vector

Vectors 395
Example 30. Find the directional derivative of ( f) at the point (1, – 2, 1) in the direction
of the normal to the surface xy 2 z = 3x + z 2 , where f = 2x 3 y 2 z 4 . (U.P., I Semester, Dec 2008)
Solution. Here, we have
f = 2x 3 y 2 z 4
3 2 4
f = i j k
(2 x y z ) 2 2 4 3 4
x y z
= 3 2 3
6 x y z i 4 xyz j 8x y z k
2 2 4 3 4 3 2 3
(f) = i j k (6x y z
i 4x yz j 8 x y z k)
x y z
= 12xy 2 z 4 + 4x 3 z 4 + 24x 3 y 2 z 2
Directional derivative of ( f )
2 4 3 4 3 2 2
= i j k (12xy z 4xz 24 x y z )
x y z
2 4 2 4 2 2 2
= (12y z 12x z 72 x y z ) i (24xyz4
48x3yz2)
j
+ (48xy 2 z 3 + 16x 3 z 3 + 48x 3 y 2 z) k
Directional derivative at (1, – 2, 1) = (48 + 12 + 288) i + (– 48 – 96) j + (192 + 16 + 192) k
= 348 i
– 144 j 400k
Normal to(xy 2 z – 3x – z 2 ) = (xy 2 z – 3x – z 2 )
2 2
= i j k
( xy z –3 x – z )
x y z
2
= 2
( y z – 3) i (2 xyz) j ( xy – 2 z)
k
Normal at(1, – 2, 1) = i
– 4 j 2k
i –4j 2k
Unit Normal Vector =
= 1 ( – 4 2 )
116
4 21 i j k
Directional derivative in the direction of normal
1
= (348 i –144 j 400 k) ( i –4j 2 k)
21
1
= (348 576 800) = 1724
Ans.
21
21
Example 31. If the directional derivative of = a x 2 y + b y 2 z + c z 2 x at the point
x 1 y 3
z
(1, 1, 1) has maximum magnitude 15 in the direction parallel to the line ,
2 2 1
find the values of a, b and c. (U.P. I semester, Winter 2001)
Solution. Given = a x 2 y + b y 2 z + c z 2 x
= i j k
x y z (a x2 y + b y 2 z + c z 2 x)
=
2 2 2
i(2 axy cz ) jax ( 2 byz) kby ( 2 czx)
at the point (1, 1, 1) = i(2 a c) j( a 2 b) kb ( 2 c)
...(1)
We know that the maximum value of the directional derivative is in the direction of .
i.e. || = 15 (2a + c) 2 + (2b + a) 2 + (2c + b) 2 = (15) 2
But, the directional derivative is given to be maximum parallel to the line