The brachistochrone, and how to differentiate with respect to a curve

Idea. Each part is the same two moves — compute \(F_y\) and \(F_{y'}\), or the Beltrami combination when \(x\) is absent — and then a question about what the equation does not say.

1. \(F_y = 2k^2y\), \(F_{y'} = 2y'\), so Euler–Lagrange is \(y'' = k^2y\) (and \(F_{y'y'} = 2 \ne 0\), so the extremal is \(C^2\) by P3). Multiply by \(y'\) and integrate, as in unit 00: \(y'^2 = k^2y^2 + E\) for a constant \(E\), and at \(x = 0\) this says \(E = y'(0)^2 \ge 0\). Try \(E = k^2b^2\) with \(b \gt 0\); then \(y'\) never vanishes, and is positive since \(y(1) \gt y(0)\). Separate and substitute \(y = b\sinh s\): then \(\mathrm dy/\sqrt{k^2y^2 + k^2b^2} = \mathrm ds/k = \mathrm dx\), so \(s = kx\) (as \(s = 0\) at \(x = 0\)), and \(y(1) = 1\) fixes \(b = 1/\sinh k\): \[ y(x) = \frac{\sinh kx}{\sinh k}. \] This is the only extremal. If \(y_1\), \(y_2\) are two, \(w = y_1 - y_2\) has \(w'' = k^2w\) and \(w(0) = w(1) = 0\); integrating \(w'^2\) by parts, \(\int_0^1(w'^2 + k^2w^2)\,\mathrm dx = [ww']_0^1 - \int_0^1 w(w'' - k^2w)\,\mathrm dx = 0\), so \(w \equiv 0\). So the other choice of constant, \(E = 0\), needs no separate treatment: whatever it would give, it cannot be a second extremal. At \(k = 2\): \(y(\tfrac12) = \sinh 1/\sinh 2 = 1.17520/3.62686 = 0.32403\), below the straight line's \(0.5\): the \(k^2y^2\) term penalises height, so the curve sags towards zero. As \(k\to0\), \(\sinh kx/\sinh k = (kx + O(k^3))/(k + O(k^3)) \to x\): the straight line, the extremal of \(\int y'^2\).

2. \(F = \sqrt{1+p^2}/y\) has no \(x\). \(F_{p} = p/(y\sqrt{1+p^2})\), and \(F - pF_p = \big((1+p^2) - p^2\big)/(y\sqrt{1+p^2}) = 1/(y\sqrt{1+p^2})\). Setting it equal to \(1/C\) with \(C \gt 0\): \(y^2(1 + y'^2) = C^2\), that is \(y^2 + (yy')^2 = C^2\). Since \(y^2 \le C^2\), write \(yy' = \pm\sqrt{C^2 - y^2}\); then \(\frac{\mathrm d}{\mathrm dx}\sqrt{C^2-y^2} = -yy'/\sqrt{C^2-y^2} = \mp1\), so \(\sqrt{C^2 - y^2} = \mp(x - x_0)\), and squaring, \[ (x - x_0)^2 + y^2 = C^2 : \] a circle of radius \(C\) centred at \((x_0, 0)\) on the axis. Check: differentiating, \(yy' = -(x-x_0)\), and \(y^2(1 + y'^2) = y^2 + (x-x_0)^2 = C^2\) ✓. The graph formulation misses the vertical lines \(x = \text{const}\), which are extremals too — the limit \(C\to\infty\) of the circles. Each upper semicircle is itself a graph over \((x_0 - C, x_0 + C)\), its vertical tangents occurring only where it meets the axis, outside the region \(y \gt 0\). (This is the upper half-plane model of hyperbolic geometry; the circles are its geodesics.)

3. \(F = yy'\): \(F_y = y'\) and \(F_{y'} = y\), so \(\frac{\mathrm d}{\mathrm dx}F_{y'} - F_y = y' - y' = 0\) for every curve. The reason: \(yy' = \big(\tfrac12y^2\big)'\), so \(J[y] = \tfrac12\big(y(b)^2 - y(a)^2\big)\) depends only on the fixed endpoint values, and its first variation vanishes in every direction. An integrand that is a total derivative is a null Lagrangian; adding one to \(F\) changes no Euler–Lagrange equation. Unit 12 uses exactly this freedom.

4. Up to the constant \(2\pi\), \(F = y\sqrt{1+p^2}\) has no \(x\), and \(F - pF_p = y/\sqrt{1+p^2} = c\). This is the deck's catenary computation with \(\mu = 0\) and \(\rho g = 1\), so \(y = c\cosh((x-x_0)/c)\), and \(x_0 = 0\) by the symmetry of the rings. The endpoint condition is \(R = c\cosh(d/c)\). With \(t = d/c\), this reads \(R/d = \cosh t/t =: m(t)\). Now \(m(t)\to\infty\) as \(t\to0\) and as \(t\to\infty\), and \(m'(t) = (t\sinh t - \cosh t)/t^2\) vanishes where \(t\tanh t = 1\), which has the single positive root \(t_* = 1.19968\) (the left side increases from \(0\) to \(\infty\)). There \(m(t_*) = 1.50888\). So a root exists if and only if \(R/d \ge 1.50888\), i.e. \(d/R \le 1/1.50888 = 0.66274\).

