introduction to vector calculus (29) · introduction to vector calculus (31) = 90° so a is...
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Introduction to Vector Calculus (29)
SOLVED EXAMPLESQ.1 If vector ˆ ˆ ˆ ˆ ˆ ˆˆ ˆA 2i j 2k, B 2i j, C 2i 3j k
then find
(a) A B
(b) A B
(c) A . B C
(d) B. C A
(e) A B C
(f) a unit vector perpendicular to both B and C
(g) Component of A along B
.
Solution:
(a) A B
= ˆ ˆ ˆ ˆ ˆ2i j 2k 2i j
= ˆ ˆ ˆ4i 2 j 2k
(b) A B
= ˆ ˆ ˆ ˆ ˆ2i j 2b 2i j
= ˆ2k
(c) A. B C
=
2 1 22 1 02 3 1
= –2 + 2 – 8 = –8
(d) B. C A
=
2 1 02 3 12 1 2
= –10 + 2= –8
(e) A B C
= B A.C C A.B
= ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2i j 2i j 2k . 2i 3j k
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2i 3j k 2i j 2k . 2i j
= ˆ ˆ12i 3k
(30) Introduction to Vector Calculus
(f) n = 2 2 2
ˆ ˆ ˆ ˆ ˆ2i j 2i 3j kB CB C 1 2 4
=ˆ ˆ ˆi 2 j 4b
21
(g) AB = ˆA cos B
= 2
A.B Bˆ ˆA.B BB
as B = BB
=
22
ˆ ˆ ˆ ˆ ˆ ˆ2i j 2k . 2i j 2i j
2 1
=ˆ ˆ10i 5j5
= ˆ ˆ2i j
Q.2: Find the angle between ˆ ˆ ˆA 2i j k
and ˆ ˆ ˆB i j 3k
.
Solution :
as A.B
= A B cos
cos = 22 2 2 2 2
ˆ ˆ ˆ ˆ ˆ ˆ2i j k . i j 3kA.BA B 2 1 1 1 1 3
=1 4cos
6 11
= 60.5° approx.
Q.3: If ˆ ˆ ˆ ˆˆ ˆA 2i 3j 7k and B 2i j k
, then show that A and B
areperpendicular to each other.
Solution :
A.B
= ˆ ˆ ˆ ˆ ˆ ˆ2i 3j 7k . 2i j k
= – 4 – 3 + 7 = 0
so A.B
= A B cos 0
cos = 0
Introduction to Vector Calculus (31)
= 90°
so A is perpendicular to B
.
Q.4: Find the unit vector perpendicular to bothˆ ˆ ˆA 2i 3j 5k
and ˆ ˆ ˆB 2i 3j k
. Also find the angle between them.
Solution:
As A B
=
ˆ ˆ ˆi j k2 3 52 3 1
= ˆ ˆ12i 8j
unit vector perpendicular to both A and B
n =A BA B
= 22
ˆ ˆ ˆ ˆ12i 8j 12i 8j20812 8
Again sin =A B
A B
=208 208
4 9 25 4 9 1 38 14
=1 208sin
38 14
= 38.7° Approx.
Q.5: A particle is acted upon by two constant forces 1ˆ ˆ ˆF i 4j 3k
and
2ˆ ˆ ˆF 3i j k
due to which particle is displaced from ˆ ˆ ˆ ˆˆ ˆi 2j 3k to 4i 5j k . Calculate
the total work done.Solution:Displacement of the particle
r = ˆ ˆ ˆ ˆ ˆ ˆ4i 5j k i 2j 3k
(32) Introduction to Vector Calculus
= ˆ ˆ ˆ3i 3j 2k Hence total work done
= Total force . displacement
= 1 2F F .r
= ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi 4 j 3k 3i j k . 3i 3j 2k
= ˆ ˆ ˆ ˆ ˆ ˆ4i 5j 4k . 3i 3j 2k 12 15 8
= 35 units.Q.6: A rigid body is rotating with angular velocity of 5 rad/s about an axis parallel
to ˆ ˆ3j k and passing through the point ˆ ˆ ˆi j 3k . Find the velocity vector of the
particle, when it is at the point ˆ ˆ ˆ2i 4 j k .
