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

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160 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS 11. n 1 + 1 ! " # \$ 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. 4 16 64 256 1024 4096 (1 + ) 1/ 1.495349 1.193722 1.067399 1.021913 1.006793 1.002033 lim !" (1 + ) 1/ = 1 15. The graphs of y 1 = 1 + 1 ! " # \$ , 2 = 2.718281828 ! , and 3 = 1 + 1 ! " # \$ +1 for 0 # # 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: = 10,000 (0.0551)10 = 10,000 0.551 = \$17,349.87 (B) = \$15,000, = \$10,000, r = 0.0551: 15,000 = 10,000 0.0551 0.0551 = 1.5 0.0551 = ln(1.5) = ln(1.5) 0.0551 ! 7.36 years 19. = ; = \$20,000, = 0.052, t = 10: 20,000 = (0.052)10 = 0.52 = 20,000 0.52 = 20,000 -0.52 ! \$11,890.41 21. 30,000 = 20,000 5 5 = 1.5 5 = ln(1.5) = 5 ! 0.0811 or 8.11%
EXERCISE 4-1 161 23. P = 10,000 e -0.08 t , 0 # # 50 (A) 0 10 20 30 40 50 10,000 4493.30 2019 907.18 407.62 183.16 The graph of is shown at the right. (B) lim !" 10,000 -0.08 = 0 25. 2 = Pe 0.07 0.07 = 2 0.07 = ln 2 = ln 2 0.07 ! 9.9 years 27. 2 = r (8) 8 = 2 8 = ln 2 = ln 2 8 ! 0.0866 or 8.66% 29. The total investment in the two accounts is given by A = 10,000 0.072 + 10,000(1 + 0.084) = 10,000[ 0.072 + (1.084) ] On a graphing utility, locate the intersection point of y 1 = 10,000[ 0.072 x + (1.084) x ] and 2 = 35,000. The result is: = ! 7.3 years. 10 0 0 50,000 31. (A) = rt ; set = 2 (B) 2 = = 2 = ln 2 = ln 2 In theory, could be any positive number. However, the restrictions on 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 = 10%; = ln 2 0.10 ! 6.93 years = 15%; = ln 2 0.15 ! 4.62 years = 20%; = ln 2 0.20 ! 3.47 years = 25%; = ln 2 0.25 ! 2.77 years = 30%; = ln 2 0.30 ! 2.31 years 33. Q = 0 e -0.0004332 1 2 0 = 0 -0.0004332 -0.0004332 = 1 2 ln( -0.0004332 ) = ln 1 2 ! " # \$ = ln 1 - ln 2 -0.0004332 = -ln 2 (ln 1 = 0) = ln 2 0.0004332 ! 0.6931 0.0004332 ! 1599.95 Thus, the half-life of radium is approximately 1600 years. 35. = 0 rt ( < 0) 1 2 0 = 0 r(30) 30 = 1 2 ln( 30 ) = ln 1 2 " # \$ % & = ln 1 – ln 2 30 = -ln 2 (ln 1 = 0) = " ln 2 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 = 0 0.013 0.013 = 2 0.013 = ln 2 = ln 2 0.013 ! 53.3 It will take approximately 53.3 years. 39. 2 0 = 0 (50) 50 = 2 50 = ln 2 = ln 2 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 = (b) b = ln 2 . LOGARITHMIC FUNCTIONS The inverse of an exponential function is called a LOGARITHMIC FUNCTION. For > 0, \$ 1, Logarithmic form Exponential form y = log x is equivalent to = y Domain: (0, " ) Domain: (- " , " ) Range: (- " , " ) Range: (0, " ) The graphs of = log and = are symmetric with respect to the line = . The two most commonly used logarithmic functions are: log = log 10 Common logarithm (base 10) ln = log Natural logarithm (base ) 3 . DERIVATIVES OF LOGARITHMIC FUNCTIONS (a) ln = 1 (b) log = 1 ln · 1 1. f ( ) = 5 + 3 + 1 f’ ( ) = 5 + 3 3. ( ) = -2 ln + 2 – 4 ( ) = - 2 + 2 5. ( ) = 3 – 6 ( ) = 3 2 – 6 7. ( ) = + – ln ( ) = + 1 - 1 9. ( ) = ln 3 = 3 ln ( ) = 3 11. ( ) = 5 – ln 5 = 5 – 5 ln ( ) = 5 - 5 13. ( ) = ln 2 + 4 = 2 ln + 4 ( ) = 2 + 4 15. ( ) = 3 + ln ; (1) = 3 ( ) = 1 ; (1) = 1 Equation of the tangent line: – 3 = 1( – 1) or = + 2

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164 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS 17.
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## 4odd - 4 ADDITIONAL DERIVATIVE TOPICS EXERCISE 4-1 Things...

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