FORMUlAS FOR DERIVATIVES FORMUlAS FOR INTEGRALS

D"c "" 0 fudv=w, - fCdu 2 D,,(u + 11)= DJtu + D"v

3 D .. (uv) = II D"II + II D"u 2 fu~du= _'_ 'I'''1 +c'n#= - I

n+!

('4) _ II Dx II - II Dx II D - -

'" II v2

5 D. /(g(X)) ~ J'(g(.,))g'(x)

3 fldu = lnlul +c "

v

f e"du=e"+C

D" II' = nu" - I Ox II 5 Sa~ du= -'- a"+C

In a

7 D" e" "" e" D~ II f Si n udu = -cos u + C

D"o" = a"ln (l D" u 7 f cos II du = sin u + C V I

D.ln l u l~ - D.u . u f sec l udu = tan u + C

I I. D. 'o .. l u l ~ -, - D.u

una Jesel udu = -COl u + C 11 D" sin II = cos II Ox II I. f sec II tan u du = sec u + C 12 DJ/cos u = -sinuD"u

11 f CSC liCOludu = -cscu+C 13 Dlt tan II = sec1 110" II I. D"cotu= - csc1u D"u 12 f Ian II till = - In I cos II I + C 15 D" sec II = sec II tan II D" II

13 J cot u du = In I sin u I + C I. D"csc u= -csc ucotuDJ

You S e $ At FORMULAS FROM GEOM

FIFTH EDITION

EARL W. SWOKOWSKI Marquette University

Australia ' Canada Mexico Singapore Spain United Kingdom United States

L

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-1\()()-7,,()-22I ,~ 1-i\()()-7"O-221-1

Prrnted in the [Inited States or :\Il1cnca

I09C:76'i,+':'1

Lihrary of Congress Cataloging-in-Puhlication Data

SWO"o\\:;"I, Earl William, Caiculu,/Llrl \\, SIIOJ,;O\\SJ,;I-'i,h cd,

I', erli, Include, index, ISB\I tJ-'i.'4-92-192-1 I, Cllculm. 2 C;ellIl1etn, Anal\ tic L S\\()kow:;J,;i, Earl William, Cliculu., \11th anaivtlc ~e()Il1etry II. Title,

Q.-\"O"S)-I 1l)()1 l)()--I II Xn 'i1'i.IS-dl'20

Dedicated to the memory of

my mother and father,

Sophia and John Swokowski

CONTENTS

PREFACE

FOR THE STUDENT

CHAPTER 1

CHAPTER 2

PRECALCULUS REVIEW

1.1

1.2

1.3

Algebra

Functions

Trigonometry

LIMITS OF FUNCTIONS

2.1

2.2

2.3

2.4

2.5

2.6

Introduction to Limits

Definition of Limit

Techniques for Finding Limits

Limits Involving Infinity

Continuous Functions

Review Exercises

xv

xx iii

2

1 4

26

39

40

5 1

58

68

77

88

vii i

CHAPTER 3

CHAPTER 4 ---

CHAPTER 5

THE DERIVATIVE

3.1

3.2

3.3

3.4

3.5

3.6

3.7

3.8

3.9

Tangent Lines and Rates of Change

Definition of Derivative

Techniques of Differentiation

Derivatives of the Trigonometric Functions

Increments and Differentials

The Chain Rule

Implicit Differentiation

Related Rates

Review Exercises

APPLICATIONS OF THE DERIVATIVE -- -

4.1 Extrema of Functions

4.2 The Mean Value Theorem

4.3 The First Derivative Test

4.4 Concavity and the Second Derivative Test

4.5 Summary of Graphical Methods

4.6 Optimization Problems

4.7 Rectilinear Motion and Other Applications

4.8 Newton's Method

4.9 Review Exercises

INTEGRALS

5.1 Antiderivatives and Indefinite Integrals

5.2 Change of Variables in Indefinite Integrals

5.3 Summation Notation and Area 5.4 The Definite Integral

5.5 Properties of the Definite Integral

5.6 The Fundamental Theorem of Calcu lus 5.7 Numerical Integration

5.8 Review Exercises

CONTENTS

89

90

98

109

1 1 8

127

137

146

152

162

165

166

177

183

191

199

207

220

232

236

239

240

250

257

266

275

282

292

300

CONTENTS

CHAPTER 6

CHAPTER 7

CHAPTER B

ix

APPLICATIONS OF THE DEFINITE INTEGRAL 303 -------------------------

6.1

6.2

6.3

6.4

6.5

6.6

6.7

6.8

6.9

Area

Solids of Revolution

Volumes by Cylindrical Shells

Vol umes by Cross Sections

Arc Length and Surfaces of Revolution

Work

Moments and Centers of Mass

Other Applications

Review Exercises

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

7.1

7.2

7.3

7.4

7.5

7.6

7.7

------

Inverse Functions

The Natural Logarithmic Function

The Natural Exponential Function

Integration

General Exponential and Logarithmic Functions

Laws of Growth and Decay

Review Exercises

INVERSE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS

8.1

8.2

8.3

8.4

8.5

Inverse Trigonometric Functions

Derivatives and Integrals

Hyperbolic Functions

Inverse Hyperbolic Functions

Review Exercises

304

313

322

328

333

342

349

358

370

373

374

381

392

399

408

416

423

425

426

433

440

447

452

x

CHAPTER 9 TECHNIQUES OF INTEGRATION

9.1

9.2

9.3

9.4

9.5

9.6

9.7

9.8

Integration by Parts

Trigonometric Integrals

Trigonometric Substitutions

Integrals of Rational Functions

Integrals Involving Quadratic Expressions Miscellaneous Substitutions

Tables of Integrals

Review Exercises

C HAP T E R 10 INDETERMINATE FORMS AND IMPROPER INTEGRALS

CHAPTER 11

10.1

10.2

10.3

10.4

10.5

INFINITE SERIES

11. 1

11.2

11.3

11.4

11.5

11.6

11.7

11.8

11.9

11.10

11. 11

The Indeterminate Forms % and :xJ :r

Other Indeterminate Forms

Integrals with Infinite Limits of Integration

I ntegrals with Discontinuous T ntegrands

Review Exercises

Sequences

Convergent or Divergent Series

Positive-Term Series

The Ratio and Root Tests

Alternating Series and Absolute Convergence

Power Series

Power Series Representations of Functions

Maclaurin and Taylor Series

Applications of Taylor Polynomials

The Binomial Series

Review Exercises

CONTENTS

455

456

463

468

472

479

482

485

488

491

492

499

504

510

5 1 7

519

520

532

543

553

557

566

573

580

590

596

599

CONTENTS

C HAP T E R 12 TOI"C) F I~OM ANALYTIC GEOMETRY ~~~~~-------12.1 Parabolas

12.2 Ellipses

12.3 Hyperbolas

12.4 Rotation of Axes

12.5 Review Exercises

C HAP T E R 13 PLANE (URVE-) AND POLAR COORDINArES

13.1

13.2

13.3

13.4

13.5

13.6

Plane Curves

Tangent Lines and Arc Length

Polar Coordinates

Integrals in Polar Coordin~ltes

Polar Equations of Conics

Review Exercises

C HAP T E R 14 VfCTOR

l

xii

C HA P T E R 15 VECTOR-VALUED FUNCTIONS

CHAPTER 16

CHAPTER 17

15.1

15.2

15.3

15.4

15.5

15.6

15.7

Vector-Valued Functions and Space Curves

Limits , Derivatives, and Integrals

Motion

Curvature

Tangential and Normal Components of Acceleration

Kepler's Laws

Review Exercises

PARTIAL DIFFERENTIATION

16.1 Functions of Several Variables

16.2 Limits and Continuity

16.3 Partial Derivatives

16.4 Increments and Differentials

16.5 Chain Rules

16.6 Directional Derivatives

16.7 Tangent Planes and Normal Lines

16.8 Extrema of Functions of Several Variables

16.9 Lagrange Multipliers

16.10 Review Exercises

MUL TIPLE INTEGRALS - -

17.1 Double Integrals

17.2 Area and Volume

17.3 Double Integrals in Polar Coordinates

17.4 Surface Area

17.5 Triple Integrals

17.6 Moments and Center of Mass

17.7 Cylindrical Coordinates

17.8 Spherical Coordinates

17.9 Change of Variables and lacobians

17.10 Review Exercises

CONTENTS

747

748

754

761

769

780

785

790

793

794

805

8 I 4

823

835

844

854

861

872

881

BBS

886

896

905

9 I I

9 I 5

926

934

942

948

961

CONTENTS

C HAP T E R 18 VECTOR CALCULUS

18.1 Vector Fields

18.2 Line Integrals

18.3 I ndependence of Path

18.4 Green's Theorem

18.5 Surface Integrals

18.6 The Divergence Theorem

18.7 Stokes' Theorem

18.8 Review Exercises

C H AP T E R 19 DIFFERENTIAL EOUATIONS

A PPENDICES

19.1

19.2

19.3

19.4

19.5

19.6

19.7

Separable Differential Equations

First-Order Linear Differential Equations

Second-Order Linear DifTerential Equations

Nonhomogeneous Linear Differential Equations

Vibrations

Series Solutions

Review Exercises

Mathematical Induction

II Theorems on Limits and Integrals

"' Tables A Trigonometric Functions

B Exponential Functions

C atural Logarithms

,V Table of Integrals

A N SWERS TO O DD - NUMBERED EXERCISES

INDEX

xii i

963

964

972

982

991

997

I a a 6

I a I I

I a I 9

102 1

I a 2 2

I a 28

I a 3 5

I a 3 9

I a 4 5

I a 5 a

I a 52

A1

AI

AS

AI9

A I 9

A2a

A2a

A2 I

A27

A97

ADDITIONAL CALCULUS TEXTBOOKS BY EARL W. SWOKOWSKI

CALCULUS, FIFTH EDITION, LATE TRIGONOMETRY VERSION This text provides a review of trigonometry and com-plete coverage of the trigonometric functions in Chapter 8. Consequently, there is slightly different trcatmcnt of limits, derivatives, and integrals.

CALCULUS OF A SINGLE VARIABLE Designed for a two-semester course , this volume consists of the first thirteen chapters of Calculus, Fifth Edition.

PREFACE

This revision of what was previously called Calculus lI'i( h A Ilaly( ic Ge()l1le( ry, A I(emate Edition was written with three objectives in mind. The first was to make the book more student-oriented by expanding discus-sions and providing more examples and figures to help clarify concepts. To further aid students, guidelines for solving problems were added in many sections of the text. The second objective was to stress the use-fulness of calculus by means of modern applications of derivatives and integrals. The third objective, to make the text as error-free as possible, was accom-plished by a careful examination of the exposition, combined with a thorough checking of each example and exercise.

