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Level surfaces of a function

  1. Feb 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Describe the level surfaces of f(x,y,z) = z + sqrt(x^2 + y^2)

    3. The attempt at a solution

    First of all, what is actually a level surface? Just a normal surface in space?

    I followed an example I found on the internet, and this is my attempt at a solution:

    First replace f(x,y,z) with a constant

    k = z + sqrt(x^2 + y^2)

    Then square (k is now another constant)

    k = z^2 + x^2 + y^2

    This is an ellipsoid, so the level surfaces are ellipsoids centered at the origin.

    Is this the right solution? If so, is it possible to say more about the ellipsoids?
     
  2. jcsd
  3. Feb 7, 2007 #2

    matt grime

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    The definition of a level surface of function of f(x,y,z) is the solutions to f(x,y,z)=k for a constant k.

    Now, please don't tell me that you think (a+b)^2=a^2+b^2, as you wrote above....
     
  4. Feb 7, 2007 #3
    OK, so

    k = z^2 + 2sqrt(x^2+y^2) + x^2 +y^2

    then.

    Is this one easy to recognize as a 3D-figure?
     
    Last edited: Feb 7, 2007
  5. Feb 7, 2007 #4

    Dick

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    Get the radical all by itself on one side of the equation before you square. It will be MUCH easier to recognize.
     
  6. Feb 7, 2007 #5
    Then i get x^2 + y^2 - z^2 = k^2 - 2kz

    Still doesn't resemble anything I'm familiar with.
     
  7. Feb 7, 2007 #6

    Dick

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    Leave it as x^2+y^2=(k-z)^2. x^2+y^2 is often called r^2, right? So write this as r=|z-k|. Can you describe it now?
     
  8. Feb 7, 2007 #7
    A sphere, isn't is?
     
  9. Feb 7, 2007 #8

    Dick

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    Nope. Look, r is the radius in the xy plane, right? If r=0 then z=k. What is z for r=1? Note k-z must be positive. Why?
     
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