Unit – 5

Vector Calculus

5.1 Vector differentiation

Vector function- A vector function can be defined as below-

If a vector ‘r’ is a function of a scalar variable ‘t’, then-

We define the derivative of a vector function as-

We can denote it as-

Similarly is the second order derivative of

Note- gives the velocity and gives acceleration.

Rules for differentiation-

Example-1: A particle moves along the curve , here ‘t’  is the time. Find its velocity and acceleration at t = 2.

Sol. Here we have-

Velocity at t = 2,

                                  =

Acceleration =

Acceleration at t = 2,

Example-2: If and then find-

1.

2.

Sol. 1. We know that-

2.

5.2. Gradient

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-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.

Which is

Hence proved.

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.

Key takeaways-

1. If a vector ‘r’ is a function of a scalar variable ‘t’, then-

We define the derivative of a vector function as-

2.    

3.    

4.     Gradient of a constant (

5.     If f and g are two scalar point functions, then

6.     If f and g are two scalar point functions, then

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

5.3  Divergence, curl and vector identities

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

graduv = u grad v + v grad u

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-

Example-6: Find the curl of F = ()i +4zj +

Solution:

Curl F=

             =

=(0-4)i-(2x-0)j+(0+1)k

 =(-4)i – (2x)j+1k

 =(-4,-2x,1)

Key takeaways-

5.4  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

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-

Key takeaways-

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

5.5  Solenoidal & Irrotational field

Irrotational field-

An irrotatonal field F is characterised by the following conditions-

1.

2. circulation along every closed surface is zero.

3.

Note- In an irrotational field for which , the vector F can always be expressed as the gradient of a scalar function provided the domain is simply connected.

So that-

Here the scalar function is called the potential.

Solenoidal field-

A  solenoidal field F is characterised by the following conditions-

1.

2. Flux along every closed surface is zero.

3.

Note- In an solenoidal field for which , the vector F can always be expressed as the curl of a vector function V.

So that-

Example-1: Prove that the vector field is irrotational and find its scalar potential.

Sol. As we know that if then field is irrotational.

So that-

So that the field is irrotational and the vector F can be expressed as the gradient of a scalar potential,

That means-

Now-

   ………………… (1)

  ……………………. (2)

Integrating (1) with respect to x, keep ‘y’ as constant-

We get-

      …………….. (3)

Integrating (1) with respect to y, keep ‘x’ as constant-

We get-

      …………….. (4)

Equating (3) and (4)-

and

So that-

Thus, the vector field F is irrotational.

Example-3: Show that the vector field is irrotational and find the scalar potential function.

Sol. Now for irrotational field we need prove- 

So that-

So that the vector field is irrotational.

Now in order to find the scalar potential function-

Key takeaways-

1. An irrotational field F is characterised by the following conditions-

1. 
2. circulation along every closed surface is zero.
3. 

2. A solenoidal field F is characterised by the following conditions-

1.

2. Flux along every closed surface is zero.

3.

5.6  Line surface & 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

; F is conservative.

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

F = ȸ

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-

5.7  Green’s lemma

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-

Now,

Along

Along

Put these values in (2), we get-

               L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

5.8  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

Because 

Example – 2 Show that 

Sol

By divergence theorem,  ..…(1)

Comparing this with  the  given  problem  let 

Hence, by (1)

                                    ………….(2)

Hence,from (2), Weget,

5.9  Stoke’s theorem

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.

References

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers