---
title: "Plane Curves"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 24
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/24-plane-curves
---

# Chapter 24 — Plane Curves

The calculus of Chapters [14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) and [16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) was built for functions $y = f(x)$; most curves of geometry and mechanics — trajectories, circles rolled on circles, orbits — refuse that form and come instead as *parametrized* curves $t \mapsto (x(t),
y(t))$ or as *polar* curves $r = r(\theta)$. This classical chapter studies both, and closes with the [conics](#def-b1-curves-conics).

## 24.1 Parametrized curves

**Definition 24.1.**

A *parametrized curve* is a [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $f
\colon I \to \R^2$, $t \mapsto M(t) = (x(t), y(t))$, with $x, y$ of class $C^1$ (at least) on the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $I$. The *velocity vector* is $f'(t) = (x'(t), y'(t))$; the point $M(t_0)$ is *regular* when $f'(t_0) \neq (0,0)$, and the *tangent* there is the line through $M(t_0)$ directed by $f'(t_0)$.

**Why $f'$ directs the tangent.** By Taylor–Young componentwise, $M(t) = M(t_0) + (t - t_0) f'(t_0) +
o(t - t_0)$: the chord direction $\frac{M(t) - M(t_0)}{t - t_0}$ tends to $f'(t_0)$. The regularity hypothesis is what makes the limit a *direction*: if $f'(t_0) = 0$, the display collapses to $M(t) = M(t_0) + o(t - t_0)$ and says nothing about how the curve leaves the point — whence the separate treatment of singular points in the method below, where the first *nonvanishing* [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) takes over the role of $f'$. Note also that the tangent line is a geometric object: any reparametrization changes $f'$ by a nonzero scalar factor and leaves the line unchanged. ∎

**Method 24.2 (Studying a parametrized curve).**

1. *Reduce the domain* using symmetries: relations between $M(-t)$ , $M(t + T)$ , … and $M(t)$ (reflections, translations, periodicity), then draw only the reduced part.
2. *Variations* : tabulate signs of $x'$ and $y'$ jointly; the curve moves right/left as $x'$ , up/down as $y'$ .
3. *Remarkable points* : horizontal tangent ( $y' = 0 \neq  x'$ ), vertical tangent ( $x' = 0 \neq y'$ ); at a *singular* point ( $f' = 0$ ), examine higher [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) for the tangent direction.
4. *Asymptotic behavior* at the ends of $I$ , then sketch.

**Example 24.3 (Asymptotic branches, worked).**

$x(t) = t$, $y(t) = t + \dfrac1t$ on $\intoo{0}{+\infty}$. As $t
\to 0^{+}$: $x \to 0$ while $y \to +\infty$ — the curve climbs along the *vertical asymptote* $x = 0$ (finite limit for one coordinate, infinite for the other). As $t \to +\infty$: both [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) blow up, so test a line: the difference

$$
y(t) - x(t) = \frac1t \longrightarrow 0^{+}
$$

exhibits the *oblique asymptote* $y = x$, approached from above. In between, $y' = 1 - \frac1{t^2}$ vanishes at $t = 1$: the point $(1, 2)$ is the low point of the branch, and $x' = 1 >
0$ throughout, so the curve always advances rightward. One arc, two asymptotes, one minimum: a complete picture from three computations — with the general recipe visible underneath: when $x, y \to \infty$ together, examine $y/x$ for a candidate slope (here $\to 1$), then $y - (\text{slope})\,x$ for the intercept and the side of approach.

**Example 24.4 (The astroid).**

$x(t) = \cos^3 t$, $y(t) = \sin^3 t$. Symmetries: $M(t + 2\pi) = M(t)$ (study a period); $M(-t)$ is the reflection of $M(t)$ in the $x$-axis; $M(\pi - t)$ in the $y$-axis; $M(\frac\pi2 -
t)$ in the diagonal $y = x$: it suffices to study $t \in
\intcc{0}{\frac\pi4}$ and unfold.

Velocity: $f'(t) = 3\sin t\cos t\,(-\cos t, \sin t)$. On $\intoo{0}{\frac\pi2}$ all points are regular with tangent directed by $(-\cos t, \sin t)$; at $t = 0$ (the point $(1,0)$) the velocity vanishes: a *cusp*, where the curve reverses along the tangent direction $(-1, 0)\cdot(\pm)$ — by symmetry the four [cusps](#ex-b1-curves-astroid) sit at $(\pm1, 0), (0, \pm1)$.

**Example 24.5 (A figure-eight, studied in full).**

$x(t) = \sin t$, $y(t) = \sin 2t$ (a *Lissajous curve*). *Symmetries*: $M(t + \pi) = (-x(t), y(t))$ (reflection in the $y$-axis), $M(-t) = (-x(t), -y(t))$ (central symmetry), $M(\pi - t) = (x(t), -y(t))$ (reflection in the $x$-axis): it suffices to study $t \in \intcc{0}{\frac\pi2}$ and unfold. *Variations*: $x' = \cos t \geq 0$ throughout, while $y' =
2\cos 2t$ is positive before $t = \frac\pi4$ and negative after: the arc climbs rightward to the summit $\bigl(\frac{\sqrt2}2,\ 1\bigr)$ at $t = \frac\pi4$ (horizontal tangent), then descends rightward to $(1, 0)$ at $t =
\frac\pi2$, where $x' = 0 \neq y'$: vertical tangent. *Double point*: $M(0) = M(\pi) = (0,0)$, with two different velocities

$$
f'(0) = (1,\ 2), \qquad f'(\pi) = (-1,\ 2) :
$$

two *regular* branches crossing at the origin at distinct angles — a double point, not a singular point: each passage is perfectly smooth, the two passages merely share their location. The whole curve is the figure-eight below.

![The Lissajous curve ( t, 2t): a double point at the origin, where two regular branches cross with velocities (1, 2) and (-1, 2), and horizontal tangents at the four summits. Symmetry reduced all the work to a quarter-period.](https://one-course.com/images/onecourse/chapters/math-3/b1-curves/fig-53dc09dad46f.svg)

*The Lissajous curve $(\sin t, \sin 2t)$: a double point at the origin, where two regular branches cross with velocities $(1, 2)$ and $(-1, 2)$, and horizontal tangents at the four summits. Symmetry reduced all the work to a quarter-period.*

![The astroid ( 3 t, 3 t): four arcs meeting at four cusps. It is the curve traced by a point of a circle of radius 1/4 rolling inside the unit circle.](https://one-course.com/images/onecourse/chapters/math-3/b1-curves/fig-8dd6fa9b9d24.svg)

*The astroid $(\cos^3 t, \sin^3 t)$: four arcs meeting at four [cusps](#ex-b1-curves-astroid). It is the curve traced by a point of a circle of radius $\frac14$ rolling inside the unit circle.*

## 24.2 Polar curves

**Definition 24.6.**

A *polar curve* is given by $r = r(\theta)$: the point of parameter $\theta$ is

$$
M(\theta) = r(\theta)\,\vec u(\theta),
\qquad \vec u(\theta) = (\cos\theta, \sin\theta).
$$

With $\vec v(\theta) = (-\sin\theta, \cos\theta) = \vec
u\,'(\theta)$, the velocity is

$$
M'(\theta) = r'(\theta)\, \vec u(\theta) + r(\theta)\, \vec
v(\theta) .
$$

Consequences: where $r \neq 0$, the point is regular, and the tangent makes with the ray the angle $V$ given by $\tan V =
\frac{r}{r'}$ (angle between $M'$ and $\vec u$); where $r(\theta_0)
= 0$, the curve passes through the origin *with tangent the ray $\theta = \theta_0$* (direction $\vec u(\theta_0)$, read off $M'(\theta_0) = r'(\theta_0)\vec u(\theta_0)$, or from the limit chord: the chord from $O$ to $M(\theta)$ is carried by $\vec
u(\theta)$ itself, which tends to $\vec u(\theta_0)$ — so the rule holds even when $r'(\theta_0) = 0$ and the point is singular, which is why polar [cusps](#ex-b1-curves-astroid) at the origin, like the cardioid’s, get their tangent for free).

