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Solution Manual for Calculus and Its Applications, 11th Edition

Solution Manual for Calculus and Its Applications, 11th Edition breaks down difficult textbook problems into simple solutions, making your study time more effective.

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     ’ S S OLUTIONS M ANUAL D AVID D UBRISKE University of Arkansas for Medical Sciences C ALCULUS AND ITS A PPLICATIONS E LEVENTH E DITION Marvin L. Bittinger Indiana University Purdue University Indianapolis David J. Ellenbogen Community College of Vermont Scott A. Surgent Arizona State University

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    TABLE OF CONTENTS Chapter R Functions, Graphs, and Models ..................................................................................................... 1 R.1 Graphs and Equations ........................................................................................................................... 1 R.2 Functions and Models ......................................................................................................................... 12 R.3 Finding Domain and Range ................................................................................................................ 25 R.4. Slope and Linear Functions ............................................................................................................... 32 R.5 Nonlinear Functions and Models ........................................................................................................ 46 R.6 Mathematical Modeling and Curve Fitting ......................................................................................... 66 Chapter 1 Differentiation ............................................................................................................................ 73 1.1 Limits: A Numerical and Graphical Approach .................................................................................... 73 1.2 Algebraic Limits and Continuity ......................................................................................................... 86 1.3 Average Rates of Change .................................................................................................................... 97 1.4 Differentiation Using Limits of Difference Quotients ....................................................................... 116 1.5 The Power and Sum–Difference Rules.............................................................................................. 132 1.6 The Product and Quotient Rules ........................................................................................................ 155 1.7 The Chain Rule .................................................................................................................................. 178 1.8 Higher-Order Derivatives .................................................................................................................. 201 Chapter 2 Applications of Differentiation ................................................................................................217 2.1 Using First Derivatives to Classify Maximum and Minimum Values and Sketch Graphs................ 217 2.2 Using Second Derivatives to Classify Maximum and Minimum Values and Sketch Graphs ........... 245 2.3 Graph Sketching: Asymptotes and Rational Functions ..................................................................... 290 2.4 Using Derivatives to Find Absolute Maximum and Minimum Values ............................................. 334 2.5 Maximum–Minimum Problems; Business, Economics and General Applications ........................... 363 2.6 Marginals and Differentials ............................................................................................................... 398 2.7 Elasticity of Demand ......................................................................................................................... 412 2.8 Implicit Differentiation and Related Rates ........................................................................................ 418 Chapter 3 Exponential and Logarithmic Functions ................................................................................434 3.1 Exponential Functions ....................................................................................................................... 434 3.2 Logarithmic Functions....................................................................................................................... 439 3.3 Applications: Uninhibited and Limited Growth Models ................................................................... 473 3.4 Applications: Decay .......................................................................................................................... 492 3.5 The Derivatives of a x and log a x ......................................................................................................... 505 3.6 A Business Application: Amortization .............................................................................................. 514 Chapter 4 Integration .................................................................................................................................525 4.1 Antidifferentiation ............................................................................................................................. 525 4.2 Antiderivatives as Areas .................................................................................................................... 538 4.3 Area and Definite Integrals................................................................................................................ 552 4.4 Properties of Definite Integrals.......................................................................................................... 569 4.5 Integration Techniques: Substitution ................................................................................................. 588 4.6 Integration Techniques: Integration by Parts ..................................................................................... 609 4.7 Integration Techniques: Tables.......................................................................................................... 628 Chapter 5 Applications of Integration ......................................................................................................637 5.1 Consumer Surplus and Producer Surplus .......................................................................................... 637 5.2 Integrating Growth and Decay Models.............................................................................................. 651 5.3 Improper Integrals ............................................................................................................................. 663 5.4 Probability ......................................................................................................................................... 674 5.5 Probability: Expected Value; The Normal Distribution .................................................................... 685 5.6 Volume .............................................................................................................................................. 699 5.7 Differential Equations ....................................................................................................................... 712