At \(d/R = 0.5\) (take \(R = 1\)), \(1 = c\cosh(0.5/c)\) has two roots: \(c = 0.84834\) and \(c = 0.23509\). Their areas, \(2\pi\int_{-0.5}^{0.5}c\cosh^2(x/c)\,\mathrm dx\), are \(5.99180\) and \(6.84566\) (in units of \(R^2\)), against \(2\pi = 6.28319\) for the two discs. The shallow catenary beats the discs; the deep one is an extremal that loses even to them — a first sign, before section 5 of the deck, that extremal does not mean minimal.

At \(d/R = 1\) there is no extremal at all. Yet the area can be made as close to \(2\pi R^2\) as desired: take a curve that drops almost vertically from \(R\) to nearly \(0\) near each ring and runs along close to the axis in between. The limit — two discs joined by a segment of the axis, the Goldschmidt curve — is not a graph with \(y \gt 0\), so the Euler–Lagrange equation, which only sees admissible graphs, cannot find it. Physically the film pops and leaves two discs.

Check. Dimensions in part 4: \(c\) is a length, \(d/c\) is a number ✓. Limit in part 1: \(k\to0\) recovers the line ✓. Limit in part 4: as \(d\to0\) the shallow root has \(c\to R\), the film becomes a short cylinder of area \(\approx 2\pi R\cdot 2d \to 0\), far below the discs' \(2\pi R^2\) ✓.

Idea. Energy gives the speed as a function of depth, the functional has no \(x\), Beltrami gives a first-order equation, and a half-angle substitution integrates it.

1. The wire's reaction is normal to the velocity and does no work, so \(\tfrac12mv^2 = mgy\) and \(v = \sqrt{2gy}\). Arc length along a graph is \(\mathrm ds = \sqrt{1+y'^2}\,\mathrm dx\). So \[ T[y] = \int_0^{1}\sqrt{\frac{1+y'^2}{2gy}}\;\mathrm dx,\qquad y(0) = 0,\ y(1) = 1 . \]

2. \(F = \sqrt{1+p^2}/\sqrt{2gy}\) has no \(x\). \(F_p = p/(\sqrt{1+p^2}\sqrt{2gy})\) and \(F - pF_p = 1/(\sqrt{1+p^2}\sqrt{2gy})\), a positive constant, which we call \(1/\sqrt{4ga}\). Squaring and inverting: \(2gy(1+y'^2) = 4ga\), i.e. \(y(1+y'^2) = 2a\).

3. \(y'^2 = (2a-y)/y\); on the descending part \(\mathrm dx = \sqrt{y/(2a-y)}\,\mathrm dy\). With \(y = 2a\sin^2(\theta/2) = a(1-\cos\theta)\): \(\sqrt{y/(2a-y)} = \tan(\theta/2)\) and \(\mathrm dy = 2a\sin(\theta/2)\cos(\theta/2)\,\mathrm d\theta\), so \(\mathrm dx = 2a\sin^2(\theta/2)\,\mathrm d\theta = a(1-\cos\theta)\,\mathrm d\theta\). Integrating from \(A\), where \(\theta = 0\): \(x = a(\theta - \sin\theta)\).

4. At \(B\), \(x/y = 1\): \((\theta_B - \sin\theta_B)/(1-\cos\theta_B) = 1\), whose unique root in \((0, 2\pi)\) (the left side is strictly increasing — the deck's constants slide) is \(\theta_B = 2.41201\), i.e. \(138.20^\circ\). Then \(a = 1/(1-\cos\theta_B) = 0.57292\) m. Along the cycloid \(\mathrm dt = \sqrt{a/g}\,\mathrm d\theta\), so \[ T = \theta_B\sqrt{a/g} = 2.41201\times0.24167 = 0.58290\ \mathrm s . \] The straight wire has length \(\sqrt2\) m and constant acceleration \(g/\sqrt2\), so \(T = \sqrt{2\cdot 2/(g\cdot1)} = 0.63855\) s: the cycloid is \(8.7\%\) quicker. Since \(\theta_B \lt \pi\), the bead is still descending at \(B\): this cycloid does not dip below \(B\) (its lowest point, \(2a = 1.14583\) m deep, lies beyond).

5. Free fall through \(1\) m takes \(\sqrt{2/g} = 0.45152\) s, but it ends at \((0, 1)\), not at \(B\). It is not an admissible curve, so it is not a competitor. It is the limit \(X/Y\to0\) of the brachistochrone, checked on the deck.

Check. Dimensions: \(a\) in metres, \(\sqrt{a/g}\) in seconds ✓. Back-substitution: \(y' = \cot(\theta/2)\) and \(y(1+y'^2) = a(1-\cos\theta)/\sin^2(\theta/2) = 2a\) ✓. Ordering: \(0.45152 \lt 0.58290 \lt 0.63855\) s, free fall fastest, line slowest ✓.

