6.1 Vector Fields - Calculus Volume 3 | OpenStax (2024)

Examples of Vector Fields

How can we model the gravitational force exerted by multiple astronomical objects? How can we model the velocity of water particles on the surface of a river? Figure 6.2 gives visual representations of such phenomena.

Figure 6.2(a) shows a gravitational field exerted by two astronomical objects, such as a star and a planet or a planet and a moon. At any point in the figure, the vector associated with a point gives the net gravitational force exerted by the two objects on an object of unit mass. The vectors of largest magnitude in the figure are the vectors closest to the larger object. The larger object has greater mass, so it exerts a gravitational force of greater magnitude than the smaller object.

Figure 6.2(b) shows the velocity of a river at points on its surface. The vector associated with a given point on the river’s surface gives the velocity of the water at that point. Since the vectors to the left of the figure are small in magnitude, the water is flowing slowly on that part of the surface. As the water moves from left to right, it encounters some rapids around a rock. The speed of the water increases, and a whirlpool occurs in part of the rapids.

Figure 6.2 (a) The gravitational field exerted by two astronomical bodies on a small object. (b) The vector velocity field of water on the surface of a river shows the varied speeds of water. Red indicates that the magnitude of the vector is greater, so the water flows more quickly; blue indicates a lesser magnitude and a slower speed of water flow.

Each figure illustrates an example of a vector field. Intuitively, a vector field is a map of vectors. In this section, we study vector fields in ${ℝ}^2$ and ${ℝ}^3.$

Drawing a Vector Field

We can now represent a vector field in terms of its components of functions or unit vectors, but representing it visually by sketching it is more complex because the domain of a vector field is in ${ℝ}^2,$ as is the range. Therefore the “graph” of a vector field in ${ℝ}^2$ lives in four-dimensional space. Since we cannot represent four-dimensional space visually, we instead draw vector fields in ${ℝ}^2$ in a plane itself. To do this, draw the vector associated with a given point at the point in a plane. For example, suppose the vector associated with point $\left(4,\mathrm{-1}\right)$ is $\langle 3,1\rangle .$ Then, we would draw vector $\langle 3,1\rangle$ at point $\left(4,\mathrm{-1}\right).$

We should plot enough vectors to see the general shape, but not so many that the sketch becomes a jumbled mess. If we were to plot the image vector at each point in the region, it would fill the region completely and is useless. Instead, we can choose points at the intersections of grid lines and plot a sample of several vectors from each quadrant of a rectangular coordinate system in ${ℝ}^2.$

There are two types of vector fields in ${ℝ}^2$ on which this chapter focuses: radial fields and rotational fields. Radial fields model certain gravitational fields and energy source fields, and rotational fields model the movement of a fluid in a vortex. In a radial field, all vectors either point directly toward or directly away from the origin. Furthermore, the magnitude of any vector depends only on its distance from the origin. In a radial field, the vector located at point $\left(x,y\right)$ is perpendicular to the circle centered at the origin that contains point $\left(x,y\right),$ and all other vectors on this circle have the same magnitude.

Example 6.2

Drawing a Radial Vector Field

Sketch the vector field $\text{F}(x,y)=\frac{x}{2}\text{i}+\frac{y}{2}\text{j}.$

Solution

To sketch this vector field, choose a sample of points from each quadrant and compute the corresponding vector. The following table gives a representative sample of points in a plane and the corresponding vectors.

$\left(x,y\right)$	$\text{F}\left(x,y\right)$	$\left(x,y\right)$	$\text{F}\left(x,y\right)$	$\left(x,y\right)$	$\text{F}\left(x,y\right)$
$\left(1,0\right)$	$\langle \frac{1}{2},0\rangle$	$\left(2,0\right)$	$\langle 1,0\rangle$	$\left(1,1\right)$	$\langle \frac{1}{2},\frac{1}{2}\rangle$
$\left(0,1\right)$	$\langle 0,\frac{1}{2}\rangle$	$\left(0,2\right)$	$\langle 0,1\rangle$	$\left(\mathrm{-1},1\right)$	$\langle -\frac{1}{2},\frac{1}{2}\rangle$
$\left(\mathrm{-1},0\right)$	$\langle -\frac{1}{2},0\rangle$	$\left(\mathrm{-2},0\right)$	$\langle \mathrm{-1},0\rangle$	$\left(\mathrm{-1},\mathrm{-1}\right)$	$\langle -\frac{1}{2},-\frac{1}{2}\rangle$
$\left(0,\mathrm{-1}\right)$	$\langle 0,-\frac{1}{2}\rangle$	$\left(0,\mathrm{-2}\right)$	$\langle 0,\mathrm{-1}\rangle$	$\left(1,\mathrm{-1}\right)$	$\langle \frac{1}{2},-\frac{1}{2}\rangle$

