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 .
V(1,0) = (0,1)
Work Area
Sketch
Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .
V(1,0) = (0,1) V(1,1) = (–1,1)
Work Area
Sketch
Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .
V(1,0) = (0,1) V(1,1) = (–1,1) V(0,1) = (–1,0)
Work Area
Sketch
Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .
V(1,0) = (0,1) V(1,1) = (–1,1) V(0,1) = (–1,0) V(–1,1) = (–1,–1)
Work Area
Sketch
Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .
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)
Work Area
Sketch
Sketch the vector field V(x,y) = (– y , x ) = – yi + xj .
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)
Work Area
Sketch
Continue with the sketch.
Compare with Figure 4.3.3 on page 286.
y
–x
yi – xj
Sketch the vector field V(x,y) = ( ——— , ——— ) = ——— .
x2 + y2 x2 + y2
x 2 + 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.
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) = x /  x2 + y2
and
fy(x,y) = y /  x2 + y2 .
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) = ( x /  x2 + y2 , y /  x2 + y2 ) is a gradient vector
field. To actually find f(x,y), we observe that
f(x,y) = (x2 + y2)1/2 + k1(y)
and
f(x,y) = (x2 + y2)1/2 + k2(x) .
Then, we must have that f(x,y) = (x2 + y2)1/2 + k .
Determine whether or not the vector field
V(x,y) = ( y , – x )
is a gradient vector field.
In order for V(x,y) = ( y , – x ) to be a gradient, we must find f(x,y) so that
fx(x,y) = y
and
fy(x,y) = – x
.
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) = 1 and fyx(x,y) = – 1 .
Consequently, V(x,y) = ( y , – x ) is 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 ?
c1  (t) = – (10sin t)i + (10cos t)j
F(c1(t)) = – (10sin t)i + (10cos t)j
Consequently, c1(t) = (10cos t)i + (10sin t)j is
a flow line of the vector field F .
(b) Is c2(t) = (10cos(2 – t) , 10sin(2 – t)) =
(10cos(2 – t))i + (10sin(2 – t))j a flow line of F ?
c2  (t) = (10sin(2 – t))i – (10cos(2 – t))j
F(c2(t)) = (– 10sin(2 – t))i + (10cos(2 – t))j
Consequently, c2(t) = (10cos(2 – t))i + (10sin(2 – t))j is
not a flow line of the vector field F .
(c) Is c3(t) = (et, e–t) = eti + e–tj a flow line of F ?
c3  (t) = eti – e–tj
F(c3(t)) = – e–ti + etj
Consequently, c3(t) = eti + e–tj is not a flow line of the vector field F .
(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.