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# 4even - 4 ADDITIONAL DERIVATIVE TOPICS EXERCISE 4-1 2 A =...

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EXERCISE 4-1 93 4 ADDITIONAL DERIVATIVE TOPICS EXERCISE 4-1 2. A = \$5,000 e 0.08 t When t = 1, A = \$5,000 e (0.08)1 = \$5,000 e 0.08 = \$5,416.44. When t = 4, A = \$5,000 e (0.08)4 = \$5,000 e 0.32 = \$6,885.64. When t = 10, A = \$5,000 e (0.08)10 = \$5,000 e 0.8 = \$11,127.70. 4. 1 2 3 4 5 6 1,000 2,000 3,000 4,000 5,000 6,000 \$7,000 A t A = 4,000e 0.08t 0 6. 2 = e 0.03 t Take the natural log of both sides of this equation ln( e 0.03 t ) = ln 2 0.03 t ln e = ln 2 0.03 t = ln 2 (ln e = 1) t = ln 2 0.03 ! 23.10 8. 3 = e 0.25 t 10. 3 = e 10 r ln( e 0.25 t ) = ln 3 ln( e 10 r ) = ln 3 0.25 t = ln 3 10 r = ln 3 t = ln3 0.25 ! 4.39 r = ln3 10 ! 0.11 12. s (1 + s ) 1/ s 0.01 2.70481 -0.01 2.73200 0.001 2.71692 -0.001 2.71964 0.0001 2.71815 -0.0001 2.71842 0.00001 2.71827 -0.00001 2.71830 ! ! 0 e = 2.7182818… 14. s 0.1 0.01 0.001 0.0001 1 + 1 s ! " # \$ s 1.270982 1.047232 1.006933 1.000921 lim s ! 0 + 1 + 1 s ! " # \$ s = 1

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94 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS 16. The graphs of y 1 = 1 + 2 n " # \$ % & n , y 2 = 7.3890560999 ! e 2 for 1 " n " 50 are given at the right. 50 0 8 0 18. (A) A = Pe rt = \$10,000 e 0.0528(3) = \$10,000 e 0.1584 = \$11,716.35 (B) 11,000 = 10,000 e 0.0528 t e 0.0528 t = 1.1 0.0528 t = ln 1.1 t = ln1.1 0.0528 ! 1.81 years 20. A = Pe rt \$50,000 = Pe 0.064(5) = Pe 0.32 Therefore, P = \$50,000 e 0.32 = \$50,000 e -0.32 ! \$36,307.45 22. 100,000 = 40,000 e 15 r e 15 r = 2.5 r = ln(2.5) 15 ! 0.0611 or 6.11% 24. P = 10,000 e -0.08 t = 5,000 e -0.08 t = 0.5 -0.08 t = ln(0.5) t = - ln(0.5) 0.08 ! 8.66 years 26. 2 P = Pe 0.05 t e 0.05 t = 2 ln( e 0.05 t ) = ln 2 0.05 t = ln 2 t = ln 2 0.05 ! 13.86 years 28. 2 P = Pe r (10) ln( e 10 r ) = ln 2 10 r = ln 2 r = ln2 10 ! 0.0693 or 6.93% 30. The total investment in the two accounts is given by A = 5,000 e 0.088 t + 7,000(1 + 0.096) t On a graphing utility, locate the intersection point of y 1 = 5,000 e 0.088 x + 7,000(1 + 0.096) x and y 2 = 20,000. The result is: x = t ! 5.7 years. 0 10 0 30,000
EXERCISE 4-1 95 32. (A) A = Pe rt ; 2 P = Pe rt 2 = e rt ; rt = ln 2; r = ln2 t (B) r t 5 10 15 20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 r = ln 2 t Although t could be any positive number, the restrictions on t are reasonable in the sense that the doubling times for most investments would be expected to be between 1 and 20 years. (C) t = 2; r = ln2 2 ! 0.347 or 34.7% t = 4; r = ln2 4 ! 0.173 or 17.3% t = 6; r = ln2 6 ! 0.116 or 11.6% t = 8; r = ln2 8 ! 0.087 or 8.7% t = 10; r = ln2 10 ! 0.069 or 6.9% t = 12; r = ln2 12 ! 0.058 or 5.8% 34. Q = Q 0 e -0.0001238 t 1 2 Q 0 = Q 0 e -0.0001238 t e -0.0001238 t = 1 2 ln( e -0.0001238 t ) = ln 1 2 ! " # \$ = ln 1 - ln 2 -0.0001238 t = -ln 2 (ln 1 = 0) t = ln2 0.0001238 ! 5,599 years 36. Q = Q 0 e rt ( r < 0) 1 2 Q 0 = Q 0 e r (90) e 90 r = 1 2 ln( e 90 r ) = ln 1 2 ! " # \$ = ln 1 - ln 2 90 r = -ln 2 (ln 1 = 0) r = - ln2 90 ! -0.0077 Thus, the continuous compound rate of decay of the strontium isotope is approximately -0.0077.

