The volume of a spherical balloon is increasing at the rate of $6 ~cm^3 / sec$. The rate of change of its surface area when its radius is 4 cm is:

$3 ~cm^2 / sec$

The correct answer is Option (4) → $3 ~cm^2 / sec$

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

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

$6 = 4\pi (4)^2 \frac{dr}{dt}$

$6 = 64\pi \frac{dr}{dt}$

$\frac{dr}{dt} = \frac{6}{64\pi} = \frac{3}{32\pi}$

$S = 4\pi r^2$

$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

$= 8\pi (4) \cdot \frac{3}{32\pi}$

$= \frac{96\pi}{32\pi} = 3$

$\text{Rate of change of surface area} = 3\ \text{cm}^2/\text{sec}$