Unit 6, lesson 85

Sketching Polar Curves

A reliable way to sketch a polar curve by hand, using the sign of the radius, symmetry tests, the zeros of the radius and the right interval of angles so that every piece is drawn once.

  • Theory part 169
  • Practice part 170
  • Polar Curves

Plotting a polar curve point by point works, but it is slow and easy to get wrong when the radius turns negative. A better plan is to read the curve from the graph of rr against θ\theta. Where rr is large the curve is far out, where rr is zero it touches the pole, and where rr is negative it is drawn on the other side.

Key terms

Angular interval

Definition The angular interval of a sketch is an interval II of θ\theta values such that the points (f(θ),θ)(f(\theta), \theta) for θ∈I\theta \in I make up the whole curve r=f(θ)r = f(\theta). If ff has period 2π2\pi, any interval of length 2π2\pi works. If moreover f(θ+π)=−f(θ)f(\theta + \pi) = -f(\theta) for all θ\theta, any interval of length π\pi already gives the whole curve, because the name

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

is the same point as (f(θ),θ)(f(\theta), \theta).

In plain words Before you plot, decide how far θ\theta has to go. Going too short misses parts of the curve, and going too far draws parts again.

Example For r=sin⁡3θr = \sin 3\theta we have sin⁡(3θ+3π)=−sin⁡3θ\sin(3\theta + 3\pi) = -\sin 3\theta, so 0≤θ≤π0 \le \theta \le \pi gives the whole rose. For r=1+cos⁡θr = 1 + \cos\theta that relation fails, and you need a full interval 0≤θ≤2π0 \le \theta \le 2\pi.

Sign of the radius

Definition On an interval where f(θ)>0f(\theta) > 0 the point (f(θ),θ)(f(\theta), \theta) lies on the ray of angle θ\theta. On an interval where f(θ)<0f(\theta) < 0 it lies on the opposite ray, of angle θ+π\theta + \pi, at distance ∣f(θ)∣\lvert f(\theta) \rvert.

In plain words Make a sign chart for rr. The negative stretches are drawn in the opposite quadrant from where θ\theta points.

Example For r=1+2cos⁡θr = 1 + 2\cos\theta the radius is negative for 2π3<θ<4π3\frac{2\pi}{3} < \theta < \frac{4\pi}{3}. At θ=π\theta = \pi we get r=−1r = -1, which is the point (1,0)(1, 0) on the right. So the inner loop lies on the right side of the pole even though these angles point left.

Symmetry tests

Definition The polar curve r=f(θ)r = f(\theta) has the following symmetries if its equation is unchanged by at least one of the listed substitutions.

Symmetric aboutReplace (r,θ)(r, \theta) by
the xx-axis(r,−θ)(r, -\theta) or (−r,π−θ)(-r, \pi - \theta)
the yy-axis(r,π−θ)(r, \pi - \theta) or (−r,−θ)(-r, -\theta)
the pole(−r,θ)(-r, \theta) or (r,θ+π)(r, \theta + \pi)

These tests are sufficient but not necessary. A curve can have a symmetry that a particular substitution fails to show.

In plain words Each substitution gives a name of the reflected point. If the equation does not notice the change, the reflected point is on the curve too.

Example For r=sin⁡2θr = \sin 2\theta, replacing (r,θ)(r, \theta) by (r,−θ)(r, -\theta) gives r=−sin⁡2θr = -\sin 2\theta, which is a different equation. But replacing by (−r,π−θ)(-r, \pi - \theta) gives

−r=sin⁡(2π−2θ)=−sin⁡2θ,-r = \sin(2\pi - 2\theta) = -\sin 2\theta,

which is the original equation. So the four petal rose is symmetric about the xx-axis, and only the second test shows it.

Tracing direction

Definition For a curve r=f(θ)r = f(\theta), as θ\theta increases the polar angle of the point (f(θ),θ)(f(\theta), \theta) is θ\theta when f(θ)>0f(\theta) > 0 and θ+π\theta + \pi when f(θ)<0f(\theta) < 0. In both cases it increases, so away from the pole the point always turns counterclockwise around the pole.

In plain words Follow the curve as θ\theta grows and the point sweeps around the pole counterclockwise, even when rr is negative.

Example For r=cos⁡θr = \cos\theta on [0,π][0, \pi], the point starts at (1,0)(1, 0) when θ=0\theta = 0, passes (12,12)(\frac{1}{2}, \frac{1}{2}) at θ=π4\theta = \frac{\pi}{4}, reaches the pole at θ=π2\theta = \frac{\pi}{2}, then passes (12,−12)(\frac{1}{2}, -\frac{1}{2}) at θ=3π4\theta = \frac{3\pi}{4} and returns to (1,0)(1, 0). The circle is traced once, counterclockwise.