Solution : Suppose r is the position vector
then r = ˆ ˆ ˆ ˆ ˆ ˆ2i 4j k i j 3k
= ˆ ˆ ˆi 5j 4k angular velocity
=
ˆ ˆ3j k 5 ˆ ˆ5 3j kˆ ˆ 103j k
linear velocity
v = 5 ˆ ˆ ˆ ˆ ˆr 3j k i 5 j 4k10
=
ˆ ˆ ˆi j k5 0 3 110 1 5 4
= 5 ˆ ˆ ˆ7i j 3k10
units.
Q.7 : Calculate the torque of a force ˆ ˆ ˆ2i 2j 5k about the point ˆ8j acting
through the point ˆ ˆ ˆ6i 4j 2k .
Solution : Here
Introduction to Vector Calculus (33)
r = ˆ ˆ ˆ ˆ8j 6i 4j 2k
= ˆ ˆ ˆ6i 4 j 2k
torque = ˆ ˆ ˆ ˆ ˆ ˆr F 6i 4j 2k 2i 2j 5k
=
ˆ ˆ ˆi j k6 4 22 2 5
= ˆ ˆ ˆ24 i 34 j 4k
Q.8: A force vector ˆ ˆ ˆ10 i 25 j 35k passes through a point (2, 5, 7). Prove thatforce is also passing through the origin.
Solution: The position vector
r = ˆ ˆ ˆ2i 5 j 7k
and moment of the form about this point i.e. torque
= r F
= ˆ ˆ ˆ ˆ ˆ ˆ2i 5j 7k 10i 25j 35k
=
ˆ ˆ ˆi j k2 5 7
10 25 35
= 0As the moment is zero, which shows that forces is passing through the origin.
Q.9: A force ˆ ˆ ˆ4i 3j 2k passes through the point (–9, 2, 1). Find the componentof moment of the force about the axis of reference.
Sol.: Here
r = ˆ ˆ ˆ9i 2 j k
so moment of force i.e. torque
r F = ˆ ˆ ˆ ˆ ˆ ˆ9i 2j k 4i 3j 2k
(34) Introduction to Vector Calculus
=
ˆ ˆ ˆi j k9 2 1
4 3 2
= ˆ ˆ ˆ7i 22 j 19k
Hence components of moment of force are 7 unit, 22 units and 19 units in x, y and zdirection respectively.
Q.10: A proton is moving with velocity 108 cm/s along z-axis through an electricfield of intensity 3 × 104 volt/cm along x-axis and magnetic field of intensity 2000 gaussalong y-axis. Calculate the magnitude and direction of total force.
Solution: Intensity of electric field
E = 4 ˆ ˆ3 10 i volt / cm 100 i esu/cm
Proton charge = 19 101.6 10 C 4.8 10 esu
Magnetic field B = ˆ2000 j gauss
velocity v = 8 ˆ10 k cm / s
so total force acting on the proton
F =
v Bq EC
= 10 810
1ˆ ˆ ˆ4.8 10 100i 10 k 2000 j3 10
= 8 ˆ4.47 10 i dyne
Hence total force acting on the proton has magnitude +4.47 × 10–8 dyne along the +ve x-direction.
Q.11: Find the value of the constant p so that
ˆ ˆ ˆ ˆˆ ˆA 2i j 3k, B 2i 3j k
and ˆ ˆ ˆC 3i pj k
are coplanar..
Solution: We know that three vectors are said to be coplanar if A. B C 0
2 1 32 3 13 p 1
= 0
Introduction to Vector Calculus (35)
4p – 38 = 0or 4p = 38
p =38 9.54
Q.12: Evaluate A B C
where
ˆ ˆ ˆ ˆ ˆA 2i j, B i j k
and ˆ ˆ ˆC 5i 3j k
.
Solution:
A B C
= B A.C C A.B
= ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k 2i j 0k . 5i 3j k
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ5i 3j k 2i j 0k . i j k
= ˆ ˆ ˆ ˆ ˆ ˆ7 i j k 1 5i 3j k
= ˆ ˆ ˆ2i 4j 8k
Q.13: If r si the position vector of any point (x, y, z) and A is a constant vector
then show that
(i) r .A . A 0
is the equation of a constant plane.
