The directional derivative of a function measures the rate of change of the function in a specific direction. It provides information about how the function changes as we move along a certain vector.


The function f(x, y, z) = xyz^2 represents a three-dimensional surface. We need to find the directional derivative of this function at the point (1, 0, 3) in the direction of the vector i - j + k.


To calculate the directional derivative, we need to use the gradient of the function. The gradient is a vector that points in the direction of the steepest increase of the function.

1. Calculate the gradient of f(x, y, z):
The gradient of f(x, y, z) = (∂f/∂x, ∂f/∂y, ∂f/∂z) is given by:
∂f/∂x = yz^2
∂f/∂y = xz^2
∂f/∂z = 2xyz

2. Evaluate the gradient at the given point:
Substituting (1, 0, 3) into the gradient, we get:
∂f/∂x = 0
∂f/∂y = 9
∂f/∂z = 0

3. Calculate the dot product of the gradient and the direction vector:
The dot product of two vectors is given by the formula:
a · b = |a| |b| cos(θ)
where a and b are vectors, |a| and |b| are their magnitudes, and θ is the angle between them.

In this case, the dot product is:
(0, 9, 0) · (1, -1, 1) = 0(1) + 9(-1) + 0(1) = -9

4. Find the magnitude of the direction vector:
The magnitude of the direction vector is given by the formula:
|v| = sqrt(a^2 + b^2 + c^2)
where v = (a, b, c) is the direction vector.

In this case, the magnitude of the direction vector is:
|v| = sqrt(1^2 + (-1)^2 + 1^2) = sqrt(3)

5. Calculate the directional derivative:
The directional derivative is given by the formula:
D_v f(x, y, z) = |v| cos(θ)
where |v| is the magnitude of the direction vector and θ is the angle between the gradient vector and the direction vector.

In this case, the directional derivative is:
D_v f(x, y, z) = (-9) / sqrt(3) = -3sqrt(3) ≈ -5.196

Therefore, the directional derivative of the function f = xyz^2 at (1, 0, 3) in the direction of the vector i - j + k is approximately -5.196.