Arc Length Formula in Calculus: Derivation and Examples

The arc length of a smooth curve \( y = f(x) \) from \( x = a \) to \( x = b \) is

$$L = \int_a^b\sqrt{1 + \big[f'(x)\big]^2}\,dx$$

“Smooth” means \( f' \) is continuous on \( [a, b] \). That condition keeps the integrand continuous, so the integral exists. In this guide you will see where the formula comes from, how to use it step by step, the handful of curves that give clean answers, and the versions for parametric and polar curves.

Where the formula comes from

Chop the curve into tiny pieces. Each piece is almost a straight segment with horizontal run \( \Delta x \) and vertical rise \( \Delta y \), so by Pythagoras its length is

$$\sqrt{\Delta x^2 + \Delta y^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}\,\Delta x$$

As the pieces shrink, \( \frac{\Delta y}{\Delta x} \to f'(x) \) and the sum becomes the integral.

Derivation, step by step

Here is the same idea written carefully, the way most textbooks present it.

Step 1: split the interval. Divide \( [a, b] \) into \( n \) subintervals of width \( \Delta x \), with grid points \( x_0 = a, x_1, \dots, x_n = b \). Connect the matching points on the curve with straight segments. This polygonal path hugs the curve, and its length gets closer to the true length as \( n \) grows.

Step 2: measure one segment. The segment over \( [x_{i-1}, x_i] \) has horizontal change \( \Delta x \) and vertical change \( \Delta y_i = f(x_i) - f(x_{i-1}) \). Its length is \( \sqrt{\Delta x^2 + \Delta y_i^2} \).

Step 3: replace the rise with a derivative. By the Mean Value Theorem, there is a point \( x_i^* \) in that subinterval where \( \Delta y_i = f'(x_i^*)\,\Delta x \). Substituting and factoring out \( \Delta x \):

$$\sqrt{\Delta x^2 + \Delta y_i^2} = \sqrt{1 + \big[f'(x_i^*)\big]^2}\,\Delta x$$

Step 4: add and take the limit. The total polygon length is a Riemann sum of the function \( \sqrt{1 + [f']^2} \). As \( n \to \infty \), a Riemann sum of a continuous function converges to its integral, which gives the formula.

Intuition: the stretch factor

Think of the integrand \( \sqrt{1 + [f'(x)]^2} \) as a stretch factor. When you move a tiny distance \( dx \) to the right, you travel a distance \( ds \) along the curve, where

$$ds = \sqrt{1 + \big[f'(x)\big]^2}\,dx$$

Where the curve is flat, \( f' = 0 \) and \( ds = dx \): no stretching. Where it is steep, each step to the right costs a lot of curve. If the tangent line makes angle \( \theta \) with the horizontal, then \( f' = \tan\theta \) and the stretch factor is \( \sec\theta \). Arc length simply adds up these stretched steps.

How to find arc length

1. Compute \( f'(x) \), using the chain rule where needed.
2. Square it and add 1. Simplify as far as you can; look for a perfect square.
3. Take the square root and set up the integral from \( a \) to \( b \).
4. Integrate exactly if possible. Otherwise, evaluate numerically.

Example 1: y = x^(3/2) on [0, 4]

This is the classic textbook example, chosen because it can be done by hand.

\( f'(x) = \frac32x^{1/2} \), so \( 1 + [f'(x)]^2 = 1 + \frac94x \).

$$\begin{aligned} L &= \int_0^4\sqrt{1 + \tfrac94x}\,dx \\ &= \frac{8}{27}\left(1 + \tfrac94x\right)^{3/2}\Big|_0^4 \\ &= \frac{8}{27}\left(10^{3/2} - 1\right) \approx 9.0734 \end{aligned}$$

(The antiderivative is a u-substitution with \( u = 1 + \frac94 x \).)

Example 2: a straight line (sanity check)

\( y = 4x \) on \( [0, 3] \): \( L = \int_0^3\sqrt{1 + 16}\,dx = 3\sqrt{17} \), which matches the distance formula between \( (0,0) \) and \( (3,12) \).