(ii) r A .r
is the equation of a sphere.
Also show that result of (i) is of the form Ax By Cz D 0 where
2 2 2D A B C and that of (ii) is of the from 2 2 2 2x y z r . [RU 2005]
Solution: (i) Suppose A = A, B, C and r x, y, z
r A . A
= x A A y B B z C C
= 2 2 2xA A yB B zC C
= 2 2 2xA yB zC A B C
= x y zA B C D
(36) Introduction to Vector Calculus
where D = 2 2 2A B C
so r A .A
= 0 x y zA B C D 0
which is an equation of a plane.
(ii) r A .A
= x A x y B y z C z
if r A .A
= 0 then
2 2 2x y zx y z A B C = 0
Which is the equation of sphere whose surface touches the origin.
Q.14: A particle moves on the curve 2 2x 2t , y t 4t, z 3t 5 where t is thetime. Find the components of velocity and acceleration at time t = 1 in the directionˆ ˆ ˆi 3j 2k .
Solution: Position vector
r = 2 2ˆ ˆ ˆ2t i t 4t j 3t 5 k
so velocity vector
drvdt
= 2 2d d dˆ ˆ ˆ2t i t 4t j 3t 5 kdt dt dt
= ˆ ˆ ˆ4t i 2t 4 j 3k
acceleration a =dvdt
= d d dˆ ˆ ˆ4t i 2t 4 j 3 kdt dt dt
= ˆ ˆ4i 2 j 0
at` t = 1, velocity ˆ ˆ ˆv 4i 2 j 3k
acceleration a = ˆ ˆ4i 2 j
and the component of ˆ ˆ ˆv along i 3j 2k
Introduction to Vector Calculus (37)
is
22 2
ˆ ˆ ˆ ˆ ˆ ˆ4i 2 j 3k . i 3j 2k
1 3 2
=16 8 14
714
and component of ˆ ˆa along i 3j 2k is
=
22 2
ˆ ˆ ˆ ˆ ˆ4i 2 j . i 3j 2k 2141 3 2
=14
7
Q.15: Calculate the unit vector, which is normal to the surface
= 2 2x y xy 3xyz at the point (1, 1, –1).
Solution : Here
= 2 2ˆ ˆ ˆi j k x y xy 3xyzx y z
= 2 2 2 2ˆ ˆx y xy 3xyz i x y xy 3xyz jx y
2 2 ˆx y xy 3xyz kz
= 2 2ˆ ˆ2xy y 3yz i x 2xy 3xz j 3xy k
At (1, 1, –1),
= ˆ ˆ ˆ ˆ2 1 3 i 1 2 3 j 3k 3k
so the unit vector normal to the surface at (1, 1, –1) is
2
ˆ3k
3 =
ˆ3k3
= k
Q.16: Find the direction derivative of 2 2x, y,z x y xy at the point (2, –1, –4)along the direction of the vector (1, 2, –1).
Solution: as
(38) Introduction to Vector Calculus
= 2 2x y xy
= ˆ ˆ ˆi j k x, y,zx y z
= 2 2 2 2 2 2ˆ ˆ ˆx y xy i x y xy j x y xy kx y z
= 2 2ˆ ˆ2xy y i x 2xy j
2, 1, 4
= ˆ3i
Position vector r = ˆ ˆ ˆi 2 j k
and unit vector along this position vector
n =ˆ ˆ ˆ ˆ ˆ ˆi 2 j k i 2 j k1 4 1 6
and direction derivative n
= ˆ ˆ ˆi 2j kˆ3i .
6
=3 6
26
Q.17: Find the equation of the tangent plane and normal line to the surface2 22x y 2z 3 at the point (2, 1, –3).