Another way

Because \(\theta_B \lt \pi\), the curve is also a graph \(x(y)\), and \(T = \int_0^1\sqrt{(1+x'^2)/(2gy)}\,\mathrm dy\) has no \(x\) at all. So the Euler–Lagrange equation says at once that \(F_{x'} = x'/\sqrt{2gy(1+x'^2)}\) is constant — the same first integral, with no Beltrami needed. Better: \(F\) is strictly convex in \(x'\) (\(F_{x'x'} = (2gy)^{-1/2}(1+x'^2)^{-3/2} \gt 0\)) and does not depend on \(x\), so for any admissible \(\eta\) convexity gives \(F(y, x'+\eta') \ge F(y,x') + F_{x'}\eta'\) pointwise, and integrating, \(T[x+\eta] \ge T[x] + \delta T[x;\eta] = T[x]\). The extremal is the global minimum among curves that are graphs over \(y\) — a complete proof, available because the endpoint is before the bottom of the arch.

Idea. Integrate by parts in the direction that needs no unknown derivative; then the slope of the extremal is defined implicitly by a \(C^1\) equation, and the implicit function theorem makes it \(C^1\).

1. Let \(A(x) = \int_a^xM(t)\,\mathrm dt\), which is \(C^1\) with \(A' = M\). For \(\eta\in C^1_0\), \(\int_a^bM\eta\,\mathrm dx = [A\eta]_a^b - \int_a^bA\eta'\,\mathrm dx = -\int_a^bA\eta'\,\mathrm dx\), since \(\eta(a) = \eta(b) = 0\). The hypothesis becomes \(\int_a^b(N - A)\eta'\,\mathrm dx = 0\) for all \(\eta\in C^1_0\). \(N - A\) is continuous, so the du Bois-Reymond lemma makes it a constant \(k\). Hence \(N = A + k\) is \(C^1\) with \(N' = A' = M\). ∎

2. The first variation of an extremal vanishes: \(\int(F_y\eta + F_{y'}\eta') = 0\) for all \(\eta\in C^1_0\), with \(F_y\) and \(F_{y'}\) evaluated along \(y\) and so continuous in \(x\). By part 1, \(\Pi(x) = F_{y'}(x,y(x),y'(x))\) is \(C^1\). Define \(G(x,p) = F_{y'}(x, y(x), p) - \Pi(x)\) near \((x_0, p_0)\), \(p_0 = y'(x_0)\). Since \(F_{y'}\) is \(C^1\) (as \(F\) is \(C^2\)) and \(y\), \(\Pi\) are \(C^1\), \(G\) is \(C^1\). It satisfies \(G(x, y'(x)) = 0\) for every \(x\), and \(\partial_pG(x_0,p_0) = F_{y'y'} \ne 0\). The implicit function theorem gives neighbourhoods \(U\ni x_0\), \(V\ni p_0\) and a \(C^1\) function \(\pi : U\to V\) such that, for \(x\in U\) and \(p\in V\), \(G(x,p) = 0\) if and only if \(p = \pi(x)\). Because \(y'\) is continuous, \(y'(x)\in V\) for \(x\) near \(x_0\), and \(G(x,y'(x)) = 0\); by uniqueness \(y'(x) = \pi(x)\) there. So \(y'\) is \(C^1\), and \(y\) is \(C^2\) near \(x_0\). ∎

3. The integrand is a square times a square, so \(J \ge 0\) always. For the given curve: on \([-1,0]\), \(y = 0\) kills the integrand; on \((0,1]\), \(y' = 2x\) makes \(2x - y' = 0\). So \(J = 0\), the least possible value, and the curve is a global minimiser. It is \(C^1\): \(y'(0^-) = 0 = y'(0^+)\). It is not \(C^2\): \(y''\) jumps from \(0\) to \(2\) at \(x = 0\). Part 2's hypothesis fails exactly there: \(F_{y'y'} = 2y^2\), which is \(0\) on the whole of \([-1,0]\), including at the kink. Where \(y \gt 0\) the hypothesis holds and \(y\) is indeed smooth.

Check. Endpoints: \(y(-1) = 0\), \(y(1) = 1\) ✓. Consistency: along the minimiser, \(\Pi = F_{y'} = -2y^2(2x - y') = 0\) identically, which is \(C^1\) as part 1 says it must be, even though \(y\) is not \(C^2\) — so part 1's conclusion is about \(F_{y'}\), not about \(y'\).

Idea. Stationary optical path, with no explicit \(x\), is exactly the Beltrami situation, and the Beltrami constant is Snell's invariant.

1. As on the deck's Snell slide: from \(P = (0, h_1)\) to \(Q = (d, -h_2)\) via \((x, 0)\), the optical path is \(n_1\sqrt{x^2+h_1^2} + n_2\sqrt{(d-x)^2+h_2^2}\); its derivative is \(n_1\sin\theta_1 - n_2\sin\theta_2\), angles measured from the normal. Setting it to zero gives Snell's law.

2. \(F = n(y)\sqrt{1+p^2}\) has no \(x\). \(F - pF_p = n(y)/\sqrt{1+p^2}\), which is therefore constant along a ray. With \(\tan\alpha = y'\), \(1/\sqrt{1+y'^2} = \cos\alpha\), so \(n\cos\alpha = \text{const}\). The angle from the vertical normal to the layers is \(\theta = \pi/2 - \alpha\), so this is \(n\sin\theta = \text{const}\): part 1 at every interface between thin layers, with the scaffolding removed — exactly Johann Bernoulli's argument for the bead, run in reverse.

