Arc Length Function Calculator & Formula

This calculator finds the length of a curve defined by a function y = f(x) between two x-values. You enter the function, the lower and upper limits, and the number of intervals for the numerical approximation; it returns the approximate arc length. The result is given in the same units as the x and y coordinates.

Formula

The exact arc length of a smooth curve y = f(x) from x = a to x = b is given by the integral:

L = ∫_a^b √(1 + [f′(x)]²) dx

where f′(x) is the derivative of the function with respect to x, and dx is an infinitesimal change in x. The formula comes from summing tiny straight-line segments: each segment has length √(dx² + dy²) = √(1 + (dy/dx)²) dx.

Because many functions do not have elementary antiderivatives for this integrand, the calculator uses a numerical method (the trapezoidal rule) on the sampled derivative. The user enters the function as a mathematical expression using x as the variable, the limits a and b, and the number of subintervals n. The calculator computes the derivative numerically by finite differences and sums the segment lengths.

Worked Example

Take the default function f(x) = x², from a = 0 to b = 1, with n = 100 intervals.

1. Divide [0,1] into 100 equal subintervals, each of width Δx = (1-0)/100 = 0.01.
2. At each point x_i = 0, 0.01, 0.02, …, 1.0, compute f(x_i) = x_i².
3. For each interval, approximate the derivative as (f(x_{i+1}) – f(x_i)) / Δx.
4. Compute the segment length: √(1 + (derivative)²) × Δx.
5. Sum all 100 segment lengths.

The exact arc length of y = x² from 0 to 1 is (1/4)(2√5 + ln(2+√5)) ≈ 1.4789. The numerical result with 100 intervals is approximately 1.4789 as well, matching to four decimal places.

When to Use an Arc Length Calculator

Arc length calculations appear in physics (distance along a curved path), engineering (length of cables, wires, or conveyor belts), computer graphics (curve rendering), and geometry (perimeter of irregular shapes). Use this calculator when you have a function in Cartesian coordinates and need the length of its graph between two points. If the curve is given parametrically or in polar coordinates, a different formula is required.

How to Choose the Number of Intervals

The accuracy of the numerical approximation improves as n increases, but so does computation time. For smooth functions, n = 100 often gives results accurate to 3-4 decimal places. For functions with steep slopes or rapid changes, use a larger n (e.g., 1000 or more). The error is roughly proportional to 1/n² for the trapezoidal rule, so doubling n reduces the error by a factor of four. However, if the function has a vertical tangent or is not differentiable at some point, the formula may fail or give misleading results.

Common Mistakes and Pitfalls

• Forgetting the derivative: The formula uses f′(x), not f(x). If you accidentally use f(x) inside the square root, you get a different (usually larger) value.
• Unit mix-ups: The result is in the same units as the x and y axes. If x is in meters and y is in meters, the arc length is in meters. If the axes have different units (e.g., x in seconds, y in meters), the result is a hybrid unit and may not be physically meaningful.
• Using the wrong limits: The limits a and b are x-values, not y-values. Double-check that you have entered the correct interval.
• Numerical noise: For very large n (e.g., 10^6), floating-point rounding can accumulate. Use a reasonable n and check stability by increasing it slightly.

Limits and Alternatives

This calculator works only for functions y = f(x) in Cartesian coordinates. For parametric curves (x(t), y(t)), use the formula L = ∫ √((dx/dt)² + (dy/dt)²) dt. For polar curves r(θ), use L = ∫ √(r² + (dr/dθ)²) dθ. If you need the arc length of a circle, simply use the circumference formula 2πr for a full circle, or rθ for a sector. For polygonal paths, sum the straight-line distances between vertices.

Frequently Asked Questions

What is the arc length formula for a function?

The arc length L of a function y = f(x) from x = a to x = b is L = ∫_a^b √(1 + [f′(x)]²) dx. You need the derivative of the function.

Can I use this calculator for parametric curves?

No, this calculator is specifically for Cartesian functions y = f(x). For parametric curves, you need a different formula and a separate calculator that accepts x(t) and y(t).

Why does the calculator ask for the number of intervals?

Because most arc length integrals cannot be solved analytically, the calculator uses a numerical approximation. The number of intervals determines how finely the curve is divided into small segments; more intervals give a more accurate result but take longer to compute.

What if my function has a vertical tangent?

If the derivative becomes infinite (e.g., at x = 0 for f(x) = √x), the formula may still work if the integral converges, but the numerical method might be inaccurate. In such cases, consider splitting the curve at the problematic point or using a different parameterization.

How accurate is the result?

For smooth functions, with n = 100, the result is usually accurate to about 3-4 decimal places. The error decreases as n increases, but floating-point rounding can become an issue for extremely large n.

Can I use this for 3D curves?

No, this calculator is for 2D functions only. For a 3D curve defined by parametric equations, you would need a formula involving the sum of squares of derivatives in all three coordinates.

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in f | Function f(x) | - | x^2 | 0
in a | Lower limit a | - | 0 | 0
in b | Upper limit b | - | 1 | 0
in n | Intervals n | - | 100 | 1
out arc | Arc length | units | sum(sqrt(1 + ((f(x_i+1)-f(x_i))/(x_i+1-x_i))^2)*(x_i+1-x_i)) | 4