sigo of k) - I(a, b). By Equation (3), !bis is the same as the sigo of

+ h, b +

Q(c) = (h 2lxx

+ 2hkl"1 + k2Iyy)l(a+do.b+ck)'

Now, if Q(O) oF 0, the sigo of Q(c) will be the same as the sigo of Q(O) for sufficiently small values of h and k. We can predict the sigo of

Q(O)

2

= h Ixx(a, b)

+ 2hkl"1(a, b) + k 2Iyy(a, b)

(4)

from the signs of Ixx and fulyy - 1"12 at (a, b). Multiply both sides of Equation (4) by Ixx and rearrange the right-band side to get (5)

From Equation (5) we see that

1. If Ixx < 0 and Ixxlyy - 1"12 > 0 at (a, b), then Q(O) < 0 for all sufficiently small nonzero values of h and Ie, and I has a local maximum value at (a, b). 2. If Ixx > 0 and Ixxlyy - 1"12 > 0 at (a, b), then Q(O) > 0 for all sufficiently small nonzero values of h and Ie, and I has a local minimum value at (a, b). 3. If Ixxlyy - 1"12 < 0 at (a, b), there are combinations of arbitrarily small nonzero values of h and k for which Q( 0) > 0, and other values for which Q( 0) < O. Arbitrarily close to the point Po(a, b, I(a, b)) on the surface z = I(x,y) there are points above Po and points below Po, so I has a saddle point at (a, b). 4. If I xxlyy - 1"12 = 0, another test is needed. The possibility that Q(O) equals zero prevents us from drawing conclusions about the sigo of Q(c).

The Error Formula for Linear Approximations We want to show that the difference E(x, y), between the values of a function I(x, y), and its linearization L(x, y) at (xo, Yo) satisfies the inequality IE(x,yli:5

~M(lx -

xol

+ Iy

- yoll'·

The function I is assumed to have continuous second partial derivatives throughout an open set containing a closed rectangular region R centered at (xo, yo). The number M is an upperbouod forl/xxl, I/yyl, and 1/"11 onR. The inequality we want comes from Equation (2). We substitute Xo and Yo for a and b, and x - Xo and y - Yo for h and k, respectively, and rearrange the result as I(x,y) = l(xo,Yo)

+ fx{xO,yo)(x

- xo)

+ Iy(xo,yo)(y

- Yo)

linearization L(x, y) 1

+ -2 (x

- Xol'lxx

+ 2(x

- xo)(y - yo)I"1

+ (y

error E(x, y)

- Yol'lyy)l(Xo + c (X-Xo.Yo ) + c (Y-Yo » .

822


TIris equation reveals that

IEI:5 !Ox - xo1 2 1/",1 + 21x - xolly - yoll/.,,1 + Iy -

YOI21/yy!).

Hence, if Mis an upper bound for the values ofl I", I, 1/.,,1, andl/yyl onR,

IEI:5 !Ox -

xol 2M

!M; ~ 4x - "- - xy 2

",- 0

~ 6 - 2y,

(4)

x- O

The volume of the entire solid is therefore

Volum, ~

y

,

,

\ 1.'.' ,·0

A(y) _

A(y) dy ~

1',-0·'

(6 - 2y) dy ~ [6y - y']: ~ 5,

in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing

(4-x-)')d.t

FIGURE 15.5 To obtaW. the cross-sectional. area A(y), we hold y fDWd and integrate with respect to x.

1,-0,·,

Volume =

11102.(4 - x -

y) duly.

The expression on the right says we can find the volume by intcgIating 4 - x - y with respect tox from x - 0 tax - 2 as in Equation (4) and integrating the result with respect to y from y - 0 to Y - 1. In this. iterated integral, the order of integration is fll'S1: x and then y, the reverse of the order in Equation (3). What do these two volume calculations with iterated integrals have to do with the double integral

jj(4 - x - y) dA R H:rsWRICAL BIOGRAPHY

Guido Fubini (1879-1943)

over the rectangle R: 0 ~ x ~ 2, 0 ~ y ~ I? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integml measures the volume of the smne region as the two iterated integrals. A theorem published in 1907 by Guido Fubini says that the double integral of any continuous functioo aver a rectangle can be calculated as an itemted integral in either order of integration. (Fubini proved his theorem in greater generality, but this is what it says in our setting.)

THEOREM l-Fubini's Theorem ;dy ~ 1'1' /(x,y) dy

d>;,

R

Fubini's Theorem says that double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini's Theorem also says that we may calculate the double integral by integrating in either order, a genuine convenience. When we calculate a volume by slicing, we may use either planes perpendicular to the x-axis or planes perpendicular to the y-axis.

EXAMPLE 1

Calculate .O'R/(x,y)

f(x,y) ~ 100 -

6x'y

dA for and

R:

O::Sx::S2,

-1::Sy::sl.

840

Chapter 15: Multiple Integrals

,

.: -100 -6.1:7

Solution

Figw-e 15.6 displays the volume beneath the surface. By Fubini's Theorem,

~ 1\200 -, - 16y) dy ~

[201ly - 8y']:,

~ 400.

Reversing the order of integration gives the same answer: x

R

,

1 ,

{' {\100 -

Jo J-l

6x'y)dydx

~ ('[lOlly - 3'V~::, dx

Jo

~ 1,\(100 -

FIGURE 15.6 The double integral f(x, y) cIA. gives the volume under this surface aver the n:ctangular region R (Example 1).

DR

3x') - (-100 - 3x')] dx

-£2

200 dx _ 400.

EXAM PLE 2

Find the volume of the region bounded above by the ellipitical paraboloid

z - 10 + x2 + 3y 2 andbelowbytherectangleR:0::S; x S 1,0

Sy::s;

2.

Solution The surface and volume a:re shown in Figure 15.7. The volume is given by the double integral

R

ff + -l' [lOy

FIGURE 15.7 The double inWgral f(x, y) gives the volume under this smfaoc aver the n:ctangular region R

DR

•

(10

V=

dA

x 2 + 3y2)dA =

11 1

2

(10

+ x 2 + 3y2) dydx

R

+ x'y + y'~:: dx

(Example 2).

=

Jor (20 + 2x

2

+ 8)dx = [2Ox + ~X3 + 8x]1 3

•

= 86.

3

0

Exercises 15.1 ~~tingH8~In~~

11.

In Exercises 1-12, evaluate the iterated integral.

1.

J.'110f'

3.1:1: la la 3

5.

2. ].' {' (x - y)dydr.

2xydydr.

(x

+y +

l

(4 -

l)dxdy

y~dydx

Y 7. ].0 '].' 0 1 +xydxdy

9. ].0 "'11[b' e2>:+y dy dx

01-1 x' -+ Y') 4. ]. '].' (1 - dr.dy o 0

12·1 la" lfr

y sin;t dx dy

(sin;t

+ cosy)dxdy

Evaluating Double Integra~ over Rectangles In Exercises 13-20, evaluate the double integral. aver t1u: given regionR.

2

13.

f' (x'y 01-2

6. ].'

1-,'],."'

2xy) dy dr.

jj(6y2 -2x)cIA.,

R:

O:!;';;t:!;'; I,

0

:!;';y:!;'; 2

R

14.

jj (v:)cIA., R

15.

R:

jjxycosYdA, R: R

0 S;t S 4,

1 Sy S 2

Y

-IS;tSI,

OSys",.

15.2 Double Integrals over General Regions

16.

JJ JJ JJ

y sin (x

+ y) dA,

R:

-1T':S

x :s 0,

O:s y ::s

1T'

R

17.

.'-Y dA,

0 '" x '" In 2,

R:

0 '" Y '" In 2

R

18.

xyexi' dA,

0 '" x '" 2, 0 '" Y '" I

R:

R

19.

ff

,xy3

}} x + I

dA,

0 '" x '" I,

R:

0 '" Y '" 2

R

20.

JJ X'y: +

I dA,

R:

0 '" x '" I,

0 '" Y '" I

R

In Exercises 21 and 22, integrate f over the given region.

21. Square f(x,y) = l/(xy) over the square I '" x'" 2, I'" Y '" 2 22. Rectangle f(x,y) = y cosxy over the rectangle 0 '" x '" 'fr,

O"'y"'l

841

Volume Beneath a Surface 23. Find the volume of the region bounded above by the paraboloid z = x' + y' and below by the square R: -I '" x '" I, -I "'y"'i. 24. Find the volume of the region bounded above by the ellipiticai paraboloid z = 16 - x' - y' and below by the square R: 0 :s x :s 2,0 :s y :s 2. 25. Find the volume of the region bounded above by the plane z = 2 - x - y and below by the square R: 0 '" x '" 1,

o '" Y '" i. 26. Find the volume of the region bounded above by the plane z = y/2 and below by the rectangle R: 0 '" x '" 4,0 '" Y '" 2. 27. Find the volume of the region bounded above by the surfaco z = 2 sin x cos y and below by the rectangle R: 0 '" x '" 'fr/2, o '" Y '" 'fr/4. 28. Find the volume of the region bounded above by the surfaco z = 4 - y' and below by the rectangle R: 0 '" x '" 1, o '" Y '" 2.

z= f(x,y)

Double Integrals over General Regions

15.2

In this section we detme and evaluate double integrals over bounded regions in the plane which are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that of determining the limits of integration. Since the region of integration may have boundaries other than line segments parallel to the coordinate axes, the limits of integration often involve variables, not just constants.

Double Integrals over Bounded, Nonrectangular Regions

"

0, the width and height of each enclosed rectangle goes to zero and their number goes to intmity. If f(x, y) is a continuous function, then these Riemann sums converge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of f(x,y) over R:

842


The nature of the boundary of R introduces issues not found in integrals over an interval. When R has a curved boundary, the n rectangles of a partition lie inside R but do not cover all of R. In order for a partition to approximate R well, the parts of R covered by small rectangles lying partly outside R must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, ellipses, and from continuous graphs over an interval, joined end to end. A curve with a "fractal" type of shape would be problematic, but such curves arise rarely in most applications. A careful discussion of which type of regions R can be used for computing double integrals is left to a more advanced text.

Volumes If I(x, y) is positive and continuous over R, we define the volume of the solid region betweenR and the surface z = I(x,y) to be JJRI(x,y) dA, as before (Figure 15.9). If R is a region like the one shown in the xy-plane in Figure 15.10, bounded "above" and ''below'' by the curves y = g2(X) and y = gl (x) and on the sides by the lines x = a, x = b, we may again calculate the volume by the method of slicing. We first calculate the cross-sectional area A(x) =

l

y =g2(X)

I(x,y) dy

y=gl(X)

and then integrate A(x) from x

V=

l

=

a to x = b to get the volume as an iterated integral:

bA(x)dx= lblg2(X) I(x,y)dydx.

a

a

(1)

gl(X)

z z =f(x,y)

o

y

x

Volume

=

lim !fiXk, Yk) aAk =

If f(x, y) dA R

FIGURE 15.9 We define the volumes of solids with curved bases as a limit of approximating rectangular boxes.

FIGURE 15.10 The area of the vertical slice shown here is A(x). To calculate the volume of the solid, we integrate this area from x = a to x = b:

l

a

b

A(x)dx =

lblg2(X) a

g,(x)

f(x,y)dydx.


843

Similarly, if R is a region like the one shown in Figure 15.11, bounded by the curves x = h2(y) and x = hl(y) and the lines y = c and y = d, then the volume calculated by slicing is given by the iterated integral

z z =f(x,y)

x

Il d

Volume =

y

hb )

(2)

I(x,y) dxdy.

h,(y)

c

That the iterated integrals in Equations (1) and (2) both give the volume that we defined to be the double integral of lover R is a consequence of the following stronger form of Fubini 's Theorem.

FIGURE 15.11 shown here is

I

c

The volume of the solid

dA(y)dy = Idlh2(v) j(x,y)dxdy. h,(v) c

TH EOREM 2-Fubini's Theorem (Stronger Form)

Let I(x, y) be continuous on a

regionR .

1. If R is defined by a on [a , b] , then

:5

x

:5

b, gl (x)

Y

:5

:5

r iirrI(x, y) dA = laig,(x)

For a given solid, Theorem 2 says we can calculate the volume as in Figure 15.10, or in the way shown here. Both calculations have the same result.

g2

b

g2(X) , with gl and g2 continuous

I(x, y) dy dx.

(x)

R

2. If R is defined by c on [c, d], then

:5

Y

:5

d, hl(y)

iJif I(x,y) dA = I

x

:5

:5

h2(y), with hI and h2 continuous

dlh2(y)

I(x,y) dxdy. hICy)

c

R

EXAMPLE 1 Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = I(x, y) = 3 - x - y.

Solution See Figure 15.12. For any x between 0 and l,y may vary from y = 0 to y = x (Figure 15.12b). Hence, V=

=

11 1

x

o

(3 - x - y) dy dx

11 [

=

0

1

1 (

o

3y - xy - -y2]Y=X dx 2 y =O

0

3

3x - ~2) dx = 2

[3~ 2

-

2

:!.-3 ]X=I 2

= 1.

x=o

When the order of integration is reversed (Figure 15.12c), the integral for the volume is

V =

= =

11 [ 1111 11 (3 - ~ 11 (% - ~ 2) [% o

(3 - x - y) dx dy

3x - ~2 - xy ]X=I dy

=

0

y

y - 3y

4y

+

Y

x=y

+ Y22 + y2) dy

dy =

The two integrals are equal, as they should be.

y - 2y2

+ ~3 I:~ =

1.

•

844

Chapter 15: Multiple Integrals z y

(3,0,0)

FIGURE 15.12 (a) Prism with a triangular base in the xy-plane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may integrate first with respect to y and then with respect to x, or the other way around (Example I). (b) Integration limits of

l

x=11Y=x f(x, y) dy dx. x=o y=o

If we integrate first with respect to y, we integrate along a vertical line through R and then integrate from left to right to include all the vertical lines in R. (c) Integration limits of

l

Y=11X=1 f(x,y) dx dy. y=o X=Y

If we integrate first with respect to x, we integrate along a horizontal line through R and then integrate from bottom to top to include all the horizontal lines in R.

Although Fubini's Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen.

EXAMPLE 2

Calculate

11 si~x

dA,

R

where R is the triangle in the xy-plane bounded by the x-axis, the line y = x, and the line x = 1.

15.2 y x

Double Integrals over General Regions

845

Solution The region of integration is shown in Figure 15.13. Ifwe integrate IlISt with respect to y and then with respect to x, we Imd

=1

l' (lS~XdY)dx l' &s~x IJdx =

= -cos (I) ~~------~--~X

o

+

I

RJ

=

1'SinXdx

0.46.

If we reverse the order of integration and attempt to calculate

J.

o

FIGURE 15.13 The region of integration in Example 2.

111. y

s~x dx dy,

we run into a problem because J«sinx)/x) dx cannot be expressed in terms of elementary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you IlISt choose doesn't work, try the other. Some_ times neither order will work, and then we need to use numerical approximations.

y

Finding Limits of Integration We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works. Using Vertical Cross-sections When faced with evaluating lfR f(x,y)dA, integrating IlISt with respect to y and then with respect to x, do the following three steps:

(a)

y

/

Leaves at y~~ Enters at y=l-x

1.

Sketch. Sketch the region of integration and label the bounding curves (Figure 15.14a).

2.

Find the y-limits of integration. Imagine a vertical line L cutting through R in the direction of increasing y. Mark the y-values where L enters and leaves. These are the y-limits of integration and are usually functions of x (instead of constants) (Figure 15.14b).

o

3. Find the x-limits of integration. Choose x-limits that include all the vertical lines x

throughR. The integral shown here (see Figure 15.14c) is

ff

(b)

f(x,y)dA

=

E~IL~~ f(x,y)dydx.

R

y

Leaves at

/y~~ Enters at

y=l-x

Using Horizontal Cross-sections To evaluate the same double integral as an iterated integral with the order of integration reversed, use horizontal lines instead of vertical lines in Steps 2 and 3 (see Figure 15.15). The integral is

JJrr f(x,y)dA R

=

J.lkr~ f(x,y)dxdy. 0

y

~t---~--~~--~x

l

x

/

Smallest x

Largest x

isx=O

isx=l

Largesty Y isy = 1 ~1

Enters at

x=l-y

(c)

FIGURE 15.14 Finding the limits of integration when integrating first with respect to y and then with respect to x.

Y --r---~~~--~

Smallest y isy ~ 0

"

~ Leaves at x

~

v'l=Y'

--;;ct-------~~--~x

o

FIGURE 15.15 Finding lhe limits of integration when integrating first with respect to x and then with respect to y.

846


EXAMPLE 3

Sketch the region of integration for the integral

1212x(4x + 2)dydx and write an equivalent integral with the order of integration reversed.

Solution The region of integration is given by the inequalities x 2 :5 y :5 2x and o :5 x :5 2. It is therefore the region bounded by the curves y = x 2 and y = 2x between x = 0 and x = 2 (Figure 15.16a). To find limits for integrating in the reverse order, we imagine a horizontal line passing from left to right through the region. It enters at x = y/2 and leaves at x = To include all such lines, we lety run from y = 0 to Y = 4 (Figure 15.16b). The integral is

...;y.

y

r

rv,.(4x + 2) dx dy. 101Y/2

•

The common value of these integrals is 8.

Properties of Double Integrals ~~----~----~X

Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications.

If f(x, y) and g(x, y) are continuous on the bounded region R, then the following properties hold

FIGURE 15.16 Region of integration for Example 3.

1. constantMUltiPle:ff cf(x,y)dA = cff f(x,y)dA R

(any number c)

R

2. Sum and Difference: ffU(X,Y) ± g(x,y)) dA

ff f(x,y) dA ± ff g(x,y) dA

=

R

R

R

3. Domination: y

(a)

if

ff f(x,y) dA ;" 0

f(x,y) ;" 0 onR

R

(b)

ff f(x,y)dA;" ff g(x,y)dA R

o

II f(x, R

y) dA

II

II

~ f(x, y) dA + f(x, y) dA RI

f(x,y) ;" g(x,y) onR

R

4. Additivity: ff f(x,y) dA --~--*-----~---------->x

if

R

=

ff f(x,y) dA + ff f(x,y) dA Rl

•

if R is the union of two nonoverlapping regions R, and R2

R2

FIGURE 15.17 The Additivity Property for rectangular regioos holds for regions bounded by continuous curves.

Property 4 assumes that the region of integration R is decomposed into nonoverlap· ping regions R, and R2 with boundaries consisting of a finite number of line segments or smooth curves. Figure 15.17 illustrates an example of this property.


847

The idea behind these properties is that integrals behave like swns. If the function f(x, y) is replaced by its constant multiple cf(x, y), then a Riemann swn for f n

Sn =

z

L f(xk, Yk) flAk k=!

is replaced by a Riemann swn for cf n

L

k=!

c

=

L

k=!

f(xk, Yk) flAk

",, \

\... I I

(a)

y

y

=

4x- 2

2

~r-----~~------L-~x

(b)

FIGURE 15.18 (a) The solid "wedgelike" region whose volume is found in Example 4. (b) The region of integration R showing the order dx dy.

Find the volwne of the wedgelike solid that lies beneath the surface z = 16 - x 2 - y2 and above the region R bounded by the curve Y = 2Vx, the line Y = 4x - 2, and the x-axis. Solution Figure 15.18a shows the surface and the "wedgelike" solid whose volwne we want to calculate. Figure 15.18b shows the region of integration in thexy-plane. If we integrate in the order dy ax (first with respect to y and then with respect to x), two integrations will be required because y varies from y = 0 to y = 2vX for 0 :::; x :::; 0.5, and then varies from y = 4x - 2 to Y = 2vX for 0.5 :::; x :::; 1. So we choose to integrate in the order ax dy, which requires only one double integral whose limits of integration are indicated in Figure 15 .18b. The volwne is then calculated as the iterated integral:

11

(16 - x 2 - y2) dA

= =

1 1

2 1(Y+2)!4

o

2 [

o

(16 - x 2 - y2)dxdy

y./4 x3 ]X=(Y+2)!4 16x - - - xy2 dx 3 x=y./4

t [

= Jo

4(y

_ [191Y

+

2) -

63y2

- ---z4 + ~ -

(y + 2)3 3 . 64 145y 3 ~

-

(y

+ 2)y2 - 4y2

4

49y4 768

+

y5 20

+

y6 3' 64

+ o

-

Sketching Regions of Integration In Exercises 1- 8, sketch the described regions of integration. 1. 0 :::; x :::; 3, 0:::; y ~ 2x

x-I ~ y ~ x 2

7. 0 ~ y ~ 1,

0 ~ x ~ sin- I Y

8. 0 ~ y ~ 8,

i

3. -2 ~ y ~ 2, y2 ~ X ~ 4 ~

y

~

1, y

~

x

~

2y

y

~ x ~ Y 1/3

y4]

+ 4 dy

7 ]2 _ 20803

Y 1344

Exercises 15.2

4. 0

cSn .

EXAMPLE 4

R

2. -1 ~ x ~ 2,

=

Taking limits as n ~ 00 shows that c limn-><Xl Sn = C ffR f dA and limn-><Xl cSn = fJR C f dA are equal. It follows that the constant multiple property carries over from swns to double integrals. Y The other properties are also easy to verify for Riemann sums, and carryover to double integrals for the same reason. While this discussion gives the idea, an actual proof that these properties hold requires a more careful analysis of how Riemann swns converge.

.J......., _

x

n

c f(xk, Yk) flAk

1680 ~ 12.4.

•

848


Finding Limits of Integration In Exercises 1}-18, write an iterated integral for lfR dA over the described region R using ( a) vertical cross-sections, (b) horizontal cross-

sections.

29.1:1"

2 dp dv

(the pv-plane)

{'{~

8t dt dx

30. Jo Jo

(the st-plane)

10.

9.

31.1~/' {E"3 cos t du dt

y

(the tu-plane)

-1r/3io

32.

1,o

'/2~4'2' 4 2 ~dvdu 1 v

(theuv-plane)

Reversing the Order of Integration --~~------~--~X

In Exercises 33-46, sketch the region of integration and write an equivalent double integral with the order of integration reversed.

33. 12.

11.

y

35.

11 1, o

4

'2.
:,y = x, aodx =

15. 16. 17.

°

18. Bounded by y = x 2 andy = x

+2

r

20. Jo Jo 21.

1, 1vG' -VI=7 Jl Jo

45.

1,'1"

y dy dx

Jtr'r' Jo e +Y dx dy X

(' p'

25. Quadrilateral I(x, y) = x/y over the region in the flIst quadrant bounded by the lines y = x, y = 2>:, x = I, and x = 2

26. Triaogle I(x, y) = x2 + y2 over the ttiaogular region with vertices (0, 0), (I, 0), aod (0, I)

27. Triaogle I(u, v) = v - Vu over the ttiangular region cut from the flISt quadraot of the uv-plane by the line u + v = I 28. Curved region I(s, t) = e' In t over the region in the flISt quadraot of the st-plane that lies above the curve s = In t from

t=llot=2 Each of Exercises 21}-32 gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral.

3ydxdy

42.

ydxdy

1v'k'

2

44.

1,o .v'k' ~/6L1/2

1,o

6x dydx

xy 2 dydx

sinr

{V3 {,m,-"

+ y) dx dy

46. Jo

Jo

v;Jdxdy

In Exercises 47-56, sketch the region of integration, reverse the order of integration, aod evaluate the integral.

r Jxrsiny --y dy dx

48.

47. Jo

1,111 2y'j;31y'j;3 51. 1, ,/2

x 2e"" dx dy

o

50.

1,2122y2 sinxy dy dx 2j,4'X' xe ~

1,o

4 _

0

52. {' {' Jo

ex' dx dy

JWi

Y

dy dx

.r dy dx

53. {1/16 {1/2 cos (I&n-x') dx dy

Jo

23. Jo Jo 3y'e"" dx dy In Exercises 25-28, integrate 1 over the giveo region.

dy dx

dx dy

{' r·r

xy dy dx (x

J"

40. Jo Jo

{' {Inx

43.

49.

{.;nx

1-x

16x dy dx

Jo

o

finding Regions of Integration and Double Integrals In Exercises 11}-24, sketch the region of integration and evaluate the integral.

0

dx dy

I,I.il'X'

1

41.

13. Bounded by y 14.

y-2

38. Jo

{'/2 {"4x' 39. Jo

0

{1n2 {'

{' {" 37. Jo

1,2:£0

54.

),1/4

l,'h o

2

-dydx -+I

~ y4

55. Square region lfR (y - 2>:2) dA where R is the region boundedbythesquarelxl + Iyl = I 56. Triaogular region

thelinesy

= ;t,Y =

lfRXY dA where R is the region hounded by +y = 2

a,andx

Volume Beneath a Surface z

=

/(x, y)

57. Find the volume of the region bounded above by the paraboloid z = x2 + y2 and below by the ttiangle enclosed by the lines y = x,x = 0, and x + y = 2 in thexy-plane. 58. Find the volume of the solid that is hounded above by the cylinder z = x 2 aod below by the region enclosed by the parabola y = 2 - x 2 and the line y = x in the xy-plane.

15.2 Double Integrals over General Regions 59. Find the volume of the solid whose base is the region in the xyplane that is bounded by the parabola y ~ 4 - x 2 and the line y = 3x, while the top of the solid is bounded by the plane z = x + 4. 60. Find the volume of the solid in the fIrSt octant bounded by the coordinate planes, the cylinder x 2 + y2 = 4, aod the plaoe z + y = 3. 61. Find the volume of the solid in the flIst octant bounded by the coordinate planes, the plaoe x = 3, aod the parabolic cylinder z ~ 4 - y2. 62. Find the volume of the solid cot froro the fIrSt octant by the surfacez = 4 - x 2 - y.

63. Find the volume of the wedge cut from the fIrSt octaot by the cylinder z ~ 12 - 3y2 and the plaoe x + y ~ 2. 64. Find the volume of the solid cot froro the square colunm Ixl + Iyl'" I by the planes z = Oaod3x + z = 3. 65. Find the volume of the solid that is bounded 00 the froot aod back by the plaoes x ~ 2 aod x = I, 00 the sides by the cylioders y = ± I/x, aod above aod below by the plaoes z = x + I aod z ~ O. 66. Find the volume of the solid bounded 00 the froot aod back by the planes x = ±'1T/3, on the sides by the cylinders y = ±secx, above by the cylinder z ~ I + y2, aod below by the xy-plaoe. In Exercises 67 and 68, sketch the region of integratioo aod the solid whose volume is giveo by the double integral. 67.

/.J-

/'

':£'116- v'25 /.o y Y

68.

(I - tx -

2r

-v'16-

'

l

Integrals over Unbounded Regions Improper double integrals cao often be coroputed similarly to improper integrals of ooe vatiable. The flIst iteratioo of the following improper integrals is cooducted just as if they were proper integrals. One thea evaluates ao improper integral of a single vatiable by taking appropriate limits, as in Sectioo 8.7. Evaluate the improper integrals in Exercises 6')..72 as iterated integrals. 69.

1

00'1. '

I

70.

74. f(x,y) = x + 2y over the regioo R inside the circle (x - 2)' + (y - 3)' = I using the partition x = 1,3/2,2,5/2, 3 aody ~ 2, 5/2, 3, 7/2, 4 with (Xk,Yk) the center (centroid) in the kth suhrectangle (provided the suhrectaogle lies within R)

Theory and Examples

v'

75. Cin:ular sector Integrate f(x, y) ~ 4 - x 2 over the smaller 2 sector cut from the disk x + y2 S 4 by the rays 8 = '1T/6 aod 8 = '1T/2. 76. Unbounded region Integrate f(x,y) ~ 1/[(x 2 - x)(y - 1)2/3]

aver the infmite rectangle 2 ::s x

-1

89.

3\IJ -

x 2 - y2dydx

x

1:1.

' _+1 dydx lOX Y

1

2].4-Y' 0

eX)! dx dy

2~s y'

I v'x2

+ y2

dxdy

0

Area by Double Integration In this section we show how to use double integrals to calculate the areas of bounded regions in the plane, and to find the average value of a function of two variables.

Areas of Bounded Regions in the Plane If we take !(x, y) = I in the defmition of the double integral over a region R in the preceding section, the Riemarm sums reduce to

S. =

• • L !(XhY.) aA. = L aA•. i- I i- I

(1)

11ris is simply the swn of the areas of the small rectangles in the partition of R, and approximates what we would like to call the area of R. As the norm of a partition of R approaches zero, the height and width of all rectangles in the partition approach zero, and the coverage of R becomes increasingly complete (Figure 15.8). We define the area of R to be the limit lim . aA. = 1IJ1I~o~

DEFINITION

iJif dA.

(2)

R

The area of a closed, bounded plane region R is

As with the other defmitions in this chapter, the definition here applies to a greater variety of regions than does the earlier single-variable definition of area, but it agrees with the earlier defmition on regions to which they both apply. To evaluate the integral in the definition of area, we integrate the constant function !(x, y) = I over R.

EXAMPLE 1 quadrant.

Find the area of the region R bounded by Y = x and y = x 2 in the frrst

15.3

851

Area by Double Integration

Solution

y

We sketch the region (Figure IS.19), noting where the two curves intersect at the origin and (1, 1), and calcolate the area as

A = =

!oJ dyax 10f' (x -

x

2 )

=

[~]:ax

ax = [; 2-

3]'0

~

=

1 6'

ax,

FIGURE 15.19

The region in Example I.

Notice that the single-variable integral Jo' (x - x 2 ) obtained from evaluating the inside iterated integral, is the integral for the area between these two curves using the method of Section S.6. • Find the area of the regionR enclosed by the parabolay = x 2 and the line

EXAMPLE 2 y = x + 2. Solution

Ifwe divide culate the area as

y

Rinto the regions R, and R2 shown in Figure IS.20a, we may cal-

On the other hand, reversing the order of integration (Figure IS.20b) gives A =

----=~~~x o

o

x

(b)

FIGURE 15.20 Calculating this area ta1res (a) two double integrals if the rust integration is with respect to x, but (b) only one if the ftrSt integration is with respect to y (Example 2).

The average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defmed on a bounded region in the plane, the average value is the integral over the region divided by the area of the region. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of R. We are led to define the average value of an integrable function f over a region R as follows:

10fRJJ IdA.

Averagevalueof/overR = area

(3)

R

If I is the temperatore of a thin piate covering R, then the double integral of lover R divided by the area of R is the plate's average temperature. If I(x, y) is the distance from the point (x, y) to a flXed point P, then the average value of lover R is the average distance of points inR fromP.

852


EXAMPLE 3 R: 0 :5 x :5

Find the average value of 'IT,

0

:5

f(x,y)

= xcosxy

over the rectsngle

f

= sinxy + C

Y :5 1.

The value of the integral of f over R is

Solution

1"1' o

xcosxy dydx =

0

1"[

sinxy

0

]Y-l dx

xcosxydy

y=O

= l"(SinX - 0)

I

dx = -cosx

= I

+

I = 2.

•

The area of R is 7f. The average value of f over R is 2/7f.

Exercises 15.3 Area by Double Integrals In Exercises 1-12, sketch 1he region bounded by 1he given lines and curves. Then express the region's area as an iterated double integral and evaluate 1he integral.

1. The coordinate axes and 1he line x

+y

= 2

2. The lines x = O,y = 2x,andy = 4

3. The parabola x

_y2 and 1he line y

=

x

=

+2

4. The parabola x = y - y2 and 1he line y = -x 5. The curve y = eX and 1he lines y = 0, x = 0, and x = In 2 6. Thecurvesy = lox andy = 2loxand1helinex = e,in1hefrrst quadrant

Find;ng Average Values 19. Find 1he average value of f(x, y) = sin (x

a. the rectangle 0 ::s x :s

1r.

b. 1he rectangle 0 ,;; x ,;; 7f,

O:s y :s

+ y) over 1r.

0,;; y ,;; 7f/2.

20. Which do you think will be larger, 1he average value of f(x,y) = xy over 1he square 0 '" x '" 1,0 '" Y '" I, or 1he average value of f over 1he quarter circle x 2 + y2 ,;; I in 1he first quadrant? Calculate 1hern to fmd out 21. Find 1he average height of 1he paraboloid z = x 2 + y2 over 1he

square 0 :s x :s 2, 0 :s y :s 2. 22. Find 1he average value of f(x,y) = I/(xy) over 1he square 102'" x '" 2102,102'" Y '" 2102.

7. The parabolas x = y2 and x = 2y _ y2 Theory and Examples 8. The parabolas x = y2 - I and x = 2y2 - 2 9. The linesy = x,y = x/3, andy = 2 10. The lines y = I - x and y = 2 and 1he curve y = eX 11. Thelinesy = 2x,y = x/2,andy = 3 - x 12. The lines y = x - 2 and y = -x and 1he curve y =

Vx

Identlfytng the Region of Integration The integrals and sums of integrals in Exercises 13-18 give 1he areas of regions in 1he xy-plane. Sketch each region, label each bounding curve with its equati~ and give the coordinates of the points where 1he curves intersect Then find 1he area of1he region.

11 1"1'1. 1°1 '6

II

o

2Y

K

dxtty

ylj3

0

o

0%

16.

ttydx

sinx X

17.

-1

18.

ttydx

+

Ox 2 -4

12:£'-x ttydx -x/2 0

-2;(

12:£0

-x Y+2

00

15.

131X(2-X) ttydx

dydx

+

14:£Vx 00

ttydx

- y. 1~

dx tty

23. Baeterium population If f(x,y) = (IO,OOOe Y )/(\ + IxV2) represents 1he "population density" of a certain bactetiun3 on 1he xy-plane, where x and y are measured in centimeters, fmd the tota! population of bactetia wi1hin 1he rectangle -5 '" x '" 5 and -2 '" Y '" O. 24. Regional population If f(x, y) = 100 (y + 1) represents 1he population density of a planar region on Earth, where x and y are measured in miles, find the number of people in the region bouoded by 1he curves x = y2 aod x = 2y _ y2. 25. Average temperature in Teus According to 1he Texas Almanac, Texas has 254 counties and a National Weather Service station in each county. Assume that at time to, each of the 254 wea1her stations recorded 1he local teroperature. Find a formula that would give a reasonable approximation of the average temperature in Texas at time to. Your answer should involve information that you would expect to be readily available in 1he Texas Almanac. 26. If Y = f(x) is a nonnegative contiouous function over 1he closed interval a '" x '" b, show that 1he double integral defInition of area for 1he closed plane region bounded by the graph of f, 1he vertical lines x = a and x = b, and the x-axis agrees with the defmition for area beuea1h 1he curve in Section 5.3.

853

15.4 Double Integrals in Polar Form

15.4

~I_D_ou_b_Le__ In_~~g~~_~_l_ 'n_P_o_ ~_ r_ ~_r_ m___________________________ Integrals are sometimes easier to evaluate if we change to polar coordinates. This section shows how to accomplish the change and how to evaluate integrals over regions whose boundaries are given by polar equations.

Integrals in Polar Coordinates When we dermed the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natoralshapes to use because their sides have either constant x-values or constant y-values. In polar coordinates, the natural shape is a ''polar rectangle" whose sides have constant r- and O-values. Suppose that a function I(r, 0) is dermed over a region R that is bounded by the rays = a and 0 = {3 and by the continuous curves r = gM) and r = g2(0). Suppose also that 0 :5 gl(O) :5 g2(O) :5 a for every value of 0 between a and {3. Then R lies in a fanshaped region Q dermed by the inequalities 0 :5 r :5 a and a :5 0 :5 {3. See Figure 15.21.

o

r=a

o~

,,--"---------------''''''''-----------~o ~

o

0

FIGURE 15.21 TheregionR:g,(8) '" r s g2(8),a s 8 s ,B,is contained in the fanshaped region Q: 0 :5 r S a, a s 8 s ,B. The partition of Q by circular arcs aod rays induces a partition of R.

We cover Q by a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii !J.r, 2!J.r, ... , m!J.r, where !J.r = aim. The rays are given by () =

Q,

0= a + M,

0= a + 2M,

... ,

o=

a + m'!J.O

= {3,

where !J.O = ({3 - a)lm'. The arcs and rays partition Q into small patches called ''polar rectangles." We number the polar rectangles that lie inside R (the order does not matter), calling their areas !J.A" !J.A 2, ... , !J.A •. We let (rk, Ok) be any point in the polar rectangle whose area is !J.Ak. We then form the sum

If I is continuous throughout R, this sum will approach a limit as we refine the grid to make !J.r and !J.O go to zero. The limit is called the double integral of lover R. In symbols, lim S. =

n~OO

JJRrr I(r, 0) dA.

854

Chapter 15: Multiple Integrals To evaluate this limit, we first have to write the sum S. in a way that expresses aA. in terms of!!J.r and!!J.lI. For convenience we choose r. to be the average of the radii of the inner and outer arcs bounding the kth polar rectangle aA•. The radius of the inner arc bounding aA. is then r. - (!!J.r/2) (Figure 15.22). The radius of the outer arc is r. + (!!J.r/2). The area of a wedge-shaped sector of a circle having radius r and angle II is

A = Large sector

o FIGURE 15.22 aA k =

The observation lbat

1 11 • r2 2

'

as can be seen by multiplying 7Tr 2 , the area of the circle, by 1I/27T, the fraction of the circle's area contained in the wedge. So the areas of the circular sectors subtended by these arcs at the origin are

Car::e:!m) -(sn:::!or)

leads to lbe formula aA, = r, !!J.r !!J.O.

)2 M

Inner mdius:

!!J. "21 ( r.---f

Outer radius:

"21 ( r. + T!!J.r)2 !!J.II.

Therefore,

aA.

= area oflarge sector - area of small sector =

y

~II [ (r. + ~r)' -

(r. -

~)'] = ~O (2r. !!J.r) =

r. ar M.

Combining this result with the sum defming S. gives

•

S. = ~ j(r., lI.h ar M.

~~----------->x

As n --> integral

00

and the values of ar and ao approach zero, these sums converge to the double

o

lim S. =

(a)

n_OO

rr

JJ R

y

Leaves at r

~

2 r sin 8 : : ~ r =

j(r, II) r dr dll.

A version of Fubini's Theorem says that the limit approached by these sums can be evaluated by repeated single integmtions with respect to r and II as

