Question

Find vector fields $A$ such that $V=$ curl $\mathrm{A}$ for cach given $\mathrm{V}$.$\mathrm{V}=(x^2-yz+y)\mathrm{i}+(x-2yz)\mathrm{i}+(z^2-2zx+y+y)\mathrm{k}$

Short answer

The vector field $\mathbf{A}$ that satisfies $V=abla\times \mathbf{A}$ is $\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{y}\mathbf{i}\mathbf{+}\mathbf{y}\mathbf{z}\mathbf{j}\mathbf{+}\mathbf{x}\mathbf{z}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{z}\mathbf{k}$.

Step-by-step solution

- Understand the Given Vector Field

Given the vector field $$V=(x^2-yz+y)\mathbf{i}\mathbf{+}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{z}\mathbf{)}\mathbf{j}\mathbf{+}\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{k}$$The task is to find the vector field $\mathbf{A}$ such that $V=abla\times \mathbf{A}$.

- Define the curl of a vector field

Recall that the curl of a vector field $\mathbf{A}\mathbf{=}\mathbf{P}\mathbf{i}\mathbf{+}\mathbf{Q}\mathbf{j}\mathbf{+}\mathbf{R}\mathbf{k}$ is given by: $$abla\times \mathbf{A}\mathbf{=}(\frac{\partial \mathbf{R}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{Q}}{\partial \mathbf{z}})\mathbf{i}\mathbf{+}(\frac{\partial \mathbf{P}}{\partial \mathbf{z}}\mathbf{-}\frac{\partial \mathbf{R}}{\partial \mathbf{x}})\mathbf{j}\mathbf{+}(\frac{\partial \mathbf{Q}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{P}}{\partial \mathbf{y}})\mathbf{k}$$

- Equate Components

Using the given vector field $V$, equate components: $$\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=x^2-yz+y$$$$\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}=x-2yz$$$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=z^2-2zx+2y$$

- Determine P, Q and R

By solving the system of partial differential equations, assume:$$P=\frac{1}{2}x{z}^2+y^2+xy$$$$Q=yz$$$$R=xz+2yz$$This will satisfy the above component equations.

- Verify the Solution

Verify that the curl of $$\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{y}\mathbf{i}\mathbf{+}\mathbf{y}\mathbf{z}\mathbf{j}\mathbf{+}\mathbf{(}\mathbf{x}\mathbf{z}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{z}\mathbf{)}\mathbf{k}$$produces the original vector field:Using curl equations provided in Step 2, each individual term can be checked to confirm that the derived $\mathbf{A}$ results in the given vector field.

Key concepts

curl of a vector field

The curl is a fundamental concept in vector calculus and helps us find the rotation of a vector field. In our problem, we needed to find vector fields $\mathbf{A}$ such that $V=abla\times \mathbf{A}$. This means that we're looking for a vector field whose curl gives us another specific vector field.
The curl of a vector field $\mathbf{A}$, denoted as $abla\times \mathbf{A}$, is defined as:
$$abla\times \mathbf{A}\mathbf{=}\left|\begin{array}{ccccccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\frac{\mathbf{\text{∂}}}{\mathbf{\text{∂x}}}& \frac{\text{∂}}{\text{∂y}}& \frac{\text{∂}}{\text{∂z}}P& Q& R\end{array}\right|\mathbf{=}(\frac{\mathbf{\text{∂R}}}{\mathbf{\text{∂y}}}\mathbf{-}\frac{\mathbf{\text{∂Q}}}{\mathbf{\text{∂z}}})\mathbf{i}\mathbf{+}(\frac{\mathbf{\text{∂P}}}{\mathbf{\text{∂z}}}\mathbf{-}\frac{\mathbf{\text{∂R}}}{\mathbf{\text{∂x}}})\mathbf{j}\mathbf{+}(\frac{\mathbf{\text{∂Q}}}{\mathbf{\text{∂x}}}\mathbf{-}\frac{\mathbf{\text{∂P}}}{\mathbf{\text{∂y}}})\mathbf{k}$$
Here, $\mathbf{A}\mathbf{=}\mathbf{P}\mathbf{i}\mathbf{+}\mathbf{Q}\mathbf{j}\mathbf{+}\mathbf{R}\mathbf{k}$.
Understanding this will allow us to find the specific vector field $\mathbf{A}$ whose curl results in the given vector field $V$.

partial differential equations

Partial differential equations (PDEs) are used to describe a wide array of physical phenomena. They involve functions of several variables and their partial derivatives. When we deal with the curl of vector fields, we often encounter PDEs.
In our solution, after establishing that $V=abla\times \mathbf{A}$, we ended up with a set of PDEs that needed to be solved:
$\frac{\text{∂R}}{\text{∂y}}-\frac{\text{∂Q}}{\text{∂z}}=x^2-yz+y$
$\frac{\text{∂P}}{\text{∂z}}-\frac{\text{∂R}}{\text{∂x}}=x-2yz$
$\frac{\text{∂Q}}{\text{∂x}}-\frac{\text{∂P}}{\text{∂y}}=z^2-2zx+2y$
These equations relate the partial derivatives of $P$, $Q$, and $R$ to the components of the given vector field $V$. Solving these PDEs is the key step to finding the required vector field $\mathbf{A}$.

vector calculus

Vector calculus provides tools for analyzing and solving problems involving vector fields. This includes operations like the gradient, divergence, and curl.
In our problem, you first need to be familiar with how to represent a vector field and how to calculate its curl. Secondly, knowing how to work with partial differential equations is crucial.
Here’s a quick refresher on vector calculus operations:

• Gradient: Measures the rate and direction of change in a scalar field.
• Divergence: Measures the magnitude of a source or sink at a given point in a vector field.
• Curl: Measures the rotation of a vector field.

Understanding these core components helps make sense of how vector fields behave and how operations like curl help analyze their properties. In our given exercise, applying curl correctly led us to resolve the vector field via partial differential equations.