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Find vector fields \(\mathbf{A}\) such that \(\mathbf{V}=\) curl \(\mathbf{A}\) for each given \(\mathbf{V}\) $$\mathbf{V}=-\mathbf{k}$$

Short Answer

Expert verified
A = -y\textbf{i}

Step by step solution

01

Recall the definition of the curl of a vector field

Given a vector field \(\textbf{A}=(A_x, A_y, A_z)\), the curl is defined as \(\text{curl} \textbf{A} = abla \times \textbf{A}\). Utilize this in subsequent steps.
02

Express the curl operator in component form

The curl of \(\textbf{A}\) can be expressed as: \(abla \times \textbf{A} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ \frac{\rightarrow}{\rightarrow x} & \frac{\rightarrow}{\rightarrow y} & \frac{\rightarrow}{\rightarrow z} \ A_x & A_y & A_z \ \right\rangle\).
03

Setup the vector equation \(\text{curl}\textbf{A} = -\textbf{k}\)

From the given \(\textbf{V} = -\textbf{k}\), set up the component-wise equations: \(abla \times \textbf{A} = -\textbf{k}\). This yields: \( \frac{\rightarrow A_z}{\rightarrow y} - \frac{\rightarrow A_y}{\rightarrow z} = -1, \frac{\rightarrow A_x}{\rightarrow z} - \frac{\rightarrow A_z}{\rightarrow x} = 0, \frac{\rightarrow A_y}{\rightarrow x} - \frac{\rightarrow A_x}{\rightarrow y} = 0\).
04

Solve the component equations

Solve \( \frac{\rightarrow A_z}{\rightarrow y} - \frac{\rightarrow A_y}{\rightarrow z} = -1\): Set \(A_z \rightarrow 0\) and \( -\frac{\rightarrow A_y}{\rightarrow z} = -1 \rightarrow A_y = z\). Solve \( \frac{\rightarrow A_x}{\rightarrow z} -= \frac{\rightarrow A_z}{\rightarrow x} = 0\) to yield constant \(A_x \rightarrow f(x,y)\). Then solve \( \frac{\rightarrow A_y}{\rightarrow x} - \frac{\rightarrow A_x}{\rightarrow y} = 0\) for \( A_y = - \frac{\rightarrow f}{\rightarrow y ^ -1},\) with substitution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

curl of a vector field
The curl of a vector field is a fundamental operator in vector calculus. It operates on a three-dimensional vector field and results in another vector field. The curl measures the rotation or 'circulation' of the field at each point. Mathematically, for a vector field \textbf{A} = (A_x, A_y, A_z)\, the curl is defined by the cross product of the del operator \( abla \times\ \textbf{A} \). In simpler terms, the curl gives us a way to describe how a vector field rotates around any given point in space.
component form of curl
To understand the curl of a vector field, it's crucial to break it down into its component form. This helps make solving real problems more manageable. The component form of the curl of a vector field \( \textbf{A}\) is given by:

\( abla \times\ \textbf{A} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ \frac{\rightarrow}{\rightarrow x} & \frac{\rightarrow}{\rightarrow y} & \frac{\rightarrow}{\rightarrow z} \ A_x & A_y & A_z \end{vmatrix} \).

This determinant expands into the following:

\( \textnormal{i} \( \frac{\rightarrow A_z}{\rightarrow y} - \frac{\rightarrow A_y}{\rightarrow z} \) - \textnormal{j} \( \frac{\rightarrow A_z}{\rightarrow x} - \frac{\rightarrow A_x}{\rightarrow z} \) + \textnormal{k} \( \frac{\rightarrow A_y}{\rightarrow x} - \frac{\rightarrow A_x}{\rightarrow y} \) \).

This represents each component of the curl of \textbf{A} in terms of partial derivatives.
solving vector component equations
When tasked with finding a vector field \(\textbf{A}\) where the curl matches a given vector field \(\textbf{V}\), you need to break it into its component equations. Given the overall equation \( abla \times \textbf{A} = - \textbf{k}\), you establish the following component equations:
  • \( \frac{\rightarrow A_z}{\rightarrow y} - \frac{\rightarrow A_y}{\rightarrow z} = -1 \)
  • \( \frac{\rightarrow A_x}{\rightarrow z} - \frac{\rightarrow A_z}{\rightarrow x} = 0 \)
  • \( \frac{\rightarrow A_y}{\rightarrow x} - \frac{\rightarrow A_x}{\rightarrow y} = 0 \)`

Each of these expressions comes from corresponding partial derivatives equating to specified components of \(\textbf{V}\). Solving these step-by-step helps find the original vector field which matches the required curl.
vector calculus
Vector calculus is a branch of mathematics that deals with vector fields and the differentiation and integration of vector functions. It includes operations like the curl, divergence, and gradient, and is essential in fields such as physics and engineering.

Key Applications are:
  • Fluid Dynamics: Describing fluid flow.
  • Electromagnetism: Analyzing electric and magnetic fields.

Understanding these operations helps us solve complex real-world problems. It provides the tools to work with field quantities that vary over time and space. So, getting a grip on concepts like curl provides a foundation for diving deeper into topics like electromagnetic fields or fluid mechanics.

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Most popular questions from this chapter

Show that \(\nabla \cdot(\mathbf{U} \times \mathbf{r})=\mathbf{r} \cdot(\nabla \times \mathbf{U})\) where \(\mathbf{U}\) is a vector function of \(x, y, z,\) and \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\).

Let \(\mathbf{r}=\mathbf{r}(t)\) be a vector whose length is always 1 (it may vary in direction). Prove that either \(\mathbf{r}\) is a constant vector or \(d \mathbf{r} / d t\) is perpendicular to \(\mathbf{r} .\) Hint: Differentiate \(\mathbf{r} \cdot \mathbf{r}.\)

For motion near the surface of the earth, we usually assume that the gravitational force on a mass \(m\) is $$\mathbf{F}=-m g \mathbf{k}$$ but for motion involving an appreciable variation in distance \(r\) from the center of the earth, we must use $$\mathbf{F}=-\frac{C}{r^{2}} \mathbf{e}_{r}=-\frac{C}{r^{2}} \frac{\mathbf{r}}{|\mathbf{r}|}=-\frac{C}{r^{3}} \mathbf{r}$$ where \(C\) is a constant. Show that both these \(\mathbf{F}\) 's are conservative, and find the potential for each.

If \(\mathbf{F}=x \mathbf{i}+y \mathbf{j},\) calculate \(\iint \mathbf{F} \cdot \mathbf{n} d \sigma\) over the part of the surface \(z=4-x^{2}-y^{2}\) that is above the \((x, y)\) plane, by applying the divergence theorem to the volume bounded by the surface and the piece that it cuts out of the \((x, y)\) plane. Hint: What is \(\mathbf{F} \cdot \mathbf{n}\) on the \((x, y)\) plane?

\(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the entire surface of a cube in the first octant with edges of length 2 along the coordinate axes, where $$\mathbf{V}=\left(x^{2}-y^{2}\right) \mathbf{i}+3 y \mathbf{j}-2 x z \mathbf{k}$$

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