Thomas' Calculus, Single Variable (12th Edition)

THOMAS' CALCULUS Twelfth Edition Single Variable Based on the original work by George B. Thomas, Jr. Massachusetts I...

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

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THOMAS'

CALCULUS Twelfth Edition

Single Variable 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

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For permission to use copyrighted matetial, grateful acknowleclgmeot is made to the copyright bolders on page C-I, which is hereby made part of this copyright page. Many ofthe designstions used by manufacturers and sellers to distiogoish their products are claimed as trsdemarks. 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 978-0-321-63742-0 I. CaicolUll-Textbooks. I. Hass, Joel. II. Thomas, George B. (George Brinton), 1914--2006. III. Thoroas, George B. (George Brinton), 1914--2006. Calcolus. Iv. Title V. Title: Calcolus. QA303.2.W452009b 515-dc22

2009023069

Copyright 0 2010, 2005, 2001 Pearson Education, Inc. All rights reserved. No part oflitis publication may be reproduced, stated in a retrieval system, or transnti!ted, in any form or by any means, electrortic, mechaeical, 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

-

PEARSON www.pearsoned.com

ISBN-10: 0-321-63742-9 ISBN-13: 978-0-321-63742-0

CONTENTS Preface

1

I

ix

1

Functions 1.1 1.2 1.3 1.4

Functions and Their Graphs I Combining Functions; Shifting and Scaling Graphs Trigonometric Functions 22 Graphing with Calco1ators and Computers 30 QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

34

35 37

AoDffiONAL 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

AoDffiONAL 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 ofTrigonometric Functions 135 The Chain Rule 142

iii

iv

Contents

3.7 3.8 3.9

Implicit Differentiation 149 Related Rates 155 Linearization and Differentials

164

QuESTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

ADDITIONAL AND ADvANCED EXERCISES

4

184

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

198

I Integration

246 5.1 5.2 5.3 5.4 5.5 5.6

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

6

180

Applications of Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7

5

175

176

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 VHopital's Rule 396 Inverse Trigonometric Functions 404 Hyperbolic Functions 416 Relative Rates of Growth 424 QuEsTIONS TO GUIDE YOUR REVIEW 429 PRACTICE EXERCISES 430 ADOffiONAL AND ADvANCED EXERCISES 433

Integration by Parts 436 Trigonometric Integrals 444 Trigonometric Substitutions 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 ADOffiONAL AND ADvANCED EXERCISES 491

453

496

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

Infinite Sequences and Series 10.1 10.2 10.3 10.4 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 '!bsts

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 ofPlane 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 ofTaylor Series 589 The Binomial Series and Applications ofTaylor Series 596 QuESTIONS TO GUIDE YOUR REVIEW 605 PRACTICE EXERCISES 605 ADDITIONAL AND ADvANCED EXERCISES 607

Curves in Space and Their Tangents 707 Integrals ofVector Functions; Projectile Motion 715 Arc Length in Space 724 Curvatore andNorma1 Vectors ofa Curve 728 Tangential and Normal Components ofAcceleration 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 ofMass 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 Al A.2 A3 AA A.5 A6 A7 A.8 A9

Real Numbers and the Real Line AP-I Mathematical Induction AP-6 Lines, Circles, and Parabolas AP-1O Proofs of Limit Theorems AP-18 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 1'H6pital'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 organization of the eleventh edition in placing antiderivatives as the rmal 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 ofthe 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.com/thomas. • Series We retain the organizational structure and content ofthe 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 ofvectors 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 coonections of multivariable ideas with their single-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.

No matter what Y positive number £ is, the graph enters this band at x = and stays.

1

y=l £

I---~.,----------''--'-----

-=-----.---------" 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 I), the range is {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 realvalued. The domains and ranges ofmany real-valued functions of a real variable are intervals or combinations ofintervals. The intervals may be open, closed, or half open, and may be finite or infinite. The range of a function is not always 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 ofa function as a machine. For instance, the Y!;;key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the y!;; key. A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element ofthe 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).

element in D.

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 = y!;;

y=~ y =

'\IT=7

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

[0, I]

Solution The formula y = x 2 gives a real 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;;,: 0. The formula y = I/x gives a real y-value for every x except x = 0. For consistency in the rules ofarithmetic, we cannot divide any number by zero. The range of y = I/x, the set ofreciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = I/(I/y). That is, for yolO the number x = I/y is the input assigned to the output valuey. The formula y = '\Ii; gives a real y-value only if x ;;,: 0. The range of y = '\Ii; 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 = '\f"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 '\f"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 - I to I. Outside this domain, I - x 2 is negative and its square root is not a real number. The values of I - x 2 vary from 0 to I on the given domain, and the square roots ofthese values do the same. The range of ~ is [0, I]. •

Graphs of Functions If j is a function with domain D, its graph consists of the pnints 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 ofpoints 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 1.4). y

y

---;01---'-;---'-;;--\--I-;---x

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

EXAMPLE 2

Graph the function y

=

FIGURE 1.4 If(x,y) lies on the graph of f, theo the value y = f(x) is the height of the graph above the point x (or below x if f(x) is uegative).

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). • How do we know that the graph of y = x 2 doesn't look like one ofthese curves? y

----'---'---~....-:---'---'---~x

-2

-I

FIGURE 1.5

Example 2.

0

2

Graph ofthe function in

y

4

Chapter 1: Functions 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 numerieally, 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 tuning 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 plotred points gives a graph of the pressure function represeoted 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

Functions and Their Graphs

y

5

y

-I

-_C-f---~f---+----jl--x

---!---+--~_x

-I

1

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 grapb ofa function f(x) = ~. (c) The lower semicircle is the grapb ofa functiong(x) ~ -~.

y

y=x

Piecewise-Defined Functions __!____c:--!--c?I''----~__!____c:___x

-3 -2 -I

0

2

3

FIGURE 1.8 The absolute value function bas domain (- 00, 00) and range [0, 00). y

Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value function

Ixl =

x;;"

°

x < 0,

The function

f(x) =

y

x, -x,

whose graph is given in Figure 1.8. The right-hand side of the equation means that the function equals x if x ;;" 0, and equals -x if x < 0. Here are some other examples.

EXAMPLE 4

FIGURE 1.9 To grapb the function y = f(x) shown here, we apply different formulas to different parts ofits domain (Example 4).

{

{-:~, I,

x I

is defined on the entire real line but has values given by different formulas depending on the position of x. The values of f are given by y = -x when x < 0, Y = x 2 when :5 x :5 I, and y = I when x > I. The function, however, is just one jUnction whose • domain is the entire set ofreal numbers (Figure 1.9).

°

,

,'Y=x 3

,

,~

,':"""'"

2

~

y~LxJ

"

-_-!2~_~I-,~'-~-c2e--!3~--x

,

"

......-0

"

EXAMPLE 5 The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer 1100r function. It is denoted l x J. Figure 1.10 shows the graph. Observe that

l2.4J = 2, l2J = 2,

-2

FIGURE 1.10 The graph of the greatest integer function y = l x J lies on or below the line y = x, so it provides an integer floor for x (Example 5).

ll.9 J lo.2J

I, = 0, =

lOJ = 0, l-0.3 J = -I

l-1.2J = -2, l-2J = -2.

•

EXAMPLE 6 The function whose value at any number x is the smallest integer greater than or e9uul to x is called the least integer function or the integer ceiling function. It is Figure 1.11 shows the graph. For positive values of x, this function might denoted I represent, for example, the cost of parking x hours in a parking lot which charges $1 for • each hour or part of an hour.

xl.

6

Chapter 1: Functions y

Increasing and Decreasing Functions

3

/ 'y=x

7

2

0---;-' ,

y

~

If the graph of a function climbs or rises as you move from left to right, we say that the function is increasing. If the gmph descends or falls as you move from left to right, the function is decreasing.

rxl

/

---:----9----,If--:---7---;:-~x

,

-2 - I ,

I

2

DEFINmONS Let f be a function defmed on an interval I and let X, and X2 be any two points in I.

3

e--;" -I

7/

-2

FIGURE 1.11 TIre graph ofthe leasl integer function y = rxl lies on or above the line y = x. so it provides an integer ceiling for x (Example 6).

1.

If f(X2)

2.

If f(X2)

>
1 + x·

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

D 68.

64. The accompanying figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long.

a. Express the y-coordinate of P in terms of x. (You might start by writing an equation for the line AB.)

b. Express the area of the rectangle in terms of x. y

a. Graphthefunctionsj(x)

= 3/(x - l)andg(x) = 2/(x + 1) together to identify the values of x for which

_3_0

0

1, the graph is scaled:

y

=

cf(x)

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

y

=

cI f(x)

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

y

=

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) f( -x)

= -

y =

EXAMPLE 4

Reflects the graph of f across the x-axis. Reflects the graph of f across the y-axis. Here we scale and reflect the graph of y =

y/;;.

(a) Vertical: Multiplying the right-hand side of y = y/;; by 3 to get y = 3 y/;; 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 = 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 = ~ = 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.

v'3

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

h

y

y/;; across the x-axis, and • y

y 4

3 y~

2

1 ~~:::::::::~::::::"-YF

Yx

----"----=-F-~-_o____cL---'--~x

FIGURE 1.32 Vertically stretching and compressing the graph y = Vi hy a factor of3 (Example 4a).

