Question

5. (2 points) Find the equation of the tangent plane to the given surface ation of the tangent plane to the given surface at point (2. -1,0): sin(xyz) = x + 2y + 3z

Answer 1

Given that sin (792) = x+2y+32 - point (2,-1,0) we know that.. The tangent plane ecuation is given by Fu (u-Ho] + Fy (y-yo) + fz ez-Zo)=0 from ecuation, written as sin(xyz) -x-2y-3zzo to a y and z partial derivative with respect side (sin(xyz) - -24~32) - sin(xyz)_21-2242 37) - on Ted Sinx Cuen 2) Li-o-odou Cosca "- Cosa 97). dr = Coscaya).yz. dr I. ..d Constant . dr = uzcos(ryz) (1) - nai Fu = yz cos(xyz)-1 (:42 - Constant) oy fy = (Sin (692)) - - -887 = Concay zu den 2–0-2011 -0

(x,z-Constant) = cos(xyz). Az - 2 1 = XZ COB CHYZ) (1) -2 fy = xz Cos(xyz)-2 Similarly fa= d sin(xyz) _ 06. 224)_ d(37) oz dz dz dz = cos (ny z), aby z)=0-0-3+") Gry-constant) = cos(xyz) ay 3 2 2 3 3 fz = xy Cos (xyz)-3 At point (2,-1,0) fu= uz Coscayz) - fx(2,-1,0) = 61) Col Cos (21-100) - | -0-1 for (2,-1,0) = -1 for = uz Cars (492)-2 fy (2,-1,0) = 2.0 cos (2.-1.0)-2 = 0-2 fy (2,-1,0)=-2 Similarly fz = xy cos (ayz) -3 f₂ (2,-1,0)=2 (-1). Cus (2.-1.05-3 =-2. Cos Co") 3 5-2 (1) – 3 fz(2,-1,0) = -5 : Cogozi1

from esuation (2,01,0) = (20,9, 20) (-1) (0-2)+(-2) (y=(-1)) + (-5)(z-o)=0 -M+2-214 Hag (Z) =0 -1+2-24-t-52=0 -n-24-52=0 (X+24+55750 = x+2y+52=O