How Can I Parametrize This Curve Using Polar Coordinates?

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In summary, the conversation discusses parametrizing the equation $(x^2+y^2)^2 = r^2 (x^2 - y^2)$ using polar coordinates. The attempt at a solution involves trying $x=r\cos{\theta}$ and $y = r\sin{\theta}$, but this does not work. The conversation then explores other potential solutions, such as $r = 0$ and $\cos(2\theta) = 1$, which would result in a line in the polar axis. However, it is noted that this does not have a defined period. The conversation also mentions that the original equation can be rewritten as $2y^2(x^2+y^2) = 0
  • #1
jakey
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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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  • #2
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:
 
  • #3
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?
 
  • #4
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:
 
  • #5
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...
 
  • #6
solve it anyway :smile:
 
  • #7
tiny-tim said:
solve it anyway :smile:

Well, r = 0 or cos (2\theta) = 1. how is this going to help, tiny-tim?
 
  • #8
jakey said:
Well, r = 0 or cos (2\theta) = 1.

and what curve is that? :wink:
 
  • #9
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.
 

What does it mean to "parametrize" a curve?

Parametrizing a curve means to represent the curve using a set of equations or parameters. This allows for a more precise description of the curve and makes it easier to calculate various properties of the curve.

Why is it important to parametrize a curve?

Parametrization allows for a more detailed and accurate understanding of the curve, which is crucial in many scientific and mathematical applications. It also allows for easier calculations of derivatives, integrals, and other properties of the curve.

How do you parametrize a curve?

The process of parametrizing a curve involves choosing one or more parameters and expressing the coordinates of points on the curve as a function of these parameters. This can be done using various methods, such as using trigonometric functions, polynomial functions, or parametric equations.

What are the benefits of using a parametric representation for a curve?

Parametric representation allows for a more versatile and flexible description of a curve, as it is not limited to a single equation. This makes it easier to manipulate and analyze the curve, and also allows for the representation of more complex and non-linear curves.

Can any curve be parametrized?

Yes, any curve can be parametrized. However, the choice of parameters and the method of parametrization may vary depending on the curve and its properties. In some cases, it may be more challenging to find a suitable parametrization, but it is always possible.

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