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Unformatted text preview: MAT1332 Spring/Summer 2009 Midterm Exam 1 Solutions. Problem 1 (6 points) Suppose the quantity v of a radioactive isotope decays according to the law dv dt = kv (1) with k > 0 a constant, v measured in moles and t measured in years. Assume that at time t = there are 20 moles of the isotope and at t = 5 there 19.25 moles remaining. (a) Solve the differential equation (1), with the initial condition given above, using separation of variables. (b) Find the value of k using your solution to (a). (c) Determine the halflife of the isotope using your solutions to (a) and (b). (d) How much of the isotope remains after 20 years? For (a) we separate variables 1 v dv =  kdt and integrate to obtain R 1 v dv =  k R dt . The left hand side of this equation is ln  v + C 1 and the righthand side is kt + C 2 , where C 1 and C 2 are constants of integration. Thus, combining the constants of integration C : = C 2 C 1 , we have ln  v = kt + C . Exponentiating gives  v  = e ln  v  = e kt + C = Ce kt where C : = e C . We are only considering the case where v takes nonnegative values and there fore, in the range of values we consider, v = Ce kt . The initial condition v (0) = 20 gives 20 = Ce = C and we have, in the range of values we consider, v = 20 e kt . For (b), note that we are given that v (5) = 19.25. As such, we have, using our solution from (a), 19.25 = 20 e 5 k . Dividing gives 77 80 = e 5 k . Taking the logarithm of both sides then yields 5 k = ln ( 77 80 ) and k =  1 5 ln 77 80 0.0076. For (c), we must fine the time t such that v ( t ) = 1 2 v (0). Using our solutions to (a) and (b) we see that this equation holds if and only if 20 e 0.0076 t = 10. I.e., if and only if e 0.0076 t = 1 2 . Taking the logarithm of both sides yields 0.0076 t = ln( 1 2 ) and t =  ln( 1 2 ) 0.0076 = ln(2) 0.0076 91.2036. Thus the halflife of the isotope is approximately 91.2036 years. For (d), we calculate v (20) = 20 e ( 0.0076)(20) 17.1798. Thus, after 20 years approximately 17.1798 moles remain. 1 Problem 2 (3 points) Does the following improper integral converge or diverge? If it converges, give its value....
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This note was uploaded on 01/16/2011 for the course MAT 1332 taught by Professor Munteanu during the Fall '07 term at University of Ottawa.
 Fall '07
 MUNTEANU
 Calculus

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