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    Chapter 6 Functions of Several Variables ................................................................................................731 6.1 Functions of Several Variables .......................................................................................................... 731 6.2 Partial Derivatives ............................................................................................................................. 737 6.3 Maximum-Minimum Problems ......................................................................................................... 755 6.4 An Application: The Least-Squares Technique ................................................................................. 772 6.5 Constrained Optimization .................................................................................................................. 787 6.6 Double Integrals ................................................................................................................................ 819

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    Chapter R Functions, Graphs, and Models Exercise Set R.1 1. Graph 1 y x   . We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 2. Graph 4 y x   . 3. Graph 1 4 y x   . We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 4. Graph 3 y x   . 5. Graph 5 3 3 y x    . We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 6. Graph 2 4 3 y x   . 7. Graph 5 x y   . We solve for y first. subtract from both sides commutative property 5 5 5 x x y y x y x        Next, we choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 8. Graph 4 x y   . x y   , x y 2  3    2, 3   0 1    0, 1  3 2   3, 2 x y   , x y 4  1   4,1  0 0   0, 0 4 1    4, 1  x y   , x y 3  8   3,8  0 3   0, 3 3 2    3, 2  x y   , x y 1  6   1, 6  0 5   0,5 2 3   2, 3

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    2 Chapter R Functions, Graphs, and Models 9. Graph 6 3 9 x y    . We solve for y first. 6 from both sides divide both sides by 3 subtract 6 3 9 3 6 9 6 9 3 3 2 3 x x y y x y x y x             Next, we choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 10. Graph 8 2 4 y x   . 11. Graph 2 5 10 x y   . We solve for y first.   subtract 2 from both sides divide both sides by 5 2 5 10 5 10 2 1 10 2 5 2 2 5 2 2 5 x x y y x y x y x y x            We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y   , x y 5  4   5, 4  0 2   0, 2 5 0   5, 0 12. Graph 5 6 12 x y   . 13. Graph 2 5 y x   . We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y   , x y 2  1    2, 1   1  4    1, 4   0 5    0, 5  1 4    1, 4  2 1    2, 1  14. Graph 2 3 y x   . 15. Graph 2 2 x y   . Since x is expressed in terms of y we first choose values for y and then compute x . Then we plot the points that are found and connect them with a smooth curve. x y   , x y 6 2    2, 2   3 1    1, 1  2 0   2, 0 3 1   1,1  6 2   2, 2  x y   , x y 2  1   2,1  0 3    0, 3  2 7    2, 7 

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    Exercise Set R.1 3 16. Graph 2 2 x y   . 17. Graph 4 y x   . We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a curve. x y   , x y 1  5   1, 5  0 4   0, 4 4 0   4, 0 5 1   5,1 6 2   6, 2 18. Graph y x  . 19. Graph 2 7 y x   . We choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y   , x y 2  3   2, 3  1  6   1, 6  0 7   0, 7 1 6   1, 6 2 3   2, 3 20. Graph 2 5 y x   . 21. Graph 3 7 y x   . First we solve for y . 3 3 7 7 y x y x     Next, we choose some x -values and calculate the corresponding y -values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y   , x y 2  1    2, 1   1  6   1, 6  0 7   0, 7 1 8   1,8 2 15   2,15 22. Graph 3 1 y x   . 23. 4 3 2 0.5 3.45 96.65 347.7 , 0 6 A t t t t t       To determine the amount of ibuprofen in the blood stream that is left after 2 hours, we substitute 2 t  into the equation and solve for A .         4 3 2 0.5 2 3.45 2 96.65 2 347.7 2 344.4 A      According to the model, approximately 344.4 milligrams of ibuprofen will remain in the blood stream 2 hours after 400 mg have been swallowed.

