Arc Length

How long is a curved path? Approximate it with short straight segments, find their lengths with the Pythagorean theorem, and let the segments shrink. The result is an integral, and evaluating it usually takes the substitution skills you already have.

Why it works

Over a small step $\Delta x$, the curve $y = f(x)$ rises $\Delta y$. The chord across that step has length $$\sqrt{(\Delta x)^{2} + (\Delta y)^{2}} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^{2}}\;\Delta x.$$ By the Mean Value Theorem, $\frac{\Delta y}{\Delta x} = f'(c)$ for some $c$ in the step. So the total length of the chords is a Riemann sum for $\int_a^b \sqrt{1 + f'(x)^{2}}\,dx$, and the chords hug the curve better and better as $\Delta x \to 0$.

Key idea

If $f'$ is continuous on $[a, b]$, the length of $y = f(x)$ from $x = a$ to $x = b$ is $$L = \int_a^b \sqrt{1 + \left[f'(x)\right]^{2}}\,dx.$$

The curve $y = \frac{x^{2}}{2}$ on $[0, 2]$ and $n$ straight chords. Each chord has length $\Delta x\sqrt{1 + (\Delta y/\Delta x)^{2}}$. Raise $n$: the total creeps up toward the exact length $\int_0^2\sqrt{1 + x^{2}}\,dx \approx 2.958$.

Worked example: A root that simplifies

Find the length of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$.

$f'(x) = x^{1/2}$, so $1 + f'(x)^{2} = 1 + x$.
$L = \int_0^3 \sqrt{1 + x}\,dx = \left[\frac{2}{3}(1 + x)^{3/2}\right]_0^3$.
$= \frac{2}{3}(8 - 1) = \frac{14}{3}$.

Answer: $\frac{14}{3}$

Worked example: A hidden perfect square

Find the length of $y = \frac{x^{3}}{6} + \frac{1}{2x}$ from $x = 1$ to $x = 2$.

$f'(x) = \frac{x^{2}}{2} - \frac{1}{2x^{2}}$, so $f'(x)^{2} = \frac{x^{4}}{4} - \frac{1}{2} + \frac{1}{4x^{4}}$.
Adding 1 flips the middle sign: $$\begin{aligned} 1 + f'(x)^{2} &= \frac{x^{4}}{4} + \frac{1}{2} + \frac{1}{4x^{4}} \\ &= \left(\frac{x^{2}}{2} + \frac{1}{2x^{2}}\right)^{2}. \end{aligned}$$
$$\begin{aligned} L &= \int_1^2 \left(\frac{x^{2}}{2} + \frac{1}{2x^{2}}\right)dx \\ &= \left[\frac{x^{3}}{6} - \frac{1}{2x}\right]_1^2 \\ &= \left(\frac{4}{3} - \frac{1}{4}\right) - \left(\frac{1}{6} - \frac{1}{2}\right) \\ &= \frac{17}{12}. \end{aligned}$$

Answer: $\frac{17}{12}$

Worked example: A trig curve

Find the length of $y = \ln(\cos x)$ from $x = 0$ to $x = \frac{\pi}{4}$.

$f'(x) = \frac{-\sin x}{\cos x} = -\tan x$, so $1 + f'(x)^{2} = 1 + \tan^{2} x = \sec^{2} x$.
On $\left[0, \frac{\pi}{4}\right]$, $\sec x > 0$, so the root is $\sec x$.
$$\begin{aligned} L &= \int_0^{\pi/4} \sec x\,dx \\ &= \Big[\ln\left|\sec x + \tan x\right|\Big]_0^{\pi/4} \\ &= \ln\left(\sqrt{2} + 1\right) - \ln 1. \end{aligned}$$

Answer: $\ln\left(\sqrt{2} + 1\right) \approx 0.881$, a bit longer than the straight-line distance from $(0, 0)$ to $\left(\frac{\pi}{4}, -\frac{\ln 2}{2}\right)$, about $0.86$.

Common mistake

Forgetting to square the derivative: writing $\sqrt{1 + f'(x)}$. Test the formula on a straight line, where you know the answer. For $y = 2x$ from 0 to 1 the length is $\sqrt{1^{2} + 2^{2}} = \sqrt{5}$. The correct formula gives $\int_0^1\sqrt{1 + 2^{2}}\,dx = \sqrt{5}$; the unsquared one gives $\sqrt{3}$.

Letting the upper limit vary gives the arc length function $s(x) = \int_a^x \sqrt{1 + f'(t)^{2}}\,dt$. By the Fundamental Theorem, $\frac{ds}{dx} = \sqrt{1 + f'(x)^{2}}$, which is often written as $ds^{2} = dx^{2} + dy^{2}$: a tiny step along the curve is the hypotenuse of a tiny right triangle. That picture will return for parametric curves and polar curves.

Most arc length integrals have no elementary antiderivative; even $y = x^{2}$ leads to a hard one. Textbook problems are chosen so that $1 + f'^{2}$ is a perfect square. In practice you set up the integral and evaluate it with the trapezoid or Simpson's rule.

Arc Length

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