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Unformatted text preview: Solutions to selected problems in Section 2.4 8. Let Q ( t ) be the amount at time t . We have the following information to set up the IVP: Q (30) = 100 , Q (120) = 30 . The IVP is: Q ( t ) = kQ ( t ) , Q (0) = Q , where Q is unknown. Solving the ODE to have Q ( t ) = Ce kt . Applying the initial condition to have C = Q , and the solution to IVP is Q ( t ) = Q e kt . (a). Then from Q (30) = 100 , Q (120) = 30, we have Q e 30 k = 100 , Q e 120 k = 30 . Solving these to have k = 1 90 ln(10 / 3) . 01338 , Q = C = 100 e 30 k 149 . 38 . (b). The half life is = ln 2 /k 51 . 83 days. (c). The time until 1% remains is determined by Q ( t ) = 0 . 01 Q , Q e kt = 0 . 01 Q , e kt = 0 . 01 , t = ln(100) k 344 . 3 . 13. The key here is to recall that the ODE to model the radioactive decay is Q ( t ) = kQ ( t ) whose general solution is Q ( t ) = Ce kt . 1 (a). From the given halflife, we have ln(2) k = 5730 , and this gives k = ln(2) 5730 . 0001209680943385594 ....
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 '06
 EDeSturler
 Equations

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