多元函数微分的几何应用

一元向量值函数及其导数

一元向量值函数

由空间解析几何知道，空间曲线$\Gamma$的参数方程为

\begin{align}\left\{\begin{array}{l}x=\varphi(t), \\y=\psi(t), \quad t \in[\alpha, \beta] \\z=\omega(t),\end{array}\right.\end{align}

方程也可以写成向量形式。若记

\begin{align}\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}, \quad \vec{f}(t)=\varphi(t) \vec {i}+\psi(t) \vec{j}+\omega(t) \vec{k}\end{align}

则方程就成为向量方程

\begin{align}\vec{r}=\vec{f}(t), t \in[\alpha, \boldsymbol{\beta}] \end{align}

定义

设数集$D\subset\mathbf{R}$，则称映射$f:D\rightarrow\mathbf{R}^{n}$为一元向量值函数，通常记为

\begin{align}\vec{r}=\vec{f}(t),t\in D\end{align}

其中数集$D$称为函数的定义域，$t$称为自变量，$\vec{r}$称为因变量

向量值函数与数量函数的关系

数量函数

• 一元函数：$y=f(x) \quad x \in D$
• 多元函数：$y=f(x, y)\quad(x, y) \in D$

向量值函数

• 一元向量值函数：$\vec{r}=\vec{j}(t) \quad t \in D$
• 多元向量值函数：$\vec{r}=\vec{f}\left(x_{1}, x_{1}, \cdots, x_{n}\right)$

一元向量值函数的极限

定义

设向量值函数$\vec{f}(t)$在点$t_{0}$的某一去心邻域内有定义，如果存在一个常向量$\vec{r}_{0}$，对于任意给定的正数$\varepsilon$，总存在正数$\delta$，使得当$t$满足$0<\left|t-t_{0}\right|<\delta$时，对应的函数值$\vec{f}(t)$都满足不等式

\begin{align}|\vec{f}(t)-\vec{r}_{0}|<\varepsilon\end{align}

那么，常向量$\vec{r}_{0}$就叫做向量值函数$\vec{f}(t)$当$t \rightarrow t_{0}$时的极限，记作

\begin{align}\lim_{t \to t_0} \vec{f}(t)=\vec{r}_0\end{align}

一元向量值函数的导数

定义

设向量值函数$\vec{r}=\vec{f}(t)$在点$t_{0}$的某一邻域内有定义，如果

\begin{align}\lim _{\Delta t \rightarrow 0} \frac{\Delta \vec{r}}{\Delta t}=\lim _{\Delta t \rightarrow 0} \frac{\vec{f}\left(t_{0}+\Delta t\right)-\vec{f}\left(t_{0}\right)}{\Delta t}\end{align}

存在，那么就称这个极限向量为向量值函数$\vec{r}=\vec{f}(t)$在$t_{0}$处的导数或导向量，记作$\vec{f}^{\prime}(t_{0})$或$\left.\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right|_{t=t_{0}}$

运算法则

设$\vec{u}(t)$、$\vec{v}(t)$是可导的向量值函数，$\overrightarrow{C}$是常向量,$c$是任一常数，$\varphi(t)$是可导的数量函数，则

• $\frac{\mathrm{d}}{\mathrm{d} t} \overrightarrow{C}=\vec{0}$
• $\frac{\mathrm{d}}{\mathrm{d} t}[c\vec{ u}(t)]=c \vec{u}^{\prime}(t)$
• $\frac{\mathrm{d}}{\mathrm{d} t}[\vec{u}(t) \pm \vec{v}(t)]=\vec{u}^{\prime}(t) \pm \vec{v}^{\prime}(t)$
• $\frac{\mathrm{d}}{\mathrm{d} t}[\varphi(t) \vec{u}(t)]=\varphi^{\prime}(t) \vec{u}(t)+\varphi(t) \vec{u}^{\prime}(t)$
• $\frac{\mathrm{d}}{\mathrm{d} t}[\vec{u}(t) \cdot \vec{v}(t)]=\vec{u}^{\prime}(t) \cdot \vec{v}(t)+\vec{u}(t) \cdot \vec{v}^{\prime}(t)$
• $\frac{\mathrm{d}}{\mathrm{d} t}[\vec{u}(t) \times \vec{v}(t)]=\vec{u}^{\prime}(t) \times \vec{v}(t)+\vec{u}(t) \times \vec{v}^{\prime}(t)$
• $\frac{\mathrm{d}}{\mathrm{d} t} \vec{u}[\varphi(t)]=\varphi^{\prime}(t) \vec{u}^{\prime}[\varphi(t)]$