3. \(n_0(1+\kappa y)/\sqrt{1+y'^2} = C\). Put \(u = 1 + \kappa y\) and \(A = C/n_0\): \(u = A\sqrt{1+y'^2}\), and since \(u' = \kappa y'\), \(u^2 = A^2(1 + u'^2/\kappa^2)\). Try \(u = A\cosh s\): then \(u' = A\sinh s\,s'\), and the equation becomes \(A^2\cosh^2 s = A^2 + A^4\sinh^2 s\,s'^2/\kappa^2\), i.e. \(s'^2 = \kappa^2/A^2\). So \(s = \kappa(x - x_0)/A\) and \[ 1 + \kappa y = A\cosh\!\Big(\frac{\kappa(x - x_0)}{A}\Big) : \] the rays are catenaries, bending upwards, towards the denser air. Check: \(y' = \sinh(\kappa(x-x_0)/A)\), \(\sqrt{1+y'^2} = \cosh(\cdot) = u/A\) ✓.

4. The grazing ray has its vertex on the road: \(y = 0\), \(y' = 0\) at \(x_0\), so \(A = 1\). Put \(x_0 = 0\): \(1 + \kappa y = \cosh(\kappa x)\). It reaches height \(h = 1.5\) m at \[ x = \frac{\operatorname{arcosh}(1 + \kappa h)}{\kappa} = \frac{\operatorname{arcosh}(1.00003)}{2.0\times10^{-5}} = 387.30\ \mathrm m, \] with slope \(\sinh(\kappa x) = 7.746\times10^{-3}\), i.e. the ray arrives travelling upwards at \(0.444^\circ\), so the eye receives it from \(0.444^\circ\) below the horizontal. Rays reaching the eye from steeper below than this left the road surface; rays from shallower below never touched the road — traced back, they bottom out above it and came from the sky. So the road looks like a sheet of reflected sky from about \(387\) m ahead: the mirage.

Check. Dimensions: \(\kappa x\) and \(\kappa h\) are pure numbers ✓. Small-angle limit: \(\cosh\kappa x\approx1 + \kappa^2x^2/2\) gives the parabola \(y = \kappa x^2/2\) and \(x = \sqrt{2h/\kappa} = 387.30\) m, agreeing to five figures ✓. Limit \(\kappa\to0\): \(x\to\infty\), no mirage in uniform air ✓.

Idea. A free end contributes one boundary condition that nobody imposed: the boundary term of the first variation must vanish by itself.

1. \(F = \tfrac12\tau y'^2 - fy\): \(F_y = -f\), \(F_{y'} = \tau y'\). Euler–Lagrange: \(\tau y'' = -f\). Natural condition at \(x = L\): \(\tau y'(L) = 0\). Integrating, \(\tau y' = f(L - x)\) (the constant chosen so that \(y'(L) = 0\)), and with \(y(0) = 0\), \[ y(x) = \frac{f}{\tau}\Big(Lx - \frac{x^2}{2}\Big),\qquad y(L) = \frac{fL^2}{2\tau} . \] With the numbers, \(y(L) = 2\times1/(2\times50) = 0.020\) m: the ring sits \(2.0\) cm down.

2. In the small-slope approximation the vertical component of the tension at the ring is \(\tau y'(L)\). The ring is massless and the rod frictionless, so the rod can push only horizontally: nothing can balance a vertical component, and it must vanish. The minimisation discovers the force balance on its own.

3. The quickest wire is still a cycloid from \(A\) (the Euler–Lagrange equation is unchanged), and the natural condition \(F_{y'} = 0\) at \(x = 1\) forces \(y' = 0\) there: the bead arrives horizontally, at the bottom of the arch, \(\theta = \pi\). So \(x = a\pi = 1\), \(a = 1/\pi = 0.31831\) m, arrival depth \(2a = 0.63662\) m, and \[ T = \pi\sqrt{a/g} = \sqrt{\pi\cdot 1/g} = 0.56590\ \mathrm s , \] against \(0.58290\) s for the fixed end \(B = (1,1)\) of P2. Freeing the end can only help, because the old answer is still a competitor; here it helps by \(0.0170\) s.

Check. Dimensions: \(fL^2/\tau\) is \(\mathrm{N\,m^{-1}}\cdot\mathrm{m^2}/\mathrm N = \mathrm m\) ✓; \(\sqrt{L/g}\) is seconds ✓. Symmetry check on part 1: a string of length \(2L\) fixed at both ends under the same load sags as \(\frac{f}{2\tau}x(2L - x)\), whose slope is zero at the middle \(x = L\) and whose left half is exactly the answer — the ring on its rod is the mirror plane ✓. Limit: \(f\to0\) gives \(y\equiv0\) ✓.

Idea. A multiplier turns a fixed-length problem into a free one; its value is the exchange rate between the constraint and the objective.

