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Math Help - Classifying Critical Points when 2nd Derivative test FAILS

  1. #1
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    Question Classifying Critical Points when 2nd Derivative test FAILS

    I'm having some trouble classifying critical points when the 2nd derivative test fails.

    The function is,

    f(x,y) = x^4 - 3x^2y^2 + y^4

    The only critical point I've found is (0,0).

    The second derivative test fails for this point, so I've got to classify it another way.

    I can try to get a rough idea of what the surface looks like by drawing cross sections, and maybe I can get a good enough picture to classify my point.

    Set x=0,

    z = y^4

    Set y=0,

    Z = x^4

    So at the bottom sits my critical point and I have to parabolas in the y-z plane and x-z plane.

    This makes me think my point may be a relative min.

    How can I figure out whether all values of Z are positive or not?

    Any ideas/suggestions?
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  2. #2
    Senior Member roninpro's Avatar
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    Wait - the second derivative test seems to work here. You have to calculate D(0,0)=f_{xx}(0,0)^2 f_{yy}(0,0)^2-(f_{xy}(0,0))^2=-4. Whenever D<0, the point in question is a saddle point.

    Classifying Critical Points when 2nd Derivative test FAILS-plot.pdf
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  3. #3
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    Quote Originally Posted by roninpro View Post
    Wait - the second derivative test seems to work here. You have to calculate D(0,0)=f_{xx}(0,0)^2 f_{yy}(0,0)^2-(f_{xy}(0,0))^2=-4. Whenever D<0, the point in question is a saddle point.

    Click image for larger version. 

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    I keep finding that,

    D(0,0)=f_{xx}(0,0)^2 f_{yy}(0,0)^2-(f_{xy}(0,0))^2=0.

    Since,

    f_{xx} = 12x^{2} -6y^{2}

    f_{xy} = -12xy

    f_{yy} = -6x^{2} + 12y^{2}

    What am I doing wrong?
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  4. #4
    Senior Member roninpro's Avatar
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    Your partial derivatives are fine. My mistake!

    Since your second derivative test fails, you might want to see what happens to some paths containing (0,0). You can show that taking the path x=0 gives a function increasing away from (0,0). On the other hand, taking the path y=x gives a function decreasing away from (0,0). Therefore, you must be on a saddle.
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  5. #5
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    Quote Originally Posted by roninpro View Post
    Your partial derivatives are fine. My mistake!

    Since your second derivative test fails, you might want to see what happens to some paths containing (0,0). You can show that taking the path x=0 gives a function increasing away from (0,0). On the other hand, taking the path y=x gives a function decreasing away from (0,0). Therefore, you must be on a saddle.
    This makes perfect sense, the only part I'm confused about is what made you choose y = x in particular? What thought process are you going through to obtain this? Hopefully it's not merely a guess!
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  6. #6
    Senior Member roninpro's Avatar
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    I actually had the benefit of a computer-generated contour plot to see which paths to take. If you were to do this purely by hand, you would probably have to stare at your function and toy with different paths and see what comes out. I would imagine that the situation becomes much more difficult when you are actually sitting on a maximum or minimum.
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