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:

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

\(=\)

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

\(\ds \)

\(=\)

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

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

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds \map \kappa t\)

\(=\)

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

\(\ds \)

\(=\)

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

$\blacksquare$

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