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Question

(A) 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}

Vedclass · Accepted Answer

$\frac{4}{3} \pi \ cm^2/sec$

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(A) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given $\frac{dV}{dt} = 4 \pi \ cm^3/sec$,we have $4 \pi r^2 \frac{dr}{dt} = 4 \pi$,which implies $\frac{dr}{dt} = \frac{1}{r^2}$.When the volume $V = 288 \pi \ cm^3$,we have $\frac{4}{3} \pi r^3 = 288 \pi$,so $r^3 = 216$,which means $r = 6 \ cm$.Now,the surface area $A = 4 \pi r^2$. Differentiating with respect to $t$,we get $\frac{dA}{dt} = 8 \pi r \frac{dr}{dt}$.Substituting $r = 6$ and $\frac{dr}{dt} = \frac{1}{r^2}$,we get $\frac{dA}{dt} = 8 \pi r 	imes \frac{1}{r^2} = \frac{8 \pi}{r}$.For $r = 6$,$\frac{dA}{dt} = \frac{8 \pi}{6} = \frac{4}{3} \pi \ cm^2/sec$.

If the volume of a spherical ball is increasing at the rate of $4 \pi \ cm^3/sec$,then the rate of change of its surface area when the volume is $288 \pi \ cm^3$ is:

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