plane-curves.tex

\chapter{Signed curvature}

On the plane, it makes sense to think about left (right) turns as positive (respectively negative).
This leads to the so-called \textit{signed curvature} of plane curves.
If you drive a car along a plane curve, then the signed curvature describes the position of the steering wheel at a given %point. 
time.

\label{chap:signed-curvature}

\section{Definitions}\label{sec:def(skur)}

Suppose $\gamma$ is a smooth unit-speed plane curve,
so $\tan(s)=\gamma'(s)$ is its unit tangent vector for any~$s$.

Let us rotate $\tan(s)$ by the angle $\tfrac\pi 2$ counterclockwise; 
denote the obtained vector by $\norm(s)$.
The pair $\tan(s),\norm(s)$ is an oriented orthonormal frame in the plane which is analogous to the Frenet frame
defined in Section~\ref{sec:frenet-frame};
we will keep the name \index{Frenet frame}\emph{Frenet frame} for it.

Recall that $\gamma''(s)\perp \gamma'(s)$ (see \ref{prop:a'-pertp-a''}).
Therefore, \index{10k@$\skur$ (signed curvature)}
\[\tan'(s)=\skur(s)\cdot \norm(s).\eqlbl{eq:tau'}\]
for a real number $\skur(s)$;
the value $\skur(s)$ is called \index{curvature}\index{signed curvature}\emph{signed curvature} of $\gamma$ at~$s$.
We may use the notation $\skur(s)_\gamma$ if we need to specify the curve~$\gamma$.

Note that 
\[\kur(s)=|\skur(s)|;\]
that is, up to sign, the signed curvature $\skur(s)$ equals the curvature $\kur(s)$  of $\gamma$ at $s$ defined in Section~\ref{sec:curvature};
the sign prescribes the direction of turn --- if $\gamma$ turns left at time $s$, then $\skur (s)>0$.
If we want to emphasize that we are working with the \textit{non-signed} curvature of a curve, 
we call it \index{curvature}\index{absolute curvature}\emph{absolute curvature}.

If we reverse the parametrization of the curve or change the orientation of the plane, then
the signed curvature changes its sign.

Since $\tan(s),\norm(s)$ is an orthonormal frame, we have 
\begin{align*}
\langle\tan,\tan\rangle&=1,
&
\langle\norm,\norm\rangle&=1, 
&
\langle\tan,\norm\rangle&=0.
\end{align*}
Differentiating these identities we get 
\begin{align*}
\langle\tan',\tan\rangle&=0,
&
\langle\norm',\norm\rangle&=0,
&
\langle\tan',\norm\rangle+\langle\tan,\norm'\rangle&=0.
\end{align*}
By \ref{eq:tau'}, $\langle\tan',\norm\rangle=\skur$. 
Therefore $\langle\tan,\norm'\rangle=-\skur$.
Hence, we get 
\[\norm'(s)=-\skur(s)\cdot \tan(s).\eqlbl{eq:nu'}\]
The equations \ref{eq:tau'} and \ref{eq:nu'} are the Frenet formulas for plane curves. 
They can be written in a matrix form as:
\[
\begin{pmatrix}
\tan'
\\
\norm'
\end{pmatrix}
=
\begin{pmatrix}
0&\skur
\\
-\skur&0
\end{pmatrix}
\cdot
\begin{pmatrix}
\tan
\\
\norm
\end{pmatrix}.
\]

\begin{thm}{Exercise}\label{ex:bike}
Let $\gamma_0\:[a,b]\to\mathbb{R}^2$ be a smooth curve and $\tan$ its tangent indicatrix.
Consider another curve $\gamma_1\:[a,b]\to\mathbb{R}^2$ defined by $\gamma_1(t)\z\df\gamma_0(t)+\tan(t)$.
Show that

\begin{minipage}{.47\textwidth}
\begin{subthm}{ex:bike:length}$\length\gamma_0\le \length\gamma_1$;
\end{subthm}
\end{minipage}
\hfill
\begin{minipage}{.47\textwidth}
\begin{subthm}{ex:bike:tc}$\tc{\gamma_0}\le \length\gamma_1$.
\end{subthm}
\end{minipage}

\end{thm}

The curves $\gamma_0$ and $\gamma_1$ in the exercise above describe the tracks of an idealized bicycle with  distance 1 from the rear to the front wheel.
Thus, by the exercise, the front wheel must have a longer track.
For more on the geometry of bicycle tracks, see the survey of Robert Foote, Mark Levi, and Serge Tabachnikov \cite{foote-levi-tabachnikov} and the references therein.

