\section{Problem 4}
\subsection{Statement}
The most critical stage of a 100m race is as the athlete accelerates
out of the blocks. The athlete will be able to maintain the speed they
manage to reach in this phase. Below are the values of an athlete's
acceleration in the first 0.8 seconds of a practice run. What is the
athlete's speed 0.8 seconds into the run?
\begin{center}
\begin{tabular}{|c|c|}
\hline
\textbf{Time (sec)} & \textbf{Acceleration (m/s$^2$)}\\\hline
0 &20.4\\
0.1 & 20.3\\
0.2 & 19.4\\
0.3 & 17.3\\
0.4 & 14.0\\
0.5 & 9.8\\
0.6 & 5.3\\
0.7 & 1.6\\
0.8 & 0\\
\hline
\end{tabular}
\end{center}

\subsection{Fitting a four parameter}
Suppose we want to fit a cubic model to the data, ie. a model of the
form $a(t)=at^3 +bt^2+ct+d$, then we will follow the same proceedure
as before. We define the total distance function, and find where this
function is a minimum.In this example if used the $V_2$ total distance
function it would be defined as
\begin{multline*}
V_2(a,b,c,d)=[(20.4-a(0))^2+
  (20.3-a(0.1))^2+(19.4-a(0.2))^2+\\
(17.3-a(0.3))^2+(14-a(0.4))^2+(9.8-a(0.5))^2+\\
(5.3-a(0.6))^2+(1.6-a(0.7))^2+(0-a(0.8))^2]^{\frac 12}
\end{multline*}
To minimize this function we find all four partial derivatives of
$V_2$, and find the locations when all four are simultaneously equal
to zero. That is solve:
\[
\pdif{V_2}{a}==0\,,\;\pdif{V_2}{b}==0\,,\;\pdif{V_2}{c}==0\,,\;\pdif{V_2}{d}==0
\]
We now need to run some type of test to classify the type of critical
points we have found. Unfortunately the equivalent of the second
derivative test is too complex for this course. We can't even examine
the graph of $V_2$ to see if the location looks like a minimum,
because the graph of $V_2$ lies in five dimensional space. The only
check we can perform is to plot the data and the new model and see if
it is a reasonable fit.
\subsection{Solution}
When each of the stages is done on Mathematica (using file:
\verb+M3_P4_acceleration.nb+) we find that the model that minimizes
the $V_2$ function is $a(t)=80.8923t^3-120.209t^2+18.8938t+20.0525$
and if this plotted next to the data we find that it is a reasonable
fit:
\begin{center}
\includegraphics[width=0.75\textwidth]{Problem4/acc_cubic.eps}
\end{center}
So we cross our fingers and hope that this really is a minimum and not
some four dimensional equivalent of a saddle point.

