Math of Finance Answer Key

Section 2.6 Applications: Growth and Decay; Mathematics of Finance 9. r = 8% compounded continuously 2.6 Applications: Growth and Decay; Mathematics of Finance rE = er ¡ 1 = e0:08 ¡ 1 = 0:0833 = 8:33% 2. y0 represents the initial quantity; k represents the rate of growth or decay. 4. The half-life of a quantity is the time period for the quantity to decay to one-half of the initial amount. 5. Assume that y = y0 ekt represents the amount remaining of a radioactive substance decaying with a half-life of T. Since y = y0 is the amount of the substance at time t = 0; then y = y20 is the amount at time t = T: Therefore, y20 = y0 ekT ; and solving for k yields 1 = ekT 2 μ ¶ 1 ln = kT 2 ¡ ¢ ln 12 k= T ln(2¡1 ) = T ln 2 =¡ : T ¡t=T ) y = y0 e(¡ ln 2=T )t = y0 e¡(t=T ) ln 2 = y0 eln(2 "μ ¶¡1 #¡t=T μ ¶t=T 1 1 ¡t=T = y0 2 = y0 = y0 2 2 8. r = 6% compounded monthly, m = 12 μ ¶12 0:06 ¡1 rE = 1 + 12 ¼ 0:0617 = 6:17% rE = er ¡ 1 = e0:05 ¡ 1 ¼ 0:0513 = 5:13% 11. A = $10; 000; r = 6%; m = 4; t = 8 ³ r ´¡tm P =A 1+ m μ ¶¡8(4) 0:06 = 10; 000 1 + 4 ¼ $6209:93 6. Assume that y = y0 ekt is the amount left of a radioactive substance decaying with a half-life of T: From Exercise 5, we know k = ¡ Tln 2 ; so 7. r = 4% compounded quarterly, m=4 ³ r ´m ¡1 rE = 1 + m μ ¶4 0:04 ¡1 = 1+ 4 ¼ 0:0406 ¼ 4:06% 10. r = 5% compounded continuously 12. A = $45,678.93; r = 7:2%; m = 12; t = 11 months ³ r ´¡tm P = A 1+ m μ ¶¡(11=12)(12) 0:072 = 45,678.93 1 + 12 ¼ $42,769.89 13. A = $7300; r = 5% compounded continuously, t=3 A = P ert A ert 7300 = 0:5(3) e P = ¼ $6283:17 14. A = $25,000; r = 4:6% compounded continuously, t = 8 A = P ert A ert 25,000 = 0:046(8) e P = ¼ $17,302.93 Chapter 2 15. r = 9% compounded semiannually μ ¶2 0:09 ¡1 rE = 1 + 2 ¼ 0:0920 ¼ 9:20% 16. r = 6:2%; m = 4 μ ¶4 0:062 ¡1 rE = 1 + 4 ¼ 0:0635 = 6:35% 17. r = 6% compounded monthly μ ¶12 0:06 rE = 1 + ¡1 12 ¼ 0:0617 ¼ 6:17% 18. A = $20,000; t = 4; r = 6:5%; m = 12 ³ r ´mt A =P 1+ m μ ¶12¢4 0:065 20,000 = P 1 + 12 20,000 =P (1:05416)48 $15,431.86 = P 19. (a) A = $307,000; t = 3; r = 6%; m = 2 ³ r ´mt A=P 1+ m μ ¶3(2) 0:06 307,000 = P 1 + 2 307,000 = P (1:03)6 307,000 =P (1:03)6 $257,107.67 = P (b) Interest = 307,000 ¡ 257,107.67 = $49,892.33 (c) P = $200,000 A = 200,000(1.03)6 = 238,810.46 The additional amount needed is 307,000 ¡ 238,810.46 = $68,189.54. NONLINEAR FUNCTIONS 20. A = $40,000; t = 5 (a) r = 0:064; m = 4 ³ r ´mt A=P 1+ m μ ¶5(4) 0:064 40,000 = P 1 + 1 40,000 =P (1:016)20 $29,119.63 = P (b) Interest = $40,000 ¡ $29,119.63 = $10,880.37 (c) P = $20,000 ¶5(4) μ 0:064 A = 20,000 1 + 4 = $27,472.88 The amount needed will be $40,000 ¡$27,472.88 = $12,527.12. 