vector

394 Vectors
Solution. Directional derivative =
x
x y z
x y
= i j k 2 2
2 2
y x 2xy
= i j
2 2 2 2 2 2
( x y ) ( x y )
Directional derivative at the point (0, 2)
4
0 2(0) (2) i
(0 4) (0 4) 4
= i j
2 2
1 x(2 x) x(2 y)
= i
j
2 2 2 2 2
2 2 2
x y ( x y ) ( x y )
Directional derivative at the point (0, 2) in the direction CA — 3 1
i.e. i j
2 2
CA OB BA i cos 30 j sin 30
i 3 1
= . i j
3 1
4 2 2
i j
2 2
3
=
Ans.
8
Example 29. Find the directional derivative of V where 2 2 2
V xy i zy j xz k, at the
point (2, 0, 3) in the direction of the outward normal to the sphere x 2 + y 2 + z 2 = 14 at the
point (3, 2, 1). (A.M.I.E.T.E., Dec. 2007)
Solution. V 2 = VV .
2 2 2 2 2 2
2
,
= ( xy i zy j xz k).( x y i zy j xz k)
= x 2 y 4 + z 2 y 4 + x 2 z 4
Directional derivative = 2
V
=
2 4 2 4 2 4
i j k ( x y z y x z )
x y z
=
4 4 2 3 3 2 4 2 3
(2xy 2 xz ) i (4x y 4 y z ) j (2y z 4 xz)
k
Directional derivative at (2, 0, 3) = (0 2281) i (0 0) j (0 4427)
k
= 324 i 432k 108 (3 i 4 k)
...(1)
Normal to x 2 + y 2 + z 2 – 14 =
2 2 2
= i j k ( x y z 14)
x y z
= (2xi 2yj
2 zk)
Normal vector at (3, 2, 1) = 6i 4 j 2 k
...(2)
Unit normal vector = 6 4 2 2(3 2 ) 3
i j k i j k i 2
j k
36 16 4 2 14 14
3i 2 j k
Directional derivative along the normal = 108(3 i 4 k). .
14
108 (9 4) 1404
=
14 14
j
—2
1
(0, 2) 30°
i
C —2
3
i
1
A
[From (1), (2)]
j
Ans.