Solution :
Here x,y,z = 2 22x y 2z 3
x
= 2 22x y 2z 4xx
y = 2 22x y 2z 2y
y
z
= 2 22x y 2z 2z
so the components , andx y z at the point (2, 1, –3) will be
Introduction to Vector Calculus (39)
x
= 4 2 8, 2 1 2, 2y z
Hence the equation of the tangent plane to the surface at the point (2, 1, –3) is
X 2 8 Y 1 2 Z 3 2 = 0
or 4X + y + Z = 6so the equation of normal to the surface at (2, 1, –3) is
X 28
=Y 1 Z 3
2 2
orX 2
4
= Y – 1 = Z + 3
Q.18: Find the angle between the surfaces 2 2 2x y z 9 and 2 2x y z 3 atthe (1, 2, 2)
Solution:
Suppose 1 = 2 2 2 2 22x y z and x y z
so 1
= ˆ ˆ ˆ2x i 2y j 2z k
and 2 = ˆ ˆ ˆ2x i 2y j k
and 1 1,2,2
= ˆ ˆ ˆ2i 4 j 4k
2 1,2,2
= ˆ ˆ ˆ2i 4 j k
since 1 2and
are normal to 1 and 2
then 1 2.
= 1 2 cos
where is the angle between thesurfaces 1 and 2.
so =1 1 2
1 2
.cos
=1 14 16 4 16cos cos
36 21 6 21
= 54.41° approx.
Q.19 (i) Provle that 1 1 2 2ˆ ˆ ˆ ˆP cos i sin j and cos i sin j
are unit vectors in
(40) Introduction to Vector Calculus
the xy-plane respectively making 1 and 2 with the x-axis.
(ii) By means of dot product, obtain the formula for 2 1cos . by similarly
formulating P and Q, obtain the formula for 2 1cos .
(iii) If is the angle between P and Q find 1 P Q2
in terms of .
Solution:
(i) Given P = 1 1
ˆ ˆcos i sin j
Q
= 2 2ˆ ˆcos i sin j
y
1
2
Q
P
P
= 2 21 1cos sin 1
Q
= 2 22 2cos sin 1
hence P and Q
are unit vectors.
(ii) P.Q
= 2 1P Q cos
= 2 11.1 cos ...(1)
But P.Q
= 1 1 2 2ˆ ˆ ˆ ˆcos i sin j . cos i sin j
= 1 2 1 2cos cos sin sin ...(2)
so 2 1cos = 1 2 1 2cos cos sin sin
let 1P P
= 1 1ˆ ˆcos i sin j
and 1Q
= 2 2ˆ ˆcos i sin j
then 1 1P .Q
= 1 21.1 cos
Introduction to Vector Calculus (41)
= 1 2 1 2cos cos sin sin
(iii) 1 1P and Q
are unit vectors.
so 1 P Q2
= 2 21 Q 2PQ cos2
=1 1 1 2cos2
= 1 cos
= 22sin2
Q.20: A vector field is given as 2 2ˆ ˆ ˆW 4x y i 7x 2z j 4xy 2z k
(i) What is the magnitude of the field at point (2. –3, 4).(ii) At what point on z-axis is the magnitude of W equal to unity? [RU 2002]Solution: (i)
W = 2 2ˆ ˆ ˆ4x y i 7x 2z j 4xy 2z k
at P(2, –3, 4), W = 2 2ˆ ˆ ˆ4 2 3 i 7 2 4 2 j 4 2 3 2 4 k
= ˆ ˆ ˆ48i 22 j 8k
W
= 2 2 248 22 8 53.4
(ii) As the required point is on z-axis so x = 0, y = 0
W = 2 2ˆ2z j 2z k for that point.
W
= 22 2 2 42z 2z 4z 4z 1
so 4 24z 4z 1 = 0
z2 = 4 16 16 1 1 1
8 2 2
z2 = –1.207 and 0.207taking z as positive z2 = 0.207
z = ± 0.455Q.21: Calculate the differential volume to obtain the expression for volume of the(i) sphere of radius 'b'
(42) Introduction to Vector Calculus
(ii) Semispherical shell of inner radius 'a' and outer radius 'b'.(iii) Cylinder of radius 'b' and height 'h'.Solution:(i) Differential volume in spherical coordinates
dV = 2r sin dr d d
here r = 0 to b, = 0 to , = 0 to 2.So volume of sphere
V =b 2
2
V 0 0 0
dV r sin dr d d
=b 2
2
0 0 0
r dr sin d d
= b
2
0 0
2 r dr cos
= b
2
0
2 1 1 r dr
=3b4
3
so Vsphere =34 b
3
(ii) For semispherical shellr1 =a, r2 = b
so dV = 2r sin dr d d
here r = a to b, 0 to , 0 to
so V =b
2
a 0 0
r sin dr d d
Introduction to Vector Calculus (43)
=b
2
a 0 0
r dr sin d d
= b
2
a 0
r dr cos
=
b2
a
r23
= 3 32 b a3
(iii) Differential volume for a cylinder
dV = r dr d dz
here r = 0 to b, z = 0 to h, and = 0 to 2.
