\begin{frame} \frametitle{Related (Dependent) Rates} \begin{exampleblock}{} Air is pumped into a spherical balloon: \begin{itemize} \item the volume increases with $100\;\text{cm}^3/\text{s}$ \end{itemize} \pause Find: rate of change of the radius when the diameter is $50cm$. \pause\medskip First step: introduce suggestive notation \begin{itemize} \pause \item let $V(t)$ be the volume after time $t$ \pause \item let $r(t)$ be the radius after time $t$ \end{itemize} \pause Then the given problem translates to \begin{talign} &V'(t) = \mpause[1]{ 100\;\text{cm}^3/\text{s} } & &\mpause[2]{ \text{Find $r'(t)$ when $r = \mpause[3]{\text{25cm}}$. } } \end{talign} \pause\pause\pause\pause How are the volume of a sphere and its radius related? \pause \begin{talign} V = \frac{4}{3}\pi r^3 &&\mpause[1]{\text{thus}} &&\mpause[1]{V'(t) = \frac{d}{dt} \left(\frac{4}{3}\pi r(t)^3\right) } \mpause[2]{ = \frac{4}{3}\pi\cdot 3r(t)^2 r'(t) } \end{talign} \pause\pause\pause We solve for $r'(t)$:\vspace{-1ex} \pause \begin{talign} r'(t) = \frac{V'(t)}{4\pi\cdot r(t)^2} %\mpause[1]{= \frac{100}{4\pi\cdot r(t)^2}} &&\mpause[1]{r'(t) = \frac{100}{4\pi\cdot 25^2} \mpause[2]{= \alert{\frac{1}{25\pi} \mpause[3]{\;\text{cm$/$s}}} } } \end{talign} \end{exampleblock} \end{frame}