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    4 Chapter R Functions, Graphs, and Models 24. 0.006 15.714 R x    We substitute 1954 in for x to get   0.006 1954 15.714 3.99 R     According to this model, the world record for the mile in 1954 is approximately 3.99 minutes. To convert this to traditional minutes-seconds we multiply the decimal part by 60 seconds.   0.99 60 59.4  Therefore the world record for the mile in 1954 was 3:59.4. Likewise, we substitute 2000 in for x to get   0.006 2000 15.714 3.714 R     According to the model, the world record for the mile in 2008 will be approximately 3.71 minutes or 3:42.8. Finally, we substitute 2015 in for x to get   0.006 2015 15.714 3.624 R     According to the model, the world record for the mile in 2015 will be approximately 3.62 minutes or 3:37.44. 25.   10.9 v t t  We substitute 2.5 in for t to get     2.5 10.9 2.5 27.25 v   Shaun White was traveling at 27.25 miles per hour when he reentered the half pipe. 26.   2 16 s t t    2 2 2 28 28 16 28 16 28 16 1.3228 s t t t t t      Danny took approximately 1.32 seconds to hit the ramp. 27. a) Locate 20 on the horizontal axis and go directly up to the graph. Then move left to the vertical axis and read the value there. We estimate the number of hearing-impaired Americans of age 20 is about 1.8 million. Follow the same process for 40, 50, and 60 to determine the number of hearing- impaired Americans at each of those ages. We estimate the number of hearing-impaired Americans of age 40 is about 4.1 million. We estimate the number of hearing-impaired Americans of age 50 is about 5.3 million. We estimate the number of hearing-impaired Americans of age 60 is about 6.0 million. b) Locate 4 on the vertical axis and move horizontally across to the graph. There are two x -values that correspond to the y -value of 4. They are 39 and 82, so there are approximately 4 million Americans age 39 who are hearing-impaired and approximately 4 million Americans age 82 who are hearing-impaired. c) The highest point on the graph appears to correspond to the x -value of 63. Therefore, age 58 appears to be the age at which the greatest number of Americans are hearing- impaired. d) Visually, we cannot tell precisely which point is the highest point on the graph or which x -value corresponds exactly to that point. The graph is not detailed enough to make that determination. 28. a) We estimate the incidence of breast cancer in 40-yr-old women is about 100 per 100,000 women. b) We estimate that at age 67 and at age 88 the incidence of breast cancer is about 400 per 100,000 women. c) The largest incidence of breast cancer occurs in woman who are approximately 79 years of age. d) Visually, we cannot tell precisely which point is the highest point on the graph or which x -value corresponds exactly to that point. The graph is not detailed enough to make that determination. 29. a)   1 t A P r       1 100,000 1 0.028 100,000 1.028 102,800 A     At the end of 1 year, the investment is worth $102,800.

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    Exercise Set R.1 5 b) 1 nt r A P n             2 1 2 2 0.028 100,000 1 2 100,000 1 0.014 100,000 1.014 A               100,000 1.028196 102,819.60 A   At the end of 1 year, the investment is worth $102,819.60. c) 1 nt r A P n               4 1 4 4 0.028 100,000 1 4 100,000 1 0.07 100,000 1.07 100,000 1.0282953744 102,829.537 102,829.54 A                At the end of 1 year, the investment is worth $102,829.54. d) 1 nt r A P n               365 1 365 365 0.028 100,000 1 365 100,000 1 0.00076712329 100,000 1.00076712329 100,000 1.02839458002 102,839.458002 102,839.46 A                At the end of 1 year, the investment is worth $102,839.46. e) There are 24 365 8760   hours in one year. 1 nt r A P n               8760 1 8760 8760 0.028 100, 000 1 8760 100, 000 1 0.000003196347 100, 000 1.000003196347 100, 000 1.02839563811 102,839.563811 102,839.56 A                At the end of 1 year, the investment is worth $102,839.56. 30. a)   1 t A P r       1 300,000 1 0.022 300,000 1.022 306, 600.00 A     At the end of 1 year, the investment is worth $306,600.00. b) 1 nt r A P n         2 1 0.022 300, 000 1 2 306, 636.30 A           At the end of 1 year, the investment is worth $306,636.30. c) 1 nt r A P n           4 1 4 0.022 300, 000 1 4 300, 000 1.0055 306, 654.65 A            At the end of 1 year, the investment is worth $306,654.65. d) 1 nt r A P n           365 1 365 0.022 300,000 1 365 300,000 1.000060273973 306,672.93 A            At the end of 1 year, the investment is worth $306,672.93.

