Problem 5 - Calculus: Early Transcendental Functions

Question

Find the curl and divergence of the given vector field. $$\left\langle x y, y z, x^{2}\right\rangle$$

Step-by-Step Solution

Answer

The curl of the given vector field is \(<-y, -2x, z - x>\), and the divergence of the vector field is \(y + z\).

Step 1: Identify the Vector Field

The given vector field is \(F = \left\langle x y, y z, x^{2}\right\rangle\). Where \(F_{x}= x y\), \(F_{y}= y z\), \(F_{z}= x^{2}\)

Step 2: Calculate the Curl of the Vector Field

The formula for the curl of a vector field F = \(\) is given by \Curl(F) = \(<\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z},\frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x},\frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y}>\)\So, \Curl(F) = \(<\frac{\partial (x^2)}{\partial y}-\frac{\partial (y z)}{\partial z},\frac{\partial (x y)}{\partial z}-\frac{\partial (x^2)}{\partial x},\frac{\partial (y z)}{\partial x}-\frac{\partial (x y)}{\partial y}>\) = \(<0-y, 0-2x, z-x>\) = \(<-y, -2x, z-x>\)

Step 3: Calculate the Divergence of the Vector Field

The formula of the divergence of a vector field F = \(\) is given by: \div(F) = \(\frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z}\)\So, \div(F) = \(\frac{\partial (x y)}{\partial x} + \frac{\partial (y z)}{\partial y} + \frac{\partial (x^2)}{\partial z}\) = \(y + z + 0\) = \(y + z\)

Key Concepts

Curl of a Vector Field

In vector calculus, the concept of curl is crucial for understanding the rotation or swirling of a vector field. Imagine a tiny paddle wheel placed in the flow of the vector field, the curl helps determine how the wheel would rotate. The curl itself is a vector that points in the direction of the axis of rotation, and its magnitude depends on how fast the object would spin.
For a vector field \( F = \langle F_x, F_y, F_z \rangle \), the curl is calculated using the formula:

• \( abla \times F = \left\langle \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right\rangle \)

In our exercise, where \( F = \langle xy, yz, x^2 \rangle \), the curl is computed as:

• \( \left\langle 0 - y, 0 - 2x, z - x \right\rangle = \langle -y, -2x, z - x \rangle \)

This result tells us about the direction and amount of rotation around each axis.
Understanding the curl provides insight into how fluids or air circulate around fixed points within the field.

Divergence of a Vector Field

The divergence of a vector field is a concept used to quantify the field's tendency to originate from or converge toward a point. Think of it as measuring how much a force spreads out or pinches in space.
For a vector field \( F = \langle F_x, F_y, F_z \rangle \), the divergence is calculated by:

• \( abla \cdot F = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \)

In the problem \( F = \langle xy, yz, x^2 \rangle \), the divergence is:

• \( \frac{\partial (xy)}{\partial x} + \frac{\partial (yz)}{\partial y} + \frac{\partial (x^2)}{\partial z} = y + z + 0 = y + z \)

This gives us a scalar field that indicates the net outward flow of the vector field from any point. Positive divergence means a source (expanding), while a negative divergence implies a sink (contracting).
Divergence is extremely useful in physics and engineering to describe phenomena such as fluid flow, electric fields, and more.

Partial Derivatives

Partial derivatives are fundamental for working with functions of multiple variables. They measure how a function changes as one of its variables is altered, keeping others constant.
Given a function \( f(x, y, z) \), the partial derivatives are represented as:

• \( \frac{\partial f}{\partial x} \)
• \( \frac{\partial f}{\partial y} \)
• \( \frac{\partial f}{\partial z} \)

These derivatives are crucial in calculating the curl and divergence of vector fields. For example, in the vector field \( F = \langle xy, yz, x^2 \rangle \), we find:

• \( \frac{\partial (xy)}{\partial x} = y \)
• \( \frac{\partial (yz)}{\partial y} = z \)
• \( \frac{\partial (x^2)}{\partial z} = 0 \)

Each partial derivative is focused on a single dimension, offering insights into how the vector field behaves in each direction.
The power of partial derivatives lies in their simplicity and ability to reduce complex multivariable problems into more manageable one-variable issues.