**Example 24.7 (The cardioid).**

$r(\theta) = 1 + \cos\theta$. Symmetry: $r(-\theta)
= r(\theta)$: reflection in the $x$-axis; study $\theta \in
\intcc{0}{\pi}$. $r$ decreases from $2$ to $0$; at $\theta = \pi$, $r = 0$: the curve reaches the origin tangentially to the ray $\theta = \pi$ (the $x$-axis), forming a [cusp](#ex-b1-curves-astroid) there — the heart’s point. Tangent at $\theta = 0$: $r' = 0$, so $\tan V = \infty$: perpendicular to the axis.

The angle $V$ deserves one more reading. At $\theta =
\frac\pi2$: $r = 1$ and $r' = -1$, so $\tan V = \frac{r}{r'} =
-1$: the tangent makes three-quarters of a right angle with the outgoing ray — the curve is already bending back toward its [cusp](#ex-b1-curves-astroid). The formula $\tan V = r/r'$ delivers tangent directions along the whole curve with no computation of $M'(\theta)$ whatsoever: it is the polar analogue of reading a slope.

**Example 24.8 (Polar to cartesian: a hidden circle).**

What is the [polar curve](#def-b1-curves-polar) $r = 2\cos\theta$? Multiply by $r$: $r^2 = 2r\cos\theta$, i.e. $x^2 + y^2 = 2x$, i.e.

$$
(x - 1)^2 + y^2 = 1 :
$$

the circle of center $(1, 0)$ and radius $1$, passing through the origin. Bookkeeping matters: as $\theta$ runs over $\intoc{-\frac\pi2}{\frac\pi2}$ the whole circle is swept exactly once ($r$ vanishes at both ends), and at $\theta =
\pm\frac\pi2$ the rule “tangent at the origin along the ray $\theta = \theta_0$” gives a *vertical* tangent there — matching the geometry, since the vertical axis is indeed tangent to this circle at $O$. For $\theta$ beyond that range, $r < 0$ retraces the same circle: a reminder that a [polar curve](#def-b1-curves-polar) is a *parametrized* object, allowed to pass over itself.

![The four-petaled rose r = 2 (). The petals along the y-axis are traced with r < 0 (the point plots on the ray opposite to ); the dashed diagonals = ± π4 are the tangents at the origin, where r vanishes.](https://one-course.com/images/onecourse/chapters/math-3/b1-curves/fig-8c1ed6913fd9.svg)

*The four-petaled rose $r = \cos 2\theta$ ([Exercise 24.4](#exo-b1-curves-4)). The petals along the $y$-axis are traced with $r < 0$ (the point plots on the ray opposite to $\theta$); the dashed diagonals $\theta = \pm\frac\pi4$ are the tangents at the origin, where $r$ vanishes.*

![The cardioid r = 1 +. Polar curves are read by sweeping the angle: the radius swells and shrinks as turns.](https://one-course.com/images/onecourse/chapters/math-3/b1-curves/fig-de2f8edc9e4e.svg)

*The cardioid $r = 1 + \cos\theta$. [Polar curves](#def-b1-curves-polar) are read by sweeping the angle: the radius swells and shrinks as $\theta$ turns.*

## 24.3 Arc length

**Definition 24.9 (Length of an arc).**

The *length* of a $C^1$ arc $f \colon
\intcc{a}{b} \to \R^2$ is the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) of the speed:

$$
L = \int_a^b \norm{f'(t)}\,\dd t
= \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\;\dd t .
$$

(Motivation: on a small [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals), $M(t + h) \approx M(t) + h
f'(t)$, so the arc is close to a polygon whose segment lengths sum to a Riemann sum of $\norm{f'}$, [Theorem 15.20](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-riemann).) For a [polar curve](#def-b1-curves-polar) $r =
r(\theta)$, the velocity $r'\vec u + r\vec v$ has [orthogonal](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#def-b1-euclid-orthogonal) components, so

$$
L = \int_{\theta_1}^{\theta_2}
\sqrt{r'(\theta)^2 + r(\theta)^2}\;\dd\theta .
$$

The length does not depend on the (monotone, $C^1$) parametrization chosen: substituting $t = \varphi(s)$ in the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) ([Theorem 15.15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-parts)) multiplies $f'$ by $\varphi'$ and $\dd t$ by $\varphi'^{-1}$.

**Example 24.10 (Sanity check: the circle).**

For $f(t) = (R\cos t, R\sin t)$ on $\intcc{0}{2\pi}$: $\norm{f'} =
R$, so $L = 2\pi R$ — the definition returns the circumference. Reparametrization test: the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $g(t) = (R\cos 2t, R\sin 2t)$ on $\intcc{0}{\pi}$ draws the *same* circle at doubled speed $\norm{g'} = 2R$, and

$$
\int_0^{\pi} 2R\,\dd t = 2\pi R
$$

again: half the time, twice the speed, same [length](#def-b1-curves-arclength) — the invariance promised in the definition, watched once on numbers. (Running $g$ on all of $\intcc{0}{2\pi}$ would give $4\pi R$: a curve traversed twice is twice as long *as a journey*; [length](#def-b1-curves-arclength) measures the parametrized path, and honest bookkeeping of the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) is part of the computation.) For the astroid $(\cos^3t, \sin^3t)$: $\norm{f'} = 3\abs{\sin
t\cos t} = \tfrac32\abs{\sin 2t}$, and by symmetry $L = 4\int_0^{\pi/2}
\tfrac32\sin 2t\,\dd t = 6$: a curve drawn inside the unit circle, of [length](#def-b1-curves-arclength) $6 < 2\pi$. The weekend problem measures the most famous arch of all.

![One arch of the cycloid (radius R = 1), the rolling circle at t = 2, and the two guide lines of the weekend problem: the chord MC to the contact point is normal to the curve, the chord MT to the top of the circle is tangent.](https://one-course.com/images/onecourse/chapters/math-3/b1-curves/fig-5f3d2a84c663.svg)

*One arch of the cycloid (radius $R = 1$), the rolling circle at $t = 2$, and the two guide lines of the weekend problem: the chord $MC$ to the contact point is *normal* to the curve, the chord $MT$ to the top of the circle is *tangent*.*

## 24.4 Conics

**Definition 24.11 (Focus–directrix definition).**

Fix a point $F$ (*focus*), a line $D$ not through $F$ (*directrix*) and $e > 0$ (*eccentricity*). The *conic* of these data is

$$
\mathcal{C} = \{M : d(M, F) = e\; d(M, D)\}:
$$

an *ellipse* for $e < 1$, a *parabola* for $e = 1$, a *hyperbola* for $e > 1$. (The circle appears as a degenerate limit $e \to 0$.)

**Example 24.12 (The definition, checked on a parabola).**

Take the parabola $y^2 = 4x$, i.e. $2p = 4$: focus $F = (1, 0)$ and directrix $D : x = -1$ (the reduced form [Theorem 24.13](#thm-b1-curves-reduced) puts them at $\pm\frac p2$). At the point $M = (1, 2)$ of the curve:

$$
MF = \sqrt{(1-1)^2 + 2^2} = 2,
\qquad
d(M, D) = 1 - (-1) = 2 :
$$

equal, as $e = 1$ demands. At $M' = (4, 4)$: $MF' = \sqrt{9 +
16} = 5$ and $d(M', D) = 5$ again. The focus–directrix definition is not an abstraction: it is a pair of distances one can measure on any point, and the algebra of the reduced equations is nothing but this measurement done once and for all.

**Theorem 24.13 (Reduced equations).**

In a well-chosen [orthonormal](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#def-b1-euclid-orthogonal) frame:

$$
\text{ellipse: } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\quad (a \geq b > 0),
\qquad
\text{hyperbola: } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,
\qquad
\text{parabola: } y^2 = 2px,
$$

with, for the ellipse: foci at $(\pm c, 0)$, $c = \sqrt{a^2 - b^2}$, $e = \frac ca$, and the bifocal characterization $MF + MF' = 2a$; for the hyperbola: $c = \sqrt{a^2 + b^2}$, $e = \frac ca$, $\abs{MF - MF'} = 2a$, asymptotes $y = \pm\frac ba x$.

