Finding Curvature

Question

I have came across $2$ formulas to find curvature:

1. $$k(s)=\|\alpha''(s)\|$$
2. $$k(t)=\frac{\det(\alpha'(t),\alpha''(t))}{\|\alpha'(t)\|^3}$$

What are the difference between the two formulas?

Answer 1

The first one applies to the case where $\Vert\alpha'(s)\Vert=1$.

The second one applies generally.

Let's use the first one, $\alpha''(s)$, as the definition of the curvature.

If $\Vert\alpha'(s)\Vert=1$, look at $\alpha(s(t))$ for some $s(t)$. Then $$ \frac{d}{dt}\alpha(s(t)) = s'(t)\cdot\alpha'(s(t)) $$ and $$ \left(\frac{d}{dt}\right)^2\alpha(s(t)) = s''(t)\cdot\alpha'(s(t)) + s'(t)^2\cdot\alpha''(s(t)). $$ In the plane, $$ \det\left(\frac{d}{dt}\alpha(s(t)),\left(\frac{d}{dt}\right)^2\alpha(s(t))\right) = |s'(t)|^3\cdot\Vert\alpha''(s(t))\Vert, $$ while in 3-dimensional space you'd express the same using the cross-product, $$ \left\Vert\frac{d}{dt}\alpha(s(t))\times\left(\frac{d}{dt}\right)^2\alpha(s(t))\right\Vert = |s'(t)|^3\cdot\Vert\alpha''(s(t))\Vert, $$ and then divide this by $|s'(t)|^3$ where $|s'(t)|=\Vert d\alpha(s(t))/dt\Vert$.