Radius of Curvature in Cartesian Form

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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 $y$ with respect to $x$ at $P$
$y' ' = \dfrac {\d^2 y} {\d x^2}$ is the second derivative of $y$ with respect to $x$ 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.

$\blacksquare$


Sources

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