**Proof.** Take the focus at the origin and the directrix vertical, $x = -h$ ($h > 0$): the condition $MF^2 = e^2 d(M, D)^2$ reads $x^2 + y^2 =
e^2 (x + h)^2$. For $e = 1$: $y^2 = 2hx + h^2$, a parabola after the shift $x \mapsto x - \frac h2$ (so $p = h$). For $e \neq 1$: completing the square in $x$,

$$
(1 - e^2)\Bigl(x - \frac{e^2 h}{1 - e^2}\Bigr)^{\!2} + y^2
= e^2h^2 + \frac{e^4 h^2}{1 - e^2}
= \frac{e^2 h^2}{1 - e^2} .
$$

Set $X = x - \frac{e^2h}{1-e^2}$ (a shift of frame). When $e < 1$, divide by the positive right-hand side: $\frac{X^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a = \frac{eh}{1 -
e^2}$, $b = \frac{eh}{\sqrt{1-e^2}}$; when $e > 1$, both sides of the display flip sign appropriately and the same division gives $\frac{X^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a = \frac{eh}{e^2 -
1}$, $b = \frac{eh}{\sqrt{e^2 - 1}}$. The stated values of $c$ follow ($c^2 = a^2 - b^2$ or $a^2 + b^2$ gives $c = ea$ in both cases, placing the focus correctly), and the bifocal properties are direct verifications on the reduced equations. In detail for the ellipse, with $F' = (c, 0)$ and $M = (X, y)$ on the curve:

$$
MF'^2 = (X - c)^2 + y^2
= X^2 - 2cX + c^2 + b^2\Bigl(1 - \frac{X^2}{a^2}\Bigr)
= \frac{c^2}{a^2}X^2 - 2cX + a^2
= (a - eX)^2 ,
$$

using $c^2 + b^2 = a^2$ and $e = \frac ca$; since $\abs X \leq
a$ and $e < 1$, $a - eX > 0$, so $MF' = a - eX$ with no square root ever extracted. The mirror computation gives $MF = a + eX$, whence $MF + MF' = 2a$, constant. For the hyperbola the same algebra yields $MF = \abs{a + eX}$ and $MF' = \abs{eX - a}$, with difference $\pm2a$ according to the branch. ∎

**Example 24.14.**

$\frac{x^2}{25} + \frac{y^2}{9} = 1$: ellipse, $a = 5$, $b = 3$, $c
= 4$: foci $(\pm4, 0)$, eccentricity $\frac45$. Its parametrization: $(5\cos t, 3\sin t)$ — a circle stretched anisotropically; the sum of distances to the foci of any of its points is $10$.

**Example 24.15 (Reading an orbit from its polar equation).**

The polar [conic](#def-b1-curves-conics) $r = \dfrac{1}{1 + \frac12\cos\theta}$ ([Exercise 24.7](#exo-b1-curves-7) with $p = 1$, $e = \frac12$) is an ellipse with a *focus at the origin* — the geometry of a planetary orbit with the sun at $O$. Extract everything from $p$ and $e$: from the reduction of [Theorem 24.13](#thm-b1-curves-reduced), $a = \dfrac{p}{1 - e^2} =
\dfrac{1}{3/4} = \dfrac43$ and $c = ea = \dfrac23$. The two apsides check it with no theory at all:

$$
r(0) = \frac{1}{3/2} = \frac23 = a - c
\quad (\text{perihelion}),
\qquad
r(\pi) = \frac{1}{1/2} = 2 = a + c
\quad (\text{aphelion}),
$$

and their sum $\frac23 + 2 = \frac83 = 2a$ recovers the major axis. The polar form is the natural one whenever a focus is physically distinguished; the reduced cartesian form, whenever the symmetry axes are. Converting between the two is exactly what the completed-square computation of the theorem does.

Eccentricity as a dial, finally: keep $p = 1$ and turn $e$ in $r
= \frac{1}{1 + e\cos\theta}$. At $e = 0$: the circle $r = 1$. At $e = \frac12$: the ellipse just studied, $r$ oscillating between $\frac23$ and $2$. At $e = 1$: $r(\theta) \to \infty$ as $\theta
\to \pi$ — the curve no longer closes: a parabola, its farthest point pushed to infinity. At $e = 2$: the denominator vanishes at $\cos\theta = -\frac12$, and only $\theta \in
\intoo{-\frac{2\pi}3}{\frac{2\pi}3}$ survives: one branch of a hyperbola, escaping along two asymptotic directions. One formula, the entire [conic](#def-b1-curves-conics) family, and the transition points $e =
1$ visible as the moment the denominator first reaches zero.

**Remark 24.16 (Common pitfalls).**

*Double points are not singular points*: at a self-crossing, each branch is regular; “singular” refers to $f'(t_0) = 0$ for one value of the parameter ([Example 24.5](#ex-b1-curves-lissajous) vs the astroid’s [cusps](#ex-b1-curves-astroid)). *Vertical tangent versus [cusp](#ex-b1-curves-astroid)*: $x'(t_0) = 0 \neq y'(t_0)$ is a regular point with vertical tangent; only $x' = y' = 0$ demands the higher-order expansion. *Negative radii plot on the opposite ray*: for $r(\theta) <
0$ the point is $-\abs{r}\,\vec u(\theta)$, at angle $\theta +
\pi$ ([Exercise 24.5](#exo-b1-curves-5)); forgetting this loses inner loops or doubles curves. *[Arc length](#def-b1-curves-arclength) integrates $\norm{f'}$, not $f'$*: split the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) at the zeros of the speed — for the astroid, integrating $\frac32\sin 2t$ over a full period without absolute values gives $0$, not $6$. *[Conics](#def-b1-curves-conics) must be reduced before being read*: in $9x^2 +
25y^2 - 36x - 50y - 164 = 0$, neither the axes, nor the center, nor the eccentricity is visible until the squares are completed ([Exercise 24.6](#exo-b1-curves-6)); and a vanishing quadratic part on one variable means parabola, not “degenerate ellipse”.

**Remark 24.17 (Where these curves go).**

[Parametrized curves](#def-b1-curves-def) are the language of mechanics: trajectories are curves, velocity vectors are tangent vectors, and the [arc length](#def-b1-curves-arclength) of [Definition 24.9](#def-b1-curves-arclength) is the distance run. The [conics](#def-b1-curves-conics) reappear wherever an inverse-square law acts — planetary orbits are ellipses with the sun at a focus. In [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar), curves become the level [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of functions of two variables, and the tangent of this chapter meets the gradient of the next. The Year 2 volume adds curvature and the local canonical form of a curve; the weekend problem below already extracts, with this year’s tools only, everything the seventeenth century knew about its most celebrated curve.

**Remark 24.18 (Perspectives inside Book 3).**

This chapter consumes the whole first half of the volume and feeds the last chapter. *Consumed*: tangent vectors are [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) ([Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)), [arc length](#def-b1-curves-arclength) and areas are [integrals](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)), [cusps](#ex-b1-curves-astroid) are settled by Taylor expansions ([Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor)), the tautochrone is a [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) differential equation ([Chapter 5](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#ch-b1-diffeq)), and every distance and angle is Euclidean ([Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid)). *Fed*: in [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar), a curve $t \mapsto (x(t),
y(t))$ drawn inside a level [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\{f = c\}$ is differentiated by the chain rule, and the resulting identity $\langle \nabla f,\
f'(t)\rangle = 0$ marries this chapter’s tangent vectors to the next one’s gradients — the tangent line of the curve and the normal direction of the surface are the same computation seen from both banks.

## 24.5 Exercises

**Exercise 24.1 ★.**

Study and sketch the curve $x(t) = t^2$, $y(t) = t^3$ (symmetries, variations, behavior at the singular point $t = 0$).

