\begin{frame}
  \frametitle{Newtons Law of Cooling/Warming}

  \begin{block}{}
    \begin{malign}
    T'(t) = k(T(t) - T_s)
    \end{malign}
  \end{block}
  \begin{exampleblock}{}
    A bottle of water is placed in the refrigerator:
    \begin{itemize}
    \pause
      \item bottle has temperature $60\textdegree$F,
    \pause
      \item refrigerator has temperature $20\textdegree$F
    \end{itemize}
    \pause
    After $2$ minutes the bottle has cooled down to $30\textdegree$F.
    \begin{itemize}
    \pause
      \item Find a formula for the temperature.
%     \pause
%       \item What is the temperature of the bottle after $1$ hour?
%     \pause
%       \item How long does it take the bottle to cool to $50\textdegree$F?
    \end{itemize}
    \pause
    \begin{talign}
      T'(t) = k(T(t) - T_s) \mpause[1]{= k(T(t) - 20)}
    \end{talign}
    \pause\pause
    We let \alert{$y(t) = T(t) - 20$}\pause, then 
    \begin{talign}
      &y(0) = \mpause[1]{ T(0) - 20 } \mpause[2]{= 60 - 20 = 40} \\
      &\mpause[3]{ y(t) = y(0)e^{kt}} \mpause[4]{ = 40 e^{kt}} \\[-1.5ex]
      &\mpause[5]{ y(2)= 40 e^{k2}} \mpause[6]{ = T(2) - 20} \mpause[7]{ = 10} 
%       \mpause[8]{\;\implies\; 30k = \ln \frac{17}{28}} 
      \mpause[8]{\;\implies\; k = \frac{\ln \frac{10}{40}}{2} = \ln \frac{1}{2}}\\[-4ex]
    \end{talign}
    \pause\pause\pause\pause\pause\pause\pause\pause\pause
    Thus \alert{$T(t) \;=\; \mpause[1]{y(t) + 20} \mpause[2]{ \;=\; 40e^{t\cdot \ln \frac{1}{2}} + 20}$}
  \end{exampleblock}
\end{frame}