UNIT 5

VECTOR CALCULUS

5.1           Scalar and vector fields

Vector function-

Suppose be a function of a scalar variable t, then-

Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.

Any vector   can be expressed as-

Here , , are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =

2. Acceleration =

Example: A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

Sol. Suppose

Now,

At t = 0	4
At t = π/2	-4
At t = 0	|v|=
At t = π/2	|v|=

Again acceleration-

Now-

At t = 0
At t = π/2
At t = 0	|a|=
At t = π/2	|a|=

Scalar point function-

If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.

Note-

Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

Vector point function-

If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.

Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Note-

Del operator-

The del operated is defined as-

Example: show that  where

Sol. Here it is given-

=

Therefore-

         Note-

Hence proved.

5.2           Gradient of a scalar field

Suppose f(x, y, z) be the scalar function and it is continuously differentiable then the vector-

Is called gradient of f and we can write is as grad f.

So that-

Here is a vector which has three components

Properties of gradient-

Property-1:

Proof:

First we will take left hand side

L.H.S =

=

=

=

Now taking R.H.S,

R.H.S. =

=

=

Here-                     L.H.S. = R.H.S.

Hence proved.

Property-2: Gradient of a constant (

Proof:

Suppose

Then

We know that the gradient-

= 0

Property-3: Gradient of the sum and difference of two functions-

If f and g are two scalar point functions, then

Proof:   

L.H.S                                

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

Proof:

So that-

Hence proved.

Property-5: Gradient of the quotient of two functions-

If f and g are two scalar point functions, then-

Proof:

So that-

Example-1: If , then show that

1.

2.

Sol.

Suppose   and

Now taking L.H.S,

Which is

Hence proved.

2.

So that

Example: If then find grad f at the point (1,-2,-1).

Sol.

Now grad f at (1 , -2, -1) will be-       

Example: If then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-                     

Now-

Apply

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

5.3           Directional derivative

Let ϕ be a scalar point function and let ϕ(P) and ϕ(Q) be the values of ϕ at two neighbouring points P and Q in the field. Then,

,  are the directional derivative of ϕ in the direction of the coordinate axes at P.

The directional derivative of ϕ in the direction l, m, n=l   + m+

The directional derivative of ϕ in the direction of =

Example: Find the directional derivative of 1/r in the direction where

Sol. Here

Now,

And

We know that-

So that-

Now,

Directional derivative =

Example: Find the directional derivative of

At the points (3, 1, 2) in the direction of the vector .

Sol. Here it is given that-

Now at the point (3, 1, 2)-

Let   be the unit vector in the given direction, then

  at (3, 1, 2)

Now,

Example: Find the directional derivatives of at the point P(1, 1, 1) in the direction of the line

Sol. Here

Direction ratio of the line are 2, -2, 1

Now directions cosines of the line are-

Which are

Directional derivative in the direction of the line-

5.4           Divergence and Curl of a vector field

Divergence (Definition)-

Suppose is a given continuous differentiable vector function then the divergence of this function can be defined as-

Curl (Definition)-

Curl of a vector function can be defined as-

Note- Irrotational vector-

If then the vector is said to be irrotational.

Vector identities:

Identity-1:  grad uv = u grad v + v grad u

Proof: 

So that

Grad uv = u grad v + v grad u

Identity-2: 

Proof:

Interchanging , we get-

We get by using above equations-

Identity-3   

Proof:   

So that-

Identity-4   

Proof:

So that,

Identity-5                  curl (u

Proof:

So that

Curl (u

Identity-6:

Proof:

So that-

Identity-7:

Proof:

So that-

Example-1: Show that-

1.

2.

Sol. We know that-

2. We know that-

= 0

Example-2: If then find the divergence and curl of .

Sol.  we know that-

Now-

Example-3: Prove that

Note- here is a constant vector and

Sol. Here    and

So that

Now-

So that-

5.5           Line, Surface and Volume integrals

The Line Integral

Let- F be vector function defined throughout some region of space and let C be any curve in that region. ṝis the position vector of a point p (x,y,z) on C then the integral ƪ F .dṝ is called the line integral of  F taken over

Now, since ṝ =xi+yi+zk          

And if F͞ =F1i + F2 j+ F3  K

Q1. Evaluate where F= cos y.i-x siny j and C is the curve y= in the xy plae from (1,0) to (0,1)

Solution: The curve y= i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

=

=

=       =-1

Q2. Evaluate where = (2xy +z2) I +x2j +3xz2 k along the curve x=t, y=t2, z= t3 from (0,0,0) to (1,1,1).

Solution :  F x dr =

Put x=t, y=t2, z= t3

Dx=dt ,dy=2tdt,                dz=3t2dt.

F x dr =

=(3t4-6t8) dt i  – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

=t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

=

=+

Example 3: Prove that   ͞͞͞F = [y2 cos x +z3] i +(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞F  (ii) the work done in moving an object in this field from (0, 1, -1) to (/ 2,-1, 2)

Sol. : (a) The fleld is conservative if cur͞͞͞͞͞͞F = 0.

Now, curl͞͞͞F = ̷̷ X                    / y            / z

Y2COS X +Z3        2y sin x-4     3xz2 + 2

; Cur = (0-0) – (3z2 – 3z2) j + (2y cos x- 2y cos x) k = 0

; F is conservative.

