Question

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

Answer 1

Answer:

Definition:

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

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

Given: $x=6e^{-3r}\sin 2\theta$ and  $y=e^{3r}\cos 2\theta$

then,  

$\frac{\partial x}{\partial r} = -18e^{-3r}\sin 2\theta$

$\frac{\partial x}{\partial \theta} = 6e^{-3r}(2 \cos 2\theta) = 12e^{-3r}\cos 2\theta$  

$\frac{\partial y}{\partial r} = 3e^{3r}\cos 2\theta$

and      

$\frac{\partial y}{\partial \theta} = e^{3r}(-2 \sin 2\theta) = -2e^{3r}\sin 2\theta$

Substitute these value in [1] ;  

$\begin{vmatrix}\frac{\partial x }{\partial r} & \frac{\partial x }{\partial \theta} \\ \frac{\partial y }{\partial r} & \frac{\partial y }{\partial \theta}\end{vmatrix}=\begin{vmatrix}-18e^{-3r}\sin 2\theta & 12e^{-3r}\cos 2\theta\\ 3e^{3r}\cos 2\theta &- 2e^{3r}\sin 2\theta\end{vmatrix}$

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

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

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; $-36 (\cos 4\theta)$