Chapter 4 Analyzing Change: Applications of Derivatives

Chapter 4 Analyzing Change: Applications of Derivatives Section 4.1 Approximating Change 1. 3% (4 percentage points per hour) 1 ( ) = 1 1 hour 30 % 3 3. 300 mph + (00 mph per hour) ( hour ) 316 3. f (3.5) f (3) + f (3)(0.5) = 17 + 4.6(0.5) = 19.3 4. g(7.5) g(7) + g (7)(0.5) = 4 + ( 1.9)(0.5) = 0.775 = mph 5. a. Increasing production from 500 to 501 units will increase total cost by approximately $17. b. If sales increase from 150 to 151 units, then profit will increase by approximately $4.75. 6. a. Increasing sales from 500 to 501 units will increase revenue by approximately $10.00 and cost by approximately $13.00. b. Increasing sales from 10 to 11 units will decrease profit by $3.46. 7. A marginal profit of $4 per shirt means that at this point the fraternity s profit is decreasing by $4 for each additional shirt sold. The fraternity should consider selling fewer shirts or increasing the sales price. 8. No. If the marginal profit at x is negative, then as x is increased the profit will decrease, but it will not necessarily be negative. 3 9. Using the two points (70,8000 and 54,0) $8000 Slope of tangent line = $500 per year 16 years of age of age Annual premium for 70-year-old $8000 Annual premium $500 for 7-year-old $8000 + ( years) = $9000 year (Estimates will vary.) 141

14 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 10. Slope of tangent line 10 years =.5 years per decade 4 decades Life expectancy in 1990 65 years Life expectancy in 000 65 years + (.5 years/decade)(1 decade) = 67.5 years (Estimates will vary.) 11. a. 97 billion dollars Slope of tangent line is approximately = 97 billion dollars per billion dollars 1 billion dollars (revenue dollars per sales dollars). Revenue is approximately $614 billion when $6 billion is spent on advertising. Revenue is approximately 614 + 0.5(97) = 66. 5 billion dollars when $6.5 billion is spent on advertising. (Discussion will vary.) b. R(6.5) $658 billion c. The solution from the model is more accurate than that which is derived from an interpretation of the graph, because it is difficult to accurately draw a tangent line on so small a graph.

Calculus Concepts Section 4.1: Approximating Change 143 1. a. Slope of tangent line 50 million kg 4 years = 1.5 million kg per year CFC-11 releases in 199 170 million kg CFC-11 releases in 1993 170 million kg + ( 1.5 million kg per year)(1 year) = 157.5 million kg b. C(5) 165 million kg c. The tangent line estimate is closer to the actual amount, because it is an underestimate of the model value, and the model overestimates the actual value. 13. a. P ( x) = 68.79(1.013087 ) thousand people in year x x x P '( x) = 68.79(ln1.013087)(1.013087 ) thousand people per year in year x In 000 the population of South Carolina was increasing by 53.6 thousand people per year. b. Between 000 and 003, the population increased by approximately 160.8 thousand people. c. By finding the slope of the tangent line at 000 and multiplying by 3, we determine the change in the tangent line from 000 through 003 and use that change to estimate the change in the population function. 14. a. A(t) = 10(1.16 t ) thousand dollars. A (t) = 10(ln 1.16)(1.16 t ) thousand dollars/year. b. A (10) = 46.656. The investment, at ten years, is changing at a rate of $46,656/year. c. Growth in the 1 st half of the 11 th year will be 1 or $3,38. (46.656) d. The percentage rate of change after 10 years is A'( t) A'(10) A( t) 100% = A(10) 100% i i = 11.9%/year. e. The percentage rate of change and the perdentage change, while both constants, are not equal. The percentage change is (b-1)i 100% = 1.6%. The percentage rate of change is the log of bi100%= 11.9%. 15. a. The population was growing at a rate of.5 million people per year in 1998. b. Between 1998 and 1999, the population of Mexico increased by approximately.5 million people.

144 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 16. a. 3 G( t) = 0.044t + 0.9t + 38 points after t hours of study G(11) 90.8 points G ( t) = 0.13t + 1.84t i. G (11) 4.7 points per hour of study ii. G(1) 94.4 points b. G(1) = G(11 + 1) G(11) + G (11)(1) (90.76 points) + (4.7 points per hour)(1 hour) = 95.03 points This is an overestimate, because the graph of the model is concave down for values of t between 11 and 1. 17. a. In 1998 the amount was increasing by 1.15 million pieces per year. b. We would expect an increase of approximately 1.15 million pieces between 1998 and 1999. c. p(4) p(3) 1.3 million pieces d. 101.9 100.4 = 1.5 million pieces e. As long as the data in part d were correctly reported, the answer to part d is the most accurate one. 18. a. Production Costs = hourly. 3 C( p) = 0.16p 8.7 p + 17 p + 69.4 dollars when p units are produced b. C ( p) = 0.48p + 17.4 p + 17 dollars per unit when p units are produced hour. C (5) = 97 dollars per unit C (0) = 16 dollars per unit C (30) = 8 dollars per unit When 5 units are produced hourly, the hourly cost is increasing at a rate of $97 per additional unit produced in an hour. When 0 units are produced hourly, the hourly cost is increasing at a rate of $16 per additional unit produced in an hour. When 30 units are produced hourly, the hourly cost is increasing at a rate of $8 per additional unit produced in an hour. c. Cost of 6th unit: C(6) C(5) $8.76 $731.90 = $90.86 Cost of 1st unit: C(1) C(0) $136.46 $1309.40 $17.06 Cost of 31st unit: C(31) C(30) $1807.6 $1719.40 = $87.86 d. The model is concave down at x = 5, but it is concave up at x = 0 and x = 30. d. C( p) 1 A( p) = = 0.16p 8.7 p + 17 p + 69.4 p dollars per unit p when p units are produced hourly