1. The arc is not a straight line (\(\ell \gt 2d\)), so it is not an extremal of length and the multiplier theorem applies: for some constant \(\lambda\), \(y\) is an extremal of \(\tilde F = y - \lambda\sqrt{1+p^2}\). With no \(x\), \(\tilde F - p\tilde F_p = y - \lambda\sqrt{1+p^2} + \lambda p^2/\sqrt{1+p^2} = y - \lambda/\sqrt{1+p^2} = C\). So \((y - C)^2(1 + y'^2) = \lambda^2\), which is P1 part 2's equation shifted by \(C\): its solutions are \((x - x_0)^2 + (y - C)^2 = \lambda^2\). The extremal is an arc of a circle of radius \(\lvert\lambda\rvert\), centre \((0, C)\) by symmetry; and \(y - C = \lambda/\sqrt{1+y'^2}\) is positive on the upper arc, so \(\lambda = R \gt 0\).

2. Chord \(2 = 2R\sin\alpha\) and length \(3 = 2R\alpha\) give \(\sin\alpha/\alpha = 2/3\), so \(\alpha = 1.49578\) rad \(= 85.70^\circ\), \(R = 1/\sin\alpha = 1.00282\) m, and the centre sits \(R\cos\alpha = 0.07516\) m below the chord. The area is the circular segment \(A^* = R^2(\alpha - \sin\alpha\cos\alpha) = 1.42907\ \mathrm m^2\).

3. Parametrise the optimal arcs by \(\alpha\), with \(R = d/\sin\alpha\): \(\ell = 2d\alpha/\sin\alpha\) and \(A^* = d^2(\alpha - \sin\alpha\cos\alpha)/\sin^2\alpha\). Differentiating, \(\mathrm d\ell/\mathrm d\alpha = 2d(\sin\alpha - \alpha\cos\alpha)/\sin^2\alpha\), and, using \(1 - \cos2\alpha = 2\sin^2\alpha\), \(\mathrm dA^*/\mathrm d\alpha = 2d^2\sin\alpha\,(\sin\alpha - \alpha\cos\alpha)/\sin^4\alpha\). The ratio is \[ \frac{\mathrm dA^*}{\mathrm d\ell} = \frac{d}{\sin\alpha} = R = \lambda, \] as the multiplier theorem promised; numerically, differentiating \(A^*(\ell)\) at \(\ell = 3\) gives \(1.00282\) m. Geometrically: lengthening the arc by \(\delta\ell\) while keeping it optimal is, to first order, pushing it outward, and a boundary of curvature radius \(R\) sweeps area \(R\) per unit of added length. The restriction \(\ell \lt \pi d\) keeps the arc less than a semicircle, so that it is a graph \(y(x)\) with no vertical tangent in \((-d, d)\); longer arcs are still circular, but only a parametric formulation can express them.

Check. Dimensions: \(\lambda\) is area per length, a length, and it is a radius ✓. Limit \(\ell\downarrow2d\): \(\sin\alpha/\alpha\to1\), \(\alpha\to0\), \(R\to\infty\), the arc flattens to the chord and the area to \(0\) ✓. Limit \(\ell\uparrow\pi d\): \(\alpha\to\pi/2\), \(R\to d\), a semicircle of area \(\pi d^2/2\) ✓.

Idea. The shape is fixed by one transcendental equation; the tension then comes for free, and the multiplier is it.

1. \(2c\sinh(5/c) = 12\) gives \(c = 4.69542\) m. Then \(\cosh(5/c) = 1.62261\) and the sag is \(c(\cosh(d/c) - 1) = 2.92344\) m. The horizontal tension is \(H = \rho gc = 1.5\times9.81\times4.69542 = 69.093\) N.

2. Force balance: the vertical component at each support carries half the weight, \(\rho g\ell/2 = 88.29\) N, and the horizontal component is \(H\), so \(T = \sqrt{69.093^2 + 88.29^2} = 112.111\) N. Multiplier: \(\mu = -\rho gc\cosh(d/c) = -112.111\) N, and the deck's claim is \(T_{\text{support}} = -\mu\) ✓. (Differentiating the minimum energy \(U^*(\ell) = -327.203\) J numerically at \(\ell = 12\) m gives \(\mathrm dU^*/\mathrm d\ell = -112.111\) N too.)

3. Along the chain \(T = \rho g c\cosh(x/c)\) (the deck's tension slide), so at \(x = \pm d\), \(T = \rho g\,c\cosh(d/c) = \rho g\big(c + c(\cosh(d/c) - 1)\big) = \rho g(c + \text{sag})\). Numerically \(1.5\times9.81\times(4.69542 + 2.92344) = 112.111\) N ✓. The reading: the tension at any point equals the weight of a length \(c + (\text{height above the lowest point})\) of cable — the directrix of the catenary lies a distance \(c\) below its lowest point.

Check. Dimensions: \(\rho g c\) is \(\mathrm{kg\,m^{-1}\,m\,s^{-2}\,m} = \mathrm N\) ✓. Two independent routes to \(112.111\) N agree ✓. Limit: a longer cable sags more and \(c\) falls, so \(H\) falls — slack cables pull less sideways ✓.

Idea. Rotational symmetry makes \(\phi\) absent from the length integrand, so its conjugate \(F_{\phi'}\) is conserved — and that conserved quantity is \(r\sin\psi\). Unit 13 will call this Noether's theorem for rotations; unit 36 re-derives it from a Killing vector.

