Exercises — Brachistochrone problem

What we recall. The founding result of the Calculus of Variations is that the fastest descent curve is the cycloid — answer (c). The equations (a point on a wheel of radius $r$ rolling along the top): $x=r(\theta -\sin \theta ),y=r(1-\cos \theta ).$ Why not the straight line? The straight line is shortest, but time is $\int ds/v$, not $\int ds$. The cycloid dives steeply first, buying speed early — this is exactly the yellow-vs-dashed-blue comparison labelled in figure s01.

Law: Conservation of Energy — kinetic energy gained equals potential energy lost (frictionless wire). $\frac{1}{2}m{v}^2=mgy\Rightarrow v=\sqrt{2gy}.$ What it looks like: deeper $y$ ⇒ bigger $v$. The mass $m$ cancels, so a marble and a bowling ball descend identically. Here $y$ points downward so $y>0$ and the square root is real.

What we use: the cusp-to-bottom formula for a bead released from rest at the cusp, $T=\pi \sqrt{\frac{r}{g}}.$ Why it is that clean: with the bead released from rest at the cusp ($y_0=0$), the speed is $v=\sqrt{2gy}=\sqrt{2gr(1-\cos \theta )}$ and the arc length is $ds=r\sqrt{2(1-\cos \theta )}d\theta$. The factor $(1-\cos \theta )$ cancels between them, leaving a constant integrand $\sqrt{r/g}$ over $\theta \in [0,\pi ]$, so $T={\int }_0^{\pi }\sqrt{r/g}d\theta =\pi \sqrt{r/g}$. Plug in: $T=\pi \sqrt{2/10}=\pi \sqrt{0.2}\approx \pi (0.4472)\approx 1.405\text{s}.$

On the dropped factor. The full integrand is $\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}=\frac{1}{\sqrt{2g}}\cdot \sqrt{\frac{1+y^{\prime 2}}{y}}$. The constant we pull out is the multiplicative factor $\frac{1}{\sqrt{2g}}$ (not $2g$ itself); dropping a positive constant multiplier does not change which curve is extremal, so we may work with $F=\sqrt{(1+y^{\prime 2})/y}$. Why Beltrami and not Euler–Lagrange? $F$ has no explicit $x$, i.e. $\partial F/\partial x=0$. The Beltrami identity stated at the top of this page then gives a conserved quantity and a first-order ODE, saving us from a messy second-order Euler–Lagrange chase. Compute $F_{y^{\prime }}=\frac{1}{\sqrt{y}}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$. Beltrami: $F-y^{\prime }{F}_{y^{\prime }}=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{y}}-\frac{y^{\prime 2}}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=\frac{(1+y^{\prime 2})-y^{\prime 2}}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=\frac{1}{\sqrt{y(1+y^{\prime 2})}}=C.$ Square and rearrange: $y(1+y^{\prime 2})=\frac{1}{C^2}\equiv k.$ What it means: small $y$ (near the top) forces large $y^{\prime }$ (steep), so the bead is nearly vertical at the start — exactly the "plunge early" behaviour.

Set up: we need $x=r(\theta -\sin \theta )=\pi$ and $y=r(1-\cos \theta )=2$ simultaneously. Guess the natural landmark: the bottom of an arch is $\theta =\pi$ (where $\sin \theta =0$, $\cos \theta =-1$). Test it: $x=r(\pi -0)=r\pi ,y=r(1-(-1))=2r.$ Matching $y=2$ gives $r=1$; then $x=\pi \cdot 1=\pi$ ✓. So $r=1,{\theta }_B=\pi$. This is precisely the endpoint $B$ drawn in figure s01. Why the test worked: the point $(\pi ,2)$ is precisely the lowest point of a unit cycloid; recognizing that landmark avoids solving a transcendental system numerically.

What we do: differentiate the parametric equations w.r.t. $\theta$ and divide (chain rule): $\frac{dx}{d\theta }=r(1-\cos \theta ),\frac{dy}{d\theta }=r\sin \theta .$ $y^{\prime }=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin \theta }{1-\cos \theta }.$ Why this simplifies to $\cot (\theta /2)$: use $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ and $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$: $y^{\prime }=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}=\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=\cot \frac{\theta }{2}.$ All cases:

• Cusp $\theta \to 0^{+}$: $\cot (\theta /2)\to +\infty$ ⇒ slope infinite ⇒ tangent is vertical (the pink double-arrow in s02). The bead starts by dropping straight down. This is why speed is gained instantly.
• Bottom $\theta =\pi$: $\cot (\pi /2)=0$ ⇒ slope zero ⇒ tangent is horizontal (the blue double-arrow in s02). The curve is flat at the lowest point, as any minimum should be.
• In between $0<\theta <\pi$: $\cot (\theta /2)>0$ and decreasing, so the curve steadily flattens — no kinks, no surprises.