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

0

2

3

-3

-2

-1

v;j3

~~~=;='x -1 0 1 2 3 4

-1

-_;;____c:-~_+___cc-_;;____c:-~x

4

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

(Example 4c).

18

Chapter 1: Functions Given the functioof(x) = x 4

EXAMPLE 5

-

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

20

-1

x

0 -10

=

16x4 + 32x 3 + 10 y

\

y

20

-2

10 x

1

0

-1

-10

-20

+ 2x3 - 5

y = _!x 4 2

x

0 -10

-20 (a)

(b)

(c)

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

c 2x 2 y

+ y2

= r 2.

(1)

y

y

r

-+---OO+-----f-~x

r

-r

r C

-r

(a) circle

FIGURE 1.36

-r

-r

(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 gmph 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 (±r/c, 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 (±r/c, 0).1n both cases, the major axis is the longer line segment. Ifwe divide both sides of Equation (I) by r 2 , we obtain

y

b

--+---4>::----+-----> X -a

19

a

-b (2)

FIGURE 1.37 2 X

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

Graph of tire ellipse

y2

+ 2 = l.a > b,wherethemajor

a b axis is horizontal.

(x - h)2

+

-'----a"2-'--

(y - k)' b2 = 1.

(3)

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 Algobraic Combinations f·g· I. f(x) = x, 2. f(x) =

=

g(x) = v;=J v;+l, g(x) = v;=J

I, g(x)

=

I

9. f(x) = vx

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

- x

Let f(x) = x - 3, g(x) = \/;;, h(x) = x', and j(x) = lx. Express each of the functions in Exercises 11 and 12 as a composite involving one or more of f,g, h, andj.

11. a. y

c. y

+ \/;;

= \/;; -

3

= x l/ 4

12.a.y=lx-3 2

5. If f(x) = x + 5 andg(x) = x - 3, fmd the following. a. f(g(O» b. g(f(O»

c. f(g(x» e. f(f( -5))

d. g(f(x»

g. f(f(x»

b. g(g(x»

r.

b. y =

Z\/;;

d. y

4x

r.

e. y = V(x - 3)'

Composites of Functions

=

y = (lx - 6)'

b.y=x'j2

LY=~

d.y=x-6

e. y = Z~ r. y = W-=3 13. Copy and coroplore the following table.

g(g(Z»

(f

g)(x)

g(x)

f(x)

I/(x + I), fmd the following.

a. x - 7

\/;;

?

a. f(g(l/Z» c. f(g(x» e. f(f(Z»

b. g(f(I/Z»

b. x + Z

3x

?

d. g(f(x))

c.

~

~

r.

b. g(g(x»

x x - I

?

g. f(f(x»

x d. -I x-

e.

?

1+ 1 x

x

f.

x

?

x

6. If f(x)

=

x - I andg(x)

=

g(g(Z»

In Exercises 7-10, write a formula for fog

0

h.

+ I, g(x) = 3x, h(x) = 4 - x 8. f(x) = 3x + 4, g(x) = lx - I, h(x) = x 2

xI

x2 .~ g(x) = - 2 - - ' h(x) = V Z - x x +I

x+Z 10. f(x) = -3- ,

In Exercises 3 and 4, fmd the domains and ranges of f, g, fig, and glf. 3. f(x) = Z, g(x) = x 2 + I

4. f(x)

I

• ,---c-;

In Exercises I and Z, fmd tire domains and ranges of f, g, f + g, and

7. f(x) = x

?

I

0

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)

f(x)

a. -I-

Ixl

x - I

(f

0

g)(x)

?

b.

?

x - I x

c.

?

V;

d.

V;

?

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.

a. j(g( -I))

g(j(O» e. g(j( -2))

b.

d. g(g(2»

a. y

j(j( -I)) f. j(g(l))

c.

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

g(j(3» e. g(j(O»

a. j(g(O» d.

b.

j(j(2»

~ ~,

g(x)

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

~

e. g(g(-l» f.

j(g(1/2)) 0

f and

t

V;

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

x-

(f

0

d. Y = (x

-2sx Period:

'1T

(c) y

y - cotx

\v

----,t:-=-tn+-+-c":-±-~x

Odd

_3

sin( -x) tan(-x) csc( -x) cot (-x)

= = =

-sinx -tanx -cscx

= -COlx

0

Domain'• X "" -2'-2 + 'Ir + 31T •... Range: y:s -1 ory ~ 1 Period: 2'lT (d)

Domain: x "'" 0, ±1f, ±21f, ••• Range: y::5 -lory 2:: 1 Period: 27l'

Domain: x "'" 0, ±'1f, ±2'1f, ... Range: _00 < Y < 00 Period: '1T (I)

(oj

FIGURE 1.46 Graphs ofthe six hasic trigonometric functions using radian measure. The shading for each trigonometric function indicates its periodicity. y

Trigonometric Identities The coordinates of any point P(x, y) in the plane can be expressed in terms of the point's distance r from the origin and the angle fJ that ray OP makes with the positive x-axis (Figure 1.42). Since xlr = cosfJ andylr = sinfJ, we have -t--~""","b~--t-~x

x = rcosfJ,

y

=

rsinfJ.

When r = I we can apply the Pythagorean theorem to the reference right triangle in Figure 1.47 and obtain the equation

FIGURE 1.47 The referee.. triangle for a general angle (}.

(3)

26

Chapter 1: Functions TIris equation, true for all values of IJ, is the most frequently used identity in trigonometry. Dividing this identity in turn by cos2 1J and sin2 1J gives

I I

+ tan2 IJ + cot' IJ

= sec2 IJ = csc2 IJ

The following formulas hold for all angles A and B (Exercise 58).

Addition Formulas cos(A sin(A

+ B) + B)

= cosA cosB - sinA sinB = sinA cosB

+ cosA sinB

(4)

There are similar formulas for cos (A - B) and sin (A - B) (Exercises 35 and 36). All the trigonometric identities needed in this book derive from Equations (3) and (4). For example, substituting IJ for both A and B in the addition formulas gives

Double-Angle Formulas cos 21J = cos2 1J - sin2 1J sin 21J = 2 sin IJ cos IJ

(5)

Additional formulas come from combining the equations cos2 1J

+

sin2 1J = I,

cos2 1J - sin2 1J = cos 21J.

We add the two equations to get 2 cos2 1J = I + cos 21J and subtract the second from the first to get 2 sin2 IJ = I - cos 21J. TIris results in the following identities, which are useful in integral calculus.

HaIf-Angie Formulas cos2 1J = I

+

cos 21J 2

(6) (7)

The Law of Cosines If a, b, and c are sides of a triangle ABC and if IJ is the angle opposite c, then

(8)

TIris equation is called the law of cosines.

1.3

B(a cos 8, a sin 8)

c 2 = (acosll- b)2 = a 2(cos211 C

27

We can see why the law holds if we introduce coordinate axes with the origin at C and the positive x-axis along one side of the triangle, as in Figure 1.48. The coordinates of A are (b, 0); the coordinates of B are (a cos II, a sin II). The square ofthe distance between A and B is therefore

y

---~

Trigonometric Functions

+

(a sin 11)2

sin211)

+ b2 -

Zabcosll

I

......--:---?~;C-~X b A(b,O) a2

=

FIGURE 1.48 The square ofthe distance betweeo A and B gives the law of cosines.

+

+ b2

-

Zabcosll.

The law of cosines generalizes the Pythagorean theorem. If II = andc 2 = a 2 + b 2 •

7T/Z,

then cos II = 0

Transformations of Trigonometric Graphs The rules for shifting, stretching, compressing, and reflecting the graph of a function summarized in the following diagram apply to the trigonometric functions we have discussed in this section.

/Vcrtical

Vertical stIm:h or compression.; reflection about x-axis ifnCgative~

y = af(b(x

+ c)) + d

~

Horizontal stretch or CODIJlI"Ssion; / reflection abouty-axis if negative

shift

Horizontal shift

The transformation rules applied to the sine function give the general sine function or sinusoid formula

f(x) = A Sin(Z; (x -

C») + D,

where IA I is the amplitude, IB I is the period, C is the horizontal shift, and D is the vertical shift. A graphical interpretation ofthe various terms is revealing and given below. y

y = A sin

(2;<x - C») + D This axis is the liney=D.

o Two Special Inequalities For any angle II measured in radians,

-1111 :5 sinll:5 1111

and

-1111 :5 I - cos II :5 1111.

28

Chapter 1: Functions

To establish these ioequalities, we picture fJ as a nonzero angle io standard position (Figure 1.49). The circle io the figure is a unit circle, solfJl equals the length ofthe circular arc AP. The length oflioe segmentAP is therefore less than IfJl. Triangle APQ is a right triangle with sides of length

y p

QP = IsiofJl,

From the Pythagorean theorem and the fact that AP

+--___co++~-+___c~X

A(l,O)

cos 6

AQ = I - cosfJ.

2

sio fJ

I-cosO

+

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

6'

and

(I - cos fJ)2 :5

6'.

By taking square roots, this is equivalent to sayiog that Isio fJl :5 IfJl

and

II - cosfJl :5 IfJl,

-lfJl :5 siofJ:5 IfJl

and

-lfJl :5 I - cosfJ :5 IfJl.

so

These inequalities will be useful io the next chapter.

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

4. If you roll a I-m-diameter wheel forward 30 em over level ground, through what aogle will the wheel tum? 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'fr/2

-2'fr/3

o

'fr/2

-'fr/6

'fr/4

S1f/6

sinO

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

xe[f,?T] 9ocosx=t, xe[-f,o] t, ['If, 7.

Sinx=~,

11. tanx =

fJ

-'fr/3

XE

3;]

8. tanx=2,

xe[o,f]

5 10ocosx=-1 3' xe[f,'T1'] 12. sinx =

-t,

XE

['If, 3;]

3'fr/4

sinO

cosO tan 8 cot 8 sec 0 esc 8 6. Copy aod complete the following table of function values. If the function is und.efmed 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

-1)

18. - cos 21TX

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"..

- 2

Graph the functions in Exercises 23-26 in the Is-plane (I-axis horizontal, s-axis vertical). What is the period of each function? What symmetries do the graphs have?

23. s

= cot 21

24. s

= -tao".1

25. s

= seC(~I)

26. s

= csc(f)

D 27.

D 28.

Graph y = taox and 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 and range of l sin x J?

= l sin x J together. What are the domain

30. Graph y = sinx and y and range of sin x 1?

= rsinx 1together. What are the domain

I-) + f)

= sinx

33. Sin(x

=

32. cos(x

~

+

34. Sin(x -

cosx

35. cos(A - B) = cosA cosB different derivation.)

36. sin(A - B)

1-) = -sinx f) -cosx

+ sinA sinB (Exercise 57 provides a

sinA cosB - cosA sinB

In Exercises 39-42, express the given quantity in terms of sin

cosx.

+ x)

40. sin(2". - x)

x)

42. cose;

+

x)

. 7". . (". ".) 43· Evaluatesffi12assm 4+3.

44.

Evaluate cos

45. Evaluate cos

If; ~.

~

0

tao(A

54. cos20

+ cosO

~

0

+ B)

taoA+taoB ~ I - taoA taoB'

Derive the formula.

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

as cos

x --+--=------'""""'---'-L--+--_+ o

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

38. What happeos if you take B = 2". in the addition formulas? Do the results agree with something you already koow?

41. Sin(3; -

53. sin20 - cosO

=

37. What happens if you take B = A in the trigonoroetric identity cos(A - B) = cosA cosB + sinA sinB? Does the result agree with something you already koow?

39. cos(".

52. sin2 0 = cos2 0

1

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

31. cos(x -

t

Theory and Examples 55. The tangent SDID formula The staodard formula for the taogent of the sum of two sogles is

a. Graph y = cos x and y = sec x together for - 3"./2 ,;; x ,;; 3"./2. Comment on the behavior of sec x in relation to the signs and values of cos x. b. Graphy = 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.

r

51. sin2 0 =

x and

0)

to obtain the addition formula for sin(A

+ B).

h. Derive the formula for cos (A + B) by suhstitoting - B for B in the formula for cos (A - B) froro Exercise 35. 59. A triangle has sides a = 2 and b = 3 and angle C = 60°. Find the length of side c.

60. A triangle has sides a

~

2 aod b

~

3 aod angle C

~

40°. Find

the length ofside c.

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

61. The law of sines

sinA

8inB

sin C

abe .

Use the accoropanying figures and the identity sin(". - 0) sin 0 , if required, to derive the law.

(-;f + 2:).

=

A

=

A

46. Evaluate sin ; ; .

Using the Douhle-Angle Formulas Find the function values in Exercises 47-50. 47• COS2 '! 8

48

'2'" 49• sm 12

" 2 3'1T 50• sm 8

2

• cos

5".

B L..--"'a--'-----' C

B £..----:a,.--J

~

62. A triangle has sides a = 2 aod b = 3 and aogle C = 60° (as in Exercise 59). Find the sine of angle 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'lDllerically small values ofx. In Sectioo 3.9, we will see why the approximatioo holds. The approximatioo error is less than I in 5000 if Ixl < 0.1. a. With your grapher in radian mode, graph y = sinx and y = x together in a viewiog window ahout 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

identify A, B, C, and D for the sine functioos 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·lxl'/'.

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

Exercises 1.4 Choosing a Viewing Window

o In

19. f(x)

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.

~

2 x +2 x2 + I x - I

23. f(x) = 6x

[-2, 2] by [-5, 5]

[-5,5] by [-25, 15] [-3, 3] by [-10, 10]

[-S,S]by[-IO, 10]

8. f(x) ~ 4x' - x' 10. f(x) = x 2(6 - x')

12. y 14. y

Ix

16. y

x 17. y = x

-

+3 +2

Chapter

II

~ -2X - 9

x2 - 3 24. f(x) = - - 2

x-

= 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)

x 2 +2x=4+4y- y 2.

18. y

= x ' /'(x 2 - 8) = x 2/'(5 - x) ~ Ix 2 - xl =I

- x

I

+3

= 1.

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

[-10, 10] by [-10, 10]

13. y = 5x 2/' - 2x 15. y ~

+I

8

31. Graph the lower half of the circle dermed by the equation

[-2,6]by[-I,4]

11. y = 2x - 3x 2/' 2

1

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

Exercises 5-30, rmd an appropriate viewing window for the given functioo and use it to display its grsph. x2 x3 5. f(x) = x' - 4x' + IS 6. f(x) = T - 2 - 2x + I

9. f(x) = x~

22. f(x)

-

25. y

[-4,S]by[-IS,25]

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

o In

+ 10

+6

- 15x 4x 2 - lOx

[-20,20] by [-100,100]

Finding a Viewing Window

7. f(x) ~ x' - 5x'

2

x2

~ -2-

X

21. f(x) ~ -x2o"_--x--'_~6

1. f(x) ~ x' - 7x 2 + 6x

a. [-I, I]by[-I, I] b. e. [-10, 10] by [-10, 10] d. 2. f(x) ~ x' - 4x 2 - 4x + 16 a. [-I, I] by [-5,5] b. e. [-S, S] by [-10,20] d. 3. f(x) = S + 12x - x' a. [-I, I]by[-I, I] b. e. [-4,4] by [-20,20] d. 4. f(x) ~ v'S + 4x - x 2 a. [-2,2] by [-2,2] b. e. [-3, 7] by [0, 10] d.

20. f(x)

35. Graph the function f(x) ~ sin 2x

I + 1.

+ cos 3x.

36. Graph the function f(x) ~ sin' x.

o

Graphing in Dot Mode Another way to avoid incorrect conoections 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 37. y ~ x - 3

39. y ~

xlxJ

38. y

~ sin~

40. y

~ -2--

x3

-

X

-

1 I

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 grsph of a real-valued functioo of a real varisble? What is the vertical line test?

3. What is a piecewiSl>-dermed functioo? Give examples.

4. What are the important types of functions frequently encouotered 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 j and g are real-valued functions, how are the domains of j + g, j - g, jg, and j / g related to the domains of j and g? Give examples.

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 = j(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 = j(x) to compress or stretch the graph by a factor c > I? Reflect the graph across a coordinate axis? Give examples.

Chapter

21

13. Graph the six basic trigonometric functions. What symmetries do the graphs have? 14. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions? 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 j(x) = A sin «(2'tT/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 ofa sphere as a function ofthe 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 tracked 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.

10 Exercises 19. y =

= x2

-

2x - I

8. y

= e-"

10 Exercises ~16, determine whether the function is even, odd, orneitheJ: 9. y = x 2 + I 10. y = x' - x 3 - x 11. y

=I

12. y

= sec x tsnx

13 .y

~ L±-L x 3 -2x

14. y

~x

15. y

= x + cosx

16. y

= xcosx

- cosx

- sinx

23. y

~ 2e~

25. y

~

27. y

= 10 (x

Ig

b.

18. If j(a - x) function.

13

~

j(a

c. I(sinx)

d. g(sec x)

+ x), show that g(x)

~

j(x

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

2

20. y = -2

x

2

22. y =

- 3

2sin(3x

24. y

+ 'tT)

- 3)

- I

+I

~

+~ +I

32 -,

tsn(2x - 'tT)

26. y ~ x 2/s 28. y = -I

+ "0/2 - x

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

c. Height above Earth's sealevel as a function ofa1mospheric pressure (assumed nonzero) d. Kinetic energy as a function ofa particle's velocity

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

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

= (x +

I)' d. R(x)~~

b. j(x)

Piecewise-Defined Functions 10 Exercises 31 and 32, fmd the (a) domain and (b) range.

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

a.

Ixl -

21. y = Vl6 -

10 Exercises 5--1l, determine whether the graph of the function is symmetric about the y-axis, the origin, or neither. 5. y = x'i' 6. y = x 2/'

7. y

1~28,

e.

31

_

{h,

Igl

+ a) is an even

-4 S x S 0 0<xs4

.y-~,

-x - 2 32.y=

x,

{

,

-x + 2,

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

36


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

e. S1retch vertically by a 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).

-c<J---~--~x

o

4

Composition of Functions In Exercises 35 and 36, f"md

a. (jog)(-l).

b. (gof)(2).

c. (j ° flex).

d. (g

I x'

35. f(x) =

° g)(x).

I

g(x) = • ~ Yx

b. Y = f(4x)

c. Y = f(-3x)

d. Y = f(2x + I)

e. y = f(t) - 4

f. Y = -3f(x)

In Exercises 37 and 38, (a) write formulas for fog and g ° f and find the (b) domain and (c) range of each.

37. f(x) = 2 - x 2 , g(x) = v;+2 38. f(x) ~ V';;, g(x) ~ ~

51.

y~ -~I +~

53. y

~ ~2 +

39. f(x)

40.

f (x)

~ ~

{

+ I, x _ I,

X

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

-2:5 x < 0 0:5x:52

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

Ii(x) 41. x

Ixl

42. x 2

Ixl

54. Y

~

I-t (-5x)1/3

55. Y

= cos2x

56• y

. x = sm 2

57. y

~

58. Y

~ COST

sin1fX

1fx

(x - f). + (x + f ).

59. Sketch the graph y = 2 cos 60. Sketch the graph y = I

sin

In Exercises 61--{j4, 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

Ix'i

44.x 2 +x

Ix

2

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

+ xl

14 - x

46.

xI

I Ixl

47.

V';;

vfxj

48. sinx

sinlxl

2

1

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~

y~

63. a. Express a in terms ofBand b.

43. x'

a. Up

±

Trigonometly In Exercises 55-58, sketch the graph ofthe given function. Whst is the pcriod of the function?

hex)

45.4-x2

52.

I

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

-x - 2 -I, , { x - 2,

+

In Exercises 51-54, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14-1.17 and applying an appropriate 1ransformation.

+2

g(x) = ~

36. f(x) = 2 - x,

a. y = f(x - 5)

65. Heigbt of a pole 1\vo wires s1retch from the top T of a vertical pole to points B and C on the grouod, where C is 10 m closer to the base of the pole than is B. If wire BTmakes an angle of 35° with the horizontal and wire CT makes an angle of 50° with the horizontal, how high is the pole? 66. Height of a weather balloon Observers at positions A and B 2 Ian apart simultaneously measure the angle of elevation of a weather balloon to be 40° and 70°, respectively. If the balloon is directly above a point on the line segroent between A and B, find the height of the balloon.

D 67.

a. Graphthefunctionf(x) ~ sinx

+ cos(x/2).

b. Whst appears to be the period of litis function? right 3

b. Down 2 units, left ~ c. Reflect about the y-axis

d. Reflect about the x-axis

c. ConIrrm your f"mding in part (b) algebraically.

D 68.

a. Graphf(x) = sin (ilx). b. Whst are the domain and range off?

c. Is f periodic? Give reasons for your answer.

Chapter 1

Chapter

= g I? 0

Give

reasons for your answer. 2. Are there two functions 1 and g with the following property? The graphs of 1 and g are not straight lines but the graph of log is a

straight line. Give reasons for your answer. 3. If I(x) is odd, can anything he said of g(x)

~

I(x) - 2? What if

f is even instead? Give reasons for your answer. 4. Ifg(x) is an odd function defined for all values of x, can anything he said ahout g(O)? Give reasons for your answer. 5. Graph the equation Ix I + Iy I = I 6. Graph the equationy

+ Iyl

+ x. = x + 14

sin x

1

b. Uniqueness Show that there is only one W!I:j to write 1 as the sura of an even and an odd function. (Hint: One W!I:j is given in part (a). If also I(x) = E,(x) + 01(X) where E 1 is even and 01 is odd, show that E - El ~ 01 - O. Then use Exercise 11 to show that E = E1 and 0 = 0 1 .) Grapher Explorations-Effects of Parameters 13. What happens to the graph of y = ax 2 + bx

+ c as

b. b changes (a and c fixed, a # OJ? c. c changes (a and b fixed, a # OJ? 14. What happens to the graph of y = a(x

1 - cosx b. 1 + cosx

+ cosx

where E is an even function and 0 is an odd function. (Hint: LetE(x) = (f(x) + 1(-x»/2. Show that E(-x) = E(x), so thatE is even. Then show that o (x) = I(x) - E(x) is odd.)

a. a changes while b and c remain fixed?

Derivations and Proofs 7. Prove the following identities. 1 - cosx sin x

37

Additional and Advanced Exercises

Functfons and Graphs 1. Are there two functions 1 and g such that log

a.


2

tan

+ b)3 +

x

a. a changes while b and c remain fixed?

2

b. b changes (a and c flXCd, a # OJ?

8. Explain the following ''proof without words" of the law of cosines. (Source: "Proof without Words: The Law of Cosines," Sidney H. Kung, Mathematics Magazine, Vol. 63, No.5, Dec. 1990, p. 342.)

cas

c. c changes (a and b flXCd, a # OJ? Geometry

15. An object's center ofmass moves at a constant velocity v along a straight line past the origin. The accompanying figure shows the coordinate system and the line ofmotion. The dots show positions

a

that are I sec apart. Why are the areas A j, A 2 , ... , As in the figure all equal? As in Kepler's equal area law (see Section 13.6), the line that joins the object's center of mass to the origin sweeps out equal areas in equal times. y

a a

10

9. Show that the area of triangle ABC is given by (1/2)ab sin C = (1/2)bc sin A = (1/2)ca sinB.

c ,,

o

5

A L------,,..-----'B

10. Show that the area of triangle ABC is given by

Vs(s - a)(s - b)(s - c)wheres = (a

+ b + c)/2isthe

10 Kilometers

15

16. a. 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). y

secoiperimeter of the triangle. 11. Show that if1 is hoth even and odd, then I(x) ~ 0 for every x in the domain of I. 12. a. Even-odd decompositions

Let

1 he

a function whose do-

main is symmetric about the origin, that is, -x belongs to the domain whenever x does. Show that 1 is the sura of an even function and an odd function:

I(x) = E(x)

+ O(x),

-o;i~-----~A"(a-,COO):>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 " IsmOcosO

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?

I sinO

0

< I < I cosO"

y (0, I)

-c-ILL8"'----_ _'c-+:=D--=-_~x A

Chapter

E

(1,0)

Technology Application Projects

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

MathematicafMaple Module: ModeIiIIg Change: Springs, Drivillg Safety, RJuJioactivity, 1Tees, 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 obj ect 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, Ga1i1eo 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 fall. It assumes negligible air resistance to slow the obj ect 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 Ga1i1eo'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 of time 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.com/thomas.

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

P from ei1her side.

Find the slope of the parabola y = x 2 at the pointP(2, 4). Write an equation for the tangent to the parabola at this point.

EXAMPLE 3

Solution We begin with a secant line through P(2, 4) and Q(2 + h, (2 + h)2) nearby. We then write an expression for the slope ofthe secant PQ and investigate what happens to the slope as Q approacbes P along the curve: ~y (2 Secant slope = =

+ h)2 - 22

~

h

h2 + 4h

+4- 4

= '"--'-='0-'--"--2

h

=

h

2

~

4h = h

+ 4.

If h > 0, then Q lies above and to the right of P, as in Figure 2.4. If h < 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope h + 4 approaches 4. We take 4 to be the parabola's slope atP. y . (2

Secant slope IS

Tangent slope

=

+ h)2 h

4

=

h + 4.

4

I I I

,Ay ~ (2

Ll

P(2,4) - -

+ h)' -

4

I Ax = h I +""'7'~l' --~--c----~x

------=.....

2+h

NOTTOSCALB

FIGURE 2.4 Finding 1he slope of1he parabola y

limit of secant slopes (Example 3).

= x 2 at 1he point P(2, 4) as 1he

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 = I and t = 2 are called instantaneous rates ofchange. 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 /ip 340 - 150 190 . Averagerateofchange:~ = 45 23 = 22 '" 8.6 flIes/day. p /

350 ~

V

300

1/ -!:J.-p-=r 190 / / /;.p V /;., - 8. ilie day /

" 250

'S

.ll

/ ' Q(45 340)

200

PI ,23,150)

.~ ISO Z

100 /

50

'/

~

&t =

22

./

o

10

20

30

40

50

Time (days)

FIGURE 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of chaoge over 22 days is the slope /ip/ /it ofthe 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 changing 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 ofExample 4 growing

on day 23?

Solution To answer this question, we examine the average rates of change over increasingly short time interva1s 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

= ap/at

Q

Slope of PQ (flies / day)

(45,340)

340 - 150 '" 8 6 45 - 23 .

(40,330)

3~~ ::: ~~O

(35,310)

310 - 150 '" 13 3 35 - 23 .

~

h

/

Q(45,340)

~ '/ P; V

250

'a

i: z

350) // ;-;:-- / ff

300

P(23,

100 /

50

265 - 150 '" 164 30 - 23 .

(30,265) FIGURE 2.6

'" 10.6

B(35,

350

o

/

~~ /llJ

~

150) J rY

~

\

~

20

30

40

50

A(14,0) Time (days)

The positioos and slopes offour secants through the pointP 00 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 t-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

~;O_-I~

= 16.7 flies/day (approximately).

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

•

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 oflength 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 t approached the values t = I and t = 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 limit.

Exercises 2.1 Average Rate. of Change In Exercises I-j), fmd the average rate of chaoge ofthe functioo over the given interval or intervals. 1. f(x) ~ x 3

+

b.[-I,I]

2. g(x)

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

v'49+l;

6. P(O) = 03 - 40'

[0,2]

+ 50;

[1,2]

Slope of a Curve at a Point

I

a.[2,3] = x2

S. R(O) ~

b. [-2,0]

In Exercises 7-14, use the method in Exsmple 3 to find o

. 1

sm

x'

x>O

2.2 Limit of a Function and Lim;t Laws

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, U(x) = O. For positive values of x arbitrarily close to zero, U(x) = I. 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 2.1Ob). (c) It oscillates too much to have a limit: f(x) has no limit as x --> 0 because the function's values oscillate between + I aod -I in every open interval containing O. The values • do not stay close to aoy one number as x --> 0 (Figure 2.IOc).

The Limit Laws When discussing limits, sometimes we use the notation 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 partico1ar discussion or example). Other times, such as in the statements ofthe 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 on the x-axis that may or may not belong to the domain of the function involved. To calculate limits of functions that are arithmetic combinations of functions having known limits, we cao use several easy rules.

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

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

x~c

1. Sum Rule:

lim g(x) = M,

then

x~c

lim(j(x)

+ g(x))

= L

+M

x~c

2. Difference Rule:

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

7. Root Rule:

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 "!' 0

lim[f(x))" =

L", n a positive integer

lim ~ =

"ifi =

x-c

x~c

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

>

L 1/", n a positive integer

0.)

In words, the Sum Ru1e 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 constant times a function is the constant times the limit of the function; the limit of a product is the product of the limits; the limit ofa quotient is the quotient ofthe limits (provided that the limit ofthe denominator is not 0); the limit ofa positive integer power (or root) ofa function is the integer power (or root) ofthe limit (provided that the root ofthe limit is a real number). It is reasonable that the properties in Theorem I 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'mition 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 Ru1e in Section 2.3, based on a precise def'mition oflimit. Ru1es 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....... c k = k and limx~c x = c (Example 3) and the properties oflimits to f"md the following limits. (a) lim(x' x~c

2 lim x '+X -1

+ 4x 2

-

3)

+ 4x 2

-

3) = lim x'

(b)

x~c

x2 + 5

(c) lim V4x 2 x--2

-

3

Solution (a) lim(x' x-c

+

lim 4x 2

=

c' + 40

lim(x' x~c

2

+x

-

x-c

X--i'C

3

-

2

lim 3

x-c

Sum and Difference Rules

Power and Multiple Rules

I)

-

Quotient Rule

lim(x 2 + 5)

x-e

lim x'

x-c

+

c' + c2 c2 (c)

lim V4x 2

x--2

-

-

lim x 2

lim I

-

x-c

x-c

I

Power or Product Rule

+5

3 =

Vx--2 lim (4x 2 -

=

Vx--2 lim 4x 2 -

=

V4(-2)2 - 3

=

Vi6=3

=

Vi3

Sum and Difference Rules

3)

Root Rule with Il

lim 3

Difference Rule

x--2

=2

Product and Multiple Rules

•

Two consequences ofTheorem I further simplify the task of calculating limits of polynomials and rational functions. To evaluate the limit of a polynomial function as x approaches c, merely substitute c for x in the formula for the function. To evaluate the limit of a rational function as x approaches a point c at which the denominator is not zero, substitute c for x in the formula for the function. (See Examples 5a and 5b.) We state these results formally as theorems.

THEOREM 2-Limits of Polynomials If P(x) = anx n + an_lx n- 1 + ... + ao, then lim P(x) = P(c) = ancn

x-c

+ an_lc n- 1 + ... + ao.

THEOREM 3-L1mlts of Rattonal Functions IfP(x) and Q(x) are polynomials and Q(c) i' 0, then . P(x) x-c Q(x)

P(c) Q(c)"

lim - - = - -

2.2

EXAMPLE 6

.

x3

x~-l

Identifying Common Factors It can be shown that if Q(x) is a polynomial and Q(c) = 0, then (x - c) is a factor of Q(x). Thus, if

the numerator and denominator of a rational function of x are both zero atx = c,theybave(x - c)asa common factor.

+ 4x 2 - 3 x2 + 5

=

(-1)3 + 4( -1)2 - 3 0 =-=0 (-If + 5 6

Theorem 3 applies only if the denominator of the rational function is not zero at the limit point c. If the denominator is zero, canceling common factors in the numerator and denominator may reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can fmd the limit by substitution in the simplified fraction.

Evaluate lim x 2

+

X -

x2 -

x---+l

2

X

•

Solutton We cannot substitute x = I because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = 1. It is, so it has a factor of (x - I) in common with the denominator. Canceling the (x - I)'s gives a simpler fraction with the same values as the original for x i' I: (x - I)(x

(a)

+ 2)

x

x(x - I)

y y=x+2 x

-

Eliminating Zero Denominators Algebraically

EXAMPLE 7

---"_:l-2c----..ot-+--------> X

51

The following calculation illustrates Theorems 2 and 3: lim

I

Limit of a Function and Lim;t Laws

+2 x

if x i' 1.

'

Using the simpler fraction, we find the limit of these values as x ---> I by substitution:

3

lim x x--+l

-----=":l-c_---..ot-+-------->x

2

+x

x2

-

- 2 = lim x

+2

X

X

x--+l

= I

+2

= 3.

1

-

See Figure 2.11.

(b)

FIGURE 2.11 The graph of f(x) ~ (x' + x - 2)/(x' - x) in part (a) is the same as the graph of g(x) = (x + 2)/x in part (b) except at x ~ I, where f is undefmed. The functions have the same lintit as x ---. I (Example 7).

Using Calculators and Computers to Estimate Limits When we cannot use the Quotient Rwe in Theorero I because the limit ofthe denominator is zero, we can try using a calcwator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example I, but calcu1ators and computers can sometimes give false values and misleading impressions for functions that are undefmed at a point or fail to have a limit there, as we now illustrate.

EXAMPLE 8

Estimate the value of lim

x-o

Yx' + 1,00 x

10.

Solutton Table 2.3 lists values of the function for several values near x = O. As x approaches 0 through the values ±I, ±0.5, ±O.IO, and ±0.01, the function seems to approach the number 0.05. As we take even smaller values of x, ±0.OO05, ±0.0001, ±0.00001, and ±O.OOOOOI, the function appears to approach the value O. Is the answer 0.05 or 0, or some other value? We resolve this question in the next example. _

52


TABLE 2.3 Computer values of f(x) ~

x

v'x' + 100 ,

10

x

near x

=

0

!(x)

±I ±0.5 ±O.I ±0.01

0.049876) 0.049969 h 0 OS? 0.049999 approac es. . 0.050000

±0.0005 ±O.OOOI ±O.OOOOI ±O.OOOOOI

0.080000) h? 0.000000 0.000000 approac es 0 . 0.000000

Using a computer or calculator may give ambiguous results, as in the last example. We cannot substitute x = 0 in the problem, and the numerator and deoominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.

EXAMPLE 9

Evaluate

" Yx2 + 100 - 10 Iun 2 •

x-o

X

Solution This is the limit we considered in Example 8. We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression Yx 2 + 100 + 10 (obtained by changing the sigu after the square root). The preliminary algebra rationalizes the numerator:

Yx2 + 100 - 10 x2

Yx 2 + 100 - 10 Yx 2 + 100 + 10 x2 • v'x 2 + 100 + 10 x 2 + 100 - 100 X

2

(Yx 2 + 100 + x

X

2

(Yx 2 +

10)

2

100 + 10)

I Yx 2 + 100 + 10'

Common

fuctor"

Cancel x2 for x

'¢

0

Therefore,

· Yx2 IlID x~o

+ 100 - 10 x2

=

I'lID r~==!,I=----:~ ~ x~o Yx 2 + 100 + 10 I

Y02 + 100 + 10 =

I

20

=

Denominator not 0 at x = 0; substitute

0.05.

This calculation provides the correct answer, in contrast to the ambiguous computer results in Example 8. • We cannot always algebraically resolve the problem of rmding the limit of a quotient where the deoominator becomes zero. In some cases the limit might then be found with the

2.2

Limit of a Function and Limit Laws

53

aid of some geometry applied to the problem (see the proof ofTheorem 7 in Section 2.4), or through methods of calculus (illustrated in Section 7.5). The next theorem is also useful.

y

The Sandwich Theorem g ~t------O--------->x

o

c

The following theorem enables us to calculate a variety oflimits. It is called the Sandwich Theorem because it refers to a function f whose values are sandwiched between the values of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of f must also approach L (Figure 2.12). You will fmd a proof in Appendix 4.

The graph of f is saodwiched between the graphs of g aod h.

FIGURE 2.12

THEOREM 4-The Sandwich Theorem

Suppose that g(x) :5 f(x) :5 h(x) for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim g(x) = lim h(x) = L. X---"'c

y

x-c

Thenlim.....d(x) = L.

The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem. y~

1-

4x'

EXAMPLE 10

Given that 2

--'-------:--~--+----x

-I

I -

0

FIGURE 2.13 Aoy function u(x) whose graph lies io the region between y = I + (x'/2) andy = I - (x'/4) has

4x

:5 u(x) :5 I

Solution

* 0,

Since lim(1 - (x 2/4)) = I

and

x~O

lim(I

x~O

+ (x 2j2»

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

EXAM PLE 11

= I,

•

The Sandwich Theorem helps us establish several important limit rules:

(a) lim sin 0 = 0 ~------')IL----~8 -1r

for all x

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

limit I as x ---> 0 (Example 10).

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

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

lim sinO = O.

8~O

--_-!2c---c-I:-""~O¥-~Ic--2;\--8

(h) From Section 1.3,0:5 I - cosO :5101 for all 0 (see Figure 2.14b), and we have lims~o(1 - cosO) = 0 or

lim cos 0 = 1.

(b)

FIGURE 2.14 The Saodwich Theorem conflflDS 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~c 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

TIf(x) :5 g(x) for allxin some open interval containingc, except and the limits of f and g both exist as x approaches c,

= c itself,

then

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 2.14a 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 lhe function g(x) graphed here, fmd lhe following limits or explaia why lhey do not exist. 8.

lim g(x)

x-I

b. lim g(x) %-2

c. lim g(x) %-3

y

d. lim g(x) x-2.S

y

4. Which of lhe following statements about 1he function y graphed here are 1rue, aod which are false? 8.

lim f(x) docs not exist.

x~2

2. For lhc function f(/) graphed here, fmd 1he following limits or explaia why lhey do not exist. 8. lim f(/) b. lim f(/) c. lim f(/) d. lim f(/) 1--2

t--I

t-O

t---{}.S

s

= f(x)

b. lim f(x) x~2

=2

c. lim f(x) docs not exist. x~l

d. lim f(x) exists at every pointxo in (-I, I). x~

...

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

...

y s

~ j(r)

o ....-&-1

3. Which of lhe following slarements about lhe function y = f(x) graphed here are 1rue, aod which are false? 8.

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 (-I, I). x~

...

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

Extstence of Limits In Exercises 5 and 6, explaia why lhe limits do not exist.

5lim~ . x~o Ixl

61" I . x~ x - I

7. Suppose lhat a function f(x) is defined for all real values ofx except x = xo. Can anything he said about lhe existence of lim~... f(x)? Give reasons for your answer. 8. Suppose iliat a function f(x) is defmed for all x in [-I, I]. Cao any1hing he said about the existence of limx_o f(x)? Give reasons for your answer.

2.2 9. Iflimr-l f(x) ~ 5, must f be defmed at x ~ I 7 If it is, must f( I) = 57 Can we conclude anything about the values of f at x ~ I 7 Explain, 10. If f(l) ~ 5, must lim,~1 f(x) exist? If it does, then must limr-l f(x) = 57 Can we cooclude anything about limr-l f(x)7 Explain.

Limit of a Function and Limit Laws

Using Limit Rulos 51. Suppose limr-o f(x) = I and lim,~o g(x) = -5, Name the rules in Theorem I that are used to accomplish steps (a), (b), and (c) of the following calcu1atioo. lim 2f(x) - g(x) ~ ,~o (j(x)

+ 7)2/3

,~o

%--7 1-+6

,~o

14. lim

%-+-2

lx 2 + 4x + 8)

(x 3 -

x-o

x-o

z~o

y--+-3

21. lim

3

v'3h+1 + I

h~O

22. lim

v5h+4 -

4-0

2

limY5h(x)

27.lim

t2

I-I

+ t - 2 t2 - 1

x+3

+

32. lim

x-I

x-

33. lim

"4 u3 -

I

34. lim if - 8

1

16

%-9

37.lim

'\."x -

x - 9

x-o

3

36. lim

38. lim

x-I

39. lim

W+12 -

x-2

x -

41. lim 2 %-+-3

2

4

+3

c. lim (j(x)

,~,

3

%-4

x-c

+ 3g(x)) ~

d

48. lim (x 2

+ 3)

c. lim (g(x»2

-

1)(2 - cosx)

yI;+4 cos (x + 'IT) SO. lim Y 7 + seC> X x-o

~

- 3. Find

b. lim xf(x) %-4

r

d

%-4

•

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) a. lim (j(x) + g(x)) x-b

x-b

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

x-b

x-o

r f(x) • ,~ f(x) - g(x)

0 and lim.~4 g (x)

c. lim 4g(x)

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

5 2

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

54. Suppose lim.~d(x) a. lim (g(x)

x-I

5andlimr-,g(x) = -2. Find

x-c

,~o

x-o

Supposelim,~d(x) =

a. lim f(x)g(x)

Find the limits in Exercises

46. lim tan x

}&-+-'II"

53.

4-x '~45-~

45. lim sec x

x-o

V(5)(5)

42.lim

44. lim sin2 x

(c)

x-I

(1)(4 - 2)

2

x + 1

43. lim (2 sin x - I) ,~o

x-I

40. lim x+2 '~-2W+5 -3

Limits with trigonometric functions 43-50,

49. lim

,~1

1

W+8 -

x-I

y51im h(x)

X+1

x

(b)

( lim p(x»)( lim 4 - lim r(x»)

'\."x

%-+-1

-~ X

4 -

x ,~4 2 -

y';+J - 2

x-I

X

v-2 v 4 -

,~1

,~1

( lim p(x»)( lim (4 - r(x»))

16y 2

-

_1_ x - I

31. lim -'--I

35. lim

vlim 5h(x)

' x2 - 7x + 10 26. lun 2 %-2 x 2 28. lim t + 3t + 2 1-+-1 t 2 - t - 2

y-o 3y 4

! - I

(a)

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

,~1

5y 3 + 8y 2 30. lim --'-;c-----co

29. lim -lx - 4 %-+-2 x 3 + 2x 2

u-l

,~1

lim'VShW ,~1 p(x)(4 - r(x))

+ 4x + 3

%-+-3 Xl

x2+3x-1O lim 25. %--5 X + 5

7 4

h

24. lim

25

%-5 Xl -

(c)

lim 7»/3

x-o

52. Let limr-l h(x) = 5, lim'~IP(X) = I, and lim,~1 r(x) = 2. Name the rules in Theorem I that are used to accomplish steps (a), (b), and (c) of the following calcu1atioo.

Limits of quotients Find the limits in Exercises 23-42. 23.lim x - 5

+

(2)(1) - (-5) (I + 7j2/3

18.lim

%--1

x-a

(lim f(x)

y+2 y-2 y2 + 5y + 6 20. lim (Zz - 8)1/3

(b)

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

s-2/3

19. lim (5 - y)4/3

x-o

+ 7))2/3

( lim (J(x)

16. lim 3s(28 - I) 17. lim 3(lx - 1)2

(a)

+ 7j2/3

x-o %-2

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

g(x))

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

12. lim (-x 2 + 5x - 2)

+ 5)

l!'!'o (2f(x) lim (j(x)

Calculating Limits Find the limits in Exercises 11-22. 11. lim (lx

55

x-b

56. Suppose that limr--2 p(x) = 4,limr--2 r(x) = 0, and lim'~-2s(x) = -3. Find a. lim (P(x) + r(x) + s(x» %-+-2

b.

lim p(x)· r(x)' s(x)

%-+-2

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

+ 5r(x))/s(x)

56


Umits of Average Rates of Change Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form

.. f(,,-x _+~h):--_f,-,(x-,-)

lim-

h occur frequently in calculus. In Exercises 57--fJ2, evaluste this limit for the given value of x and function f. b-O

57. f(x) = x 2• X = I 59. f(x) = 3x - 4, x = 2 61. f(x) =

v;,.

58. f(x) = x 2• X = -2 60. f(x) = I/x. x = -2

Using the Sandwich Theorem 63. If v'5 - 2x lim,-+o f(x).

2

'"

f(x) '" ~ fur -I'" x '" I. fmd

2

.

X

o

----'c-t----';C"---> x

o

Xo

New challenge: I E = 100,000

--'-j---~l----~x

o

Xo

Response:

Ix - Xo I < EXAMPLE 2

Xo

New challenge: E= ...

81/100,000

Show that lim (5x - 3) = 2.

x->I

Solution Set Xo = I ,f(x) = 5x - 3, and L = 2 in the definition oflimit. For any given > 0, we have to find a suitable [; > 0 so that if x '" 1 and x is within distance [; of Xo = 1, that is, whenever E

o
0 such that for all x 0< Ix - xol < Il can be accomplished in two steps.

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

NOT TO SCALE

FIGURE 2.22

*

*

The function and intervals

in Example 4.

EXAMPLE 5

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

{xI,

2

f(x) Solution

=

x*2

,

x

Our task is to show that given E > 0 there exists all> 0 such that for all x

If(x) - 41
0 such that for all x

=

I(j(x) - L)

0< Ix - cl < Il,

If(x) - LI < e/2.

Similarly, sioce lim~c g(x) = M, there exists a number 112

o

Therefore, for any e

I(g(x)

0, there exists Il

>

63

The Precise Definition of a Limit

0 such that

- f(x» - (M - L)I < e

whenever

0

< Ix - cl
0 by hypothesis, we take e = L - M in particular and we have a number Il > 0 such that

I(g(x)

- f(x» - (M - L)I < L - M

whenever

0

< Ix - cl
0 such that for an x 0< Ix-xol 0 satisfies the following condition: For all x,

0 < Ix - 11 < 8

That is, for each 8 that

+

R

If(x) - 21 < 1/2.

> 0 show that there is a value of x such

0< Ix - II < 8

v

~

If(x) - 21 '" 1/2.

and

This will show that limx_ d(x)

b. Show that lim_1 f(x) ~ I. c. Show that1im_ 1 f(x) ~ 1.5.

~

2.

66


58. Leth(x)

~

{

X',

X

y

2 2 2.

•

2

y

4

3

---1'---'------+--+ x

o

2

COMPUTER EXPLORATIONS -;;1..L-----:c-----+ x

o

In Exercises 61--{j6, you will further explore rmding deltas graphically. Use a CAS to perform the followiug steps:

a. Plot the functioo y = f(x) near the point Xo being approached. Show that a. lim h(x) #' 4

b. Guess the value of the limit L aod then evaluate the limit symbolically to see if you guessed correctly.

b. lim h(x) #' 3

c. Using the value. = 0.2, graph the haoding lines Yl = L - • aod Y2 = L + • together with the functioo f near Xo.

x~2 x~2

c. lim h(x) #' 2 x~2

59. For the functioo graphed here, explain why

d. From your graph in part (c), estimate a 8

0
c) only.

One-Sided Limits To have a limit L as x approaches c, a function f must be dermed on both sides of c and its values f(x) must approach L as x approaches c from either side. Because of this, ordinary limits are called two-sided.

2.4 One-Sided Limits y

J

•

F~

•

0

67

If j fails to have a tw-sided limit at c, it may still have a one-sided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a right-hand limit. From the left, it is a left-hand limit. The function j(x) = x/lxl (Figure 2.24) has limit I as x approaches 0 from the right, and limit - I as x approaches 0 from the left. Since these one-sided limit values are not the same, there is no single number that j(x) approaches as x approaches O. So j(x) does not have a (two-sided) limit at O• Intuitively, if j(x) is defmed on an interval (c, b), where c < b, and approaches arbitrarily close to L as x approaches c from within that interval, then j has right-hand limit L at c. We write lim j(x) = L. x-c+

FIGURE 2.24 Different right-hand and left-hand limits at tire origin.

The symbol "x --> c+" means that we consider only values of x greater than c. Similarly, if j(x) is defined on an interval (a, c), where a < c and approaches arbitrarily close to M as x approaches c from within that interval, then j has left-hand limit M at c. We write lim j(x)

x-e-

= M.

The symbol "x --> c-" means that we consider only x values less than c. These informal defmitions of one-sided limits are illustrated in Figure 2.25. For the functionj(x) = x/lxl in Figure 2.24 we have lim j(x)

x-o+

=

I

and

lim j(x)

x-o-

y

=

-1.

y

f(·)

L

o (a)

-C:+---~"'-----"-'------>.

o

•

c

lim f(.)

~c'"

M

f(·)

~

•

c

L

FIGURE 2.25 approaches c. (b) Left-hand limit as> approaches c. y

EXAMPLE 1 The domain of j(x) Figure 2.26. We have

=

V4=7 is [-2, 2]; its graph is the semicircle in and

~~---;;t---+--.

-2

0

FIGURE 2.26 lim

x--2+

2

lim

x-2-

v'4=7 =

v'4

-

.2 =

0 and

0 (Example I).

The function does not have a left-hand limit at x = -2 or a right-hand limit at x does not have ordinary two-sided limits at either - 2 or 2.

=

2. It •

One-sided limits have all the properties listed in Theorem I in Section 2.2. The righthand limit ofthe sum oftwo functions is the sum oftheir right-hand limits, and so on. The theorems for limits of polynomials and mtional functions hold with one-sided limits, as does the Sandwich Theorem and Theorem 5. One-sided limits are related to limits in the following way. THEOREM 6 A function j(x) has a limit asx approaches c if and only if it has left-hand and right-hand limits there and these one-sided limits are equal: lim j(x) .~c

= L

limj(x) .~c

=

L

and

lim j(x)

x-c+

= L.

68


EXAMPLE 2

y

For the function graphed in Figure 2.27, limx~o+ f(x) =

Atx = 0: 2

1,

limx~o- f(x) and limx~o f(x) do not exist. The function is not defined to the left of x = O. limx~l-

Atx = 1:

•

-=+-~-----o-----o----'---x

o

2

3

f(x)

=

limx~l+ f(x) =

4

0 even though f(l) = 1, 1,

limx~l f(x) does not exist. The right- and left-hand limits are not

equal.

FIGURE 2.27 Graph of the function in Example 2.

limx~'- f(x) =

1, limx~2+ f(x) = 1, limx~2f(x) = 1 even though f(2) = 2. limx~r f(x) = limx~3+ f(x) = lim~3 f(x) = f(3) = 2. limx~4- f(x) = 1 even though f( 4) oF 1, limx~4+ f(x) and limx~4 f(x) do not exist. The function is not defined to the right of x = 4.

Atx = 2:

Atx = 3:

y

Atx = 4:

•

At every other point c in [0, 4], f(x) has limit f(c). L+
0 be given. Here Xo = 0 and L = 0, so we want to find all> 0 such that for all x

O<x .

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

vx,

Find (a) lim......o' j(x) and (h) lim,_o- j(x); tlIen use lintit definitions to verilY your tmdings. (0) Based on your cooclusions in parts :

l<xX

"'¢'3x -

~

I?

31.

}~sin(x -

sinx)

32.

~Sin(fcos(tant»)

33. lim sec (y sec' y - tan' Y - I) y~1

34.

%~ tan(~cos (Sinx 1/3»)

35. lim cos ( t-O

V 19 -

1f

3 sec 2t

)

36.

lim v'csc'x x-1T16

+ 5V3tanx

2.5

Continuous Extensions

Theory and Examples

37. Defmeg(3) in a way that extends g(x) = (x 2 continuous at x = 3.

-

9)/(x - 3) to be

= (1 2 + 31 -

38. Defmeh(2) in a way that extends h(l) to be continuous at 1 = 2.

10)/(1 - 2)

39. Defme f(l) in a way that extends f(s) ~ (s' - 1)/(s2 - I) to be continuous at s = 1. 40. Defme g(4) in a way that extends

g(x)

= (x 2 -

to be continuous at x

16)/(x 2

4.

=

53. Roots of a cubic has three solutions in the interval [-4,4].

55. Solving an equation If f(x) = x' - 8x + 10, show that there are values c for which f(c) equals (a) 'IT; (b) (c) 5,000,000.

~

2 -

I, x < 3

2ax.

x~3

{x

56. Explain why the following fIVe statements ask for the same infor-

mation. a. Find the roots of f(x)

continuous at every x? 42. For what value of b is X

g(x) =

{

b~2,

X '"

-2

~ {~~~ -

e. Solve the equation x~2

2a,

continuous at x

x__-"-'----

Limits Involving Infinity; Asymptotes of Graphs

x>M The implication will hold if M = 1/E or any larger positive number (Figure 2.50). This proves limx~oo (I/x) = O. (b> Let E > 0 be given. We must find a number N such that for all x

x 2 (where the denominator is zero), as shown in the graph. • Notice in Example 8 that if the degree of the nwnerator in a rational function is greater than the degree ofthe denominator, then the limit as Ix Ibecomes large is +00 or -00, depending on the signs asswned by the nwnerator and denominator.

Infinite Limits

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

~o

You can get as low

you want by taking X close enough to O.

FIGURE 2.57 lim 1= 00 x_lt X

x

No matter how low -B is, the graph goes lower.

j8\

-B

Let us look again at the function f(x) = I/x. As x ---> 0+, the values of f grow without bound, eventually reaching and surpassing every positive real nwnher. That is, given any positive real nwnber B, however large, the values of f become larger still (Figure 2.57). Thus, f has no limit as x ---> 0+. It is nevertheless convenient to describe the behavior of f by saying that f(x) approaches 00 as x ---> 0+. We write lim f(x) x-o+

One-sided inf"mite limits: and lim 1 = -00. x_o-x

=

lim x-o+

i

= 00

In writing this equation, we are not saying that the limit exists. Nor are we saying that there

is a real nwnber 00, for there is no such nwnber. Rather, we are saying that limx~o+ (I/x) does not exist because I/x becomes arbitrarily large andpositive as x ---> 0+ . As x ---> 0-, the values of f(x) = I/x become arbitrarily large and negative. Given any negative real nwnber -B, the values of f eventually lie below -B. (See Figure 2.57.) We write

y

1 Y=x-t

lim f(x) x-o-

=

lim

x-o-

i

= -00.

Again, we are not saying that the limit exists and equais the nwnber - 00 . There is no real nwnber - 00 . We are describing the behavior of a function whose limit as x ---> 0- does not

1 ----:-----;;+---;1--2~----;;3---x

exist because its values become arbitrarily large and negative.

EXAMPLE 9

FIGURE 2.58 Near x = I, the function y = I/(x - I) behaves the way the function y = I/x behaves near x = o. Its grapb is the grapb of y = I/x sbifted I unit to the right (Example 9).

x-l+ x - I

· nI u

I x_rx - I"

and

Geometric Solution The graph of y = I/(x - I) is the graph of y = I/x shifted I unit to the right (Figure 2.58). Therefore, y = I/(x - I) behaves near I exactly the way y = I/x behaves near 0:

lim _1_ 1

= 00

x_l+X -

lim _1_ 1

and

=-00

x_l-X -

Analytic Solution Think about the nwnber x - I and its reciprocal. As x ---> 1+, we have (x - I) ---> 0+ and I/(x - I) ---> oo.Asx---> 1-, we have (x - I) --->0- and I/(x - 1)--->

y

-00.

} x 0

Find lim _1_

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) in Example 10 approaches inf"mity as x --+ O.

as

x ---> o.

Solution 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

lim f(x) x-o

=

lim \ x-Ox

= 00.

•

90

Chapter 2: Limits and Continuity The function y = l/x shows no consistent behavior as x --> o. We have l/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 lim._ o (l/x 2) = 00.

EXAMPLE 11

These examples illustrate that rational functions can behave in various

ways near zeros of the denominator.

.-2 .-2

.-2 .-2 +

(a) lim (x - 2)' = lim (x - 2)2 = lim x - 2 = 0 .-2 X 2 - 4 (x - 2)(x + 2) X + 2

(b) lim x - 2 = lim .-2 X 2 -

(c) y

4

x - 2 = lim _1_ = ! (x - 2)(x + 2) X 2 4

lim x - 3 = lim

.-2+ X 2 -

4

.-2+

(d) lim x - 3 = lim • -2- X 2 -

4

.-T

.-2

(e) lim x - 3 = lim • -2 X 2 -

Bf------/-----j----'\---

--;;-I;f----'c~--o:---C;\---~x

Xo

+8

.-2

(f) lim

4

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

The values are negative for x > 2,xnear2.

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

The values are positive forx < 2,xnear2.

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

See pans (c) and (d).

2-x = lim -(x-2)=lim -I =-00 (x - 2)3 .-2 (x - 2)3 .-2 (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 requiring f(x) to lie arbitrarily close to a finite number L for all x sufficiently close to xo, the defInitions ofinfinite limits require f(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seen before. Figures 2.60 and 2.61 accompany these definitions.

DEFINmONS 1. We say that f(x) approaches infinity as x approaches xo, and write lim f(x) = 00, x-xo if for every positive real number B there exists a corresponding /l that for all x O' Horizontal

lim 1 = x---+o-x

-00

;;;;;

asympto"'. y~O

We say that the line x = 0 (the y-axis) is a vertical asymptote of the graph of f(x) = I/x. Observe that the denominator is zero at x = 0 and the function is undefined there.

Vertieal asymptote,

x=O

The coordina'" axes are asymptotes ofboth branches ofthe hyperbola y ~ 1/•. FIGURE 2.&2

DEFINmON A line x y = f(x) if either

=

a is a vertical asymptote of the graph ofa function or

EXAMPLE 13

limj(x) .-a

=

±OO.

Find the horizontal and vertical asymptotes of the curve

x+3

y = x + 2'

Solution We are interested in the behavior as x ---> ± 00 and the behavior as x ---> - 2, where the denominator is zero.

92


The asymptotes are quickly revealed ifwe recast the rational function as a polynomial with a remainder, by dividing (x + 2) into (x + 3):

y

Vertical asymptote, x

=-2

6 5 4

I

x+3

y= x+2

~ 1

Horizontal

x + 2)" + 3 x+2 I

+ _1_

x+2

asymptote,

y- 1

-~-;--c";-':k-±--;-~--7----;;----;;---X -5 -4 -3\-2 -1 0 - \ -1

1

2

This result enables us to rewrite y as:

3

I

y=l+ x + 2 '

-2

As x ---> ± 00, the curve approaches the horizontal asymptote y = I; as x ---> - 2, the curve approaches the vertical asymptote x = -2. We see that the curve in question is the graph of I(x) = I/x shifted I unit up and 2 uuits left (Figure 2.63). The asymptotes, instead of being the coordinate axes, are now the lines y = I and x = - 2. •

-3 -4

FIGURE 2.63 The lines y = I and x = - 2 are asymptotes ofthe curve in Example 13.

EXAMPLE 14

Find the horizontal and vertical asymptotes ofthe graph of

I(x) y y~

Vertical

asymptote, x=-2

=

2

Horizontal asymptote, y

8 x2 _ 4'

Solution We are interested in the behavior as x ---> ± 00 and as x ---> ±2, where the denominator is zero. Notice that I is an even function of x, so its graph is symmetric with respect to the y-axis.

__ 8_ x2 _ 4

Vertical asymptote, x

= -

=

0

(a) The behavior as x---> ±OO. Since limx~oo/(x) = 0, the line y = 0 is a horizontal asymptote of the graph to the right. By symmetry it is an asymptote to the left as well (Figure 2.64). Notice that the curve approaches the x-axis from ouly the negative side (or from below). Also, 1(0) = 2.

(b) The behavior as x---> ±2. Since

and the line x = 2 is a vertical asymptote both from the right and from the left. By symmetry, the line x = - 2 is also a vertical asymptote. FIGURE 2.64 Graph ofthe function in Example 14. Notice that the curve approaches the x-axis from only one side.

There are no other asymptotes because I has a finite limit at every other point.

EXAMPLE 15

Asymptotes do not have to he two-sided.

The curves

Y

=

secx

=

I

cosx

and

y

=

tanx

=

sin x cosx

both have vertical asymptotes at odd-integer multiples oftr12, where cos x y

y

=

y

secx

-+----'-+-~-+_--'--t-~x

3

•

=

0 (Figure 2.65).

y = tanx

-+~'----+-~Oo-+_~-t-~x

3

FIGURE 2.65 The graphs of sec x and tan x have infmitely many vertical asymptotes (Example 15).

•

2.6

Limits Involving Infinity; Asymptotes of Graphs

93

Dominant Terms In Example 8 we saw that by long division we could rewrite the function

x2 - 3 f(x) = 2>; - 4 as a linear function plus a remainder tenn:

This tells us immediately that

f(x) '"

f+ I

I f(x) '" 2>; - 4

y

For x numerically large, 2x

~ 4 is near O.

For x near 2, this term is very large.

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 tota1 value of f is insignificant. It behaves like 1/(2); - 4) when x is so close to 2 that 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.

If we want to know how

----;2!---~1 ~""'O'+--""-!----C:-2 ~x

-5 Let f(x) = 3x 4 - 2>;3 + 3x 2 - 5x + 6 and g(x) = 3x 4 • 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

± 00 • We fmd that 2Oc O---;_loO:O"-""'O::+-~"OIOcO---;2f.oO~x

-_-C0

-100,000 (bJ

FIGURE 2.66 The graphs off and g are (a) distinct for Ixl small, and (b) nearly identical forlxllarge (Example 16).

lim f(x) = lim 3x 4 - 2>;3 + 3x 2 - 5x .~±OO g(x) .~±oo 3x4 = =

+6

. ( 1 -2- +1 - - 5 + 2) lim 3x x2 3x 3 x4

x_±OO

1,

which means that f and g appear nearly identical for Ix Ilarge.

•

94


Exercises 2.6 Finding Limits

Limits of Rational Funettons

1. For the function f whose graph is given, determine the following limits.

a. lim f(x)

b.

d. lim f(x)

e. lim f(x)

r.

h. lim f(x)

i.

x~,

lim f(x)

%-0+

%-+-3

g. lim f(x) x~O

lim _ f(x)

e.

%--3+

x~oo

%-+-3

lim_ f(x)

In Exercises 13-22, f"md the limit of each rational function (a) as

x ---+

00

and (b) as x ---+

13. f(x)

=

+3 +7

14. f(x)

2x' + 7 = x-,~-"x"",~+'--'x'--+-7

15. f(x)

= -"'--±-l 2

16. f(x)

=~+7

17. h(x)

7x' =,x-3x+6:< ,

18. g(x) = -x~'---4'-x-+-1

2x Sx

x~o

lim f(x)

%_-00

y

x

+3

3

,-,/

19. g(x) = lOx'

1/

f

r

-6 - -4 - -i/o

,

s 6

3

2. For the function f whose graph is given, determine the following limits.

a. lim f(x)

b. lim f(x)

d. lim f(x)

e.

c. lim_ f(x)

x-2+

x~4

x~,

x~,

r.

lim f(x)

x--3+

lim_ f(x)

%-+-3

g. lim f(x)

h. lim f(x)

i. lim_ f(x)

j. lim f(x)

k. lim f(x)

L

%-0+

%-+-3 x~o

21. h(x)

=

-2x' - 2x + 3 3x' + 3: ±oo Find the limits in Exercises 80-86.

(v;+9 - v;+4) lim (Vx 2 + 25 - w-=!) lim (w-+3 + x) x_-oo lim (2x + Vrc 4X"2-+--c3x -_--c 2) x_-oo

80. lim

a. x-+O+

(x -

~

00,

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

t---+ 0+

I

and lim f(x)

lim f(x) =

%_-00

X-±OO

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

~/3 x

~ -00,

%_-00

c. x-+O-

(:/5 +

-I, and

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

57 r x -3x+2 • un x 3 -2x 2 as

59. lim(2 -

~

.1'--1+ -00

%-1-

and lim f(x) =

];_00

1 + 4 as

c.x-+l+

62. lim(

~

2, f( -I)

lim f(x) =

-

Lx-+-2+

x

~

71. flO) = 0, lim f(x) = 0, lim f(x) = b. x-+O-

61. lim(

~

x~"-

c. x --> V'2

a.

~

lim f(x)

a. x-+O+

60. lim

2x 68·y=x+1

0, f(l) lim f(x) = I

69. flO)

c. x---+ -1+

a.

x+3

=x +2

x~oo

a. x-+ 1+

. x2 56. lun 2x

-3

Inventing Graphs and Functions In Exercises 6 ±oo to establish the limits in Exercises 87 and 88. 87. Iff baa the coostant value f(x) = k, thea lim f(x) = k.

xl ~

95. lim

00

96. lim _1_ ~

-00

98. lim _1_ ~ %-1-1 - x 2

00

2

x_2-X -

1~

-00

x_o-x

97. lim _1_ ~

2

x_2+X -

00

X~OO

88. Iff baa the coostant value f(x) = k, thea lim f(x) = k. X~-OO

Use formal defmitions to prove the lintit statemeats in Exercises 8l}-92.

-I 89. lim -2 =

x-o

91. lim

x~3

90. lim

-00

-2 = (x - 3)'

x-

I -I1 =

00

x-o x

X

92. lim

-00

X~-,

(x

Oblique Asymptotes Grapb the rational functioos in Exercises 9l}-104. Include the grapbs and equations of the asymptotes. x2 x 2 +1 99. y ~ - - I 100. y ~ - - I

I

+ 5)2

=

101. y ~

x2

x-

x 2 -1

-4

-.t=T

102. y ~ 2x

+4

00

93. Here is the defmition oflnflnite rigbt-band limit.

x2

-1

103. y = - x -

104. y =

x 3 +1 --2X

Additional Graphing Exerdses

D Grapb We say that f(x) approacbes infmity as x approacbes Xo from the right, and write lim f(x) =

x-xo+

x

106. y

105. y = _ ,.-------, v4 - x 2

00,

if. for every positive real number B. there exists a corresponding nomber 8 > 0 sucb that for all x xo<x<xo+8

the curves in Exercises 105-108. Explain the relationship betweea the curve's formula and what you see.

f(x) > B.

107. y =

x 2/ 3

-I = _,.-------, 2

v4 - x

+ ~f3

108. y

x

= sin

(+-) x + I

D Grapb the functions in Exercises 109 and 110. Thea answer the following questioos.

a. How does the graph bebave as x ---> o+? Modify the defmition to cover the following cases.

a.

lim_f(x)

b. How does the graph behave as x ---> ± oo?

~ 00

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

X-Xo

lim f(x)

= -00

Give reasons for your answers.

c. limj(x)

= -00

109. y

b.

x-xo+ %-Xo

Chapter

r

~ &(x - ~

3(

X )2/3

110'Y~2 x - I


1. What is the average rate of change ofthe 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 calculatiog lintits? Give exam-

2. What lintit must be calculated to fmd the rate of cbaoge 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 does not exist? Give examples.

3. Give an informal or inmitive defmition of the lintit lim f(x) = L. x~X

Why is the definition "informal''? Give examples. 4. Does the existence and value of the limit of a function f(x) as x approaches Xo ever depead on what bappeos at x ~ xo? Explain and give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples.

ples ofbow the theorems are used

8. What is the value oflim._o «(sinO)/O)? Does it matter whether 0 is measured in degrees or radians? Explain. 9. What exactly does lim~x f(x) ~ L mean? Give an example in whicb you fmd a 8 > 0 for a givea f, L, xo, and. > 0 in the precise defmition oflintit.

10. Give precise defmitions of the following statements. a. lim~2- f(x) = 5

b. lim~2' f(x) = 5

c. lim~2 f(x) =

d. lim~2 f(x) = -

00

00

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 function is continuous? 13. What does it mean for a function to be right-continuous at a point? Left-continuous? How are continuity and onl>-sided continuity related?

14. What does it mean for a function 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 grapbingandsolvingtheequationj(x) ~ O? 17. Under what circumstances can you extend a function j(x) to be continuous at a point x = c? Give an example. 18. What exact1y do limr _ oo j(x) = L and limr-~ j(x) = L mean? Give examples. 19. What are limr_± k (k a constant) and limr-± (I/x)? How do you extend these results to other functions? Give examples. 20. How do you rmd the limit of a rational function as x ---+ ± oo? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.

Practice Exerdses

Limns and Continuity 1. Graph the function

In Exercises 5 and 6, rmd the value that limr-Og(x) must have if the given lintit statements hold.

I,

j(x) =

I

-~:

-x, I,

x:5

-I

5. lim (4 - xg(x)) = I

-1

<x
-1 and a '¢ O. one positive and one negative: _ -I T+ ( a ) -

+~ a

'

r_(a)

~

-I-~ a .

Generalized Limits involving

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

c. Support your conclusions by graphing r+(a) andr_(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, and 0.05.

+ 2 cos x

Ii:

8

The formula lime_o(sin 8)/8 = I can be genera1ized. If limx _, f(x) = 0 and f(x) is never zero in an open interval containing the point x = c, except possibly c itself, then lim sinf(x) = I f(x) .

x~,

Here are several examples.

.

a. lim. 8mx

,

=

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 lun • x-+-l X + 1 x-+-l (x 2 - X - 2) . (x' - x - 2) . (x + I)(x - 2) lim =1' lim =-3 x

+1

x--l

x

+1

. sin (I - '\!'x) . sin (I - '\!'x) I - '\!'x d. lun = lun = x-I x-I x-I 1_ ~ x-I

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

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

'\!'x)

Find the limits in Exercises 25-30. sin (I - cosx) 25. lim x

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

22. Root of so equation Show that the equation x has at least one solution.

2

b. Find a similar expression for min {a, h}, the smaller of a andb.

x--l

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

la - bl

equals a if a '" b and equals h if h '" a. In other words, max {a, hI gives the larger of the two numbers a andh.

a. Show that f is discontinuous 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 fmitely manyratiooa1 numbers rin [0, I] such thatf(r) '" E.)

+

= 0

23. Bounded function. A real-valued function f is bounded from above on a set D if there exists a number N such that f(x) '" N for an x in D. We call N, when it exists, an upper bound for f on D and say that f is bounded from above by N. In a similar manner, we say that f is bounded from below on D if there exists a number M such that f(x) '" M for an x in D. We call M, when it

x~o

27. lim

sin (sinx)

x

x~o

29. lim

x-2

28. lim

sin (x'

x

x~o

+ x)

sin ( '\!'x

sin (x' - 4) 2

30. lim

x

x-9

x

- 3) 9

Oblique Asymptote. Find all possible oblique asymptotes 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

101


Mathematica,tMap1e Modules: 1IIke It to the Limit Part I Part II (Zero Raised to the Power Zero: What Does it Mean?) Part III (One-Sided Limits) Visualize and interpret the limit concept through graphical and numerical explorations.

Part IV (What a Difference a Power 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 (Rates of Growth) 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 function, 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.

Tangents and the Derivative at a Point

3.1

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). Ifthe 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. (xo, f(xo))

: _______ J

~-~',

h

: I

~-------''------......I.....--------~

o

FIGURE 3.1 The slope of the tangent r t p. r f(xo + h) - f(xo) Ine a IS h~ h

X

DEFINITIONS

The slope of the curve y

==

f{x) at the point P{xo, f{xo)) is the

number

. m==hm

h~O

f{xo

+ h) - f{xo) h

(provided the limit exists).

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

y=l slope is-~

I .'

(a) Find the slope of the curve y = I/x at any point x = a oF 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

I slope is-l

(a) Here fix) = I/x. The slope at (a, I/a) is

I

atx =-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).

-h

I·

h~ ha(a + h)

lim

1 a - (a + h)

h~O

h a(a + h)

lim

=

-I

h~O a(a + h)

I

a

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 4

=

I

+ hh - a =

a

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 = -I, the slope is -1/(-1)2 = -I (Figure 3.2).

y

slopels. 1

= lim h~O

(2. ~)

~'----t-~x sl . 1 opels-"4

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

I a2

I 4'

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 oF 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 f 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 f 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 = fix) 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 1sec, 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? We let f(l) = 161 2. The average speed of the rock over the interval between 1 = 1 and 1 = 1 + h seconds, for h > 0, was found tu be

Solution

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

Tangent Lines with Spedfied Slopes

In Exercises 1-4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points PI

At what points do the graphs of the functions in Exercises 23 and 24 have horizontal tangeats?

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 tangeot to the curve y ~ I/(x - I). 26. Find an equatioo of the straight line having slope 1/4 that is tangeat to the curve y = ~.

1 I-

Rates of Change

r

27. Object dropped from a tower An o!!iec! is dropped from the top of a 100-m-high tower. Its height above ground after 1 sec is 100 - 4.91' m. How fast is it falling 2 sec after it is dropped? 28. Speed of a rocket At 1 sec after liftoff, the height of a rocket is 31' ft. How fast is the rocket climbing 10 sec after liftoff?

3.

4.

y

29. Circle's changing area What is the rate of chauge ofthe area of a circle (A ~ 'lTr') with respect to the radius whea the radius is r = 3?

y

30. Ball's changing volume What is the rate of change of the volume ofa ball (V ~ (4/3)'lTr') with respect to the radius whea the radius is r = 2?

3

31. Show that the line y = point (xo, nlXo + b).

nIX

+b

is its cwo tangeat line at any

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

o

Testtng for Tangents 33. Does the graph of

In Exercises 5-10, fmd an equatioo for the tangeat to the curve at the givea point. Thea sketch the curve and tangeot together.

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

= 2~,

9. y

= x',

(-1,3) (1,2)

(-2, -8)

6. y ~ (x - I)' 8. y 10. y

=

1"x

+

(I, I)

I,

34. Does the graph of

(-2,-t)

= :',

+ I,

(2,5)

x 13. g(x) = x _ 2' 15. h(l) ~ I',

(3,3)

(2,8)

17. f(x) =~, (4,2)

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

8

14. g(x) = x"

=

5x 2 ,

X =

I 21. y = x _ I'

-1 x = 3

20. y

=

+ 31, (1,4) v;+l, (8,3)

1 - x 2,

x - I 22'Y=x+I'

x ~ 0 x =0

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

Vertical Tangents (2,2)

In Exercises 19-22, find the slope of the curve at the point indicated. 19. y

g(x) ~ {x sin (1Ix), 0,

(1,-1)

16. h(l) ~ I' 18. f(x) =

x ~ 0 x= 0

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

(-I, I)

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

f(x) ~ {x' sin (1Ix), 0,

X =

2

x=O

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

+ h) - f(O) h = lim _1_ = 4-0 h2j3

00

.

106

Chapter 3: Differentiation y

36. Does lbe graph of

{O,1,

U(x) =

x
x -I 0 2

Y Y ~J(x) D: -3s x s 3

3.3

D 60.

Theory and Examples In Exercises 4~52,

115

Graph y ~ 3x in a window that has - 2 '" x '" 2, 0 '" Y '" 3. 2

Then, on the same screen, graph

a. Find the derivative j'(x) of the given function y = f(x). b. Graph Y = f(x) and y = j'(x) side by side using separate sets of coordinate axes, and answer the following questions. c. For what values of x, if any, is i' positive? Zero? Negative? d. Over what intervals ofx-values, if any, does the function y = j(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more ahout this relationship in Section 4.3.)

49. Y

Differentiation Rules

= -x 2

SO. Y

51. y ~ x'/3

= -I/x

52. Y ~ x 4/4

53. Tangent to a parabola Does the parabola y = 2x 2 - 13x + 5 have a tangent wbose slope is -I ? If so, find an equation for the line and the point of tangency. Ifnot, wby not?

=

54. Tangent to y Vi Does any tangent to the curve y = Vi cross the x-axis at x = -I? If so. f'md an equation for the line and the point of tangency. Ifnot, wbynot?

(x + h)' - x' Y =

h

for h = 2, 1,0.2. Then try h = -2, -I, -0.2. Explain wbat is goingon. 61. Derivative of y = Ixl Graph the derivative of f(x) = 14 Then graphy = Gxl- O)/(x - 0) = lxi/x. What can you conclude?

D 62.

Weierstrass's nowhere differentiable continu.ous function The som of the rltst eight terms of the Weierstrass function f(x) = ~::".o (2/3Y' cos (9"'lTx) is

+ (2/3)1 cos (9'ITx) + (2/3)2COS(9''lTx) + (2/3)' cos (9''lTx) + ... + (2/3)7 cos (9 7'ITx).

g(x) = cos ('lTx)

Graph this snm. Zoom in several times. How wiggly and bnmpy is this graph? SpecifY a viewing window in which the displayed portion of the graph is smooth.

COMPUTER EXPLORATIONS

55. Derivative of - f Does knowing that a function f(x) is difThrentiable at x = Xo tell you anything about the differentiability of the function -fat x ~ xo? Give reasons for your answer.

Use a CAS to perform the following steps for the functions in Exercises 63--{j8.

56. Derivative of multiples Does knowing that a function g(t) is differentiable at t = 7 tell you anything ahout the differentiability of the function 3g at t ~ 7? Give reasons for your answer.

b. Derme the difference quotient q at a general point x, with general step size h.

57. Limit of a quotient Suppose that functions g(t) and h(t) are dermed for all values of t and g(O) = h(O) = O. Can Itm,....o (g(t))/(h(t)) exist? If it does exist, must it equal zero?

d. Substi1ute the value x = Xo and plot the function y = f(x) together with its tangent line at that point.

Give reasons for your answers. 2

58. a. Let f(x) be a function satisfYing If(x) I '" x for -I '" x '" I. Show that f is differentiable atx ~ 0 and rmdj'(O).

X2 Sin}, X::F 0

f(x) = {

Graphy ~ 1/(2Vi) in a window that has 0'" the SaJ13e screeo, graph

v'XTI -

for h ~ 1,0.5,0.1. Then try h is going on.

3.3

e. Substitute various values for x larger and smaller than Xo into the formnla obtained in part (c). Do the nombers make sense with your piclure?

x'" 2. Theo, on

65. f(x) =

-2--'

66. f(x) ~

Vi

h ~

64. f(x) =

+ x 2 - x, Xo ~ I x 1/3 + x 2/3, Xo = I

63. f(x) ~ x'

0, x = 0 is differentiable atx = 0 and rmdj'(O).

y=

e. Take the limit as h .... O. What formula does this give?

f. Graph the formnla obtained in part (c). What does it mean wben its values are negative? 'hro? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer.

b. Sbowthat

D 59.

a. Plot y = f(x) to see that function~ global behavior.

-I, -0.5, -0.1. Explain what

4x

Xo = 2

x +I

x - I 3x

2

+

I

'

Xo ~ -I

67. f(x) = sin 2x, Xo = 'IT/2 68. f(x) = x 2 cos x, Xo = 'IT/4

Differentiation Rules This section introduces several rules that allow us to differentiate constant functions, power functions, polynomials, rational functions, and certain combinations of them, simply and directly, without having to take limits each time.

Powers, Multiples, Sums, and Differences A simple rule of differentiation is that the derivative of every constant function is zero.

116

Chapter 3: Differentiation

y

Derivative of a Constant Function

c

If f has the constant value f(x) = c, then

df d dx = dx (c) = O. ~I-------"---"-_--'-c~-~x

o

x

x+h

FIGURE 3.9 The rule (d/dx)(c) = 0 is another way to say that the values of constant functions never change and that the slope ofa horizontal line is zero at every point.

Proof We apply the definition of the derivative to f(x) = c, the function whose outputs have the constant value c (Figure 3.9). At every value of x, we rmd that

t'(x)

lim f(x + h) - f(x) h

=

h-O

= lim c h-O

- c h

= limO = h-O

o.

•

From Section 3.1, we know that

From Example 2 of the last section we also know that

These two examples illustrate a general rule for differentiating a power x'. We first prove the rule when n is a positive integer.

Power Rule for Positive Integers: If n is a positive integer, then

H!SlORICAL BIOGRAPHY

Richard Courant (1888-1972)

Proof of the PoSitive Integer Power Rule

z' - x·

=

The formula

(z - x)(z·-l + z·-2 X + ... + zx·- 2 + x·- 1)

can be verified by multiplying out the right-hand side. Then from the alternative formula for the definition of the derivative,

t'(x)

= lim z_x

f(z) - f(x) zx

= lim(z·-l

= lim z' - x· z_x z x

+ z'-":,; + ... + zx·- 2 + x·- 1)

• terms

z~x

• The Power Rule is actoally valid for all real numbers n. We have seen examples for a negative integer and fractional power, but n could be an irrational number as well. To apply the Power Rule, we subtract 1 from the original exponent n and multiply the result by n. Here we state the general version of the rule, but postpone its proof until Chapter 7.

3.3


117

Power Rule (General Version) If n is any real number, then

~xn

=

dx

nx n- t '

for all x where the powers x' and X,-l are defined.

EXAMPLE 1 (a)

Differentiate the following powers ofx.

x'

(d)

~4

(f) Yx2+w

(e) x-4/'

x

Solution (8) :fx(x') = 3x'-1 = 3x 2

(c) :fx(xv'i)

=

(e) ~(x-4/') = dx

(b) :fx(x 2/') = t x (2f3)-l = tx-I/' (d) :fx84) = :fx(x-4) = -4x-4-1 = -4x-s = -:s

v'2xv'i-l

_ix -(4/')-1 3

_i x - 7/'

=

3

(f) :fx(Yx2+w) = :fx(X 1+(W/2») =

(I +

;)x 1+(W/2)-1 =

t(2 +

1T)W

•

The next rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.

Derivative Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then :fx(cu) = c:. y

In particular, if n is any real number, then

:fx (ex') = cnx,-t. Slope /

3

(I, 3)

~

3(:!x) -00; ~

6(1)

~

6

Proof d

.

cu(x

~_cu = hm

2

Slope

~

:!x

~

2(1)

~

2

+

h-O

UA.

h) - cu(x) h

Derivative def'mition with j(x) ~ ",,(x)

=

. u(x + h) - u(x) dun ---'-----;----'~ h~O h

Constant Multiple Limit Property

=

du c dx

u is differenti.able.

1(1, I)

•

1

EXAMPLE 2

1 1

1

----'~r='----__:_--____!--->x

o

1

2

FIGURE 3.10 Thegraphsofy ~ x'and y = 3x'. Tripling the y-coordinate triples the slope (Example 2).

(8) The derivative formula

~(3X2) dx

= 3·2:< = 6x

says that if we rescale the graph of y = x'bymultiplying eachy-coordinate by 3, then we multiply the slope at each point by 3 (Figure 3.10).

118


(b) Negative ora function The derivative ofthe negative of a differentiable function u is the negative ofthe function's derivative. The Constant Multiple Rule with c = -I gives

•

l£(-u) = l£(-Iou) = -I.l£(u) = _du ~

I

Denoting Fnnctions by u and v The functions we are working with when we need a differentiation formula are likely to be denoted by letters like f aod g.

~

~~.

The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.

Derivative Sum Rule If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable. At such points,

We do not want to use these same letters when statiog general differentiation rules, so we use letters like u aod v instead that are not likely to be already in use.

d du dv ~(u+v)=~+~.

For example, if y = x 4 + 12x, then y is the sum of u(x) = x 4 and v(x) = 12x. We thenbave

dy d ~ = ~(X4)

d

d ~(I2x) = 4x 3 + 12.

We apply the def"mition of the derivative to f(x) = u(x) + v(x):

Proof ~_

+

() [u x

+v

( ) . [u(x + h) + v(x + h)] - [u(x) + v(x)] x ] = lim h h--i'O

UA,

= lim [u(x + h) - u(x) + v(x + h) - v(x)] h~O

h h . u(x + h) - u(x) . v(x + h) - v(x) = lim h +1Jm h h--i'O

h~O

du

dv

UA,

UA,

=~_+~_ . •

Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives:

l£(u _ v) = l£[u + (-I)v] = du + (-I) dv = du _ dv ~

~

~

~

~

~.

The Sum Rule also extends to finite sums of more than two functions. If Ul, "2, ... , Un

are differentiable at x, then so is "1 + d

~ (Ul + U2 + ... + un) =

dUl

Iii

U2

+

+ ... + Un, and

dU2

Iii

+ ... +

dUn

Iii'

For instance, to see that the rule holds for three functions we compute

d

d

d

~ (Ul + U2 + u,) = ~ ((Ul + U2) + u,) = ~ (Ul + U2) +

du,

Iii

=

dUl

Iii

+

dU2

Iii

+

du,

Iii'

A proofhy mathematical induction for any finite number of terms is given in Appendix 2.

EXAMPLE 3 Solution

:

Find the derivative of the polynomial y = x 3 +

= ;"x

3

= 3x 2 +

+ ;"

2 (jcx ) - ;" (5x)

i. 2x 3

+ ;" (I)

- 5 + 0 = 3x 2 +

llx 3

jc x 2 -

5x + 1.

Smn and Difference Rules

5

•

3.3

119


We can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 3. All polynomials are differentiable at all values of x.

EXAMPLE 4 where? Solution

Does the curve y

=

x4

-

2x 2 + 2 have any horizontal tangents? If so,

The horizontal tangents, if any, occur where the slope dy/dx is zero. We have

dy

Now solve the equation dx

=

0 for x: 4x 3 - 4x = 0

(-I, 1)

-1

(I, 1)

4x(x 2

I) = 0

-

x=O,I,-1.

0

FIGURE 3.11 The curve in Example 4 and its horizontal tangents.

The curve y = x 4 - 2x 2 + 2 has horizontal tangents at x = 0, I, and -1. The corresponding points on the curve are (0, 2), (I, I) and (-I, I). See Figure 3.11. We will see in Chapter 4 that finding the values of x where the derivative of a function is equal to zero is an important and useful procedure. _

Products and Quotients While the derivative of the sum of two functions is the sum oftheir derivatives, the derivative of the product of two functions is not the product oftheir derivatives. For instance,

d

d

dx (x), dx (x) = I, I = 1.

while

The derivative of a product oftwo functions is the sum of two products, as we now explain.

Derivative Product Rule If u and v are differentiable at x, then so is their product uv, and d() dv du dx uv =udx+v dx '

The derivative of the product uv is u times the derivative ofv plus v times the derivative ofu. Inprime notation, (uv)' = uv' + vu'. In fimction notation,

:fx [f(x)g(x)] = EXAMPLE 5

Find the derivative ofy

+ g(x)j'(x).

f(x)g'(x) = (x 2

+

1)(x 3

+ 3).

Solution

(a) From the Product Rille with u

:fx [(x 2 + l)(x 3 + 3)] =

=

x2

(x 2 4

+ I and v

+

1)(3x 2 ) 2

=

+ 4

x3

(x 3

+ 3, we f'md + 3)(2x)

=3x +3x +2x +fix = 5x 4

+ 3x 2 + fix.

d

-(uv) dx

~

O.

v(x

(b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial: y = (x 2 + l)(x 3 + 3) = x 5 + x 3 + 3x 2 + 3

+ h)

= 5x

:

f---------, u(x+h)..1.v /+------'--'---'----. ---

4

+

3x 2

+ 6x.

&v

u(x)v(x)

o

Iv(x) 4u

Proof of the Derivative Product Rule

it (uv)

/

4u \ u(x) u(x + h)

Then the change in the product uv is the difference in areas ofthe larger and smaller "squares," which is the sum ofthe

upper and right-hand reddish-shaded rectangles. That is, .i(uv)

~ ~

u(x u(x

.i(uv) -h-

it( ) _ ' dx

uv -11m

h-O

[( h) v(x ux+

+ h)

,

(

v(x

h-O

4u

u(x + h)v(x + h) - u(x + h)v(x) + u(x + h)v(x) - u(x)v(x) h

=

= I'

= I=u

= u(x + h)T + v(x)T'

The limit as h .... 0+ gives the Product

h) - u(x)v(x)

h

h-O

To change this fraction into an equivalent one that contains difference quotients for the derivatives ofu and v, we subtract and add u(x + h)v(x) in the numerator:

+ h)v(x + h) - u(x)v(x) + h).iv + v(x)4u. 4V

+ h)v(x +

lim u(x

=

dx

Divisioo by h gives

Rille.

•

This is in agreement with our flISt calculation.

v(x)

h-O

x+

h) I'

.

=

- v(x)

h

+

( ) u(x +vx

h) - v(x)

h

()'

+ v x ·1=

h

h-O

+ h)

- U(X)] u(x

+

h) - u(x)

h

h-O

As h approaches zero, u(x + h) approaches u(x) because u, being differentiable at x, is continuous atx. The two fractions approach the values of dv/dx atx and du/dx atx. In short,

d() dv dx uv = u dx

du

+ v dx'

•

The derivative ofthe quotient oftwo functions is given by the Quotient Rule.

Derivative Quotient Rule If u and v are differentiable at x and if v(x) '" 0, then the quotient u/v is differentiable at x, and

it dx

('!)

du =

dv

vfiX - ufiX

if'

V

In function notation,

it [f(X)]

= g(x)t'(x) - f(x)g'(x)

dx g(x)

EXAMPLE 6

Solution

Find the derivative of y

g2(x)'

t2 - I t + I

= -3--'

We apply the Quotient Rule with u dy dt

=

t2

-

I and v

(t 3 + I)' 2t - (t 2 - I). 3t 2 (t 3 + 1)2

+ 2t - 3t 4 + 3t 2 (t 3 + 1)2 -t 4 + 3t 2 + 2t (t 3 + I)'

=

t 3 + I:

~

(Il)

dt v

~ v(du/dt) - u(du/dt)

if

2t 4

•

3.3


121

Proof of the Derivative Quotient Rule u(x + h)

('!)

~

lim v(x + h)

=

:' - lOx - 5x-'

27. Y

26.r=2(.vo+VO)

u(O)

2

- 7t'(x

+ 5)

= 5,

u'(O)

= -3,

v(O)

= -I,

v'(O)

=0

= 2.

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

a. dx (uv)

d d. dx (7v - 2u)

3.3

passing through the point (3, 8).

u(i) = 2, u'(I) = 0, v(l) = 5, v'(I) = -I.

d

~

123

52. Find all points (x,y) on the graph of f(x) ~ x' with tangent lines

42. Suppose u and v are differentiable functioos of x and that

Find the values of the following derivatives at x

Oifferentiation Rules

f(x) ~ x'

I.

d d. dx (7v - 2u)

a. dx (uv)

Slopes and rangents 43. a. Normal to aeurve Find an equation for the line perpendicular to the tangent to the curve y = x' - 4x + I at the point (2, I). b. Smallest slope What 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 equatioos for the horizontal tangents to the curve y = x' - 3x - 2. Also !md equations for the lines that are perpendicular to these tangents at the points of tangency.

b. Smallest slope What 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" serpentine (graphed here) at the origin and the point (I, 2). y

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 tangent line together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordiuates. Dc. ConImn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

54. a. Find an equation for the line that is tangent to the curve y = x 3 - 6x 2 + Sx at the origin. D b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom. and Trace to estimate the point~ coordiuates. Dc. ConImn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

Theory and Examples 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 converting 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.

g(x)

x

47. Quadratic tangent to identity function The curve y = ax' + bx + c passes through the point (I, 2) and is tangent to the line y = x at the origin. Find a, b, and c. 48. Quadratics having a common tangent The curves y = x 2 + ax + b and y = ex - x 2 have a common tangent line at the point (I, 0). 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.

~

ax { x" - 3x,

if x < 0 if x '" 0

58. Find the values of a and b that make the following function differentiable for all x-values.

ax + b, () = { bx'-3, fx

x> x:5

-I -1

59. The general polynomial of degree n has the fonn

P(x) where a.

¢'

=

anx n + an_tXn-1

+ ... + a2x 2 + alX + ao

O. Find P' (x).

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

124


where C is a positive constant and M is the amount ofmedicine 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 dR/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? What does this say about the Derivative Constant Multiple Rule?

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

where m is a positive integer.

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 P =

62. The Reciprocal Rule

dx

'

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

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

!£

2

nRT _ an V - nb V2

(1) __ -.L v

-

dv v 2dx ·

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 gives the formula d dv -(uv) = u dx dx

The Derivative Product Rule

+

du vdx

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 A(q) =

ofx.

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 1U2 u3 U4 offour differentiable functions of x?

c. What is the formula for the derivative of a product Ul U2 U3 ... 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 dA/dq and d 2A/dq 2.

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 (f(x + h) - f(x))/h as the average rate of change in f over the interval from x to x + h, we can interpret its limit as h ~ 0 as the rate at which f is changing at the point x.

3.4

DEFINmON

The Derivative as a Rate of Change

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 ofchange.

EXAMPLE 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 m 2/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

11s--->!'1

~

/(t).

The displacement ofthe object over the time interval from tto t

+ At

•

=

tis

) S

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

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

=

+ tit (Figure 3.12) is

/(t + tit) - /(t) ,

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

=

displacement tis /(t travel time = tit =

+ tit) - /(t) tit

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 + tit as tit 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 + tit) - /(t) ti 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 ofmotion. When the object is moving forward (s increasing), the velocity is positive; when the object is moving backward (s 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

I ~------_)I

o

s increasing:

s decreasing:

positive slope so

negative slope so moving downward

moving upward

FIGURE 3.13 For motioo s ~ 1(1) along a straight line (the vertical axis), v = dsldt is positive when s increases and negative whens decreases. The blue curves represent positioo along the lioe over time; they do not portray the path of motion, which lies along the s-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 speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.

DEFINmON

Speed is the absolute value ofvelocity. Speed = Iv(I)1 =

I~I

EXAMPLE 2 Figure 3.14 shows the graph of the velocity v = 1'(1) ofa particle moving along a horizontal line (as opposed to showing a position function s = 1(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. •

3.4 The Derivative as a Rate of Change

127

v MOVES FORWARD

I

I

I

I FORWARD I I AGAIN I

I I

(v> 0)

I I

Speeds~Steady~ Slows

up

:(v =

consO:

Velocity v

I

I

down

J

~ 1'(')

