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\bf{Question}

\quad Let $\gamma(t)$ be a regular plane curve, and let $d$ be a
positive constant.  The curve $\delta(t)$ given by
$$\delta(t)=\gamma(t)+dN(t)$$ is called the {\it parallel} to
$\gamma$ at distance $d$.  Show that $\delta(t)$ is a regular
curve {\it except} where $\delta$ intersects the evolute of
$\gamma$.

Taking $\gamma$ to be a parabola, sketch $\delta$ for increasing
values of $d$.



\bf{Answer}

$\delta(t) = \gamma(t) + dN(t)$, where $\gamma$ is unit speed, so
$\dot{\gamma}=\gamma'$, $\dot{N}=N$.

$\dot{\delta}(t) = \gamma'(t) + d N'(t) = T - d.KT = (1-dK)T$.

Thus $\dot{\delta} \ne 0 $ except when $1-dK=0$, i.e. $d= \rho =
K^{-1}$;

then $\delta(t)=$ centre of curve $\Rightarrow$ lies on evolute.

\begin{center}
\epsfig{file=342-5A-1.eps, width=50mm}

Note how the singular points of the parallels trace out the
evolute.
\end{center}



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