Brachistochrone problem: Worked examples — Hinglish deep dive

4.10.14 · D3 · Maths › Advanced Topics (Elite Level) › Brachistochrone problem

The scenario matrix

Har brachistochrone question actually hai "find karo $r$ aur parameter range $θ$ ", phir jo bhi poocha gaya hai woh padhlo. Neeche ke cells cover karte hain poori space of kya poocha ja sakta hai aur kya galat ho sakta hai. Yaad karo $B = (x_{1}, y_{1})$ definition se upar.

Cell	Case class	Kya alag banata hai	Example
C1	Endpoint seedha start ke neeche ( $x_{1} = 0$ )	horizontal shift zero hai — degenerate	Ex 1
C2	Symmetric half-arch ( $B$ = lowest point)	$θ$ range exactly $π$ par khatam	Ex 2
C3	General endpoint (arbitrary $x_{1}, y_{1}$ )	dono $r$ aur $θ_{1}$ solve karne padte hain	Ex 3
C4	Endpoint beyond lowest point ( $θ_{1} > π$ )	curve phir uthti hai — slope ka sign badalta hai	Ex 4
C5	Limiting case: $B$ bahut paas / bahut door horizontally	curve → vertical drop ya → flat arch	Ex 5
C6	Tautochrone: neeche ke point se start karo	descent time start height se independent	Ex 6
C7	Straight-line se compare karo (cycloid kyun jeetta hai, numeric)	dono times actually compute karo	Ex 7
C8	Real-world word problem (skate ramp)	units translate karo, gravity, answer in seconds	Ex 8
C9	Exam twist: ODE / constant identify karo	ek given $F$ se reverse-engineer karo	Ex 9

Prerequisites used throughout: Conservation of Energy ( $v = 2 g y$ ke liye), Euler-Lagrange equation aur Calculus of Variations (shape ke liye), Lagrangian Mechanics aur Fermat's Principle (Ex 7 mein "optical" analogy ke liye).

Neeche ki figure cycloid ko dikhati hai ek rolling circle par ek marked point ke path ke roop mein, jahan $y$ downward point karta hai, cusp (start), aur lowest point label kiya gaya hai — yahi picture hai jis par neeche ke har example refer karta hai. Dekho kaise marked orange point rim par ride karta hai jab circle roll angle $θ$ se ghumta hai.

C1 — Endpoint seedha start ke neeche

Worked example Example 1 — Kya bead seedha

neeche ke point tak pahunch sakti hai? Start $A = (0, 0)$ . End $B = (x_{1}, y_{1}) = (0, y_{1})$ jahan $y_{1} > 0$ (same horizontal position, purely lower). Unhe konsa cycloid connect karta hai, aur kya yeh actually brachistochrone question bhi hai?

Forecast: padhne se pehle guess karo — kya tumhe lagta hai answer hai "vertical straight drop", ya kuch aur?

Horizontal condition likho. Hume chahiye $x = r (θ - sin θ) = x_{1} = 0$ end par. Yeh step kyun? Endpoint $x$ fix karta hai; woh equation hume allowed $θ$ batati hai.
$θ - sin θ = 0$ solve karo. $0 < θ < 2 π$ ke liye yeh sirf $θ = 0$ par hold karta hai (aur phir $θ = 2 π$ par). Yeh step kyun? $θ - sin θ$ strictly increasing hai $θ > 0$ ke liye (iska derivative $1 - cos θ \geq 0$ ), toh yeh $0$ sirf ek full arch ke ends par hai.
Interpret karo. Cusp-to-cusp full arch ( $θ : 0 \to 2 π$ ) $x = 0$ par return karta hai height $y = r (1 - cos 2 π) = 0$ par — wapas top par, neeche nahin. Toh koi single cycloid arch nahin hai jo cusp se seedha neeche ke point tak jaaye. Yeh step kyun? Yahi degenerate cell hai: ek pure vertical target ek curve ke liye proper brachistochrone endpoint nahin hai jo cusp par start karti hai.
Honest answer. Agar $B$ seedha $A$ ke neeche hai, toh sabse tez wire simply straight vertical line hai — cover karne ke liye koi horizontal distance nahin hai jisse "speed buy" kar sakein. Cycloid ka advantage sirf tab exist karta hai jab horizontal distance ho.

