Unit - 4

Vector calculus

4.1 Differentiation of vectors

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-

We can denote it as-

Similarly, is the second order derivative of

Note- gives the velocity and gives acceleration.

Note-

1. Velocity =

2. Acceleration =

The derivative of the velocity vector V(t) is called acceleration A(t), and it is given as-

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,

Again acceleration-

Now-

Key takeaways-

1. Any vector  can be expressed as-

Here , , are the scalar functions of t.

2. Velocity =

3. Acceleration =

4.2 Scalar and vector point functions Gradient of a scalar field and directional derivative

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

Gradient-

The vector differential operator is written as-

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.

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-

Key takeaways-

1. 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).

2.  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).

Gradient of a constant (

4.3 Divergence and curl of a vector field and their physical interpretations

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-

Key takeaways-

If

then the vector is said to be irrotational.

curl (u

4.4 Integration of vectors, line integral, surface integral, volume integral

The Line Integral

Let- F be vector function defined throughout some region of space and let C be any curve in that region. ṝ is 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) dti – (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 = [y2cos 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 field is conservative if cur͞͞͞͞͞͞F = 0.

Now, curl͞͞͞ F = ̷̷ 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

(y2cos x=z3) i + (2y sin x-4) j + (3xz2 + 2) k =   i +   j + k

= y2cos x + z3, = 2y sin x – 4, = 3xz2 + 2

Now, = dx + dy + dz

= (y2cos x + z3) dx +(2y sin x – 4) dy + (3xz2 + 2) dz

= (y2cos 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 is 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-

Key takeaways-

When the path of integration is closed curve then we use the notation for integration-

  in place of

Surface integral =

The volume integral is denoted by

5.      If in a conservative field 

Then this is the condition for independence of path

4.5 Stoke's and Gauss theorems (without proof) and their simple applications

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.

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), We get,

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,

=

= =

.[

=

References: