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Pre-Calc Homework Solutions 287

# Pre-Calc Homework Solutions 287 - Chapter 6 Review 51 Using...

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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 n 0 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 ) 5 2 } k V A } t 2 C ) c 2 y ) 5 e 2 ( kA / V ) t 2 C c 2 y 5 6 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 2 c ) e 2 ( kA / V ) t (b) lim t y ( t ) 5 lim t [ c 1 ( y 0 2 c ) e 2 ( kA / V ) t ] 5 c 53. (a) P ( t ) 5 } 1 1 15 e 0 4.3 2 t } 5 } 1 1 1 e 5 4 0 .3 e 2 t } This is P 5 } 1 1 M Ae 2 kt } where M 5 150, A 5 e 4.3 , and k 5 1. Therefore, it is a solution of the logistic differential equation.
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