Fluid Dynamics, free surface boundary condition derivation. I'm stuck on part of the derivation for a boundary condition for the free surface of a fluid. The fluid concerns a rectangular trough of water whose surface is at a height $$ z=h(x,t) $$ (Problem in 2D)
Now another condition stated earlier (which I understand), states the surface moves with the fluid. The surface given by $ S =z-h(x,t) $.
This means $$\frac{D}{Dt}(z-h)=0$$
Evaluated on the surface $z=h$ , where I'm referring to the material derivative.
Fully expanding this out i get (with subscript derivative notation):
$$z_t-h_t+uz_x+vz_y+wz_z-uh_x-vh_y-wh_z=0$$
Here u,v and w are the x, y and z components of the flow respectively.
The final condition is said to be. 
$$-h_t-uh_x+w=0$$
I can't work out how so many of the derivatives are zero. The derivatives of z and what exactly z refer to seems to be my main area of confusion. Is it the coordinate system itself or the height of the fluid or something else. 
 A: The variables $x$,$y$ and $z$ are Cartesian coordinates and are independent variables -- both spatially and temporally
Thus, 
$$z_t = \frac{\partial z}{\partial t} = 0, \,\,\,z_x = \frac{\partial z}{\partial x} = 0, \,\,\, z_y = \frac{\partial z}{\partial y} = 0, \,\,\, z_z = \frac{\partial z}{\partial z} = 1$$
Since $h$ is a function of only $t$ and $x$, the only nonzero partial derivatives are $h_t$ and $h_x$.
Substituting into your general expression we see that only three nonzero terms remain in the free-surface (kinematic) condition:
$$\frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$
A: $z=h(x,t)$ is the equation of the free surface . each fluid particle on the free surface has two coordinate $x , z$ and since the fluid particle moves with time . clearly $z$ is a function of $x , t$
The first condition on the free surface is called impermeability condition or no slip condition 
$$\frac{dz}{dt}=\frac{dh(x,t)}{dt}=\frac{\partial h}{\partial t} + \frac{d h}{dx}\frac{dx}{dt}$$
which implies that $$w = \ h_t \ + u h_x$$