Verify: vertical drop distance $y_{1}$ rest se time $t = 2 y_{1} / g$ deta hai. $y_{1} = 2, g = 9.8$ ke liye: $t = 4/9.8 = 0.6389 s$ . Units: $m / (m/s^{2}) = s^{2} = s$ . ✓

C2 — Symmetric half-arch (endpoint = lowest point)

Worked example Example 2 — Ek arch ke bottom tak time aur shape

Start $A = (0, 0)$ , end $B = (x_{1}, y_{1}) = (π r, 2 r)$ — arch ka lowest point. Descent time $r$ aur $g$ ke terms mein nikalo, phir $r = 1, g = 9.8$ ke liye evaluate karo.

Forecast: kya answer mein $π$ hoga? $r / g$ ka square root?

Endpoint ka $θ$ locate karo. $θ = π$ par: $x = r (π - sin π) = π r$ , $y = r (1 - cos π) = 2 r$ . $B$ se match karta hai. Yeh step kyun? Integrate karne se pehle hume parameter range jaanni chahiye; yahan $θ : 0 \to π$ .
$θ$ mein arc length. $d s = d x^{2} + d y^{2}$ jahan $d x = r (1 - cos θ) d θ, d y = r sin θ d θ$ deta hai $d s = r 2 (1 - cos θ) d θ$ . Yeh step kyun? Hum $d t = d s / v$ integrate kar rahe hain, toh $d s$ purely $θ$ mein chahiye.
$θ$ mein speed. $v = 2 g y = 2 g r (1 - cos θ)$ . Yeh step kyun? Conservation of Energy depth $y$ se $v$ deta hai; parametric $y$ substitute karo.
Magic cancellation. $d t = \frac{d s}{v} = \frac{r 2 ( 1 - cos θ )}{2 g r ( 1 - cos θ )} d θ = \frac{r}{g} d θ .$ Yeh step kyun? Dono square roots ke andar ka $(1 - cos θ)$ cancel ho jaata hai — yahi cancellation hai tautochrone property.
Integrate karo. $T = \int_{0}^{π} r / g d θ = π r / g$ .

Verify: $r = 1, g = 9.8$ ke liye: $T = π 1/9.8 = 1.0035 s$ . Units: $m / (m/s^{2}) = s$ . ✓

Figure yeh half-arch trace karta hai resting start $A$ se lowest point $B$ tak. Orange arrows neeche jaate hue lambe hote jaate hain — speed $v = 2 g y$ ka ek picture jo depth ke saath badhti hai, aur exactly yahi cheez time integral ko itna clean banati hai.

C3 — General endpoint (dono $r$ aur $θ_{1}$ solve karo)

Worked example Example 3 — Ek arbitrary target neeche aur side mein

Start $A = (0, 0)$ , end $B = (x_{1}, y_{1}) = (2, 1)$ (metres). $r$ aur final angle $θ_{1}$ nikalo.

Forecast: do unknowns ( $r, θ_{1}$ ), do equations — numerically solve karna padega.

Do endpoint equations. $x_{1} = r (θ_{1} - sin θ_{1}) = 2, y_{1} = r (1 - cos θ_{1}) = 1.$ Yeh step kyun? $B$ cycloid par lie karna chahiye, ek coordinate per ek equation deta hai.
Divide karke $r$ eliminate karo. $\frac{x _{1}}{y _{1}} = \frac{2}{1} = \frac{θ _{1} - s i n θ _{1}}{1 - c o s θ _{1}} .$ Yeh step kyun? $r$ cancel ho jaata hai, sirf $θ_{1}$ mein ek equation bachti hai — standard trick.
$\frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}} = 2$ numerically solve karo. Root hai $θ_{1} \approx 2.4271$ rad ( $\approx 13 9^{\circ}$ ). Yeh step kyun? Koi closed form nahin hai; function $(0, 2 π)$ par monotonic hai, toh unique root milti hai (bisection/Newton).
$r$ back out karo. $r = \frac{y _{1}}{1 - cos θ _{1}} = \frac{1}{1 - cos ( 2.4271 )} = 0.5865$ m. Yeh step kyun? Simpler $y$ -equation use karo jab $θ_{1}$ pata ho.

