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Math Help - Curve of intersection with 2 surfaces

  1. #1
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    Curve of intersection with 2 surfaces

    Show that the curve with vector equation  r(t)=<2cos^{2} t, sin(2t), 2sint>
    is the curve of intersection of the surfaces and .

    Attempt:

    (1)
    (x-1)^2+y^2=1
    y^2 = 1-(x-1)^2

    (2)

    x^2+y^2+z^2=4

    plug (1) into (2) for y


    x^2+1 -(x-1)^2+z^2=4

    Expanding and canceling


    x^2+1-x^2+2x-1+z^2=4

    (3)
    2x+z^2=4

    vector equation  r(t)=<2cos^{2} t, sin(2t), 2sint>

    (4)
    x= 2cos^{2} t
    y= sin(2t)
    z= 2sint

    plug (4) into (3)

    2(2cos^2(t)) +(2sint)^2=4
    4cos^2(t) +4sin^{2}t=4

    divide out the 4

    \frac{4cos^2(t) +4sin^{2}(t)=4}{4}

    cos^2(t) +sin^{2}(t)=1

    identity
    cos^2(t) +sin^{2}(t) = 1
    therefore 1= 1

    Did i even do this correct?
    If not where did i go wrong?
    Thank you
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  2. #2
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    Athens, OH, USA
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    Re: Curve of intersection with 2 surfaces

    Hi,
    Here are some helpful comments, I hope.

    Curve of intersection with 2 surfaces-mhfcalc9.png
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