Q. 66 - Precalculus Enhanced with Graphing Utilities

Question

The volume $V$ (in cubic meters) of the hot-air balloon is given by $V\left(r\right)=\frac{4}{3}{\mathrm{\pi r}}^3.$If the radius $r$ is  function of $t$ as $r\left(t\right)=\frac{2}{3}{t}^3,$ find the volume $V$ as a function of the time $t$. 

Step-by-Step Solution

Answer

Hence, $\mathrm{V}\left(\mathrm{t}\right)=\frac{32}{81}{\mathrm{\pi t}}^9$

Step 1. Given Information

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

Step 2. Solution

We have to find $V$ as a function of time $t.$

Here, $V\left(r\right)=\frac{4}{3}{\mathrm{\pi r}}^3$ and $r\left(t\right)=\frac{2}{3}{t}^3.$

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

$$\begin{gathered} V\circ r\left(t\right)=V\left(r\right(t\left)\right) \\ V\circ r\left(t\right)=\frac{4}{3}\mathrm{\pi }{\left\{\mathrm{r}\right(\mathrm{t}\left)\right\}}^3\mathrm{V}\circ \mathrm{r}\left(\mathrm{t}\right)=\frac{4}{3}\mathrm{\pi }{\left\{\frac{2}{3}{\mathrm{t}}^3\right\}}^3 \\ \mathrm{V}\circ \mathrm{r}\left(\mathrm{t}\right)=\frac{4}{3}\mathrm{\pi }\times \frac{8}{27}{\mathrm{t}}^9 \\ \mathrm{V}\circ \mathrm{r}\left(\mathrm{t}\right)=\frac{32}{81}{\mathrm{\pi t}}^9 \\ \mathrm{V}\left(\mathrm{t}\right)=\frac{32}{81}{\mathrm{\pi t}}^9 \end{gathered}$$

Step 3. Final answer

Hence, $V$ as a function of $t$ is $\mathrm{V}\left(\mathrm{t}\right)=\frac{32}{81}{\mathrm{\pi t}}^9$