# Thread: Expressing a equation in polar coordinates.

1. ## Expressing a equation in polar coordinates.

I have to express y = 6x^2 in this fashion. here is my work.

y = r sin(theta) = 6x^2

x = r cos(theta) -> x^2 = r^2*(cos(theta))^2

rsin(theta) = 6x^2

rsin(theta) = 6( r^2) (cos(theta))^2

divide both by r...

sin(theta) = 6r (cos(theta))^2

divided the (cos(theta))^2 and the 6 to leave r alone...

(tan(theta)sec(theta))/6 = r

This is the part where i got lost, how do you find theta in this situation?

2. You already seem to have your answer: namely that the function $y=6x^2$ in polar form is $r=\frac{1}{6}\tan\theta\sec\theta$.

3. Originally Posted by driver327
I have to express y = 6x^2 in this fashion. here is my work.

y = r sin(theta) = 6x^2

x = r cos(theta) -> x^2 = r^2*(cos(theta))^2

rsin(theta) = 6x^2

rsin(theta) = 6( r^2) (cos(theta))^2

divide both by r...

sin(theta) = 6r (cos(theta))^2

divided the (cos(theta))^2 and the 6 to leave r alone...

(tan(theta)sec(theta))/6 = r

This is the part where i got lost, how do you find theta in this situation?
That would be difficult, but why do you ask? The problem, you said, was to change the formula to polar coordinates and, as Tikoloshe says, you have done that.

4. Ok, i will keep it as is. I just assumed that if I had to find r, I had to find theta as well. thank you both!