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Unformatted text preview: 1 MATH 141B QUIZ#8 Solutions Date: Tues March 25, 2008 1. Separating and integrating gives − 20 integraltext c − 2 dc = integraltext dt , or 20 c = t + K, where K is a constant. Since c (0) = 0 . 1 = 1 / 10 , we get K = 200. Therefore, 20 c = t + 200 and hence c 20 = 1 t + 200 . This gives c ( t ) = 20 t + 200 . Setting this equal to 0 . 05 = 1 / 20 , gives 20 t + 200 = 1 20 . Solving for t , we get t = 200 seconds, which is equal to 3 minutes and 20 seconds. Therefore, A) is the answer. 2. We need to solve N ′ ( t ) = 0 . The fast way of solving this is to realize we are given the derivative dN dt = kN · sin(2 πt ) . Since N negationslash = 0 , this happens when sin(2 πt ) = 0 . This means 2 πt = 0 or t = 2 πt = π. In other words, t = 0 or t = 1 / 2 . Recall from last semester, that any t for which N ′ ( t ) = 0 can be a max, a min, or neither. From the graph of sin, we conclude that t = 0 gives a minimum, and t = 1 / 2 gives rise to a maximum....
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This note was uploaded on 03/30/2008 for the course MATH 141B taught by Professor Marcfabbri during the Spring '08 term at Penn State.
 Spring '08
 MARCFABBRI
 Math, Logic

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