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\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}