Calculus 2

Arc length

A curve is not made of straight pieces, so no ruler fits along it. How long is it?

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Area between curvesCalculus 1 The Pythagorean theoremMath 8

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Arc length and speed on parametric and polar curvesCalculus 2 Surface area of a solid of revolutionCalculus 2

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Arc length

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Why this exists

A straight segment is measured by the Pythagorean theorem: a run of \Delta x and a rise of \Delta y give \displaystyle \sqrt{(\Delta x)^2 + (\Delta y)^2}. A circle’s arcs are a fraction of the circumference, but only because a circle bends the same amount everywhere. A parabola or a hanging cable bends differently at every point, and no formula from geometry measures it.

The way through is the usual one for integrals. Cut the curve into short pieces, replace each by a straight segment (the hypotenuse of a tiny right triangle), add, and let the pieces shrink. As with area, the sum is a Riemann sum and its limit is an integral.

Definition

Let f have a derivative f' that is continuous on [a, b]. Such a function is called smooth on [a, b]: its graph has no corners and no vertical tangents there.

Split [a, b] into n pieces of width \Delta x = \dfrac{b - a}{n}, with endpoints a = x_0 < x_1 < \dots < x_n = b, and mark the points P_i = (x_i, f(x_i)) on the graph. Joining P_0, P_1, \dots, P_n by straight segments gives a path inscribed in the curve (Figure 1). The arc length L of the graph from x = a to x = b is the limit of the lengths of these paths:

L = \lim_{n \to \infty} \sum_{i=1}^{n} \lvert P_{i-1} P_i \rvert

Here \lvert P_{i-1} P_i \rvert is the length of the segment from P_{i-1} to P_i.

Figure 1An inscribed path. Each segment is the hypotenuse of a small right triangle.

This is a definition: it says what “the length of a curve” means. The work below turns it into an integral.

Intuition

Zoom in on a smooth curve and a short piece is nearly its own tangent line. Over a run of dx the tangent rises f'(x)\,dx, so the piece has length about

\begin{aligned} &\sqrt{(dx)^2 + \big(f'(x)\,dx\big)^2} \\ &\quad = \sqrt{1 + [f'(x)]^2}\,dx \end{aligned}

The factor \displaystyle \sqrt{1 + [f'(x)]^2} is a stretch factor: how many times longer the curve is than the stretch of x-axis beneath it. It is 1 where the graph is flat and large where it is steep. Arc length adds it up across [a, b].

Results

Arc length of a graph y = f(x). If f' is continuous on [a, b], the length of the graph of f from x = a to x = b is

L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx

Arc length of a graph x = g(y). If g' is continuous on [c, d], the length of the curve x = g(y) from y = c to y = d is

L = \int_c^d \sqrt{1 + [g'(y)]^2}\,dy

The second form is the first with x and y exchanged. Both come from the arc length element

ds = \sqrt{(dx)^2 + (dy)^2}

factored as \displaystyle \sqrt{1 + (dy/dx)^2}\,dx or as \displaystyle \sqrt{1 + (dx/dy)^2}\,dy, whichever variable the curve is written in.

The length is at least the chord. By the triangle inequality, every inscribed path is at least as long as the single segment from P_0 to P_n, so \displaystyle L \ge \sqrt{(b - a)^2 + (f(b) - f(a))^2}. Since the stretch factor is at least 1, also L \ge b - a.

Most of these integrals have no elementary antiderivative. Already y = x^3 and y = \sin x give integrands whose antiderivatives cannot be built from polynomials, roots, exponentials, logarithms and trigonometric functions (stated here without proof). For such curves a decimal from numerical integration, by computer or by the trapezoidal rule or Simpson’s rule, is the normal answer. Exact lengths come mostly from curves for which 1 + [f'(x)]^2 is a perfect square, as in Examples 2 and 3.