Calculus Concepts Section 4.1: Approximating Change 145 e. 1 A ( p) = 0.3 p 8.7 + 17 p + 69.4 p dollars per unit per unit when p units are produced hourly When the production level is at 5, the average cost is changing at a rate of A (5) 30.08 dollars per unit per unit produced hourly. When the production level is at 0, the average cost is changing at a rate of A (0) 6.47 dollars per unit per unit produced hourly. When the production level is at 30, the average cost is changing at a rate of A (30) 6.71 dollars per unit per unit produced hourly. When 5 units are produced hourly, the per-unit cost increases by $30.08 per additional unit produced hourly. When 0 units are produced hourly, the per-unit cost decreases by $6.47 per additional unit produced hourly. When 30 units are produced hourly, the per-unit cost decreases by $6.71 per additional unit produced hourly. 19. a. A = 300(1 + ).065 1 (1 t) b. A = 300(1.06697) t c. A() = $341.53 d. A () = 300(ln 1.06697)(1.06697) =.14 dollars/year. e. A(.5) A() +.5 A'() = $347.07 0. a. A = 000(1 + ).03 1 (1 t) b. A = 000(1.035) t c. A(5) = $346.5 d. A (5) = 000(ln 1.035)(1.035) 5 = 74.99 dollars/year. e. A(5.5) A(5) +.5 A'(5) = $384.0 1. Sales a. ( ) 4 R( x) = 7.03 10 x + 1.666x + 47.130 dollars when x hot dogs are sold, 100 < x < 1500. b. Cost: C( x) = 0. 5x dollars when x hot dogs are sold, 100 < x < 1500. 4 Profit: P( x) = R( x) C( x) = ( 7.03 10 ) x + 1.166x + 47. 130 dollars when x hot dogs are sold, 100 < x < 1500. c. Marginal Revenue = R (x) = -.0014x + 1.666 dollars/hot dog. x (hot dogs) R ( x ) c ( x ) (dollars per hot dog) (dollars per hot dog) 00 1.38 0.50 0.88 800 0.54 0.50 0.04 1100 0.1 0.50 0.38 p ( x ) (dollars per hot dog) 1400 0.30 0.50 0.80

146 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts If the number of hot dogs sold increases from 00 to 01, the revenue increases by approximately $1.38 and the profit increases by approximately $0.88. If the number increases from 800 to 801, the revenue increases by 0.54, but the profit sees almost no increase (4 cents). If the number increases from 1100 to 1100, the increase in revenue is only approximately 1 cents. Because this marginal revenue is less than the marginal cost at a sales level of 1100, the result of the sales increase from 1100 to 1101 is a decrease of $0.38 in profit. If the number of hot dogs increases from 1400 to 1401, revenue declines by approximately 30 cents and profit declines by approximately 80 cents. d. The marginal values in part c are the slopes of the graphs shown here. For example, at x = 800, the slope of the revenue graph is $0.54 per hot dog, the slope of the cost graph is $0.50 per hot dog, and the slope of the profit graph is $0.04 per hot dog. We see from the graph that maximum profit is realized when approximately 800 hot dogs are sold. Revenue is greatest near x = 1100, so the marginal revenue there is small. However, once costs are factored in, the profit is actually declining at this sales level. This is illustrated by the graph.. Production a. b. 3 C( x) = 0.068x.933x + 55.69x + 146.983dollars when x hundred balls are produced hourly C ( x) = 0.04x 5.865x + 55.69 dollars per hundred balls when x hundred balls are produced hourly C (10) 16.99 dollars per hundred balls or $0.17 per ball The hourly cost will increase by approximately $0.17 for each additional ball produced in an hour. c. C (3) 39.51dollars per hundred balls or $0.395 per ball d. e. C (1) $1.94 dollars per hundred balls or $0. per ball When 300 balls are produced hourly, the hourly cost will increase by approximately $0.395 for each additional ball produced in an hour. When 100 balls are produced hourly, the hourly cost will increase by approximately $0. for each additional ball produced in an hour. C( x) 1 A( x) = = 0.068x.933x + 55.69 + 149.983x dollars per hundred balls x when x hundred balls are produced hourly A ( x) = 0.136x.933 146.983x dollars per hundred balls per hundred balls when x hundred balls are produced hourly (3) 18.86 A dollars per hundred balls per additional hundred balls produced hourly or $0.00189 per ball per additional ball produced hourly

Calculus Concepts Section 4.1: Approximating Change 147 A (17) 1.13 dollars per hundred balls per additional hundred balls produced hourly or $0.000113 per ball per additional ball produced hourly When 300 balls are produced hourly, the cost per ball will decrease by approximately $0.00189 for each additional ball produced in an hour. When 1700 balls are produced hourly, the cost per ball will decrease by approximately $0.000113 for each additional ball produced in an hour. 3. CPI United States: a. 3 b. A( t) = 0.109t 1.555t + 10.97t + 100.30 t years after 1980 A ( t) = 0.37t 3.111t + 10.97 index points per year t years after 1980 A (7) 5. index points per year c. 1988 CPI estimate: (CPI in 1987) + A (7) (1 year) 137.9 + (5. index points per year)(1 year) = 143.1 Note: The estimate can also be calculated using the value of A(7) instead of the actual CPI in 1987. Because the model closely agrees with the actual value in 1987, the value of this estimate is not significantly affected by this choice. Canada: a. 3 b. C( t) = 0.150t.171t + 15.814t + 99.650 t years after 1980 C ( t) = 0.450t 4.343t + 15.814 index points per year t years after 1980 C (7) 7.5 index points per year c. 1988 CPI estimate: (CPI in 1987) + C (7) (1 year) 155.4 + (7.5 index points per year)(1 year) = 16.9 Note: The estimate can also be calculated using the value of C(7) instead of the actual CPI in 1987. Because the model closely agrees with the actual CPI in 1987, the value of this estimate is not significantly affected by this choice. Peru: a. t P( t ) = 85.11(.0135 ) t years after 1980 b. P ( t) = 85.11(ln.0135)(.0135 ) t 59.558(.0135 ) index points per year t years after 1980 P (7) 7984 index points per year t c. 1988 CPI estimate: (CPI in 1987) + (7) P (1 year) 11,150 + (7984 index points per year)(1 year) = 19,134 Note: The estimate can also be calculated using the value of P(7) instead of the actual CPI in 1987. If this is done, the estimate will be approximately 19,394.

