Radius of Curvature in Cartesian Form

Theorem
Let $C$ be a curve defined by a real function which is twice differentiable.

Let $C$ be embedded in a cartesian plane.

The radius of curvature $\rho$ of $C$ at a point $P = \tuple {x, y}$ is given by:


 * $\rho = \dfrac {\paren {1 + y'^2}^{3/2} } {\size {y''} }$

where:
 * $y' = \dfrac {\d y} {\d x}$ is the derivative of $y$ $x$ at $P$
 * $y'' = \dfrac {\d^2 y} {\d x^2}$ is the second derivative of $y$ $x$ at $P$.

Proof
By definition, the radius of curvature $\rho$ is given by:
 * $\rho = \dfrac 1 {\size \kappa}$

where $\kappa$ is the curvature, given in Cartesian form as:
 * $\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$

As $\paren {1 + y'^2}^{3/2}$ is positive, it follows that:
 * $\size {\dfrac {y} {\paren {1 + y'^2}^{3/2} } } = \dfrac {\size {y} } {\paren {1 + y'^2}^{3/2} }$

Hence the result.