---
title: "Curves"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 18
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/18-curves
---

# Chapter 18 — Curves

Geometry now turns differential. A curve is a point moving through space; calculus gives us its velocity and acceleration, and geometry asks what is intrinsic — independent of how fast we traverse the trajectory. The answers are *[arc length](#def-b2-curves-length)*, which measures the trajectory itself, and *[curvature](#thm-b2-curves-frenet2d)*, which measures how it bends. In dimension $3$ a second invariant, *[torsion](#thm-b2-curves-frenet3d)*, measures how the curve twists out of its plane. The bookkeeping device for all of this is the moving *Frenet frame*.

## 18.1 Parametrized arcs

**Definition 18.1 (Parametrized arc).**

A *parametrized arc* of class $\mathcal{C}^k$ ($k \geq 1$) is a map $\gamma \colon I \to \R^n$ of class $\mathcal{C}^k$ on an interval $I$. A point $\gamma(t)$ is *regular* if $\gamma'(t) \neq 0$, and the arc is *regular* if all its points are. The line through $\gamma(t)$ directed by $\gamma'(t)$ is the *tangent line* at a regular point.

**Definition 18.2 (Change of parameter).**

A *change of parameter* of class $\mathcal{C}^k$ is a $\mathcal{C}^k$ diffeomorphism $\theta \colon J \to I$ between intervals ($\theta' \neq 0$ everywhere). The arcs $\gamma$ and $\gamma \circ \theta$ are called *equivalent*; a *geometric arc* (or curve) is an equivalence class. Notions invariant under change of parameter — the trajectory, the [tangent line](#def-b2-curves-arc), [arc length](#def-b2-curves-length), [curvature](#thm-b2-curves-frenet2d) — are called *geometric*.

**Remark 18.3.**

The trajectory alone does not determine the [geometric arc](#def-b2-curves-reparam): the parametrizations $t \mapsto (\cos t, \sin t)$ on $[0, 2\pi]$ and on $[0, 4\pi]$ have the same image but travel the circle once and twice. A [geometric arc](#def-b2-curves-reparam) remembers the multiplicity and orientation of the traversal, not its speed.

**Example 18.4 (Speed changes nothing geometric).**

Parametrize the unit circle by

$$
\gamma(t) = (\cos t^2,\ \sin t^2),
\qquad t \in \intcc0{\sqrt{2\pi}} .
$$

The speed $\norm{\gamma'(t)} = 2t$ grows linearly, yet

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

the same [length](#def-b2-curves-length) as at constant speed — as [Theorem 18.7](#thm-b2-curves-lengthinv) promises, via the [change of parameter](#def-b2-curves-reparam) $t \mapsto t^2$. The [tangent line](#def-b2-curves-arc), the [curvature](#thm-b2-curves-frenet2d) computed from [Proposition 18.17](#prop-b2-curves-kappaformula), and every other geometric quantity agree as well; only $t = 0$ deserves a glance, where $\gamma'(0) = 0$ makes *this parametrization* irregular although the trajectory is a perfect circle. Geometric statements tolerate bad parametrizations badly: reparametrize first, conclude second.

**Example 18.5.**

The arc $\gamma(t) = (t^2, t^3)$ is $\mathcal{C}^\infty$ but not regular: $\gamma'(0) = (0, 0)$. Its trajectory, the semicubical parabola $y^2 = x^3$, has a *cusp* at the origin: smoothness of the parametrization does not prevent a geometric singularity where the velocity vanishes. This is why the regularity hypothesis $\gamma' \neq 0$ is not cosmetic.

## 18.2 Arc length

**Definition 18.6 (Arc length).**

Let $\gamma \colon [a, b] \to \R^n$ be a $\mathcal{C}^1$ arc. Its *length* is

$$
L(\gamma) = \int_a^b \norm{\gamma'(t)}\, \dd t ,
$$

where $\norm{\cdot}$ is the Euclidean [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm). The *arc-length function* based at $t_0$ is $s(t) = \int_{t_0}^t
\norm{\gamma'(u)}\,\dd u$.

**Theorem 18.7 (Length is geometric; polygonal characterization).**

1. If $\theta \colon [c, d] \to [a, b]$ is a $\mathcal{C}^1$ [change of parameter](#def-b2-curves-reparam) , then $L(\gamma \circ \theta) = L(\gamma)$ .
2. $L(\gamma)$ is the supremum of the [lengths](#def-b2-curves-length) of inscribed polygons: $$L(\gamma) = \sup\Bigl\{\, \sum_{i=1}^{m} \norm{\gamma(t_i) - \gamma(t_{i-1})} \;:\; a = t_0 < t_1 < \dots < t_m = b \,\Bigr\}.$$

**Proof.** *1.* By the change of variables $t = \theta(u)$ (Year 1 volume, valid since $\theta$ is $\mathcal{C}^1$ monotone),

$$
\int_c^d \norm{(\gamma\circ\theta)'(u)}\,\dd u
= \int_c^d \norm{\gamma'(\theta(u))}\,\abs{\theta'(u)}\,\dd u
= \int_a^b \norm{\gamma'(t)}\,\dd t ,
$$

where we used $(\gamma\circ\theta)' = \theta'\cdot
(\gamma'\circ\theta)$ and, if $\theta$ is decreasing, the sign of $\theta'$ is absorbed by the reversal of the bounds.

*2.* For any subdivision, $\gamma(t_i) - \gamma(t_{i-1}) =
\int_{t_{i-1}}^{t_i}\gamma'(t)\,\dd t$, so by the triangle inequality for integrals $\norm{\gamma(t_i) - \gamma(t_{i-1})}
\leq \int_{t_{i-1}}^{t_i}\norm{\gamma'}$: every polygon is shorter than $L(\gamma)$, hence $\sup \leq L(\gamma)$.

For the reverse inequality, let $\varepsilon > 0$. Since $\gamma'$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on the [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) $[a,b]$, it is uniformly continuous: there is $\delta > 0$ with $\norm{\gamma'(t) - \gamma'(u)} \leq
\varepsilon$ whenever $\abs{t - u} \leq \delta$. Take a subdivision of step $\leq \delta$. On each piece, for $t \in [t_{i-1}, t_i]$,

$$
\gamma(t_i) - \gamma(t_{i-1})
= \int_{t_{i-1}}^{t_i}\gamma'(t)\,\dd t
= (t_i - t_{i-1})\,\gamma'(t_{i-1}) + R_i,
\qquad
\norm{R_i} \leq \varepsilon\,(t_i - t_{i-1}),
$$

since $R_i = \int_{t_{i-1}}^{t_i}(\gamma'(t) -
\gamma'(t_{i-1}))\,\dd t$. Hence

$$
\norm{\gamma(t_i) - \gamma(t_{i-1})}
\geq (t_i - t_{i-1})\norm{\gamma'(t_{i-1})}
- \varepsilon (t_i - t_{i-1}) .
$$

Summing, and comparing $\sum (t_i -
t_{i-1})\norm{\gamma'(t_{i-1})}$ with $\int_a^b\norm{\gamma'}$ (a Riemann sum of the [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) function $\norm{\gamma'}$, within $\varepsilon(b - a)$ of the integral for $\delta$ small enough by uniform [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) again), we get a polygon of [length](#def-b2-curves-length) $\geq
L(\gamma) - 2\varepsilon(b - a)$. Letting $\varepsilon \to 0$ proves the claim. ∎

**Example 18.8 (Archimedes and the inscribed polygons).**

For the unit circle, the inscribed regular $n$-gon has [length](#def-b2-curves-length) $L_n = 2n\sin\frac\pi n$, and the expansion $\sin x = x -
\frac{x^3}6 + O(x^5)$ gives

$$
L_n = 2\pi - \frac{\pi^3}{3n^2} + O\Bigl(\frac1{n^4}\Bigr) :
$$

the polygonal [lengths](#def-b2-curves-length) of [Theorem 18.7](#thm-b2-curves-lengthinv) converge *quadratically*. Numerically: $L_6 = 6$ (the hexagon, giving the crude $\pi > 3$), while $L_{96} =
192\sin\frac{\pi}{96} \approx 6.28206$ against $2\pi \approx
6.28319$ — the error $0.00113$ agrees with the predicted $\pi^3/(3\cdot96^2) \approx 0.00112$. This is why Archimedes, doubling the hexagon five times to $96$ sides, could bracket $\pi$ to three digits by hand: each doubling divides the error by four. The supremum in the polygonal characterization is not just attained in the limit; it is attained *fast*, because a smooth curve separates from its chords only at second order.

**Theorem 18.9 (Arc-length parametrization).**

Let $\gamma \colon I \to \R^n$ be a regular $\mathcal{C}^k$ arc ($k \geq 1$). The [arc-length function](#def-b2-curves-length) $s$ is a $\mathcal{C}^k$ diffeomorphism from $I$ onto an interval $J$, and $\tilde\gamma =
\gamma \circ s^{-1}$ satisfies $\norm{\tilde\gamma'} = 1$ everywhere. Up to translation of the parameter and orientation, this *[arc-length](#def-b2-curves-length)* (or *unit-speed*) parametrization is unique.

**Proof.** $s'(t) = \norm{\gamma'(t)} > 0$ and $s'$ is $\mathcal{C}^{k-1}$ (composition of the $\mathcal{C}^{k-1}$ map $\gamma'$ with the [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm), smooth away from $0$), so $s$ is $\mathcal{C}^k$, strictly increasing, a bijection onto $J = s(I)$, and its inverse is $\mathcal{C}^k$ by the inverse function theorem in dimension $1$ (Year 1 volume). Then

$$
\tilde\gamma'(\sigma)
= \frac{\gamma'(t)}{s'(t)}
= \frac{\gamma'(t)}{\norm{\gamma'(t)}},
\qquad t = s^{-1}(\sigma),
$$

a unit vector. If $\hat\gamma = \gamma\circ\theta$ is another unit-speed parametrization, then $\abs{\theta'} = 1$, so $\theta'
= \pm 1$ constant ([continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)), i.e. $\theta(u) = \pm u + c$. ∎

**Remark 18.10.**

[Arc length](#def-b2-curves-length) is the parameter that separates geometry from dynamics. A trajectory can be traversed with any speed profile — the physics of the motion — but every parametrization-invariant question (shape, bending, osculation) has a canonical clock, the distance travelled. This is why all [curvature](#thm-b2-curves-frenet2d) formulas below are *defined* at unit speed and then *translated* to arbitrary parametrizations by the chain rule: the translation factors are powers of $v = s'$, and tracking them correctly is the entire content of [Proposition 18.17](#prop-b2-curves-kappaformula).

**Example 18.11 (Circle and helix).**

For the circle $\gamma(t) = (R\cos t, R\sin t)$, $\norm{\gamma'} =
R$, so $s = Rt$ and the [length](#def-b2-curves-length) of a full turn is $2\pi R$. For the helix $\gamma(t) = (a\cos t,\ a\sin t,\ bt)$ with $a > 0$, $\norm{\gamma'(t)} = \sqrt{a^2 + b^2}$ is constant: the helix is traversed at constant speed, and $s = t\sqrt{a^2 + b^2}$.