\section{Fundamental theorem of plane curves}

\begin{thm}{Theorem}\label{thm:fund-curves-2D}
Let $s\mapsto \skur(s)$ be a smooth real-valued function defined on a real interval $\mathbb{I}$.
Then there is a smooth unit-speed curve $\gamma\:\mathbb{I}\to\mathbb{R}^2$ with signed curvature $\skur(s)$.
Moreover, $\gamma$ is uniquely defined up to a orientation-preserving motion of the plane.
\end{thm}

This theorem is a partial case of its 3-dimensional analog (\ref{thm:fund-curves}), but we present a direct proof.

\parbf{Proof.} 
Fix $s_0\in\mathbb{I}$.
Consider the function
\[\theta(s)
=
\int_{s_0}^s\skur(t)\cdot dt.\]
By the fundamental theorem of calculus, we have $\theta'(s)\z=\skur(s)$ for all~$s$.

Set 
$\tan(s)=(\cos[\theta(s)],\sin[\theta(s)])$,
and let $\norm(s)$ be its counterclockwise rotation by angle $\tfrac\pi2$; so 
$\norm(s)=(-\sin[\theta(s)],\cos[\theta(s)])$.
Consider the curve 
\[\gamma(s)=\int_{s_0}^s\tan(s)\cdot ds.\]
Since $|\gamma'|=|\tan|=1$, the curve $\gamma$ is unit-speed, and its Frenet frame is~$\tan,\norm$. 

Note that
\begin{align*}
\gamma''(s)&=\tan'(s)=
\\
&=(\cos[\theta(s)]',\sin[\theta(s)]')=
\\
&=\theta'(s)\cdot (-\sin[\theta(s)],\cos[\theta(s)])=
\\
&=\skur(s)\cdot \norm(s).
\end{align*}
So, $\skur(s)$ is the signed curvature of $\gamma$ at~$s$. 

This proves the existence;
it remains to prove the uniqueness.

Assume $\gamma_1$ and $\gamma_2$ are two curves that satisfy the assumptions of the theorem.
Applying a rigid motion, we can assume that $\gamma_1(s_0)\z=\gamma_2(s_0)$, and both curves have the same Frenet frame at $s_0$.
Let us denote by $\tan_1,\norm_1$ and $\tan_2,\norm_2$ the Frenet frames of $\gamma_1$ and $\gamma_2$, respectively.
Both triples $\gamma_i,\tan_i,\norm_i$ satisfy the following system of ordinary differential equations 
\[
\begin{cases}
\gamma_i'=\tan_i,
\\
\tan_i'=\skur\cdot\norm_i,
\\
\norm_i'=-\skur\cdot\tan_i.
\end{cases}
\]
Moreover, they have the same initial values at $s_0$.
By the uniqueness of solutions of ordinary differential equations (\ref{thm:ODE}), we have $\gamma_1=\gamma_2$.
\qeds

Suppose $\gamma\:\mathbb{I}\to\mathbb{R}^2$ is a unit-speed curve.
A continuous function $\theta\:\mathbb{I}\to\mathbb{R}$ is called \index{continuous!argument}\emph{continuous argument} of $\gamma$ if 
\[\gamma'(s)=(\cos [\theta(s)],\sin[\theta(s)])\]
for any $s$.
The proof of the theorem implies the following.



\begin{thm}{Corollary}\label{cor:2D-angle}
For any smooth curve unit-speed curve $\gamma\:\mathbb{I}\to\mathbb{R}^2$ there is a continuous argument $\theta\:\mathbb{I}\to\mathbb{R}$.
Moreover 
\[\theta'(s)=\skur(s),\]
where $\skur$ denotes the signed curvature of~$\gamma$.
\end{thm}


