Solutions for Calculus Concepts And Contexts Enhanced Edition 4th Us Edition by Stewart by 6alsm

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COMPLETE SOLUTIONS MANUAL

for Stewart’s

SINGLE VARIABLE CALCULUS

CONCEPTS AND CONTEXTS

FOURTH EDITION

JEFFERY A. COLE Anoka Ramsey Community

TIMOTHY J. FLAHERTY Carnegie

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2010 Brooks/Cole, Cengage Learning

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PREFACE

This Complete Solutions Manual contains solutions to all exercises in the texts Single Variable Calculus:Concepts and Contexts, Fourth Edition,and Chapters 1–8 of Calculus:Concepts and Contexts, Fourth Edition,by James Stewart. A student version of this manual is also available; it contains solutions to the odd-numbered exercises in each chapter section,the review sections,the True-False Quizzes,and the Focus on Problem Solving sections,as well as solutions to all the exercises in the Concept Checks. No solutions to the Projects appear in the student version. It is our hope that by browsing through the solutions,professors will save time in determining appropriate assignments for their particular classes.

Some nonstandard notation is used in order to save space. If you see a symbol that you don’t recognize,refer to the Table of Abbreviations and Symbols on page v.

We appreciate feedback concerning errors,solution correctness or style,and manual style. Any comments may be sent directly to us at jeff.cole@anokaramsey.edu or tim@andrew.cmu.edu,or in care of the publisher:Cengage Learning Brooks/Cole,0 Davis Drive,Belmont,CA 94002.

We would like to thank Jim Stewart,for his guidance; Brian Betsill,Kathi Townes,and Rebekah Million,of TECH-arts,for their production services; and Richard Stratton and Jeannine Lawless,of Cengage Learning Brooks/Cole,for entrusting us with this project as well as for their patience and support.

Jeffery A. Cole Anoka Ramsey Community College

Timothy J. Flaherty Carnegie Mellon University

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ABBREVIATIONS AND SYMBOLS

CDconcavedownward

CUconcaveupward

Dthedomainof

FDTFirstDerivativeTest

HAhorizontalasymptote(s)

Iintervalofconvergence

I/DIncreasing/DecreasingTest

IPin ectionpoint(s)

Rradiusofconvergence

VAverticalasymptote(s)

CAS = indicatestheuseofacomputeralgebrasystem.

H = indicatestheuseofl’Hospital’sRule.

= indicatestheuseofFormula intheTableofIntegralsinthebackendpapers.

s = indicatestheuseofthesubstitution { =sin =cos }

c = indicatestheuseofthesubstitution { =cos = sin }.

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CONTENTS ■ DIAGNOSTIC TESTS1

1 ■ FUNCTIONS AND MODELS9

1.1 Four Ways to Represent a Function9

1.2 Mathematical Models:A Catalog of Essential Functions20

1.3 New Functions from Old Functions26

1.4 Graphing Calculators and Computers38

1.5 Exponential Functions45

1.6 Inverse Functions and Logarithms51

1.7 Parametric Curves59

Laboratory Project ■ Running Circles Around Circles71 Review73

Principles of Problem Solving83

2 ■ LIMITS AND DERIVATIVES 87

2.1 The Tangent and Velocity Problems87

2.2 The Limit of a Function90

2.3 Calculating Limits Using the Limit Laws96

2.4 Continuity104

2.5 Limits Involving Inﬁnity 113

2.6 Derivatives and Rates of Change123

2.7 The Derivative as a Function134

2.8 What Does Say about ?146 Review152

Focus on Problem Solving163

3 ■ DIFFERENTIATION RULES 167

3.1 Derivatives of Polynomials and Exponential Functions167 Applied Project ■ Building a Better Roller Coaster177

3.2 The Product and Quotient Rules179 f f

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3.3 Derivatives of Trigonometric Functions188

3.4 The Chain Rule194

Laboratory Project ■ Bézier Curves207

Applied Project ■ Where Should a Pilot Start Descent?208

3.5 Implicit Differentiation209

3.6 Inverse Trigonometric Functions and Their Derivatives219

3.7 Derivatives of Logarithmic Functions225

Discovery Project ■ Hyperbolic Functions230

3.8 Rates of Change in the Natural and Social Sciences232

3.9 Linear Approximations and Differentials241

Laboratory Project ■ Taylor Polynomials247 Review249

Focus on Problem Solving261

4 ■ APPLICATIONS OF DIFFERENTIATION 271

4.1 Related Rates271

4.2 Maximum and Minimum Values278

Applied Project ■ The Calculus of Rainbows288

4.3 Derivatives and the Shapes of Curves289

4.4 Graphing with Calculus and Calculators310

4.5 Indeterminate Forms and l’Hospital’s Rule328

4.6 Optimization Problems340

Applied Project ■ The Shape of a Can358

4.7 Newton’s Method359

4.8 Antiderivatives368 Review375

Focus on Problem Solving393

5 ■ INTEGRALS 403

5.1 Areas and Distances403

5.2 The Deﬁnite Integral412

5.3 Evaluating Deﬁnite Integrals421

Discovery Project ■ Area Functions428

5.4 The Fundamental Theorem of Calculus430

5.5 The Substitution Rule436

5.6 Integration by Parts444

Parts444

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5.7 Additional Techniques of Integration452

5.8 Integration Using Tables and Computer Algebra Systems459

Discovery Project ■ Patterns in Integrals465

5.9 Approximate Integration467

5.10 Improper Integrals479

Review491

Focus on Problem Solving505

6 ■ APPLICATIONS OF INTEGRATION 513

6.1 More about Areas513

6.2 Volumes523

Discovery Project ■ Rotating on a Slant537

6.3 Volumes by Cylindrical Shells538

6.4 Arc Length546

Discovery Project ■ Arc Length Contest554

6.5 Average Value of a Function554

Applied Project ■ Where To Sit at the Movies557

6.6 Applications to Physics and Engineering557

Discovery Project ■ Complementary Coffee Cups568

6.7 Applications to Economics and Biology569

6.8 Probability572

Review575

Focus on Problem Solving585

7 ■ DIFFERENTIAL EQUATIONS 593

7.1 Modeling with Differential Equations593

7.2 Direction Fields and Euler’s Method596

7.3 Separable Equations604

Applied Project ■ How Fast Does a Tank Drain?616

Applied Project ■ Which Is Faster, Going Up or Coming Down?618

7.4 Exponential Growth and Decay619

Applied Project ■ Calculus and Baseball624

7.5 The Logistic Equation625

7.6 Predator-Prey Systems635 Review640

Focus on Problem Solving647

F ocus on Problem Solvin g 647Solving6

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8 ■ INFINITE SEQUENCES AND SERIES 653

8.1 Sequences653

Laboratory Project ■ Logistic Sequences661

8.2 Series665

8.3 The Integral and Comparison Tests; Estimating Sums678

8.4 Other Convergence Tests684

8.5 Power Series691

8.6 Representations of Functions as Power Series698

8.7 Taylor and Maclaurin Series707

Laboratory Project ■ An Elusive Limit721

8.8 Applications of Taylor Polynomials722

Applied Project ■ Radiation from the Stars733 Review735

Focus on Problem Solving747

■ APPENDIXES 753

A Intervals,Inequalities,and Absolute Values753

B Coordinate Geometry756

C Trigonometry762

D Precise Deﬁnitions of Limits766

F Sigma Notation771

G Integration of Rational Functions by Partial Fractions774

H Polar Coordinates785

Discovery Project ■ Conic Sections in Polar Coordinates808

I Complex Numbers809

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DIAGNOSTICTESTS

TestAAlgebra

1. (a) ( 3)4 =( 3)( 3)( 3)( 3)=81 (b) 34 = (3)(3)(3)(3)= 81 (c) 3 4 = 1 34 = 1

2. (a)Notethat

Or: Usetheformulaforthedifferenceoftwosquarestoseethat +

Note: Aquickerwaytoexpandthisbinomialistousetheformula ( + )

(e)SeeReferencePage1forthebinomialformula ( + )3 = 3

.Usingit,weget (

(a)Usingthedifferenceoftwosquaresformula,

Thislastexpressionwasobtainedusingthesumoftwocubesformula,

and =3.[SeeReferencePage1inthetextbook.]

2 ,sowewillfactorout

,wehave

Inintervalnotation,theansweris [ 4 3). (b)

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.Now, (

4) willchangesignatthecritical values = 2 and =4.Thusthepossibleintervalsofsolutionare ( 2), ( 2 4),and (4 ).Bychoosinga singletestvaluefromeachinterval,weseethat ( 2 4) istheonlyintervalthatsatis estheinequality.

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(c)Theinequality ( 1)( +2) 0 hascriticalvaluesof 2 0 and 1.Thecorrespondingpossibleintervalsofsolution are ( 2), ( 2 0), (0 1) and (1 ).Bychoosingasingletestvaluefromeachinterval,weseethatbothintervals ( 2 0) and (1 ) satisfytheinequality.Thus,thesolutionistheunionofthesetwointervals: ( 2 0) (1 )

(d) | 4| 3 3 4 3 1 7.Inintervalnotation,theansweris (1 7)

(e) 2 3 +1 1 2 3

Now,theexpression 4 +1 maychangesignsatthecriticalvalues = 1 and =4,sothepossibleintervalsofsolution are ( 1), ( 1 4],and [4 ).Bychoosingasingletestvaluefromeachinterval,weseethat ( 1 4] istheonly intervalthatsatis estheinequality.

10. (a)False.Inorderforthestatementtobetrue,itmustholdforallrealnumbers,so,toshowthatthestatementisfalse,pick =1 and =2 andobservethat (1+2)2 =12 +22 .Ingeneral, ( + )2 = 2 +2 + 2

(b)Trueaslongas and arenonnegativerealnumbers.Toseethis,thinkintermsofthelawsofexponents: =( )1

(c)False.Toseethis,let =1 and =2,then 1

(d)False.Toseethis,let =1 and =2,then 1+1(2)

(e)False.Toseethis,let

and

(f)Truesince 1 = 1 ,aslongas =0 and =0

TestBAnalyticGeometry

1. (a)Usingthepoint (2 5) and = 3 inthepoint-slopeequationofaline, 1 = ( 1 ),weget ( 5)= 3( 2) +5= 3 +6 = 3 +1

(b)Alineparalleltothe -axismustbehorizontalandthushaveaslopeof 0.Sincethelinepassesthroughthepoint (2 5), the -coordinateofeverypointonthelineis 5,sotheequationis = 5.

(c)Alineparalleltothe -axisisverticalwithunde nedslope.Sothe -coordinateofeverypointonthelineis2andsothe equationis =2

(d)Notethat 2 4 =3 4 = 2 +3 = 1 2 3 4 .Thustheslopeofthegivenlineis = 1 2 .Hence,the slopeofthelinewe’relookingforisalso 1 2 (sincethelinewe’relookingforisrequiredtobeparalleltothegivenline).

Sotheequationofthelineis ( 5)= 1 2 ( 2) +5= 1 2 1 = 1 2 6

2. Firstwe’ll ndthedistancebetweenthetwogivenpointsinordertoobtaintheradius, ,ofthecircle: = [3 ( 1)]2 +( 2 4)2 = 42 +( 6)2 = 52.Nextusethestandardequationofacircle, ( )2 +( )2 = 2 ,where ( ) isthecenter,toget ( +1)2 +( 4)2 =52.

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3. Wemustrewritetheequationinstandardforminordertoidentifythecenterandradius.Notethat 2 + 2 6 +10 +9=0 2 6 +9+ 2 +10 =0.Fortheleft-handsideofthelatterequation,we factorthe rstthreetermsandcompletethesquareonthelasttwotermsasfollows: 2 6 +9+ 2 +10 =0 ( 3)2 + 2 +10 +25=25 ( 3)2 +( +5)2 =25.Thus,thecenterofthecircleis (3 5) andtheradiusis 5.

4. (a) ( 7 4) and (5 12) = 12 4 5 ( 7) = 16 12 = 4 3 (b) 4= 4 3 [ ( 7)] 4= 4 3 28 3 3 12= 4 28 4 +3 +16=0.Putting =0, weget 4 +16=0,sothe -interceptis 4,andsubstituting 0 for resultsina -interceptof 16 3 .

(c)Themidpointisobtainedbyaveragingthecorrespondingcoordinatesofbothpoints: 7+5 2 4+( 12) 2 =( 1 4). (d) = [5 ( 7)]2 +( 12 4)2 = 122 +( 16)2 = 144+256= 400=20

(e)Theperpendicularbisectoristhelinethatintersectsthelinesegment atarightanglethroughitsmidpoint.Thusthe perpendicularbisectorpassesthrough ( 1 4) andhasslope 3 4 [theslopeisobtainedbytakingthenegativereciprocalof theanswerfrompart(a)].Sotheperpendicularbisectorisgivenby +4= 3 4 [ ( 1)] or 3 4 =13.

(f)Thecenteroftherequiredcircleisthemidpointof ,andtheradiusishalfthelengthof ,whichis 10.Thus,the equationis ( +1)2 +( +4)2 =100.

5. (a)Graphthecorrespondinghorizontallines(givenbytheequations = 1 and =3)assolidlines.Theinequality 1 describesthepoints ( ) thatlie onor above theline = 1.Theinequality 3 describesthepoints ( ) thatlieonor below theline =3.Sothepairofinequalities 1 3 describesthepointsthatlieonor between thelines = 1 and =3

(b)Notethatthegiveninequalitiescanbewrittenas 4 4 and 2 2, respectively.Sotheregionliesbetweentheverticallines = 4 and =4 and betweenthehorizontallines = 2 and =2.Asshowninthegraph,the regioncommontobothgraphsisarectangle(minusitsedges)centeredatthe origin.

(c)We rstgraph =1 1 2 asadottedline.Since 1 1 2 ,thepointsinthe regionlie below thisline.

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(d)We rstgraphtheparabola = 2 1 usingasolidcurve.Since 2 1, thepointsintheregionlieonor above theparabola.

(e)Wegraphthecircle 2 + 2 =4 usingadottedcurve.Since 2 + 2 2,the regionconsistsofpointswhosedistancefromtheoriginislessthan2,thatis, thepointsthatlie inside thecircle.

(f)Theequation 9 2 +16 2 =144 isanellipsecenteredat (0 0).Weputitin standardformbydividingby 144 andget 2 16 + 2 9 =1.The -interceptsare locatedatadistanceof 16=4 fromthecenterwhilethe -interceptsarea distanceof 9=3 fromthecenter(seethegraph).

TestCFunctions

1. (a)Locate 1 onthe -axisandthengodowntothepointonthegraphwithan -coordinateof 1.Thecorresponding -coordinateisthevalueofthefunctionat = 1,whichis 2.So, ( 1)= 2

(b)Usingthesametechniqueasinpart(a),weget (2) 2 8

(c)Locate 2 onthe -axisandthengoleftandrightto ndallpointsonthegraphwitha -coordinateof 2.Thecorresponding -coordinatesarethe -valueswearesearchingfor.So = 3 and =1

(d)Usingthesametechniqueasinpart(c),weget 2 5 and 0 3

(e)Thedomainisallthe -valuesforwhichthegraphexists,andtherangeisallthe -valuesforwhichthegraphexists. Thus,thedomainis [ 3 3],andtherangeis [ 2 3]

2. Notethat (2+ )=(2+ )3 and (2)=23 =8.Sothedifferencequotientbecomes (2+ ) (2) = (2+ )3 8 =

= (12+6 + 2 ) =12+6 + 2 .

