The rate of change of the volume of a sphere | vedclass

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(A) Let $r$ be the radius of the sphere. 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

$5/2 \ m$

Vedclass · Answer

(A) Let $r$ be the radius of the sphere. 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 $S = 4 \pi r^2$. We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dS}$. Using the chain rule,$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$. Differentiating $V$ with respect to $r$: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3} \pi r^3) = 4 \pi r^2$. Differentiating $S$ with respect to $r$: $\frac{dS}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r$. Therefore,$\frac{dV}{dS} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$. When the radius $r = 5 \ m$,the rate of change is $\frac{5}{2} \ m$.

The rate of change of the volume of a sphere with respect to its surface area,when the radius is $5 \ m$ is

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