---
title: The Calculus of Variations
module: Historical Notes and the Calculus of Variations
moduleNumber: 10
lessonNumber: 1
order: 1001
summary: >
  Ordinary calculus finds the point where a function is stationary; the calculus
  of variations finds the whole curve where an integral is stationary. Euler's
  differential equation is the necessary condition for an extremal, and it becomes
  integrable in three cases, solving the shortest-path, minimal-surface, and
  brachistochrone problems. Lagrange multipliers extend the method to isoperimetric
  constraints, and Hamilton's principle recovers Newton's law from a single
  stationary integral.
topics: [Historical Notes and the Calculus of Variations]
sources:
  - book: Simmons
    ref: "Ch. 12 — The Calculus of Variations; §66 Some Typical Problems; §67 Euler's Differential Equation for an Extremal"
  - book: Simmons
    ref: "Ch. 12 §68 Isoperimetric Problems; Appendix B Hamilton's Principle and Its Implications"
  - book: Simmons
    ref: "Ch. 1 §6 The Brachistochrone; Fermat and the Bernoullis"
---

Elementary calculus locates the points at which a function of one variable is
largest or smallest: differentiate, set the derivative to zero, solve. The
calculus of variations poses the same question one level up. The unknown is no
longer a number but a whole function, and the quantity to be made stationary is
not a function value but an **integral** whose integrand depends on that function
and its derivative. The output of the method is a differential equation: many of
the equations solved elsewhere in these notes — the cycloid of the
[brachistochrone](/differential-equations/first-order/separable-and-exact), the
catenary, the equations of motion of a mechanical
[oscillator](/differential-equations/second-order-linear/mechanical-electrical-vibrations)
— arise first as the condition that some integral be as small as possible.[^s66]

## Problems that depend on a whole curve

Fix two points $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ in the plane and consider the
family of functions

$$
y = y(x), \qquad y(x_1) = y_1, \quad y(x_2) = y_2,
$$

whose graphs join $P$ to $Q$. Three classical questions single out one member of
this family by a minimizing property:

- **Shortest path.** Which curve joining $P$ and $Q$ has the least arc length? The
  length of $y(x)$ is $\int_{x_1}^{x_2} \sqrt{1 + (y')^2}\, \d x$.
- **Minimal surface of revolution.** Which curve, revolved about the $x$-axis,
  sweeps out the surface of least area? The area is
  $\int_{x_1}^{x_2} 2\pi y \sqrt{1 + (y')^2}\, \d x$.
- **Brachistochrone.** Down which frictionless wire from $P$ to a lower point $Q$
  does a bead slide in the least time? With speed $v = \sqrt{2gy}$ from energy
  conservation, the descent time is
  $\int_{x_1}^{x_2} \sqrt{1 + (y')^2}\big/\sqrt{2gy}\; \d x$.

Each is a special case of one general problem: among all admissible functions
$y(x)$ joining $P$ and $Q$, find the one that gives a stationary value to the
integral

$$
I(y) = \int_{x_1}^{x_2} f(x, y, y')\, \d x.
$$

The integrand $f$ carries the geometry or physics of the problem; the three
questions above correspond to $f = \sqrt{1+(y')^2}$, $f = 2\pi y\sqrt{1+(y')^2}$,
and $f = \sqrt{1+(y')^2}\big/\sqrt{2gy}$.

$$
% caption: Infinitely many admissible curves join $P$ to $Q$; the integral $I(y)$
% assigns a number (length, area, or time) to each, and the problem is to select
% the curve on which that number is stationary.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (P) at (0,0);
\coordinate (Q) at (7,2);
\draw[black] (P) .. controls (2.0,2.3) and (4.5,2.7) .. (Q);
\draw[black] (P) .. controls (1.5,1.7) and (5.2,3.0) .. (Q);
\draw[black] (P) .. controls (3.0,1.3) and (5.6,2.5) .. (Q);
\draw[acc, very thick] (P) -- (Q);
\fill[black] (P) circle (2.4pt);
\fill[black] (Q) circle (2.4pt);
\node[anchor=north east] at (P) {$P$};
\node[anchor=south west] at (Q) {$Q$};
\node[anchor=south, black] at (3.4,2.55) {admissible curves};
\node[anchor=north, acc] at (4.6,0.6) {a candidate extremal};
\end{tikzpicture}
$$

> **Definition (Functional).** A rule $I$ that assigns a real number to each
> function in some admissible class. Here $I(y) = \int_{x_1}^{x_2} f(x,y,y')\,\d x$.
> The admissible functions are those with continuous second derivatives that
> satisfy the prescribed boundary conditions $y(x_1)=y_1$, $y(x_2)=y_2$.

