Check whether the curve repeats
A circle drawn twice
The orange point traces the circle \(r = 5\sin\theta\), which passes through the pole. The teal region shows the area that the turning ray covers.
Drag \(\theta\) from 0 to \(\pi\). Here \(\sin\theta \ge 0\), so the point is on the ray. It goes once round the circle and returns to the pole at \(\theta = \pi\), and the disk is complete.
Drag \(\theta\) from \(\pi\) to \(2\pi\). The radius is negative, so the point is behind the pole. It traces the circle again, and the disk has a second, darker layer of teal.
An area integral over \(0 \le \theta \le 2\pi\) counts the disk twice, so this circle needs only \(0 \le \theta \le \pi\).
Two points, half a turn apart
A curve \(r = f(\theta)\) repeats when it returns to points that it drew before.
The orange point A is on the curve at the angle \(\theta\). The orange point B is half a turn later, at the angle \(\theta + \pi\) and the distance \(f(\theta + \pi)\) from the pole. When A and B are different points, a red segment joins them.
For the circle, A and B are the same point at every angle, so the circle repeats after half a turn.
Press the first button to change the curve to a rose with four petals. B is now on the other side of the pole from A, so the rose does not repeat after half a turn.
The pairs \((r, \theta)\) and \((-r, \theta + \pi)\) name the same point, so A and B meet when \(f(\theta + \pi) = -f(\theta)\).
Notation
A full turn points the ray in the same direction. The curve repeats after \(2\pi\) if the radius is the same.
\[f(\theta + 2\pi) = f(\theta)\]This holds for the sine or the cosine of a whole-number multiple of \(\theta\).
A half turn points the ray the opposite way. The curve repeats after \(\pi\) if the radius changes sign, because then B is at A.
\[f(\theta + \pi) = -f(\theta)\]If \(f(\theta + \pi) = f(\theta)\), B is a new point on the opposite side of the pole.
Use the shorter interval that works, starting at any angle \(\theta_0\).
\[\begin{aligned} \theta_0 &\le \theta \le \theta_0 + \pi \\ \theta_0 &\le \theta \le \theta_0 + 2\pi \end{aligned}\]For a rose \(r = a\sin n\theta\) or \(r = a\cos n\theta\), the test depends on \(n\).
\[\sin(n\theta + n\pi) = (-1)^n \sin n\theta\]If \(n\) is odd, half a turn is enough, and the rose has \(n\) petals. If \(n\) is even, a full turn is needed, and the rose has \(2n\) petals.
When it fails
The spiral \(r = \frac{\theta}{3}\) does not repeat, because its radius grows on every turn, so B never meets A.
Some curves repeat only after more than a full turn. The curve \(r = \sin\frac{\theta}{2}\) repeats after \(4\pi\), because \(\sin\frac{\theta + 4\pi}{2} = \sin\frac{\theta}{2}\).
Example 1: a rose with an odd number
Find the shortest interval \(0 \le \theta \le T\) that traces \(r = 4\sin 5\theta\) once. Count the petals.
Compare the radius half a turn later with the radius now. The orange points A and B, at the angles \(\theta\) and \(\theta + \pi\), are the same point.
\[\begin{aligned} r(\theta + \pi) &= 4\sin(5\theta + 5\pi) \\ &= -4\sin 5\theta = -r(\theta) \end{aligned}\]The radius changes sign, so the curve repeats after half a turn. The teal region shows that the ray covers the rose once for \(0 \le \theta \le \pi\).
\[T = \pi\]Because \(n = 5\) is odd, the rose has 5 petals. Each petal runs between two neighboring zeros of \(r\), \(\frac{\pi}{5}\) apart, so the five petals use \(5\cdot\frac{\pi}{5} = \pi\).
Example 2: a rose with an even number
Find the shortest interval \(0 \le \theta \le T\) that traces \(r = 3\sin 4\theta\) once.
Compare the radius half a turn later with the radius now. They are equal, so B is on the opposite side of the pole from A.
\[r(\theta + \pi) = 3\sin(4\theta + 4\pi) = r(\theta)\]A tempting wrong move: The radius repeats, so the curve repeats after half a turn. Use \(0 \le \theta \le \pi\).
\[T = \pi\]The figure shows that \(0 \le \theta \le \pi\) traces only four of the eight petals. The next half turn draws four new petals in the gaps.
The test for a repeat is \(r(\theta + \pi) = -r(\theta)\). Here the sign is the same, so B is at a new point.
Test a full turn. The ray points the same way and the radius is the same, so the curve repeats after \(2\pi\), and \(T = 2\pi\). Because \(n = 4\) is even, the rose has 8 petals.
\[r(\theta + 2\pi) = 3\sin(4\theta + 8\pi) = r(\theta)\]
Practice
Find the shortest interval \(0 \le \theta \le T\) that traces each curve once: \(r = 6\cos 3\theta\) and \(r = 3 + 2\cos\theta\).
Compare \(f(\theta + \pi)\) with \(-f(\theta)\) first. Then drag \(\theta\) and stop where the teal region is complete, before any part turns darker.
Solution
For \(6\cos 3\theta\), the radius half a turn later changes sign, so the rose repeats after half a turn and \(T = \pi\).
\[6\cos(3\theta + 3\pi) = -6\cos 3\theta\]For \(3 + 2\cos\theta\), the radius is at least 1, so it cannot change sign. A full turn repeats the curve because \(\cos(\theta + 2\pi) = \cos\theta\), so \(T = 2\pi\).
Questions
In the figures, the orange ray turns from \(\theta = 0\), and the teal region shows what it covers. Questions 1 and 2 use the rose \(r = 3\sin 2\theta\), with the ray at \(\theta = \pi\).
How many petals has the ray drawn so far?
- 2
- 4
- 1
Answer
2
How far must \(\theta\) go to trace the whole rose once?
- To \(2\pi\)
- To \(\pi\)
- To \(4\pi\)
Answer
To \(2\pi\)
Which interval traces \(r = 4 + 3\cos\theta\) once?
- \(0 \le \theta \le 2\pi\)
- \(0 \le \theta \le \pi\)
- \(0 \le \theta \le 4\pi\)
Answer
\(0 \le \theta \le 2\pi\)
Summary
You compare \(f(\theta + \pi)\) with \(-f(\theta)\) and \(f(\theta + 2\pi)\) with \(f(\theta)\), and you use the shortest interval that traces the curve once.
This works because the pairs \((r, \theta)\) and \((-r, \theta + \pi)\) name the same point, so a change of sign after half a turn repeats the curve.
You use it before you integrate, so that you count no region twice.
This is one of four skills in topic 9.8 of Which & Why: AP Calculus.