Point on surface closest to origin.

[mathman44]

Homework Statement

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

z² - xy = 1

that is closest to the origin.

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!

 

[Mark44]

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

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]

Tyvm.

 

[HallsofIvy]

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

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

Here that becomes 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(-y\vec{i}- x\vec{j}+ 2z\vec{k})

 

[lanedance]

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

And \nabla F 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...