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Implicit Function Thm Application

  1. Jul 6, 2012 #1
    Investigate the possibility of solving x2-4x+2y2-yz = 1 for each of its variables in terms of the other two near the point (2,-1,3).

    Attempt:

    Ok so using the IFT I was able to determine that I can only slze for y,z. But in the question they ask me to solve for y and z. z was not problem, but y, they got a solution like this:

    y = (z - [z2+8(1-x2+4x)]1/2) / 4

    as well they said x = 2 (+ or -) [5-2y2+yz]1/2 and the square root vanishes at (y,z) = (-1,3) so there are two values of “ for some nearby values of and
    and none for others.

    Question: How did they solve for y? (What bloody algebra trick did they use that I'm not seeing? and how did those conditions on x appear?
     
  2. jcsd
  3. Jul 6, 2012 #2

    tiny-tim

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    hi trap101! :smile:
    that's the standard -b ± √etc formula :wink:
    your typing hasn't come out :redface:
     
  4. Jul 6, 2012 #3
    there are two values of x for some nearby values of y and z
    and none for others.



    Well after doing the mechanics for solving for y, I guess in that sort of scenario I should assume the other "variables" in this case x, z are just constants?
     
    Last edited: Jul 6, 2012
  5. Jul 7, 2012 #4

    tiny-tim

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    yup! :biggrin:
     
  6. Jul 7, 2012 #5
    thnks again and won't be the last time, unless I get struck by lightning and all of a sudden I become a genius
     
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