Question

If the volume of a sphere increases at the rate of $2\pi c{m}^3/sec$, then the rate of increase of its radius (in cm/sec), when the volume is $288\pi c{m}^3$ is

• A. $\frac{1}{36}$
• B. $\frac{1}{72}$
• C. $\frac{1}{18}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{1}{72}$

$\frac{dV}{dt}=2\pi {cm}^3/sec$

$V=\frac{4}{3}\pi r^3$

$288\pi =\frac{4}{3}\pi r^3\Rightarrow r^3=216\Rightarrow r=6$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$\frac{2\pi }{4\pi \times 6\times 6}=\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{72}cm/sec$