例题

设$\vec{f}(t)=(\cos t)\vec{i}+(\sin t)\vec{j}+t\vec{k}$，求$\lim_{t\rightarrow\frac{\pi}{4}}\vec{f}(t)$及$\vec{f}'(t)$

解

\begin{align} \lim _{t \rightarrow \frac{\pi}{4}} \vec{f}(t) & =\left(\lim _{t \rightarrow \frac{\pi}{4}} \cos t\right) \vec{i}+\left(\lim _{t \rightarrow \frac{\pi}{4}} \sin t\right) \vec{j}+\left(\lim _{t \rightarrow \frac{\pi}{4}} t\right) \vec{k} \\ & =\frac{\sqrt{2}}{2} \vec{i}+\frac{\sqrt{2}}{2} \vec{j}+\frac{\pi}{4} \vec{k} \\\vec{f}'(t)&=(\cos t)' \vec{i}+(\sin t)'\vec{j}+(t)'\vec{k}\\&=(-\sin t)\vec{i}+(\cot)\vec{j}+\vec{k}\end{align}

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设空间曲线$\Gamma$的向量方程为：$\vec{r}=\vec{f}(t)=\left(t^{2}+1,4 t-3,2 t^{2}-6 t\right), t \in R$，求曲线$\Gamma$在与$t=2$相应的点处的单位切向量。

\begin{align}&\vec{f}^{\prime}(t)=(2 t, 4,4 t-6) , t \in R\\&\vec{f}^{\prime}(2)=(4,4,2)=2(2,2,1) \\ &\overrightarrow{T}=(2,2,1),|\overrightarrow{T}|=\sqrt{2^{2}+2^{2}+1}=3\end{align}

由导向量的几何意义知，曲线$\Gamma$在与$t=2$相应的点处的一个单位切向量是$\left(\frac{2}{3},\frac{2}{3},\frac{1}{3}\right)$，其指向与t的增长方向一致；另一个单位切向量是$\left(-\frac{2}{3},-\frac{2}{3}\right.,\left.-\frac{1}{3}\right)$，其指向与t的增长方向相反

空间曲线的切线与法平面

第一种形式

求曲线$\Gamma$与点$M(x_0,y_0,z_0)$处的切线与法平面，记$\vec f(t)=(\varphi(t),\psi(t),\omega(t)),t\in[\alpha,\beta]$。点$M(x_0,y_0,z_0)$对应参数$t_0$

切向量

\begin{align}\overrightarrow{T}=\vec {f}^{\prime}\left(t_{0}\right)=\left(\varphi^{\prime}\left(t_{0}\right),\psi^{\prime}\left(t_{0}\right),\omega^{\prime}\left(t_{0}\right)\right)\end{align}

切线方程

\begin{align}\frac{x-x_{0}}{\varphi^{\prime}\left(t_{0}\right)}=\frac{y-y_{0}}{\psi^{\prime}\left(t_{0}\right)}=\frac{z-z_{0}}{\omega^{\prime}\left(t_{0}\right)}\end{align}

法平面方程

\begin{align}\varphi^{\prime}\left(t_{0}\right)\left(x-x_{0}\right)+\psi^{\prime}\left(t_{0}\right)\left(y-y_{0}\right)+\omega^{\prime}\left(t_{0}\right)\left(z-z_{0}\right)=0\end{align}

例题

求曲线$x=t,y=t^{2},z=t^{3}$在点$(1,1,1)$处的切线及法平面方程

因为$x_{t}^{\prime}=1,y_{t}^{\prime}=2t,z_{t}^{\prime}=3t^{2}$，而点$(1,1,1)$所对应的参数$t_{0}=1$。所以切向量

\begin{align}\overrightarrow{T}=(1,2,3)\end{align}

切线方程为

\begin{align}\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{3}\end{align}