Throughout, $f$ is assumed to have continuous second-order partial derivatives in
its three arguments $x$, $y$, $y'$, which are treated as independent when the
partials $\partial f/\partial y$ and $\partial f/\partial y'$ are formed.

## Euler's equation for an extremal

Suppose an admissible function $y(x)$ makes $I$ stationary. To turn that
supposition into an equation, compare $y$ with nearby admissible functions.
Choose any **variation** $\eta(x)$ with continuous second derivative that vanishes
at the endpoints,

$$
\eta(x_1) = \eta(x_2) = 0,
$$

and form the one-parameter family $\bar{y}(x) = y(x) + \alpha\,\eta(x)$. Every
member joins $P$ to $Q$, and the whole family collapses onto the minimizing curve
$y$ at $\alpha = 0$. The deviation of a neighbor from $y$ is $\alpha\,\eta(x)$.

$$
% caption: The minimizing curve $y$ and a neighbor $\bar y = y + \alpha\eta$ that
% shares its endpoints; the vertical gap between them is $\alpha\,\eta(x)$, which
% vanishes at both ends.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (A) at (0,0);
\coordinate (B) at (7,0.4);
\draw[acc, very thick] (A) .. controls (2.5,1.8) and (4.5,1.6) .. (B);
\draw[black, thick, dashed] (A) .. controls (2.2,2.6) and (5,2.1) .. (B);
\fill[black] (A) circle (2.2pt);
\fill[black] (B) circle (2.2pt);
\node[anchor=north east] at (A) {$(x_1,y_1)$};
\node[anchor=north west] at (B) {$(x_2,y_2)$};
\node[acc, anchor=east] at (2.5,1.35) {$y(x)$};
\node[black, anchor=south] at (3.0,2.45) {neighbor $\bar y$};
\draw[<->, black] (3.5,1.53) -- (3.5,2.28);
\node[anchor=west, black] at (3.78,1.9) {variation};
\end{tikzpicture}
$$

Substituting $\bar{y}$ and $\bar{y}' = y' + \alpha\eta'$ into the integral turns
$I$ into an ordinary function of the single number $\alpha$,

$$
I(\alpha) = \int_{x_1}^{x_2} f\big(x,\; y + \alpha\eta,\; y' + \alpha\eta'\big)\, \d x.
$$

Since $y$ minimizes the integral, $I(\alpha)$ has a minimum at $\alpha = 0$, so
$I'(0) = 0$. Differentiating under the integral sign and using the chain rule,

$$
I'(\alpha) = \int_{x_1}^{x_2} \parens{\frac{\partial f}{\partial y}\,\eta(x) + \frac{\partial f}{\partial y'}\,\eta'(x)}\, \d x,
$$

and setting $\alpha = 0$ gives

$$
\int_{x_1}^{x_2} \parens{\frac{\partial f}{\partial y}\,\eta + \frac{\partial f}{\partial y'}\,\eta'}\, \d x = 0.
$$

The derivative $\eta'$ can be removed by integrating the second term by parts.
Because $\eta$ vanishes at both endpoints the boundary term drops, leaving

$$
\int_{x_1}^{x_2} \eta(x)\parens{\frac{\partial f}{\partial y} - \frac{\d}{\d x}\frac{\partial f}{\partial y'}}\, \d x = 0.
$$

This holds for **every** admissible variation $\eta$. If the bracketed factor were
nonzero — say positive — at some interior point, continuity would keep it positive
on a small subinterval, and choosing an $\eta$ that is positive there and zero
elsewhere would make the integral positive, a contradiction. The bracket must
therefore vanish identically. That conclusion is the **fundamental lemma of the
calculus of variations**, and it yields Euler's equation.[^s67]

