[tex]x=t^2-s^2, y=ts,u=x,v=-y[/tex]
a) compute derivative matrices
[tex]\vec{D}f(x,y) = \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right][/tex]
[tex]\vec{D}f(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right][/tex]
b) express (u,v) in terms of (t,s)
[tex]f(u(x,y),v(x,y) = (t^2-s2,-(ts))[/tex]
c) Evaluate [tex]\vec{D}(u,v)[/tex]
[tex]\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][/tex]
[tex]= \left[\begin{array}{cc}2t&-2s\\-s&-t\end{array}\right][/tex]
d) verify if chain rule holds
need help with this last part, also need to know if I even did the rest correctly
a) compute derivative matrices
[tex]\vec{D}f(x,y) = \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right][/tex]
[tex]\vec{D}f(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right][/tex]
b) express (u,v) in terms of (t,s)
[tex]f(u(x,y),v(x,y) = (t^2-s2,-(ts))[/tex]
c) Evaluate [tex]\vec{D}(u,v)[/tex]
[tex]\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][/tex]
[tex]= \left[\begin{array}{cc}2t&-2s\\-s&-t\end{array}\right][/tex]
d) verify if chain rule holds
need help with this last part, also need to know if I even did the rest correctly
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