1. \(\gamma(z) = (r\cos\phi, r\sin\phi, z)\) has \(\gamma' = (r'\cos\phi - r\phi'\sin\phi,\ r'\sin\phi + r\phi'\cos\phi,\ 1)\). Squaring and adding, the cross terms \(\mp2rr'\phi'\sin\phi\cos\phi\) cancel, leaving \(\lvert\gamma'\rvert^2 = r'^2 + r^2\phi'^2 + 1\). Unit 03's arc length gives the formula.

2. \(F = \sqrt{1 + r'^2 + r^2\phi'^2}\) depends on \(z\) and \(\phi'\) but not on \(\phi\), so Euler–Lagrange for \(\phi\) reads \(\frac{\mathrm d}{\mathrm dz}F_{\phi'} = 0\): \[ F_{\phi'} = \frac{r^2\phi'}{\sqrt{1 + r'^2 + r^2\phi'^2}} = c . \] Now the unit tangent is \(\gamma'/\lvert\gamma'\rvert\) and \(e_\phi = (-\sin\phi, \cos\phi, 0)\) is the unit vector along the parallel, perpendicular to the meridian. Their dot product is \(r\phi'/\lvert\gamma'\rvert = \sin\psi\). So \(F_{\phi'} = r\cdot r\phi'/\lvert\gamma'\rvert = r\sin\psi\), and the first integral is \(r\sin\psi = c\).

3. \(\lvert\sin\psi\rvert\le1\) gives \(r \ge \lvert c\rvert\) all along the geodesic. At \(r = c\), \(\sin\psi = 1\): the geodesic runs along the parallel momentarily and turns back. (There \(\phi(z)\) has infinite slope, so the description by \(\phi(z)\) breaks down; the relation holds on each side and, by continuity along the curve, at the turning point.)

4. On the cone, \(z = r/\tan\alpha\), so the meridian element is \(\mathrm dm = \sqrt{\mathrm dr^2 + \mathrm dz^2} = \mathrm dr/\sin\alpha\). By part 1, \(\mathrm ds^2 = \mathrm dm^2 + r^2\mathrm d\phi^2\), and the angle \(\psi\) to the meridian splits the step as \(r\,\mathrm d\phi = \sin\psi\,\mathrm ds\), \(\mathrm dm = \cos\psi\,\mathrm ds\). Hence on the outgoing half \[ \frac{\mathrm d\phi}{\mathrm dr} = \frac{\tan\psi}{r\sin\alpha} = \frac{c}{r\sin\alpha\sqrt{r^2-c^2}},\qquad \sin\psi = c/r . \] With \(r = c\sec t\), \(t\in[0,\pi/2)\): \(\mathrm dr = c\sec t\tan t\,\mathrm dt\), \(\sqrt{r^2-c^2} = c\tan t\), so \(\int_c^\infty\frac{c\,\mathrm dr}{r\sqrt{r^2-c^2}} = \int_0^{\pi/2}\mathrm dt = \frac\pi2\). Incoming and outgoing halves each contribute \(\pi/(2\sin\alpha)\), so \[ \Delta\phi = \frac{\pi}{\sin\alpha} . \] At \(\alpha = 20^\circ\), \(\Delta\phi = 9.1854\) rad \(= 526.3^\circ \gt 360^\circ\): the geodesic wraps more than once round the cone and crosses itself. The threshold is \(\sin\alpha = \tfrac12\), \(\alpha = 30^\circ\), where \(\Delta\phi = 2\pi\) exactly.

Check. Limit \(\alpha\to\pi/2\): the cone flattens to a plane, \(\Delta\phi\to\pi\), and a straight line in the plane does turn through exactly \(\pi\) as seen from a point off it ✓. Independent check: cut the cone along a meridian and unroll it; it becomes a flat sector of angle \(2\pi\sin\alpha\) (circumference \(2\pi r\) laid out at slant distance \(r/\sin\alpha\)), a geodesic becomes a straight line turning through \(\pi\) in the plane, and a plane angle \(\beta\) corresponds to \(\phi = \beta/\sin\alpha\) — so \(\Delta\phi = \pi/\sin\alpha\) ✓. Dimensions: \(c\) is a length, \(\Delta\phi\) a pure number ✓.

Idea. The equator past the antipode is a great circle, hence stationary; but the great circles from the start all re-meet at the antipode, and past a conjugate point the second variation can be made negative.

1. The point is \((\cos\lambda\cos\phi, \cos\lambda\sin\phi, \sin\lambda)\). Its \(\phi\)-derivative is \((-\lambda'\sin\lambda\cos\phi - \cos\lambda\sin\phi,\ -\lambda'\sin\lambda\sin\phi + \cos\lambda\cos\phi,\ \lambda'\cos\lambda)\); the squared norm is \(\lambda'^2(\sin^2\lambda + \cos^2\lambda) + \cos^2\lambda\), the cross terms cancelling. So \(F = \sqrt{p^2 + \cos^2\lambda}\). \(F_\lambda = -\sin\lambda\cos\lambda/F\) and \(F_p = p/F\) both vanish identically on \(\lambda\equiv0\), so Euler–Lagrange, \((F_p)' = F_\lambda\), holds: the equator is an extremal for every \(\ell\).

