Solution Manual for Calculus and Its Applications, 11th Edition

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Solution Manual for Calculus and Its Applications, 11th Edition - Page 1 preview image

’SSOLUTIONSMANUALDAVIDDUBRISKEUniversity of Arkansas for Medical SciencesCALCULUSAND ITSAPPLICATIONSELEVENTHEDITIONMarvin L. BittingerIndiana University Purdue University IndianapolisDavid J. EllenbogenCommunity College of VermontScott A. SurgentArizona State University

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TABLE OF CONTENTSChapter R Functions, Graphs, and Models..................................................................................................... 1R.1 Graphs and Equations ...........................................................................................................................1R.2 Functions and Models.........................................................................................................................12R.3 Finding Domain and Range ................................................................................................................25R.4. Slope and Linear Functions ...............................................................................................................32R.5 Nonlinear Functions and Models ........................................................................................................46R.6 Mathematical Modeling and Curve Fitting .........................................................................................66Chapter 1Differentiation ............................................................................................................................ 731.1 Limits: A Numerical and Graphical Approach....................................................................................731.2 Algebraic Limits and Continuity .........................................................................................................861.3 Average Rates of Change ....................................................................................................................971.4 Differentiation Using Limits of Difference Quotients.......................................................................1161.5 The Power and Sum–Difference Rules..............................................................................................1321.6 The Product and Quotient Rules........................................................................................................1551.7 The Chain Rule..................................................................................................................................1781.8 Higher-Order Derivatives ..................................................................................................................201Chapter 2Applications of Differentiation ................................................................................................2172.1 Using First Derivatives to Classify Maximum and Minimum Values and Sketch Graphs................2172.2 Using Second Derivatives to Classify Maximum and Minimum Values and Sketch Graphs ...........2452.3 Graph Sketching: Asymptotes and Rational Functions .....................................................................2902.4 Using Derivatives to Find Absolute Maximum and Minimum Values .............................................3342.5 Maximum–Minimum Problems; Business, Economics and General Applications ...........................3632.6 Marginals and Differentials ...............................................................................................................3982.7 Elasticity of Demand .........................................................................................................................4122.8 Implicit Differentiation and Related Rates ........................................................................................418Chapter 3Exponential and Logarithmic Functions ................................................................................4343.1 Exponential Functions .......................................................................................................................4343.2 Logarithmic Functions.......................................................................................................................4393.3 Applications: Uninhibited and Limited Growth Models ...................................................................4733.4 Applications: Decay ..........................................................................................................................4923.5 The Derivatives ofaxand logax.........................................................................................................5053.6 A Business Application: Amortization ..............................................................................................514Chapter 4Integration .................................................................................................................................5254.1 Antidifferentiation .............................................................................................................................5254.2 Antiderivatives as Areas....................................................................................................................5384.3 Area and Definite Integrals................................................................................................................5524.4 Properties of Definite Integrals..........................................................................................................5694.5 Integration Techniques: Substitution.................................................................................................5884.6 Integration Techniques: Integration by Parts.....................................................................................6094.7 Integration Techniques: Tables..........................................................................................................628Chapter 5Applications of Integration ......................................................................................................6375.1 Consumer Surplus and Producer Surplus ..........................................................................................6375.2 Integrating Growth and Decay Models..............................................................................................6515.3 Improper Integrals .............................................................................................................................6635.4 Probability .........................................................................................................................................6745.5 Probability: Expected Value; The Normal Distribution ....................................................................6855.6 Volume ..............................................................................................................................................6995.7 Differential Equations .......................................................................................................................712

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Chapter 6Functions of Several Variables ................................................................................................7316.1 Functions of Several Variables..........................................................................................................7316.2 Partial Derivatives .............................................................................................................................7376.3 Maximum-Minimum Problems .........................................................................................................7556.4 An Application: The Least-Squares Technique.................................................................................7726.5 Constrained Optimization..................................................................................................................7876.6 Double Integrals ................................................................................................................................819

