Question

1. For each of following vector functions determine whether the equation Vø=F possesses a solution, and determine that solution if it exists. (a) F=2xyz'i- (x²z' +2y)}+3x²yz°k (b) F=2xyî + (x² +2yz)j+(y² +1)&

Answer 1

Solution * According to Coltece from the data * let us assume that the equations are * To determine the vector function should be a equations of To=f and its possesers of the solution are @ F = any zsi - (m?z? +24)j + 3n+ y zk. O TO=F Comite dy je k ] 6 = F – delenit olej + Do k = aryż?– (*+23+2y) ; + 3x*yzik then e os - 2xyz? - =-(022424) –0, = 32ºyz" — from @ we get Now we have to solve from ③ we get (heteze 6) == ho you a {@way... ? (he tę? *) = f up to ² qt € → ficy,z) = -(292423+ yu+hcz) it is to should be functions are y&z only and its 70=F. let us F= 2nyit(**+ 2yz)j +9+1)— © o$=F → Cosmitón it k] $ - F >> 2xy it (atay zji+6+1) Now the gay -, ele = a +292, +®, y=y71 →

from 6 We get Solve. 0= my +f169,2) —® Now we have to solve them from @ oping = a + thing → out being = x +2yz => sofligin ay z then fily,z) = yʻzth(2) Now we have to substacle the above equations all filyr) = y 2+h(z) and p=ry + y 2 thee) delea yt olen → you that y71 = tolong 1 >> dh=dz then integrating above esuation h(z)=z+C and its substuted Ø (my, z) = x'y +y+z+z+c it is Ċ has arbitary Constant # Henica Prowed * +z +c