Arc Length of Curves: Complete Guide

30 September 2026Mathematics Try the interactive 3D model →

Arc length is the exact length of a curve measured along the curve itself, like laying a thread along a winding road and then straightening the thread against a ruler. For a smooth function y = f(x), calculus computes this length with a single integral.

This matters whenever the length of something curved is needed: the cable hanging between two towers, the track of a roller coaster, the border of a curved garden plot, or the path of a robot arm. Straight-line distance always underestimates these, sometimes badly.

Key Concepts

Infinitesimal hypotenuse: a tiny piece of curve is nearly straight, with length ds = √(dx² + dy²). Factoring out dx gives ds = √(1 + (dy/dx)²) dx, which is what we integrate.
Smoothness: the formula needs dy/dx to exist and be continuous on the interval; sharp corners must be handled piece by piece.
Steepness penalty: where the curve is steep, (dy/dx)² is large and each dx contributes much more than dx to the length, exactly as intuition suggests.
Parametric form: for a curve given by x(t), y(t), the length is ∫ √( (dx/dt)² + (dy/dt)² ) dt, the same hypotenuse idea with t as the parameter.

Formulas

As a function of x: L = ∫√(1 + (dy/dx)²) dx from x = a to x = b, in length units.
As a function of y: L = ∫√(1 + (dx/dy)²) dy from y = c to y = d, useful when dy/dx is messy but dx/dy is simple.
Parametric: L = ∫√((dx/dt)² + (dy/dt)²) dt from t = t₁ to t = t₂.

Worked Example

Find the arc length of y = (2/3)x^(3/2) from x = 0 to x = 4.

Differentiate: dy/dx = (2/3)·(3/2)x^(1/2) = √x.
Form the integrand: 1 + (dy/dx)² = 1 + x, so L = ∫₀⁴ √(1 + x) dx.
Substitute: let u = 1 + x, so du = dx; when x = 0, u = 1, and when x = 4, u = 5. Then L = ∫₁⁵ √u du.
Integrate: ∫ √u du = (2/3)u^(3/2). Evaluating: (2/3)(5^(3/2) − 1^(3/2)) = (2/3)(5√5 − 1).
Compute numerically: √5 ≈ 2.2361, so 5√5 ≈ 11.1803; minus 1 gives 10.1803; times 2/3 gives ≈ 6.7869, about 6.79 units. The straight-line distance between the endpoints (0, 0) and (4, 16/3) is √(16 + 28.44) ≈ 6.67, slightly less, as expected.

Common Misconceptions

The chord is close enough. For gentle curves it is, but steep or wavy curves are far longer than their endpoints suggest; only the integral is exact.
Forgetting the square root. Integrating 1 + (dy/dx)² instead of its square root hugely overestimates the length.
Integrating dy/dx itself. ∫ dy/dx dx = f(b) − f(a) is just the net vertical change, not the length along the curve.

Real-World Applications

Civil engineering: lengths of suspension-bridge cables, arches, and curved road or rail alignments for material estimates.
Manufacturing: cutting lengths for bent pipes, mouldings, and wires shaped along a designed curve.
Computer graphics: spacing objects evenly along a path needs the true curve length, not the straight-line gap.
Surveying: measuring winding boundaries and coastlines from mathematical models.

Arc length treats a curve as infinitely many tiny hypotenuses and adds them with the integral of √(1 + (dy/dx)²). Differentiate carefully, keep the square root, and the result is the true length a thread would measure along the curve.