Question

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

• (A) $80\pi \mathrm{cum}/\mathrm{s}$
• (B) $144\pi \mathrm{cum}/\mathrm{s}$
• (C) $80\mathrm{cum}/\mathrm{s}$
• (D) $64\mathrm{cum}/\mathrm{s}$

Answer 1

The Correct Option is A

Explanation:
Let $h$ and $r$ be the height and radius of cylinder. Given that $\frac{dr}{dt}=3\mathrm{m}/\mathrm{s},\frac{dh}{dt}=-4\mathrm{m}/\mathrm{s}$Also, $V=\pi r^{2}h$On differentiating w.r.t. $\frac{dV}{dt}=\pi [r^2\frac{dh}{dt}+h\cdot 2r\frac{dr}{dt}]$At $r=4\mathrm{m}$ and $h=6\mathrm{m}$$\frac{dV}{dt}=\pi [-64+144]\mathrm{cum}/\mathrm{s}$$\frac{dV}{dt}=80\pi \mathrm{cu}\mathrm{m}/\mathrm{s}$