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:

\(\ds \dfrac {\d t} {\d s}\)

\(=\)

\(\ds \dfrac 1 {\paren {x'^2 + y'^2}^{1/2} }\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\paren {r^2 \theta'^2 + r'2}^{1/2} }\)

\(\ds \dfrac {\d g} {\d t}\)

\(=\)

\(\ds \map {\dfrac \d {\d t} } {\dfrac {y'} {x'} }\)

\(\ds \)

\(=\)

\(\ds \map {\dfrac \d {\d t} } {\dfrac {r' \sin \theta + r \theta' \cos \theta} {r' \cos \theta - r \theta' \sin \theta} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {x' y'' - y' x''} {\paren {x'}^2}\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 r'^2 \theta' + r r' \theta'' - r r'' \theta' + r^2 \theta'^3} {\paren {x'}^2}\)

\(\ds \)

\(=\)

\(\ds \dfrac {r' \paren {r \theta'}' - r r'' \theta' + \paren {r'^2 + \paren {r \theta'}^2} \theta'} {\paren {x'}^2}\)

\(\ds \leadsto \ \ \)

\(\ds \map \kappa t\)

\(=\)

\(\ds \dfrac {x'^2} {s'^2} \cdot \dfrac {2 r'^2 \theta' + r r' \theta'' - r r'' \theta' + r^2 \theta'^3} {\paren {x'}^2} \cdot \dfrac 1 {\paren {r'^2 + \paren {r \theta'}^2}^{1/2} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 r'^2 \theta' + r r'' \theta' + r r' \theta'' + r^2 \theta'^3} {\paren {r'^2 + \paren {r \theta'}^2}^{1/2} }\)

$\blacksquare$

This article needs proofreading. In particular: I found this on a scrap of paper in my notes which I am about to shred. Have no clue about its rigor or accuracy. Please feel free to give this whatever workover is needed. If you believe all issues are dealt with, please remove {{Proofread}} from the code. 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 {{Proofread}} from the code.