1. [-1- r" 16lT- {Ll *, ir a > of du | /i "' l/a +Tn + lal-'2"0'. Jf -u--:/:ar +n - lZr^n_r r@_ q6 i f a

1. a[1 t"r,r,-,L+cl a al l c o t r , - , L + cl a ai f l u l < ai f l u l > aIn formulas 27 through 38, we may replaceln(z * t/FRl by sinh-lf,lnlu * trt41 by cosh-'f,. la+ tfi4lt"l , bY sinh-'f,I,, I#:rnlu +{F+l+c,8. t/itTV Idu:l 1ffif +$,non,)ovl. cerrt.D=n$U" + vFEA + cf _2e.Juz/uz + 02 d" --t(2u' + a2) r/uz:; . tffil + ctoI. r y : fr*' n-z " " lLPl*,,t. I =-: tfi, - a2-4 sec-,..1'ltAzA ' f I/FTF au FtA , ,-r-. l-ff:-T*hlu+vTfr+c1 u'z ilu u - 1n2 zz. J ffi:it/F tF -;hlu * tffiv1 + c7"C "*[n-k3st.L :1r " . - , lg*l . ""' J uy@-- oz a"-- lal' -,uI+:_YT*,ftr.J(uz- + szytroz" :t (2u,+ 'Sazl)i rE7+{r'r1"+ tFTTl+c*q'R I d " : - L L cJ (uz-rszlstz -raz1/ffi -" I#:sin-l i * '*. I 1/o:utur:l tF7+f"in-,I*,f - nt Juz1ff-i uzo ":t (2u,- sz1/ F= uz+ f; sin-'I *,' 42. t/F=F - ' J 1au: {a,77 - a tnl'* *, u 1 Yal u I: {a'7 u' - a cosh-l X *,-L"?. 1 t lF=Ta u : _ l ; ' - u ' _ s i n - lL +cJ u2 u ---- a

2. 44.45.i46.49.50.t%:-tt m+o: sin-rL+cJ t/az-uz 2 2 nlL:-1 , n l o * f f i ! * .J u ! a 2 - u 2 a I u I: -+cosh1-l* 'f - du E-u' , - ,J uz tfaz - u2 azuFsftrl*,C: ont4futingL, wu- n3f . - , u- a a' ^ ^ - - ,Itrau-F du:T t/2au-uz +7 cos-r[ " r m d u : 2 u 2 - T - t o ' f f i r e47.Ioz - uz)stdzu : -[ t "'- sq')4sI#:#+caAa . _, Ir , r {az - uz +?sin-l;*t(r-+)*,* t c o s - l ( r - ) * ,sl I ry : tffi * ac os(-r'- :)+ cszI.r y:-rY-cos--,:()1* ,s3 I#:cos-r(1-T)*,Forms Containing Trigonometric Functionsff.'''59. f sinudu cosu+CJf60. f c o s u d u : s i n u * CJf61. I t a t u d u : l n l s e c u l +CJ62. | .o. u du- lnlsin ul + cJf63. I t". u du - lnl secu * tan ul + Cr: lnltan (*n + iu)l+ C64. ["r.u du- lnlcsc t/ - cot ul + CJ- lnltan iul + C6 5 . I r u . ' u d u : t a n u * CJf56. f csc2udu cotu+CJr57.Jsecil tanudu: secil+C58. f.r. u cotu duJf69.Isin2 u du : Lrtt- *sin 2u * Cf70. I cos' u du: *u * *sin 2u * CJf71.Jtanz u du: tan tt - u+ Cvsr '4 .t + t t raua *acos - ' ( r -g) +cJ l f f i u - u z' - - a /5trt5r .[ uz du (" |l')- ER +3o-' cos-l 166rFo, . f du @- ,-/-J f f i : - *T 'f d u : -u - a , - .J (2au - u2)st2 a2!2au - uzvs'L8r'. 1 , udu. -++cJ (Zau - u2)3t2 atffi=Er72.cot2u du: - cot u - u+ CIt 73.J t t : - l , i n n - r u c o s t l+ n " 1 l . , i r r ' - ' u d u J s i n n u t n- r i l c o s u + -n Ju r ^ rf 74. Jcosnu du:+cos' - ra sin"*+J.o'"-2 u duu du:#tan'-r "- Itann-27s.J .ur,"u ilu76/..ot" u du : -i+cotn-'"-Icotu-uz i lu77.J r".' u du :#secn-2 u tan" +#/'"t'-2 u du78.Icscu' du #csc'-z u cot"+ #J t""-' u du7s./ri^ mu sinn ud u- - -W + W+c- f , sin (m * n)u , sr(m --n)u-t80.c o s n ud u : - ; f f i +@ t Lc lcosmuf a cos(rn * n)u cos(m - n)u -L .81.Jsinmu cosnu du- -@ r Lf82.Iu sin u du: sin u - tt cos ll + Cf83.Iil cos u du: cos u * u sin z + C84. I u'sin z du - 2u srn u * (2 - u2) cos u * CJ 3. WITH ANALYTIC GEOMETRYthird edition g((I{{!itLoulsLelthold UNIVERSITY OF SOUTHERN CALIFORNIAHARPER & ROW, PUBLISHERSNew York, HagerstoutnS, anF ranciscoL, ondon 4. Sponsoring Editor: George J. TeleckiProject Editor: Karen A. fuddDesigner: Rita NaughtonProduction Supervisor: Francis X. GiordanoCompositor: Progressive TypographersPrinter and Binder: Kingsport PressArt Studio: J & R Technical Services Inc.Chapter opening art: "Study Light,, by patrick CaulfieldTHE CALCULUS WITH ANALYTIC GEOMETRY,Copyright @ 1.9681, ,972t,9 76 by Louis LeitholdAll rights reserved. Printed in the United States of America. No part of this book may beused or reproduced in any manner whatsoever without written permission except in thecase of brief quotations embodied in critical articles and reviews. For information addressHarper & Row, Publishers, Inc., 10 East 53rd street, New york, N.y. 10022.Library of Congress Cataloging in Publication DataLeithold, Louis.The calculus, with analytic geometry.Includes index.L. Calculus. 2. Geometry, Analytic, L Title.QA303.L428 7975b 515'.1s 75-26639ISBN 0-06-043951-3Edition 5. To Gordon Marc 6. GontentsChapter 1Preface )cu,REAL NUMBERS,INTRODUCTION TOANALYTIC GEOMETRY,AND FUNCTIONSpage 1Chapter 2LIMITS AND CONTINUIT}paSe ocChapter 3THE DERIVATIVEpage 1101.1 Sets, Real Numbers, and Inequalities 21.2 Absolute Value 141.3 The Number Plane and Graphs of Equations 2L1.4 Distance Formula and Midpoint Formula 281.5 Equations of a Line 331.6 The Circle 43L.7 Functions and Their Graphs 481.8 Function Notation, Operations on Functions, and Types of Func-tions562.1, The Limit of a Function 662.2 Theorems on Limits of Functions 742.3 One-Sided Limits 852.4 Infinite Limits 882.5 Continuity of a Function at a Number 972.6 Theorems on Continuity 1-0L3.1 The Tangent Line 1-1-i-3.2 Instantaneous Velocity in Rectilinear Motion 11,53.3 The Derivative of a Function L21-3.4 Differentiability and Continuity L263.5 Some Theorems on Differentiation of Algebraic Functions 1.303.6 The Derivative of a Composite Function L383.7 The Derivative of the Power Function for Rational Exponents L423.8 Implicit Differentiation 1-453.9 The Derivative as a Rate of Change 1.503.10 Related Rates 1543.1L Derivatives of Higher Order 157 7. -TtIViii CONTENTSChapter 4TOPICS ON LIMITS,CONTINUITY, AND THEDERIVATIVEpage 164Chapter 5ADDITIONAT APPLICATIONSOF THE DERIVATIVEpage 204Chapter 5THE DIFFERENTIAL ANDANTIDIFFERENTIATIONpage 243Chapter 7THE DEFINITE INTEGRALpage 275Chapter 8APPLICATIONS OF THEDEFINITE INTEGRALpage 3234.L Limits at Infinity 7654.2 Horizontal and Vertical Asymptotes L714.3 Additional Theorems on Limits of Functions 1.744.4 Continuity on an Interval 1774.5 Maximum and Minimum Values of a Function 19L4.6 Applications Involving an Absolute Extremum on a Closed In-tervalL894.7 Rolle's Theorem and the Mean-Value Theorem lgs5.1 Increasing and Decreasing Functions and the First-DerivativeTest 2055.2 The second-Derivative Test for Relative Extrema 2i.15.3 Additional Problems Involving Absolute Extrema 2135.4 Concavity and Points of Inflection 2205.5 Applications to Drawing a Sketch of the Graph of a Function 2275.6 An Application of the Derivative in Economics 2805.1 The Differential 2446.2 Differential Formulas 2495.3 The Inverse of Differentiation 2536.4 Differential Equations with Variables Separable6.5 Antidifferentiation and Rectilinear Motion 2656.6 Applications of Antidifferentiation in Economics2612697.L The Sigma Notation 2767.2 Area 28L7.3 The Definite Integral 2887.4 Properties of the Definite Integral 2gG7.5 The Mean-Value Theorem for Integrals 3067.6 The Fundamental Theorem of the Calculus SLi.8.1 Area of a Region in a Plane 3248.2 Volume of a Solid of Revolution: Circular-Disk and Circular-RingMethods 3308.3 Volume of a Solid of Revolution: Cylindrical-Shell Method 3J68.4 Volume of a Solid Having Known Parallel Plane Sections 3418.5 Work 3448.6 Liquid Pressure 3488.7 Center of Mass of a Rod 35L8.8 Center of Mass of a Plane Region 3568.9 Center of Mass of a Solid of Revolution 3668.10 Length of Arc of a Plane Curve 372 8. Chapter 9 9.1LOGARITHMIC AND 9.2EXPONENTIAL FUNCTIONS 9.3page 38L 9.49.59.6CONTENTSThe Natural Logarithmic Function 382The Graph of the Natural Logarithmic Function 391'The Inverse of a Function 395The Exponential Function 405Other Exponential and Logarithmic FunctionsLaws of Growth and DecaY 420The Sine and Cosine Functions 431Derivatives of the Sine and Cosine Functions 438Integrals Involving Powers of Sine and Cosine 447The Tangent, Cotangent, Secant, and Cosecant Functions 452An Application of the Tangent Function to the Slope of aLine 461,10.5 Integrals Involving the Tangent, Cotangent, Secant, and Co-secant4664L4ChaPter L0TRIGONOMETRICFUNCTIONSpage 430Chapter LLTECHNIQUES OFINTEGRATIONpage 49710.1t0.210.310.410.sChapter L2 Lz.LHYPERBOTIC FUNCTIONS I2,2page5 35 12.3Chapter 13 13.1POLAR COORDINATES T3.2page5 54 13.313.413.5Chapter 1.4THE CONIC SECTIONSpage 578InverseTrigonometricFunctions 47LDerivatives of the Inverse Trigonometric FunctionsIntegrals Yielding Inverse Trigonometric FunctionsIntroduction 492Integration by Parts 493Integration by Trigonometric Substitution 498Integration of Rational Functions by Partial Fractions.2: The Denominator Has Otly Linear Factors 504Integration of Rational Functions by Partial Fractions.4: The Denominator Contains Quadratic Factors 51.2Integration ofRational Functions of Sine and CosineMiscellaneous Substitutions 51-9The Trapezoidal Rule 521,Simpson's Rule 526The Hyperbolic Functions 536The Inverse Hyperbolic Functions 543Integrals Yielding Inverse Hyperbolic Functions 548The Polar Coordinate System 555Graphs of Equations in Polar Coordinates 560Intersection of Graphs in Polar Coordinates 567Tangent Lines of Polar Curves 571-Area of a Region in Polar CoordinatesL0.710.810.911.1LL.211.3Lt.411.5LL.6'Ll..71.L.8'n.9477484Cases L andCases 3 and516LA.l The Parabola 579L4.2 Translation of Axes 58314.3 Some Properties of Conics 588L4.4 Polar Equations of the Conics 592573 9. II {:LINi1X CONTENTSL4.5 Cartesian Equations of the Conics Sgg14.5 The Ellipse 606L4.7 The Hyperbola 61314.8 Rotation of Axes 620chapter L5 15.1 The Indeterminate Form 0/0 629INDETERMINATE FORMS, 1,5.2 Other Indeterminate Forms 636IMPROPER INTEGRALS, AND 15.3 Improper Integrals with Infinite Limits of IntegrationTAYLOR'S FORMULA lS.4 Other Improper Integrals 647Pag6e2 8 15.5 Taylor's Formula 6s7Chapter 15 1,5.1 Sequences 5G0INFINITE SERIES 1'6.2 Monotonic and Bounded Sequencse 667 page6 5e 16.g Infinite Serieso f Constant Term s 6731,6.4 Infinite Series of Positive Terms 6g41,6.5 The Integral Test 69415.6 Infinite series of Positive and Negative Terms 697L6 7 Power Series 707L6.8 Differentiation of Power Series 71316.9 Integration of Power Series 72215.10 Taylor Series 729L6.ll The Binomial Series 738641Chapter 17VECTORS IN THE PLANEAND PARAMETRICEQUATIONSpage 745Chapter 18VECTORS IN THREE-DIMENSIONALSPACE ANDSOLID ANALYTICGEOMETRYpage 81077.1, Vectors in the Plane 74677.2 Properties of Vector Addition and Scalar Multiplication 751L7.3 Dot Product 75617.4 vector-Valued Functions and Parametric Equations 76377.5 Calculus of Vector-Valued Function s 77277.6 Length of Arc 779t7.7 Plane Motion 785I7.8 The Unit Tangent and Unit Normal Vectors and Arc Length as aParameter 792L7.9 Curvature 79617.10 Tangential and Normal Components of Acceleration g0418.1 R3, The Three-Dimensional Number Space 8L1L8.2 Vectors in Three-Dimensional Space 81818.3 The Dot Product in V, 82518.4 Planes 82918.5 Lines in R3 836L8.6 Cross Product 8421,8.7 Cylinders and Surfaces of Revolution BS218.8 Quadric Surfaces 85818.9 Curves in R3 86418.10 Cylindrical and Spherical Coordinates 872 10. Chapter 19DIFFERENTIAL CALCULUSOF FUNCTIONS OFSEVERAL VARIABLESpage 880Chapter 20DIRECTIONAL DERIVATIVES,GRADIENTS, APPTICATIONSOF PARTIAL DERIVATIVES,AND LINE INTEGRALSpage 944ChaPter 2|MULTIPTE INTEGRATIONpage 1001APPENDIXpage A-1CONTENTSL9.']