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43. continued (b) Note that we choose a small interval of x -values because the y -values decrease very rapidly and our calculator cannot handle the calculations for x #2 1. (This occurs because the analytic solution is y 52 2 1 ln (2 2 e 2 x ), which has an asymptote at x ln 2 < 2 0.69. Obviously, the Euler approximations are misleading for x 0.7.) [ 2 1, 0.2] by [ 2 10, 2] 44. Set y 1 52} x e 2 y 1 1 x y } and use IMPEULG with initial values x 5 0 and y 5 0 and step sizes 0.1 and 2 0.1. (a) [ 2 0.2, 4.5] by [ 2 5, 1] (b) [ 2 4.5, 0.2] by [ 2 1, 5] 45. (a) Half-life 5 } ln k 2 } 2.645 5 } ln k 2 } k 5 } 2 l . n 64 2 5 } < 0.262059 (b) Mean life 5 } 1 k } < 3.81593 years 46. T 2 T s 5 ( T 0 2 T s ) e 2 kt T 2 40 5 (220 2 40) e 2 kt Use the fact that T 5 180 and t 5 15 to find k . 180 2 40 5 (220 2 40) e 2 ( k )(15) e 15 k 5 } 1 1 8 4 0 0 } 5 } 9 7 } k 5 } 1 1 5 } ln } 9 7 } T 2 40 5 (220 2 40) e 2 ((1/15) ln (9/7)) t 70 2 40 5 (220 2 40) e 2 ((1/15) ln (9/7)) t e ((1/15) ln (9/7)) t 5 } 1 3 8 0 0 } 5 6 1 } 1 1 5 } ln } 9 7 } 2 t 5 ln 6 t 5 } l 1 n 5 ( l 9 n /7 6 ) } < 107 min It took a total of about 107 minutes to cool from 220 8 F to 70 8 F. Therefore, the time to cool from 180 8 F to 70 8 F was about 92 minutes. 47. T 2 T s 5 ( T 0 2 T s ) e 2 kt We have the system: 5 Thus, } 4 3 6 9 2 2 T T s s } 5 e 2 10 k and } 3 4 3 6 2 2 T T s s } 5 e 2 20 k . Since ( e 2 10 k ) 2 5 e 2 20 k , this means: ( } 4 3 6 9 2 2 T T s s } ) 2 5 } 3 4 3 6 2 2 T T s s } (39 2 T s ) 2 5 (33 2 T s )(46 2 T s ) 1521 2 78 T s 1 T s 2 5 1518 2 79 T s 1 T s 2 T s 3 The refrigerator temperature was 2 3 8 C. 48. Use the method of Example 3 in Section 6.4. e 2 kt 5 0.995 2 kt 5 ln 0.995 t 1 k } ln 0.995 5 l 7 n 0 2 0 } ln 0.995 < 41.2 The painting is about 41.2 years old. 49. Use the method of Example 3 in Section 6.4. Since 90% of the carbon-14 has decayed, 10% remains. e 2 kt 5 0.1 2 kt 5 ln 0.1 t 1 k } ln 0.1 5 l 7 n 0 2 0 } ln 0.1 < 18,935 The charcoal sample is about 18,935 years old. 50. Use t 5 1988 2 1924 5 64 years. 250 e rt 5 7500 e rt 5 30 rt 5 ln 30 r 5 } ln t 30 } 5 } ln 64 30 } < 0.053 The rate of appreciation is about 0.053, or 5.3%. 39 2 T s 5 (46 2 T s ) e 2 10 k 33 2 T s 5 (46 2 T s ) e 2 20 k 286 Chapter 6 Review

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51. Using the Law of Exponential Change in Section 6.4 with appropriate changes of variables, the solution to the differential equation is L ( x ) 5 L 0 e 2 kx , where L 0 5 L (0) is the surface intensity. We know 0.5 5 e 2 18 k , so k 5 } l 2 n0 1 . 8 5 } and our equation becomes L ( x ) 5 L 0 e (ln 0.5)( x /18) 5 L 0 1 } 1 2 } 2 x /18 . We now find the depth where the intensity is one-tenth of the surface value. 0.1 5 1 } 1 2 } 2 x /18 ln 0.1 5 } 1 x 8 } ln 1 } 1 2 } 2 x 5 } 18 ln ln 0. 0 5 .1 } < 59.8 ft You can work without artificial light to a depth of about 59.8 feet. 52. (a) } d d y t } 5 } k V A } ( c 2 y ) E } c d 2 y y } 5 E } k V A } dt 2 ln ) c 2 y ) 5 } k V A } t 1 C ln ) c 2 y ) 52} k V A } t 2 C ) c 2 y ) 5 e 2 ( kA / V ) t 2 C c 2 y 56 e 2 ( kA / V ) t 2 C y 5 c 6 e 2 ( kA / V ) t 2 C y 5 c 1 De 2 ( kA / V ) t Initial condition y 5 y 0 when t 5 0 y 0 5 c 1 D y 0 2 c 5 D Solution: y 5 c 1 ( y 0
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