Calculus 2

Arc length and speed on parametric and polar curves

A pen plotter’s pen is steered by two formulas of time, one for how far right it is and one for how far up. How fast is the pen moving along its line at a given moment, and how long is the line it draws, when that line loops and doubles back and is not the graph of any function?

You need

Arc lengthCalculus 2 Derivatives of parametric curvesCalculus 2

Unlocks

Vector valued functions and motion in the planeCalculus 2

Where you are 5 in Parametric, polar and vector valued functions

Calculus8

Parametric, polar and vector valued functions5

Area in polar coordinates Derivatives of parametric curves Slope of a polar curve

Arc length and speed on parametric and polar curves

Vector valued functions and motion in the plane

Why this exists

Kawira programs a pen plotter: at each time t (in seconds) the pen sits at x(t) centimeters right of a corner and y(t) centimeters up. Muthoni and Kamande draw with a spirograph, a toothed wheel rolling inside a ring with a pen through one of its holes; the pen’s position is again a pair of formulas in one parameter. Two questions come up at once. How long is the line on the paper? How fast is the pen moving along it at a given instant?

Arc length answered the first question for a graph y = f(x): add up tiny hypotenuses and take a limit. But a spirograph star loops back on itself and fails the vertical line test, so it is no graph, and even a curve that is a graph can be easier to handle in its parametric form. Derivatives of parametric curves took slopes straight from \displaystyle \frac{dx}{dt} and \displaystyle \frac{dy}{dt}; this lesson takes length and speed from the same two rates. Day 1 does curves given by a parameter. Day 2 does polar curves r = f(\theta) (Polar coordinates), which are parametric curves in disguise.

Definition

Let x = x(t), y = y(t) for a \le t \le b, with x' and y' continuous on [a, b]. As t runs from a to b, the point \big(x(t), y(t)\big) moves along a path (Parametric equations).

The arc length traced from t = a to t = b is the limit of the lengths of inscribed polygons: choose a = t_0 < t_1 < \dots < t_n = b, join consecutive points by straight chords, add the chord lengths, and let the longest step t_i - t_{i-1} shrink to 0. This is the same definition as for a graph.

Write s(t) for the arc length traced from a up to time t. The speed is \displaystyle \frac{ds}{dt}, the rate at which arc length is swept out as the parameter advances. It is never negative: s only grows, whichever way the point moves.

If the parameter traces the curve once (no stretch of the curve is covered twice), the arc length from a to b is the length of the curve. If some stretch is covered twice, the same integral is the distance traveled by the moving point, and the curve itself is shorter.

A polar curve r = f(\theta), \alpha \le \theta \le \beta, is the parametric curve x = f(\theta)\cos\theta, y = f(\theta)\sin\theta with \theta as the parameter. Its arc length is defined exactly as above.

Intuition

Over a short time step \Delta t the pen moves \displaystyle \Delta x \approx \frac{dx}{dt}\Delta t across and \displaystyle \Delta y \approx \frac{dy}{dt}\Delta t up. Over so short a step the curve is nearly straight, so by the Pythagorean theorem (the The distance formula) the pen covers about

\begin{aligned} \Delta s &\approx \sqrt{(\Delta x)^2 + (\Delta y)^2} \\ &\approx \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,\Delta t. \end{aligned}

The square root is the speed: the two rates are the legs of a right triangle and the speed is its hypotenuse. Adding up \Delta s over [a, b] and letting \Delta t \to 0 turns the sum into an integral. Length is the integral of speed, just as distance traveled is the integral of speed for a car on a road.

Figure 1Kawira’s plotter line x = 6t^2, y = 4t^3, 0 \le t \le 2, and four chords. Each chord has length \displaystyle \sqrt{(\Delta x)^2 + (\Delta y)^2}; as the chords shorten, their total approaches the length of the curve.

Two things follow from the picture. The length of a curve does not depend on how fast it is drawn: a pen that moves twice as fast covers the same line in half the time. And the speed is 0 exactly when both rates are 0, which is where a curve can have a sharp point, a cusp.

Results

Speed and arc length of a parametric curve. If x' and y' are continuous on [a, b], then

\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2},

L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt.

L is the length of the curve when t traces it once, and the distance traveled otherwise.

