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Thread: Is this right? differential

  1. #1
    Senior Member
    May 2010

    Is this right? differential

    Hi there, I have this exercise, I'd like to know what you think, if I did this right:

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

    So I choose $\displaystyle P_0(0,-1,0,1)$ which satisfies the system

    Then I've verified that $\displaystyle F_1,F_2$ satisfies the implicit function theorem, both functions are polynomials of class $\displaystyle C^k$ y

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

    Then I consider:

    $\displaystyle dF=\frac{{\partial F}}{{\partial x}}dx+\frac{{\partial F}}{{\partial y}}dy$

    $\displaystyle \frac{{\partial F}}{{\partial x}}=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial x}}+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial x}}$

    $\displaystyle \frac{{\partial F}}{{\partial y}}=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial y}}+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial y}}$

    And I set:

    $\displaystyle \frac{{\partial F_i}}{{\partial x}}=\begin{Bmatrix} 1+3u^2\frac{{\partial u}}{{\partial x}}+\frac{{\partial v}}{{\partial x}}=0\\3x^2-4\frac{{\partial u}}{{\partial x}}+4v^3\frac{{\partial v}}{{\partial x}}=0\end{matrix}$

    From where I get:
    $\displaystyle \frac{{\partial u}}{{\partial x}}=0$
    $\displaystyle \frac{{\partial v}}{{\partial x}} \right |_{P_0}=-1$

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

    $\displaystyle dF=\frac{{\partial F}}{{\partial u}}\frac{{\partial u}}{{\partial x}}dx+\frac{{\partial F}}{{\partial v}}\frac{{\partial v}}{{\partial y}}dy$

    Is this right?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Nov 2010
    Madrid, Spain

    Re: Is this right? differential

    Quote Originally Posted by Ulysses View Post
    Is this right?
    Yes, it is correct (only one mistake: $\displaystyle \frac{\partial u}{\partial x}(0,-1)$ it is $\displaystyle 1$ ).
    Last edited by FernandoRevilla; Aug 15th 2011 at 11:45 PM.
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