Question

If the rate of change of volume of the sphere is equal to the rate of change of its radius, find the radius of the sphere.

Answer 1

So here we first assume volume as V and radius as R and after applying formula of vol of sphere $V=\dfrac{4}{3}\pi {{R}^{3}}$ now rate of change is given so we will differentiate it which gives expression as $dV=\dfrac{4}{3}\pi 3{{R}^{2}}dR$ but rate of change of volume is equal to the rate of change of radius means $dV=dR$ putting it in equation to get the correct solution.

Complete step by step answer:
Given a sphere whose rate of change of volume is equal to the rate of change of its radius and we have to find the radius of sphere
So first we assume volume of sphere as V and radius of that sphere as R
Now we know that formula of volume of sphere is $V=\dfrac{4}{3}\pi {{R}^{3}}$ and given condition is related to derivative so we should differentiate this expression, which gives
$dV=\dfrac{4}{3}\pi 3{{R}^{2}}dR$ (using differentiation property $d({{x}^{3}})=3{{x}^{2}}dx$)
But according to given condition rate of change of volume is equal to the rate of change of its radius which means $dV=dR$ so on putting this condition in equation $dV=\dfrac{4}{3}\pi 3{{R}^{2}}dR$
We get expression as $1=\dfrac{4}{3}\pi 3{{R}^{2}}$ which on solving looks $\dfrac{3}{4\pi \times 3}={{R}^{2}}$ further solving looks
$\dfrac{1}{4\pi }={{R}^{2}}$ now taking square root and we got value of R as $R=\dfrac{1}{2\sqrt{\pi }}$

Hence radius of given sphere is $R=\dfrac{1}{2\sqrt{\pi }}$

Note: Most of the students do mistake while taking derivative for example they do derivative as $d({{x}^{3}})=3{{x}^{3}}dx$, forgets to decrease the power by 1 it should be $d({{x}^{3}})=3{{x}^{2}}dx$, some of students remember wrong formula $V=\dfrac{4}{3}\pi {{R}^{2}}$ which should be $V=\dfrac{4}{3}\pi {{R}^{3}}$