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

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


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}


Volume of a Sphere


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.


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}

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Gurleen Kaur

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