Conservative vector fields — potential functions: Question bank

4.4.26 · D5 · Maths › Multivariable Calculus › Conservative vector fields — potential functions

Shuru karne se pehle, teen words jo hum baar baar use karte hain:

Ek conservative field woh hoti hai jo $\nabla f$ ke barabar hoti hai (kisi scalar hill $f$ ka gradient) — dekho Gradient and directional derivative.
Ek loop integral $\oint_{C} F \cdot d r$ ek closed path par total "push along" hota hai — dekho Line integrals of vector fields.
Curl local swirl measure karta hai; curl $= 0$ cross-partial test $P_{y} = Q_{x}$ hai — dekho Curl of a vector field.

Error dhundho

"Maine

P

x

mein integrate kiya aur

f = x^{2} y + 3 x

mila, toh yahi potential hai."

Error: ==

+ g (y)

== drop kar diya.

x

ke respect mein integrate karne par

y

ka ek unknown function bachta hai, jise equation

f_{y} = Q

se pin down karna hota hai.

P_{y} = Q_{x}

hold hua, toh field conservative hai — domain check karne ki zarurat nahi."

Error: simply-connected requirement skip kar di. Hole wale domain par test pass ho sakta hai jabki hole ke around ek loop nonzero integral de sakta hai.

"Har vector field kisi na kisi cheez ka gradient hoti hai, toh main bas

f

dhundh lunga."

Error: sirf curl-free fields ko nice domains par potentials hoti hain.

(y, - x)

ke liye,

P_{y} = 1 \neq = - 1 = Q_{x}

, toh koi

f

exist nahi karta.

f

dhundhne ke liye maine

P

x

mein integrate kiya, phir

Q

y

mein integrate kiya, phir unhe add kar diya."

Error: isse shared terms double-count ho jaate hain. Sahi method

P

ko ek baar integrate karta hai, phir result ko

y

mein differentiate karke

Q

se match karta hai taaki sirf missing piece

g (y)

recover ho sake.

A

B

tak

C

par work hard tha, toh maine ek straight-line path choose kiya asaan karne ke liye — lekin sirf kyunki main lucky tha."

Error nahi hai, lekin reasoning galat hai: yeh luck nahi hai. Ek baar field conservative confirm ho jaaye, har path ek hi answer deta hai, isliye easy path choose karna hamesha legal hai.

g^{'} (y) = - 4 y

toh

g (y) = - 4 y

Error: integrate karna bhool gaye.

g (y) = \int - 4 y d y = - 2 y^{2} + C

; tumhe antidifferentiate karna hota hai, copy nahi.

"2D mein Curl zero matlab single number

P_{y} - Q_{x}

zero hai, toh

F

automatically kahin bhi conservative hai."

Error: "kahin bhi" domain assumption phir se chhupa leta hai; is implication ke liye zaruri hai ki jinpe tumhe care hai un loops se enclosed koi holes na hon.

Why questions

Cross-partial test mein sirf mixed partials kyun hote hain,

P_{x}

Q_{y}

kyun nahi?

Kyunki conservativeness

f_{x y}

f_{y x}

se compare karta hai; "straight" partials

f_{xx}, f_{y y}

mein koi cross-consistency information nahi hoti aur woh unconstrained hote hain.

Clairaut's theorem (equality of mixed partials) test ka engine kyun hai?

Yeh guarantee karta hai ki smooth

f

ke liye

f_{x y} = f_{y x}

; iska ulta padhne par, agar

F = \nabla f

toh

P_{y} = Q_{x}

zaroor hold karna chahiye, isliye

P_{y} = Q_{x}

ki failure kisi bhi potential ko rule out kar deti hai.

Domain mein hole hone se "curl-zero ⟹ conservative" step kyun toot jaata hai?

Hole ek loop ko ek aisi point ke around ghoomne deta hai jahan field defined nahi hai;

f

ko recover karne ke liye us missing point ko fill karna padta hai, aur koi single-valued

f

yeh nahi kar sakta, isliye path-independence fail ho jaati hai — yahi punctured-plane counterexample hai.

