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Math Help - stationary points revisited

  1. #1
    Member billym's Avatar
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    stationary points revisited

    I have this function and I am looking for the stationary points:

    <br />
f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6<br />

    I have the two partial derivatives:

    <br />
4x(x^2 + y^2 + 1) = 0,  (E.1)<br />
    <br />
4y(y^2 + x^2 - 1) = 0,  (E.2)<br />

    I have already found the stationary points: (0,0), (0,1) and (0,-1)

    For (E.2) can we set:

    <br />
y^2 + x^2 = 1<br />

    and thus have:

    <br />
( \sqrt{1/2}, \sqrt{1/2} )<br />

    or is this useless because it doesn't satisfy (E.1) ?
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  2. #2
    MHF Contributor
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    Two things:

    1) I'd have to agree that it is of no value because it does not satisfy (E.1)

    2) Why didn't you wonder about the other three really obvious suspects (not to mention the infinitely many others) from that last errant exercise?
    Last edited by TKHunny; May 5th 2008 at 08:56 AM. Reason: Rephrase
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  3. #3
    Member billym's Avatar
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    1. Yeah i figured it was a dumb question, just making sure.

    2. Huh?
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  4. #4
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    Not a dumb question. It shows you were on the right track.

    x^{2} + y^{2} = 1 defines a circle. How many points are there on a circle?
    Last edited by TKHunny; May 5th 2008 at 12:04 PM.
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  5. #5
    Member billym's Avatar
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    oooohh... does that mean (0,0) isn't a solution?
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  6. #6
    Member billym's Avatar
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    or wait no. sorry. whoops.
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  7. #7
    Member billym's Avatar
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    now i am confused
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  8. #8
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    Please be unconfused. You got it right in the first place. It was only talking about the circle that became confusing. Rule out the circle and you're done.

    Of course, (0,1) and (0,-1) were identified before you started playing with the circle, so keep those two.
    Last edited by TKHunny; May 5th 2008 at 12:04 PM. Reason: Expansion
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