I

i

:

I

I

I

I

I

I

I

I

r

Stands

,

(v

I

I I

till

,S

I I

2

: (v> 0)

~ Speeds ~ I up I

I I

o

I

5

3,

~

I

I

~I

I 0) ,

6

, , , , , , , ,

I ,

l

I

Speeds

4- Slows ~ down

I

MOVES BACKWARD

, I

IUp

I

,

I

, I

I

:

(v

< 0)

,

:

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

Bernard Balzano (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 1is s = /(1), then the body's acceleration at time 1is 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 significant. The value of g in the equation s = (1/2)gI2 depends on the units used to measure 1and s. With 1in secoods (the usua1 unit), the value ofg determined by measurement at sea level is approximately 32 ft./sec 2 (feet per second squared) in Eoglish units, and g = 9.8 m/sec2

128


r (seconds)

=0 t=1 t

•.. ' ,

'-'

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

s (meters)

0 5 10

t

~

2

15

..

' , '-'

j =

30

~

3

'-'

O.

EXAMPLE 3 Figure 3.15 shows the free fail of a beavy bail bearing released from rest at time I = Osee.

35 t

=

An object does not exhibit jerkiness during free fail.

25

'..,

~ (g)

20

40

-

50 0 -50 -100 0

2

4 6 8 10 Time after launch (sec)

! ! ! ! '!

-I

12

c. When did the rocket reach its highest point? What was its velocity then? d. When did the parachute pop out? How fast was the rocket falling then? e. How long did the rocket fall before the parachute opened? f. When was the rocket's acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)?

J I a. How long did it take the balls to fall the first 160 cm? What was their average velocity for the period? b. How fast were the balls falling when they reached the l60-cm mark? What was their acceleration then? c. About how fast was the light flashing (flashes per second)?

134


20. A traveling trud< The accompanying graph shows the positioo s of a truck traveling 00 a highway. The truck starts at t ~ 0 aod returns 15 h later at t = 15.

Economics 23. Marginal co.t Suppo.e that the dollar co.t of prodocing x washing machines is c(x) ~ 2000 + 10(1,; - 0.lx 2 •

a. Use the technique desctihed in Sectioo 3.2, Example 3, to

a. Find the average cost per machine of producing the rlrst 100 washing machines.

graphthetruck~velocityv ~

0:

dy dx

lim ay 4x~O

ax

lim ay, au

ax

4x~O au

lim ay, lim au

=

4x~O

4x~O

au

ax (Note that..:1u --+ 0 as ~- 0 since g is continuous.)

lim ay, lim au ..:1u_o.6.u .:1%-0 ax dy du =du'dx· =

The problem with this argument is that it could be true that au = 0 even when ax o¢' 0, so the cancellation of au in Equation (I) would be invalid. A proof requires a different approach that avoids this flaw, and we give one such proof in Section 3.9. •

EXAMPLE 2 An object moves along the x-axis so that its position at any time t ;;,: 0 is given by x(t) = cos(t 2 + I). Find the velocity ofthe object as a function oft. Solution We know that the velocity is dx/dt. In this instance, x is a composite function: x = cos(u) andu = t 2 + I. Webave

dx du du dt

= -sin(u)

x ~ cos(u)

2t.

=

By the Chain Rule,

dx dx du dt=du'dt tb:

= -sin(u)' 2t = -sin(t 2 + 1)' 2t

du evaluated at u

•

= -2tsin(t 2 + I).

UOutside-lnside u Rule A difficulty with the Leibniz notation is that it doesn't state specifically where the derivatives in the Chain Rule are supposed to be evaluated. So it sometimes helps to think about the Chain Rule using functional notation. If y = f(g(x)), then

dy dx

=

f'(g(x))' g'(x).

In words, differentiate the "outside" function f and evaluate it at the "inside" function g(x) left alone; then multiply by the derivative of the "inside function."

EXAMPLE 3 Solution

Differentiate sin (x 2

+ x) with respect to x.

We apply the Chain Rule directly and find

:Ix sin (x 2 + x) = inside

cos (x 2

+ x), (2x +

inside left alone

1).

derivative of the inside

•

3.6 The Chain Rule

145

Repeated Use of the Chain Rule We sometimes have to use the Chain Rule two or more times to fmd a derivative. HIsTORICAL BIOGRAPHY

Johann Bernoulli (1667-1748)

EXAMPLE 4

Find the derivative ofg(t) = tan(5 - sin2t).

Solution Notice here that the tangent is a function of 5 - sin 2t, whereas the sine is a function of2t, which is itself a function oft. Therefore, by the Chain Rule,

g'(t)

~(tan (5

- sin 2t))

Derivati:~ of tan u with u=5-sin2t

= sec2 (5 -

sin2t)· :t(5 - sin2t)

Derivative of 5 - sin u with u = 2t

2 = sec (5 -

sin2t)·

=

(0 - cos2t· ~(2t))

= sec2 (5 -

sin2t)'(-cos2t)'2 -2(cos2t) se~(5 - sin2t).

=

•

The Chain Rule with Powers of a Function If j is a differentiable function of u and if u is a differentiable function of x, then substitoting y = j(u) into the Chain Rule fonnula

dy dy du dx=du'dx leads to the fonnula

~j(u) =

f'(u)a;.

Ifn is any real number and j is a power function, j(u) = un, the Power Rule tells us thatf'(u) = nun-I. Ifu is a differentiable function ofx, then we can use the Chain Rule to extend this to the Power Chain Rule: ~( n) _

dx u

EXAMPLE 5

- nu

n-I

du dx'

The Power Chain Rule simplifies computing the derivative of a power of

an expression. (a) ~(5x3 - x 4)7 = 7(5x 3 - x4)6~(5x3 - x 4) dx dx = 7(5x 3 - x 4)6(5' 3x 2 - 4x 3) =

(b)

~ (_I-) = dx 3x-2

Power Chain Rule with u

= 5;t3

- X 4 ,1I

=7

7(5x 3 - x 4)6(15x 2 - 4x 3) 4/dtwheo 8 ~ 3'1f/2 ifs ~ cos8andd8/dt ~ 5. 76. Find eIy/dt wheox ~ I ify ~ x 2

+ 7x

- 5andd>/dt ~ 1/3.

What happens if you can write a function as a composite in different

b. y

= x by using lhe Chain Rule withy as a compos-

= (u/5) + 7 = I + (I/u)

u and u and

= 5x - 35 = I/(x - I).

78. Find eIy/dx if y = x'/2 by using the Chain Rule wilh y as a composite of

a. y = u' b. y

= 'VU

and and

101)] + 25

and is graphed in the accompanying figure.

a. On what day is the temperature increasing the fastest? ing when it is increasing at its fastest? y

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.

a. y

[;t7s (x -

b. About how many degrees per day is the temperature increas-

Theory and Examples

77. Find dy/dx ify ite of

= A cos (2'1fbt) ,

u = vX

u =x'.

79. Findthetangeottoy = «x - I)/(x 80. Find the tangeot to y =

y'x 2

-

X

+

1))2 atx = O.

+ 7 at x

= 2.

81. a. Find lhe tangeot to lhe 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

a. Find equations for the tangeots to the curves y = sin 2x and y ~ -sin(x/2) at lhe origin. Is lhere anything special about how the tangeots are related? Give reasons for your answer. b. Can anything be said about lhe tangeots to lhe curves y ~ sinmxandy ~ -sin(x/m) at the origin (m a constant '¢' O)? Give reasons for your answer. c. For a giveo m, what are the largest values the slopes of lhe 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, 2'1f1, lhe 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.

60 H-+-+-+--bo>""i-.,..d-+-+-+-+-+-H .... fi:'

i

I....'

c... 40 HH-+~'-+-+-+-+' "~-+--+--+-+-+-I

~

--~,-----~ - - - - 1'\ ----~--I>x

20 - - - V0

-til

-20

,,~
tion is constant. 87. Falling meteorite The velocity of a heavy meteorite eotering Earth's atmosphere is inversely proportional to VS wheo it is s km from Earth's ceoter. Show that the meteorite's acceleration is inversely proportional to S2 • 88. Particle acceleration A particle moves along lhe x-axis wilh velocity dx/dt ~ f(x). Show that lhe particle's acceleration is

f(x)1'(x). 89. Temperature and the period of a pendnIum For oscillations of small amplitude (short swings), we may safely model the relationship betweeo the period T and the leogth L of a simple peodulum with the equation

T

~ 2'1f~,

where g is the constant acceleration of gravity at the peodulum's

location. If we measure g in centimeters per second squared, we measure L in ceotimeters and T in seconds. If lhe peodulum is

Implicit Differentiation

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

a. Graph dg/dt (where defmed) over [-"., ".].

b. Findd//dt. c. Graph d// dt. Where does the approximation of dg/dt by d//dt seem to be best? Least good? Approximations by 1rigonome1ric polynomials are important in the theories of heat and oscillation, but we must not expect too much of them, as we see in the next exercise.

constant, tiL du = kL. 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 composites

(j

0

g)(x) =

lxi' =

are both differentiable at x

o

and

(g

0

/)(x) =

Ix'i =

91. The derivative of sin 2x Graph the function y = 2 cos 2x for -2 '" x '" 3.5. Then, on the same screen, graph

+ 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. 92. The derivative of cos (x') Graph y = -2x sin (x') -2 '" x '" 3. Then, on the same screen, graph cos ((x

y=

+ h)') -

~

g(t)

'~f(t)

/"

o

-"

= 0 even though g itself is not differ-

sin 2(x

, /

x'

entiable at x = O. Does Ibis contradict the Chain Rule? Explain.

y=

o

x'

,

Ixl. Then the

Suppose that /(x) = x' and g(x) =

149

"

94. (Continuation ofExercise 93.) In Exercise 93, the 1rigonome1ric polynomial /(t) that approximated the sawtooth function g(t) on [-"., ".j 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 approximating 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.

cos (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.

,


I

Trigonometric Polynomials

-"

93. As the accompanying figure shows, the 1rigonometric "polynomial"

k(t)

~h(t)

1

-2"

~

"2

0

"

-1

s = /(t) = 0.78540 - 0.63662 cos 21 - 0.07074cos6t - 0.02546 cos lOt - 0.01299 cos 14t

a. Graph dk/dt (where defmed) over [-"., ".j.

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 fann 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 technique.

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 usua1 way. Then we differentiate the equations inJplicitly, 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 implicitly as a differentiable function of x so that dyjax exists.

EXAMPLE 1 FIGURE 3.26 The curve + y' - 9xy ~ Dis not the graph of

x'

anyone function of x. The curve can,

however, be divided into separate arcs that are the graphs offunctions ofx. This particular curve, called afolium, dates to Descartes in 1638.

Find dyjax if y 2 = x.

The equation y2 = x defmes two differentiable functions of x that we can actoally find, namely y, = YX and Y2 = -YX (Figure 3.27). We know how to calculate the derivative of each of these for x > 0:

Solution

dYl ax

I 2YX

dY2 ax

and

1 2YX .

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 dyjax, 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

= -t

2Y2

x

The Chain Rule give. ::.

ax

-Yx

dy ax

t

=--

2Yx

FIGURE 3.27 The equationy2 - x = 0, or y2 = X as it is usually written, dermes two differentiable functions ofx on the interval x > O. Example I shows how to find the derivatives of these functions without solving the equation y2 = x for y.

(y 2 ) ~

d 2 dy dx [f(x)] ~ 2f(x)f'(x) ~ 2y dx'

2y dy = 1

Q(x, -vX)

Y2~

/

Slope

=

I 2y'

This one fannula gives the derivatives we calculated for both explicit solutions y, = YX andY2 = -YX:

dYl ax

EXAMPLE 2

I 2Yl

I 2YX

and

d)i2 ax

I

2( -YX)

I 2YX .

•

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 Y2, 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 2V25=9

3 4'

3.7 Implicit Differentiation y

151

We can solve this problem more easily by differentiating the given equation of the cITcle implicitly with respect to x:

~(X2) + ~(y2) dx

dx

=

dx

dy 2x+ 2y dx=O

--_5*---~Ot----f;5;--~X

dy dx The slope at (3, -4) is--!I = Y (3, ....) FIGURE 3.28 The circle combines the graphs of two functions. The graph of Y2 is the lower semicircle and passes through (3, -4).

~(25)

x y'

-~ = 1. -4 4

Notice that unlike the slope formula for dy,)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 ofother implicitly defmed functions, we proceed as in Examples I and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides ofthe defming equation.

Implldt Differentiation 1. Differentiate both sides ofthe 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.

EXAMPLE 3

y

Find dy/dx if y2 = x 2

+ sinxy

(Figure 3.29).

T~i'+sinxy

Solution

We differentiate the equation implicitly.

y2 :. (y2)

FIGURE 3.29 The graph of y 2 = x 2 + sin>;)' in Example 3.

2Y: -

(cosxy)

=

x 2 + sinxy

= :.

(x 2) + :. (sinXY)

2y dy dx = 2x

+

d (xy ) (cosxy) dx

2y: = 2x

+

(cosxy)&

(x :)

=

2x + (cosxy)y

dy (2y - xcosxy) dx = 2x dy dx

+ x :)

Differentiate both sides with respect to x ...

. .. treatingY as a function of and using the Chain Rule.

x

Treatxy as a product

Collect terms with dy/dx.

+ ycosxy

2x + Y cosxy 2y - x cosxy

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 also be used to find higher derivatives.

EXAMPLE 4

Find d'y/dx 2 if 2x' - 3y 2 = 8.

Solution To start, we differentiate both sides ofthe equation with respect to x in order to findy' = dy/dx.

JI (2x' dx

JI (8)

- 3y 2) =

dx

1leatyasa functianofx.

6y'y' = 0

6>;2 -

2

y'

x y,

=

whenyoF 0

Solvefory'.

We now apply the Quotient Rille to fmd y" . y" =

Finally, we substitute y'

=

fx (~)

y ;,x2 ' =

= 2xy

~ - ;~.y'

x 2/y to express y" in terms ofx andy. 2

2

_ 2x x (x- ) x' _ 2x y" ----Y

Curve of lens smface

y2

Y

Y

•

wheny oF 0

y"

Lenses, Tangents, and Normal Lines In the law that describes how light changes direction as it enters a lens, the important an-

gles 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 to 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 perpendicu1ar to the tangent ofthe profile curve at the point of entry. FIGURE 3.30 TIre profile ofa leos, showing the beoding (refraction) ofa ray oflight as it passes through the leos

surface.

y

EXAMPLE 5 Show that the point (2, 4) lies on the curve x' + y' - 9xy find the tangent and normal tu the curve there (Figure 3.31).

=

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 firat use implicit differentiation to fmd a fonnu1a for dy/dx: x'+y'-9xy=0

3x

2

JI (x') + JI (y') dx

dx

-

JI (9xy)

=

JI (0)

dy + 3y 2 dx

- 9 (d x dx

Y

+ Y dx) dx

=

0

dy (3y 2 - 9x) dx

FIGURE 3.31 Example 5 shows how to fmd equations for the taogeot and normal to the folium ofDescartes at (2, 4).

dx

+ 3x 2

-

dx

9y = 0

3(y2 - 3x) dy = 9y - 3x 2 dx dy 3y-x 2 dx

y2 - 3x'

Differentiate both sides with respect to x. TreatX)' as a product andy

as 8 function of x.

Solve for dy/d

X

(2,3)

(3, -4)

(-1,3)

32. y2 - 2x - 4y - I = 0, 2

35. 2xy + 'If siny = 2'1f,

44. The folium of Deseartes

(-2, I)

+ 3xy + 2y 2 + 17y - 6 = 0, 2 34. x - V3xy + 2y 2 = 5, ( V3, 2 ) 33. 6x

y

(-2, I) aod(-2,-1)

(-1,0)

(I, 'If/2)

36. xsin2y = ycos2x, ('If/4,'If/2) 37. y ~ 2 sin ('IfX - y), (1,0) 38. x 2 cos2 y - siny ~ 0,

(See Figure 3.26.)

a. Find the slope of the folium ofDescartes x'

+ y'

- 9xy ~ 0

at the points (4, 2) aod (2, 4). b. At what point other thao the origin does the folium have a horizontal tsngent?

c. Find the coordinates of the point A in Figure 3.26, where the folium has a vertical tsngent.

(0,'If)

39. Parallel tangents Find the two points where the curve x 2 + xy + y2 = 7 crosses the x-axis, aod show that the taogents to the curve at tlrese points are parallel. What is the common slope of these taogents?

Theory and Examples 45. Intersecting normal The line that is normal to tire curve x 2 + 2xy - 3y 2 = 0 at (I, I) intersects the curve at what other point?

40. Normals parallel to a line Find the normals to the curve xy + 2x - y ~ 0 that are parallel to the line 2x + y ~ O.

46. Power rule for rational exponents 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,

41. The eight curve Find the slopes of the curve y' ~ y2 - x 2 at the two points shown here.

~ dl;Xpl.

y

-----'~------>x

47. Normals to a parabola Show that if it is possible to draw three normals from the point (a, 0) to the parabola x = y2 shown in the accompaoying diagram, then a must be greater thao 1/2. One of the normals is the x-axis. For what value of a are the other two normals perpendicular? y

-1

42. The cissoid ofDioeles (from about 200 B.c.) Find equations for the tsngent and normal to tire cissoid of Diocles y2(2 - x) ~ x' at (1,1). y

= EqX "'1.)-1 .

3.8

Related Rates

155

x' D 10 Exercises 51 and 52, fmd both dy/dx (treatingy as a differeotiable

48. Is there anything special about the tangents to the curves y' ~ and 2;x' + 3y 2 = 5 at the points (I, ± I)? Give reasoos for your

answer. y

function of x) and dx/dy (treating x as a differeotiable function ofy). How do dy/dx and dx/dy seern to be related? Explain the relationship geometrically in terms of the graphs. 51. xy'

+ x'y

52. x'

= 6

+ y2

= sin2 y

COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises 53-{)(). a. Plot the equation with the implicit plotter ofa CAS. Check to see that the given point P satisfies the equation.

-f----=+ 0, the difference quotient

I(a

+ Ilx) - I(a) Ax

approaches I'(a) (remember the definition of I'(a», so the quantity in parentheses becomes a very small number (which is why we called it e). In fact, e --> 0 as Ax--> O. When Ax is small, the approximation error e Ax is smaller still.

l'(a)llx

+ e Ax

true

estimated

error

change

change

III

=

Although we do not know the exact size of the error, it is the product e • Ilx of two small quantities that both approach zero as Ax --> O. For many common functions, whenever Ilx is small, the error is still smaller.

Change In Y

= f(x) near x = II

If y = I(x) is differentiable at x = a and x changes from a to a change Ily in I is given by

Ily in which e --> 0 as Ilx --> O.

=

I'(a) Ax + e Ax

+ Ilx,

the (1)

170

Chapter 3: Differentiation In Example 6 we found that

aA

'IT(10.1)2 - 'IT(10)2 = (102.01 - 100)'IT = (2'IT

=

-dA

-

+ 0.01 'IT) m2

so the approximation error is aA - dA = E/1r = 0.01'IT and O.OI'ITjO.1 = O.I'IT m.

error E

=

0.01 'lTj /1r =

Proof of the Chain Rule Equation (I) enables us to prove the Chain Rule correctly. Our goal is to show that if I( u) is a differentiable function of u and u = g(x) is a differentiable function of x, then the composite y = l(g(x)) is a differentiable function ofx. Since a function is differentiable if and only if it bas a derivative at each point in its domain, we must show that whenever g is differentiable at Xo and 1 is differentiable at g(xo) , then the composite is differentiable at Xo and the derivative ofthe composite satisfies the equation

: Ix~xo =

I'(g(xo))' g'(xo).

Let /1x be an increment in x and let /1u and /1y be the corresponding increments in u andy. Applying Equation (I) we have

/1u where El

--->

=

+ El /1x

=

(g'(xo)

l'(uo)/1u + E2/1u

=

(I'(uo) + E2)/1u,

g'(xo)/1x

+ E,J/1x,

0 as /1x ---> o. Similarly,

/1y

=

where E2 ---> 0 as /1u ---> O. Notice also that /1u ---> 0 as /1x ---> O. Combining the equations for /1u and /1y gives

/1y

=

(j'(uo) + E2)(g'(Xo) + El)/1x,

so

Since El and E2 go to zero as /1x goes to zero, three ofthe four terms on the right vanish in the limit, leaving

dyl

-

ax

= X=Xo

lim -/1y = I'(uo)g'(xo) = I'(g(xo))' g'(xo).

J1x~O 4x

•

Sensitivity to Change The equation df = I'(x) dx tells how sensitive the output of 1 is to a cbange in input at different values of x. The larger the value of I' at x, the greater the effect of a given cbange dx. As we move from a to a nearby point a + dx, we can describe the change in 1 in three ways:

Absolute change Relative change Percentage change

True

Estimated

/11 = I(a + dx) - I(a) /11 I(a)

dl = dl

/11 I(a) X 100

I'(a)dx

I(a)

dl

I(a) X 100

3.9 Linearization and Differentials

171

You want to calculate the depth of a well from the equation s = 16t 2 by timing how long it takes a heavy stone you drop to splash into the water below. How sensitive will your calculations be to a O.I-sec error in measuring the time?

EXAMPLE 7

Solution

The size of ds in the equation

ds = 32tdt depends on how big t is. If t = 2 sec, the change caused by dt = 0.1 is about

ds = 32(2)(0.1) = 6.4 ft. 1bree seconds later at t = 5 sec, the change caused by the same dt is

ds = 32(5)(0.1) = 16 ft. For a fixed error in the time measurement, the error in using ds to estimate the depth is _ larger when the time it takes until the stone splashes into the water is longer.

EXAMPLE 8 In the late 1830s, French physiologist Jean Poiseuille (''pwa-ZOY'') discovered the formula we use today to predict how much the radius of a partially clogged artery decreases the normal volume of flow. His formula, V = /cr', says that the volume V of fluid flowing through a small pipe or tube in a unit of time at a rIXed pressure is a constant times the fourth power of the tube's radius r. How does a 10% decrease in r affect V? (See Figure 3.45.)

Angiography

Angioplasty

FIGURE 3.45 To uoblock a clogged artery, an opaque dye is injected into it to make the inside visible under X-rays. Then a balloontipped catheter is inflated inside the artery to widen it at the blockage site. Solution

The differentials of r and V are related by the equation

dV =

a;

dr = 4/cr 3 dr.

The relative change in V is 3

dV = 4/cr dr = 4dr V /cr' r . The relative change in V is 4 times the relative change in r, so a 10% decrease in r will _ result in a 40% decrease in the flow.

EXAMPLE 9

Newton's second law,

d dv F = dt (mv) = m dt = ma,

172

Chapter 3: Differentiation is stated with the assumption that mass is constant, but we know this is not strictly true because the mass ofa body increases with velocity. In Einstein's corrected formula, mass bas the value

mo

m = -:-Vrl=_~if~/c=ii2' where the ''rest mass" mo represents the mass of a body that is not moving and c is the speed oflight, which is about 300,000 lan/sec. Use the approximation

1 • ~2 '" 1 vi - x

1

+ -2x

2

(2)

to estimate the increase ~m in mass resulting from the added velocity v.

When v is very small compared with c, if/ c 2 is close to zero and it is safe to use the approximation

Solution

Eq, (2) with x

~

*

to obtain

or

m

Ri

ma

1 2(1)

+ Zmov

c2 .

Equation (3) expresses the increase in mass that results from the added velocity v.

(3)

•

Converting Mass to Energy Equation (3) derived in Example 9 bas an important interpretation. In Newtonian physics, (1/2)m oif is the kinetic energy (KE) of the body, and if we rewrite Equation (3) in the form

we see that

or (~m)c2 '" ~(KE).

So the change in kinetic energy ~(KE) in going from velocity 0 to velocity v is approximately equal to (~m)c2, the cbange in mass times the square of the speed oflight. Using c '" 3 X 10 8 m/sec, we see that a small change in mass can create a large cbange in energy.

3.9

173

Linearization and Differentials

Exercises 3.9 finding L1nearlzattons

Approxlmatton Error

In Exercises 1-5, find the linearizationL(x) of f(x) atx = a.

In Exercises 29-34, each function f(x) changes value when x changes from xo to Xo + dx. Find

1. f(x) = x' - 2x

2. f(x) ~

+ 3, a

W+9, 1

3. f(x)=x+x'

= 2

a ~ -4

a. the change 4f

~

a

tanx,

(b) cosx

+ 2x, Xo

xo

8. f(x) = X-I, 9. f(x) =

Find the lineariza-

lif ~ f(xo + dx)

(c) tanx

10. f(x)

~

1

+ x, Xo

3,

f(xo)

--c-I----L-----'--~------>x

Xo

xo+dx

~

+ 2x, Xo = I, 2x + 4x - 3, Xo

~ l' Xo

30. f(x) = = -I, dx = 0.1 31. f(X) = x' - x, Xo = 1, dx = 0.1

8.1

32. f(X) = x" Xo = 1, dx = 0.1 33. f(X) = x-I, Xo = 0.5, dx = 0.1

= 1.3

+ x)'

at x ~ 0 is

14. Use the linear approximation (I + xl' '" 1 + Ax to find an approximation for the functionf(x) for values ofxnear zero.

a. f(x) = (1 - x)'

~ +x (4 + 3X)'/3

c. f(x) = •

dx = 0.1

2

13. Show that the linearization of f(x) ~ (I L(x) = 1 + Ax.

d. f(x) =

v'2+7 "

f. f(x) =

34. f(x) ~ x' - 2x

+ 3, xo

~ 2,

dx ~ 0.1

Dlfferenttal Estimates of Change In Exercises 35-40, write a differential formula that estimates the

given change in volume or s'Wface area. 35. The change in the volume V = (4/3)'lTr' of a sphere when the radius changes from ro to ro + dr

h. f(x) = 1 :. x

V1

e. f(x) =

-

fix

29. f(x) = x 2

Xo = -0.9

11. f(x) =~, Xo = 8.5

12. f(x) = x

L

Tangent

o

~ 0.1

df.,~ f'(xo) fix

(xo,f(xo)

= 0.9

+ 4x -

2x 2

- f(xo);

y

~ 'IT

L1nearlzatton for Approxlmatton In Exercises 7-12, fmd a linearization at a suitably chosen integer near Xo at which the given function and its derivative are easy to evaluate. 7. f(x) ~ x 2

+ dx)

c. the approximation error 14f - dfl.

6. Common Hnear appnIlimations at x = 0 tions of the following functions at x = O.

(a) sinx

f(xo

h. the value ofthe estimate df = 1'(xo) dx; and

a= 1

4. f(x) =~, a = -8 5. f(x)

~

1(1 -

,,,

2

~ x)'

36. The change in the volume V = lengths change from xo to Xo + dx

x' of a

cube when the edge

37. The change in the surface area S ~ 6x 2 of a cube when the edge lengths change from xo to Xo + dx

+ xl' '"

38. The change in the lateral surface area S = 'lTrv'r 2 + h 2 of a right circular cone when the radius changes from ro to ro + dr and the height does not change

16. Find the linearization of f(x) ~ y';+J + sinx atx ~ O. How is it related to the individual linearizations of y';+J and sin x atx = O?

39. The change in the volume V = 'lTr 2 h of a right circular cylinder when the radius changes from ro to ro + dr and the height does notchange

Derivatives in Differenttal Form

40. The change in the lateral surface area S = 2'lTrh of a right circular cylinder when the height changes from ho to ho + dh and the

15. Faster than a calculator Use the approximation (I 1 + Ax to estimate the following.

a. (1.0002)"

h. '\7"1.009

In Exercises 17-28, find dy.

17. y ~

x' -

3v'i

2x l+x 21. 2y'/2 + -9' - x = 0

19.

Y~--2

18. y ~ 20.y~

24. y

25. y

26. y

= cos (x ) = sec(x 2 -

27. y

= 3 csc (1

28. y

=2

- 2v'i)

Applicattons

2v'i

( ./C) 3 I + vX 22. -9'2 - 4x'/2 - y

= sin (5v'i) = 4 tan (x'/3)

23. y

radius does not change

xv'l=7

41. The radius of a circle is increased from 2.00 to 2.02 m.

=0

a. Estimate the resulting change in area.

h. Express the estimate as a percentage of the circle's original area.

2

I)

cot(Jx)

42. The diameter of a tree was lOin. Doring the following year, the circumference increased 2 in. About how much did the tree's

diameter increase? The tree's cross-section area?

174


43. Estimating volume Estimate the volume ofmaterial in a cylindrical shell with length 30 in., radius 6 in., and shell thickness 0.5 in. 0.5 in. """'--/

6in.L.....

~30in.-'>i

44. Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%? 45. Tolerance The radius r of a circle is measured with an error of at most 2%. What is the maximum corresponding percentage error in computing the circle's

As a member of NASA's medical team, you want to know how sensitive W is to apparent changes in g caused by flight maneuvers, and this depends on the initial value of g. As part ofyour investigation, you decide to compare the effect on W of a given change dg on the moon, where g = 5.2 ft/sec2 , with the effect the same change dg would have on Earth, where g = 32 ft/sec 2 • Use the simplified equation above to find the ratio of dWmoon to dWEarth· 52. Measuring acceleration of gravity When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum's period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of Ii.T, we can estimate the variation in g from the equation T = 27T(L/g)I/2 that relates T, g, andL. a. With L held constant and g as the independent variable, calculate dT and use it to answer parts (b) and (c).

a. circumference? b. area? 46. Tolerance The edge x of a cube is measured with an error of at most 0.5%. What is the maximum corresponding percentage error in computing the cube's a. surface area? b. volume? 47. Tolerance The height and radius of a right circular cylinder are equal, so the cylinder's volume is V = 7Th 3 . The volume is to be calculated with an error of no more than I % of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h. 48. Tolerance a. About how accurately must the interior diameter of a 10-mhigh cylindrical storage tank be measured to calculate the tank's volume to within 1% of its true value? b. About how accurately must the tank's exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5% of the true amount? 49. The diameter of a sphere is measured as 100 ± 1 em and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation. 50. Estimate the allowable percentage error in measuring the diameter D of a sphere if the volume is to be calculated correctly to within 3%. 51. The effect of flight maneuvers on the heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation

W= PV+

b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 1DO-em pendulum is moved from a location where g = 980 cm/sec2 to a new location. This increases the period by dT = 0.001 sec. Find dg and estimate the value of g at the new location. 53. The linearization is the best linear approximation Suppose that y = f(x) is differentiable at x = a and that g(x) = m(x - a) + c is a linear function in which m and c are constants. If the error E(x) = f(x) - g(x) were small enough near x = a, we might think of using g as a linear approximation of f instead ofthelinearizationL(x) = f(a) + f'(a)(x - a). Show that if we impose on g the conditions 1. E(a) = 0

The approximation error is zero at x

. E(x) 2.hm x - a =0 x->a

The error is negligible when compared withx - a.

The linearization, L(x): y = f(a) + f'(a)(x - a)

\

Some other linear approximation, g(x): y = m(x - a) + c y =f(x)

-------'-------~)

+ gb (a, b constant).

I

x

54. Quadratic approximations a. Let Q(x) = bo + bl(x - a) + Mx - a? be a quadratic approximation to f(x) at x = a with the properties:

i) Q(a) = f(a) ii) Q'(a) = f'(a) iii) Q"(a) = j"(a).

W = a

a.

then g(x) = f(a) + f'(a)(x - a). Thus, the linearization L(x) gives the only linear approximation whose error is both zero at x = a and negligible in comparison with x - a.

vat? 2g ,

where W is the work per unit time, P is the average blood pressure, V is the volume ofblood pumped out during the unit of time, a ("delta") is the weight density of the blood, v is the average velocity ofthe exiting blood, and g is the acceleration of gravity. When P, V, a, and v remain constant, W becomes a function ofg, and the equation takes the simplified form

=

Determine the coefficients bo, bl' and b2.

Chapter 3 Questions to Guide Your Review b. Find the quadratic approximation to j(x) x =

D c.

o.

~

a. Plot the function j over 1.

1/(1 - x) at

b. Fiod the linearization L of the function at the point o.

Grapb j(x)

= 1/( I

- x) and its quadratic approximation at

c. Plot j and L together on a single graph.

x ~ O. Then zoom in on the two graphs at the point (0, I). Comment on what you see.

D d.

Find the quadratic approximation to g(x) = I/x at x = I. Graph g and its quadratic approximation together. Comment on what you see.

D e.

Find the quadratic approximation to h(x) = ~ at x ~ O. Graph h and its quadratic approximation together. Comment on what you see.

f. What are the linearizations of j, g, and h at the respective points in parts (b), (d), and (e)?

COMPUTER EXPLORATIONS In Exercises 55-58, use a CAS to estimate the magnitude of the error

in using the lineatization in place of the function over a specified interval 1. Perform the following steps:

Chapter

2. What role does the derivative play in defming slopes, tangents, and rates of change? 3. How can you sOl13etimes graph the derivative of a function when all you have is a table of the function's values? 4. What does it mean for a function to be differentiable on an open interval? On a closed interval? 5. How are derivatives and one-sided derivatives related?

6. Describe geOI13Oltica11y when a function typically does not have a derivative at a point 7. How is a function's differentiability at a point related to its continuity there, ifat all? 8. What rules do you know for calculating derivatives? Give sOl13e examples. 9. Explain how the three formulas

tL (x') ~ dx

d

(UI

e. From your graph in part (d), estimate as large a 8 > 0 as you can, satisfying

Ix -

Ij(x) - L(x)1
O.

The minus sign indicates that the volume is decreasing. We assume that the proportiona1ity factor k is constant. (It probably depends on many things, such as the relative homidity of the surrouoding air, the air temperatore, and the incidence or absence of sunlight, to name only a few.) Assume a particular set of conditions in which the cobe lost 1/4 of its volume during the first hour, and that the volume is Vowhen 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. Derivatives, Slopes, Tangent Lines, and Making Movies Part. I-ill. You will visualize thc derivative at a point, the linearization of a function, and thc 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 Movie.). Parts N and V of the module can he used to animate 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. ModonAlong II Straight Line: Position .... Velocity .... Acceleration Ohserve 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)

~

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

----I---.....---~x

FIGURE 4.1 Absolute extrema for the sine and cosine functions on [ -'IT/2, 'IT /2] . These values can depend on the domain of a function.

184

~

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 ex1rema 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.

•

----"x

a

FIGURE 4.5 ~ x ~ b.

: :

e

d

b

How to identifY types ofmaxima and minima for a functioo willi domain

a

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 ofits domain, and if I' is defined at c, then

THEOREM 2-The First Derivative Theorem for Local Extreme Values I

I I I I

I'{c) = O.

,

Secant slopes

2:

0

Secant slopes =s 0 (never po~itive)

(never ,negative) I I I I

I I I I

I

!

)

e

X

FIGURE 4.6 A curve with a local maximum. value. The slope at c,

simultaneously the limit ofnoopositive

Proof To prove that I' (c) is zero at a local extremum, we show first that I' (c) cannot be positive and second that I' (c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what I' (c) must be. To begin, suppose that f has a local maximum value at x = c (Figure 4.6) so that f{x) - ftc) ,;; 0 for all values of x near enough to c. Since c is an interior point of !'s domain, I' (c) is derfied by the tw-sided limit

r

numbers and nonnegative numbers, is zero.

f{x) - ftc)

x~

xc'

This means that the right-hand and left-hand limits both exist at x When we examine these limits separately, we find that

0

-.

Because (:::t - c) < 0 and f(x) " f{e)

(2)

Together, Equations (I) and (2) imply I'{c) = O. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use f{x) 2: ftc}, which reverses the inequalities in Equations (1) and (2). •

188

Chapter 4: Applications of Derivatives y

----'--

+-"=------'---~x

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 I can possibly have an extreme value (local or global) are

1. 2. 3.

interior points where f' = 0, interior points where f' is undefined, endpoints of the domain of I.

The following definition helps us to summarize. (al

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 I 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 I at all critical points and endpoints. 2. Take the largest and smallest of these values.

EXAMPLE 2 [-2,1].

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 = 1: Critical point value:

1(0) = 0

Endpoint values:

1(-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

7

Find the absolute maximum and minimum values of g(l) = 81 - 14 on

EXAMPLE 3 [-2,1].

y (1,7)

189


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- t4

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 /(x) interval [-2,3]. FIGURE 4.8 The extreme values of = 81 - 14 on [-2, I] (Example 3).

g(l)

We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The 1rrst derivative

j'(x) = ~x-I/3 = _2_ 3 3\Yx

y =x 2l3 , -2sxs 3

Absolute maximum; also a local maximum.

has no zeros but is undefined at the interior point x = O. The values of / at this one critical point and at the endpoints are

-o--"---~ x o c b

Geometrically, the Mean Value Theorem says that somewhere b&weeo a and b the curve has at least ooe tangeot parallel to chord AB.

f(bl

=~(a)

= I'(c).

(1)

FIGURE 4.13

Proof We picture the graph of f and draw a line through the points A(a, f(a» and B(b,

feb»~.

(See Figure 4.14.) The line is the graph of the function

g(x) = f(a)

+

f(bl

=~(a)

(x - a)

(2)

(point-slope equation). The vertical difference between the graphs of f and g at x is HIsTORICAL BIOGRAPHY

Joseph-Louis Lagrange (1736-1813)

hex) = f(x) - g(x) = f(x) - f(a) -

feb) - f(a) b (x - a).

-a

(3)

Figure 4.15 shows the graphs of f, g, and h together. The function h satisfies the bypotheses of Rolle's Theorem on [a, b]. It is continuous on [a, b] and differentiable on (a, b) because both f and g are. Also, heal = h(b) = 0 because the graphs of f and g both pass through A and B. Therefore h'(c) = 0 at some point c E (a, b). This is the point we want for Equation (1).

194

Chapter 4: Applications of Derivatives

B(b,f(b))

B

I I I I I I I I

a

y

t

y ~ Y1=7,-I" ,,5 I I

'x

b

FIGURE 4.14 The graph of j and the chordAB over the interval [a, b].

h(x)

~

f(x) - g(x)

/

:............... i ............

a

x

b

'x

FIGURE 4.15 The chordAB is the graph of the functiong(x). The function h(x) = j(x) - g (x) gives the vertical distance between the graphs of j and g at x.

-f---C:+--~~x

-I

0

I

FIGURE 4.16 The function j(X) = ~ satisfies the hypotheses (and conclusion) of the Mean Value Theorem on [-I, I] even though j is not differentiable at-land!.

To verify Equation (1), we differentiate both sides of Equation (3) with respect to x and then set x = e:

h'(x)

=

I'(x) _ I(b) - I(a) b-a

Derivative ofEq. (3) ...

h'(e)

=

I'(e) _ I(b) - I(a) b a

... withx=c

y B(2,4)

4

;J

3

y

0= I'(e) _ I(b) - I(a) b-a

= x2

I'(e)

=

I(bl-

~(a),

h'(c)

~

0

Rcarnmged

which is what we set out to prove. The hypotheses of the Mean Value Theorem do not require either a or b. Continuity at a and b is enough (Figure 4.16). FIGURE 4.17 As we fmd in Example 2, c = 1 is where the tangent is parallel to the chord. s

I

-

to he differentiable at

The function I(x) = x 2 (Figure 4.17) is continuous for 0 :s; x :s; 2 and differentiable for 0 < x < 2. Since 1(0) = 0 and 1(2) = 4, the Mean Value Theorem says that at some point e in the interval, the derivative I' (x) = 2x must have the value (4 - 0)/(2 - 0) = 2. In this case we can identify e by solving the equation 2c = 2 to get e = I. However, it is not always easy to find e algebraically, even though we know it always exists. _

EXAMPLE 2

400

g 320

A Physical Interpretation

u

240

H i3 160

We can think of the number (f(b) - I(a»/(b - a) as the average change in I over [a, b] and I' (c) as an instantaneous change. Then the Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.

&J~

o ~~~-;5~~-TIme (sec)

FIGURE 4.18 Distance versus elapsed time for the car in Example 3.

EXAMPLE 3 If a car accelerating ftmn zero takes 8 sec to go 352 ft, its average velocity for the 8-sec interval is 352/8 = 44 ft/ sec. The Mean Value Theorem says that at some point during the accelemtion the speedometer must read exactly 30 mph (44 ft/ sec) (Figure 4.18). _

4.2

The Mean Value Theorem

195

Mathematical Consequences At the beginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollary of the Mean Value Theorem provides the answer that only constant functions have zero derivatives.

COROLLARY 1 If j'(x) = 0 at each point x of an open interval (a, b), then j(x) = C for all x E (a, b), where C is a constant.

Proof We want to show that j has a constant vaIue on the interval (a, b). We do so by showing that if Xl and X2 are any two points in (a, b) with Xl < X2, then j(x,) = j(X2). Now j satisfies the hypotheses of the Mean Value Theorem on [Xl, X2]: It is differentiable at every point of [Xl. X2] and hence continuous at every point as well. Therefore,

at some point c between Xl and X2. Since j' = 0 throughout (a, b), this equation implies successively that

j(X2) - j(XI) X2 Xl = 0,

j(X2) - j(XI) = 0,

and

j(XI) = j(X2).

•

At the beginning of this section, we also asked about the relationship between two functions that have identical derivatives over an interval. The next corollary tells us that their vaIues on the interval have a constant difference.

y

COROLLARY 2 Ifj'(x) = g'(x) at each point X in an open interval (a, b), then there exists a constant C such that j(x) = g(x) + C for all X E (a, b). That is, j - g is a constant function on (a, b).

Proof At each point X E (a, b) the derivative ofthe difference function h = j - g is h'(x) = j'(x) - g'(x) = O. Thus, h(x) = C on (a, b) by Corollary I. That is, j(x) - g(x) = C on (a, b), so j(x) = g(x) + C. • ----\c-\---"-+-"'--f+----->X

-2

FIGURE 4.19 From a geometric point of view, Corollary 2 of the Meao Value Theorem says that the graphs offunctions with identical derivatives on an interval cao differ only by a vertical shift there. The graphs ofthe functions with derivative Z. are the parabolas y ~ x' + C, shown here for selected values of C.

Corollaries 1 and 2 are also true if the open interval (a, b) fails to be finite. That is, they remain true if the interval is (a, 00), (-00, b), or (-00, 00). Corollary 2 plays an important role when we discuss antiderivatives in Section 4.7. It tells us, for instance, that since the derivative of j(x) = x 2 on (-00, 00) is z., any other function with derivative 2x on ( - 00, 00) must have the fonnula x 2 + C for some vaIue of C (Figure 4.19).

EXAMPLE 4 Find the function j(x) whose derivative is sin X and whose graph passes through the point (0, 2). Solution Since the derivative of g(x) = -cosx is g'(x) = sinx, we see that j and g have the same derivative. Corollary 2 then says that j(x) = -cosx + C for some

196

Chapter 4: Applications of Derivatives constant C. Since the graph of j passes through the point (0, 2), the value of C is determined from the condition that j( 0) = 2:

j(O)

= -cos

The function isj(x) = -cosx

+

(0)

+C

=

2,

C = 3.

so

•

3.

Finding Velodty and Position from Acceleration We can use Corollary 2 to find the velocity and position functions of an object moving along a vertical line. Assume the object or body is falling freely from rest with acceleration 9.8 m/sec2. We assume the position 8(/) of the body is measured positive downward from the rest position (so the vertical coordinate line points downward, in the direction of the motion, with the rest position at 0). We know that the velocity V(/) is some function whose derivative is 9.8. We also know that the derivative of g(/) = 9.81 is 9.8. By Corollary 2,

+

v(/) = 9.81

C

for some constant C. Since the body falls from rest, v(O) = O. Thus 9.8(0)

+

C = 0,

and

C = O.

The velocity function must be v(/) = 9.81. What about the position function 8(/)? We know that 8(/) is some function whose derivative is 9.8/. We also know that the derivative of j(/) = 4.9/2 is 9.8/. By Corollary 2,

8(1) = 4.9/ 2 + C for some constant C. Since 8(0) = 0, 4.9(0)'

+

C = 0,

and

C = O.

The position function is 8(/) = 4.9/ 2 until the body hits the ground The ability to find functions from their rates of change is one of the very powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.

Exercises 4.2 Checldng the Mean Yalue Theoll!m

Find the value or values of c that satisfY the equation f(b) - f(a) ~ f'(c) b-a

Which of the functions in Exercises 7-12 satisfY the hypotheses of the Mean Value Theorem on the giveo interval, aod which do not? Give reasons for your answers. 7. f(x) = x 2/3 , [-1,8] [0, I]

in the conclusion of the Mean Value Theorem for the functions and in-

8. f(x) = x'I',

tervals in Exercises l--{j. 1. f(x) ~ x 2 + Zl: - I,

9. f(x) = v'x(1 - x),

2. f(x) = x 2/3 ,

3. f(x)

4. f(x) =

[t,2]

v;=J,

S.f(x)~x3-x2,

6. g(x)

~

X3 {

x2:

SinX

[0, I]

=x+~,

[1,3] [-1,2]

-2::=:; x::=:; 0 < x '" 2

°

[0, I]

[0, I] 10. f(x) =

{

-'T1' ~x

x'

0,

0

x=o

x2 - x 11. f(x) ~ { Zl:2 _ ~x - 3,

12. f(x) = {Zl: - 3,2 6x - x

0 throughout an interval [a, b], then f' bas at most one zero in [a, b]. What iff" < 0 throughout [a, b] instead? 18. Sbow that a cubic polynontial can have at most three real zeros. Sbow that the functions in Exercises 19-26 have exactly one zero in the given interval.

20. f(x)

~

21. g(l) =

x3

41. v ~ 9.81

+ 5,

42. v = 321 - 2,

44. v

= IjTcos W'

23. 7(0) = 0

+

sin2

(~)

24.7(0) = 20 - oos2 0 I 25. 7(0) = sec 0 - 03

(0,00) 3.1,

(-I, I)

sin 'frl, 2

8,

+ v2,

+ 5,

tanO - cotO - 0,

(-00,00) (-00,00)

(0, 'fr/2) (0, 'fr/2)

Find;ng Functions from Derivatives 27. Suppose that f( -I) ~ 3 and that f'(x) ~ 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.

,(0) 21

~

0

,

s('7T""")

=

1

Exercises 45-48 give the acceleration a = d 2,/dI 2 , initial velocity, and initial position of a body moving on a coordinate line. Find the body's position at time I. 45. a = 32,

-

,(0) ~ 10 ,(0.5) = 4

(-00,0)

~ I + v'1+t -

~

F;nd;ng Position from velocity or Acceleration

+ ~ + 7,

4,

P(O, 0)

Exercises 41-44 give the velocity v ~ ds/dl and initial position ofa body moving along a coordinate line. Find the body's position at time t.

~

vt + v'1+t -

(-;r, 0)

40. 7'(1) = secltanl - I,

43. v

x

p(-I, I)

csc'O,

[-2, -I]

22. g(l) = I

26. 7(0)

x

I,

+ 3x +

19. f(x) ~ x'

I, P(O, 0)

v(O) = 20, v(O) ~ -3,

,(0) = 5

46. a

~

9.8,

47. a

~

-4 sin 21,

v(O) ~ 2,

,(0)

9 31 48. a = "" cos 'fr'

v(O) = 0,

,(0) = -I

,(0)

~

0 ~

-3

Applkations 49. Temperature change It took 14 sec for a mercury thermometer to rise from -19'C to lOO'C when it was taken from a freezer and placed in boiling water. Show that somewbere along the way the mercury was rising at the rate of 8.5°C/sec.

198


50. A trucker handed in a ticket at a toll booth showing that in 2 bours she had covered 159 mi 00 a toll road with speed limit 65 mph. The trucker was cited for speeding. Why? 51. Classical accounts tell us that a 17 0 on (1,00).

71. Let

a. Show that I(x) '" I for all x. b. Mustl'(l) = O?Explain.

72. Let I(x) = px2 + qx + r be a quadratic function defined on a closed interval [a, b]. Show that there is exactly ooe 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 differentiable 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 3 (a, h). If I'(x)

Suppose that

I

is continuous on [a, h] and differentiable on

> 0 at each point x E (a, h), then I

is increasing on [a, h].

Ifl'(x) < 0 at each point x E (a, h), then/is decreasing on [a, h].

Proof Let XI andx2 be any two points in [a, h] with XI < X2. The Mean Value Theorem applied to I on [Xh X2] says that

I(X2) - l(xI)

=

l'(e)(x2 - XI)

for some e between XI and X2. The sign of the right-hand side of this equation is the same as the sign of I'(e) because X2 - X, is positive. Therefore, I(X2) > l(xI) if I' is positive • on (a, h) and/(x2) < I(x,) 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 e in (a, h). Ifl'(e) > 0, thenl'(x) > 0 for all x in (a, h) so/isincreasing on [a, h] by Corollary 3; if I'(e) < 0, then I is decreasing on [a, h]. The next example illustrates how we use this procedore. y y~x'-I2x - 5

Find the critical points of I(x) = x 3 - 12x - 5 and identify the intervals on which I is increasing and on which I is decreasing.

EXAMPLE 1

20

Solution

The function I is everywhere continuous and differentiable. The f"mt derivative

I'(x)

FIGURE 4.20 The function fix) ~ - 12x - 5 is monotonic on three separate intervals (Example 1).

=

3x 2 - 12

=

3(x + 2)(x - 2)

=

3(x2 - 4)

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 gmph of I is given in Figure 4.20.

x3

Interval I' evaluated Sign ofl' Behaviorofl

-00 < X 0 immediately to the left and t' < 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 t' (x) changes. Absolute max

f' undefined 1

1

:

No extremum f'~ 0

1

1

11' 0 implies that I is increasing on [c, b]. Therefore, I(x) ;", I(c) for every XE (a, b). By definition, I has a local minito positive at c, there are numbers a and b such that a

t' >

mum ate. Parts (2) and (3) are proved similarly.

•

4.3

EXAMPLE 2

201

Monotonic Functions and the First Derivative Test

Find the critical points of I(x) = x l /3(x - 4) = x'/3 - 4x l / 3.

Identify the intervals on which I is increasing and decreasing. Find the function's local and absolute extreme values. Solution The function I is continuous at all x since it is the product of two continuous functions, x 1/3 and (x - 4). The first derivative f'(x) =

~ (X 4/3 - 4xl/3) ~

_i

-3x

=

i xl/3 _ i 3

3

2/

x- 3

-2/3( _ ) _ 4(x - 1) x 1 2/3 3x

is zero at x = 1 and undefined at x = O. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where I might have an extreme value. The critical points partition the x-axis into intervals on which f' is either positive or negative. The sign pattern off' reveals the behavior of I between and at the critical points, as summarized in the following table.

y 4

Interval Sign off' Behavior of f

FIGURE 4.22 The functionf(x) ~ x 1/3 (x - 4) decreases when x < I and increases when x > I (Example 2).

x Oforx < landf'(x) < Oforx> I;

39. h(x) = x /'(x 2 '

40. k(x) = x 2/'(x 2

b. f'(x)

4)

-

-

4)

a. IdentifY the functioo's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute?

Support your fmdings with a graphing calculator or compurer grapher.

41. j(x)

=

2x - x 2 ,

-00

42. j(x) = (x

+ I)',

43. g(x) = x 2

-

44. g(x) = -x 2 45. j(l) = 121 -

48. k(x)

=

x' T x'

:5 2

0 for x oF I;

In Exercises 41-52:

D c.

21f

00 as x ---> 0+;

00

as x ---> 0-,

b. h(O) = 0, -2 :5 h(x) :5 0 for all x, h'(x) --->

00

as x ---> 0-,

and h'(x) ---+ -00 as x ---+ 0+. 00

67. Discuss the extreme-value behavior of the functioo j(x) = x sin (I/x), x oF 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). 70. Determine the values of constants a, b, c, and d so that f(x) = ax 3 + bx2 + ex + d has a local maximum at the point (0,0) and a local minimum at the point (I, -I).

absolute minimum? An absolute maximum? (See Exercise 49 in

Sectioo 2.3.) 68. Find the intervals on which the functioo f(x) a

-=F

203

~ ax 2 + bx + c, 0, is increasing and decreasing. Describe the reasoning be-

hind your answer.

Concavity and Curve Sketching

4.4 y

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 aboot 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 features of functions. Identifying and knowing the locations of these features 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 coocave down on (- 00, 0) and concave up 00 (0,00) (Example la).

As you can see in Figure 4.23, the curve y = x 3 rises as x increases, but the portions de-

fmed 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 ofthe 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 (a)

The graph of a differentiable function y

= j(x)

is

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 1.

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/" > 0 on I, and decreases if/" < o.

The Second Derivative Test for Concavity Let y = j(x) be twice-differentiable on an interval 1. 1. If/" 2. If/"

> 0 on I, the graph of j over I is concave up. < 0 on I, the graph of j over I is concave down.

If y = j(x) is twice-differentiable, we will use the notations /" and y" interchangeably when denoting the second derivative.

204


EXAMPLE 1

y

x 3 (Figure 4.23) is concave down on (-00,0) where y' = 6x < 0 and concave up on (0, 00) where y' = 6x > o. (b) The curve y = x 2 (Figure 4.24) is concave up on (-00, 00) because its second derivative y" = 2 is always positive. _ (8) The curve

EXAMPLE 2

y

=

Detennine the concavity of y = 3

+ sin x on [0, 21T].

-+-+~:+ 0:

EXAMPLE 5

y

------"+----~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

s(l)

=

21 3

+ 221 -

141 2

-

5,

12: O.

Find the velocity and acceleration, and describe the motion of the particle. Solutton

The velocity is

V(I)

=

=

61 2

V'(I)

=

S'(I)

-

281 + 22 = 2(1 - 1)(31 - 11),

and the acceleration is

a(l)

=

S"(I)

=

121 - 28 = 4(31 - 7).

When the function S(I) is increasing, the particle is moving to the right; when S(I) is decreasing, the particle is moving to the left. Notice that the first derivative (v = s') is zero at the critical points 1 = I and 1 = 11/3.

0 O,thenfhasalocalminimumatx = c. 3. If f' (c) = 0 and j" (c) = 0, then the test fails. The function f may have a local maximum, a local minimum, or neither.

!\V f' ~ O.!" < 0 ~localmax

f'~O.!,,>O ~

local min

Proof Part (1). If I"(c) < 0, then I"(x) < 0 on some open interval I containing the point c, since I" is continuous. Therefore, f' is decreasing on 1. Since f' (c) = 0, the sign of f' changes from positive to negative at c so f has a local maximum at c by the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions y = x',y = -x" and y = x 3 • For each function, the frrst and second derivatives are zero at x = O. Yet the function y = x' has a local minimum there, y = -x' has a local maximum, and y = x 3 is increasing in any open interval containing x = 0 (having neither a maximum nor a minimum there). Thus the test fails. _ This test requires us to know I" only al c ilself and not in an interval about c. This makes the test easy to apply. That's the good news. The bad news is that the test is inconclusive if I" = 0 or if j" does not exist at x = c. When this happens, use the First Derivative Test for local extreme values. Together f' and j" tell us the shape of the function's graph-that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is toming or bending as defined by its concavity. We use this information to sketch a graph ofthe function that captures its key features.

EXAMPLE 7

Sketch a graph ofthe function

using the following steps. (8) Identify where the extrema of f occur. (b) Find the intervals on which f is increasing and the intervals on which f is decreasing.

(c) Find where the graph of f is concave up and where it is concave down. (d) Sketch the general shape ofthe graph for j. (e) Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. Then sketch the curve.

4,4


207

The function f is continuous since j'(x) = 4x' - 12x 2 exists. The domain of f is (- 00, 00), and the domain of j' is also (- 00, 00). Thus, the critical points of f occur only at the zeros of j' . Since Solutton

j'(x)

4x' - 12x 2

=

the flISt derivative is zero at x = 0 and x vals where f is increasing or decreasing.

Interval

=

=

4x 2(x - 3)

3. We use these critical points to define inter-

x 0 yielding a relative minimum by the Second Derivative Test. At x = 1,1"(1) = -I < 0 yielding a relative maximum by the Second Derivative test.

4. Increasing and decreasing. We see that on the interval (-00, -I) the derivative j' (x) < 0, and the curve is decreasing. On the interval (-I, I), j' (x) > 0 and the curve is increasing; it is decreasing on (I, 00) where j'(x) < 0 again.

4.4 Concavity and Curve Sketching

209

5. Inflection points. Notice that the denominator of the second derivative (Step 2) is always positive. The second derivative I" is zero when x = - V3, 0, and V3. The second derivative changes sign at each of these points: negative on ( - 00, - V3) , positive on ( - V3, 0), negative on (0, V3), and positive again on ( V3, 00 ). Thus each point is a point of inflection. The curve is concave down on the interval ( - 00, - V3), concave up on ( - V3, 0), concave down on (0, V3), and concave up again on ( V3, 00 ) . 6. Asymptotes. Expanding the numerator of I(x) and then dividing both numerator and denominator by x' gives

+ I)' + x'

(x) = (x I

1

I

+

Expanding numerator

(2/x) (I/x')

Point of inflection wherex ~ 0 2 (1,2)/ Y

We see that I(x) --+ 1+ as x --+ 00 and that I(x) --+ 1- as x --+ - 00. Thus, the line y = I is a horizontal asymptote. Since 1 decreases on ( - 00, - I) and then increases on ( - I, I), we know that I( -I) = 0 is a local minimum. Although 1 decreases on (I, 00), it never crosses the horizontal asymptote y = I on that interval (it approaches the asymptote from above). So the graph never becomes negative, and I( -I) = 0 is an absolute minimum as well. Likewise, 1(1) = 2 is an absolute maximum because the graph never crosses the asymptote y = I on the interval (- 00, - I), approaching it from below. Therefore, there are no vertical asymptotes (the range of 1 is 0,,;; y,,;; 2).

---~~_"II'-i----;----_x

Point of inflection wherex ~ -0

FIGURE 4.30

The graph of y

=

(x + I)' , l+x

+ (I/x') +I

7.

(Example 8).

The graph of 1 is sketched in Figure 4.30. Notice how the graph is concave down as it approaches the horizontal asymptote y = I as x --+ - 00, and concave up in its approach to y = I asx --+ 00. •

EXAMPLE 9

Sketch the graph of I(x) =

x' + 4 ----zx-.

Solution

1.

2.

The domain of 1 is all nonzero real numbers. There are no intercepts because neither x nor I(x) can be zero. Since I(-x) = - I(x), we note that 1 is an odd function, so the graph of 1 is symmetric about the origin. We calculate the derivatives ofthe function, but rITst rewrite it in order to simplify our computations:

I(x)

=

I'(x)

=

x' ~ 4

t-:'

=

t+~

=

x'z:,

4

Function simplified for differm.tiation

Combine fractions to solve easily /,(x)

=

o.

Exists throughout the entire domain off

3.

The critical points occur at x = ±2 where I'(x) = O. Since 1"(-2) < 0 and 1"(2) > 0, we see from the Second Derivative Thst that a relative maximum occurs at x = - 2 with 1(-2) = - 2, and a relative minimum occurs at x = 2 with 1(2) = 2.

210

Chapter 4: Applications of Derivatives 4.

On the interval (- 00, - 2) the derivative f' is positive because x 2 - 4 > 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

There are no points of inflection because I"(x) < 0 whenever x < 0, f"(x) > 0 whenever x > 0, and f" 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-

6.

y=! 2

From the rewritten formula for f(x), we see that

_L-_L-~¥=-----L_--'--~x

-4

-2

2

~ FIGURE 4.31

lim (~+~) x-o+ 2 x

4

The graph ofy =

=

+00

and

lim (~+~) x-o2 x

00

-00' 00,

so the y-axis is a vertical asymptote. Also, as x -+ or as x -+ the graph of f(x) approaches the line y = x/2. Thus y = x/2 is an oblique asymptote.

2

x +4 ----zx-

=

7.

The graph of f is sketched in Figure 4.31.

•

(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 first 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 determine 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 following figure summarizes how the derivative and second derivative affect the shape of a graph.

y

rJd" 7

> 0 ~ rises ftom.

Differentiable ~ smooth, connected; graph may rise and fall

y'

,/

(

y" > 0

or

"

~ 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

or

y" < 0

left to righ~ may be wavy

\ :/

~ concave down throughout; no waves; graph may rise or fall

f\

and y"O at a point; graph has local mjnjmum

y'~

4,4


211

Exercises 4.4 Analyzing Functions from Graphs Identify the inflectioo points and local maxima and minima of the functioos graphed in Exercises 1-8. IdentifY the intervals on which the functions are concave up and concave down.

1.

2.

16. y = I - (x + I)' 17. y = x 4 - a 2 = x 2(x 2 - 2) 18. y = -x 4 + 6x 2 - 4 = x 2(6 - x 2) - 4 19. y ~ 4x' - x 4 ~ x'(4 - x) 20. y ~ x 4 + a' ~ x'(x + 2) 21. y ~ x' - 5x 4 ~ x 4(x - 5) 22. y

= x(~ -

5)'

= x + sinx, O::s x ::s 21T 24. y = x - sinx, O::s x ::s 2w

23. Y

-I--lt----+--.x

25. y

= V3x -

2 cos x,

4 26. y = 3x - tanx,

3.

1T

2 <xO

93. The accompanying figure shows a portioo of the graph of a twicedifferentiable functioo y = f(x). At esch of the five labeled points, classifY y' and y" as positive, negative, or zero.

x:s; 0

x2,

x = -,--

x> 0

y

Sketching y from Graphs of y' and y" Each of Exercises 71-74 shows the graphs of the first and second derivatives of a functioo y ~ f(x). Copy the picture and add to it a sketch of the approximate graph of f, given that the graph passes through the point P. 71.

72.

s

~f(X)

~ R

p

T

Q

Y

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

o

94. Sketch a smooth coooected curve y x

Y ~r(x)

73.

Y

f(-2) = 8,

1'(2) = 1'( -2) = 0,

flO) = 4,

I'(x) < 0 for

f(2) = 0,

r(x) < 0

I'(x)

P

= f(x) with

>

0

for

Ixl > 2,

r(x)

>

0

Ixl
O.

95. Sketch the graph of a twice-differentiable function y = f(x) with the following properties. !.abel coordinates where possible.

x 74.

Y

y

Derivatives

x

0

>

> 0,

X

20 40 60 80100120 Thousands of ooits produced

100. The aeeompanying graph shows the moothly reveoue of the Widget Corporatioo for thc last 12 yeas>. Doring approximately what time intervals was the marginal revenue increasing? Decreasing?

111. Find the values of coostants a, b, and c so that the graph of y = ax' + bx2 + ex has a local maximum at x = 3, local minimumatx = -I, and inflectioo point at (I, II). 112. Find the values of coostants a, b, and c so that the graph of y = (x 2 + a)/(bx + c) has a local minimum at x = 3 and a 1cal maximum at (-I, -2).


y

10 Exercises 113-116, fmd the inflectioo points (if any) 00 the graph of the function and the coordinates of the points on the graph where the functioo has a local maximum or local minimum value. Thou graph the function in a region large eoough to show all these points simultanously. Add to your picture the graphs of the functioo's fll'St and secood derivatives. How are the values at which these graphs intersect the x-axis related to the graph of the function? 10 what other ways are the grapha ofthe derivatives related to the graph ofthe functioo? 101. Suppose the derivative of the functioo y = f(x) is

y'

= (x

- I)'(x - 2).

At what points, if any, does the graph of f have a local minimum, local maximum, or point of inflectioo? (Hint: Draw the sign pattern for y' .) 102. Suppose the derivative of the functioo y ~ f(x) is

y'

= (x

- 1)2(X - 2)(x - 4).

At what points, if any, does the graph of f have a local mini-

mum, local maximum., or point ofinflection?

113. y ~ x' - 5x' - 240 115. Y

= ~x' +

116. Y

x4 x 3 =4 - 3 -

16x 2

-

4x 2

114. Y ~ x' - 12x 2

25

+ 12x +

20

117. Graph f(x) = 2x' - 4x 2 + I and its frrst two derivatives lgether. Commeot on the behavior off in relatioo to the signs and values of j' and f" . 118. Graph f(x) = x cos x and its second derivative together for o,;; x ,;; 2".. Commeot 00 the behavior of the graph of fin rlatioo to the signs and values of f" .

214

Chapter 4: AppLications of Derivatives

Applied Optimization

4.5

What are the dimensions of a rectangle with lured 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. Dnnv a picture. Label lIllY part that may be important to the problem.. 3. h1troduce 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 o/the unknown. 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 corners 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 ofthis variable:

tin What size comers maximize the box's volmne (Example I)?

, y - x(12 -

2x'f,

O:Sx:S6

o

"""/ , 6

FIGURE 4.33 The volume of the box in Figure 4.32 graphed as a fimction of x.

Y=IIlw

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 miDimum. value of 0 at x = 0 and x = 6 and a maximum near x = 2. To learn more, we examine the first derivative of V with respect to x: :

~

144 - 9&

+ 120' ~

12(12 - 8x

+ x') ~

12(2 - x)(6 - x).

Of the two zeros, x - 2 and x - 6, only x - 21ies in the interior of the function's domain

and :makes the critical-point list. The values of Yat this one critical point and two endpoints are Critical-point value: Endpoint values:

Y(2) - 128 Y(O) ~ 0,

Y(6)

~

O.

The:JDaXimum 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? SoLution Volume ofcan: 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 = 27Tr 2

+ 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

=

27fr2

Tall and thin

+

2000, r r

/> 0

o Short and wide

FIGURE 4.35

The graph of A = 27fr 2

+ 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 27Tr 2 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 rrrst derivative is zero.

dA dr

=

4'ITr _ 2000 r2

o=

4'ITr _ 2000

SeldA/dr ~

r2

4'1Tr 3 = 2000 r =

o.

Multiply by r'.

~5~0

'" 5.42

Solve forr.

What happens at r = "¥500/'IT? The second derivative 2

d A = 4'IT dr 2

+

4000

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.

=

7T

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, Y/4 -

X

2)

2 ---cL-L-----cciL----"--c~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 -

Area:2x~.

x 2,

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~ = ~

-2x 2 + 2(4 - x 2)

0

= 0

2

8-4x =0 x 2 = 20rx

=

±v'2.

4.5 Applied Optimization

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:

2V2v14=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

Medium 1 -oo+---x----""''k---+--~x

Medium 2

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 .!)I-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

d-x B

Time = distance rate

FIGURE 4.37 Alightrayrefracted (deflected from its palh) 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

= C1

=

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: I = 11

+ 12

=

Va 2 + x 2 C,

+

Vb 2 + (d - x)' C2

This equation expresses I as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of I on this closed interval. We find the derivative

dl dx dtIdx negative

I

~-

o

-

dtIdx zero

-

-

-

~x

Xo

d-x

+ x2

and observe that it is continuous. In terms of the angles 61 and 62 in Figure 4.37,

dtldx positive

~ ++++ +++( •

x c,Va2

dl dx

sin 6,

sin 62

-----

Cl

C2·

d

FIGURE 4.38 The siga pattero of dt/dx in Example 4.

The function I 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 ofx (Exercise 62). At this unique point we then have sin II,

c,

sin 112 C2

•

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 ofthe 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 value 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) = r'(x)

--;cl""'-L-------:c---L----c-------->x

o

Items produced

FIGURE 4.39 The graph of a typical cost function starts concave dowo BOd laler turns concave 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 occurring where c' (x) = r' (x). Farther to the right, cost exceeds revenue (perllaps because of a combination ofrising labor and material costs BOd market saturation) and production levels become unprofitable again.

4.5 Applied Optimization

219

Supposethatr(x) = 9xandc(x) = x 3 - 6x 2 + 15x, where x represents millions ofMP3 players produced. Is there a production level that maximizes profit? If so, what is it?

EXAMPLE 5

y

Solution

Notice that r'(x) = 9 and c'(x) = 3x2 -

3x 2 -

IZl; + IS.

IZl; + IS = 9

Selc'(x) ~ T'(X).

2

3x -1Zl;+6=0 The two solutions of the quadratic equation are

vTi

= 2 -

V2 '" 0.586

vTi

= 2 +

V2 '" 3.414.

12 -6 : Maximum : for profit

12 +6

I I I

Local maximum for loss "---'---;=----J---'----;cc---~ 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"(xl since r"(x) is everywhere zero. Thus, p"(x) = 6(2 - x), which is negative atx = 2 + v'2

V2.

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. •

Exercises 4.5 Mathematical Applications Whenever you are maximizing or minimizing a function ofa single variable, we wge 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 uoits long. &.

Express the y-mial fimctioo y ~ ax' + fu;2 + ex + d, where y (- L) ~ H and

y(O) = O. a. What is dy/dx atx = O? b. What is dy/dx atx = -L? c. Use the values forth'/dx atx ~ oand x y(O) = Oandy(-L) = Htoshowthat

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 margiuaI cost 55. Showthatifr(x) = 6xandc(x) =

x' - 6x

2

+

15xareyourrev-

enue and cost functions, then the best you can do is break. even ~

-Ltogetherwith

(have revenue equal cost). 56. Production level Suppose that c(x) = x' - 20>:2 + 20,000>: is the cost of maoufacturing x iterus. Find a productioo level that

will minimize the average cost ofmaking x items. 57. You are to construct an open rectangular box with a square base y

Landing path

H = Cruising altitude

Airport

II Business and Economics 51. It costs you c dollars each to maoufacture and distribute hackpacks. If the backpacks sell at x dollars each, the number sold is given by

n = x ~ c

and a volume of48 ft'. Ifmaterial for the bottom costs $6/ft' and materisl for the sides costs $4/ft', what dimensions will resolt in

+ b(100

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? 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: $200 per persoo if 50 people (the minimum number to book the tour) go on the tour.

a. When we cough, the trachea (windpipe) contracts to increase the velocity ofthe 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 persoo is reduced by $2.

velocity and whether it really cootraets that much when we cough.

Under reasonable assumptions about the elasticity ofthe

It costs $6000 (a fixed cost) plus $32 per person 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 frictioo, 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

hq

length of the tracbea. Show that v is greatest when r = (2/3)ro; that is, when the trachea is about 33% coottaeted. The remarkable fact is that X-ray photographs conf'mn that the trachea cootracts 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 he), k is the cost of

placing an order (the same, no matter how often you order), c is

(a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to fmd the quantity that will minimize A(q). What is it? (The formula you get for the answer is called the Wilson lot sizefimnula.)

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?

< r - roo

where ro is the rest radius ofthe trachea in centimeters and c is a positive constant whose value depends in part on the

q + cm + T'

the cost of ooe item (a constant), m is the number of items sold each week (a constant), and h is the weekly holding cost per item

70 < 2: -

v = c(ro - r)r 2 em/sec,

D

h. Take ro to he 0.5 and c to he I and graph v over the interval r '" 0.5. Compare what you see with the claim that v is at a maximum when r ~ (2/3 )ro.

o'"

Theory and Examples 61. An inequality for positive integers

Show that if a, b, c, and d

are positive integers, then (a 2

+

1)(b 2 + 1)(c 2

abed

+

1)(d2

+

I) > 16.

4.6

62. The derivative dt/dx in Example 4

b. Is f ever negative? Explain.

f(x)~~

+x

mum value on the interval 0 < x
0, applying Newton's method to

f(x) =

'\Ix,

x '" 0

{ ~,

x =

rbjdtinw

pIC\'iouI cquatioo.

s(0) - 80

Initial condition:

We solve this initial value problem to fmd the height as a function of t.

1.

Solve the differential equation: Finding the gmeral antiderivative of - 32t s - -16t

2.

2

+

12t

+ 12 gives

+ C.

Evaluate C: 80 - -16(0)' C

~

+ 12(0) + C

Initial cmditiou(O) - 80

80.

The package's height above ground at time t is

s

=

-16t 2

+

12t

+ 80.

Use the solution: To fmd how long it takes the package to reach the ground, we set s equal to 0 and solve for t: -16t 2

+

-4t2

12t+80-0

+3t+20=0

-3 ± v329 t8 t ~ -1.89, t ~ 2.64. The package hits the ground about 2.64 sec after it is dropped from the balloon. (The negative root has no physical meaning.) •

235

4.7 Antiderivatives

Indefinite Integrals A special symbol is used to denote the collection of all antiderivatives of a function j.

DEFINmON The collection of all antiderivatives of f is called the indefinite integral of f with respect to x, and is denoted by J f(x)dx.

J

The symbol is an integral sign. The function gral, and x is the variable of integration.

f

is the integrand of the inte-

After the integral sign in the notation we just defined, the integrand function is always followed by a differential to indicate the variable of integration. We will have more to say about why this is important in Chapter 5. Using this notation, we restate the solutions of Example I, as follows:

2 2xdx=x +C,

J

J J

cosxdx = sinx

(2x

+ cos x) dx

+ =

C,

x2

+ sinx +

C.

This notation is related to the main application of antiderivatives, which will be explored in Chapter 5. Antiderivatives play a key role in computing limits of certain infinite sums, an unexpected and wonderfully useful role that is described in a central result of Chapter 5, called the Fundamental Theorem of Calculus.

EXAMPLE 6

Evaluate

J

(x 2

-

2x

Solutton If we recognize that (x 3/3) - x 2 we can evaluate the integral as

+ 5)dx.

+ 5x is

an antiderivative of x 2

-

2x

+

5,

antiderivati~

J

(x 2

-

2x

x3

+ 5)dx = 3 -

x2

+

5x

+ £.

lUbitrary constant

If we do not recognize the antiderivative right away, we can generate it term-by-term with the Sum, Difference, and Constant Multiple Rules:

J (x

2

-

2x

+ 5) dx

=

2

J x dx - J 2x dx

+ J 5 dx

2 = JX dx-ZJXdx+5Jldx =

(;3 +

=

x 3 + C, - x 2 - ZC2 + 5x + 5C3.

3

C,) _ Z (x;

+ C2) +

5(x

+ C3)

236

Chapter 4: Applications of Derivatives This formula is more complicated than it needs to be. Ifwe combine C" -2C" and 5C3 into a single arbitrary constant C = C, - 2C, + 5C3, the formula simplifies to

x3 --x'+5x+C 3 and still gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term-by-term. Write

J

(x' - 2x + 5) dx

=

J J J x' dx -

+

2x dx

5 dx

3

=

x 3 - x' + 5x + C.

Find the simplest antiderivative you can for each part and add the arbitrary constant of integration at the end _

Exercises 4.7 finding Antlderlvatlves

finding Indefinite Integrals

10 Exercises 1-16, f"md an antiderivative for each function. Do as many as you ean mentally. Check your answers by differentiation.

10 Exercises 17-54, find 1he most general antiderivative or indeflnite integral. Check your answers by differentiation.

1.a.2x 2. a. 6x

e.x'-2x+1 e. x 7 - 6x + 8

b.

3. a. -3x-4

b. x-4

b.x' x7

e. x-4

-3

+ 2x + 3

b. ~ +x 2

C.

I 5. a. 2

b.~ x'

5 c. 2- 2 x

2 6. a. - x3

b. _1_

I x e. y!; + _1_

4. a. 2x-3

X

2

3

b. 1x-'/3

e. -tx-4/3

10. a. 12x -112

b. _lx-3j2

e. -~x-5j2

11. a. -w sin '11'X

b. 3 sinx

c. sin 71'X

12. a.

'T1'

cos '7TX

e. ~+_I-

2

'1fX

'T1'X

c. COST

2

+

e. -see' 3x 2

14. a. ese' x

b. _lese' 3x

e. 1-8ese'2x

15. a. cscxcotx

b. -esc 5x cot 5x

c. -w esc 1TX cot '7TX 2 2

16. a. secxtanx

b. 4 see 3x tan 3x

c. sccTtanT

=

-

5x

+ 7)dx

27. 29.

1(8Y-)I')dy

x- 1/3 dx (Y!;

+

33.

1

tV;;t;

20.

~) dx

V; dt

'1fX

37.

17Sin~

dO

39·/(-3eSC'x)dx 41.

1

ese °2eot 0 dO

1(~

+ 4t 3 ) dt

22.1(1 - x' 1(t - :3 +

3x')dx

24.

26.

35. 1(-2 eost) dl

13. a. see' x

2

3

t) dt

18·/(5-6x)dx

Ix -'I'

2x) dx

dx

I(Yf + ~)dx 30. I(t- )I')dy 32. IX-3(x+ I)dx 28.

34.

wooSX'

2 e' -x b. -se 3 3 2

1(2x

+

31. 12x(l - x -3) dx 3 sin 3x

-

1 1 1

25.

~

2

+ I)dx

23·/8,-x'-t)dx

y!;

3

b. 1rcos

(3t'

c. x - 3

9. a. 3x 2 -1/3

8.a. 4~ x

19. 21.

3

~ y!;

(x

-x-3 + x-I

2x 3 b. _1_ 2Y!; b. _1_ 3~

7. a.

17.1

36.

1 ~3V; 4

1(-5

dt

sin I) dt

38. 13 eos 50 dO 40.

42.

1(_~'X)dx

1~seeOtanOdO

4.7

I

43. 1(4seextmx - 2see'x)dx 44.

45.

47. 49.

I I l(i I I

(sin 2x - esc' x) dx I

cos 2x - 3 sin3x) dx

48·/1-~s6tdt

+ tm'B) dB

50./(2

+ tm' B =

sec' B)

eot?-x)dx

+ cot?- x = esc' x)

eosB(tmB

+ sec B) dB

+ tm'B) dB

l(i I

52.

eot?-xdx

(Hint: I

53.

1(2

46.

+ ~oS4t dt

(Hint: I

51.

k (esc' x - esex eotx) dx

54.

55. 56.

57.

58.

59. 60.

I I I I I I

'

(7x - 2) dx

=

+ 5)-'dx

(3x

(7x - 2)4 28

(3x

~

+

(x ~

I) dx = -3 cot

I

(x

+ I)' dx

(x

+ I)' dx

= (2x

+ I)' + C

c. 16(2x

+ I)' dx

= (2x

+ I)' + C

64. Right, or wrong? Say which for each formula aod give a briefrea-

son for each answer.

a.

I~dx=

b.

l~dx=~+C

c.

I~dx=t(~)'+c

+C

(x ~

+C

I)

3)' +c

66. Right, or wrong? Give a briefreason why.

_ ')-c-c-s_in_(_x') _ _ sin (x') _x_co_s_(_x__ dxx , x

+c

Initial Yalue Problems 67. Which of the following graphs shows the solution of the initial value problero

dy dx=2x, y=4whenx=l?

+C y

y

+I +C

x

I

+I +C

= x

4

61. Right, or wrong? Say which for each formula aod give a briefreason for each answer.

a. IXSinXdx =

~2 sinx + C

(1,4)

3

3 2

2 1 -1 0

b. IXSinXdx=-XCOSX+C

c. IXSinXdx

_-",15",(x,--+---,:"3)_'_ dx~ (x -+(X-2)4 x-2

I

I

= - x

v'x'+x+ C

65. Right, or wrong? Give a briefreason why.

I

sec'(5x - I)dx = ttm(5x - I) esc'

+ I)' dx

C

+ 5)-1 3

b. 13(2x

eseB. dB cscO - 8m8

Checking Antiderivative Formulas VerifY the formulas in Exercises 55--j)O by differentiation.

237

Antiderivatives

=

-xcosx

x

1

(a)

+ sinx + C

62. Right, or wrong? Say which for each formula and give a briefrea-

-1 0

x

x

(b)

(c)

Give reasons for your answer. 68. Which of the following graphs shows the solution of the initial value problero

son for each answer.

a. b. c.

I I I

tmBsee'BdB

~

tmBsee'BdB

~ ktm'B + C

""':J'B

tmBsee'BdB = ksec'B

+C

63. Right, or wrong? Say which for each formula aod give a briefreason for each answer.

a.

I

(2x

+ I)' dx

=

(2x

+ I)' 3

dy dx = -x, Y = I whenx = -I?

+C

+C

y

,~. o

(a)

y

x

x

(b)

Give reasons for your answer.

Hy y

o

(c)

x

238


Solve the initial value problems in Exercises 69-1l8. dy 69. dx = 2x - 7,

Solution (Integral) Curves Exercises 91-94 show solution curves of differential equations. 10 each exercise, fmd an equation for the curve through the labeled point

y(2) = 0

dy 70. dx = 10 - x, y(O) = -I dy 1 71. dx = x 2

+ X,

dy

72. dx = 9x 2

~

73• dy dx

I • /'

dy dx

~

77.

+ 5, y(-I)

+ COSI,

cos I

~; ~

+ sinl,

dr

78. dO ~ cos,,-o,

d'y

82.

-2

dx

d'y dx 2

83. ddl r2 2

84 d s 'd1 2

87.

~

0

I

v(O) = I

y'(O)

y'(O)

= 2,

~

4, y(O)

y(O)

~

tis dl

r(l)

=I

~

tis I

~

s(4)

~4

8'

= 0;

dlt~4

3 '

Applications 95. Finding displacement from an antiderivative of velocity

a. Suppose that the velocity of a body moving along the s-axis is

= I•

31.

I

=0

-drl dt 1=1

85. dx' ~ 6; d'O

~

2 = -I';

d'y

86. dl'

s(,,-)

0"(0)

= -2,

0'(0)

-sin t + cos t; ym(o) = 7, y"(O) = y'(O)

=

88. y(4) = -cosx + 8 sin2x; ym(o) ~ 0, y"(O) ~ y'(O)

~

-z'I0(0). = v2

r.;

y(O)

I, y(O)

~

=0 3

~

9.81 - 3.

90. Uniqneness of solutions If differentiable functions y = F(x) and y = G(x) both solve the initial value problem y(xo) = Yo.

on an interval I, must F(x) = G(x) for every x in

~

I to I

~

3 given

iii) Now find the body's displacement from I = 1 to I given that s ~ So when I ~ O.

=3

b. Suppose that the position s of a body moving along a coordinate line is a differentiable function oftime t. Is it true that once you know an antiderivative of the velocity function tis/dlyou can fmd the body's displacement from I = a to I ~ b even if you do not know the body's exact position at

either of those times? Give reasons for your answer.

89. Find the curve y ~ f(x) in the xy-plane that passes through the point (9, 4) and whose slope at each point is 3 \1';;.

dy dx = f(x),

v

i1) Find the body's displacement from I thats = -2 when 1= O.

y(4) =

= -I,

~

i) Find the body's displacement aver the time interval from I ~ I tol ~ 3giventhats ~ 5 when I ~ O.

y"(O) ~ -8, y'(O) ~ 0, y(O) ~ 5

for your answer.

94. y

r(O) ~ 1

2 - 6x;

= 0;

fly = sinx _ cos x dx y

-5

r(O)

dv I 79. dl = Zsecltanl,

-2 ~

~

= 0

s(O) ~ 4

-,,-sin,,-O,

81.

93.

92.

y(4) ~ 0

2vx

75. dl ~ I

y

0; y(2) = I

3x -2/3 , y(-I)

74. dx ~

76. :

4x

-

>

X

d2~ 1_ ~x1J3 dx 3

91.

n Give reasons

96. Liftoff from Earth A rocket lifts offthe surface of Earth with a constant acceleration of 20 rnjsec'. How fast will the rocket be going I min later? 97. Stopping a car in time You are driving along a highway at a steady 60 mph (88 ft/sec) when you see an accident ahead and slan3 on the brakes. What constant deceleration is required to stop your car in 242 ft? To fmd out, carry out the following steps.

239

Chapter 4 Questions to Guide Your Review 1. Solve the initial value problem Differential equation:

d 2s dt 2 = -k

(k constant)

101. Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is a

S=21

tis ~ 88ands ~ Owheot ~ 0 dt .

Initial conditions:

~

Differential equation:

O. (The answer will

3. Find the value of k that makes s = 242 for the value of t you found in Step 2. 98. Stopping a motorcycle The State of lllinois Cycle Rider Safety Program requires riders to be able to brake from 30 mph (44 ft/sec) to 0 in 45 ft. What constant deceleration does it take to do that?

Chapter

tis ~ Oands ~ 4wheot ~ 0 dt

VO and s

=

So when t

=

O.

where s is the body's height above 1he sorface. The equation has a minus sign because the acceleration acts downward, in the direction of decreasing s. The velocity Vo is positive if the object is rising at time t = 0 and negative if the object is falling. Instead of using 1he result of Exercise 101, you can derive Equation (2) directly by solving an appropriste initial value problem. What initial value problem? Solve it to be sure you have 1he right one, explaining the solution steps as you go along.

o 100. The hammer aod the feather

Initial conditions:

=

(2)

a. the velocity v = tis/dt in terms oft

d 2s - 2 ~ -5.2 ft/sec 2 dt

tis dt

102. Free fall near the surface of a plaoet For free fall near the sorface of a planet where 1he acceleration due to gravity has a constant msgnitude ofg leng1!l-units/sec', Equation (I) in Exercise 10I takes the form

b. the position s in terms of t.

Differential equation:

d 2s dt 2 = a

Initial conditions:

99. Motion along a coordinate line A particle moves on a coordinate line with acceleration a ~ d 2s/dt 2 ~ 15vt - (3/vt) , subject to the conditions that tis/dt = 4 and s = 0 wheo t = I. Find

WheoApollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonsttate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the eveot shows the hammer and 1he feather falling more slowly than on Ear1h, where, in a vacuum, 1hey would have taken only half a second to fall1he 4 ft. How long did it take the hammer and fea1her to fall 4 ft on 1he moon? To fmd out, solve the following initial value problem for s as a function of t. Then fmd 1he value of t that makes s equal to O.

(1)

+vot+so.

where Vo and So are the body's velocity and position at time t = O. Derive this equation by solving the initial value problem

Measuring time and distance from when the brakes are applied

2. Find the value of t that makes tis/dt involve k.)

2

COMPlITER EXPLORATIONS Use a CAS to solve the initial problems in Exercises 103-106. Plot the solution curves.

103. y' = cos2 x + sinx, y(".) = I 104·y'=t+ x, y(I)=-1 105. y' = 106. y"

I

V4 - x 2

~ ~ +~,

,y(O) = 2 y(l)

~

0,

y'(l)

~

0


1. What can be said about 1he extreme values of a function that is continuous on a closed interval? 2. What does it mean for a function to have a local extreme value on its domain? An absolute extreme value? How are local and absolute extreme values related, if at all? Give examples. 3. How do you find the absolute extrema of a continuous function on a closed interval? Give examples. 4. What are the hypotheses and conclusion of Rolle's Theorem? Are the bypotheses really necessary? Explain. S. What are the hypotheses and conclusion of the Mean Value Theorem? What pbysical interpretations might 1he theorem have?

6. State 1he Mean Value Theorem's three corollaries. 7. How can you sometimes identify a function f(x) by knowing f' and knowing 1he value of f at a point x = xo? Give an example. 8. What is 1he First Derivative Test for Local Extreme Values? Give examples of how it is applied. 9. How do you test a twice-differentiable function to determine where its graph is concave up or concave down? Give examples. 10. What is an inflection point? Give an example. What pbysical significance do inflection points sometimes have? 11. What is 1he Second Derivative Test for Local Extreme Values? Give examples of how it is applied.

240


12. What do the derivatives of a functioo tell you about the shape of its graph?

18. Can a function have more than one antiderivative? If so, how are the antiderivatives related? Explain.

13. List the steps you would take to graph a polynomial functioo. Illustrate with an example.

19. What is an indefinite integral? How do yeu evaluate ooe? What general formulas do yeu know for fmding indefmite integrals?

14. What is a cusp? Give examples.

20. How can you sometimes solve a differential equation of the form

15. List the steps you would take to graph a ratiooal functioo. Illustrate with an example. 16. Outline a general strategy for solving max-min problems. Give examples. 17. Describe Newton's method for solving equatioos. Give an example. What is the theory behind the method? What are some of the things to watch out for when you use the method?

Chapter

dy/dx

Practice Exercises

= cscx + 2 cotx have

b. Now factor I'(x) and show that j has a local maxinnun atx = 1.70998 andlocalminirnaatx = ±v3 '" ±1.73205.

"\7'5 '"

c. Zoom in on the graph to fmd a viewing window that shows the and x = v3. presence of the extreme values at x =

"\7'5

any local maximum values?

Give reasons for your answer. 3. Does j(x) = (7 + x)(11 - 3X)I/3 have an absolute minimum

The moral here is that without calculus the existence of two of the three extreme values would probably have gone unnoticed. On any normal graph of the function, the values would lie close enough together to fall within the dimensions of a single pixel on the screen. (Source: Uses o/Technology in the Mathematics Curriculum, hy Benny Evans and Jerry Johuson, Oklaboma State University, published in 1990 under Natiooal Science Foundation Grant USE-8950044.)

value? An absolute maximum? If so. fmd them or give reasons why they fail to exist. List all critical points off. 4. Find values of a and b such that the function

has a local extreme value of 1 at x = 3. Is this extreme value a local maximum, or a local minimum? Give reasons for your answer.

J,

S. The greatest integer functioo j(x) = l x defmed for all values of x, assumes a local maximum value of 0 at each point of [0, I). Could any of these local maximum values also be local minimum values of f7 Give reasons for your answer.

6. a. Give an example of a differentiable function j whose fIrst derivative is zero at some point c even though f has neither a local maximum nor a local minimum at c.

b. How is this consistent with Theorem. 2 in Section 4.1? Give

reasons for your answer. 7. The function y ~ I/x does not take on either a maximum or a minimum on the interval 0 < x < 1 even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for continuous functions? Why?

8. What are the maximum and minimum values of the function y ~ Ix I on the imerval -I :5 x < I? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why?

D

j(x)?

22. Ifyeu know the acceleration of a body moving along a coordinate line as a function of time, what more do you need to know to find the body's position function? Give an example.

Extreme Values 1. Does j(x) = x 3 + 2>: + tan x have any local maximum or minimum. values? Give reasons for your answer. 2. Does g(x)

~

21. What is an initial value problem? How do yeu solve one? Give an example.

9. A graph that is large enough to show a function's global behavior may fail to reveal important local features. The graph of j(x) = (x'/8) - (x 6/2) - x' + 5x 3 is a case in point.

a. Graph j over the interval -2.5

:5

x

:5 2.5. Where does the

graph appear to have local extreme values or points of inflection?

D 10.

(Continuation ofExercise 9.) a. Graph j(x) = (x'/8) - (2/5)x' - 5x - (5/x 2 ) + II over the interval - 2 :5 x :5 2. Where does the graph appear to have local extreme values or points of inflection? b. Show that j has a local maximum value at x = ~ '" 1.2585 1.2599. and a local minimum value at x ~

--rz '" c. Zoom in to fmd a viewing window that shows the presence of the extreme values at x = ~ and x = --rz.

The Mea. Value Theo",m 11. a. Show that g(l) = sin2 1 - 31 decreases on every interval in its domain.

b. How many solutions does the equation sin2 t - 3t = 5 have? Give reasons for your answer. 12. a. Show that y

~

