Exploring Polar Rose: Where Did I Go Wrong?

[JanClaesen]

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?

 

[arildno]

"(2x)^2 + (1-2y)^2 = 1" This is an equation for a circle.

Your parametrization is not.

 

[tiny-tim]

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]

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]

Hint: multiply both sides by r².

 

[HallsofIvy]

And use the identity $sin(\theta)= 2sin(\theta)cos(\theta)$.

 

[tiny-tim]

And use the identity $sin(\theta)= 2sin(\theta)cos(\theta)$.

He knows that …

Since sin t * cos t = (1/2) * sin 2t …

(and have a theta: θ )​

 

[JanClaesen]

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]

(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]

(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]

Is there a human way to do this also for sin(3θ)?

Hint: try it for cos(3θ) + isin(3θ)