Multivariable-Calculus · Unit 5 · Video 3 · Interactive Practice

The Cycloid: Choosing a Parameter That Is Not Time

IKey Formulas

Formula	Name	What you need
$\big(x(t),\ y(t),\ z(t)\big)$	Any trajectory of a moving point	Each coordinate as a function of one parameter
$Q(t) = Q_0 + t\,\overrightarrow{Q_0Q_1}$	The line, with time as the parameter	Two points $Q_0$ and $Q_1$
$\big(x(2t),\ y(2t)\big)$	The same curve at double speed	The original motion: same points, same order, half the time
$P = \big(x(\theta),\ y(\theta)\big)$	The cycloid, with the angle turned as the parameter	Wheel radius $a$ , $P$ starting at the origin; $0 \le \theta \le 2\pi$ is one arch

Key Insight: How fast the wheel rolls is not part of the curve, so time is the wrong parameter. The angle turned and the distance rolled fix each other, and the angle wins because the turning is the complicated part of the motion.

IITwo Motions at Once

Rolling carries $P$ forward and swings it around the centre at the same time; neither motion alone makes arches.

Motion θ turned 120°

IIISame Curve, Different Schedule

Changing the rolling speed changes when $P$ reaches each point of the arch, never which points it reaches.

Wheel B Clock t 2 s

IVThe Angle Places the Point

Knowing only the angle $\theta$ the wheel has turned, where on its rim is $P$ ?

VQuiz Questions

Problem 1 · Half a Turn

Given: a wheel of radius $a$ rolls to the right along the $x$ -axis, and the painted point $P$ starts at the origin — where is $P$ once the wheel has turned through half a turn, $\theta = \pi$ ?

✅ Correct! Half a turn swings the painted radius from straight down to straight up while the wheel rolls forward, so

P

is at the top of the wheel: the top of the first arch.

❌ Not yet. Coming back down to the road, where the first arch ends, takes a full turn,

\theta = 2\pi

. Half a turn is only halfway through the arch.

❌ That is a quarter turn. Level with the centre and behind the contact point is where

P

is at

\theta = \tfrac{\pi}{2}

. Half a turn swings the painted radius all the way from straight down to straight up.

❌ Close, but the wheel has also moved. Above the origin is where

P

would be if the wheel spun in place. Rolling carries the whole wheel forward, so the top of the wheel sits above the new contact point.

Show solution

The angle $\theta$ is measured at the centre, from the downward vertical to the painted radius. At $\theta = 0$ the painted radius points straight down, at the origin.

Rolling to the right turns the wheel clockwise: at $\theta = \tfrac{\pi}{2}$ the painted radius points backwards, and at $\theta = \pi$ it points straight up.

Both motions happen at once, so meanwhile the wheel has rolled forward, and its contact point is always directly below its centre. Hence $P$ is directly above the contact point, at the very top of the wheel: the highest point of the first arch. It is not above the origin; that would be a wheel spinning in place, with no forward motion.

Problem 2 · Double Speed

Given: two identical wheels, A and B, start together with their painted points at the origin, and B rolls twice as fast as A. The point on A reaches the top of its first arch at $t = 6$ s.

How does B's curve compare with A's?

When did B's point pass that same top?

✅ Correct! Speed changes the schedule, not the curve: B visits every point of A's cycloid in the same order, each at half of A's clock reading.

❌ Not quite. A faster wheel covers more road per second, not more road per turn. One full turn still makes one arch, and the wheel is the same size.

❌ Not quite. At the top of an arch the point is at the top of the wheel, and B's wheel is the same size as A's, so its arches are exactly as high.

❌ Not quite. B's point has less time, but B's wheel also turns faster: at any moment B has reached the place A reaches at twice that time, so B passes through exactly the same points.

❌ That is the wrong direction.

12

s would suit a slower wheel. B is faster, so it reaches the same point sooner:

2t = 6

gives

t = 3

❌ Same curve, different schedule. B passes the same points as A, but it gets to each one in half the time A takes.

❌ Not quite. The two traces lie exactly on top of each other, so B passes every point A passes, the top of the first arch included.

Show solution

Double speed means that at any moment B has reached the place A reaches at twice that time:

\big(x_B(t),\ y_B(t)\big) = \big(x_A(2t),\ y_A(2t)\big)

The curve. As $t$ runs through all times, so does $2t$ : B visits every point A visits, in the same order. The two traces coincide exactly; only the clock readings attached to the points differ.

The time. A is at the top of its first arch at $t = 6$ s. B is there when $2t = 6$ , that is at $t = 3$ s. At every shared point the slower wheel's clock reads twice the faster one's.

