MISSION · Calculus
Arc Length of a Curve
Straighten the curve into infinitely many hypotenuses and add them up
LEVEL 1 · EXPLORE
Make the idea move
Change the controls and watch what happens. Follow your curiosity and see what you notice.
Loading the activity…
LEVEL 2 · LEARN
Unlock the explanation
Open the chapters in any order. Try the examples as you go.
The idea in one sentence
Chop the curve into pieces so short they are effectively straight, use Pythagoras on each, and integrate.
Building the formula
Take a tiny piece of curve. Over a short enough stretch it is indistinguishable from its own tangent line, so it is the hypotenuse of a right triangle with legs \(dx\) and \(dy\):
\[ds^2 = dx^2 + dy^2\]Factor out \(dx^2\) and take the root:
\[ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\;dx\]Integrating that from \(a\) to \(b\) adds all the little hypotenuses together:
\[L = \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\,dx\]The whole formula is Pythagoras plus the observation that a smooth curve is locally straight. There is nothing else in it.
Checking it on something you already know
A formula this abstract deserves a test where the answer is known independently.
Take \(y = \frac{4}{3}x\) from \(x = 0\) to \(x = 3\). The derivative is \(\frac43\), so
\[L = \int_0^3 \sqrt{1 + \tfrac{16}{9}}\;dx = \int_0^3 \tfrac53\,dx = 5\]And the straight-line distance from \((0,0)\) to \((3,4)\) is indeed 5. The integral has reproduced the 3-4-5 triangle, which is a good sign.
Notice the length is always at least \(b - a\), since the square root is never below 1. A curve is never shorter than the interval it spans, and it equals it only when the curve is horizontal.
Why the integrals are usually horrible
Arc length is conceptually the simplest thing in this chapter and computationally the worst.
\(y = x^2\) gives \(\int \sqrt{1 + 4x^2}\,dx\), which needs trigonometric substitution and produces logarithms. The circumference of an ellipse has no elementary closed form at all — it defines a function, the elliptic integral, invented for that purpose.
So most real arc lengths are found numerically, with Riemann sums or better. This is a topic where the formula is the lesson and the antiderivative is usually somebody else’s problem.
The exceptions that do work out neatly are worth recognising, because those are the ones that appear on exams: straight lines, \(y = x^{3/2}\), the catenary \(y = \cosh x\), and a handful of others engineered so the expression under the root becomes a perfect square.
The parametric and polar versions
When \(x\) and \(y\) both depend on \(t\), divide \(ds^2 = dx^2 + dy^2\) by \(dt^2\) instead:
\[L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt\]This version is often the better one, because it handles curves that double back or are not functions of \(x\) at all — a full circle, for instance.
It also has a physical reading. If \(t\) is time, the square root is the speed, and arc length is speed integrated over time, which is total distance travelled. Same formula, two subjects.
In polar form it becomes \(\int \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta\).
A related use
Spin the curve about an axis and each \(ds\) sweeps out a thin band of circumference \(2\pi y\), giving the surface area of revolution:
\[S = \int 2\pi y \sqrt{1 + (y')^2}\,dx\]Note the contrast with volumes of revolution, which use \(dx\) because a disk’s thickness is horizontal. Surface area uses \(ds\) because the surface follows the slanted edge. Using \(dx\) for a surface is the standard error, and it systematically undercounts.
Common mix-ups
- Forgetting the 1. The integrand is \(\sqrt{1 + (y')^2}\), not \(\sqrt{(y')^2}\).
- Squaring the function instead of the derivative.
- Splitting the root. \(\sqrt{1 + (y')^2} \ne 1 + y'\).
- Using \(dx\) for a surface of revolution where \(ds\) is needed.
- Expecting a tidy antiderivative. Most do not have one.
Check yourself
What is inside the square root, how long is \(y = \frac43 x\) on \([0,3]\), and what is \(ds^2\) equal to?
The integrand is \(\sqrt{1 + (dy/dx)^2}\). For the line, \(\int_0^3 \frac53 dx = 5\) — the 3-4-5 triangle. And \(ds^2 = dx^2 + dy^2\), straight from Pythagoras.
LEVEL 3 · REFLECT
Add this quest to your trail
When you feel ready, mark this idea as explored. You can always come back and try it again.
EXPLORER STRATEGIES