Connection between cesaro equation and polar coordinates

[Jhenrique]

First, I'd like that you read this littler article (http://mathworld.wolfram.com/NaturalEquation.html). The solution given by Euler that coonects the system cartesian (x, y) with the curvature κ of the "cesaro system" (s, κ), is that the derivative of the cartesian tangential angle φ* wrt arc length s results the curvature κ.

However, in the 2D plane is definied the polar tangential angle ψ* too. Thus if I want express a curve given s and κ in terms of r and θ (or vice-versa) I need to establish a connection between the curvature κ and the polar system. So, would be correct to say that the derivative of ψ wrt s is equal to κ?

*
https://en.wikipedia.org/wiki/Tangential_angle
https://en.wikipedia.org/wiki/Subtangent
https://en.wikipedia.org/wiki/List_...rtesian_coordinates_from_Ces.C3.A0ro_equation

EDIT:

 

[Chestermiller]

In polar coordinates, the unit tangent to a curve that is parameterized as r(s) and θ(s) is given by:

$$\vec{u}=\frac{dr}{ds}\vec{i}_r+r\frac{dθ}{ds}\vec{i}_θ$$

So, the curvature vector is given by$$κ\vec{n}=\frac{d\vec{u}}{ds}=\left(\frac{d^2r}{ds^2}-r(\frac{dθ}{ds})^2\right)\vec{i}_r+\left(r\frac{d^2θ}{ds^2}+2\frac{dr}{ds}\frac{dθ}{ds}\right)\vec{i}_θ$$
where $\vec{n}$ is the unit normal to the curve.

 

[Jhenrique]

I didn't understood how your answer answers my ask...

 

[Chestermiller]

I didn't understood how your answer answers my ask...

I must have misunderstood your question. I thought you were just interested in getting an equation for the curvature in terms of the parameterized r(s) and θ(s). Apparently not.

Chet