Find the directional derivative of f(x, y, z) 5y x + z = at P(4,1,-3)
in the direction from P to Q(3,2,-2).
NOTE: Enter the exact answer.
We have the function f(x, y, z) = 5yx + z. We want to find the directional derivative at the point P(4, 1, -3) in the direction of the vector PQ. First, let's find the vector PQ.
PQ = Q - P = (3 - 4, 2 - 1, -2 - (-3)) = (-1, 1, 1)
Now, we need to normalize this vector to get the unit direction vector. The magnitude of PQ is:
|PQ| = √((-1)^2 + 1^2 + 1^2) = √3
So the unit direction vector is:
u = PQ/|PQ| = (-1/√3, 1/√3, 1/√3)
Next, we need to find the gradient of the function f(x, y, z) at point P. To do this, we'll find the partial derivatives of f with respect to x, y, and z and evaluate them at P(4, 1, -3):
∂f/∂x = 5y = 5(1) = 5
∂f/∂y = 5x = 5(4) = 20
∂f/∂z = 1
So the gradient of f at P is ∇f(P) = (5, 20, 1).
Now, to find the directional derivative of f at point P in the direction of u, we need to take the dot product of the gradient vector and the unit direction vector:
D_uf(P) = ∇f(P) · u = (5, 20, 1) · (-1/√3, 1/√3, 1/√3) = 5(-1/√3) + 20(1/√3) + 1(1/√3)
D_uf(P) = (-5 + 20 + 1)/√3 = 16/√3
Therefore, the directional derivative of f(x, y, z) = 5yx + z at P(4, 1, -3) in the direction from P to Q(3, 2, -2) is 16/√3.