tan 8 increases on every interval in its domain.

b. If the conclusion in part (a) is really correct, how do you explain the fact that tan". = 0 is less than tan ("./4) = I? 13. a. Show that the equation lution on [0, I].

D b.

x' + 2>:2 -

2 ~ 0 has exactly one so-

Find the solution to as many decimal places as yeu can.

14. a. Show that j(x) = x/(x domain. b. Show that j(x) = x 3 mum values.

+

I) increases on every interval in its

+ 2>:

has no local maximum or mini-

Chapter 4

15. Water in a reservoir As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acre-ft in 24 hours. Show that at some instant during that period the reservoir's volume was increasing at a rate in excess of225,000 gal/min. (An acre-foot is 43,560 ft' , the volume that would cover I acre to the depth of I ft. A cubic foot holds 7.48 gal.)

Practice Exercises

241

body's (8) velocity equal to zero? (b) acceleration equal to zero? Dur-

ing approximately what time intervals does the body move (c) forward? (d) backward?

21.

s

16. The formula F(x) ~ 3x + C gives a different function for each value of C. All of these functions, however, have the same derivative with respect to x, namely F' (x) ~ 3. Are these the only differentiable functions whose derivative is 3? Could there be any

others? Give reasons for your answers. 17. Show that

Graphs and Graphing Graph the curves in Exercises 23-32.

even though

I

x

x+I"-x+I' Doesn't this contradict Corollary 2 of the Mean Value Theorem?

23. Y = x' - (x'/6) 24.y=x'-3x'+3 25. y= -x'+6x'-9x+3

Give reasons for your answer. 18. Calculate the ftrst derivatives of f(x) = x'/(x' + I) and g(x) ~ -I/(x' + I). What can you conclude about the graphs

of these functions?

26. Y ~ (l/8)(x'

+ 3x' - 9x -

27. Y ~ x'(8 - x) 28. y 29. Y

= x'(2x' - 9) = x - 3x'/3 = x l /3(x - 4)

Analyzing Graphs In Exercises 19 and 20, use the graph to answer the questions.

30. y

19. IdentifY any global extreme values of f and the values of x at which they occur.

31.y~x~ 32. Y

y y

~

f(x)

(1, 1)

"-....H)

-"O+----~x

20. Estimate the intervals on which the function y = f(x) is

27)

= xv'4=7

Each of Exercises 33-38 gives the first derivative of a function y ~ f(x). (8) At what points, if any, does the graph of f have a local maximum, local mininuon, or inflection point? (b) Sketch the general abape of the graph.

x'

33. y' ~ 16 -

34. y' =x 2 -x-6

35. y'

~

6x(x

+ I)(x -

37. y'

=

x4

2x 2

-

2)

36. y' ~ x'(6 - 4x) 38. y'

=

4x 2

-

x4

In Exercises 39-42, graph each function. Then use the function's ftrst derivative to explain what you see.

s. increasing. b. decreasing. c. Use the given graph off' to indicate where any local extreme

values of the function OCCllI, and whether each extreme is a relative maximum or minimum. y

39. y = x'/3 + (x - 1)1/3 41. Y = x l /3

1)1/3

40. Y = x'/3

+ (x -

1)'/3

42. Y = x'/3 - (x - 1)1/3

Sketch the grapha of the rational functions in Exercises 43-50.

x

(2, 3)

+ (x -

+

I

43,y~x-3

45• y

~x'+1 x

2x

44,y~x+5

46 ~x'-x+1 • y x

3

x 4 -1

47. Y =

x +2 ----zx-

48. Y = ~

~

x2 - 4 -,-x-3

50. Y

49. Y

~

x2 -,-x-4

-2 OptImization 51. The sum of two noonegative numbers is 36. Find the numbers if Each of the grapha in Exercises 21 and 22 is the graph of the position function s = f(l) of a body moving on a coordinate line (I represents time). At approximately what times (if any) is each

a. the difference oftheir square roots is to be as large as possible.

b. the sum oftheir square roots is to be as large as possible.

242


52. The sum of two nonnegative numbers is 20. Find the numbers

of the corridor shown here? Round your answer down to the nearest foot.

a. if the product of one number and the square root of the other is to be as large as possible.

y

b. if one number plus the square root of the other is to be as large as possible. 53. An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 - x 2 • Find the largest area the triangle can have. 54. A customer has asked you to design an open-top rectangular stainless steel vat. It is to have a square base and a volume of 32 ft3, to be welded from quarter-inch plate, and to weigh no more than necessary. What dimensions do you recommend? 55. Find the height and radius of the largest right circular cylinder that can be put in a sphere of radius

v3.

56. The figure here shows two right circular cones, one upside down inside the other. The two bases are parallel, and the vertex of the smaller cone lies at the center of the larger cone's base. What values of rand h will give the smaller cone the largest possible volume?

o '--------'8'------~x Newton's Method 61. Let f(x) = 3x - x 3 • Show that the equation f(x) = -4 has a solution in the interval [2, 3] and use Newton's method to find it. 62. Let f(x) = x 4 - x 3 . Show that the equation f(x) = 75 has a solution in the interval [3, 4] and use Newton's method to find it.

Finding Indefinite IntegraLs Find the indefinite integrals (most general antiderivatives) in Exercises 63-78. Check your answers by differentiation.

r

63.

1 2'

65.

1

67

•

J J J

3

(x + 5x - 7) dx

~) dt

(3Vt + dr

57. Manufacturing tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where o ~ x ~ 4 and

73.

75. 40 - lOx

5-x·

Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make? 58. Particle motion The positions of two particles on the s-axis are SI = cos t and S2 = cos (t + 7r/4).

77. 78.

60. The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the comer

J ~dB 2

+ x 4)-1/4 dx

72.

2

sec :0 ds

74.

esc \l2B cot \l2B dB

76.

sin2

~ dx

2

I dx

cos

3

+ t) dt

-

6 dr

(r -

J J J

(Hint: sin2 B = 1 -

7

+

0

(2 - x)3/5 dx

csc

2

sec

7TS

ds

~ tan ~ dB

~os 2B)

Solve the initial value problems in Exercises 79-82. dy

79. -d =

x

o 59. Open-top box

An open-top rectangular box is constructed from a 10-in.-by-16-in. piece of cardboard by cutting squares of equal side length from the comers and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume. Support your answers graphically.

70.

68

(8t

InitiaL VaLue ProbLems

a. What is the farthest apart the particles ever get? b. When do the particles collide?

J J J J J

x 3(1

•

66.

+ 5)2

(r

J ~ JC!;; -:4) J \12)3

64.

69. J30W+! dO 71.

y=

(8,6)

6

=

80. : 2

81.

d r

-2

dt

82.

d 3r -3 dt

x2

+

1

- - 2- ,

x

y(l) = -1

(x + } Y.

y(1) = 1

" r 3 = 15vt + "r; V t

= -cos t;

r'(l) = 8,

r"(O) = r'(O) = 0,

r(l) = 0

r(O) = -1

dt

Chapter 4 Additional and Advanced Exercises


Chapter

1. What can you say about a function whose maximum and minimum values on an interval are equal? Give reasons for your answer. 2. Is it true that a discontinuous function cannot have both an absolute maximum and an absolute minimum value on a closed interval? Give reasons for your answer.

3. Can you conclude anything about the extreme values of a continuous function on an open interval? On a half-open interval? Give reasons for your answer.

4. Local extrema

243

Use the sign pattern for the derivative

11. Finding a function Use the following information to find the values of a, b, and e in the formula I(x) = (x + a)/ (bx 2

+

ex

+

2).

i) The values of a, b, and e are either 0 or 1.

ii) The graph of I passes through the point ( - 1, 0) . iii) The line y = 1 is an asymptote of the graph of I.

12. Horizontal tangent will the curve y = x 3 tal tangent?

For what value or values of the constant k 3x - 4 have exactly one horizon-

+ kx 2 +

13. Largest inscribed triangle

I

to identify the points where values.

has local maximum and minimum

Points A and B lie at the ends of a diameter of a unit circle and point C lies on the circumference. Is it true that the area of triangle AB C is largest when the triangle is isosceles? How do you know?

14. Proving the second derivative test

5. Local extrema a. Suppose that the first derivative of y = I(x) is y' = 6(x

The Second Derivative Test for Local Maxima and Minima (Section 4.4) says:

a.

+ 1)(x - 2)2.

At what points, if any, does the graph of I have a local maximum, local minimum, or point of inflection?

b. Suppose that the first derivative of y = I(x) is y' = 6x(x

+ 1)(x -

2).

At what points, if any, does the graph of I have a local maximum, local minimum, or point of inflection? 6. If I' (x) ~ 2 for all x, what is the most the values of I can increase on [0, 6]? Give reasons for your answer. Suppose that I is continuous on [a, b] and that e is an interior point of the interval. Show that if I'(x) ~ 0 on [a, e) and I'(x) ~ 0 on (e, b], then I(x) is never less than I(e) on [a, b].

7. Bounding a function

I has a local maximum value at x

= e if I' (e) = 0 and

I"(e) < 0 b.

I has a local minimum value at x

= e if I' (e) = 0 and

I"(e) > O. To prove statement (a), let e = (1/2) II" (e) I.Then use the fact that .

I"(e) = hm

+ h) -

I'(e

h~O

I'(e

+ h)

h

> 0,

I '(e + h

S

.

= hm - - - h~O

h

to conclude that for some S

o < Ihl
0 feet from the right goal post. (See the accompanying figure.) Find the distance h from the goal post line that gives the kicker his largest angle f3 . Assume that the football field is flat.

a tax of t dollars per item sold? Comment on the difference between this price and the price before the tax. 20. Estimating reciprocals without division You can estimate the value ofthe reciprocal of a number a without ever dividing by a if youapplyNewton'smethodtothefunctionj(x) = (1/x) - a. For example, if a = 3, the function involved is j(x) = (1/x) - 3. a. Graphy = (1/x) - 3. Where does the graph cross the x-axis? b. Show that the recursion formula in this case is

Goal post line

so there is no need for division.

h

21. To find x = ~, we apply Newton's method to j(x) = x q - a. Here we assume that a is a positive real number and q is a positive integer. Show that Xl is a "weighted average" of Xo and a/x oq - l , and find the coefficients mo, ml such that

mo > O,ml > 0, mo + ml = 1. Football

17. A max-min problem with a variable answer Sometimes the solution of a max-min problem depends on the proportions of the shapes involved. As a case in point, suppose that a right circular cylinder of radius r and height h is inscribed in a right circular cone of radius R and height H, as shown here. Find the value of r (in terms of R and H) that maximizes the total surface area of the cylinder (including top and bottom). As you will see, the solution depends on whether H ::::; 2R or H > 2R.

What conclusion would you reach if Xo and a/x oq - l were equal? What would be the value of Xl in that case? 22. The family of straight lines y = ax + b (a, b arbitrary constants) can be characterized by the relation y" = O. Find a similar relation satisfied by the family of all circles

where h and r are arbitrary constants. (Hint: Eliminate h and r from the set of three equations including the given one and two obtained by successive differentiation.) 23. Assume that the brakes of an automobile produce a constant deceleration of k ft/sec 2 • (a) Determine what k must be to bring an automobile traveling 60 mi/hr (88 ft/sec) to rest in a distance of 100 ft from the point where the brakes are applied. (b) With the same k, how far would a car traveling 30 mi/hr travel before being brought to a stop? 24. Let j(x), g(x) be two continuously differentiable functions satisfying the relationships j'(x) = g(x) and j"(x) = - j(x). Let h(x) = f2(x) + g2(x). Ifh(O) = 5, findh(IO). 25. Can there be a curve satisfying the following conditions? d 2y/dx 2 is everywhere equal to zero and, when x = 0, y = 0 and dy/dx = 1. Give a reason for your answer. 26. Find the equation for the curve in the xy-plane that passes through the point ( 1, - 1) if its slope at x is always 3x 2 + 2.

18. Minimizing a parameter Find the smallest value of the positive constant m that will make mx - 1 + (1/x) greater than or equal to zero for all positive values of x. 19. Suppose that it costs a company y = a + bx dollars to produce x units per week. It can sell x units per week at a price of p = c - ex dollars per unit. Each of a, b, c, and e represents a positive constant. (a) What production level maximizes the profit? (b) What is the corresponding price? (c) What is the weekly profit at this level of production? (d) At what price should each item be sold to maximize profits if the government imposes

27. A particle moves along the x-axis. Its acceleration is a = -t 2 • At t = 0, the particle is at the origin. In the course of its motion, it reaches the point x = b, where b > 0, but no point beyond b. Determine its velocity at t = O.

Vi - (l/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' + 2hx + c with 0 > O. By coosidering the minimum, prove that f(x) '" 0 for all real x if, and ooly if, b 2 - ac :s; O. 30. Schwarz's inequality

and deduce Schwarz's inequality: (01 h,

+ o,h, + ... + a"h.)' '" (0" + 0,' + ... + a,,')(h,' + h,' + ... + h.').

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.

a. In Exercise 29, let

Chapter

245


MathematicatMap1e Modules: ModonAlong 0 Straight Line: Position .... Velocity .... Acceleration You will observe the shape of a grapb through dramatic animated visualizatioos of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated

Newton's Method: Estimate 'fr to How Many Places? 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.

Area and Estimating with Finite Sums

5.1

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.

y

0.5 R

~---_.L...--_-_....:L....-_~X

a

0.5

FIGURE 5.1 The area of the region R cannot be found by a simple formula.

246

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 !,

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

-oI---~=----'!--~x

o

o

0.5

0.25

0.5

(a)

0.75

(b)

FIGURE 5.2 l

13 I)

3

t

250

Chapter 5: Integration With / evaluated at t = 0, I, and 2, we have

D '" /(It) at

+ /(t2)

+ /(t3) at + [160 - 9.8(1)](1) +

at

= [160 - 9.8(0)](1)

[160 - 9.8(2)](1)

= 450.6.

(b) Three subintervals oflength I, with / evaluated at right endpoints giving a lower sum:

o

11

t2

•

•

13

.) t

1

2

3

~--_-

x

o

o

U

M

15

2

b x

a

The lengths of the subintervals are Ax! = 0.2, Ax, = 0.4, Ax, = 0.4, Ax, = 0.5, and Axs = 0.5. The longest subinterval length is 0.5, so the norm of the partition is Ilpll = 0.5. In this example, there are two subintervals of this length. _ (b)

The curve of Figure 5.9 with rectangles from rmer partitioos of [a, b]. Finer partitioos create collections of rectangles with thinner bases that FIGURE 5.10

approximate the regioo betweeo the grapb off and the x-axis with increasing

accuracy.

Any Riensann sum associated with a partition of a closed interval [a, b] dermes rectangles that approximate the region between the graph of a continuous function f and the x-axis. Partitions with norm approaching zero lead to collections ofrectangles that approximate this region with increasing accuracy, as suggested by Figore 5.10. We will see in the next section that if the function f is continuous over the closed interval [a, b], then no matter how we choose the partition P and the points Ck in its subintervals to construct a Riemann sum, a single limiting value is approached as the subinterval widths, controlled by the norm of the partition, approach zero.

Exercises 5.2 Sigma Notation Write the sums in Exercises 1--6 without sigma notation. Then evaluate them. ,

6k

l'~k+1

'-I, 3. ~ coskw '-I,

5. ~(_1)k+1 Sinf

2

7. Which of the following express I + 2 + 4 + 8 + 16 + 32 in sigma notation?

±

. '-I,

k-I k

4. ~ Sink1T

'-I,

6. ~(-I)'coskw

8. Which of the following express I - 2 sigma notation?

,

a. ~(-2)k-l k=1

, b. ~(-l)'2' k=O

+ 4 - 8 + 16 - 32 in

, c. ~(-1)k+12k+' k=-2

262

Chapter 5: Integration 5

9. Which formula is not equivalent to the other two? 4 (-I)k-! ~--

a.

b.

~ k- I

2 (-I)' ~-

1 (-I)' c.~-

~k+ I

'~1 k + 2

10. Which formula is not equivalent to the other two?

,

4

'-I

+ I)'

11. I + 2 + 3 + 4 + 5 + 6

12. I + 4 + 9 + 16

I I I I 13. 2 + 4 + 8 + 16

14. 2 + 4 + 6 + 8 + 10

I

2

3

16'-:5+:5-:5+:5-:5

1-1

k-l

•

• b,

b. ~b'1 6

1=1

•

c. ~(a, + b,) 1=1

" e. ~(b, - 2a,)

d. ~(a, - b,) k=l

~

" 0 and ~b, ~

1=1

I. Fiod the values of

"

c.

'-I

'-I " d. ~(b, '-I

I)

I)

Evaluate the sums io Exercises 19-32. 10

19. a. ~k k-I 13

20. a. ~k k-I

7

21. ~(-2k) ...1

•

23. ~(3 - k 2 ) ...1

5.3

264

500

30. a. ~k

b. ~k2

c. ~k(k - 1)

" 31. a. ~4

b. ~c

i=1 17

k=3 71

k=18

k=3

b"

"

" c.~(k-I)

'-I " ~(k+2n) b. '-I ~*

'-I

" k

c. ~2 k-l

n

RIemann Sums Partition the ioterval ioto four suhiotervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum ~~-d(c,) .1. 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 IYb 39.],

,w

31.

x'dx

40.

Jo[" x'dx

Use the rules in Table 5,4 aod Equations (1)-(3) to evaluate the integrals in Exercises 41-50.

41. ]" 7 dx

42.],'sxdx

43.],'(21 - 3) dl

44.

Jo (I - v'2) dl

45. j,' (I + I) dz

46.

],0(2z - 3) dz

47·1'3U'dU

48.

f' 24u' du J'/2

49.

],\3x' + x - 5) dx

[0

50.1°(3X'

+x -

5) dx

FInding Area by Definite Integrals 10 Exercises 51-54, use a definite integral to rmd the area of the region between the giveo curve aod the x-axis on the interval [0, b].

51. Y = b. _],'h(U)dU

>

26.1'3Idl, O

28. f(x) = 3x

29.

b. /,' [f(x)

e. /,'[2f(x) - 3h(x)] dx

4

b

['/2

Use the rules in Table 5,4 to rmd

271


3x'

52. Y = ,,-x'

272

Chapter 5: Integration

Finding Average value In Exercises 55--{j2, graph the function and fmd its average value over the given interval.

55. f(x) = x' - I 56. f(x) =

x' -T

on

[0,0]

10' sin (x') ax cannot possibly be 2.

76. Show that the value of fo' and 3.

~ ax lies between 2 v'2

77. Integrals of nonnegative functions

57. f(x) = -3x' - I

[0,3]

on

75. Show that the value of

on

[0, I]

58. f(x) ~ 3x' - 3 on [0, I]

'" 2.8

Use the Max-Min inequal-

ity to show that iff is integrable then

f(x) '" 0 on

=*

[a, b]

J.bf(x) ax '" O.

59. f(t) = (t - I)' on [0,3] 60. f(t) ~ t' - t on 61. g(x) =

62. h(x) =

Ixl -Ixl

[-2, I] on a. [-1,1], h. [I, 3], and c. [-1,3]

I

78. Integrals of nonpositi"" functions then

Show that if f is integrable

on a. [-1,0], h. [O,I],and c. [-1,1]

f(x) '" 0 on

=*

[a, b]

J.bf(X) ax '" O.

Definite Integrals as Limits Use the method of Example 4a to evaluate the defmite integrals in Exercises 63-70.

63. J.b C ax

65. J.b x'ax,

67.1;

66.

(3x' - 2x + I) ax

1:

68.1:

10'

80. The inequality sec x '" I + (x'/2) holds on (-"./2, ",/2). Use it to find a lower bound for the value of fo' sec x

(x - x') ax

81. Ifav(f) rea11y is a typical value of the integrshle function f(x) on [a, b], then the constant function av(f) should have the same integral over [a, b] as f. Does it? That is, does

+

ax.

x'ax J.b av(j) ax

69. J.b x' ax,

a< b

ax.

I) ax

64. /,' (2x

a< b

79. Use the inequality sin x '" x, which holds for x '" 0, to fmd an upper bound for the value of sin x

~ J.bf(x) ax?

70. /,' (3x - x') ax


Theory and Examples 71. What values of a and b maximize the value of

82. It would be nice if average values of integrable functions obeyed the following rules on an interval [a, b].

a. av(j + g) = av(j) + av(g) h. av(kf)

= k av(j)

c. av(j) '" av(g)

(Hint: Where is the integrand positive?) 72. What values of a and b miuimize the value of

(any nornber k) if f(x) '" g(x)

on

[a, b].

Do these rules ever hold? Give reasons for your answers. 83. Upper and lower sums for increasing functions

a. Suppose the graph of a continuous function f(x) rises steadily as x moves from left to right across an interval [a, b]. Let P be a partition of [a, b] into n subintervals oflength ~ (b - alin. Show by referring to the accompanying figure that the difference between the upper and lower soros for f on this partition can be represented graphically as the area of a rectangleR whose dimensiona are [f(b) - f(a)] by (Hint: The difference U - L is the sum of areas of rectangles whose diagonals QOQb Q,Q" ... , Q._,Q.lie along the

ax

73. Use the Max-Min Inequality to fmd upper and lower bounds for the value of

' I /,o 1 + x 2

--ax•

74. (Continuation of Exercise 73.) Use the Max-Min Inequality to fmd upper and lower bounds for 0,' - -I a2 x /o , 1 x

+

and

ax.

curve. There is no overlapping when these rectangles are shifted horizontally onto R.)

/,' - -Ia x . 2 0.5 1 + x

ax.

h. Suppose that instead ofbeing equal, the lengths of the subintervals ofthe partition of [a, b] vary in size. Show that

Add these to arrive at an improved estimate of

' - -I a x /,o 1 + x 2 •

U- L '" If(b) - j(a)lax.....

where

ax,... is the norm of P, and hence that lirn~~o = O.

(U - L)

5.3


273

y

y

1

f (b) - f(a)

Rj --->I~I 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 contiuuous. Use this and the figure for Exercise 86 to show that iff is contiuuous and E > 0 is given, it is possible to make U - L '" E' (b - a) by making the largest of the sufficiently small.

ax,'s

88. If you average 30 mijh on a 150-mi trip and then return over the same 150 mi at the rate of 50 mifh, what is your average speed for the trip? Give reasons for your answer. COMPUTER EXPLORATIONS !fyour CAS can draw rectangles associated with Riemann sums, use it to drsw rectangles associated with Riemann sums that converge to the integrals in Exercises 8~94. Use n ~ 4, 10,20, and 50 subintervals of equallengtb in each case. {' I 89. Jo (I - x) dx ~ 2

274


90.j,'<X' + I)

dl; =

91.

J~ cos x dl; =

92.

Jo

93.

[iX I

1

b. Partition the interval into n = 100, 200, and 1000 subintervals of equallengtb, and evaluate the function at the midpoint of each subinterval.

0

c. Compute the average value of the function values generated in part (b).

[~/4

d. Solve the equationf(x) ~ (average value) for x using the average value calculated in part (c) for the n = 1000 partitioning.

sec' x dl; = I dl;

94.1' ~dl;

~

95. f(x) = sinx

I

(The integra1'svalue is about 0.693.)

10 Exercises 95-98, use a CAS to perform the following steps:

a. Plot the functions over the given interval.

5.4

I

y

f(e), average heighl --.1--'----~-----O~x

a

b

----->I

FIGURE 5.16 The value f(e) in the Mean Value Theorem. is, in a sense, the average (or mean) height off on [a, b]. When f '" 0, the area of the rectangle is the area under the graph of f from a to b, f(e)(b - a) =

98.f(x)=xsin'~

[~, 'IT]

on

on

[~,'IT]

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 than by taking limits of Riemann sums as we did iu Section 5.3. Leibniz and Newton exploited this relationship and started mathematical 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.

Sir Isaac Newton (1642-1727)

c ~b- a

~

[0, 'IT]

on

The Fundamental Theorem of Calculus

HISTORICAL BIOGRAPHY

o

97. f(x) = x sin

[0, 'IT]

on

96. f(x) ~ sin' x

l'

f(x) dl;.

Mean Value Theorem for Definite Integrals In the previous section we defiued the average value of a continuous function over a I(x) dx divided by the length or width closed iuterva1 [a, hI as the def'mite iutegral 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, h]. 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.

J:

THEOREM 3-The Mean Value Theorem for Definite Integrals ous on [a, hI, then at some poiut c iu [a, hI,

l

If 1 is continu-

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

Avm~evw~ln ~~~-------~notassumed

"t---t----;;------x o 2 FIGURE 5.17

A discontinuous function

275

Since 1 is continuous, the Intermediate Value Theorem for Continuous Functions (Section 2.5) says that 1 must assume every value between min 1 and max f. It must therefore assume the value (l/(b - a))J:/(x) dx at some pointe in [a, b]. • The continuity of 1 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 asswne its average value.

b/(x) dx

=

0,

then I(x) = 0 at least once in [a, b]. Solution

The average value of 1 on [a, b] is

1 b

avU) = b _I a a I(x) dx = b _I a . 0 = O. By the Mean Value Theorem, 1 assumes this value at some point e

E

[a, b].

•

Fundamental Theorem, Part 1 area

y

~

If 1(/) is an integrable function over a finite interval I, then the integral from any f'1xed number a E I to another number x E I defines a new function F whose value at x is

F(x)

F(x) = [ / ( / ) dl.

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 > Q.

(1)

For example, if 1 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 ofintegration of an integral, but F is just like any other real-valued function ofa real variable. For each value ofthe input x, there is a well-defmed numerical output, in this case the defmite integral of1 from a to x. Equation (I) gives a way to defme new functions (as we will see in Section 7.2), but its importance now is the connection it makes between integrals and derivatives. If1 is any continuous function, then the Fundamental Theorem asserts that F is a differentiable function of x whose derivative is 1 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 1 ;;,: 0 on [a, b], then the computation of F'(x) from the defmition of the derivative means taking the limit as h ---> 0 ofthe difference quotient F(x b In Equation (I), F(x) is the area to the left of x. Also, F(x + h) is the area to the left of

FIGURE 5.19

+ h. The difference quotient [F(x + h) - F(x)l/h is then

x

approximately equal to f(x), the height ofthe rectangle shown here.

+ h) - F(x) h

For h > 0, the numerator is obtained by subtracting two areas, so it is the area under the graph of1 from x to x + h (Figure 5.19). If h is small, this area is approximately equal to the area ofthe rectangle 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 ofthis 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 1 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:

di

F'(x) = ax

X

(2)

I(t) dt = I(x).

a

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 Theorero to tmd dyJax if

X

(t 3

+ I)dt

(b) y = 153tSintdt

(c) Y =

1'"

costdt

We calculate the derivatives with respect to the independent variable x.

i

X

(a) -dy = -d (t 3 + I)dt = x 3 + I ax ax a 5 dy = ax d1x 3tsintdt = ax d ( f ' 3tsintdt ) (b) ax

-15

= -

d (X ax15 3tsintdt

= -

3x sin X

Eq.(2)withf(r) ~ r'

+

1

Table 5.4, Rule 1

Eq. (2) with f(t) ~ 31Sin t

(C) The upper limit of integration is not x but

x 2• This

makes y a composite of the two

functions,

y =

1"

and

costdt

We must therefore apply the Chain Rule when finding dyJax.

dy dy du ax=du'ax =

(~1" costdt), du 1

= casu'

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 f(x) for each x in (a, b). Thus, Fi,,-(x_+~h)'----_Fi-"(x-'-)

F'(x) = lim _ h~O

=

h

l~d [[+hf(t) dt l

= lim -h h--+O

l

[f(t) dt]

x h

+ f(t) dt

Table 5.4. Rule 5

x

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 f in the interval between x and x + h. That is, for some number c in this interval,

I1

h

x

X

+h f(t) dt = ftc).

(4)

All h --+ 0, x + h approaches x, forcing c to approach x also (because c is trapped between x and x + h). Since f is continuous atx, f(c) approaches f(x): lim ftc) = f(x).

(5)

h~O

In conclusion, we have l

F'(x) = lim -h h-O

l

x h

+ f(t) dt

x

= lim ftc)

Eq. (4)

= f(x).

Eq. (5)

h~O

Ifx = a or b, then the limit ofEquation (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 ofthe 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.

If f is continuous at every point in [a, b] and F is any antiderivative of f on [a, b], then

THEOREM 4 (Continued)-The Fundamental Theorem of Calculus, Part 2

l

bf

(x) dx

=

F(b) - F(a).

Proof Part I of the Fundamental Theorem tells us that an antiderivative of f exists, namely G(x) = [f(t) dt.

Thus, if F is allY antiderivative of f, 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.bI(t) dt - J."/(t) dt

=

J.bl(t) dt - 0

=

J.bI(t) dt.

•

The Evaluation Theorem is important because it says that to calculate the definite integral of lover an interval [a, b] we need do only two things: 1.

Find an antiderivative F of I, and

2.

Calculate the number F(b) - F(a), which is equal to

1: I(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 1 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.

0

sin (I') dl

j,' - ,+- - dl 2 1

,

1

4

+ 1)10 dl)' dl

7f

Ixl < -2

v l - t2

o

J,['I,dl,

40.y= [0 ~ Jtau 1 + t Area Ia Exercises 41-44, fmd the total area between the region and the

x-axis. 41. y = -x 2 42. y

= 3x 2 -

-

lx,

3,

43. y=x'-3x 2 44. y

= x ' /' - x,

-3:5 x

-2:S; x

+2x,

:5

:5

2

2

0"'x"'2

-I '" x '" 8

5.4 The Fundamental Theorem of Calculus

Find 1he areas of1he shaded regions in Exercises 45-48. 45.

283

56. Show that if k is a positive constant, then the area between the x-axis and one arch of1he curve y = sin kx is 2/k.

y

57. Co.tfrom marginal eo.t The marginal cost ofprintiog a poster when x posters have been printed is x = 1T ~f-------""-L~x

2v'i

58. Revenue from marginal revenue Suppose that a company's marginal revenue from 1he manufaeture and sale of eggbeaters is

y

~ .. 6

47.

I

dx

dollars. Find e(100) - e(l), 1he eost ofprintiog posters 2-100.

o

46.

de

:

48.

+

2 - 2/(x

1)2,

where r is measured in thousands of dollars and x in thousands of units. How much money should 1he company expect from a produetion run of x = 3 1housand eggbeater>? To fmd out, integrate 1he marginal revenue from x = 0 to x = 3. 59. The temperature T(OF) ofaroom at time I minutes is given by

6

y

=

y

2

T = 85 - 3'\!25=t for 0 '"

25.

a. Find 1he room's temperature when 1= 0, 1= 16, and 1= 25.

= sec 8 tan 8

y

I '"

b. Find 1he room's average temperature for 0 '" I '" 25. --;;;;f-----Jl~--:oor:----> 6

60. The height H (It) of a palm tree after growing for I years is given

4

by H

-V2

=

1

'1

-dl - 3

d. Y =

-I

I

=

x'

for 0 '" I '" 8.

Y('IT)

=

b. Find 1he tree's average height for 0 '" I '" 8. f(l) dl = x 2 - 2x + I. Find f(x).

61. Suppose 1hat

J:

50. y'

-3

=

f(x) = 2 -

' I

w

/.

=

4

["

g(x) = 3 + JI

atx Express 1he solutions of1he initial value problems in Exercises 53 and 54 in terms ofintegrals. dy 53. dx

=

seex, y(2)

=

54. :

=

'\1'1+7,

y(l)

9

I

I + dl

64. Find 1he linearization of

I 52. y' = x' y(l) = -3

51. y' = sec x, y(O) = 4

['+1

J2

atx = I.

t dl - 3

seex, y(-I)

Jt

62. Findf(4)if f(l)dl = xeoS'ITX. 63. Find 1he linearization of

b. Y = /.' secldl + 4

1 I

e. y = 1'seeldl+4 dy 49. dx

51 1/3

a. Find 1he tree's height when I = 0, I = 4, and I = 8.

Initial Value Problems Each of 1he following functions solves one of 1he initial value problems in Exercises 4l}-52. Whieb function solves which problem? Give briefreasons for your answers.

a. y

'\It+1 +

=

3

sec (I - I)dl

= -1.

65. Suppose that f has a positive derivative for all values of x and that f( I) = O. Which of 1he following statements must be true of 1he function g(x) = /.' f(l) dl?

=

-2 Give reasons for your answers.

Theory and Examples 55. Archimede.' area formula for parabolie arebe. Archimedes (287-212 B.C.), inventor, ntilitary engineer, pbysicist, and 1he greatest mathematician ofclassical times in the Western world, discovered that the area under a parabolic arch is two-thirds the base times 1he height. Sketch 1he parabolic arch y = h - (4hlb 2 )x 2, -b/2 '" x '" b/2, assunting that h and b are positive. Then use calculus to fmd 1he area of the region enclosed between 1he arch and the x-axis.

a. g is a differentiable function of x. b. g is a continuous function of x.

e. The graph ofg has a horizontal taogent at x d. g has a local maximum at x

=

I.

e. g has a local minimum at x

=

1.

f. The graph ofg has an inflection point at x

=

=

I.

I.

g. The graph of dgl dx erosses 1he x-axis at x = I.

284


66. Another proof of The Evaluation Theorem

= Xo < Xl

< X2 ••• < X n = b be any partition of [a, b], and letFbe any antiderivative of j. Show that

a. Let a

, F(b) - F(a)

= "5'. t:1

[F(x,) - F(X'-l)]'

67. j(x) ~ x' - 4x'

e. From part (b) and the defmition of the defmite integral, show that

F(b) - F(a) COMPUTER EXPLORATIONS

~ /.' j(x) dx.

J:

In Exercises 67-70, letF(x) = j(t) dt for the specified function j and interval [a, b]. Use a CAS to perform the following steps and

answer the questions posed. &.

69. j(x) = sin2xcos~,

b. SolvetheequationF'(x) = O. What can you see to be true about the graphs of j and F at points where F' (x) ~ O? Is your observation borne out by Part I of the Fundamental Theorem coupled with information provided by the ftrst derivative? Explain your answer.

[O,~]

[0,21f]

[0,21f]

In Exercises 71-74, letF(x) =

J:V 25.

u

~ 2x'

27.

30.

~ I- cosf

u

+ 2y)
~ x 3f2 -

13. /

v'X sin' (x 3f2

14. /

:,cOS'(~)dx, u=-~

- 1) dx,

u

= y'

/

.'x 26. / tan7X 2 sec2 dx

28. Jr' (7 - ~~)' dr

dr

(V-

secztanz dz

33. /

~COS(+-I)dt

34. /

~ cos(Vt + 3) dt

35. /

~ sin~cos~dO

36. /

I

37. / t 3(1 39. / b. Using u = esc 20

(V -

esc - 2 1T) - cot - 2 1T) - dv 32. /

/

+ 4y 2 + I

15. / esc' 20 cot 20 dO a. Using u = cot 20

24. / tan' x sec' x dx

. 'x x dx sm 3 cos 3

Jr' (~; -I)'

+ 4) dz

sin(2t+ I) dt cos' (2t + I)

u ~ I - r3 1)'(y3

+ 2) dx

22. /cos (3z

29. / x 112 sin (x 3f2 + I) dx

~ 2t

31.

+ 4y 2 +

/

If'dO

v'Xx(l I+ v'X' dx x)

23. / sec' (3x

8. / x sin (2x') dx,

(I - cosf)' sinfdt, .~,

~ 3x' + 4x

u~I+~

dx,

b. Usingu= ~

17.

21. /

I

+ 2)(3x' + 4x)' dx, u

(1 + ~)1/3 /

+

+8

a. Usingu = 5x

Evaluate the integrals in Exercises 17-50.

u = 7x - I

4. /

6.

+4

= 2x

16./~ 5x + 8

41.

+ t')3 dt

:,~2-~dx

/ "----;Il dx ~

38. /

40.

~

cos\18

\18 sin' \18

~x ;, I dx

/1.~x2 3 x

42.

dO

/~x

X 3 '

- I dx x2

- I

dx

5.6 Substitution and Area Between Curves

43. 45. 47. 49.

! ! ! x'W+1 ! ~

x(x - 1)'· dt

44.

(x + 1)'(1 - x)' dt

46.

(x 2

dt

48.

4)' dt

SO.

!X~dt

! ! 3x'W+! ! ~

(x + 5)(x - 5)'/3 dt

(x

dt

4)' dt

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplilY 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.

51.

!

2

2

18 tan X sec x dt (2 + tan' x)2 a. u = tan x , followed byv = u', then byw = 2

b. u ~ tan'x,followedbyv ~ 2

+v

58.

~; =

59.

~;~ ~

(~- 0)'

2

300s

-4 sin (21 -

!

+U

VI + sin (x - I) sin (x - I) cos (x - I)dt

b.

!

57.

~ = 8sin2 (I + ~).

5.6

-2u du

!

-u 2 +

C2

u = cosx = -cos2 x

sin2xdx

+ C2

2sin:rcosx=sin2x

a. Show by evaiuatiog the integral in the expression I (1/60) -

I)', .(1) = 3

~ 4x (x 2 + 8r'/',

=

+ Ct

Can all three integrations be correct'l Give reasons for your answer. 64. (Continuation afExample 9.)

Solve the initial value problems in Exercises 55-60.

56. :

= 8in2 X

+ Ct

c. !2 sin x COS X ax = !

In;tlal Value Problems -

2 sin x cos x dx

u= sinx

= u2

=

sin YO dO VOcos'v1i

= 121 (31

~0

~ _ 00~2x + C,.

!

~

100, .(0)

a. !2SinXCOSXdx = !2UdU

Evaluate the integrals in Exercises 53 and 54. (2r - I)cos V3(2r - I)' + 6 53. dr V3(2r - I)' + 6

55.

~

63. It looks as if we can integrate 2 sin x cosx with respect to x in three different ways:

2

2

.'(0)

f),

Theory and Examples 61. The velocity of a particle moving back and fotth on a line is v ~ tis/dl ~ 6 sin 21 m/sec for all I. If. ~ 0 when I ~ 0, fmd the value of. when I = 'IT/2 sec. 62. The acceleration of a patticle moving back and fotth on a line is a ~ d 2./dI 2 ~ ~ cos 'lT1 m/sec' for all I. If • ~ 0 and v ~ 8 m/sec when I = 0, fmd. when I = I sec.

a. u = x - I, followed by v = sinu,thenbyw = I + if b. u = sin (x - I), followed by v = I + u 2 C. U = I + sin2 (x - I)

54.!

f

d 2y 60. dt 2 = 4 sec2 2xtan 2x, y'(O) = 4, y(O) = -I

c.u=2+tan3 x

52.

r(O) =

291

y(O)

['/60

oj.

V~sinI20'ITldl

that the average value of V = is zero.

V~sin

120 'lT1 over a full cycle

b. The circoit that rons your electric stove is rated 240 volts rms.

What is the peak value of the allowable voltage?

~0

c. Showthat

['/60

J.

.(0) = 8

2

(V.".)'sin 120'ITldl=

(V,

)'

;;; .

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-Substftutlon In Definite Integrals If g' is continuous on the interval [a, b] and 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 Itt F(g(x)) ~

tf(g(x)). g'(x) dx = F(g(x)) [ :

~

=

F(g(b)) - F(g(a))

=

F(u)

]

.~g(b) .~g(a)

9(b)

=

1

F(g(x))g'(x) f(g(x))g'(x)

Fundamental

f(u) duo

Theorem, Part 2

g(al

•

To use the formula, make the same u-substitotion u = g(x) and du = g'(x) dx you would use to evaluate the corresponding indefinite integral. Then integrate the transformed integral with respect to u from the value g(a) (the value ofu at x = a) to the value g(b)(thevalueofuatx = b).

EXAMPLE 1 Solution

Evaluate

1:

3x 2W+1 dx.

We have two choices.

Method 1: Transform the integral and evaluate the transformed integral with the transformed limits given in Theorem 7. Let" = x 3 + l,du = 3,x 2 dt.

[3x

2

W+1dx

=

[VUdU

=

Z 3/ 2 _u 3

When x ~ -I,. ~ (-I)' + I ~ O. Whenx ~ I,. ~ (I)' + I ~ 2.

]2

Evaluate the new deImite integral.

0

= Hz3/2 -

0

3/2] = HzV2] =

4'(2

Method 2: Transform the integral as an indefinite integral, integrate, change back to x, and use the original x-limits.

J W+1 dx JVU 3x

2

=

=

'2. u 3/2 + C

=

'2. (x 3 + 3

=

3

+

= H «(1)3

= HZ3/2 -

1)3/2

+C

1)3/2L

+

=

x 3 + l,du

=

3x 2 dx.

Integrate with respect to u.

3

[3x W+1 dx t(x 2

Letu

du

Replace. by

Usc the integral just found, with limits of intcgmtion for x.

1)3/2 - «-1)3

0

x' + l.

+

3/ 2] = HZV2] =

1)3/2]

4'(2

•

5.6 Substitution and Area Between Curves

293

Which method is better----evaluating the transformed definite integral with transfanned limits using Theorem 7, or transfanning the integral, integrating, and transforming back to use the original limits of integration? In Example 1, the flISt method seems easier, but that is not always the case. Generally, it is best to know both methods and to use whichever one seems better at the time.

EXAMPLE 2

We use the method of transfanning the limits of integration. Let u ~ cot 0, du ~

("/2 cotlicsc'lidli

=

lOU' (-du)

J"/4

1

=

WhenO

~

When 0

~

-cae' 0 dO,

-du ~ esc' 0 dB. ",/4,u ~ cot (",/4) ",/2, u ~ cot (",/2)

~

I.

~

O.

-lOUdU

• Definite Integrals of Symmetric Functions

y

The Substitotion Fannula in Theorem 7 simplifies the calculation of definite integrals of even and odd functions (Section 1.1) over a synuuetric interval [-a, a] (Figure 5.24).

THEOREM 8 -a

o

a

(a)

Let f be continuous on the aymmetric interval [ -a, a].

Iff is even, then 1>(X) dx

(a)

y

--!--+--"f-----\c----r.ao-~x

(b) Iff is odd, then

FIGURE 5.24

(a)f"""", J~f(x):k.

Riemann sum

~l

1:

(c•• /(c.)

T

A

/(c.) - g(c.)

I

a

b

H

n

As Ilpll ...... 0, the sums on the right approach the limit [/(x) - g(x dx because f and g are continuous. We take the area of the region to be the value of this integral. That is,

y

(c•• g(c.)

ax.

FIGURE 5.27 The area ,uk of the kth rectangle is the product of its height, f(q) - g(c.) , and its width, 4>:•.

X

=

.

lim L[/(ck) - g(Ck)] 4>:k =

11P1I~o k~l

l

b

[((x) - g(x)] dx.

a

DEFINmON If f and g are continuous with f(x) ;" g(x) throughout [a, b], then the area of the region between the curves y = f(x) and y = g(x) from a to b is the integral of (f - g) from a to b:

A

=

l

b

[/(x) - g(x)] dx.

When applying this definition it is helpful to graph the curves. The graph reveals which curve is the upper curve f and which is the lower curve g. It also helps you f'md the limits of integration if they are not given. You may need to find where the curves intersect to

5.6 Suhstitution and Area Between Curves y

295

detennine the limits of integration, and this may involve solving the equation f(x) = g(x) for vslues of x. Then you can integrate the function f - g for the area between the intersections.

EXAMPLE 4

Find the area of the region enclosed by the parabola y = 2 - x 2 andthe

liney = -x. Solution First we sketch the two curves (Figure 5.28). The limits of integration are found by solving Y = 2 - x 2 and y = -x simultaneously for x. 2 - x 2 = -x

Equate f(x) and g(x).

2

x -x-2=O

FIGURE 5.28 The region in Example 4 with a typical approximating rectangle.

(x

+ I)(x - 2)

Rewrite.

0 x=2.

x=-l,

=

Factor. Solve.

The region runs from x = - I to x = 2. The limits of integration are a = - I, b = 2. The area between the curves is

A = =

t 2 1

[f(x) - g(x)]

-1

HIsTORICAL BIOGRAPHY

=

Richard Dedekind (1831-1916)

EXAMPLE 5

Area

2

~ {(Yx - x + 2)dx J,'

' l

Area ~ 0 Yxdx

(x,f(x» I

\

+x

- x 2)

x 2)

(-x)]

-

ax = [ 2x 2 + ~ - ~3]2 2 3_

(4 + ~ - ~) - (-2 + ! + t)

=

ax

1

~

•

If the formula for a bounding curve changes at one or more points, we subdivide the region into subregions that correspond to the fonnula changes and apply the fonnula for the area between curves to each subregion.

y =

y

(2

ax = 1~[(2 -

(x,f(x»

y~Yx \ (4,2) B

Find the area of the region in the first quadrant that is bounded above by = x - 2.

vX and below by the x-axis and the line y

Solution The sketch (Figure 5.29) shows that the region's upper boundary is the graph of f(x) = vX. The lower boundary changes from g(x) = 0 for 0 :5 X :5 2 to g(x) = x - 2 for 2 :5 X :5 4 (both fonnulas agree at x = 2). We subdivide the region at x = 2 into subregions A and B, shown in Figure 5.29. The limits of integration for region A are a = 0 and b = 2. The left-hand limit for region B is a = 2. To f'md the right-hand limit, we solve the equations y = vX and y = x - 2 simultaneously for x: vX=x-2

\

Equate f(x) andg(x).

x = (x - 2)2 = x 2

y~x-2

-

4x

+4

x 2 -5x+4=O (x - I)(x - 4) = 0

e+-~---oc--!:-------!:-->x o / y~O 2 4

x

(x, g(x»

FIGURE 5.29 When the fonnu1a for a bounding curve changes, the area integral changes to become the sum ofintegrals to match, one integral for each ofthe shaded regions shown here for Example 5.

= 1,

Square both sides. Rewrite. Factor.

x

=

4.

Solve.

Only the vslue x = 4 satisfies the equation vX = x - 2. The vslue x = I is an extraneous root introduced by squaring. The right-hand limit is b = 4. ForO For2

:5 X :5 :5

x

:5

2: 4:

f(x) - g(x) = f(x) - g(x) =

vX vX -

0 =

vX

(x - 2) =

vX -

x

+2

296

Chapter 5: Integration We add the areas of subregions A and B to find the tota1 area: 2 Vi dx Tota1 area = ]. 0

+ {'(Vi - x + 2) dx

J2

.... orB

area of A

=

['1. X3/2]2 3

+

0

['1. X3/2 _ 3

2 x 2

+ 2x]4 2

=

t(2)3/2 - 0

+ (t(4)3/2 - 8 + 8) - (t(2)3/2 - 2 + 4)

=

t (8) -

~o.

2 =

•

Integration with Respect to y Ifa region's bounding curves are described by functions ofy, the approximating rectangles are horizontal instead of vertical and the basic formula hasy in place ofx. For regions like these: y

y

'1

x ~ g(Y)~===i

use the fannula

A = y

2

(4,2) (g(y), y)

l

d

[f(y) - g(y)] dy.

In this equation f always denotes the right-hand curve and g the left-hand curve, so f(y) - g(y) is nonnegative.

x=y+2

EXAMPLE 6

-;;f--::---:!:----:--x o y~O 2 4 FIGURE 5.30

It talres two

integrations to fmd the area ofthis region ifwe integrate with respect to x. It talres only one lfwe integrate with respect to y (Example 6).

Find the area ofthe region in Example 5 by integrating with respect toy.

Solution We first sketch the region and a typical horizontal rectangle based on a partition of an interval ofy-values (Figure 5.30). The region's right-hand boundary is the line x = y + 2, so f(y) = y + 2. The left-hand boundary is the curve x = y2,so g(y) = y2. The lower limit ofintegration is y = O. We find the upper limit by solving x = y + 2 and x = y2 sinJultaneously for y: y + 2 = y2 y2 _ Y _ 2 = 0 (y

+

I)(y - 2) = 0

y=-l,

y=2

Equa1o/(y) ~ y

+ 2 and

g(y) ~ y'.

Rcwri1o. Factor.

Solve.

The upper limit of integration is b = 2. (The value y = - I gives a point of intersection below the x-axis.)

5.6

297

Substitution and Area Between Curves

The area of the region is

A

=

l

d

[f(y) - g{y)] dy

= ].2[y + Z _ y2] dy =

].\z +

=

[Zy + y; - ~3J:

y - y2] dy

=4+!-~= ~O.

•

This is the result of Example 5, found with less work.

Exercises 5.6 17.

Jor/ oos-3 26sin26d6 Jor 5(5 -

6

Evaluating Definite Integrals

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises 1-24.

1. ..

t v':Y+l

dy

1~ v':Y+l dy Lr~dr

21. J.' (4y - y2 + 4y 3 + 1)-2/3 (12y 2 - 2y + 4) dy

I>.

3...

Jor/

1>.1

tan x sec' x dl;

-fr/4

4. a. J.'" 3 cos2 x sin x dx

b.

1

4 cos 1)'/4 sin I dl 20.

sin 21)3/2 cos 21 dl

22. J.\y3 + 6y 2 - 12y + 9t'/2 (y2 + 4y - 4) dy

0

x x

Jo{~/4 (I -

I>.

J.'r~dr tan sec' dl;

13~/2 ~ cot' (6) 6 sec' (6) 6 d6

19.

2...

4

18.

3~

3 cos2 x sinx dx

23. J.Y/,;iv'8 cos2 (63/2) d6

24.1~'/21-2 sin2 (I + +) dl

2~

5

J.'r

6

J.

3

Vi

7.

8... {'

Jo

9

4

{v'3

1>.1

-Vi

I>.

5r 2)' dr

+r lOy,;

(I + v 3/2)2

... Jo

0

I(1 2 + 1)'/3 dl

,,1'( -1

I>. 1>3(1 + 14)3 dl

+ 14)3 dl

(1

4x

W+1

dv

I>.

dl;

I>. I>.

11

I>.

12 13. a. 14. ..

E/2(2 {2'"

Jo

1°

5r

J.o (4 + r 2 )'

y

dr y = (1 - cau) sinx

j4

lOy,; dv , (I + ,,'/2)2 v'3

I

4x

0

dl;

-_C02t---+..c------;c-~·

x

2 + tanf)sec fd' I>.

cos z dz \14 + 3 sinz

b.

I>.

dw

15. J.' v't'+2t (514 + 2) dl

16

1 (I 1-~/2~/2(Z 1-~ \14 J.

o

"

dl;

cos31)sin3ldl

I) I + tan z sec'zdl

~

cosz dz + 3 sinz ~/2 . smw dw o (3+2cosw)'

j4

~-"--------'---->.

3

1-,\10+9 ~/6

sinw

26. y

~/3

cos31)sin3ldl

~/2 (3 + 2 cos w)'

25.

-v'3W+1

10... {' k d l ; 4 +9

Jo x 6 Jor/ (I -

'

1(12 + 1)'/3dl

Area Find the total areas of the sbaded regions in Exercises 25-40.

dy

. , 2yY(I + yY)'

28.

27. y

y = ~(COSX)(sin(1T + 1Tsinx»

y

298


29.

36.

30.

Y

y

y

y=!sec 2 t 2

y

1

~

1

2

"2

0

"

o

x

37.

38. Y

sf

(-3,5)

-4

y=x 2 -4 y

31. (-2,8)

(2,8)

8

39.

40. Y

NOTTOSCAIE

32.

Y

y=~

3

(3, 1)

~lf-~oi: 44.y=x2 -2x and y=x 45. y=x2 and y= -x 2 +4x 46. y~7-2x2 and y~x2+4 47. y ~ x' - 4x 2 + 4 and y = x 2 48. Y

=

xv'a 2

49. Y

=

v'R

x 2,

a

>

0,

and

5y

=

x

-

and y = 0

+6

(How many intersection points

are there?)

50. y

= 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 ~ y2

and x ~ y

+2

y

53. y2 - 4x = 4

and

4x - y = 16

54. x - y2 = 0

and x

+ 2y 2 = 3

55. x = y2 - Y

and x = 2y 2 - 2y - 6

56. x - y2j3 ~ 0 57. x = y2 - I

and x

+ y4

~ 2

and x = lyl'V'l=Y'

58. x = y' - y2

and x = 2y

-oi-""'-:,--!---+-_x

o

Find the areas ofthe regions enclosed by the curves in Exercises 59-{j2. 59.4x 2 +y=4 and x 4 _y= I 60. x'-y~O 61. x 62. x

299

Substitution and Area Between Curves

3x2_y~4

and

+ 4y 2 ~ 4 + y2 ~ 3

and x and 4x

+ y4 + y2

~ I, ~

for

x "" 0

0

2

79. The figrrre here shows triangle AOC inscribed in the region cut from the parabola y ~ x 2 by the line y ~ a 2 • Find the limit of the ratio of the area of the triangle to the area of the parabolic region as a approaches zero. y

Find the areas of the regions enclosed by the lines and curves in Exercises 63-70.

63. y

= 2 sin x and

64. y ~ 8 cosx

y

= sin2x, o::=:;

x::=:;

'IT

and y ~ sec' x,

65. y = cos ('1Tx/2)

-'1T/3:5 x :5 '1T/3 and y = I - x 2

66. y = sin ('1Tx/2)

and y = x

67. y ~ sec' x, 2

68. x = tan y

y ~ tan2 x,

X

~ -'1T/4, 2

and x = -tan y,

69. x = 3 siny Vcosy 70. y ~ sec' ('1Tx/3)

-'1T/4:5 y:5 '1T/4

and x = 0,

and y ~ xli',

and x ~ '1T/4

0:5 Y :5 '1T/2 -1:5 x :5 I

71. Find the area ofthe 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 - ylj3 ~ o and x - yl/5 ~ o.

80. Suppose the area of the region between 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 = f(x) and y = 2f(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 frrst quadrant bounded by the liney

= x,the line x = 2,thecurvey = l/x2 ,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. 75. The region bounded below by the parabola y ~ x 2 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.

a. Sketch the region and draw a line y

~ 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.

82. True, sometimes true, or never true? The area of the region between the graphs of the continuous functions y = f(x) and y = g(x)andtheverticallinesx = 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.)

l\f(x) - g(x)] dx.


2

76. Find the area ofthe region between the curve y = 3 - x and the line y ~ -I by integrating with respect to (8) x, (b) y. 77. Find the area of the region in the fJrst quadrant bounded on the left by the y-axis, below by the line y = x/4, above left by the curvey ~ I + v'X,andaboverightbythecurvey ~ 2/v'X. 78. Find the area of the region in the fJrst quadrant bounded on the left by they-axis, below by the curve x = 2yY, above left by the curve x ~ (y - 1)2, and above right by the line x ~ 3 - y.

Theory and Examples 83. Suppose thatF(x) is an antiderivative of f(x) Express {' sin2x dx x

11 in terms ofF.

= (sinx)/x,

x

> O.

300


84. Show that iff is continuous, then

l

1f (X) dx =

l

y

1f (1 - x) dx.

85. Suppose that

Find 1>(X)dx

if(a) f is odd, (h) f is even. 86. a. Show that if f is odd on [-a, a], then

89. Use a substitution to verify Equation (I). 90. For each of the following functions, graph f(x) over [a, b] and f(x + c) over [a - c, b - c] to convince yourself that Equation (I) is reasonable.

L:f(X)dx = O. 1>. Test the result in part (a) with f(x)

87.

~

sinxanda

~

'fr/2.

a. f(x) ~ x 2 , a ~ 0, b ~ I,

Iff is a continuous function, fmd the value ofthe integral ('

1=

Jo

by making the substitution u integral to 1.

~

%

a - x and adding the resulting

COMPUTER EXPLORATIONS Io Exercises 91-94, you will find the area hetween curves in the plane when you cannot fmd their points of intersection using siruple algebra. Use a CAS to perform the following steps:

a. Plot the curves together to see what they look like and how many points ofintersection they have. 1>. Use the numerical equation solver in your CAS to fmd all the points ofintersection. c. Iotegrate If(x) - g(x) I over consecutive pairs ofintersection values. d. Sum together the integrals found in part (c).

""1t dt = 1'1t dt . 1

The Shift Property for Definite Integral. A basic property of definite integrals is their invariance under translation, as expressed by the equation [f(x) dx =

t~'f(x + c) dx.

(1)

The equation holds wbeoever f is integrable and defmed for the necessary values ofx. For example, in the accompanying figure show that -1(x 1 -2

+ 2)' dx

=

(l x 'dx

Jo

because the areas of the shaded regions are congruent.

Chapter

v;=4,

c. f(x) =

88. By using a substitution, prove that for all positive numbers x andy,

1

a = 0, b = 'fr, C = 'fr/2 a = 4, b = 8, c = 5

1>. f(x) = sinx,

f(x) dx f(x) + f(a - x)

c ~ I

3

2

91. f(x) =

x x T - 2 -

92. f(x) ~

2x· -

3x'

2x

+

1 + 3'

10,

g(x) = x - I

g(x) ~ 8 - 12x

93. f(x) ~ x

+ sin (2x),

94. f(x) ~

cos x, g(x) ~ x' - x

x2

g(x) ~ x'


I. How can you sometimes estimate quantities like distance traveled, area, and average value with 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'lDll?

4. What is the norm of a partition of a closed interval? 5. What is the defmite integral ofa function f over a closed interval [a, b]? When can you he sure it exists?

6. What is the 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 with defntite integrals (Table 5.4). Give examples. 9. What is the Fundamental Theorem of Calculus? Why is it so iruportant? Illustrate each part of the theorem with an example. 10. What is the Net Change Theorem? 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" ofeach 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.

13. How is inregration by substitution relared to the Chain Rule?

Chapter

16. How do you defme and calculare the area of the region between the graphs of two continuous functions? Give an example.

Practice Exerdses

Finite Sums and Estimates 1. The accompanying figure shows the graph ofthe velocity (fr/sec) ofa model rocket for the first 8 sec after launch. The rocket accelerared straight up for the fIrSt 2 sec and then coasred to reach its maximum height at t = 8 sec.

-

'0

20

20

20

a. ~3al

b. ~ (al

k-l

0

50 0

2

Definite Integrals 6

4

8

Time after launch (sec)

I

!;- -..J

I

+-H'

1"\

I

+ 1-+ 2

+ f-\ ++

4

6

8

10

Time (sec) 10

10

3. Suppose that ~al = -2 and ~bl = 25. Find the value of k-I

•

lim ~(2el-I)-If2.1>;1,wherePisapartitionof[I,5]

1V'1I~0 1~1

•

6. lim ~e.<e.' -1)1/3.1>;.,wherePisapartitionof[I,3] 1V'1I~0 1-1 7. 8.

+

lim

1V'1I~0 t:1

111 (cos ("--2 aXl, where P is a partition of [-"IT, 0]

))

•

lim ~(sinel)(cos el) .1>;1, whereP is a partition of[O, "lT12]

1V'1I~0 1~1

t,

9. If 3f(x) dx = 12, J~,f(x) dx = 6, and J~,g(x) dx = 2, fmd the values of the following.

a.J:

\ 1\

1/1

5.

f(x) dx

e. j,-'g(x) dx

1-

o

In Exercises 5-8, express each limit as a defutire inregral. Then evaluare the integral to fmd the value of the limit. In each case, P is a partition of the given interval and the numbers Ck are chosen from the

subinrervals of P.

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 doting the time inrerval from t ~ 0 to t ~ 10 sec. About how far did the body travel doting those 10 sec? b. Sketch a graph of s as a function of t for 0 ,;; t ,;; 10 assumings(O) ~ O. 5

+ b1 )

d. ~(al- 2)

••

.~

>

k-l

'0

~100

OJ

10 - - b d. ~ 1 i-I 2

I)

4. Suppose that ~al = 0 and ~bl = 7. Find the values of

200 150 1 ill

(5 )

10

e. ~(al + b1 1-1

e.

l

b.l'f(X)dx d.1:(-"lTg (X))dx

'(f(x) + g(x)) 5 dx

-,

10. If Jo'f(x)dx = "IT, fo'7g(x)dx = 7, and the values of the following.

101g(x)dx =

a. J.'g(x) dx

b. j,'g(x) dx

e.l°f (x) dx

d. J.'

i-I

e. J.'(g(x) - 3f(x)) dx

V2 f(x) dx

2, fmd

302


Area Io Exercises 11-14, rmd the total area ofthe region between the graph of f and the x-axis. 11. f(x) ~ x 2 - 4x + 3, 0:5 x :5 3

32. Find the total area of the regioo between the curves y = sin x and y = cosx for 0 ::5 x :s; 3w/2.

12. f(x) = I - (x 2/4),

33. Show that y

-2:5 x :5 3

13. f(x) ~ 5 - 5x 2/',

dx 2

0:5 x :5 4

Find the areas of the regioos enclosed by the curves and lines in Exercises 15-26. 15. y

= x,

y

16. y ~ x,

17.

Vi +

=

l/x2,

I/Vi,

y ~

vY = I,

x

x ~

= 0,

l'+

dl solves the initial value problen3

I - x 2 ; y'(I)

J;( I

=0

= 3,

y(l)

= I.

+ 2~) dl solves the initial value

d 2y -2 ~ Vsecxtanx; dx

2 y

=2

34. Show that y = problen3

=2

X

~ x2 +

d'y

-1:5 x:5 8

Vi,

14. f(x) ~ I -

Initial Value Problems

y'(O) ~ 3, y(O) ~ O.

Express the solutions of the initial value probleD3S in Exercises 35 and 36 in terms of integrals.

y

sin x y () 35. dy dx = -x-, 5 =-3 = V2 - sin2 x,

36. :

y(-I) = 2

Evaluating Indefinite Integrals Evaluate the integrals in Exercises 37-44. 37. /2(COSX)-1/2 sinx dx

--:cr-----=--:-~x

o

+ I + 2 cos (28 +

39. /(28 18. x'

+

vY = I,

x

= 0,

y

= 0,

for

0:5 x :5 I

y

1

x3

= 2y 2,

X

= 0,

+ 4,

y

=3

20. x

=4

- y2,

X

=0

y = 4x - 16

23. y = sinx, y = x,

(~ + 2 sec2 (28 -

41. /

(I -

43. /

Vtsin(2I'/2) dl

t) + t) (I

dl

'If)) d8

+

~~2 -

42. /

(I

I dl

44. /

sec 8 tJln 8 VI

+ sec 8 d8

Evaluating Definite Integrals Evaluate the integrals in Exercises 45-70.

21. y2 ~ 4x, y ~ 4x - 2 22. y2 = 4x

l))d8

40. /

+ vY = I, 0 =s x =s 1

o 19. x

38. / (tJlnxt'/2 sec2x dx

0:5 x :5 'If/4

24. y = Isinxl, y = I, -'If/2:5 x :5 'If/2 25. y=2sinx, y=sin2x, O::5x::5'11' 26. y = 8 cosx, y = see> x,

-'If/3 :5 x :5 'If/3

2 45. 1>3X - 4x

{2

47. 49. 51 •

J1

+ 7) dx

48.

{'~

1.

50.

IVt

1

36 dx

(2) + I)'

0

27. Find the area of the ''triangular'' regioo hounded 00 the left by x + y = 2, on the right by y = x 2,and above byy = 2.

53. {' x-1/3(1 - x 2/3)'/2 dx

28. Find the area of the ''triangular'' regioo bounded 00 the left by y ~ Vi, 00 the right by y ~ 6 - x,andbelowbyy ~ I. 29. Find the extren3e values of f(x) ~ x' - 3x 2and rmd the area of

55.

the region enclosed by the graph off and the x-axis. 30. Find the area ofthe regioo cut frOO3 the rlrst quadrant by the curve x 1/2 + y1/2 ~ a1/2. 31. Find the total area of the region enclosed by the curve x = y2/3 and the linesx = y and y = -1.

il/8

57. 59.

J.1J'

J. 1~

8in2 5r

61.1

'~

dr

•

54. 56.

sec> 8 d8

58.

(88' - 12>2

+

J1 X

-01/3

5) dx

'(I l 1. 1. 1

cof ~dx

60.

secxtanxdx

62.

dx

+ VU)l/2 •r vu

du

dr

V'(7 - 5r)2 1/2 0 x'(1 + 9x')-'/2 dx 0

J.~/' cos2 (41 -

1

'~/'

1

'~/'

csc2 x

f)

ax

~/'

0

-/3

1

1

52

~/'

0

1.

{27

4

if dv

J1

46.

f tJln2~d8 ~/'

csczcotzdz

dl


63.

r~/2

Jo

79. Y =

S(sinx)'/2cosxdx

' l %

1 1o 69. 1o

~/2

1S sin' 3x cos 3x dx

65. 67.

3 .

smxcosx dx

VI+3sin2 x

~/3

tan 8 dB V2""'-8

press

71. rmd the average value of I(x)

= nu; + b

y

answer. 83. Find dy/dx if Y = 84. Find dy/dx if Y = in your caleulation.

72. rmd the average value of

v'1+7. Ex-

fal v'1+7 dx in terms of F md give a reason for your

your caleulation.

over[-l,l]

b. over [-k, k] L

ICC&"

I

-,-- dt t +1

82. Suppose that F(x) is m antiderlvative of I(x) =

Average Valu.. L

, 1

80. Y =

Theory and Examples 81. Is it true that every function Y = I(x) that is differentiable on [a, b] is itself the derivative of some function on [a, b]? Give reaSODS for your answer.

~/2 ~/2

- -6, d l 3 +t

303

t

V1+I'dl. Explain the main steps in

t..% (1/(1 - 12»

dt. Explain the main steps

85. A new parking lot To meet Ibe demand for parking, your town has allocated the area shown here. As the town engineer, you have been asked by the town council to find out if lbe lot can be buih

= "\I'h aver [0, 3]

b. y= ~aver[O,a] 73. Let I be a function that is differentiable on [a, b]. In Cbapter 2 we defmed the average rate ofcbange of lover [a, b] to bc

I(b) - I(a) b-a

for $10,000. The cost to elearlbe land will be $0.10 a square foot, and lbe lot will cost $2.00 a square foot to pave. Can lbe job be done for $1O,000? Use a lower sum estimate to see. (Answers may vary slightly, depending on lbe estimate used.)

md thc instantaneous rate ofcbange of I at.< to be I'(x). In this chapter we defmed the average va1ue ofa function. For the new definition ofaverage to be consistent wilb !be old one, we sbould have

Oft

I(b) - I(a) b_ a = average value of I' on [a, b]. 51 ft

Is this lbe ease? Give reasons for your answer. 74. Is it true that !be average value of an integrable function over an interval oflengtb 2 is half the function's integral aver tbe interval? Give reasons for your answer.

D 75.

49.5 ft Vertical spacing = IS ft

Compute !be average value of!be temperature function

I(x) = 37 sin

U:s

64.4ft

(x - 101») + 2S

67.5ft

for a 36s-day year. This is one way to estimate tbe annual meBD air temperature in Fairbanks, Alaska. The National Weatber Service's official figure, a numerical average of lbe daily normal mean air temperatures for the year, is 2S.7"F, whicb is slightly higher 1han the average value of!be approximating function/(x).

D 76.

Specific beat of. gas Specific heat C. is the amount ofbeat required to raise the temperature of a given mass of gas with COIlstant volume by I"C, measured in units ofea1/deg-mole (calories per degree gr.un molecule). The specific beat of oxygen depends on its temperature T md satisfies !be formula

C.

= 8.27

+ 10-' (UT -

J.87T2 ).

rmd the average value of C. for 20" :5 T :5 67S·C and the ternperatun: at which it is attained.

54ft

42ft

86. Skydivers A md B are in a helicopter havering at 6400 It Skydiver A jumps and descends for 4 sec befure opening her parachute. The helicopter then climbs to 7000 It md hovers there. Forty-fIVe seconds after A leaves lbe aircraft, B jumps and descends for l3 sec before opening his parachute. Both skydivers descend at 16 f1/sec with parachutes open. Assume that the skydivers f.ill freely (no effective air ",sistance) before their parachutes open. •• At what altitude does A's parachute open? b. At what altitude does B's parachute open?

DHf~adnglnt~b

c. Which skydiver lands fJ",t?

In Exercises 77-80, find dy/dx.

77. y =

1% V2 + cos'ldt

78. Y =

J2r'" V2 + cos'tdt

304


Chapter

Additional and Advanced Exercises 9. Finding a curve Find the equation for the curve in the xy-plane if its slope at x is always that passes through the point (I,

Theory and Examples

1. a. If 1'7f(X) dx

b. If 1 'f (x) dx

~ 7,

= 4 andf(x)

[v7Wdx =

~ 17

does 1 'f (x) dx

V4 =

3: x

2, x = 0 about

a. the x-axis

b. the y-axis

e. the line x = 4

d. the Hoe y = 2

33. The region in the rITst quadrant bounded by the curve x = y - y' and the y-axis about

a. the x-axis

b. the Hoe y

~

I

-2

0

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 rmd 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 a. the x-axis

b. the y-axis

e. the line x = I

d. the Hoe y = I

3S. The region bounded by y ~

--;cf"""':....-+.----> x

o

b. the y-axis

x2

and y = x about a. the y-axis b. the Hoe x = I 37. The region in the rITSt quadrant that is bounded above by the curvey = I/x'/"ontheleftbythelinex = 1/16, and below by 36. The region bounded by y = 2x -

the line y

y

v-;; and y ~ x 2/8 about

a. the x-axis

= 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 drill bit ofradius 3.

Find the volume ofthe solid by

a. Find the volume ofthe bead.

a. the washer method. b. the shell method. 38. The region in the rITSt quadrant that is bounded above by the curve y = I/V-;;, on the left by the Hoe x = 1/4, and below by the line y

= 1 is revolved about the y-axis to generate a solid.

Find the volume ofthe solid by

a. the washer 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 ofthe removed portion ofthe sphere.

42. A Bondt cake, well known for having a ringed shape, is formed by revolving aroond the y-axis the region boonded 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 JITCcise 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 many 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 x = a to x = b. Ie order to derive an integral formula for the length ofthe curve, we assume that fhas a continuous derivative at every point of [a, bl. Such a function is called smooth, and its gmph is a smooth eurve because it does not have any breaks, corners, or cusps.

y

327

Arc Length

6.3

lJ.-,

-f---~-,----,----rlf-----'---'------"-----c---"----_X

FIGURE 6.22 The length of the polygonal path PoP,P2 ••• p. approximates the length ofthe curve y = f(x) from point A to pointB. y

AYk

I I

I

L1

r------I AXl

P,

I I I

We partition the interval [a, bj into n subintervals with a = Xo < XI < X2 < ... < x. = b. IfY, = f(x,), then the corresponding point P,(x" y,) 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 ofthe curve (Figure 6.22). If Axk = Xk - Xl-I and ~Yk = Yk - Yk-lo then a representative line segment in the path bas length (see Figure 6.23)

-ccI-cc'--------cL--> X

o

Xl_l

Xk

FIGURE 6.23 The arc P'_IP, ofthe curve y = f(x) is approximated by the straight line segment shown here, which has length L, ~ v'(ax,)2 + (ay,)2.

so the length of the curve is approximated by the sum (1)

We expect the approximation to improve as the partition of [a, bj becomes rmer. Now, by the Mean Value Theorem, there is a point c'" with Xk-I < Ck < X", such that ~Yk =

f'(et) Axk.

With this substitution for ~Y'" the sums in Equation (I) take the form

•

~ Lk

=

•

~ V(Axk)2 + (f'(Ck)~Xk)2

=

•

~ VI + [('(Ck»)' ~Xk'

(2)

Because VI + [('(x»)' is continuous on [a, bj, the limit of the Riemann sum on the righthand side ofEquation (2) exists as the norm of the partition goes to zero, giving lim

±

n_ OO k=1

Lk = lim

±

n_ OO k=l

VI

+

[('(et»)' Axk =

r

Ja

VI

+ [('(x»)' dx.

We derme the value ofthis limiting integral to be the length ofthe curve.

DEFINmON If j' is continuous on [a, bj, then the length (arc length) of the curve Y = f(x) from the point A = (a,j(a» to the pointB = (b,j(b» is the value ofthe integral

L

=

t

VI

+

[{'(x)]'dx =

t ~I (:y +

dx.

(3)

328

Chapter 6: Applications of Definite Integrals

y

EXAMPLE 1

Find the length of the curve (Figure 6.24)

Y =

Solution

4V2 3 x3 / 2 -

I ,O";x";l.

We use Equation (3) with a = 0, b = I, and

4V2 x3/ 2 _

y =

I

~ =

3

l,y

R:

0.89

dy = 4V2 • lx'/2 = 2V2x1/2 dx 3 2

FIGURE 6.24 The length of the curve is slightly larger than the length of the line segment joining points A and B (Example 1).

(:Y

Y

2

(2V2x1/

=

&.

=

The length of the curve over x = 0 to x = I is

L= =

[)1+

(:Ydx=

t· t(I +

&)3/2

[~dx

J: Ii '"

Eq. (3) with a=O,b=1

Let" = 1 + ~. integrate, and replace u by

2.17.

=

1

+ 8.l'I

.........L_x

0

Work and FLuid Forces

6.5

In everyday life, work means an activity that requires muscular or mental effort. In science, the term refers specifically to a force acting on a body (or object) and the body's subsequent displacement. This section shows how to calculate work. The applications run from compressing nrilroad car springs and emptying subterranean tanks to forcing electrons together and lifting satellites into orbit.

Work Done by a Constant Force When a body moves a distance d along a stnright line as a result of being acted on by a force of constant magnitude F in the direction of motion, we define the work W done by the force on the body with the formula

W=Fd

(Constant-force formula for work).

(1)

From Equation (I) we see that the unit OfWOlK in any system is the unit of force multiplied by the unit ofdistance. In SI units (SI stands for Systeme International, or International System), the unit offorce is a newton, the unit of distance is a meter, and the unit of work is a newton-meter (Nom). This combination appears so often it has a special name, the joule. In the British system, the unit of WOIk is the foot-pound, a unit frequently used by engineers.

I

Joules The joule, ahbreviated J and pronounced ''jewel,'' is named after the English physicist James Prescott Joule (181l>-1889). The defming equation is 1joule

~

(I newton)(I meter).

Insymbols,IJ~

IN·m.

EXAMPLE 1 Suppose you jack up the side of a 2000-lb car 1.25 ft to change a tire. The jack applies a constant vertical force of about 1000 Ib in lifting the side of the car (but because of the mechanical advantage of the jack, the force you apply to the jack itself is ouly about 30 Ib). The total work performed by the jack on the car is 1000 X 1.25 = 1250 ft-Ib. In SI units, the jack has applied a force of 4448 N through a distance of 0.381 m to do 4448 X 0.381 '" 1695 J of work. _

338


Work Done by a Variable Force Along a Line If the force you apply varies along the way, as it will if you are compressing a spring, the fonnula W = Fd has to be replaced by an integral fonnuia that takes the variation in F into account. Suppose that the force performing the work acts on an object moving along a straight line, which we take to be the x-axis. We assurne that the magnitode of the force is a continuous functionF of the object's positionx. We want to f'md the work done over the interval from x = a to x = b. We partition [a, b] in the usual way and choose an arbitrary point Cl in each subinterval [Xl- h xii. If the subinterval is short enough, the continuous function F will not vary much from Xl-I to Xl. The amount of work done across the interval will be about F(cl) times the distance the same as it wouid be if F were constant and we couid apply Equation (I). The total work done from a to b is therefore approximated by the Riemannsurn

axh

n

Work

L F(Cl) axl. 1-1

RJ

We expect the approximation to improve as the nonn of the partition goes to zero, so we define the work done by the force from a to b to be the integral of F from a to b: n

n~ ~ F(cl)

axl =

lb

a F(x) dx.

DEFINITION The work done by a variable force F(x) in the direction of motion along the x-axis from x = a to x = b is W =

l

b

F(x) dx.

(2)

The units of the integral arejouies ifF is in newtons and x is in meters, and foot-pounds if F is in pounds and x is in feet. So the work done by a force of F(x) = l/x 2 newtons in moving an object along the x-axis from x = I m to x = 10 m is

W=

1'0 I 1

x2

1]'01 =

dx = - X

I

- 10

+

I = 0.9 J.

Hooke's Law for Springs: F = Ioc F

Hooke's Law says that the force required to hold a stretched or compressed spring x units from its natoral (unstressed) length is proportional to x. In symbols, F= kx.

j 4f--.3,-----::+--"""71:-+. (ft) 3

33. Rectangular plate In a pool filled with water to a depth of 10ft, calculate the fluid force on one side of a 3 ft by 4 ft rectangular plate ifthe plate rests vertically at the bottom of the pool L

on its 4-ft edge

b. on its 3-ft edge.

34. Semldn:ulor plate Calculate the fluid force on one side of a semicin:ular plate of nuIius 5 ft that rests vertically on its diameter at the bottom ofa pool filled with water to a depth of6 it. y Surface of water

6

5

Work and Fluid Forces

345

38. Semicircular plate A semieircular plate 2 ft in diameter sticks straight down into freshwater with the diameter along the surface. Find the force exerted by the water on one side of the plate.

39. 1iIled plate Calculate the fluid force 00 one side ofa 5 ft by 5 ft square plate if the plate is at the bottom ofa pool ftlled with water to a depth of g ft and L 1yina flat on its 5 ft by 5 ft face. b. restiog vertically on a 5-ft edge. c. restiog on a 5-ft edge and tilted at 45' to the bottom ofthe pool. 40. lilted plate Calculate the fluid force 00 one side of a righttriangular plate with edges 3 ft, 4 ft, and 5 ft if the plate sits at the

bottom of a pool ftlled with water to a depth of 6 ft on its 3-ft edge and tilted at 60' to the bottom of the pool. 41. The cubical metal tank shown here has a parabolic gate held in place by bolts and designed to withs1lmd a fluid force of 160 lb without rupturing. The liquid you plan to store has a weightdensity of 50 Ib/tt' . a. What is the fluid force on the gate when the liquid is 2 ft deep? b. What is the maximwn height to which the container can be filled without exceeding the gate's design limitation? y (ft)

_ _L-_ _--"--_ _---l_ _ •

(-1, 1)

(I, I)

35. Triangular plate The isosceles triangular plate shown here is submerged vertically I ft below the surface of a freshwater lake. L

----:-----'''':0:+-'''------:--. (ft)

Find the fluid force against one face of the plate.

b. What wculd be the fluid force on one side of the plate if the water were seawater instead of freshwater? Surface level

Parabolic gate

EDlarged view of

plUBbolic gate

42. The end plates of the trough shown here were designed to withs1lmd a fluid force of 6667 lb. How many cubic fi:et of water can the tank hold without exceeding this limitation? Round down to the nearest cubic fool y(ft) (-4, 10) (4, 10)

36. Rotated triangular plate The plate in Exercise 35 is revolved 180' about line AB so that part of the plate sticks out of the lake, as shown here. What force does the water exert on one face of the plate now?

---cc'of----+. (ft) End view of trough

43. A vertical rectangular plate a uuits long by b uuits wide is sui>metged in a fluid of weight-density w with its long edges parallel to the fluid's surface. Find the average value of the pressure along the vertical dimension ofthe plate. Explain your answer. 37. New EngbmdAquarium The viewing portion of the rectsngular glass window in a typical fIsh tank at the New England Aquarium in Boston is 63 in. wide and runs from 0.5 in. below the water's surface to 33.5 in. below the surface. Find the fluid force against this portion of the window. The \Wight-density of seawater is 64 Ib/tt' . (In case you were wcndering, the glass is 3/4 in. thick and the tank wal1s extend 4 in. above the water to keep the fISh fromjumpina oul)

44. (Continuation oJfuerciae 43.) Show that the force exerted by the fluid 00 one side of the plate is the average value of the pressure (found in Exercise 43) times the area of the plate.

45. Water pours into the tank shown here at the rate of 4 tt'/min. The tank's cross-sections are 4-ft.:I-1o::---:--- x

o

(X, y)

center ofmass (c.m.): length: width: area: mass:

= (x,! [((x)

+ g(x)])

j(x) - g(x)

ax

dA = [((x) - g(x)] ax dm = Il dA = Il[{(x) - g(x)]

ax.

b

The moment ofthe plate about the y-axis is FIGURE 6.52 Modeling the plate bounded by _ curves with vertical strips. The strip

c.m. is halfway, so 'ji

=

t

[f(x)

J 1 b

My =

+ g(x)].

xdm =

xll[{(x) - g(x)]

ax,

+ g(x)]'Il[{(x)

- g(x)]

and the moment about the x-axis is

Jy 1t b

Mx =

dm =

b

=

1 a

2Il [f(x)

[((x)

- g2(x)]

ax

ax.

These moments give the formulas

11

b

X= M

a

Ilx [((x) - g(x)]

ax

(6)

(7)

EXAMPLE 3 Find the center ofmass for the thin plate bounded by the curves g(x) and j(x) = V;,O :5

X

:5

I, (Figure 6.53) using Equations (6) and (7) with the density

function Il(x) = x 2 •

Solutton We first compute the mass of the plate, where dm = Il[{(x) - g(x)] ax: M =

t 10

(V; _~)2 ax

x2

= x/2

=

t 10

3 (:x:'/2 _ x ) 2

ax

=

[2.7 x 7/2 _ Ix-]! =~. 8 0 56

352

Chapter 6: Applications of Definite Integrals Then from Equations (6) and (7) we get

x= =

y 1

56 9

[x2.{v'~ -~) dx

(x7/2 _~4)

56 [ 9

= 56

[2. x9/2 _

9 9

~ xs]l 10

0

dx 308

405'

and g(x) ~ ~

--o-\------------:--x

o

1

252

FIGURE 6.53

405'

The region in Example 3.

•

The center of mass is shown in Figure 6.53.

Centroids When the density function is constant, it cancels out of the numerator and denominator of the formulas for x and y. Thus, when the density is constant, the location of the center of mass is a feature of the geometry of the object and not of the material from which it is made. In such cases, engineers may call the center of mass the centroid of the shape, as in "Find the centroid of a triangle or a solid cone." To do so, just set [; equal to I and proceed to fmd x and y as before, by dividing moments by masses. y A typical small segment of wire has

tim

~

a tis ~ aad8.

a

-a (a)

EXAMPLE 4 Find the center of mass (centroid) of a thin wire of constant density [; shaped like a semicircle of radius a.

Va

length:

y

mass: distance ofc.m. to x-axis:

a c.m.

-a

ds = a dO dm = [; tis = [;a dO y = a sin O.

Mass per unit length times 1ength

Hence,

(0. ~a)

0

x

2 2 (Figure 6.54). The Solution We model the wire with the semicircle y = distribution of mass is symmetric about the y-axis, so x = O. To find y, we imagine the wire divided into short subarc segments. If (x, y) is the center of mass of a subarc and 0 is the angle between the x-axis and the radial line joining the origin to (x, y), then y = a sin 0 is a function of the angle 0 measured in radians (see Figure 6.54a). The length ds of the subarc containing (x, y) subtends an angle of dO radians, so tis = a dO. Thus a typical subarc segment has these relevant data for calculating y:

a

(b)

FIGURE 6.54 The semicircular wire in Example 4. (a) The dimensions and variables used in rmding the center of mass. (b) The center of mass does not lie on the wire.

_ fydm fo"a sinO'[;a dO y= = f dm fo" [;a dO

[;a2[ -cos 0]; [;a7f

2 = 7f

a.

The center of mass lies on the axis of symmetry at the point (0, 2a/7f), about two-thirds of the way up from the origin (Figure 6.54b). Notice how [; cancels in the equation for y, so we could have set [; = I everywhere and obtained the same value for y. • In Example 4 we found the center of mass of a thin wire lying along the graph of a differentiable function in the xy-plane. In Chapter 16 we willleam how to find the center of mass of wires lying along more general smooth curves in the plane (or in space).

6.6

Moments and Centers of Mass

353

Auid Forces and Centroids

Surface level of fluid 11 = centroid depth

Ifwe know the location ofthe centroid of a submerged flat vertical plate (Figure 6.55), We can take a shortcut to find the force against one side of the plate. From Equation (7) in Section 6.5,

l l

b

Plate centroid

F =

= W

The force against one side ofthe plate is w·1i . plate area. FIGURE 6.55

w X (strip depth) X L(y) dy b

(strip depth) X L(y) dy

= w X (moment about surface level line of region occupied by plate) = w X (depth of plate's centroid) X (area of plate).

Fluid Forces and Centroids The force ofa fluid of weight-density w against one side ofa submerged flat vertical plate is the product ofw, the distance Ii from the plate's centroid to the fluid surface, and the plate's area: F=

whA.

(8)

EXAMPLE 5 A flat isosceles triangular plate with base 6 ft and height 3 ft is submerged vertically, base up with its vertex at the origin, so that the base is 2 ft below the surface ofa swimming pool. (fbis is Example 6, Section 6.5.) Use Equation (8) to find the force exerted by the water against one side ofthe plate. Solutton The centroid of the triangle (Figure 6.43) lies on the y-axis, one-third of the way from the base to the vertex, so Ii = 3 (where y = 2) since the pool's surface is y = 5. The triangle's area is A = t(base)(height) = t(6)(3) = 9. Hence,

•

F = whA = (62.4)(3)(9) = 1684.8 lb. y

d

The Theorems of Pappus -----~-=---~

In the third century, an Alexandrian Greek named Pappas discovered two formulas that relate centroids to surfaces and solids of revolution. The formulas provide shortcuts to a number of otherwise lengthy calculations.

L(y)

c

- - -- -

-~-"-~---~

Centroid

-O:+--------·~x FIGURE 6.56 The region R is to be revolved (once) about the x-axis to generate a solid. A 1700-year-old theorem says that the solid's volume can be calculated by multiplying the region's area by the distance traveled by its centroid during the revolution.

THEOREM 1

Pappus's Theorem for Volumes

If a plane region is revolved onCe about a line in the plane that does not cut through the region's interior, then the volume of the solid it generates is equal to the region's area times the distance traveled by the region's centroid during the revolution. If p is the distance from the axis of revolution to the centroid, then

v=

27TpA.

(9)

Proof We draw the axis of revolution as the x-axis with the region R in the first quadrant (Figure 6.56). We let L(y) denote the length of the cross-section of R perpendicular to the y-axis at y. We assume L(y) to be continuous.

354

Chapter 6: AppLications of Definite Integrals By the method of cylindrical shells, the volume of the solid generated by revolving the region about the x-axis is

v

~

l' 2,,(sbell

nldius)(shell height) dy

~ 2,,1' y L(y) dy,

(10)

The y-coordinate ofR's centroid is

,

_l'

Distance from Will of revoluUOII to centroid

I

'ji 0 in the equation y = x'/(4p), the y-coonlinate of the centroid of the parabolic segment abown here is ji = (3/5}a. y

Use the result in Exercise 17 to fmel the centroids of the triangles wbose vertices appear in Exercises 111-22. Assume a, b > D. 18. (-1,0),(1,0),(0,3) 19. (0,0), (I, 0), (0, I) ZO. (0,0), (a, 0), (0, a) 21. (0,0), (a, 0), (0, b) 11. (0,0), (a, 0), (a/2, b)

Tlrin Wires 23. Constant density Find the moment about Ibe x-axis of a wire of constant density that lies along the curve y = V; from x = 0 tox = 2. 24. Constant density Find the moment about Ibe x-axis ofa wire of constant density that lies along the corve y = x' from x = 0 tox = 1.

Th. Theorems of Pappus 33. The square region with vertices (0, 2), (2, D), (4, 2), and (2, 4) is revolved about the x-axis to generate a solid. Fmel the volume and surface area of the solid. 34. Use a theorem of Pappus to fmd the volume generated by revolving about Ibe line x = 5 Ibe triangular region bounded by the coordinate axes and the line 2x + y = 6 (see Exercise 17). 35. Find lbe volume of lbe torus generated by revolving the circle (x - 2)' + y' = I about lbe y-axis.

Chapter 6 Practice Exercises 36. Use the theorems of Pappus to fmd the lateral surface area and the volume ofa right-circular cone. 37. Use Pappus's Theorem for sorface area and the fact that the surface area ofa sphere ofradius a is 4'1Ta 2 to fmd the centroid ofthe 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 point (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 point (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

ofthe given triangle. Use the fact that the volume of a cone ofradius r and height h is V ~ ~ 'lTr 2h. 43.

y (0, b)

=;;;t----~~x (0,0) (a, 0)

44.

y

(a, c)

41. The region ofExercise 40 is revolved about the line y = x - a to generate a solid. Find the volume ofthe solid.

Chapter


1. How do 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 calculationa by these methods. 3. Describe the method of cylindrical sbells. Give an example. 4. How do you fmd the length of the grapb of a smooth function over a closed interval? 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

~

6. How do you defme and calculate the work done by a variable force directed along a portion of the x-axis? How do you calculate the work it takes to pump a liquid from a tank? Give examples. 7. How do you calculate the force exerted by a liquid against a portion ofa flat vertical wall? Give an example. 8. What is a center ofmass? a centroid? 9. How do you locate the center of mass ofa thin flat plate ofmatrial? Give an example. 10. How do 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 ofthe 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-sectiona perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y ~ x 2 to the parabola y ~ ~.

3. The solid lies between planes perpendicular to the.HXis at x = 'IT/4 and x ~ 5'IT/4. The cross-sectiona between these planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y=2sinx.

2. The base of the solid is the region in the fust quadrant between the line y = x and the parabola y = 2 ~. The cross-sections of

4. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 6. The cross-sectiona between these planes

358

Chapter 6: Applications of Definite Integrals y

are squares whose bases run from the x-axis up to the curve x'f2 + y'f2 =

v'6.

y ~ll;t-----:+--+'ll~X

6

-2

2

Lengths of Curves Find the leogths of the curves in Exercises 17-20.

17. y

5. The solid lies betweeo planes perpeodicular to the x-axis at x = D and x ~ 4. The cross-sections of the solid perpeodicular to the x-axis between these planes are circular disks whose diameters run from the curve x' ~ 4y to the curve y' ~ 4x.

= x 'f2

- (1/3)x'f2,

I '" x '" 4

18. x = y'/3, I '" y '" 8 19. y = (5/12)x 6/' - (5/8)x 4/', 20. x ~ (y'/12)

+ (l/y),

I '" y '" 2

Areas of Surfaces of Revolution In Exercises 21-24, fmd the areas ofthe surfaces geoerated hy revolv-

6. The base of the solid is the region bounded by the parabola y2 = 4x and the line x = 1 in the xy-plane. Each cross-section perpendicular to the x-axis is an equilateral triangle with one edge in the plane. (The triangles all lie on the same side of the plane.)

ing the curves about the given axes.

7. Find the volume of the solid geoerated by revolving the region bounded by the x-axis, the curve y = 3x4, and the lines x = 1 and x = -I about (a) the x-axis; (b) the y-axis; (c) the line x = I; (d) the liney = 3.

24. x =

8. Find the volume of the solid geoerated by revolving the "triangular" region bounded by the curve y = 4/x' and the lines x = I and y ~ 1/2 about (a) the x-axis; (b) the y-axis; (c) the line x = 2; (d) the liney = 4. 9. Find the volume of the solid geoerated by revolving the region bounded on the left by the parabola x ~ y' + I and on the right by the line x = 5 about (a) thex-axis; (b)they-axis; (c) the line x = 5. 10. Find the volume of the solid geoerated by revolving the region bouoded by the parabola y' = 4x and the line y = x about (a) the x-axis; (b) they-axis; (c) the line x ~ 4; (d) the liney ~ 4.

11. Find the volume of the solid geoerated by revolving the "triangular" region bouoded by the x-axis, the line x = 'If/3, and the curve y ~ tan x in the fust quadrant about the x-axis. 12. Find the volume of the solid geoerated by revolving the region bounded by the curve y = sin x and the lines x = 0, x = '1T', and y = 2 about the line y = 2. 13. Find the volume of the solid geoerated by revolving the region be2x about (a) the x-axis; tween the x-axis and the curve y ~ (b) the liney = -I; (c) the line x = 2; (d) the line y = 2.

x' -

14. Find the volume of the solid geoerated by revolving about the x-axis the region bouoded by y = 2tanx,y = O,x = -'If/4,and x = 'If/4. (The region lies in the frrst and third quadrants and resembles a skewed bowtie.)

V3

ft is 15. Volume of a solid sphere hole A rouod hole of radius bored throngb the cooter of a solid sphere of a radius 2 ft. Find the volume of material removed from the sphere.

_haD

16. Voiome of a The prome of a fuothall resembles the ellipse shown here. Find the football's volume to the nearest cubic inch.

I '" x '" 32

21. y ~~, 0 '" x '" 3; x-axis 22. y = x 3/3, 0:5 x:5 1; x-axis 23. x ~

v'4y -

YY,

y',

I '" y '" 2;

2 '" y '" 6;

y-axis

y-axis

Work 25. Lifting equipmeot A rock climber is about to baul up 100 N (about 22.5 Ib) of equipmeot that bas beeo banging beoeath her on 40 m of rope that weighs 0.8 newtoo per meter. How much work will it take? (Hint: Solve for the rope and equipmeot separately, theo add.) 26. Leaky tank truck You drove an 800-gal tank truck of water from the base of Mt. Washington to the summit and discovered on arrival that the tank was only balffull. You started with a full tank, clin3bed at a steady rate, and accomplisbed the 475D-ft elevation cbange in 50 min. Assuming that the water leaked out at a steady rate, how much work was speat in carrying water to the top? Do not couot the work done in getting yourself and the truck there. Water weighs 8 Ib/U.S. gal. 27. Stretching a spring If a force of20 Ib is required to hold a spring I ft beyond its unstressed leogth, how much work does it take to stretch the spring this far? An additional lOot? 28. Garage door spring A force of 200 N will stretch a garage door spring 0.8 m beyond its uostressed length. How far will a 300-N force stretch the spring? How much work does it take to stretch the spriog this far from its unstressed length? 29. Pumping a reservoir A reservoir sbaped like a right-circular cone, point down, 20 ft across the top and 8 ft deep, is full of water. How much work does it take to pump the water to a level 6 ft above the top? 30. Pumping a reservoir (Continuation of Exercise 29.) The reservoir is filled to a depth of5 ft, and the water is to be pumped to the same level as the top. How much work does it take? 31. Pumping a couical tank A right-circular conical tank, point down, with top radius 5 ft and height 10 ft is filled with a liquid whose weight-deosity is 60 Ib/ft' . How much work does it take to pump the liquid to a point 2 ft above the tank? If the pump is

Chapter 6 Additional and Advanced Exercises driven by a motor rated at 275 fl:-1b/sec (1/2 hp), how long will it take to empty the tank?

y

32. Pumping a cylindrical tank A storage tank is a right-circular cylinder 20 ft long and 8 ft in diameter with its axis horizonta1. If the tank is half full of olive oil weighing 57 lb/ft3 , find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank. Centers of Mass and Centroids 33. Find the centroid of a thin, flat plate covering the region enclosed by the parabolas y = 2x 2 and y = 3 - x 2 •

34. Find the centroid ofa thin, flat plate covering the region enclosed by the x-axis, the lines x = 2 and x = -2, and the parabola y = x 2•

2 --'----~.:::::....------'-~x

FLuid Force 39. Trough of water The vertical triangular plate shown here is the end plate of a trough full of water (w = 62.4). What is the fluid force against the plate?

Chapter

4

UNITS IN FEET

40. Trough of maple syrup The vertical trapezoidal plate shown here is the end plate of a trough full of maple syrup weighing 75lb/ft3 . What is the force exerted by the syrup against the end plate ofthe trough when the syrup is 10 in. deep? y y=x-2// ..,----------"t--/ ----->-"-2--0+---/-}/2"--- x

36. Find the centroid of a thin, flat plate covering the region enclosed by the parabola y2 = x and the line x = 2y.

38. a. Find the center ofmass of a thin plate of constant density covering the region between the curve y = 3/x 3/ 2 and the x-axis fromx = 1 tox = 9. b. Find the plate's center ofmass if, instead ofbeing constant, the density is 8(x) = x. (Use vertical strips.)

