Math Problem Statement

Find the directional derivative of f(x,y,z)=z3−x2y at the point (-2, 5, 2) in the direction of the vector v=⟨−3,−1,2⟩.

Solution

To find the directional derivative of the function f(x,y,z)=z3x2yf(x, y, z) = z^3 - x^2 y at the point (2,5,2)(-2, 5, 2) in the direction of the vector v=3,1,2\mathbf{v} = \langle -3, -1, 2 \rangle, follow these steps:

Step 1: Find the gradient of the function f(x,y,z)f(x, y, z)

The gradient of f(x,y,z)f(x, y, z), denoted by f(x,y,z)\nabla f(x, y, z), is the vector of partial derivatives:

f(x,y,z)=(fx,fy,fz)\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)

  1. Partial derivative with respect to xx: fx=2xy\frac{\partial f}{\partial x} = -2xy
  2. Partial derivative with respect to yy: fy=x2\frac{\partial f}{\partial y} = -x^2
  3. Partial derivative with respect to zz: fz=3z2\frac{\partial f}{\partial z} = 3z^2

Thus, the gradient is:

f(x,y,z)=2xy,x2,3z2\nabla f(x, y, z) = \langle -2xy, -x^2, 3z^2 \rangle

Step 2: Evaluate the gradient at the point (2,5,2)(-2, 5, 2)

Substitute x=2x = -2, y=5y = 5, and z=2z = 2 into the gradient:

f(2,5,2)=2(2)(5),(2)2,3(2)2\nabla f(-2, 5, 2) = \langle -2(-2)(5), -(-2)^2, 3(2)^2 \rangle f(2,5,2)=20,4,12\nabla f(-2, 5, 2) = \langle 20, -4, 12 \rangle

Step 3: Normalize the direction vector v\mathbf{v}

The direction vector v=3,1,2\mathbf{v} = \langle -3, -1, 2 \rangle needs to be normalized:

v=(3)2+(1)2+22=9+1+4=14|\mathbf{v}| = \sqrt{(-3)^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}

The normalized vector is:

u=1143,1,2=314,114,214\mathbf{u} = \frac{1}{\sqrt{14}} \langle -3, -1, 2 \rangle = \langle \frac{-3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \rangle

Step 4: Compute the directional derivative

The directional derivative of f(x,y,z)f(x, y, z) at the point (2,5,2)(-2, 5, 2) in the direction of v\mathbf{v} is given by the dot product of the gradient and the normalized vector:

Dvf(2,5,2)=f(2,5,2)uD_{\mathbf{v}} f(-2, 5, 2) = \nabla f(-2, 5, 2) \cdot \mathbf{u} Dvf(2,5,2)=20,4,12314,114,214D_{\mathbf{v}} f(-2, 5, 2) = \langle 20, -4, 12 \rangle \cdot \langle \frac{-3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \rangle Dvf(2,5,2)=114(20(3)+(4)(1)+12(2))D_{\mathbf{v}} f(-2, 5, 2) = \frac{1}{\sqrt{14}} (20(-3) + (-4)(-1) + 12(2)) Dvf(2,5,2)=114(60+4+24)D_{\mathbf{v}} f(-2, 5, 2) = \frac{1}{\sqrt{14}} (-60 + 4 + 24) Dvf(2,5,2)=114(32)D_{\mathbf{v}} f(-2, 5, 2) = \frac{1}{\sqrt{14}} (-32) Dvf(2,5,2)=3214D_{\mathbf{v}} f(-2, 5, 2) = \frac{-32}{\sqrt{14}}

Thus, the directional derivative is:

Dvf(2,5,2)=3214D_{\mathbf{v}} f(-2, 5, 2) = \frac{-32}{\sqrt{14}}

Would you like more details on any part of this solution?

Here are 5 related questions:

  1. How do you interpret the magnitude of the directional derivative?
  2. What does it mean if the directional derivative is positive or negative?
  3. How would the directional derivative change if the direction vector were different?
  4. Can you find the maximum rate of change of the function at a given point?
  5. How do gradient vectors relate to level curves or surfaces?

Tip: The magnitude of the gradient indicates the steepest ascent direction and rate of change at a point.

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Math Problem Analysis

Mathematical Concepts

Directional Derivatives
Gradient Vectors
Multivariable Calculus
Vector Normalization
Dot Product

Formulas

Gradient of f: ∇f(x, y, z) = (∂f/∂x, ∂f/∂y, ∂f/∂z)
Directional derivative: D_𝑣 f = ∇f ⋅ u
Vector normalization: u = v / |v|
Dot product: A ⋅ B = A₁B₁ + A₂B₂ + A₃B₃

Theorems

Gradient Theorem
Properties of Dot Product

Suitable Grade Level

Grades 11-12 or Early College