scalar fields and vector fields - iit ropar

Scalar Fields and Vector fields

Definition

• A scalar field is an assignment of a scalar toeach point in region in the space. E.g. thetemperature at a point on the earth is a scalarfield.

• A vector field is an assignment of a vector toeach point in a region in the space. e.g. thevelocity field of a moving fluid is a vector fieldas it associates a velocity vector to each pointin the fluid.

Definition

• A scalar field is a map from D to ℜ, where D is

a subset of ℜn.

• A vector field is a map from D to ℜn, where D

is a subset of ℜn.is a subset of ℜn.

• For n=2: vector field in plane,

• for n=3: vector field in space

• Example: Gradient field

Line integral

• Line integral in a scalar field

• Line integral in a vector field

LINE INTEGRAL IN A SCALAR FIELD

MOTIVATION

A rescue team follows a path in a danger area where for each

position the degree of radiation is defined. Compute the total

amount of radiation gathered by the rescue team along the

path.path.

RESCUE

BASE

Piecewise Smooth Curves


A classic property of gravitational fields is that, subject to

certain physical constraints, the work done by gravity on an

object moving between two points in the field is

independent of the path taken by the object.

One of the constraints is that the path must be a piecewise

smooth curve. Recall that a plane curve C given bysmooth curve. Recall that a plane curve C given by

r(t) = x(t)i + y(t)j, a ≤ t ≤ b

is smooth if

are continuous on [a, b] and not simultaneously 0 on (a, b).


Similarly, a space curve C given by

r(t) = x(t)i + y(t)j + z(t)k, a ≤ t ≤ b

is smooth ifis smooth if

are continuous on [a, b] and not simultaneously 0 on (a, b).

A curve C is piecewise smooth if the interval [a, b] can be

partitioned into a finite number of subintervals, on each of

which C is smooth.

Example 1 – Finding a Piecewise Smooth Parametrization

Find a piecewise smooth parametrization of the

graph of C shown in Figure.

Because C consists of three line segments C1, C2,

and C3, you can construct a smooth parametrization

for each segment and piece them together by

making the last t-value in Ci correspond to the first t-

value in Ci + 1, as follows.

Example 1 – Solution

value in Ci + 1, as follows.

So, C is given by


Because C1, C2, and C3 are smooth, it follows that C is

piecewise smooth.

Parametrization of a curve induces an

orientation to the curve.

For instance, in Example 1, the curve is


For instance, in Example 1, the curve is

oriented such that the positive direction is

from (0, 0, 0), following the curve to (1, 2, 1).

Line Integrals

You will study a new type of integral called a line integral

for which you integrate over a piecewise smooth curve C.

To introduce the concept of a line integral, consider themass of a wire of finite length, given by a curve C inspace.

The density (mass per unit length) of the wire at the point(x, y, z) is given by f(x, y, z).

Line Integrals

Partition the curve C by the points

P0, P1, …, Pn

producing n subarcs, as shown in Figure.producing n subarcs, as shown in Figure.

Line Integrals

The length of the ith subarc is given by ∆si.

Next, choose a point (xi, yi, zi) in each subarc.

If the length of each subarc is small, the total massof the wire can be approximated by the sumof the wire can be approximated by the sum

If you let ||∆|| denote the length of the longestsubarc and let ||∆|| approach 0, it seemsreasonable that the limit of this sum approachesthe mass of the wire.

Line Integrals

Line Integrals

To evaluate a line integral over a plane curve C

given by r(t) = x(t)i + y(t)j, use the fact that

A similar formula holds for a space curve.

Line Integrals

Note that if f(x, y, z) = 1, the line integral gives the arc length of the

curve C. That is,

Evaluate

where C is the line segment shown in Figure.

Example 2 – Evaluating a Line Integral

Begin by writing a parametric form of the

equation of the line segment:

x = t, y = 2t, and z = t, 0 ≤ t ≤ 1.


Therefore, x'(t) = 1, y'(t) = 2, and z'(t) = 1, which

implies that

So, the line integral takes the following form.


Line Integrals

For parametrizations given by r(t) = x(t)i +

y(t)j + z(t)k, it is helpful to remember the form of ds as

• Just as for an ordinary single integral, we can

interpret the line integral of a positive

function as an area.

• In fact, if f(x, y) ≥ 0, represents the area of one side of the “fence” or “curtain” shown here,

( ),Cf x y ds∫

shown here, whose:

– Base is C.

– Height above the point (x, y) is f(x, y).

• Now, let C be a piecewise-smooth curve.

– That is, C is a union of a finite number of smooth

curves C1, C2, …, Cn, where the initial point of Ci+1

is the terminal point of Ci.

• Then, we define the integral of f along C

as the sum of the integrals of f along each

of the smooth pieces of C:

( ),f x y ds∫ ( )

( ) ( )

( )1 2

,

, ,

... ,n

C

C C

C

f x y ds

f x y ds f x y ds

f x y ds

= +

+ +

∫∫ ∫

∫

LINE INTEGRAL IN A VECTOR FIELD

MOTIVATION

A ship sails from an island to another one along a fixed

route. Knowing all the sea currents, how much fuel will

be needed ?

One of the most important physical applications

of line integrals is that of finding the work done

on an object moving in a force field.

For example, Figure shows an inverse square

force field similar to the gravitational field of the

sun.sun.

Line Integrals of Vector Fields

To see how a line integral can be used to find

work done in a force field F, consider an object

moving along a path C in the field, as shown in

Figure.

Figure15.13


To determine the work done by the force, you needconsider only that part of the force that is acting in thesame direction as that in which the object is moving.

This means that at each point on C, you can consider theprojection F � T of the force vector F onto the unit tangentprojection F � T of the force vector F onto the unit tangentvector T.