2

=

L

R

V2 t-.....,,\t-~

Y2 esc 9

Enters atr

=

V2 esc 8

8

~~~--------->x

o

Largest 6 is

I'

L

2

/

// Y

=x

V2 I--+-....:/~ / / / /

/

R

1

,~plr~"'(.)

8=a

j(r,lI) r dr dll. r=g,(8)

Finding Limits of Integration The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. To evaluate lfR j(r, II) dA over a region R in polar coordinates, integmting fITst with respect to r and then with respect to II, take the following steps.

(b) y /

iJif

j(r,lI) dA =

~ Smallest 6 is ~.

1. Sketch. Sketch the region and label the bounding curves (Figure 15.23a). 2. Find the r-limits of integration. Imagine a my L from the origin cutting through R in the direction of increasing r. Mark the r-values where L enters and leaves R. These are the r-limits of integmtion. They usually depend on the angle II that L makes with the positive x-axis (Figure 15.23b).

/

~~-L--------->x

o

(c)

FIGURE 15.23 Finding !be limits of integration in polar coordinates.

3.

Find the II-limits of integration. Find the smallest and largest II-values that bound R. These are the II-limits of integmtion (Figure 15.23c). The polar iterated integral is

iJif R

j(r, II) dA =

1

·~"/21r~2

9='1fj4

r=V2csc6

j(r, II) r dr dll.


855

EXAMPLE 1 Find the limits of integration for integrating fir, 0) over the region R that lies inside the cardioid r = I + cos 0 and outside the circle r = I.

y

Solutton

O=-I

L

Enters at r=1

Leaves at

1. We flfst sketch the region and label the bounding curves (Figure 15.24). 2. Next we fmd the r-limits of integration. A typical ray from the origin enters R where r = I and leaves where r = I + cos O. 3. Finally we find the O-limits ofintegration. The rays from the origin that intersectR run from 0 = -7f/2 to 0 = 7f/2. The integral is

r=I+0089

1

-fTj2

FIGURE 15.24 Finding the limits of

integratioo in polar coordinates for the regioo in Example 1.

~/211+'0"

•

fir, 0) r dr dO.

1

If fir, 0) is the constant function whose value is I, then the integral of f over R is the areaofR.

Area in Polar Coordinates

I Area Differential in Polar Coordinate.

The area of a closed and bounded region R in the polar coordinate plane is

dA = rdrd6

A = IfrdrdO. R

This fonnula for area is consistent with all earlier fonnulas, although we do not prove this fact. y

Leaves at r=

V4 co, 28

" I. ~------~~------+---~X

Enters at r= 0

r 2 =400829

To integrate over the shaded regioo, we run r from 0 to y'4 cos 26 and 6 from 0 to ",/4 (Exarople 2). FIGURE 15.25

EXAMPLE 2

Find the area enclosed by the lemniscate r2

=

4 cos 20.

Solutton We graph the lemniscate to determine the limits of integration (Figure 15.25) and see from the symmetry of the region that the tota1 area is 4 times the first-quadrant portion.

J.

A=4

~/4:£V4"'26

J.~/4

00

=

J.

4

0

[r2]'=V4'0'2'

rdrdO=4"2 0

~/4

2 cos 20 dO

]~/4

=

4 sin 20

0

dO

r=O

=

4.

•

Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integralifR fix, y) dx dy into a polar integral has two steps. First substitute x = r cos 0 and y = r sin 0, and replace dx dy by r dr dO in the Cartesian integral. Then supply polar limits of integration for the boundary of R. The Cartesian integral then becomes

II R

fix, y) dx dy =

II

fir cos 0, r sin 0) r dr dO,

G

where G denotes the same region of integration now described in polar coordinates. This is like the substitution method in Chapter 5 except that there are now two variables to substitute for instead of one. Notice that the area differential dx dy is not replaced by dr dO but by r dr dO. A more general discussion of changes of variables (substitutions) in multiple integrals is given in Section 15.8.

856


EXAMPLE 3

Evaluate

II

e

x2 y2

+ dy dx,

R

where R is the semicircular region bounded by the x-axis and the curve y = ~ (Figure 15.26). y

y=~

SoLution In Cartesian coordinates, the integral in question is a nonelementary integral and there is no direct way to integrate e x2 +y2 with respect to either x or y. Yet this integral and others like it are important in mathematics--in statistics, for example--and we need to find a way to evaluate it. Polar coordinates save the day. Substituting x = r cos e, y = r sin e and replacing dy dx by r dr de enables us to evaluate the integral as

0=0 --~~------~~------~

__ x

= Jot'l"2 (e - 1) de

FIGURE 15.26 The semicircular region in Example 3 is the region

o :s r :s

1,

o :s ()

:s

7T.

= ;

(e - 1).

The r in the r dr de was just what we needed to integrate e r2 • Without it, we would have been unable to find an antiderivative for the first (innermost) iterated integral. _

EXAMPLE 4

SoLution

Evaluate the integral

Integration with respect to y gives

t (x2~+

Jo

(1 - X2) 3/2) 3 dx,

an integral difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. The region of integration in Cartesian coordinates is given by the inequalites 0 ::; y ::; ~ and o ::; x ::; 1, which correspond to the interior of the unit quarter circle x 2 + y2 = 1 in the first quadrant. (See Figure 15.26, first quadrant.) Substituting the polar coordinates x = r cos e, y = r sin e, 0 ::; e ::; 7T/2 and 0 ::; r ::; 1, and replacing dx dy by r dr de in the double integral, we get

z

_1

-

o

7T /

2

_1

[r4 ]r=1 4 de -

r=O

7T /

2

!4 de -_ 1T8·

0

Why is the polar coordinate transformation so effective here? One reason is that x 2 simplifies to r2. Another is that the limits of integration become constants.

EXAMPLE 5 z x

FIGURE 15.27 Example 5.

_

Find the volume of the solid region bounded above by the paraboloid

= 9 - x 2 - y2 and below by the unit circle in the xy-plane.

The region of integration R is the unit circle x 2 + y2 = 1, which is described in polar coordinates by r = 1,0 ::; e ::; 21T. The solid region is shown in Figure 15.27. The volume is given by the double integral

SoLution The solid region in

+ y2


=

10f'''10{' (9r -

r

3

)

857

dr dO

• EXAMPLE 6 Using polar integration, find the area of the regionR in the xy-plane eoclosed by the circle x 2 + y2 + 4, above the line y = 1, and below the line y = 0x. Solutton

A sketch of the region R is showo in Figure 15.28. First we note that the line y = 0x has slope 0 = tan 6, so 6 = 7T/3. Next we observe that the line y = 1 intersects the circle x 2 + y2 = 4 wheo x 2 + 1 = 4, or x = 0. Moreover, the radial line from the origin through the point (0,1) has slope 1/0 = tan 0, giving its angle of inclination as 6 = 7T/6. This information is showo in Figure 15.28. Now, for the regionR, as 6 varies from 7T/6 to 7T/3, the polar coordinate r varies from the horizontal line y = 1 to the circle x 2 + y2 = 4. Substituting r sin 0 for Y in the equation for the horizontal line, we have r sin 6 = 1, or r = esc 6, which is the polar equation of the line. The polar equation for the circle is r = 2. So in polar coordinates, for 7T/6 :5 6 :5 7T/3, r varies from r = esc 0 to r = 2. It follows that the iterated integral for the area theo gives

y

~~--~~------~~X

o

2


1 2

/3

{{ dA = {" Jrr/6

JJR

The region R in

- {"/3 - Jrr/6

rdrd6

esc (J

[1 r2l'~2 2

d6

r- csc9

,,/3 1

csc20] dO

=

1

=

21 [46 + cot 6 ],,/ "/6

=

1 (47T +

,,/6

2 [4 -

3

_1_) _1(47T + 0)

23026

= 7T -

Exercises 15.4 Regions in Polar Coordinates

4.

3.

In Exercises 1-1l, describe the given region in polar coordinates.

L

y

1 y --~--~~----~->x

-1

--4-----+-----,--> x 9

--------+-~--4-->x

o

4

0

1

0

3'

•

858

Chapter 15: Multiple Integrals Area in Polar Coordinates

6.

5. y

27. Find the area of the region cut from the fIrst quadrant by the curve r = 2(2 - sin 26)112.

y

28. Cardioid overlapping a clrcle Find the area of the region that lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1.

2t--- - - - - ,

29. One leaf of a rose r = 12 cos 36.

30. SnaR shen Find the area of the region eoclosed by the positive x-axis aod spiral r = 46/3,0 '" 6 '" 21T. The region looks like a anall shell.

2

7. The region enclosed by the circle x 2

8.

+ y2 = 21:. Theregionenclosedbythesemicirclex 2 + y2 =

2y,y '" O.

In Exercises 9-22, change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

1 11,~

11. Jo Jo

I

2

-~ 1

+

~

-Vd'-x'

22

YX

+ y2

/,v22/,'v4=Y'

Find the awrage height of

Ya 2 - x2 -

y2 above the disk

34. Average height ofa cone Find the average height of the (single) dxdy

dydx

conez = Yx 2

+ y2 above the diskx 2 + y2

'" a 2 inthexy-plane.

35. Average distance from interior of disk to center Find the average distance from a point P(x, y) in the disk x 2 + y2 '" a 2 to the origin.

2

2 X + y2f

+

dydx

°

1 1Vi-v'I=Yl =Y'

f(r, 6) r dr d6.

R

33. Awrage height of a hemisphere

1n2:£V(ln2)'-" • ~ • vx'+,' dx dy

1,°

JJ

36. Average distance squared from a point in a disk to a point in

-~ (I

1

20.

1

16.

dydx

18.111~ 19.

12.

Are~(R)

a1~ dydx

the hemispherical surface z = x 2 + y2 ~ a 2 in the xy-plane.

17.1°1° -1

+ y2) dx dy

+ y2) dx dy

Average values In polar coordinates, the average value of a function over a region R (Section 15.3) is giveo by

0

x dx dy

V3r Jl

-1

(x 2

0

(x

13. 1,61,' 1

1,11,Vi=Y'

0

{'{VH

15.

10.

dy dx

-1

31. Cardioid in the r"st quadrant Find the area of the region cut from the frrst quadrant by the cardioid r = 1 + sin 6. 32. Overlapping cardioids Find the area of the region conunon to the interiors of the cardioids r = I + cos 6 and r = I - cos 6.

Evaluating Polar Integrals

9.

Find the area enclosed by one leaf of the rose

its houndary Find the average value of the square of the distance from the point p(x, y) in the disk x 2 + y2 '" I to the boundary pointA(I, 0).

Theory and Examples In (x 2

+ y2 + I) dx dy

-1

37. Converting to a polar integral Integrate f(x,y) = [In (x 2 + y2)]fYx 2 + y2 OWl the region 1 '" x 2 + y2 S e.

21.1,11~ (x + 2y)dydx

38. Converting to a polar integral Integrate f(x, y) = [In (x 2 + y2)]/(x 2 + y2) 0'"' the region I '" x 2 + y2 S .2.

22. 121,~

39. Volume of noncircular right cylinder The region that lies inside the cardioid r = 1 + cos 8 and outside the circle r = I is the base of a solid right cylinder. The top of the cylioder lies in the plane z = x. Find the cylinder's volume.

1

0

1 (X 2

+ y2)2

dydx

In Exercises 23-26, sketch the region of integration and convert each polar integral or sum of integrals to a Cartesian integral or sum of in· tegrals. Do not evaluate the integrals. 12 {' 23. Jo Jo r3 sin 6 cos 6 dr d6

r

24.

r121='

r2 cos 6 dr d6

i1r/6 I /4 {''''''' 25. Jo Jo

r

40. Volume of noncircular right cylinder The region enclosed by the leoutiscate r2 = 2 cos 26 is the base of a solid tight cylinder whose top is bounded by the sphere z = ~. Find the cylinder's volume. 41. Converting to polar integrals

a. The usual way to evaluate the improper integral 00 I = e --;f1 dx is frrst to calculate its square:

10

r' sin2 6 dr d6 Evaluate the last integral using polar coordinates and solve the resulting equation for 1.

15.5 Triple Integrals in Rectangular Coordinates b. Evaluate lim erf(x) =

x-HX)

lim

!O0

X

2 - /' • e/ dt.

X-H)()

46. Area Suppose that the area of a region in the polar coordinate plane is

1

3Tr/412sin6

A =

V 11'

Tr/4

42. Converting to a polar integral


43. Existence Integrate the function j(x, y) = 1/(1 - x 2 - y2) over the disk x 2 + y2 ~ 3/4. Does the integral of j(x, y) over

rdrdO. esc 6

Sketch the region and find its area. COMPUTER EXPLORATIONS In Exercises 47- 50, use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise.

a. Plot the Cartesian region of integration in the xy-plane.

the disk x 2 + y2 ~ I exist? Give reasons for your answer.

b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for r andO.

44. Area formula in polar coordinates Use the double integral in polar coordinates to derive the formula

1

859

.6 1

2

c. Using the results in part (b), plot the polar region of integration in the rO-plane.

for the area of the fan-shaped region between the origin and polar curve r = j(O), a ~ 0 ~ {3.

d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility.

A =

a

- r 2 dO

45. Average distance to a given point inside a disk Let Po be a point inside a circle of radius a and let h denote the distance from Po to the center of the circle. Let d denote the distance from an arbitrary point P to Po. Find the average value of d 2 over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and Po on the x-axis.)

15.5

47.

49.

1111 1l o

x

1

o

Y

- 2 - - 2 dy

+Y

x

Y 3 /

-y/3

dx

Y

Yx 2 + y2

48.

l1 1

o

x 2 x / ---dydx

0

x 2 + y2

dxdy

Triple Integrals in Rectangular Coordinates Just as double integrals allow us to deal with more general situations than could be handled by single integrals, triple integrals enable us to solve still more general problems. We use triple integrals to calculate the volumes of three-dimensional shapes and the average value of a function over a three-dimensional region. Triple integrals also arise in the study of vector fields and fluid flow in three dimensions, as we will see in Chapter 16.

Triple Integrals If F(x, y, z) is a function defined on a closed, bounded region D in space, such as the region occupied by a solid ball or a lump of clay, then the integral of F over D may be defined in the following way. We partition a rectangular boxlike region containing D into rectangular cells by planes parallel to the coordinate axes (Figure 15.29). We number the cells that lie completely inside D from 1 to n in some order, the kth cell having dimensions ~Xk by ~Yk by ~Zk and volume ~ Vk = ~k~Yk~Zk. We choose a point (Xk, Yk, Zk) in each cell and form the sum n

Sn = ~ F(xk, Yk , Zk) ~ Vk.

(1)

k=l

FIGURE 15.29 Partitioning a solid with rectangular cells of volume ~ Vk •

We are interested in what happens as D is partitioned by smaller and smaller cells, so that ~k, ~Yk, ~Zk and the norm of the partition Ilpll, the largest value among ~k, ~Yk, ~Zk, all approach zero. When a single limiting value is attained, no matter how the partitions and points (Xk, Yk, Zk) are chosen, we say that F is integrable over D. As before, it can be

860

Chapter 15: Multiple Integrals shown that when F is continuous and the bounding surface of D is fonned from tmitely many smooth surfaces joined together along tmitely many smooth curves, then F is integrable. As Ilpll --> 0 and the number of cells n goes to 00, the sums S. approach a limit. We call this limit the triple integral of F over D and write

•

~S. =

iiiF(x,y,Z)dV

lim S. =

or

IWII~O

D

rrr F(x,y,z)

111 D

dxdy dz .

The regions D over which continuous functions are integrable are those having ''reasonably smooth" boundaries.

Volume of a Region in Space If Fis the constant function whose value is 1, then the sums in Equation (1) reduce to

As dxk, t.Yk, and t.Zk approach zero, the cells t. V. become smaller and more numerous and ml up more and more of D. We therefore define the volume of D to be the triple integral

DEFINITION

The volume of a closed, bounded region D in space is

V= iff dV. D

TIris detmition is in agreement with our previous detmitions of volume, although we omit the verification of this fact. As we see in a moment, this integral enables us to calculate the volumes of solids enclosed by curved surfaces.

Finding Limits of Integration in the Order dz dy dx We evaluate a triple integral by applying a three-dimensional version of Fubini's Theorem (Section 15.2) to evaluate it by three repeated single integrations. As with double integrals, there is a geometric procedure for finding the limits of integration for these single integrals. To evaluate

iff F(x,y,z)dV D

over a region D, integrate first with respect to z, then with respect to y, and finally with respect to x. (You might choose a different order of integration, but the procedure is similar, as we illustrate in Example 2.)

1.

Sketch. Sketch the region D along with its "shadow" R (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of D and the upper and lower bounding curves of R.

15.5 Triple Integrals in Rectangular Coordinates

861

z z = fz(x, y)

D

y

x

2.

Find the z-limits a/integration . Draw a line M passing through a typical point (x,y) in R parallel to the z-axis. As z increases, M enters D at z = f 1(x, y) and leaves at z = h(x,y). These are the z-limits of integration. z

D

y

a b

--- ---

x

(x,y)

3.

Find the y-limits a/integration. Draw a line L through (x,y) parallel to the y-axis. As y increases, L enters Rat y = gl(X) and leaves at y = g2(X). These are the y-limits of integration. z

D

Enters at y = g,(x) a

y

x b

x

Leaves at y = g2(x)

862


4.

Find the x-limits of integration. Choose x-limits that include all lines through R parallel to the y-axis (x = a and x = b in the preceding figure). These are the x-limits of

integration. The integral is

l

x=bly=g2(X)lz= h(x,y)

x=a

F(x, y, z) dz dy dx. y=gl(X)

z= !J(x, y)

Follow similar procedures if you change the order of integration. The "shadow" of region D lies in the plane of the last two variables with respect to which the iterated integration takes place. The preceding procedure applies whenever a solid region D is bounded above and below by a surface, and when the "shadow" region R is bounded by a lower and upper curve. It does not apply to regions with complicated holes through them, although sometimes such regions can be subdivided into simpler regions for which the procedure does apply.

EXAMPLE 1

Find the volume of the region D enclosed by the surfaces z = x 2

+

3y2

and z = 8 - x 2 - y2. SoLution

The volume is

v=

111

dzdydx,

D

the integral of F(x, y, z) = lover D. To find the limits of integration for evaluating the integral, we first sketch the region. The surfaces (Figure 15.30) intersect on the elliptical cylinder x 2 + 3y2 = 8 - x 2 - y2 or x 2 + 2y2 = 4, z > O. The boundary of the region R, the projection of D onto the xy-plane, is an ellipse with the same equation: x 2 + 2y2 = 4. The ''upper'' boundary of R is the curve y = V(4 - x 2)/2. The lower boundary is the curve y = -V(4 - x 2 )/2. Now we find the z-limits of integration. The line M passing through a typical point (x,y) in R parallel to the z-axis enters D at z = x 2 + 3y2 and leaves at z = 8 - x 2 - y2. M

The curve of intersection

'-(-2, 0,4)

I I

Enters at

z =x 2 +

3y2

I

I

J-----

-- I I , I '.,,1

Enters at~-----=_ y = -V(4 - x 2 )/2

x

R

Leavesat

~

L

y

y = V(4 - x 2 )/2

FIGURE 15.30 The volume of the region enclosed by two paraboloids, calculated in Example 1.


863

Next we find the y-limits of integration. The line L through (x, y) parallel to the y-axis enters Rat y = - Y(4 - x 2 )/2 and leaves at y = Y(4 - x 2 )/2. Finally we find the x-limits of integration. As L sweeps across R, the value of x varies from x = -2 at (-2,0,0) to x = 2 at (2, 0, 0). The volume of Dis

v=

JJJ

dzdydx

D

1 11 1

21V(4 - X2)/ 218 - X2 - y2

=

- 2 - V(4 - x 2)/ 2 x2 +3y2

V (4 - X2)/ 2

2

=

dzdydx

(8 - 2x2 - 4y2) dy dx

- 2 - V(4 - x 2)/ 2 2 [

=

-2

=

4 ]y=V(4 - X2)/ 2 (8 - 2x 2)y - _y3 dx 3 y= - V(4 - x 2)/ 2

81TY2.

•

After integration with the substitution x = 2 sin u

In the next example, we project D onto the xz-plane instead of the xy-plane, to show how to use a different order of integration. z __________ (0, 1,1)

EXAMPLE 2 Set up the limits of integration for evaluating the triple integral of a function F(x, y, z) over the tetrahedron D with vertices (0, 0, 0), (1 , 1,0), (0, 1, 0), and (0, 1, 1). Use the order of integration dy dz dx. Solution

We sketch D along with its "shadow" R in the xz-plane (Figure 15.31). The upper (right-hand) bounding surface of D lies in the plane y = 1. The lower (left-hand) bounding surface lies in the plane y = x + z. The upper boundary of R is the line z = 1 - x. The lower boundary is the line z = 0. First we find the y-limits of integration. The line through a typical point (x, z) in R parallel to the y-axis enters D at y = x + z and leaves at y = 1. Next we find the z-limits of integration. The line L through (x, z) parallel to the z-axis enters R at z = and leaves at z = 1 - x. Finally we find the x-limits of integration. As L sweeps across R, the value of x varies from x = to x = 1. The integral is

°

(1,1,0)

°

ll i

x

Finding the limits of integration for evaluating the triple integral of a function defined over the tetrahedron D (Examples 2 and 3).

l

o

FIGURE 15.31

l x -

0

l

•

F(x,y, z) dy dz dx.

x+ z

EXAM PLE 3 Integrate F(x, y , z) = 1 over the tetrahedron D in Example 2 in the order dz dy dx, and then integrate in the order dy dz dx. Solution

First we find the z-limits of integration. A line M parallel to the z-axis through a typical point (x, y) in the xy-plane "shadow" enters the tetrahedron at z = and exits through the upper plane where z = y - x (Figure 15.32). Next we find the y-limits of integration. On the xy-plane, where z = 0, the sloped side of the tetrahedron crosses the plane along the line y = x . A line L through (x, y) parallel to the y-axis enters the shadow in the xy-plane at y = x and exits at y = 1 (Figure 15.32).

°

864


z

Finally we find the x-limits of integration. As the line L parallel to the y-axis in the previous step sweeps out the shadow, the value of x varies from x = 0 to x = 1 at the point (1, 1,0) (see Figure 15.32). The integral is

ri

Jo

x

(y-x

l

Jo

F(x,y,z) dzdydx.

For example, if F(x,y, z) = 1, we would find the volume of the tetrahedron to be

v=

1

lillY- Xdzdydx l

= l i\y-x)dY dx x

FIGURE 15.32 The tetrahedron in Example 3 showing how the limits of integration are found for the order dz dy dx.

1

6' We get the same result by integrating with the order dy dz dx. From Example 2,

V

=

l i o

l l X l dydzdx l 0

r r-

x+z

x

=

Jo Jo

=

1 --(1 - x)3

6

(1 - x - z) dz dx

]1

•

0

Average Value of a Function in Space The average value of a function F over a region D in space is defined by the formula Average value of F over D =

vol~e of D

Jff

F dV.

(2)

D

For example, if F(x,y, z) = Yx 2 + y2 + z2, then the average value of F over D is the average distance of points inD from the origin. If F(x,y, z) is the temperature at (x,y, z) on a solid that occupies a region D in space, then the average value of F over D is the average temperature of the solid.

865


EXAMPLE 4 Find the average value of F(x,y, z) = xyz throughout the cubical region D bounded by the coordinate planes and the planes x = 2, Y = 2, and z = 2 in the tIrst octant. Solution We sketch the cube with enough detail to show the limits of integration (Figure 15.33). We then use Equation (2) to calculate the average value of F over the cube. The volume of the region Dis (2)(2)(2) = 8. The value of the integral of F over the cube is

D

121212

xyzdxdydz =

000

FIGURE 15.33 in Example 4.

The region of integration

1212 [2 12 12

~ yz ]X~2 dydz

00

=

[]Y~2 y2z dz=

o

= 12122yzdydz

x- O

y=O

00

[]2

4zdz= 2z2

0

=8.

0

With these values, Equation (2) gives / Average value of = lYz over the cube vo ume

rrr

111 cube

(-81 )(8)

xyz dV =

=

I.

In evaluating the integral, we chose the order dx dy dz, but any of the other five possible orders would have done as well. •

Properties of Triple Integrals Triple integrals have the same algebraic properties as double and single integrals. Simply replace the double integrals in the four properties given in Section 15.2, page 846, with triple integrals.

Exercises 15.5 Triple Integrals In Different Iteration Orders 1. Evaluate 1he integral in Example 2 taking F(x,y, z) = 1 to Imd 1he volmne of1he tetrahedron in 1he order dz dx dy. 2. Volume of rectangular .olid Write six different iterated triple

Evaluating Triple Iterated Integrals Evaluate 1he integrals in Exercises 7-20.

7. 1, '1, '1,

'<x 2

+ y2 + z2) dz ely dx

integrals for 1he volume of 1he rectangular solid in 1he fIrst octant bounded by 1he coordinate planes aod 1he planes x = I, Y = 2, and z = 3. Evaluate one of1he integrals.

9.

3. Volume of tetrahedron Write six different iterated triple inre. grals for 1he volmne of 1he tetrahedron cut from 1he IlrSt octant by 1he plane 6x + 3y + 2z = 6. Evaluate one of 1he integrals.

4. Volume of solid Write six different iterated triple integrals for 1he volmne of 1he region in 1he first octant enclosed by 1he cylinder x 2 + z2 = 4 and 1he plane y = 3. Evaluate one of 1he integrals.

10.

1,'1,'-"'1.'-"'-'

12.1'Jo{'Jo

{2(x

13.

{i' {

872


TABLE 15.2

Moments of inertia (second moments) formulas

THREE-DIMENSIONAL SOLID About the x-axis:

Ix =

fJJ(y2 + Z2) /l dV

About the y-axis:

Iy =

fJJ(X 2+ z2) /l dV

h

fJJr 2/l dV

6

~

6(z,y,z)

About the z-axis:

About a line L:

=

rex, y, z) = distance from. the point (x, y, z) to line L

TWO-DIMENSIONAL PLATE About the x-axis:

Ix =

About the y-axis:

Iy =

About a line L:

h

ff ff ff r2(x,y) 2

6

y /ldA

~

6(z,y)

2

x /l dA

=

rex, y) = distance from (x,y)toL

/l dA

10 = ff(x 2 + y2)/ldA = Ix + Iy

About the origin (polar moment):

Similarly, and

Iz

_ M (2 a + b2) . -T2

•

EXAMPLE 4 A thin plate covers the triangular region bouoded by the x-axis and the lines x = I and y = 2x in the first quadrant. The plate's density at the point (x, y) is /l(x, y) = 6x + 6y + 6. Find the plate's moments of inertia about the coordinate axes and the origin. y 2

(1,2)

Solution We sketch the plate and put in enough detail to determine the limits of integration for the integrals we bave to evaluate (Figure 15.40). The moment of inertia about the

x-axis is

nf----+--~x

=

J.' [

2xy3

3 y' + 2y 3 lY~2z dx +_

o

FIGURE 15.40 The triangular region covered by the plate in Example 4,

=

[s.x S + 4x'l~

y~O

2

=

12.

=

J.' 0

(4Ox'

+

16x 3) dx

15.6

Moments and Centers of Mass

873

Similarly, the moment of inertia about the y-axis is

Beam A

Iy =

Axis

39 10t 10(2x x 2S(x,y) dy dx = s·

Notice that we integrate y2 times density in calculating Ix and x 2 times density to find IySince we know Ix and Iy, we do not need to evaluate an integral to tmd 10; we can use the equation 10 = Ix + Iy from Table 15.2 instead: BeamB

10 = 12

+ 39 5

=

60

+ 39 5

99 5'

•

Axis

The moment of inertia also plays a role in detennining how much a horizontal metal beam will bend under a load. The stiffness of the beam is a constant times I, the moment of inertia of a typical cross-section of the beam about the beam's longitudinal axis. The greater the value of I, the stiffer the beam and the less it will bend under a given load. That FIGURE 15.41 The greater the polar is why we use I-beams instead of beams whose cross-sections are square. The flanges at moment of inertia of the cross-section of a the top and bottom of the beam hold most of the beam's mass away from the longitudinal beam about the beam's longitudinal axis, the axis to increase the value of I (Figure 15.41). stiffer the beam. Beams A aod B have the same cross-sectional area, but A is stiffer.

Exercises 15.6 Plates of Constant Density 1. Finding a center of mass Find the center of mass of a thin plate of density B = 3 bounded by the lines x = 0, y = x, and the parabola y = 2 - x 2 in the tITst quadrant 2. Finding moments of inertia Find the moments of inertia about the coordinate axes of a thin rectangular plate of constant density B bounded by the lines x = 3 and y = 3 in the rITst quadraot. 3. Finding a centroid Fiod the centroid of the region in the first quadrant bounded by the x·axis, the parabola y2 = 2x, and the lioex+y=4. 4. Finding a centroid Find the centroid of the triangular region cut from the rITSt quadrant by the line x + y = 3. 5. Finding a centroid Fiod the centroid of the region cut from the rITSt quadrant by the circle x 2 + y2 = a 2. 6. Finding a centroid Find the centroid of the region b&ween the x-axis and the archy = sin x, 0:5 x :s; 71'. 7. Finding moments of ioertia Find the moment of inertia about the x·axis of a thio plate of deosity B = I bounded by the circle x 2 + y2 = 4. Then use your result to rmd 1, and 10 for the plate. 8. Finding a moment of inertia Find the moment of inertia with respect to the y·axis of a thin sheet of constant density B = I bouoded by the curve y = (sin2 x)/x 2 and the interval 'TT' :s; x :s; 2w of the x-axis.

9. The centroid of an infinite region Find the centroid of the inrmite region io the second quadrant enclosed by the coordinate axes and the curve y = e~. (Use improper integrals in the massmoment formulas.)

10. The fin! moment of an ioImite plate Find the rITSt moment about the y-axis of a thin plate of density B(x, y) = I covering the 2 infmite region under the curve y = e -x j2 in the first quadrant.

Plates with Varying Density 11. Finding a moment of inertia Find the moment of inertia about the x·axis of a thio plate bounded by the parabola x = y - y2 and the line x + y = if B(x,y) = x + y.

ma..

o

12. Finding Find the mass of a thin plate occupying the smaller region cut from the ellipse x 2 + 4y2 = 12 by the parabola x = 4y2 if B(x, y) = 5x. 13. Finding a center of mass Find the center of mass of a thin triangular plate bounded by the y-axis aod the lines y = x and y = 2 -xifB(x,y) = 6>:+ 3y+ 3. 14. Finding a center of mass and moment of inertia Find the center of mass and moment of inertia about the x-axis of a thin plate bounded by the curves x = y2 and x = 2y - y2 if the density at the point (x. y) is B(x,y) = y + 1. 15. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin rectangular plate cut from the rITSt quadrant by the lioes x = 6 and y = I if B(x,y) = x + Y + 1. 16. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin plate bounded by the line y = I and the parabola y = x 2 if the density is B(x,y) = y + 1. 17. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin plate bounded by the x·axis, the lines x = ± I, and the parabola y = x 2 if B(x,y) = 7y + 1.

874


18. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the x-axis of a thin rectangular plate bounded by the linesx = 0, x = 20, Y = -I, and y = I if B(x,y) = I + (x/20).

19. Center of mass, moments of inertia Find the center of mass, the moment of inertia about the coordinate axes, and the polar moment of inertia of a thin triangular plate bounded by the linesy = x,y = -x, andy = I ifB(x,y) = Y + I. 20. Center of mass, moments of inertia B(x,y) = 3x 2 + I.

Repeat Exercise 19 for

l---..

y

Solids with Constant Den';ty 21. Moments of inertia Find the moments of inertia of the rectangular solid shown bere with respect to its edges by calculating Ix, [Y' and /z·

25. a. Center of mass Find the center of mass of a solid of constant density bounded below by the paraboloid z = x 2 + y2 and above by the plane z = 4. b. Find the plane z = c that divides the solid into "'" parts of equal

volume. This plane does not pass through the center of mass. 26. Moments A solid cube, 2 units on a side, is bounded by the

c I I

-- b

)-/

/ /

planes x

= ±I,z = ±I,y = 3,

and y

= 5.

Find the center of

mass and the moments of inertia about the coordinate axes.

Y

27. Moment of inertia about aline A wedge like the one in Exercise 22 bas a = 4, b = 6, and c = 3. Make a quick sketch to

a

22. Moments of inertia

The coordinate axes in the figure run through the centroid of a solid wedge parallel to the labeled

edges. Find Ix, Iy, and Iz if a

=

b

=

6 and c

=

4.

check for yourself that the square of the distance from a typical point (x, y, z) of the wedge to the line L: z = O,y = 6 is r2 = (y - 6)2 + z2 Then calculate the moment of inertia of the wedge about L. 28. Moment of inertia about aline A wedge like the one in Exercise 22 bas a = 4, b = 6, and c = 3. Make a quick sketch to check for yourself that the square of the distance from a typical is point (x, y, z) of the wedge to the line L: x = 4, Y = r2 = (x - 4)2 + y2. Then calculate the moment of inertia of the wedge about L.

°

Centroid 81(0,0,0)

c

~

3

a

23. Center of mas. and moment. of inertia A solid ''trough'' of constant density is bounded below by the surface z = 4y2, above by the plane z = 4, and on the ends by the planes x = I and x = -I. Find the center of mass and the moments of inertia with

Solids with varying Den';ty In Exercises 29 and 30, fmd

a. the mass of the solid. b. the center of mass. 29. A solid region in the flIst octant is bounded by the coordinate planes and the plane x + y + z = 2. The density of the solid is B(x,y,z) = 2x.

°

respect to the three axes. 24. Center of ma.. A solid of constant density is bounded below by the plane z = 0, on the sides by the elliptical cylinder x 2 + 4y2 = 4, and above by the plane z = 2 - x (see the accompanying figure).

a. Find>, andy. b. Evaluate the integral Mxy =

1'J -

,,- /

(1j2)~J.2-x

-(1/2)~ 0

zdzdydx

using integral tables to carry out the fmal integration with respect to x. Then divide Mxy by M to verify that z = 5/4.

°

30. A solid in the flIst octant is boWlded by the planes y = and z = andbythesurfacesz = 4 - x 2 andx = y2 (seethe accompanying figure). Its density function is B(x, y, z) = ir:xy, /r; a constant

///

2 /

,/

//

'

X=y2

~

} (2, Yz,O)

Y

15.7 Triple Integrals in Cylindrical and Spherical Coordinates In Exercises 31 and 32, fmel

31.

32.

33.

34.

a. the mass of the solid. b. the center of mass. c. the moments of inertia about the coordinate axes. A solid cube in the fIrSt octant is bounded by the coordinate planes and by the planes x = I,y = I,andz = l.Thedensityof the cube is 8(x,y,z) = x + Y + z + l. A wedge like the one in Exercise 22 has dimensions a = 2, b = 6, and c = 3. The density is 8(x,y,z) = x + l. Notice that if the density is constant, the center of mass will be (0, 0, 0). Mass Find the mass of the solid bounded by the planes x + z = I, x - z = -I, y = 0 and the surface y = The density of the solid is 8(x,y,z) = 2y + 5. Mass Find the mass of the solid region bounded by the parabolic surfaces z = 16 - 2x 2 - 2y2 and z = 2x 2 + 2y2 if the density of the solid is 8(x,y,z) = v'x 2 + y2.

875

~1

L

I

.........v=xi+yJ

- ',

(x,y, z)

-

~ ~~/ tcJn. /

I I I

(h, 0, 0)

Yz.

.y D

I

b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line L c.m . along the z-axis and the line L perpeodicular to the xy-plaoe at the point (h, 0, 0). Let D be the region of space occupied by the body. Then, in

the notation of the fIgure, Theory and Examples The PIII'IIlle1 ADs Theorem Let L,m. be a line through the center of mass of a body of mass m aud let L be a parallel line h units away from L c.m • The Parallel Axis Theorem says that the moments of inertia lc.m. aud h of the body about L,m. aud L satisfY the equation

(2) As in the two-dimensional case, the theorem. gives a quick way to calculate one moment when the other moment and the mass are

known. 35. Proof of the PIII'IIi1e1 ADs Theorem a. Show that the fIrst moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the yz-plane. What does the formula x = M,,/M then tell you?)

15.7

=

ifflv -hil

2

dm.

D

Expand the integrand in this integral and complete the proof. 36. The moment of inertia about a dismeter of a solid sphere of conslant density and radius a is (2/5)ma 2, where m is the mass of the sphere. Find the moment of inertia about a line tangent to the sphere. 37. The moment of inertia of the solid in Exercise 21 about the z-axis is I, = abc(a 2 + b 2 )/3.

•• Use Equation (2) to fmd the moment of inertia of the solid about 1he line parallel to 1he z-axis through the solid's center of mass. b. Use Equation (2) and 1he result in part (a) to fmd the moment of inertia of 1he solid about the line x = 0, y = 2b. 38. If a = b = 6 and c = 4, 1he moment of inertia of the solid wedge in Exercise 22 about the x-axis is I, = 208. Find 1he moment of inertia of the wedge about 1he line y = 4,z = -4/3 (1he edge of the wedge's narrow end).

Triple Integrals in Cylindrical and Spherical Coordinates When a calculation in physics, engineering, or geometry involves a cylinder, cone, or sphere, we can often simplify our work by using cylindrical or spherical coordinates, which are introduced in this section. The procedure for transfonning to these coordinates and evaluating the resulting triple integrals is similar to the transformation to polar coordinates in the plane studied in Section 15.4.

z P(r, 8, z)

o

h

z

Integration in Cylindrical Coordinates We obtain cylindrical coordinates for space by combining polar coordinates in the xy-plane with the usual z-axis. This assigns to every point in space one or more coordinate triples of the form (r, fJ, z), as shown in Figure 15.42.

DEFINmON

Cylindrical coordinates represent a point P in space by ordered

triples (r, fJ, z) in which The cylindrical coordinates of a point in space are r, 6, andz. FIGURE 15.42

1. r and fJ are polar coordinates for the vertical projection of P on the xy-plane 2. z is the rectangular vertical coordinate.

876


The values of x, y, r, and () in rectangular and cylindrical coordinates are related by the usual equations. Equations Relating Rectangular (x,y, z) and Cylindrical (r, 8, z) Coordinates x

= rcos(),

y

= rsin(),

z = z,

tan() = y/x

z 0=00 ,

randzvary-

zoo

Z=

rand 0 vary

I

In cylindrical coordinates, the equation r = a describes not just a circle in the xy-plane but an entire cylinder about the z-axis (Figure 15.43). The z-axis is given by r = o. The equation () = ()o describes the plane that contains the z-axis and makes an angle ()o with the positive x-axis. And, just as in rectangular coordinates, the equation z = Zo describes a plane perpendicular to the z-axis. Cylindrical coordinates are good for describing cylinders whose axes run along the z-axis and planes that either contain the z-axis or lie perpendicular to the z-axis. Surfaces like these have equations of constant coordinate value:

r=4 y

() =

7T

Plane containing the z-axis

3 z = 2.

r = a,

oand z vary

FIGURE 15.43 Constant-coordinate equations in cylindrical coordinates yield cylinders and planes.

Cylinder. radius 4. axis the z-axis

Plane perpendicular to the z-axis

When computing triple integrals over a region D in cylindrical coordinates, we partition the region into n small cylindrical wedges, rather than into rectangular boxes. In the kth cylindrical wedge, r, () and z change by tJ..rk, tJ..()k, and tJ..Zk, and the largest of these numbers among all the cylindrical wedges is called the norm of the partition. We define the triple integral as a limit of Riemann sums using these wedges. The volume of such a cylindrical wedge tJ.. Vk is obtained by taking the area !J..A k of its base in the rO-plane and multiplying by the height tJ..z (Figure 15.44). For a point (rk, ()k, Zk) in the center of the kth wedge, we calculated in polar coordinates that !J..A k = rk tJ..rk tJ..()k. So tJ.. Vk = tJ..Zk rk tJ..rk tJ..()k and a Riemann sum for lover D has the form n

Sn =

I Volume Differential in Cylindrical Coordinates

dV

=

dzrdrdO

'L I(rk, ()k, Zk) tJ..Zk rk tJ..rk tJ..()k. k=!

The triple integral of a function lover D is obtained by taking a limit of such Riemann sums with partitions whose norms approach zero:

n~~Sn =

Iff D

IdV=

1IIIdzrdrd(). D

Triple integrals in cylindrical coordinates are then evaluated as iterated integrals, as in the following example.

z r !:J.r!:J.O

rM

L

T!:J.z

~~ ~!:JL . __ ~----J ~ - - -

r------1~ !:J.r

EXAMPLE 1 Find the limits of integration in cylindrical coordinates for integrating a function I(r, (), z) over the region D bounded below by the plane z = 0, laterally by the circular cylinder x 2 + (y - 1)2 = 1, and above by the paraboloid z = x 2 + y2. SoLution The base of D is also the region's projection R on the xy-plane. The boundary of R is the circle x 2 + (y - 1f = 1. Its polar coordinate equation is

+ (y - 1)2 = 1 2 x + y2 - 2y + 1 = 1 r2 - 2r sin () = 0 r = 2 sin(). x2

FIGURE 15.44 In cylindrical coordinates the volume of the wedge is approximated by the product a v = az r ar ao.

15.7 Triple Integrals in Cylindrical and Spherical Coordinates

Top Cartesian: z = .x2 Cylindrical: z = ,2

877

The region is sketched in Figure 15.45. We find the limits of integration, starting with the z-limits. A line M through a typical point (r, 0) in R parallel to the z-axis enters D at z = 0 and leaves at z = x 2 + y2 = r2. Next we find the r-limits of integration. A ray L through (r, 0) from the origin enters R at r = 0 and leaves at r = 2 sin O. Finally we find the O-limits of integration. As L sweeps across R, the angle 0 it makes with the positive x-axis runs from 0 = 0 to 0 = 7T. The integral is

+I

iii

fer, 0, z) dV = l1r12sinolr2 fer, 0, z) dz r dr dO.

•

D i-=----~y

L

x

Example 1 illustrates a good procedure for finding limits of integration in cylindrical coordinates. The procedure is summarized as follows.

How to Integrate in Cylindrical Coordinates To evaluate

iff

FIGURE 15.45 Finding the limits of integration for evaluating an integral in cylindrical coordinates (Example l).

fer, 0, z) dV

D

over a region D in space in cylindrical coordinates, integrating first with respect to z, then with respect to r, and finally with respect to 0, take the following steps.

1.

Sketch . Sketch the region D along with its projection R on the xy-plane. Label the surfaces and curves that bound D and R. z

,. - -...... /

z = g2(r, 0)

y x r = h2(O)

2.

Find the z-limits of integration. Draw a line M through a typical point (r, 0) of R parallel to the z-axis. As z increases, M enters D at z = g\(r, 0) and leaves at z = g2(r, 0). These are the z-limits of integration. z z = gir, 0)

y x

878


3.

Find the r-limits of integration. Draw a ray L through (r, 0) from the origin. The ray enters R at r = hI (0) and leaves at r = h2( 0) . These are the r-limits of integration. z

4.

Find the O-limits of integration. As L sweeps across R, the angle 0 it makes with the positivex-axis runs from 0 = a to 0 = (3. These are the O-limits of integration. The integral is

hrJ!

f(r,O,z)dV=

l

O=f31r=h2(O)lz=g2(r, o)

f(r,O,z)dzrdrdO .

O=a

D

r=h,(O)

z=gk, O)

EXAMPLE 2 Find the centroid (8 = 1) of the solid enclosed by the cylinder x 2 + y2 = 4, bounded above by the paraboloid z = x 2 + y2, and bounded below by the xy-plane. 4

Centroid

R

x

FIGURE 15.46 Example 2 shows how to find the centroid of this solid.

SoLution We sketch the solid, bounded above by the paraboloid z = r2 and below by the plane z = 0 (Figure 15.46). Its base R is the disk 0 ::; r ::; 2 in the xy-plane. The solid's centroid (x, y, :z) lies on its axis of symmetry, here the z-axis. This makes x = y = O. To find :z , we divide the first moment Mxy by the mass M. To find the limits of integration for the mass and moment integrals, we continue with the four basic steps. We completed our initial sketch. The remaining steps give the limits of integration. The z-limits. A line M through a typical point (r, 0) in the base parallel to the z-axis enters the solid at z = 0 and leaves at z = r2. The r-limits. A ray L through (r,O) from the origin enters R at r = 0 and leaves at r = 2. The O-limits. As L sweeps over the base like a clock hand, the angle 0 it makes with the positive x-axis runs from 0 = 0 to 0 = 27T. The value of Mxy is

Mxy

r7T {2 r

-1 1

{27T {2 [ 2]r2

2

Jo Jo Jo

=

z dz r dr dO

_1

27T 2r 5

-

o

0

2 dr dO -

=

Jo Jo

27T [~]2 12 dO

z2

0

_1

-

r dr dO

27T l§.

000

3 dO -_ 327T 3 .

The value of Mis

M

r7T {2 r

=

Jo Jo Jo

=

Jo Jo

{27T {2

{27T {2 [

2

dz r dr dO

r3 dr dO =

=

Jo Jo

z

]r2 0

{27T [ 4]2 ~ 0 dO =

Jo

r dr dO

r7T

Jo

4 dO = 87T.

879

15.7 Triple Integrals in Cylindrical and Spherical Coordinates Therefore,

z P(p,

_ Mxy 321T 1 4 z = M = -3- 81T = "3 '

cP, 0)

•

and the centroid is (0, 0, 4/3). Notice that the centroid lies outside the solid.

z=pcoscp

7----y /

x FIGURE 15.47 The spherical coordinates 1>, and (J and their relation to x, y, z, and r.

Spherical Coordinates and Integration Spherical coordinates locate points in space with two angles and one distance, as shown in Figure 15.47. The first coordinate, p = 1oFl, is the point's distance from t~ origin. Unlike r, the variable p is never negative. The second coordinate, 4>, is the angle OP makes with the positive z-axis. It is required to lie in the interval [0, 1T]. The third coordinate is the angle 8 as measured in cylindrical coordinates.

p,

DEFINITION Spherical coordinates represent a point P in space by ordered triples (p, 4>, 8) in which 1. p is the distance from P to the origin. 2. 4> is the angle oF makes with the positive z-axis (0 ::; 4> ::; 1T). 3. 8 is the angle from cylindrical coordinates (0 ::; 8 ::; 21T).

z

cp

=

CPo,

p and 0 vary

~

On maps of the Earth, 8 is related to the meridian of a point on the Earth and 4> to its latitude, while p is related to elevation above the Earth's surface. The equation p = a describes the sphere of radius a centered at the origin (Figure 15.48). The equation 4> = 4>0 describes a single cone whose vertex lies at the origin and whose axis lies along the z-axis. (We broaden our interpretation to include the xy-plane as the cone 4> = 1T/2.) If 4>0 is greater than 1T/2, the cone 4> = 4>0 opens downward. The equation 8 = 80 describes the half-plane that contains the z-axis and makes an angle 80 with the positive x-axis.

Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates

= p sin 4>, z = p cos 4>, r

FIGURE 15.48 Constant-coordinate equations in spherical coordinates yield spheres, single cones, and half-planes.

EXAMPLE 3 Solution

x = rcos8 = psin4>cos8, y = rsin8 = psin4>sin8,

Find a spherical coordinate equation for the sphere x 2 + y2

(1)

+ (z

- 1)2

We use Equations (1) to substitute for x , y , and z:

x 2 + y2 + (z - 1)2

=

1

+ p2 sin2 4> sin2 8 + (p cos 4> - 1)2 = 1 p2 sin2 4>( cos2 8 + sin2 8) + p2 cos2 4> - 2p cos 4> + 1 = 1 p2 sin2 4> cos2 8

p2

=

Eqs. (I)

2pcos 4>

p = 2 cos 4> .

p> 0

=

1.

880

Chapter 15: Multiple Integrals z x2

+ y2 + (z

2

- 1)2 = 1 p=2cose/>

The angle cP varies from 0 at the north pole of the sphere to 1T/2 at the south pole; the angle (J does not appear in the expression for p, reflecting the symmetry about the z-axis (see Figure 15.49). •

EXAMPLE 4

Find a spherical coordinate equation for the cone z = v'x 2 + y2.

SoLution 1 Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z = y. The angle between the cone and the positive z-axis is therefore 1T/4 radians. The cone consists of the points whose spherical coordinates have cP equal to 1T/4, so its equation is cP = 1T/4. (See Figure 15.50.)

p

Use algebra. Ifwe use Equations (1) to substitute for x,y, andz we obtain the

SoLution 2

same result:

z FIGURE 15.49

The sphere in Example 3.

=

P cos cP =

v'x2

+ y2

v'p2 sin2 cP

P cos cP = P sin cP

z

Example 3 p

>

0, sine/> ;;, 0

cos cP = sin cP

e/>=JI 4

cP

•

1T

=

4·

Spherical coordinates are useful for describing spheres centered at the origin, half-planes hinged along the z-axis, and cones whose vertices lie at the origin and whose axes lie along the z-axis. Surfaces like these have equations of constant coordinate value:

y

x

p=4

Sphere, radius 4, center at origin

cp=1T 3

Cone opening up from the origin, making an angle of 71/3 radians with the positive z-axis

(J = FIGURE 15.50

1T 3.

The cone in Example 4.

I Volume Differential in Spherical Coordinates

dV = p2 sin e/> dp de/> dO

z p sin e/>

Half-plane, hinged along the z-axis, making an angle of 71/3 radians with the positive x-axis

When computing triple integrals over a region D in spherical coordinates, we partition the region into n spherical wedges. The size of the kth spherical wedge, which contains a point (Ph CPh (h), is given by the changes liPh li(Jh and liCPk in p, (J, and cpo Such a spherical wedge has one edge a circular arc of length Pk liCPh another edge a circular arc of length Pk sin CPk li(Jh and thickness lipk. The spherical wedge closely approximates a cube of these dimensions when liPh li(Jh and liCPk are all small (Figure 15.51). It can be shown that the volume of this spherical wedge Ii Vk is Ii Vk = pl sin CPk liPk liCPk li(Jk for (Ph CPh (h) a point chosen inside the wedge. The corresponding Riemann sum for a function f(p, cP, (J) is n

Sn

= ~ f(Ph CPh (Jk) Pk2 sin CPk liPk liCPk li(Jk. k=l

As the norm of a partition approaches zero, and the spherical wedges get smaller, the Riemann sums have a limit when f is continuous:

n~ Sn =

111

f(p,

cP, (J) dV =

D

111

f(p,

cP, (J) p2 sin cP dp dcp d(J.

D

In spherical coordinates, we have dV = p2 sin cP dp dcp d(J. FIGURE 15.51

dV

= =

In spherical coordinates

dp· p de/> • p sin e/> dO p2 sin e/> dp de/> dO.

To evaluate integrals in spherical coordinates, we usually integrate first with respect to p. The procedure for finding the limits of integration is as follows. We restrict our attention to integrating over domains that are solids of revolution about the z-axis (or portions thereof) and for which the limits for (J and cP are constant.


881

How to Integrate in Spherical Coordinates To evaluate

111

f(p,

cP, 0) dV

D

over a region D in space in spherical coordinates, integrating first with respect to p, then with respect to cp, and finally with respect to 0, take the following steps.

1.

Sketch . Sketch the region D along with its projection R on the ~-plane. Label the surfaces that bound D. z

y x

2.

Find the p-limits of integration. Draw a ray M from the origin through D making an angle cp with the positive z-axis. Also draw the projection of M on the xy-plane (call the projection L). The ray L makes an angle 0 with the positive x-axis. As p increases, M enters D at p = gl(CP, 0) and leaves at p = g2(CP, 0). These are the p-limits of integration. z

(J =

0.-----

y

x

3.

Find the cp-limits of integration. For any given 0, the angle cP that M makes with the z-axis runs from cP = CPmin to cP = CPmax. These are the cP -limits of integration.

882


Find the a-limits of integration. The ray L sweeps over R as a runs from a to (3. These are the a-limits of integration. The integral is

4.

hrJ!

f(p, cp, a) dV

lI= a

D

p

R


4>=4>mm

II)

f(p, cp, a) p2 sin cp dp dcp dO.

p=g,(4), II)

EXAMPLE 5

z

x

1

1l=f314>=4>maxlP=gz(q"

=

L

Y

The ice cream cone in

::5

Find the volume of the "ice cream cone" D cut from the solid sphere I by the cone cp = 1T/3.

SoLution The volume is V = fffvp2 sin cp dp dcp dO, the integral of f(p, cp, a) = I over D . To find the limits of integration for evaluating the integral, we begin by sketching D and its projection R on the xy-plane (Figure 15.52). The p-limits of integration . We draw a ray M from the origin through D making an angle cp with the positive z-axis. We also draw L, the projection of M on the xy-plane, along with the angle athat L makes with the positive x-axis. Ray M enters D at p = 0 and leaves at p = 1. The cp-limits of integration. The cone cp = 1T/3 makes an angle of 1T/3 with the positive z-axis. For any given a, the angle cp can run from cp = 0 to cp = 1T/3. The a-limits of integration. The ray L sweeps over R as a runs from 0 to 21T. The volume is

r

7T

{7T/3 [p3]1

{27T {7T/31 '3 sin cp dcp dO

=

Jo Jo

=

/ 1 + -1) dO = --coscp ]7T 3dO = 127T (-10 27T [1 3 63 0 0

3

0

sin cp dcp dO =

Jo Jo

1 -(21T) = 1T .

6

3

•

EXAMPLE 6 A solid of constant density 8 = 1 occupies the region D in Example 5. Find the solid's moment of inertia about the z-axis. SoLution

In rectangular coordinates, the moment is

In spherical coordinates, x 2 + y2 = (p sin cp cos 0)2 Hence,

+ (p sin cp sin a?

=

p2 sin2 cpo

For the region in Example 5, this becomes

=

1r sJo

7T

(

1

-2 + 1 +

SJo{27T 245 dO = 241 (21T)

1 1) 1 24 - '3 dO =

=

1T

12'

•


883

Coordinate Conversion Formulas C'YuNDRICAL TO

SPHERICAL TO

SPHERICAL TO

REcTANGULAR

RECTANGULAR

CYLINDRICAL

X = rcosO y = rsinO z=z

= psincpcosO

r=psincp

Y = psincpsinO

z=pcoscp

z=pcoscp

0=0

X

Corresponding fonnulas for dV in triple integrals:

dV= dxdydz = dzrdrdO = ; sin cp dp dcp dO In the next section we offer a more general procedure for detennining dV in cylindrical and spherical coordinates. The results, of course, will he the same.

Exercises 15.7 Evaluating Integrals in Cylindrical Coordinates Evaluate the cylindricsl coordinate integrals in Exercises l--{j.

1.

J.

'~J.1J~ dzrdrd6

2.

J.'~ {'/2~ {3+24,.' dz r dr d6

4.

OOr

3. 5.

Jo Jo J.'~J.'Jl~3 dzrdrd6

6.

J.'~

00,.1/3

o

1 {11 o Jo -1/2

/2(r'sin'6

J.'~J.'f,v'i8::;> dzrdrd6 J.~ {'/~ {'~ z dz r dr d6 0

Jo l-v'h'-

Changtng the Order of Integration In Cylindrical Coordinates

9. 10.

J.'~ {' {'/3 r'drdzd6 o

Jo Jo

0

0

J.

1J.V,Jo{'~ r-2

I 1J.'wJo{1+~"4rdrd6dz -I 0

(r' cos' 6 + z')r d6 dr dz

J.'l~/"~ o

8.

(rsin6

•• dzdrd6

b. drdzd6

Finding Iterated Integrals in Cylindrical Coordinates 13. Give the limits of integratioo for evaluatiog the integral

JJJ

f(r, 6, z) dz r dr d6

+ I)rd6dzdr

14. Convert the integrsl

[J.vq[(x' +

a. dz drd6

y') dz dx dy

to an equivalent integrsl in cylindricsl coordinates and evaluate

the result In Exercises 15--20, set up the iterated integral for evaluatiog fffnf(r, 6, z) dz r dr d6 over the given regionD.

15. D is the right circular cylinder whose base is the circle r = 2 sin 6 in the xy.plane and whose top lies in the plane z = 4 - y.

z

0

11. Let D be the region bounded below by the plane z = 0, above by + y' + z' = 4, aod on the sides by the cylinder the sphere + y' = 1. Set up the triple integrals in cylindricsl coordi· nates that give the volume of D using the following orders of in· tegration.

x'

c. d6dzdr

as an iterated integral over the region that is bounded below by the plaoe z = 0, on the side by the cytioder r = cos 6, aod on top by the paraboloid z = 3r'.

+ z')dzrdrd6

The integrals we have seen so far suggest that there are preferred or· ders of integration for cylindricsl coordinates, but ot1rer orders usually work well and are occasionally easier to evaluate. Evaluate the integrsls in Exercises 7-10.

7.

integrals in cylindricsl coordinates that give the volwne of D using the following orders of integration.

z=4-y

x'

b. dr dz d6

c. d6 dz dr

12. Let D be the regioo bounded below by the cone z = y'x' + y' and above by the paraboloid z = 2 y'. Set up the triple

x' -

/

884


16. D is the right circular cylinder whose base is the circle r = 3 cos 0 and whose top lies in the plane z = 5 - x .

z

z z=5-x

y

"r=3cos8

x

x

y=x

17. D is the solid right cylinder whose base is the region in the xy-plane that lies inside the cardioid r = I + cos 0 and outside the circle r = I and whose top lies in the plane z = 4.

z

x

Evaluating Integrals in Spherical Coordinates Evaluate the spherical coordinate integrals in Exercises 21- 26.

rrr {7r {2sin'"

21.

Jo Jo Jo

24.

Jo Jo Jo

25.

1211'111'/312

r=1+cos8

18. D is the solid right cylinder whose base is the region between the circles r = cos 0 and r = 2 cos 0 and whose top lies in the plane z = 3 - y.

r 7r/2 {7r e5p3sin 1> dp d1> dO o

z

p2 sin 1> dp d1> dO

3

3p2 sin 1> dp d1> dO

sec'"

0

{27r {7r/4 {sec'"

26.

Jo Jo Jo

(p cos 1» p2 sin 1> dp d1> dO

Changing the Order of Integration in Spherical Coordinates The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals in Exercises 27- 30. 27.

121-11' 111'11'/4/2 0

p3 sin 21> d1> dO dp

o

19. D is the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = I and whose top lies in the plane z = 2 - y .

z

1 11'/312 esc "'1211'p2 sin 1> dO dp d1> 11'/6 esc'" 0

29.1 11 1 7r

z=2-y

2

28.

ix I

7r /\2P sin 3 1> d1> dO dp

30. {7r/2 {7r/212 5p4 sin31> dp dO d1> J7r/6 J- 7r/2 esc'" 31. Let D be the region in Exercise 11. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.

a. dp d1> dO x

20. D is the prism whose base is the triangle in the xy-plane bounded by the y-axis and the lines y = x and y = I and whose top lies in the plane z = 2 - x.

b. d1> dp dO

v'

32. Let D be the region bounded below by the cone z = x 2 + y2 and above by the plane z = 1. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.

a. dp d1> dO

b. d1> dp dO

885

15.7 Triple Integrals in Cylindrical and Spherical Coordinates Finding Iterated Integrals in Spherical Coordinates In Exercises 33- 38, (a) find the spherical coordinate limits for the in-

volume of D as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find V.

tegral that calculates the volume of the given solid and then (b) evaluate the integral.

41. Let D be the smaller cap cut from a solid ball of radius 2 units by a plane I unit from the center of the sphere. Express the volume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals.

33. The solid between the sphere p = cos 4> and the hemisphere p = 2,z ~

°

z p

= coscf> _

42. Express the moment of inertia Iz of the solid hemisphere x 2 + y2 + z2 ::s I, z ~ 0, as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find I z •

_ _

1-

Volumes Find the volumes of the solids in Exercises 43-48. 43.

x

y

34. The solid bounded below by the hemisphere p = I, Z above by the cardioid of revolution p = I + cos 4>

44.

z

z ~

0, and

z p

=

1

x x

45.

46.

z

x

35. The solid enclosed by the cardioid of revolution p = I - cos 4> 36. The upper portion cut from the solid in Exercise 35 by the xy-plane

r= 3cos(J"-..:

37. The solid bounded below by the sphere p = 2 cos 4> and above by 2 + y2 the cone Z =

y

x

' r = -3 cos (J

x

Yx

47.

y

48.

z

z

y

x

x

38. The solid bounded below by the xy-plane, on the sides by the sphere p = 2, and above by the cone 4> = 1T/3

x

y

r

z

=

cos (J

49. Sphere and cones Find the volume of the portion of the solid sphere p::S a that lies between the cones 4> = 1T/3 and 4> = 21T/3 . p=2 x

y

50. Sphere and half-planes Find the volume of the region cut from the solid sphere p ::s a by the half-planes () = and () = 1T/6 in the first octant.

°

51. Sphere and plane Find the volume of the smaller region cut from the solid sphere p ::s 2 by the plane z = 1.

Finding Triple Integrals 39. Set up triple integrals for the volume of the sphere p = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. 40. Let D be the region in the first octant that is bounded below by the cone 4> = 1T/4 and above by the sphere p = 3. Express the

52. Cone and planes Find the volume of the solid enclosed by the x 2 + y2 between the planes z = I and z = 2. cone z =

Y

53. Cylinder and paraboloid Find the volume of the region bounded below by the plane z = 0, laterally by the cylinder x 2 + y2 = I, and above by the paraboloid z = x 2 + y2.

886


54. Cylinder and p .....boloids Find 1he volwne of the regioo bouoded below by the paraboloid z ~ + y', laterally by the cylinder x' + y' = l,andaboveby1l1eparaboloidz =x' + y' + 1.

73. Moment of inertia of solid cone Find the moment of inertia of a right circular cooe of base radius I and height I about an axis 1hrough 1I1e vertex parallel to 1I1e base. (Take 8 = 1.)

55. Cylinder and cones Find the volume of the solid cut from the thick-walled cylinder 1:5 + y' :5 2 by the cooes z ~

x'

74. Moment of inertia of solid sphere Find the moment of inertia ofa solid sphere of radius a about a diameter. (Take 8 = 1.)

56. Sphere and cylinder Find 1he volume of the regioo that lies inside the sp\rere + y' + z, ~ 2 and outside the cylinder

75. Moment of inertia of solid cone Find the moment of inertia of a right circular cone of base radius a and height h about its asis. (Hint: Place the cone with its vertex at the origin and its axis aloog 1I1e z-axis.)

x'

±Vx' + y'.

x'

x'+y'=1. 57. Cylinder and planes Find the volume of the regioo enclosed by the cylinder + y' ~ 4 and the planes z ~ 0 and y + z ~ 4.

x'

58. Cylinder and planes Find the volume of the regioo enclosed by the cylinder + y' = 4 and the planes z = 0 and x + y + z = 4.

x'

76. Variable density A solid is bounded on 1I1e top by 1I1e paraboloidz = r',oothebottombytheplanez = O,andonthesidesby 1I1e cylinder r = 1. Find 1I1e center of mass and 1I1e moment of inertia about the z-axis if the density is

a. 8(r, 0, z) = z

b. 8(r, 0, z) = r.

59. Region trapped by paraboloids Find the volume of the regioo bounded above by 1he paraboloid z = 5 y' and below by the paraboloid z = 4x' + 4y'.

77. Variable density A solid is bounded below by the cone z = + y' and above by the plane z = I. Find 1I1e center of mass and the moment of inertia about the z-axis if the density is

60. Paraboloid and cylinder Find the volume of the regioo bounded above by the paraboloid z = 9 y', below by the xy-plane, and lying outside the cylinder x' + y' ~ 1.

78. Variable density A solid ball is bounded by the sphere p Find the moment of inertia about the z-axis if the density is

x' -

x' -

61. Cylinder and sphere Find 1he volume of the region cut from the solid cylinder + y' :5 I by the sphere + y' + z, = 4.

x'

x'

62. Sphere and paraboloid Find the volume of the region bounded above by the sphere + y' + z, ~ 2 and below by the paraboloidz = x 2 + y2.

x'

Average Values 63. Find the average value of the functioo I(r, 0, z) = r over the regioo bounded by the cylinder r ~ I between the planes z ~ -I andz = 1. 64. Find the average value of the function I(r, 0, z) = r over the solid ball bounded by the sphere + z, = 1. (This is 1he sphere + y' + = 1.)

x'

r'

z'

65. Find the average value of the function I(p, y-equations for the boundary of R

= =

x y/2 x (y/2) y=O y=4

+

1

Corresponding uv-equations for the boundary of G

Simplified uv-equations

+v =2v/2 =v u + =(2v/2) + 1=v + 1 2v =0 2v =4

u=O u v=O v=2

u v

=1

The Jacobian of the transformation (again from Equations (5» is

ax au J(u, v) = ay au

ax av ay av

a au(u + v)

a av(u + v)

a au (2v)

a av (2v)

=

I~ ~I

=

2.

We now have everything we need to apply Equation (I):

:L

4

1 o

v

X

J

- (Y/2 +I

2x - y -2-dxdy

1 1"-1 2

V

=

-

x- yf2

ul.J(u,v)ldudv

v- O

u=O

=f[(U)(2) du dv =f

v=u

I

[u

2

]:

dv

=f dv =2.

--c1{--+--~U

• EXAMPLE 3

v=-2u

Evaluate

{' {'-x

-2

10 10

~ (y

- 2x)' dy dx.

Solution

We sketch the region R of integration in the xy-plane and identify its boundaries (Figure 15.56). The integrand suggests the transformation u = x + y and v = y - 2x. Routine algebra produces x and y as functions of u and v:

y

x

=

u

v

3 - 3'

y

2u

=

v

3 + 3'

(6)

From Equations (6), we can find the boundaries of the uv-region G (Figure 15.56).

x=o R -o ~--' \--~~X

.>y-equations for the boundary of R

Corresponding uv-equations for the boundary of G

Simplified uv-equations

y~O

FIGURE 15.56 The equations x ~ (u/3) - (v/3) andy ~ (2u/3) + (v/3) transform G into R. Reversing the transformation by the equations u ~ x + y and v ~ y - 2x transforms R into G (Example 3).

x+y=1 x=O y=O

(1-¥)+(2;+¥)=1 '!_"=O 3 3 2u+,,=0 3 3

u

=I

v=u

v

=-2u

890

Chapter 15: Multiple Integrals The Jacobian of the transfonnation in Equations (6) is

J(u, v) =

ax au

ax

I

I

av

3

3

ay au

ay av

2 3

3

I 3.

I

Applying Equation (I), we evaluate the integral:

=

1'1" o

-2u

(I)

1'£' [I

u'(2v 2 3

dv du = 3

0

u'(2 -v3 3

]V-"

du

v - -2u

• In the next example we illustrate a nonlinear transfonnation of coordinates resulting from simplifying the fonn of the integrand Like the polar coordinatea' transfonnation, nonlinear transfonnations can map a straight line boundary of a region into a curved boundary (or vice versa with the inverse transfonnation). In general, nonlinear transfunnations are more complex to ana1yze than linear ooes, and a complete treatment is left to a more advanced course. y

EXAMPLE 4


2 1----\~-"'--_;,f___:

1, l1/r VX~eVxYaxdy. 2

y

1

Solution

The square root terms in the integt'lll!d suggest that we might simplify the inteand v = vYf;:. Squaring these equatioos, we readily gratioo by substituting u = have u 2 = xy and v 2 = y/x, which imply that u 2if = y2 and u 2/if = x 2. So we obtain the transfonnation (in the same ordering of the variables as discussed before)

~~--~----~--~.x

o

1

2

'\IxY

u x=v

FIGURE 15.57 The region of integration R in Example 4.

y = uv.

and

Let's firat see what happens to the integrand itself under this transfonnatioo. The Jacobian of the transfonnation is

u=1

J-----

J(u, v) =

xy=l

/

v=l

1

2

ax av

ay au

ay av

I

-u

v

u

v if

2u v'

If G is the region of integration in the uv-plane, then by Equation (I) the transfonned

I

y=x ,

double integral under the substitution is

I

0

ax au

U

FIGURE 15.58 The boundaries of the region G correspond to those of region R in Figure 15.57. Notice as we move counterclockwise around the region R, we also move counterclockwise around the region G. The inverse transformation equations u ~ v'i,Y, v ~ vY7i prodoce the regioo G from the region R.

11 ~ R

e VxY ax dy =

11

ve"

~ du dv =

G

11

2ue" du dv.

G

The transfonned integrand function is easier to integrate than the original one, so we proceed to detennine the limits of integratioo for the transfonned integral. The regioo of integratioo R of the original integral in the xy-plane is shown in Figore and v = vYf;:, we see that the image of 15.57. From the substitution equations u = the left-hand boundary xy = I for R is the vertical line segment u = 1,2 ;;,: v ;;,: I, in G (see Figure 15.58). Likewise, the right-hand boundary y = x of R maps to the horizontal line segment v = I, I :5 U :5 2, in G. Finally, the horizontal top boundary Y = 2 of R

'\IxY

15.8 Substitutions in Multiple Integrals

891

maps to uv = 2, 1 ::; v ::; 2, in G. As we move counterclockwise around the boundary of the region R, we also move counterclockwise around the boundary of G, as shown in Figure 15.58. Knowing the region of integration G in the uv-plane, we can now write equivalent iterated integrals:

1P ~ 2

I

ll /y \jx

e vXY dx dy

=

11 2

I

2U /

2ue U dv duo

Note the order of integration.

I

We now evaluate the transformed integral on the right-hand side,

11 2

2U /

u

2ue dv du

=

212 ]~:~/u 212 212 (2 -

=

2[(2 - u)e U

=

2(e 2

= =

vue U

(2e U

du

ue U ) du

-

u)e U du

(e

-

+

+ et:~

Integrate by parts.

e» = 2e(e - 2).

•

Substitutions in Triple Integrals The cylindrical and spherical coordinate substitutions in Section 15.7 are special cases of a substitution method that pictures changes of variables in triple integrals as transformations of three-dimensional regions. The method is like the method for double integrals except that now we work in three dimensions instead of two. Suppose that a region G in uvw-space is transformed one-to-one into the region D in xyz-space by differentiable equations of the form

x

=

g(u, v, w),

y

= h(u, v, w),

z = k(u, v, w),

as suggested in Figure 15.59. Then any function F(x, y, z) defined on D can be thought of as a function F(g(u, v, w), h(u, v, w), k(u, v, w»

= H(u, v, w)

defined on G. If g, h, and k have continuous first partial derivatives, then the integral of F(x, y, z) over D is related to the integral of H(u , v , w) over G by the equation

fff

F(x, y, z) dx dy dz =

D

fff

H(u, v, w) IJ(u , v, w) Idu dv dw.

G

z

w x = g(u, v, w) y = h(u, v, w) z = k(u, v, w) )

u

Cartesian uvw-space

x

Cartesian xyz-space

FIGURE 15.59 The equations x = g(u, v, w),y = h(u, v, w), and z = k(u, v, w) allow us to change an integral over a region D in Cartesian xyz-space into an integral over a region G in Cartesian uyw-space using Equation (7).

(7)

892


z

Cube with sides parallel to the coordinate axes

The factor J(u, v, w), whose absolute value appears in this equation, is the Jacobian

determinant

J(u, v, w) =

()

r

iJx iJu

iJx iJv

iJx iJw

iJy iJu

iJy iJv

iJy iJw

iJz iJu

iJz iJv

iJz iJw

iJ(x,y,z) iJ(u, v, w)·

Cartesian r()z-space

x = rcos ()

= I yz=z

rsin ()

This determinant measures how much the volume near a point in G is being expanded or contracted by the transformation from (u, v, w) to (x, y, z) coordinates. As in the twodimensional case, the derivation of the change-of-variable formula in Equation (7) is omitted. For cylindrical coordinates, r, 8 , and z take the place of u, v, and w. The transformation from Cartesian r8z-space to Cartesian ~z-space is given by the equations

z

x

= rcos8,

y

=

z=z

rsin 8,

(Figure 15.60). The Jacobian of the transformation is

J(r, 8, z) = x

Cartesian xyz-space

FIGURE 15.60 The equations x = r cos 8, y = r sin 8, and z = z transform the cube G into a cylindrical wedgeD.

iJx iJr

iJx iJ8

iJx iJz

iJy iJr

iJy iJ8

iJy iJz

cos 8 sin 8 0

iJz iJr

iJz iJ8

= r cos2 8

-r sin 8

0

r cos 8

0

0

iJz iJz

+

r sin2 8 = r.

The corresponding version of Equation (7) is

fff

F(x,y,z)dxdydz

fff

=

H(r,8,z)lrldrd8dz.

G

D

We can drop the absolute value signs whenever r ;:::: O. For spherical coordinates, p, 0, b > 0, in the xy-plane. Find the frrst moment of the plate about the origin. (Hint: Use the transformationx = arcos8,y = brsin8.) 12. The area of an ellipse The area '/Tab of the ellipse x'/a' + y'/b' = 1 can be found by integrating the function f(x, y) = I over the region bounded by the ellipse in the xy-plane. Evaluating the integral directly requires a trigonometric substi1ution. Au easier way to evaluate the integral is to use the transformation x = au, Y = bv and evaluate the transformed integral over the disk G: + s 1 in the uv-plaue. Find the area this way.

22. Volume of an ellipsoid

= au, y = 00, and z = cwo Then find the volume of an appropriate region in uvw-space.)

23. Evaluate

over the solid ellipsoid

x2

=

(x 2

24. Let D be the region in xyz-space defmed by the inequalities

1

15. Use the transformation x = u/v,y = uv to evaluate the integral som

1

+ y2) dx dy + 1414~ (x 2 + y2) dx dy. 2

y/4

= cw. Then integrate over an

v to evaluate the

by lust writing it as an integral over a region G in the uv-plane.

Y

abc

= au, y = bv, and z appropriate region in uvw-space.)

' [,,+4)/2 y'(2x _ y)e(2r-y )' dx dy ]. o }yf2

11l/y

z2

(Hint: Let x

by frrst writing it as an integral over a region G in the uv-plane.

+ (i/2)v,y

y2

2+2+2~1.

13. Use the transformation in Exercise 2 to evaluate the integral

].'/3l'-2y(x + 2y)e,,-r) dx dy

Find the volume of the ellipsoid

(Hint: Let x

u' v'

14. Use the transformation x = u integral

Yx' + y'dydx.

17. Find the Jacobiana(x,y)/a(u, v) of the transformation

//2(X - y)dxdy.

2

{2VFx

10 10

R

for the region R in the fIrst quadrant bounded by the lines y = -(3/2)x + I, y = -(3/2)x + 3, y = -(I/4)x, and y = -(1/4)x + I.

895

~

x

~

2,

0

~

xy

~

2,

0

~

z

~

1.

Evaluate

ff/(x'y

+ 3xyz) dx dy dz

D

by applying the transfonnation

u

= x,

v

= xy,

w

= 3z

and integrating over an appropriate region G in uvw-space.

896


25. Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base toward the top, show, by traosforming the appropriate integrals, that the center of mass of a solid semiellipsoid (x 2/a 2) + (y2/b 2) + (z2/e2) '" I, Z '" 0, lies on the z-axis three-eighths of the way from the base toward the top. (You cao do this without evaluating aoy of the integrals.)

Chapter

iI!"3

26. Cylindrical shell. In Section 6.2, we learned how to rmd the volume of a solid of revolution using the shell method; namely, if the region between the curve y = f(x) aod the x-axis from a to b (0 < a < b) is revolved about the y-axis, the volume of the 2'1fxf(x) dx. Prove that rmding volumes by resulting solid is using triple integrals gives the same result. (Hint: Use cylindrical coordinates with the roles ofy andz cbanged.)

1:


1. Derme the double integral of a function of two vatiables over a

7. How are double and triple integrals in rectangular coordinates

2. How are double integrals evaluated as iterated integrals? Does the

used to calculate volumes, average values, masses, moments, and centers of mass? Give examples.

order of integration matter? How are the limits of integration de-

8. How are triple integrals dermed in cylindrical and sphetical coor-

bounded region in the coordinate plane.

termined? Give examples.

3. How are double integrals used to calculate areas and average values. Give examples.

dinates? Why might one prefer working in one of these coordinate systems to worldng in rectangular coordinates? 9. How are triple integrals in cylindrical and sphetical coordinates evaluated? How are the limits of integration fouod? Give examples.

4. How can you change a double integral in rectangular coordinates into a double integral in polar coordinates? Why might it be worthwhile to do so? Give an example.

10. How are substitutions in double integrals pictured as transformations of two-dimensional regions? Give a sample calculation.

S. Derme the triple integral of a function f(x, y, z) over a bounded region in space.

11. How are substitutions in triple integrals pictured as transformations ofthree-dimensiona\ regions? Give a sample calculation.

6. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example.

Chapter

ilM

Practice Exercises

Evaluating Double Iterated Integrals

Areas and Volumes Using Double Integrals

In Exercises 1-4, sketch the region of integration and evaluate the double integral.

13. Area between line and parabola Find the area of the region enclosed by the line y = 2x + 4 and the parabola y = 4 - x 2 in the xy-plane.

['0 {1/Y

1.

3.

J1 Jo ye" dx dy f2:Lv9=4t' ' tdsdt /, o

-V9-4t1

2./,1/," 4.

eY/r dydx

1J2/,ovY

vY xydxdy

In Exercises 5-8, sketch the region of integration and write an equivalent integral with the order of integration reversed. Then evaluate both integrals.

4:Lr,-4)f2

s. /, o -V4=Y 7.

' /,o

dxdy

6.

-V9-4,,'

Jo iX2

{2 {,-r'

f2

:Lv9=4Y' ydxdy

[' {\.;;. dy dx

8.

Jo Jo

2xdydx

Evaluate the integrals in Exercises ~12.

rz h~

9. {'

8

11. /, o

4 cos (x 2 ) dx dy

dydx ~ y4 + 1 2

h

10. 12.

{2 (1 er' dx dy

h~

/,1h 1

2'If sm . 'IfX 2 dx dy o"o/Y x2

14. Area bonnded by lines and parabola Find the area of the ''triangular" region in the xy-plane that is bounded on the right by the parabola y = x 2 , on the left by the line x + y = 2, and above by the line y = 4. 15. Volume of the region under a p ...aboloid Find the volume under the psraholoid Z = x 2 + y2 above the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xy-plane.

16. Volume of the region under p ...abolic cylinder Find the volume under the parabolic cylinder z = x 2 above the region enclosed by the parabola y = 6 - x 2 and the line y = x in the xy-plane.

Average Values Find the average value of f(x, y) and 18.

= xy over the regions in Exercises 17

17. The square bounded by the lines x

=

I, Y

=

I in the first

quadrant 18. The quarter circle x 2 + y2 '" I in the fIrst quadrant

Chapter 15

Polar Coordinates Evaluate the integrals in Exercises 19 and 20 by changing to polar coordinates.

19.

2dydx 1 11~ _~ (I + x2 + y2)2

In (x 2 + y2

+

31. Cylindrical to rectangular coordinates

21. Integrating over lemniscate Integrate the function f(x, y) = I/O + x 2 + y2)2 over the region enclosed by one loop of the lemniscate (x 2 + y2)2 - (x 2 - y2) = O. 22. Integrate f(x,y) = I/O + x 2 + y2)2 over and

The triangle with vertices (0, 0), (1, 0),

(I, v3).

b. First quadrant

The first quadrant of the xy-plane.

cos (x

l 710ln 21ln4 25.1 1'1 in

24.

26.

In 5

In6

0

1

x

j I

r ~ 0

r

to (a) rectangular coordinates with the order of integration dz dx dy and (b) spherical coordinates. Then (c) evaluate one of the integrals. 32. Rectangular to cylindrical coordinates (a) Convert to cylindrical coordinates. Then (b) evaluate the new integral.

33. Rectangular to spherical coordinates (a) Convert to spherical coordinates. Then (b) evaluate the new integral.

+ Y + z) dx dy dz

r dzdydx 1 11~ -~ JVx'+y'

e(x+y+z) dz dy dx

- I

x y + (2x - y - z) dz dy dx

0

io(2"ior./2!~3 dz r dr dO,

-~

elxlz2Y 3dydzdx I

Convert

21xy2 dz dy dx 1o 11~1(X'+Y') - (x'+y')

Evaluating Triple Iterated Integrals Evaluate the integrals in Exercises 23- 26.

23.1"1"1"

29. Average value Find the average value of f(x,y,z) = 30xz ~ over the rectangular solid in the first octant bounded by the coordinate planes and the planes x = I, y = 3, z = I.

Cylindrical and Spherical Coordinates

1) dx dy

- I

a. Triangular region

897

30. Average value Find the average value of p over the solid sphere p ::s a (spherical coordinates).

- I

20.111v'-v'JJ=Y2 =Y2

Practice Exercises

Z

Volumes and Average Values Using Triple Integrals 27. Volume Find the volume of the wedge-shaped region enclosed on the side by the cylinder x = -cosy, -7r/2 ::s y ::s 7r/2, on the top by the plane z = - 2x, and below by the xy-plane.

z

z=-2x

34. Rectangular, cylindrical, and spherical coordinates Write an iterated triple integral for the integral of f(x, y, z) = 6 + 4yover the region in the first octant bounded by the cone z = v'x 2 + y2 , the cylinder x 2 + y2 = I, and the coordinate planes in (a) rectangular coordinates, (b) cylindrical coordinates, and (c) spherical coordinates. Then (d) find the integral of f by evaluating one of the triple integrals. 35. Cylindrical to rectangular coordinates Set up an integral in rectangular coordinates equivalent to the integral

{"/2

r.f3 {~ r 3(sin 0 cos

io il il

O)z2 dz dr dO.

Arrange the order of integration to be z first, then y, then x. 36. Rectangular to cylindrical coordinates

1

2iva- x'iv4- x' - y'

o

2 7T

'2

x

y

28. Volume Find the volume of the solid that is bounded above by the cylinder z = 4 - x 2, on the sides by the cylinder x 2 + y2 = 4, and below by the xy-plane.

z

0

The volume of a solid is

dzdydx.

- V4 - x'- y'

a. Describe the solid by giving equations for the surfaces that form its boundary. b. Convert the integral to cylindrical coordinates but do not evaluate the integral. 37. Spherical versus cylindrical coordinates Triple integrals involving spherical shapes do not always require spherical coordinates for convenient evaluation. Some calculations may be accomplished more easily with cylindrical coordinates. As a case in point, find the volume of the region bounded above by the sphere x 2 + y2 + z2 = 8 and below by the plane z = 2 by using (a) cylindrical coordinates and (b) spherical coordinates.

Masses and Moments 38. Finding /z in spherical coordinates Find the moment of inertia about the z-axis of a solid of constant density 8 = I that is bounded above by the sphere p = 2 and below by the cone cf> = 7r/3 (spherical coordinates).

898


39. Moment of inertia of. "thick" sphere Find the moment of inertia of a solid of constant density 8 bounded by two concentric spheres of radii a and b (a < b) about a diameter. 40. Moment of inertia of an apple Find the moment of inertia about the z-axis of a solid of density 8 ~ I enclosed by the spherical coordinate surface p ~ 1 - cos"'. The solid is the red curve rotated about the z-axis in the accompanying figure.

z

bounded by the line y ~ x and the parabola y ~ x 2 in the xy-plane if the density is 8(x, y) ~ x + 1 . 47. Plate witb variable density Find the mass and first moments about the coordinate axes of a thin square plate bounded by the linesx ~ ± I, Y ~ ± I in the xy-plane if the density is 8(x, y) ~ x 2 + y2 + 1/3. 48. Triangles with same inertial moment Find the moment of inertia about the x-axis of a thin triangular plate of constant density 8 whose base lies along the interval [0, b1on the x-axis and whose vertex lies on the line y = h above the x-axis. As you will see, it

does not matter where on the line this vertex lies. All such triangles have the same moment of inertia about the x-axis. 49. Centroid Find the centroid of the region in the polar coordinate plane defined by the inequalities 0 :5 r :5 3, -7T/3 :5 9 :5 7T/3. 50. Centroid Find the centroid of the region in the first quadrant bounded by the rays 9 ~ 0 and 9 ~ 7T/2 and the circles r ~ I andr ~ 3.

51. a. Centroid Find the centroid of the region in the polar coordinate plane that lies inside the cardioid r = 1 + cos (} and outside the circle r 41. Centroid Find the centroid of the "triangular" region bounded by the lines x ~ 2, Y ~ 2 and the byperbola xy ~ 2 in the xy-plane. 42. Centroid Find the centroid of the region between the parabola x + y2 - 2y ~ 0 and the line x + 2y ~ 0 in the xy-plane. 43. Polar moment Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 ~ 3 bounded by the y-axis and the lines y ~ 2x and y ~ 4 in the xy-plane. 44. Polar moment Find the polar moment of inertia about the center of a thin rectangular sheet of constant density 8 ~ 1 bounded by the lines

a. x

~

1>. x

~

±2, Y ±a, y

~

~

± I in the xy-plane ±b in the xy-plane.

(Hint: Find Ix. Then use the formula for Ix to fmd Iy and add the two to find 10).

~

1.

b. Sketch the region and show the centroid in your sketch.

52. a. Centroid Find the centroid of the plane region defined by the polar coordinate inequalities 0 ~ r ~ a, -a ~ (J ~ a (0 < a::;; '7T).Howdoesthecentroidmoveasa~'7T-? b. Sketch the region fora ~ 57T/6 and show the centroid in your sketch.

Substitutions S3. Show that if u

~

x - y and v

~

y, then

54. What relationship must hold between the constants a, b, and c to mske

45. Inertial moment Find the moment of inertia about the x-axis of a thin plate of constant density 8 covering the triangle with vertices (0, 0), (3, 0), and (3, 2) in thexy-plane. 46. Plate with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin plate

Chapter

iI5

(Hint: l.et s ~ ax + ,By and t ~ oyx ac - b 2. Thenax2 + 2bxy + cy2 =

+ 8y, where (a8 - ,Boy)' 2 2 8 + t .)

~

Additional and Advanced Exerdses

Volumes 1. Sand pile: double and triple integrals The base of a sand pile covers the region in the xy-plane that is bounded by the parabola x 2 + y ~ 6 and the line y ~ x . The height of the sand above the point (x, y) is x 2• Express the volume of sand as (a) a double integral, (b) a triple integral. Then (0) fmd the volume.

3. Solid eylindrioal region between two planes Find the volume of the portion of the solideylinderx 2 + y2:5 I that lies between the planes z ~ o and x + y + z ~ 2.

2. Water in a hemispherioal bowl A hemispherical bowl of radius 5 em is filled with water to within 3 em of the top. Find the volume of water in the bowl.

5. Two paraboloid. Find the volume of the region bounded above by the paraboloid z ~ 3 - x 2 - y2 and below by the paraboloid z ~ 2x 2 + 2y2.

4. Sphere and paraboloid Find the volume of the region bounded above by the sphere x 2 loidz = x 2 + y2.

+ y2 + z2

~ 2 and below by the parabo-

Chapter 15 Additional and Advanced Exercises 6. Spherical coordinates Find the volume of the region enclosed by the spherical coordinate surface p = 2 sin cP (see accompanying figure).

z p=2sincp

Similarly, it can be shown that

1l"

X

V

em(x-t) f(t) dt du dv = lx (x - t)2 em(x-t) f(t) dt. 2 l0 0 0 0

14. Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that

11

y

X

f(x) ( l g(x - y)f(y) dY ) dx =

7. Hole in sphere A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is

=

11 (1

1

f(y)

g(x - y)f(x) dx) dy

Ire Jo

g( Ix - yl )f(x)f(y) dx dy.

2Jo

Masses and Moments

[2" [v'3 [~

Jo Jo Jl

V= 2

899

15. Minimizing polar inertia A thin plate of constant density is to occupy the triangular region in the first quadrant of the xy-plane having vertices (0, 0), (a, 0), and (a, l/a). What value of a will minimize the plate's polar moment of inertia about the origin?

r dr dz dO.

a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral.

8. Sphere and cylinder Find the volume of material cut from the solid sphere r2 + z2 ::::; 9 by the cylinder r = 3 sin o. 9. Two paraboloids Find the volume of the region enclosed by the surfaces z = x 2 + y2 and z = (x 2 + y2 + 1)/2.

16. Polar inertia of triangular plate Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 = 3 bounded by the y-axis and the lines y = 2x and y = 4 in the xy-plane.

10. Cylinder and surface z = xy Find the volume of the region in the first octant that lies between the cylinders r = 1 and r = 2 and that is bounded below by the xy-plane and above by the surfacez = xy.

17. Mass and polar inertia of a counterweight The counterweight of a flywheel of constant density 1 has the form of the smaller segment cut from a circle of radius a by a chord at a distance b from the center (b < a). Find the mass of the counterweight and its polar moment of inertia about the center of the wheel.

Changing the Order of Integration

18. Centroid of boomerang Find the centroid of the boomerangshaped region between the parabolas y2 = -4(x - 1) and y2 = -2(x - 2) in thexy-plane.

11. Evaluate the integral

1

00

e-ax -

o

dx. Theory and Examples

(Hint: Use the relation

e -ax - e -bx x

e-bx

x

=

l

19. Evaluate b

e-XYdy

a

to form a double integral and evaluate the integral by changing the order of integration.) 12. a. Polar coordinates that

ra Jo

sin {3

Show, by changing to polar coordinates,

[Y';-I- In (x 2 + y2) dx dy = a 2{3 (Ina -

Jy cot{3

where a

where a and b are positive numbers and

t),

20. Show that

> 0 and 0 < (3 < 'TT" /2.

b. Rewrite the Cartesian integral with the order of integration reversed.

over the rectangle Xo ::::; x ::::; xl. Yo ::::; Y ::::; yl. is

13. Reducing a double to a single integral By changing the order of integration, show that the following double integral can be reduced to a single integral: l

x l

u

em(x-t) f(t) dt du = l

x (x - t)em(x-t) f(t) dt.

21. Suppose that f(x,y) can be written as a product f(x, y) = F(x )G(y) of a function of x and a function of y. Then

900


the integral off over the rectangle R: a ~ x ~ b, c be evaluated as a product as well, by the formula

ff

dx)

!(x,y) dA = ( [ F(x)

(t

~

y

G(y) dy ).

~

d can

a. If you have not yet done Exercise 41 in Section 15.4, do it now to show that

(1)

1=

Joroo e f dy = Y; ;rr.

R

The argument is that

ii

!(x,y) dA

=

R

= =

t ([ dx) t (G(Y{ t ([ dx )G(Y) F(x)G(y)

b. Substitute y = Vi in Equation (2) to show that r( 1/2) = 2I = Y-ii. dy

F(x)dx) dy

F(x)

dy

= ( [ F(X)dx)t G(y)dy. L

(i)

(ii) (iii)

(iv)

Give reasons for steps (i) through (iv).

vertical.

When it applies, Equation (I) can be a time-saver. Vse it to evaluate the following integrals. b.

{1n2 {~/2 eX cosy dy dx

Jo Jo

c.

24. Total electriclll chorge over circul... plate The electrical charge distribution on a circular plate of radius R meters is O"(r, 0) = kr(1 - sinO) coulomb/m2 (k a constant). Integrate 0" over the plate to fmd the total charge Q. 25. A parabolic rain gauge A bowl is in the shape of the graph of z = x 2 + y2 from z = 0 to z = 10 in. You plan to calibrate the bowl to make it into a rain gauge. What height in the bowl would correspond to 1 in. of rain? 3 in. of rain? 26. Water in ...temte dish A parabolic satellite dish is 2 m wide and 1/2 m deep. Its axis of symmetry is tilted 30 degrees from the

{2 {I x dx dy

11 i-I Y 2

22. Let Du! denote the derivative of !(x,y) = (x 2 + y2)/2 in the direction of the unit vector u = U1 i + U2j. Finding average value Find the average value of Du! over the triangular region cut from the fIrSt quadrant by the line x+y=l. b. Average value and centroid Show in general that the average value of Du! over a region in the lJ'-plane is the value of D.! at the cen1roid of the region. 23. The value ofr(I/2) The gamma function, L

a. Set up, but do not evaluate, a triple integral in rectangular coordinates that gives the amount ofwator the satellite dish will hold. (Hint: Put your coordinate system so that the satellite dish is in "standsrd position" and the plane of the water level is slanted) (Caution: The limits of integration are not "nice.") b. What would be the anta1lest tilt of the satellite dish so that it holds no water? 27. An infinite half-cylinder Let D be the interior of the infmite right circular half-cylinder of radius I with its single-end face suspended I unit above the coigin and its axis the ray from (0, 0, I) to 00. Vse cylindrical coordinates to evaluate

iff

z(r2

+ z 2t'/2 dV.

D

rex) = 1,00 1"-1 e-t dt,

I: dx

extends the factorial function from the noonegative integers to other real values. Of particular interest in the theery of differential equations is the number

r(1)

=

2

Chapter

n~

roo t(I/2)-1 e-' dt = roo e-t dt.

Jo

Jo Vt

(2)

28. Hypervolume We have learned that I is the length of the interval [a, b] on the number line (one-dimensional space), ffR I dA is the area ofregionR in thexy-plane (two.. The curved path C,: ret) = ti

+ t'j + t 4k,

1. f(x,y,z)

=

(x' + y' + z')-l/2

2. f(x,y,z) = InVx' + y' + z' 3. g(x,y, z) = e' - In (x' + y') 4. g(x,y, z) = xy + yz + xz 5. Give a fonnula F = M(x, y)i + N(x, y)j for the vector field in the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from (x, y) to the origin. (The field is not defined at (0, 0).) 6. Give a fonnula F = M(x, y)i + N(x, y li for the vector field in the plane that has the properties that F = 0 at (0, 0) and that at any other point (a, b), F is tangent to the circle x' + y' = + b' and points in the clockwise direction with magnitode IFI = Va' + b'.

a'

c. The path C, U C4 consisting of the line segment from (0, 0, 0) to (I, 1,0) followed by the segment from (I, 1,0) to (I, I, I)

7. F = 3yi + 2xj + 4zk 9. F = Vzi - 2xj + Vyk

8. F = [1/(x' + l)]j 10. F = xyi + yzj + X2k

11. F = (3x' - 3x)i + 3zj + k 12. F

= (y + z)i +

(z

+ xli + (x + y)k z

(0,

o. 0)

~

Line Integrals nf Vector Fields In Exercises 7-12, fmd the line integrals of F from (0, 0, 0) to (I, I, I) over each of the following paths in the accompanying figore.

0 s t '" I 0 '" t '" I

(I, 1.0)

918

Chapter 16: Integration in Vector Fields

Une Integrals with Respect to x, y, and z

Une Integrals in the Plane

In Exercises 13-16, fmd the line integrals along the given path C.

23. Evaluate Ie xy d< + (x + y) dy along the curve y ~ x' from (-I,l)to(2,4).

13.1 (x -

y)d Part 2 We want tu show that for any two points A and Bin D, the integral of F • dr has the same value over any two paths C 1 and C2 from A to B. We reverse the direction on C2 to make a path -C2 fromB toA (Figure 16.24). Together, C, and -C2 make a closed loop C, and by assumption,

FIGURE 16.24 Ifwe have two p.ths from A to B, one of them cao be reversed to make.loop.

Thus, the integrals over C , and C2 give the same value. Note that the definition of F' dr shows that changing the direction along a curve reverses the sign of the line integral.

16.3 B

B

925

Path Independence, Conservative Fields, and Potential Functions

Proof that Part 2 => Part 1 We want to show that the integral of F • dr is zero over any closed loop C. We pick two points A and B on C and use them to break C into two pieces: Cl from A to B followed by C2 from B back to A (Figure 16.25). Then

f

F ' dr =

c

r

lei

F.dr

+

r

icz

F.dr =

r F.dr - JA-r F·dr

JA.

=

O.

•

The following diagram summarizes the results of Theorems 2 and 3. A

A

Theorem 2

If A andB lie on a loop, we cao reverse part of the loop to lIIlIke two patha from A to B. FIGURE16.25

F = VIon D

=

Theorem 3

= fF'dr

F conservative onD

=

0

c

over any loop in D

Two questions arise:

1.

How do we know whether a given vector field F is conservative?

2.

If F is in fact conservative, how do we f"md a potential function I (so that F = V/)?

Finding Potentials for Conservative Fields The test for a vector field being conservative involves the equivalence of certain partial derivatives of the field components.

Component Test for Conservative Fields Let F = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k be a field on a connected and simply connected domain whose component functions have continuous first partial derivatives. Then, F is conservative if and only if

oP oy

ON

oM

OP

oz'

iJz

ax'

oN oX

and

oM oy'

(2)

Proof that Equations (2) hold if F is conservative There is a potential function I such that

F = Mi

+ Nj + Pk

01

= oX

i

+

01

oyj

+

01

oz k.

Hence, 2

oP _ l.. (01) _ 0 1 oy - oy OZ - oyoz

all

Mixed Derivative Theorem, Section 14.3

ozoy

=:z (:;)

=

~~.

The others in Equations (2) are proved similarly.

•

The second half of the proof, that Equations (2) imply that F is conservative, is a consequence of Stokes' Theorem, taken up in Section 16.7, and requires our assumption that the domain of F be simply connected.

926

Chapter 16: Integration in Vector Fields Once we know that F is conservative, we usually want to find a potential function for F. This requires solving the equation Vj = F or

oj i ox

+ oj. + oj k oyJ

OZ

Mi

=

+ N· + Pk J

for j. We accomplish this by integrating the three equations

oj oX =M,

oj oy =N,

oj oz =P,

as illustrated in the next example.

EXAMPLE 3 Show that F = (eX cosy + yz)i + (xz - eX siny)j servative over its natural domain and find a potential function for it.

+ (xy + z)k is con-

Solution The natoral domain of F is all of space, which is connected and simply connected. We apply the test in Equations (2) to M = eXcosy

+ yz,

N

=

xz - eXsiny,

P=xy+z

and calculate

OP oN oy =x= oz'

oM

iiZ

OP

oN = -exsiny oX

= y = ox'

+z

= oM.

oy

The partial derivatives are continuous, so these equalities tell us that F is conservative, so there is a function j with Vj = F (Theorem 2). We fmd j by integrating the equations

oj = eXcosy oX

-

+ yz

'

oj oy = xz - eX siny,

oj oz = xy

+ z.

(3)

We integrate the first equation with respect to x, holding y and z fIXed, to get

j(x,y, z) = eX cosy

+ xyz + g(y, z).

We write the constant of integration as a function of y and z because its value may depend ony andz, though not onx. We then calculate ojjoy from this equation and match it with the expression for ojjoy in Equations (3). This gives

-eX siny + xz +

:~ =

xz - eX siny,

so ogjoy = O. Therefore, g is a function of z alone, and

j(x,y, z) = eX cosy

+ xyz + h(z).

We now calculate ojjoz from this equation and match it to the formula forojjoz in Equations (3). This gives

dh xy+ dz =xy+z,

or

so Z2

h(z) =

2 + c.

dh dz

=

z,

16.3

Path Independence, Conservative Fields, and Potential Functions

927

Hence,

f(x,y,z) = e'cosy

Z2

+ ~z + 2 + c.

-

We have infinitely many potential functions ofF, one for each value of C.

EXAMPLE 4

+ (cosz)k is not conservative.

Show that F = (h - 3)i - zj

We apply the component test in Equations (2) and find immediately that

Solution

oP 0 oy = oy (cosz) = 0,

oN = -(-z) 0 = -\. oz oz

-

-

The two are unequal, so F is not conservative. No further testing is required.

EXAMPLE 5

Show that the vectnr field F =

2 -y 2 i x +y

+

2 X 2j x +y

+ Ok

satisfies the equations in the Component Test, but is not conservative over its natural domain. Explain why this is possible. Solution We have M = -y/(x 2 Component Test, we find

oP=O=oN oy oz'

+ y2), N

=

x/(x 2 + y2), and P = O. If we apply the

oP=O=oM ax

and

iJz'

oM oy

y2 - x 2 (x 2 + y2)2

oN oX .

So it may appear that the field F passes the Component Test. However, the test assumes that the domain of F is simply connected, which is not the case. Since x 2 + y2 cannot equal zero, the natoral domain is the complement of the z-axis and contains loops that cannot be contracted In a point. One such loop is the unit circle C in the ~-plane. The circle is parametrized by r(l) = (COSI)i + (sinl)j,O :5 I :5 2,.. This loop wraps around the z-axis and cannot be contracted to a point while staying within the complement of the z-axis. To show that F is not conservative, we compute the line integral F • dr around the loop C. First we write the field in terms of the parameter I: C

f

F =

x

2

- y2.1 +Y

+ x

2

x. 2J =

+Y

Next we f'md dr/dl = (-sinl)i

f C

F·dr =

-sint. __ 2 1 + cos-t

. 2 sm I

+

+

. 2 sm I

cost . ( . ). _2 J = -sml 1 + co.-t

+ (cosl).l·

(COSI)j, and then calculate the line integral as

f F'~~ 1 dl

=

2 2 " (sin t

+ cos2 t) dt

=

2,..

C

Since the line integral ofF around the loop C is not zero, the field F is not conservative, by Theorem 3. _ Example 5 shows that the Component Test does not apply when the domain of the field is not simply connected. However, if we change the domain in the example so that it is restricted to the ball of radius 1 centered at the point (2, 2, 2), orin any similar ball-shaped region which does not contain a piece of the z-axis, then this new domain D is simply connected Now the partial derivative Equations (2), as well as all the assumptions of the Component Test, are satiafied In this new situation, the field F in Example 5 is conservative on D.

928

Chapter 16: Integration in Vector Fields Just as we must be careful with a function when determining if it satisfies a property throughout its domain (like continuity or the intennediate value property), so must we also be careful with a vector field in determining the properties it mayor may not have over its assigned domain.

Exact Differential Forms It is often conveuient to express work and ciroulation integrals in the differential form

lMdx+Ndy+Pdz discussed in Section 16.2. Such line integrals are relatively easy to evaluate if M dx + N dy + P dz is the total differential of a function f and C is any path joining the two points from A to B. For then

{

( af ax dx

Jc M dx + N dy + P dz = Jc =

1B

+

af ay dy

Vf'dr

= f(B) - f(A).

+

af az dz

Vfisconservative. Theorem 1

Thus,

1B

df= f(B) - f(A),

just as with differentiable functions of a single variable.

DEFINmONS Any expression M(x,y,z)dx + N(x,y,z)dy + P(x,y,z)dz is a differential form. A differential form is exaet on a domain D in space if af

af

af

Mdx+Ndy+Pdz=~dx+~dy+~dz=~

for some scalar function f throughout D.

Notice that if M dx + N dy + P dz = df on D, then F = Mi + Nj + Pk is the gra. dient field of f on D. Conversely, if F = V/, then the form M dx + N dy + P dz is exact. The test for the form's being exact is therefore the same as the test for F being conservative.

Component Test for Exactness of M dx + N dy + p dz The differential form M dx + N dy + P dz is exact on a connected and simply connected domain if and only if

ap ay

aN az'

aM

ap

iJz

ax'

and

This is equivalent to saying that the field F = Mi

aN ax

aM ay'

+ Nj + Pk is conservative.

16.3

EXAMPLE 6

Path Independence, (onservative Fields, and Potential Functions

929

+ x dy + 4 dz is exact and evaluate the integral

Show that y dx

OJ

16.4 Green's Theorem in the Plane 31. Evaluating a work integral two ways Let F ~ V(x'y2) and let C be the path in the xy·plane from (-I, I) to (I, I) that consists of the line segment from (-I, 1) to (0, 0) followed by the line segment from (0, 0) to (I, I). Evaluate leF'dr in two ways.

34. Gradient of a line integral vative vector field aod

32. Integral along different path. Evaluate the line integral Ie 2x cos y dx - x 2 sin y dy along the following paths C in the xy-plane.

a. The parabola y ~ (x - 1)2 from (1,0) to (0, I) b. The line segment from (-I,,,.) to (I, 0) c. The x-axis from (-I, 0) to (I, 0) d. The astroid r(l) ~ (eos'l)i + (sin' I)j, 0 ,;; I';; 2"., counterclockwise from (I, 0) back to (I, 0) y

Vj is a conser-

~

L

F·dr.

(0.0.0)

ShowthatVg

b. Usej(x,y) ~ x'y2 as a potential function forF.

~

«3.)

g(x,y,z)

a. Find parametrizations for the segments that make up C and evaluate the integral.

Suppose that F

931

~

F.

35. Path of lea.t work You have beeo asked to find tlte patlt along which a force field F will perform tlte least work in moving a particle betweeo two locations. A quick calculation on your part shows F to be conservative. How should you respond? Give rea-

sons for your answer. 36. A revealing experiment By experiment, you fmd that a force field F performs only half as much work in moving an object along patlt C1 from A to B as it does in moving tlte object along path C2 from A to B. What can you conclude about F? Give rea-

sons for your answer. 37. Work by a constant force Show that tlte work done by a constant force field F ~ ai + bj + ck in moving a particle along anypatltfromA toB is W ~ F·AB.

(0. 1)

38. Gravitational field (-1,0)

(1,0)

~-E~--~--~~~x

a. Find a potential function for the gravitational field F

~

-GmM

_~:nc.'_+-,y~j_+~zk~ (x2 + y2 + z2)'/2

(0, -1)

(G, m, aodM are eoostants). 33, a. Exact differential form How are tlte constants a, b, and c related if the following differeotial form is exact? (ay2 + 2czx) dx + y(bx + cz) dy + (ay2 + ex 2) dz For what values of b and c will F ~ (y2 + 2=)i + y(bx + cz)j + (y2 + ex 2 )k

b. Gradient field

b. Let P1 aod P2 be points at distance 81 and 82 from the origin. Show that tlte work done by tlte gravitational field in part (a) in moving a particle from PI to Pz is

GmM(1. _1.). 82

Sl

be a gradient field?

16.4

Green's Theorem in the Plane If F is a conservative field, then we know F = Vf for a differentiable function f, and we can calculate the line integral of F over any path C joining point A to B as fcF'dr = f(B) - f(A). In this section we derive a method for computing a work or flux integral over a closed curve C in the plane when the field F is not conservative. This method, known as Green's Theorem, allows us to convert the line integral into a double integral over the region enclosed by C. The discussion is given in terms of velocity fields of fluid flows (a fluid is a liquid or a gas) because they are easy to visualize. However, Green's Theorem applies to any vector field, independent of any particular interpretation of the field, provided the assumptions of the theorem are satisfied. We introduce two new ideas for Green's Theorem: divergence and circulation density around an axis perpendicular to the plane.

Divergence Suppose that F(x, y) = M(x, y)i + N(x, y)i is the velocity field of a fluid flowing in the plane and that the first partial derivatives of M and N are continuous at each point of a region R. Let (x, y) be a point in R and let A be a small rectangle with one corner at (x, y) that, along with its interior, lies entirely in R. The sides of the rectangle, parallel to the coordinate axes, have lengths of dx and 0 A gas expanding at the point (%". Yo).

~r/

/1",>

0,

= 2c: If c

:x (y)

+ ~ (ex)

= 0: The gas is neither expanding nor compressing.

= 0: The gas is neither expanding nor compressing.

( -y ) + ~( x ) = 2xy 2xy = (d) divF = ~ oX x 2 + y2 oy x2 + y2 (x2 + y2)2 (x2 + y2)2 the divergence is zero at all points in the domain of the velocity field.

o.

. Again,

• Cases (b), (c), and (d) of Figure 16.28 are plausible models for the two-dimensional flow of a liquid. In fluid dynamics, when the velocity field of a flowing liquid always bas divergence equal to zero, as in those cases, the liquid is said to be incompressihle.

Spin Around an Axis: The k-Component of Curl The second idea we need for Green's Theorem bas to do with measuring how a floating paddle wheel, with axis perpendicular to the plane, spins at a point in a fluid flowing in a plane region. This idea gives some sense of how the fluid is circulating around axes located at different points and perpendicular to the region. Physicists sometimes refer to this as the circulation density of a vector field F at a point. To obtain it, we retorn to the velocity field

F(x,y)

=

M(x,y)i + N(x,y)j

and consider the rectangle A in Figure 16.29 (where we assume both components ofF are positive). (x.y

+ Oy)

(x

+ ox.y + Oy)

~

F·(-I) 0 Counterclockwise circulation

We sum opposite pairs to get

Top and bottom:

Vertical axis

kt

cb

+----

F(x,y + Jly)' (-i) Jlx = -M(x,y + Jly)Jlx F(x,y)·i Jlx = M(x,y)Jlx F(x + Jlx,y). j Jly = N(x + Jlx,y)Jly F(x,y)' (-j) Jly = -N(x,y)Jly.

Right and left:

-(M(x,y (N(x

+ Jly) - M(x,y»Jlx '" -

+ Jlx,y) - N(x,y))Jly '"

(~~ JlY)Jlx

(~~ Jlx )Jly.

Adding these last two equations gives the net circulation relative to the counterclockwise orientation, and dividing by JlxJly gives an estimate of the circulation density for the rectangle:

CurlF (xo,Yo)' k < 0 Clockwise circulation

Circulation around rectangle aN aM rectangle area '" ax - ay'

FIGURE 16.30 In the flow of an incompressible fluid over a plane region, the k-component of the curl measures the rate of the fluid's rotation at a point. The k-component of the curl is positive at points where the rotation is counterclockwise and negative where the rotation is clockwise.

We let Jlx and Jly approach zero to define the circulation density of F at the point (x,y). If we see a counterclockwise rotation looking downward onto the xy-plane from the tip of the unit k vector, then the circulation density is positive (Figure 16.30). The value of the circulation density is the k-component of a more general circulation vector field we derme in Section 16.7, called the curl of the vector field F. For Green's Theorem, we need only this k-component.

DEFINmON The circulation density of a vector field F = Mi point (x, y) is the scalar expression aN ax

aM ay'

+ Nj at the (2)

---

This expression is also called the k-component of the curl, denoted by (curl F) • k.

If water is moving about a region in the :ry-plane in a thin layer, then the k-component of the curl at a point (xo, YO) gives a way to measure how fast and in what direction a small paddle wheel spins if it is put into the water at (xo, YO) with its axis perpendicular to the plane, parallel to k (Figure 16.30).

EXAMPLE 2 Find the circulation density, and interpret what it means, for each vector field in Example 1. Solution (8) Uniform expansion: (curl F)' k = at very small scales.

:x (ey) -

~ (ex)

= O. The gas is not circulating

936

Chapter 16: Integration in Vector Fields y

(b) Rotation: (curl F)' k =

:x

(ex) -

~ (-cy)

indicates rotation at every point. If c the rotation is clockwise.

(0) Shear: (curl F)' k = -

:o ~

--- -

~ (y)

>

= 2c. The constant circulation denaity

0, the rotation is counterclockwise; if c

X

FIGURE 16.36 Green's Theorem may be applied to the aonular region R by sununing the line integrals along the houodaries Ct

and C, in the directions shown.

Exercises 16.4 Verifying Green's Theorem 10 Exercises 1-4, verifY the conclusion of Green's Theorem by evaluating both sides of Equations (3) and (4) for the fieldF = Mi + Nj. Take the domains of integration in each case to be the disk R: x 2 + y2 ,,; a 2 and its boundiog circle C: r = (a cos t)i + (a sin t)j, t '" 2,.. l.F=-yi+xj 2.F=yi

9. F

=

(xy

+ y2ji + (x

- y)j

10. F

=

+ 3yji + (2x

(x

- y)j

y

y

°'"

3. F=2xi-3yj

~~----~----~~X

4. F = -x'yi + xy2j

-2

2

-I

Clrculatton and Flux 10 Exercises 5-14, use Green's Theorem to fmd the counterclockwise circolation and outward flux for the field F and curve C. S. F = (x - y)i + (y - x)j C: The square bounded by x = O,X = I,y = O,y = I

11. F =X 3y 2 i

+ kX4yj

12. F = ~i+ (tan-ty)j I+y

y

y

6. F = (x 2 + 4y)i + (x + y2)j C: The square bounded by x = O,X = I,y = O,y = I

(2,2)

7. F = (y2 - x 2)i + (x2 + y2)j C: The triangle houoded by y = 0, x = 3 , and y = x 8. F = (x + y)i - (x 2 + y2)j C: The triangle houoded by y = 0, x = I , and y = x

-I