CHANGES FOR THE FIFTH EDITION Suggestions for improvements from instructors and reviewers resulted in a great deal of rewriting and reorganization. The principal changes are highlighted in the following list.

CHAPTER 1 The number of review sections has been reduced from six to three, with proofs of results from precalculus replaced by examples on inequalities, equations, and graphs.

CHAPTER 2 There is more emphasis on the graphical significance of limits. A physical application and tol-erance statements are used to motivate the E-() defini-tion. Limits involving CfJ (formerly in Chapter 4) are discussed in Section 2.4.

CHAPTER 3 The interpretations of the derivative as the slope of a tangent line and as the rate of change of a function are considered simultaneously instead of

in separate sections. The power rule for rational num-bers and the concept of higher derivatives are intro-duced in Section 3.2. Greater emphasis is given to the use of differentials for linear approximations of func-tion values.

CHAPTER 4 The definition of concavity has been changed to make it easier to relate the sign of a second derivative to the shape of a graph. A new section titled SUlllmary of Graphical Methods includes a list of guide-lines for sketching the graph of a function.

CHAPTER 5 Antiderivatives and indefinite integrals are discussed in the first two sections instead of in different chapters. There are fifteen new examples per-taining to definite integrals.

CHAPTER 6 Almost every example on applications of definite integrals has been rewritten so as to replace formal limits of sums with a more intuitive method using differentials. Guidelines are stated that contain strategies for finding areas and volumes.

CHAPTER 7 The formula for the derivative of an 111-verse function is proved in the first section instead of the last. Integrals of the tangent, cotangent, secant, and cosecant functions (formerly in Chapter 8) are discussed in Section 7.4.

CHAPTER 8 The topics considered are restricted to inverse trigonometric and hyperbolic functions.

CHAPTER 9 Explanations and guidelines for methods of integration have been improved.

CHAPTER 10 For easy reference, definitions and no-tations for indeterminate forms are stated in tables. The discussion of Taylor's formula has been moved to Chapter 11.

x vi

CHAPTER II More emphasis is given to the differences between sequences, partial sums, sums of infinite series, and the fact that tests for convergence do not determine the sum of a series. The material on power series representations of functions and Taylor series has been completely reorganized.

CHAPTER 12 Additional calculus applications in-volving conic sections are included so that the discus-sion is not simply a review of precalculus topics .

CHAPTER 13 The concept of orientation of a curve has been added and incorporated into examples and exercIses.

CHAPTER 14 To help readers visualize and sketch surfaces, many examples contain charts that show the trace on each coordinate plane. The discussion of cylindrical and spherical coordinates has been moved to Chapter 17.

CHAPTER 15 The introduction to vector-valued func-tions has been rewritten and integrated with the no-tion of space curves. The significance of using an arc length parameter has been given more prominence.

CHAPTER 16 Sixteen new figures provide a greater emphasis on graphs and geometric interpretations of functions of several variables. Newton 's method for a system of two nonlinear equations is explained in Sec-tion 16.4. The discussion of Lagrange multipliers has been expanded.

CHAPTER 17 The definition of double integral and methods of evaluation appear in one section instead of two. The discussions of triple integrals in cylindrical coordinates and in spherical coordinates are presented in two separate sections.

CHAPTER 18 There is a more detailed explanation of conservative vector fields and of the technique of de-termining a potential function from its gradient. Two new examples apply Stokes' theorem and the concept of circulation to the analysis of winds inside a tornado.

CHAPTER 19 Series solutions of differential equations are discussed in the last section.

FEATURES OF THE TEXT APPLICATIONS The previous edition contained applied problems from fields such as engineering, physics , chemistry, biology , economics , physiology, sociology, psychology , ecology, oceanography , meteorology, radiotherapy, astronautics, and transportation. This

PREFACE

already extensive list was augmented with examples a nd exercises that include modern applications of cal-culus to the design of computers , the analysis of tem-perature grids, and the measurement of the thickness of the ozone layer , the greenhouse effect, vertical wind shear, the circulation of winds inside tornados and hurricanes, the energy released by earthquakes, the density of the atmosphere , the movement of robot arms, and the health hazards of radon gas.

EXAMPLES Well-structured and graded examples pro-vide detailed solutions of problems similar to those that appear in exercise sets . Many examples contain graphs, charts, or tables to help readers understand procedures and solutions. There are also labeled illus-trations , which are brief demonstrations of the use of definitions , laws , or theorems. Whenever feasible, ap-plications are included that indicate the usefulness of a topic.

EXERCISES Exercise sets begin with routine drill prob-lems and progress gradually to more difficult types. Many exercises containing graphs were added to this edition. Applied problems generally appear near the end of a set, to allow students to gain confidence in manipulations and new ideas before attempting ques-tions that require analyses of practical situations.

A new feature in this edition is over 300 exer-cises, designated by the symbol ~, spec ifically de-signed to be solved with the aid of a scientific calculator or computer. Graphics capabilities are required for some of these exercises (see the remarks that follow under Calculators).

Review exercises at the end of each chapter except the first may be used to prepare for examinations.

ANSWERS The answer sec tion at the end of the text provides answers for most of the odd-numbered exer-cises. Considerable thought and effort were devoted to making this section a learning device instead of merely a place to check answers. To illustrate , if an answer is the area of a region or the volume of a solid, a suitable (set-up) definite integral is often given along with its value. Numerical answers for many exercises are stated in both an exact and an approximate form. Graphs , proofs, and hints are included whenever appropriate.

CALCULATORS Since students may have access to a variety of calculators or computers, no attempt was made to categorize the exercises that are designated by 0 . The statement of a problem should provide

PREFACE

sufficient information for determining whether a par-ticular calculator or available computer software can be used to obtain a numerical solution. For example, if an exercise states that the trapezoidal rule with n = 4 should be applied, then almost any calculator is suit-able, provided the function is not too complicated. However , for n = 20, a programmable calculator or computer is recommended. If the solution of an exer-cise involves a graph, a graphics calculator may be adequate; however , complicated functions or surfaces may require sophisticated computer software. Since numerical accuracy is also dependent on the type of computational device used , some answers are rounded to two decimal places , but in other cases eight-decimal-place accuracy may be given.

TEXT DESIGN AND FIGURES The text has been com-pletely redesigned to make discussions easier to follow

x vii

and to highlight important concepts. All graphics have been redone. Graphs of functions of one or two vari-ables were computer-generated and plotted to a high degree of accuracy using the latest technology. Colors are used to distinguish between different parts of fig-ures. For example, the graph of a function may be shown in blue and a tangent line to the graph in red . Labels are in the same color as the parts of the figure they identify. The use of several colors should help readers visualize surfaces and solids more readily than in previous editions.

FLEXIBILITY Syllabi from schools that used previous editions attcst to the flexibility of the text. Sections and chapters can be rearranged in different ways, depending on the objectives and the length of the course.

Earl JJ. Swo/.:ow\/.:i

,

ACKNOWLEDGMENTS

I am indebted to JerTery Cole, of Anoka-Ramsey Com-munity College, for his excellent work on the exercise sets, the answer section , and artwork preparation for the figures. Jeff also carefully read the manuscript and offered many suggestions that helped make discus-sions and examples more student-oriented. He often reminded me of the motto I adopted when I started on thi s revision: "Teach the Class."

I wish to thank Gary Rockswold , of Mankato State University, for supplying many of the new ap-plied problems and calculator exerciscs.

I am also grateful to the following individuals, who reviewed the manuscript:

Stephen Andrilli, LaSal/e Unirersitj' Robert Beezer, UnirCl'silY of Puyel Soulld Clifton Clarridge, Santa Monica Col/eye Andrew Demetropoulos, Monle/air Stare Col/eye Richard Grassl, M uhlellhery Col/eye Harvey B. Keynes, Unil'ersity 01' Minnesota Karl R. Klose, Susquehanna Unil'ersitj' James E. McKenna, SUN Y Col/eye at Fredonia James R. McKinney, Cali/(JI'Ilia Slate Polytechnic

U niversit.\', POlllona Riehard A. Quint, Velltura Col/ege Leon F. Sagan, Anne Arundel Comllluility Col/eye David J. Sprows, Vil/allo['{/ Ullirersit.\' Ronald Stoltenberg, Sam Houston Slate Ullicersity Jan Vandever, South Dakota State Ullil'ersily Kenneth L. Wiggins, Wal/a Wal/a Col/eye

The following mathematies educators met with me and represcntatives from PWS KENT for several days in the summer of 1989 and latcr reviewed the manuscript:

Riehard D. Armstrong, SI. Louis COllllllUlliTY Col/eye at Florissalll Valley

Paul W. Britt. Louisiana State Unicersit.l' Jeffery A. Cole, Anoka-Ramsey COIll/llllllilY Colleye William James Lewis. Ullil'ersily o{Nehraska- Lillcoln Michael A. Penna, illdiallu UniL'ersit)' Purdue

U niL'CI'sit,l' at indianapolis

Their comments on pedagogy and their specific rec-ommendations about the content of calculus courses helped me to improve the book.

I am thankful for the excellent eooperation of the starT of PWS KENT. Two people in the company de-serve special mention. Managing Editor Dave Geggis supervised the project, contacted many reviewers and users of the text, and was a constant source of in-formation and advice. Dave is also responsible for finding the authors who developed the excellent ancil-laries that accompany this book. Elisc Kaiser rede-signed thc book, coordinated the production of the text, and kept the schedule running smoothly and on time. Sally Liftand, of LiOand et aI., Bookmakers, took exceptional care in seeing that no editorial inconsis-tencies occurred.

Finally , I am deeply grateful to my wife Maureen for her understanding, patience, support, and love .