**Example 18.12 (Arc length in polar coordinates).**

A polar curve $r = r(\theta)$ is the arc $\gamma(\theta) =
(r\cos\theta,\ r\sin\theta)$, with

$$
\gamma'(\theta) = (r'\cos\theta - r\sin\theta,\
r'\sin\theta + r\cos\theta),
\qquad
\norm{\gamma'(\theta)}^2 = r'^2 + r^2
$$

(the cross terms cancel): the polar [length](#def-b2-curves-length) element is $\sqrt{r^2 + r'^2}\,\dd\theta$. For the cardioid $r = 1 +
\cos\theta$:

$$
r^2 + r'^2 = (1 + \cos\theta)^2 + \sin^2\theta = 2 +
2\cos\theta = 4\cos^2\tfrac\theta2,
$$

and on $\intcc{-\pi}{\pi}$ the half-angle cosine is nonnegative, so

$$
L = \int_{-\pi}^{\pi}2\cos\tfrac\theta2\,\dd\theta
= \Bigl[4\sin\tfrac\theta2\Bigr]_{-\pi}^{\pi}
= 4 - (-4) = 8 :
$$

like the cycloid arch of [Exercise 18.1](#exo-b2-curves-1), a curve built from circles has a rational [length](#def-b2-curves-length), with no $\pi$ anywhere. The half-angle factorization is the standard trick for [lengths](#def-b2-curves-length) of circle-generated curves; when it fails (the ellipse), the [length](#def-b2-curves-length) is a genuinely new function — an elliptic integral, beyond elementary closed forms.

## 18.3 Curvature in the plane

Throughout this section, arcs are $\mathcal{C}^2$ and regular in the oriented Euclidean plane. We parametrize by [arc length](#def-b2-curves-length) and write $T(s) = \tilde\gamma'(s)$ for the unit tangent, and $N(s)$ for the unit vector directly orthogonal to $T(s)$ (rotation of $T$ by $+\pi/2$).

**Theorem 18.13 (Plane Frenet formulas).**

Let $\tilde\gamma$ be a unit-speed $\mathcal{C}^2$ arc in the oriented plane. There is a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) function $\kappa$, the (algebraic) *curvature*, such that

$$
T'(s) = \kappa(s)\, N(s),
\qquad
N'(s) = -\kappa(s)\, T(s).
$$

**Proof.** Since $\norm{T(s)}^2 = 1$ for all $s$, differentiating the scalar product gives $2\langle T'(s), T(s)\rangle = 0$: $T'(s)$ is orthogonal to $T(s)$, hence collinear with $N(s)$ (dimension $2$); write $T'(s)
= \kappa(s) N(s)$ with $\kappa(s) = \langle T'(s), N(s)\rangle$, [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity). Likewise $N' \perp N$, so $N' = \lambda T$; and differentiating $\langle T, N\rangle = 0$ gives $\langle T', N\rangle +
\langle T, N'\rangle = \kappa + \lambda = 0$. ∎

**Definition 18.14.**

When $\kappa(s) \neq 0$, the *radius of curvature* is $R(s) =
1/\abs{\kappa(s)}$ and the *center of curvature* is $\tilde\gamma(s) + \frac{1}{\kappa(s)} N(s)$; the circle with that center and radius $R(s)$ is the *osculating circle*, the best circular approximation of the curve at $\tilde\gamma(s)$.

**Example 18.15 (The osculating circle of the exponential).**

For $y = \eu^x$ at the point $(0, 1)$: $f'(0) = f''(0) = 1$, so by the graph formula below,

$$
\kappa(0) = \frac{1}{(1 + 1)^{3/2}} = \frac1{2\sqrt2},
\qquad R = 2\sqrt2 .
$$

The unit tangent is $T = \frac{(1, 1)}{\sqrt2}$, the direct normal $N = \frac{(-1, 1)}{\sqrt2}$, and the [center of curvature](#def-b2-curves-curvature) is

$$
(0, 1) + 2\sqrt2\cdot\frac{(-1, 1)}{\sqrt2} = (-2,\ 3) :
$$

the [osculating circle](#def-b2-curves-curvature) has equation $(x + 2)^2 + (y - 3)^2 =
8$. As a check of the “best circular approximation” claim: solving the circle’s equation for $y$ near $(0,1)$ and expanding gives $y = 1 + x + \frac{x^2}2 + O(x^3)$ — exactly the second-order Taylor expansion of $\eu^x$. The [osculating circle](#def-b2-curves-curvature) matches value, slope *and* second derivative; an ordinary tangent circle would match only the first two.

**Example 18.16 (The evolute of a circle is its center).**

For the circle of radius $R$ traversed counterclockwise, $\kappa = 1/R$ and $N$ points toward the center, so the [center of curvature](#def-b2-curves-curvature) $\tilde\gamma + \frac1\kappa N$ is the center of the circle, for every $s$: the [osculating circle](#def-b2-curves-curvature) of a circle is the circle itself, and the locus of centers of [curvature](#thm-b2-curves-frenet2d) collapses to a point. This degenerate case calibrates [Exercise 18.6](#exo-b2-curves-6): there the evolute’s velocity is $-\frac{\kappa'}{\kappa^2}N$, which vanishes identically precisely when $\kappa$ is constant.

**Proposition 18.17 (Curvature in an arbitrary parametrization).**

For a regular $\mathcal{C}^2$ plane arc $\gamma(t) = (x(t), y(t))$,

$$
\kappa(t)
= \frac{x'(t)\,y''(t) - y'(t)\,x''(t)}
       {\bigl(x'(t)^2 + y'(t)^2\bigr)^{3/2}} ,
$$

in particular $\kappa = \dfrac{y''}{(1 + y'^2)^{3/2}}$ for a graph $y = f(x)$.

**Proof.** Write $v(t) = \norm{\gamma'(t)} = s'(t)$, so $\gamma' = vT$ (composing the unit-speed data with $s$). Differentiating,

$$
\gamma'' = v'T + v\,T'\cdot s' = v'T + v^2\kappa N .
$$

Now take the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) (in the canonical oriented basis) of $(\gamma', \gamma'')$: since $\det(T, T) = 0$ and $\det(T, N) = 1$,

$$
\det(\gamma', \gamma'') = \det(vT,\ v'T + v^2\kappa N)
= v^3\kappa .
$$

The left side is $x'y'' - y'x''$, and $v^3 = (x'^2 +
y'^2)^{3/2}$. The graph case is the parametrization $t \mapsto (t,
f(t))$. ∎

**Example 18.18 (Circle, line, parabola).**

A line has $\kappa = 0$ (and conversely: $T' = 0$ means $T$ constant, so $\tilde\gamma(s) = \tilde\gamma(0) + sT$, a line). The circle of radius $R$ traversed counterclockwise has $\kappa = 1/R$: with $\gamma(t) = (R\cos t, R\sin t)$, the formula gives $\kappa =
R^2/R^3$. For the parabola $y = x^2/2$: $\kappa(x) = 1/(1 +
x^2)^{3/2}$, maximal at the vertex — the parabola is most sharply bent where it turns around.

**Remark 18.19 (Common pitfalls around curvature).**

(i) The algebraic [curvature](#thm-b2-curves-frenet2d) of a plane arc changes sign when the orientation of the arc or of the plane is reversed: only $\abs\kappa$ and $R = 1/\abs\kappa$ are purely geometric. A circle traversed clockwise has $\kappa = -1/R$. (ii) The graph formula $\kappa = f''/(1 + f'^2)^{3/2}$ silently chooses the parametrization by $x$; applying it to a curve that is not a graph near the point (vertical tangent) is the classical blunder. (iii) At a point where $\gamma' = 0$ nothing is defined — neither $T$ nor $\kappa$ — and the trajectory may genuinely break ([Example 18.5](#ex-b2-curves-cusp)); always check regularity before differentiating the unit tangent. (iv) In space, $\kappa =
\norm{T'} \geq 0$ by convention: there is no sign to get wrong, but also no sign to exploit — inflection-type information moves into the [torsion](#thm-b2-curves-frenet3d). (v) Finally, $\kappa$ is a derivative *with respect to [arc length](#def-b2-curves-length)*: for a non-unit-speed parametrization, forgetting the factor $v^3$ in [Proposition 18.17](#prop-b2-curves-kappaformula) is the most frequent error in practice.

![The parabola y = x2/2, its moving Frenet frame (T, N), and the osculating circle at the vertex (radius 1, since (0) = 1). The frame turns as the point moves; curvature is the rate of that turning per unit of arc length.](https://one-course.com/images/onecourse/chapters/math-4/b2-curves/fig-3bd00bac0056.svg)

***Figure 18.1.** The parabola $y = x^2/2$, its moving Frenet frame $(T,
N)$, and the [osculating circle](#def-b2-curves-curvature) at the vertex (radius $1$, since $\kappa(0) = 1$). The frame turns as the point moves; [curvature](#thm-b2-curves-frenet2d) is the rate of that turning per unit of [arc length](#def-b2-curves-length).*

**Theorem 18.20 (Curvature determines the curve).**

Let $\kappa \colon J \to \R$ be [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity). There exists a unit-speed $\mathcal{C}^2$ arc in the plane with [curvature](#thm-b2-curves-frenet2d) $\kappa$, and it is unique up to a direct isometry (rotation followed by translation).

**Proof.** *Existence.* Fix $s_0 \in J$ and set $\varphi(s) =
\int_{s_0}^s \kappa(u)\,\dd u$, then

$$
\tilde\gamma(s) = \Bigl(\int_{s_0}^s \cos\varphi(u)\,\dd u,\
\int_{s_0}^s \sin\varphi(u)\,\dd u\Bigr).
$$

Then $T(s) = (\cos\varphi(s), \sin\varphi(s))$ is a unit vector, $N(s) = (-\sin\varphi, \cos\varphi)$, and

$$
T'(s) = \varphi'(s)\,(-\sin\varphi, \cos\varphi)
= \kappa(s)\,N(s) :
$$

the arc is unit-speed with [curvature](#thm-b2-curves-frenet2d) $\kappa$.

*Uniqueness.* Let $\gamma_1, \gamma_2$ be unit-speed arcs with the same [curvature](#thm-b2-curves-frenet2d). Each unit tangent lifts to an angle function, by an explicit construction: view $T_j$ as the complex number $z_j = a_j + \iu b_j$ of modulus $1$, pick $\varphi_j(0)$ with $z_j(0) = \eu^{\iu\varphi_j(0)}$, and set

$$
\varphi_j(s) = \varphi_j(0) + \int_0^s\det\bigl(T_j,
T_j'\bigr)(u)\,\dd u .
$$

From $\abs{z_j} = 1$: $\operatorname{Re}(\conj{z_j}\,z_j') =
0$, so $\conj{z_j}\,z_j' = \iu\det(T_j, T_j') =
\iu\,\varphi_j'$, i.e. $z_j' = \iu\varphi_j'z_j$; then

$$
\bigl(z_j\,\eu^{-\iu\varphi_j}\bigr)'
= \eu^{-\iu\varphi_j}\bigl(z_j' - \iu\varphi_j'z_j\bigr) = 0,
$$

so $z_j = \eu^{\iu\varphi_j}$ throughout: $T_j =
(\cos\varphi_j, \sin\varphi_j)$ with $\varphi_j$ of class $\mathcal C^1$. Moreover $\det(T_j, T_j') = \det(T_j,
\kappa N_j) = \kappa$, so $\varphi_j' = \kappa$. Hence $\varphi_2 = \varphi_1 + c$ for a constant $c$: $T_2$ is $T_1$ rotated by the fixed angle $c$, so integrating, $\gamma_2 = \rho(\gamma_1) + w$ where $\rho$ is the rotation of angle $c$ and $w$ a constant vector. ∎

**Remark 18.21.**

This is the one-dimensional prototype of a *fundamental theorem of geometry*: a [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) set of local invariants (here, one function) classifies the object up to rigid motion. The three-dimensional version below needs two invariants.