Arc length of a polar curve. If f' is continuous on [\alpha, \beta] and r = f(\theta), then

L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta.

Graphs are a special case. With x = t, y = f(t) the parametric formula becomes \displaystyle \int_a^b \sqrt{1 + f'(x)^2}\,dx, the formula of Arc length.

The length does not depend on the parametrization. Tracing the same curve once by another parameter, at any pace and in either direction, gives the same L.

Why it is true

The parametric formula. Take a partition a = t_0 < \dots < t_n = b with steps \Delta t_i. The chord from t_{i-1} to t_i has length \displaystyle \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}. By the mean value theorem applied to x and to y separately, \Delta x_i = x'(p_i)\,\Delta t_i and \Delta y_i = y'(q_i)\,\Delta t_i for some p_i and q_i in [t_{i-1}, t_i]. So the polygon has length

\sum_{i=1}^{n} \sqrt{x'(p_i)^2 + y'(q_i)^2}\;\Delta t_i.

If q_i were p_i, this would be a Riemann sum for the speed. They may differ, but y' is continuous on the closed interval [a, b], hence uniformly continuous (stated here without proof): for any \varepsilon > 0, once every step is small enough, \lvert y'(q_i) - y'(p_i) \rvert < \varepsilon. The triangle inequality gives

\left\lvert \sqrt{A^2 + B^2} - \sqrt{A^2 + C^2} \right\rvert \le \lvert B - C \rvert,

so the polygon length differs from a true Riemann sum by less than \varepsilon(b - a). As the steps shrink, the Riemann sums approach the integral of the speed and the difference goes to 0, so the polygon lengths approach L.

Speed. The arc length up to time t is \displaystyle s(t) = \int_a^t \sqrt{x'(u)^2 + y'(u)^2}\,du, and the integrand is continuous, so the The fundamental theorem of calculus gives \displaystyle \frac{ds}{dt} = \sqrt{x'(t)^2 + y'(t)^2}.

Independence of the parametrization. Suppose t = \varphi(u) with \varphi' continuous and \varphi' > 0, carrying [c, d] onto [a, b], and set X(u) = x\big(\varphi(u)\big), Y(u) = y\big(\varphi(u)\big). By the chain rule the new speed is \displaystyle \sqrt{x'(\varphi(u))^2 + y'(\varphi(u))^2}\;\varphi'(u), and substituting t = \varphi(u) turns \int_c^d of it into \int_a^b of the old speed. (Exercise 14 asks for the details and the case \varphi' < 0.)

The polar formula. With x = r\cos\theta, y = r\sin\theta and r = f(\theta), the product rule gives

\begin{aligned} \frac{dx}{d\theta} &= \frac{dr}{d\theta}\cos\theta - r\sin\theta, \\ \frac{dy}{d\theta} &= \frac{dr}{d\theta}\sin\theta + r\cos\theta. \end{aligned}

Squaring and adding, the cross terms \displaystyle \mp 2r\frac{dr}{d\theta}\sin\theta\cos\theta cancel, and \sin^2\theta + \cos^2\theta = 1 collects the rest:

\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = \left(\frac{dr}{d\theta}\right)^2 + r^2.

The parametric formula with \theta as the parameter is now the polar one. In words: \displaystyle \frac{dr}{d\theta} is the rate of moving straight away from the pole, r is the rate of moving around it (an angle step \Delta\theta sweeps an arc of about r\,\Delta\theta), and the two directions are perpendicular.

Worked examples

Example 1

Day 1: Kawira’s plotter line

Kawira’s plotter moves its pen along x = 6t^2, y = 4t^3 for 0 \le t \le 2, in centimeters and seconds. Find the pen’s speed at t = 1 and at t = 2, and the length of the line it draws.

Solution

The rates are \displaystyle \frac{dx}{dt} = 12t and \displaystyle \frac{dy}{dt} = 12t^2, both continuous. Their squares add to

144t^2 + 144t^4 = 144t^2(1 + t^2),

so the speed is \displaystyle \sqrt{144t^2(1 + t^2)} = 12\lvert t \rvert\sqrt{1 + t^2} = 12t\sqrt{1 + t^2}, since t \ge 0 on this interval. At t = 1 the speed is \displaystyle 12\sqrt{2} \approx 16.97 centimeters per second, and at t = 2 it is \displaystyle 24\sqrt{5} \approx 53.67. The pen speeds up as it goes.

