Gradient of arctan(y/x): The Polar Angle Field

Chain rule, then clear the compound fraction

Use $\frac{d}{du}\arctan u = \frac{1}{1+u^{2}}$ with $u = \frac{y}{x}$. For the $x$ partial, $y$ is frozen, so $u$ is a constant divided by $x$ and $\frac{\partial u}{\partial x} = -\frac{y}{x^{2}}$.

The compound fraction is the step people abandon. Multiply the top and bottom of the first factor by $x^{2}$.

The $y$ partial is easier, since $\frac{\partial u}{\partial y} = \frac{1}{x}$.

Both $x^{2}$ factors cancel completely, which is the sign that the answer wants to be written in terms of $r^{2} = x^{2}+y^{2}$ rather than in terms of the quotient you started with.

This function is the polar angle

For $x > 0$, $\arctan\left(\frac{y}{x}\right)$ is exactly the polar angle $\theta$. Its level curves are rays out of the origin, and the gradient must be perpendicular to those rays, so it circles the origin.

The dot product with the position vector is zero at every point, confirming the circulation. The length is again $\frac{1}{r}$, since the numerator vector $\left\langle -y, x\right\rangle$ has length $r$.

At $(1,1)$ the gradient is $\left\langle -\frac{1}{2}, \frac{1}{2}\right\rangle$, pointing counterclockwise, which is the direction in which the angle $\theta$ increases.

This field returns later in the course as the standard example that a curl-free field on a region with a hole need not be conservative. Its circulation around a loop enclosing the origin is $2\pi$, not zero, because $\theta$ itself cannot be defined continuously all the way round.

The mistake: mishandling the inner quotient

Two errors cover most wrong answers.

• Writing $\frac{\partial}{\partial x}\left(\frac{y}{x}\right) = \frac{-y}{x}$, losing the square in the denominator. Rewrite the quotient as $y x^{-1}$ and the power rule hands you $-y x^{-2}$ with no ambiguity.
• Leaving the answer as $\frac{-y/x^{2}}{1 + y^{2}/x^{2}}$ and treating it as finished. It is correct but unusable, and it hides the fact that the two components differ only by a swap and a sign.

Check at $(1,0)$, on the positive $x$ axis: the true gradient is $\left\langle 0, 1 \right\rangle$. Moving in $y$ from there starts sweeping the angle upward at rate $1$, while moving along the axis does not change the angle at all. Any candidate answer with a nonzero first component at that point is wrong.

One boundary condition to keep stated: everything here requires $x \neq 0$, since the formula $\arctan\left(\frac{y}{x}\right)$ is undefined on the $y$ axis even though the angle itself is perfectly well behaved there.