Drop of water sliding on a surface (curve of *steepest descent* ) I've always heard that if you put a water droplet on a surface, it will slide following the curve of steepest descent (a curve perpendicular everywhere to the surface level curves). The usual explanation of this fact is given by means of Vector Calculus. 
Recently I've been learning about fluid dynamics and a little about surface phenomena. Now I know that for a real droplet, the dynamics of its motion on an arbitrary surface is much more difficult to analyze. 
I want to know if there is a way to show that for some particular conditions(maybe some approximation) of the droplet and the surface, so that one can show that the droplet will slide following the curve of steepest descent. Some references are welcome.
 A: The following argument assumes the droplet is a point particle as mentioned by @m3tro in his comment.  I think your question also assumed a point particle because you ask about "...THE curve of steepest descent...".  Were it not a point particle, it would straddle an infinite number of curves.
There are two forces acting on the droplet at any location on the surface.


*

*The force due to gravity.  This force is perpendicular to the surface level curve at the droplet location by definition.

*The normal force acting between the droplet and the surface.  This force is also perpendicular to the surface level curve at the droplet location because the force is perpendicular to the surface (by definition) and the surface level curve is on the surface.
So we have two forces acting on the droplet, both perpendicular to the surface level curve.  Therefore, the resultant force is also perpendicular to the surface level curve.  Therefore, the acceleration of the droplet must also be perpendicular to the surface level curve and along the curve of steepest descent.
You have specified that there is no inertia effect.  I assume that this means the droplet has no "memory" of its current motion and will always move in the direction of instantaneous acceleration.  This will also be the case when the speed of the droplet is very, very small.
I think this proves that the droplet will move along the curve of steepest descent in the absence of inertia.  With inertia, the path will be dependent on the speed and therefore dependent on the "friction" force of the droplet on the surface.
The paper When Does Water Find the Shortest Path Downhill? The Geometry of Steepest Descent Curves Seems to support my conclusion, although I don't have access to the full paper.  In a footnote to the paper preview it says...

The steepest descent model is not accurate for all situations.  One problem is that it accounts for potential energy, but ignores kinetic energy.  When water flows quickly, it tends to keep flowing in the same direction, even if it has to go somewhat uphill to do that.

A: I will attempt to prove or disprove the claim using analytical mechanics:

A droplet of water will follow the steepest path of descent along any curved surface.

There are a number of assumptions I will take:


*

*Friction does not apply

*The droplet is of a small enough size that I can approximate its motion using a point mass


Using Vector Calculus, the path of steepest descent of any particle on any surface given by a function of two variables will have a velocity vector that is proportional to the negative gradient at any point. In other words
$$\dot{\bf{r}}=-k\nabla f$$
This equation is our basis for proving the statement. If our equations of motion match this, then your statement is true. I will use Cartesian Coordinates for simplicity; our equation becomes:
$$\dot{x}=-\frac{\partial f(x,y)}{\partial x}, \space \dot{y}=-\frac{\partial f(x,y)}{\partial y}$$
$$z=f(x,y)$$ The statement for z is holonomic constraint that is more appropriately written as:
$$\lambda (f(x,y)-z)=0$$
($\lambda$ is a lagrange multiplier). We are now prepared to obtain our equations of motion. The Lagrangian is 
$$L=T-U$$
where
$$T = \frac 1 2 m (\dot{x}^2+\dot{y}^2+\dot{z}^2)$$
and
$$U=m g z$$
Let's add $\lambda (f(x,y)-z)$ to our Lagrangian (in other words $0$).
$$L=\frac 1 2 m (\dot{x}^2+\dot{y}^2+\dot{z}^2))+m g z+\lambda (f(x,y)-z)$$
Now we can use the Euler-Lagrange Equations to get the equations of motion.
$$\frac {\partial{L}} {\partial{q_i}}- \frac{d}{dt} \frac {\partial{L}} {\partial{\dot{q_i}}}=0$$
Plugging in our Lagrangian and coordinates ($x$,$y$,$z$,$\lambda$) we get:
$$-m \ddot{x}+\lambda \frac{\partial{f(x,y)}}{\partial{x}}=0, \space -m \ddot{y}+\lambda \frac{\partial{f(x,y)}}{\partial{y}}=0$$ 
in other words,
$$\ddot{\bf{r}}=\frac{\lambda}{m}\nabla f$$
In summary, this means that the droplet will accelerate along the path of steepest descent, but it will not necessarily travel the path of steepest descent.
A: Certainly, it is possible to show that for some particular conditions the droplet will slide following the curve of steepest descent. One just has to make sure that the surface and the initial position of the droplet have the right symmetries. For example, a droplet on an inclined plane will certainly fall along the curve of steepest descent (if it has no initial velocity). So will a droplet placed in the "middle" of an inclined cylindrical pipe. In general, if the surface has a mirror symmetry plane that is perpendicular to the ground, and you place the droplet at the intersection curve between this symmetry plane and the surface, it will follow this intersection curve, which is also the curve of steepest descent unless it is flat. (This intersection curve will be either the top of a "mountain ridge" or the bottom of a "valley". I assume that a droplet placed on top of a "mountain ridge" won't break into two droplets)
Added example: If you'd rather think in more mathematic terms, imagine a surface given by
$$z = f(x,y)$$
where $f(x,y)$ is an even function of $x$. If you place the droplet so that its center of mass is at $x=0$, it will flow down the curve of steepest descent given by $x=0$. See e.g. this valley or this ridge.
A: Let's call the formula of the hill the drop is traveling on $z=f(x,y)$. 
In addition, let's approximate the drop using a point mass. To further simplify the problem, let's ignore friction. 
The net force on the particle is the sum of the normal force and the point's weight. 
We can find a vector parallel to the normal vector by computing the following cross product: 
$$\left(1,0,\frac{\partial f}{\partial x}\right)\times\left(0,1,\frac{\partial f}{\partial y}\right)=\left(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1\right)$$
Let's call this vector $N$. Then, the unit normal vector is 
$$\frac{1}{\left|\left|N\right|\right|}\left(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1\right)$$
So, the normal force, $F_n$, is
$$\frac{\left|\left|F_n\right|\right|}{\left|\left|N\right|\right|}\left(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1\right)$$
We already know that the point's weight is $$\left(0,0,-mg\right)$$
So, the net force is 
$$\left(-\frac{\left|\left|F_n\right|\right|}{\left|\left|N\right|\right|}\frac{\partial f}{\partial x},-\frac{\left|\left|F_n\right|\right|}{\left|\left|N\right|\right|}\frac{\partial f}{\partial y},\frac{\left|\left|F_n\right|\right|}{\left|\left|N\right|\right|}-mg\right)$$
We can see that the projection of this vector on the $xy$ plane is pointing in the same direction as the negative gradient vector, which is 
$$-\nabla f(x,y)=\left(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y}\right)$$
This implies that the point accelerates towards the path of steepest descent.