so V =V
dV
V =b h 2
a 0 0
r dr d dz
=b h 2
0 0 0
r dr dz d
=
b2
0
r2 .h .2
so cyl.V = 2b h
Q.22 : For positive x, y, z let = 40 xyz c/m3. Find the total charge within theregion bounded by x = 0, y = 0, 0 2x 3y 10 and 0 z 2 .
Solution : Here
Q =y5 z
0 0 0
40 xyz dx dydz
(44) Introduction to Vector Calculus
=
10 2x 25 2 23
0 0 0
y z40x dx2 2
= 5
2 3
0
40 100x 40x 4x dx9
=
52 3 4
0
40 100x 40x 4x9 2 3 4
Q = 925.926 CQ.23: Given point P in Cartesian coordinate system as P(1, 2, 3). Calculate its
coordinates in cylindrical system.Solution: As given x = 1, y = 2, z = 3
= 2 2 1 yx y tan , z zx
so = 2 21 2 5 2.236
= 1 2tan 63.431
z = 3so Pcyl. = (2.236, 63.43°, 3)Q.24: The cooridnate of a point P in cylindrical system is P(1, 45°, 2). find its
equivalent in cartesion system.Solution:Here = 1, = 45°, z = 2
and x = cos , y sin , z z
x =11. cos 45 0.7072
y = 0.707z = 2
so Pcart. = (0.707, 0.707,2)Q.25: Find the constant m such that the vector
Introduction to Vector Calculus (45)
= ˆ ˆ ˆx 3y i y 2z j x mz k is solenoidal.
Solution: The vector will be solenoidal if
so ˆ ˆ ˆ ˆ ˆ ˆi j k x 3y i y 2z j x mz k 0x y z
i.e. x 3y y 2z x mz 0x y z
or 1 + 1 + m = 0m = –2
Q.26: Find 3 3 3div F and curl F if F grad x y z 3xyz
. [WBUT 2001]
Solution: Here F = grad 3 3 3x y z 3xyz
= 3 3 3 3 3 3ˆ ˆx y z 3xyz i x y z 3xyz jx y
3 3 3 ˆx y z 3xyz kz
= 2 2 2ˆ ˆ ˆ3x 3yz i 3y 3xz j 3z 3xy k
div F
= 2 2 23x 3yz 3y 3xz 3z 3xyx y z
= 6 x y z
F =
2 2 2
ˆ ˆ ˆi j k
x y z
3x 3yz 3y 3xz 3z 3xy
= ˆ ˆ ˆ3x 3x i 3y 3y j 3z 3z k 0
Q.27 Show that curl grad f = 0 where f = x2y + 2xy + z2.
(46) Introduction to Vector Calculus
Solution: grad f =f f fˆ ˆ ˆi j kx y z
= 2ˆ ˆ ˆ2xy 2y i x 2x j 2z k
curl grad f =2
ˆ ˆ ˆi j k
x y z
2xy 2y x 2x 2z
= ˆ0 0 2x 2 2x 2 k 0
Q.28: If the scalar function 2x,y,z 2xy z ,. is its corresponding scalar fieldis solenoidal or irrotational?
Solution : Let
F = ˆ ˆ ˆ2y i 2x j 2zk
so .F
= 2y 2x 2zx y z
= 0 +0 + 2 = 2 0
So field is not solenoidal.
