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June 22nd, 2018, 11:29 PM  #1 
Newbie Joined: Jun 2018 From: India Posts: 1 Thanks: 0  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. Thanks! Last edited by skipjack; June 24th, 2018 at 05:29 PM. 
June 23rd, 2018, 12:28 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
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: $\displaystyle x(t)=\cos(k\theta)\cos(t)$ $\displaystyle y(t)=\cos(k\theta)\sin(t)$ 
June 23rd, 2018, 10:40 PM  #3 
Member Joined: Oct 2017 From: Japan Posts: 60 Thanks: 2 
The following video may be of interest. 
June 24th, 2018, 05:34 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,191 Thanks: 1649  
June 24th, 2018, 06:06 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
Oops...I meant: $\displaystyle x(t)=\cos(kt)\cos(t)$ $\displaystyle y(t)=\cos(kt)\sin(t)$ 
June 24th, 2018, 06:45 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
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conversion, curve, parametric, polar, rose 
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