# Polar coordinates

1. Feb 7, 2006

### Steel_City82

Heres where Im struggling, I cant seem to change equations from rectangular to polar and vice versa

an example

x^2+y^2-2ax=0

heres what I got when I tried
r=2a cos theta
and thats a graph of a rose curve, I think, Im about 10% sure on that answer

heres an example of one I have no clue on

(x^2+y^2)(arctan(y/x))^2=a^2

heres what im thinkin on this one

the x^2+y^2 can = r^2 and the arctan (y/x) can = theta
so you would have (r^2)(theta^2)=a^2

I dont know, I just cant get this

2. Feb 8, 2006

### Tide

What, exactly, are you having doubts about? It looks fine to me.

3. Feb 8, 2006

### HallsofIvy

Staff Emeritus
x^2+y^2-2ax=0
Okay, obviously $x^2+ y^2= r^2$ and $2ax= 2ar cos(\theta)$ so the is $r^2- 2ar cos(\theta)= 0$ which you can write as $r^2= 2ar cos(\theta)$ and, as long as r is not 0, divide by r to get $r= 2a cos(\theta)$ as you have.

(x^2+y^2)(arctan(y/x))^2=a^2
Again $r^2= x^2+ y^2$ and, essentially by definition, $arctan(y/x)= \theta$ so this is simply $r2\theta^2= a^2$ as you have.
Assuming everything is positive, you can reduce that to $r\theta= a$.