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Thread: Jacobian matrix exercise

  1. #1
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    Jacobian matrix exercise

    Hi everyone,

    I'm trying to do an exercise about jacobian matrix but I don't really know how to do it can you help me? The exercise says:

    H:R \rightarrow R^3 is a C^1 vectorial field in R and G:R^2 \rightarrow R^3 is defined as follow:

    G(x,y) = H(x^2+3y)

    Calculate JH(6) knowing that JG(0,2) = \begin{pmatrix} 0 & 3 \\ 0 & 0 \\ 0 & -6  \end{pmatrix}
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  2. #2
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    Quote Originally Posted by jollysa87 View Post
    Hi everyone,

    I'm trying to do an exercise about jacobian matrix but I don't really know how to do it can you help me? The exercise says:

    H:R \rightarrow R^3 is a C^1 vectorial field in R and G:R^2 \rightarrow R^3 is defined as follow:

    G(x,y) = H(x^2+3y)

    Calculate JH(6) knowing that JG(0,2) = \begin{pmatrix} 0 & 3 \\ 0 & 0 \\ 0 & -6  \end{pmatrix}
    Define K:\mathbb{R}^2\to\mathbb{R} by K(x,y) = x^2+3y. Jacobians respect matrix multiplication (that's just a way of expressing the chain rule), so JG(0,2) = J(H\circ K)(0,2) = JH(6).JK(0,2) (since K(0,2) = 6).
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  3. #3
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    Thank you for the explanation... So I have that:

    JG(0,2) = JH(6) \cdot JK(0,2) \Rightarrow \begin{pmatrix} 0 & 3 \\ 0 & 0 \\ 0 & -6  \end{pmatrix} = \begin{pmatrix} D_xH_1(6) \\ D_xH_2(6) \\ D_xH_3(6)  \end{pmatrix} \cdot \begin{pmatrix} 0 & 3  \end{pmatrix}

    I can see that:

     D_xH1(6)*3 = 3  \Rightarrow D_xH1(6) = 1
     D_xH2(6)*3 = 0  \Rightarrow D_xH2(6) = 0
     D_xH3(6)*3 = -6 \Rightarrow D_xH3(6) = -2

    So the solution is:

    JH(6)=\begin{pmatrix} 1 \\ 0 \\ -2  \end{pmatrix}

    Am I right?
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  4. #4
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    Quote Originally Posted by jollysa87 View Post
    So the solution is:

    JH(6)=\begin{pmatrix} 1 \\ 0 \\ -2  \end{pmatrix}

    Am I right?
    Yes.
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