148 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts Brazil: a. t B( t ) = 73, 430(.615939 ) t years after 1980 b. B ( t) = 73.430(ln.615939)(.615939 ) t 70.61(.615939 ) index points per year t years after 1980 B (7) 59,193 index points per year t c. 1988 CPI estimate: (CPI in 1987) + B (7) (1 year) 77,58 + (59,193 index points per year)(1 year) = 136,451 Note: The estimate can also be calculated using the value of B(7) instead of the actual CPI in 1987. If this is done, the estimate will be approximately 10,748. 4. Revenue a. Revenue = R( x) = 1.16x + 54.8x 105.60 dollars, where x = the price for a large one-topping pizza, in dollars. 9.5 < x < 14.5. b. R '( x) = 4.3x + 54.8 dollars per dollar of pizza price. R (9.5) = $9.3 When the price of a large one-topping pizza is $9.5, the revenue is increasing by $9.3 for every additional dollar in the price of pizza. c. Revenue = (10.5 9.5)(9.3) = $9.3 d. R (11.50) = -$5.4 When the price of a large one-topping pizza is $11.50, the revenue is decreasing by $5.40 for every additional dollar in the price of pizza. e. Revenue = (11.50 1.50)( 5.4) = $5.40 f. In both cases the graph is concave down. 5. Advertising Note: This Activity can be solved using either a cubic model or a logistic model. The following solution uses a cubic model. a. b. 3 R( A) = 0.158A + 5.35A 3.056A + 154.884 thousand dollars of revenue when A thousand dollars is spent on advertising. 5 < A < 19. 3 R ( A) = 0.473A + 10.471A 3.056 thousand dollars of revenue per thousand dollars of advertising when A thousand dollars is spent on advertising R (10) 34.3 thousand dollars of revenue per thousand dollars of advertising When $10,000 is spent on advertising, revenue is increasing by $34.3 thousand per thousand advertising dollars. If advertising is increased from $10,000 to $11,000, the car dealership can expect an approximate increase in revenue of $34,300. c. (18) 1.0 R thousand dollars of revenue per thousand of dollars of advertising When $18,000 is spent on advertising, revenue is increasing by $1.0 thousand per thousand advertising dollars. If advertising is increased from $18,000 to $19,000, the car dealership can expect an approximate increase in revenue of $1,000.

Calculus Concepts Section 4.1: Approximating Change 149 6. Newspapers 3 a. n ( x) = 0.008x 0.31x + 3.457x + 51. 588 million newspapers x years after 1980. The curvature of the scatter plot suggests either a logistic or a cubic function, except that the data point for 1986 is lower than that for 1988. A cubic would be the better model for the max/min behavior. b. n(7) 67.1 million newspapers c. n '( x) = 0.04x 0.64x + 3. 457 million newspapers per year n (18) -0.4 million newspapers per year d. n (0) 0.1 million newspapers per year. We approximate the change in circulation between 1990 and 1991 to be an increase of 0.1 million newspapers. 7. One possible answer: Close to the point of tangency, a tangent line and a curve are close to one another. The farther away from the point of tangency we move, the more the tangent line deviates from the curve. Thus the tangent line near the point of tangency will usually produce a good estimate, but the tangent line farther away from the point of tangency will produce a poor estimate. 8. One possible answer: Because a line tangent to a point on a concave-up portion of a curve lies below the curve near the point of tangency, estimates taken from it will be under approximations (unless the tangent line cuts through the curve at some point). Because a line tangent to a point on a concave-down portion of a curve lies above the curve near the point of tangency, estimates taken from it will be over approximations (unless the tangent line cuts through the curve at some point). f ( x + h) f ( x) 9. One possible answer: By definition f '( x) = lim. Assuming h is relatively close h 0 h f ( x + h) f ( x) to zero, f '( x). Multiplying both sides of this approximation by h yields h h f '( x) f ( x + h) f ( x). Section 4. Relative and Absolute Extreme Points 1. Quadratic, cubic, and many product, quotient, and composite functions could have relative maxima or minima.. A graph of the function can be used to find the approximate values of the relative maxima and minima. (If technology is used, very accurate approximations can be obtained.) The exact values can be obtained by determining the exact output values where the derivative is zero, provided that the derivative graph crosses (not just touches) the input axis at that value. Additional relative minima or maxima may occur at the breakpoints of a piecewise continuous function.

150 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 3. The derivative is zero at the absolute maximum point. 4. The derivative is zero at the relative minimum point and at the relative maximum point. 5. The derivative is zero at the absolute maximum point marked with an X. The derivative where the graph is broken, at the relative minimum, is undefined. 6. The derivative is undefined at the absolute maximum point.

Calculus Concepts Section 4.: Relative and Absolute Extreme Points 151 7. 8. The derivative is zero at both absolute maximum points. The derivative does not exist at the relative minimum point. The derivative is undefined at the relative minimum point. 9. One possible answer: One such graph is y = x 3, which does not have a relative minimum or maximum at x = 0 even though the derivative is zero at this point. 10. One possible answer: One such graph is shown to the right. 11. a. All statements are true. b. The derivative does not exist at x = because f is not continuous there, so the third statement is false. c. The slope of the graph is negative, f ( x) < 0, to the left of x = because the graph is decreasing, so the second statement is false. d. The derivative does not exist at x = because f is not smooth there, so the third statement is not true.

15 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 1.a. The slope of this graph is always positive, so the second statement is not true. b. All statements are true. c. All statements are true. d. The third statement is not true because the derivative does not exist at x =. 13. One possible graph: 14. One possible graph: 15. One possible graph: 16. One possible graph: 17. a. The derivative formula is f (x) = x +.5 b. Using technology, the relative minimum value is approximately -7.565, which occurs at x -1.5. c.

Calculus Concepts Section 4.: Relative and Absolute Extreme Points 153 18. a. g (x) = -6x + 14.1 b. The relative maximum is 0.3675, which occurs at x =.35 19. a. h (x) = 3x 16x - 6 b. The relative maximum is 1.077, which occurs at x -0.35; The relative minimum is -108.99, which occurs at x 5.685 0. a. j (x) = 0.9x +.4x - 6 b. The relative maximum is 8.146, which occurs at x -4.39; The relative minimum is -1.301, which occurs at x 1.573 1. a. f (t) = 1(ln 1.5)(1.5 t ) + 1(ln 0.5)(0.5 t ) b. This function is always increasing, so does not have a relative maximum nor relative minimum.. a. j (t) = -5e -t + 1 t b. The relative maximum is.508 which occurs at t 0.59 3. a. g (x) =.1x 1.76x + 4.81 b. The relative maximum is 19.888, which occurs at x 3.633; The relative minimum is 11.779, which occurs at x 11.034 c. On the closed interval [ 0, 14.5 ], The absolute minimum is found at the point (11.034, 11.779); The absolute maximum is found at the point (3.633, 19.888) 4. a. The relative maximum is.86, which occurs at x 0.51; There is no relative minimum. b. On the closed interval [ 0, 10 ], the absolute minimum is found at the point (10, -9.30); the absolute maximum is found at the point (.51,.86) c. The graph of the derivative crosses the x-axis at the point x = 0.51 and no other time. 5. Grasshoppers a. At 9.449 C, the greatest percentage of eggs, 95.598%, eggs hatch. b. 9.449 C corresponds to 49 F. 6. Population The absolute minimum is found at the point (3.754, 6.156), which corresponds to 004. The absolute maximum is found at the point (7.116, 10.46), which corresponds to 08. 7. River Rate a. The flow rate for h = 0 was 13.0 cfs; for h = 11 it was 331.305 cfs. b. The absolute minimum is found at the point (.388, 11.311), or when h 0.4 hours The absolute maximum is found at the point (8.900, 387.975) or h 8.9 hours

