Question

normal to the surface xy^2+xy+2=0 is​

Answer 1

Answer:

(a) f = xyz + 3x

2 − 4 x0 = (1, 1, 1)

∇f = (yz + 6x, xz, xy) ∇f|x0 = (7, 1, 1) n = ± √

1

51 (7, 1, 1)

(b) f = 3yz

2 + 2x

2 − 4xy2 − 3 x0 = (0, 1, 1)

∇f = (4x − 4y

2

, 3z

2 − 4x, 6yz) ∇f|x0 = (−4, 3, 6)

Equation of tangent plane is (x − x0) · n = 0

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n

x − x0

−4(x − 0) + 3(y − 1) + 6(z − 1) = 0

− 4x + 3y + 6z = 9

(c) f1 = 2x

2 + y

2 + z

2 − 4 f2 = x

2 + y

2 − z − 1 x0 = (1, 1, 1)

∇f1 = (4x, 2y, 2z) n1 = √

1

24 (4, 2, 2) = √

1

6

(2, 1, 1)

∇f2 = (2x, 2y, −z) n2 =

1

3

(4, 2, 2)

Angle between surfaces = angle between normals = θ

cos θ = n1 · n2 = √

1

6

1

3

5

θ = cos−1

³

5

3

√

6

´