~~~~O

-4

35. Find the centroid of a thin, flat plate covering the "triangular" region in the first quadrant bounded by the y-axis, the parabola y = x 2/4, and the line y = 4.

37. Find the center of mass of a thin, flat plate covering the region enclosed by the parabola y2 = x and the line x = 2y if the density function is 8(y) = 1 + y. (Use horizontal strips.)

359

/ / / /

UNITS IN FEET

41. Force on a parabolic gate A flat vertical gate in the face of a dam is shaped like the parabolic region between the curve y = 4x 2 and the line y = 4, with measurements in feet. The top of the gate lies 5 ft below the surface of the water. Find the force exerted by the water against the gate (w = 62.4).

D 42.

You plan to store mercury (w = 849lb/nJ) in a vertical rectangular tank with a 1 ft square base side whose interior side wall can withstand a total fluid force of 40,000 lb. About how many cubic feet of mercury can you store in the tank at anyone time?


VoLume and Length 1. A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y = f(x), the x-axis, and the fixed line x = a and the variable line x = b, b > a. Its volume, for all b, is b 2 - ab. Find f(x).

2. A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y = f(x) , the x-axis, and the lines x = 0 and x = a. Its volume, for all a > 0, is a 2 + a. Find f(x).

5. Find the volume of the solid formed by revolving the region bounded by the graphs of y = x and y = x 2 about the line y = x. 6. Consider a right-circular cylinder of diameter 1. Form a wedge by making one slice parallel to the base of the cylinder completely through the cylinder, and another slice at an angle of 45° to the first slice and intersecting the first slice at the opposite edge of the cylinder (see accompanying diagram). Find the volume of the wedge.

3. Suppose that the increasing function f(x) is smooth for x ~ 0 and that f( 0) = a. Let s(x) denote the length of the graph of f from (0, a) to (x, f(x», x > O. Find f(x) if s(x) = Cx for some constant C. What are the allowable values for C? 4. a. ShowthatforO

b. Generalize the result in part (a).

/

/

I

I

Tr/2,

a

/

/

/

/

Va2 + sin2 a.

360


Surface Area 7. At points on the curve y = 2 ~, line segments of length h = y are drawn perpendicular to the xy-plane. (See accompanying figure.) Find the area of the surface formed by these perpendiculars from (0, 0) to (3, 2\13). y

10. Work and kinetic energy Suppose a 1.6-oz golf ball is placed on a vertical spring with force constant k = 2 lb/ in. The spring is compressed 6 in. and released. About how high does the ball go (measured from the spring's rest position)? Centers of Mass 11. Find the centroid of the region bounded below by the x-axis and above by the curve y = 1 - x n , n an even positive integer. What is the limiting position of the centroid as n ~ oo?

'7

/ /

/

2\13

2~

(3,2\13) ~

"

y=2~

x

8. At points on a circle of radius a, line segments are drawn perpendicular to the plane of the circle, the perpendicular at each point P being of length ks, where s is the length of the arc of the circle measured counterclockwise from (a, 0) to P and k is a positive constant, as shown here. Find the area of the surface formed by the perpendiculars along the arc beginning at (a, 0) and extending once around the circle.

12. If you haul a telephone pole on a two-wheeled carriage behind a truck, you want the wheels to be 3 ft or so behind the pole's center of mass to provide an adequate "tongue" weight. The 40-ft wooden telephone poles used by Verizon have a 27-in. circumference at the top and a 43.5-in. circumference at the base. About how far from the top is the center of mass? 13. Suppose that a thin metal plate of area A and constant density S occupies a region R in the xy-plane, and let My be the plate's moment about the y-axis. Show that the plate's moment about the line x = b is a. My - bSA if the plate lies to the right of the line, and b. bSA - My if the plate lies to the left of the line.

14. Find the center ofmass ofa thin plate covering the region bounded by the curve y 2 = 4ax and the line x = a, a = positive constant, if the density at (x,y) is directly proportional to (a) x, (b) IYI. 15. a. Find the centroid of the region in the first quadrant bounded by two concentric circles and the coordinate axes, if the circles have radii a and b, 0 < a < b, and their centers are at the origin. b. Find the limits of the coordinates of the centroid as a approaches b and discuss the meaning of the result.

16. A triangular comer is cut from a square 1 ft on a side. The area of the triangle removed is 36 in2 . If the centroid of the remaining region is 7 in. from one side of the original square, how far is it from the remaining sides?

l a x

Work 9. A particle of mass m starts from rest at time [ = 0 and is moved along the x-axis with constant acceleration a from x = 0 to x = h against a variable force of magnitude F([) = [2. Find the work done.

Chapter •

FLuid Force 17. A triangular plate ABC is submerged in water with its plane vertical. The side AB, 4 ft long, is 6 ft below the surface of the water, while the vertex C is 2 ft below the surface. Find the force exerted by the water on one side of the plate.

18. A vertical rectangular plate is submerged in a fluid with its top edge parallel to the fluid's surface. Show that the force exerted by the fluid on one side of the plate equals the average value of the pressure up and down the plate times the area of the plate.


Mathematica/Maple Modules: Using Riemann Sums to Estimate Areas, Volumes, and Lengths of Curves Visualize and approximate areas and volumes in Part I and Part II: Volumes of Revolution; and Part III: Lengths of Curves. Modeling a Bungee Cord Jump Collect data (or use data previously collected) to build and refine a model for the force exerted by a jumper's bungee cord. Use the work-energy theorem to compute the distance fallen for a given jumper and a given length of bungee cord.

7 TRANSCENDENTAL FUNCTIONS OVERVIEW Functions can be classified into two broad complementary groups called algebraic functions and transcendental functions (see Section 1.1). Except for the trigonometric functions, our studies so far have concentrated on the algebraic functions. In this chapter we investigate the calculus of important transcendental functions, including the logarithmic, exponential, inverse trigonometric, and hyperbolic functions. They occur frequently in many mathematical settings and scientific applications.

7.1

Inverse Functions and Their Derivatives A function that undoes, or inverts, the effect of a function f is called the inverse of f. Many common functions, though not all, are paired with an inverse. Important inverse functions often show up in applications. Inverse functions also playa key role in the development and properties of the logarithmic and exponential functions, studied in Section 7.3.

One-to-One Functions A function is a rule that assigns a value from its range to each element in its domain. Some functions assign the same range value to more than one element in the domain. The function f(x) = x 2 assigns the same value, 1, to both ofthe numbers -1 and + 1; the sines of 7T/3 and 27T/3 are both 0/2. Other functions assume each value in their range no more than once. The square roots and cubes of different numbers are always different. A function that has distinct values at distinct elements in its domain is called one-to-one. These functions take on anyone value in their range exactly once.

DEFINITION A function f(x) is one-to-one on a domain D if f(Xl) #- f(X2) whenever Xl #- X2 inDo EXAMPLE 1 Some functions are one-to-one on their entire natural domain. Other functions are not one-to-one on their entire domain, but by restricting the function to a smaller domain we can create a function that is one-to-one. The original and restricted functions are not the same functions, because they have different domains. However, the two functions have the same values on the smaller domain, so the original function is an extension ofthe restricted function from its smaller domain to the larger domain. (a) f(x) = \IX is one-to-one on any domain of nonnegative numbers because vx;: whenever Xl #- X2·

v.x;- #-

(b) g(x) = sin x is not one-to-one on the interval [0,7T] because sin (7T/6) = sin(57T/6). In fact, for each element Xl in the subinterval [0, 7T/2) there is a corresponding element X2 in the subinterval (7T/2, 7T] satisfying sinxl = Sinx2' so distinct elements in

361

362

Chapter 7: Transcendental Functions y

the domain are assigned to the same value in the range. The sine function is one-toone on [0, ,,/2], however, because it is an increasing function on [0, ,,/2] giving distinct outputs for distinct inputs. • The graph of a one-to-one function y = f(x) can intersect a given horizontal line at most once. If the function intersects the line more than once, it assumes the same y-value for at least two different x-values and is therefore not one-to-one (Fignre 7.1).

(a) One-to-one: Graph meets each horizontal line at most once.

The Horizontal Une Test for One-to-One Fnnctions A function y = fix) is one-to-one if and only if its graph intersects each horizontalline at most once.

y

Same y-value

0.5 --!--'~'---!--

-1

0

I

\

,

5:;

, Inverse Functions

y = sinx (b) Not one-to-one: Graph meets one or more horizonta1lines more than once.

=,'

=

FIGURE 7.1 (e)y andy v'Xare one-to-one on their domains (-00,00) and [O,oo).(b)y =x2 andy = sin x are not one-la-one on their domains (-00,00).

Since each output of a one-to-one function comes from just one input, the effect of the function can be inverted to send an output back to the input from which it came.

DEFINmON Suppose that f is a one-to-one function on a domain D with rnnge R. The inverse function 1 is defmed by

r

rl(b) = The domain of

r

0

if f(o) = b.

l is R and the rnnge of

r

l

is D.

r

r

l for the inverse of f is read "f inverse." The "-I" in l is not an The symbol exponent; rl(x) does not mean 1/fix). Notice that the domains and ranges of f and l are interchanged.

EXAMPLE 2

r

Suppose a one-to-one function y = fix) is given by a table of values

x fix)

r

A table for the values of x = l(y) can then be obtained by simply interchanging the values in the columns of the table for f:

y

•

rl(y)

r

If we apply f to send an input x to the outputf(x) and follow by applying l to fix) we get right back to x, just where we started Similarly, if we take some number y in the range of f, apply l to it, and then apply f to the resulting value rl(y), we get back the value y with which we began. Composing a function and its inverse has the same effect as

r

doing nothing.

U-l U

0

0

f)(x) = x,

rl)(y) = y,

for all x in the domain of f for all y in the domain of f- l (or range of f)

Only a one-to-one function can have an inverse. The reason is that if f(xI) = y and f(X2) = y for two distinct inputs Xl and X2, then there is no way to assign a value to rl(y) that satisfies both rlU(Xl)) = Xl and r l U(X2)) = X2.

7.1

Inverse Functions and Their Derivatives

363

A function that is increasing on an interval so it satisfies the inequality f(X2) > 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 i' 0 and flO) = 0, defined on (- 00, 00) and passing the horizontal line test.

Finding Inverses The graphs ofa 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

°

x DOMAINOF

f

(a) To :lind the value off at x, we start at X, go up to the curve, and then over to the y-axis.

°

X RANGE OF

/-1

r'

(b) The graph of ia Ibe graph off, hot 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 ofj-l is the range off, The range ofj-l is the domain off.

y

y ~r'(x)

--L---'O";/01"-_....,I-_ _---->, y /

----onl--""""l~---~x

/ / / / / / /

(c) To draw !he graph ofr' in Ibe

(d) Then we interchange the letters x and y. 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.

FIGURE 7.2 Detennining the graph ofy = r'(x) from the graph ofy = j(x). The graph of is obtained 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

r'

to the y-axis to get the value of (x). Figure 7.2 indicates the relationship between the graphs of 1 and 1 • The graphs are interchanged by reflection through the line y = x. 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 ofpassing from 1 to 1-1 can be summarized as a !w-step procedure.

r

1.

2. y

Solve the equation y = I(x) for x. This gives a formula x = r1(y) where x is expressed as a function ofy. Interchange x and y, obtaining a formula y = r1(x) where 1 is expressed in the conventional format with x as the independent variable and y as the dependent variable.

r

y~2x-2

EXAMPLE 3

Find the inverse of y

~ x + I, expressed as a function ofx.

=

Solution 1.

Solve for x in terms ofy:

y =

2y = x

~joo"C~jL-f-----~x

-2

/

/

x

/

// -i /

FIGURE 7.3 Graphingf(x) = (1/2)x + I and r'(x) = 2x - 2 together shows the

2I x +

2.

Interchange x and y:

I

+2

= 2y -

2.

y = 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:

graphs' symmetry with respect to the line

r1(f(x))

=

2(~X + I) - 2 =

I(r (x)) '

=

~(2x -

y = x (Example 3).

2)

+

+2- 2

x

I = x - I

=

x

+

I = x.

;;"

0, expressed as a function

•

y

EXAMPLE 4 ofx. Solution

Find the inverse of the function y

=

x 2, X

We flISt solve for x in terms of y: y = x2

l:tl ~""'-------~x

another (Example 4).

because x

2:

0

We then interchange x andy, obtaining

~

FIGURE 7,4 The functions y ~ Vi and y = xl, X ~ 0, are inverses of one

= x

y= vX. = x 2,x;;"

The inverse ofthe functiony O,isthefunctiony = vX (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 function y = x 2, with no domain restrictions, is not one-to-one (Figure 7.lb) and therefore has _ oo~e.

Derivatives of Inverses of Differentiable Functions If we calculate the derivatives of I(x) = (1/2)x + I and its inverse r1(x) from Example 3, we see that

~/(x)=~(~x+I)=~ ~rl(x)

=

~(2x

- 2)

=

2.

= 2x -

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 # 0, the reflected line has slope I/m.

The slopes are reciprocal: (r')'(b) ~ _I_or (f-l)'(b) ~ I 1'(0) f'(f '(b»

FIGURE 7.5 The graphs of inverse functions hsve reciprocal slopes at corresponding points.

r

The reciprocal relationship between the slopes of I and 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = I(x) at the point (a, I(a» is j'(a) and j'(a) # 0, then the slope of y = r'(x) at the point (f(a), a) is the reciprocal 1/ j'(a) (Figure 7.5). Ifwe set b = I(a), then

(f-I)'(b)

=

j'~a)

=

j'(f \b»'

r'

If y = I(x) has a horizontal tangent line at (a, I(a» then the inverse function has a is not differentiable vertical tangent line at (f(a), a), and this wmite slope implies that at I(a). Theorem I gives the conditions under which is differentiable in its domain (which is the same as the range of I).

r'

THEOREM I-The Derivative Rule for Inverses and j'(x) exists and is never zero on I, then its domain (the range of f). The value of

r'

If I has an interval I as domain

r is differentiable at every point in (r ')' at a point b in the domain of r ' isthereciprocalofthevalueofj' at the point a = r'(b): 1

I (/ -1)'(b) _ - j'(f '(b»

(1)

or

drl I dx

x~b

-

_1_

d/l dx

x-r'(b)

Theorem I makes two assertions. The tITSt of these has to do with the conditions under which I-I is differentiable; the second assertion is a formula for the derivative of

366


I -1 when it exists. While we omit the proof of the first assertion, the second one is proved in the following way:

l(r 1(x»

=

x

Inverse function relationship

:x l(r (x»

=

I

Differentiating both sides

=

I

Chain Rule

1

/'(r 1(x»' ~ r dx d

dx

1

(x)

1-1(

I

) _

x-

/'(f

Solving for the derivative

'(x»"

Thefunction/(x) = x 2 ,x;" oand its inverse r'(x) = v'Xhavederivatives/,(x) = 2xand(r 1)'(x) = 1/(2v'X). Let's verify that Theorem I gives the same fannula for the derivative of 1(x):

y

EXAMPLE 5

r

4

(r')'(x)

/'(f \x»

=

3

rex)

I

2(f l(x»

2

~

2x with x replaced

by r'(xl

I

I/J

2(v'X)"

--ccfL----!--~___!---"-----~ x

o

t

2

3

4

FIGURE 7.6 The derivative of r'(x) ~ Vi at the point (4, 2) is the

Theorem I gives a derivative that agrees with the known derivative of the square root function. Let's examine Theorem I at a specific point. We pick x = 2 (the number a) and = 4 (the value b). Theorem I says that the derivative of I at 2, /,(2) = 4, and the derivative ofr l at 1(2), (f-1 )'(4), are reciprocals. It states that

1(2)

reciprocal ofthe derivative of I(x) = x 2 at (2, 4) (Example 5).

(/

- 1)'(4)

_

I

_

I

_ I

- /'(f 1(4» - /,(2) - 2x

I.-2 -_4'I

•

See Figure 7.6.

y

6

y=x 3 -

(2, 6)

2 Slope 3:2

= 12

x=2

I x=/(2) =

I

I

dl dx

I

I 12

Eq. (I)

x=2

•

7.1

367

Inverse Functions and Their Derivatives

Exercises 7.1 Identifying One-to-One Functions Graphically Which of the functions graphed in Exercises l--{j are one-to-one, and which are not? y y 2. 1. y

W..

= -3x3 X

y 1

o

13. y

y

= intx

14.

y ~f(x)

...

~

y

sin x,

-¥sxs¥

0-0 0-0 0-0

y ~ f(x)

"

0-0 0-0 0-0

6. y=

l x

\

::r..

one-to-one. 7. f(x) =

G,-

x,

x I

10. f(x)

'2

15.

x

16. y

6

f(x)

f(x) ~ 6 - lx, O=SxS3

~

X+I, -ISxSO 2 { -2 + "3x, 0 < x < 3

-"--c+-----fJ-->x -I 3

--:ct------'::c-->x

o

3

17. a. Graph the functionf(x) ~ ~, 0 S x S I. What symmetry does the graph have?

In Exercises 7-10, determine from its graph if the function is

8. f(x)

"

x

y

y

taox,

2

0

y

~

-¥<xo

o

I

y

12.

y

~f---j'''-------

y=x 4 _x 2

3.

11.

~

4,

r

x,

v?

b. Show that f is its own inverse. (Rernernher that = x if 0.) 18. a. Graph the functionf(x) ~ I/x. What symmetry does the graph have?

x'"

b. Show that f is its own inverse.

x s-3 x> -3

Fonnulas for Inverse Functions Each of Exercises 19--24 gives a formula for a function y ~ f(x) and shows the grapha of f andr'. Find a formula for r l in each case. 19. f(x) = x 2

x>O

+ I, x '" 0

y

Graphing Inverse Functions Each of Exercises 11-16 shows the graph of a function y ~ f(x). Copy the graph and draw in the line y = x. Then use symmetry with to your sketch. (It is reapect to the line y ~ x to add the graph of not necessary to rmd a formula for r l .) Identify the domain and rangeofr'.

r'

20. f(x) = x 2,

x

S

0

y

-----'~---l--~x

--:ct-----:-------> x

o

368


22. j(x) ~ x 2 - lx

21. j(x) ~ x' - I

+

I, x "" I

40. a. Showthath(x) = x'/4andk(x) = (4x)'/3 are inverses of one

another.

y

y

h. Graph h and k over an x-inrervallarge enough to show the graphs inrersecting at (2, 2) and (-2, -2). Be sure the picture shows the required symmetry about the line y = x. c. Find the slopes of the tangents to the graphs of h and k at (2,2) and (-2, -2). -----jf-c-hl---~x --c-+-----'""-----~x

o

23. j(x) ~ (x

+ 1)2,

X "" -I 24. j(x) ~ x 2/3,

y

x "" 0

y

d. What lines are tangent to the corves at the origin? 41. Let j(x) ~ x' - 3x 2 - I, x "" 2. Find the value of dr'/dx at the point x = -I =j(3). 42. Let j(x) = x 2 - 4x - 5,x > 2. Find the value of dr'/dx at the point x ~ 0 ~ j(5).

43. Suppose that the differentiable function y = j(x) bas an inverse and that the graph of j passes through the point (2, 4) and hss a slope ofl/3 there. Find the value of dr'/dx atx ~ 4. 44. Suppose that the differentiable function y = g(x) bas an inverse and that the graph of g passes through the origin with slope 2. Find the slope of the graph of g-' at the origin. Inverses of Lines 45. a. Find the inverse of the function j(x) stant different from zero.

~

mx:, where m is a con-

b. What can you conclude about the inverse of a function y ~ j(x) whose graph is a line through the origin with a Derlvattves of Inverse Functtons Each of Exercises 25-34 gives a fannula for a function y = j(x). In each case, fmd r'(x) and identilY the domain and range of r'. Ail a check, showthatJ0

+3

---=2

x 33. j(x) = x 2 - lx, x '" I (Hint: Complete the square.)

30. j(x) = I/x', 32. j(x) = •

x¢'O

y; r

VX - 3

46. Show that the graph of the inverse of j(x) ~ mx: + h, where m and h are constants and m ¢' 0, is a line with slope I/m and y-intercept -h/m. 47. a. Find the inverse ofj(x) = x + I.Graphjanditsinverse together. Add the line y = x to your sketch, drawing it with dashes or dots for contrast b. Find the inverse of j(x) = x + h(h constant). How is the reJared to the graph of f? graph of

r'

e. What can you conclude about the inverses of functions whose graphs are lines parallel to the line y = x? 48. a. Find the inverse ofj(x) = -x + I. Graph the line y ~ -x + I together with the line y ~ x. At whst angle do

+ 1)'/'

34. j(x) = (lx'

nonzero slope m?

the lines intersect? In Exercises 35-38:

a. Findr'(x). b. Graph j and

b. Find the inverse of j(x) = -x + h (h constant). What angle does the line y ~ -x + h mske with the line y ~ x?

r' together.

e. Evaluaredj/dxatx ~ aanddr'/dxatx ~ j(a) to show that at thesepointsdr'/dx = I/(dj/dx).

+

3,

a = -I

37. j(x) ~ 5 - 4x,

a ~ 1/2

35. j(x) = lx

36. j(x) = (1/5)x + 7, a = -I 38. j(x) ~ lx 2, X "" 0, a ~ 5

39. a. Show that j(x) ~ x' and g(x) ~

'Ii'X

are inverses of one

e. What can you conclude about the inverses of functions whose graphs are lines perpendicular to the line y = x? Increasing and Decreasing Functions 49. Show that increasing functions and decreasing functions are aneto-one. That is, show that for any Xl and X2 in I, X2 =F Xl implies j(X2) ¢' j(x,).

another. b. Graph j and g over an x-inrervallarge enough to show the graphs in_ecting at (I, I) and ( -I, -I). Be sure the pictore shows the required symmetry about the line y = x. e. Find the slopes of the tangents to the graphs of j and g at (I, I) and (-I, -I) (four tangents in all).

d. What lines are tangent to the carves at the origin?

Use the results of Exercise 49 to show that the functions in Exercises 5(}-54 hsve inverses over their domains. Find a formula for dr'/dx using Theorem I.

50. j(x) = (1/3)x

+ (5/6)

52. j(x) = I - 8x' 54. j(x) =

x S/'

51. j(x) = 27x'

53. j(x) = (I - x)'

7.2

Theory and Appl;cations 55. If f(x) is on-to-one, can anything be said about g(x) = - f(x)?

Is it also one-to-one? Give reasons for your answer. 56. If f(x) is one-to-one and f(x) is never zero, can anything be said about h(x) = 1/f(x)? Is it also one-to-one? Give reasons for your answer. 57. Suppose that the range of g lies in the domain of f so that the composite fog is dermed. If f and g are one-ro-one, can anything be said about fog? Give reasons for your answer.

Natural Logarithms

369

COMPlITER EXPLORATIONS 10 Exercises 61--{j8, you will explore some functions and their inverses together wilb their derivatives and linear approximating functions at specified points. Perform the followiog steps using your CAS: a. Plot the function y = f(x) together with its derivative over the giveo interval. Explain why you know that f is on-to-<me over lbe interval. b. Solve the equationy = f(x) for x as a function ofy, and name lbe resulting inverse function g.

58. If a composite fog is one-to-one, must g be one-to-one? Give reasons for your answer.

c. Find the equation for the tangeot line to f at the specified point

59. Assome that f and g are differentiable functions that are inverses of one another so that (g 0 f)(x) = x. Differentiate both sides of this equation with respect to x using the Chain Rule to express (g 0 f)'(x) as a product of derivatives of g and f. What do you fmd? (This is not a proof of Theorem 1 because we assume here is differentiable.) the theorem ~ conclusion that g =

d. Find the equation for the tangent line to g at the point (f(Xo), xo) located symmetrically across the 45 0 line y = x (which is the graph ofthe ideotity function). Use Theorem I to find the slope of this tangeot line.

r'

60. Equivalence of the washer and sheIl methods for rmding volume Let f be differeotiable and increasing on the interval a '" x '" b, with a > 0, and suppose that f bas a differeotiable inverse, Revolve about the y-axis the region bounded by the graph of f and the lines x = a and y = f(b) to geoerate a solid. Theo the values of the integrals giveo by the washer and shell methods for the volume have ideotiea1 values:

r '.

i

f('J 1T«r'(y»' -

a') t(y =

f(a)

l'

21TX(f(b) - f(x)) dx.

To prove this equality, derme

i

61. Y

l'

1T«r (y))' - a') t(y '

21TX(f(t) - f(x)) dx.

4x

63. y = - , - - ,

+

I

x'

64. y = - , - - , x + I 65. y

= x3

66. y

=

-

t '" x '" 4,

2,

- 3x+2 62.y-2x-11'

3x 2

Xo = 3

-2< _ x
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 ~ 1f/6 to x ~ 1f/2 is revolved about the x-axis to generate a solid. Find the volume of the solid.

Evaluate the integrals in Exercises 37-54.

37.

~

I)

b. Using part (a), show that Inx

32

\Yr

35. y = {'" In vldt

/12

70. a. Provethatf(x) = x -Inxisincressingforx

2

33. y-In

•

Jor

69. Locate and identifY the absolute extreme values of

= I + Int 29 • y I - Int

31. y = In (sec (In 8))

cottdt

Theory and Applications

In (sec 8 + tan8)

I 27. y=ln . ~ xvx + I

54

v'i+3 sin 8

xlnx

Inx 21. y ~ I + Inx

26. y

3

55. y

16. y

'0

1

~/2

Logarithmic Differentiation

14. y = (lnx)3

-t-

52.

secytany 2 + secy
In Exercises 55- 0, bave 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 general logarithmic functions, log.,x.

y y = In-1x

or 7

+ sinx) for a

Exponential Functions

7.3 8

a. Grapb y = sin x and the curves y = In (a 4, 8, 20, and 50 together for 0 ,;; x ,;; 23.

b. Why do the curves flatten as a increases? (Hint: Find an

D 88.

y(O) ~ 0

'" x at x ~ O.

c. Graph 10 (I + x) and x togcther for 0 ,;; x,;; O.5.Use differeot colors, ifavailable. At what points does the approximation ofln (I + x) seem best? !.cast good? By reading coordinstes from the graphs, find as good an upper bound for the error as your grapher will allow.

Solve the initial value problems in Exercises 83 and 84. dy I dx ~ I + x'

+ x)

b. Estimate to five decimal places the error involved in replacing 10 (I + x) by x on the interval [0, 0.1].

a. Find the ceotroid of the region b _ the curve y = I/x and the x-axis from x ~ I to x ~ 2. Give the coordinates to two decimal places. b. Sketch the region and show the centroid in your sketch. 80. a. Find the ceoter ofmass ofa thin plate of constant deosity covering the region b _ the curve y = I/Vi and the x-axis from x = ltox = 16. b. Find the ceoter of mass if, instead of being constant, the densityfunctionisS(x) = 4/Vi. 81. Use a derivative to show that f(x) = 10 (x' - I) is one-to-one.

83.

The linearization of In (1 + x) atx ~ 0 Instead of approximating In x near x = I, we approximate In (I + x) near x = O. We get a simpler formula this way. a. Derive the linearization 10 (l

D 79.

377

X =

lny

6

The Inverse of In x and the Number e The function In x, being an increasing function of x with domain (0, 00) and range (-00,00), has an inverse In-1x with domain (-00,00) and range (0, 00). The graph of In-I x is the graph of In x reflected across the line y = x. As you can see in Figure 7.10, lim In-I x = 00

lim In-I x

and

x_-oo

X--i'OO

= O.

The function In-I X is usually denoted as exp x. We now show that exp x is an exponential function with base e. The number e was deimed tu satislY the equation In (e) = I, so e = exp (I). We can raise the number e to a rational power r in the usual algebraic way:

e 2 =e'e,

e-2

=l e 2'

el/2=~,

and so on. Since e is positive, e' is positive too, so we can take the logarithm of e'. When

we do, we find that for r rational FIGURE 7.10 Thegrapbsofy ~ lox andy = In-I x = expx. Thenumbereis 10-1 1 = exp (I).

Ine T

=

rlne

=

r-1

=

r.

Theorem 2, Rule 4

Then applying the function In-I to both sides of the equation In e' = r, we find that e' = exp r

for r rational.

cxp i.In-I .

(1)

We have not yet found a way to give an obvious meaning to eX for x irrational. But In-I x has meaning for any x, rational or irrational. So Equation (I) provides a way to extend the

378

Chapter 7: Transcendental Functions definition of eX to irrational values of x. The function exp x is defined for all x, so we use it to assign a value to eX at every point.

DEFINmON For every real number x, we define the natural exponential function to be eX = exp x. Typical values of

ex

x

eX (rounded)

-I

0.37

o

I

I

2.72

2

7.39

10 100

For the first time we bave a precise meaning for an irrational exponent-we are nrising a specific number e to any real power x, rational or irrational. Since the functions In x and eX are inverses of one another, we bave the following relationships.

Inverse Equations for eX and In JC

e lnx = x

22026 2.6881 X 1043

In(eX)=x

EXAMPLE 1

(all x

> 0)

(all x)

Solve the equation e2>-6 = 4 for x.

Solution We take the natorallogaritbrn of both sides of the equation and use the second inverse equation:

In (e2>-6)

=

In 4

2x-6=ln4 2x = 6

Invorse n:lationship

+ 1n4

x = 3

+ Iln4 = 3 + 1n4 ' /2 2

x = 3

+ 1n2

•

The Derivative and Integral of eX According to Theorem I, the natural exponential function is differentiable because it is the inverse of a differentiable function whose derivative is never zero. We calculate its derivative using the inverse relationship and Chain Rule:

In (eX)

x

Inverse n:lationship

!L ax In (eX)

= 1

Differentiate both sides.

:x· :Ix (eX)

= I

Eq. (2), Section 72, with

:x eX y

=

=

iT.

u~ ...

Solve for the derivative.

dyJax

That is, for = eX, we find that = eX so the natural exponential function eX is its own derivative. Moreover, if f(x) = eX, then t' (0) = eO = 1. This means that the natoral exponential function eX bas slope I as it crosses the y-axis at x = O. The Chain Rule extends the derivative result for the natural exponential function to a more general form involving a function u(x):

If u is any differentiable function of x, then

d u_ .du axe-eax'

(2)

7.3 Exponential Functions

EXAMPLE 2

379

We fmd derivatives of the exponential using Equation (2).

(a) iL(Se x) = SiLex = Sex

ax

ax

(b) iLe-x

ax

= e-xiL(-x) = e-C«-I) = -e-C
0 and X,

the exponential function with

base a is

Theorem 3 is also valid for aX, the exponential function with base a. For example, aXl. aX2

= eXtlna. e X2lna

Def'mition of a~

= eXtlna+x2lna

Law I

= e(J:l+ xz)lna =

a X1 +X2 •

In particular, an. a-I = a n- l for any real number n.

Factor In a Def"mition of a~


381

Proof of the Power Rule (General Version) The definition of the general exponential function enables us to make sense of raising any positive number to a real power n, rational or irrational. That is, we can define the power function y = x n for any exponent n. For any x > 0 and for any real number n,

DEFINmON

Because the logaritbm and exponential functions are inverses of each other, the definition gives

In x n = n In x,

for all real numbers n.

That is, the Power Rule for the naturallogaritbm holds for all real exponents n, not just for rational exponents as previously stated in Theorem 2. The definition of the power function also enables us to establish the derivative Power Rule for any real power n, as stated in Section 3.3. General Power Rule for Derivatives For x > 0 and any real number n,

If x :5 0, then the formula holds whenever the derivative, x n, and x n- 1 all exist.

Proof Differentiating x n with respect to x gives Deflnition of x", x

>0

Chain Rule for eO, Eq. (2)

Dcf'mition and derivative of In x

In short, whenever x

For x < 0, if y

> 0,

=

x n, y' , and x n - 1 all exist, then

Inlyl

=

Inlxl n

=

n InIx!-

Using implicit differentiation (which assumes the existence ofthe derivative y') and Equation (4) in Section 7.2, we have

y'

n

y

x'

Solving for the derivative,

y xn y' = n x = n x = nx n-

1

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

Differentiate f(x) = xx, x

o.

>

Solution We cannot apply the power rule here because the exponent is the variable x rather than being a constant value n (rational or irrational). However, from the def"mition ofthe general exponential function we note that f(x) = XX = exlnx , and differentiation gives /,(x) =

~ (e xInX ) dx

=

e xlnx ~ (x In x) dx

=

e

=

XX (in x + I).

Xlnx ( Inx

Eq. (2) with U

~

.In.

+ x·i)

•

.>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. THEOREM 4-The Number. as a Untit limit

e

Proof Iff(x)

=

=

The number e can be calculated as the

lim (I

x-o

+ x)l/x.

Inx,then/,(x) = I/x,so/,(I) = I. But,bythedef"mitionofderivative,

/,(1) = lim f(1 h-O

=

lim In (I

x-o =

lim In (I

x-o

+ h) - f(l)

=

lim f(1

h

x-o

+ x)

- In I = lim

x-o

X

+ x)!/x

=

In [lim(I

x-o

+ x) - f(l) x

lin (I + x)

lnl~O

X

+ x)!/x]

In is continuous;

use Theorem lOin Chapter 2.

Because /,(1) = I, we have In [lim(1

.-0

+ X)!/x]

=

I

Therefore, exponentiating both sides we get lim (I + x)!/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 = e xlna . Then we have

383


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 ofthe general exponential function.

If a > 0 and u is a differentiable function of x, then a U is a differentiable function ofx and

%x aU

=

aUlna :,.

(3)

The integral equivalent ofthis last result gives the general antiderivative

J

aUdu

y~ F E-----'--

2

a>

1.Asx---+oo,wehaveax---+Oif

O 1.As x-+ -00, we have aX -+ 00 if 0 < a < 1 andax-+Oifa> 1.

aU

Ina

+ C.

(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 d -(aX) dx 2

FIGURE 7.11 Exponential functions decrease if 0 < a < 1 and increase if

= -

=

d -(axIna) dx

=

(Inaj2a x

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

We fmd derivatives and integrals using Equations (3) and (4).

(a)

~3x

(b)

~rx = rX(ln3)~(-x)

dx

=

3x ln3

dx

dx

=

- rx ln3

d· . d . (C) dx 3""X = 3,mx(ln3) dx (sinx) = 3,mx(ln3) cosx

(d) (e)

J J

2x dx =

~x2 + C

2 .... cosxdx =

Eq. (3) with a

~

Eq.(3)witha

~ 3,_ ~-.

... ,_

~

2u du =

2sinx

= In2

+

C

~U2 + C

_

Il

~

•

,in.

Eq.(4)witha

J

3, _

~ 2,_ ~.

~ ,in',d_ ~ cosx 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 natura110garithms ... . .. divided by In. ... . .. gives Rule 1 for base a logarithms.

Quotient Rule: log" ~ = log" x - log" y

Derivatives and Integrals Involving log. x

Reciprocal Rule:

To rmd 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) = Ina'u tU'

d I du -(Iog"u) = -I , tU In a u tU

7.3

Exponential Functions

385

EXAMPLE 6 I

I

(8)

axd loglO(3x +

(b)

r°'6)X ax = ~zJ ~x ax =

axd (3x +

I) = In 10 • 3x + I

log,x

~zJUdU

3

I) = (In 1O)(3x + I)

~ :;

u~lnx, du~itU

I u2 I (lnx)' (In x)' =lnzT+C=lnz-z-+ C = ZlnZ +C

•

Exercises 7.3

1. a.

e--()·3t =

b.

27

2. a. • - 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 ofthe form dy/dx = ky for some nonzero constant k. More generally, suppose we have a differential equation ofthe form

dy dx = f(x,y),

(3)

7.4 Exponential Change and Separable Differential Equations

389

where j 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) = j(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 j can be expressed as a product of a function of x and a function of y. The differential equation then has the fann dy dx = g(x)H(y).

g is a function of:t; His a function of y.

When we rewrite this equation in the form dy dx

g(x) h(y) ,

1 H(y) ~ hey)

its differential fann 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 ofthis 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

g(x) dx.

Solve the differential equation dy dx = (I + y)e", y

Solution Since I the variables.

+ y is never zero for y >

-I, we can solve the equation by separating

dy = (I dx

-

dy

=

> -1.

+ y)e"

(I + y)e" dx

~=e"dx 1+ Y 1+ Y J ~=Je"dx

In(l+y)=e"+C The last equation gives y as an implicit function of x.

Treat dyldx 88 a quotient of

diffi:rentials and multiply both sides by dx. Divide by (1

+ y).

Integrafe both sides. C represents the combined

constants of integration.

•

390


EXAMPLE 2 Solution

Solve the equation y(x

+

I):

= X(y2

+

I).

We change to differential form, separate the variables, and integrate:

y(x

+

I)dy = X(y2

ydy y2+1

+

I)dx

xdx x+1

- -

- -

x =F

1 J ~-J (1 __ l+y2-

tln(1

+ y2)

x+1

=x -lnlx

)dx

-1

Dividcxbyx + 1.

+ II + c.

The last equation gives the solution y as an implicit function of x.

•

The initial value problem

dy dt = ky,

y(O) = Yo

involves a separable differential equation, and the solution y = yoe kt expresses exponential change. We now present several examples of such change.

Unlimited Population Growth Strictly speaking, the nwnber of individuals in a population (of people, plants, animals, or bacteria, for example) is a discontinuous function oftime because it takes on discrete values. However, when the nwnber of individuals becomes large enough, the population can be approximated by a continuous function. Differentiability of the approximating function is another reasonable hypothesis in many settings, allowing for the use of calculus to model and predict population sizes. If we asswne that the proportion of reproducing individuals remains constant and asswne a constant fertility, then at any instant t the birth rate is proportional to the nwnber y(t) of individuals present. Let's asswne, too, that the death rate of the population is stable and proportioual to y(t). If, further, we neglect departures and arrivals, the growth rate dy/dt is the birth rate minus the death rate, which is the difference of the two proportioualities under our asswnptions. In other words, dy/ dt = ky so that y = yoe kt , where Yo is the size of the population at time t = O. As with all kinds of growth, there may be limitations imposed by the surrounding environment, but we will not go into these here. The proportiouality dy/dt = ky models unlimitedpopulation growth. In the following example we asswne this population model to look at how the nwnber of individuals infected by a disease within a given population decreases as the disease is appropriately treated.

EXAMPLE 3 One model for the way diseases die out when properly treated asswnes that the rate dy/ dt at which the nwnber of infected people changes is proportioual to the nwnber y. The nwnber of people cured is proportioual to the nwnber y that are infected with the disease. Suppose that in the course of any given year the nwnber of cases of a disease is reduced by 20%. If there are 10,000 cases today, how many years will it take to reduce the nwnber to 1000? We use the equation y = yoe kt • There are three things to find: the value of Yo, the value oft, and the time t when y = 1000. The value of Yo. We are free to count time beginning anywhere we want. If we count fromtoday,theny = 10,000 when t = O,soYo = 10,000. Our equation is now

Solution

y = 1O,000e kt •

(5)

7.4

Exponential Change and Separable Differential Equations

391

The value of k. When t = I year, the number of cases will be 80% of its present value, or 8000. Hence, 8000 = 1O,000e.\(1)

e k = 0.8 In (e k ) = In 0.8 k= 1n0.8

Eq. (5) with t ~ I and y ~ 8000 Logs ofboth sides

< O.

At any given time t,

y = 10,000e(lno.')t.

(6)

The value oft that makes y = 1000. We set y equal to 1000 in Equation (6) and solve for t: 1000 = 1O,000e (In O.'ll 0.1

e(lno.')1 =

(lnO.8)t

= 1n0.1

t = :

~:~

Logs of both sides

'" 10.32 years.

It will take a little more than I 0 years to reduce the number of ca,e, to 1000.

•

Radioactivity

I

For radon-222 gas, t is measured in days and k = 0.18. For radiwn-226, which used to be painted on watch dials to make them glow at night (a dangerous practice), t is measured in years and k ~ 4.3 X 10-4.

Some atoms are unstable and can spontaneously emit mass or radiation. This process is called radioactive decay, and an element whose atoms go spontaneously through this process is called radioactive. Sometimes when an atom emits some of its mass through this process of radioactivity, the remainder of the atom re-forms to make an atom of some new element. For example, radioactive carbon-14 decays into nitrogen; radium, through a number ofintermediate radioactive steps, decays into lead. Experiments have shown that at any given time the rate at which a radioactive element decays (as measured by the number of nuclei that change per unit time) is approximately proportional to the number of radioactive nuclei present. Thus, the decay of a radioactive element is described by the equation dy/ dt = - ky, k > O. It is conventional to use -k, with k > 0, to emphasize thaty is decreasing. If Yo is the number ofradioactive nuclei present at time zero, the number still present at any later time t will be

k> O. The half-life of a radioactive element is the time required for half of the radioactive nuclei present in a sample to decay. It is an interesting fact that the half-life is a constant that does not depend on the number of radioactive nuclei initially present in the sample, but only on the radioactive substance. To see wby, let Yo be the number of radioactive nuclei initially present in the sample. Then the number ypresent at any later time twill be y = yoe-kt • We seek the value oft at which the number of radioactive nuclei present equals half the original number:

yoe-kl =

e-kl = -kt

!YO

2

!

