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How to parametrize inersection of 2 paraboloids?

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to parametrize the equations z = x^2 and x = y^2 in order to find the unit tangent vector, normal vector, and binormal vector.

    2. Relevant equations

    z = x^2 and x = y^2

    3. The attempt at a solution

    I tried setting x = t, so then y = [itex]\sqrt{}t[/itex] and then z = t^2.
    Using that equation r(t) = < t, [itex]\sqrt{}t[/itex] , t^2> I would get completely wrong answers from the book, and if I tried another function like r(t) = < t^2 , t , t^4> , I would get different results from my previous vector. Any ideas?
     
  2. jcsd
  3. Jul 19, 2012 #2

    jbunniii

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    You can see that your first parameterization is not going to give you the full intersection, because there won't be any points with negative y values.

    The second parameterization looks more promising. Can you show your calculations for the tangent and normal vectors?
     
  4. Jul 19, 2012 #3
    Well I needed to find the unit tangent and normal vectors to find the normal and osculating plane equations, and the question only asked for it at the point < 1,1,1> so y isn't negative. However, with either function it looks like I am not getting the correct answer.

    When I used r(t) = < t, [itex]\sqrt{}t[/itex] , t^2> I got that

    r'(t) = <1 , 1/2sqrt(t) , 2t> and if I found the magnitude of that, I get [itex]\sqrt{}5.25[/itex]

    From that, my unit tangent vector is 1/sqrt(5.25) <1 , 1/2sqrt(t) , 2t >

    Something already looks wrong, so when I kept going, I obviously didn't get the right answer
     
  5. Jul 19, 2012 #4

    jbunniii

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    Well, you have to evaluate this at t = 1, but aside from that, I don't see anything wrong. What answer does the book give?

    By the way, I get the same answer using either of your two parameterizations.
     
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