~~~--~-------+X

(0,0)

-I

16.4 Green's Theorem in the Plane 13. F

~

(x

+ eX siny)i + (x + eX cosy)j

C: The right·hand loop of the lemniscate r2 ~ cos 28 14. F

Green'. Theorem Area Formula

~ (tan-l~} + In(x 2 + y2)j

AreaofR

tf

~

15. Find the counterclockwise circulation and outward flux of the field F ~ xyi + y2j around and over the boundary of the region enclosed by the curves y ~ x 2 and y ~ x in the frrst quadrant

The reason is that by Equation (3), run backward,

R

~

17. Find the outward flux of the field

~

f

(y2 dx

(y - 2)2 ~ 4

(3ydx

0, x

~

+y

~

I,y

~

0

(6y

(2x

7T,

0 :s y :s sinx

+ x)dx + (y + 2x)dy

C: The circle (x - 2)2

f

.. f f

j(x) dx

I - cost

+ g(y) dy

ky dx

+ hx dy (k and h constants).

c 30. Integral dependent only on .....

Show that the value of

+ (x2y + 2x) dy

xy2 dx

around any square depends ouly on the area of the square and not on its location in the plane. 31. What is special about the integral

f

4x 3y dx

+ x' dy?

Give reasons for your answer.

f

+ 2xdy)

c

24.

~

32. What is special about the integral

C: The boundary of 0 :s x :s

f

I - sinl, Y

29. Let C be the boundary of a region on which Green's Theorem holds. Use Green~ Theorem to calculate

c

c

23.

~

28. One arch of the cycloid x

0 s I :5 2".

c

+ x 2 dy)

C: The mangle bounded by x

f

0:5 I :5 2".

f

4y)j

C

22.

+ (b sin I)j, + (sin3 1)j,

b.

Using Green's Theorem Apply Green~ Theorem to evaluate the integrals in Exercises 21-24.

21.

27. The astroid r(l) ~ (COS3 1)i

(a cos I)i

c

C: The boundary of the ''triangular'' region in the frrst quadrant enclosed by the x-axis, the line x = 1, and the curve y = x 3

C: The circle (x

0:5 1 :5 2".

~

26. The ellipse r(l) ~ (a cos I)i

once counterclockwise around the given curve. 19. F ~ lxy':i + 4x2y2j

+ (2x - 2)2 +

+ (a sin l)j,

25. The circle r(l)

+ cos8),a > O.

Work In Exercises 19 and 20, fmd the vrork done by F in moving a particle

(4x - 2y)i

txdy - tydx.

Use the Green's Theorem area formula given above to fmd the areas of the regions enclosed by the curves in Exercises 25-28.

18. Find the counterclockwise circulation of F ~ (y + eX In y)i + (eX/y)j around the boundary of the region that is bounded above by the curve y ~ 3 - x 2 and below by the curve y ~ x' + 1.

