\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}