Problem 3 · Two and a Half Turns

Given: a bicycle wheel of radius $35$ cm rolls along a flat road, with a reflector on its rim that starts touching the road. The wheel turns through $\theta = 5\pi$ — where is the reflector, and how high above the road?

Where is the reflector?

How high above the road is it?

✅ Correct!

5\pi

is two full turns and a half: two complete arches, then half of the third, which leaves the reflector at the very top of the wheel,

2 \times 35 = 70

cm up.

❌ Check the size of a turn. One full turn is

2\pi

, not

\pi

, so

5\pi

is two and a half turns, not five.

❌ You stopped too early. The reflector is back on the road where arch 2 ends at

\theta = 4\pi

. The extra

\pi

is half a turn more, which lifts it to the top of the next arch.

❌ Count the arches again. The two full turns complete arches 1 and 2; the half turn after them is spent on arch 3.

❌ That is the height of the centre. The arch rises higher than that: at its top the reflector is at the very top of the wheel, one more radius up.

❌ Not at this angle. The reflector touches the road only after whole turns,

\theta = 2\pi, 4\pi, \dots

;

5\pi

is half a turn past

4\pi

❌ Too high.

140

cm is twice the wheel's full height. The top of the wheel is one diameter above the road:

2 \times 35 = 70

cm.

Show solution

Step 1 — count whole turns. One full turn is $\theta = 2\pi$ and traces one arch. Split the angle:

5\pi = 2 \cdot 2\pi + \pi

That is two full turns, then half a turn more.

Step 2 — after the whole turns. At $\theta = 4\pi$ the reflector is back on the road where arch 2 ends, in the same arrangement as at the start, only further along the road.

Step 3 — the extra half turn. Half a turn swings the painted radius from straight down to straight up, so the reflector is at the very top of the wheel: the top of arch 3.

Step 4 — the height. The centre is one radius above the road, and the top of the wheel is one more radius above the centre:

35 + 35 = 70 \text{ cm}

Problem 4 · Same Line, New Schedule

Given: the line through $Q_0 = (-1, 2, 2)$ and $Q_1 = (1, 3, -1)$ , traced as $x = -1 + 2t$ , $y = 2 + t$ , $z = 2 - 3t$ . Run the motion at double speed: at time $s$ the new point is where the old one was at time $2s$ .

Which equations describe the double-speed motion?

At what $s$ does the new point reach $Q_1$ ?

✅ Correct! Substituting

t = 2s

doubles every coefficient of the parameter and keeps the constants: the same line and the same start

Q_0

, at twice the velocity, so

Q_1

arrives at

s = \tfrac{1}{2}

❌ The constants must stay. Doubling them moves the start to

(-2, 4, 4)

, which is not on the line: only the parameter changes,

t = 2s

, so only the coefficients of

s

double.

❌ That is half speed. These equations come from

t = \tfrac{s}{2}

. Double speed means the new point at time

s

is where the old one was at time

2s

, so substitute

t = 2s

❌ Same line, old schedule. These are the original equations with the letter renamed; the timing must change, so that at

s = 1

the point is where the old one was at

t = 2

❌ The faster motion arrives sooner. At

s = 2

the new point is where the old one was at

t = 4

, far beyond

Q_1

. It reaches

Q_1

, the old

t = 1

, when

2s = 1

❌ That is the old schedule. The old point reaches

Q_1

t = 1

; the double-speed point gets there in half the time.

❌ Not quite. Double speed traces the same line through

Q_0

and

Q_1

, only on a faster schedule, so it certainly reaches

Q_1

Show solution

Step 1 — substitute $t = 2s$ .

x = -1 + 2(2s) = -1 + 4s, \qquad y = 2 + 2s, \qquad z = 2 - 3(2s) = 2 - 6s

Step 2 — check the endpoints. $s = 0$ gives $(-1, 2, 2) = Q_0$ , and $s = \tfrac{1}{2}$ gives $(-1 + 2,\ 2 + 1,\ 2 - 3) = (1, 3, -1) = Q_1$ . The coefficients of $s$ form $\langle 4, 2, -6\rangle = 2\langle 2, 1, -3\rangle$ : a vector along the same line, now twice as long, so the point moves twice as fast.

Step 3 — why not double everything? The equations $x = -2 + 4s$ , $y = 4 + 2s$ , $z = 4 - 6s$ start at $(-2, 4, 4)$ . On the original line, $-1 + 2t = -2$ forces $t = -\tfrac{1}{2}$ , and then $y = 2 - \tfrac{1}{2} = \tfrac{3}{2} \neq 4$ . That start is off the line, so doubling everything gives a parallel line, not the same one.

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