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\documentclass[12pt,thmsa]{article}
\usepackage{amssymb}

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\usepackage{sw20jart}

%TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart}

%TCIDATA{<META NAME="ViewSettings" CONTENT="17">}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{Created=Sat Nov 16 20:56:01 1996}
%TCIDATA{LastRevised=Thu Feb 13 11:19:31 1997}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=Exam.cst}

\input{tcilatex}
\begin{document}


\section{Quiz}

\subsection{Text}

\section{Calculus, 2nd Semester}

\subsection{Growth Problems and Separable Differential Equations}

Your Name: 
%TCIMACRO{\HTML{<Input TYPE="Text" NAME="STUDENTNAME" VALUE=" " SIZE=0>}}
%BeginExpansion
%%
%EndExpansion

\vspace{1pt}

\subsubsection{Setup}

Choice Space: 2

Title:Growth Problems and Separable Differential Equations

Choices: No Break, Radio buttons, Permute

\section{Question}

\subsection{Setup}

$T_{1}:=\limfunc{rand}(150,1000)$

$T_{2}:=\func{rand}(40,\frac{T_{1}}{3})$

$T_{half}:=\func{rand}(T_{2}+1,T_{1}-1)$

Condition:\ $(T_{1}-T_{2}>2)$

\subsection{Statement}

A small object had temperature $%
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluate}}
%BeginExpansion
T_{1}%
%EndExpansion
\unit{\UNICODE{0x2109}}$ when it was dropped into a large pool of water at
temperature $%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
\unit{\UNICODE{0x2109}}$. Five minutes after it was put into the water, the
temperature of the object was $%
%TCIMACRO{\FORMULA{T_{half}}{T_{half}}{evaluate}}
%BeginExpansion
T_{half}%
%EndExpansion
$ $\unit{\UNICODE{0x2109}}$ degrees. Newton's law of cooling says that the
rate of change of temperature of the object is directly proportional to the
difference between it's own temperature and the temperature of the
surroundings.

What was the temperature of the object 10 minutes after it was put in the
water?\bigskip

\subsection{Choices}

\begin{itemize}
\item  $%
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{\left( T_{1}-T_{2}\right) \frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}+T_{2}}{evaluate}}
%BeginExpansion
\left( T_{1}-T_{2}\right) \frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}+T_{2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{\left( T_{1}-1.0T_{2}\right) \frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}+T_{2}}{evaluatenum}}
%BeginExpansion
\left( T_{1}-1.0T_{2}\right) \frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}+T_{2}%
%EndExpansion
\unit{\UNICODE{0x2109}}\correctchoice{}$

\item  $%
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }}{\left( T_{1}-T_{2}\right) \frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}}{evaluate}}
%BeginExpansion
\left( T_{1}-T_{2}\right) \frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }}{\left( T_{1}-1.0T_{2}\right) \frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}}{evaluatenum}}
%BeginExpansion
\left( T_{1}-1.0T_{2}\right) \frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}%
%EndExpansion
\unit{\UNICODE{0x2109}}$

\item  $%
%TCIMACRO{
%\FORMULA{e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{\frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}+T_{2}}{evaluate}}
%BeginExpansion
\frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}+T_{2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{\frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}+T_{2}}{evaluatenum}}
%BeginExpansion
\frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}+T_{2}%
%EndExpansion
\unit{\UNICODE{0x2109}}$

\item  $%
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{-\left( T_{1}-T_{2}\right) \frac{T_{half}-T_{2}}{-T_{1}+T_{2}}+T_{2}}{evaluate}}
%BeginExpansion
-\left( T_{1}-T_{2}\right) \frac{T_{half}-T_{2}}{-T_{1}+T_{2}}+T_{2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{-1.0\left( T_{1}-1.0T_{2}\right) \frac{T_{half}-1.0T_{2}}{-1.0T_{1}+T_{2}}+T_{2}}{evaluatenum}}
%BeginExpansion
-1.0\left( T_{1}-1.0T_{2}\right) \frac{T_{half}-1.0T_{2}}{-1.0T_{1}+T_{2}}+T_{2}%
%EndExpansion
\unit{\UNICODE{0x2109}}$
\end{itemize}

