Arc Length and Distance Traveled

Part 1 · Hook

Why this matters

Lay a string along a curve, straighten it, and measure it: that is the curve's length. Calculus gets the same number without a string. Approximate the curve by short straight chords, find each chord's length with the Pythagorean theorem, and let the chords shrink. The sum becomes an integral.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the distance between $(1, 2)$ and $(4, 6)$ ?

5
7
25
$\sqrt{7}$

Show the answer

$\sqrt{3^2 + 4^2} = 5$ .

Correct: 5:
7:
25:
$\sqrt{7}$ :

2. If $y = \frac{2}{3}x^{3/2}$ , then $\frac{dy}{dx} =$

$x^{1/2}$
$\frac{2}{3}x^{1/2}$
$\frac{4}{15}x^{5/2}$
$x^{3/2}$

Show the answer

$\frac{2}{3}\cdot\frac{3}{2}x^{1/2}$ $= x^{1/2}$ .

Correct: $x^{1/2}$ :
$\frac{2}{3}x^{1/2}$ :
$\frac{4}{15}x^{5/2}$ :
$x^{3/2}$ :

Part 4 · See it

See it first

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. — Three chords along y = (2/3)x^(3/2) on [0, 3] total about 4.647, a little less than the curve’s length, 14/3 ≈ 4.667. Each chord is a hypotenuse; with more chords the total becomes the arc length integral. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Replace a short piece of the curve by its chord, with run $\Delta x$ and rise $\Delta y$ the chord's length is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ $= \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}\,\Delta x$
By the Mean Value Theorem, $\frac{\Delta y}{\Delta x} = f'(c)$ for some $c$ in the piecethe chord total is a Riemann sum of $\sqrt{1 + (f'(x))^2}$
Let the pieces shrinkthe sum becomes $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ , the arc length
The curve is given as $x = g(y)$ swap roles: $\int_c^d \sqrt{1 + (g'(y))^2}\,dy$ with $y$ -limits

Part 6 · Key ideas

Key ideas

Arc length of $y = f(x)$ from $x = a$ to $x = b$ : $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ , for $f'$ continuous on $[a, b]$ .
Square the derivative, not the function; keep the square root and the 1.
Few arc length integrals have exact values; most are calculator questions. Write the setup, then the value to three decimal places.
From a table of $f'$ , estimate the length with a Riemann or trapezoidal sum of $\sqrt{1 + (f')^2}$ .

Part 7 · Misconception

A common mistake

The wrong idea: The length of $y = x^3$ from 0 to 1 is $\int_0^1 \sqrt{1 + x^6}\,dx$ .

What actually happens: $x^6$ squares the function. The formula squares the derivative: $y' = 3x^2$ , so the length is $\int_0^1 \sqrt{1 + 9x^4}\,dx$ $\approx 1.548$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Chords along $y = \frac{x^2}{2}$

The graph of

f(x) = \frac{x^2}{2}

for

0 \le x \le 2

is shown with two chords, from

(0, 0)

to

\left(1, \frac{1}{2}\right)

and from

\left(1, \frac{1}{2}\right)

to

(2, 2)

.

1. Find the length of the chord from $(0, 0)$ to $\left(1, \frac{1}{2}\right)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

By the Pythagorean theorem: $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ $= \sqrt{1 + \frac{1}{4}}$ $= \frac{\sqrt{5}}{2}$ .

Answer: $\frac{\sqrt{5}}{2}$ ≈ 1.118

2. Find the total length of the two chords.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\sqrt{1 + \frac{1}{4}}$ $+ \sqrt{1 + \frac{9}{4}}$ $= \frac{\sqrt{5}}{2}$ $+ \frac{\sqrt{13}}{2}$ $\approx 2.921$ .

