Exercises — Brachistochrone problem

Hum kya recall karte hain. Calculus of Variations ka founding result yeh hai ki fastest descent curve cycloid hai — answer (c). Equations (radius $r$ ke wheel ki rim par ek point jo upar se roll karta hai): $x=r(\theta -\sin \theta ),y=r(1-\cos \theta ).$ Straight line kyun nahi? Straight line shortest hoti hai, lekin time $\int ds/v$ hai, na ki $\int ds$. Cycloid pehle steeply dive karta hai, jis se speed jaldi milti hai — yahi figure s01 mein yellow-vs-dashed-blue comparison hai.

Law: Conservation of Energy — gain ki gayi kinetic energy equals potential energy lost (frictionless wire). $\frac{1}{2}m{v}^2=mgy\Rightarrow v=\sqrt{2gy}.$ Yeh kaisa dikhta hai: zyada $y$ ⇒ zyada $v$. Mass $m$ cancel ho jaata hai, toh ek marble aur bowling ball ek jaise descend karte hain. Yahan $y$ neeche point karta hai toh $y>0$ hai aur square root real hai.

Hum kya use karte hain: cusp se bottom tak bead ke liye formula, jab bead cusp se rest par release ho: $T=\pi \sqrt{\frac{r}{g}}.$ Yeh itna clean kyun hai: cusp par rest se release hone par ($y_0=0$), speed $v=\sqrt{2gy}=\sqrt{2gr(1-\cos \theta )}$ hai aur arc length $ds=r\sqrt{2(1-\cos \theta )}d\theta$ hai. $(1-\cos \theta )$ factor unke beech cancel ho jaata hai, ek constant integrand $\sqrt{r/g}$ bacha deta hai $\theta \in [0,\pi ]$ par, toh $T={\int }_0^{\pi }\sqrt{r/g}d\theta =\pi \sqrt{r/g}$. Plug in: $T=\pi \sqrt{2/10}=\pi \sqrt{0.2}\approx \pi (0.4472)\approx 1.405\text{s}.$

Dropped factor ke baare mein. Full integrand $\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}=\frac{1}{\sqrt{2g}}\cdot \sqrt{\frac{1+y^{\prime 2}}{y}}$ hai. Jo constant hum bahar nikalte hain woh multiplicative factor $\frac{1}{\sqrt{2g}}$ hai (na ki $2g$ khud); ek positive constant multiplier drop karne se koi fark nahi padta ki kaunsa curve extremal hai, toh hum $F=\sqrt{(1+y^{\prime 2})/y}$ ke saath kaam kar sakte hain. Beltrami kyun aur Euler–Lagrange kyun nahi? $F$ mein koi explicit $x$ nahi hai, yaani $\partial F/\partial x=0$. Is page ke upar stated Beltrami identity phir ek conserved quantity aur ek first-order ODE deti hai, jo hume ek messy second-order Euler–Lagrange chase se bachata hai. Compute karo $F_{y^{\prime }}=\frac{1}{\sqrt{y}}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$. 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 karo aur rearrange karo: $y(1+y^{\prime 2})=\frac{1}{C^2}\equiv k.$ Iska matlab: chhota $y$ (top ke paas) bada $y^{\prime }$ force karta hai (steep), toh bead shuru mein almost vertical hota hai — bilkul wahi "pehle plunge karo" behaviour.

Setup karo: hume $x=r(\theta -\sin \theta )=\pi$ aur $y=r(1-\cos \theta )=2$ simultaneously chahiye. Natural landmark guess karo: ek arch ka bottom $\theta =\pi$ hai (jahan $\sin \theta =0$, $\cos \theta =-1$). Test karo: $x=r(\pi -0)=r\pi ,y=r(1-(-1))=2r.$ $y=2$ match karne se $r=1$ milta hai; phir $x=\pi \cdot 1=\pi$ ✓. Toh $r=1,{\theta }_B=\pi$. Yahi woh endpoint $B$ hai jo figure s01 mein drawn hai. Test kyun kaam kiya: point $(\pi ,2)$ precisely ek unit cycloid ka lowest point hai; us landmark ko recognize karne se hume ek transcendental system numerically solve karne ki zaroorat nahi padti.

Hum kya karte hain: parametric equations ko $\theta$ ke saath differentiate karo aur divide karo (chain rule): $\frac{dx}{d\theta }=r(1-\cos \theta ),\frac{dy}{d\theta }=r\sin \theta .$ $y^{\prime }=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin \theta }{1-\cos \theta }.$ Yeh $\cot (\theta /2)$ mein simplify kyun hota hai: $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ aur $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$ use karo: $y^{\prime }=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}=\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=\cot \frac{\theta }{2}.$ Saare cases:

• Cusp $\theta \to 0^{+}$: $\cot (\theta /2)\to +\infty$ ⇒ slope infinite ⇒ tangent vertical hai (s02 mein pink double-arrow). Bead seedha neeche girke start karta hai. Yahi wajah hai ki speed turant gain hoti hai.
• Bottom $\theta =\pi$: $\cot (\pi /2)=0$ ⇒ slope zero ⇒ tangent horizontal hai (s02 mein blue double-arrow). Curve lowest point par flat hai, jaisa koi bhi minimum hona chahiye.
• Beech mein $0<\theta <\pi$: $\cot (\theta /2)>0$ aur decreasing, toh curve steadily flatten hoti hai — koi kinks nahi, koi surprises nahi.

