I am reading Do Carmo's Differential Geometry of Curves and Surfaces, and a passage on page $18$ is confusing me.
He writes:
In what fallows we shall restrict ourselves to curves $\alpha(s)$ parametrized by arc length without singular points of order $1$. We shall denote $t(s) = \alpha'(s)$, the unit tangent vector of $\alpha$ at $s$. Thus $t'(s) = k(s)n(s)$ (where $k(s)$ is the curvature of $\alpha$). The unit vector $b(s) = t(s) \times n(s)$ is normal to the osculating plane and will be called the binormal vector at $x$. Since $b(s)$ is a unit vector, the length of $|b'(s)|$ measures the rate of change of the neighboring osculating planes with the osculating plane at $s$; that is, $b'(s)$ measures how rapidly the curve pulls away from the osculating plane at $s$ in a neighborhood of $s$.
I see that because $t(s), b(s)$ are both of norm $1$ and perpendicular that their cross product has length $1$. I am not sure why since $b(s)$ is a unit vector, that the length $|b'(s)$| mesasures how rapdidly the curve pulls away from the osculating plane at $s$. How does the norm of $b'(s)$ play into things?
If I look at what $b'(s)$ is, well its $\frac{d}{dt} (t(s) \times n(s)) = \frac{dt}{ds}\times n(s) + t(s) \times \frac{dn}{ds} $, and well from here I'm not really sure how to proceed. Any insights appreciated.