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    6 Chapter R Functions, Graphs, and Models e) There are 24 365 8760   hours in one year. 1 nt r A P n           8760 1 8760 0.022 300,000 1 8760 300,000 1.000002511416 306,673.13 A            At the end of 1 year, the investment is worth $306,673.13. 31. a)   1 t A P r       3 3 30, 000 1 0.04 30, 000 1.04 33, 745.92 A     At the end of 3 years, the investment is worth $33,745.92. b) 1 nt r A P n             2 3 6 0.04 30, 000 1 2 30, 000 1.02 30, 000 1.1262 33, 784.87 A             At the end of 3 years, the investment is worth $33,784.87. c) 1 nt r A P n             4 3 12 0.04 30, 000 1 4 30, 000 1.01 30, 000 1.1268 33,804.7509 33,804.75 A              At the end of 3 years, the investment is worth $33,804.75. d) 1 nt r A P n             365 3 1095 0.04 30, 000 1 365 30, 000 1.000109589 30, 000 1.1274893877 33,824.68163 33,824.68 A              At the end of 3 years, the investment is worth $33,824.68. e) There are 24 365 8760   hours in one year. 1 nt r A P n             8760 3 26,280 0.04 30, 000 1 8760 30, 000 1.000004566210046 30, 000 1.127496542677 33,824.89628031334 33,824.90 A              At the end of 3 years, the investment is worth $33,824.90. 32. a)   1 t A P r     4 1000 1 0.05 1215.51 A    At the end of 4 years, the investment is worth $1215.51. b) 1 nt r A P n           2 4 8 0.05 1000 1 2 1000 1.025 1218.40 A            At the end of 4 years, the investment is worth $1218.40. c) 1 nt r A P n           4 4 16 0.05 1000 1 4 1000 1.0125 1219.89 A            At the end of 4 years, the investment is worth $1219.89.

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    Exercise Set R.1 7 d) 1 nt r A P n           365 4 1460 0.05 1000 1 365 1000 1.000136986 1221.39 A            At the end of 4 years, the investment is worth $1221.39. e) There are 24 365 8760   hours in one year. 1 nt r A P n           8760 4 35040 0.05 1000 1 8760 1000 1.000005708 1221.40 A            At the end of 4 years, the investment is worth $1221.40. 33. Using the formula: 1 12 12 1 1 12 n n r r M P r                               We substitute 18,000 for P , 0.064   6.4% 0.064  for r , and 36   3 12 36   for n . Then we use a calculator to perform the computation. 36 36 0.064 0.064 1 12 12 18,000 0.064 1 1 12 550.86 M                                The monthly payment on the loan will be approximately $550.86. 34. 30 years 30 12 360 months    100, 000; 0.048 P r   360 360 0.048 0.048 1 12 12 100,000 0.048 1 1 12 524.67 M                                The monthly payment on the loan will be approximately $524.67. 35.   1 1 n r W P r            We substitute 3000 for P , 0.0657   6.57% 0.0657  for r , and 18 for n .   18 1 0.0657 1 3000 0.0657 97,881.97 W             Rounded to the nearest cent, Kate’s annuity will be worth $97,881.97 after 18 years. 36. We substitute 50,000 for W , 0.0725 1 7 % 0.0725 4        in for r , and 20 for n . Then we proceed to solve for P.   20 1 0.0725 1 50, 000 0.0725 P              20 50,000 1 0.0725 1 0.0725 1186.74 P P             Paulo will need to invest $1186.74 annually to reach a goal of $50,000 after 20 years. 37. a) Locate 230 on the vertical axis and then think of a horizontal line extending across the graph from this point. The years for which the graph lies above this line are the years for which the condor population was at or above 230. We determine that the condor population is above 230 for the period from 2007 to 2012. b) Locate 200 on the vertical axis and then think of a horizontal line extending across the graph from this point. The years for which the graph touches this line are the years for which the population was at 200. The condor population was at 200 in 2005. c) Locate the highest point on the graph and extend a line vertically to the horizontal axis. The year which the condor population was the highest was 2012. d) Locate the lowest point on the graph and extend a line vertically to the horizontal axis. The year when the condor population was the lowest was 2002.

