Curvature of a plane curve in Cartesian coordinates Let $\alpha:I\to R^2, t\mapsto (x(t),y(t))$ be a plane curve, not neccessarily parameterized by arc length.
We want to compute its signed curvature, defined at each point as the signed curvature of its arc length reparameterization.
We want to compute the curvature of $\alpha$ in function of $x,y$ and their derivatives.
So I thought that I would take $\beta = \alpha \circ \phi$, where $\phi$ is an arc reparameterization.
Then $\beta'=\frac{\alpha' \circ \phi}{|\alpha' \circ \phi|}$.
But my joy ends there, since trying to compute the second derivative $\beta''$ is nightmarish. What is a clean way of proceeding from here?
 A: 
Use the notation $\dot{\beta}$ for derivative of $\beta$ with respect to arc length $s$ and $\alpha'$ for the derivative of $\alpha$ with respect to parameter $t.$  Then
$$
  \dot{\beta}(s) = \dot{(\alpha \circ \phi)}(s) = \alpha'(t) \dot{\phi}(s). \nonumber
$$
Since $\phi(s) = t,$ $\dot{\phi} = dt/ds = 1/(ds/dt) = 1/|\alpha'|.$  So
$$
  \dot{\beta}(s) = \alpha'(t) / |\alpha'(t)|. \tag{1}\label{ref1}
$$

We can use a trick to make computing the second derivative easier: multiply both sides
of \eqref{ref1} by $|\alpha'(t)|$ so that we can use the product rule instead of the quotient rule to differentiate.  So \eqref{ref1} becomes
$$
  \dot{\beta}(s)(|\alpha'|\circ\phi)(s) = (\alpha'\circ\phi)(s). \tag{2}\label{ref2}
$$
When we differentiate \eqref{ref2}, we will need to differentiate $(|\alpha'|\circ\phi)(s).$  To do this, write $(|\alpha'|\circ\phi)(s) = (\sqrt{\alpha' \cdot \alpha'} \circ \phi)(s)$ (where the dot between $\alpha'$ and $\alpha'$ denotes vector dot
product).  Then we can easily compute the derivative:
$$
  \frac{d}{ds}(|\alpha'| \circ \phi) = \frac{\alpha'' \cdot \alpha'}{|\alpha'|}\dot{\phi} = \frac{\alpha'' \cdot \alpha'}{|\alpha'|^2}.
$$


Now we can compute the second derivate of $\beta$ with respect to $s$ by 
differentiating \eqref{ref2}:
$$
  \ddot{\beta}|\alpha'| + \dot{\beta}(\alpha'' \cdot \alpha')/|\alpha'|^2 = 
  \alpha'' \dot{\phi} = \alpha''/|\alpha'|.
$$

Solving for $\ddot{\beta}$ yields 
$$
  \ddot{\beta} = \frac{\alpha''|\alpha'| - \alpha'(\alpha'' \cdot \alpha')/|\alpha'|}
                {|\alpha'|^3} \tag{3}\label{ref3}.  
$$


Equation \eqref{ref3} is valid for arbitrary dimensional curves.  The squared norm of the numerator of \eqref{ref3} is $(\alpha''\cdot \alpha'')(\alpha'\cdot \alpha') - (\alpha'\cdot \alpha'')^2.$  In two dimensions, this is the perfect square $(x''y' - x'y'')^2.$

