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R = cos(theta) in polar coordinates? 
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#1
Nov1310, 05:46 PM

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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
Nov1410, 08:11 AM

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hullo merry!
it's going fine, thanks for asking! yes, it is a circle, with diameter from (0,0) to (0,1) … a little bit of geometry will enable you to confirm that your mistake was that you can't have negative values of r!! 


#3
Nov1410, 11:37 AM

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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 yaxis. 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 xaxis. 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. 


#4
Nov1410, 11:46 AM

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R = cos(theta) in polar coordinates?
HallsofIvy likes coordinate equations, and I like geometry!



#5
Nov1510, 03:16 PM

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


#6
Nov3010, 06:20 PM

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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: I am confused with arildno's statement as well @.@ 


#7
Dec110, 04:15 AM

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hi merry!
(just got up …) 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) but r = cosθ is two circles, touching at the origin 


#8
Dec110, 05:51 AM

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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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