21. P = $60,000 (a) r = 8% compounded quarterly: ³ r ´tm A=P 1+ m μ ¶5(4) 0:08 = 60; 000 1 + 4 ¼ $89,156.84 r = 7:75% compounded continuously A = P ert = 60; 000e0:0775(5) ¼ $88; 397:58 Linda will earn more money at 8% compounded quarterly. (b) She will earn $759.26 more. (c) r = 8%; m = 4 : ³ r ´m ¡1 rE = 1 + m μ ¶4 0:08 ¡1 = 1+ 4 ¼ 0:0824 = 8:24% Section 2.6 Applications: Growth and Decay; Mathematics of Finance r = 7.75% compounded continuously: (b) 11,000 = 5000e0:063t 11 = e0:063t 5 rE = er ¡ 1 = e0:0775 ¡ 1 ¼ 0:0806 = 8:06% 0:063t = ln ¡ ¢ ln 11 5 t= 0:063 t ¼ 12:52 (d) A = $80,000 A = P ert 80,000 = 60,000e0:0775t 4 = e0:0775t 3 4 ln = ln e0:0775t 3 ln 4 ¡ ln 3 = 0:0775t ln 4 ¡ ln 3 =t 0:0775 3:71 = t $60,000 will grow to $80,000 in about 3.71 years. μ ¶4x 0:08 (e) 60,000 1 + ¸ 80,000 4 80,000 (1:02)4x ¸ 60,000 4 (1:02)4x ¸ 3 μ ¶ 4 4x log (1:02) ¸ log 3 μ ¶ 4 4x log (1:02) ¸ log 3 ¡ ¢ log 43 ¼ 3:63 x¸ 4 log (1:02) It will take about 3.63 years. 22. (a) μ ¶4t 0:063 11,000 = 5000 1 + 4 11 = (1:01575)4t 5 μ ¶ 11 ln = 4t ln 1:01575 5 ¡ ¢ ln 11 5 t= 4 ln 1:01575 t ¼ 12:61 Since the interest is only added at the end of the quarter, it will take 12.75 years. μ 11 5 ¶ It will take about 12.52 years. 23. The …gure is not correct. (1 + 0:09)(1 + 0:08)(1 + 0:07) = 1:2596 This is a 25.96% increase. 24. The statement is not correct. (1 + r)15 = 1:94 + 1 1 + r = (2:94)1=15 r = (2:94)1=15 ¡ 1 r ¼ 0:0745 The required annual percent increase is 7:45%: 25. S(x) = 1000 ¡ 800e¡x (a) S(0) = 1000 ¡ 800e0 = 1000 ¡ 800 = 200 (b) S(x) = 500 500 = 1000 ¡ 800e¡x ¡500 = ¡800e¡x 5 = e¡x 8 5 ln = ln e¡x 8 5 ¡ ln = x 8 0:47 ¼ x Sales reach 500 in about 1 2 year: (c) Since 800e¡x will never actually be zero, S(x) = 1000 ¡ 800e¡x will never be 1000. (d) Graphing the function y = S(x) on a graphing calculator will show that there is a horizontal asymptote at y = 1000: This indicates that the limit on sales is 1000 units. Chapter 2 26. S(x) = 5000 ¡ 4000e¡x (c) No, the answer in part (b) is too small. Exponential growth does not accurately describe population growth for the world over a long period of time. (a) S(0) = 5000 ¡ 4000e0 = 1000 Since S(x) represents sales in thousands, in year 0 the sales are $1,000,000. (b) Let S(x) = 4500: 4500 = 5000 ¡ 4000e¡x ¡500 = ¡4000e¡x 0:125 = e¡x ¡ ln 0:125 = x x¼2 28. (a) The function giving the number after t hours is y = y0 2t=12 (b) For 10 days, t = 10 ¢ 24 or 240. y = (1)2240=12 = 220 = 1,048,576 It will take about 2 years for sales to reach $4,500,000. (c) Graphing the function y = S(x) on a graphing calculator will show that there is a horizontal asymptote at y = 5000: Since this represents $5000 thousand or $5,000,000, the limit on sales is $5,000,000. For 15 days, t = 15 ¢ 24 or 360. y = (1)2360=12 = 230 = 1,073,741,824 29. From 1960 to 2005 is an interval of t = 45 years. 