154 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 8. Lake Level Because 1996 was a leap year, the number of days from October 1, 1995, to July 31, 1996, was 31 + 30 + 31 + 31 + 9 + 31 + 30 + 31 + 30 + 31 = 305 days. We using technology to solve the equation 7 4 L ( d) = 3( 5.345 10 ) d + (.543 10 ) d 0.019 = 0, and we obtain the solutions d 43.8, 73.4. To find the maximum height of the lake for 0 x 305, we compare the outputs at the endpoints with the outputs corresponding to the places where the derivative is 0: (0, 66.19) (43.8, 65.794) (73.4, 69.08) (305, 68.87) The highest lake level was 69.08 feet above sea level after approximately 73.4 days, which is below 69.1 feet above sea level. Yes, the lake remained below the maximum level. 9. Swim Time a. S( x) = 0.181x 8.463x + 147.376 seconds at age x years. b. The model gives a minimum time of 48.5 seconds occurring at 3.4 years. c. The minimum time in the table is 49 seconds, which occurs at 4 years of age. 30. Costs a. Hourly Cost = C(x) = 0.0198x 3 1.779x + 58.4x + 15.079 dollars, when the production level is x units per hour, (1 < x < 61). b. The marginal cost function is C (x) = 0.0594x 3.558x + 58.4 dollars. At the production level of 40 units per hour, the marginal cost is $11.14 dollars per unit. c. Average Hourly Cost = A(x) = 0.0198x 1.779x + 58.4 + 15.079x -1 dollars, when the production level is x units per hour, (1 < x < 61). d. Using technology, the minimum average hourly cost occurs at (46.69, 1.783) the point where 46.69 units per hour are being produced. The average hourly cost is $1.78, and the total cost is A(46.69) = $1017.08. 31. Sales a. A quadratic or exponential model can be used to model the data, but the exponential model may be a better choice because it does not predict that demand will increase for prices above $40. An exponential model for the data is R( p ) = 316.765(0.949 ) dozen roses when the price per dozen is p dollars. b. Multiply R(p) by the price, p. The consumer expenditure is E( p) = 316.765 p(0.949 ) dollars spent on roses each week when the price per dozen is p dollars. c. Using technology, we find that E(p) is maximized at p 19.16 dollars. A price of $19.16 per dozen maximizes consumer expenditure. d. Profit is given by F( p) = E( p) 6 R( p) = 316.765( p 6)(0.949 ). Using technology, F(p) is maximized at p 5.16 dollars. A price of $5.16 per dozen maximizes profit. e. Marginal values are with respect to the number of units sold or produced. In this activity, the input is price, so derivatives are with respect to price and are not marginals. p p p

Calculus Concepts Section 4.: Relative and Absolute Extreme Points 155 3. Demand a. A quadratic model for the data is p( x) = 0.0866x 10.15x + 99.710 dollars when the demand is x, that is, when x tenants desire the facilities. (Note that an exponential model does not fit the data well. This is a rare case in which shifting the data up actually produces a better-fitting exponential model.) b. To find the revenue, multiply the price, p(x), by the demand, x. 3 R( x) = 0.0866x 10.15x + 99.710x dollars when the demand is x tenants c. Examining a graph of R, we see that the maximum occurs between x = 15 and x = 5. On this interval, the derivative is zero at x 19.86. Because the number of tenants must be an integer, we compare the values of R(x) for x = 0 and x = 19. The greater output value occurs at x = 0. The price corresponding to this demand is p(0) 131.85. Thus the revenue is maximized at a price of approximately $13 and a demand of 0 tenants. The marginal revenue at the maximum point is near zero. 33. Refuse a. b. 3 G( t) = 0.008t 0.347t + 6.108t + 79.690 million tons of garbage taken to a landfill t years after 1975 G ( t) = 0.05t 0.693t + 6.108 million tons of garbage per year t years after 1975 c. In 005 the amount of garbage was increasing by G (30) 8.1million tons per year. d. Because the derivative graph exists for all input values and never crosses the horizontal axis, G(t) has no relative maxima. 34. Price p a. Exponential model: A( p ) = 568.074(0.96558 ) dollars Quadratic model: price is p dollars tickets sold on average when the price is p A( p) = 0.15p 16.007 p + 543.86 tickets sold on average when the The exponential model probably better reflects the probable attendance if the price is raised beyond $35 because attendance is likely to continue to decline. (The quadratic model predicts that attendance will begin to increase around $53.) b. Multiply the exponential ticket function by the price, p, to obtain the revenue function. p R( p) = 568.074 p(0.96558 ) dollars of revenue when the ticket price is p dollars. c. Using technology, the maximum point on the revenue graph is approximately (8.55, 5966.86). This corresponds to a ticket price of $8.55, which results in revenue of approximately $5967. The resulting average attendance is A(8.55) 09.

156 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 35. One possible answer: The graph shown below indicates there is an absolute maximum to the right of -3 and an absolute minimum to the left of 1. A view of the graph showing more of the horizontal axis indicates that y = is a horizontal asymptote for the graph. Use technology to find the absolute extrema, or solve the equation x(x x + 3) 4x 1 y = + = 0 ( x + ) x + In either case you should find the absolute minimum point of approximately (0.73, 1.317) and the absolute maximum point of approximately (.73,.183). Thus the absolute maximum is approximately.18, and the absolute minimum is approximately 1.3. 36. One possible answer: The derivative of y is zero for three values of x: x = 3.5, x 1.049, and x 1.549. A graph indicates that x = 3.5 and x 1.549 correspond to relative minima and x 1.049 correspond- ing to a relative maximum. There are no places where the derivative is not defined. Observing the end behavior of the graph (rising infinitely on both sides) and comparing the values of y for x = 3.5 and x 1.549, we conclude that there is no absolute maximum and the absolute y = 3( 1549. ) + ( 1549. ) ( 35. + 1549. ) 6.31. minimum is [ ] Section 4.3 Inflection Points 1. Production a. One visual estimate of the inflection points is (198, 5) and (018, 5). Note: There are also smaller inflection points at approximately (191,.5), (197, ), (1930, ), and (1935, 3). b. The input values of the inflection points are the years in which the rate of crude oil production is estimated to be increasing and decreasing most rapidly. We estimate that the rate of production was increasing most rapidly in 198, when production was approximately 5 billion barrels per year, and that it will be decreasing most rapidly in 018, when production is estimated to be approximately 5 billion barrels per year.