In addition to all the persons named here, I ex-press my sincere appreciation to the many students and teachers who have helped shape my views on how calculus should be presented in the classroom. I am grateful to many people who have reviewed the va-rious editions of this book. My thanks go to these reviewers:

Alfred D. Andrew Georyia Institute of Technoloyy Jan Frederick Andrus University 01' Nell' Orleans

ACKNOWLEDGMENTS

Jacquelinc E, Harab [lIil'el'sity of Seuula, Las Veyas Phillip W, Bean ,\lel'cer Uilirersil.\' Daniel D, Benice At on I (} 0 1111' r,l' Col/eye Delmar L. Rovar UIJit'ersily of Texas U Pel.lo Chri,tian C. Braunschwcigcr MU!'(jllt'lte Lllil'ersil\' Robert M, Brooks Ulliu'rsi1\' (}f ['tuil Ronald F. Bruck ['lIilersitl of Soulilern Cali/emlia David C. Buchthal ['lIirersil\' of Akroll Dawson Carr SUlldhil/s Communily Co/lC'ye James L. Cornelle Irma Stull' Unircl'S/t] John F. Derwent Unirel'sil)' of NoIre DUll/(' Danicl Drucker Wayne ,

x x

Norman K. Nystrom American River College James Osterburg University of Cincinnati Richard R. Patterson Indiana University- Purdue University at Indianapolis Charles V. Peele Marshall University Ada Peluso CU NY-Hunter College David A. Petrie Cypress College Neal C. Raber University of Akron William H. Robinson Ventura Col/eye Jean E. Rubin Purdue University John T. Scheick Ohio State University Eugene P. Schlereth U niversit y of Tennessee Jon W. Scott Montgomery College Richard D. Semmler Northern Virginia Community College- Annandale Leonard Shapiro University of Minnesota

ACKNOWLEDGMENTS

Donald Sherbert University of Illinois Eugene R. Speer Rutgers U niversit y Monty J. Strauss Texas Tech University John Tung Miami University Charles Van Gordon Millersville State Colleye Richard G. Vinson University of South Alabama Roman W . Voronka New Jersey Institute of Technoloyy Dale E. Walston University of Texas at Austin Frederick R. Ward Boise State University Alan Wiederhold San Jacinto College Loyd V. Wilcox Golden West College T. J. Worosz Metropolitan State College Dennis Wortman U niversit y of Massachusetts - Boston

EHS

SUPPLEMENTS

STUDENT STUDY GUIDE Richard M. Grassl's Student Study Guide leads students through all major topics in the text and reinforces their understanding through review sections, drills , and self-tests .

STUDENT SOLUTIONS MANUAL, VOLUMES I AND" Jeffery A. Cole and Gary K. Rockswold's Student Solutions Manual contains solutions, worked out in detail , to a subset of the odd-numbered exercises from the text.

INSTRUCTOR'S SOLUTIONS MANUAL, VOLUMES I AND " Jeffery A. Cole and Gary K. Rockswold 's Instructor's Solutions Manual gives solutions or answers for all exercises in the text.

EVEN-NUMBERED ANSWER BOOK Jeffery A. Cole and Gary K. Rockswold's Even-Numhered Answer Book is for both instructor and student use.

PRINTED TEST BANK Available to instructors, the Printed Test Bank contains sample tests for each chapter.

TRANSPARENCIES Also available to instructors are four-color acetate transparencies of selected figures from the text.

SOFTWARE MATHEMATICS PLOTTING PACKAGE Available for IBM-PCs and compatibles, the Mathematics Plotting Package combines superb plotting and graphing soft-ware. This public domain program developed at the United States Naval Academy is accompanied by a text-specific Instructor's Resource Guide to MPP, by Howard Penn and Craig Bailey. The guide provides instructors with notes on how to teach key concepts by using MPP to illustrate examples and problems from the text. Also included are Example Files for

MPP-disks containing examples and problems from the Resource Guide.

PC:SOLVE Available for IBM-PCs and compatibles, PC:SOL VE is an interactive mathematical language with scratch pad that supports calculations and graph-ics, combined with a calculus curriculum library of key concepts to promote critical thinking and problem solving. A course license for PC:SOL VE is available upon adoption of this text.

GRAPHER Steve Scarborough's Grapher , for the Macintosh, is a flexible program that can be used to generate several types of graphs. In addition to plot-ting rectangular and polar curves and interpolating polynomials, it handles parametric equations, systems of two first-order differential equations, series, and direction fields.

TrueBASIC CALCULUS Available for IBM-PCs and compatibles, TrueBASIC Calculus is a disk and manual package that is equally useful for self-study, free exploration of topics, and solutions of problems. The record and playback feature make the package ideal for classroom demonstrations.

COMPUTERIZED TESTING EXPTest This testing program for IBM-PCs and com-patibles allows users to view and edit all tests, adding to, deleting from, and modifying existing questions. Any number of student tests can be created- multiple forms for larger sections or single tests for individual use. Users can create a question bank using mathe-matical symbols and notation. A graphics importa-tion feature permits display and printing of graphs , diagrams , and maps provided with the test banks. The

x X I i

package includes easy-to-fo llow documentation, wit h a quick-start guide. A demonstration disk is ava ilab le.

LXRTest This Macintosh testing program a llows users to create, view, and edit tests. Qucstions can be stored by objectives, a nd the user can crea te questions using multiple-choice , true/ false, fi ll-in-the-blank, essay, and matching formats. The order of a lternatives and the ordcr of questions can be scrambled. A demonstration disk is available (the user must have HyperCard to run the demo disk).

SUPPLEMENTS

Rftpulr rAI r "ATOR SUPPLEMENTS

CALCULUS ACTIVITIES FOR GRAPHIC CALCULATORS Dennis Pence's Calculus Activities/or Graphic Calcu-lators, for students and instructors, offers exercises a nd examples utilizing the graphing ca lculator (Sharp, Casio, a nd HP2SS). Classroom tested , these varied ac-tivities will enhance understanding of calculus topics for those using graphing calculators.

CALCULUS ACTIVITIES FOR THE TI-81 GRAPHIC CALCULATOR For students and instructors using the TI-Sl graphic calculator, Dennis Pence provides calculus exercises and examples in Calculus Activities for the T/-8 / Graphic Calculator.

FOR THE STUDENT

Calculus was invented in the seventeenth century as a tool for investigating problems that involve motion . Algebra and trigonometry may be used to study ob-jects that move at constant speeds along linear or cir-cular paths, but calculus is needed if the speed varies or if the path is irregular. An accurate description of motion requires precise definitions of velocity (the rate at which distance changes per unit time) and accelera-tion (the rate at which velocity changes). These defini-tions may be obtained by using one of the fundamental concepts of calculus-the derivative.

Although calculus was developed to so lve prob-lems in physics , its power and versa tility have led to uses in many diverse fields of study. Modern-day a p-plications of the derivative include investigating the rate of growth of bacteria in a culture, predicting the outcome of a chemical reaction , measuring instan-taneous changes in electrical current , describing the behavior of atomic particles, estimating tumor shrink-age in radiation therapy , forecasting economic profits and losses , and analyzing vibrations in mechanical systems.

The derivative is also useful in so lving problems tha t involve maximum or minimum va lues, such as manufacturing the least expensive rectangular box that has a given volume, calculating the greatest dis-tance a rocket will travel , obta ining the maximum safe flow of traffic across a long bridge , determining the number of wells to drill in an o il fie ld for the most efficient production, finding the point between two light sources at which illumination will be greatest, and maximizing corporate revenue for a particular product. Mathematicians often employ deriva tives to find tangent lines to curves and to help analyze graphs of complicated functions.

Another fundamental concept of calculus- the definite integral- is motivated by the problem of find-ing areas of regions that have curved boundaries. Definite integrals are employed as extensively as de-rivatives and in as many different fields. Some appli-cations are finding the center of mass or moment of inertia of a so lid , determining the work required to send a space probe to another planet , calculating the blood flow through an arteriole, estimating deprecia-tion of eq uipment in a manufacturing plant , and inter-preting the amount of dye dilution in physiological tests that involve tracer methods . We can also use definite integrals to investigate mathematical concepts such as a rea of a curved surface, volume of a geometric solid, or length of a curve.

The concepts of derivative and definite integral are defined by limitmg processes. The notion of limit is the initia l idea that separates calculus from elemen-tary mathematics. Sir Isaac Newton (1642 - 1727) and Gottfried Wilhelm Leibniz (1646 - 1716) independently discovered the connection between derivatives and integrals and are both credited with the invention of calculus. Many other mathematicia ns have added greatly to its development in the las t 300 years.

The applications of calculus mentioned here repre-sent just a few of the many considered in this book . We can't possibly discuss all the uses of calculus, and more are being developed with every advance in tech-nology . Whatever your field of interest , calculus is probably used in some pure or applied investigat ions. Perhaps you will discover a new application for thi s branch of science.

CHAPTER

1

PRECALCULUS REVIEW

INTRODUCTION

In this chapter we review topics from precalculus mathematics that are essential for the study of calcu-lus. After a brief discussion of inequalities , equations, absolute values , and graphs , we turn our attention to functions. To say that the concept of function is im-portant in mathematics is an understatement. It is literally the foundation of calculus and the backbone of the entire subject. You will find the word funcrion and the symbol f or f(x) used frequently on almost every page of this text.

In precalculus courses we study properties of func-tions by using algebra and graphical methods that include plotting points, determining symmetry, and making horizontal or vertical shifts. These techniques are adequate for obtaining a rough sketch of a graph; however , calculus is required to find precisely where graphs of functions rise or fall , exact coordinates of high or low points , slopes of tangent lines , and many other useful facts. Applied problems that cannot be solved by means of algebra, geometry , or trigonom-etry can often be attacked by representing physical quantities in terms of functions and then applying the tools developed in calculus.

With the preceding remarks in mind , read Section 1.2 carefully. A good understanding of this material is required before beginning the next chapter.

2 CHAPTER 1 PRECALCULUS REVIEW

1.1 ALGEBRA

ILLUSTRATION

Properties of inequalities (1. I)

This section contains a review of topics from algebra th at a re prerequisites for calculus. We shall state important fact s and work examples without supplying detailed reasons to justify our work. A more extensive coverage of this material can be found in texts on precalculus ma thematics.

All concepts in calculus are based on properties of the set IR of real numbers . There is a one-to-one correspondence between IR and points on a coordinale line (or rea/line) / as illust rated in Figure 1.1, where 0 is the origin. The number 0 (zero) is neither positive nor nega tive.

FIGURE 1.1 () B A

.. - 3 - 2 I-I

I () I

I 2

I 31 4 23 5 h a

-'

- 1.5 I , v'2 2.33 TT Negative real Positive real ~ numbe rs >1< numbers

If a and b are real numbers , then a > h (a is greater than b) if a - h is positive. An equivalent statement is 17 < a (b is less than a ). Referring to the coordinate line in Figure 1.1, we see that a > b if and only if the point A corresponding to a lies to the right of the point B corresponding to h. Other types of inequality symbols include a S 17, which means a < b or a = b, and a < b s c, which means a < hand b s c.

- 5 > 3 - - 7 < - 2 - (_ 3)2> 0 - a2 ~ 0 for every real number a

The following properties can be proved.

(i) If a > band b > c, then a > c. (ii) If a > b, then a + c > b + c.

(iii) If a > b, then a - c > b - c. (iv) If a > band c is positive, then ac > bc. (v) If a > band c is negative , then ac < bc .

l I

I

Analogous properties are true if the inequality signs a re reversed. Thus , if a < band b < c, then a < c; if a < /7, then a + c < h + c; and so on .

