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\begin{document}

\newpage
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\title{\ifnum\TeacherEdition=1 TEACHER VERSION %\footnote{DELETE THIS FOOTNOTE BEFORE SUBMISSION.  In the \TeX \, source code there is a toggle/option to produce a Teacher Version or a Student Version by ``carving'' out with  $\backslash$teacher\{\} that material which should only appear in the Teacher version. See the source code.}
 \else STUDENT VERSION \fi\\At what age do people get married?}



\markboth{Marriage Age}{Marriage Age}

\author{Tracy Weyand\\
Rose-Hulman Institute of Technology\\
Terre Haute IN 47803 USA\\
weyand@rose-hulman.edu
}

\makeStitlePDFLaTex


\begin{abstract}
Students will build and analyze a model of the fraction of people who are married (for the first time) by a certain age. This model comes from a paper by Hernes\cite{Hernes} and, in this project, is compared to another model used by Coale\cite{Coale}. 

These models are first-order ordinary differential equations (which are also separable and autonomous). Students are asked to set up and solve these differential equations. They are then expected to analyze the behavior of these models, both mathematically (for example, by analyzing the stability of equilibrium points) and physically (determining whether the solution makes sense or not). Reasonable initial conditions are discussed. Finally, an alternatively scaled model is presented, and students are asked to relate this solution to the previous one.
\end{abstract}


\section*{SCENARIO DESCRIPTION}

There are many reasons someone may make the decision to or not to get married. Similarly there are many factors that cause people to get married at a wide variety of ages. Here's my story: I was fresh out of college when I got the phone call from my friend that I've had since preschool asking me to be in her wedding. I excitedly agreed. Fast forward a year: the wedding just happened and I'm telling my best friend from high school that she better wait a while until my bank account is replenished before she asks me to be in her wedding. It's only a month later when I get a phone call from her that begins ``Please hear me out before you hang up,"   and I find myself getting ready to be in another wedding.

Is there a correlation? Is that when my friends were at the ``common age" for (a first) marriage? Let's see if we can find an age at which people marry the most quickly; i.e. an age at which the fraction of people who are married increases the most rapidly.


Define the following functions:
\begin{itemize}
\item $p(t)$: the number of people in the United States who are $t$ years old
\item $M(t)$: the number of people in the United States who are $t$ years old and have been married
\item $m(t) := \dfrac{M(t)}{p(t)}$: the fraction of people of age $t$ in the United States who have been married
\end{itemize}
We will be considering differential equations in terms of $m(t)$, because we are looking for the age at which $m(t)$ is increasing the fastest.

\section*{The Peer Pressure Model}
We assume that a person's chances of getting married (in some small time interval $\Delta t$) is proportional to the fraction of people who are already married by that age; this can be considered peer pressure\cite{Bender} .

\begin{enumerate}
\item \label{first} We first assume that the rate of change of $m(t)$ is proportional to the number of interactions (the fraction of) people married (for a first time) by age $t$ have with (the fraction of) people of age $t$ who have not yet married. 
\begin{enumerate}
\item \label{start} Using our notation above, what fraction of people are married by age $t$? What fraction are unmarried at age $t$? How many interactions are there between these groups of people?
\item  \label{main} Write and solve a differential equation modeling the situation with these assumptions.
\item \label{long-term} What is the long-term behavior of this solution? Does this make sense for the scenario?\\
\emph{Hint: Based on the situation, is the proportionality constant positive or negative? Why?}
\item \label{equilibrium} This differential equation is autonomous.
\begin{enumerate}
\item Find the equilibrium point(s).
\item Analyze the stability of the equilibrium point(s).
\end{enumerate}

\item \label{Maple} Your general solution to Problem \ref{main} should have two unknown constants in it; the proportionality constant and another constant that arose when solving the first-order differential equation. Choose three different combinations of values for these constants. Plot the solution with these values plugged in. 
\begin{enumerate}
\item Do these graphs support your answers from Problem \ref{long-term} and \ref{equilibrium}?
\item What do you notice about solution at $t=0$? Does this make sense?
\end{enumerate}

\item\label{criticism} One criticism of this model is that it assumes people experience that peer pressure to the same degree at every age. Come up with at least two other criticisms.
\end{enumerate}
\end{enumerate}



\noindent Let's try to compensate for the criticism mentioned in Problem \ref{criticism} (that people probably don't experience peer pressure to the same extent at every age). We can do this by making a proportionality function $f(t)$ (rather than a proportionality constant), and leaving everything else unchanged. 


\begin{enumerate}
\setcounter{enumi}{1}
\item \label{second}  Write and solve a differential equation (as much as possible) modeling the situation with these assumptions.
\end{enumerate}

\noindent The first proportionality function used by G. Hernes \cite{Hernes} in an attempt to match numerical data was
\[f(t) = Ab^t\]
where $b<1$ and $A$ is the the initial fraction of people married.

\begin{enumerate}
\setcounter{enumi}{2}
\item \label{conditions} How should the proportionality function $f(t)$ behave for the model to make sense? Therefore, why is this a reasonable guess? 

