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

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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