Question

The radius of a cylinder is increasing at the rate of $3m/s$ and its altitude is decreasing at the rate of $4m/s$. The rate of change of volume when radius is $4m/s$. The rate of change of volume when radius is $4m$ and altitude is $6m$, is

• A. $-64\pi +144m^3/s$
• B. $\pi -80m^3/s$
• C. $80m^3/s$
• D. $64m^3/s$

Answer 1

Answer: (A)

$-64\pi +144m^3/s$

Let hand r be the height and radius of cylinder
Given that, $\frac{dr}{dt}=3m/s,\frac{dh}{dt}=-4m/s$
Also, $V=\pi r^{2}h$
On differentiating with respect to t, we get
$\frac{dv}{dt}=\pi r^2\frac{dh}{dt}+h.2r\frac{dr}{dt}$
At $r=4m$ and $h=6$ m
$\therefore \frac{DV}{dt}=\pi -64+144=80\pi m^3/s$