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Homework Help: Level curves

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to sketch level curves of [itex]T(x, y) = 50(1 + x^2 + 3y^2)^{-1}[/itex] and [itex] V(x, y) = \sqrt{1 - 9x^2 -4y^2}[/itex]

    3. The attempt at a solution

    Is it correct that they are ellipses?

    ie [tex] 1 = \frac{9}{1 - c^2} x^2 + \frac{4}{1 - c^2}y^2[/itex]

    for V(x, y) = c = constant
    I feel so rusty going back to school :s
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2


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    That's for V? Setting [itex]V(x,y)= \sqrt{1- 9x^2- 4y^2}= c[/itex], then [itex]1- 9x^2- 4y^2= c^2[/itex], [itex]9x^2+ 4y^2= 1-c^2[/itex],
    [tex]\frac{9}{1- c^2}x^2+ \frac{4}{1-c^2}y^2= 1[/tex]
    just as you say. Yes, that's an ellipse. It might be easier to recognise if you wrote it
    [tex]\frac{x^2}{\left(\frac{\sqrt{1-c^2}}{3}\right)^2}+ \frac{y^2}{\left(\frac{\sqrt{1-c^2}}{2}\right)^2}= 1[/tex]
    an ellipse with center at (0,0) and semi-axes of length
    [tex]\frac{\sqrt{1-c^2}}{3} [/tex]

    Similarly, [itex]T(x,y)= 50(1+ x^2+ 3y^2)^{-1}= c[/itex] gives [itex]c(1+ x^2+ 3y^2)= 50[/itex] so [itex]1+ x^2+ 3y^2= 50/c[/itex], [itex]x^2+ 3y^2= (50/c- 1)[/itex]. Now divide both sides by 50/c- 1:
    [tex]\frac{x^2}{50/c-1}+ \frac{y^2}{\frac{50/c-1}{3}}= 1[/tex]
    again, an ellipse with center at (0,0), semi-axes of length
    [tex]\sqrt{50/c- 1}[/tex]
    [tex]\sqrt{\frac{50/c- 1}{3}}[/tex]
  4. Sep 11, 2007 #3
    That does make it easier to understand.

    Thanks for your help.
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