Module 3

Vector

3.1Vector function of scalar Quantity

If t is a scalar quantity and if to each value of t is an interval there corresponds a vector then we say that is a vector function of a scalar variable t and we denote it as

3.2 Vector operator Del ∇

We define a new operator ∇ by

∇=i + j+ k

3.3 Gradient

For a real-valued function f(x,y,z)f(x,y,z) on R3R3, the gradient ∇f(x,y,z)∇f(x,y,z) is a vector-valued function on R3R3 , that is, its value at a point (x,y,z)(x,y,z) is the vector

∇f(x,y,z)=(∂f/∂x,∂f/∂y,∂f/∂z)=∂f/∂xi+∂f/∂yj+∂f/∂zk

3.4 Divergence

For example, it is often convenient to write the divergence div f as ∇⋅f∇⋅f, since for a vector field f(x,y,z)=f1(x,y,z)i+f2(x,y,z)j+f3(x,y,z)kf(x,y,z)=f1(x,y,z)i+f2(x,y,z)j+f3(x,y,z)k, the dot product of f with ∇∇ (thought of as a vector) makes sense:

∇⋅f=(∂/∂xi+∂/∂yj+∂/∂zk)⋅(f1(x,y,z)i+f2(x,y,z)j+f3(x,y,z)k)

                         =(∂/∂x)(f1)+(∂/∂y)(f2)+(∂/∂z)(f3)

                         =∂f1/∂x+∂f2/∂y+∂f3/ ∂z

                         =div f

3.5 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 =

3.6 Solenoidal

A solenoidal vector field satisfies

for every vector , where  is the divergence. If this condition is satisfied, there exists a vector , known as the vector potential, such that

where ∇ X A is the curl. This follows from the vector identity

If A is an irrotational field, then

is solenoidal. If  and  are irrotational, then

is solenoidal. The quantity

Where  is the gradient, is always solenoidal. For a function  satisfying Laplace's equation

it follows that ∇ϕ is solenoidal (and also irrotational).

3.7 Irrotational Field

A vector field  for which the curl vanishes

3.8 Scalar Potential

A conservative vector field (for which the curl ) may be assigned a scalar potential

where  is a line integral.

3.9 Sums

Q1. Calculate the curl for the following vector field.

        F⃗ =x3y2 i⃗ +x2y3z4 j⃗ +x2z2 k⃗

Solution:  In order to calculate the curl, we need to recall the formula.

where P, Q, and R correspond to the components of a given vector field: F⃗ =Pi⃗ +Qj⃗ +Rk⃗ 

=((x2z2)−(x2y3z4) )i⃗ +((x3y2)−(xz2) )j⃗ +((x2y3z4)−(x3y2) )k⃗ 

=(0−4x2y3z3)i⃗ +(0−2xz2)j⃗ +(2xy3z4−2x3y)k⃗ 

Thus the curl is

=(−4x2y3z3)i⃗ +(−2xz2)j⃗ +(2xy3z4−2x3y) k⃗

Q2. Find the directional derivative of Θ=x2y cos z at (1,2,π/2) in the direction of a = 2i+3j+2k.

Solution : ∇ϕ = i + k

= 2xy cos zi+ x2 cos zj -x2y sin zk

At (1,2,π/2)        ∇ϕ = 0i +0j-2k

Directional directive in the direction of 2i+3j+2k.

=(0i+0j-2k).     =-

Q3. In what direction from the point (2,1,-1) is the directional derivative of ϕ=x2yz3 maximum? What is its magnitude?

Solution : ∇ϕ = i + k

                         = -4i-4j+12k

Directional derivative is maximum in the direction of ∇Θ. Hence, directional derivative is maximum in the direction of -4i-4j+12k

Its magnitude =    =4

Example 4: Prove that   ͞͞͞F = [y2 cos x +z3] i +(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for  ͞͞͞F  (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 cur ͞͞͞ ͞͞͞F = 0.

