The radius of a sphere is changing. At an ins | vedclass

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Question

(B) Given that, at any instant of time, the rate of change in volume with respect to time is equal to the rate of change in surface area with respect

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given that, at any instant of time, the rate of change in volume with respect to time is equal to the rate of change in surface area with respect to time, i.e.,$\frac{dV}{dt} = \frac{dS}{dt}$ $\ldots(i)$ The volume of a sphere of radius $r$ is $V = \frac{4}{3} \pi r^3$. Differentiating with respect to $t$, we get $\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3} \pi r^3) = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$ $\ldots(ii)$ The surface area of a sphere of radius $r$ is $S = 4 \pi r^2$. Differentiating with respect to $t$, we get $\frac{dS}{dt} = \frac{d}{dt}(4 \pi r^2) = 4 \pi (2r) \frac{dr}{dt} = 8 \pi r \frac{dr}{dt}$ $\ldots(iii)$ Substituting the values from equations $(ii)$ and $(iii)$ into equation $(i)$, we get $4 \pi r^2 \frac{dr}{dt} = 8 \pi r \frac{dr}{dt}$. Assuming $\frac{dr}{dt} 
eq 0$, we divide both sides by $4 \pi r \frac{dr}{dt}$ to obtain $r = 2$.

The radius of a sphere is changing. At an instant of time, the rate of change in its volume and its surface area are equal. Then the value of the radius at that instant is?

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