Vector Fields

Learning Objectives

• Recognize a vector field in a plane or in space.
• Sketch a vector field from a given equation.
• Identify a conservative field and its associated potential function.

Vector fields are an important tool for describing many physical concepts, such as gravitation and electromagnetism, which affect the behavior of objects over a large region of a plane or of space. They are also useful for dealing with large-scale behavior such as atmospheric storms or deep-sea ocean currents. In this section, we examine the basic definitions and graphs of vector fields so we can study them in more detail in the rest of this chapter.

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 (PageIndex{1}) gives visual representations of such phenomena.

Figure (PageIndex{1a}) 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 (PageIndex{1b}) 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.

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).

DEFINITION: vector field

• A vector field (vecs{F}) in (ℝ^2) is an assignment of a two-dimensional vector (vecs{F}(x,y)) to each point ((x,y)) of a subset (D) of (ℝ^2). The subset (D) is the domain of the vector field.
• A vector field (vecs{F}) in (ℝ^3) is an assignment of a three-dimensional vector (vecs{F}(x,y,z)) to each point ((x,y,z)) of a subset (D) of (ℝ^3). The subset (D) is the domain of the vector field.

Vector Fields in (ℝ^2)

A vector field in (ℝ^2) can be represented in either of two equivalent ways. The first way is to use a vector with components that are two-variable functions:

[vecs{F}(x,y)=⟨P(x,y),Q(x,y)⟩]

The second way is to use the standard unit vectors:

[vecs{F}(x,y)=P(x,y) ,hat{mathbf i}+Q(x,y) ,hat{mathbf j}.]

A vector field is said to be continuous if its component functions are continuous.

Example (PageIndex{1}): Finding a Vector Associated with a Given Point

Let (vecs{F} (x,y)=(2y^2+x−4),hat{mathbf i}+cos(x),hat{mathbf j}) be a vector field in (ℝ^2). Note that this is an example of a continuous vector field since both component functions are continuous. What vector is associated with point ((0,−1))?

Solution

Substitute the point values for (x) and (y):

[egin{align*} vecs{F} (0,-1) &=(2{(−1)}^2+0−4) ,hat{mathbf i}+cos(0) ,hat{mathbf j} [4pt] &=−2 ,hat{mathbf i} + hat{mathbf j}. end{align*}]

Exercise (PageIndex{1})

Let (vecs{G}(x,y)=x^2y,hat{mathbf i}−(x+y),hat{mathbf j}) be a vector field in (ℝ^2). What vector is associated with the point ((−2,3))?

Hint

Substitute the point values into the vector function.

(vecs{G}(−2,3)=12hat{mathbf i}−hat{mathbf j})

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 ((4,−1)) is (⟨3,1⟩). Then, we would draw vector (⟨3,1⟩) at point ((4,−1)).

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 ((x,y)) is perpendicular to the circle centered at the origin that contains point ((x,y)), and all other vectors on this circle have the same magnitude.

Example (PageIndex{2}): Drawing a Radial Vector Field

Sketch the vector field (vecs{F} (x,y)=dfrac{x}{2}hat{mathbf i}+dfrac{y}{2}hat{mathbf 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.

Table (PageIndex{1})
((x,y))(vecs{F}(x,y))((x,y))(vecs{F}(x,y))((x,y))(vecs{F}(x,y))
((1,0))(⟨dfrac{1}{2},0⟩)((2,0))(⟨1,0⟩)((1,1))(⟨dfrac{1}{2},dfrac{1}{2}⟩)
((0,1))(⟨0,dfrac{1}{2}⟩)((0,2))(⟨0,1⟩)((−1,1))(⟨−dfrac{1}{2},dfrac{1}{2}⟩)
((−1,0))(⟨−dfrac{1}{2},0⟩)((−2,0))(⟨−1,0⟩)((−1,−1))(⟨−dfrac{1}{2},−dfrac{1}{2}⟩)
((0,−1))(⟨0,−dfrac{1}{2}⟩)((0,−2))(⟨0,−1⟩)((1,−1))(⟨dfrac{1}{2},−dfrac{1}{2}⟩)

Figure (PageIndex{2a}) shows the vector field. To see that each vector is perpendicular to the corresponding circle, Figure (PageIndex{2b}) shows circles overlain on the vector field.

Exercise (PageIndex{2})

Draw the radial field (vecs{F} (x,y)=−dfrac{x}{3}hat{mathbf i}−dfrac{y}{3}hat{mathbf j}).

Hint

Sketch enough vectors to get an idea of the shape.

In contrast to radial fields, in a rotational field, the vector at point ((x,y)) 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 (PageIndex{3}): Drawing a Rotational Vector Field

Sketch the vector field (vecs{F} (x,y)=⟨y,,−x⟩).

Solution

Create a table (see the one that follows) using a representative sample of points in a plane and their corresponding vectors. Figure (PageIndex{3}) shows the resulting vector field.