(b) Since F is conservative there exists a scalar potential ȸ such that

F = ȸ

(y2 cos x=z3) i + (2y sin x-4) j + (3xz2 + 2) k =   i +   j + k

= y2 cos x + z3, = 2y sin x – 4, = 3xz2 + 2

Now, = dx + dy + dz

= (y2 cos x + z3) dx +(2y sin x – 4)dy + (3xz2 + 2)dz

= (y2 cos x dx + 2y sin x dy) +(z3dx +3xz2dz) +(- 4 dy) + (2 dz)

=d(y2 sin x + z3x – 4y -2z)

ȸ = y2 sin x +z3x – 4y -2z

(c)  now, work done = .d   ͞r

=  dx + (2y sin x – 4) dy + ( 3xz2 + 2) dz

=   (y2 sin x + z3x – 4y + 2z)    (as shown above)

= [ y2 sin x + z3x – 4y + 2z ]( /2, -1, 2)

= [ 1 +8 + 4 + 4 ] – { - 4 – 2} =4 + 15 

Sums Based on Line Integral

1. Evaluate where =yz i+zx  j+xy k and C is the position of the curve.

= (a cost)i+(b sint)j+ct k , from y=0 to t=π/4.

Soln.  = (a cost)i+(b sint)j+ct k

The parametric eqn. Of the curve are x= a cost, y=b sint, z=ct              (i)

=

Putting values of x,y,z from (i),

Dx=-a sint

Dy=b cost

Dz=c dt

=

=

==

2. Find the circulation of around the curve C where =yi+zj+xk and C is circle .

Soln. Parametric eqn of circle are:

x=a cos

y=a sin

z=0

=xi+yj+zk = a cosi + b cos + 0 k

d=(-a sin i + a cos j)d

Circulation = = +zj+xk). d

=-a sin i + a cos j)d

= =

Surface integrals-

An integral which we evaluate over a surface is called a surface integral.

Surface integral =

Volume integrals-

The volume integral is denoted by

And defined as-

If   , then

Note-

If in a conservative field 

Then this is the condition for independence of path.

Example: Evaluate , where S is the surface of the sphere in the first octant.

Sol. Here-

Which becomes-

Example: Evaluate , where and V is the closed reason bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.

Sol.

Here-   4x + 2y + z = 8

Put y = 0 and z = 0 in this, we get

4x = 8 or x = 2

Limit of x varies from 0 to 2 and y varies from 0 to 4 – 2x

And z varies from 0 to 8 – 4x – 2y

So that-

So that-

Example: Evaluate if V is the region in the first octant bounded by and the plane x = 2 and .

Sol.

x varies from 0 to 2

The volume will be-

5.6           Green’s theorem in a plane

If C be a regular closed curve in the xy-plane and S is the region bounded by C then,

Where P and Q are the continuously differentiable functions inside and on C.

Green’s theorem in vector form-

Example-1: Apply Green’s theorem to evaluate where C is the boundary of the area enclosed by the x-axis and the upper half of circle

Sol. We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P =   and   Q =

Now-

  and

So that by Green’s theorem, we have the following integral-

Example-2: Evaluate by using Green’s theorem, where C is a triangle formed by

Sol. First we will draw the figure-

Here the vertices of triangle OED are (0,0), (

Now by using Green’s theorem-

Here P = y – sinx, and Q  = cosx

So that-

  and

Now-

=

Which is the required answer.

Example-3: Verify green’s theorem in xy-plane for where C is the boundary of the region enclosed by

Sol.

On comparing with green’s theorem,

We get-

P = and Q =

  and

By using Green’s theorem-

  ………….. (1)

And left hand side=

………….. (2)

Now,

Along

Along

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

5.7           Gauss’s divergence theorem

Gauss divergence theorem

If V is the volume bounded by a closed surface S and is a vector point function with continuous derivative-

Then it can be written as-

Where unit vector to the surface S.

Example-1: Prove the following by using Gauss divergence theorem-

1.

2.

Where S is any closed surface having volume V and

Sol. Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

= 3V

2.

Because 

Example – 2 Show that 

Sol

By divergence theorem,  ..…(1)

Comparing  this  with  the  given  problem  let 

Hence, by (1)

                                    ………….(2)

Now ,  

Hence,from (2), Weget,

Example Based on Gauss Divergence Theorem

1. Show that

Soln.  We have Gauss Divergence Theorem

By data, F=  

=(n+3)

2 Prove that  =

Soln. By Gauss Divergence Theorem,

=

= =

.[

=

5.8           Stoke’s theorem (without proofs) and their verification

If is any continuously differentiable vector point function and S is a surface bounded by a curve C, then-

Example-1: Verify stoke’s theorem when and surface S is the part of sphere , above the xy-plane.

Sol.

We know that by stoke’s theorem,

Here C is the unit circle-

So that-

Now again on the unit circle C, z = 0

Dz = 0

Suppose, 

And           

Now

  ……………… (1)

Now-

Curl

Using spherical polar coordinates-

  ………………… (2)

From equation (1) and (2), stoke’s theorem is verified.

Example-2: If and C is the boundary of the triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0), then evaluate by using Stoke’s theorem.

Sol. Here we see that z-coordinates of each vertex of the triangle is zero, so that the triangle lies in the xy-plane and

Now,

Curl

Curl

The equation of the line OB is y = x

Now by stoke’s theorem,

Example-3: Verify Stoke’s theorem for the given function-

Where C is the unit circle in the xy-plane.

Sol. Suppose-

Here

We know that unit circle in xy-plane-

Or

So that,

Now

Curl

Now,

Hence the Stoke’s theorem is verified.