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Is this Differential right?

  1. Aug 15, 2011 #1
    Hi there, I have this exercise, I'd like to know what you think, if I did this right:

    If [tex]z=F(u,v)=5u^2+4v-7[/tex], with [tex]\begin{Bmatrix} x+y^3+u^3+v=0\\x^3+y-4u+v^4=0\end{matrix}[/tex]. Determine if its possible the dF over a point adequately chosen.

    So I choose [tex]P_0(0,-1,0,1)[/tex] which satisfies the system

    Then I've verified that [tex]F_1,F_2[/tex] satisfies the implicit function theorem, both functions are polynomials of class [tex]C^k[/tex] y

    [tex]\frac{{\partial (F_1,F_2)}}{{\partial (u,v)}}=4\neq{0}[/tex] then [tex]\exists{ E_r(P_0)}:\begin{Bmatrix} u=u(x,y)\\v=v(x,y)\end{matrix}[/tex]

    Then I consider:

    [tex]dF=\frac{{\partial F}}{{\partial x}}dx+\frac{{\partial F}}{{\partial y}}dy[/tex]

    [tex]\frac{{\partial F}}{{\partial x}}=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial x}}+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial x}}[/tex]

    [tex]\frac{{\partial F}}{{\partial y}}=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial y}}+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial y}}[/tex]

    And I set:

    [tex]\frac{{\partial F_i}}{{\partial x}}= 1+3u^2\frac{{\partial u}}{{\partial x}}+\frac{{\partial v}}{{\partial x}}=0\\3x^2-4\frac{{\partial u}}{{\partial x}}+4v^3\frac{{\partial v}}{{\partial x}}=0\end{matrix}[/tex]

    From where I get:
    [tex]\frac{{\partial u}}{{\partial x}}=0[/tex]
    [tex]\frac{{\partial v}}{{\partial x}}_{P_0}=-1[/tex]

    Analogous procedure for the derivatives with respect to y, and:

    [tex]dF=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial x}}dx+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial y}}dy[/tex]

    Is this right?
    Bye.
     
  2. jcsd
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