Equivalence of Definitions of Curvature/Whewell Form to Parametric Polar 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 $C$.
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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