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Thread: 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 $\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>$
    is the curve of intersection of the surfaces and .

    Attempt:

    (1)
    $\displaystyle (x-1)^2+y^2=1$
    $\displaystyle y^2 = 1-(x-1)^2$

    (2)

    $\displaystyle x^2+y^2+z^2=4$

    plug (1) into (2) for y


    $\displaystyle x^2+1 -(x-1)^2+z^2=4$

    Expanding and canceling


    $\displaystyle x^2+1-x^2+2x-1+z^2=4$

    (3)
    $\displaystyle 2x+z^2=4$

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

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

    plug (4) into (3)

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

    divide out the 4

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

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

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

    Did i even do this correct?
    If not where did i go wrong?
    Thank you
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  2. #2
    MHF Contributor
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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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