Question

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 3 million barrels of oil in the well; six years later 1,500,000 barrels remain.

Answer 1

If $A(t)$ is the amount of oil in the well (let's say in millions of barrels) at time $t[tex] (years), then [tex]\dfrac{\mathrm dA}{\mathrm dt}=kA$

for some negative value of $k$. You're told that the well started off with 3 million barrels, so $A(0)=3$, and that after 6 years, there are 1.5 million barrels, so $A(6)=1.5$.

Solve the differential equation above:

$\dfrac{\mathrm dA}{\mathrm dt}=kA\implies\dfrac{\mathrm dA}A=k\,\mathrm dt$

Integrate both sides.

$\displaystyle\int\frac{\mathrm dA}A=k\int\mathrm dt$
$\ln|A|=kt+C$
$A=e^{kt+C}$
$A=Ce^{kt}$

Given that $A(0)=3$ and $A(6)=1.5$, you have

$\begin{cases}3=Ce^{0k}\\1.5=Ce^{6k}\end{cases}\implies C=3,\,k=-\dfrac{\ln2}6$

So the amount of oil in the well is given by the function

$A(t)=3e^{-\frac{\ln2}6t}=3\times2^{-t/6}$