Connection between curvature and arc-length

Question

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?

Answer 1

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

Answer 2

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.