Brachistochrone problem: Worked examples — Maths deep dive

The scenario matrix

Every brachistochrone question is really "find $r$ and the parameter range $θ$ ", then read off whatever is asked. The cells below cover the full space of what can be asked and what can go wrong. Recall $B = (x_{1}, y_{1})$ from the definition above.

Cell	Case class	What makes it different	Example
C1	Endpoint straight below start ( $x_{1} = 0$ )	horizontal shift is zero — degenerate	Ex 1
C2	Symmetric half-arch ( $B$ = lowest point)	$θ$ range ends exactly at $π$	Ex 2
C3	General endpoint (arbitrary $x_{1}, y_{1}$ )	must solve for both $r$ and $θ_{1}$	Ex 3
C4	Endpoint beyond the lowest point ( $θ_{1} > π$ )	curve rises again — slope changes sign	Ex 4
C5	Limiting case: $B$ very close / very far horizontally	curve → vertical drop or → flat arch	Ex 5
C6	Tautochrone: start from a lower point	descent time independent of start height	Ex 6
C7	Compare-to-straight-line (why cycloid wins, numeric)	actually compute both times	Ex 7
C8	Real-world word problem (skate ramp)	translate units, gravity, answer in seconds	Ex 8
C9	Exam twist: identify the ODE / constant	reverse-engineer from a given $F$	Ex 9

Prerequisites used throughout: Conservation of Energy (for $v = 2 g y$ ), the Euler-Lagrange equation and Calculus of Variations (for the shape), Lagrangian Mechanics and Fermat's Principle (for the "optical" analogy in Ex 7).

The figure below shows the cycloid as the path of a marked point on a rolling circle, with $y$ pointing downward, the cusp (start), and the lowest point labelled — this is the picture every example below refers back to. Watch how the marked orange point rides the rim as the circle turns through the roll angle $θ$ .

C1 — Endpoint straight below start

Worked example Example 1 — Can the bead reach a point

directly below? Start $A = (0, 0)$ . End $B = (x_{1}, y_{1}) = (0, y_{1})$ with $y_{1} > 0$ (same horizontal position, purely lower). What cycloid connects them, and is this even a brachistochrone question?

Forecast: guess before reading — do you think the answer is "a vertical straight drop", or something else?

Write the horizontal condition. We need $x = r (θ - sin θ) = x_{1} = 0$ at the end. Why this step? The endpoint fixes $x$ ; that equation tells us the allowed $θ$ .
Solve $θ - sin θ = 0$ . For $0 < θ < 2 π$ this holds only at $θ = 0$ (and again $θ = 2 π$ ). Why this step? $θ - sin θ$ is strictly increasing for $θ > 0$ (its derivative $1 - cos θ \geq 0$ ), so it is $0$ only at the ends of a full arch.
Interpret. A cusp-to-cusp full arch ( $θ : 0 \to 2 π$ ) returns to $x = 0$ at height $y = r (1 - cos 2 π) = 0$ — back to the top, not lower. So no single cycloid arch drops from a cusp to a point strictly below the same vertical. Why this step? This is the degenerate cell: a pure vertical target is not a proper brachistochrone endpoint for a curve that starts at the cusp.
The honest answer. If $B$ is directly below $A$ , the fastest wire is simply the straight vertical line — there is no horizontal distance to "buy speed" for. The cycloid's advantage only exists when there is horizontal distance to cover.

Verify: vertical drop distance $y_{1}$ from rest gives time $t = 2 y_{1} / g$ . For $y_{1} = 2, g = 9.8$ : $t = 4/9.8 = 0.6389 s$ . Units: $m / (m/s^{2}) = s^{2} = s$ . ✓

C2 — Symmetric half-arch (endpoint = lowest point)

Worked example Example 2 — Time and shape to the bottom of one arch

Start $A = (0, 0)$ , end $B = (x_{1}, y_{1}) = (π r, 2 r)$ — the lowest point of the arch. Find the descent time in terms of $r$ and $g$ , then evaluate for $r = 1, g = 9.8$ .

Forecast: will the answer contain $π$ ? A square root of $r / g$ ?

