Arc length, explained with examples
Arc length is the length of a curve between two points. Calculus finds it by adding up tiny straight segments, and AP Calculus BC uses it for graphs of functions, parametric curves and the total distance travelled by a moving particle.
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Where the formula comes from
Zoom in far enough on a smooth curve and a tiny piece of it looks straight. If that tiny piece covers a horizontal change dx and a vertical change dy, Pythagoras gives its length as √(dx² + dy²). Factor out dx: the length is √(1 + (dy/dx)²) dx. Adding up all those tiny lengths with an integral gives the length of the whole curve. The same idea, with a different variable, gives the parametric formula.
The formulas
- For y = f(x) from x = a to x = b: L = ∫ab √(1 + [f'(x)]²) dx
- For a parametric curve x(t), y(t) from t = a to t = b: L = ∫ab √((dx/dt)² + (dy/dt)²) dt
Arc length is always positive. The integrand is a square root of a sum of squares, so it is never negative.
Worked examples you can do by hand
Most arc length integrals cannot be done exactly, which is why AP questions usually allow a calculator. A few are designed to simplify neatly.
Example 1: a power function
Find the length of y = (2/3)x3/2 from x = 0 to x = 3.
- f'(x) = x1/2, so 1 + [f'(x)]² = 1 + x.
- L = ∫03 √(1 + x) dx = [(2/3)(1 + x)3/2]03 = (2/3)(8 − 1).
- L = 14/3 ≈ 4.67.
Example 2: a circle
Find the length of the parametric curve x = 3cos t, y = 3sin t for 0 ≤ t ≤ 2π.
- dx/dt = −3sin t and dy/dt = 3cos t, so the integrand is √(9sin²t + 9cos²t) = 3.
- L = ∫02π 3 dt = 6π, the circumference of a circle of radius 3, as expected.
Example 3: a logarithm
Find the length of y = ln(cos x) from x = 0 to x = π/4.
- f'(x) = −tan x, so 1 + tan²x = sec²x and the integrand is sec x.
- L = ∫0π/4 sec x dx = [ln|sec x + tan x|]0π/4 = ln(√2 + 1).
- L ≈ 0.881.
Example 4: a parametric curve
Find the length of x = t², y = t³ for 0 ≤ t ≤ 1.
- dx/dt = 2t, dy/dt = 3t². The integrand is √(4t² + 9t4) = t√(4 + 9t²) for t ≥ 0.
- Substitute u = 4 + 9t²: L = (1/27)[(4 + 9t²)3/2]01 = (133/2 − 8)/27.
- L ≈ 1.44.
Calculator questions
On the calculator-allowed sections, you will usually be asked to write the integral and evaluate it numerically. Set it up carefully, with the derivative inside the square root, and give the answer to three decimal places.
Example 5: with a calculator
Find the length of y = x² from x = 0 to x = 1.
- f'(x) = 2x, so L = ∫01 √(1 + 4x²) dx.
- Using a calculator: L ≈ 1.479. Write the integral in full before the number; that setup earns a point by itself.
Speed and total distance
For a particle moving in the plane with position (x(t), y(t)), its speed at time t is √((dx/dt)² + (dy/dt)²), the magnitude of the velocity vector. The total distance travelled from t = a to t = b is the integral of speed, which is exactly the parametric arc length formula. This is how AP BC free-response questions on particle motion in two dimensions use arc length.
Example 6: particle motion
A particle has velocity components dx/dt = 3 and dy/dt = 4t for 0 ≤ t ≤ 2. Write and evaluate an integral for the total distance travelled.
- Speed = √(9 + 16t²).
- Distance = ∫02 √(9 + 16t²) dt ≈ 10.465 units, using a calculator.
Why arc length integrals are often hard
The square root in the formula usually blocks exact integration. Even a simple curve like y = x² leads to ∫√(1 + 4x²) dx, which needs advanced techniques, and y = sin x gives an integral with no elementary answer at all. That is why exam questions designed for working by hand use special functions, such as x3/2 or ln(cos x), where 1 + [f'(x)]² turns into a perfect square. If an integral looks impossible by hand on a non-calculator question, check whether the expression under the root simplifies.
Free-response tips
When an AP question says “find the total distance travelled by the particle”, write the speed integral with its limits before using your calculator, and give units if the question uses them. When it asks for the length of a curve given as y = f(x), make sure you differentiate first and square the derivative inside the root. Readers award a point for a correct integral expression even if the numerical answer is wrong, so never skip the setup.
Displacement versus distance
Do not confuse total distance travelled with displacement. Displacement is the straight-line change in position, found from the change in x and the change in y separately. Distance travelled follows the actual path, so it is always at least as large as the size of the displacement. A particle that goes round a full circle has travelled 2πr but has zero displacement.
Common mistakes
- Forgetting the 1 inside the square root for y = f(x).
- Squaring the derivative after integrating, instead of before.
- Using the arc length formula for displacement.
- Leaving the integral without limits, or with the wrong variable, in a parametric question.
- Writing √(1 + f'(x))² instead of √(1 + [f'(x)]²).
Practice questions
Try each one before opening the answer.
Q1. Write the integral for the length of y = sin x from 0 to π.
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Q2. Find the length of the line y = 2x + 1 from x = 0 to x = 3 using the formula.
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Q3. A particle has dx/dt = cos t and dy/dt = sin t. What is its speed?
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Q4. Is arc length ever negative?
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Frequently asked questions
Is arc length on AP Calculus AB?
No. Arc length for functions and parametric curves is part of AP Calculus BC only, in Units 8 and 9.
Is arc length in A level Maths?
Not in A level Maths. It appears in some A level Further Maths options.
Key takeaways
- Arc length adds up tiny straight segments using Pythagoras.
- y = f(x): L = ∫ √(1 + [f'(x)]²) dx.
- Parametric: L = ∫ √((dx/dt)² + (dy/dt)²) dt.
- Total distance travelled is the integral of speed.
- Most arc length integrals need a calculator; set them up carefully.