
#1
Oct2509, 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: z^{2}  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 d^{2} = (z^{2}1)/y + (z^{2}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
Oct2509, 08:13 PM

Mentor
P: 20,997

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 z^{2} with an expression it is equal to, because of the definition of the surface. Equivalently, you can minimize the distance squared, which is d^{2} = f(x, y) = x^{2} + y^{2} + 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. 



#3
Oct2509, 09:34 PM

P: 210

Tyvm.




#4
Oct2609, 05:12 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

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] 



#5
Oct2609, 05:49 AM

HW Helper
P: 3,309

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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