R = cos(theta) in polar coordinates?

[merry]

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!

 

[tiny-tim]

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]

If you multiply both sides of r= cos(\theta) by r you get
r^2= rcos(\theta) which is the same as x^2+ y^2= x or
x^2- x+ 1/4+ y^2= 1/4
(x- 1/2)^2+ y^2= 1/4
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 \theta= \pi, r= -1 which gives the point (1, 0) on the positive x-axis. Note that cos(\theta) goes from 0 to 0 as \theta goes from 0 to \pi and again as \theta goes from \pi to 2\pi.

As \theta goes from 0 to 2\pi, the point goes around the circle twice.

 

[tiny-tim]

HallsofIvy likes coordinate equations, and I like geometry!

 

[arildno]

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

As \theta goes from 0 to 2\pi, 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:
0\leq{r}\leq{1},-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}

If we instead have the equation,
r=|\cos\theta|, 0\leq\theta\leq{2\pi}
then this is a double circle, joining at the origin.

 

[merry]

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 don't quite understand what this means tho:

Note that cos(\theta) goes from 0 to 0 as \theta goes from 0 to \pi and again as \theta goes from \pi to 2\pi.

As \theta goes from 0 to 2\pi, the point goes around the circle twice.

HallsofIvy, doesn't cos(\theta) go from 1 to -1 as \theta goes from 0 to \pi?

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

 

[tiny-tim]

hi merry!

(just got up :zzz: …)

HallsofIvy, doesn't cos(\theta) go from 1 to -1 as \theta goes from 0 to \pi?

(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]

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