**Example 18.22 (Constant curvature means circle).**

Take $\kappa \equiv \kappa_0 > 0$ in the existence formula: $\varphi(s) = \kappa_0 s$ and

$$
\tilde\gamma(s) = \Bigl(\frac{\sin\kappa_0s}{\kappa_0},\
\frac{1 - \cos\kappa_0s}{\kappa_0}\Bigr) :
$$

the circle of radius $1/\kappa_0$ centered at $(0,
1/\kappa_0)$, traversed at unit speed. By the uniqueness half of the theorem, *every* unit-speed arc of constant [curvature](#thm-b2-curves-frenet2d) $\kappa_0$ is a piece of a circle of radius $1/\kappa_0$ (or a line if $\kappa_0 = 0$) — the converse of the computation in [Example 18.18](#ex-b2-curves-kappaexamples), and the plane case of [Exercise 18.9](#exo-b2-curves-9).

**Example 18.23 (Reconstructing a curve from its curvature).**

Which unit-speed curve has [radius of curvature](#def-b2-curves-curvature) $R(s) = 1 +
s^2$? Following the existence proof with $\kappa(s) =
\frac1{1+s^2}$ and $s_0 = 0$: $\varphi(s) = \arctan s$, so

$$
T(s) = (\cos\arctan s,\ \sin\arctan s)
= \Bigl(\frac{1}{\sqrt{1+s^2}},\
\frac{s}{\sqrt{1+s^2}}\Bigr),
$$

and integrating,

$$
\tilde\gamma(s) = \Bigl(\ln\bigl(s + \sqrt{1 + s^2}\bigr),\
\sqrt{1 + s^2} - 1\Bigr).
$$

Setting $x = \ln(s + \sqrt{1+s^2})$, i.e. $s = \sinh x$, the second coordinate is $\cosh x - 1$: the curve is the *catenary* $y = \cosh x - 1$. This closes the loop with [Exercise 18.3](#exo-b2-curves-3), where we computed $R = \cosh^2 x = 1 +
\sinh^2 x = 1 + s^2$ directly: the fundamental theorem guarantees the catenary is the *only* curve with this [curvature](#thm-b2-curves-frenet2d) profile, up to a direct isometry.

**Remark 18.24 (Where curvature is used next).**

The decomposition $\gamma'' = v'T + v^2\kappa N$ obtained in the proof of [Proposition 18.17](#prop-b2-curves-kappaformula) is the kinematics of every curved motion: tangential versus centripetal acceleration. [Curvature](#thm-b2-curves-frenet2d) returns for surfaces ([Chapter 19](https://one-course.com/books/math/4/en/chapter/19-surfaces#ch-b2-surfaces)) through the [curvature](#thm-b2-curves-frenet2d) of curves drawn on them, and the [envelope](#pb-b2-curves-1) calculus of this chapter’s weekend problem — evolutes, caustics — is the geometric optics of wavefronts. The Year 3 volume takes the intrinsic point of view up again for submanifolds of $\R^n$.

## 18.4 Frenet frame in space

Now let $\tilde\gamma \colon J \to \R^3$ be a unit-speed $\mathcal{C}^3$ arc that is *biregular*: $T'(s) \neq 0$ for all $s$. Then $\kappa(s) = \norm{T'(s)} > 0$ defines the *[curvature](#thm-b2-curves-frenet2d)* (no sign in space: there is no preferred normal orientation), and we set:

$$
N(s) = \frac{T'(s)}{\kappa(s)}
\quad\text{(principal normal)},
\qquad
B(s) = T(s) \wedge N(s)
\quad\text{(binormal)} ,
$$

so that $(T, N, B)$ is a direct orthonormal frame, the *Frenet frame*. The plane through $\tilde\gamma(s)$ spanned by $T, N$ is the *osculating plane*.

**Theorem 18.25 (Frenet formulas in space).**

There is a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) function $\tau$, the *torsion*, with

$$
T' = \kappa N, \qquad
N' = -\kappa T + \tau B, \qquad
B' = -\tau N .
$$

**Proof.** The first formula is the definition of $N$. Each of the vectors $T, N, B$ has constant [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $1$ and they are pairwise orthogonal; differentiating the six relations $\langle X, Y\rangle = \delta_{XY}$ shows the matrix of $(T', N', B')$ in the basis $(T, N, B)$ is antisymmetric: indeed $\langle X', Y\rangle + \langle X, Y'\rangle = 0$ and $\langle X', X\rangle = 0$. Its $(N, T)$ entry is $\langle N', T\rangle =
-\langle N, T'\rangle = -\kappa$, and its $(T, B)$-column entry $\langle T', B\rangle = \kappa\langle N, B\rangle = 0$. Naming the remaining free entry $\tau = \langle N', B\rangle$ gives exactly the three displayed formulas: antisymmetry fills in $\langle B', N\rangle = -\tau$ and $\langle B', T\rangle = 0$. [Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $\tau = \langle N', B\rangle$ is clear since $N'$ and $B$ are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ($\tilde\gamma$ is $\mathcal{C}^3$, so $N = T'/\kappa$ is $\mathcal{C}^1$). ∎

**Example 18.26 (The Darboux vector).**

The three Frenet formulas compress into one. Set $\omega(s) =
\tau\,T + \kappa\,B$ (the *Darboux vector*). Using $B
\wedge T = N$, $T \wedge N = B$, $N \wedge B = T$:

$$
\omega \wedge T = \kappa\,N = T', \qquad
\omega \wedge N = \tau\,B - \kappa\,T = N', \qquad
\omega \wedge B = -\tau\,N = B' :
$$

each frame vector evolves by $X' = \omega \wedge X$, the kinematic signature of an *instantaneous rotation* with angular velocity vector $\omega$. The frame spins at rate $\norm\omega = \sqrt{\kappa^2 + \tau^2}$ about the moving axis $\omega$; [curvature](#thm-b2-curves-frenet2d) is the component of the spin about the binormal, [torsion](#thm-b2-curves-frenet3d) the component about the tangent. For the helix, $\omega$ is a constant vector along the cylinder’s axis — which is exactly why the helix’s frame precesses evenly. The antisymmetry of the Frenet matrix, exploited in [Exercise 18.7](#exo-b2-curves-7), is the matrix form of this single geometric fact.

**Proposition 18.27 (Torsion measures planarity).**

A biregular arc is contained in a plane if and only if $\tau
\equiv 0$; in that case the plane is the (constant) osculating plane.

**Proof.** If $\tau \equiv 0$, then $B' = 0$, so $B$ is a constant unit vector $B_0$, and

$$
\frac{\dd}{\dd s}\langle \tilde\gamma(s), B_0\rangle
= \langle T(s), B_0\rangle = 0 :
$$

$\langle \tilde\gamma, B_0\rangle$ is constant, so the arc lies in a plane orthogonal to $B_0$. Conversely, if the arc lies in a plane $P$, then $T$ and $T'$ (hence $N$) are parallel to the direction of $P$ for all $s$; so $B = T \wedge N$ is one of the two unit normals of $P$, and being [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) it is constant; then $0 = B'
= -\tau N$ with $N \neq 0$ forces $\tau \equiv 0$. ∎

**Example 18.28 (A tilted circle has zero torsion).**

The arc $\gamma(t) = \bigl(\cos t,\ \tfrac{\sin t}{\sqrt2},\
\tfrac{\sin t}{\sqrt2}\bigr)$ lies in the plane $y = z$, and is the unit circle of that plane (check: $\norm{\gamma(t)} =
1$ and the plane’s orthonormal basis $(1,0,0)$, $(0, \tfrac1{\sqrt2}, \tfrac1{\sqrt2})$ exhibits the standard parametrization). Without any Frenet computation, [Proposition 18.27](#prop-b2-curves-torsion) predicts $\tau \equiv 0$, and the fixed binormal must be the plane’s unit normal $\pm(0,
\tfrac1{\sqrt2}, -\tfrac1{\sqrt2})$. [Torsion](#thm-b2-curves-frenet3d) does not measure being “tilted in space”; it measures *leaving* a plane. Only the helix’s nonzero $b$ below produces genuine [torsion](#thm-b2-curves-frenet3d).

**Example 18.29 (The helix).**

For the helix $\gamma(t) = (a\cos t, a\sin t, bt)$, $a > 0$, we computed $s = ct$ with $c = \sqrt{a^2 + b^2}$. Then

$$
T = \frac1c(-a\sin t,\ a\cos t,\ b),
\qquad
T' \cdot \frac{\dd t}{\dd s}
= \frac{1}{c^2}(-a\cos t, -a\sin t, 0),
$$

so $\kappa = a/c^2 = a/(a^2 + b^2)$ and $N = (-\cos t, -\sin t,
0)$: the principal normal points horizontally toward the axis. Next $B = T \wedge N = \frac1c(b\sin t, -b\cos t, a)$, and $B'
\frac{\dd t}{\dd s} = \frac{b}{c^2}(\cos t, \sin t, 0) = -\tau N$ gives

$$
\boxed{\ \kappa = \frac{a}{a^2 + b^2}, \qquad
\tau = \frac{b}{a^2 + b^2}. \ }
$$

Both invariants are constant — and one can show, conversely, that the only biregular curves with constant $\kappa > 0$ and constant $\tau$ are helices (circles when $\tau = 0$). Note the signs: $b >
0$ gives a right-handed helix with positive [torsion](#thm-b2-curves-frenet3d).

**Example 18.30 (Curvature and torsion without arc length; the twisted cubic).**

Reparametrizing by [arc length](#def-b2-curves-length) is usually impossible in closed form, so the invariants must be extracted from the raw derivatives. Write $v = \norm{\gamma'} = s'$; then $\gamma' =
vT$ and, as in the proof of [Proposition 18.17](#prop-b2-curves-kappaformula),

$$
\gamma'' = v'T + v^2\kappa N,
\qquad
\gamma' \wedge \gamma'' = v^3\kappa\,(T \wedge N) =
v^3\kappa\,B .
$$

Taking [norms](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) ($\kappa \geq 0$ in space):

$$
\kappa = \frac{\norm{\gamma' \wedge \gamma''}}{v^3} .
$$

Differentiating $\gamma''$ once more and converting $N' =
v(-\kappa T + \tau B)$ (chain rule through $s$), the only $B$-component comes from the last term:

$$
\gamma''' = \bigl(v'' - v^3\kappa^2\bigr)T +
\bigl(v'v\kappa + (v^2\kappa)'\bigr)N + v^3\kappa\tau\,B,
$$

so that, pairing with $\gamma' \wedge \gamma'' = v^3\kappa B$,

$$
\det(\gamma', \gamma'', \gamma''') = \langle\gamma' \wedge
\gamma'',\ \gamma'''\rangle = v^6\kappa^2\tau,
\qquad\text{i.e.}\qquad
\tau = \frac{\det(\gamma', \gamma'',
\gamma''')}{\norm{\gamma' \wedge \gamma''}^2} .
$$

Application to the *twisted cubic* $\gamma(t) = (t,\
t^2,\ t^3)$ at $t = 0$: $\gamma' = (1, 0, 0)$, $\gamma'' =
(0, 2, 0)$, $\gamma''' = (0, 0, 6)$, so $v = 1$,

$$
\gamma' \wedge \gamma'' = (0, 0, 2), \qquad \kappa(0) = 2,
\qquad
\det(\gamma', \gamma'', \gamma''') = 12, \qquad \tau(0) =
\frac{12}{4} = 3 .
$$

Closing insight: both formulas are ratios in which the speed $v$ cancels to the exact degree needed — $\kappa$ scales like a second derivative per unit [length](#def-b2-curves-length), $\tau$ like the mixed volume of three derivatives per squared area — which is *why* they are geometric while $\gamma''$ itself is not.

