vector

Vectors 423
1 3 4 2
= (2 t 3) (2 dt ) (2)( t t ) dt ( t – 2)(3 t t dt ) =
0
2
t 2 5 3 7 6 4
= 4 6 t t t – t
2 5 7 4
1
0
1 4 6 3
(4t 6 2t 3 t – 6 t ) dt
0
2 2 5 3 7 3 4
= 2t 6 t t t – t
5 7 2
2 3 3
= 2 6 – = 7.32857. Ans.
5 7 2
Example 68. The acceleration of a particle at time t is given by
a
= 18 cos 3tiˆ
8sin 2tj ˆ 6 tk ˆ .
If the velocity v and displacement r be zero at t = 0, find v and r at any point t.
2
Solution. Here, a d r
=
2
dt
On integrating, we have
= 18 cos 3tiˆ
8sin 2tj ˆ 6 tkˆ
.
v = dr
iˆ
18 cos 3t dt ˆj 8sin 2t dt k ˆ 6t dt
dt
2
v = 6sin 3tiˆ 4cos 2tj ˆ 3t kˆ
c ...(1)
At t = 0, v = 0
Putting t = 0 and
v = 0 in (1), we get
0 = 4ˆj
c c 4ˆj
dr
2
v = 6sin 3tiˆ 4(cos 2t 1) ˆj 3tkˆ
dt
Again integrating, we have
2
r
= iˆ
6sin 3 ˆ 4(cos 2 1) ˆ
t dt j t dt k3t dt
3
r = 2cos 3 tiˆ(2sin 2t 4 t)
ˆj t kˆ
c
...(2)
At, t = 0, r = 0
Putting t = 0 and r = 0 in (2), we get
0 = 2iˆC ˆ
1 C1
2i
Hence, r
3
= 2(1cos 3) t iˆ
2(sin 2t 2 t)
ˆj t kˆ
Ans.
Example 69. If A 2 2
(3x 6 y) iˆ
–14yzj ˆ 20 xz kˆ
, evaluate the line integral Adr . from
(0, 0, 0) to (1, 1, 1) along the curve C.
x = t, y = t 2 , z = t 3 . (Uttarakhand, I Semester, Dec. 2006)
Solution. We have,
A . dr =
C
=
C
C
2 2
[(3x 6 y) iˆ –14yzj ˆ 20 xz kˆ].[ iˆ dx ˆjdy kˆ
dz]
2 2
[(3x 6 y) dx –14yzdy 20 xz dz]
If x = t, y = t 2 , z = t 3 , then points (0, 0, 0) and (1, 1, 1) correspond to t = 0 and t = 1 respectively.
Now, A . dr =
C
=
t 1 2 2 2 3 2 3 2 3
t 0
[(3t 6 t ) d ( t)–14 t t d ( t ) 20 t ( t ) d ( t )]
t 1 2 5 7 2
[9 t dt –14 t .2tdt 20 t .3 t dt]
=
t 0
1
1 2 6 9
0
1
0
(9 t –28 t 60 t ) dt