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Sketching level curves of f(x,y)

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Sketch the level curve of the surface [tex]z = \frac{x^2 - 2y + 6}{3x^2 + y}[/tex] belonging to height z = 1 indicating the points at which the curves cut the y−axis.


    2. Relevant equations



    3. The attempt at a solution

    I put [tex]1 = \frac{x^2 - 2y + 6}{3x^2 + y}[/tex] but then don't know how to proceed.

    The answer shows an inverted parabola at y = 2, but I don't know how to get that.
     
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  3. Jun 12, 2009 #2

    CompuChip

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    From
    [tex]
    1 = \frac{x^2 - 2y + 6}{3x^2 + y}
    [/tex]
    it follows that
    [tex]
    3x^2 + y = x^2 - 2y + 6
    [/tex]

    Can you cast this in the form
    y = f(x) ?
     
  4. Jun 12, 2009 #3
    Do you mean like this...

    [tex]3x^2 + y = x^2 - 2y + 6[/tex]
    [tex]3y = -2x^2 + 6[/tex]
    [tex]y = \frac{-2x^2 + 6}{3}[/tex]
    [tex]y = \frac{-2x^2}{3} + 2[/tex]

    ???
     
  5. Jun 12, 2009 #4

    CompuChip

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    Yes.
    Now, that's a parabola, isn't it? It's of the general form y = ax^2 + bx + c with a = -2/3, b = 0, c = 2; you can see that it is inverted (like a mountain top) because a < 0.
     
  6. Jun 12, 2009 #5
    Ok, that explanation helped, thankyou.
     
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