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Sketching Level Curves

  1. Dec 28, 2008 #1
    1. The problem statement, all variables and given/known data
    z=f(x,y)= -[tex]\sqrt{9-2x^2-y^2}[/tex]

    Sketch the level curves for f(x,y)

    2. Relevant equations

    3. The attempt at a solution

    I am really poor at this. I let z = c (a constant) . Substituting c into the equation and rearranging it, I got this 9-c[tex]^{2}[/tex]=2x[tex]^{2}[/tex]+y[tex]^{2}[/tex]

    From this, I know i will get a ellipse. But I am stuck here. How do I find the domain for c and the radius? Some hints will be helpful. Thanks
     
  2. jcsd
  3. Dec 28, 2008 #2

    tiny-tim

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    Hi jeffreylze! :smile:
    Hint: first, put 9 - c2 = a, so it's 2x2 + y2 = a …

    i] what values can a have?

    ii] yes, it's an ellipse, and obviously its centre is at the origin, so just find the points where it intersects the x and y axes. :smile:
     
  4. Dec 28, 2008 #3
    i) Since ellipse has a general equation, (x-h)^2/a^2 + (y-k)^2/b^2 = 1, hence the a you were referring to would have the values >= 0 ? I dont know, I am quite confused.
     
  5. Dec 28, 2008 #4

    HallsofIvy

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    No, a2 must be non-negative but a itself can be any (non-zero) number. Bacause of the symmetry, it really doesn't matter.
     
  6. Dec 28, 2008 #5
    Hence, 9-c^2 >= 0 ?
     
  7. Dec 28, 2008 #6

    tiny-tim

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    Nooo … 9 - c2 = a,

    and the only restriction is that c2 ≥ 0 …

    so a … ? :smile:
     
  8. Dec 28, 2008 #7
    so a[tex]\leq[/tex]9 ? Am i right this time?
     
  9. Dec 28, 2008 #8

    tiny-tim

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    Yup! :biggrin:

    oops … sorry … i forgot what this was all about :redface: … i forgot about the √(9 - 2x2 + y2), so your first answer, a ≥ 0, is also correct: 0 ≤ a ≤ 9.

    And now … find the points where it intersects the x and y axes. :smile:
     
  10. Dec 28, 2008 #9
    but the i still couldn't get the answer -3 ≤ c ≤ 0. This is what i did, for 0 ≤ a ≤ 9, I subbed a= 9 - c^2 into it to get 0 ≤ 9 - c^2 ≤ 9 , rearranged and i got 0≤|c|≤3 . Where did i go wrong?
     
  11. Dec 29, 2008 #10

    tiny-tim

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    Well, basically, you're right,

    but you dropped a condition when you started to use c2 instead of c …

    when you go back from c2 to c, you have to remember that c to c2 is single-valued, but c2 to c is double-valued, and you have to select the right square-root (of c2) …

    in this case, the original equation for f specified the negative square-root …

    so you combine the condition c ≤ 0 with your result 0≤|c|≤3 to give …

    da-daa! … -3 ≤ c ≤ 0. :smile:

    (actually, it would have been a lot quicker, and just as accurate, to say, just by looking, that the square-root of nine-minus-something-positive must be between 0 and 3 :wink:)
     
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