Why it is true

Take one piece of the path, from P_{i-1} to P_i. Its run is \Delta x and its rise is \Delta y_i = f(x_i) - f(x_{i-1}), so by the Pythagorean theorem

\lvert P_{i-1} P_i \rvert = \sqrt{(\Delta x)^2 + (\Delta y_i)^2}

The rise is a difference of function values, and the mean value theorem turns it into a derivative. f is continuous on [x_{i-1}, x_i] and differentiable inside it, so there is a point x_i^* strictly between x_{i-1} and x_i where

f(x_i) - f(x_{i-1}) = f'(x_i^*)\,\Delta x

In words: somewhere above the piece, the tangent line is parallel to the segment. Substitute, and factor (\Delta x)^2 out of the root (its square root is \Delta x, which is positive):

\begin{aligned} \lvert P_{i-1} P_i \rvert &= \sqrt{(\Delta x)^2 + [f'(x_i^*)]^2 (\Delta x)^2} \\ &= \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x \end{aligned}

So the whole path has length

\sum_{i=1}^{n} \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x

This is a Riemann sum for the function \displaystyle \sqrt{1 + [f'(x)]^2} on [a, b], with the sample point x_i^* in the ith piece. That function is continuous, because f' is, so its Riemann sums converge to its integral as n \to \infty whatever sample points are used. The limit that defines L is therefore \displaystyle \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.

The hypothesis is used twice: differentiability for the mean value theorem, and continuity of f' for the limit. For x = g(y) the same argument runs with [c, d] split into pieces.

Worked examples

Example 1

A curve whose integrand simplifies at once

Find the length of y = \tfrac{2}{3}x^{3/2} from x = 0 to x = 3.

Solution

The derivative is \displaystyle f'(x) = \tfrac{2}{3} \cdot \tfrac{3}{2} x^{1/2} = \sqrt{x}, continuous on [0, 3]. Its square is x, so the stretch factor is \displaystyle \sqrt{1 + x}, and the power rule in reverse finishes the job:

\begin{aligned} L &= \int_0^3 \sqrt{1 + x}\,dx = \frac{2}{3}(1 + x)^{3/2}\Big|_0^3 \\ &= \frac{2}{3}(8 - 1) = \frac{14}{3} \end{aligned}

Check. The chord runs from (0, 0) to \displaystyle (3, 2\sqrt{3}) and has length \displaystyle \sqrt{9 + 12} = \sqrt{21} \approx 4.58. The curve should be a little longer, and \tfrac{14}{3} \approx 4.67 is.

Example 2

A perfect square under the root

Find the length of y = \dfrac{x^3}{12} + \dfrac{1}{x} from x = 1 to x = 3.

Solution

Here f'(x) = \dfrac{x^2}{4} - \dfrac{1}{x^2}. When it is squared, the cross term is 2 \cdot \dfrac{x^2}{4} \cdot \dfrac{1}{x^2} = \dfrac{1}{2}:

[f'(x)]^2 = \frac{x^4}{16} - \frac{1}{2} + \frac{1}{x^4}

Adding 1 changes only the middle term, from -\tfrac12 to +\tfrac12, and that turns a difference squared into a sum squared:

\begin{aligned} 1 + [f'(x)]^2 &= \frac{x^4}{16} + \frac{1}{2} + \frac{1}{x^4} \\ &= \left(\frac{x^2}{4} + \frac{1}{x^2}\right)^2 \end{aligned}

The quantity in parentheses is positive, so the square root gives it back unchanged:

\begin{aligned} L &= \int_1^3 \left(\frac{x^2}{4} + \frac{1}{x^2}\right)dx \\ &= \left[\frac{x^3}{12} - \frac{1}{x}\right]_1^3 \\ &= \left(\frac{9}{4} - \frac{1}{3}\right) - \left(\frac{1}{12} - 1\right) \\ &= \frac{23}{12} + \frac{11}{12} = \frac{17}{6} \end{aligned}

The pattern: if f' = u - v with uv = \tfrac14, then 1 + (u - v)^2 = (u + v)^2.

Example 3

The power line

Parvaneh is surveying a power line strung between two towers 60 meters apart on level ground. With the origin on the ground halfway between the towers and distances in meters, the cable hangs along

y = 15\left(e^{x/30} + e^{-x/30}\right), \qquad -30 \le x \le 30

as in Figure 2. How much cable is there between the towers, and how far does it sag?

