关于偏微分

来源:百度知道 编辑:UC知道 时间:2024/09/21 15:26:14
只要答案就行,我想看看自己有没有算错
偏微分和常微分是不一样的~~~题目的答案还是方程~~~~用u,r,θ和偏微分表示

偏导符号打不出,我用 d 来代替吧

r = 根号下(x^2 + y^2)
θ = arc tg(y/x)

dr/dx = …… = x/r
dr/dy = …… = y/r
dθ/dx = …… = -y/r^2
dθ/dy = …… = x/r^2

du/dy = (du/dr)(dr/dy) + (du/dθ)(dθ/dy)
= (du/dr)(y/r) + (du/dθ)(x/r^2)
= (du/dr)sinθ + (du/dθ)(cosθ/r)
du/dx = (du/dr)(dr/dx) + (du/dθ)(dθ/dx)
= (du/dr)(x/r) + (du/dθ)(-y/r^2)
= (du/dr)cosθ - (du/dθ)(sinθ/r)

x du/dy + y du/dx = 0
rcosθ *[(du/dr)sinθ + (du/dθ)(cosθ/r)] + rsinθ * [(du/dr)cosθ - (du/dθ)(sinθ/r)] = 0
2sinθcosθ(du/dr) + {[(cosθ)^2 - (sinθ)^2]/r } (du/dθ) = 0

sin(2θ) (du/dr) + [cos(2θ)/r] (du/dθ) = 0

补充:把du写成dx,dy以及dr,dθ,算出来结果是1/r*(du/dr),看起来还是不错的。