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Vector calculus, surfaces, and planes.

  1. Sep 20, 2014 #1
    I have attached an image... Okay, so I have been stuck on this problem for like 2 hours now and I have no idea how to find r(x). I know the trace is the intersection of the plane and the surface. My first attempt was to substitute the plane y+2x=0 equation for the surface equation by solving for y getting y=-2x. So I got 3x^2-(-2x)^2+2x+2(-2x) = z = f(x,y).
    So I got an equation which is only a function of x e.g f(x) = -x^2-2x. However, how do I go from this to a vector equation r(x) so I can differentiate to get the tangent vector equation.
     

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  3. Sep 20, 2014 #2

    LCKurtz

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    You could use the vector parameterization of the surface ##x = x, y = y, z = 3x^3 - y^2 +2x+2y## so for the surface$$
    \vec R(x,y) = \langle x,y,3x^3 - y^2 +2x+2y\rangle$$Now put ##y = -2x## in that and you will have a vector equation for the required curve.

    [Edit] Fixed typo in parameterization ##x^2## should have been ##x^3##.
     
    Last edited: Sep 20, 2014
  4. Sep 20, 2014 #3
    okay, I haven't solved it yet using what you've told me, but why would that work? How would I know if using x and not x^2 or 4x would be the correct parametrization? Just seems arbitrary how your define the position vector of the surface. mhmmm....
     
  5. Sep 20, 2014 #4

    LCKurtz

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    Parameterizations of surfaces need two variables. Since ##z## is given in terms of ##x## and ##y##, it is most natural to use them as the parameters.
     
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