Q. 65 - Precalculus Enhanced with Graphing Utilities

Question

The surface area S (in square meters) of a hot-air balloon is given by $S\left(r\right)=4{\mathrm{\pi r}}^2$, where $r$ is the radius of the balloon (in meters).If the radius $r$ is increasing with time $t$ (in seconds) according to the formula $r\left(t\right)=\frac{2}{3}{t}^3, t\ge 0,$ find the surface area S of the balloon as a function of the time $t.$

Step-by-Step Solution

Answer

$S\left(t\right)=\frac{16}{9}{\mathrm{\pi t}}^6.$

Step 1. Given Information

Given that $S\left(r\right)=4{\mathrm{\pi r}}^2$ and $r\left(t\right)=\frac{2}{3}{t}^3, t\ge 0.$

Step 2. Solution

We have to find surface area $S$ of the balloon as a function of time $t.$

We have $S\left(r\right)$ is function of $r$ and $r\left(t\right)$ is function of $t$.

So, we have to find $S\circ r\left(t\right).$

Now,

$$\begin{gathered} S\circ r\left(t\right)=S\left(r\right(t\left)\right) \\ S\circ r\left(t\right)=4\mathrm{\pi }{\left\{\mathrm{r}\right(\mathrm{t}\left)\right\}}^2\mathrm{S}\circ \mathrm{r}\left(\mathrm{t}\right)=4\mathrm{\pi }{\left\{\frac{2}{3}{\mathrm{t}}^3\right\}}^2 \\ \mathrm{S}\circ \mathrm{r}\left(\mathrm{t}\right)=4\mathrm{\pi }\left(\frac{4}{9}{\mathrm{t}}^6\right) \\ \mathrm{S}\circ \mathrm{r}\left(\mathrm{t}\right)=\frac{16}{9}{\mathrm{\pi t}}^6. \end{gathered}$$

Step3. Final Answer

Hence, S as a function of t is $S\left(t\right)=\frac{16}{9}{\mathrm{\pi t}}^6.$