-. Functions of More Than One Variable 88Llg.2 Limits of Functions of More Than one Variable 889tg.3 Continuity of Functions of More Than One Variable 90019.4 Partial Derivatives 90519.5 Differentiability and the Total Differential 91'319.6 The Chain Rule 92619.7 Higher-Order Partial Derivatives 93420.L Directional Derivatives and Gradients 94520.2 Tangent Planes and Normals to Surfaces 95320.3 Extrema of Functions of Two Variables 95620.4 Some Applications of Partial Derivatives to Economics 96720.5 Obtaining a Function from Its Gradient 97520.6 Line Integrals 98120.7 Line Integrals Independent of the Path 98921.1, The Double Integral L0022'1..2 Evaluation of Double Integrals and Iterated Integrals 100821.3 Center of Mass and Moments of Inertia 1'0L621.4 The Double Integral in Polar Coordinates 1'02221.5 Area of a Surface 102821.6 The Triple Integral 103421.7 The Triple Integral in Cylindrical and Spherical Coordinates L039Table 1.Table 2Table 3Table 4Table 5Table 6Table 7Powers and Roots A-2Natural Logarithms A-3Exponential Functions A-5Hyperbolic Functions A-L2Trigonometric Functions A-13Common Logarithms A-1-4The Greek Alphabet A-15ANSWERS TO ODD-NUMBERED EXERCISES A.1.7INDEX 4.45 11. ACKNOWLEDGMENTS Revieweorsf rheC atcutuProfessor William D.Professor Archie D. lProfessor Phillip ClaProfessor Reuben WProfessor Jacob GolilProfessor Robert K. Goodrich, University of ColoradoProfessor Albert Herr, Drexel UniversityProfessor James F. Hurley, University of ConnecticutProfessor Gordon L. Miller, Wisconsin State UniversitvProfessor William W. Mitchell, Jr., phoenix CollegeProfessor Roger B. Nelsen, Lewis and Clark Colle-geProfessor Robert A. Nowlan, southern connectictit state collegeSister Madeleine Rose, Holy Names CollegeProfessor George W. Schultz, St. petersbuig Junior CollegeProfessor Donald R. Sherbert, Universitv of IllinoisProfessor] ohn V_adneyF, ulton-Montgomery community collegeProfessor David Whitman, San Diego State CollegeProduction Staff at Harper & RowGeorge Telecki, Mathematics EditorKaren fudd, Project EditorRita Naughton, DesignerAssistants for Answers to ExercisesJacqueline Dewar, Loyola Marymount UniversityKen Kast, Logicon, Inc.|ean Kilmer, West Covina Unified School DistrictCover and Chapter Openins ArtistPatrick Cadlfield, Londin, EnglandTo these _people and to all the users of the first and second editions who have suggestedchanges, I express my deep appreciation.L. L. 12. r l ] FTETAGEThis third edition of THE CALCULUS WITH ANALYTIC GEOMETRY,like the other two, is designed for prospective mathematics majorsas well as for students whose primary interest is in engineering, thephysical sciences, or nontechnical fields. A knowledge of high-schoolalgebra and geometry is assumed.The text is available either in one volume or in two parts: Part I con-sistsof the-first sixteen chapters, and Part II comprises Chapters 16through 21 (Chapter L6 on Infinite Series is included in both parts to makethe use of the two-volume set more flexible). The material in Part I con-sistsof the differential and integral calculus of functions of a singlevariable and plane analytic geometrlz, and it may be covered in a one-yearcourse of nine or ten semester hours or twelve quarter hours. The secondpart is suitable for a course consisting of five or six semester hours oreight quarter hours. It includes the calculus of several variables and atreatment of vectors in the plane, as well as in three dimensions, with avector approach to solid analytic geometry.The objectives of the previous editions have been maintained. I haveendeavored to achieve a healthy balance between the presentation ofelementary calculus from a rigorous approach and that from the older,intuitive, and computational point of view. Bearing in mind that a text-bookshould be written for the student, I have attempted to keep the pre-sentationgeared to a beginner's experience and maturity and to leave nostep unexplained or omitted. I desire that the reader be aware that proofsof theorems are necessary and that these proofs be well motivated andcarefully explained so that they are understandable to the student who hasachieved an average mastery of the preceding sections of the book. If atheorem is stated without proof ,I have generally augmented the discus-sionby both figures and examples, and in such cases I have alwaysstressed that what is presented is an illustration of the content of thetheorem and is not a proof.Changes in the third edition occur in the first five chapters. The first 13. xlv PREFACEsection of Chapter I has been rewritten to give a more detailed expositionof the real-number system. The introduction to analytic geometry in thischapter includes the traditional material on straight lines as well as thatof the circle, but a discussion of the parabola is postponed to Chapter 14,The Conic Sections. Functions are now introduced in Chapter 1. I havedefined a function as a set of ordered pairs and have used this idea topoint up the concept of a function as a correspondence between sets ofreal numbers.The treatment of limits and continuity which formerly consisted often sections in Chapter 2 is now in fwo chapters (2 and 4), with the chap-teron the derivative placed between them. The concepts of limit and con-tinuityare at the heart of any first course in the calculus. The notion of alimit of a function is first given a step-by-step motivation, which bringsthe discussion from computing the value of a function near a number,through an intuitive treatment of the limiting process, up to a rigorousepsilon-delta definition. A sequence of examples progressively graded indifficulty is included. All the limit theorems are stated, and some proofsare presented in the text, while other proofs have been outlined in theexercises. In the discussion of continuity, I have used as examples andcounterexamples "common, everyday" functions and have avoided thosethat would have little intuitive meaning.In Chapter 3, before giving the formal definition of a derivative, Ihave defined the tangent line to a curve and instantaneous velocity inrectilinear motion in order to demonstrate in advance that the concept ofa derivative is of wide application, both geometrical and physical. The-oremson differentiation are proved and illustrated by examples. Ap-plicationof the derivative to related rates is included.Additional topics on limits and continuity are given in Chapter 4.Continuity on a closed interwal is defined and discussed, followed bythe introduction of the Extreme-Value Theorem, which involves suchfunctions. Then the Extreme-Value Theorem is used to find the absoluteextrema of functions continuous on a closed interval. Chapter 4 concludeswith Rolle's Theorem and the Mean-Value Theorem. Chapter 5 givesadditional applications of the derivative, including problems on curvesketching as well as some related to business and economics.The antiderivative is treated in Chapter 6. I use the term "antidif-ferentiation"instead of indefinite integration, but the standard notation! f (x) dx is retained so that you are not given a bizarre new notation thatwould make the reading of standard references difficult. This notation willsuggest that some relation must exist between definite integrals, intro-ducedin Chapter 7, and antiderivatives, but I see no harm in this as longas the presentation gives the theoretically proper view of the definiteintegral as the limit of sums. Exercises involving the evaluation of defi-niteintegrals by finding limits of sums are given in Chapter 7 to stressthat this is how they are calculated. The introduction of the definite inte- 14. PREFACEgral follows the definition of the measure of the area under a curye as alimit of sums. Elementary properties of the definite integral are derivedand the fundamental theorem of the calculus is proved. It is emphasizedthat this is a theorem, and an important one, because it provides us withan alternative to computing limits of sums. It is also emphasized that thedefinite integral is in no sense some special type of antiderivative. InChapter 8 I have given numerous applications of definite integrals. Thepresentation highlights not only the manipulative techniques but alsothe fundamental principles involved. In each application, the definitionsof the new terms are intuitively motivated and explained.The treatment of logarithmic and exponential functions in Chapter 9is the modern approach. The natural logarithm is defined as an integral,and after the discussion of the inverse of a function, the exponentialfunction is defined as the inverse of the natural logarithmic function. Anirrational power of a real number is then defined. The trigonometricfunctions are defined in Chapter 10 as functions assigning numbers tonumbers. The important trigonometric identities are derived and usedto obtain the formulas for the derivatives and integrals of these functions.Following are sections on the differentiation and integration of the trig-onometricfunctions as well as of the inverse trigonometric functions.Chapter LL, on techniques of integration, involves one of the mostimportant computational aspects of the calculus. I have explained thetheoretical backgrounds of each different method after an introductorymotivation. The mastery of integration techniques depends upon theexamples, and I have used as illustrations problems that the student willcertainly meet in practice, those which require patience and persistenceto solve. The material on the approximation of definite integrals includesthe statement of theorems for computing the bounds of the error involvedin these approximations. The theorems and the problems that go withthem, being self-contained, can be omitted from a course if the instructorso wishes.A self-contained treatment of hyperbolic functions is in Chapter 12.This chapter may be studied immediately following the discussion of thecircular trigonometric functions in Chapter L0, if so desired. The geo-metricinterpretation of the hyperbolic functions is postponed untilChapter L7 because it involves the use of parametric equations.Polar coordinates and some of their applications are given in Chap-ter13. In Chapter 1.4, conics are treated as a unified subject to stress theirnatural and close relationship to each other. The parabola is discussed inthe first two sections. Then equations of the conics in polar coordinatesare treated, and the cartesian equations of the ellipse and the hyperbolaare derived from the polar equations. The topics of indeterminate forms,improper integrals, and Taylor's formula, and the computational tech-niquesinvolved, are presented in Chapter L5.I have attempted in Chapter 16 to give as complete a treatment of 15. xvr PREFACEinfinite series as is feasible in an elementary calculus text. In addition tothe customary computational material, I have included the proof of theequivalence of convergence and boundedness of monotonic sequencesbased on the completeness property of the real numbers and the proofsof the computational processes involving differentiation and integrationof power series.The first five sections of Chapter L7 on vectors in the plane can betaken up after Chapter 5 if it is desired to introduce vectors earlier in thecourse. The approach to vectors is modern, and it serves both as an intro-ductionto the viewpoint of linear algebra and to that of classical vectoranalysis. The applications are to physics and geometry. Chapter 18 treatsvectors in three-dimensional space, and, if desired, the topics in the firstthree sections of this chapter may be studied concurrently with the corre-spondingtopics in Chapter 17.Limits, continuity, and differentiation of functions of several variablesare considered in Chapter L9. The discussion and examples are appliedmainly to functions of two and three variables; however, statements ofmost of the definitions and theorems are extended to functions of nvariables.In Chaptet 20, a section on directional derivatives and gradients isfollowed by a section that shows the application of the gradient to findingan equation of the tangent plane to a surface. Applications of partialderivatives to the solution of extrema problems and an introduction toLagrange multipliers are presented, as well as a section on applications ofpartial derivatives in economics. Three sections, new in the third edition,are devoted to line integrals and related topics. The double integral of afunction of two variables and the triple integral of a function of threevariables, along with some applications to physics, engineering, andgeometry/ are given in Chapter 2/...New to this edition is a short table of integrals appearing on the frontand back endpapers. However, as stated in Chapter L1, you are advisedto use a table of integrals only after you have mastered integration.Louis Leithold 16. Real numberq, introductionto analytic geometryand functions 17. