Problem 16 - Advanced Engineering Mathematics

Question

Find the curl and the divergence of the given vector field. $$ \mathbf{F}(x, y, z)=x^{2} \sin y z \mathbf{i}+z \cos x z^{3} \mathbf{j}+y e^{5 x y} \mathbf{k} $$

Step-by-Step Solution

Answer

Curl: \(\nabla \times \mathbf{F} = \left( e^{5xy} + 5xy e^{5xy} - \cos(xz^3) + 3x z^2 \sin(xz^3) \right)\mathbf{i} + \left( x^2 y \cos(yz) - 5y^2 e^{5xy} \right)\mathbf{j} + \left( -z^4 \sin(xz^3) - x^2 z \cos(yz) \right)\mathbf{k} \). Divergence: \( \nabla \cdot \mathbf{F} = 2x \sin(yz) \).

Step 1: Define the Curl

The curl of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k} \). We will apply this formula to find the curl of the given vector field.

Step 2: Calculate Partial Derivatives for Curl

For the given \( \mathbf{F}(x, y, z) = x^2 \sin(yz)\mathbf{i} + z \cos(xz^3)\mathbf{j} + y e^{5xy}\mathbf{k} \), we need the partial derivatives: \( \frac{\partial R}{\partial y} = e^{5xy} + 5xy e^{5xy} \), \( \frac{\partial Q}{\partial z} = \cos(xz^3) - 3x z^2 \sin(xz^3) \), \( \frac{\partial P}{\partial z} = x^2 y \cos(yz) \), \( \frac{\partial R}{\partial x} = 5y^2 e^{5xy} \), \( \frac{\partial Q}{\partial x} = -z^4 \sin(xz^3) \), and \( \frac{\partial P}{\partial y} = x^2 z \cos(yz) \).

Step 3: Substitute Derivatives into Curl Formula

Substituting the derivatives into the curl formula, we have: \( abla \times \mathbf{F} = \left( e^{5xy} + 5xy e^{5xy} - \cos(xz^3) + 3x z^2 \sin(xz^3) \right)\mathbf{i} + \left( x^2 y \cos(yz) - 5y^2 e^{5xy} \right)\mathbf{j} + \left( -z^4 \sin(xz^3) - x^2 z \cos(yz) \right)\mathbf{k} \).

Step 4: Define the Divergence

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

Step 5: Calculate Partial Derivatives for Divergence

For the vector field \( \mathbf{F} \), the partial derivatives are: \( \frac{\partial P}{\partial x} = 2x \sin(yz) \), \( \frac{\partial Q}{\partial y} = 0 \), and \( \frac{\partial R}{\partial z} = 0 \).

Step 6: Substitute Derivatives into Divergence Formula

Substituting the derived terms into the divergence formula, we have: \( abla \cdot \mathbf{F} = 2x \sin(yz) + 0 + 0 = 2x \sin(yz) \).

Key Concepts

Curl of a Vector Field

To understand the curl of a vector field, imagine a vector field as a flowing liquid, and the curl describes the rotation of this fluid. Specifically, the curl measures how much the flow is rotating around a given point.

In mathematical terms, for a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), the curl is computed using partial derivatives and is given by the formula:

• \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k} \)

Here's how it works with our example vector field \( \mathbf{F}(x, y, z)= x^2 \sin(yz)\mathbf{i} + z \cos(xz^3)\mathbf{j} + y e^{5xy}\mathbf{k} \):

We start by calculating the necessary partial derivatives needed for the curl formula. These derivatives represent the rates of change of the vector field components relative to the variables \( x, y, \) and \( z \).

• \( \frac{\partial R}{\partial y}, \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}, \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}, \frac{\partial P}{\partial y} \)

After computing these derivatives, we substitute them back into the curl formula to find the rotation of the vector field at each point. This operation is fundamental in physics and engineering, particularly in electromagnetism, where the curl can describe the rotational aspect of magnetic and electric fields.

Divergence of a Vector Field

The divergence of a vector field describes how much the field behaves like a source or sink at a particular point. It's like looking at whether more of the field flows into a point or away from it.

Mathematically, for a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), the divergence is given by:

• \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \)

Why do we care about divergence? It's a key concept in fluid dynamics where it helps describe how fluids expand or compress. In solid mechanics, it's important in assessing how materials deform.

Applying this to our vector field example, we need to find the partial derivatives of each component concerning its respective variable (\(x\) for \(P\), \(y\) for \(Q\), and \(z\) for \(R\)):

• \( \frac{\partial P}{\partial x} \)
• \( \frac{\partial Q}{\partial y} \)
• \( \frac{\partial R}{\partial z} \)

When you sum these derivatives, you get the divergence: \( 2x \sin(yz) \). This result tells us about the volumetric density of the vector field sources or sinks at a point.

Partial Derivatives

Partial derivatives allow us to study how a function changes with respect to one of its variables while keeping the other variables constant. They are especially useful in multivariable calculus when dealing with functions that have more than one input.

Imagine you're in a landscape where you want to measure the slope (derivative) when moving straight north (without changing east-west position). This illustration explains what a partial derivative does.

In the context of our vector field \( \mathbf{F}(x, y, z)= x^2 \sin(yz)\mathbf{i} + z \cos(xz^3)\mathbf{j} + y e^{5xy}\mathbf{k} \), each component of \( \mathbf{F} \) depends on multiple variables. Here's how we manage partial derivatives:

• For \( P = x^2 \sin(yz) \), partial derivatives like \( \frac{\partial P}{\partial x} \) and \( \frac{\partial P}{\partial y} \) respect that \( P \) changes as \( x \) and \( y \) change independently.
• Similarly, we compute \( \frac{\partial Q}{\partial z} \) or \( \frac{\partial R}{\partial y} \) for the other components.

This technique helps us find how small changes in a single variable affect the overall system, providing critical insights in fields like physics, engineering, and economics.

Understanding partial derivatives is vital for analyzing and solving problems involving vector fields, like calculating curls and divergences.