How Can I Parametrize This Curve Using Polar Coordinates?

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Homework Help Overview

The discussion revolves around parametrizing the equation \((x^2+y^2)^2 = r^2 (x^2 - y^2)\) using polar coordinates. Participants are exploring the implications of this equation and how to express it in terms of polar coordinates.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial attempts to use \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\) but express confusion over the equality of both sides of the equation. There are questions about the implications of \(r = 0\) and the conditions under which \(\cos(2\theta) = 1\) holds.

Discussion Status

The conversation is active, with participants sharing their thoughts on the parametrization and questioning the validity of their approaches. Some participants suggest that the equation simplifies under certain conditions, while others express uncertainty about how to proceed with the parametrization for the line integral.

Contextual Notes

There is a mention of needing the parametrization for evaluating a line integral, which adds a layer of complexity to the discussion. Participants are also considering the implications of using Cartesian coordinates versus polar coordinates in their analysis.

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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?
 
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hi jakey! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)
jakey said:
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 :smile:​
 
tiny-tim said:
hi jakey! :smile:

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


yes it does …

show us how far you've got :smile:​


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?
 
hi jakey! :smile:
jakey said:
but these two don't equal...right?

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

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

what are the solutions to that, and what curve does it represent? :wink:​
 
tiny-tim said:
hi jakey! :smile:


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

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

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

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...
 
solve it anyway :smile:
 
tiny-tim said:
solve it anyway :smile:

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

and what curve is that? :wink:
 
tiny-tim said:
and what curve is that? :wink:

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.
 
  • #10
jakey said:
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 :wink:
 
  • #11
tiny-tim said:
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 :wink:

WOW, really?? But I couldn't find a period for this...? Or, is it t \in (-\infty, \infty)?
 
  • #12
period? :confused:

what's the question?​
 
  • #13
tiny-tim said:
period? :confused:

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.
 
  • #14
jakey said:
… 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 :wink:)

∫ |y| ds ? …

well that's 0 :confused:
 
  • #15
tiny-tim said:
(btw, if you don't use polar coordinates, the original equation becomes 2y2(x2+y2) = 0, or y = 0 :wink:)

∫ |y| ds ? …

well that's 0 :confused:


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.
 

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