@vv 
Yes, the last option is the solution.

@C_R 
"Does this mean that the trajectory is given? By that, do you mean the trajectory of the contact point?"
Yes, of course. These are just graphs that appear in turn, reminding  of kinematics.
A small curved torus (or paraboloid) is tied with a set of points on its surface (intersection line), this set of points rolls along another set of points (trajectorie) on a stationary surface. In each frame, the equation of the moving surface is transformed based on the position of the points, and we get another graph and, accordingly, a frame.

@Rouben Rostamian  
You explained much better than me, and in a few words, how the algorithm works.
And special thanks to you for the  "Драгилев".

@Rouben Rostamian  
In several of my old posts there was a proposal to apply the Method to solving the inverse kinematics of manipulators (if you want, I will try to find all these messages). There we proceed as follows: we impose additional connections on the links of the manipulator. We transform it "mathematically" into a spatial lever mechanism with one degree of freedom. After which the inverse problem is solved automatically. And we choose the type of constraints from the simplest ones, so that the working point can easily move along its trajectory.
Here we use a similar technique for rolling a figure on a surface. For example, we take a small torus, intersect it with a plane. In the same way, we obtain a trajectory on a stationary surface, intersecting it with some surface. We make the step along both intersection curves the same there and there. We tie the equation of the small torus to the points of the section line and roll the curve along the curve. But we do not show these curves, we only show the graphs of our surfaces. The plane of the points of the section curve on the small torus is directed approximately along the fixed curve in the osculating plane of this fixed curve and is additionally corrected by the normal to the fixed surface at each point. We can say that we leave it 1 degree of freedom, which is what we see in the figure.
(I keep using the letter "h" in his last name, but now I don't know if it's right or not.)

@C_R  
Did you like it? Thank you, I thought a lot on the algorithm. In fact, nothing rolls in my algorithm. The rolling surface is in its initial state before each new frame. Then we look at where its current location on the trajectory should be, based on the very first point of the trajectory,
we apply Euler angles with an offset, and after memorizing the corresponding graph, we send the rolling surface to its initial state. And from which side to roll, we determine using the normal to the stationary surface.
Honestly, I have not yet encountered such 3D animations with surfaces, and it would be interesting to see rolling surfaces without slipping performed by MapleSim.
In any case, these are just toys. The only thing I will draw your attention to is that the Dragнilev method is also used here.☺

And just a little bit more

Adding images on this topic

@acer 
Are you talking about G1[iii]? I couldn't think of anything else. After all, each frame produces a new equation, and it needs to be displayed on the graph. And even then, the animation is shown after 2 frames due to resource limitations.

@Rouben Rostamian  No, it's much simpler. This is computational geometry. A set of points is rigidly attached to a curved (small) torus, and a trajectory is laid out on the large torus, along which the small torus supposedly rolls. At each point of the trajectory, calculations are made so that the small torus imitates rolling without slipping. This is done using Euler angles: we recalculate the coordinates of the attached points, and then, based on this, we transform the equation of the small torus and draw another graph of motion.
I'll never be able to handle physics equations.

@Rouben Rostamian  
I thought  if we move along the graph line itself, then the proximity between the roots has a weaker effect on the situation with their possible omission. That is, we are talking about the values ​​y(s) and x(s), where s is a natural parameter (arc length). And then we can, for example, use a cycle "to count" harmonics based on their average length. And each time start counting from the last point on the curve. Then we should equally successfully pass "empty" ones, where y does not change sign, and places with a cluster of roots. In this case, most likely, it is necessary to additionally clarify each last point on the line at each step of the supposed cycle. 

(edited)
No, it looks like it won't be more reliable than your proposal. To move along the curve, an unacceptable number of significant digits would be required. "My" method is already very poorly controlled beyond 10^5

@Rouben Rostamian  

I don't need all the solutions of the cluster. I want to make sure that at a significant distance, for example, at a distance of 10^5 - 10^6 from the origin, not a single root out of about 50 roots in a row is guaranteed to be skipped. I'm just comparing with the capabilities of that very method of continuation by parameter. Here, I recently picked up an example. I'm already embarrassed that I attracted so much attention to myself, but it is difficult for me to satisfy my curiosity on my own due to my knowledge of the package.

@Rouben Rostamian  
You understood everything correctly. The only thing is, I wanted to see not the closest, in your words, cluster, but the most distant one, where it is still possible to find solutions in a row.
(Well, I use this very translator.☺)

@mmcdara  
Everything is correct, only if the following condition is met: at least 50 roots in a row (not a single missing root) in its environment.

@mmcdara  
Would I really suggest this example if the solutions were to be sought at such a "significant" distance from 0? ☺ It is need to find not the first 52 solutions, but the last ones that you can. 
By the way, if you decide to try again, please consider the capabilities of Maple 17.

Confirmation of problems with the Schatz mechanism.