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Point on surface closest to origin.

by mathman44
Tags: closest, origin, point, surface
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mathman44
#1
Oct25-09, 07:11 PM
P: 210
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!
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Mark44
#2
Oct25-09, 08:13 PM
Mentor
P: 21,281
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.
mathman44
#3
Oct25-09, 09:34 PM
P: 210
Tyvm.

HallsofIvy
#4
Oct26-09, 05:12 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,544
Point on surface closest to origin.

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]
lanedance
#5
Oct26-09, 05:49 AM
HW Helper
P: 3,307
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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