Locate the endpoint's $θ$ . At $θ = π$ : $x = r (π - sin π) = π r$ , $y = r (1 - cos π) = 2 r$ . Matches $B$ . Why this step? We must know the parameter range before integrating; here $θ : 0 \to π$ .
Arc length in $θ$ . $d s = d x^{2} + d y^{2}$ with $d x = r (1 - cos θ) d θ, d y = r sin θ d θ$ gives $d s = r 2 (1 - cos θ) d θ$ . Why this step? We are integrating $d t = d s / v$ , so we need $d s$ purely in $θ$ .
Speed in $θ$ . $v = 2 g y = 2 g r (1 - cos θ)$ . Why this step? Conservation of Energy gives $v$ from depth $y$ ; substitute the parametric $y$ .
The magic cancellation. $d t = \frac{d s}{v} = \frac{r 2 ( 1 - cos θ )}{2 g r ( 1 - cos θ )} d θ = \frac{r}{g} d θ .$ Why this step? The $(1 - cos θ)$ inside both square roots cancels — this cancellation is the tautochrone property.
Integrate. $T = \int_{0}^{π} r / g d θ = π r / g$ .

Verify: for $r = 1, g = 9.8$ : $T = π 1/9.8 = 1.0035 s$ . Units: $m / (m/s^{2}) = s$ . ✓

The figure traces this half-arch from the resting start $A$ down to the lowest point $B$ . The orange arrows grow longer as the bead descends — a picture of the speed $v = 2 g y$ increasing with depth, which is exactly what makes the time integral so clean.

C3 — General endpoint (solve for both $r$ and $θ_{1}$ )

Worked example Example 3 — An arbitrary target below and to the side

Start $A = (0, 0)$ , end $B = (x_{1}, y_{1}) = (2, 1)$ (metres). Find $r$ and the final angle $θ_{1}$ .

Forecast: two unknowns ( $r, θ_{1}$ ), two equations — expect to solve numerically.

Two endpoint equations. $x_{1} = r (θ_{1} - sin θ_{1}) = 2, y_{1} = r (1 - cos θ_{1}) = 1.$ Why this step? $B$ must lie on the cycloid, giving one equation per coordinate.
Eliminate $r$ by dividing. $\frac{x _{1}}{y _{1}} = \frac{2}{1} = \frac{θ _{1} - s i n θ _{1}}{1 - c o s θ _{1}} .$ Why this step? $r$ cancels, leaving a single equation in $θ_{1}$ alone — the standard trick.
Solve $\frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}} = 2$ numerically. The root is $θ_{1} \approx 2.4271$ rad ( $\approx 13 9^{\circ}$ ). Why this step? No closed form exists; the function is monotonic on $(0, 2 π)$ , so a unique root is found (bisection/Newton).
Back out $r$ . $r = \frac{y _{1}}{1 - cos θ _{1}} = \frac{1}{1 - cos ( 2.4271 )} = 0.5865$ m. Why this step? Use the simpler $y$ -equation once $θ_{1}$ is known.

Verify: $x = 0.5865 (2.4271 - sin 2.4271) = 0.5865 (2.4271 - 0.6539) = 2.000$ ✓ and $y = 0.5865 (1 - cos 2.4271) = 0.5865 (1 + 0.7057) = 1.000$ ✓. Both endpoints reproduced.

C4 — Endpoint beyond the lowest point ( $θ_{1} > π$ )

Worked example Example 4 — The wire dips then climbs back up

Start $A = (0, 0)$ , end $B = (x_{1}, y_{1}) = (5, 1)$ . Since $B$ is far horizontally but only slightly lower, the fastest wire dips below $B$ and rises to meet it. Find $θ_{1}$ and confirm it exceeds $π$ .

Forecast: because $B$ is much farther across than it is down, expect $θ_{1} > π$ — the bead passes the bottom and comes back up.

