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)?
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.