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96 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS 38. P = P 0 e rt 2 P 0 = P 0 e 0.0085 t or e 0.0085 t = 2 Thus, ln( e 0.0085 t ) = ln 2 and 0.0085 t = ln 2 Therefore, t = ln2 0.0085 ! 81.57 years 40. 2 P 0 = P 0 e r (200) e 200 r = 2 ln( e 200 r ) = ln 2 200 r = ln 2 r = ln2 200 ! 0.0035 or 0.35% EXERCISE 4-2 2. f ( x ) = -7 e x – 2 x + 5 f ’( x ) = -7 e x - 2 4. f ( x ) = 6 ln x x 3 + 2 f ’( x ) = 6 1 x " # \$ % & - 3 x 2 = 6 x - 3 x 2 6. f ( x ) = 9 e x + 2 x 2 f ’( x ) = 9 e x + 4 x 8. f ( x ) = ln x + 2 e x – 3 x 2 f ’( x ) = 1 x + 2 e x – 6 x 10. f ( x ) = ln x 8 = 8 ln x f ’( x ) = 8 1 x " # \$ % & = 8 x 12. f ( x ) = 4 + ln x 9 = 4 + 9 ln x f ’( x ) = 9 1 x " # \$ % & = 9 x 14. f ( x ) = ln x 10 + 2 ln x = 10 ln x + 2 ln x = 12 ln x f ’( x ) = 12 1 x " # \$ % & = 12 x 16. f ( x ) = 2 ln x f ’( x ) = 2 1 x " # \$ % & = 2 x For x = 1, the slope of the tangent line is m = f ’(1) = 2 1 = 2, and f (1) = 2 ln 1 = 2(0) = 0. So, the equation of the tangent line at x = 1 is: y – 0 = 2( x – 1) or y = 2 x – 2. 18. f ( x ) = e x + 1 f ’( x ) = e x For x = 0, m = f ’(0) = e 0 = 1 and f (0) = 2, so the equation of the tangent line at x = 0 is: y – 2 = 1( x – 0) or y = x + 2. 20. f ( x ) = 1 + ln x 4 = 1 + 4 ln x f ’( x ) = 4 1 x " # \$ % & = 4 x For x = e , m = f ’( e ) = 4 e and f ( e ) = 1 + 4 ln e = 5, so the equation of the tangent line at x = e is: y – 5 = 4 e ( x – e ) = 4 e " # \$ % & x – 4 or y = (4 e -1 ) x + 1
EXERCISE 4-2 97 22. f ( x ) = 5 e x f ’( x ) = 5 e x For x = 1, m = f ’(1) = 5 e and f (1) = 5 e , so the equation of the tangent line at x = 1 is: y – 5 e = 5 e ( x – 1) or y = (5 e ) x 28. f ( x ) = 2 + 3 ln 1

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4even - 4 ADDITIONAL DERIVATIVE TOPICS EXERCISE 4-1 2 A =...

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