3. (a)Setthedenominatorequalto0andsolveto ndrestrictionsonthedomain: 2 + 2=0 ( 1)( +2)=0 =1 or = 2.Thus,thedomainisallrealnumbersexcept 1 or 2 or,ininterval notation, ( 2) ( 2 1) (1 )

(b)Notethatthedenominatorisalwaysgreaterthanorequalto 1,andthenumeratorisde nedforallrealnumbers.Thus,the domainis ( )

(c)Notethatthefunction isthesumoftworootfunctions.So isde nedontheintersectionofthedomainsofthesetwo rootfunctions.Thedomainofasquarerootfunctionisfoundbysettingitsradicandgreaterthanorequalto 0.Now,

4 0 4 and 2 1 0 ( 1)( +1) 0 1 or 1.Thus,thedomainof is ( 1] [1 4]

4. (a)Re ectthegraphof aboutthe -axis.

(b)Stretchthegraphof verticallybyafactorof 2,thenshift 1 unitdownward.

(c)Shiftthegraphof right 3 units,thenup 2 units.

5. (a)Makeatableandthenconnectthepointswithasmoothcurve:

2 1 0 1 2 8 1 0 1 8

(b)Shiftthegraphfrompart(a)left 1 unit.

(c)Shiftthegraphfrompart(a)right 2 unitsandup 3 units.

(d)Firstplot = 2 .Next,togetthegraphof ( )=4 2 , re ect aboutthe x-axisandthenshiftitupward 4 units.

(e)Makeatableandthenconnectthepointswithasmoothcurve: 0 1 4 9 0 1 2 3

(f)Stretchthegraphfrompart(e)verticallybyafactoroftwo.

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(g)Firstplot =2 .Next,getthegraphof = 2 byre ectingthegraphof =2 aboutthe x-axis.

(h)Notethat =1+ 1 =1+1 .So rstplot =1 andthenshiftit upward 1 unit.

6. (a) ( 2)=1 ( 2)2 = 3 and (1)=2(1)+1=3

(b)For 0 plot ( )=1 2 and,onthesameplane,for 0 plotthegraph of ( )=2 +1

( )( )= ( ( ))= ( 2 +2 1)=2( 2 +2

( )(

TestDTrigonometry

(a)

2. (a) 5 6 =

3. Wewillusethearclengthformula, = ,where isarclength, istheradiusofthecircle,and isthemeasureofthe centralangleinradians.First,notethat 30 =30 180 = 6 .So =(12) 6 =2 cm.

4. (a) tan( 3)= 3 Youcanreadthevaluefromarighttrianglewithsides1,2,and 3

(b)Notethat 7 6 canbethoughtofasanangleinthethirdquadrantwithreferenceangle 6.Thus, sin(7 6)= 1 2 , sincethesinefunctionisnegativeinthethirdquadrant.

(c)Notethat 5 3 canbethoughtofasanangleinthefourthquadrantwithreferenceangle 3.Thus, sec(5 3)= 1 cos(5 3) = 1 1 2 =2,sincethecosinefunctionispositiveinthefourthquadrant.

5. sin = 24 =24sin and cos = 24 =24cos

= 24 sin =24

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So,usingthesumidentityforthesine,wehave

9. We rstgraph =sin2 (bycompressingthegraphof sin byafactorof2)andthenshiftitupward 1 unit.

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1 FUNCTIONSANDMODELS

1.1FourWaysToRepresentaFunction

Inexercisesrequiringestimationsorapproximations,youranswersmayvaryslightlyfromtheanswersgivenhere.

1. (a)Thepoint (1 3) isonthegraphof ,so (1)=3

(b)When = 1, isabout 0 2,so ( 1) 0 2.

(d)Areasonableestimatefor when =0 is = 0 8

(e)Thedomainof consistsofall -valuesonthegraphof .Forthisfunction,thedomainis 2 4,or [ 2 4]

Therangeof consistsofall -valuesonthegraphof .Forthisfunction,therangeis 1 3,or [ 1 3].

(f)As increasesfrom 2 to 1, increasesfrom 1 to 3.Thus, isincreasingontheinterval [ 2 1]

2. (a)Thepoint ( 4 2) isonthegraphof ,so ( 4)= 2.Thepoint (3 4) isonthegraphof ,so (3)=4

(b)Wearelookingforthevaluesof forwhichthe -valuesareequal.The -valuesfor and areequalatthepoints ( 2 1) and (2 2),sothedesiredvaluesof are 2 and 2

(d)As increasesfrom 0 to 4, decreasesfrom 3 to 1.Thus, isdecreasingontheinterval [0 4]

(e)Thedomainof consistsofall -valuesonthegraphof .Forthisfunction,thedomainis 4 4,or [ 4 4]

Therangeof consistsofall -valuesonthegraphof .Forthisfunction,therangeis 2 3,or [ 2 3]

(f)Thedomainof is [ 4 3] andtherangeis [0 5 4]

3. FromFigure1inthetext,thelowestpointoccursatabout ( )=(12 85).Thehighestpointoccursatabout (17 115). Thus,therangeoftheverticalgroundaccelerationis 85 115.Writteninintervalnotation,weget [ 85 115]

4. Example1: Acarisdrivenat 60 mi hfor 2 hours.Thedistance traveledbythecarisafunctionofthetime .Thedomainofthe functionis { | 0 2},where ismeasuredinhours.Therange ofthefunctionis { | 0 120},where ismeasuredinmiles.

Example2: Atacertainuniversity,thenumberofstudents on campusatanytimeonaparticulardayisafunctionofthetime after midnight.Thedomainofthefunctionis { | 0 24},where is measuredinhours.Therangeofthefunctionis { | 0 }, where isanintegerand isthelargestnumberofstudentson campusatonce.

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Example3: Acertainemployeeispaid $8 00 perhourandworksa maximumof 30 hoursperweek.Thenumberofhoursworkedis roundeddowntothenearestquarterofanhour.Thisemployee’s grossweeklypay isafunctionofthenumberofhoursworked Thedomainofthefunctionis [0 30] andtherangeofthefunctionis

5. No,thecurveisnotthegraphofafunctionbecauseaverticallineintersectsthecurvemorethanonce.Hence,thecurvefails theVerticalLineTest.

6. Yes,thecurveisthegraphofafunctionbecauseitpassestheVerticalLineTest.Thedomainis [ 2 2] andtherange is [ 1 2].

7. Yes,thecurveisthegraphofafunctionbecauseitpassestheVerticalLineTest.Thedomainis [ 3 2] andtherange is [ 3 2) [ 1 3].

8. No,thecurveisnotthegraphofafunctionsincefor =0, ±1,and ±2,therearein nitelymanypointsonthecurve.

9. Theperson’sweightincreasedtoabout 160 poundsatage 20 andstayedfairlysteadyfor 10 years.Theperson’sweight droppedtoabout 120 poundsforthenext 5 years,thenincreasedrapidlytoabout 170 pounds.Thenext 30 yearssawagradual increaseto 190 pounds.Possiblereasonsforthedropinweightat 30 yearsofage:diet,exercise,healthproblems.

10. First,thetubwas lledwithwatertoaheightof15 in.Thenapersongotintothetub,raisingthewaterlevelto20 in.At around12minutes,thepersonstoodupinthetubbutthenimmediatelysatdown.Finally,ataround17minutes,thepersongot outofthetub,andthendrainedthewater.

11. Thewaterwillcooldownalmosttofreezingastheicemelts.Then,when theicehasmelted,thewaterwillslowlywarmuptoroomtemperature.

12. RunnerAwontherace,reachingthe nishlineat 100 metersinabout 15 seconds,followedbyrunnerBwithatimeofabout 19 seconds,andthenbyrunnerCwho nishedinaround 23 seconds.Binitiallyledtherace,followedbyC,andthenA. CthenpassedBtoleadforawhile.ThenApassed rstB,andthenpassedCtotaketheleadand nish rst.Finally, BpassedCto nishinsecondplace.Allthreerunnerscompletedtherace.

13. (a)Thepowerconsumptionat6 AM is 500MW whichisobtainedbyreadingthevalueofpower when =6 fromthe graph.At6 PM wereadthevalueof when =18 obtainingapproximately 730MW

(b)Theminimumpowerconsumptionisdeterminedby ndingthetimeforthelowestpointonthegraph, =4 or 4 AM.The maximumpowerconsumptioncorrespondstothehighestpointonthegraph,whichoccursjustbefore =12 orright beforenoon.Thesetimesarereasonable,consideringthepowerconsumptionschedulesofmostindividualsand businesses.

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14. Thesummersolstice(thelongestdayoftheyear)is aroundJune21,andthewintersolstice(theshortestday) isaroundDecember22.(Exchangethedatesforthe southernhemisphere.)

16. Thevalueofthecardecreasesfairlyrapidlyinitially,then somewhatlessrapidly.

15. Ofcourse,thisgraphdependsstronglyonthe geographicallocation!

18. Thetemperatureofthepiewouldincreaserapidly,level offtooventemperature,decreaserapidly,andthenlevel offtoroomtemperature.

17. Asthepriceincreases,theamountsolddecreases.

19.

20. (a)

(b)

(c)

(d)

21. (a)

(b)Fromthegraph,weestimatethenumberofUScell-phone subscriberstobeabout126millionin2001and207million in2005.

22. (a)

(b)Fromthegraphinpart(a),weestimatethetemperatureat 11:00 AM tobeabout81 F.

)

24. Asphericalballoonwithradius +1 hasvolume ( +1)= 4 3 ( +1)3 =

.Wewishto ndthe amountofairneededtoin atetheballoonfromaradiusof to +1.Hence,weneedto ndthedifference ( +1) ( )=

. 25. ( )=4+3 2 ,so (3+ )=4+3(3+ ) (3+ )2 =4+9+3 (9+6 + 2 )=4 3 2 , and (3+ ) (3) = (4 3 2 ) 4 = ( 3 ) = 3

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26. ( )= 3 ,so ( + )=( + )3 = 3 +3 2 +3 2 + 3 , and ( + ) ( ) = ( 3 +3 2 +3 2 + 3 ) 3 = (3 2 +3 + 2 ) =3 2 +3 + 2

27. ( ) ( ) = 1 1 = = ( ) = 1( ) ( ) = 1

28. ( ) (1) 1 =

= +1 ( +1)( 1) = ( 1) ( +1)( 1) = 1 +1

29. ( )=( +4) ( 2 9) isde nedforall exceptwhen 0= 2 9 0=( +3)( 3) = 3 or 3,sothe domainis { R | = 3 3} =( 3) ( 3 3) (3 ).

30. ( )=(2 3 5) ( 2 + 6) isde nedforall exceptwhen 0= 2 + 6 0=( +3)( 2) = 3 or 2,sothedomainis { R | = 3 2} =( 3) ( 3 2) (2 )

31. ( )= 3 2 1 isde nedforallrealnumbres.Infact 3 ( ),where ( ) isapolynomial,isde nedforallrealnumbers. Thus,thedomainis R or ( )

32. ( )= 3 2+ isde nedwhen 3 0 3 and 2+ 0 2.Thus,thedomainis 2 3,or [ 2 3]

33. ( )=1 4 2 5 isde nedwhen 2 5 0 ( 5) 0.Notethat 2 5 =0 sincethatwouldresultin divisionbyzero.Theexpression ( 5) ispositiveif 0 or 5.(SeeAppendixAformethodsforsolving inequalities.)Thus,thedomainis ( 0) (5 )

34. ( )= 4 2 .Now = 4 2 2 =4 2 2 + 2 =4,so thegraphisthetophalfofacircleofradius 2 withcenterattheorigin.Thedomain is | 4 2 0 = | 4 2 = { | 2 | |} =[ 2 2].Fromthegraph, therangeis 0 2,or [0 2]

35. ( )=2 0 4 isde nedforallrealnumbers,sothedomainis R, or ( ) Thegraphof isalinewithslope 0 4 and -intercept 2

36. ( )= 2 2 +1=( 1)2 isde nedforallrealnumbers,sothe domainis R,or ( ).Thegraphof isaparabolawithvertex (1 0).

NOT FOR SALE

37. ( )=2 + 2 isde nedforallrealnumbers,sothedomainis R,or ( ).Thegraphof isaparabolaopeningupwardsincethe coef cientof 2 ispositive.To ndthe -intercepts,let =0 andsolve for 0=2 + 2 = (2+ ) =0 or = 2.The -coordinateof thevertexishalfwaybetweenthe -intercepts,thatis,at = 1.Since ( 1)=2( 1)+( 1)2 = 2+1= 1,thevertexis ( 1 1)

38. ( )= 4 2 2 = (2+ )(2 ) 2 ,sofor =2, ( )=2+ .Thedomain is { | =2}.Sothegraphof isthesameasthegraphofthefunction ( )= +2 (aline)exceptfortheholeat (2 4)

39. ( )= 5 isde nedwhen 5 0 or 5,sothedomainis [5 )

Since = 5 2 = 5 = 2 +5,weseethat isthe tophalfofaparabola.

40. ( )= |2 +1| = 2 +1 (2 +1) if 2 +1 0 if 2 +1 1 = 2 +1 2 1 if 1 2 if 1 2

Thedomainis R,or ( )

41. ( )= 3 + | | .Since | | = if 0 if 0 ,wehave ( )=

Notethat isnotde nedfor =0.Thedomainis ( 0) (0 )

42. ( )= | | = if 0 if 0 = 0 if 0 2 if 0 .

Thedomainis R,or ( )

43. ( )= +2 if 0 1 if 0

Thedomainis R.

44. ( )= 3 1 2 if 2 2 5 if 2

Thedomainis R.

45. ( )= +2 if 1 2 if 1

Notethatfor = 1,both +2 and 2 areequalto1.Thedomainis R

46. ( )= +9 if 3 2 if | | 3 6 if 3

Notethatfor = 3,both +9 and 2 areequalto 6;andfor =3,both 2 and 6 areequalto 6.Thedomainis R.

47. Recallthattheslope ofalinebetweenthetwopoints ( 1 1 ) and ( 2 2 ) is = 2 1 2 1 andanequationoftheline connectingthosetwopointsis 1 = ( 1 ).Theslopeofthelinesegmentjoiningthepoints (1 3) and (5 7) is 7 ( 3) 5 1 = 5 2 ,soanequationis ( 3)= 5 2 ( 1).Thefunctionis ( )= 5 2 11 2 , 1 5

48. Theslopeofthelinesegmentjoiningthepoints ( 5 10) and (7 10) is 10 10 7 ( 5) = 5 3 ,soanequationis 10= 5 3 [ ( 5)].Thefunctionis ( )= 5 3 + 5 3 , 5 7

49. Weneedtosolvethegivenequationfor . +( 1)2 =0 ( 1)2 = 1= ± =1 ± .Theexpressionwiththepositiveradicalrepresentsthetophalfoftheparabola,andtheonewiththenegative radicalrepresentsthebottomhalf.Hence,wewant ( )=1 .Notethatthedomainis 0

NOT FOR SALE

50. 2 +( 2)2 =4 ( 2)2 =4 2 2= ± 4 2 =2 ± 4 2 .Thetophalfisgivenby thefunction ( )=2+ 4 2 , 2 2.

