Question

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 $cm^3/min$, when the radius is $2\, cm$ and the height is $3\, cm$ is

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

Answer 1

Answer: (D)

$\frac{2\pi}{5}$

Answer 2

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 $h = 3$,
$\frac{dV}{dt}=\frac{\pi\left(2\right)}{5}\left(-2+3\right)=\frac{2\pi}{5}.$