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

The Cycloid's Bottom Point Is a Cusp: Taylor Terms Settle It

IKey Formulas

Formula	Name	What you need
$x(\theta) = \theta - \sin\theta,\quad y(\theta) = 1 - \cos\theta$	Cycloid with lengths measured in radii	$x = a\theta - a\sin\theta$ , $y = a - a\cos\theta$ divided by $a$
$f(t) \approx f(0) + t\,f'(0) + \tfrac{1}{2}t^2 f''(0) + \tfrac{1}{6}t^3 f'''(0)$	Taylor approximation for small $t$	The derivatives of $f$ at $0$
$\sin\theta \approx \theta - \tfrac{\theta^3}{6},\quad \cos\theta \approx 1 - \tfrac{\theta^2}{2}$	The first corrections	$\sin$ : $0, 1, 0, -1$ ; $\cos$ : $1, 0, -1, 0$
$x \approx \tfrac{\theta^3}{6},\ \ y \approx \tfrac{\theta^2}{2},\ \ \dfrac{y}{x} \approx \dfrac{3}{\theta} \to \infty$	Vertical tangent: a cusp	The terms that survive the cancellation

Key Insight: $\sin\theta \approx \theta$ and $\cos\theta \approx 1$ give $x \approx 0$ , $y \approx 0$ — true of every candidate picture: a corner, a horizontal tangent, a loop, a vertical tangent. The leading terms cancel exactly, so the shape is decided by the first terms that survive: $\theta^3$ in $x$ and $\theta^2$ in $y$ .

IITaylor Terms, One Order at a Time

The linear terms cancel exactly, so the first terms that survive decide the shape of the bottom.

Terms through θ¹ θ 0.6

Through $\theta^1$

\sin\theta \approx \theta, \qquad \cos\theta \approx 1

x \approx \theta - \theta = 0

y \approx 1 - 1 = 0

Every $\theta$ lands on the origin, which all four candidate pictures pass through: this level decides nothing.

Through $\theta^2$

\sin\theta \approx \theta \quad \big(\sin''(0) = 0\big)

\cos\theta \approx 1 - \tfrac{\theta^2}{2}

x \approx \theta - \theta = 0

y \approx 1 - \big(1 - \tfrac{\theta^2}{2}\big) = \tfrac{\theta^2}{2}

The vertical motion appears first; there is no forward motion at this order.

Through $\theta^3$

\sin\theta \approx \theta - \tfrac{\theta^3}{6}

\cos\theta \approx 1 - \tfrac{\theta^2}{2} \quad \big(\cos'''(0) = 0\big)

x \approx \theta - \big(\theta - \tfrac{\theta^3}{6}\big) = \tfrac{\theta^3}{6}

y \approx 1 - \big(1 - \tfrac{\theta^2}{2}\big) = \tfrac{\theta^2}{2}

The forward creep $\tfrac{\theta^3}{6}$ is far smaller than the rise $\tfrac{\theta^2}{2}$ : the two branches meet at the origin in a cusp.

IIIThe Chord Stands Up

The chord from the bottom to the point at $\theta$ has slope $y/x \approx 3/\theta$ , which grows without bound as $\theta \to 0$ .

θ → 0 θ = 1

IVOnly the Rim Makes a Cusp

A point at distance $b$ from the hub traces $x = \theta - b\sin\theta$ , $y = 1 - b\cos\theta$ ; only $b = 1$ cancels the linear terms.

b 1

VQuiz Questions

Problem 1 · The Terms That Survive

Given: the cycloid $x(\theta) = \theta - \sin\theta$ , $y(\theta) = 1 - \cos\theta$ — find the leading term of each coordinate near $\theta = 0$ , using Taylor approximations of $\sin\theta$ and $\cos\theta$ through the $\theta^3$ term.

First: the horizontal coordinate

Then: the vertical coordinate

✅ Correct! The

\theta

's cancel in

x

and the

1

's cancel in

y

; the first corrections survive:

x \approx \tfrac{\theta^3}{6}

y \approx \tfrac{\theta^2}{2}

❌ That is the first-order answer.

\sin\theta \approx \theta

gives

x \approx \theta - \theta = 0

, which every candidate picture satisfies; keep the

-\tfrac{\theta^3}{6}

\sin\theta

❌ Check the sign.

\theta - \big(\theta - \tfrac{\theta^3}{6}\big) = \theta - \theta + \tfrac{\theta^3}{6}

: subtracting the negative correction leaves it positive.

