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 $P = \left({x, y}\right)$ is given by:


 * $\rho = \dfrac {\left({1 + y'^2}\right)^{3/2} } {\left\lvert{y''}\right\rvert}$

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

Proof
By definition, the radius of curvature $\rho$ is given by:
 * $\rho = \dfrac 1 {\left\lvert{k}\right\rvert}$

where $k$ is the curvature, given in Cartesian form as:
 * $k = \dfrac {y''} {\left({1 + y'^2}\right)^{3/2} }$

As $\left({1 + y'^2}\right)^{3/2}$ is positive, it follows that:
 * $\left\lvert{\dfrac {y} {\left({1 + y'^2}\right)^{3/2} } }\right\rvert = \dfrac {\left\lvert{y}\right\rvert} {\left({1 + y'^2}\right)^{3/2} }$

Hence the result.