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080311quiz7solution

# 080311quiz7solution - same rate What is the amount of...

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MATH 126 QUIZ #7 Solution 03/13/08 Mathias Knape Last Name : Circle discussion section First Name: 11 am 12 pm 1 pm Note: In order to receive full credit, please show all your work (conceptual steps, deﬁnitions you use, . ..) 1. Determine whether the sequence converges of diverges. If it converges ﬁnd the limit. (a) a n = n +1 3 n - 1 lim n →∞ n + 1 3 n - 1 = lim n →∞ 1 + 1 n 3 - 1 n = 1 3 (b) a n = ( - 1) n ln n n Observe that - ln n n a n ln n n Furthermore, we can show by using L’Hospital that lim x →∞ ln x x = lim x →∞ 1 x 1 2 x = lim x →∞ 2 x = 0 . Thus it follows by the Squeeze Theorem that lim n →∞ a n = 0. 2. Find the solution of dL dt = kL 2 ln t when L (1) = - 1. Separating variables gives us Z 1 L 2 dL = Z k ln tdt. Using integration by parts and solving for L leads to L = 1 kt - kt ln t - C From the initial condition L (1) = - 1 it then follows that C = k + 1. Hence L = 1 kt - kt ln t - k - 1 . 3. A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6 % alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the

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Unformatted text preview: same rate. What is the amount of alcohol in the vat at time t ? Let y ( t ) be the amount of alcohol at time t . Then dy ( t ) dt = rate in-rate out = . 06 × 5 gal min-y ( t ) 500 × 5 gal min = 30-y ( t ) 100 gal min . Separating variables we obtain Z 1 30-y ( t ) dy = Z 1 100 dt. Thus-ln | 30-y | = 1 100 t + C. The initial condition is y (0) = 0 . 04 × 500 = 20 gal. Plugging this in the equation above allows us to ﬁnd that C =-ln 10. Therefore we have that ln | 30-y | =-1 100 t + ln 10 . Taking the exponent of this equation we get that | 30-y | = 10 e-1 100 t . Now observe that the y (0) = 0, y ( t ) is continuous and the right hand side is always positive. Thus y = 30-10 e-1 100 t ....
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080311quiz7solution - same rate What is the amount of...

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