~

f

+ t)dydx

R

c

(3xy - ~)i + (eX + tan-I y)j I +y

aerossthe cardioidr ~ a(1

II (t

~ JJ dydx ~

AreaofR

16. Find the counterclockwise circulation and the outward flux of the field F ~ (-siny)i + (xcosy)j around and over the square cut from the frrst quadrant by the lines x ~ "./2 and y ~ "./2.

F

xdy - ydx

c

C: The boundary of the region defmed by the polar coordinate inequalities 1 :s r :s 2, 0 ::s 8 ::s 11'

20. F

941

+ y2) dx +

+ (y -

(lxy

_y3dy+x 3 dx?

c Give reasons for your answer. 33. Are. as a Une integral Show that if R is a region in the plane bounded by • piecewise smooth, simple closed curve C, then

3)' ~ 4

+ 3y) dy

c C: Any simple closed curve in the plane for which Green's Theorem holds Calculating A .... with Green" Theo...m If a simple closed curve C in the plane and the region R it encloses satisfy the hypotheses of Green's Theorem, the area of R is given by

AreaofR

~

f

xdy

~

c

-

f

c

ydx.

34. Definite integral .s • Une integral Suppose that a nonnegative functiony ~ j(x) has a continuous frrstderivative on [a, b]. Let C be the boundary of the region in the xy-plane that is bounded below by the x-axis, above by the graph of j, and on the sides by the lines x ~ a and x ~ b. Show that

/.b

j(x)dx

~

-

f

c

ydx.

942


35. Area and the centroid Let A be the area and xthe x-coordinate of the centroid of a region R that is bounded by a piecewise

a. Let f(x,y) ~ In (x 2 + y2) and let Cbe the circle x 2 + y2 = a 2. Evaluate the flux integral

smooth, simple closed curve C in the ~-p1me. Show that

f

Vf·nds.

c 36. Moment of inertia Let Iy be the moment of ioertia about the y-axis of the region ioExercise 35. Show that

37. Green's Theorem and Laplace's equation Assuming that all the necessary derivatives exist and are continuous, show that if f(x,y) satisfies the Laplace equation

a2f a2f '+-2=0. ax ay then

b. Let Kbe an arbitrary smooth, simple closed curve in the plane that does not pass through (0, 0). Use Green's Theorem to show that

hss two possible values, depending on whether (0, 0) lies inside K or outside K.

40. Bendi::J:80n's criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors F ~ M(x, y)i + N(x, y)j of the flow's velocity field are the tsngent vectors of the streamlines. Show that if the flow takes place over a simply connected region R (no holes or missiog points) and that if M, + Ny ¢ 0 throughout R, then none of the streamlines in R is closed In other words, no particle of fluid ever has a closed trajectory in R. The criterion M, + Ny ¢ 0

is called Bendi:J:son's criterion for the nonexistence of closed trajectories. 41. Establish Equation (1) to rmish the proof of the special case of for all closed curves C to which Green's Theorem applies. (The converse is also true: If the line integral is always zero, !ben f satisfies the Laplace equation.)

38. Mnjrnjzjng work Among all smooth, simple closed curves in the plane, oriented counterclockwise, find the one along which the work done by

Green's Theorem. 42. Curl component of conservative fields

Can anytbiog be said

about the curl component of a conservative two-dimensional vector field? Give reasons for your answer.

COMPUTER EXPLORATIONS In Exercises 43-46, use a CAS and Green's Theorem to rmd the counterclockwise circulation of the field F around the simple closed curve C. Perform the followiog CAS steps.

a. Plot C in the xy-p1me. is greatest. (Hint: Where is (curl F) • k positive?) 39. Regions with many holes Green's Theorem holds for a region R with any rmite number of holes as long as the bounding curves

are smooth, simple, and closed and we integrate over each component of the boundary io the direction that keeps R on our immediate left as we go along (see accompanying figure).

b. Determine the integrand (aN/ax) - (aM/ay) for the curl form of Green's Theorem. c. Determioe the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. 43. F ~ (2x - y)i + (x + 3y)j, C: The ellipse x 2 + 4y2 ~ 4 44. F ~ (2x' - y')i

45. F ~ x-'e"i

+ (x' + y')j,

2 C: The ellipse ~

I

+ (e"lnx + 2x)j,

C: The boundary of the region dermed by y ~ 1 andy ~ 2 (above) 46. F ~ xeYi

y2

+"9 ~

+ x' (below)

+ (4x 2 Iny)j,

C: The triangle with vertices (0, 0), (2, 0), and (0, 4)

16.5

16.5

I

Surfaces and Area

g43

Surfaces and Area We have dermed curves in the plane in three different ways:

11

II -

constant

__+,::-::;_ --"11 _ constant (II, v)

f(x)

Explicitform:

y

Implicitfonn:

F(x,yl

Paramc1ric vector fonn:

r(tl

=

~

0

~ f(t~

+ g(tll,

a" t "

h.

We have analogous def'mitions of surfaces in space:

R

;;!---+---. o , Curve 11 _ CODStInt

Explicitfonn:

z ~ f(x, y)

Implicitform:

F(x,y,z) -

o.

There is also a parametric form for surfaces that gives the position of a point on the surface as a vector function of two variables. We discuss this new form in this section and apply the form to obtain the area of a surface as a double integral. Double integral formulas fur areas of surfaces given in implicit and explicit forms IU'e then obtained as special cases

of the more general parametric formula.

Curve" _ constaD1 %

r(M_ 11) _ f(M, 11)1 + g(lI, v)J + 11(11,1/)11:, Position vector to surface point

~y FIGURE 16.37 A parametrized swfacc S ~sscd IS a vector function of two variables dcfmed on a region R.

Parametrizations of Surfaces Suppose

r(a,v) - f(a,v)1

+ g(a,v)l + h(a,v)k

(1)

is a continuous vector function that is dermed on a region R in the uv-plane and one-toone on the interior of R (Figure 16.37). We call the range of r the .urface S defined or traced by r. Equation (1) together with the domain R constitute a parametrization of the surface. The variables u and v are the parameten, and R is the parameter domain. To s:implify our discussion, we take R to be a rectangle def"med by inequalities of the form a SUS b, c S v S d. The requirement that r be one-to-one on the interior of R eDSW'eS that S does not cross itself. Notice that Equation (1) is the vector equivalent of three parametric equations:

x = /(u, v), EXAMPLE 1 Cone:

y = g(u,v),

z = h(u, v).

Find a parametrization of the cone

J;

OSzS 1.

.t - W+"J2

Solution Here, cylindrical coordinates provide a parametrization. A ~t (x,y, z) on the cone (Figure 16.38) has x = rcos6,y = rsinl1, and z = Yx 2 + y2 = r, with S r S 1 and 0 S 8 S 2'IT. Thking u = r and v = 6 in Equation (1) gives the parametrization

o

.

"

r(r, 6)

(,"os 6)1 + (r sin 6)1 + rk,

o :S r :S

1,

O:s 11

S

2'IT.

The parametrization is one-to-one on the interior of the domain R, though not on the boundary tip of its cone where r = O. • y

EXAMPLE 2 FIGURE 16.38 The cone in Example I can be parametrized using cylindrical coordinates.

~

Find a parametrization of the sphere x 2

+ y2 + z2

= a 2.

Solution Spherical coordinates provide what we need. A typical point (x, y, z) on the sphere (Figure 16.39) has x=asin~cos8, y=asinq,sin8, and z=acosq"

944


o ~ t/J ~ 7T, Q ~ 0 ~ 27T. Taking U = t/J and v

= 0 in Equation (1) gives the parame-

1rization r(, 9) - (a sin co, 9)i

+ (asin.p,in9)J + (ao .. (sin2cf> + cos2cf»

=

y'a 4 sin4 cf>cos /1

=

y'a 4

=

a 2 y'sin2 cf>

4

4

4

=

a 2 sin cf>,

4

4

4

•

16.5 Surfaces and Area

since sin cf>

;:::: 0 for 0 :::; cf> :::; A= =

947

1T. Therefore, the area of the sphere is

121rl1r a 2sin cf> dcf> de 121r [-a 2cos cf>

I

de

=

121r 2a 2de =

41Ta

2

units squared.

•

This agrees with the well-known formula for the surface area of a sphere.

z

EXAMPLE 6 Let S be the "football" surface formed by rotating the curve x = cos z, y = 0, -1T/2 :::; z :::; 1T/2 around the z-axis (see Figure 16.44). Find a parametrization for S and compute its surface area.

A

~

/ () -

x

r = cos z is the radius of a circle

J. at height z ________ (x, y, z)

y

Solution Example 2 suggests finding a parametrization of S based on its rotation around the z-axis. If we rotate a point (x, 0, z) on the curve x = cos z, y = 0 about the z-axis, we obtain a circle at height z above the xy-plane that is centered on the z-axis and has radius r = cos z (see Figure 16.44). The point sweeps out the circle through an angle of rotation e,o :::; e :::; 21T. We let (x, y, z) be an arbitrary point on this circle, and define the parameters u = z and v = e. Then we have x = r cos e = cos u cos v, y = r sin e = cos u sin v, and z = u giving a parametrization for S as r (u, v )

. v J. + u, k = cos u cos V I.+ cos u sm

_1T
In Figure 16.47, we see how a subdivision of R (considered as a rectangle for simplicity) divides the surface S into corresponding curved surface elements, or patches, of area

tJ.u""

Rl

Ir.. X rvl

dudv.

As we did for the subdivisions when defming double integrals in Section 15.2, we number the surface element patches in some order with their areas given by dul. tJ.uz, ••• , tJ.ull• To form a Riemann sum over S, we choose a point (X},y,.. z.t) in the kth patch, multiply the value of the function G at that point by the area tJ.U,b and add together the products:

•

~

G(Xob Job

z.t) tJ.u,t.

Depending on how we pick (Xob Job z.t) in the kth patch, we may get different values for this Riemann sum. Then we take the limit as the number of surface patches increases, their areas shrink to zero, and both 4u --+ 0 and tJ.v --+ O. This limit, whenever it exists independent of all choices made, delmes the .urface integral of G over the Ianace 80s (!)

954

Chapter 16: Integration in Vector Fields Notice the analogy with the definition of the double integral (Section 15.2) and with the line integral (Section 16.1). If S is a piecewise smooth surface, and G is continuous over S, then the surface integral defined by Equation (I) can be shown to exist. The formula for evaluating the surface integral depends on the manner in which S is described, parametrically, implicitly or explicitly, as discussed in Section 16.5.

Formulas for a Surface Integral

1. For a smooth surface S defined parametrically as r(u, v) = I(u, v)i + g(u, v)j + h(u, v)k, (u, v) eR, and a continuous function G(x, y, z) defined on S, the surface integral of Gover S is given by the double integral over R,

JJ G(x,y,Z)dCT

=

JJ G(/(u, v),g(u, v),h(u, v)) Iru X rvl dudv.

S

(2)

R

2. For a surface S given implicitly by F(x, y, z) = c, where F is a continuously differentiable function, with S lying above its closed and hounded shadow region R in the coordinate plane beneath it, the surface integral of the continuous function G over S is given by the double integral over R,

JJ

G(x,y,Z)dCT

S

=

JJ G(x,y'Z)I~;~~1

(3)

dA,

R

where p is a unit vector normal to R and VF' p

'¢'

O.

3. For a surface S given explicitly as the graph of Z = I(x, y), where I is a continuously differentiable function over a region R in the xy-plane, the surface integral of the continuous function G over S is given by the double integral over R,

JJ G(x, y,z) dCT S

=

JJ G(x, y,f(x, y))

V/. 2 + I; + I dxdy.

(4)

R

The surface integral in Equation (I) takes on different meanings in different applications. If G has the constant vaIue I, the integral gives the area of S. If G gives the mass density of a thin shell of material modeled by S, the integral gives the mass of the shell. If G gives the charge density of a thin shell, then the integral gives the total charge.

EXAMPLE 1

Integrate G(x, y, z) = x 2 over the cone Z =

Vx 2 + y2, 0 :5 Z

:5

I.

Solution Using Equation (2) and the calculations from Example 4 in Section 16.5, we have Irr X rei = Vir and

JJX 2 dCT= /.2w/"(r 2 cos2 6)(Vir)drd6 S

x~rcos9

955

16.6 Surface Integrals

Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum of their integrals and so on. The domain Additivity Property takes the form

11

Gda =

11

Gda

+

s]

S

11

Gda

11

+ ... +

Gda.

Sn

S2

When S is partitioned by smooth curves into a finite number of smooth patches with nonoverlapping interiors (i.e., if S is piecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of a function over the surface of a cube is the sum of the integrals over the faces of the cube. We integrate over a turtle shell of welded plates by integrating over one plate at a time and adding the results.

EXAMPLE 2 Integrate G(x, y, z) = xyz over the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1 (Figure 16.48).

z

J :

I I I I I

1-_

/ 0

x

SoLution We integrate xyz over each of the six sides and add the results. Since xyz = 0 on the sides that lie in the coordinate planes, the integral over the surface of the cube reduces to

~

--_

///

/ /f

SMd

/

11 --- ~ i idec

Y

SideB

FIGURE 16.48

11

xyz da =

xyz da

11

+

Side A

Cube surface

xyz da

+

SideB

11

xyz da.

SideC

Side A is the surface f(x, y , z) = z = lover the square region Rxy: 0 1, in the xy-plane. For this surface and region,

:5 X :5

o :5 Y :5

p = k,

Vf

=

k,

The cube in Example 2.

da =

IVfl

=

1,

IVf'pl

IVfl IVf ' pl dA

=

1 Tdxdy = dxdy

=

Ik'kl

=

1,

1

xyz = xy(1) = xy and

Symmetry tells us that the integrals ofxyz over sides Band C are also 1/4. Hence,

ff

}}

xyz da =

1

1

1

"4 + "4 + "4

=

3

•

"4 '

Cube surface

EXAMPLE 3 Integrate G (x, y , z) = VI - x 2 - y2 over the "football" surface S formed by rotating the curve x = cos z, y = 0 , -1T/2 :5 z :5 1T/2, around the z-axis. SoLution The surface is displayed in Figure 16.44, and in Example 6 of Section 16.5 we found the parametrization X

=

cos u cos v,

y = cos u sin v,

z = u,

1T

1T

-"2:5 U :5"2

and

0:5

v

:5 21T,

where v represents the angle of rotation from the xz-plane about the z-axis. Substituting this parametrization into the expression for G gives

The surface area differential for the parametrization was found to be (Example 6, Section 16.5) da = cos u

VI

+ sin2 u du dv.

956

Chapter 16: Integration in Vector FieLds

These calculations give the surface integral

fJif

\11 -

x2

-

y2

du

=

1

21T 11T/2

o

S

-1T/2

Isin u Icos u \11 + sin2 ududv

{21T {1T/2 Jo sin u cos u

= 2 Jo

_1

-

w=1+sin2 u

00

21T

12 ~

o

1

I d d Vw W v

2 32 = 21T 3 w / 0

\11 + sin2 u du dv dw = 2 sin u c;s u du Whenu=Ow=l. ' When u = 'TT/2, w = 2.

]2 = 4; (2V2 - 1).

•

1

Orientation

FIGURE 16049 Smooth closed surfaces in space are orientable. The outward unit normal vector defines the positive direction at each point. c

d

a

b

Start

We call a smooth surface S orientable or two-sided if it is possible to define a field n of unit normal vectors on S that varies continuously with position. Any patch or subportion of an orientable surface is orientable. Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable. By convention, we choose n on a closed surface to point outward. Once n has been chosen, we say that we have oriented the surface, and we call the surface together with its normal field an oriented surface. The vector n at any point is called the positive direction at that point (Figure 16.49). The Mobius band in Figure 16.50 is not orientable. No matter where you start to construct a continuous unit normal field (shown as the shaft of a thumbtack in the figure), moving the vector continuously around the surface in the manner shown will return it to the starting point with a direction opposite to the one it had when it started out. The vector at that point cannot point both ways and yet it must if the field is to be continuous. We conclude that no such field exists.

Surface Integral for Flux Suppose that F is a continuous vector field defined over an oriented surface S and that n is the chosen unit normal field on the surface. We call the integral of F n over S the flux of F across S in the positive direction. Thus, the flux is the integral over S of the scalar component of F in the direction of n. 0

FIGURE 16050 To make a Mobius band, take a rectangular strip of paper abed, give the end be a single twist, and paste the ends of the strip together to match a with e and b with d. The Mobius band is a nonorientable or one-sided surface.

DEFINITION The flux of a three-dimensional vector field F across an oriented surface S in the direction of n is Flux =

11

Fondu.

(5)

s

The definition is analogous to the flux of a two-dimensional field F across a plane curve C. In the plane (Section 16.2), the flux is

1

Fonds,

the integral of the scalar component ofF normal to the curve. If F is the velocity field of a three-dimensional fluid flow, the flux of F across S is the net rate at which fluid is crossing S in the chosen positive direction. We discuss such flows in more detail in Section 16.7.


EXAMPLE 4

z

y

=

x

2

,

0

:5 X

957

Find the flux of F = yzi + xj - z 2 k through the parabolic cylinder :5 1,0 :5 z :5 4, in the direction n indicated in Figure 16.51.

SoLution On the surface we have x = x, y = x 2 , and z = z, so we automatically have the parametrization rex, z) = xi + x 2j + zk,O :5 x :5 1,0 :5 z :5 4. The cross product of tangent vectors is k

rx X rz = 1

j 2x

o

0

1

I

n"-- ~-. 1

__~ /'

,1

, y

j.

The unit normal vectors pointing outward from the surface as indicated in Figure 16.51 are n =

x

FIGURE 16051 Finding the flux through the surface of a parabolic cylinder (Example 4).

o = 2xi -

rx X rz Irx X rzl

2xi - j

= -----;====;::=====

\/4x 2 + 1·

On the surface, y = x 2 , so the vector field there is

Thus, F n 0

= \/4X; + 1 ((x 2z)(2x) + (x)( -1) + (-z2)(O)) 2x 3z - x

\/4x 2 + 1 . The flux of F outward through the surface is

1 ]X=1x=o dz z - -x 2 2 1o 10 (2x z - x) dxdz = 14 [1-x 4 1

=

3

4

0

=

41- (z -

1

o 2

1

2

0

1) dz

1]4 (z - 1)2

= -

4

0

1

= 4(9) - 4(1) = 2. If S is part of a level surface g(x, y, z) fields

•

= c, then n may be taken to be one of the two Vg

(6)

n = ±IVgl'

depending on which one gives the preferred direction. The corresponding flux is Flux =

11

Fondu

s

=

rr (F0 ±Vg) IVgopl IVgl dA

11

1Vg1

Eqs. (6) and (3)

R

=

rr FOIVgopl ±Vg dA.

11 R

(7)

958

Chapter 16: Integration in Vector FieLds EXAMPLE 5 Find the flux of F = yzj + z 2k outward through the surface S cut from the cylinder y2 + z2 = 1, z 2':: 0, by the planes x = and x = 1.

I;

Rx> t

(1, -1, 0)

°

SoLution The outward normal field on S (Figure 16.52) may be calculated from the gradient of g(x,y, z) = y2 + z2 to be

s

1 ----- y

(1, 1,0)

Vg +IVgl

n =

With P

=

=

2yj

+ 2zk

~ I.

2vl

•

= Yl + zk.

k, we also have

x

FIGURE 16.52 Calculating the flux ofa vector field outward through the surface S. The area of the shadow region Rry is 2 (Example 5).

2yj + 2zk Y4y2 + 4z2

du =

IVgl IVg.kl

2

We can drop the absolute value bars because z The value of F • n on the surface is F • n = (yzj =

y2z

=

z.

12z1

dA =

2'::

1 dA = zdA.

°on S.

+ z 2k) • (yj + zk) + z3 = z(y2 + z2)

The surface projects onto the shadow region Rxy, which is the rectangle in the xy-plane shown in Figure 16.52. Therefore, the flux ofF outward through S is

• Moments and Masses of Thin Shells Thin shells of material like bowls, metal drums, and domes are modeled with surfaces. Their moments and masses are calculated with the formulas in Table 16.3. The derivations are similar to those in Section 6.6. The formulas are like those for line integrals in Table 16.1, Section 16.1. TABLE 16.3 Mass and moment formulas for very thin shells

Mass:

M =

11

8 du

8 = 8(x,y,z) = density at (x,y,z) as mass per unit area

s

First moments about the coordinate planes:

Myz

=

11

Mxz

x8du,

s

=

11

y 8 du,

s

Coordinates of center of mass:

Moments of inertia about coordinate axes:

r(x,y, z)

=

distance from point (x,y, z) to line L


EXAMPLE 6

959

Find the center ofmass of a thin hemispherical shell of radius a and con-

""'" density 6.

Solution We model the shell with the hemisphere

s

•

•

f(x,y, z) = x" y

x

FIGURE 16.53 Thecentl:rofmassofa thin hemispherical shell of constant density lies on the axis of symmetry haI:fway from the base to the ~ (Example 6).

+ y2 + z2 =

a 2,

z"o

(Figure 16.53). The symmetry of the surface about the z-axis tells us thatj = mains only to findi from the fonnulaz = Mxy/M• The mass of the shell is

JJs

M -

JJs

6 du - 6

O. Itre-

du - (6)(are. of S) - 2"0'6.

To evaluate the integral for MJiY' we take p

=

k and calculate

IVII

~

12xi + 2yj + 2m1

IVI'pl

~

IVI'kl ~ 12z1 ~ 2z IVII 0 IVI'pl dA ~ zdA.

da ~

Y=

~ 2V,'

+ y' + z,

~ 20

Then

Mxy -

JJ

Jf z~dA JJ

z6 du -

- 8a

I)

S

R

2 3 dA. - &(7Ta ) - &iTa

R

z--M --21Ta 8 --

_

Mxy

'1f'a 3l) 2

a

2'

•

The shell's C"",", ofDUlllS is the paID! (0, 0, 0/2).

EXAMPLE 7 Find the center of mass of a thin shell of density 8 cone z = Yx 2 + y2 by the planes z = 1 andz = 2 (Figure 16.54).

=

Solution The symmetry of the surface about the z-axis tells us that Z = Mxy/M. Working as in Example 4 of Section 16.5, we have

x= y =

r(r,6) = (rc088)i

+ (rsine)j + rk.

1

~

r

:S

2,

O:s

and

FIGURE 16.54 The cone frustmn formed 2 + y 2 iscutby whentheconez the planes z - 1 !IIld z - 2 (Example 7).

Therefon:,

Yx:

- VZJ,'"[lnr];d6 - VZJ,'"1n2d6 ~ 2..VZ1n2,

1/z2 cut from the

O. We :fmd

(J :S 217',

960


Mxy =

ff

Bzdu = 12'1r12 :2rV2rdrd6

S

=

Vi12'1r12 drdf)

1,"

~ v'? _

M,.

z~-=

M

dO

~ 2..v'?,

2..v'? v'? In 2

2..

1

=-

1n2'

•

The shell's center ofmass is the point (0, 0, I/ln2).

Exercises 16.6 Surface Integrals In Exercises 1--8, integrare the given function over the given surface.

1. Parabolie cy6nder G(x,y,z) = x, over the parabolic cylinder y = Xl, 0 :S X :S 2,0 :S z :S 3

2. Circalar cyIID.der

y2 +z2 =

G(x,y,z) = z, am: the cylindricalsuxfacc

I, 1)

4,z~ O,I:Sx:S4

3. Spbere G(x,y,z) =x 2,ovcrthe1lllitsphcR:x2 + yl +z2 = 1 4. Hemisphere G(x,y,z) = z2, aver the hcmisphcIe x 2 + y2 + z2=a 2,z2:0 5. Portion of plaDe F(x,y, z) = z, am: the portion of the plane x + y + z = 4 that lies .baw: the square 0 s x s I, o s y s I, in the xy-plane 6. Coae F(x,y, z) = z - x, am: the cone z = Yx 2 + y2, OSzsl

(0,0, y

yS - 4z, over the parabolic

7. Parabolie dome H(r,y,z) - x 2 domez - 1 - x 2 - y2,z 2: 0

8. SpberiuJ. eap H(x,y,z) - yz, over the part of the sphere x2 + y2 + 1'2 _ 4 that lies above the cone z _ Yx2 + y2 9. Integrate G(x,y,z) - x + Y + z over the surface of the cube cut from the fust octant by the planes x - a, y - a, Z - a. 10. Intqp-ate G(r,y, z) = y + I' aver the surl'ace of the wedge in the frrst octant bomded by the coordinate planes and the planes x - 2andy+z - 1. 11. Intqp-ate G(x,y,z) = X)'% over the surface of the rectangular solid cut from the frrst octant by the planes x - a, y - b, and z - c. 12. Intqp-ate G(x,y,z) = X)'% over the surface of the rectangular solid bounded by the planes x - ±a, y - ±b, and z - ±c.

(I, I, 0)

16. Integrate G(x,y,z) = x am: the !IUIfaoe given by

z=X2+y for Osxsl, -lSysl. 17. IntegrateG(x,y,z) - xyz am: the triangular surface with vertices (1, 0, 0), (0, 2, 0), om! (0, 1, 1).

I, 1)

13. Intqp-ate G(x,y,z) = x + Y + z aver the portion of the plane b + 2y + z = 2 that lies in the fmt octant

14. Intqp-atc G(x,y, z) = xVyl + 4 aver the surface cut from the parabolic cylinder y2 + 4z = 16 by the planes x = 0, x = I, andz = O. 15. Intqp-atc G(x,y,z) = z - x over the portion of the graph of z = x + y2 above the triangle in the xy-plane having vertices (0. o. 0). (1. 1.0). om! (0. 1. 0). (s.. " " _ figure.)

18. Integrate G(x,y,z) - x - y - z aver the portion of the plllllC x + Y = 1 in the fmt octant between z = 0 and z = 1 (lICe the accompanying f1jUIC).

16.6

Surface Integrals

961

+k + zyj + z 2k xi + yj + zk xi + yj + zk V2 2 x + y2 + z

33. F(x,y,z) = yi - xj

z

34. F(x,y,z) = zxi 35. F(x,y,z)

=

(1,0, 1)

36. F(x,y,z) = - ~"""-- y

x

Finding FLux Across a Surface In Exercises 19-28, use a parametrization to find the flux ffsF· n du across the surface in the given direction. 19. Parabolic cylinder F = z 2i + xj - 3zk outward (normal away from the x-axis) through the surface cut from the parabolic cylinder z = 4 - y2 by the planes x = 0, x = 1 , and z = 0 20. Parabolic cylinder F = x 2j - xzk outward (normal away from the yz-plane) through the surface cut from the parabolic cylinder y = x 2 , -1 :::; x :::; 1 , by the planes z = 0 and z = 2 21. Sphere F = zk across the portion of the sphere x 2 + y2 + z2 = a 2 in the first octant in the direction away from the origin zk across the sphere x 2

22. Sphere F = xi + yj + in the direction away from the origin

24. Cylinder F = xi + yj + zk outward through the portion of the cylinder x 2 + y2 = 1 cut by the planes z = 0 and z = a 25. Cone F = xyi - zk outward (normal away from the z-axis) through the cone z = Yx 2 + y2, 0 :::; Z :::; 1 26. Cone F = y 2i + xzj - k outward (normal away from the z-axis) through the cone z = 2Yx 2 + y2, 0 :::; Z :::; 2 27. Cone frustum F = -xi - yj + z 2k outward (normal away from the z-axis) through the portion of the cone z = Yx 2 + y2 between the planes z = 1 and z = 2 28. Paraboloid F = 4xi + 4yj + 2k outward (normal away from the z-axis) through the surface cut from the bottom of the paraboloid z = x 2 + y2 by the plane z = 1 In Exercises 29 and 30, find the flux of the field F across the portion of the given surface in the specified direction. =

-i

+ 2j + 3k

S: rectangular surface z = 0, direction k 30. F(x, y, z) = yx 2i - 2j + xzk S: rectangular surface y = 0, direction - j

0:::; x :::; 2,

0:::; y :::; 3,

=

zk

32. F(x, y, z)

=

-yi

+ xj

39. Let Sbe the portion of the cylinder y = eX in the first octant that projects parallel to the x-axis onto the rectangle Ryz: 1 :::; y :::; 2, o :::; z :::; 1 in the yz-plane (see the accompanying figure). Let n be the unit vector normal to S that points away from the yz-plane. Find the flux of the field F(x, y, z) = -2i + 2yj + zk across S in the direction of n.

z 1 --_

1 --x S

~y / ' "

40. Let S be the portion of the cylinder y = In x in the first octant whose projection parallel to the y-axis onto the xz-plane is the rectangle Rxz: 1 :::; x :::; e, 0 :::; z :::; 1. Let n be the unit vector normal to S that points away from the xz-plane. Find the flux of F = 2yj + zk through S in the direction ofn. 41. Find the outward flux of the field F = 2xyi + 2yzj + 2xzk across the surface of the cube cut from the first octant by the planes x = a,y = a,z = a. 42. Find the outward flux of the field F = xzi + yzj + k across the surface of the upper cap cut from the solid sphere x 2 + y2 + z2 :::; 25 by the plane z = 3. Moments and Masses 43. Centroid Find the centroid of the portion of the sphere x 2 + y2 + z2 = a 2 that lies in the first octant. 44. Centroid Find the centroid of the surface cut from the cylinder y2 + z2 = 9, z ~ 0, by the planes x = 0 and x = 3 (resembles the surface in Example 5).

-1:::; x :::; 2,

2:::; z :::; 7,

In Exercises 31-36, find the flux of the field F across the portion of the sphere x 2 + y2 + z2 = a 2 in the first octant in the direction away from the origin.

31. F(x,y,z)

38. Find the flux of the field F(x,y,z) = 4xi + 4yj + 2k outward (away from the z-axis) through the surface cut from the bottom of the paraboloid z = x 2 + y2 by the plane z = 1.

+ y2 + z2 = a 2

23. Plane F = 2xyi + 2yzj + 2xzk upward across the portion of the plane x + y + z = 2a that lies above the square o :::; x :::; a, 0 :::; y :::; a , in the xy-plane

29. F(x,y, z)

37. Find the flux of the field F(x,y, z) = z2j + xj - 3zk outward through the surface cut from the parabolic cylinder z = 4 - y2 by the planes x = O,X = 1, andz = O.

45. Thin shell of constant density Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density {) cut from the cone x 2 + y2 - z2 = 0 by the planes z = 1 andz = 2. 46. Conical surface of constant density Find the moment of inertia about the z-axis of a thin shell of constant density {) cut from the cone 4x 2 + 4y2 - z2 = 0, Z ~ 0, by the circular cylinder x 2 + y2 = 2x (see the accompanying figure).

962

Chapter 16: Integration in Vector FieLds 47. Spherical shells

z

a. Find the moment of inertia about a diameter of a thin spherical shell of radius a and constant density 8. (Work with a hemispherical shell and double the result.) b. Use the Parallel Axis Theorem (Exercises 15.6) and the result in part (a) to find the moment of inertia about a line tangent to the shell. 48. Conical Surface Find the centroid of the lateral surface of a solid cone of base radius a and height h (cone surface minus the base). y

x

16.7

or r=2cosO

Stokes' Theorem

-/CurlF «" ffJ FIGURE 16.55 The circulation vector at a point (x, y, z) in a plane in a tbreedimensional fluid flow. Notice its right-hand relation to the rotating particles in the fluid.

As we saw in Section 16.4, the circulation density or curl component of a two-dimensional field F = Mi + Nj at a point (x, y) is described by the scalar quantity (aN/ax - aM/ay). In three dimensions, circulation is described with a vector. Suppose that F is the velocity field of a fluid flowing in space. Particles near the point (x,y, z) in the fluid tend to rotate around an axis through (x,y, z) that is parallel to a certain vector we are about to define. This vector points in the direction for which the rotation is counterclockwise when viewed looking down onto the plane of the circulation from the tip of the arrow representing the vector. This is the direction your right-hand thumb points when your fingers curl around the axis of rotation in the way consistent with the rotating motion of the particles in the fluid (see Figure 16.55). The length ofthe vector measures the rate of rotation. The vector is called the curl vector and for the vector field F = Mi + Nj + Pk it is defined to be curl F =

(ap + (aM _ ap)j + (aN _ aM)k. ay _ aN)i az az ax ax ay

(1)

This information is a consequence of Stokes' Theorem, the generalization to space ofthe circulation-curl form of Green's Theorem and the subject of this section. Notice that (curl F)· k = (aN/ax - aM/ay) is consistent with our definition in Section 16.4 when F = M(x,y)i + N(x,y)j. The formula for curl F in Equation (1) is often written using the symbolic operator

V

.a +.a =lax Jay

+k

a az·

(The symbol V is pronounced "de1.") The curl of F is V X F : j

k

a VXF= ax

a a ay az M N P

_ aN)i + (aM _ ap). + (aN _ aM)k ( ap ay az az ax J ax ay =

curl F.

(2)

16.7

Stokes'Theorem

curl F = V X F

EXAMPLE 1 SoLution

Find the curl ofF = (x 2 - z)i

963

(3)

+ xeZj + xyk.

We use Equation (3) and the determinant form, so curlF = V X F j

k

a

a

a

ax 2 x - z

ay

az

xe z

xy

- ~(xeZ))i - (~(xy) - ~(x2 (~(xy) ay az ax az + (~(xeZ) - ~(x2 - Z))k ax ay

z))j

= (x - xeZ)i - (y + l)j + (e Z - O)k =

x(I - e)i - (y

+

I)j

+

•

e~

As we will see, the operator V has a number of other applications. For instance, when applied to a scalar function f(x, y, z), it gives the gradient of f:

Vf =

af. ax.

af.

af

+ ayl + azk.

It is sometimes read as "del f' as well as "grad f."

Stokes' Theorem

s - ~-

x) G

...... - - - - - -

n

'/

FIGURE 16.56 The orientation of the bounding curve C gives it a right-handed relation to the normal field n. If the thumb of a right hand points along n, the fingers curl in the direction of C.

Stokes' Theorem generalizes Green's Theorem to three dimensions. The circulation-curl form of Green's Theorem relates the counterclockwise circulation of a vector field around a simple closed curve C in the xy-plane to a double integral over the plane region R enclosed by C. Stokes' Theorem relates the circulation of a vector field around the boundary C of an oriented surface S in space (Figure 16.56) to a surface integral over the surface S. We require that the surface be piecewise smooth, which means that it is a finite union of smooth surfaces joining along smooth curves.

THEOREM 6-Stokes'Theorem

Let S be a piecewise smooth oriented surface having a piecewise smooth boundary curve C. Let F = Mi + Nj + Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counterclockwise with respect to the surface's unit normal vector n equals the integral of V X F· n over S.

f

F·dr =

c

Counterclockwise circulation

11

V X F·nda-

s

Curl integral

(4)

964

Chapter 16: Integration in Vector Fields Notice from Equation (4) that if two different oriented surfaces SI and S2 have the same boundary C, their curl integrals are equal:

Both curl integrals equal the counterclockwise circulation integral on the left side ofEquation (4) as long as the unit normal vectors 01 and 02 correctly orient the surfaces. If C is a curve in the xy-plane, oriented counterclockwise, and R is the region in the xy-plane bounded by C, then du = dx dy and

(V

x

F)' 0 = (V

x

F)' k = ( -aN - -aM) . ax ay

Under these conditions, Stokes' equation becomes

f

F'dr

=

11 (~~ -~~)

dxdy,

R which is the circulation-curl form of the equation in Green's Theorem. Conversely, by reversing these steps we can rewrite the circulation-curl form of Green's Theorem for two-dimensional fields in del notation as C

f

Circulation

F·dr =

11

V

(5)

X F·kdA.

R

C

See Figure 16.57. Evaluate Equation (4) for the hemisphere S: x 2 + y2 its bounding circle C: x 2 + y2 = 9, z = 0, and the field F = yi - xj.

EXAMPLE 2

+ z2 =

9, z

2:

0,

Solution

FIGURE 16.57 Comparison of Green's Theorem and Stokes' Theorem.

The hemisphere looks much like the surface in Figure 16.56 with the bounding circle C in the xy-plane (see Figure 16.58). We calculate the counterclockwise circulation around C (as viewed from above) using the parametrization reO) = (3 cos O)i + (3 sin O)j, :5 0 :5 21T:

°

dr = (-3 sin 0 dO)i

+ (3 cos 0 dO)j

F = yi - xj = (3 sin O)i - (3 cos O)j

f

F'dr = -9 sin2 0 dO - 9 cos2 0 dO = -9 dO F • dr = 127T-9dO=-181T.

C

For the curl integral ofF, we have

V X F

= (ap _ aN)i + (aM _ ap). + (aN _ aM)k ay

=

y 0=

x

du FIGURE 16.58 A hemisphere and a disk, each with boundary C (Examples 2 and 3).

V

az

+ (0 + yj + zk Yx 2 + y2 + z2

(0 - O)i xi

az

O)j =

3

= zdA

X F'odu = _ 2zl dA = -2dA

3 z

ax J

ax

+ (-1 - l)k + yj + zk

=

xi

ay

-2k Outer unit nonnal

3 Section 16.6, Example 6, with a = 3

16.7 Stokes'Theorem and

II

II

V X F'nder =

s

965

-2dA = -18,..

.r+y2.s9

The circulation around the circle equals the integral of the curl over the hemisphere, as it should. _ The surface integral in Stokes' Theorem can be computed using any surface having boundary curve C, provided the surface is properly oriented and lies within the domain of the field F. The next example illustrates this fact for the circulation around the curve C in Example 2.

EXAMPLE 3 Calculate the circulation around the bounding circle C in Example 2 using the disk of radius 3 centered at the origin in the xy-plane as the surface S (instead of the hemisphere). See Figure 16.58. Solution Ai; in Example 2, V X F = - 2k. For the surface being the described disk in the xy-plane, we have the normal vector n = k so that V X F'nder = -2k'kdA = -2dA and

II

II

V X F'nder =

s

-2dA = -18,.,

.r+y2.s9

a simpler calculation than before. z

_

Find the circulation of the field F = (x 2 - y)i + 4zj + x 2k around the curve C in which the plane z = 2 meets the cone z = Vx 2 + y2, counterclockwise as viewed from above (Figure 16.59).

EXAMPLE 4

C:X 2 +y2=4. z=2

/

Solution Stokes' Theorem enables us to fmd the circulation by integrating over the surface of the cone. Traversing C in the counterclockwise direction viewed from above corresponds to tsking the inner normal n to the cone, the normal with a positive k-component. We parametrize the cone as r(r, 0) = (r cos O)i + (r sin O)j + rk,

s: r(t) ~ (r cos 6)1 + (r sin 6)J + rk x

FIGURE 16.59 in Example 4.

o :5 r :5 2,

0:5 0 :5 2,..

We then have

"-. y

n=

The curve C and cone S

=

Vz der

=

r, X r.

-(rcosO)i - (rsinO)j + rk

Ir, X r.1

rVz

Section 16.5, Example 4

(-(COSO)i - (sinO)j + k)

rVz drdO

Section 16.5, Example 4

VXF=-4i-2xj+k

Example 1

= -4i - 2rcosOj + k.

x = rcos(J

Accordingly, V X F'n =

=

Vz Vz

(4 cosO + 2r cos 0 sinO + I) (4 cosO + rsin20 + I)

966

Chapter 16: Integration in Vector Fields and the circulation is

f

F·dr =

e

II

V X F'ndu

Stokes'Theorem,Eq.(4)

S =

/.2"/.2 Vz (4

cosO

+ rsin20 + l)(rV2drdO)

= 41T.

•

EXAMPLE 5 The cone used in Example 4 is not the easiest surface to use for calculating the circulation around the bounding circle Clying in the plane z = 3.lfinstead we use the flat disk of radius 3 centered on the z-axis and lying in the plane z = 3, then the normal vector to the surface S is n = k. Just as in the computation for Example 4, we still bave V X F = -4i - 2xj + k. However, now we get V X F· n = 1, so that

II S

V X F· n du =

II

The shadow is the disk of radius 2 in the xy-plane.

1 dA = 41T.

x2+y2S4

TIris result agrees with the circulation value found in Example 4.

Paddle Wheel Interpretation of V

•

F

X

Suppose that F is the velocity field of a fluid moving in a region R in space containing the closed curve C. Then

f

F·dr

e

is the circulation of the fluid around C. By Stokes' Theorem, the circulation is equal to the flux of V X F through any suitably oriented surface S with boundary C:

f

F·dr =

II

V X F·ndu.

s

e

Suppose we fIX a point Q in the region R and a direction u at Q. Take C to be a circle of radius p, with center at Q, whose plane is normal to u. If V X F is continuous at Q, the average value of the u-component of V X F over the circular disk S bounded by C approaches the u-component of V X F at Q as the radius p --> 0: (V X F'u)g = lim

-l,JJrr V X F·udu.

p-o trp

s

If we apply Stokes' Theorem and replace the surface integral by a line integral over C, we get CurlF _ _

(V X F'u)g = lim

l..2f F·dr.

p-O'lrp

(6)

e

The left-hand side of Equation (6) has its maximum value when u is the direction of V X F. When p is small, the liruit on the right-hand side of Equation (6) is approximately -l-fF'dr 7f'p2

FIGURE 16.60 The paddle wheel interpretation of curl F.

e

'

which is the circulation around C divided by the area of the disk (circulation density). Suppose that a small paddle wheel of radius p is introduced into the fluid at Q, with its axle directed along u (Figure 16.60). The circulation of the fluid around C affects the rate

16.7 Stokes'Theorem

967

of spin of the paddle wheel. The wheel spins fastest when the circulation integral is maximized; therefore it spins fastest when the axle of the paddle wheel points in the direction of V X F.

EXAMPLE 6 A fluid of constant density rotates around the z-axis with velocity F = w( -yi + xj), where w is a positive constant called the angular velocity ofthe rotation (Figure 16.61). Find V X F and relate it to the circulation density. SoLution

+

With F = -wyi

wxj, we find the curl

P(x,y, z)

V X F = (ap _ aN)i ay az =

(0 - O)i

+

+

(aM _ ap). az ax J

+ (w -

(0 - O)j

+

(aN _ aM)k ax ay

(-w»k = 2wk.

By Stokes' Theorem, the circulation of F around a circle C of radius p bounding a disk S in a plane normal to V X F , say the xy-plane, is F = w(-yl + xj)

x

/

P(x,y, O)

f

F·dr =

e

y

11

V X F·odO" =

s

11 s

Thus solving this last equation for 2w, we have

FIGURE 16.61 A steady rotational flow parallel to the xy-plane, with constant angular velocity w in the positive (counterclockwise) direction (Example 6).

2wk·kdxdy = (2w)( 1Tp2).

(V X F)· k = 2w =

-1-f 1Tp2

F· dr,

e

•

consistent with Equation (6) whenu = k.

EXAMPLE 7 Use Stokes' Theorem to evaluate Je F· dr, ifF = xzi + xyj + 3xzk and C is the boundary of the portion of the plane 2x + Y + z = 2 in the first octant, traversed counterclockwise as viewed from above (Figure 16.62). z

SoLution

y

The plane is the level surface I(x, y, z) = 2 of the function I(x, y, z) = 2x

+

+ z. The unit normal vector o =

(0,0,2)

VI IVI I =

(2i 12i

+ j + k) + j + kI =

1 (. 21

V6

•

+J +

)

k

is consistent with the counterclockwise motion around C. To apply Stokes' Theorem, we find

j

:x

ay

:z

xz

xy

3xz

V X F = (x - 3(2 - 2x - y»j

+ yk

=

(7x

6 + y)

=

~ (7X + 4y -

curl F = V X F =

x y


k

a

=

(x - 3z)j

+ yk.

On the plane, z equals 2 - 2x - y, so

The planar surface in

+

3y - 6)j

+ yk

and

V X F· 0

=

~ (7X + 3y -

The surface area element is dO" =

IV/I IVI. kl

V6

dA = -1- dx dy.

6).

968

Chapter 16: Integration in Vector Fields The circulation :is

f

ff

F·dr =

C

V X Fonda

Stnkcs''Ibcarem, Eq. (4)

S

11

1 2 2x -

=

~ (7X + 4y -

6)\16 dydx

f' {'-'" (7x+4y-6)dydx~-1.

•

~ JoJo

Let the surface S be the ellipitica1 paraboloid z = x 2 + 4y21ying beneath the plane z = 1 (Figure 16.63). We define the orientation of Sby taking the innernonnal vector n to the surface, which is the normal having a positive k-component. Find the flux of the curl V X F across S in the direction n for the vector field F = yI - xzj + xz2JL

EXAM PLE 8

Solution We use Stokes'Theorem to calculate the curl integral by finding the equivalent counterclockwise circulation of F around the cmve of intersection C of the paraboloid z - x 2 + 4y2 and the plane z - I, as shown in Figure 16.63. Note that the orientation of S is consistent with tntversiDg C in a counterclockwise direction around the z-axis. The cmve Cis the ellipse Xl + 4yl - 1 in the plane z - 1. We can parametrize the ellipse by x - cos t,y - sin t, z - 1 for 0 S t S 2v, so C is given by

!

y

FIGURE 1&.&3 The portion of the ellipiticai paraboloid in Example 8, showing its curve of intersection C with the p1.am: z = I and its inner normal orientation by n.

r(')

~

(cost)1

t

To compute the circulation integral iF dr, we evaluate F along C and fmd the velocity vectDr dr/ dt: 0

F(r('» -

t

(sin ')1 - (..st)J

-:;;

~

-(sin ')1

and

•

+

t

+ (cost)k

(cos Ill·

Thco,

f

f2.

F·dr - Jo

c D

c

B

o :S t :S 21T.

+ (sin 'll + k,

dr F(r(,»· dt dt

f2'(1-"lsin t - "lco;t 1)

=

Jo

=

-t 12w

2

dt

dJ = -1T.

(. )

Therefore the flux of the curl across S in the direction n for the field F is

ff

V X F'nd"

~ -v.

s

•

Proof of Stokes' Theorem for Polyhedral Surfaces Let S be • polyItednU surlioce consmting of • ,mire mnobee of plane regions '" fuceo. (See Figure 16.64 for examples.) We apply Green's Theorem to each separate face of S. There

are two types of faces: (b)

FIGURE 18.84 (a)Partofapolyhedral surface. (b) Other polyhedral. surfaces.

1. 2.

Those that are surrounded on all sides by other faces. Those that have one or more edges that are not adjacent to other faces.

969

16.7 Stokes'Theorem

The boundary Jl of S consists of those edges of the type 2 faces that are not adjacent to other faces. In Figure 16.64a, the triangles EAB, BCE, and CDE represent a part of S, with ABCD part of the boundary Jl. We apply a generalized tangential form of Green's Theorem to the three triangles ofFigore 16.64a in tum and add the results to get

In the generalized form, the line integral of F around the curve enclosing the plane region R normal to n equals the double integral of (curl F) • n over R. The three line integrals on the left-hand side of Equation (7) combine into a single line integral taken around the periphery ABCDE because the integrals along interior segments cancel in pairs. For example, the integral along segment BE in triangle ABE is opposite in sign to the integral along the same segment in triangle EBC. The same holds for segment CEo Hence, Equation (7) reduces to

f

F·dr

=

AllCDE

II

VX

F·nda-.

AllCDE

When we apply the generalized form of Green's Theorem to all the faces and add the results, we get

f

F ' dr =

IlvxF.nda-. s

~

This is Stokes' Theorem for the polyhedral surface S in Figore 16.64a. More general polyhedral surfaces are shown in Figore 16.64b and the proof can be extended to them. General smooth surfaces can be obtained as limits of polyhedral surfaces.

Stokes' Theorem for Surfaces with Holes Stokes' Theorem holds for an oriented surface S that has one or more holes (Figore 16.65). The surface integral over S of the normal component of V X F equals the sum of the line integrals around all the boundary curves of the tangential component of F, where the curves are to be traced in the direction induced by the orientation of S. For such surfaces the theorem is unchanged, but C is considered as a union of simple closed curves.

An Important Identity The following identity arises frequently in mathematics and the physical sciences.

curl grad I = 0

FIGURE 16.65 Stokes' Theorem also holds for oriented surfaces with holes.

or

V X Vf= 0

(8)

This identity holds for any function I(x, y, z) whose second partial derivatives are continuous. The proof goes like this:

i

a V X VI = ax al ax

j

k

a ay al ay

a az al az

= (fry - Iyz)i - (fzx -

Ixzli + (fp;

- Iry)k.

If the second partial derivatives are continuous, the mixed second derivatives in parentheses are equal (Theorem 2, Section 14.3) and the vector is zero.

970


Conservative Fields and Stokes' Theorem In Section 16.3, we found that a field F being conaervative in an open region D in space is equivalent to the integral of F around every closed loop in D being zero. This, in turn, is equivalent in simply connected open regions to saying that V X F = 0 (which gives a test for determining if F is conservative for such regions).

(a)

THEOREM 7-Curl F = 0 Related to the Closed-Loop Property

If V X F = 0 at every point of a simply connected open region D in space, then on any piecewisesmooth closed path C in D,

f

•

Sketch of a Proof Theorem 7 can be proved in two steps. The first step is for simple closed curves (loops that do not cross themselves), like the one in Figure 16.66a. A theorem from topology, a branch of advanced mathematics, states that every smooth simple closed curve C in a simply connected open region D is the boundary of a smooth two-sided surface S that also lies in D. Hence, by Stokes' Theorem,

f

(b)

0

31. y" + Y = 2 cosx 32. y"

+ y'

+ Bxe~

+ sinx, Yp = A cosX' + B sinx + sinx, YP = Ax cosX' + Bx sinx

- 2y = xe x , YP = Ax 2e x

+ Bxe x

coefficients. dy

33 - - = eX • dx' dx

+ e-r

d'y dy 35. dx,-4 -5y =ex +4

dx

=

cscx,

39. y" - 8y' =

o<x<w

43. y"

+

45. y"

+Y

42. y"

2y' = x 2 =

46. y" - 3y'

e~

-

44. y"

secxtanx,

+

+ 4y = sinx + 4y' + 5y = x + 2 + 9y = 9x - cosX'

40. y"

e!b:

41. y" - y' = x 3

_'!! 2

< x < '!! 2

2y = eX - e'h

The method of undetermined coefficients can sometimes be used to solve flISt-order ordinary differential equations. Use the method to solve the equations in Exercises 47-50.

47. y' - 3y = eX

48. y'

+ 4y = x

49. y' - 3y = 5e'"

SO. y'

+Y

d'y 51. dx'

+ Y = sec x, -2 < x < 2;

d'y 52. dx'

+ Y = e2x;

,1f

y(O)

= 0,

1f

= sinx

y(O)

= y'(O) = I

2

=5

y'(O)

d'y

53. y"

x' - x, y(0) = 0, + y' = x, YP = T

54. y"

+ Y = x, YP = 2 sin x + x,

55.

