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{\bf Question}

Water flows into a reservoir from a river at the rate of $R \,
{\rm kgs}^{-1}.$ The outflow from the reservoir is $kd \, {\rm
kgs}^{-1},$ where $d$ is the depth of the reservoir and $k (&gt;0)$
is a constant.  Assuming that the reservoir is a cylinder with
cross-sectional area $A$ and depth $d$ find the depth of the
reservoir as a function of time if it initially has a depth $d_0.$

\vspace{.25in}

{\bf Answer}

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Mass is conserved, so $\ds \frac{dm}{dt} =$ net inflow; $m = Ad
\rho$

$\ds A \rho \dot d = R - kd$

This differential equation for $d$ has the solution:

$\ds d(t) = \frac{R}{k} - B\exp\left(-\frac{k}{A\rho}t\right)$

Now d(0) = d$_0$ therefore $\ds d = \frac{R}{k} + \left( d_0 -
\frac{R}{k} \right) \exp\left(- \frac{kt}{A\rho}\right)$



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