法平面方程为

\begin{align}(x-1)+2(y-1)+3(z-1)=0\end{align}

化简后得

\begin{align}x+2y+3z=6\end{align}

第二种形式

如果空间曲线$\Gamma$的方程以下列的形式给出

\begin{align}\left\{\begin{array}{l}y=\varphi(x)\\z=\psi(x)\end{array}\right.\end{align}

设$M(x_0,y_0,z_0)$为曲线方程上$\Gamma$的一点求过$M$点的切线及其法平面方程，取$x$为参数，它就可以表示为参数方程的形式

\begin{align}\left\{\begin{array}{l}x=x\\y=\varphi(x)\\z=\psi(x)\end{array}\right.\end{align}

点$M$对应参数$x=x_{0}$。则根据上面的讨论可知，$\overrightarrow{T}=\left(1,\varphi^{\prime}\left(x_{0}\right)\right.,\left.\psi^{\prime}\left(x_{0}\right)\right)$，因此曲线$\Gamma$在点$M\left(x_{0},y_{0},z_{0}\right)$处的切线方程为

\begin{align}\frac{x-x_{0}}{1}=\frac{y-y_{0}}{\varphi^{\prime}\left(x_{0}\right)}=\frac{z-z_{0}}{\psi^{\prime}\left(x_{0}\right)}\end{align}

在点$M\left(x_{0},y_{0},z_{0}\right)$处的法平面方程为

\begin{align}\left(x-x_{0}\right)+\varphi^{\prime}\left(x_{0}\right)\left(y-y_{0}\right)+\psi^{\prime}\left(x_{0}\right)\left(z-z_{0}\right)=0\end{align}

第三种形式

设空间曲线$\Gamma$的方程以下列的形式给出

\begin{align}\left\{\begin{array}{l}F(x,y,z)=0\\G(x,y,z)=0\end{array}\right.\end{align}

$M\left(x_{0},y_{0},z_{0}\right)$是曲线$\Gamma$上的一个点，求过$M$的切线和法平面方程。设$F$和$G$有对各个变量的连续偏导数，且在点$M\left(x_{0},y_{0},z_{0}\right)$的某一邻域内确定了一组函数$y=\varphi(x),z=\psi(x)$则

\begin{align}\begin{array}{l}F[x,\varphi(x),\psi(x)]\equiv0\\G[x,\varphi(x),\psi(x)]\equiv0\end{array}\end{align}

两边分别对$x$求全导数

\begin{align}\left\{\begin{array}{l} \frac{\partial F}{\partial x}+\frac{\partial F}{\partial y} \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\partial F}{\partial z} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ \frac{\partial G}{\partial x}+\frac{\partial G}{\partial y} \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\partial G}{\partial z} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.\end{align}

由假设可知，在点$M$的某个邻域内

\begin{align}J=\frac{\partial(F, G)}{\partial(y, z)} \neq 0\end{align}

故可解得

\begin{align}\frac{\mathrm{d} y}{\mathrm{~d} x}=\varphi^{\prime}(x)=\frac{\left|\begin{array}{ll}F_{z} & F_{x} \\G_{z} & G_{x}\end{array}\right|}{\left|\begin{array}{ll}F_{y} & F_{z} \\G_{y} & G_{z}\end{array}\right|}, \frac{\mathrm{d} z}{\mathrm{~d} x}=\psi^{\prime}(x)=\frac{\left|\begin{array}{ll}F_{x} & F_{y} \\G_{x} & G_{y}\end{array}\right|}{\left|\begin{array}{ll}F_{y} & F_{z} \\G_{y} & G_{z}\end{array}\right|}\end{align}

切向量

\begin{align}\overrightarrow{T}=\left(1,\varphi^{\prime}\left(x_{0}\right),\psi^{\prime}\left(x_{0}\right)\right)\end{align}

