Unit 6, lesson 84

Some Polar Curves

The standard polar curves, circles, cardioids, limaçons, roses and spirals, with how to rotate them, spot their symmetry, find where two of them meet and see how they pass through the pole.

  • Theory part 167
  • Practice part 168
  • Polar Curves

Some shapes are awkward in xx and yy but almost trivial in polar form. A flower with four petals is r=cos⁡2θr = \cos 2\theta. A heart shape is r=1+cos⁡θr = 1 + \cos\theta. Knowing a small family of these curves by sight lets you read a polar equation the way you read y=x2y = x^2 as a parabola.

Key terms

Circles

Definition For a real number a≠0a \neq 0,

r=ais the circle x2+y2=a2,r = a \quad \text{is the circle } x^2 + y^2 = a^2,r=2acos⁡θis the circle (x−a)2+y2=a2,r = 2a\cos\theta \quad \text{is the circle } (x - a)^2 + y^2 = a^2,r=2asin⁡θis the circle x2+(y−a)2=a2.r = 2a\sin\theta \quad \text{is the circle } x^2 + (y - a)^2 = a^2.

The first is traced once for 0≤θ<2π0 \le \theta < 2\pi. The last two pass through the pole, and each of them is traced exactly once for 0≤θ<π0 \le \theta < \pi.

In plain words A constant radius gives a circle around the pole. A radius that follows cos⁡θ\cos\theta or sin⁡θ\sin\theta gives a circle that touches the pole and sits to the side.

Example For r=4cos⁡θr = 4\cos\theta multiply by rr to get x2+y2=4xx^2 + y^2 = 4x, so (x−2)2+y2=4(x - 2)^2 + y^2 = 4. This is the circle of radius 22 centered at (2,0)(2, 0).

Limaçons

Definition A limaçon is a curve r=b+acos⁡θr = b + a\cos\theta or r=b+asin⁡θr = b + a\sin\theta with a≠0a \neq 0 and b≠0b \neq 0. Its shape depends on the ratio ∣b∣/∣a∣\lvert b \rvert / \lvert a \rvert.

Ratio ∣b∣/∣a∣\lvert b \rvert / \lvert a \rvertShape
less than 11inner loop
equal to 11cardioid
between 11 and 22dimpled
22 or moreconvex

In plain words The constant bb pushes the curve away from the pole and the term acos⁡θa\cos\theta pulls one side in. If the pull is stronger than the push, rr turns negative for a while and the curve makes a small loop inside.

Example For r=1+2cos⁡θr = 1 + 2\cos\theta the ratio is 12\frac{1}{2}, so there is an inner loop. The radius is 00 when cos⁡θ=−12\cos\theta = -\frac{1}{2}, at θ=2π3\theta = \frac{2\pi}{3} and θ=4π3\theta = \frac{4\pi}{3}, and negative between them, where the small loop is drawn. The farthest point is (3,0)(3, 0) at θ=0\theta = 0.

Cardioids

Definition A cardioid is a limaçon with ∣b∣=∣a∣\lvert b \rvert = \lvert a \rvert, for example

r=a(1+cos⁡θ),a>0,0≤θ<2π.r = a(1 + \cos\theta), \qquad a > 0, \quad 0 \le \theta < 2\pi.

It passes through the pole exactly once, at θ=π\theta = \pi, where it has a cusp.

In plain words The heart shaped member of the limaçon family. The radius just reaches zero and comes back without going negative.

Example For r=1+cos⁡θr = 1 + \cos\theta a few values fix the shape.

θ\theta00π2\frac{\pi}{2}π\pi3π2\frac{3\pi}{2}
rr22110011

The curve reaches out to (2,0)(2, 0), crosses the yy-axis at (0,1)(0, 1) and (0,−1)(0, -1), and comes to a point at the pole on the left.

