Q. 14 - Calculus

Question

Suppose the radius r, volume V, and surface area S of a sphere are functions of time t. How are $\frac{dV}{dt}$ and $\frac{dS}{dt}$ related? 

Step-by-Step Solution

Answer

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

Step 1. Formula used.

The volume of the sphere is given by $V=\frac{4}{3}{\mathrm{\pi r}}^3$ cu. units.

The surface area of the sphere is given by $S=4{\mathrm{\pi r}}^2$ sq. units.

Step 2. Apply the differentiation to V .

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

Apply the differentiation to $V=\frac{4}{3}{\mathrm{\pi r}}^3$ with respect to t as follows.

$$\begin{gathered} \frac{dV}{dt}=\frac{4}{3}\mathrm{\pi }\left(3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}\right) \\ \frac{\mathrm{dV}}{\mathrm{dt}}=4{\mathrm{\pi r}}^2\frac{\mathrm{dr}}{\mathrm{dt}} \\ \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4{\mathrm{\pi r}}^2}\frac{\mathrm{dV}}{\mathrm{dt}} \end{gathered}$$

It is found that $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4{\mathrm{\pi r}}^2}\frac{\mathrm{dV}}{\mathrm{dt}}$.

Step 3. Apply the differentiation to S .

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

Apply the differentiation to $S=4{\mathrm{\pi r}}^2$ with respect to t as follows.

$$\begin{gathered} \frac{dS}{dt}=4\mathrm{\pi }\left(2\mathrm{r}\frac{\mathrm{dr}}{\mathrm{dt}}\right) \\ \frac{\mathrm{dS}}{\mathrm{dt}}=8\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}} \\ \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{8\mathrm{\pi r}}\frac{dS}{\mathrm{dt}} \end{gathered}$$

It is found that $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{8\mathrm{\pi r}}\frac{dS}{\mathrm{dt}}$.

Step 4. Equating the derivative.

From step 2, $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4{\mathrm{\pi r}}^2}\frac{\mathrm{dV}}{\mathrm{dt}}$.

From step 3, $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{8\mathrm{\pi r}}\frac{dS}{\mathrm{dt}}$.

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

$$\begin{gathered} \frac{1}{4{\mathrm{\pi r}}^2}\frac{dV}{dt}=\frac{1}{8\mathrm{\pi r}}\frac{dS}{\mathrm{dt}} \\ \frac{dV}{dt}=\frac{4{\mathrm{\pi r}}^2}{8\mathrm{\pi r}}\frac{dS}{\mathrm{dt}} \\ \frac{dV}{dt}=\frac{r}{2}\frac{dS}{\mathrm{dt}} \end{gathered}$$

Step 5. Conclusion.

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