Table (PageIndex{2})
((x,y))(vecs{F}(x,y))((x,y))(vecs{F}(x,y))((x,y))(vecs{F}(x,y))
((1,0))(⟨0,−1⟩)((2,0))(⟨0,−2⟩)((1,1))(⟨1,−1⟩)
((0,1))(⟨1,0⟩)((0,2))(⟨2,0⟩)((−1,1))(⟨1,1⟩)
((−1,0))(⟨0,1⟩)((−2,0))(⟨0,2⟩)((−1,−1))(⟨−1,1⟩)
((0,−1))(⟨−1,0⟩)((0,−2))(⟨−2,0⟩)((1,−1))(⟨−1,−1⟩)

Analysis

Note that vector (vecs{F}(a,b)=⟨b,−a⟩) points clockwise and is perpendicular to radial vector (⟨a,b⟩). (We can verify this assertion by computing the dot product of the two vectors: (⟨a,b⟩·⟨−b,a⟩=−ab+ab=0).) Furthermore, vector (⟨b,−a⟩) has length (r=sqrt{a^2+b^2}). Thus, we have a complete description of this rotational vector field: the vector associated with point ((a,b)) 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 (PageIndex{3}) 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 (PageIndex{4}): Sketching a Vector Field

Sketch vector field (vecs{F}(x,y)=dfrac{y}{x^2+y^2}hat{mathbf i}, -dfrac{x}{x^2+y^2}hat{mathbf j}).

Solution

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

(||vecs{F}(a,b)||=sqrt{dfrac{a^2+b^2}{ {(a^2+b^2)}^2 }} =sqrt{dfrac{1}{a^2+b^2}}).

Since (dfrac{1}{a^2+b^2} ightarrow infty) as ((a,b) ightarrow (0,0)), then (||vecs F(a,b)|| ightarrow infty) as ((a,b) ightarrow (0,0)). This vector field looks similar to the vector field in Example (PageIndex{3}), but in this case the magnitudes of the vectors close to the origin are large. Table (PageIndex{3}) shows a sample of points and the corresponding vectors, and Figure (PageIndex{5}) shows the vector field. Note that this vector field models the whirlpool motion of the river in Figure (PageIndex{5})(b). The domain of this vector field is all of (ℝ^2) except for point ((0,0)).

Table (PageIndex{3})
((x,y))(vecs{F}(x,y))((x,y))(vecs{F}(x,y))((x,y))(vecs{F}(x,y))
((1,0))(⟨0,−1⟩)((2,0))(⟨0,−dfrac{1}{2}⟩)((1,1))(⟨dfrac{1}{2},−dfrac{1}{2}⟩)
((0,1))(⟨1,0⟩)((0,2))(⟨dfrac{1}{2},0⟩)((−1,1))(⟨dfrac{1}{2},dfrac{1}{2}⟩)
((−1,0))(⟨0,1⟩)((−2,0))(⟨0,dfrac{1}{2}⟩)((−1,−1))(⟨−dfrac{1}{2},dfrac{1}{2}⟩)
((0,−1))(⟨−1,0⟩)((0,−2))(⟨−dfrac{1}{2},0⟩)((1,−1))(⟨−dfrac{1}{2},−dfrac{1}{2}⟩)

Exercise (PageIndex{4})

Sketch vector field (vecs{F}(x,y)=⟨−2y,,2x⟩). Is the vector field radial, rotational, or neither?

Hint

Substitute enough points into (vecs{F}) to get an idea of the shape.

Rotational

Example (PageIndex{5}): Velocity Field of a Fluid

Suppose that (vecs{v} (x,y)=−dfrac{2y}{x^2+y^2}hat{mathbf i}+dfrac{2x}{x^2+y^2}hat{mathbf j}) is the velocity field of a fluid. How fast is the fluid moving at point ((1,−1))? (Assume the units of speed are meters per second.)

Solution

To find the velocity of the fluid at point ((1,−1)), substitute the point into (vecs{v} ):

(vecs{v}(1,−1)=dfrac{−2(−1)}{1+1}hat{mathbf i}+dfrac{2(1)}{1+1}hat{mathbf j}=hat{mathbf i}+hat{mathbf j}).

The speed of the fluid at ((1,−1)) is the magnitude of this vector. Therefore, the speed is (||hat{mathbf i}+hat{mathbf j}||=sqrt{2}) m/sec.

Exercise (PageIndex{5})

Vector field (vecs{v} (x,y)=⟨4|x|,,1⟩) models the velocity of water on the surface of a river. What is the speed of the water at point ((2,3))? Use meters per second as the units.

Hint

Remember, speed is the magnitude of velocity.

(sqrt{65}) m/sec

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 (vecs{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.

Example (PageIndex{6}): A Unit Vector Field

Show that vector field (vecs{F} (x,y)=leftlangledfrac{y}{sqrt{x^2+y^2}},−dfrac{x}{sqrt{x^2+y^2}} ight angle) is a unit vector field.

Solution

To show that (vecs{F}) is a unit field, we must show that the magnitude of each vector is (1). Note that

[egin{align*} sqrt{ left(dfrac{y}{sqrt{x^2+y^2}} ight)^2+left(−dfrac{x}{sqrt{x^2+y^2}} ight)^2} &=sqrt{ dfrac{y^2}{x^2+y^2}+dfrac{x^2}{x^2+y^2}} [4pt] &=sqrt{dfrac{x^2+y^2}{x^2+y^2}} [4pt] &=1 end{align*}]

Therefore, (vecs{F} ) is a unit vector field.

Exercise (PageIndex{6})

Is vector field (vecs{F} (x,y)=⟨−y,,x⟩) a unit vector field?

Hint

Calculate the magnitude of (vecs{F} ) at an arbitrary point ((x,y)).

No.

