The rate of change of the volume of a sphere with respect to its surface area, when its radius is 2 cm , is

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Aicharya · Accepted Answer

Volume of sphere $(V)=\frac{4}{3} \pi r^3$ Surface area of sphere $(A)=4 \pi r^2$ $\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^2 	ext { and } \frac{\mathrm{dA}}{\mathrm{dr}}=8 \pi \mathrm{r}$ $\begin{aligned} & 	herefore \quad\left(\frac{d V}{d A}ight)=\frac{\left(\frac{d V}{d r}ight)}{\left(\frac{d A}{d r}ight)}=\frac{4 \pi r^2}{8 \pi r}=\frac{r}{2} \ & 	herefore \quad\left(\frac{d V}{d A}ight)_{r=2}=1 \mathrm{~cm}^3 / \mathrm{cm}^2\end{aligned}$

The rate of change of the volume of a sphere with respect to its surface area, when its radius is 2 cm , is

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