Neither rate is 0 for t > 0, and x = 6t^2 is increasing, so the line is traced once. With u = 1 + t^2, du = 2t\,dt, an antiderivative of \displaystyle 12t\sqrt{1 + t^2} is 4(1 + t^2)^{3/2}; differentiating it back gives \displaystyle 4 \cdot \frac{3}{2}(1 + t^2)^{1/2} \cdot 2t = 12t\sqrt{1 + t^2}. So

\begin{aligned} L &= \int_0^2 12t\sqrt{1 + t^2}\,dt \\ &= \Big[4(1 + t^2)^{3/2}\Big]_0^2 = 4\big(5\sqrt{5} - 1\big) \\ &= 20\sqrt{5} - 4 \approx 40.72 \text{ cm}. \end{aligned}

Check: the line runs from (0, 0) to (24, 32), and the straight chord between those points has length \displaystyle \sqrt{24^2 + 32^2} = 40. The curve is a little longer than its chord, as it must be.

Example 2

Day 1: one circle drawn three ways

Muthoni draws a circle of radius 4 centimeters three ways: (a) x = 4\cos t, y = 4\sin t, 0 \le t \le 2\pi; (b) x = 4\cos(t^2), y = 4\sin(t^2), \displaystyle 0 \le t \le \sqrt{2\pi}; (c) x = 4\cos 3t, y = 4\sin 3t, 0 \le t \le 2\pi. Find the speed and the integral of the speed in each case, and say what each integral measures.

Solution

(a) The rates are -4\sin t and 4\cos t, so the speed is \displaystyle \sqrt{16\sin^2 t + 16\cos^2 t} = 4, constant. The integral is \int_0^{2\pi} 4\,dt = 8\pi, the circumference 2\pi \cdot 4.

(b) By the chain rule the rates are -8t\sin(t^2) and 8t\cos(t^2), so the speed is 8t (with t \ge 0): the pen starts at rest and speeds up. The angle t^2 runs from 0 to 2\pi, so the circle is traced once, and

\int_0^{\sqrt{2\pi}} 8t\,dt = \Big[4t^2\Big]_0^{\sqrt{2\pi}} = 8\pi.

Same curve, different pace, same length.

(c) The rates are -12\sin 3t and 12\cos 3t, so the speed is 12 and the integral is 24\pi. The angle 3t runs from 0 to 6\pi: the pen goes around three times. The integral 24\pi is the distance the pen traveled; the circle on the paper is still 8\pi long.

Example 3

Day 1: the spirograph star

Kamande’s ring has radius 10 centimeters and his wheel radius 2, with the pen on the rim of the wheel. The pen follows

\begin{aligned} x &= 8\cos t + 2\cos 4t, \\ y &= 8\sin t - 2\sin 4t, \end{aligned}

for 0 \le t \le 2\pi. Find the speed, the times the pen stops, and the length of the star.

Solution

The rates are \displaystyle \frac{dx}{dt} = -8\sin t - 8\sin 4t and \displaystyle \frac{dy}{dt} = 8\cos t - 8\cos 4t. Square and add; \sin^2 + \cos^2 = 1 twice and the identity \cos t\cos 4t - \sin t\sin 4t = \cos 5t give

\begin{aligned} \left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2 &= 64\big(2 - 2\cos 5t\big) \\ &= 256\sin^2\tfrac{5t}{2}, \end{aligned}

using the power-reducing formula 1 - \cos 2A = 2\sin^2 A with \displaystyle A = \frac{5t}{2}. So the speed is \displaystyle 16\left\lvert \sin\frac{5t}{2} \right\rvert. The absolute value is needed because a square root is never negative while \displaystyle \sin\frac{5t}{2} changes sign.

The pen stops where \displaystyle \sin\frac{5t}{2} = 0: \displaystyle t = 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5} (and 2\pi, the start again). There x^2 + y^2 = 100: these are the five points of the star, on the ring, where the rim of the wheel touches the ring. Between them the speed rises to 16 centimeters per second, at \displaystyle t = \frac{\pi}{5}, \frac{3\pi}{5} and so on.

