A map f : Rn R defined by f(x1,x2,…,xn) is called a scalar field.

A map F : R2 R2 defined by F(x,y) = ( F1(x,y) , F2(x,y) ) is called a vector field in the plane.

A map F : R3 R3 defined by F(x,y,z) = ( F1(x,y,z) , F2(x,y,z), F3(x,y,z) ) is called a vector field in space.

The definition of a vector field can be extended to Rn for any n (as on page 285 of the textbook). Often, we will denote a vector field as F(x), where x = (x,y), or x = (x,y,z), etc.

Vector fields can be used to model the flow of fluid through a pipe, heat conductivity, gravitational force fields, etc. (Note that each of the component functions of a vector field is a scalar field.)

We shall consider (unless otherwise stated) vector fields with component functions that have continuous partial derivatives of at least the first order.

Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .

Work Area Sketch

V(1,0) = (0,1)

Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .

Work Area Sketch

V(1,0) = (0,1) V(1,1) = (–1,1)

Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .

Work Area Sketch

V(1,0) = (0,1) V(1,1) = (–1,1) V(0,1) = (–1,0)

Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .

Work Area Sketch

V(1,0) = (0,1) V(1,1) = (–1,1) V(0,1) = (–1,0) V(–1,1) = (–1,–1)

Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .

Work Area Sketch

V(1,0) = (0,1) V(1,1) = (–1,1) V(0,1) = (–1,0) V(–1,1) = (–1,–1)

V(–1,0) = (0,–1)

Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .

Work Area Sketch

V(1,0) = (0,1) V(1,1) = (–1,1) V(0,1) = (–1,0) V(–1,1) = (–1,–1)

V(–1,0) = (0,–1) V(–1,–1) = (1,–1) V(0,–1) = (1,0) V(1,–1) = (1,1)

Continue with the sketch.

Compare with Figure 4.3.3 on page 286.

Sketch the vector field V(x,y) = ( ) = y – x——— , ———x2 + y2 x2 + y2

V(1,0) = (0, –1)

V(0,1) = (1,0)

V(–1,0) = (0,1)

V(0,–1) = (–1,0)

V(2,0) = (0,–1/2)

V(0,2) = (1/2,0)

V(–2,0) = (0,1/2)

V(0,–2) = (–1/2,0)

Compare with Figure 4.3.4 on page 287.

yi – xj——— .x2 + y2

A vector field F(x) is a gradient vector field, if we can find a function f(x) such that f(x) = F(x) .

Determine whether or not the vector field

F(x,y) = ( x / x2 + y2 , y / x2 + y2 ) is a gradient vector field.

In order for F(x,y) = (x / x2 + y2 , y / x2 + y2 ) to be a gradient, we must find f(x,y) so that

fx(x,y) = and fy(x,y) =

If this were possible, then it must be true that fxy(x,y) = fyx(x,y).

It is easy to verify that fxy(x,y) = fyx(x,y) = – xy / (x2 + y2)3/2 .

Consequently, F(x,y) = is a gradient vector field. To actually find f(x,y), we observe thatf(x,y) = and f(x,y) =(x2 + y2)1/2 + k1(y) (x2 + y2)1/2 + k2(x) .

Then, we must have that f(x,y) = (x2 + y2)1/2 + k .

x / x2 + y2 y / x2 + y2 .

( x / x2 + y2 , y / x2 + y2 )

In order for V(x,y) = ( y , – x ) to be a gradient, we must find f(x,y) so that

fx(x,y) = and fy(x,y) = .If this were possible, then it must be true that fxy(x,y) = fyx(x,y).

It is easy to verify that fxy(x,y) = and fyx(x,y) =1 – 1 .

Consequently, V(x,y) = ( y , – x ) is

Determine whether or not the vector field

V(x,y) = ( y , – x ) is a gradient vector field.

– xy

not a gradient vector field.

If F is a vector field, then a flow line for F is a path c(t) such that c (t) = F(c(t)) , that is, F yields the velocity field of the path c(t).

Consider the vector field F(x,y) = – yi + xj .

(a) Is c1(t) = (10cos t , 10sin t) = (10cos t)i + (10sin t)j a flow line of F ?

(b) Is c2(t) = (10cos(2 – t) , 10sin(2 – t)) = (10cos(2 – t))i + (10sin(2 – t))j a flow line of F ?

c1 (t) = F(c1(t)) =

c2 (t) =

F(c2(t)) =

Consequently, c1(t) = (10cos t)i + (10sin t)j is

Consequently, c2(t) = (10cos(2 – t))i + (10sin(2 – t))j is

– (10sin t)i + (10cos t)j – (10sin t)i + (10cos t)j

a flow line of the vector field F .

not a flow line of the vector field F .

(10sin(2 – t))i – (10cos(2 – t))j

(– 10sin(2 – t))i + (10cos(2 – t))j

(d) What other flow lines of the vector field F(x,y) can be found?

In order for c(t) = x(t)i + y(t)j to be a flow line, we must have c (t) = F(c(t)), that is,

x (t)i + y (t)j = – y(t)i + x(t)j .

Solutions to such problems can be suggested by examining a picture of the vector field (or solutions can be found by solving the corresponding system of differential equations).

Paths of the form c(t) = ( r0cos(t + t0) , r0sin(t + t0) ) will work for any constants r0 and t0.

(c) Is c3(t) = (et, e–t) = eti + e–tj a flow line of F ?

c3 (t) = F(c3(t)) =

Consequently, c3(t) = eti + e–tj is not a flow line of the vector field F .

eti – e–tj – e–ti + etj