设函数z=z(x,y)由方程x^2+y^2+z^2=xf(y/x)确定,且f可微求,z对x,y的偏

Question

设函数z=z(x,y)由方程x^2+y^2+z^2=xf(y/x)确定,且f可微求,z对x,y的偏导

Answer 1

x/y=u，f(x/y)=f(u) 2xdx+2ydy+2zdz=f(u)dy+yf'(u)*(ydx-xdy)/y^2=f(u)dy+f'(u)*(ydx-xdy)/y 2xydx+2y^2dy+2yzdz=yf(u)dy+f'(u)*(ydx-xdy)=(x^2+y^2+z^2)dy+f'(u)*(ydx-xdy) y[2x-f'(u)]dx+[y^2-x^2-z^2+xf'(u)]dy+2yzdz=0 f'(u)抵消不掉