#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 29 maths textbook solution

Answer: $\frac{d s}{d t}=10 \mathrm{~cm}^{2} / \mathrm{sec}$

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

Given: The volume of spherical balloon is increasing at the rate of 25  cm3/sec

Solution: Let the radius of the given spherical balloon be $r$ cm and  $V$  be its volume at any instant time.

Then according to the given criteria,

The rate of volume of the spherical balloon is increasing is $\frac{d V}{d t}=25 \: cm^{3}/sec$        ....…(i)

But volume of balloon is $\frac{4}{3} \pi r^{3}$

Applying derivative

\begin{aligned} &\frac{d V}{d t}=\frac{d\left[\frac{4}{3} \pi r^{3}\right]}{d t} \\\\ &\frac{d V}{d t}=\frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t} \end{aligned}

Substituting the value from equation (i)

\begin{aligned} &25=\frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t}\\\\ &\frac{d r}{d t}=\frac{25}{4 \pi r^{2}} \end{aligned}.......(ii)

Now the surface area of spherical balloon at any time $t$  will be $S=4 \pi r^{2} \mathrm{~cm}^{2}$

Applying derivative,

$\frac{d S}{d t}=\frac{d\left(4 \pi r^{2}\right)}{d t}=4 \pi \times 2 r \frac{d r}{d t}$

Substituting the value from equation (ii) we get,

$\frac{d S}{d t}=4 \pi \times 2 r \times \frac{25}{4 \pi r^{2}}$

So, when the radius is 5 cm,

$\frac{d S}{d t}=\frac{25 \times 2}{5}=10 \mathrm{~cm}^{2} / \mathrm{sec}$