2

= In! = -In2

2

t = 1n2 k

R;2

~ = yer' + 2vY er'

dy 20. fix=xy+3x-2y-6

27. Working underwater The intensity L(x) oflightx feet beoeaili the surface of the ocean satisfies the differential equation

tiL

22. :

= eX - Y

+ eX +

-0.6y

called ''inversion'' that changes the sugar's molecular s1ructW'e. Once the process has begun, lhe rate of chaage of lhe amount of raw sugar is proportional to the amount of raw sugar remaining. If 1000 kg ofraw sugar reduces to 800 kg ofraw sugar during lhe IlrSt 10 hours, how much raw sugar will remain after aoother 14 hours?

dy e 2r - y 18. dx = eX+Y 19. y2 ~ =

~

wheo t is measured in hours. If there are 100 grams of B-giucono lactoue present wheo t = 0, how many grams will be left after ilie first hour? 26. The inversion of sngar The processing of raw sugar has a step

x> 0 16. (secx) ~ = e Y+'"

= eY+v',;,

Differential equation: dp/dh = kp (k a constant) Initial condition: p = Po wheo h = 0

c. At what altitode does ilie pressure equal 900 millibars? 25. Fint-order chemical reactions In some chemical reactions,

Separable Differential Equations Solve the differential equation in Exercises g....22. • ,.. dy _ 9. 2 vxy fix - I, x,y > 0

now (as a perceotage of our preseot tooili size)? 24. Atmospheric pressure The earth's atmospheric pressme p is ofteo modeled by assuruing that lhe rate dp/dh at which p chaages wiili the altitode h above sea level is proportional to p. Suppose that ilie pressure at sea level is 1013 millibars (about 14.7 pounds per square inch) aod that the pressure at ao altitode of20 km is 90 millibars. &. Solve lhe initial value problem

to express p in terms of h. Determine ilie values of po aod k from ilie giveo altitude-pressure data. b. What is the atmospheric pressure at h ~ 50 km?

x>O 8.

value of k in the equation y = yoe lt . Then use this value of k

to aoswer the following questions. b. In about how maoy years will human teeth be 90% of their

Initial value Prohlems

Differential equation

23. Human ....Iution continues The aoalysis of tooth shrinkage by C. Loring Brace aod colleagues at the University of Michigao's Museom ofAothropology indicates that human tooth size is continuing to decrease aod that ilie 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. a. If t represents time in years and y represents tooth size, use theconditionthaty ~ 0.99yowheot ~ 1000 to find the

e-Y

+1

Applications and Examples D The answers to most of the following exercises are in terms of logarithms aod exponeotials. A calculator cao be helpful, enabling you to express the aoswers in decimal form.

fix

~

-kL.

As a diver, you know from experieoce that diving to 18 ft in ilie Caribbean Sea cuts ilie intensity in half. You csunot work wiiliout artificial light wheo the intensity falls below on.,..tenth of the surface value. About how deep cao you expect to work wiiliout artificiallight?

7.4 28. Voltage in a discharging capacitor Suppose that electricity is draining 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. Cholera hacteria Suppose that the bactetia in a colony can grow unchecked, by the law of exponential change. The colony starts with I bacterium and doubles every balf-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 an infected perliOll, many bacteris are destroyed, but this example helps explain why a person who 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? 31. The incidence of a disea.e (Continuation ofExample 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 ofcases to less than I? 32. The U.S. population The U.S. Census Bureau keeps a running clock totaling the U.S. population. On March 26, 2008, the tota1 was increasing at the rate of I perlion 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 constant rate, fmd the rate constant for the population's growth (people per 365-day year). b. At this rate, what will the U.S. population be at 2:31 P.M. EST on March 26, 2015? 33. Oil depletion Suppose the amount of oil pumped from one ofthe canyon wells in Whittier, California, decreases at the continuous 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 ordeni, your firm's .ale. department applies a continuous discount that makes the unit price a functionp(x) of the numher 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 unit price P(90) for a 90-unit order. c. The sales department has asked you to find out if it is discounting so much that the fInD's revenue, r(x) ~ x· p(x) , will actually be less for a 100-unit order than, say, for a

Exponential Change and Separable Differential Equations

395

90-unit order. Reassure them by showing that r has its maximum value atx ~ 100. d. Graph the revenue function r(x)

~

xp(x) for 0 :5 x :5 200.

35. Plutooium-239 The half-life of the plutonium 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 half-life of polonium is 139 days, but your sample will not be useful to you after 95% of the radioactive nuclei present 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. The mean life of a radioactive 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 mean life ofa carbon-14 nucleus is more than 8000 years. Show that 95% ofthe radioactive nuclei origiually present in a sample will disintegrate within three mean lifetimes, i.e., by time t = 3/k. Thus, the mean life ofanucleus gives a quick way to estimate how long the radioactivity of a sample will last. 38. Californium-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 westem world since its discovery by Glenn Seaburg in 1950. The balf-life of the isotope is 2.645 years--Iong enough for a usful 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 i. the value of k in the decay equation for this isotope? b. What is the isotope's mean life? (See Exercise 37.) c. How long will it take 95% of a sample's radioactive nuclei to disintegrate? 39. Cooling .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 answer the following questions. a. How much longer would it take the soup to cool to 35'C? b. Instead ofbeing left to stand in the room, the cup of90'C soup is put in a freezer whose temperature is -15°C. How long will it take the .oup 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 temperliturewas 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. Silver cooling in air The temperliture of an ingot of silver is 60'C above room temperliture right now. Twenty minutes ago, it

was 70D e above room temperature. How far above room temperature will the silver be

a. 15 min from now?

b. 2 hours from now?

c. When will the silver be 1DoC 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 contained 44.5% of the carbon-14 found in living matter. About how old is Crater Lake? 44. The sensitivity of earhon-14 dating to measurement To see the effect ofa relatively small error in the estimate ofthe amount of carhon-14 in a sample heing dated, consider this hypothetical situation:

a. A fossilized bone found in centrallllinois in the year A.D. 2000 contains 17% ofits original carhon-14 content. Estimate the year the animal died.

45. Carhon-14 The oldest known frozen humao mummy, discovered in the Schnalstal glacier of the Italian Alps in 1991 and called Olzi, was found wearing straw shoes and a leather coat with goat fur, and holdiog a copper ax and stone dagger. It was estimated that Otzi died 5000 years hefore he was discovered in the melting glacier. How much of the origins! carhon-14 remained in Otzi at the time of his discovery? 46. Art forgery A painting attrihuted to Vermeer (1632-1675), which should contain no more than 96.2% of its originsl carhon14, contains 99.5% instead. Ahouthow old is the forgery?

I>. Repeat part (a) assuming 18% instead of 17%.

e. 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'H6pitai (1661-1704) Johano Bernoulli (1667-1748)

John (Johann) 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'Hopital" Rule, after Guillaume de I'HOpitai. He was a French nobleman who wrote the first introductory differential calculus text, where the rule f"ITSt 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 cannot apply the Quotient Rille 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 undef"med. 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 Rille, 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 cannot evaluate. We use % as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as 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 manipillations. 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

'( ) = li f(x) - f(a) f a m x 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:Hopital's Rule enables us to draw on our success with derivatives to evaluate limits that otherwise lead to indetenninate fonns. THEOREM 5- L'H6pital's Rule Supposetbatf(a) = g(a) = 0, that f and g are differentiable on an open interval I containing a, and that g' (x) oF 0 on fif x oF a. Then lim f(x) g(x)

=

x~.

lim f'(x) g'(x) ,

x~.

assuming that the limit on the right side ofthis equation exists. We give a proof ofTheorem 5 at the end of this section.

I Caution

To apply I'H6pital's Rille to Ilg, divide the derivative of I by the derivative ofg. Do not fall into the trap oftaking the detivative of IIg. The quotient to use is ng', not <JIg)'.

EXAMPLE 1 The following limits involve % indeterminate forms, so we apply I'H5pital's Rule. In some cases, it must be applied repeatedly. (a) lim 3x - sinx = lim 3 - cosx = 3 - cos x l

x-o

x

x-o

1

1

= 2

x=o

1 (b) lim

x-o

(c) lim

yIJh - I

=

x

yIJh -

lim 2y1Jh

x-o

1

I 2

1 - x/2

o

2

x-o

o

X

=

=

xr'/ x-o 2x . -(1/4)(1 + xr lim lim (1/2)(1 +

2

-

32 /

x-a

2

1/2 Still

I 8

=--

Not~; limit is found

o

(d) lim x - sinx

x-o

~; differentiate again.

o

x3

=

lim I - cosx

x-o =

3x 2

lim sinx

x-o 6x

Not~; limit is found. Here is a summary of the procedure we followed in Example I. Using UH6pital's Rule

Tofmd lim f(x) g(x)

x~.

by I'HOpital's Rule, continue to differentiate f 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:H5pital'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+x 2 =

. 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'HOpitai '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 inde• terminate 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 oc and + oc mean the same thing.

Positive for x

o

. sin x (b) Iun-2 .%-0-

o

x

=

> 0

lim cosx x-o- 2x

Indetenninate Fonns

= -00

Negative for x < 0

OI:J / OI:J, OI:J •

0,

OI:J -

•

OI: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 we11 as to 0/0. Iff(x) --+ ±oo andg(x) --+ ±oo asx --+ a, then lim fix) = lim f'(x) g(x) x~. g'(x)

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 -l)

In (esc x)

x~w/2 (x - ('If/2l)2

lim 0 - sinO cosO ,~o

x-I+

%-1+

53. lim (lnx)l/x

54. lim (In x) 1/(>-.) x-e+

x~oo

24. lim

I sin I 1_01 - cost

56. lim.

X IJlu

X~OO

26.

lim

x-(n/2)-

28. ,~o lim

("-2 -

(1/2)' - I 6

x) tan x

57. lim (1

+ 2x)I/(2lnx)

58. lim (eX x~o

x~oo

59.

li.s x'

x~o

a>O

38. lim (lnx - In sin x) x-o+ 40. lim x-o+

x~+ In (sinx)

• %-1+

t- 3

36. lim

y~O

X~OO

I' - I

1-14t 3 -

(-SInt

x-(7rf2)-

.~o

8.

2

Inx

%-0+

y

37. lim (In 2x - In(x

Applying l'H6pital's Rule Use I'Hllpital's rule to fmd the limits in Exercises 7-50.

34. lim In (eX - I)

+ 25 - 5

y~O

~oox3+x+l

x2

lo~x

33. lim 35. lim

-

32 r log, x • x~log,(x + 3)

31. lim In(x + I) %_00

3. lim 5x - 3x x_OO 7x 2 + 1 ~o

x-o~

2. lim sin 5x

4

2

, 3x - I 30. lun~1

60.

lim

%-0+

+ x)!/x

(1 + l)X X

7.5 Indeterminate Forms and rH6pitars Rule

61. lim %_00

63.

65.

(Xx+- I2)'

lim

62.

2

lim x 1nx

x_oo

X)

lim xtan(" x-o+ 2

X

lim sin x oinx

66.

X~

...

x_ OO

69.

v.;-+I

lim cotx escx

70

lim sec x x-(7ff2t tanx

o 81. oc -

oc Form

a. Estimate the value of X~OO

=x

hy graphing f(x) val ofx-values.

• %-0+

. ZX - 3x 71. x~3x + 4%

82. Find lim

x~oo

(W+i - -...h).

o 83. 0/0 Form

i' 73• x~xex I· e

Estimate the value of lim 2x' - (3x

75. Which one is correct, and which one is wrong? Give reasons for

your answers.

=

b. lim x-3 %-3 x 2 - 3

=Q=o 6

and the limit

. (1+"I)X =e.

your answers.

lim

a. lim x '-2x x_ox2 - sinx

2x ~lim-

x' -

X~OO

2

a. Use I'H6pital's Rule to show that

x-O 2x - cosX'

= lim

lim x_o2

2

+ sin x

2x = lim 2x - 2 sinx x-O 2x cosX'

-2

d.

lim Inx

-00

= x~o+ (I/x) = ()O = -

Db. Graph

f(x)

~

(I +

1

x~"'(1

=

lim x-O+

lim (I +

k--+oo

x)

(I/x)

(-l/x 2)

=

lim (-x)

=0

x-o+

78. Find all values of c thst satisfY the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.

a. f(x) ~ x, g(x) ~ x', (a, b) ~ (-2,0) b. f(x) ~ x, g(x) ~ x', (a, b) arbitrary c. f(x) = x'/3 - 4x, g(x) = x', (a, b) = (0,3) 79. Continuous extension Find a value ofc thst makes the function f(x) =

and g(x)

~

(I

+~)'

85. Show thst

lim x In x ~ lim In/x

X~...

:S

toge1her for x '" O. How does the behavior off compare with thst ofg? Estimate the value oflim~oo f(x). c. Confmn your estimate oflim~oo f(x) by calculatiog it with I'Hllpital's Rule.

b. lim xlnx = 0'(-00) = -00 x-o+ x

X~OO

0 - 1 = 2

%-0+

In

lim

2+0

others wrong? Give reasons for your answers. a. lim xlnx ~ 0·(-00) ~ 0

r

. (1+"I)X =e.

= _2_ = I

77. Only one of these calculations is correct Which one? Why are the

c. x-'?J+ X

I)-...h + 2

X

76. Which one is correct, and which one is wrong? Give reasons for

x_ox2

+

1 hy graphing. Then confmn your estimate with I'Hllpital's Rule. x-I

84. This exercise explores the difference between the limit

lim ~=l x-3 2x 6

a. lim x-3 x-3 x 2 - 3

b.

- Yx' + x over a suitably large inter-

b. Now confirm your estimate hy fmding the limit with I'Hllpital's Rule. Ail the fIrSt step, multiply f(x) hy the fraction (x + ~)/(x + Yx' + x) and simplifY the new numerator.

68. lim -...h x-o+ Ysinx

'\!9x+1

bX) = O?

lim (tan 2x + "- + sin x-a x3 x2 X

lim (x-~)

Theory and Applications I:Hllpital'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

80. For what values of a and b is

x+2

lim (In x)' x-o+

64.

x-o+

(X' + I )'IX

403

9X - 3sin3x, x ~ 0 5x' { c, x= 0

continuous at x = O. Explain why your value of c works.

86. Given that x a. Xl/X b. xiii'

1')' k

=

e'.

> 0, fmd the maximum value, if any, of

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.

(I)

a. y~xtan "

2x

3x + e b. y~ 2x+e'"

e-Vi' 88. Findf'(O) for f(x) ~ { 0, '

x ~ 0 x = O.

404

D 89.


D 90.

The continuons extension of(sinxf to [0, 'IT]

The function (sin x)...• (Continuation ofExercise 89.)

= (sinx)""'on1heinterva1-7:5X:5 7. How do you account for the gaps in the graph? How wide are the gaps?

a. Graphf(x) ~ (sinxf on the interval 0 :5 x:5 'IT. What value would you assign to f to make it continuous at x = O?

a. Graphf(x)

b. VerilY your conclusion in part (a) by rmding lim,....,. f(x) with I'H6pital~ Rule.

b. Now graph f on the interval 0 :5 x :5 '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)+.)

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 crosse. the x-axis. To simplilY your work, you might want to delete the exponential factor from the expression for f' aod 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:

Y = sin-Ix

'IT'

-1';;x,;;1

Range: -TTI2 :S Y :S 'lT12

_~

1

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.

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 -1 at x = -1t/2 to + 1 at x = 1t/2. By restricting its domain to the interval [-1t/2, 1t/2], we make it one-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

sin x

= sinx Domain: [-'IT/2, 'IT/2]

y

Range: [-1,1]

y = cosx Domain: [0, 'IT] Range: [-1, 1] y

sec x

-o;;t---,,;;---;;,,~x

2

i( y = cotx Domain: (0, 'IT) Range: (-00, 00)

y = tao x Domain: (-'IT/2, 'IT/2) Range: (-00, 00)

y = secx Domain: [0, 'IT/2) U ('IT/2, 'IT] Range: (-00, -1] U [I, 00)

----,;,,;--c;cot---c,,;;-~x

-~

2

y = cscx Domain: [-'IT/2, 0) U (0, 'IT/2] Range: (-00, -1] U [I, 00)

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 and y = cos-1 x for radian angles in the frrst quadrant. For a unit circle, lhe = 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 x, 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

or

y

= cos- 1 x

or

y

= tan-I

x

or

= coC l X y = sec- t x

equation s

whose sine is x."

y = sin-Ix

y y

Y = arcsinx Y = arccosx y = arctanx

or

Y = arccotx y = arcsecx

or

y

or

= csc- l x

= arccscx

These equations are read "y equals the arcsine of x" or "y equals arcsin x' and so OD. Caution The - I in the expressions for the inverse means ''inverse.'' It does not mean reciprocal. For example, the reciprocal of sin x is (sinx)-l = I/sinx = cscx.

y

x2

+ y2 = 1

Arc whose sine is x

~-+--

Angle whose sine is x -+-----::r\---'-~-t:_-~x

o

x

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 tske a closer look at these functions and their derivatives. Domain: -1 =S x =S 1 Range" - J! :s y s '!

Domain: -1:s x:s 1 Range: 0 :S Y :S '1r

y

y

.

1

2

Anglewho!le

2

Domain: -00 < x < Range:

OQ

-I 1 >0

4

•

xV25x s - I·

Derivatives of the Other Three We cou1d use the same techniques to find the derivatives ofthe other three inverse trigonometric fimctions--arccosine, arccotangent, and arccosecant---but there is an easier way,

thanks to the following identities.

Inverse Function-Inverse Cofunction Identities cos-I x

=

TT/2 - sin- l x

coC I X csc- I x

=

TT/2 -

=

TT/2 - sec-I x

tan-I

x

We saw the tmt of these identities in Equation (4). The others are derived in a similar way. It follows easily that the derivatives ofthe inverse cofunctions are the negatives ofthe

7.6 Inverse Trigonometric Functions

411

derivatives ofthe corresponding inverse functions. For example, the derivative of COS-I x is calculated as follows: d (COS-1) d (1T . -1 x ) dx x =dx 2- sm

Identity

1

Derivative of arcsine

v"1=7

The derivatives of the inverse trigonometric functions are summarized in Table 7.3.

TABLE

7.3 Derivatives of the inverse trigonometric functions

d(sin- 1 u)

I

dx

1.

du

d(cos- I u)

2.

1

dx

I

lui

I

Integration Formulas The derivative formulas in Table 7.3 yield three useful integration formulas in Table 704. The formulas are readily verified by differentiating the functions on the right-band sides.

TABLE

7.4 Integrals evaLuated with inverse trigonometric functions

The following formulas hold for any constant a

1.

2. 3.

Jv'a~u_ J~ J a2

u2 =

du

(*) + ~tan-l (*) +

u 2 = sin-

uv'u2 _ a 2

=

1

a

2

(Valid for u

C

(Valid for all u) (Valid for

2

x -I,

, , , ,

y

l ,

x

, ,

, ,, ,

(a)

TABLE 7.9

Identities for inverse

hyperbolic functions sech-1x = cosh-11 x csch-1X = sinh-11 x

(c)

FIGURE 7.26 The graphs oflbe inverse hyperbolic tangent, cotangent, and cosecant of x.

Useful Identities We use the identities in Table 7.9 to calculate the values ofsech- 1 x, csch- I x, and coth- I x on calculators that give only cosh-1 x, sinh-1 x, and tanh-I x. These identities are direct consequences of the defmitions. For example, if 0 < x ,,;; 1, then

420

Chapter 7: Transcendental Functions We also know that sech (sech- I 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- I u)

dx d(cosh- I u)

dx d(tanh- I u)

Derivatives of inverse hyperbolic functions

I du ~dx I du W-=1dx'

u

du 2dx l-u '

lui
I by applying Theorem I with f(x) = coshx andr'(x) = COsh-I x. Theorem I can be applied because the derivative of cosh x is positive for 0 < x.

(r')'(x)

=

The=ml

f'(j \ (x)) I sinh (cosh

1 x)

I Ycosh2 (cosh 1 x) - I I

1'(-)

~

coshl

U -

sinhu

=

sinh_

sinhl

U =

I,

Vcosh2 u - 1

421

7.7 Hyperbolic Functions HIsTORICAL BIOGRAPHY

The Chain Rule gives the final result:

Sonya Kova1evsky (185a>O

r-'(')+c ~coth-l

du

a

4.

Integrals leading to inverse hyperbolic functions

a

u2

'

+ u2

a2

u2 > a 2

(*) + C,

u2

because = lim 2>;2 =

00

I'H6pital's Rule

%---+00

00

for any positive integer n because

lim Inx = lim

I/x

,_00 x l / n ,_00 (I/n)x(l/n)-I

!'H6pilal's Rule

= lim --"-= O. x---+oo x 1/ n

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 : log" x

.

.

In x/In a

hm--=hm ,_00 10llbX ,_00 Inx/lnb

Inb =Ina'

•

The limiting ratio is always finite and never zero.

x --->

I

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 ,_00

= L1

and

together imply

If L, andL2 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+ x25 x x_OO

(2VX -

x-oo

1)2 = lim

x

x-oo

(2VX Vi

=1

'

1)2 = lim x-oo

(2 __Vi

1_)2 = 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.

Here we use little-oh notation.

EXAMPLE 3

(a) lnx = o(x) as x ---> (b) x 2 = 0(x 3

+

00

because

lim lnx = 0 x---+ OO X 2

1) as x --->

00

because

•

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 /

Here we use big-oh notation.

EXAMPLE 4 (a) x (b) eX

+

sinx = O(x) as x --->

+ x2

= O(g) ("I is big-oh of g").

=

O( eX) as x --->

00

00

because because

x

+ xsinx ,;;

eX

+x x 2 ~lasx--+oo

.

2 for x sufficiently large.

e

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 rate, thoo I = O(g) and g = OU) (Exercise ll). (c) x = O( eX) as x --->

00

because

Sequential vs. Binary Search Computer sciootists often measure the efficioocy of an algorithm by counting the number of steps a computer must take to execute the algorithm. There can be significant differooces in how efficiently algorithms perform, even if they are designed to accomplish the sarae task. These differooces are often described in big-oh notation. Here is an exarople. Webster. 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 arrangemoot. 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, thoo 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 ) That certainly beats a possible 26,000 steps.

< I.

428


For a list of length n, a sequential search algorithm takes on the order of n steps to find a word or detennine 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(log2 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 Compartsons with the Exponentfal eX

1. Which of the following functions grow faster than eX as x --+ 00 ? Which grow at the same rate as eX? Which grow slower? Lx-3 b.x 3 +sin2x c. v'X c. (3/2Y'

d.

4X

r.

eX/2

g. eX/2

h. log"x

g. In (In x)

2. Which of the following functions grow faster than eX as x --+ Which grow at the same rate as eX? Which grow slower?

a. lOx' + 30x c. vT+7

+

I

00

?

d. (5/2Y' f. xeX"

3. Which ofthe following functions grow faster than x 2 as x --+ 00 ? Which grow at the same rate as x'? Which grow slower? a.x 2 +4x b.x S -x 2 c. v'x' + x 3 d. (x + 3)2 e. xlnx f. 2X" 3 g. x e -x h. &x 2 2

4. Which ofthe following functions grow faster than x as x --+

00

?

Which grow at the same rate as x'? Which grow slower? b. lOx'

r.

(I/IOY' h. x 2 + IOOx

Compartsons with the Logartthm In x 5. Which of the following functions grow faster than In x as x ---+ oo? Which grow at the same rate as In x? Which grow

c. In v'X e. x g.

1/x

Ordering Functions by Growth Rates

a. eX

b.x'

c. (lnxY'

d. e X / 2

a. 2x c. (ln2Y'

d. v'X

r.

5lnx

b. x 2 d. eX

Big-oh and Little-oh; Order

9. True, or false? As x ---+ 00. a. x = o(x) c. x = O(x + 5) e. eX = o(e2%) g. Inx = 0(1n 2x) 10. True, or false? As x

d. log" (x')

b. In 2x

+ 5)

8. Order the following functions from slowest growing to fastest growing as x ---+ 00.

Compartsons with the Power i'

slower? a. 10g)x

h. In (2x

7. Order the following functions from slowest growing to fastest growing as x ---+ 00.

b. x Inx - x

h • ex-I

a. x' + v'X c. x'e ~ e. x 3 - x' g. (l.l Y'

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. log, (x 2 ) b. log" lOx c. 1/v'X d. I/x' f. e-x e. x - 2lnx

a. x

=

d. x = 0(2x)

f. x + Inx = O(x) W+5 = O(x)

h. ---+ 00,

~ 3 = O(})

c. } - :'

b.x=0(x+5)

o(})

b. }

+ :' =

O(})

d. 2

+ cosx

= 0(2)

e. eX + x = O(e X) f. x Inx = 0(x 2 ) g. In (In x) = O(lnx) h. In (x) = 0(1n (x 2 + I)) 11. Show that ifpositive functions f(x) and g(x) grow at the same rate asx--+ 00, theof= O(g) andg = 0(/). 12. Wheo is a polynomial f(x) of smaller order than a polynomial g(x) as x ---+ oo? Give reasons for your answer. 13. Wheo 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 ofrational functions tell us about the relative growth ofpolynomia1sasx~ oo?

D

Other Comparisons DIS. Investigate

lim In (x + I) Inx

.%_00

and

.

11m

];_00

In (x + 999) In x .

X'/I,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. c. Even x '/10 takes a long time to overtake In x. Experiment with a calculator to fmd the value of x at which the graphs of x 1/10 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.

D 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 1han 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

x_OO

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 + I and y';+J 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. a. Suppose you have three different algorithms for solving the same prohlem and each algorithm takes a number of steps that is of the order ofone of the functions listed here: n lo~ n,

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 lJ for any positive integer n, even x 1,000,000 . (Hint: What is the nth derivative of XII?) 20. The function c" outgrows any polynomial Show that e" grows faster as x ---> 00 than any polynomial

21. a. Show that Inx grows slower as x ---> 00 than xii' for any positive integer n, even x 1/1,000,000 •

Db.

n 3j2 , n(lo~ n? .

Which of the algorithms is the most efficient 10 the long roo? Give reasons for your answer.

D 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,

Y;; log, n,

(log, n? .

D 25.

Suppose you are looking for an item in an ordered list one million iterus long. How many steps might it take to fmd that item with a sequential search? A binary search?

D 26.

You are looking for an item in an ordered list 450,000 iterus long (the length of Webster. Third New International Dictionary). How many steps might it take to fmd the item with a sequential search? A binary search?

Although the values ofx'/I,OOll,OOO eventually overtake the values 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

429


1. What functions have inverses? How do you know if two functions f and g are ioverses of one another? Give examples of functions that are (are not) ioverses of one another. 2. How are the domains, raoges, and graphs of functions and their ioverses related? Give an example. 3. How can you sometimes express the inverse ofa function of x as a function ofx? 4. Under what circumstances can you be sure that the ioverse of a function f is differentiable? How are the derivatives of f and related?

r'

5. What is the natoral 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 iotegrals of tan x and cot x? 8. How is the exponential function c' 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 truth is there 10 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 chaoge? How can it he derived from an initial value problem? What are some of the applications of the law? 12. Descrihe 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 - OO? Give examples.

14. How can you sometimes handle limits that lead to indeterminate forms Xl, 00, and 00 001 Give examples.

r

15. How are the inverse trigooometric 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 functioos? How do suhstitutioo and completing the square broaden the applicatioo 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

19. What are the derivatives of the six basic hyperbolic functioos? What are the correspooding integral formulas? What similarities do you see here with the six basic trigonometric functions? 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-' x, and coth-I x using a calculator's keys for cosh- 1 X, sinh-1 x, and tanh-I x? 21. What integrals lead natorally to inverse hyperbolic functioos? 22. How do you compare the growth rates of positive functions as X~OO?

23. What roles do the functioos e% and In x play in growth comparisons? 24. Describe big-oh and little-oh notatioo. Give examples. 25. Which is more efficient--- 0)

31. / e% sin (e") dx

2

7. y = 10!l2 (x /2) 9. y = 8--1 11. y = 5x,·6

12. y = y'2,,-v2

+ 2)"+2

14. y = 2(lnx)"/2

13. y = (x

10. y = 9"

15. y = sin-I~, 16. y

33. / e% sec2 (e% - 7) dx

1 17. y

~

Incos-Ix

18. y~zcos-Iz- ~

39.

>1

41.

sec-lVi

w tan

40.

1~

v

dv

(lnx)-'

/

42.

44.

- x - dx

46.

1 rcsC> (I

48.

1

'3

1'~ 2 cot 'ffX dx

1 W

tan (In v)

/

1'~ - x - dx I

~3 dx

49. / x3%' dx 51.

38.

__ 21_ dl o t 2 - 25

/

csc2 x e OOl %dx

36. /

, xdx

+ Inr) dr

6

cos I dt -wf2 1 - smt /

/

dv vlnv

/

In (x - 5) x 5 dx

/cos(I-lnv)

v

SO. / 2""% seC> x dx 52.

1'2 1

I

5x dx

dv

Chapter7

53.

54.1' (~ -~)

l' (~+ ~)dx

56.1

1

55.1- e --«+1) dx 57.

/.,"5

e'(3e'

59.1' ~ ' l 61.

(I

+ 1)-'/2 dr

+ 7lnxt l/'

58.

dx

60

(In (v + I)? v+1 dv

1

1'/ 1

"

65.

4

'19 -

4x 2

66.

6712~2

68

69 ·

J

dy yY4y 2

-

2/3

dy

70.

I

•

J

72. 74

dx

Y-2x-x 2

1-

•

1

75 •

77 •

-2

J

(t

105. lim

(i _1)t

4 _ 25x2

~

2

+

2dv

76

+ 4v + 5

•

--=-r~== - 16

dt I)Yt 2 + 2t - 8

78

•

:-::,~== Iy IV5y 2 - 3

Y-x 2

(3t

dx +4x-1

3 dv

+ 4v + 4

+

dt I)Y9t 2 + 6t

Solving Equations In Exerci,e, 79-84, solve for y.

79. 3" =

80. 4-> = 3"+2

81. ge'" ~ x 2

82. 3"

= x + Iny

~

84. In (lOlny)

= 1n5x

86.1im x ' - 1 x-I x b - 1

85.limx2+3x-4 x-I x 1

88. lim

x-o

2 8in x

+ smx

lim sec 7x cos 3x

x-n/2-

94. lim(1

93. lim (esc x - eotx)

x_a\;4

x~o

2

95. lim (Yx + X~OO

96

x

tan~

, sinmx 90• Iun . x-o smnx

89. 11m - ( x-o tan x 2) 91.

G'

X

+ I - Yx

lim _x 'x_oo

2-1

,)

x_ x2

+t

108. lim x-o+

2

-

x)

(I

3 )X + _x

= log,x,

g(x) = log,x I g(x) = x + X

e. f(x) ~ x/IOO, d. f(x) ~ x,

g(x)

e. f(x) = ese- l x,

g(x) = I/x

r.

g(x)

f(x)

~

~ xe~

g(x) ~ tan-I x

sinhx,

~

eX

110. Does f grow faster, slower, or at the same rate as g as x -+ oo?

Give reasons for your answers. .. f(x)

~ 3~,

g(x)

b. f(x) ~ 1n2x,

~ 2~

g(x) ~ Inx 2

e. f(x) = lOx' + d. f(x) ~ tan- l(1jx),

2x 2,

g(x) = eX g(x) ~ I/x

e. f(x) = sin- (l/x),

g(x) = l/x 2

r.

g(x) =

f(x) = seehx,

e~

111. True, or false? Give reasons for your answers.

3 Inx

Apply;ng L:H6pUal's Rule Use I'HOpital's Rule to find the limits in Exercises 85-108.

•

+ I )'"x

l

2"+1

83. In (y - I)

y-o+

b. f(x) = x,

dy

-21'15

4v 2

106. lim e-l/Y lny

Comparing Growth Rates of Functions 109. Does f grow faster, slower, or at the same rate as g as x -+ oo? .. f(x)

24dy

yVy 2

-1

+3_x

Give reasons for your answers.

1

V

2

.t_4ex4

eX - 1

x_OO

2

sin ('ll'x)

104. lim

t

107. lim (eX

I

6dx

1Y J I J 1 J

t-O+

102. lim x smx x-o tan3 x

+ 2t)

+ Int)tlntdt

-1/5

•

101. lim 5 - 5 eosx 1 x_oe x x t-O+

z-m" -

x-o eX-l

t - In (I 2 t

-V.M

71. /, :-::-r~= Yz/3ly IV9y2 - I 73

1)1/2 d6

8ln31og, 6 d6 6

{'

100. lim

x_oex-l

· lv'33 + t 2

• _24+3t

98. lim 3' - I 6-0 8

x

103. lim

1/5

6dx

x-o

dx

x~

64'1

-:;:~~

-'/4

l"-I-

97. lim lOX - I 99. lim 2'"" - I

dw

/.,"9 e'(e' _

' l

-6- d6

1

2w

e

62. j,4(1

'IOg., 6

63.

dx

0

-'"2

-2

x2

3x

1

431

Practice Exercises

-~) x2

.. ~2 + ~4 = O(~) 2 x e. x

~

b.

x x o(x + Inx)

~2 + ~4 = O(~) 4

x x d. In (lnx)

e. tan-I x ~ 0(1)

~

x o(lnx)

f. eoshx ~ O(e X )

112. Tme, or false? Give reasons for your answers.

"~~O(~+~) x4 x2 x4

b.~~o(~+~) x4 x2 x4

c. Inx = o(x + I)

d. 1n2x = O(lnx)

e. see-l x = 0(1)

f. sinhx = O(e X )

Theory and Applications 113. Thefimctioof(x) = eX + x,heingdifferen1ishleaodone-to-one, has a differeotiable inverse (x) , Find the value of /dx at the point f(ln 2),

rl

drl

114. Find the inverse of the fimctioof(x) = I + (I/x),x # O.Thea showthatr l (f(x)) ~ f(r l (x)) ~ x and that

drll dx

__1_ ft. 0).

119. A particle is traveling upward and to the right aloog the curve y = Inx. Its x-coordinare is increasing at the rate (fix/dt) = VX 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. -xl' . Her y-coordinare is changiog at the rate dy/dt = ( -1/4)~ ft/sec. At approximarely what rate is her x-coordinate changiog when she reaches the bottom ofthe slide at x = 9 ft? (Take.' to be 20 and rouod your answer to the nearest ft/sec.) 121. 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 = e -r. What dimensions give the rectangle its largest area, and what is that area? y

In Exercises 129-132 solve the initial value problem.

.-'-Y-',

129. :

=

dy 130. fix

=-

ylny

I

+x

"

y(O) = -2

=•

y(0)

,

131. xdy - (y

+ vY) fix

132. y-' dyfix =

,f-, y(O) = • +I

= 0, y(l) = I

1

133. What is the age ofa sample ofcharcoal in which 90% of the carbon-14 origioally present has decayed? 134. Cooling a pie A deep-dish apple pie, whose inremal 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 teruperatore was 180°F. How long did it take the pie to cool from there to 70°F? 135. Locating a .olar .tation You are under cootract to build a solar .tation at grouod level on the east-we.t 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

cot

x

C1 50 -

60 - co

x

~.

Then fmd the value of x that maximizes 8.

DO

~+------'---'---=-. 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 dimensions give the rectangle its largest area, and what is that area?

::~

0.2 0.1

o

,_

..

~

~I

II

>x

D 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)/VX b. y ~ .-.' e. y ~ (l + x).~

D 124. Graph f(x)

= x In x. Does the functioo appear to have an absolure minimum value? Cnofmn your answer with calculus.

DD

DO DO

o

x

SOm

136. A rouod underwater transmissioo cable consists ofa core ofcopper wire. surrounded by nonconducting insulation. 1f x denores the ratio of the radius of the core to the thickness of the insulation, it is known that the speed ofthe transmission signal is given by the equation v = x'in (I/x). Ifthe radius ofthe core is I em, what insulation thickness h will allow the greatest transmission speed? Insulation x= !.h

Chapter 7 Additional and Advanced Exercises

'f}

Chapter


Limns

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.

1. lim

[b

b-l-)O

2. lim} {' tan-I r dt

dx

~

%_00

%-0+

Jo

+ .')'/'

4. lim (x

3. lim (cos VX)I/,

%_00

5. lim (_1_ 11_00 n+ 1

+n+2 +... +~) 2n

• I) g(x

x

6. lim 1 (ellA + n_oon

+ ... + e(n-l)/II + en/II)

volving the region about the x-axis. Find the following limits.

b. lim V(t)/A(t)

1_00

c. lim V(t)/A(t) 1-0+

/-+00

8. Varying a logarithm's base a. Findlimlog.,2aso .... 0+, 1-, 1+, and

0< a

00.

= log., 2 as a function of a over the interval

:5

4.

of the larger area to the smaller?

010. Graph j(x) = tan-I x + tan-l(l/x) for -5

x :5 5. Then use calculus to explain what you see. How would you expect j to behave beyond the interval [-5,5]? Give reasons for your answer. :5

0 does x(r) = (x')'? Give reasons for your

answer.

012.

a. Showthatg(O) = O.

b. Showthatg'(x) = I

+ [g(x)]'.

c. Find g(x) by solving the differential equation in part (b).

17. Center of mass Find the center of mass of a thin plate of constant density covering the region in the firat and fourth quadnlnts enclosed by the curves y ~ 1/(1 + x') and y ~ -1/(1 + x') and by the lines x = 0 and x = I. 18. Solid of revolution The region between the curve y = 1/(2VX) and the x-axis from x ~ 1/4 to x ~ 4 is revolved about the x-axis to generate a solid.

b. Find the centroid ofthe region.

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 = •. What is the ratio

>

=I

h

a. Find the volume of the solid.

Theory and Examples

11. For what x

+ g(y)

il) lim g(h) = 0 iii) lim g(h)

7. LelA(t) be the area ofthe region in the IITst quadrant enclosed by the coordinate axes, the curve y = e -x, and the vertical line x ~ t, t > O. Let V(t) be the volume of the solid generated by rea. lim A(t)

g(x)

= I _ g(x)g(y) forallrealnumbers x,y, and

.~o

_1_

e2/A

+ y)

+ y in the domain ofg

h-O

o b. Graph y

433

Graphj(x) = (sinx)"nxover[O, 31T]. Explain what you see.

13. Findj'(2) if j(x)

~ • .<x) andg(x) ~

],'_t-4 dt. , I +t

14. a. Find dfjdx if j(x) =

19. The best hranching angles for hlood vessels and pipes 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 SOD2C energysaving point of view. We might require, for instance, that energy loss due to friction be minimized aloog the sectionAOB showo in the accompaoylng figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe upstream from B, and 0 is the point where the branchiog occurs. A law due to Poiseui11e states that the loss of energy due to friction in nontorbulent flow is proportional to the length of the path and inversely proportional to the fourth power of the radius. Thus, the loss along AO is (kd l )/R 4 and along OB is (kd,)/r 4 , where k 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:

['2~tdt.

dl

d,

L~k4+k4' R r

b. Find j(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) = ofor allxandh(x) = ofor allx. b. If j(x) = h (x) + j 0 (x) is the sum of an even function h(x) and an odd function jo(x) , then show that

h(x) = j(x)

+ j( -x) 2

and

jo(x) = j(x) - j( -x) 2

c. What is the sigoificance of the result in part (b)?

A l d _

1 22.

29.

cos' 4x dx

10.

1.

'"/2

3 sffi 3 dx

sin' 2x cos 2x dx

cos3 xdx

7

. x

J J f sins~dx

8.

sin yth>

21.

[~

sinS xdx

[~/2

19.

27.

4tan'xdx

46.

~/'

J J J J J J

sec x tan2 X dx sec' x tan' x dx

sec' x tan2 X dx

eX sec' eX dx

3 sec' 3x dx 6

sec

1

~/'

X

dx

6tan'xdx

~/'

J J

cot' 2x dx

8 cot" tdt

8.3 Products of Sines and Cosines Evaluate the integrals in Exercises 51-56.

J

52.

J

53.

J~ sin3xsin3xdx

54.

J.1T/2 sinxcosxdx

55.

sin 11: cos 2x dx

J

cos 11: cos 4x dx

56.

sin 2x cos 11: dx

59.

61.

J J J

sin' 8 cos 38 d8

58.

cos' 8 sin 28 d8

60.

sin 8 cos 8 cos 38 d8

62.

63

•

J

sec x dx tan x

8.3

64.

esc x dx

66.

x sin' x dx

68.

J J

449

cotx dx

cos2 x

x cos' x dx

Applications

J~; cos x cos 7x dx

69. Arc length

Find the length of the curve

y

~

In (secx),

0:5 x:5 ",/4.

cos' 28 sin 8 d8

70. Center of gravity Find the center of gravity of the region bounded by the x-axis, the curve y = sec x, and the lines x = -",/4,x = ",/4.

sin' 8 cos 28 d8

71. Volume Find the volume generated by revolving one arch of the curve y = sinx about the x-axis.

J J J

72. Area

VI

sin 8 sin 28 sin 38 d8

J

3

sin

X

cos4 X

+

Find the area between the x-axis and the curve y cos4x,O ~ x ~

~

'IT.

73. Centroid Find the centroid of the region bounded by the graphs ofy = x + cos x andy = o forO ::=:; x::=:; 2'7T.

Assorted Integrations Use any method to evaluate the integrals in Exercises 63--{)8. 3

tan' x

67.

Exercises 57--{)2 require the use of various trigonometric identities before you evaluate the integrals.

57.

J J

65.

51.

Trigonometric Substitutions

74. Volume Find the volume ofthe solid formed by revolving the region bounded by the graphs of y = sin x + sec x, Y = 0, x = 0, and x ~ ",/3 about the x-axis.

dx

Trigonometric Substitutions Trigonometric substitutions occur when we replace the variable of integration by a trigonometric function. The most common substitutions are x = a tan IJ, x = a sin IJ, and

x = a sec IJ. These substitutions are effective in transfurming integrals involving Va 2 + x 2 , Va' - x 2 , and Vx' - a 2 into integrals we can evaluate directly since they come from the reference right triangles in Figure 8.2. Withx = atanlJ,

a2

+ x2

=

a2

+ a 2 tan2 1J

x'

=

a2

-

= a 2 (1

+ tan2 1J)

= a 2 sec2 1J.

Withx = asinlJ,

a2

-

a 2 sin2 1J

=

a'(l - sin'lJ)

=

a' cos2 1J.

a x=atan8

x=asin8

x=asecO

Va' + x' ~ aisecOI

Va' - x' ~ aleosOI

Vx'- a'~ aitanOI

FIGURE 8.2 Reference triaogles for the three basic substitutions identifying the sides labeled x and a for each substitution.

450

Chapter 8: Techniques of Integration Withx = asecO,

x2 6

"2

-

a2

= a 2 sec'0 -

a2

= a 2 (sec2 0 -

I) = a 2 tan2 0.

We want any substitution we use in an integration to be reversible so that we can change back to the original variable afterward. For example, if x = a tan 0, we want to be able to set = tan-I (x/a) after the integration takes place. If x = a sin 0, we want to be able to set 0= sin-I (xja) when we're done, and similarly fur x = asecO. As we know from Section 7.6, the functions in these substitutions bave inverses only for selected values of 0 (Figure 8.3). For reversibility,

o

x

a

-2 "

x

=

(~)

with

-2 " < 0 < 2' "

sin-I ("') a

with

-2';; 0 ,;; 2'

with

1:Ofor-~5'

452

Chapter 8: Techniques of Integration to put the radicand in the form x 2

x 2 x -

-

a 2 • We then substitute

2

2

dx = Ssec 0 lanO dO,

= Ssec 0

(~)' = =

is

- I) =

25

~lan20 25

~X2 (~)' ~ ItanOI ~lanO. =

-

j

dx

v'25x 2 -

FIGURE 8.6 Ifx = (2/5)sec6, 0< 6 < 'IT/2,then6 ~ sec- 1(5x/2), and we can read the values ofthe other 1rigonometric functions of6 from this right 1riangle (Example 3).

4 = =

=

j

dx

5v'x 2 -

(4/25) =

tj secOdO tin tin I~ + v'25~2

(2/5) sec 0 lan 0 dO 5' (2/5) lanO

IsecO

=

> ofor

tan 6

0< 0 < ",/2

=

With these substitutions, we have

j

Vy2 - 49 Y dy, Y

fix 13. • ~, 1 X 2 VX 2 - 1

1VI

3fix

21.

+ 9x 2

2 4.1. fix 2 oS+2x

3.12~2 -24+x 5. 1.0

•

x

-7 2

10.

> 7 12.

>

1

14.

1. ' /2Vi 0

1 1 1 1

2fix

5fix , x V25x 2 - 9

> -3 5

dy, Y

>5 31.

x

>

1

Assorted Integrattons Use any method to evaluate the integrals in Exercises 15-34. Most will require 1rigonome1ric substitutions, but some can be evaluated by other methods. x2 16. --2fix 4+x 9 - x2

15·1~fix . 1W+4

17

x'fix

18.

1 1

29

•

• 2fix ~, 2 X 3 Vx - 1

fix X2~

0

2 fix '/2' (x - 1)

33 ·

1 1 1 1

x

6

Sfix

(4x 2

22.

4x2 fix (1 _ x 2 )'/2

(1 - x 2 )'/2

27.

20·1~dw

w2

100 fix 36 + 25x 2

25.1

v'l=9?dt

y'

Sdw

w2

V3/2

23.

VI _ 4x2

V y 2 - 25

1 V4 1 1. +

x> 1

1

x Vx 2 - 4 fix

1 fix 24. 1.0 (4 _ x2)'/2

26.

1(x,x~~)'/2' 1 1 1 1 (I-x 2 )1/2

fix

1)2

x' fix x2 - 1

28.

30 • 32 •

4

34.

fix

6dt (9t 2 + 1)2

25

xfix 2 + 4x

(1 - r 2 )'/2

2

v dv (I - v 2 )'/2

x

x> 1

r

8

dr

In Exercises 35-4S, use an appropriate substitution and then a 1rigonome1ric substitution to evaluate the integrals.

35 •

1.

ln4

37.

I dt

Ve'"

0

['/

e

4

+9

2 dt

11/12 vi + 4tvl

36

•

lin

(4/')

In ('/4)

t dt

_-"e~~

(I

+ e,")'/2

38 {' dy . yVI + (lny)'

1,

8.4 Integration of RationaL Functions by PartiaL Fractions

.. J J · '\17=1 J · 'V1+? . . J)4 ;x JVx~

I ) : = 2V3,

+ 21): ~

+ I):

= I

2x

= x2

+2

(I,x

2),

x(3) = 0

0),

x(l)

+ I (I > -I), x(O)

~

5

59. Social diffusion Sociologists s0n3ctimes use the phrase "social diffusion" to describe the way information spreads through a population. The information might be a rumor, a cultural fad, or news about a tecbnica1 innovation. In a sufficiently large population, the nU03ber of people x wbo bave the information is treated as a differentiable function oftime I, and the rate of diffusion, dx/dl, is asSU03ed to be proportional to the number of people who bave the information times the number of people wbo do not. This leads to the equation

dx ~ kx(N - x) dl ' where N is the nmober of people in the population. Suppose I is in days, k = 1/250, and two people start a TUmor at time I = 0 in a population of N = 1000 people.

x(1) = -1fV3/4

>

3

a. Find x as a function of t. I

b. When will balf the population bave heard the rumor? (This is when the rumor will be spreading the fastest.)

o 60. Second-order chemical reactions

= 0

Appllcatfons and Examples In Exercises 55 and 56, fmd the volwne of the solid generated volving the shaded region about the indicated axis.

by re-

Many chemical reactions are the result of the interaction of two molecules that undergo a cbange to produce a new product. The rate of the reaction typically depends on the coocentrations of the two kinds of molecules. If a is the amount of substance A and b is the amount of substance B at time I = 0, and ifx is the amount ofproduct at time I, then the rate of formation of x may be given the differential equation

by

55. The x-axis y~

y

~

k(a - x)(b - x),

or

I dx ~ k (a - x)(b - x) dl '

2

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

o

dx dl

3

Y3x-x 2 (0.5, 2.68) (2.5, 2.68)

0.5

2.5

where k is a constant for the reaction. Integrate both sides of this equation to obtain a relation betweeo x and I j:=

x A2 * sqrt (a A2

+ x A2);

Then you use the integrate command on j, identifYing the variable of integration:

>

int(j,x);

Maple retorns the answer

t

x(a 2 + x 2)3/2 - ta2xYa2 + x 2 - t a4ln (x + Ya 2 + x 2).

If you want to see ifthe answer can be simplified, enter

>

simplify(%);

Maple retorns

ta2xYa2 + x 2 + tx3Ya2 + x 2 - ta4ln (x + Ya 2 + x 2). If you want the def'mite integralfor 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 4ln(2) 64

32

8

- ta4ln(1T + Y4a 2 +,.2) + l~a4ln(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 + tln(YZ EXAMPLE 6

Solution

•

I).

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

+ I~

cos(xj2sin(x)

+

2 15 sin(x).

•

Compoter algebra systems vary in how they process integrations. We used Maple in Examples 5 and 6. Mathematica would have returned somewhat different results:

466

Chapter 8: Techniques of Integration

1.

In Example 5, given

In [IJ:= 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]'2' 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 ofrandom 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

x

dx ,

f

e (.'j d!:,

f

f~

d!:,

VI - k'sin2 xd!:,

f

In (In x) d!:,

0

f

sinx d!: x '

< k < I,

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 I, because they are continuous. However, none ofthe antiderivatives are elementary. None ofthe integrals you are asked to evaluate in the present chapter fall into this category, but you may encounter nonelementary integrals in your other work.

8.5 Integral Tables and Computer Algebra Systems

467

Exercises 8.5 Using Integral Tables

2

Use Ibe table of integrals at Ibe back oflbe book to evaluate Ibe integrals in Exercises 1-26.

1/ ·

2/ · xy';+4

dx x~

38.

Yx 2 -

/

4x

+5

dx

Y5 - 4x - x 2 dx

39. /

x 2 Y2x - x 2 dx

40. /

dx

x

Using Reduction Fonnulas

xdx

4 /

xdx (2x + 3)'/2

Use reduction fonnulas to evaluate Ibe integrals in Exercises 41-50.

5./x~dx

6. /

x(7x + 5)'/2 dx

41. /

7·/7

8 / dx • x 2Y4x - 9

3 /

·

9. /

11 /

·

13.

\Ix=2

•

dx

xY4x -x2 dx

10.

/~dx dx x~

12 /

dx xv7+7-

·

/~ dx

14.

/~dx

15. /

e 2t cos 3t dt

16. /

e-3t sin 4t dt

17. /

X

cos-1 xdx

18. /

xtao-l xdx

19. / 21. /

X

2 tao- l xdx

sin 3x cos 2x dx 8 sin 4t sin

25.

8 8 coS:3 cos 4 d8

/

24•

. t . t d

Sffi

/

3

S1n"6

t

cOS~COS78d8

26. /

31. / 33. /

34./ 36. /

2 sin2 t sec' t dt

45. /4tao'2xdx

46. /8COt'tdt

47. /

2 sec' 7fX dx

48. / 3 sec' 3x dx

49. /

csc' x dx

50. /

e'sec' (e' - I)dt

l' 0

(x 2 + 1)2 dx

28.

l

sin- ...;xdx

...;x

~

dx

cottYI - sin2 tdt,

0

/X2+6x (x2

+ 3)' dx

30. /

cos-:rx...;x dx

32. /

~ dx ...;x

oo

The value we assign to the area under the curve from 0 to 00 is J.o

oo e--./2dx =

lim J.b e--./2 dx

=

2.

b_oo 0

y

DEnNmON Integrals with inf"mite limits of integration are improper integrals ofType I. 1. If f(x) is continuous on [a, 00), then

---+-----====--=-=.x

1

00

a

1 and diverges if p :5 1.

EXAMPLE 3

For what values ofp does the integral tegral does converge, what is its value?

Solution

If p

I

1

'¢'

bdx -

xP

J,"" dx/x

P

converge? When the in-

1,

=

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' { 00,

P

< 1

because lim _1_ = {O,

b-oo bP - 1

00,

p>1 p 0+:

Therefore the area under the curve from 0 10 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 ofType II. 1. If fix) is continuous on (a, bl and discontinuous at a, then b b fix) ax = lim+i fix) ax.

i i a

C-Q

c

2. If fix) is continuous on [a, b) and discontinuous at b, then b

a

fix)

ax =

c

lim_i fix)

c-b

a

3. If fix) is discontinuous at e, where a [a, c) U (e, bl, then [

fix)

ax =

[f(X)

ax +

evalf(%, 6);

The symbol % instructa the computer to evaluate the last expression on the screen, in this case (-1/2)". + In (5) + arctan (2). Maple returns 1.14579. Using Mathematica, entering

In [1]:= Integrate [(x

+ 3)/«x

- l)(x'2

+

I)), {x, 2, Infinity}]

returns

Out [1]=

-".

2

+ ArcTan [2] + Log [5].

To obtain a numerical result with six digits, use the command ''N[%, 6]"; it also yields

1.14579.

Tests for Convergence and Divergence

y

When we cannot evaluate an improper integral directly, we try to determine whether it converges or diverges. If the integral diverges, that's the end of the story. If it converges, we can use numerical methods to approximate its value. The principal tests for convergence or divergence are the Direct Comparison Test and the Limit Comparison Test.

EXAMPLE 6 Solutfon

--;cr---:---~~-~x

FIGURE 8.19 The graph of e~' lies below the graph of e ~ for x > I (Example 6).

Does the integral

.hoo e --i' ax converge?

By definition,

We cannot evaluate this integral directly because it is nonelementary. But we can show that its limit as b --> 00 is f"mite. We know that fIb e --i' ax is an increasing function of b. Therefore either it becomes infinite as b --> 00 or it has a finite limit as b --> 00. It does not become infinite: For every value ofx 2: I, we have e--i' :5 e-x (Figure 8.19) so that

l

b

e-i' ax:5

l

b

e-x ax =

-e-b + e-1 < e- 1

'"

0.36788.

Hence,

converges to some definite f"mite value. We do not know exactly what the value is except that it is something positive and less than 0.37. Here we are relying on the completeness • property ofthe real numbers, discussed in Appendix 6.

484


The comparison of e --i' and e-x in Example 6 is a special case of the following test.

THEOREM 2-Dtred Comparison Test Let with 0 ,;; f(x) ,;; g(x) for all x ;" a. Then


Karl Weierstrass (1815-1897)

1.

to

f(x) dx

converges if

diverges if

2. [ " g(x) dx

f

and g be continuous on [a, 00)

converges.

[ " g(x) dx

diverges.

[ ' f(x)dx

Proof The reasoning behind the argument establishing Theorem 2 is similar to that in Example 6. If 0 ,;; f(x) ,;; g(x) for x ;" a, then from Rule 7 in Theorem 2 of Section 5.3 we have

b

> a.

From this it can be argued, as in Example 6, that

to to

f(x) dx

[X'

converges if

Turning this around says that

EXAMPLE 7

g(x)dx

to

diverges if

g(x)dx

converges.

f(x)dx

diverges.

•

These examples illustrate how we use Theorem 2.

2

(a) JOO sin x dx

converges because

x2

1

on

[1,00)

OOI

J

and

2dx

x

1

converges.

Example 3

diverges.

Example 3

OO

(b)

J Vx 1

1

Vx 2 -

I

2 -

dx

diverges because

0.1 >

0.1 -

1

x

on

OO I

[1,00)

J

and

1

If the positive functions f and g are con-

THEOREM 3-Llmlt Comparison Test tinuous on [a, 00), and if lim f(x) = L g(x) ,

X~OO

then

1

xdx

00

O -00

1

3

Lower endpoint

j

-dx-

a

Y~r;: 1

~-I)'"

+ x2

1

3

dx (X - 1)'/3

r

=

e

dx

d~+1d (X - 1)'/3

Y Y~

Y

1

_

1

1 (x -

I)'"

Y- 1 +-x

~=-C+----=-"""',,->x

o

6. Interior point

3. Both limits

e

10

dx (x - 1)2/3 =

t

10

dx (x - 1)'/3

+

j3 1

Y Y

1

Y~

1 (x -

I)'"

~O+--~---~x

dx (X - 1)2/3

8.7 Improper Integrals

487

Exercises 8.7 Evaluating Improper Integrals Evaluate lhe integrals in Exercises 1-34 without using tables.

1 •

ax

{I

Jo

5.

1

ax

13

6.

-1 X

ax

8

Jo VI==7

100 ], 100

2ax

-2

-oox-l 2 -2--dv 2 v - v

2xax 2

17

8

Jo

19

ax

roo

· Jo

(I + x)Vi

dv

roo

· Jo (I + v 2)(1 + tan Iv) 21. J~ 8e'd8 23.

J~ e-i>1ax 2

27 •

ax

2

29 •

1, 1 v'fxT 1

tis

~

31 •

•

-1

B'

. 11roo ~ x3 + 1

e

12.],00~ 2

51

roo

ax

52],00

ax

2

100

t

+4

'2~

1

-

+ I tis

Jo v'4=7

18.

roo

Jl

ax

I

+ Xl

22.],00 2e -IJ sin 8 d8 -x'

ax

30 •

34

roo

00

100 1

· Jo

62.

Vex - x

ax

00

~W+1

0

Jl x(inx)"

1>.

2X dx

ax

e'+ e ~

],00 ax

x(inx)"

2

,~oo

roo

J!,/(x)dx

Show that

2xax

Jo x2 + 1 diverges and bence that

00 2xax

1 l' 2xax ~ o. --00

x2

+

1

diverges. Then show lhat lim

b_oo -b x 2

+1

Exercises 67-70 are about lhe infmite region in lhe first quadrant between the curve y = e ---x and the x-axis.

cot8d8

cos 8 d8

Vi

~

66. J~oo/(x)dxmaynotequal lim

{~/2

],1 _e--Vi ax

1

64.

(' ax

~/2 (". - 28) /3

40.

itroo eX _I

I

.~ax

(x + 1)(x + I)

Jo

x

I

],2 inx ax 60. [00 in (in x) ax

59. Jl ~ ax 61.

00

58.

roo ,

a.

ax 2

38.1~/2

],o

2 dt t 3/2-1

'o~

sin8d8

e-1/, ax

+;osx dx

65. Find lhe values ofp for which eacb integral cooverges.

t'\J'12=4 2 ax ],

37. r

-2

54],00 xax • 2 W-=1 00 56.1 I + ~inx ax

Theory and Examples

2

36.

ln2 x

2

/.00

4rdr

tan8d8

39.

57.

63.

Jo v'1-=-7 ],' dt

Jo

~

00

1

J:" 2xe

'2~

xl

,

20. roo 16 tan-I x ax 1

it 55.1

roo ~ 1+ 6

~

x~

10

. Jo W+1

53. roo~ax

xax

16. {2 s

r/2

Jo

· , Vi-I

· Jo

x2

Testing for Convergence In Exercises 35-64, use integration, the Direct Comparison Test, or lhe Limit Coroparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. 35.

/.00 ax

48

50

32

+6

46.1:-xinlxlax

dv

-1

d8 + 58

Jol - x

49],00

--00

00

33

{' ax

44·

o 1- x

47

28. {'

Is~

ax

--2

],

~

26.],1 (-inx) ax

v'4=7

], 0

(Hint:t",=sintfort"'= 0)

10.1 ~

24.

25.],1 xinxax

dt.

t-smt

45. 1: in Ixlax

• ~ (x 2 + 4)3/2

+ I d8 vB' + 28

{I

sint

X

• ]0, rO. 999

14

• ~ (x + I)'

15.

ax

2

-2--

Vi +

2

43.

1;3

-s

1

10

ax

1

dt

Jo

42. {I

~

1

2j3

{I

11.

{'

4· Jo

Vi 1

9.

2-. 1 xl. OOI

+1

x

0

3.

7.

100 ax

ax],00 2

41. {~

1

67. Find the area of the region. 68. Find lhe centroid of the region. 69. Find the volume of the solid generated by revolving the region

about the y-axis.

488


70. Find the volume of the solid generated by revolving the region about the x-axis.

D

71. Find the area of the region that lies between the curves y = sec x andy = tan x from x = Otox = Tr/2.

b. Explore the convergence of

72. The region in Exercise 71 is revolved about the x-axis to generate a solid. a. Find the volume of the solid. b. Show that the inner and outer surfaces ofthe solid have infinite area. 73. Estimating the value of a convergent improper integral whose domain is infinite a. Show that

1

00

and hence that

e-3x dx =

te-9
1,

12Tri~1

b

2Tr

=

The function

- -1- 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

1

b

+ :4dx >

2e-t2

o

74. The infinite paint can or Gabriel's hom As Example 3 shows, the integral f,oo (dx/x) diverges. This means that the integral

1

2

2 -t V;dt,

a. Plot the error function for 0 ,.;; x ,.;; 25.

x2

00

r

Jo

b. Explore the convergence of

2

e- dx numerically.

,

The function erf(x) =

dx without introducing an error ofmagnitude greater than 0.000042.

D

dt.

called the error function, has important applications in probability and statistics.

2

2

1 s~t

0.000042,

hoo e -x dx < 0.000042. Explain why this

roo means that Joe -x

a. Plot the integrand (sin t)/t for t > 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.

idx.

1:

y

f(x) dx

=

1.

In what follows, let f.L = 0 and u = 1.

o

D

--...I -_ I

1

---

------------

----

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 ofrevolution 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

> 1, and for b > 1,

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)
00 -->:

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 lineA" - e.

;tl1c~tdt x-->o+ t

sintdt

8. lim

%

Evaluate the limits in Exercises 9 and 10 by identifying them with definite integrals and evaluating the integrals.

9. lim

±1n41

n-->00 1-1

+~

Applfcattons 11. Finding an: length

10. lim

~ Vn 2I-

.......00 .1;-0

t2

Find the length of the curve

y -1% Vcos2tdt,

O:!;';;t:!;';

'lr/4.

21. Centmid of a region Find the centroid of the region in the plane enclosed by the curves y - ±( I - ;t2t1/2 and the lines ;t - Oand;t - 1. 23. Length of. curve ;t-lto;t-e.

Find the length of the curve y - In;t from

24. Finding mrfaee area Find the area of the smface generated by revolving the curve in Exercise 23 about the y-axis. 25. The.urfaee poenrted by an astroid The gnlph of the equation ;t2/3 + y2/3 _ I is an a.ftroid (see accompanying ftgure). Find the area ofthe surface generatM by revolving the curve about the x-mds.

,

11. Finding an: lengtb Find the length ofthe graph of the fimction y -In(l _;t2), O:!;'; x:!;'; 1/2.

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. 14. FIndIng volume The region in the first quadrant that is enclosed by the".ui>, the """" ~ '/(x~). ud the tim", ~ 1 and ;t = 4 is revolved about the »-axis to gwerate a solid. Find the volume of the solid.

y

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 tile 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 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.

17. Finding volume Let R be the ''t:riangular" 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 volume of the solid generated by revolving R about L the ;t-axUt. 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-lIXis and the curve

is revolved about the »-axis to generate the solid shown here. L

Show that f is continuous at;t = O.

-1

26. Lengthofaclll'ft Find the leug1h of the curve

y= [%v'Vt-ldt.

l:!;';xsI6.

27. For what value or values of a does

[G::,-~)dt converge? Evaluate the corresponding integral(s). 28. For each ;t > O. let G(;t) for each x > O.

10

00

e ~ dt. Prove that ;tG(;t) - 1

29. IDf'Ulite area lIDd (mite wlume What values of p have the following 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 :follow.. iog 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 in:finile but the voluIne of the solid genented by revolving the region about one of the COOIt1ina1J; axes is finite. The S.mm. Function .nd Stlrlfng's Fonnula Euler's gamma ftmction f(;t) ("gamma ofr'; f is a Greek capital g) uses an integral to extend the factorial function ftom. the nonnegative integers to otheI' real. values. The formula is

b. Find the volume of the solid.

,

_1 r 00

o

,

f(;t)

1e-t dt,

;t> O.

For each positive;t. the number f{x) is the integral oft r - 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

"'\Yn! '" ~.

o b. 2

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 refmement of Equation (2) gives

--+-±---+---;;+----:--:-----;;-~x

o

3

-I

f(x) =

-2 -3

(~Y l#e1/(l"')(1

FIGURE 8.21 Euler's gamma function f(x) is a continuous function of> whose value at each positive integer n + 1 is n!' The defIning 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 + I»/x, which is the subject of Exercise 31. 31. H n is a nonnegative integer, f(n

+ 1)

(~Y l#e1/(l"'), (6)

Cmnpare the values given for IO! by your calculator, Stirling's approximation, and Equation (6). Tabular Integratton The technique of tabular integration also applies to integrals of the fonn f(x )g(x) fix when neither function can be differentiated repeatedly to become zero. For example, to evaluate

J

= n!

b. Then apply integration by parts to the integral for f(x show that f(x + I) = This gives

xf(x).

+

J

I) to

eh"cosxdt

f(2)

~

If(l)

~

I

we begin as before with a table listing successive derivatives of e'lx

f(3)

~

2f(2)

~

2

and integrals of cos x:

f(4)

~

3f(3)

~

6

+ I)

= nf(n) = n!

(1)

Use mathematical induction to verify Equation (I) for every nonnegative integer n.

32. Stirling's formula Scottish mathematician James Stirling (1692-1770) showed that

e2" and its

cos x and its

derivatives

integrals

e"'~

coox

2o"'~inX (+) ~

4e2¥

-cosx

<E-

so, for large x, .(x) --+ 0 aox --+

DC.

(2)

f1rst row except for multi-

(~)" VZ;;;

J

~ +(e"'sinx) +

(Stirling's formula).

(3)

a. Stirling's approsimation for n! Use Equation (3) and the fact that n! = nf(n) to show that

n! '"

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 interpret the table as saying

e"'cooxfix

Dropping .(x) leads to the approximation

(21i (ex)X "l/x

Stophere:Rowissameas

plicative constants (4 on the left, -I on the right).

x~('iYfIrf(X)= I,

f(x) '"

.(x»

which tells us that

a. Show that f( I) = I.

c.

+

or f(x) '"

r(n

(5)

(Stirling's approIimation).

(4)

(2o"'(-coox»

J

(4e"')( -cos x) fix.

We take signed products from the diagonal arrows and a signed int.,. gra1 for the last horizontal arrow. Transposing the integral on the righthand side over to the left-hand side now gives

5

J

e 2x cosxdx

=

e2t"sinx

+ 2e2xcosx

494


or

.

