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Polar rose

by JanClaesen
Tags: polar, rose
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JanClaesen
#1
Nov7-09, 03:19 PM
P: 56
I'm trying to express the polar rose as an implicit function:
r(t)=sin t

x = sin t * cos t
y = sin^2 t

Since sin t * cos t = (1/2) * sin 2t and sin^2 t = (1/2) * (1-cos 2t)

(2x)^2 + (1-2y)^2 = 1
4x^2 -4y + 4y^2 = 0

When I plot this, Maple plots a circle, where have I gone wrong?
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arildno
#2
Nov7-09, 04:17 PM
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"(2x)^2 + (1-2y)^2 = 1" This is an equation for a circle.

Your parametrization is not.
tiny-tim
#3
Nov7-09, 04:42 PM
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Hi JanClaesen!

(have a theta: θ )

A rose is usually r = ksinθ or r = kcosθ see http://en.wikipedia.org/wiki/Rose_(mathematics).

For k = 1, it is a circle.

(But you could have got the same equation if you'd just made it r2 = y )

JanClaesen
#4
Nov7-09, 06:58 PM
P: 56
Polar rose

Do you have any hints on how to find the Cartesian equation for r(θ)=sin(2θ), I really can't seem to find it. :)
tiny-tim
#5
Nov8-09, 02:49 AM
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Hint: multiply both sides by r2.
HallsofIvy
#6
Nov8-09, 04:29 AM
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And use the identity [itex]sin(\theta)= 2sin(\theta)cos(\theta)[/itex].
tiny-tim
#7
Nov8-09, 04:38 AM
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Quote Quote by HallsofIvy View Post
And use the identity [itex]sin(\theta)= 2sin(\theta)cos(\theta)[/itex].
He knows that
Quote Quote by JanClaesen View Post
Since sin t * cos t = (1/2) * sin 2t
(and have a theta: θ )
JanClaesen
#8
Nov8-09, 05:29 AM
P: 56
Wow, that was clever, thank you
For those interested:

xy = 0.5(x^2+y^2)(x^2+y^2)^(1/2)

(where x^2+y^2 = sin^2 (2θ) )
tiny-tim
#9
Nov8-09, 05:37 AM
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(try using the X2 tag just above the Reply box )

That's it!

And then expand it , and put it all on the left:
(x2 + y2)3 - 2xy = 0.
JanClaesen
#10
Nov8-09, 06:33 AM
P: 56
Quote Quote by tiny-tim View Post
(try using the X2 tag just above the Reply box )

That's it!

And then expand it , and put it all on the left:
(x2 + y2)3 - 2xy = 0.
Yep, thanks again

Is there a human way to do this also for sin(3θ)? Or would that be a computer job?
I'm trying to do this now, but I have a feeling it's quite tough.
tiny-tim
#11
Nov8-09, 06:44 AM
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Quote Quote by JanClaesen View Post
Is there a human way to do this also for sin(3θ)?
Hint: try it for cos(3θ) + isin(3θ)


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