Discussion Forum

The radius of a right circular cylinder increases at the rate of 0.1 cm/min, and the height decreases at the rate of 0.2 cm/min. The rate of change of the volume of the cylinder, in cm³/min, when the radius is 2 cm and the height is 3 cm is

Answer:

Explanation:

$V=\pi r^{2}h$

Differentiating both sides, we get

$\frac{dV}{dt}=\pi\left(r^{2}\frac{dh}{dt}+2r\frac{dr}{dt}h\right)$

= $\pi r\left(r\frac{dh}{dt}+2h\frac{dr}{dt}\right)$

$\frac{dr}{dt}=\frac{1}{10}and\frac{dh}{dt}=-\frac{2}{10}$

$\frac{dV}{dt}=\pi r\left(r\left(\frac{-2}{10}+2h\frac{1}{10}\right)\right)$

= $\frac{\pi\ r}{5}\left(-r+h\right)$

Thus, when r = 2 and h = 3,

 $\frac{dV}{dt}=\frac{\pi (2)}{5}(-2+3)=\frac{2\pi}{5}$