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Lagrange multipliers

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data

    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
     
  2. jcsd
  3. Jun 12, 2010 #2

    lanedance

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    i would check your grad g

    also i think you mean closest to (0,0,0)?
     
  4. Jun 12, 2010 #3

    HallsofIvy

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    This s wrong. You have dropped a sign.

     
  5. Jun 12, 2010 #4
    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
     
  6. Jun 12, 2010 #5

    lanedance

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    (0,0,0) is not in your surface
     
  7. Jun 12, 2010 #6

    lanedance

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    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....
     
  8. Jun 12, 2010 #7

    cronxeh

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    Just curious, shouldn't d = sqrt(x^2+y^2+z^2)
     
  9. Jun 12, 2010 #8

    lanedance

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    yeah, but minimising x^2 will minimise |x| so its ok - but i suppose you should show it
     
    Last edited: Jun 12, 2010
  10. Jun 12, 2010 #9
    the problem states by itself that it shouldnt be sqrt(x^2+y^2+z^2) I think i understand the solution now,

    thanks for your help :)
     
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