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{\large\bf Semester Project Part 1}.  Help is available by telephone, office visit or email, or visit
the Math Center in LCB to ask for assistance on the problems.

{\bf Problem 3}. Three applications for the Newton cooling equation
$y' = -h(y-y_1)$ are considered, where $h,y_1$ are constants.


\begin{itemize}
\item[\BF{(a)}] {\bf Cooling}. An apple initially $23$ degrees Celsius is placed in a refrigerator at $2$ degrees Celsius. The exponential model is the apple temperature $u(t)=2+21 e^{-ht}$. Display the differential equation and the initial condition.

\item[\BF{(b)}] {\bf Heating}. A beef roast initially $8$ degrees Celsius is placed in an oven at $190$ degrees Celsius. The exponential model is the roast temperature $u(t)=190 - 182 e^{-ht}$. Display the differential equation and the initial condition.

\item[\BF{(c)}] {\bf Fish Length}. K.\ L.\ von Bertalanffy  in 1934 modeled the growth of fish using the equation $\frac{dL}{dt} = h(L_\infty -L(t))$. The fish has mature length $L_\infty$ inches, length $L(t)$ while growing, $t$ is in months and $h$ is the growth rate. Given growth data of $L(0)=0$, $L(1)=5$, $L(2)=7$, find the mature length $L_\infty$, the growth rate $h$ and the months to grow to $95$\% of mature length.

\end{itemize}

{\bf References}. Edwards-Penney section 1.5.
Course notes on \BF{Newton's linear drag model}:\newline
\href{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/2250kinetics.pdf}
\newline
Course notes on \BF{Newton cooling}:\newline
\href{http://www.math.utah.edu/~gustafso/s2019/2280/lectureslides/linearapplications2-5.pdf}
\newline
Wikipedia biography of \BF{Ludwig von Bertalanffy}:\newline
\href{http://en.wikipedia.org/wiki/Ludwig_von_Bertalanffy}. 
\newline
Pisces Conservation Ltd {\BFGrowth Models}, especially Gompertz, logistic and von Bertalanffy.\newline
\href{http://www.pisces-conservation.com/growthhelp/index.html}
\newline
Serway and Vuille, {\em College Physics 9/E}, Brooks-Cole (2011), ISBN-10: 0840062060.\newline
The \BF{Coffee Cooling Problem}, a Wolfram Demonstration by S.M. Binder.\newline
\href{http://demonstrations.wolfram.com/TheCoffeeCoolingProblem/}

\vspace{1pc}

{\bf Problem 4}.
Logistic growth $F(x)=rx(1-x/M)$ 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 $M$.

\begin{itemize}
\item[{\bf (a)}.] Assume a pond has carrying capacity $M=780$ thousand fish.  If $92$\% of the the fish survive to maturity, then $r=0.92$. Display the no-harvesting model $x'(t)=F(x(t))$, using only symbols $x$ and $t$.

\item[{\bf (b)}.] Assume constant harvesting $H \ge 0$ to give the model $x'(t)=F(x(t))-H$. Use the college algebra quadratic formula to find the equilibrium points in terms of symbols $r,M,H$. Then verify facts {\bf A, B, C} from your answer.
     \begin{itemize}
     \item[{\bf b1}.] If $H=\frac{rM}{4}$, then there is one equilibrium point $x=\frac{M}{2}$ (a double real root).
     \item[{\bf b2}.] If $H>\frac{rM}{4}$, then there is no equilibrium point.
     \item[{\bf b3}.] If $0<H<\frac{rM}{4}$, then there are two equilibrium points.
     \end{itemize}

 \item[{\bf (c)}.] Replace symbols $r,M$ by $0.92$ and $780$. Create a short filmstrip of 5 hand-drawn phase diagrams for the
 equation $x'(t)=F(x(t))-H$ using the successive harvest values
 $$H=0,\frac{1}{4}\left(\frac{rM}{4}\right),\frac{1}{2}\left(\frac{rM}{4}\right), \frac{1}{1}\left(\frac{rM}{4}\right), \frac{11}{10}\left(\frac{rM}{4}\right).$$
  Each phase diagram shows the equilibria and at least 5 threaded solutions, with labels for funnel, spout and node. The graph window is
  $t=0$ to $36$ months and $x=0$ to $2M$.

\item[{\bf (d)}.] Justify a guess for the {\bf maximum sustainable harvest}, based on your 5 diagrams. This is an approximate value for the largest catch $H$ that can be taken over 36 months.
\end{itemize}


{\bf References}. Edwards-Penney sections 2.1, 2.2.
Course document on the \BF{Logistic Equation}:\newline
\href{http://www.math.utah.edu/~gustafso/s2015/2280/2250logistic.pdf}
\newline
Course document on \BF{Stability}:\newline
\href{http://www.math.utah.edu/~gustafso/s2015/2280/2250phaseline.pdf}
\newline
Course document on \BF{Fish Farming}:\newline
\href{http://www.math.utah.edu/~gustafso/s2015/2280/fishFarming2014.pdf}. 
\newline
A logistic investigation in Malaysia by \BF{M.F. Laham 2012}:\newline
\href{http://www.ukm.my/jsm/pdf_files/SM-PDF-41-2-2012/04%20Mohamed%20Faris.pdf}


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r:=0.92:M:=780:q:=r*M/4;H:=1*q/2; # H=q is a bifurcation point
F:=P->r*(1-P/M)*P;a:=M/2;b:=M/2;
if (H<=q) then EQUIL:=[solve(F(P)-H=0,P)];a:=EQUIL[1]:b:=EQUIL[2]: fi;
de:=diff(P(t),t)=F(P(t))-H;
DEtools[DEplot](de,P(t),t=0..36,P=0..2*M,
[[0,M/8],[0,M/4],[P(0)=M/2],[0,M],[0,3*M/2],[0,2*M],[P(0)=a],[P(0)=b] ]);
unassign('P','M','r','H');F:=P->r*(1-P/M)*P;expand(F(x)-H);
solve(r^2-4*(-r/M)*(-H)=0,H);
solve(F(P)=H,P);


















