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{\bf Chapters 1,2. Sample Problem 3}.

Suppose a cup of hot chocolate has an initial temperature of $185^\circ F$ when freshly poured and the desired drinking temperature is $160^\circ F$.  After $50$ seconds in a room at $68^\circ F$, the temperature has cooled to $181^\circ F$. {\bf Newton's Law of Cooling} applies to model the temperature $u(t)$ of the chocolate by the initial value problem
$$\frac{du}{dt} = -h\left( u(t)-68  \right),\quad u(0)=185,$$
where $h>0$ is the cooling constant, to be determined from supplied information.


\begin{itemize}
\item[{\bf 1}.] Find an equation for the temperature $u(t)$ at any time $t$.

\item[{\bf 2}.] Find the Newton cooling constant $h$.

\item[{\bf 3}.] Determine the time required for the chocolate to cool to $160^\circ F$.

\end{itemize}
{\bf References}. Edwards-Penney section 1.5.
Serway and Vuille, {\em College Physics 9/E}, Brooks-Cole (2011), ISBN-10: 0840062060. An answer check might use {\em The Coffee Cooling Problem}, a Wolfram Demonstration Project contributed by S.M. Binder, which can be found at \\
\href{http://demonstrations.wolfram.com/TheCoffeeCoolingProblem/}
{http://demonstrations.wolfram.com/TheCoffeeCoolingProblem/}.

{\bf Credits}. Created by Rebecca Terry, January 2014.

\vspace{3pc}

{\bf Chapters 1,2. Sample Problem 4}. Logistic growth $F(x)=rx(1-x/K)$ can be used to describe the annual natural growth of a fish stock. Symbol $x(t)$ is the stock biomass in number of fish at the start of month $t$. The intrinsic growth rate is symbol $r$. The environmental carrying capacity in stock biomass terms is symbol $K$.

\begin{itemize}
\item[{\bf 1}.] Assume a fish pond has carrying capacity $K=780500$ and that $80$\% of the the fish survive to maturity. We'll assume 6 months to maturity and $r=0.8$. Write in detail the no-harvesting model $x'(t)=F(x(t))$ and find the equilibrium values.

\item[{\bf 2}.] Assume constant harvesting $H$ to give the model $x'(t)=F(x(t))-H$. Use the quadratic formula from algebra to find the equilibrium points as a function of symbol $H \ge 0$.
    Then verify the following results.

     \begin{itemize}\normalsize\sf
     \item[] If $H=156100$, then there are two states: extinction for $x(0)<390250$
         and limiting population $390250$ otherwise.
         % Laham uses 389482 instead of K/2=390250

     \item[] If $H > 156100$, then the extinction state is the only possibility.

    \item[] If $H < 156100$, then there are two equilibria. The larger equilibrium population size is stable and the smaller is unstable. These numbers imply sustainable harvest for certain population sizes, but not all.
     \end{itemize}


 \item[{\bf 3}.] Assume a constant harvest rate $H$. Create two graphics of the population $x(t)$ over $36$ months. The first uses a harvesting size $H$ to show sustainable harvest. The second uses a different size $H$ to show non-sustainable harvest. Handwritten plots are expected, or a computer plot, if you know how.

\end{itemize}


{\bf References}. Edwards-Penney sections 2.1, 2.2.
Course documents:\newline
{\scriptsize\url{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/2250logistic.pdf}} \quad \BF{Logistic Equation},\newline
{\scriptsize\url{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/2250phaseline.pdf}}\quad \BF{Stability},\newline
{\scriptsize\url{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/fishFarming2014.pdf}}  \quad\BF{Fish Farming}\newline
and a logistic investigation in Malaysia by \BF{M.F. Laham 2012}:\newline
\url{http://www.ukm.my/jsm/pdf_files/SM-PDF-41-2-2012/04%20Mohamed%20Faris.pdf}
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