Unit 5, lesson 67

Arc Length of Function Graphs

How the length of a curved graph is defined through inscribed polygons, why it equals the integral of the square root of one plus the squared derivative, and what to do when that integral has no formula.

  • Theory part 133
  • Practice part 134
  • Applications of Integration

Lay a piece of string along a curve, then straighten it and measure it. That is what arc length means. To compute it, approximate the curve by a chain of short straight segments, whose lengths the distance formula gives. More and shorter segments hug the curve better, and the limit of their total length is an integral.

Key terms

Arc length as a limit of polygonal lengths

Definition Let ff be continuous on [a,b][a, b]. For a partition a=x0<x1<⋯<xn=ba = x_0 < x_1 < \dots < x_n = b, let Pi=(xi,f(xi))P_i = (x_i, f(x_i)). The length of the inscribed polygon is

L(P)=∑i=1n∣Pi−1Pi∣=∑i=1n(xi−xi−1)2+(f(xi)−f(xi−1))2.L(P) = \sum_{i=1}^n \lvert P_{i-1} P_i \rvert = \sum_{i=1}^n \sqrt{(x_i - x_{i-1})^2 + \left(f(x_i) - f(x_{i-1})\right)^2}.

The arc length of the graph is the least upper bound of L(P)L(P) over all partitions, when this bound is finite, and the graph is then called rectifiable. Adding a point to a partition never decreases L(P)L(P), by the triangle inequality. Because ff is continuous, the least upper bound also equals lim⁡L(P)\lim L(P) as the longest subinterval width tends to 00. This is a theorem of analysis, and for the smooth functions of the next section it follows directly from the Riemann sum argument given there.

In plain words Join points on the curve with straight segments and add their lengths. Adding more points never makes the total shorter, and the arc length is the value those totals approach.

Example For f(x)=x2f(x) = x^2 on [0,1][0, 1], with nn equal subintervals,

nnPolygon length
111.41421.4142
221.46041.4604
441.47431.4743
881.47781.4778
16161.47871.4787

The exact length, from the formula below, is about 1.47891.4789.

Smooth and piecewise smooth graphs

Definition A function ff is smooth on [a,b][a, b] if f′f' exists and is continuous on [a,b][a, b], using one-sided derivatives at the endpoints. If ff is smooth on [a,b][a, b], the length of its graph is

L=∫ab1+f′(x)2 dx.L = \int_a^b \sqrt{1 + f'(x)^2} \, dx.

This follows from the mean value theorem. On each subinterval, f(xi)−f(xi−1)=f′(xi∗) Δxif(x_i) - f(x_{i-1}) = f'(x_i^*)\,\Delta x_i for some xi∗x_i^* in (xi−1,xi)(x_{i-1}, x_i), so

L(P)=∑i=1n1+f′(xi∗)2 Δxi.L(P) = \sum_{i=1}^n \sqrt{1 + f'(x_i^*)^2}\,\Delta x_i.

This is a Riemann sum of a continuous function, so it tends to the integral as the longest subinterval width tends to 00. If ff is continuous on [a,b][a, b] and smooth on each of finitely many subintervals that cover [a,b][a, b], then ff is piecewise smooth and its length is the sum of the lengths of the pieces.

In plain words A tiny step dxdx along the axis comes with a rise f′(x) dxf'(x)\,dx, so the tiny piece of curve has length 1+f′(x)2 dx\sqrt{1 + f'(x)^2}\,dx by the theorem of Pythagoras. Corners are fine as long as you split there.

Example For f(x)=x3/2f(x) = x^{3/2} on [0,4][0, 4], f′(x)=32x1/2f'(x) = \frac{3}{2}x^{1/2} and

L=∫041+9x4 dx=827[(1+9x4)3/2]04=827(1010−1)≈9.073.L = \int_0^4 \sqrt{1 + \frac{9x}{4}} \, dx = \frac{8}{27}\left[\left(1 + \frac{9x}{4}\right)^{3/2}\right]_0^4 = \frac{8}{27}\left(10\sqrt{10} - 1\right) \approx 9.073.

For the piecewise smooth g(x)=∣x∣g(x) = \lvert x \rvert on [−1,2][-1, 2], split at 00. The pieces have lengths 2\sqrt{2} and 222\sqrt{2}, for a total of 323\sqrt{2}.

The dx and dy formulas

Definition If the curve is described as x=g(y)x = g(y) for c≤y≤dc \le y \le d, with g′g' continuous on [c,d][c, d], its length is

L=∫cd1+g′(y)2 dy.L = \int_c^d \sqrt{1 + g'(y)^2} \, dy.

