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Curve of intersection of a plane and function

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    f(x,y) = x^2 + xy + y^2 and z=2x+y intersect, find a parameterization of the curve where they intersect.
    2. Relevant equations


    3. The attempt at a solution
    I am lost. I know that z is the partial derivative of the original function, if that's of any use. I can visualize it but not quite sure how to begin calculating.
     
    Last edited: Jan 25, 2015
  2. jcsd
  3. Jan 25, 2015 #2

    Mark44

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    It would help to know what is the original function, as well as the actual problem you're working on. From the equations above, f(x, y) is a z value, so you could equate x2 + xy + y2 with 2x + y. I'm not sure this leads to anything informative, so I'm wondering if you made a mistake somewhere.
     
  4. Jan 25, 2015 #3

    Ray Vickson

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    You seem to know something that is wrong: z is NOT the partial derivative of the original function (whatever that function is, since you do not say).
     
  5. Jan 25, 2015 #4

    Stephen Tashi

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    A brute force approach is to let the parameter be [itex] t [/itex]. Set [itex] x = t [/itex]. Substitute for [itex] t [/itex] for [itex] x [/itex] in the equation [itex] x^2 + xy + y^2 = 2x + y [/itex]. Solve for [itex] y [/itex] in terms of [itex] t [/itex].

    There might be a clever way. Solving a quadratic equation for y whose coefficients are expressions in t isn't very elegant.

    Consider the surface [itex] z = x^2 + xy + y^2 - (2x + y) [/itex]. The requested intersection is a level curve of this surface for [itex] z = 0 [/itex]. Do you know any slick ways to parameterize the level curves of a surface? ( I haven't solved the problem myself, so I'm just speculating.) The value of the directional derivative of the surface along tangents to the level curve should be zero. Perhaps that's a useful fact.
     
  6. Jan 25, 2015 #5
    Let f(x,y)=x^2 + xy + y^2 for all x in R2.
    Find a parameterization of the curve given by the intersection of f(x,y) and the plane z=2x+y. That's the problem exactly as on paper.
     
  7. Jan 25, 2015 #6

    Ray Vickson

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    It should have said "find a parametrization of the curve of intersection of the two surfaces ##z = f(x,y) ## and ##z = 2x + y##."
     
  8. Jan 25, 2015 #7
    I think I solved it.
     
  9. Jan 26, 2015 #8

    LCKurtz

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    In case anyone is still interested, I think the "natural" way to parameterize that intersection is to take the equation$$
    x^2+xy + y^2 - 2x -y = 0$$which is the projection of the intersection curve in the xy plane. Rotate and translate that to get its center and the standard form of the equation of the ellipse$$\frac {(x-x_0)^2}{a^2} + \frac {(y-y_0)^2}{b^2} = 1$$. Then let $$
    x =x_0+a\cos t,~~y = y_0 + b\sin t, z = 2x+y$$With a bit of help from Maple you get$$
    \vec r(t) = \langle 1 +\frac 1 {\sqrt 3} \cos t -\sin t,\frac 1 {\sqrt 3} \cos t +\sin t,2 + \sqrt 3\cos t -\sin t\rangle$$
    Here's a picture:

    ellipse.jpg
     
  10. Jan 27, 2015 #9

    HallsofIvy

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    I presume you mean the intersection or z= x^2+ xy+ y^2 and z= 2x+ y. That intersection is given by x^2+ xy+ y^2- 2x- y= 0. It is possible, by rotating the axes, to remove the "xy" term getting an equation of an ellipse or other conic section for which it is easy to find parameters. Then use z= 2x+ y to get a parametric equation for z.

     
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