Unit 6, lesson 89

Arc Lengths of Polar Curves

How the parametric length formula turns into the integral of the square root of r squared plus the square of its derivative, and how to choose the angles so a polar curve is measured exactly once.

  • Theory part 177
  • Practice part 178
  • Calculus of Polar Curves

How long is the edge of a cardioid? There is no ruler for curved lines, but there is the same idea used for parametric curves. Add up the distance covered in tiny steps of θ\theta. In polar form the step has a particularly tidy size, and many classic curves give exact answers.

Key terms

The polar speed

Definition Let ff be differentiable on an interval and write the curve r=f(θ)r = f(\theta) as x=f(θ)cos⁡θx = f(\theta)\cos\theta, y=f(θ)sin⁡θy = f(\theta)\sin\theta. Then

(dxdθ)2+(dydθ)2=f(θ)2+f′(θ)2,\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = f(\theta)^2 + f'(\theta)^2,

so the point moves along the curve at the rate

dsdθ=r2+(drdθ)2.\frac{ds}{d\theta} = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}.

In plain words A small step dθd\theta moves the point r dθr\,d\theta around the pole and drdr away from it. These two motions are at right angles, so the step has length r2+(dr/dθ)2 dθ\sqrt{r^2 + (dr/d\theta)^2}\,d\theta.

Example The cross terms cancel, since

(f′cos⁡θ−fsin⁡θ)2+(f′sin⁡θ+fcos⁡θ)2=f′2(cos⁡2θ+sin⁡2θ)+f2(sin⁡2θ+cos⁡2θ)=f2+f′2.(f'\cos\theta - f\sin\theta)^2 + (f'\sin\theta + f\cos\theta)^2 = f'^2(\cos^2\theta + \sin^2\theta) + f^2(\sin^2\theta + \cos^2\theta) = f^2 + f'^2.

For the spiral r=θr = \theta the speed is θ2+1\sqrt{\theta^2 + 1}, so at θ=2π\theta = 2\pi it is about 6.366.36.

Polar arc length

Definition Let f′f' be continuous on [a,b][a, b] and suppose the curve r=f(θ)r = f(\theta) is traced exactly once as θ\theta increases from aa to bb. Then its length is

L=∫abf(θ)2+f′(θ)2 dθ.L = \int_a^b \sqrt{f(\theta)^2 + f'(\theta)^2}\, d\theta.

In plain words Length equals the integral of the polar speed over an interval that traces the curve once.

Example For the cardioid r=1+cos⁡θr = 1 + \cos\theta we have f′(θ)=−sin⁡θf'(\theta) = -\sin\theta and

f2+f′2=1+2cos⁡θ+cos⁡2θ+sin⁡2θ=2+2cos⁡θ=4cos⁡2θ2.f^2 + f'^2 = 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta = 2 + 2\cos\theta = 4\cos^2\frac{\theta}{2}.

The square root is 2∣cos⁡θ2∣2\left\lvert\cos\frac{\theta}{2}\right\rvert, and the absolute value matters. On [0,2π][0, 2\pi],

L=∫0π2cos⁡θ2 dθ+∫π2π(−2cos⁡θ2)dθ=4+4=8.L = \int_0^{\pi} 2\cos\frac{\theta}{2}\, d\theta + \int_{\pi}^{2\pi} \left(-2\cos\frac{\theta}{2}\right) d\theta = 4 + 4 = 8.

Smoothness

Definition The length formula above assumes that ff is continuously differentiable on [a,b][a, b], that is, f′f' exists and is continuous there, with one-sided derivatives at the endpoints. The parametrization is regular where the speed f2+f′2\sqrt{f^2 + f'^2} is positive, which fails only where f(θ)=0f(\theta) = 0 and f′(θ)=0f'(\theta) = 0 at once, so only at the pole. Regularity is not needed for the length formula. The integrand f2+f′2\sqrt{f^2 + f'^2} is continuous whenever f′f' is, so the integral still gives the length when the point stops for an instant at the pole.

In plain words You need f′f' to be continuous. The curve itself may still have a cusp at the pole, where the point comes to rest for an instant, and the integral handles it without trouble.

Example The cardioid r=1+cos⁡θr = 1 + \cos\theta has f(π)=0f(\pi) = 0 and f′(π)=−sin⁡π=0f'(\pi) = -\sin\pi = 0, so the speed is 00 at θ=π\theta = \pi, which is the cusp at the pole. Since f′(θ)=−sin⁡θf'(\theta) = -\sin\theta is continuous, the formula applies, and it gave the length 88 above.

Parameter interval

Definition The parameter interval for a length computation must trace each part of the curve exactly once. If an interval runs over the whole curve exactly kk times, the integral equals kk times the length.

In plain words Find a range of θ\theta that draws the whole curve exactly once, and integrate over that.

Example For the circle r=2cos⁡θr = 2\cos\theta the speed squared is 4cos⁡2θ+4sin⁡2θ=44\cos^2\theta + 4\sin^2\theta = 4. The circle is traced once on [0,π][0, \pi], so

L=∫0π2 dθ=2π,L = \int_0^{\pi} 2\, d\theta = 2\pi,

the circumference of a circle of radius 11. Integrating over [0,2π][0, 2\pi] would give 4π4\pi, twice the true length.

Piecewise curves

Definition If a path is made of arcs r=fk(θ)r = f_k(\theta), θk−1≤θ≤θk\theta_{k-1} \le \theta \le \theta_k, with each fkf_k continuously differentiable on its interval and each arc traced once, then the length of the path is

L=∑k∫θk−1θkfk(θ)2+fk′(θ)2 dθ.L = \sum_k \int_{\theta_{k-1}}^{\theta_k} \sqrt{f_k(\theta)^2 + f_k'(\theta)^2}\, d\theta.

The same applies to a single curve r=f(θ)r = f(\theta) whose derivative has finitely many jumps, by splitting at the jumps.

In plain words Boundaries made of pieces of different curves are measured piece by piece, then added.

Example The region inside both r=sin⁡θr = \sin\theta and r=cos⁡θr = \cos\theta is bounded by r=sin⁡θr = \sin\theta for 0≤θ≤π40 \le \theta \le \frac{\pi}{4} and r=cos⁡θr = \cos\theta for π4≤θ≤π2\frac{\pi}{4} \le \theta \le \frac{\pi}{2}. Each piece has speed sin⁡2θ+cos⁡2θ=1\sqrt{\sin^2\theta + \cos^2\theta} = 1, so the boundary has length π4+π4=π2\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.

Check yourself

  1. Find the length of the spiral r=eθr = e^{\theta} for 0≤θ≤10 \le \theta \le 1.
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Here f=f′=eθf = f' = e^{\theta}, so f2+f′2=2e2θf^2 + f'^2 = 2e^{2\theta} and

L=∫012 eθ dθ=2(e−1)≈2.43.L = \int_0^1 \sqrt{2}\,e^{\theta}\, d\theta = \sqrt{2}(e - 1) \approx 2.43.
  1. Find the length of r=θ2r = \theta^2 for 0≤θ≤50 \le \theta \le \sqrt{5}.
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With f′=2θf' = 2\theta the speed is θ4+4θ2=θθ2+4\sqrt{\theta^4 + 4\theta^2} = \theta\sqrt{\theta^2 + 4} for θ≥0\theta \ge 0, so

L=∫05θθ2+4 dθ=[13(θ2+4)3/2]05=27−83=193.L = \int_0^{\sqrt{5}} \theta\sqrt{\theta^2 + 4}\, d\theta = \left[\frac{1}{3}(\theta^2 + 4)^{3/2}\right]_0^{\sqrt{5}} = \frac{27 - 8}{3} = \frac{19}{3}.
  1. Someone finds the length of r=3sin⁡θr = 3\sin\theta by integrating over [0,2π][0, 2\pi]. What do they get, and what is the right answer?
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The speed is 33, so they get 6π6\pi. The circle is traced once on [0,π][0, \pi], so its length is 3π3\pi.

How this lesson connects