# Polar coordinates

1. Jan 26, 2008

### ggcheck

1. The problem statement, all variables and given/known data
change the following equation into polar form:

y=x^2

3. The attempt at a solution
r*sin(t) = r^2 * cos(t)^2

stuck after this... my friend suggested that I cancel an r, but won't that get rid of one of the solutions?

I'm not really sure how to proceed

2. Jan 26, 2008

### Fresh_Angel

Well, from what I remember, I would say that your friend is right. Cancelling an r is the way to go, mostly because it's nice to have the answer in the "r=" form. It's really the same as if you had x^2=xy, to solve for y you'd simple cancel an x to get y=x.

3. Jan 26, 2008

### ggcheck

I thought that since "r" was a variable we couldn't cancel it out because we would be losing one of the solutions...

but assuming we can, how should I proceed?

4. Jan 27, 2008

### Fresh_Angel

I believe dividing both sides by r* cos(t)^2 gives you the equation in polar coordinates solved for r. Does the question ask you to do anything else?

5. Jan 27, 2008

### sutupidmath

take care this part is not right, maybe it is just a typo, but it shoul read like this
$$(cos(t))^{2}$$

6. Jan 27, 2008

### sutupidmath

By just dividing both sides by r you will defenitely loose one solution, the one when x takes negative values.

7. Jan 27, 2008

### ggcheck

yes, that is how it should read

8. Jan 27, 2008

### ggcheck

any tips on how to get started? :)

9. Jan 27, 2008

### sutupidmath

well one solution is going to be zero. I think there are two ways of going about this
r*sin(t) = r^2 * (cos(t))^2
r^2*(cos(t))^2 -r*sin(t)=0 now factor a r out
r( r*(cos(t))^2 -sin(t))=0, so which are the two solutions here?
or you might want to solve this r^2*(cos(t))^2 -r*sin(t)=0 as a quadratic equation.