> **Theorem (Euler's equation).** If an admissible function $y(x)$ gives a
> stationary value to $I(y) = \int_{x_1}^{x_2} f(x,y,y')\,\d x$, then $y$ satisfies
>
> $$
> \frac{\d}{\d x}\parens{\frac{\partial f}{\partial y'}} - \frac{\partial f}{\partial y} = 0.
> $$

$$
% caption: The logic of the derivation: disturb the minimizing curve, demand that
% the disturbed integral be stationary in alpha, integrate by parts to strip the
% eta-prime term, and invoke the fundamental lemma.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=33mm, minimum height=11mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[b] (a) at (0,0) {embed $y$ in a\\family of neighbors};
\node[b] (b) at (4.2,0) {f\/irst variation\\vanishes};
\node[b] (c) at (8.4,0) {integrate by parts\\(boundary term $=0$)};
\node[b] (d) at (8.4,-2.1) {fundamental lemma:\\bracket $=0$};
\node[b, draw=acc, text=acc, thick] (e) at (4.2,-2.1) {Euler's equation};
\draw[->, thick] (a) -- (b);
\draw[->, thick] (b) -- (c);
\draw[->, thick] (c) -- (d);
\draw[->, thick] (d) -- (e);
\end{tikzpicture}
$$

Euler's equation is a **necessary** condition, not a sufficient one. Exactly as
$g'(x_0) = 0$ in ordinary calculus can signal a maximum, a minimum, or an
inflection, a solution of Euler's equation only makes the integral **stationary**.
Such a solution is called a **stationary function** or, when the boundary
conditions are left off, an **extremal**; whether it actually minimizes must be
settled by other means, usually the geometry or physics of the problem.
Expanding the total derivative shows the equation is second order in general,

$$
f_{y'y'}\,\frac{\d^2 y}{\d x^2} + f_{y'y}\,\frac{\d y}{\d x} + \big(f_{y'x} - f_y\big) = 0,
$$

so its extremals form a two-parameter family, and the two constants are fixed by
the boundary conditions. A second-order nonlinear equation of this kind is rarely
solvable in closed form, but three recurring special forms of $f$ collapse it to
something integrable.

## Three solvable cases

The structure of $f$ determines which terms of Euler's equation remain.

| Case | Missing from $f$ | Euler's equation reduces to | Extremals |
| --- | --- | --- | --- |
| A | $x$ and $y$ | $f_{y'y'}\,y'' = 0$, so $y'' = 0$ | straight lines $y = c_1 x + c_2$ |
| B | $y$ | $\dfrac{\d}{\d x}\dfrac{\partial f}{\partial y'} = 0$ | first integral $\dfrac{\partial f}{\partial y'} = c_1$ |
| C | $x$ | — | first integral $y'\dfrac{\partial f}{\partial y'} - f = c_1$ |

The first integral in Case C follows from a total-derivative computation. When
$x$ is absent from $f$, the combination $y'\,f_{y'} - f$ has zero total derivative
along an extremal, because

$$
\frac{\d}{\d x}\parens{y'\frac{\partial f}{\partial y'} - f} = y'\parens{\frac{\d}{\d x}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y}} - \frac{\partial f}{\partial x},
$$

and both the bracket (by Euler's equation) and $\partial f/\partial x$ (by
assumption) vanish. The surviving first integral $y'\,f_{y'} - f = c_1$ is the
**Beltrami identity**, and it turns a second-order problem into a first-order one.

> **Worked example (shortest path, Case A).** For arc length
> $f(y') = \sqrt{1 + (y')^2}$ neither $x$ nor $y$ appears, so Case A applies. Since
>
> $$
> f_{y'y'} = \frac{1}{\big[1 + (y')^2\big]^{3/2}} \neq 0,
> $$
>
> the equation forces $y'' = 0$ and the extremals are all straight lines. The
> boundary conditions pick out
>
> $$
> y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}\,(x - x_1),
> $$
>
> the segment $PQ$. The analysis alone shows only that if a stationary curve exists
> it must be this line; that the line genuinely minimizes follows because arc length
> has a smallest value but no largest one.