The absolute va lue I a I of a real number a is defined as follows: if a ~ 0 if a < 0

If a is the coordinate of the point A on the coordinate line in Fig-ure 1.1, then I a I is the number of units (that is, the distance) between A and the origin o.

1 1 ALGEBRA

ILLUSTRATION

IJr(Jpertle~ r f ,t'

4

Intervals (1.3)

FIGURE 1.2 1 1 1 (I 1 1 I ] 1

o x

CHAPTER 1 PRECALCULU S REVIEW

NOTATION DEFINITION GRAPH

(a , b) {x:a < x < b} ~ ~ a h

[a, b] {x: a ::;; x ::;; b} E 3 ~ a h

[a, b) {x:a::;;x a} a [a , 00 ) {x:x ?: a} r [

a

(-oo,b) {x:x < b} j ~ b

(-oo, b] {x:x ::;; b} 3 ~ b

( -00,00) IR

EXAMPLE 2 Solve each inequality a nd sketch the graph orits solution. 4 - 3x

la) - 5 ::;; - 2- < l ib) x 2 - 10 > 3x

SOLUTION 4 - 3x

la) - 5::;; - 2- < 1 (given)

- 10 ::;; 4 - 3x < 2 (multiply by 2) - 14 ::;; - 3x < -2 (subtract 4)

134 ?: x > 1 (divide by - 3) 1 < x ::;; 134 (equivalent inequality)

Hence the solutions are the numbers in the half-open interval (1, In. The graph is sketched in F igure 1.2.

Ib) x 2 - 10 > 3x (given) x 2 - 3x - 10 > 0 (subtract 3.\) (x - 5)(x + 2) > 0 (factor)

FIGURE 1.3 Sign of factor

x + 2: - - - + + + + + x - 5: -

+ + + + + + + +

1 I 1 1 1 II! I I 1 1 I I o

I I 1 1 1 I 1 I ( I 1 1 () 'i

x

x

1.1 ALGEBRA

FIGURE 1.4

1

o

FIGURE 1.5

1 1 - 1 0

( I ) 1 234 5 x

)1 1 1 x

5

We next examine the signs of the factors x - 5 and x + 2, as shown in Figure 1.3. Since (x - 5)(x + 2) > 0 if both factors have the same sign, the solutions are the real numbers in the union ( - x, - 2) u (5 , co ), as illustrated in Figure 1.3.

Inequalities involving abso lute value occur frequent ly in calculus.

EXAMPLE 3 Solve each inequality and sketch the graph of its solution. (a) 1 x - 3 1 < 0.5 (bJ 12x - 7 1 > 3

SOLUTIO N (a) 1 x - 3 1 < 0.5 (given)

- 0.5 < x - 3 < 0.5 (property of absolute value) 2.5 < x < 3.5 (add 3)

The solutions are the real numbers in the open interval (2.5, 3.5), as shown in Figure lA.

(b) 12.x - 7 1 > 3 (given) 2.\' - 7 < - 3 or 2x - 7 > 3 (property of absolute value)

2.\' < 4 or 2x > 10 (add 7) .\' < 2 or x > 5 (divide by 2)

The solutions are given by ( - x, 2) u (5 , co ). The graph is sketched in Figure 1.5.

A rectangular coordinate system is an assignment of ordered pairs (a, b) to points in a plane, as illustrated in Figure 1.6. The plane is ca lled a coordinate plane, or an xy-plane. Note that in this context (a, b) is not an open interval. It should always be clear from our discussion whether (a , iJ) represents a point or an interval.

FIGURE 1.6 y

(0. 5) -+.3) h -- .. (a. h) I

: (5. 2) -+. (}) I

o II x

(0. 3) S. 3) (5. 3)

.,

6

Distance formula (l.4)

/\/Iidpolnt formula (l.S)

FIGURE J.7 y

I(

x

'I

CHAPTER 1 PRECALCULUS REVIEW

The following formulas can be proved.

The distance between PI and P z is ~----~--------~

d(P I, Pz) = -/(x z - x1)Z + (yz - YI)Z. y

I' ( \ \)

x

The midpoint M of segment PIP z is

M(XI + Xz YI + Yz) 2 ' 2 .

y

I' (\ . \ )

x

EXAMPLE 4 Given A( - 2, 3) and B(4 , - 2) , find fal d(A, B) fbI the midpoint M of segment AB SOLUTION The points are plotted in Figure 1.7 . Using the formulas in (1.4) and (1.5) , we obtain the following: fal d(A, B) = (4 + 2)Z + (- 2 - W = -/36 + 25 = $I

M ( - 2 + 4 3 + ( - 2)) = M( 1 .1) fbI 2' 2 ' z

1.1 ALGEBRA

Symmetries of graprls (1.6)

/il y-axis y

x

Test (i): Substitution of - x for x leads to the same equation.

7

An equation in x and y is an equality such as

2x + 3y = 5, Y = x 2 - 5x + 2, or )"2 + si n x = 8.

A solution is an ordered pair (a, h) that produces a true statement when x = a and J = h. The graph of the equation consists of all points (a, h) in a plane that correspond to the solutions. We shall assume that you have experience in sketching graphs of basic equations in x and y. Certain graphs have symmetries as indicated in (1.6), which states tests that can be applied to an equation in x and J' to determine a symmetry.

/iil x-axis y

x

Test (ii): Substitution of - y for y leads to the same equation.

/iiil Origin y

I

iG') x

\.

Test (iii): Substitution of both - x for x and - y for y leads to the same equation.

Tests for symmetry are useful in the next example, because they enable us to ske tch only half of a graph and then reflect that half through an axis or the origin as shown in (1.6). We shall plot several points on each graph to illustrate solutions of the equation; however , a principal ohjective ill yraphing is 10 obtain (Ill accurate sketch \\ithour plottiny many (or any) points.

EXAMPLE 5 Sketch the graph of /el 4)' = x 3

SOLUTION lal By symmetry test (i), the graph of y = ~X2 is symmetric with respect to the y-axis. Some points (x, y) on the graph are listed in the following table.

x

y

o

o

2

2

3 9 2

4

8

,

8

FIGURE 1.8 y

\'

( I , I x

FIGURE 1.11 y

/'(.1 , \)

x

Equation of a circle (1.71 FIGURE 1 12

y

f)(..( , 5)

x

CHAPTER 1 PRECALCULUS REVIEW

FIGURE 1.9 Y

(0 , 0) x

FIGURE 1.10 y

x

Plotting, drawing a smooth curve through the points , and then using symmetry gives us the sketch in Figure 1.8. The graph is a parabola , with r el'tex (0 , 0) and axis along the y-axis. Parabolas are discussed in detail in Chapter 12. (bl By symmetry test (ii) , the graph of )'2 = X is symmetric with respect to the x-axis. Points above the x-axis are given by y = Fx. Several such points are (0 , 0) , (I , I), (4 , 2) , and (9 , 3). Plotting and using symmetry gives us Figure 1.9. The graph is a parabola with vertex (0,0) and axis along the x-axis. (el By symmetry test (iii), the graph of 4)' = x 3 is symmetric with respect to the origin. Several points on the graph are (0,0), (1 , t ), and (2,2). Plotting and using symmetry gives us the sketch in Figure 1.10,

A circle with center C(h , k) and radius I' is illustrated in Figure 1.11. If P(x , y) is any point on the circle, then by the distance formula (1.4), d(P , C) = r, or [d(P , C)J2 = 1'2. This leads us to the following equation.

If the radius r is I, then the circle is called a unit circle. A unit circle U with center at the origin has the equation x 2 + )'2 = I.

EXAMPLE 6 Find an equation of the circle that has center C( - 2, 3) and contains the point D(4, 5).

SOLUTION The circle is illustrated in Figure 1.12. Since D is on the circle, the radius r is d(C , D). By the distance formula ,

r = ( - 2 - 4) 2 + (3 - 5)2 =

1.1 ALGEBRA

lines /1.8/

/II Slope m: Y2 - YI

m= ---X2 - XI

y

/' 1\ . \ )

x

Special Ilne~ /1.9/

(il Vertical: m undefined Horizontal: m =

y

y = b (a. b)

,I

x

9

Using the equation of a circle with h = - 2, k = 3, and r = y40 gives us (x + 2)1 + (y - W = 40,

or x 2 + y2 + 4x - 6y - 27 = 0.

[n calculus we oftcn consider lines in a coordinate planc. The following formulas are used for finding thcir equa tions.

(iiI Point-Slope Form: Y-Yl= m(x-x 1)

y

Slope 11/

x

(iiil Slope-Intercept Form : y= mx+b

y

Siopc m ~

((l , / l

x

Somc specia l types of lines and properties of thci slo pes a re given in (1.9).

(iiI Parallel: m 1 = m2 (iiil Perpendicula r: m1 m2 = - 1

y y

x x

1 0 CHAPTER 1 PRECALCULUS REVIEW

FIGURE 1.13 la) m = -i

y

Ib) m = ~

B( 2. - I)

EXAMPLE 7 Sketch the line through each pair of points and find its slope . lal A( - 1,4) and B(3. 2) leI A(4, 3) and B( - 2. 3)

Ibl A(2, 5) and B( - 2, - I) Idl A(4, - I) and B(4 , 4)

SOLUTION The lines are sketched in Figure 1.13.

Ie) m = 0 Id) m undefined y y y

8( 2.3) A(4 . 3)

x x

8(-+ . -+)

x

A (.f . I)

From the slope formula (1.8)(i): 2 - 4 - 2 1 lal 111 = = - = --3 - ( - 1) 4 2

5 - ( - 1) 6 3 Ibl I1J = = - = -2 - ( - 2) 4 2

3 - 3 0 leI I1J = = - = 0 4 - ( - 2) 6 4 - ( - 1) 5

Idl m = = -0' which is undefined. Note that the line is vertical. 4 - 4

A linear equation in x and y is an equation of the form ax + hy = c (or ax + hy + d = 0) , with a and h not both zero. The graph of a linear equation is a line .

EXAMPLE 8 Find a linear equation for the line through A(I , 7) and B( - 3,2). SOLUTION The slope I1J of the line is

7 - 2 m=

1 - ( - 3) 5 4

We may use the coordinates of either A or B for (x l' .lit) in the point-slope form (1.8)(ii). Using A(1 , 7) gives us

y - 7 = -i-(x - 1), which is equivalent to

4y - 28 = 5x - 5, or 5x - 4y = - 23 .

1.1 ALGEBRA 1 1

EXAMPLE 9 (a) Find the slope of the line 1 with equation 2x - 5y = 9. (b) Find linear equations for thc lines through P(3 , - 4) that are parallel to 1 and perpendicular to I.

