#### Need solution for RD Sharma maths class 12 chapter Derivative as a Rate Measure exercise 12.1 question 3

Hint:

Here we have to differentiate volume of sphere with respect to surface area of sphere

$V\! olume\; o\! f\; spher\! e, V=\frac{4}{3}\pi r^{2}$

$\therefore S\! ur\! f\! ace\; area\; o\! f\; spher\! e, A=4\pi r^{2}$

Given:

Radius of sphere, $r=2cm$

Solution:

Here we have,

Radius, $r=2cm$

Area, $A=4\pi r^{2}$

Let’s differentiate it with respect to area ($A$)

$\Rightarrow \frac{d(A)}{dA}=4(2)\pi r\frac{dr}{dA}$                                    $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\Rightarrow 1=8\pi r\frac{dr}{dA}$

$\therefore \frac{dr}{dA}=\frac{1}{8\pi r}\; \; \; \; \; \; \; \; \; \; \; ......(i)$

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

Let’s differentiate with respect to Area (A)

$\Rightarrow \frac{d V}{d A}=4\pi r^{2}\frac{dr}{d A}$

$\Rightarrow \frac{d V}{d A}=\frac{4\pi r^{2}}{8\pi r}$                                                        {From the Equation (i)}

$\therefore \frac{d V}{d A}=\frac{r}{2}$

To putting value of $r=2cm$

$\therefore \frac{d V}{d A}=1cm$

Note:

We cannot put Area directly in formula of volume,

$V=\frac{4}{3}\pi r^{2}$

$S\! o, we\; can\; not\; write,V=\frac{A.r}{3}$