204k views
3 votes
Find an equation for a tangent plane z=e-x2-y2 at the point (0,0,1).

User Darkless
by
8.0k points

1 Answer

2 votes

Answer: equation of the tangent plane is z = 1

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

User Jamik
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories