Line integrals — scalar and vector, work done: Question bank

4.4.27 · D5 · Maths › Multivariable Calculus › Line integrals — scalar and vector, work done

Sahi hai ya galat — justify karo

A scalar line integral

\int_{C} f d s

kabhi sign nahi badalta jab curve ko reverse karo.

Sahi —

d s = ∣ r^{'} (t) ∣ d t

ek unsigned length hai, isliye

f \cdot d s

ka total same rehta hai chahe tum wire par kis bhi direction mein chalo.

A vector line integral

\int_{C} F \cdot d r

kabhi sign nahi badalta jab curve ko reverse karo.

Galat — reversing karne se

d r = r^{'} d t

flip ho jaata hai, isliye har dot product flip ho jaata hai, aur result

- W

milta hai. Work ka sign hota hai kyunki field ke against kheenchne se energy wapas milti hai.

Agar do alag parametrisations same curve ko same direction mein trace karti hain, to dono same vector line integral deti hain.

Sahi — value sirf geometric path aur uski orientation par depend karti hai, is par nahi ki parameter kitni tezi se chalta hai; speed factor dono tareekon mein sahi se cancel ho jaata hai.

\int_{C} f d s

negative ho sakta hai agar

f

kahi

C

par zero se neeche jaaye.

Sahi —

d s > 0

hamesha hota hai, lekin

f

negative ho sakta hai, isliye integrand

f d s

negative ho sakta hai. (Physical density ke liye

f \geq 0

hoga, lekin abstractly ho sakta hai.)

Kisi bhi closed curve

C

ke liye,

\oint_{C} F \cdot d r = 0

Galat — yeh tabhi guaranteed hai jab

F

conservative ho (dekho Gradient and conservative fields). Ek rotational field jaise

F = (- y, x)

nonzero loop integral deta hai.

Agar

F = \nabla ϕ

, to

\int_{C} F \cdot d r

sirf

C

ke endpoints par depend karta hai.

Sahi — Fundamental Theorem for Line Integrals ke according,

W = ϕ (end) - ϕ (start)

, isliye beech mein path ki shape irrelevant hai.

\int_{C} f d s

mein

f = 1

rakhne se

C

ki arc length milti hai.

Sahi — tum

1 \cdot d s = d s

ko curve par sum kar rahe ho, jo bilkul Arc length ki definition hai.

\int_{C} F \cdot d r

equals

\int_{C} (F \cdot T) d s

same orientation ke liye.

Sahi — kyunki

d r = T d s

jahan

T = r^{'} /∣ r^{'} ∣

, to vector integral literally

F

ke tangential component ka scalar integral hai.

Agar

F

har jagah curve ke perpendicular hai, to work zero hai.

Sahi — tangential component

F \cdot T = 0

har jagah hai, isliye kuch accumulate nahi hota; sideways force koi work nahi karta.

Error dhundo

"Wire ki mass

= \int_{a}^{b} f (r (t)) d t

." Kya galat hai?

∣ r^{'} (t) ∣

factor missing hai;

d t

parameter mein ek step hai, length mein nahi. Sahi hai

\int_{a}^{b} f (r (t)) ∣ r^{'} (t) ∣ d t

d r

aur

d s

same tiny quantity hain, isliye main inhe swap kar sakta hoon." Kyun galat hai?

d s = ∣ r^{'} ∣ d t

ek positive scalar hai (length);

d r = r^{'} d t

ek vector hai (displacement). Ye

d r = T d s

se related hain, equal nahi.

"Maine curve reverse kiya, isliye apne mass integral par sign flip karoon." Error?

Mass ek scalar line integral hai — yeh orientation-independent hai, isliye koi sign flip nahi. Sirf vector/work integral flip hota hai.

F \cdot d r

ka matlab hai

F

ki length ko

d r

ki length se multiply karo." Error?

Dot product sirf

F

ka woh component rakhta hai jo

d r

ke along ho; yeh

∣ F ∣∣ d r ∣ cos θ

hai, plain product nahi. (Dekho Dot product.)

"Kyunki unit circle ke liye

∣ r^{'} (t) ∣ = 1

hai, main yeh factor hamesha drop kar sakta hoon." Error?

Yeh

1

sirf isliye hai kyunki us particular parametrisation ki speed unit hai. Ek reparametrisation jaise

r (2 t)

ki speed

2

hogi; tumhe har baar

∣ r^{'} ∣

recompute karna hoga.

"Work nikalne ke liye main

∣ F ∣

ko arc length par integrate karta hoon." Error?