**Remark 18.31 (Fundamental theorem for space curves).**

As in the plane, the pair $(\kappa, \tau)$ with $\kappa > 0$ determines a biregular arc up to direct isometry of $\R^3$: the Frenet formulas form a linear differential system for the frame $(T, N, B)$, to which the Cauchy–Lipschitz theory of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq) applies; orthonormality of the solution frame is preserved because the coefficient matrix is antisymmetric (same Gram-matrix argument as in [Exercise 18.7](#exo-b2-curves-7)), and the curve is recovered by integrating $T$. We leave the details to the reader as a substantial but instructive exercise.

## 18.5 Local study: position with respect to the tangent

**Proposition 18.32 (Local shape at a regular point).**

Let $\gamma$ be a plane arc of class $\mathcal{C}^k$ at $t_0$, with $p$ the smallest index with $\gamma^{(p)}(t_0) \neq 0$ and $q$ the smallest index $> p$ with $\gamma^{(q)}(t_0)$ not collinear with $\gamma^{(p)}(t_0)$ (assuming both exist, $q \leq k$). In the basis $(u, v) = (\gamma^{(p)}(t_0), \gamma^{(q)}(t_0))$ centered at $\gamma(t_0)$, Taylor–Young gives coordinates

$$
X(t) \sim \frac{(t - t_0)^p}{p!},
\qquad
Y(t) \sim \frac{(t - t_0)^q}{q!} .
$$

The local picture depends only on the parities of $p$ and $q$:

| $p$ odd, $q$ even | *ordinary point* | curve crosses no line, stays on one side of tangent |
| --- | --- | --- |
| $p$ odd, $q$ odd | *inflection point* | curve crosses its tangent |
| $p$ even, $q$ odd | *cusp of the first kind* | both branches on opposite sides of tangent |
| $p$ even, $q$ even | *cusp of the second kind* | both branches on the same side |

**Proof.** Taylor–Young at order $q$ (the function $\gamma$ is $\mathcal{C}^q$ near $t_0$):

$$
\gamma(t) - \gamma(t_0)
= \sum_{j=p}^{q} \frac{(t-t_0)^j}{j!}\,\gamma^{(j)}(t_0)
+ o\bigl((t-t_0)^q\bigr).
$$

By the choice of $p$ and $q$, each $\gamma^{(j)}(t_0)$ with $p
\leq j < q$ is collinear with $u$; collecting components in the basis $(u, v)$: $X(t) = \frac{(t-t_0)^p}{p!}(1 + o(1))$ and $Y(t) = \frac{(t-t_0)^q}{q!}(1 + o(1))$. The sign table of $X$ and $Y$ for $t \gtrless t_0$ — governed exactly by the parities — gives the four pictures: for instance if $p$ is even, $X > 0$ on both sides (both branches leave in the direction $+u$: a cusp), and the side of the [tangent line](#def-b2-curves-arc) ($\operatorname{sign} Y$) flips with $q$ odd. ∎

**Example 18.33.**

For $\gamma(t) = (t^2, t^3)$ at $t_0 = 0$ ([Example 18.5](#ex-b2-curves-cusp)): $\gamma'' (0)= (2, 0)$, $\gamma'''(0) =
(0, 6)$, so $p = 2$, $q = 3$: a cusp of the first kind, the familiar picture of the semicubical parabola. For $\gamma(t) = (t,
t^3)$ at $0$: $p = 1$, $q = 3$: inflection — the cubic crosses its tangent.

**Remark 18.34 (Perspectives within this volume).**

Curves feed the next chapters in three ways. Drawn on a surface, they define its tangent planes and its first fundamental form ([Chapter 19](https://one-course.com/books/math/4/en/chapter/19-surfaces#ch-b2-surfaces)), and their [lengths](#def-b2-curves-length) are computed by restricting the ambient metric — the chapter ahead is largely this chapter relativized. The [envelope](#pb-b2-curves-1) calculus of the weekend problem meets double integrals in [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint), where the [astroid](#pb-b2-curves-1)’s area is recomputed by Green’s formula ([Exercise 20.5](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#exo-b2-multint-5)) — one curve, two theories, matching answers. And the Frenet system already used the linear differential equations of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq) (existence, uniqueness, and the orthogonality-preservation argument of [Exercise 18.7](#exo-b2-curves-7)): the fundamental theorem of curves is a differential equations theorem wearing geometric clothes.

**Remark 18.35 (Method: running the local study).**

In practice the classification is a four-step routine. *One*, differentiate at $t_0$ until the first nonzero derivative appears: its index is $p$, its value the vector $u$. *Two*, keep differentiating until a derivative not collinear with $u$ appears: index $q$, vector $v$. *Three*, read the parities $(p, q)$ in the table. *Four*, draw: the curve leaves along $+u$ if $p$ is odd (along $u$ then back along $u$ if $p$ is even), on the side of $v$ dictated by the sign of $Y$. Two cautions. The frame $(u, v)$ is generally *not* orthonormal — the table describes positions relative to the [tangent line](#def-b2-curves-arc), not angles or distances, so do not read [curvature](#thm-b2-curves-frenet2d) off the picture. And intermediate derivatives collinear with $u$ are allowed between ranks $p$ and $q$ (they only shift the expansion of $X$); what must *not* happen is stopping at the first nonzero derivative and guessing $q = p + 1$: for $\gamma(t)
= (t^2, t^4 + t^5)$ the naive guess $q = 3$ is wrong, as $\gamma^{(3)}(0)$ is still collinear with $\gamma''(0)$ — this is precisely [Exercise 18.5](#exo-b2-curves-5).

## 18.6 Exercises

**Exercise 18.1 ★.**

Compute the [length](#def-b2-curves-length) of one arch of the cycloid $\gamma(t) = (t -
\sin t,\ 1 - \cos t)$, $t \in [0, 2\pi]$. *(Use $1 - \cos t =
2\sin^2(t/2)$.)*

**Solution of Exercise 18.1.**

$\gamma'(t) = (1 - \cos t,\ \sin t)$, so

$$
\norm{\gamma'(t)}^2 = (1 - \cos t)^2 + \sin^2 t
= 2 - 2\cos t = 4\sin^2\tfrac t2 ,
$$

and $\norm{\gamma'(t)} = 2\sin\frac t2$ (nonnegative on $[0,
2\pi]$). Hence

$$
L = \int_0^{2\pi} 2\sin\tfrac t2\,\dd t
= \Bigl[-4\cos\tfrac t2\Bigr]_0^{2\pi} = 8 :
$$

one arch of the cycloid has [length](#def-b2-curves-length) $8$ (for a wheel of radius $1$) — a famous result of Wren, with no $\pi$ in sight.

**Exercise 18.2 ★.**

Compute the [curvature](#thm-b2-curves-frenet2d) of the ellipse $\gamma(t) = (a\cos t,\ b\sin
t)$ ($a > b > 0$) and locate the points of maximal and minimal [curvature](#thm-b2-curves-frenet2d).

**Solution of Exercise 18.2.**

With $x = a\cos t$, $y = b\sin t$: $x' = -a\sin t$, $y' = b\cos
t$, $x'' = -a\cos t$, $y'' = -b\sin t$, so by [Proposition 18.17](#prop-b2-curves-kappaformula)

$$
\kappa(t)
= \frac{x'y'' - y'x''}{(x'^2 + y'^2)^{3/2}}
= \frac{ab\sin^2 t + ab\cos^2 t}
       {(a^2\sin^2 t + b^2\cos^2 t)^{3/2}}
= \frac{ab}{(a^2\sin^2 t + b^2\cos^2 t)^{3/2}} .
$$

The denominator is minimal when $\sin t = 0$ (value $b^3$, points $(\pm a, 0)$) and maximal when $\cos t = 0$ (value $a^3$, points $(0, \pm b)$), since $a > b$. Hence $\kappa$ is maximal at the ends of the major axis, $\kappa_{\max} = a/b^2$, and minimal at the ends of the minor axis, $\kappa_{\min} = b/a^2$: the ellipse bends most sharply at the tips of its long axis.

**Exercise 18.3 ★.**

Show that the [arc length](#def-b2-curves-length) of the graph of $f(x) = \cosh x$ over $[0, x]$ equals $\sinh x$, and compute the [curvature](#thm-b2-curves-frenet2d) of this curve (the *catenary*). Verify that $R(x) = 1/\kappa(x) =
\cosh^2 x$.

**Solution of Exercise 18.3.**

For the graph $\gamma(x) = (x, \cosh x)$: $\norm{\gamma'(x)} =
\sqrt{1 + \sinh^2 x} = \cosh x$, so the [arc length](#def-b2-curves-length) from $0$ to $x$ is $\int_0^x \cosh u\,\dd u = \sinh x$. [Curvature](#thm-b2-curves-frenet2d) of a graph ([Proposition 18.17](#prop-b2-curves-kappaformula)):

$$
\kappa(x) = \frac{f''(x)}{(1 + f'(x)^2)^{3/2}}
= \frac{\cosh x}{\cosh^3 x} = \frac{1}{\cosh^2 x} ,
$$

so $R(x) = \cosh^2 x$, as announced. Note the neat coincidence $R(x) = 1 + s(x)^2$ with $s = \sinh x$ the [arc length](#def-b2-curves-length): the catenary’s [radius of curvature](#def-b2-curves-curvature) grows with the square of the [arc length](#def-b2-curves-length) from the vertex.

**Exercise 18.4 ★★.**

(Logarithmic spiral) Let $\gamma(t) = e^{t}(\cos t,\ \sin t)$, $t
\in \R$. Show that the angle between $\gamma(t)$ and $\gamma'(t)$ is constant, compute the [arc length](#def-b2-curves-length) of $\gamma$ on $(-\infty, 0]$ (finite!), and the [curvature](#thm-b2-curves-frenet2d).

**Solution of Exercise 18.4.**

$\gamma'(t) = e^t(\cos t - \sin t,\ \sin t + \cos t)$, so

$$
\langle \gamma(t), \gamma'(t)\rangle = e^{2t}
\bigl(\cos t(\cos t - \sin t) + \sin t(\sin t + \cos t)\bigr)
= e^{2t},
$$

while $\norm{\gamma(t)} = e^t$ and $\norm{\gamma'(t)} =
e^t\sqrt 2$. Hence

$$
\cos\angle\bigl(\gamma, \gamma'\bigr)
= \frac{e^{2t}}{e^t \cdot e^t\sqrt2} = \frac{1}{\sqrt2} :
$$

the tangent always makes the angle $\pi/4$ with the radius — the *equiangular* property of the logarithmic spiral. [Arc length](#def-b2-curves-length) on $(-\infty, 0]$:

$$
\int_{-\infty}^0 \norm{\gamma'(t)}\,\dd t
= \sqrt2\int_{-\infty}^0 e^t\,\dd t = \sqrt 2 ,
$$

finite although the spiral winds infinitely many times around the origin. [Curvature](#thm-b2-curves-frenet2d): with $x'y'' - y'x''$ computed from $\gamma'' =
e^t(-2\sin t,\ 2\cos t)$,

$$
x'y'' - y'x'' = e^{2t}\bigl(2\cos t(\cos t - \sin t)
+ 2\sin t(\sin t + \cos t)\bigr) = 2e^{2t},
$$

so $\kappa(t) = \dfrac{2e^{2t}}{(e^t\sqrt2)^3} =
\dfrac{1}{e^t\sqrt2}$: the [curvature](#thm-b2-curves-frenet2d) is $1/(\sqrt2\,
\norm{\gamma})$, decaying as the spiral grows.

