3-25 微积分笔记

多个方程组确定隐函数

$F(x,y,u,v),G(x,y,u,v),p_0(x_0,y_0,u_0,v_0)$在邻域$D$有连续偏导，$F|_{p_0}=0,G|_{p_0}=0,\frac{\partial(F,G)}{\partial(u,v)} \neq 0$，则$u=u(x,y),v=v(x,y)$在$(x_0,y_0)$附近存在且具有连续偏导。

对方程组

$$\begin{cases} F(x,y,u,v)=0\\G(x,y,u,v)=0 \end{cases}$$
求偏导，得到
$$\frac{\partial u}{\partial x}=-\frac{\partial (F,G)}{\partial (x,v)}/\frac{\partial (F,G)}{\partial (u,v)},\qquad \frac{\partial v}{\partial x}=-\frac{\partial (F,G)}{\partial (u,x)}/\frac{\partial (F,G)}{\partial (u,v)}$$

例2、设$z=z(x,y)$由$x=e^{u+v},y=e^{u-v},z=uv$确定。求$z'_x,z'_y$.
解：设

$$\begin{cases}F(x,y,u,v,z)=\exp(u+v)-x=0 \\ G(x,y,u,v,z)=\exp(u-v)-y=0 \\ H(x,y,u,v,z)=z-uv = 0\end{cases}$$
然后根据公式搞出来。

或者直接针对原方程组求偏导，然后消元，把$z'_x$表达出来。

例3、设$y=g(x,z),f(x-z,y)=0$，其中$f,g$具有连续偏导数。它们确定了函数$y=y(x)$，求$y'$.
解：由

$$\begin{cases}F(x,y,z)=y-g(x,z)=0 \\ G(x,y,z)=f(x-z,y) = 0\end{cases}$$
得
$$y'=-\frac{\partial (F,G)}{\partial (x,z)}/\frac{\partial (F,G)}{\partial (y,z)}$$
即
$$y' = -\left|\begin{array}{qwq} -g'_x & -g'_z \\f'_1&-f'_1 \end{array}\right|/\left|\begin{array}{qwq} 1 & -g'_z \\f'_2&-f'_1 \end{array}\right|$$

解法二、由

$$\begin{cases}y=g(x,z) \\ f(x-z,y)=0\end{cases}$$ 对$x$求导：
$$\begin{cases} y'=g'_x+g'_z\cdot z'\\f'_1(1-z')+f'_2y'=0 \end{cases}$$
即
$$\begin{cases} y'-g'_zz' = g_x' \\ f_2'y' - f_1'z'=-f'_1 \end{cases}$$

微分法的几何应用

空间曲线的切线和法平面

（一）设空间曲线$\Gamma:x=x(t),y=y(t),z=z(t)$，$t\in I$（$I$为某区间），$p_0(x_0,y_0,z_0)\in \Gamma$对应$t=t_0$，$x(t),y(t),z(t)$可导，且在$t_0$处$(x'(t)+y'(t)+z'(t))|_{t_0}\neq 0$.
考虑割线$M_0M$：$M_0(x_0,y_0,z_0),M(x_0+\Delta x,y_0+\Delta y,z_0+\Delta z)$，分别对应$t_0,t_0+\Delta t$.那么

$$\overrightarrow{MM_0} =\{\Delta x,\Delta y,\Delta z\} \parallel \{\frac{\Delta x}{\Delta t},\frac{\Delta y}{\Delta t},\frac{\Delta z}{\Delta t}\}$$

切线方向向量：

$$\vec{\tau} = \{x'(t),y'(t),z'(t)\}|_{t_0}$$
切线方程：
$$\frac{x-x_0}{x'(t_0)} = \frac{y-y_0}{y'(t_0)} = \frac{z-z_0}{z'(t_0)}$$
法平面方程：
$$x'(t_0)(x-x_0)+y'(t_0)(y-y_0)+z'(t_0)(z-z_0) = 0$$

（二）设空间曲线$\Gamma:\begin{cases}F(x,y,z)=0 \\ G(x,y,z)=0\end{cases}$在$p_0$附近$F,G$具有连续偏导数，且

$$\Bigg [\frac{\partial (F,G)}{\partial (y,z)}\Bigg |_{p_0}\Bigg]^2+\Bigg [\frac{\partial (F,G)}{\partial (z,x)}\Bigg |_{p_0}\Bigg]^2+\Bigg [\frac{\partial (F,G)}{\partial (x,y)}\Bigg |_{p_0}\Bigg]^2 \neq 0$$

不妨设$\frac{\partial (F,g)}{\partial (y,z)}\Big |_{p_0}\neq 0$，则空间曲线$\Gamma:x=x,y=y(x),z=z(x)$

则切线：

$$\vec{\tau} = \{1,y',z'\}$$
其中
$$y'=-\frac{\partial (F,G)}{\partial (x,z)}/\frac{\partial (F,G)}{\partial (y,z)},\qquad z=-\frac{\partial (F,G)}{\partial (y,x)}/\frac{\partial (F,G)}{\partial (y,z)}$$

可取

$$\vec \tau = \Bigg\{ \frac{\partial (F,G)}{\partial (y,z)},\frac{\partial (F,G)}{\partial (z,x)},\frac{\partial (F,G)}{\partial (x,y)} \Bigg\}\Bigg | _{p_0}$$

例1、求螺旋线$x=a\cos \theta,y=a\sin \theta,z=b\theta$在$(a,0,0)$处的切线和法平面。

解：易求切线方向向量$\tau = \{-a\sin \theta,a\cos \theta ,b\}|_{\theta_0=0}=\{0,a,b\}$

切线：

$$\frac{x-a}{0}=\frac ya = \frac zb$$
法平面：
$$ay+bz=0$$

例2、求曲线$\begin{cases}z^2=3x^2+y^2 \\ 2x^2+3y^2+z^2 = 9\end{cases}$在$M_0(1,-1,2)$处的切线方程。

解：易知

$$\begin{cases}F(x,y,z) = z^2 -3x^2 - y^2 = 0\\ G(x,y,z) = 2x^2 + 3y^2 +z^2 -9 = 0\end{cases}$$
切线
$$\vec \tau = \Bigg\{ \frac{\partial (F,G)}{\partial (y,z)},\frac{\partial (F,G)}{\partial (z,x)},\frac{\partial (F,G)}{\partial (x,y)} \Bigg\}\Bigg | _{p_0}$$
可取$\vec\tau=(8,10,7)$.由此可以搞出切线方程。