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    What's the difference between a line and a half-line with polar coordinates.

    For example, which of these equations would be half-lines:
    a)  \theta=3

    b)  rcos\theta=3

    c)  r(cos\theta - \frac{\pi}{6})=3

    I'm a bit confused here, as I would put a & c as lines as b as a half-line, my reasoning being:
    a) is just an angle, hence the line goes on forever
    c) Based on the text-book I'm following, this defines it the line AP
    (where A is perpendicular to O and P is a point on the line). I'm guessing that AP is not just the chord AP, but continues on to infinity
    b) has an adjacent of 3, so will always stop

    Can someone clear this up for me - thnx

    I would say that:

    a) is a half-line, because that's a line coming out from the origin. Remember if you want a (full-)line you'd have to specify theta=3 or theta=2pi-3.

    b) is a line. Just think about polar-cartesian transforms. That's just the cartesian line x=3 in polar form.

    c) ... Not sure, to be honest. I suspect it's a line again, but I can't rigorously justify that. Yet.

    For c) Expand out the cosine using the addition formula and remember

    r \cos \theta = x and r \sin \theta = y

    OP: You've made a typo for (c). It should say r\cos(\theta-\frac \pi 6) = 3.

    I'd agree that a) is a half-line because theta is defined as the angle counter-clockwise about the line of origin , and hence the line can only extend in one direction.If it were a full line , you'd also need a definition of the angle for 2\pi - \theta as well.

    b) is a full line because rcos\theta = 3 for all theta values.It's equivalent with the line x = 3 due to Pythagoras.

    c) - is a full line because after multiplying out the compound angle you'd end up with constants of cos (pi/6) and sin (pi/6) , and converting to cartesian format gives you the equation of a straight line which can be defined below the line of origin.

    (Original post by Operator)
    OP: You've made a typo for (c). It should say r\cos(\theta-\frac \pi 6) = 3.
    Yeahhh I wondered that. It'd make sense anyway, because according to my calculator (c) isn't even a straight line.

    You don't even really need to use the angle formulae to see it's a line. (c) is exactly the same as (b) but where the angle varies slightly diffrently. So effectively throwing in the "- pi/6" into the cosine just rotates the line x=3 a bit (however finding the precise equation of the line will obviously be by using the angle formulae)
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Updated: September 22, 2008

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