gistfile1.tex

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\documentclass[oneside,english]{amsart}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{graphicx}
\usepackage{esint}

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 \theoremstyle{plain}
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  \newtheorem{prop}[thm]{Proposition}
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\begin{document}
By $AB$ denote the arc of the bicycle from back wheel to front. By
$\Gamma$ denote the path of the front wheel and by $\gamma$ the
path of the back wheel. Let $s$ be the arc length parameter for $\Gamma$
and $t$ the arc length parameter for $\gamma$. Let $\alpha(s)$
be the angle between $\Gamma'(s)$ and $BA$. We will denote by $v$
the tangent vector to $BA$ at $\Gamma(s)$ and by $\tilde{v}$ the
tangent vector to $BA$ at $\gamma(s)$. By $\kappa$ denote the geodesic
curvature of $\Gamma$ and by $k$ that of $\gamma$.

%
\begin{figure}[h]
\includegraphics[scale=0.5]{figures/sphere}

\caption{Vectors on a sphere}

\end{figure}


\begin{prop}
The condition $T_{l}(\gamma)=\Gamma$is equivalent to the differential
equation on the function $\alpha(s)$:

\[
\frac{d\alpha(s)}{ds}+\kappa(s)=\frac{\cos(l)}{\sin(l)}\sin(\alpha)\]

\end{prop}
\begin{proof}
First we write down an expression for $v$ and use it to obtain an
expression for $\gamma$:

\[
v=\sin(\alpha)(\Gamma(s)\wedge\Gamma'(s))+\cos(\alpha)\Gamma'(s)\]


\[
\gamma(s)=\cos(l)\Gamma(s)+\sin(l)v\]


Note that $\Gamma''(s)=-\Gamma(s)+\kappa(\Gamma(s)\wedge\Gamma'(s))$.
We can use this to find the derivative $\gamma'=\frac{d\gamma}{ds}$:

\begin{eqnarray*}
\gamma'(s) & = & (\sin(l)\cos(\alpha)(\kappa+\alpha')(\Gamma(s)\wedge\Gamma'(s))+\\
 &  & (\cos(l)-\sin(l)\sin(\alpha)(\kappa+\alpha'))(\Gamma'(s))+\\
 &  & (-\sin(l)\cos(\alpha))\Gamma(s)\end{eqnarray*}


We can also obtain an expression for $\tilde{v}$ by transporting
$v$ along $AB$:

\[
\tilde{v}=\cos(l)v-\sin(l)\Gamma(s)\]


Now, $\tilde{v}$ and $\gamma'(s)$ are parallel so we equate their
wedge product to zero to obtain:

\[
0=\tilde{v}\wedge\gamma'(s)=A(\Gamma(s))+B(\Gamma'(s))+C(\Gamma(s)\wedge\Gamma'(s))\]


Since these three vectors are orthogonal and non-zero, all of A, B,
and C must be equal to 0. This gives us that either:

\[
\frac{d\alpha(s)}{ds}+\kappa(s)=\frac{\cos(l)}{\sin(l)}\sin(\alpha)\]


or

\begin{eqnarray*}
\sin(l)\cos(\alpha) & = & 0\\
\cos(l) & = & 0\\
\sin(\alpha) & = & 0\end{eqnarray*}


But this system cannot be satisfied for any pair of $\alpha$ and
$l$ so we are done.
\end{proof}
\begin{cor}
$\left|\frac{dt}{ds}\right|=\left|\cos(\alpha)\right|$
\end{cor}
\begin{proof}
We obtain an expression for $\gamma'(s)$ as above, then take its
magnitude and substitute in $\kappa(s)=\frac{\cos(l)}{\sin(l)}\sin(\alpha)-\frac{d\alpha(s)}{ds}$.
Simplification yields the result. 
\end{proof}
\begin{cor}
$k=\frac{\tan(\tilde{\alpha})}{\sin(l)}$where $\tilde{\alpha}$ denotes
the angle function $\alpha$ parameterized by $t$ instead of $s$. 
\end{cor}
\begin{proof}
Let $\sigma$ denote the orientation of the rear wheel ($+1$ if the
same direction as $\gamma'$and $-1$ otherwise). Note

\[
\Gamma(t)=\cos(l)\gamma(t)+\sigma\sin(l)\gamma'(t)\]


\[
\Gamma'(t)=-\sigma\sin(l)(\gamma)+\cos(l)(\gamma'(t))+\sigma\sin(l)k(t)(\gamma\wedge\gamma'(t))\]


\[
v=-(-\sin(l)\gamma+\sigma\cos(l)\gamma')=\sin(l)\gamma-\sigma\cos(l)\gamma'\]


\[
\Gamma'\wedge v=\cos(l)\sin(l)k(t)(\gamma)+\sigma\sin^{2}(l)k(t)(\gamma')\]


However, we can also write

\[
\Gamma'\wedge v=(\sin(\alpha)(||v||)(||\Gamma'||))\Gamma=(\sin(\tilde{\alpha})\sqrt{1+(\sin(l)k)^{2}})\Gamma\]


Plugging in our formula for $\Gamma$ from above and equating the
two expressions we find that

\[
\sin(l)k=\sin(\tilde{\alpha})\sqrt{1+(\sin(l)k)^{2}}\]


Substituting we see this is solved for 

\[
\sin(l)k=\tan(\tilde{\alpha})\]

\end{proof}
\begin{thm}
\textup{$\int\kappa(s)ds=\cos(l)\int k(t)dt$}
\end{thm}
\begin{proof}
We first rewrite the integral of curvature of $\gamma$\begin{eqnarray*}
\int k(t)dt & = & \frac{1}{\sin(l)}\int\tan(\tilde{\alpha}(t))dt\\
 & = & \frac{1}{\sin(l)}\int\tan(\alpha(s))\frac{dt}{ds}ds\\
 & = & \frac{1}{\sin(l)}\int\tan(\alpha(s))\cos(\alpha(s))ds\\
 & = & \frac{1}{\sin(l)}\int\sin(\alpha(s))ds\end{eqnarray*}


However, integrating both sides of our above differential equation
we get:

\begin{eqnarray*}
\int(\frac{d\alpha(s)}{ds}+\kappa(s))ds & = & \int\frac{\cos(l)}{\sin(l)}\sin(\alpha(s))ds\\
\int\kappa(s)ds & = & \frac{\cos(l)}{\sin(l)}\int\sin(\alpha(s))ds\\
 & = & \cos(l)\int k(t)dt\end{eqnarray*}

\end{proof}

\end{document}