91Ó°ÊÓ

The volume of the sphere is given by $V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$cu. units.

The surface area of the sphere is given by$S=4{\mathrm{Ï€°ù}}^2$sq. units.

Given that radius(r), volume(V), surface area(S) are functions of t.

Apply the differentiation to $V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$ with respect to tas follows.

role="math" localid="1648726072573" $$\begin{gathered} \frac{dV}{dt}=\frac{4}{3}\mathrm{Ï€}\left(3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}\right) \\ \frac{\mathrm{dV}}{\mathrm{dt}}=4{\mathrm{Ï€°ù}}^2\frac{\mathrm{dr}}{\mathrm{dt}} \\ \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4{\mathrm{Ï€°ù}}^2}\frac{\mathrm{dV}}{\mathrm{dt}} \end{gathered}$$

It is found that $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4{\mathrm{Ï€°ù}}^2}\frac{\mathrm{dV}}{\mathrm{dt}}$.

Given that radius(r), volume(V), surface area(S) are functions of t.

Apply the differentiation to $S=4{\mathrm{Ï€°ù}}^2$ with respect to t as follows.

role="math" localid="1648726391483" $$\begin{gathered} \frac{dS}{dt}=4\mathrm{Ï€}\left(2\mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}\right) \\ \frac{\mathrm{dS}}{\mathrm{dt}}=8\mathrm{Ï€°ù}\frac{\mathrm{dr}}{\mathrm{dt}} \\ \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{8\mathrm{Ï€°ù}}\frac{dS}{\mathrm{dt}} \end{gathered}$$

It is found that$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{8\mathrm{Ï€°ù}}\frac{dS}{\mathrm{dt}}$.

From step 2, $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4{\mathrm{Ï€°ù}}^2}\frac{\mathrm{dV}}{\mathrm{dt}}$.

From step 3, $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{8\mathrm{Ï€°ù}}\frac{dS}{\mathrm{dt}}$.

Equating $\frac{dr}{dt}$obtained in both step 2 and step 3 as follows.

$$\begin{gathered} \frac{1}{4{\mathrm{Ï€°ù}}^2}\frac{dV}{dt}=\frac{1}{8\mathrm{Ï€°ù}}\frac{dS}{\mathrm{dt}} \\ \frac{dV}{dt}=\frac{4{\mathrm{Ï€°ù}}^2}{8\mathrm{Ï€°ù}}\frac{dS}{\mathrm{dt}} \\ \frac{dV}{dt}=\frac{r}{2}\frac{dS}{\mathrm{dt}} \end{gathered}$$

The derivatives$\frac{dV}{dt}$and$\frac{dS}{dt}$are related by$\frac{dV}{dt}=\frac{r}{2}\frac{dS}{\mathrm{dt}}$.