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Domain and how it relates to a Level Curve

  1. Feb 13, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Find the domain of the function f(x,y) = 1/sqrt(x^2-y). Sketch the level curves f=1, f=1/2, f=1/3.

    2. Relevant equations


    3. The attempt at a solution

    Domain is rather simple to find: x^2>y AND x^2-y cannot equal 0.

    Level curves are also simple to find. They are simply parabolas that are shifted down according to what f =1. For f=1 shift the parabola down 1. For f=1/2 shift the parabola down 4. For f=3 shift the parabola down 9.

    My simple question is this: Upon looking at my level curve graph, I notice that anything inside the y=x^2-1 parabola is NOT consistent with the domain.
    I would then 'mark' this section off in my level curve? (say, darken it out and clarify that it is NOT in the domain).
    or
    Would I leave it as is as it isn't needed in the level curve?

    I can't seem to find a clear quick answer on google. Seems logical that I would clarify that it is not part of the graph.
     
  2. jcsd
  3. Feb 13, 2015 #2

    Mark44

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    More simply, this is y < x2, all the points in the plane that are below the graph of y = x2.
    How are you getting this? The point (0, -1/2) is "inside" the parabola y = x2 + 1, and f(0, -1/2) = ##\frac{1}{\sqrt{0 - (-1/2)}}## is defined.
     
  4. Feb 13, 2015 #3

    BvU

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    No need to go inside. Your level curve is that parabola itself, and it is completely inside the domain (i.e. below y = x2)
     
  5. Feb 13, 2015 #4

    LCKurtz

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    It would be much clearer if you wrote how you got those (correct) statements. When you set ##f(x,y)=c## you have ##\frac 1 {\sqrt{x^2-y}} = c##. This immediately tells you that ##c>0##, there are no level curves for ##c\le 0##, and the simplified equations of the level curves are ##y=x^2-\frac 1 {c^2}##.
     
  6. Feb 13, 2015 #5

    RJLiberator

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    Okay, guys:
    Thank you for the tips. I see that I was wrong to think that anything above y=x^2-1 is out of the domain.

    However, this conversation leads me to believe that y=x^2 is the part that needs to be shaded out, correct?
     
  7. Feb 14, 2015 #6

    Mark44

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    And all of the points inside this parabola. Maybe that's what you meant, but it isn't what you said.
     
  8. Feb 14, 2015 #7

    RJLiberator

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    I thought it was y=x^2-1. I see my error thanks to your observations. I see that it is y=x^2 as the shaded part. :)
     
  9. Feb 14, 2015 #8

    BvU

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    He ho, you do it again: y=x2 is a line, you can't shade that !
     
  10. Feb 14, 2015 #9

    RJLiberator

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    What do you mean? y=x^2 is a parabola and the shaded part is the region above y=x^2.
     
  11. Feb 14, 2015 #10

    BvU

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    Yes, so the wording "I see that it is y=x^2 as the shaded part" keeps triggering folks like Mark and me to point out that that can't be done.

    And you want to shade ##y\ge x^2## although it's difficult to distinguish from ## y > x^2 ## :)

    (Not all lines are straight lines...)
     
  12. Feb 14, 2015 #11

    RJLiberator

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    Ahhhhhh
    The inequality is the one that we are looking for. I understand. Thank you for pointing out the error in my wording.
     
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