\section{Total signed curvature}\label{sec:Total signed curvature}

Let $\gamma\:\mathbb{I}\to\mathbb{R}^2$ be a smooth unit-speed plane curve.
The \index{curvature}\index{total!signed curvature}\emph{total signed curvature} of $\gamma$, denoted by $\tgc\gamma$, is defined as the integral \index{10psi@$\tgc\gamma$ (total signed curvature)}
\[\tgc\gamma
=
\int_\mathbb{I} \skur(s)\cdot ds,\eqlbl{eq:tsc-k}\]
where $\skur$ denotes the signed curvature of~$\gamma$.

If $\mathbb{I}=[a,b]$, then 
\[\tgc\gamma=\theta(b)-\theta(a),\eqlbl{eq:tsc-theta}\]
where $\theta$ is a continuous argument of $\gamma$ (see \ref{cor:2D-angle}).

If $\gamma$ is a piecewise smooth plane curve, then we define its total signed curvature as the sum of the total signed curvatures of its arcs plus the sum of the \textit{signed} external angles at its joints;
they are positive where $\gamma$ turns left, negative where $\gamma$ turns right, and 0 where $\gamma$ goes straight.
It is undefined if $\gamma$ turns exactly backwards;
that is, if it has a cusp.

In other words, if $\gamma$ is a concatenation of smooth arcs $\gamma_1,\dots,\gamma_n$, then 
\[\tgc\gamma=\tgc{\gamma_1}+\dots+\tgc{\gamma_n}+\theta_1+\dots+\theta_{n-1},\]
where $\theta_i$ is the signed external angle at the joint between $\gamma_i$ and $\gamma_{i+1}$.
If $\gamma$ is closed, then the concatenation is cyclic, and
\[\tgc\gamma=\tgc{\gamma_1}+\dots+\tgc{\gamma_n}+\theta_1+\dots+\theta_{n},\]
where $\theta_n$ is the signed external angle at the joint between $\gamma_n$ and $\gamma_1$.

Since $\left|\int \skur(s)\cdot ds\right|\le \int|\skur(s)|\cdot ds$, we have
\[|\tgc\gamma|\le \tc\gamma\eqlbl{eq:tsc-tc}\] 
for any smooth plane curve $\gamma$;
that is, the total signed curvature $\tgc{}$ cannot exceed the total curvature $\tc{}$ in absolute value.
The equality holds if and only if the signed curvature does not change the sign.

\begin{thm}{Exercise}\label{ex:trochoids}
A trochoid is a curve traced out by a point fixed to a wheel as it rolls along a straight line.
\begin{figure}[!ht]
\centering
\begin{lpic}[t(-0mm),b(0mm),r(0mm),l(0mm)]{asy/trochoids}
\lbl[l]{4,0;{\footnotesize $-\tfrac32$}}
\lbl[l]{4,6;{\footnotesize $-1$}}
\lbl[tl]{10,15;{\footnotesize $-\tfrac12$}}
\lbl[t]{22,17;{\footnotesize $0$}}
\lbl[r]{3,23.6;{\footnotesize $\tfrac12$}}
\lbl[r]{3,29.4;{\footnotesize $1$}}
\lbl[r]{3,35.2;{\footnotesize $\tfrac32$}}

\end{lpic}
\end{figure}
A family of \index{trochoid}\emph{trochoids} $\gamma_a\:[0,2\cdot\pi]\to \mathbb{R}^2$ (see the picture) can be parametrized as
\[\gamma_a(t)=(t+a\cdot \sin t, a\cdot \cos t).\]
\begin{enumerate}[(a)]
\item Given $a\in \mathbb{R}$, find $\tgc{\gamma_a}$ if it is defined.
\item Given $a\in \mathbb{R}$, find $\tc{\gamma_a}$.
\end{enumerate}
\end{thm}

\begin{thm}{Proposition}\label{prop:total-signed-curvature}
Any simple closed smooth plane curve $\gamma$ has total signed curvature  $\pm2\cdot\pi$; it is $+2\cdot\pi$
if the region bounded by $\gamma$ lies on the left from it and  $-2\cdot\pi$ otherwise.

