# Quick Question about Converting Polar cordinates

Quatros

## Homework Statement

I'm suppose to convert Sqrt[12x-2x^2] into a polar equation.

## The Attempt at a Solution

I went from that equation to r(sin(theta)^2 + 2cos(theta)^2)= 12cos(theta), I really don't know where to go from there.

Mentor

## Homework Statement

I'm suppose to convert Sqrt[12x-2x^2] into a polar equation.

## The Attempt at a Solution

I went from that equation to r(sin(theta)^2 + 2cos(theta)^2)= 12cos(theta), I really don't know where to go from there.
Rectangular to polar conversion usually would involve a 2-dimensional quantity. Can you post an image of the full question along with any figure?

Homework Helper
Gold Member
Could you please show a complete equation in part 1. The problem statement is incomplete.

Quatros
All I got was to convert y = sqrt[12x-2x^2] to polar form.

Homework Helper
Gold Member
When you square both sides, you get ## r^2 ##. You can let ## sin^2(\theta)+cos^2(\theta)=1 ##, leaving one ## r^2 cos^2(\theta) ##. ## \\ ## Additional comment=the equation ## y^2=12x-2x^2 ## looks like an ellipse that is translated in the x-direction. In this case, it would only be taking the positive values of ## y ##. ## \\ ## Additional note: Except for simple graphs, polar coordinate expressions can often be somewhat clumsy to work with.

Quatros
When you square both sides, you get ## r^2 ##. You can let ## sin^2(\theta)+cos^2(\theta)=1 ##, leaving one ## r^2 cos^2(\theta) ##. ## \\ ## Additional comment=the equation ## y^2=12x-2x^2 ## looks like an ellipse that is translated in the x-direction. In this case, it would only be taking the positive values of ## y ##. ## \\ ## Additional note: Except for simple graphs, polar coordinate expressions can often be somewhat clumsy to work with.

I'm a bit confused,
I went from y^2+2x^2 = 12x

then I converted to

r^2cos(theta) ^2- 2r^2sin(theta ) = 12rsin(theta)

r^2(cos(theta)^2-2r^2sin(theta )) = r(12sin(theta)

Homework Helper
Gold Member
## x=r \cos(\theta) ##. You have it reversed.

Mentor
I'm a bit confused,
I went from y^2+2x^2 = 12x

then I converted to

r^2cos(theta) ^2- 2r^2sin(theta ) = 12rsin(theta)
This is incorrect. The relationship is ##y = r\sin(\theta)## and ##x = r\cos(\theta)##, not the other way around, as you have it.
Quatros said:
r^2(cos(theta)^2-2r^2sin(theta )) = r(12sin(theta)

Quatros
## x=r \cos(\theta) ##. You have it reversed.

r^2(sin(theta)^2-2r^2cos(theta )^2) = r(12cos(theta))
12cos(theta) = (sin(theta)^2 +2cos(theta)^2)r ,

which is where i get stuck.

Homework Helper
Gold Member
r^2(sin(theta)^2-2r^2cos(theta )^2) = r(12cos(theta))
12cos(theta) = (sin(theta)^2 +2cos(theta)^2)r ,

which is where i get stuck.
Can you see that ## sin^2(\theta)+2 cos^2(\theta)=[sin^2(\theta)+cos^2(\theta)]+cos^2(\theta) ##? That should be obvious.

Quatros
Ohhh, I see then i get 12cos(theta) = 1+ rcos^2(theta), then i just solve for r?

Homework Helper
Gold Member
Ohhh, I see then i get 12cos(theta) = 1+ rcos^2(theta), then i just solve for r?
## 12 \cos(\theta)=r(1+\cos^2(\theta) ) ##. You can then solve for ## r ##.