If the surface area of a sphere of radius r i | vedclass

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(D) The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{d

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proportional to $r$

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(D) The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \dots (i)$. The surface area of a sphere is $S = 4\pi r^2$. Differentiating with respect to time $t$,we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$. Given that $\frac{dS}{dt} = 8 \, cm^2/s$,we have $8 = 8\pi r \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{1}{\pi r}$. Substituting the value of $\frac{dr}{dt}$ into equation $(i)$,we get $\frac{dV}{dt} = 4\pi r^2 	imes \frac{1}{\pi r} = 4r$. Thus,the rate of change of volume $\frac{dV}{dt}$ is proportional to $r$.

If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8 \, cm^2/s$,then the rate of change of its volume is

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