Final answer:
The linear approximation of the function at point (2, 4, 4) is used to find the approximate value at (2.03, 3.99, 3.98), which rounds to 5.9933.
Step-by-step explanation:
The given function f(x, y, z) = √(x^2 + y^2 + z^2) represents the distance from the origin to a point in three-dimensional space. To find the linear approximation, we calculate the gradient of the function at point (2, 4, 4).
First, let's find the partial derivatives at this point:
- f_x = rac{x}{√(x^2 + y^2 + z^2)}
- f_y = rac{y}{√(x^2 + y^2 + z^2)}
- f_z = rac{z}{√(x^2 + y^2 + z^2)}
At (2, 4, 4), the gradient is (1/√(36)) ⋅ (2, 4, 4) = (1/6) ⋅ (2, 4, 4) = (1/3, 2/3, 2/3).
The linear approximation formula is L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c), where (a, b, c) is the point of approximation and (x, y, z) is the point we're approximating.
Applying this to the given problem: L(2.03, 3.99, 3.98) ≈ √(36) + (1/3)(2.03 - 2) + (2/3)(3.99 - 4) + (2/3)(3.98 - 4) ≈ 6 + 0.01 + 0 - 0.02/3 ≈ 6 - 0.0067 ≈ 5.9933. Therefore, our approximation of f(2.03, 3.99, 3.98) is approximately 5.9933.