Exercise 1.3.60

Question

A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of $2 \mathrm{~cm}/	ext{s}.$
\begin{enumerate}[label=(\alph*)]
\item Express the radius $r$ of the balloon as a function of the time $t$ (in seconds)
\item If $V$ is the volume of the balloon as a function of the radius, find $V \circ r$ and interpret it.
\end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)]\item Since the radius of the balloon increases by $2 \mathrm{~cm}$ with every second, the function $r$ of the time should be defined as follows:
$$
\forall t \in\left[0, \frac{r_{\max }}{2}ight] ; \quad r(t)=2 t 	ext {, }
$$
where $r_{	ext {max }}$ is the largest radius a baboon can attain.

\item First, we should define $V$ as the function of the radius of the balloon to compose the functions $V$ and $r$. Thus, we have.
$$
V:\left[0, r_{	ext {max }}ight] ightarrow\left[0, V_{	ext {max }}ight]: V(r)=\frac{4}{3} \pi r^{3}
$$
where $r_{	ext {max }}$ is the maximal value of the radius of the baton, and $V_{	ext {max }}$ is the maximal volume of the balloon. Thus, we have:
$$
\forall t \in\left[0, \frac{r_{	ext {max }}}{2}ight]: V\circ r(t)=\frac{32}{3} \pi t^{3} 	ext {, }
$$
The formula above represents the value of the balloon at a particular time $t$. 
\end{enumerate}

Exercise 1.3.60

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Exercise 1.3.60

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