Moreover, the same statement holds for any simple closed piecewise smooth plane curve $\gamma$ if its total signed curvature is defined.
\end{thm}

This proposition is called sometimes {}\emph{Umlaufsatz}; it is a differential-geometric analog of the theorem about the sum of the internal angles of a polygon (\ref{thm:sum=(n-2)pi}) which we use in the proof.
A more conceptual proof was given by Heinz Hopf \cite{hopf1935}, \cite[p. 42]{hopf1989}.

\parbf{Proof.}
Without loss of generality, we may assume that $\gamma$ is oriented in such a way that the region bounded by $\gamma$ lies on its left.
We can also assume that $\gamma$ is unit-speed.

Consider a closed polygonal line $p_1\dots p_n$ inscribed in~$\gamma$.
We can assume that the arcs between the vertices are sufficiently small 
so that the polygonal line is simple, and each arc $\gamma_i$ from $p_i$ to $p_{i+1}$ has small total absolute curvature, say  $\tc{\gamma_i}<\pi$ for each~$i$.


Assume $p_i=\gamma(t_i)$.
As usual, we use indexes modulo $n$; in particular, $p_{n+1}\z=p_1$.
Set 
\begin{align*}
\vec w_i&=p_{i+1}-p_i,& \vec v_i&=\gamma'(t_i),
\\
\alpha_i&=\measuredangle(\vec v_i,\vec w_i),&\beta_i&=\measuredangle(\vec w_{i-1},\vec v_i),
\end{align*}
where $\alpha_i,\beta_i\in(-\pi,\pi)$ are signed angles --- $\alpha_i$ is positive if $\vec w_i$ points to the left from~$\vec v_i$.

{

\begin{wrapfigure}[13]{o}{43 mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-59}
\vskip0mm
\end{wrapfigure}

By \ref{eq:tsc-theta}, the value
\[\tgc{\gamma_i}-\alpha_i-\beta_{i+1}\eqlbl{eq:Psi-alpha-beta}\]
is a multiple of $2\cdot\pi$.
Since $\tc{\gamma_i}<\pi$, the chord lemma (\ref{lem:chord}) implies that $|\alpha_i|\z+|\beta_i|<\pi$.
By \ref{eq:tsc-tc}, we have that $|\tgc{\gamma_i}|\z\le\tc{\gamma_i}$;
therefore the value in \ref{eq:Psi-alpha-beta} vanishes.
In other words, for each $i$ we have
\[\tgc{\gamma_i}=\alpha_i+\beta_{i+1}.\]

}

Note that 
\[\delta_i=\pi-\alpha_i-\beta_i\eqlbl{eq:delta=pi-alpha-beta}\] 
is the internal angle of the polygonal line at $p_i$;
$\delta_i\in (0,2\cdot\pi)$ for each~$i$.
Recall that the sum of the internal angles of an $n$-gon is $(n-2)\cdot \pi$ (see \ref{thm:sum=(n-2)pi}); that is,
\[\delta_1+\dots+\delta_n=(n-2)\cdot \pi.\]
Therefore, 
\[
\begin{aligned}
\tgc\gamma&=\tgc{\gamma_1}+\dots+\tgc{\gamma_n}=
\\
&=(\alpha_1+\beta_2)+\dots+(\alpha_n+\beta_1)=
\\
&=(\beta_1+\alpha_1)+\dots+(\beta_n+\alpha_n)=
\\
&=(\pi-\delta_1)+\dots+(\pi-\delta_n)=
\\
&=n\cdot\pi-(n-2)\cdot \pi=
\\
&=2\cdot\pi.
\end{aligned}\eqlbl{eq:delta=pi-alpha-beta-sum}\]

