Register to reply

Parametrize this curve

by jakey
Tags: curve, parametrize
Share this thread:
jakey
#1
Apr3-11, 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?
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
tiny-tim
#2
Apr3-11, 07:38 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
hi jakey!

(have a theta: θ and try using the X2 icon just above the Reply box )
Quote Quote by jakey View Post
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
#3
Apr3-11, 07:42 AM
P: 51
Quote Quote by tiny-tim View Post
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
#4
Apr3-11, 07:47 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
Parametrize this curve

hi jakey!
Quote Quote by jakey View Post
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
#5
Apr3-11, 07:51 AM
P: 51
Quote Quote by tiny-tim View Post
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
#6
Apr3-11, 07:57 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
solve it anyway
jakey
#7
Apr3-11, 08:01 AM
P: 51
Quote Quote by tiny-tim View Post
solve it anyway
Well, r = 0 or cos (2\theta) = 1. how is this gonna help, tiny-tim?
tiny-tim
#8
Apr3-11, 08:02 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
Quote Quote by jakey View Post
Well, r = 0 or cos (2\theta) = 1.
and what curve is that?
jakey
#9
Apr3-11, 08:09 AM
P: 51
Quote Quote by tiny-tim View Post
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
#10
Apr3-11, 08:17 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
Quote Quote by jakey View Post
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
#11
Apr3-11, 08:21 AM
P: 51
Quote Quote by tiny-tim View Post
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
#12
Apr3-11, 08:23 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
period?
what's the question?
jakey
#13
Apr3-11, 08:28 AM
P: 51
Quote Quote by tiny-tim View Post
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
#14
Apr3-11, 08:36 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
Quote Quote by jakey View Post
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
#15
Apr3-11, 08:47 AM
P: 51
Quote Quote by tiny-tim View Post
(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.


Register to reply

Related Discussions
Parametrize surfaces Calculus & Beyond Homework 0
Parametrize trajectory of a hocjey puck Calculus & Beyond Homework 2
Parametrize a surface Calculus & Beyond Homework 4
Parametrize a cylinder Calculus & Beyond Homework 4
Find a vector parametrization for: y^2+2x^2-2x=10 Calculus & Beyond Homework 1