# Equivalence of Definitions of Curvature/Whewell Form to Parametric Polar Form

## 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 $C$.

### Parametric Polar Form

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

$\begin{cases} r = \map r t \\ \theta = \map \theta t \end{cases}$

The curvature $\kappa$ of $C$ at a point $P = \polar {r, \theta}$ is given by:

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

where:

$r' = \dfrac {\d r} {\d t}$ is the derivative of $r$ with respect to $t$ at $P$
$\theta' = \dfrac {\d \theta} {\d t}$ is the derivative of $\theta$ with respect to $t$ at $P$
$r''$ and $\theta''$ are the second derivatives of $r$ 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 $C$.

Let us consider $C$ expressed in cartesian form:

 $\ds x$ $=$ $\ds r \cos \theta$ $\ds y$ $=$ $\ds r \sin \theta$

Then:

 $\ds \dfrac {\d x} {\d t}$ $=$ $\ds r' \cos \theta - r \theta' \sin \theta$ Chain Rule for Derivatives, Derivative of Cosine Function $\ds \dfrac {\d y} {\d t}$ $=$ $\ds r' \sin \theta + r \theta' \cos \theta$ Chain Rule for Derivatives, Derivative of Sine Function

In Whewell form:

 $\ds \tan \psi$ $=$ $\ds \dfrac {y'} {x'}$ $\ds s^2$ $=$ $\ds x'^2 + y'^2$ $\ds$ $=$ $\ds r^2 \theta'^2 + r'2$ $\ds \leadsto \ \$ $\ds \map \kappa t$ $=$ $\ds \dfrac {\map \arctan {\dfrac {y'} {x'} }' } {s'}$

Let:

 $\ds g$ $=$ $\ds \dfrac {y'} {x'}$ $\ds$ $=$ $\ds \tan \psi$ $\ds \leadsto \ \$ $\ds \dfrac \d {\d \psi}$ $=$ $\ds \dfrac {\d g} {\d \psi}$ $\ds \leadsto \ \$ $\ds \dfrac {\d \psi} {\d g}$ $=$ $\ds \dfrac 1 {1 + g^2}$ $\ds$ $=$ $\ds \dfrac {x'^2} {x'^2 + y'^2}$

We have:

$\map \kappa t = \dfrac {\d \psi} {\d g} \dfrac {\d g} {\d t} \dfrac {\d t} {\d s}$

Then:

 $\ds \dfrac {\d t} {\d s}$ $=$ $\ds \dfrac 1 {\paren {x'^2 + y'^2}^{1/2} }$ $\ds$ $=$ $\ds \dfrac 1 {\paren {r^2 \theta'^2 + r'2}^{1/2} }$

 $\ds \dfrac {\d g} {\d t}$ $=$ $\ds \map {\dfrac \d {\d t} } {\dfrac {y'} {x'} }$ $\ds$ $=$ $\ds \map {\dfrac \d {\d t} } {\dfrac {r' \sin \theta + r \theta' \cos \theta} {r' \cos \theta - r \theta' \sin \theta} }$ $\ds$ $=$ $\ds \dfrac {x' y'' - y' x''} {\paren {x'}^2}$ $\ds$ $=$ $\ds \dfrac {2 r'^2 \theta' + r r' \theta'' - r r'' \theta' + r^2 \theta'^3} {\paren {x'}^2}$ $\ds$ $=$ $\ds \dfrac {r' \paren {r \theta'}' - r r'' \theta' + \paren {r'^2 + \paren {r \theta'}^2} \theta'} {\paren {x'}^2}$ $\ds \leadsto \ \$ $\ds \map \kappa t$ $=$ $\ds \dfrac {x'^2} {s'^2} \cdot \dfrac {2 r'^2 \theta' + r r' \theta'' - r r'' \theta' + r^2 \theta'^3} {\paren {x'}^2} \cdot \dfrac 1 {\paren {r'^2 + \paren {r \theta'}^2}^{1/2} }$ $\ds$ $=$ $\ds \dfrac {2 r'^2 \theta' + r r'' \theta' + r r' \theta'' + r^2 \theta'^3} {\paren {r'^2 + \paren {r \theta'}^2}^{1/2} }$

$\blacksquare$