多元函数微积分期中复习复盘笔记

多元微积分期中

极限与连续

定义

对于函数 $f:U\to V$
重极限：

\forall ε > 0, \exists δ > 0, 当 x \in U, ‖ x - x_{0} ‖ < 0 时, ‖ f (x) - A ‖ < ε ⇓ lim_{x \to x_{0}} f (x) = A

$\forall \varepsilon >0,\exists \delta>0,当\bm x\in U,\|\bm x-\bm x_0\|<0时,\|f(\bm x)-A\|<\varepsilon\\\Downarrow\\\lim_{\bm x\to\bm x_0}f(\bm x)=A\\$

连续：

lim_{x \to x_{0}} f (x) = f (x_{0})

$\lim_{\bm x\to\bm x_0}f(\bm x)=f(\bm x_0)\\$

累次极限：

形 如 lim_{x \to x_{0}} lim_{y \to y_{0}} f (x, y) 、 lim_{y \to y_{0}} lim_{x \to x_{0}} f (x, y)

$形如\lim_{x\to x_0}\lim_{y\to y_0}f(x,y)、\lim_{y\to y_0}\lim_{x\to x_0}f(x,y)\\$

重极限与累次极限的关系

累次极限存在但不相等 $\to$ 重极限不存在
重极限存在、累次极限存在 $\to$ 累次极限相等

例题

例1：求极限:

lim_{x \to 0, y \to 0} \frac{\sqrt{x y + 1} - 1}{x y}

$\lim_{x→0,y→0}\frac{\sqrt{xy+1}-1}{xy}$

解：

lim_{x \to 0, y \to 0} \frac{\sqrt{x y + 1} - 1}{x y} = lim_{x \to 0, y \to 0} \frac{x y}{x y \sqrt{x y + 1} + 1} = lim_{x \to 0, y \to 0} \frac{1}{\sqrt{x y + 1} + 1} = \frac{1}{2}

$\lim_{x→0,y→0}\frac{\sqrt{xy+1}-1}{xy}=\lim_{x→0,y→0} \dfrac{xy}{xy\sqrt{xy+1}+1}=\lim_{x→0,y→0}\dfrac{1}{\sqrt{xy+1}+1}=\frac{1}{2}$

例2：求极限

lim_{x \to \infty, y \to \infty} \frac{x + y}{x^{2} - x y + y^{2}}

$\lim_{x→∞,y→∞} \dfrac{x+y}{x^2-xy+y^2}$

解：由于 $|\dfrac{x+y}{x^2-xy+y^2}|≤|\dfrac{\frac{1}{y}+\frac{1}{x}}{\frac{x}{y}+\frac{y}{x}-1}|≤|\dfrac{1}{y}+\dfrac{1}{x}|→0$ ，故由夹逼准则知极限为0。

例3：证明

lim_{x \to 0, y \to 0} \frac{x y^{2}}{x^{2} + y^{2}} = 0

$\lim_{x\to0,y\to 0}\dfrac{xy^2}{x^2+y^2}=0\\$

解：

| \frac{x y^{2}}{x^{2} + y^{2}} | = | y | \cdot \frac{| x y |}{x^{2} + y^{2}} \leq \frac{1}{2} | y | \to 0

$\left|\dfrac{xy^2}{x^2+y^2}\right|=|y|\cdot\dfrac{|xy|}{x^2+y^2}\le \frac{1}{2}|y|\to 0\\$

例4：求极限

lim_{x \to 0, y \to 0} \frac{\sin (x^{2} y + y^{4})}{x^{2} + y^{2}}

$\lim_{x\to0,y\to0}\dfrac{\sin(x^2y+y^4)}{x^2+y^2}\\$

解：
由 $|\sin x|\le |x|$ ，则

\begin{aligned} | \frac{\sin (x^{2} y + y^{4})}{x^{2} + y^{2}} | & \leq | \frac{x^{2} y + y^{4}}{x^{2} + y^{2}} | \\ \leq | \frac{x^{2} y}{x^{2} + y^{2}} | + | \frac{y^{4}}{x^{2} + y^{2}} | = | y | \cdot \frac{x^{2}}{x^{2} + y^{2}} + y^{2} \cdot \frac{y^{2}}{x^{2} + y^{2}} \\ \leq | y | + y^{2} \to 0 \end{aligned}

$\begin{aligned}\left|\dfrac{\sin(x^2y+y^4)}{x^2+y^2}\right|&\le \left|\dfrac{x^2y+y^4}{x^2+y^2}\right|\\&\le\left|\dfrac{x^2y}{x^2+y^2}\right|+\left|\dfrac{y^4}{x^2+y^2}\right|=|y|\cdot\dfrac{x^2}{x^2+y^2}+y^2\cdot\dfrac{y^2}{x^2+y^2}\\&\le |y|+y^2\to 0\end{aligned}\\$