**Solution of Exercise 24.1.**

$M(-t) = (t^2, -t^3)$: reflection in the $x$-axis; study $t \geq 0$. Both $x' = 2t$ and $y' = 3t^2$ are $\geq 0$: the branch moves right and up, from $(0,0)$ to infinity. At $t = 0$ the velocity vanishes; expansions $x = t^2$, $y = t^3$ show the tangent is the $x$-axis ($y/x = t \to 0$) with $y$ changing sign while $x \geq 0$: a *[cusp](#ex-b1-curves-astroid)* pointing left. The curve is the semi-cubical parabola $y^2 = x^3$.

**Exercise 24.2 ★.**

For the cycloid $x(t) = t - \sin t$, $y(t) = 1 - \cos t$ (the trajectory of a point of a rolling wheel of radius $1$): identify the period-translation symmetry, the singular points, and the tangent direction at $t = 0$ *(expand $x$ and $y$ to the first nonzero orders)*.

**Solution of Exercise 24.2.**

$M(t + 2\pi) = M(t) + (2\pi, 0)$: the curve repeats, translated by one wheel circumference; study one period. $x'(t) = 1 - \cos t \geq
0$, $y'(t) = \sin t$: the arch rises on $\intoo{0}{\pi}$, falls on $\intoo{\pi}{2\pi}$, culminating at $(\pi, 2)$. Singular points where $x' = y' = 0$: $t \in 2\pi\Z$, on the ground. Near $t = 0$:

$$
x(t) = \frac{t^3}{6} + o(t^3),
\qquad
y(t) = \frac{t^2}{2} + o(t^2):
$$

$x$ changes sign, $y \geq 0$, and $\frac{x}{y^{3/2}}$ bounded: the tangent is *vertical* (direction $(0,1)$), a [cusp](#ex-b1-curves-astroid) where the tracked point momentarily has zero speed — the physical signature of rolling without slipping.

**Exercise 24.3 ★.**

Give the tangent line to the curve $(\cos^3 t, \sin^3 t)$ at $t =
\frac\pi4$, and check that the segment of this tangent cut by the axes has [length](#def-b1-curves-arclength) $1$ — a famous property of the astroid (true at every regular point).

**Solution of Exercise 24.3.**

At $t = \frac\pi4$: $M = \bigl(\frac{\sqrt2}{4},
\frac{\sqrt2}{4}\bigr)$, tangent direction $(-\cos t, \sin t) =
\frac{1}{\sqrt2}(-1, 1)$ ([Example 24.4](#ex-b1-curves-astroid)). Tangent line: $y - \frac{\sqrt2}{4} = -(x - \frac{\sqrt2}{4})$, i.e. $x +
y = \frac{\sqrt2}{2}$. Intercepts: $\bigl(\frac{\sqrt2}{2},
0\bigr)$ and $\bigl(0, \frac{\sqrt2}{2}\bigr)$; the segment between them has [length](#def-b1-curves-arclength) $\sqrt{\frac12 + \frac12} = 1$.

(General point $t$: the tangent at $(\cos^3 t, \sin^3 t)$ cuts the axes at $(\cos t, 0)$ and $(0, \sin t)$ — check that the line through these points has direction $(-\cos t, \sin t)$ and passes through $M(t)$ — and the cut segment has [length](#def-b1-curves-arclength) $\sqrt{\cos^2 t +
\sin^2 t} = 1$.)

**Exercise 24.4 ★.**

Sketch the [polar curves](#def-b1-curves-polar) $r = \cos 2\theta$ (four-petaled rose) and $r = \frac{1}{\cos\theta}$ on $\intoo{-\frac\pi2}{\frac\pi2}$ (recognize a line).

**Solution of Exercise 24.4.**

$r = \cos 2\theta$: period $\pi$ in $\theta$, symmetric in both axes; $r$ vanishes at $\theta = \pm\frac\pi4$ (tangents at the origin along the diagonals) and is negative for $\theta \in
\intoo{\frac\pi4}{\frac{3\pi}{4}}$, where the points plot on the opposite ray — producing four petals along the axes directions $\theta = 0, \frac\pi2, \pi, \frac{3\pi}{2}$, each of maximal radius $1$.

$r\cos\theta = 1$ is the equation $x = 1$: the [polar curve](#def-b1-curves-polar) $r =
\frac{1}{\cos\theta}$ is the vertical line $x = 1$ (swept once for $\theta \in \intoo{-\frac\pi2}{\frac\pi2}$).

**Exercise 24.5 ★★.**

Study the [polar curve](#def-b1-curves-polar) $r = 1 + 2\cos\theta$ (a limaçon): domain where $r \geq 0$ vs $r < 0$ (points plotted with negative radius sit on the opposite ray), passage through the origin, inner loop, sketch.

**Solution of Exercise 24.5.**

$r(\theta) = 1 + 2\cos\theta$ vanishes for $\cos\theta = -\frac12$: $\theta = \pm\frac{2\pi}{3}$. Symmetry in the $x$-axis; study $\theta \in \intcc{0}{\pi}$. For $\theta \in
\intco{0}{\frac{2\pi}{3}}$: $r > 0$, decreasing from $3$ to $0$: outer arc, entering the origin tangent to the ray $\theta =
\frac{2\pi}{3}$. For $\theta \in \intoc{\frac{2\pi}{3}}{\pi}$: $r <
0$: the points $r\vec u(\theta)$ sit on the opposite ray (angle $\theta - \pi \in \intoc{-\frac\pi3}{0}$), with distance $\abs r$ growing from $0$ to $1$: this draws a small *inner loop* through the origin. At $\theta = \pi$, $r = -1$ and the point is $-\vec u(\pi) = (1, 0)$: the loop closes on the positive $x$-axis. Sketch: a big heart-like outer curve with maximum reach $3$ at $\theta = 0$, plus a loop inside it through $O$ and $(1,0)$.

**Exercise 24.6 ★★.**

Identify the [conic](#def-b1-curves-conics) $9x^2 + 25y^2 - 36x - 50y - 164 = 0$: reduce by completing squares, give center, semi-axes, foci, eccentricity.

**Solution of Exercise 24.6.**

Complete squares:

$$
9(x^2 - 4x) + 25(y^2 - 2y) = 164
\iff 9(x-2)^2 + 25(y-1)^2 = 164 + 36 + 25 = 225 .
$$

Dividing by $225$: $\frac{(x-2)^2}{25} + \frac{(y-1)^2}{9} = 1$: an ellipse of center $(2, 1)$, semi-axes $a = 5$ (horizontal), $b =
3$; $c = \sqrt{25 - 9} = 4$: foci $(2 \pm 4,\, 1) = (-2, 1)$ and $(6, 1)$; eccentricity $e = \frac45$.

**Exercise 24.7 ★★.**

Prove that the polar equation of a [conic](#def-b1-curves-conics) with focus at the origin is

$$
r = \frac{p}{1 + e\cos\theta}
$$

(directrix vertical at distance $\frac pe$ right of the focus). Which values of $\theta$ are allowed when $e > 1$?

**Solution of Exercise 24.7.**

Focus at the origin, directrix $D: x = d$ with $d = \frac pe > 0$. For a point $M = (r\cos\theta, r\sin\theta)$ with $r > 0$:

$$
MF = r,
\qquad
d(M, D) = \abs{d - r\cos\theta},
$$

and the [conic](#def-b1-curves-conics) condition $MF = e\,d(M, D)$, in the regime where $M$ is on the focus side of the directrix ($r\cos\theta < d$), reads $r
= e(d - r\cos\theta)$, i.e.

$$
r(1 + e\cos\theta) = ed = p,
\qquad
r = \frac{p}{1 + e\cos\theta} .
$$

For $e < 1$ the denominator never vanishes: all $\theta$ allowed (ellipse). For $e = 1$: $\theta \neq \pi$ (parabola, open towards the directrix’s far side). For $e > 1$: need $1 + e\cos\theta > 0$, i.e. $\theta \in \intoo{-\theta_0}{\theta_0}$ with $\theta_0 =
\arccos\bigl(-\frac1e\bigr)$: one *branch* of the hyperbola (the other branch corresponds to the sign choice $r < 0$, or to the second focus).