SOLUTION (a) If we rewrite the equation as 5y = 2x - 9 and divide both sides by 5, we obtain

Y = ix-l Comparing this equation with the slope-intercept form y = I17X + b, we see that the slope is 111 = l (b) 8y (ii) and (iii) of (1.9). the line through P(3 , - 4) parallel to 1 has slope

~ and the line perpendicular to 1 has slope -l The corresponding equa-tions are

and

y+ 4=i(x-3) , or 2x-5y=26 )" + 4 = -!(x - 3), or 5x + 2y = 7.

EXAMPLE 10 Sketch the graphs of 4x + 3y = 5 and 3x - 2y = 8, and find their point of intersection.

SOLUTION 80th equations are linear, and hence the graphs are lines. To sketch the graphs (see Figure 1.14), we can use the x-intercepts and the y-intercepts, found by letting y = 0 and x = 0, respectively.

The coordinates of the point P of intersection of the lines are obtained from the solution of the following system of equations:

FIGURE 1.14 y

{4X + 3y = 5 3x - 2y = 8

.\.1 - 2\' ~

x

4x + 3y = 5

I 2 CHAPTER 1 PRECALCULUS REVIEW

To eliminate y from the system, we begin by multiplying the first equation by 2 and the second by 3:

{8.\ + 6y = 10 9.\ - 6y = 24

Next we add both sides of the equations, to obtain 17x = 34, or x = 2.

This is the x-coordinate of the point of intersection. To find the y-coordinate of P, we let .\ = 2 in 4.\ + 3y = 5, obtaining

4(2)+3)' = 5, or y= - 1. Hence P has coordinates (2, - I).

EXERCISES 1. 1

Exer. 1-8: Rewrite without using the absolute value symbol.

1 lal( - 5) 13 - 61 Ib) 1- 61 /( - 2) Ie) I - 71 + 14 1 2 la) (4) 16 - 71 Ib) 5/1- 21 Ie) 1- I I + 1- 9 1 3 la) 14 - n I Ib) In - 4 1 Ie) I J2 - 1.5 1 4 la) I J3 - 1.7 1 Ib) I 1.7 - J31 Ie) I! - 11 ) 3 5 13 + x I if x < - 3 6 15 - x I if x > 5 7 12 - x I if x < 2 8 17 + x I if x ~ - 7

Exer. 9-12: Solve the equation by factoring.

9 15x 2 - 12 =-8x

11 2x(4x + 15) = 27 10 15x2 -14=29x

12 x(3x + 10) = 77 Exer. 13-16: Solve the equation by using the quadratic formula.

13 x2+4x+2=0

15 2x 2 -3x-4=0

14 x2-6x-3=0

16 3x 2 + 5x + 1 = 0

Exer. 17-38: Solve the inequality and express the solu-tion in terms of intervals whenever possible.

1 7 2x + 5 < 3x - 7

2x - 3 19 3< --< 7

- 5

21 x 2 - x - 6 < 0

23 x 2 - 2x - 5 > 3

25 x(2x + 3) ~ 5 x+1

27 --> 2 2x - 3

I 3 29 - - >--

x-2 - x + 1

31 I x + 3 1 < 0.01

18 x - 8 > 5x + 3

4x + I 20 -2 0.002 36 13x - 71 ~ 5 38 1- II - 7x I> 6

Exer. 39-40: Describe the set of all points P(x, y) in a coordinate plane that satisfy the given condition. 39 la) x = -2 Ib) )' = 3 Ie) x ~ 0 Id) xy > 0

Ie) )' < 0 (f) I x I ::; 2 and I y I ::; I 40 la) y=-2 (b) x= - 4 Ie) xIJ < 0 Id) X.l" = 0

(e) y > I (f) I x I ~ 2 and I y I ~ 3 Exer. 41-42: Find la) d(A, B) and (b) the midpoint of AB. 41 A(4 , - 3), B(6 , 2) 42 A( - 2, - 5) , B(4, 6) 43 Show that the triangle with vertices A(8, 5) , B( I, - 2),

and C( - 3, 2) is a right triangle , and find its area. 44 Show that the points A( - 4, 2). B( 1.4), C(3, - I) , and

D(-2, - 3) are vertices ofa square. Exer. 45-56: Sketch the graph of the equation. 45 Y = 2X2 - I

47 x = h 2 49 Y = x 3 - 8

51 Y = fx - 4 53 (x + W + (y - 2)2 = 9 55 )'= -JI6-x 2

46 y = _x 2 +2

48 X= _ 2.1'2

50 )' = _x 3 + I 52 Y = \Ix - 4

54 x 2 + (.l' - 2)2 = 25 56 Y = J 4 - x 2

Exer. 57-60: Find an equation of the circle that satisfies the given conditions.

57 Center C(2 , - 3); radius 5 58 Center C( - 4,6); passing through P(I , 2) 59 Tangent to both axes; center in the second quadrant;

radius 4

EXERCISES 1.1

60 Endpoints of a diameter A(4, - 3) and B( - 2, 7) Exer. 61-66: Find an equation of the line that satisfies the given conditions.

61 Through A(S, -3); slope - 4 62 Through A( - I, 4); slope ~ 63 x-intercept 4; y-intercept - 3

64 Through A(S, 2) and B( - 1,4) 65 Through A(2, -4); parallel to the line Sx - 2y = 4 66 ThroughA(7 , -3); perpendicular to the line 2x - S)' = 8 Exer. 67-68: Find an equation for the perpendicular bisector of AB.

67 A(3, - I) , B( -2, 6) 68 A(4 , 2), B( - 2, 10) Exer. 69-72: Sketch the graphs of the lines and find their point of intersection.

69 2x + 3y = 2; x - 2y = 8

70 4x + S)' = 13 ; 3x + y = -4 71 2x + S;- = 16; 3x - 7;- = 24

72 7x - 8.r = 9; 4x + 3;- = - 10

o 73 Approximate the coordinates of the point of intersection of the lines

(JITS - O.I)x + (0.11) 2 3 y = 1/ .j5 (2.Sl)2f3 x + (6.27 - J3)y = 12.

o 74 Approximate the smallest root of the following equa-tion: x 2 - (6.7 X 106)x + 1.08 = O. To avoid calculating a zero value for this root , rewrite the quadratic formula as

2c x= -------;~==

- b b2 - 4ac

75 The rate at which a tablet of vitamin C begins to dissolve depends on the surface area of the tablet. One brand of tablet is 2 centimeters long and is in the shape of a cylin-der with hemispheres of diameter O.S centimeter attached to both ends (see figure) . A second brand of tablet is to be manufactured in the shape of a right circular cylinder of altitude O.S centimeter. EXERCISE 75

~----2cm >1 I ,

='=1o.s cm

~~JO.S cm

1 3

(aj Find the diameter of the second tablet so that its surface area is equal to that of the first tablet.

(bj Find the vol ume of each tablet. 76 A manufacturer of tin cans wishes to construct a right

circular cylindrical can of height 20 centimeters and of capacity 3000 cm 3 (see figure). Find the inner radius r of the can.

EXERCISE 76

I , I

~r~

77 Shown in the figure is a simple magnifier consisting of a convex lens. The object to be magnified is positioned so that its distance p from the lens is less than the focal length f. The linear magnification M is the ratio of the image size to the object size. It is shown in physics that M = f jU - pl If f = 6 cm, how far shou ld the object be placed from the lens so that its image appears at least three times as large?

EXERCISE 77

78 As the altitude of a space shuttle increases, an astro-naut's weight decreases until a state of weightlessness is achieved. The weight of a 12S-pound astronaut at an alt itude of x kilometers above sea level is given by

W = 12S( 6400 )2 6400 + x

At what altitudes is the astronaut's weight less than S pounds?

79 The braking distance d (in feet) of a car traveling u mijhr is approximated by d = v + (u 2/20). Determine velocities that result in braking distances of less than 7S feet.

80 For a drug to have a beneficial effect, its concentration in the bloodstream must exceed a certain value, the

1 4

minimum therapeutic lerel. Suppose that the concentra-tion c of a drug 1 hours after it is taken orally is given by c = 20( ((2 + 4) mg/ L. If the minimum therapeutic level is 4 mg/ L, determine when this level is exceeded.

CHAPTER 1 PRECALCULUS REVIEW

82 Pharmacological products must specify recommended dosages for adults and chi ldren. Two formulas for modi-fication of adult dosage levels for young children are

Cowling's rule: y = 214(1 + I)a 81 The electrical resistance R (in ohms) for a pure metal

wire is related to its temperature T (in 0c) by the formula R = Ro( I + aT) for positive constants a and Ro.

Friend's rule: y = lsta where a denotes the adult dose (in milligrams) and 1 denotes the age of the child (in years).

lal For what temperature is R = Ro? Ibl Assuming that the resistance is 0 if T = - 273 C

(absolute zero) , find a.

lal If a = 100. graph the two linear equations on the same axes for 0 :0; 1 :0; 12.

Ibl For what age do the two formulas specify the same dosage? lei Silver wire has a resistance of 1.25 ohms at 0 C. At

what temperature is the resistance 2 ohms?

1.2 FUNCTIONS

Definition (1.10)

FIGURE 1.15

z~;(W) D a~ ~(Z)

~ f(x) f(a)

E

The notion of/unction is basic for much of our work in calculus. We may define a function as follows.

A function f from a set D to a set E is a correspondence that assigns to each element x of D exactly one element y of E.

The element y of E is the value of f at x and is denoted by f(x) , read f ofx . The set D is the domain of the function. The range of f is the subset of E consisting of all possible function values J(x) for x in D.

We sometimes depict functions as in Figure 1.15, where the sets D and E are represented by points within regions in a plane. The curved arrows indicate that the elements J(x), J(\\') , J(::) , and J(a) of E correspond to the elements x, W, Z, and a of D. It is important to remember that to each x in D there is assigned exactly one Junction value J(x) in E; however , different elements of D, such as wand z in Figure 1.15 , may yield the same function value in E. Throughout Chapters 1- 14, the phrase J is a Junction will mean that the domain and range of J are sets of real numbers .

We usually define a function J by stating a formula or rule for finding J(x), such as f(x) = J.x - 2. The domain is then assumed to be the set of all real numbers such that f(x) is real. Thus , for J(x) =~, the domain is the infinite interval [2 , ex) ). If x is in the domain , we say that f is defined at x, or that f(x) exists. If S is a subset of the domain , then J is defined on S. The terminology J is undefined at x means that x is not in the domain of f.

