Curvature of a curve

Remarks

Example: the circle $\gamma(t) = (r \cos t, r \sin t)$ with $r > 0$ has $\norm{\gamma'(t)} = r$ and $T(t) = (-\sin t, \cos t)$, so $\norm{T'(t)} = 1$ and $\kappa(t) = 1/r$. Non-example: for $\gamma(t) = (t^2, t^3)$ the formula is not defined at $t = 0$, because $\gamma'(0) = 0$; the curve is not regular there. The curvature does not depend on the regular parametrisation chosen. In $\R^3$ it is given by $\kappa = \norm{\gamma' \times \gamma''} / \norm{\gamma'}^3$, where $\times$ is the cross product. For a plane curve $\gamma = (x, y)$ one also uses the signed curvature $$k = \frac{x' y'' - y' x''}{(x'^2 + y'^2)^{3/2}},$$ which is positive where the curve turns left; then $\kappa = \abs{k}$.