About the Curve Arc Length Calculator
This arc length calculator measures how long the curve y = f(x) is between two x values — the distance an ant would walk along the graph, not the straight-line gap between the endpoints. Type a function such as x^2, sin(x), x^(3/2) or ln(x), set the start and end of the interval, and it returns the arc length together with the derivative it used and the chord length for comparison.
It is meant for calculus students checking arc length integrals (many of which have no neat antiderivative), and for anyone who needs the real length of a curved profile: a cable hanging between two posts, a curved ramp, a road or a cut along a parabola. The chart plots the curve and the running length so you can see where most of the length accumulates.
The derivative is found symbolically and the integral is evaluated numerically with adaptive Gaussian quadrature, so results are accurate to many decimal places for smooth functions. The function must be continuous and differentiable on the interval; use radians for trigonometric functions and ln(x) for the natural logarithm.
With the default inputs, the arc length l is 1.478943. Change any value above to recalculate instantly.
How to use the curve arc length calculator
- 1Type the function using x as the variable, e.g. x^2 or sin(x).
- 2Enter the starting x value a and the ending x value b (pi and fractions are allowed).
- 3Read the arc length, the derivative used and the integrand.
- 4Compare with the chord length to see how much extra length the curvature adds.
- 5Use the chart to see where along the interval the length builds up fastest.
Formula and method
Split the curve into tiny pieces. Over a small step dx the curve rises by dy = f′(x)·dx, so by Pythagoras each piece has length √(dx² + dy²) = √(1 + f′(x)²)·dx. Adding the pieces from x = a to x = b gives the arc length integral above.
The calculator differentiates f symbolically (falling back to a central-difference derivative when needed), then evaluates the integral with adaptive 5-point Gauss–Legendre quadrature (pieces are halved until the estimate stops changing) after the substitution x = a + (b − a)(3t² − 2t³), which packs nodes near both ends. Gauss nodes never touch the endpoints, so curves like √x whose derivative is infinite at an end are still handled accurately. If f itself becomes infinite in the interval, or the adaptive refinement cannot make the estimate settle, the length is reported as infinite or non-convergent instead of printing a meaningless number. The chord is the straight-line distance √((b − a)² + (f(b) − f(a))²); the arc is always at least that long.
- L
- Arc length of the curve
- f′(x)
- Derivative (slope) of the curve
- a, b
- Start and end x values
Worked examples
Parabola y = x² from 0 to 1
Here f′(x) = 2x, so L = ∫₀¹ √(1 + 4x²) dx = (2√5 + asinh 2) ÷ 4 ≈ 1.478943. The straight line from (0, 0) to (1, 1) is √2 ≈ 1.414214, so the curve is about 4.6% longer.
y = x^(3/2) from 0 to 4
f′(x) = 1.5√x, so the integrand is √(1 + 9x/4). This one has an exact antiderivative: L = (8/27)(10^(3/2) − 1) ≈ 9.073415 — a classic textbook arc length problem.
One arch of sin(x) from 0 to π
With f′(x) = cos x, L = ∫₀^π √(1 + cos²x) dx, an elliptic integral with no elementary answer. Numerically it is ≈ 3.820198, about 21.6% longer than the flat distance π.
Vertical tangent: y = √x from 0 to 1
Here f′(x) = 1/(2√x) is infinite at x = 0, but the integral still converges. y = √x on [0, 1] is the mirror image of y = x² on [0, 1], so its length is the same ≈ 1.478943.
A straight line y = 2x + 1 from 0 to 3
For a line the slope is constant, so L = 3√(1 + 2²) = 3√5 ≈ 6.708204, exactly the chord length — the ratio is 1.
Frequently asked questions
What is the arc length formula?+
For a curve y = f(x) from x = a to x = b, the arc length is L = ∫ₐᵇ √(1 + f′(x)²) dx. It comes from applying Pythagoras to tiny straight segments of the curve and adding them up.
Why can’t I find an exact answer for most arc lengths?+
The square root √(1 + f′(x)²) rarely has an elementary antiderivative. Even simple curves like sin(x) or an ellipse lead to elliptic integrals, which is why arc lengths are usually computed numerically, as this calculator does.
How do I find the arc length of a parametric or polar curve?+
For x(t), y(t) use L = ∫ √(x′(t)² + y′(t)²) dt; for a polar curve r(θ) use L = ∫ √(r² + (dr/dθ)²) dθ. This tool handles curves written as y = f(x); rewrite simple parametric curves in that form where possible.
Is arc length always longer than the straight-line distance?+
Yes. The straight line is the shortest path between two points, so the arc length is always greater than or equal to the chord. They are equal only when the curve is a straight line on the interval.
What if the derivative is infinite at an endpoint?+
Curves such as y = √x have a vertical tangent at x = 0, so f′ blows up there. The integral can still converge. This calculator packs its quadrature nodes towards both ends and never evaluates the exact endpoints, so it handles such curves well — √x from 0 to 1 gives 1.478943, matching the exact value.