❌ That is the expansion of $\sin\theta$ itself. The coordinate is

\theta

minus it:

\theta - \big(\theta - \tfrac{\theta^3}{6}\big)

❌ The second-order Taylor term carries a factor $\tfrac{1}{2}$ .

\tfrac{1}{2}\theta^2\cos''(0) = -\tfrac{\theta^2}{2}

, so

\cos\theta \approx 1 - \tfrac{\theta^2}{2}

, not

1 - \theta^2

❌ That is $\cos\theta$ itself. The coordinate is

1

minus it:

1 - \big(1 - \tfrac{\theta^2}{2}\big)

❌ That is the first-order answer $1 - 1 = 0$ . The

\theta^2

term of

\cos\theta

survives the cancellation.

Show solution

Step 1 — derivatives at $0$ . For $\sin$ : $0, 1, 0, -1$ . For $\cos$ : $1, 0, -1, 0$ . Feeding them into $f(0) + t f'(0) + \tfrac{1}{2}t^2 f''(0) + \tfrac{1}{6}t^3 f'''(0)$ :

\sin\theta \approx \theta - \frac{\theta^3}{6}, \qquad \cos\theta \approx 1 - \frac{\theta^2}{2}

Step 2 — substitute and cancel.

x \approx \theta - \Big(\theta - \frac{\theta^3}{6}\Big) = \frac{\theta^3}{6}, \qquad y \approx 1 - \Big(1 - \frac{\theta^2}{2}\Big) = \frac{\theta^2}{2}

Each coordinate is a difference of two nearly equal quantities. The leading terms cancel exactly, and what survives — the first correction of each expansion — is what decides the shape.

Problem 2 · Rise Over Run

Given: a classmate divides the leading terms as $\dfrac{x}{y} \approx \dfrac{\theta^3/6}{\theta^2/2} = \dfrac{\theta}{3} \to 0$ and concludes that the cycloid meets the ground with a horizontal tangent — which statement is right?

✅ Correct!

x/y \to 0

says the run becomes negligible next to the rise — the same vertical tangent, read the other way up.

❌ A slope is rise over run, $y/x$ . The classmate's

x/y

is run over rise; it tending to

0

means the run vanishes relative to the rise, so the tangent is vertical.

❌ The algebra is fine.

\dfrac{\theta^3}{6} \cdot \dfrac{2}{\theta^2} = \dfrac{\theta}{3}

. The mistake is reading

x/y

as the slope.

❌ Both coordinates do tend to $0$ , but their leading terms settle the ratio.

y/x \approx 3/\theta

— deciding such a

0/0

is exactly what the extra Taylor terms are for.

Show solution

Divide the leading terms both ways:

\frac{x}{y} \approx \frac{\theta^3/6}{\theta^2/2} = \frac{\theta}{3}, \qquad \frac{y}{x} \approx \frac{\theta^2/2}{\theta^3/6} = \frac{3}{\theta}

The slope of the chord from the origin to the point is rise over run, $y/x$ , and $3/\theta \to \infty$ as $\theta \to 0$ : the curve reaches the origin along a vertical tangent, so the bottom is a cusp.

The classmate's $x/y \to 0$ says the same thing — the run is negligible next to the rise — but it is the reciprocal of the slope, so reading it as a slope turns vertical into horizontal.

Problem 3 · Numbers at One Tenth

Given: $\theta = \tfrac{1}{10}$ , so $\theta^2 = \tfrac{1}{100}$ and $\theta^3 = \tfrac{1}{1000}$ — estimate the coordinate $x$ and the chord slope $y/x$ from the leading terms.

First: the horizontal coordinate

Then: the chord slope

✅ Correct!

x \approx \tfrac{1}{6000}

and

y \approx \tfrac{1}{200}

: the point has risen thirty times as far as it has moved forward.

❌ $\tfrac{1}{1000}$ is the power $\theta^3$ , not the coordinate.

x \approx \tfrac{\theta^3}{6} = \tfrac{1}{6} \cdot \tfrac{1}{1000}

❌ The coordinate $x$ goes with $\theta^3$ , not $\theta^2$ .

\tfrac{\theta^2}{6} = \tfrac{1}{600}

mixes the two expansions.

❌ $\tfrac{1}{200} = \tfrac{\theta^2}{2}$ is the height $y$ . The horizontal coordinate is

\tfrac{\theta^3}{6}

❌ $\tfrac{\theta^2}{\theta^3} = \tfrac{1}{\theta} = 10$ is only the ratio of the powers. The constants contribute

\tfrac{1/2}{1/6} = 3

as well.