\subsection{Answer}

$\left( 
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluatenum}}
%BeginExpansion
T_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluatenum}}
%BeginExpansion
T_{2}%
%EndExpansion
\right) \left( -\frac{%
%TCIMACRO{\FORMULA{T_{half}-T_{2}}{T_{half}-T_{2}}{evaluate} }
%BeginExpansion
T_{half}-T_{2}%
%EndExpansion
}{-%
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluate} }
%BeginExpansion
T_{1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate} }
%BeginExpansion
T_{2}%
%EndExpansion
}\right) ^{10}+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluatenum}}
%BeginExpansion
T_{2}%
%EndExpansion
=%
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{\left( T_{1}-T_{2}\right) \frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}+T_{2}}{evaluate}}
%BeginExpansion
\left( T_{1}-T_{2}\right) \frac{\left( T_{half}-T_{2}\right) ^{2}}{\left( -T_{1}+T_{2}\right) ^{2}}+T_{2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{(T_{1}-T_{2})e^{\frac{10}{5}\ln \left( -\frac{T_{half}-T_{2}}{-T_{1}+T_{2}}\right) }+T_{2}}{\left( T_{1}-1.0T_{2}\right) \frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}+T_{2}}{evaluatenum}}
%BeginExpansion
\left( T_{1}-1.0T_{2}\right) \frac{\left( T_{half}-1.0T_{2}\right) ^{2}}{\left( -1.0T_{1}+T_{2}\right) ^{2}}+T_{2}%
%EndExpansion
\unit{\UNICODE{0x2109}}$

\subsection{Solution}

\begin{enumerate}
\item  Start with $\frac{dT}{dt}=k(T-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
)$.

\item  Then $\frac{dT}{T-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate} }
%BeginExpansion
T_{2}%
%EndExpansion
}=kdt$ and $\int \frac{dT}{T-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate} }
%BeginExpansion
T_{2}%
%EndExpansion
}=\int kdt$.

\item  Integrate to get $\allowbreak \ln \left( T-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
\right) =kt+C$.

\item  Solve for $T=e^{kt+C}+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
$.

\item  Rewrite as $T=e^{C}e^{kt}+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
$.

\item  Clearly, $e^{C}=%
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluate}}
%BeginExpansion
T_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
$.

\item  Then you get $%
%TCIMACRO{\FORMULA{T_{half}}{T_{half}}{evaluate}}
%BeginExpansion
T_{half}%
%EndExpansion
=\left( 
%TCIMACRO{\FORMULA{T_{1}-T_{2}}{T_{1}-T_{2}}{evaluate}}
%BeginExpansion
T_{1}-T_{2}%
%EndExpansion
\right) e^{k5}+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate}}
%BeginExpansion
T_{2}%
%EndExpansion
$.

\item  Solve, $k=\frac{1}{5}\ln \left( -\frac{%
%TCIMACRO{\FORMULA{T_{half}-T_{2}}{T_{half}-T_{2}}{evaluate} }
%BeginExpansion
T_{half}-T_{2}%
%EndExpansion
}{-%
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluate} }
%BeginExpansion
T_{1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate} }
%BeginExpansion
T_{2}%
%EndExpansion
}\right) \bigskip \thickapprox 
%TCIMACRO{
%\FORMULA{\frac{1}{5}\ln \left( -\frac{\FORMULA{T_{half}-T_{2}}{T_{half}-T_{2}}{evaluate}}{-\FORMULA{T_{1}}{T_{1}}{evaluate}+\FORMULA{T_{2}}{T_{2}}{evaluate}}\right) }{.\,\allowbreak 2\ln \left( -1.0\frac{T_{half}-1.0T_{2}}{-1.0T_{1}+T_{2}}\right) }{evaluatenum}}
%BeginExpansion
.\,\allowbreak 2\ln \left( -1.0\frac{T_{half}-1.0T_{2}}{-1.0T_{1}+T_{2}}\right) %
%EndExpansion
$.

