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Intersection of surface and plane.

  1. Sep 20, 2010 #1
    Parameterizing vector function for intersection of cylinder and plane

    1. The problem statement, all variables and given/known data

    Problem asks us to find the vector function of the curve which is created when the plane y= 5/2 intersects the ellptic cyl. (x^2)/4 + (z^2)/6 = 5



    2. Relevant equations




    3. The attempt at a solution

    I know its going to be an ellipse formed....

    I took the given ellptic cyl. equation, and divided by 5 to get (x^2)/20 + (z^2)/30 = 1.

    ***I parameterized by using x=cos(t) and z=sin(t) and got ((cost)^2)/20 + ((sint)^2)/30 =1.

    Now, by looking at examples that are somewhat similar, I could tell the answer by looking at number relationships. However, I am unsure of the true way to go about getting my final solution.

    My "made up way" of solving was to set either x or z to zero before parameterizing.

    My final answers are x=(sqrt[20])cos(t) y=5/2 z=(sqrt[30])sin(t)

    I checked my answer using a graphing program, and it is correct, but I am just unsure about going about the TRUE way of solving once I get to the part labeled *** above.

    Thanks.
     
    Last edited: Sep 20, 2010
  2. jcsd
  3. Sep 21, 2010 #2
    Any takers? I'm sure I have the answer right, just not sure of the "correct" last couple steps to get to that answer so that I can show work properly.
     
  4. Sep 21, 2010 #3

    Mark44

    Staff: Mentor

    This step seems flaky to me.

    The correct parametrization is, I believe, x = a*cos(t), z = b*sin(t). Then you have
    [tex]\frac{a^2 cos^2(t)}{20} + \frac{b^2 sin^2(t)}{30} = 1[/tex]

    For this equation to be identically true for all t, it must be that a2 = 20 and b2 = 30, which makes the parametric form of the ellipse (in the x-z plane)
    x = [itex]\sqrt{20}[/itex] cos(t)
    z = [itex]\sqrt{30}[/itex] sin(t)

    I think this is the right way to go about it.
     
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