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Chapter RFunctions, Graphs, and ModelsExercise Set R.11.Graph1yx.We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.2.Graph4yx.3.Graph14yx .We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.4.Graph3yx .5.Graph533yx .We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.6.Graph243yx.7.Graph5xy.We solve foryfirst.subtractfrom both sidescommutative property555xxyyxyx Next, we choose somex-values and calculatethe correspondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.8.Graph4xy.xy,xy232,3010,1323, 2xy,xy414,1000, 0414,1xy,xy383,8030, 3323,2xy,xy161, 6050,5232, 3

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2Chapter R Functions, Graphs, and Models9.Graph639xy .We solve foryfirst.6from both sidesdivide both sides by 3subtract639369693323xxyyxyxyx   Next, we choose somex-values and calculatethe correspondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.10.Graph824yx.11.Graph2510xy.We solve foryfirst.subtract 2from both sidesdivide both sides by 5251051021 1025225225xxyyxyxyxyx We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.xy,xy545, 4020, 2505, 012.Graph5612xy.13.Graph25yx.We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.xy,xy212,1141,4050,5141,4212,114.Graph23yx.15.Graph22xy.Sincexis expressed in terms ofywe firstchoose values foryand then computex. Thenwe plot the points that are found and connectthem with a smooth curve.xy,xy622,2311,1202, 0311,1622, 2xy,xy212,1030,3272,7

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Exercise Set R.1316.Graph22xy.17.Graph4yx.We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with acurve.xy,xy151, 5040, 4404, 0515,1626, 218.Graphyx.19.Graph27yx.We choose somex-values and calculate thecorrespondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.xy,xy232, 3161, 6070, 7161, 6232, 320.Graph25yx.21.Graph37yx.First we solve fory.3377yxyxNext, we choose somex-values and calculatethe correspondingy-values to find some orderedpairs that are solutions of the equation. Thenwe plot the points and connect them with asmooth curve.xy,xy212,1161, 6070, 7181,82152,1522.Graph31yx.23.4320.53.4596.65347.7 , 06AtttttTo determine the amount of ibuprofen in theblood stream that is left after 2 hours, wesubstitute2tinto the equation and solve forA.    4320.5 23.45 296.65 2347.7 2344.4AAccording to the model, approximately 344.4milligrams of ibuprofen will remain in the bloodstream 2 hours after 400 mg have beenswallowed.

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4Chapter R Functions, Graphs, and Models24.0.00615.714Rx We substitute 1954 in forxto get0.006 195415.7143.99R According to this model, the world record forthe mile in 1954 is approximately 3.99 minutes.To convert this to traditional minutes-secondswe multiply the decimal part by 60 seconds.0.99 6059.4Therefore the world record for the mile in 1954was 3:59.4.Likewise, we substitute 2000 in forxto get0.006 200015.7143.714R According to the model, the world record forthe mile in 2008 will be approximately 3.71minutes or 3:42.8.Finally, we substitute 2015 in forxto get0.006 201515.7143.624R According to the model, the world record forthe mile in 2015 will be approximately 3.62minutes or 3:37.44.25.10.9v ttWe substitute 2.5 in fortto get2.510.9 2.527.25vShaun White was traveling at 27.25 miles perhour when he reentered the half pipe.26. 216stt 222282816281628161.3228s tttttDanny took approximately 1.32 seconds to hitthe ramp.27.a)Locate 20 on the horizontal axis and godirectly up to the graph. Then move left tothe vertical axis and read the value there.We estimate the number of hearing-impairedAmericans of age 20 is about 1.8 million.Follow the same process for 40, 50, and 60to determine the number of hearing-impaired Americans at each of those ages.We estimate the number of hearing-impairedAmericans of age 40 is about 4.1 million.We estimate the number of hearing-impairedAmericans of age 50 is about 5.3 million.We estimate the number of hearing-impairedAmericans of age 60 is about 6.0 million.b)Locate 4 on the vertical axis and movehorizontally across to the graph. There aretwox-values that correspond to they-valueof 4. They are 39 and 82, so there areapproximately 4 million Americans age 39who are hearing-impaired andapproximately 4 million Americans age 82who are hearing-impaired.c)The highest point on the graph appears tocorrespond to thex-value of 63. Therefore,age 58 appears to be the age at which thegreatest number of Americans are hearing-impaired.d)Visually, we cannot tell precisely whichpoint is the highest point on the graph orwhichx-value corresponds exactly to thatpoint. The graph is not detailed enough tomake that determination.28.a)We estimate the incidence of breast cancerin 40-yr-old women is about 100 per100,000 women.b)We estimate that at age 67 and at age 88 theincidence of breast cancer is about 400 per100,000 women.c)The largest incidence of breast cancer occursin woman who are approximately 79 yearsof age.d)Visually, we cannot tell precisely whichpoint is the highest point on the graph orwhichx-value corresponds exactly to thatpoint. The graph is not detailed enough tomake that determination.29.a)1tAPr1100,000 10.028100,000 1.028102,800AAt the end of 1 year, the investment is worth$102,800.

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Exercise Set R.15b)1ntrAPn2 1220.028100,000 12100,000 10.014100,000 1.014A100,000 1.028196102,819.60AAt the end of 1 year, the investment is worth$102,819.60.c)1ntrAPn4 1440.028100,000 14100,000 10.07100,000 1.07100,000 1.0282953744102,829.537102,829.54AAt the end of 1 year, the investment is worth$102,829.54.d)1ntrAPn365 13653650.028100,000 1365100,000 10.00076712329100,000 1.00076712329100,000 1.02839458002102,839.458002102,839.46AAt the end of 1 year, the investment is worth$102,839.46.e)There are24 3658760hours in one year.1ntrAPn8760 1876087600.028100, 000 18760100, 000 10.000003196347100, 000 1.000003196347100, 000 1.02839563811102,839.563811102,839.56AAt the end of 1 year, the investment is worth$102,839.56.30.a)1tAPr1300,000 10.022300,000 1.022306, 600.00AAt the end of 1 year, the investment is worth$306,600.00.b)1ntrAPn2 10.022300, 000 12306, 636.30AAt the end of 1 year, the investment is worth$306,636.30.c)1ntrAPn4 140.022300, 000 14300, 000 1.0055306, 654.65AAt the end of 1 year, the investment is worth$306,654.65.d)1ntrAPn365 13650.022300,000 1365300,000 1.000060273973306,672.93AAt the end of 1 year, the investment is worth$306,672.93.

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6Chapter R Functions, Graphs, and Modelse)There are24 3658760hours in one year.1ntrAPn8760 187600.022300,000 18760300,000 1.000002511416306,673.13AAt the end of 1 year, the investment is worth$306,673.13.31.a)1tAPr3330, 000 10.0430, 000 1.0433, 745.92AAt the end of 3 years, the investment isworth $33,745.92.b)1ntrAPn2 360.0430, 000 1230, 000 1.0230, 000 1.126233, 784.87AAt the end of 3 years, the investment isworth $33,784.87.c)1ntrAPn4 3120.0430, 000 1430, 000 1.0130, 000 1.126833,804.750933,804.75AAt the end of 3 years, the investment isworth $33,804.75.d)1ntrAPn365 310950.0430, 000 136530, 000 1.00010958930, 000 1.127489387733,824.6816333,824.68AAt the end of 3 years, the investment isworth $33,824.68.e)There are24 3658760hours in one year.1ntrAPn8760 326,2800.0430, 000 1876030, 000 1.00000456621004630, 000 1.12749654267733,824.8962803133433,824.90AAt the end of 3 years, the investment isworth $33,824.90.32.a)1tAPr41000 10.051215.51AAt the end of 4 years, the investment isworth $1215.51.b)1ntrAPn2 480.051000 121000 1.0251218.40AAt the end of 4 years, the investment isworth $1218.40.c)1ntrAPn4 4160.051000 141000 1.01251219.89AAt the end of 4 years, the investment isworth $1219.89.

