What does $r'(t) ⋅r''(t)=0$ mean geometrically?

Question

I am trying to understand dot and cross products from a physics perspective. If a space curve $r(t)$ satisfies the equation $r'(t)×r''(t)=0$ for all $t$, I understand that the derivative of $r'(t)$ is parallel to $r'(t)$, so the velocity vector $r'(t)$ does not change direction. Thus, this curve moves along a line.

However, I'm not sure what $r'(t) ⋅r''(t)=0$ means. My intuition is that it represents motion along a circle, or part of a circle, since the velocity and acceleration vectors should be perpendicular in that case.

Any guidance is greatly appreciated!

Answer 1

Observe the identity $$\frac{d}{dt}|r'(t)|^2=2r'(t)\cdot r''(t)$$

Thus, $r'(t)\cdot r''(t)\equiv 0$ if and only if the motion $r$ has constant speed.

P.S. The circle you mentioned is somewhat relevant to this, since the velocity vector $r'(t)$ would lie on a fixed circle centered at the origin.

P.S.(again) Actually, $r'(t)\times r''(t)\equiv 0$ does not imply that $r(t)$ lies on a fixed line in general. You may consider $r$ that stalls at a point, stays at that point for a while(say 1 second), then changes direction, and re-accelerates. -- Still, your intuition is correct if we add condition that $r'(t)$ is never zero. (the proof is clear; the notion of direction you mentioned now makes sense)

Answer 2

You can integrate both sides:

$$ \int_{t=0}^{t=t_f} \vec{r} \cdot \dot{\vec{r} } dt = 0$$

But, $ \frac{d \vec{r} \cdot \vec{r} }{dt} = 2 ( \vec{r} \cdot \dot{\vec{r}})$:

$$ [\vec{r} \cdot \vec{r}]_{t=0}^{t=t_f} =0$$

Or,

$$\vec{r} \cdot \vec{r}|_{t_f} = \vec{r} \cdot \vec{r}|_{t=0} = C$$

Meaning that the length of vector is constant. If the length of vector is constant, we can imagine it to be spinning about a origin. A physics example would be to keep origin at the center of a circle and track how a particle spins around a circular loop centered at the same origin.