On finding geodesics of an arbitrary surface. I have almost no knowledge about geodesics neither differential geometry, but I've read about it and I know some of its ideas. Because of this, I was trying to derive the parametric equation of a particle moving on any surface in 2D. I'm just seeking for advice to see if my "experiments" make sense.
For the things I have read about the topic, I know the particle acceleration is perpendicular to the surface and not in any other direction.
Procedure:
I start by finding two arbitrary basis vectors tangent to the surface:
$
\vec{e_1} = \vec{i} + \frac{\partial f}{\partial x}\vec{k} \\
\vec{e_2} = \vec{j} + \frac{\partial f}{\partial y}\vec{k}
$
Find a perpendicular vector to the surface (gradient):
$
\frac{\mathrm d^2 \vec{r}}{\mathrm d t^2} \propto \nabla \vec{F} \\
\nabla \vec{F} = \vec{e_1} \times \vec{e_2} = -\frac{\partial f}{\partial x}\vec{i} -\frac{\partial f}{\partial y}\vec{j} + \vec{k}
$
Negating the gradient should give us the acceleration vector we were seeking for:
$
\frac{\mathrm d^2 \vec{r}}{\mathrm d t^2} = -\nabla \vec{F} \\
\frac{\mathrm d^2 \vec{r}}{\mathrm d t^2} = \frac{\partial f}{\partial x}\vec{i} +\frac{\partial f}{\partial y}\vec{j} - \vec{k}
$
I don't know if this has anything to do with the actual formulation of geodesics, but I've tried to simulate it and it make sense, I get the results I was expecting.
The thing is, it is possible to solve this differential equation and find the parametric equation of the trajectory that a particle would follow? And for any particular function like: $f(x, y) = R x y$ or $f(x, y) = -\frac{R}{\sqrt{x^2 + y^2}}$ (where R is a constant)? If so, how would I get the equation?
If there is something I'm missing or I have wrong, feel free to correct me.
Edit: I know this is not the rigorous formulation of geodesics, that requires the metric tensor, it is just for learning the concepts and to try to learn a little on my own.
 A: There is a very simple way to construct a geodesic (but not between two end point you want), take a surface, take sticky tape and unroll the sticky tape along the surface, cut it off after you get a geodesic of required length. Boom you got a geodesic.
The things you really need to characterize geodesic is something known as curvature. Curvature intuitively is like how much you need to steer your car to keep on the curve. We can split total curvature of a geodesic into two parts:
$$ \kappa= \kappa_g + \kappa_n$$
The $\kappa_g$ and $\kappa_n$ magnitude's are the  projection of the acceleration vector into the tangent plane and a plane spanned by normal of the surface and tangent of the curve at that pointrespectively for a particle moving at unit speed along said curve.
For a geodesic, $\kappa_g=0$ meaning there is only acceleration along the normal direction as we move along the curve on the surface. This means physically that if we could peel the geodesic curve of the surface, we could paste it onto a paper as a flat line.
Explanation based on Tristan Needham's Visual differential geometry, above section is based of first chapter and chapter-11.
note: $\kappa_g$ and $\kappa_n$ are vectors
