Problem 246 - Calculus Volume 3

Question

Find the divergence of \(\mathbf{F}\) at the given point. $$ \mathbf{F}(x, y, z)=x y z \mathbf{i}+y \mathbf{j}+z \mathbf{k} \text { at }(1,2,3) $$

Step-by-Step Solution

Answer

The divergence at the point (1, 2, 3) is 8.

Step 1: Understand the Problem

We need to find the divergence of the vector field \( \mathbf{F}(x, y, z) = x y z \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) at the point \( (1, 2, 3) \).

Step 2: Recall the Formula for Divergence

The divergence of a vector field \( \mathbf{F}(x, y, z) = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by the formula \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).

Step 3: Identify Components of \( \mathbf{F} \)

In the vector field \( \mathbf{F}(x, y, z) = x y z \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), the components are \( P = xyz \), \( Q = y \), and \( R = z \).

Step 4: Compute Partial Derivative with respect to \( x \)

Find \( \frac{\partial P}{\partial x} \). Since \( P = xyz \), we treat \( y \) and \( z \) as constants, giving \( \frac{\partial (xyz)}{\partial x} = yz \).

Step 5: Compute Partial Derivative with respect to \( y \)

Find \( \frac{\partial Q}{\partial y} \). Since \( Q = y \), this gives us \( \frac{\partial y}{\partial y} = 1 \).

Step 6: Compute Partial Derivative with respect to \( z \)

Find \( \frac{\partial R}{\partial z} \). Since \( R = z \), we have \( \frac{\partial z}{\partial z} = 1 \).

Step 7: Sum the Partial Derivatives

Add the partial derivatives: \( abla \cdot \mathbf{F} = yz + 1 + 1 \).

Step 8: Evaluate at the Given Point \( (1, 2, 3) \)

Substitute \( y = 2 \) and \( z = 3 \) into the expression obtained in Step 7: \( abla \cdot \mathbf{F} = 2 \cdot 3 + 1 + 1 = 6 + 1 + 1 = 8 \).

Key Concepts

Vector Fields

A vector field is a mathematical construction representing a function that provides a vector for each point in space. This can be visualized as an array of arrows where each arrow has a direction and magnitude specific to that location. In physics, vector fields often represent quantities like velocity or force fields, indicating how these quantities vary over space.
In the context of the original exercise, the vector field \( \mathbf{F}(x, y, z) = x y z \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) is a function of three variables (\( x, y, z \)), where each point in the space has an associated vector.

• \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) represent the unit vectors in the direction of the x, y, and z axes, respectively.
• The components of the vector field are not constant; they depend on the position \((x, y, z)\).

Understanding vector fields is crucial for visualizing various physical phenomena and is a core part of vector calculus.

Partial Derivatives

Partial derivatives are an essential part of multivariable calculus. They measure the rate at which a function changes with respect to one of its variables while keeping the other variables constant. This concept is fundamental when dealing with fields or functions of multiple variables.
In our original exercise, each component of the vector field \( \mathbf{F} \) is differentiated partially:

• For \( P = xyz \), the partial derivative with respect to \( x \) is \( \frac{\partial (xyz)}{\partial x} = yz \), treating \( y \) and \( z \) as constants.
• For \( Q = y \), the partial derivative with respect to \( y \) is \( \frac{\partial y}{\partial y} = 1 \), an indication that \( Q \) changes linearly.
• For \( R = z \), \( \frac{\partial z}{\partial z} = 1 \), as it also changes linearly in the direction of \( z \).

Partial derivatives form the backbone of differential calculus applied to functions of several variables, allowing us to analyze how each variable impacts the function's behavior independently.

Calculus

Calculus is the branch of mathematics dealing with the study of change and motion. It includes techniques like differentiation and integration. In the realm of vector fields and physics, calculus provides us with tools like divergence, curl, and various integral theorems which help in understanding and quantifying spatial phenomena.
In this exercise, we used calculus to find the divergence of a given vector field. Divergence is a scalar value that gives the magnitude of a field's source at a given point, influencing how much the field is emanating from or converging at that point.
The divergence for the vector field \( \mathbf{F}(x, y, z) = x y z \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) was computed as:

• Calculate partial derivatives for each component.
• Sum these to find \( abla \cdot \mathbf{F} = yz + 1 + 1 \).

By evaluating \( abla \cdot \mathbf{F} \) at the point \( (1, 2, 3) \), we used calculus to quantify the divergence and found it to be 8, showcasing how calculus allows us to obtain precise and meaningful insights about mathematical models.