I have been studying differential geometry lately, and in my text (DoCarmo), he (at least at the moment) is only addressing the properties of a curve when they are regular, and hasn't really mentioned why nor what changes when the curve is not regular.
To much dismay, there is a problem in the text which reads as follows
A regular parametrized curve $\alpha$ has the property that all its tangent lines pass through a fixed point.
a) Prove that the trace of $\alpha$ is a (segment of a) straight line
b) Does the conclusion still hold if $\alpha$ is not regular?
I proved a) as follows:
WLOG, Assume $\alpha$ is parametrized by arc-length.
Say the point which the tangent lines pass through is $c$. Then $\exists \lambda: \mathbb{R} \to \mathbb{R} \ $ such that $$\alpha(s) + \lambda(s) \alpha'(s) = c$$ I assumed $\lambda(s)$ was differentiable, which made sense to me on an intuitive level, though I have no real justification for. Differentiating:
$$\alpha'(s)(1 + \lambda'(s)) + \lambda(s)\alpha''(s) = 0$$ Since $\alpha'(s)$ and $\alpha''(s)$ are orthogonal, this requires $$\lambda'(s) = -1 \ \ \ \ \text{ and } \ \ \ \ \ \alpha''(s) = 0 \implies k(s) = 0$$
Using either of these will give me that $\alpha(s)$ is the parametrization of a straight line.
However, I fail to see exactly where I implemented the regularity. I feel as though the conclusion likely fails for b), but am unsure where exactly it fails.
So what exactly does regularity give us here that being singular does not?
If a curve is not regular, it cannot -in general- be parametrized by the arc length. For instance, consider a $C^ 2$ curve c, which for $t\in [-1,0]$ describe the horizontal segement $[-1,0]\times \{0\}$, by the law $(t^3,0)$ then for $t>0$ the vertical segment $ \{0\}\times [0,1]$, by the law $(0,t^3)$. This curve is $C^2$, not regular, and all its tangent lines pass through the origin.