The speed repeats with period \displaystyle \frac{2\pi}{5}, so the five arms have equal length, and on the first arm \displaystyle \sin\frac{5t}{2} \ge 0:

\begin{aligned} L &= 5\int_0^{2\pi/5} 16\sin\tfrac{5t}{2}\,dt \\ &= 5 \cdot \tfrac{32}{5}\Big[-\cos\tfrac{5t}{2}\Big]_0^{2\pi/5} \\ &= 32\big(1 - \cos\pi\big) = 64 \text{ cm}. \end{aligned}

The absolute value matters: \displaystyle \int_0^{2\pi} 16\sin\frac{5t}{2}\,dt is only \displaystyle \frac{64}{5}, because on alternate arms the integrand is negative and cancels.

Figure 2Kamande’s spirograph with a ring of radius 10 and a wheel of radius 2. Left, the pen on the rim of the wheel (Example 3), with five cusps where the pen stops. Right, the pen in a hole 1 centimeter from the wheel’s center (Example 4), never stopping.

Example 4

Day 1: the pen in a hole, a length that needs a number

Kamande moves the pen to a hole 1 centimeter from the wheel’s center, so the path becomes x = 8\cos t + \cos 4t, y = 8\sin t - \sin 4t, 0 \le t \le 2\pi. Find the speed, its smallest and largest values, and the length of the path.

Solution

Now \displaystyle \frac{dx}{dt} = -8\sin t - 4\sin 4t and \displaystyle \frac{dy}{dt} = 8\cos t - 4\cos 4t, and the same identities give

\begin{aligned} \left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2 &= 64 + 16 - 64\cos 5t \\ &= 16\big(5 - 4\cos 5t\big). \end{aligned}

The speed is \displaystyle 4\sqrt{5 - 4\cos 5t}. It is smallest, \displaystyle 4\sqrt{1} = 4, when \cos 5t = 1, and largest, \displaystyle 4\sqrt{9} = 12, when \cos 5t = -1. It is never 0, so this curve has no cusps; the figure shows smooth lobes. The path closes up once over [0, 2\pi], so

L = \int_0^{2\pi} 4\sqrt{5 - 4\cos 5t}\,dt.

This integrand has no elementary antiderivative (it leads to an elliptic integral), which is the usual situation for arc length. A numerical method, Simpson’s rule with a few hundred subintervals or a calculator’s integrator, gives L \approx 53.46 centimeters. A rough check: the speed stays between 4 and 12, so L lies between 4 \cdot 2\pi \approx 25.1 and 12 \cdot 2\pi \approx 75.4. The path is shorter than the star’s 64 centimeters even though it takes the same time.

Example 5

Day 2: a straight line in polar form

On Day 2 Kamande switches the plotter to polar mode. He enters r = 4\sec\theta for \displaystyle 0 \le \theta \le \frac{\pi}{3}. Find the length of the line drawn, and check it without calculus.

Solution

Here \displaystyle \frac{dr}{d\theta} = 4\sec\theta\tan\theta, so

\begin{aligned} r^2 + \left(\tfrac{dr}{d\theta}\right)^2 &= 16\sec^2\theta\,(1 + \tan^2\theta) \\ &= 16\sec^4\theta, \end{aligned}

by the identity 1 + \tan^2\theta = \sec^2\theta. The square root is 4\sec^2\theta, positive on the interval, and \displaystyle \frac{d}{d\theta}(4\tan\theta) = 4\sec^2\theta, so

L = \int_0^{\pi/3} 4\sec^2\theta\,d\theta = \Big[4\tan\theta\Big]_0^{\pi/3} = 4\sqrt{3}.

Check: r\cos\theta = 4 means x = 4 (Converting an equation between the two systems), a vertical line. At \theta = 0 the point is (4, 0), and at \displaystyle \theta = \frac{\pi}{3} it is r = 8 at angle \displaystyle \frac{\pi}{3}, the point \displaystyle (4, 4\sqrt{3}). The segment between them has length \displaystyle 4\sqrt{3}.

Example 6

Day 2: a curve that closes after three half turns

Find the length of the curve \displaystyle r = 6\sin^3\frac{\theta}{3}, and explain why 0 \le \theta \le 3\pi is the right interval.

