TITLE: Why is the derivative of a vector orthogonal to the vector itself?
QUESTION [24 upvotes]: $R(t) \cdot R'(t) = 0$, 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?

REPLY [4 votes]: Perhaps extended comment... I just find the answer here to be short and sweet:
A vector of constant magnitude will be orthogonal to its derivative because:
$$r(t)\cdot r(t) = \Vert r(t)\Vert^2 = C$$ 
for some $C$ independent of $t.$ Therefore, differentiating with respect to $t$ we obtain
$$2 r(t) \cdot r'(t) = 0$$