\item  Thus $T=\left( 
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluatenum}}
%BeginExpansion
T_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluatenum}}
%BeginExpansion
T_{2}%
%EndExpansion
\right) e^{\frac{1}{5}\ln \left( -\frac{%
%TCIMACRO{\FORMULA{T_{half}-T_{2}}{T_{half}-T_{2}}{evaluate} }
%BeginExpansion
T_{half}-T_{2}%
%EndExpansion
}{-%
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluate} }
%BeginExpansion
T_{1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate} }
%BeginExpansion
T_{2}%
%EndExpansion
}\right) t}+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluatenum}}
%BeginExpansion
T_{2}%
%EndExpansion
$

\item  i.e. $T=\left( 
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluatenum}}
%BeginExpansion
T_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluatenum}}
%BeginExpansion
T_{2}%
%EndExpansion
\right) \left( -\frac{%
%TCIMACRO{\FORMULA{T_{half}-T_{2}}{T_{half}-T_{2}}{evaluate} }
%BeginExpansion
T_{half}-T_{2}%
%EndExpansion
}{-%
%TCIMACRO{\FORMULA{T_{1}}{T_{1}}{evaluate} }
%BeginExpansion
T_{1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluate} }
%BeginExpansion
T_{2}%
%EndExpansion
}\right) ^{\frac{t}{5}}+%
%TCIMACRO{\FORMULA{T_{2}}{T_{2}}{evaluatenum}}
%BeginExpansion
T_{2}%
%EndExpansion
$
\end{enumerate}

\section{Question}

\subsection{Setup}

$m:=\limfunc{rand}(2,10)$

$n:=\func{rand}(2,50)$

$k:=\func{rand}(2,8)$

\subsection{Statement}

A bacteria colony is observed to increase from $%
%TCIMACRO{\FORMULA{n\cdot 100}{100n}{evaluate}}
%BeginExpansion
100n%
%EndExpansion
$ to $%
%TCIMACRO{\FORMULA{mn\cdot 100}{100mn}{evaluate}}
%BeginExpansion
100mn%
%EndExpansion
$ in $%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}
%BeginExpansion
k%
%EndExpansion
$ hours. Assuming exponential growth, after how many more hours will the
colony grow to $%
%TCIMACRO{\FORMULA{(k+m)n\cdot 100}{100\left( k+m\right) n}{evaluate}}
%BeginExpansion
100\left( k+m\right) n%
%EndExpansion
$?

\subsection{Choices}

\begin{itemize}
\item  $%
%TCIMACRO{
%\FORMULA{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) }{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) }{evaluate}}
%BeginExpansion
k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) %
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) }{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1.0\right) }{evaluatenum}}
%BeginExpansion
k\left( \frac{\ln \left( k+m\right) }{\ln m}-1.0\right) %
%EndExpansion
\unit{h}\correctchoice{}$

\item  $%
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( k+m\right) }{\ln m}}{k\frac{\ln \left( k+m\right) }{\ln m}}{evaluate}}
%BeginExpansion
k\frac{\ln \left( k+m\right) }{\ln m}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( k+m\right) }{\ln m}}{k\frac{\ln \left( k+m\right) }{\ln m}}{evaluatenum}}
%BeginExpansion
k\frac{\ln \left( k+m\right) }{\ln m}%
%EndExpansion
\unit{h}$

\item  $%
%TCIMACRO{
%\FORMULA{k\left( \frac{\ln \left( k\right) }{\ln m}-1\right) }{k\left( \frac{\ln k}{\ln m}-1\right) }{evaluate}}
%BeginExpansion
k\left( \frac{\ln k}{\ln m}-1\right) %
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\left( \frac{\ln \left( k\right) }{\ln m}-1\right) }{k\left( \frac{\ln k}{\ln m}-1.0\right) }{evaluatenum}}
%BeginExpansion
k\left( \frac{\ln k}{\ln m}-1.0\right) %
%EndExpansion
\unit{h}$

\item  $%
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( k+m\right) }{\ln m}-1}{k\frac{\ln \left( k+m\right) }{\ln m}-1}{evaluate}}
%BeginExpansion
k\frac{\ln \left( k+m\right) }{\ln m}-1%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( k+m\right) }{\ln m}-1}{k\frac{\ln \left( k+m\right) }{\ln m}-1.0}{evaluatenum}}
%BeginExpansion
k\frac{\ln \left( k+m\right) }{\ln m}-1.0%
%EndExpansion
\unit{h}$
\end{itemize}

\subsection{Answer}

$%
%TCIMACRO{
%\FORMULA{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) }{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) }{evaluate}}
%BeginExpansion
k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) %
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1\right) }{k\left( \frac{\ln \left( k+m\right) }{\ln m}-1.0\right) }{evaluatenum}}
%BeginExpansion
k\left( \frac{\ln \left( k+m\right) }{\ln m}-1.0\right) %
%EndExpansion
\unit{h}$

\subsection{Solution}

\begin{enumerate}
\item  Set up $P=Ce^{rt}$.