❌ That is $x/y$ , run over rise. The slope is rise over run:

\tfrac{1/200}{1/6000}

❌ $\tfrac{1/2}{1/6} = 3$ is only the ratio of the constants. The powers contribute

\tfrac{\theta^2}{\theta^3} = \tfrac{1}{\theta} = 10

as well.

Show solution

Step 1 — the coordinates.

x \approx \frac{\theta^3}{6} = \frac{1}{6} \cdot \frac{1}{1000} = \frac{1}{6000}, \qquad y \approx \frac{\theta^2}{2} = \frac{1}{2} \cdot \frac{1}{100} = \frac{1}{200}

Step 2 — the slope.

\frac{y}{x} \approx \frac{1/200}{1/6000} = \frac{6000}{200} = 30 = \frac{3}{\theta}

Check against the exact coordinates: $x = 0.1 - \sin 0.1 \approx 0.00016658$ and $y = 1 - \cos 0.1 \approx 0.0049958$ , whose ratio is $\approx 29.99$ . Note that $\theta^2$ and $\theta^3$ are not the coordinates: each carries its constant, $\tfrac{1}{2}$ or $\tfrac{1}{6}$ .

Problem 4 · A Reflector on a Spoke

Given: a reflector halfway between the hub and the rim of a rolling wheel of radius $1$ traces $x = \theta - \tfrac{1}{2}\sin\theta$ , $y = 1 - \tfrac{1}{2}\cos\theta$ , with its lowest point $\big(0, \tfrac{1}{2}\big)$ at $\theta = 0$ — find the leading term of $x$ and the kind of point the lowest point is.

First: the leading term of the horizontal coordinate

Then: the lowest point is

✅ Correct! Off the rim the linear terms no longer cancel:

x \approx \tfrac{\theta}{2}

while

y - \tfrac{1}{2} \approx \tfrac{\theta^2}{4}

, so the chord slope is

\approx \tfrac{\theta}{2} \to 0

❌ Only half of the $\theta$ cancels.

\theta - \tfrac{1}{2}\theta = \tfrac{\theta}{2}

survives, and near

0

it swamps the

\tfrac{\theta^3}{12}

term.

❌ Check the sign.

\theta - \tfrac{1}{2}\sin\theta \approx \theta - \tfrac{1}{2}\theta = \tfrac{\theta}{2}

❌ That is the rim point's answer. Here

\sin\theta

carries a factor

\tfrac{1}{2}

, so the linear terms do not cancel:

\theta - \tfrac{1}{2}\theta = \tfrac{\theta}{2}

❌ A corner needs different slopes on the two sides. Here

\big(y - \tfrac{1}{2}\big)/x \approx \tfrac{\theta}{2}

tends to

0

from both sides: one horizontal tangent.

❌ A vertical tangent needs the rise to dwarf the run. Here it is the other way round: the run

\tfrac{\theta}{2}

dwarfs the rise

\tfrac{\theta^2}{4}

❌ A loop needs the point to move backward at some stage. Here

x'(\theta) = 1 - \tfrac{1}{2}\cos\theta \geq \tfrac{1}{2}

, so

x

always increases.

Show solution

Step 1 — expand. With $\sin\theta \approx \theta - \tfrac{\theta^3}{6}$ and $\cos\theta \approx 1 - \tfrac{\theta^2}{2}$ :

x \approx \theta - \frac{1}{2}\Big(\theta - \frac{\theta^3}{6}\Big) = \frac{\theta}{2} + \frac{\theta^3}{12}, \qquad y \approx 1 - \frac{1}{2}\Big(1 - \frac{\theta^2}{2}\Big) = \frac{1}{2} + \frac{\theta^2}{4}

Step 2 — measure from the lowest point. The run is $x \approx \tfrac{\theta}{2}$ — the linear term survives, and swamps $\tfrac{\theta^3}{12}$ — and the rise is $y - \tfrac{1}{2} \approx \tfrac{\theta^2}{4}$ :

\frac{y - \tfrac{1}{2}}{x} \approx \frac{\theta^2/4}{\theta/2} = \frac{\theta}{2} \to 0

Step 3 — classify. The chord slope tends to $0$ from both sides, so there is a single horizontal tangent, and $x$ increases steadily, so the path never doubles back into a loop: the lowest point is smooth. The cusp belongs to the rim alone, where $\theta - \sin\theta$ loses its linear term completely.

Solved: 0 / 4