Idea: ${\theta }_0$ par rest se release hue bead ke liye, $\theta =\pi$ tak slide karte hue $T=\int ds/v$ compute karo, aur dikhao ki answer mein ${\theta }_0$ dependence nahi hai. General start ke saath speed. Depth $y_0=r(1-\cos {\theta }_0)$ par rest se release hone par, depth $y$ par Conservation of Energy deta hai: $v=\sqrt{2g(y-y_0)}=\sqrt{2gr(\cos {\theta }_0-\cos \theta )}.$ 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 karo. $T={\int }_{{\theta }_0}^{\pi }\frac{r\sqrt{2(1-\cos \theta )}}{\sqrt{2gr(\cos {\theta }_0-\cos \theta )}}d\theta =\sqrt{\frac{r}{g}}{\int }_{{\theta }_0}^{\pi }\sqrt{\frac{1-\cos \theta }{\cos {\theta }_0-\cos \theta }}d\theta .$ Dono half-angle pieces rewrite karo. Identities use karo: $1-\cos \theta =2{\sin }^2\frac{\theta }{2},\cos {\theta }_0-\cos \theta =2{\cos }^2\frac{{\theta }_0}{2}-2{\cos }^2\frac{\theta }{2}$ (doosra $\cos \varphi =2{\cos }^2\frac{\varphi }{2}-1$ ko har angle par apply karke aur subtract karke: $\cos {\theta }_0-\cos \theta =(2{\cos }^2\frac{{\theta }_0}{2}-1)-(2{\cos }^2\frac{\theta }{2}-1)=2{\cos }^2\frac{{\theta }_0}{2}-2{\cos }^2\frac{\theta }{2}$). Substitute karne par ratio ban jaata hai: $\frac{1-\cos \theta }{\cos {\theta }_0-\cos \theta }=\frac{2{\sin }^2\frac{\theta }{2}}{2({\cos }^2\frac{{\theta }_0}{2}-{\cos }^2\frac{\theta }{2})}=\frac{{\sin }^2\frac{\theta }{2}}{{\cos }^2\frac{{\theta }_0}{2}-{\cos }^2\frac{\theta }{2}}.$ Clever substitution — 2's kahan se aate hain aur kahan jaate hain. Let karo: $u=\cos \frac{\theta }{2},du=-\frac{1}{2}\sin \frac{\theta }{2}d\theta \Rightarrow \sin \frac{\theta }{2}d\theta =-2du.$ Numerator $\sqrt{{\sin }^2\frac{\theta }{2}}=\sin \frac{\theta }{2}$, $d\theta$ ke saath combine hokar exactly $\sin \frac{\theta }{2}d\theta =-2du$ deta hai — yahi factor $2$ ka origin hai. $a=\cos \frac{{\theta }_0}{2}$ likhte hue: $T=\sqrt{\frac{r}{g}}{\int }_{{\theta }_0}^{\pi }\frac{\sin \frac{\theta }{2}d\theta }{\sqrt{a^2-{\cos }^2\frac{\theta }{2}}}=\sqrt{\frac{r}{g}}{\int }_{u=a}^{u=0}\frac{-2du}{\sqrt{a^2-u^2}}=2\sqrt{\frac{r}{g}}{\int }_0^a\frac{du}{\sqrt{a^2-u^2}}.$ Limits note karo: $\theta ={\theta }_0\Rightarrow u=\cos \frac{{\theta }_0}{2}=a$, aur $\theta =\pi \Rightarrow u=\cos \frac{\pi }{2}=0$; limits flip karne se minus sign khatam hota hai, aur factor $2$ bachta hai (cancel nahi hota). Evaluate karo. ${\int }_0^a\frac{du}{\sqrt{a^2-u^2}}=[\arcsin \frac{u}{a}{]}_0^a=\arcsin 1-\arcsin 0=\frac{\pi }{2}.$ Result. $\boxed{T=2\sqrt{\frac{r}{g}}\cdot \frac{\pi }{2}=\pi \sqrt{\frac{r}{g}}}$ Punchline: upper limit $a=\cos ({\theta }_0/2)$ $\arcsin$ ke andar cancel ho gaya — answer $\pi \sqrt{r/g}$ mein koi ${\theta }_0$ nahi hai. Har release height bottom tak same time mein pahunchti hai. Yahi Cycloid ki tautochrone (equal-time) property hai — Huygens' pendulum clock ka principle. Level 2 ke saath consistency. ${\theta }_0=0$ set karo (cusp par rest se release): formula $T=\pi \sqrt{r/g}$ deta hai, jo Exercise 2.1 se exactly match karta hai. Koi factor-of-2 discrepancy nahi — dono results agree karte hain kyunki dono rest se bead describe karte hain jo bottom tak slide karta hai.

