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

  • #1
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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

  • #2
Math_QED
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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?
 
  • #3
40
1
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.
 
  • #4
pasmith
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To write [itex]z = g(x,y)[/itex] you need [itex]z[/itex] to be uniquely determined by [itex](x,y)[/itex]. In other words, [itex]f(x,y,z) = C[/itex] must have a unique solution for [itex]z[/itex] given [itex](x,y)[/itex]. This can be guaranteed by insisting that [itex]\frac{\partial f}{\partial z} >0[/itex] so that given [itex](x,y)[/itex] there is at most one [itex]z[/itex] such that [itex]f(x,y,z) = C[/itex].
 
  • #5
Ray Vickson
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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
[tex] dz = -\frac{f_z}{f_z} dx - \frac{f_y}{f_z} dy. [/tex]
This is a partial differential equation to determine ##z## in terms of ##x## and ##y##.
 

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