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Unformatted text preview: M (15) = ( . 93) M (0), which implies that Qe15 k = ( . 93) Q15 k = ln( . 93) k =ln( . 93) 15 Now, we want to nd M (30) M (0) 100 = Qe30 k Q 100 = 100 e30ln( . 93) 15 = 100 e 2 ln( . 93) = 100( . 93) 2 (b). How long will it take until the moisture is reduced to 50% of its original quantity? We want to nd t so that M ( t ) = 1 2 M (0) = 1 2 Q . Thus, we want t so that Qekt = 1 2 Q , i.e., so that ekt = 1 2 . Thus, we want t so thatkt = ln 1 2 , i.e., t = ln 1 2k = 15 ln 1 2 ln( . 93) 4. Plot f ( x ) = 2 sin( x )1. Be sure to mark important points on the axes! The plot is a simple sine wave, oscillating between 3 and +1, with period 2 , since it has amplitude 2 and has been shifted downwards by 1....
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Zhang

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