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% 26/01/2022 10AM

\begin{example*}
    Let $C$ be the helix $\bf{x}(t) = (\cos t,
    \sin t, t)$. Then $\dot{\bf{x}}(t) = (-\sin t,
    \cos t, 1)$.
    \[ \implies \dfrac{s}{t} = \left|
    \dfrac{\bf{x}}{t} \right| = \sqrt{2} \implies
    s = \sqrt{2}t .\]
    The distance along the curve between
    $\bf{x}(0) = (1, 0, 0)$ and $\bf{x}(2\pi) =
    (1, 0, 2\pi)$ is
    \[ s = \int_0^{2\pi} \dd t |\dot{\bf{x}}| =
    \sqrt{2} \times 2\pi = \sqrt{8} \pi \]
    \[ \bf{x}(s) = (\cos(s/\sqrt{2}),
    \sin(s/\sqrt{2}, s/\sqrt{2}) \]
    \[ \bf{t} = \dfrac{\bf{x}}{s} =
    \frac{1}{\sqrt{2}} \left( - \sin \left(
    \frac{s}{\sqrt{2}} \right), \cos \left(
    \frac{s}{\sqrt{2}} \right), 1 \right) \]
    \[ \dfrac{\bf{t}}{s} = \ub{\half}_{\kappa}
    \ub{\left( - \cos
    \left( \frac{s}{\sqrt{2}} \right), -\sin
    \left( \frac{s}{\sqrt{2}} \right), 0 \right)
    }_{\bf{n}} .\]
\end{example*}

\noindent
For curves in $\RR^3$, define the \emph{binormal}
\[ \bf{b} = \bf{t} \times \bf{n} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c39e482082bf11ec.png}
\end{center}
\begin{note*}
    $\bf{t}$, $\bf{n}$ and $\bf{b}$ are an
    orthonormal basis for each $s$ (at least with
    $\kappa(s) \neq 0$).
\end{note*}

\noindent
Because $|\bf{b}| = 1$ we have $\bf{b} \cdot
\dfrac{\bf{b}}{s} = 0$. Moreover,
\[ \bf{t} \cdot \bf{b} = 0 \implies
\dfrac{\bf{t}}{s} \cdot \bf{b} + \bf{t} \cdot
\dfrac{\bf{b}}{s} = 0 \]
\[ \implies \kappa \bf{n} \cdot \bf{b} + \bf{t}
\cdot \dfrac{\bf{b}}{s} = 0 \]
\[ \implies \bf{t} \cdot \dfrac{\bf{b}}{s} = 0 \]
so $\dfrac{\bf{b}}{s}$ is $\perp$ to $\bf{b}$ and
$\bf{t}$ hence $\dfrac{\bf{b}}{s}$ is parallel to
$\bf{n}$. \myskip
Define the \emph{torsion}, $\tau(s)$ as
\[ \dfrac{\bf{b}}{s} = -\tau(s) \bf{n} \]
The torsion measures how much the curve twists out
of the plane. (It vanishes for planar curves.)

\begin{note*}
    \[ \dfrac{\bf{t}}{s} = \kappa(s) (\bf{b}
    \times \bf{t}) \]
    \[ \dfrac{\bf{b}}{s} = \tau(s) (\bf{t} \times
    \bf{b}) \]
    These are six first order DEs in six unknowns
    $\bf{t}$ and $\bf{b}$. For fixed $\kappa(s)$
    and $\tau(s)$, there is a unique solution if
    we're given $\bf{t}(0)$ and $\bf{b}(0)$. \\
    i.e. $\kappa$ and $\tau$ specify $C$ up to
    translations / rotations.
\end{note*}

\subsection{Line Integrals}

A scalar field $\phi(\bf{x})$ is a map
\[ \phi : \RR^n \to R \]
We would like to integrate $\phi(\bf{x})$ along a
curve $C$ given by $\bf{x}(t)$ in a way that is
\emph{independent} of the parametrisation. \myskip
We work with the arc length. Let $\bf{x}(s)$ be a
curve $C$ that urns from $\bf{x} = \bf{a}$ to
$\bf{x} = \bf{b}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/a9f9ca9282c011ec.png}
\end{center}
We define the \emph{line integral} from $\bf{a}$
to $\bf{b}$
\[ \int_C \phi \dd s = \int_{s_a}^{s_b}
\phi(\bf{x}(s)) \dd s \]
where we take $s_a < s_b$.

\begin{note*}
    This is defined so that $\int_C \dd s$ is the
    length of $C$ and is always positive. In other
    words, the line integral from $\bf{a}$ to
    $\bf{b}$ gives the same answer as $\bf{b}$ to
    $\bf{a}$.
\end{note*}

If you're given the curve $\bf{x}(t)$ using some
other parameter, with $\bf{x}(t_a) = \bf{a}$ and
$\bf{x}(t_b) = \bf{b}$ and $t_a < t_b$ then
\begin{align*}
    \int_C \phi \dd s
    &= \int_{b_a}^{b_b} \phi(\bf{x}(t))
    \dfrac{s}{t} \dd t \\
    &= \int_{t_a}^{t_b} \phi(\bf{x}(t))
    |\dot{\bf{x}}| \dd t
\end{align*}
(using $\dfrac{s}{t} = |\dot{\bf{x}}|$. This
factor $|\dot{\bf{x}}|$ ensures independence of
parametrisation.) \myskip
A vector field $\bf{F}(\bf{x})$ is a map
\[ \bf{F} : \RR^m \to \RR^n .\]
The \emph{line integral} of a vector field
$\bf{F}(\bf{x})$ along a curve $C$, parametrised
by $\bf{x}(t)$, from $\bf{x}(t_a) = \bf{a}$ to
$\bf{x}(t_b) = \bf{b}$ is
\[ \int_C \bf{F} \cdot \dd \bf{x} =
\int_{t_a}^{t_b} \bf{F}(\bf{x}(t)) \cdot
\dot{\bf{x}}(t) \dd t \]
This is the integral of $\bf{F}$ tangent to the
curve (for example $\tau = \tau(t)$).

\begin{note*}
    This time the direction of the integral
    matters. The integral from $\bf{a}$ to
    $\bf{b}$ is the negative of the integral from
    $\bf{b}$ to $\bf{a}$.
\end{note*}

\noindent
The choice of direction along $C$ is called an
\emph{orientation}. \myskip
Again: the line integral of a scalar field does
\emph{not} depend on the orientation of $C$; the
line integral of a vector field does depend on the
orientation.
