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Math Help - Curve of intersection of surfaces

  1. #1
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    Curve of intersection of surfaces

    The problem statement, all variables and given/known data

    See first figure attached

    The attempt at a solution

    I was able to sketch the two curves individually to get an idea of what I'm looking at, but I still can't really visualize how the two curves would intersect each other in the first octant.

    Is this crucial to answering this question?

    If I can write parametric equations for x, y and z then I could express the curve of intersection in that form. Then I'd simply have to look at values of t for which x=0, y=0 and z=0, right? Those would be the endpoints of the curve when they are exiting the first octant.

    Any ideas/suggestions/tips?

    Thanks again!
    Attached Thumbnails Attached Thumbnails Curve of intersection of surfaces-foqa1.jpg   Curve of intersection of surfaces-foq.jpg  
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  2. #2
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    You have the two surfaces defined by x^2+ \frac{z^2}{4}= 1, a cylinder, and z= \sqrt{4x^2+ y^2}, the "upper half" of the cone given by z^2= 4x^2+ y^2.

    If you replace the z^2 in the first equation by 4x^2+ y^2, you get x^2+ x^2+ \frac{y^2}{4}= 1 or 2x^2+ \frac{y^2}{4}= 1, an ellipse with major semi-axis along the y axis with length 2 and minor semi-axis along the x-axis, of length \sqrt{2}{2}. That is the equation of the projection into the xy-plane.

    I don't think the problem actually requires you to find parametric equations for the curve, but if you want to, a standard parameterization of the ellipse 2x^2+ \frac{y^2}{4}= 1 would be x= \frac{\sqrt{2}}{2}cos(t), y= 2 sin(t). Then z= \sqrt{4x^2+ y^2}= \sqrt{2cos^2(t)+ 4 sin^2(t)}= \sqrt{2+ 2sin^2(t)}.

    The boundaries of the first octant are the coordinate planes, x= 0, y= 0, and z= 0. If z= 0, then the first equation becomes x^2= 1 and since we are in the first octant, x= 1. But then the second equation becomes 0= \sqrt{4+ y^2} which is impossible. The curve of intersection does not cross the xz-plane.

    If x= 0, then the first equation becomes \frac{z^2}{4}= 1 so z^2= 4 and, since we are in the first octant, z= 2. then the second equation becomes 2= \sqrt{y^2} so y= 2. The endpoint in the yz-plane is (0, 2, 2).

    If y= 0, then the second equation becomes z= \sqrt{4x^2} or z= 2x. Then the first equation becomes x^2+ x^2= 2x^2= 1 so that x= \sqrt{2}{2}, z= \sqrt{2} and the endpoint in the xz-plane is (\sqrt{2}{2}, 0, \sqrt{2}).
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