1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the point on the curve y = x^2 that is closest to the point (0, b).

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the point on the curve y = x2 that is closest to the point (0, b).

    2. Relevant equations
    Lagrange multipliers

    3. The attempt at a solution
    L(x,y,λ) = x2 + (y-b)2 - λ(x2-y)
    Lx = 2x - 2λx = 0
    Ly = 2y - 2b + λ = 0
    Lλ = x2 - y = 0

    Then solve for x and y? Is this right?
     
    Last edited: Nov 27, 2009
  2. jcsd
  3. Nov 27, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's ok so far. Keep going.
     
  4. Nov 27, 2009 #3
    2x = 2λx -> λ = 1
    2y = 2b + λ -> y = (2b+1)/2
    x^2 = y -> x = ((2b+1)/2)^1/2
     
  5. Nov 27, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've got a sign error in your y equation. And you've also ignored a possible solution of the first equation. How about x=0? And in the last equation there are two solutions for x.
     
    Last edited: Nov 27, 2009
  6. Nov 27, 2009 #5
    So basically, there are three points on the curve that are equally closest to (0,b)?
     
  7. Nov 27, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Depends on b. The lagrange multiplier method only gives you the critical points. There might be one or three of them depending on the value of b. You then have to check them to figure out which is closest.
     
  8. Nov 27, 2009 #7
    Well, I was given the point (0, b), without any information about b, so I guess I should just put down all three points then...
     
  9. Nov 27, 2009 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, whatever b is you should check to see which points are minimum. You have only shown that they are critical points. It is quite possible that one of the points is a local maximum for the distance.
     
  10. Nov 27, 2009 #9
    Lxx = 2 - 2λ
    Lyy = 2
    Lxy = 0

    if λ = 1, then discriminant = LxxLyy - Lxy2 = 0...
     
  11. Nov 27, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Why don't you just plug the points you get into the distance formula and figure out which one is closest?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook