Determine the directional derivative of phi =xy^z in the direction i^-2j^+2k^ at the point (3, 1, -1)

To determine the directional derivative of the function φ = xy^z in the direction i^-2j^+2k^ at the point (3, 1, -1), you'll need to follow these steps:

Step 1: Find the gradient of φ.
The gradient of φ, denoted as ∇φ, is a vector that consists of the partial derivatives of φ with respect to each variable. In this case, we have φ = xy^z, so the gradient is given by:
∇φ = (∂φ/∂x)i^ + (∂φ/∂y)j^ + (∂φ/∂z)k^

Step 2: Evaluate the gradient at the given point.
To find the gradient at the point (3, 1, -1), substitute the values of x, y, and z into the partial derivatives:
∂φ/∂x = y^z, evaluated at (3, 1, -1)
∂φ/∂y = x(y^(z-1)), evaluated at (3, 1, -1)
∂φ/∂z = xy^z ln(y), evaluated at (3, 1, -1)

Step 3: Compute the directional derivative.
The directional derivative, denoted as Dφ, is found by taking the dot product of the gradient and the direction vector. In this case, the direction vector is i^-2j^+2k^, which can be rewritten as (-2i^ + 2k^). Thus,
Dφ = ∇φ · (-2i^ + 2k^)

Step 4: Substitute the computed partial derivatives into the directional derivative expression.
Multiplying the gradient components by the corresponding direction vector components, you can calculate the directional derivative of φ.

Dφ = (∂φ/∂x)(-2) + (∂φ/∂y)(0) + (∂φ/∂z)(2)

Note that the y-component is zero because there is no contribution from the j^ direction in the given direction vector.

Finally, substitute the evaluated partial derivatives from Step 2 into the equation above. Make sure to evaluate them at the point (3, 1, -1).