例5：证明极限 $\displaystyle \lim_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2y^2+(x-y)^2}$ 不存在
解：
当 $(x,y)$ 沿着 $y=x$ 趋近于 $(0,0)$ 时

lim_{x \to 0, y = x} \frac{x^{2} y^{2}}{x^{2} y^{2} + (x - y)^{2}} = lim_{x \to 0} \frac{x^{4}}{x^{4} + (x - x)^{2}} = 1

$\lim_{x\to 0,y=x}\dfrac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x\to 0}\dfrac{x^4}{x^4+(x-x)^2}=1\\$

当 $(x,y)$ 沿着 $y=0$ 趋近于 $(0,0)$ 时

lim_{x \to 0, y = 0} \frac{x^{2} y^{2}}{x^{2} y^{2} + (x - y)^{2}} = 0

$\lim_{x\to 0,y=0}\dfrac{x^2y^2}{x^2y^2+(x-y)^2}=0\\$

因此极限不存在

多元函数的泰勒公式

对于函数 $f(x,y)$ ,将其在 $(x_0,y_0)$ 处进行展开

设

g (t) = f (x_{0} + t (x - x_{0}), y_{0} + t (y - y_{0}))

$g(t)=f(x_0+t(x-x_0),y_0+t(y-y_0))$

则

g (1) = f (x, y)

$g(1)=f(x,y)\\$

由泰勒公式，有

g (t) = g (0) + g^{'} (0) t + \frac{1}{2} g^{″} (0) t^{2} + \frac{1}{6} g^{‴} (ξ) t^{2} ξ \in (0, t)

$g(t)=g(0)+g'(0)t+\frac{1}{2}g''(0)t^2+\cfrac{1}{6}g'''(\xi)t^2\quad \xi\in(0,t)\\$

则

g (1) = g (0) + g^{'} (0) + \frac{1}{2} g^{″} (0) + \frac{1}{6} g^{‴} (ξ) ξ \in (0, 1)

$g(1)=g(0)+g'(0)+\cfrac{1}{2}g''(0)+\cfrac{1}{6}g'''(\xi)\quad\xi\in(0,1)\\$

又

\begin{aligned} g (0) & = f (x_{0}, y_{0}) \\ g^{'} (0) & = (x - x_{0}) \frac{\partial f}{\partial x} |_{(x_{0}, y_{0})} + (y - y_{0}) \frac{\partial f}{\partial y} |_{(x_{0}, y_{0})} \\ g^{″} (0) & = (x - x_{0})^{2} \frac{\partial^{2} f}{\partial x^{2}} |_{(x_{0}, y_{0})} + 2 (x - x_{0}) (y - y_{0}) \frac{\partial^{2} f}{\partial x \partial y} |_{(x_{0}, y_{0})} + (y - y_{0})^{2} \frac{\partial^{2} f}{\partial y^{2}} \end{aligned}

$\begin{aligned}g(0)&=f(x_0,y_0)\\g'(0)&=(x-x_0)\cfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}+(y-y_0)\cfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\\g''(0)&=(x-x_0)^2\cfrac{\partial^2 f}{\partial x^2}\bigg|_{(x_0,y_0)}+2(x-x_0)(y-y_0)\cfrac{\partial^2 f}{\partial x\partial y}\bigg|_{(x_0,y_0)}+(y-y_0)^2\cfrac{\partial ^2f}{\partial y^2}\end{aligned}\\$

则

\begin{aligned} f (x, y) & = f (x_{0}, y_{0}) \\ + (x - x_{0}) \frac{\partial f}{\partial x} |_{(x_{0}, y_{0})} + (y - y_{0}) \frac{\partial f}{\partial y} |_{(x_{0}, y_{0})} \\ + \frac{1}{2} (x - x_{0})^{2} \frac{\partial^{2} f}{\partial x^{2}} |_{(x_{0}, y_{0})} + (x - x_{0}) (y - y_{0}) \frac{\partial^{2} f}{\partial x \partial y} |_{(x_{0}, y_{0})} + \frac{1}{2} (y - y_{0})^{2} \frac{\partial^{2} f}{\partial y^{2}} \\ + o (x^{2} + y^{2}) \end{aligned}

$\begin{aligned}f(x,y)&=f(x_0,y_0)\\&+(x-x_0)\cfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}+(y-y_0)\cfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\\&+\cfrac{1}{2}(x-x_0)^2\cfrac{\partial^2 f}{\partial x^2}\bigg|_{(x_0,y_0)}+(x-x_0)(y-y_0)\cfrac{\partial^2 f}{\partial x\partial y}\bigg|_{(x_0,y_0)}+\cfrac{1}{2}(y-y_0)^2\cfrac{\partial ^2f}{\partial y^2}\\&+o(x^2+y^2)\end{aligned}\\$

