<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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 Verify that if the tangent to a plane curve $\gamma$ at the
point $\gamma(s)$ makes an angle $\psi(s)$ with the $x$-axis
(turning from the positive $x$-direction to the tangent vector)
then $\kappa(s)={d\psi\over ds}(s)$ (where $s={\rm arc length}$ as
usual).
\smallskip
Suppose $y'(s)\ne0$ and let the tangent line at $\gamma(s)$ meet
the $x$-axis at $x=c(s)$, say.  Show that if $y(s)\ne0$ (so the
point $\gamma(s)$ is not on the $x$-axis itself) then
$\kappa(s)=0$ if and only if $c'(s)=0$.



\bf{Answer}

Very easy: we see that

\begin{eqnarray*}
T(s) &amp; = &amp; (\cos \phi, \sin \phi)\\ \Rightarrow  T'(s) &amp; = &amp;
(-\sin \phi, \cos \phi)\\ \textrm{But }N(s) &amp; = &amp; (-\sin \phi,
\cos \phi)\\ \Rightarrow  K(s) &amp; = &amp; \frac{d \phi}{d s}.
\end{eqnarray*}

\begin{center}
\epsfig{file=342-4A-12.eps, width=50mm}
\end{center}

$\gamma(s)=(x(s), y(s))$, unit speed.

Equation of tangent to curve $\gamma$ at $(x(s), y(s))$ is
$$(y-y(s))x'(s) = (x-x(s))y'(s).$$

This meets the $x$-axis at $(c(s), 0)$ so we have $$-y(s)x'(s) =
(c(s) - x(s))y'(s). \ \longleftarrow (1)$$

Differentiate: $-y'x' =yx'' = (c'-x')y' + (c-x)y''$, (dropping the
$s$).

Giving: $$-yx'' = c'y' + (c-x) y''$$ Substitute $(c-x)$ from $(1)$
to get (with $y' \ne 0$)
\begin{eqnarray*}
-yx'' &amp; = &amp; c'y' + \frac{-yx'}{y'}.y''\\ \textrm{i.e. } c'(y')^2 &amp;
= &amp; y(x'y''-y'x'') = yK\\ \textrm{as }x'(s)^2 + y'(s)^2 &amp; = &amp; 1
\end{eqnarray*}

Hence since $y \ne 0$ we see $K=0$ precisely when $c'=0$.


\end{document}
</pre></body></html>