Find the directional derivative of f(x,y,z)=z3−x2y
at the point (-2, 5, 1) in the direction of the vector v=⟨−5,−5,−3⟩
.
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.
To find the directional derivative of the function f(x, y, z) = z^3 − x^2y at the point (-2, 5, 1) in the direction of the vector v = ⟨−5, −5, −3⟩, follow these steps:
Step 1: Calculate the gradient of the function.
The gradient of the function f(x, y, z) is given by ∇f = ⟨∂f/∂x, ∂f/∂y, ∂f/∂z⟩.
∂f/∂x = -2xy, ∂f/∂y = -x^2, and ∂f/∂z = 3z^2.
Therefore, the gradient of f(x, y, z) is ∇f = ⟨-2xy, -x^2, 3z^2⟩.
Step 2: Evaluate the gradient at the given point.
Substitute the values x = -2, y = 5, and z = 1 into the gradient expression:
∇f(-2, 5, 1) = ⟨-2(-2)(5), -(-2)^2, 3(1)^2⟩
= ⟨20, -4, 3⟩.
Step 3: Normalize the vector v.
To find the directional derivative, we need to normalize the vector v by dividing it by its magnitude. The magnitude of the vector v = ⟨-5, -5, -3⟩ is given by ||v|| = sqrt((-5)^2 + (-5)^2 + (-3)^2) = sqrt(59).
Therefore, the normalized vector v̂ is v̂ = v/||v|| = ⟨-5/sqrt(59), -5/sqrt(59), -3/sqrt(59)⟩.
Step 4: Calculate the dot product.
To find the directional derivative, we need to take the dot product of the normalized vector v̂ and the gradient ∇f at the given point.
v̂ ⋅ ∇f(-2, 5, 1) = (-5/sqrt(59))(20) + (-5/sqrt(59))(-4) + (-3/sqrt(59))(3)
= (-100/sqrt(59)) + (20/sqrt(59)) - (9/sqrt(59))
= -89/sqrt(59).
Therefore, 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⟩ is -89/sqrt(59).