In Exercises 33-36, solve the given differential equations (a) by variation of parameters and (h) by the method of undetermined d'y

cotx, O<X 0 in each exercise. 59. x7" + 2xy' - 2y = x 2, Yl = x- 2 , Y2 = x 60.

x7" + xy'

- Y=

X,

YI = x-I, Y2 = x

17.3

I ,,,

Applications

17-17

AppLications In this section we apply second-order differential equations to the study of vibrating springs and electric circuits.

Vibrations A spring has its upper end fastened to a rigid support, as shown in Figure 17.2. An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an equilibrium position. According to Hooke's Law (Section 6.5), the tension force in the spring is ks, where k is the spring constant. The force due to gravity pulling down on the spring is mg, and equilibrium requires that massm

ks = mg.

at equilibrium

y

FIGURE 17.2

Massm stretches a spring by length s to the equilibrium position at y = O.

(1)

Suppose that the object is pulled down an additionsl amount Yo beyond the equilibrium position and then released We want to study the object's motion, that is, the vertical position of its center of mass at any future time. Let y, with positive direction downward, denote the displacement position of the object away from the equilibrium position y = 0 at any time t after the motion has started. Then the forces acting on the object are (see Figure 17.3)

Fp

=

mg,

F, = k(s

F,=ll

the propulsion force due to gravity,

+ y),

the restoring force of the spring's tension,

dy , dt

a frictional force assumed proportional to velocity.

The frictional force tends to retard the motion of the object. The resultant of these forces is F = Fp - F. - F" and by Newton's second law F = ma, we must then have

a"ly m dt 2

=

dy mg - ks - ky - Il dt'

By Equation (1), mg - ks = 0, so this last equation becomes

dly m dt 2 y

Yo

start

position y

FIGURE 17.3 Thepropulsion force ("";ght) Fp pulls iIle mass downward, but !he spring restoring force F, aod fric1iooa1 force F, pullille mass upward. The motion starts at Y = Yo with iIle mass vihratiog up aod down.

+

dy Il dt

+ ky

=

(2)

0,

subject to the initial conditions y(O) = Yo and y' (0) = O. (Here we use the prime notation to denote differentiation with respect to time t.) You might expect that the motion predicted by Equation (2) will be oscillatory about the equilibrium position y = 0 and eventually dsmp to zero because of the retarding frictional force. This is indeed the case, and we will show how the constants m, Il, and k determine the nature of the dsmping. You will also see that if there is no friction (so Il = 0), then the object will simply oscillate indefinitely.

Simple Harmonic Motion Suppose first that there is no retarding frictional force. Then Il = 0 and there is no dsmping. lfwe substitote w = Vkj;. to simplify our calculations, then the second-order equation (2) becomes with

y(O) = Yo

and

y'(O) = O.

17 -18 Chapter 17: Second-Order Differential Equations The auxiliary equation is r2

+ if

0,

=

having the imaginary roots r = ±wi. The general solution to the differential equation in (2) is

Y =

Cl

cos rut

+

sin wt.

C2

(3)

To fit the initial conditions, we compute

y'

-Cl Cd sin

=

rut +

C2W cos

wt

and then substitute the conditions. This yields c, = Yo and C2 =

o. The particular solution

Y=Yocoswl

(4)

describes the motion of the object. Equation (4) represents simple harmonic motion of amplitude Yo and period T = 27T/W. The general solution given by Equation (3) can be combined into a single term by using the trigonometric identity sin(WI

+ cf»

+ sin wi cos cf>.

wi sin cf>

= cos

To apply the identity, we take (see Figure 17.4)

c,

=

C sin cf>

and

C2

=

C cos cf>,

where

C= Vc,2 FIGURE 17_4 = Ceosc/>.

C2

c, =

C sin c/> and

+

co'

_

and

-1 Cl

cf> - tan

C2·

Then the general solution in Equation (3) can be written in the alternative form

Y = C sin (WI

+ cf».

(5)

Here C and cf> may he taken as two new arbitrary constants, replacing the two constants c, and C2. Equation (5) represents simple harmonic motion of amplitude C and period T = 27T/W. The angle wi + cf> is called the phase angie, and cf> may be interpreted as its initial value. A graph of the simple harmonic motion represented by Equation (5) is given in Figure 17.5.

y

Period

~ T=~ ~

C Csin."

I I

I I

o -C y=Csin(w,+.,,)

FIGURE 17.5 Simple hannonie motion of amplitude C and period T wi1h initial phase angle c/> (Equation 5).

17.3 Applications

17 -19

Damped Motion Assume now that there is friction in the spring system, so [; "" O. If we substitute w = ~ and 2b = [;/m, then the differential equation (2) is

y"

+ 2by' + .Jy

=

O.

(6)

The auxiliary equation is ,2+2b,+w2 =O,

with roots , = -b ± yCb"2-_-w'''. Three cases now present themselves, depending upon the relative sizes of b and w. Case 1: b = Cd. The double root of the auxiliary equation is real and equals, = w. The general solution to Equation (6) is

y = (Cl

+ c2t)e-"".

This situation of motion is called critical damping and is not oscillatory. Figure 17.6a shows an example of this kind of damped motion. Case 2:

b>

-b + is given by

'1 =

The roots of the auxiliary equation are real and unequal, given by 2 Vb - w' and '2 = -b - Vb 2 - w'. The general solution to Equation (6) Cd.

y = Cl e (-b+v'b'-w'p

+ C2e (-b-v'b'-w'Jt.

Here again the motion is not oscillatory and both '1 and '2 are negative. Thus y approaches zero as time goes on. This motion is referred to as overdamping (see Figure l7.6b). Case 3: b , =

O.

(l)

17-24

Chapter 17: Second-Order Differential Equations To solve Equation (I), we rlISt make the change of variables

z = Inx

y(x) = Y(z).

and

We next use the chain rule to find the derivatives y'(x) and y"(x):

y' (x) =

fx Y(z)

=

~Y(z):

= Y' (z){

and

-~2 Y'(z) + ! y"(z)dz

y"(x) = i£y'(x) = i£y'(z)! x = dx dx

x

x

dx

=

-~2 Y'(z) + ~2 Y"(z). x x

Substituting these two derivatives into the left-hand side of Equation (I), we rmd

ax"y"

+ bxy' +

:2

cy = ax 2 ( = aY"(z)

Y'(z)

+ (b

:2

+

+ bX({ Y'(Z)) +

Y"(Z))

- a)Y'(z)

cY(z)

+ cY(z).

Therefore, the substitutions give us the second-order linear clifferential equation with constant coefficients

aY"(z)

+ (b

- a)Y'(z)

+ cY(z)

=

o.

(2)

We can solve Equation (2) using the method of Section 17.1. That is, we find the roots to the associated auxiliary equation

ar2

+ (b

- a)r

+C

= 0

(3)

to find the general solution for Y(z). After rmding Y(z), we can detennine y(x) from the substitutionz = Inx.

EXAMPLE 1

Find the general solution of the equation x 2y"

+

2xy' - 2y =

o.

Solution This is an Euler equation with a = I, b = 2, and C = -2. The auxiliary equation (3) for Y(z) is r2

+ (2

+ 2)

- I)r - 2 = (r - I)(r

=

0,

with roots r = -2 and r = 1. The solution for Y(z) is given by

Y(z) = Ct e -2z

+ C2ez.

Substituting z = In x gives the general solution for y(x):

y(x) = cte-2lnx

EXAMPLE 2

+ c2e lnx

= CtX- 2

Solve the Euler equation x"y" - 5xy'

+

+ C2X

•

9y = O.

Solution Since a = I, b = -5, and C = 9, the auxiliary equation (3) for Y(z) is r2

+ (-5

- I)r

+9

= (r -

3)2 = O.

The auxiliary equation has the double root r = 3 giving

Y(z) = cte 3z

+

c2ze3~

Substituting z = In x into this expression gives the general solution

y(x) = Cte3lnx

+ c2lnxe3lnx

= CtX3

+ C2x3lnx

•

17.4 Euler Equations

EXAMPLE 3 Find the particular solution to x"y" - 3xy' initial conditions y(1) = 0 andy'(I) = 1. Solution Here a = I, b = -3, and gives

C

+ 68y

17-25

= 0 that satisfies the

= 68 substitoted into the auxiliary equation (3)

r 2 -4r+68=0. The roots are r = 2

+

8i and r = 2 - 8i giving the solution

Y(z) = e2z(cl cos 8z

+ C2 sin 8z).

Substitoting z = In x into this expression gives

y(x) =

e

21nx

( Cl

cos (8 In x)

+ C2 sin (8 In x»).

From the initial condition y(l) = 0, we see that Cl = 0 and

y(x) = C2x2 sin (8 In x).

y

To fit the second initial condition, we need the derivative

10

y'(x) = c2(8x cos (8 In x) 5

Since y'(I) = I, we immediately obtain C2 = 1/8. Therefore, the particular solution satisfying both initial conditions is

o

2

10

y(x) =

2

-5

+ 2xsin(8lnx»).

t

x 2 sin (8 In x).

y= "ssin(8lnx)

Since -I :5 sin (8 In x) :5 I, the solution satisfies -10

x2

-8 :5 y(x) FIGURE 17.8 Example 3.

:5

x2

8'

Graph of the solution to

A graph of the solution is shown in Figure 17.8.

EXERCISES 17.4 In Exercises 1-24, rmd the general solution to the given Euler equation. Assume x > 0 throughout

1. x"y" + 2xy' - 2y

=

0

=

0

3.x"y"-6y=0 5. xV - Sxy'

+ 8y

=

0

4.x"y"+xy'-y=0

7. 3xV + 4xy' = 0 9. x"y" - xy' + Y = 0 11. x"y" - xy' + Sy = 0

+ 3xy' + lOy = 0 15. 4xV + 8xy' + Sy = 0 17. x"y" + 3xy' + Y = 0 19. xV + xy' = 0 13. x"y"

2. x"y" + xy' - 4y

+ 7xy' + 2y = 0 8. xV + 6xy' + 4y = 0 10. x"y" - xy' + 2y = 0 12. x"y" + 7xy' + l3y = 0 14. x"y" - Sxy' + lOy = 0 16. 4x"y" - 4xy' + Sy = 0 18. x"y" - 3xy' + 9y = 0 20. 4xV + Y = 0 6. 2xV

21. 9x"y' + ISxy' 22. 16xV - 8xy'

+Y = 0 + 9y = 0 23. 16xV + S6xy' + 2Sy = 0 24. 4x"y' - 16xy' + 2Sy = 0 In Exercises 25-30, solve the given initial value problem.

25. xV 26. 6x"y' 27. xV

+ 3xy' - 3y = 0, y(l) = I, y'(I) = -I + 7xy' - 2y = 0, y(1) = 0, y'(I) = I - xy' + Y = 0, y(l) = I, y'(I) = I

28. xV + 7xy' + 9y = 0, y(l) = I, y'(I) = 0 29. xV - xy' + 2y = 0, y(1) = -I, y'(I) = I 30. xV + 3xy' + Sy = 0, y(l) = I, y'(I) = 0

•

17 -26

Chapter 17: Second-Order Differential Equations

Power-Series SoLutions In this section we exteod our stody of second-order linear homogeoeous equations with variable coefficieots. With the Euler equations in Section 17.4, the power of the variable x in the nonconstant coefficieot had to match the order of the derivative with which it was paired: x 2 with y', Xl with y', and x O(=I) with y. Here we drop that requiremeot so we ean solve more geoeral equations.

Method of Solution The poweMeries method for solving a second-order homogeneous differential equation consists of finding the coefficieots of a power series 00

y(x) = ~ c.x' = Co

.-0

+ ClX + C2x2 + ...

(1)

which solves the equation. To apply the method we substitote the series and its derivatives into the differential equation to determine the coefficieots Co, Ct, C2, .... The technique for finding the coefficieots is similar to that used in the method of uodetermined coefficieots preseoted in Section 17.2. In our first example we demonstrate the method in the setting of a simple equation whose general solution we already know. This is to help you become more comfortable with solutions expressed in series form.

EXAMPLE 1

Solve the equation y'

+Y

= 0

by the power-series method.

Solution We assume the series solution takes the form of

and calculate the derivatives 00

y'

=

~

00

ncll x n -

1

y"

and

n(n - 1)c.x·- 2.

= ~

11=1

11=2

Substitotion of these forms into the second-order equation gives us 00

00

11 - 2

11 - 0

~ n(n - l)c.x·- 2 + ~ c.x· = O. Next, we equate the coefficients of each power of x to zero as summarized in the followiog table.

Powerofx

Coefficient Equation

1

xo

2(I)C2

+ Co

=

0

or

C2 = -ZCo

xl

3(2)C3

+ Cl

=

0

or

C3 = -3-2 Cl

x2

4(3)C4

+ C2

=

0

or

C4 = -4'3 C2

x3

5(4)cs

+ C3

=

0

or

Cs = -5'4 C3

x4

6(5)C6

+ C4

=

0

or

C6 = -6'5 C4

+ C.-2

=

0

or

en =

X,,-2

n(n - l)c.

1 1 1 1

1 n(n - 1) C.-2

17.5 Power-Series Solutions

17-27

From the table we notice that the coefficients with even indices (n = 2k, k = 1,2,3, ... ) are related to each other and the coefficients with odd indices (n = 2k + 1) are also interrelated. We treat each group in turn. Even indices: Here n = 2k, so the power is x 2k--2. From tlie last line of the table, we have 2k(2k - l)c2k

+ C2k-2

=

0

or

C2k =

I 2k(2k _ I) C2k-2·

From this recursive relation we find

c2k = [ 2k(2;- 1)][ (2k - 2;(2k - 3)]-" [-

=

4(~)][-!]co

(-1)1 (2k)! co·

Odd indices: Here n = 2k line of the table yields (2k

+ 1, so the power is X2k-I. Substituting this into the last

+ 1)(2k)c2k+ I + C2k-1

=

0

or

c2k+ I =

(2k

I

+ I )(2k) C2k-I·

Thus,

C2k+1 = [ (2k+ \)(2k)] [ (2k -

=

1)~2k -

2J . ht4)] ht2)]cl

(_1)1 (2k + I)! CI·

Writing the power series by grouping its even and odd powers together and substituting for the coefficients yields

00

=

~

;Y' + 6y

l)y"

+ y'

+ 2)y' + 2y = 0 2 13. (x - l)y" + 4'>;y' + 2y = 0 14. y" - 2xy' + 4y = 0

4. y" - 3y'

xV -

-

10. y"

12. .>;y" - (x

3.y"+4y=O

5.

9. (x 2

16. (I - x 2)y" - .>;y'

=0

=

0

+ 6y

=

17. y" - xy' + 3y 18.

xV -

4.>;Y'

+ 4y 0

= 0

ApPENDICES Real Numbers and the Real Line

A.1

This section reviews real numbers, inequalities, intervals, and absolute values.

Real Numbers Much of calculus is based on properties of the real number system. Real numbers are numbers that can be expressed as decimals, such as

- 4"3 =

"31 =

v2 =

-0.75000 . .. 0.33333 . . . 1.4142 .. .

The dots ... in each case indicate that the sequence of decimal digits goes on forever. Every conceivable decimal expansion represents a real number, although some numbers have two representations. For instance, the infinite decimals .999 . .. and 1.000 ... represent the same real number 1. A similar statement holds for any number with an infinite tail of9's. The real numbers can be represented geometrically as points on a number line called the real line.

Rules for inequalities I

If a, b, and e are real numbers, then: l.a 0

~

7i> 0

1

6. If a and b are both positive or both

. then a < b negatIve,

~

b1 < 7i1

-2

I

o

1

3

2

I I

I

37T

4

)

The symbollR denotes either the real number system or, equivalently, the real line. The properties of the real number system fall into three categories: algebraic properties, order properties, and completeness. The algebraic properties say that the real numbers can be added, subtracted, multiplied, and divided (except by 0) to produce more real numbers under the usual rules of arithmetic. You can never divide by O. The order properties of real numbers are given in Appendix 6. The useful rwes at the left can be derived from them, where the symbol ~ means "implies." Notice the rules for multiplying an inequality by a number. Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign. For example, 2 < 5 but -2 > -5 and 1/2 > 1/5. The completeness property of the real number system is deeper and harder to defme precisely. However, the property is essential to the idea ofa limit (Chapter 2). Roughly speaking, it says that there are enough real numbers to "complete" the real number line, in the sense that there are no "holes" or "gaps" in it. Many theorems of calcwus would fail if the real number system were not complete. The topic is best saved for a more advanced course, but Appendix 6 hints about what is involved and how the real numbers are constructed.

AP-l

AP-2

Appendices

We distinguish three special subsets of real numbers.

1. 2.

The natural numbers, namely 1,2,3,4, ...

3.

The rational numbers, namely tbe numbers tbat can be expressed in tbe form of a fraction min, where m and n are integers and n oF o. Examples are

The integers, namely 0, ± 1, ±2, ±3, ...

1 3'

4 9

-4 9

4 -9'

200 13'

57 = 57 1

and

The rational numbers are precisely tbe real numbers witb decimal expansions tbat are eitber

(a) terminating (ending in an infinite string of zeros), for example,

43

=

0.75000 ... = 0.75

or

(b) eventually repeating (ending witb a block of digits tbat repeats over and over), for example 23

IT

=

2.090909... = 2.09

The bar indicates the block of repeating digits.

A terminating decimal expansion is a special type of repeating decimal, since tbe ending zeros repeat. The set of rational numbers has all tbe algebraic and order properties of tbe real numbers but lacks tbe completeness property. For example, tbere is no rational number whose should be. square is 2; tbere is a ''hole'' in tbe rational line where Real numbers tbat are not rational are called irrational numbers. They are characterized by having nonterminating and nomepeating decimal expansions. Examples are 'fr, and 10g!O 3. Since every decimal expansion represents a real number, it should be clear tbat tbere are infmitely many irrational numbers. Botb rational and irrational numbers are found arbitrarily close to any point on tbe realline. Set notation is very useful for specifying a particular subset of real numbers. A set is a collection of objects, and tbese objects are tbe elements oftbe set. If S is a set, tbe notation a E S means tbat a is an element of S, and a I'! S means tbat a is not an element of S. If S and T are sets, tben S U T is tbeir union and consists of all elements belonging eitber to S or T (or to botb S and T). The intersection S n T consists of all elements belonging to botb S and T. The empty set 0 is tbe set tbat contains no elements. For example, tbe intersection of tbe rational numbers and tbe irrational numbers is tbe empty set. Some sets can be described by listing tbeir elements in braces. For instance, tbe set A consisting of tbe natural numbers (or positive integers) less than 6 can be expressed as

v2

v2, ¥S,

A = {1,2,3,4,5}. The entire set of integers is written as {O, ± 1, ±2, ±3, ... } . Anotber way to describe a set is to enclose in braces a rule tbat generates all tbe elements of tbe set. For instance, tbe set

A = {xix is an integer and 0 is tbe set of positive integers less tban 6.

<x
6 is an interval, as is the set of all x such that - 2 :5 X :5 5. The set of all nonzero real numbers is not an interval; since 0 is absent, the set fails to contain every real number between -I and I (for example). Geometrically, intervals correspond to rays and line segments on the real line, along with the real line itself. Intervals of numbers corresponding to line segments are finite intervals; intervals corresponding to rays and the real line are infinite intervals. A finite interval is said to be closed if it contains both of its endpoints, half-open if it contains one endpoint but not the other, and open if it contains neither endpoint. The endpoints are also called boundary points; they make up the interval's boundary. The remaining points of the interval are interior points and together comprise the interval's interior. Infinite intervals are closed if they contain a finite endpoint, and open otherwise. The entire real line n;l is an infinite interval that is both open and closed Table A.I summarizes the various types of intervals.

TABLE A.l Types of intervals Notation

Set description

Type

(a, b)

{xla < x < b}

Open

[a, b]

{xla

:5

x

Closed

[a, b)

{xla

:5

x < b}

(a, b]

{xla < x

(a, 00)

{xix> a}

Open

: a

[a, 00)

{xix;;,: a}

Closed

a

(-00, b)

{xlx x T.

OrX:s~l.

The solution set is the half-open interval (1,11/5] (Figure Alc).

Absolute Value

-

The absolute value of a number x, denoted by Ix I, is dermed by the fonnula Ixl = {

EXAMPLE 2

131 = 3,

101 = 0,

x, -x,

x2:0 x < O.

I-51 = -(-5) = 5,

Hall = lal

-

Geometrically, the absolute value of x is the distance from x to 0 on the real number line. Since distances are always positive or 0, we see that Ixl 2: 0 for every real number x, andlxl = oif and ouly if x = O.Also, Ix - yl = the distance between x andy ~ 1-51~5 I

-5

) ,

131 ---->

I

I

o

3

on the real line (Figure A2). Since the symbol Va always denotes the nonnegative square root of a, an alternate definition of Ix Iis

+---14 - II ~ II - 41 ~ 3 ~

I

I

4

FIGURE A.2 Absolute values give distances between points on the number line.

Ixl=W. It is important to remember that W = Ia I. Do not write W = a unless you already know that a 2: O. The absolute value has the following properties. (You are asked to prove these properties in the exercises.)

Appendix 1 Real Numbers and the Real Line

AP-5

Absolute Value Properties 1.1-al = lal

A number and its additive inverse or negative have the same absolute value.

2. labl = lallbl

The absolute value of a product is the product of the absolute values.

lal bal =jbj l 4. la + bl ,;; lal

The absolute value of a quotient is the quotient of the absolute values.

3.

~ a --~)¥I,--- a ~ ~~~.~----~I------~~

-a

x 0 ~ Ixl --->l

a

FIGUREA.3 Ixl < a means x lies between -a and a.

+ Ibl

Note that I-al oF -Ial. For example, 1-31 = 3, wbereas -131 = -3. Ifa and b differ in sign, then la + bl is less than lal + Ibl. In all other cases, la + bl equals Ia I + Ib I. Absolute value bars in expressions like 1-3 + 51 work like parentheses: We do the arithmetic inside before taking the absolute value. EXAMPLE 3 1-3 13

AbsoLute vaLues and intervals Ixl = a

~

6. Ixl < a

~

7.lxl>a

=

8. Ix I ,;; a 9. Ix I ;;" a

=

~

x = ±a -a < x < a x>aorx I

A.2

I

14.1 3 -}I < t 17.lr~

(2) 2

la + bl '" lal + Ibl 9. 18 - 391 =

(3)

= (Ial + Ibl)'

~

Solve the inequalities in Exercises 10--17, expressing the solution sets as intervals or unions of intervals. Also, show each solution set on the real line. 10. Ixl < 2

2

2

Solve the equations in Exercises 7-9. 7. Iyl = 3

(1)

=a 2 +2ab+b 2

4.5x-3"'7-3x

I

x< 0

23. Solve the equation Ix - II = I-x.

In Exercises 3--6, solve the inequalities and show the solution sets on the rea1line. 3. -2x

-

22. Do not fall into the trap of thinking I-al = a. For what real numbers a is this equation true? For what real numbers is it false?

12. 13y - 71 < 4 15. 1291'" 4

(4)

25. Prove that Iab I = Ia II b I for any nwnbers a and b. 26. Ifl xl '" 3 and x > -1/2, what can you say abootx? 27. Graph the inequalitylxl + Iyl '" 1. 28. For any number a, pruve that I-a I = Ia I. 29. Let a he any positive nwnber. Prove that Ix I > a if and only if x>aorx 3k for some k 20 7. Then

It looks as if n!

n, =

(k + I)! = (k + I)(k!)

Thus, for k

20

>

(k + 1)3k

>

7.3 k

>

3k+l.

7,

k! > 3k implies

(k + I)! > 3k+l.

The mathematical induction principle now guarantees n!

20

3" for all n

Proof of the Derivative Sum Rule for Sums of Finitely Many Functions We prove the statement

20

7.

•

AP-9

Appendix 2 Mathematical Induction

by mathematical induction. The statement is true for n = 2, as was proved in Section 3.3. This is Step 1 of the induction proof. Step 2 is to show that if the statement is true for any positive integer n = k, where k ;", no = 2, then it is also true for n = k + 1. So suppose that d

dx (U1

+ U2 + ... + Uk)

=

dU1 dU2 ax + ax +

(1)

Then d dx

(U1

+ U2 + ... + uk + UH 1) ~

Call this function v.

Call the function delmed by this sum u.

+ Uk) + dU1

dU2

dUk

=dx -++ dx

d

Sum Rule for ax (u + v)

dUk+1

dx

dUk+1

+dx- + dx

Eq. (1)

With these steps verified, the mathematical induction principle now guarantees the Sum Rille for every integer n ;", 2.

Exercises A.2 1. Assuming that the triangle inequality Ia for aoy two numbers a and b, show that IX1

+ X2 + ... + x,1

:5 Ixd

+ b I :5

Ia I + Ib I holds

6. Showthatn! 7. Show that 2'

+ IX21 + ... + lx, I

8. Show that 2" '" 1/8 for n '" -3. 9. Sum. of .quare.

for aoy n numbers.

positive integers is

2. Show that if r #' I, then

for every positive integer n.

d

dv

3. Use the Product Rnie, dx (uv) = u dx = I to show that

:! (x')

du

+ v dx' and the fact that

= nx,-l for every positive inte-

gern.

4. Suppose that a function fIx) has the property that f(xlxz) = f(xl) + f(x2) for aoytwo positive nurohers Xl andxz. Show that

Show that the sum of the squares ofilie fmt n

+ t}n +

1 - r"+l 1 +r+r2+"'+r"=~--"- I r

:! (x)

> n'ifnislargeenough. > n Z ifn is large enuugh.

+

I)

3

10. Sum. of cub.. Show that ilie sum of the cubes of the I!rst n positive integers is (n(n + 1)/2),. 11. Rules for Imite .um. Show that ilie following Imite sum rules hold for every positive integer n. (See Section 5.2.)

a.

" "" L(a,+ b,) = La,+ Lb, k=i

k=i

k=l

f(xlxz" ·x,) = f(xl) + f(xz) + ... + fIx,)

for the product ofany n positive numbers Xl. X2 •••• , xn.

(aoy number c)

S. Showthat

2 2 2 I -+-+···+-=1-1 2

3

3

for all positive integers n.

3"

(if

a, has ilie constant value c)

3"

12. Show that Ix"1 = Ix I" for every positive integer n and every real numberX'.

AP-l0

Appendices

Lines, Circles, and Parabolas

A.3

TIris section reviews coordinates, lines, distance, circles, and parabolas in the plane. The notion of increment is also discussed. y

b

-------o P(a,b) I I I I I I I I I

Positive y-axis 3

~ 2 Origin

Negative x-axis

:

~--~"~~~~~~~~~x -3 -2 -1 0 1 \2 a 3 -1

Positive x-axis

Negative y-axis 2

~ -3

FIGURE A.4 Cartesian coordinates in the plane are based on two perpendicular axes intersecting at the origin.


Rene Descartes (1596-1650)

y

(1,3) 3

Second quadrant (-, +)

•

•

First 2

quadrant (+, +)

•

1

(-2, 1)

-2

(0,0) -1

(2, 1)

~

(1,0)

0

2

x

(-2,-1)

• Third

-1

quadrant (-, -)

-2

•

Fourth quadrant (+, -)

(1, -2)

FIGURE A.S Points labeled in the xy-coordinate or Cartesiao plane, The points on the axes all have coordinate pairs but are usually labeled with single real numbers, (so (I, 0) on the x-axis is labeled as I). Notice the coordinate sign patterns of the quadrants.

Cartesian Coordinates in the Plane In Appendix I we identified the points on the line with real numbers by assigning them coordinates. Points in the plane can be identified with ordered pairs of real numbers. To begin, we draw two perpendicular coordinate lines that intersect at the O-point of each line. These lines are called coordinate axes in the plane. On the horizontal x-axis, numbers are denoted by x and increase to the right. On the vertical y-axis, numbers are denoted by y and increase upward (Figure AA). Thus ''upward'' and "to the right" are posi-

tive directions, whereas "downward" and ''to the left" are considered as negative. The origin 0, also labeled 0, of the coordinate system is the point in the plane where x and y are both zero. IfP is any point in the plane, it can be located by exactly one ordered pair of real numbers in the following way. Draw lines through P perpendicular to the two coordinate axes. These lines intersect the axes at points with coordinates a and b (Figure AA). The ordered pair (a, b) is assigned to the pointP and is called its coordinate pair. The first number a is the ;»-coordinate (or abscissa) of P; the second number b is the y-eoordinate (or ordinate) of P. The x-coordinate of every point on the y-axis is O. The y-coordinate of every point on the x-axis is O. The origin is the point (0, 0). Starting with an ordered pair (a, b), we can reverse the process and arrive at a correspondingpointPin the plane. Often we identifY Pwith the ordered pair and write P(a, b). We sometimes also refer to "the point (a, b)" and it will be clear from the context when (a, b) refers to a point in the plane and not to an open interval on the real line. Several points labeled by their coordinates are shown in Figure A.5. TIris coordinate system is called the rectangular coordinate system or Cartesian coordinate system (after the sixteenth-century French mathematician Rene Descartes). The coordinate axes of this coordinate or Cartesian plane divide the plane into four regions called quadrants, numbered counterclockwise as shown in Figure A.5. The graph of an equation or inequality in the variables x and y is the set of all points P(x, y) in the plane whose coordinates satisfY the equation or inequality. When we plot data in the coordinate plane or graph formulas whose variables have different units of measure, we do not need to use the same scale on the two axes. If we plot time vs. thruat for a rocket motor, for example, there is no reason to place the mark that shows I sec on the time axis the same distance from the origin as the mark that shows I Ib on the throst axis. Usually when we graph functions whose variables do not represent pbysical measurements and when we draw figures in the coordinate plane to study their geometry and trigonometry, we try to make the scales on the axes identical. A vertical unit of distance then looks the same as a horizontal unit. As on a surveyor's map or a scale drawing, line segments that are supposed to have the same length will look as if they do and angles that are supposed to be congruent will look congruent. Computer displays and calculator displays are another matter. The vertical and horizontal scales on machine-generated graphs usually differ, and there are corresponding distortions in distances, slopes, and angles. Circles may look like ellipses, rectangles may look like squares, right angles may appear to be acute or obtuse, and so on. We discuss these displays and distortions in greater detail in Section 1.4.

Increments and Straight Lines When a particle moves from one point in the plane to another, the net changes in its coordinates are called increments. They are calculated by subtracting the coordinates of the

Appendix 3 Lines. Circles, and Parabolas

starting point from the coordinates of the ending point. If x changes from x, to X2, the increment in x is

y C(5,6)

6

5

I

B(2, 5)

4

EXAMPLE 1 In going from the point A( 4, -3) to the point B(2, 5) the increments in the x- and y-coordinates are

';';~j

3

~x =

2 1

AP-ll

~y=5-(-3)=8.

2 - 4 = -2,

From C(5, 6) to D(5, 1) the coordinate increments are

D(5, I)

~O+-~I--~~~~--5~-->X

~=5-5=0,

~Y =

1- 6

=

-5.

-I

See Figure A6.

-2

Given two points PI(Xt.YI) and P2(X2,Y2) in the plane, we call the increments ~ = X2 - XI and ~Y = Y2 - YI the run and the rise, respectively, between PI and P2. Two such points always determine a unique straight line (usually called simply a line) passing throogh them both. We call the line PIP2. Any nonverticalline in the plane has the property that the mtio

-3

~~_ A(4,-3)

(2,-3)

I1x =-2

FIGURE A.6 Coordinate increments may be positive, negative, or zero (Example I).

•

~Y

rise

Y2 - YI

m=run=~=X2

y

XI

has the same value for every choice of the two points PI(Xt.YI) and P2(X2,n) on the line (Figure A 7). This is because the mtios of corresponding sides for similar triangles are equal.

!!.y

(rise)

!!.y'

PI (xl' Yl)

<JI-----:--~

!!.x

Q(x"y,)

DEFINITION

The constant ratio rise

m = run =

~Y ~

Y2 - YI

= X2

XI

(run) -----------~ Q'

is the slope of the nonverticalline PIP2.

!!.x' ----~~--------------~x

o

FIGURE A.7 Triangles P,QP, and Pl'Q'P2' are similar, so the ratio of their sides has the same value for any two

points on the line. Ibis common value is the line's slope.

The slope tells us the direction (uphill, downhill) and steepness of a line. A line with positive slope rises uphill to the right; one with negative slope falls downhill to the right (Figure A8). The greater the absolute value of the slope, the more mpid the rise or fall. The slope of a vertical line is undefined. Since the run ~ is zero for a vertical line, we cannot form the slope ratio m. The direction and steepness of a line can also be measured with an angle. The angle of inclination of a line that crosses the x-axis is the smallest counterclockwise angle from the x-axis to the line (Figure A9). The inclination of a horizontal line is 0°. The inclination of a vertical line is 90°. If c/> (the Greek letter phi) is the inclination of a line, then o :5 c/> < 180°. The relationship between the slope m of a nonverticalline and the line's angle of inclination c/> is shown in Figure A.10:

m

= tanc/>.

Straight lines have relatively simple equations. All points on the vertical line through the point a on the x-axis have x-coordinates equal to a. Thus, X = a is an equation for the vertical line. Similarly, Y = b is an equation for the horizon/alline meeting the y-axis at b. (See Figure All.) We can write an equation for a nonvertical straight line L if we know its slope m and the coordinates of one point PI(Xt. YI) on it. If Pix, y) is any other point on L, then we can

AP-12

Appendices

use the two points PI and P to compute the slope,

y

Y - YI

m = x -

Xl

so that

Y - YI

=

mix - Xl),

Y =YI + mix -Xl).

or

The equation

Y

= YI

+ mix -

XI)

is the point-slope equation of the line that passes through the point (Xj,YI) and has slope m. FIGURE A.S The slope of LI is dy 6 - (-2) 8 m=dX= 3 0 3'

y

That is, y increases 8 units every time

x increases 3 units. The slope of L2 is dy 2- 5 -3 m = dx = 4 - 0 = T' That is, y decreases 3 units every time

~.s

x increases 4 units. this

--~~--~~x I

X

I

. . . __ ~//Jnotthis

--~----------~x

Inotthis

FIGURE A.9 Angles of inclination are measured counterclockwise from the x-axis.

EXAMPLE 2

FIGURE A.I0 The slope ofanonver1ical line is the tangent of its aog1e of inclination.

Write an equation for the line through the point (2, 3) with slope - 3/2.

Solution We substitute XI = 2,YI = 3, and m = -3/2 into the point-slope equation and obtain

When X = 0, Y = 6

EXAMPLE 3

y

6

Along this line, x=2

Solution

3 -Zx + 6.

=

80

the line intersects the y-axis at Y = 6.

or

Y

=

•

Write an equation for the line through (-2, -I) and (3,4).

The line's slope is

-I - 4

5 4

3

3 - z(x - 2),

Y

-5

m = -2 _ 3 = -5 = I. Along this line, y~3

+ f-- +-'-- - -

We can use this slope with either of the two given points in the point-slope equation:

(2,3)

2

With (x"yU = (-2, -1)

Y = -I

+ I'(x - (-2))

y=-I+x+2 FIGURE A.II The standard equations for the vertical and horizontal lines through (2,3) are x = 2 andy = 3.

Y = 4

+ I'(x - 3)

y=4+x-3

y=x+l -..............Sam.....u1t

Either way, Y = x

With (x"yU = (3,4)

+

/ y=x+l

1 is an equation for the line (Figure A.12).

•

Appendix 3 Lines, Circles, and Parabolass

AP-13

The y-coordinate of the point where a nonverticalline intersects the y-axis is called the y-intercept of the line. Similarly, the x-intercept of a nonhorizootalline is the x-coordinate of the point where it crosses the x-axis (Figure A. 13). A line with slope m and y-intercept b passes through the point (0, b), so it has equation

y = b

+ m(x

- 0),

or, more simply,

y

=

mx + b.

The equation

y=mx+b FIGUREA.12 The line in Example 3.

is called the slope-intercept equation of the line with slope mandy-intercept b.

Lines with equations of the form y = mx have y-intercept 0 and so pass through the origin. Equations oflines are called linear equations. The equation

(A and B not both 0)

Ax+By=C

is called the generailinear equation in x and y because its graph always represents a line and every line has an equation in this form (including lines with undefined slope).

Parallel and Perpendicular Lines

FIGURE A.13 LineLhasx-intercept a and y-intercept h.

Lines that are parallel have equal angles of inclination, so they have the same slope (if they are not vertical). Conversely, lines with equal slopes have equal angles of inclination and so are parallel. If two nonverticallines Ll and L2 are perpendicular, their slopes ml and m2 satisfY ml m2 = -I, so each slope is the negative reciprocal of the other: mt =

Y

I -m2'

To see this, notice by inspecting similar triangles in Figure A.14 that ml = a/h, and m2 = -h/a. Hence, mlm2 = (a/h)( -h/a) = -\.

Distance and Circles in the Plane The distance between points in the plane is calculated with a formula that comes from the Pythagorean theorem (Figure A.15).

FIGURE A.14 4ADCissimi1arto !!J.CDB. Hence 1 is also the upper angle in !!J.CDB. From the sides of !!J.CDB, we read tan 1 ~ a/h.

Y

Y2

This distance is

d~

Vrel'-2---'1~12C-+-cI-Y2---Y~112

~

V ('2-'1)2+ (Y2-Yl)2

\ Yl

o

'2

FIGURE A.15 To calculate the distance between P(XloYl) and Q(X2,Y2) , apply the Pythagorean theorem to triaogle PCQ.

AP-14

Appendices

Distance Formula for Points in the Plane The distance between P(Xl ,Yl) and Q(X2,Y2) is

+ (~y)2

d = Y(Jlx)2

=

Y(X2 - Xl)'

+ (Yo - Yl)'.

EXAMPLE 4

y

(al The distance between P( -1,2) and Q(3, 4) is

Y(3 - (-1)' + (4 - 2)'

a

=

Y(4)' + (2)'

v20 = v'4=5 =

=

2v'5.

(bl The distance from the origin to P(x, y) is

C(h, k)

• (x - h)'

+ (y -

k)' ~ a'

~--~--~~--~----~x

o

By def'mition, a circle of radius a is the set of all points P(x, y) whose distance from some center C(h, k) equals a (FigureA.16). From the distance formula, P lies on the circle if and only if

FIGURE A.16 A circle of radius a in the xy-plane, with center at (h, k).

Y(X - h)2 + (y - k)2

a,

=

so

(X - h)2 + (y - k)2

=

a 2.

{ll

Equation (1) is the standard equation ofa circle with center (h, k) and radius a. The circle of radius a = 1 and centered at the origin is the unit circle with equation

X2+y2=1.

EXAMPLE 5 (al The standard equation for the circle ofmdius 2 centered at (3, 4) is

(X - 3)2 + (y - 4)2

=

22

=

=

3

4.

(bl The circle

(X - 1)'

+ (y + 5)'

hash = 1,k= -5, and a = v'3.The center is the point (h,k) = (1,-5) and the • radius is a = v'3.

If an equation for a circle is not in standard form, we can find the circle's center and radius by first converting the equation to standard fonn. The algehmic technique for doing so is completing the square.

EXAMPLE 6

Find the center and mdius of the circle

X2

+ y2 + 4x

- 6y - 3 = O.

Appendix 3 Lines, Circles, and Parabolas

AP-15

Solution We convert the equation to standard form by completing the squares in x and y: x 2 + y2 + 4x - 6y - 3 = 0 Start with the given equation. (x 2 + 4x)

+ (y2 - 6y)

=

Gather terms. Move the con_ t to the right-hand side.

3

(X2+4X+ (~Y) + &2_ 6 Y+ (~6Y) =

3+ (~y + (~6y (x 2 + 4x

(x

+ 4) + (y2 - 6y + 9)

+ 2)' +

=

3

Add the square of half the coefficient of x to each side of the equation. Do the same for y. The parenthetical expressions on the left-hand side are now perfect

squares.

+4+9 Write each quadratic as a squared

(y - 3)2 = 16

linear expression.

•

The center is ( - 2, 3) and the radius is a = 4. The points (x, y) satisfYing the inequality Interior: (x - h)2

+ (y

_ k)2

< .2

~--------~----------~x

o

h

(x - h)2

make up the interior region of the circle with center (h, k) and redius a (Figure A.17). The circle's exterior consists of the points (x,y) satisfying

FIGURE A_17 The interior and exterior of the crrcle (x - h)' + (y - k)' = a 2.

y y =x2 (2,4)

+ (y - k)' < a 2

(x - h)2

+ (y - k)' > a 2.

Parabolas The geometric defmition and properties of general parabolas are reviewed in Section 11.6. Here we look at parabolas arising as the gmphs of equations of the furm y = ax 2 + bx + c. EXAMPLE 7

Considertheequationy = x 2 • Some points whose coordinates satisfy this

equation are (0, 0), (I, I),

(~, ~). (-I, I), (2,4), and (-2,4). These points (and all oth-

ers satisfying the equation) make up a smooth curve called a parabola (Figure A.IB).

•

--~--~~~~--~~X

-2

-1

FIGURE A.18 (Example 7).

0

2

The gmph of an equation of the form

y

Theparabolay = ,2

= ax 2

is a parabola whose axis (axis of symmetry) is the y-axis. The parabola's vertex (point where the parabola and axis cross) lies at the origin. The parabola opens upward if a > 0 and downward if a < O. The larger the value ofla I, the narrower the parabola (Figure A. 19). Generally, the gmph of y = ax 2 + bx + c is a shifted and scaled version of the parabola y = x 2 • We discuss shifting and scaling of gmphs in more detail in Section 1.2.

TheGraphofy=ax2 +bx+c,

a*'O

The graph of the equation y = ax 2 + bx + c, a oF 0, is a parabola. The parabola opens upward if a > 0 and downward if a < O. The axis is the line (2)

The vertex of the parabola is the point where the axis and parabola intersect. Its x-coordinate is x = -b/2a; itsy-coordinate is found by substitoting x = -b/2a in the parabola's equation.

AP-16

Appendices Notice that if a = 0, 1hen we have y = bx + c, which is an equation for a line. The axis, given by Equation (2), can be found by completing 1he square.

EXAMPLE 8

Graph the equation y = -

I a =--

-

+ 4.

X

+

+ c we see that

bx

b = -1,

2'

c = 4.

< 0, the parabola opens downward. From Equation (2) the axis is the vertical b

x=--=

2a

FIGURE A.19 Besides determining 1he direction in which 1he parabola y = ax 2 opens, the number a is a scaling factor. The parabola widens as a approacbes zero aod narrows as Ia Ibecomes large.

2

Comparing the equation with y = ax 2

Solution

Since a line

tx

( -1) 2(-1/2)

=

-1.

Whenx = -1, we have

y = -t(_1)2 - (-I)

+ 4 =~.

The vertex is (-1,9/2). The x-intercepts are where y = 0:

_lx 2 2

-

x+4

=

0

x 2 +2x-8=O (x - 2)(x + 4) = 0 x = 2, x =-4

-k-;;--;;---;--;;i---;-~t---~x

,\2

We plot some points, sketch the axis, and use the direction of opening to complete the graph in Figure A.20. •

i

Intercepts at x=-4andx=2

FIGURE A.20 The parabola in Example 8.

Exercises A.3 Distance#, Slopes, and Lines In Exercises 1 aod 2, a particle moves from A to B in 1he coordinate plaoe. Find 1he incremeots dx and .1y in 1he particle's coordinates. Also rmd 1he distance from A toB.

1. A( -3,2), B( -1, -2)

+ y2 =

I

S. A( -1,2), B( -2, -I)

6. A(2,3), B( -1,3)

2. A( -3.2, -2), B( -8.1, -2)

Describe 1he graphs of 1he equations in Exercises 3 and 4.

3. x 2

Plot 1he points in Exercises 5 and 6 and find 1he slope (if any) of 1he line 1hey detennine. Also rmd 1he coramon slope (if any) of1he lines perpeodicular to line AB.

4. x 2

+ y2

'"

3

In Exercises 7 aod 8, find ao equation for ;y-ptane. Include the circle's center in your skeroh. Also, label the circle's x- and y-intercepts, if any, with their coordinate pairs.

21. C(0,2),

a= 2

23. C( -v3, -2), a

22. C( -1,5), a = =

v'iO

10' 0'

o

1

2

2 3 4 5 Distance lbrough wall (incbes)

6

7

The ten2perature changes in the wall in Exercises 41 and 42.

Graph the circles whose equations are given in Exercises 24-26. Label each circle's center and intercepts (if any) with their coordinate pairs. 24. x 2 + y2 + 4x - 4y + 4 = 0

25. x 2 + y2 - 3y - 4

=

0

26. x 2 + y2 - 4x

+ 4y

=

0

Parabolas Graph the parabolas in Exercises 27-30. Label the vertex, axis, and intercepts in each case. 27.y=x2 -2x-3 28.y=-x 2 +4x

29. y

=

-x 2

-

6x - 5

30. y

=

I 2x2

+X +4

Inequalities Describe the regions defmed by the inequalities and pairs of inequalities in Exercises 31-34. 32. (x - 1)2 + y2 :5 4 31. x 2 + y2 > 7

33. x 2 + y2 34. x 2 + y2

42. Insulation According to the figure in Exercise 41, which of the materials is the best insulator? The poorest? Explain. 43. Pressure under water The pressure p experienced by a diver under water is related to the diver's depth d by an equation of the form p = kd + I (k a constant). At the surface, the pressure is I atmosphere. The pressure at 100 meters is about 10.94 atmospheres. Find the pressure at 50 meters. 44. Reflected light A ray of light comes in along the line x + Y = I from the second quadrant and reflects off the x-axis (see the accompanying figure). The angle of incidence is equal to the angle of reflection. Write an equation for the line along which the dparting light travels. y

x+y=l

> I, x 2 + y2 < 4 + 6y < 0, y>-3

35. Write an inequality that describes the points that lie inside the circle with center ( - 2, I) and radius

v'6.

36. Write a pair of inequalities that describe the points that lie inside or on the circle with center (0, 0) and radius and on or to the right of the vertical line through (I, 0).

V2,

--~------~L-----------~x

o

The path of the light ray in Exercise 44. Angles of incidence and reflection are measured from the perpendicular.

AP-18

Appendices

45. Fahrenheit vo. Celsius

47. By ealeulatiog the lengths ofits sides, show that the triangle with vertices at the points A(I, 2), B(5, 5), and C(4, -2) is isosceles but not equilateral.

In the FC·plane, sketch the grnph of the

equation C =

~(F - 32)

48. Show that the triangle with verticesA(O, 0), B( I, is equilatera1.

9

linking Fahrenheit and Celsius temperatures. On the same graph

49. Show that the poiots A(2, -I), B(I, 3), and C( - 3, 2) are vertices of a square, and fmd the fourth vertex.

sketch the line C = F. Is there a temperature at which a Celsius thermometer gives the same numerical reading as a Fahrenheit

50. Three differeot parallelograms have vertices at (-I, I), (2, 0), and (2, 3). Sketch them and fmd the coordinates of the fourth ver-

thermometer? If so, fmd it. 46. The Mt. Washington Cog Ranway Civil eogioeers calculate the slope of roadhed as the ratio of the distance it rises or falls to the distance it ruos horizootally. They eall1his ratio the grade of the roadbed, usually wtitteo as a percentage. Aloog the coast, commercial rallroad grades are usually less than 2%. In the mountaios, they may go as high as 4%. Highway grades are usually less than 5%. The steepest part of the Mt Washingtoo Cog Railway in New Hampshire has an exceptiooa1 37.1 % grade. Aloog this part of the track, the seats in the froot of the car are 14 ft above those in the rear. About how far apart are the front and rear rows of seats?

A.4

v'3), and C(2, 0)

tex of each.

51. For what value of /r; is the line 2x + ~ = 3 perpeodicular to the lioe 4x + y = I? For what value of k are the lioes parallel? 52. Midpoint of a line segment nates Xl (

Show that the point with coordi-

+ X2 2

+ .Y2)

Yl

'

2

is the midpoint of the line segment joining P(Xl. Yl) to Q(X2, Y2).

Proofs of Limit Theorems This appendix proves Theorem I, Parts 2-5, and Theorem 4 from Section 2.2.

THEOREM l-Llmlt Laws

If L, M, c, and k are real numbers and

lim f(x) = L

and

x--"'c

lim g(x) = M,

then

x--"'c

1. Sum Rule:

lim (f(x) x-c

+ g(x»

= L

2. Difference Rule:

lim (f(x) x-c

- g(x»

= L - M

3. Constant Multiple Rule:

lim (k' x-c

4. Product Rule:

lim (f(x) x-c

f(x))

=

-g(x»

+M

k' L = L· M

M .. 0

5. Quotient Rule: 6. Power Rule:

lim [j(x)]' = L', n a positive integer x-c

7. Root Rule:

lim x-c

ifiW = "'?'L =

(If n is even, we assume that lim f(x) = L x-c

>

L 11" n a positive integer

0.)

We proved the Sum Rule in Section 2.3 and the Power and Root Rules are proved in more advanced texts. We obtain the Difference Rule by replacing g(x) by -g(x) and M by -Min the Sum Rule. The Constant Multiple Rule is the special case g(x) = k of the Product Rule. This leaves only the Product and Quotient Rules.

Procrf crf the Limit Product Rule We show that for any E > 0 there exists a lJ that for all x in the intersection D of the domains of f and g,

0< Ix - cl

If(x)g(x) - LMI

0 such

Appendix 4 Proofs of Limit Theorems Suppose then that

E

AP-19

is a positive number, and write f(x) and g(x) as

f(x)

L + (f(x) - L),

=

g(x)

=

M + (g(x) - M).

Multiply these expressions together and subtract LM:

f(x)· g(x) - LM

= =

=

(L + (f(x) - L»)(M + (g(x) - M)) - LM LM + L(g(x) - M) + M(f(x) - L) + (f(x) - L)(g(x) - M) - LM L(g(x) - M) + M(f(x) - L) + (f(x) - L)(g(x) - M).

(l)

Since f and g have limits L and M as x --> c, there exist positive numbers Ill, 1l2, 1l3, and Il. such that for allx in D

o< o< o
=> =>

If(x) - LI < yI;f3 Ig(x) - MI < yI;f3 If(x) - LI < E/(3(1 + IMI» Ig(x) - MI < E/(3(i + ILl».

(2)

lfwe take Il to be the smallest numbers III through Il., the inequalities on the right-hand side of the Implications (2) will hold simultaneously for 0 < Ix - cl < Il. Therefore, for all x inD, 0 < Ix - cl < Il implies

::~~tqualityaPPlied

If(x)· g(x) - LMI

+ IMllf(x) - LI + If(x) - Lllg(x) - MI ,;; (I + ILI)lg(x) - MI + (i + IMI)lf(x) - LI + If(x) - Lllg(x) - MI ,;; ILIIg(x) - MI

Values from (2)

•

This completes the proof of the Limit Product Rule.

Proof of the Limit Quotient Rule We show that lim..... ,(I/g(x))

=

l/M. We can then

conclude that

lim f(x) g(x)

.~,

= lim

.~,

(f(X)

._1_) g(x)

= lim

.~,

f(x) • lim

.~,

_1_ g(x)

=

L.l M

=

£

M

by the Limit Product Rule. Let E > 0 be given. To show that lim.~,(l/g(x)) = I/M, we need to show that there exists all> 0 such that for all x

0< Ix - cl < Il

=>

Ig(~) -

11

0, there exists a positive number III such that for all x

0< Ix - cl < III

=>

M

Ig(x) - MI < 2.