Figure 2The cable of Example 3, drawn to scale. It meets the towers about 46.3 meters up.

Solution

By the chain rule, y' = \tfrac12\left(e^{x/30} - e^{-x/30}\right). Square it, using e^{x/30} \cdot e^{-x/30} = 1 for the cross term:

(y')^2 = \tfrac14\left(e^{x/15} - 2 + e^{-x/15}\right)

Adding 1 = \tfrac44 flips the sign of the middle term, exactly as in Example 2:

\begin{aligned} 1 + (y')^2 &= \tfrac14\left(e^{x/15} + 2 + e^{-x/15}\right) \\ &= \left[\tfrac12\left(e^{x/30} + e^{-x/30}\right)\right]^2 \end{aligned}

The bracket is positive, so

\begin{aligned} L &= \int_{-30}^{30} \tfrac12\left(e^{x/30} + e^{-x/30}\right)dx \\ &= 15\left[e^{x/30} - e^{-x/30}\right]_{-30}^{30} \\ &= 15\left[\left(e - e^{-1}\right) - \left(e^{-1} - e\right)\right] \\ &= 30\left(e - e^{-1}\right) \approx 70.51 \end{aligned}

That is about 70.5 meters of cable for a 60 meter span. The lowest point is at x = 0, at height 15(1 + 1) = 30 meters. The cable meets the towers at height 15\left(e + e^{-1}\right) \approx 46.29 meters, so it sags about 16.3 meters.

The stretch factor here is \tfrac{1}{30} of the height y itself, which Exercise 14 turns into a formula.

Example 4

Integrating in y when the slope blows up

Find the length of y = x^{2/3} from (0, 0) to (8, 4).

Solution

In x, f'(x) = \tfrac{2}{3}x^{-1/3}, which is undefined at x = 0. The curve has a vertical tangent at the origin, so f' is not continuous on [0, 8] and the x form of the formula does not apply as stated.

Turn the curve around. For x \ge 0 and y \ge 0, the equation y = x^{2/3} says the same as x = y^{3/2}, with y running from 0 to 4. Now g'(y) = \tfrac32 y^{1/2} is continuous on [0, 4], and [g'(y)]^2 = \tfrac94 y:

L = \int_0^4 \sqrt{1 + \tfrac94 y}\,dy

Substitute u = 1 + \tfrac94 y, so du = \tfrac94\,dy, and u runs from 1 to 10:

\begin{aligned} L &= \frac49 \int_1^{10} u^{1/2}\,du = \frac49 \cdot \frac23 \left[u^{3/2}\right]_1^{10} \\ &= \frac{8}{27}\left(10\sqrt{10} - 1\right) \approx 9.07 \end{aligned}

Check. The chord from (0, 0) to (8, 4) has length \displaystyle \sqrt{80} \approx 8.94, just under 9.07.

Example 5

A parabola, by the definition and by integration

Find the length of y = x^2 from x = 0 to x = 1.

Solution

f'(x) = 2x, so

L = \int_0^1 \sqrt{1 + 4x^2}\,dx

This time 1 + 4x^2 is not a perfect square. A first estimate comes straight from the definition. With n = 4 the points are (0, 0), \left(\tfrac14, \tfrac{1}{16}\right), \left(\tfrac12, \tfrac14\right), \left(\tfrac34, \tfrac{9}{16}\right) and (1, 1). Each run is \tfrac14 and the rises are \tfrac{1}{16}, \tfrac{3}{16}, \tfrac{5}{16}, \tfrac{7}{16}, so the four segments have lengths

\begin{aligned} &\tfrac{1}{16}\left(\sqrt{17} + \sqrt{25} + \sqrt{41} + \sqrt{65}\right) \\ &\approx 0.2577 + 0.3125 + 0.4002 + 0.5039 \\ &= 1.4743 \end{aligned}

since \displaystyle \sqrt{\tfrac{16}{256} + \tfrac{1}{256}} = \tfrac{\sqrt{17}}{16}, and so on. An inscribed path is never longer than the curve, so L \ge 1.4743. Numerical integration of \displaystyle \sqrt{1 + 4x^2} gives

L \approx 1.4789

Here an exact antiderivative does exist: the trigonometric substitution 2x = \tan\theta produces it, and

L = \frac{\sqrt5}{2} + \frac14 \ln\left(2 + \sqrt5\right) \approx 1.4789

For most curves, though, the numerical value is the only one there is.

