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# 高等微積分((

Q:Suppose that the functionh:R3->R is continuously differentiable.Let p be a point in R3 at which ∇h(p) ≠0 and define c= h(p).Use the Implicit Function Theoremto show that there is a neighborhood N of p such that S={(x,y,z)| in N| h(x,y,z)=c}is a surface.

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- mathmanliuLv 71 decade agoFavorite Answer
(∂h/∂x, ∂h/∂y, ∂h/∂z)≠(0,0,0)

不失一般性,可設∂h/∂z ≠0

因 h連續可微分=>存在 P之neighborhood N, 使∂h/∂z≠0

由 implicit function thm: 在N內可視z為x, y之函數=> z=f(x,y)

即得曲面 z= f(x,y) (z 為 x, y之顯函數)

Note:若 ∂h/∂y≠ 0 => y可為 z, x之函數 y= g(x,z)亦得曲面

∂h/∂x≠0 => x=g(y,z)亦得曲面

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