(3)

For any numbers A andB it can be shown that IAI-IBI,;; IA - BI and IBI-IAI,;; IA - BI, from which it follows that IIAI-IBII ,;; IA - BI. With A = g(x) and B = M, this becomes Ilg(x) I - IMII ,;; Ig(x) - MI,

AP-20

Appendices

which can he combined with the inequality on the right in Implication (3) to get, in turn,

IMI IIg(x)I-IMII 0 there exists all> 0 such that for all x the interval c < x < c + Il is contained in I and the inequality implies L -

E

O. We can then apply the Sandwich Theorem for Sequences (Section 10.1, Theorem 2) to conclude that x"/n! --> O. The first step in showing that Ixr/n! --> 0 is to choose an integer M > lxi, so that (Ixl/M) < I. By Limit 4, just proved, we then have (Ixl/M)" --> O. We then restrict our attention to values of n > M. For these values of n, we can write Ixr n!

Ixr

1·2· ... 'M'(M+ I)'(M+ 2)' ... 'n (. - M) factors

Ixl" = Ixl"MM = MM (I~)" - MlM" M MlM" Ml \.M .

0 because (Ix I/M)" --> O. •

Appendix 6 Theory of the Real Numbers

A.6

AP-23

Theory of the Real Numbers A rigorous development of calculus is based on properties of!be real numbers. Many results about functions, derivatives, and integrals would be false if stated for functions dermed ouly on the rational numbers. In this appendix we briefly examine some basic concepts of the theory of the reals that bint at wbat might be learned in a deeper, more theoretical study of calculus. Three types of properties make the real numbers wbat they are. These are the algebraic, order, and completeness properties. The algebraic properties involve addition and multiplication, subtraction and division. They apply to rational or complex numbers as well as to the reals. The structure of numbers is built around a set with addition and multiplication operations. The following properties are required of addition and multiplication. Al

A2 A3 A4

+ (b + c) = (a + b) + c for all a, b, c. + b = b + a for all a, b. There is a number called "0" such that a + 0 = a for all a. For each number a, there is a b such that a + b = O. a a

M2

a(bc) = (ab)cforalla,b,c. ab = ba for all a, b.

Ml M3

There is a number called "1" such that a' 1 = a for all a.

M4

For each nonzero a, there is a b such that ab = 1.

D

a(b

+

c) = ab

+ bc for all a, b, c.

Al and Ml are associative laws, A2 and M2 are commutativity laws, A3 and M3 are identity laws, and D is the distributive law. Sets that bave these algebraic properties are examples of fields, and are studied in depth in the area of theoretical mathematics called abstract algebra. The order properties allow us to compare the size of any two numbers. The order properties are 01

For any a and b, either a :5 b or b :5 a or both.

02

lfa:5 lfa:5 lfa:5 lfa:5

03 04

OS

bandb:5 athena = b. bandb:5 c then a :5 c.

bthena

+

c:5 b

+

c.

bandO:5 c then ac :5 bc.

03 is the transitivity law, and 04 and 05 relate ordering to addition and multiplication. We can order the reals, the integers, and the rational numbers, but we cannot order the complex numbers. There is no reasonable way to decide whether a number like i = yCl is bigger or smaller than zero. A field in which the size of any two elements can be compared as above is called an ordered field. Both the rational numbers and the real numbers are ordered fields, and there are many others. We can think of real numbers geometrically, lining them up as points on a line. The completeness property says that !be real numbers correspond to all points on the line, with no ~'holes" or "gaps." The rationals, in contrast, omit points such as and 1T' , and the integers even leave out fractions like 1/2. The reals, baving the completeness property, omit no points. What exactly do we mean by this vague idea of missing holes? To answer this we must give a more precise description of completeness. A number M is an upper bound for a set of numbers if all numbers in the set are smaller than or equal to M. M is a least upper bound if it is the smallest upper bound. For example, M = 2 is an upper bound for the

Vi

AP-24 Appendices negative nmnbers. So is M = 1, showing that 2 is not a least upper bound. The least upper bound for the set of negative nmnbers is M = O. We define a complete ordered field to be one in which every nonempty set bounded above has a least upper bound. If we work with just the mtional nmnbers, the set of nmnbers less than v'2 is bounded, but it does not have a mtionalleast upper bound, since any mtional upper bound M can be replaced by a slightly smaller mtional nmnber that is still larger than So the mtionals are not complete. In the real nmnbers, a set that is bounded above always has a least upper bound. The reals are a complete ordered field. The completeness property is at the heart of many results in calculus. One example occurs when searching for a maxin3mn value for a function on a closed interval [a, bI, as in Section 4.1. The function y = , - x' has a maximmn value on [0, II at the point x satis· fYing I - 3x 2 = 0, or x = Vf13. If we limited our considemtion to functions defined only on mtional nmnbers, we would have to conclude that the function has no maximmn, since Vf13 is irrational (Figure A.22). The Extreme Value Theorem (Section 4.1), which in3plies that continuous functions on closed intervals [a, bI have a maxm.mn value, is not true for functions dermed only on the mtionals. The Intermediate Value Theorem in3plies that a continuous function 1 on an interval [a,blwith/(a) < Oand/(b) > omust be zero somewhere in [a, bl.The function values cannot jmnp from negative to positive without there being some point x in [a, bI where I(x) = O. The Intermediate Value Theorem also relies on the completeness of the real nmnbers and is false for continuous functions defined only on the mtionals. The function I(x) = 3x 2 - 1 has 1(0) = -I and 1(1) = 2, but if we consider 1 only on the mtional an nmnbers, it never equals zero. The only value ofx for which I(x) = 0 is x = irrational nmnber. We have captored the desired properties of the reals by saying that the real nmnbers are a complete ordered field. But we're not quite finished. Greek mathematicians in the school of Pytbagoms tried to in3pose another property on the nmnbers of the real line, the condition that all nmnbers are mtios of integers. They learned that their effort was doomed when they discovered irrational nmnbers such as \12. How do we know that our efforts to specny the real nmnbers are not also flawed, for some unseen reason? The artist Escher drew optical illusions of spiral staircases that went up and up until they rejoined them· selves at the bottom. An engineer trying to build such a staircase would find that no strue· tore realized the plans the architect had dmwn. Could it be that our design for the reals contains some subtle contradiction, and that no construction of such a nmnber system can be made? We resolve this issue by giving a specific description of the real nmnbers and verify· ing that the algebraic, order, and completeness properties are satisfied in this model. This is called a construction of the reals, and just as stairs can be built with wood, stone, or steel, there are several approaches to constructing the reals. One construction treats the reals as all the inf'mite decimals,

Vz.

