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\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}}
\end{talign}
}
\end{exampleblock}
\vspace{10cm}
\end{frame}