Idea: compute $T=\int ds/v$ for a bead released from rest at ${\theta }_0$, sliding to $\theta =\pi$, and show the answer has no ${\theta }_0$ dependence. Speed with a general start. Released from rest at depth $y_0=r(1-\cos {\theta }_0)$, at depth $y$ the Conservation of Energy gives $v=\sqrt{2g(y-y_0)}=\sqrt{2gr(\cos {\theta }_0-\cos \theta )}.$ Arc length. $ds=\sqrt{(dx{)}^2+(dy{)}^2}=r\sqrt{(1-\cos \theta {)}^2+{\sin }^2\theta }d\theta =r\sqrt{2(1-\cos \theta )}d\theta .$ Assemble. $T={\int }_{{\theta }_0}^{\pi }\frac{r\sqrt{2(1-\cos \theta )}}{\sqrt{2gr(\cos {\theta }_0-\cos \theta )}}d\theta =\sqrt{\frac{r}{g}}{\int }_{{\theta }_0}^{\pi }\sqrt{\frac{1-\cos \theta }{\cos {\theta }_0-\cos \theta }}d\theta .$ Rewrite both half-angle pieces. Use the identities $1-\cos \theta =2{\sin }^2\frac{\theta }{2},\cos {\theta }_0-\cos \theta =2{\cos }^2\frac{{\theta }_0}{2}-2{\cos }^2\frac{\theta }{2}$ (the second from $\cos \varphi =2{\cos }^2\frac{\varphi }{2}-1$ applied to each angle and subtracting: $\cos {\theta }_0-\cos \theta =(2{\cos }^2\frac{{\theta }_0}{2}-1)-(2{\cos }^2\frac{\theta }{2}-1)=2{\cos }^2\frac{{\theta }_0}{2}-2{\cos }^2\frac{\theta }{2}$). Substituting, the ratio becomes $\frac{1-\cos \theta }{\cos {\theta }_0-\cos \theta }=\frac{2{\sin }^2\frac{\theta }{2}}{2({\cos }^2\frac{{\theta }_0}{2}-{\cos }^2\frac{\theta }{2})}=\frac{{\sin }^2\frac{\theta }{2}}{{\cos }^2\frac{{\theta }_0}{2}-{\cos }^2\frac{\theta }{2}}.$ The clever substitution — where the 2's come from and go. Let $u=\cos \frac{\theta }{2},du=-\frac{1}{2}\sin \frac{\theta }{2}d\theta \Rightarrow \sin \frac{\theta }{2}d\theta =-2du.$ The numerator $\sqrt{{\sin }^2\frac{\theta }{2}}=\sin \frac{\theta }{2}$ combines with $d\theta$ to give exactly $\sin \frac{\theta }{2}d\theta =-2du$ — that is the origin of the factor $2$. Writing $a=\cos \frac{{\theta }_0}{2}$, $T=\sqrt{\frac{r}{g}}{\int }_{{\theta }_0}^{\pi }\frac{\sin \frac{\theta }{2}d\theta }{\sqrt{a^2-{\cos }^2\frac{\theta }{2}}}=\sqrt{\frac{r}{g}}{\int }_{u=a}^{u=0}\frac{-2du}{\sqrt{a^2-u^2}}=2\sqrt{\frac{r}{g}}{\int }_0^a\frac{du}{\sqrt{a^2-u^2}}.$ Note the limits: $\theta ={\theta }_0\Rightarrow u=\cos \frac{{\theta }_0}{2}=a$, and $\theta =\pi \Rightarrow u=\cos \frac{\pi }{2}=0$; flipping the limits kills the minus sign, and the factor $2$ survives (it does not cancel). Evaluate. ${\int }_0^a\frac{du}{\sqrt{a^2-u^2}}=[\arcsin \frac{u}{a}{]}_0^a=\arcsin 1-\arcsin 0=\frac{\pi }{2}.$ Result. $\boxed{T=2\sqrt{\frac{r}{g}}\cdot \frac{\pi }{2}=\pi \sqrt{\frac{r}{g}}}$ Punchline: the upper limit $a=\cos ({\theta }_0/2)$ cancelled inside the $\arcsin$ — the answer $\pi \sqrt{r/g}$ contains no ${\theta }_0$. Every release height reaches the bottom in the same time. That is the tautochrone (equal-time) property of the Cycloid — the principle behind Huygens' pendulum clock. Consistency with Level 2. Set ${\theta }_0=0$ (release from rest at the cusp): the formula gives $T=\pi \sqrt{r/g}$, matching Exercise 2.1 exactly. No factor-of-2 discrepancy — the two results agree because both describe a bead released from rest and sliding to the bottom.

