The volume of a sphere is increasing at the rate of $8c{m}^3/s$. Find the rate at which its surface area is increasing when the radius of the sphere is $12cm$.

Volume and surface area of sphere :

$V=\frac{4}{3}\pi r^{3}S=4\pi r^2$

Rate of volume and surface area :

$\frac{dV}{dt}=\frac{4}{3}\pi \times 3r^2\times dr/dt\frac{dS}{dt}=4\pi \times 2r\times dr/dt$

$\frac{dV}{dt}=8c{m}^3/s\Rightarrow \frac{4}{3}\pi \times 3r^2\times \frac{dr}{dt}=8\Rightarrow \frac{dr}{dt}=\frac{8\times 7}{22\times 4\times {12}^2}=0.00441$

Now $\frac{dS}{dt}=4\pi \times 2r\times \frac{dr}{dt}=\frac{4\times 22\times 2\times 12}{7}\times 0.00441=65.33{cm}^2/s$