Now F
=
ˆ ˆ ˆi j k
x y z2y 2x 2z
= 0so field is irrotational.Q.29: Verify the divergence theorem for the vector function
F = 2ˆ ˆ ˆ4xz i y j yz k
taken over the cube bounded by x = 0, 1 y = 0, 1, z = 0, 1.[WBUT (math) 2002]
Solution.:
Introduction to Vector Calculus (47)
x
y
z
3 2
5
1
67
0
4
for face 4567; ˆn i and x = 1
1
ˆF . n ds =
1 1
0 0
4z dydz 2
2
ˆF . n ds = i dy dz 0 for 1230
for ˆ2561 n = j, y 1
3
ˆF. n ds = –1
and for 30744
ˆF. n ds = 0
for face 2345,5
ˆF. n ds =
12
and6
ˆF. n ds = 0 for 0167
totalS
ˆF. n ds = 1 32 0 1 0 0
2 2
Again F
= 24xz y yzx y z
= 4z – y
(48) Introduction to Vector Calculus
1 1 1
0 0 0
.F dV
= 1 1 1
0 0 0
4z y dx dy dz
=
11 1 2
0 0 0
4z yz dx dy2
=32
s
ˆF. n ds =
V
F dV
Hence divergence theorem is verified.
Q.30: Calculate the line integral of ˆ ˆA cos z sin z
around the edge L of thewedge defined by 0 4 , 0 30 , z 0.
Solution:
Given A = ˆ ˆcos z sin z
differential length
dl
= ˆˆ ˆd d dzz
x
y
0
(3)(2)
(1)
Circulation of A around path is
L
A.dl
=1 2 3
A.dl A.dl A.dl
for (1) 0, d 0, z 0, dz 0
1
A.dl
= 1
ˆˆ ˆ ˆcos zsin d d dz z
Introduction to Vector Calculus (49)
=
44 2
1 0 0
cos d d2
=16 82
for (2) 4, d = 0
z = 0, dz = 0
2
A. dl
=2
cos d zsin dz 0
for (3) / 6, d = 0, z = 0, dz = 0
3
A. dl
=3
cos d
=0
4
cos d6
=00 2
4 4
3 d 0.8662 2
= –6.93
So totalL
A. dl
= 8 + 0 – 6.93
= 1.07
Q.31 Given 2A x xy
, calculate A . ds
over the region y = x2, 0 < x < 2.
Solution:So y = x2, ds = dx dy
(50) Introduction to Vector Calculus
x
y
0
2
2
A. ds
= 2x xy dx dy
= 2 22 x 2 x
2
0 y 0 0 y 0
x dx dy xy dx dy
=2 2 5
4
x 0 x 0
xx dx dx2
=32 645 12
= 11.7
Q.32 For a scalar function zx ysin sin e2 3
. Calculate the magnitude
direction of maximum rate of increase of at the point (1, 1, 1).Solution: As gradient of a scalar function gives the magnitude and direction of max. rate
of change of that
So = ˆ ˆ ˆi j kx y z
=z zx y x yˆ ˆsin sin e i sin sin e j
x 2 3 y 2 3
zx y ˆsin sin e kz 2 3
=z zx x yˆ ˆe sin j sin sin e k
6 2 2 3
at (1, 1, 1)
Introduction to Vector Calculus (51)
1,1,1 =1 1ˆ ˆe sin j sin sin e k
6 2 2 3
= ˆ ˆ0.367 j 0.866 0.367 k6
= ˆ ˆ0.192 j 0.318 k
and 1,1,1 = 1/22 20.192 0.318
= 0.37
and j =ˆ ˆ0.192j 0.318k0.37
= ˆ ˆ0.52j 0.86k
Q.33 : Determine the divergence of the following vector fields at given points–
(i) ˆ ˆ ˆA yzi 4 x y j xyz k at 1, 2,1
(ii) ˆˆ ˆB z sin 5 z cos z z at 5, , 12
(iii) ˆ ˆˆC 2rsin cos r cos r at 1, ,3 3
Solution: In cartesion
(a) .A
= yz 4x 4y xyzx y z
= 0 + 4 + xy
.A
= 4 + xy
1, 2,1at 1, 2,1 , .A
= 4 + 1 × (–2)
= 2(b) In cylindrical
.B
= z1 1B B B
z
(52) Introduction to Vector Calculus
given B = 1 zz sin B 5 z cos , B z
.B
= 2 1z sin 5. z sin 1
= 1 3z sin
at 5, , 12
.B
= 1 – (3 × 1 × 1)
= –2(c) In spherical system
.C
= 2r2
C1 1 1r C sin Cr r sin rsinr
= 322 cossin cos r sin
r r sin rsinr
=cos cot6sin cos
r
at 1, ,3 3
.C
=cos cot
3 36sin cos3 3 1
= 2.6 + 0.288= 2.88
Q.34: Find the nature of the vector 2ˆ ˆ ˆF 30i 2xy j 5xz k
. [RU 2003]
Solution :
.F
= 30 2xy 5xzx y y
= 2x + 10xz 0
Introduction to Vector Calculus (53)
Curl F
= F =
2
ˆ ˆ ˆi j k
x y z
30 2xy 5xz
= 2 ˆ ˆ5z j 2yk
0
as .F 0
fields is not solenoidal
and F 0
field is rotational.