Example 3: the catenary y = cosh x

On \( [0, 1] \): \( 1 + \sinh^2 x = \cosh^2 x \), so the root simplifies perfectly:

$$L = \int_0^1\cosh x\,dx = \sinh 1 \approx 1.1752$$

The catenary is the shape of a hanging chain or cable, so this calculation tells you how much cable you need between two supports.

Example 4: a parabola (needs numerics)

\( y = x^2 \) on \( [0, 1] \): \( L = \int_0^1\sqrt{1 + 4x^2}\,dx \approx 1.4789 \). It has a closed form, but it’s messy, and most arc length integrals don’t have nice antiderivatives at all. That’s why a numerical calculator is so useful here.

Example 5: a designed perfect square

Exam problems often use curves built so that \( 1 + [f']^2 \) becomes a perfect square. Take \( y = \frac{x^3}{6} + \frac{1}{2x} \) on \( [1, 2] \). The derivative is

$$f'(x) = \frac{x^2}{2} - \frac{1}{2x^2}$$

Squaring gives \( \frac{x^4}{4} - \frac12 + \frac{1}{4x^4} \). Adding 1 flips the middle sign:

$$1 + \big[f'(x)\big]^2 = \frac{x^4}{4} + \frac12 + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2$$

The square root cancels the square, leaving a polynomial-style integral:

$$\begin{aligned} L &= \int_1^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)dx \\ &= \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^2 \\ &= \left(\frac43 - \frac14\right) - \left(\frac16 - \frac12\right) = \frac{17}{12} \end{aligned}$$

Whenever the derivative looks like “something minus something,” check whether adding 1 turns it into “something plus something” squared.

Semicircle check

For \( y = \sqrt{1 - x^2} \) on \( [-1, 1] \), \( f'(x) = \frac{-x}{\sqrt{1-x^2}} \) and the integral gives \( \pi \), half the circumference of the unit circle, as it should. (It’s an improper integral because \( f' \) blows up at the ends.)

Parametric and polar curves

For \( x = x(t),\ y = y(t) \), \( \alpha\le t\le\beta \):

$$L = \int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$$

For a polar curve \( r = r(\theta) \): \( L = \int\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \).

If the curve is easier to write as \( x = g(y) \) for \( c \le y \le d \), swap the roles of the variables: \( L = \int_c^d\sqrt{1 + [g'(y)]^2}\,dy \).

Example 6: one arch of a cycloid

A point on the rim of a rolling wheel of radius 1 traces \( x = t - \sin t \), \( y = 1 - \cos t \). One arch runs from \( t = 0 \) to \( t = 2\pi \). The derivatives are \( 1 - \cos t \) and \( \sin t \), so

$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 2 - 2\cos t = 4\sin^2\frac t2$$

using the half-angle identity. On \( [0, 2\pi] \), \( \sin\frac t2 \ge 0 \), so the root is \( 2\sin\frac t2 \) and

$$L = \int_0^{2\pi}2\sin\frac t2\,dt = \Big[-4\cos\frac t2\Big]_0^{2\pi} = 8$$

One arch is exactly 8 times the wheel’s radius, a neat result that needs no approximation.

Example 7: a cardioid in polar form

For \( r = 1 + \cos\theta \), \( \frac{dr}{d\theta} = -\sin\theta \) and \( r^2 + (r')^2 = 2 + 2\cos\theta = 4\cos^2\frac\theta2 \). The root is \( 2\left|\cos\frac\theta2\right| \), and integrating over \( [0, 2\pi] \) gives a total length of 8. Keep the absolute value: \( \cos\frac\theta2 \) is negative on the second half of the interval, and dropping the bars would make the answer 0.

The arc length function

If you let the upper limit vary, you get the distance traveled along the curve from \( a \) to \( x \):

$$s(x) = \int_a^x\sqrt{1 + \big[f'(t)\big]^2}\,dt$$

By the Fundamental Theorem of Calculus, \( s'(x) = \sqrt{1 + [f'(x)]^2} \), which is exactly the stretch factor from before. This function is used to reparametrize curves by arc length, a key step in studying curvature.

