\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}