The rate of change of the surface are of a sphere of radius r

Let $r$ be the radius of sphere at any time $t$.

It is given that,

$\frac{d r}{d t}=2 \mathrm{~cm} / \mathrm{sec}$

Surface area of the sphere, $S=4 \pi r^{2}$

$S=4 \pi r^{2}$

Differentiating both sides with respect to $t$, we get

$\frac{d S}{d t}=4 \pi \frac{d}{d t} r^{2}$

$\Rightarrow \frac{d S}{d t}=4 \pi \times 2 r \frac{d r}{d t}$

$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$

Putting $\frac{d r}{d t}=2 \mathrm{~cm} / \mathrm{sec}$, we get

$\frac{d S}{d t}=8 \pi r \times 2=16 \pi r \mathrm{~cm}^{2} / \mathrm{sec}$

Thus, the rate of change of the surface are of a sphere of radius $r$ when the radius is increasing at the rate of $2 \mathrm{~cm} / \mathrm{sec}$ is $16 \pi r \mathrm{~cm}^{2} / \mathrm{sec}$.

The rate of change of the surface are of a sphere of radius $r$ when the radius is increasing at the rate of $2 \mathrm{~cm} / \mathrm{sec}$ is __16π">ππr cm2/sec___.