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA. ND FUNCTIONS1.1 SETS, REAL NUMBERS, The idea of "set" is used extensively in mathematics and is such a ba-AND INEQUALITIES sic concept that it is not given a formal definition. We can say that aset is a collection of objects, and the objects in a set are called the elementsof a set. We may speak of the set of books in the New York Public Library,the set of citizens of the United States, the set of trees in Golden GatePark, and so on. In calculus, we are concerned with the set of real numbers.Before discussing this set, we introduce some notation and definitions.We want every set to be weII defined; that is, there should be somerule or property that enables one to decide whether a given object is oris not an element of a specific set. A pair of braces { } used with wordsor symbols can describe a set.If S is the set of natural numbers less than 6,we can write the set S as{ L , 2 , 3 , 4 , 5 }We can also write the set S as{r, such that r is a natural number less than 6}where the symbol. "x" is called a"variable." A aariable is a symbol usedto represent any element of a given set. Another way of writing the aboveset S is to use what is called set-builder notation, where a vertical bar isused in place of the words "such that." Using set-builder notation todescribe the set S, we have{rlr is a natural number less than 6}which is read "the set of all r such that r is a natural number less than G."The set of natural numbers will be denoted by N. Therefore, we maywrite the set N as{ 7 , 2 , 3 , I. lwhere the three dots are used to indicate that the list goes on and on withno last number. With set-builder notation the set N may be written as{rlr is a natural number}.The symbol " e " is used to indicate that a specific element belongsto a set. Hence, we may write 8 N, which is read "8 is an element of N.,,The notation a,b e S indicates that both a andb are elements of S. Thesymbol fi is read "is not an element of." Thus, we read * w as "+ is notan element of N."We denote the set of all integers by /. Because every element of N isalso an element of / (that is, every natural number is an integer), we saythat N is a "subset" of /, written N E /.:',.tDefinition The set S is a subset of the set T, written S e T,if and only if every elementof S is also an element of T.If , in addition, there is at least one element of T-J 18. 1.1 SETS,R EALN UMBERSA, ND INEQUALITIESwhich is not an element of S, then S is a proper subset of T, and it is writ-tenSC T.Observe from the definition that every set is a subset of itself, but a setis not a proper subset of itself.In Definition 1.1.1, the "if and only if" qualification is used to com-binetwo statements: (i) "the set S is a subset of the set T if every elementof S is also an element of T"; and (ii) "the set S is a subset of set T only ifevery element of S is also an element of 7," which is logically equivalentto the statement"if. S is a subset of.T, then every element of S is also anelement of. T."o rLLUsrRArroN 1: Let N be the set of natural numbers and let M be theset denoted by {rlr is a natural number less than 1,0}. Because everyelement of M is also an element of N, M is a subset of N and we writeM e N. Also, there is at least one element of N which is not an elementof M, and so M is a proper subset of N and we may write M C N. Further-more,because {5} is the set consisting of the number 5, {6} C M, whichstates that the set consisting of the single element 5 is a Proper subset ofthe set M. We may also write 5 M, which states that the number 6 isan element of the set M. oConsider the set {xlzx * L:0, and x e I}. This set contains no ele-mentsbecause there is no integer solution of the equation 2x * 1. : 0.Such a set is called the "empty set" or the "null set."1.t.2 Definition The empty set (or null sef) is the set that contains no elements. The emPtyset is denoted by the symbol A.The concept of "subset" may be used to define what is meant by twosets being "equal."1.1.3 Definition Two setsA and B are said to be equal,w ritten A: B, if and only if A e BandB e A.Essentially, this definition states that the two sets A and B are equalif and only if every element of A is an element of B and every element of Bis an element of A, that is, if the sets A and B have identical elements.There are two operations on sets that we shall find useful as weproceed. These operations are given in Definitions 1.1.4 and 1.1.5.1.1.4 Definition Let A and B be two sets. The union of A and B, denoted by A U B and read"A union 8," is the set of all elements that are in A or in B or in both Aand B. 19. REAL NUMBERS,I NTRODUCTIONT O ANALYTICG EOMETRYA, ND FUNCTIONSExAMPLE1 : Let A: {2, 4, 6,8,10, LzI , B : {1, 4, 9, '1.5}, andC - {2,10} .F ind(a)AuB(c)BUCSOLUTION:(a)Au(b)Au(c)BU(d)Au1..1..D5 efinition Let A and Band tead " Aand B.B - {1,2,4 , 6,9,9, "1.01,, 21, ,6}C: {2, 4, 6,8, 1,01, 2}C : {1,2, 4,9, 1 , 0 , 1 , 6 }A: {2, 4, 6, 8, I0,12} : 4be two sets.T he intersectiono f A and B, denoted by A n Bintersection 8," is the set of all elements that are in both A(b)Auc(d)AuAnxelvrpr.E2: If A, B, and C arethe sets defined in Example 1,find@)AnB(c)BnCSOLUTION:(a) A n g : { 4 }(c) B f i C:AC: {2, 1.0}A: { 2 ,4 , 6 , 8 , t 0 , 7 2 1: 4(b)An(d)An(b)Anc@)AnA1.1.6 Axiom(Closure and Uniqueness Laws)1.1.7 Axiom(Commutatiae Laws)L.L.8 Axiom(Associatiae Laws)1."1,.A9 xiom(DistributiaeL aw)The real number system consists of a set of elements called real numbersand two operations called addition and multiplication The set of real num-bersis denoted by Rt. The operation of addition is denoted by the symbol"*", and the operation of multiplication is denoted by the symbol',.',.rf a, b c Rt, a * b denotes the sum of a and b, and a - b (or ab) denotestheir product.We now present seven axioms that give laws governing the operationsof addition and multiplication on the set Rl. The word axiom is used toindicate a formal statement that is assumed to be true without proof.rf a,b e Rt, then sib is a unique real number, andab is a unique realnumber.If a, b e Rr, thena*b:b*a a n d ab:baIf. a, b, c Rl, thena + (b* c) : (a + b ) + c and a(bc) : (ab)clf a,b, c C R1, thena ( b + c ) : a b * a c1..L.10 Axiom There exist two distinct real numbers 0 and 1 such that for anv real num- J (Existencoef IdentityE lements) ber A,a i l : a and A . t : a 20. 1.1 SETS,R EAL NI.JMBERSA,N D INEOUALTTIESL.1.11 Axiom For every real number a, there exists(Existencoef N egatiaoer Additiaeln aerse) of a (or additiae inaerse of a), denotedsuch thata * ( - s ) : Q'1,.1.12A xiom(Existenceo f Reciprocaol rMultiplica tiae lnue rse)1..1..1D3e finition'i,.1.14 Definitiona real number called the negatiaeby -a (read "the negative of a"),For every real number a, except 0, there exists a real number called thereciprocat of a (or multiplicatiae inaerse of a), denoted by a-t, such thata ' a - r , : tAxioms 1,.1,.6th rough 1,.I.L2 are called field axioms because if theseaxioms are satisfied by a set of elements, then the set is called afield underthe two operations involved. Hence, the set R1 is a field under additionand multiplication. For the set / of integers, each of the axioms t.1.6through 1.1.11 is satisfied, but Axiom L.1,.1,i2s not satisfied (for instance,the integer 2 has no multiplicative inverse it /). Therefore, the set ofintegers is not a field under addition and multiplication.lf a, b Rl, the operation of subtraction assigns to a andb areal number,denoted by a - b (read "a minus b"), called the difference of a and b, wherea - b : A * ( - b ) ( 1 )Equality (1) is read "A minus b equals a plus the negatle of- b."I f . a , b G Rl , a n d b * 0 , t h e o p e r a t i o no f d i a i s i o na s s i g n s t o a a n d b a r c a lnumber, denoted by o + b (read "a divided by b"), called the quotient ofa and b, wherea : b : A ' b - rOther notations for the quotient of a and b ate! and albbBy using the field axioms and Definitions 1.1.13 and 1.1.14, we canderive properties of the real numbers from which follow the familiaralgebraic operations as well as the techniques of solving equations,factoring, and so forth. In this book we are not concerned with showinghow such properties are derived from the axioms.Properties that can be shown to be logical consequences of axioms aretheorems. In the statement of most theorems there are two parts: the "if"part, called the hypothesis, and the "then" part, called the conclusion. Theargument verifying a theorem is a proof. A proof consists of showingthat the conclusion follows from the assumed truth of the hypothesis. 21. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONSL.1..L5 Axiom(OrdeAr xiom)The concept of a real number being "positive" is given in the fol-lowingaxiom.In the set of real numbers there exists a subset called the positiae numberssuch that(i) if a Rr, exactly one of the following three statements holds:a: 0 a is positive -a is positive.(ii) the sum of two positive numbers is positive.