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(ðz/ðy)=1
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/21 05:43:37
![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(ðz/ðy)=1](/uploads/image/z/5272371-27-1.jpg?t=%E8%AF%81%E6%98%8E%E7%94%B1%E6%96%B9%E7%A8%8BF%EF%BC%88x-az%2Cy-bz%EF%BC%89%3D0%E7%A1%AE%E5%AE%9A%E7%9A%84%E5%87%BD%E6%95%B0z%3Dz%EF%BC%88x%2Cy%EF%BC%89%E5%BA%94%E6%BB%A1%E8%B6%B3a%EF%BC%88%26%23240%3Bz%2F%26%23240%3Bx%EF%BC%89%2Bb%EF%BC%88%26%23240%3Bz%2F%26%23240%3By%EF%BC%89%3D1)
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(ðz/ðy)=1
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(ðz/ðy)=1
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(ðz/ðy)=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
不知道 不好意思
用代入法求