y 0.5

~~~~~~~~~~~~~x

0.1

0.3

0.5

I

0.7

0.9 1

vV3 FIGURE A.22 The maximum value of = , - ,'on [0, I] occurs at 1he irrational number x = V1j3. y

V1!3,

a.dt d2d,d4 • •• In this approach a real nmnber is an integer a followed by a sequence of decimal digits dlo d2, d" ... , each between 0 and 9. This sequence may stop, or repeat in a periodic pattern, or keep going forever with no pattern. In this form, 2.00, 0.3333333 . .. and 3.1415926535898 ... represent three familiar real nmnbers. The real meaning of the dots " ... " following these digits requires development of the theory of sequences and series, as in Chapter 10. Each real nmnber is constructed as the limit of a sequence ofmtional nmnbers given by its rmite decin3al approxin3ations. An inf'mite decimal is then the same

as a series dl d2 a+-+-+ .. · 10 100 This decirnal construction of the real nmnbers is not entirely straightforward. It's easy enough to check that it gives nmnbers that satisfy the completeness and order properties,

Appendix 7 Complex Numbers

AP-25

but verifying the algebraic properties is rather involved. Even adding or multiplying two numbers requires an inf"mite number of operations. Making sense of division requires a careful argument involving limits of rational approximations to infinite decimals. A different approach was taken by Richard Dedekind (1831-1916), a German mathematician, who gave the first rigorous construction of the real numbers in 1872. Given any real number x, we can divide the rational numbers into two sets: those less !ban or equal to x and those greater. Dedekind cleverly reversed tIris reasoning and defined a real number to be a division of the rational numbers into two such sets. This seems like a strange approach, but such indirect methods of constructing new structures from old are common in theoretical mathematics. These and other approaches can be used to construct a system of numbers having the desired algebraic, order, and completeness properties. A final issue that arises is whether all the constructions give the same tiring. Is it possible that different constructions result in different number systems satisfying all the required properties? If yes, which of these is the real numbers? Fortunately, the answer turns out to be no. The reals are the only number system satisfying the algebraic, order, and completeness properties. Confusion about the nature of the numbers and about limits caused considerable controversy in the early development of calculus. Calculus pioneers such as Newton, Leibniz, and their successors, when looking at what happens to the difference quotient ~y ~

f(x +

~x)

- f(x)

~

as each of ~y and ~ approach zero, talked about the resulting derivative being a quotient of two inf"mitely small quantities. These "inf"mitesimals;' written dx and dy, were thought to be some new kind of number, smaller !ban any fixed number but not zero. Similarly, a def"mite integral was thought of as a sum of an infinite number of inf"mitesimals

f(x)-dx as x varied over a closed interval. While the approximating difference quotients ~y/ ~x were understood much as today, it was the quotient of infinitesimal quantities, rather than a limit, that was thought to encapsulate the meaning of the derivative. This way of thinking led to logical difficulties, as attempted def"mitions and manipulations of infinitesimals ran into contradictions and inconsistencies. The more concrete and computable difference quotients did not cause such trouble, but they were thought of merely as useful calculation tools. Difference quotients were used to work out the numerical value of the derivative and to derive general formulas for calculation, but were not considered to be at the heart of the question of what the derivative actually was. Today we realize that the logical problems associated with infinitesimals can be avoided by defining the derivative to be the limit of its approximating difference quotients. The ambiguities of the old approach are no longer present, and in the standard theory of calculus, infinitesimals are neither needed nor used.

A.7

Complex Numbers Complex numbers are expressions of the form a + ib, where a and b are real numbers and i is a symbol for v'=I. Unfortunately, the words ''real'' and ''imaginary'' have connotations that somehow place v'=I in a less favorable position in our minds !ban v'2. As a matter of fact, a good deal of imagination, in the sense of inventiveness, has been required to construct the real number system, which forms the basis of the calculus (see Appendix A.6). In this appendix we review the various stages of this invention. The further invention of a complex number system is then presented.

AP- 26

Appendices

The Development of the Real Numbers The earliest stage of number development was the recognition of the counting numbers 1,2,3, ... , which we now call the natural numbers or the positive integers. Certain simple arithmetical operations coo be performed with these numbers without getting out· side the system. That is, the system of positive integers is closed under the operations of addition ood multiplication. By this we meoo that if m ood n are any positive integers, then

m+n

=

P

ood

mn

= q

(1)

are also positive integers. Given the two positive integers on the left side of either equation in (1), we coo find the corresponding positive integer on the right side. More thoo this, we cao sometimes specify the positive integers m ood P ood tmd a positive integer n such that m + n = p. For instance, 3 + n = 7 coo be solved when the ouly numbers we know are the positive integers. But the equation 7 + n = 3 cannot be solved unless the number system is enlarged. The number zero ood the negative integers were invented to solve equations like 7 + n = 3. In a civilization that recognizes all the integers

... , -3, -2, -1, 0,1,2,3, ... ,

(2)

educated person coo always find the missing integer that solves the equation = P when given the other two integers in the equation. Suppose our educated people also know how to multiply any two of the integers in the list (2). If, in Equations (1), they are given m ood q, they discover that sometimes they coo find n ood sometimes they cannot. Using their imagination, they may be inspired to invent still more numbers ood introduce fractions, which are just ordered pairs min of integers m ood n. The number zero has special properties that may bother them for a while, but they ultimately discover that it is haody to have all ratios of inte· gers min, excluding only those having zero in the denominator. This system, called the set of rational numbers, is now rich enough for them to perform the rational operations of arithmetic: 00

m

+n

1. (a) addition (b) subtraction

1

FIGURE A.23 With a straightedge and compass, it is possible to construct a segment of irrationalleogth.

2. (a) multiplication (b) division

on ooy two numbers in the system, except that they cannot divide by zero because it is meaoingless. The geometry of the unit square (Figure A.23) ood the Pythagoreoo theorem showed that they could construct a geometric lioe segment that, in terms of some basic unit of length, bas length equal to Thus they could solve the equation

Vz.

x2

= 2

by a geometric constroction. But then they discovered that the lioe segment representing is 00 incommensurable quaotity. This meoos that cannot be expressed as the ratio of two integer multiples of some unit oflength. That is, our educated people could not tmd a rational number solution of the equation x 2 = 2. There is no rational number whose square is 2. To see why, suppose that there were such a rational number. Then we could find integers p ood q with no common factor other tbao I, ood such that

Vz

v'2

(3)

Since p ood q are integers, p must be even; otherwise its product with itself would be odd. In ayrobols, p = 2PI, where PI is 00 integer. This leads to 2pl2 = q2 which says q must be even, say q = 2qj, where ql is 00 integer. This makes 2 a factor of both p ood q, contrary to our choice ofp ood q as integers with no common factor other tbao 1. Hence there is no rational number whose square is 2.

AP-28

Appendices

and are familiar with the further fact that when the discriminant, b 2 - 4ac, is negative, the solutions in Equation (9) do not belong to any of the systems discussed above. In fact, the very simple quadratic equation

x2 + 1

=

0

is impossible to solve if the only number systems that can be used are the three invented systems mentioned so far. Thus we come to the fourth invented system, the set of all complex numbers a + ib. We could dispense entirely with the symbol i and use the ordered pair notation (a, b). Since, under algebraic operations, the numbers a and b are treated somewhat differently, it is essential to keep the order straight. We therefore might say that the complex number system consists of the set of all ordered pairs of real numbers (a, b), together with the rules by which they are to be equated, added, multiplied, and so on, listed below. We will use both the (a, b) notation and the notation a + ib in the discussion that follows. We call a the real part and b the imaginary part of the complex number (a, b). We make the following definitions. Equality a+ib=c+id if and only if a = candb = d.

Two complex numbers (a, b) and (c, d) are equal if and only if a = candb = d.

Addition (a + ib) =

+ (c + id) (a + c) + i(b + d)

The sum of the two complex numbers (a, b) and (c, d) is the complex number (a + c, b + d).

Multiplication (a + ib)(c + id) = (ac - bd) + i(ad c(a

+ ib)

=

ac

The product of two complex numbers (a, b) and (c, d) is the complex number (ac - bd, ad

+ bc)

+ i(bc)

+ be).

The product of a real number c and the complex number (a, b) is the complex number (ac, be).

The set of all complex numbers (a, b) in which the second number b is zero has all the properties of the set of real numbers a. For example, addition and multiplication of (a, 0) and (c, 0) give

(a,O)

+ (e,O)

=

(a

+ c,O),

(a, 0)' (e, 0) = (ae, 0), which are numbers of the same type with imaginary part equal to zero. Also, if we multiply a "real number" (a, 0) and the complex number (c, d), we get (a, 0)' (c, d) = (ac, ad) = a(c, d). In particular, the complex number (0, 0) plays the role of zero in the complex number

system, and the complex number (1, 0) plays the role of unity or one. The number pair (0, 1), which has real part equal to zero and imaginary part equal to one, has the property that its square,

(0,1)(0,1) = (-1,0), has real part equal to minus one and imaginary part equal to zero. Therefore, in the system of complex numbers (a, b) there is a number x = (0,1) whose square can be added to unity = (1,0) to produce zero = (O,O),thatis,

(0, 1)2

+ (1,0)

=

(0,0).


AP-27

Although our educated people could not f"md a rational solution of the equation x 2 = 2, they could get a sequence of rational numbers

I I'

7 5'

41 29'

239 169'

... ,

(4)

whose squares form a sequence

I I'

49 25'

57,121 28,561'

1681 841 '

... ,

(5)

that converges to 2 as its limit. This time their imagination suggested that they needed the concept of a limit of a sequence of rational numbers. If we accept the fact that an increasing sequence that is bounded from above always approaches a limit (Theorem 6, Section 10.1) and observe that the sequence in (4) has these properties, then we want it to have a limit L. This would also mean, from (5), that L 2 = 2, and hence L is not one of our rational numbers. If to the rational numbers we further add the limits of all bounded increasing sequences of rational numbers, we arrive at the system of all ''real'' numbers. The word real is placed in quotes because there is nothing that is either "more real" or "less real" about this system than there is about any other mathematical system.

The Complex Numbers lmagioation was called upon at many stsges during the development of the real number system. In fact, the art of invention was needed at least three times in constructing the systems we have discussed so far:

1.

The first invented system: the set of all integers as constructed from the counting numbers.

2.

The second invented system: the set of rational numbers min as constructed from the integers.

3.

The third invented system: the set of all real numbers x as constructed from the rational numbers.

These invented systems form a hierarchy in which each system contsins the previous system. Each system is also richer than its predecessor in that it permits additional opera· tions to be performed without going outside the system:

1.

In the system of all integers, we can solve all equations of the form

x

+a

= 0,

(6)

where a can be any integer. 2.

In the system of all rational numbers, we can solve all equations of the form

ax + b

=

0,

(7)

provided a and b are rational numbers and a .. O. 3.

In the system of all real numbers, we can solve all of Equations (6) and (7) and, in ad· dition, all quadratic equations

ax 2

+ bx + c

= 0

having

a" 0

and

b2

-

4ac ;" O.

(8)

You are probably familiar with the formula that gives the solutions of Equation (8), namely,

x=

-b ± Vb 2 2a

-

4ac

(9)


The equation

x

2

+I

=

AP-29

°

therefore bas a solution x = (0, I) in this new number system. You are probably more familiar with the a + ib notation than you are with the notation (a, b). And since the laws of algebra for the ordered pairs enable us to write

(a, b) = (a,O)

+ (0, b)

= a(l,

0)

+ b(O, I),

while (I, 0) behaves like unity and (0, I) behaves like a square root of minus one, we need not hesitate to write a + ib in place of (a, b). The i associated with b is like a 1Iacer element that tags the imaginary part of a + ib. We can pass at will from the rea1m of ordered pairs (a, b) to the realm of expressions a + ib, and conversely. But there is nothing less "real" about the symbol (0, 1) = i than there is about the symbol (1, 0) = I, once we have learned the laws of algebra in the complex number system of ordered pairs (a, b). To reduce any rational combination of complex numbers to a single complex number, we apply the laws of elementary algebra, replacing i 2 wherever it appears by -I. Of course, we cannot divide by the complex number (0, 0) = + iO. But if a + ib oF 0, then we may carry out a division as follows:

°

+ id a + ib e

(e (a

+ iri)(a + ib)(a

- ib) - ib)

(ae

+

bd)

+

i(ad - be)

a2 + b 2

The result is a complex number x + iy with

ae

x

= a2

+ bd + b2 '

y

=

ad - be a2 + b 2 '

anda 2 + b 2 oF O,sincea + ib = (a, b) oF (0,0). The number a - ib that is used as multiplier to clear the i from the denominator is called the complex conjugate of a + ib. It is customary to use (read "z bar") to denote the complex conjugate of z; thus

z

z

=

a + ib,

z=

a - ib.

Multiplying the numerator and denominator of the fraction (e + iri)/(a + ib) by the complex conjugate of the denominator will alwsys replace the denominator by a real number.

EXAMPLE 1

We give some illustrations of the arithmetic operations with complex

numbers. (8) (2 + 3i) + (6 - 2i) = (2 + 6) + (3 - 2)i = 8 + i

(b) (2 + 3i) - (6 - 2i) = (2 - 6) + (3 - (-2»i = -4 + 5i (c) (2 + 3i)(6 - 2i) = (2)(6) + (2)( -2i) + (3i)(6) + (3i)( -2i) = 12 - 4i + 18i - 6i 2 = 12 + 14i + 6 = 18 + 14i

(d) 2 + 3i = 2 + 3i 6 + 2i 6-2i 6-2i6+2i 12 + 4i + 18i + 6i2 36 + 12i - 12i - 4i2 _6+22i_l+!l. 40 -20 20'

Argand Diagrams There are two geometric representations of the complex number z = x + iy: L

as the point P(x, y) in the xy-plane

2_

as the vector

oF from the origin to P.

•

AP-30

Appendices

y

In each representation, the x-axis is called the real axis and the y-axis is the imaginary axis. Both representations are Argand diagrams for x + iy (Figure A.24). In tenns of the polar coordinates ofx andy, we have

x

rcosB,

=

y

rsin8,

=

and

+ isinll). (10) We define the absolute value of a coroplex number x + iy to be the length r of a vector z = x

FIGURE A.24 This Argand diagram represents z = x + iy both as a point P(x, y) and as a vector liP .

+ iy

= r(cosll

oP from the origin to P(x, y). We denote the absolute value by vertical bars; thus, Ix

+

iyl = Yx 2

+ y2.

If we always choose the polar coordinates r and II so that r is nonnegative, then

r

= Ix

+

iyl.

The polar angle II is called the argument of z and is written II = arg z. Of course, any integer multiple of 2'IT may be added to II to produce another appropriate angle. The following equation gives a useful formula connecting a complex number z, its conjugate z, and its absolute value Izl, namely,

z·z =

Iz12.

Eulefs Formula The identity ei(J =

cos 8 + isin8,

called Euler'. formula, enables us to rewrite Equation (10) as

This formula, in tum, leads to the following rules for calculating products, quotients, powers, and roots of coroplex numbers. It also leads to AIgand diagrams for ,/9. Since cos II + i sin II is what we get from Equation (10) by taking r = I, we can say that ,i9 is represented by a unit vector that makes an angle II with the positive x-axis, as shown in Figure A.25.

Yei6=cos8+isin6

Yeil':l=cos6+isin8 \

-t----~o~-L---f~x

(a)

(cos 6, sin 6)

-t----~o~~~-t->x

(b)

FIGURE A.25 AIgand diagrams for eil = cos 6 vector and (b) as a poio!.

+ i sio 6 (a) as a

Products To multiply two complex numbers, we multiply their absolute values and add their angles. Let

(11)


AP-31

so that

y

Then

and hence IZIZ,I = rlr, = Izd'lz,1

+ 9,

arg (ZIZ,) = 91 FIGURE A.26 When ZI and z, are multiplied,lzlz,1 = rl'r,and arg (ZIZ,) = 81 + 8,.

= arg ZI

+ arg z,.

(12)

Thus, the product of two complex numbers is represented by a vector whose length is the product oftbe lengths oftbe two factors and whose argument is tbe sum of their arguments (Figure A.26). In particular, from Equation (12) a vector may be rotated counterclockwise through an angle 9 by multiplying it by e i'. Multiplication by i rotates 90°, by -I rotates 180°, by -i rotates 270°, and so on.

v'3 -

EXAMPLE 2 Let ZI = I + i, z, = i. We plot these complex numbers in an Argand diagram (Figure A.27) from which we read off the polar representations

y I I

Z2 =

:v3-1

2e -i'lf/6 .

I

~~--L-r-~~'-----~-r~xThen

1+v3

ZIZ, = 2v'Zexpe: - i n =

2v'ZexpG~)

-I

= 2v'Z

FIGURE A.27

To multiply two complex

numbers, multiply tbeir absolute values and add tbeir arguments.

(COS;; + i sin ;;) '" 2.73 + 0.73i. •

The notation exp(A) stands fore A •

Quotients Suppose,., '" 0 in Equation (11). Then

Hence

7,

ZII = 71 = Iz,l IZII z, I

and

arg (ZI) z, = 91 - 9, = argzi - argz,.

That is, we divide lengths and subtract angles for the quotient of complex numbers.

EXAMPLE 3

Letzi = 1

1+i v'3 _i

_ -

+ iandz,

=

v'3 -

i,asinExample2.Then

v'Ze'w/4 _ v'Z ''''/1' 0707 ( 57T + . . 57T) 2e 'w/6 2 e "'. cos 12 I sm 12

'" 0.183

+ 0.683i.

•

AP-32

Appendices

Powers If n is a positive integer, we may apply the product fonnulas in Equation (12) to fmd zll = z·z· ... ·Z.

With z

=

rej(J, we

11

factors

obtain • summands

(13) The length r = Iz Iis raised to the nth power and the angle 0 = arg z is multiplied by n. If we take r = I in Equation (13), we obtain De Moivre's Theorem.

De Mcrivre's Theorem (cosO

+ isinO)n

= cosnO

+ isinnO.

(14)

If we expand the left side of De Moivre's equation above by the Binomial Theorem and reduce it to the form a + ib, we obtain formulas for cos nO and sin n6 as polynomials of degree n in cos 6 and sin 6.

EXAMPLE 4

Ifn = 3 in Equation (14), we have (cosO

+

isin(J)3 = cos 36

+ isin30.

The left side of this equation expands to cos3 6

+

3i cos2 6 sin 6 - 3 cos 6 sin2 6 - i sin3 6.

The real part of this must equal cos 36 and the imaginary part must equal sin 36. Therefore, cos 36 = cos3 6 - 3 cos 6 sin2 6, sin 36 = 3 cos2 6 sin 6 - sin3 6.

•

Roots If z = rei' is a complex number different from zero and n is a positive integer, then there are precisely n different complex numbers Wo, WI, ... , Wn-l, that are nth roots ofz. To see why, let W = peia be an nth root of z = rei', so that wI! =

z

or

Then

p =

"0

is the real, positive nth root of r. For the argument, although we cannot say that na and 6 must be equal, we can say that they may differ only by an integer multiple of2'lT. That is,

na

=

6 + 217r,

k

=

0, ±I, ±2, ....

Therefore,

a

6

2'lT

=,,+ kn.


AP-33

Hence, all the nth roots of z = re iO are given by

y

It = 0, ±1, ±2, ....

----j---~r-=~-_;;_t_~x

There might appear to be infmitely many different answers corresponding to the infinitely many possible values of k, but k = n + m gives the same answer as k = m in Equation (15). Thus, we need only take n consecutive values for k to obtain all the different nth roots of z. For convenience, we take

k W2

FIGURE A.28 z = re iIJ ,

The three cube roots of

= 0, 1,2, ... ,n -

1.

All the nth roots of re iO lie on a circle centered at the origin and having radius equal to the real, positive nth root of r. One of them has argument a = (I/n. The others are uniformly spaced around the circle, each being separated from its neighbors by an angle equal to 27T/n. Figure A.28 illustrates the placement of the three cube roots, wo, WI> W2, of the complex number z = re llJ •

EXAMPLE 5

y

(15)

Find the four fourth roots of - 16.

Solution As our first step, we plot the number -16 in an Argand diagram (Figure A.29) and determine its polar representation re iO • Here, z = -16, r = + 16, and (I = 7T. One of the fourth roots of 16e i " is 2e i"/4. We obtain others by successive additions of 27T/4 = 7T/2 to the argument of this first one. Hence, '¢'16

.

expz1T -

2

.(7T 37T 57T 77T)

expz

4'4'4'4 '

and the four roots are ~---- +----~-~--r~X

-16

Wo = 2 [cos

FIGURE A.29 of-16.

The four fourth roots

~ + isin ~ 1= Vz(1 + i)

Wj

= 2 [cos

3; + isin 3;] = Vz(-1 + i)

W2

= 2 [cos

5; + isin 5; 1= Vz(-1 -

W3

= 2 [cos

7; + isin 7; 1= Vz(1 -

i)

i).

•

The Fundamental Theorem of Algebra

v=1

One might say that the invention of is all well and good and leads to a number system that is richer than the real number system alone; but where will this process end? Are we also going to invent still more systems so as to obtain ~, ~,and so on? But it turns out this is not necessary. These numbers are already expressible in terms of the complex number system a + ib. In fact, the Fundamental Theorem of Algebra says that with the introduction of the complex numbers we now have enough numbers to factor every polynomial into a product of linear factors and so enough numbers to solve every possible polynomial equation.

AP-34

Appendices

The Fundamental Theorem of Algebra Every polynomial equation of the form anzll

+ Cln_lZIl - 1 + ... +

alZ

+ ao

=

0,

in which the coefficients 110, aj, ... , an are any complex numbers, whose degree n is greater than or equal to one, and whose leading coefficient a. is not zero, has exactly n roots in the complex number system, provided each multiple root of multiplicity m is counted as m roots.

A proof of this theorem can be found in almost any text on the theory of functions of a complex variable.

Exercises A.7 Operations with Complex Numbers 1. How computers multiply compleJ[ numbers Find (a, b)· (e, d)

+ be). (2,3)'(4,-2) c. (-I, -2)·(2, 1)

= (ae - bd,ad

b. (2,-1)'(-2,3)

L

15. cos 48

2. Solve the following equations for the real numbers, x andy. L

(3

b.

(II-I + ~)' + _1_. I + x+lY

- 2(x - iy) = x

+ iy

=

+ 4 = O. 22. Find the six solutions of the equation z6 + 2z3 + 2 = O. 23. Find all solutions of the equatioo x' + 4x' + 16 = O. 24. Solve the equatioo x' + I = O.

b. (-z)

c. -z d. l/z 4. Show that the distance b&ween the two points ZI and z, in an Argand diagram is IZI - z,l. In Exercises 5-10, graph the points z

=

x

+ iy that satisfy the given

conditions. 5.

L

Izl = 2

11 = 2 Iz + 11 = Iz - 11 Iz + 11'" Izl

2

8.

9.

.. Extend the results of Exercise 26 to show that f(Z) = f(z) if II

\1'=3)'

13. I + fV'3 1- fV3

f(z)

=

a,.zn

+ a,._tz,.-l + ... + atZ + ao

is a polynomial with real coefficients ao•...• a,..

Express the complex numbers in Exercises 11-14 in the form re i6 , with r '" 0 and -'IT < 8 '" 'IT. Draw an Argand diagram for each calculation.

+

25. Complex numben and vecton in the plane Show with an Argand diagram that the law for adding complex numbers is the same as the para1lelogram law for adding vectors. 26. Complex arithmetic with conjugate. Show that the coojugate of the sum (product, or quotient) of two complex numbers, ZI and z" is the same as the sum (product, or quotient) of their conjugates. complex-conjugate pairs

11 = I Iz + il = Iz -

7. Iz +

11. (1

Theory and Examples

27. Complex roots of polynomials with real coefficients come in b. Izl

6. Iz -

10.

19. Find the three cube mots of -8i.

21. Find the four solutioos of the equation z· - 2z'

i

Graph;ng and Geometry 3. How may the followIDg complex numbers be obtained from z = x + iy geometrically? Sketch.

z

18. Find the two square roots of i. 20. Find the six sixth roots of 64.

c. (3 - 2i)(x + iy) = 2(x - 2iy) + 2i - I

L

16. sin 48

17. Find the three cube roots of I.

(This is how complcx numbers are multiplied by computers.)

+ 4iJ>

Powers and Roots Use De Moivre's Theorem to express the trigonometric functions in Exercises 15 and 16 in tenDs of cos 8 and sin 8.

12L±l . I - i 14. (2

+ 3i)(1

- 2i)

= O,wheref(z) is a polynomial with real coefficients as in part (a), show that the conjugate z is also a root of the equation. (Hint: Let f(z) = u + iv = 0; then both u and v are zero. Use the fact that f(z) = f(z) = u - iv.) 28. Absolutevalueofaconjngate Showthatlzl = Izi. 29. When z = i If z and z are equal, what can you say about the location of the point z in the complex plane? b. Ifzis a root of the equatioof(z)

Appendix 8 The Distributive Law for Vector Cross Products 30. Real and imaginary parts Let Re(z) denote the real part of z and Im(z) the imaginary part. Show that the following relations hold for any complex nwnbers z, Zl, and Z2. a. z

+Z=

b. z -

z=

A.8

c. IRe(z)I :s Izl + z21 2 = Iztl 2 + IZ212

d. IZI e. IZI

2Re(z)

+ z21 :s

Iztl

AP-35

+ 2Re(zlz2)

+ IZ21

2iIm(z)

The Distributive Law for Vector Cross Products In this appendix we prove the Distributive Law u X (v

+ w)

=

u X v

+u

X w,

which is Property 2 in Section 12.4.

Proof To derive the Distributive Law, we construct u X v a new way. We draw u and v from the common point 0 and construct a plane M perpendicular to u at 0 (Figure A.30). We then project v orthogonally onto M, yielding a vector v' with length Iv Isin 8 . We rotate v' 90° about u in the positive sense to produce a vector v" . Finally, we multiply v" by the length of u. The resulting vector Iu Iv" is equal to u X v since v" has the same direction as u X v by its construction (Figure A.30) and lullv"l = lullv'l = lullvlsin8 = lu X vi.

M

FIGURE A.30

As explained in the text, u X v = Iu Iv" .

Now each of these three operations, namely,

1.

projection onto M

2.

rotation about u through 90°

3.

multiplication by the scalar Iu I

when applied to a triangle whose plane is not parallel to u, will produce another triangle. If we start with the triangle whose sides are v, w, and v + w (Figure A.31) and apply these three steps, we successively obtain the following:

1.

2.

+ w)' satisfying the vector equation v' + w' = (v + w)' A triangle whose sides are v", w" , and (v + w)" satisfying the vector equation v" + w" = (v + w)" A triangle whose sides are v', w', and (v

AP-36

Appendices

(the double prime on each vector has the same meaning as in Figure A.30)

M

FIGURE A.31 The vectors, v, W, v + w, and their projections ooto a plane perpendicular to u.

3. A triangle whose sides are lulv",lulw", andlul(v + w)" satisfying the vectorequation lulv"

+

lulw" = lul(v

+

w)".

Substituting lulv" = u X v, lulw' = u X w, and lul(v from our discussion above into this last equation gives u X v

+u

X w = u X (v

+ w)"

= u X (v

+ w)

+ w),

•

which is the law we wanted to establish.

A.9

The Mixed Derivative Theorem and the Increment Theorem This appendix derives the Mixed Derivative Theorem (Theorem 2, Section 14.3) and the Increment Theorem for Functions of Two Variables (Theorem 3, Section 14.3). Euler rlISt published the Mixed Derivative Theorem in 1734, in a series of papers he wrote on hydrodynamics.

THEOREM 2-Tbe Mixed Derivative Theorem If I(x, y) and its partial derivatives Ix. Iy, I"", and Iyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then

I",,(a, b)

=

Iyx(a, b).

Proof The equality of 1",,( a, b) and Iyx( a, b) can be established hy four applications of the Mean Value Theorem (Theorem 4, Section 4.2). By hypothesis, the point (a, b) lies in the interior of a rectangle R in the xy-plane on which I, Ix. Iy, I"", and Iyx are all defined. We let h and k be the numbers such that the point (a + h, b + k) also lies in R, and we consider the difference ~ =

F(a + h) - F(a),

(1)

I(x, b + k) - I(x, b).

(2)

where

F(x)

=

Appendix 9 The Mixed Derivative Theorem and the Increment Theorem

AP-37

We apply the Mean Value Theorem to F, which is continuous because it is differentiable, Then Equation (1) becomes

a where

=

hF'(c,),

(3)

c, lies between a and a + h, From Equation (2), F'(x)

fx(x,b + k) - fx(x,b),

=

so Equation (3) becomes

a

=

h[j.(cl, b + k) - fx(cl, b)l,

(4)

Now we apply the Mean Value Theorem to the functiong(y) = f.(c"y) and have

g(b

+ k) - g(b)

kg'(d,),

=

or

fx(cl, b + k) - fx(c!, b) for some d, between b and b

=

kfry(c!, d,)

+ k, By substituting this into Equation (4), we get

a

=

hkfry(cl, d,)

(5)

for some point (c), d,) in the rectangle R' whose vertices are the four points (a, b), (a + h, b), (a + h,b + k), and (a, b + k). (See FigureA.32.) By substituting from Equation (2) into Equation (1), we may also write

y

R

•

(a, b)

a=

R'

k

= =

h

f(a + h,b + k) - f(a + h,b) - f(a,b + k) + f(a,b) [j(a + h,b + k) - f(a,b + k)l - [f(a + h,b) - f(a,b)l cf>(b + k) - cf>(b),

(6)

where -o~------------------->x

FIGURE A.32 The key to proving fry(a, b) = f",(a, b) is that no matter how small R' is, fry and f", tske on equal values somewhere inside R' (although not necessarily at the same point).

cf>(y)

=

f(a + h,y) - f(a,y).

(7)

The Mean Value Theorem applied to Equation (6) now gives

a for some d2 between b and b

=

k¢'(d2 )

(8)

+ k. By Equation (7),

cf>'(y)

=

Na + h,y) - fy(a,y).

(9)

Substitoting from Equation (9) into Equation (8) gives

a=

k[jy(a + h, d2) - fy(a, d2)].

Finally, we apply the Mean Value Theorem to the expression in brackets and get (10)

for some C2 between a and a + h. Together, Equations (5) and (10) show that

fry(c), d,)

=

f",(c2, d2),

(11)

where (c), d,) and (C2, d2) both lie in the rectangle R' (Figure A.32). Equation (11) is not quite the result we want, since it says only that fry has the same value at (c), d,) that f yx has at (C2, d2 ). The numbers h and k in our discussion, however, may be made as small as we wish. The hypothesis that fry and fyx are both continuous at (a, b) means that fry(c),d,) = fry(a,b) + €,andfyx(c2,d2) = fyx(a,b) + €2,whereeachof€),€2-->Oas both h, k--> O. Hence, if we let hand k--+ 0, we have fry(a, b) = fyx(a, b). • The eqnality of f ry( a, b) and f yx( a, b) can be proved with hypotheses weaker than the ones we assumed. For example, it is enough for f, fx> and fy to exist in R and for fry to be continuous at (a, b). Then fyx will exist at (a, b) and eqnal fry at that point.

AP-38

Appendices

THEOREM 3-The Increment Theorem for Functions of Two variables Suppose that the first partial derivatives of f(x, y) are defined throughout an open region R cootaining the point (xo, YO) and that fx and fy are cootiouous at (xo, yo). Then the change

o.

+ '\'x, Yo + ,\.y)

,~~. B(xo

/

1

+ ax, Yo)

Proof We work within a rectangle T centered at A(xo, YO) and lying within R, and we assume that At and ay are already so small that the line segment joining A to B(xo + At, YO) and the line segment joining B to C(xo + At, Yo + ay) lie in the interior of T (Figure A.33). We may think of 0 to make the new graph synnne1ric with respect to the y-axis.

A-6

Chapter 2: Answers to Odd-Numbered Exercises

43_ Reflects the portion for y < 0 across the x-axis 45_ Reflects the portion for y < 0 across the x-axis 47_ Adds the mirror image of the portion for x > 0 to make the new graph symmetric with respect to the y-axis 49_ (_) y = g (x - 3) (0) y

= g(-x)

+

t

(d) y

(b) y = g (x

= -g(x)

,

' --VI +~

, -,

~ -4 -3 -2-1

-1

55_ Period 'IT

Y=~+1

1

,

3 •

+h "\1'1+1; ("\1'1+1; - 1)/h

Y _ 2col(x_~)

I

-, b

I

b

63_ (-) a = tanB 65_ '" 16.98 m

= V3

(b) a = 2V3/3 a (b) c = sinA

67. (b)

c

= 4V3/3

1. Yes.Forinstancc:f(x) = I/xandg(x) = I/x,orf(x) = 2x andg(x) = x/2, or f(x) = eX andg(x) = lnx. 3. Iff(x) is odd, theng(x) = f(x) - 2 is not odd. Norisg(x) evcn, unless f(x) = a for all x. If f is even, theo g(x) = f(x) - 2 is

also even. ,

I)/h

(b) g(x) =

Vx

1.1

1.01

1.001

1.0001

1.04880

1.004987

1.0004998

1.0000499

0.4880

0.4987

0.4998

0.499

1.00001

1.000001

1.000005

1.0000005

0.5

0.5 (0) 0.5

("\1'1+1; -

(d) 0.5

21. (8) IS mph, 3.3 mph, 10 mph (0) 20 mph when t = 3.5 br

(b) 10 mph, 0 mph, 4 mph

4'IT

Additional and Advanced Exerdses, pp. 37-38

5.

(b) '" 50 m/sec or 180 km/h 17. (8)

(b) '" $56,000/year (0) '" $42,000/year 19. (_) 0.414213,0.449489,

=

are m/sec.

,

,

61_ (_) a

5. I

x

,

y _ cos2l'

3~

(b) y = 4x - 7 (b) Y = 2x - 7 (b) y = 12x - 16 (b) Y = -9x - 2 15. Your estimates may not completely agree with these. (8) PQ, PQ, PQ, PQ. The appropriate units 43 46 49 50

57_ Period2

,

59_

x

(b) -

7. (_) 4 9. (8) 2 11. (_) 12 13. (8) -9

,

53_

(b) I

!

3. (_) -

= 5g(x)

(f) y = g(5x) 51_

Section 2.1, pp. 44-46 1. (8) 19

+ ~) - 2

(0) y

CHAPTER 2

Section 2.2, pp. 54-57 1. (8) Does not exist As x approaches I from the right, g(x) approaches o. As x approaches I from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x --+ 1. (b) I (0) a (d) 1/2 3. (8) True (b) True (0) False (d) False (0) False (f) True (g) True 5. Asx approaches a from the left, x/lxl approaches -1. Asx approaches a from the right, x/lxl approaches 1. There is no single number L that the function values all get arbitrarily close to as x---+ O. 7. Nothing can be said. 9. No; no; no 11. -9 13.-8 15. 5/8 17. 27 19. 16 21. 3/2 23. 1/10 25.-7 27. 3/2 29. -1/2 31. -I 33. 4/3 35. 1/6 37. 4 39. 1/2 41.3/2 43. -I 45. I 47. 1/3 49. ~


51. (s) (c) 53. (8) 55. (s) 57. 2 65. (8) 67. (s)

QuotientRule (b) DitferenceandPowerRules Sum and Constant Multiple Rules -10 (b) -20 (c) -I (d) 5/7 4 (b) -21 (c) -12 (d) -7/3 59. 3 61. 1/(2V7) 63. The limit is 1. f(x) = (x 2 - 9)/(x + 3)

Vs

Vi?),

25. (v'i5,

2 27. (2 _ 0.03 m' 29.

Vi? -

8 =

+

0.03)

•

m'

(k-,i",i,+t),

A-7

4 ... 0.12 0.03

=

m

U

8=,i,

31.L=-3,

8=0.01

33. L

= 4, 8 = 0.05 35. L = 4, 8 = 0.75 55. [3.384, 3.387]. To be safe, the left endpoint was rounded up aod the right endpoint rounded down. 59. The limit does not exist as x approaclres 3.

Section 2.4, pp. 71-73 lim f(x)

(c)

69. (s) G(x)

=

+ 6)/(x 2 + 4x

(x

1. (a) (f) (1 0 such that for all x

xo - S < x < Xo => f(x) > B. (b) For every negative real number -B there exists a corresponding nomber S > a such that for all x Xo

Xo - S

<x
f(x) < -B.

Xo

101. y , , , , ,

6

,--',-,

,

+ S => f(x) < -B.

99.

4

10

Xo

(0) For every negative real number -B there exists a corresponding nomber S > 0 such that for all x

65.

,

<x
;'+1 (x' - 1)'(x' + x + I)'

~

y(')

=

0

yen) =

33. y' 35 dr 'd8 37. ':;;

=

a for n '" 5

3x 2

+ 4x

- S,y" Oforn O?:: 4

2x - 7x-2 ,y"

=

6x

=

2

+

+ 4,ym

=

6,

14x-3

'dB'

-z-' - I, ~';

dpi

39. dq ~ 7;q

I

~

,

4

6

,

10

~ 1.-' d'PII zq--4 - 5q-
; - l)eot(J>; - I) 63. 16(z" + 1)2(5x + I) 65. 5/2 67. -'1f/4 69. a 71.-5 73. (a) 2/3 (b) 2'1f + 5 (c) 15 - 8'1f (d) 37/6 (.)-1

_2xy _ y2

10(z" + I)' 11. Witb u

55. 6 tan (sin' I) sec2 (sin' I) sin2 I cos I

I sin (12) v'1 + cos (12)

-67 x -67 2

2

(b) Y ~ 'if x - 'if

'1f

+2

x I (b) Y ~ - 2'1f + 2'1f

(Y?, 0), Slope:

-2

m~-Iat (~,~). m~V3at (~,t)


43. (-3,2):m

(3, -2): m 45. (3, -I) dy y' 51. dx

= _2;; (-3, -2): m = 2;; (3,2): m = 2;;

17. (3x

27

= - If

+ 2xy

= x2 + 3xy2'

=

13. (a)

dx

x2

+ 3xy2

dx

dy

y' + 2xy'

dy

21. I dy/dx

1T72 ~;

(b)

dV 2dh (0) dt = 1T7 dt

15. (a) I volt/sec

dR

(0) dt =

+

f' is eveu

23. h' is defmed but not continuous at x = 0; k' is def'med and continuous at x = O. 27.
.

xsmx

:ix (x; sinx + C) = ~ sinx + ~2

COSX

=

x2

+ TeosX'

(b) Wrong:

:ix (-x cos:< + C) = -cou + x sinx

(0) Right::ix (-x cos:< + sinx + C) = -cos:< + x sin x +

cosX' = x sin x . A. (2x + I)' ) _ 3(2x + 1)'(2) _ 63. (a) Wrong. dx 3 + C 3 Z(2x + I)' (b) Wrong:

:ix «2x + I)' + C) = 3(2x + 1)'(2) =

6(2x + I)' (0) Right:

:ix «2x +

I)' + C) = 6(2x + 1)2

A-22

Chapter 4: Answers to Odd-Numbered Exercises (a) Local maximum at x = 4; local minimum at x = -4; infIectioo point at x = 0 (b) x=4

Right

67_ (b)

69_ y=x'-7x+ 10

I x' I 71_ Y=-X+T-2

73_ y

=

75_ s

9xl/' + 4

77_ r = cos("O) - I

t + sin t + 4 I I 79_ v = 2 sect + 2 =

83_r=~+2t-2

81_ y=-x'+x'+4x+1 85_ y = x' - 4x' 87_ y 89_ 93. 95_ 97.

=

+

x - -4

(a) Local maximum at x = 0; local minima at x = -I and x = 2; infIectioo points at x = (1 ± v7)/3

5

-sint+ cost + t ' - I

y=2x'/'-50

(b)

y = -sinx - cosx - 2 (a) (i) 33.2 units, (Ii) 33.2 units, (Iii) 33.2 units

t= 88/k,k= 16 99. (a) v = IOt'/' - 6t l /2

(b) True

x=-1

37. (a) Localmaximumatx = -v'2;localminimumatx = v'2; infIectioo points at x = ± I aod 0

(b) s = 4t'/' - 4t'/2

(b)

Practice Exerdses, pp. 240-242 1. No 3. No minimum; absolute maximum: 1(1) 5. Yes,exceptatx = points: x = I aodll/3 7. No 11. (b) One 13. (b) 0.8555996772

=

a

""-

16; critical

19. GlObaiminimumvalueortatx = 2

(b) t = 3,9 21. (a) t = 0,6,12 (d) a < t < 6, 12 < t < 14

23.

""-

91_y=x-x 4/'+t

< t < 12

(c) 6

43. 25.

45. y

y- x+, -J~·L ,

y

x-3

y. x2;l :

x-3:

-H {

:

, ----- i ---

3

,

:

------ I

-~~~~+-~~~~_x

--=~_"'I.Jc-"-:-+:-"',--_x

-, 27.

-4 -3 -2

-v ,'-I

//A~~

y

47.

49. Y

Y

,,

4

,~/

"

2

x=-V3 :,

:x=~ :

-,

-,

y

51. (a) 0,36

(b) 18,18

53. 54 sqUlll"O units

55. Height = 2, radius = v'2

Vs hundred'" 276 tires, 2(5 - Vs) hundred'" 553 tires

57. x = 5 y =

59. Dimensions: base is 6 in. by 12 in., height maximum volume = 144 in3

),=1

----T----

1

31.

~-4

i'= x2-3

,

, ,

~~!:-~ -"-~~~~I~'~'~x

y

:

4

,'~ - ~ :'

29.

1 2 3 4

=

2 in.;


5 x, = 2.195823345 63. "4 + Ix2 - 7x + C 65. 21'/2 - t + C 67. - ~ 5 + C 69. (82 + 1)'/2 + C X4

61.

r

71.

t

75. -

5•

· SID'11' -

. 8m

'" + sm . '"

2

JzC'C

73. 10 tan :0 + C

\128 + C

79. y = x - } - I

77.

kX - .in I + C

81. r = 41'/2 + 41'/2 - 81

=

V32-

2

7• All 0 fthem

9. b

6 4 '

11.

~k

13.

.t-l

(1 + x 4)'/4 + C

3

A-23

17. 19. 21. 29. 31. 33.

~~ i'

15.

k- I

(s) -15 (b) I (s) 55 (b) 385

-56

~ (_1)1+11.

(s)

(b) y

y

,

(2,3)

1. The function is constant on the interval. 3. The extreme points will not be at the end of an open interval. 5. (s) A local minimwn atx = -I; points of inflection at x = 0 and x = 2 (b) A local maximum at x = 0 and local minima

(0) 16

(0) I (d) -11 (0) 3025

23. -73 25. 240 27. 3376 (b) 3500 (0) 2620 (b) en (0) (n 2 - n)/2

(s) 21 (s) 4n

Additional and Advanced Exercises, pp. 243-245

k

k- l

~) _ ;J;2_1.

OSl::S2

O:Sx:S2

Left·....

2

(~')

--!(x) . Xl-I, :

2

finfl' . atx=- I and x= 2 ;pomtso ectionatx= I ± 3v'7 9. No 11. a = I, b = 0, e = I 15. Dril1 the hole at y = h/2. 17. r =

19. (s)

2(H~ R) forH > 2R,r = e-b --ze

(d) e

e+b

(b) -2-

13. Yes RifHS 2R 2

2

(0) b -2be +e +4ae

4e

(0)

y

,

+b+1 2

21. ma = 1 -

I

7i' ml

23. (s) k = -38.72 25. Yce.,y = x

=

(2.3)

-

f~) _ x2_1,

O:Sx:!i:2

I 7j

2

(b) 25 ft

27• Vo = 2\12 3 b'/4

o '" c: "2 -:

"3

"4

,

CHAPTER 5

-1

Section 5.1, pp. 253-255 1. 3. 5. 9. 13.

(s) 0.125 (b) 0.21875 (0) 0.625 (d) 0.46875 (s) 1.066667 (b) 1.283333 (0) 2.666667 (d) 2.083333

0.3125,0.328125 7. 1.5,1.574603 (s) 87 in. (b) 87 in. 11. (s) 3490 ft (b) 3840 ft (s) 74.65 ft/sec (b) 45.28 ft/sec (0) 146.59 ft 31 15. 16 17. I 19. (s) Upper = 758 gal, lower = 543 gal (b) Upper = 2363 gal, lower = 1693 gal (0) '" 31.4h, '" 32.4 h 21. (s) 2 (0) 8

35. (s)

(b) y

f~) - sin::t, -'IT

l y

f(; 0, so g" never cbanges sign. (g) True, since g' (I) ~ 1(1) ~ 0 aod g' (x) ~ I(x) is an increasing func1ioo ofx (because I'(x) > 0).

Section 5.5, pp. 290-291 1.

+C

4)'

13.

= bin and right-<mdpoint

t

11. -6(1 - r')'/2 + C

i

(x'/2 - I) -

15. (a) -

77. J.'/(X)dx '" J.'Odx

~0

1/4

69. b 4/4 - a 4/4

63. c(b - a) 65. b 3/3 - a 3/3 67. 9 71. a ~ 0 aod b ~ I maximize the integral. 73. Upperbound ~ I, lower bound ~ 1/2 75. For example, ].' sin (x') dx '" ].' dx

~

~I

79. Upper bound

13. -,,/4

15. I -

21. -3/4

23.

29. (cosVi)

43. 1/2

f

v'2 -

C~)

35. -tx-'/2sinx 45."

49. d,sincey'

~

7. I

5. 8

fg + I 31. 41'

~ 1/2

11.

33.

a

19. -8/3

25. -I

39. I

37. 0 47.

9. 2V3

17. 2 -4 v'2

27. 16

V1+7

25. tSin'

~

41.28/3

+C

+

C

. (I) +

33. -sm t - I I (I 16

I (x - 1)12 43 . IT

47.

~ (x'

I 31. 2 cos (21 + I) + C sin' (1/8) 2 + C

+C

I + rr(x - I) 11 + C

+ 1)'/2 -

t (x'

+ 1)3/' + C

6

(b)

+ tan3 x + C 6

3

2+tanx

6

2

-I , ,+C 4(x - 4)

49.

+ tan3 x

+C

+C

-3

59. s

59. a. 1'(0) = 70'F, 1'(16) = 76'F 1'(25) = 85'F b. av(T) ~ 75'F 61. 2x - 2 63. -]x + 5

(~: - I)' + C

1)'/2 + C 39. 32(2 - :x

57. s = 41 - 2 sin (21 +

2 55. 3 bh

I 23. 3tan(3x + 2) + C

53. isin Y3(2r - I)' + 6 + C ~

+ C

-~(I - 8')'/4 + C

35. -

C

4I (csc' 28)

I 4 2 45. -g(1 - x)' + 7(1 - x)' - 3(1 - x)' + C

51. b,sineey' = secxaody(O) = ]'°seCldl+ 4 = 4

r 53. y = J, sec 1dl + 3

19.

27.

44 + C + I)

2( I - x3)'/2 41. 27 3

(c) -

J.~ I ~ tdl - 3

(t)

51. (a) - 2

VJ:"

I :xaody(,,)

(b) -

2 ('/2 ) 29. -3cOS x + I + C

Section 5.4, pp. 282-284 3. -10/3

+ C

21. (-2/(1 + Vi» + C

37.

1. 6

sin (2x'/2 - 2) + C

4I (cot' 28)

17. -t(3 - 2x)'/2 55. av(f) ~ 0 57. av(f) ~ -2 59. av(f) ~ I 61. (a) av(g) ~ -1/2 (b) av(g) ~ I (c) av(g)

-t(x' + 5)-3 + C

3.

5. 110 (3x' + 4x)' + C 9. t sec 21 + C

~ ].']X' dx ~ b'

53. Using n subintervals ofleogth!u values:

i (2x +

~ sin (21- f)

~)

55. s

~

t(31' - 1)4 - 5

+ 9

+ 1001+ I

61.6m

57. $9.00

Section 5.6, pp. 297-300 1. (a) 14/3 (b) 2/3 5. (a) 15/16 (b) 0 11. (a) 1/6 (b) 1/2

3. (a) 1/2 (b) -1/2 7. (a) 0 (b) 1/8 9. (a) 4 13. (a) 0 (b) 0 15. 2V3

(b) 0

A-25


19. 3'/2 - I 21. 3 23. 'IT/3 25. 16/3 29. 'IT/2 31. 128/15 33. 4/3 37. 38/3 39. 49/6 41. 32/3 43. 48/5 47. 8 49. 5/3 (There are three intersection points.) 53. 243/8 55. 125/6 57.2 59. 104/15 4 4 63.4 65. 3- 17 67. 'IT/2 61. 56/15

17. 3/4 27. 2'/2 35. 5/6 45. 8/3 51. 18

y

,

_~'-~_I~-+-~-~'~X

69. 2 71. 1/2 73. I 75. (8) (±Vc, c) (b) c ~ 4'/3 77. 11/3 79. 3/4

17. 1/2

Practice Exercises, pp. 301-303 1. (_) About 680 ft

15. 13/3

25. (a) 0 (0) Y

(b) h (feet) 600

23. (b) 'lTr'

(b) -I (0) -'IT (d) x ~ I 2x + 2 - 'IT (I) X ~ -I,x ~ 2 sin4y siny 29 - - - -

(g) [-2'IT,0]

~

27. 2/x

700

21. /.If(x) dx

19. 1/6

. Vy

soo

2Vy

400

300 200

100 ----;;0t-""-.;----7-;---;-~ t

3. (8) -1/2 5.

l'(2x -

9. (_) 4

11. 8/3

(b) 31

~2

1)-1/2 dx

(b) 2

(0) 13

(d) -2'IT

V2 1 23. ~ 32 +-2--

17. 1/6

29. Min: -4, max: 0, area: 27/4

35. Y~ J,X(S~I)dl - 3 39.

t cos (21'/2) + C

53. 27V3/160

61. -I 63.2 65. -2 71. (8) b (b) b 75. 25'F

83.

41.

77. V~2-+-c-o~s"x

-v'\+?

~3 + i

57. V3

7. f(x) ~ ~ x2 + 1

-6

79. - - . 3 +x

9. Y ~ x'

+ 2x - 4

I 2 13'2- 17 y

_~8--_';-~0~~~X

-.

y=-4

5. (a) 2V3 13. (8) s'h

19. 3;'IT

51. 8

49. (a)

16'1T

15

53. (a) V ~

7'IT

6

(b)

117'IT 37. -5-

56'1T

(0)

29. 2'IT

64'IT

15

31. 3'IT

39. 'IT ('IT - 2) (0) ~

51. V

8'IT

T

4'IT 41. T (d)

, , 2a b'1T

I (b) 120'IT m/sec

'lTh'(3a - h) 3

~ 3308 em'

15. 2:;

32'IT (b) -5-

47. (8) 8'IT

(b) 36

23. 'IT

27. 2'IT

15

7. (8) 60

(b) s'h

21. 36'1T

2'IT 35. T

45.

(b) 8

59. 4 - ~ + a

81. Yes

\o/l2

(b)

y

17. 4 - 'IT

57. V

5. (a) 1/4

11. 36/5

11. 10

69.V2-1

67. I

Additional and Advanced Exerdses, pp. 304-307 (b) No

9. 8'IT

43. 8'IT

59. 6V3 - 2'IT

85. Cost '" $10,899 using a lower sum estimate

1. (_) Yes

6 3. 13

--' - 2'IT 33. .,.

+ C 49. I

47.2

1. 16

1 25. 'IT(f + 2V2 - 13 )

37. -4(oosx)l/2 + C

45. 16

55. 'IT/2

21. 9/8

31. 6/5

0' + 6 + sin (26 + 1) + C

43. -

(e) 8/5

19. 18

8~-7

27.

25.4

CHAPTER 6 Section 6.1, pp. 316-319

(d) 0

J~COSIdx ~ 2

7.

(0) -2

15. I

13. 62

(sec)

Section 6.2, pp. 324-326 1. 6'1T

11.

i~

17. 8:;

3. 2'IT

5•

14'1T 3

7. 8'IT

13. (b) 4'IT

15.

19. 4:;

I~

21.

I~; (3V2 +

(d) 241r

(b) 32'IT (e) 60'1T

(0) 28'IT (I) 48'IT

2~'IT

(b) 27'IT

(0) 72'IT

23. (8) 16'1T 25. (8)

27. (8) 6;

2 (b)

7

5'IT 9• 6

5 (0) 2'IT

5)

(d) 108'IT

5

(d) 2'IT

29. (8) About the x-axis: V

~ ~~; about the y-axis: V ~ ~

(b) About the x-axis: V

~ ~~; about the y-axis: V ~ ~

2241r

15

A-26 31_ (a)


5;

(b) 4". 3

(d) 2".

(c) 2".

3

5. (a) 2'lf1'(3 - x ' /2j2VI

4". 15

(b) 7". 30 24". (b) 48". 35_ (a) -55 9". 37- (a) 16 (b) 9". 16 39_ Disk: 2 integrals; washer: 2 integrals; shell: I integral

33_ (a)

41_ (a) 25:".

(b)

+ (I -

3x l/2j2dx

(0) S'" 63.37

(b) ,

,

~". .~--;----;;---"'.~.,

Section 6.3, pp. 330-332 3• 53 6

5 123 • 32

11. (a)

1>11

+ 4x2 dx

13. (a)

/,W,Ii + cos2y t(y

1• 12

15. (a) 17. (a)

1:

V I

[wi'

Jo

+

(y

secxdx

+

7 99 • 8

9 2 •

(b)

~

±X

(0) '" 3.82

0.8

(0) '" 9.29

.I: -!oJ Uintdt

(0) '" 0.55

31. {7 (\03/2 - I)

o '--coO'".l""0~2""0~3""0'",""o~,""0.'".""0.~7... ,

9. 4".Ys

11. 3'lfYs

17. 'If(Vs - 1)/9 23. 253"./20

Jo[wi' (tau) VI

+ sec' xdx

1. 5. 7. 15.

,

25. 31. 35. 39.

(b)

~ VI

27. Order 226.2 liters of each color.

3. 4 em, 0.08 J (b) 905 in.-Ib, 2714 in.-Ib 780 J 9. 72,900 ft-Ib 13. 160 ft-Ib (a) 1,497,600 ft-Ib (b) I hr, 40 min (d) At 62.26Ib/ft': a) 1,494,240 ft-Ib b) I hr, 40 min At 62.59Ib/ft': a) 1,502,160 ft-Ib b) I hr, 40.1 min 400 N/m

+ y-4 t(y

19. 23;611" ft-Ib

85.1 ft-Ib 27. 1684.81b 33. 1164.8 Ib 37. (a) 12,4801b

41. (a) 93.33 Ib

3. (a) 2".1

15. 2'lf

21. (2'lf/3)(2Vz - I)

(a) 7238 Ib/in.

17. 37,306 ft-Ib

2

13. 98'lf/81

19. 35".Ys/3

Section 6.5, pp. 342-346

Section 6.4, pp. 335-337

(b)

,

0.' 1)2 t(y

+ C where C is any real number

1. (a) 2'lf

(0) S'" 2.08

(0) '" 6.13

19. (a) y ~ Vi from (I, I) to (4, 2) (b) Only one. We know the derivative of the function and the value of the function at one value ofx. 21. I 27. Yes, f(x)

/,w/3 (J,' tan t dt) sec y dy

7. (a) 2'lf

21. 7,238,229.47 ft-Ib

98.35 ft-Ib 29. 91.32 in.-oz (a) 6364.8Ib (b) 5990.4lb 1309 Ib (b) 8580lb (0) 9722.3 Ib

(b) 3 ft

wb 43 . 2

45. No. The tank will overflow because the movable end will have moved only 3! ft by the time the tank is full.

(0) S'" 5.02

, 2

Section 6.6, pp. 355-357 1. x ~ O,y ~ 12/5 3. x ~ I,y 5. 9.

13. 7-~~~~~~~' 0.6 0.7 0.8 0.9

15.

~

-3/5

x ~ 16/105,y ~ 8/15 7. x ~ O,y ~ 'If/8 x ~ IS ~ -2/5 11. x ~ Y ~ _2_ 4-". x ~ 3/2,y ~ 1/2 (a) 2~ (b) x ~ 2,y ~ 0


,

(0)

11. D: (0, I]

R: [0,

15. D: [0,6]

R: [0,3]

,

4

A-27

13. D: [-I, I] R: [-,../2,,../2]

00)

y. ---±."

,

0

(2,0)

r

4

y. --±."

-4

35.

= Y = 1/3 21. j = a/3,y = b/3 23. 0- mr x= ,Y =4 j = 1/2,y = 4 29. j = 6/5,y = 8/7 V = 32,.., S = 32v'2,.. 4,..' 37. j = O,y = ~ 39• x- = 0,Y =

41.

v'2,..a 3(4 + 3,..)/6

19. 25 . 27.

33.

j

43.

=

j

~:o

3.,..'

7. (8) 2,.. 9. (a) 8,..

6

4b

3'17"

~,y = ~

~----~----~6~r

19. r'(x)

23. r'(x)

5. 7;;

25. r'(x)

(b),.. (0) 12,../5 (d) 26?r/5 (b) 1088,../15 (0) 512,../15

27. r'(x) range:

11. ,..(3\13 - ,..)/3

13. (a) 16,../15 15. 21. 27. 31. 35. 39.

2~'" It'

(b) 8,../5

0 17. 13

(0) 8,../3

29. r'(x)

(d) 32,../5

range:

33. r'(x) range:

Additional and Advanced Exemses, pp. 359-360

5 -,..'300

11.

j

7. 28/3

I

109. -Inx

6 89' ln7

101. 3 ~2


111. (x + I)'

(X ~

I + In (x +

1))

115. (sinx)'(lnsinx + xcotx) ~

119. Maximum: I at x

evt)' (~I + k)

113.

121. (a) Absmax: }atx

~

~

~

(b) Y

I

~~

0, Y

~

89. (a) We should assigo the value I to I(x) continuous at x = o.

117. cosxx.x-'(1 + Inx)

0, minimum: 2 - 2 10 2 at x

~

87. (a) Y

83. -I

10 2

A-29

(sin x)' to make it

y

I

(b) (2, :,)

123. Abs max ofl/(2e) assumed at x ~ 125. 2 127. Y ~ e X/ ' - I

1jVe

e' 2~

I 131. In (\12 + I) d I 133. (a) tb;(xlnx -x + C) ~x·x + Inx - I + 0 ~ Inx

129.

I (b) e - I

(e) The maximum value of I(x) is close to I near the point x '" 1.55 (see the graph in part (a».

135. (b) Ierror I '" 0.02140 (e) L(x) ~ x + I never overestimates eX.

Section 7.6, pp. 413-416 1. 3. 5. 7.

(a) (a) (a) (a)

'1f/4 (b) -'1f/3 -'1f/6 (b) '1f/4 '1f/3 (b) 3'1f/4 3'1f/4 (b) '1f/6

9. 1/\12 19.0 137. 21n 5

139. x '" -0.76666

141. (a) L(x)

~

25.

I + (102).< '" 0.69x + I

29.

1

~C

y'/' - x'/2

17. y

~ sin (x' +

21. 41o(Vy 23. 25. 31. 33. 39. 41.

+ 2) ~

15. e-Y + 2eV. ~ C 19. tlnlY' - 21

C) ex'

13. '1f/2

15. '1f/2

17. '1f/2

~ 23.~ I-x' 1-21'

21. I

.~

Ils + IIv8' + 8 -I

31.

v!=7

-2s

27.

(x' + I)Vx' + 2s'

-I 2Vt (1 + I)

33

I • (tan 1x)(1 + x')

35.

11. eY - eX ~ C

13. -x + 2tanvY ~ C

11. -1/Y3

-e' -I 37 -ls" • ".------; • .,.-------, 39. 0 le'IV(e')' - I ve" - I v I - 8' 41. sin- 1x 43. sin- 1 ~3 + C 45. _1_ tan-I _x_ + C

Section 7.4, pp. 394-396 9.

(e) '1f/6 (e) -'1f/3 (e) '1f/6 (e) 2'1f/3

47.

~ x' + C

~sec-ll~1

53. -7f/12

+C

(a) -0.00001 (b) 10,536 years (e) 82% 54.88 g 27. 59.8 ft 29. 2.8147498 X 10 14 (a) 8 years (b) 32.02 years

55.

v'i7

+ C

49.2'1f/3

51. '1f/16

~sin-12(r - I) + C

57. ,?tan-1(xvi)+c

15.28 years 35. 56,562 years (a) 17.5 min (b) 13.26 min -3°C 43. Aboot 6658 years 45. 54.44%

v'i7

59.tsec-112X;II+c

61. 7f

63. 7f/12

65. k sin-1y' + C

69. '1f

71. ktan-1

f;

I) + C

67. sin-1(x - 2) + C

73. 27f

75.11n(x'+4)+2tan-l~+C 2

Section 7.5, pp. 402-404 1. 1/4 3. 5/7 5. 1/2 7. 1/4 9. -23/7 15. -16 17. -2 19. 1/4 21. 2 23. 3 27. In 3 39.

-00

51. I/e 63. 0

I

31. In 2

29. In 2 41. -1/2 53. I 65. +1

75. (b) is correct

43. -I

55. I/e 67. 3

33. I

69. I

77. (d) is correct.

71. 0

0 77. x + 10 (x' + 9) - 13 tan-I 79. sec-1 Ix +

37. 1n2

47. 0 59. I

13. 0

25.-1

35. 1/2

45. I

57. e 1/2

11. 5/7

2

83. t(sin- ' xl' + C

49.2 61.

73. 00 27

79.c~TO

e' 81. (b)

11 +

89. 1tan-1 -I

T

97. I

(tan-~

103. y

C

j

+ C

81. e""-'x + C 85. Inltan-1yl + C

Vi) + C

~ sin-1x 105.

91. 5 y

87. Y3 - I

93. 2

~ sec-1x +

95. I 2;,x

>I

A-30


107_ (b) y = 3Ys

109_ 8 = COS-I

(~) "" 54.7'

3. sinhx

121. w2/2 123. (a) w2/2 (b) 2" 125. (a) 0.84107 (b) -0.72973 (0) 0.46365 127. (a) Domain: all real numbers except those having the fonn

f + k1r where !is an integer Range:

= 8/15, tanh x = 8/17, cothx = 17/8, sechx = 15/17,

cschx = 15/8 I 5. x + x

7.

0 s.

r. tanhVt 15. sech v t + Vt 2-

25.

,

I 2Vx(1 + x)

35.

<x
, Yl ~ 4.2, Y2 ~ 6.216, y, ~ 9.697

- ~ (I + Yx)'/2 + ~ (I + Yx)3/2 + C

~I + C

+

x - y;y(I)

11. y(exact)

l

v'3

~

I I

+

+C

Yzi Yz

45.

+C

35. Euler's method gives y '" 3.45835; the exact solution is y ~ I + , '" 3.71828. 37. y '" 1.5000; exact value is 1.5275.

Section 9.2, pp. 508-510 39.xln(ax)-x+C

1.

eX

+C

y~-x-'

cosx' x 3• y -- C - 3 x

x>O

v393'"

7. y

9. y

~

11••

~

I'

/

C

(/ - I)' (/ - I)' 13. r ~ (csc 8)(lnlsec 81 + C), 0 < 8 < ",/2 _31_ I '" 15. y -22 e 2 ' 1 7. y -_-Ocos8 + 28 21.y~yoekt

23. (b) is correct, but 0.,,'>0

1'0

(0)

Y

1.5lb/gai

27.8 min

Section 9.4, pp. 522-523 1. y' = (y + 2)(y - 3) (a) y = -2 is a stable equilibrium value and y = 3 is an unstable equilibrium.

(b)

y'

,

-4

=

2(y

+ 2)&

,>.

-

t).

)",'0,,">0

6

•

,','>0

1~

, ,

:,'>0:

,

2) (y

,'0

,

4

Y


. d2

,

(0)

(k)

A-37

(k) (g - /iiV k 2)

ConCavIty: dt v2 = -2 /iiv o .2"

~0

"" 0

~3~5)4P P'>O

P">O

As

t~ 00, i(t) ---+ isteadysta.te

=

~.

P 4

F:::::---===:::='~P'>O'P' >0 2

0.4

0.5

-1

0.6

0.7 P'>O,P' ; and y: fix

dt = (a -

-2

13. Before tire catastrophe, the population exhibits logistic growth and P(t) increases toward M o, the stable equilibrium. After tire catastrophe, the population declines logistically and p(t) decreases toward M the new stable equilibrium. " p

p

, , ,

--~

----t----

,

dv _ k 2 15. dt - g - /iiv, Equilibrium' dv .~

----

~,

'--

g,k,m > Oandv(t) '" 0

=g

-

~ v 2 = 0 ~ v = rmg m ~~

dy =

bylx,

a

< 0,

m(1 _l')y - nxy y(m - my - nx) =

dt MM' Rest points are (0, 0), unstable, and (0, M), stable. 5. (8) Logistic growth occurs in the absence of the competitor, and involves a simple intemction between the species: growth dontinales the competition when either population is sma11, so it is difficult to drive either species to extinction. (b) a: per capita growth rate for trout m: per capita growth rate for bass b: intensity of competition to the trout n: intensity of competition to the bass k,: enviroomental carrying capacity for the trout k2: enviroomental carrying capacity for tire bass

~: growth versus competition or net growth of trout

W: relative survival of bass

A-38

Chapter 10: Answers to Odd-Numbered Exercises dx

(0) dt ~ 0

when

ely dt ~ 0 when

x~o

or y

y~O

or y

a

33. (a) y = -I is stable and y = I is unstable.

a bk,x,

~

b-

~

k2n k2 - /ilx.

