Water evaporates from an open bowl of unspecified shape at a rate proportional to the area ofthe water surface; that is, $$\frac{dV}{dt} = -cA(h)$$ where V is the volume of water, A(h) is the area of the water surface when the depth is h, and c is a positive constant.
Show that $$\frac{dh}{dt} = -c$$
It gives a hint $$V=\int_{0}^{h}A(x)dx$$
I can't see how dh/dt is brought up and related to the first formula, so I have no idea what to do.
$\endgroup$ 11 Answer
$\begingroup$From the hint: (and wlog) $$\frac{dV}{dh}=A$$
By the chain rule; $$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}$$ There is now sufficient information in the question for you to proceed and prove that $$\frac{dh}{dt}=-c$$
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