Math Core

How long is a curve? A ruler only measures straight segments, but calculus lets you break a curve into tiny pieces that are almost straight, measure each piece with the Pythagorean theorem, and add them up. Arc length is a BC-only topic, and the same idea returns in the next unit for parametric and polar curves.

Building the formula

Suppose $y = f(x)$ has a continuous derivative on $[a, b]$ . Zoom in on a tiny piece of the curve. It looks like the hypotenuse of a right triangle with horizontal leg $dx$ and vertical leg $dy$ , so its length is

$ds = \sqrt{(dx)^2 + (dy)^2}.$

Factor $(dx)^2$ out of the square root:

$ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.$

Adding all the pieces from $x = a$ to $x = b$ gives the length of the whole curve.

Arc length

If $f'$ is continuous on $[a, b]$ , the length of the curve $y = f(x)$ from $x = a$ to $x = b$ is

$L = \int_a^b \sqrt{1 + \big(f'(x)\big)^2}\,dx.$

If the curve is given as $x = g(y)$ for $c \le y \le d$ , the length is

$L = \int_c^d \sqrt{1 + \big(g'(y)\big)^2}\,dy.$

A quick check: for a line $y = mx + b$ , the integrand is the constant $\sqrt{1 + m^2}$ , so the length from $x = a$ to $x = b$ is $(b - a)\sqrt{1 + m^2}$ . That is exactly the distance formula, which is a good sign the formula is right.

When can you integrate by hand?

The square root makes most arc length integrals impossible to find with elementary antiderivatives. Even $y = x^2$ from $0$ to $1$ gives $\int_0^1 \sqrt{1 + 4x^2}\,dx$ , which needs a trig substitution. On the AP exam, arc length appears in two ways:

Calculator questions: set up the integral and evaluate it numerically.
Hand questions: the function is chosen so that $1 + (f')^2$ is a perfect square, or so a simple $u$ -substitution works.

The perfect-square trick shows up often. If $f'(x) = A - B$ where $4AB = 1$ , then

$1 + (A - B)^2 = A^2 - 2AB + B^2 + 1 = A^2 + 2AB + B^2 = (A + B)^2,$

and the square root disappears.

The curve y = (2/3)x^(3/2) from (0, 0) to (3, 2√3). Its length is 14/3 ≈ 4.67, a bit more than the straight-line distance √21 ≈ 4.58.Open in grapher →

Worked example: A u-substitution

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

Here $y' = x^{1/2}$ , so $1 + (y')^2 = 1 + x$ .

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

Worked example: A perfect square

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

$y' = \frac{x^2}{2} - \frac{1}{2x^2}.$

With $A = \dfrac{x^2}{2}$ and $B = \dfrac{1}{2x^2}$ , we have $4AB = 1$ . So

$1 + (y')^2 = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2.$

$L = \int_1^2 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^2 = \left(\frac{4}{3} - \frac{1}{4}\right) - \left(\frac{1}{6} - \frac{1}{2}\right) = \frac{13}{12} + \frac{1}{3} = \frac{17}{12}.$

Worked example: A trig identity

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

$y' = \frac{-\sin x}{\cos x} = -\tan x, \qquad 1 + (y')^2 = 1 + \tan^2 x = \sec^2 x.$

On $\left[0, \tfrac{\pi}{4}\right]$ , $\sec x > 0$ , so $\sqrt{\sec^2 x} = \sec x$ .

$L = \int_0^{\pi/4} \sec x\,dx = \Big[\ln|\sec x + \tan x|\Big]_0^{\pi/4} = \ln\big(\sqrt{2} + 1\big) - \ln 1 = \ln\big(\sqrt{2} + 1\big).$

Worked example: Calculator active

Find the length of $y = x^2$ from $x = 0$ to $x = 1$ , to three decimal places.

$L = \int_0^1 \sqrt{1 + (2x)^2}\,dx = \int_0^1 \sqrt{1 + 4x^2}\,dx \approx 1.479.$

On a calculator question, the setup is worth as much as the number. Write the integral, then evaluate.

Common mistake

The most common mistake is forgetting to square the derivative, or forgetting the $1$ . The integrand is $\sqrt{1 + (f')^2}$ , not $\sqrt{1 + f'}$ and not $\sqrt{(f')^2}$ . Another slip is taking the square root of $1 + (f')^2$ term by term: $\sqrt{1 + u^2}$ is not $1 + u$ .

Tip

Arc length is always at least the straight-line distance between the endpoints. If your answer is shorter than the chord, something is wrong.

Curves given as x = g(y)

Sometimes $x$ is the easier function. For example, the curve $x = \tfrac{1}{2}y^2$ from $y = 0$ to $y = 1$ has $\dfrac{dx}{dy} = y$ , so its length is $\int_0^1 \sqrt{1 + y^2}\,dy$ . Use whichever variable makes the derivative simpler, and make sure the limits match the variable you integrate in.