Arc Length and Distance Traveled — AP® Calculus BC Notes

Area under a curve and volume of a solid came from adding up thin slices. The arc length of a curve, its length measured along the curve, comes the same way: from adding up short straight pieces. Each piece is the hypotenuse of a small right triangle, so the Pythagorean theorem does the work.

Chords first

Split $[a, b]$ into pieces. Over a piece with run $\Delta x$ , the curve rises $\Delta y$ , and the chord joining the two points has length

$\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}\,\Delta x$

The figure shows three chords along $y = \frac{2}{3}x^{3/2}$ on $[0, 3]$ . Their total, about 4.647, is a little less than the curve's length: a chord is the shortest path between its endpoints. More, shorter chords bring the total closer.

The curve y = (2/3)x to the 3/2 from x = 0 to x = 3, in violet, rising from (0, 0) to (3, 2√3), about 3.464. Three dashed blue chords join the points at x = 0, 1, 2 and 3, with dotted right triangles showing each run and rise. The chords measure 1.202, 1.577 and 1.869, a total of 4.647, a little less than the curve’s length 14/3, about 4.667. With more chords the total becomes the integral of √(1 + (dy/dx)²) dx, here the integral from 0 to 3 of √(1 + x) dx. — **Figure 1.** Each chord is a hypotenuse; their total approaches the length of the curve. LevlPrep original diagram.

From chords to an integral

If $f$ is differentiable, the Mean Value Theorem gives a point $c$ in each piece with $\frac{\Delta y}{\Delta x} = f'(c)$ . So the chord total is $\sum \sqrt{1 + (f'(c))^2}\,\Delta x$ , a Riemann sum. As the pieces shrink it becomes an integral:

$\text{arc length} = \int_a^b \sqrt{1 + (f'(x))^2}\,dx,$

for a function $f$ whose derivative is continuous on $[a, b]$ .

Worked example 1. Find the length of the graph of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$ .

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

Setup: $\int_0^3 \sqrt{1 + x}\,dx$ .

Evaluate: $\frac{2}{3}(1 + x)^{3/2}\Big|_0^3$ $= \frac{2}{3}(8 - 1)$ $= \frac{14}{3}$ $\approx 4.667$ .

Exact answers like this are rare: the square root usually has no elementary antiderivative. Most arc length questions are calculator questions, where the setup earns a point and the value to three decimal places earns another.

Worked example 2 (calculator). Find the length of one arch of $y = \sin x$ , from $x = 0$ to $x = \pi$ .

$y' = \cos x$ , so the length is $\int_0^\pi \sqrt{1 + \cos^2 x}\,dx$ $\approx 3.820$ (radian mode).

Sanity check: the arch is longer than its base, $\pi \approx 3.142$ , and longer than the two chords from $(0, 0)$ to $\left(\frac{\pi}{2}, 1\right)$ to $(\pi, 0)$ , which total about 3.724.

Setups that lose the point

Curve and interval	Right setup	Common wrong setup
$y = x^3$ , $[0, 1]$	$\int_0^1 \sqrt{1 + 9x^4}\,dx$	$\int_0^1 \sqrt{1 + x^6}\,dx$ (squared $f$ , not $f'$ )
$y = \sqrt{x}$ , $[1, 4]$	$\int_1^4 \sqrt{1 + \frac{1}{4x}}\,dx$	$\int_1^4 \sqrt{1 + \frac{1}{2\sqrt{x}}}\,dx$ (derivative not squared)
$y = \frac{x^2}{2}$ , $[0, 2]$	$\int_0^2 \sqrt{1 + x^2}\,dx$	$\int_0^2 (1 + x^2)\,dx$ (no square root)

The scoring guidelines give the setup point only for an integrand in the right form, with the right limits. Write $\sqrt{1 + (f'(x))^2}$ with your derivative substituted, then evaluate.

Curves given as x = g(y)

When $x$ is a function of $y$ , swap the roles: the length from $y = c$ to $y = d$ is $\int_c^d \sqrt{1 + (g'(y))^2}\,dy$ . For $x = y^2$ from $y = 0$ to $y = 2$ : $\int_0^2 \sqrt{1 + 4y^2}\,dy$ $\approx 4.647$ . The limits are $y$ -values.

Arc length from a table

If only values of $f'$ are known, estimate $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ with a Riemann or trapezoidal sum of the values of $\sqrt{1 + (f'(x))^2}$ , using the table's widths. For example, with $f'(0) = 0$ , $f'(2) = \frac{3}{4}$ and $f'(4) = \frac{12}{5}$ , the left sum on $[0, 6]$ is $2\left(1 + \frac{5}{4} + \frac{13}{5}\right) = 9.7$ .

Reading an arc length in context

If $h(x)$ is the height of a cable in meters, $x$ meters from a pole, then $\int_0^5 \sqrt{1 + (h'(x))^2}\,dx$ is the length of cable between $x = 0$ and $x = 5$ , in meters. Name the curve, the interval and the unit. The length does not depend on where the curve sits: moving a graph up or down leaves $f'$ , and so its length, unchanged.