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Find the equation for the rate of change of the volume V, where $V=\frac{1}{3}\pi r^2 h$ and the radius r and the height h are both functions of time t.

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1 Answer

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Guess you're looking for this... $$\frac{dV}{dt}=\frac{1}{3}\pi\frac{d(r^2h)}{dt}$$ $$\frac{dV}{dt}=\frac{1}{3}\pi(2rh\frac{dr}{dt}+r^2\frac{dh}{dt})$$

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