The case of piecewise smooth curves is done the same way;
we need to subdivide the arcs in the cyclic concatenation further to meet the requirement above, and, instead of equation \ref{eq:delta=pi-alpha-beta}, we have 
\[\delta_i=\pi-\alpha_i-\beta_i-\theta_i,\]
where $\theta_i$ is the signed external angle of $\gamma$ at $p_i$; it vanishes if the curve $\gamma$ is smooth at $p_i$.
Therefore, instead of equation \ref{eq:delta=pi-alpha-beta-sum}, we have
\begin{align*}
\tgc\gamma&=\tgc{\gamma_1}+\dots+\tgc{\gamma_n}+\theta_1+\dots+\theta_n=
\\
&=(\alpha_1+\beta_2)+\dots+(\alpha_n+\beta_1)+\theta_1+\dots+\theta_n=
\\
&=(\beta_1+\alpha_1+\theta_1)+\dots+(\beta_n+\alpha_n+\theta_n)=
\\
&=(\pi-\delta_1)+\dots+(\pi-\delta_n)=
\\
&=n\cdot\pi-(n-2)\cdot \pi=
\\
&=2\cdot\pi.
\end{align*}
\qedsf

\begin{thm}{Exercise}\label{ex:zero-tsc}
Draw a smooth closed plane curve $\gamma$ such that 

\begin{subthm}{ex:zero-tsc:0}
$\tgc\gamma=0$;
\end{subthm}
 
\begin{subthm}{ex:zero-tsc:5}
$\tgc\gamma=\tc\gamma=10\cdot\pi$;
\end{subthm}

\begin{subthm}{ex:zero-tsc:2-4}
$\tgc\gamma=2\cdot \pi$ and $\tc\gamma=4\cdot\pi$.
\end{subthm}

\end{thm}

\begin{thm}{Exercise}\label{ex:length'}
Let $\gamma\:[a,b]\to\mathbb{R}^2$ be a smooth plane curve with Frenet frame $\tan,\norm$.
Given a real parameter $\ell$, consider
the curve $\gamma_\ell(t)\z=\gamma(t)\z+\ell\cdot\norm(t)$; it is called a \index{parallel!curve}\emph{parallel curve} of $\gamma$ at signed distance~$\ell$.

\begin{subthm}{ex:length':reg}
Show that the parametrization $\gamma_\ell$ is a regular if $\ell\cdot \skur(t)_\gamma\ne 1$ for all~$t$.
\end{subthm}
 
\begin{subthm}{ex:length':formula}
Set $L(\ell)=\length\gamma_\ell$.
Show that 
\[L(\ell)=L(0)-\ell\cdot\tgc\gamma\eqlbl{eq:length(parallel-curve)}\]
for all $\ell$ sufficiently close to $0$. 
\end{subthm}

\begin{subthm}{ex:length':antiformula}
Describe an example showing that formula \ref{eq:length(parallel-curve)} may not hold for some values~$\ell$. 
\end{subthm}

\end{thm}


\section{Osculating circle}

\begin{thm}{Proposition}\label{prop:circline}
Given a point $p\in\mathbb{R}^2$,
a unit vector $\tan$ 
and a real number $\skur$, there is a unique smooth unit-speed curve $\sigma\:\mathbb{R}\to\mathbb{R}^2$ 
that starts at $p$ in the direction of $\tan$ and signed curvature $\skur$.

Moreover, if $\skur=0$, then it is a line $\sigma(s)=p+s\cdot \tan$;
if $\skur\ne 0$, then $\sigma$ runs around a circle of radius $\tfrac1{|\skur|}$ with center at $p+\tfrac1\skur\cdot \norm$, where $\tan,\norm$ is an oriented orthonormal frame.
\end{thm}

\parbf{Proof.}
Choose a coordinate system such that $p$ is its origin and $\tan$ points in the direction of the $x$-axis.
Therefore, $\norm$ points in the direction of the $y$-axis.
Then
\begin{align*}\theta(s)&=\int_{0}^s\skur\cdot dt=\skur\cdot s
\end{align*}
is a continuous argument of $\gamma$, see \ref{cor:2D-angle}.
Therefore,
\[\sigma'(s)=(\cos[\skur\cdot s],\sin[\skur\cdot s]).\]
It remains to integrate the last identity.
If $\skur=0$, we get $\sigma(s)=(s,0)$
which describes the line $\sigma(s)=p+s\cdot \tan$.

