\begin{frame} \frametitle{Related (Dependent) Rates} \vspace{-.5ex} \begin{exampleblock}{} A water tank has the shape of an inverted circular cone: \begin{itemize} \item base radius $2m$ and the height is $4m$, \item water is pumped into the tank at a rate of $2\text{m}^3/\text{min}$. \end{itemize} At what rate is the water rising when the water is $3$m deep? \begin{minipage}{.39\textwidth} \begin{tikzpicture}[default,scale=1] \draw [draw=none,fill=cblue!20] (0,0) -- (1,2) -- (-1,2) -- cycle; \draw [draw=cblue!30,fill=cblue!15] (0,2) ellipse (1 and .27); \begin{scope}[line width=.5pt] \draw (1.5,3) -- (0,0) -- (-1.5,3); \draw (0,3) ellipse (1.5 and .4); \draw[<->] (1.3,0) -- node[right] {$h$} (1.3,2); \draw[<->] (1.8,0) -- node[right] {$4$} (1.8,3); \draw (1.2,0) -- (1.4,0); \draw (1.2,2) -- (1.4,2); \draw (1.7,0) -- (1.9,0); \draw (1.7,3) -- (1.9,3); \draw[dashed] (0,0) -- (0,3) -- node[above] {$2$} (1.5,3); \draw[dashed] (0,2) -- node[above] {$r$} (1,2); \end{scope} \end{tikzpicture} \end{minipage} \begin{minipage}{.59\textwidth} \pause \begin{talign} V = \frac{1}{3}\pi r^2 h \end{talign} \pause How is $r$ related to $h$? \pause \begin{talign} \frac{r}{h} = \frac{2}{4} \mpause[1]{\quad\implies\quad r = \frac{1}{2}h} \end{talign}\vspace{-2ex} \pause\pause \begin{talign} V = \frac{1}{3}\pi (\frac{1}{2}h)^2 h \mpause[1]{= \frac{1}{12}\pi h^3} \end{talign} \end{minipage} \pause\pause\medskip We differentiate both sides with respect to $t$: \begin{talign} \frac{dV}{dt} = \frac{d}{dt} (\frac{1}{12}\pi h^3) \mpause[1]{= \frac{1}{12}\pi 3h^2 \frac{dh}{dt} } \mpause[2]{\;\implies\; \frac{dh}{dt} = \frac{4}{\pi h^2}\frac{dV}{dt} } \mpause[3]{\stackrel{h=3}{=} \frac{4}{\pi 9} \cdot 2} \end{talign} \pause\pause\pause\pause Thus the water rises with $8/(\pi 9)$m/min when its is $3$m deep. \end{exampleblock} \end{frame}