2. On \(\lambda = 0\), \(p = 0\): \(F = 1\), and \(F_{pp} = \cos^2\lambda/F^3 = 1\); \(F_{\lambda p} = \partial_\lambda(p/F) = -p\,F_\lambda/F^2 = 0\); \(F_{\lambda\lambda} = \partial_\lambda(-\sin\lambda\cos\lambda/F) = -\cos2\lambda/F + \sin\lambda\cos\lambda\,F_\lambda/F^2 = -1\). So \(\delta^2L = \int(F_{\lambda\lambda}\eta^2 + 2F_{\lambda p}\eta\eta' + F_{pp}\eta'^2) = \int(\eta'^2 - \eta^2)\). \(P = 1 \gt 0\): Legendre holds, strictly.

3. \(P = 1\), \(Q = -1\), so the accessory equation is \(\eta'' + \eta = 0\); with \(\eta(0) = 0\), \(\eta = \sin\phi\) is a solution, and it vanishes next at \(\phi = \pi\). No solution \(h \not\equiv 0\) vanishing at \(0\) can vanish at some \(c \in (0,\pi)\): multiplying \(h'' + h = 0\) by \(h\) and integrating by parts gives \(\int_0^c(h'^2 - h^2) = [hh']_0^c = 0\), which part 5's strict inequality on \([0,c]\) forbids (and \(h\) cannot vanish on all of \([0,c]\), since \(h'^2 + h^2\) is constant along a solution). So \(\pi\) is the first conjugate point: \((\cos\pi, \sin\pi, 0) = (-1,0,0)\), the antipode of the start. The family: the great circle through \((1,0,0)\) tilted by \(\beta\) about the \(x\)-axis lies in the plane \(z = \tan\beta\,y\); in coordinates \(\tan\lambda = \tan\beta\sin\phi\). Then \(\partial_\beta\lambda\big|_{\beta=0} = \sin\phi\) — the Jacobi field — and every member passes through the antipode, where \(\sin\phi = 0\).

4. \(\eta = \sin(2\phi/3)\) vanishes at \(0\) and \(3\pi/2\). Then \(\delta^2L = \int_0^{3\pi/2}\big(\tfrac49\cos^2\tfrac{2\phi}{3} - \sin^2\tfrac{2\phi}{3}\big)\mathrm d\phi = \tfrac{3\pi}{4}\big(\tfrac49 - 1\big) = -\tfrac{5\pi}{12} = -1.30900\). Taylor predicts \(L[\varepsilon\eta]\approx\tfrac{3\pi}{2} + \tfrac{\varepsilon^2}{2}\delta^2L\), i.e. shorter by \(0.0065450\) at \(\varepsilon = 0.1\). Quadrature of the exact integrand gives \(L = 4.705854\) against \(3\pi/2 = 4.712389\): shorter by \(0.0065354\). A curve within \(0.1\) rad of the equator, with slope within \(0.067\), is shorter: the \(270^\circ\) arc is stationary and not even a weak local minimum.

5. For \(\ell \lt \pi\), \(u(\phi) = \sin\big(\phi + (\pi-\ell)/2\big)\) solves \(u'' + u = 0\) and is positive on \([0,\ell]\), since its argument stays in \(\big[(\pi-\ell)/2, (\pi+\ell)/2\big]\subset(0,\pi)\). The Picone identity (deck, Jacobi's theorem) gives \(\int_0^\ell(\eta'^2 - \eta^2) = \int_0^\ell(\eta' - \eta u'/u)^2 \gt 0\) for \(\eta\ne0\). ∎

Check. The two ways of computing the shortening agree to \(0.15\%\), the difference being fourth order in \(\varepsilon\) ✓. Dimensions: all angles, all pure numbers ✓. Special case \(\ell = \pi\): \(\sin\phi\) itself gives \(\delta^2L = 0\) — the half great circle ties with its tilted neighbours, as it must, since all of them have length \(\pi\) ✓.

Idea. Two integrations by parts leave an interior term, which gives the equation, and two boundary terms at the free end, which give one condition each.

1. \(\delta J = \int_0^L(EIy''\eta'' + w\eta)\,\mathrm dx\). Integrating by parts twice, \(\int_0^LEIy''\eta'' = [EIy''\eta']_0^L - [EIy'''\eta]_0^L + \int_0^LEIy''''\eta\), and the terms at \(0\) vanish. So \(\delta J = \int_0^L(EIy'''' + w)\eta\,\mathrm dx + EIy''(L)\eta'(L) - EIy'''(L)\eta(L)\). Variations vanishing near both ends give the Euler–Poisson equation \(EIy'''' = -w\) (fundamental lemma). That leaves \(\delta J = EIy''(L)\eta'(L) - EIy'''(L)\eta(L)\) for every admissible \(\eta\). Take \(\eta = x^2(x - L)\), which has \(\eta(0) = \eta'(0) = 0\), \(\eta(L) = 0\) and \(\eta'(L) = L^2\): it isolates the first term. Then \(\eta = x^2\), with \(\eta(L) = L^2\), isolates the second. So \[ EIy''(L) = 0,\qquad EIy'''(L) = 0 . \] \(EIy''\) is the bending moment and \(EIy'''\) the shear force: a free end carries neither.