The same problem posed on a curved surface, where the shortest joining curve is
called a **geodesic**, is the seed of differential geometry.

> **Worked example (minimal surface of revolution, Case C).** For
> $f(y, y') = 2\pi y\sqrt{1 + (y')^2}$ the variable $x$ is missing, so the Beltrami
> identity applies. Working it out,
>
> $$
> \frac{y(y')^2}{\sqrt{1+(y')^2}} - y\sqrt{1+(y')^2} = c_1
> \quad\Longrightarrow\quad
> c_1\, y' = \sqrt{y^2 - c_1^2}.
> $$
>
> Separating variables and integrating gives
> $x = c_1\log\!\big(y + \sqrt{y^2 - c_1^2}\big) + c_2'$, and solving for $y$,
>
> $$
> y = c_1 \cosh\!\parens{\frac{x - c_2}{c_1}}.
> $$
>
> The extremals are **catenaries**, and the minimal surface — when it exists — is
> the **catenoid** swept out by revolving one.

Whether the two constants can be chosen to thread both endpoints is delicate: for
some placements of $Q$ there are two catenaries, for others one, and for others
none at all, in which case the smallest surface degenerates to two disks joined by
a thread (the Goldschmidt solution).

> **Worked example (brachistochrone, Case C).** The descent-time integrand
> $f(y,y') = \sqrt{1 + (y')^2}\big/\sqrt{2gy}$ again omits $x$. The Beltrami identity
> reduces, after simplification, to
>
> $$
> y\big[1 + (y')^2\big] = c,
> $$
>
> the very equation John Bernoulli reached in 1696 by an optical analogy. Its
> solution is the **cycloid**
>
> $$
> x = a(\theta - \sin\theta), \qquad y = a(1 - \cos\theta),
> $$
>
> traced by a point on a circle of radius $a = c/2$ rolling beneath the $x$-axis,
> the radius chosen so the first inverted arch reaches $Q$.[^s6]

The straight wire is the shortest path but not the fastest: a curve that plunges
steeply at first trades extra length for early speed, giving a shorter descent
time.

$$
% caption: From A to a lower point B, three wires: the straight segment, a
% circular arc, and the cycloid. The bead reaches B soonest along the cycloid,
% even though that path is the longest of the three.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (A) at (0,3);
\coordinate (B) at (5,0);
% straight
\draw[black, thick] (A) -- (B);
% circular arc (shallow)
\draw[black, thick] (A) .. controls (1.2,2.0) and (3.2,0.55) .. (B);
% cycloid (steep drop then flatten) -- fastest
\draw[acc, very thick] (A) .. controls (0.35,1.2) and (2.3,0.05) .. (B);
\fill[black] (A) circle (2.4pt);
\fill[black] (B) circle (2.4pt);
\node[anchor=south east] at (A) {$A$};
\node[anchor=north west] at (B) {$B$};
\node[acc, anchor=west] at (1.3,0.55) {cycloid (fastest)};
\node[black, anchor=west] at (3.15,1.95) {straight};
\node[black, anchor=west] at (4.15,1.2) {arc};
\draw[->, black] (-0.5,3) -- (-0.5,0) node[midway, left, black] {$g$};
\end{tikzpicture}
$$

The cycloid also carries the **tautochrone** property discovered by Huygens: a
bead released from any point of an inverted cycloidal arch reaches the bottom in
the same time. Bernoulli exulted that the tautochrone of Huygens and his own
brachistochrone were the same curve.

## Several unknown functions

When the integrand depends on two functions $y(x)$ and $z(x)$,

$$
I = \int_{x_1}^{x_2} f(x, y, z, y', z')\, \d x,
$$

the same argument with two independent variations $\eta_1$, $\eta_2$ produces a
**system** of Euler equations, one per unknown,

$$
\frac{\d}{\d x}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y} = 0,
\qquad
\frac{\d}{\d x}\frac{\partial f}{\partial z'} - \frac{\partial f}{\partial z} = 0.
$$

This extension carries the method into dynamics, where a configuration needs
several coordinates.

