# Function f(x,y,z) of three variables becomes z = g(x,y)

Hello everyone,

I have a theoretical calculus question. I am working on a exercise where you have to consider f(x,y,z) and express the variable z as a function of x and y on a certain level surface around a certain (x0,y0,z0).

I found out that the condition for this to be able is that the triple (x0,y0,z0) is such that ∂f/∂z > 0. I am able to do the rest of the exercise because I know the rules as to how to find derivatives of this g(x,y) and other things, but I wanna know exactly why the condition stated above works.

What I can think of myself: There has to be a positive change in f when z changes is the meaning of the condition. I must be overlooking something, because I don't see why this makes z expressable in terms of x and y. If anyone could clarify, thank you very much!

Kind regards,
Tom

## Answers and Replies

member 587159
Hello everyone,

I have a theoretical calculus question. I am working on a exercise where you have to consider f(x,y,z) and express the variable z as a function of x and y on a certain level surface around a certain (x0,y0,z0).

I found out that the condition for this to be able is that the triple (x0,y0,z0) is such that ∂f/∂z > 0. I am able to do the rest of the exercise because I know the rules as to how to find derivatives of this g(x,y) and other things, but I wanna know exactly why the condition stated above works.

What I can think of myself: There has to be a positive change in f when z changes is the meaning of the condition. I must be overlooking something, because I don't see why this makes z expressable in terms of x and y. If anyone could clarify, thank you very much!

Kind regards,
Tom

No need to write in bold. Could you maybe give the question?

Sorry I didn't see I was writing in bold.
The question is: Why does the condition ∂f/∂z > 0 make g(x,y) exist on the level surface? I don't understand the theory behind it.

pasmith
Homework Helper
To write $z = g(x,y)$ you need $z$ to be uniquely determined by $(x,y)$. In other words, $f(x,y,z) = C$ must have a unique solution for $z$ given $(x,y)$. This can be guaranteed by insisting that $\frac{\partial f}{\partial z} >0$ so that given $(x,y)$ there is at most one $z$ such that $f(x,y,z) = C$.

Ray Vickson
$$dz = -\frac{f_z}{f_z} dx - \frac{f_y}{f_z} dy.$$