Straight line. Yahan $y=x$ on $0\le x\le 2$, toh $y^{\prime }=1$ aur $\sqrt{1+y^{\prime 2}}=\sqrt{2}$. 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. Hume $(0,0)$ aur $(2,2)$ se jaane waali cycloid chahiye. $r({\theta }_B-\sin {\theta }_B)=2$ aur $r(1-\cos {\theta }_B)=2$ simultaneously solve karo. Divide karne se $r$ eliminate hota hai: $\frac{{\theta }_B-\sin {\theta }_B}{1-\cos {\theta }_B}=1.$ Numerically yeh ${\theta }_B\approx 2.4120$ deta hai, phir $r=\frac{2}{1-\cos {\theta }_B}\approx 1.1454$. Cycloid time (parametric form, arc/speed with $g=1$, bead $A$ par rest se release hota hai jahan $\theta =0$): $T_{\text{cyc}}={\int }_0^{{\theta }_B}\frac{r\sqrt{2(1-\cos \theta )}}{\sqrt{2r(1-\cos \theta )}}d\theta =\sqrt{r}{\int }_0^{{\theta }_B}d\theta =\sqrt{r}{\theta }_B\approx \sqrt{1.1454}\cdot 2.4120\approx \mathrm{2.581.}$ Compare karo aur ratio report karo. $T_{\text{line}}\approx 2.828$ vs $T_{\text{cyc}}\approx 2.581$, toh cycloid faster hai aur requested ratio hai: $\frac{T_{\text{line}}}{T_{\text{cyc}}}\approx \frac{2.828}{2.581}\approx \mathrm{1.096.}$ Iska matlab: bhale hi cycloid straight chord ke neeche dip karta hai aur geometrically zyada lamba hai (bilkul jaisa s01 mein drawn hai), phir bhi yeh line ko time mein roughly $9.6\%$ se beat karta hai — "farther but faster" principle, quantified.

EK angle aur uska reference fix karo — switching mat karo. Maano $\varphi$ woh angle hai jo path ki tangent horizontal (layers ki direction) ke saath banati hai. Path ke along ek chhota step $(dx,dy)$ hai, toh right triangle se jiska hypotenuse $ds=\sqrt{d{x}^2+d{y}^2}$ hai: $\cos \varphi =\frac{dx}{ds}=\frac{1}{\sqrt{1+y^{\prime 2}}},\sin \varphi =\frac{dy}{ds}=\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}.$ Is $\varphi$ se Beltrami rewrite karo. Kyunki $v=\sqrt{2gy}$, $\frac{1}{\sqrt{2gy}\sqrt{1+y^{\prime 2}}}=\frac{1}{v}\cdot \frac{1}{\sqrt{1+y^{\prime 2}}}=\frac{\cos \varphi }{v}=C.$ Ab Snell's law se map karo — complementary-angle bridge. Optics mein Fermat's Principle least-time paths deta hai, aur speed $v$ wali horizontal layers ke across ek light ray obey karta hai: $\frac{\sin \vartheta }{v}=\text{const},$ jahan $\vartheta$ layers ke normal se measured incidence angle hai — aur horizontal layer ka normal vertical hota hai. Hamara $\varphi$ horizontal se measure hota hai, toh $\varphi$ aur $\vartheta$ complementary hain: $\vartheta =90^{\circ }-\varphi$. Isliye: $\sin \vartheta =\sin (90^{\circ }-\varphi )=\cos \varphi ,$ aur hamara conserved quantity $\frac{\cos \varphi }{v}$ literally $\frac{\sin \vartheta }{v}$ hai — Snell's invariant. Toh brachistochrone ek light ray ki tarah bend karta hai ek aisi medium mein jo depth ke saath speed up hoti hai. Yahi Johann Bernoulli ka original 1696 optical solution hai: dono "least-time" principles ek hi law hain. Consistency check. Cusp par tangent vertical hai, toh $\varphi =90^{\circ }$ (horizontal se) aur $\vartheta =0^{\circ }$ (vertical normal se). Phir $\cos \varphi =0=\sin \vartheta$, $y^{\prime }\to \infty$ ke saath consistent. Complementary bookkeeping boundary par hold karta hai.

Tautochrone proof kaunsa exercise number test karta hai? ::: 4.1 Kaunsa exercise is problem ko Snell's law / Fermat se connect karta hai? ::: 5.2 Kaunse do exercises mein endpoint se $r$ solve karna padta hai? ::: 3.1 and 5.1