Q.35: Given the vector field 2ˆ ˆ ˆG 16xy 3 i 8x j xk
.
(i) Is G irrotational (or conservative)?(ii) Find the net flux of G over the cube 0<x, y, z < 1.(iii) Determine the circulation of G around the edge of the square z = 0, 0 < x,
y < 1. Assume anticlockwise direction. [RU 2003]Solution:
(i) G
=2
ˆ ˆ ˆi j k
x y z
16xy z 8x x
= ˆ ˆ ˆ0 i 1 1 j 16x 16x k 0
So G is irrotational.
(ii) Net flux of G over the cube
= V
. G dV
. G
= 216xy z 8x xx y z
= 16y 0 0 16y
so V
. G dV
=1 1 1
0 0 0
16y dx dydz 16 dx dz y dy
(54) Introduction to Vector Calculus
=
12
0
y16.1.1 82
(iii)0
G . dl
= y 0 x 1
1 12
x 0 y 0z 0 z 0
16xy z dx 8x dy
y 1 x 0
0 02
x 1 y 1z 0 z 0
16xy z dx 8x dy
= 021
01
x0 8 1 y 16 1 02
= 8 – 8 = 0.
SUMMARY• A vector is with magnitude and direction.• In space a quantity is specified by a function.• When the result of multiplication of two vectors is a scalar then it is called scalar
product or dot product.• When product is a vector then it is called vector product.
• Multiplication of three vectors can give scalar A . B C
or a vector A B C
.
• Vector differentiation is done using dal () operator the gradient of a scalar field is
, divergence as . A
and curl by A
and laplacian by 2A.
• In Cartesion coordinate system
dl
= ˆ ˆ ˆdx i dy j dz k, dV dx dy dz
Gradient = ˆ ˆ ˆi j kx y z
Divergences . A
=yx zAA A
x y z
Introduction to Vector Calculus (55)
Curl A
=x y z
ˆ ˆ ˆi j k
x x zA A A
Laplacian =2 2 2
2 2 2x y z
• In cylindrical system
dl
= ˆˆ ˆd d dz z, dV d d dz
gradient T =T 1 T Tˆˆ z
z
divergence . A
= zA A1 1Az
Curl A
=
z
ˆˆ z1
zA A A
Laplacian =2
2 2 21 1
z
• In spherical system
dl
= ˆ ˆˆdr r rd r sin d
dV = 2r sin d dr d
gradient T =T 1 T 1 Tˆ ˆrr r r sin
divergence
= 2r2
A1 1 1r A A sinr r sin rsinr
(56) Introduction to Vector Calculus
Curl A
=
r
ˆ ˆr r sin1
r sin rA rA rsin A
Laplacian =2
21 1r sin
r r r sin
2
2 2 21
sin
• Gauss Divergence theorem
s
A . ds
= V
. A dV
• Stoke's theorem
L
A . dl
= V
A . ds
• A vector field is solenoidal if
Irrotational or conservative if
• Triple products
A . B C B C A C A B
A B C B A . C C A . B
• Second derivatives
. A 0
f 0
2A . A A
EXERCISE
Introduction to Vector Calculus (57)
1. Give the basic concepts of transformation of one coordinate system to another. Derivenecessary relations for rectangular, cylindrical and spherical systems. [RU 2003]
2. Write short note on "Physical significance of curl, divergence and gradient".3. State-Gauss divergence theorem. Write its applications, advantages and limitations.
