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 ${\text{cm}}^3/\text{min},$ 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

The correct option is D $\frac{2\pi }{5}$

Given that $\frac{dr}{dt}=\frac{1}{10}$ and $\frac{dh}{dt}=\frac{-2}{10}$
Volume of the cylinder $V=\pi r^{2}h$
Differentiating both sides
$\frac{dV}{dt}=\pi (r^2\frac{dh}{dt}+2r\frac{dr}{dt}h)$
$\Rightarrow \frac{dV}{dt}=\pi r(r\frac{dh}{dt}+2h\frac{dr}{dt})$
$\Rightarrow \frac{dV}{dt}=\pi r(r\left(\frac{-2}{10}\right)+2h\left(\frac{1}{10}\right))$
$\Rightarrow \frac{dV}{dt}=\frac{\pi r}{5}(-r+h)$
Thus, when $r=2$ and $h=3,$
$\frac{dV}{dt}=\frac{\pi \left(2\right)}{5}(-2+3)=\frac{2\pi }{5}$