The discussion revolves around converting the expression Sqrt[12x-2x^2] into polar coordinates. The subject area is primarily focused on polar coordinate transformations in the context of algebra and geometry.
Several participants have provided insights and suggestions regarding the conversion process, including the use of trigonometric identities and relationships between x, y, r, and θ. There is an ongoing exploration of different interpretations and approaches, with no clear consensus reached yet.
Some participants note that the original problem statement may be incomplete, and there are discussions about the implications of working with polar coordinates, particularly regarding the complexity of expressions and the nature of the graph being represented.
Rectangular to polar conversion usually would involve a 2-dimensional quantity. Can you post an image of the full question along with any figure?Quatros said:Homework Statement
I'm suppose to convert Sqrt[12x-2x^2] into a polar equation.
Homework Equations
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.
Charles Link said: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.
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: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)
Quatros said:r^2(cos(theta)^2-2r^2sin(theta )) = r(12sin(theta)
Charles Link said:## x=r \cos(\theta) ##. You have it reversed.
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 said: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.
## 12 \cos(\theta)=r(1+\cos^2(\theta) ) ##. You can then solve for ## r ##.Quatros said:Ohhh, I see then i get 12cos(theta) = 1+ rcos^2(theta), then i just solve for r?