Curvature ~ dT/ds or dn/ds?

1. Jun 29, 2010

Living_Dog

curvature ~ dT/ds or dn/ds??

In Riemannian Geometry: A Beginner's Guide on pg. 5, Chap. 2, the curvature, $$\kappa$$, is defined as the rate of change of the tangent vector, T, along the curve, x(t), where 't' is a parameter along the curve. (A physical analogy is the tangential velocity and centripetal acceleration in circular motion.)

$$\overrightarrow{\kappa} \equiv \f{d\overrightarrow{T}}/{ds}$$

However, then the author says that the magnitude of the curvature is equal to the magnitude of the rate of change of the normal.

$$\kappa \doteq |\f{d\overrightarrow{n}}/{ds}|$$

I don't understand how it can be both.

Souldn't $$\kappa \propto |\overrightarrow{n}|$$ ?

IOW, the magnitude of a vector, in this case the curvature, is the magnitude of the rate of change of the tangential vector. No?