51. For 0 3,thegraphisthelinewithslope 1 and -intercept 3,thatis, = +3.For 3 5,thegraphistheline withslope 2 passingthrough (3 0);thatis, 0=2( 3),or =2 6.Sothefunctionis ( )= +3 if 0 3 2 6 if 3 5

52. For 4 2,thegraphisthelinewithslope 3 2 passingthrough ( 2 0);thatis, 0= 3 2 [ ( 2)],or = 3 2 3.For 2 2,thegraphisthetophalfofthecirclewithcenter (0 0) andradius 2.Anequationofthecircle is 2 + 2 =4,soanequationofthetophalfis = 4 2 .For 2 4,thegraphisthelinewithslope 3 2 passing through (2 0);thatis, 0= 3 2 ( 2),or = 3 2 3.Sothefunctionis ( )= 3 2 3 if 4 2

53. Letthelengthandwidthoftherectanglebe and .Thentheperimeteris 2 +2 =20 andtheareais = . Solvingthe rstequationfor intermsof gives = 20 2 2 =10 .Thus, ( )= (10 )=10 2 .Since lengthsarepositive,thedomainof is 0 10.Ifwefurtherrestrict tobelargerthan ,then 5 10 wouldbe thedomain.

54. Letthelengthandwidthoftherectanglebe and .Thentheareais =16,sothat =16 .Theperimeteris =2 +2 ,so ( )=2 +2(16 )=2 +32 ,andthedomainof is 0,sincelengthsmustbepositive quantities.Ifwefurtherrestrict tobelargerthan ,then 4 wouldbethedomain.

55. Letthelengthofasideoftheequilateraltrianglebe .ThenbythePythagoreanTheorem,theheight ofthetrianglesatis es 2 + 1 2 2 = 2 ,sothat 2 = 2 1 4 2 = 3 4 2 and = 3 2 .Usingtheformulaforthearea ofatriangle, = 1 2 (base)(height),weobtain ( )= 1 2 ( ) 3 2 = 3 4 2 ,withdomain 0

56. Letthevolumeofthecubebe andthelengthofanedgebe .Then = 3 so = 3 ,andthesurfaceareais ( )=6 3 2 =6 2 3 ,withdomain 0.

57. Leteachsideofthebaseoftheboxhavelength ,andlettheheightoftheboxbe .Sincethevolumeis 2,weknowthat 2= 2 ,sothat =2 2 ,andthesurfaceareais = 2 +4 .Thus, ( )= 2 +4 (2 2 )= 2 +(8 ),with domain 0.

NOT FOR SALE

58. Theareaofthewindowis

,where istheheightoftherectangularportionofthewindow.

Sincethelengths and mustbepositivequantities,wehave 0 and 0.For 0,wehave 2 0

59. Theheightoftheboxis andthelengthandwidthare =20 2 , =12 2 .Then = andso ( )=(20 2 )(12 2 )( )=4(10 )(6 )(

Thesides , ,and mustbepositive.Thus, 0

;and

60. Forthe rst 1200kWh, ( )=10+0 06

.Combiningtheserestrictionsgivesusthedomain 0 6

Forusageover 1200kWh,thecostis ( )=10+0 06(1200)+0 07( 1200)=82+0 07( 1200)

Thus, ( )= 10+0 06 if 0 1200 82+0 07( 1200) if 1200

61. (a)

(b)On $14,000,taxisassessedon $4000,and 10%($4000)=$400 On $26,000,taxisassessedon $16,000,and 10%($10,000)+15%($6000)=$1000+$900=$1900

(c)Asinpart(b),thereis$1000taxassessedon$20,000ofincome,so thegraphof isalinesegmentfrom (10,000 0) to (20,000 1000)

Thetaxon$30,000is$2500,sothegraphof for 20,000 is theraywithinitialpoint (20,000 1000) thatpassesthrough (30,000 2500).

62. Oneexampleistheamountpaidforcableortelephonesystemrepairinthehome,usuallymeasuredtothenearestquarterhour. Anotherexampleistheamountpaidbyastudentintuitionfees,ifthefeesvaryaccordingtothenumberofcreditsforwhich thestudenthasregistered.

63. isanoddfunctionbecauseitsgraphissymmetricabouttheorigin. isanevenfunctionbecauseitsgraphissymmetricwith respecttothe -axis.

64. isnotanevenfunctionsinceitisnotsymmetricwithrespecttothe -axis. isnotanoddfunctionsinceitisnotsymmetric abouttheorigin.Hence, is neither evennorodd. isanevenfunctionbecauseitsgraphissymmetricwithrespecttothe -axis.

65. (a)Becauseanevenfunctionissymmetricwithrespecttothe -axis,andthepoint (5 3) isonthegraphofthisevenfunction, thepoint ( 5 3) mustalsobeonitsgraph.

(b)Becauseanoddfunctionissymmetricwithrespecttotheorigin,andthepoint (5 3) isonthegraphofthisoddfunction, thepoint ( 5 3) mustalsobeonitsgraph.

66. (a)If iseven,wegettherestofthegraphbyre ecting aboutthe -axis.

(b)If isodd,wegettherestofthegraphbyrotating 180 abouttheorigin.

NOT FOR SALE

69. ( )= +1 ,so ( )= +1 = 1

Sincethisisneither ( ) nor ( ),thefunction is neitherevennorodd.

71. ( )=1+3 2 4 .

( )=1+3( )2 ( )4 =1+3 2 4 = ( ).

So isanevenfunction.

70. ( )= | | ( )=( ) | | =( ) | | = ( | |) = ( )

So isanoddfunction.

72. ( )=1+3 3 5 ,so ( )=1+3( )3 ( )5 =1+3( 3 ) ( 5 ) =1 3 3 + 5

Sincethisisneither ( ) nor ( ),thefunction is neitherevennorodd.

73. (i)If and arebothevenfunctions,then ( )= ( ) and ( )= ( ).Now

( + )( )= ( )+ ( )= ( )+ ( )=( + )( ),so + isan even function.

(ii)If and arebothoddfunctions,then ( )= ( ) and ( )= ( ).Now

( + )( )= ( )+ ( )= ( )+[ ( )]= [ ( )+ ( )] = ( + )( ),so + isan odd function.

(iii)If isanevenfunctionand isanoddfunction,then ( + )( )= ( )+ ( )= ( )+[ ( )]= ( ) ( ), whichisnot ( + )( ) nor ( + )( ),so + is neither evennorodd.(Exception:if isthezerofunction,then + willbe odd.If isthezerofunction,then + willbe even.)

74. (i)If and arebothevenfunctions,then ( )= ( ) and ( )= ( ).Now ( )( )= ( ) ( )= ( ) ( )=( )( ),so isan even function.

(ii)If and arebothoddfunctions,then ( )= ( ) and ( )= ( ).Now ( )( )= ( ) ( )=[ ( )][ ( )]= ( ) ( )= ( )( ),so isan even function.

(iii)If isanevenfunctionand isanoddfunction,then

( )( )= ( ) ( )= ( )[ ( )]= [ ( ) ( )]= ( )( ),so isan odd function.

NOT FOR SALE

1.2MathematicalModels:A CatalogofEssentialFunctions

1. (a) ( )=log 2 isalogarithmicfunction.

(b) ( )= 4 isarootfunctionwith =4

(d) ( )=1 1 1 +2 54 2 isapolynomialofdegree 2 (alsocalleda quadraticfunction).

(e) ( )=5 isanexponentialfunction.

(f) ( )=sin cos 2 isatrigonometricfunction.

2. (a) = isanexponentialfunction(noticethat isthe exponent).

(b) = isapowerfunction(noticethat isthe base).

(d) =tan cos isatrigonometricfunction.

(e) = (1+ ) isarationalfunctionbecauseitisaratioofpolynomials.

(f) = 3 1 (1+ 3 ) isanalgebraicfunctionbecauseitinvolvespolynomialsandrootsofpolynomials.

3. Wenoticefromthe gurethat and areevenfunctions(symmetricwithrespecttothe -axis)andthat isanoddfunction (symmetricwithrespecttotheorigin).So(b) = 5 mustbe .Since is atterthan neartheorigin,wemusthave

4. (a)Thegraphof =3 isaline(choice ).

(b) =3 isanexponentialfunction(choice ).

(d) = 3 = 1 3 isarootfunction(choice ).

5. (a)Anequationforthefamilyoflinearfunctionswithslope 2 is = ( )=2 + ,where isthe -intercept.

NOT FOR SALE

(b) (2)=1 meansthatthepoint (2 1) isonthegraphof .Wecanusethe point-slopeformofalinetoobtainanequationforthefamilyoflinear functionsthroughthepoint (2 1) 1= ( 2),whichisequivalent to = +(1 2 ) inslope-interceptform.

(c)Tobelongtobothfamilies,anequationmusthaveslope =2,sotheequationinpart(b), = +(1 2 ), becomes =2 3.Itisthe only functionthatbelongstobothfamilies.

6. Allmembersofthefamilyoflinearfunctions ( )=1+ ( +3) have graphsthatarelinespassingthroughthepoint ( 3 1).

7. Allmembersofthefamilyoflinearfunctions ( )= havegraphs thatarelineswithslope 1.The -interceptis

8. Thevertexoftheparabolaontheleftis (3 0),soanequationis = ( 3)2 +0.Sincethepoint (4 2) isonthe parabola,we’llsubstitute 4 for and 2 for to nd . 2= (4 3)2 =2,soanequationis ( )=2( 3)2 . The -interceptoftheparabolaontherightis (0 1),soanequationis = 2 + +1.Sincethepoints ( 2 2) and (1 2 5) areontheparabola,we’llsubstitute 2 for and 2 for aswellas 1 for and 2 5 for toobtaintwoequations withtheunknowns and ( 2 2): 2=4 2 +1 4 2 =1 (1) (1 2 5): 2 5= + +1 + = 3 5 (2) 2 · (2) + (1) givesus 6 = 6 = 1.From (2), 1+ = 3 5 = 2 5,soanequation is ( )= 2 2 5 +1.

9. Since ( 1)= (0)= (2)=0, haszerosof 1, 0,and 2,soanequationfor is ( )= [ ( 1)]( 0)( 2), or ( )= ( +1)( 2).Because (1)=6,we’llsubstitute 1 for and 6 for ( ). 6= (1)(2)( 1) 2 =6 = 3,soanequationfor is ( )= 3 ( +1)( 2)

NOT FOR SALE

10. (a)For =0 02 +8 50,theslopeis 0 02,whichmeansthattheaveragesurfacetemperatureoftheworldisincreasingata rateof 0 02 C peryear.The -interceptis 8 50,whichrepresentstheaveragesurfacetemperaturein C intheyear1900.

(b) =2100 1900=200 =0 02(200)+8 50=12 50 C

11. (a) =200,so =0 0417 ( +1)=0 0417(200)( +1)=8 34 +8 34.Theslopeis 8 34,whichrepresentsthe changeinmgofthedosageforachildforeachchangeof1yearinage.

(b)Foranewborn, =0,so =8 34 mg.

12. (a)

13. (a)

(b)Theslopeof 4 meansthatforeachincreaseof 1 dollarfora rentalspace,thenumberofspacesrented decreases by 4.The -interceptof 200 isthenumberofspacesthatwouldbeoccupied iftherewerenochargeforeachspace.The -interceptof 50 isthe smallestrentalfeethatresultsinnospacesrented.

(b)Theslopeof 9 5 meansthat increases 9 5 degreesforeachincrease of 1 C.(Equivalently, increasesby 9 when increasesby 5 and decreasesby 9 when decreasesby 5.)The -interceptof 32 istheFahrenheittemperaturecorrespondingtoaCelsius temperatureof 0

14. (a)Let = distancetraveled(inmiles)and = timeelapsed(inhours).At =0, =0 andat =50 minutes =50 · 1 60 = 5 6 h, =40.Thuswe havetwopoints: (0 0) and 5 6 40 ,so = 40 0 5 6 0 =48 andso =48 . (b)

(c)Theslopeis 48 andrepresentsthecar’sspeedinmi h.

15. (a)Using inplaceof and inplaceof ,we ndtheslopetobe

.Soalinear equationis

(b)Theslopeof 1 6 meansthatthetemperatureinFahrenheitdegreesincreasesone-sixthasrapidlyasthenumberofcricket chirpsperminute.Saiddifferently,eachincreaseof 6 cricketchirpsperminutecorrespondstoanincreaseof 1 F.

(c)When =150,thetemperatureisgivenapproximatelyby = 1 6 (150)+ 307 6 =76 16 F 76 F

NOT FOR SALE

16. (a)Let denotethenumberofchairsproducedinonedayand theassociated cost.Usingthepoints (100 2200) and (300 4800),wegettheslope

4800 2200 300 100 = 2600 200 =13.So 2200=13( 100) =13 +900

(b)Theslopeofthelineinpart(a)is 13 anditrepresentsthecost(indollars) ofproducingeachadditionalchair.

(c)The -interceptis 900 anditrepresentsthe xeddailycostsofoperating thefactory.

17. (a)Wearegiven changeinpressure 10 feetchangeindepth = 4 34 10 =0 434.Using forpressureand fordepthwiththepoint ( )=(0 15),wehavetheslope-interceptformoftheline, =0 434 +15

(b)When =100,then 100=0 434 +15 0 434 =85 = 85 0 434 195 85 feet.Thus,thepressureis 100lb in2 atadepthofapproximately 196 feet.

18. (a)Using inplaceof and inplaceof ,we ndtheslopetobe

Soalinearequationis 460= 1 4 ( 800)

(b)Letting =1500 weget = 1 4 (1500)+260=635

Thecostofdriving1500milesis$635.

(d)The -interceptrepresentsthe xedcost,$260.

(e)Alinearfunctiongivesasuitablemodelinthissituationbecauseyouhave xedmonthlycostssuchasinsuranceandcar payments,aswellascoststhatincreaseasyoudrive,suchasgasoline,oil,andtires,andthecostoftheseforeach additionalmiledrivenisaconstant.

19. (a)Thedataappeartobeperiodicandasineorcosinefunctionwouldmakethebestmodel.Amodeloftheform ( )= cos( )+ seemsappropriate.

(b)Thedataappeartobedecreasinginalinearfashion.Amodeloftheform ( )= + seemsappropriate.

20. (a)Thedataappeartobeincreasingexponentially.Amodeloftheform ( )= or ( )= + seemsappropriate.

(b)Thedataappeartobedecreasingsimilarlytothevaluesofthereciprocalfunction.Amodeloftheform ( )= seems appropriate.

INSTRUCTOR USE ONLY

NOT FOR SALE

Exercises21 – 24:Somevaluesaregiventomanydecimalplaces.Thesearetheresultsgivenbyseveralcomputeralgebrasystems roundingisleft tothereader.

21. (a)

(b)Usingthepoints (4000 14 1) and (60,000 8 2),weobtain

14 1= 8 2 14 1 60,000 4000 ( 4000) or,equivalently, 0 000105357 +14 521429

Alinearmodeldoesseemappropriate.

(c)Usingacomputingdevice,weobtaintheleastsquaresregressionline = 0 0000997855 +13 950764

Thefollowingcommandsandscreensillustratehowto ndtheleastsquaresregressionlineonaTI-84Plus. Enterthedataintolistone(L1)andlisttwo(L2).Press toentertheeditor.

FindtheregessionlineandstoreitinY1 .Press .

