What is the polar formula for $y=x$? $y=x$ is a basic cartesian equation, but I'm at a loss as to what it is in polar form. It seems the only way I've found to express it is with $r$ on both side of the equation, but is there a way of writing it with $r$ on only one side?
 A: One way to see it is the following:
Knowing that $x=r\cos(\theta)$ and that $y=r\sin(\theta)$, the equation $x=y$ implies the equation:
$$r\cos(\theta)=r\sin(\theta)$$
Subtracting to one side, you get:
$$r\cos(\theta)-r\sin(\theta) = 0$$
Factoring, we have:
$$r(\cos(\theta)-\sin(\theta))=0$$
Now, two things multiplied equal zero if and only if at least one of them is zero.  This is true when either $r=0$ or when $(\cos(\theta)-\sin(\theta))=0$
This is true whenever $r=0$ and $\theta$ is anything, or when $r$ is anything and $\theta=\frac{\pi}{4}$ or $\theta=\frac{5\pi}{4}$
A: The polar coordinates $\rho,~\theta$ describe $x,~y$ in terms of angle (begining from the positive $x$ axis) and distance (from the zero element). Your curve is $y=x$, the diagonal, which has constant angle $\pi/4$ or $5\pi/4$ (in the case of the first and the third quadrant, respectively). These expressions are exactly the traslation of your curve into polar coordinates. Besides, you can apply $x=\rho\cos\theta, ~y=\rho\sin\theta$ to $y=x$ and get $\tan\theta=1$, meaning that $\theta=\pi/4$ or $5\pi/4$.
A: Imagine $(r,\theta)$ to be a set of marching instructions: you turn $\theta$ and march a distance of $r$.
In the case of $y=x$, this line makes a $\frac{\pi}{4}$ angle with the positive $x$ axis. So turn $\theta = \frac{\pi}{4}$ and march, well, march however far you'd like. Since we can move any $r$, we leave it unspecified.
$$\theta = \pi/4$$
As mentioned elsewhere, $5\pi/4$ would be equivalent.
A: Any line through origin in cartesian form has the equation $y=mx$; in polar form it will be given by $\theta = \tan^{-1} (m)$. Exceptionally the $y$-axis is given by $\theta=\pi/2$. Only the lines avoiding the origin will have equations involving both $r$ and $\theta$.