分子分母中带下标$M$的行列式表示行列式在点$M\left(x_{0},y_{0},z_{0}\right)$的值。把上面的切向量$\overrightarrow{T}$乘$\left|\begin{array}{ll}F_{y}&F_{z}\\G_{y}&G_{z}\end{array}\right|_{M}$，得

\begin{align}\overrightarrow{T}_{1}=\left(\left|\begin{array}{ll} F_{y} & F_{z} \\ G_{y} & G_{z} \end{array}\right|_{M},\left|\begin{array}{ll} F_{z} & F_{x} \\ G_{z} & G_{x} \end{array}\right|_{M},\left|\begin{array}{ll} F_{x} & F_{y} \\ G_{x} & G_{y} \end{array}\right|_{M}\right)\end{align}

这也是曲线$\Gamma$在点$M$处的一个切向量。

由此可写出曲线$\Gamma$在点$M\left(x_{0},y_{0},z_{0}\right)$处的切线方程为

\begin{align}\frac{x-x_{0}}{\left|\begin{array}{cc} F_{y} & F_{z} \\ G_{y} & G_{z} \end{array}\right|_{M}}=\frac{y-y_{0}}{\left|\begin{array}{cc} F_{z} & F_{x} \\ G_{z} & G_{x} \end{array}\right|_{M}}=\frac{z-z_{0}}{\left|\begin{array}{cc} F_{x} & F_{y} \\ G_{x} & G_{y} \end{array}\right|_{M}}\end{align}

曲线$\Gamma$在点$M\left(x_{0},y_{0},z_{0}\right)$处的法平面方程为

\begin{align}\left|\begin{array}{ll} F_{y} & F_{z} \\ G_{y} & G_{z} \end{array}\right|_{M}\left(x-x_{0}\right)+\left|\begin{array}{ll} F_{z} & F_{x} \\ G_{z} & G_{x} \end{array}\right|_{M}\left(y-y_{0}\right)+\left|\begin{array}{ll} F_{x} & F_{y} \\ G_{x} & G_{y} \end{array}\right|_{M}\left(z-z_{0}\right)=0\end{align}

例题

求曲线$x^{2}+y^{2}+z^{2}=6,x+y+z=0$在点$(1,-2,1)$处的切线及法平面方程

将所给方程的两边对$x$求导

\begin{align}\left\{\begin{array}{l} 2 x+2 y \cdot \frac{d y}{d x}+2 z \cdot \frac{d z}{d x}=0 \\ 1+\frac{d y}{d x}+\frac{d z}{d x}=0 \end{array}\right.\end{align}

移项化简后为

\begin{align}\left\{\begin{array}{l} y \frac{\mathrm{d} y}{\mathrm{~d} x}+z \frac{\mathrm{d} z}{\mathrm{~d} x}=-x \\ \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\mathrm{d} z}{\mathrm{~d} x}=-1 \end{array}\right.\end{align}

由此得

\begin{align} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left|\begin{array}{cc} -x & z \\ -1 & 1 \end{array}\right|}{\left|\begin{array}{ll} y & z \\ 1 & 1 \end{array}\right|}=\frac{z-x}{y-z}, \frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{\left|\begin{array}{cc} y & -x \\ 1 & -1 \end{array}\right|}{\left|\begin{array}{ll} y & z \\ 1 & 1 \end{array}\right|}=\frac{x-y}{y-z} \\ \left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{(1,-2,1)}=0,\left.\frac{\mathrm{d} z}{\mathrm{~d} x}\right|_{(1,-2,1)}=-1 \\ \end{align}

从而切向量为

\begin{align}\overrightarrow{T}=(1,0,-1)\end{align}

切线方程为

\begin{align}\frac{x-1}{1}=\frac{y+2}{0}=\frac{z-1}{-1}\end{align}

法平面方程为

\begin{align}(x-1)+0 \cdot(y+2)-(z-1)=0\end{align}

化简后得

\begin{align}x-z=0\end{align}

曲面的切平面与法线

曲面为$F(x,y,z)=0$的情形

设$M(\left.x_{0},y_{0},b_{0}\right)$为曲面$\Sigma$上任意一点，在曲面$\Sigma$上，通过点$M$任取一条曲线

\begin{align}\Gamma:\left\{\begin{array}{l}x=\varphi(t)\\y=\psi(t)\\z=w(t)\end{array}\quad\alpha\leqslant t\leqslant\beta\right.\end{align}

