My Math Forum Parametric to Polar conversion of Rose Curve

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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: 12,204 Thanks: 511 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: $\displaystyle x(t)=\cos(k\theta)\cos(t)$ $\displaystyle y(t)=\cos(k\theta)\sin(t)$ Thanks from Isbell
 June 23rd, 2018, 10:40 PM #3 Member   Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3 The following video may be of interest.
June 24th, 2018, 05:34 PM   #4
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Quote:
 Originally Posted by MarkFL I would use the parameter $\displaystyle t$ and . . .
Eh?

 June 24th, 2018, 06:06 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Oops...I meant: $\displaystyle x(t)=\cos(kt)\cos(t)$ $\displaystyle y(t)=\cos(kt)\sin(t)$ Thanks from Isbell
 June 24th, 2018, 06:45 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Like this: Thanks from Isbell

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