Notefromthelast gurethattheregressionlinehasbeenstoredinY1 andthatPlot1hasbeenturnedon(Plot1is highlighted).YoucanturnonPlot1fromtheY=menubyplacingthecursoronPlot1andpressing orby pressing

Nowpress toproduceagraphofthedataandtheregression line.Notethatchoice9oftheZOOMmenuautomaticallyselectsawindow thatdisplaysallofthedata.

(d)When =25,000, 11 456;orabout 11 5 per 100 population.

NOT FOR SALE

(e)When =80,000, 5 968;orabouta 6% chance.

(f)When =200,000, isnegative,sothemodeldoesnotapply.

22. (a) (b)

(c)When =100 F, =264 7 265 chirps min.

23. (a)Alinearmodelseemsappropriateoverthetimeinterval considered.

Usingacomputingdevice,weobtaintheleastsquares regressionline =4 856 220 96

(b)Usingacomputingdevice,weobtaintheleastsquares regressionline 0 027 47 758

(c)When =2004, =6 35,whichishigherthantheactualwinningheightof 5 95m

(d)No,sincethetimesappeartobelevelingoffandgettingfurtherawayfromthemodel.

24. Bylookingatthescatterplotofthedata,weruleout thepowerandlogarithmicmodels.

Wetryvariousmodels:

Linear = 0 4305454545 +870 1836364

Quadratic: =0 0048939394 2 19 78607576 +20006 95485

Scatterplot

Cubic: = 0 00007319347 3 +0 4391142191 2 878 4298718 +585960 983

Quartic: =0 0000079020979 4 0 0625787879 3 +185 8422838 2 245290 9304 +121409472 7

Exponential: =2 6182302 × 1021 (0 9767893094) [continued]

25.

Linearmodel

NOT FOR SALE

Quarticmodel Exponentialmodel

Cubicmodel

Afterexaminingthegraphsofthese models,weseethatallthemodels aregoodandthequarticmodelis thebest.

Usingthismodel,weobtainestimates 13 6% and 10 2% fortheruralpercentagesin1988and2002respectively.

Usingacomputingdevice,weobtainthecubicfunction

= 3 + 2 + + with =0 0012937, = 7 06142, =12,823,and = 7,743,770.When =1925, 1914 (million).

26. (a) =1 000431227 1 499528750

(b)Thepowermodelinpart(a)isapproximately = 1 5 .Squaringbothsidesgivesus 2 = 3 ,sothemodelmatches

Kepler’sThirdLaw, 2 = 3

1.3NewFunctionsfromOldFunctions

1. (a)Ifthegraphof isshifted 3 unitsupward,itsequationbecomes = ( )+3

(b)Ifthegraphof isshifted 3 unitsdownward,itsequationbecomes = ( ) 3

(c)Ifthegraphof isshifted 3 unitstotheright,itsequationbecomes = ( 3).

(d)Ifthegraphof isshifted 3 unitstotheleft,itsequationbecomes = ( +3).

(e)Ifthegraphof isre ectedaboutthe -axis,itsequationbecomes = ( )

(f)Ifthegraphof isre ectedaboutthe -axis,itsequationbecomes = ( )

(g)Ifthegraphof isstretchedverticallybyafactorof 3,itsequationbecomes =3 ( )

(h)Ifthegraphof isshrunkverticallybyafactorof 3,itsequationbecomes = 1 3 ( )

INSTRUCTOR USE ONLY

NOT FOR SALE

2. (a)Toobtainthegraphof = ( )+8 fromthegraphof = ( ),shiftthegraph 8 unitsupward.

(b)Toobtainthegraphof = ( +8) fromthegraphof = ( ),shiftthegraph 8 unitstotheleft.

(c)Toobtainthegraphof =8 ( ) fromthegraphof = ( ),stretchthegraphverticallybyafactorof 8

(d)Toobtainthegraphof = (8 ) fromthegraphof = ( ),shrinkthegraphhorizontallybyafactorof 8

(e)Toobtainthegraphof = ( ) 1 fromthegraphof = ( ), rstre ectthegraphaboutthe -axis,andthenshiftit 1 unitdownward.

(f)Toobtainthegraphof =8 ( 1 8 ) fromthegraphof = ( ),stretchthegraphhorizontallyandverticallybyafactor of 8.

3. (a)(graph3)Thegraphof isshifted 4 unitstotherightandhasequation = ( 4).

(b)(graph1)Thegraphof isshifted 3 unitsupwardandhasequation = ( )+3

(c)(graph4)Thegraphof isshrunkverticallybyafactorof 3 andhasequation = 1 3 ( ).

(d)(graph5)Thegraphof isshifted 4 unitstotheleftandre ectedaboutthe -axis.Itsequationis = ( +4)

(e)(graph2)Thegraphof isshifted 6 unitstotheleftandstretchedverticallybyafactorof 2.Itsequationis =2 ( +6).

4. (a)Tograph = ( ) 2,weshiftthegraphof , 2 unitsdownward.Thepoint (1 2) onthegraphof correspondstothepoint (1 2 2)=(1 0).

(b)Tograph = ( 2),weshiftthegraphof , 2 unitstotheright.Thepoint (1 2) onthegraphof correspondstothepoint (1+2 2)=(3 2).

(c)Tograph = 2 ( ),were ectthegraphaboutthe -axisandstretchthegraphverticallybyafactorof 2 Thepoint (1 2) onthegraphof correspondstothe point (1 2 2)=(1 4).

(d)Tograph = ( 1 3 )+1,westretchthegraph horizontallybyafactorof 3 andshiftit 1 unitupward. Thepoint (1 2) onthegraphof correspondstothe point (1 3 2+1)=(3 3).

NOT FOR SALE

5. (a)Tograph = (2 ) weshrinkthegraphof horizontallybyafactorof 2

(b)Tograph = 1 2 westretchthegraphof horizontallybyafactorof 2.

Thepoint (4 1) onthegraphof correspondstothe point 1 2 · 4 1 =(2 1)

(c)Tograph = ( ) were ectthegraphof about the -axis.

Thepoint (4 1) onthegraphof correspondstothe point (2 · 4 1)=(8 1)

(d)Tograph = ( ) were ectthegraphof about the -axis,thenaboutthe -axis.

Thepoint (4 1) onthegraphof correspondstothe point ( 1 · 4 1)=( 4 1)

Thepoint (4 1) onthegraphof correspondstothe point ( 1 4 1 1)=( 4 1)

6. Thegraphof = ( )= 3 2 hasbeenshifted 2 unitstotherightandstretchedverticallybyafactorof 2. Thus,afunctiondescribingthegraphis

7. Thegraphof = ( )= 3 2 hasbeenshifted 4 unitstotheleft,re ectedaboutthe -axis,andshifteddownward 1 unit.Thus,afunctiondescribingthegraphis

Thisfunctioncanbewrittenas

8. (a)Thegraphof =2sin canbeobtainedfromthegraph of =sin bystretchingitverticallybyafactorof 2

(b)Thegraphof =1+ canbeobtainedfrom thegraphof = byshiftingitupward 1 unit.

NOT FOR SALE

9. = 3 :Startwiththegraphof = 3 andre ectaboutthe -axis.Note:Re ectingaboutthe -axisgivesthesameresult sincesubstituting for givesus =( )3 = 3 .

10. =1 2 = 2 +1:Startwiththegraphof = 2 ,re ectaboutthe -axis,andthenshift 1 unitupward.

11. =( +1)2 :Startwiththegraphof = 2 andshift 1 unittotheleft.

12. = 2 4 +3=( 2 4 +4) 1=( 2)2 1:Startwiththegraphof = 2 ,shift 2 unitstotheright, andthenshift 1 unitdownward.

13. =1+2cos :Startwiththegraphof =cos ,stretchverticallybyafactorof 2,andthenshift 1 unitupward.

14. =4sin3 :Startwiththegraphof =sin ,compresshorizontallybyafactorof 3,andthenstretchverticallybya factorof 4.

15. =sin( 2):Startwiththegraphof =sin andstretchhorizontallybyafactorof 2

16. =1 ( 4):Startwiththegraphof =1 andshift 4 unitstotheright.

17. = +3 :Startwiththegraphof = andshift 3 unitstotheleft. 18. = | | 2:Startwiththegraphof = | | andshift 2 unitsdownward.

NOT FOR SALE

19. = 1 2 ( 2 +8 )= 1 2 ( 2 +8 +16) 8= 1 2 ( +4)2 8:Startwiththegraphof = 2 ,compressverticallybya factorof 2,shift 4 unitstotheleft,andthenshift 8 unitsdownward. 0 0 0 0

20. =1+ 3 1 :Startwiththegraphof = 3 ,shift 1 unittotheright,andthenshift 1 unitupward.

21. = | 2|:Startwiththegraphof = | | andshift 2 unitstotheright.

22. = 1 4 tan( 4 ):Startwiththegraphof =tan ,shift 4 unitstotheright,andthencompressverticallybyafactorof 4

23. = | 1|:Startwiththegraphof = ,shiftit 1 unitdownward,andthenre ecttheportionofthegraphbelowthe -axisaboutthe -axis.

NOT FOR SALE

24. = |cos |:Startwiththegraphof =cos ,shrinkithorizontallybyafactorof ,andre ectallthepartsofthegraph belowthe -axisaboutthe -axis.

25. ThisisjustlikethesolutiontoExample4excepttheamplitudeofthecurve(the30 NcurveinFigure9onJune21)is

14 12=2.Sothefunctionis ( )=12+2sin 2 365 ( 80) .March31isthe 90thdayoftheyear,sothemodelgives (90) 12 34 h.Thedaylighttime(5:51 AM to6:18 PM)is 12 hoursand 27 minutes,or 12 45 h.Themodelvaluediffers fromtheactualvalueby 12 45 12 34 12 45 0 009,lessthan 1%.

26. UsingasinefunctiontomodelthebrightnessofDeltaCepheiasafunctionoftime,wetakeitsperiodtobe 5 4 days,its amplitudetobe 0 35 (onthescaleofmagnitude),anditsaveragemagnitudetobe 4 0.Ifwetake =0 atatimeofaverage brightness,thenthemagnitude(brightness)asafunctionoftime indayscanbemodeledbytheformula

( )=4 0+0 35sin 2 5 4

28. Themostimportantfeaturesofthegivengrapharethe -interceptsandthemaximum andminimumpoints.Thegraphof =1 ( ) hasverticalasymptotesatthe -values wherethereare -interceptsonthegraphof = ( ).Themaximumof 1 onthegraph of = ( ) correspondstoaminimumof 1 1=1 on =1 ( ).Similarly,the minimumonthegraphof = ( ) correspondstoamaximumonthegraphof =1 ( ).Asthevaluesof getlarge(positivelyornegatively)onthegraphof = ( ),thevaluesof getclosetozeroonthegraphof =1 ( ).

29. ( )= 3 +2 2 ; ( )=3 2 1 = R forboth and (a) ( + )( )=( 3 +2 2 )+(3 2 1)= 3 +5

( )(

( )( )=(

( )= 3 +2 2 3 2 1 , = | = ± 1 3 since 3 2 1 =0.

NOT FOR SALE

30. ( )= 3 , =( 3]; ( )= 2 1, =( 1] [1 )

(a) ( + )( )= 3 + 2 1, =( 1] [1 3],whichistheintersectionofthedomainsof and (b) ( )( )= 3 2 1, =( 1] [1 3]

(d) ( )= 3 2 1 , =( 1) (1 3].Wemustexclude = ±1 sincethesevalueswouldmake unde ned.

31. ( )= 2 1, = R; ( )=2 +1, = R

(a) ( )( )= ( ( ))= (2 +1)=(2 +1)2 1=(4 2 +4 +1) 1=4 2 +4 , = R

(b) ( )( )= ( ( ))= ( 2 1)=2( 2 1)+1=(2 2 2)+1=2 2 1, = R

(d) ( )( )= ( ( ))= (2 +1)=2(2 +1)+1=(4 +2)+1=4 +3, = R

32. ( )= 2; ( )= 2 +3 +4. = R forboth and ,andhencefortheircomposites.

(a) ( )( )= ( ( ))= ( 2 +3 +4)=( 2 +3 +4) 2= 2 +3 +2 (b) ( )( )= ( ( ))= ( 2)=( 2)2 +3( 2)+4= 2 4 +4+3 6+4= 2 +2

(d) ( )( )= ( ( ))= ( 2 +3 +4)=( 2 +3 +4)2 +3(

33. ( )=1 3 ; ( )=cos = R forboth and ,andhencefortheircomposites.

(a) ( )( )= ( ( ))= (cos )=1 3cos .

(b) ( )( )= ( ( ))= (1 3 )=cos(1 3 ).

(d) ( )( )= ( ( ))= (cos )=cos(cos ) [Notethatthisis not cos cos .]

34. ( )= , =[0 ); ( )= 3 1 , = R.

(a) ( )( )= ( ( ))= 3 1 = 3 1 = 6 1 .

Thedomainof is { | 3 1 0} = { | 1 0} = { | 1} =( 1]

(b) ( )( )= ( ( ))= ( )= 3 1 .

Thedomainof is { | isinthedomainof and ( ) isinthedomainof }.Thisisthedomainof , thatis, [0 ).

(d) ( )( )= ( ( ))= 3 1 = 3 1 3 1 ,andthedomainis ( )

35. ( )= + 1 , = { | =0}; ( )= +1 +2 , = { | = 2}

(a) ( )( )= ( ( ))= +1

Since ( ) isnotde nedfor = 2 and ( ( )) isnotde nedfor = 2 and = 1, thedomainof( )( ) is = { | = 2 1}.

(b) ( )( )= ( ( ))= + 1 =

Since ( ) isnotde nedfor =0 and ( ( )) isnotde nedfor = 1, thedomainof ( )( ) is = { | = 1 0}

4 +3 2 +1 ( 2 +1) = { | =0}

(d) ( )( )= ( ( ))= +1 +2 = +1 +2 +1 +1 +2 +2 = +1+1( +2) +2 +1+2( +2) +2 = +1+ +2

Since ( ) isnotde nedfor = 2 and ( ( )) isnotde nedfor = 5 3 , thedomainof ( )( ) is = | = 2 5 3

36. ( )= 1+ , = { | = 1}; ( )=sin2 , = R.

(a) ( )( )= ( ( ))= (sin2 )= sin2 1+sin2

Domain: 1+sin2 =0 sin2 = 1 2 = 3 2 +2 = 3 4 + [ aninteger].

(b) ( )( )= ( ( ))= 1+ =sin 2 1+ .

