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Finding the tangent line of the curve of intersection

  1. Nov 6, 2011 #1
    Given the paraboloid z = 6 - x - x2 -2y2 and the plane x = 1, find curve of intersection and the parametric equations of the tangent line to this curve at point (1,2,-4).


    So I plugged x=1 into the paraboloid equation and got z = 4-2y2.

    Then I take the derivative of the curve of intersection:
    dz/dy = -4y

    Then I plug in 2 for t since y=t=2 at the given point and get the slope -8.

    The parametric equation of a line is r = r0 +tv
    I have r0 = (1,2,-4) and I have a slope -8. I'm not sure how to get a vector.
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    Simon Bridge

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    Consider in general:

    dy/dx=m ... means that every 1 unit you go in x, you go m units in y right?
    The vector that does that must be (1,m)t right?
     
  4. Nov 6, 2011 #3
    yes, so if -8 is dz/dy then for every unit in y I go down 8 in z. so the vector would be <0, 1, -8>? then the parametric equation is

    x = 1
    y = 2 + t
    z = -4 -8t
     
  5. Nov 6, 2011 #4

    Simon Bridge

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    There you go :)
     
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