$R(t) \cdot R'(t) = 0$, which is what every source I can find tells me. Even though I understand the proof I don't understand the underlying concept. If $R(t)\cdot R'(t) = 0$, then $R'(t)$ is orthogonal to $R(t)$, right?
But you use the same derivative to find the tangent of a curve. Then somehow if you differentiate the tangent itself, you get the normal to the curve.
I really can't wrap my head around this. Could someone help me understand?
Best Answer
I'm going to assume that you mean to ask why the derivative of a fixed length vector is perpendicular to the vector itself. Here's the idea:
$$\vec{r}(t)\cdot\vec{r}'(t) = \frac{1}{2}\frac{d}{dt}(\vec{r}(t)\cdot\vec{r}(t)) = 0.$$
If you want to think about why it is true, think about an object rotating on a circle around the origin. Draw some diagrams to figure out what the position and velocity vectors are. Position points outward radially if we imagine the center of the circle is the origin of the $xy$ plane. Velocity is the rate of change of position. If you consider the average velocity over some interval of time, it would be a secant vector (if you want to call it that). As you take a limit, the secant vector will become a vector tangent to the circle. See the images below. (Note that there are length considerations for $v_{\text{avg}}$ that I am ignoring, but that's not super important for this heuristic argument.)