# 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 = 1, = \$5,000 (0.08)1 = \$5,000 0.08 = \$5,416.44. When = 4, = \$5,000 (0.08)4 = \$5,000 0.32 = \$6,885.64. When = 10, = \$5,000 (0.08)10 = \$5,000 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 = 0.03 Take the natural log of both sides of this equation ln( 0.03 ) = ln 2 0.03 ln = ln 2 0.03 = ln 2 (ln e = 1) = ln 2 0.03 ! 23.10 8. 3 = 0.25 10. 3 = 10 r ln( 0.25 ) = ln 3 ln( 10 ) = ln 3 0.25 = ln 3 10 = ln 3 = ln 3 0.25 ! 4.39 = ln 3 10 ! 0.11 12. s (1 + s ) 1/ 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 = 2.7182818… 14. 0.1 0.01 0.001 0.0001 1 + 1 ! " # \$ 1.270982 1.047232 1.006933 1.000921 lim ! 0 + 1 + 1 ! " # \$ = 1

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94 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS 16. The graphs of y 1 = 1 + 2 n " # \$ % & , 2 = 7.3890560999 ! e 2 for 1 " " 50 are given at the right. 50 0 8 0 18. (A) A = Pe rt = \$10,000 0.0528(3) = \$10,000 0.1584 = \$11,716.35 (B) 11,000 = 10,000 0.0528 t 0.0528 = 1.1 0.0528 = ln 1.1 = ln 1.1 0.0528 ! 1.81 years 20. = \$50,000 = 0.064(5) = 0.32 Therefore, P = \$50,000 0.32 = \$50,000 -0.32 ! \$36,307.45 22. 100,000 = 40,000 15 r 15 = 2.5 = ln(2.5) 15 ! 0.0611 or 6.11% 24. = 10,000 -0.08 = 5,000 -0.08 = 0.5 -0.08 = ln(0.5) = - ln(0.5) 0.08 ! 8.66 years 26. 2 = 0.05 0.05 = 2 ln( 0.05 ) = ln 2 0.05 = ln 2 = ln 2 0.05 ! 13.86 years 28. 2 = (10) ln( 10 ) = ln 2 10 = ln 2 = ln 2 10 ! 0.0693 or 6.93% 30. The total investment in the two accounts is given by = 5,000 0.088 + 7,000(1 + 0.096) On a graphing utility, locate the intersection point of 1 = 5,000 0.088 x + 7,000(1 + 0.096) x and 2 = 20,000. The result is: = ! 5.7 years. 0 10 0 30,000
EXERCISE 4-1 95 32. (A) A = Pe rt ; 2 P = 2 = e ; = ln 2; r = ln 2 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 could be any positive number, the restrictions on are reasonable in the sense that the doubling times for most investments would be expected to be between 1 and 20 years. (C) = 2; = ln 2 2 ! 0.347 or 34.7% = 4; = ln 2 4 ! 0.173 or 17.3% = 6; = ln 2 6 ! 0.116 or 11.6% = 8; = ln 2 8 ! 0.087 or 8.7% = 10; = ln 2 10 ! 0.069 or 6.9% = 12; = ln 2 12 ! 0.058 or 5.8% 34. Q = 0 -0.0001238 1 2 0 = 0 -0.0001238 -0.0001238 = 1 2 ln( -0.0001238 ) = ln 1 2 ! " # \$ = ln 1 - ln 2 -0.0001238 = -ln 2 (ln 1 = 0) = ln 2 0.0001238 ! 5,599 years 36. = 0 ( < 0) 1 2 0 = 0 (90) 90 = 1 2 ln( 90 ) = ln 1 2 ! " # \$ = ln 1 - ln 2 90 = -ln 2 (ln 1 = 0) = - ln 2 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 0 = 0 0.0085 t or 0.0085 = 2 Thus, ln( 0.0085 ) = ln 2 and 0.0085 = ln 2 Therefore, = ln 2 0.0085 ! 81.57 years 40. 2 0 = 0 r (200) 200 = 2 ln( 200 ) = ln 2 200 = ln 2 = ln 2 200 ! 0.0035 or 0.35% EXERCISE 4-2 2. f ( x ) = -7 – 2 + 5 ’( ) = -7 - 2 4. ( ) = 6 ln 3 + 2 ’( ) = 6 1 " # \$ % & - 3 2 = 6 - 3 2 6. ( ) = 9 + 2 2 ’( ) = 9 + 4 8. ( ) = ln + 2 – 3 2 ’( ) = 1 + 2 – 6 10. ( ) = ln 8 = 8 ln ’( ) = 8 1 " # \$ % & = 8 12. ( ) = 4 + ln 9 = 4 + 9 ln ’( ) = 9 1 " # \$ % & = 9 14. ( ) = ln 10 + 2 ln = 10 ln + 2 ln = 12 ln ’( ) = 12 1 " # \$ % & = 12 16. ( ) = 2 ln ’( ) = 2 1 " # \$ % & = 2 For = 1, the slope of the tangent line is m = ’(1) = 2 1 = 2, and (1) = 2 ln 1 = 2(0) = 0. So, the equation of the tangent line at = 1 is: y – 0 = 2( – 1) or = 2 – 2. 18. ( ) = + 1 ’( ) = For = 0, = ’(0) = 0 = 1 and (0) = 2, so the equation of the tangent line at = 0 is: – 2 = 1( – 0) or = + 2. 20. ( ) = 1 + ln 4 = 1 + 4 ln ’( ) = 4 1 " # \$ % & = 4 For = , = ’( ) = 4 and ( ) = 1 + 4 ln = 5, so the equation of the tangent line at = is: – 5 = 4 ( x – e ) = 4 " # \$ % & – 4 or = (4 -1 ) + 1
EXERCISE 4-2 97 22. f ( x ) = 5 e ’( ) = 5 For = 1, m = ’(1) = 5 and (1) = 5 , so the equation of the tangent line at = 1 is: y – 5 = 5 ( – 1) or = (5 ) 28. ( ) = 2 + 3 ln 1 = 2 + 3 ln -1 = 2 + 3(-1) ln = 2 – 3 ln ’( ) = -3 1 " # \$ % & = - 3 30. ( ) = + 5 ln 6 = + 5(ln 6 + ln ) = + 5 ln 6 + 5 ln ’( ) = 1 + 5 1 " # \$ % & = 1 + 5 = + 5 32. = 3 log 5 dy dx = 3 1 ln 5 " 1 # \$ % & ( = 3 ln 5 34. = 4 = 4 ln 4 = (2 2 ) ln 2 2 = (2 2 )2 ln 2 = 2 2 +1 ln 2 36. = log + 4 2 + 1 = 1 ln 10 · 1 + 8 = 1 + 8 2 ln 10 ln 10 38. = 5 – 5 = 5 4 – 5 ln 5 40. = -log 2 + 10 ln = - 1 ln 2 · 1 + 10 1 " # \$ % & = 10 " 1 ln 2 # \$ % & ( 1 42. = 3 – 3 = -3 ln 3 44. On a graphing utility, graph 1 = and 2 = 5 . Rounded off to two decimal places, the points of intersection are: (1.30, 3.65), (12.71, 332,105.11).

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## This note was uploaded on 04/21/2009 for the course MATH 121 taught by Professor Hamidy during the Spring '09 term at Miramar College.

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

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