Directional Derivative of f(x, y, z) = x^2 - y^2 + 2z^2

To find the directional derivative of the function f(x, y, z) = x^2 - y^2 + 2z^2 at the point (1, 2, 3) in the direction 4i - 2j + k, we need to follow these steps:

Step 1: Calculate the Gradient Vector
The gradient vector of a function is a vector that points in the direction of the greatest rate of increase of the function at any given point. It is calculated by taking the partial derivatives of the function with respect to each variable.

Given the function f(x, y, z) = x^2 - y^2 + 2z^2, we can find the gradient vector as follows:

∇f(x, y, z) = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

∂f/∂x = 2x
∂f/∂y = -2y
∂f/∂z = 4z

Therefore, the gradient vector is:
∇f(x, y, z) = 2xi - 2yj + 4zk

Step 2: Normalize the Direction Vector
To find the directional derivative, we need to normalize the direction vector (4i - 2j + k) by dividing it by its magnitude. The magnitude of a vector is calculated using the formula:
|v| = sqrt(vx^2 + vy^2 + vz^2)

In this case, the magnitude of the direction vector is:
|4i - 2j + k| = sqrt(4^2 + (-2)^2 + 1^2) = sqrt(21)

Therefore, the normalized direction vector is:
v = (4i - 2j + k) / sqrt(21) = (4/sqrt(21))i - (2/sqrt(21))j + (1/sqrt(21))k

Step 3: Calculate the Dot Product
The directional derivative of f(x, y, z) in the direction of v is given by the dot product of the gradient vector and the normalized direction vector.

Dv(f) = ∇f(x, y, z) · v

Dv(f) = (2xi - 2yj + 4zk) · ((4/sqrt(21))i - (2/sqrt(21))j + (1/sqrt(21))k)

Dv(f) = (2 * (4/sqrt(21))) + (-2 * (2/sqrt(21))) + (4 * (1/sqrt(21)))

Dv(f) = (8/sqrt(21)) - (4/sqrt(21)) + (4/sqrt(21))

Dv(f) = 8/sqrt(21)

Therefore, the directional derivative of f(x, y, z) = x^2 - y^2 + 2z^2 at the point (1, 2, 3) in the direction 4i - 2j + k is 8/sqrt(