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M I

Unit - 4

Vector calculus

 

Q1) Define vector function.

A1)

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.

 

Q2) A particle moves along the curve , here ‘t’ is the time. Find its velocity and acceleration at t = 2.

A2)

Here we have-

Then, velocity

Velocity at t = 2,

=

Acceleration =

Acceleration at t = 2,

Q3) If and then find-

1.

2.

 

A3)

We know that-

1.

 

2.

 

Q4) If , then show that

1.

2.

A4)

1. Suppose   and

Now taking L.H.S,

Which is

Hence proved.

 

2.

So that

 

Q5) If then find grad f at the point (1, -2, -1).

A5)

 

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

 

Q6) If then prove that grad u, grad v and grad w are coplanar.

A6)

Here-                     

 

Now-

Apply

 

Which becomes zero.

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

 

Q7) Find the directional derivative of 1/r in the direction where

A7)

Here

Now,

And

We know that-

So that-

Now,

Directional derivative-

 

Q8) Find the directional derivative of

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

A8)

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,

 

Q9) Show that

A9)

Interchanging , we get-

We get by using above equations-

 

Q10) Prove that

A10)

So that-

 

Q11) Prove that

A11)

So that-

 

Q12) If then find the divergence and curl of .

A12)

We know that-

Now-

 

Q13) Prove that

A13)

here    and

So that

Now-

So that-

 

 

Q14) 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)

A14)

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

=

=

=       =-1

Q15) 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).

A15)

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

=

=+

 

Q16) Prove that   ͞͞͞F = [y2cos x +z3] i+(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞ F (ii) the work done in moving an object in this field from (0, 1, -1) to (/ 2, -1, 2)

A16)

(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

 

Q17) Evaluate , where S is the surface of the sphere in the first octant.

A17)

Here-

 

Which becomes-

 

Q18) Verify stoke’s theorem when and surface S is the part of sphere , above the xy-plane.

A18)

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.

 

Q19) Verify Stoke’s theorem for the given function-

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

A19)

Suppose-

Here

We know that unit circle in xy-plane-

Or

So that,

Now

Curl

 

Now,

Hence the Stoke’s theorem is verified.

 

Q20) Show that 

A20)

By divergence theorem,  ...… (1)

Comparing this with the given problem let 

Hence, by (1)

                                    …………. (2)

Now,

Hence, from (2), We get,