Q: Kya F=(y,−x) conservative hai? Pehle predict karo, phir check karo.
Forecast: "Rotational-looking lag raha hai, shayad nahi."
Verify:Py=1, Qx=−1. Equal nahi ⟹ conservative nahi. Unit circle ke around loop integral =−2π=0. Forecast confirm ho gaya.
Ek scalar potential f exist karta hai jaise ki F=∇f.
Fundamental Theorem of Line Integrals ka statement?
∫C∇f⋅dr=f(B)−f(A) — sirf endpoints pe depend karta hai.
Conservative field ka loop integral?
∮CF⋅dr=0.
Conservativeness ke liye 2D test?
∂P/∂y=∂Q/∂x (simply connected domain pe).
Cross-partial test kaam kyun karta hai?
Kyunki Py=fxy, Qx=fyx, aur Clairaut's theorem deta hai fxy=fyx.
"Simply connected" kyun required hai?
Holes curl-free fields ko nonzero loop integrals ke saath allow karte hain (jaise (−y,x)/(x2+y2)).
F=(P,Q) se f recover karne ka method?
P ko x mein integrate karo f=∫Pdx+g(y) pane ke liye, phir fy=Q set karo g solve karne ke liye.
3D mein conservative hone ki condition?
curlF=0, yaani Py=Qx,Pz=Rx,Qz=Ry.
Kya (y,−x) conservative hai?
Nahi; Py=1=−1=Qx.
Recall Feynman: ek 12-saal ke bachche ko samjhao
Ek pahadi park imagine karo. Har jagah ek arrow hai jo bata raha hai ki ball kidhar roll karegi — hamesha neeche ki taraf. Woh arrow map ek gradient field hai, aur pahadi ka height mappotential function hai. Yahan cool part hai: agar tum bench se swings tak jaate ho, toh tumhe jo total "downhill help" milti hai woh sirf bench aur swings ke height difference pe depend karti hai — is baat pe nahi ki tumne lamba winding raasta liya ya chhota wala. Aur agar tum loop mein chalo aur usi bench pe wapas aao, toh tum same height pe ho, isliye net help zero hai. Aisi tarah kaam karne wale fields "conservative" hain. Aisi fields jaise ek spinning whirlpool, jahan tum chakkar lagaate raho aur energy lete raho, woh NOT conservative hain.