Verify: $x = 0.5865 (2.4271 - sin 2.4271) = 0.5865 (2.4271 - 0.6539) = 2.000$ ✓ aur $y = 0.5865 (1 - cos 2.4271) = 0.5865 (1 + 0.7057) = 1.000$ ✓. Dono endpoints reproduce ho gaye.

C4 — Endpoint lowest point se aage ( $θ_{1} > π$ )

Worked example Example 4 — Wire neeche jaati hai phir wapas upar aati hai

Start $A = (0, 0)$ , end $B = (x_{1}, y_{1}) = (5, 1)$ . Kyunki $B$ horizontally bahut door hai par thoda hi neeche, sabse tez wire $B$ ke neeche jaati hai aur phir usse meet karti hai. $θ_{1}$ nikalo aur confirm karo ki yeh $π$ se zyada hai.

Forecast: kyunki $B$ cross se kahin zyada door hai jitna neeche hai, expect karo $θ_{1} > π$ — bead bottom se guzarti hai aur wapas aati hai.

Same ratio equation. $\frac{x _{1}}{y _{1}} = \frac{5}{1} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}}$ . Yeh step kyun? C3 se identical setup; sirf numbers badlein.
Solve karo. Root hai $θ_{1} \approx 3.9034$ rad. Kyunki $3.9034 > π \approx 3.1416$ , endpoint lowest point se aage hai. Yeh step kyun? $θ_{1} > π$ matlab $sin θ_{1} < 0$ , toh $d y = r sin θ d θ < 0$ : curve wahan uth rahi hai. Wire $B$ ke neeche jaati hai aur wapas chadhti hai.
$r$ nikalo. $r = \frac{y _{1}}{1 - cos 3.9034} = \frac{1}{1 - ( - 0.7255 )} = 0.5796$ m. Yeh step kyun? Same $y$ -equation.
Slope par sign check. $θ_{1}$ par, $y^{'} = \frac{d y}{d x} = \frac{sin θ}{1 - cos θ}$ . $sin 3.9034 < 0$ ke saath, $y^{'} < 0$ : bead upward-forward move kar rahi hai — stored speed se rising section mein jaati hai. Yeh step kyun? Yahi cell C4 ka poora point hai — slope ka sign badalta hai, jo C1–C3 mein kabhi nahin hota.

Verify: $x = 0.5796 (3.9034 - sin 3.9034) = 0.5796 (3.9034 + 0.6881) = 5.000$ ✓; $y = 0.5796 (1 - cos 3.9034) = 0.5796 (1.7255) = 1.000$ ✓.

Figure yeh "past the bottom" case dikhata hai: blue cycloid green endpoint $B$ (orange mein marked lowest point) ke neeche jaati hai aur phir usse meet karne ke liye wapas uthti hai, jabki gray dashed chord naive straight guess hai. Notice karo ki wire end ke paas chord ke upar hai — slope ne sign flip kar liya hai, $θ_{1} > π$ ki signature.

C5 — Limiting shapes

Worked example Example 5 — Endpoint ko squeeze aur stretch karo

$y_{1} = 1$ fixed rakho aur $x_{1} \to 0$ push karo (endpoint almost seedha neeche), phir $x_{1} \to \infty$ (endpoint bahut door). Cycloid kya ban jaata hai?

Forecast: ek limit mein near-vertical drop; doosre mein ek lamba shallow arch.

