Unit - 5
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,
