Conversion of parametric form to polar for the rose curve

In summary, the conversation is discussing the conversion from the general parametric form to the general polar form for the Rhodonea curve, also known as the rose curve. The parametric equation is given as x=cos(k(theta))cos(theta) and y=cos(k(theta))sin(theta), and the main question is how this was converted to the polar equation r=cos(k). It is noted that a few different equations can also work for this conversion. The conversation also mentions that parametric descriptions are not unique.
  • #1
Alphonso2001
2
0
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!
 
Last edited:
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  • #2
Alphonso2001 said:
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.
 
  • #3
Hi
mfb said:
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
 
  • #4
##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
 

Related to Conversion of parametric form to polar for the rose curve

1. How do I convert a parametric form to polar for the rose curve?

To convert a parametric form to polar for the rose curve, you can use the following equations:

x = r * cos(theta)

y = r * sin(theta)

where r is the radial distance and theta is the angle in radians.

2. What is the purpose of converting parametric form to polar for the rose curve?

The purpose of converting parametric form to polar for the rose curve is to represent the curve in terms of its polar coordinates, which can provide a more intuitive understanding of its shape and behavior.

3. Can you provide an example of converting parametric form to polar for the rose curve?

Sure, let's take the parametric form x = sin(3t) and y = sin(4t). To convert this to polar form, we can use the equations mentioned in the first question. This will result in r = sin(3t) and theta = 4t. So the polar form of the rose curve would be r = sin(3theta) with 0 <= theta <= 2pi.

4. Is it possible to convert any parametric form to polar for the rose curve?

Yes, it is possible to convert any parametric form to polar for the rose curve as long as the equations for x and y can be expressed in terms of r and theta.

5. Can polar equations be converted back to parametric form for the rose curve?

Yes, polar equations can be converted back to parametric form for the rose curve by using the inverse equations:

x = r * cos(theta)

y = r * sin(theta)

where r is the radial distance and theta is the angle in radians.

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