Question

证明由方程F（x-az,y-bz）=0确定的函数z=z（x,y）应满足a（ðz/ðx）+b（

Answer 1

设u=x-az,v=y-bz

则,原方程写为 F(u,v)=0

方程F(u,v)=0 两端分别对x,y求偏导得

ðF/ðx=ðF/ðu*(ðu/ðx+ðu/ðz*ðz/ðx)+ðF/ðv*(ðv/ðz*ðz/ðx)

=ðF/ðu*(1-a*ðz/ðx)+ðF/ðv*(-b*ðz/ðx)

=ðF/ðu-a*ðF/ðu*ðz/ðx-b*ðF/ðv*ðz/ðx

=ðF/ðu-(a*ðF/ðu+b*ðF/ðv)ðz/ðx

=0

得：ðz/ðx=a*(ðF/ðu)/(a*ðF/ðu+b*ðF/ðv)

ðF/ðy=ðF/ðu*(ðu/ðz*ðz/ðy)+ðF/ðv*(ðv/ðy+ðv/ðz*ðz/ðy)

=ðF/ðu*(-a*ðz/ðy)+ðF/ðv*(1-b*ðz/ðy)

=-a*ðF/ðu*ðz/ðy+ðF/ðv-b*ðF/ðv*ðz/ðy

=ðF/ðv-(a*ðF/ðu+b*ðF/ðv)*ðz/ðy

得：ðz/ðy=a*(ðF/ðv)/(a*ðF/ðu+b*ðF/ðv)

则,a(ðz/ðx)+b(ðz/ðy)

=a*(ðF/ðu)/(a*ðF/ðu+b*ðF/ðv)+b*(ðF/ðv)/(a*ðF/ðu+b*ðF/ðv)

=(a*ðF/ðu+b*ðF/ðv)/(a*ðF/ðu+b*ðF/ðv)

=1