Domain: { | = 1}

Since ( ) isnotde nedfor = 1,and ( ( )) isnotde nedfor = 1 2 , thedomainof ( )( ) is = { | = 1 1 2 }

(d) ( )( )= ( ( ))= (sin2 )=sin(2sin2 )

Domain: R

INSTRUCTOR USE ONLY

NOT FOR SALE

37. ( )( )= ( ( ( )))= ( ( 1))= (2( 1))=2( 1)+1=2 1

38. ( )( )= ( ( ( )))= ( (1 ))= ((1 )2 )=2(1 )2 1=2 2 4 +1

39. ( )( )= ( ( ( )))= ( ( 3 +2))= [( 3 +2)2 ] = ( 6 +4 3 +4)= ( 6 +4 3 +4) 3= 6

40. ( )( )= ( ( ( )))= ( ( 3 ))= 3 3 1 =tan 3 3 1

41. Let ( )=2 + 2 and ( )= 4 .Then ( )( )= ( ( ))= (2 + 2 )=(2 + 2 )4 = ( ).

42. Let ( )=cos and ( )= 2 .Then ( )( )= ( ( ))= (cos )=(cos )2 =cos 2 = ( ).

43. Let ( )= 3 and ( )= 1+ .Then ( )( )= ( ( ))= ( 3 )= 3 1+ 3 = ( )

44. Let ( )= 1+ and ( )= 3 .Then ( )( )= ( ( ))= 1+ = 3 1+ = ( )

45. Let ( )=cos and ( )= .Then ( )( )= ( ( ))= (cos )= cos = ( )

46. Let ( )=tan and ( )= 1+ .Then ( )( )= ( ( ))= (tan )= tan 1+tan = ( )

47. Let ( )= 2 , ( )=3 ,and ( )=1 .Then ( )( )= ( ( ( )))= ( ( 2 ))= 3 2 =1 3 2 = ( )

48. Let ( )= | |, ( )=2+ ,and ( )= 8 .Then ( )( )= ( ( ( )))= ( (| |))= (2+ | |)= 8 2+ | | = ( ).

49. Let ( )= , ( )=sec ,and ( )= 4 .Then ( )( )= ( ( ( )))= ( ( ))= (sec )=(sec )4 =sec 4 ( )= ( ).

50. (a) ( (1))= (6)=5

(b) ( (1))= (3)=2

(e) ( )(3)= ( (3))= (4)=1 (f) ( )(6)= ( (6))= (3)=4

51. (a) (2)=5,becausethepoint (2 5) isonthegraphof .Thus, ( (2))= (5)=4,becausethepoint (5 4) isonthe graphof .

(b) ( (0))= (0)=3

(d) ( )(6)= ( (6))= (6).Thisvalueisnotde ned,becausethereisnopointonthegraphof thathas -coordinate 6.

(e) ( )( 2)= ( ( 2))= (1)=4

(f) ( )(4)= ( (4))= (2)= 2

INSTRUCTOR

NOT FOR SALE

52. To ndaparticularvalueof ( ( )),sayfor =0,wenotefromthegraphthat (0) 2 8 and (2 8) 0 5.Thus, ( (0)) (2 8) 0 5.Theothervalueslistedinthetablewereobtainedinasimilarfashion.

( ) ( ( )) 5 0 2 4

53. (a)Usingtherelationship distance = rate · time withtheradius asthedistance,wehave ( )=60

(b) = 2 ( )( )= ( ( ))= (60 )2 =3600 2 .Thisformulagivesustheextentoftherippledarea (incm2 )atanytime

54. (a)Theradius oftheballoonisincreasingatarateof 2cm s,so ( )=(2cm s)( s)=2 (in cm).

(b)Using = 4 3 3 ,weget ( )( )= ( ( ))= (2 )= 4 3 (2 )3 = 32 3 3 Theresult, = 32 3 3 ,givesthevolumeoftheballoon(in cm 3 )asafunctionoftime(in s).

55. (a)Fromthe gure,wehavearighttrianglewithlegs 6 and ,andhypotenuse BythePythagoreanTheorem, 2 +62 = 2 = ( )= 2 +36

(b)Using = ,weget =(30 km h)( hours)=30 (inkm).Thus, = ( )=30

(c) ( )( )= ( ( ))= (30 )= (30 )2 +36= 900 2 +36.Thisfunctionrepresentsthedistancebetweenthe lighthouseandtheshipasafunctionofthetimeelapsedsincenoon.

56. (a) = ( )=350

(b)ThereisaPythagoreanrelationshipinvolvingthelegswithlengths and 1 andthehypotenusewithlength : 2 +12 = 2 .Thus, ( )= 2 +1

57. (a) ( )= 0 if 0 1 if 0 (b) ( )= 0 if 0 120 if 0 so ( )=120 ( )

(c) Startingwiththeformulainpart(b),wereplace 120 with 240 tore ectthe differentvoltage.Also,becausewearestarting 5 unitstotherightof =0, wereplace with 5.Thus,theformulais ( )=240 ( 5)

NOT FOR SALE

58. (a) ( )= ( ) = 0 if 0 if 0 (b) ( )= 0 if 0 2 if 0 60 so ( )=2 ( ), 60. (c) ( )= 0 if 7 4( 7) if 7 32 so ( )=4( 7) ( 7), 32.

59. If ( )= 1 + 1 and ( )= 2 + 2 ,then

( )( )= ( ( ))= ( 2 + 2 )= 1 ( 2 + 2 )+ 1 = 1 2 + 1 2 + 1

So isalinearfunctionwithslope 1 2

60. If ( )=1 04 ,then

( )( )= ( ( ))= (1 04 )=1 04(1 04 )=(1 04)2 , ( )( )= (( )( ))= ((1 04)2 )=1 04(1 04)2 =(1 04)3 ,and ( )( )= (( )( ))= ((1 04)3 )=1 04(1 04)3 =(1 04)4

Thesecompositionsrepresenttheamountoftheinvestmentafter2,3,and4years. Basedonthispattern,whenwecompose copiesof ,wegettheformula ( ··· ) 0 s ( )=(1 04)

61. (a)Byexaminingthevariabletermsin and ,wededucethatwemustsquare togettheterms 4 2 and 4 in .Ifwelet ( )= 2 + ,then ( )( )= ( ( ))= (2 +1)=(2 +1)2 + =4 2 +4 +(1+ ).Since ( )=4 2 +4 +7,wemusthave 1+ =7.So =6 and ( )= 2 +6.

(b)Weneedafunction sothat ( ( ))=3( ( ))+5= ( ).But ( )=3 2 +3 +2=3( 2 + )+2=3( 2 + 1)+5,soweseethat ( )= 2 + 1

62. Weneedafunction sothat ( ( ))= ( +4)= ( )=4 1=4( +4) 17.Soweseethatthefunction mustbe ( )=4 17

63. Weneedtoexamine ( ). ( )=( )( )= ( ( ))= ( ( )) [because iseven] = ( ) Because ( )= ( ), isanevenfunction.

64. ( )= ( ( ))= ( ( )).Atthispoint,wecan’tsimplifytheexpression,sowemighttryto ndacounterexampleto showthat isnotanoddfunction.Let ( )= ,anoddfunction,and ( )= 2 + .Then ( )= 2 + whichisneither evennorodd.

Nowsuppose isanoddfunction.Then ( ( ))= ( ( ))= ( ).Hence, ( )= ( ),andso isoddif both and areodd.

Nowsuppose isanevenfunction.Then ( ( ))= ( ( ))= ( ).Hence, ( )= ( ),andso isevenif is oddand iseven.

INSTRUCTOR

NOT FOR SALE

1.4GraphingCalculatorsandComputers

1. ( )= 3 5 2

(a) [ 5 5] by [ 5 5] (Thereisnographshown.)

(b) [0 10] by [0 2] (c) [0 10] by [0 10]

Themostappropriategraphisproducedinviewingrectangle(c).

2. ( )= 4 16 2 +20

(a) [ 3 3] by [ 3 3]

(b) [ 10 10] by [ 10 10]

(d) [ 5 5] by [ 50 50]

Themostappropriategraphisproducedinviewingrectangle(d).

3. Sincethegraphof ( )= 2 36 +32 isaparabolaopeningupward, anappropriateviewingrectangleshouldincludetheminimumpoint.

Completingthesquare,weget ( )=( 18)2 292,andsothe minimumpointis (18 292)

4. Anappropriateviewingrectanglefor ( )= 3 +15 2 +65 should includethehighandlowpoints.

5. ( )= 4 81 4 isde nedwhen 81 4 0 4 81 | | 3,sothedomainof is [ 3 3].Also 0 4 81 4 4 81=3, sotherangeis [0 3]

6. ( )= 0 1 +20 isde nedwhen 0 1 +20 0 200, sothedomainof is [ 200 ).

7. Thegraphof ( )= 3 225 issymmetricwithrespecttotheorigin.

Since ( )= 3 225 = ( 2 225)= ( +15)( 15),there are -interceptsat 0, 15,and 15. (20)=3500.

8. Thegraphof ( )= ( 2 +100) issymmetricwithrespecttothe origin.

9. Theperiodof ( )=sin(1000 ) is 2 1000 0 0063 anditsrangeis [ 1 1].Since ( )=sin2 (1000 ) isthesquareof ,itsrangeis [0 1] andaviewingrectangleof [ 0 01 0 01] by [0 1 1] seems appropriate.

10. Theperiodof ( )=cos(0 001 ) is 2 0 001 6300 anditsrange is [ 1 1],soaviewingrectangleof [ 10,000 10,000] by [ 1 5 1 5] seemsappropriate.

11. Thedomainof = is 0,sothedomainof ( )=sin is [0 ) andtherangeis [ 1 1].Withalittletrial-and-errorexperimentation,we nd thatanXmaxof100illustratesthegeneralshapeof ,soanappropriate viewingrectangleis [0 100] by [ 1 5 1 5]

NOT FOR SALE

12. Oneperiodof =sec occursontheinterval 2 2 2 3 2 2 20 3 2 1 40 3 40 ,orequivalently, 0 025 0 075.

13. The rstterm, 10sin ,hasperiod 2 andrange [ 10 10].Itwillbethedominantterminany“large”graphof =10sin +sin100 ,asshowninthe rst gure.Thesecondterm, sin100 ,hasperiod 2 100 = 50 andrange [ 1 1]

Itcausesthebumpsinthe rst gureandwillbethedominantterminany“small”graph,asshownintheviewnearthe origininthesecond gure.

14. = 2 +0 02sin(50 )

15. (a)The rst gureshowsthe"big picture"for ( )=( 10)3 2 . Thesecond gureshowsamaximum near =10

(b)Youneedmorethanonewindowbecausenosinglewindowcanshowwhatthefunctionlookslikeglobally andthedetailofthefunctionnear =10

16. Thefunction ( )= 2 30 hasdomain ( 30].Itsgraphisvery steepnear =30,sopartofthegraphmayappeartobemissing.

INSTRUCTOR

NOT FOR SALE

COMPLETE SOLUTIONS MANUAL for Stewart’s

MULTIVARIABLE CALCULUS: CONCEPTS AND CONTEXTS

FOURTH EDITION

2010 Brooks/Cole, Cengage Learning

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ISBN-13: 978-0-495-56056-2

ISBN-10: 0-495-56056-1

Brooks/Cole 10 Davis Drive Belmont, CA 94002-3098 USA

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PREFACE

This Complete Solutions Manual contains detailed solutions to all exercises in the text Multivariable Calculus:Concepts and Contexts, Fourth Edition (Chapters 8–13 of Calculus: Concepts and Contexts, Fourth Edition) by James Stewart. A Student Solutions Manual is also available,which contains solutions to the odd-numbered exercises in each chapter section,review section,True-False Quiz,and Focus on Problem Solving section as well as all solutions to the Concept Check questions. (It does not,however,include solutions to any of the projects.)

While I have extended every effort to ensure the accuracy of the solutions presented,I would appreciate correspondence regarding any errors that may exist. Other suggestions or comments are also welcome,and can be sent to me at the email address or mailing address below.

I would like to thank James Stewart for entrusting me with the writing of this manual and offering suggestions,Kathi Townes,Stephanie Kuhns,and Rebekah Steele of TECH-arts for typesetting and producing this manual,and Brian Betsill of TECH-arts for creating the illustrations. Brian Karasek prepared solutions for comparison of accuracy and style in addition to proofreading manuscript; his assistance and suggestions were very helpful and much appreciated. Finally, I would like to thank Richard Stratton and Elizabeth Neustaetter of Brooks/Cole,Cengage Learning for their trust,assistance,and patience.

DANCLEGG

dclegg@palomar.edu

Palomar College Department of Mathematics 1140 West Mission Road San Marcos,CA 92069

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8 ■ INFINITE SEQUENCES AND SERIES 1

8.1 Sequences1

Laboratory Project ■ Logistic Sequences9

8.2 Series13

8.3 The Integral and Comparison Tests; Estimating Sums26

8.4 Other Convergence Tests32

8.5 Power Series39

8.6 Representations of Functions as Power Series46

8.7 Taylor and Maclaurin Series55

Laboratory Project ■ An Elusive Limit69

8.8 Applications of Taylor Polynomials70

Applied Project ■ Radiation from the Stars81 Review83

Focus on Problem Solving95

9 ■ VECTORS AND THE GEOMETRY OF SPACE101

9.1 Three-Dimensional Coordinate Systems101

9.2 Vectors108

9.3 The Dot Product115

9.4 The Cross Product122

Discovery Project ■ The Geometry of a Tetrahedron130

9.5 Equations of Lines and Planes132

Laboratory Project ■ Putting 3D in Perspective141

9.6 Functions and Surfaces143

9.7 Cylindrical and Spherical Coordinates151

Laboratory Project ■ Families of Surfaces156 Review158

Focus on Problem Solving169

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10 ■ VECTOR FUNCTIONS175

10.1 Vector Functions and Space Curves175

10.2 Derivatives and Integrals of Vector Functions185

10.3 Arc Length and Curvature195

10.4 Motion in Space:Velocity and Acceleration208

Applied Project ■ Kepler’s Laws218

10.5 Parametric Surfaces219 Review225

Focus on Problem Solving231

11 ■ PARTIAL DERIVATIVES239

11.1 Functions of Several Variables239

11.2 Limits and Continuity249

11.3 Partial Derivatives256

11.4 Tangent Planes and Linear Approximations272

11.5 The Chain Rule280

11.6 Directional Derivatives and the Gradient Vector290

11.7 Maximum and Minimum Values302

Applied Project ■ Designing a Dumpster318

Discovery Project ■ Quadratic Approximations and Critical Points320

11.8 Lagrange Multipliers323

Applied Project ■ Rocket Science333

Applied Project ■ Hydro-Turbine Optimization335 Review338

Focus on Problem Solving351

12 ■ MULTIPLE INTEGRALS357

12.1 Double Integrals over Rectangles357

12.2 Iterated Integrals362

12.3 Double Integrals over General Regions368

12.4 Double Integrals in Polar Coordinates380

12.5 Applications of Double Integrals386

12.6 Surface Area395

NOT FOR SALE

12.7 Triple Integrals400

Discovery Project ■ Volumes of Hyperspheres416

12.8 Triple Integrals in Cylindrical and Spherical Coordinates417

Applied Project ■ Roller Derby425

Discovery Project ■ The Intersection of Three Cylinders427

12.9 Change of Variables in Multiple Integrals429 Review435

Focus on Problem Solving447

13 ■ VECTOR CALCULUS453

13.1 Vector Fields453

13.2 Line Integrals458

13.3 The Fundamental Theorem for Line Integrals466

13.4 Green’s Theorem471

13.5 Curl and Divergence478

13.6 Surface Integrals486

13.7 Stokes’Theorem497

13.8 The Divergence Theorem501 Review505

Focus on Problem Solving515

■ APPENDIXES 519

D Precise Deﬁnitions of Limits519

H Polar Coordinates519

Discovery Project ■ Conic Sections in Polar Coordinates543

I Complex Numbers544

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INFINITESEQUENCESANDSERIES

8.1Sequences

1. (a)Asequenceisanorderedlistofnumbers.Itcanalsobedeﬁnedasafunctionwhosedomainisthesetofpositiveintegers.

(b)Theterms approach 8 as becomeslarge.Infact,wecanmake ascloseto 8 aswelikebytaking sufﬁciently large.

