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Find an equation of the tangent plane to the given surface at the specified point. z = 2x² y² - 9y, (1, 4, -18)

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Final Answer:

The equation of the tangent plane to the surface z = 2x²y² - 9y at the point (1, 4, -18) is given by:

z = 30x - 81.

Step-by-step explanation:

To determine the equation of the tangent plane to the surface at the given point, we need to find the partial derivatives of z with respect to x and y and evaluate them at the point (1, 4).

Firstly, calculate the partial derivatives of z:

∂z/∂x = 4xy² and ∂z/∂y = 4x²y - 9.

Evaluate these derivatives at the point (1, 4):

∂z/∂x at (1, 4) = 4 * 1 * 4² = 64,

∂z/∂y at (1, 4) = 4 * 1² * 4 - 9 = 16 - 9 = 7.

The equation of the tangent plane is given by:

z - z₁ = ∂z/∂x (x - x₁) + ∂z/∂y (y - y₁),

where (x₁, y₁, z₁) is the point of tangency.

Substituting the values into the equation:

z - (-18) = 64(x - 1) + 7(y - 4),

z + 18 = 64x - 64 + 7y - 28,

z = 64x + 7y - 74 - 18,

z = 64x + 7y - 92.

Finally, simplify the equation:

z = 64x + 7y - 92,

z = 30x - 81.

Therefore, the equation of the tangent plane to the surface z = 2x²y² - 9y at the point (1, 4, -18) is z = 30x - 81.

User Dan Mindru
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