Pre-Calc Homework Solutions 264

Pre-Calc Homework Solutions 264 - 38. (a) To simplify...

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36. continued (e) 3 y 0 5 y 0 e rt 3 5 e rt ln 3 5 rt t 5} ln r 3 } Since ln 3 < 1.099, a suitable rule is } 1 1 0 0 0 8 r } or } 10 i 8 } . (We choose 108 instead of 110 because 108 has more factors.) 37. (a) x 1 1 1 } 1 x } 2 x 10 2.5937 100 2.7048 1000 2.7169 10,000 2.7181 100,000 2.7183 e < 2.7183 Graphical support: y 1 5 1 1 1 } 1 x } 2 x , y 2 5 e [0, 50] by [0, 4] (b) r 5 2 x 1 1 1 } 2 x } 2 x 10 6.1917 100 7.2446 1000 7.3743 10,000 7.3876 100,000 7.3889 e 2 < 7.389 Graphical support: y 1 5 1 1 1 } 2 x } 2 x , y 2 5 e 2 [0, 500] by [0, 10] r 5 0.5 x 1 1 1 } 0 x .5 } 2 x 10 1.6289 100 1.6467 1000 1.6485 10,000 1.6487 100,000 1.6487 e 0.5 < 1.6487 Graphical support: y 1 5 1 1 1 } 0 x .5 } 2 x , y 2 5 e 0.5 [0, 10] by [0, 3] (c) As we compound more times, the increment of time between compounding approaches 0. Continuous compounding is based on an instantaneous rate of change which is a limit of average rates as the increment in time approaches 0.
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Unformatted text preview: 38. (a) To simplify calculations somewhat, we may write: v ( t ) 5 ! } m k g } } e e a a t t 1 2 e e 2 2 a a t t }} e e a a t t } 5 ! } m k g } } e e 2 2 a a t t 2 1 1 1 } 5 ! } m k g } 5 ! } m k g } 1 1 2 } e 2 at 2 1 1 } 2 The left side of the differential equation is: m } d d v t } 5 m ! } m k g } (2)( e 2 at 1 1) 2 2 (2 ae 2 at ) 5 4 ma ! } m k g } ( e 2 at 1 1) 2 2 ( e 2 at ) 5 4 m ! } g m k } ! } m k g } ( e 2 at 1 1) 2 2 ( e 2 at ) 5 } ( e 4 2 m at g 1 e 2 1 at ) 2 } ( e 2 at 1 1) 2 2 }} e 2 at 1 1 264 Section 6.4...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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