Ek conservative force "energy conserve" kyun karta hai?

Kiya gaya kaam sirf endpoints ke beech potential difference ke barabar hota hai, isliye ek round trip tumhe same potential par wapas le aata hai — looping se koi energy create ya nikali nahi jaati, yeh Conservation of energy in physics se match karta hai.

Green's theorem ko ek curl-free field ke liye filled region par loop integral zero confirm karne ke liye kyun use kiya ja sakta hai?

Green's theorem loop integral ko enclosed area par

Q_{x} - P_{y}

ke double integral mein convert kar deta hai; agar woh integrand puri hole-free region mein zero hai, toh loop integral zero hai.

Fundamental Theorem of Line Integrals ko ek potential ki zarurat kyun hai, sirf koi bhi scalar function kyun nahi chalega?

Derivation chain rule

\frac{d}{d t} f (r (t)) = \nabla f \cdot r^{'}

use karta hai; yeh integrand ko perfect derivative mein tab hi turn karta hai jab

F

actually

\nabla f

ke barabar ho.

f

jaanne par

C

ki parametrization bilkul kyun skip ho jaati hai?

Kyunki integral sirf

f (B) - f (A)

tak collapse ho jaata hai, jo sirf do endpoints par depend karta hai — curve ki shape aur speed kabhi nahi aati.

Edge cases

Kya zero field

F = 0

conservative hai?

Haan — yeh kisi bhi constant

f

ke liye

\nabla f

hai; trivially curl-free aur har integral zero hai.

Punctured field

(- y, x) / (x^{2} + y^{2})

ka loop integral ek aisi loop ke around kya hai jo origin ko enclose nahi karti?

Zero — hole ke bahar hone par, woh region effectively wahan simply connected hai, isliye curl-free ek vanishing loop deta hai, aur sirf origin ke around ghoomne wale loops

2 π

dete hain.

Kya ek field ek region par conservative ho sakti hai lekin ek bade region par nahi jo hole contain kare?

Haan — hole-free patch tak restrict karne par yeh

\nabla f

ho sakti hai, lekin puncture ke across extend karne par potential multivalued ho sakta hai aur conservativeness globally toot sakti hai.

Agar

P_{y} = Q_{x}

sirf ek point par fail ho, toh kya field phir bhi non-conservative hai?

Haan — ek genuine gradient field ko

P_{y} = Q_{x}

har jagah satisfy karna chahiye jahan woh defined aur smooth ho, isliye ek violating point kisi bhi region par jo use contain kare, kisi bhi potential ko rule out kar deta hai.

Potential mein constant

C

add karne se koi physically measurable quantity badlti hai?

Nahi — work aur loop integrals differences

f (B) - f (A)

par depend karte hain, aur

C

cancel ho jaata hai; isliye potentials sirf ek constant tak hi defined hote hain.

Recovery method mein kya hota hai agar field conservative nahi hai — woh kahan toot ti hai?

Step

f_{y} = Q

impossible ho jaata hai:

P

integrate karne par tumhe ek aisa

f_{y}

milta hai jo abhi bhi

x

(ya koi aur variable jو nahi hona chahiye) contain karta hai, isliye

g^{'} (y)

ek honest function of

y

alone nahi ban sakta, jo signal karta hai ki koi potential exist nahi karta.

Kya ek field jo sirf ek akele curve par defined ho (2D region par nahi) meaningfully "conservative" hai?

Concept ko partial derivatives lene aur paths compare karne ke liye ek open domain chahiye; ek akele curve par koi alternative paths nahi hote, isliye cross-partial test defined bhi nahi hota.

Recall Poore page ka ek-line litmus

Curl zero aur koi holes nahi ⟹ conservative ⟹ path-independent ⟹ loop integrals vanish ⟹ ek potential $f$ exist karta hai, unique up to $+ C$ .

Question bank — Conservative vector fields — potential functions

True or false — justify karo

Error dhundho

Why questions

Edge cases