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:
| \(\ds y'\) | \(=\) | \(\ds \dfrac {\d y} {\d x}\) | is the derivative of $y$ with respect to $x$ at $P$ | |||||||||||
| \(\ds y' '\) | \(=\) | \(\ds \dfrac {\d^2 y} {\d x^2}\) | 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.
$\blacksquare$