Arc Length Formula
Calculate the length of a curve using the arc length integral formula for Cartesian and parametric forms.
The Formula
The arc length formula computes the exact length of a curve between two points. Unlike a straight line where distance is trivial to calculate, a curved path requires calculus because its direction changes continuously. The formula works by breaking the curve into infinitesimally small straight segments, computing each tiny length using the Pythagorean theorem, and summing them all via integration.
The expression inside the square root, 1 + (dy/dx)2, comes directly from the Pythagorean theorem applied to an infinitesimal right triangle. As you move a tiny distance dx along the x-axis, the curve rises by dy. The actual distance traveled along the curve is √(dx2 + dy2), which factors to √(1 + (dy/dx)2) dx.
For curves described parametrically as x = x(t) and y = y(t), the formula takes a different but equivalent form:
The parametric version is especially useful for curves that loop back on themselves or cannot be expressed as a single function y = f(x). Circles, ellipses, spirals, and cycloids are all naturally described in parametric form.
In polar coordinates r = f(θ), the arc length becomes:
Arc length integrals are often difficult or impossible to evaluate analytically. Many practical curves produce integrands that have no closed-form antiderivative. In such cases, numerical integration methods like Simpson's rule or Gaussian quadrature are used to approximate the result. This is perfectly acceptable in engineering and scientific applications where a decimal answer is what you need.
Variables
| Symbol | Meaning |
|---|---|
| L | Arc length of the curve (same units as x and y) |
| a, b | Limits of integration along the x-axis |
| dy/dx | Derivative of the curve (slope at each point) |
| t₁, t₂ | Parameter limits (for parametric form) |
| dx/dt, dy/dt | Parametric derivatives with respect to parameter t |
Example 1
Find the arc length of y = x3/2 from x = 0 to x = 4.
Compute derivative: dy/dx = (3/2)x1/2
(dy/dx)2 = (9/4)x
L = ∫04 √(1 + 9x/4) dx
Let u = 1 + 9x/4, du = 9/4 dx → dx = 4/9 du
L = (4/9) × (2/3) × [u3/2] from u=1 to u=10
L = (8/27) × (103/2 − 1)
L = (8/27)(10√10 − 1) ≈ 9.07 units
Example 2
Find the circumference of a circle of radius 3 using the parametric arc length formula.
Parametrize: x = 3 cos(t), y = 3 sin(t), with t from 0 to 2π
dx/dt = −3 sin(t), dy/dt = 3 cos(t)
(dx/dt)2 + (dy/dt)2 = 9 sin2(t) + 9 cos2(t) = 9
L = ∫02π √9 dt = ∫02π 3 dt
L = 6π ≈ 18.85 units — matching the familiar formula C = 2πr
When to Use It
The arc length formula is needed whenever you must measure the actual distance along a curved path.
- Calculating the length of a road, railroad track, or cable following a curved route
- Finding the perimeter of irregular shapes in engineering design
- Computing distances traveled along trajectories in physics simulations
- Determining the amount of material needed to cover a curved surface (e.g., pipe insulation)
- Measuring coastline lengths or river distances in geography