Length of Arc of Evolute equals Difference in Radii of Curvature

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $C$ be a curve defined by a real function which is twice differentiable.

Let the curvature of $C$ be non-constant.

The length of arc of the evolute $E$ of $C$ between any two points $Q_1$ and $Q_2$ of $C$ is equal to the difference between the radii of curvature at the corresponding points $P_1$ and $P_2$ of $C$.


Proof

CenterOfCurvature.png


Let $P = \tuple {x, y}$ be a general point on $C$.

Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.


From the above diagram:

$(1): \quad \begin {cases} x - X = \pm \rho \sin \psi \\ Y - y = \pm \rho \cos \psi \end {cases}$

where:

$\rho$ is the radius of curvature of $C$ at $P$
$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.


Whether the sign is plus or minus depends on whether the curve is convex or concave.

For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive.

The case for $k < 0$ can then be treated similarly.


Thus we have $k > 0$ and so $(1)$ can be written:

$(2): \quad \begin {cases} X = x - \rho \sin \psi \\ Y = y +\rho \cos \psi \end {cases}$

By definition of curvature:

$k = \dfrac {\d \psi} {\d s}$

and:

$\rho = \dfrac 1 k = \dfrac {\d s} {\d \psi}$


Hence:

\(\ds \rho \sin \psi\) \(=\) \(\ds \dfrac {\d s} {\d \psi} \dfrac {\d y} {\d s}\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\d y} {\d \psi}\)

and:

\(\ds \rho \cos \psi\) \(=\) \(\ds \dfrac {\d s} {\d \psi} \dfrac {\d x} {\d s}\)
\(\text {(4)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\d x} {\d \psi}\)


Differentiating $(2)$ with respect to $\psi$:

\(\ds \dfrac {\d X} {\d \psi}\) \(=\) \(\ds \dfrac {\d x} {\d \psi} - \rho \cos \psi - \dfrac {\d \rho} {\d \psi} \sin \psi\)
\(\text {(5)}: \quad\) \(\ds \) \(=\) \(\ds - \dfrac {\d \rho} {\d \psi} \sin \psi\) from $(3)$

and:

\(\ds \dfrac {\d Y} {\d \psi}\) \(=\) \(\ds \dfrac {\d y} {\d \psi} - \rho \sin \psi + \dfrac {\d \rho} {\d \psi} \cos \psi\)
\(\text {(6)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\d \rho} {\d \psi} \cos \psi\) from $(4)$


Let $S$ be the length of arc of the evolute $E$ from a fixed point $Q_0$ on $E$ to the variable point $Q$ corresponding to $P$.

We have that:

$\d S^2 = \d X^2 + \d Y^2$

and so:

\(\ds \paren {\dfrac {\d S} {\d \psi} }^2\) \(=\) \(\ds \paren {\dfrac {\d X} {\d \psi} }^2 + \paren {\dfrac {\d Y} {\d \psi} }^2\)
\(\ds \) \(=\) \(\ds \paren {- \dfrac {\d \rho} {\d \psi} \sin \psi}^2 + \paren {\dfrac {\d \rho} {\d \psi} \cos \psi}^2\) from $(5)$ and $(6)$
\(\ds \) \(=\) \(\ds \paren {\dfrac {\d \rho} {\d \psi} }^2 \paren {\sin^2 \psi + \cos^2 \psi}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac {\d \rho} {\d \psi} }^2\) Sum of Squares of Sine and Cosine


Choosing the direction of $S$ so that $S$ and $\rho$ increase together, this tells us:

$\dfrac {\d S} {\d \psi} = \dfrac {\d \rho} {\d \psi}$

Integrating with respect to $\psi$ gives:

$S = \rho + c$

where $c$ is a constant.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Length_of_Arc_of_Evolute_equals_Difference_in_Radii_of_Curvature&oldid=740223"