Isse direction ki parwah kiye bina poori force magnitude count ho jaati. Work sirf tangential part use karta hai:

\int_{C} (F \cdot T) d s

, isliye motion ke perpendicular force kuch contribute nahi karta.

"Green's Theorem se main kisi bhi

\int_{C} F \cdot d r

ko double integral mein convert kar sakta hoon." Error?

Green's Theorem ek closed curve par apply hota hai jo kisi region ko bound kare (aur

F

uspar smooth honi chahiye); yeh open path par apply nahi hota.

Why questions

Speed factor

∣ r^{'} (t) ∣

scalar integral mein kyun aata hai lekin raw velocity

r^{'} (t)

vector wale mein kyun aata hai?

Scalar integral ko sirf kitni length har step cover karti hai yeh chahiye (ek magnitude), jabki vector integral ko har step ki direction chahiye taaki

F

se dot kar sake, isliye woh poora velocity vector rakhta hai.

\int_{C} F \cdot d r

parametrise karne ki speed se independent kyun hai, jabki

r^{'}

tez speed par bada hota jaata hai?

Ek tez parametrisation

t

-interval ko exactly utne hi factor se shrink karti hai jitna woh

r^{'}

grow karti hai; dono effects

\int F \cdot r^{'} d t

mein cancel ho jaate hain, aur geometric value bachti hai.

Conservative field loop integrals ko zero kyun karta hai?

Kyunki

W = ϕ (end) - ϕ (start)

, aur closed loop par start aur end coincide karte hain, isliye difference zero ho jaata hai.

Work ke liye magnitudes multiply karne ki jagah dot product kyun lete hain?

Sirf force ka woh component jo motion ki direction mein ho energy transfer karta hai; dot product

F \cdot T

exactly woh forward-pushing part extract karta hai aur sideways part discard kar deta hai.

Ek curve par line integral ko hum hamesha

t

mein ordinary single integral mein kyun collapse kar sakte hain?

Ek curve 1-dimensional hai — ek number

t

tumhari position fix kar deta hai — isliye parametrisation substitute karne se sab kuch standard integral ban jaata hai

[a, b]

interval par.

Orientation reverse karna vector integral ko kyun flip karta hai lekin direction-preserving reparametrisation nahi?

Reversal literally har displacement

d r

ko negate karta hai, sum ko negate karta hai; same-direction reparametrisation sirf speed ko relabel karta hai, jo cancel out ho jaata hai aur answer same rehta hai.

Edge cases

\int_{C} f d s

kya hoga agar

C

ek single point hai (degenerate, zero length)?

Zero — integrate karne ke liye koi arc length hi nahi, isliye

f \cdot d s

ka sum empty hai.

Agar

F = 0

har jagah ho, to work kya hai?

Zero — har

F \cdot d r = 0

hai, path ya uski length se koi farak nahi padta.

Ek closed curve ke liye,

\oint_{C} 1 d s

kya compute karta hai, aur kya yeh zero ho sakta hai?

Yeh loop ka total perimeter (arc length) compute karta hai; yeh zero tabhi hoga jab loop degenerate ho (ek point), warna strictly positive hota hai.

Agar

F

conservative hai, to

\int_{C} F \cdot d r

ek aise path par kya hoga jo same point se start aur end hota hai lekin beech mein kaafi door ghoom aata hai?

Phir bhi zero — path-independence sirf endpoints ke coincide hone par depend karti hai, path kitna dur bhatka is par nahi.

\int_{C} f d s

ka kya hoga agar tum same curve ko do baar traverse karo (ek baar ghoom ke, phir dobara)?

Yeh double ho jaata hai — arc-length contribution doosri baar phir accumulate hoti hai;

d s

hamesha positive hai isliye koi cancellation nahi hoti.

\oint_{C} F \cdot d r

(non-conservative

F

) ka kya hoga agar closed curve ko do baar loop karo?

Yeh double ho jaata hai — har lap par same nonzero circulation accumulate hoti hai kyunki dono baar orientation preserved rehta hai.

Agar curve par ek corner ho (ek point par velocity undefined ho), to kya line integrals phir bhi defined ho sakte hain?

Haan —

C

ko smooth pieces mein tod do aur integrals add karo; zero length ka ek single non-smooth point kuch contribute nahi karta.

7. Connections

Line integrals — scalar and vector, work done — woh parent topic jise yeh bank drill karta hai.
Dot product — har work question ke peeche "component along the path" engine.
Parametric curves — kyun ek number $t$ ek curve integral ko collapse karta hai.
Arc length — $f = 1$ edge case.
Gradient and conservative fields aur Fundamental Theorem for Line Integrals — path-independence traps.
Green's Theorem — closed-curve conversion caveat.