Problem: Find the directional derivative of the function f(x, y, z) = x^2 - y^2 + 2z^2 at the point P(1, 2, 3) in the direction of the line PQ, where Q has the coordinates (5, 0, 4).

Solution:

To find the directional derivative, we need to compute the dot product between the gradient vector of the function and the unit vector in the direction of PQ. Let's break down the solution into steps.

Step 1: Find the gradient vector:
The gradient vector of a function f(x, y, z) is given by:

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

Here, f(x, y, z) = x^2 - y^2 + 2z^2. Taking partial derivatives with respect to x, y, and z, we get:

∂f/∂x = 2x
∂f/∂y = -2y
∂f/∂z = 4z

So the gradient vector ∇f is:

∇f = (2x, -2y, 4z)

Step 2: Find the unit vector in the direction of PQ:
To find the unit vector in the direction of PQ, we need to calculate the direction vector PQ and then normalize it.

Direction vector PQ = Q - P = (5, 0, 4) - (1, 2, 3) = (4, -2, 1)

To normalize this vector, we divide it by its magnitude:

||PQ|| = √(4^2 + (-2)^2 + 1^2) = √21

So, the unit vector in the direction of PQ is:

u = (4/√21, -2/√21, 1/√21)

Step 3: Compute the directional derivative:
The directional derivative Df(P, u) is given by the dot product of the gradient vector ∇f and the unit vector u:

Df(P, u) = ∇f · u

Substituting the values, we get:

Df(P, u) = (2x, -2y, 4z) · (4/√21, -2/√21, 1/√21)
= (2x)(4/√21) + (-2y)(-2/√21) + (4z)(1/√21)
= (8x/√21) + (4y/√21) + (4z/√21)

Now, substituting the coordinates of point P(1, 2, 3) into the equation, we have:

Df(P, u) = (8(1)/√21) + (4(2)/√21) + (4(3)/√21)
= (8/√21) + (8/√21) + (12/√21)
= (24/√21)

Therefore, the directional derivative of the function f(x, y