(c)Theterms becomelargeas becomeslarge.Infact,wecanmake aslargeaswelikebytaking sufﬁcientlylarge.

2. (a)FromDeﬁnition1,aconvergentsequenceisasequenceforwhich lim exists.Examples: {1 }, {1 2 }

(b)Adivergentsequenceisasequenceforwhich lim doesnot exist.Examples: { }, {sin }

3. Theﬁrstsixtermsof =

4. {cos( 3)}

1 .Thesequencedoesnotappeartohavealimit.Thevalueswillcycle throughtheﬁrstsixnumbersinthesequence—neverapproachingaparticularnumber.

5. 1 1 3 1 5 1 7 1 9 .Thedenominatorofthe nthtermisthe nthpositiveoddinteger,so = 1 2 1

7. {2 7 12 17 }.Eachtermislargerthantheprecedingoneby 5,so = 1 + ( 1)=2+5( 1)=5 3.

8. 1 4 2 9 3 16 4 25 .Thenumeratorofthe thtermis anditsdenominatoris ( +1)2 .Includingthealternatingsigns, weget =( 1) ( +1)2 . 9.

10. {5 1 5 1 5 1 }.Theaverageof 5 and 1 is 3,sowecanthinkofthesequenceasalternatelyadding 2 and 2 to 3. Thus, =3+( 1) +1 2

13. =1 (0 2) ,so lim =1 0=1 by(7).Converges

NOT FOR SALE

14. = 3 +1 = 3 ( +1) = 2 1+1 2 ,so as since lim 2 = and lim (1+1 2 )=1.Diverges

15. Becausethenaturalexponentialfunctioniscontinuousat 0,Theorem5enablesustowrite lim =lim 1 = lim (1 ) = 0 =1 Converges

.Converges

17. If = 2 1+8 ,then lim =lim (2 ) (1+8 ) =lim 2 1 +8 = 2 8 = 4 .Since tan iscontinuousat 4 ,by Theorem5, lim tan 2 1+8

.Converges

18. Usingthelastlimitlawforsequencesandthecontinuityofthesquarerootfunction,

19. = ( 1) 1 2 +1 = ( 1) 1 +1 ,so 0 | | = 1 +1 1

Theorem4.Converges

,butthetermsofthesequence { } alternateinsign,sothesequence 1 3 5 convergesto 1 andthesequence 2 4 6 convergesto +1 Thisshowsthatthegivensequencedivergessinceitstermsdon’tapproachasinglerealnumber.

21. = + 2 1

because 1+ 2

and .Converges

=cos(2 ).As , 2 0,so cos(2 ) cos0=1 because cos iscontinuous.Converges

= 2 = 2 .Since lim 2 H =lim 2

=lim

=0,itfollowsfromTheorem2that lim =0.Converges

2 as ,sosince lim arctan = 2 ,wehave lim arctan2 = 2 .Converges

2 2 1 2 [since 0 cos 2 1],sosince lim 1 2 =0, cos 2 2 convergesto 0 bytheSqueezeTheorem. 26. = cos = ( 1) .Since | | = as ,thegivensequencediverges.

lim 1+ 2 =lim ln = 2 ,sobyTheorem2, lim 1+ 2 = 2 .Convergent

29.

NOT FOR SALE

,so lim =8lim 21 =8 2lim (1 ) =8 20 =8 byTheorem5,sincethefunction ( )=2 iscontinuousat 0. Convergent

.Converges

=0 bythe SqueezeTheorem.Converges

31. {0 1 0 0 1 0 0 0 1 } divergessincethesequencetakesononlytwovalues,0and1,andneverstaysarbitrarilycloseto eitherone(oranyothervalue)for sufﬁcientlylarge.

34. 0 | | = 3 ! = 3 1 3 2 3 3 3 ( 1) 3 3 1 3 2 3 [for 2] = 27 2 0 as ,sobytheSqueeze

TheoremandTheorem4, {( 3) !} convergesto 0

35. Fromthegraph,itappearsthatthesequenceconvergesto 1 {( 2 ) } convergesto 0 by(7),andhence {1+( 2 ) } convergesto 1+0=1

36. Fromthegraph,itappearsthatthesequenceconvergestoanumber greaterthan 3 lim =lim sin =lim sin =lim 0+ sin = =1 =

37.

NOT FOR SALE

Fromthegraph,itappearsthatthesequenceconvergesto 1 2 As , =

so lim = 1 2

38.

Fromthegraph,itappearsthatthesequenceconvergesto 5

Hence, 5 bytheSqueezeTheorem.

39.

Alternatesolution: Let =(3 +5 )1 .Then

lim ln =lim ln(3 +5 ) H =lim 3 ln3+5 ln5 3 +5 =lim 3 5 ln3+ln5 3 5 +1 =ln5,

so lim = ln5 =5,andso 3 +5 convergesto 5

Fromthegraph,itappearsthatthesequence { } = 2 cos 1+ 2 is divergent,sinceitoscillatesbetween 1 and 1 (approximately).To provethis,supposethat { } convergesto .If = 2 1+ 2 ,then { } convergesto 1,and lim = 1 = .But =cos ,so lim doesnotexist.Thiscontradictionshowsthat { } diverges. 40.

Fromthegraph,itappearsthatthesequenceapproaches 0

(a)

convergesto 0.

,and

. (b) lim =1000lim (1 06) ,sothesequencedivergesby(7)with =1 06 1.

NOT FOR SALE

42. (a)Substitute 1 to 6 for in =100 1 0025 1 0 0025 toget 1 =$0, 2 =$0 25, 3 =$0 75, 4 =$1 50, 5 =$2 51,and 6 =$3 76

(b)Fortwoyears,use 2 12=24 for toget $70 28

43. (a)Wearegiventhattheinitialpopulationis 5000,so 0 =5000.Thenumberofcatﬁshincreasesby 8% permonthandis decreasedby 300 permonth,so 1 = 0 +8% 0 300=1 08 0 300, 2 =1 08 1 300,andsoon.Thus, =1 08 1 300.

(b)Usingtherecursiveformulawith 0 =5000,weget 1 =5100, 2 =5208, 3 =5325 (roundinganyportionofa catﬁsh), 4 =5451, 5 =5587,and 6 =5734,whichisthenumberofcatﬁshinthepondaftersixmonths.

44. +1 = 1 2 if isanevennumber 3 +1 if isanoddnumber When 1 =11,theﬁrst 40 termsare 11, 34, 17, 52, 26,

ThefamousCollatzconjectureisthatthissequencealwaysreaches 1,regardlessofthestartingpoint 1 .

45. (a) 1 =1,

3=1, 4 =4 3 =4 1=3, 5 =4 4 =4 3=1.Sincethetermsofthesequencealternatebetween 1 and 3, thesequenceisdivergent.

(b)

2=2.Sinceallofthetermsare 2, lim =2 andhence,the sequenceisconvergent.

46. (a)Since lim = ,theterms approach as becomeslarge.Becausewecanmake ascloseto aswewish, +1 willalsobeclose,andso lim +1 = (b) 1 =1,

(c)If =lim then lim +1 = also,so mustsatisfy

;hence,thesequenceisconvergent.

(since hastobenon-negativeifitexists).

NOT FOR SALE

47. (a)Let bethenumberofrabbitpairsinthe nthmonth.Clearly 1 =1= 2 .Inthe nthmonth,eachpairthatis 2 ormoremonthsold(thatis, 2 pairs)willproduceanewpairtoaddtothe 1 pairsalreadypresent.Thus, = 1 + 2 ,sothat { } = { },theFibonaccisequence. (b)

then =lim 1 and =lim 2 ,so mustsatisfy

[since mustbepositive]. 48. For 2, 2 2, 2 2 2, ,

Alternatesolution:Let =lim .(Wecouldshowthelimitexistsbyshowingthat { } isboundedandincreasing.)

Then mustsatisfy = 2 2 =2 ( 2)=0. =0 sincethesequenceincreases,so =2.

50. = 2 3 3 +4 deﬁnesanincreasingsequencesincefor ( )=

3 3 +4 , 0 ( )= (3 +4)(2) (2 3)(3) (3 +4)2 = 17 (3 +4)2 0.Thesequenceisboundedsince 1 = 1 7 for 1, and 2 3 3 2 3 = 2 3 for 1.

51. Thetermsof = ( 1) alternateinsign,sothesequenceisnotmonotonic.Theﬁrstﬁvetermsare 1, 2, 3, 4,and 5. Since lim | | =lim = ,thesequenceisnotbounded.

52. = + 1 deﬁnesanincreasingsequencesincethefunction ( )= + 1 isincreasingfor 1 [ 0 ( )=1 1 2 0 for 1.]Thesequenceisunboundedsince as .(Itis,however,boundedbelowby 1 =2.)

53. Since { } isadecreasingsequence, +1 forall 1.Becauseallofitstermsliebetween 5 and 8, { } isa boundedsequence.BytheMonotonicSequenceTheorem, { } isconvergent;thatis, { } hasalimit mustbelessthan 8 since { } isdecreasing,so 5 8

54. (a)Let bethestatementthat +1 and 3 1 isobviouslytrue.Wewillassumethat istrueand thenshowthatasaconsequence +1 mustalsobetrue. +2 +1 2+ +1 2+

2+ +1 2+ +1 ,whichistheinductionhypothesis. +1 3 2+ 3 2+ 9 7,whichiscertainlytruebecauseweareassumingthat 3.So istrueforall ,andso 1 3 (showingthatthesequenceisbounded),andhencebytheMonotonicSequenceTheorem, lim exists.

(b)If =lim ,then lim +1 = also,so = 2+ 2 =2+ 2 2=0 ( +1)( 2)=0 =2 [since can’tbenegative].

55. 1 =1, +1 =3 1 .Weshowbyinductionthat { } isincreasingandboundedaboveby 3.Let betheproposition that +1 and 0 3.Clearly 1 istrue.Assumethat istrue.Then +1

1 1 +1 1 .Now

3 1 = +1 +1 .Thisprovesthat { } isincreasingandbounded aboveby 3,so 1= 1 3,thatis, { } isbounded,andhenceconvergentbytheMonotonicSequenceTheorem. If =lim ,then lim +1 = also,so mustsatisfy =3 1 2 3 +1=0

But 1,so = 3+ 5 2 .

56. 1 =2, +1 = 1 3 .Weuseinduction.Let bethestatementthat 0 +1 2.Clearly 1 istrue,since 2 =1 (3 2)=1.Nowassumethat istrue.Then

bytheinduction hypothesis,so +1 istrue.Toﬁndthelimit,weusethefactthat

,sowemusthave

57. (0 8) 0 000001 ln(0 8) ln(0 000001) ln(0 8)

000001)

8) 61 9,so mustbeatleast 62 tosatisfythegiveninequality.

58. (a)If iscontinuous,then ( )= lim =lim ( )=lim +1 =lim = byExercise46(a). (b)Byrepeatedlypressingthecosinekeyonthecalculator(thatis,takingcosineofthepreviousanswer)untilthedisplayed valuestabilizes,weseethat 0 73909

59. (a)Suppose { } convergesto .Then +1 = + lim +1 =

+lim = + 2 + = ( + )=0 =0 or = . (b) +1 = + = 1+ since 1+ 1

INSTRUCTOR

NOT FOR SALE

(c)Bypart(b), 1 0 , 2 1 2 0 , 3 2 3 0 ,etc.Ingeneral, 0 ,

so lim lim · 0 =0 since By(7) lim =0 if 1 1.Here = (0 1)

(d)Let .Weﬁrstshow,byinduction,thatif 0 ,then and +1 For =0,wehave 1 0 = 0 +

Nowwesupposetheassertionistruefor = ,thatis, and +1 .Then +1 = + = ( )+ + = ( ) + 0 because .So +1 .And +2

( +1 ) + +1 0 since +1 .Therefore, +2 +1 .Thus,theassertionistruefor = +1.Itisthereforetrueforall bymathematicalinduction.

Asimilarproofbyinductionshowsthatif 0 ,then and { } isdecreasing.

Ineithercasethesequence { } isboundedandmonotonic,soitisconvergentbytheMonotonicSequenceTheorem. Itthenfollowsfrompart(a)that lim =

.Noticethat

8 .Itappearsthattheoddterms areincreasingandtheeventermsaredecreasing.Let’sprovethat 2

bymathematical induction.Supposethat

Wehavethusshown,byinduction,that theoddtermsareincreasingandtheeventermsaredecreasing.Alsoalltermslie between 1 and 2,soboth { } and { } areboundedmonotonicsequencesandthereforeconvergentbythe MonotonicSequenceTheorem.Let

2 .Takinglimitsofboth sides,weget =

[since 0].Thus, lim 2 = 2

Similarly,weﬁndthat lim 2

= 2.Sincetheeventermsapproach 2 andtheoddtermsalsoapproach 2,it followsthatthesequenceasawholeapproaches 2,thatis, lim = 2

NOT FOR SALE

LABORATORYPROJECTLogisticSequences

1. TowritesuchaprograminMapleitisbesttocalculateallthepointsﬁrstandthengraphthem.Onepossiblesequenceof commands[taking 0 = 1 2 and =1 5 forthedifferenceequation]is

t:=’t’;p(0):=1/2;k:=1.5; forjfrom1to20dop(j):=k*p(j-1)*(1-p(j-1))od; plot([seq([t,p(t)]t=0..20)],t=0..20,p=0..0.5,style=point);

InMathematica,wecanusethefollowingprogram:

p[0]=1/2

k=1.5

p[j ]:=k*p[j-1]*(1-p[j-1])

P=Table[p[t],{t,20}]

ListPlot[P]

With 0 = 1 2 and =1 5:

0 0 5 7 0 3338465076 14 0 3333373303

1 0 375 8 0 3335895255 15 0 3333353318

2 0 3515625 9 0 3334613309 16 0 3333343326

3 0 3419494629 10 0 3333973076 17 0 3333338329

4 0 3375300416 11 0 3333653143 18 0 3333335831

5 0 3354052689 12 0 3333493223 19 0 3333334582

6 0 3343628617 13 0 3333413274 20 0 3333333958

With 0 = 1 2 and =2 5:

0 0 5 7 0 6004164790 14 0 5999967417

1 0 625 8 0 5997913269 15 0 6000016291

2 0 5859375 9 0 6001042277 16 0 5999991854

3 0 6065368651 10 0 5999478590 17 0 6000004073

4 0 5966247409 11 0 6000260637 18 0 5999997964

5 0 6016591486 12 0 5999869664 19 0 6000001018

6 0 5991635437 13 0 6000065164 20 0 5999999491

Bothofthesesequencesseemtoconverge theﬁrsttoabout 1 3 ,thesecondtoabout0.60 .