Figure 6.3(a) shows the vector field. To see that each vector is perpendicular to the corresponding circle, Figure 6.3(b) shows circles overlain on the vector field.

Figure 6.3 (a) A visual representation of the radial vector field $\text{F}(x,y)=\frac{x}{2}\text{i}+\frac{y}{2}\text{j}.$ (b) The radial vector field $\text{F}(x,y)=\frac{x}{2}\text{i}+\frac{y}{2}\text{j}$ with overlaid circles. Notice that each vector is perpendicular to the circle on which it is located.

In contrast to radial fields, in a rotational field, the vector at point $\left(x,y\right)$ is tangent (not perpendicular) to a circle with radius $r=\sqrt{x^2+y^2}.$ In a standard rotational field, all vectors point either in a clockwise direction or in a counterclockwise direction, and the magnitude of a vector depends only on its distance from the origin. Both of the following examples are clockwise rotational fields, and we see from their visual representations that the vectors appear to rotate around the origin.

Example 6.3

Chapter Opener: Drawing a Rotational Vector Field

Figure 6.4 (credit: modification of work by NASA)

Sketch the vector field $\text{F}\left(x,y\right)=\langle y,\text{−}x\rangle .$

Solution

Create a table (see the one that follows) using a representative sample of points in a plane and their corresponding vectors. Figure 6.6 shows the resulting vector field.

$\left(x,y\right)$	$\text{F}\left(x,y\right)$	$\left(x,y\right)$	$\text{F}\left(x,y\right)$	$\left(x,y\right)$	$\text{F}\left(x,y\right)$
$\left(1,0\right)$	$\langle 0,\mathrm{-1}\rangle$	$\left(2,0\right)$	$\langle 0,\mathrm{-2}\rangle$	$\left(1,1\right)$	$\langle 1,\mathrm{-1}\rangle$
$\left(0,1\right)$	$\langle 1,0\rangle$	$\left(0,2\right)$	$\langle 2,0\rangle$	$\left(\mathrm{-1},1\right)$	$\langle 1,1\rangle$
$\left(\mathrm{-1},0\right)$	$\langle 0,1\rangle$	$\left(\mathrm{-2},0\right)$	$\langle 0,2\rangle$	$\left(\mathrm{-1},\mathrm{-1}\right)$	$\langle \mathrm{-1},1\rangle$
$\left(0,\mathrm{-1}\right)$	$\langle \mathrm{-1},0\rangle$	$\left(0,\mathrm{-2}\right)$	$\langle \mathrm{-2},0\rangle$	$\left(1,\mathrm{-1}\right)$	$\langle \mathrm{-1},\mathrm{-1}\rangle$

Figure 6.5 (a) A visual representation of vector field $\text{F}\left(x,y\right)=\langle y,\text{−}x\rangle .$ (b) Vector field $\text{F}\left(x,y\right)=\langle y,\text{−}x\rangle$ with circles centered at the origin. (c) Vector $\text{F}\left(a,b\right)$ is perpendicular to radial vector $\langle a,b\rangle$ at point $\left(a,b\right).$

Analysis

Note that vector $\text{F}\left(a,b\right)=\langle b,\text{−}a\rangle$ points clockwise and is perpendicular to radial vector $\langle a,b\rangle .$ (We can verify this assertion by computing the dot product of the two vectors: $\langle a,b\rangle ·\langle \text{−}b,a\rangle =\text{−}ab+ab=0.)$ Furthermore, vector $\langle b,\text{−}a\rangle$ has length $r=\sqrt{a^2+b^2}.$ Thus, we have a complete description of this rotational vector field: the vector associated with point $\left(a,b\right)$ is the vector with length r tangent to the circle with radius r, and it points in the clockwise direction.