Solution

By the chain rule, \displaystyle \frac{dr}{d\theta} = 6 \cdot 3\sin^2\frac{\theta}{3}\cos\frac{\theta}{3} \cdot \frac{1}{3} = 6\sin^2\frac{\theta}{3}\cos\frac{\theta}{3}. Then

\begin{aligned} r^2 + \left(\tfrac{dr}{d\theta}\right)^2 &= 36\sin^4\tfrac{\theta}{3}\Big(\sin^2\tfrac{\theta}{3} + \cos^2\tfrac{\theta}{3}\Big) \\ &= 36\sin^4\tfrac{\theta}{3}, \end{aligned}

a perfect square, with root \displaystyle 6\sin^2\frac{\theta}{3}.

The interval: on 0 \le \theta \le 3\pi, \displaystyle \frac{\theta}{3} runs over [0, \pi], so r \ge 0, starting and ending at the pole. For the next stretch, \displaystyle \sin\frac{\theta + 3\pi}{3} = -\sin\frac{\theta}{3}, so r(\theta + 3\pi) = -r(\theta). A negative r at the angle \theta + 3\pi points opposite to the direction \theta + 3\pi, which is the direction \theta itself: the point is the same. So [3\pi, 6\pi] retraces the curve, and [0, 3\pi] traces it once.

With \displaystyle \sin^2 u = \frac{1 - \cos 2u}{2},

\begin{aligned} L &= \int_0^{3\pi} 6\sin^2\tfrac{\theta}{3}\,d\theta \\ &= \int_0^{3\pi} 3\Big(1 - \cos\tfrac{2\theta}{3}\Big)\,d\theta \\ &= \Big[3\theta - \tfrac{9}{2}\sin\tfrac{2\theta}{3}\Big]_0^{3\pi} \\ &= 9\pi \approx 28.27. \end{aligned}

Integrating over [0, 6\pi] would give 18\pi, the distance a pen travels going around twice.

Figure 3The curve \displaystyle r = 6\sin^3\frac{\theta}{3}, traced once as \theta runs from 0 to 3\pi (Example 6). It is farthest from the pole at \displaystyle \theta = \frac{3\pi}{2}, where r = 6.

Example 7

Day 2: a limaçon, with bounds to trust the number

Muthoni draws the limaçon r = 3 + 2\cos\theta, 0 \le \theta \le 2\pi, in centimeters. Find its length, and give two bounds that confirm the value.

Solution

Since 3 + 2\cos\theta \ge 1 > 0, the curve never reaches the pole and is traced once over [0, 2\pi]. With \displaystyle \frac{dr}{d\theta} = -2\sin\theta,

\begin{aligned} r^2 + \left(\tfrac{dr}{d\theta}\right)^2 &= 9 + 12\cos\theta + 4\cos^2\theta + 4\sin^2\theta \\ &= 13 + 12\cos\theta. \end{aligned}

This is not a perfect square, and \displaystyle \int_0^{2\pi} \sqrt{13 + 12\cos\theta}\,d\theta has no elementary antiderivative. Numerically, L \approx 21.01 centimeters.

Bounds: since \displaystyle \left(\frac{dr}{d\theta}\right)^2 \ge 0, the integrand is at least r, so L \ge \int_0^{2\pi} (3 + 2\cos\theta)\,d\theta = 6\pi \approx 18.85. And 13 + 12\cos\theta \le 25, so L \le 5 \cdot 2\pi = 10\pi \approx 31.42. The value 21.01 sits between them, much nearer the lower bound, because \displaystyle \frac{dr}{d\theta} is small next to r for most of the trip.

Boundaries

The formulas need x' and y' (or f') continuous on the closed interval. A finite number of points where a rate jumps can be handled by splitting the interval; a rate that blows up, as \displaystyle \frac{dy}{dt} = \frac{1}{2\sqrt{t}} does at t = 0, makes the integral improper.
The integral measures the curve’s length only when the parameter traces it once. A curve traced twice counts its length twice (Example 2c, and [0, 6\pi] in Example 6); the integral is then the distance traveled.
The speed is a square root and never negative. When it simplifies to an expression that changes sign, such as \displaystyle 16\sin\frac{5t}{2}, the absolute value must be kept and the interval split where the sign changes.
A speed of 0 does not stop the formula from working, but it is where the curve can have a cusp (Example 3); the length stays finite.
Most arc length integrals have no elementary antiderivative; a closed form is the exception, and a numerical value is a full answer.

