Question

Find equations for the tangent plane and the normal line at point P P. (Xo. No: 2) (5,1,0) on the surface - Cos (2x) + 6x2y+3e* + 2y2= 154. The equation for the tangent plane is a

Answer 1

V We equation of surface on x2 have to find the the tangent plane the $14,9,2) = COS (STN) + 6x'y+ 3 e +242-154 at the point Po Cho, Yo, 2) = 15,1,0 equation of the tangent at any point (ho, 9o,2o) f14, 5,2) (4-40) 21 (1) + (9-%) 1 / 1 (Po) + (2-2) 24 (2) 50 f(0, 1, 2) = cosuTht 68²y + 3 e42 +242 - 154 zo. af = -stsinine +1264 +3ę ek? (2) - (0) on als 3,0) = -sesinst + 12:5:1+3 é o+ 60 +3.0 = 60 plane to the surface an 5.0 of a 6x²4.22 Selis,i,o, - 65242.0= 6x25=150 27 - 11/ 15,00 3 e12. 2+2y. at I az (5,1,0) = 3-e 5+21 2 75+2=1). equation of of the tangent blane at Po (5,1,ol is. af Po (345). (X- Houd en +(yys) et 90-5) 60+ (4-1)/50 + (2-0) 19 co 2 boutsoy+ 172 - 450 50

Q. we have to find line the equation of at Po(5,1,o) to the normal the given surface. equation OF the normal line at But P (Ho, Yo, 2o) is. H-Yo yoso z- zo at hulle 2 1/2 21 help port; we knoce from the previous at allpo 60 2 1 - Ipor iso It willpo=13 equation the required normal liners. x-s. 's 60 y-1 150 2-0 12 or 1-5 60 Is = 20 = tlet Any x=600+ 5.1 y = 150 ft 2:17t Any doubt then comment. 9 will help you ASAP. СТНАЕСү00.