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

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

Answers (1)

Answer:

\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}

Posted by

Gurleen Kaur

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