What condition is fulfilled by the path of a mass sliding down a lumpy hill? Suppose I have a hill that goes up and down unevenly.  It's frictionless, and I want to slide a point mass down the hill.  I am interested in the path it takes.  (By "path" I mean the trail it leaves behind, not the time-dependent trajectory.)
I can find the equations of motion of the point mass and, from  there, find the path by finding the entire time-dependent motion.  Is there a way to find the path directly, given the initial velocity of the particle?  For example, does the correct path extremize something?
 A: I think you're simply looking for the shortest path between the starting point and the arrival point under the constraint that the path should be on the surface.
So as Mark pointed out, this can't be right. But what is still correct is the principle of least action: 
$$S=\int \; \left[ \frac{1}{2}m v^2 - mgz - \lambda (z-f(x,y)) \right] \; dt $$
I chose a function of the form $z=f(x,y)$ to describe the landscape, but more general forms are possible. 
Extremizing this action leads to the following equations of motion:
$$\begin{align}
m\frac{d^2 x}{dt^2} & = \lambda \frac{\partial f}{\partial x} \\\\

m\frac{d^2 y}{dt^2} & = \lambda \frac{\partial f}{\partial y} \\\\

m\frac{d^2 z}{dt^2} & = -mg -\lambda \\\\

z=f(x,y)
\end{align}$$
A: I think what Raskolnikov says is incorrect (I'd post this as a comment but can't because of low reputation). What he says would be correct in the absence of gravity, so it is just about finding geodesics on some manifold given by your surface.
Your problem is harder, though, because what you are interested in are not geodesics, but trajectories that correspond to non-zero force acting on the particle. I am not sure what more could I write about this, because your question is too general: it is equivalent to solving equations of motion on (almost) arbitrary two-dimensional manifold with (almost) arbitrary potential.
A: If you know the path $\vec{r}(t)$ then you know the acceleration $\vec{a}=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ and from that you get the force needed. $\vec{F}=m\cdot\vec{a}$ .This is also known as a forward dynamics problem.
Given the hill shape $\vec{r}(s)$ where $s$ is a parameter and the
derivatives $\vec{r}'=\frac{\mathrm{d}\vec{r}}{\mathrm{d}s}$, $\vec{r}''=\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}s^{2}}$  
fully describe the hill properties. To follow the hill, the particle
velocity has to be tangent to the hill $\vec{v}=\upsilon\vec{r}'$  
and the acceleration $\vec{a}=\dot{\upsilon}\vec{r}'+\upsilon^{2}\vec{r}''$.
The equations of motion are $\sum\vec{F}=m\,\vec{a}$ and the reaction
forces have to provide no work (or power) making them normal to the
path. Why? $\vec{F}_{\mathrm{reaction}}\cdot\vec{v}=\vec{F}_{\mathrm{reaction}}\cdot\upsilon\vec{r}'=0$  only if $\vec{F}_{\mathrm{reaction}}$ is perpendicular to $\vec{r}'$.  
If you can define the instanteneous tangential and normal unit vectors
at each point $s$ then the equations of motion are decomposed nicely
along those directions allowing to solve for the tangential acceleration
$\dot{\upsilon}$ and the reaction force $F_{\mathrm{reaction}}$.
I am not fully understanding your question of, given a shape (path) what is the path without the equations of motion? You either know the path and get the forces, or you know the forces and get the path.
