Arc Length of a Curve: Formula & Examples

The arc length of a curve y = f(x) from x = a to x = b is $L = \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\,dx$ . The formula adds up tiny straight pieces of the curve, each found with the Pythagorean theorem. It works whenever f′ is continuous on [a, b].

Before this lesson: Distance Formula, Power Rule and Definite Integrals

Measuring a curve with tiny straight pieces

You can’t lay a ruler along a curve, but you can along a short straight piece. Break the curve into many short pieces. Each piece is the long side of a small right triangle with legs Δx and Δy.

By the Pythagorean theorem, a piece has length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ . Factor out Δx and use Δy/Δx ≈ f′(x). Each piece is about $\sqrt{1 + (f'(x))^2}\,\Delta x$ . As the pieces shrink, the sum becomes an integral.

The curve y = (2/3)x^(3/2) from x = 0 to x = 3, with four straight chords joining points on it. One chord is magnified as the long side of a right triangle with horizontal leg Δx, vertical leg Δy and slanted side Δs. — Each tiny piece of the curve is close to a straight chord of length √(Δx² + Δy²).

Arc length formula

L = \int_a^b \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2}\,dx

If the curve is written as x = g(y) from y = c to y = d, use the same idea with the roles switched:

L = \int_c^d \sqrt{1 + \left(\dfrac{dx}{dy}\right)^2}\,dy

Find the derivative f′(x).
Square it and add 1.
Simplify the square root, if you can. Many textbook problems turn into a perfect square.
Integrate from a to b. If there is no antiderivative you can find, estimate the integral numerically.

Arc length examples

Example 1: A power curve

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

f'(x) = \dfrac{2}{3}\cdot\dfrac{3}{2}x^{1/2} = x^{1/2}

Use the power rule.

1 + \left(x^{1/2}\right)^2 = 1 + x

Square and add 1.

L = \int_0^3 \sqrt{1 + x}\,dx

Set up the integral.

= \Big[\dfrac{2}{3}(1 + x)^{3/2}\Big]_0^3

Integrate with u = 1 + x (see u-substitution).

= \dfrac{2}{3}\left(4^{3/2} - 1^{3/2}\right) = \dfrac{2}{3}(8 - 1)

Plug in; 4^(3/2) = 8.

= \dfrac{14}{3} \approx 4.667

Multiply.

Answer: $L = \dfrac{14}{3}$ .

Example 2: A line, as a check

Find the length of y = 2x + 1 from x = 0 to x = 3, and compare with the distance formula.

f'(x) = 2, \; 1 + 2^2 = 5

Differentiate, square and add 1.

L = \int_0^3 \sqrt{5}\,dx = 3\sqrt{5}

The integrand is constant.

\sqrt{(3 - 0)^2 + (7 - 1)^2} = \sqrt{45} = 3\sqrt{5}

Distance from (0, 1) to (3, 7) (see simplifying square roots).

Answer: $L = 3\sqrt{5} \approx 6.708$ . Both methods agree, as they should for a straight line.

Example 3: A perfect square under the root

Find the arc length of $y = \dfrac{x^3}{6} + \dfrac{1}{2x}$ from x = 1 to x = 2.

f'(x) = \dfrac{x^2}{2} - \dfrac{1}{2x^2}

Use the power rule; 1/(2x) = (1/2)x⁻¹.

\left(f'(x)\right)^2 = \dfrac{x^4}{4} - \dfrac{1}{2} + \dfrac{1}{4x^4}

Square the binomial; the middle term is 2 · (x²/2) · (1/(2x²)) = 1/2 (see special products of binomials).

1 + \left(f'(x)\right)^2 = \dfrac{x^4}{4} + \dfrac{1}{2} + \dfrac{1}{4x^4}

Add 1; −1/2 + 1 = 1/2.

= \left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)^2

Recognize a perfect square (see perfect square trinomials).

L = \int_1^2 \left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)dx

The square root undoes the square; the inside is positive.

= \Big[\dfrac{x^3}{6} - \dfrac{1}{2x}\Big]_1^2

Integrate.

= \left(\dfrac{4}{3} - \dfrac{1}{4}\right) - \left(\dfrac{1}{6} - \dfrac{1}{2}\right)

Plug in 2 and 1.

= \dfrac{13}{12} + \dfrac{1}{3} = \dfrac{17}{12}

Simplify each part, then add.

Answer: $L = \dfrac{17}{12} \approx 1.417$ .

Example 4: Estimating with Simpson’s rule

Estimate the arc length of y = x² from x = 0 to x = 1 with Simpson’s rule, n = 4.

f'(x) = 2x, \; L = \int_0^1 \sqrt{1 + 4x^2}\,dx

Set up the integral.

g(0) = 1, \; g(0.25) \approx 1.118034, \; g(0.5) \approx 1.414214

Let g(x) = √(1 + 4x²) and find the first three values.

g(0.75) \approx 1.802776, \; g(1) \approx 2.236068

Find the last two.

