Answer: equation of the tangent plane is z = 1
Step-by-step explanation:
Given equation
z = e^(-x²-y²) at point (0,0,1)
now let z = f(x,y)
Δf(x,y) = [ fx, fy ]
= (-2xe^(-x²-y²)), (-2ye^(-x²-y²))
now
Δf (0,0) = [ 0, 0 ] = [ a, b ]
equation of the tangent plane therefore will be
z - z₀ = a(x-x₀) + b(y-y₀)
z - 1 = 0(x-0) + 0(y-0)
z - 1 = 0 + 0
z = 1
Therefore equation of the tangent plane is z = 1