Answer: $\frac{\sqrt{5} + \sqrt{13}}{2}$ ≈ 2.921

3. Which integral gives the exact length of the graph of $f$ from $x = 0$ to $x = 2$ ?

$\int_0^2 \sqrt{1 + \frac{x^4}{4}}\,dx$
$\int_0^2 \left(1 + x^2\right)dx$
$\int_0^2 \sqrt{1 + x}\,dx$
$\int_0^2 \sqrt{1 + x^2}\,dx$

Show the answer

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

$\int_0^2 \sqrt{1 + \frac{x^4}{4}}\,dx$ : That squares $f(x)$ itself. The formula uses the derivative: $f'(x) = x$ .
$\int_0^2 \left(1 + x^2\right)dx$ : The square root is missing: the integrand comes from a hypotenuse, $\sqrt{1 + (f')^2}$ .
$\int_0^2 \sqrt{1 + x}\,dx$ : The derivative must be squared: $\sqrt{1 + (f'(x))^2}$ $= \sqrt{1 + x^2}$ .
Correct: $\int_0^2 \sqrt{1 + x^2}\,dx$ : Right: $f'(x) = x$ , so the integrand is $\sqrt{1 + (f'(x))^2}$ $= \sqrt{1 + x^2}$ .

4. How does the two-chord total compare with the length of the graph?

It is less: a chord is the shortest path between its endpoints
It is greater, because $f$ is concave up
It is equal, because the chords meet the curve at its endpoints
No comparison is possible without evaluating the integral

Show the answer

Each chord is the straight path between two points of the curve, so the chord total, about 2.921, is less than the arc length, about 2.958. More chords raise the total toward it.

Correct: It is less: a chord is the shortest path between its endpoints: Right: each piece of the curve is at least as long as its chord, and here strictly longer, since the curve bends.
It is greater, because $f$ is concave up: Concavity decides whether chords lie above or below the curve, not whether they are shorter. A straight segment is never longer than a curved path.
It is equal, because the chords meet the curve at its endpoints: Sharing endpoints does not make the lengths equal: the curve bends between them.
No comparison is possible without evaluating the integral: The comparison follows from geometry alone: the straight path is the shortest.

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

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$y' = x^{1/2}$ , so the length is $\int_0^3 \sqrt{1 + x}\,dx$ $= \frac{2}{3}(1 + x)^{3/2}\Big|_0^3$ $= \frac{2}{3}(8 - 1)$ $= \frac{14}{3}$ .

Answer: $\frac{14}{3}$ ≈ 4.667

6. A cable hangs between two poles 5 meters apart; $h(x)$ is its height, in meters, above the point $x$ meters from the first pole. What does $\int_0^5 \sqrt{1 + (h'(x))^2}\,dx$ $= 6.2$ mean?

The cable's height changes by 6.2 meters between the poles
The area between the cable and the ground is 6.2 square meters
The average height of the cable is 6.2 meters
The cable is 6.2 meters long between the two poles

Show the answer

$\int_a^b \sqrt{1 + (h'(x))^2}\,dx$ is the length of the graph of $h$ on $[a, b]$ , here the cable from one pole to the other, in meters.

The cable's height changes by 6.2 meters between the poles: A change in height is $\int_0^5 h'(x)\,dx$ $= h(5) - h(0)$ , without the root.
The area between the cable and the ground is 6.2 square meters: An area would be $\int_0^5 h(x)\,dx$ , in square meters.
The average height of the cable is 6.2 meters: An average height would divide $\int_0^5 h(x)\,dx$ by 5.
Correct: The cable is 6.2 meters long between the two poles: Right: the arc length integral of the graph of $h$ from $x = 0$ to $x = 5$ is the cable's length, in meters.

Part 9 · Summary

Summary

The length of the graph of

y = f(x)

from

x = a

to

x = b

is

\int_a^b \sqrt{1 + (f'(x))^2}\,dx

: the limit of sums of chord lengths, each a hypotenuse. Square the derivative, keep the root, use the right limits, and expect to need a calculator for most values.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Chords along y=x22y = \frac{x^2}{2}

Summary

What comes next

Connections

Chords along $y = \frac{x^2}{2}$