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 (vecs{F} =⟨P,Q,R⟩) is a vector field, then the corresponding unit vector field is (iglangle frac{P}{||vecs F||}, frac{Q}{||vecs F||}, frac{R}{||vecs F||}ig angle). Notice that if (vecs{F}(x,y)=⟨y,,−x⟩) is the vector field from Example (PageIndex{6}), then the magnitude of (vecs{F} ) is (sqrt{x^2+y^2}), and therefore the corresponding unit vector field is the field (vecs{G} ) from the previous example.

If (vecs{F} ) is a vector field, then the process of dividing (vecs{F} ) by its magnitude to form unit vector field (vecs{F}/||vecs{F}||) is called normalizing the field (vecs{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

[vecs{F}(x,y,z)=⟨P(x,y,z),Q(x,y,z),R(x,y,z)⟩]

or

[vecs{F}(x,y,z)=P(x,y,z)hat{mathbf i}+Q(x,y,z)hat{mathbf j}+R(x,y,z)hat{mathbf k}.]

Example (PageIndex{7}): Sketching a Vector Field in Three Dimensions

Describe vector field (vecs{F}(x,y,z)=⟨1,,1,,z⟩).

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 (vecs{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 ((a,b,0)) has vector (⟨1,1,0⟩) 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 (PageIndex{6}) shows this vector field.

Figure (PageIndex{6}): A visual representation of vector field (vecs{F}(x,y,z)=⟨1,1,z⟩).

Exercise (PageIndex{7})

Sketch vector field (vecs{G}(x,y,z)=⟨2,,dfrac{z}{2},,1⟩).

Hint

Substitute enough points into the vector field to get an idea of the general shape.

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

Example (PageIndex{8}): Describing a Gravitational Vector Field

Newton’s law of gravitation states that (vecs{F}=−Gdfrac{m_1m_2}{r^2}hat{mathbf 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 ((x,y,z)). Field (vecs{F}) denotes the gravitational force that object 1 exerts on object 2, (r) is the distance between the two objects, and (hat{mathbf 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 (hat{mathbf r}=dfrac{⟨x,y,z⟩}{||⟨x,y,z⟩||}), and hence (hat{mathbf r}=iglangledfrac{x}{r},dfrac{y}{r},dfrac{z}{r}ig angle). Therefore, gravitational vector field (vecs{F}) exerted by object 1 on object 2 is

[ vecs{F}=−Gm_1m_2iglangledfrac{x}{r^3},dfrac{y}{r^3},dfrac{z}{r^3}ig angle. onumber]

This is an example of a radial vector field in (ℝ^3).

Figure (PageIndex{7}) 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.

Exercise (PageIndex{8})

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 ((15,−5,10)), measured in units of 10 to the eighth power kilometers. Given that the universal gravitational constant is (G=6.67384×10^{−11}m^3{kg}^{−1}s^{−2}), find the gravitational force vector that asteroid 1 exerts on asteroid 2.

Hint

Follow Example (PageIndex{8}) and first compute the distance between the asteroids.

(1.49063×{10}^{−18}), (4.96876×{10}^{−19}), (9.93752×{10}^{−19}) N

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

[ ext{grad}, f =vecs abla f(x,y) =f_x(x,y) hat{mathbf i} +f_y(x,y) hat{mathbf j}. ]

We can see from the form in which the gradient is written that (vecs abla 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

[ ext{grad}, f =vecs abla f(x,y,z) = f_x(x,y,z) hat{mathbf i}+f_y(x,y,z) hat{mathbf j}+f_z(x,y,z)hat{mathbf 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.

A vector field (vecs{F}) in (ℝ^2) or in (ℝ^3) is a gradient field if there exists a scalar function (f) such that (vecs abla f=vecs{F}).

Example (PageIndex{9}): Sketching a Gradient Vector Field

Use technology to plot the gradient vector field of (f(x,y)=x^2y^2).

Solution

The gradient of (f) is (vecs abla f(x,y)=⟨2xy^2,,2x^2y⟩). To sketch the vector field, use a computer algebra system such as Mathematica. Figure (PageIndex{8}) shows (vecs abla f).

Exercise (PageIndex{9})

Use technology to plot the gradient vector field of (f(x,y)=sin xcos y).

Hint

Consider the function (f(x,y)=x^2y^2) from Example (PageIndex{9}). Figure (PageIndex{9}) 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.

As we learned earlier, a vector field (vecs{F}) is a conservative vector field, or a gradient field if there exists a scalar function (f) such that (vecs abla f=vecs{F}). In this situation, (f) is called a potential function for (vecs{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 (vecs{F}) is defined instead as a function such that (−vecs abla f=vecs{F}). This is the case for certain contexts in physics, for example.

Example (PageIndex{10}): Verifying a Potential Function

Is (f(x,y,z)=x^2yz−sin(xy)) a potential function for vector field

(vecs{F}(x,y,z)=⟨2xyz−ycos(xy),x^2z−xcos(xy),x^2y⟩)?

Solution

We need to confirm whether (vecs abla f=vecs{F}). We have

[ egin{align*} f_x(x,y) =2xyz−ycos(xy) [4pt] f_y(x,y) =x^2z−xcos(xy) [4pt] f_z(x,y) =x^2y end{align*}.]

Therefore, (vecs abla f=vecs{F}) and (f) is a potential function for (vecs{F}).

