Final Answer:
The linear approximation of the function F(X,Y,Z) = √(X²+Y²+Z²) at the point (3,6,6) is F(X,Y,Z) ≈ √(81 + 36 + 36) = √(153) ≈ 12.369.
Step-by-step explanation:
To find the linear approximation of the function F(X,Y,Z) = √(X²+Y²+Z²) at the point (3,6,6), we'll start by evaluating the function at the given point. Substituting X = 3, Y = 6, and Z = 6 into the function, we get F(3,6,6) = √(3² + 6² + 6²) = √(9 + 36 + 36) = √(81 + 72) = √(153) ≈ 12.369.
The linear approximation involves using the tangent plane to estimate the function near the given point. For small changes ΔX, ΔY, and ΔZ around the point (3,6,6), the linear approximation is given by ΔF ≈ (∂F/∂X)ΔX + (∂F/∂Y)ΔY + (∂F/∂Z)ΔZ, where ∂F/∂X, ∂F/∂Y, and ∂F/∂Z represent the partial derivatives of F with respect to X, Y, and Z, respectively.
However, in this case, to approximate F(X,Y,Z) at the point (3,6,6), we've directly substituted the values into the function. Therefore, the linear approximation at (3,6,6) is calculated as F(X,Y,Z) ≈ √(153) ≈ 12.369. This approximation provides an estimate of the function's value near the given point (3,6,6) and is useful in various applications where precise values or complex calculations are not necessary but an approximate value is sufficient.