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Exercise Set R.17d)1ntrAPn365 414600.051000 13651000 1.0001369861221.39AAt the end of 4 years, the investment isworth $1221.39.e)There are24 3658760hours in one year.1ntrAPn8760 4350400.051000 187601000 1.0000057081221.40AAt the end of 4 years, the investment isworth $1221.40.33.Using the formula:112121112nnrrMPrWe substitute 18,000 forP, 0.0646.4%0.064forr, and 363 1236forn.Then we use a calculator to perform thecomputation.36360.0640.0641121218,0000.0641112550.86MThe monthly payment on the loan will beapproximately $550.86.34.30 years30 12360 months100, 000;0.048Pr3603600.0480.04811212100,0000.0481112524.67MThe monthly payment on the loan will beapproximately $524.67.35.11nrWPrWe substitute 3000 forP, 0.06576.57%0.0657forr, and 18 forn.1810.0657130000.065797,881.97WRounded to the nearest cent, Kate’s annuity willbe worth $97,881.97 after 18 years.36.We substitute 50,000 forW, 0.072517%0.07254in forr, and 20 forn. Thenwe proceed to solve forP.2010.0725150, 0000.0725P2050,00010.072510.07251186.74PPPaulo will need to invest $1186.74 annually toreach a goal of $50,000 after 20 years.37.a)Locate 230 on the vertical axis and thenthink of a horizontal line extending acrossthe graph from this point. The years forwhich the graph lies above this line are theyears for which the condor population wasat or above 230. We determine that thecondor population is above 230 for theperiod from 2007 to 2012.b)Locate 200 on the vertical axis and thenthink of a horizontal line extending acrossthe graph from this point. The years forwhich the graph touches this line are theyears for which the population was at 200.The condor population was at 200 in 2005.c)Locate the highest point on the graph andextend a line vertically to the horizontalaxis. The year which the condor populationwas the highest was 2012.d)Locate the lowest point on the graph andextend a line vertically to the horizontalaxis. The year when the condor populationwas the lowest was 2002.

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Solution Manual for Calculus and Its Applications, 11th Edition - Page 12 preview image

8Chapter R Functions, Graphs, and Models38.Answers will vary, but might includereferences to increased conservation efforts,Lower mortality rates of hatchlings,reduced number of predators etc…39.a) Using the formula11nrWPrwesubstitute 1200 forP, 0.044%0.04forrand 35 forn.3510.04112000.0488,382.67WSally will have approximately $88,382.67in her account when she retires.b) Sally invested $1200 per year for 35 years.Therefore, the total amount of her originalpayments is:$1200 35$42, 000. Sincethe total amount in the account was$88,382.67, the interest earned over the 35years is:$88, 382.67$42, 000$46, 382.67Therefore, $42,000 was the total amount ofSally’s payments and $46,382.67 was thetotal amount of her interest.40.a)Using the formula:112121112nnrrMPrWe substitute $88,382.67 forP, 0.044%0.04forr, and 15806515forn. Then we use a calculator to perform thecomputation.1801800.040.041121288,382.670.041112653.76MSally should take a monthly payment of$653.76.b)Sally received 180 payments of $653.76,therefore she received a total of$653.76180$117, 676.80during the 15years. Of that 42,000 was what sheoriginally contributed, leaving $75,676.80 ininterest.41.Substituting the information into the formula forannual yield give us120.053110.0543.12YThus, the annual percentage yield is 5.43%42.Substituting the information into the formula forannual yield give us40.041110.0416.4YThus, the annual percentage yield is 4.16%43.Substituting the information into the formula forannual yield give us520.0375110.0382.52YThus, the annual percentage yield is 3.82%44.Substituting the information into the formula forannual yield give us3650.04110.0408.365YThus, the annual percentage yield is 4.08%45.a)The annual yield for Western Bank is10.045110.045.1WBYThus the annual yield for Western Bank is4.5%.The annual yield for CommonwealthSavings is120.0443110.0452.12CWYThus the annual yield for Commonwealthsavings is 4.52%.b) Commonwealth savings has the higherannual yield.46.a)The annual yield for Sierra Savings is10.05110.05.1SSYThus the annual yield for Sierra Savings is5%.The annual yield for Foothill bank is520.0488110.0499860.05.52FBYThus the annual yield for Foothill bank is5% (4.9986%).b) Sierra Savings offers a slightly higherannual yield.

