| Line 3: | Line 3: | ||
<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. | ||
| Line 10: | Line 10: | ||
-Dat Tran | -Dat Tran | ||
| + | |||
| + | Here's another way to think about it. 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: | ||
| + | |||
| + | <math>\frac{dA}{dt}=-kA</math> | ||
| + | |||
| + | Then we know that <math>A(t)=A_0e^{-kt}</math>. If 10% of the oil is gone after one year, then | ||
| + | |||
| + | <math>A(1)=A_0e^{-k\cdot 1}=\frac{9}{10}A_0e^{-k}</math>, | ||
| + | |||
| + | and so <math>e^{k}=\frac{9}{10}</math>. | ||
Revision as of 10:34, 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 another way to think about it. 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, then
$ A(1)=A_0e^{-k\cdot 1}=\frac{9}{10}A_0e^{-k} $,
and so $ e^{k}=\frac{9}{10} $.
