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\begin{frame}
\frametitle{Related (Dependent) Rates}

\begin{exampleblock}{}
Air is pumped into a spherical balloon:
\begin{itemize}
\item the volume increases with $100\;\text{cm}^3/\text{s}$
\end{itemize}
\pause
Find: rate of change of the radius when the diameter is $50cm$.
\pause\medskip

First step: introduce suggestive notation
\begin{itemize}
\pause
\item let $V(t)$ be the volume after time $t$
\pause
\item let $r(t)$ be the radius after time $t$
\end{itemize}
\pause
Then the given problem translates to
\begin{talign}
&V'(t) = \mpause[1]{ 100\;\text{cm}^3/\text{s} } &
&\mpause[2]{ \text{Find $r'(t)$ when $r = \mpause[3]{\text{25cm}}$. } }
\end{talign}
\pause\pause\pause\pause
How are the volume of a sphere and its radius related?
\pause
\begin{talign}
V = \frac{4}{3}\pi r^3 &&\mpause[1]{\text{thus}}
&&\mpause[1]{V'(t) = \frac{d}{dt} \left(\frac{4}{3}\pi r(t)^3\right)  }
\mpause[2]{ = \frac{4}{3}\pi\cdot  3r(t)^2 r'(t)  }
\end{talign}
\pause\pause\pause
We solve for $r'(t)$:\vspace{-1ex}
\pause
\begin{talign}
r'(t) = \frac{V'(t)}{4\pi\cdot r(t)^2} %\mpause[1]{= \frac{100}{4\pi\cdot r(t)^2}}
&&\mpause[1]{r'(t) = \frac{100}{4\pi\cdot 25^2} \mpause[2]{= \alert{\frac{1}{25\pi} \mpause[3]{\;\text{cm$/$s}}} } }
\end{talign}
\end{exampleblock}
\end{frame}