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Two-dimensional vector fields Sketch the following vector fields. $$\mathbf{F}=\langle x, y-x\rangle$$

Short Answer

Expert verified
Answer: The most important part of sketching a vector field is to visualize the vector field and understand how the components of the vector affect the direction and magnitude.

Step by step solution

01

Understand the vector field components

The given vector field is \(\mathbf{F} = \langle x, y-x \rangle\). Here, the x-component is x, and the y-component is y-x. These components will determine the direction and magnitude of vectors in the vector field.
02

Create a grid

Create a grid of points (x, y) in the two-dimensional space. The points chosen should cover the region of interest for the vector field sketch. For this example, let's consider a grid with points ranging from -4 to 4 for both x and y axes, with an interval of 1.
03

Calculate vector F for each point in the grid

For each point (x, y) in the grid, compute the vector \(\mathbf{F}\) at that point by plugging in the x and y values into the components formula: \(\mathbf{F} = \langle x, y-x \rangle\). For example, for the point (0,0), the vector F would be \(\mathbf{F} = \langle 0, 0-0 \rangle = \langle 0, 0 \rangle\). Repeat this for all points in the grid.
04

Draw vector arrows

At each point (x, y) in the grid, draw a vector arrow starting from that point and pointing in the direction of the vector \(\mathbf{F}\) calculated in the previous step. The length of the arrow should represent the magnitude of vector F as well. Note that longer arrows indicate stronger magnitude and vice versa.
05

Observe the pattern

Once all the vector arrows are drawn, observe the overall pattern of the vector field. This will give an idea of how the vector field behaves in the two-dimensional space. Notice the regions where vector magnitudes are stronger and where the direction of vectors change gradually or abruptly. By following these steps, the vector field sketch for \(\mathbf{F} = \langle x, y-x \rangle\) can be obtained, giving an insight into the behavior of this two-dimensional vector field.

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Key Concepts

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

Vector Field Components
When we talk about vector field components, we're discussing the individual parts of a vector field that inform us about the direction and strength of the field at any given point. In our example of the vector field \( \mathbf{F} = \langle x, y-x \rangle \), the components are essentially instructions on how to build each vector at a given point.

Think of vector field components like a recipe; for each position in space, specified by the coordinates (x, y), the components tell us how much to move along the x-axis and the y-axis. The x-component of our vector field is simply 'x', indicating that the vector's horizontal component is proportional to its x-coordinate. The y-component, 'y-x', implies that the vertical component depends on both y-coordinate and x-coordinate. To visualize how these components work together, you can picture a grid with arrows at each intersection point: each arrow starts at the grid point and extends in the direction and magnitude prescribed by the components at that specific location.

This nuanced 'combination of ingredients' helps determine the overall look and behavior of the vector field, revealing patterns such as swirls, sinks, sources, and uniform flow when visualized.
Two-Dimensional Vector Space
The two-dimensional vector space is a plane where each point can be represented by a pair of numbers, typically (x, y), that correspond to positions on the x-axis and y-axis. In the context of vector fields, a two-dimensional space serves as a 'canvas' where we can draw and analyze vectors. The beauty of working within such a plane is the intuitive grasp it provides since every position is just a hop and a skip away—horizontal for x, vertical for y.

You may be familiar with coordinate graphing from math class—the two-dimensional vector space is like an elaborate coordinate graph where, instead of plotting points, we're plotting vectors with both magnitude and direction. Creating a grid in this space allows us to systematically explore the vector field by looking at how vectors change from point to point. For example, in our exercise with the vector field \( \mathbf{F} = \langle x, y-x \rangle \), this space allows us to see whether the vectors point inward, outward, align with certain curves, or form distinctive patterns such as whirlpools or radiating lines, providing a visual insight into the field's behavior.
Magnitude and Direction of Vectors
Delving into magnitude and direction of vectors unveils how strong vectors are and where they're headed. Magnitude represents the length or size of the vector, commonly interpreted as the 'force' or 'intensity'. Direction, on the other hand, tells us which way the vector is pointing.