Calculus Concepts Section 4.3: Inflection Points 157. Advertising a. b. The inflection point occurs when the rate at which revenue is increasing with respect to the amount spent on advertising is greatest. It can be regarded as the point of diminishing returns. c. Answers may vary. It is important to note that additional dollars spent after the inflection point are generating diminishing returns. 3. For polynomial functions, as these appear to be, you can identify the function and its derivative by noticing the number of inflection points. Because a derivative has a power one less than the original function, it will also have one less inflection point. Thus graph b with two inflection points is the function. Graph a with one inflection point is the derivative, and graph c with no inflection points is the second derivative. 4. Using the same reasoning as in Activity 3, we conclude that graph a is the function, graph c is the derivative, and graph b is the second derivative. 5. Graph c appears to have a minimum at 1 and an inflection point at. Graph b crosses the horizontal axis at 1 and graph a crosses it at. Thus graph c is the function, graph b is the derivative, and graph a is the second derivative. 6. Graph a appears to have a minimum at 0, a maximum at, and inflection points at 3 and between 1 and 0. Graph b crosses the horizontal axis at and 0 and graph c crosses it at 3.4 and between 1 and 0. Thus graph a is the function, graph b is the derivative, and graph c is the second derivative. 7. f (x) = -3; f (x) = 0 8. g (t) = e t ; g (t) = e t 9. c (u) = 6u 7; c (u) = 6 10. k (t) = -4.t + 7; k (t) = -4. 11. p (u) = -6.3u + 7u; p (u) = -1.6u + 7 1. f (s) = 96s 4.s +7; f (s) = 19s 4. 13. g (t) = 37(ln 1.05)(1.05 t ); g (t) = 37(ln 1.05) (1.05 t )

158 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 14. h (t) = -3(ln.0)(.0 t ); h (t) = -3(ln.0) (.0 t ) 15. f (x) = 3.x -1 ; f (x) = -3.x - 16. g (x) = 3e 3x x -1 ; g (x) = -9e 3x x - 17. 18. 3.9t 3.9t L'( t) = 131.04 e (1 +.1 e ) ; L (t) = 0.0t 0.0t L'( t) = 199. e (1 + 99.6 e ) ; 146.435e 511.056e L"( t) = + L"( t) = 7.8t 3.9t 3.9t 3.9t ( 1 +.1e ) ( 1 +.1e ) 3 7.936e 3.984e +.04 t.0 t.0 t.0t ( 1 + 99.6e ) ( 1 + 99.6e ) 3 19. f '( x) = 3x 1x + ; f "( x) = 6x 1 = 0 at x = 0. g '( t) = -0.3t +.4t + 3.6 ; g"( t) = -0.6t +.4 = 0 at t = 4 1. Using technology, the second derivative = 0 at the point x = 3.356. Using technology, the second derivative = 0 at the point t = -17.918 3. There is no value for t for which the second derivative is zero, therefore there is no inflection point. 4. There is no value for t for which the second derivative is zero, therefore there is no inflection point. 5. a. g( x) 3 = 0.04x 0.88x + 4.81x + 1.11 g ( x) = 0.1x 1.76x + 4.81

Calculus Concepts Section 4.3: Inflection Points 159 b. g ( x) = 0. 4x 176. The inflection point on the graph of g is approximately (7.333, 15.834). This is a point of most rapid decline. 0 Bx 6. a. f ( x) = LABe + 0. 5 x. Using the formula 1 19e Bx (1 + Ae ) 0.5x ( e ) 0.5x ( 1+ 19e ) 0(1)(0.5) 19 f ( x) = = 190e 1+ 19e x ( ) 0.5x 0.5 for the derivative, we have 0. 5x 0. 5x 0. 5x ( ) ( + ) + 0. 5x ( + e ) d f x = dx e e e d ( ) 190 1 19 190 1 19 dx ( 0.5 x ) ( 0.5 x 3 ) 0.5 x ( 0.5 x ) ( 0.5 x e e e e e ) = 190 ( 0.5) 1+ 19 + ( ) 1+ 19 19 ( 0.5) 0. 5x 0. 5x x 0. 5x 3 = 95e 1 + 19e + 3610e 1 + 19e ( ) ( )

160 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts b. The inflection point of f is approximately (5.889, 10). This is a point of most rapid increase. 7. Study Time 45 0. 96915t 1 a. P( t) = = 45 + 0. 96915 t ( 1 + 594. e ) 1 5. 94e 0. 96915t 0. 96915t P t e e ( ) ( ) ( ) = 45( 1) 1 + 594. 594. ( 0. 96915) t ( e ) 0.96915t 0.96915 percent after studying for t hours = 59.047115e 1+ 5.94 percentage points per hour after studying for t hours 0. 96915t ( ) ( + e 0. 96915t ) d P ( t) = 59. 047115 dx e 1 5. 94 + 59 047115 0.. e 96915 d t 0. 96915t ( 1 + 5. 94e ) dx ( 0. 96915 t ) ( 0. 96915 t e e ). t 0. 96915t 3 0. 96915t e ( e ) ( e ) = 59. 047115 ( 0. 96915) 1 + 5. 94 + 59. 047115 0 96915 ( ) 1 + 5. 94 5. 94 ( 0. 96915)

