## Find an equation of the tangent plane to the given surface at the specified point. z = 3(x − 1)2 + 2(y + 3)2 + 7, (5, 1, 87)

Question

Find an equation of the tangent plane to the given surface at the specified point. z = 3(x − 1)2 + 2(y + 3)2 + 7, (5, 1, 87)

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6 months 2021-08-10T01:26:31+00:00 1 Answers 9 views 0

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

= 3×2- 6x + 2y2 + 12y + 28

Rearrange the equation of the surface into form f(x, y, z) = 0

f(x, y, z) = 3×2- 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 (3×2- 6x + 2y2 + 12y + 28 – z) i + ∂/∂y (3×2- 6x + 2y2 + 12y + 28 – z) j +∂/∂z (3×2- 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