141Bquiz8csols

141Bquiz8csols - 1 MATH 141B QUIZ#8 Solutions Date: Tues...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 2

141Bquiz8csols - 1 MATH 141B QUIZ#8 Solutions Date: Tues...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online