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Math Help - Please Help (Vector Functions)

  1. #1
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    Please Help (Vector Functions)

    Hi again, I need to find the the vector function that represents the curve of intersection of the two surfaces listed below.

    Paraboloid: Z = 4x^2 + y^2
    Parabolic Cylinder: y = x^2

    I had a very difficult time following my instructor and office hours with the TA next Weds but i really want to understand this before then to start my homework. Please help thanks again!
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  2. #2
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    Quote Originally Posted by multivariablecalc View Post
    Hi again, I need to find the the vector function that represents the curve of intersection of the two surfaces listed below.

    Paraboloid: Z = 4x^2 + y^2
    Parabolic Cylinder: y = x^2

    I had a very difficult time following my instructor and office hours with the TA next Weds but i really want to understand this before then to start my homework. Please help thanks again!
    If you let y = x^2 then z = 4x^2 + y^2 = 4x^2 + x^4. If you let  x = t then y = t^2 and z = 4t^2 + t^4. This is the curve of intersection. If you want the vector which represents this space curve then

     <br />
\vec{r} = \,< t, t^2, 4t^2+t^4>.<br />
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    Thanks for the response Danny. I figured it out! but i just stumbled on to one more of the same problem.

    Hyperboloid z = x^2-y^2
    Cylinder x^2+y^2 = 1. I understand the fx and y part of the answer but i do not understand how Z comes out to be cos2t if you can help out with this extra one i would really appreciate it. Thanks a bunch!
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  4. #4
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    Quote Originally Posted by multivariablecalc View Post
    Thanks for the response Danny. I figured it out! but i just stumbled on to one more of the same problem.

    Hyperboloid z = x^2-y^2
    Cylinder x^2+y^2 = 1. I understand the fx and y part of the answer but i do not understand how Z comes out to be cos2t if you can help out with this extra one i would really appreciate it. Thanks a bunch!
    It you let x = \cos t and y = \sin t then

    x^2+y^2 = \cos^2 t + \sin^2 t = 1 so the cylinder is automatically satisfied. Now substituting these into your surface gives

     <br />
z = x^2 - y^2 = \cos^2 t - \sin^2 t = \cos 2t.<br />
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    Danny thanks again!
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