R = cos(theta) in polar coordinates?

  1. r = cos(theta) in polar coordinates??

    Hullo everyone!

    Hows it going?
    I am confused with how to interpret the graph of r = cos(theta) in polar coordinates.
    I tried graphing it manually. and this is how I interpreted it:

    r(0) = cos(0) = 1
    r(pi/2) = 0
    r(-pi) = -1
    r(3pi/2) = 0
    r(2pi) = 1

    This gives me three points in a line @.@ with one on the negative x axis, one on the origin, and one on the positive x axis. But apparently, its supposed to be a circle; how so?
    Could someone please explain where I am going wrong?
    Thanks v. much!
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,043
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    hullo merry! :smile:

    it's going fine, thanks for asking! :biggrin:

    yes, it is a circle, with diameter from (0,0) to (0,1) …

    a little bit of geometry will enable you to confirm that :wink:

    your mistake was that you can't have negative values of r!! :redface:
     
  4. HallsofIvy

    HallsofIvy 40,674
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    Re: r = cos(theta) in polar coordinates??

    If you multiply both sides of [itex]r= cos(\theta)[/itex] by r you get
    [itex]r^2= rcos(\theta)[/itex] which is the same as [itex]x^2+ y^2= x[/itex] or
    [tex]x^2- x+ 1/4+ y^2= 1/4[/tex]
    [tex](x- 1/2)^2+ y^2= 1/4[/tex]
    a circle with center at (1/2, 0) and radius 1/2 so it is tangent to the y-axis.

    While you can have r negative in the equation, that is interpreted in polar coordinates as the radius going the other way. With [itex]\theta= \pi[/itex], [itex]r= -1[/itex] which gives the point (1, 0) on the positive x-axis. Note that [itex]cos(\theta)[/itex] goes from 0 to 0 as [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex] and again as [itex]\theta[/itex] goes from [itex]\pi[/itex] to [itex]2\pi[/itex].

    As [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex], the point goes around the circle twice.
     
  5. tiny-tim

    tiny-tim 26,043
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    HallsofIvy likes coordinate equations, and I like geometry! :biggrin:
     
  6. arildno

    arildno 12,015
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    Re: r = cos(theta) in polar coordinates??

    I totally disagree.

    "r", in this case is measured from the origin, theta the angle the radial vector makes with the positiv x-axis.

    Thus, we have the bounds:
    [tex]0\leq{r}\leq{1},-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}[/tex]

    If we instead have the equation,
    [tex]r=|\cos\theta|, 0\leq\theta\leq{2\pi}[/tex]
    then this is a double circle, joining at the origin.
     
  7. Re: r = cos(theta) in polar coordinates??

    Thanks guys! It makes more sense now =D
    Sorry for not replying! I didnt think my question would get noticed, so I gave up!

    I dont quite understand what this means tho:

    HallsofIvy, doesnt [itex]cos(\theta)[/itex] go from 1 to -1 as [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex]?

    I am confused with arildno's statement as well @.@
     
  8. tiny-tim

    tiny-tim 26,043
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    hi merry! :smile:

    (just got up :zzz: …)
    (have a theta: θ and a pi: π :wink:)

    not quite … θ goes from -π/2 to π/2 as r goes from 0 to 1 and back to 0 (as arildno said);

    between π/2 and -π/2, r doesn't exist (because r can't be negative)
    arildno is saying that r = cosθ is one circle

    but r = |cosθ| is two circles, touching at the origin :smile:
     
  9. HallsofIvy

    HallsofIvy 40,674
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    Re: r = cos(theta) in polar coordinates??

    Notice the absolute value in arildno's "[itex]r= |cos\theta|[/itex]". If [math]-\pi/2\le \theta\le \pi/2[/math] then [itex]cos(\theta)[/itex] is positive so [itex]r= cos(\theta)[/itex]. That is the circle with center at (1, 0) and radius 1. If [itex]\pi/2\le \theta\le 3\pi/2[/itex] then [itex]|cos(\theta)|[/itex] is still positive so we have the circle with center at (-1, 0) and radius 1. The area tangent at (0, 0).

    I said earlier that r< 0 is interpreted as 'the opposite direction'. Arildno disagrees with that but I consider it a matter of one convention rather than another.
     
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