Problem 217 - Calculus Volume 3

Question

For the following exercises, find the curl of \(\mathbf{F}\). \(\mathbf{F}(x, y, z)=(x-y) \mathbf{i}+(y-z) \mathbf{j}+(z-x) \mathbf{k}\)

Step-by-Step Solution

Answer

The curl is \( \langle 1, 1, 1 \rangle \).

Step 1: Understand the Curl Formula

The curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by the cross product \( abla \times \mathbf{F} \). In component form, the curl is computed as:\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}. \]

Step 2: Identify Component Functions

Given \( \mathbf{F}(x, y, z) = (x-y) \mathbf{i} + (y-z) \mathbf{j} + (z-x) \mathbf{k} \), identify the component functions as \( P = x-y \), \( Q = y-z \), \( R = z-x \).

Step 3: Compute Partial Derivatives

Calculate the required partial derivatives:- \( \frac{\partial R}{\partial y} = \frac{\partial (z-x)}{\partial y} = 0 \)- \( \frac{\partial Q}{\partial z} = \frac{\partial (y-z)}{\partial z} = -1 \)- \( \frac{\partial P}{\partial z} = \frac{\partial (x-y)}{\partial z} = 0 \)- \( \frac{\partial R}{\partial x} = \frac{\partial (z-x)}{\partial x} = -1 \)- \( \frac{\partial Q}{\partial x} = \frac{\partial (y-z)}{\partial x} = 0 \)- \( \frac{\partial P}{\partial y} = \frac{\partial (x-y)}{\partial y} = -1 \).

Step 4: Assemble the Curl Expression

Substitute the derivatives into the curl formula:\[ abla \times \mathbf{F} = \left( 0 - (-1) \right) \mathbf{i} + \left( 0 - (-1) \right) \mathbf{j} + \left( 0 - (-1) \right) \mathbf{k}. \]

Step 5: Simplify the Result

Simplifying the expression, we get:\( abla \times \mathbf{F} = 1 \mathbf{i} + 1 \mathbf{j} + 1 \mathbf{k} \). So, the curl of \( \mathbf{F} \) is \( \langle 1, 1, 1 \rangle \).

Key Concepts

curl of a vector field

The curl of a vector field is an important concept in vector calculus that indicates the field's rotational tendency at a particular point in space. For a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the curl is represented by the expression \( abla \times \mathbf{F} \). This is equivalent to finding the circulation of the field per unit area as we shrink the area to zero.

The mathematical interpretation of the curl is akin to the amount of 'twisting' or 'spinning' a vector field induces at any point. If you visualize tiny paddles placed in the field, the curl shows how likely they are to spin.

It’s crucial to understand that the curl is different from divergence, which measures a field's magnitude of expansion or compression rather than its rotation.

In practical terms, the curl is computed using a three-dimensional cross product involving the del operator \( abla \) and the vector field \( \mathbf{F} \). This leads to a new vector that denotes the curl direction and its magnitude.

By calculating the curl, we can understand the dynamics of vector fields in physical systems, such as the circulation in fluid flow or magnetic field rotation around a wire carrying an electric current.

partial derivatives

Partial derivatives are fundamental in multivariable calculus, allowing us to find the rate at which a function changes with respect to one variable while keeping others constant. In the context of vector fields, partial derivatives help us calculate the curl by showing how each component of a vector field changes independently.

Consider a function \( f(x, y, z) \). The partial derivative of \( f \) with respect to \( x \), noted as \( \frac{\partial f}{\partial x} \), measures how \( f \) changes as \( x \) changes, while \( y \) and \( z \) remain fixed. Similarly, partial derivatives with respect to \( y \) and \( z \) help in capturing the directional changes along those axes.

In the exercise, partial derivatives are used to evaluate changes in each component of the vector field \( \mathbf{F} \) before assembling them into the curl formula. For example:

• \( \frac{\partial R}{\partial y} = 0 \) because \( R = z-x \) doesn't depend on \( y \).
• \( \frac{\partial Q}{\partial z} = -1 \) because \( Q = y-z \) changes linearly with \( z \).

Mastering the computation of partial derivatives is essential, as they are building blocks for not only curl but various other operations in calculus, such as gradient and divergence.

cross product

The cross product, also known as the vector product, is an algebraic operation that returns a vector perpendicular to two given vectors in three-dimensional space. For a vector field \( \mathbf{F} \), finding the curl involves taking the cross product of the del operator \( abla \) and \( \mathbf{F} \).

Let’s break down the basic idea: If \( \mathbf{a} \) and \( \mathbf{b} \) are two vectors, their cross product \( \mathbf{a} \times \mathbf{b} \) is a vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). The direction of this new vector can be determined using the right-hand rule, and its magnitude is given by \( |\mathbf{a}| \times |\mathbf{b}| \times \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).

In the context of the curl operation:

• The cross product allows the computation of rotational characteristics within the vector field.
• It combines the interactions of each partial derivative obtained from the vector field components.

The vector calculated from the cross product encodes information about the rotational movement of the field in a way that scalar products or element-wise multiplications cannot. This is why the cross product is instrumental in defining physical concepts such as torque and angular momentum, beside its role in calculating the curl of a vector field.