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Point on surface closest to origin.

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Hello all. The question asked here is to find the point on the surface:

    z2 - xy = 1

    that is closest to the origin.

    3. The attempt at a solution

    I only have experience in doing this with 2 variables. I begin by trying to find "d", I get that d = sqrt[...] but I have three variables to deal with. I end up with

    d2 = (z2-1)/y + (z2-1)/x + xy + 1

    The problem is that the partial derivatives with respect to x, y and z are quite messy and they don't look right... Does this look correct so far? Thanks all!
  2. jcsd
  3. Oct 25, 2009 #2


    Staff: Mentor

    You want the points on this surface whose distance is smallest. This means you want to minimize
    [tex]d~=~\sqrt{x^2 + y^2 + z^2}~=~\sqrt{x^2 + y^2 + xy + 1} [/tex]

    In the second square root, I replaced z2 with an expression it is equal to, because of the definition of the surface.
    Equivalently, you can minimize the distance squared, which is
    d2 = f(x, y) = x2 + y2 + xy + 1

    You need to keep in mind that there is a domain here, {(x, y) | xy + 1 >= 0}. It seems very likely to me that you'll have a minimum point on the boundary of this domain.
  4. Oct 25, 2009 #3
  5. Oct 26, 2009 #4


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    Science Advisor

    Or you could use Lagrange multipliers. Since the same point that minimizes distance will minimize distance square you can take [itex]F(x,y,z)= x^2+ y^2+ z^2[/itex] as the function to be minimized subject to the constraint that [itex]G(x,y,z)= z^2- xy= 1[/itex].

    The max or min will occur where [itex]\nabla F[/itex] and [itex]\nabla G[/itex] are parallel- that is, that [itex]\nabla F= \lambda\nabla G[/itex] for some number [itex]\lambda[/itex], the "Lagrange multiplier".

    Here that becomes [itex]2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(-y\vec{i}- x\vec{j}+ 2z\vec{k})[/itex]
  6. Oct 26, 2009 #5


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    Homework Helper

    thats interesting geomtrically as well, as [itex]\nabla G[/itex] is perpindicular to the level surfaces of G(x,y,z).

    And [itex]\nabla F[/itex] will always point in the radial direction of the position vector

    so i think you could probably use this to show that the vector connecting a point with the closest point on a given surface, will be normal to the tangent plane of the surface...
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