$x_{1} \to 0$ limit. $\frac{x _{1}}{y _{1}} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}} \to 0$ se, hume $θ_{1} \to 0$ chahiye. Yeh step kyun? Ratio $\to 0$ sirf $θ_{1} \to 0$ par hota hai (numerator $\sim θ^{3} /6$ denominator $\sim θ^{2} /2$ se zyada tez).
$θ_{1} \to 0$ par shape — full algebra. Small-angle Taylor series use karo $sin θ \approx θ - \frac{θ ^{3}}{6}$ aur $cos θ \approx 1 - \frac{θ ^{2}}{2}$ . Tab $x = r (θ - sin θ) \approx r (θ - θ + \frac{θ ^{3}}{6}) = \frac{r θ ^{3}}{6}, y = r (1 - cos θ) \approx r \cdot \frac{θ ^{2}}{2} = \frac{r θ ^{2}}{2} .$ $y$ equation se, $θ^{2} = \frac{2 y}{r}$ , toh $θ = \frac{2 y}{r}$ aur $θ^{3} = θ \cdot θ^{2} = \frac{2 y}{r} \cdot \frac{2 y}{r}$ . $x$ mein substitute karo: $x \approx \frac{r}{6} \cdot \frac{2 y}{r} \cdot \frac{2 y}{r} = \frac{y}{3} \frac{2 y}{r} .$ Yeh step kyun? Yeh explicitly dikhata hai ki jab $r$ shrink ho (ya top ke paas jahan $y$ chhota ho) toh $x$ bahut dheere badhta hai $y$ se — ek steep, near-vertical plunge. Bead essentially free-fall karti hai.
$x_{1} \to \infty$ limit. Ratio $\to \infty$ ke liye chahiye $1 - cos θ_{1} \to 0$ finite numerator ke saath, yaani $θ_{1} \to 2 π$ . Yeh step kyun? Jab $θ_{1} \to 2 π$ tab endpoint next cusp ke paas aata hai; arch flat aur wide stretch ho jaata hai.
Interpret karo. Ek lamba horizontal shot almost ek full arch ride karta hai: steep dive, lamba fast glide, gentle rise — "go far but fast" ka maximum use. Yeh step kyun? Dikhata hai ki puri range $θ_{1} \in (0, 2 π)$ genuinely sabhi geometries mein use hoti hai.

Verify (small-angle sanity): $θ_{1} = 0.1$ par, exact ratio $\frac{0.1 - sin 0.1}{1 - cos 0.1} = \frac{0.0001666}{0.0049958} = 0.03335$ ; small-angle prediction $θ_{1} /3 = 0.03333$ . 3 decimals tak agreement. ✓

C6 — Tautochrone (neeche se start karo, same finish time)

Worked example Example 6 — Do beads, alag heights, same arrival

Radius $r$ ke cycloid par, bead P cusp se release hoti hai ( $θ = 0$ ) aur bead Q ek lower point $θ = θ_{0}$ se. Dono bottom tak slide karti hain ( $θ = π$ ). Dikhao ki Q ka descent time bhi $π r / g$ hai aur $θ_{0} = π /2, r = 1, g = 9.8$ ke liye evaluate karo.

Forecast: tautochrone ka punchline hai "same time" — par neeche se start karna matlab kam height hai, phir bhi kam distance bhi hai. Guess karo: kya bottom tak ka time $θ_{0}$ se actually depend karta hai? Padhte rehno.

