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# A map f : R n  R defined by f ( x 1 ,x 2 ,…, x n ) is called a scalar field . - PowerPoint PPT Presentation

A map f : R n  R defined by f ( x 1 ,x 2 ,…, x n ) is called a scalar field. A map F : R 2  R 2 defined by F ( x,y ) = ( F 1 ( x,y ) , F 2 ( x,y ) ) is called a vector field in the plane.

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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

——— , ———

x2 + y2 x2 + y2

yi – xj

——— .

x2 + y2

Sketch the vector field V(x,y) = ( ) =

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) = and fy(x,y) =

x /  x2 + y2y /  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) = is a gradient vector field. To actually find f(x,y), we observe that

f(x,y) = and f(x,y) =

( x /  x2 + y2 , y /  x2 + y2 )

(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 .

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) = and fy(x,y) = .

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) = and fyx(x,y) =

1

– 1 .

Consequently, V(x,y) = ( y , – x ) is

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) =

– (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 .

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.