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

[tomkoolen]

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 want to 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

 

[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 want to 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?

 

[tomkoolen]

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]

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]

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.

First of all: it works equally well if $\partial f /\partial z < 0$ instead. All you need is $\partial f /\partial z \neq 0$.

Denote the partial derivatives by subscripts, so that $f_x = \partial f/\partial x$, etc. Geometrically: the gradient vector $\nabla f = (f_x, f_y, f_z)$ is perpendicular to the tangent plane of $f$ at $(x,y,z)$, so if $f_z(x_0,y_0,z_0) = 0$ the tangent plane is vertical (parallel to the $z$-axis) at $(x_0,y_0,z_0)$. That means that $z$ could not be a nice, smooth, single-valued function of $x$ and $y$ in the immediate vicinity of $(x_0,y_0)$.

Alternatively, from $f(x,,y,z) =0$ we have $f_x dx + f_y dy + f_z dz = 0$, so if $f_z \neq 0$ we can divide through by it to get
$$dz = -\frac{f_z}{f_z} dx - \frac{f_y}{f_z} dy.$$
This is a partial differential equation to determine $z$ in terms of $x$ and $y$.