Q. 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 and altitude is 6m, is
A) $80\pi\ cu\ m/s$
B) $144\pi\ cu\ m/s$
C) $80\ cu\ m/s$
D) $64\ cu\ m/s$
Explanation
(a)
Let $h\$and r be the height and radius of cylinder.
Given that ,$\frac{dr}{dt} = 3m/s,\$ $\frac{dh}{dt} = - 4m/s$
Let volume of cylinder, $V = \pi r^{2}h$
$\Rightarrow \ \ \ \frac{dV}{dt} = \pi\left\lbrack r^{2}\frac{dh}{dt} + h.2r\frac{dr}{dt}\ \right\rbrack$
At $r = 4m$
And $h = 6m$
$\therefore\ \ \frac{dV}{dt} = \pi\lbrack - 64 + 144\rbrack = 80\pi cu\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.