The calculation of magnitude involves a bit of Pythagorean theorem, where the length of a vector \( \mathbf{v} = \langle a, b \rangle \) is the square root of \( a^2 + b^2 \). For instance, in our vector field if we have a vector \( \mathbf{F} = \langle 2, 1 \rangle \), its magnitude would be \( \sqrt{2^2 + 1^2} = \sqrt{5} \), signifying a moderate strength in context. The direction, given by the angle of the vector in relation to the x-axis, can be found using trigonometry, with the angle \( \theta \) determined by \( \arctan(\frac{b}{a}) \).

In our example vector field, arrows are drawn with different lengths and orientations, mirroring the calculated magnitudes and directions for their respective grid points. Hence, the sketch of the vector field collectively illustrates the variation in magnitudes and direction as we scan across the vector space. The learner must note that while magnitude relates to the 'push' of the vector, direction is vital in understanding the 'pull' in which the dynamics of the field develop.

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Most popular questions from this chapter

Consider the potential function \(\varphi(x, y, z)=G(\rho),\) where \(G\) is any twice differentiable function and \(\rho=\sqrt{x^{2}+y^{2}+z^{2}} ;\) therefore, \(G\) depends only on the distance from the origin. a. Show that the gradient vector field associated with \(\varphi\) is \(\mathbf{F}=\nabla \varphi=G^{\prime}(\rho) \frac{\mathbf{r}}{\rho},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(\rho=|\mathbf{r}|\) b. Let \(S\) be the sphere of radius \(a\) centered at the origin and let \(D\) be the region enclosed by \(S\). Show that the flux of \(\mathbf{F}\) across \(S\) is $$\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi a^{2} G^{\prime}(a) $$ c. Show that \(\nabla \cdot \mathbf{F}=\nabla \cdot \nabla \varphi=\frac{2 G^{\prime}(\rho)}{\rho}+G^{\prime \prime}(\rho)\) d. Use part (c) to show that the flux across \(S\) (as given in part (b)) is also obtained by the volume integral \(\iiint_{D} \nabla \cdot \mathbf{F} d V\). (Hint: use spherical coordinates and integrate by parts.)

The potential function for the force field due to a charge \(q\) at the origin is \(\varphi=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{|\mathbf{r}|},\) where \(\mathbf{r}=\langle x, y, z\rangle\) is the position vector of a point in the field and \(\varepsilon_{0}\) is the permittivity of free space. a. Compute the force field \(\mathbf{F}=-\nabla \varphi\). b. Show that the field is irrotational; that is \(\nabla \times \mathbf{F}=\mathbf{0}\).

Find the work required to move an object in the following force fields along a line segment between the given points. Check to see whether the force is conservative. $$\mathbf{F}=\langle x, 2\rangle \text { from } A(0,0) \text { to } B(2,4)$$

The French physicist André-Marie Ampère \((1775-1836)\) discovered that an electrical current \(I\) in a wire produces a magnetic field \(\mathbf{B} .\) A special case of Ampère's Law relates the current to the magnetic field through the equation \(\oint_{C} \mathbf{B} \cdot d \mathbf{r}=\mu I,\) where \(C\) is any closed curve through which the wire passes and \(\mu\) is a physical constant. Assume that the current \(I\) is given in terms of the current density \(\mathbf{J}\) as \(I=\iint_{S} \mathbf{J} \cdot \mathbf{n} d S\) where \(S\) is an oriented surface with \(C\) as a boundary. Use Stokes' Theorem to show that an equivalent form of Ampère's Law is \(\nabla \times \mathbf{B}=\mu \mathbf{J}\)

a. Let \(\mathbf{a}=\langle 0,1,0\rangle, \mathbf{r}=\langle x, y, z\rangle,\) and consider the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r} .\) Use the right-hand rule for cross products to find the direction of \(\mathbf{F}\) at the points (0,1,1),(1,1,0),(0,1,-1), and (-1,1,0). b. With \(\mathbf{a}=\langle 0,1,0\rangle,\) explain why the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r}\) circles the \(y\) -axis in the counterclockwise direction looking along a from head to tail (that is, in the negative \(y\) -direction).

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