Question

Find the linear) equation of the tangent plane to the surface z = 2y2 – 4x2 + x at the point (4,-1, -58). Your answer should be in the form of an equation, e.g., something like ax + by + cz = dor a(x – 20) + b(y – yo) + c(z – zo) = 0 would work. Equation:

Answer 1

z + 31 x + 4y = 62.

Explanation:

Equation of the 3-d surface S: z = 2y² - 4 x² + x.

Point A = (4,-1,-58) = (x₀, y₀, z₀)

S: z = f(x,y).

fₓ = -8 x + 1, => fₓ (x₀, y₀) = -8×4+1 = -31.

fᵧ = 4 y, => fᵧ (x₀, y₀) = -4.

So the normal vector at A is given by (fₓ, fᵧ, -1) or (-31, -4, -1).

Equation of the plane tangential to the surface is given by:

z - z₀ = fₓ (x₀, y₀) × (x - x₀) + fᵧ (x₀, y₀) × (y - y₀) .

=> z - (-58) = -31×(x - 4) - 4 (y - (-1))

=> z +58 +31 x -124 + 4y + 4 = 0.

=> z + 31 x + 4y = 62.