Where this goes next

The speed \displaystyle \sqrt{x'^2 + y'^2} is the length of the vector \displaystyle \left(\frac{dx}{dt}, \frac{dy}{dt}\right). In Vector valued functions and motion in the plane that vector is the velocity of a moving particle, and distance traveled is the integral of its length. Polar curves get their slopes from the same derivatives of x = r\cos\theta and y = r\sin\theta in Slope of a polar curve, and their areas in Area in polar coordinates. Curves already met in cardioids and limaçons and Polar circles and symmetry tests can now be measured, as in the exercises.

Exercises

Day 1: lengths of parametric curves

Find the length of x = 1 + 3t, y = 2 - 4t, 0 \le t \le 2, and check it with the distance formula.
Find the length of \displaystyle x = \frac{1}{2}e^{2t} - t, y = 2e^t, 0 \le t \le \ln 2.
Kawira’s plotter draws the curve traced by the end of a string unwound from a spool of radius 3 centimeters: x = 3\cos t + 3t\sin t, y = 3\sin t - 3t\cos t, starting at t = 0. Find the speed, the length drawn from t = 0 to t = T, and the time T at which the line drawn is 24 centimeters long.
For x = t^3 - 3t, y = 3t^2: (a) show the speed is 3(t^2 + 1) and find the length for 0 \le t \le 2; (b) the curve crosses itself; find the point and the length of the loop.

Day 1: speed, and when the integral is not the length

Muthoni’s plotter draws the ellipse x = 6\cos t, y = 2\sin t. Find the speed at t = 0, \displaystyle t = \frac{\pi}{4} and \displaystyle t = \frac{\pi}{2}, and the points where the pen moves fastest and slowest.
Find the length of the ellipse in Exercise 5 to two decimal places.
For x = 3\cos^2 t, y = 4\sin^2 t, 0 \le t \le \pi, find the integral of the speed and the length of the curve, and explain the difference.

Day 2: lengths of polar curves

For the circle r = 10\cos\theta, find the integral of \displaystyle \sqrt{r^2 + (dr/d\theta)^2} over 0 \le \theta \le \pi and over 0 \le \theta \le 2\pi, and say which is the circumference.
Find the length of the part of the cardioid r = 5 + 5\cos\theta with \displaystyle 0 \le \theta \le \frac{\pi}{3}.
The spiral r = 4e^{-3\theta/4}, \theta \ge 0, winds around the pole infinitely often. Find its length as an improper integral, and say why the answer is finite.
The limaçon r = 1 + 2\cos\theta has an inner loop, traced for \displaystyle \frac{2\pi}{3} \le \theta \le \frac{4\pi}{3}. Explain why, and find the length of the inner loop to two decimal places.
Kamande’s polar plotter turns its arm one radian per second, so \theta = t, and draws r = 2 + \sin\theta centimeters. Find the pen’s speed at \theta = 0, \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}, and where it is fastest and slowest.

Reasoning

Show that for x = t, y = f(t) the parametric arc length formula is the arc length formula for the graph y = f(x).
Let t = \varphi(u), c \le u \le d, with \varphi' continuous and never 0, carry [c, d] onto [a, b]. Show that X(u) = x(\varphi(u)), Y(u) = y(\varphi(u)) has the same arc length as x(t), y(t) on [a, b], whether \varphi increases or decreases.
A polar plotter turns at a steady \omega radians per second, so \theta = \omega t. Show that the pen’s speed is \displaystyle \omega\sqrt{r^2 + (dr/d\theta)^2}, and find it for the circle r = a.
Show that the length L of the ellipse x = 6\cos t, y = 2\sin t satisfies \displaystyle 8\sqrt{10} < L < 32, and compare with Exercise 6.
Show that a polar curve on [\alpha, \beta] has length at least \int_\alpha^\beta \lvert r \rvert\,d\theta, with equality exactly when r is constant there.
Someone claims: if the speed is never 0 on [a, b], then \displaystyle \int_a^b \frac{ds}{dt}\,dt is the length of the curve. Refute the claim with x = 2\cos t, y = 2\sin t, 0 \le t \le 4\pi.

