I am beginning to learn about polar coordinates. Of course, one of the first shapes I learned was the rose curve. I am getting stuck on finding the highest and lowest values of [MATH]\theta[/MATH] that a petal can be. I know that for the endpoints you find the maximum of r. Do you simply find the max and min of [MATH]\theta[/MATH] in order to find out this answer?
Finding the area of a polar curve
- Thread starter burt
- Start date
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,750
To make sure we are talking about the same thing, please tell us what you mean by "the rose curve". In particular, this is actually a family of different curves, and it is possible you have only been shown one example. See https://en.wikipedia.org/wiki/Rose_(mathematics)
Also, how are you defining "a petal"? I would not think of it as delimited by maximum values of r but by r=0.
Finally, what do you mean by "find the max and min of θ"? It may be best if you show your actual work so we can be sure.
Also, how are you defining "a petal"? I would not think of it as delimited by maximum values of r but by r=0.
Finally, what do you mean by "find the max and min of θ"? It may be best if you show your actual work so we can be sure.
- Joined
- Nov 24, 2012
- Messages
- 3,021
Let's consider the polar equation:
[MATH]r=a\cos(n\theta)[/MATH] where \(0<a\) and \(n\in\mathbb{N}\).
The petals will begin and terminate for values of \(n\theta\) for which \(r=0\). So, we want to look at:
[MATH]\cos(n\theta)=0\implies n\theta=\frac{\pi}{2}+k\pi=\frac{\pi}{2}(2k+1)\implies \theta=\frac{\pi}{2n}(2k+1)[/MATH] where \(k\in\mathbb{Z}\).
For a particular petal, we want two "adjacent" angles beginning with an odd value for \(k\). So we could write these as:
[MATH]\theta\in\left\{\frac{\pi}{2n}(4k-1),\frac{\pi}{2n}(4k+1)\right\}[/MATH]
[MATH]r=a\cos(n\theta)[/MATH] where \(0<a\) and \(n\in\mathbb{N}\).
The petals will begin and terminate for values of \(n\theta\) for which \(r=0\). So, we want to look at:
[MATH]\cos(n\theta)=0\implies n\theta=\frac{\pi}{2}+k\pi=\frac{\pi}{2}(2k+1)\implies \theta=\frac{\pi}{2n}(2k+1)[/MATH] where \(k\in\mathbb{Z}\).
For a particular petal, we want two "adjacent" angles beginning with an odd value for \(k\). So we could write these as:
[MATH]\theta\in\left\{\frac{\pi}{2n}(4k-1),\frac{\pi}{2n}(4k+1)\right\}[/MATH]