What is the intuition behind the unit normal vector being the derivative of the unit tangent vector? I've seen the math, but... It just doesn't make sense to me. How is the slope going to point perpendicular to the vector that is clearly a straight line not going in that direction?
Take some point $P_1 = C(s_1)$ on a curve $C$. Let $T_1$ be the unit tangent at this point. Now move a tiny distance along the curve, to get to the point $P_2 = C(s_2)$, and let $T_2$ be the unit tangent at this point. Now draw the vector $V = T_2 - T_1$. If $P_1$ and $P_2$ are very close together, $T_1$ and $T_2$ will be almost equal, and $V$ will be roughly perpendicular to both of them (because the triangle with sides $T_1$ and $T_2$ will be almost isosceles and very skinny). Here's a picture.

But, by the definition of derivative,
$$
\frac{dT}{ds} = \lim_{s_2 \to s1} \frac{T_2 - T_1}{s_2 - s_1}
$$
and this limit will also be roughly perpendicular to $T_1$.
This can be very confusing. A simple explanation is as follows: The (principle) unit Normal vector (dT/dt) is always orthogonal to the curve and, therefore, also to the unit Tangent vector T, and is contained in the vector subspace defined by the two unit vectors T and N. The acceleration of a particle on a curve is partioned between the unit Tangent and the unit Normal vector. Thus, acceleration = kT + mN. For example, a stright line has k=0, and there is  no acceleration in the normal direction. However, the unit Normal vector to the line is still well defined. The point is that the unit Normal vector is part of a coordinate system that may be used to define a component of acceleration.