1 + 4(1.118034) + 2(1.414214) + 4(1.802776) + 2.236068 \approx 17.747735

Apply the weights 1, 4, 2, 4, 1 (see Simpson’s rule).

L \approx \dfrac{0.25}{3}(17.747735) \approx 1.478978

Multiply by Δx/3.

Answer: about 1.4790. The exact value, found with trig substitution, is (2√5 + ln(2 + √5))/4 ≈ 1.478943.

Example 5: Length of a hanging cable

A power line between two poles 60 feet apart hangs in the shape $y = 20\cosh\dfrac{x}{20}$ , for x from −30 to 30 (in feet). How long is the cable?

y' = \sinh\dfrac{x}{20}

Chain rule: 20 · (1/20) sinh(x/20) (see hyperbolic functions).

1 + \sinh^2\dfrac{x}{20} = \cosh^2\dfrac{x}{20}

Use cosh²u − sinh²u = 1.

L = \int_{-30}^{30} \cosh\dfrac{x}{20}\,dx

cosh is positive, so the root is cosh(x/20).

= \Big[20\sinh\dfrac{x}{20}\Big]_{-30}^{30} = 20\sinh 1.5 - 20\sinh(-1.5)

Integrate and plug in.

= 40\sinh 1.5 \approx 40(2.129279) \approx 85.17

sinh is odd, so sinh(−1.5) = −sinh 1.5.

Answer: the cable is about 85.2 feet long, about 25 feet more than the straight distance between the poles.

Common mistakes

Mistake	Example of the error	Fix
Forgetting to square the derivative	$\int_0^3 \sqrt{1 + x^{1/2}}\,dx$ for y = (2/3)x^(3/2)	Square f′ first: 1 + (x^(1/2))² = 1 + x.
Leaving out the 1	$\int_a^b \sqrt{(f'(x))^2}\,dx$	The 1 comes from Δx². Always use 1 + (f′)².
Taking the root term by term	$\sqrt{1 + 4x^2} = 1 + 2x$	A square root does not split over a sum. Simplify only when the inside is a perfect square.
Using f instead of f′	$\int_0^1 \sqrt{1 + (x^2)^2}\,dx$ for y = x²	Use the derivative: $\int_0^1 \sqrt{1 + (2x)^2}\,dx$ .
Mixing up the variables	Using x-limits with the dy formula	With dx/dy, the limits are y-values, from c to d.

Related methods

The arc length function

Measuring from a fixed point a, the length up to x is $s(x) = \int_a^x \sqrt{1 + (f'(t))^2}\,dt$ . By the fundamental theorem, $\dfrac{ds}{dx} = \sqrt{1 + (f'(x))^2}$ , which gives the arc length differential $ds = \sqrt{dx^2 + dy^2}$ .

Arc length of a circle

For a circle, the geometry formula s = rθ is easier. See arc length in Geometry. The calculus formula gives the same answer, as practice problem 8 shows.

Parametric and polar curves

For a curve given by x(t) and y(t), the length is $\int \sqrt{(x')^2 + (y')^2}\,dt$ . That comes later in this course, with parametric and polar curves.

Practice problems

Find the arc length of each curve.

y = 3x − 2, from x = 1 to x = 4
$y = \dfrac{2}{3}x^{3/2}$ , from x = 0 to x = 8
$y = \dfrac{1}{3}(x^2 + 2)^{3/2}$ , from x = 0 to x = 3
$y = \dfrac{x^4}{8} + \dfrac{1}{4x^2}$ , from x = 1 to x = 2
y = ln(cos x), from x = 0 to x = π/4
$x = \dfrac{2}{3}(y - 1)^{3/2}$ , from y = 1 to y = 4
Set up the arc length integral for y = sin x from 0 to π, and estimate it.
Use the formula to find the length of the upper half of the circle x² + y² = 9.
y = cosh x, from x = 0 to x = ln 2
Estimate the length of y = x³ from x = 0 to x = 1 with Simpson’s rule, n = 2.

Show answers

f'(x) = 3, \; \sqrt{1 + 9} = \sqrt{10}

Differentiate, square, add 1.

L = \int_1^4 \sqrt{10}\,dx = 3\sqrt{10}

Constant integrand; 4 − 1 = 3.

Answer: $3\sqrt{10} \approx 9.487$

1 + (x^{1/2})^2 = 1 + x

f′(x) = x^(1/2), as in Example 1.

L = \Big[\dfrac{2}{3}(1 + x)^{3/2}\Big]_0^8 = \dfrac{2}{3}(27 - 1)

9^(3/2) = 27.

= \dfrac{52}{3}

Multiply.

Answer: $\dfrac{52}{3} \approx 17.33$

f'(x) = \dfrac{1}{3}\cdot\dfrac{3}{2}(x^2 + 2)^{1/2}\cdot 2x = x\sqrt{x^2 + 2}

Chain rule (see chain rule).

