Find an equation of the tangent plane to the given surface at the specified point. z = 4(x - 1)² + 3(y + 3)² + 8; (2, -2, 15)

Solution:

The surface given is

z = 4(x - 1)² + 3(y + 3)² + 8

We can write it as

F(x, y, z) = z - 4(x - 1)² + 3(y + 3)² + 8 = 0

We should find the equation of tangent which passes through (2, -2, 15)

The equation of a tangent is

F_x(x₀, y₀, z₀).(x - x₀) + F_y(x₀, y₀, z₀), (y - y₀) + F_z(x₀, y₀, z₀).(z - z₀) = 0

Here

F_x = -8(x - 1)

F_z(2, -2, 15) = -8(2 - 1) = -8 (1) = -8

F_y = 6(y + 3)

F_y(2, -2, 15) = 6 (-2 + 3) = 6 (1) = 6

F_z = 1

F_z(2, -2, 15) = 1

Now substituting all these values

-8(x - 2) + 6 (y + 2) + 1 (z - 15) = 0

Using the distributive property

-8x + 16 + 6y + 12 + z - 15 = 0

By further calculation

-8x + 6y + z + 13 = 0

Multiplying -1

8x - 6y - z - 13 = 0

Therefore, the equation of the tangent plane is 8x - 6y - z - 13 = 0.

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Find an equation of the tangent plane to the given surface at the specified point. z = 4(x - 1)² + 3(y + 3)² + 8; (2, -2, 15)

Summary:

An equation of the tangent plane to the given surface at the specified point z= 4(x - 1)² + 3(y + 3)² + 8; (2, -2, 15) is 8x - 6y - z - 13 = 0.