Roses

Definition For a≠0a \neq 0 and an integer n≥2n \ge 2, the curves r=acos⁡nθr = a\cos n\theta and r=asin⁡nθr = a\sin n\theta are roses. Each petal has length ∣a∣\lvert a \rvert. The rose has nn petals when nn is odd and 2n2n petals when nn is even.

In plain words As θ\theta goes around, the radius swings out and back 2n2n times. For even nn each swing makes a new petal. For odd nn the negative swings land exactly on top of the positive ones, so you see only nn petals.

Example The rose r=cos⁡2θr = \cos 2\theta has 44 petals of length 11, pointing in the four directions along the two coordinate axes. The rose r=sin⁡3θr = \sin 3\theta has 33 petals, with tips in the directions π6\frac{\pi}{6}, 5π6\frac{5\pi}{6} and 3π2\frac{3\pi}{2}.

Spirals

Definition The spiral of Archimedes is r=aθr = a\theta with a>0a > 0 and θ≥0\theta \ge 0. The logarithmic spiral is r=aebθr = ae^{b\theta} with a>0a > 0 and b≠0b \neq 0, for all real θ\theta.

In plain words In an Archimedean spiral the gap between turns stays the same. In a logarithmic spiral each turn is a fixed multiple larger than the last.

Example On r=θr = \theta the points on the positive xx-axis occur at θ=2π,4π,6π\theta = 2\pi, 4\pi, 6\pi, at distances 2π2\pi, 4π4\pi, 6π6\pi from the pole, so consecutive turns are 2π2\pi apart. On r=e0.1θr = e^{0.1\theta} each full turn multiplies the distance by e0.2π≈1.87e^{0.2\pi} \approx 1.87.

Rotations

Definition Let α\alpha be a real number. The graph of r=f(θ−α)r = f(\theta - \alpha) is the graph of r=f(θ)r = f(\theta) rotated counterclockwise about the pole through the angle α\alpha. Indeed, (r,θ)(r, \theta) satisfies r=f(θ)r = f(\theta) exactly when (r,θ+α)(r, \theta + \alpha) satisfies r=f((θ+α)−α)r = f((\theta + \alpha) - \alpha).

In plain words Subtracting α\alpha inside the function turns the whole picture by α\alpha, in the positive direction.

Example Since cos⁡(θ−π2)=sin⁡θ\cos(\theta - \frac{\pi}{2}) = \sin\theta, the cardioid r=1+sin⁡θr = 1 + \sin\theta is r=1+cos⁡θr = 1 + \cos\theta turned a quarter turn counterclockwise. Its far point moves from (2,0)(2, 0) to (0,2)(0, 2) and its cusp moves to the bottom.

Symmetry

Definition A polar curve is symmetric about the xx-axis if whenever a point (x,y)(x, y) is on it, so is (x,−y)(x, -y), and similarly for the yy-axis and the pole. For r=f(θ)r = f(\theta) with ff even, f(−θ)=f(θ)f(-\theta) = f(\theta), the curve is symmetric about the xx-axis, because (r,−θ)(r, -\theta) is the mirror image of (r,θ)(r, \theta) in that axis.

In plain words Curves built from cos⁡θ\cos\theta are mirror images of themselves across the xx-axis, and curves built from sin⁡θ\sin\theta are usually mirror images across the yy-axis. You only need to plot half and reflect.

Example For r=1+2cos⁡θr = 1 + 2\cos\theta, f(−θ)=1+2cos⁡θ=f(θ)f(-\theta) = 1 + 2\cos\theta = f(\theta), so the limaçon is symmetric about the xx-axis. Plotting 0≤θ≤π0 \le \theta \le \pi and reflecting gives the whole curve.

