This is the work that I've done.

The transformation between polar and rectangular coordinates can be expressed as: $x = r \cos \theta$, $y = r \sin \theta$.

Hence, $\frac{\partial x}{\partial r} = \cos \theta$.

Furthermore, we can write $r = \sqrt{x^2 + y^2}$ as part of the reverse transformation. Taking the partial derivative of this with respect to $x$, we have $\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} = \frac{r \cos \theta}{r} = \cos \theta$.

Have I made some mistake or is it correct that both of these partial derivatives are equal to $\cos \theta$? How can one interpret this fact?

Thank you for any insight!

  • 3
    $\begingroup$ The "problem" is that the two partial derivatives are holding two different things constant, they are not reciprocals of each other. The first one holds $y$ constant but the second one holds $\theta$ constant. I have problem in quotes because it is not really a problem. $\endgroup$ Jun 4, 2020 at 21:05
  • $\begingroup$ In addition to what @NinadMunshi said, it is only a "problem" if you incorrectly believe the nonsensical statement that $\dfrac{\partial x}{\partial r} = \dfrac{1}{\frac{\partial r}{\partial x}}$. $\endgroup$
    – peek-a-boo
    Jun 4, 2020 at 21:07
  • $\begingroup$ I don't know for intuition, but this property can easily be understood geometrically. $\endgroup$
    – Winther
    Jun 4, 2020 at 21:09
  • $\begingroup$ @peek-a-boo, that is certainly not nonsensical; in fact, it is often true in the context of differential geometry. $\endgroup$ Jun 4, 2020 at 21:10
  • $\begingroup$ @Carlo I said nonsensical because under the typical interpretation of these variables, using $\dfrac{\partial}{\partial r}$ means we're considering everything as functions of $r, \theta$ and performing differentiation while keep $\theta$ constant, while if we say $\dfrac{\partial}{\partial x}$, means we keep $y$ constant. If one intends anything other than this "standard" interpretation, then it should be clearly mentioned (like you did in your answer). But under the "standard" interpretation of the notation, that equation is definitely false. $\endgroup$
    – peek-a-boo
    Jun 4, 2020 at 21:16

2 Answers 2


Unfortunately, in the derivation of $\frac{\partial r}{\partial x},$ it does not make sense to substitute $x = r \cos \theta$ because you are viewing $r$ and $x$ as functions of $x$ and $y.$ In fact, it is actually true that for fixed $\theta,$ we have that$$\frac{\partial x}{\partial r} = \frac{1}{\frac{\partial r}{\partial x}}.$$

Proof. Using implicit differentiation on $x = r \cos \theta,$ we have that $$1 = \frac{\partial}{\partial x} x = \frac{\partial}{\partial x} r \cos \theta = \cos \theta \frac{\partial}{\partial x} r = \cos \theta\frac{\partial r}{\partial x}.$$ Crucially, we are viewing $x$ and $r = r(x, y)$ as functions of $x$ and $y$ with $\theta$ constant. QED.


For illustrative purposes the geometric view.

enter image description here


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