/

. " dx -- e" smX' +5 2" e cosX' e cosx

2z

(9)

smx=--2' I +z

+ C'

Finally, x = 2tan-1 z,so

after dividing by 5 and adding the constant of integration. Use tabular integration to evaluate the integrals in Exercises 33-40. 33. /

e"cos3xdx

34. /

35. / sin 3x sinxca

(to)

e"'sin4xdx

Examples

s/

36. / cos5xsin4xdt

37. /

e~sinbxdx

38. /

e~cosbxdx

39. /

In (ax) dx

40. / x 2 1n (ax) dx

.

I dx~/I+z'~ I + cosx 2 I + z'

=/dz=Z+C = tan

The Substitution z

~

(f)

+ C

tan (x/2)

The substitution

z = tan~ 2

(7)

b. /

I dx = / 2 + sinx

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 accompanying figure

I

+ z2

2dz

2 + 2z + 2z 2 1 + z2 z' +

~+ I = /

(z +

(I/~)2 + 3/4

=/,;~,; =

~tan-l (~)

+ C

= ~tan-12z + I + C y!3 y!3 2 _II + 2 tan (x/2) =y!3tan y!3 +C

Use the substitutions in Equations (7}-{IO) to evaluate the integrals in Exercises 41-48. Integrals like these arise in calculating the average aogu1ar velocity of the output shaft of a uoiversal joint when 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

W/2

43. cosx

~ 2cos2 (f) - I ~ sec2~X/2) 1-

o /.

I

w/2

45. /. 0

2 _1=_2_ _ 1 I + tan2 (x/2) I + z2 cosX'

I

z2

= --2' I +z

(8)

dx.

- smX'

dx . 1 + smx dO 2+cosO

47. / . dt smt - cost

~I~~. __dx"'"-c-_ _ + SInX' + cosX'

42. /

44.

46

l

W/2

,"/3

1

dx 1 - cosX'

2w /3 cos 0 dO • '"/2 sinO cos 0 + sinO

48. /

costdt 1 - cost

Use the substitution z = tan (0/2) to evaluate the integrals in Exercises 49 and 50.

and

. . x x sin (x/2) 2 (x) smx = 2 sm 2cos2 = 2 cos (x/2) • cos 2 = 2tan~. I 2 sec2 (x/2)

2 tan (x/2) I + tan2 (x/2)

49. /

secOdO

50. /

cscOdO


Chapter


Mathematica/Maple Modules: Riemann, Trapezoidal, and SimpsonApproximations Part I: Visualize the error involved in using Riemann sums to approximate the area under a curve.

Part II: Build a table of values and compute the relative magnitude of the error as a function of!be step size 4>:. Part III: Investigate !be effect of the detivative function on !be error. Part. IV and V: Trapezoidal Rule approximations. PartVI: Simpson'. Rule approximations.

Games o/Chance: Exploring the Monte Carlo Probabilistic Technique/or Numerkal Integration Graphically explore the Monte Carlo method for approximating def"mite integrals. Computing Probabilities with Improper Integral. More explorations of the Monte Carlo method for approximating definite integral•.

495

9 FIRST-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In Section 4.7 we introduced differential equations of the form dyjdx = f(x), where f is given and y is an unknown function of x. When f is continuous over some interval, we found the general solution y(x) by integration, y = f(x) dx. In Section 7.4 we solved separable differential equations. Such equations arise when investigating exponential growth or decay, for example. In this chapter we study some other types offirst-order differential equations. They involve only first derivatives of the unknown function.

J

9.1

Solutions, Slope Fields, and Euler's Method We begin this section by defining general differential equations involving first derivatives. We then look at slope fields, which give a geometric picture ofthe solutions to such equations. Many differential equations cannot be solved by obtaining an explicit formula for the solution. However, we can often find numerical approximations to solutions. We present one such method here, called Euler s method, upon which many other numerical methods are based.

General First-Order Differential Equations and Solutions A first-order differential equation is an equation

dy dx = f(x,y)

(1)

in which f(x, y) is a function of two variables defined on a region in the xy-plane. The equation is of first order because it involves only the first derivative dyjdx (and not higher-order derivatives). We point out that the equations

y' = f(x,y)

and

d dx Y = f(x,y)

are equivalent to Equation (I) and all three forms will be used interchangeably in the text. A solution of Equation (I) is a differentiable function y = y(x) defined on an interval I ofx-values (perhaps infinite) such that d dxy(x) = f(x,y(x))

on that interval. That is, wheny(x) and its derivative y'(x) are substituted into Equation (1), the resulting equation is true for all x over the interval I. The general solution to a firstorder differential equation is a solution that contains all possible solutions. The general solution always contains an arbitrary constant, but having this property doesn't mean a solution is the general solution. That is, a solution may contain an arbitrary constant without being the general solution. Establishing that a solution is the general solution may

496

9.1

Solutions, Slope Fields, and Euler's Method

497

require deeper results from the theory of differential equations and is best studied in a more advanced course.

EXAMPLE 1

Show that every member of the family of functions

y=t+ 2 is a solution of the first-order differential equation

dy I dx = x(2 - y) on the interval (0, 00), where C is any constant. Solution Differentiating y = Cjx :

=

+ 2 gives C

~ (i)

+0

= -

~.

We need to show that the differential equation is satisfied when we substitute into it the expressions (Cjx) + 2 for y, and -Cjx 2 for dy/dx. That is, we need to verifY that for all

XE(O,OO),

This last equation follows immediately by expanding the expression on the right-hand side:

1x [2 _(cx + 2)] 1(_c) C x =

= _

X

Therefore, for every value of C, the function y = C/x equation.

x2·

+ 2 is a solution of the differential _

As was the case in finding antiderivatives, we often need a particular mther than the general solution to a first-order differential equation y' = f(x, y). The particular solution satisfying the initial condition y(xo) = Yo is the solution y = y(x) whose value is Yo when x = XC. Thus the gmph of the particular solution passes through the point (xo, Yo) in the ~-plane. A first...rder initial value problem is a differential equation y' = f(x, y) whose solution must satisfY an initial condition y(xo) = Yo.

EXAMPLE 2

Show that the function

Y=

(x + I) -

I -ex 3

is a solution to the first-order initial value problem

dy dx=y-x, Solution

The equation

dy dx=y-x is a first-order differential equation with f(x, y) = y - x. On the left side ofthe equation:

dy = ~ dxdx

(x +

I

_lex) 3

=

I _

leX

3'

498

Chapter 9: First-Order Oifferential Equations y

On the right side ofthe equation: y

=

(x

+ 1) -

!e x

3

y - x = (x

+ I) - tex - x

=

I - tex.

-"-----!----"+-----J;\--L-~x

4

The function satisfies the initial condition because

y(O) FIGURE 9.1 Graph ofthe solution to the initial value problem in Example 2.

=

I X] I 2 =1--=-. [(x+I)--e 3 x-o 3 3

The graph ofthe function is shown in Figure 9.1.

•

Slope Fields: Viewing Solution Curves Each time we specify an initial condition y(xo) = Yo for the solution of a differential equation y' = I(x, y), the solution curve (graph ofthe solution) is required to pass through the point (xo, Yo) and to have slope I(xo, Yo) there. We can picture these slopes graphically by drawing short line segments of slope I(x, y) at selected points (x, y) in the region of the '!y-plane that constitutes the domain of f. Each segment has the same slope as the solution curve through (x, y) and so is tangent to the curve there. The resuiting picture is called a slope field (or direction field) and gives a visualization of the general shape of the solution curves. Figure 9.2a shows a slope field, with a particular solution sketched into it in Figure 9.2b. We see how these line segments indicate the direction the solution curve tskes at each point it passes through.

FIGURE 9.2 ±ooy(x) = O. We will see that knowing the overall behavior of the solution curves is often critical to understanding and predicting outcomes in a real-world system modeled by a differential equation.

9.1 Solutions, Slope Fields, and Euler's Method

499

(c) y'~ (I-x)y+ ~

y

y

~

L(x)

~

Yo + f(x.. yoJ(x - xo)

FIGURE 9.3 Slope fields (top row) and selected solution curves (bottom row). In computer renditions, slope segments are sometimes portrayed with arrows, as they are here. This is not to be taken as an indication that slopes have directions, however, for they do not.

Yo

o FIGURE 9.4 The linearization L(x) of y

~

y(x) atx

~

xo.

Constructing a slope field with pencil and paper can be quite tedious. All our examples were generated by a computer.

Euler's Method If we do not require or cannot immediately tmd an exact solution giving an explicit formula for an initial value problemy' = f(x,y),y(xo) = Yo, we can often use a computer to generate a table of approximate numerical values of y for values of x in an appropriate interval. Such a table is called a numerical solution of the problem, and the method by which we generate the table is called a numerical method. Given a differential equation dyl ax = fix, y) and an initial condition y(xo) = Yo, we can approximate the solution y = y(x) by its linearization

L(x) FIGURE 9.5 The fIrst Euler srep

= L(x,).

approximates y(x,) with y,

Y

Euler approximation (x2.

Yv

=

y(xo)

+ y'(xo)(x - xo)

I ,

I I ,

Troe solution ClU'Ve y ~ y(x) : I I

I I

FIGURE 9.6 Three sreps in the Euler

approximation to the solution ofthe initial value problem y' = f(x, y), y (xo) = Yo. As we take more steps, the errors involved

usually accumulate, but not in the exaggerated way shown here.

L(x)

=

Yo

+ f(xo,yo)(x - xo).

The function L(x) gives a good approximation to the solution y(x) in a short interval about Xo (Figure 9.4). The basis ofEuler's method is to patch together a string oflinearizations to approximate the curve over a longer stretch. Here is how the method works. We know the point (xo, Yo) lies on the solution curve. Suppose that we specify a new value for the independent variable to be x, = Xo + ax. (Recall that ax = ax in the definition of differentials.) If the increment ax is small, then

y, I

or

=

L(x,)

=

Yo + f(xo,Yo)

ax

is a good approximation to the exact solution value y = y(x,). So from the point (xo, Yo), which lies exactly on the solution curve, we have obtained the point (Xlo yd, which lies very close to the point (XloY(X,)) on the solution curve (Figure 9.5). Using the point (xloyd and the slope f(xloY,) of the solution curve through (XloY,), we take a second step. Setting X2 = + dx, we use the linearization ofthe solution curve through (Xlo Yt) to calculate

x,

Y2

=

y, + f(xloY,)

ax.

This gives the next approximation (X2, Y2) to values along the solution curve Y = y(x) (Figure 9.6). Continuing in this fashion, we take a third step from the point (",Y2) with slope f(x2, Y2) to obtain the third approximation

Y3

=

Y2

+ f(x2' Yo) ax,

500

Chapter 9: First-Order Oifferential Equations and so 00. We are literally building an approximatioo to ooe of the solutioos by following the directioo ofthe slope field ofthe differential equatioo. The steps in Figure 9.6 are drawn large to illustrate the constroctioo process, so the approximatioo looks crude. In practice, would be small enough to make the red curve hug the blue one and give a good approximatioo throughout.

ax

EXAMPLE 3 Find the first three approximatioos Yl,Y2,Y3 using Euler's method for the initial value problem y(O) = I, y' = I + Y, starting at Xo = 0 with

ax =

0.1.

Solution We have the starting values Xo = 0 and Yo = 1. Next we determine the values of x at which the Euler approximations will take place: Xl = Xo + = 0.1, X2 = Xo + 2 = 0.2, and X3 = Xo + 3 = 0.3. Then we find

ax

ax

ax

First:

Second:

Yl

Yo + /(xo, Yo)

=

Yo + (I

=

I

+ (I + 1)(0.1)

1.2

=

+ /(Xl, Yl) ax Yl + (I + Yl) ax 1.2 + (I + 1.2)(0.1)

Y2 = Yl = =

Third:

ax + Yo) ax

=

=

+ /(X2, Y2) ax Yo + (I + Y2) ax 1.42 + (I + 1.42)(0.1)

1.42

Y3 = Y2 = =

=

1.662

•

The step-by-step process used in Example 3 can be continued easily. Using equally spaced values for the independent variable in the table for the numerical solutioo, and generating n ofthem, set

=

Xo + Xl +

ax ax

=

Xn-l

+ dx.

Xl X2

=

Xn

Then calculate the approximations to the solutioo,

Yl Yo

= =

Yo + /(xo,yo) ax Yl + /(Xl, Yl) ax

Yn = Yn-l


Leonhard Euler (1703-1783)

+

/(xn-l,Yn-l)

ax.

The number of steps n can be as large as we like, but errors can accumulate if n is too large. Euler's method is easy to implement 00 a computer or calculator. A computer program generates a table ofnumerical solutions to an initial value problem, allowing us to input Xo and Yo, the number of steps n, and the step size It then calculates the approximate solution values Yl, Y2, ... ,Yn in iterative fashion, as just described Solving the separable equatioo in Example 3, we find that the exact solutioo to the initial value problem is Y = 2e x - I. We use this informatioo in Exarople 4.

ax.

9.1 Solutions, Slope Fields, and Euler's Method

EXAMPLE 4

501

Use Euler's method to solve

y' = I

+ y,

y(O)

=

I,

on the interval 0 :5 x :5 I, starting at Xo = 0 and taking (a) dx = 0.1 and (b) dx = 0.05. Compare the approximations with the values ofthe exact solution y = 2e x - I. Solution

(a) We used a computer to generate the approximate values in Table 9.1. The "error" column is obtained by subtracting the unrounded Euler values from the unrounded values found using the exact solution. All entries are then rounded to four decimal places.

TABLE 9.1 Euler solution of y' = 1 step size dx = 0.1

y 4

3

2

y=2e x - l

+ y, y( 0)

=

1,

x

y (Euler)

y (exact)

Error

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

I 1.2 1.42 1.662 1.9282 2.2210 2.5431 2.8974 3.2872 3.7159 4.1875

I 1.2103 1.4428 1.6997 1.9836 2.2974 2.6442 3.0275 3.4511 3.9192 4.4366

0 0.0103 0.0228 0.0377 0.0554 0.0764 0.1011 0.1301 0.1639 0.2033 0.2491

---oJ--------:--~x

o

FIGURE 9.7 Thegraphofy = 2e' - I superimposed on a scatterplot of the Euler approximations shown in Table 9.1 (Example 4).

By the time we reach x = I (after 10 steps), the error is about 5.6% of the exact solution. A plot of the exact solution curve with the scatterplot of Euler solution points from Table 9.1 is shown in Figure 9.7.

(b) One way to try to reduce the error is to decrease the step size. Table 9.2 shows the results and their comparisons with the exact solutions when we decrease the step size to 0.05, doubling the number of steps to 20. As in Table 9.1, all computations are performed before rounding. This time when we reach x = I, the relative error is only about 2.9%. _ It might be tempting to reduce the step size even further in Exaruple 4 to obtain greater accuracy. Each additional calculation, however, not ouly requires additional computer time but more importantly adds to the buildup of round-off errors due to the approximate representations of numbers inside the computer. The analysis of error and the investigation of methods to reduce it when making numerical calculations are important but are appropriate for a more advanced course. There are numerical methods more accurate than Euler's method, usually presented in a further study of differential equations.

502

Chapter 9: First-Order Oifferential Equations

Euler solution of y' step size dx = 0.05

TABLE 9.2

x 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1

=

+ y, y(O)

=

1,

y(Euler)

y (eIllet)

Error

I

I

1.1 1.205 1.3153 1.4310 1.5526 1.6802 1.8142 1.9549 2.1027 2.2578 2.4207 2.5917 2.7713 2.9599 3.1579 3.3657 3.5840 3.8132 4.0539 4.3066

1.1025 1.2103 1.3237 1.4428 1.5681 1.6997 1.8381 1.9836 2.1366 2.2974 2.4665 2.6442 2.8311 3.0275 3.2340 3.4511 3.6793 3.9192 4.1714 4.4366

0 0.0025 0.0053 0.0084 0.0118 0.0155 0.0195 0.0239 0.0287 0.0340 0.0397 0.0458 0.0525 0.0598 0.0676 0.0761 0.0853 0.0953 0.1060 0.1175 0.1300

Exercises 9.1 Slope Fields In Exercises 1-4, match the differential equations with their slope fields, graphed here. y

y

!!WlH I!!!!!!! IIIIIIII IIIIIIII I111I111 III III

........................................................

•

.......

~~~~~~~~ ~~~~~~~~ ~~~~~~~~ ~~~~~~~~

\\\\\\H \\\\\\\\ (e)

1.y'=x+y (a)

(b)

3.yl=_~

2.y'=y+1

9.1

In Exercises 5 and 6, copy the slope fields and sketch in some of the

solution curves. 5. y'

Solutions, Slope Fields, and Euler's Method

503

19. Use the Euler method with dx = 0.5 to estimate y(5) if y' = y2/~andy(l) = -I. What is the exact value ofy(5)? 20. Use the Euler method with dx ~ 1/3 to estimate y(2) if y' = xsinyandy(O) = I. What is the exact value ofy(2)?

~(y+2)(y-2)

y

21. Show that the solution of the initial value problem

lllllllt llllllll

y'

~x+y,

y(XO)~Yo

is y = -I - x

x

= y(y +

COMPUTER EXPLORATIONS In Exercises 23-28, obtain a slope field and add to it graphs of the

solution curves passing through the given points.

I)(y - I)

23. y' = y with

y

a. (0, I) h. (0,2) 24. y' = 2(y - 4) with

11111\1'-.... 11111111-.... -....-...-. ...; .....; ..,..............

-4'"............4 ---.............

a. (0, I)

h. (0,4)

x

a. (0, I) a. (0, I)

27. y' = (y - I)(x

Integral Equations In Exercises 7-10, write an equivalent rust-order differential equation and initial condition for y.

8. y ~ 9. y

l'

c. (0,5)

h. (0, -2)

c. (0, 1/4)

d. (-I, -I)

h. (0,2)

c. (0, -I)

d. (0,0)

c. (0,3)

d. (I, -I)

26. y' ~ y2 with

a. (0, -I)

+

c. (0, -I)

25. y' = y(x + y) with

......-....-....-.,.,.-.................

:./............Z. . . . ---.. . . .4-

IIIIII tt 11I111I1 7. y = -I

(I - y(I»dl

28. y'

+ 2) with h. (0, I)

~~ with 2

x +4 a. (0,2)

h. (0, -6)

c.

29. A logistic equation

~ 2 - / , ' (I + y(l))sinldl

O::s;x::S;4,

y' = y(2 - y), y(O) = 1/2;

O~y~3

30. y' = (sinx)(siny), y(O) = 2;

+ /,' y(l) dl

Using Euler's Method

13. y' 14. y'

DIs. D 16.

y'

y'

= x(1 - y), y(l) = 0, dx = 0.2 = 2xy + 2y, y(O) = 3, dx = 0.2 = y2(l + 2x), y( -I) = I, dx = 0.5 = 2xe", y(O) = 2, dx = 0.1 = ye', y(O) = 2, dx = 0.5

17. Use the Euler method with dx ~ 0.2 to estimate y(l) if y' andy(O) = I. What is the exact value ofy(I)?

-6 '" x '" 6,

-6 '" y '" 6

Exercises 31 and 32 have no explicit solution in tenns of elementary functions. Use a CAS to explore graphically each of the differential equations.

In Exercises 11-16, use Euler's method to calculate the rust three approximations to the given initial value prohlem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. y 11. y' ~ I - x' y(2) ~ -I, dx ~ 0.5

12. y'

(-2V3, -4)

In Exercises 29 and 30, obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to rmd the general solution of the differential equation.

['I Jl t dl

10. y = I

".

22. What integral equation is equivalent to the initial value problem y' = f(x), y(xo) = Yo?

iiifiiu ifiiifif 6. y'

+ (I + Xo + Yo) er

31. y' = cos (2x - y), y(O) = 2;

0 '" x '" 5,

0 '" y '" 5

32. A Gompertz equation y' = y(I/2 - loy), 0:5x:54, Q:5y:53

y(O) = 1/3;

33. Use a CAS to rmd the solutions of y' initial conditiony(O) = 0, if f(x) is

a. 2x

b. sin 2x

c. 3e x/2

+y

~

f(x) subjcctto the

d 2e -x/2 cos 2x.

Graph all four solutions over the interval -2 '" x '" 6 to compare the results. 34. a. Use a CAS to plot the slope field of the differential equation

~

y

18. Use the Euler method with dx = 0.2 to estimate y(2) if y' = y/x and y(l) ~ 2. What is the exact value ofy(2)?

,_3x2 +4x+2 y 2(y - I) over the region -3 :5 x :5 3 and -3 :5 Y :5 3. h. Separate the variables and use a CAS integrator to rmd the general solution in implicit fonn.

504


c. Using a CAS implicit function grapher, plot solution CUIveS for the arbitrary constsnt values C = -6, -4, -2, 0, 2, 4, 6.

d. Find and graph the solution that satisfies the specified initial condition over the interval [0, b].

d. Find and graph the solution that satisfies the initial condition y(O) ~ -I.

e. Find the Euler nurnetical approximation to the solution of the initial value problem with 4 subintervals of the x-interva1 and plot the Euler approximation supetimposed on the graph produced in part (d).

In Exercises 35-38, use Euler~ method with the specified step size to estimate the value ofthe solution at the given point x.... Find the value ofthe exact solution at x" . 35. y' ~ 2xe"', y(O) ~ 2, dx ~ 0.1, x· ~ I 36. y' ~ 2y 2(x - I), y(2) ~ -1/2,

dx ~ 0.1,

g. Find the error (y(exact) - y(Euler)) at the specified point x = b for each ofyour four Euler approximations. Discuss the improvement in the percentage error.

x· ~ 3

37. y' ~ -Vx/y, y> 0, y(O) ~ I, dx ~ 0.1, x· ~ I 38. y' ~ I

+ y2, y(O)

~ 0,

dx ~ 0.1,

f. Repcatpart (c) for 8,16, and 32 subinterva1s. Plot these three Euler approximations supetimposed on the graph from part (e).

x· ~ I

Use a CAS to explore graphically each of the differential equations in Exercises 39-42. Perform the followiog steps to help with your explo-

rations.

39. y' = x b~ I

+ y,

y(O) = -7/10;

40. y' = -x/y, y(O) = 2; 41. y' = y(2 - y),

a. Plot a slope field for the differential equation in the given >;y-wiodow.

-4'" x '" 4,

-3 '" x'" 3,

y(O) = 1/2;

-4'" y '" 4;

-3 '" y '" 3; b = 2

0 '" x '" 4, 0 '" y '" 3;

b~3

b. Find the general solution of the differential equation using your CAS DE solver.

42. y' = (sinx)(siny), y(O) = 2; b ~ 37f/2

-6 '" x '" 6,

-6 '" y '" 6;

c. Graph the solutions for the values of the arbitrary constaot C ~ -2, -I, 0, I, 2 superimposed on your slope field plot.

9.2

First-Order Linear Equations A first-order linear differential equation is one that can be written in the form

dy

dx

+ P(x)y

=

Q(x),

(1)

where P and Q are continuous functions of x. Equation (I) is the linear equation's standard form. Since the exponential growth/decay equation dy/dx = ky (Section 7.4) can be put in the standard form

dy --ky=O dx '

we see it is a linear equation with P(x) = -k and Q(x) = O. Equation (I) is lillear (iny) because y and its derivative dy/ dx occur only to the fJrst power, they are not multiplied together, nor do they appear as the argument ofa function (such as siny,

EXAMPLE 1

e", or V dy/dx).

Put the following equation in standard form:

dy xdx

=

x 2 + 3y

'

x > O.

Solution

Divide by~. Standsrd funn withP(x) and Q(x) ~ x

~

-3/x

9.2

First-Order Linear Equations

Notice thatP(x) is -3/x, not +3/x. The standard form is y' + P(x)y nus sign is part ofthe formula for p(x).

=

505

Q(x) , so the mi•

Solving Linear Equations We solve the equation

dy dx + P(x)y

=

Q(x)

by multiplying both sides by a positive function v(x) that transforms the left-hand side into the derivative ofthe product v(x) •y. We will show how to find v in a moment, but fITst we want to show how, once found, it provides the solution we seek. Here is why multiplying by v(x) works:

dy dx + P(x)y

=

Q(x)

dy v(x) dx + P(x)v(x)y

=

v(x)Q(x)

Multiply by posiliw> v(.

0,

o.

•

9.2

Solution

With x

>


507

0, we write the equation in standard form:

,

lnx

I

Y - 3x Y

=

+

I

3x

Then the integrating factor is given by

v =

eJ -dx/3.

=

e H / 3)ln. =

X- I / 3 .

•

>0

Thus Left-hand side is try.

Integration by parts of the right-hand side gives x- I / 3y =

-x- I /3(lnx

+ 1) +

J

x-4/3

dx

+ C.

Therefore

or, solving for y,

y = -(lnx

+ 4) + ex!/3.

When x = I and y = -2 this last equation becomes

-2

=

-(0 + 4) + C,

so

C = 2. Substitution into the equation for y gives the particular solution

y = 2x!/3 - lnx - 4.

•

In solving the linear equation in Example 2, we integrated both sides of the equation after multiplying each side by the integrating factor. However, we can shorten the amount of work, as in Example 3, by remembering that the left-hand side always integrates into the product v(x) •Y ofthe integrating factor times the solution function. From Equation (2) this means that

v(x)y =

J

v(x)Q(x) dx.

(4)

We need only integrate the product of the integrating factor v(x) with Q(x) on the righthand side ofEquation (I) and then equate the result with v(x)y to obtain the general solution. Nevertheless, to emphasize the role of v(x) in the solution process, we sometimes follow the complete procedure as illustrated in Example 2. Observe that if the function Q(x) is identically zero in the standard form given by Equation (I), the linear equation is separable and can be solved by the method of Section 7.4:

dy dx

+ P(x)y

= Q(x)

dy dx

+ P(x)y

=

t

0

= - P(x) dx

Q(x) - 0 Sepamting the variables

508


RL Circuits The diagram in Figure 9.8 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 he closed to connect a constant electrical source of V volts. Ohm's Law, V = RI, has to be augmented for such a circuit. The correct equation accounting for hoth resistance and inductance is

v +,---11---,2> I

6

I

a

Switch

b

di L dt

L

R

+ R'I

= V,

(5)

where i is the current in amperes and t is the time in seconds. By solving this equation, we can predict how the current will flow after the switch is closed.

FIGURE 9.8 The RL circuit in Example 4.

EXAMPLE 4

The switch in the RL circuit in Figure 9.8 is closed at time t = O. How will the current flow as a function of time?

Solution Equation (5) is a flIst-order linear differential equation for i as a function of t. Its standard form is (6)

and the corresponding solution, given that i = 0 when t = 0, is (7)

f:We leave the calculation of the solution for you to do in Exercise 28.) Since R and L are positive, -(R/L) is negative and e-(RILl' --> 0 as t --> 00. Thus,

(1' -

lim i = lim 1_00

o

2~

L R

R

R

1_00

1'e- x

4. y' + (tanx)y = cos' x,

2. eX :+2eX y= I

5.x

dy I +2y=ldx

x,

6. (I

+ x)y' + y ~ y;

8. e 2x y'

0

dy

-'ff/2

< x < 'ff/2

x>O

+ 2e 2x y cosx

~ 2x

10. x dx = -x- - 2y, x > 0

7. 2y' ~ exf2

+y

9. xy' - Y ~ 2xlnx

9.2 11. (I - I)' ~ 12. (I

~

2s

3(1

13. sin 6 ~;

+ (cos6)r =

14. tan 6 ~;

+r

+ I) + ( I )" 1+ I

tan6,

2

1 > -I

0 < 6 < 1r/2

Ldi+R'=O dt l ,

0 < 6 < 1r/2

= sin 6,

which is Equatioo (5) wilb V = O.

a. Solve the equation to express i as a function of t.

Solving Inmal Value Problems

b. How loog after lbe switch is throwo will it take lbe current to fall to half its original value?

Solve lbe initial value problems in Exercises 15-20.

dy 15. dl

+ 2y

16. 1dy dl

c. Showtbat the value oflbe current wheo 1 = L/R is I/e. (The significance of this time is explained in lbe next exercise.)

y(O) = I

= 3,

, 1 > 0, + 2y - I,

dy . 17. 6 d6 + Y = sm6,

509

26. Current in an open RL circuit If lbe switch is thrown open after lbe current in an RL circuit has built up to its steady-state value I = VIR, the decaying current (see accompanying figure) obeys the equatioo

+ 4(1 - 1)2s ~ 1 + I, 1 > I

+ I) tls + d1


Y() 2

=I

6> 0, Y(1r/2) = I

dy _, 18. 6 d6 - 2y - 6 sec 6 tan 6, dy

6> 0,

y(1r/3)

~

2

x'

19. (x + I) tis - 2(x 2 + x)y = x e+ I' x > -I, y(O) = 5 20.

dxdy + xy

~

3!-

x, y(O) ~ -6

R

21. Solve lbe expooeotial growth/decay initial value probleoJ for y as a functioo of 1 by lbinking of lbe differential equatioo as a rrrstorderlineareqoatioowilbP(x) = -k and Q(x) = 0:

dy dl

~

ky

(kconstant),

y(O)

~

Yo

22. Solve lbe following initial value probleoJ for u as a functioo of t.

~~ + ~u

= 0

(k and m positive coostants),

u(O) = Uo

27. TIme constants Engineers call !be number L/R lbe time constant of the RL circuit in Figure 9.9. The sigoificance oflbe time coostant is lbat 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.

a. Find the value of i in Equatioo (7) tbat correspoods to 1 = 3L/R and show tbat it is about 95% of the steady-state value I ~ V/ R. b. Approxiruately what perceotage oflbe 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 tbat the solutioo of lbe equation

Theory and Examples 23. Is eilber oflbe following eqoatioos correct? Give reasoos for your

answers. b.

x/ tdx ~ xlnlxl

+ ex

24. Is eilber oflbe following eqoatioos correct? Give reasoos for your

answers. a.

co~x/ cosX' dx =

1/

b. cosX'

is i ~

tanx

cos x dx = tanx

l' + Ce-(RlL)'

+C

C

+ cosX'

25. Current in a closed BL circuit How many seconds after lbe switch in an RL circuit is closed will it take lbe current i to reach half of its steady-state value? Notice tbat lbe time depeods 00 R and L and not 00 how much voltage is applied.

R

b. Theo use the initial cooditioo i(O) ~ 0 to determine lbe value of C. This will complete lbe detivatioo ofEquatioo (7). c. Showtbati = VIR is a solution ofEqoatioo (6) and tbat i = Ce -{RIL)t satisfies the equation

di dl

R.

+ I'

=

o.

510

Chapter 9: First-Order Oifferential 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

+ P(x)y

~

Q(x)y". Of,

equivalently,

Observe that, if n = 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'

31. xy'

+y

Y -2

32. x"y'

~

_y~xy2

+ 2xy

~ y'

AppLications We now look at four applications of frrst-order differential equations. The frrst application analyzes an object moving along a straight line while subject to a fOrce opposing its motion. The second is a model ofpopulation growth. The third application considers a curve or curves intersecting each curve in a second family ofcurves orthogonally (that is, at right angles). The fmal application analyzes chemical concentrations entering and leaving a container. The various models involve separable or linear frrst-rder equations.

Motion with Resistance Proportional to Velocity In sonJe 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 line with position function s and velocity vat time t. From Newton's second law ofmotion, the resisting force opposing the motion is Force = mass 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 mass 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 oftime 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 (I) 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

=

+ C.

0 when I = 0 gives vom 0=--+ C k

vom C=-

and

k

The body's position at time I is therefore () sl

_ m _ vom ( -vom T e -(klm)t +VoT - T I - e -(kim)t) .

(2)

To find how far the body will coast, we fmd the limit ofs(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 k

=

t_OO

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: ifm is large, the body will coast a long way.

lIn 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 ft/sec (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-tllS = e-tllS =

I

EQ.(I)withk~ 1/3.

m

1/11

-1/18

=

In (1/11)

I

=

18ln 11 '" 43 sec.

=

=

6, Va

=

11, v

=

1

-In 11

We answer the second question with Equation (3): . Distance coasted

vom

=

T

=

198 ft.

=

11·6

1/3

•

Inaccuracy of the Exponential Population Growth Model In Section 7.4 we modeled population growth with the Law of Exponential Change: dP di

=

kP,

P(O)

=

Po

whereP is the population at time I, k > 0 is a constant growth mte, and Po is the size ofthe 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 dP;dl = k

(4)

512


TABLE

World populatioo (1980-2008)

I

)

30

t

FIGURE 9.10 Notice that the value of the solutionP ~ 4454e°.0 17• is 7169 when t = 28, which is nearly 7% more than the actual population in 2008.

Orthogooal trajectory

9.3 World population (midyear)

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 ofthe 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 '" !1P, 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 orthogonal trajectory intersects the family ofcurves at right angles, or orthogonally.

!1P/P

0.017P,

P(O) = 4454.

The solution to this initial value problem gives the population function P = 4454.°.017/. In 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 affecting 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

-+--+-+~+-+--+~X

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 x 2 + y2 = a 2 , centered at the origin (Figure 9.12). Such mutoa1ly orthogonal systems ofcurves 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.

=

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/ dx values. Differentiating xy = a implicitly gives

FIGURE 9.12 Every straight line through the origin is orthogonal to the family of circles centered at the origin.

dy xdx+y=O

or

dy dx

Y 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 y

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.

Each curve is orthogonal to every curve it meets in the other family (Example 2).

(5)

FIGURE 9.13

where b = 2C is an arbitrary constant. The orthogonal trajectories are the family ofhyper• bolas 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 mixture is kept uniform by stirring and flows out of the container at a known rate. 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 (rate at WhiCh) (rate 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 rate of the chemical at time t is y(t) Departure rate = V{t) -(outflowrate)

=

concentration in ) ( container at time t - (outflow rate) .

(7)

Accordingly, Equation (6) becomes

dy . . y(t) dt = (chenncal'samvalrate) - V(t) . {outflow rate).

(B)

If, say, y is measured in pounds, V in gallons, and t in minutes, the units in Equation (8) are pounds

minutes

pounds pounds gallons minutes - gallons - minutes'

EXAMPLE 3 In an oil refinery, a storage tank contains 2000 gal of gasoline that initially has 100 Ib of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 Ib of additive per gallon is pumped into the tank at a rate of 40 gal/min.

514

Chapter 9: First-Drder Differential Equations

The wt:ll-mixed solution is pumped out at a mte of 45 gal/min. How much of the additive is in the tank 20 min after the pumping process begins (Figure 9.14)? ) 40 ga1Imin containing 2 IbIgal

4~ gaJJmi.n c~taining

2 lblgal V

FIGURE 9.14 The storage tank in Example 3 mixes input liquid with stored liquid to produce an output liquid.

Lety be the 8JD01mt (in pounds) of additive in the tank at time I. We know that y = 100 when I = O. The number of gallons of gasoline and additive in solution in the tank at any time I is

Solution

~ 2000 gal + (40: -

V(I)

~

45 : ) (tmin)

(2000 - 51) gal.

Therefore, Rate out

=

y(1) Y(I)' outflow rate

Eq.(7) OutflOW' rue ia4S pJ/min

- (200: ~

51) 45

muh - 2000 - St.

45y Ib 2000 51 min .

Also, Rate in

~

(2

~)(40 : )

~801b. mID

The differen:ti.al equation modeling the mixture process is

dy 45y dt ~ 80 - 2000 51

Eq. (8)

in pmmds per minute. To solve this differeIltial equation, wt: Ill'St write it in standard linear £ann: dy dt

45

51 Y ~ 80.

+ 2000

Thus, P(I) - 45/(2000 - 51) and Q{I) - 80. The integnUing factnr is v(l)

=

eJPdt

=

eJ~dl

= e-9In(2000-St)

~ (2000 -

5Ir'.

9.3

Applications

515

Multiplying both sides ofthe standard equation by v(l) and integrating both sides gives 9 (2000 - 51r •

(2000 - 51r9

(t + 200~5_ Y) 51

t+

= 80(2000 - 51r

9

45(2000 - 51)-10 Y = 80(2000 - 51r9

: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(2oo0 - 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. Coasting bicycle A 66-kg cyclist on a 7-kg bicycle starts coasting on level ground at 9 m/scc. The k in Equation (I) is about 3.9 kgfsec.

a. About how far will tire cyclist coast before reaching a complete stop?

b. How long will it take tire cyclist~ speed to drop to I m/sec? 2. Coasting battleship Suppose that an Iowa class battleship has mass around 51,000 melric 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 tire ship~ speed to drop to I mfsec? 3. The data io Table 9.4 were collected with a motion detector and a CBLTM by Valerie Sharritts, a mathematics teacher at St. Fraucis DeSaies High School in Columbus, Ohio. The table shows tire dislance s (meters) coasted on io-lioe skates io t sec by her daughter Ashley when she was 10 years old. Fiod a model for Ashley's

516


position given by the data in Table 9.4 in the form ofEquation (2). Her initial velocity was Vo ~ 2.75 mlsec, her mass m ~ 39.92 kg

In Exercises 5-10, fmd the orthogonal 1rajectories of the family of

(she weighed 88Ib), and her total coasting distaoce was 4.91 m.

curves. Sketch several members ofeach family.

TABLE 9.4 Ashley Sharritts skating data

t (see)

s(m)

t (see)

s(m)

t (see)

s(m)

0

0

2.24

3.05

4.48

4.77

0.16

0.31

2.40

3.22

4.64

4.82

0.32

0.57

2.56

3.38

4.80

4.84

0.48

0.80

2.72

3.52

4.96

4.86

0.64

1.05

2.88

3.67

5.12

4.88

Orthogonal Trajectories

5.y=mx 7. kx' + y' ~ I 9. y = ce ---x

6.y=cx'

8. 2x' + y' ~ 10. y = e b

11. Show that the curves 2x 2

+ 3y 2

= 5 and y2 =

c'

x' are orthogonal.

12. Find the family of solutions of the given differential equation and the fantily of orthogonal trajectories. Sketch both fantilies.

a. xlix + ydy

=0

b. xdy - 2ylix

=a

MIxture Problems 13. Salt mixture A tank initially contains 100 gal of brine in which 50 Ib of salt are dissolved. A brine containing 2 Ib/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at the rate of4 gal/min.

0.80

1.28

3.04

3.82

5.28

4.89

0.96

1.50

3.20

3.96

5.44

4.90

1.12

1.72

3.36

4.08

5.60

4.90

1.28

1.93

3.52

4.18

5.76

4.91

a. At what rate (pounda per minute) does salt enter the tank at time I?

1.44

2.09

3.68

4.31

5.92

4.90

b. What is the volume ofbrine in the tank at time t?

1.60

2.30

3.84

4.41

6.08

4.91

1.76

2.53

4.00

4.52

6.24

4.90

c. At what rate (pounda per minute) does salt leave the tank at time t?

1.92

2.73

4.16

4.63

6.40

4.91

d. Write down and solve the initial value problem describing the

2.08

2.89

4.32

4.69

6.56

4.91

4. Coasting to a stop Thhle 9.5 shows the distaoce s (meters) coasU:d on in-line skates in terms of time t (seconds) by Kelly Schmitzer. Find a model for her position in the form ofEquation (2). Her initial velocity was Va ~ 0.80 mfsec, her mass m ~ 49.90 kg (110 Ib), and her total coasting distaoce was 1.32 m.

s(m)

0

0

1.5

0.89

3.1

1.30

0.1

0.07

1.7

0.97

3.3

1.31

t (see)

s(m)

0.3

0.22

1.9

1.05

3.5

1.32

0.5

0.36

2.1

1.11

3.7

1.32

0.7

0.49

2.3

1.17

3.9

1.32

0.9

0.60

2.5

1.22

4.1

1.32

1.1

0.71

2.7

1.25

4.3

1.32

1.3

0.81

2.9

1.28

4.5

1.32

9.4

a. At what time will the tank be full? will it contain?

t (see)

s(m)

14. Mixture problem A 200-gal tank is half full of distilled water. At time t = 0, a solution containing 0.5Ib/gal ofconcen1rate enters the tank at the rate of 5 gal/min, and the well-stirred mixture is withdrawn at the rate 00 gal/min. b. At the time the tank is full, how many pounds ofconcen1rate

TABLE 9.5 Kelly Schmitzer skating data

t (see)

mixing process.

e. Find the concentration ofsalt in the tank 25 min afler the process starts.

15. Fertilizer mixture A tank contains 100 gal offresh water. A solution containing I Ib/gal ofsoluble lawn fertilizer runs into the tank at the rate of I gal/min, and the mixture is pomped out of the tank at the rate 00 gal/min Find the maxinuun amount of fertilizer in the tank and the time required to reach the maxinuun. An executive conference room. of a corporation contains 4500 ft' of air initially free ofcarbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the racon at the rate of 0.3 ft'/min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft'/min. Find the time when the concentration ofcarbon monoxide in the room. reaches 0.01 %.

16. Carbon monOlide pollution

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 graphs is a powerful tool in understanding real-world systems. The starting ideas for a graphical 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 Ifdy/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. ------l@)------------i@~--->,

-1

2

Y

518

Chapter 9: First-Order Oifferential Equations 2.

IdentifY and label the intervals where y' > 0 and y' < O. 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 = -1. Solution curves between the lines y = -I and y = 2 fall away from y = 2 toward Y = - 1. 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 inJplicit differentiation. y' = (y

+ I)(y - 2)

= y2 - Y -

2

< O. To find y", we differ-

Formula for y' ...

y" = ~(Y') = ~(y2 _ Y _ 2)

dx

y

=

y'>O y">O

dx

differentiated implicitly

2y'y' - y'

= (2y -

I)y'

= (2y -

I)(y

with respect to x

+

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


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 object is equal to the net force applied to it. In mathematical terms,

F

=

d

(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 object 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 and Equation (2) takes the simpler form

F= ma,

or

0,

(3)

known as Newton s 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

ypositive

Fp FIGURE 9.19

=

An object falling under the

influence ofgravity with a resistive force assumed to be proportional to the velocity.

=

mg,

the force due to gravity. If, however, we think of a real body falling through the air----1lay, 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 m-=mg-ku dt dv k dt =g- m V '

-

dv>O dt __

I

I I

dv 0 when v > mg/k. Figure 9.21 adds tIris infonnation to the phase 11oe. Notice the similarity to the phase line for Newton's law of cooling (Figure 9.17). The solution curves are similar as well (Figure 9.22). Figure 9.22 shows two typical solution curves. Regardless of the initial velocity, we see the body's velocity tending toward the limiting value v = mg/k. This value, a stable equilibrium point, is called the body's terminal velocity. Skydivers can vary their terminal velocity from 95 mph to 180 mph by changing the amount of body area opposing the fall, which affects the value of k.

Logistic Population Growth In Section 9.3 we examined population growth using the model of exponential change. That is, if P represents the number of individuals and we neglect departures and arrivals, then

dP

'JYpical velocity curves for a falling body encountering resistance. The value v = mg/k is the tenninal velocity.

lit

FIGURE 9.22

=

kP,

(5)

where k > 0 is the birth rate minus the death rate per individual per unit time. Because the natural environment has only a limited number ofresources to sustain life, it is reasonable to assume that only a maximum population M can be accommodated. As the population approaches this limiting population or carrying capacity, resources become less abundant and the growth rate k decreases. A simple relationship exhibiting this behavior is

k

=

r(M - P),

where r > 0 is a constant. Notice that k decreases as P increases toward M and that k is negative if P is greater than M. Substituting r(M - P) for k in Equation (5) gives the differential equation

';J; I I I I

@ 0

dP >0 dt

I I I I

@

•M •

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


dP >0 dt I I

aZp

I

dt 2

@, o

>0 : aZp < 0 I

J'

dt2

I,

I

M

I I I I

ZP) and tiP/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 llrst quadrant of the tP-plane into horizontal bands in which we know the signs of both tiP/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

If P

tit0 I dt 2 @. '>P I

M

1:

FIGURE 9.24 The completed phase line for logistic growth (Equation 6).

> M, then (M -

P

Limiting

f----====:;:::;~~""'''':::::::'''''....~--population

TIme

FIGURE 9.25

Population corves for logistic growth.

Exercises 9.4 Phase Unes and Solution Curves 10 Exercises 1-8,

a. IdentifY the equilibrium values. Which are stahle and which are unstable? b. Construct a phase line. IdentifY the sigos of y' and y'.

c. Sketch several solution curves. dy dy 2. dx 1. dx = (y + 2)(y - 3)

= y2

- 4

dy 3. dx ~ y' - Y

4. dx ~ y' - 2y

dy

5.y'=vy, y>O 7. y' = (y - I)(y - 2)(y - 3)

6. y' = y 8. y' = y3 _ y2

vy,

y > 0

Models of Population Growth The autonomous diffi:rentia1 equations in Exercises 9--12 represent models for population 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, and which are unstable?

dP 9. dt = I - 2P 11. dP dt ~ 2P(P - 3)

dP 10. dt = P( I - 2P)

(I) 2

12. dP dt ~ 3P{I - P) P -

13. Catastrophic change in logistic growth Suppose that a healthy population of some species is growing in a limited environment

and that the current population Po is fairly close to the carrying capacity Mo. You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddeo1y a catastrophe such as the Mount St Helens volcauic eruption contaminates the lake and destroys a siguificant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity M, considerably less than M o and, in fact, less than the current population Po. Starting at some time before the catastrophe, s1retch a "before-and-after" curve that shows how the fish population responds to the change in environment 14. ControUing a population The fish and game department in a certain state is planning to issue huntiog permits to control the deerpopulation (one deer per permit). It is known that if the deer population falls below a certain level m, the deer will become extioct. It is also known that if the deer population rises above the carrying capacity M, the population will decrease back to M through disease and ma1nutrition.

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),

where P is the population of the deer and r is a positive constant ofproportionality. 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

c. ShowthatifP

> Mforallt,thenlim,_ooP(t) < m for all t?

~

M.

d. What happens if P

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 permits should he issued? Applications and Examples 15. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance 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. enrreot in ao RL-clrcuit The accompanying 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

where k is a constant 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 a160-lb skydiver (mg = 160) and with time in seconds and distance in feet, a typical value of k is 0.005. What is the

Switch

diver's terminal velocity?

y;,

16. Resistance proportional to A body ofmass m is projected vertically downWlltd with initial velocity vo. Assume that the resisting force is proportional to the square root of the velocity and fmd the terminal velocity from a graphical analysis. 17. Sailing A sailboat is runuing along a s1raight course with the wind providing a constant fotWlltd 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 I fr/sec. What is the maximum velocity in feet per second of the boat under this wind? 18. The spread ofinformation Sociologists recognize a phenomenon called social diffitsion, which is the spreading of a piece ofinformation, technological innovation, or cu1toral fad among a population. The members of the population can he divided into _ classes: those who have the information and those who do not In a ftxed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it IfX denotes the number 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 booyant force eqoals 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 descends, complete the following steps. a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Fignrc 9.19. b. Using v(t) for the pearl's velocity as a function 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 UN.

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 hass living in a common pond.

524


Phase Planes A general system of two first-order differential 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 y(t) 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), y(t» in the xy-plane starting at some specified point. Therefore the solution functions define a solution curve through the specified point, called a trajectory 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 y(t) 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 differential equations. Several factors affect the mtes 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 bass in the pond:

dx dt = (a - by)x,

(la)

dy dt = (m - nx)y.

(lb)

Here x(t) represents the trout population, y(t) 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 y(t) 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 gmphical 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 population x(t) is constant. Similarly, if x = min, theo Equation (Ib) implies dy/dt = 0, and the bass population y(t) is constant. This information is recorded in Figure 9.26. y Bass

y Bass

y Bass I

.!!

hunter mode~ show that each trajectory is traversed in a counterclockwise direction as time t increases. Along each trajectory, both the rabbit and fox populations fluctuate between their maximum and minimum levels. The maximum and

utinimum levels for the rabbit population occur where the trajectory intersects the horizontal line y = alb. For the fox population, they occur where the trajectory intersects the vertical line x = c/d. When the rabbit population is at its maximum, the fox population is below its maximum value. As the rabbit population declines from this point in time, we move counterclockwise around the trajectory, and the fox population grows until it reaches its maximum value.

At this point the rabbit population has declined to x

~

cld and is

no longer at its peak value. We see that the fox population reaches

its maximum value at a later time than the rabbits. The predator population lags behind that of the prey in achieving its maximum

values. This lag effect is showe in Figure 9.38, which grapha both x(t) and y(t).


529

14. At some time during a trajectory cycle, a wolf invades the rabbit-fox territory, eats some rabbits, and then leaves. Does this mean that the fox population will from then on have a lower IIUlXimwn value? Explain your answer.

>,y

+------+ Lag time

FIGURE 9.38 The fox and rabbit populations oscillate periodically, with the lIUlXimwn fox population lagging the IIUlXimwn rabbit population.


Chapter •

1. What is a frrst-order differential equation? When is a function a

solution of such an equation? 2. What is a general solution? A particular solution? 3. What is the slope field of a differential equation y' What can we learn from such fields?

= f(x, y)?

4. Describe Euler's method for solving the initial value problem y' = f(x, y), y(xo) = Yo numerically. Give an example. Comment on the method's accuracy. Why might you want to solve an initial

value problem numerically? 5. How do you solve linear fIrSt-order differential equations? 6. What is an orthogonal trajectory of a family of curves? Deseribe how one is found for a given farnily of curves.

Chapter

1. y' = xe"~

5. y' =

+ xcos

2

ydx ~ 0

~

9. Why is the exponential model unrealistic for predicting long-term population growth? How does the logistic model correct for the deficiency in the exponential model for population growth? What is the logistic differential equation? What is the fonn ofits solution? Desctibe the graph of the logistic solution. 10. What is an autonomous system of differential equations? What is a solution to such a system? What is a trajectory of the system?

xj2 = xe

Initial Value Problems In Exercises 17-22 solve the initial value problem.

2. y' = xye" 4.

a 2 dx -

6. y' =

7. x(x - I) th> - ydx = 0 9. 2y' - y

8. How do you construct the phase line for an autonomous differential equation? How does the 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. secxth>

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?

3vYcscxth> ~ 0

xe~-Ycscy

th>

17.(x+I)dx+2y~x,

dy

+ 2y

18. x dx

_ 2 - x

+ I,

x>-I, x

> 0,

15. 16.

~

y(l)

I

I)x- I

8. y' = (y2 y' 10. 2: + Y = e-xsinx

+ 2y ~ I - x-I 12. xy' - y ~ alnx x (l + eX) th> + (ye + e~) dx ~ 0 e~ th> + (e~y - 4x) dx ~ 0 (x + 3y 2) th> + y dx ~ 0 (Hint: d(xy) ~ y dx + x th» xth> + (3y - x-2 cosx) dx = 0, x> 0

th> 19. dx

+ h). = x 2, y(O)

= -I

11. xy'

13. 14.

y(O)~1

20. xth> 21. xy'

+ (y -

+ (x

22. y dx

cosx)dx = 0,

y(~)

= 0

- 2)y ~ h3e~, y(l) ~ 0

+ (3x

- xy

+ 2) dy

= 0,

y(2) = -I, y < 0

530


Euler's Method

023.

+

y' = y

cosx,

0,;; x ,;; 2;

y(O) = 0;

Xo

Massm

mgR 2

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.

F~---

,2

~

I I

= 0

I I I

024. y' ~ (2 - y)(2x + 3), y(-3) ~ I; -3:sx:s -1; Xo =-3

dy

025.c~3; dx ~

dy

026.c=4;

dx

=

x-2y

X+!, x2

x

y(l)

'

center

=I

In Exercises 27 and 28, use Euler's method to solve the initial value problem graphically, starting at Xo ~ 0 with

a. dx

027. dxdy =

~

; +2'

eX Y

x2

dy

b. dx

0.1.

028. dx ~ - eY

~

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

-0.1.

v2

2gR2 =

-s-

+ VOl

2gR.

-

y(O) = -2

Thus, the velocity remains positive as loog as Vo '"

+Y

+ x'

v'2iii

y(O) ~ 0

Slope Fields

29.y'=x

30. y' = I/x

= xy

v'2iii,

The velocity Vo = is the moon~ escape velocity. A body projected upward with this velocity or a greater one will escape from the moon's gravitational pull.

In Exercises 2!1-32, sketcb 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. 31. y'

!

y(O) ~ I

-2y+1

iI

M~'sill

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.

32. y' = I/y

Autonomous Differential Equations and Phase Lines In Exercises 33 and 34:

v'2iii, then

b. Show that if Vo =

s=RI+ (

3vo

2R

)2/3

t.

36. Coasting to a stop Table 9.6 shows the dislaDce s (meters) coasted 00 in-line skates in t sec by Johnathon Krueger. Find a model for his position in the form of Equatioo (2) ofSectioo 9.3. His initial velocity was Vo = 0.86 mlsec, his mass m = 30.84 kg (he weighed 68 Ib), and his total coastiog dislaDce 0.97 m.

a. IdentifY the equilibrium values. Which are stable and which are unstable?

TABLE 9.6 Johnathon Krueger skating data

b. Construct a phase line. IdentifY the signs of y' and y'. c. Sketch a representative selection of solution curves. 33.

dy dx

=y2-1

34.

dy dx

=y_y2

Applications 35. Escape velocity

The gravitationsl attraction F exerted by an

airless moon on a body ofmass 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 figure). The force F is negative because it acts in the direction of decreasing s.

Chapter

I (see)

s(m)

I (see)

s(m)

I (see)

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 ofthe movement ofa dissolved substance across a cell's membrane is described by the equation

dy A = k-(c - y) dt V .

-

In this equation, y is the concentration of the substance inside the cell and dy/dtis the rate at whichy changes over time. The letters

L

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

a. Vetify that y(x) problem y'

substance outside the cell. The equation says that the rate at which the conceotration changes within the cell is proportionsl to the difference between it and the outside concentration.

~

531

Yl (x) - Y2(X) satisfies the initial value

+ P(x)y = 0,

y(xo) = O.

b. For the integrating factor v(x) =

eJP«)"', show that

d tb: (v(x )[Yl (x) - Y2(X)]) = O.

a. Solve the equation for y(t), using Yo to deoote y(O). Conclude that v(x )[Yl (x) - Y2(X)]

b. Find the steady-state conceotration, Iim,-..ooy(t).

2. Height of a rocket If an exteroal force F acts upon a system whose mass varies with time, Newton's law of motion is d(mv) tim ----;j/ = F + (v + u)d/' 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 aystem at the rate dm/dt. Suppose that a rocket of initial mass mo starts from rest, but is driven upward by ruing some of its mass directly backward at the constant rate of dmldt = -b units per second and at constant speed relative to the rocket U = -c. The only external force acting on the rocket is F ~ -mg due to gravity. Under these assumptions, sbow that the height of the rocket above the ground at the end of t seconds (t small compared to molb) is _ [ mo - bt In mo - btl _ y-ct+ b mo

1 2 zgt·

3. a. Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Fundanleotal Theorem of Calculus, Part I to show that soy function y satisfYing the equation v(x)y for v(x) =

=

J

v(x)Q(x) tb:

+C

eJ p(.) '" is a solution to the first-rder linear equstion dy tb:

+ P(x)y

= Q(x).

J:'

b. If C ~ Yov(xo) v(t)Q(t) dt, theo sbow that any solution Y in part (a) satisfies the initial condition y(xo) = Yo.

4. (Continuation ofExercise 3.) Assmne the hypotheses ofExercise 3, and assume that Yl (x) and Y2(X) are both solutions to the firstorder linear equation satisfYing the initial condition y(xo) = Yo.

Chapter ID

constant.

a

c. From part (a), we have Yl (xo) - Y2(XO) = O. Since v(x) > 0 fora < x < b, usc part (b) to cs1ablisb that Yl(X) - Y2(x) = 0 on the interval (a, b). ThusYl(X) ~ Y2(X) for all a < x < b.

Homogeneous Equatfons A lITst-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 = ylx. Then, y ~ vxand dy dv tb:=v+xtb:.

Substitution into the original differential equstion and collecting tenus with like vatiables theo gives the separable equstion tb:

dv F(v) = O.

x +v_

Afler solving this separable equstion, the solution ofthe original equation is obtained when we replace v by ylx .

Solve the bomogeoeous equstions in Exercises 5-10. First put the equstion in the funn ofa homogeneous equstion.

5. (x 2 + y 2)tb: + xydy 6. x 2 dy

+ (y2

7. (xeY/·

+ y)tb:

- xy)tb:

=0 =0

- xdy = 0

8. (x +y)dy+ (x -y)tb: y y-x 9. y' = x + cos - x -

10.

~

0

(x sin ~ - ycos~) tb: + xcos ~ dy


Mathematica/Maple Modules: Drug Dosages:Are They Effective?Are They Safe? Formulate and solve an initial value model for the absorption ofa drug in the bloodstream. First-Order Differential Equations and Slope Fields Plot slope fields and solution curves for various initial conditions to selected lltSt-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 X n 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 this 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 ofthe sequence be any integer. In the sequence above, {an} starts with a, while {b n } starts withb 6 • Sequences can be described by writing rules that specify their terms, such as an =

V;,

bn = (_I)n+l{,

en = n ; I,

dn = (_I)n+l,

or by listing !enns: {an} =

{Vi, Vi, V3, ... , V;, ... }

{b n }

{1,-!,t,-i,.. ·,Hr' {, ... }

=

1234

n-I

{en} = { 0'2'3'4'5'''''-n-''''

}

{dn} = {I, -I, I, -I, I, -I, ... , (_I)n+\ ... }.

We also sometimes write

Figure 10.1 shows two ways to represent sequences graphically. The fIrst marks the rlrst few points from a" a2, a3,"" an,'" on the real axis. The second method shows the graph ofthe function defIning the sequence. The function is derIDed only on integer inputs, and the graph consists of some points in the xy-plane located at (I, a,), (2, a2), ... , (n, a,,), .... a" a,

•

"2 03 a4 a ,

1

0

•

• •

2

a" ~

•

••• •

0

a"

a,

'" a2

0

vn

3 2 1

1

1 an =n

~t

_L---L---Ll_ll_~, n

2

3

4

5

a"

a, a.

• •

a,

a, a3 0

•

••

1

aJ! =

FIGURE 10.1

(_1)1I+1~

~f

Sequences can be represented as points on the real line or as

points in the plane where the horizontal axis n is the index number ofthe term and the vertical axis a" 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, V2, V3, ... , V;;, ... } L-E

o

L L+E

•• •••• (•• -I- )

f----------~L

L

have tenDs that get larger than any number as n increases, and sequences like

+E

------------(n,a,,)-~-.-- f------~~--~L

-

E

{I, -I, I, -I, I, -I, ... , (_I).+I, ... }

bounce back and forth between I and - I , never converging to a single value. The following definition captures the meaning of having a sequence converge to a limiting value. It says that if we go far enough out in the sequence, by taking the index n to be larger than some value N, the difference between a. and the limit of the sequence becomes less than any preselected number E > 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,

c:+---:----:----:-----:':~~---n

o

2 3

N

n

n>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 {(n, a.)}. In this figure, all the a,/s after aN lie within e of L.

FIGURE 10.2

la. - LI

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 ofthe sequence (Figure 10.2).

The definition is very similar to the definition ofthe limit ofa function f(x) asx tends to 00 (limx~oof(x) in Section 2.6). We will exploit this connection to calculate limits of HISTORICAL BIOGRAPHY

Nicole Oresme (ca. 132

0 be given. We must show that there exists an integer N such that for all n,

n>N This implication will hold if (I/n) < E or n > I/E. If N is any integer greater than I/E, the implication will hold for all n > N. This proves that lim.~oo(l/n) = O.

(h) Let E

>

0 be given. We must show that there exists an integer N such that for all n,

n>N

=>

Ik - kl
N we have an < m, then we say {a.} diverges to negative infinity and write

lim n~OO

an

= -00

or

an ~ -00.

chosen number m.

A sequence may diverge without diverging to infinity or negative inf"mity, as we saw in Example 2. The sequences {I, -2, 3, -4,5, -6,7, -8, ... } and {I, 0, 2, 0, 3, O, ... } are also examples of such divergence.

Calculating Limits of Sequences Since sequences are functions with domain restricted to the positive integers, it is not surprising that the theorems on limits offunctions given in Chapter 2 have versions for sequences.

THEOREM 1 Let {an} and ibn} be sequences of real numbers, and let A and B be real numbers. The following rules hold if Iim.~oo a. = A and limn_ oo bn = B. 1. Sum Rule:

lim,,~oo(an

2. Difference Rule:

lim,,~oo(an

+ bn)

= A

+B

3. Constant Multiple Rule:

- bn) = A - B lim,,~oo(k'bn) = k·B (any number k)

4. Product Rule:

lim,,~oo(an'

5. Quotient Rule:

Iim,,~oo b

bn) = A' B an A n

= Jj

The proof is similar to that ofTheorem I of Section 2.2 and is omitted.

536

Chapter 10: Infinite Sequences and Series

EXAMPLE 3 (a)

lim

,,---+00

By combining Theorem I with the limits of Example I, we have:

(_1)

= -I- lim

n

11--+ 00

1 n

= -1·0 = 0

ConstanlMultipleRuleandExamplela

I) n - I) = lim ( (b) lim ( -nI = lim I 11---+00 11---+00 11---+00

n

lim

11---+ 00

nI =

I - 0 = I

Difference Rule andExamplela

Product Rule

- 7 6 (4/n 6 ) - 7 0 - 7 (d) lim 4 6 n=lim =--=-7. n~OO n + 3 n~OO I + (3/n 6 ) I +0

Sum and Quotient Rules

•

Be cautious in applying Theorem 1. It does not say, for example, that each of the sequences {an} and ibn} have limits if their sum {an + bn} has a limit. For instance, {an} = {1,2,3, ... } and {bn} = {-1,-2,-3, ... } both diverge, but their sum {an + bn} = {O, 0, O, ... } clearly converges to o. One consequence ofTheorem I is that every nonzero multiple of a divergent sequence {an} diverges. For suppose, 10 the contrary, that {can} convetges fur some number e oF o. Then, by1llking k = I/e in the Constant Multiple Rule in Theorem I, we see that the sequence

... .... .. .... a; fn

L

b~.~T~.~~:::f~r/I

•• -

c:f--"----------->.

o

FIGURE 10.4 The terms of sequence {bn } are sandwiched betweeo those of {a,,} and {cn}, forcing them to the same common limit L.

converges. Thus, {ea.} cannot converge unless {a.} also converges. If {an} does not converge, then {can} does not converge. The next theorem is the sequence version of the Sandwich Theorem in Section 2.2. You are asked to prove the theorem in Exercise 109. (See Figure 10.4.)

THEOREM 2-The Sandwich Theorem for Sequences Let {an}, {bn} , and {en} be sequences ofreal numbers. If a. :5 bn :5 en holds for all n beyond some index N, and if lUnn-+oo an = lilIln-+oo en = L, then limn_ oo b" = L also. An immediate consequence of Theorem 2 is that, if Ibnl :5 en and en --+ 0, then bn ---+ 0 because -en ~ bn ~ Cn " We use this fact in the next example.

EXAMPLE 4

Since I/n --+ 0, we know that because

_l
(c) (- I)nk--+ 0

because

_1n 0)

n~OO

5. n_ limOO

In

(1+"-)"=e n

(lxl

n 6. lim "-- = 0

(any x)

n~OO

X

(any x)

< I)

4. limxn=O n_OO

Formulas (3) through (6), x remains rlXed as n --->

n!

00.

Proof The first limit was computed in Example 7. The next two can be proved by taking logarithms and applying Theorem 4 (Exercises 107 and 108). The remaining proofs are • given in Appendix 5.

EXAMPLE 9 (a)

In ~n2)

(b)

W

=

These are examples ofthe limits in Theorem 5.

2 ~n

---> 2. 0 =

0

= n2/n = (n lln )'---> (1)2 = I

(c) '{i3;; = 3 /n(n lln ) ---> 1·1 = I ' (d)

(-t)"---> 0

Formula 1 Formula 2

Formula 3 with x

=

Formula 4 with x

= _!

3 and Formula 2

2

10.1 Sequences

I Factorial Notatioo

(e)

The notation n! (Un factorial") means the product I . 2 . 3 ... n ofthe integers from I to n. Notice that (n + I)! = (n + 1). n!. Thus, 4! ~ 1'2'3'4 ~ 24 and 5! = 1·2·3·4'5 = 5'4! = 120. We derme OJ to be 1. Factorials grow even faster than exponentials, as the table suggests. The values in the table are roooded.

It

I 5 10 20

en 3 148 22,026 4.9 X 108

It!

I 120 3,628,800 2.4 X 10 18

(f)

(n ~ 2)" (I + --rnn~e-2 =

loon

-~O

n!

Fonnula 5 with x

=

-2

Fonnula 6 with x

=

100

539

•

Recursive Definitions So far, we have calculated each a" directly from the value of n. But sequences are often defmed recursively by giving

1. The value(s) of the initial tenn or tenns, and 2. A rule, called a recursion formula, for calculating any later tenn from tenns that precede it.

EXAMPLE 10

a,

(a) The statements = I and a" = an-l + I for n > I define the sequence 1,2,3, ... , n, ... ofpositive integers. With = I, we have a2 = + I = 2, a3 = a2 + I = 3, and so on. (h) The statements = I and an = n'an-l for n> I defme the sequence 1,2,6,24, ... , n!,... of factorials. With a, = I, we have a2 = 2· a, = 2, a3 = 3'a2 = 6,,,. = 4'a3 = 24, and so on.

a,

a,

a,

(c) The statements a, = I, a2 = I, and an+l = a" + an-l for n > 2 define the sequence I, 1,2,3,5, ... of Fibonacci numbers. With = I and a2 = I, we have a3 = I + I = 2, a4 = 2 + I = 3, as = 3 + 2 = 5, and so on.

a,

(d) As we can see by applying Newton's method (see Exercise 133), the statements 2 XQ = landxn+l = Xn -- [(sinxn -- xn )J(cosxn -- 2xn)]forn > o define 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 {an} is bounded from above if there exists a number M such that an ,,; M for all n. The number M is an upper bound for {an}. If M is an upper bound for {an} but no number less than M is an upper bound for {an}, then M is the least upper bound for {a,,}. A sequence {a,,} is bounded from below if there exists a number m such that an ;" 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 {an}, then m is the greatest lower bound for {a,,}. If {an} is bounded from above and below, the {an} is bonnded. If {an} is not bounded, then we say that {an} is an nnbounded sequence. EXAMPLE 11 (a) The sequence 1,2,3, ... , n, ... has no upper bound since it eventnally surpasses every number M. However, it is bounded below by every real number less than or equal to 1. The number m = I is the greatest lower bound ofthe sequence. 123 n . (h) Thesequence Z'3'4"'" n + I"" lSbounded above by every real number greater than or equal to I. The upper bound M = I 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


I

Convergent sequences are bounded

If a sequence {a,,} converges to the number L, then by def"mition there is a number N such that Ia" - L I < 1 if n > N. That is,

L-lO

• 1 _ 1 sm1j

= 2n

n

I 3 7 17 a a + 2b --I' 2' S' 12'···· b' a + b •....

----

n2

77. all

85. all

(In n)'

a. VerilY that X1 2 - 2Y12 ~ -I,xi - 2yi ~ +1 and, more generally, that if a 2 - 2b 2 = -I or + I , then

(a + 2b)2 - 2(a +

I (3I)' + vzn

b)'

X n+l

90. an

1

=

/,' pdx, I

!'(x )· n

Do the sequences converge? If so, to what value? In each case, begin by identifYing the functioo f that generates the sequence.

p>1

1 X

Recursively Deftned Sequences In Exercises 91-98, assume that each sequence converges and fmd its

limit 91. at = 2,

Cln+l = 1

92. at = -1, a ll +1 93. al = -4,

72

+ all

95. al = 5, 96. al ~ 3,

I

an + 6 = all + 2

-

2

X n+l = X n -

~ =

b. Xo = I,

X n +l = X n -

---'0;--

c.

X n+l = XII -

1

Xo =

1,

V8 + Za, V8 + Za,

a,+1 =

I

x n2

a. :to = I,

x"

1

2: + Xn

tanxn - 1

sec2 X n

102. a. Suppose that f(x) is differentisble for all X in [0, I] and that f(O) ~ O. Define sequence {a,} by the rule a, ~ nf(l/n). Show that lim,_ooa, ~ 1'(0). Use the result in part (a) to fmd the limits of the following sequences {a,}.

""+1 = ""+1 = vSa;; ""+1 ~ 12 - v;;.

94. al = 0,

-I,

f(x,)

= Xn -

I

1/,'1xdx

or

101. Newton'. method The following sequences come from the recursioo formula for Newtoo's method,

= --n-

• ,,------; •~ vn 2 - 1 - vn 2 +n

= +1

b. The fractions Tn = xJYn approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that r,2 - 2 ~ ±(lly,)2 and that y, is not less than n.)

87. an=n-~

y;;

89. an = 11

b. an = n tan-I

d. a,

I

97.2,2+ , 2 + - -,2+ I , ... 2 1 2+-2 2+-2+ 1 2 98.

Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third. sequence. Let XII and y, he, respectively, the numerator and the denominator of the nth fractioo r, = xJy,.

respectively.

(Inn)"')

84.a,=~ 86. a, =

X n+1 =Xl +X2 +"'+X/I'

=

+ 5,)1/,

1-1

n!

10 611

71. a,

76. a, = sinh (In n)

82• a,

99. The first tenn of a sequence is Xl = 1. Each succeeding term is the sum of all those that come hefore it:

~ nn~ (Hint: Compare with lin.)

all = - - , -

64.

Theory and Examples

62. a, = vt32n + 1

VI, VI + VI, VI + VI + VI, VI + VI + VI + VI, ...

k

c.

an

= n(e 1jn

-

1)

~ nln(1 +~)

103. Pythagorean triple. A triple of positive integers a, b, and c is called a Pythagorean triple if a 2 + b 2 = c 2. Let a he an odd positive integer and let

be, respectively, the integer floor and ceiling for a2/2.

10.1

Sequences

543

2' - I

117. a" ~ ~ 119. a"

l~J

~ «(-I)' + l)(n ~

120. The fl!St tenn of a sequence is Xl = cos (I). The next terms are X2 = XI or cos (2), whichever is larger; and X3 = X2 or cos (3), whichever is larger (farther to the right). In general,

x.+1 6

121. a" = I

a

a. Show that a 2 + b 2 ~ c 2 • (Hint: Let a ~ 2n band c in tenns ofn.)

+

104. The nth root of n! a. Showthat1im.~oo(2mr)1/(2n)= I and hence, using Stirling's approximation (Chapter 8, Additional Exercise 32a), that

D b.

for large values of n.

Test the approximation in part (a) for n ~ 40,50, 60, ... , as far as your calculator will allow.

b. Prove that lim.~oo(l/n') = 0 if c is any positive constant (Hint: If. = 0.001 andc = 0.04, how large shouldNbeto ensure that II/n' - 01 < • ifn > N?)

106. The zipper theorem Prove the "zipper theorem" for sequences: If {a.} and {b.} both converge to L, then the sequence

126. Uniquene•• of least upper bounds Show that if M 1 and M2 are least upper bounds for the sequence {a.}, then M1 ~ M2.

That is, a sequence cannot have two different least upper bounds. 127. Is it true that a sequence {a.} of positive numbers must converge if it is bounded from above? Give reasons for your answer. 128. Prove that if {a.} is a convergent sequence, then to every positive number E there corresponds an integer N such that for all m

andn,

= I.

108. Prove that limn~ooX1/. = I, (x

> 0).

110. Prove Theorem 3.

In Exercises 111-114, detennine ifthe sequence is monotonic and ifit is bounded.

3n + I n+T

2"311 113. an =-, n.

(2n + 3)! 112. a" = (n + I)!

n>N

~

Ia", -

a" 1
N and

131. For a sequence {a,,} the tenns of even index are denored by au and the terms of odd index by au+ 1. Prove that if au ---+ L and a2k+1 ~ L, then an ~ L.

converges to L.

111. a. =

n+ I 122. an = - n -

129. Uniquene•• of limit. Prove that limits of sequences are unique. That is, show that if L 1 and L, are numbers such that a" -+ LI and an -+ L2. then Ll = L,..

if c is any positive constant.

109. Prove Theorem 2.

+..j,2.'

+ I)}.

{x.. cos (n

124. at = 1, an+l = 2a" - 3 125. The sequence {n/(n + I)} bas a least upper houud of 1 Show that if M is a number less than I, then the terms of {n/(n + I)} eventually exceed M. That is, if M < I there is an inreger N such that n/(n + I) > M whenever n > N. Since n/(n + I) < I for every n, this proves that I is a least upper bound for {n/(n + I)}.

105. a. Assuming that limn~oo(l/n') = 0 if c is any positive constant, show that

107. Prove that limn~oo'Cf,;

~ max

I and express

b. By direct calculation, or by appealing to the accompanying figure, fmd

vt,;! ~ ~

I)

D

= Xn -

f'(x )· n

a. Show that the recursion formula for f(x) ~ x 2 - a, a > 0, can be written as X.+1 = (x. + a/x.)/2. b. Starting with Xo = I and a = 3, calculate successive tenns of the sequence until the display begins to repeat. What number is being approximared? Explain.

544


x,

a. Calculate and theo plot the flISt 25 terms of the sequeoce. Does the sequeoce appear to be bouoded from above or below? Does it appear to converge or diverge? If it does cooverge, what is the limit L? b. If the sequence converges, find an integer N such that Ia" - L I :5 0.01 for n '" N. How far in the sequence do you have to get for the terms to lie within 0.0001 ofL?

0134. Arecursivedefioi1ionof"./2 Ifyou starl with = I anddefme the subsequent terms of {x,} by the rule x, = X,-1 + COSX,-I, you geoerate a sequence that converges rapidly to 'If/2. (a) Try it. (b) Use the accompaoying figure to explain why the convergence is so rapid. y

135. a, =

COSXn_l

%

136. a" = I

= an + 5"

137. at = I,

Qn+1

138.

an+! = an

at =

139. an

=

1,

+ (-2)" 140. Cln

sinn

sinn


Use a CAS to perform the followiog steps for the sequeoces in Exercises 135-1%.

10.2

(I + 0~5)"

=

n sin

k

Inn

141. a, = -n-

142. a" = 11

143. a, = (0.9999)'

144. a" = (123456)'/'

145. an

=

n 41 1%. a" ~ 19"

8' I

n.

Infinite Series An infinite series is the sum of an infmite sequence of numbers

a, + 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 + ... + an

is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n gets 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

I +

I

I

I

I

2 + 4: + 8 + 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

I

First:

SI =

Second:

s2=1+ 2I

Third:

I I s,=I+-+2 4

nth:

I I I s, =1+-+-+ .. ·+-2 4 2,-1

Suggestive expression for partial sum

2 - I

I 3 2 7 4

2 - -I 2 2 - -I 4

2' - I 2,-1

I22,-1

10.2

Infinite Series

545

Indeed there is a pattern. The partial sums form a sequence whose nth tenn is SrI

=

1 2 - 2n - 1 •

This sequence of partial sums converges to 2 because 1im,,~00(l/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 ofpartial 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.

I

1/4

~

o

III

FIGURE 10.8 As the lengths I,

,/2, ,/4, '/" ...

1/8

2

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 + a2 + ,.. + 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 a, + a2 + ... + 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 shorthand when summ.a.tion from 1 to 00 is

understood

546


Geometric Series Geometric series are series of the form 00

a+

+

aT

ar 2

+ ... + ar n- t + ...

=

L ar n- t 11=1

in which a and r are fixed real numbers and a oF O. The series can also be written as ~:o ar". The ratio r can be positive, as in I I (I)a-l 1+-+-+···+ + ...

2

4

2

r~1/2.a~1

'

or negative, as in

I-t+~- .. ·+

r~-1/3.a~1

( - t f ' +....

If r = I, the nth partial sum of the geometric series is Sa = a + a(l) + a(I)2 + ... + a(I)a-l = na,

and the series diverges because lima~oo Sa = ± 00, depending on the sign of a. If r = -I, the series diverges because the nth partial sums alternate between a and O. If 1r 1 oF I. we can detennine the convergence or divergence ofthe series in the following way:

rSn

=

+ aT + ar 2 + ... + ar",-l aT + ar 2 + ... + arn- 1 + ar

Sn - rSn

=

a - ar"

8n = a

ll

Multiply So by r. Subtract rSn from Sn' Most of the terms on the right cancel.

s.(1 - r) = a(1 - r a) 8n =

Factor.

a(1 - r a ) 1- r '

(r oF I).

We can solve for 811 if r #:- 1.

If 1 rl < I, then r a --> 0 as n --> 00 (as in Section 10.1) and Sa --> a/(I - r). Iflrl then 1ral--> 00 and the series diverges.

> I,

Iflrl < I, the geometric series a + ar + ar2 + ... + ar a- I + ... converges to a/(I - r): 00

~ 11=1

ar n -

t =

_a_, 1- r

Irl

•

which converges to kA by the Constant Multiple Rule for sequences.

All corollaries ofTheorem 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 diverge.

Caution

~bn

diverges, then ~(an + bn) and ~(a.

Remember that ~(an + bn) can converge when ~a. and ~bn both diverge. I + I + I + "'and~bn = (-I) + (-I) + (-1) +'''diverge, = 0 + 0 + 0 + ... converges to O.

Forexample,~an = whereas ~(an + bn)

EXAMPLE 9 00

(s)

L

n=1

- bn) both

Find the sums of the following series.

3n-' _ 1 6n ,

00

=

L

n=1 00

=

L

11=1

(

1 1 ) zn-' - 6n-'

1 zn-' -

I 1 - (I/Z)

4 = Z -6- = 5

5

00

L

11=1

I 6n-' 1

1 - (1/6)

Difference Rule

Geometric series with a ~ 1 andr ~ 1/2,1/6

550


(b)

Constant 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

00

Lan

+

= at

a2

+ ... +

ak-l

n=1

+ ~ an" 1I=k

Conversely, if~:;"_ka. converges for any k

>

1, 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 n in the fannula fora, by n - h: 00

00

~ an =

11=1

~ an-h

= at

+

a2

+ Q3 + ....

n=-l+h

To lower the starting value ofthe index h units, replace the n in the fannula for a. by n 00

+ h:

00

:L an = n=l-lt :L an+h = 11=1

at

+

a2

+

Q3

+ ....

We saw this reindexing in starting a geometric series with the index n = 0 instead of the index n = 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

1 1

+ 2 + 4 + ...

as 00

1

~2n,

00

1

~ 2,-5' n=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 Find;ng nth Partial Sums In 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.

Using the nth-Term Test In Exercises 27-34, use the nth-Term Test for divergence to show that

the series is divergent, or state that the test is inconclusive.

1. 2+~+~+l+ ... +~+ ... 3 9 27 3'-1

27.

~n

n + 10

2 ~ + _9_ + _9_ + ... + _9_ + ... • 100 1002 100' 100'

00

I

29.

~~ 11-0 n

3.1_

00

1 + 1 _ 1 +"'+(_1),-1_1_+ ... 2

4

8

2,-1

~

Series with Geometric Terms In Exercises 7-14, write out the fITst few terms ofeach series to show how the series starts. Then fwd the sum of the series. 00 I 00 (-I)'

7. ~-4'

8. ,-2 ~4'

,-0

00 7

9. 11. 13.

00 5 10. ~(-I)' 4'

~4'

,-0

(5 + I) (I + -----sn (-I)') ~ 00

~

2"

00

00

~

14.

2"

m (~Y +

32.

,-I

1&

(tY

e;:I)

+

(~)' + (~)' +

.

(~2 Y+ (~2)' + (~2)' + (~2)' + (~2)' + ...

Repeating Declmals 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 ...

22. O.d ~ O.dddd 23. 0.06

= 0.06666

24. 1.414

.

, where d is a digit .

1.414414414 ...

~

25. 1.24123

~

26. 3.142857

~+-+ e n

11-0

00 34. ~ cosmr

,-0

Telescop;ng Series In 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. 00 35. ~

n-I

(In-~I I) n

00 37. ~ (In v;;-+l

-

36.~(3

,-I \;;2

3) (n + I)'

In y';;)

00 38. ~ (tan (n) - tan (n - I)) 11=1

1.24123 123 123

= 3.142857 142857

. .

00 40. ~

,-I

(v;;-+4 -

~)

Find the sum ofeach series in Exercises 41-48. 41

~. 4 • .;:1 (4n - 3)(4n

43

~ 40n • ,~1 (2n - 1)'(2n + 1)2

.

+ (t)' + (t)4 + (t)' + ...

21. 0.7 = 0.7777

,

(5 I)

16. 1+ (-3) + (-3)2 + (-3)' + (-3)4 + 17. (t) +

~+11-1 n + 3

12.~ 2"-3"

3"

In Exercises 15-18, determine if the geometric series converges or diverges. If a series converges. fmd its sum.

15. 1+

30.

00

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) + ...

,-I (n 00

00 I 31. ~ cos n

4.1- 2 +4 - 8 + ... + (-1)'-12'-1 + ...

n(n + I) + 2)(n + 3)

28.

45.

+ I)

00(1 I) ~ y';; - v;;-+l

42

•

~

~

6 (2n - 1)(2n + I)

44.~2n+1 2 ,~1 n (n

46.

00(1 ~ 21/,

+ 1)2 I) - 21/(,+1)

~. ( I _ I ) '.;:1 In(n+2) In(n+l) 00 48. ~(tan-l (n) - tan-I (n + I)) 47

,-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 00(1)' SO. ~(v'2r 49. ~ • r;:: ,-0 11=0 v2 00 51. ~ (-1)'+1 3, 52. ~ ( -1)'+1 n 11=1 2 11=1 00 00 54. ~ cos:'11' 53. ~ cosmr n=O 5 ,-0

552


00

00

57. 59.

61.

,-I

2

00

58.

211 _ 1

~3" 00

83. Show by example that ~(aJb,) may diverge even though ~a" and ~b, converge and no b, equals O.

56. ~ In 3'

~ 10' ,-I 00

I

00

e-'" ,-0

55. ~

00 60.~ 00

1

illustrate the fact that toAB.

I

I)'

,

62. ~"IFI n!

~ I~O'

2"+ 3" 63. ~11=1 4"

00

00

64.

21'.

~a"b,

85. Show by example that ~(a,/b,) may cooverge to something other than A/Beven whoo A ~ ~a" B ~ ~b, # 0, and no b, equals O.

I-Ii

(

~a, and B = ~b, that may converge without being equal

84. Find convergoot geometric series A =

I

~"' Ixl> 11=0 X

86. If ~a, converges and a, > 0 for all n, can anything be said about ~(I/a,,)? Give reasons for your answer.

+ 4/1

~311+411

87. What happeos if you add a fmite number of terms to a divergent series or delete a ftnite number of tenns :from a divergent series?

65.~1n( n ~I)


11""1

Geometric Sertes with a Vartable 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'"

a. a = 2 b. a = 13/2. 90. Find the value of b for which 1 + e b + e 2JJ + e 3b + ... = 9. 91. For whst values of r does the infinite series

,-0

,-0

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 ~arll-1 that converges to the number 5if

00 (-I)' ( I )' 72. ~-2- 3 + sinx

converge? Find the sum ofthe series when it converges. In Exercises 73-78, find the values of x for which the given geometric senes converges. Also, fmd the sum of the series (as a functioo of x) for those values of x. 00

00

73. ~2'x'

74. ~(-I)Y'"

,-0

,-0

00

75. ~(-I)'(x

,-0

92. Show that the error (L - so) obtained by replacing a convergoot geometric series with one of its partial sums Sn is arn/(l - r). 93. The accompanying figure shows the first five of a sequeuce of squares. The outermost square hss an area of 4 m'. 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.

00 ( - I)"(x - 3)' 76. ~ 11=0 2

+ I)'

00

00

77. ~ sin"x

78. ~(lnx)'

,-

Theory and Examples 79. The series in Exercise 5 can also be written as

~

,-I (n

I

~

d

+ I)(n + 2) an

,--I (n

Write it as a sum beginning with (a) n (c) n = 5.

~

I

+ 3)(n + 4) . -2, (b) n

~

0,

80. The series in Exercise 6 can also be written as 00

5

~ n(n + I)

00

and

5

~ (n + I)(n + 2) .

Write it as a sum beginning with (a) n = -I, (b) n = 3, (c) n ~ 20. 81. Make up an infinite series ofnonzero terms whose sum. is a. 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.

94. Helga von Koch's snowflake enrve Helga voo Koch's snowflake is a curve of infmite leogth that oocloses a regioo of fmite area. To see why this is so, suppose the curve is generated by starting with an equilateral triangle whose sides have leogth I.

a. Find the loogth L, of the nth curve C, and show that

limn.. . . oo Ln =

00.

b. Find the area A, of the region oocloscd by C, and show that lim,~ooA, = (8/5)A 1 •

6000 c,

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 ~~1

00 •

11

If f(x) dx is [mite, the right-hand inequality shows that ~a" is [mite. If .!too f(x) dx is infmite, the left-hand inequality shows that ~an is infmite. Hence the series 00

and the integral are both finite or both infinite.

_

10.3 The Integral Test

EXAMPLE 3

555

Show that the p-series

~~= I~+~+~+···+~+··· (p a real constant) converges if p

Solution

If p

>

>

I, and diverges if p

I, then f(x) = l/xP is a positive decreasing function ofx. Since

1

00

1

-1 dx = xP

1

00

x-P dx = lim

1

converges if p

> I, diverges if p

:5 I.

[ ~x

-p+l ]b ~

-P + 1

b_oo

=

_1_ lim ( _ I1 - I)

=

_1_(0 _ I)

b~oo

1- P

1

bP=

I-p

I ~ 1 I The p-series 1I~1 nP

I.

:5

_1_

p-I'

bP - 1 --+ 00 as b --+ 00 because p - 1 > 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

foo ~dx =

11

xP

_1_ lim (b1-p - I) 1 - Pb_OO

= 00.

The series diverges by the Integral Test. If p = I, we have the (divergent) hannonic series

1 1 ... +-+ 1 ... 1 +-+-+ 2 3 n We have convergence for p

>

1 but divergence for all other values ofp.

•

The p-series with p = 1 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.) The series ~~~l (1/(n 2 Integral Test. The function f(x) = 1/(x 2 x ~ 1,and

EXAMPLE 4

00

1 1

+ +

I)) is not ap-series, but it converges by the I) is positive, continuous, and decreasing for

b~oo

_1_ dx = lim arctan x b +I [ ],

x2

=

lim [arctan b - arctan I]

b~oo

1T - 1T -

2

4

1T

-

4·

Again we emphasize that 1T/4 is not the sum ofthe 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 -

Sn

= 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

+

cz,,+3

+ ... ;;,:

f(x) fix.

• +1

Similarly, from Figure 10.11b, we fmd an upper bound with

1

00

R.

= cz,,+1

+

a.+2

+

cz,,+3

+ ... ,,;;

f(x) fix.

These comparisons prove the following result giving bounds on the size ofthe 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 ~cz" converges to 8. Then the remainder R. = 8 - s. satisfies the inequalities

1

00

1

00

f(x) fix ,,;; R. ,,;;

n+l

f(x) fix.

(1)

n

Ifwe add the partial sum s. to each side of the inequalities in (I), we get

1

00

s.

+

1

00

f(x) fix ,,;; 8 ,,;; s.

+

n+l

f(x) fix

(2)

n

since s. + R. = 8. 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 ofthe interval containing 8, as given by (2).

EXAMPLE 5 n = 10.

Estimate the sum of the series ~(I/n2) 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

1N.

for all (a) If ~c" converges, then ~a" also converges.

(b) If

y

~dn

diverges, then

~a"

also diverges.

Proof In Part (a), the partial sums of ~a" are bounded above by 00

M

= at

+

a2

+ ... +

aN

+

:L n=N+l

Cn ·

They therefore form a nondecreasing sequence with a limit L ,;; M. That is, if ~c" converges, then so does ~a". Figure 10.12 depicts this result, where each term of each series is interpreted as the area ofa rectangle Gust like we did for the integral test in Figure 10.11). In Part (b), the partial sums of ~an are not bounded from above. Ifthey were, the partial sums for ~dn would be bounded by FIGURE 10.12 If tire total area ~c" ofthe taller en rectangles is f'mite, then so is the total area ~all of the shorter a" rectangles.

00

M' = dl + d2 + ... + dN +

~ a" n=N+l

•

and ~d" would have to converge instead of diverge. EXAMPLE 1

We apply Theorem 10 to several series.

(a) The series 00

~ 5n

5 - I

diverges because its nth term

5

5n - I

=_1_>1 I n n-

S

HisTORICAL BIOGRAPHY

Albert of Saxony (Co. 1316-1390)

is greater than the nth term of the divergent harmonic series. (b) The series

~l=

I + l + l + l + ... I! 2! 3! converges because its terms are all positive and less than or equal to the corresponding terms of 00 I I I 1+~n=I+I+-+2+"" n~O 2 2 2 The geometric series on the left converges and we have "~O n!

00

I +

I

~ 2n = I + I _

I (1/2) = 3.

The fact that 3 is an upper bound for the partial sums of~;::'o (lin!) does not mean that the series converges to 3. As we will see in Section 10.9, the series converges to e. (C) The series

5+~+1+1+ I + I + I + ... + I + ... 37 2+V1 4+V2 8+V3 2"+Y!;; converges. To see this, we ignore the first three terms and compare the remaining terms with those ofthe convergent geometric series ~:;'~o(l/2n). The term I/{ 2n + V;;) of

560

Chapter 10: Infinite Sequences and Series the truncated sequence is less than the corresponding term 1/2' of the geometric series. We see that lenD by term we have the comparison

1+

1 + I + I + ... ,,;1+1+1+1+ .... 2+V1 4+0 8+\13 2 4 8

So the truncated series and the original series converge by an application of the Comparison Test. _

The Limit Comparison Test We now introduce a comparison test that is particularly useful for series in which tin is a rational function of n.

THEOREM l1-Llmtt Comparison Test all n

2:

Suppose that a,

>

0 and b,

>

0 for

N (N an integer).

1. If lim ban = c n

11_00

>

0, then

2. If lim ban = 0 and 11- 00

11

3. If lim ban =

00

n

11_00

~a, and ~b, hoth converge or hoth diverge.

~b, converges, then ~a, converges.

and ~b, diverges, then

~tIn diverges.

Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 55a and b. Since c/2 > 0, there exists an integer N such that for all n n

Thus, for n

>

>

~ I~ -cl

N

0 and lim an

= 00.

.~oo

> 0 and lim n'a" .~oo

Prove that ~an diverges.

= O. Prove that ~a. con-

61. Show that ~::,,~, «In n)q/nP) converges for p>l.

-00

Does your CAS fmd a closed form answer for this limit?

2

n (n

+ 1)

•

e. Explain why taking the fITst M terms in the series in part (b) gives a better approximation to S than taking the fIrst M terms in the otigina1 series ~::'-1 (I/n 2 ).

00 ?

b. Plot the fITst 100 points (I4 (n + I)(n + I) n+ I .

=

The series diverges because p = 4 is greater than I. (c) Ifa, = 4'n!n!/(2n)!, then

4'+I(n + I)!(n + I)! (2n)! (2n + 2)(2n + 1)(2n)!' 4'n!n!

a.+l

a. =

4(n + I)(n + I) 2(n + I) (2n + 2)(2n + I) = 2n + I ---> 1.

Because the limit is p = I, we cannot decide from the Ratio Test whether the series converges. When we notice that a.+l/a, = (2n + 2)/(2n + I), we conclude that a.+l is always greater than a. because (2n + 2)/(2n + I) is always greater than 1. Therefore, all tenns are greater than or equal to a, = 2, and the nth tenn does not ap_ proach zero as n ---> 00. The series diverges.

The Root Test The convergence tests we have so far for ~a, work best when the fonnula for a, is relatively simple. However, consider the series with the tenns

n/2" a, = { 1/2',

nodd

n even.

To investigate convergence we write out several terms ofthe series:

fa=l+l+~+l+~+l+2+ ... 2 3 4 5 6 7

'~1'

2

2 2 ' 113

2

2

15

2

2

17

= 2 + 4 + 8 + 16 + 32 + 64 + 128 + ... Clearly, this is not a geometric series. The nth tenn approaches zero as n ---> 00, so the nth-Tenn Test does not tell us if the series diverges. The Integral Test does not look promising. The Ratio Test produces

a,+1 =

a.

12~'

n odd

n+ I 2 '

n even.

All n ---> 00, the ratio is alternately small and large and has no limit. However, we will see that the following test establishes that the series converges.

THEOREM 13-The Root Test

Let

~a,

be a series with

a. ;", 0 for n ;", N,

and suppose that

Then (8) the series converges if p < I, (b) the series diverges if p finite, (0) the test is inconclusive if p = I.

>

lor P is in-

566


Proof (8) P < 1. Choose an e > 0 so small that p + e < I. Since -xYa: ---> p, the terms -xYa: eventually get closer than e to p. In other words, there exists an index M 2: N such that -xYa: I for n > M. The terms of the series do not converge to zero. The series diverges by the nth-Term Test. (0) p = 1. The series ~:1 (lin) and ~:;C~1 (1/n 2) show that the test is not conclusive when p = I. The first series diverges and the second converges, but in both cases -xYa: ---> 1. _

/2" {n1I' 2 ,

. the senes · · WI t h tenus a. = Consl'der agam

EXAMPLE 2

nodd 11 even.

Does ~a. converge?

Solution We apply the Root Test, finding that

/2, {'fin1/2,

%: = a.

nodd n even.

Therefore,

t ,;

~.

-xYa: ,;

I

Since 'fin ---> (Section 10.1, Theorem 5), we have Iim"~",, -xYa: = 1/2 by the Sandwich Theorem. The limit is less than 1, so the series converges by the Root Test. _

EXAMPLE 3 ""

(8)

Solution (8) (b)

(0)

2

~ ~.

Which of the following series converge, and which diverge? (b)

~ 2:

n=-1 11

(0)

"" (

~

I ). 1+ n

We apply the Root Ths! to each series.

"" n 2 1f.n2 W L 2' converges because 2' = •• r.:;. n=1 v~

2' . ~2' L"" --, diverges because --, = 11

,,=111

=

('finY

--2- --->

1.

2 2 r), ---> --, > 1. n 1

( •• V

L""(I)' -+1 converges because ~(I)' -+1 = 11 n

,,=1

12

2
0 11

< 1.

-

10.5 The Ratio and Root Tests

567

Exercises 10.5 Using the Ratio Test

00

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. 4

4

~

~n~2

II_I

3

00

21l +1

37. 39.

~

41

311 +2

6·L11=2 Inn

n2 (n n!

+ 2)!

8 ~ • ,~I

32/1

(20

n5' 3) In (n

+

L

11. ~ (4n + 3)' 3n - 5

13 ~ • ~ (3

15.

L

00

,-I

(

+

8 (l/n))2o

12. 14.

i i

(In(e sin'

+

I)

2

+

k) r'

(~)

I)'"

,~OO

= e")

I

00

~n

1+,

Determining Convergence or Divergence

00

19.

21.

23.

20. ~ InOIl

I.

&1 ~O' 00

~

2

22.

+ (-I)'

24.

1.25'

3)'

,-I

31.

OO(

28.

n

(I I)

~. ~n ~ (n

/I-I

+

/I-I 00

(I I)'

~_ n~n

.-=1

l)!

~n

L

11=1

L n . ,-2 (In n )('/2)

40

(Inn )' n lin n. n n(n + 2)!

3' L--,;; n2 00

42.

11-1

~ (20

(n!? (20)'•

45. al = 2,

44.~

+ 3)(2' + 3) 3'+2

,-I

all +l =

46.

al =

1, °n+l

47.

a, =

3'

48.

al

49.

al =

2,

SO.

a, =

5,

51.

a,

52.

al =

I

= 3,

n

= I,

a"

n

3n - I 2n + 5 ""

n

= n

+ 1 an

I

+ Inn

n

+ Inn

0n+l =

I

2'

an

1 + tan-I n =

""+I = an+!

1 + sinn

Cln+1 =

n

n

+ 10

an

alJ

a, ~ t, ""+1 ~ ~ 54. a, t, ""+I (",,)'+1 =

=

Convergence or Divergence Which of the series io Exercises 55-62 converge, and which diverge? Give reasons for your answers. 00 (3n)! 00 2' I I 55 L~ 56. n!(n + I)!(n + 2)! • ,~I (2o)!

&1

00

57.

(n!)'

00

:.:1 00

,

60.

1.3 .... '(20 - l) 4lJ 2lJn!

62.

~ [2 -4-

1.3

(20-1)

(2n)](3'

+ I)

(n!)'

LlJ-l n(lJ1) 00

~ 00

58.

n( ') 2 lJ

61.

+ 2)

00

L ( ')2 lJ-l n

59. Y

I

n.

,

Recursively Def"med Terms Which of the series ~;l a" defmed by the formulas in Exercises 45-54 converge, and which diverge? Give reasons for your answers.

(Inn)'

30. ~ Ii - n 2

32. I)(n

I)'

3n

L-,n 00

.-=1

33. ~

L-

,-I /1""1

29. ~ li- n 2 00

(-2)' , 3

26·LI--

~ In;

/I-I

(n; 2)' ,-I ~ 00

00 ( 25·LI-"

27.

00

53.

I

00

~n!e-II 00

L L

11=2 00

In Exercises 17-44, use any method tc determine if tire series con-

verges or diverges. Give reasons for your answer. 00 Vi 00 17. ~ nZII 18. ~n2e-

~

+ I)!

3'n.,

38. "5' n;

1-"

(Hint: lim (I + x/n)' 16. "5'.

,

n2'(n

11'-

00

. ,_I

43.

In Exercises 9-16, use the Roo! Ths! tc detennine if each series converges or diverges. 00 7 00 4' 10. (3 )" 9. /I-I (2n + 5 )' /1""1 n

L

oo

36.

3!n!3 11

~ (20 :

00

Using the Root Test

/I-I

+ 3)!

(n

00

. ~n3"-1 00

~:II 00

~ 00

L, 00

35.

,

~ (;,)'

568


Theory and Examples 63. Neither the Ratio Test nor the Root Thst helps with p-senes. Try

65. Let a, = {

them on

Does I ~n=l nP 00

~an

n/ 2'

1/2':

ifn is a prime numher otherwise.

converge? Give reasons for your answer.

66. Show that ~:_I 2(i'J/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

-2

+

1- 2

1

1

1

2 + 3 - 4 + 5 - ... +

I -

1-

1

1

1

2+4- 8+

+3- 4+5- 6+

(_1)'+1 n

+ ...

(1)

(-1)'4

+ ~ +...

(2)

+ (-1)'+ In + ...

(3)

We see from these examples that the nth term of an alternating series is of the fonn 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)n+11 = I n~l n

_1 + 1_1 + ... 2

3

4

clearly satisfies the tbree requirements of Theorem 14 with N = I; it therefore converges. _ Rather tban directly verifying the defInition Un ;;,: Un +1o a second way to show that the sequence {un} is nonincreasing is to define a differentiable function f{x) satisfying f{n) = Un' That is, the values of f 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 f{x) is nonincreasing for x ;;,: N. It follows that f{n) ;;,: f{n + I), or Un ;;,: Un + 10 for n ;;,: N.

EXAMPLE 2 Consider the sequence where Un = IOn/{n 2 + 16). Detme f{x) 2 IOx/{x + 16). Then from the Derivative Quotient Rule,

f '() x

=

2

=

1O{16 - x ) 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. ~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 In Exercises 1-14, detennine if the alternating series converges or diverges. Some of the aeries do not satislY the conditions of the Alternating Series Test.

1. 3.

~(_I)'+I_I-

,~1

V;;

~(-I)'+I~ IFI n3

2.

11-1

9.

13.

n,n ++ 45

~(_I)'+I I~n

12. i(-I)'In(1 +

~(_I)'+I

V;; + I n+ 1

14.

Inn

k)

V;;

+

I

Absolute and ConcUtional Convergence Which of the aeries in Exercises 15-48 converge absolutely, which converge, and which diverge? Give reasons for your answers. 00

15. L(_I),+I(O.I)' 17.

18.

V;;

00

n

00

23. L(-I),+l 25.

3 5

+n

+n

~(_I)'+II ~ n

11=1

(-I)' ,~I I + V;; 0 0 ,

19. L(-I)'+I+IFI n +1 00 1 21. ~(-I)' + 3

n-l

~

n

20. L(-I),+I 11=1

n;

00

24. L

s:,n

(-2)'+1 + 5'

11-1 n

26.

~(_I)'+I(\Ylo)

11=1

32. Y(-5t'

:=1

n (-100)'

00

34.

1

n.

L

(_1)'-1

~,c'-='--- 11-1 n + 2n + 1 00

36. ,~_. CO~n1T ~ 00 (-I)'+I(n!)'

+ I)'

~

(2n)'

00

,(In)!

39.

~(-I) 2"nln

38.

~

40.

~(-I)' (In'+ 1)1

(In)l

00

00

00

11=1

11=1

(nl)'3'

00

+ vn -

43. L(-I)'(Yn /1=1

(-I)'

00

44.

n)

V;;) 00

L.r .~ vn+ vn+ 1

45. L(-I)' sechn

,-I

11=1 00

46. L(-I)'cschn

,-I

47.

I

I

I

4: - 6 + 8 -

48. I

I 10

I

I

+ IT - 14 + ...

1111111

+ 4: - 9 - 16 + 25 + 36 - 49 - 64 + ...

2

00.

22. ~(-I)'

Inn n - Inn

00

37.

n-I

~(-I),_I-

,~1

(0 I)'

00

16. L(-I)'+I---il

n-l

~(_I)'

41. L(-I)'(~ - V;;)42. L(-l)'(~-

~(_I)'+I 3~

:=1

L 11=1

nlnn

11-1

~ I

35. ~_ cos mr ~ nV;; 00 (-I)'(n

~(_I)'+I_I_

11-2

30.

n + 1

00

~(-I)' (n + I)!

10.

~

33.

10"

00

8.

00

,

11=1

10

:=1

:.:1

~(_I)'+I_I-

11=2

00

4. L ( - I ) ' - )(' 11=2 In n 6. L(-I),+I

28.

,-I ~(-I)' ~-1 n

31. Y(-I)'

4

~(_I)'+I (~)'

II-I

11.

~(-I)'+I+ n f2 00

7. ~(_I)'+I n2:

29.

L( -1)'n'(2/3)'

n-I

11=1 00

00

S'L(-I)'--,"IFI n + 1

00

27.

Error Estimation In Exercises 49-52, estimate the magnitude of the error involved in using the sum of the first four tenns to approximate the sum of the en-

tire series.

574


51. f(-I),+1 (O.~I)"

N. you will see in Section 10.7, the sum is In(I.Ot).

,-I

52. I

~I~

i.(

is the same as the sum of the ftrSt n terms of the series I

I

I

I

I

1·2 + 2·3 + 3·4 + 4'5 + 5·6 + .... -1)'1',

0< I < I Do these series converge? What is the sum of the fIrSt 2n

+

I

tenns of the fIrst series? If the series converge, what is their swn? 63. Show that if~:1 a, diverges, then ~:;"_lla,1 diverges.

In Exercises 53-56, determine how many terms should be used to estimate the sum ofthe entire series with an error ofless than 0.001. 1 53. f(-l)' -2--

55.

54. f(_I),+1

n +3

11=1

IFI

~(_I)'+1 (n + ~y;;y

56.

64. Show that if~:1 a, converges absolutely, then

z"n +1

~(-I)' In(ln(~ + 2))

65. Show that if~:1 a" and ~:1 b, brrth converge shsolutely, then so do the following.

D Approximate the sums in Exercises 57 and 58 with an error of magnitude less than 5 X 10..... n .

11=0

58.

f (-I)' 1,

As you will see in Section 10.9,

the sum. is e-1 .

n.

n-O

D

Theory and Examples 59. a. The series

1_1 +

l _ l + l _ l + ... + l _ l + ... 32942783'2'

does not meet one of the conditions ofTheorem 14. Which

one? b. Use Theorem 17 to find the sum of the series in part (a).

D 60.

c. ~

As you will see in Section 10.9, the sum is cos 1, the cosine oflradian.

1

00

57. ~(-I)'(2 )'

The 1intit L of an alternating series that satisfies the conditions of Theorem 14 lies between the values of any two consecutive par-

tial sums. This suggests using the average Sn

+23,,+1 -_

Sn

+ 12 (-1)'+2an+!

ka" (k any number)

66. Show ~ example that ~~=1 a"bll may diverge even if ~~=1 an and ~,-1 b, brrth converge. 67. In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to -1/2. Start the new arrangement with the fifllt negative term, which is -1/2. Whenever you have a sum that is less than or equal to -1/2, start introducing positive terms, taken in order, until the new total is greater than -1/2. Then add negative terms until the total is less than or equal to -1/2 again. Continue this process until your partial sums have been above the larget at least three times and fInish at or below it. If 9" is the sum of the fIrst n terms of your new series, plot the points (n, s,) to illustrate how the sums are behaving. 68. OutHne of the proof of the Rearrangement Theorem (Theorem!?)

a. Let e be a positive real number, let L = ~~-1 an, and let St

= ~:=1 an· Show that for some index Nt and for some

index N2

O?::

Nt,

to estimate L. Compute

I

I

s20+2'21 as an approximation to the sum of the alternating harmonic series. The exact sum is ln2 = 0.69314718 ....

61. The sign of the remainder of an alternating series that satisfies the conditions of Theorem 14 Prove the assertion in Theorem IS that whenever an alternating series satisfying the conditions ofTheorem 14 is approximated with one ofits partial sums, then the remainder (sum of the unused terms) has the same sign as the first unused term. (Hinl: Group the remainder's terms in consecutive pairs.)

62. Show that the sum of the flfst 2n terms ofthe series 111111111

1-2+2-J+J-~+~-J+J-~+'"

Since all the terms aI, a2, ... , aN: appear somewhere in the sequence {b,}, there is an index N, '" N2 such that if n ~ N3 , then (:~k-l bk ) - SN: is at most a sum. of terms elm withm O?:: Nt. Therefore,ifn O?:: N3,

:5

~la.1

k-Nt

+ ISN, - LI
1 because the nth tenn does not converge to zero. At x = I, we get the alternating hannonic series 1 - 1/2 + 1/3 - 1/4 + "', which converges. At x = -1, we get -1 - 1/2 1/3 - 1/4 - "', the negative of the hannonic series; it diverges. Series (a) converges for - 1 < x ,;; 1 and diverges elsewhere.

o

-1

x2n+l

(b)

I2n + 1

.2n -

11

2(.

x 2n - 1

+ 1) - 1 ~ 2. + 1

The series converges absolutely for x 2 < 1. It diverges for x 2 > 1 because the nth tenn does not converge to zero. At x = 1 the series becomes 1 - 1/3 + 1/5 - 1/7 + "', which converges by the Alternating Series Theorem. It also converges at x = -1 because it is again an alternating series that satisfies the conditions for convergence. The value at x = - 1 is the negative of the value at x = 1. Series (b) converges for - 1 ,;; x ,;; 1 and diverges elsewhere.

(c)

>x

o

-1

IU~:'1 I(n + 1)!'x'n! I = X·+I

Ixl

n + 1--+0foreveryx.

.1

(. + I)!

The series converges absolutely for all x.

o

IU~:'1 I

(n

(d)

+

1)!x'+

nix'

l

l=

(n + 1)lxl--+oounlessx

The series diverges for all values of x except x =

•

o

>x

=

o.

o.

•

The previous example illustrated how a power series might converge. The next result shows that if a power series converges at more than one value, then it converges over an entire interval of values. The interval might be finite or inf'mite and contain one, both, or none of its endpoints. We will see that each endpoint of a finite interval must be tested independently for convergence or divergence.

578


THEOREM 18-The Convergence Theorem for Power Serfes If the power series co La,x' = ao + a,x + a2x2 + ... converges at x = c oF 0, then it converges ,-0 absolutely for all x withlxl < lei- If the series diverges atx = d, then it diverges for all x with Ixl > Idl.

Proof The proof uses the Comparison Test, with the given series compared to a converging geometric series. Suppose the series ~:o anc' converges. Then lim,~co anc' = 0 by the nth-Term Test. Hence, there is an integer N such that Ianc'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 ofconvergence R '" O.

Since Ix/c I < 1, it follows that the geometric series ~:o Ix/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 ~an(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 hut 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 coX" 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 cod" diverges. Let S be the set of values of x for which ~:o coX" 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 pcint 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)l" 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 (f"mite)

R (it does not even converge conditionally) because the

interval of convergence.

test for convergence or divergence at each endpoint, as in Examples 3a and b. Use a Comparison Test, the Integral Test, or the Alternating Series Test.

nth term does not approach zero for those values of x.

Operations on Power Series On the intersection of their intervals of convergence, two pcwer series can be added and subtracted term by term just like series of constants (Theorem 8). They can be multiplied just as we multiply pclynomials, 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

