We take partial derivatives of the radius with respect to Cartesian coordinates all the time, especially in physics. While we can very easily do so with the quotient / chain rule, I’ve always been annoyed by that process (e.g. dealing with fractional exponents / cancelling out the 2s). As such, below is a proof by picture which is hopefully more illuminating / less frustrating.

Suppose we have the vector \(\mathbf{r}\) living in \(\mathbb{R}^2\) decomposed into its components \(\mathbf{r} = x \mathbf{e}_x + y \mathbf{e}_y\), where \(\mathbf{e}_x = (1,0)\) and \(\mathbf{e}_y = (0,1)\). Let’s define \(r = \sqrt{x^2 + y^2} = \|\mathbf{r}\|\), the length of \(\mathbf{r}\).

To find \(\partial_x r\), instead of using calculus, we can draw the following picture.

We notice that as \(h \to 0\), we have \(\|\mathbf{r} + h \mathbf{e_x}\| \to r + h \partial_x(r)\). Taking this approximation, we have

\[\cos \theta = \frac{x}{r} = \partial_x(r)\]

Similarly, we can conclude

\[\sin \theta = \frac{y}{r} = \partial_y(r)\]

These results further inform us of the gradient of the radius.

\[\begin{align*} x\mathbf{e_x} + y \mathbf{e_y} &= r (\partial_x(r) \mathbf{e_x} + \partial_y(r) \mathbf{e_y}) \\ \therefore \quad \mathbf{r} &= r \nabla r \end{align*}\] Next maths post Previous maths post

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