27. (a) P = P0 e kt P = P0 ert 1500 = 59e45r When t = 1650; P = 470: When t = 2005; P = 6451: 1500 = e45r 59 1500 45r = ln 59 1500 1 ln r= 45 59 ¼ 0:0719 1650k 470 = P0 e 6451 = P0 e2005k P0 e2005k 6451 = 470 P0 e1650k 6451 = e355k 470 μ ¶ 6451 355k = ln 470 ¡ 6451 ¢ ln 470 k= 355 k ¼ 0:007378 Substitute this value into 470 = P0 e1650k to …nd P0 : 470 = P0 e1650(0:007378) P0 = NONLINEAR FUNCTIONS 470 This is an annual increase of 7.19%. 30. y = y0 ekt y = 40,000; y0 = 25,000; t = 10 (a) 40,000 = 25,000ek(10) 1:6 = e10k ln 1:6 = 10k 0:047 = k The equation is e1650(0:007378) y = 25,000e0:047t : P0 ¼ 0:002427 Therefore, P (t) = 0:002427e0:007378t : (b) P (1) = 0:002427e0:007378 ¼ 0:002445 million or 2445 The exponential equation gives a world population of only 2445 in the year 1. (b) y = 60,000 60,000 = 25,000e0:047t 2:4 = e0:047t ln 2:4 = 0:047t 18:6 = t There will be 60,000 bacteria in about 18.6 hours. Section 2.6 Applications: Growth and Decay; Mathematics of Finance 31. y = y0 ekt (a) y = 20,000; y0 = 50,000; t = 9 20,000 = 50,000e9k 0:4 = e9k ln 0:4 = 9k ¡0:102 = k 34. (a) From the graph, the risks of chromosomal abnormality per 1000 at ages 20, 35, 42, and 49 are 2, 5, 29, and 125, respectively. (Note: It is di¢cult to read the graph accurately. If you read di¤erent values from the graph, your answers to parts (b)-(e) may di¤er from those given here.) (b) y = Cekt The equation is When t = 20; y = 2; and when t = 35; y = 5: ¡0:102t y = 50,000e (b) : 1 (50,000) = 25,000 2 25,000 = 50,000e¡0:102t 0:5 = e¡0:102t ln 0:5 = ¡0:102t 6:8 = t 2 = Ce20k 5 = Ce35k Ce35k 5 = 2 Ce20k 2:5 = e15k 15k = ln 2:5 k= Half the bacteria remain after about 6.8 hours. 32. f (t) = 500e0:1t (a) ln 2:5 15 k ¼ 0:061 (c) y = Cekt When t = 42; y = 29; and when t = 49; y = 125: f (t) = 3000 3000 = 500e0:1t 6 = e0:1t ln 6 = 0:1t 17:9 ¼ t 29 = Ce42k 125 = Ce49k Ce49k 125 = 29 Ce42k 125 = e7k 29 μ ¶ 125 7k = ln 29 ¡ 125 ¢ ln 29 k= 7 k ¼ 0:21 It will take 17.9 days. (b) If t = 0 corresponds to January 1, the date January 17 should be placed on the product. January 18 would be more than 17.9 days. 33. Use y = y0 e¡kt : (d) Since the values of k are di¤erent, we cannot assume the graph is of the form y = Cekt : When t = 5; y = 0:37y0 : 0:37y0 = y0 e¡5k 0:37 = e¡5k ¡5k = ln (0:37) k= ln (0:37) ¡5 k ¼ 0:1989 (e) The results are summarized in the following table. n 2 3 4 Value of k for [20, 35] 0:0011 2:6 £ 10¡5 6:8 £ 10¡7 Value of k for [42, 49] 0:0023 3:4 £ 10¡5 5:5 £ 10¡7 The value of n should be somewhere between 3 and 4. Chapter 2 35. A(t) = A0 ekt 0:60 A0 = A0 e(¡ ln 2=5600)t 0:60 = e(¡ ln 2=5600)t ln 2 t ln 0:60 = ¡ 5600 5600(ln 0:60) =t ¡ ln 2 4127 ¼ t The sample was about 4100 years old. 36. 1 A0 = A0 e¡0:053t 2 1 = e¡0:053t 2 1 ln = ¡0:053t 2 ln 1 ¡ ln 2 = ¡0:053t NONLINEAR FUNCTIONS μ ¶t=13 1 38. (a) A(t) = A0 2 μ ¶100=13 1 A(100) = 4:0 2 A(100) ¼ 0:0193 After 100 years, about 0:0193 gram will remain. μ ¶t=13 1 (b) 0:1 = 4:0 2 μ ¶t=13 1 0:1 = 4:0 2 μ ¶ 1 t ln ln 0:025 = 13 2 t= 13 ln 0:025 ¡ ¢ ln 12 t ¼ 69:19 It will take 69 years. 