# Parametrize this curve

by jakey
Tags: curve, parametrize
 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?
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P: 26,148
hi jakey!

(have a theta: θ and try using the X2 icon just above the Reply box )
 Quote by jakey 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
P: 51
 Quote by tiny-tim 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?

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P: 26,148
Parametrize this curve

hi jakey!
 Quote by 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?
P: 51
 Quote by tiny-tim 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...
 Sci Advisor HW Helper Thanks P: 26,148 solve it anyway
P: 51
 Quote by tiny-tim solve it anyway
Well, r = 0 or cos (2\theta) = 1. how is this gonna help, tiny-tim?
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P: 26,148
 Quote by jakey Well, r = 0 or cos (2\theta) = 1.
and what curve is that?
P: 51
 Quote by tiny-tim 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.
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P: 26,148
 Quote by jakey 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
P: 51
 Quote by tiny-tim 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)?
 Sci Advisor HW Helper Thanks P: 26,148 period? what's the question?
P: 51
 Quote by tiny-tim 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.
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P: 26,148
 Quote by jakey … 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
P: 51
 Quote by tiny-tim (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.

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