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Curve of intersection of a plane and curve

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data
    I have to find the slope of the tangent line at (-1,1,5) to the curve of intersection of the surface z = x2 + 4y2 and the plane x = -1

    2. The attempt at a solution
    I really am having trouble figuring out where to start. The question is even numbered, and the only one like it in the chapter, so I have been having trouble figuring out what to do.
    I tried googling it different ways, the explanation of a problem that is kinda like this one told me to take the gradient
    2xi + 8yj - k
    and then plug in (-1,1,5) and that would be the normal n1
    And I was supposed to do the cross of n1 and the normal of the plane, and then some more stuff but their problem was different, and I am not really sure where to go from here, or if this is even right...
     
  2. jcsd
  3. Jul 14, 2012 #2

    LCKurtz

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    You wouldn't usually use the term "slope" for the tangent to a 3D curve. You would ask for a direction vector. If you plug in ##x=-1## you get ##z=1+4y^2##. With these you can parameterize the curve ##\vec R = \langle -1,y,1+4y^2\rangle##. Now it should be easy to find the direction vector for the tangent to the curve.
     
  4. Jul 15, 2012 #3
    Thank you, that helped greatly, though I am still not the best at getting that curve.

    A later problem has z=√(x2 + 4y2) and 3z = x + 2y + 8
    I was trying to make the two Zs equal to each other, and solve for x or y, but I couldn't get any of them separate.
    Same thing, finding the tangent line at a point (3,2,5)
     
  5. Jul 15, 2012 #4

    LCKurtz

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    Forum policy is that you start a new thread if you have a new problem. You are more likely to get an answer.
     
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