Solve Polar Coordinates: y=x^2

[ggcheck]

Homework Statement

change the following equation into polar form:

y=x^2

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

 

[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.

 

[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?

 

[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?

 

[sutupidmath]

Homework Statement

cos(t)^2

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

 

[sutupidmath]

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

 

[ggcheck]

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

yes, that is how it should read

 

[ggcheck]

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

any tips on how to get started? :)

 

[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.