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  "body": "\n## Definition\n\nLet $n \\ge 2$, let $I \\subseteq \\R$ be an interval, and let $\\gamma \\colon I \\to \\R^n$ be a [[regular|06JMR2XWXW]] [[differentiable curve|C57T562NKB]] of class $C^2$.\nLet $T(t) = \\gamma'(t) / \\norm{\\gamma'(t)}$ be its unit [[tangent vector|0D98771SM0]].\nThe *curvature* of $\\gamma$ at $t \\in I$ is\n$$\\kappa(t) = \\frac{\\norm{T'(t)}}{\\norm{\\gamma'(t)}}.$$\nIf $\\gamma$ is parametrised by [[arc length|00R1JDDV5X]] $s$, this reads $\\kappa(s) = \\norm{\\gamma''(s)}$.\n\n## Remarks\n\nExample: the circle $\\gamma(t) = (r \\cos t, r \\sin t)$ with $r \u003e 0$ has $\\norm{\\gamma'(t)} = r$ and $T(t) = (-\\sin t, \\cos t)$, so $\\norm{T'(t)} = 1$ and $\\kappa(t) = 1/r$.\nNon-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.\nThe curvature does not depend on the regular parametrisation chosen.\nIn $\\R^3$ it is given by $\\kappa = \\norm{\\gamma' \\times \\gamma''} / \\norm{\\gamma'}^3$, where $\\times$ is the [[cross product|0SEDGXG9S3]].\nFor a plane curve $\\gamma = (x, y)$ one also uses the *signed curvature*\n$$k = \\frac{x' y'' - y' x''}{(x'^2 + y'^2)^{3/2}},$$\nwhich is positive where the curve turns left; then $\\kappa = \\abs{k}$.\n",
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