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Solution Manual for Calculus and Its Applications, 11th Edition - Page 13 preview image

Exercise Set R.1947.The annual yield for Stockman’s Bank is10.042110.042.1SBYThus the annual yield for Stockman’s Bank is4.2%.In order to compete, the annual yield forMesalands Savings must be 4.2%. We substituteinto the annual yield formula and solve for theinterest rate1121121121212110.0421211.0421211.042121.042112121.04210.0412.rrrrrrMesalands Savings needs to offer at least 4.12%compounded monthly to be competitive.48.The annual yield for Belltown Bank is10.0375110.0375.1SBYThus the annual yield for Belltown Bank is3.75%.In order to compete, the annual yield forShea Savings must be 3.75%. We substituteinto the annual yield formula and solve for theinterest rate:14141444110.0375411.0375411.037541.03751441.037510.03698.rrrrrrShea Savings needs to offer at least 3.7%(3.698%) compounded quarterly to becompetitive.49.Graph150yxWe use the following window.Next, we type the equation into the calculator.The resulting graph is:50.Graph25yxUsing the following window:The resulting graph is

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Solution Manual for Calculus and Its Applications, 11th Edition - Page 14 preview image

10Chapter R Functions, Graphs, and Models51.Graph322413yxxxWe use the following window:Next, we type the equation in to the calculator.The resulting graph is:52.Graph237yx.Using a standard window.Results in the graph:53.Graph9.64.2100xy .First, we solve fory.subtract 9.6from both sides9.64.21004.21009.69.61004.2xxyyxxy  Next, we set the window to be:Next, we type the equation into the calculator.The resulting graph is:54.Graph22.34.89yxx .We use the following window:The resulting graph is:

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Solution Manual for Calculus and Its Applications, 11th Edition - Page 15 preview image

Exercise Set R.11155.Graph24xy.First we solve fory.2subtracting 42from both sidestaking the square rootof both sides444xyxyxyNext, we set the window to the standardwindow:It is important to remember that we must graphboth the positive root and the negative root.We type both equations into the calculator.This resulting graph is:56.Graph28xy.Solving foryyields:8yx We use the standard window.The resulting graph is:

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Solution Manual for Calculus and Its Applications, 11th Edition - Page 16 preview image

12Chapter R Functions, Graphs, and ModelsExercise Set R.21.The correspondence is a function because eachmember of the domain corresponds to only onemember of the range.2.The correspondence is a function because eachmember of the domain corresponds to only onemember of the range.3.The correspondence is a function because eachmember of the domain corresponds to only onemember of the range.4.The correspondence isnota function becauseone member of the domain, 6, corresponds totwo members of the range, – 6 and – 7 .5.The correspondence is a function because eachmember of the domain corresponds to only onemember of the range, even though two membersof the domain, 10 pc. Chicken McNuggets andthe Crispy Chicken both correspond to $4.29.6.The correspondence is a function because eachmember of the domain corresponds to only onemember of the range.7.The correspondence is a function because eachiPod has exactly one amount of memory.8.The correspondence is a function because eachiPod has exactly one serial number.9.The correspondence is a function because eachiPod has exactly one number of songs at anygiven time.10.The correspondence is a function because eachiPod has exactly one number of Avril Lavignesongs at any given time.11.The correspondence is a function because anynumber squared and then increased by 8,corresponds to exactly one number greater thanor equal to 8.12.The correspondence is a function because anynumber raised to the fourth power correspondsto exactly one nonnegative number.13.The correspondence is a function because everyfemale has exactly one biological mother.14.The correspondence is a function because everymale has exactly one biological father.15.This correspondence isnota function, becauseit is reasonable to assume at least one avenue isintersected by more than one cross street.16.This correspondence isnota function, becauseit is reasonable to assume that a textbook hasmore than one even-numbered page.17.The correspondence is a function because eachshape has exactly one area.18.The correspondence is a function because eachshape has exactly one perimeter.19.a)43fxx  5.14 5.1317.45.014 5.01317.045.0014 5.001317.00454 5317ffffx5.15.015.0015fx17.417.0417.00417b)43fxx    44 431334 339242311434314 134434143443ffffkkkfttttfxhxhxh 20.a)32fxxx4.14.014.0014fx14.314.0314.00314b)32fxx  53 521713121323213 123323532332fffkkkfttttfxhxhxh 

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