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\markright{Solution Keys\hfill Page }


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\n {\bf \textcolor{red}{Note:}} \textcolor{blue}{Solutions to
Sample Test 3}

%\begin{enumerate}
\n {\bf \textcolor{red}{[\#1]}} (a) Let $f(y)=-y^2(y-3)$.
Equilibrium solutions are zero points of $f(y)=0$. They are
$y=0,y=3$. (b) For any initial point between the to equilibrium
points $y=0, y=3$, $dy/dt>0$. So the graph of the solution
increases, approaching $y=3$ asymptotically. (c) By Euler's
method, $y_{n+1}=y_n+\Delta x f(y_n)=y_n+0.5(-y_n^2(y_n-3))$

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\begin{table}[h]%\label{table1}
%\caption{Trophic Characteristic in Reproductive Rate Ratios}
%\hskip 2in
\begin{tabular}{c|c|c|c}\hline
$n$ & $t_n$ & $y_n$ & $\Delta xf(y_n)$ \\
\hline
0 & 0 & 1 & $0.5(-1^2(1-3))=1$ \\
1 & 0.5 & $1+1=2$ & $0.5(-2^2(2-3))=2$ \\
2 & 1 & $2+2=4$ & -- \\
\hline
\end{tabular}
\end{table}
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\n Hence $y(1)~4$.


\n {\bf \textcolor{red}{[\#2]}} (a) $\begin{cases}
\displaystyle{\frac{dy}{dt}=5\text{g/L}\times 100\text{L/min}-
\frac{y(t)}{1000}\text{g/L}\times 100\text{L/min}=500-\frac{y(t)}{10}}& \text{} \\
y(0)=0 & \text{}
\end{cases}$

(b) By the method of separation of variables,
\begin{equation*}
\begin{split}
&\frac{dy}{dt}=\frac{5000-y(t)}{10}\\
&\frac{dy}{5000-y}=\frac{1}{10}dt\\
&\int\frac{dy}{5000-y}=\int\frac{1}{10}dt=\frac{t}{10}+C_1\\
&-\ln|5000-y|=\frac{t}{10}+C_1\\
&|5000-y|=e^{-\frac{t}{10}-C_1}=C_2e^{-\frac{t}{10}}\\
&5000-y=\pm C_2e^{-\frac{t}{10}}=C_3e^{-\frac{t}{10}}\\
&y(t)=5000-C_3e^{-\frac{t}{10}}=5000+Ce^{-\frac{t}{10}}\\
\end{split}
\end{equation*}
Use the initial condition $y(0)=0$, we have
\[
0=y(0)=5000+Ce^{-\frac{0}{10}}=5000+C\Longrightarrow C=-5000,
\]
and
\[
y(t)=5000(1-e^{-\frac{t}{10}}).
\]
(c) The time $t$ that happens satisfies
\[
3000=y(t)=5000(1-e^{-\frac{t}{10}}).
\]
Solving for $t$ to have: $3/5=1-e^{-\frac{t}{10}}$,
$e^{-\frac{t}{10}}=1-3/5=2/5$, and $t=-10\ln 2/5~9.2$min.

\n {\bf \textcolor{red}{[\#3]}}
\begin{equation*}
\begin{split}
&\frac{dy}{dx}=\frac{2xy}{x^2+1}\\
&\frac{dy}{y}=\frac{2x}{x^2+1}dx\\
&\int\frac{dy}{y}=\int\frac{2x}{x^2+1}dx\\
&\ln|y|=\ln(x^2+1)+C_1\\
&|y|=e^{\ln(x^2+1)+C_1}=C_2(x^2+1)\\
&y=\pm C_2(x^2+1)=C(x^2+1)\\
\end{split}
\end{equation*}
With $3=y(0)=C(0^2+1)=C$ we have $y(x)=3(x^2+1)$.


\n {\bf \textcolor{red}{[\#4]}} $A=(-1,1,2),B=(1,-1,0),C=(2,1,1)$.
(a) $\vec{AB}=(2,-2,-2), \vec{BC}=(1,2,1)$.
(b)$\vec{AB}\cdot\vec{BC}=2(1)-2(2)-2(1)=-4$.
$\vec{AB}\times\vec{BC}=\left|\begin{array}{rrr}
\vec i & \vec j & \vec k \\ 2&-2&-2\\
1&2&2 \end{array}\right|=2\vec i-4\vec j+6\vec k=(2,-4,6)$. (d)
No. Vectors parallel to $(-1,1,0)$ have the form
$k(-1,1,0)=(-k,k,0)$. Since $\vec{AB}\cdot
(-k,k,0)=2(-k)-2(k)-2(0)=-4k\ne 0$ unless $k=0$. (e)
$(1/2)||\vec{AB}\times\vec{BC}||=\sqrt{14}$. (f) $\cos\theta
=\frac{\vec{AB}\cdot\vec{BC}}{||\vec{AB}||\cdot||\vec{BC}||}=-\frac{\sqrt
2 }{3}$ and $\theta= \arccos -\frac{\sqrt 2 }{3}=2.06$ radians.
(g) $(x-(-1),y-1,z-2)\times (\vec{AB}\times\vec{BC})=0$ which
simplifies to $2(x+1)-4(y-1)+6(z-2)=0 \Longrightarrow x-2y+3z=3$.
(h)
$Proj_{\vec{BC}}\vec{AB}=\frac{\vec{AB}\cdot\vec{BC}}{||\vec{BC}
||^2}\vec{BC}=-\frac{2}{3}\langle 1,2,1\rangle$.

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