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    8 Chapter R Functions, Graphs, and Models 38. Answers will vary, but might include references to increased conservation efforts, Lower mortality rates of hatchlings, reduced number of predators etc… 39. a) Using the formula   1 1 n r W P r            we substitute 1200 for P , 0.04   4% 0.04  for r and 35 for n .   35 1 0.04 1 1200 0.04 88,382.67 W             Sally will have approximately $88,382.67 in her account when she retires. b) Sally invested $1200 per year for 35 years. Therefore, the total amount of her original payments is: $1200 35 $42, 000   . Since the total amount in the account was $88,382.67, the interest earned over the 35 years is: $88, 382.67 $42, 000 $46, 382.67   Therefore, $42,000 was the total amount of Sally’s payments and $46,382.67 was the total amount of her interest. 40. a) Using the formula: 1 12 12 1 1 12 n n r r M P r                               We substitute $88,382.67 for P , 0.04   4% 0.04  for r , and 15   80 65 15   for n . Then we use a calculator to perform the computation. 180 180 0.04 0.04 1 12 12 88,382.67 0.04 1 1 12 653.76 M                                Sally should take a monthly payment of $653.76. b) Sally received 180 payments of $653.76, therefore she received a total of $653.76 180 $117, 676.80   during the 15 years. Of that 42,000 was what she originally contributed, leaving $75,676.80 in interest. 41. Substituting the information into the formula for annual yield give us 12 0.053 1 1 0.0543. 12 Y           Thus, the annual percentage yield is 5.43% 42. Substituting the information into the formula for annual yield give us 4 0.041 1 1 0.0416. 4 Y           Thus, the annual percentage yield is 4.16% 43. Substituting the information into the formula for annual yield give us 52 0.0375 1 1 0.0382. 52 Y           Thus, the annual percentage yield is 3.82% 44. Substituting the information into the formula for annual yield give us 365 0.04 1 1 0.0408. 365 Y           Thus, the annual percentage yield is 4.08% 45. a) The annual yield for Western Bank is 1 0.045 1 1 0.045. 1 WB Y           Thus the annual yield for Western Bank is 4.5%. The annual yield for Commonwealth Savings is 12 0.0443 1 1 0.0452. 12 CW Y           Thus the annual yield for Commonwealth savings is 4.52%. b) Commonwealth savings has the higher annual yield. 46. a) The annual yield for Sierra Savings is 1 0.05 1 1 0.05. 1 SS Y           Thus the annual yield for Sierra Savings is 5%. The annual yield for Foothill bank is 52 0.0488 1 1 0.049986 0.05. 52 FB Y            Thus the annual yield for Foothill bank is 5% (4.9986%). b) Sierra Savings offers a slightly higher annual yield.

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    Exercise Set R.1 9 47. The annual yield for Stockman’s Bank is 1 0.042 1 1 0.042. 1 SB Y           Thus the annual yield for Stockman’s Bank is 4.2%. In order to compete, the annual yield for Mesalands Savings must be 4.2%. We substitute into the annual yield formula and solve for the interest rate       1 12 1 12 1 12 12 12 1 1 0.042 12 1 1.042 12 1 1.042 12 1.042 1 12 12 1.042 1 0.0412. r r r r r r                               Mesalands Savings needs to offer at least 4.12% compounded monthly to be competitive. 48. The annual yield for Belltown Bank is 1 0.0375 1 1 0.0375. 1 SB Y           Thus the annual yield for Belltown Bank is 3.75%. In order to compete, the annual yield for Shea Savings must be 3.75%. We substitute into the annual yield formula and solve for the interest rate:       1 4 1 4 1 4 4 4 1 1 0.0375 4 1 1.0375 4 1 1.0375 4 1.0375 1 4 4 1.0375 1 0.03698. r r r r r r                               Shea Savings needs to offer at least 3.7% (3.698%) compounded quarterly to be competitive. 49. Graph 150 y x   We use the following window. Next, we type the equation into the calculator. The resulting graph is: 50. Graph 25 y x   Using the following window: The resulting graph is