Straight line. Here $y=x$ on $0\le x\le 2$, so $y^{\prime }=1$ and $\sqrt{1+y^{\prime 2}}=\sqrt{2}$. Speed $v=\sqrt{2gy}=\sqrt{2y}$ (with $g=1$). $T_{\text{line}}={\int }_0^2\frac{\sqrt{2}}{\sqrt{2y}}dx={\int }_0^2\frac{\sqrt{2}}{\sqrt{2x}}dx={\int }_0^2{x}^{-1/2}dx=[2\sqrt{x}{]}_0^2=2\sqrt{2}\approx \mathrm{2.828.}$ Cycloid. We need the cycloid through $(0,0)$ and $(2,2)$. Solve $r({\theta }_B-\sin {\theta }_B)=2$ and $r(1-\cos {\theta }_B)=2$ simultaneously. Dividing eliminates $r$: $\frac{{\theta }_B-\sin {\theta }_B}{1-\cos {\theta }_B}=1.$ Numerically this gives ${\theta }_B\approx 2.4120$, then $r=\frac{2}{1-\cos {\theta }_B}\approx 1.1454$. Cycloid time (parametric form, arc/speed with $g=1$, bead released from rest at $A$ where $\theta =0$): $T_{\text{cyc}}={\int }_0^{{\theta }_B}\frac{r\sqrt{2(1-\cos \theta )}}{\sqrt{2r(1-\cos \theta )}}d\theta =\sqrt{r}{\int }_0^{{\theta }_B}d\theta =\sqrt{r}{\theta }_B\approx \sqrt{1.1454}\cdot 2.4120\approx \mathrm{2.581.}$ Compare and report the ratio. $T_{\text{line}}\approx 2.828$ vs $T_{\text{cyc}}\approx 2.581$, so the cycloid is faster and the requested ratio is $\frac{T_{\text{line}}}{T_{\text{cyc}}}\approx \frac{2.828}{2.581}\approx \mathrm{1.096.}$ What it means: even though the cycloid dips below the straight chord and is geometrically longer (exactly as drawn in s01), it beats the line by roughly $9.6\%$ in time — the "farther but faster" principle, quantified.

Fix ONE angle and its reference — no switching. Let $\varphi$ be the angle the tangent to the path makes with the horizontal (the direction of the layers). A tiny step along the path is $(dx,dy)$, so from the right triangle with hypotenuse $ds=\sqrt{d{x}^2+d{y}^2}$: $\cos \varphi =\frac{dx}{ds}=\frac{1}{\sqrt{1+y^{\prime 2}}},\sin \varphi =\frac{dy}{ds}=\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}.$ Rewrite Beltrami with this $\varphi$. Since $v=\sqrt{2gy}$, $\frac{1}{\sqrt{2gy}\sqrt{1+y^{\prime 2}}}=\frac{1}{v}\cdot \frac{1}{\sqrt{1+y^{\prime 2}}}=\frac{\cos \varphi }{v}=C.$ Now map to Snell's law — the complementary-angle bridge. In optics Fermat's Principle gives least-time paths, and across horizontal layers of speed $v$ a light ray obeys $\frac{\sin \vartheta }{v}=\text{const},$ where $\vartheta$ is the angle of incidence measured from the normal to the layers — and the normal to a horizontal layer is vertical. Our $\varphi$ is measured from the horizontal, so $\varphi$ and $\vartheta$ are complementary: $\vartheta =90^{\circ }-\varphi$. Therefore $\sin \vartheta =\sin (90^{\circ }-\varphi )=\cos \varphi ,$ and our conserved quantity $\frac{\cos \varphi }{v}$ is literally $\frac{\sin \vartheta }{v}$ — Snell's invariant. So the brachistochrone bends like a light ray through a medium that speeds up with depth. This is Johann Bernoulli's original 1696 optical solution: the two "least-time" principles are the same law. Consistency check. At the cusp the tangent is vertical, so $\varphi =90^{\circ }$ (from horizontal) and $\vartheta =0^{\circ }$ (from the vertical normal). Then $\cos \varphi =0=\sin \vartheta$, consistent with $y^{\prime }\to \infty$. The complementary bookkeeping holds at the boundary.

Which exercise number tests the tautochrone proof? ::: 4.1 Which exercise connects the problem to Snell's law / Fermat? ::: 5.2 Which two exercises require solving for $r$ from an endpoint? ::: 3.1 and 5.1