d"'y ely (b) dx 2 = 2y dx = 2y(y2 - I) )' _,_1

By picking alb > k2 and min > k" we insure that an equilibrium pcint exists inside the fIrst quadrant

Y

I

~>o dx

•

: d7>O : d2y;2+6 19. y = 40..,') 17. y = 6(x + I)' 21. y ~ .~(3x' - 1>;2) 11. y=

t(I -

23.

-2

Additional and Advanced Exerdses, pp. 530-531 x

y

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0 0.1000 0.2095 0.3285 0.4568 0.5946 0.7418 0.8986 1.0649 1.2411 1.4273

1.1

1.6241 1.8319 2.0513 2.2832 2.5285 2.7884 3.0643 3.3579 3.6709 4.0057

1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

c + (YO - c).-k(AjY)1 (b) Steady-state solution: yoo = c 5. x 2(x 2 + 2y2) = C

1. (a) y =

7. In Ixl + ."/' = C 9. In Ixl -In lsec(y/x - I)

+ tan(y/x - 1)1

CHAPTER 10 Section 10.1, pp. 541-544 1. a, = 0, a2 = -1/4, a, = -2/9, flo, = -3/16 3. a, = l,a2 = -1/3,a, = 1/5,,,,- = -1/7 5. a, = 1/2, a2 = 1/2, a, = 1/2, flo, = 1/2 3 7 15 31 63 127 255 511 1023 7. 1'1' 4' 8' 16' 32' 64' 128' 256' 512

25. y(3) '" 0.8981 27.

(.)~

IIIIIIII 9. 2, I, -1'-4' 8'16'- 32' - 64' 128' 256

11. I, 1,2,3,5,8, 13,21,34,55 [ 0.2, 4.5] by [ 25,O.S]

(b) Note that we choose a small interval ofx-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x :5 -I. (This occurs because the analytic solution is y = -2 + 1n(2 - .~), which has an asymptote atx = -In 2 '" -0.69. Obviously, the Euler approximations are misleading for x :5 -0.7.)

~ .

~

..

[-1,0.2] by [-la, 2]

( ) 29. yexact 31. y(exact)

123

.1

= IX -I;y(2) '" 0.4; exact value IS I' = -.(>'-')/2;y(2) '" -3.4192; exact value is

-.'/2 '" -4.4817.

= C

15. a. = (- I)·+l(n)2, n '" I

19. an =

n2 -

1, n ~ 1

_ 3n + 2 > 23.an n! ,n_l 27. 33. 39. 45. 51. 57.

13. a.

= (-I)'+',n 211 -

'" I

1

17. a. = 3(n + 2)'n '" I

21. an = 4n - 3, n ~ 1 I + (_1)'+' 25.an = 2 ,n~l

29. Converges, -I 31. Converges, -5 Diverges 37. Converges, 1/2 41. Converges, 43. Converges, I 47. Converges, 0 49. Converges, 0 53. Converges,. 7 55. Converges, I 59. Diverges 61. Converges, 4 63. Converges, 0 65. Diverges 67. Converges, e-1 69. Converges, .2/' 71. Converges, x (x > 0) 73. Converges, 0 75. Converges, I 77. Converges, 1/2 79. Converges, I 81. Converges, 'If/2 83. Converges, 0 85. Converges, 0 87. Converges, 1/2 89. Converges, 0 Converges, 2 Diverges 35. Converges, 0 Converges, 0 Converges, I Converges, I

91. 8

93. 4

95. 5

v'2

97. I

+

v'2

99. x.

= 2.- 2

A-39


101. (a) f(x) = x' - 2,1.414213562 "" v'2 (b) f(x) = tan (x) - 1,0.7853981635 "" '1T/4 (0) f(x) = eX, diverges 103. (b) 1 111. Nondecreasing, bounded 113. Not nondecreasing, bounded 115. Converges, nondecreasing sequence theorem 117. Converges, nondecreasing sequence theorem 119. Diverges, defmition of divergence 121. Converges 123. Converges 133. (b) v'3

17. Converges; geometric series, r 19. Diverges; Integral Test 21. Converges; geometric series, r 23. Diverges; Integral Test

2(1 - (1/3)') 1-(1/3),3

7

IJ_con

Converges; Integral Thst Converges; Integral Test Converges; Integral Thst (a) y

~2 >

1

33. Diverges; ntb-Thnm Thst 37. Converges; Integral Test 41. a = 1

1114 7. 1 - 4 + 16 - 64 + ..., 5

G t) +

35. s. = 1 - n

s. f =

~

(n ~

45. 1

37. s, =

2}

21. 7/9

In Vn+1; diverges (b) "" 41.55 45. True 47. (b) n '" 251,415

converges,-~ ~2

47. -

49. Converges, 2 +

v'2

55. Converges, - ,e'- e - 1 57. Converges, 2/9 59. Converges, 3/2 61. Diverges 63. Converges, 4 65. Diverges 67. Converges, '11' ~ e 51. Converges, 1

,

+

1; converges, 1

oos-1

43. 5

53. Diverges

49.

s.

=

• I ~.,

,,- 1 n

""

1.195

51. 10 60

59. (a) 1.20166 s S s 1.20253

(b) S"" 1.2021, error < 0.0005

Section 10.4, pp. 562-563

=

1. Converges; compare witb ~ (I/n')

=

3. Diverges; compare witb ~

1, r = -x; converges to 1/(1 + x) forlxl < 1 3, r = (x - 1)/2; converges to 6/(3 - x) for x in (-1,3) 1 1 1 73.lxl;2 + 4x' + X4 (I - 2x)' ~ I - 6>; + 12x2 - 8x' 0.00267 17. 0.10000 19. 0.09994 21. 0.10000 1 x3 x' xU 13'6! '" 0.00011 25. T - 7'3! + 11'5!

5.I-x+T-2 9 · 11. 13. 15.

13 5 3. 1+"2 x + gX 2 + 16 x 3 + ...

A-42


27_ (a)

x2

x4

T -

(b)

13. Wi1h a = 22'1f, cosx = I -

12

x2

x4

T -

3-4

+

29. 1/2

31. -1/24

39. 3/2

41. e

49.

I

x'

(b)

x8

+ ... + (-I)" 31.32

5-6 - 7-8

35. -I

37. 2

45. V3/2

47. I

33. 1/3

43. cos ~

I - 6! (x - 22".)'

x 32

~x

23. a

x3

,-I

2 - x - 6 - 40 - ill

61.1-2x+3x 2 -4x'+··· 67. (a) -I

71 x + •

(b)

(1/v2)(1 + ,)

(o)-j

,-0

~

( _ 1),,.z.+IX2n+l

(2n

(x

00

+ I)!

+

1) 65. 2 - - 2 I' • .

61.

3(x

+ I?

+, 2 ·2!

~

(-I)'x,o.j3 (2n)!

00

63.

(n ;;,

85. (a) 00

(0) I/q

(b) 6

~

Section 11.1, pp. 616-618 1.

3.

,

5.

7.

, 2

n!

+ I)' + 2, ·3! + ...

I): the seties converges to In (b) a = I, b = 0

"'-.L . , 2

t=~ -2

t=O

-,

2

,

-2

(t) . 9.

,

+

Y.f

(x - '1f/3)'

3. Diverges; n1h-Term Test 7. Diverges; nth-Term Test

V3 I -2- (x - '1f/3) - 4 (x - '1f/3?

+ ... 2

11. Wi1h a = 0, e' = I

3

+ x + x2! + x3! + ...

0

t='IT

-,

75. 1/12

87. It converges.

1. Converges; Comparison Test 5. Converges; Comparison Test

2I -

,

~2+-i _ l


9. Wi1h a = '1f/3, cosx =

,

------~o~~--~,

«'1fx)/2)'

l_l(x 2

83. In

27. (b) Yes

9(x

3) +l(x - 3? _l(x - 3)' 4 4 4' 4' 69. 0.4849171431 71. 0.4872223583 73. 7/2 77. -2 79. r = -3, s = 9/2 81. 2/3 67.

-7/6

±sI

21. b =

CHAPTER 11

Converges to I 3. Converges to -I 5. Diverges Converges to 0 9. Converges to I 11. Converges to e-s Converges to 3 15. Converges to In 2 17. Diverges 1/6 21. 3/2 23. e/(e - I) 25. Diverges Converges conditiona11y 29. Converges conditionally Converges absolutely 33. Converges absolutely Converges absolutely 37. Converges absolutely Converges absolutely (0) x = -7 (a) 3, -7 '" x < -I (b) -7 < x < -I (0) None (a) 1/3,0 '" x '" 2/3 (b) 0 '" x '" 2/3 (a) 00, for all x (b) For all x (0) None (a) V3, -V3 < x < V3 (b) -V3 < x < V3 (0) None 49. (a) e, -e < x < e (b) -e < x < e (0) Empty set 00 I I 4 X 57. ~2'x' 51. 1 +x'4'5 53. sinx,'11',O 55. e ,ln2,2 1. 7. 13. 19. 27. 31. 35. 39. 41. 43. 45. 47.

59.

17. "./2

~xs + ...' forallx 30

+3

1x 3 -

Practice Exercises, pp. 605-607

00

+ 1! (x - 22'1f)'

33. (a) R, = Coe-"'(I - e"""")/(I - e-Ito ), R = Co(e -"')/(1 - e -Ito) = Co/(e lto - 1) (b) R, = I/e '" 0.368, RIO = R(I - e- ZO ) '" R(0.9999546) '" 0.58195; R '" 0.58198; 0 < (R - Rzo)/R < 0.0001 (0) 7

5x 7

3x 5

=

31. (a) ~nx'-1

-I 55. 500 terms 57. 4 terms (I + X)2 . x3 3x 5 5x 7 6 + 40 + ill' radius of convergence = I

'IT

2,L

(x - 22'1f?

00

•

+

x2

=

t

+ ...

15. Converges, limit = b

51

+ x2

59. (a) x

x6

--~f---+---~~~,

11.

,

"

4

,


13.

15. y

3

-- ,

19. I

y=-b -?

t=O

29.

"

\

\

-,

\

21. 3a'1T

23. ab1T

31. 8~

521T 33. -3-

,

'1r

37. (j,y)

r

•

d"y

fix' = 108

13. Y = 9x - I,

y

=

39. (j,y) =

(~

-

(t,1T -1)

43. (a) x = I, Y = 0,

.. •

, I

,

I

,,/

45. r

\ \

\

\

\

,, ,,

(~, I).

y = -a sin I, 0 s I S 21T y = asint, O:s; t:S; 2'I'T y = -asint, O:s; t:S; 4'1T y = a sin I, Os IS 41T x = -I + 51, y = -3 + 41, x = I' + I, Y = I, I s 0

35. x=2cott, y=2sin2 t, asin'ltanl, y

=

a sin'l,

Os I

d"y

fix' =

1T

r

n an integer (d) (2,(2n + 1)1T) and (-2,2n1T),n an integer

5. (a) (3,0) O:s; t :s; 2w

(0)(3,0)

< 1T/Z

39. (I, I)

7. (a)

(b) (-3,0) (f)

(V2, f) 1~1T)

d"y

11.

I 2

j'3+z,

(I, \13) (-1,\13)

(d) (b)

1

1)

(b) (-1,0)

(d) (-5, 1T - tan-I

Y

13.

~) Y .~

+-C+---t-~r

Z=-4

(g)(-3,0)

(d) (5'1T - tan-

(a)

(0) (-Z, 5;)

(-I, \13)

(b) (3, 1T)

9. (-3V2, 5:)

V2

(0)

(1,\13)

+ 4' fix' = -2 7. Y = lx - \1'3, fix' = -3\1'3

11.y=\I'3x-

__

(~O)

(b) (Z,2n1T)and(-Z,(2n + 1)1T),n an integer

-4

d"y

d"y y=x-4, fix'

-,

(a) (2'f + 2n1T) and (-z,f + (2n + 1)1T ),naninteger

Ox

,,

v(nl2,o)

,

11.

x _ _3

),=2

2

----_,~~+-~~-->x

,,

2

-~rl-2---->x

31. Asymptotes: y

,

~

±2x

33. Asymptotes: y

,

,,

,,

,,

,

~ ±X/2

A-46

Chapter 11: Answers to Odd·Numbered Exercises

57. C(-2, 0), a = 4 59. V(-I,I), F(-I,O) . . (x+2)2 2 __ 61. Ellipse. 5 + Y - I, C( 2, 0), F(O, 0) and 39. (a) Vertex: (I, -2); focus: (3, -2); directrix: x = -I (b) ,

V(v'5 -

F(-4, 0),

2,O)andV(-v'5 - 2,0)

. (x - 1)2 63. Ellipse: 2 2

F(O, I),

~O++'I~2~'~--------~' -2

2

+ (y - I) = I, C(I, I), F(2, V(v'2 + 1,1) and V( -v'2 + 1,1)

65. Hyperbola: (x - 1)2 - (y - 2)2 = I,

"'- _F(3,_2)

I) and

C(I,2),

F( I + v'2, 2) andF(1

- v'2,2), V(2, 2) and asymptotes: (y - 2) = ±(x - I) (y - 3)' x2 67. Hyperbola: 6 - T = I, C(O, 3), F(O, 6) and

V(1,-2)

·4

V(O, 2);

41. (a) Foci: (4 ± (b) ,

0, 3); vertices: (8, 3) and (0, 3); center: (4,3)

F(O, 0),

+ 3)andV(O,-v'6 + 3); v'2x + 3 or y = - v'2x + 3

V(O, v'6

asymptotes: y =

69. (b) I: I 73. Length = 2 v'2, width = v'2, area = 4 75. 241T 77. x = O,y = O:y = -2x;x = O,y = 2:y = 2x + 2; x = 4,y = O:y = 2x - 8

79. .>;=0,

y=

j!

Section 11.7, pp. 653-654 43. (a) Center: (2, 0); foci: (7, 0) and (-3, 0); vertices: (6,0) and

(-2,0); asymptotes: y = (b)

±~ (x

1. e

=

3

S'

. directrices are x =

- 2)

3. e

F(±3,O);

25

±T'

=

I "1/2;

F(O, ± I);

directrices arey = ±2.

,

,

-s .........-/2

5. e 45. (y + 3)2 = 4(x + 2), V(-2, -3), F(-I, -3), directrix: x = -3 47. (x - 1)2 = 8(y + 7), V(1, -7), F(I, -5), directrix: y = -9 (x + 2)' (y + 1)2 49. 6 + 9 = I, F( -2, ± v3 - I), V(-2, ±3 - I),

51.

53.

(x - 2)2 3

4

I y'3; F(O, ±I);

directrices arey

,

= ±3.

7. e

=

~;

F(±v3, 0);

directrices are x = ±3v3.

,

C(-2, -I)

(y - 3)2

+ 2 V(±v3 + 2,3),

(x - 2)'

=

-

= I, F(3, 3) andF(I, 3), C(2,3)

(y - 2)2 5 = I,

.-16 .-13

C(2,2),

F(5, 2) and F( -1,2),

v'5

V(4, 2) and V(O, 2); asymptotes: (y - 2) = ± T (x - 2)

55. (y + 1)2 - (x + 1)2 = I, C(-I,-I), F(-I, v'2 -I) andF(-I,-v'2 - I), V(-I,O)andV(-I,-2); asymptotes (y + I) = ±(x + I)

A-47


17. e =

v'2;

F(±v'2,O);

~,__ . Ull~ctrices

are x =

19. e =

1

± v'2 .

v'2;

F(O, ±4);

directrices are y

=

±2.

y

y

y

f-~ _ I --------~

6

, -./8-

-- 2

,

__ ____

,_

--~--~~~--~--~->x ~ :~~/~J 2 4 ---2 --./8,

.

-4 F,

49. rcos(o 21. e

= Vs;

~,__

F( n/iii, 0);

.

UllOctriceS are x

23. e =

2

Vs;

F(O,

±vio);

53.

.;f)

=

3

51. rcos(o

+

f)

55.

y

5

= y

directrices arey = ±2 V'TIr

= ± V'TIr

r=-2005 8

, .

------~~~--_f--->x

Radin!

--::J~~~~: ~~~,_~___

•

.

=1

0/2 ---~----

RadinI=2

--~--~~~·~·--~--_7->x

-4

c->~---.----+-------_>x

->

4

57. r

=

12 cosO

59. r

=

108inO

y

•

y

r= IOIinB

J-+(y-Sl - 2S

(;x

10 5 - sinO

39. y

y

61. r = -2 cos 0

x- I

63. r

= -sin (J y

y

,= --'-1+0018 ;+~+t)2 -

(.;+li+l=l r _ _20016

t

r _ -sinO

--~--.---t--------->x

--------__~._------~.x

65.

67.

43.

41. y

y

y

y 4

r=4sinO

------~~~~------>x

A-48


71-

,

,

11. (a) y

~

±lxl'/2

(b)

-8- - 1

y~ ±~

~~--_~,--~-~l>. ---~-1~~---~'

,--'-

,

,

-1

x=2

4+WI!I9

, ---~--~~-~.

27. y

~ -~

29. x 2 + (y + 2)2 ~ 4

,

,

_ 1 1 2sin8

r -

2

75_ (b) - - - - - - - - - - - - - -

Planet

Perihelion

Aphelion

Mercury

0.3075 AU 0.7184 AU 0.9833 AU 1.3817 AU 4.9512 AU 9.0210 AU 18.2977 AU 29.8135 AU

0.4667 AU 0.7282 AU 1.0167 AU 1.6663 AU 5.4548 AU 10.0570 AU 20.0623 AU 30.3065 AU

Venus Eartb Mars

Jupiter Saturn Uranus Neptune

•

,

r _ -4!in8

:s?-+(y+2r- - 4

•

2

>'=-1 31.

(x - v'2)' + y2 =

,

33. r = -5 sin 8

2

,

r _ _.5sinS

Practice Exercises, pp.6SS-6S7 1.

T=

3.

,

Moo! 8

;.+~+tY=-¥

,

x

41-4..'=1

1 1=0

2

----~o+----~·

35.r=3cos8

37.

,

,

5. \

\

7. x

= 3 cos t,

y'3 9. y ~2x

I

Y = 4 sin t, 1 1

+4'4

---~r----~6~'

", ,,

), _ .1:

\ \

,

0

~ t ~

2'1T

39. d

41. 1

51. 8

53.

'IT -

43. k

3

45. i

47.

29 'IT

49.2

'IT

+4


55. Focus is (0, -I), ~

directrix is y

57. Focus is

1.

(~, 0).

directrix is x

,

,

4 81.r~I+2cose

-l

~

85. (a) 241T

,_1

2 83.r~2+sine

(b) 161T

Additional and Advanced Exercises, pp. 657-658

,

61. e

~

2; the asymptotes

arey ~ ±V3x.

,

3.

3x2+3y2-8y+4~O

7. (a)

(y - 1)2 x2 16 - 48 ~ I

5. F(O,±I)

&+~y

(b)

(i~)

,

--r---~----+~c-~'

11. x2+4,r-4 - 0

,

13.

,2->,-1 - .

\

-4

63. (x - 2)2 ~ -12(y - 3), V(2, 3), F(2, 0), directrix is y ~ 6.

65.

(x + 3)2

9

+

(y + 5)2

25

~

1,C(-3,-5),F(-3,-I)and

F(-3, -9), V(-3, -10) and V(-3, 0).

67.

(y -

2Yz)' 8

A-49

(x - 2)2 2 ~ I, c( 2, 2Yz),

F( 2, 2 Yz ± Vio), v( 2, 4 Yz) and V(2, 0), the asymptotes

,

15.

arey ~ 2x - 4 + 2Yzandy ~ -2x + 4 + 2Yz.

9X2+4y2-36=O

69. Hyperbola: C(2, 0), V(O, 0) and V(4, 0), the foci are

F( 2 ± Vs, 0) and the asymptotes are y

~ ±x

""

; 2.

71. Parabola: V(-3,1),F(-7,I)aodthedirectrixisx

~

1.

73. Ellipse: C(-3,2),F(-3 ± 0,2), V(I,2)aodV(-7,2) 75. Circle: C(I, I) and radius ~ v'2 77. V(I,O) 79. V(2, 'II') aod V(6,1T)

,

,

17. (a) r ~ e 2B

,=_2_ l+cosO

----~~~----~, (1,0)

,- --'-1-20019

(b)

4 19.r~I+2cose

Vs (e"" 2

I)

2 21.r~2+sine

23. x ~ (a + b)cose - bCOS(a; be), y~ (a + b)sine - bSin(a; be) 27.

1T

2

x2

---~

(~)

,

I

A-50

Chapter 12: Answers to Odd-Numbered Exercises 23. The vector v is horizontal and 1 in. long. The vectors u and w

:~ in. loog.

CHAPTER 12

are

Section 12.1, pp. 663-664

horizootal. All vectors must be drawn ta scale.

1. The line through the paint (2, 3, 0) parallel to the z-axis 3. The x-axis 5. The circle x 2 + y2 = 4 in the xy-plane 2 7. The circle x + z2 = 4 in the xz-plane 9. The circle y2 + z2 = I in the yz-plane 11. The circle x 2 + y2 = 16 in the xy-plane 13. The ellipse formed by the intersectian afthe cylinder x 2 + y2 = 4 and the plane z = y 15. The parabala y = x 2 in the xy-plane 17. (a) The first quadrant afthexy-plane (b) The fourth quadrant afthe xy-plane 19. (a) The ball afradius I centered at the arigin (b) All points more than I unit from the origin 21. (a) The ball afradius 2 centered at the arigin with the interior af the ball af radius I ceotered at the origin removed (b) The solid upper hemisphere af radius I ceotered at the origin 23. (a) The regioo on or inside the parabala y = x 2 in the xy-plane and all points above this region (b) The regioo 00 or to the left afthe parabala x = y2 in the xyplane and all paints above it that are 2 units or less away from the xy-plane 25. (a) x = 3 (b) Y = -I (0) z = -2 27. (a) z = I (b) x = 3 (0) Y = -I 29. (a) x 2 + (y - 2)' = 4, z = 0 (b) (y - 2)' + z2 = 4, x = a (0) x 2 + z2 = 4, Y = 2 31. (a) y = 3, z = -I (b) x = I, z = -I (0) x = I, Y = 3 2 33. x + y2 +z2 = 25,z= 3 35. a I 37. a

(a)~

"'z '"

z'"

< I

39. (a) (x - 1)2 + (y - I)' + (z - I)'

>

(b) (x - 1)2 + (y - 1)2 + (z - 1)2

&-t)'

53. (x + I)' +

55. C(-2,0,2),a

= VB

59. (a) v'y2 + z2

61. v'i7 + V:l3 65. (a) (0,3, -3)

t)' ~~ c(-i,-i,-i).a = 5~

+ (z + 57.

=

(b) v'x 2 + z2

(0)

3(ti+

29.

~(~i - ~j - ~k)

31. (a) 2i

+6

(12, -19)

9. (I, -4)

.=

(b) V 505

11. (-2, -3)

15. \ -

Yj,-t)

21. 3i

+ 5j - 8k

3. (a) (1,3) 7. (a)

13. \

~

•• __

27.5(k)

(b) -\13k

(0) l.j +.2.k 10 5

(d) 6i - 2j + 3k

7

33. 13 (12i - 5k) 35. (a) _3_ i + _4_ j - _I_k 5V2 5V2 V2 37. (a) -

(b) (1/2,3,5/2)

~i - ~j - ~k

39. A(4, -3, 5)

3

2' b

41. a =

=

G,H)

(b)

I

2

43. "'(-338.095,725.046) 1()() cos 45° 45. IFd = sin 75' '" 73.205 N, =

I~;~~~O'

'" 89.658 N,

FI = (-IFd cas 30', IFd sin 30') '" (-63.397,36.603), F2 = (IF2 Icas45',IF2 Isin45') '" (63.397,63.397) 100 sin 75' 47. W = cas 40' '" 126.093 N, IFd =

ws:;;~'

'" 106.933 N

(t,

5r)

(b)

/ I 14) \5'"5

2

'

2 (b) i + j - 2k

(0) (2,2, I)

v'iO (b)

-t, Yj)

17. -3i + 2j - k

=

5 + 1OV2 5\13 - 1OV2)

51. (a) ti + t j - 3k

Section 12.2, pp. 672-674

5. (a)

-w

(b) (5 cas 60' + IOcas315',5sin60' + IOsin315')

63. Y = I (b) (0, 5, -5)

(b) 3v'i3

tk)

tj -

49. (a) (5 cas 60', 5 sin 60') =

(0) v'x 2 + y2

w

~~ (d)

(

1. (a) (9, -6)

8+1'+"

25.

IF21

51. (x - 1)2 + (y - 2)2 + (z - 3)' = 14

(b)

u

I

41. 3 43. 7 45. 2\13 47. C( -2,0,2), a = 2V2 49. C( V2, V2, - V2), a = V2

w is vertical and u makes a 45' angle with the

19. -3i + 16j

vm 5

Section 12.3, pp. 680-682 1. (a) -25,5,5

(b) -I

3. (a) 25, IS, 5

(b)

.=.1::

5. (a) 2, v34, v3 (d) 1\ (5j -

3k)

(0) -5

t

(0)

t

(b)

2 \l3V34

(d)

(d) -2i + 4j - Vsk

~ (lOi + (0)

2 V34

llj - 2k)


7. (8) 10 (0) 10

+ vTI, V26, V21 +

vTI

V26

9. 0.75 rad

(b) 10

9.

+ vTI

(d)

11.

,

V546 vTI 10 +26 (5i + j)

___ j+k

11. 1.77 rad

13. AngleatA = cos-1 B = cos- 1

G) '"

C = cos- 1

(~)

A-51

-=-:1'---~-----+,

(~)

'"

63.435degrees,aogleat

53.130 degrees, aogle at '"

13.

63.435 degrees.

23. Horizootal compooeot '" 1188 fi/sec, vertical componeot: '" 167 fi/sec 25. (8) Since Icos 81 '" I,wehavelu·vl = lullvllcos81 s lullvl (I) = lullvl· (b) We have equality precisely wheu Icos 81 = I or wheu ooe or both ofu and v are O. In the case of nonzero vectors, we have equality wheu 8 = 0 or 'IT , that is, whoo the vectors are parallel. 27. a 35. -2x + Y = -3 33.x+2y=4

,

,

17. (a)

-21+ J i+2j

15. (8) 2v6

--~~~--~o~--*~--~, 2

v'2 2

(b) ± _1_ (2i v6

+ j + k)

(b) ± _1_ (i - j)

v'2

19. 8 21. 7 23. (8) None (b) u aod w 25. 1OV3 fi-Ib 27. (8) True (b) Not always true (0) True (d) True (e) Not always true (f) True (g) True (b) True o·v 29. (a) projyu= v.vv (b) ±uXv (oj ±(uXv)Xw

-2.;+y - -3

(d) 1(0 X v)·wl

~O4-~--------~4~~'

37. x+y=-I

39. 2x - y = 0

,

,

P(l,2)

1'(-2, 1)

------f-----~.,

--~2--~~--~~--~, -I -

x+y=-1

43. 3464 J

(I) lu l :

l 1 (b) No (0) Yes (d) No 31. (8) Yes 33. No, v need not equal w. For example, i + j '" -i + j, but i X (i + j) = i X i + i X j = 0 + k = k and i X (-i + j) = -i X i + i X j = 0 + k = k. • ,.;;:;; 11 25 35. 2 37. 13 39. v 129 41. T 43. T 45.

t

47.

vp

49. If A = ali + a2j aod B = hli + h2j, thoo

2J

i -I

41. 5 J

(e) (u X v) X (u X w)

j

k

I-J

45.

'IT

"4

47.

'IT

"6

aod the triangle's area is

49. 0.14

tlA BI ±t I:: ::1· X

Section 12.4, pp. 686-688 1. 10 x vi = 3,directionisti +

tj

7.IOXVI=6Vs,directinois~idirection is - _1_ i + ....2.... k Vs Vs

The applicable sign is ( + ) if the acute angle from A to B runs couoterclockwise in the xy-plane, and ( - ) if it runs clockwise.

+ tk;lv x ul = 3,

directioois-.2. i _Ij -.2.k 3 3 3 3. lu x vi = 0, no direclino; Iv X ul = 0, no direction 5. 10 X vi = 6, direction is -k; Iv X 01 = 6, directioo is k .vsk;lvxol=6Vs,

=

Section 12.5, pp. 694-696 1. 3. 5. 7.

x=3+1, y=-4+1, z=-I+I x = -2 + 51, y = 51, z = 3 - 51 x=O, y=21, z=1 x = I, y = I, z = I + I 9. x = I, Y = -7 + 21, z = 21 11. x = t, Y = 0, z = 0

A-52 13•

Chapter 12: Answers to Odd·Numbered Exercises

X=I~y=l.

O S I S

Z

= 2' I,

15. x

I

=

I. Y

z=O.

(I,l, l)

I I I

17.

= I + I•

-I SISO

-7iL-----r----+,

,

/

/ / /

I I I

~c_=_>----4il, 1, 0)

'-"-"*"""- :-~/~-~" I I

,

(1.0.0)

/

_____ J/

23.

21.

/

, 17. x

= 0,

z = I.

19. x = 2 - 2t. Y = 21. z=2-21. OSlsl

y = I - 21.

0

,

S

I '" I

O.

_.2. i

D.I = 7; D..,.I = -7; D• .! = 7 where Ul = 41. '1f/Vz 43. (8) lil,2) = 1,(1,2) = 2 , 45.

(b) 14/5

vfl(O,_l,I) - J+211:

, vfl(o,_l,_lj - J-2k

t

j

-

v TVi

+ ~ k;

.3. j

-

.6: k;

777

v TVi


47. Tangent: 4x - y - 5z = 4; normal line: x=2+4t,y= -l-t,z= 1-51 49. 2y - z - 2 = 0 51. Tangent: x + y = 'IT + I; normal line: y = x Y

17. f(x,y) =

~ + 4,g(x,y)

=

A-63

I +~

19. y = 2lnlsinxl + ln2 'IT

+ I

, - -.;+'11"+ I

b

21. (8) _ (2i + 7j) v53 1 2 23. w = e -C -rr t sin 1TX

(b)

~ (98i -

127j + 58k)

29,097

CHAPTER 15 53. x= 1-21,y= I,z= 1/2+21 55. Answers will depend 00 the upper bound used for

57. 59. 61.

63. 67. 69.

71. 73. 75. 77. 79. 81.

Section 15.1, pp. 840-841

Ifni, Ifryl, If", I· WithM = v2/2, lEI'" 0.0142. With M = I, lEI'" 0.02. L(x,y,z) = y - 3z,L(x,y,z) = x + y - z - I Be more careful with the diameter. dI = 0.038, % change in 1= 15.83%, more sensitive to voltage change (8) 5% 65. Local minimum of -8 at( -2, -2) Saddle point at (0, O),f(O, 0) = 0; local maximum of 1/4 at (-1/2, -1/2) Saddle point at (0, 0), f(O, 0) = 0; local minimum of -4 at

0

---co,¥-- ---i---» o

25. ~ln2 2 29. 8

27. -1/10 53. 1/(80".)

31. 2".

•

u

-,

v _ _p

u_p

-,

/,2

p

y

(-11'/3,2)

(O.5.0.~

(~-'l)

4/.(4-Y)/2 dxdy

,•

-,•

o

>

0.'

-'1:-), _ 1

-x-]= 1 -1

y (1,1)

4 -,=4-11: 2

57. 4/3 67.

y=x

y-'"

(1,2)

61. 16

II I I I

/

(l,t)

z=l-

tx-t,

,L-----

16>; dx dy

y

+ In2)

•

/

'

,

/

~------------~~,~>

/

/ /

9

/

3

(1,1)

---::ot---;----»

-01-o +~ -----»

41.1' [~3Ydydx

43. /.'

-1)0

y

__

65. 2(1

['[(~ 39. Jo Jo

Jl 1my

~~

63. 20

2

37. [' [' dx dy

,+ ,-e"

59. 625/12

y

-.ot-+----------»

~+-

0

,

L'

xy dx dy

69.

1 71.".'

77.

/. '1'->

(x'

o

x

73. -

;2

+ y')dydx

75.

20r

=4 -

3

y

y

____

-L~>

~----_f.--------~-»

45.

>

-1

35. lL>dydx

0

y

-1

x+y - l

-x+y - I

(-2,-2)

33.

,

55. -2/3

y

[" [' (x+y)dydx

it Jinx

y

47.2 y (11",'11")

'IT

y- >

o ~+.,------------~-»

.'

•

79. Risthe set of points (x,y) such that x' + 2y' < 4. 81. No, by Fubini's Theorem, the two orders of integration must give the same result. 85. 0.603 87. 0.233


Section 15.3, p. 852 1.

1,'1,'-' dydx = I,'I,'-Y dxdy =

13. 12

2 or

3

1'1-Y' dxdy=Jt, 2

'-2y-2

2

15.

,

-12

(12,6)

,2=3%

y=~

, _ linx

12

x ~

0

NOf TO SCALE

•

y - x+2

x- -y'J(-4,-2)

-,

17.

1

(11'/4, -v2J2)

2

0

---~4----~----~0~'

,

,

y= COlI X

,

(-1,1)

,

V2 -

A-65

3 2

x

,

--oJ o ------~,--~X

{'"'r 5. Jo Jo dy dx

=

7.

1

'1,y-r dxdy-3 _1 I, o

,

' \ (2,-1)

r

(1, I)

---cojL-----~----+x

--;Cot--:"lnco,--------+x

------~~~~,--~x

y - -~

19. (a) a (b) 4/~ 21. 8/3 23. 40,000(1 - .-')In (7/2) '" 43,329

Section 15.4, pp. 857-859 1.

9.I,'1'Y 1 dxdy = 4

I

11.

'1" 1'Y

1,o x/2 1 dydx + ' 1,o y/2 Idxdy+

,

, _2

,

~r~

9

3.

*" ~ (} ~ 3:. 0 ~ r ~ esc 8

;;/2

x

1

)'/2

1

~

'"

2' a '" r

13. 36

23.

= 3

~

'" 2 cos 0

v'3

15. 2 -

19. (2 In 2 - I) (~/2)

/,'l'-x dydx 2 /,'l'-Y Idxdy=2 1

-2 '" 0

11. 2~

I I

,

2w, 0

~

7.

y·ll:

y_x

~

5. 0 '" 0 '"

{' r 1 dydx + i2f' )x/3(' 1 dydx = 4 ,

(J

~, 1 '" r '" 2 v'3 sec 0; 671' 0 (x2 + y2)' (c) 9/2 (0) 0

/2i -

+ y2)'

(a) 9/2 (b) 13/3 (a) 1/3 (b) -1/5 (a) 2 (b) 3/2 (0) 1/2 -15/2 15. 36 17. (a) -5/6

(b) 0 (0) -7/12 1/2 21. -17 23. 69/4 25. -39/2 27. 25/6 (a) Cire, = 0, eire, = 217, flux, = 217, flux2 = 0 (b) Cire1 = 0, cirC2 = 8'I'T, flUXI = 8w, flUX2 = 0 31. Circ = 0, flux = a 217 33. Cire = a 217, flux = 0

1"

35. (a) 39.

.,

(0) I

(b) 0

37. (a) 32

(b) 32

(0) 32

Y


11 %' 2

I. (a) _,

6

%

-

11-%'].%' 2

2

(b)

x dy fix

_,

6

0

%

dz dy ~ --

(0) 125/4

3. 217

7. (a) Hole radius = I, sphere radius = 2

5. 317/2

(b) 4V317

9. 17/4

11.

In(£)

15.

I/~

41. (a) G 43. F

17. Mass 10 19.

= a 2 oos-'(£)

= ~4 cos-'(£)

';b (eo'h' -

25. h =

I)

- bv'a 2 - b 2,

~ Va2 _

_

21. (b) I

=

(t) '-?J

Section 16.1, pp. 90S-907

11.~

9. Vi

(£)

17. V3ln 21.

!~ (e16

13. 3\1 14

31.

i

(17'/2 - I)

27.

\O~ -

= 217Vi~

(b) Vi (b) I,

+

51. 0

53.

"2I

+2+2z +C

15. -16 =

5. Not couservative

2

+C

V(x2; I)

+ tau (x + y) + tln(y2 + z2) + C

17. 1

19.9ln2

29. (a) I

21.0

(b) 1

23.-3

(0) I

f(B) - f(A) = a(xb - xa)

+ b(yb

- ya)

+ c(zb

- za) =

Section 16.4, pp. 940-942

2

29. 8

In (I + Vi)

= 417Vi~

+ y2F

F·lk.

33. 2Vi - I

35. (a) 4Vi - 2 39. (a) I,

+ 9)

23. 17 (40'/2 - 13 3/2)

25. i(5 3/2 - 7Vi - 1)

v'x 2

31. (a) 2 (b) 2 33. (a) c = b = 2a (b) c = b = 2 35. It does not matter what path you use. The work will be the same on any path because the field is conservative. 37. The force F is couservative because all partial derivatives of M, N, andP are zero. f(x,y, z) = ax + by + cz + C; A = (xa,ya, za) andB = (xb,yb, zb). Therefore, JF' dr =

(b) .1..(173/2 - I) 12

19. (a) 4Vs

- e64 )

7. Graph (f)

15. i(5Vs

3y2

11. f(x,y,z) = xlnx - x

CHAPTER 16 5. Graph (d)

=

3. Not couservative 2

9. f(x, y, z) = xeY+:z.

27. F

1

47.48

7. f(x,y,z)=x

27. 217[t -

3. Graph (g)

xi+yj v'x2 + y2

1. Conservative

13. 49

1. Graph (c)

(b) G

-yi

Section 16.3, pp. 929-931

b2 _ b: (a2 _ b2)'/2

(c) 0

V20 in., h = V60 in.

+ xi

=

37. I,

41. 1%

=

= 217

217~a3 - 2

1. 5. 9. 13. 17. 29.

Flux = 0, eire = 2'7Ta 2 3. Flux = -71'a 2 , eire = 0 Flux = 2, eire = 0 7. Flux = -9, eire = 9 Flux = -11/60, eire = -7/60 11. Flux = 64/9, eire = 0 Flux = 1/2, eire = 1/2 15. Flux = 1/5, eire = -1/12 0 19. 2/33 21. 0 23. -1617 25. l7a 2 27. 317/8 (a) 0 if C is traversed counterclockwise (b) (h - k)(area oftheregiou) 39. (a) 0

A-69


Section 16.5, pp. 951-953 1. r(r, 0) ~ (r cas O)i

+ (r sin O)j + r'I casO)i + (V3sin4>sinO)j +

~ 1'~1~ sin' 4> cas' 0 d4> dO ~ ~

3. II x' du

s

~ 1'1' (4 -

5. IlzdU (falX

~

U,Y

~

17.

[IV;

~

rdrdO

19.1'~1' TV's drdO ~

7. II x'v'S - 4zdu

21.1'~14 I dudv ~ 61T

8",vS

23. Jo Jo uv'4u' + I du dv

['~J.~ 2 sin 4> d4> dO ~

Jo

~ 1'1'~ u'cas'v·v'4u' + I·

s

+ Idvdu~ [' ['~ u'(4u' +

uv'4u'

~~

11.

9. 9a'

1 15. 30 (Vi 19. -32

abe 4 (ab +

+ 6v'6)

,

21. "';

27. -73,,/6

ae

+ be)

I) cos' v dvdu

~

11" 12

13. 2

17. v'6/30 23. 13a 4/6

,

29. 18

31. "'; 39. -4

25. 2",/3

,

33. "': 41. 3a 4

"'r

['~ ['

25.

~ 3V3

v)

(V3 cos 4»1 :5 2,,/3,0 :5 0 :5 2",

9. r(x,y) ~ xi + yj + (4 - y')k, 0 :5 X :5 2, -2 '" Y '" 2 11. r(u, v) ~ ui + (3 cas v)j + (3 sin v)k, 0 :5 u :5 3, o s v ::s 21T 13. (a) r(r, 0) ~ (r cas O)i + (r sin O)j + (1 - rcosO - r sin 0)1olic functions, 416-421

def'mition of, 416-417 derivatives of, 417-418, 420-421 graphs of, 419 identities for, 416-417, 419-420 integrals of, 417-418 """",",418-419,420-421 six basic, 417 Hyperl>olic paraboloid, 698, 699 Hyperl>oloids, 697, 699 i-component of vector, 669 Ideal gas law, 824 Identity function, 7, 48, 969 Image, 887 Implicitdiff=ntiation, 149-153,779-781 furmuia for, 780

Implicit Function Theorem, 781, 948 Implicit surfaces, 948-950 Improper integrals, 478-486 approximations to, 485 of Type I, 478, 486 ofJYpe II, 481, 486 Increasing function, 6, 199--200 Increment Theorem for Functions of Two Variable., 771,AP-38-AP-40 Increments, AP-IO-AP-13 Indef'mite integrals, 235-236 See also Antiderivatives defmition of, 235, 284, 285 evaluation with substitution rule, 284--290 Indepeodentvariableoffunction, 1,747, 824-825 Indeterminate furms oflimils, 396-402, 600-@1 Indeterminate powers, 400 Index of sequence, 532 Index of summation, 256, 550 Induction, mathematical, AP-6--AP-9 Inequalities, rules for, AP-l solving of, AP-~AP-4 Inertia, moments of, 870-873 Infinite discontinuities, 75 Iofinite (unboonded) interwia, 6, AP-3 Iofinite limits, 89-90 def"mition of, precise, 90-93 of integration, 478-480 Infinite sequence, 532-533. See also Sequences Infinite series. 544-550 Iofinitesimals, AP-19 Infinity, divergence of sequence to, 535 limits at, 84-93 of rational functions, 86

Inflection, point of, 188, 204-206, 209 Initial poin~ of curve, 6 \0 of vector, 665

Initial ray in polar coordinates, 627 Initial speed in projectile motion, 718 Initial value problems, 233, 390, 497 Inner products_ See Dot product Input variable of function, 1,747 Instantaneous rates of change, 43--44 derivative as, 124-125 Instantaneous speed, 39-41 Instantaneous velocity, 125-126 Integer ceiling function (Least integer function), 5 Integer floor function (Greatest integer function), 5 Integers, AP-26 positive, ~rule for, 116-117 starting, AP-8 Integrable functions, 264-265, 837, 859 Integral furm, product rule in, 436-439 Integral sigo, 235 Integral tables, 46~64, T-I-T-6 Integral test, 55~556 error estimation, 555-556 remainder in, 555-556 Integral1heorcms for vector fields, 980-981 Integrals, approximation of, by Riemann sum, 261 by lower swns, 248 by midpoint rule, 248, 469 by Simpson's Rule, 470-472 by Trapezoidal Rule, 469-470 by upper sums, 248 Brief Table of, 435 def'mite. See Defmite integrals double. See Double integrals evaluated with inverse trigonometric functiona, 411-412 f(l/u) du, 373-374 for fluid foree against vortical flat plate, 342 of hyperbolic functions, 417-418 improper, 478-486 approximations to, 485 of Type I, 478, 486 of Type II, 481, 486 indef'Inite, 235-236, 284-285 involving lo&zx, 384-385 iterated, 838 line. See Line integrals logaritlun def'med as, 370 multiple, 836-1!37 substitution in. 887-894 nonelemeotary, 466, 598-599 polar, changing Cartesian integrals into, 855--1!57 in polar coordinates, 853-854 of powers of tan x and soc x, 446-447 of rate, 278-280 repeated, 838 of sin2 x and cos2 x, 289-290

surface, 953-960, 965 table of, 463-464, T-I-T-6 of tan x, cotx, sec x and cscx, 374-375 trigonometric, 444-448 triple. See Triple integrals

1-7

of vector fields, 910 of vector functions, 715-717 work, 337-338, 727-728, 912-914 Integraods, 235 with vortical asymptote~ 481-482 Integrate connnand (CAS), 465 Integrating factor, 505-507 Integration, 246-307 basic fannulas, 435 with CAS, 465-466 in cylindrical coordinates, 875-879 and diffi:rentiation, relationship between, 280 formulas, 411-413 limits of. See Limits, of integration numerical,468-475 by parts, 436-441 by parts formula, 436-437 of rational functions by partial fractions, 45~61

with respect to y, area between CID'VeS, 296-297 in spbctical coordinates, 881--1!83

tabular, 440-441 techniques of, 435-495 term-by-term forpower selie., 581 by trigonometric substitution, 449-452 variable of, 235, 264 in vector fields, 901-981 of vector function, 716-717 Intetior poin~ 806, AP-3 continuity at, 74 for regions in plane, 749 for regions in space, 751 Intermediate Value Property, 80 Intermediate Value Theorem, AP-24 for continuous functions, 80-81, 217, 274-275 Intermediate variable, 776-777 Intersection,lines of, 692-693 Intersection of sets, AP-2 Interval of convergence, 579 Intervals, AP-3 differentiable on, 109-110 parameter, 6HJ-{i11 types of, AP-3 Inverse equations, 378, 384 Inverse function-inverse cofunction identities, 410-411 Inverse functions, 11,361,362-363 derivative rule for, 365 and derivatives, 364-366 fmding, 36~364 bypeIi>olic,418-419,420-421 one-to-one, 361 trigonometric. See Inverse trigonometric functions Inverse trigonometric functions, 404-413 cofunctionidentities, 410-411 def'mition of, 404-405 derivatives of, 411 integrals evaluated with, 411-412 Inverses, def'Ining of, 404-405 integration and differentiation operations, 280 oflnx and number e, 377-378 of tan x, cob, sec x, and csc x, 407-408

1-8

Index

Inational numbers, AP-2 as exponents, 377-378

Limit circle, 526

Jacobi, Carl Gustav Iacob, 887

Limit Comparison Ths~ 483. 484. 560-561 Limit Laws. 46-54. 757 llieorem, 49 Limit Power Rule, 49 Limit Product Rule, proof of, AP-18--AP-19 Limit Quotient Rule. proof of. AP-I 7. j + j + j + j eotxdx + j + j a2 dx+ x2 ltan-l~ + a a j Va dx+ x sinb-l~ + 3.

=

5.

aX

In Ixl

C (a

=

In Isinxl

13.

=

15.

eosbxdx = sinbx

17.

2

2

10.

C

12. 14.

C

C

16.

C

=

19.

8.

C

esexeolxdx = -esex

11.

6.

0, a "" 1)

C

ese2 xdx = -eolx

=

j

xn+l

x"dx=n+I+C (n "" -I)

4.jeX dx=e X +C

C

eosxdx = sinx

9.

2.

18.

C (a> 0)

20.

j sinxdx j j j tanxdx j sinbxdx

-eosx

=

see2 xdx = tanx

+C

+C

seextanxdx = seex In Iseexl

=

j

• / 2dx va

x2

-

j V dx j Vx dx a x

x2 2

-

eosbx

=

-

a2 2

+C +C

+C

=. sm-1 ~ a + = =

C

~ see-1 I~ I + C eosh- 1 ~

+

C (x

> a > 0)

Forms Involving ax + b 21. 22.

j (ax + b)" dx

(ax + b)"+1 a(n + 1) + C, n "" -1

=

1

j

" _ (ax + b)"+1 [ax + b b x(ax + b) dx a2 n + 2 - n + 1 + C,

23. j(ax + bt 1dx = 25.

27.

j x(ax + W 2dx j

=

~In lax + bl

+C

1

:2 [ In Iax + b I + ax : b + C

" 2 (Vax + b (Vax + b) dx = a n+2

(2

+ C, n "" -2

n "" -1,-2

24.

jx(ax+b)-ldx=~- :21nIax+bl

26.

j x(ax~ b) tin lax: bl + C

28.

+C

=

j~ dx=2~+bj xVaxdx + b T-l

T-2

A BriefTable of Integrals

29. (a) / 30.

ax

x~

Vb

2

X

/

34. /

a

2

ax +x

0

2

- Vb I + C

Vax+b+Vb

/~ ax=- ~ x

Forms Involving 32.

_I_In I~

=

+'!/ 2

ax

x~

(b) /

+ C

_ atan I -1 x a+ C

ax

Va + x

2

sinh-l~ + C =

=

35. /Va2+x2ax=IVa2+x2+

31 /

·

ax x 2Vax + b

33 /

ax (a 2 + x 2)2

•

In (x + Va 2 +

x2 )

Va 2 + x 2 _ a In Ia + V: + x21 + C

38. /

2 Va: 2+ x ax

=

In (x + Va 2 + x 2) _ Va /

39./

2 2 2 x ax=_a 1n (x+Va 2 +x2)+xVa +x2+c Va2 + x2 2 2

42.

/

44. /

0

2

2

41. /

43/

Ix

·

sin-l~ + C

46. /x 2Va 2 - x 2 ax =

~sin-l~ -

ax

(a 2 _ x 2)2 2

2

x2

Forms Involving il 52. / 53. /

V. 2ax x

-

a2

=

Vx 2 - a 2 ax

Va 2 - x 2 2 +C ax 0

2

In Ix + Vx 2 - a 2 + C 1

=

,+

2

Va x +C ax

IVx2 - a2 -

=

x

2a2(a2 _ x2) 2

+_I_Inlx+al +C 4a3

2+

x- a

~2 sin-l~ + C

kXVa 2 - x 2 (a 2 - 2x2) + C

2

=

= _

45. / Va - x ax = IVa - x

50. /

.

+C

x + _I-tan-l~ + C 2a 2(a 2 + x 2) 2a3 a

2

x 2Va + x 2

2

ax x 2Va 2 -

=

ax2

47. /Va -x ax=Va2_x2_alnla+V:2-x21+c 48. /Va: -x

51 /

ax

xVax + b

il

-

I In x + a 2 ax _ x 2 = 2a - al a +C

2 x

2b

2 x +C

-!In la + V:2 + x21 + C

=

Va:X- x 2 =

-"-/

C

2

=

Forms Involving

~ bx

+

ag' In (x + Va 2 + X2) + C

2 2 Va x + x ax

xVa + x 2

= -

b

+C

37. /

ax2

~tan-l~ax Vb b

~1n(x+Va2+x2)+C

~(a2 + 2x 2)Va 2 + x 2 -

36. /x 2Va 2 + x 2 ax =

40. /

=

+ il

2 -

2

ax

x~

~21n Ix +

Vx 2 - a 2 + C 1

2

ax=-sin-l~-

2 2 Va -x +C x

ax = _! In Ia + V:2 - x21 + C xVa 2 - x 2

A BriefTable ofIntegrals

54.

(Yx 2 - a 2)" dx =

J

55J dx • (Yx 2 -a 2)"

x(Yx2 - a 2)n n+ I -

(v' 2 =xx-a

2t

(2-n)a 2

nn~

J

2

I

(Yx 2 - a2)"-2 dx,

n'¢' -I

n

_n-3J dx (n-2)a 2 (Yx 2 -a 2)"

56. Jx(YX 2 - a 2)" dx

(Yx2 - a 2 )"+2 = n+2

57. J x 2Yx 2 - a 2 dx

i(2x 2 - a 2)Yx 2 - a 2 - as'1n Ix + Yx 2 - a 21 + C

58.

2

J

Yx2x - a dx = v'x 2

2 59. J YX: 2- a dx 60. 61.

=

J J "Ii

Yx2 _ a2

x

-a2

dx

x2

dx

-

-

a sec-I Ixl a +C

1- YX 2x-

2 a + C

=

In Ix + Yx 2 _ a 2

=

2 a 1n Ix + Yx 2 - a 21 + "-Yx 2 - a 2 + C 2 2

2

x

a2

-

+ C, n'¢'-2

62.

aI sec-llxl a

Trigonometric Forms 63. JSinaxdx

65. J sin2axdx 67.

68•

J J

sin" ax ax

J

(b)

J

(c)

72. 74.

J

. J J J

64.

C

=

~ - s~~ax + C

=

smn-I : cos ax

.

· ax + n dx -- cosn axI sm cosaX' na

69. (a)

70.

-~cosax +

=

IJ "--=---IJ

+

n ;

n

2

-

x

C

sin2ax

cosn-2 ax dx

a2

4

b2

r

2 2 a '¢' b

cos2ax 4a + C

~In Isin axl

2

cosaxdx-2+~+C

sin(a - b)x sin(a + b)x 2(a _ b) + 2(a + b) + C,

C?sax dx = smax

~sinax +

cos ax dx =

2 2 a ,¢,b

sm ax cos ax dx = -

71.

+ C

73.

sinax I cosaxdx = -aln Icosaxl + C

J

J J

.. sin(a - b)x sin(a + b)x smaxsmbxdx= 2(a-b) - 2(a+b) +C, cosaxcosbxdx =

2

dx = Yx - a + C 2 x 2Yx 2 -a 2 ax

sin,,-2 ax dx

cos(a + b)x cos(a - b)x . smaxcosbxdx = - 2(a + b) - 2(a _ b) + C,

. n-I ax cosm+1 ax sm am ( +n) 76.

66.

J

J J

+ -n + - I m

sin",+l ax COS",-l ax m- 1 sinn ax cosmax dx = aim + n) + m+ n

n

.

,,-2

SID

J J

On _sinn+lax sm axcosaxdx - (n + I)a + C, PI·

-

n'¢'-I

cosn+1ax

cos axsmaxdx - - (n + I)a + C,

m dx axcos ax , n '¢' -m

sinn ax cosm-2 ax dx,

m '¢' -n

(reduces sinn ax) (reduces cosmax)

n'¢'-I

T-3

T-4


77. J b + :inax

=

78. J

=

79.

81.

dx. b + csmax

J J

1 + dx sinax =

dx

=

dx b + ccosax

=

83. J 1+ %sax

89. J tanaxdx

~In Isecaxl

X,-I cos ax dx

~In Isinaxl

x + C

92. J cot'axdx

=

94. J cot" ax dx

= -

:'~ '0 - J ~In Isec ax +

n" 1

tan,-2 ax dx,

96. Jcscaxdx

tanaxl + C

J ~tanax 2J " J cs~~:~c~ax if , ----.uzJ sec,,-2 ax tan ax n a{n _ 1) + n _ 1

100. J csc'axdx

sec ax tan ax dx -

sec"ax

,-2

sec

+ : ::::

=

x: sin ax -

~sinax + ~J

=

X,-I sin ax dx

+ C

-~cotax -

x + C

:(:~ '0 - J

-~In Icscax +

=

C

-~cotax +

Cot"-2 ax dx, n" 1

cotaxl + C

C

ax dx, n" 1

n" 1

csc"-2 ax dx,

n"O

+ C,

=

98. J csc2 axdx

+ C

sec ax dx -

:2 cos ax +

=

90. J cotaxdx

=

101.

86. J xcosaxdx

-~cota; + C

=

+ C

93. J tan' ax dx

99.

%sax

=

~tanax -

sec2 axdx =

b2 c

+ C,

v'

aVc2 _ b2

=

T)]

2 c2 In Ic + bsinax + - b cosaxl + C, aVc2 _ b2 b + csmax

-1

-a1 tan (7T 4 - 2ax)

b+ccosax

82.J

aVb~2_ c 2 tan-I [~~ ~ ~ tan(; -

102.

, J

- ----.uz-

csc ax cot ax dx - -

csc"ax

+ C,

n"O

Inverse Trigonometric Forms 103. J sin-Iaxdx

=

X sin-I ax +

~Vl

105. J tan-I ax dx

=

Xtan-I ax -

2~ In (1

J J J

11

•

-1

-

x ll + 1

• -1

- a 2x2 + C + a2x 2 ) + C a

J J J

106.

x sm axdx--+1 sm ax--+1 n n

107.

X,+I a x'cos-Iaxdx = n + 1 cos-lax + n + 1

108. x' tan-I ax dx

=

x"+l

104. J cos-Iaxdx = X COS-I ax -

a

n + 1 tan-I ax - n + 1

x"+ 1 dx

,I

'

vI - a 2x 2 X,+I dx VI _ a2x 2 '

x"+l dx 1 + a 2x 2'

n"-1 n"-1

n"-1

~Vl

- a2x 2 + C

A BriefTable ofIntegrals

Exponential and Logarithmic Forms 109.

111.

j eax dx j

=

xe ax dx

~eax + c

=

eax

-;;0 (ax

110.

-

114. je ax sinbx dx =

/ax 2(acosbx+bsinbx)+C

116.jlnaxdx=Xlnax-x+c

a +b

n+1

x '(lnax)mdx

b > 0, b oF 1

b > O,b oF 1

117. jX"(lnax)mdx = x"+l(lnax)m - ----"'----jX"(lnax)m-' dx

j

Ibax

a In b + C,

2eax 2 (a sinbx - b cos bx) + C a +b

l1s.jeaxcosbxdx=

118.

j

=

- I) + C

~i:,: a~bjX"-lbaxdx,

113. jX"baxdx =

bax dx

n+1

(lnax)m+l m + 1 + C,

=

m oF -I

n oF-I

'

119.

dx xlnax

j

=

lnllnaxl + C

Forms Involving Y2ax - xl, a> 0 120. 121.

j Y2axdx - x sin-1 (x ~ a) + C j Y2ax - x 2 dx x ; a y2ax - x 2 + 'fsin-1 (x ~ a) + C 2

=

=

"

2 122. j ( Y2ax - x ) dx = 123

dx

•j

124. 125.

(Y2ax - x2)"

=

na 2 j ( (x - a)( Y2ax - X2)" 2)"-2 n+ 1 + n+ 1 Y2ax - x dx

(x - a)(Y2ax - x2)2-" n - 3 j + ----""-~2 2 (n - 2)a (n - 2)a

dx

(Y2ax _ x2)" 2

j xY2ax _ x 2 dx (x + a)(2x - ~a)Y2ax - x 2 + ~ sin-1 (x ~ a) + C j Y2~ - x 2 dx Y2ax _ x 2 + a sin-1 (x ~ a) + C j Y2a;2 - dx -2 ~2a ; sin-1 (9) + C j x Y2axdx -x j Y2axxdx- a sin-l (x ~ a) - Y2ax - x 2 + C =

=

2

126. 127.

x

x -

=

x2

=

128.

~coshax + C

130.

= 2

_1~2a a

- x +C x

Hyperbolic Forms 129. jSinhaxdx = 131. 133.

j sinh2axdx j sinh" ax dx

=

S~2ax -

=

sinh"-l ~ cosh ax - n -;; 1

I+C

j cosh ax dx j cosh2ax dx n °

132.

j sinh"-2 ax dx,

~ sinh ax + C

=

oF

=

s~;ax + I + C

T-5

T-6


134. J cosh" ax dx

cosh"-l:: sinh ax

=

J cosh"-2 ax dx,

+ n -;;

1

135.Jx sinh ax dx ~ cosh ax - :2 sinh ax + C 137.Jx" sinh ax dx x; cosh ax - ~Jx"-l cosh ax dx 139.Jtanhaxdx ~ln(coshax) + C 141.J tanh2 ax dx x - ~ tanh ax + C =

=

~~-;;;:

+

144.] coth" ax dx

=

(~th~-;;;:

+

145'] sechaxdx

J J

tanh"-2 ax dx,

n i' 1

coth"-2 ax dx,

n i' 1

=

sech"-2 ax tanh ax (n _ l)a

=

=

2eax

[ ebx e-bx ] a + b - a _ b

2eax

[ ebx a +b

+

bx ea _ b

~ sinh ax -

:2 cosh ax

x" cosh ax dx = x; sinh ax -

~ln Isinh axl =

x -

X

=

w/2

1

157. o

f(n) = (n - 1)1,

1W

/2

sin" x dx =

0

cos" x dx =

n

x"-l sinh ax dx

+

C

-~cothax + C

2 1

4

T-

T-

n i' 0

+

152.

J

csch" ax coth ax dx = -

csch" ax

na

C,

1+ C,

>

°

156.1""e-ax ' dx

1.3.5 ..... (n - 1) 7r •2·4·6· .... n 2' 2 • 4 • 6 ••••• (n - 1) 3.5.7 •...• n '

1

J

~

~ coth ax + C

Some Definite Integrals

155.1""x"-le- dx

C

+C

~ln Itanh~1

cschaxdx =

+

J sech"-2 ax, dx 1 n n - 2J h"-2 dx 41 ax, n n- 1 esc

+ nn -_

= -

. 153.Je ax sinhbxdx 154.Jeaxcoshbxdx

=

csch2 axdx =

ax coth ax (n1 )a 50J esch" ax dx CSCh"-2 151.J sech" ax tanh ax dx sec~= ax + C,

1 .

J

146.

~ tanh ax + C

147.J sech2 ax dx 149.J sech" ax dx --

138.

J 148. J

~sin-l(tanhax) + C

=

Jx cosh ax dx =

=

=

136.

140. J cothaxdx 142. J coth2 ax dx

=

143'] tanh" ax dx

n i' 0

=

t.J%,

if n is an even integer ;,,2 ifn is an odd integer;,,3

a

>0

+

C,

ni'O

CREDITS Page ii, photo, Forest Edge, Hokuto, Hokkaido, Japan 2004 © Michael Kenna; Page 1, photo, Getty Images; Page 39, photo, Getty Images; Page 102, photo, Getty Images; Page 133, Section 3.4, photo for Exercise 19, PSSC Physics, 2nd ed., DC Heath & Co. with Education Development Center, Inc.; Page 178, Chapter 3 Practice Exercises, graphs for Exercise 94, NCPMF "Differentiation" by W. U. Walton et al., Project CALC, Education Development Center, Inc.; Page 181, Chapter 3 Additional and Advanced Exercises, photo for Exercise 9, APIWide World Photos; Page 184, photo, Getty Images; Page 246, photo, Getty Images; Page 308, photo, Getty Images; Page 347, Figure 6.44, PSSC Physics, 2nd ed., DC Heath & Co. with Education Development Center, Inc.; Page 361, photo, Getty Images; Page 435, photo, Getty Images; Page 496, photo, Getty Images; Page 532, photo, Getty Images; Page 547, Figure 10.9b, PSSC Physics, 2nd ed., DC Heath & Co. with Edocation Development Center, Inc.; Page 610, photo, Getty Images; Page 660, photo, Getty Images; Page 707, photo, Getty Images; Page 722, Section 13.2, photo for Exercise 35, PSSC Physics, 2nd ed, DC Heath & Co. with Education Development Center, Inc.; Page 747, photo, Getty Images; Page 836, photo, Getty Images; Page 901, photo, Getty Images; Page 908, Figures 16.6 and 16.7, NCFMF Book ofFilm Notes, 1974, MIT Press with Education Development Center, Inc.; Page 909, Figure 16.15, InterNetwork Media, Inc., and NASAlJPL; Page AP-l, photo, Getty Images.

C-1

6653_AWLThomas_EPp01-07 7/29/09 7:30 PM Page 3

BASIC ALGEBRA FORMULAS Arithmetic Operations asb + cd = ab + ac,

ac a#c = b d bd

a c ad + bc + = , b d bd

a>b a d = #c b c>d

Laws of Signs a a -a = - = b b -b

-s - ad = a, Zero Division by zero is not defined.

0 If a Z 0: a = 0,

a 0 = 1,

0a = 0

For any number a: a # 0 = 0 # a = 0 Laws of Exponents a ma n = a m + n,

sabdm = a mb m,

sa m dn = a mn,

n

a m>n = 2a m =

n aBm A2

If a Z 0, am = a m - n, an

a 0 = 1,

a -m =

1 . am

The Binomial Theorem For any positive integer n, sa + bdn = a n + na n - 1b + +

nsn - 1d n - 2 2 a b 1#2

nsn - 1dsn - 2d n - 3 3 a b + Á + nab n - 1 + b n . 1#2#3

For instance, sa + bd2 = a 2 + 2ab + b 2,

sa - bd2 = a 2 - 2ab + b 2

sa + bd3 = a 3 + 3a 2b + 3ab 2 + b 3,

sa - bd3 = a 3 - 3a 2b + 3ab 2 - b 3.

Factoring the Difference of Like Integer Powers, n>1 a n - b n = sa - bdsa n - 1 + a n - 2b + a n - 3b 2 + Á + ab n - 2 + b n - 1 d For instance, a 2 - b 2 = sa - bdsa + bd, a 3 - b 3 = sa - bdsa 2 + ab + b 2 d, a 4 - b 4 = sa - bdsa 3 + a 2b + ab 2 + b 3 d. Completing the Square If a Z 0, ax 2 + bx + c = au 2 + C The Quadratic Formula

au = x + sb>2ad, C = c -

If a Z 0 and ax 2 + bx + c = 0, then x =

-b ; 2b 2 - 4ac . 2a

b2 b 4a


GEOMETRY FORMULAS A = area, B = area of base, C = circumference, S = lateral area or surface area, V = volume Triangle

Similar Triangles c'

c h

a'

Pythagorean Theorem a

c

b

b' b

b

a' b' c' abc

A 1 bh 2

Parallelogram

Trapezoid

a a2 b2 c2

Circle a

h h

A ␲r 2, C 2␲r

r

b b A bh

A 1 (a b)h 2

Any Cylinder or Prism with Parallel Bases

Right Circular Cylinder r

h

h h

V Bh

B

B V ␲r2h S 2␲rh Area of side

Any Cone or Pyramid

Right Circular Cone

h h

Sphere

h s r

V 1 Bh B

3

B

V 1 ␲r2h 3 S ␲rs Area of side

V 43 ␲r3, S 4␲r2


LIMITS General Laws

Specific Formulas

If L, M, c, and k are real numbers and

If Psxd = an x n + an - 1 x n - 1 + Á + a0 , then

lim ƒsxd = L

and

x:c

lim gsxd = M,

x:c

then

lim Psxd = Pscd = an c n + an - 1 c n - 1 + Á + a0 .

x:c

lim sƒsxd + gsxdd = L + M

Sum Rule:

x:c

lim sƒsxd - gsxdd = L - M

Difference Rule:

x:c

If P(x) and Q(x) are polynomials and Qscd Z 0, then

lim sƒsxd # gsxdd = L # M

Product Rule:

Psxd Pscd = . Qscd x:c Qsxd lim

x:c

Constant Multiple Rule: Quotient Rule:

lim sk # ƒsxdd = k # L

x:c

lim

x:c

ƒsxd L = , M gsxd

M Z 0 If ƒ(x) is continuous at x = c, then

The Sandwich Theorem

lim ƒsxd = ƒscd.

x:c

If gsxd … ƒsxd … hsxd in an open interval containing c, except possibly at x = c, and if lim gsxd = lim hsxd = L,

x:c

x:c

lim

x:0

then limx:c ƒsxd = L.

sin x x = 1

and

lim

x:0

1 - cos x = 0 x

Inequalities

L’Hôpital’s Rule

If ƒsxd … gsxd in an open interval containing c, except possibly at x = c, and both limits exist, then

If ƒsad = gsad = 0, both ƒ¿ and g¿ exist in an open interval I containing a, and g¿sxd Z 0 on I if x Z a, then

lim ƒsxd … lim gsxd.

x:c

x:c

lim

x:a

Continuity

ƒsxd ƒ¿sxd = lim , gsxd x:a g¿sxd

assuming the limit on the right side exists.

If g is continuous at L and limx:c ƒsxd = L, then lim g(ƒsxdd = gsLd.

x:c


DIFFERENTIATION RULES General Formulas

Inverse Trigonometric Functions

Assume u and y are differentiable functions of x. d Constant: scd = 0 dx dy d du Sum: + su + yd = dx dx dx dy d du Difference: su - yd = dx dx dx d du Constant Multiple: scud = c dx dx d dy du Product: + y suyd = u dx dx dx

d 1 ssin-1 xd = dx 21 - x 2

d 1 scos-1 xd = dx 21 - x 2

d 1 stan-1 xd = dx 1 + x2

d 1 ssec-1 xd = dx ƒ x ƒ 2x 2 - 1

d 1 scot-1 xd = dx 1 + x2

d 1 scsc-1 xd = dx ƒ x ƒ 2x 2 - 1

du dy y - u d u dx dx a b = dx y y2 d n x = nx n - 1 dx d sƒsgsxdd = ƒ¿sgsxdd # g¿sxd dx

Quotient: Power: Chain Rule:

Hyperbolic Functions d ssinh xd = cosh x dx d stanh xd = sech2 x dx d scoth xd = - csch2 x dx

d scosh xd = sinh x dx d ssech xd = - sech x tanh x dx d scsch xd = - csch x coth x dx

Inverse Hyperbolic Functions Trigonometric Functions d ssin xd = cos x dx d stan xd = sec2 x dx d scot xd = - csc2 x dx

d scos xd = - sin x dx d ssec xd = sec x tan x dx d scsc xd = - csc x cot x dx

d 1 ssinh-1 xd = dx 21 + x 2

d 1 scosh-1 xd = 2 dx 2x - 1

d 1 stanh-1 xd = dx 1 - x2

d 1 ssech-1 xd = dx x21 - x 2

d 1 scoth-1 xd = dx 1 - x2

d 1 scsch-1 xd = dx ƒ x ƒ 21 + x 2

Exponential and Logarithmic Functions

Parametric Equations

d x e = ex dx d x a = a x ln a dx

If x = ƒstd and y = gstd are differentiable, then

d 1 ln x = x dx d 1 sloga xd = dx x ln a

y¿ =

dy>dt dy = dx dx>dt

and

d 2y dx

2

=

dy¿>dt . dx>dt


INTEGRATION RULES General Formulas a

Zero:

ƒsxd dx = 0

La a

Order of Integration:

b

ƒsxd dx = -

Lb b

Constant Multiples:

ƒsxd dx

La

b

kƒsxd dx = k

La

La

b

b

-ƒsxd dx = -

La

La

b

sƒsxd ; gsxdd dx =

La b

Additivity:

sk = - 1d

ƒsxd dx

b

Sums and Differences:

sAny number kd

ƒsxd dx

La

b

ƒsxd dx ;

c

ƒsxd dx +

La

gsxd dx

c

ƒsxd dx =

ƒsxd dx La Lb La Max-Min Inequality: If max ƒ and min ƒ are the maximum and minimum values of ƒ on [a, b], then b

min ƒ # sb - ad …

La

ƒsxd dx … max ƒ # sb - ad.

b

Domination:

ƒsxd Ú gsxd

on

[a, b]

implies

La

b

ƒsxd dx Ú

La

gsxd dx

b

ƒsxd Ú 0

on

[a, b]

implies

La

ƒsxd dx Ú 0

The Fundamental Theorem of Calculus x

Part 1 If ƒ is continuous on [a, b], then Fsxd = 1a ƒstd dt is continuous on [a, b] and differentiable on (a, b) and its derivative is ƒ(x); x

F¿(x) =

d ƒstd dt = ƒsxd. dx La

Part 2 If ƒ is continuous at every point of [a, b] and F is any antiderivative of ƒ on [a, b], then b

La

ƒsxd dx = Fsbd - Fsad.

Substitution in Definite Integrals b

La

Integration by Parts gsbd

ƒsgsxdd # g¿sxd dx =

Lgsad

ƒsxd g¿sxd dx = ƒsxdgsxd D a -

b

ƒsud du

La

b

b

La

ƒ¿sxdgsxd dx

6653_AWLThomas_Last3Bkpgs 7/29/09 7:31 PM Page 2

tan A + tan B 1 - tan A tan B tan A - tan B tan sA - Bd = 1 + tan A tan B

Trigonometry Formulas

tan sA + Bd = y

1. Definitions and Fundamental Identities y 1 sin u = r = Sine: csc u Cosine:

x 1 cos u = r = sec u

Tangent:

y 1 tan u = x = cot u

P(x, y) r 0

y

␪ x

x

sin aA +

p b = cos A, 2

sec2 u = 1 + tan2 u,

sin 2u = 2 sin u cos u, 1 + cos 2u , 2

csc2 u = 1 + cot2 u

cos 2u = cos2 u - sin2 u sin2 u =

1 - cos 2u 2

sin A cos B =

1 1 sin sA - Bd + sin sA + Bd 2 2

sin A + sin B = 2 sin

1 1 sA + Bd cos sA - Bd 2 2

sin A - sin B = 2 cos

1 1 sA + Bd sin sA - Bd 2 2 1 1 sA + Bd cos sA - Bd 2 2

cos A - cos B = - 2 sin

cos sA - Bd = cos A cos B + sin A sin B

1 1 sA + Bd sin sA - Bd 2 2

y

y y sin x

y cos x

Trigonometric Functions Radian Measure

Degrees

1

兹2 45

C ir

兹2

1

␲

3␲ 2␲ 2

x

–␲ – ␲ 2

␲ 2

0

␲

3␲ 2␲ 2

x

Domain: (–, ) Range: [–1, 1]

1

␲ 4

90

r

␲ 2

y

1

y

y tan x

it cir cl

y sec x

e

Un

1

θ

␲ 2

0

y sinx Domain: (–, ) Range: [–1, 1]

␲ 4

45 s

–␲ – ␲ 2

Radians

p b = - sin A 2

cos aA +

1 1 cos sA - Bd + cos sA + Bd 2 2

cos A + cos B = 2 cos

cos sA + Bd = cos A cos B - sin A sin B

p b = sin A 2

1 1 cos sA - Bd - cos sA + Bd 2 2

sin sA + Bd = sin A cos B + cos A sin B sin sA - Bd = sin A cos B - cos A sin B

cos aA -

cos A cos B = cos s -ud = cos u

sin2 u + cos2 u = 1,

cos2 u =

p b = - cos A, 2

sin A sin B =

2. Identities sin s -ud = - sin u,

sin aA -

s cle of ra diu

1

r ␲ 6

30 兹3

2

60

u s s or u = r , r = 1 = u 180° = p radians .

90 1

– 3␲ –␲ – ␲ 2 2 兹3

2 ␲ 3

␲ 2

0 ␲ ␲ 3␲ 2 2

x

Domain: All real numbers except odd integer multiples of ␲/2 Range: (–, )

– 3␲ –␲ – ␲ 0 2 2

␲ ␲ 3␲ 2 2

x

Domain: All real numbers except odd integer multiples of ␲/2 Range: (–, –1] h [1, )

1 y

y

y csc x

The angles of two common triangles, in degrees and radians. 1 –␲ – ␲ 0 2

y cot x

1 ␲ 2

␲ 3␲ 2␲ 2

Domain: x 0, ␲, 2␲, . . . Range: (–, –1] h [1, )

x

–␲ – ␲ 0 2

␲ 2

␲ 3␲ 2␲ 2

Domain: x 0, ␲, 2␲, . . . Range: (–, )

x


SERIES Tests for Convergence of Infinite Series 1. The nth-Term Test: Unless an : 0, the series diverges. 2. Geometric series: g ar n converges if ƒ r ƒ 6 1; otherwise it diverges. 3. p-series: g 1>n p converges if p 7 1; otherwise it diverges. 4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test or the Limit Comparison Test.

5. Series with some negative terms: Does g ƒ an ƒ converge? If yes, so does gan since absolute convergence implies convergence. 6. Alternating series: gan converges if the series satisfies the conditions of the Alternating Series Test.

Taylor Series q

1 = 1 + x + x 2 + Á + x n + Á = a x n, 1 - x n=0

ƒxƒ 6 1

q

1 = 1 - x + x 2 - Á + s -xdn + Á = a s - 1dnx n, 1 + x n=0

ƒxƒ 6 1

q

ex = 1 + x + sin x = x cos x = 1 -

xn x2 xn + Á + + Á = a , 2! n! n = 0 n!

ƒxƒ 6 q

q s - 1dnx 2n + 1 x5 x3 x 2n + 1 + - Á + s -1dn + Á = a , 3! 5! s2n + 1d! n = 0 s2n + 1d! q s - 1dnx 2n x4 x2 x 2n + - Á + s -1dn + Á = a , 2! 4! s2nd! s2nd! n=0

ln s1 + xd = x -

ƒxƒ 6 q

ƒxƒ 6 q

q s - 1dn - 1x n x2 x3 xn + - Á + s -1dn - 1 n + Á = a , n 2 3 n=1

-1 6 x … 1

x 2n + 1 x5 1 + x x3 x 2n + 1 + Á + = 2 tanh-1 x = 2 ax + + + Á b = 2a , 5 1 - x 3 2n + 1 n = 0 2n + 1 q

ln

tan-1 x = x -

q s - 1dnx 2n + 1 x5 x3 x 2n + 1 - Á + s -1dn + + Á = a , 5 3 2n + 1 2n + 1 n=0

ƒxƒ 6 1

ƒxƒ … 1

Binomial Series s1 + xdm = 1 + mx +

msm - 1dsm - 2dx 3 msm - 1dx 2 msm - 1dsm - 2d Á sm - k + 1dx k + Á + + Á + 2! 3! k!

q m = 1 + a a b x k, k=1 k

ƒ x ƒ 6 1,

where m a b = m, 1

msm - 1d m a b = , 2! 2

msm - 1d Á sm - k + 1d m a b = k! k

for k Ú 3.


VECTOR OPERATOR FORMULAS (CARTESIAN FORM) Formulas for Grad, Div, Curl, and the Laplacian Cartesian (x, y, z) i, j, and k are unit vectors in the directions of increasing x, y, and z. M, N, and P are the scalar components of F(x, y, z) in these directions.

The Fundamental Theorem of Line Integrals 1.

Let F = Mi + Nj + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then there exists a differentiable function ƒ such that 0ƒ 0ƒ 0ƒ i + j + k F = §ƒ = 0x 0y 0z B

2.

if and only if for all points A and B in D the value of 1A F # dr is independent of the path joining A to B in D. If the integral is independent of the path from A to B, its value is

0ƒ 0ƒ 0ƒ §ƒ = i + j + k 0x 0y 0z

Gradient

B

LA Divergence

§#F =

F # dr = ƒsBd - ƒsAd.

0N 0P 0M + + 0x 0y 0z Green’s Theorem and Its Generalization to Three Dimensions

Curl

Laplacian

i

j

k

0 § * F = 4 0x

0 0y

0 4 0z

M

N

P

§2ƒ =

0 2ƒ 0x

2

0 2ƒ + 0y

2

Normal form of Green’s Theorem:

C

Divergence Theorem:

0 2ƒ + 0z

2

F # n ds = § # F dA F 6 6

Tangential form of Green’s Theorem:

su *

= sv *

u * sv * wd =

su # wdv -

= sw *

ud # v

Stokes’ Theorem:

9

§ # F dV

D

F # dr = § * F # k dA F 6

C

wd # u

F # n ds =

S

Vector Triple Products vd # w

R

R

F # dr = § * F # n ds F 6

C

S

su # vdw

Vector Identities In the identities here, ƒ and g are differentiable scalar functions, F, F1 , and F2 are differentiable vector fields, and a and b are real constants. § * s§ƒd = 0 §sƒgd = ƒ§g + g§ƒ § # sgFd = g§ # F + §g # F § * sgFd = g§ * F + §g * F § # saF1 + bF2 d = a§ # F1 + b§ # F2 § * saF1 + bF2 d = a§ * F1 + b§ * F2 §sF1 # F2 d = sF1 # §dF2 + sF2 # §dF1 + F1 * s§ * F2 d + F2 * s§ * F1 d

§ # sF1 * F2 d = F2 # § * F1 - F1 # § * F2 § * sF1 * F2 d = sF2 # §dF1 - sF1 # §dF2 + s§ # F2 dF1 - s § # F1 dF2 § * s§ * Fd = §s § # Fd - s § # §dF = §s§ # Fd - §2F 1 s§ * Fd * F = sF # §dF - §sF # Fd 2

Thomas' Calculus (12th Edition)

Thomas' Calculus, Multivariable (12th Edition)

Thomas' Calculus, Single Variable (12th Edition)

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Thomas' Calculus, 11th Edition

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Solution Manual Thomas' Calculus, 11th Edition

Thomas' Calculus. SOLUTION MANUAL

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Social Psychology 12th Edition

Managerial Accounting, 12th edition

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Lonely Planet Mexico, 12th Edition

Engineering Mechanics: Statics (12th Edition)

Builder's Reference Book 12th Edition

Simpson's Forensic Medicine, 12th edition

Diseases of Poultry 12th Edition

Langman's Medical Embryology, 12th Edition

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Principles of Marketing, 12th Edition

Calculus, 4th Edition

Calculus, 3rd Edition

Calculus, 7th Edition

Calculus, 7th Edition

Thomas' Calculus (12th Edition)

Thomas' Calculus, Multivariable (12th Edition)

Thomas' Calculus, Single Variable (12th Edition)

Thomas' Calculus, 11th Edition

Thomas' Calculus, 11th Edition

Thomas' Calculus, 11th Edition

Thomas' calculus

Thomas Calculus

Solution Manual Thomas' Calculus, 11th Edition

Thomas' Calculus. SOLUTION MANUAL

International Economics, 12th Edition

Human Biology, 12th Edition

Understanding Nutrition (12th edition)

Social Psychology 12th Edition

Managerial Accounting, 12th edition

Fundamentals of Selling, 12th Edition

Lonely Planet Mexico, 12th Edition

Engineering Mechanics: Statics (12th Edition)

Builder's Reference Book 12th Edition

Simpson's Forensic Medicine, 12th edition

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