Question

Use the Chain Rule to find the indicated partial derivatives. и = = х2+ yz, x = pr cos e, y = pr sin , 2 = p+r ди др au ar au де when p = 2, r= 1, = 0 ди др au ar ди де

Answer 1

Solution : Given that U = x3+yz , X = browse , y = brsîne , Z = +r = 4(37,2) =x+yz , x(br,e) = þrcose, y (er; e) = þrsing and z (B730=6+r, X u = x3 +32 → au 2X 3x0² +0 of xo 个 au 2x – 3X2 (4. Theorem: Let u(67,z) be a frenetion X, Y, Z and xCBre), y(bre), 7(b5e) Then by chain Ruleg weget au ag au ap ax ap ay ap az and u = x3+yz » au = 0 + Z au au axt he (1) → au Z Z e 2u24+ -(5) ne ele au ax ax ar + all az ƏZ ar (cli ay u=x² x3 + YZ and (2) → au az 0+y ne 24 24+24 24+24 az 2x 20 20 az ao oe → ne ze =y L(3) 6) Since x = prcose Then ㅋ ax rocoso хе ele prose JX -- br sine de (8) 9

Sinee, y = prsine Then ay=psine 24 = $ sino/; ia he esmed ab 20 2p (10) (12) Sinee Z=b+p Then 2Z 1 a Z 1 = a Z 30 ao (13) (14) (5) Therefore from (16), we get au au at + az ap ne From (1), we get au ax au az ax ap 38 to From (4,5,6), we get au 3226 au =Z] de = 3x2 ro caso + Z rosino + ty ab ne = haze хе Re From (7),(0),13), we get 24 = 3(prcose) ne ErcosA + (p+r).rsine + prsine Since x = prcose, y = prsing and Z = p+r 2x = rose, ay el= singh de az ap=1 au - Bu 3 ?-?oset Frsin 6+2Sim Then + prsine (3x2)rcose) +(z)rSine) +($) (1) Now cohen þ=2, r=1, 0 = 0 de he au TT = 3 x?rcos +z rsino ty de (16) 3(2)?(1)?(05(0) 3+ (2) (1) sin() +(2)? Sin (0) +2).(0) Sino) → au ap = 3.4.1 + 2.1.0 +4.0 + 2.1.0 since COS(O) = 1 and Sin (O) = 0

de rette = 4.3 +0+0+0 → ne = 12 ap (Answer) From (17), weget 9 24. = 3x2bcose + z bsino + y From (2 we get 24 22 agar au zu ta n elle au az az an ele au ar = 3 3(prcose) pcose + (p+r) þsine + prsing Since x=br cose, y = bosine, z =b+r From (4,5), (6), we get lau = 3x25|24 = zl,342 - y ax when yo = 1; Ⓡ = 2, 0 = 0 , From (8), (11),(4), we get C ne ele 31 2:1. cosa = 10503(2)0056) 3(2. xe de posel, 2 * = psine and ә +(2+1).2-Sing(0) + 2.1 Sin (0) ze T = ele dhe me 3 (2.1)?2. -2.1 + (3).2.0 +2.1.0 Since cos (O) = 1 and Sinco) = 0 au ap Therefore y = (@x2)pose) +(z) Chine) +8)(1) 3x2 pcose + z p sine +y -(17) du → = ele → 2u = 3:23 +0+ 0 ele = 3.8 จริง he ele he ele 24 (Answer)

From (18), we get Ze ne au 20 au an ax 20 + - + Өe De ze = ne ax au 20 From (3), we get au au ay --3px?nsin 0 + z prcoso ay From (1), (5), (6), we get au 3x3 = Z1; 24- y =-3Cþrcose)ksine + (p+r) þrcos From (9), (12), (5), weget 3X -- brsinel24 - brose 22 - 0 = -3 3p3 cos?e sin o + b2r cose +brzcoset 24 = 3x2(-þrsine) + z (brcose) + y(0) a = -353p3 cos?8 Sine + b2po2cose+br_cose 24 - 3px? Sine + Z prcos 8 +0 brcoso de 3 bel ne Өe Then 介 E ao when p = 1, b=2, 0 = 0 → ne Өe – 3ppxz sine + Zprcose (18) he Өe. --3(2)(1) 3605? (0) Sin (0) + 2212 cos(0) +2:12 Cos (0) he Өe = -3.8.10 + 4.1.1 +2.1.1 since cos(O)=1 and sin(o)=0 he = 0 + 4 +2 au 20 = 6 lo Answer