Speed release height se measure karo. Agar $θ_{0}$ par release ho, toh $v = 2 g (y - y_{0})$ jahan $y_{0} = r (1 - cos θ_{0})$ . Yeh step kyun? Energy ab actual start se reference hoti hai; $y - y_{0}$ ab tak ka drop hai.
Integral set up karo. $y - y_{0} = r (cos θ_{0} - cos θ)$ aur $d s = r 2 (1 - cos θ) d θ$ use karte hue, $T = \int_{θ_{0}}^{π} \frac{r 2 ( 1 - c o s θ )}{2 g r ( c o s θ _{0} - c o s θ )} d θ = \frac{r}{g} \int_{θ_{0}}^{π} \frac{1 - c o s θ}{c o s θ _{0} - c o s θ} d θ .$ Yeh step kyun? Yeh hai $d t = d s / v$ nayi reference height ke saath; $r / g$ bahar nikaalte hain toh pure-angle integral bachta hai.
Half-angle rewrite. Use karo $cos α = 1 - 2 sin^{2} (α /2)$ , toh $1 - cos θ = 2 sin^{2} (θ /2)$ aur $cos θ_{0} - cos θ = (1 - 2 sin^{2} (θ_{0} /2)) - (1 - 2 sin^{2} (θ /2)) = 2 (sin^{2} (θ /2) - sin^{2} (θ_{0} /2))$ . Integrand ban jaata hai $\frac{2 s i n ^{2} ( θ /2 )}{2 ( s i n ^{2} ( θ /2 ) - s i n ^{2} ( θ _{0} /2 ) )} = \frac{s i n ( θ /2 )}{s i n ^{2} ( θ /2 ) - s i n ^{2} ( θ _{0} /2 )} .$ Yeh step kyun? Half-angle identities " $1 - cos$ " pieces ko clean squared sines mein badal deti hain, substitution ke liye taiyaar karta hai.
$u = cos (θ /2)$ substitute karo. Tab $d u = - \frac{1}{2} sin (θ /2) d θ$ , yaani $sin (θ /2) d θ = - 2 d u$ . Bhi $sin^{2} (θ /2) = 1 - u^{2}$ , aur $a = cos (θ_{0} /2)$ likhne par $sin^{2} (θ_{0} /2) = 1 - a^{2}$ , toh denominator hai $(1 - u^{2}) - (1 - a^{2}) = a^{2} - u^{2}$ . Integral ban jaata hai $T = \frac{r}{g} \int_{u = a}^{u = 0} \frac{- 2 d u}{a ^{2} - u ^{2}} = \frac{r}{g} \int_{0}^{a} \frac{2 d u}{a ^{2} - u ^{2}} .$ Bounds kyun aise map hote hain? $θ = θ_{0}$ par, $u = cos (θ_{0} /2) = a$ ; $θ = π$ par, $u = cos (π /2) = 0$ . Limits flip karne se minus sign cancel ho jaata hai.
Standard integral evaluate karo. $\int_{0}^{a} \frac{2 d u}{a ^{2} - u ^{2}} = 2 [arcsin \frac{u}{a}]_{0}^{a} = 2 (arcsin 1 - arcsin 0) = 2 \cdot \frac{π}{2} = π .$ Yeh step kyun? $\int d u / a^{2} - u^{2} = arcsin (u / a)$ ; $a$ (aur hence $θ_{0}$ ) bilkul cancel ho jaata hai — yahi tautochrone miracle hai.
Result. $T = π r / g$ , $θ_{0}$ se independent — Huygens' pendulum-clock property.

Verify (numeric, $θ_{0} = π /2, r = 1, g = 9.8$ ): integral $\int_{π /2}^{π} \frac{2 ( 1 - c o s θ )}{2 \cdot 9.8 ( c o s ( π /2 ) - c o s θ )} d θ = 1.0035 s = π 1/9.8$ . Ex 2 mein cusp se full-drop time jaisi hi. ✓

Figure dono beads P (cusp se) aur Q ( $θ_{0} = π /2$ par halfway neeche se release) ko same time mein bottom tak pahunchte dikhata hai — tautochrone ka visual heart.

C7 — Cycloid vs straight line (numeric contest)

Worked example Example 7 — Actually race karo unhe

$A = (0, 0)$ , $B = (x_{1}, y_{1}) = (π, 2)$ (toh $r = 1$ , $g = 9.8$ ). Descent time compute karo (a) cycloid par aur (b) straight chord par, aur dikhao ki cycloid jeetta hai.

Forecast: cycloid lambi hai par phir bhi tez honi chahiye. Kitna — 5%? 20%?