If $\skur\ne 0$, we get
\[\sigma(s)=(\tfrac1\skur\cdot\sin[\skur\cdot s],
\tfrac1\skur\cdot(1-\cos[\skur\cdot s]));\]
it is the circle of radius $r=\tfrac1{|\skur|}$ centered at $(0,\tfrac1\skur)=p+\tfrac1\skur\cdot\norm$.
\qeds

\begin{thm}{Definition}
Let $\gamma$ be a smooth unit-speed plane curve;
denote by $\skur(s)$ the signed curvature of $\gamma$ at~$s$.

The unit-speed curve $\sigma_s$ of constant signed curvature $\skur(s)$ that starts at $\gamma(s)$ in the direction $\gamma'(s)$ is called the \index{osculating!circle}\emph{osculating circle} of $\gamma$ at~$s$.

The center and radius of the osculating circle at a given point are called the \index{center of curvature}\emph{center of curvature} and \index{radius of curvature}\emph{radius of curvature} of the curve at that point.
\end{thm}

{

\begin{wrapfigure}{o}{31 mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-21}
\vskip0mm
\end{wrapfigure}

The \textit{osculating circle} might be a circle or a line.
In the latter case, the center of curvature is undefined and the radius of curvature is infinite. 


The osculating circle $\sigma_s$ can be also defined as the unique circle (or line) that has \index{order of contact}\emph{second-order contact} with $\gamma$ at $s$;
that is, $\rho(\ell)\z=o(\ell^2)$, where $\rho(\ell)$ denotes the distance from $\gamma(s+\ell)$ to $\sigma_s$.

The following exercise is recommended to the reader familiar with the notion of \index{inversion}\emph{inversion}.

}

\begin{thm}{Advanced exercise}\label{ex:inverse}
Suppose $\gamma$ is a smooth plane curve that does not pass thru the origin.
Let $\hat \gamma$ be the inversion of $\gamma$ with respect to a circle centered at the origin.
Show that the osculating circle of $\hat\gamma$ at $s$ is the inversion of the osculating circle of $\gamma$ at~$s$.
\end{thm}

\section{Spiral lemma}
\label{spiral}
\index{spiral lemma}

The following lemma was proved by Peter Tait \cite{tait}
and later rediscovered by Adolf Kneser \cite{kneser}.

\begin{thm}{Lemma}\label{lem:spiral}
Assume $\gamma$ is a smooth plane curve with strictly decreasing positive signed curvature. Then the osculating circles of $\gamma$ are nested; that is, if $\sigma_s$ denotes the osculating circle of $\gamma$ at $s$,
then $\sigma_{s_0}$ lies in the open disc bounded by $\sigma_{s_1}$ for any $s_0<s_1$. \index{10sigma@$\sigma_s$ (osculating circle)}
\end{thm}

{

\begin{wrapfigure}{r}{31 mm}
\vskip-6mm
\includegraphics{mppics/pic-61}
\end{wrapfigure}


The picture shows a curve $\gamma$ as in the theorem and its osculating circles.
They form a peculiar foliation of the annulus; it meets the following property:
if a smooth function is constant on each osculating circle, then it must be constant in the annulus \cite[Lecture 10]{fuchs-tabachnikov}.
Also, the curve $\gamma$ is tangent to a circle of the foliation at each of its points.
However, it does not run along any of those circles.

}

\parbf{Proof.}
Let $\tan(s),\norm(s)$ be the Frenet frame,
$\omega(s)$, $r(s)$
the center and radius of curvature of~$\gamma$.
By \ref{prop:circline},  we have
\[\omega(s)=\gamma(s)+r(s)\cdot \norm(s).\]

Since $\skur>0$, we have that $r(s)\cdot\skur(s)=1$.
Therefore, applying Frenet formula \ref{eq:nu'}, we get that
\begin{align*}
\omega'(s)&=\gamma'(s)+r'(s)\cdot \norm(s)+r(s)\cdot \norm'(s)=
\\
&=\tan(s)+r'(s)\cdot \norm(s)-r(s)\cdot \skur(s)\cdot \tan(s)=
\\
&=r'(s)\cdot \norm(s).
\end{align*}
Since $\skur(s)$ is decreasing, $r(s)$ is increasing;
therefore $r'\ge 0$.
It follows that $|\omega'(s)|= r'(s)$ and $\omega'(s)$ points in the direction of $\norm(s)$.

