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} $.
