Directional derivative of xy²+yz³

The directional derivative of φ at a point in the direction of a vector d is ∇φ · d̂, where d̂ is the unit vector along d. It is the rate of change of φ per unit distance in that direction.

Gradient

φ = xy² + yz³

∂φ/∂x = y²
∂φ/∂y = 2xy + z³
∂φ/∂z = 3yz²

At (2, −1, 1):

∂φ/∂x = 1
∂φ/∂y = 2(2)(−1) + 1 = −3
∂φ/∂z = 3(−1)(1) = −3
∇φ = i − 3j − 3k

Unit vector

d = i + 2j + 2k, |d| = √(1 + 4 + 4) = 3, so d̂ = (i + 2j + 2k)/3.

Directional derivative

∇φ · d̂ = (1·1 + (−3)(2) + (−3)(2))/3 = (1 − 6 − 6)/3 = −11/3

Answer: the directional derivative is −11/3 ≈ −3.67 (φ decreases in that direction).