Converting to Polar co-ordinates

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October 19th 2009, 12:42 AM
Peleus

Converting to Polar co-ordinates

Hi all,

I've got to convert this equation to polar co-ordinates.

$ x^2 + 2y^2 = y(y + 3x) $

My working so far is as follows,

$ x^2 + 2y^2 = y^2 + 3xy $

$ x^2 - 3xy + y^2 = 0 $

Let $y = (r \cos\theta)^2$ and $x = (r \sin\theta)^2$

$ (r \cos \theta)^2 - 3(r \cos \theta)(r \sin \theta) + (r \sin \theta)^2 = 0 $

$ r^2 \cos ^2\theta - 3(r \cos \theta)(r \sin \theta) + r^2 \sin ^2\theta = 0 $

$ r^2 (\cos ^2\theta + \sin ^2\theta) = 3(r \cos\theta)(r \sin\theta) $

$ r = \sqrt{3(r \cos\theta)(r \sin\theta)} $

I can't see this being right with r on the RHS though, anyone able to help? Cheers.
October 19th 2009, 01:02 AM
mr fantastic

Quote:

Originally Posted by Peleus

Hi all,

I've got to convert this equation to polar co-ordinates.

$ x^2 + 2y^2 = y(y + 3x) $

My working so far is as follows,

$ x^2 + 2y^2 = y^2 + 3xy $

$ x^2 - 3xy + y^2 = 0 $

Let $y = (r \cos\theta)^2$ and $x = (r \sin\theta)^2$

$ (r \cos \theta)^2 - 3(r \cos \theta)(r \sin \theta) + (r \sin \theta)^2 = 0 $

$ r^2 \cos ^2\theta - 3(r \cos \theta)(r \sin \theta) + r^2 \sin ^2\theta = 0 $

$ r^2 (\cos ^2\theta + \sin ^2\theta) = 3(r \cos\theta)(r \sin\theta) $

$ r = \sqrt{3(r \cos\theta)(r \sin\theta)} $

I can't see this being right with r on the RHS though, anyone able to help? Cheers.

Your equation simplifies to

Case 1: $r = 0$

Case 2: $1 = 3 \cos \theta \sin \theta = \frac{3}{2} \sin (2 \theta) \Rightarrow \sin (2 \theta) = \frac{2}{3}$

so you have four rays, all with terminus at the origin (alternatively, two lines passing through the origin). See http://www.wolframalpha.com/input/?i...%28y+%2B+3x%29
October 19th 2009, 01:26 AM
Peleus

Any hints as to how it actually simplifies down to that? I'm not sure at what point I should be manipulating it to produce that result.
October 19th 2009, 01:28 AM
mr fantastic

Quote:

Originally Posted by Peleus

Any hints as to how it actually simplifies down to that? I'm not sure at what point I should be manipulating it to produce that result.

Delete the last line of your calcultation (which should have had a $\pm \sqrt{}$ in it by the way) and simplify the second last line.