Problem 14 - Advanced Engineering Mathematics

Question

Find the curl and the divergence of the given vector field. $$ \mathbf{F}(x, y, z)=y z \ln x \mathbf{i}+(2 x-3 y z) \mathbf{j}+x y^{2} z^{3} \mathbf{k} $$

Step-by-Step Solution

Answer

Curl: \( (2xyz^3 + 3y) \mathbf{i} - (y^2z^3 - y\ln x) \mathbf{j} + (2 - z\ln x) \mathbf{k} \); Divergence: \( \frac{yz}{x} - 3z + 3xy^2z^2 \)."}

Step 1: Identify the Vector Field Components

The given vector field \( \mathbf{F} \) has three components: \( F_1 = yz \ln x \), \( F_2 = 2x - 3yz \), and \( F_3 = xy^2z^3 \). These correspond to the \( i \), \( j \), and \( k \) components, respectively.

Step 2: Compute the Curl of \( \mathbf{F} \)

The curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k} \).

Step 3: Calculate Partial Derivatives for Curl

Calculate the necessary partial derivatives: - \( \frac{\partial F_3}{\partial y} = 2xyz^3 \)- \( \frac{\partial F_2}{\partial z} = -3y \)- \( \frac{\partial F_3}{\partial x} = y^2z^3 \)- \( \frac{\partial F_1}{\partial z} = y\ln x \)- \( \frac{\partial F_2}{\partial x} = 2 \)- \( \frac{\partial F_1}{\partial y} = z\ln x \)

Step 4: Substitute Partials into Curl Formula

Substitute these into the curl formula:\[ abla \times \mathbf{F} = (2xyz^3 + 3y) \mathbf{i} - (y^2z^3 - y\ln x) \mathbf{j} + (2 - z\ln x) \mathbf{k}. \]

Step 5: Simplify Curl Expression

Simplify further if needed:\[ abla \times \mathbf{F} = (2xyz^3 + 3y) \mathbf{i} - (y^2z^3 - y\ln x) \mathbf{j} + (2 - z\ln x) \mathbf{k}. \]

Step 6: Compute the Divergence of \( \mathbf{F} \)

The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by \( abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \).

Step 7: Calculate Partial Derivatives for Divergence

Find the necessary partial derivatives:- \( \frac{\partial F_1}{\partial x} = \frac{yz}{x} \)- \( \frac{\partial F_2}{\partial y} = -3z \)- \( \frac{\partial F_3}{\partial z} = 3xy^2z^2 \)

Step 8: Substitute Partials into Divergence Formula

Substitute these into the divergence formula:\[ abla \cdot \mathbf{F} = \frac{yz}{x} - 3z + 3xy^2z^2. \]

Step 9: Simplify Divergence Expression

Simplify the expression if needed:\[ abla \cdot \mathbf{F} = \frac{yz}{x} - 3z + 3xy^2z^2. \]

Key Concepts

Curl of a Vector Field

In vector calculus, the curl of a vector field is a fascinating concept that helps us understand the rotation or the "swirl" at a point in a three-dimensional field. If you imagine a small paddle wheel in a fluid flow, the curl measures how much the wheel would rotate.

To find the curl of a vector field \( \mathbf{F}(x, y, z) \), we employ the formula:

• \( abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k} \).

This formula utilizes partial derivatives to calculate the components of the curl, assessing how the field components change with respect to the coordinates.

For example, evaluating the curl for the vector field \( \mathbf{F}(x, y, z) = yz \ln x \mathbf{i} + (2x - 3yz) \mathbf{j} + xy^2z^3 \mathbf{k} \), we first determine the needed partial derivatives:

• \( \frac{\partial F_3}{\partial y} = 2xyz^3 \)
• \( \frac{\partial F_2}{\partial z} = -3y \)
• \( \frac{\partial F_3}{\partial x} = y^2z^3 \)
• \( \frac{\partial F_1}{\partial z} = y\ln x \)
• \( \frac{\partial F_2}{\partial x} = 2 \)
• \( \frac{\partial F_1}{\partial y} = z\ln x \)

Substituting these into the curl formula gives us:

• \( abla \times \mathbf{F} = (2xyz^3 + 3y) \mathbf{i} - (y^2z^3 - y\ln x) \mathbf{j} + (2 - z\ln x) \mathbf{k} \).

The result describes how the vector field rotates and varies in three dimensions.

Divergence of a Vector Field

Divergence is another key concept in vector calculus that describes the magnitude of a source or sink at a given point in a vector field. Essentially, it measures how much a vector field spreads out from (or converges to) a particular point.

The formula for divergence is relatively straightforward:

• \( abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \).

This involves adding up the partial derivatives of each component of the vector field with respect to its corresponding variable.

Considering the example vector field \( \mathbf{F}(x, y, z) = yz \ln x \mathbf{i} + (2x - 3yz) \mathbf{j} + xy^2z^3 \mathbf{k} \), let's compute its divergence. The necessary derivatives are as follows:

• \( \frac{\partial F_1}{\partial x} = \frac{yz}{x} \)
• \( \frac{\partial F_2}{\partial y} = -3z \)
• \( \frac{\partial F_3}{\partial z} = 3xy^2z^2 \)

By plugging these into our divergence formula, we get:

• \( abla \cdot \mathbf{F} = \frac{yz}{x} - 3z + 3xy^2z^2 \).

This value indicates how much the vector field diverges or converges at any point \( (x, y, z) \). High divergence means strong source strength, whereas negative values can indicate a sink behavior.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of several variables. They allow us to understand how a function changes as one of the variables is varied, while keeping the other variables constant.

To compute a partial derivative, we differentiate with respect to one variable, treating all other variables as constants. This concept is crucial in vector calculus, particularly when calculating the curl and divergence of vector fields.

In the context of our example vector field \( \mathbf{F}(x, y, z) = yz \ln x \mathbf{i} + (2x - 3yz) \mathbf{j} + xy^2z^3 \mathbf{k} \), let's examine how partial derivatives are utilized:

• For the curl: we calculate
  • \( \frac{\partial F_3}{\partial y} \), \( \frac{\partial F_2}{\partial z} \), \( \frac{\partial F_3}{\partial x} \), \( \frac{\partial F_1}{\partial z} \), \( \frac{\partial F_2}{\partial x} \), \( \frac{\partial F_1}{\partial y} \)
  to see how each component interacts with the others.
• For the divergence: we calculate
  • \( \frac{\partial F_1}{\partial x} \), \( \frac{\partial F_2}{\partial y} \), \( \frac{\partial F_3}{\partial z} \)
  to add up how each component contributes to the total divergence.

By finding these partial derivatives, we can extract vital information from the equations and understand how changes in one variable affect the whole vector field.