# Line from origin with elevation theta in polar form?

1. Sep 17, 2007

### BlackWyvern

1. The problem statement, all variables and given/known data
I need to get y = mx in polar form. I looked at it logically, and I figured I needed a function f(x) that will resolve 0 when x =/= 0, and will resolve 1 when x = 0. I thought for a while, and then realized that sine sorta does that, and just needs a tweak to do it exactly:

$$r = cos^\infty(\theta - c)$$

But this function, although great, is cumbersome. I can't find any other formulas, and my attempts to derive one only equal in cancellations on both sides.

2. Relevant equations
x = rcos(theta)
y = rsin(theta)
m = tan(theta)
Anything else you can think of.

Just to make it clear, I would like it in r form, I know of the $$\theta = \varphi$$ equation.

Last edited: Sep 17, 2007
2. Sep 17, 2007

### BlackWyvern

$$r = \lim_{\substack{x\rightarrow 0}} x\sec(\theta-c) \,$$

This also seems to make a line from the origin.

3. Sep 17, 2007

### HallsofIvy

Staff Emeritus
Just as you cannot get a vertical line in y= mx+ b form, you cannot cannot get a line through the origin in $r= f(\theta)$ form. A line through the origin at fixed "angle of elevation" $\phi$ (better not to use $\theta$ for this constant) is $\theta= \phi$. r, of course, can be number, positive, zero, or negative.

4. Sep 17, 2007

### BlackWyvern

I see, thanks for that.