## Isoperimetric problems and Lagrange multipliers

A different class of problem fixes a second integral while making the first
stationary. The prototype is the **isoperimetric problem** the Greeks posed: among
all closed plane curves of a given length, which encloses the greatest area? The
answer is a circle, and the general pattern — extremize one integral subject to a
constraint that a second integral equal a prescribed value $c$ — takes its name
from it.

The tool is the same **Lagrange multiplier** that handles constrained extrema in
ordinary calculus. To find where $f(x,y)$ is stationary subject to $g(x,y) = 0$,
one forms $F = f + \lambda g$ and treats its stationary values as unconstrained; the
multiplier $\lambda$ removes the constraint at the cost of one extra unknown, and
it keeps the symmetry of $x$ and $y$ that eliminating a variable would destroy.[^slag]

> **Theorem (Euler's equation with an integral constraint).** To give a stationary
> value to $I = \int_{x_1}^{x_2} f(x,y,y')\,\d x$ subject to
> $J = \int_{x_1}^{x_2} g(x,y,y')\,\d x = c$, the stationary function must satisfy
> Euler's equation for $F = f + \lambda g$,
>
> $$
> \frac{\d}{\d x}\frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = 0.
> $$
>
> Its extremals carry three constants — two of integration and the multiplier
> $\lambda$ — fixed by the two boundary conditions and the constraint $J = c$.

The derivation mirrors the unconstrained one but disturbs $y$ with a
**two-parameter** family $\bar y = y + \alpha_1\eta_1 + \alpha_2\eta_2$, because a
single variation would in general break the side condition $J = c$. The parameters
$\alpha_1, \alpha_2$ are then linked by the constraint — the setting
Lagrange multipliers were built for.

> **Worked example (maximum-area arc).** To find the curve of fixed length $L$
> joining $(0,0)$ and $(1,0)$ that encloses the greatest area above the $x$-axis,
> maximize $\int_0^1 y\,\d x$ subject to $\int_0^1 \sqrt{1+(y')^2}\,\d x = L$. Here
> $F = y + \lambda\sqrt{1+(y')^2}$, and Euler's equation for $F$ works out to
> constant curvature,
>
> $$
> \frac{y''}{\big[1+(y')^2\big]^{3/2}} = \frac{1}{\lambda}.
> $$
>
> A curve of constant curvature $1/\lambda$ is a circular arc of radius $\lambda$;
> integrating twice confirms $(x - c_1)^2 + (y - c_2)^2 = \lambda^2$. Treating the
> full closed curve parametrically recovers the classical result in sharp form: a
> closed curve of length $L$ enclosing area $A$ obeys the **isoperimetric
> inequality**
>
> $$
> A \le \frac{L^2}{4\pi},
> $$
>
> with equality only for the circle.

$$
% caption: Among closed curves of a fixed perimeter $L$, area grows as the shape
% rounds out; the circle attains the isoperimetric bound $A = L^2/4\pi$, and every
% other shape of the same length encloses strictly less.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% squished ellipse
\draw[black, thick] (0,0) ellipse (1.35 and 0.5);
\node[black, anchor=north] at (0,-0.62) {small area};
% rounder ellipse
\draw[black, thick] (3.6,0) ellipse (1.1 and 0.78);
\node[black, anchor=north] at (3.6,-0.9) {larger};
% circle -- optimal
\draw[acc, very thick] (7.1,0) circle (0.95);
\node[acc, anchor=north] at (7.1,-1.05) {circle: largest area};
\draw[->, black] (1.6,1.15) -- (5.9,1.15);
\node[black, anchor=south] at (3.75,1.15) {same perimeter $L$, more area};
\end{tikzpicture}
$$

## Hamilton's principle

The method applies to mechanics, where it recasts Newton's law as
the condition that a certain integral be stationary. Consider a particle of mass
$m$ moving under a conservative force $\mathbf{F}$ derived from a potential energy
$V$, so that $\mathbf{F} = -\nabla V$. With kinetic energy $T = \tfrac{1}{2}mv^2$,
define the **action** over the motion from time $t_1$ to $t_2$ as

$$
A = \int_{t_1}^{t_2} (T - V)\, \d t.
$$

The integrand $L = T - V$ is the **Lagrangian**. For a single particle in
rectangular coordinates,

