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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 half-life 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 right-hand 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 non-negative 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 half-life 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