**Exercise 24.8 ★★.**

From the bifocal property $MF + MF' = 2a$, deduce the “gardener’s construction” of the ellipse, and prove that the tangent at $M$ makes equal angles with $MF$ and $MF'$ *(reflection property; differentiate $\norm{M(t) - F} + \norm{M(t) - F'} = 2a$ and interpret the vanishing sum of unit-vector products)*.

**Solution of Exercise 24.8.**

*Gardener:* attach a string of [length](#def-b1-curves-arclength) $2a$ to two stakes $F,
F'$ and keep it taut with the tracing point $M$: the constraint is exactly $MF + MF' = 2a$, so the traced curve is the ellipse.

*Reflection property:* let $t \mapsto M(t)$ be a regular parametrization and $u(t) = \frac{M(t) - F}{\norm{M(t) - F}}$, $u'(t)$ the analogous unit vector to $F'$. Differentiating $\norm{M - F} + \norm{M - F'} = 2a$, using $\frac{\dd}{\dd
t}\norm{M - F} = \bigl\langle \frac{M - F}{\norm{M-F}},\,
M'\bigr\rangle$ (chain rule on $\sqrt{\langle\cdot,\cdot\rangle}$):

$$
\bigl\langle u + u',\, M'\bigr\rangle = 0 .
$$

So the tangent direction $M'$ is [orthogonal](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#def-b1-euclid-orthogonal) to the bisector direction $u + u'$ of the two focal rays: the tangent makes equal angles with $MF$ and $MF'$. (A light ray from one focus reflects off the ellipse to the other focus.)

**Exercise 24.9 ★★★.**

The *lemniscate of Bernoulli* is the [polar curve](#def-b1-curves-polar) $r^2 = \cos
2\theta$ (take $r = \sqrt{\cos2\theta}$ where defined).

1. Give its domain, symmetries, tangents at the origin, and sketch it.
2. Prove that it is the locus of points $M$ with $MF \cdot MF'  = \frac12$ where $F, F' = \bigl(\pm\frac{1}{\sqrt2},  0\bigr)$ . *(Compute $MF^2\,MF'^2$ in polar [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates).)*

**Solution of Exercise 24.9.**

1. Domain: $\cos 2\theta \geq 0$ : $\theta \in  \intcc{-\frac\pi4}{\frac\pi4} \cup  \intcc{\frac{3\pi}{4}}{\frac{5\pi}{4}}$ . Symmetries: $\theta  \mapsto -\theta$ ( $x$ -axis) and $\theta \mapsto \pi -  \theta$ ( $y$ -axis): study $\intcc{0}{\frac\pi4}$ ; $r$ decreases from $1$ to $0$ . At $\theta = \pm\frac\pi4$ : $r =  0$ , tangents at the origin along the diagonals. The curve is the $\infty$ symbol: two symmetric loops meeting at $O$ , reaching $(\pm 1, 0)$ .
2. With $F, F' = (\pm c, 0)$, $c = \frac{1}{\sqrt 2}$, and $M  = (r\cos\theta, r\sin\theta)$: $$MF^2\, MF'^2  = \bigl((r\cos\theta - c)^2 + r^2\sin^2\theta\bigr)  \bigl((r\cos\theta + c)^2 + r^2\sin^2\theta\bigr)  = (r^2 + c^2)^2 - (2rc\cos\theta)^2 ,$$ by the identity $(A - B)(A + B) = A^2 - B^2$ with $A = r^2  + c^2$, $B = 2rc\cos\theta$. With $c^2 = \frac12$: $$MF^2 MF'^2 = \Bigl(r^2 + \frac12\Bigr)^2 - 2r^2\cos^2\theta  = r^4 + r^2\bigl(1 - 2\cos^2\theta\bigr) + \frac14  = r^4 - r^2\cos 2\theta + \frac14 .$$ On the lemniscate, $r^2 = \cos 2\theta$: the first two terms cancel, leaving $MF^2MF'^2 = \frac14$, i.e. $MF \cdot MF' =  \frac12$. Conversely, the computation read backwards shows the locus equation $MF\,MF' = \frac12$ is $r^2 = \cos  2\theta$ (for $r \neq 0$; and $O$ satisfies both).

**Exercise 24.10 ★★.**

Compute the total [length](#def-b1-curves-arclength) of the cardioid $r = 1 + \cos\theta$ using the polar formula of [Definition 24.9](#def-b1-curves-arclength) *(the identity $1 + \cos\theta = 2\cos^2\frac\theta2$ turns the square root into $2\abs{\cos\frac\theta2}$)*.

**Solution of Exercise 24.10.**

$r = 1 + \cos\theta$, $r' = -\sin\theta$:

$$
r^2 + r'^2 = 1 + 2\cos\theta + \cos^2\theta + \sin^2\theta
= 2 + 2\cos\theta = 4\cos^2\frac\theta2 ,
$$

so $\sqrt{r^2 + r'^2} = 2\abs{\cos\frac\theta2}$ and, by the symmetry in the $x$-axis,

$$
L = \int_0^{2\pi} 2\Bigl|\cos\frac\theta2\Bigr|\,\dd\theta
= 2\int_0^{\pi} 2\cos\frac\theta2\,\dd\theta
= 2\Bigl[4\sin\frac\theta2\Bigr]_0^{\pi} = 8 .
$$

Another algebraic, $\pi$-free perimeter.

**Exercise 24.11 ★★.**

Show that the tangent to the ellipse $(a\cos t, b\sin t)$ at the point of parameter $t$ has equation

$$
\frac{x\cos t}{a} + \frac{y\sin t}{b} = 1 ,
$$

and deduce the tangent to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $(x_0, y_0)$ of the ellipse: $\frac{x\,x_0}{a^2} +
\frac{y\,y_0}{b^2} = 1$ (the “splitting of the squares” rule).

**Solution of Exercise 24.11.**

The point $(a\cos t, b\sin t)$ satisfies the equation: $\cos^2 t + \sin^2 t = 1$. The line’s normal vector is $\bigl(\frac{\cos t}a, \frac{\sin t}b\bigr)$, and its product with the velocity $(-a\sin t, b\cos t)$ is $-\sin t\cos t +
\sin t\cos t = 0$: the line passes through the point with the tangent direction — it is the tangent. For $(x_0, y_0)$ on the ellipse, write $\cos t = \frac{x_0}a$, $\sin t = \frac{y_0}b$ and substitute:

$$
\frac{x\,x_0}{a^2} + \frac{y\,y_0}{b^2} = 1 ,
$$

obtained from the ellipse equation by “splitting” $x^2 \mapsto
x\,x_0$ and $y^2 \mapsto y\,y_0$.

**Exercise 24.12 ★★★.**

The *logarithmic spiral* is the [polar curve](#def-b1-curves-polar) $r =
\eu^{k\theta}$ ($k > 0$ fixed, $\theta \in \R$).

1. Show that the angle $V$ between the radius and the tangent is *constant* ( $\tan V = \frac1k$ ) — the spiral crosses every ray at the same angle.
2. Show that rotating the spiral by an angle $c$ [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) it onto its scaling by the factor $\eu^{-kc}$ : every rotation of the spiral is a magnification of it (self-similarity).
3. Compute the [length](#def-b1-curves-arclength) of the arc $\theta \in  \intoc{-\infty}{\theta_0}$ (as a limit of [lengths](#def-b1-curves-arclength) on $\intcc{A}{\theta_0}$ , $A \to -\infty$ ) and observe that it is finite: a curve spiraling infinitely many times around the origin, of finite [length](#def-b1-curves-arclength) .

