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Question: Find the directional derivative of f(x,y,z) = xy+yz+zx at P(1,-1,3) in the direction from P to Q(2,4,5).
Using the formula for the directional derivative of f at P in the direction of a unit vector u, the directional derivative is: F'(P, u) = u_x * ∂f/∂x + u_y * ∂f/∂y + u_z * ∂f/∂z Where u_x, u_y and u_z are the components of the unit vector u. P = (1,-1,3) and Q = (2,4,5) and the unit vector u = (Q - P)/|Q - P| = (1,5,2)/√36 So, the directional derivative = u_x * ∂f/∂x + u_y * ∂f/∂y + u_z * ∂f/∂z = (1/√36)*∂f/∂x + (5/√36)*∂f/∂y + (2/√36)*∂f/∂z = (1/√36)* y + (5/√36)* z + (2/√36)* x = (1/√36)*(-1) + (5/√36)*(5) + (2/√36)*(1) = (-1/√36) + (25/√36) + (2/√36) = (22/√36) ≈ 3.02.
March 3, 2023, 9:07 p.m.
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