vector

Vectors 389
Since the scalar product of grad u, grad v and grad w are zero, hence these vectors are
coplanar vectors.
Proved.
Example 18. Find the directional derivative of x 2 y 2 z 2 at the point (1, 1, –1) in the direction
of the tangent to the curve x = e t , y = sin 2t + 1, z = 1 – cos t at t = 0.
(Nagpur University, Summer 2005)
Solution. Let = x 2 y 2 z 2
Directional Derivative of
2 2 2
= = i j k ( x y z )
x y z
2 2 2 2 2 2
= 2xy z i 2yx z j 2zx y k
Directional Derivative of at (1, 1, –1)
Tangent vector,
=
Tangent(at t = 0) =
2 2 2 2 2 2
2(1)(1) ( 1) i 2(1)(1) ( 1) j 2( 1)(1) (1) k
= 2 i 2 j 2k
...(1)
t
r
= xi yj zk e i (sin 2t 1) j (1
cos t)
k
T dr
t
= e i 2cos 2tj
sin tk
dt
0
e i 2(cos 0) j (sin 0) k i 2 j
...(2)
( i 2 j)
Required directional derivative along tangent = (2 i 2 j 2 k)
1
4
[From (1), (2)]
= 2 4 0 6 Ans.
5 5
Example 19. Find the unit normal to the surface xy 3 z 2 = 4 at (–1, –1, 2). (M.U. 2008)
Solution. Let (x, y, z) = xy 3 z 2 = 4
We know that is the vector normal to the surface (x, y, z) = c.
Normal vector = = i
x j
y k
z
Now =
Normal vector =
i ( xyz) j ( xyz) k ( xyz)
x y z
3 2 3 2 3 2
3 2 2 2 3
yz i 3xy z j 2xyzk
Normal vector at (–1, –1, 2) = 4i 12 j 4k
Unit vector normal to the surface at (–1, –1, 2).
4i 12 j 4k
1
= ( i 3 j k ) Ans.
| | 16 144 16 11
Example 20. Find the rate of change of = xyz in the direction normal to the surface
x 2 y + y 2 x + yz 2 = 3 at the point (1, 1, 1). (Nagpur University, Summer 2001)
Solution. Rate of change of =
= i j k ( x yz)
iyz jxz kxy
x y z