# Geometric intuition behind simple identities of derivatives of polar coordinates

Consider planar coordinates, given (say in the first quadrant to avoid any range issues) by

$$\left\lbrace \begin{matrix}x=r\cos\theta\\y=r\sin\theta\\\end{matrix}\right.\qquad\text{and}\qquad\left\lbrace \begin{matrix}r=\sqrt{x^2+y^2}\\\theta=\arctan \frac yx\ \ \\\end{matrix}\right.$$

One straightforwardly computes that ${\partial x\over\partial r}=\cos\theta$ and ${\partial y\over\partial r}=\sin\theta$. It turns out that computing the derivatives the other way around yields the same results: $${\partial r\over\partial x}={\partial \sqrt{x^2+y^2}\over\partial x}=\frac12\frac{2x}{\sqrt{x^2+y^2}} =\cos\theta\ \ \left(={\partial x\over\partial r}\right)$$ and similarly ${\partial r\over\partial y}={\partial y\over\partial r}$.

Apart from direct computation, one could also argue that the inverse of the jacobian matrix $\left(\matrix{\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\\}\right)$ is $\left(\matrix{\cos\theta&\sin\theta\\-\frac{\sin\theta}{r}&\frac{\cos\theta}{r}\\}\right)$ where one can see that the entries $\cos\theta$ and $\sin\theta$ appear twice.

Is there yet a deeper, possibly geometrical, explanation for the identities $\displaystyle{\partial x\over\partial r}={\partial r\over\partial x}$ and $\displaystyle{\partial y\over\partial r}={\partial r\over\partial y}$?

To which extent can one regard this as an intrinsic natural definition of $\cos$ and $\sin$?

• Why isn't $\frac{\partial r}{\partial x} = \frac{1}{\cos\theta}?$ – saulspatz May 15 '18 at 18:20
• @saulspatz Because a partial derivative is not a fraction even though the notation looks like it. Compute it, you will see (use the chain rule). – Arnaud Mortier May 15 '18 at 18:38
• I did, starting from $r=x/\cos\theta$ – saulspatz May 15 '18 at 19:43
• @saulspatz I edited to clarify. Your computation is wrong because you treat $\theta$ as being independent from $x$ which it is not. That is, ${\partial \theta\over\partial x}\neq 0$. – Arnaud Mortier May 15 '18 at 19:46
• Partial derivative notation is ambiguous. The calculation is "wrong" because you and @saulspatz assign different meanings to the notation. – Hurkyl May 15 '18 at 20:29