J4 + x EXAMPLE 1 Let J(x) = . I -x lal Find the domain of f. Ibl Find J(5), J( - 2) , f( - a), and -J(a). SOLUTION lal Note that J(x) is a real number if and only if the radicand 4 + x is nonnegative and the denominator 1 - x is not equal to O. Thus, J(x) exists if and only if

4 + x ;?: 0 and 1 - x#-O

1.2 FUNCTIONS

FIGURE 1.16

1 5

or, equivalently, x:;::: - 4 and X", 1.

Hence the domain is [ -4,1) u (I , (0). (b) To find values of f , we substitute for x :

f (5) = fl+5 = J9 = _~ I - 5 - 4 4

[( - 2) = J 4 + (- 2) = j2 . I - (-2) 3 f( _ a)= J 4+( - a)= 4 - a

l-(- a) I + a

- f (a) = _ ~ = J4+"G I - a a - I

Ma ny formul as that occur in mathematics and the sciences determine functions. For insta nce , the formula A = nr2 for the area A of a circle of radius r assigns to each positive real number r exactly one value of A. The letter r, which represents an arbitrary number from the domain, is an independent variable. The letter A , which represents a number from the range , is a dependent variable , since its value depends on the number assigned to r . If two variables r and A are related in this manner, we say that A is a junction of r. As another example, if an automobile travels at a unifo rm rate of 50 mi/hr , then the distance d (in miles) traveled in time t (in hours) is given by d = SOt , and hence the distance d is a function of time t .

EXAMPLE 2 A steel storage tank for propane gas is to be constructed in the shape of a right circular cylinder of altitude 10 feet with a hemisphere attached to each end . The radius r is yet to be determined. Express the vo lume V of the tank as a function of r .

SOLUTION The tank is sketched in Figure 1.16. We may find the volume of the cylindrical part of the tank by multiplying the altitude 10 by the a rea nr2 of the base of the cylinder:

volume of cylinder = 10(nr2) = lOnr 2

The two hemispherical ends , taken together, form a sphere of radius r. Using the formula for the volume of a sphere, we obtain

volume of the two ends = 4nr3 . Thus, the volume V of the tank is

V = 4nr3 + IOnr2 = 1nr2(2r + 15). This formula expresses Vas a function of r.

If f is a function , we may use a graph to illustrate the change in the function value f( x ) as x varies through the domain of f. By definition,

1 6

FIGURE 1.17 \'

T-Range of J

i_

\

I k-Domain of J ~

CHAPTER 1 PRECALCULUS REVIEW

the graph of a function I is the graph of the equation J = I(x) for x in the domain of f. As shown in Figure 1.17. we often attach the label y = f(x)

J(x) to a sketch of the graph. Note that if P(a, h) is on the graph, then the y-coordinate b is the func tion value I(a). The figure exhibits the domain off' (the set of possible va lues of x) and the range off (the corresponding values of y), Although we have pictured the domain and range as closed intervals, they may be infinite intervals or other sets of real numbers,

It is important to note that since there is exactly one value f(a) for x each a in the domain, only one point on the graph has x-coordinate a,

Thus. erery rerrical line intersects the graph of' a function in at most one point. Consequently, the graph of a function cannot be a figure such as a circle. which can be intersected by a vertical line in more than one point.

The x-intercepts of the graph of a function I a re the solutions of the equation f(x) = O. These numbers are the zeros of the function. The y-intercept of the graph is f(O), if it exists,

If f is an even function - that is , if I( -x) = I(x) for every x in the domain of f - then the graph off is symmetric with respect to the y-axis, by symmetry test (i) of (\.6). If f is an odd function that is , if f( -x) = -I(x) for every x in the domain of f - then the graph of I is symmetric with respect to the origin, by symmetry test (iii). Most functions in calculus are neither even nor odd.

The next illustration contains sketches of graphs of some common funct ions. You should check each for the indicated symmetry, domain, and range.

ILLUSTRATION

FUNCTION f

- I(x)=~

- f(x) = x 2

- I(x) = x 3

GRAPH

y

y

y

x

x

x

SYMMETRY

none

y-axis (even function)

origin (odd function)

DOMAIN D, RANGE R

D = [0, (0) R = [0, (0 )

D = (-00 , (0 ) R = [0, (0 )

D=( -oo,oo) R = (- 00, (0 )

(COlli iI/lied)

1.2 FUNCTIONS

FIGURE 1.18 y

1 7

FUNCTION f _ fix) = X 2 3

GRAPH

y SYMMETRY

y-axIs DOMAIN D, RANGE R

D = (-x, 00 ) R = [0, x) (even function)

x

_ fix) = X l3 y OrIgll1 D=(- x . x ) R=(- x . x )

- fix) = Ix I

. 1 - fix) = -

x

x

x

)'

x

y

x

(odd function)

y-axIs (even function)

orIgin (odd function)

D = (-x, 00 ) R = [0, 00 )

D = ( - 00, 0) u (0, (0) R = (-00,0) u (0, (0)

Functions that are described by more than one expression, as in the next example, are called piecewise-defined functions.

EXAMPLE 3 Sketch the graph of the function f defined as follows:

{

2X + 3 if x < 0 fix) = Xl if 0 ::; x < 2

1 if x;:::: 2

SOLUTION If x < 0, then fix) = 2x + 3. and the graph of f is part of the line)' = 2x + 3, as indicated in Figure 1.18. The small circle indicates that (0,3) is not on the graph.

If 0 ::; x < 2, then fix) = Xl, and the graph of f is part of the parabola y = Xl. Note that (2,4) is not on the graph .

If x ;:::: 2, the function values are always 1, and the graph is a horizontal half-line with endpoint (2, 1).

(

1 8

ILLUSTRATION

FIGURE 1.19 y

~

I I I I I I ~-

I I I I I . x

.......... ! ~

CHAPTER I PRECALCULUS REVIEW

If x is a real number, we define the symbol [x] as follows: [x] = n, where n is the greatest integer such that n ~ x

If we identify IR with points on a coordinate line, then n is the first integer to the left of (or equal to) x .

-[0.5] = 0

-[ 1.8] = I

-[yIs] = 2

-[3] = 3

-[ - 3] = - 3

-[ - 2.7] = - 3

-[ -)3] =-2

-[- 0.5] = - 1

The greatest integer function f is defined by f(x) = [xl

EXAMPLE 4 Sketch the graph of the greatest integer function.

SOLUTION The x- and y-coordinates of some points on the graph may be listed as follows:

Values of x

- 2:-S;x

1.2 FUNCTIONS

FIGURE 1.20 Vertical shifts, c > 0

FIGURE 1,22

Y

FIGURE 1.24

y = x 2 + 4

\' .r-

y = x 2 - 2

y

x

1 0 Y = :jx-

x

FIGURE 1.21 Horizontal shifts, c > 0

y = f(x + c)

a - c

FIGURE 1,23 y

y

\' - f(x)

a

y = (x + 2)" Y x 2 Y = (x - 4)"

FIGURE 1.25

Y ,

\ ' = x-

x

x

1 9

y = f(x - c)

a + c x

20

....

CHAPTER 1 PRECALCULUS REVIEW

If f and g are functions , we define the sum f + g, difference f - g, product fg, and quotient j i g as follows:

(f + g)(x) = f(x) + g(x) (f - g)(x) = f(x) - g(x)

(fg)(x) = f(x)g(x)

( L )(x) = f(x) g g(x) The domain of f + g, f - g, and fg is the intersection of the domains

of f and g- that is , the numbers that are common to both domains. The domain of f i g consists of all numbers x in the intersection such that g(x) #- O.

EXAMPLE 5 Let f(x) = .}4 - x 2 and g(x) = 3x + I. Find the sum , difference, product, and quotient of f and g, and specify the domain of each.

SOLUTION The domain of f is the closed interval [ -2, 2J , and the domain of g is IR. Consequently, the intersection of their domains is [ -2, 2J, and we obtain the following:

(f + g)(x) = .}4 - x 2 + (3x + 1), - 2 :s; x :s; 2 (f - g)(x) = .}4 - x 2 - (3x + 1),

(fg)(x) = .}4 - x 2 (3x + 1),

(f) J4=7 g (x) = 3x + I ' - 2:s;x:s;2

- 2 :s;x:s; 2

- 2 :s; x :s; 2 and x#--

A function f is a polynomial function if f(x) is a polynomial- that is , if

where the coefficients ao, aI' . .. , a" are real numbers and the exponents are nonnegative integers. If a" #- 0, then f has degree n. The following are special cases, where a =1= 0:

degree 0: f(x) = a (constant function) degree 1: f(x) = ax + b (linear function) degree 2: f(x) = ax 2 + bx + c (quadratic function)

A rational function is a quotient of two polynomial functions. Later in the text we shall use methods of calculus to investigate graphs of poly-nomial and rational functions.

An algebraic function is a function that can be expressed in terms of sums, differences , products, quotients, or rational powers of polynomials. For example, if

1.2 FUNCTIONS

Definition (1.11)

FIGURE 1.26

g

x

Domain of g

Domain of f f(g(x))

2 1

thenf is an algebraic function. Functions that are not algebraic are termed transcendental. The trigonometric, exponential. and logarithmic functions considered later are examples of transcendental functions.

In the remainder of this sect ion we shall discuss how two functions f and 9 may be used to obtain the compositefitnctionsf 9 and 9 f (read f circle y and 9 circle f, respectively). The function f r 9 is defined as follows.

l The composite function f o g of f and 9 is defined by (.f 0 g)(x) = f(g(x)). The domain of f o g is the set of all x in the domain of 9 such that g(x) is in the domain of f.

Figure 1.26 illustrates relationships between f, g, and f g. Note that for x in the domain of y, we first .find y(x) (which must be in the domain off) and then, secol1d,findf(g(x)).

For the composite function y ' f, we reverse this order, first finding I(x) and then finding g(.f(x)). The domain of g o f is the set of a ll x in the domain of I such that f(x) is in the domain of y.

EXAMPLE 6 If f(x) = x 2 - 1 and g(x) = 3x + 5, find (aJ U y)(x) and the domain of f y (bJ (y I)(x) and the domain of 9 ' f SOLUTION

(aJ U ' y)(x) = f(g(x)) (definition off If) = f(3x + 5) (definition of y) = (3x + 5f - 1 (definition off) = 9x 2 + 30x + 24 (simplifying)

The domain of both I and 9 is ~ . Since for each x in ~ (the domain of 9) the function value 9(X) is in ~ (the domain of j), the domain of f o g is also R (bJ (y f)(x) = gU(x))

= g(X2 - I ) = 3(x 2 - I) + 5 = 3x 2 + 2

(definition of If f) (definition of .n (definition of y) (simplifying)

Since for each x in ~ (the domain of j) the function value f(x) is in ~ (the domain of y), the domain of 9 f is ~ .