**Solution of Exercise 24.12.**

1. $r' = k\eu^{k\theta}$ , so $\tan V = \frac{r}{r'} =  \frac1k$ : constant. The spiral cuts every ray from the origin at the same angle $V = \arctan\frac1k$ .
2. In complex notation the spiral is $\{\eu^{k\theta}  \eu^{\iu\theta} : \theta \in \R\}$. Rotating by $c$ multiplies by $\eu^{\iu c}$: $$\eu^{k\theta}\eu^{\iu(\theta + c)}  = \eu^{-kc}\,\eu^{k(\theta + c)}\eu^{\iu(\theta+c)} ,$$ and as $\theta + c$ runs over $\R$ this describes $\eu^{-kc}$ times the spiral: rotation $=$ scaling. No other smooth curve but lines and circles has this property.
3. $\sqrt{r^2 + r'^2} = \sqrt{1 + k^2}\,\eu^{k\theta}$, so on $\intcc{A}{\theta_0}$ the [length](#def-b1-curves-arclength) is $\frac{\sqrt{1+k^2}}{k}\bigl(\eu^{k\theta_0} -  \eu^{kA}\bigr)$, and as $A \to -\infty$: $$L = \frac{\sqrt{1 + k^2}}{k}\,\eu^{k\theta_0} < \infty :$$ infinitely many turns around the origin, finite total [length](#def-b1-curves-arclength) (the turns shrink geometrically).

## 24.6 Problem: the cycloid, queen of curves

**Problem 24.1.**

A wheel of radius $R$ rolls without slipping along the $x$-axis; the point of the rim initially at the origin [traces](https://one-course.com/books/math/3/en/chapter/21-matrices#def-b1-matrices-transpose) the *cycloid*. The seventeenth century fought over this curve — Galileo weighed paper cutouts of it, Wren measured it, Roberval computed its area, Huygens built clocks on it — and every one of their results is within reach of this chapter. We prove the four classics: the tangent construction, Wren’s [length](#def-b1-curves-arclength) $8R$, the area $3\pi R^2$, and Huygens’ tautochrone property.

**Part I — Rolling and the tangent.**

1. After the wheel has turned by an angle $t$, its center sits at $\Omega(t) = (Rt, R)$ (rolling without slipping: contact distance $=$ arc rolled). Show that the marked point is at $$M(t) = \bigl(R(t - \sin t),\; R(1 - \cos t)\bigr).$$
2. Compute $f'(t)$ and show $\norm{f'(t)} =  2R\,\abs{\sin\frac t2}$ ; locate the singular points ( [cusps](#ex-b1-curves-astroid) — cf. [Exercise 24.2](#exo-b1-curves-2) ) and the top of each arch.
3. Let $C(t) = (Rt, 0)$ be the contact point and $T(t) =  (Rt, 2R)$ the top of the wheel. Prove that for $0 < t <  2\pi$ the vector $M - C$ is *normal* to the curve at $M(t)$ and the vector $T - M$ is *tangent* : to draw the tangent to a cycloid, join the point to the top of its rolling circle. *(Factor everything through $\sin\frac t2$ and $\cos\frac t2$.)*
4. Interpret question 3 kinematically: the contact point is the *instantaneous center of rotation* , and the speed of $M$ equals its distance to $C$ (for unit angular velocity). Verify $\norm{M - C} = 2R\abs{\sin\frac t2}$ .
5. Prove the height–speed relation $$\norm{f'(t)}^2 = 2R\,y(t) :$$ on a cycloid traversed at unit angular velocity, the speed at each point is exactly the free-fall speed for a drop equal to the current height. (Keep this for Part IV.)

**Part II — Wren’s theorem: the arch has [length](#def-b1-curves-arclength) $8R$.**

6. Using [Definition 24.9](#def-b1-curves-arclength), compute the [length](#def-b1-curves-arclength) of one arch: $$L = \int_0^{2\pi} 2R\sin\frac t2\,\dd t = 8R$$ (Wren’s theorem, 1658). Four wheel diameters, and no $\pi$ anywhere.
7. Compute the [arc length](#def-b1-curves-arclength) from the [cusp](#ex-b1-curves-astroid) : $s(t) =  4R\bigl(1 - \cos\frac t2\bigr)$ , and check $s(2\pi) =  8R$ .
8. Now measure the arc from the *apex* $t = \pi$: $\sigma(t) = \abs{4R\cos\frac t2}$. Prove the intrinsic relation $$\sigma^2 = 8R\,\bigl(2R - y\bigr) :$$ the squared arc distance from the top is proportional to the height drop below the top.
9. Sanity checks: recover from question 8 that the half-arch from apex to [cusp](#ex-b1-curves-astroid) has [length](#def-b1-curves-arclength) $4R$ , and compare with the astroid computation of [Example 24.10](#ex-b1-curves-circlelength) — both curves have algebraic, $\pi$ -free [lengths](#def-b1-curves-arclength) ; explain what makes this possible even though both are built from circles. *(Look at the form of $\norm{f'}$.)*

**Part III — Roberval’s area: $3\pi R^2$.**

10. Justify that the area between one arch and the ground is $A = \int_0^{2\pi} y(t)\,x'(t)\,\dd t$ *(the substitution $x = x(t)$ in $\int y\,\dd x$, [Theorem 15.15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-parts); $x$ is increasing)* .
11. Compute $$A = R^2\int_0^{2\pi}(1 - \cos t)^2\,\dd t = 3\pi R^2 :$$ exactly *three* times the area of the wheel — the ratio Galileo had guessed by weighing.
12. Same method for the astroid $(\cos^3t, \sin^3t)$ : show that the enclosed area is $\frac{3\pi}8$ *(linearize $\sin^4 t\cos^2 t$; only the constant term survives over a full period)* .
13. Sanity check the formula of question 10 on the upper unit semicircle $(\cos t, \sin t)$ , $t$ from $\pi$ to $0$ : does it return $\frac\pi2$ ?

**Part IV — Huygens’ tautochrone.** Flip the arch: a frictionless bead slides, under gravity $g$, inside the cycloidal bowl

$$
x(t) = R(t + \sin t), \qquad y(t) = R(1 - \cos t)
\qquad (t \in \intcc{-\pi}{\pi}),
$$

whose lowest point is the origin ($y$ measured upward).

14. Compute the speed $\norm{f'(t)} = 2R\cos\frac t2$, the [arc length](#def-b1-curves-arclength) from the bottom $s(t) = 4R\sin\frac t2$, and prove the key identity $$y = \frac{s^2}{8R} .$$
15. The bead released at rest from the point of parameter $t_0 > 0$ obeys energy conservation: if $s(\tau)$ denotes its arc position at time $\tau$, then $\frac12\bigl(  \frac{\dd s}{\dd\tau}\bigr)^2 + g\,y = g\,y_0$. Rewrite this, using question 14, as $$\Bigl(\frac{\dd s}{\dd\tau}\Bigr)^{\!2}  = \frac{g}{4R}\,\bigl(s_0^2 - s^2\bigr),  \qquad s_0 = s(t_0).$$
16. Differentiate with respect to $\tau$ and obtain the harmonic oscillator $$\frac{\dd^2 s}{\dd\tau^2} = -\frac{g}{4R}\,s ;$$ solve it with [Theorem 5.10](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#thm-b1-diffeq-homogeneous2) and the initial conditions: $s(\tau) = s_0\cos(\omega\tau)$, $\omega = \sqrt{g/4R}$.
17. Deduce the *tautochrone property* (Huygens, 1659): the time to reach the bottom, $$T_{\downarrow} = \frac{\pi}{2\omega}  = \pi\sqrt{\frac Rg}\,,$$ does not depend on the release point — beads released together from any two heights of the bowl arrive together.
18. Verify by substitution that $s(\tau) =  s_0\cos\omega\tau$ satisfies the first-order energy equation of question 15 exactly (not only the differentiated one), and explain in one sentence why a circular pendulum is only *approximately* isochronous while the cycloid is exactly so.
19. Numerically: what radius $R$ makes the descent time exactly one second ( $g = 9.81$ )? Note how close the answer is to one meter, and how the full oscillation period $2\pi\sqrt{4R/g}$ compares with the small-angle pendulum formula $2\pi\sqrt{\ell/g}$ for $\ell = 4R$ .