Both formulas are cases of ds=dx2+dy2ds = \sqrt{dx^2 + dy^2}. Use whichever variable makes the derivative continuous and the integral simpler.

In plain words If the curve is a nicer function of yy than of xx, measure it along the yy-axis instead.

Example The graph of y=x2/3y = x^{2/3} from (0,0)(0, 0) to (8,4)(8, 4) has dydx=23x−1/3\frac{dy}{dx} = \frac{2}{3}x^{-1/3}, which is not defined at x=0x = 0, so the dxdx formula is not covered by the theorem there. As x=y3/2x = y^{3/2} for 0≤y≤40 \le y \le 4, the dydy formula has g′(y)=32y1/2g'(y) = \frac{3}{2}y^{1/2}, which is continuous, and

L=∫041+9y4 dy=827(1010−1)≈9.073.L = \int_0^4 \sqrt{1 + \frac{9y}{4}} \, dy = \frac{8}{27}\left(10\sqrt{10} - 1\right) \approx 9.073.

It is the same curve as the previous example with the axes swapped, so the same length is no surprise.

Arc length function

Definition Let ff be smooth on [a,b][a, b]. The arc length function measured from aa is

s(x)=∫ax1+f′(t)2 dt,a≤x≤b.s(x) = \int_a^x \sqrt{1 + f'(t)^2} \, dt, \qquad a \le x \le b.

By the fundamental theorem of calculus, s′(x)=1+f′(x)2≥1s'(x) = \sqrt{1 + f'(x)^2} \ge 1, so ss is strictly increasing, and in differential form ds2=dx2+dy2ds^2 = dx^2 + dy^2.

In plain words The value s(x)s(x) is how far you have walked along the curve from its starting point when you reach the point above xx.

Example For f(x)=cosh⁡xf(x) = \cosh x measured from a=0a = 0, the integrand is 1+sinh⁡2t=cosh⁡t\sqrt{1 + \sinh^2 t} = \cosh t, so s(x)=sinh⁡xs(x) = \sinh x. At x=ln⁡2x = \ln 2,

s(ln⁡2)=sinh⁡(ln⁡2)=2−122=34.s(\ln 2) = \sinh(\ln 2) = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4}.

Non-elementary length integrals

Definition For many smooth functions ff, even simple ones, the integrand 1+f′(x)2\sqrt{1 + f'(x)^2} has no elementary antiderivative. The arc length still exists as a definite integral of a continuous function, and it is found by numerical integration, such as Simpson’s rule, to any desired accuracy.

In plain words Curves with neat length formulas are the exception. The square root usually defeats every technique, and a numerical value is the honest answer.

Example One arch of y=sin⁡xy = \sin x on [0,π][0, \pi] has length

L=∫0π1+cos⁡2x dx≈3.8202,L = \int_0^\pi \sqrt{1 + \cos^2 x} \, dx \approx 3.8202,

an integral with no elementary antiderivative. As a sanity check, it is longer than both the straight chord of length π≈3.1416\pi \approx 3.1416 and the path up and down through the peak, 2(π/2)2+1≈3.72422\sqrt{(\pi/2)^2 + 1} \approx 3.7242. That path is an inscribed polygon, so the arc must be at least as long.

Check yourself

  1. Find the length of y=23x3/2y = \frac{2}{3}x^{3/2} for 0≤x≤30 \le x \le 3.
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The derivative is y′=x1/2y' = x^{1/2}, so

L=∫031+x dx=23(8−1)=143.L = \int_0^3 \sqrt{1 + x} \, dx = \frac{2}{3}\left(8 - 1\right) = \frac{14}{3}.
  1. Find the length of y=ln⁡(cos⁡x)y = \ln(\cos x) for 0≤x≤π40 \le x \le \frac{\pi}{4}.
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The derivative is y′=−tan⁡xy' = -\tan x, so 1+tan⁡2x=sec⁡x\sqrt{1 + \tan^2 x} = \sec x on this interval, where sec⁡x>0\sec x > 0. Then

L=∫0π/4sec⁡x dx=ln⁡(2+1)≈0.8814.L = \int_0^{\pi/4} \sec x \, dx = \ln\left(\sqrt{2} + 1\right) \approx 0.8814.
  1. Use the integral formula to find the length of y=2x+1y = 2x + 1 on [0,3][0, 3], and check it with the distance formula.
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The slope is 22, so

L=∫031+4 dx=35.L = \int_0^3 \sqrt{1 + 4} \, dx = 3\sqrt{5}.

The endpoints (0,1)(0, 1) and (3,7)(3, 7) are 9+36=45=35\sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} apart.

How this lesson connects

It also uses distance formula.