Point on surface closest to origin.

  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. Mark44

    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. Tyvm.
     
  5. HallsofIvy

    HallsofIvy 40,658
    Staff Emeritus
    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. lanedance

    lanedance 3,307
    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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