5
$\begingroup$

Let $r(t)=(x(t),y(t))$ be a curve in the plane. We can find its arc-length using the formula $$\ell=\int \sqrt{x^\prime (t)^2+y^\prime(t)^2}dt=\int\Vert r^\prime(t)\Vert dt.$$

We also can find the curvature of this curve. Denote by $v(t)=r^\prime(t)$ the tangent vector, one can express it as $$v(t)=\exp(i\theta(t))$$ getting equivalently $$\kappa_r(t)=|r^{\prime\prime}(t)|=\bigg\vert \frac{d\theta}{dt}\bigg\vert=\frac{|x'y''-y'x''|}{|r^\prime(t)|^3}.$$

In the special case of $ r(t)=(t,y(t)\, )$ we get the formula

$$\kappa=\dfrac{|y''|}{(1+y^{'2})^{1.5}}$$

That seems like the connection between the arc-length and the curvature has to be very strong. Can one express (even if $r(t)=(t,f(t))$) the curvature in terms of the arc-length?

$\endgroup$
  • 3
    $\begingroup$ Note that curvature cannot be expressed purely in terms of arc length. One way to see this is to note that any two curves of the same length are diffeomorphic to each other by an arc-length-preserving diffeomorphism. The formula given by @DavidQuinn expresses curvature in terms of two quantities -- the arc length and the tangent angle. $\endgroup$ – Jack Lee Sep 13 '15 at 15:36
  • $\begingroup$ For the connection see math.stackexchange.com/questions/1272514/… $\endgroup$ – Michael Hoppe May 15 '18 at 22:10
5
$\begingroup$

Using intrinsic coordinates, yes. Curvature is then $$\kappa=\frac{d\psi}{ds}$$ where $\psi$ is the tangential angle and $s$ is arc length.

$\endgroup$
  • $\begingroup$ Thanks. What do you mean by intrinsic coordinates? Can one also write the signed curvature in intrinsic coordinates? If so: does it have similar expression? $\endgroup$ – Wenchao Shang Sep 13 '15 at 12:24
  • $\begingroup$ Intrinsic coordinates are when the curve is expressed as $s=f(\psi)$. $\kappa$ changes sign according to the shape of the curve (convex to concave). For example, $s=c\tan \psi$ is the equation of a catenary. $\endgroup$ – David Quinn Sep 13 '15 at 12:32
  • $\begingroup$ Thanks. just to verify: $s(\tau)=\int_0^\tau \Vert r^\prime(t)\Vert dt$? $\endgroup$ – Wenchao Shang Sep 13 '15 at 12:37
  • 2
    $\begingroup$ Yes, that's right $\endgroup$ – David Quinn Sep 13 '15 at 12:48
  • $\begingroup$ To add: the Cesàro equation for a curve expresses curvature as a function of arclength, while the Whewell equation for a curve expresses the tangential angle as a function of arclength. $\endgroup$ – J. M. is a poor mathematician Mar 1 '18 at 11:00
2
$\begingroup$

Denoting rotation by $\phi$ and arc by $t$,

$$\kappa= \frac{ d\phi}{dt}$$

By "strong" you mean $ natural, intrinsic, isometrically\, invariant $ etc.These depend upon first fundamental form and are therefore independent of euclidean motions of translation and rotation .. $x,y, \theta$. But all the above three $ \phi,t $ and its derivative $\kappa$ are strong.

Examples of natural equation are $\kappa = 1$ ( unit Circle ) , $\kappa = t $ ( Cornu spiral) $ ,\kappa = \cos \phi $, (the Cycloid), movable anywhere in the $x-y$ plane, constants of integration are Euclidean motions.

In this notation $ \theta,\psi $ (polar coordinate rotation around origin and angle arc to radius vector ) are individually not "strong", but $ \theta + \psi $ is strong.

EDIT 1:

Only now it occurs to me. Instead of strong we could use rigid, as curvature and torsion are invariant in all Euclidean motions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.