Smoothness of the asymptotic parametrization of a ruled surface

Question

Let $S$ be a smooth developable surface in $\mathbb{R}^{3}$. It is well known that, if $S$ is free of planar points, then it admits a local parametrization of the form

$$\begin{align} \sigma \colon I \times U &\to \mathbb{R}^{3}\\ (t,u) &\mapsto \gamma(t) + u X(t), \end{align}$$

where both the curve $\gamma \colon I \to \mathbb{R}^{3}$ and the unit vector field $X \colon I \to \mathbb{R}^{3}$ along $\gamma$ are smooth maps.

Surprisingly, the same result does not hold if planar points are present. Indeed, if we merely require $S$ to be free of planar neighborhoods (i.e., the set of planar points is nowhere dense), then $X$ is generally only continuous, not differentiable; see V. Ushakov, Parameterisation of developable surfaces by asymptotic lines, Bull. Austral. Math. Soc. 54 (1996), 411–421.

I am wondering whether the converse statement holds. Namely, given a smooth curve $\gamma$ in $\mathbb{R}^{3}$ and a continuous vector field $X$ along $\gamma$ that is smooth in a dense subset, is then the image of $\sigma$ a smooth ruled surface?

Answer 1

No. E.g., let $I=(0,2\pi)$, $U=(-1,1)$, $\gamma(t)=(t,0,0)$, and $X(t)=(0,\cos t,|\sin t|)$:

Here the unit vector $X(t)$ rotates counterclockwise from $(0,1,0)$ to $(0,0,1)$ to $(0,-1,0)$ in the $yz$-plane (orthogonal to the straight line $\gamma$) during the time interval $(0,\pi]$ and then back, clockwise, from $(0,-1,0)$ to $(0,0,1)$ to $(0,1,0)$ during the time interval $[\pi,2\pi)$.