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Taking the derivative

  1. Aug 2, 2005 #1
    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 dont 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?
     
  2. jcsd
  3. Aug 2, 2005 #2

    matt grime

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    becuase we are taking a level curve (when F=0), or if you like, F, a fucntion of three variables, and f are completetly different things, F=0 implicitly defines z as a function of x and y (f).
     
    Last edited: Aug 2, 2005
  4. Aug 2, 2005 #3
    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 dont quite know what you mean in this part. z is implicitely defined as a function of x, and y. I dont 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 explination, sorry.
     
  5. Aug 2, 2005 #4

    matt grime

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    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 fucntion 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 differnt 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
     
    Last edited: Aug 2, 2005
  6. Aug 2, 2005 #5
    Hmmmm, I like your explination, becuase 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 truely 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?
     
    Last edited: Aug 2, 2005
  7. Aug 2, 2005 #6

    matt grime

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    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 fucntion 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 fucntion 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?
     
  8. Aug 2, 2005 #7
    "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 dependance on one another.
     
  9. Aug 2, 2005 #8

    Hurkyl

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    It might help you to realize this:

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

    means exactly the same thing as

    defining [itex]F(\spadesuit, q, \xi) = f(\spadesuit, q) - \xi[/itex].
     
  10. Aug 4, 2005 #9
    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?
     
  11. Aug 4, 2005 #10

    matt grime

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    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.
     
  12. Aug 4, 2005 #11
    I see, could I write it as F(x,y,f(x,y))=0 ?
     
  13. Aug 4, 2005 #12

    matt grime

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    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 fucntion

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

    that is the fucntion. OK?

    then we take the level curve of that fucntion given by requiring F=0.
     
    Last edited: Aug 4, 2005
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