vector

402 Vectors
2i 4 j 4k
Directional derivative along the outer normal = (12i 48 j 48 k).
416 16
[From (1), (2)]
24 192 192
= = 68 Ans.
6
Example 40. Show that div (grad r n ) = n (n + 1)r n – 2 , where
2 2 2
r = x y z
2 1
Hence, show that = 0. (U.P. I Semester, Dec. 2004, Winter 2002)
r
Solution. grad (r n
n
n n
) = i r j r k r by definition
x y z
n 1 r
n 1 r
n
1r
n 1 r r
r
= inr j nr knr . = nr i j k
x y z
x y z
n 1 x y z
n2 n 2
= nr i j k nr ( xi y j zk) nr r.
r r r
2 2 2 2 r r x
r x y z 2r 2x etc.
x x r
Thus, grad (r n ) = n 2 n 2 n 2
nr xi nr yj nr zk
...(1)
div grad r n = div [ n2 n2 n 2
nr xi nr yj nr zk ]
=
=
=
n2 n 2 n 2
i j k .( nr xi nr y j nr zk)
x y z
n 2 n 2 n 2
( nr
x) ( nr
y) ( nr z)
x y z
n2 n3 r
n2 n 3
r
nr nx ( n 2) r nr ny ( n 2) r
x y
nr nz ( n
2) r
n n r r r
3 nr nn ( 2) r
x y z
x y z
= 2 3
[From (1)]
(By definition)
z
n2 n3
r
n 2 n 3 x y z
= 3 nr nn ( 2) r
x y
z
r r r
2 2 2 2 r r x
r x y z 2r 2x etc.
x x r
= 3nr n – 2 + n (n – 2)r n – 4 [x 2 + y 2 + z 2 ]
= 3nr n – 2 + n (n – 2) r n – 4 .r 2 ( r 2 = x 2 + y 2 + z 2 )
= r n – 2 [3n + n 2 – 2n] = r n – 2 (n 2 + n) = n(n + 1) r n – 2
If we put n = –1
div grad (r – 1 ) = –1 (–1 + 1) r – 1 – 2
2 1
r
= 0
r
1
Ques. If r xi y j zk,
and r = |r| find div .
2 (U.P. I Sem., Dec. 2006) Ans. 2
r
r