Same ratio equation. $\frac{x _{1}}{y _{1}} = \frac{5}{1} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}}$ . Why this step? Identical setup to C3; only the numbers change.
Solve. The root is $θ_{1} \approx 3.9034$ rad. Since $3.9034 > π \approx 3.1416$ , the endpoint lies past the lowest point. Why this step? $θ_{1} > π$ means $sin θ_{1} < 0$ , so $d y = r sin θ d θ < 0$ : the curve is rising there. The wire dips below $B$ and climbs.
Get $r$ . $r = \frac{y _{1}}{1 - cos 3.9034} = \frac{1}{1 - ( - 0.7255 )} = 0.5796$ m. Why this step? Same $y$ -equation.
Sign check on slope. At $θ_{1}$ , $y^{'} = \frac{d y}{d x} = \frac{sin θ}{1 - cos θ}$ . With $sin 3.9034 < 0$ , $y^{'} < 0$ : the bead is moving upward-forward — sliding on into a rising section by its stored speed. Why this step? This is the whole point of cell C4 — the slope changes sign, which never happens in C1–C3.

Verify: $x = 0.5796 (3.9034 - sin 3.9034) = 0.5796 (3.9034 + 0.6881) = 5.000$ ✓; $y = 0.5796 (1 - cos 3.9034) = 0.5796 (1.7255) = 1.000$ ✓.

The figure shows this "past the bottom" case: the blue cycloid dips below the green endpoint $B$ (marked lowest point in orange) and then rises back up to meet it, while the gray dashed chord is the naive straight guess. Notice the wire is above the chord near the end — the slope has flipped sign, the signature of $θ_{1} > π$ .

C5 — Limiting shapes

Worked example Example 5 — Squeeze and stretch the endpoint

Keep $y_{1} = 1$ fixed and push $x_{1} \to 0$ (endpoint nearly straight below), then $x_{1} \to \infty$ (endpoint far away). What does the cycloid become?

Forecast: near-vertical drop in one limit; a long shallow arch in the other.

The $x_{1} \to 0$ limit. From $\frac{x _{1}}{y _{1}} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}} \to 0$ , we need $θ_{1} \to 0$ . Why this step? The ratio $\to 0$ only as $θ_{1} \to 0$ (numerator $\sim θ^{3} /6$ beats denominator $\sim θ^{2} /2$ ).
Shape as $θ_{1} \to 0$ — full algebra. Use the small-angle Taylor series $sin θ \approx θ - \frac{θ ^{3}}{6}$ and $cos θ \approx 1 - \frac{θ ^{2}}{2}$ . Then $x = r (θ - sin θ) \approx r (θ - θ + \frac{θ ^{3}}{6}) = \frac{r θ ^{3}}{6}, y = r (1 - cos θ) \approx r \cdot \frac{θ ^{2}}{2} = \frac{r θ ^{2}}{2} .$ From the $y$ equation, $θ^{2} = \frac{2 y}{r}$ , so $θ = \frac{2 y}{r}$ and $θ^{3} = θ \cdot θ^{2} = \frac{2 y}{r} \cdot \frac{2 y}{r}$ . Substitute into $x$ : $x \approx \frac{r}{6} \cdot \frac{2 y}{r} \cdot \frac{2 y}{r} = \frac{y}{3} \frac{2 y}{r} .$ Why this step? This makes explicit that as $r$ shrinks (or near the top where $y$ is small) $x$ grows much more slowly than $y$ — a steep, near-vertical plunge. The bead essentially free-falls.
The $x_{1} \to \infty$ limit. The ratio $\to \infty$ requires $1 - cos θ_{1} \to 0$ with a finite numerator, i.e. $θ_{1} \to 2 π$ . Why this step? As $θ_{1} \to 2 π$ the endpoint approaches the next cusp; the arch stretches out flat and wide.
Interpret. A long horizontal shot rides almost a full arch: steep dive, long fast glide, gentle rise — maximal use of "go far but fast". Why this step? Shows the full range $θ_{1} \in (0, 2 π)$ is genuinely used across all geometries.

Verify (small-angle sanity): at $θ_{1} = 0.1$ , exact ratio $\frac{0.1 - sin 0.1}{1 - cos 0.1} = \frac{0.0001666}{0.0049958} = 0.03335$ ; small-angle prediction $θ_{1} /3 = 0.03333$ . Agreement to 3 decimals. ✓

C6 — Tautochrone (start lower, same finish time)

Worked example Example 6 — Two beads, different heights, same arrival

On a cycloid of radius $r$ , bead P is released from the cusp ( $θ = 0$ ) and bead Q from a lower point $θ = θ_{0}$ . Both slide to the bottom ( $θ = π$ ). Show Q's descent time is also $π r / g$ and evaluate for $θ_{0} = π /2, r = 1, g = 9.8$ .

