\documentclass{amsart}
\usepackage{graphicx}

%THIS IS A SAMPLE LAB. NOTE THE WAY FIGURES ARE PLACED AND REFERENCED.
%  ALSO NOTE HOW EQUATIONS ARE REFERENCED.
% TYPOS WERE INTRODUCED TO CHECK ISPELL

\title{ The Bicycle Race (A Sample Write Up) }

\author{ Albert Schueller }

\begin{document}

\maketitle


\section{Introduction}

We will consider the situation of a cyclist passing a refreshment station
in a bicycle race and the relative positions of the cyclist and her chasing
support car.  The cylcist rides at a constant velocity with her distance in
meters from the refreshment station given by $kt$ where $t$ is time in
seconds after passing the station.  The pursuing support car's distance in
meters from the station is given by $\frac{1}{3}(10t^2-t^3)$.  There are
several aspects of the situation we will analyze.

We are given the position of the cyclist relative to the location of the
refreshment station as a function of time.  We denote by $s_c(t)$ the
position of the cyclist at time $t$.  Denote by $s_a(t)$ the
position function of the auto.  From the discussion in \S2.3 of
Calculus, Stewart 3rd ed, we know that the velocity of the cyclist is
$s'_c(t)$ and the velocity of the auto is $s'_a(t)$.  In
this article we compute the cyclist's velocity and the time at which the
chase car catches the cyclist.  We then explure the effect
of different cyclist velocities on meeting times.  Following this, we
attempt to illustrate the various scenarios graphically.  Finally, in
we connect a theortical result about cubic polynomials with
the situation of the cyclist and the chase car.

\section{Velocity and meeting time}

In this section we determine the speed of the the cyclist and how long it takes
the chase car to catch her.  We have the additional information that when
the chase car catches the cyclist their velocities are the same.  This
provides us with two algebraic equations.  When the chase car catches the
cyclist their positions are the same, thus $$s_c(t)=s_a(t).$$  Furthermore,
when their positions coincide, by design, their velocities are the same,
thus $$s'_c(t)=s'_a(t).$$  Together, we have two equations in the two
unknowns $k$ and $t$.
%
\begin{eqnarray}
kt & = & \frac{1}{3}(10t^2-t^3) \label{posn_eqn} \\
k & = & \frac{1}{3}(20t-3t^2) \label{deriv_eqn} 
\end{eqnarray}
%
Using the equation (\ref{deriv_eqn}), we eliminate $k$ from the equation
(\ref{posn_eqn}) and solve for $t$.  We find two solutions, $t=0$s and
$t=5$s.  The first solution is consistent with the fact that their
positions coincided at the refreshment station.  The second time, 5
seconds, is when the support car catches the cyclist.

The velocity of the cyclist as a function of time is given by $s'_c(t) =
k$, where $k$ is a constant.  Determining $k$ will give the velocity of the
cyclist.  We know that at $t=5$s equation (\ref{deriv_eqn}) holds.  We can
determine $k$ by substituting $t=5$s into equation (\ref{deriv_eqn}).
Doing so we find $k=25/3$ m/s.  Thus the cyclist is traveling at a rate of
$25/3$ m/s.

\section{Different velocities}

We will now consider what happens if $k$ is allowed to vary.  We will
derive a formula for $t$ in terms of $k$ for the times when the position of
the cyclist and the car coincide.

By setting the two position functions equal to each other and solving for
$t$, we can determine the length of time it takes for the pursuing car to
catch the cyclist.
%
\begin{eqnarray*}
s_c(t) & = & s_a(t) \\
kt & = & \frac{1}{3}(10t^2-t^3) \\
3kt & = & 10t^2-t^3 \\
3k & = & 10t-t^2,~t\neq 0 \\
t^2 - 10t + 3k & = & 0,~t\neq 0 \\
\end{eqnarray*}

From this calculation we see that $t=0$ is one solution, which is
consistent with the fact that their positions are the same when the cyclist
passes the refreshment station.  We see that there are potentially two more
times where the car and the cyclist have the same position.  To determine
these times, we must apply the quadratic formula, 
$$
t=\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}
$$ 
where $a=1$, $b=-10$ and $c=3k$.  Doing so we have
$$
t=\frac{10 \pm \sqrt{100-12k}}{2}
$$  In order for the solutions to be
real--valued, we must have $k \leq 25/3$.  Assuming $k>0$, we see that the
expression for $t$ is always positive and yields two distinct times.  From
this we may conclude that if she rides faster than $25/3$ m/s, the support
car will not catch her.  If she rides slower than $25/3$ m/s, the support
car will meet her twice.

\section{Graphical illustration}

In this section we provide graphs to illustrate what happens in the first
case where the car and the cyclist meet and have the same velocity and in
the second case where the velocity may change.  Figure \ref{CycleAuto}, shows
the position functions of the cyclist and the car on the same axes with
$k=25/3$ m/s--the optimal speed.  We see that after $t=0$s the two
functions intersect only once at $t=5$s.  

\begin{figure}[h]
\centering
\includegraphics[height=2.5in]{cycle1} $\quad$
\includegraphics[height=2.5in]{cycle2}
\includegraphics[height=2.5in]{cycle3}
\caption{A plot of $s_a(t)=\frac{1}{3}(10t^2-t^3)$ (the position
of the auto, shown as the solid curve) and $s_a(t)=kt$
(the position of the cyclist, shown as a dotted line).  The figures
shown are for $k=25/3, k=26/3$ and $k=20/3$, for the upper left,
upper right and lower graphs respectively.}
\label{CycleAuto}
\end{figure}


The second graph in Figure \ref{CycleAuto} shows the cyclist's velocity 
slightly
faster at $k=26/3$ m/s.  In this case we see that the two plots fail to
intersect after $t=0$s.  The third graph,
Figure \ref{CycleAuto} shows the cyclist's velocity slightly
slower at $k=20/3$ m/s.  In this case we see that the two plots intersect
twice after $t=0$s.  

In the situation depicted in the second graph, it would be difficult for
the support car to give refreshment since there would still be some
distance between the car the cyclist.

In the situation depicted in the third graph, it would be difficult for
the car to give refreshment because the velocities of the cyclist and the
car would be too different.

\section{Cubic polynomials}

We will now show that if a cubic polynomial, $P(t)$, has a repeated real
root at $t=a$, then $P'(a)=0$.  We will then use this result to interpret
the relationship between the cyclist and the auto.  Let $t=a$ be a double
root.  If $b$ is the remaining root of $P(t)$, then we may write $$P(t) =
K(t-a)^2(t-b)$$ where $K$ is the coefficient on the $t^3$ term.  Thus,
$$P'(t) = K(t-a)^2 + 2K(t-a)(t-b).$$  Hence, $P'(a) = 0$.

This result is relevant to the first situation in which the car and the
cyclist are traveling at the same speed when they meet.  The cubic
polynomial in question is $P(t) = s_c(t) - s_a(t)$ and represents the
difference in the positions of the cyclist and the car.  Using the {\bf 
factor()} call in maple with $k=25/3$ m/s we have $$P(t)
=\frac{1}{3}t(t-5)^2.$$  We see that $t=5$ is a double root of $P$.  Notice
further that $P'$ gives the difference in the velocities of the cyclist and
the auto.  The above double root result states that $P'(5)=0$ which is
consistent with the fact that at the time that they meet the cyclist and
the auto are traveling at the same speed.

In the later examples where $k$ is allowed to vary, we find that $P$ has
not double roots and the delicate condition of matching position and speed
is destroyed.  




\end{document}