$$
L = \tfrac{1}{2}m\big(\dot x^2 + \dot y^2 + \dot z^2\big) - V(x,y,z),
$$

and requiring the action to be stationary means the Euler equations must hold in
each coordinate. Carrying out $\frac{\d}{\d t}\frac{\partial L}{\partial \dot x} - \frac{\partial L}{\partial x} = 0$ and its partners gives

$$
m\ddot x = -\frac{\partial V}{\partial x}, \qquad
m\ddot y = -\frac{\partial V}{\partial y}, \qquad
m\ddot z = -\frac{\partial V}{\partial z},
$$

that is, $m\,\ddot{\mathbf r} = \mathbf F$ — Newton's second law.[^shp]

> **Theorem (Hamilton's principle).** A particle moving from $P_1$ at time $t_1$ to
> $P_2$ at time $t_2$ under conservative forces follows the path for which the
> action $A = \int_{t_1}^{t_2}(T - V)\,\d t$ is stationary. Over a short enough time
> interval the stationary value is a minimum, and the statement is called the
> **principle of least action**.

The equivalence runs both ways: assuming Newton's law yields Hamilton's principle,
and assuming the principle yields Newton's law, so the vectorial and variational
formulations of mechanics are the same physics in two languages. The variational
form has one decisive advantage — it is written in energies alone, with no
reference to any coordinate system.

$$
% caption: The two equivalent formulations of particle dynamics: Newton's
% vectorial law and Hamilton's stationary-action principle each imply the other
% through Euler's equation applied to the Lagrangian.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=32mm, minimum height=13mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[b, draw=acc, text=acc] (n) at (0,0) {Newton\\$m\,\mathbf a = \mathbf F$};
\node[b, draw=acc, text=acc] (h) at (6.4,0) {Hamilton\\action $A$ stationary};
\node[b] (e) at (3.2,-2.0) {Euler's equation\\for the Lagrangian};
\draw[<->, thick] (n) -- (h) node[midway, above, black, font=\scriptsize] {equivalent};
\draw[->, black] (h) -- (e);
\draw[->, black] (e) -- (n);
\end{tikzpicture}
$$

For a system with constraints one drops rectangular coordinates in favor of
**generalized coordinates** $q_1,\dots,q_m$, one per degree of freedom. Hamilton's
principle then reads off **Lagrange's equations of motion**,

$$
\frac{\d}{\d t}\parens{\frac{\partial L}{\partial \dot q_j}} - \frac{\partial L}{\partial q_j} = 0,
\qquad j = 1,\dots,m,
$$

a system of second-order equations whose solution is the motion. Conservation of
energy $T + V = E$ falls out of them when $L$ has no explicit time dependence,
using Euler's theorem on the homogeneous quadratic $T$. Simmons records Planck's
verdict that among all the laws of physics the principle of least action comes
nearest to the ideal of condensing every natural phenomenon into a single
statement; it extends unchanged from one particle to systems, rigid bodies, and
continuous media, and reappears in electromagnetism, relativity, and quantum
theory.

[^s66]: **Simmons**, _Differential Equations with Applications and Historical Notes_, §66 — the shortest-path, minimal-surface-of-revolution, and brachistochrone problems as instances of minimizing $\int f(x,y,y')\,\d x$, and the definition of admissible functions.
[^s67]: **Simmons**, §67 — the variation $\bar y = y + \alpha\eta$, the condition $I'(0)=0$, integration by parts, the fundamental lemma, Euler's equation, and Cases A–C with the shortest-path, catenary, and cycloid examples.
[^s6]: **Simmons**, §6 — Bernoulli's optical (Snell's-law) solution of the brachistochrone leading to $y[1+(y')^2] = c$ and the cycloid, and Huygens's tautochrone property.
[^slag]: **Simmons**, §68 — Lagrange multipliers for finite constraints, Euler's equation for $F = f + \lambda g$ under an integral side condition, the maximum-area arc, and the isoperimetric inequality $A \le L^2/4\pi$.
[^shp]: **Simmons**, Ch. 12 Appendix B — the action $\int(T-V)\,\d t$, the Lagrangian $L = T - V$, the derivation of Newton's law from stationary action, Lagrange's equations in generalized coordinates, and conservation of energy.