Math Problem Statement

Find the directional derivative of 0 = 682+92722372 at the point (1,-1,1) in the direction of @ =î+§+k.

Solution

To find the directional derivative of the function f(x,y,z)=682+92722372f(x, y, z) = 682 + 92722372 at the point (1,1,1)(1, -1, 1) in the direction of the vector v=i+j+k\mathbf{v} = \mathbf{i} + \mathbf{j} + \mathbf{k}, let us solve this step by step:


1. General Formula for the Directional Derivative

The directional derivative of f(x,y,z)f(x, y, z) in the direction of a unit vector u\mathbf{u} is given by: Duf(x,y,z)=fuD_{\mathbf{u}}f(x, y, z) = \nabla f \cdot \mathbf{u} Where:

  • f=(fx,fy,fz)\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) is the gradient vector of ff.
  • u=vv\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} is the unit vector in the direction of v\mathbf{v}.

2. Step-by-Step Solution

Step 1: Calculate the gradient f\nabla f

The function f(x,y,z)=682+92722372f(x, y, z) = 682 + 92722372 is a constant function. The gradient of a constant function is: f=(fx,fy,fz)=(0,0,0)\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (0, 0, 0)

Step 2: Normalize the direction vector v=i+j+k\mathbf{v} = \mathbf{i} + \mathbf{j} + \mathbf{k}

The magnitude of v\mathbf{v} is: v=12+12+12=3\|\mathbf{v}\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} The unit vector is: u=vv=(13,13,13)\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)

Step 3: Compute the directional derivative

The directional derivative is given by: Duf(x,y,z)=fuD_{\mathbf{u}}f(x, y, z) = \nabla f \cdot \mathbf{u} Substituting f=(0,0,0)\nabla f = (0, 0, 0) and u=(13,13,13)\mathbf{u} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right), we get: Duf(x,y,z)=(0,0,0)(13,13,13)=0D_{\mathbf{u}}f(x, y, z) = (0, 0, 0) \cdot \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) = 0


Final Answer:

The directional derivative of the given function at the point (1,1,1)(1, -1, 1) in the direction of i+j+k\mathbf{i} + \mathbf{j} + \mathbf{k} is: 0\boxed{0}

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Related Questions:

  1. What is the meaning of a zero directional derivative in a given direction?
  2. How does the gradient relate to the steepest direction of change?
  3. What happens to the directional derivative if the vector v\mathbf{v} is not normalized?
  4. How would the directional derivative change if f(x,y,z)f(x, y, z) were not constant?
  5. Can the directional derivative ever exceed the magnitude of the gradient?

Tip:

The gradient vector f\nabla f always points in the direction of the steepest ascent, and its magnitude represents the rate of change in that direction!

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Math Problem Analysis

Mathematical Concepts

Directional Derivative
Gradient Vector
Vector Normalization

Formulas

Directional derivative formula: D_u f(x, y, z) = ∇f · u
Gradient: ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)
Unit vector: u = v / ||v||

Theorems

Properties of Constant Functions

Suitable Grade Level

Grades 11-12