@C_R  

I didn't see your message because there was no request to me. I came across this place completely by accident. Click "Reply" and then I will know because the flag in the upper right corner of the page will turn red for me.
b[i] is a way to make Maple do natural parameterization when I use internal procedures. Such actions are much easier to perform if you write the code for solving the ODE yourself.

Yes, that's right: we get successive solutions and do this based on solving the Cauchy problem. That is, we solve a system of nonlinear equations using a numerical solution of the ODE.
In this case, it is the Draghilev method.

Very good, +. I really like pictures like this. And I myself sometimes have fun this way.   
Are there any plans for examples of rolling in 3D and plans for rolling of surfaces too?

@vv  OK, thank you. 

@vv  
I don't know, maybe it's all about the accuracy of the ODE solution, but in all positions of the circle, the fsolve procedure shows, in addition to the point of tangency, the presence of intersection points with the transcendental curve.

@vv 
I don't agree with you, because near your solution, in addition to the point of tangency, there immediately appears a point of intersection of the circle with the curve. After all, I suggested an approach to finding a solution, or not?
I would like it without intersection and with one point of contact.

@vv 
As always, everything is great with you, +. Only I already said that I know this solution. I wanted to get exactly the case when the point of tangency is unique on each curve. 

@mmcdara   @C_R  
What was meant is that each curve touches the circle at only one point. But, of course, formally this is a solution.
Your solution is contained in the set of solutions that are presented in the animation. Here is approximately that same frame.
(The animation itself is done with a rough step.)

It is clear that there may be moments when the circle formally touches both curves, but simultaneously intersects one of them. And in this case the same situation arises.
Let's look at the lower right corner of the overall graph.

To find a solution to this problem, we can, for example, do the following: starting to construct inscribed circles, watch the moment when the inscribed circle crosses the transcendental curve. And as soon as the intersection occurs, fix the previous position of the circle. If you do this with a sufficiently small integration step, then you can talk about a good approximation to the solution.
The coordinates of the inscribed circle are approximately [0.584307201828654, -1.27793596947425], and the maximum radius is somewhere around 0.656227713070106.
IN_EL_SIN_1.mw

@janhardo  
Thank you for your appreciation. If you have any questions, I will always be happy to answer.

@dharr, in fact, this is an underdetermined system of equations: 4 equations and 6 variables.

Maple17, Windows7, immediately (without specifying the range).

Simply an idea for obtaining an approximate result. This is almost a manual solution plus using the same algorithm to reduce the distance between points on the surface.
It seemed to me that this surface is very sly, and simple techniques work very poorly on it. If, as an initial approximation to the solution, we take the section of the surface by the plane 1.470702751*x1-2.941405502*x2-1.470702751*x3 = 0, which passes through the given points and the middle of the edge, then we get a plane curve (of blue color).
 

By replacing the resulting curve with a finite set of segments, we can slightly change its shape, simultaneously turning this curve into a spatial one. The figure shows a place due to which the overall length of the curve is slightly reduced.

For example, if the plane curve had a length of approximately 3.9762, then with the same number of segments the spatial curve would be approximately 3.9409 (the length of the segment of the plane curve is 0.030000, and the spatial one is  from 0.026501 to 0.029995). All this is rather rough, but the decrease in length occurs with any number of segments (within reason, of course).

Edited 31.07.2024

@dharr   This is a numerical test of the blue curve to be geodesic.
(The text is very amateurish.)
WITH_NORMAL_2.mw

@dharr  Thank you very much for checking the length of the curve. I’ll still have to deal with Ruben’s code, since I’m not very familiar with the theory, and Maple is are still a mystery for me.
As for the blue line, this is also a geodesic. This is the simplest solution, but not the shortest curve between points. That, that this is a geodesic can be verified numerically by constructing osculating planes. The normals to the surface at each point of the curve will lie in the osculating  plane.
For example, for the second curve I did such a selective check.(If I find my code, I'll try to do the same for the blue curve. But I think it's obvious.)

@Rouben Rostamian  Yes, great. 
I've been playing around with  this algorithm.
I got two geodesics. One has a length of approximately 3.8437,



 and the short one has a length of approximately 3.7288.