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Seems like everytime I pick a random problem for review I get a hard one, calc 3ish

  1. Oct 3, 2006 #1
    find the minimum distance between the origin and the surface z^2=[8-y^2-x^4*y^2]/2

    so d=sqrt(x^2+y^2+z^2)^(1/2) where for z^2 I sub in that expression in x and y?

    so then I take the partials of d wrt to x and y and set those equal to 0, then I solved that system of equations and got x=+-1 and y=+-1 but I'm not sure about that. I also got x=0 and y=0 but I guess that's the max distance since it's the initial point? I dunno. I got the ultimate distance being sqrt(5) by plugging 1 back into the expression for d, you're welcome to practice on it too but I just need to know if the procedure was right
  2. jcsd
  3. Oct 3, 2006 #2


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    Homework Helper

    Proceedure is correct, but it can be simplified since the point at which the minimum distance occurs is also where the square ofthe distance is minimized (so you can drop the square root on d and get the same location). Another method is to use the method of Lagrange multipliers: I'll use it to check your work.

    We want to minimize [tex]d^2=x^2+y^2+z^2[/tex] subject to the constraint [tex]2z^2+y^2+x^4y^2=8[/tex]. According to the method of Lagrange multipliers we should solve this system of equations:

    [tex]\left< 2x,2y,2z\right> = \lambda\left< 4x^3y^2,2y+2x^4y,4z \right>[/tex] and [tex]2z^2+y^2+x^4y^2=8[/tex]

    this simplifies to

    [tex]x=2\lambda x^3y^2[tex]
    [tex]y=\lambda y(1+x^4)[tex]
    [tex]z=2\lambda z[tex]

    according to the third equation, either [tex]z=0[/tex] or [tex]\lambda =\frac{1}{2}[/tex]. Assume presently that [tex]z=0[/tex], then the fourth equations becomes [tex]y^2(1+x^4)=8[/tex] so that multipling the second equation by y gives [tex]y^2=\lambda y^2(1+x^4) = 8\lambda[tex] and putting this into the first equation gives [tex]x=16\lambda ^2 x^3[tex]

    finish later perhaps... (seems like the other method would be easier).
    Last edited: Oct 4, 2006
  4. Oct 3, 2006 #3
    yah I noticed that, when you work it out you get

    Fx=1/2(original expression to the -1/2)(another expression)=0

    Fy=1/2(original to the -1/2)(another)=0

    so you can just chuck the original expressions and solve the two "others" for x and y, which is where you end up if you start from d^2

    I couldn't quite convince myself that the minimum of the distance would be the mininum of the d^2 function, I might just be thinking too hard
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