Arc Length Calculator

The arc length calculator computes the exact length of a curve between two points by integrating the arc length formula. It is used by calculus students, engineers designing curved structures, and anyone who needs the true distance along a curve rather than the straight-line distance between endpoints. Enter your function and interval to get the arc length with working steps.

How to Use the Arc Length Calculator

1. Enter your function f(x) in the Arc Length Calculator input field. Set the lower bound a and upper bound b of the interval.
2. Click Calculate to evaluate the arc length integral.
3. Review the derivative f'(x) and the integrand shown in the working.
4. For parametric curves, switch to parametric mode and enter x(t) and y(t) with the parameter range.

The Formula

For a function y = f(x) on the interval [a, b], the arc length L is:

L = Integral from a to b of sqrt(1 + [f'(x)]^2) dx

Here, f'(x) is the derivative of the function and the square root term accounts for the actual distance along the curve rather than just the horizontal distance. For a parametric curve x = x(t), y = y(t):

L = Integral from t1 to t2 of sqrt([x'(t)]^2 + [y'(t)]^2) dt

This second form is the Pythagorean theorem applied to infinitely small segments of the curve.

Real-World Example

Find the arc length of f(x) = (2/3)x^(3/2) on the interval [0, 3].

1. Find f'(x) = x^(1/2) = sqrt(x)
2. Set up the integral: L = Integral from 0 to 3 of sqrt(1 + x) dx
3. Let u = 1 + x, so du = dx
4. L = Integral from 1 to 4 of sqrt(u) du = [(2/3)u^(3/2)] from 1 to 4
5. L = (2/3)(8) - (2/3)(1) = 16/3 - 2/3 = 14/3 approximately 4.667

The arc length along the curve from x = 0 to x = 3 is 14/3 units, which is longer than the straight-line distance of sqrt(9 + f(3)^2) = sqrt(9 + (2/3)(3sqrt(3))^2).

Where Arc Length Appears in Practice

Structural engineers use arc length when designing arches, cables, and curved road profiles. The length of cable needed to span a suspension bridge follows the catenary arc length formula. In manufacturing, the arc length of a spline determines how much material is needed to form a curved part. In computer graphics, arc length parameterisation is used to move objects at constant speed along curved paths regardless of how the curve is defined. Knowing the true path length rather than just the endpoint distance is critical in all these contexts.

Frequently Asked Questions

Why is the formula sqrt(1 + [f'(x)]^2) and not just the integral of f(x)? The formula comes from applying the Pythagorean theorem to infinitely small pieces of the curve. Each tiny segment has horizontal length dx and vertical height f'(x)dx, so its true length is sqrt(dx^2 + [f'(x)dx]^2), which simplifies to sqrt(1 + [f'(x)]^2) dx after factoring out dx.

Can arc length always be computed exactly? No. Many arc length integrals have no closed-form solution and must be evaluated numerically. For example, the arc length of an ellipse involves elliptic integrals with no simple expression. Numerical integration methods such as Simpson's rule are commonly used in these cases.

Is the arc length formula different for polar curves? Yes. For a polar curve r = f(theta) from theta = a to theta = b, the arc length is the integral of sqrt(r^2 + [dr/d(theta)]^2) d(theta). This accounts for the fact that distance in polar coordinates involves both radial and angular components.

What is the difference between arc length and arc length parameterisation? Arc length is a scalar value giving the total length of a curve segment. Arc length parameterisation re-expresses the curve so that the parameter t directly equals the distance travelled along the curve from the starting point, making it useful for animation and physics simulations.

Also try these free tools:

• Definite Integral Calculator
• Area Between Curves Calculator
• Derivative Calculator

Worked examples with closed-form answers

Example 1, a closed form you can check. Take f(x) = x squared on [0, 1]. Then f'(x) = 2x and the integrand is sqrt(1 plus 4x squared). The antiderivative is (x divided by 2) times sqrt(1 plus 4x squared) plus (1 divided by 4) times ln(2x plus sqrt(1 plus 4x squared)).

Evaluating at the bounds:

• At x = 1: 0.5 times 2.236068 plus 0.25 times 1.443635 = 1.118034 plus 0.360909 = 1.478943
• At x = 0: 0 plus 0.25 times ln(1) = 0

So L = 1.478943 units. The straight-line distance from (0, 0) to (1, 1) is sqrt(2) = 1.414214, which makes the curve 0.064729 units longer than the chord. That gap is the reason the integral exists: the distance formula takes the shortcut, the arc length integral walks the path.

Example 2, a power function. Take f(x) = x to the power 3/2 on [0, 4]. Then f'(x) = 1.5 times sqrt(x) and 1 plus the square of that is 1 plus 2.25x. Substituting u = 1 plus 2.25x gives L = (2 divided by 9) times (2 divided by 3) times the change in u to the power 3/2 from 1 to 10, which is 0.296296 times 30.622776 = 9.073415 units. The chord from (0, 0) to (4, 8) is sqrt(80) = 8.944272, so the arc is longer again.

Example 3, the same curve over wider intervals. f(x) = x squared, evaluated exactly, so the widening gap between path and shortcut is visible:

Numerical evaluation and how fast it converges

Most arc length integrals have no elementary antiderivative, so the value comes from quadrature. The two rules used most often behave very differently on the same problem. This is Example 1 again, computed both ways, against the exact value 1.47894286.

Every doubling of the subinterval count cuts the trapezoidal error by about four, which is its second-order behaviour, and the Simpson error by about sixteen, which is fourth order. At 16 subintervals Simpson's rule is already correct to eight decimal places. That is why a calculator settles on an answer quickly, and why the interval count matters far less than the rule chosen.

Method and assumptions

The formula is built by cutting the curve into pieces, applying the Pythagorean theorem to each piece, and taking the limit. On a piece running from x to x plus delta x, the horizontal run is delta x and the rise is f'(x) delta x by the mean value theorem, so the piece length is sqrt(delta x squared plus the square of f'(x) delta x), which factors to sqrt(1 plus the square of f'(x)) delta x. Summing and letting delta x go to zero gives the integral.

The calculator assumes:

• The interval runs low to high. An integral with a above b returns the negative of the length, and the tool reports the magnitude instead.
• f is differentiable with a continuous derivative on the interval. A vertical tangent makes the integrand unbounded and turns the integral improper.
• The curve is traversed once. A path that loops back on itself is counted each time it passes, which is correct for distance travelled but not for the length of the set of points.
• Both axes use the same unit. Arc length comes out in axis units, so an x axis in metres against a y axis in millimetres returns a number in neither.
• The geometry is flat. Arc length on a sphere is a different formula with a cosine term in the integrand.

The parametric form, checked against the function form

The same curve can be written either way, and the two integrals have to agree. For y = f(x) = x squared on [0, 1] the function form integrates sqrt(1 plus 4x squared) dx. Writing that same curve parametrically as x(t) = t and y(t) = t squared over the same range gives the parametric integrand sqrt(1 squared plus (2t) squared) = sqrt(1 plus 4t squared), integrated against dt. After the substitution x = t the two integrands are identical, so both return 1.478943. Running a curve through both modes is a quick way to confirm the tool is wired correctly before trusting it on a problem that has no closed form.

Source

The integral definition follows from the mean value theorem argument above and appears in that form in every standard single-variable calculus text. Where a closed form exists, the antiderivative used in Example 1 is the standard table entry for the square root of a quadratic. Curves with no elementary antiderivative, of which the ellipse is the usual example, reduce to elliptic integrals and are evaluated numerically, as the convergence table shows.