#### Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  2

Answer: $3\; cm^{2}/sec$

Hint: Here we all know the volume of sphere is $\mathrm({v})=\frac{4}{3} \pi r^{3}$

Given: $r=2cm$ and the volume of sphere is increasing at the rate of $\text { 3 cubic cm/second }$

Solution:  Let r be the radius and v be the volume of then

$\mathrm({v})=\frac{4}{3} \pi r^{3}$

$\frac{d v}{d t}=\frac{4}{3} \pi \times 3 r^{2} \times \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi r^{2}} \times \frac{d V}{d t}$

\begin{aligned} &\frac{d r}{d t}=\frac{3}{4 \pi(2)^{2}}[r=2 \mathrm{~cm}] \\\\ &\frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}

Now,let S be the surface area of sphere

So,         $S=4 \pi r^{2}$

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

$\Rightarrow \frac{d S}{d t}=8 \pi \times 2 \times \frac{3}{16 \pi}$                            (because r=2)

$\Rightarrow \frac{d S}{d t}=3 \mathrm{~cm}^{2} / \mathrm{sec}$