I am struggling to understand the definition of biregular curve.
Definition
A curve $I \ni t \mapsto P(t) \in \mathbb{R}^3$ is said to be biregular when the velocity and accelleration vectors are linearly independent, that is when their cross product is different from zero $$dP/dt \land d^2P/d^2t \neq 0$$
Q.
I read that this is equivalent to the fact that $$d \underline{t}/ds \neq 0$$ and this is what I am having trouble understanding. Why is so?
Recall that $$s= \int_0^t \|dP/dt\| \, dt,$$ is the curvilinean coordinate, and $$\underline{t}(s)=dP/ds$$ so that $$dP/dt= dP/ds \ \dot{s}= \underline{t} \ \dot{s}$$ where $$\dot{s}=ds/dt.$$
Moreover $$d^2P/d^2t= d\underline{t}/ds \ \dot{s}^2 + \underline{t} \ \ddot{s}.$$ Hence we have
$$dP/dt \land d^2P/d^2t = \underline{t} \ \dot{s} \land (d\underline{t}/ds \ \dot{s}^2 + \underline{t} \ \ddot{s})$$$$ = \dot{s}^3(\underline{t}\land d\underline{t}/ds)+\ddot{s}(\underline{t}\land \underline{t})=\dot{s}^3(\underline{t}\land d\underline{t}/ds).$$
So, the curve is biregular $\iff$ $\dot{s}=\|dP/dt\|\neq 0$ and $\underline{t}\land d\underline{t}/ds \neq 0.$
How is this equivalent to the fact that $d \underline{t}/ds \neq 0$?
Why is it not possible that $d\underline{t}/ds$ and $\underline{t}$ are both different from zero and yet linearly dependent?
I can't bring myself to use little letters for these vectors when we already have the time parameter $t$.
Because $\|T(s)\| = 1$ for all $s$, it follows that $dT/ds$ is orthogonal to $T$, so they will always be linearly independent provided $dT/ds\ne 0$.