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Chapter 15: Problem 6

Sketch the vector field by drawing some representative nonintersecting vectors. The vectors need not be drawn to scale, but they should be in reasonably correct proportion relative to each other.\(\mathbf{F}(x, y)=y \mathbf{j}, \quad y>0\)

Short Answer

Expert verified
Vectors point upward, increasing in magnitude with \( y \), for \( y>0 \).

Step by step solution

01

Understand the Vector Field Components

The vector field is given by \( \mathbf{F}(x, y)=y \mathbf{j} \). Here, \( \mathbf{j} \) is the unit vector in the direction of the positive \( y \)-axis. This means that the vector field has no \( i \)-component (no horizontal component). The magnitude and direction of each vector depends only on the \( y \)-value and it points upwards because of the positive \( \mathbf{j} \) component. For \( y > 0 \), the vectors will always have a positive magnitude.
02

Determine the Vector Magnitudes

For any point \((x, y)\), the vector has the form \( \mathbf{F}(x, y) = y \mathbf{j} \). This means at any point \((x, y)\), the magnitude of the vector is simply \( y \). So, as \( y \) increases, the length of the vector increases. When \( y = 1 \), the vector's length is 1; when \( y = 2 \), the vector's length is 2, and so on.
03

Sketch the Vector Field

To sketch this vector field, draw vectors originating at several points in the plane where \( y > 0 \). For example, choose points where \( y = 1, 2, 3, \ldots \). At each chosen point \((x, y)\), draw a vector pointing directly upwards (along the positive \( y \)-axis) with a length proportional to \( y \). For \( y=1 \), draw short vectors at points like \((0, 1), (1, 1),\) etc. For \( y=2 \), draw longer vectors at points like \((0, 2), (1, 2),\) etc. Ensure that vectors do not intersect, and the pattern shows increasing length as \( y \) increases.
04

Analyze the Sketch for Accuracy

Check the sketch to verify that vectors are aligned along the positive \( y \)-axis with increasing length as \( y \) increases. Make sure the overall pattern shows an array of upward-pointing vectors that convey the sense of flow in the direction of increasing \( y \), confirming that the vector field is consistent with the formula \( \mathbf{F}(x, y) = y \mathbf{j} \).

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Key Concepts

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

Vector Magnitude
In a vector field, the magnitude of a vector provides an understanding of its strength or size at a given point. For the vector field \( \mathbf{F}(x, y) = y \mathbf{j} \),the magnitude is dictated solely by the \( y \)-component. The formula shows that each vector's magnitude is equal to the value of \( y \).

To comprehend this, consider these key points:
  • When \( y = 1 \), the magnitude of the vector is 1, representing a basic unit length in the vector field.
  • As \( y \) increases, the magnitude of the vector directly proportionally increases. For instance, if \( y = 3 \), the magnitude becomes 3, making the vector three times longer than when \( y = 1 \).
  • This linear relationship between the vector magnitude and the \( y \)-value results in vectors lengthening as they move upwards in the plane.
By understanding vector magnitudes, students can better predict and sketch vector fields by visualizing how the magnitude changes with \( y \).
Sketching Vectors
Sketching vectors begins with recognizing how length and direction are displayed. For \( \mathbf{F}(x, y) = y \mathbf{j} \),all vectors point upwards because the direction is determined by \( \mathbf{j} \), which is along the positive \( y \)-axis.

Here's how to effectively sketch vectors:
  • Start by selecting several points where \( y > 0 \); points such as \( (0, 1), (1, 1), (0, 2), (1, 2) \) work well.
  • Draw arrows originating from these points; their length should represent the \( y \)-value of the point.
  • Keep vectors non-intersecting and proportional to each other to maintain clarity in the sketch.
The aim is to provide a visual pattern where vectors appear as an array of arrows increasing in size, clearly indicating how the vector field behaves with higher \( y \)-values.
Coordinate System
A coordinate system serves as the backdrop when sketching vector fields. In our given exercise, the coordinate system uses the Cartesian plane, consisting of the \( x \)-axis (horizontal) and \( y \)-axis (vertical).

Understanding this system is crucial for accurately representing vectors:
  • The Cartesian coordinate system allows us to locate points precisely, expressed as \((x, y)\).
  • In the vector field \( \mathbf{F}(x, y) = y \mathbf{j} \),vectors are constructed at coordinates where \( y \) is positive.
  • Vectors have no horizontal component, making the \( x \)-coordinate flexible without affecting the vector's magnitude or direction.
By using the coordinate system efficiently, one can identify points to draw vectors and truly understand how the vector field manifests visually, especially when variables like \( y \) change.

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Most popular questions from this chapter

Determine whether \(\mathbf{F}\) is a conservative vector field. If so, find a potential function for it.$$ \mathbf{F}(x, y)=x \mathbf{i}+y \mathbf{j} $$

Verify Formula (2) in Stokes' Theorem by evaluating the line integral and the surface integral. Assume that the surface has an upward orientation.\(\mathbf{F}(x, y, z)=(x-y) \mathbf{i}+(y-z) \mathbf{j}+(z-x) \mathbf{k} ; \sigma\) is the por- tion of the plane \(x+y+z=1\) in the first octant.

Use a CAS to evaluate the line integrals along the given curves.$$ \begin{aligned} &\text { (a) } \int_{C}\left(x^{3}+y^{3}\right) d s \\ &\quad C: \mathbf{r}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j} \quad(0 \leq t \leq \ln 2) \\ &\text { (b) } \int_{C} x e^{z} d x+(x-z) d y+\left(x^{2}+y^{2}+z^{2}\right) d z \\ &C: x=\sin t, y=\cos t, \quad z=t \quad(0 \leq t \leq \pi / 2) \end{aligned} $$

Let \(C\) be the line segment from \((0,0)\) to \((0,1) .\) In each part, evaluate the line integral along \(C\) by inspection, and explain your reasoning. (a) \(\int_{C} d s\) (b) \(\int_{C} \sin x y d y\)

Let \(k\) be a constant, \(\mathbf{F}=\mathbf{F}(x, y, z), \mathbf{G}=\mathbf{G}(x, y, z)\), and \(\phi=\phi(x, y, z)\). Prove the following identities, assuming that all derivatives involved exist and are continuous.\(\operatorname{curl}(\phi \mathbf{F})=\phi \operatorname{curl} \mathbf{F}+\nabla \phi \times \mathbf{F}\)
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