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<math>A(t)=A_{0}(9/10)^t</math> | <math>A(t)=A_{0}(9/10)^t</math> | ||
| − | Then, solve for when A(t)/A_{0} is 1/5. | + | Then, solve for when <math>A(t)/A_{0}</math> is 1/5. |
Sorry for the confusion I caused in class about this problem. | Sorry for the confusion I caused in class about this problem. | ||
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-Dat Tran | -Dat Tran | ||
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| + | Here's a few more words about this problem. We assume that the oil <math>A(t)</math> satisfies the standard differential equation of decay that is proportional to the amount present at any given time: | ||
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| + | <math>\frac{dA}{dt}=-kA.</math> | ||
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| + | Then we know that <math>A(t)=A_0e^{-kt}</math>. If 10% of the oil is gone after one year, that means | ||
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| + | <math>A(1)=A_0e^{-k}=\frac{9}{10}A_0</math>, | ||
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| + | and so <math>e^{-k}=\frac{9}{10}</math>. Finally, note that <math>e^{-kt}=(e^{-k})^t</math> and you get Mr. Tran's formula, | ||
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| + | <math>A(t)=A_{0}(9/10)^t</math>. | ||
Latest revision as of 10:37, 3 October 2008
Think about the problem similar to the half-life problem, but instead of one half, we have 9/10.
$ A(t)=A_{0}(9/10)^t $
Then, solve for when $ A(t)/A_{0} $ is 1/5.
Sorry for the confusion I caused in class about this problem.
Your TA,
-Dat Tran
Here's a few more words about this problem. We assume that the oil $ A(t) $ satisfies the standard differential equation of decay that is proportional to the amount present at any given time:
$ \frac{dA}{dt}=-kA. $
Then we know that $ A(t)=A_0e^{-kt} $. If 10% of the oil is gone after one year, that means
$ A(1)=A_0e^{-k}=\frac{9}{10}A_0 $,
and so $ e^{-k}=\frac{9}{10} $. Finally, note that $ e^{-kt}=(e^{-k})^t $ and you get Mr. Tran's formula,
$ A(t)=A_{0}(9/10)^t $.