**Exercise 18.5 ★★.**

Determine $p$, $q$ and the local shape (ordinary, inflection, cusp) of $\gamma(t) = (t^2,\ t^4 + t^5)$ at $t = 0$, and of $\gamma(t) = (t^3,\ t^4)$ at $t = 0$.

**Solution of Exercise 18.5.**

*First arc:* $\gamma(t) = (t^2,\ t^4 + t^5)$. Derivatives at $0$: $\gamma'' = (2, 0) \neq 0$, so $p = 2$. Then $\gamma^{(3)}(0)
= (0, 0)$, $\gamma^{(4)}(0) = (0, 24)$, not collinear with $(2,
0)$: $q = 4$. Both even: *cusp of the second kind* — both branches leave in the direction $+u = (1,0)$ and stay on the same side of the tangent. (Indeed $y = x^2 \pm x^{5/2}$ on the two branches: same sign for small $x$.)

*Second arc:* $\gamma(t) = (t^3, t^4)$. $\gamma'(0) =
\gamma''(0) = 0$, $\gamma^{(3)}(0) = (6, 0)$: $p = 3$, odd. Next $\gamma^{(4)}(0) = (0, 24)$: $q = 4$, even. Odd–even: *ordinary point* — despite the vanishing velocity, the trajectory $y = x^{4/3}$ crosses the origin smoothly, staying above its tangent $y = 0$.

**Exercise 18.6 ★★.**

Let $\gamma$ be a unit-speed plane arc with $\kappa(s) > 0$ for all $s$, and let $c(s) = \gamma(s) + \frac{1}{\kappa(s)}N(s)$ be the [center of curvature](#def-b2-curves-curvature) (the curve $c$ is the *evolute*). Assuming $\kappa$ is $\mathcal{C}^1$, show that $c'(s) =
-\frac{\kappa'(s)}{\kappa(s)^2}N(s)$: the evolute is tangent to the normal lines of $\gamma$.

**Solution of Exercise 18.6.**

Differentiate $c(s) = \gamma(s) + \dfrac{1}{\kappa(s)}N(s)$ using the plane Frenet formulas ([Theorem 18.13](#thm-b2-curves-frenet2d)):

$$
c'(s) = T(s) - \frac{\kappa'(s)}{\kappa(s)^2}N(s)
+ \frac{1}{\kappa(s)}\,\bigl(-\kappa(s)T(s)\bigr)
= -\frac{\kappa'(s)}{\kappa(s)^2}\,N(s) ,
$$

the tangent terms cancelling exactly. So the velocity of the evolute is carried by $N(s)$, which directs the normal line of $\gamma$ at $\gamma(s)$ — and the point $c(s)$ lies on that very normal line: the evolute is the *[envelope](#pb-b2-curves-1) of the normals*. (Where $\kappa' = 0$ the evolute has a singular point; this is what produces the cusps of the evolute of an ellipse.)

**Exercise 18.7 ★★★.**

Let $A(s)$ be a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) family of antisymmetric $3 \times 3$ matrices and $F' = F A$ a matrix solution with $F(s_0)$ orthogonal. Show that $F(s)$ is orthogonal for all $s$. *(Differentiate $G = F F^{\mathsf T}$ and use uniqueness in Cauchy–Lipschitz.)* Explain the relevance to the Frenet system.

**Solution of Exercise 18.7.**

Let $G(s) = F(s)F(s)^{\mathsf T}$. Then, using $F' = FA$ and $(F^{\mathsf T})' = (F')^{\mathsf T} = A^{\mathsf T}F^{\mathsf T}$,

$$
G' = F'F^{\mathsf T} + F(F^{\mathsf T})'
= FAF^{\mathsf T} + FA^{\mathsf T}F^{\mathsf T}
= F(A + A^{\mathsf T})F^{\mathsf T} = 0
$$

by antisymmetry. So $G$ is constant on the interval, equal to $G(s_0) = F(s_0)F(s_0)^{\mathsf T} = I$: $F(s)$ is orthogonal for every $s$. (Alternatively, without computing $G'$ to zero: both $G$ and the constant $I$ solve the linear system $Y' = YA +
A^{\mathsf T}Y$ with the same initial value, and Cauchy–Lipschitz uniqueness for linear systems, [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq), forces $G
\equiv I$.)

*Relevance:* the Frenet system $(T, N, B)' = (T, N, B)\,A(s)$ has the antisymmetric coefficient matrix

$$
A = \begin{pmatrix} 0 & -\kappa & 0\\ \kappa & 0 & -\tau\\
0 & \tau & 0\end{pmatrix}
$$

(columns expressing $T', N', B'$). The computation above shows that a solution frame that starts orthonormal *stays* orthonormal — the key step in the fundamental theorem reconstructing a curve from $(\kappa, \tau)$.

**Exercise 18.8 ★★★.**

(Total [curvature](#thm-b2-curves-frenet2d) of a closed convex curve) Let $\tilde\gamma$ be a unit-speed $\mathcal{C}^2$ closed plane arc of [length](#def-b2-curves-length) $L$ (so $\tilde\gamma(s + L) = \tilde\gamma(s)$), traversed once counterclockwise. Using the angle function $\varphi$ with $T =
(\cos\varphi, \sin\varphi)$ from [Theorem 18.20](#thm-b2-curves-fundamental), explain why $\varphi(L) -
\varphi(0)$ is a multiple of $2\pi$, and show that $\int_0^L \kappa(s)\,\dd s = \varphi(L) - \varphi(0)$. (For a circle of radius $R$: $\int \kappa = \frac1R \cdot 2\pi R = 2\pi$. The theorem of turning tangents asserts the value $2\pi$ for every simple closed curve; you are not asked to prove that.)

**Solution of Exercise 18.8.**

By [Theorem 18.20](#thm-b2-curves-fundamental) (uniqueness part), there is a $\mathcal{C}^1$ angle function $\varphi$ with $T(s) =
(\cos\varphi(s), \sin\varphi(s))$ and $\varphi' = \kappa$. Hence

$$
\int_0^L \kappa(s)\,\dd s = \varphi(L) - \varphi(0) .
$$

Since the arc is closed of period $L$, $T(L) = T(0)$: $(\cos\varphi(L), \sin\varphi(L)) = (\cos\varphi(0),
\sin\varphi(0))$, so $\varphi(L) - \varphi(0) \in 2\pi\Z$. The total [curvature](#thm-b2-curves-frenet2d) of a closed curve is therefore always an integer multiple of $2\pi$ — the integer being the *winding number* of the tangent (the number of full turns $T$ makes). For the circle of radius $R$: $\kappa = 1/R$ and $L = 2\pi R$, total [curvature](#thm-b2-curves-frenet2d) $2\pi$, winding number $1$; the theorem of turning tangents states this value holds for every simple closed curve.

**Exercise 18.9 ★★★.**

Show that a biregular space curve with constant $\kappa > 0$ and $\tau = 0$ is (an arc of) a circle of radius $1/\kappa$. *(Use [Proposition 18.27](#prop-b2-curves-torsion), then show the center $\gamma + \frac1\kappa N$ is constant.)*

**Solution of Exercise 18.9.**

Since $\tau \equiv 0$, the curve lies in a plane ([Proposition 18.27](#prop-b2-curves-torsion)); work in that plane. Consider the candidate center

$$
c(s) = \gamma(s) + \frac{1}{\kappa}N(s)
\qquad (\kappa \text{ constant}).
$$

Differentiating with the Frenet formulas ($N' = -\kappa T + \tau B
= -\kappa T$ here):

$$
c'(s) = T + \frac1\kappa(-\kappa T) = 0 ,
$$

so $c$ is a constant point $\Omega$. Then $\norm{\gamma(s) -
\Omega} = \norm{-\frac1\kappa N(s)} = \frac1\kappa$ for all $s$: the curve lies on the circle of center $\Omega$ and radius $1/\kappa$ (in its plane), and being a nonconstant arc of it, it is an arc of that circle.

**Exercise 18.10 ★.**

Compute the [arc length](#def-b2-curves-length) of the parabola $y = x^2/2$ over $\intcc0a$ and show that it equals

$$
\tfrac12\Bigl(a\sqrt{1 + a^2} + \ln\bigl(a + \sqrt{1 +
a^2}\bigr)\Bigr).
$$

**Solution of Exercise 18.10.**

For the graph $\gamma(x) = (x, x^2/2)$, $\norm{\gamma'(x)} =
\sqrt{1 + x^2}$, so $L = \int_0^a\sqrt{1+x^2}\,\dd x$. Substituting $x = \sinh u$ ($\dd x = \cosh u\,\dd u$, $u$ from $0$ to $u_a = \ln(a + \sqrt{1+a^2})$):

$$
L = \int_0^{u_a}\cosh^2 u\,\dd u
= \frac12\bigl[u + \sinh u\cosh u\bigr]_0^{u_a}
= \frac12\Bigl(\ln\bigl(a + \sqrt{1+a^2}\bigr) +
a\sqrt{1+a^2}\Bigr),
$$

using $\cosh^2 u = \frac{1 + \cosh 2u}2$ and $\sinh u_a = a$, $\cosh u_a = \sqrt{1 + a^2}$.

**Exercise 18.11 ★★.**

Let $\gamma$ be a regular $\mathcal C^2$ arc in $\R^n$ all of whose [tangent lines](#def-b2-curves-arc) pass through a fixed point $P$. Prove that the trajectory of $\gamma$ is contained in a straight line. *(Parametrize by [arc length](#def-b2-curves-length), write $\gamma(s) +
\lambda(s)T(s) = P$ and differentiate.)*

**Solution of Exercise 18.11.**

Parametrize by [arc length](#def-b2-curves-length) ([Theorem 18.9](#thm-b2-curves-arclength)) and set $\lambda(s) =
\langle P - \gamma(s), T(s)\rangle$, a $\mathcal C^1$ function; since $P$ lies on the [tangent line](#def-b2-curves-arc) at $\gamma(s)$, the vector $P - \gamma(s)$ is collinear with $T(s)$, so $P =
\gamma(s) + \lambda(s)T(s)$. Differentiating,

$$
0 = T(s) + \lambda'(s)T(s) + \lambda(s)T'(s)
= \bigl(1 + \lambda'(s)\bigr)T(s) + \lambda(s)T'(s),
$$

and $T'(s) \perp T(s)$ (differentiate $\norm T^2 = 1$), so both components vanish: $\lambda' = -1$ and $\lambda T' = 0$. Then $\lambda(s) = c - s$ vanishes at most once, so $T' = 0$ on a dense set, hence everywhere by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity): $T$ is a constant unit vector and $\gamma(s) = \gamma(s_0) + (s -
s_0)T$: a straight line (through $P$, as it must be).

