推 aacvbn:真的拿證明題束手無側呀~~ 太感謝你了!! 11/15 20:19
※ 引述《aacvbn ( 姿勢+)》之銘言：
: ->
: for a saclar function f and a vector G , prove that
: -> -> ->
: ▽╳(fG ) = f▽╳G + (▽f) ╳ G in Cartesian coordinates.
: 謝謝!
勤勞點......就能K.O.這類的題目了= =
▽╳(fG ) = f▽╳G + (▽f) ╳ G
let f=g(x,y,z)
G=Ai+Bj+Ck A,B,C is x,y,z function
▽╳(fG=Agi+Bgj+Cgk)=| i j k |
| |
| @ @ @ |
| --- --- ----|
| @x @y @z |
| |
| Ag Bg Cg |
=i{(Cyg+Cgy)-(Bzg+Bgz)}+
j{(Azg+Agz)-(Cxg+Cgx)}+
k{(Bxg+Bgx)-(Ayg+Agy)} ...(1)
f▽╳G =gi{Cy-Bz}+gj{Az-Cx}+gk{Bx-Ay} ....(2)
(▽f) ╳ G={gxi+gyj+gzk}╳ {Ai+Bj+Ck}
=i{Cgy-Bgz}+j{Agz-Cgx}+k{Bgx-Agy} ....(3)
(2)+(3)=(1)
▽╳(fG ) = f▽╳G + (▽f) ╳ G Q.E.D.
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