Find the directional derivative of f(x,y,z)=z3−x2y

Question

at the point (-2, 5, 1) in the direction of the vector v=⟨−5,−5,−3⟩
.

Answer 1

To find the directional derivative of f(x,y,z)= z^3−x^2y at the point (-2, 5, 1) in the direction of the vector v=⟨−5,−5,−3⟩, we need to calculate the gradient of f(x,y,z) at the given point and then dot it with the direction vector (normalized).

First, find the gradient of f(x,y,z) by taking the partial derivatives of each variable:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)
∂f/∂x = -2xy
∂f/∂y = -x^2
∂f/∂z = 3z^2

At the point (-2, 5, 1):
∂f/∂x = -2(5)(-2) = 20
∂f/∂y = -(-2)^2 = -4
∂f/∂z = 3(1)^2 = 3

Therefore, the gradient of f(x,y,z) at (-2, 5, 1) is ∇f = (20, -4, 3).

Next, normalize the direction vector v:
||v|| = √((-5)^2 + (-5)^2 + (-3)^2) = √(25 + 25 + 9) = √59
v_normalized = (-5/√59, -5/√59, -3/√59)

Finally, calculate the directional derivative by taking the dot product of the gradient and the normalized direction vector:
d_f = ∇f · v_normalized
= (20, -4, 3) · (-5/√59, -5/√59, -3/√59)
= (20)(-5/√59) + (-4)(-5/√59) + (3)(-3/√59)
= -100/√59 + 20/√59 - 9/√59
= (-100 + 20 - 9) / √59
= -89/√59

Therefore, the directional derivative of f(x,y,z) at the point (-2, 5, 1) in the direction of the vector v=⟨−5,−5,−3⟩ is -89/√59.