Radius of Curvature in Parametric Cartesian Form

Definition

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

Let $C$ be embedded in a cartesian plane and defined by the parametric equations:

$\begin{cases} x = \map x t \\ y = \map y t \end{cases}$

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

$\rho = \size {\dfrac {\tuple {x'^2 + y'^2}^{3/2} } {x' y - y' x} }$

where:

$x' = \dfrac {\d x} {\d t}$ is the derivative of $x$ with respect to $t$ at $P$

$y' = \dfrac {\d y} {\d t}$ is the derivative of $y$ with respect to $t$ at $P$

$x$ and $y$ are the second derivatives of $x$ and $y$ with respect to $t$ at $P$.

$\size {\, \cdot \,}$ denotes the absolute value function.

$\kappa = \dfrac {x' y - y' x} {\tuple {x'^2 + y'^2}^{3/2} }$

where:

$x' = \dfrac {\d x} {\d t}$ is the derivative of $x$ with respect to $t$ at $P$

$y' = \dfrac {\d y} {\d t}$ is the derivative of $y$ with respect to $t$ at $P$

$x$ and $y$ are the second derivatives of $x$ and $y$ with respect to $t$ at $P$.

By definition, the radius of curvature $\rho$ is given by:

$\rho = \dfrac 1 {\size \kappa}$

where $\kappa$ is the curvature, given in parametric Cartesian form as:

$\kappa = \dfrac {x' y - y' x} {\tuple {x'^2 + y'^2}^{3/2} }$

As $\tuple {x'^2 + y'^2}^{3/2}$ is positive, it follows that:

$\size {\dfrac {x' y - y' x} {\tuple {x'^2 + y'^2}^{3/2} } } = \dfrac {\size {x' y - y' x} } {\tuple {x'^2 + y'^2}^{3/2} }$

Hence the result.

$\blacksquare$