The radius of a cylinder is increasing at the rate of 2 cm/s and its altitude is decreasing at the rate of 3 cm/s. Find the rate of change of volume, when the radius is 3 cm and altitude is 5 cm.
Let the radius of cylinder $= r$ cm
Let the height (altitude) of cylinder $= h$ cm
Rate of increase of radius of cylinder $=\dfrac{dr}{dt} = 2 \; cm/s$ $\;\; \cdots$ (1)
Rate of decrease of height of cylinder $= \dfrac{dh}{dt} = -3 \; cm/s$ $\;\; \cdots$ (2)
Volume of cylinder $=V=\pi r^2 h$ $\;\; \cdots$ (3)
Differentiating equation (3) with respect to time t gives
$\dfrac{dV}{dt} = \pi \left(2rh\dfrac{dr}{dt} + r^2 \dfrac{dh}{dt}\right)$ $\;\; \cdots$ (4)
When $r=3 \; cm$ and $h = 5 \; cm$, in view of equations (1) and (2), equation (4) becomes
$\dfrac{dV}{dt} = \pi \left(2 \times 3 \times 5 \times 2 - 3^2 \times 3\right)= 33 \pi$
$\therefore$ The volume of the cylinder is increasing at the rate of $33 \pi \; cm^3/s$.