$M\left(x_{0},y_{0},z_{0}\right)$对应参数为$t=t_{0}$

则过$M$的切线方程为

\begin{align}\frac{x-x_{0}}{\varphi^{\prime}\left(t_{0}\right)}=\frac{y-y_{0}}{\psi^{\prime}\left(t_{0}\right)}=\frac{z-z_{0}}{\omega^{\prime}\left(t_{0}\right)}\end{align}

切平面方程

\begin{align}F_{x}\left(x_{0}, y_{0}, z_{0}\right)\left(x-x_{0}\right)+F_{y}\left(x_{0}, y_{0}, z_{0}\right)\left(y-y_{0}\right)+ F_{z}\left(x_{0}, y_{0}, z_{0}\right)\left(z-z_{0}\right)=0\end{align}

法线方程

\begin{align}\frac{x-x_{0}}{F_{x}\left(x_{0}, y_{0}, z_{0}\right)}=\frac{y-y_{0}}{F_{y}\left(x_{0}, y_{0}, z_{0}\right)}=\frac{z-z_{0}}{F_{z}\left(x_{0}, y_{0}, z_{0}\right)}\end{align}

曲面法向量

\begin{align}\vec n=\left(F_{x}\left(x_{0}, y_{0}, z_{0}\right), F_{y}\left(x_{0}, y_{0}, z_{0}\right), F_{z}\left(x_{0}, y_{0}, z_{0}\right)\right)\end{align}

曲面$z=f(x,y)$的情形

令$F(x,y,z)=f(x,y)-z=0$可见

\begin{align}F_{x}(x, y, z)=f_{x}(x, y), F_{y}(x, y, z)=f_{y}(x, y), F_{z}(x, y, z)=-1\end{align}

切平面方程为

\begin{align}f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)-\left(z-z_{0}\right)=0\end{align}

法线方程为

\begin{align}\frac{x-x_{0}}{f_{x}\left(x_{0}, y_{0}\right)}=\frac{y-y_{0}}{f_{y}\left(x_{0}, y_{0}\right)}=\frac{z-z_{0}}{-1}\end{align}

方向余弦

\begin{align}\cos \alpha=\frac{-f_{x}}{\sqrt{1+f_{x}^{2}+f_{y}^{2}}}, \quad \cos \beta=\frac{-f_{y}}{\sqrt{1+f_{x}^{2}+f_{y}^{2}}}, \quad \cos \gamma=\frac{1}{\sqrt{1+f_{x}^{2}+f_{y}^{2}}}\end{align}

例题

求球面$x^{2}+y^{2}+z^{2}=14$在点$(1,2,3)$处的切平面及法线方程

\begin{align}F(x,y,z)=x^{2}+y^{2}+z^{2}-14\end{align}

\begin{align}F_x=2x,F_y=2y,F_z=2z\end{align}

所以

\begin{align}\left.\vec{n}\right|_{(1,2,3)}=(2,4,6)\end{align}

所以在点$(1,2,3)$处此球面的切平面方程为

\begin{align}2(x-1)+4(y-2)+6(z-3)=0\end{align}

即

\begin{align}x+2y+3z-14=0\end{align}

法线方程为

\begin{align}\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\end{align}

即

\begin{align}\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\end{align}

由此可见,法线经过原点(即球心)

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求旋转抛物面$z=x^{2}+y^{2}-1$在点$(2,1,4)$处的切平面及法线方程

\begin{align}f(x,y)=x^{2}+y^{2}-1\end{align}

\begin{align}&\vec{n}=\left(f_{x},f_{y},-1\right)=(2x,2y,-1)\\&\left.\vec{n}\right|_{(2,1,4)}=(4,2,-1)\end{align}

所以在点$(2,1,4)$处的切平面方程为

\begin{align}4(x-2)+2(y-1)-(z-4)=0\end{align}

即

\begin{align}4x+2y-z-6=0\end{align}

法线方程为

\begin{align}\frac{x-2}{4}=\frac{y-1}{2}=\frac{z-4}{-1}\end{align}