# Equivalence of Definitions of Curvature

## Theorem

The following definitions of the concept of Curvature are equivalent:

### Whewell Form

The curvature $\kappa$ of $C$ at a point $P$ can be expressed in the form of a Whewell equation as:

$\kappa = \dfrac {\d \psi} {\d s}$

where:

$\psi$ is the turning angle of $C$
$s$ is the arc length of $C$.

### Cartesian Form

Let $C$ be embedded in a cartesian plane.

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

$\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$

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 $P$.

### Parametric Form

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

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

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 $P$.

## Proof

Consider the curvature of a curve $C$ at a point $P$ expressed as a Whewell equation:

$\kappa = \dfrac {\d \psi} {\d s}$

where:

$\psi$ is the turning angle of $C$
$s$ is the arc length of $C$.

The derivative of the tangent of the turning angle $\psi$ at a point $P$ with respect to $\psi$ is also the derivative of the tangent to $C$ at $P$, again with respect to $\psi$.

That is:

 $\displaystyle \frac {\d} {\d \psi} \tan \psi$ $=$ $\displaystyle \frac {\d} {\d \psi} \frac {\d y} {\d x}$ $\displaystyle \leadsto \ \$ $\displaystyle \sec^2 \psi$ $=$ $\displaystyle \frac {\d y'} {\d \psi}$ Derivative of Tangent Function $\displaystyle \leadsto \ \$ $\displaystyle 1 + \tan^2 \psi$ $=$ $\displaystyle \frac {\d y'} {\d \psi}$ Difference of Squares of Secant and Tangent $\displaystyle \leadsto \ \$ $\displaystyle 1 + y'^2$ $=$ $\displaystyle \frac {\d y'} {\d \psi}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac 1 {1 + y'^2}$ $=$ $\displaystyle \frac {\d \psi} {\d y'}$

We also have that:

 $\displaystyle \d s$ $=$ $\displaystyle \sqrt {\d x^2 + \d y^2}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d s} {\d x}$ $=$ $\displaystyle \sqrt {\paren {\frac {\d x} {\d x} }^2 + \paren {\frac {\d y} {\d x} }^2}$ $\displaystyle$ $=$ $\displaystyle \sqrt {1 + y'^2}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d x} {\d s}$ $=$ $\displaystyle \frac 1 {\paren {1 + y'^2}^{1/2} }$

Then:

 $\displaystyle \kappa$ $=$ $\displaystyle \dfrac {\d \psi} {\d s}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\d \psi} {\d y'} \dfrac {\d y'} {\d x} \dfrac {\d x} {\d s}$ Chain Rule for Derivatives $\displaystyle$ $=$ $\displaystyle \frac 1 {1 + y'^2} y'' \frac 1 {\paren {1 + y'^2}^{1/2} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$

which is the Cartesian form of curvature as required.

$\blacksquare$