Example 6

A curve with no elementary length

Set up and evaluate the length of one arch of y = \sin x, from x = 0 to x = \pi.

Solution

f'(x) = \cos x, so

L = \int_0^{\pi} \sqrt{1 + \cos^2 x}\,dx

The antiderivative of \displaystyle \sqrt{1 + \cos^2 x} is not elementary. It is an elliptic integral, and no substitution will produce it (stated here without proof). The definite integral is still a number, and numerical integration gives

L \approx 3.8202

Two bounds show the value is reasonable. The chord from (0, 0) to (\pi, 0) has length \pi, so L \ge \pi \approx 3.1416. And \displaystyle \sqrt{1 + \cos^2 x} \le 1 + \lvert \cos x \rvert, because squaring both sides gives 1 + \cos^2 x \le 1 + 2\lvert\cos x\rvert + \cos^2 x. So L \le \pi + \int_0^{\pi} \lvert \cos x \rvert\,dx = \pi + 2 \approx 5.1416. The value 3.8202 sits between the two.

Boundaries

f' must be continuous on the closed interval. At a vertical tangent, as for y = x^{2/3} at the origin, f' is unbounded and the x form becomes an improper integral. Writing the curve as x = g(y) often removes the trouble (Example 4).

A corner splits the interval. y = \lvert x \rvert has no derivative at 0; its length on [-2, 2] is \displaystyle 2\sqrt2 + 2\sqrt2 = 4\sqrt2, from the formula on each side of the corner.

The curve must be a graph in the variable of integration. A whole circle has two heights for most x and two x values for most heights. Split it into graphs, or use a parameter, which is arc length for parametric curves.

Continuity alone does not give a finite length. The graph of y = x\sin(1/x) on (0, 1], with y = 0 at x = 0, is continuous, yet its inscribed paths grow without bound (stated without proof). Smoothness rules this out.

An exact answer is the exception. A decimal from numerical integration is a complete answer.

Where this goes next

Turn the graph of f about the x-axis and each piece ds sweeps out a thin band; adding their areas gives the area of a surface of revolution, built on ds the way the disk method is built on slices. When a point traces a curve over time t, ds divided by dt is its speed: arc length and speed on parametric and polar curves. And the length out to a moving endpoint, \displaystyle s(x) = \int_a^x \sqrt{1 + [f'(t)]^2}\,dt, has derivative \displaystyle \sqrt{1 + [f'(x)]^2} by the first fundamental theorem of calculus: the stretch factor is the rate at which length piles up.

Exercises

Exact lengths

Find the length of each curve.

y = x^{3/2} from x = 0 to x = \tfrac43.
y = \dfrac{x^4}{8} + \dfrac{1}{4x^2} from x = 1 to x = 2.
y = \dfrac{x^2}{2} - \dfrac{\ln x}{4} from x = 1 to x = e.
y = \ln(\cos x) from x = 0 to x = \tfrac{\pi}{4}. (Use \int \sec x\,dx = \ln\lvert \sec x + \tan x \rvert + C.)

Integrating in y

Find the length of x = \tfrac23 (y - 1)^{3/2} from y = 1 to y = 9.
Find the length of x = \dfrac{y^4}{4} + \dfrac{1}{8y^2} from y = 1 to y = 2.
Set up the length of y = \ln x from x = 1 to x = e twice, once as an integral in x and once as an integral in y, and evaluate it numerically.

Setting up and estimating

Set up the length of y = x^3 from x = 0 to x = 1. Without integrating, explain why the length lies between \displaystyle \sqrt2 and 2. Then evaluate it numerically.
Estimate the length of y = \dfrac{1}{x} from x = 1 to x = 2 with an inscribed path of two segments, and compare with the value from numerical integration.
Sahkyo is building a garden arch in the shape of y = 4 - x^2 above the ground y = 0, with distances in feet. Set up the length of the arch and evaluate it numerically.