Note that in Example 6, f(y(x)) and g(.f(x)) are not always the same; that is, I y i= 9 f.

If two functions f and y both have domain ~ , then the domain of f 0 y and y , f is also ~ . This was illustrated in Example 6. The next example shows that the domain of a composite function may differ from those of the two given functions.

2 2 CHAPTER I PRECALCULUS REVIEW

EXAMPLE 7 If {(x) = X 2 - 16 and g(x) = Fx, find fal (f 0 g)(x) and the domain of { 0 9 fbI (g {)(x) and the domain of 9 0 f

SOLUTION We first note that the domain of f is IR and the domain of 9 is the set of all nonnegative real numbers - that is , the interval [0 , CX) ). We may proceed as follows.

fal (f 0 g)(x) = f(g(x)) = f(Fx) = (Fxf - 16 = x - 16

(definition of f 9) (definition of g) (definition off) (sim pi ifyi ng)

If we considered only the final expression x - 16, we might be led to believe that the domain of f o g was IR, since x - 16 is defined for every real number x. However , this is not the case. By definition, the domain of f o g is the set of all x in [0, CJJ ) (the domain of g) such that g(x) is in IR (the domain of f) . Since g(x) = Fx is in IR for every x in [0, CX) ), it follows that the domain of f o g is [0, CX) ). fbI (g 0 f)(x) = g(f(x))

= g(x 2 - 16) = Jx2 - 16

(definition of 9 Il (definition off) (definition of y)

By definition, the domain of 9 0 f is the set of all x in IR (the domain of f) such that f(x) = x 2 - 16 is in [0, CX) ) (the domain of g). The state-ment x 2 - 16 is in [0, CX) ) is eq uiva1ent to each of the inequalities

x2

- 16 ?: 0, x 2 ?: 16, and Ix l?: 4.

Thus, the domain of 9 0 f is the union ( - CX) , - 4J u [4, CJJ ). Note that this is different from the domains of both f and g.

If f and 9 are functions such that y = f(u) and u = g(x) ,

then substituting for u in y = f(u) yields y = f(g(x)) .

For certain problems in calculus we reverse this procedure; that is , given y = h(x) for some function h, we find a composite function form y = f(u) and u = g(x) such that h(x) = f(g(x)).

EXAMPLE 8 Express y = (2x + 5)B in a composite function form. SOLUTION Suppose, for a real number x, we wanted to evaluate (2x + 5)B by using a calculator. We would first calculate 2x + 5 and then raise the result to the eighth power. This suggests that we let

u = 2x + 5 and y = uB,

which is a composite function form for y = (2x + 5)B.

EXERCISES 1.2

ILLUSTRATION

EXERCISES 1.2

23

The method used in the preceding example can be extended to other functions. In general, suppose we are given y = h(x). To choose the inside expression u = g(x) in a composite function form , ask the following ques-tion: If a calculator were being used , which part of the expression h(x) would be evaluated first? This often leads to a suitable choice for LI = g(x). After choosing u, refer to h(x) to determine y = f(u). The following illustra-tion contains typical problems.

FUNCTION VALUE CHOICE FOR u = g(x) CHOICE FOR Y = feu)

-y = (x 3 - 5x + 1)" II = x3 - 5x + .I' = u"

-J' = JX2 - 4 II = x 2 - 4 Y = J;t

2 2

-y= -- II = 3x + 7 .1'= -

3x + 7 u

The composite function form is never unique. For example, consider the first expression in the preceding illustration:

y = (x 3 _ 5x + 1)4 If n is any nonzero integer, we could choose

u={x3 -5x+l)" and y=U 411 . Thus , there are an unlimited number of composite function forms. Gen-erally, our goal is to choose a form such that the expression for y is simple , as we did in the illustration.

1 If f(x) = yX - 4 - 3x, find f(4), f(8), and f(13). Exer. 11-12: Determine whether f is even, odd, or nei-ther even nor odd.

x 2 If fix) = x _ 3' find f( - 2), flO), and f(3.0 1).

Exer. 3-6: If a and h are real numbers, find and simplify fa) f(a) , fbI f( - a), fe) - f(a), fd) f(a + h), fe) f(a) + f(h),

f(a + h) - f(a) and If) h ' provided h #. O.

3 fix) = 5x - 2 4 fix) = 3 - 4x 5 f(x) = x 2 - X + 3 6 fix) = 2X" + 3x - 7

Exer. 7-10: Find the domain of f. x+l

7 fix) = - 3-4-x - x

.J2x - 3 9 fix) = --=----

x2 - 5x + 4

4x 8 fix) = 6x2 + 13x _ 5

J 4x - 3 10 fix) = 2 4

x -

II fa) fix) = 5x J + 2x fbI fix) = Ixl - 3 fe) fIx) = (8x 3 - 3X 2 )3

12 fa) fIx) = 3x4 + 2X" - 5 fbI fIx) = 6x 5 - 4x 3 + 2x Ie) fix) = xix - 5)

Exer. 13-22: Sketch, on the same coordinate plane, the graphs of f for the given values of c. (Make use of sym-metry, vertical shifts, horizontal shifts, stretching, or reflecting. ) 13 fix) = 1 x 1 + c; c = 0, I, - 3 14 f(x)=lx-cl; c=0,2,-3 15 f(x)=2 J;+c; c=0,3,-2

24

16 f (x) = , 9 - x 2 + c; 17 f (x) = 2Jx - e;

e = 0, I , -3

c = 0, I , - 2

c = 0, I , -2

e = 1,3, -2

e = 0, I , -2

18 f(x) = -2(x-e)2; 19 f(x) = eJ4 - X 2 ; !O f (x) = (x + e)3;

21 f( x ) = (X - e)2/3 + 2; e = 0,4, - 3 22 f (x) = (x - 1) 13 - e; e = 0, 2, - I

ExeL 23-24: The graph of a function f with doma in o ::5 x ::5 4 is shown in the figure. Sketch the graph of the given equation.

23

y

24

y

x

r = f(x + 3) (b) )' = f(x - 3) (e) y = f(x) + 3 (d) )' = I(x) - 3 (e) )' = - 3{(x) (f) .r = - ,\((.x) (9) )' = -I(x + 2) - 3 (h) Y = I(x - 2) + 3

(a) .\ = I(x - 2) (b) )' = I(x + 2) (e) )' = I(x) - 2 (a) )' = f(x) + 2 (el )' = - 2I(x) (f )' = -U(x)

x (9) Y = -fIx + 4) - 2 (h )' = I(x - 4) + 2

ExeL 25-30: Sketch the graph of f.

r' if x:s; - I 25 f(x) = x3 if lxl < I -x+ 3 if x;:::: 1 r- 3 if x:s; -2 26 I (x) = _x 2 if -2

EXERCISES ).2

[] 52 If f(x) = ,x3 + I - I , approximate f(O.OOOI). [n order to avoid calculating a zero value for f(O .OOOI), rewrite the formula for f as

53 An open box is to be made from a rectangular piece of cardboard 20 inches x 30 inches by cutting out identical sq uares of area x 2 from each corner and turning up the sides (see figure) . Express the volume V of the box as a function of x.

EXERCISE 53 IE 30 )1 I x xl I .:;. IE') " I

! I---------~;~lll ----------' ___ il

54 An open-top aquarium of height 1.5 feet is to have a volume of 6 ft 3 Let x denote the length of the base, and let )' denote the width (see figure). laj Express J as a function of x. Ibj Express the total number of square feet S of glass

needed as a function of x.

EXERCISE 54

25

55 A hot-air ba ll oon is released at I :00 P.M. and rises ver-tica lly at a rate of 2 m sec. An observation point is situ-ated 100 meters from a point on the gro und directly below the balloon (see figure). [f r denotes the time (in seconds) after I : 00 P.M. , express the distance d be-tween the balloon and the observa ti on point as a func-tion of r.

EXERCISE 55

11 II

I I r---------------~/~I--------~

I I ( I Observation / I point / P>

1-,~100 m

56 Refer to Example 2. A steel storage tank for propane gas is to be constructed in the shape of a right circular cylinder of al titude 10 feet wi th a hemisphere attached to each end. The radius r is ye t to be determined. Express the surface area S of the tank as a function of r.

57 From an ex terior point P that is h units from a circle of radius r, a tangent line is drawn to the circle (see figure) . Let J denote the distance from the point P to the point of tangency T. laj Express y as a fu nction of h. (Hint : [f C is the center

of the circle, then PT is perpendicular to CT.) Ibj If r is the radius of the earth and h is the altitude of

a space shuttle, then we can derive a formula for the maximum distance (to the earth) that an astronaut can see from the shuttle . In particular, if h = 200 mi and r ~ 4000 mi , approximate y .

EXERCISE 57

.1'//

T /"

/" /

~-----II r

58 Triangle A BC is inscribed in a semicircle of diameter 15 (see the figure on the foll owing page). laj [f x denotes the length of side AC, express the length

)' of side BC as a function of x, and state its domain. (Hint: Angle ACB is a right angle.)

Ibj Express the area of triangle ABC as a function of x.

26

EXERCISE 58 c

,/

A~--------11 B

The relative positions of an airport runway and a 20-foot-tall control tower are shown in the figure. The beginning of the runway is at a perpendicular distance of 300 feet from the base of the tower. If x denotes the distance an airplane has moved down the runway, ex-press the distance d between the airplane and the control booth as a function of x.

EXERCISE 59

60 An open rectangular storage shelter consisting of two vertical sides , 4 feet wide , and a Aat roof is to be attached to an existing structure as illustrated in the figure. The Aat roof is made of tin that costs $5 per square foot , and the other two sides are made of plywood costing $2 per square foot.

If $400 is available for construction , express the length y as a function of the height x.

EXERCISE 60

1.3 fRIGONOMf: TRY

CHAPTER 1 PRECALCULUS REVIEW

Ib) Express the volume V inside the shelter as a function of x.

61 The shape of the first spacecraft in the Apollo program was a frustum of a right circular cone, a solid formed by truncating a cone by a plane parallel to its base. For the frustum shown in the figure , the radii a and b have already been determined. la) Use similar triangles to express)' as a function of h. Ib) Express the volume of the frustum as a function of h. Ie) If a = 6 ft and b = 3 ft , for what value of h is the

volume of the frustum 600 ft 3?

EXERCISE 61 I b ~ I I t------ "7t, :

y / I ' I / I " t---- / ! \I

h I

l_ i I I I

~a~

62 A right circular cylinder of radius r and height h is in-scribed in a cone of altitude J 2 and base radius 4, as illustrated in the figure. la) Express h as a function of r . (Hint: Use similar

triangles.) Ib) Express the volume V of the cylinder as a function

of r.

