Question

The radius of a right circular cylinder increases at the rate of $0.1\text{cm/min}$, and the height decreases at the rate of $0.2\text{cm/min}$. The rate of change of the volume of the cylinder, in ${\text{cm}}^3/\text{min}$, when the radius is $2\text{cm}$ and the height is $3\text{cm}$ is
(The negative sign(-) indicates that volume decreases)

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

Answer 1

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

Let $r$ and $h$ be the radius and height of the cylinder respectively.
Given, $r=2\text{cm},h=3\text{cm}$
$\frac{dr}{dt}=0.1\text{cm/min}$
$\frac{dh}{dt}=-0.2\text{cm/min}$

$V=\pi r^{2}h$
Differentiating both sides
$\frac{dV}{dt}=\pi (r^2\frac{dh}{dt}+2r\frac{dr}{dt}h)=\pi r(r\frac{dh}{dt}+2\frac{dr}{dt}h)$
$=\pi r\left(r\right(-0.2)+2h(0.1\left)\right)=\frac{\pi r}{5}(-r+h)$
Thus, when $r=2$ and $h=3$,
$\frac{dV}{dt}=\frac{\pi \left(2\right)}{5}(-2+3)=\frac{2\pi }{5}{\text{cm}}^3/\text{min}$