Exercise (PageIndex{10})

Is (f(x,y,z)=x^2cos(yz)+y^2z^2) a potential function for (vecs{F}(x,y,z)=⟨2xcos(yz),−x^2z sin(yz)+2yz^2,y^2⟩)?

Hint

No

Example (PageIndex{11}): Verifying a Potential Function

The velocity of a fluid is modeled by field (vecs v(x,y)=⟨xy, frac{x^2}{2}−y⟩). Verify that (f(x,y)=dfrac{x^2y}{2}−dfrac{y^2}{2}) is a potential function for (vecs{v}).

Solution

To show that (f) is a potential function, we must show that (vecs abla f=vecs v). Note that (f_x(x,y)=xy) and (f_y(x,y)=dfrac{x^2}{2}−y). Therefore, (vecs abla f(x,y)=⟨xy, frac{x^2}{2}−y⟩) and (f) is a potential function for (vecs{v}) (Figure (PageIndex{10})).

Exercise (PageIndex{11})

Verify that (f(x,y)=x^2y^2+x) is a potential function for velocity field (vecs{v}(x,y)=⟨3x^2y^2+1,2x^3y⟩).

Hint

(vecs abla f(x,y)=vecs{v}(x,y))

If (vecs{F}) is a conservative vector field, then there is at least one potential function (f) such that (vecs abla f=vecs{F}). But, could there be more than one potential function? If so, is there any relationship between two potential functions for the same vector field? Before answering these questions, let’s recall some facts from single-variable calculus to guide our intuition. Recall that if (k(x)) is an integrable function, then (k) has infinitely many antiderivatives. Furthermore, if (vecs{F}) and (vecs{G}) are both antiderivatives of (k), then (vecs{F}) and (vecs{G}) differ only by a constant. That is, there is some number (C) such that (vecs{F}(x)=vecs{G}(x)+C).

Now let (vecs{F}) be a conservative vector field and let (f) and (g) be potential functions for (vecs{F}). Since the gradient is like a derivative, (vecs{F}) being conservative means that (vecs{F}) is “integrable” with “antiderivatives” (f) and (g). Therefore, if the analogy with single-variable calculus is valid, we expect there is some constant (C) such that (f(x)=g(x)+C). The next theorem says that this is indeed the case.

To state the next theorem with precision, we need to assume the domain of the vector field is connected and open. To be connected means if (P_1) and (P_2) are any two points in the domain, then you can walk from (P_1) to (P_2) along a path that stays entirely inside the domain.

UNIQUENESS OF POTENTIAL FUNCTIONS

Let (vecs{F}) be a conservative vector field on an open and connected domain and let (f) and (g) be functions such that (vecs abla f=vecs{F}) and (vecs abla g=vecs{G}). Then, there is a constant (C) such that (f=g+C).

Proof

Since (f) and (g) are both potential functions for (vecs{F}), then (vecs abla (f−g)=vecs abla f−vecs abla g=vecs{F}−vecs{F}=vecs 0). Let (h=f−g), then we have (vecs abla h=vecs 0).We would like to show that (h) is a constant function.

Assume (h) is a function of (x) and (y) (the logic of this proof extends to any number of independent variables). Since (vecs abla h=vecs 0), we have (h_x(x,y)=0) and (h_y(x,y)=0). The expression (h_x(x,y)=0) implies that (h) is a constant function with respect to (x)—that is, (h(x,y)=k_1(y)) for some function (k_1). Similarly, (h_y(x,y)=0) implies (h(x,y)=k_2(x)) for some function (k_2). Therefore, function (h) depends only on (y) and also depends only on (x). Thus, (h(x,y)=C) for some constant (C) on the connected domain of (vecs{F}). Note that we really do need connectedness at this point; if the domain of (vecs{F}) came in two separate pieces, then (k) could be a constant (C_1) on one piece but could be a different constant (C_2) on the other piece. Since (f−g=h=C), we have that (f=g+C), as desired.

(square)

Conservative vector fields also have a special property called the cross-partial property. This property helps test whether a given vector field is conservative.

THE CROSS-PARTIAL PROPERTY OF CONSERVATIVE VECTOR FIELDS

Let (vecs{F}) be a vector field in two or three dimensions such that the component functions of (vecs{F}) have continuous second-order mixed-partial derivatives on the domain of (vecs{F}).

If (vecs{F}(x,y)=⟨P(x,y),Q(x,y)⟩) is a conservative vector field in (ℝ^2), then

[dfrac{partial P}{partial y}=dfrac{partial Q}{partial x}.]

If (vecs{F}(x,y,z)=⟨P(x,y,z),Q(x,y,z),R(x,y,z)⟩) is a conservative vector field in ({mathbb{R}}^3), then

[ egin{align*} dfrac{partial P}{partial y} =dfrac{partial Q}{partial x} [4pt] dfrac{partial Q}{partial z} =dfrac{partial R}{partial y} [4pt] dfrac{partial R}{partial x} =dfrac{partial P}{partial z}. end{align*}]

Proof

Since (vecs{F}) is conservative, there is a function (f(x,y)) such that (vecs abla f=vecs{F}). Therefore, by the definition of the gradient, (f_x=P) and (f_y=Q). By Clairaut’s theorem, (f_{xy}=f_{yx}), But, (f_{xy}=P_y) and (f_{yx}=Q_{x}), and thus (P_y=Q_x).

