Jacobian for Conversion Between Polar and Cartesian Resembles a Rotation Matrix In matrix form, supposing $z=f(x,y)$ where x,y represent the cartesian coordinate system, one can write the chain rule:
$ \begin{bmatrix}
z_x \\
z_y 
\end{bmatrix} = \begin{bmatrix}
r_x & \theta_x \\
r_y & \theta_y 
\end{bmatrix}   
 \begin{bmatrix}
z_r \\
z_\theta 
\end{bmatrix}$
Which turns out to be
$ \begin{bmatrix}
z_x \\
z_y 
\end{bmatrix} =  \begin{bmatrix}
\cos\theta & -\frac{\sin\theta}{r} \\
\sin\theta & \frac{\cos\theta}{r} 
\end{bmatrix}  
 \begin{bmatrix}
z_r \\
z_\theta 
\end{bmatrix}$
Why does this middle matrix resemble a rotation and a scaling by 1/r so much?
 A: Because that is what it is in a way.  The action of a rotation matrix is specified by the action of it on basis vectors. Under rotation of a given angle, e_x gets sent to r hat of that angle and e_y to theta hat of that angle.
A: Because the Jacobian represents the best linear approximation or differential $dF$ of the change of coordinates map $F(r,\theta) = r(\cos \theta, \sin \theta)$ that stretches a rectangle in the $(r,\theta)$ plane to a "polar rectangle" in the $(x,y)$ plane. (Screenshot below shows a relevant drawing I made when teaching Calc 3 online.)

If you remember that the columns of the matrix tell you where the original basis vectors (in this case, the canonical basis in the $(r,\theta)$ plane) get mapped to, then it is simple to understand geometrically why the the first column vector points radially outwards, and why the second column vector is rotated 90 degrees and also stretched by the radial factor (points further away from the origin in the $(x,y)$ plane must rotate faster as you move vertically in the $(r,\theta)$ plane, since the rectangles get bigger).

