Question

compute \(\dfrac{\partial x}{\partial u}\) , \(\dfrac{\partial x}{\partial v}\), \(\dfrac{\partial y}{\partial u}\), \(\dfrac{\partial y}{\partial v }\)
if \(u(x,y) = x^2-y^2\) and \(v(x,y) = 2xy\)
please, help

Answer 1

I got it

Answer 2

Step1: Jacobian of f
\[Jf =\left[\begin{matrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}\end{matrix}\right]= \left[\begin{matrix}2x&-2y\\2y&2x\end{matrix}\right]\]

Answer 3

now \(det Jf= 4x^2 +4y^2\)

Answer 4

\((Jf)^{-} =\dfrac{1}{det Jf}(cofactor~~Jf)\) = \(\dfrac{1}{4x^2+4y^2}\left[\begin{matrix}2x&-2y\\-2y&-2x\end{matrix}\right]\)

Answer 5

this inverse is Jacobian of f inverse, that is
\[(Jf)^- =\left[\begin{matrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{matrix}\right]\]

Answer 6

hence
\[(Jf)^- =\left[\begin{matrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{matrix}\right]=\dfrac{1}{4x^2+4y^2}\left[\begin{matrix}2x&-2y\\-2y&-2x\end{matrix}\right]\]

Answer 7

Therefore:
\(\dfrac{\partial x}{\partial u}=\dfrac{2x}{4x^2+4y^2}\)

\(\dfrac{\partial x}{\partial v}=\dfrac{-2y}{4x^2+4y^2}\)

\(\dfrac{\partial y}{\partial u}=\dfrac{-2y}{4x^2+4y^2}\)

\(\dfrac{\partial y}{\partial v}=\dfrac{-2x}{4x^2+4y^2}\)

Answer 8

Lalala... life is beautiful. Math is beautiful.