\begin{frame}
  \frametitle{Review - Midterm Exam 3}
  
  \begin{exampleblock}{}
    Two people are standing together. One begins to walk east at a rate of 2 miles per hour,
    and at the same time the second begins to walk north at a rate of 3 miles per hour.
    How fast is their distance growing when the first has walked 4 miles?
    \pause\medskip
    
    \begin{minipage}{.2\textwidth}
      \begin{tikzpicture}[default,ultra thick]
        \draw[->] (0,0) -- node[below] {$x$} (1,0);
        \draw[->] (0,0) -- node[left] {$y$} (0,1);
        \mpause[1]{
        \draw[dashed,shorten <= 2mm, shorten >= 2mm] (0,1) -- node[above right] {$z$} (1,0);
        }
      \end{tikzpicture}
    \end{minipage}
    \begin{minipage}{.79\textwidth}
      \pause\pause
      We know that
      \pause
      \begin{talign}
        \frac{d}{dx} = 2 &&
        \frac{d}{dy} = 3 &&
        \mpause[1]{z^2 = }\mpause{x^2 + y^2}
      \end{talign}
    \end{minipage}
    \pause\pause\pause\smallskip
    
    The first has walked $4$ miles when $t = \pause 4/2 = 2h$.
    \pause
    At time $t = 2h$:\vspace{-.5ex}
    \begin{talign}
      x = 4 && y = \mpause[1]{2*3 = 6} && z = \mpause{\sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}}
    \end{talign}
    \pause\pause\pause
    We use implicit differentiation:\vspace{-.5ex}
    \begin{talign}
      &\frac{d}{dt} z^2 = \frac{d}{dt}(x^2 + y^2)
      \mpause[1]{\;\implies\; 2zz' = 2xx' + 2yy'} \\
      &\mpause{z' = \frac{2xx' + 2yy'}{2z}}
      \mpause{= \frac{2 \cdot 4 \cdot 2 + 2\cdot 6 \cdot 3}{2\cdot 2\sqrt{13}}}
      \mpause{= \frac{13}{\sqrt{13}}}
      \mpause{= \sqrt{13}}
      \mpause{\approx 3.6}
    \end{talign}\vspace{-1.25ex}%
    \pause\pause\pause\pause\pause\pause\pause
    
    Their distance increases with \alert{$\sqrt{13}$ miles per hour}.\vspace{-.6ex}
  \end{exampleblock}
\end{frame}