/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Sketch the following vector fiel... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 23

Sketch the following vector fields. $$\mathbf{F}=\langle y,-x, 0\rangle$$

Short Answer

Expert verified
Question: Sketch the vector field F(x, y, z) = (y, -x, 0). Answer: To sketch the vector field, divide the xy-plane into four quadrants based on the signs of x and y. Draw vectors in each quadrant as follows: In the first quadrant, vectors point towards the third quadrant; In the second quadrant, vectors point towards the second quadrant; In the third quadrant, vectors point towards the fourth quadrant; In the fourth quadrant, vectors point towards the first quadrant. Ensure that the vector lengths are proportional to their magnitudes, and remember that all vectors lie in the xy-plane.

Step by step solution

01

Understand the vector components

The components of the given vector field are F_x = y, F_y = -x, and F_z = 0. The x-component of the vector (F_x) depends on the y-coordinate, while the y-component (F_y) depends on the x-coordinate. The z-component (F_z) is always 0, meaning that all vectors in the vector field will reside in the xy-plane.
02

Identify directions of the vectors

Since the x-component of the vector depends on the y-coordinate, and the y-component depends on the x-coordinate, the directions of the vectors in the vector field will be influenced by the signs of x and y. There are four regions to consider based on the signs of x and y: 1. x > 0 and y > 0: In this region, F_x = y > 0 and F_y = -x < 0. This means that the vectors will point towards the third quadrant. 2. x < 0 and y > 0: In this region, F_x = y > 0 and F_y = -x > 0. This means that the vectors will point towards the second quadrant. 3. x < 0 and y < 0: In this region, F_x = y < 0 and F_y = -x < 0. This means that the vectors will point towards the fourth quadrant. 4. x > 0 and y < 0: In this region, F_x = y < 0 and F_y = -x > 0. This means that the vectors will point towards the first quadrant.
03

Draw the vector field

Now that we understand how the vector components determine the directions of the vectors, we can draw the vector field. First, draw the xy-plane and divide it into four quadrants using the x and y axes. Then, sketch vectors in each quadrant based on the information from Step 2. Remember that all vectors have a z-component of 0, so they will lie in the xy-plane. - In the first quadrant (x > 0 and y > 0), draw vectors pointing towards the third quadrant. - In the second quadrant (x < 0 and y > 0), draw vectors pointing towards the second quadrant. - In the third quadrant (x < 0 and y < 0), draw vectors pointing towards the fourth quadrant. - In the fourth quadrant (x > 0 and y < 0), draw vectors pointing towards the first quadrant. Ensure that the lengths of the vectors are proportional to their magnitudes which depend on the x and y coordinates. This sketch will represent the given vector field.

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Key Concepts

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

Vector Components
When discussing vector fields, it's crucial to understand the vector components involved. A vector in two-dimensional space is represented by its x and y components, denoted as \( \mathbf{F} = \langle F_x, F_y \rangle \). In this particular exercise, we have the vector field \( \mathbf{F} = \langle y, -x, 0 \rangle \).

Here, the components are as follows:

  • \( F_x = y \): This implies that the x-component of the vector relies on its position along the y-axis.
  • \( F_y = -x \): This shows that the y-component is governed by the x-axis position, changing direction depending on whether x is positive or negative.
  • \( F_z = 0 \): This indicates that there's no component along the z-axis, keeping the vectors entirely in the xy-plane.
This composition of the vector is essential because vectors are direction-sensitive and their behavior is derived from their components.
Quadrants in the xy-plane
The xy-plane is divided into four distinct quadrants, each characterized by the signs of the x and y coordinates:

  • First Quadrant where \( x > 0 \) and \( y > 0 \).
  • Second Quadrant where \( x < 0 \) and \( y > 0 \).
  • Third Quadrant where \( x < 0 \) and \( y < 0 \).
  • Fourth Quadrant where \( x > 0 \) and \( y < 0 \).

Understanding these quadrants is integral when sketching or analyzing vector fields as the direction and magnitude of vectors are affected by the quadrant they lie in.

For example, in this exercise, the vector components \( F_x = y \) and \( F_y = -x \) result in unique directions based on their positions:

  • In the first quadrant, vectors lean towards the third quadrant operations because \( F_x \) is influenced by a positive \( y \) and \( F_y \) by a negative \( x \).
  • Each quadrant behaves similarly based on the respective signs of x and y.
Vectors Direction and Magnitude
A vector's direction and magnitude tell us how and where it points and how intense that pointing is. The direction is determined by its components, which can be considered analogous to components creating an arrow.

For the vector field \( \mathbf{F} = \langle y, -x, 0 \rangle \), the direction alters according to values of \( x \) and \( y \).

Here is how you determine direction in various quadrants:

  • Direction: Use vector components to analyze direction automatically affected by the sign of x and y. For instance, a positive \( y \) and negative \( x \) in the first quadrant point the vector towards a specific direction (usually opposite to the fourth quadrant).
  • Magnitude: Calculated as \( \sqrt{F_x^2 + F_y^2} = \sqrt{y^2 + (-x)^2} \), showing how strong or long the vector is. This magnitude helps depict the proportional length of vectors when constructing the field.
This exercise focuses on understanding these directional tendencies by sketching the vector field across the xy-plane.

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Most popular questions from this chapter

Let \(f\) be differentiable and positive on the interval \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. Use Theorem 17.14 to show that the area of \(S\) (as given in Section 6.6 ) is $$ \int_{a}^{b} 2 \pi f(x) \sqrt{1+f^{\prime}(x)^{2}} d x $$

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Cone and sphere The cone \(z^{2}=x^{2}+y^{2},\) for \(z \geq 0,\) cuts the sphere \(x^{2}+y^{2}+z^{2}=16\) along a curve \(C\) a. Find the surface area of the sphere below \(C,\) for \(z \geq 0\) b. Find the surface area of the sphere above \(C\) c. Find the surface area of the cone below \(C\), for \(z \geq 0\)

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