Question

The radius of a cylinder is increasing at the rate $2$ cm/sec. and its altitude is decreasing at the rate of $3$ cm/sec. Find the rate of change of volume when radius is $3$ cm and altitude $5$ cm.

Answer 1

We have, $V=\pi r^{2}h,\frac{dr}{dt}=2$ and $\frac{dh}{dt}=-3$
$\therefore V=\pi R^{2}h\Rightarrow \frac{dV}{dt}=\pi \{2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\}\Rightarrow \frac{dV}{dt}=\pi (4rh-3r^2)$
When r = 3, h = 5,we obtain
$\frac{dV}{dt}=\pi (60-27)=33\pi$$c{m}^3/sec$