Intersections of polar curves

Definition A point lies on both r=f(θ)r = f(\theta) and r=g(θ)r = g(\theta) when some name of it satisfies the first equation and some, possibly different, name satisfies the second. When ff and gg have period 2π2\pi, the intersection points other than the pole come from solving

f(θ)=g(θ),f(θ)=−g(θ+π),f(\theta) = g(\theta), \qquad f(\theta) = -g(\theta + \pi),

for θ\theta in [0,2π)[0, 2\pi), and you must check separately whether the pole lies on both curves.

In plain words Setting the two equations equal finds only the points that both curves reach at the same angle with the same radius. Points they reach under different names, including the pole, must be hunted down separately.

Example Take r=cos⁡2θr = \cos 2\theta and the circle r=12r = \frac{1}{2}. Solving cos⁡2θ=12\cos 2\theta = \frac{1}{2} gives θ=±π6,±5π6\theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}, four points. The circle is also r=−12r = -\frac{1}{2} under the names (−12,θ+π)(-\frac{1}{2}, \theta + \pi), and solving cos⁡2θ=−12\cos 2\theta = -\frac{1}{2} gives θ=π3,2π3,4π3,5π3\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, four more points. The pole is not on the circle, so there is nothing more to check. In all there are eight points, two on each of the four petals, and a sketch confirms it.

Behavior at the pole

Definition A curve r=f(θ)r = f(\theta) passes through the pole at every θ0\theta_0 with f(θ0)=0f(\theta_0) = 0. If ff is differentiable at θ0\theta_0 with f(θ0)=0f(\theta_0) = 0 and f′(θ0)≠0f'(\theta_0) \neq 0, then the curve is tangent at the pole to the line θ=θ0\theta = \theta_0.

In plain words A curve enters the pole heading in the direction of the angle at which rr becomes zero.

Example For r=sin⁡3θr = \sin 3\theta the zeros in [0,π)[0, \pi) are θ=0,π3,2π3\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, and f′(θ)=3cos⁡3θ=±3≠0f'(\theta) = 3\cos 3\theta = \pm 3 \neq 0 there. So the petals meet at the pole tangent to the lines y=0y = 0, y=3 xy = \sqrt{3}\,x and y=−3 xy = -\sqrt{3}\,x.

Check yourself

  1. Name the shape of r=3+2sin⁡θr = 3 + 2\sin\theta and give its largest and smallest distance from the pole.
Show the answer

The ratio 32\frac{3}{2} is between 11 and 22, so it is a dimpled limaçon. The largest distance is 55 at θ=π2\theta = \frac{\pi}{2} and the smallest is 11 at θ=3π2\theta = \frac{3\pi}{2}.

  1. How many petals does r=2cos⁡4θr = 2\cos 4\theta have, and how long is each?
Show the answer

Here n=4n = 4 is even, so there are 88 petals, each of length 22.

  1. Find all intersection points of r=cos⁡θr = \cos\theta and r=1−cos⁡θr = 1 - \cos\theta.
Show the answer

Setting the radii equal, cos⁡θ=1−cos⁡θ\cos\theta = 1 - \cos\theta gives cos⁡θ=12\cos\theta = \frac{1}{2}, so the points (12,π3)(\frac{1}{2}, \frac{\pi}{3}) and (12,−π3)(\frac{1}{2}, -\frac{\pi}{3}) are on both curves. The second equation

cos⁡θ=−(1−cos⁡(θ+π))\cos\theta = -\bigl(1 - \cos(\theta + \pi)\bigr)

simplifies to cos⁡θ=−12\cos\theta = -\frac{1}{2}. Its solutions θ=2π3\theta = \frac{2\pi}{3} and θ=4π3\theta = \frac{4\pi}{3} give the names (−12,2π3)(-\frac{1}{2}, \frac{2\pi}{3}) and (−12,4π3)(-\frac{1}{2}, \frac{4\pi}{3}), which are the two points already found. The pole is on both curves too, at θ=π2\theta = \frac{\pi}{2} on the first and θ=0\theta = 0 on the second, so there are three points in all.

How this lesson connects

It also uses trigonometric graphs.