(iii) the product of two positive numbers is positive.Axiom 1.1.15 is called the order axiom because it enables us to orderthe elements of the set Rr. In Definitions 'J,.7.77and 1.1.18 we use thisaxiom to define the relations of "greater than" and "less than" on Rr.The negatives of the elements of the set of positive numbers form theset of "negative" numbers, as given in the following definition.The real number a is negatiae iI and only if -a is positive.From Axiom 1.1.15a nd Definition 7.1..76it follows that a real numberis either a positive number, a negative number, or zero. Ary real numbercan be classified as a rqtional number or an irrational number. A rationalnumber is any number that can be expressed as the ratio of two integers.That is, a rational number is a number of the form plq, where p a:nd, qare integers and q + 0.The rational numbers consist of the following:The integers (positive, negative, and zero). , -5, -4, -3, -2, -L, 0, 1, 2, 3, 4, 5, .The positive and negative fractions such as+-#+The positive and negative terminating decimar.s uch as1*1.16 Definition2s6:ffi -0.0032:513,251L,000,000The positive and negative nonterminatingre peatingd ecimalss uch as0.333. :+ -0549s49s49...:-ftThe real numbers which are not rational numbers are called irrationalnumbers. These are positive and negative nonterminating, nonrepeatingdecimals, for example,rt: t.732. n : 3.L4'1,59. tan 140o: -0.839L.The field axioms do not imply any ordering of the real numbers. Thatis, by means of the field axioms alone we cannot state that 1 is less than 22. 1.1 SETS,R EALN UMBERSA, ND INEQUALITIES2,2isless than 3, and so on. However, we have introduced the order axiom(Axiom 1.1.15), and because the set Rr of real numbers satisfies the orderaxiom and the field axioms, we say that Rl is an ordered field.We use the concept of a positive number given in the order axiomto define what we mean by one real number being "less than" another.11*7 Definition I f a,b Rt, then a is less than b (written ab) i f andonlyif b islessthan a; with symbols we writea l b i f a n d o n l y i f b < al.l.l9 Definition The symbols = (" is less than or equal to") and = (" is greater than orequal to") are defined as follows:(1) a - b i f and only i f either a < b or a:b.(ii a > b if. and only if either a > b ot a: b-The statements a 1b, a) b, a 8,2> -4, -3> - 6 ,5 = 3, -10 < -7.In particular, a 1b and a > b are called strict inequal-ities,whereas a -b anda= b are called nonstrict inequalities..The following theorems can be proved by using L.L.6 through 1.1'.19."1..1.20T heorem (i) a > 0 if and only if a is positivq.(ii) a < 0 if and only if a is negative.(1ii) a > 0 if and only if -a < 0.(iv) a < 0 if and only if -a > 0.1 . 7 . 2 1T h e o r em I f . a < b a n d b 1 c , t h e n a 1 c .o rLLUsrRArroN 3: 3 < 5 and 6 2 a n d 3 > - 5 ; s o 7 * 3 > 2 + ( - 5 ) . oIf a > b and if c is any positive number, then ac ) bc.o rLLUsrRArroN 12: -3 > -7; so (-3) > eDA.L.1.30 Theorem 24. n1.1.31 Theorem1.1.32 Theorem7 91.L.33 Definition1.1 SETS,R EALN UMBERSA, ND INEQUALITIESlf. a > b and if c is any negative number, then ac I bc.o rLLUsrRArroN 13: -3 > -7; so (-3)(-4) < (-7)(-4). oIf a > b > }and c > d > 0, then ac > bd.o r L LUSrRAr r o1N4 : 4 > 3 > 0 a n d 7 > 5 ) 0 ; s o 4 ( 7 ) > 3 ( 6 ) . oSo far we have required the set Rl of real numbers to satisfy the fieldaxioms and the order axiom, and we have stated that because of this re-quirementRr is an ordered field. There is one more condition that is im-posedupon the set R1. This condition is called the axiom of completeness(Axiom 16.2.5). We defer the statement of this axiom until Section 16.2because it requires some terminology that is best introduced and dis-cussedlater. However, we now give a geometric interpretation to the setof real numbers by associating them with the points on a horizontal line,called an axis. The axiom of completeness guarantees that there is a one-to-one correspondence between the set Rl and the set of points on an axis.Refer to Figure 1.1.1. A point on the axis is chosen to represent thenumber 0. This point is called the origin A unit of distance is selected.Then each positive number x is represented by the point at a distance of runits to the right of the origin, and each negative number x is representedby the point at a distance of -x units to the left of the origin (it should benoted that if r is negative, then -r is positive). To each real number therecorresponds a unique point on the axis, and with each point on the axisthere is associated only one real number; hence, we have a one-to-onecorrespondence between Rl and the points on the axis. So the points onthe axis are identified with the numbers they represent, and we shalluse the same symbol for both the number and the point representingthat number on the axis. We identify Rl with the axis, and we call Rl thereal number line.We see that a < b if and only if the point representing the number ais to the left of the point representing the number b. Similarly, a > b ifand only if the point representing a is to the right of the point repre-sentingb. For instance, the number 2 is less than the number 5 and thepoint 2 is to the left of the point 5. We could also write 5 ) 2 and saythat the point 5 is to the right of the point 2.A numberx i s between a andbif a ( rand x a}(ii) (-.o, b): {xlx < b}(iii) la, +*7 : {xlx > a}(iv) (-m, bl: {xlx - b}(v) (-*, **) : RlaFigure1 .1.6-bFigure1 .1.7b" 1.1.35 Definition'1,.1.3D6 efinition-a bFigure1 .1.4f---La bFigure1 .1.5Figure 1,.1,.6il lustrates the interval (a, *a), and Fig. r.7.T illustratesthe interval (-*, b). Note that (-m,1oo) denotes the set of all real numbers.For each of the intervals (a,b), [a, b], lq, b), and (a, b] the numbers a 26. EXAMPL3E: Find the solutionset of the inequality2*3x 0; that is, x ) 3.Multiplying on both sides of the inequality by x- 3, we getx14x-12Adding -4x to both members, we obtain-3x < -12Dividing on both sides by-3 and reversing the direction of the inequality,we havex ) 4Thus the solution set of Case 1 is {rlx > 3} n {xlr > 4} or, equivalently,{xlx > 4}, which is the interval (4, +oo;.Case2: x-3 < 0 ; that is,x -12or, equivalently,x14Therefore, .x must be less than 4 and also less than 3. Thus, the solutionset of Case 2 is the interval (-*, 3).If the solution sets for Cases 1 and 2 are combined, we obtain(--, 3) U (4, **), which is illustrated in Fig. 1.1.11.solurroN: The inequality will be satisfied when both factors have thesames ign, that iisf, x t 3 > 0andx + 4> 0,ori f x t 3 < 0andr + 4 < 0.Let us consider the two cases.Casel-: xt3 > 0 and x t 4 ) 0. T h a t i s ,x>-3 and x ) - 4Thus, both inequalities hold if x > -3, which is the interval (-3, +*;.Case2: x-13 < 0 and x * 4 < 0 . T h a t i s ,x 1-3 and x 1-4Both inequalities hold if x < -4, which is the interval (-*,-4).If we combine the solution sets for Cases 1 and 2,we have (-*,-4) U(-3, +-;. 29. 14 REAL NUMBERS,I NTRODUCTIONT O ANALYTICG EOMETRYA, ND FUNCTIONSExercises7 .LIn Exercises1 through 10,l ist the elementso f the given set if A: {0,2,4,6,8}, B :{0,3, 6 , 9 } .I,.AUB 2.CUD 3 .4.CnD 5.BUD 6 .7.BnD 8.AnC g .10.(AuB)n(cuD){1, z , 4, B}, C : {1, i,s,7,9}, and D :(Bnc)l1"t*T"-:11ANBAUC@n D) u-ilP';3,t"i ,T:tyo.n set of the siven inequatity and illustrate the solution on the real number1 . ' 1 . . 5 x + 2 > x - 6? , l - x 1 4 . 3 x - 5- < i x +4 ' - ' 317.2 > - 3 - 3 x > - 7r )20r i- = 1. - r L - x2 3 . ( x - 3 ) ( r + s ) > 02 6 . * * 3 r * 1 > 012. 3- r 2x-3>5,f8I. .9i 1 42 1 ,*.> 424. x2 -3r* 2> 02 7 . 4 f I 9 x < 9?n x * ' j , - x" - ' 2-r29. -31,.3 * x32.35.36.37.38.39.?x-t Z-7"x x22.f 02 f - 6 x * 3 < 0743x-7-3 - 2 xProve Theorem 1,.1,.22.33. Prove Theorem 1,.1.27. 94.Prove Theorems 1.1,.24a nd 1.1.25.lf n > b = O,prove that a2 > W.lf. a and b are nonnegative numbers, and az ) br, prove that a > b.Prove that if a > 0 and b > 0, then A2 : b2 if and only if a: b.Pncve that if b > a > 0 and c ) 0, thena* c ab + t ' b1.2 ABSOLUTE VALUE40. Prove that if x 1 !, then r < t(x + y) < y.The concept of the "absolute value" ofa number is used in some importantdefinitions in the study of calculus. Furtherrnore, you will need. tb workwith inequalities involving "absolute value."The absolute aalue of x, denoted by lrl, is defined byl * l - * * i f x > 0l t l : - t i f x < ol0l :01.2|t Definition 30. 1.2A BSOLUTVEA LUEThus, the absolute value of a positive number or zero is equal tothe number itself. The absolute value of a nggative numloer is the cor-respondingpositive number because the negative of a negative numberis positive.. ILLUsTRATI1o:N l3 l :3; l -51: - ( -5) :5; 18- L4l : l -61: - ( -6) :6.We see from the definition that the absolute value of a number iseither a positive number or zero; that is, it is nonnegative.In terms of geometry, the absolute value of a number x is its distancefrom 0, without regard to direction. In general, lo - bl is the distancebetween a and b without regard to direction, that is, without regard towhich is the larger number. Refer to Fig. 1..2.1'.We have the following properties of absolute values.l"l < a lf. and only if -a 1x I a,where a > 0.Irl = a if. and only if -a < )c < a, where a > 0.ltl > a if andonly if x > a or x I -a,where a > 0.ltl = a if andonly if x > a or x s -a,where a ) 0.The proof of a theorem that has ant "if and only if " qualification re-quirestwo parts, as illustrated in the following proof of Theorem 1.2.2.plnr l.: Prove that lrl < a if.-a < x I a,where a ) O.Here,we havetoconsidert wo cases:. x> 0 and r < 0.Casel: x=0.Then ltl : x. Becauser < a,we concludet hat lrl < a.Case2: x 0. Here wemust show that whenever the inequality ltl < a holds, the inequality-a < x 1a also holds. Assume that lrl < a and consider the two casesx > 0 a n d r ( 0 .CaseT: x > 0 .Then ltl : r. Because lxl < a,we conclude that r < a. Also, becausea ) 0, it follows from Theorem 1.1.31 that-a < 0. Thus/ we have -a 10< x < a , o t - a < x 1 a .Case2: x 0i f r * 3 < 0Figure1 .3.5soLUTroN: Fromhavey:x*3 i fandy--(x+3)or, equivalently,y:x*3 i f x > - 3andy - - ( x + 3 ) i f x 1 - 3Table 1.3.3 gives some values of x and y satisfying Eq. (Z).Table1 .3.3x 0 L 2 3 -L -2 -3 -4 -5 -6 -7 -8 -9v 3 4 5 6 2 1 0 1, 2 3 4 5 6A sketch of the graph of Eq. (7) is shown in Fig. 1,.9.6.soLUTroN: By the property of real numbers that ab:O if anda: 0 or b :0, we have from Eq. (8)x - 2 y * 3 : 0 40. 1.3 THE NUMBER PLANE AND GRAPHS OF EQUATIONS(10)y-xz:oThe coordinates of all points that satisfy Eq. (8) will satisfy eitherEq. (9) or Eq. (L0), and the coordinates of any point that satisfies eitherEq. (9) or (10) will satisfy Eq. (8). Therefore, the graph of Eq. (8) will con-sistof the graphs of Eqs. (9) and (10).T able 1..3.4g ives some valueso f rand y satisfying Eq. (9), and Table 1.3.5 gives some values of x and ysatisfying Eq. (10). A sketch of the graph of Eq. (8) is shown in Fig. 7.3.7.TableL .3.4x 0L23-1-2-3-4-5v 821r31,+o-+-1Table1 .3.5x 01,23-7-2-3v 0L491491.3.3 Definition An equation of a graph is an equation which is satisfied by the coordinatesof those, and only those, points on the graph.o ILLUSTRATIoN 1: In R2, y:8 is an equation whose graph consists ofthose points having an ordinate of 8. This is a line which is parallel to thex axis, and 8 units above the r axis. oIn drawing a sketch of the graph of an equation, it is often helpfulto consider properties of symmetry of a graph.1.3.4 Definition Two points P and Q are said to be symmetric with respect to a line if andonly if the line is the perpendicular bisector of the line segment PQ. Twopoints P and Q are said to be symmetric with respect to a thirdpoint if andonly if the third point is the midpoint of the line segment PQ.