Calculus Concepts Section 4.3: Inflection Points 161 51. 049033e 1 + 5. 94 t ( e ) 0. 96915 t 0. 96915 1. 9385t 0. 96915t 3 + 98. 46511e ( 1 + 5. 94e ) percentage points per hour per hour after studying for t hours Solving P ( t) = 0 for t gives t 1838.. The inflection point on P is approximately (1.838,.5). After approximately 1.8 hours of study (1 hour and 50 minutes), the percentage of new material being retained is increasing most rapidly. At that time, approximately.5% of the material has been retained. b. The answer agrees with that given in the discussion at the end of the section. 8. Population 5 3 3 a. p ( x) = 4( 1619. 10 ) x 3( 1675. 10 ) x + ( 0. 050) x 0. 308 percentage points per year x years after the end of 000 5 3 p ( x) = 1(1.619 10 ) x 6(1.675 10 ) x + (0.050) percentage points per year per year x years after the end of 000 Solving for x in the equation p ( x) = 0 gives x 13.44, 38.9. The lower value corresponds to a maximum slope value. To make sure it is the absolute maximum for the years between 000 and 050, we compare the slope value at x 13.44 with the slope values at the endpoints: p (0) = 0.308 and p (50) 0.5. Both of these values are less than the slope at x 13.44. The percentage will be increasing most rapidly in 014, when the percentage will be approximately 8.05% and will be increasing at a rate of 0.855 percentage point per year. b. The second solution found in part a, x 38.9, corresponds to a minimum on the derivative graph. Again, we compare the slope at that inflection point with the slopes at the endpoints to determine the year in which the most rapid decrease is expected to take place. We find that the slope in 000 has a greater magnitude than the slope for x 38.9. Thus the percentage will be decreasing most rapidly in 000, when the percentage will be approximately 6.69% and will be decreasing at a rate of 0.308 percentage point per year. 9. Grasshoppers a. 4 3 P( t) = 0.00645t + 0.488t 1.991t + 136.560t 395.154 percent when the temperature is t C 3 P ( t) = 0.058t + 1.464t 5.98t + 136.560 percentage points per C when the temperature is t C P ( t) = 0.0774t +.98t 5.98 percentage points per C per C when the temperature is t C

16 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts b. Because the graph of P crosses the t-axis twice, there are two inflection points. These are approximately (14., 59.4) and (3.6, 5.8). The point of most rapid decrease on the graph of P is (14., 59.4). (The other inflection point is a point of least rapid decrease.) The most rapid decrease occurs at 14. C, when 59.4% of eggs hatch. At this temperature, P (14.) 11.1, so the percentage of eggs hatching is declining by 11.1 percentage points per C. A small increase in temperature will result in a relatively large increase in the percentage of eggs not hatching. 30. Home Sale a. We solve the equation H ( x) = 0 and obtain the solution x 14. 11. A look at the graph indicates that this point corresponds to the least rapid increase. The median house size was increasing least rapidly in 1994 ( x 14. 11). At the time, the median house size was approximately H (14.11) 1938 square feet and was increasing at a rate of approximately H (14.11) 7.3 square feet per year. b. Square 00 footage 100 000 1900 1800 1700 7 9 11 13 15 17 19 1

Calculus Concepts Section 4.3: Inflection Points 163 80 60 H'(x) 40 0 0 7 9 11 13 15 17 19 1 0 10 H''(x) 0 7-10 9 11 13 15 17 19 1-0 c. We compare the slopes of the graph of H at the endpoints corresponding to x = 7 and x = 1: H ( 7) 61.7 square feet per year H (1) 58.4 square feet per year Thus the median house size was increasing most rapidly in 1987. 31. Price a. The relative maximum point on the derivative graph between the values x = 4 and x = 10 occurs where x 5.785. This is the inflection point on the original function. The relative minimum point on the derivative graph between the values x = 4 and x = 10 occurs where x 8.115. This is the other inflection point on the original function. b. The relative max and min on the first derivative graph correspond to the x-intercepts on the second derivative graph. c. According to the model, between 1990 and 001, the gas prices were declining most rapidly in 1990, and they were increasing most rapidly in 001. This is easiest to see if one examines the first derivative graph. d. According to the model, between 1994 and 000, the gas price was decreasing most rapidly in 1999 (approximately where x = 8.115) and it was increasing most rapidly in 1996 (approximately where x = 5.785). 3. Cable TV 6.7 P( x) = 6 + percent x years after the end of 1970 0.58x 1 + 38.7e Using the formula LABe Bx Bx (1 + Ae ) 0.58x 0.58x ( 1 + 38.7e ) for the derivative, we have 66.034e P( x) = percentage points per year x years after the end of 1970

164 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts The inflection point occurs where P ( x) is a maximum, at x 14.1. The percentage was increasing most rapidly in the first half of 1985, when the percentage was P(14.1) 31.1% and the rate of change of the percentage was P(14.1) 4.04 percentage points per year. 33. Donors 3 D( t) = 10.47t + 08.114t 168.805t + 9775.035 donors t years after 1975 D ( t) = 30.741t + 416.88t 168.805 donors per year t years after 1975 a. Using technology, we find that (0.418, 9740.089) is the approximate relative minimum point, and (13.11, 0,4.033) is the approximate relative maximum point on the cubic model. b. The inflection point occurs where D ( t) has its maximum, at t 6.770. The inflection point is approximately (6.8, 14,991.1). c. i. Because 6.8 is between t = 6 (the end of 1981) and t = 7 (the end of 198), the inflection point occurs during 198, shortly after the team won the National Championship. This is when the number of donors was increasing most rapidly. ii. The relative maximum occurred around the same time that a new coach was hired. After this time, the number of donors declined. 34. Cable TV a. 3 A( x) = 0. 16x + 1596. x + 180. x + 40. 930 annual dollars per person x years after 1984 A ( x) = 0.378x + 3.19x + 1.80 annual dollars per person per year x years after 1984 A ( x) = 0.756x + 3.19 annual dollars per person per year per year x years after 1984 The inflection point on the graph of A is approximately (4., 67.507). The corresponding point on the graph of A is the relative maximum, approximately (4., 8.541). The corresponding point on the graph of A is the x-intercept, approximately (4., 0).

Calculus Concepts Section 4.3: Inflection Points 165 b. For integer values of x, the maximum value of A ( x) is A (4) 8.5. In 1988, the rate of change of the annual amount spent per person was $8.5 per year. 35. Labor a. 6 N( h) = 0.654h 1 + 11.49e components after h hours. Using the formula LABe Bx (1 + Ae ) derivative, we have 0.654h ( e ) 0.654h ( 1+ 11.49e ) 6( 0.654)(11.49) N ( h) = = 465.8965e 1+ 11.49e components per hour after h hours. h ( ) 0.654h 0.654 Bx for the The greatest rate occurs when N ( h) is maximized, at h 3.733 hours, or approximately 3 hours and 44 minutes after she began working. b. Her employer may wish to give her a break after 4 hours to prevent a decline in her productivity.

