# 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.

$\blacksquare$