(3'2oo rLLUsrRArroN 2: The points (3,2) and (3,-2) are symmetric withrespect to the x axis, the points (3, 2) and (-3, 2) are symmetric withrespect to the y axis, and the points (3,2) and (-3, -2) are symmetricwithrespect to the origin (see Fig. 1.3.8). oIn general, the points (x, y) and (x, -y) are symmetric with respectto the x axis, the points (r, y) and (-x, y) are symmetric with respectto the y axLSa, nd the points (x,y) and (-r, -y) arc symmetric with respectto the origin.Figure1 .3.7Figure 1.3.8 41. REALN UMBERSI,N TRODUCTION1.3.5 DefinitionL.3.6 Theorem(Tests t'or Symmetry)rxlrvrpr.r 5: Draw a sketch of thegraph of the equationTO ANALYTICG EOMETRYA, ND FUNCTIONSxy:1 ( 1 1 )are symmetric with respect to R.From Definition 1.3.5 it follows that if a point (x, y) is on a graphwhich is symmetric with respect to the x axis, then the point (x,-y) alsomust be on the graph. And, if both the points (x,y) and (x,-y) are on thegraph, then the graph is symmetric with respect to the x axis. Therefore,the coordinates of the point (x,-y) as well as (x, y) must satisfy an equa-tionof the graph. Hence, we may conclude that the graph of an equationin r and y is symmetric with respect to the r axis if and only if an equiv-alentequation is obtained when y is replaced by -y in the equation. Wehave thus proved part (i) in the following theorem. The proofs of parts(ii) and (iii) are similar.The graph of an equation in r and y is(i) symmetric with respect to the x axis if and only if an equivalentequation is obtained when y is replaced by -y in the equation;(ii) symmetric with respect to the y axis if and only if an equivalentequation is obtained when x is replaced by -x in the equation;(iii) symmetric with respect to the origin if and only if an equivalentequation is obtained when r is replaced by -x and y is replacedby -y in the equation.The graph in Fig. 1.3.2 is symmetric with respect to the y axis, andfor Eq. (1) an equivalent equation is obtained when x is replaced by -r.In Example 1 we have Eq. (2) for which an equivalent equation is ob-tainedwhen y is replaced by-y, and its graph sketched inFig. L.3.3 issymmetric with respect to the r axis. The following example gives a graphwhich is symmetric with respect to the origin.solurroN: We see that if in Eq. (1,1)x is replaced by -x and y is replacedby -y,an equivalent equation is obtained; hence,by Theorem 1.3.6(iii)the graph is symmetric with respect to the origin. Table L.3.6 gives somevalues of x and y satisfying Eq. (11).TableL .3.6 42. 1.3 THE NUMBER PLANE AND GRAPHS OF EQUATIONSFrom Eq. (11)w e obtain y : 'l.lx.W e seet hat as r increasest hroughpositive values, y decreasesth rough positive values and gets closer andclosert o zero.A s x decreasesth rough positive values,y increasesth roughpositive values and gets larger and larger. As I increases through nega-tivevalues (i.e., r takes on the values -4, -3, -2, -t, -+, etc.), y takeson negative values having larger and larger absolute values. A sketch ofthe graph is shown in Fig. 1..3.9.Figure 1.3.9Exercise1s.3In Exercises 1 through 6, plot the dven Point P and such of the following PoinB as may aPPly:(a) The point O suitr ttrit ttre tnJ through Q and P is perpendicular to the r axis and is bisected by it. Give the coordi-natesof Q.(b) The poiniR such that the line through P and R is perpmdicular to and i6 bisected by the y axis. Give the coordinatesof R.(c) The point s such that the line through P and 5 iE bisected by the origrn. Give the_coordin"F-"-9f s.-1ai fn" poi"t T such that the line *Eough P and 7 is perpendicular to and is bisected by the als' line though the originbisecting the first and thtd quadrants. Give the coordinates of T.1 . P( 1, - 2 )4. P(-2, -2)In Exercises 7 through2S, draw a sketch of the graph of the equation.7'Y:2x*5l0.y:-nffi13. x: -J1 5 . y : - l x + 2 119. 4 f *9Y2:362. P(-2,2)5. P(-1 ,-3)8 ' Y : 4 x - 31 1 . y ' : x - 3L4. x: y2 * 'I..17. y: l r l - 520. 4x2-9 Y' :363. P(2,2)5. P(0, -3)Y: 81y:5y:lx-51y - - l x l + zA:4x33x2 - L3xy - L0y2: gxa - 5x2y I 4y2: g(c) y' :2x(c y' : -2N22. Y' : 41cs 23. 4x2 - Az :02 5 . x z * y ' , : O 26. ( 2 x + V - l ) ( 4 V + r 2 ) : 02 8 .( y 2- x * z ) (V + tE- + ) :O29. Draw a sketch of the graph of each of the following equations:(a) V: lzx (b v : -{2x30. Draw a sketch of the graph of each of the following equations:(a) Y: !-2x (b y : -!-2x 43. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONS3L. Draw a sketch of the graph of each of the following equations:(a) r * 3 y : 0 (b) r - 3 y : 0 (c) r' - 9y' :032. (a) Write an equation whose gtaPh is the r axis. (b) Write an equation whose graph is the y axis. (c) Write an equafionwhose graph is the set of all points on either the .x axis or the y axis.33. (a) Write a,n__equ{iol whose graph consists of all points having an abscissa of 4. O) Write an equation whose graphconsistE of all points having an ordinate of -3.34. Prove that a graPh that i5 symmetric with respect to both coordinate axes is also symmetric with rcspect to the origin.35' Pro-ee that a graPh that is symmeFic with rcspect to any two perpendicular lines is also sym.rretric with respect to theirpoint of intersection-1.4 DISTANCE FORMULAAND MIDPOINT FORMULAD ( - 2 , 4 )C(-2, -g)CD:7(b)vCD: - 6(a)If A is the point (x,.,y) and B is the point (xr,y) (i.e., A and B havethesame ordinate but different abscissas), then the directed distance fromA to B, denoted by AB, is defined ds x2 - x1.o ILLUSTRATTo1N: Refer to Fig. 1.4.1(a),( b), and (c). uA B : ' / . , 4(b)Figur1e. 4.1If .4 is the point (3, 4) and B is the point (9, 4), then AB :9 - 3: 6.rf A is the point (-8, 0) and B is the point (6,0), then B :6 - (-s) : 14.If A is the point (4, 2) and B is the point (L,2), then AB : 1. - 4: -3. wesee that AB is positive if B is to the right of A, and AB is negative if B isto the left of A. oIf C is the point (x,yr) and D is the point (x, yr), then the directeddistance from C to D, denoted by CD, is defined as!z- !r.o rLLUsrRArroN 2: Refer to Fig. La.2@) and (b).If C is the point (L, -2) and D is the point (1, -8), then CD : -g-J-2)--6. If C is the point (-2, -3) and D is the point (-2,4), thenCD:4- (-3) :7. Tlr.e number Cp is positiveif D is above C, and CDis negative if D is below C. oWe consider a directed distance AB as the signed distance traveledby u particle that starts at A(rr, y) and travels to B(x2, !). ln such a case,the abscissa of the particle changes from 11 to x2, dfld we use the notationAr ("delta x") to denote this change; that is,A,x: xz- xtTherefore, AB: Lx.B(r, z) A(4,2)Figure 1.4.2 44. Pr(x", vz)Figure 1.4.3EXAMPLEl .: If a point P(x, U) issuch that its distance fromA(3,2) is always twice its dis-tancefrom B(-4, L), find anequation which the coordinatesof P must satisfy.1.4 DISTANCEF ORMULAA ND MIDPOINTF ORMULAIt is important to note that the symbol Ar denotes the difference be-tweenthe abscissa of B and the abscissa of A, and it does not mean "deltamultiplied by x."Similarly, if we consider a particle moving along a line parallel tothe y axis from a point C(x, yr) to a point D(x, yr), then the ordinate ofthe particle changes from Ar to Az.We denote this change by Ly otLy: a z - arThus, CD: Ly.Now let Pr(r1, yr) and Pr(xr, a) be any two points in the plane. wewish to obtain a formula for finding the nonrtegative distance betweenthese two points. We shatl denote this distance by lPtPrl. W" use abso-lute-value bars because we are concerned only with the length, which isa nonnegative number, of the line segment between the two points P1 andPr. To derive the formula, we note that lffil is the length of the hypot-"rrrrr"of a right triangle PLMP2T. his is illustrated in Fig. 1.4.3f or P1a ndPr, both of which are in the first quadrant.Using the Pythagorean theorem, we haveEF"f: lArl2+ llvl'lF-nl:MV+WPThat is,lF:P,l: (1)Formula (1) holds for all possible positions of Pt and P, in all fourqqadrants. The length of the hypotenuse will always be lFfrl, and thelengths of the two legs will always be [Ar[ and lAyl (see Exercises L and2). We state this result as a theorem.'1,.4.T1 heorem The undirected distanceb etween the two points Pr(xr, Ar) and Pr(xr, yr)is given bysolurroN: From the statement of the problemlPTl :2lPBlUsing formula (1), we have@:z@Squaring on both sides, we havex2 - 6x+ 9 + y' - 4y I 4: 4(xz *8r * t6 + y2 - 2y * 1)Soor, equivalently,3x2* 3y ' * 38r- 4y + 55:0 45. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONSExAMPLE2 : Show that the tri-anglewith vertices at A(-2, 4),B(-5, L), and C(-6,5) isisosceles.Figure 1.4.4nxervrrrn 3: Prove analyticallythat the lengths of the diagonalsof a rectangle are equal.The triangle is shown in Fig. L.4.4.@:4+16:fr@:tr6+1:{lTW:/N:jfiSOLUTION:lBe:lla-e:;IEAI:Therefore,Hence, triangle ABC is isosceles.SOLUTIOND: raw a generalr ectangle.B ecausew e can chooset he coordi-nateaxes anywhere in the plane , and because the choice of the positionof the axes does not affect the truth of the theorem, we take the origin atone vertex, the x axis along one side, and the y axis along another side.This procedure simplifies the coordinates of the vertices on the two axes.Refer to Fig. 1..4.5.Now the hypothesis and the conclusion of the theorem can be stated.HypothesisO: ABC is a rectanglew ith diagonalsO B and AC.Co n c lus io n : IO -B| : lrel .loB:lf f i : r t+62lAC:lW :f r ]AtTherefore,lml : ld-clLet P, (xr, yr) and P, (xr, yr) be the endpoints of a line segment. Weshall denote this line segment by PtPr. This is not to be confused with thenotation PrPr, which denotes the directed distance from P, to Pr. That is,PrP, denotes a number, whereas PrP2 is a line segment. Let P(x, y) be themidpoint of the line segment P,Pr. Refer to Fig. L.4.6.In Fig. I.4.6 we see that triangles P'RP andPTPrare congruent. There-fore,l-PtRl : lPTl, and so r - xr: xz- x, giving usA(-2,4)B(-5, i.)C(0, b) B(a,b)(0,0)Figure 1.4.5 46. 1.4 DISTANCEF ORMULAA ND MIDPOINTF ORMULAPz (xz, v z)P(x,y)T(xr, Y)u__ _ _ J S ( r r , y r )R(t, yt)Figure1 .4.6In the derivation of formulas (2) and (3) it was assumed that x, ) xtand y, ) yr.The same formulas are obtained by using any orderings ofthese numbers (see Exercises 3 and 4).solurroN: Draw a general quadrilateral. Take the origin at one vertexand the x axis along one side. This method simplifies the coordinates ofthe two vertices on the r axis. See Fig. L.4.7.Hypothesis: OABC is a quadrilateral. M is the midpoint of OA, N isthe midpoint of CB, R is the midpoint of OC, and S is the midpoint of AB.Conclusion: MN and RS bisect each other.pRooF: To prove that two line segments bisect each other, we show thatthey have the same midpoint. Using formulas (2) and (3), we obtain thecoordinates of M, N, R, and S. M is the point Ga, 0), N is the point(+(b+ d),tr(c* e ) ) , R i s the point 1!d,te), and S i s thepoint G@+b),ic).The abscissao f the midpoint of MNis i lLa ++(b + d) l : * (a+b + d) .The ordinate of the midpoint of MN is +[0 + Lk + e)f : ik + e).Therefore, the midpoint of MN is the point (ifu + b + d), ik + e)).The abscissao f the midpoint of RS is i l ia+*@ + b) l : * (a+b + d) .The ordinate of the midpoint of RS is tlie + *cf : Ik + e).Therefore, the midpoint of RS is the point (ifu + b + d), i(c + e)).Thus, the midpoint of MN is the same point as the midpoint of RS.Therefore, MN and RS bisect each other. Inxaurr,E 4: Prove analyticallythat the line segments ioining themidpoints of the opposite sidesof any quadrilateral bisect eachother.B(b, c)Similarly, lRPl : lTPzl.T hen A - Ar: Az- A, andthereforeHence, the coordinates of the midpoint oftively, the average of the abscissas and thethe endpoints of the line segment.