166 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 36. Lake Level 7 3 4 a. L( d) ( 5.345 10 ) d (.543 10 ) d 0.019d 66.19 6 4 d days after September 30, 1995; ( ) ( ) = + + feet above sea level L ( d) = 1.6035 10 d + 5.086 10 d 0.019 feet per day d days after September 30, 1995; Because 1996 was a leap year, the number of days from October 1, 1995, to July 31, 1996, was 31 + 30 + 31 + 31 + 9 + 31 + 30 + 31 + 30 + 31 = 305 days. The lake level was rising most rapidly when L ( d) was a maximized at 159 days after the end of September 30, 1995 which occurred on March 7, 1996. b. One possible answer: A spring thaw (maybe in conjunction with rain) probably caused the lake level to rise rapidly. c. One possible answer: Yes, because melting snow and rainfall are likely to follow a similar pattern each year. 37. Labor 10, 11110. a. H( w) = + 0. 77966 w total labor-hours after w weeks 1 1153. e 0. 77966w 0. 77966w b. H w ( e ) ( e ) c. ( ) = 10, 11110. ( 1) 1 + 1153. 1153. ( 0. 77966) w ( e ) 0.77966w 0.77966 8,488,330.433e 1+ 1153. labor-hours per week after w weeks The derivative gives the manager information approximately the number of labor-hours spent each week d. The maximum point on the graph of H is approximately (9.685, 1840.134). Keeping in mind that the model must be discretely interpreted, we conclude that in the tenth week, the most labor-hours are needed. That number is H (10) 1816 labor-hours. d e. H w = ( ) ( + ) dx e 0. 77966w e 0. 77966w ( ) 8, 488, 330. 433 1 1153. + 8 488 330 433( 0. 77966 w d,,. ) ( 1 + 1153. 0. e 77966 w e ) dx = 8, 488, 330. 433( 0. 77966 w e )( 0. 77966) ( 1 + 1153. e 0. 77966 w ) + 8, 488, 330. 433( 0. 77966 w e )( ) ( 1 + 1153. e 0. 77966 w ) 0. 77966w ( 1153. e ) ( 0.77966) 3

Calculus Concepts Section 4.3: Inflection Points 167 ( 0. 77966 w )( 0. 77966 w e e ) ( 145 10 10. ) ( e 1. 45593 w w )( + e 0. 1 1153. 7766 ) 6, 179, 14. 51 1 + 1153. + Use technology to find the maximum of H ( w), which occurs at w 7.876. The point of most rapid increase on the graph of H is (7.876, 16.756). This occurs approximately 8 weeks into the job, and the number of labor-hours per week is increasing by approximately H (8) 513 labor-hours per week per week. f. Use technology to find the minimum of the graph of H, which occurs at w 11.494. The point of most rapid decrease on the graph of H is (11.494, 16.756). This occurs approximately 1 weeks into the job when the number of labor-hours per week is changing by approximately H (1) = 486 labor-hours per week per week. g. By solving the equation H ( w) = 0, we can find the input values that correspond to a maximum or minimum point on the graph of H, which corresponds to inflection points on the graph of H, the weekly labor-hour curve. h. Since the minimum of H ( w) occurs approximately 4 weeks after the maximum of H ( w), the second job should begin approximately 4 weeks into the first job. 3 38. Advertising 57454.18 a. P( x) = 0.090864x 1 + 31.876e dollars where x is the number of labor hours b. Using the formula P ( x) 166410.3e LABe Bx Bx (1 + Ae ) 0.090864x 0.090864x ( 1+ 31.876e ) for the derivative, we have dollars per labor hour where x is the number of labor hours. The inflection point occurs when P ( x) is a maximum at x 38 labor hours. The profit is increasing most rapidly at 38 labor hours, the profit is P(38) 8,77 dollars, and the rate of change is P (38) 1305 dollars per labor hour. c. Answers may vary. 39. Refuse a. Between 1980 and 1985, the average rate of change was smallest at 1 117 = 1 million tons per year. 1985 1980 b. 3 g( t) = 0.008t 0.347t + 6.108t + 79.690 million tons t years after 1970 c. g ( t) = 0.05t 0.693t + 6.108 million tons per year t years after 1970 g ( t) = 0.051t 0.693 million tons per year per year t years after 1970

168 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts d. g ( t) = 0 e. 0.0507t 0.693 = 0 0.0507t = 0.693 t 13.684 Solving g ( t) = 0 gives t 13.684, which corresponds to mid-1984. The corresponding amount of garbage is g(13.684) 10 million tons and the corresponding rate of increase is g (13.684) 1.4 million tons per year. Because the graph of g crosses the t-axis at 13.68, we know that input corresponds to an inflection point of the graph of g. Because the graph of g has a minimum at that same value, we know that it corresponds to a point of slowest increase on the graph of g. f. The year with the smallest rate of change is 1984, with g(14) 10.4 million tons of garbage, increasing at a rate of g ( 14) 137. million tons per year. 40. Revenue a. Using symmetric difference quotients to estimate rates of change, we see that revenue was growing most rapidly in 1998, when the rate of change was approximately 1713.1 1180.4 = 66.35 million dollars per year. 1999 1997 3 b. R( t) = 5.85t + 11.667t 689.840t + 191. 979 million dollars of revenue t years after 1990 c. R '( t) = 15.854t + 43.334t 689. 840 million dollars per year t years after 1990 R ''( t) = 31.708t + 43.334 million dollars per year per year t years after 1990 d. Solving R ( t) = 0 gives t 7.7. For integer values of t, R ( t) has its maximum at t = 18, in 008. According to the model, the revenue were R(18) 1475.4 million, and the rate of change was R (18) 4.19 million dollars per year. 41. Reaction a. The first differences are greatest between 6 and 10 minutes, indicating the most rapid increase in activity.

Calculus Concepts Section 4.3: Inflection Points 169 1930. b. A( m) =. 1 + 3170. e 0 439118m U / 100 µ L m minutes after the mixture reaches 95 C The inflection point, whose input is found using technology to locate the maximum point on a graph of A, is approximately (7.87, 0.965). After approximately 7.9 minutes, the activity was approximately 0.97 U/100 µl and was increasing most rapidly at a rate of approximately 0.1 U/100 µl/min. 4. Emissions 4 3 a. ( ) ( ) ( ) N( t) = 3.611 10 t +.83 10 t + 1.349 10 t + 6.990 millions of metric tons t years after 1940 b. ( 3 ) ( N ( t) 1.083 10 t 4.567 10 ) t ( 1.349 10 ) = + + millions of metric tons per year t years after 1940 c. Using technology, we find that N ( t) is maximized at t 1.1. For integer values of t, the maximum value of N ( t) is obtained at t = 1, which corresponds to the year 1961. The amount of emissions was N(1) 14.0 million metric tons, and emissions were increasing at the rate of N (1) 0.5 million metric tons per year. 43. The graph of f is always concave up. A parabola that opens upward fits this description. 44. Possible graphs are parabolas or lines. 45. a. The graph is concave up between x = 0 and x =, has an inflection point at x = and is concave down between x = and x = 4. b. 46. a. The graph is concave up between x = 0 and x =, has inflection points at x = 0 and x = and is concave down to the left of x = 0 and to the right of x =. b. One possible graph is shown.