**Exercise 18.12 ★★★.**

(Fundamental theorem for space curves) Let $\kappa > 0$ and $\tau$ be [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) functions on an interval $J$. Carry out the program of the remark following [Example 18.29](#ex-b2-curves-helixfrenet): (a) show that the linear system $F' = FA(s)$, with $A(s)$ the antisymmetric Frenet matrix built from $\kappa, \tau$ and $F(s_0)$ a direct orthonormal frame, has a unique global solution, which remains a direct orthonormal frame; (b) construct a unit-speed biregular curve with [curvature](#thm-b2-curves-frenet2d) $\kappa$ and [torsion](#thm-b2-curves-frenet3d) $\tau$; (c) prove uniqueness up to a direct isometry of $\R^3$.

**Solution of Exercise 18.12.**

(a) The Frenet matrix

$$
A(s) = \begin{pmatrix} 0 & -\kappa & 0\\ \kappa & 0 & -\tau\\
0 & \tau & 0\end{pmatrix}
$$

has [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) entries, so the linear system $F' = FA(s)$, $F(s_0) = F_0$ (a direct orthonormal matrix) has a unique solution on all of $J$ ([Theorem 16.4](https://one-course.com/books/math/4/en/chapter/16-differential-equations#thm-b2-diffeq-linear)). By [Exercise 18.7](#exo-b2-curves-7), $F(s)$ is orthogonal for every $s$; $\det F$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) with values in $\{\pm1\}$ and equals $1$ at $s_0$, so $F(s)$ is direct for all $s$.

(b) Read off the rows $T, N, B$ of $F$ (so that $T' = \kappa
N$, $N' = -\kappa T + \tau B$, $B' = -\tau N$) and set $\gamma(s) = \gamma_0 + \int_{s_0}^s T(u)\,\dd u$. Then $\gamma' = T$ is a unit vector: unit speed; $T' = \kappa N$ with $\kappa > 0$ and $N$ unit orthogonal to $T$, so $\gamma$ is biregular with [curvature](#thm-b2-curves-frenet2d) $\norm{T'} = \kappa$ and principal normal $N$; the binormal is $T \wedge N = B$ (direct orthonormal frame), and $B' = -\tau N$ identifies the [torsion](#thm-b2-curves-frenet3d) as $\tau$.

(c) Let $\gamma_1, \gamma_2$ be unit-speed biregular curves with the same $(\kappa, \tau)$. There is a unique direct isometry $\Phi = \rho + w$ ($\rho \in SO(3)$) sending $\gamma_1(s_0)$ to $\gamma_2(s_0)$ and the Frenet frame of $\gamma_1$ at $s_0$ to that of $\gamma_2$ at $s_0$. The curve $\Phi\circ\gamma_1$ is unit-speed with the same invariants (its frame is $\rho$ applied to that of $\gamma_1$, and $\rho$ preserves cross products, being direct). Now the frames of $\Phi\circ\gamma_1$ and $\gamma_2$ both solve $F' =
FA(s)$ with the same initial value, so they coincide by uniqueness; in particular the tangents agree, and integrating from the common point $s_0$: $\Phi\circ\gamma_1 = \gamma_2$.

## 18.7 Problem: envelopes — the astroid, two evolutes, and a caustic

![A ladder of length 1 sliding down a wall (blue positions) never crosses the astroid x2/3 + y2/3 = 1 (red): the astroid is the envelope of the family of segments, tangent to every one of them.](https://one-course.com/images/onecourse/chapters/math-4/b2-curves/fig-86eeb596a0c4.svg)

*A ladder of [length](#def-b2-curves-length) $1$ sliding down a wall (blue positions) never crosses the [astroid](#pb-b2-curves-1) $x^{2/3} + y^{2/3} = 1$ (red): the [astroid](#pb-b2-curves-1) is the *[envelope](#pb-b2-curves-1)* of the family of segments, tangent to every one of them.*

**Problem 18.1.**

Weekend problem — the envelope machine and four classical curves

A one-parameter family of lines usually fails to cover the plane evenly: the lines pile up along a curve tangent to all of them, their *envelope*. Light rays make [envelopes](#pb-b2-curves-1) visible as *caustics* — the bright cusped curve in a mug of coffee. This problem builds the general [envelope](#pb-b2-curves-1) machine, then runs it four times: the sliding ladder ([astroid](#pb-b2-curves-1)), the normals of the parabola and of the cycloid (evolutes, with Huygens’ pendulum at the end), and the coffee-cup caustic ([nephroid](#pb-b2-curves-1)). Throughout, $D_t$ denotes the line of equation $a(t)\,x + b(t)\,y = c(t)$, where $a, b, c$ are $\mathcal C^2$ functions with $(a(t), b(t)) \neq (0,0)$, and $\Delta(t) = a(t)b'(t) - a'(t)b(t)$.

**Part I — The [envelope](#pb-b2-curves-1) machine.**

1. Suppose $\Delta(t) \neq 0$. Show that the *characteristic system* $$\begin{cases} a(t)\,x + b(t)\,y = c(t)\\  a'(t)\,x + b'(t)\,y = c'(t)\end{cases}$$ has a unique solution $E(t) = (x(t), y(t))$, given by $x = \dfrac{cb' - c'b}{\Delta}$, $y = \dfrac{ac' -  a'c}{\Delta}$.
2. Assume moreover that $E$ is $\mathcal C^1$ near $t$ with $E'(t) \neq 0$ . Differentiating the first equation of the system, show $a(t)\,x'(t) +  b(t)\,y'(t) = 0$ , and conclude that the curve $E$ passes through a point of $D_t$ *with the direction of* $D_t$ : the family is tangent to $E$ , which is called its *[envelope](#pb-b2-curves-1)* .
3. Sanity check: the [tangent lines](#def-b2-curves-arc) of the parabola $y =  x^2/2$ at the points $(t, t^2/2)$ are $tx - y =  t^2/2$ . Verify that the [envelope](#pb-b2-curves-1) machine returns the parabola itself.
4. ( [Envelope](#pb-b2-curves-1) of the normals) Let $\gamma$ be unit-speed with $\kappa(s) \neq 0$ . The normal line at $\gamma(s)$ is $\{M : \langle M - \gamma(s), T(s)  \rangle = 0\}$ . Show that its characteristic system forces $\langle M - \gamma(s), N(s)\rangle =  1/\kappa(s)$ , hence that the characteristic point is the [center of curvature](#def-b2-curves-curvature) : *the [envelope](#pb-b2-curves-1) of the normals is the evolute* , recovering [Exercise 18.6](#exo-b2-curves-6) . Check $\Delta(s) =  \kappa(s)$ .
5. Two degenerations. For the pencil $D_\theta : x\cos  \theta + y\sin\theta = 0$ , show that the characteristic point is the origin for every $\theta$ (the “ [envelope](#pb-b2-curves-1) ” collapses to a point, and $E' = 0$ : question 2 does not apply). For a family of parallel lines ( $a, b$ constant), show $\Delta \equiv 0$ and that the characteristic system is in general inconsistent: no [envelope](#pb-b2-curves-1) .

**Part II — The sliding ladder and the [astroid](#pb-b2-curves-1).** A segment of [length](#def-b2-curves-length) $1$ slides with one end $P_t =
(\cos t, 0)$ on the floor and the other $Q_t = (0, \sin t)$ on the wall, $t \in \intoo0{\pi/2}$.

6. Show that the line $(P_tQ_t)$ has equation $x\sin t +  y\cos t = \sin t\cos t$, and that the [envelope](#pb-b2-curves-1) machine gives the characteristic point $$E(t) = (\cos^3 t,\ \sin^3 t) :$$ the *astroid*, of implicit equation $x^{2/3} + y^{2/3} = 1$ (extended to the other quadrants by symmetry).
7. Show that $E'(0) = 0$ and, using the local classification ( [Proposition 18.32](#prop-b2-curves-local) ), that the [astroid](#pb-b2-curves-1) has a cusp of the first kind at $(1, 0)$ — and likewise at its four axis points.
8. Compute $\norm{E'(t)} = \tfrac32\abs{\sin 2t}$ and deduce that the total [length](#def-b2-curves-length) of the [astroid](#pb-b2-curves-1) is $6$ .
9. Where does the ladder touch the [astroid](#pb-b2-curves-1) ? Show $E(t) =  P_t + \sin^2 t\,(Q_t - P_t)$ : the contact point divides the ladder in the ratio $\sin^2 t : \cos^2  t$ , sweeping it from one end to the other as the ladder slides.
10. Compute the area enclosed by the [astroid](#pb-b2-curves-1) : show that the first-quadrant area is $3\int_0^{\pi/2}\sin^4  t\cos^2 t\,\dd t$ , evaluate the integral by linearization ( $\sin^2 2t = \tfrac{1 - \cos 4t}2$ ), and conclude that the total area is $3\pi/8$ .

**Part III — The evolute of the parabola.** Let $\gamma(t) = (t, t^2/2)$.

11. Show that the normal line at $\gamma(t)$ has equation $x + t\,y = t + t^3/2$ .
12. Run the [envelope](#pb-b2-curves-1) machine: show that the [envelope](#pb-b2-curves-1) of the normals is $$E(t) = \Bigl(-t^3,\ 1 + \tfrac32 t^2\Bigr),$$ with implicit equation $x^2 = \tfrac8{27}(y - 1)^3$: a semicubical parabola.
13. Cross-check with question 4: compute the [center of curvature](#def-b2-curves-curvature) $\gamma(t) + \frac1{\kappa(t)}N(t)$ from $\kappa(t) = (1 + t^2)^{-3/2}$ ( [Example 18.18](#ex-b2-curves-kappaexamples) ) and recover the same point.
14. Show that the evolute has a cusp of the first kind at $(0, 1)$ , the [center of curvature](#def-b2-curves-curvature) at the vertex — the point where $\kappa$ is extremal, as predicted by the formula $c' = -\frac{\kappa'}{\kappa^2}N$ of [Exercise 18.6](#exo-b2-curves-6) .
15. How many normals of the parabola pass through a given point $(x_0, y_0)$ ? Show the answer is governed by the cubic $\tfrac{t^3}2 + (1 - y_0)\,t - x_0 = 0$ ; treat the axis case $x_0 = 0$ completely (one normal for $y_0 < 1$ , three for $y_0 > 1$ ), and interpret the evolute as the transition curve.

**Part IV — The coffee-cup caustic.** Parallel rays of direction $(1, 0)$ strike the inside of the mirror circle $x^2 + y^2 = 1$; the ray hitting $P_\theta =
(\cos\theta, \sin\theta)$ reflects according to the law of reflection.

