How Can I Parametrize This Curve Using Polar Coordinates?

[jakey]

Homework Statement

Parametrize the following equation using polar coordinates:

Homework Equations

$(x^2+y^2)^2 = r^2 (x^2 - y^2)$

The Attempt at a Solution

It seems that $x=r\cos{\theta}$ and $y = r\sin{\theta}$ don't work. any suggestions?

 

[tiny-tim]

hi jakey!

(have a theta: θ and try using the X² icon just above the Reply box )

It seems that $x=r\cos{\theta}$ and $y = r\sin{\theta}$ don't work. any suggestions?

yes it does …

show us how far you've got ​

 

[jakey]

hi jakey!

(have a theta: θ and try using the X² icon just above the Reply box )


yes it does …

show us how far you've got ​

hi tiny-tim, thanks for the reply :)

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?

 

[tiny-tim]

hi jakey!

but these two don't equal...right?

d'oh! if the question says they're equal, then they're equal!

the equation has become r⁴ = r⁴(cos²θ - sin²θ) …

what are the solutions to that, and what curve does it represent? ​

 

[jakey]

hi jakey!


d'oh! if the question says they're equal, then they're equal!

the equation has become r⁴ = r⁴(cos²θ - sin²θ) …

what are the solutions to that, and what curve does it represent? ​

Hi tiny-tim, I need the parametrization for "x" and "y" so that the equation holds. I need this to evaluate a line integral that's why I am not looking for the corresponding polar equation...

 

[tiny-tim]

solve it anyway

 

[jakey]

solve it anyway

Well, r = 0 or cos (2\theta) = 1. how is this going to help, tiny-tim?

 

[tiny-tim]

Well, r = 0 or cos (2\theta) = 1.

and what curve is that?

 

[jakey]

and what curve is that?

hi tiny-tim, I'm stuck. for one, r can't be 0 as the formula I'm dealing with has r=22. i just typed it as r as a generalization.

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.

 

[tiny-tim]

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.

well that's the parametrisation, isn't it? …

0 ≤ r < ∞, along the line θ = 0 or π …

in Cartesian coordinates: (x(t),y(t)) = (t, 0) for any t

 

[jakey]

well that's the parametrisation, isn't it? …

0 ≤ r < ∞, along the line θ = 0 or π …

in Cartesian coordinates: (x(t),y(t)) = (t, 0) for any t

WOW, really?? But I couldn't find a period for this...? Or, is it t \in (-\infty, \infty)?

 

[tiny-tim]

period?

what's the question?​

 

[jakey]

period?

what's the question?​

Hmm, this is the hint given:

"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.

 

[tiny-tim]

… the equation i gave above is the curve for the line integral of \int |y| ds.

(btw, if you don't use polar coordinates, the original equation becomes 2y²(x²+y²) = 0, or y = 0 )

∫ |y| ds ? …

well that's 0

 

[jakey]

(btw, if you don't use polar coordinates, the original equation becomes 2y²(x²+y²) = 0, or y = 0 )

∫ |y| ds ? …

well that's 0

it can't be. btw, it's ∫_C |y| ds where C is the curve I gave above. I need to parametrize it so I could use ds = ||r'(t)|| dt.