\item  From the initial condition, $%
%TCIMACRO{\FORMULA{100n}{100n}{evaluate}}
%BeginExpansion
100n%
%EndExpansion
=C.$

\item  Thus we have $P=%
%TCIMACRO{\FORMULA{100n}{100n}{evaluate}}
%BeginExpansion
100n%
%EndExpansion
e^{rt}$.

\item  Then, $%
%TCIMACRO{\FORMULA{100mn}{100mn}{evaluate}}
%BeginExpansion
100mn%
%EndExpansion
=%
%TCIMACRO{\FORMULA{100n}{100n}{evaluate}}
%BeginExpansion
100n%
%EndExpansion
e^{r%
%TCIMACRO{\FORMULA{k}{k}{evaluate} }
%BeginExpansion
k%
%EndExpansion
}$ gives $r=\frac{\ln \left( 
%TCIMACRO{\FORMULA{m}{m}{evaluate} }
%BeginExpansion
m%
%EndExpansion
\right) }{%
%TCIMACRO{\FORMULA{k}{k}{evaluate} }
%BeginExpansion
k%
%EndExpansion
}$.

\item  Solve $%
%TCIMACRO{\FORMULA{(k+m)n\cdot 100}{100\left( k+m\right) n}{evaluate}}
%BeginExpansion
100\left( k+m\right) n%
%EndExpansion
=%
%TCIMACRO{\FORMULA{100n}{100n}{evaluate}}
%BeginExpansion
100n%
%EndExpansion
e^{\frac{\ln \left( 
%TCIMACRO{\FORMULA{m}{m}{evaluate} }
%BeginExpansion
m%
%EndExpansion
\right) }{%
%TCIMACRO{\FORMULA{k}{k}{evaluate} }
%BeginExpansion
k%
%EndExpansion
}\cdot t}$ to get $t=%
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( k+m\right) }{\ln m}}{k\frac{\ln \left( k+m\right) }{\ln m}}{evaluate}}
%BeginExpansion
k\frac{\ln \left( k+m\right) }{\ln m}%
%EndExpansion
$.\bigskip
\end{enumerate}

\section{Question}

\subsection{Setup}

$m:=\limfunc{rand}(500,5000)$

$n:=\func{rand}(m,\frac{15}{8}m)$

$k:=\func{rand}(2,5)$

\subsection{Statement}

The number of cells in a culture increases exponentially. Today we have $%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}
%BeginExpansion
m%
%EndExpansion
$ cells, and in $%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}
%BeginExpansion
k%
%EndExpansion
$ days, the number will grow to $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
$. Find the doubling time.\bigskip

\subsection{Choices}

\begin{itemize}
\item  $%
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( \frac{n}{m}\right) }{\ln 2}}{k\frac{\ln \frac{n}{m}}{\ln 2}}{evaluate}}
%BeginExpansion
k\frac{\ln \frac{n}{m}}{\ln 2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( \frac{n}{m}\right) }{\ln 2}}{1.\,\allowbreak 442\,7k\ln \frac{n}{m}}{evaluatenum}}
%BeginExpansion
1.\,\allowbreak 442\,7k\ln \frac{n}{m}%
%EndExpansion
\unit{d}\correctchoice{}$

\item  $%
%TCIMACRO{
%\FORMULA{k\ln \left( \frac{n}{m}\right) }{k\ln \frac{n}{m}}{evaluate}}
%BeginExpansion
k\ln \frac{n}{m}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\ln \left( \frac{n}{m}\right) }{k\ln \frac{n}{m}}{evaluatenum}}
%BeginExpansion
k\ln \frac{n}{m}%
%EndExpansion
\unit{d}$

