GDC-12
3
34 Solution (a) Rewrite as 2 2 14 y x z 14 2 2 2 z y x Thus, we obtain 2 2 2 ) , , ( z y x z y x Then, ) ( 2 2 2 2 k j i k j i z y x z y x At the point (−1,3,2), The unit normal vector is 14 2 3 ˆ k j i n 14 2 2 3 ) 1 ( 2 2 2 2 and ) 2 3 ( 2 k j i
-
Upload
gthulasi787126 -
Category
Documents
-
view
214 -
download
0
description
GDC
Transcript of GDC-12
-
34
Solution
(a) Rewrite as2214 yxz 14222 zyx
Thus, we obtain222),,( zyxzyx
Then, )(2222 kjikji zyxzyx
At the point (1,3,2),
The unit normal vector is14
23
kjin
14223)1(2 222 and)23(2 kji
-
35
(b) Rewrite as
Thus, we obtain
Then,
At the point (1,0,1),
The unit normal vector is6
2
kjin
6211 222 and
23),,( zyxezyx y
kji zyxee yy 2)3( 2
23 zyxey 023 zyxey
kjikji 2)1(2])0(31[ 200 ee
-
36
Divergence of Vector Fields
The divergence is an operator that
measures the magnitude of a vector field's
source or sink at a given point
The divergence of a vector field is a scalar
(x,y,z)V (x dx,y,z)V
x x+dx