How long is a curve? You can't lay a ruler along a bend — but you can chop it into tiny straight
pieces, and each piece is just the hypotenuse of a little right triangle. Arc length is Pythagoras, integrated.
1 · A curve is a chain of tiny hypotenuses
Start with one straight chord from end to end — useless, it ignores every bend. Add more chords
and the chain starts to hug the curve. Each chord is the hypotenuse of a right triangle with legs Δx and Δy.
Segments: 1
polygon length = —true arc length = ≈ 7.640
Each little hypotenuse has length Δs = √(Δx² + Δy²)
by the Pythagorean theorem. The curve's length is the sum of all of them — and as the pieces shrink, that sum becomes
an integral. The only question left is how to turn that √ into something you can integrate.
2 · Where the formula comes from
The famous arc-length formula looks intimidating, but it's nothing more than that little triangle,
tidied up. Three steps:
1
Pythagoras on one piece. A tiny segment of curve is the hypotenuse of a right triangle with horizontal leg Δx and vertical leg Δy.
2
Factor out Δx. Pull Δx out of the square root so the inside is written with the slope Δy/Δx.
3
Δy/Δx becomes the derivative. As the piece shrinks, the slope of the chord becomes f′(x), the sum becomes an integral, and Δx becomes dx.
That's the whole formula — and now you can see it's not arbitrary. The 1 under the
root is the horizontal leg (Δx² factored out), and the [f′(x)]² is the vertical leg (the
slope, squared). It's the Pythagorean theorem wearing an integral sign.
3 · A note on curves given as x = g(y)
Almost every problem you'll see is y = f(x) — that's the formula from the last
section, and it's what gets tested. But the same Pythagorean idea works no matter which variable a curve is written in.
If a curve is more natural as x = g(y) (a sideways curve that fails the vertical-line
test, say), you just swap the roles of x and y: differentiate x with respect to y, and integrate over the y-interval.
Nothing about the logic changes — you're still adding up little hypotenuses, only now you factored out Δy instead of Δx.
Coming in Unit 9.
When x and y both depend on a parameter t, the same idea gives a curve's length (and a particle's distance traveled)
from its speed. You'll meet that parametric form once parametric curves are introduced in Unit 9.
4 · Worked examples
Here's the catch the formula hides: that square root is almost never integrable by hand. Exam
problems are rigged so the inside becomes a perfect square. Spotting that is the real skill.
A clean setup
Problem. Find the length of from
x = 0 to x = 3.
1
Differentiate. Bring down the power.
2
Build the integrand. 1 + (f′)² = 1 + x — and the messy root collapses to √(1+x).
3
Integrate. A clean power rule on (1+x).
L = 14/3
The perfect-square trick
This one looks hopeless until you notice 1 + (f′)² is a perfect square. That's
not luck — it's how every by-hand arc-length problem is built.
Problem. Find the length of from
x = 1 to x = 2.
1
Differentiate. f′(x) = x²/2 − 1/(2x²).
2
Spot the perfect square. Adding 1 turns the difference into a sum-squared — the cross terms cancel perfectly. This is the move to look for.
3
Root vanishes, then integrate. √ of a perfect square is just the bracket.
L = 17/12
Exam tip. If 1 + (f′)² doesn't become a perfect square, you're not meant to integrate
by hand — it's a calculator question, or the task is only to set up the integral. Always write the integral
first; you'll get credit for the correct setup even when the antiderivative is out of reach.
5 · Does the formula actually work?
A formula you can't sanity-check is just a thing to memorize. So let's aim it at a shape whose
length we already know — a circle — and see if it gives the right answer.
Take the top half of the unit circle, y = √(1 − x²) on
[−1, 1]. We already know the answer: a full unit circle has circumference
2πr = 2π, so the top half should come out to exactly π. Does the formula agree?
Check. Find the arc length of on
[−1, 1] — and compare to the known semicircle length π.
1
Differentiate. The derivative of √(1 − x²).
2
Build the integrand — and watch it simplify. The 1 and the fraction combine over a common denominator, and almost everything cancels.
3
Integrate. That's the arcsine integral — one worth recognizing.
L = π ✓ — exactly the semicircle length we expected
It works. The formula, built from nothing but Pythagoras on tiny segments, reproduces the circumference
of a circle that geometry handed us thousands of years ago. That's the sign of a formula you can trust — not just
one you memorized.
6 · Drill it
Set up the integral, spot the perfect square, pick the right variable. Five quick checks.
Score: 0 / 0
That wraps the methods of Unit 8. Put it all together in the Exam Practice
capstone, grab the cheat sheet for quick review — or head on to Unit 9,
where arc length returns for parametric curves and motion.