#### Provide solution for RD Sharma maths class 12 chapter Derivative as a Rate Measure exercise 12.1 question 2

$\frac{\pi D^{2}}{2}, D\; \; is\; the\; diameter\; o\! f\; sphere.$

Hint:

To find rate of change of volume we have to find the differentiation of volume of sphere with respect to diameter

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

Here,$R$= Radius,

Relation between Radius $R$ and diameter $D$ is $D=2R$

$\therefore R=\frac{D}{2}$

$\therefore V\! olume\; o\! f spher\! e, V=\frac{4}{3}\pi \left ( \frac{D}{2} \right )^{3} = \frac{4}{3}\pi \frac{D^{3}}{8}$

$\therefore V=\frac{\pi D^{3}}{6}$

Given:

Volume of a Sphere

Solution:

Here we have,

$D$= Diameter of sphere
$V$= Volume of sphere

Volume of sphere in terms of diameter,

$\therefore V=\frac{\pi D^{3}}{6}$

Let’s differentiate equation (i) with respect to diameter,

$\therefore \frac{d V}{d D}=\frac{d}{d D}\left ( \frac{\pi D^{3}}{6} \right )$

$\therefore \frac{d V}{d D}= \frac{\pi D^{2}}{2}$                                            $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

This is the rate of change of the volume of sphere with respect to diameter.

Note:

Here answer is with respect to diameter but with respect to radius, rate of change of volume of sphere is

$\therefore \frac{d V}{d R}= 4\pi R^{2}$