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Math Help - Am I doing this right?

  1. #1
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    Am I doing this right?

    Find a vector function that represents the curve of intersection of the two surfaces.

    The cylinder x^2 + y^2 = 4 and the surface z = xy.


    What I do is make z = 0, then x^2 + y^2 = 4 is a circle and
    xy = 0.

    Then I make these equations equal each other.

    x^2 + y^2 - 4 = xy and simplify

    x^2 + y^2 - xy - 4 = 0

    But now I can't seem to complete the square.

    Any help would be appreciated.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    Find a vector function that represents the curve of intersection of the two surfaces.

    The cylinder x^2 + y^2 = 4 and the surface z = xy.


    What I do is make z = 0, then x^2 + y^2 = 4 is a circle and
    xy = 0.

    Then I make these equations equal each other.

    x^2 + y^2 - 4 = xy and simplify

    x^2 + y^2 - xy - 4 = 0

    But now I can't seem to complete the square.

    Any help would be appreciated.
    You can finish this by rotating the coordinate system about the z axis by some angle to eliminate the xy term.

    However, I'm confused about something. Why are you specifically setting z = 0?

    -Dan
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  3. #3
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    I'm setting z = 0 to simplify the expression so I can complete the square. It appears that this is not the way to go. Could you write out in mathematical terms the rotation about the z - axis here for me please. I guess when you deal with surfaces its not the same as curves.
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  4. #4
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    Quote Originally Posted by Undefdisfigure View Post
    Find a vector function that represents the curve of intersection of the two surfaces.

    The cylinder x^2 + y^2 = 4 and the surface z = xy.


    What I do is make z = 0, then x^2 + y^2 = 4 is a circle and
    xy = 0.

    Then I make these equations equal each other.

    x^2 + y^2 - 4 = xy and simplify

    x^2 + y^2 - xy - 4 = 0

    But now I can't seem to complete the square.

    Any help would be appreciated.
    It's easier than you think:

    The curve is the set of points that satisfy x^2 + y^2 = 4 and z = xy. Using parametric equations:

    To satisfy x^2 + y^2 = 4, x = 2 \cos t\, and \,y = 2 \sin t\, (note: clearly other choices are possible).

    Therefore, to satisfy z = xy it is required that \,z = (2 \cos t) \, (2 \sin t) = 2 \sin(2 t).

    Note that -\infty < t < \infty.

    Therefore the vector equation of the curve is:

    r = (\,2 \cos t\,) i + (\,2 \sin t\,) j + (\,2 \sin(2t)\,) k, where -\infty < t < \infty.
    Last edited by mr fantastic; March 22nd 2008 at 05:06 AM. Reason: Added the red stuff
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    x^2 + y^2 - xy - 4 = 0
    Since you asked:
    Let
    x = x^{\prime}~cos(\theta) - y^{\prime}~sin(\theta)
    and
    y = x^{\prime}~sin(\theta) + y^{\prime}~cos(\theta)

    (This represents a rotational change of coordinates.)

    Then
    x^2 + y^2 - xy - 4 = 0
    becomes
    (x^{\prime}~cos(\theta) - y^{\prime}~sin(\theta))^2 + (x^{\prime}~sin(\theta) + y^{\prime}~cos(\theta))^2 -  (x^{\prime}~cos(\theta) - y^{\prime}~sin(\theta))(x^{\prime}~sin(\theta) + y^{\prime}~cos(\theta)) - 4 = 0

    Now expand this and find the angle \theta that makes the x^{\prime}y^{\prime} coefficient 0. (It's a bit of work and more than I wish to type out at the moment.)

    -Dan
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