Show that a curve $r(t)$ is a straight line if $r'(t)$ and $r''(t)$ are linearly dependent for all $t$.

Question

Show that a curve $r=r(t)$ of class $C^m \; (m\geq 2)$, where $t$ is arbitrary, is a straight line if $r'(t)$ and $r''(t)$ are linearly dependent for all $t$.

So if $r'$ and $r''$ are linearly dependent, that means that there exists a real number $a$ so that $r'(t)=a\cdot r''(t)$.

If $r$ is parametrised by arc length, we get that $r'(s)=r'(s(t))\cdot s'(t)$, thus we have

$$r''(s)=r''(s(t))\cdot s'(t) + r'(s(t))s''(t)=r''(s)s'(t)+a r''(s)s''(t)$$

From here we have that $r''(s)(1-s'(t)-s''(t))=0$, so by direct integration of $r''(s)=0$ we get that $r(s)=c_1 t+c_2$.

Is my way of proving correct?

Answer 1

Counter example: $r(t) = (1,2,3)$.

Linear dependency: $$ \lambda_1 \dot{r} + \lambda_2 \ddot{r} = 0 \wedge (\lambda_1, \lambda_2) \ne 0 $$

Answer 2

$$r''(t)=\lambda(t)r'(t)$$ is equivalent to

$$\frac{x''(t)}{x'(t)}=\frac{y''(t)}{y'(t)}=\frac{z''(t)}{z'(t)}=\lambda(t),$$ or by logarithmic integration

$$C_xx'(t)=C_yy'(t)=C_yy'(t)=\mu(t).$$

Integrating once again,

$$C_xx(t)+C'_x=C_yy(t)+C'_y=C_zz(t)+C'_z=\nu(t).$$

This is the equation of a straight line traversed at speed $\nu'(t)\sqrt{\frac1{C_x^2}+\frac1{C_y^2}+\frac1{C_z^2}}$.

Answer 3

You can use the curvature, if $r'(t)=ar''(t)$ then for any t : $$\kappa(t)=\frac{||r'(t)\times r''(t)||}{||r'(t)||^3}=\frac{||r'(t)\times ar'(t)||}{||r'(t)||^3}=|a|\frac{||r'(t)\times r'(t)||}{||r'(t)||^3}=0$$ so $r(t)$ is a parametrization of a straight line.