Pole crossings

Definition The curve r=f(θ)r = f(\theta) meets the pole at each θ0\theta_0 with f(θ0)=0f(\theta_0) = 0. If ff is continuous and changes sign at θ0\theta_0, the curve passes through the pole and continues on the opposite side. If ff does not change sign at θ0\theta_0, the curve reaches the pole and turns back to the same side, as at a cusp.

In plain words Zeros of rr are where the curve touches the center. Whether rr changes sign there tells you if the curve goes straight through or bounces back.

Example For r=cos⁡2θr = \cos 2\theta the zero at θ=π4\theta = \frac{\pi}{4} is a sign change from positive to negative, so the curve goes through the pole there. For r=1+cos⁡θr = 1 + \cos\theta the zero at θ=π\theta = \pi has r≥0r \ge 0 on both sides, so the curve comes in and goes back out, making the cusp of the cardioid.

Petals and loops

Definition Let θ1<θ2\theta_1 < \theta_2 be consecutive zeros of a continuous ff, with ff of one sign on (θ1,θ2)(\theta_1, \theta_2). The arc made of the points

(f(θ),θ),θ1≤θ≤θ2,(f(\theta), \theta), \qquad \theta_1 \le \theta \le \theta_2,

starts and ends at the pole and is a closed loop. For a rose it is called a petal, and its tip points in the direction θ\theta where ∣f∣\lvert f \rvert is largest, or the opposite direction if f<0f < 0.

In plain words Between two neighboring zeros of rr the curve leaves the pole, goes out and comes back. That is one petal or one loop.

Example The zeros of r=cos⁡2θr = \cos 2\theta are ±π4,±3π4\pm\frac{\pi}{4}, \pm\frac{3\pi}{4}. On [−π4,π4][-\frac{\pi}{4}, \frac{\pi}{4}] the radius is positive with peak 11 at θ=0\theta = 0, a petal pointing right. On [π4,3π4][\frac{\pi}{4}, \frac{3\pi}{4}] it is negative with peak size 11 at θ=π2\theta = \frac{\pi}{2}, so this petal points in the direction 3π2\frac{3\pi}{2}, down. Four such arcs make the four petals.

Avoiding repeated tracing

Definition An interval II of angles traces the curve r=f(θ)r = f(\theta) once if distinct θ\theta in II give distinct points, apart from finitely many crossing points such as the pole. For calculations of area and length, use an interval that gives the whole curve, or the part you want, and traces it once.

In plain words A nice drawing does not care if you pass over the same petal twice, but an integral does.

Example For r=sin⁡3θr = \sin 3\theta, the interval [0,π][0, \pi] traces the three petals once. On [π,2π][\pi, 2\pi] write θ=φ+π\theta = \varphi + \pi with 0≤φ≤π0 \le \varphi \le \pi. Then

(sin⁡3θ,θ)=(−sin⁡3φ,φ+π),(\sin 3\theta, \theta) = (-\sin 3\varphi, \varphi + \pi),

which is the point (sin⁡3φ,φ)(\sin 3\varphi, \varphi) already drawn. So [0,2π][0, 2\pi] traces every petal twice.

Check yourself

  1. On which interval of θ\theta is the circle r=2cos⁡θr = 2\cos\theta traced exactly once?
Show the answer

It is traced once on 0≤θ<π0 \le \theta < \pi. Since 2cos⁡(θ+π)=−2cos⁡θ2\cos(\theta + \pi) = -2\cos\theta, the interval [π,2π)[\pi, 2\pi) repeats the same points.

  1. For r=1−2sin⁡θr = 1 - 2\sin\theta, find the angles where r<0r < 0. What part of the curve do they draw?
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The radius is negative when sin⁡θ>12\sin\theta > \frac{1}{2}, that is for π6<θ<5π6\frac{\pi}{6} < \theta < \frac{5\pi}{6}. These angles draw the inner loop, which lies below the pole because the points are reflected through it.

  1. Use the tests to show that r2=cos⁡2θr^2 = \cos 2\theta is symmetric about the pole.
Show the answer

Replacing (r,θ)(r, \theta) by (−r,θ)(-r, \theta) gives (−r)2=cos⁡2θ(-r)^2 = \cos 2\theta, the same equation.

How this lesson connects