Answers

10. The speed is \displaystyle \sqrt{9 + 16} = 5, so L = 5 \cdot 2 = 10. The endpoints are (1, 2) and (7, -6), at distance \displaystyle \sqrt{36 + 64} = 10.
\displaystyle \frac{3}{2} + \ln 2 \approx 2.19. \left(e^{2t} - 1\right)^2 + \left(2e^t\right)^2 = e^{4t} + 2e^{2t} + 1 = \left(e^{2t} + 1\right)^2, so the speed is e^{2t} + 1 and \displaystyle L = \Big[\frac{1}{2}e^{2t} + t\Big]_0^{\ln 2} = 2 + \ln 2 - \frac{1}{2}.
Speed 3t; length \displaystyle \frac{3}{2}T^2; T = 4 seconds. \displaystyle \frac{dx}{dt} = -3\sin t + 3\sin t + 3t\cos t = 3t\cos t and \displaystyle \frac{dy}{dt} = 3t\sin t, so the speed is 3t for t \ge 0, and \displaystyle \int_0^T 3t\,dt = \frac{3}{2}T^2. Setting \displaystyle \frac{3}{2}T^2 = 24 gives T^2 = 16, and T > 0.
(a) L = 14. (3t^2 - 3)^2 + (6t)^2 is 9(t^4 + 2t^2 + 1), which is 9(t^2 + 1)^2, and \displaystyle \int_0^2 3(t^2 + 1)\,dt = 3\left(\frac{8}{3} + 2\right). (b) The point (0, 9), at \displaystyle t = \pm\sqrt{3}; the loop has length \displaystyle 12\sqrt{3} \approx 20.78. Equal y values with s \ne t force s = -t, and then x(-t) = -x(t) = x(t) gives t^3 - 3t = 0, so \displaystyle t = \pm\sqrt{3}; \displaystyle \int_{-\sqrt{3}}^{\sqrt{3}} 3(t^2 + 1)\,dt = 6\sqrt{3} + 6\sqrt{3}.
2, \displaystyle 2\sqrt{5} \approx 4.47 and 6. The speed is \displaystyle \sqrt{36\sin^2 t + 4\cos^2 t} = \sqrt{4 + 32\sin^2 t}. It is largest, 6, at (0, 2) and (0, -2), where \sin^2 t = 1, and smallest, 2, at (6, 0) and (-6, 0), where \sin t = 0.
L \approx 26.73. The ellipse is traced once on [0, 2\pi], and \displaystyle \int_0^{2\pi} \sqrt{4 + 32\sin^2 t}\,dt has no elementary antiderivative; a numerical integrator gives 26.730.
The integral is 10; the curve has length 5. \displaystyle \frac{dx}{dt} = -3\sin 2t and \displaystyle \frac{dy}{dt} = 4\sin 2t, so the speed is 5\lvert\sin 2t\rvert and \int_0^\pi 5\lvert\sin 2t\rvert\,dt = 2\int_0^{\pi/2} 5\sin 2t\,dt = 10. Since \displaystyle \frac{x}{3} + \frac{y}{4} = 1 with x, y \ge 0, the curve is the segment from (3, 0) to (0, 4), of length 5; the point goes out at \displaystyle t = \frac{\pi}{2} and back by t = \pi, covering it twice.
10\pi over [0, \pi], the circumference; 20\pi over [0, 2\pi]. \displaystyle r^2 + \left(\frac{dr}{d\theta}\right)^2 = 100\cos^2\theta + 100\sin^2\theta = 100. The circle has diameter 10 and is traced once on [0, \pi], twice on [0, 2\pi].
10. The sum r^2 + (dr/d\theta)^2 is 25(1 + \cos\theta)^2 + 25\sin^2\theta, which simplifies to \displaystyle 50(1 + \cos\theta) = 100\cos^2\frac{\theta}{2}, with root \displaystyle 10\cos\frac{\theta}{2} on this interval; \displaystyle \Big[20\sin\frac{\theta}{2}\Big]_0^{\pi/3} = 20 \cdot \frac{1}{2}.
\displaystyle \frac{20}{3}. \displaystyle \frac{dr}{d\theta} = -3e^{-3\theta/4}, so \displaystyle r^2 + \left(\frac{dr}{d\theta}\right)^2 = (16 + 9)e^{-3\theta/2}, with root 5e^{-3\theta/4}. Then \displaystyle \int_0^b 5e^{-3\theta/4}\,d\theta = \frac{20}{3}\left(1 - e^{-3b/4}\right), and as b \to \infty this approaches \displaystyle \frac{20}{3}: the improper integral converges. Each turn is shorter than the one before by the factor e^{-3\pi/2} \approx 0.009, so the turns add up like a convergent geometric series.