INSTRUCTOR USE ONLY

Both of thesesequencesseemtoconverge the ﬁrsttoabout rsttoaboutR,thesecond toabout0.60 esecondtoabout0.60

With 0 = 7 8 and =1 5:

NOT FOR SALE

0 0 875 7 0 3239166554 14 0 3332554829

1 0 1640625 8 0 3284919837 15 0 3332943990

2 0 2057189941 9 0 3308775005 16 0 3333138639

3 0 2450980344 10 0 3320963702 17 0 3333235980

4 0 2775374819 11 0 3327125567 18 0 3333284655

5 0 3007656421 12 0 3330223670 19 0 3333308994

6 0 3154585059 13 0 3331777051 20 0 3333321164

With 0 = 7 8 and =2 5:

0 0 875 7 0 6016572368 14 0 5999869815

1 0 2734375 8 0 5991645155 15 0 6000065088

2 0 4966735840 9 0 6004159972 16 0 5999967455

3 0 6249723374 10 0 5997915688 17 0 6000016272

4 0 5859547872 11 0 6001041070 18 0 5999991864

5 0 6065294364 12 0 5999479194 19 0 6000004068

6 0 5966286980 13 0 6000260335 20 0 5999997966

Thelimitofthesequenceseemstodependon ,butnoton 0

2. With 0 = 7 8 and =3 2:

0 0 875 7 0 5830728495 14 0 7990633827

1 0 35 8 0 7779164854 15 0 5137954979

2 0 728 9 0 5528397669 16 0 7993909896

3 0 6336512 10 0 7910654689 17 0 5131681132

4 0 7428395416 11 0 5288988570 18 0 7994451225

5 0 6112926626 12 0 7973275394 19 0 5130643795

6 0 7603646184 13 0 5171082698 20 0 7994538304

Itseemsthateventuallytheterms ﬂuctuatebetweentwovalues(about 0 5 and 0 8 inthiscase).

NOT FOR SALE

3. With 0 = 7 8 and =3 42:

0 0 875 7 0 4523028596 14 0 8442074951

1 0 3740625 8 0 8472194412 15 0 4498025048

2 0 8007579316 9 0 4426802161 16 0 8463823232

3 0 5456427596 10 0 8437633929 17 0 4446659586

4 0 8478752457 11 0 4508474156 18 0 8445284520

5 0 4411212220 12 0 8467373602 19 0 4490464985

6 0 8431438501 13 0 4438243545 20 0 8461207931

With 0 = 7 8 and =3 45:

0 0 875 7 0 4670259170 14 0 8403376122

1 0 37734375 8 0 8587488490 15 0 4628875685

2 0 8105962830 9 0 4184824586 16 0 8577482026

3 0 5296783241 10 0 8395743720 17 0 4209559716

4 0 8594612299 11 0 4646778983 18 0 8409445432

5 0 4167173034 12 0 8581956045 19 0 4614610237

6 0 8385707740 13 0 4198508858 20 0 8573758782

Fromthegraphsabove,itseemsthatfor between 3 4 and 3 5,thetermseventuallyfluctuatebetweenfourvalues.Inthe graphbelow,thepatternfollowedbythetermsis 0 395 0 832 0 487 0 869 0 395 .Notethatevenfor =3 42 (asinthe firstgraph),therearefourdistinct“branches”;evenafter 1000 terms,thefirstandthirdtermsinthepatterndifferbyabout 2 × 10 9 ,whilethefirstandfifthtermsdifferbyonly 2 × 10 10 .With 0 = 7 8 and =3 48:

NOT FOR SALE

Fromthegraphs,itseemsthatif 0 ischangedby 0 001,thewholegraphchangescompletely.(Note,however,thatthismight bepartiallyduetoaccumulatedround-offerrorintheCAS.ThesegraphsweregeneratedbyMaplewith100-digitaccuracy, anddifferentdegreesofaccuracygivedifferentgraphs.)Thereseemtobesomesomeﬂeetingpatternsinthesegraphs,buton thewholetheyarecertainlyverychaotic.As increases,thegraphspreadsoutvertically,withmoreextremevaluescloseto 0 or 1

NOT FOR SALE

8.2Series

1. (a)Asequenceisanorderedlistofnumberswhereasaseriesisthe sum ofalistofnumbers.

(b)Aseriesisconvergentifthesequenceofpartialsumsisaconvergentsequence.Aseriesisdivergentifitisnotconvergent.

2. =1 =5 meansthatbyaddingsufﬁcientlymanytermsoftheserieswecangetascloseasweliketothenumber 5

Inotherwords,itmeansthat lim =5,where isthe thpartialsum,thatis, =1

1 2 40000

2 1 92000

3 2 01600

4 1 99680

5 2 00064

6 1 99987

7 2 00003

8 1 99999

9 2 00000

10 2 00000

Fromthegraphandthetable,itseemsthattheseriesconvergesto 2.Infact,itisageometric

serieswith = 2 4 and = 1 5 ,soitssumis

Notethatthedotcorrespondingto =1 ispartofboth { } and { }

TI-86Note: Tograph { } and { },setyourcalculatortoParammodeandDrawDotmode.(DrawDotisunder

GRAPH,MORE,FORMT(F3).)Nowunder E(t)= maketheassignments: xt1=t,yt1=12/(-5)ˆt,xt2=t, yt2=sumseq(yt1,t,1,t,1). (sumandseqareunderLIST,OPS(F5),MORE.)UnderWINDuse 1,10,1,0,10,1,-3,1,1 toobtainagraphsimilartotheoneabove.ThenuseTRACE(F4)toseethevalues.

1 0 54030

2 0 12416

3 0 86584

4 1 51948

5 1 23582

6 0 27565

7 0 47825

8 0 33275

9 0 57838

10 1 41745

Theseries =1 cos diverges,sinceitstermsdonotapproach 0

1 0 44721

2 1 15432

3 1 98637

4 2 88080

5 3 80927

6 4 75796

7 5 71948

8 6 68962

9 7 66581

10 8 64639

NOT FOR SALE

Theseries =1 2 +4 diverges,sinceitstermsdonotapproach 0

1 4 90000

2 8 33000 3 10 73100

4 12 41170

5 13 58819

6 14 41173

7 14 98821

8 15 39175

9 15 67422

10 15 87196

Fromthegraphandthetable,weseethatthetermsaregettingsmallerandmayapproach 0, andthattheseriesapproachesavaluenear 16.Theseriesisgeometricwith 1 =4 9 and =0 7,soitssumis

1 0 29289

2 0 42265

3 0 50000

4 0 55279

5 0 59175

6 0 62204

7 0 64645

8 0 66667

9 0 68377

10 0 69849

Fromthegraphandthetable,itseemsthattheseriesconverges.

NOT FOR SALE

2 0 12500

3 0 19167

4 0 23333

5 0 26190

6 0 28274

7 0 29861

8 0 31111

9 0 32121

10 0 32955

11 0 33654

Fromthegraphandthetable,itseemsthattheseriesconverges.

9. (a) lim =lim 2 3 +1 = 2 3 ,sothe sequence { } isconvergentby(8.1.1).

(b)Since lim = 2 3 =0,the series =1 isdivergentbytheTestforDivergence.

10. (a)Both =1 and =1 representthesumoftheﬁrst termsofthesequence { },thatis,the thpartialsum. (b) =1 = + + + terms = ,which,ingeneral,isnotthesameas =1 =

11. 3 4+ 16 3 64 9 + isageometricserieswithratio = 4 3 .Since | | = 4 3 1,theseriesdiverges.

12. 4+3+ 9 4 + 27 16 + isageometricserieswithratio 3 4 .Since | | = 3 4 1,theseriesconvergesto 1 = 4 1 3 4 =16

13. 10 2+0 4 0 08+ isageometricserieswithratio 2 10 = 1 5 .Since | | = 1 5 1,theseriesconvergesto 1 = 10 1 ( 1 5) = 10 6 5 = 50 6 = 25 3

14. 1+0 4+0 16+0 064+ ··· isageometricserieswithratio =0 4= 2 5 .Since | | = 2 5 1,theseriesconvergesto 1 = 1 1 2 5 = 5 3 .

15. =1 6(0 9) 1 isageometricserieswithﬁrstterm =6 andratio =0 9.Since | | =0 9 1,theseriesconvergesto 1 = 6 1 0 9 = 6 0 1 =60

Converges.

NOT FOR SALE

.Theﬁrstseriesisaconvergent geometricseries (| | = 1 2 1),butthesecondseriesisadivergentgeometricseries (| | = 3 2 1),sotheoriginalseries isdivergent. 25. =1 2=2+ 2+ 3 2+ 4 2+ divergesbytheTestforDivergencesince lim =lim 2=lim 21 =2

=1 =0 26. =1 (cos1) isageometricserieswithratio =cos1 0 540302.Itconvergesbecause | | 1.Itssumis

divergesbytheTestforDivergencesince lim =lim arctan = 2 =0

NOT FOR SALE INSTRUCTOR

28. =1 (0 8) 1 (0 3) = =1 (0 8) 1 =1 (0 3) [differenceoftwoconvergentgeometricseries] = 1 1 0 8 0 3 1

29. =1 1 = =1 1 isageometricserieswithﬁrstterm = 1 andratio = 1 .Since | | = 1 1,theseriesconverges to 1 1 1 = 1 1 1 = 1 1 .ByExample6, =1 1 ( +1) =1.Thus,byTheorem8(ii), =1 1 + 1 ( +1) = =1

30. =1 3 5 + 2 divergesbecause =1 2 =2 =1 1 diverges.(Ifitconverged,then

2 · 2 =1 1 wouldalsoconvergeby

Theorem8(i),butweknowfromExample7thattheharmonicseries =1 1 diverges.)Ifthegivenseriesconverges,thenthe difference =1 3 5 + 2 =1 3 5 mustconverge(since =1 3 5 isaconvergentgeometricseries)andequal =1 2 ,but wehavejustseenthat =1 2 diverges,sothegivenseriesmustalsodiverge.

31. Usingpartialfractions,thepartialsumsoftheseries =2

32. Fortheseries

33. Fortheseries

[usingpartialfractions].Thelattersumis

NOT FOR SALE

34. Fortheseries =1 ln +1 , =(ln1 ln2)+(ln2 ln3)+(ln3 ln4)+ +[ln ln( +1)]=ln1 ln( +1)= ln( +1) [telescopingseries]

Thus, lim = ,sotheseriesisdivergent.

35. (a)Manypeoplewouldguessthat 1,butnotethat consistsofaninﬁnitenumberof 9s.

(b) =0 99999 = 9 10 + 9 100 + 9 1000 + 9 10,000 + = =1 9 10 ,whichisageometricserieswith 1 =0 9 and =0 1.Itssumis 0 9 1 0 1 = 0 9 0 9 =1,thatis, =1

(c)Thenumber 1 hastwodecimalrepresentations, 1 00000 and 0 99999

(d)Exceptfor 0,allrationalnumbersthathaveaterminatingdecimalrepresentationcanbewritteninmorethanoneway.For example, 0 5 canbewrittenas 0 49999 aswellas 0 50000

36.

NOT FOR SALE

43. =0 cos 2 isageometricserieswithﬁrstterm1andratio = cos 2 ,soitconverges | | 1.But | | = |cos | 2

forall .Thus,theseriesconvergesforallrealvaluesof andthesumoftheseriesis 1 1 (cos ) 2 = 2 2 cos

44. Because 1 0 and ln iscontinuous,wehave lim ln 1+ 1 =ln1=0

Wenowshowthattheseries =1 ln 1+ 1 = =1 ln +1 = =1 [ln( +1) ln ] diverges. =(ln2 ln1)+(ln3 ln2)+ +(ln( +1) ln )=ln( +1) ln1=ln( +1)

As , =ln( +1) ,sotheseriesdiverges.

45. Afterdeﬁning ,Weuse convert(f,parfrac); inMaple, Apart inMathematica,or ExpandRational and Simplify inDerivetoﬁndthatthegeneraltermis

3 1 (

3 .Sothe nthpartialsumis

Theseriesconvergesto lim =1.Thiscanbeconﬁrmedbydirectlycomputingthesumusing sum(f,1..infinity); (inMaple), Sum[f,{n,1,Infinity}] (inMathematica),or CalculusSum (from 1 to )and Simplify (inDerive).

46. SeeExercise45forspeciﬁcCAScommands.

Theseriesconvergesto lim = 1 4 .

NOT FOR SALE

49. (a)Aftertheﬁrstpillistaken, 100mg ofthedrugisinthebody.Afterthesecondpillistaken, 100mg plus 100(5%)mg remainsinthebody.Afterthethirdpillistaken 100mg plus 100(5%)mg plus [100(5%)](5%)mg remainsinthebody. Thisgivesus 100+100(0 05)+100(0 05)2 =105 25mg ofthedrugremaininginthebodyafterthepatienttakesthree pills.

(b)Continuingthepatternestablishedinpart(a),weget 100+100(0 05)+100(0 05)2 + +100(0 05) 1 mg after pillsaretaken.By(3),thissumis 100(1 0 05 ) 1 0 05 .

(c)Theamountofthedrugremaininginthebodyinthelongruncanbeapproximatedbysummingtheinﬁniteserieswith 1 =100 and =0 05.Thissumis 100 1 0 05 = 100 0 95 105 26mg

50. (a)Westartwith fliesbeingreleased.After 1 day,wehave 1 fliesplusanewreleaseof flies.After 2 days,wehave 2 + 1 + flies.After days,wehave + 1 + + 1 + = =0 = (1 +1 ) 1 by(3).In thelongrun,wehaveageometricserieswith 1 = and = ,soitssumis 1

(b)Wewantthesumtobe 10,000,so 10,000= 1 0 9 [frompart(a)] =10,000(0 1)=1000

51. (a)Theﬁrststepinthechainoccurswhenthelocalgovernmentspends dollars.Thepeoplewhoreceiveitspendafraction ofthose dollars,thatis, dollars.Thosewhoreceivethe dollarsspendafraction ofit,thatis, 2 dollars. Continuinginthisway,weseethatthetotalspendingafter transactionsis = + + 2 + + –1

by(3).

(b)

= [since + =1] = [since =1 ] If =0 8,then =1 =0 2 andthemultiplieris =1 =5

52. (a)Initially,theballfallsadistance ,thenreboundsadistance ,falls ,rebounds 2 ,falls 2 ,etc.Thetotal distanceittravelsis

(b)FromExample3inSection2.1,weknowthataballfalls 1 2 2 metersin seconds,where isthegravitational acceleration.Thus,aballfalls metersin = 2 seconds.Thetotaltraveltimeinsecondsis

NOT FOR SALE

(c)Itwillhelptomakeachartofthetimeforeachdescentandeachreboundoftheball,togetherwiththevelocityjustbefore andjustaftereachbounce.Recallthatthetimeinsecondsneededtofall metersis 2 .Theballhitsthegroundwith velocity 2 = 2 (takingtheupwarddirectiontobepositive)andreboundswithvelocity 2 = 2 ,takingtime 2 toreachthetopofitsbounce,whereitsvelocityis 0.Atthatpoint, itsheightis 2 .Alltheseresultsfollowfromtheformulasforverticalmotionwithgravitationalacceleration :

Thetotaltraveltimeinsecondsis

Anothermethod: Wecouldusepart(b).Atthetopofthebounce,theheightis 2 = ,so = andtheresultfollows frompart(b).

53. =2 (1+ ) isageometricserieswith =(1+ ) 2 and =(1+ ) 1 ,sotheseriesconvergeswhen (1+ ) 1

.Wecalculatethesumofthe

.However,thenegativerootisinadmissiblebecause

54. =0 = =0 ( ) isageometricserieswith =( )0 =1 and = .If 1,ithassum 1 1 ,so 1 1 =10 1 10 =1 = 9 10 =ln 9 10 .

NOT FOR SALE

55. = 1+ 1 2 + 1 3 + + 1 = 1 1 2

Thus, +1 and lim = .Since { } isincreasing, lim = ,implyingthattheharmonicseriesis divergent.