Since $\norm'(s)=-\skur(s)\cdot\tan(s)$, the direction of $\omega'(s)$ cannot be constant on a nontrivial interval;
in other words, the curve $s\mapsto \omega(s)$ contains no line segments.

\begin{wrapfigure}[5]{o}{35 mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-84}
\end{wrapfigure}

In particular, $|\omega(s_1)-\omega(s_0)|\z<\length(\omega|_{[s_0,s_1]})$ for any $s_0<s_1$.
Therefore, 
\begin{align*}
|\omega(s_1)-\omega(s_0)|&<\length(\omega|_{[s_0,s_1]})=
\\
&=\int_{s_0}^{s_1}|\omega'(s)|\cdot ds=
\\
&=\int_{s_0}^{s_1}r'(s)\cdot ds=
\\
&=r(s_1)-r(s_0).
\end{align*}
In other words, the distance between the centers of $\sigma_{s_1}$ and $\sigma_{s_0}$
is strictly less than the difference between their radii, hence the result.
\qeds


The curve $s\mapsto \omega(s)$ is called the \index{evolute}\emph{evolute} of $\gamma$; 
it traces the centers of curvature of the curve. 
The evolute of $\gamma$ can be written as 
\[\omega(t)=\gamma(t)+\tfrac1{\skur(t)}\cdot \norm(t);\] 
in the proof, we showed that $(\tfrac1{\skur})'\cdot\norm$ is its velocity vector.


\begin{thm}{Exercise}\label{ex:evolute}
Let $\omega$ be the evolute of a smooth plane curve~$\gamma$.
Suppose $\gamma$ has positive signed curvature $\skur$ and $\skur' \neq 0$ at all points.
%%%%%%% Vertices haven't been defined yet.
Find the Frenet frame and curvature of \(\omega\) in terms of $\skur$ and Frenet frame $(\tan,\norm)$ of $\gamma$.
\end{thm}

The following theorem states formally that 
\textit{if you drive on the plane and turn the steering wheel to the left all the time,
then you will not be able to come back to the place you started.}


\begin{thm}{Theorem}\label{thm:spiral}
Assume $\gamma$ is a smooth plane curve with positive strictly monotone signed curvature. 
Then $\gamma$ is simple.
\end{thm}

The same statement holds true without assuming positivity of curvature; the proof requires only minor modifications.

\parbf{Proof.}
Note that $\gamma(s)$ lies on the osculating circle $\sigma_s$ of $\gamma$ at~$s$.
If $s_1\ne s_0$, then by \ref{lem:spiral}, $\sigma_{s_0}$ does not intersect $\sigma_{s_1}$.
Therefore, $\gamma(s_1)\ne \gamma(s_0)$,
hence the result.\qeds

\begin{thm}{Advanced exercise}\label{ex:3D-spiral}
Show that a 3-dimensional analog of the theorem does not hold.
That is, there are self-intersecting smooth space curves with strictly monotone curvature.
\end{thm}

\begin{thm}{Exercise}\label{ex:double-tangent}
Assume $\gamma$ is a smooth plane curve with positive strictly monotone signed curvature.

\begin{subthm}{ex:double-tangent:a}Show that no line can be tangent to $\gamma$ at two distinct points.
\end{subthm}

\begin{subthm}{ex:double-tangent:b}Show that no circle can be tangent to $\gamma$ at three distinct points. 
\end{subthm}

\end{thm}

{

\begin{wrapfigure}{o}{25 mm}
\vskip-6mm
\centering
\includegraphics{mppics/pic-25}
\vskip0mm
\end{wrapfigure}

Part \ref{SHORT.ex:double-tangent:a} does not hold if we allow the curvature to be negative; an example is shown on the picture.

}

\begin{thm}{Advanced exercise}\label{ex:spherical-spiral}
Show that a smooth spherical curve with nonvanishing torsion has no self-intersections.
\end{thm}