P9-P13多元函数微积分

2019-10-01 本文已影响0人陈文瑜

解析几何

多元函数基本概念

二元函数
$z = f(x,y)$
几何意义就是一个曲面
二元函数极限
$\lim_{x \rightarrow x_0,y \rightarrow y_0}f(x,y)$
以任意方式逼近 $(x_0,y_0)$

偏导数

导数定义
$\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = \lim _{\Delta x \rightarrow 0} \frac {f(x_0 + \Delta x)-f(x_0)}{\Delta x}$
对 $x$ 的偏导数( $y$ 不变)
$\lim_{\Delta x \rightarrow 0} \frac {\Delta_x Z}{\Delta x} = \lim _{\Delta x \rightarrow 0} \frac {f(x_0 +\Delta x,y_0) - f(x_0,y_0)}{\Delta x}$
$f_x^\prime (x_0,y_0) \quad \frac {\partial f(x_0,y_0)}{\partial x}$
二阶偏导
$\frac {\partial}{\partial x}({\frac {\partial z}{\partial x})} = \frac {\partial^2 z}{\partial x^2} = z_{xx}^{\prime \prime} = f _{xx}^{\prime \prime}(x,y)$

全微分

偏导数
偏增量
全增量
$\Delta z = f(x+\Delta x,y+\Delta y)-f(x,y)$
全增量面积近似
$\Delta S = (x+\Delta x)(y +\Delta y)-xy \approx y\Delta x + x\Delta y$
全微分
$dz = f^\prime_x dx +f^\prime _y dy$
偏微分
$d_x z \quad d_y z$
三元 $u = f(x,y,z)$
$du = f^\prime _x dx +f^\prime _y dy +f^\prime _z dz$

多元复合函数求导

$z = f(x,y) \quad z_x^\prime \quad z_y^\prime$
$z=f(u,v) \quad u = \phi (x,y) \quad v=\phi(x,y)$
$z \rightarrow u\rightarrow x \quad \quad z \rightarrow u \rightarrow y$
$z \rightarrow v\rightarrow x \quad \quad z \rightarrow v \rightarrow y$

$\frac {\partial z}{\partial x} = \frac {\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$

$\frac {\partial z}{\partial y} = \frac {\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$

举一个栗子
$z = uv +sint \quad u=e^t \quad v=cost$
分析
$z \rightarrow u\rightarrow t \quad \quad z \rightarrow v \rightarrow t \quad z \rightarrow t \rightarrow t$
求解
$\frac{dz}{dt} = \frac {\partial z}{\partial u}\frac {du}{dt} +\frac {\partial z}{\partial v} \frac{\partial v}{\partial t} + \frac {\partial z}{\partial t}$
理解 $z=f(u,x,y) \quad u=\phi (x,y)$
$\frac {\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac {\partial f}{\partial x}$

二阶偏导

表达方式
$\frac {\partial ^2 z}{\partial x^2} \quad \frac {\partial ^2 z}{\partial x \partial y}$
栗子
$z=f(xy,x^2+y^2) \quad z =f(u,v) \quad u=xy \quad v = x^2+y^2$
注意事项
$f^\prime _u \quad f^\prime _v 任然是u \quad v 的函数$
也就是
$\frac {\partial f_u^\prime}{\partial x} = f_{uu}^{\prime \prime} \frac {\partial u}{\partial x} + f_{uv}^{\prime \prime} \frac {\partial v}{\partial x}$
求解
$\frac {\partial z}{\partial x} = f_u^\prime \frac {\partial u}{\partial x} + f_v^\prime \frac {\partial v}{\partial x} = f_u ^\prime y + f_v ^\prime 2x$
$\frac {\partial ^2 z}{\partial x^2} = ....$

隐函数求导

$F(x,y) = 0$ 右边为0

求解过程
$F(x,f(x)) =0$
$\frac {\partial F}{\partial x} + \frac {\partial F}{\partial y} \frac {dy}{dx} = 0$
$\frac {dy}{dx} = - \frac {\frac {\partial F}{\partial x}}{\frac {\partial F}{\partial y}} = - \frac {F_x ^\prime}{F _y ^\prime}$
再来一个三元的
$F(x,y,z) = 0$
$\frac {\partial z}{\partial x} = - \frac {F _x ^\prime}{F_z ^\prime}$