Line from origin with elevation theta in polar form?

[BlackWyvern]

Homework Statement

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 sort of 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.

Homework 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.

 

[BlackWyvern]

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

This also seems to make a line from the origin.

 

[HallsofIvy]

Homework Statement

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 sort of 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 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.

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.

 

[BlackWyvern]

I see, thanks for that.