Multivariable-Calculus · Unit 5 · Video 4 · Interactive Practice

One Vector Split Into Three: The Cycloid's Equations

IKey Formulas

Formula	Piece	Read off from
$\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AB} + \overrightarrow{BP}$	Head-to-tail split	$A$ = contact point, $B$ = centre; any chain from $O$ to $P$ adds up to $\overrightarrow{OP}$
$\overrightarrow{OA} = \langle a\theta,\ 0\rangle, \quad \overrightarrow{AB} = \langle 0,\ a\rangle$	Along the road, up to the centre	Rolling without slipping: road $=$ arc $= a\theta$ , with $\theta$ in radians
$\overrightarrow{BP} = \langle -a\sin\theta,\ -a\cos\theta\rangle$	Centre out to the rim	Right triangle with hypotenuse $a$ and $\theta$ measured from the downward vertical
$x(\theta) = a\theta - a\sin\theta, \quad y(\theta) = a - a\cos\theta$	The cycloid	The three added component by component

Key Insight: The sine lands in the horizontal slot because $\theta$ is measured from the downward vertical: the vertical leg is adjacent to $\theta$ , the horizontal leg opposite it.

IIVisualization 1 — Rolled Distance Equals Arc Length

Rolling without slipping matches road to rim: the arc from $P$ to $A$ has exactly the length $|\overrightarrow{OA}|$ .

θ (rad) 2.50 Rolled back 0%

IIIVisualization 2 — Why the Sine Goes Across

With $\theta$ measured from the downward vertical, the vertical leg is adjacent to $\theta$ and the horizontal leg opposite.

💡 The vertical is not a free choice: $P$ began at the bottom of the wheel, so the turning angle — the same $\theta$ that gives $|\overrightarrow{OA}| = a\theta$ — is measured from the downward vertical.

IVVisualization 3 — Adding the Three Vectors

Laid head to tail, $\overrightarrow{OA}$ , $\overrightarrow{AB}$ and $\overrightarrow{BP}$ end at $P$ for every $\theta$ , and $P$ traces the cycloid.

θ (rad) 1.20

💡 Where the arch comes down to the road at $\theta = 2\pi$ , the picture cannot tell a corner from a smooth landing — the formulas $x(\theta)$ and $y(\theta)$ can.

VQuiz Questions

Problem 1 · A Quarter Turn

Given: a wheel of radius $a = 3$ rolls along the $x$ -axis with the rim point $P$ starting at the origin, and has made a quarter turn, $\theta = \tfrac{\pi}{2}$ — find the position of $P$ .

✅ Correct!

\overrightarrow{OA} = \langle \tfrac{3\pi}{2}, 0\rangle

\overrightarrow{AB} = \langle 0, 3\rangle

\overrightarrow{BP} = \langle -3, 0\rangle

: after a quarter turn

P

is level with the centre, one radius behind it.

❌ Sign slip on $\overrightarrow{BP}$ . The wheel turns clockwise as it rolls right, so

P

swings up the back of the wheel: at

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

it is behind the centre, and the horizontal component is

-a\sin\theta = -3

, not

+3

❌ Sine and cosine swapped. That would leave

P

on the road after a quarter turn. With

\theta

from the downward vertical the horizontal slot takes the sine:

x = a\theta - a\sin\theta

y = a - a\cos\theta

❌ Degrees.

3 \times 90

is no length on this wheel: arc

= a\theta

holds only with

\theta

in radians, so a quarter turn rolls the wheel

3 \cdot \tfrac{\pi}{2} = \tfrac{3\pi}{2}

Show solution

Write down the three vectors with $a = 3$ and $\theta = \tfrac{\pi}{2}$ (so $\sin\theta = 1$ , $\cos\theta = 0$ ):

\overrightarrow{OA} = \langle a\theta,\ 0\rangle = \left\langle \tfrac{3\pi}{2},\ 0\right\rangle, \qquad \overrightarrow{AB} = \langle 0,\ 3\rangle, \qquad \overrightarrow{BP} = \langle -3\sin\tfrac{\pi}{2},\ -3\cos\tfrac{\pi}{2}\rangle = \langle -3,\ 0\rangle

Add component by component:

\overrightarrow{OP} = \left\langle \tfrac{3\pi}{2} + 0 - 3,\ \ 0 + 3 + 0\right\rangle = \left\langle \tfrac{3\pi}{2} - 3,\ 3\right\rangle \approx \langle 1.71,\ 3\rangle

So $P = \left(\tfrac{3\pi}{2} - 3,\ 3\right)$ : at the height of the centre, $3$ units behind it.

Problem 2 · Where Does the Sine Go?

Given: a wheel of radius $a = 2$ that has turned through $\theta = \tfrac{\pi}{6}$ , measured from the downward vertical as in the cycloid — find $\overrightarrow{BP}$ , from the centre to the rim point.

✅ Correct! The short horizontal leg is opposite the small angle

\theta

, so it is

a\sin\theta = 1

; the long vertical leg is adjacent,

a\cos\theta = \sqrt{3}

; both point left and down.

