Brachistochrone problem

4.10.14 · Maths › Advanced Topics (Elite Level)

KYA poocha ja raha hai

Humein ek functional — ek number jo poore function par depend karta hai — ko minimize karne ka tool chahiye.

KAISE time functional set up karte hain (first principles)

Step 1 — Time = distance / speed, jod ke. $ds$ length ke ek tiny arc ke saath bead speed $v$ se travel karti hai, toh time lagta hai $dt=ds/v$. Total: $T={\int }_A^B\frac{ds}{v}$

Yeh step kyun? Time locally accumulate hota hai; local $dt$ ko integrate karne se total time milta hai.

Step 2 — Arc length curve ke terms mein. $ds=\sqrt{d{x}^2+d{y}^2}=\sqrt{1+y^{\prime 2}}dx,y^{\prime }=\frac{dy}{dx}$

Kyun? Ek infinitesimal step $(dx,dy)$ par Pythagoras.

Step 3 — Energy conservation se speed. Maano bead height $y=0$ par rest se shuru hoti hai aur $y$ ko neeche ki taraf positive maapte hain. Energy conservation (bina friction ke): $\frac{1}{2}m{v}^2=mgy\Rightarrow v=\sqrt{2gy}$

Kyun? Gain ki gayi kinetic energy = khoyi gayi potential energy. Mass cancel ho jaata hai — answer mass-independent hai.

Step 4 — Functional assemble karo. $\boxed{T[y]={\int }_0^{x_1}\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}dx}$

Toh hamara integrand ("Lagrangian") hai $F(y,y^{\prime })=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}.$

KAISE minimize karein: Euler–Lagrange + Beltrami

Key shortcut: yahan $F$ mein koi explicit $x$ nahi hai. Aise cases ke liye ek conserved quantity hai (Beltrami identity):

KAISE Beltrami equation solve karein

$F_{y^{\prime }}$ compute karo: $F_{y^{\prime }}=\frac{1}{\sqrt{2gy}}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}.$

Phir $F-y^{\prime }{F}_{y^{\prime }}=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}-\frac{y^{\prime 2}}{\sqrt{2gy}\sqrt{1+y^{\prime 2}}}=\frac{1}{\sqrt{2gy}\sqrt{1+y^{\prime 2}}}=C.$

Yeh simplify kyun hota hai? Dono terms ka denominator $\sqrt{2gy}\sqrt{1+y^{\prime 2}}$ share hota hai; numerator ban jaata hai $(1+y^{\prime 2})-y^{\prime 2}=1$.

Square karke aur $2g$ ko constant mein absorb karke: $y(1+y^{\prime 2})=k(k=\frac{1}{2g{C}^2},\text{ ek positive constant}).$

Yeh brachistochrone ODE hai. Separation se solve karo: $y^{\prime }=\sqrt{\frac{k-y}{y}}\Rightarrow dx=\sqrt{\frac{y}{k-y}}dy.$

Substitution $y=\frac{k}{2}(1-\cos \theta )$ (toh $dy=\frac{k}{2}\sin \theta d\theta$):

• $k-y=\frac{k}{2}(1+\cos \theta )$, aur $\frac{y}{k-y}=\frac{1-\cos \theta }{1+\cos \theta }={\tan }^2(\theta /2)$.
• Toh $dx=\tan (\theta /2)\cdot \frac{k}{2}\sin \theta d\theta =\frac{k}{2}(1-\cos \theta )d\theta$ ($\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ use karke).
• Integrate karo: $x=\frac{k}{2}(\theta -\sin \theta )$.

KYUN cycloid jeetta hai (intuition)

Worked examples

Common mistakes

Connections

• Calculus of Variations — parent framework; yeh iska founding problem hai.
• Euler-Lagrange equation — woh master condition jo humne Beltrami se reduce ki.
• Cycloid — solution curve; tautochrone bhi.
• Conservation of Energy — $v=\sqrt{2gy}$ deta hai.
• Fermat's Principle — optics mein analogous "least-time" idea (Snell's law).
• Lagrangian Mechanics — physical motion ke liye wahi variational machinery.

Concept Map