The Path of a Pebble on a Rolling Wheel

A pebble stuck in a tire traces a cycloid: it stops dead each time it touches the road, then swings forward at twice the car's speed. Integrating with the parametric equations shows each arch encloses exactly three times the wheel's area and is exactly eight radii long.

How it works

Area and arc length with parametric equations

In one sentence

For $x = f(t)$ , $y = g(t)$ , $\alpha \le t \le \beta$ , traced once left to right, the area under the curve is $\int_\alpha^\beta g(t)f'(t)\,dt$ and its length is $\int_\alpha^\beta\sqrt{f'(t)^2 + g'(t)^2}\,dt$ .

General formula

{\htmlData{sym=0}{x}} = {\htmlData{sym=2}{r}}({\htmlData{sym=3}{t}} - \sin {\htmlData{sym=3}{t}}),\; {\htmlData{sym=1}{y}} = {\htmlData{sym=2}{r}}(1 - \cos {\htmlData{sym=3}{t}}), \qquad {\htmlData{sym=4}{A}} = \int_0^{2\pi} {\htmlData{sym=1}{y}}\,\frac{d{\htmlData{sym=0}{x}}}{d{\htmlData{sym=3}{t}}}\,d{\htmlData{sym=3}{t}} = 3\pi {\htmlData{sym=2}{r}}^2, \qquad {\htmlData{sym=5}{L}} = \int_0^{2\pi}\sqrt{\left(\frac{d{\htmlData{sym=0}{x}}}{d{\htmlData{sym=3}{t}}}\right)^2 + \left(\frac{d{\htmlData{sym=1}{y}}}{d{\htmlData{sym=3}{t}}}\right)^2}\,d{\htmlData{sym=3}{t}} = 8{\htmlData{sym=2}{r}}

Swap $dx$ for $\frac{dx}{dt}dt$ and you can integrate along the path by time instead of by $x$ . Speed times time is distance, so integrating speed gives length.

What the symbols mean

$x$
Horizontal position
Where the pebble is along the road.
$y$
Height
How high the pebble is.
$r$
Wheel radius
Size of the rolling wheel.
$t$
Angle rolled
How far the wheel has turned, in radians.
$A$
Area under one arch
Exactly three times the wheel's area.
$L$
Arch length
Exactly eight radii.

As the wheel rolls through angle $t$ , its center moves $rt$ forward and the pebble sits at angle $t$ around the center: $x = r(t - \sin t)$ , $y = r(1 - \cos t)$ .
Area under one arch: trade $dx$ for $\frac{dx}{dt}dt$ and integrate $y\,\frac{dx}{dt} = r^2(1 - \cos t)^2$ from $0$ to $2\pi$ : $3\pi r^2$ .
Arc length: the pebble's speed is $\sqrt{x'^2 + y'^2} = 2r\left|\sin\frac{t}{2}\right|$ ; integrating gives $8r$ .
The pebble touches the road with zero speed (a cusp) and moves at twice the car's speed at the top.

Learn the math

Area with Parametric EquationsCalc II · Lesson 18, Parametric Equations and Polar Coordinates Arc Length and Surface Area with Parametric EquationsCalc II · Lesson 19, Parametric Equations and Polar Coordinates