微分与偏导的几何应用

空间曲面的切平面与法向量

首先何谓空间曲面，空间曲面有两种表现形式，一种以显函数形式表示
$$y=f(x,y)$$一种则以参数方程的形式表示
$$\{\begin{array}{l}x=x(u,v)\\ y=y(u,v)\\ z=z(u,v)\end{array}$$如果以显函数形式表示，空间曲面在某点的切平面就表示为其全微分的超平面
$$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$问题是：如果以参数方程的形式表示，切平面该如何求得呢？我们当然是要化成显函数的形式了，但是显函数不总是那么容易求得，这时，我们想到隐函数存在定理。假设$\frac{\partial (x,y)}{\partial (u,v)}\ne 0$，由隐函数存在定理2，在局部$x_0=x(u_0,v_0),y_0=y(u_0,v_0)$就存在一个连续可微的隐函数组
$$\{\begin{array}{l}u=u(x,y)\\ v=v(x,y)\end{array}$$代入第三个方程中，就得到显函数
$$z=z(u(x,y),v(x,y))$$于是
$$\{\begin{array}{l}\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}\end{array}$$只需要我们求出$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$
自然地，我们联想到Cramer法则：$$\{\begin{array}{l}\frac{\partial u}{\partial x}=\frac{\frac{\partial y}{\partial v}}{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\\ \frac{\partial v}{\partial x}=\frac{-\frac{\partial y}{\partial u}}{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\\ \frac{\partial u}{\partial y}=\frac{-\frac{\partial x}{\partial v}}{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\\ \frac{\partial v}{\partial y}=\frac{\frac{\partial x}{\partial u}}{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}\end{array}$$这样，我们就求得了所有偏导数，就可以得到切平面的方程，切平面的法向量就是$(-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1)$如果是参数方程形式$$\frac{\partial z}{\partial x}=\frac{\det \left[\begin{array}{cc}\frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right]}{\det \left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right]}$$$$\frac{\partial z}{\partial y}=\frac{\det \left[\begin{array}{cc}\frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\end{array}\right]}{\det \left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right]}$$
因此，参数方程形式下，法向量为：
$$(\det \left[\begin{array}{cc}\frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}\end{array}\right],\det \left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}\end{array}\right],\det \left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right])$$于是，参数方程形式下，只要以上三个行列式之一不为0，切平面就可以表为$$\frac{\partial (y,z)}{\partial (u,v)}(x-x_0)+\frac{\partial (x,z)}{\partial (u,v)}(y-y_0)+\frac{\partial (x,y)}{\partial (u,v)}(y-y_0)=0$$

空间曲线的切线与法平面

空间曲线则为以下的参数方程形式
$$\{\begin{array}{l}x=x(t)\\ y=y(t)\\ z=z(t)\end{array}$$其中$t\in [\alpha ,\beta ]$，切线的方向向量该如何求解呢？$t\in [\alpha ,\beta ]$，$t+\Delta t\in [\alpha ,\beta ]$，假设$x(t),y(t),z(t)$都可微，那么：方向向量就可以通过逼近的方式得到
$$\frac{1}{\Delta t}[(x(t+\Delta t),y(t+\Delta t),z(t+\Delta t))-(x(t),y(t),z(t))]\to (x^{\prime }(t),y^{\prime }(t),z^{\prime }(t))$$与之正交的一切向量组成其法平面：
$$x^{\prime }(t)(x-x(t))+y^{\prime }(t)(y-y(t))+z^{\prime }(z-z(t))=0$$当然，空间曲线可以由两个空间曲面相交得到
$$\{\begin{array}{l}F(x,y,z)=0\\ G(x,y,z)=0\end{array}$$在这种情况下，如何求得方向向量和法平面呢？我们不妨先讨论如下的情形：
$$\{\begin{array}{l}z=F(x,y)\\ z=G(x,y)\end{array}$$相交得到的空间曲线应当如何求解法平面，假设$\frac{\partial (F,G)}{\partial (x,y)}\ne 0$，那么，由隐函数存在定理，就可以反解出
$$\{\begin{array}{l}x=x(z)\\ y=y(z)\end{array}$$再补充
$$z=z$$就成了参数方程形式
$$\{\begin{array}{l}x=x(z)\\ y=y(z)\\ z=z\end{array}$$同样地，对于更一般的形式，不妨假设$\frac{\partial (F,G)}{\partial (x,y)},\frac{\partial (F,G)}{\partial (x,z)},\frac{\partial (F,G)}{\partial (y,z)}$不全为0，假设$\frac{\partial (F,G)}{\partial (x,y)}\ne 0$，由隐函数存在定理，可以反解出
$$\{\begin{array}{l}x=x(z)\\ y=y(z)\end{array}$$再补充
$$z=z$$就成了参数方程形式
$$\{\begin{array}{l}x=x(z)\\ y=y(z)\\ z=z\end{array}$$下面我们对一般形式进行讨论：
$$x^{\prime }(z)=\frac{\frac{\partial (F,G)}{\partial (y,z)}}{\frac{\partial (F,G)}{\partial (x,y)}}$$$$y^{\prime }(z)=\frac{\frac{\partial (F,G)}{\partial (x,z)}}{\frac{\partial (F,G)}{\partial (x,y)}}$$因此，法平面方程就是：
$$\frac{\partial (F,G)}{\partial (y,z)}(x-x_0)+\frac{\partial (F,G)}{\partial (x,z)}(y-y_0)+\frac{\partial (F,G)}{\partial (x,y)}(z-z_0)=0$$