The power line

Find the length of y = \tfrac12\left(e^{x} + e^{-x}\right) from x = 0 to x = \ln 2.
A longer span hangs along y = 20\left(e^{x/40} + e^{-x/40}\right) between towers at x = -40 and x = 40, in meters. Find the length of cable and how far it sags.
Ulviyya’s towers stand on a slope, so the cable of Example 3, y = 15\left(e^{x/30} + e^{-x/30}\right), runs from x = -15 to x = 45. Find the length of cable.
For the cable y = \dfrac{c}{2}\left(e^{x/c} + e^{-x/c}\right) with c > 0, show that the length from the lowest point (0, c) out to the point (x, y), x > 0, is \displaystyle \sqrt{y^2 - c^2}.

Reasoning

Show that the formula gives the length of y = mx + k from x = a to x = b, and that the result agrees with the distance formula.
Apply the formula to the upper half of the circle x^2 + y^2 = r^2 from x = 0 to x = \tfrac{r}{2}, and confirm the result against the length of an arc as a fraction of the circumference.
Explain without integrating why \displaystyle y = \sqrt{x} from x = 0 to x = 1 has the same length as the parabola of Example 5. Why is the x form awkward for \displaystyle y = \sqrt{x} when the y form is not?
Refute the claim that multiplying a function by 2 doubles the length of its graph, using y = x and y = 2x on [0, 1].
Explain why adding one more point to an inscribed path can never make it shorter, and why that means no inscribed path is longer than L.
Show that L \ge b - a for every smooth f on [a, b], with equality only when f is constant.