56. Theareabetween = 1 and = for 0 1 is

Wecanseefromthediagramthatas ,thesumoftheareas betweenthesuccessivecurvesapproachestheareaoftheunitsquare, thatis, 1.So =1 1 ( +1) =1

57. Let bethediameterof .Wedrawlinesfromthecentersofthe to thecenterof (or ),andusingthePythagoreanTheorem,wecanwrite

Similarly,

,andingeneral,

.Ifweactuallycalculate 2 and 3 fromtheformulasabove,weﬁndthattheyare 1 6 = 1 2 · 3 and 1 12 = 1 3 · 4 respectively,sowesuspectthatingeneral, = 1 ( +1) .Toprovethis,weuseinduction:Assumethatforall , =

[telescopingsum].Substitutingthisintoour formulafor +1 ,weget

,andtheinductioniscomplete.

Now,weobservethatthepartialsums =1 ofthediametersofthecirclesapproach 1 as ;thatis, =1 = =1 1 ( +1) =1,whichiswhatwewantedtoprove.

NOT FOR SALE

58. | | = sin , | | = | | sin = sin2 , | | = | | sin = sin3 , .Therefore, | | + | | + | | + | | + = =1 sin = sin 1 sin sincethisisageometricserieswith =sin and |sin | 1 because 0 2 .

59. Theseries 1 1+1 1+1 1+ diverges(geometricserieswith = 1)sowecannotsaythat 0=1 1+1 1+1 1+ .

60. If =1 isconvergent,then lim =0 byTheorem6,so lim 1 =0,andso =1 1 isdivergentbytheTestfor Divergence.

61. Supposeonthecontrarythat ( + ) converges.Then ( + ) and areconvergentseries.Soby Theorem8(iii), [( + ) ] wouldalsobeconvergent.But [( + ) ]= ,acontradiction,since isgiventobedivergent.

62. No.Forexample,take = and = ( ),whichbothdiverge,yet ( + )= 0,whichconverges withsum 0

63. Thepartialsums { } formanincreasingsequence,since 1 = 0 forall .Also,thesequence { } isbounded since 1000 forall .SobytheMonotonicSequenceTheorem,thesequenceofpartialsumsconverges,thatis,theseries isconvergent.

64. (a)RHS

[frompart(a)]

[asabove]

NOT FOR SALE

65. (a)Attheﬁrststep,onlytheinterval 1 3 2 3 (length 1 3 )isremoved.Atthesecondstep,weremovetheintervals 1 9 2 9 and 7 9 8 9 ,whichhaveatotallengthof 2 · 1 3 2 .Atthethirdstep,weremove 22 intervals,eachoflength 1 3 3 .Ingeneral, atthe nthstepweremove 2 1 intervals,eachoflength 1 3 ,foralengthof 2 1 · 1 3 = 1 3 2 3 1 .Thus,thetotal lengthofallremovedintervalsis =1 1 3 2 3 1 = 1 3 1 2 3 =1 geometricserieswith = 1 3 and = 2 3 .Noticethatat the thstep,theleftmostintervalthatisremovedis 1 3 2 3 ,soweneverremove 0,and 0 isintheCantorset.Also, therightmostintervalremovedis 1

,so 1 isneverremoved.SomeothernumbersintheCantorset are

(b)Thearearemovedattheﬁrststepis 1 9 ;atthesecondstep, 8 · 1 9 2 ;atthethirdstep, (8)2 · 1 9 3 .Ingeneral,thearea removedatthe thstepis (8) 1 1 9 = 1 9 8 9 1 ,sothetotalareaofallremovedsquaresis

Thelimitsseemtobe 5 3 , 8 3 , 2, 3, 667,and 334.Notethatthelimitsappeartobe“weighted”moretoward 2 .Ingeneral,we guessthatthelimitis

atotalof 1 timesinthiscalculation,onceforeach between 3 and +1.Nowwecanwrite

NOT FOR SALE

67. (a)For =1 ( +1)! , 1 = 1 1 2 = 1 2 , 2 = 1 2 + 2 1 2 3 = 5 6 , 3 = 5 6 + 3 1 2 3 4 = 23 24 , 4 = 23 24 + 4 1 2 3 4 5 = 119 120 .Thedenominatorsare ( +1)!,soaguesswouldbe = ( +1)! 1 ( +1)!

(b)For =1, 1 = 1 2 = 2! 1 2! ,sotheformulaholdsfor =1.Assume = ( +1)! 1 ( +1)! .Then +1 = ( +1)! 1 ( +1)! + +1 ( +2)! = ( +1)! 1 ( +1)! + +1 ( +1)!( +2) = ( +2)! ( +2)+ +1 ( +2)! = ( +2)! 1 ( +2)!

Thus,theformulaistruefor = +1.Sobyinduction,theguessiscorrect.

68. Let 1 = radiusofthelargecircle, 2 = radiusofnextcircle,andsoon. Fromtheﬁgurewehave =60 and cos60 = 1 | |,so | | =2 1 and | | =2 2 .Therefore,

1 =3 2 .Ingeneral,wehave +1 = 1 3 ,sothetotalareais

Sincethesidesofthetrianglehavelength

.Theareaofthetriangleis 3 4 ,sothecirclesoccupyabout 83 1% oftheareaofthetriangle.

NOT FOR SALE

8.3TheIntegralandComparison Tests;EstimatingSums

1. Thepictureshowsthat 2 = 1 21 3 2 1 1 1 3 , 3 = 1 31 3 3 2 1 1 3 ,andsoon,so =2 1 1 3 1 1 1 3 .The integralconvergesby(5.10.2)with =1 3 1,sotheseriesconverges.

2. Fromtheﬁrstﬁgure,weseethat

6 1 ( ) 5 =1 .Fromthesecondﬁgure, weseethat 6 =2 6 1 ( ) .Thus,we have 6 =2 6 1 ( ) 5 =1

3. (a)Wecannotsayanythingabout .If forall and isconvergent,then couldbeconvergentor divergent.(SeethenoteafterExample2.)

(b)If forall ,then isconvergent.[Thisispart(i)oftheComparisonTest.]

4. (a)If forall ,then isdivergent.[Thisispart(ii)oftheComparisonTest.]

(b)Wecannotsayanythingabout .If forall and isdivergent,then couldbeconvergentor divergent.

5. =1 isa -serieswith = =1 isageometricseries.By(1),the -seriesisconvergentif 1.Inthiscase, =1 = =1 1 ,so 1 1 arethevaluesforwhichtheseriesconverge.Ageometricseries

6. Thefunction ( )=1

iscontinuous,positive,anddecreasingon [1 ),sotheIntegralTestapplies.

Sincethisimproperintegralisconvergent,theseries =1 1 5 isalsoconvergentbytheIntegralTest.

7. Thefunction ( )=1 5 = 1 5 iscontinuous,positive,anddecreasingon [1 ),sotheIntegralTestapplies.

NOT FOR SALE

SECTION8.3 THEINTEGRALANDCOMPARISONTESTS;ESTIMATINGSUMS

8. Thefunction ( )=1 +4=( +4) 1 2 iscontinuous,positive,anddecreasingon [1 ),sotheIntegralTestapplies. 1 ( +4) 1 2 =lim 1 ( +4) 1 2 =lim 2( +4)1 2 1 =lim 2 +4 2 5 = ,sotheseries =1 1 +4 diverges.

9. 2 3 +1 2 3 = 1 2 2 1 2 forall 1,so =1 2 3 +1 convergesbycomparisonwith =1 1 2 ,whichconverges becauseitisa p-serieswith =2 1.

10. 3 4 1 3 4 = 1 forall 2,so =2 3 4 1 divergesbycomparisonwith =2 1 ,whichdivergesbecauseitisa p-series with =1 1 (theharmonicseries).

11. Theseries =1 1 0 85 isa -serieswith =0 85 1,soitdivergesby(1).Therefore,theseries =1 2 0 85 mustalsodiverge, forifitconverged,then =1 1 0 85 wouldhavetoconverge[byTheorem8(i)inSection8.2].

12. =1 1 4 and =1 1 2 are -serieswith 1,sotheyconvergeby(1).Thus, =1 3 1 2 convergesbyTheorem8(i)in

Section8.2.ItfollowsfromTheorem8(ii)thatthegivenseries =1 ( 1 4 +3 1 2 ) alsoconverges.

13. 1+ 1 8 + 1 27 + 1 64 + 1 125 + = =1 1 3 .Thisisa -serieswith =3 1,soitconvergesby(1).

14. 1+

3 2 .Thisisa -serieswith = 3 2 1,soitconvergesby(1).

15. ( )= iscontinuousandpositiveon [1 ). 0 ( )= + = (1 ) 0 for 1,so isdecreasing on [1 ).Thus,theIntegralTestapplies. 1 =lim 1 =lim 1 [byparts] =lim [ + 1 + 1 ]=2 since lim =lim ( ) H =lim (1 )=0 and lim =0.Thus, =1 converges.

16. ( )= 2 3 +1 iscontinuousandpositiveon [2 ),andalsodecreasingsince 0 ( )= (2 3 ) ( 3 +1)2 0 for 2, sowecanusetheIntegralTest[notethat is not decreasingon [1 )] 2

ln9 = ,sotheseries =2 2 3 +1 diverges,andsodoes thegivenseries,

NOT FOR SALE

17. ( )= 1 ln iscontinuousandpositiveon [2 ),andalsodecreasingsince 0 ( )= 1+ln 2 (ln )2 0 for 2,sowecan

usetheIntegralTest. 2 1 ln =lim [ln(ln )]2 =lim [ln(ln ) ln(ln2)]= ,sotheseries =2 1 ln diverges.

Anothersolution: Thefunction ( )=1 ( 2 +9) iscontinuous,positive,anddecreasingon [1 ),sotheIntegralTest applies.

.Sincetheintegralconverges,sodoestheseries.

aconstantmultipleofaconvergent -series[ =2 1].Thetermsofthegivenseriesarepositivefor 1,whichisgood enough.

NOT FOR SALE

25. UsetheLimitComparisonTestwith = 1+4 1+3 and = 4 3 :

diverges,sodoes =1

orusetheTestforDivergence.

.Alternatively,usetheComparisonTestwith

convergesbycomparisonwiththeconvergent -series

27. 2+( 1) 3 ,and =1 3 convergesbecauseitisaconstantmultipleoftheconvergent -series =1 1 = 3 2 1 ,sothegivenseriesconvergesbytheComparisonTest.

28.

,sothegivenseriesconvergesbycomparisonwithaconstantmultipleofa convergentgeometricseries.

29. UsetheLimitComparisonTestwith =sin 1 and = 1 .Then and areserieswithpositivetermsand lim =lim sin(1 ) 1 =lim 0 sin =1 0.Since =1 isthedivergentharmonicseries, =1 sin(1 ) alsodiverges.[Notethatwecouldalsousel’Hospital’sRuletoevaluatethelimit: lim sin(1 ) 1 H =lim cos(1 ) · 1 2 1 2 =lim cos 1 =cos0=1.]

30. If =

and =

,then

, so =1 2 5 3 + +1 divergesbytheLimitComparisonTestwiththedivergentharmonicseries =1 1 (Notethat 0 for 6.)

31. Wehavealreadyshown(inExercise17)thatwhen =1 theseries =2 1 (ln ) diverges,soassumethat =1. ( )= 1 (ln ) iscontinuousandpositiveon [2 ),and 0 ( )= +ln 2 (ln ) +1 0 if ,sothat iseventually decreasingandwecanusetheIntegralTest. 2 1 (ln ) =lim (ln )1

Thislimitexistswhenever 1 0 1,sotheseriesconvergesfor 1

NOT FOR SALE

32. (a) ( )=1 4 ispositiveandcontinuousand 0 ( )= 4 5 isnegativefor 0,andsotheIntegralTestapplies.

1 082037+0 000250=1 082287 1 082037+0 000333=1 082370,soweget 1 08233 with error 0 00005 (c) 1 4 =

thatis,for 32.

33. (a) ( )= 1 2 ispositiveandcontinuousand 0 ( )= 2 3 isnegativefor 0,andsotheIntegralTestapplies.

1 549768+0 090909=1 640677 1 549768+0 1=1 649768,soweget 1 64522 (theaverageof 1 640677 and 1 649768)witherror 0 005 (themaximumof 1 649768 1 64522 and 1 64522 1 640677,roundedup). (c) 1 2 =

35. ( )=1 (2 +1)6 iscontinuous,positive,anddecreasingon [1 ),sotheIntegralTestapplies.Using(3), (2 +1) 6 =lim 1 10(2 +1)5 = 1 10(2 +1)5 .Tobecorrecttoﬁvedecimalplaces,wewant

NOT FOR SALE

36. ( )= 1 (ln )2 ispositiveandcontinuousand 0 ( )= ln

)3 isnegativefor 1,sotheIntegralTestapplies. Using(3),weneed 0

,sowewouldhavetotakethis manyterms,whichwouldbeproblematicbecause

(b)Bypart(a), 106 1+ln106 14 82 15 and 109 1+ln10

41. Since 10 9

foreach ,andsince =1 9

willalwaysconvergebytheComparisonTest.

isaconvergentgeometricseries |

42. Firstweobservethat,byl’Hospital’sRule, lim 0 ln(1+ ) =lim 0 1 1+ =1.Also,if converges,then lim =0 by

Theorem8.2.6.Therefore, lim ln(1+ ) =lim 0 ln(1+ ) =1 0.Wearegiventhat isconvergentand 0

Thus, ln(1+ ) isconvergentbytheLimitComparisonTest.

NOT FOR SALE

43. Yes.Since isaconvergentserieswithpositiveterms, lim =0 byTheorem8.2.6,and = sin( ) isa serieswithpositiveterms(forlargeenough ).Wehave lim =lim sin( ) =1 0 byTheorem3.3.2.Thus, isalsoconvergentbytheLimitComparisonTest.

44. ln = ln ln = ln ln = ln = 1 ln .Thisisa -series,whichconvergesforall suchthat ln 1 ln 1 1 1 [with 0].

45. lim =lim 1 ,soweapplytheLimitComparisonTestwith = 1 .Since lim 0 weknowthateitherboth

seriesconvergeorbothseriesdiverge,andwealsoknowthat =1 1 diverges[ -serieswith =1].Therefore, mustbe divergent.

46. Fortheseries

Thus, =1 1 +1 =lim

.Sinceaconstantmultipleofadivergentseries isdivergent,thelastlimitexistsonlyif 1=0,sotheoriginalseriesconvergesonlyif =1

8.4OtherConvergenceTests

1. (a)Analternatingseriesisaserieswhosetermsarealternatelypositiveandnegative.

(b)Analternatingseries =1 = =1 ( 1) 1 ,where = | |,convergesif 0 +1 forall and lim =0. (ThisistheAlternatingSeriesTest.)

(c)Theerrorinvolvedinusingthepartialsum asanapproximationtothetotalsum istheremainder = andthe sizeoftheerrorissmallerthan +1 ;thatis, | | +1 .(ThisistheAlternatingSeriesEstimationTheorem.)

2. (a)Since lim +1 =8 1,part(b)oftheRatioTesttellsusthattheseries isdivergent.

(b)Since lim +1 =0 8 1,part(a)oftheRatioTesttellsusthattheseries isabsolutelyconvergent(and thereforeconvergent).

(c)Since lim +1 =1,theRatioTestfailsandtheseries mightconvergeoritmightdiverge.