(3)a line segment are, respec-averageof the ordinates ofFigure 1.4.7 47. 92 REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONSExercise7s. 41. Derive distance formula (1) if Pr is in the third quadrant and P, is in the second quadrant. Draw a figure.2. Derive distance formula (1) if Pr is in the second quadrant and P, is in the fourth quadrant. Draw a figure.3. Derive midpoint formulas (2) and (3) if Pr is in the first quadrant and p, is in the third quadrant.4. Derive midpoint formulas (2) and (3) iI4 (rr, yr) and &(rr, yr) are both in the second quadrant an d x, > xrand,y, > yr.5. Find the length of the medians of the rriangle having vertices A(2,3),8(3,-g), and C(-1,-1).5. Find the midpoints of the diagonals of the quadrilateral whose vertices are (0,0), (0,4), (3,5), and (3, 1).(.Prove that the trianSle with verticesA (3,-6), B(8,-2), and C(-1, -1) is a right triangle. Find the areao f the triangle. '-l(HrNr: Use the converse of the Pythagorean theorem.)8. Prove that the Points A(6, -lS) , B(-2, 2), C(13, 10), and D(21, -5) are the vertices of a square. Find the tength of9.10.a diagonal.By using distance formula (1), prove that the points (-3, 2) , (1, -2), and (9, -10) lie on a line.If one end of a line segment is the point (-4, 2) and the midpoint is (3, -L), find the coordinates of the other end ofthe line segment.11. The abscissa of a Point iB -6, and its distance from the point (1, 3) is V-74. Find the ordinate of the point.12. Determine whether or not the poifis (14,7) , (2, 2) , and, (-4, -1) lie on a line by using distance formula (1).'/;3:.llf two vertices of an equilateral hiangle are (-4,3) and (0,0), find the third vertex.d Findan equation that must be satisfied by the coordinates of any point that is equidistant from the two points (-3, 2)and (4,6).15. Find an equation that must be satisfied by the coordinates of any point whose distance from the point (5, 3) is atwaystwo units Feater than it8 distance frcm the point (-4, -2).16. Giver the two Points d(-3,4) andB(2,5), find the coordinates of a point P on the line through A and B such thatpis (a) twice as far from 4 as frcm B, and (b) twice as far frcm B as from _A.17. Find the coordinates of the three points that divide the tine segment ftom l(-5, 3) to 8(6, g) into four equal parts.18. If rr. and r, are positiv_in tgg:ers,-proveth at the coordinateso f the point p(r, y), which divides the line segment plp,in the ratio rr/r2-that i3, IP-,,PUlPrE:I r,/rr-are given by,:(t" - rt)x, + rrxuur4 , : (r, - ,r)y, + ,rv"f z " t 2In Exercises 19 through 23, use the formulas of Exercise 18 to find the coordinates oI point p.19' The Point P is on the line segment between points Pr (1 ,3) ?flrdP, 26,2) and is three times as far from P, as it b ftom pr.20. The Point P is on the line segment between points Pl(1 ,3) and Pr(6,2) and is three times as far from Pr as it is frcmpr.21. The Point P i8 on the line through P' and P, and is three times as far ftom Pr(6, 2) as it is from P, (1,3) but is not be-tweenPr and Pr.22. The Point P is on the Une though P, and P, and is three times as far from P1(1, 3) as it is ftom Pr(6,2) but is not be-tweenP1 and P2.The point P is on the line through Pr(-3,5) and Pr(-L,2) so that ppr : + . p.,pr.Find an equation whose graph is the circle that is the set of all points that are at a distance of 4 units from the point(1, 3).23.24. 48. 1.5 EQUATIONSO F A LINE25. (a) Find an equation whose graph consist8 of all points equidistant frcm the Points (-1,2) and (3,4). (b) Draw a skekhof the gnph of the equation found in (a)'26. prove analytically that the sum of the squares of the distanc$ of any Point from two opposite vertics of any rcctangleis equal to the sum of the squares of its distances from the other two verticeE'22. prcve analytically that the line segment ioining the midpoints of two opposite sides of any quadrilateral and the linesegment joining the midpoints of the diagonala of the quadrilateral bieect each other'28. prove analytically that the midpoint of the hypotenuse of any right triangle is equidistant frcm each of the three vettices.29. Prove analytically that if the lengths of two of the medians of a triangle arc equal, the triangle is isoeceles.1..s EQUATIONS OF A LINE Letl be a nonvertical line and Pr(x' U) andP"(xr, A)be any two distinctpoints on l. Figure L.5.1 shows such a line. In the figure, R is the pointQc, Ur), and the points Pr, Pu and R are vertices of a right triangle; further-more,P,.R: xz-xr and R-Pr: Az-Ar The number Az-Ar gives themeasure of the change in the ordinate from Pr to Pr, and it may be posi-tive,negative , or zero. The number x2 - xl gives the measure of the changein the abscissa from P, to Pr, and it may be positive or negative. The num-berx, - rr may not be zero becaus? x2 * r, since the line I is not vertical.For all choices of the points P, and P, on l, the quotientFigure 1.5.1Az- Ur, Xz- xris constanq this quotient is called the "slope" of the line. Following is theformal definition.1.S.1 Definition If Pr(xr, Ar) and Pr(xr, yr) arc any two distinct points on line l, which isnot parallel to the y axis, then the slope of l, denoted by m, is given by;:yE (1)Xz * ?CrIn Eq. (L), x, * x1 since / is not parallel to the y axis.The value of m computed from Eq.(t) is independent of the choiceof the two pointr_P, and P2 on L To show this, suppose we choose two-at different points, tr(n, yr) and Pr(ir,yr), and compute a number nr fromEe. (1).7z,Ai*- Az- ar m : - -x z - x tWe shall show tlnilt n - m. Refer to Fig. 1.5.2. Triangles PrRPz and P1RP2are similar, so the lengths of corresponding sides are proportional.Therefore,V:- Y_-t Az-ArXz- Xr Xz- Xtorm:mR(rz, yJP, (xr, y,Pr(7r,V,)R(xr,Yr)Figure 1.5.2x 49. 34 REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONSHence, we conclude that the value of m computed from Eq. (1) is the samenumber no matter what two points on I are selected.In Fig. 1.5.2, x2 ) x1, Uz ) Ar, x, > n, and y2 > fr. The discussionabove is valid for any ordering of these pairs of numbers since Definition1.5.1 holds for any ordering.In sec. 1.4 we defined Ly:uz-ur and Ar:xz rr. substitutingthese values into Eq. (L), we haveMultiplying on both sides of this equation by Lx, we obtainLy: m Lx e)It is seen from Eq. (2) that if we consider a particle moving alongline l, the change in the ordinate of the particle is proportional to thechange in the abscissa, and the constant of proportionality is the slopeof the line.If the slope of a line is positive, then as the abscissa of a point on theline increases, the ordinate increases. Such a line is shown in Fig. 1.5.3.In Fig. "1..5.4w, e have a line whose slope is negative. For this line, as theabscissa of a point on the line increases, the ordinate decreases. Note thatif the line is parallel to the x axis, then Uz: Ur and so m:0.If the line is parallel to the y axis, xz: xr, thus, Eq. (1) is meaninglessbecause we cannot divide by zerc. This is the reason that lines paiallelto the y axis, or vertical lines, are excluded in Definition 1.5.L. We Jay thata vertical line does not have a slope.. rLLUSrRArroN l.: Let I be the line through the points P,,(2,3) and p2(4,7).The slope of I, by Definition "!..5.'!i.s, given byRefert o Fig. 1.5.5.r f P(x,y ) andQ @+ L*,y + Lil area ny two pointson I, thenLA _ . ,A x - 'Ay :2 A,xThus, if a particle is moving along the line l, the change in the ordinateis twice the change in the abscissa. That is, if the particle is at pr(4,7)and the abscissa is increased by one unit, then the ordinate is increasedby two units, and the particle is at the point Pr(5,9). Similarly, if the par-ticleis at Pt(2,3) and the abscissa is decreased by three units, then the or-dinateis decreased by six units, and the particle is at Pr(-L, -3). oSince two points Pr(xr, yr) and Pr(xr, y) determine a unique line, weFigure1 .5.3Figure 1.5.4P + ( - 1P r $ , 9 )*:7n!:zLP:-(2,3)Figure 1.5.5 50. 1.5 EQUATIONSO F A LINEshould be able to obtain an equation of the line through these two points.Consider P(x,y) any point on the line. We want an equation that is satis-fiedby r and y if and only rf P(x, D is on the line through P1(rr, yr) andPr(xr, Ar).We distinguish two cases.Case 7: xz: xr.In this case the line through P, and P, is parallel to the y axis, and allpoints on this line have the same abscissa. So P(x,, U) ts any point on theline if and only ifX : Xr ( 3 )Equation (3) is an equation of a line parallel to the y axis. Note that thisequation is independent of y; that is, the ordinate may have any valuewhatsoever, and the point P(x,y) is on the line whenever the abscissa is rr.Case 2: x2 * x1.The slope of the line through Pt and P, is given byry:Uz- A rxz- x rlf P(x, g) isgiven by1--A-A tx- x rThe point P will be on the line through P, and Pz if and only if the valueof m fuom Eq. (4) is the same as the value of m ftom Eq. (5), that is, if andonly ifU - At -Az- ArX- X r X z - XrMultiplying on both sides of this equation by (x - xr),we obtainEquation (5) is satisfied by the coordinates of P, as well as by the coordi-natesof any other point on the line through P1 and Pr.Equation (6) is called tkte two-point fonn of an equation of the line.It gives an equation of the line if two points on the line are known.o rLLUsrRArroN 2: An equation of the line through the two points (5, -3)and (-2,3) isy - ( - 3 ) : * ( r - 5 )y+3--2@-6)3x*4Y:6(4)any point on the line except (, !r), the slope is also(s)(6) 51. REAL NUMBERSI,N TRODUCTIONTO ANALYTICG EOMETRYA, ND FUNCTIONSIf in Eq. (6) we replace (y, - !r)l(x, - x) by *, we getI i . : r ) i ' I . ' t I + l- o : '; . . . i Yfl:gqt{,&Es'}* "atu e)Equation (7) is called the point-slope form of an equation of the line.It gives an equation of the line if a point Pr(xr, a) on the line and theslope m of the line are known.. rLLUSrRArroN 3: An equation of the line through the point (-4,-5)and having a slope of 2 isy - (-s) :Zlx - (-4) l2x-y*3:0 oIf we choose the particular point (0, b) (i.e., the point where the lineintersects the y axis) for the point (h, A) in Eq. (7), we havey-b:m(x-0)or, equivalently,+bThe number b, which is the ordinate of the point where the lineintersects the y axis, is called the y intercept of the line. Consequently,Eq. (8) is called the slope-intercepfto rm of an equation of the line. This_form is especially important because it enables us to find the slope of aline from its equation. It is also important becausei t expressest tre y co-ordinateexplicitly in terms of the r coordinate.(8)rxevrpln L: Given the linehaving the equation 3r I 4y :7,find the slope of the line.sol.urroN: Solving the equation f.or !, we have, : - f i x * IComparing this equation with Eq. (8), we see thatTherefore, the slope is -*.and b : * .Another form of an equation of a line is the one involving the inter-ceptsof a line. We define the r intercept of a line as the abscissa of thepoint at which the line intersects the r axis. The r intercept is denotedby o. If the r intercept a and the y intercept b arc given, we have twopoints (a, 0) and (0, b) on the line. Applying Eq. (6), the two-point form,we haveh - 0 . Y-o:6ik- a)-aY:bx-abbx * aA: ab 52. 