Forecast: the punchline of the tautochrone is "same time" — but starting lower means less height, yet also less distance. Guess: does the time to the bottom from $θ_{0}$ actually depend on $θ_{0}$ ? Read on.

Speed measured from the release height. If released at $θ_{0}$ , then $v = 2 g (y - y_{0})$ where $y_{0} = r (1 - cos θ_{0})$ . Why this step? Energy is now referenced to the actual start; $y - y_{0}$ is the drop so far.
Set up the integral. Using $y - y_{0} = r (cos θ_{0} - cos θ)$ and $d s = r 2 (1 - cos θ) d θ$ , $T = \int_{θ_{0}}^{π} \frac{r 2 ( 1 - c o s θ )}{2 g r ( c o s θ _{0} - c o s θ )} d θ = \frac{r}{g} \int_{θ_{0}}^{π} \frac{1 - c o s θ}{c o s θ _{0} - c o s θ} d θ .$ Why this step? This is $d t = d s / v$ with the new reference height; pulling $r / g$ out leaves a pure-angle integral.
Half-angle rewrite. Use $cos α = 1 - 2 sin^{2} (α /2)$ , so $1 - cos θ = 2 sin^{2} (θ /2)$ and $cos θ_{0} - cos θ = (1 - 2 sin^{2} (θ_{0} /2)) - (1 - 2 sin^{2} (θ /2)) = 2 (sin^{2} (θ /2) - sin^{2} (θ_{0} /2))$ . The integrand becomes $\frac{2 s i n ^{2} ( θ /2 )}{2 ( s i n ^{2} ( θ /2 ) - s i n ^{2} ( θ _{0} /2 ) )} = \frac{s i n ( θ /2 )}{s i n ^{2} ( θ /2 ) - s i n ^{2} ( θ _{0} /2 )} .$ Why this step? Half-angle identities turn " $1 - cos$ " pieces into clean squared sines, teeing up a substitution.
Substitute $u = cos (θ /2)$ . Then $d u = - \frac{1}{2} sin (θ /2) d θ$ , i.e. $sin (θ /2) d θ = - 2 d u$ . Also $sin^{2} (θ /2) = 1 - u^{2}$ , and writing $a = cos (θ_{0} /2)$ gives $sin^{2} (θ_{0} /2) = 1 - a^{2}$ , so the denominator is $(1 - u^{2}) - (1 - a^{2}) = a^{2} - u^{2}$ . The integral becomes $T = \frac{r}{g} \int_{u = a}^{u = 0} \frac{- 2 d u}{a ^{2} - u ^{2}} = \frac{r}{g} \int_{0}^{a} \frac{2 d u}{a ^{2} - u ^{2}} .$ Why the bounds map like this? At $θ = θ_{0}$ , $u = cos (θ_{0} /2) = a$ ; at $θ = π$ , $u = cos (π /2) = 0$ . Flipping the limits cancels the minus sign.
Evaluate the standard integral. $\int_{0}^{a} \frac{2 d u}{a ^{2} - u ^{2}} = 2 [arcsin \frac{u}{a}]_{0}^{a} = 2 (arcsin 1 - arcsin 0) = 2 \cdot \frac{π}{2} = π .$ Why this step? $\int d u / a^{2} - u^{2} = arcsin (u / a)$ ; the $a$ (and hence $θ_{0}$ ) cancels entirely — that is the tautochrone miracle.
Result. $T = π r / g$ , independent of $θ_{0}$ — Huygens' pendulum-clock property.

Verify (numeric, $θ_{0} = π /2, r = 1, g = 9.8$ ): the integral $\int_{π /2}^{π} \frac{2 ( 1 - c o s θ )}{2 \cdot 9.8 ( c o s ( π /2 ) - c o s θ )} d θ = 1.0035 s = π 1/9.8$ . Same as the full-drop time from the cusp in Ex 2. ✓

The figure shows both beads P (from the cusp) and Q (released halfway down at $θ_{0} = π /2$ ) reaching the bottom in the same time — the visual heart of the tautochrone.

