Arc Length of Curve Calculator
Calculate the exact and numerical arc length $L$ of any differentiable curve $y = f(x)$ over an interval $[a, b]$ using the fundamental calculus arc length integral $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$.
Cartesian Curve & Subtended Arc Plot
Interval [a, b]Calculus Derivation & Infinitesimal Pythagorean Hypotenuse (dL) Formula: dL = √(dx² + dy²) = √(1 + [f'(x)]²) dx
How to Find the Arc Length of a Curve Using Calculus
To find the arc length of a differentiable function y = f(x) on the interval [a, b], compute the first derivative f'(x), square it, add 1, take the square root to form the differential element dL = √(1 + [f'(x)]²) dx, and evaluate the definite integral: L = ∫ₐᵇ √(1 + [f'(x)]²) dx. For example, for the parabola y = x² on [0, 2], f'(x) = 2x, giving L = ∫₀² √(1 + 4x²) dx ≈ 4.6468 units (compared to straight-line distance 4.4721 units).
Anatomy of Curve Arc Length & Infinitesimal Pythagorean Theorem
The calculus arc length formula solves a fundamental problem: how do we measure the distance along a curved path that constantly changes direction?
The core insight comes from zooming in on an infinitesimal segment of the curve: over an infinitesimally small horizontal displacement $dx$ and vertical displacement $dy$, the curve resembles the hypotenuse $dL$ of a tiny right triangle:
(dL)² = (dx)² + (dy)² ⇒ dL = √(1 + [dy/dx]²) dx
Summing these infinite infinitesimal hypotenuses from $x = a$ to $x = b$ produces the Riemann integral for total length $L$.
Hypotenuse of infinitesimal step.
Accumulated curve distance on [a, b].
Requires a smooth $C^1$ differentiable curve.
Cartesian Arc Length Formulas (y = f(x) and x = g(y))
Standard form integrating with respect to x.
Useful when dy/dx has vertical tangents.
Parametric & Polar Arc Length Formulations
Speed integrated over parameter time t.
Spiral and rose curve arc length.
Numerical Simpson Quadrature for Non-Elementary Integrals
Because the square root term $\sqrt{1 + [f'(x)]^2}$ rarely yields closed-form antiderivatives (except for carefully constructed textbook examples like $x^{3/2}$ or $\ln(\cos x)$), our interactive calculator employs a 1,000-subinterval Composite Simpson's $1/3$ Rule:
L ≈ (h / 3) [ g(x₀) + 4g(x₁) + 2g(x₂) + 4g(x₃) + ... + g(x_n) ] where g(x) = √(1 + [f'(x)]²)
Exact Integrals for Benchmark Curves (Parabola, Catenary, Astroid)
| Curve Name | Equation | Derivative f'(x) | Exact Arc Length Closed Form |
|---|---|---|---|
| Semi-Cubical Parabola | y = x^(3/2) on [0, 4] | (3/2)x^(1/2) | (8/27)(10√10 − 1) ≈ 9.0734 |
| Catenary (Hanging Chain) | y = a⋅cosh(x/a) on [0, b] | sinh(x/a) | a⋅sinh(b/a) |
| Log Cosine | y = ln(cos x) on [0, π/4] | −tan x | ln(1 + √2) ≈ 0.8814 |
Step-by-Step Worked Calculus Solutions
Find the arc length of y = (2/3)(x² + 1)^(3/2) on the interval [0, 1].
1. Compute derivative via chain rule: f'(x) = (2/3)(3/2)(x² + 1)^(1/2)(2x) = 2x√(x² + 1).
2. Form 1 + [f'(x)]² = 1 + 4x²(x² + 1) = 4x⁴ + 4x² + 1 = (2x² + 1)².
3. Take square root: √( (2x² + 1)² ) = 2x² + 1.
4. Integrate: L = ∫₀¹ (2x² + 1) dx = [ (2/3)x³ + x ]₀¹ = (2/3)(1) + 1 = 5/3 ≈ 1.6667 units.
Common Pitfalls: Squaring Derivatives & Interval Limits
Integrating $\sqrt{[f'(x)]^2} dx = |f'(x)| dx$ calculates the total vertical variation $\int |dy|$, NOT the actual hypotenuse length $dL$. The $+1$ represents $(dx)^2$ and is essential.
If the curve has a sharp cusp or corner (like $y = |x|$ at $x = 0$), split the integral into continuous smooth pieces at the corner point.
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