Question

If the volume of spherical ball is increasing at the rate of $4\pi cc/sec$ then the rate of change of its surface area when the volume is $288\pi cc$ is

• A. $\frac{4}{3}\pi c{m}^2/sec$
• B. $\frac{2}{3}\pi c{m}^2/sec$
• C. $4\pi c{m}^2/sec$
• D. $2\pi c{m}^2/sec$

Answer 1

Answer: (A)

$\frac{4}{3}\pi c{m}^2/sec$

$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$
When $V=288\pi$
$⟹288\pi =\frac{4}{3}\pi r^3\Rightarrow r=6$
Given, $\frac{dv}{dt}=4\pi$
$\therefore 4\pi r^2\frac{dr}{dt}=4\pi =\frac{dr}{dt}=\frac{1}{r^2}$
$A=$ Surface area $=4\pi r^2$
$\therefore \frac{dA}{dt}=8\pi r\frac{dr}{dt}=8\pi r\times \frac{1}{r^2}=\frac{8\pi }{r}=\frac{8\pi }{6}=\frac{4\pi }{3}$

Hence, $\frac{dA}{dt}=\frac{4\pi }{3}c{m}^2/sec$