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Line from origin with elevation theta in polar form?

  1. Sep 17, 2007 #1
    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:

    [tex] r = cos^\infty(\theta - c) [/tex]

    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 [tex]\theta = \varphi[/tex] equation.
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2
    [tex]r = \lim_{\substack{x\rightarrow 0}} x\sec(\theta-c) \,[/tex]

    This also seems to make a line from the origin.
  4. Sep 17, 2007 #3


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    Just as you cannot get a vertical line in y= mx+ b form, you cannot cannot get a line through the origin in [itex]r= f(\theta)[/itex] form. A line through the origin at fixed "angle of elevation" [itex]\phi[/itex] (better not to use [itex]\theta[/itex] for this constant) is [itex]\theta= \phi[/itex]. r, of course, can be number, positive, zero, or negative.
  5. Sep 17, 2007 #4
    I see, thanks for that.
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