Unit 5, lesson 33

Vector Fields

What a vector field is, how velocity and force fields model flowing fluids and pulling forces, and how gradient fields, potentials and field lines depend on where the field is defined.

  • Theory part 65
  • Practice part 66
  • Vector Fields and Line Integrals

A weather map of wind shows a small arrow at each city, pointing where the air moves and drawn longer where it moves faster. Nobody draws one arrow for the whole country. The wind is a different vector at every point. That assignment of a vector to each point is a vector field, and it is the main object of the rest of this course.

Key terms

Plane and space vector fields

Definition Let DD be a set in R2\mathbb{R}^2. A vector field on DD is a function F\mathbf{F} that assigns to each point (x,y)(x, y) in DD a vector in R2\mathbb{R}^2,

F(x,y)=⟨P(x,y), Q(x,y)⟩.\mathbf{F}(x, y) = \langle P(x, y),\, Q(x, y) \rangle.

A vector field on a set EE in R3\mathbb{R}^3 assigns to each (x,y,z)(x, y, z) in EE a vector ⟨P,Q,R⟩\langle P, Q, R \rangle in R3\mathbb{R}^3. The scalar functions PP, QQ and RR are the component functions. The field is continuous, or C1C^1, on an open set when all its component functions are continuous, or have continuous first partial derivatives, there. To picture F\mathbf{F}, draw the arrow F(x,y)\mathbf{F}(x, y) with its tail at the point (x,y)(x, y).

In plain words A vector field is a rule that hands you an arrow at every point. The input is a point, the output is a vector, and the arrow is drawn starting at the input point.

Example Take F(x,y)=⟨−y,x⟩\mathbf{F}(x, y) = \langle -y, x \rangle on the whole plane. A few values are

PointVector attached there
(1,0)(1, 0)⟨0,1⟩\langle 0, 1 \rangle
(0,1)(0, 1)⟨−1,0⟩\langle -1, 0 \rangle
(1,2)(1, 2)⟨−2,1⟩\langle -2, 1 \rangle
(−2,−2)(-2, -2)⟨2,−2⟩\langle 2, -2 \rangle

At (1,2)(1, 2) the arrow ⟨−2,1⟩\langle -2, 1 \rangle is perpendicular to the position vector ⟨1,2⟩\langle 1, 2 \rangle, since their dot product is −2+2=0-2 + 2 = 0. The same is true at every point, so the arrows circle the origin counterclockwise, and their length x2+y2\sqrt{x^2 + y^2} grows with the distance from the origin. When arrows are drawn shorter than their true length so that a picture stays readable, scaling every arrow by the same factor keeps their relative lengths honest. Some pictures also cap the longest arrows, which keeps every direction correct but no longer shows lengths to scale.

Velocity field

Definition A steady fluid flow in a region EE is described by its velocity field v\mathbf{v}, where v(x,y,z)\mathbf{v}(x, y, z) is the velocity of the fluid particle passing through (x,y,z)(x, y, z), the same at every moment. A particle carried by the flow moves along a path r(t)\mathbf{r}(t) with

r′(t)=v(r(t))\mathbf{r}'(t) = \mathbf{v}\big( \mathbf{r}(t) \big)

for every tt in an interval. Its speed at time tt is ∣v(r(t))∣\lvert \mathbf{v}(\mathbf{r}(t)) \rvert.

In plain words The field tells each drop of fluid how fast and in which direction to move at the place where it is right now.

Example Water in a round tank turns with velocity field v=⟨−y,x⟩\mathbf{v} = \langle -y, x \rangle, in meters per second. Does a particle starting at (1,0)(1, 0) follow r(t)=⟨cos⁡t,sin⁡t⟩\mathbf{r}(t) = \langle \cos t, \sin t \rangle? Differentiate and compare,

r′(t)=⟨−sin⁡t,cos⁡t⟩,v(r(t))=⟨−sin⁡t,cos⁡t⟩.\begin{aligned} \mathbf{r}'(t) &= \langle -\sin t, \cos t \rangle, \\[0.4em] \mathbf{v}\big( \mathbf{r}(t) \big) &= \langle -\sin t, \cos t \rangle. \end{aligned}

They agree for all tt, so yes. The particle moves around the unit circle at speed 1 and completes one lap in 2π≈6.282\pi \approx 6.28 seconds.

Force field

Definition A force field assigns to each point the force that a given test object would feel if placed there. The standard example is the gravitational field of a mass MM at the origin acting on a mass mm. With x=⟨x,y,z⟩\mathbf{x} = \langle x, y, z \rangle and k=GMm>0k = GMm > 0, where GG is the gravitational constant,

F(x)=−k∣x∣3 x,x≠0.\mathbf{F}(\mathbf{x}) = -\frac{k}{\lvert \mathbf{x} \rvert^3} \, \mathbf{x}, \qquad \mathbf{x} \ne \mathbf{0}.