0 ¡ ln 2 =t ¡0:053 13 ¼ t 39. (a) The half-life of plutonium 241 is about 13 years. μ ¶t=1620 1 2 μ ¶100=1620 1 A(100) = 4:0 2 A(t) = A0 A(100) ¼ 3:8 After 100 years, about 3.8 grams will remain. 37. 1 A0 = A0 e¡0:00043t 2 1 = e¡0:00043t 2 1 ln = ¡0:00043t 2 ln 1 ¡ ln 2 = ¡0:00043t 0 ¡ ln 2 =t ¡0:00043 1612 ¼ t (b) μ ¶t=1620 1 2 μ ¶t=1620 1 0:1 = 4 2 1 t ln ln 0:025 = 1620 2 1620 ln 0:025 ¡ ¢ t= ln 12 0:1 = 4:0 t ¼ 8600 The half-life of radium 226 is about 1600 years. The half-life is about 8600 years. Section 2.6 Applications: Growth and Decay; Mathematics of Finance 42. y = 40e¡0:004t 40. (a) y = y0 ekt When t = 0; y = 500; so y0 = 500: When t = 3; y = 386: (a) t = 180 y = 40e¡0:004(180) = 40e¡0:72 ¼ 19:5 watts 386 = 500e3k 386 = e3k 500 1 = e¡0:004t 2 1 ln = ¡0:004t 2 ln 1 ¡ ln 2 =t ¡0:004 e3k = 0:772 3k = ln 0:772 k= ln 0:772 3 k ¼ ¡0:0863 y = 500e¡0:0863t (b) 1 y0 = y0 e¡0:0863t 2 1 ln = ¡0:0863t 2 ¡ ¢ ln 12 t= ¡0:0863 t ¼ 8:0 173 ¼ t It will take about 173 days. (c) The power will never be completely gone. The power will approach 0 watts but will never be exactly 0. 43. A(t) = A0 41. (a) y = y0 ekt When t = 0; y = 25:0; so y0 = 25:0: When t = 50; y = 19:5: 19:5 = 25:0e50k 19:5 = e50k 25:0 μ ¶ 19:5 50k = ln 25:0 ¡ 19:5 ¢ ln 25:0 k= 50 k ¼ ¡0:00497 y = 25:0e¡0:00497t 1 y0 = y0 e¡0:00497t 2 1 = e¡0:00497t 2 μ ¶ 1 ¡0:00497t = ln 2 ¡ ¢ ln 12 t= ¡0:00497 t ¼ 139 The half-life is about 139 days. μ ¶t=5600 1 2 μ ¶43,000=5600 1 A(43,000) = A0 2 The half-life is about 8.0 days. (b) 20 = 40e¡0:0004t (b) ¼ 0:005A0 About 0.5% of the original carbon 14 was present. 44. P (t) = 100e¡0:1t (a) P (4) = 100e¡0:1(4) ¼ 67% (b) P (10) = 100e¡0:1(10) ¼ 37% (c) 10 = 100e¡0:1t 0:1 = e¡0:1t ln (0:1) = ¡0:1t ¡ ln (0:1) =t 0:1 23 ¼ t It would take about 23 days. (d) 1 = 100e¡0:1t 0:01 = e¡0:1t ln (0:01) = ¡0:1t ¡ ln (0:01) =t 0:1 46 ¼ t It would take about 46 days. Chapter 2 45. (a) Let t = the number of degrees Celsius. y = y0 ¢ ekt y0 = 10 when t = 0± : To …nd k; let y = 11 when t = 10± : 11 = 10e10k 11 10 10k = ln 1:1 e10k = ln 1:1 10 ¼ 0:0095 k= The equation is y = 10e0:0095t : (b) Let y = 15; solve for t: 15 = 10e0:0095t ln 1:5 = 0:0095t ln 1:5 0:0095 ¼ 42:7 t= 15 grams will dissolve at 42.7± C. 46. t = 9; T0 = 18; C = 5; k = :6 f(t) = T0 + Ce¡kt f(t) = 18 + 5e¡0:6(9) = 18 + 5e¡5:4 ¼ 18:02 The temperature is about 18.02± : 47. f (t) = T0 + Ce¡kt 25 = 20 + 100e¡0:1t 5 = 100e¡0:1t ¡0:1t e = 0:05 ¡0:1t = ln 0:05 ln 0:05 ¡0:1 ¼ 30 t= It will take about 30 min. NONLINEAR FUNCTIONS 48. C = ¡14:6; k = 0:6; T0 = 18± ; f(t) = 10± f(t) = T0 + Ce¡kt f (t) = 18 + (¡14:6)e¡0:6t ¡8 = ¡14:6e¡0:6t 0:5479 = e¡0:6t ln 0:5479 = ¡0:6t ¡ ln 0:5479 =t 0:6 1¼t It would take about 1 hour for the pizza to thaw.