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    10 Chapter R Functions, Graphs, and Models 51. Graph 3 2 2 4 13 y x x x     We use the following window: Next, we type the equation in to the calculator. The resulting graph is: 52. Graph 23 7 y x   . Using a standard window. Results in the graph: 53. Graph 9.6 4.2 100 x y    . First, we solve for y . subtract 9.6 from both sides 9.6 4.2 100 4.2 100 9.6 9.6 100 4.2 x x y y x x y          Next, we set the window to be: Next, we type the equation into the calculator. The resulting graph is: 54. Graph 2 2.3 4.8 9 y x x     . We use the following window: The resulting graph is:

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    Exercise Set R.1 11 55. Graph 2 4 x y   . First we solve for y . 2 subtracting 4 2 from both sides taking the square root of both sides 4 4 4 x y x y x y        Next, we set the window to the standard window: It is important to remember that we must graph both the positive root and the negative root. We type both equations into the calculator. This resulting graph is: 56. Graph 2 8 x y   . Solving for y yields: 8 y x    We use the standard window. The resulting graph is:

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    12 Chapter R Functions, Graphs, and Models Exercise Set R.2 1. The correspondence is a function because each member of the domain corresponds to only one member of the range. 2. The correspondence is a function because each member of the domain corresponds to only one member of the range. 3. The correspondence is a function because each member of the domain corresponds to only one member of the range. 4. The correspondence is not a function because one member of the domain, 6, corresponds to two members of the range, – 6 and – 7 . 5. The correspondence is a function because each member of the domain corresponds to only one member of the range, even though two members of the domain, 10 pc. Chicken McNuggets and the Crispy Chicken both correspond to $4.29. 6. The correspondence is a function because each member of the domain corresponds to only one member of the range. 7. The correspondence is a function because each iPod has exactly one amount of memory. 8. The correspondence is a function because each iPod has exactly one serial number. 9. The correspondence is a function because each iPod has exactly one number of songs at any given time. 10. The correspondence is a function because each iPod has exactly one number of Avril Lavigne songs at any given time. 11. The correspondence is a function because any number squared and then increased by 8, corresponds to exactly one number greater than or equal to 8. 12. The correspondence is a function because any number raised to the fourth power corresponds to exactly one nonnegative number. 13. The correspondence is a function because every female has exactly one biological mother. 14. The correspondence is a function because every male has exactly one biological father. 15. This correspondence is not a function, because it is reasonable to assume at least one avenue is intersected by more than one cross street. 16. This correspondence is not a function, because it is reasonable to assume that a textbook has more than one even-numbered page. 17. The correspondence is a function because each shape has exactly one area. 18. The correspondence is a function because each shape has exactly one perimeter. 19. a)   4 3 f x x                   5.1 4 5.1 3 17.4 5.01 4 5.01 3 17.04 5.001 4 5.001 3 17.004 5 4 5 3 17 f f f f             x 5.1 5.01 5.001 5   f x 17.4 17.04 17.004 17 b)   4 3 f x x                           4 4 4 3 13 3 4 3 3 9 2 4 2 3 11 4 3 4 3 1 4 1 3 4 4 3 4 1 4 3 4 4 3 f f f f k k k f t t t t f x h x h x h                                 20. a)   3 2 f x x   x 4.1 4.01 4.001 4   f x 14.3 14.03 14.003 14 b)   3 2 f x x                       5 3 5 2 17 1 3 1 2 1 3 2 3 2 1 3 1 2 3 3 2 3 5 3 2 3 3 2 f f f k k k f t t t t f x h x h x h                             

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