16. From the mirror symmetry in the normal (the radius), justify that the reflected direction is $v = u -  2\langle u, n\rangle n$ with $u = (1,0)$ , $n =  (\cos\theta, \sin\theta)$ , and compute $v =  -(\cos2\theta, \sin2\theta)$ .
17. Show that the reflected ray lies on the line $$x\sin 2\theta - y\cos 2\theta = \sin\theta .$$
18. Run the [envelope](#pb-b2-curves-1) machine ($\Delta = 2$): show that the caustic is $$E(\theta) = \Bigl(\tfrac{3\cos\theta -  \cos3\theta}4,\ \tfrac{3\sin\theta -  \sin3\theta}4\Bigr),$$ the *nephroid*.
19. Compute $E'(\theta) = \tfrac32\sin\theta\,  (\cos2\theta, \sin2\theta)$ ; check that the tangent direction is the reflected-ray direction (question 16), locate the two cusps $(\pm\tfrac12, 0)$ , and show that the reflected ray crosses the axis $y = 0$ at $x = \frac1{2\cos\theta}$ — so nearly axial rays focus at $x = \tfrac12$ : the focal [length](#def-b2-curves-length) $R/2$ of a mirror of radius $R$ .
20. Show that the [nephroid](#pb-b2-curves-1) has total [length](#def-b2-curves-length) $6$ and that near $\theta = 0$, $$E(\theta) - \bigl(\tfrac12, 0\bigr) =  \bigl(\tfrac34\theta^2 + o(\theta^2),\ \theta^3 +  o(\theta^3)\bigr) :$$ a cusp of the first kind, pointing along the axis.
21. Explain in one paragraph why the caustic is bright: through every point just outside the caustic pass two reflected rays, through every point on it the rays are “infinitely concentrated” (the map $(\theta,  \text{distance along ray}) \mapsto \R^2$ has a critical point exactly on the [envelope](#pb-b2-curves-1) ).

**Part V — Huygens: the cycloid is its own evolute.** Let $\gamma(t) = (t - \sin t,\ 1 - \cos t)$, $t \in
\intoo0{2\pi}$, one arch of the cycloid.

22. Compute $\kappa(t) = -\dfrac1{4\sin(t/2)}$ and the [center of curvature](#def-b2-curves-curvature); show that the evolute is $$c(t) = (t + \sin t,\ \cos t - 1),$$ and that the substitution $t = u + \pi$ exhibits it as the original cycloid translated by $(\pi, -2)$: *the evolute of a cycloid is a congruent cycloid* (Huygens).
23. Verify that the [radius of curvature](#def-b2-curves-curvature) at the apex $t =  \pi$ equals $4$ , which is half the [length](#def-b2-curves-length) $8$ of one arch ( [Exercise 18.1](#exo-b2-curves-1) ); locate the cusp of the evolute directly below the apex, at distance $4$ .
24. (The string property) Let $\gamma$ be unit-speed with $\kappa > 0$ , $\kappa$ of class $\mathcal C^1$ and $R = 1/\kappa$ strictly monotone. Using $c' = R'N$ , show that the [arc length](#def-b2-curves-length) of the evolute between $c(s_0)$ and $c(s_1)$ is $\abs{R(s_1) - R(s_0)}$ . Interpret: a taut string unwound from the evolute, of [length](#def-b2-curves-length) $R(s_0)$ at the start, has its free end trace the original curve — so a pendulum swinging between two cycloidal cheeks of Huygens’ clock describes a cycloid.
25. Synthesis. The machine of Part I produced the [astroid](#pb-b2-curves-1) , a semicubical parabola, a [nephroid](#pb-b2-curves-1) and a cycloid. For each of the four families, state in one sentence where the hypotheses $\Delta \neq 0$ and $E' \neq 0$ held or failed, and what geometric event (cusp, focus, degeneracy) each failure of $E' \neq 0$ signalled. Where must extrema of [curvature](#thm-b2-curves-frenet2d) appear on the [envelope](#pb-b2-curves-1) of the normals, and why?

**Solution of Problem 18.1.**

**1.** The system is linear in $(x, y)$ with [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) $\Delta(t) = a b' - a'b \neq 0$: Cramer’s rule gives the unique solution

$$
x = \frac{c b' - c' b}{\Delta}, \qquad
y = \frac{a c' - a' c}{\Delta}.
$$

**2.** Since $a(t)x(t) + b(t)y(t) = c(t)$ identically, differentiating gives $a'x + b'y + ax' + by' = c'$; the second characteristic equation kills $a'x + b'y - c'$, so $a(t)x'(t) + b(t)y'(t) = 0$: $E'(t)$ is orthogonal to $(a,
b)$, hence parallel to $(-b, a)$, the direction of $D_t$. As $E(t) \in D_t$ (first equation) and $E'(t) \neq 0$, the line $D_t$ is exactly the [tangent line](#def-b2-curves-arc) of the curve $E$ at $E(t)$.

**3.** Here $(a, b, c) = (t, -1, t^2/2)$, so $\Delta =
t\cdot0 - 1\cdot(-1) = 1$ and

$$
x = \frac{c b' - c' b}{\Delta} = \tfrac{t^2}2\cdot 0 +
t = t, \qquad
y = \frac{a c' - a' c}{\Delta} = t\cdot t - \tfrac{t^2}2 =
\tfrac{t^2}2 :
$$

the [envelope](#pb-b2-curves-1) of the [tangent lines](#def-b2-curves-arc) of the parabola is the parabola, as it should be.

**4.** The normal line is $\langle M, T(s)\rangle =
\langle\gamma(s), T(s)\rangle$: coefficients $a = T_1$, $b =
T_2$, $c = \langle\gamma, T\rangle$. Differentiating with Frenet ($T' = \kappa N$): $a' = \kappa N_1$, $b' = \kappa
N_2$, and $c' = \langle T, T\rangle + \langle\gamma, \kappa
N\rangle = 1 + \kappa\langle\gamma, N\rangle$. The second characteristic equation $\kappa\langle M, N\rangle = 1 +
\kappa\langle\gamma, N\rangle$ reads $\kappa\langle M -
\gamma, N\rangle = 1$. The first says $M - \gamma \perp T$, so $M - \gamma = \mu N$ with $\mu = 1/\kappa$: the characteristic point is $\gamma + \frac1\kappa N$, the [center of curvature](#def-b2-curves-curvature), and the [envelope](#pb-b2-curves-1) of the normals is the evolute of [Exercise 18.6](#exo-b2-curves-6). Finally $\Delta = T_1\kappa N_2 -
\kappa N_1 T_2 = \kappa\det(T, N) = \kappa \neq 0$.

**5.** Pencil: the system $x\cos\theta + y\sin\theta =
0$, $-x\sin\theta + y\cos\theta = 0$ has [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) $1$ and solution $(0,0)$ for every $\theta$: $E \equiv (0,0)$, $E'
\equiv 0$, and there is no curve — merely the common point of all the lines. Parallel family: $a' = b' = 0$ gives $\Delta \equiv 0$ and the second equation $0 = c'(t)$, which fails as soon as the family actually moves: no characteristic point, and indeed a family of parallel lines touches no curve along all its members.

**6.** The line through $(\cos t, 0)$ and $(0, \sin t)$ is $\frac x{\cos t} + \frac y{\sin t} = 1$, i.e. $x\sin t +
y\cos t = \sin t\cos t$. With $(a, b, c) = (\sin t, \cos t,
\sin t\cos t)$: $a' = \cos t$, $b' = -\sin t$, $c' = \cos
2t$, $\Delta = -\sin^2 t - \cos^2 t = -1$. Cramer:

$$
\begin{align*}
x &= \frac{cb' - c'b}{-1} = \sin^2 t\cos t + \cos 2t\cos t
= \cos t\,(\sin^2 t + \cos^2 t - \sin^2 t) = \cos^3 t,\\
y &= \frac{ac' - a'c}{-1} = \sin t\cos^2 t - \sin t\cos 2t
= \sin t\,(\cos^2 t - \cos^2 t + \sin^2 t) = \sin^3 t .
\end{align*}
$$

And $(\cos^3t)^{2/3} + (\sin^3t)^{2/3} = 1$: the [astroid](#pb-b2-curves-1).

**7.** $E'(t) = 3(-\cos^2 t\sin t,\ \sin^2 t\cos t)$ vanishes at $t = 0$. There, $E''(0) = (-3, 0) \neq 0$ gives $p = 2$; the $x$-component of $E$ is even in $t$, so $E'''(0) = (0, 6)$, not collinear: $q = 3$. Even–odd: cusp of the first kind at $(1, 0)$ ([Proposition 18.32](#prop-b2-curves-local)), with tangent along the $x$-axis. The symmetries $x \mapsto -x$, $y \mapsto -y$, $(x,
y) \mapsto (y, x)$ of the [astroid](#pb-b2-curves-1) transport the cusp to $(-1,
0)$ and $(0, \pm1)$.

**8.** $E'(t) = 3\sin t\cos t\,(-\cos t, \sin t)$, so $\norm{E'(t)} = 3\abs{\sin t\cos t} = \tfrac32\abs{\sin 2t}$. One quadrant: $\int_0^{\pi/2}\tfrac32\sin 2t\,\dd t =
\tfrac32$, and by symmetry the total [length](#def-b2-curves-length) is $4 \cdot
\tfrac32 = 6$.

**9.** $E(t) - P_t = (\cos^3 t - \cos t,\ \sin^3 t) =
\sin^2 t\,(-\cos t,\ \sin t) = \sin^2 t\,(Q_t - P_t)$. So the contact point is the [barycenter](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-barycenter) of $(P_t, \cos^2 t)$ and $(Q_t, \sin^2 t)$: as $t$ runs from $0$ to $\pi/2$ it slides from the floor end to the wall end of the ladder.

**10.** In the first quadrant the region under the [astroid](#pb-b2-curves-1) has area $\int_0^1 y\,\dd x$ with $x = \cos^3 t$ decreasing from $1$ to $0$ as $t$ goes from $0$ to $\pi/2$:

$$
\int_0^1 y\,\dd x
= \int_{\pi/2}^{0}\sin^3 t\,(-3\cos^2 t\sin t)\,\dd t
= 3\int_0^{\pi/2}\sin^4 t\cos^2 t\,\dd t .
$$

Linearize: $\sin^4 t\cos^2 t = (\sin t\cos t)^2\sin^2 t =
\tfrac18\bigl(\sin^2 2t - \sin^2 2t\cos 2t\bigr)$, and $\int_0^{\pi/2}\sin^2 2t\,\dd t = \tfrac\pi4$ while $\int_0^{\pi/2}\sin^2 2t\cos 2t\,\dd t =
\bigl[\tfrac{\sin^3 2t}6\bigr]_0^{\pi/2} = 0$. So the integral is $\tfrac\pi{32}$, the quadrant area $\tfrac{3\pi}{32}$, and the enclosed area $4 \cdot
\tfrac{3\pi}{32} = \tfrac{3\pi}8$.

**11.** The tangent at $\gamma(t) = (t, t^2/2)$ is directed by $(1, t)$, so the normal line is $\{(x, y) : (x -
t) + t\,(y - t^2/2) = 0\}$, i.e. $x + t\,y = t +
\tfrac{t^3}2$.

**12.** $(a, b, c) = (1, t, t + t^3/2)$: $a' = 0$, $b' =
1$, $c' = 1 + \tfrac32 t^2$, $\Delta = 1$. Then $y = ac' -
a'c = 1 + \tfrac32t^2$ and

$$
x = cb' - c'b = t + \tfrac{t^3}2 - t\Bigl(1 +
\tfrac32t^2\Bigr) = -t^3 .
$$

Eliminating $t$: $t^2 = \tfrac23(y - 1)$ and $x^2 = t^6 =
\tfrac8{27}(y - 1)^3$: a semicubical parabola with vertex $(0, 1)$.

**13.** $T = (1, t)/\sqrt{1+t^2}$, $N = (-t,
1)/\sqrt{1+t^2}$, and $\kappa = (1+t^2)^{-3/2}$, so

$$
\gamma + \frac1\kappa N = (t, \tfrac{t^2}2) +
(1+t^2)\,(-t, 1) = \Bigl(-t^3,\ 1 + \tfrac32t^2\Bigr),
$$

the same curve: the [envelope](#pb-b2-curves-1) of the normals is the locus of the centers of [curvature](#thm-b2-curves-frenet2d), as question 4 promised.

