Final answer:
The directional derivative of the function f(x, y) = x² - y² at the point (1,1) in the direction of the vector A = 3i + 2j is 2/√13.
Step-by-step explanation:
The directional derivative of the function f(x, y) = x² - y² at the point P₀ = (1,1) in the direction of the vector A = 3i + 2j. Since the function is two-dimensional and the vector A provided has a third component (-6k), we ignore this component as it is irrelevant to the calculation of the directional derivative in the xy-plane. To find the directional derivative, we first need to compute the gradient of f, ∇f, at the point (1,1), and then find the unit vector in the direction of A. The gradient is given by ∇f = (2x, -2y), so ∇f at (1,1) is (2, -2). The unit vector in the direction of A is u_A = A/|A| where |A| is the magnitude of A, computed as |A| = √(3² + 2²) = √13. The unit vector u_A = (3/√13, 2/√13). The directional derivative D_u f at P₀ is the dot product of ∇f and u_A, which is (2, -2) · (3/√13, 2/√13) = (6/√13 - 4/√13), which simplifies to 2/√13. Thus, the directional derivative of f at (1,1) in the direction A is 2/√13.