**Part V — Dividends, and synthesis.**

20. Apply the tangent rule of question 3 at $t =  \frac\pi2$ (take $R = 1$ ): compute $M$ , $T$ , the direction of $MT$ , and check it against $f'(\frac\pi2)$ .
21. (Trochoids) Mark instead a point at distance $d$ from the center ( $d \neq R$ ): the curve is $x = Rt -  d\sin t$ , $y = R - d\cos t$ . Show that for $d < R$ the curve is regular everywhere and is the graph of no singular behavior ( $x' > 0$ : it advances), while for $d  > R$ the abscissa $x'$ changes sign and the curve makes loops — the flanged railway wheel whose rim points travel *backwards* .
22. (Brachistochrone teaser) From the [cusp](#ex-b1-curves-astroid) $(\pi R, 2R)$ of the bowl to the bottom, compare the cycloid descent time $\pi\sqrt{R/g}$ with the time along the straight chute joining the same points *(constant acceleration $g\sin\alpha$ along the chord)* : show the chord takes $\sqrt{\pi^2 + 4}\,\sqrt{R/g} \approx 3.72\sqrt{R/g}$ . The curve beats the line — it is, in fact, the fastest of all curves, a result of the calculus of variations.
23. Show that on the arch ($0 < t < 2\pi$), $$\frac{\dd y}{\dd x} = \cot\frac t2,  \qquad  \frac{\dd^2 y}{\dd x^2} =  -\frac{1}{4R\sin^4\frac t2} < 0 :$$ the arch is concave, with vertical tangents exactly at the [cusps](#ex-b1-curves-astroid).
24. Recover the vertical [cusp](#ex-b1-curves-astroid) tangent of [Exercise 24.2](#exo-b1-curves-2) geometrically: compute the limit direction of the chord $MT$ of question 3 as $t \to  0^{+}$ , with no expansions at all.
25. Synthesis, in four sentences: which three named theorems this problem proved (with their numbers $8R$ , $3\pi  R^2$ , $\pi\sqrt{R/g}$ and their authors); which single computational device ( $\sin\frac t2$ , $\cos\frac t2$ factorizations) powered all of Parts I, II and IV; how the intrinsic relation $y = s^2/8R$ converted geometry into a [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) differential equation; and which chapter of this book each Part leaned on.

**Solution of Problem 24.1.**

**1.** Rolling without slipping means the contact point has traveled a distance equal to the arc of wheel unrolled: after turning by $t$, the center is at $(Rt, R)$. The marked point sits on the rim at angle $t$ *behind* the downward vertical (the wheel turns clockwise while advancing):

$$
M(t) = \Omega(t) + R(-\sin t, -\cos t)
= \bigl(R(t - \sin t),\ R(1 - \cos t)\bigr),
$$

which is correct at $t = 0$ ($M = (0,0)$) and at $t = \pi$ ($M =
(\pi R, 2R)$, the top).

**2.** $f'(t) = R(1 - \cos t,\ \sin t)$ and

$$
\norm{f'}^2 = R^2\bigl((1-\cos t)^2 + \sin^2 t\bigr)
= 2R^2(1 - \cos t) = 4R^2\sin^2\frac t2 ,
$$

so $\norm{f'} = 2R\abs{\sin\frac t2}$. Singular points at $t \in
2\pi\Z$: the [cusps](#ex-b1-curves-astroid) on the ground ([Exercise 24.2](#exo-b1-curves-2)); the top of the arch is $t = \pi$, where the speed $2R$ is maximal.

**3.** Half-angle factorizations:

$$
f'(t) = 2R\sin\frac t2\,\Bigl(\sin\frac t2,\ \cos\frac t2\Bigr),
\quad
M - C = 2R\sin\frac t2\,\Bigl(-\cos\frac t2,\ \sin\frac t2\Bigr),
$$

$$
T - M = \bigl(R\sin t,\ R(1 + \cos t)\bigr)
= 2R\cos\frac t2\,\Bigl(\sin\frac t2,\ \cos\frac t2\Bigr).
$$

For $0 < t < 2\pi$, $\sin\frac t2 \neq 0$: $M - C$ is [orthogonal](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#def-b1-euclid-orthogonal) to the tangent direction $\bigl(\sin\frac t2, \cos\frac
t2\bigr)$ (their product is $-\sin\frac t2\cos\frac t2 +
\sin\frac t2\cos\frac t2 = 0$), and $T - M$ is parallel to it. The chord to the top of the wheel *is* the tangent; the chord to the contact point is the normal.

**4.** At each instant the wheel pivots about its contact point (that point has zero velocity: rolling without slipping), so every rigid point of the wheel moves [orthogonally](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#def-b1-euclid-orthogonal) to the line joining it to $C$, with speed (angular velocity $1$) equal to that distance. Check: $\norm{M - C} = 2R\abs{\sin\frac t2} =
\norm{f'(t)}$.

**5.** $2R\,y(t) = 2R\cdot 2R\sin^2\frac t2 =
4R^2\sin^2\frac t2 = \norm{f'(t)}^2$. The speed at height $y$ is $\sqrt{2R\,y}$ — formally the law $v = \sqrt{2gh}$ of [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) fall, with the ground playing the ceiling; Part IV turns this observation into clockwork.

**6.** By [Definition 24.9](#def-b1-curves-arclength) and question 2 ($\sin\frac t2 \geq 0$ on $\intcc{0}{2\pi}$):

$$
L = \int_0^{2\pi} 2R\sin\frac t2\,\dd t
= 2R\Bigl[-2\cos\frac t2\Bigr]_0^{2\pi} = 2R(2 + 2) = 8R .
$$

Wren’s theorem: four diameters exactly.

**7.** $s(t) = \int_0^t 2R\sin\frac u2\,\dd u =
4R\bigl(1 - \cos\frac t2\bigr)$; $s(2\pi) = 4R(1+1) = 8R$, consistent.

**8.** $\sigma(t) = \abs{s(t) - s(\pi)} = \abs{4R(1 -
\cos\frac t2) - 4R} = 4R\abs{\cos\frac t2}$, so $\sigma^2 =
16R^2\cos^2\frac t2$; and $2R - y = R(1 + \cos t) =
2R\cos^2\frac t2$, whence $8R(2R - y) = 16R^2\cos^2\frac t2 =
\sigma^2$.

**9.** At the [cusp](#ex-b1-curves-astroid) $t = 0$ (or $2\pi$): $\sigma = 4R$, half of $8R$: the apex halves the arch. In both computations the speed is $\abs{\text{trigonometric polynomial in } t/2}$, whose antiderivative is again trigonometric: the [length](#def-b1-curves-arclength) is a difference of *values* of cosines — rational numbers times $R$ — with no arc of circle to measure, hence no $\pi$. The circle itself has constant speed, so its [length](#def-b1-curves-arclength) [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) produces the full [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) [length](#def-b1-curves-arclength) $2\pi$; the cycloid’s speed vanishes at the ends and integrates algebraically.

**10.** The arch is swept with $x$ increasing from $0$ to $2\pi R$ ($x' = R(1 - \cos t) \geq 0$, vanishing only at isolated points). Substituting $x = x(t)$ in the area [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) $\int_0^{2\pi R} y\,\dd x$ ([Theorem 15.15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-parts)) gives $A = \int_0^{2\pi} y(t)\,x'(t)\,\dd t$.

**11.**

$$
A = \int_0^{2\pi} R(1 - \cos t)\cdot R(1 - \cos t)\,\dd t
= R^2\int_0^{2\pi}\bigl(1 - 2\cos t + \cos^2 t\bigr)\dd t
= R^2\Bigl(2\pi - 0 + \pi\Bigr) = 3\pi R^2 ,
$$

using $\int_0^{2\pi}\cos^2 = \pi$. Exactly three wheel areas: Galileo’s balance said “about $3$”; Roberval’s computation says “exactly”.

