Is there anything more I can say about a vector function that is parallel to its derivative?

Question

The problem is to find the set of curves in $\mathbb{R}^3$ given by a vector equation $\mathbf{r}(t)$ with the property that the vector $\mathbf{r}'(t)$ is parallel to $\mathbf{r}(t)$ for all $t$ in the domain. My solution is that we must have $\mathbf{r}'(t)=f(t)\mathbf{r}(t)$, and this implies that $\mathbf{r}(t)=F(t)\mathbf{r}_0$ for some $\mathbf{r}_0\in\mathbb{R}^3$ and differentiable $F(t)$. In other words, they are curves traced by the vector function $\mathbf{r}(t)=\langle c_1F(t),c_2F(t),c_3F(t) \rangle$ for constants $c_1,c_2,c_3$. My question is, is there anything more descriptive I can say about these curves? Is there a name for these kinds of curves?

Answer 1

As you have a variable scalar $F(t)$ times a fixed vector ${\bf r}_0$, the curve will lie along a straight line through the origin. If the range of $F$ is all of $\Bbb R$ it will be the whole line, if not then it will be only a part of the line such as a ray or a segment.

Answer 2

Assuming $\dot{r}(t) \ne 0$, for any $t$, as $\dot{r}(t)= f(t) r(t)$ one have that the $\ddot{r}(t) = ({f}^2(t)+\dot{f}(t))r(t)$ and therefore $\dot{r}(t)\times \ddot{r}(t)=0$. So the $\kappa=0$, i.e. the curvature is zero and the curve is a straight line.