Arc Length

A small piece is almost straight

Zoom in on a short piece of a smooth curve and it looks straight. Across the piece x changes by dx and y by dy, so the piece is the hypotenuse of a right triangle with legs dx and dy.

By Pythagoras its length ds satisfies $d{s}^2=d{x}^2+d{y}^2$ds² = dx² + dy², so $ds=\sqrt{d{x}^2+d{y}^2}$ds = √(dx² + dy²).

A short piece of curve, ds, as the hypotenuse of a right triangle with legs dx across and dy up.

Take dx out of the root

Factor $d{x}^2$dx² out of the root: $\sqrt{d{x}^2+d{y}^2}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$√(dx² + dy²) = √(1 + (dy/dx)²) dx. Adding every piece from x = a to x = b is integrating, so the length of the curve is $L=\underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$L = ∫ √(1 + (dy/dx)²) dx from a to b.

A straight line checks it. For $y=\frac{4x}{3}$y = 4x/3 from 0 to 3, $\frac{dy}{dx}=\frac{4}{3}$dy/dx = 4/3, so the root is $\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac{5}{3}$√(1 + 16/9) = √(25/9) = 5/3 at every x, and $L=\frac{5}{3}\times 3=5$L = 5/3 × 3 = 5: the hypotenuse of a 3, 4, 5 triangle.

The parabola $y={\left(x-2\right)}^2/1.5$y = (x − 2)² / 1.5 from 0 to 4, followed by 4 straight chords, each $\sqrt{\Delta x^2+\Delta y^2}$√(Δx² + Δy²). The chords add to 6.876, short of the true length 6.977, because each chord cuts across a bend. Add chords and the sum climbs: 8 give 6.951, and 13 come within 0.01.

A curve that integrates exactly

Take $y=\frac{2}{3}x\sqrt{x}$y = ⅔x√x from x = 0 to 3. Its gradient is $\frac{dy}{dx}=\sqrt{x}$dy/dx = √x, so $1+{\left(\frac{dy}{dx}\right)}^2=1+x$1 + (dy/dx)² = 1 + x, and L is the integral of $\sqrt{1+x}$√(1 + x) from 0 to 3.

An antiderivative is $\frac{2}{3}\left(1+x\right)\sqrt{1+x}$⅔(1 + x)√(1 + x). At x = 3 that is $\frac{2}{3}\times 4\times 2=\frac{16}{3}$⅔ × 4 × 2 = 16/3, and at x = 0 it is ⅔, so $L=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$L = 16/3 − 2/3 = 14/3, about 4.667.

The straight chord between the ends, (0, 0) and $\left(3,2\sqrt{3}\right)$(3, 2√3), is $\sqrt{9+12}=\sqrt{21}$√(9 + 12) = √21, about 4.583. The curve is a little longer, as any curve between two points must be.

$y=\frac{2}{3}x\sqrt{x}$y = ⅔x√x from (0, 0) to $\left(3,2\sqrt{3}\right)$(3, 2√3), with the straight chord between the ends. The curve is $\frac{14}{3}$14/3 long, about 4.667, and the chord is $\sqrt{21}$√21, about 4.583.

Most lengths are found numerically

For most curves the root $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$√(1 + (dy/dx)²) has no simple antiderivative, and the length is found numerically, by adding many short chords as the figure above does.

For $y=x^2$y = x² from 0 to 1, $\frac{dy}{dx}=2x$dy/dx = 2x and L is the integral of $\sqrt{1+4x^2}$√(1 + 4x²) from 0 to 1, which comes to about 1.479. Two bounds check it: the chord from (0, 0) to (1, 1) is $\sqrt{2}$√2, about 1.414, and the path 1 across and then 1 up is 2, so the curve’s length lies between them.

This one does have an exact value, through the substitution 2x = sinh u: $\frac{\sqrt{5}}{2}+\frac{1}{4}\mathrm{arsinh}2$√5/2 + ¼ arsinh 2, which is the same 1.479.

A curve given by a parameter

When x and y are both given in terms of a parameter t, take dt out of the root instead: $ds=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$ds = √((dx/dt)² + (dy/dt)²) dt, and L is its integral between the two values of t.

A circle checks it. With x = r cos t and y = r sin t, $\frac{dx}{dt}=-r\sin t$dx/dt = −r sin t and $\frac{dy}{dt}=r\cos t$dy/dt = r cos t, so the root is $\sqrt{r^2{\sin }^{2}t+r^2{\cos }^{2}t}=r$√(r²sin²t + r²cos²t) = r. From t = 0 to $2\pi$2π the length is $2\pi r$2πr, the circumference.

For $x=t^2$x = t² and $y=t^3$y = t³ from t = 0 to 1, $\frac{dx}{dt}=2t$dx/dt = 2t and $\frac{dy}{dt}=3t^2$dy/dt = 3t², so the root is $\sqrt{4t^2+9t^4}=t\sqrt{4+9t^2}$√(4t² + 9t⁴) = t√(4 + 9t²). An antiderivative is $\frac{\left(4+9t^2\right)\sqrt{4+9t^2}}{27}$(4 + 9t²)√(4 + 9t²)/27, so $L=\frac{13\sqrt{13}-8}{27}$L = (13√13 − 8)/27, about 1.440. The chord from (0, 0) to (1, 1) is $\sqrt{2}$√2, about 1.414.

The usual mistakes

Integrating $\frac{dy}{dx}$dy/dx. That adds up the rises and gives the change in y, not the length.

Dropping the 1 under the root. The 1 is the dx side of the triangle; without it the integral again measures only the rise.

Giving the chord. The straight distance between the ends is shorter than the curve.

Using $\frac{dy}{dx}$dy/dx for a curve given by a parameter. Take dt out of the root, so that both $\frac{dx}{dt}$dx/dt and $\frac{dy}{dt}$dy/dt are squared and added.

A roller coaster

In the application below, a roller coaster’s first drop has gradient $\sqrt{\frac{x}{96}}$√(x/96), so $1+{\left(\frac{dy}{dx}\right)}^2=1+\frac{x}{96}$1 + (dy/dx)² = 1 + x/96 and the root integrates exactly. The integral of the gradient alone gives how far the track drops, and the straight ramp is the chord.