Question bank — Brachistochrone problem

False. Time hai $\int ds/v$, sirf $\int ds$ nahi; cycloid pehle steep neeche jaata hai isliye $v$ already badi hoti hai lambe stretch par. Shorter-but-slower, longer-but-faster se haarta hai.

False. Mass $\frac{1}{2}m{v}^2=mgy$ mein cancel ho jaata hai, isliye $v=\sqrt{2gy}$ mass-free hai; poora functional $\int \sqrt{1+y^{\prime 2}}/\sqrt{2gy}dx$ mein koi $m$ nahi hai.

False (shape ke liye), true (time ke liye). $g$ ek overall constant $\sqrt{2g}$ ke roop mein functional ko multiply karta hai, isliye har path ka time equally scale hota hai aur yeh change nahi kar sakta kaun sa path minimal hai. Shape ek cycloid hoti hai chahe gravity ki strength kuch bhi ho.

False. Euler–Lagrange sirf time ko stationary banata hai (first variation zero); yeh candidates ko flag karta hai. Cycloid ko genuinely minimum check kiya jaata hai alag se — E–L akela min ko max ya saddle se distinguish nahi kar sakta.

False. Beltrami, E–L ka ek consequence hai, sirf tab valid hai jab $F$ mein koi explicit $x$ na ho; yeh ek first integral (conserved quantity) hai jo E–L ki second-order equation ki jagah ek first-order equation deta hai.

False. Yeh tautochrone property hai: dono ek hi time mein bottom tak pahunchti hain, $\pi \sqrt{r/g}$ lowest point tak, starting height se independent.

False. Agar $B$ directly $A$ ke neeche hai, toh free vertical fall hi sabse fast path hai — cycloid us vertical line mein degenerate ho jaata hai (uska steepest possible start). Koi horizontal distance nahi matlab koi advantage trade karne ko nahi hai.

False. Hum total time minimize karte hain $\int ds/v$. Average speed ek summary number hai jo discard karta hai ki speed kahan fast hai aur kahan slow — lekin kahan matter karta hai, isliye yeh optimization target nahi ho sakta.

False. $T[y]$ ek functional hai: iska input ek poora curve $y(x)$ hai aur output ek number hai (time). Numbers par optimize karne se functions par optimize karne ka yeh shift hi woh cheez hai jisne Calculus of Variations ko launch kiya.

Error sign convention mein hai. Yahan $y$ neeche ki taraf positive measure kiya gaya hai, isliye girte hue bead ka $y>0$ hai aur $v=\sqrt{2gy}$ real hai. $y$ upar choose karne se sign flip ho jaata — pehle apna convention batao.

Galat — Beltrami ke liye $\partial F/\partial x=0$ chahiye (koi explicit $x$-dependence nahi). Agar $F$ directly $x$ par depend karta hai, toh quantity $F-y^{\prime }{F}_{y^{\prime }}$ conserved nahi hoti aur poora Euler–Lagrange equation use karna padega.

Ulta hai. Jab $y\to 0$, product force karta hai $y^{\prime 2}\to \infty$, isliye $y^{\prime }\to \infty$: curve vertical shuru hoti hai, flat nahi. Woh steep plunge exactly woh hai jisse bead pehle speed gain karti hai.

Start par $y=0$ pe integrand $1/\sqrt{y}$ ki tarah blow up karta hai. Yeh ek improper integral hai lekin phir bhi converge karta hai ($1/\sqrt{y}$ singularity integrable hai) — total time finite hai chahe integrand bounded na ho.

Square karne se $y^{\prime }$ ki sign chhup jaati hai: descending part par $y^{\prime }>0$ hai, lekin ek poore arch ke lowest point ke baad bead upar jaati hai aur $y^{\prime }<0$ hota hai. Single ODE $y(1+y^{\prime 2})=k$ dono branches cover karti hai; har segment par sahi root choose karni padegi.

$C=0$ require karta $\sqrt{2gy}\sqrt{1+y^{\prime 2}}=\infty$ har jagah — degenerate, koi curve nahi. $C$ ek positive constant hai jo cycloid ko $B$ se pass karwaane ke liye fix hoti hai; yeh kisi real brachistochrone ke liye kabhi zero nahi hoti.

Nahi. Sine curve mein $y$ ek explicit function of $x$ hoti hai; cycloid genuinely parametric hai aur apne cusp ke paas slope infinite ho jaata hai — koi single-valued sine yeh nahi kar sakta. Yeh ek rolling circle ke rim-point ka path hai (dekho Cycloid).

Kyunki speed path ke along continuously vary karti hai; sirf local infinitesimal times $ds/v$ ko sum karna hi slow regions (top) ko fast regions (bottom) ke against sahi weight deta hai.

Kyunki $F$ mein koi explicit $x$ nahi hai, Beltrami seedha ek first-order ODE ($y(1+y^{\prime 2})=k$) deta hai, instead of messy second-order Euler–Lagrange equation ke jo otherwise do baar integrate karni padti.

Substitution $y=\frac{k}{2}(1-\cos \theta )$ jo $y(1+y^{\prime 2})=k$ solve karti hai, exactly $x=r(\theta -\sin \theta )$ produce karti hai — algebra force karta hai rolling-circle curve; koi aur shape ODE satisfy nahi karta.

Factor $(1-\cos \theta )$ arc length aur speed dono mein identically appear hota hai, isliye $ds/v$ mein cancel ho jaata hai. Woh cancellation simultaneously woh cheez hai jo total time minimize karti hai aur woh cheez bhi jo descent time ko starting height se independent banati hai.

Yeh pehla famous case tha jahan optimization ek poori function (curve) par thi naa ki numbers par, jisme nayi machinery ki zaroorat thi — Euler-Lagrange equation — ordinary calculus se aage.

Fermat kehta hai light least-time path leta hai; ek aisa medium jismein speed depth ke saath $\sqrt{2gy}$ ki tarah badhti hai light ko cycloid mein bend karta hai, isliye bead aur light ray dono ek hi "least time" law follow karte hain.

Tab $B$ neeche nahi hai, isliye rest se start karne wala bead kabhi wahan pahunch nahi sakta (convert karne ke liye koi potential energy nahi). Problem jaise pose ki gayi hai, $B$ strictly $A$ ke neeche hona chahiye.

Cycloid phir bhi apply hoti hai lekin ek bada radius $r$ aur chhota $\theta$-range use karta hai; curve chord ke neeche dip karti hai aur $B$ ki height se bhi neeche ja sakti hai phir wapas upar aati hai — long horizontal run ke liye speed gain karne ke liye zaroorat se zyada gehre jaana.

Horizontal distance $\to 0$, cycloid ka arch ek vertical line par collapse ho jaata hai — bead simply free-fall karta hai, jo ki genuinely fastest descent hai jab cover karne ke liye kuch horizontal na ho.

Haan. Ek poore arch ke lowest point ke baad bead $B$ ki taraf upar jaati hai (wahi cycloid par). Woh pehle deep plunge karne ke liye thoda sacrifice karti hai, phir thoda climb — net time phir bhi minimal hota hai.

Lowest point tak time $\pi \sqrt{r/g}$ hai, isliye $r$ double karne se time $\sqrt{2}$ se multiply ho jaata hai — ek bada, gentle arch zyada time leta hai chahe bead higher speeds reach kare.

Bead wahan rest se start karti hai $v=0$ ke saath, aur vertical tangent ka matlab sirf yeh hai ki woh ek instant ke liye seedha neeche girti hai; time contribution finite rehta hai kyunki singular integrand integrable hai.