\item \label{third} Solve your differential equation from Problem \ref{second} with this proportionality function.

\item  \label{conditions 2} Recall that $A$ is the initial fraction of people married. Why doesn't this make sense, and what problem does it cause with your solution from Problem \ref{third}? %How could this problem be avoided?

\end{enumerate} 

\noindent  To work around this initial condition problem, A. J. Coale \cite{Coale} made the following linear transformation of the age axis:
\[x:=at-c.\]

\noindent The other axis was scaled to the number of people whose first marriage was at time $x$ (i.e. our function $M(x)$).
Upon doing this, it was found that the numerical data closely fits the curve
\[M(x) = e^{-e^x}.\]

\begin{enumerate}
\setcounter{enumi}{5}
\item \label{comparison} Does this make sense given your solution in Problem 4? Explain and justify your reasoning.
\item \label{question} Time to go back to the original question. Was this to be expected? Is there a ``peak" age at which people get married?
\end{enumerate}


\begin{thebibliography}{98}




\bibitem{Bender} Bender, E. 1978. {\it Introduction to Mathematical Modeling\/}. New York: John Wiley \& Sons.


\end{thebibliography}



\teacher{
\section*{NOTES FOR TEACHERS}

I would expect this project to take students 90 to 120 minutes to complete. The mathematics (of setting up and solving the differential equations) will not take anywhere near that long, but many questions require thoughtful interpretations of the physical situation and solution forms; this is what will be the time consuming part for students. If you intend this to be in done in class, you can have students work on Problem 1 during one fifty minute class and the remainder the following class.

 This project can easily be shortened. For example, one could 
\begin{itemize}
\item provide the models; this will delete problem \ref{start} and shorten \ref{main} to just ``solve",
\item skip Problem \ref{second} and only consider the proportionality function given above Problem \ref{conditions}, and/or
\item skip Problem \ref{comparison} (comparison with a differently scaled solution), which is interesting but not crucial. 
\end{itemize}
\noindent Many of the other questions could also be skipped depending on what you have/have not covered in class. It is also possible to introduce this problem at the very beginning of the course as a simple model, and then subsequent questions could be asked, either in class as examples or in homework, as each relevant mathematics topic is introduced. The prerequisites for each question are outlined below.

It could also be used as an out-of-class group project. This modeling scenario would be a good way to introduce the idea of equilibrium points and stability; see $^\dagger$ below. The references contain pages of numerical data that could be used if desired.\\

\underline{\textbf{Prerequisites:}}
\begin{itemize}
\item \ref{start}:* Basic modeling 
\item \ref{main}:* Basic modeling (including the idea of ``proportional to") and solving separable first-order differential equations
\item \ref{long-term}: Calculus (taking a limit)
\item \ref{equilibrium}$^\dagger$ Autonomous equations (including equilibrium points and stability)
\item \ref{Maple}: Initial conditions and graphing solutions. Technology (a graphing calculator, Mathematica, Maple, etc.) is required only for Problem \ref{Maple}, and can easily be cut if necessary.
\item \ref{criticism}: None
\item \ref{second}: Solve a first-order separable differential equation
\item \ref{conditions}: None
\item \ref{third}: Solve a first-order separable differential equation
\item \ref{conditions 2}: Initial conditions
\item \ref{comparison}: None (besides knowing properties of logarithmic and exponential functions)
\item \ref{question}: Calculus (maximum rate of change)
\end{itemize}


\noindent As an alternative, one could give the model:
\[m'(t) = c\cdot m(t)(1-m(t)).\]

\noindent This question could easily be reworded to introduce the concept of an equilibrium point/solution and stability as follows:

\ref{equilibrium} Notice that we missed a solution!\
\begin{adjustwidth}{1cm}{}
(i) For what constant value(s) $c$ is $m(t) = c$ a solution? These are called \emph{equilibrium solutions}. \\
(ii) Relate these solutions to Problem \ref{long-term}. What do you notice? Due to this, the larger equilibrium solution is referred to as \emph{stable} whereas the smaller one is called \emph{unstable}.\\
\end{adjustwidth}




