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Homework Help: Find the critical points of (x^2+y^2-4)(x+y) and their nature

  1. Aug 7, 2011 #1

    I am studying for my Analysis final and came across this problem I just can't get my head around:
    1. The problem statement, all variables and given/known data
    Find the critical points of [itex]f(x,y) = (x^2+y^2-4)(x+y)[/itex] and their nature.

    2. Relevant equations
    [itex]\vec{\nabla} f(x,y) = \vec{0}[/itex]

    3. The attempt at a solution
    [itex]\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0[/itex]
    [itex]\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0[/itex]
    These are both ellipses with center (0,0) and angles [itex]\frac{5\pi}{8}[/itex] and [itex]\frac{\pi}{8}[/itex] respectively.
    According to Wolfram Alpha, there are four intersection points. (Click here to view) However, I need to solve this problem analytically and I wish to understand how it is done.
    I have tried to eliminate a variable from the equations, but it is not possible due to the [itex]2xy[/itex] term. Another solution would be to write the ellipses in a parametric form, but this I feel would be overly complicated.

    Could anyone please shed some light on this?
    Thanks in advance!
  2. jcsd
  3. Aug 7, 2011 #2


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    The first thing I would do is subtract one equation from the other to get
    [tex]3(x^2- y^2)+ (y^2- x^2)= 2(x^2- y^2)= 0[/tex]
    so that either y= x or y= -x.

  4. Aug 7, 2011 #3


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    Subtract the second equation from the first one and factorise.

  5. Aug 7, 2011 #4
    Thanks for your advice.
    However, this doesn't give me the four points of intersection. If I take the solutions [itex]y=\pm x[/itex], I get the solutions [itex] (0,0) \wedge (\pm \sqrt{2},\pm \sqrt{2})[/itex].
    However, the points [itex] (\pm \sqrt{\frac{2}{3}},\pm \sqrt{\frac{2}{3}})[/itex] are also solutions of the equation. And the point [itex](0,0)[/itex] is not a solution at all.
    How would I get the other two points?
    Thanks in advance!
  6. Aug 7, 2011 #5


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    Your values are not correct. Give the solutions as pairs.
    (√2, √2) is not a solution, but (√2, -√2) is.
    (0,0) is not a solution. How did you get it? If you substitute x=y into one of the original equation, you will get the pair with √(2/3)

    Last edited: Aug 7, 2011
  7. Aug 7, 2011 #6
    I got this:
    [itex]y = \pm x[/itex]
    [itex]x = 0 \vee x=\pm \sqrt{2}[/itex]
    According to the solutions manual, which only states the values, not the method, the solutions are:
    [itex]x = \pm \sqrt{2} \vee x = \pm \sqrt{\frac{2}{3}}[/itex]
  8. Aug 7, 2011 #7
    Oops... I see it now. I used f(x,y) instead of the partial derivatives.
    I will try again!
  9. Aug 7, 2011 #8
    Yes! I got it!
    Thank you for your help, I appreciate it! :D
  10. Aug 7, 2011 #9


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    You see, it was worth to write it again...:smile:

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