\item  $%
%TCIMACRO{\FORMULA{k\ln n}{k\ln n}{evaluate}}
%BeginExpansion
k\ln n%
%EndExpansion
\approx 
%TCIMACRO{\FORMULA{k\ln n}{k\ln n}{evaluatenum}}
%BeginExpansion
k\ln n%
%EndExpansion
\unit{d}$

\item  $%
%TCIMACRO{
%\FORMULA{k\ln 2\ln \left( \frac{n}{m}\right) }{k\ln 2\ln \frac{n}{m}}{evaluate}}
%BeginExpansion
k\ln 2\ln \frac{n}{m}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\ln 2\ln \left( \frac{n}{m}\right) }{.\,\allowbreak 693\,15k\ln \frac{n}{m}}{evaluatenum}}
%BeginExpansion
.\,\allowbreak 693\,15k\ln \frac{n}{m}%
%EndExpansion
\unit{d}$
\end{itemize}

\subsection{Answer}

$%
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( \frac{n}{m}\right) }{\ln 2}}{k\frac{\ln \frac{n}{m}}{\ln 2}}{evaluate}}
%BeginExpansion
k\frac{\ln \frac{n}{m}}{\ln 2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( \frac{n}{m}\right) }{\ln 2}}{1.\,\allowbreak 442\,7k\ln \frac{n}{m}}{evaluatenum}}
%BeginExpansion
1.\,\allowbreak 442\,7k\ln \frac{n}{m}%
%EndExpansion
\unit{d}$

\subsection{Solution}

\begin{enumerate}
\item  Set up $P=Ce^{rt}$.

\item  From the initial condition, $C=%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}
%BeginExpansion
m%
%EndExpansion
$

\item  Thus we have $P=%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}
%BeginExpansion
m%
%EndExpansion
e^{rt}$.

\item  Then, $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
=%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}
%BeginExpansion
m%
%EndExpansion
e^{r%
%TCIMACRO{\FORMULA{k}{k}{evaluate} }
%BeginExpansion
k%
%EndExpansion
}$ gives $r=\frac{\ln \left( 
%TCIMACRO{\FORMULA{\frac{n}{m}}{\frac{n}{m}}{evaluate} }
%BeginExpansion
\frac{n}{m}%
%EndExpansion
\right) }{%
%TCIMACRO{\FORMULA{k}{k}{evaluate} }
%BeginExpansion
k%
%EndExpansion
}$.

\item  Solve $%
%TCIMACRO{\FORMULA{2m}{2m}{evaluate}}
%BeginExpansion
2m%
%EndExpansion
=%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}
%BeginExpansion
m%
%EndExpansion
e^{\frac{\ln \left( 
%TCIMACRO{\FORMULA{\frac{n}{m}}{\frac{n}{m}}{evaluate} }
%BeginExpansion
\frac{n}{m}%
%EndExpansion
\right) }{%
%TCIMACRO{\FORMULA{k}{k}{evaluate} }
%BeginExpansion
k%
%EndExpansion
}\cdot t}$ to get $t=%
%TCIMACRO{
%\FORMULA{k\frac{\ln \left( \frac{n}{m}\right) }{\ln 2}}{k\frac{\ln \frac{n}{m}}{\ln 2}}{evaluate}}
%BeginExpansion
k\frac{\ln \frac{n}{m}}{\ln 2}%
%EndExpansion
$.\bigskip
\end{enumerate}

\section{Question}

\subsection{Setup}

$n:=\func{rand}(9)+1$

$b:=\func{rand}(2,99)$

\subsection{Statement}

$%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
$\% of a certain radioactive substance decays in $%
%TCIMACRO{\FORMULA{b}{b}{evaluate}}
%BeginExpansion
b%
%EndExpansion
$ days. What is the half life of the substance?\bigskip

\subsection{Choices}

\begin{itemize}
\item  $%
%TCIMACRO{
%\FORMULA{\frac{-b\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{-b\frac{\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{evaluate}}
%BeginExpansion
-b\frac{\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{\frac{-b\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{-.\,\allowbreak 693\,15\frac{b}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }}{evaluatenum}}
%BeginExpansion
-.\,\allowbreak 693\,15\frac{b}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }%
%EndExpansion
\unit{d}\correctchoice{}$