**12.** With $x = \cos^3 t$, $y = \sin^3 t$: $y\,x' =
-3\sin^4 t\cos^2 t$. Linearize:

$$
\sin^4 t\cos^2 t = \frac{1 - \cos 2t}{2}\cdot\frac{\sin^2
2t}{4} = \frac{\sin^2 2t}{8} - \frac{\sin^2 2t\cos 2t}{8} ,
$$

and over $\intcc{0}{2\pi}$: $\int\sin^2 2t = \pi$, $\int \sin^2
2t\cos 2t = \bigl[\frac{\sin^3 2t}{6}\bigr] = 0$. So $\oint
y\,\dd x = -3\cdot\frac\pi8$, and the enclosed area is $\frac{3\pi}8$ (the sign records the counterclockwise orientation).

**13.** For $(\cos t, \sin t)$ with $t$ from $\pi$ to $0$, $x$ increases from $-1$ to $1$ and

$$
\int_\pi^0 \sin t\cdot(-\sin t)\,\dd t
= \int_0^\pi \sin^2 t\,\dd t = \frac\pi2 :
$$

the formula returns the area of the upper half-disc, as it must.

**14.** $x' = R(1 + \cos t) = 2R\cos^2\frac t2$, $y' =
R\sin t = 2R\sin\frac t2\cos\frac t2$, so $\norm{f'} =
2R\cos\frac t2$ (nonnegative on $\intcc{-\pi}{\pi}$). Arc from the bottom: $s(t) = \int_0^t 2R\cos\frac u2\,\dd u =
4R\sin\frac t2$. Then

$$
y = R(1 - \cos t) = 2R\sin^2\frac t2
= 2R\Bigl(\frac{s}{4R}\Bigr)^{\!2} = \frac{s^2}{8R} .
$$

**15.** Energy conservation with $v = \frac{\dd s}{\dd\tau}$ and $y = \frac{s^2}{8R}$, $y_0 = \frac{s_0^2}{8R}$:

$$
\Bigl(\frac{\dd s}{\dd\tau}\Bigr)^{\!2} = 2g(y_0 - y)
= \frac{2g}{8R}\bigl(s_0^2 - s^2\bigr)
= \frac{g}{4R}\bigl(s_0^2 - s^2\bigr).
$$

**16.** Differentiating in $\tau$: $2s's'' =
-\frac{g}{4R}\,2ss'$, so wherever $s' \neq 0$ (hence everywhere by [continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous)), $s'' = -\frac{g}{4R}s$: the harmonic oscillator. By [Theorem 5.10](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#thm-b1-diffeq-homogeneous2), $s(\tau) =
A\cos\omega\tau + B\sin\omega\tau$ with $\omega =
\sqrt{g/(4R)}$; the initial conditions $s(0) = s_0$, $s'(0) = 0$ give $s(\tau) = s_0\cos\omega\tau$.

**17.** The bead reaches the bottom when $s = 0$, i.e. at $\omega\tau = \frac\pi2$:

$$
T_{\downarrow} = \frac{\pi}{2\omega}
= \frac\pi2\sqrt{\frac{4R}{g}} = \pi\sqrt{\frac Rg}\,,
$$

independent of $s_0$: released from anywhere in the bowl, all beads arrive at the same instant — the tautochrone.

**18.** Substituting $s = s_0\cos\omega\tau$: the left side is $s_0^2\omega^2\sin^2\omega\tau$ and the right side is $\frac{g}{4R}s_0^2(1 - \cos^2\omega\tau) =
s_0^2\omega^2\sin^2\omega\tau$: exact equality, so no spurious solution was introduced. For a circular arc, $y$ is not proportional to $s^2$ ($y = \ell(1 - \cos\frac s\ell) =
\frac{s^2}{2\ell} - \frac{s^4}{24\ell^3} + \dots$): the restoring term is only *approximately* [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def), so the period of a circular pendulum drifts with amplitude, while the cycloid’s is rigorously constant.

**19.** $\pi\sqrt{R/g} = 1$ gives $R = \frac{g}{\pi^2}
\approx \frac{9.81}{9.87} \approx 0.994$ m — almost exactly one meter. The full oscillation takes $\frac{2\pi}\omega =
2\pi\sqrt{4R/g}$, which is precisely the small-angle formula $2\pi\sqrt{\ell/g}$ for a pendulum of [length](#def-b1-curves-arclength) $\ell = 4R$: Huygens suspended his pendulum from cycloidal cheeks of exactly that proportion.

**20.** At $t = \frac\pi2$, $R = 1$: $M = \bigl(\frac\pi2 -
1,\ 1\bigr)$, $T = \bigl(\frac\pi2,\ 2\bigr)$, so $T - M = (1,
1)$. And $f'(\frac\pi2) = (1 - 0,\ 1) = (1, 1)$: the chord to the top is the velocity, as promised.

**21.** For the trochoid, $x'(t) = R - d\cos t$ and $y'(t)
= d\sin t$. If $d < R$: $x' \geq R - d > 0$, the point always advances; the velocity never vanishes (its first component is positive): no singular point, a regular wave. If $d > R$: $x'(0)
= R - d < 0 < R + d = x'(\pi)$, so the point moves backwards near the contact instants and forwards elsewhere: the curve crosses itself in loops. A point on the flange of a railway wheel (below the rail head, $d > R$) travels backwards at every turn.

**22.** The chord from the [cusp](#ex-b1-curves-astroid) $(\pi R, 2R)$ to the origin has [length](#def-b1-curves-arclength) $L = R\sqrt{\pi^2 + 4}$ and slope angle $\alpha$ with $\sin\alpha = \frac{2R}{L}$. Sliding from rest with constant acceleration $g\sin\alpha$: $L = \frac12 g\sin\alpha\,T^2$, so

$$
T = \sqrt{\frac{2L}{g\sin\alpha}} = \sqrt{\frac{2L^2}{2Rg}}
= \frac{L}{\sqrt{Rg}} = \sqrt{\pi^2 + 4}\,\sqrt{\frac Rg}
\approx 3.72\sqrt{\frac Rg}\,,
$$

against $\pi\sqrt{R/g} \approx 3.14\sqrt{R/g}$ for the cycloid: the curved path is faster. It is in fact the fastest possible — the brachistochrone — a theorem of the calculus of variations, beyond this volume.

**23.** $\dfrac{\dd y}{\dd x} = \dfrac{y'}{x'} =
\dfrac{\sin t}{1 - \cos t} = \cot\dfrac t2$ (half-angle formulas). Then

$$
\frac{\dd^2y}{\dd x^2}
= \frac{\dd}{\dd t}\Bigl(\cot\frac t2\Bigr)\cdot\frac{1}{x'(t)}
= -\frac{1}{2\sin^2\frac t2}\cdot\frac{1}{2R\sin^2\frac t2}
= -\frac{1}{4R\sin^4\frac t2} < 0 :
$$

concave throughout the arch; as $t \to 0^+$ or $2\pi^-$ the slope $\cot\frac t2 \to \pm\infty$: vertical tangents at the [cusps](#ex-b1-curves-astroid).

**24.** The chord direction is $T - M \parallel
\bigl(\sin\frac t2, \cos\frac t2\bigr)$, which tends to $(0, 1)$ as $t \to 0^{+}$: the tangent at the [cusp](#ex-b1-curves-astroid) is vertical — recovered from pure geometry, with no Taylor expansion.

**25.** (i) Wren’s theorem, $L = 8R$; Roberval’s area, $A =
3\pi R^2$ (Galileo’s conjectured ratio $3$); Huygens’ tautochrone, $T_\downarrow = \pi\sqrt{R/g}$. (ii) Every computation ran on the half-angle factorizations $f' =
2R\sin\frac t2(\sin\frac t2, \cos\frac t2)$ and their bowl analogue — one identity powering tangent, [length](#def-b1-curves-arclength) and clock alike. (iii) The intrinsic relation $y = s^2/8R$ converted the energy equation into $s'' = -\frac{g}{4R}s$, a [linear](https://one-course.com/books/math/3/en/chapter/20-linear-maps#def-b1-linmaps-def) equation with constant coefficients whose solutions are exactly isochronous. (iv) Part I used the differential calculus of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative), Part II–III the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) of [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration), Part IV the differential equations of [Chapter 5](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#ch-b1-diffeq) — the cycloid is this book’s curriculum rolled into one curve.