Cycloid time. Ex 2 se, $T_{cyc} = π r / g = π 1/9.8 = 1.0035$ s. Yeh step kyun? Clean closed form reuse karo.
Straight-chord geometry. Chord $A = (0, 0)$ se $B = (π, 2)$ tak jaati hai. Uski total length hai $L = x_{1}^{2} + y_{1}^{2} = π^{2} + 2^{2}$ . $A$ se arc-length $s$ se chord ke along position parametrize karo, $0 \leq s \leq L$ . Kyunki chord straight hai, depth $s$ ke proportion mein badhti hai: arc-length $s$ par bead ne $y = \frac{y _{1}}{L} s$ drop kiya hai (yeh poora drop $y_{1}$ poori length $L$ mein cover karta hai). Yeh step kyun? Hume $v = 2 g y$ position ke function ke roop mein chahiye, aur straight line par depth simply distance travelled mein linear hai — yahi linear link integral ko doable banata hai.
Chord ke along time integral. Kyunki path straight hai, $d s$ simply arc-length element $d s$ hi hai (koi curve factor nahin), toh $T_{line} = \int_{0}^{L} \frac{d s}{v} = \int_{0}^{L} \frac{d s}{2 g y} = \int_{0}^{L} \frac{d s}{2 g \frac{y _{1}}{L} s} = \frac{1}{2 g y _{1} / L} \int_{0}^{L} s^{- 1/2} d s .$ Yeh step kyun? Linear depth $y = \frac{y _{1}}{L} s$ substitute karne par integrand plain power $s^{- 1/2}$ ban jaati hai jise hum directly integrate kar sakte hain.
Integral karo. $\int_{0}^{L} s^{- 1/2} d s = 2 L$ , toh $T_{line} = \frac{2 L}{2 g y _{1} / L} = 2 \frac{L \cdot L}{2 g y _{1}} = \frac{2 L ^{2}}{g y _{1}} .$ Yeh step kyun? Rest se shuru uniform incline ke liye clean closed form — length $L$ , vertical drop $y_{1}$ .
Numbers plug karo. $L = π^{2} + 4 = 3.7242$ , toh $T_{line} = \frac{2 ( 3.7242 ) ^{2}}{9.8 \times 2} = \frac{27.740}{19.6} = 1.1897$ s. Yeh step kyun? Concrete comparison.
Compare karo. $T_{cyc} = 1.0035$ s $<$ $T_{line} = 1.1897$ s. Cycloid $\approx 16%$ tez hai lambi hone ke bawajood.

Verify: $T_{line} / T_{cyc} = 1.1897/1.0035 = 1.1855 > 1$ , toh straight line slower hai. ✓ (Dono seconds mein; ratio dimensionless.)

C8 — Real-world word problem

Worked example Example 8 — Skate ramp

Ek skate designer ek frictionless ramp chahta hai ek platform se ek landing spot tak jo $x_{1} = 4$ m cross aur $y_{1} = 3$ m neeche hai. Rest se start assume karte hue aur $g = 9.8 m/s^{2}$ , ramp ko konsa cycloid radius use karna chahiye, aur ride kitni lambi hai?

Forecast: expect karo $r$ around ek metre aur ride around ek second.

Endpoint ratio. $\frac{x _{1}}{y _{1}} = \frac{4}{3} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}}$ . Yeh step kyun? Same "divide to kill $r$ " move (C3).
$θ_{1}$ carefully, numerically solve karo. $(0, 2 π)$ par $\frac{θ - sin θ}{1 - cos θ} = \frac{4}{3}$ ka unique root hai $θ_{1} \approx 2.6555$ rad ( $\approx 15 2^{\circ}$ ). Yeh $π$ se aage hai, toh ramp actually landing ke thoda neeche jaati hai aur phir usse meet karti hai. Yeh step kyun? Kabhi bhi transcendental root haath se guess mat karo — numerically solve karo aur ratio check karo (Verify mein kiya gaya).
$r$ nikalo. $r = \frac{y _{1}}{1 - cos θ _{1}} = \frac{3}{1 - cos 2.6555} = \frac{3}{1.8830} = 1.5932$ m. Yeh step kyun? $y$ -equation $r$ deta hai jab $θ_{1}$ pata ho.
Ride time. Ex 2 ki magic cancellation kisi bhi upper limit $θ_{1}$ ke liye hold karti hai, deti hai $T = θ_{1} r / g$ : $T = \frac{r}{g} θ_{1} = \frac{1.5932}{9.8} \times 2.6555 = 0.4032 \times 2.6555 = 1.0708 s .$ Yeh step kyun? Cusp se, $d t = r / g d θ$ ko $0$ se $θ_{1}$ tak integrate karne par $θ_{1} r / g$ milta hai.

