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Homework Help: Vector Calculus - Closest point to the origin

  1. May 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A curve C in space is defined implicitly on the cylinder x^2+y^2=1 by the additional equation: x^2-xy+y^2-z^2=1. Find the point or points on C closest to the origin.


    2. Relevant equations

    d = ((x-x0)+(y-y0)+(z-z0))^(1/2) - This is the distance formula.
    Please note that I did NOT learn Lagrange Multipliers, yet - it is the next section in my math book.


    3. The attempt at a solution

    First, I used the distance formula: ((x-x0)^2+(y-y0)^2+(z-z0)^2)^(1/2).
    where (x0,y0,z0) = (0,0,0) - the origin.

    I solved for z from the additional equation: x^2-xy+y^2-z^2=1.

    So the equation now looks like this: ((x^2+y^2+(x^2-xy+y^2-1))^(1/2)

    I then square the entire equation (so that I can derive easier) because the extrema points remained the same if the equation were not squared.

    I solved the partials for x and y. And for some reason, I got fx=4x-y and fy=4y-x. Thus the point of intersection, which I believe to be incorrect, is (0,0).

    I plugged (0,0) back into the additional equation and I have square root by a negative number!

    And now... I am stuck.

    Can anyone please show me how to approach this problem using the derivation of the distance formula and NOT using Lagrange Multipliers?
     
  2. jcsd
  3. May 14, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi number0! Welcome to PF! :smile:

    (have a square-root: √ and try using the X2 and X2 tags just above the Reply box :wink:)

    Hint: you know x2 + y2 = 1, so you really only need to minimise z2, and you can use x2 + y2 = 1 in the definition of C. :wink:
     
  4. May 14, 2010 #3
    Thanks tiny-tim!
     
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