Vector Analysis (Parametized curve)

Question

The question is find a familiar parameterized curve that has the property $r(t) \times\dfrac{dr}{dt}=0$. The only curve that I can see that works is the line through the origin. I was just wondering is there others?

Answer 1

Let $r=(x,y,z)$, then $$ r\times \dot{r}=0\Longleftrightarrow \left|\begin{matrix}x&x'& i\\y&y'& j\\ z&z'&k\end{matrix}\right|=0 $$ or equivalently $$ xy'=x'y,\quad xz'=x'z,\quad yz'=y'z. $$ Assuming that $x(t)\ne 0$, in some interval $I=(t_1,t_2)$, and setting $a(t)=x'(t)/x(t)$, you obtain $$ y'=a(t)y, \quad z'=a(t)z, $$ hence $$ y(t)=\exp\Big(\int_{t_0}^t a(s)\,ds\Big)y(t_0), \quad z(t)=\exp\Big(\int_{t_0}^t a(s)\,ds\Big)z(t_0), \quad t\in (t_1,t_2) $$ and of course $$ x(t)=x(t)/x(t_0) \times x(t_0)=\exp\Big(\int_{t_0}^t a(s)\,ds\Big)x(t_0). $$ Thus $$ \big(x(t),y(t),z(t)\big)=\exp\Big(\int_{t_0}^t a(s)\,ds\Big)\big(x(t_0),y(t_0),z(t_0)\big), $$ i.e., $r=(x,y,z)$ is on a straight line passing from the origin.