Question

The radius of a cylinder is increasing at the rate of $2m/s$ and its height is decreasing at the rate of $3m/s$. When the radius is $3m$ and height is $5m$, then the volume of the cylinder would change at the rate of

• A. $87\pi m^3/s$
• B. $33\pi m^3/s$
• C. $27\pi m^3/s$
• D. $15\pi m^3/s$

Answer 1

Answer: (B)

Given, $\frac{dr}{dt}=2m/s$
$\frac{dh}{dt}=-3m/s$
$\therefore$ Volume of cylinder, $V=\pi r^{2}h$
$\therefore \frac{dV}{dt}=\pi \left[2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right]$
At $r=3m$ and $h=5m$,
$\frac{dV}{dt}=\pi [2\times 3\times 5\times 2-9\times 3]$ [using Eq. (i)]
At $r=3m$ and $h=5m$,
$\frac{dV}{dt}=\pi [2\times 3\times 5\times 2-9\times 3]$ [using Eq. (i)]
$\Rightarrow \frac{dV}{dt}=\pi [60-27]=33\pi$
$\Rightarrow \frac{dV}{dt}=33\pi m^3/s$