In this Quora Answer, a very intuitive explanation for the Jacobian is provided, there is however a step I don't understand:

He takes this square:

Original square

And via a polar coordinates transformation transforms it into this parallelogram:

Transformed parallelogram

I get the idea, what I don't get however is how the new coordinates are calculated. He says

This square transforms into a parallelogram with the coordinate change. The point $(x,y)$ transforms to the point $(r,\theta)$. The point $(x + \mathrm{d}x,y)$ becomes $( r(x + \mathrm{d}x,y) , \theta(x + \mathrm{d}x,y))$. Because it's a small square, we can find these coordinates using calculus. The new point is $\left(r + \frac{\partial r}{\partial x}\mathrm{d}x, \theta + \frac{\partial \theta}{\partial x}\mathrm{d}x\right)$

But if the polar transformation is $ x = r\cdot cos(\theta) , y = r\cdot sin(\theta) $, shouldn't $(x,y) \rightarrow ( r\cdot cos(\theta), r\cdot sin(\theta)) $ ? Why is it $(x,y) \rightarrow (r,\theta) $ instead? And what about the partial derivatives?

  • 1
    $\begingroup$ If you have a point with polar coordinates $(r, \theta)$, then the same point has the cartesian coordinates $x = r\cos(\theta), y = r\sin(\theta)$. However, in this case they're forgoing writing down the direct relationship between the coordinates, and instead they just say "The cartesian point $(x, y)$ transforms to the polar point $(r, \theta)$." For a concrete example, we could say that "The cartesian point $(3, 3)$ transforms to the polar point $(3\sqrt 2, \pi/4)$." $\endgroup$
    – Arthur
    Jun 11, 2016 at 11:47

1 Answer 1


The formula you gave below describes the relationship between Cartesian coordinate (x,y) and polar coordinate $(r, \theta)$. They are not supposed to be written in this way, i.e. in bracket.

$$(x,y) \rightarrow ( r\cdot cos(\theta), r\cdot sin(\theta))$$

The correct way to write this is,

$$x=r\cdot cos(\theta)$$ $$y=r\cdot sin(\theta)$$

Reversely, you can write the relationship between polar coordinate $(r,\theta)$ and Cartesian coordinate (x,y) as below. note: care should be taken for calculating $\theta$.

$$r=\sqrt {x^2+y^2}$$ $$\theta = \tan^{-1}{\frac yx}$$

As you can see polar coordinate r and $\theta$ are functions of x and y. Theoretically, you can do partial derivative and then calculate Jacobin matrix.

You may ask yourself again if you got the idea of transforming a square to a parallelogram. A better way to comprehend this is to assume coordinates of the square's vertices can transform them to polar coordinate system and draw it in $(r, \theta)$ plane.


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