Common mistakes

• Forgetting to square the derivative. The integrand is \( \sqrt{1 + [f'(x)]^2} \), not \( \sqrt{1 + f'(x)} \).
• Using f instead of f’. Arc length depends on the slope, not the height. Shifting a curve up does not change its length.
• Splitting the square root. \( \sqrt{1 + u^2} \) is not \( 1 + u \). Only simplify the root after you have written the inside as a perfect square.
• Dropping absolute values. \( \sqrt{g^2} = |g| \). If \( g \) changes sign on the interval, split the integral, as in the cardioid example.
• Mixing variables in parametric problems. When integrating in \( t \), use \( t \)-limits and \( t \)-derivatives throughout, not \( \frac{dy}{dx} \) with \( x \)-limits.

Where arc length is used

• Engineering and design: lengths of cables, pipes, roads and roller-coaster tracks.
• Physics: the distance an object travels along a path is the integral of its speed, which is the parametric arc length formula.
• Surface area of revolution: rotating \( y = f(x) \) about the x-axis gives surface area \( S = 2\pi\int_a^b f(x)\,ds \). For example, rotating \( y = x^3 \) on \( [0, 1] \) gives \( \frac{\pi}{27}\left(10^{3/2} - 1\right) \approx 3.563 \).
• Volumes vs lengths: the same slicing idea gives volumes in the disk and washer method; arc length slices a curve instead of a solid.

Arc length calculator

The gadget differentiates \( f \) symbolically and evaluates the integral with adaptive Gauss–Kronrod quadrature:

For a full-page version that shows \( f'(x) \) and the integral it sets up, use the arc length calculator.

Practice problems

Try these before checking the answers.

1. \( y = \tfrac23x^{3/2},\ 0 \le x \le 3 \)
2. \( y = 2x + 1,\ 0\le x\le 2 \)
3. \( y = \ln(\cos x),\ 0 \le x \le \frac\pi4 \)
4. \( y = \tfrac13(x^2 + 2)^{3/2},\ 0 \le x \le 3 \)
5. \( y = \frac{x^2}{4} - \frac{\ln x}{2},\ 1 \le x \le e \)
6. The circle \( x = 3\cos t,\ y = 3\sin t,\ 0 \le t \le 2\pi \)

Answers: (1) \( \frac{14}{3} \); (2) \( 2\sqrt5 \); (3) \( \ln\left(1 + \sqrt2\right) \); (4) \( 12 \), since \( 1 + [f']^2 = (x^2 + 1)^2 \); (5) \( \frac{e^2 + 1}{4} \approx 2.097 \); (6) \( 6\pi \).

FAQ

Why is the arc length always at least b − a?

Because \( \sqrt{1 + (f')^2} \ge 1 \). A curve can’t be shorter than the horizontal distance it covers.

Is arc length the same as displacement?

No. Arc length is the distance traveled along the curve; displacement is the straight-line distance between the endpoints.

Why do so few arc length integrals have a closed form?

The square root of \( 1 + [f']^2 \) rarely simplifies. Even the parabola needs a trig substitution, and an ellipse leads to an elliptic integral with no elementary antiderivative. In practice you evaluate most of them numerically, for example with a definite integral calculator.

What does ds mean?

\( ds \) is the length of an infinitesimal piece of curve. In Cartesian form \( ds = \sqrt{dx^2 + dy^2} \), and every arc length formula is just this expression rewritten in terms of \( dx \), \( dy \), \( dt \) or \( d\theta \).

Does the direction I trace the curve change the length?

No. As long as you integrate from the smaller limit to the larger one and trace the curve once, the length is the same in either direction. Be careful with parametric curves that retrace part of their path, which would count that part twice.

Further reading

• Arc Length (Paul’s Online Math Notes) — both forms of the formula with several worked examples, including trig substitution
• Arc Length of a Curve and Surface Area (OpenStax Calculus Volume 1) — a textbook treatment that also develops the surface area formula