I saw a similar problem from Standford, but I couldn't figure out how to apply Standford's answer to this problem.
Show that the depth of water in the tank at any time satisfies the below differential equation
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ordinary-differential-equations
mathematical-modeling
1 Answers
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The volume $V(t)$ of water in the tank at time $t$ is given by
$$V(t)=\frac{\pi h^3(t)}{3}\,\tan^2(\phi)$$
Hence, the rate of change in $V(t)$, $\frac{dv(t)}{dt}=\overbrace{k}^{\text{Inward flow rate}}-\overbrace{\alpha \sqrt{h(t)}}^{\text{Outward flow rate}}$, is given by
$$\begin{align} \frac{dV(t)}{dt}&=\pi h^2(t)\tan^2(\phi)\frac{dh(t)}{dt}\\\\ &=k-\alpha \sqrt {h(t)} \end{align}$$
whereupon solving for $\frac{dh(t)}{dt}$ yields
$$\bbox[5px,border:2px solid #C0A000]{\frac{dh(t)}{dt}=\frac{k-\alpha \sqrt{h(t)}}{\pi h^2(t)\tan^2(\phi)}}$$
as was to be shown!
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0How did you know that V(t)=(πh^3(t)/3)(tan^2(ϕ))? – 2017-02-22
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1The volume can be found by integrating $\int_0^h \pi^2 r^2(\tilde h)\,d\tilde h=\int_0^h \pi (\tilde h \tan(\phi))^2\,d\tilde h$ where we used $r=h\tan(\phi)$ – 2017-02-22