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Parametrize this curve 
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#1
Apr311, 07:01 AM

P: 51

1. The problem statement, all variables and given/known data
Parametrize the following equation using polar coordinates: 2. Relevant equations $(x^2+y^2)^2 = r^2 (x^2  y^2)$ 3. The attempt at a solution It seems that $x=r\cos{\theta}$ and $y = r\sin{\theta}$ don't work. any suggestions? 


#2
Apr311, 07:38 AM

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hi jakey!
(have a theta: θ and try using the X^{2} icon just above the Reply box ) show us how far you've got 


#3
Apr311, 07:42 AM

P: 51

well, the left hand side would equal to (r^2)^2= r^4. the right hand side would equal to r^2 (r^2 cos^2 θ  r^2 sin^2 θ) but these two don't equal...right? 


#4
Apr311, 07:47 AM

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Parametrize this curve
hi jakey!
the equation has become r^{4} = r^{4}(cos^{2}θ  sin^{2}θ) … what are the solutions to that, and what curve does it represent? 


#5
Apr311, 07:51 AM

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#7
Apr311, 08:01 AM

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#9
Apr311, 08:09 AM

P: 51

well,for cos 2\theta = 1, a solution, for example would be theta = 0 or theta = \pi. that would just be a line in the polar axis. 


#10
Apr311, 08:17 AM

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0 ≤ r < ∞, along the line θ = 0 or π … in Cartesian coordinates: (x(t),y(t)) = (t, 0) for any t 


#11
Apr311, 08:21 AM

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#13
Apr311, 08:28 AM

P: 51

"Parametrize the curve first using polar coordinates. Next, find the period which is to be done in Cartersian coordinates." you see, the equation i gave above is the curve for the line integral of \int y ds. 


#14
Apr311, 08:36 AM

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∫ y ds ? … well that's 0 


#15
Apr311, 08:47 AM

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