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Math Help - Jacobian matrix and chain rule

  1. #1
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    Jacobian matrix and chain rule

    question gives

    x=t^2-s^2, y=ts,u=x,v=-y

    a) compute derivative matrix (jacobian matrix)

    f(u,v)=f(t^2-s^2,-(ts))

    \vec{D}f(u,v) = \left[ \begin{array}{cc}2t&-2s\\-s&-t\end{array}\right]

    b) express (u,v) in terms of (t,s)

    (u,v) = (t^2-s^2, -(ts))

    c) calculate \vec{D}(u,v)

    \left[\begin{array}{cc}1&-1\end{array}\right] \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right] = \left[\begin{array}{cc}2t-s&-2s-t\end{array}\right]

    d) verify chain rule holds

    f(t^2-s^2,-(ts)) not sure how to go about verifying this when im not sure if the rest is correct
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  2. #2
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    Re: Jacobian matrix and chain rule

    re-did this one

    a) compute derivative matrices

    \vec{D}f(x,y) = \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]


    \vec{D}f(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right]

    b) express (u,v) in terms of (t,s)

    f(u(x,y),v(x,y) = (t^2-s2,-(ts))

    c) evaluate \vec{D}(u,v)

    \vec{D}(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right] \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]

    = \left[\begin{array}{cc}2t&-2s\\-s&-t\end{array}\right]

    d) verify chain rule holds

    this one I am still not sure how to show if it holds
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  3. #3
    Junior Member Dark Sun's Avatar
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    Re: Jacobian matrix and chain rule

    You have already practically shown d)

    Df(u(x,y),v(x,y))=\left[\begin{array}{cc}2t & -2s \\ -s & -t \end{array} \right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1 \end{array} \right]\cdot \left[\begin{array}{cc}2t & -2s \\ s & t \end{array} \right]=Df(u,y)\cdot Df(x,y)
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