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Solving for variables as functions of other variables

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the equations

    can be solved for (x,y,z) as functions of (u,v) near the point (x,y,z,u,v)=(1,1,1,1,1) and find dy/du at (u,v)=(1,1)

    2. Relevant equations
    Multivariable calculus differentiation

    3. The attempt at a solution

    I want to know if I'm doing this correctly (I don't think I am). What I did was find the Jacobian and plug the point into that to get a numerical matrix. Then I take determinants of matrices consisting of [x/y/z u v] (or is it [x y z]?) and show that none of them are zero.
    Then I find the inverse of [u v] and do [u v]^-1 * [x y z]
    I'm not very good at this.
  2. jcsd
  3. Feb 25, 2015 #2


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    What did you get ? Why do this ?

    What did you get ? Why do this ? Is that enough to show that the original three equations can be solved simultaneously for (x,y,z) as functions of (u,v) near the point (x,y,z,u,v)=(1,1,1,1,1) ? Aren't you mixing up with things that apply to linear equations only ? And why the question mark ?

    What did you get ? How do you do this at all ?

    How whould you show that the original equations can be solved for (u, v) = (1,1) ?
  4. Feb 25, 2015 #3


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    The last equation, xu+ yu- xyz= 1 is the same as (x+ y)u= xyz+ 1 so u= (xyx+ 1)/(x+ y). Putting that into the second equation gives
    x^3z+ 2y- [(xyz+ 1)/(x+ y)]v= 2 or v= (x^2z+ 2y- 2)(x+y)/(xyz+1).
    You should to see if, with those, xy^2+ zu+ v^2 is equal to 3. If not, the equation cannot "be solved for (x,y,z) as functions of (u,v)".

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