2. Integrate \(y'''' = -w/EI\) four times, fixing one constant at each stage: \(y''' = -w(x - L)/EI\) (so \(y'''(L) = 0\)); \(y'' = -w(x-L)^2/(2EI)\) (so \(y''(L) = 0\)); \(y' = -w\big((x-L)^3 + L^3\big)/(6EI)\) (so \(y'(0) = 0\)); \(y = -w\big((x-L)^4 - L^4 + 4L^3x\big)/(24EI)\) (so \(y(0) = 0\)). Expanding \((x-L)^4\), \[ y(x) = -\frac{w\,x^2\,(x^2 - 4Lx + 6L^2)}{24EI},\qquad y(L) = -\frac{wL^4}{8EI} . \]

3. \(I = 0.025\times(0.001)^3/12 = 2.0833\times10^{-12}\ \mathrm m^4\); \(w = 7850\times0.025\times0.001\times9.81 = 1.9252\ \mathrm{N\,m^{-1}}\). Then \(wL^4/(8EI) = 1.9252\times0.0081/(8\times2.0\times10^{11}\times2.0833\times10^{-12}) = 4.68\times10^{-3}\) m. The tip droops about \(4.7\) mm.

Check. Dimensions: \(wL^4/(EI)\) is \(\mathrm{N\,m^{-1}}\cdot\mathrm m^4/(\mathrm{N\,m^{-2}}\cdot\mathrm m^4) = \mathrm m\) ✓. Limits: \(E\to\infty\) gives no droop ✓; \(L^4\) scaling means a \(60\) cm overhang droops sixteen times as far, \(75\) mm — at which point small slopes are no longer a fair assumption. Sign: the tip goes down, \(y(L) \lt 0\) ✓.

Idea. Curves can make both terms small by oscillating ever faster, but the limit of the oscillation is \(y = 0\), where the first term is at its worst. Non-convexity in \(y'\) is what lets the limit be worse than the sequence.

1. Both terms are non-negative and continuous, so \(J[y] = 0\) forces both to vanish identically (a continuous non-negative function with zero integral is zero). Then \(y\equiv0\), so \(y'\equiv0\), and the first term is \(1\), not \(0\). Contradiction; hence \(J[y] \gt 0\). ∎

2. Let \(z_n\) rise with slope \(1\) on \([k/n, k/n + 1/(2n)]\) and fall with slope \(-1\) on \([k/n + 1/(2n), (k+1)/n]\), for \(k = 0,\dots,n-1\). It vanishes at both ends and is piecewise \(C^1\) with \(2n - 1\) interior corners. Where differentiable, \((z_n'^2 - 1)^2 = 0\), and on one tooth \(\int z_n^2 = 2\int_0^{1/(2n)}t^2\,\mathrm dt = 1/(12n^3)\); so \(J[z_n] = 1/(12n^2)\), e.g. \(8.33\times10^{-4}\) at \(n = 10\).

Repair: at each corner \(x_k\), with slope \(s = \pm1\) before and \(-s\) after, replace the slope on \([x_k - \delta, x_k + \delta]\) by the linear function going from \(s\) to \(-s\). The new slope is continuous, so its integral \(y\) is \(C^1\); and since the linear slope, like the old one, has zero integral over the interval (it is odd about \(x_k\)), \(y\) agrees with \(z_n\) outside the repair intervals. Inside them \(\lvert y'\rvert\le1\) and \(\lvert y\rvert\le1/(2n)\), so the integrand is at most \(1 + 1/(4n^2)\le2\). Hence \(J[y]\le1/(12n^2) + (2n-1)\cdot2\delta\cdot2\). Choosing \(\delta = 1/(48n^3)\) (small enough that the intervals do not overlap) gives \(J[y]\le1/(12n^2) + 1/(6n^2) = 1/(4n^2)\to0\). With part 1, \(\inf J = 0\) and it is not attained. ∎

3. \(F = (p^2 - 1)^2 + y^2\): \(F_y = 2y\), \(F_p = 4p(p^2 - 1)\). On \(y\equiv0\) both vanish, so Euler–Lagrange holds, and \(J[0] = 1\). The Legendre coefficient is \(F_{pp} = 12p^2 - 4\), which at \(p = 0\) is \(-4 \lt 0\). By the Legendre condition, \(y\equiv0\) is not a weak local minimum: a squeezed bump lowers \(J\).

4. Step 3, lower semicontinuity. The minimising sequence converges uniformly to \(0\) (step 2 succeeds for positions), but \(J[0] = 1 \gt 0 = \lim J[y_n]\). The failure is caused by \(F\) not being convex in \(p\) — it has two wells at \(p = \pm1\) — so the slopes of the sequence can keep oscillating between \(+1\) and \(-1\), costing nothing, while their local averages — and hence the limit curve's slope — tend to \(0\), where the cost is highest. Tonelli's convexity hypothesis is exactly what excludes this.

Check. Numbers: \(J[z_n] = 1/(12n^2)\) is \(1/12 = 0.0833\) at \(n = 1\), and a single tooth of height \(\tfrac12\) has \(\int z^2 = 2\int_0^{1/2}t^2 = 1/12\) ✓. Dimensions: the example is dimensionless. Limit: the bound \(1/(4n^2)\to0\) as \(n\to\infty\) ✓.