On a small subarc of length ∆si, the increment of work is

∆Wi = (force)(distance)

≈ [F(xi, yi, zi) � T(xi, yi, zi)] ∆si

where (xi, yi, zi) is a point in the ith subarc.


Consequently, the total work done is given by the

following integral.

This line integral appears in other contexts and is

the basis of the following definition of the line

integral of a vector field.

Note in the definition that

Find the work done by the force field

on a particle as it moves along the helix given by

Example – Work Done by a Force

from the point (1, 0, 0) to (–1, 0, 3π),

as shown in Figure.

Because

r(t) = x(t)i + y(t)j + z(t)k

= cos ti + sin tj + tk

it follows that x(t) = cos t, y(t) = sin t, and z(t) = t.

Example – Solution

So, the force field can be written as

To find the work done by the force field in moving a

particle along the curve C, use the fact that

r'(t) = –sin ti + cos tj + k

and write the following.



For line integrals of vector functions, the

orientation of the curve C is important.

If the orientation of the curve is reversed, the If the orientation of the curve is reversed, the

unit tangent vector T(t) is changed to –T(t), and

you obtain

Line Integrals in Differential FormLine Integrals in Differential Form

Line Integrals in Differential Form

A second commonly used form of line integrals is

derived from the vector field notation used in the

preceding section.

If F is a vector field of the form F(x, y) = Mi + Nj, and C is

given by r(t) = x(t)i + y(t)j, then F • dr is often written as given by r(t) = x(t)i + y(t)j, then F • dr is often written as

M dx + N dy.

Line Integrals in Differential Form

This differential form can be extended to three

variables. The parentheses are often omitted, as

follows.

Example – Evaluating a Line Integral in Differential Form

Let C be the circle of radius 3 given by

r(t) = 3 cos ti + 3 sin tj, 0 ≤ t ≤ 2π

as shown in Figure. Evaluate the line integral


Because x = 3 cos t and y = 3 sin t, you have dx = –3 sin t

dt and dy = 3 cos t dt. So, the line integral is

Suppose instead of being a force field, suppose that F represents

the velocity field of a fluid flowing through a region in space.

Under these circumstances, the integral of F .T along a curve in

the region gives the fluid’s flow along the curve.

EXAMPLE: Finding Circulation Around a Circle

Flux Across a Plane Curve

To find the rate at which a fluid is entering or leaving a

region enclosed by a smooth curve C in the xy-plane, we

calculate the line integral over C of F.n, the scalar

component of the fluid’s velocity field in the direction of

the curve’s outward-pointing normal vector.

Notice the difference between flux and

circulation: Flux is the integral of the normal

component of F; circulation is the integral of

the tangential component of F.the tangential component of F.

How to evaluate Flux of F across C

we choose a smoothparameterization

that traces the curve C exactlyonce as t increases from a to b. Wecan find the outward unit normalvector n by crossing the curve’svector n by crossing the curve’sunit tangent vector T with thevector k.

• If the motion is clockwise, k×Tpoints outward;

• if the motion is counterclockwise, T×k points outward

We choose: n = T × k

Now,

Here the circle on the integral shows that the integration around the closedcurve C is to be in the counterclockwise direction.

EXAMPLE: Finding Flux Across a Circle

Note that the flux of F across the circle is positive, implies the net flow across the curve

is outward. A net inward flow would have given a negative flux.

Path Independence

Under differentiability conditions, a field F is conservative iff it is the

gradient field of a scalar function ƒ; i.e., iff for some ƒ. The

function ƒ then has a special name.

once we have found a potential function ƒ for a field F,

we can evaluate all the work integrals in the domain of

F over any path between A and B by

Connectivity and Simple Connectivity

• All curves are piecewise smooth, that is,

made up of finitely many smooth pieces

connected end to end.

• The components of F have continuous first • The components of F have continuous first

partial derivatives implies that when

this continuity requirement guarantees that

the mixed second derivatives of the potential

function ƒ are equal.

Simple curve: A curve that doesn’t intersect

itself anywhere between its endpoints.

r(a) = r(b) for a simple closed curve

But r(a) ≠ r(b) when a < t1 < t2 < b

Simply-connected region: A simply-connected region in the

plane is a connected region D such that every simple closed

curve in D encloses only points that are in D.

Intuitively speaking, a simply-connected

region contains no hole and can’t consist

of two separate pieces.of two separate pieces.

An open connected region means that every point can be connected to every other point

by a smooth curve that lies in the region. Note that connectivity and simple connectivity

are not the same, and neither implies the other. Think of connected regions as being in

“one piece” and simply connected regions as not having any “holes that catch loops.”

INDEPENDENCE OF PATH

Suppose C1 and C2 are two piecewise-smooth curves (which are

called paths) that have the same initial point A and terminal point

B. We have

Note: F is conservative on D is equivalent to saying that

the integral of F around every closed path in D is zero. In

other words, the line integral of a conservative vector field

depends only on the initial point and terminal point of a

curve.

EXAMPLE: Finding Work Done by a Conservative Field

Proof that Part 1 Part 2Proof that Part 1 Part 2

If we have two paths from A to B, one of

them can be reversed to make a loop.

Finding Potentials for Conservative Fields

EXAMPLE: Finding a Potential Function

Integrating first equation w.r.t. ‘x’

Differentiating it w.r.t. ‘y’ and equating with the

Computing ‘g’ as a function of ‘y’ givesComputing ‘g’ as a function of ‘y’ gives

Further differentiating ‘f’ w.r.t. to ‘z’ and equating it with

Integrating, we have

EXAMPLE: Showing That a Differential Form Is Exact

scalar fields and vector fields - iit ropar

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