And because it is infinitesimally small, a normal vector becomes the inward facing vector?
For example a parametric function $\vec{r(t)=\langle{}x(t), y(t)\rangle}$
- it's tangents are $\vec{r(t)}= \langle{}x'(t), y'(t)\rangle{}$. Take it's unit vector so we only consider direction, not scale. Thus: $\vec{r(t)}_{unitVec}= \langle{}x'(t), y'(t)\rangle{}\cdot\frac{1}{\sqrt{x'(t)^2+y(t)^2}}$
To find the normal, take the tangent of the unit vector $\vec{r''(t)}_{unitVec}=\vec{r(t)}_{Normal}=$ Normal Vectors of $\vec{r(t)}$.
- where $\frac{\vec{T'(t)}}{||\vec{T'(t)}||}=N$, is the unit normal?
Is this line of thinking correct?
This then can be divided by arc length to get Curvature like so: $\frac{\vec{r(t)}_{Normal}}{MagnitudeOfCurve'sTangent}\rightarrow \frac{\vec{r(t)}_{Normal}}{\sqrt{x'(t)^2+y(t)^2}}=\frac{||d\vec{T}||}{||d\vec{S}||}=\frac{||\frac{d\vec{T(t)}}{dt}||}{||\frac{d\vec{S(t)}}{dt}||}=k$
Drawing below to be verified, technically, tangent of tangent is at the end but because it is infinitesimally small, it stems from the same point?.. Is that right?
the curvature of a curve is defined in differential geometry as the coefficient $k$ of the unit normal.
So if you have
$T'(t)=kN$
Then $k$ is the curvature.
in general if you have a unit vector
$x(t), ||x(t)||=1, \forall t$
then
$x(t)\cdot \frac{dx}{dt}\\ =\frac{1}{2}\frac{d(x(t)\cdot x(t))}{dt}\\ =\frac{1}{2}\frac{d||x(t)||^2}{dt}=0$