First and second derivative of a line parametrization are linearly dependent

Question

I got a homework problem in my differential geometry class with which I'm stuck:

Prove: A map

$X: \mathbb{R} \supseteq I \ \rightarrow \mathbb{R}^n, t \rightarrow X(t)$, with $X' \ne 0$

parametrises a line or a line segment if $X'(t)$ and $X''(t)$ are linearly dependent.

• At first I solved it for two dimensions, then an Ansatz can be made for the straight line:

$y(t) = kx(t) + d$, $y'(t) = kx'(t)$ and $y''(t) = kx''(t)$.

And if $X'= (x'(t),y'(t))^T$ and $y''= (x''(t),y''(t))^T$ are linearly dependent, then the determinant of the matrix

\begin{bmatrix} x'&x'' \\ y'&y'' \end{bmatrix}

must equal zero, which implies $x'y'' - x''y' = 0$ or $ \frac{x'}{x''}=\frac{y'}{y''}$. Then I just insert the $x' and x''$ from my Ansatz and see that it fits.

But is that a proove? I'm not sure... (Still it just works in 2 dimensions.)

• Another approach works in three dimensions, then the curvature is given by

$\kappa = \frac{||X' \times X''||}{||X'||^3}$

and if $X', X''$ are linearly dependent, then $X'=a X''$ hold with some constant $a$. So the cross product of two parallel vectors is zero and therefore the curvature is zero which means, $X$ must be a straight line. But does this hold for higher dimensions $n>3$?

So my question is, how do I prove it for n dimensions?

Answer 1

If $X'(t)$ and $X''(t)$ are linearly dependent with $X(t)\ne0$ for all $t$ then there is a function $t\mapsto\lambda (t)$, defined in a neighborhood $U$ of $t=0$, such that $$X''(t)=\lambda(t)X'(t)\qquad(t\in U)\ .$$ If $X(\cdot)\in C^2$ the function $\lambda(t)={X''(t)\cdot X'(t)\over |X'(t)|^2}$ will be automatically continuous.

Let $X'(0)=v_0\ne0$, and consider a fixed vector $e\perp v_0$. The auxiliary scalar function $$\phi(t):=e\cdot X'(t)$$ satisfies $\phi(0)=0$ and $$\phi'(t)=e\cdot X''(t)=\lambda(t)\ e\cdot X'(t)\ ,$$ hence is a solution of the IVP $$\phi'(t)=\lambda(t)\phi(t), \quad\phi(0)=0\ .\tag{1}$$ Since $\phi(t)\equiv0$ is a solution of $(1)$, the general existence and uniqueness theorem for ODEs in fact enforces $\phi(t)\equiv0$, hence $X'(t)\perp e$ for all times. As $e\perp v_0$ was chosen arbitrarily we can conclude that $$X'(t)=g(t) v_0\qquad(t\in U)$$ for a certain function $t\mapsto g(t)>0$. It is then clear that the point $t\mapsto X(t)$ moves on a line with direction $v_0$.

Answer 2

I think I got it:

$X: \mathbb{R} \rightarrow \mathbb{R}^n : t \rightarrow X(t)$ mit $X'(t) \ne 0$ $\forall t$

One has: $X''(t)=a(t) X'(t), a(t) \in \mathbb{R}$ (linearly dependent)

Choose an arbitrary component $x_i$ \begin{align} x_i''(t) &= a(t) x_i'(t) &\\ \frac{x_i''}{x_i'} &= a &| \int dt \\ \int (ln(x_i'))'dt &= \int a dt &| A := \int a dt \\ ln(x_i') + C_{i1} &= A &| e^{(...)} \\ C_{i2} x_i' &= e^{A} &| \int dt \\ C_{i2} x_i + C_{i3} &= \frac{e^{A}}{a} \\ x_i(t) &= \frac{e^{A} - aC_{i3}}{a C_{i2}} \end{align}

This holds for all components $x_1,...,x_n$. Then one has for the curve

\begin{align} X(t) &= \frac{1}{a}u \cdot (e^{A} - a v) &| u = (1/C_{12} ,..., 1/C_{n2})^T \\ &= \frac{1}{a}u e^{A} - uv &| v = (C_{13} ,..., C_{n3})^T \\ &= \frac{1}{a}u e^{A} + w &| w = (-C_{13}/C_{12} ,..., -C_{n3}/C_{n2})^T \\ &= \lambda(t) u + w &| \lambda = \frac{e^{A(t)}}{a(t)} \end{align}

This is the parametrization of a straight line, with $w$ beeing an arbitrary point on the line, $u$ a vector parallel zu the line. $\lambda(t)$ is a paramter, depending on which the curve is run through.

By parametrization to constant speed $X'(t) = T$ ($T_i = k_i =const.$ in all components) the acceleration vector $X''(t) = (T)' = 0$ vanishes. Such a parametrization can be done for any curve, for which the prove was given in the lecture.

I am glad if somebody could confirm or confute my result, thanks.