Answers

\displaystyle L = \frac{56}{27} \approx 2.07. The integrand is \displaystyle \sqrt{1 + \tfrac94 x}, so L = \tfrac{8}{27}\left[(1 + \tfrac94 x)^{3/2}\right]_0^{4/3}, which is \tfrac{8}{27}(8 - 1).
\displaystyle L = \frac{33}{16}. f'(x) = \tfrac{x^3}{2} - \tfrac{1}{2x^3}, product \tfrac14, so the integrand is \tfrac{x^3}{2} + \tfrac{1}{2x^3}. Then L = \left[\tfrac{x^4}{8} - \tfrac{1}{4x^2}\right]_1^2, which is \tfrac{31}{16} + \tfrac{2}{16}.
\displaystyle L = \frac{2e^2 - 1}{4} \approx 3.4445. f'(x) = x - \tfrac{1}{4x}, product \tfrac14, so the integrand is x + \tfrac{1}{4x} and L = \left[\tfrac{x^2}{2} + \tfrac{\ln x}{4}\right]_1^e = \tfrac{e^2}{2} - \tfrac14.
\displaystyle L = \ln\left(1 + \sqrt2\right) \approx 0.8814. 1 + \tan^2 x = \sec^2 x and \sec x > 0 here, so L = \ln\lvert \sec x + \tan x \rvert \Big|_0^{\pi/4}.
\displaystyle L = \frac{52}{3}. g'(y) = (y - 1)^{1/2}, so 1 + [g'(y)]^2 = y and \displaystyle L = \int_1^9 \sqrt{y}\,dy = \tfrac23(27 - 1).
\displaystyle L = \frac{123}{32}. g'(y) = y^3 - \tfrac{1}{4y^3}, product \tfrac14, so the integrand is y^3 + \tfrac{1}{4y^3} and L = \left[\tfrac{y^4}{4} - \tfrac{1}{8y^2}\right]_1^2 = \tfrac{127}{32} - \tfrac{4}{32}.
In x: \displaystyle \int_1^e \sqrt{1 + \tfrac{1}{x^2}}\,dx. In y: the curve is x = e^y for 0 \le y \le 1, so \displaystyle \int_0^1 \sqrt{1 + e^{2y}}\,dy. Both give L \approx 2.0035.
\displaystyle L = \int_0^1 \sqrt{1 + 9x^4}\,dx \approx 1.5479. The chord from (0, 0) to (1, 1) has length \displaystyle \sqrt2, a lower bound. Since 1 + 9x^4 \le (1 + 3x^2)^2, the integrand is at most 1 + 3x^2, and \int_0^1 (1 + 3x^2)\,dx = 2.
Path through (1, 1), \left(\tfrac32, \tfrac23\right), \left(2, \tfrac12\right): \displaystyle \tfrac{\sqrt{13}}{6} + \tfrac{\sqrt{10}}{6} \approx 1.1280. Numerical integration of \displaystyle \int_1^2 \sqrt{1 + \tfrac{1}{x^4}}\,dx gives L \approx 1.1321, slightly longer, as it must be.
The arch meets the ground at x = \pm 2 and y' = -2x, so \displaystyle L = \int_{-2}^{2} \sqrt{1 + 4x^2}\,dx \approx 9.2936 feet.
\displaystyle L = \frac34. 1 + (y')^2 = \left[\tfrac12(e^x + e^{-x})\right]^2, so L = \tfrac12\left(e^x - e^{-x}\right)\Big|_0^{\ln 2} = \tfrac12\left(2 - \tfrac12\right).
L = 40\left(e - e^{-1}\right) \approx 94.02 meters. As in Example 3, the integrand is \tfrac12\left(e^{x/40} + e^{-x/40}\right), whose antiderivative is 20\left(e^{x/40} - e^{-x/40}\right). The sag is 20\left(e + e^{-1}\right) - 40 \approx 21.72 meters.
L = 15\left(e^{3/2} - e^{-3/2} + e^{1/2} - e^{-1/2}\right) \approx 79.51 meters. The antiderivative from Example 3 is 15\left(e^{x/30} - e^{-x/30}\right); evaluate it at 45 and subtract its value at -15.
As in Example 3, the integrand is \tfrac12\left(e^{x/c} + e^{-x/c}\right), so the length is \tfrac{c}{2}\left(e^{x/c} - e^{-x/c}\right). Also y^2 - c^2 equals \tfrac{c^2}{4}\left(e^{2x/c} - 2 + e^{-2x/c}\right), the square of that length, which is positive for x > 0. So the length is \displaystyle \sqrt{y^2 - c^2}.
f' = m, so \displaystyle L = (b - a)\sqrt{1 + m^2}. The endpoints differ by b - a across and m(b - a) up, and \displaystyle \sqrt{(b - a)^2 + m^2(b - a)^2} is the same number.
\displaystyle L = \frac{\pi r}{6}. Here 1 + (y')^2 = \tfrac{r^2}{r^2 - x^2}, so \displaystyle L = \int_0^{r/2} \tfrac{r}{\sqrt{r^2 - x^2}}\,dx, which is r\arcsin\tfrac{x}{r}\Big|_0^{r/2} = r \cdot \tfrac{\pi}{6}. The arc spans a central angle of 30^\circ, and \tfrac{30}{360} \cdot 2\pi r = \tfrac{\pi r}{6}.
Both have length L \approx 1.4789. \displaystyle y = \sqrt{x} is x = y^2 for 0 \le y \le 1, the parabola of Example 5 reflected in y = x, which keeps lengths; the y form gives exactly \displaystyle \int_0^1 \sqrt{1 + 4y^2}\,dy. In x, \displaystyle f'(x) = \tfrac{1}{2\sqrt{x}} is unbounded near 0, so the x integral is improper.
y = x gives \displaystyle \sqrt2 and y = 2x gives \displaystyle \sqrt5, but twice \displaystyle \sqrt2 is \displaystyle \sqrt8. Doubling f doubles f', which turns the integrand into \displaystyle \sqrt{1 + 4[f'(x)]^2}, not \displaystyle 2\sqrt{1 + [f'(x)]^2}.
A new point R between P and Q replaces \lvert PQ \rvert by \lvert PR \rvert + \lvert RQ \rvert, at least as large by the triangle inequality. Refining a path until its pieces shrink to zero never shortens it, and the lengths approach L, so the path was at most L.
\displaystyle \sqrt{1 + [f'(x)]^2} \ge 1, so L \ge b - a. If f'(x_0) \ne 0, continuity keeps f' \ne 0 near x_0, the integrand exceeds 1 there, and L > b - a. So equality forces f' = 0 everywhere, and by the mean value theorem f is constant.