The vector points toward the origin and has magnitude k∣x∣2\frac{k}{\lvert \mathbf{x} \rvert^2}, so it obeys an inverse square law. The electric force between two charges has the same form, with a constant that may have either sign.

In plain words Put a small object at a point and the field tells you the push or pull it feels there. For gravity, the pull aims at the central mass and weakens with the square of the distance.

Example With k=8k = 8, at the point (0,0,2)(0, 0, 2) the length of x\mathbf{x} is 22, so

F(0,0,2)=−823⟨0,0,2⟩=⟨0,0,−2⟩.\begin{aligned} \mathbf{F}(0, 0, 2) &= -\frac{8}{2^3} \langle 0, 0, 2 \rangle \\[0.4em] &= \langle 0, 0, -2 \rangle. \end{aligned}

The force points straight down toward the origin with magnitude 2=8222 = \frac{8}{2^2}. At the point (0,0,4)(0, 0, 4), twice as far away, the magnitude is 816=12\frac{8}{16} = \frac{1}{2}, a quarter as much.

Gradient field

Definition If ff is a scalar function with first partial derivatives on an open set DD, its gradient field is

∇f(x,y,z)=⟨fx, fy, fz⟩,\nabla f(x, y, z) = \langle f_x,\, f_y,\, f_z \rangle,

and similarly ∇f=⟨fx,fy⟩\nabla f = \langle f_x, f_y \rangle in the plane. At a point where ∇f\nabla f is nonzero and ff is differentiable, the gradient points in the direction in which ff increases fastest and is perpendicular to the velocity of every differentiable curve that stays in the level set of ff and passes through that point. When ff is C1C^1 that level set is, near the point, a smooth curve or surface.

In plain words Every scalar function gives a vector field for free. At each point the arrow points uphill on ff, at right angles to the level curve or level surface.

Example For f(x,y,z)=x2y+zf(x, y, z) = x^2 y + z,

∇f=⟨2xy, x2, 1⟩.\nabla f = \langle 2xy,\, x^2,\, 1 \rangle.

At (1,2,0)(1, 2, 0) this is ⟨4,1,1⟩\langle 4, 1, 1 \rangle. That vector is normal to the level surface x2y+z=2x^2 y + z = 2 at the point (1,2,0)(1, 2, 0).

Potential

Definition A vector field F\mathbf{F} on an open set DD is conservative, or a gradient field, if there is a differentiable function ff on DD with ∇f=F\nabla f = \mathbf{F} at every point of DD. Such an ff is called a potential function for F\mathbf{F}. If DD is also connected, any two potentials differ by a constant. Physicists often use the potential energy V=−fV = -f, so that F=−∇V\mathbf{F} = -\nabla V. Not every field has a potential. If F=⟨P,Q⟩\mathbf{F} = \langle P, Q \rangle is C1C^1 on DD and has a potential there, then the potential has continuous second partials and Clairaut’s theorem forces Py=QxP_y = Q_x on DD.

In plain words A potential is a scalar function whose gradient is the given field. Finding one turns a vector question into a scalar one.

Example For the gravitational field with k=8k = 8, the function f(x)=8∣x∣f(\mathbf{x}) = \frac{8}{\lvert \mathbf{x} \rvert} is a potential on R3\mathbb{R}^3 without the origin. Indeed

∂∂x8x2+y2+z2=−8x∣x∣3,\frac{\partial}{\partial x} \frac{8}{\sqrt{x^2 + y^2 + z^2}} = -\frac{8x}{\lvert \mathbf{x} \rvert^3},

and the same pattern holds in yy and zz. In contrast, ⟨−y,x⟩\langle -y, x \rangle has no potential on the plane, since Py=−1P_y = -1 and Qx=1Q_x = 1 differ.

Field lines

Definition A field line, also called a flow line or integral curve, of a continuous vector field F\mathbf{F} on an open set DD is a differentiable curve r(t)\mathbf{r}(t) in DD with

r′(t)=F(r(t))\mathbf{r}'(t) = \mathbf{F}\big( \mathbf{r}(t) \big)

for every tt in an interval. In the plane this is the system x′=P(x,y)x' = P(x, y), y′=Q(x,y)y' = Q(x, y). If F\mathbf{F} is C1C^1 on DD, the existence and uniqueness theorem for differential equations gives, for each point pp of DD, exactly one maximal field line with r(0)=p\mathbf{r}(0) = p. So two field lines that share a point lie along the same curve, and different field lines never cross. At a point where F\mathbf{F} is zero the field line is constant.