~~~oa.x· andB(x) = ~:;;'o b.x· converge absolutely forlxl

=

c. = aob.

If

< R, and

•

= ~akb.-.,

+ a,b.- , + a2b.-2 + ... + an-lb! + 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:

(~x').r~ (-l)'~) \,;~o \,;~o n + I = (I

+ x + x 2 + ...)(x _

~2

+ x; - .. -)

(x _ x; + x; _ .. -) + (x2 _ x; + by!

x

2

5x

~4

Multiplysecondseries ... _ .. -)

+ (x 3

~4 + x;

_

byx 3

x

by

_ .. -) + ...

x'

4

and gather the IllS! four powers.

=x+T+T-6····

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/O - x) = ~:;;'ox' converges absolutely for Ixl < I, it follows from Theorem 20 that 1/(1 - 4x 2 ) = ~~_o(4x2)'convergesabsolutelyforI4x21 < I 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 Theonm

radius of convergence R

>

If ~c.(x - a)' has

0, it defines a function

00

j(x)

= ~c.(x

- a)'

on the interval

a - R

x

-0.5

a

0.5

1.0

The graph of f(x) and its Taylor polynomials FIGURE 10.17

PI(x) = I P2(X) = I P,(x) = I

= e'

+x + x + (x 2/2!) + x + (x 2/2!) + (x'/3!).

Notice the very close agreement near the center x = 0 (Example 2).

/"(0)

In)(o)

2!

n!

1(0) + f'(O)x + - - x 2 + ... + - - x n + ... 2

= I

x x" +x+-+ .. ·+-+ .. · 2 n! 00

=

k

L~,· k-O •

This is also the Maclaurin series for e'. In the next section we will see that the series converges to e' at every x.

10.8 Taylor and Maclaurin Series

587

The Taylor polynomial of order n at x = 0 is Xl

Xli

+".+,. n.

P.(x)=I+x+2

EXAMPLE 3

•

Find the Taylor series and Taylor polynomials generated by I(x) = cosx

atx = O.

Solution

The cosine and its derivatives are

I(x) =

cos x,

f"(x) =

-cos x,

=

-sin x,

p(x) =

sin x,

I'(x)

/o.)(x) = (-I)" cosx, At x = 0, the cosines are I and the sines are 0, so

/0.+1)(0) =

/0.)(0) = (-I)",

o.

The Taylor series generated by 1 at 0 is

1(0)

+ 1'(0)x + l'i!O) x 2 + f"~~0) x3 + ... + /:~O) x" + ... =

I

+ O'x

00

=

~

x2 - 2!

x4

x 2n

+ 0'x 3 + 4! + ... + (-I)" (2n)! + ...

(-I)kx 2k (2k)! .

This is also the Maclaurin series for cos x. Notice that only even powers of x occur in the Taylor series generated by the cosine function, which is consistent with the fact that it is an even function. In Section 10.9, we will see that the series converges to cos x at every x. Because /0.+1)(0) = 0, the Taylor polynomials oforders 2n and 2n + I are identical:

x2 Po.(x) = Po.+l(X) = I - 2!

x4

x 2n

+ 4! - ... + (-I)" (2n)!

.

Figure 10.18 shows how well these polynomials approximate I(x) = cosx near x = O. Only the right-hand portions of the graphs are given because the graphs are symmetric about the y-axis. • J

2

C:+---:-~c-:---:'---Y---r=--:---\':--~----:o~x

o

-1

-2

FIGURE 10.18

The polynomials n

Po.(x) = ~

(-I)*,;2>

(2k)!

converge to cos x as n ~ 00. We can deduce the behavior of cos x arbitrarily far away solely from knowing the values ofthe cosine and its derivatives at x = 0 (Example 3).

588


EXAMPLE 4

It can be shown (though not easily) that

y

f(x) =

o {e~I/x',

x = 0 x i' 0

(Figure 10.19) has derivatives of all orders at x = 0 and that 1')(0) = 0 for all n. This means that the Taylor series generated by f at x = 0 is -----'---~+...L-'---------'----x -2

-I

0

/"(0)

f(O) + /,(O)x + ~ x 2 + ... +

2

/,')(0)

----,;r- x'

+ ...

= 0 + O·x + 0'x 2 + ... + O·x· + ...

The graph ofthe continuous extension of y = e-1/;;- is so flat at the origin that all ofits derivatives there are zero (Example 4). Therefore its Taylor series is not the function itself. FIGURE 10.19

=0+0+···+0+···. The series converges for every x (its sum is 0) but converges to f(x) only at x = O. That is, the Taylor series generated by f(x) in this example is not equal to the function f(x) itselL •

Two questions still remain.

1.

For what values of x can we normally expect a Taylor series to converge to its generating function?

2.

How accurately do a function's Taylor polynomials approximate the function on a given interval?

The answers are provided by a theorem ofTaylor in the next section.

Exercises 10.8 + x' + 1,

FInding Taylor Polynomials

25. f(x) = x'

In Exercises 1-10, find the Taylorpolynomia1s of orders 0, 1,2, and 3 generated by f at a.

26. f(x) ~ 3x s - x'

2. f(x) = sinx,

I. f(x) = e"', a = 0 3. f(x) = Inx, a = I

4. f(x) = In (1 + x),

5. f(x) = I/x, a = 2 7. f(x) ~ sinx, a ~ 'If/4 9. f(x) =~,

a= 0

a= 4

6. f(x) = I/(x

8. f(x)

~

tanx,

+

2),

a

~

a= 0 a =0 'If/4

10. f(x) =~, a = 0

FInding Taylnr Series at x = 0 (Maclaurin Series) Find the Maclaurin series for the functions in Exercises 11-22. 11. e-x 12. xe~ 13. 1

I

16. sin ~

17. 7 cos (-x)

18. 5 cos '1fX

19. cosh x

=

eX

+ e--;t"

21. x' - lx' - 5x

+4

22. x

x'+ 1

FInding Taylor and Maclaurin Series In Exercises 23-32, fmd the Taylor senes generated by f at x 23. f(x) = x' - lx + 4, a = 2 24. f(x) = lx'

+ x' +

3x - 8,

a = I

a ~ 1

28. f(x) ~ 1/(1 - x)',

a ~ 0

29. f(x) ~ e', a ~ 2 30. f(x) = T, a = 1

31. f(x) = cos (lx + ('If/2», a = 'If/4 32. f(x) = y';+l, a = 0

Theory and Examples 37. Usc the Thylor senes generated by e' at x = a to show that

. eX-e-;; 20. sinhx = 2

2

a ~ -1

verges absolutely. 33. f(x) = cosx - (2/(1 - x» 34. f(x) = (1 - x + x') e' 35. f(x) = (sin x) In (I + x) 36. f(x) = x sin' x

1- x

15. sin 3x

- 2,

In Exercises 33-36, fmd the first three nonzero terms ofthe Maclaurin series for each function and the values of x for which the series con-

14. 2 + x

+x

27. f(x) ~ I/x',

a = -2

+ lx' + x'

e'

~

a.

= eO

[1 + (x _ a) + (x ;1 a)' + ... ].

38. (Continuation ofExercise 37.) Find the Thylor senes generated by e' at x = I. Compare your answer with the fonnu1a in Exercise 37. 39. Let f(x) have derivatives through order n at x = a. Show that the Taylor polynomial of order n and its flISt n derivatives have the same values that f and its first n derivatives have at x = a.

589

10.9 Convergence of Taylor Series 40. Approximation properties of Taylor polynomials Suppose that f(x) is differentiable on an interval centered at x ~ a and that g(x) = bo + bl(X - a) + ... + bo(x - a)O is a polynomial of degree n with constant coefficients ho, ... , b". Let E(x) = f(x) - g(x). Show that ifwe impose on g the conditions

i) E(a) = 0 ") lim II

The approximation error is zcro at:t

E(x)

x-a (x _ a)"

0,

=

Thus, the Taylor polynomial Po(x) is the only polynomial of degree less than or equal to n whose error is both zero at x = a and negligible when compared with (x - a)o. Quadratic Approsimations The Taylor polynomial of order 2 generated by a twice-differentiable function f(x) at x = a is called the quadratic approximation of f at x = a. In Exercises 41-46, fmd the (a) linearization (Taylor polynomial of order I) and (b) quadratic approximation off at x ~ O. 42. f(x) = e'"'x 41. f(x) = In (cosx)

a.

The error is neg1igible wheo compared to (x - a)II,

then g(x)

= f(a) + f'(a)(x

43. f(x) = 1/"\11=7 45. f(x) = sinx

j"(a) - a) + 2i(x - a}' +

44. f(x) = coshx 46. f(x) = tanx

f(o)(a) +--,-(x-a)o. n.

10.9

l_co_n_v_erg-",-e_n_C_e_of_l_a~YL_O_rS_e_ri_e_s

_

In the last section we asked when a Taylor series for a function can be expected to converge to that (generating) function. We answer the question in this section with the following theorem.

/0)

If f and its first n derivatives 1', /", ... , are continuous on the closed interval between a and b, and is differentiable on the open interval between a and b, then there exists a number c between a and b such that

THEOREM 23-Taylor's Theonom

I(b) = I(a)

+ I'(a)(b

+

/0)

/,,(a)

+ 2 ! (b

- a)

AO)(a)

_l"_ _ (b _

n!

a

)0 +

- a)2

Ao+l)(C)

l"

(n

+ I)!

+ ...

(b _

a

)0+1

.

Taylor's Theorem is a generalization of the Mean Value Theorem (Exercise 45). There is a proof ofTaylor's Theorem at the end of this section. When we apply Taylor's Theorem, we usually want to hold a fixed and treat b as an independent variable. Taylor's formula is easier to use in circumstances like these if we change b to x. Here is a version of the theorem with this change.

Taylor's Formula If I has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, I(x) = I(a)

+ I'(a)(x

- a)

/o)(a)

+ --, -(x n.

/,,(a)

+ 2 ! (x 0

a)

- a)2

+ ...

+ Ro(x),

(1)

where

/o+l)(c) + 1)1 (x - a)o+1

R.(x) = (n

for some c between a and x.

(2)

590

Chapter 10: Infinite Sequences and Series When we state Taylor's theorem this way, it says that for each x

f(x)

=

E

I,

P.{x) + R.(x).

1.+

1 ) at a point The function R.(x) is detennined by the value of the (n + I)st derivative c that dependa on both a and x, and that lies somewhere between them. For any value of n we want, the equation gives both a polynomial approximation of f of that order and a formula for the error involved in using that approximation over the interval I. Equation (I) is called Taylor's formula. The function R.(x) is called the remainder of order n or the error term for the approximation of f by p.(x) over I.

If R.(x) --+ 0 as n --+ 00 for all x E I, we say that the Taylor series generated by f at x = a converges to f on I, and we write 00

f(x)

=

t'k)(a)

L -k'-(x k-O •

a)k.

Often we can estimate R. without knowing the value ofc, as the following example illustrates.

EXAMPLE 1 Show that the Taylor series generated by f(x) f(x) for every real value of x.

= eX at x =

0 converges to

Solution The function has derivatives of all orders throughout the interval I (-00,00). Equations (I) and (2) withf(x) = eX and a = 0 give

x2 xn e X = I +x+-+···+-+R(x) 2! n! n

=

Polynomial from Section 10.8, Example 2

and

_ eC n+l R.(x) - (n + 1)1 x

for some c between 0 and x.

Since eX is an increasing function of x, e' lies between eO = I and eX. When x is negative, soisc,ande' < 1. When x is zero, eX = landR.(x) = O.Whenxispositive,soisc,and e' < eX. Thus, for R.(x) given as above,

Ixl·+ 1

IR.{x) I ,;; (n + I)!

whenx

~

0,

and

x n+ 1

IR.{x) I < eX (n + I)!

when x

>

O.

Finally, because X ·+

lim

.~OO (n

+

1

I)!

= 0

for every x,

Section 10.1, Theorem 5

(3)

I Tbe Number e as a Series 00

1

/FO

n!

e~ ~-

• We can use the result of Example I with x = I to write

e

=

I

+ I + -2!I + ... + -n!I + Rn (I),

10.9 Convergence of Taylor Series

591

where for some c between 0 and I, C

Rn(l) = e (n

11)1
x

o

-1

-2

FIGURE 10.20 The polyoomials n

P2n+l(X)

(-1)'x2H1

~ ~ (2k + I)!

converge to sin x as n -+ 00. Notice how closely P3(X) approximates the sine curve for x '" I (Example 5).

594


A Proof of Taylo(s Theorem We prove Taylor's theorem assuming a The Taylor polynomial

< b. The proof for a > b is nearly the same.

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 ofthe form K(x - a)n+ I , 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

"'.(x)

=

Pn(x) + K(x - a)n+1

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 = "'.(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+1

(7)

With K defined by Equation (7), the function

F(x)

=

f(x) - "'.(x)

"'n

measures the difference between the original function f and the approximating function 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 CI in (a, b).

Next,becauseF'(a) = F'(cl) = OandbothF'andF" are continuous on [a, ctJ,weknow that

F"(C2)

=

0

for some C2 in (a, CI).

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+1 in (a, cn) such that

F(n)(a)

(8)

Ifwe differentiate F(x) = f(x) - P.(x) - K(x - a)n+1 a total ofn

F(n+I)(X)

=

+I

,n+I)(x) - 0 - (n + l)!K.

times, we get (9)

Equations (8) and (9) together give

K

=

,n+I)(c) (n + I)!

for some numberc = Cn+1 in (a, b).

(10)

10.9

Convergence of Taylor Series

595

Equations (7) and (10) give

f(b)

=

Pn(b)

+

/n+1)(c) + I)! (b - a)n+l.

(n

•

This concludes the proof.

Exercises 10.9 Finding Taylor Series Use substitution (as in Example 4) to find the Taylor series at x the functions in Exercises 1-10. 2. e-x/2

1. e--5x

(~x)

4. sin

7. !n(1

8. tao-I

0 of

3. 5 sin (-x)

5. cos 5x

+ x')

~

2

6. cos

(x'/3/V2)

9 I . 1 + ~x3

(3x 4 )

10. -2-

-x

Use power series operations to fmd the Taylor series at x functions in Exercises 11-28.

.

14. smx - x

+ x' 3!

13.

15. x cos 'll"x

16. x' cos (x')

= (I + cos2x)/2.)

18. sin'x

19. 1- 2x

I (I - x)'

24. sinX" cosx 27.

t!n (I + x')

0 for the

12. x 2 sinx

17. cos'x (Hint: cos'x

21.

x' T

~

x'

I

20. xln(I

+ cosx

+ 2x)

25.ex+I~x + x)

26. cosX' - sinx

e~

sin x

!nO +x) 30.

I - x

2

32. cos x' sinx

40. The estimate ~ = I + (x/2) is used when x is small. Estimate the error wben Ix I < 0.01.

42. (Continuation of Exercise 41.) When x < 0, the series for eX is an alternating series. Use the Alternating Series Estimation TheorOll3 to estimate the error that results from replacing eX by I + x + (x'/2) when -0.1 < x < O. Compare your estimate with the one you obtained in Exercise 41. Theory and Examples 43. Use the identity sin' x ~ (I - cos 2x)/2 to obtain the Maclaurin series for 8in2 x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for

sin 2x.

45. Taylor's Theorem and the Mean Value Theorem Explain how the Mean Value Theoren3 (Section 4.2, Theorem 4) is a special

31. (tao-1 x)'

46. Linearizations at iDfIection points Show that if the graph of a twicl>-differentiable function I(x) has an inflection point at

x = a, then the linearization of f at x = a is also the quadratic approximation of 1 at x = a. This explains why taogent lines fit so well at inflection points. 47. The (second) seoond derivative test

34. sin (tan- 1 x)

I(x) ~ I(a) Error Estimates 35. Estimate the error if P,(x) ~ x - (x' /6) is used to estimate the valueofsinxatx = 0.1. 36. Esti:matetheerrorifP4(x) = I + x + (x'/2) + (x'/6) is used to estimate the value of eX at x ~ 1/2.

+ (x 4/24)

37. For approximately what values of x can you replace sin x by x - (x'/6) with an error ofmagnitude no greater tban 5 X 1O-4?


< IO-'? For

case ofTaylor's Theorem. -!n(I - x)

Find the fIrst four nonzero terms in the Maclaurin series for the functions in Exercises 29-34. 29.

39. How close is the approximation sinx = x when Ixl which of these values ofx is x < sinx?

44. (Continuation of Exercise 43.) Use the identity cos'x ~ cos 2x + sin2 X to obtain a power series for cos2 x.

2 22. ( I-x )'

28. !n(1

too small? Give reasons for your answer.

41. The approximation eX = I + x + (x'/2) is used when x is small. Use the Remainder Estimation Theorem to estimate the error whenlxl < 0.1.

4

I

38. If cos x is replaced by I - (x' /2) and Ix I < 0.5, what estimate can be made of the error? Does I - (x'/2) tend to be too large, or

Use the equatiou

/"(c,)

+ f'(a)(x - a) + -2-(x - a)'

to establish the followiug test. Let 1 have continuous fust and second derivatives and suppose thatf'(a) = O. Then a.

1 bas a local maximum at a if/"'"

0 throughout an interval

whose interior contains a; b.

1 bas a local minimum at a if f" '" 0 throughout an interval whose interior contains a.

596


48. A cubic apprmdmation Use Taylor's formula with a = 0 and n ~ 3 to fmd the standard cubic approximation of f(x) ~ 1/( I - x) at x = O. Give an upper bound for the magnitode of the error in the approximation whenlxl :5 0.1. 49. a. Use Taylor's formula withn ~ 2 to find the quadratic approximationoff(x) = (I + x)k atx = (kaconstant).

o

b. If k = 3, for approximately what values ofx in the interval [0, I] will the error in the quadratic approximation be less than 1/100? 50. Improving apprmdmations of 'If that P + sin P gives an approximation correct to 3n decimals. (Hint: LetP = 'If + x.) Try it with a calculator.

51. The Taylor .eries generated by f(x) ~ ~::'~oa.x· is ~:=o GilX' A function defined by a power series ~:o GnX" with a radius of convergence R > 0 has a Taylor series that converges to the function at every point of (-R, R). Show this by showing that the Taylor series generated by f(x) = ~::'-o a.x' is the series ~:o Cl"X n itself. An immediate consequence ofthis is that series like • 2 xsmx=x

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 (a) and (b) for the functions and intervals in Exercises

53-58. Step 1: Plot the function over the specified interval.

a. Let P be an approximation of 'IT accurate to n decimals. Show

D b.

and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of x can the function be replaced by each

4

6

8

3!

5!

7!

x + X- - -X + ... --

and

Step 2: Find the Taylor polynomials PI(x), P2(x), and P,(x) at

x = O. Step 3: Calculate the (n + I)st derivative /,'+1)(c) associated with the remainder tenn for each Taylor polynomial. Plot the derivative as a function ofc over the specified. interval and estimate its maximum absolute value, M. Step 4: Calculate the remainder R,(x) for each polynomial. Using the estimate M from Step 3 in place of /'+1)(c), plot R,(x) over the specified interval. Then estimate the values of x that answer question (a). Step 5: Compare ynur estimated error with the actual error E.(x) ~ If(x) - P,(x) I by plotting E,(x) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5.

obtained by multiplying Taylor series by powers of x, as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent. 52. Taylor serie. for even function. and odd functions (Continuation ofSection 10.7, Exercise 55.) Suppose thatf(x) ~ ~:oa.x' converges for all x in an open interval ( - R, R). Show that

a. Iff is even, then al = a, = a, = ... = 0, Le., the Taylor series for f at x

~

0 contains ouly even powers of x.

b. Iff is odd, then no ~ a2 ~ Cl4 ~ ... ~ 0, Le., the Taylor series for f at x = 0 contains ouly odd powers of x.

1 53. f(x) = .~'

vl+x

54. f(x) = (1 55. f(x) =

+ x)'/2,

Ixl:5

3

4

-t:5 x :5 2

+-, +

Ixl:5 2 x I (cosx)(sinlx), Ixl:5 2

56. f(x)

~

57. f(x)

~ e~coslx,

Ixl:5 I

58. f(x) ~ e'/3 sin lx,

Ixl:5 2

COMPUTER EXPLORATIONS Taylor's formula with n ~ I and a ~ 0 gives the linearization of a function at x = O. With n = 2 and n = 3 we obtain the standard quadratic

10.10 I--Th_e_B_in_o_rn_i_a_L_Se_n_·e_s_a_n_d_A....:p....:.p_Li_c_ati_·o_n_s_o_f_l_a=-yL_o_r_S_en_·e_s

_

In this section we introduce the binonrial series for estimating powers and roots of binomial expressions (1 + x)m. We also show how series can be used to evaluate nonelementary integrals and limits that lead to indeterminate [onns, and we provide a derivation of the Taylor series for tan-I x. This section concludes with a reference table of frequently used series.

10.10 The Binomial Series and Applications of Taylor Series

597

The Binomial Series for Powers and Roots The Taylor series generated by f(x) = (I

I

+ mx +

m(m - I) 2 2! x

+ x)m, when m is constant, is

+

m(m - I)(m - 2) 3 3! x

+

m(m - I)(m - 2)'" (m - k k!

+ ... + l)

k

x

This series, called the binomial series, converges absolutely for series, we first list the function and its derivatives:

f(x)

=

+ ...

IxI
;)'

I)'

Approxfmattons and Nonelementary Integrals

0.2 sinx

/.°

0.1 17.

o

/.o

2

42.

16.

v'1+74 dx

18.

/.0.25

1

+x

1

/.0.2 I ° "---i- dx ~

dx

0

21.

/.0.1 s~x dx

20.

0.1 v'1+7 dx /.°

22.

~ dx

4S

X

fo

24. Estimate the error if cos VI is approximated by I -

T+ ~! 4

8

in the

t + ~! 2

,

~!

10 Exercises 25-28, fmd a polynomial that will approximate F(x) throughout the given interval with an error of magnitude less than 10-' .

25. F(x) =

I

28. F(x) ~

xln(1 + I) ° I dl, /.

x-a

31. lim

x

t

1-0

4

'

Y - tan-I Y

y-o

y

33. lun

(b) [0, I] (b) [0, I]

35. lim x 2(e- 1/ x'

y-o

-

x~oo

37. lim

1n(1

+x

2

)

~'------"

x-a 1

sinB - B + (11'/6) 32. 9-0 lim "' tr

34. lun

cosX'

I)

r

1f's

1T'7

3"5!

3"7!

2

2'

2'

2'

3 - 3"3 + 3"5 - 3"7 x 3 + x 4 + X S + x 6 + ... 34x 4

3'>:2 21 XS

-

+ ...

3'x'

+ 41 - 61 +'" + x 7 - x 9 + XII - •••

2 3 22.x 4 23x S 24x 6 50. x -2>; +21-31+41-'"

51. -I +2>;-3x 2 +4x'-5x 4 + ... 52.1+

x x2 x 3 x 4 + + + +",

2

3

4

5

Theory and Examples 53. Replace x by -x in the Taylor series for In (I + x) to oblain a series for In (I - x). Thea subtract Ibis from the Thylor series for In (I + x) to sbowthatforlxl < I,

+

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