Question

Prove that the map T: D to R^2 defined by T(x,y)= (u,v) is not invertible where u=x cos y , v= x sin y and Domain D= {( x,y) | x>0.

Answer 1

T(x, y)= (u, v)

where by u = x cos y and v = x sin y

To check whether the mapping T: D $\rightarrow$ R² is invertible or not we will find the Jacobian.

$J_T(x, y)$ = $\begin{vmatrix} \dfrac{ ∂u}{ ∂x} & \dfrac{ ∂u}{ ∂y} \\ \dfrac{ ∂v}{ ∂x} & \dfrac{ ∂v}{ ∂y} \end{vmatrix}$

Now

$\dfrac{ ∂u}{ ∂x}$ = cos y

$\dfrac{ ∂u}{ ∂y}$ = - x sin y

$\dfrac{ ∂v}{ ∂x}$ = sin y

$\dfrac{ ∂v}{ ∂y}$ = x cos y

$J_T(x, y)$ = $\begin{vmatrix} cos \ y & - x \ sin \ y \\ sin \ y & x \ cos \ y \end{vmatrix}$

$J_T(x, y)$ = x cos² y + x sin² y

$J_T(x, y)$ = x

Now the inverse of T(x, y)= (u, v) will be the inverse of matrix $J_T(x, y)$. 

according to the domain of T(x, y) x can be zero. So the function will not be invertible as matrix J_T (x, y) will become singular . Hence, we say that T(x, y) = (u, v) is not invertible.