Arc length formula — derivation — Hinglish notes

4.2.16 · Maths › Calculus II — Integration

Step 1 — Curve ko line segments se approximate karo

Curve par points $a = x_{0} < x_{1} < \dots < x_{n} = b$ chuno. Consecutive points $(x_{i - 1}, y_{i - 1})$ aur $(x_{i}, y_{i})$ ko seedhe chords se connect karo. Curve ki total length inhi chord lengths ke sum ke approximately barabar hai.

Figure — Arc length formula — derivation

Har chord ek tiny right triangle ki hypotenuse hai jiska horizontal leg $Δ x_{i} = x_{i} - x_{i - 1}$ aur vertical leg $Δ y_{i} = y_{i} - y_{i - 1}$ hai:

$L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2}$

Step 2 — Square root se $Δ x_{i}$ bahar nikalo

Yeh step kyun? Humein $x$ mein integral chahiye, isliye sab kuch $Δ x_{i}$ ke terms mein likhna chahte hain.

$L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2} = (Δ x_{i})^{2} (1 + \frac{( Δ y _{i} ) ^{2}}{( Δ x _{i} ) ^{2}}) = 1 + (\frac{Δ y _{i}}{Δ x _{i}})^{2} Δ x_{i}$

( $Δ x_{i}$ ko bahar nikal sakte hain kyunki $Δ x_{i} > 0$ hai.)

Step 3 — Mean Value Theorem use karo

Yeh step kyun? Humein secant slope $\frac{Δ y _{i}}{Δ x _{i}}$ ko actual derivative $f^{'}$ mein badalna hai, taaki limit ek clean integrand ban jaaye.

Agar $f$ , $[x_{i - 1}, x_{i}]$ par continuous hai aur andar differentiable hai, to Mean Value Theorem guarantee karta hai ki koi $x_{i}^{*} \in (x_{i - 1}, x_{i})$ hoga jiske liye

$f^{'} (x_{i}^{*}) = \frac{f ( x _{i} ) - f ( x _{i - 1} )}{x _{i} - x _{i - 1}} = \frac{Δ y _{i}}{Δ x _{i}}$

Substitute karo:

$L_{i} = 1 + (f^{'} (x_{i}^{*}))^{2} Δ x_{i}$

Step 4 — Sum karo aur limit lo (Riemann sum → integral)

$L = lim_{n \to \infty} \sum_{i = 1}^{n} 1 + (f^{'} (x_{i}^{*}))^{2} Δ x_{i}$

Yeh bilkul exactly function $g (x) = 1 + (f^{'} (x))^{2}$ ka Riemann sum hai. Jab tak $f^{'}$ continuous hai, limit definite integral hoti hai:

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek saamp ek unche-neeche raaste par pada hai. Tum jaanna chahte ho saamp kitna lamba hai. Tum sirf raaste ki width measure nahi kar sakte — saamp oopar-neeche jaata hai. Toh tum saamp ke body ke saath bahut saare tiny seedhe matchsticks rakh do. Har matchstick thoda "across" aur thoda "upar" cover karti hai. Corner-triangle rule (Pythagoras) se, har matchstick ki length $across^{2} + up^{2}$ hai. Saari matchsticks jodhho. Ab aur bhi choti matchsticks use karo, aur choti, hamesha ke liye — yeh "infinitely many tiny matchsticks jodhna" wahi hai jo integral sign $\int$ ka matlab hai. Yahi arc length hai.

Flashcards

y = f (x)

ke liye arc length integrand kya hai?

1 + (f^{'} (x))^{2}

d x

mein

a

b

tak integrate karo.

Arc length formula mein "

+ 1

" kahan se aata hai?

Horizontal leg se:

d s^{2} = d x^{2} + d y^{2}

d x^{2}

se divide karne par

1 + (d y / d x)^{2}

milta hai.

Secant slope

Δ y /Δ x

f^{'} (x_{i}^{*})

mein kaun sa theorem badalta hai?

Mean Value Theorem.

Har tiny chord ki length kis geometric fact se milti hai?

Pythagoras:

L_{i} = (Δ x_{i})^{2} + (Δ y_{i})^{2}

Curve

x = g (y)

ki arc length

y = c

y = d

tak?

\int_{c}^{d} 1 + (d x / d y)^{2} d y

Arc length hamesha endpoints ke beech seedhe chord se ≥ kyun hoti hai?

Seedhi line sabse chota raasta hai; curve chords ka sum hai jo kabhi apne seedhe piece se choti nahi hoti.

d s

d x, d y

ke terms mein arc length differential kya hai?

d s = d x^{2} + d y^{2}

Common error: arc length ke liye

\int y d x

integrate karna — woh actually kya compute karta hai?

Curve ke neeche ka area, uski length nahi.

Connections

Pythagorean theorem — har chord length ka engine.
Mean Value Theorem — secant slopes ko derivatives mein convert karta hai.
Riemann sums and the definite integral — chord-sum limit ko $\int$ mein badalta hai.
Surface area of revolution — usi $d s$ element ko reuse karta hai: $S = \int 2 π y d s$ .
Parametric arc length — $L = \int (d x / d t)^{2} + (d y / d t)^{2} d t$ tak generalize karta hai.
Polar arc length — $L = \int r^{2} + (d r / d θ)^{2} d θ$ .