What Happens to z When Taking the Derivative of a Surface Equation?

[Cyrus]

In stewart, page 806 he says:

"In the special case in which the equation of a surface S is of the form z=f(x,y) (that is, S is the graph of a function f of two variables), we can write the equation as

F(x,y,z) = f(x,y) - z = 0

and regard S as a level surface (with k=0) of F. Then

Fx(x0,y0,z0) = fx(x0,y0)
Fy(x0,y0,z0) = fy(x0,y0)
Fz(x0,y0,z0)= -1 "

end quote

I understand his moving z to the other side.
But when the take the derivative W.R.T x, what about the z? z is not a variable, z is a functin of x and y, so why don't u have some dz/dx term in there, Fx(x0,y0,z0)=fx(x0,y0) -dz/dx . How did z no longer become a dependent variable on x and y?

 

[matt grime]

because we are taking a level curve (when F=0), or if you like, F, a function of three variables, and f are completetly different things, F=0 implicitly defines z as a function of x and y (f).

 

[Cyrus]

because we are taking a level curve (when F=0), or if you like, F, a function of three variables, and f are completetly different things, F=0 implicitly defines z as a function of x and y (f).

Right, F=0 is a level surface of three variables, which means that for any value of x,y or z=f(x,y) the function spits out the value zero.

" F=0 implicitly defines z as a function of x and y (f)"

I don't quite know what you mean in this part. z is implicitely defined as a function of x, and y. I don't know what you mean by y(f) though. How does making F(x,y,z) change the fact that z is no longer depended on x any y? I could not figure that out from your explanation, sorry.

 

[matt grime]

What's not to figure out? You appear to have a common theme in your threads of simply leaping to a conclusion and not thinking about it. We have two different ways of thinking about the same thing, that's all. I can either have z as a function of x and y explicitly (your dependent situation) or I can take a *different* function of three (independent) variables and use that to implicitly define z as a function of x and y.

The f in brackets was to indicate how F=0 implicitly defines z as a function of y, ie the function is z=f(x,y).

You are used to doing this for two variables, surely.

Consider the equation

xy+y^2+x=1

this defines a locus in the xy plane. Would you be happy to find me dy/dx from this?

Seeing as you are assigning too much meaning to symbols, cnosider this, let f be a fuction of x and y (look no mention of z) and now define a new function F of three variables F(x,y,z) = f(x,y)-z

now, the level surface F=0 is the set of points x,y,z satisfying f(x,y)=z. I can plot this as a surface.

See, no mention of dependent or indepenent variables, which is all a red herring anyway.

The nature of the z changes exactly because we declare them to be different in the different expressions. Again, just a notational convention that you need to learn. There is no smoke and mirrors going on.

perhaps it would help if you used completely different letters: F(u,v,w)=f(u,v)-w

 

[Cyrus]

Hmmmm, I like your explanation, because through your way, z is has no relation to f(x,y). By setting f(x,y)-z=0, we can still have any value of x or y, but it forces us to limit our scope to value of the (variable) z, so that the equation is satisfied. So in that sense, z is truly a variable now, it can be whatever it wants to be, but the equality forces a restriction on it. Is that an ok way to think about it?

Ok, here is a better question to ask. When we have z=f(x,y), then z is a function of x and y. So z is not an independent variable. If I take the derivative w.r.t x or y, id get dz/dx or dz/dy.

But when I declare F(x,y,z), does that change z from being a dependent variable into a dependent variable, but with a restirction imposed on it, since it must satisfy f(x,y)-z=0?

 

[matt grime]

Gah! Whether or not something is or is not dependent would depend upon what you're doign. IN THIS CASE x,y and z are INDEPENDENT variables of a function F, absolutely indpendent, totally independent.

From this we can create a function z=f(x,y) by NOW taking a level curve and from this we can think of z as a function of the independent variables x and y (z is not a variable now, if you like, dependent or otherwise, z is a function of x and y). this level curve also may define y as a function of x and z or even x as a function of y and z.


"when i declare F(x,y,z)". What does that mean?

 

[Cyrus]

"when i declare F(x,y,z)". What does that mean?

That F is a function of three imput variables, x,y,z which have no dependence on one another.

 

[Hurkyl]

It might help you to realize this:

defining F(x, y, z) = f(x, y) - z

means exactly the same thing as

defining $F(\spadesuit, q, \xi) = f(\spadesuit, q) - \xi$.

 

[Cyrus]

Hey matt. I thought about what you said. Could you help me out with this please.

Lets say we have, a function defined as F(x,y,z) = f(x,y) -z =0

In this situation, x,y,and z are independent variables. So I could pick any value for x and y, but then I am limited in what I can choose for z in order for the equation to equal zero. Would It be incorrect for me to say that I could pick a value for x and z, but then be limited in my choice of y, while still keeping the equation the same, f(x,y)-z=0 , or would I have to rewrite it as f(x,z)-y=0?

 

[matt grime]

No, F(x,y,z)=0 is not a function of 3 independent variables at all. it is an equality. F(x,y,z) is a function of 3 independent vairables.

 

[Cyrus]

I see, could I write it as F(x,y,f(x,y))=0 ?

 

[matt grime]

That would depend what you meant when you wrote it. If you meant a function of three inputs, then no, in fact that makes even less sense now. The whole point of this is that we define a function

F(x,y,z)=f(x,y)-z

that is the function. OK?

then we take the level curve of that function given by requiring F=0.