Curvature defined as k(t)=(norm(T'(t)))/(v(t)) , explanation of the proof. I have problems to understand how it's derived the formula for the curvature k(t)=(norm(T'(t)))/(v(t)). The derivation starts with the equality (dT)/(ds) (ds)/(dt)=(dT)/(dt) where T is the unit velocity vector for the curve and s is the arc length function for the curve. So (dT)/(ds)=(T′(t))/(v(t)) and from here it's all clear to me how to finish. Initially I supposed it was just an application of chain rule, but if I rewrite the proof in Netwon's notation I obtain [T(s(t))]′=T′(s(t))s′(t)=T′(s(t))v(t) So to get the final equality I should have [T(s(t))]′=T′(t), but is it always the case ? I don't know if it is useful the fact that T(s(t)) and T(t) have the same trajectory (in general they differ for a c

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