If X' x X''=0, is X on straight line?

Question

If a $C^2$ curve $X(t)$ in $R^3$ satisfies $X' \times X''=0$, how can I show that $X$ is a part of a straight line?

(I think there is a condition that the curve is a regular curve)

Answer 1

This is saying that $X''$ and $X'$ are always collinear. If the curve is regular, i.e. $X'$ is never $0$, then you can compute the rate of change of the unit tangent vector $T = X'/\|X'\|$ and find $$ T' = \frac{(X' \cdot X') X'' - (X' \cdot X'') X'}{\|X'\|^3} = 0$$ so $T$ is constant, and thus $X$ lies on the straight line through $X(0)$ in direction $T$.

Answer 2

This would not be true in general, at least not without some extra non-vanishing conditions. For example, suppose the curve is $$X(t) = \begin{cases} e^{-1/t} \mathbf{i}, & t > 0; \\ 0, & t = 0; \\ e^{1/t} \mathbf{j}, & t < 0. \end{cases}$$ Then $X(t)$ is even a $C^\infty$ curve, and it is clear that $X'(t) \times X''(t) = 0$ for $t \ne 0$. At $t = 0$, $X'(t) = X''(t) = 0$, so there also, $X'(t) \times X''(t) = 0$. But obviously $X$ does not lie along a single line.

Answer 3

Given that $X(t)$ is regular, that is, $X'(t)$ vanishes nowhere, I think we can show it like this:

If

$X' \times X'' = 0, \tag{1}$

then $X''$ is collinear with $X'$; that is, for each $t$ there is a $\lambda(t)$ such that

$X''(t) = \lambda(t) X'(t); \tag{2}$

since $X(t)$ is $C^2$, $X'(t)$ is $C^1$ and $X''(t)$ is $C^0$; thus $\lambda(t)$ is $C^0$ as well. We can see this even more explicitly by inspecting the components of $X(t)$; letting

$X(t) = (x_1(t), x_2(t), x_3(t)), \tag{3}$

we may write

$x_i''(t) = \lambda(t) x_i'(t), \tag{4}$

$1 \le i \le 3$, with each $x_i''(t)$ of class $C^0$ and the $x_i'(t)$ of class $C^1$; we see once again that $\lambda(t) \in C^0$. Of course, we have used the regularity of $X(t)$ in arriving at this conclusion.

The unique solution(s) to the (4) are thus of the form

$x_i'(t) = x_i'(t_0) e^{ \int_{t_0}^t \lambda(s) ds}, \tag{5}$

or $X'(t) = X'(0) e^{ \int_{t_0}^t \lambda(s) ds}. \tag{6}$

if we set

$\mu(t) = e^{ \int_{t_0}^t \lambda(s)ds}, \tag{7}$

so that

$X'(t) = \mu(t) X'(t_0), \tag{8}$

then . $X(t) = (\displaystyle \int_{t_0}^t \mu(s) ds) X'(t_0) + X_0, \tag{9}$

or in components,

$x_i(t) = (\displaystyle \int_{t_0}^t \mu(s) ds) x_i'(t_0) + x_i(t_0); \tag{10}$

taking

$s(t) = \displaystyle \int_{t_0}^t \mu(s) ds,\tag{11}$

we see that $X(t)$ lies in the line

$X(s) = sX'(0) + X_0. \tag{12}$

Nota Bene: Of course we must credit user Daniel Schepler for a fine discussion of the prospect that $X'(t) = 0$ for some $t$. End of Note.