Question bank — Brachistochrone problem

False. Time is $\int ds/v$, not $\int ds$; the cycloid dives steeply first so $v$ is already large over the long stretch. Shorter-but-slower loses to longer-but-faster.

False. Mass cancels in $\frac{1}{2}m{v}^2=mgy$, so $v=\sqrt{2gy}$ is mass-free; the whole functional $\int \sqrt{1+y^{\prime 2}}/\sqrt{2gy}dx$ has no $m$.

False (for shape), true (for time). $g$ appears as an overall constant $\sqrt{2g}$ multiplying the functional, so it scales every path's time equally and cannot change which path is minimal. The shape is a cycloid regardless of gravity's strength.

False. Euler–Lagrange only makes time stationary (first variation zero); it flags candidates. The cycloid is checked to be a genuine minimum separately — E–L alone cannot distinguish min from max or saddle.

False. Beltrami is a consequence of E–L, valid only when $F$ has no explicit $x$; it's a first integral (conserved quantity) that trades E–L's second-order equation for a first-order one.

False. This is the tautochrone property: both reach the bottom in the same time, $\pi \sqrt{r/g}$ to the lowest point, independent of starting height.

False. If $B$ is directly below $A$, free vertical fall is the fastest path — the cycloid degenerates into that vertical line (its steepest possible start). No horizontal distance means no advantage to trade for.

False. We minimize total time $\int ds/v$. Average speed is a summary number that discards where the speed is fast versus slow — but where matters, so it can't be the optimization target.

False. $T[y]$ is a functional: its input is an entire curve $y(x)$ and its output is one number (the time). This shift from optimizing over numbers to optimizing over functions is what launched the Calculus of Variations.

The error is the sign convention. Here $y$ is measured downward as positive, so a falling bead has $y>0$ and $v=\sqrt{2gy}$ is real. Choosing $y$ upward would flip the sign — state your convention first.

Wrong — Beltrami requires $\partial F/\partial x=0$ (no explicit $x$-dependence). If $F$ depends directly on $x$, the quantity $F-y^{\prime }{F}_{y^{\prime }}$ is not conserved and you must use the full Euler–Lagrange equation.

Backwards. As $y\to 0$, the product forces $y^{\prime 2}\to \infty$, so $y^{\prime }\to \infty$: the curve starts vertical, not flat. That steep plunge is exactly how the bead gains speed early.

At the start $y=0$ the integrand blows up like $1/\sqrt{y}$. It's an improper integral but still converges (the $1/\sqrt{y}$ singularity is integrable) — the total time is finite even though the integrand is not bounded.

Squaring hides the sign of $y^{\prime }$: on the descending part $y^{\prime }>0$, but after the lowest point of a full arch the bead rises and $y^{\prime }<0$. The single ODE $y(1+y^{\prime 2})=k$ covers both branches; you must pick the right root per segment.

$C=0$ would require $\sqrt{2gy}\sqrt{1+y^{\prime 2}}=\infty$ everywhere — degenerate, not a curve. $C$ is a positive constant fixed by making the cycloid pass through $B$; it is never zero for a real brachistochrone.

No. A sine curve is $y$ as an explicit function of $x$; the cycloid is genuinely parametric and near its cusp the slope becomes infinite — no single-valued sine can do that. It is the rim-point path of a rolling circle (see Cycloid).

Because speed varies continuously along the path; only summing the local infinitesimal times $ds/v$ correctly weights the slow regions (top) against the fast regions (bottom).

Because $F$ has no explicit $x$, Beltrami hands us a first-order ODE ($y(1+y^{\prime 2})=k$) directly, instead of the messy second-order Euler–Lagrange equation we'd otherwise integrate twice.

The substitution $y=\frac{k}{2}(1-\cos \theta )$ that solves $y(1+y^{\prime 2})=k$ produces exactly $x=r(\theta -\sin \theta )$ — the algebra forces the rolling-circle curve; no other shape satisfies the ODE.

The factor $(1-\cos \theta )$ appears identically in the arc length and in the speed, so it cancels in $ds/v$. That cancellation is simultaneously what minimizes total time and what makes descent time independent of start height.

It was the first famous case of minimizing over an entire function (the curve) rather than over numbers, demanding new machinery — the Euler-Lagrange equation — beyond ordinary calculus.

Fermat says light takes the least-time path; a medium whose speed grows with depth like $\sqrt{2gy}$ bends light into a cycloid, so the bead and a light ray obey the same "least time" law.

Then $B$ is not lower, so a bead starting from rest can never reach it (no potential energy to convert). The problem as posed needs $B$ strictly below $A$.

The cycloid still applies but uses a large radius $r$ and small $\theta$-range; the curve dips below the chord and can even pass below $B$'s height then rise back up — going deeper than necessary to gain speed for the long horizontal run.

The horizontal distance $\to 0$, the cycloid's arch collapses onto a vertical line — the bead simply free-falls, which is indeed the fastest descent when there's nothing horizontal to cover.

Yes. For a full arch past the lowest point the bead rises toward $B$ (still on the same cycloid). It sacrifices a little to plunge deep early, then climbs slightly — net time is still minimal.

Time to the lowest point is $\pi \sqrt{r/g}$, so doubling $r$ multiplies the time by $\sqrt{2}$ — a larger, gentler arch takes longer even though the bead reaches higher speeds.

The bead starts at rest with $v=0$ there, and the vertical tangent just means it drops straight down for an instant; the time contribution stays finite because the singular integrand is integrable.