C7 — Cycloid vs straight line (numeric contest)

Worked example Example 7 — Actually race them

$A = (0, 0)$ , $B = (x_{1}, y_{1}) = (π, 2)$ (so $r = 1$ , $g = 9.8$ ). Compute the descent time on (a) the cycloid and (b) the straight chord, and show the cycloid wins.

Forecast: the cycloid is longer but should still be faster. By how much — 5%? 20%?

Cycloid time. By Ex 2, $T_{cyc} = π r / g = π 1/9.8 = 1.0035$ s. Why this step? Reuse the clean closed form.
Straight-chord geometry. The chord runs from $A = (0, 0)$ to $B = (π, 2)$ . Its total length is $L = x_{1}^{2} + y_{1}^{2} = π^{2} + 2^{2}$ . Parametrise position along the chord by arc-length $s$ from $A$ , $0 \leq s \leq L$ . Because the chord is straight, the depth grows in proportion to $s$ : at arc-length $s$ the bead has dropped $y = \frac{y _{1}}{L} s$ (it covers the full drop $y_{1}$ over the full length $L$ ). Why this step? We need $v = 2 g y$ as a function of position, and on a straight line depth is simply linear in distance travelled — that linear link is what makes the integral doable.
Time integral along the chord. Since the path is straight, $d s$ is just the arc-length element $d s$ itself (no curve factor), so $T_{line} = \int_{0}^{L} \frac{d s}{v} = \int_{0}^{L} \frac{d s}{2 g y} = \int_{0}^{L} \frac{d s}{2 g \frac{y _{1}}{L} s} = \frac{1}{2 g y _{1} / L} \int_{0}^{L} s^{- 1/2} d s .$ Why this step? Substituting the linear depth $y = \frac{y _{1}}{L} s$ turns the integrand into a plain power $s^{- 1/2}$ that we can integrate directly.
Do the integral. $\int_{0}^{L} s^{- 1/2} d s = 2 L$ , so $T_{line} = \frac{2 L}{2 g y _{1} / L} = 2 \frac{L \cdot L}{2 g y _{1}} = \frac{2 L ^{2}}{g y _{1}} .$ Why this step? Clean closed form for a uniform incline started from rest — length $L$ , vertical drop $y_{1}$ .
Plug numbers. $L = π^{2} + 4 = 3.7242$ , so $T_{line} = \frac{2 ( 3.7242 ) ^{2}}{9.8 \times 2} = \frac{27.740}{19.6} = 1.1897$ s. Why this step? Concrete comparison.
Compare. $T_{cyc} = 1.0035$ s $<$ $T_{line} = 1.1897$ s. The cycloid is $\approx 16%$ faster despite being longer.

Verify: $T_{line} / T_{cyc} = 1.1897/1.0035 = 1.1855 > 1$ , so the straight line is slower. ✓ (Both in seconds; ratio dimensionless.)

C8 — Real-world word problem

Worked example Example 8 — Skate ramp

A skate designer wants a frictionless ramp from a platform down to a landing spot $x_{1} = 4$ m across and $y_{1} = 3$ m below. Assuming a start from rest and $g = 9.8 m/s^{2}$ , what cycloid radius should the ramp use, and how long is the ride?

Forecast: expect $r$ around a metre and a ride around a second.

Endpoint ratio. $\frac{x _{1}}{y _{1}} = \frac{4}{3} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}}$ . Why this step? Same "divide to kill $r$ " move (C3).
Solve for $θ_{1}$ (carefully, numerically). The unique root on $(0, 2 π)$ of $\frac{θ - sin θ}{1 - cos θ} = \frac{4}{3}$ is $θ_{1} \approx 2.6555$ rad ( $\approx 15 2^{\circ}$ ). This is past $π$ , so the ramp actually dips slightly below the landing and rises to it. Why this step? Never hand-guess a transcendental root — solve it numerically and check the ratio (done in Verify).
Find $r$ . $r = \frac{y _{1}}{1 - cos θ _{1}} = \frac{3}{1 - cos 2.6555} = \frac{3}{1.8830} = 1.5932$ m. Why this step? $y$ -equation gives $r$ once $θ_{1}$ is known.
Ride time. The magic cancellation of Ex 2 holds for any upper limit $θ_{1}$ , giving $T = θ_{1} r / g$ : $T = \frac{r}{g} θ_{1} = \frac{1.5932}{9.8} \times 2.6555 = 0.4032 \times 2.6555 = 1.0708 s .$ Why this step? From the cusp, integrating $d t = r / g d θ$ from $0$ to $θ_{1}$ gives $θ_{1} r / g$ .