**14.** $E'(t) = (-3t^2, 3t)$ vanishes at $t = 0$; $E''(0) = (0, 3) \neq 0$ gives $p = 2$ and $E'''(0) = (-6,
0)$ gives $q = 3$: a cusp of the first kind at $(0, 1)$. The vertex is where $\kappa = (1+t^2)^{-3/2}$ is maximal, so $\kappa'(0) = 0$ and the evolute’s velocity $-\frac{\kappa'}{\kappa^2}N$ vanishes exactly there: cusps of the evolute sit at the extrema of [curvature](#thm-b2-curves-frenet2d).

**15.** The normal at parameter $t$ passes through $(x_0, y_0)$ iff $x_0 + t\,y_0 = t + \tfrac{t^3}2$, i.e.

$$
\frac{t^3}2 + (1 - y_0)\,t - x_0 = 0 ,
$$

a cubic in $t$: one or three real roots (counted without multiplicity, for generic points). On the axis $x_0 = 0$ it factors as $t\bigl(\tfrac{t^2}2 + 1 - y_0\bigr) = 0$: the root $t = 0$ (the axis is the normal at the vertex), plus $t = \pm\sqrt{2(y_0 - 1)}$ when $y_0 > 1$. So: one normal for $y_0 < 1$, three for $y_0 > 1$, and at $y_0 = 1$ the triple root marks the cusp of the evolute. In general a double root of the cubic means the point satisfies both the line equation and its $t$-derivative — it lies *on the [envelope](#pb-b2-curves-1)*: the evolute is precisely the boundary between the one-normal and three-normal regions.

**16.** Reflection in the mirror reverses the normal component of the direction and keeps the tangential one: writing $u = \langle u, n\rangle n + u_{\mathrm{tan}}$, the reflected direction is $u_{\mathrm{tan}} - \langle u,
n\rangle n = u - 2\langle u, n\rangle n$. Here $\langle u,
n\rangle = \cos\theta$, so

$$
v = (1, 0) - 2\cos\theta\,(\cos\theta, \sin\theta)
= (1 - 2\cos^2\theta,\ -2\sin\theta\cos\theta)
= -(\cos2\theta,\ \sin2\theta).
$$

**17.** The reflected ray passes through $P_\theta =
(\cos\theta, \sin\theta)$ with direction $(\cos2\theta,
\sin2\theta)$; a normal vector is $(-\sin2\theta,
\cos2\theta)$, so the line is

$$
-\sin2\theta\,(x - \cos\theta) + \cos2\theta\,(y -
\sin\theta) = 0,
$$

and the constant is $-\sin2\theta\cos\theta +
\cos2\theta\sin\theta = -\sin\theta$: multiplying by $-1$, $x\sin2\theta - y\cos2\theta = \sin\theta$.

**18.** $(a, b, c) = (\sin2\theta, -\cos2\theta,
\sin\theta)$: $a' = 2\cos2\theta$, $b' = 2\sin2\theta$, $c' =
\cos\theta$, $\Delta = 2\sin^22\theta + 2\cos^22\theta = 2$. Cramer, then product-to-sum formulas:

$$
\begin{align*}
x &= \frac{2\sin\theta\sin2\theta +
\cos\theta\cos2\theta}{2}
= \frac{(\cos\theta - \cos3\theta) + \frac12(\cos\theta +
\cos3\theta)}{2} = \frac{3\cos\theta - \cos3\theta}4,\\
y &= \frac{\sin2\theta\cos\theta - 2\cos2\theta\sin\theta}2
= \frac{\frac12(\sin3\theta + \sin\theta) - (\sin3\theta -
\sin\theta)}2 = \frac{3\sin\theta - \sin3\theta}4 :
\end{align*}
$$

the [nephroid](#pb-b2-curves-1), a closed curve with two cusps.

**19.** Differentiating and factoring with $\sin3\theta
- \sin\theta = 2\cos2\theta\sin\theta$, $\cos\theta -
\cos3\theta = 2\sin2\theta\sin\theta$:

$$
E'(\theta) = \tfrac34\bigl(\sin3\theta - \sin\theta,\
\cos\theta - \cos3\theta\bigr)
= \tfrac32\sin\theta\,(\cos2\theta, \sin2\theta),
$$

parallel to the reflected direction of question 16: each reflected ray is tangent to the caustic, as the [envelope](#pb-b2-curves-1) property demands. $E' = 0$ exactly at $\sin\theta = 0$: $E(0) = (\tfrac12, 0)$ and $E(\pi) = (-\tfrac12, 0)$, the two cusps. Setting $y = 0$ in the line equation: $x\sin2\theta =
\sin\theta$, so $x = \frac1{2\cos\theta} \to \frac12$ as $\theta \to 0$: paraxial rays focus at distance $R/2$ from the center — the focal [length](#def-b2-curves-length) of the spherical mirror.

**20.** $\norm{E'(\theta)} = \tfrac32\abs{\sin\theta}$, so the [length](#def-b2-curves-length) is $\tfrac32\int_0^{2\pi}\abs{\sin\theta}\,
\dd\theta = \tfrac32\cdot4 = 6$. Near $\theta = 0$, with $\cos k\theta = 1 - \tfrac{k^2\theta^2}2 + O(\theta^4)$ and $\sin k\theta = k\theta - \tfrac{k^3\theta^3}6 +
O(\theta^5)$:

$$
x - \tfrac12 = \frac{3\theta^2 + O(\theta^4)}{4}
= \tfrac34\theta^2 + O(\theta^4),
\qquad
y = \frac{4\theta^3 + O(\theta^5)}{4}
= \theta^3 + O(\theta^5) :
$$

$p = 2$, $q = 3$, a cusp of the first kind pointing along the axis — the bright point of the coffee-cup caustic.

**21.** Parametrize the illuminated points by $\Phi(\theta, r) = P_\theta + r\,v_\theta$ (position along each reflected ray). The Jacobian [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) $\det(\partial_\theta\Phi, \partial_r\Phi) =
\det(P_\theta' + rv_\theta', v_\theta)$ is [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) in $r$ and vanishes for exactly one $r = r_*(\theta)$ — and $\Phi(\theta, r_*(\theta))$ is the characteristic point, since there the ray direction and the variation of the family become dependent. Off the [envelope](#pb-b2-curves-1) the map is a local diffeomorphism, and a point just inside the caustic is hit by two nearby rays (two solutions $\theta$), a point outside by none from that part of the family; on the caustic the two merge. Light intensity is inversely proportional to the Jacobian’s absolute value, so it blows up along the [envelope](#pb-b2-curves-1): the caustic is the bright curve, brightest of all at the cusp, where the degeneracy is worst.

**22.** $x' = 1 - \cos t$, $y' = \sin t$, $x'' = \sin
t$, $y'' = \cos t$, so $x'y'' - y'x'' = \cos t - 1$ and $\norm{\gamma'}^2 = 2(1 - \cos t) = 4\sin^2\tfrac t2$; by [Proposition 18.17](#prop-b2-curves-kappaformula),

$$
\kappa(t) = \frac{-(1 - \cos t)}{8\sin^3\tfrac t2}
= \frac{-2\sin^2\tfrac t2}{8\sin^3\tfrac t2}
= -\frac1{4\sin\tfrac t2}
\qquad (0 < t < 2\pi).
$$

With $T = (\sin\tfrac t2, \cos\tfrac t2)$ (divide $\gamma'$ by $2\sin\tfrac t2$) and $N = (-\cos\tfrac t2, \sin\tfrac
t2)$:

$$
\gamma + \frac1\kappa N
= \gamma - 4\sin\tfrac t2\,\Bigl(-\cos\tfrac t2,\
\sin\tfrac t2\Bigr)
= \bigl(t - \sin t + 2\sin t,\ 1 - \cos t - 2(1 - \cos
t)\bigr),
$$

i.e. $c(t) = (t + \sin t,\ \cos t - 1)$. Substituting $t = u
+ \pi$:

$$
c = \bigl(u + \pi - \sin u,\ -\cos u - 1\bigr)
= \bigl((u - \sin u) + \pi,\ (1 - \cos u) - 2\bigr) :
$$

the cycloid $\gamma(u)$ translated by $(\pi, -2)$. The evolute of a cycloid is a congruent cycloid, hanging one level below.

**23.** $R(t) = 1/\abs{\kappa(t)} = 4\sin\tfrac t2$, so $R(\pi) = 4$: half of the arch [length](#def-b2-curves-length) $8$ computed in [Exercise 18.1](#exo-b2-curves-1). The evolute’s velocity $c'(t) = (1 +
\cos t, -\sin t)$ vanishes at $t = \pi$: the cusp is $c(\pi)
= (\pi, -2)$, directly below the apex $\gamma(\pi) = (\pi,
2)$, at distance $4 = R(\pi)$, exactly the [length](#def-b2-curves-length) of the osculating radius there.

**24.** From [Exercise 18.6](#exo-b2-curves-6), $c'(s) =
-\frac{\kappa'(s)}{\kappa(s)^2}N(s) = R'(s)\,N(s)$, so $\norm{c'(s)} = \abs{R'(s)}$ and, for monotone $R$,

$$
\int_{s_0}^{s_1}\norm{c'(s)}\,\dd s =
\Bigl|\int_{s_0}^{s_1}R'(s)\,\dd s\Bigr| = \abs{R(s_1) -
R(s_0)} .
$$

Say $R$ decreases. A string laid along the evolute beyond $c(s_0)$ and prolonged by the segment from $c(s_0)$ to $\gamma(s_0)$ (which is tangent to the evolute, by question 4) has, when peeled off up to $c(s)$ and stretched taut, straight part of [length](#def-b2-curves-length) $R(s_0) - \bigl(R(s_0) - R(s)\bigr) = R(s)$ pointing from $c(s)$ along the normal — landing exactly on $\gamma(s)$: the free end traces the original curve (“involute”). Huygens hung a pendulum between two cycloidal cheeks: the cord wraps on the evolute, so the bob describes a cycloid — the tautochrone, whose oscillation period does not depend on the amplitude.

**25.** Tangents of the parabola: $\Delta = 1$ and $E' =
(1, t) \neq 0$ everywhere — smooth [envelope](#pb-b2-curves-1) (the parabola itself). Sliding ladder: $\Delta = -1$, but $E' =
\tfrac32\sin 2t\,(-\cos t, \sin t)$ vanishes at the quadrant ends — the four cusps of the [astroid](#pb-b2-curves-1). Normals of the parabola and of the cycloid: $\Delta = \kappa \neq 0$, and $E' = R'N$ vanishes exactly where the [curvature](#thm-b2-curves-frenet2d) is extremal — cusps of the evolutes at $(0,1)$ and $(\pi, -2)$. Caustic: $\Delta = 2$, and $E' = \tfrac32\sin\theta\,(\cos2\theta,
\sin2\theta)$ vanishes at $\theta = 0, \pi$ — the two cusps of the [nephroid](#pb-b2-curves-1), the focal points of the mirror. Extrema of [curvature](#thm-b2-curves-frenet2d) *must* produce cusps on the [envelope](#pb-b2-curves-1) of the normals, since the evolute’s velocity is $R'N$: this is why the evolute of the ellipse has four cusps (four vertices), and the degenerate families (pencil, parallels) are the cases where the machine outputs a point or nothing at all.