Sketches such as that in Figure 6.6 are often used to analyze major storm systems, including hurricanes and cyclones. In the northern hemisphere, storms rotate counterclockwise; in the southern hemisphere, storms rotate clockwise. (This is an effect caused by Earth’s rotation about its axis and is called the Coriolis Effect.)

Example 6.4

Sketching a Vector Field

Sketch vector field $\text{F}(x,y)=\frac{y}{x^2+y^2}\text{i}-\frac{x}{x^2+y^2}\text{j}.$

Solution

To visualize this vector field, first note that the dot product $\text{F}(a,b)·(a\text{i}+b\text{j})$ is zero for any point $\left(a,b\right).$ Therefore, each vector is tangent to the circle on which it is located. Also, as $\left(a,b\right)\to \left(0,0\right),$ the magnitude of $\text{F}\left(a,b\right)$ goes to infinity. To see this, note that

$$\left|\left|\text{F}\left(a,b\right)\right|\right|=\sqrt{\frac{a^2+b^2}{{\left(a^2+b^2\right)}^2}}=\sqrt{\frac{1}{a^2+b^2}}.$$

Since $\frac{1}{a^2+b^2}\to \infty$ as $\left(a,b\right)\to \left(0,0\right),$ then $\left|\left|\text{F}\left(a,b\right)\right|\right|\to \infty$ as $\left(a,b\right)\to \left(0,0\right).$ This vector field looks similar to the vector field in Example 6.3, but in this case the magnitudes of the vectors close to the origin are large. The table below shows a sample of points and the corresponding vectors, and Figure 6.6 shows the vector field. Note that this vector field models the whirlpool motion of the river in Figure 6.2(b). The domain of this vector field is all of ${ℝ}^2$ except for point $\left(0,0\right).$

$\left(x,y\right)$	$\text{F}\left(x,y\right)$	$\left(x,y\right)$	$\text{F}\left(x,y\right)$	$\left(x,y\right)$	$\text{F}\left(x,y\right)$
$\left(1,0\right)$	$\langle 0,\mathrm{-1}\rangle$	$\left(2,0\right)$	$\langle 0,-\frac{1}{2}\rangle$	$\left(1,1\right)$	$\langle \frac{1}{2},-\frac{1}{2}\rangle$
$\left(0,1\right)$	$\langle 1,0\rangle$	$\left(0,2\right)$	$\langle \frac{1}{2},0\rangle$	$\left(\mathrm{-1},1\right)$	$\langle \frac{1}{2},\frac{1}{2}\rangle$
$\left(\mathrm{-1},0\right)$	$\langle 0,1\rangle$	$\left(\mathrm{-2},0\right)$	$\langle 0,\frac{1}{2}\rangle$	$\left(\mathrm{-1},\mathrm{-1}\right)$	$\langle -\frac{1}{2},\frac{1}{2}\rangle$
$\left(0,\mathrm{-1}\right)$	$\langle \mathrm{-1},0\rangle$	$\left(0,\mathrm{-2}\right)$	$\langle -\frac{1}{2},0\rangle$	$\left(1,\mathrm{-1}\right)$	$\langle -\frac{1}{2},-\frac{1}{2}\rangle$

Figure 6.6 A visual representation of vector field $\text{F}(x,y)=\frac{y}{x^2+y^2}\text{i}-\frac{x}{x^2+y^2}\text{j}.$ This vector field could be used to model whirlpool motion of a fluid.

We have examined vector fields that contain vectors of various magnitudes, but just as we have unit vectors, we can also have a unit vector field. A vector field F is a unit vector field if the magnitude of each vector in the field is 1. In a unit vector field, the only relevant information is the direction of each vector.

Why are unit vector fields important? Suppose we are studying the flow of a fluid, and we care only about the direction in which the fluid is flowing at a given point. In this case, the speed of the fluid (which is the magnitude of the corresponding velocity vector) is irrelevant, because all we care about is the direction of each vector. Therefore, the unit vector field associated with velocity is the field we would study.

