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I'm trying to prove the formula to calculate the curvature of a plane curve. But I end up with the wrong sign and can't figure out why:

I want to proof $\kappa(t) = \frac{\dot c(t) \cdot \ddot c^\perp (t)}{\| \dot c \|^3}$

My proof:

Let $\tilde c(s) := c \circ \varphi(s) \Rightarrow c = \tilde c \circ \varphi^{-1}$ be parametrized by the arc length with $\varphi$ orientation preserving. $\Rightarrow \|\dot{\tilde c}\| = 1$

From the definition of curvature follows with rotating both sides by $\frac{\pi}{2}$:

$\ddot{\tilde c} = \tilde k(s) \tilde n(s) = \tilde k(s) \dot{\tilde c}^\perp(s)\\ \Rightarrow \ddot{\tilde c}^\perp = -\tilde k(s) \dot{\tilde c}(s)\\ \Rightarrow \dot{\tilde c}(s) \ddot{\tilde c}^\perp(s) = -\tilde k(s)$

Now I simply substitute:

$ \dot{ c} = \dot{\tilde c} \circ \varphi^{-1} \cdot \dot \varphi^{-1}\\ \ddot{ c} = \ddot{\tilde c} \circ \varphi \cdot (\dot \varphi^{-1})^2 + \dot{\tilde c} \circ \varphi^{-1} \cdot \ddot \varphi^{-1}$

$ \dot c(t) \cdot \ddot c^\perp (t) = \dot{\tilde c}(s) \ddot{\tilde c}^\perp(s) \cdot (\dot \varphi^{-1})^3 = \dot{\tilde c}(s) \ddot{\tilde c}^\perp(s) \cdot \|\dot {\tilde c(s)}\cdot (\dot \varphi^{-1})^3\|\\ \Rightarrow \frac{\dot c(t) \cdot \ddot c^\perp (t)}{\| \dot c \|^3} = - \tilde \kappa(s) = - \kappa(t)$

Can anyone see my error(s)?

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    $\begingroup$ Try applying the formula to a simple case first, such as a circle running clockwise or anticlockwise. That should give you some sense of what may be going wrong. $\endgroup$ – Semiclassical Aug 5 '14 at 15:39
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    $\begingroup$ In the plane, curvature does possess a sign; this does not happen for curves in three dimensions, for example. A curve that traces the standard capital letter S has positive curvature in some places, negative in others. Multiplying the curvature function by $-1$ gives another correct curvature function. You may be looking at incompatible definitions from more than one source. The requirement is just that the absolute value of the reciprocal give the radius of the best approximating circle at that point, when the curvature is nonzero. $\endgroup$ – Will Jagy Aug 5 '14 at 18:04
  • $\begingroup$ Ok, thanks. I just tried it with the unit circle and got $-1$ using the formula that is to be proven. So it seems to be a definition issue. Thanks for the help. $\endgroup$ – user44789 Aug 5 '14 at 18:34
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Hint: See Geometry of Curves and Surfaces of Mantredo P. do Carmo http://www.maths.ed.ac.uk/~aar/papers/docarmo.pdf pag. 25, as was mentioned, the curvature in two dimensions has signal and three not. And so you take the module as above.

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