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June 23rd, 2018, 12:29 AM   #1
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Parametric to Polar conversion of Rose Curve

Hi,

The main question revolves around the Rhodonea curve AKA rose curve. The polar equation given for the curve is r = cos(kθ). The parametric equation is x = cos(kθ)cos(θ), y = cos(kθ)sin(θ). Can anyone show me the conversion from the general parametric form to the general polar form? Basically, what I am looking for is the working. How did the parametric form get converted to polar?


P.S. In the case that the aforementioned doesn't happen, even if you are able to find the general rectangular coordinate form for the given polar equation above, it will work for me.


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Last edited by skipjack; June 24th, 2018 at 06:29 PM.
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June 23rd, 2018, 01:28 AM   #2
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I would use the parameter $\displaystyle t$ and:

$\displaystyle x(t)=r\cos(t)$

$\displaystyle y(t)=r\sin(t)$

We are given:

$\displaystyle r=\cos(k\theta)$

Hence:

$\displaystyle x(t)=\cos(k\theta)\cos(t)$

$\displaystyle y(t)=\cos(k\theta)\sin(t)$
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June 23rd, 2018, 11:40 PM   #3
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The following video may be of interest.

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June 24th, 2018, 06:34 PM   #4
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Quote:
Originally Posted by MarkFL View Post
I would use the parameter $\displaystyle t$ and . . .
Eh?
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June 24th, 2018, 07:06 PM   #5
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Oops...I meant:

$\displaystyle x(t)=\cos(kt)\cos(t)$

$\displaystyle y(t)=\cos(kt)\sin(t)$
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June 24th, 2018, 07:45 PM   #6
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