\item  $%
%TCIMACRO{
%\FORMULA{\frac{-b\ln 2}{\ln \left( 1-\frac{1}{100}\right) }}{-b\frac{\ln 2}{\ln \frac{99}{100}}}{evaluate}}
%BeginExpansion
-b\frac{\ln 2}{\ln \frac{99}{100}}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{\frac{-b\ln 2}{\ln \left( 1-\frac{1}{100}\right) }}{68.\,\allowbreak 968b}{evaluatenum}}
%BeginExpansion
68.\,\allowbreak 968b%
%EndExpansion
\unit{d}$

\item  $%
%TCIMACRO{
%\FORMULA{\frac{-b}{\ln \left( 1-\frac{1}{100}n\right) }}{-\frac{b}{\ln \left( 1-\frac{1}{100}n\right) }}{evaluate}}
%BeginExpansion
-\frac{b}{\ln \left( 1-\frac{1}{100}n\right) }%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{\frac{-b}{\ln \left( 1-\frac{1}{100}n\right) }}{-1.0\frac{b}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }}{evaluatenum}}
%BeginExpansion
-1.0\frac{b}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }%
%EndExpansion
\unit{d}$

\item  $%
%TCIMACRO{
%\FORMULA{\frac{-\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{-\frac{\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{evaluate}}
%BeginExpansion
-\frac{\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{\frac{-\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{-\frac{.\,\allowbreak 693\,15}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }}{evaluatenum}}
%BeginExpansion
-\frac{.\,\allowbreak 693\,15}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }%
%EndExpansion
\unit{d}$
\end{itemize}

\subsection{Answer}

$\frac{1}{%
%TCIMACRO{
%\FORMULA{-\frac{\ln \left( 1-\frac{1}{100}n\right) }{b\ln 2}}{-\frac{\ln \left( 1-\frac{1}{100}n\right) }{b\ln 2}}{evaluate} }
%BeginExpansion
-\frac{\ln \left( 1-\frac{1}{100}n\right) }{b\ln 2}%
%EndExpansion
}\approx 
%TCIMACRO{
%\FORMULA{-\frac{b\ln 2}{\ln \left( 1-\frac{1}{100}n\right) }}{-.\,\allowbreak 693\,15\frac{b}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }}{evaluatenum}}
%BeginExpansion
-.\,\allowbreak 693\,15\frac{b}{\ln \left( 1.0-.0\,\allowbreak 1n\right) }%
%EndExpansion
\unit{d}$

\subsection{Solution}

\begin{enumerate}
\item  Set up $P=\left( \frac{1}{2}\right) ^{rt}$.

\item  Then $\frac{100-%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }
%BeginExpansion
n%
%EndExpansion
}{100}=\left( \frac{1}{2}\right) ^{r%
%TCIMACRO{\FORMULA{b}{b}{evaluatenum} }
%BeginExpansion
b%
%EndExpansion
}$ so $r=-\frac{\ln \left( 1-\frac{1}{100}%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }
%BeginExpansion
n%
%EndExpansion
\right) }{%
%TCIMACRO{\FORMULA{b}{b}{evaluatenum} }
%BeginExpansion
b%
%EndExpansion
\ln 2}$.\bigskip
\end{enumerate}

\section{Question}

\subsection{Setup}

$n:=\func{rand}(10,47)$

\subsection{Statement}

Determine the age of an ancient tree that contains $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
$\% of the $^{14}C$ it contained while living. Use the fact that after a
living tree dies, the $^{14}C$ decays exponentially with half life 5750
years.\bigskip

\subsection{Choices}

\begin{itemize}
\item  $%
%TCIMACRO{
%\FORMULA{-5750\frac{\ln \frac{n}{100}}{\ln 2}}{-5750\frac{\ln \frac{1}{100}n}{\ln 2}}{evaluate}}
%BeginExpansion
-5750\frac{\ln \frac{1}{100}n}{\ln 2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{-5750\frac{\ln \frac{n}{100}}{\ln 2}}{-8295.\,\allowbreak 5\ln .0\,\allowbreak 1n}{evaluatenum}}
%BeginExpansion
-8295.\,\allowbreak 5\ln .0\,\allowbreak 1n%
%EndExpansion
\unit{y}\correctchoice{}$