[RU 2002]4. State and prove stoke's theorem. [RU 2000]5. Explain how stoke's theorem enables us to obtain the integral form of ampere circuital law.6. Explain various types of vector fields.
(i) Solenoidal and irrotational fields.(ii) Irrotational but not solenoidal fields(iii) Solenoidal but not irrotaitonal fields(iv) Neither irrotational nor solenoidal fields.
7. For the vectors ˆ ˆ ˆ ˆ ˆA i 3k and B 5i 2 j 6k
calculate
(i) A B
(ii) A B
(iii) A . B
(iv) A B
(v) Angle between A and B
(vi) A unit vector parallel to 3A B
.
(vii) Length of the projection of A on B
.
Ans.: ˆ ˆ ˆ ˆ ˆ ˆˆ ˆi 6i 2j 3k ii 4i 2j 9k iii 13 iv 6i 21j 2k
ˆ ˆ ˆ8i 2j 3kv 60 vi vii 1.6m
77
8. Use the differential volume dV to find volume of region.
(i) 0 x 1, 1 y 2, 3 z 3 [Ans.: 6]
(ii) 2 5, , 1 z 43
[Ans.: 110]
9. Find area of the region 0 on the spherical shell of radius 'b'. [Ans. 2b2]
10. Evaluate the gradient of the following scalar fields
(58) Introduction to Vector Calculus
(a) zP e sin 2x [Ans.: z zˆ ˆP 2cos 2xe i sin2xe k ]
(b) 2q zcos 2 2ˆˆAns. : q 2 zcos zsin cos z
(c) s 20r sin cos ˆAns. : r 20sin cos r 20rcos
ˆ ˆcos 20rsin sin
11. If f = xy + yz + xz then(i) Find the magnitude and direction of the maximum rate of change of the function at
point (1, 2, 3)(ii) Find the rate of change of the function at the same point in the direction of the
vector.
Ans. : i f 14
ˆ ˆ ˆf 2i 3j kˆ ff 14
(ii) df f .dl 11
12. If ˆ ˆ ˆT 2x i 3y j 4z k and V = xyz evaluate . VT
. [Ans.: 2xyz]
13. If 2 2 2U xz x y y z find div (grad U). [Ans.: 2(–y + z2 + y2)]
14. Given 2 2 2 2 3ˆ ˆ ˆD 6xyz i 3x z j 6x y k C/m
Find the total charge lying within the region
bounded by 0 < x < 1, 1 < y < 2 and z 1 by separately evaluating each side of divergencetheorem. [Ans.: 6C]
15. If ˆ ˆ ˆA x y 1 i j x y k
then prove that A . 0
16. Prove that vector ˆ ˆ ˆA x y z i 2y zx j 2z xy k
is not solenoidal.[WBUT 2005]
17. If 2 2x y 2z find .
. [WBUT 2005]
18. Show that 2 2ˆ ˆ ˆB 2xyz i x z 2y j x yk
is irrotational. [WBUT 2005]
19. find a unit vector perpendicular to 2 2 2x y z 100 at (1, 2, 3). [WBUT 2007]
Introduction to Vector Calculus (59)
ˆ ˆ ˆi 2j 3kAns. :14
20. if 2 3 23x y y z find
at (1, –2, –1). [WBUT 2004]
ˆ ˆ ˆAns. : 12i 9j 16k
21. Show that 3 2 2ˆ ˆ ˆF 2xy z i x j 3xz k
is a conservative force field. Find also the scalar
potential. [WBUT 2006, 2003] 2 3Ans. : x y z x constant
22. Evaluates s
ˆF.n ds where 2ˆ ˆ ˆF 8x z i y j yzk
and s is the surface of the cube bounded
by x = 0, 1, y = 0, 1, z = 0, 1.