vector

408 Vectors
Curl V =
=
=
2 2
i j k [2 x yzi ( x z 2 y) j x yk]
x y z
i j k
x y z
2 2
2xyz x z 2y x y
2 2
( x x ) i (2xy 2 xy) j (2xz 2 xz) k 0
Hence, V (x, y, z) is irrotational.
To find corresponding scalar function u, consider the following relations given
V
= grad (u)
or V = ( u)
...(1)
u u u
du =
dx
dy
dz (Total differential coefficient)
x y z
u u u
= i j k .( i dx jdy k dz)
x y z
= udr
. V.
d r
[From (1)]
2 2
= [2 xyzi ( xz 2 y) j x yk].( i dx jdy kdz)
= 2 x y z dx + (x 2 z + 2y) dy + x 2 y dz
= y(2x z dx + x 2 dz) + (x 2 z) dy + 2y dy
= [yd (x 2 z) + (x 2 z) dy] + 2y dy = d(x 2 yz) + 2y dy
Integrating, we get u = x 2 yz + y 2 Ans.
Example 49. A fluid motion is given by v ( y z) i ( z x) j ( x y) k.
Show that the
motion is irrotational and hence find the velocity potential.
(Uttarakhand, I Semester 2006; U.P., I Semester, Winter 2003)
Solution. Curl v = v
= i j k [( y z) i ( z x) j ( x y) k]
x y z
=
i j k
x y z
y z z x x y
= (1 1) i (1 1) j (1 1) k 0
Hence,
v is irrotational.
To find the corresponding velocity potential , consider the following relation.
v =
d = dx
dy
dz
[Total Differential coefficient]
x y z