1.5E OUATIONOSF A LINEDividing by ab, if. a * 0 and b + 0, we obtain(e)Equation (9) is called the intercept fonn of an equation of the line.Obviously it does not apply to a line through the origin, because for sucha line both a and b arc zero.nxlupln 2: The point (2,3)bisects that portion of a linewhich is cut off by the coordi-nateaxes. Find an equation ofthe line.Figure 1.5.6soLUrIoN: Refer to Fig. 1.5.6. lf. a isthe y intercept of the line, then theline segment joining (a, 0) and (0,haveandt: rythe r intercept of the line and b ispoint (2, 3) is the midPoint of theb). By the midpoint formulas/ weA:4 and b:6The intercept form, Eq.x + ! - t4 ' 6or, equivalently,3x*2y-12(9), gives ust.5.2 Theorem The graph of the equationAx*By*C:0where A, B, and C are constants and where not both A and B are zero, isa straight line.PRooF: Consider the two cases B + 0 and B : 0.CaseL: B+0.Because B + 0, we divide on both sides of Eq. (10) by B and ob-tainAC y--i*-, (11)Equation (11) is an equation of a straight line because it is in the slope-interceptform, where m: -AlB and b -- -CIB.Case2 : B :0.BecauseB :0, we may concludet hat A * 0 and thus haveAx*C:0(10)A a t l 53. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONSor/ equivalently,*:- aC (12)Equation (12) is in the form of Eq. (3), and so the graph is a straight lineparallel to the y axis. This completes the proof. rBecause the graph of Eq. (10) is a straight line, it is calle d a linearequation. Equation (10) is the general equation of the first degree in x andy.Because two points determine a line, to draw a sketch of the graph ofa straight line from its equation we need only determine the cooidit utetof two points on the line, plot the two points, and then draw the line.Aty two points will suffice, but it is usually convenient to plot the twopoints where the line intersects the two axes (which are given by theintercepts).rxeupm 3: Given line lr, havingthe equation 2x - 3y : 12, andline Ir, having the equation4x * 3y:6, draw a sketch ofeach of the lines. Then find thecoordinates of the point ofintersection of l, and Ir.Figure 1.5.7solurroN: To draw a sketch of the graph of lr, we find the intercepts aand b. In the equation of Ir, we substitute 0 for r and get b - -4.In theequation of Ir, we substitute 0 for y and get a: 6. Similarly, we obtainthe intercepts a and b for lr, and for l, we have a:8 and, b: 2. The twolines are plotted in Fig. 1..5.7.To find the coordinates of the point of intersection of /, and Ir, wesolve the two equations simultaneously. Because the point must lie onboth lines, its coordinates must satisfy both equations. If both equationsare put in the slope-intercept form, we havea:?x-4 andEliminating y gives3x-4:-gx*22x-L2:-4x#6x : 3Soy:BG)_4-tx*2Therefore, the point of intersection is (3, -2).1'5'3 Theorem If /r and l, are two distinct nonvertical lines having slopes m, and mr,respectively, then /, and l, are parallel if and only if mr: //t2.PRooF: Let an equation of /, be A : tftrx * br, and let an equation of l,be y : rnzx * br. Because there is an "if and only if " qualiiication, theproof consists of two parts.panr 1: Prove that l, and I, are parallel if ftir: tnz. 54. 1.5E QUATIONOSF A LINEAssume that I, and I, are not parallel. Let us show that this assumP-tionleads to a contradiction. If I, and I, are not parallel, then they intersect.Catl this point of intersection P(*o, /o).The coordinates of P must satisfythe equations of 11 and Ir, and so we havels: rnlxs* b1 and !o: fti2xo* b2But mr: rrt2rw hich givesls: mfis * b1 and !o: nttfio * b2from which it follows thatbl: bz. Thus, because rrtr: mrandbl: b2, bothlines 11 and /, have the same equation , A : mrx * br, and so the lines arethe same. But this contradicts the hypothesis that l, and I, are distinctlines. Therefore, our assumption is false. So we conclude that /r and 12are parallel.pARr 2: Prove that l, and I, are parallel only if mt: lftz. Here we mustshow that if Ir and l, are parallel, then rflt: rtrz.Assume that m, * mr. Solving the equations forand I, simultane-ously,we get, upon eliminating V,mtx * br: m2x * b2from which it follows that(m, - mr)x: bz - b tBecause we have assumed that m, * *r, this gives- - _ _b r - b ,L _t7- ffizHence, 11 and l, have a point of intersection, which contradicts the hy-pothesisthat l, and 12 are parallel. So our assumption is false, and there-foremr: nflz. IIn Fig. 1.5.8, the two lines Ir and l, are perpendicular. We state andprove the following theorem on the slopes of two perpendicular lines.1.5.4 Theorem If neither line /, nor line l, is vertical, then Ir and 12 are perpendicular ifand only if the product of their slopes is -1.. That is, if m, is the slope of11 and m, is the slope of. Ir, then Ir and I, are perpendicular if and only ifntfltz - -1.PROOF:Panr 1: Prove that l, and 12 are perPendicular only if mtm2 - -1'Let L, be the line through the origin parallel to l, and let L, be theline through the origin parallel to Ir. See Fig. 1.5.8. Therefore,by Theorem1..5.3, the slope of line L, ts m, and the slope of line L, is m2. Because neither 55. REAL NUMBERS,I NTRODUCTIONT O ANALYTICG EOMETRYA, ND FUNCTIONSL, nor L, rs vertical, these two lines intersect the line x: ! at points p,and Pr, respectively.T he abscissao f both P, and P, is L. Let / be the ordi-nateof Pt. Since Lr containst he points (0,0) and (1.,y ) andi ts slopei s mr,we have from Definition 1.5.1a - 0ttL:i=and so y = mr.Similarly, the ordinate of P, is shown to be mr.Applying the Pythagorean theorem to right triangle prop2, we getlot r ,+;zl@ f : lP,kl ,By applyingt he distancef ormula,w e obtainlo-4lr: (1 - 0), f- (mr- 0), :7 * m]lOtrlr: (1 - o), * (mr- o), :r * mz2lF J rP: (1- 1) , * (mr - mr ) ,- nt r z-2 mrmr t mr ,Substituting into Eq. (13) gives(l + mrz)+ (1 + m22-) ftizz- 2mrm,I .mrt2: -2mrffiztltinz - -1( 13)penr 2: Prove that I, and l, are perpendicular if. mrm" - -1. Starting withlt|fitz - -'1, we can reverse the steps of the proof of Part 1 in order toobtainlO-&l'l+1 trl': lPEl'zfrom which it follows, from the converse of the Pythagorean theorem,that Lt andL, are pe{pendicular; hence, /, and l, are perpendicular. ITheorem I.5.4 states that if two lines are perpendicular and neitherone is vertical, the slope of one of the lines is the negative reciprocal ofthe slope of the other line.o rLLUsrRArroN 4: If line /t is perpendicular to line I, and the slope of /,is 3, then the slope of /, must be -9. oFigure 1.5.8sxarvrpr,n4 : Prove by meahs ofslopes that the four pointsA(5,2), B(9, 6), C(4, g), andDQ,4) are the vertices of arectangle.solurroN: See Fig. 1.5.9. Letntr: the slope of ABm, J the slope of BC- 86- 6- 2- 8 - 64- I 56. C(4,8)D1.5 EQUATIONSO F A LINE 41B(8,6)Figure 1.5.9EXAMPLE 5: Given the line /having the equation2 x * 3 Y - 5 : 0find an equation of the lineperpendicular to line I andpassing through the pointAet,3)./fi.:the sloPoef DC:ft: ''rn.:the slopeo f.A D:=:-+.BecauseTnr: mr, AB ll DC./n2: ma, BC ll AD.ntrttrz- -'1, AB L BC.Therefore, quadrilateral ABCD has opposite sides parallel and two adja-centsides perpendicular, and we conclude that ABCD is a rectangle.solurroN: Because the required line is perpendicular to line l, its slopemust be the negative reciprocal of the slope of l. We find the slope of lbyputting its equation into the slope-intercept form. Solving the given'equation for y, we obtainy--tx**Therefore, the slope of I is -3, and the slope of the required line is 9. ge-causewe also know that the required line contains the point (-L, 3), weuse the point-slope form, which givesy - 3 : $ ( x * t )2y-6:3x*33x-2Y+9:0Exercise1s- .5In Exercises L through 4, find the slope of the line through the given points.1. (2, -3), ( - 4 , 3 )3.( ++, ),e E,& )In Exercises 5 through14, find an equation of the line satisfying the given conditions.5. The slope is 4 and through the point (2,-3).6. Through the two points (3, L) and (-5,4).7. Through the point (-3, -4) and parallel to the y axis.8. Through the point (I,-7) and parallel to the x axis.9. The x intercept is -3, and the y intercept is 4.2. (5,2) , (-2, -3)4 . ( - 2 .T, 0 . 3 ), ( 2 . 3 , 1 . 4 ) 57. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA. ND FUNCTIONS10. Through (1, 4) and parallel to the line whose equation is 2r - 5y + 7:0.11. Through (-2, -5) and having a slope of l/5.12. Through the origin and bisecting the angle between the axes in the first and third quadranb,13. Through the origin and bisecting the angle between the axes in the second and fou h quadrants.14. The slope is -2, and the r intercept is 4.15. Find an equation of the line thouth the points (1, 3) and (2, -2), and put the equation in the intercept form.dg. Fitrd"trequation of the line through the points (3, -5) and (1, -2), and put the equation in the slope-intercept form.17. Show by means of slopes that the points (-4, -1) , (3, +) , (8, -4) , and (2, -9) are the vertices of a trapezoid.18. Thrce consecutive vertices of a parallelogram are (-4, 1), (2,3),and (8,9). Find the coordinates of the fou h vertex.19. For each of the following sets of three point6, detemine by means of slopes if the points are on a line: (a) (2, 3),(-4,-7), (s,8);(b) (-3,6), (3,2), (e,-2); (cl (2, -7), (7,7), (3,t); and (d) (1,6), (1,2), (-s,-4).20. Prove by means of slopes that the three points A(3, 1), B(6,0), and C(4, ) are the vertices of a right triangle, and findthe area of the triangle./2T. Given the line I having the equation 2y - 3r : 4 and the point P(1, -3), find (a) an equation of the line through P" and perpendicul to l/(b) the shortest distance from P to line l.22. lf A,B, C, and D arc cdhstarts, show that (a) the lines Ax + By+ C:0 a dAx+By+D:0 are parallel and (b) theljmesA r+ By + C:0 and Bx- Ay * D:0 arep erpendicular.23. Given the line l, having the equation Ar + By + C: O, B + 0, find (a) the slope, (b) the y intercept, (c) the x intercept,(d) an equation of the line through the origin perpendicular to L24. Find an equation of the line which has equal intercepts and which passes throuth the point (8, -6).25. Find equations o{ the three medians of the triangle having vertices A(3, -2) , BG, a), and C(-l,1), and prove thatthey met in a point.26. Find equations of the perpendicular bisectors of the sides of the triande having vertices A(-l , -3) , B (5, -3), andC(5,5), and prove that they meet in a point.27. Find an equation of each of the lines through the point (3,2), which forms with the coordinate axes a triangle of area 12.28. Let il be the line havingt he equation.r4'*r Bly+ C1:0,andlet I'be the line having the equationA rx+ 82! + Ca:0.If ll is not parallel to L and iJ & is any constant, the equationA'x 'l Bry I C' * k(A"x * Bry * C") :0represents an unlimited number of lines. Prove that eadr of these lines contains the point of intersection df lr and t.29. Given an equation ot It is 2x 'f 3y - 5: 0 and an equation of l, is 3x + 5y - 8 : 0, by using Exercise 28 and withoutfinding the coordinates of the point of intersection of ,r and lr, find an equation of the line through this point of inter-sectionand (a) passing through the point (1, 3); (b) parallel to the r axis; (c) parallel to the y axis; (d) having slope -2;(e) perpendicular to the line having the eqtatio 2x + y :7; (f) forming an isosceles triangle with the coordinateaxes.30. Find an equation of each straight line that is perpendicular to the line having the equation 5r - y: I and that formswith the coordinate axes a triantle having an area of measure 5.31. Prove analytically that the diatonals of a rhombus are perpendicular.32. Prove analytically that the line segments joining conEecutive midpoints of the 6ide8 of any quadrilateral form aparallelogram. 