Respuesta :
Answer:
So for the point (5, 1, 87) the normal vector to surface is given by:
fx(5,1) = [6(5) – 6] , fy(5,1)= [4(1) +12]
fx(5,1) =(30 – 6) fy (5,1) = 16
fx(5,1) = 24
Step-by-step explanation:
Equation is z = 3(x -1)2 + 2(y +3)2 + 7
Z= 3(x2+1-2x) +2(y2+6y+9)+7
= 3x2- 6x + 2y2 + 12y + 28
Rearrange the equation of the surface into form f(x, y, z) = 0
f(x, y, z) = 3x2- 6x + 2y2 + 12y + 28 - z
∆f(x,y,z) = ∂f/∂x i + ∂f/∂y j+ ∂f/∂z k
Partially differentiate with respect to variable:
= ∂/∂x (3x2- 6x + 2y2 + 12y + 28 - z) i + ∂/∂y (3x2- 6x + 2y2 + 12y + 28 - z) j +∂/∂z (3x2- 6x + 2y2 + 12y + 28 – z) k
=(6x-6)i + (4y + 12 )j +(-1)k
So for the point (5, 1, 87) the normal vector to surface is given by:
fx(5,1) = [6(5) – 6] , fy(5,1)= [4(1) +12]
fx(5,1) =(30 – 6) fy (5,1) = 16
fx(5,1) = 24
