\begin{frame}<-20> \frametitle{Related (Dependent) Rates} \begin{exampleblock}{} We have a right-angled triangle of the form \begin{center}\vspace{-1ex} \begin{tikzpicture}[default,yscale=.8] \draw (0,0) -- node[above] {$x$} (3,0) -- node[below right] {$\onslide<10->{\alert{H}}$} (0,-2) -- node[left] {$20cm$} (0,0); \draw[gray] (0.2,0) -- ++(0,-.2) -- ++ (-.2,0); \node at (.25,-1.55) {$\phi$}; \end{tikzpicture}\vspace{-1.5ex} \end{center} The length $x$ increases with $4cm/s$.\\ How fast is the angle $\phi$ changing when $x = 15cm$? \pause\medskip \alt<-4>{ The quantities $x$ and $\phi$ are related by: \begin{talign} \tan \phi = \frac{x}{20} \end{talign} \pause Differentiating both sides yields: \begin{talign} \frac{d}{dt} \tan \phi = \frac{d}{dt} \frac{x}{20} \mpause[1]{\;\implies\; \frac{1}{(\cos \phi)^2}\cdot\frac{d\phi}{dt} = \frac{1}{20}\cdot \frac{dx}{dt} } \end{talign} }{ \pause[5]\vspace{-3ex} \begin{talign} &\frac{1}{(\cos \phi)^2}\cdot\frac{d\phi}{dt} = \frac{1}{20}\cdot \frac{dx}{dt}\\ &\mpause[1]{\implies \quad \frac{d\phi}{dt} = \frac{(\cos \phi)^2}{20}\cdot \frac{dx}{dt} } \mpause[2]{= \frac{(\cos \phi)^2}{20}\cdot 4 } \mpause[3]{= \frac{(\cos \phi)^2}{5} } \end{talign} \pause\pause\pause\pause We have $\cos \phi \pause= 20 / H \pause= 20 / \sqrt{15^2 + 20^2} \pause= 20/25 = 4/5$. \pause Thus \begin{talign} \frac{d\phi}{dt} = \left(\frac{4}{5}\right)^2\cdot \frac{1}{5} \mpause[1]{= \frac{4^2}{5^3}} \mpause[2]{= \frac{16}{125}\text{rad/s}} \end{talign} } \end{exampleblock} \vspace{10cm} \end{frame}