Question

G find the jacobian of the transformation. x = 6e−3r sin 2θ, y = e3r cos 2θ

Answer 1

Definition:

The Jacobian of the transformation x = f(r, θ) and y = g(r ,θ) is:

`(\partial (x,y) )/(\partial (r,\theta))=\begin{vmatrix}(\partial x )/(\partial r) & (\partial x )/(\partial \theta) \\ (\partial y )/(\partial r) & (\partial y )/(\partial \theta)\end{vmatrix}` =
`(\partial x)/(\partial r)\cdot (\partial y)/(\partial \theta)-(\partial x)/(\partial \theta)(\partial y)/(\partial r)` ......[1]

Given:
`x=6e^(-3r)\sin 2\theta` and
`y=e^(3r)\cos 2\theta`

then,

`(\partial x)/(\partial r) = -18e^(-3r)\sin 2\theta`

`(\partial x)/(\partial \theta) = 6e^(-3r)(2 \cos 2\theta) = 12e^(-3r)\cos 2\theta`

`(\partial y)/(\partial r) = 3e^(3r)\cos 2\theta`

and

`(\partial y)/(\partial \theta) = e^(3r)(-2 \sin 2\theta) = -2e^(3r)\sin 2\theta`

Substitute these value in [1] ;

`\begin{vmatrix}(\partial x )/(\partial r) & (\partial x )/(\partial \theta) \\ (\partial y )/(\partial r) & (\partial y )/(\partial \theta)\end{vmatrix}=\begin{vmatrix}-18e^(-3r)\sin 2\theta & 12e^(-3r)\cos 2\theta\\ 3e^(3r)\cos 2\theta &- 2e^(3r)\sin 2\theta\end{vmatrix}`

=
`(-18e^(-3r)\sin 2\theta)\cdot(-2e^(3r)\sin 2\theta)-(12e^(-3r)\cos 2\theta)\cdot(3e^(3r)\cos 2\theta)`

=
`(-18 \cdot -2)e^(-3r+3r) \sin^2 2\theta - (12 \cdot 3)e^(-3r+3r) \cos^2 2\theta`

On simplify:

`\begin{vmatrix}-18e^(-3r)\sin 2\theta & 12e^(-3r)\cos 2\theta\\ 3e^(3r)\cos 2\theta &- 2e^(3r)\sin 2\theta\end{vmatrix} = 36 \sin^2 2\theta -36 \cos^2 2\theta=36(\sin^2 2\theta-\cos^2 2\theta)`

=
`-36 (\cos 4\theta)`

[ Use
`\cos^2 2\theta -\sin^2 2\theta = \cos(4\theta)`]

therefore, the jacobian transformation of
`x=6e^(-3r)\sin 2\theta` and
`y=e^(3r)\cos 2\theta` is;
`-36 (\cos 4\theta)`