If $\text{F}=\langle P,Q,R\rangle$ is a vector field, then the corresponding unit vector field is $\langle \frac{P}{\left|\left|\text{F}\right|\right|},\frac{Q}{\left|\left|\text{F}\right|\right|},\frac{R}{\left|\left|\text{F}\right|\right|}\rangle .$ Notice that if $\text{F}\left(x,y\right)=\langle y,\text{−}x\rangle$ is the vector field from Example 6.3, then the magnitude of F is $\sqrt{x^2+y^2},$ and therefore the corresponding unit vector field is the field G from the previous example.

If F is a vector field, then the process of dividing F by its magnitude to form unit vector field $\text{F}\text{/}\left|\left|\text{F}\right|\right|$ is called normalizing the field F.

Vector Fields in ${ℝ}^3$

We have seen several examples of vector fields in ${ℝ}^2;$ let’s now turn our attention to vector fields in ${ℝ}^3.$ These vector fields can be used to model gravitational or electromagnetic fields, and they can also be used to model fluid flow or heat flow in three dimensions. A two-dimensional vector field can really only model the movement of water on a two-dimensional slice of a river (such as the river’s surface). Since a river flows through three spatial dimensions, to model the flow of the entire depth of the river, we need a vector field in three dimensions.

The extra dimension of a three-dimensional field can make vector fields in ${ℝ}^3$ more difficult to visualize, but the idea is the same. To visualize a vector field in ${ℝ}^3,$ plot enough vectors to show the overall shape. We can use a similar method to visualizing a vector field in ${ℝ}^2$ by choosing points in each octant.

Just as with vector fields in ${ℝ}^2,$ we can represent vector fields in ${ℝ}^3$ with component functions. We simply need an extra component function for the extra dimension. We write either

$$\text{F}\left(x,y,z\right)=\langle P\left(x,y,z\right),Q\left(x,y,z\right),R\left(x,y,z\right)\rangle$$

(6.3)

or

$$\text{F}\left(x,y,z\right)=P\left(x,y,z\right)\text{i}+Q\left(x,y,z\right)\text{j}+R\left(x,y,z\right)\text{k}.$$

(6.4)

Example 6.7

Sketching a Vector Field in Three Dimensions

Describe vector field $\text{F}\left(x,y,z\right)=\langle 1,1,z\rangle .$

Solution

For this vector field, the x and y components are constant, so every point in ${ℝ}^3$ has an associated vector with x and y components equal to one. To visualize F, we first consider what the field looks like in the xy-plane. In the xy-plane, $z=0.$ Hence, each point of the form $\left(a,b,0\right)$ has vector $\langle 1,1,0\rangle$ associated with it. For points not in the xy-plane but slightly above it, the associated vector has a small but positive z component, and therefore the associated vector points slightly upward. For points that are far above the xy-plane, the z component is large, so the vector is almost vertical. Figure 6.7 shows this vector field.

Figure 6.7 A visual representation of vector field $\text{F}\left(x,y,z\right)=\langle 1,1,z\rangle .$

Checkpoint 6.6

Sketch vector field $\text{G}\left(x,y,z\right)=\langle 2,\frac{z}{2},1\rangle .$

In the next example, we explore one of the classic cases of a three-dimensional vector field: a gravitational field.

Example 6.8

Describing a Gravitational Vector Field

Newton’s law of gravitation states that $\text{F}=\text{−}G\frac{m_1{m}_2}{r^2}\widehat{r},$ where G is the universal gravitational constant. It describes the gravitational field exerted by an object (object 1) of mass $m_1$ located at the origin on another object (object 2) of mass $m_2$ located at point $\left(x,y,z\right).$ Field F denotes the gravitational force that object 1 exerts on object 2, r is the distance between the two objects, and $\widehat{r}$ indicates the unit vector from the first object to the second. The minus sign shows that the gravitational force attracts toward the origin; that is, the force of object 1 is attractive. Sketch the vector field associated with this equation.

Solution

Since object 1 is located at the origin, the distance between the objects is given by $r=\sqrt{x^2+y^2+z^2}.$ The unit vector from object 1 to object 2 is $\widehat{r}=\frac{\langle x,y,z\rangle }{\left|\left|\langle x,y,z\rangle \right|\right|},$ and hence $\widehat{r}=\langle \frac{x}{r},\frac{y}{r},\frac{z}{r}\rangle .$ Therefore, gravitational vector field F exerted by object 1 on object 2 is

$$\text{F}=\text{−}G{m}_1{m}_2\langle \frac{x}{r^3},\frac{y}{r^3},\frac{z}{r^3}\rangle .$$

This is an example of a radial vector field in ${ℝ}^3.$

Figure 6.8 shows what this gravitational field looks like for a large mass at the origin. Note that the magnitudes of the vectors increase as the vectors get closer to the origin.

Figure 6.8 A visual representation of gravitational vector field $\text{F}=\text{−}G{m}_1{m}_2\langle \frac{x}{r^3},\frac{y}{r^3},\frac{z}{r^3}\rangle$ for a large mass at the origin.

Checkpoint 6.7

The mass of asteroid 1 is 750,000 kg and the mass of asteroid 2 is 130,000 kg. Assume asteroid 1 is located at the origin, and asteroid 2 is located at $\left(15,\mathrm{-5},10\right),$ measured in units of 10 to the eighth power kilometers. Given that the universal gravitational constant is $G=6.67384\times {10}^{\mathrm{-11}}{\text{m}}^3{\text{kg}}^{\mathrm{-1}}{\text{s}}^{\mathrm{-2}},$ find the gravitational force vector that asteroid 1 exerts on asteroid 2.

Gradient Fields

In this section, we study a special kind of vector field called a gradient field or a conservative field. These vector fields are extremely important in physics because they can be used to model physical systems in which energy is conserved. Gravitational fields and electric fields associated with a static charge are examples of gradient fields.

Recall that if $f$ is a (scalar) function of x and y, then the gradient of $f$ is

$$\text{grad}f=\text{∇}f=f_x(x,y)\text{i}+f_y(x,y)\text{j}.$$

We can see from the form in which the gradient is written that $\text{∇}f$ is a vector field in ${ℝ}^2.$ Similarly, if $f$ is a function of x, y, and z, then the gradient of $f$ is

$$\text{grad}f=\text{∇}f=f_x(x,y,z)\text{i}+f_y(x,y,z)\text{j}+f_z(x,y,z)\text{k}.$$

The gradient of a three-variable function is a vector field in ${ℝ}^3.$

A gradient field is a vector field that can be written as the gradient of a function, and we have the following definition.

Example 6.9

Sketching a Gradient Vector Field

Use technology to plot the gradient vector field of $f\left(x,y\right)=x^2{y}^2.$

Solution

The gradient of $f$ is $\text{∇}f=\langle 2x{y}^2,2x^{2}y\rangle .$ To sketch the vector field, use a computer algebra system such as Mathematica. Figure 6.9 shows $\text{∇}f.$

Figure 6.9 The gradient vector field is $\text{∇}f,$ where $f\left(x,y\right)=x^2{y}^2.$

Consider the function $f\left(x,y\right)=x^2{y}^2$ from Example 6.9. Figure 6.11 shows the level curves of this function overlaid on the function’s gradient vector field. The gradient vectors are perpendicular to the level curves, and the magnitudes of the vectors get larger as the level curves get closer together, because closely grouped level curves indicate the graph is steep, and the magnitude of the gradient vector is the largest value of the directional derivative. Therefore, you can see the local steepness of a graph by investigating the corresponding function’s gradient field.

Figure 6.10 The gradient field of $f\left(x,y\right)=x^2{y}^2$ and several level curves of $f.$ Notice that as the level curves get closer together, the magnitude of the gradient vectors increases.

As we learned earlier, a vector field $\text{F}$ is a conservative vector field, or a gradient field if there exists a scalar function $f$ such that $\text{∇}f=\text{F}.$ In this situation, $f$ is called a potential function for $\text{F}.$ Conservative vector fields arise in many applications, particularly in physics. The reason such fields are called conservative is that they model forces of physical systems in which energy is conserved. We study conservative vector fields in more detail later in this chapter.

You might notice that, in some applications, a potential function $f$ for F is defined instead as a function such that $\text{−}\text{∇}f=\text{F}.$ This is the case for certain contexts in physics, for example.