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Math Help - Intersection of functions --> vector function

  1. #1
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    Intersection of functions --> vector function

    I'm a bit confused as to how to approach this problem, the only piece of information I can decipher is the statement at the end of this post, if anyone can shed some light, I'd greatly appreciate it. Here's the question...

    A circular cylinder of radius 1 and the parabolic cylinder z = 2x^2 - 1 intersect as shown in the attached image.

    Which vector function has the curve of intersection as its graph if x(0) = 1.

    The answer choices are...

    (cost)i + (sint)j + (cos2t - 1)k
    (cost)i + (sint)j + (cos2t)k
    (cost)i + (sint)j + (1 - cos2t)k
    (sint)i + (cost)j + (cos2t - 1)k
    (sint)i + (cost)j + (cos2t)k
    (sint)i + (cost)j + (1 - cos2t)k


    I'm under the impression that since x(0) has to equal 1, then i must be cost, since cos(0) = 1. If that is correct that eliminates three of the six choices.
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  2. #2
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    Quote Originally Posted by ocryan View Post
    I'm a bit confused as to how to approach this problem, the only piece of information I can decipher is the statement at the end of this post, if anyone can shed some light, I'd greatly appreciate it. Here's the question...

    A circular cylinder of radius 1 and the parabolic cylinder z = 2x^2 - 1 intersect as shown in the attached image.

    Which vector function has the curve of intersection as its graph if x(0) = 1.

    The answer choices are...

    (cost)i + (sint)j + (cos2t - 1)k
    (cost)i + (sint)j + (cos2t)k
    (cost)i + (sint)j + (1 - cos2t)k
    (sint)i + (cost)j + (cos2t - 1)k
    (sint)i + (cost)j + (cos2t)k
    (sint)i + (cost)j + (1 - cos2t)k


    I'm under the impression that since x(0) has to equal 1, then i must be cost, since cos(0) = 1. If that is correct that eliminates three of the six choices.
    Yes and which of the remaining three choices satisfies z = 2x^2-1?
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