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Finding where two curves share tangent lines

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Find all points for which the curves [itex]x^2+y^2+z^2=3[/itex] and [itex]x^3+y^3+z^3=3[/itex] share the same tangent line.


    2. Relevant equations
    Sharing the same tangent line amounts to having the same derivative. The constraint then is that [itex]3x^2+3y^2+3z^2=2x+2y+2z[/itex]. The points must obviously also lie on the original curves.


    3. The attempt at a solution
    Combining the constraint on the derivatives ([itex]3(x^2+y^2+z^2)-2(x+y+z)=0[/itex]) with the constraint that [itex]x^2+y^2+z^2=x^3+y^3+z^3=3[/itex] we see that the constraint on the derivatives becomes [itex]3(3)-2(x+y+z)=0[/itex] which is just the planar equation [itex]2(x+y+z)=9[/itex]. This feels wrong to me; these curves should not intersect at a plane. Am I right?
     
  2. jcsd
  3. Sep 5, 2011 #2

    LCKurtz

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    Those aren't curves, they are surfaces. Tangent lines aren't usually what you talk about with surfaces although I guess there's no law against it.:uhh:

    If it's any help to you, here's a picture of the two surfaces:

    surfaces.jpg
     
  4. Sep 5, 2011 #3

    dynamicsolo

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    I believe this is not quite finished, though. You have found that your points must lie in the plane x + y + z = 9/2 , but you must still introduce a constraint, since plainly this equation alone permits "solutions" which are far from the sphere. You could, say, write z in terms of x and y using the equation for the sphere.
     
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