# Equivalence of Definitions of Curvature/Whewell Form to Cartesian Form

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## 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 $P$.

## 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:

\(\ds \frac {\d} {\d \psi} \tan \psi\) | \(=\) | \(\ds \frac {\d} {\d \psi} \frac {\d y} {\d x}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \sec^2 \psi\) | \(=\) | \(\ds \frac {\d y'} {\d \psi}\) | Derivative of Tangent Function | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 1 + \tan^2 \psi\) | \(=\) | \(\ds \frac {\d y'} {\d \psi}\) | Difference of Squares of Secant and Tangent | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 1 + y'^2\) | \(=\) | \(\ds \frac {\d y'} {\d \psi}\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac 1 {1 + y'^2}\) | \(=\) | \(\ds \frac {\d \psi} {\d y'}\) |

This article, or a section of it, needs explaining.In particular: Clarify the relationship between the tangent and turning angle. The latter is still inadequately defined.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

We also have that:

\(\ds \d s\) | \(=\) | \(\ds \sqrt {\d x^2 + \d y^2}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {\d s} {\d x}\) | \(=\) | \(\ds \sqrt {\paren {\frac {\d x} {\d x} }^2 + \paren {\frac {\d y} {\d x} }^2}\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \sqrt {1 + y'^2}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d s}\) | \(=\) | \(\ds \frac 1 {\paren {1 + y'^2}^{1/2} }\) |

Then:

\(\ds \kappa\) | \(=\) | \(\ds \dfrac {\d \psi} {\d s}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {\d \psi} {\d y'} \dfrac {\d y'} {\d x} \dfrac {\d x} {\d s}\) | Chain Rule for Derivatives | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {1 + y'^2} y'' \frac 1 {\paren {1 + y'^2}^{1/2} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {y''} {\paren {1 + y'^2}^{3/2} }\) |

which is the Cartesian form of curvature as required.

$\blacksquare$