\underline{\textbf{Expected Solutions:}}
\begin{enumerate}
\item
\begin{enumerate}
\item $m(t), 1-m(t), m(t)(1-m(t))$
\item The model described can be written as
\[m'(t) = cm(t)(1-m(t)\]
\[\mbox{Explicit Solution: } m(t) = \dfrac{Ke^{ct}}{1+Ke^{ct}}\]
\item Technically, this depends on the proportionality constant $c$. However, from the model, the number of people getting married for the first time should be increasing (especially when $m(t)$ is small), so we can assume that $c>0$. With this assumption,
\[\lim_{t\to\infty} m(t) = \lim_{t\to\infty} \dfrac{Ke^{ct}}{1+Ke^{ct}} = 1.\]
This would mean that everyone will eventually be married, which isn't reasonable; it should be less than one.
\item
\begin{enumerate}
\item The equilibrium solutions are the (constant function) solutions to
\[m'(t) = cm(t)(1-m(t)) = 0,\]
so $m(t)=0 $ or $m(t)=1$ are the equilibrium solutions (meaning 0 and 1 are the equilibrium points).
\item We note that $m'(t) > 0$ if $0<m<1$. This means that $m(t) = 1$ is a stable point and $m(t) =0$ is an unstable point.
\end{enumerate} 
\item 
\begin{enumerate}
\item This graph is the solution with $c=.2$ and $K = 1$. I plotted it from $t=0$ to 100 because that (more than) covers (most) people's lifespan. The graph Figure~\ref{fig:graph} does support the claim that $\lim_{t\to\infty}m(t) = 1$, $m=1$ is a stable point, and $m=0$ is an unstable point. Actually, $\lim_{t\to-\infty} m(t) =0$, but people can't live negative years. 
\item This does not make sense at $t=0$, as it starts at $m(0)= 0.5$, which would mean half of the people are married at birth. 
\end{enumerate} 





\item Based on the stability, this model implies that everyone will eventually be married, which obviously isn't the case (though taking $t\to\infty$ doesn't really make sense either because people don't live forever). As mentioned in \ref{Maple} ii, the solution does not make sense at $t=0$. According to the graph, approximately 90\% of people are married by age 10, which also does not make sense. The solution ``form" seems reasonable, but it increases too much too soon. There are \underline{many} other possible criticisms here.

\begin{figure}[h!]
\begin{center}\label{fig:graph}
\includegraphics[scale=.5]{1-136-SampleGraph}
\caption{Graph in support of claim.}
\end{center}
\vspace*{-.5 true cm}
\end{figure}

\end{enumerate}
\item 
\[m'(t) = f(t)m(t)(1-m(t))\]
\[\mbox{Implicit Solution: } \dfrac{m(t)}{1-m(t)} = Ke^{\int f(t)\, dt}\]
\[\mbox{Explicit Solution: } m(t) = \dfrac{Ke^{\int f(t)\, dt}}{1+Ke^{\int f(t)\, dt}}\]
%\begin{align*}
%m'(t) = f(t)m(t)(1-m(t))\\
%\dfrac{dm}{m(1-m)} = f(t)dt\\
%\int \dfrac{dm}{m(1-m)} = \int \left(\dfrac{1}{m} + \dfrac{1}{1-m}\right)\, dt = \ln(m(t))-\ln(1-m(t)) %= \ln\left(\dfrac{m(t)}{1-m(t)}\right) = \int f(t)\, dt
%\end{align*}
\item The function should be decreasing over time. People who never want to get married never will. Therefore, as $t$ increases, the number of people (left) to succumb to so-called ``peer pressure" (i.e. those considering marriage but who have never married) will be decreasing. This function makes sense because, since $b<1$, this function is decreasing with time.
\item 
\[\int f(t)\, dt = \dfrac{Ab^t}{\ln(b)}\]
\[m(t) = \dfrac{Ke^{Ab^t/\ln(b)}}{1+Ke^{Ab^t/\ln(b)}}\]
\item This does not make sense because the initial fraction of people married is 0, and if $A=0$, then the solution 
\[m(t)=\dfrac{K}{1+K},\]
is a constant. %This problem could be avoided by considering an initial-condition not a zero, but rather at, say 18, the legal minimum age to marry in the United States (without parental consent).
\item Yes, this does make sense. Recall that
\[m(t) = \dfrac{Ke^{Ab^t/\ln(b)}}{1+Ke^{Ab^t/\ln(b)}} = \dfrac{M(t)}{p(t)}\]
so assume that $M(t)$ is proportional to $Ke^{Ab^t/\ln(b)}$. Then $M(x)$ behaves like
\[Ke^{Ab^{x+c}/(a\ln(b))} = Ce^{\alpha b^x/\ln(b)}.\]
Let us look at $b^x$. We want to switch the base of this exponential function to $e$. If
\[b^x = y = e^{kx}\]
then
\[x\ln(b) = \ln(y) = kx \to k = \ln(b) < 0\]
since $b<1$. Therefore $M(x)$ does behave like
\[e^{-e^x},\]
modulo some constants.
\item The derivative of $m(t)$ (Problem \ref{third}) or $M(t)$ (Problem \ref{comparison}) is never zero. It is clearly larger in some places than others, but as there are no critical points, there is no point at which the rate is increasing the fastest.
\end{enumerate}




\begin{thebibliography}{99}


\bibitem{Bender} Bender, E. 1978. {\it Introduction to Mathematical Modeling\/}. New York: John Wiley \& Sons.

\bibitem{Hernes} Hernes, G. 1972. The process of entry into first marriage.  {\it American Sociological Review\/}. 37: 173-182.



\bibitem{Coale} Coale, A. 1971. Age Patterns of Marriage. {\it Population Studies\/}.  25(2): 193-214. 

\end{thebibliography}
}



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