Question

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

• A. $-2\pi$
• B. $-\frac{8\pi }{5}$
• C. $-\frac{3\pi }{5}$
• D. $\frac{2\pi }{5}$

Answer 1

Answer: (D)

Given $V={\pi }^{2}h.$
Differentiating both sides, we get
$\frac{dV}{dt}=\pi \left(r^2\frac{dh}{dt}+2r\frac{dr}{dr}h\right)-\pi r\left(r\frac{dh}{dt}+2h\frac{dr}{dt}\right)$
$\frac{dr}{dt}=\frac{1}{10}$and $\frac{dh}{dt}=-\frac{2}{10}$
$\frac{dV}{dt}=\pi r\left(r\left(-\frac{2}{10}\right)+2h\left(\frac{1}{10}\right)\right)=\frac{\pi r}{5}\left(-r+h\right)$
Thus, when $r=2$ and $h=3$,
$\frac{dV}{dt}=\frac{\pi \left(2\right)}{5}\left(-2+3\right)=\frac{2\pi }{5}.$