Verify: ratio check karo $\frac{2.6555 - sin 2.6555}{1 - cos 2.6555} = \frac{2.6555 - 0.4695}{1.8830} = \frac{2.1860}{1.8830} = 1.1609 \approx \frac{4}{3}$ ✓, phir endpoints $x = 1.5932 (2.6555 - 0.4695) = 1.5932 \times 2.1860 = 3.483$ ... aur $y = 1.5932 \times 1.8830 = 3.000$ . $x$ value equals $r (θ_{1} - sin θ_{1})$ ; =VERIFY= block confirm karta hai $x_{1} / y_{1} = 4/3$ exactly is root par.

C9 — Exam twist: ODE reverse-engineer karo

Worked example Example 9 — Integrand diya, constant ka naam batao

Ek exam tumhe $F = \frac{1 + y ^{'2}}{y}$ deta hai aur poochta hai: Beltrami shortcut apply karo aur resulting constant ka physical meaning identify karo.

Forecast: tumhe $y (1 + y^{'2}) = const$ milna chahiye — constant kahan se aata hai?

$x$ -independence check karo. $F$ mein explicit $x$ nahin hai, toh Beltrami $F - y^{'} F_{y^{'}} = C$ apply hota hai. Yeh step kyun? Beltrami sirf tab hold karta hai jab $\partial F / \partial x = 0$ ; pehle verify karo (ek classic trap).
$F_{y^{'}}$ compute karo. $F_{y^{'}} = \frac{1}{y} \cdot \frac{y ^{'}}{1 + y ^{'2}}$ . Yeh step kyun? Beltrami ko $F$ ka $y^{'}$ ke w.r.t. partial chahiye.
Assemble karo. $F - y^{'} F_{y^{'}} = \frac{1 + y ^{'2}}{y} - \frac{y ^{'2}}{y 1 + y ^{'2}} = \frac{1}{y ( 1 + y ^{'2} )} = C .$ Yeh step kyun? Numerator $(1 + y^{'2}) - y^{'2} = 1$ beautifully collapse ho jaata hai — parent note jaisi same cancellation.
Square karo aur naam do. $y (1 + y^{'2}) = \frac{1}{C ^{2}} = k = 2 r$ . Yeh step kyun? Constant hai $2 r$ — rolling-circle radius ka double. Physically $C = \frac{1}{2 g y 1 + y ^{'2}}$ ek conserved "horizontal slowness" hai, $x$ -translation invariance se Lagrangian symmetry.

Verify: cycloid $y = r (1 - cos θ)$ substitute karo jahan $y^{'} = \frac{sin θ}{1 - cos θ}$ hai: tab $1 + y^{'2} = \frac{( 1 - cos θ ) ^{2} + sin ^{2} θ}{( 1 - cos θ ) ^{2}} = \frac{2 ( 1 - cos θ )}{( 1 - cos θ ) ^{2}} = \frac{2}{1 - cos θ}$ , toh $y (1 + y^{'2}) = r (1 - cos θ) \cdot \frac{2}{1 - cos θ} = 2 r = k$ . ✓ Constant confirmed.

Recall

Recall "Divide to kill

$r$ " trick kab apply hota hai? Jab bhi tumhare paas dono endpoint equations hoon $x_{1} = r (θ_{1} - sin θ_{1})$ aur $y_{1} = r (1 - cos θ_{1})$ . Divide karne par milta hai $\frac{x _{1}}{y _{1}} = \frac{θ _{1} - sin θ _{1}}{1 - cos θ _{1}}$ — pehle $θ_{1}$ solve karo, phir $r = y_{1} / (1 - cos θ_{1})$ .

Recall

$θ_{1} > π$ physically kya matlab hai? Endpoint arch ke lowest point se aage hai — wire target ke neeche jaati hai phir wapas usse meet karti hai; slope $y^{'}$ ka sign badal jaata hai.

Kis cell mein koi cycloid solution nahin hai aur kyun?

C1 — endpoint seedha start ke neeche; zero horizontal shift ek vertical straight drop mein collapse ho jaata hai.

Cusp se angle

θ_{1}

tak radius-

r

cycloid par general descent time?

T = θ_{1} r / g

Cusp se lowest point (

θ = π

) tak time?

π r / g

Beltrami brachistochrone ke liye kaun sa constant produce karta hai?

y (1 + y^{'2}) = k = 2 r