\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\setlength{\textwidth}{6.5in} \setlength{\textheight}{9in}
\setlength{\topmargin}{-.25in} \setlength{\oddsidemargin}{-.25in}
\newcommand\br{\textbf{R}}
\pagestyle{empty}
\begin{document}

\parindent=0pt

\bf{Question}

\quad Let
\begin{eqnarray*}\gamma_1(t)&=&(\cosh2t+2\cosh t,2\sinh
t-\sinh2t)\\ \gamma_2(t)&=&(\cosh2t-2\cosh t,2\sinh t-\sinh2t).
\end{eqnarray*}

Show that $\gamma_2$ is a regular curve, while $\gamma_1$ has a
singularity at the point $(3,0)$.  Sketch $\gamma_1$ and
$\gamma_2$. Find the centre of curvature of $\gamma_2$ at the
point $(-1,0)$.

[{\sl Exploit symmetry, and look at the directions of the tangents
to the curves.}]




\bf{Answer}

Write $\gamma(t) = (\cosh 2t + 2 \epsilon \cosh t, 2 \sinh t -
\epsilon \sinh 2t)$

Where $\epsilon = \pm 1$ (so $\gamma_i$ has $\epsilon =
(-1)^{i+1}$)

So $\dot{\gamma}(t) = (2 sinh 2t + 2\epsilon \sinh t, 2 \cosh t -
2\epsilon \cosh 2t).$

For $\epsilon=-1$ the second component never vanishes ($\cosh t$
always $\leq 1$); for $\epsilon = +1$ we have

first component $=2 \sinh t (2 \cosh t + \epsilon)$, which $=0$
just when $\sinh t=0$, i.e. $t=0$; then $\dot{\gamma}(0) = (0,0)$,
$\gamma(0) = (3,0)$.

Write $\gamma_i(t) = (x_i(t), y_i(t))$ for $i=1,2$ and note the
following facts
\begin{itemize}
\item
$x_i(t)$ is an even function ($\to +\infty$ as $t \to \pm\infty$),
$y_i(t)$ is an odd function.

\item
$y_i(t) =0$ when $2 \sinh t = 2\epsilon \sinh t \cosh t$, i.e.
$t=0$ or $\epsilon \cosh t=1$, i.e. $t=0$ ($\epsilon = 1$). So
$\gamma_1$ crosses the $x$-axis at $(3,0)$, $\gamma_2$ crosses the
$x$-axis at $(-1,0)$.

\item
$x_i(t)=0$ when $\cosh 2t = -2\epsilon \cosh t$, which does not
happen when $\epsilon=1$.
\end{itemize}

When $\epsilon=-1$ it occurs when $c = \cosh t$ satisfies $2c^2-1
= 2c$, i.e. $c=\frac{1}{2}(1 + \sqrt{3})$ ($c$ always $\geq 1$).

For $\epsilon = +1$ we have
\begin{eqnarray*}
\gamma(t) & = & \left ( 1 + \frac{(2t)^2}{2!}+\cdots+ 2\left ( 1 +
\frac{t^2}{2!} + \cdots \right ), 2 \left ( t + \frac{t^3}{3!}+
\cdots \right ) - \left ( 2t + \frac{(2t)^3}{3!} + \cdots \right )
\right )\\ & = & (3,0) + (3t^2+ \cdots, -t^3 + \cdots)
\end{eqnarray*}
$\Rightarrow \frac{3}{2}$-power \underline{cusp}.

Also $\dot{x}_i(t)$ has the same sign as $t$, and $\dot{y}_i(t)$
is always $\leq 0$ for $i=1$ and $>0$ for $i=2$.

Finally, the curves must never meet, they lie on opposite sides of
the curve $$(\cosh 2t, \sinh 2t) = \gamma_0(t), \ \ \textrm{i.e.
}x^2-y^2=1$$.

\begin{center}
\epsfig{file=342-5A-2.eps, width=65mm}
\end{center}

When $t=0$ we find $\| \dot{\gamma}_2 \| =4$ and $\dot{x}\ddot{y}
- \dot{y}\ddot{x} = -8$, so $\rho = \frac{1}{K} = -8$.

Also $N(0)=(-1,0)$ so centre of curvature is $$\gamma_2(0) + \rho
N(0) = (7,0).$$


\end{document}