% vim: tw=50
% 24/01/2022 10AM

\noindent
We will sometimes write $\dd \bf{x} = \dot{\bf{x}}
\dd t$. \\
If we're in Cartesian coordinates then we just
differentiate vector components
\[ \bf{x}(t) = \bf{x}^i(t) \bf{e}_i \implies
\dot{\bf{x}}(t) = \dot{x}^t(t) \bf{e}_i \]

\begin{note*}
    If we have a function $f(t)$ and vectors
    $\bf{g}(t)$ and $\bf{h}(t)$ then the following
    identities hold
    \begin{align*}
        \dfrac{}{t} (\bf{f} \bf{g})
        &= \dot{\bf{f}} \bf{g} + \bf{f}
        \dot{\bf{g}} \\
        \dfrac{}{t} (\bf{g} \cdot \bf{h})
        &= \dot{\bf{g}} \cdot \bf{h} + \bf{g}
        \cdot \dot{\bf{h}} \\
        \dfrac{}{t} (\bf{g} \times \bf{h})
        &= \dot{\bf{g}} \times \bf{h} + \bf{g}
        \times \dot{\bf{h}}
    \end{align*}
    (just apply the product rule to the
    components.)
\end{note*}

\subsubsection*{Tangent Vectors}

The derivative $\dot{\bf{x}}(t)$ is the
\emph{tangent vector} to the curve
\begin{center}
\begin{tsqx}
    ! size(5cm);
    arc3 origin dir(40) dir(160)
    label $C$ @ dir(40)+(0.1,0)
    label $O$ @ origin-(0,0.1)
    (0,0)->>dir(110)
    (0,0)->>dir(80)
    label $\mathbf{x}(t)$ @ 1.1*dir(110)
    label $\mathbf{x}(t + \delta t)$ @ 1.1*dir(80)
    dir(110)->>dir(80) red
    label $\delta \mathbf{x}$ @ dir(95)-(0,0.1) +col:red
\end{tsqx}
\end{center}

\begin{note*}
    The direction $\delta \mathbf{x}(t)$ is
    independent of the parametrisation (at least
    up to a sign), while the magnitude does depend
    on parametrisation.
\end{note*}

\noindent
For example, these two maps give the same curve
$C$ in $\RR^2$
\[ \bf{x}(t) = (t, t) \implies \dot{\bf{x}} =
(1, 1) \]
\[ \bf{x}(t) = (t^3, t^3) \implies \dot{\bf{x}} =
3t^2 (1, 1) \]
$C$ is just a line in $\RR^2$. In the second case
$\dot{\bf{x}} = 0$ at $t = 0$ but this is due to
the parametrisation, not to $C$ itself. \\
A parametrisation is \emph{regular} if
$\dot{\bf{x}}(t) \neq 0 \,\,\forall\,\, t$. \\
In what follows, we'll assume regular
parametrisations.

\bigskip

\noindent
The \emph{arc length} is the distance along the
curve. For nearby points
\begin{align*}
    \delta s
    &= |\delta \bf{x}| + O(|\delta
    \bf{x}|^2) \\
    &= |\dot{\bf{x}} \delta t| + O(\delta t^2) \\
    \implies \dfrac{s}{t}
    &= \pm |\dot{\bf{x}}|
\end{align*}
($\pm$ depends on whether $s$ increases or
decreases as $t$ increases.) \\
The arc length is defined by
\[ s = \int_{t_0}^t \dd t' |\dot{\bf{x}} (t')| \]

\begin{note*}
    For $t > t_0$, $s > 0$, and for $t < t_0$, $s
    < 0$.
\end{note*}

\begin{claim*}
    $s$ is independent of our choice of
    parametrisation.
\end{claim*}

\begin{proof}
    Pick a different choice $\tau(t)$. Assume
    $\dfrac{\tau}{t} > 0$. Then
    \[ \dfrac{\bf{x}}{t} = \dfrac{\bf{x}}{\tau}
    \dfrac{\tau}{t} \]
    and
    \begin{align*}
        s
        &= \int_{t_0}^t \dd t' \left|
        \dfrac{\bf{x}}{t'} \right| \\
        &= \int_{t_0}^t \dd t'
        \dfrac{\tau'}{t'} \left|
        \frac{\bf{x}}{\tau'} \right| \\
        &= \int_{\tau_0}^\tau \dd \tau' \left|
        \dfrac{\bf{x}}{\tau'} \right|
    \end{align*}
    ($\tau_0 = \tau(t_0)$)
\end{proof}

\bigskip

\noindent
This means that $s$ itself is a natural
parametrisation of the curve. We can think of
$\bf{x}(s)$.

\bigskip

\noindent
Because $\dfrac{s}{t} = |\dot{\bf{x}}(t)|$, the
associated tangent vector $\dfrac{\bf{x}}{s}$ has
$\left| \dfrac{\bf{x}}{s} \right| = 1$.

\subsubsection*{Curvature and Torsion}

A curve $C$ parametrised by the arc length $s$,
has tangent vector
\[ \bf{t} = \dfrac{\bf{x}}{s} \]

\begin{note*}
    $\bf{t}$ is not the same thing as our previous
    parameter!
\end{note*}

\noindent
This has $|\bf{t}| = 1$. \myskip
The curvature $\kappa(S)$ is
\[ \kappa(s) = \left| \dfrac[2]{\bf{x}}{s} \right|
= \left| \dfrac{\bf{t}}{s} \right| \]
To get some intuition, consider a circle
\[ \bf{x}(t) = (R \cos t, R \sin t) \]
Use $\dfrac{s}{t} = |\dot{\bf{x}}|$ to get $s =
Rt$.
\[ \implies \bf{x}(s) = (R \cos \left( \frac{s}{R}
\right), R \sin \left( \frac{s}{R} \right) \]
\[ \implies \kappa(s) = \frac{1}{R} \]
(which is constant.) \myskip
Define the \emph{(principle) normal}
\[ \bf{n} = \frac{1}{\kappa} \dfrac[2]{x}{s} =
\frac{1}{\kappa} \dfrac{t}{s} \]
(when $\kappa(s) \neq 0$)

\begin{note*}
    $|\bf{n}| = 1$.
\end{note*}

\begin{claim*}
    If $\kappa \neq 0$, then $\bf{n} \cdot \bf{t}
    = 0$.
\end{claim*}

\begin{proof}
    $\bf{t} \cdot \bf{t} = 1 \implies
    \dfrac{}{s}(\bf{t} \cdot \bf{t}) = 2\bf{t}
    \cdot \dfrac{\bf{t}}{s} = 0$.
\end{proof}

\noindent
Hence $\bf{n}$ and $\bf{t}$ define the
\emph{oscillating plane}. \myskip
The curvature $\kappa(s)$ of a curve coincides
with that of a circle touching $C$, at $S$, lying
in the plane.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d79525187d0311ec.png}
\end{center}

\begin{note*}
    Because $\bf{t} \cdot \dfrac{\bf{t}}{s} = 0$,
    for curves in $\RR^3$, we can also compute the
    curvature as
    \[ \kappa = \left| \bf{t} \times
    \dfrac{\bf{t}}{s} \right| = \ub{|t|}_{=1} \left|
    \dfrac{\bf{t}}{s} \right| \]
\end{note*}