(square)

Clairaut’s theorem gives a fast proof of the cross-partial property of conservative vector fields in (ℝ^3), just as it did for vector fields in (ℝ^2).

The Cross-Partial Property of Conservative Vector Fields shows that most vector fields are not conservative. The cross-partial property is difficult to satisfy in general, so most vector fields won’t have equal cross-partials.

Show that rotational vector field (vecs{F}(x,y)=⟨y,,−x⟩) is not conservative.

Solution

Let (P(x,y)=y) and (Q(x,y)=−x). If (vecs{F}) is conservative, then the cross-partials would be equal—that is, (P_y) would equal (Q_x).Therefore, to show that (vecs{F}) is not conservative, check that (P_y≠Q_x). Since (P_y=1) and (Q_x=−1), the vector field is not conservative.

Exercise (PageIndex{12})

Show that vector field (vecs F(x,y)=xy,hat{mathbf i}−x^2y,hat{mathbf j}) is not conservative.

Hint

Check the cross-partials.

(P_y(x,y)=x) and (Q_x(x,y)=−2xy). Since (P_y(x,y) ≠ Q_x(x,y)), (vecs F) is not conservative.

Example (PageIndex{13}): Showing a Vector Field Is Not Conservative

Is vector field (vecs{F}(x,y,z)=⟨7,−2,x^3⟩) conservative?

Solution

Let (P(x,y,z)=7), (Q(x,y,z)=−2), and (R(x,y,z)=x^3). If (vecs{F}) is conservative, then all three cross-partial equations will be satisfied—that is, if (vecs{F}) is conservative, then (P_y) would equal (Q_x), (Q_z) would equal (R_y), and (R_x) would equal (P_z). Note that

[P_y=Q_x=R_y=Q_z=0 onumber]

so the first two necessary equalities hold. However, (R_x(x,y,z)=x^3) and (P_z(x,y,z)=0) so (R_x≠P_z). Therefore, (vecs{F}) is not conservative.

Exercise (PageIndex{13})

Is vector field (vecs{G}(x,y,z)=⟨y,,x,,xyz⟩) conservative?

Hint

Check the cross-partials.

No

We conclude this section with a word of warning: The Cross-Partial Property of Conservative Vector Fields says that if (vecs{F}) is conservative, then (vecs{F}) has the cross-partial property. The theorem does not say that, if (vecs{F}) has the cross-partial property, then (vecs{F}) is conservative (the converse of an implication is not logically equivalent to the original implication). In other words, The Cross-Partial Property of Conservative Vector Fields can only help determine that a field is not conservative; it does not let you conclude that a vector field is conservative.

For example, consider vector field (vecs{F}(x,y)=⟨x^2y,dfrac{x^3}{3}⟩). This field has the cross-partial property, so it is natural to try to use The Cross-Partial Property of Conservative Vector Fields to conclude this vector field is conservative. However, this is a misapplication of the theorem. We learn later how to conclude that (vecs F) is conservative.

Key Concepts

• A vector field assigns a vector (vecs{F}(x,y)) to each point ((x,y)) in a subset (D) of (ℝ^2) or (ℝ^3). (vecs{F}(x,y,z)) to each point ((x,y,z)) in a subset (D) of (ℝ^3).
• Vector fields can describe the distribution of vector quantities such as forces or velocities over a region of the plane or of space. They are in common use in such areas as physics, engineering, meteorology, oceanography.
• We can sketch a vector field by examining its defining equation to determine relative magnitudes in various locations and then drawing enough vectors to determine a pattern.
• A vector field (vecs{F}) is called conservative if there exists a scalar function (f) such that (vecs abla f=vecs{F}).

Key Equations

• Vector field in (ℝ^2)
(vecs{F}(x,y)=⟨P(x,y),,Q(x,y)⟩)
or
(vecs{F}(x,y)=P(x,y) ,hat{mathbf i}+Q(x,y) ,hat{mathbf j})
• Vector field in (ℝ^3)
(vecs{F}(x,y,z)=⟨P(x,y,z),,Q(x,y,z),,R(x,y,z)⟩)
or
(vecs{F}(x,y,z)=P(x,y,z) ,hat{mathbf i} +Q(x,y,z) ,hat{mathbf j}+R(x,y,z) ,hat{mathbf k})

Glossary

conservative field
a vector field for which there exists a scalar function (f) such that (vecs ∇f=vecs{F})
a vector field (vecs{F}) for which there exists a scalar function (f) such that (vecs ∇f=vecs{F}); in other words, a vector field that is the gradient of a function; such vector fields are also called conservative
potential function
a scalar function (f) such that (vecs ∇f=vecs{F})
a vector field in which all vectors either point directly toward or directly away from the origin; the magnitude of any vector depends only on its distance from the origin
rotational field
a vector field in which the vector at point ((x,y)) is tangent to a circle with radius (r=sqrt{x^2+y^2}); in a rotational field, all vectors flow either clockwise or counterclockwise, and the magnitude of a vector depends only on its distance from the origin
unit vector field
a vector field in which the magnitude of every vector is 1
vector field
measured in (ℝ^2), an assignment of a vector (vecs{F}(x,y)) to each point ((x,y)) of a subset (D) of (ℝ^2); in (ℝ^3), an assignment of a vector (vecs{F}(x,y,z)) to each point ((x,y,z)) of a subset (D) of (ℝ^3)

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Vector fields in Processing

In this article I want to program vector fields in Processing. If you are not familiar with vector fields, I suggest you watch this video from Khan Academy:

But in short, vector field is a way to visualize multivariable vector-va l ued functions. We give them several variables as an input and instead of a single number (like scalar valued functions) they return us a whole vector. Vectors have both magnitude (or length) and direction. We can think of their direction as an angle of rotation. Usually, vectors are represented with arrows. Vectors and vector fields can be in as many dimensions as we want, but now we will focus on two-dimensional vector fields.

