Re: [理工] [電磁學]-David cheng向量一題
※ 引述《aacvbn ( 姿勢+)》之銘言:
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: For a scalar function f and a vector A , prove that
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: ▽˙(fA) = f▽˙A + A ˙▽f in cartesian coordinates.
: 謝謝高手指導~
▽˙(fA) = f▽˙A + A ˙▽f in cartesian coordinates.
let f=g(x,y,z)
A=Bi+Cj+Dk B,C,D is (x,y,z) function. not const.
@Bg(x,y,z) @Cg(x,y,z) @Dg(x,y,z)
▽˙(fA)=----------- + ----------- + ----------
@x @y @z
=(Bxg+Bgx) + (Cyg+Cgy) + (Dzg +Dgz)
=g(Bx+Cy+Dz) + (Bgx+Cgy+Dgz) ...(1)
f▽˙A=g(Bx+Cy+Dz)
A ˙▽f=Bgx+Cgy+Dgz
f▽˙A + A ˙▽f =g(Bx+Cy+Dz)+Bgx+Cgy+Dgz ....(2)
(1)=(2)
=>▽˙(fA)= f▽˙A + A ˙▽f
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接下來乖孩子不要學,0分
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let ▽=▽_f+▽_A ,_x表僅對x作用
▽˙(fA)=(▽_f+▽_A )˙(fA)
=(▽_f)˙(fA)+▽_A˙(fA)
=A˙▽_ff+f▽_A˙A
=A˙▽f + f▽˙A
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◆ From: 123.193.214.165
※ 編輯: iyenn 來自: 123.193.214.165 (11/14 01:43)
※ 編輯: iyenn 來自: 123.193.214.165 (11/14 01:45)
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