1 + x^2(x^2 + 2) = x^4 + 2x^2 + 1 = (x^2 + 1)^2

Square, add 1 and factor.

L = \int_0^3 (x^2 + 1)\,dx = \Big[\dfrac{x^3}{3} + x\Big]_0^3 = 9 + 3

Take the root and integrate.

= 12

Add.

Answer: 12

f'(x) = \dfrac{x^3}{2} - \dfrac{1}{2x^3}

Power rule.

1 + (f')^2 = \dfrac{x^6}{4} + \dfrac{1}{2} + \dfrac{1}{4x^6} = \left(\dfrac{x^3}{2} + \dfrac{1}{2x^3}\right)^2

Square, add 1, and recognize a perfect square.

L = \Big[\dfrac{x^4}{8} - \dfrac{1}{4x^2}\Big]_1^2 = \left(2 - \dfrac{1}{16}\right) - \left(\dfrac{1}{8} - \dfrac{1}{4}\right)

Integrate and plug in.

= \dfrac{31}{16} + \dfrac{2}{16} = \dfrac{33}{16}

Simplify.

Answer: $\dfrac{33}{16} \approx 2.063$

f'(x) = \dfrac{-\sin x}{\cos x} = -\tan x

Derivative of ln(cos x) by the chain rule.

1 + \tan^2 x = \sec^2 x

Pythagorean identity (see fundamental trig identities).

L = \int_0^{\pi/4} \sec x\,dx = \Big[\ln|\sec x + \tan x|\Big]_0^{\pi/4}

sec x is positive here; integrate.

= \ln(\sqrt{2} + 1) - \ln 1 \approx 0.8814

sec(π/4) = √2 and tan(π/4) = 1.

Answer: $\ln(\sqrt{2} + 1) \approx 0.8814$

\dfrac{dx}{dy} = (y - 1)^{1/2}, \; 1 + (y - 1) = y

Use the dy formula.

L = \int_1^4 \sqrt{y}\,dy = \Big[\dfrac{2}{3}y^{3/2}\Big]_1^4 = \dfrac{2}{3}(8 - 1)

Integrate and plug in.

= \dfrac{14}{3}

Multiply.

Answer: $\dfrac{14}{3}$

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

f′(x) = cos x.

\approx 3.8202

This has no elementary antiderivative; a numerical method gives the value.

Answer: about 3.820

f'(x) = \dfrac{-x}{\sqrt{9 - x^2}}, \; 1 + \dfrac{x^2}{9 - x^2} = \dfrac{9}{9 - x^2}

Here y = √(9 − x²); use a common denominator.

L = \int_{-3}^{3} \dfrac{3}{\sqrt{9 - x^2}}\,dx = \Big[3\arcsin\dfrac{x}{3}\Big]_{-3}^{3}

An improper integral at both ends, but it converges.

= 3\left(\dfrac{\pi}{2}\right) - 3\left(-\dfrac{\pi}{2}\right) = 3\pi

arcsin 1 = π/2 and arcsin(−1) = −π/2.

Answer: 3π, half of the circumference 6π

1 + \sinh^2 x = \cosh^2 x

The derivative of cosh x is sinh x.

L = \int_0^{\ln 2} \cosh x\,dx = \sinh(\ln 2) = \dfrac{2 - 1/2}{2}

Integrate and use the definition of sinh.

= \dfrac{3}{4}

Simplify.

Answer: $\dfrac{3}{4}$

10.

L = \int_0^1 \sqrt{1 + 9x^4}\,dx

f′(x) = 3x².

g(0) = 1, \; g(0.5) = \sqrt{1 + \tfrac{9}{16}} = 1.25, \; g(1) = \sqrt{10} \approx 3.162278

Values at 0, 0.5 and 1.

\dfrac{0.5}{3}\big(1 + 4(1.25) + 3.162278\big) \approx 1.527046

Weights 1, 4, 1.

Answer: about 1.527 (the true value is about 1.5479)

Arc length FAQ

What is the arc length formula?

L = ∫ from a to b of √(1 + (dy/dx)²) dx. It comes from adding the lengths of tiny straight pieces with the Pythagorean theorem.

Why are arc length integrals so hard?

The square root rarely simplifies, so most arc length integrals have no elementary antiderivative. Textbook problems are chosen so 1 + (f′)² is a perfect square; otherwise, estimate numerically.

How do you find arc length with respect to y?

Write x as a function of y and use L = ∫ from c to d of √(1 + (dx/dy)²) dy, with y-limits.

Is this the same as arc length of a circle?

For a circle, both give the same answer, but s = rθ is much faster. The integral formula works for any smooth curve.

Summary

Arc length: L = ∫ₐᵇ √(1 + (f′(x))²) dx.
Find f′, square it, add 1, then simplify the root if you can.
For x = g(y), use √(1 + (dx/dy)²) dy.
Many problems turn 1 + (f′)² into a perfect square.
If not, estimate the integral with Simpson’s rule or another numerical method.

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