Conversion of parametric form to polar for the rose curve

[Alphonso2001]

Hi,

The main question revolves around the Rhodonea curve AKA rose curve. The polar equation given for the curve is r=cos(k). The parametric equation is = cos(k(theta)) cos (theta), = cos(k(theta)) sin(theta) . 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 meThanks!

 

[mfb]

The parametric equation is = cos cos (), = cos () sin

There is something missing.

You can find x and y as function of k, this gives you a natural parametrization. Every monotonous function k(t) that covers the same range will work as well, plus a few more exotic examples.

Parametric descriptions are never unique.

 

[Alphonso2001]

Hi

There is something missing.

You can find x and y as function of k, this gives you a natural parametrization. Every monotonous function k(t) that covers the same range will work as well, plus a few more exotic examples.

Parametric descriptions are never unique.

My bad. I edited it

 

[lurflurf]

$x=r\cos(\theta)=\cos(k \theta)\cos(\theta)\\ y=r\sin(\theta)=\cos(k \theta)\sin(\theta)$
either equation gives the polar equation and r can be isolated by cancellation