58. 1.6 THE CIBCLE 4333. Prove analytically that the diagonals of a parallelogram bisect each other.34. Prove analytically that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelo-gram.1.5 THE CIRCLE'1,.6.1D efinition1.6.2 TheoremThe simplest curve that is the graph of a quadratic equation in two vari-ablesis the "citcle," which we now define.A circle is the set of all points in a plane equidistant from a fixedpoint. The fixed point is called the center of the circle, and the measureof the constant equal distance is called the radius of the circle.The circle with center at the point C(h, k) and radius r has as an equation' ; l - , :i :;,(x *'h}P, * (g1 = k)2,* 7zi'1 (1)pRooF: The point P(x, y) lies on the circle if and only iflncl: ,that is, if and only if@:rThis is true if and only if( x - h ) ' + ( y - k ) z - - t zwhich is Eq. (1). Equation (1) is satisfied by the coordinates of those andonly those points which lie on the given circle. Hence, (1) is an equationof the circle. IFrom Definition L.3.2, it follows that the graph of Eq. (1) is the circlewith center at (h, k) and radius r.If the center of the circle is at the origin, then h: k: 0; therefore,its equation is x' + y' : 12.o rLLUsrRArroN 1: Figure 1.6.1 shows the circle with center at (2, -3) andradius equal to 4. For this circle h:2,k:-3, and r:4. We obtain an equa-tionof the circle by substituting these values into Eq. (1) and we obtain( x - 2 ) ' + l y - ( - 3 ) 1 2 : 4 2(x-2)'+ (y + 3 ) ' : t 6Squaring and then combining terms, we havex 2 - 4 x + 4 + y ' * 5 y * 9 : t 6' :-L-J- 'C(2, - 3)Figure 1.6.1 )c'+y'-4x+6y-3:0 59. REAL NUMBERSI,N TRODUCTIONTO ANALYTICG EOMETRYA, ND FUNCTIONSEXAMPLE 1: Given the equationx'+ y'* 6x-2y - 15: oprove that the graph of thisequation is a circle, and find itscenter and radius.solurroN: The given equation may be written as( x r + 6 x ) t ( y r - 2 y ) : t sCompleting the squares of the terms in parentheses by adding 9 and 1on both sides of the equation, we have(x' * 6x *9) + (y' - 2y I1) : 15 + 9 + 1,(r*3)2+(V-1)':25Comparing this equation with Eq. (1), we see that this is an equation ofa circle with its center at (-3, I) and a radius of 5.In Eq. (L), removing parenthesesa nd combining terms givesx' + y'- Zhx - Zky + (h, + k2 - rr) :0 Q)Equation (2) is of the form)e+y'*Dx+Ey*F:0 ( 3 )where D : -2h, E : -2k, and F : h2 * k2 - f.Equation (3) is called the general form of an equation of a circle,whereas Eq. (1) is called the center-rsdiusfo rm. Becausee very circle hasa center and a radius, its equation can be put in the center-radius form,and hence into the general form.We now consider the question of whether or not the graph of everyequation of the formx'+y,*Dx*Ey*F:0is a circle. To determine this, we shall attempt to write this equation in thecenter-iadius form. We rewrite the equation asx'+y,*Dx*Ey--Fand completet he squareso f the terms in parenthesesb y adding]nDza ndiE' on both sides, thus giving us(x'* Dxt +D')+ (y' * Ey+ +E'z-)- F + +Dz* iE,or, guivalently,(x + LD)2+ (y + +E)2: i(D' + Ez- 4F)Equation (a) is in the form of Eq. (1) if and only if+(D,+E2-4F):rzWe now consider three cases, namely, (D'+ E2 - 4F) as positive,zero, and negative.Case 7: (D, + E2 - 4F) > 0.Then r': *(D' + E2 - 4F), and so Eq. (a) is an equation of a circlehaving a radius equal to it/FE= 4F and its center at (-*D,-+E).(4) 60. 1.6 THE CIRCLECase2 : Dz+ Ez- 4F: 0.Equation ( ) is then of the form(x + tD)z + (Y * iE)z :0Because the only real values of r and y satisfying Eq. (5) are v - -L2D andy:-+E, the graph is the point (-+D, -+E). Comparing Eq. (5) withEq. (1), we see that h:-+D, k:-iE, and r - 0. Thus, this point can becalled a point-circle.Case 3: (Dt + E2 - 4F) < 0.Then Eq. (4) has a negative number on the right side and the sum ofthe squares of two real numbers on the left side. There are no real valuesof r and y that satisfy such an equation; consequently, we say the graphis the empty set.Before stating the results of these three cases as a theorem, we ob-servethat an equation of the formAxz*Ay'+Dx*Ey* F:0 where A+0can be written in the form of Eq. (3) by dividing by A, thereby obtainingrc'+y'+Dt r *,*E= A,UF +i :0Equation (6) is a special case of the general equation of the second degree:Axz * Bxy * Cy' t Dx * Ey * F : 0in which the coefficients of x2 and yz are equal and which has no xy term.We have, then, the following theorem.1.6.3 Theorem The graph of any second-degree equation in R2 in r and U, in which thecoefficients of x2 and y2 are equal and in which there is no xy term, iseither a circle, a point-circle, or the empty set.o rLLUSrRArroN 2: The equation2 x 2 * 2 y ' * L Z x - B y + 3 1 : 0is of the form (6), and therefore its graph is either a circle, a point-circle,or the empty set. If the equation is put in the form of Eq. (1), wehavex'+y't6x-4y-F#:o(x'+ 6x) + (y'- 4y):-T(x' * 6x *9) + (y' - 4y + 4) :-#+ 9 + 4(r*3)2 + (Y-2)':-EThereforet,h e graphi s the empty set.(5)(6) 61. REALN UMBERSI,N TRODUCTIOTNO ANALYTICG EOMETRYA, ND FUNCTIONSExAMPLE 2: Find an equation ofthe circle through the threepoints A(4,5), B(3, -2), andc(1,,- 4).nxeuprr 3: Find an equation ofthe circle with its center at thepoint C(1,6) and tangent to theline / having the equationx - v - 1 : 0 .soLUrIoN: The general form of an equation of the circle isx'+y,tDx-fEy*F:0Becauseth e threep oints A, B,and C must lie on the circle,t he coordi-natesof these points must satisfy the equation. So we haveL6+25+4D+5E+F:09+ 4+3D-2E*F:01+16+ D-4E*F:0or, guivalently,4D+ 5E+ F:-473D-2E+F:-L3D - 4E * F:-17Solving these three equations simultaneously, we getD:7 E:-5 F : - 4 4Thus, an equation of the circle isx'+y'*7x-5y-44:0In the following example we have a line which is tangent to a circle.The definition of the tangent line to a general curve at a specific point isgiven in Sec. 3.L. However, for a circle we use the definition from planegeometry which states that a tangent line at a point P on the circle is theline intersecting the circle at only the point P.soLUrIoN: See Fig. L.6.2. Given that h: L and k: 6, if we find r, we canobtain an equation of the circle by using the center-radius form.Let I, be the line through C and the point P, which is the point oftangency of line I with the circle.r: lPClHence, we must find the coordinates of P. We do this by finding an equa-tionof /, and then finding the point of intersection of /1 with /. Since 11is along a diameter of the circle, and / is tangent to the circle , I, is perpen-dicularto l. Because the slope of / is L, the slope of 11 is -L. Therefore,using the point-slope form of an equation of a line, we obtain as an equa-tionof I,y - 5 - - 1 ( r - 1 ) 62. Figure 1.6.2Exercise1s.6 /- )-In Exercises 1 through 4, find an equation of the circle with center at C and radius r. Write the equation in both the center-radiusforrr and the gmeral form.'r . c 1 4 , - 3 ) , r: S3. C(-5, -12), f :3In Exercises 5 through 10, find an equation of the circle satisfying the given condition$.5. Center is at (1, 2 and through the point (3, -1).5. Center is at (-2,5) and tangent to the line x:7./Z) C"nl l . f * y ' - 6 x - 8 y * 9 : 013. 3f * 3y' * 4y - 7 :015. 12* y' - 2x * LOy* L9: 01 7 . f * y ' - 1 0 r * 6 y * 3 6 : 01,6 THE CIRCLEor, equivalently,x+Y-7:0Solving this equation simultaneously with the given equation of l, namely,X- A - L:0we get x: 4 and y: 3. Thus, P is the point (4, 3). Therefore,,:l-PCl:ffior, equivalently,r : !18So, an equation of the circle is(x-l)'+ (Y-6)': (f t ) 'or, equivalently,x' + y ' - 2x- 12y * L9:0, . c(Q, 0 ) , r : 84. C ( - l , l ) , r : 21 2 . 2 f * 2 y ' - 2 x * 2 y * 7 : 01 4 . * * y ' - 10r-lOy+25:016. 4* l- 4y' * 24x - 4y t L : 01 8 . x 2 * y ' * 2 x - 4 y * 5 : 0"ris at (-3, -5) and tangent to the line l}x t 5y - 4:0.w8. Through the three points (2, 8) , (7 , 3), and (-2,0).9. Tangent to the line 3x * y * 2:0 at (-L, 1) and through the point (3, 5).10. Tangent to the line 3x * 4y - 16:0 at (4,1.) and with a radius of 5. (Two possible circles.)In Exercises L1 through 14, find the center and radius of each circle, and draw a sketch of the graph.In Exercises L5 through 20, determine whether the graph is a circle, a point-circle, or the empty set. 63. 48 REALN UMBERSIN, TRODUCTIOTNO ANALYTICG EOMETRYA,N D FUNCTIONS*+ f -{f.x+36v-719:0W19fI6x-6y+5-0Find an equation of the comrnon chord_ of the two cirdes f* 3F+rlt - 6y - 12:0 and f* l+gr- 2yrg:o.(rrrrvr: If the coordinates of a Point gatisfy two different equations, then thjcoordinates also satisfy the difierence ofthe two equatione.)Find the points of intersection of the two cirdes in Exercise 21.Find an equation of the line which is tangent to the circle .d + y, - 4t + 6y - 12: O at the point (5, 1).Findanequationofeachofthetwolineshavingslope-jgwhicharetangenttothecirclef*l*2x-W-g:0.From the origin, dbrda of the cirde t' * I * 4r: 0 are drawn. Prve that the set of midpoinb of these chords is a circle.Prove analytically that a line from th center of any circle biEecting any drord is perpendicular to the chord.Prcve analytically that an angle inscribed in a semicircle is a right angle.Given the line y: flr + b tangent to the circle rP * y, = r2, find an equation involving m, b, and t1.7 FUNCTIONS AND We intuitively consider y to be a function of r if there is some rule byTHEIR GRAPHS which a unique value is assigned to y by a corresponding value of x.Familiar examples of such relationships are given by equations such asy:2f+5 ( 1 )andY:tP-g19.20.21.22.iI24.25.26.27.28.(2)Table 1.7.1It is not necessary that r and y be related by an equation in order fora functional relationship to exist between them, as shown in the fol-lowingillustration.o ILLUsTRATIoN1 ,: lf y is the number of cents in the postage of a domesticfirst class letter, and if r is the number of ounces in the weight of the letter,then y is a function of x. For this functional relationship, there is no equa-tioninvolving r and /; however, the relationship between r and y maybe given by means of a table, such as Table 1.7.L. .x 0 such that I f(x)-51 < e whenever 0 < lx- 1l < 6, westate that the limit of f (x) as r approaches1 is equal to 5 or, exPressedinsymbols,trm/(r):s ( 4 )You will note that we state 0 < lr - 11. This condition is imposedbecause we are concerned only with values of f (x) for x close to L, but notf.or x: L. As a matter of fact, this function is not define