Tangent Plane and Nomrla Line

Question

Consider the following.

z + 6 = xe^y cos(z), (6, 0, 0)

(a)

Find an equation of the tangent plane to the given surface at the specified point.

(b)

Find an equation of the normal line to the given surface at the specified point.

(x(t), y(t), z(t)) =

Answer 1

F(x, y, z) = z + 6 - xe^ycos(z) = 0; ∇F = <- e^ycos(z), - xe^ycos(z), 1 + xe^ysin(z)> = <- 1, - 6, 0>

(a) Equation of tangent plane: <x - 6, y, z>• <- 1, - 6, 0> = 0; - 1(x - 6) - 6y = 0; x + 6y - 6 = 0

(b) Equation of normal line: (x - 6)/(- 1) = y/(- 6) = z/0 = t; x = 6 - t, y = - 6t; z = 0