User:Egm6322.s09.bit.sahin/r3

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

Remembering that $$\theta =tan^{-1}\left (\frac{y}{x} \right )$$, If we define $$s=y/x$$ then

$$\frac{\partial \theta }{\partial y}=\frac{\partial }{\partial s}tan^{-1}(s)\frac{\partial s}{\partial y}=\frac{1}{1+\left (y/x \right )^{2}}\left (\frac{1}{x} \right )=\frac{x}{x^{2}+y^{2}}=\frac{rcos\theta }{r^{2}}=\frac{cos\theta }{r}$$

As a result, we obtain Jacobian matrix as

$$\underline{J}=\begin{bmatrix} cos\theta & sin\theta \\ -\frac{sin\theta }{r}&\frac{cos\theta }{r} \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rcos\theta &rsin\theta  \\ -sin\theta &cos\theta \end{bmatrix}$$

In addition to this method, a simpler method can be used to determine the Jacobian matrix directly in polar coordinates.

$$x_{i}=x_{i}\left (\overline{x}_{1},\overline{x}_{2} \right )=x_{i}\left (r,\theta \right )$$

Instead of inversing the equation, differentiating $$\frac{\partial x_{i}}{\partial \overline{x}_{j}}$$ is simpler (e.g., $$\frac{\partial x}{\partial r},\frac{\partial x}{\partial \theta})$$.

$$\underline{A}=\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

Here $$\underline{A}=\underline{J}^{-1}$$. Therefore it can be written that

$$\underline{J} \underline{A}=\underline{I}=\left [\partial _{ij} \right ]=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$

$$\underline{J} $$ can easily be determined.

$$\underline{J}= \underline{A}^{-1}=\left (\underline{J}^{-1} \right )^{-1}=\begin{bmatrix} cos\theta & -rsin\theta \\ sin\theta & rcos\theta \end{bmatrix}^{-1}$$

$$\underline{A}^{-1}=\frac{1}{det\underline{A}}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

Eventually,

$$\underline{J}=\frac{1}{r}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

which is identical with the previous result.