MATH 2214 Midterm: MATH 2214 Virginia Tech 2.4 Exam
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1: the amount of the substance must be given by ce kt for some constants. If t is measured in days, we are told that ce 30k = 100 and. Dividing the two equations, we nd e 90k = 3/10, i. e. k = The half life is = ln 2/k 51. 83 days. The time until 1% remains is ln 100/k 344. 3 days: a) we are told that ln 2/k = 5730, hence k = ln 2/5730 0. 00012097. Next we are told that i. e. e kt = 0. 3, t = 1 k ln(10/3) 9953 years: we can repeat the calculation in part a assuming = 5700 and = 5760, respectively. The corresponding values of k are 0. 0001216 and 0. 00012034. The ages obtained are 9901 and 10005 years. Hence the uncertainty is ap- proximately 50 years: we have q(60000)/q(0) = exp( 60000k) 0. 000704.