170 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 47. One possible answer: Cubic and logistic models have inflection points, as do some product, quotient, and composite functions. Note: In the answer in the text exponential is listed twice and cubic is omitted. 48. One possible answer: Approximate values of inflection points can often be determined by viewing a graph of the function. Accurate values may be determined by finding the input values corresponding to a relative minimum or maximum of the function s derivative. The second derivative is always undefined or zero at an inflection point. Section 4.4 Interconnected Change: Related Rates df dx 1. = 3 dp ds. p = 5 dx dx dk dx 3. = 1x dy dy 4. dy = 7x dx + 4x dx + 4 dx dy ( 7 4 4 dx = x + x + ) 5. dg = 3 3e x dx 6. dg = 30xe 15x dx df x dx 7. = 6(ln1.0)(1.0 ) 8. dp 5 ds = dx 7 + s dx dh da da 9. = 6 + 6ln a dy dy dy da = 6(1 + ln a) dy

Calculus Concepts Section 4.4: Interconnected Change: Related Rates 171 10. Write v as v = π h( xw + w ) dv πh x dw w dw = + dv dw = π h( x + w) ds dh π r r h ( h) πrh dh = r + h 11. 1 ( ) 1 = + 1. 1 dh dr 0 = π r hr 3 + dh dr 0 = r + hr dh dr r = hr dh h dr = r 13. Use the Product Rule with πr as the first term and ( ) 1 1 dr dh dr 0 = πr r + h r h π r h + + + πr dr dh dr 0 = r + h + π r + h r + h 14. We apply the Product Rule twice: dw 0 = π hw + (derivative of π hw)( x + w) dw dw dh 0 = π hw + (πh + π w )( x + w) r + h as the second term. 15. a. w = 31.54 + 1.97 ln 5 5.4 gallons per day dw 1.97 dg 1.97 b. = = inches per year 0.43 gallon per day per year g 5 1 The amount of water transpired is increasing by approximately 0.43 gallon per day per year. In other words, in one year, the tree will be transpiring approximately 0.4 gallon more each day than it currently is transpiring. 16. a. 5 feet 8 inches = 68 inches = h 0.45w 0.45w B = = for a 5 8 woman weighing w pounds 0.00064516(68 ).9831984

17 Chapter 4: Analyzing Change: Applications of Derivatives Calculus Concepts 17. a. db 0.45 dw b. For the specific equation in part a, we have =..9831984 c. Because the variable w does not appear in the equation in part b, the fact that the woman weighs 160 pounds is not relevant information. We have dw = 1 pound per month, so db 0.45 = (1) 0.15 point per month..9831984 d. Because the variable B does not appear in the equation in part b, the fact that the woman has body mass index of 4 points is not relevant information. We have db = 0.1point per 0.45 dw month, so 0.1 =.9831984 dw ( 0.1).9831984 = 0.7 pound per month 0.45 0.45(100) 45 B = = 0.00064516h 0.00064516h 3 b. ( h ) points db 45 45 = dh = dh 0.00064516 3 0.00064516h c. Evaluate the equation in part b at h = 63 inches and dh = 0.5 inch per year to obtain db 0.789 point per year. 18. a. We are given h = 53% (remains constant), t = 80 F and F per hour dx = where x is time measured in hours. The question asks for A and da. We find A as dx A =.70 + 0.885(80) 78.7(0.53) + 1.0(80)(0.53) 83.7 F. Next, we treat h as a constant and find the derivative of the apparent temperature function with respect to x: da = 0.885 + 1.h dx dx dx After substituting the known values, we are able to solve for da dx : da = 0.885() + 1.(0.53)() = 3.04 F per hour dx The apparent temperature is approximately 83.7 F and is increasing at a rate of approximately 3.04 F per hour. b. We are given t = 100 F (remains constant), h = 0.3 and dh = 0.0 per hour where x is dx time measured in hours. (Note that we must convert both the humidity and the rate of change of humidity to decimals.) The question asks for A and da. We find A as dx A =.70 + 0.885(100) 78.7(0.30) + 1.0(100)(0.30) 103.6 F. Next, we treat t as a constant and find the derivative of the apparent temperature function with respect to x: da = 78.7 dh + 1.t dx dx dh dx

Calculus Concepts Section 4.4: Interconnected Change: Related Rates 173 After substituting the known values, we are able to solve for da dx : da = 78.7( 0.0) + 1.(100)( 0.0) = 0.86 F per hour dx The apparent temperature is approximately 103.6 F and is decreasing at a rate of approximately 0.83 F per hour. 19. a. We know h = 3 feet, d = 10 foot, dh = 0.5 foot per year, and we wish to find 1 dv. We treat d as a constant and find the derivative with respect to time t to obtain the related rates 1.73995 0.133187 equation dv = 0.00198d 1.133187h dh. Substituting the values given above results in dv 0.0014 cubic foot per year. 0. a. 1. a. b. We know h = 34 feet, d = 1 foot, dd = 1 foot per year, and we wish to find dv. We treat h as a constant and find the derivative with respect to time t to obtain the related rates 0.73995 1.133187 equation dv = 0.00198(1.73995 d ) h dd. Substituting the values given above results in dv 0.0347 cubic foot per year. 11.56P D = megajoules per person, where P is the number of kilograms of wheat produced K per hectare per year and K is the carrying capacity of the crop in people per hectare dd 1 dk 11.56 dp b. = 11.56P + K K c. In order to answer the question posed, we need to solve for dp in the equation 1 dk 11.56 dp = 11.56(10) +. Doing so requires that we know values of K and dk, K K which we are not given. We interpret dp as the rate of change of the crop production with respect to time. This rate of change tells us how rapidly the yearly amount of wheat grown is changing as time changes. 5 3 5 M 3 3 M L = 0.4 = 48.1035K 48.1035 5 3 5 b. ( = K 3 3 ) dl M dk 48.1035 K c. We are given K = 47 and dk = 0.5. Using the fact that L = 8 and the original equation, we dl can find the value of M corresponding to the current levels of labor and capital: M 781.39. Substituting the known values into the equation in part b gives dl 0.057 thousand worker-hours per year.