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> [!definition] Definition. ([[signed unit-normal vector to plane curve]])
> To keep track of whether a plane [[parameterized curve|curve]] is traveling counter-clockwise or clockwise, we define a notion of **signed curvature**. Let $\alpha(s)$ be a unit-speed curve in $\mathbb{R}^{2}$. Then the **signed normal** $n_\text{s}(s)$ of $\alpha$ at $s$ is defined $n_{\text{s}}(s):= \begin{bmatrix}
0 & -1 \\ 1 & 0
\end{bmatrix}T(s),$
where $T$ is the [[tangent indicatrix of a unit-speed curve|tangent indicatrix]] evaluated at $s$.
\
For the case where $\alpha=\alpha(t)$ is not unit-speed, we explicitly have $k_{\text{s}}(t)=\frac{\alpha'(t)^{\perp} \cdot \alpha''(t)}{\|\alpha'(t)\|^{3}}.$
> [!justification]
> We know that because $|T(s)| \equiv 1$, $n(s)=\frac{\alpha''(s)}{\|\alpha''(s)\|}$ is [[orthogonal]] to $T(s)$, so when we 'manually rotate' $T(s)$ to get $n_{\text{s}}(s)$ we find that either $n_{\text{s}}(s)=n(s)$ or $n_{\text{s}}(s)=-n(s)$. In the first case, the curve is turning counter-clockwise at $s$. In the second case, it is turning clockwise.
From the signed normal we can define signed curvature:
> [!definition] (Signed Curvature of a Plane Curve)
> Let $\alpha(s)$ be a [[regular curve]] with unit-speed parameterization. Then its signed curvature $\kappa_{\text{s}}(s)$ at $s$ is defined as $\kappa_{\text{s}}(s):= \alpha''(s) \cdot n_{\text{s}}(s)=\begin{cases}
\kappa(s) & \text{ if } n(s)=n_{\text{s}}(s) \\
-\kappa(s) & \text{ if } n(s) = -n_{\text{s}}(s).
\end{cases}$
> [!justification]
> In light of the above justification block: assuming $\kappa$ is never 0, when $n(s)=n_{\text{s}}(s)$ the curve is rotating counterclockwise and we get $\kappa_{\text{s}}>0$ and when $n(s)=-n_{\text{s}}(s)$ it is rotating clockwise and we get $\kappa_{\text{s}}(s)<0$.
> (For the first equality, recall that $\alpha''(s)$ and $n_{\text{s}}(s)$ are collinear, so that $\alpha''(s) \cdot n_{\text{s}}(s)= \pm \|\alpha''(s)\| \cancel{\|n_{\text{s}}(s)\|}^{1}$).
> [!basicexample]
>
>
Let $\alpha(s)$ be the circle of radius $R$ parameterized as $\alpha(s)=\left( R \cos \frac{t}{R}, R \sin \frac{t}{R} \right).$
Its tangent is given by $T(s)=\left( - \sin \frac{t}{R}, \cos \frac{t}{R} \right).$
Its signed normal vector is $\begin{bmatrix}
0 & -1 \\ 1 & 0
\end{bmatrix} \begin{bmatrix}
-\sin \frac{t}{R} \\ \cos \frac{t}{R}
\end{bmatrix} = \begin{bmatrix}
-\cos \frac{t}{R} \\ - \sin \frac{t}{R}
\end{bmatrix}.$
Its second derivative is $\alpha''(s)=\left( -\frac{1}{R} \cos \frac{t}{R}, \frac{1}{R} \sin \frac{t}{R} \right)$. Therefore, its signed curvature is $\alpha''(s) \cdot n_{\text{s}}(s)= \frac{1}{ R}.$
Now, define the curve $\beta(s)=\left( R \cos \frac{t}{R}, -R \sin \frac{t}{R} \right).$
Notice that $\beta(s)$ has the same trace as $\alpha(s)$.
This time $T_{\beta}(s)=\left( - \sin \frac{t}{R}, -\cos \frac{t}{R} \right),$
meaning that $n_{\text{s}}(s)=\begin{bmatrix}
0 & -1 \\ 1 & 0
\end{bmatrix} \begin{bmatrix}
-\sin \frac{t}{R} \\ -\cos \frac{t}{R}
\end{bmatrix} = \begin{bmatrix}
\cos \frac{t}{R} \\ - \sin \frac{t}{R}
\end{bmatrix}.$
Since $\beta''(s)=\left( -\cos \frac{t}{R}, \sin \frac{t}{R} \right),$
we have
$\kappa_{\text{s}}^{(\beta)} (s)= \beta''(s) \cdot n_{\text{s}}^{(\beta)}(s)=-\frac{1}{R}.$
This all checks out: when we traced clockwise, we got negative curvature.
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```