极值与条件极值

多元函数的极值

现在我们来讨论多元函数的极值问题，首先，同一元函数一样，如果多元函数$f(x_1,\cdots ,x_n)$在$(x_1^0,\cdots ,x_n^0)$处取极值点，并且在$(x_1^0,\cdots ,x_n^0)$连续可微，那么首先在该点对各变元的偏导数为0，即
$$f^{\prime }(x_1^0,\cdots ,x_n^0)=0$$满足这个条件的点就称为驻点，然而驻点不一定都是极值点。同样地，对极值点的判定，我们可以借助“二阶导数”，也就是海色矩阵进行。

定理15.1 $f(x_1,\cdots ,x_n)$在$(x_1^0,\cdots ,x_n^0)$的某个邻域上二阶连续可微，$f^{\prime }(x_1^0,\cdots ,x_n^0)=0$，$H_f(x_0)$为$f(x_1,\cdots ,x_n)$$(x_1^0,\cdots ,x_n^0)$处的海色矩阵，则
(1)$H_f(x_0)$正定，则$f$在$(x_1^0,\cdots ,x_n^0)$处取严格极小值
(2)$H_f(x_0)$负定，则$f$在$(x_1^0,\cdots ,x_n^0)$处取严格极大值
(3)$H_f(x_0)$不定，则$f$在$(x_1^0,\cdots ,x_n^0)$处不取极值

证：
(1)(2)我们仅证明(1)，(2)的证明是类似的
由Taylor公式$$f(x)=f(x_0)+\frac{1}{2}(x-x_0{)}^T{H}_f(x_0)(x-x_0)+o((x-x_0{)}^T(x-x_0))$$由于$H_f(x_0)$正定，存在正定矩阵$Q$，使得$$Q{H}_f(x_0)Q^T=D=diag({\lambda }_1,\cdots ,{\lambda }_n)$$其中，${\lambda }_1,\cdots ,{\lambda }_n$都是正数，是$H_f(x_0)$的特征值。对$x-x_0$作变换$\Delta y=Q(x-x_0)$，就有
$$f(x)-f(x_0)=\frac{1}{2}\Delta y^{T}D\Delta y+o(\Delta y^T\Delta y)$$于是$$\frac{f(x)-f(x_0)}{\Delta x^T\Delta x}=\frac{1}{2}\frac{\Delta y^{T}D\Delta y}{\Delta y^T\Delta y}+\frac{o(\Delta y^T\Delta y)}{\Delta y^T\Delta y}$$
$$\frac{\Delta y^{T}D\Delta y}{\Delta y^T\Delta y}=\frac{{\sum }_{k=1}^n{\lambda }_k\Delta y_k^2}{{\sum }_{k=1}^n\Delta y_k^2}\ge \min ({\lambda }_1,\cdots ,{\lambda }_n)>0\text{\hspace{1em}(1)}$$存在$\delta >0$，当$\Delta x^T\Delta x<\sqrt{\delta }$时
$$|\frac{o(\Delta y^T\Delta y)}{\Delta y^T\Delta y}|<\frac{\min ({\lambda }_1,\cdots ,{\lambda }_n)}{2}$$这样
$$\frac{f(x)-f(x_0)}{\Delta x^T\Delta x}>0$$从而$$f(x)>f(x_0)$$这样$f(x)$在$x_0$处取得严格极小值
反过来，按照公式(1)，如果$H_f(x_0)$不定，那么存在一正一负两个特征值，用相同的方法就可以证明$f(x)$在$x_0$处不取极值，就证得了(3)

