Proof curvature and torsion Given position vector $p(s)$ on the curve with $\kappa$ as curvature and $T$ as torsion. Proof these 2 equations
a) $$\kappa = \| p''(s)\| = \|p'(s) \cdot p''(s)\|$$
b) $$\kappa^2 T = (p'(s),p''(s), p'''(s))$$
 A: I believe you have mispelt your problem.
If your curve is parametrized by $p:I=(\alpha,\beta) \to \mathbb{R^2}$ where $-\infty\leq\alpha<\beta\leq+\infty$, and $|p'(t)|=1$ $\forall t\in (\alpha,\beta)$ then by the definition of curvature $\kappa=|p''(t)|$. If your curve $p$ is not unit-speed and $\beta(s(t))=p(t)$ is a unit-speed reparametrization of your original function then $$(1)\quad \kappa=|p'(t)\times p''(t)| /|p'(t)|^3$$
But since we are supposed to show that $\kappa=|p''(s)|$ I assume $p$ is unit-speed and I'll answer based on that.
If $p$ is unit-speed $|p'(s) \cdot p''(s)|=|p'(s)|\cdot|p''(s)| \cdot |cos \angle(p'(s),p''(s))|=|p''(s)|$, because if p is unit-speed $|p'(s)|=1$ and when you differenciate $p'(s) \cdot p'(s) = 1$ on both sides and divide by $2$ you get $p'(s) \cdot p''(s) = 0$, so $p'(s)$ is orthogonal to $p''(s)$ for all $s \in I$, so $|cos \angle(p'(s),p''(s))|=1$ $\forall s \in I$.
If you meant $\kappa=\|p'(s) \times p''(s)\|$ it follows by simplifying equation $(1)$.
Regarding your (b) question I think you need to edit it because it can't be something in $\mathbb{R^3}$ if you are multiplying scalars like $\kappa$ and $\tau$.
