71/87
\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}