First of all, we need a mathematical function that we will give us a 2d vector field. For example:

We give this function a point on the coordinate plane and it returns us a vector. Now, if we draw all vectors at the origin it looks very crammed. So, we translate output vector to the input point. However, if we draw them in full length it still might look jam-packed. There are a couple solutions to this problem. First, we can just ignore the magnitude and draw vectors with a fixed length. But then we lose a feeling of their magnitude completely. Second, we can use color. Lastly, we can scale down all vectors so that maximum magnitude is always the same length for all fields. This image from Wolfram alpha shows how the vector field of our example function looks like:

We should think of coordinate plane as a grid and vectors will be assigned to each square. Let’s start coding now.

Actually, we don’t need a lot of global variables:

• len is a length of a side of each square
• cols and rows are the numbers of columns and rows on the grid
• grid is a 2d array of Cell objects that represent each square on the grid.

First, I write a vector field function. Its type is PVector and it receives x and y coordinates of each square. Then I calculate components of a vector: u is x-component and v is y-component. Finally, I just return this vector.

You may think that it is strange that we calculate u and v because u is equal to x and y is equal to v and we can pass them to PVector constructor directly. But it is the case only for our example function. If the formula is different, these components are not equal and it makes sense to calculate them separately.

Now, let’s make the Cell object. It needs two indexes i and j that represent its position in 2d array that I called grid and vector associated with a vector field. I will pass only indexes to the constructor function. Then I will calculate x and y coordinates of the center of a cell and a corresponding vector. Also you can make an argument and magnitude separate variables.

Then, I create a show function that will draw an arrow corresponding to an associated vector. First of all, I check if the magnitude is 0. If it is, there is essentially no vector and we cannot draw anything. And then, I draw an arrow and rotate it by an argument of a vector. Drawing an arrow may sound simple, but it is actually a little bit tricky. Now, let me explain how we can do that.

Let l be a length of an arrow and r be the length of two smaller legs.

First, we translate to the center of a cell. So, (0, 0) is now its center. We draw an arrow so that its center is at (0, 0). Then, it rear points are (-l/2 0) and (l/2, 0).

Smaller legs are rotated relatively to the arrow. I call the angle of rotation theta. Now, we use a bit of trigonometry to calculate coordinates of their ends. We can note that their x-coordinates are the same and their y-coordinates have the same absolute value but opposite signs. Now we can derive a simple system of equations in polar coordinates. Here is my sketch to make it clear.

So, I calculate these points and draw lines. And before that, I rotate everything by the argument of an angle. But it is important to use push() and pop() functions to apply rotation to only one particular arrow. I also added a buffer to make arrows a little bit shorter and not overlap with other arrows. Code should look like this:

Now there is only a tiny step left. We need to make a setup function, there we will draw background, initiate all variables, and draw all vectors.

Vector Fields

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Mathematical Expression Editor

We introduce the idea of a vector at every point in space.

Types of functions

When we started on our journey exploring calculus, we investigated functions . Typically, we interpret these functions as being curves in the -plane:

Now we are ready for a new type of function.

Vector fields

Now we will study vector-valued functions of several variables: We interpret these functions as vector fields, meaning for each point in the -plane we have a vector.

The second choice is not a vector field.

The third choice is not a vector field.

The fourth choice is a constant vector field, and is the correct answer.

Properties of vector fields

As we will see in the chapters to come, there are two important qualities of vector fields that we are usually on the look-out for. The first is rotation and the second is expansion. In the sections to come, we will make precise what we mean by rotation and expansion. In this section we simply seek to make you aware that these are the fundamental properties of vector fields.

Very loosely speaking a radial field is one where the vectors are all pointing toward a spot, or away from a spot. Let’s see some examples of radial vector fields.

Each of the vector fields above is a radial vector field. Let’s give an explicit definition.

Fun fact: Newton’s law of gravitation defines a radial vector field.

Some fields look like they are expanding and are. Other fields look like the are expanding but they aren’t. In the sections to come, we’re going to use calculus to precisely define what we mean by a field “expanding.” This property will be called divergence.

Rotational fields

Vector fields can easily exhibit what looks like “rotation” to the human eye. Let’s show you a few examples.

At this point, we’re going to give some “spoilers.” It turns out that from a local perspective, meaning looking at points very very close to each other, only the first example exhibits “rotation.” While the second example looks like it is “rotating,” as we will see, it does not exhibit “local rotation.” Moreover, in future sections we will see that rotation (even local rotation) in three-dimensional space must always happen around some “axis” like this:

In this final section, we will talk about fields that arise as the gradient of some differentiable function. As we will see in future sections, these are some of the nicest vector fields to work with mathematically.

Let’s take a look at a gradient field.

The shape of things to come

Now we present the beginning of a big idea. By the end of this course, we hope to give you a glimpse of “what’s out there.” For this we’re going to need some notation. Think of and as sets of numbers, like or or or .

