Polar Graphs of Roses and Lemniscates

Definition Let $a \ne 0$ and let $n \ge 2$ be an integer. The graphs of

are rose curves. The curve consists of petals. Each petal starts and ends at the pole, is drawn by the angles between two consecutive zeros of the radius, which are $\frac{\pi}{n}$ apart, and reaches out to distance $\lvert a \rvert$ at its tip. For $n = 1$ the same equations give circles through the pole, not roses.

Definition In $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ with $n > 0$ the number $n$ is the angular frequency. The radius has period $\frac{2\pi}{n}$ as a function of $\theta$, so it completes $n$ full cycles while the direction $\theta$ turns once around the pole. When $n$ is a positive integer, the radius satisfies $f(\theta + 2\pi) = f(\theta)$, so after one full turn the point is back where it started and the curve closes up. When $n$ is not an integer, one turn is not enough. For a fraction such as $n = \frac{3}{2}$ the curve needs more than one turn to close, and for an irrational $n$ it never closes up. The petal rules below are for integer $n$ only.

Definition Let $a \ne 0$ and let $n \ge 2$ be an integer. The rose $r = a\cos(n\theta)$, and likewise $r = a\sin(n\theta)$, has $n$ petals when $n$ is odd and $2n$ petals when $n$ is even. The reason is the identity

which holds because adding $\pi$ to $\theta$ adds $n\pi$ to $n\theta$. For odd $n$ the name $(f(\theta + \pi), \theta + \pi) = (-f(\theta), \theta + \pi)$ is the same point as $(f(\theta), \theta)$, so the second half turn redraws the $n$ petals of the first. For even $n$ the second half turn draws the reflections of the first $n$ petals through the pole, which fall into the gaps between them. The rule applies only to an equation in exactly this form, so rewrite a disguised equation first.

Definition When $f(\theta) < 0$, the point $(f(\theta), \theta)$ lies at distance $\lvert f(\theta) \rvert$ from the pole on the ray opposite the direction $\theta$, which is the ray of angle $\theta + \pi$. So on a rose, an interval between consecutive zeros where the radius is negative draws a petal on the opposite side of the pole from the directions in that interval.

Definition Let $a > 0$. The graphs of

are lemniscates. Each is a figure eight made of two loops that meet at the pole, and each loop reaches distance $a$ from the pole. The first has its tips at $(a, 0)$ and $(-a, 0)$. The second is the first turned through $\frac{\pi}{4}$, since $\sin 2\theta = \cos 2(\theta - \frac{\pi}{4})$, and its tips lie on the line $y = x$. Multiplying the first equation by $r^2$ and using $x = r\cos\theta$ and $y = r\sin\theta$ gives its rectangular form

Multiplying by $r^2$ could only add the pole, and the pole is already on the curve, at $\theta = \frac{\pi}{4}$.

Definition For an equation $r^2 = g(\theta)$, a real radius exists only at the angles where $g(\theta) \ge 0$. At those angles the real values of $r$ are

which are the two real branches. They are different when $g(\theta) > 0$ and both equal $0$ when $g(\theta) = 0$. The graph is everything both branches draw over the angles with $g(\theta) \ge 0$, and the angles with $g(\theta) < 0$ give no points. For $r^2 = a^2\cos 2\theta$ the allowed angles in $[0, 2\pi)$ are

Definition A tracing interval for a polar equation is an interval of angles that draws the whole graph, with every real branch included when the equation fixes only $r^2$, and draws no part of it twice, apart from finitely many points such as the pole. For the rose $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ with an integer $n \ge 2$, the interval $[0, \pi)$ is a tracing interval when $n$ is odd, and the full interval $[0, 2\pi)$ is needed when $n$ is even. For the lemniscate $r^2 = a^2\cos 2\theta$, the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$ with both branches is a tracing interval.