条件极值与拉格朗日乘数法

很多情况下，我们都是给定一定的条件求极值，即如下形式的优化问题：目标函数为$$y=f(x_1,\cdots ,x_n)$$条件为
$$\{\begin{array}{l}{g}_1(x_1,\cdots ,x_n)=0\\ g_2(x_1,\cdots ,x_n)=0\\ \cdots \\ g_m(x_1,\cdots ,x_n)=0\end{array}$$其中$m<n$，在该条件下求极值或最值，我们假定$\frac{\partial (g_1,\cdots ,g_m)}{\partial (x_1,\cdots ,x_m)}\ne 0$，由隐函数存在定理，就存在隐函数组
$$\{\begin{array}{l}{x}_1=h_1(x_{m+1},\cdots ,x_n)\\ \cdots \\ x_m=h_m(x_{m+1},\cdots ,x_n)\end{array}$$代入到目标函数中，就化成无条件极值问题
$$y=F(x_{m+1},\cdots ,x_n)=f(h(x_{m+1},\cdots ,x_n),x_{m+1},\cdots ,x_n)$$由取极值的必要条件，就要求
$$\frac{\partial F}{\partial x_k}=0k=m+1,\cdots ,n$$对$k=m+1,\cdots ,n$，就有
$$\sum _{i=1}^m\frac{\partial h_i}{\partial x_k}\frac{\partial f}{\partial x_i}+\frac{\partial f}{\partial x_k}=0\text{\hspace{1em}(2)}$$由隐函数存在定理，令
$$J=\left[\begin{array}{ccc}\frac{\partial g_1}{\partial x_1}& \cdots & \frac{\partial g_1}{\partial x_m}\\ \cdots & \cdots & \cdots \\ \frac{\partial g_m}{\partial x_1}& \cdots & \frac{\partial g_m}{\partial x_m}\end{array}\right]$$由(2)，就有$$-\left[\begin{array}{ccc}\frac{\partial f}{\partial x_1}& \cdots & \frac{\partial f}{\partial x_m}\end{array}\right]J^{-1}\left[\begin{array}{c}\frac{\partial g_1}{\partial x_k}\\ \cdots \\ \frac{\partial g_m}{\partial x_k}\end{array}\right]+\frac{\partial f}{\partial x_k}=0\text{\hspace{1em}(2)}$$令
$$\lambda =\left[\begin{array}{ccc}{\lambda }_1& \cdots & {\lambda }_m\end{array}\right]=-\left[\begin{array}{ccc}\frac{\partial f}{\partial x_1}& \cdots & \frac{\partial f}{\partial x_m}\end{array}\right]J^{-1}$$代入到(3)中，就有$$\frac{\partial f}{\partial x_k}+\sum _{i=1}^m{\lambda }_i\frac{\partial g_i}{\partial x_k}=0$$当然，$k=1,\cdots ,m$时
$$J^{-1}\left[\begin{array}{c}\frac{\partial g_1}{\partial x_k}\\ \cdots \\ \frac{\partial g_m}{\partial x_k}\end{array}\right]$$是线性方程组
$$Jx=\left[\begin{array}{c}\frac{\partial g_1}{\partial x_k}\\ \cdots \\ \frac{\partial g_m}{\partial x_k}\end{array}\right]$$的解，观察$J$的构造，就有
$$J^{-1}\left[\begin{array}{c}\frac{\partial g_1}{\partial x_k}\\ \cdots \\ \frac{\partial g_m}{\partial x_k}\end{array}\right]=e_k$$即除了第$k$个变元为1外，其余变元都为0，因此(3)也成立，而(3就相当于以下函数分别对$x_1,\cdots ,x_n$求偏导并等于0$$L(x_1,\cdots ,x_n,{\lambda }_1,\cdots ,{\lambda }_m)=f(x_1,\cdots ,x_n)+\sum _{i=1}^m{\lambda }_i{g}_i(x_1,\cdots ,x_n)$$在结合条件，取条件极值的必要条件是以上函数对各变元（包括$\lambda$）的偏导数都为0，因而称为拉格朗日乘数法。但是，以上只是取条件极值的必要条件，还不能确认该点是条件极值点。确认条件极值点，一般来讲有两种方法，第一，利用条件海色矩阵，第二，利用最值的存在性。