#### need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 7

Answer$\frac{d V}{d t}=2 \pi \mathrm{cm}^{3} / \mathrm{sec}$

Hint: As we know that the volume of bubble is $V=\frac{4}{3} \pi r^{3}$ .

Given: The radius of air bubble is increasing at the rate of 0.5 cm/sec

Solution: Suppose the radius of the given air bubble be $r$ cm and let $V$  be the volume of the air bubble at any instant time.

The rate of increasing in the radius of the air bubble is $\frac{\mathrm{dr}}{\mathrm{dt}}=0.5 \mathrm{~cm} / \mathrm{sec}$

By applying derivative on volume

$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t}$                    $\therefore\left(v=\frac{4}{3} \pi r^{3}\right)$

Let’s differentiate  $d\left(\frac{4}{3} \pi r^{3}\right)$

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

Let’s  put value of $\frac{d r}{d t}=0.5$    (given)

$\frac{d V}{d t}=4 \pi r^{2} \times 0.5$

Let’s put  value of $r = 1$

\begin{aligned} &\frac{d V}{d t}=4 \pi \times 1^{2} \times 0.5 \\\\ &\frac{d V}{d t}=2 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}

The rate is increasing is $2 \pi \mathrm{cm}^{3} / \mathrm{sec}$  when the radius is 1  cm.