The rate of change of the volume of the spher | vedclass

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(B) Let the radius of the sphere be $r$. The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$. The surface area $S$ of the sphere is giv

Vedclass · Accepted Answer

$\frac{1}{4} \sqrt{\frac{S}{\pi}}$

Vedclass · Answer

(B) Let the radius of the sphere be $r$. The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$. The surface area $S$ of the sphere is given by $S = 4 \pi r^2$. From the surface area formula,we have $r^2 = \frac{S}{4 \pi}$,which implies $r = \sqrt{\frac{S}{4 \pi}} = \frac{1}{2} \sqrt{\frac{S}{\pi}}$. We need to find $\frac{dV}{dS}$. Using the chain rule,$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$. $\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} \pi r^3) = 4 \pi r^2$. $\frac{dS}{dr} = \frac{d}{dr} (4 \pi r^2) = 8 \pi r$. Therefore,$\frac{dV}{dS} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$. Substituting the value of $r$,we get $\frac{dV}{dS} = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{S}{\pi}}) = \frac{1}{4} \sqrt{\frac{S}{\pi}}$. Thus,the correct option is $B$.

The rate of change of the volume of the sphere with respect to its surface area $S$ is . . . . . . .

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