EXERCISE 62 r-------:

112 r-- !

h I __ 1_ !

I I I I

~r~ I I

k-4~

[n geometry an angle is determined by two rays, or line segments, having the same initial point 0 (the vertex of the angle). [f A and B are points on the rays II and 12 in Figure 1.27, we refer to angle ADB, or LAOB. We

1.3 TRIGONOMETRY

FIGURE 1.27

FIGURE 1.28

FIGURE 1.29 0= t radians

u

()

y

y

Initial ,ide

x

A( I. II) x

Relationships between degrees and radians (1.12)

27

often use a lower-case Greek letter, such as e, rx, or [3, to denote an angle. In trigonometry we may also interpret L AOB as a rotation of ray II in the plane determined by A, 0, and B: We rotate 11 (the initial side of the angle) about 0 , to a position specified by 12 (the terminal side). The amount or direction of rotation is not restricted; we can let 11 make several revolu-tions in either direction about 0 before stopping at 12 , Thus, many angles may have the same initial and terminal sides.

If we introduce a rectangular coordinate system, then the standard position of an angle 0 is obtained by taking the vertex at the origin and the initial side along the positive x-axis (sec Figure 1.28). The angle e is positive for a counterclockwise rotation and negative for a clockwise rotation.

The magnitude of an angle may be expressed in either degrees or ra-dians. An angle of degree measure 1 corresponds to 3~0 of a complete revolution in the counterclockwise direction. One minute (1 ') is 610 of a degree, and one second (I ") is Jo of a minute. In calculus the most im-portant unit of angular measure is the radian. To define radian measure , consider the unit circle U with center at the origin of a rectangular coor-dinate system, and let e be an angle in standard position (see Figure 1.29). If we rotate the x-axis to the terminal side of e, its point of intersection with U travels a certain distance t before arriving at its final position P(x, y). If t is considered positive for a counterclockwise rotation and negative for a clockwise rotation , then e is an angle of t radians , and we write either e = t or e = t radians. In Figure 1.29, ( is the length of the arc AP. If e = 1 (that is, if e is an angle of 1 radian), then the length of

" arc AP on U is 1 (see Figure 1.30). FIGURE 1.30

y

. \( I. II ) x

u

Since the circumference of the unit circle is 2rr, it follows that

2rr radians = 3600

This fact gives us the following relationships.

180 = rr radians rr d' 1 = 180 ra lan (180)0 1 radian = ---;-

If we approximate rr/ 180 and 180/ rr, we obtain

1 0 ~ 0.01745 radian and 1 radian ~ 57.29578.

The next theorem is a consequence of the formulas in (1.12).

l

28

l Radians

Degrees

FIGURE 1.31

()

Theorem (1.13)

n n n o

6 4 3

0 30 45 60

n

2

90

Length of a circular arc (1.14)

Area of a circular sector (1.15)

CHAPTER 1 PRECALCULUS REVIEW

IiI To change radian measure to degrees, multiply by 180/n. liil To change degree measure to radians , mu ltiply by n/ 180.

When radian measure o{an angle is used, I/O IInilS lI'ill be indicated. Thus , if an angle has radian mcasure 5, we write 0 = 5 instead of e = 5 radians. There should be no confusion as to whether radian or degree mcasurc is being used, since if 8 has degree measure 5 , we write 8 = 5, nOlO = 5.

2n 3n 3 4

120 0 135

5n 6 n

7n 6

5n 4

150 0 180 210c 225

4n 3

240

3n 2

270

5n 3

300

7n 4

315

11 n 6

2n

3300 360

T he above tablc displays the relationship between the radian and degree measures of several common angles. The entrics may be checked by using Theorcm (1.13).

A central angle of a circle is an angle 0 whose vertex is at the center of the circle, as illustrated in Figure 1.31 . We say that arc AB sub tends 0, or that e is sub/ended hy AB. The relationship betwecn thc length s of As, the radian measure of 0, and the radius r of the circle follows.

If an a rc of length s on a circle of radius I' subtends a central angle of rad ian measure 0, then

s = reo

PROOF If 05 , is the length of any other arc on thc circle and if 0, is the radian measure of the corresponding central anglc , thcn , from planc gco-metry , the ratio of the arcs is the same as the ratio of the angular measures; that is , sis I = % I' and hence s = S I o/e I' If we consider the special case in which 8 , = 2n, then S, = 2nr, and we obtain S = 2nrO/(2n) = 1'0. _

We will makc usc of the next result later in thc tcxt.

If e is the rad ian measu re of a central angle of a circle of radius rand if A is the area of the circular sector determined by 0, then

A = 1/"28 .

PROOF A typical anglc and circular sector arc shown in Figure 1.31 . If 01 is any other central angle and A) the area of thc corresponding sector, then, from plane geometry, A/ A, = 0/8

"

or A = A ,0/8" If we consider the special case 0 , = 2n, then A I = n/"2 and A = nr20/ (2n) = 1/"20. _

The six trigonometric functions are the sine, cosine, tangent , cosecant, secant, and cotangent , denotcd by sin, cos, tan, csc. sec, and cot , respec-

1.3 TRIGONOMETRY

The trigonometric functions (1.16)

FIGURE 132 )'

II

sin () > 0 A ll esc 8 > 0 positive

tan () > 0 cos () > 0 x cot () > 0 sec () > 0

III IV

29

tively. We may define the trigonometric functions in terms of either an angle e or a real number x. There are two standard methods that employ angles:

1. If 0 is acute (0 < 0 < n12), we may use a right triangle. 2. If 0 is any angle (in standard position), we may use the point P(a, b) at which the terminal side of 0 intersects the circle X2 + y2 = ,.2.

In the following definitions , the abbreviations adj, opp, and hyp are used for the lengths of the adjacent side, opposite side, and hypotenuse of a right triangle having 0 as an angle.

III Of an Acute Angle e:

(hyp) c

()

il (adj) liil Of Any Angle 0:

y

x~ + \' ::! = ,.2

b (opp)

x

11111 Of a Real , umber x:

. e b c S1l1 = - csc e = b c

a cos e = - c sec e = -

c a

b a tan e = - cot e = b

a

. e b r S1l1 = - csc e = b r

a cos e = - r sec e = -

r a

b a tan e = - cot e = b

a

The value ( r a trigonometric function at a real number x is its value at an angle of x radians.

Note that, by (iii), there is no difference between trigonometric func-tions of angles measured in radians and trigonometric functions of real numbers. For example, we can interpret sin 2 as either the sine of an angle of 2 radians or the sine of the real number 2.

The valucs of the trigonometric functions of acute angles in (i) are ratios of sides of a right triangle, and hence they are positive real numbers . For the general case (ii) , the sign of the function value depends on the quadrant containing the terminal side of e. For example, if e is in quadrant TI, then a < 0, b > 0, and hence sin 0 = blr > 0 and csc e = rib> O. The other four functions are negative. These facts are indicated schematically 111 Figure 1.32. You should check the signs in the remaining quadrants.

30

Fundamental Identities /1.17)

CHAPTER' PRECALCULUS REVIEW

We see from (ii) of Definition (1.16) that the domain of sin and cos consists of all angles O. Since tan 0 and sec 0 are undefined if a = (that is, if the terminal side of 0 is on the y-axis), the domain of tan and sec consists of all angles except those of radian measure (n/2) + nn , where n is an integer. The domain of cot and csc consists of all angles except those of radian measure nn , since cot 0 and csc 0 are undefined if b = 0.

Note from (ii) that

I sin 0 I ::;; 1, I cos 0 I ::;; 1, I csc 0 I ::2: 1, and I sec 0 I ::2: 1 for every 0 in the domains of these functions .

We shall next li st some important relationships that exist among the trigonometric functions . Recall that an expression such as sin 2 0 means (sin O)(sin 0).

I 1 csc 0 = --

sin 0

1 sec 0 = --0

cos

I cot 0 = --0

tan

O sin 0 tan = --cos 0

cos 0 cot 0 = - '- 0

sm

sin 2 0 + cos 2 0 = 1

1 + tan 2 0 = sec2 0

1 + coe 0 = csc2 0 J

Each fundamental identity can be proved by referring to (ii) of Definition (1.16). For example:

r 1 I csc 0 = - = - - = --

b (b/r) sin 0 tan 0 = ~ = (b/ r) = sin 0

a (a/ r) cos 0 b2 a2 a2 + b2 r2

sin 2 0 + cos 2 0 = - + - = = - = 1 r2 r2 r2 r2

Fundamental identities are useful for changing the form of an ex-pression that involves trigonometric functions. To illustrate , sll1ce cos 2 0 = 1 - sin 2 0,

sin 0 sin 0 tan 0 = - - = - r== =::;==

cos 0 1 - sin 2 0

r n Chapter 9 we will use trigonometric substitutions of the type illus-trated in the next example.

EXAMPLE 1 If a> 0, express -./a 2 - x 2 in terms of a trigo-nometric function of 0 without radicals by making the trigonometric substitution

x = a sin 0 n n for - - < 0 < - . 2- -2

1.3 TRIGONOMETRY 3 1

Special values of the trigonometric functions (1.18)

FIGURE 1.33

2

SOLUTION We let x = a sin 8:

,)a2 _ x 2 = ,)a 2 - (a si n 8)2 = ,)a2 - a2 sin 2 8 = ,)a2(1 - sin 2 8) = ,)a2 cos2 8 = a cos 8

The last equality is true because, first , JGi = a if a > 0, and, second , if - nl2 ~ 0 ~ n12, then cos 8 ;::: 0 and hence ,)cos 2 8 = cos 8.

There are a variety of methods for finding values of trigonometric fu nctions. For certain special cases we can refer to the right triangles in Figure 1.33. Applying (i) of Definition (1.16) gives us the following .

0 0 : SlJ:SO~ ~co /RADIANS) /DEGREES) cot 0 sec 0 n 1 13 13 13 213 30 - 2 6 2 2 3 3 n

45 J2 J2 J2 J2 4 2 2 n

60 13 1 13 13 213 - 2 3 2 2 3 3

Two reasons for stressing these special values are that (I) they are exact and (2) they occur frequently in work involving trigonometry . Because of their importance, it is a good idea either to memorize the table or to be able to find the values quickly by using the triangles in Figure 1.33.

It is possible to approximate, to any degree of accuracy, the values of the trigonometric functions for any angle. Table A in Appendix III gives four-decimal-place approximations to some values.

Scientific calculators have keys labeled I SIN I, I cos I, and ITAN I that can be used to approximate values of these functions. The values of csc, sec, and cot may then be found by means of the reciprocal key ITE] . Before using a calculator to find func