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高数小题,用全微分形式不变性怎么解呢?
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2016-06-29 · 知道合伙人教育行家
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两边同时微分,得到
F1'·d(x+z/y)+F2'·d(y+z/x)=0
F1'·(dx-z/y²·dy+1/y·dz)
+F2'·(-z/x²·dx+dy+1/x·dz)=0
整理得到:
(F1'·1/y+F2'·1/x)dz
=(F2'·z/x²-F1')dx+(F1'·z/y²-F2')dy
所以,
dz=(F2'·z/x²-F1')/(F1'·1/y+F2'·1/x)·dx
+(F1'·z/y²-F2')/(F1'·1/y+F2'·1/x)·dy
=(F2'·yz-F1'·x²y)/(F1'·x²+F2'·xy)·dx
+(F1'·xz-F2'·xy²)/(F1'·xy+F2'·y²)·dy
根据全微分的叠加原理,
两个偏导数分别为:
zx=(F2'·yz-F1'·x²y)/(F1'·x²+F2'·xy)
zy=(F1'·xz-F2'·xy²)/(F1'·xy+F2'·y²)
F1'·d(x+z/y)+F2'·d(y+z/x)=0
F1'·(dx-z/y²·dy+1/y·dz)
+F2'·(-z/x²·dx+dy+1/x·dz)=0
整理得到:
(F1'·1/y+F2'·1/x)dz
=(F2'·z/x²-F1')dx+(F1'·z/y²-F2')dy
所以,
dz=(F2'·z/x²-F1')/(F1'·1/y+F2'·1/x)·dx
+(F1'·z/y²-F2')/(F1'·1/y+F2'·1/x)·dy
=(F2'·yz-F1'·x²y)/(F1'·x²+F2'·xy)·dx
+(F1'·xz-F2'·xy²)/(F1'·xy+F2'·y²)·dy
根据全微分的叠加原理,
两个偏导数分别为:
zx=(F2'·yz-F1'·x²y)/(F1'·x²+F2'·xy)
zy=(F1'·xz-F2'·xy²)/(F1'·xy+F2'·y²)
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