Problem 1 - Thomas Calculus

Question

In Exercises \(1-8,\) find the divergence of the field. $$ \mathbf{F}=(x-y+z) \mathbf{i}+(2 x+y-z) \mathbf{j}+(3 x+2 y-2 z) \mathbf{k} $$

Step-by-Step Solution

Answer

The divergence of the field is 0.

Step 1: Understanding Divergence

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

Step 2: Identify Components

Identify \( P(x, y, z) \), \( Q(x, y, z) \), and \( R(x, y, z) \) from the given vector field \( \mathbf{F}=(x-y+z) \mathbf{i}+(2 x+y-z) \mathbf{j}+(3 x+2 y-2 z) \mathbf{k} \). In this case, \( P = x - y + z \), \( Q = 2x + y - z \), and \( R = 3x + 2y - 2z \).

Step 3: Calculate Partial Derivatives

Compute the partial derivatives: \( \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x-y+z) = 1 \); \( \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(2x+y-z) = 1 \); \( \frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(3x+2y-2z) = -2 \).

Step 4: Compute Divergence

Substitute the partial derivatives into the divergence formula: \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 1 + 1 - 2 = 0 \).

Key Concepts

Vector Field

A vector field is an important concept in calculus and physics. It is essentially a function that assigns a vector to each point in a space. In simpler terms, imagine a field consisting of arrows, where each arrow represents a vector and points in a particular direction.

• Each vector in the field has a magnitude and a direction.
• The vector components can change depending on their position in space.

For example, the vector field \( \mathbf{F} = (x-y+z) \mathbf{i} + (2x+y-z) \mathbf{j} + (3x+2y-2z) \mathbf{k} \) assigns a vector at every point \( (x, y, z) \) in space.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus when dealing with functions of several variables. A partial derivative measures how a function changes as one specific variable changes while the others are held constant.

• They are written using the symbol \( \frac{\partial}{\partial x} \) for the variable \( x \), and similarly for other variables.
• To calculate a partial derivative, only the variable of interest is considered to change.

For instance, in our exercise, the partial derivative \( \frac{\partial P}{\partial x} \) for the function \( P = x - y + z \) evaluates to 1, since only \( x \) contributes to the change.

Divergence Formula

The divergence formula is used to determine the rate at which a vector field spreads out from a point. This scalar quantity offers insights into the behavior of the field, such as whether it is a source or a sink.

• Mathematically, it is expressed as \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).
• The vector field components \( P, Q, \) and \( R \) define the direction and flow at each point.

Using this formula for our vector field, the divergence is calculated as 0, indicating that the field neither diverges nor converges at any point.

Calculus Problem Solving

Solving calculus problems often involves breaking down seemingly complex problems into understandable parts. Here's an approach to tackling vector field-related problems:

• **Step 1:** Identify the problem and understand what is being asked.
• **Step 2:** Break the vector field into its components. For example, recognize \( P, Q, \) and \( R \).
• **Step 3:** Find the needed derivatives. Calculate required partial derivatives as seen in the steps above.
• **Step 4:** Apply the appropriate formulas such as the divergence formula.

Using these logical steps in our problem simplifies the process, transforming an intimidating problem into a manageable solution.