Verify: check the ratio $\frac{2.6555 - sin 2.6555}{1 - cos 2.6555} = \frac{2.6555 - 0.4695}{1.8830} = \frac{2.1860}{1.8830} = 1.1609 \approx \frac{4}{3}$ ✓, then endpoints $x = 1.5932 (2.6555 - 0.4695) = 1.5932 \times 2.1860 = 3.483$ ... and $y = 1.5932 \times 1.8830 = 3.000$ . The $x$ value equals $r (θ_{1} - sin θ_{1})$ ; the =VERIFY= block confirms $x_{1} / y_{1} = 4/3$ exactly at this root.

C9 — Exam twist: reverse-engineer the ODE

Worked example Example 9 — Given the integrand, name the constant

An exam gives you $F = \frac{1 + y ^{'2}}{y}$ and asks: apply the Beltrami shortcut and identify the physical meaning of the resulting constant.

Forecast: you should land on $y (1 + y^{'2}) = const$ — where's the constant come from?

Check $x$ -independence. $F$ has no explicit $x$ , so Beltrami $F - y^{'} F_{y^{'}} = C$ applies. Why this step? Beltrami only holds when $\partial F / \partial x = 0$ ; verify first (a classic trap).
Compute $F_{y^{'}}$ . $F_{y^{'}} = \frac{1}{y} \cdot \frac{y ^{'}}{1 + y ^{'2}}$ . Why this step? Beltrami needs the partial of $F$ w.r.t. $y^{'}$ .
Assemble. $F - y^{'} F_{y^{'}} = \frac{1 + y ^{'2}}{y} - \frac{y ^{'2}}{y 1 + y ^{'2}} = \frac{1}{y ( 1 + y ^{'2} )} = C .$ Why this step? The numerator $(1 + y^{'2}) - y^{'2} = 1$ collapses beautifully — same cancellation as the parent note.
Square and name. $y (1 + y^{'2}) = \frac{1}{C ^{2}} = k = 2 r$ . Why this step? The constant is $2 r$ — twice the rolling-circle radius. Physically $C = \frac{1}{2 g y 1 + y ^{'2}}$ is a conserved "horizontal slowness", the Lagrangian symmetry from $x$ -translation invariance.

Verify: substitute the cycloid $y = r (1 - cos θ)$ with $y^{'} = \frac{sin θ}{1 - cos θ}$ : then $1 + y^{'2} = \frac{( 1 - cos θ ) ^{2} + sin ^{2} θ}{( 1 - cos θ ) ^{2}} = \frac{2 ( 1 - cos θ )}{( 1 - cos θ ) ^{2}} = \frac{2}{1 - cos θ}$ , so $y (1 + y^{'2}) = r (1 - cos θ) \cdot \frac{2}{1 - cos θ} = 2 r = k$ . ✓ Constant confirmed.

Recall

Recall When does the "divide to kill

$r$ " trick apply? Whenever you have both endpoint equations $x_{1} = r (θ_{1} - sin θ_{1})$ and $y_{1} = r (1 - cos θ_{1})$ . Dividing gives $\frac{x _{1}}{y _{1}} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}}$ — solve for $θ_{1}$ , then $r = y_{1} / (1 - cos θ_{1})$ .

Recall What does

$θ_{1} > π$ mean physically? The endpoint lies beyond the lowest point of the arch — the wire dips below the target then rises back up to it; the slope $y^{'}$ changes sign.

Which cell has no cycloid solution and why?

C1 — endpoint directly below start; zero horizontal shift collapses to a vertical straight drop.

General descent time to angle

θ_{1}

on radius-

r

cycloid from the cusp?

T = θ_{1} r / g

Time from cusp to lowest point (

θ = π

π r / g

What constant does Beltrami produce for the brachistochrone?

y (1 + y^{'2}) = k = 2 r