\begin{frame}
  \frametitle{Related (Dependent) Rates}
  \vspace{-.5ex}
  \begin{exampleblock}{}
    A ladder of length $10\text{ft}$ rests against a vertical wall.
    \begin{itemize}
      \item the bottom of the ladder slides away from the wall with $1\text{ft}/\text{s}$
    \end{itemize}
    How fast is the top sliding when the bottom is $6\text{ft}$ from the wall?
    \pause\smallskip
    
    \begin{minipage}{.54\textwidth}
    \begin{center}
      \begin{tikzpicture}[default]
        \draw (0,0) -- node[at end,below,anchor=north] {ground} (3,0);
        \draw (0,0) -- node[at end,left,anchor=east] {wall} (0,2.3);
        \begin{scope}[cred]
        \draw[dashed] (0,1.5) -- node[above,anchor=south west] {$10$ft ladder} node [at start,left] {$y$} node [at end,below] {$x$} (2,0);
        \end{scope}
        \begin{scope}[cblue]
        \mpause[1]{
          \draw[->] (1.7,-.5) -- node [below] {$\frac{d}{dt}x = 1$}++(.6,0);
        }
        \mpause[2]{
          \draw[->] (-.5,1.8) -- node [left] {$\frac{d}{dt}y = $?\;\;} ++(0,-.6);
        }
        \end{scope}
      \end{tikzpicture}
    \end{center}
    \end{minipage}
    \begin{minipage}{.45\textwidth}
      \ \\
      \mpause[3]{Thus}
      \begin{talign}
        &\mpause[4]{x^2 + y^2 = 10^2} \\
        &\mpause[5]{\stackrel{x=6}{\implies} 6^2 + y^2 = 10^2} \\
        &\mpause[6]{\implies y = \pm \sqrt{10^2-6^2}} \\
        &\mpause[7]{\implies y = 8} \\
      \end{talign}
    \end{minipage}
    \pause\pause\pause\pause\pause\pause\pause\vspace{-1ex}
    \begin{talign}
      &\mpause[1]{\frac{d}{dt} (x^2 + y^2) = \frac{d}{dt} 10^2} 
      \mpause[2]{\implies 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 }\\
      &\mpause[3]{\implies \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} }
      \mpause[4]{\implies \frac{dy}{dt} = -\frac{6}{8}\cdot 1 = -\frac{3}{4} }
    \end{talign}\vspace{-1.5ex}
    \pause\pause\pause\pause\pause
    
    The top slides with \alert{$\frac{3}{4}$ft/s} when the bottom is $6$ft from the wall.
  \end{exampleblock}
\end{frame}