Multivariable function notations

[medwatt]

Hello,
I've been having some trouble getting some notations straight and hence my question.
Usually when I see f(x,y) it means to me there is some variable z produced for any combination of x and y in the domain of the function. So, z=x^2+y^2 I imagine as a paraboloid.
So z=f(x,y) ... is a function of three variables with x and y being independent . . . but sometimes I see u=F(x,y,z)=f(x,y)-z . . . all of a sudden it seems as though z has suddenly become independent and there are now four variables.
I hope someone can clarify my mishap in reading functions properly or provide me with a source where I can read all various ways functions are written.
Thank You...

EDIT: Reason why I raised this question.
I have a function z=ln(xy^2).
1. First consideration . . .
If z=f(x,y) then : F(x,y,z)=f(x,y)-z
Hence grad(F)=<1/x,2/y,-1> . . . At point(1,1,0) . . . grad(F)=<1,2,-1> . . . where grad(F) is a vector normal to the surface at that point
2. Second consideration . . .
If z=f(x,y) then I can also do . . grad(z)=<1/x,2/y> . . . At point(1,1,0) . . . grad(F)=<1,2> . . . which is also normal to the surface with a rising rate of modulus(<1,2>)

So I interpreted the problem in two ways and I had two seemingly similar representations of normal vectors to the curve with one having an extra term that I cannot interpret.

 

[Whovian]

So z=f(x,y) ... is a function of three variables with x and y being independent

Nope, it's an equation in 3 variables. We really don't care too much about what's a function of what here.

u=F(x,y,z)=f(x,y)-z . . . all of a sudden it seems as though z has suddenly become independent and there are now four variables.

Some clearing up should be done by explaining what exactly we mean by a graph. A graph of some equation (not always z=f(x,y) for some f) is the set of all points (x,y,z) that satisfy the equation. Once we clear that up, it shouldn't really matter what's dependent on what. It's just convention that we usually try to write z=f(x,y).

EDIT: Reason why I raised this question.
I have a function z=ln(xy^2).
1. First consideration . . .
If z=f(x,y) then : F(x,y,z)=f(x,y)-z
Hence grad(F)=<1/x,2/y,-1> . . . At point(1,1,0) . . . grad(F)=<1,2,-1> . . . where grad(F) is a vector normal to the surface at that point
2. Second consideration . . .
If z=f(x,y) then I can also do . . grad(z)=<1/x,2/y> . . . At point(1,1,0) . . . grad(F)=<1,2> . . . which is also normal to the surface with a rising rate of modulus(<1,2>)

The second is $\nabla f$ in two-dimensional space. It's normal to the curve $f\left(x,y\right)=C$, where C is a constant, not to the surface. The first is $\nabla\left(f\left(x,y\right)-z\right)$, where z is some new variable, independent of x and y (unless we're only looking at the gradient when we're on the surface,) which is our third variable if we're taking the gradient in three-dimensional space.

 

[medwatt]

If ∇f is normal to a curve and the curve lies on the surface . . . shouldn't that mean the vector is also normal to the surface because after all a vector only acts at a point ?

 

[Whovian]

If ∇f is normal to a curve and the curve lies on the surface . . . shouldn't that mean the vector is also normal to the surface because after all a vector only acts at a point ?

Only if the surface is vertical, for instance, we could consider the plane $x=z$. For given z, the gradient of the function f, where $z=f\left(x,y\right)$, is $\left\langle1,0\right\rangle$, which sticks out from the plane, but not perpendicular to it.

Another thing to note is that $\left\langle1,0\right\rangle$ doesn't really mean anything in a three-dimensional example. So, for this to make sense, we'd "cast" it (to use the programming term) to $\left\langle1,0,0\right\rangle$ for three-dimensional problems. Note that this is just an arbitrary choice of "cast" that's the most intuitive.