Now,  cur ͞͞͞F = ̷̷ X                    / y            / z

                                i                           j                    k

                            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 +3xz2 dz) +(- 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 

Q5. If x2zi – 2y3z3j + xy2z2k find dvi and curl at (1,-1, 1)

Solution : div = ∇.= =

                            =  2xz – 6y2z3 + 2xy2z =(2-6+2) = -2

Curl =

            =i(2xyz2 + 6y3z2) – j(y2z2- x2) + k(0-0)

            =-8 at (1,-1,1)

Q6. Find the angle between the normal to the surface xy = z2 at the points (1,4,2) and (-3,-3,3)

Solution: let ϕ = xy-z2

                   ∇ϕ= i     =yi + xj -2zk =4i + j -4k

                ∇ϕ = 3i – 3j-6k

But these are the normal to the surface at given points. Angle between two vectors is given by (4i + j -4k).( 4i + j -4k)= |4i + j- 4k|.|-3i-3j -6K|cos θ.

If θ is the angle between then cos θ= = .

3.10 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 =F1 i + 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)dt i –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

 =

 =+

Q3. Prove that   ͞͞͞F = [y2 cos x +z3] i +(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for  ͞͞͞F  (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 cur ͞͞͞ ͞͞͞F = 0.

Now, curl ͞͞͞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

(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 +3xz2 dz) +(- 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 

3.11 Problems on work done

Q1. Find the work done in moving a particle once round the complete circle x2+y2=a2 , z=0 in the force filed given by = sin yi +(x + x cos y )j.

Solution: work done= = (x+x cos y )dy

                 Using parametric equation x=a cos θ, y=a sinθ

 Work done   = 

          = .

3.12. Green’s Theorem

If P and Q are two functions of x, y and their partial derivatives ,   are continuous single valued functions over the closed region bounded by a curve C then

 } dx dy.

EXAMPLE – 1     Verify green’s theorem for    and  C  is  the  triangle  having  verticles  A (0,2 ) , B (2,0 ) , C (4,2 ).

SOLUTION:       By green  theorem.

     Here ,          

(a)   Along AB , since the equation  of  AB  is 

Putting 

Along  BC , since  the  equation  of  BC , .

Along  CA , since  the  equation  of  CA,  is  y = 2 , dy = 0.

(b)     

                                                     .

From  (1)  and (2) , the  theorem  is  verified .

Example 2: Evaluate by Green ‘s theorem = - xy (xi –yj) and c is r= a (1+ cos )

Sol : By Green’s Theorem , ) dx dy

Now,  . d ṝ = 2 y i + xy2 j ) . (d xi + dy j) = 2y dx + xy2 dy )

By comparison                            p= - x2y,                           Q = xy2

2, = - x2            2 + x2) dx dy

To evaluate the integral , we put x = r cos , y = r sin for the cardioid  r = a ( 1 + cos ), we take the integral from

2 . r dr d = 2 3 dr d

= ]a ( 1 + cos )                            dθ = (1+ cos θ)4dθ

=8a4 = a4

3.13 Stokes’s Theorem

The integral of the normal component of the curl of a vector F͞ over a surface S is equal to the line integral of the tangential component of F͞ around the curve bounding S i.e

          F͞ )ds =

Q1. Use  stoke’s  theorem  to  evaluate 

SOLUTION :     We  have  by  stoke’s  theorem 

                     Now ,   

3.14 Gauss divergence theorem

The divergence theorem states that the surface integral of the normal component of a vector point function “F” over a closed surface “S” is equal to the volume integral of the divergence of F⃗ taken over the volume “V” enclosed by the surface S. Thus, the divergence theorem is symbolically denoted as:

∬v∫▽F⃗ .dV=∬sF⃗ .n⃗ .dS

Sums on surface and volume

EXAMPLE – 1      Use divergence theorem to show that  where S is any closed surface enclosing a volume V.

SOLUTION:     By divergence theorem 

                  Here ,    

                                = 6V

EXAMPLE – 2       Show that 

SOLUTION:    By divergence theorem,     ..…(1)

                          Comparing  this  with  the  given  problem  let 

Hence, by (1)

                                                                …………. (2)

Now ,  

Hence, from (2), We get,