• is the set of continuous functions from to .
• is the set of differentiable functions from to whose first-derivative is continuous.
• is the set of differentiable functions from to whose first and second derivatives are continuous.
• is the set of differentiable function from to where the first th derivatives are continuous.
• is the set of differentiable functions from to where all of the derivatives are continuous.

The gradient turns functions of several variables into vector fields.

We can write this with our new notation as:

Now we give a method to determine if a field is a gradient field.

And so we see , and thus is a gradient field. Now let’s try to find a potential function. To do this, we’ll antidifferentiate—in essence we want to “undo” the gradient. Write with me:

where is a function of . In a similar way:

We need to make this happen, we set and . From this we find our potential function is .

If a vector field v admits a vector potential A, then from the equality

(divergence of the curl is zero) one obtains

which implies that v must be a solenoidal vector field.

be a solenoidal vector field which is twice continuously differentiable. Assume that v(x) decreases sufficiently fast as ||x||→∞. Define

Then, A is a vector potential for v, that is,

A generalization of this theorem is the Helmholtz decomposition which states that any vector field can be decomposed as a sum of a solenoidal vector field and an irrotational vector field.

The vector potential admitted by a solenoidal field is not unique. If A is a vector potential for v, then so is

where f is any continuously differentiable scalar function. This follows from the fact that the curl of the gradient is zero.

This nonuniqueness leads to a degree of freedom in the formulation of electrodynamics, or gauge freedom, and requires choosing a gauge.

Conservative Vector Fields

Definition. A vector field $$is said to be conservative in a region D if = abla f for some scalar function f in D. The function f is called a scalar potential of$$ in $D.$

Example. Determine whether the vector field is conservative and if so, find a scalar potential function egin (x,y,z)=y^2+left(2x y+e^<3z> ight)+left(3y e^<3z> ight) end

Solution. If there is such a function $f$ then $f_x(x,y,z)=y^2$, $f_y(x,y,z)=2 x y+e^<3z>$, and $f_z(x,y,z)=3y e^<3z>.$ Integrating $f_x$ with respect to $x,$ $f(x,y,z)=x y^2+g(y,z).$ Then differentiating $f$ with respect to $y,$ we have $f_y(x,y,z)=2x y+g_y(y,z)$ and this yields $g_y(y,z)=e^<3z>.$ Thus $g(y,z)=y e^<3z>+h(z)$ and we have $f(x,y,z)=x y^2+y e^<3z>+h(z).$ Finally, differentiating $f$ with respect to $z$ and comparing, we obtain $h'(z)=0$ and therefore, $h(z)=K,$ a constant. The desired scalar function is $f(x,y,z)=x y^2+y e^<3z>+K$ with $= abla f.$

Definition. A region $D$ in the plane is called connected (one piece) if it has the property: (i) any two points in the region can be connected by a piecewise smooth curve lying entirely within $D$ and a simply connected region (no holes) is a connected region $D$ that has the property: (ii) every closed curve in $D$ encloses only points that are in $D.$

Theorem. (Conservative in Space) Suppose that the vector field $$and mathop are both continuous in the simply connected region D of mathbb^3. Then$$ is conservative in $D$ if and only if $mathop=<0>.$

Theorem. (Conservative in the Plane) Consider the vector field $(x,y)=u(x,y)+v(x,y)$ where $u$ and $v$ have continuous first partials in the open simply connected region $D$ in the plane. Then $(x,y)$ is conservative in $D$ if and only if egin frac=frac end on $D.$

Example. Determine whether the vector field is conservative and if so, find a scalar potential function $(x,y)=2 x y+x y^3 .$

Solution. Since $u(x,y)=2 x y$, $v(x,y)= x y ^3$, and $frac=2x eq y^3=frac,$ we see that $$is not a conservative vector field. Example. Show that the vector field egin (x,y,z)= left(frac<1+x^2>+ an ^<-1>z ight) +left( an ^<-1>x ight) +left(frac<1+z^2> ight) end is conservative and find a scalar potential function. Solution. Since  ext=0, it follows$$ is conservative. Now we set out to find the scalar potential function $f.$ Since eginfrac=frac<1+x^2>+ an ^<-1>zend we set eginf(x,y,z)=y an ^<-1>x+x an ^<-1>z+c(y,z).end Since eginfrac= an ^<-1>z=fracleft(y an ^<-1>x+x an ^ <-1>z+c ight)= an ^<-1>x+frac,end we find $frac=0$ and $c=c_1(z)$ and so we set eginf(x,y,z)=y an ^<-1>x+x an ^<-1>z+c_1(z).end Since eginfrac=frac<1+z^2>=fracleft[y an ^<-1>x+x an ^<-1>z+c_1(z) ight] = frac <1+z^2>+ c’_1(z),end we then find $c_1 ‘(z)=0,$ $c_1=0$ and so we obtain egin f(x,y,z)=y an ^<-1>x+x an ^<-1>z end as desired.

Vector Fields

The best way to introduce vector fields is with an example. Consider the two-dimensional vector field

For each point (x,y) in the xy-plane the function F (x,y) assigns a vector. The coefficient of i is the x component of the vector. The coefficient of j is the y component of the vector. Alternatively, we can use the notation <y,sin x> to denote the vector field. The figure below plots the vector field.

The length of the vector in the plot is proportional to the actual magnitude of the vector.

In general, a vector field in two dimenensions is a function that assigns to each point (x,y) of the xy-plane a two-dimensional vector F (x,y). The standard notation is

Here P(x,y) is the x-component function of the vector field and Q(x,y) is the y-component function of the vector field. In some cases the vector field is only defined for a region D of the xy-plane.

A vector field in three dimensions is a function F that assigns to each point (x,y,z) in xyz-space a three dimensional vector F (x,y,z). The notation is

Again, the vector field may only be defined in a certain region D of xyz-space.

Vector fields arise in a number of disciplines in the physical sciences including

• Mechanics: gravitational fields. At each point the vector field gives the direction and magnitude of the force on a particle.
• Electricity and Magnetism: electric and magnetic fields. At each point the vector field gives the direction and magnitude of the force on a particle.
• Fluid Mechanics: velocity fields. At each point the vector field gives the velocity of a fluid.

A type of vector field arising in a number of applications, including mechanics and electricity and magnetism, is a conservative vector field. In this case the vector field is defined in terms of the gradient of a scalar function f(x,y,z):

Vector Fields versus Vector Functions

Vector fields and vector functions are two different types of functions. Recall that a vector function in three dimensions is denoted r (t)=<f(t),g(t),h(t)>. A vector function has three components, each of which is a function of ONE variable. A vector function represents a curve in space.

A vector field in three dimensions, F (x,y,z)=<f(x,y,z),g(x,y,z),h(x,y,z)>, has three components, each of which is a function of THREE variables. A vector field assigns a vector to each point in a region in xyz space.

Vector Fields

You are about to erase your work on this activity. Are you sure you want to do this?

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Mathematical Expression Editor

We introduce the idea of a vector at every point in space.

Types of functions

When we started on our journey exploring calculus, we investigated functions . Typically, we interpret these functions as being curves in the -plane:

Now we are ready for a new type of function.

Vector fields

Now we will study vector-valued functions of several variables: We interpret these functions as vector fields, meaning for each point in the -plane we have a vector.

The second choice is not a vector field.

The third choice is not a vector field.

The fourth choice is a constant vector field, and is the correct answer.

Properties of vector fields

As we will see in the chapters to come, there are two important qualities of vector fields that we are usually on the look-out for. The first is rotation and the second is expansion. In the sections to come, we will make precise what we mean by rotation and expansion. In this section we simply seek to make you aware that these are the fundamental properties of vector fields.

Very loosely speaking a radial field is one where the vectors are all pointing toward a spot, or away from a spot. Let’s see some examples of radial vector fields.

Each of the vector fields above is a radial vector field. Let’s give an explicit definition.

Fun fact: Newton’s law of gravitation defines a radial vector field.

Some fields look like they are expanding and are. Other fields look like the are expanding but they aren’t. In the sections to come, we’re going to use calculus to precisely define what we mean by a field “expanding.” This property will be called divergence.

Rotational fields

Vector fields can easily exhibit what looks like “rotation” to the human eye. Let’s show you a few examples.

At this point, we’re going to give some “spoilers.” It turns out that from a local perspective, meaning looking at points very very close to each other, only the first example exhibits “rotation.” While the second example looks like it is “rotating,” as we will see, it does not exhibit “local rotation.” Moreover, in future sections we will see that rotation (even local rotation) in three-dimensional space must always happen around some “axis” like this:

In this final section, we will talk about fields that arise as the gradient of some differentiable function. As we will see in future sections, these are some of the nicest vector fields to work with mathematically.

Let’s take a look at a gradient field.

The shape of things to come

Now we present the beginning of a big idea. By the end of this course, we hope to give you a glimpse of “what’s out there.” For this we’re going to need some notation. Think of and as sets of numbers, like or or or .

• is the set of continuous functions from to .
• is the set of differentiable functions from to whose first-derivative is continuous.
• is the set of differentiable functions from to whose first and second derivatives are continuous.
• is the set of differentiable function from to where the first th derivatives are continuous.
• is the set of differentiable functions from to where all of the derivatives are continuous.

The gradient turns functions of several variables into vector fields.

We can write this with our new notation as:

Now we give a method to determine if a field is a gradient field.

And so we see , and thus is a gradient field. Now let’s try to find a potential function. To do this, we’ll antidifferentiate—in essence we want to “undo” the gradient. Write with me:

where is a function of . In a similar way:

We need to make this happen, we set and . From this we find our potential function is .

Vector Field Plotter

A vector function is a function that takes a number of inputs, and returns a vector. For simplicity, let's keep things in 2 dimensions and call those inputs (x) and (y). Mathematically speaking, this can be written as

Where ( hat ) and ( hat ) are unit vectors along the (x) and (y) axes respectively. Then, if we have a grid like the one above, we can systematically pick points on the grid at which to plot the corresponding vector. The end result is known as a vector field.

Our interactive demo allows you to enter any function you like for ( g(x,y) ) and ( h(x,y) ). When the page first loads, these functions are set to

The values ( a,b,c ) and ( d ) just simple constants. They are, however, linked to the sliders, which means you can adjust them and watch how the vector field changes in real time. Don't let them confuse you though, you can safely ignore these if desired.

To enter a new vector function, type your expressions in the ( hat ) and ( hat ) text boxes and press "update expression". The types of functions you can enter are explained in the table below.