Solve Lagrange Multipliers for x,y,z in Min Distance Problem

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Homework Help Overview

The problem involves finding points on the surface defined by the equation x² - z² = 1 that minimize the distance to the origin (0,0,0). The original poster is using the distance formula d = x² + y² + z² and applying Lagrange multipliers to find the minimum distance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Lagrange multipliers, questioning the correctness of gradients and signs in the equations. There is confusion regarding the values of λ and the implications for x and z. Some participants also explore the validity of points derived from the calculations.

Discussion Status

The discussion is ongoing, with participants providing corrections and suggestions for checking assumptions. There is an exploration of different cases for λ and their implications for the points on the surface. Some participants express confusion about certain points and seek clarification.

Contextual Notes

Participants note that the point (0,0,0) does not lie on the defined surface, and there is a suggestion to visualize the surface to aid understanding. The problem's constraints and the nature of the surface are under consideration.

kliker
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Homework Statement



find the points on the surface x^2-z^2 = 1 which are in minimum distance from (0,0)

i should find the points using d = x^2+y^2+z^2

first of all

gradf = λ gradg

where f = d and g = x^2-z^2

so we have (2x,2y,2z) = λ (2x,0,2z)

now
2x = λ2x
2y = 0 => y = 0
2z = λ2z

so λ=1

but now how can i find x and z?

im totally confused
 
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i would check your grad g

also i think you mean closest to (0,0,0)?
 
kliker said:

Homework Statement



find the points on the surface x^2-z^2 = 1 which are in minimum distance from (0,0)

i should find the points using d = x^2+y^2+z^2

first of all

gradf = λ gradg

where f = d and g = x^2-z^2

so we have (2x,2y,2z) = λ (2x,0,2z)
This s wrong. You have dropped a sign.

now
2x = λ2x
2y = 0 => y = 0
2z = λ2z

so λ=1

but now how can i find x and z?

im totally confused
 
oh you're right so now i have

2x = λ2x
2y = 0 => y = 0
2z = -λ2z => λ*2z+2z = 0 => 2z(1+λ)=0

hence we have λ = -1 and z = 0

but if z is 0 then x = +1 or x = -1

so for λ = -1 we have two points (1,0,0) and (-1,0,0)

is this correct? also, i would like to ask something, if these points are correct then why for λ = -1 in the first equation i get -4x = 0 hence x = 0?

but when x = 0 the point will be (0,0,0), can we check for this point too?

thanks for your help
 
(0,0,0) is not in your surface
 
its also worth drawing the surface to help understand what is going on

try drawing the curve given by y=0, x^2-z^2 = 1 in the xz plane...
 
Just curious, shouldn't d = sqrt(x^2+y^2+z^2)
 
yeah, but minimising x^2 will minimise |x| so its ok - but i suppose you should show it
 
Last edited:
the problem states by itself that it shouldn't be sqrt(x^2+y^2+z^2) I think i understand the solution now,

thanks for your help :)
 

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