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Calculus 2 - Midterm 1 Solutions

# Calculus 2 - Midterm 1 Solutions - MAT1332 Spring/Summer...

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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 = 0 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 0 : = C 2 - C 1 , we have ln | v | = - kt + C 0 . Exponentiating gives | v | = e ln | v | = e - kt + C 0 = Ce - kt where C : = e C 0 . 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 0 = 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

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Problem 2 (3 points) Does the following improper integral converge or diverge? If it converges, give its value. Z 1 ln( x ) x 2 dx . First let us find the indefinite integral R ln( x ) x 2 dx . There are several ways to find the integral. One is to use integration by parts with f ( x ) : = - 1 x , g ( x ) : = ln( x ). Instead I will show you a way to do it using the fact that ln( x
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