In plain words A field line is a curve that follows the arrows, with its velocity equal to the arrow at every point. For a velocity field it is the path of a floating particle.

Example For F(x,y)=⟨x,−y⟩\mathbf{F}(x, y) = \langle x, -y \rangle the system x′=xx' = x, y′=−yy' = -y gives

r(t)=⟨x0et, y0e−t⟩.\mathbf{r}(t) = \langle x_0 e^{t},\, y_0 e^{-t} \rangle.

Through (1,2)(1, 2) the field line is r(t)=⟨et,2e−t⟩\mathbf{r}(t) = \langle e^t, 2e^{-t} \rangle. Check that r′(t)=⟨et,−2e−t⟩\mathbf{r}'(t) = \langle e^t, -2e^{-t} \rangle equals F(et,2e−t)\mathbf{F}(e^t, 2e^{-t}). Along it the product xyxy stays equal to 22, so this field line is one branch of the hyperbola xy=2xy = 2. The origin is a constant field line, since F(0,0)=0\mathbf{F}(0, 0) = \mathbf{0}.

Domain and singularities

Definition The domain of a vector field given by a formula is the set of points where every component is defined, unless a smaller domain is stated. A singularity is a point, or a set of points such as a line, where the field is not defined but which is surrounded by points of the domain, typically because the magnitude grows without bound as you approach it. Statements about potentials and field lines always refer to a specific domain, and the answer can change when the domain changes.

In plain words Before computing anything, ask where the formula makes sense. A field that blows up at a point is not defined there, and that gap can matter for every later question.

Example The field

G(x,y)=⟨−yx2+y2, xx2+y2⟩\mathbf{G}(x, y) = \left\langle \frac{-y}{x^2 + y^2},\, \frac{x}{x^2 + y^2} \right\rangle

has domain R2\mathbb{R}^2 without the origin, and ∣G∣=1x2+y2\lvert \mathbf{G} \rvert = \frac{1}{\sqrt{x^2 + y^2}}, which grows without bound near (0,0)(0, 0). On the half plane x>0x > 0 it has the potential arctan⁡(y/x)\arctan(y/x), the polar angle. On the whole punctured plane it has no potential, as the lesson on conservative fields shows, even though Py=QxP_y = Q_x there. The gravitational field also has a singularity at the origin, where its magnitude k∣x∣2\frac{k}{\lvert \mathbf{x} \rvert^2} grows without bound.

Check yourself

  1. A fluid has velocity field v=⟨−2y,2x⟩\mathbf{v} = \langle -2y, 2x \rangle. Show that r(t)=⟨3cos⁡2t,3sin⁡2t⟩\mathbf{r}(t) = \langle 3 \cos 2t, 3 \sin 2t \rangle follows the flow, and find the particle’s speed and the time for one lap.
Show the answer

The derivative is r′(t)=⟨−6sin⁡2t,6cos⁡2t⟩\mathbf{r}'(t) = \langle -6 \sin 2t, 6 \cos 2t \rangle, and v(r(t))=⟨−6sin⁡2t,6cos⁡2t⟩\mathbf{v}(\mathbf{r}(t)) = \langle -6 \sin 2t, 6 \cos 2t \rangle, so they agree. The speed is 6. The circle has circumference 6π6\pi, so one lap takes π\pi time units, which matches the period of cos⁡2t\cos 2t.

  1. Find the gradient field of f(x,y)=ln⁡(x2+y2)f(x, y) = \ln(x^2 + y^2), its domain and its magnitude.
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The gradient is

∇f=⟨2xx2+y2, 2yx2+y2⟩,\nabla f = \left\langle \frac{2x}{x^2 + y^2},\, \frac{2y}{x^2 + y^2} \right\rangle,

defined on the plane without the origin, where it has a singularity. It points straight away from the origin with magnitude 2x2+y2\frac{2}{\sqrt{x^2 + y^2}}.

  1. Which of the fields ⟨y,x⟩\langle y, x \rangle and ⟨y,−x⟩\langle y, -x \rangle on the plane is a gradient field?
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The field ⟨y,x⟩\langle y, x \rangle is ∇(xy)\nabla (xy). For ⟨y,−x⟩\langle y, -x \rangle we get Py=1P_y = 1 and Qx=−1Q_x = -1, which differ, so by Clairaut’s theorem it has no potential.

How this lesson connects

It also uses Vectors, derivatives of curves, gradients and elementary differential equations.

Resources for this lesson