Polar curve

In polar coordinates a point is given by an angle θ, measured from the positive x-axis, and a signed distance r along that direction. A polar curve $r = f(\theta)$ is the set of points that have a name (r, θ) fitting the equation.

In the polar graph, θ and its ray are red, and r is the yellow segment from the origin to the blue pen. The r–θ graph draws r against θ as an ordinary graph, with r on the same scale. Where that graph crosses the θ-axis, the curve passes through the origin. Where it lies below the axis, the pen is on the far side of the origin, and yellow spokes mark the stretch drawn so far with r < 0.

A negative r puts the point |r| from the origin on the opposite ray, so $(-r, \theta + \pi)$ is the same point as $(r, \theta)$. A point therefore has many names, and it lies on the curve when one of them fits, not necessarily all.

A rose $r = \cos k\theta$, with k a positive integer, draws one petal between each two neighbouring zeros of $\cos k\theta$, which are π/k apart. Whether a full turn gives 2k petals or k petals drawn twice comes from one line, $\cos k(\theta + \pi) = (-1)^k \cos k\theta$. For odd k, adding π to θ reverses the ray and changes the sign of r, which leaves the point where it was. The second half-turn therefore retraces the first. Each of the k petals is drawn once with r > 0 and once with r < 0. For even k the point at θ + π is the point at θ turned half a turn about the origin, so all 2k petals are different, half of them drawn by negative r. The 1× and 2× outside the tips count how many times each whole petal has been drawn so far.

A limaçon $r = a + b\cos\theta$ with a, b > 0 takes one of four shapes, decided by b/a alone, so the page fixes a = 1 and turns b. For b ≤ 1/2 the curve is convex, for 1/2 < b < 1 it has a dimple, and at b = 1 it is the cardioid $r = 1 + \cos\theta$. For b > 1, r is negative while $\cos\theta < -1/b$, and those θ draw an inner loop on the far side of the origin.

Differentiating x and y with respect to θ gives the pen’s speed per radian. On the cardioid it is $ds/d\theta = 2\lvert\cos(\theta/2)\rvert$, which is 0 at θ = π. The pen stops there, and the curve has a cusp at the origin. A rose with k > 1, by contrast, passes through the origin at its fastest.

The Archimedean spiral $r = \theta$, here for 0 ≤ θ ≤ 4π, moves outward at a steady rate as it turns. θ and θ + 2π point along the same ray but not to the same point: each turn crosses every ray 2π farther out.