\item  $%
%TCIMACRO{\FORMULA{n5750}{5750n}{evaluate}}
%BeginExpansion
5750n%
%EndExpansion
\unit{y}$

\item  $%
%TCIMACRO{
%\FORMULA{-5750\ln \frac{n}{100}}{-5750\ln \frac{1}{100}n}{evaluate}}
%BeginExpansion
-5750\ln \frac{1}{100}n%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{-5750\ln \frac{n}{100}}{-5750.0\ln .0\,\allowbreak 1n}{evaluatenum}}
%BeginExpansion
-5750.0\ln .0\,\allowbreak 1n%
%EndExpansion
\unit{y}$

\item  $%
%TCIMACRO{
%\FORMULA{5750\frac{\ln n}{\ln 2}}{5750\frac{\ln n}{\ln 2}}{evaluate}}
%BeginExpansion
5750\frac{\ln n}{\ln 2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{5750\frac{\ln n}{\ln 2}}{8295.\,\allowbreak 5\ln n}{evaluatenum}}
%BeginExpansion
8295.\,\allowbreak 5\ln n%
%EndExpansion
\unit{y}$
\end{itemize}

\subsection{Answer}

$%
%TCIMACRO{
%\FORMULA{-5750\frac{\ln \frac{n}{100}}{\ln 2}}{-5750\frac{\ln \frac{1}{100}n}{\ln 2}}{evaluate}}
%BeginExpansion
-5750\frac{\ln \frac{1}{100}n}{\ln 2}%
%EndExpansion
\approx 
%TCIMACRO{
%\FORMULA{-5750\frac{\ln \frac{n}{100}}{\ln 2}}{-8295.\,\allowbreak 5\ln .0\,\allowbreak 1n}{evaluatenum}}
%BeginExpansion
-8295.\,\allowbreak 5\ln .0\,\allowbreak 1n%
%EndExpansion
\unit{y}$

\subsection{Solution}

\begin{enumerate}
\item  Set up $P=\left( \frac{1}{2}\right) ^{\frac{t}{5750}}$.

\item  Get $%
%TCIMACRO{\FORMULA{\frac{n}{100}}{\frac{1}{100}n}{evaluate}}
%BeginExpansion
\frac{1}{100}n%
%EndExpansion
=\left( \frac{1}{2}\right) ^{\frac{t}{5750}}$.

\item  Solve for $t$.\bigskip
\end{enumerate}

\section{Question}

\subsection{Setup}

$n:=\func{rand}(10^{5},10^{6})$

$m:=\func{rand}(10,\frac{n}{50})$

$M:=\func{rand}(10m,\frac{n}{3})$

\subsection{Statement}

An epidemic is spreading through Townville, population $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
$. The rate at which people get infected (in people/day) is directly
proportional to the number of already infected people and inversely
proportional to the number of healthy people. The Center for Disease Control
gathered the following data

\[
\begin{tabular}{|l|l|l|}
\hline
day & 3 November & 17 November \\ \hline
\# sick & $%
%TCIMACRO{\FORMULA{m}{m}{evaluate} }
%BeginExpansion
m%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{M}{M}{evaluate} }
%BeginExpansion
M%
%EndExpansion
$ \\ \hline
\end{tabular}
\]

Set up a differential equation and integrate to find a equation relating the
population and time.

\subsection{Answer}

$-I+%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
\ln I=kt+C$. You can solve for the $k$ and $C$ using the collected data if
you wish.

\subsection{Solution}

\begin{enumerate}
\item  Set up $I$, the number of infected people.

\item  Statement says $\frac{dI}{dt}=k\frac{I}{%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }
%BeginExpansion
n%
%EndExpansion
-I}$.

\item  Separate $\frac{%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }
%BeginExpansion
n%
%EndExpansion
-I}{I}dI=kdt$.

\item  Integrate $\int \frac{%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }
%BeginExpansion
n%
%EndExpansion
-I}{I}dI=\int kdt$ to get $-I+%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}
%BeginExpansion
n%
%EndExpansion
\ln I=kt+C$.
\end{enumerate}

\subsection{Response}

TextArea(30,2)

\end{document}
