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In Riemannian Geometry: A Beginner's Guide on pg. 5, Chap. 2, the curvature, [tex]\kappa[/tex], isdefinedas 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.)

[tex]\overrightarrow{\kappa} \equiv \f{d\overrightarrow{T}}/{ds}[/tex]

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

[tex]\kappa \doteq |\f{d\overrightarrow{n}}/{ds}|[/tex]

I don't understand how it can be both.

Souldn't [tex]\kappa \propto |\overrightarrow{n}|[/tex] ?

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

Thanks in advance.

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# Curvature ~ dT/ds or dn/ds?

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