Unit - 4

Vector differential calculus

4.1 Vector triple product, Product of four vectors

VECTOR PRODUCT OR CROSS PRODUCT:

The vector, or cross product of two vectors and is defined to be a vector such that

(i) Its magnitude is | || |sin , where is the angle between .

(ii) Its direction is perpendicular to both vectors and

(iii) It forms with a right handed system.

Let be a unit vector perpendicular to both the vectors and

||

Useful results:

Since , , are three mutually perpendicular unit vectors, then

SCALAR TRIPLE PRODUCT:

Let and be three vectors then their dot product is written as

If

Similarly, have the same value.

The value of the product depends upon the cyclic order of the vector, but is c independent of the position of the dot and cross. These may be interchanged.

The value of the product changes if the order is non-cyclic.

VECTOR PRODUCT OF THREE VECTORS:

Let and be three vectors then their vector product is written as

Let

Example: prove that

Sol:

Here we have

= 0 + 0 + 0

= 0

Example: prove that

Sol:

Let

Now

VECTOR PRODUCT OF FOUR VECTORS:

Let , and   be four vectors then their vector product is written as

Example: show that

Sol:

LHS

Hence proved

4.2 Scalar point function, Vector point Function

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

Tangential and normal accelerations-

It is very important to understand that the magnitude of acceleration is not always the rate of change of |V|.

Suppose be the vector-valued function which denotes the position of any object as a function of time.

Then

Then the tangential and normal component of acceleration are given as below-

Example: A object move in the path where t is the time in seconds and distance is measured in feets.

Then find and as functions of t.

Sol.

We know that-

And

Now we will use-

And now-

Key takeaways-

1. Any vector  can be expressed as-

Here , , are the scalar functions of t.

1. Velocity =

2. Acceleration =

3. 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).

4. 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).

4.3 Vector differentiation, Gradient, Divergence and Curl

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-

1.

2.

3.

4.

5.

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-

Then, velocity

Velocity at t = 2,

=

Acceleration =

Acceleration at t = 2,

Example-2: If and then find-

1.

2.

Sol. 1. We know that-

2.

Example-3: 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|=

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-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.

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-

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-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-

Example-4: Find the curl of F(x,y,z) = 3i+2zj-xk

Ans.

Curl F =

=

= i -

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

= -2i+j

Example-5: What is the curl of the vector field F= ( x +y +z ,x-y-z,)?

Solution:

Curl F =

=

=

= (2y+1)i-(2x-1)j+(1-1)k

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

= (2y+1, 1-2x,0)

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-

1. 
2. 
3. 
4. 
5. Curl (u
6. 
7. 

4.4 Directional derivatives with their physical interpretation

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-

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 =

4.5 Solenoidal and irrotational motions

Irrotational field-

An irrotational 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-

Example-2: Prove that the vector field is solenoidal and irrotational.

Sol. We know that if  then the vector field will be solenoidal.

So that-

=

So that the vector field is solenoidal.

Now for irrotational field we need prove- 

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.

References:

1. Higher Engineering Mathematics: B. S. Grewal

2. Applied Mathematics Volume I & II: J. N. Wartikar

3. Textbook of Engineering Mathematics: Bali, Iyenger (Laxmi Prakashan)