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# 4odd - 4 ADDITIONAL DERIVATIVE TOPICS EXERCISE 4-1 Things...

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EXERCISE 4-1 159 4 ADDITIONAL DERIVATIVE TOPICS EXERCISE 4-1 Things to remember: 1 . THE NUMBER e The irrational number e is defined by e = lim n ! " 1 + 1 n ! " # \$ n or alternatively, e = lim s ! 0 (1 + s ) 1/ s e = 2.7182818… 2 . CONTINUOUS COMPOUND INTEREST A = Pe rt where P = Principal r = Annual nominal interest rate compounded continuously t = Time in years A = Amount at time t 1. A = \$1000 e 0.1 t When t = 2, A = \$1000 e (0.1)2 = \$1000 e 0.2 = \$1221.40. When t = 5, A = \$1000 e (0.1)5 = \$1000 e 0.5 = \$1628.72. When t = 8, A = \$1000 e (0.1)8 = \$1000 e 0.8 = \$2225.54 3. A t \$5,000 0 1 2 \$10,000 \$15,000 A = 6,000 e 0.1 t 3 4 5 6 7 8 5. 2 = e 0.06 t Take the natural log of both sides of this equation ln( e 0.06 t ) = ln 2 0.06 t ln e = ln 2 0.06 t = ln 2 (ln e = 1) t = ln 2 0.06 ! 11.55 7. 3 = e 0.1 t 9. 2 = e 5 r ln( e 0.1 t ) = ln 3 ln( e 5 r ) = ln 2 0.1 t = ln 3 5 r = ln 2 t = ln3 0.1 ! 10.99 r = ln2 5 ! 0.14

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160 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS 11. n 1 + 1 n ! " # \$ n 10 2.59374 100 2.70481 1000 2.71692 10,000 2.71815 100,000 2.71827 1,000,000 2.71828 10,000,000 2.71828 ! ! " e = 2.7182818… 13. n 4 16 64 256 1024 4096 (1 + n ) 1/ n 1.495349 1.193722 1.067399 1.021913 1.006793 1.002033 lim n ! " (1 + n ) 1/ n = 1 15. The graphs of y 1 = 1 + 1 n ! " # \$ n , y 2 = 2.718281828 ! e , and y 3 = 1 + 1 n ! " # \$ n +1 for 0 # n # 20 are given at the right. 1 y y 2 y 3 4 0 0 20 17. (A) A = Pe rt ; P = \$10,000, r = 5.51% = 0.0551, t = 10: A = 10,000 e (0.0551)10 = 10,000 e 0.551 = \$17,349.87 (B) A = \$15,000, P = \$10,000, r = 0.0551: 15,000 = 10,000 e 0.0551 t e 0.0551 t = 1.5 0.0551 t = ln(1.5) t = ln(1.5) 0.0551 ! 7.36 years 19. A = Pe rt ; A = \$20,000, r = 0.052, t = 10: 20,000 = Pe (0.052)10 = Pe 0.52 P = 20,000 e 0.52 = 20,000 e -0.52 ! \$11,890.41 21. 30,000 = 20,000 e 5 r e 5 r = 1.5 5 r = ln(1.5) r = ln(1.5) 5 ! 0.0811 or 8.11%
EXERCISE 4-1 161 23. P = 10,000 e -0.08 t , 0 # t # 50 (A) t 0 10 20 30 40 50 P 10,000 4493.30 2019 907.18 407.62 183.16 The graph of P is shown at the right. (B) lim t ! " 10,000 e -0.08 t = 0 25. 2 P = Pe 0.07 t e 0.07 t = 2 0.07 t = ln 2 t = ln2 0.07 ! 9.9 years 27. 2 P = Pe r (8) e 8 r = 2 8 r = ln 2 r = ln2 8 ! 0.0866 or 8.66% 29. The total investment in the two accounts is given by A = 10,000 e 0.072 t + 10,000(1 + 0.084) t = 10,000[ e 0.072 t + (1.084) t ] On a graphing utility, locate the intersection point of y 1 = 10,000[ e 0.072 x + (1.084) x ] and y 2 = 35,000. The result is: x = t ! 7.3 years. 10 0 0 50,000 31. (A) A = Pe rt ; set A = 2 P (B) 2 P = Pe rt e rt = 2 rt = ln 2 t = ln2 r In theory, r could be any positive number. However, the restrictions on r are 0.1 0.2 0.3 5 10 15 20 25 30 35 r t t = ln 2 r reasonable in the sense that most investments would be expected to earn between 2% and 30%.

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162 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS (C) r = 5%; t = ln 2 0.05 ! 13.86 years r = 10%; t = ln2 0.10 ! 6.93 years r = 15%; t = ln 2 0.15 ! 4.62 years r = 20%; t = ln2 0.20 ! 3.47 years r = 25%; t = ln 2 0.25 ! 2.77 years r = 30%; t = ln2 0.30 ! 2.31 years 33. Q = Q 0 e -0.0004332 t 1 2 Q 0 = Q 0 e -0.0004332 t e -0.0004332 t = 1 2 ln( e -0.0004332 t ) = ln 1 2 ! " # \$ = ln 1 - ln 2 -0.0004332 t = -ln 2 (ln 1 = 0) t = ln2 0.0004332 ! 0.6931 0.0004332 ! 1599.95 Thus, the half-life of radium is approximately 1600 years. 35. Q = Q 0 e rt ( r < 0) 1 2 Q 0 = Q 0 e r(30) e 30 r = 1 2 ln( e 30 r ) = ln 1 2 " # \$ % & = ln 1 – ln 2 30 r = -ln 2 (ln 1 = 0) r = " ln2 30 ! " 0.6931 30 ! -0.0231 Thus, the continuous compound rate of decay of the cesium isotope is approximately -0.0231. 37. 2 P 0 = P 0 e 0.013 t e 0.013 t = 2 0.013 t = ln 2 t = ln2 0.013 ! 53.3 It will take approximately 53.3 years. 39. 2 P 0 = P 0 e r (50) e 50 r = 2 50 r = ln 2 r = ln2 50 ! 0.0139 or 1.39%
EXERCISE 4-2 163 EXERCISE 4-2 Things to remember: 1 . DERIVATIVES OF EXPONENTIAL FUNCTIONS (a) d dx e x = e x (b) d dx b x = b x ln b 2 . LOGARITHMIC FUNCTIONS The inverse of an exponential function is called a LOGARITHMIC FUNCTION. For b > 0, b \$ 1, Logarithmic form Exponential form y = log b x is equivalent to x = b y Domain: (0, " ) Domain: (- " , " ) Range: (- " , " ) Range: (0, " ) The graphs of y = log b x and y = b x are symmetric with respect to the line y = x .

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