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R = cos(theta) in polar coordinates?

by merry
Tags: circle, graph, polar coordinates, r=cos
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merry
#1
Nov13-10, 05:46 PM
P: 45
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!
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tiny-tim
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Nov14-10, 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!!
HallsofIvy
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Nov14-10, 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 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.

tiny-tim
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Nov14-10, 11:46 AM
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R = cos(theta) in polar coordinates?

HallsofIvy likes coordinate equations, and I like geometry!
arildno
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Nov15-10, 03:16 PM
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Quote Quote by HallsofIvy View Post
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.
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.
merry
#6
Nov30-10, 06:20 PM
P: 45
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:

Quote Quote by HallsofIvy View Post
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.
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 @.@
tiny-tim
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Dec1-10, 04:15 AM
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hi merry!

(just got up )
Quote Quote by merry View Post
HallsofIvy, doesnt [itex]cos(\theta)[/itex] go from 1 to -1 as [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex]?
(have a theta: θ and a pi: π )

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)
I am confused with arildno's statement as well @.@
arildno is saying that r = cosθ is one circle

but r = |cosθ| is two circles, touching at the origin
HallsofIvy
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Dec1-10, 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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