Convert the equation to polar coordinates, y=x^2

[81ttopcoupe]

First post for me, so if I'm breaking a rule be gentle. If someone can point me to how to get the theta, pi, etc, symbols that would be great also.

y=x^2
rsin(theta)=rcos(theta)rcos(theta)
rsin(theta)=r^2 x cos^2(theta)

At this point I'm kind of lost, maybe I went down the wrong path.

Help appreciated, Thanks and Regards.

 

[Deleted member 4993]

First post for me, so if I'm breaking a rule be gentle. If someone can point me to how to get the theta, pi, etc, symbols that would be great also.

y=x^2
rsin(theta)=rcos(theta)rcos(theta)
rsin(theta)=r^2 x cos^2(theta)

At this point I'm kind of lost, maybe I went down the wrong path.

Help appreciated, Thanks and Regards.

You are almost there...

\(\displaystyle r*sin(\theta) \ = \ r^2 * cos^2(\theta)\)

\(\displaystyle \frac{sin(\theta)}{cos^2(\theta)} \ = \ r\)

\(\displaystyle r \ = \ tan(\theta) * sec(\theta)\)

That's it...........

 

[81ttopcoupe]

Thank you.
One more question to polish this off. There is short cut type algebra operation you used between the first two steps you show. What is this? I can't remember what that's called or the method, or maybe I'm getting confused with something else, what I am thinking about worked because the ratios would still be the same if that makes any sense.

 

[tkhunny]

If there is such a "short cut", forget it. Learn to use algebra. Really.