(急)多元函数微积分证明题
设函数u=f(x,,y,z),x=rsinψcosθ,y=rsinψsinθ,z=rcosψ,,其中f具有连续偏导数,证明:1.如果xdu/dx+ydu/dy+zdu/d...
设函数u=f(x,,y,z),x=rsinψcosθ,y=rsinψsinθ,z=rcosψ,,其中f具有连续偏导数,证明:
1.如果xdu/dx+ydu/dy+zdu/dz=0,则u仅是ψ和θ的函数;
2.如果(du/dx)/x=(du/dy)y=(du/dz)/z,则u仅是r的函数、
偏导符号打不出来,用d代替。 展开
1.如果xdu/dx+ydu/dy+zdu/dz=0,则u仅是ψ和θ的函数;
2.如果(du/dx)/x=(du/dy)y=(du/dz)/z,则u仅是r的函数、
偏导符号打不出来,用d代替。 展开
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1. du/dr = du/dx * dx/dr + du/dy * dy/dr + du/dz * dz/dr
= du/dx sinψcosθ + du/dy sinψsinθ + du/dz cosψ
= 1/r [xdu/dx + ydu/dy +zdu/dz) =0
所以u与r无关,只是ψ和θ的函数
2.同理,
du/dθ = du/dx dx/dθ + du/dy dy/dθ + du/dz dz/dθ
= -rsinθsinψ du/dx + rsinψcosθdu/dy
= -r^2sin^2ψ sinθ cosθ[du/dx /x - du/dy /y]=0
同理可以怎么du/dψ =0
所以u只与r有关
= du/dx sinψcosθ + du/dy sinψsinθ + du/dz cosψ
= 1/r [xdu/dx + ydu/dy +zdu/dz) =0
所以u与r无关,只是ψ和θ的函数
2.同理,
du/dθ = du/dx dx/dθ + du/dy dy/dθ + du/dz dz/dθ
= -rsinθsinψ du/dx + rsinψcosθdu/dy
= -r^2sin^2ψ sinθ cosθ[du/dx /x - du/dy /y]=0
同理可以怎么du/dψ =0
所以u只与r有关
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