❌ The cosine went across. That is right only for an angle measured from the horizontal. Here

\theta

is measured from the downward vertical, so the vertical leg is adjacent (

a\cos\theta = \sqrt{3}

) and the horizontal leg opposite (

a\sin\theta = 1

❌ $P$ is behind the centre, not ahead. Rolling right turns the wheel clockwise, carrying

P

from the bottom up the left side, so the horizontal component is

-a\sin\theta = -1

❌ For $\theta = \tfrac{\pi}{6}$ , $P$ is still below the centre. Going from

B

P

goes down, so the vertical component is

-a\cos\theta = -\sqrt{3}

Show solution

Right triangle at $B$ : hypotenuse $|\overrightarrow{BP}| = a = 2$ , angle $\theta = \tfrac{\pi}{6}$ between the hypotenuse and the vertical leg.

\text{vertical leg} = a\cos\tfrac{\pi}{6} = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}, \qquad \text{horizontal leg} = a\sin\tfrac{\pi}{6} = 2 \cdot \tfrac{1}{2} = 1

From $B$ to $P$ goes left and down, so both components are negative:

\overrightarrow{BP} = \langle -a\sin\theta,\ -a\cos\theta\rangle = \langle -1,\ -\sqrt{3}\rangle

Check: $|\overrightarrow{BP}| = \sqrt{1 + 3} = 2 = a$ ✓

Problem 3 · A Reflector After One Metre

Given: a bicycle wheel of radius $a = 0.5$ m carries a reflector on its rim, touching the road at the start; the bike rolls $1$ m forward without slipping — find the angle the wheel has turned and the reflector's height.

Angle turned

Height of the reflector

✅ Correct!

\theta = 1/0.5 = 2

rad, past a quarter turn, so

\cos 2 \lt 0

and the reflector sits above the centre:

y = 0.5 - 0.5\cos 2 \approx 0.71

❌ Inverted. Road

=

arc gives

1 = a\theta

, so

\theta = 1 / a = 1/0.5

, not

a / 1

❌ The distance rolled is $a\theta$ , not $\theta$ . Divide the

1

m by the radius

0.5

❌ That counts revolutions.

1

m over a circumference of

2\pi a = \pi

m is

\tfrac{1}{\pi}

of a turn; in radians that is

2\pi \cdot \tfrac{1}{\pi} = 2

❌ Check the angle. Rolling without slipping: distance

= a\theta

with

\theta

in radians.

❌ Sine in the vertical slot. That is

0.5 - 0.5\sin 2

. The vertical leg is adjacent to

\theta

, so

y = a - a\cos\theta

❌ That is $a + a\cos\theta$ . The vertical component of

\overrightarrow{BP}

-a\cos\theta

, which is positive here because

\cos 2 \lt 0

: the reflector is above the centre, not below it.

❌ That uses $\theta = 1$ . The angle is distance over radius,

1/0.5 = 2

rad.

❌ Check the height.

y = a - a\cos\theta

with

\theta

in radians.

Show solution

Angle. Rolling without slipping makes the road covered equal the arc of rim: $a\theta = 1$ , so

\theta = \frac{1}{0.5} = 2 \text{ rad} \quad (\approx 114.6\degree,\ \text{past a quarter turn})

Height. The reflector starts at the origin, so it follows the cycloid:

y = a - a\cos\theta = 0.5 - 0.5\cos 2 \approx 0.5 - 0.5(-0.4161) = 0.5 + 0.20805 \approx 0.71 \text{ m}

Since $\cos 2 \lt 0$ , the term $-a\cos\theta$ is positive: the reflector is about $0.21$ m above the centre of the wheel.

Problem 4 · A Reflector on a Spoke

Given: a wheel of radius $a$ rolls along the $x$ -axis carrying a reflector $Q$ on a spoke, at distance $b$ from the centre with $0 \lt b \lt a$ ; $Q$ starts directly below the centre, at $(0,\ a - b)$ — find the position of $Q$ after the wheel turns through $\theta$ .

✅ Correct! The same split works:

\overrightarrow{OA}

and

\overrightarrow{AB}

belong to the wheel, and only the last vector shrinks to length

b

❌ That is the rim point's cycloid.

\overrightarrow{BQ}

has length

b

, not

a

\overrightarrow{BQ} = \langle -b\sin\theta,\ -b\cos\theta\rangle

❌ The road is covered by the rim, not by the spoke. The wheel of radius

a

rolls without slipping, so

|\overrightarrow{OA}| = a\theta

, and the centre is still

a

above the road.

❌ Check $\overrightarrow{AB}$ . It runs from the road up to the centre, so it is

\langle 0,\ a\rangle

however far out along the spoke

Q

sits.

Show solution

Split $\overrightarrow{OQ}$ through the contact point $A$ and the centre $B$ :

\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AB} + \overrightarrow{BQ}

The first two depend only on the wheel: $\overrightarrow{OA} = \langle a\theta, 0\rangle$ (rolling without slipping) and $\overrightarrow{AB} = \langle 0, a\rangle$ . The last has length $b$ and makes angle $\theta$ with the downward vertical, because $Q$ started directly below $B$ :

\overrightarrow{BQ} = \langle -b\sin\theta,\ -b\cos\theta\rangle

\overrightarrow{OQ} = \langle a\theta - b\sin\theta,\ a - b\cos\theta\rangle

Check at $\theta = 0$ : $(0,\ a - b)$ ✓. Since $b \lt a$ , the height $a - b\cos\theta \geq a - b$ is always positive: this curve never touches the road.

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