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nignag [31]
3 years ago
11

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

Mathematics
1 answer:
Nadya [2.5K]3 years ago
7 0

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

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3 years ago
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Alinara [238K]

Answer:

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3 years ago
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<em>(sorry for the late answer)</em>

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