L \approx 2.68. r = 0 when \displaystyle \cos\theta = -\frac{1}{2}, at \displaystyle \theta = \frac{2\pi}{3} and \displaystyle \frac{4\pi}{3}, and r < 0 between them, so the point leaves the pole, runs around the small loop on the side opposite its angle, and returns. With \displaystyle \frac{dr}{d\theta} = -2\sin\theta, \displaystyle r^2 + \left(\frac{dr}{d\theta}\right)^2 = 5 + 4\cos\theta (the formula uses r^2, so the sign of r does not matter), and numerically \displaystyle \int_{2\pi/3}^{4\pi/3} \sqrt{5 + 4\cos\theta}\,d\theta = 2.682.
\displaystyle \sqrt{5} \approx 2.24, 3 and 1 centimeters per second. The speed is \displaystyle \sqrt{(2 + \sin t)^2 + \cos^2 t} = \sqrt{5 + 4\sin t}, fastest (3) at \displaystyle \theta = \frac{\pi}{2} and slowest (1) at \displaystyle \theta = \frac{3\pi}{2}.
\displaystyle \frac{dx}{dt} = 1 and \displaystyle \frac{dy}{dt} = f'(t), so \displaystyle L = \int_a^b \sqrt{1 + f'(t)^2}\,dt; renaming the variable t as x gives \displaystyle \int_a^b \sqrt{1 + f'(x)^2}\,dx.
By the chain rule X'(u) = x'(\varphi(u))\varphi'(u) and Y'(u) = y'(\varphi(u))\varphi'(u), so the new speed is \displaystyle \sqrt{x'(\varphi(u))^2 + y'(\varphi(u))^2}\,\lvert\varphi'(u)\rvert. If \varphi' > 0, then \varphi(c) = a, \varphi(d) = b, and substituting t = \varphi(u) gives \displaystyle \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt. If \varphi' < 0, then \lvert\varphi'\rvert = -\varphi', \varphi(c) = b, \varphi(d) = a, and the same substitution gives -\int_b^a = \int_a^b of the same integrand.
By the chain rule, \displaystyle \frac{ds}{dt} = \frac{ds}{d\theta}\cdot\frac{d\theta}{dt}, where \displaystyle \frac{ds}{d\theta} = \sqrt{r^2 + (dr/d\theta)^2} is the speed with \theta as parameter and \displaystyle \frac{d\theta}{dt} = \omega. For r = a the speed is \displaystyle \omega\sqrt{a^2} = a\omega (with a > 0), the familiar speed of circular motion.
The four points (\pm 6, 0), (0, \pm 2) are on the ellipse, and each quarter of the ellipse is at least as long as the chord joining its ends, \displaystyle \sqrt{36 + 4} = 2\sqrt{10}; it is longer, since the quarter is not straight. On each quarter x and y are monotone and \displaystyle \sqrt{x'^2 + y'^2} < \lvert x' \rvert + \lvert y' \rvert wherever both rates are nonzero, so the quarter is shorter than 6 + 2 = 8. Hence \displaystyle 8\sqrt{10} \approx 25.30 < L < 32, and Exercise 6’s 26.73 lies between.
\displaystyle \sqrt{r^2 + (dr/d\theta)^2} \ge \sqrt{r^2} = \lvert r \rvert at every \theta, so the integrals compare the same way. Equality of the integrals of two continuous functions, one never smaller, forces equality everywhere, so \displaystyle \frac{dr}{d\theta} = 0 on the interval and r is constant; conversely a constant r gives equality.
The speed is \displaystyle \sqrt{4\sin^2 t + 4\cos^2 t} = 2, never 0, and \int_0^{4\pi} 2\,dt = 8\pi. The curve is the circle of radius 2, whose length is 4\pi: the parameter goes around twice. A nonzero speed rules out stops, not repeats.