或

\begin{aligned} f (x, y) & = f (x_{0}, y_{0}) \\ + (x - x_{0}) \frac{\partial f}{\partial x} |_{(x_{0}, y_{0})} + (y - y_{0}) \frac{\partial f}{\partial y} |_{(x_{0}, y_{0})} \\ + \frac{1}{2} (x - x_{0})^{2} \frac{\partial^{2} f}{\partial x^{2}} |_{M} + (x - x_{0}) (y - y_{0}) \frac{\partial^{2} f}{\partial x \partial y} |_{M} + \frac{1}{2} (y - y_{0})^{2} \frac{\partial^{2} f}{\partial y^{2}} |_{M} \end{aligned} M \in (x_{0}, x) \times (y_{0}, y)

$\begin{aligned}f(x,y)&=f(x_0,y_0)\\&+(x-x_0)\cfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}+(y-y_0)\cfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\\&+\cfrac{1}{2}(x-x_0)^2\cfrac{\partial^2 f}{\partial x^2}\bigg|_{M}+(x-x_0)(y-y_0)\cfrac{\partial ^2f}{\partial x\partial y}\bigg|_{M}+\cfrac{1}{2}(y-y_0)^2\cfrac{\partial^2 f}{\partial y^2}\bigg|_{M}\end{aligned}\\M\in(x_0,x)\times(y_0,y)\\$

可微、全微分与偏导数

向量值函数的可微性

对于向量值函数 $\bm f:\mathbb{R^m}\to\mathbb{R^n}$ ，如果存在线性映射 $L:\mathbb{R^m}\to\mathbb{R^n}$ 满足

lim_{h \to 0} \frac{‖ f (x_{0} + h) - f (h) - L (h) ‖}{‖ h ‖} = 0

$\lim_{\bm h\to 0}\dfrac{\|\bm f(\bm x_0+\bm h)-\bm f(\bm h)-L(\bm h)\|}{\|\bm h\|}=0\\$

则称其在 $\bm x_0$ 处可微，记微分

D f (h) = L (h)

$\mathrm{D}\bm f(\bm h)=L(\bm h)\\$

标量函数的可微性、全微分与偏导数

如果一个多元标量函数 $f$ 的全增量能够写成线性增量加上一个无穷小量，即

Δ f = A Δ x + B Δ y + o (ρ)

$\Delta f = A\Delta x + B\Delta y + o(\rho)$

其中 $\rho = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ ，那么该函数在该点可微，其中 $A$ 和 $B$ 分别是该点对 $x$ 和 $y$ 的偏导数，记为 $\dfrac{\partial f}{\partial x}=A,\dfrac{\partial f}{\partial y}=B$ 。

即

lim_{\begin{matrix} Δ x \to 0 \\ Δ y \to 0 \end{matrix}} \frac{Δ f - A Δ x - B Δ y}{\sqrt{(Δ x)^{2} + (Δ y)^{2}}} = 0

$\lim_{\begin{matrix}\small\scriptstyle\Delta x\to 0\\\small\scriptstyle\Delta y\to 0\end{matrix}}\frac{\Delta f-A\Delta x-B\Delta y}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=0$

或

lim_{x \to x_{0}, y \to y_{0}} \frac{f (x, y) - f (x_{0}, y_{0}) - A (x - x_{0}) - B (y - y_{0})}{\sqrt{(x - x_{0})^{2} + (y - y_{0})^{2}}} = 0

$\lim_{x\to x_0,y\to y_0}\dfrac{f(x,y)-f(x_0,y_0)-A(x-x_0)-B(y-y_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0\\$

记微分

d f (x, y) = A x + B y

$\mathrm{d}f(x,y)=Ax+By$

偏导数的定义如下：

f_{x} (x_{0}, y_{0}) = \frac{\partial f}{\partial x} |_{(x_{0}, y_{0})} = lim_{x \to x_{0}} \frac{f (x, y_{0}) - f (x_{0}, y_{0})}{x - x_{0}} f_{y} (x_{0}, y_{0}) = \frac{\partial f}{\partial y} |_{(x_{0}, y_{0})} = lim_{y \to y_{0}} \frac{f (x_{0}, y) - f (x_{0}, y_{0})}{y - y_{0}}

$f_x(x_0,y_0)=\dfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}=\lim_{x\to x_0}\dfrac{f(x,y_0)-f(x_0,y_0)}{x-x_0}\\f_y(x_0,y_0)=\dfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}=\lim_{y\to y_0}\dfrac{f(x_0,y)-f(x_0,y_0)}{y-y_0}$

全微分为

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y

$\mathrm{d}f=\dfrac{\partial f}{\partial x}\mathrm{d}x+\dfrac{\partial f}{\partial y}\mathrm{d}y\\$

雅各比矩阵

\frac{\partial (y_{1}, y_{2})}{\partial (x_{1}, x_{2})} = [\begin{matrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} \\ \frac{\partial y_{2}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{2}} \end{matrix}]

$\cfrac{\partial (y_1,y_2)}{\partial (x_1,x_2)}=\begin{bmatrix}\cfrac{\partial y_1}{\partial x_1}&\cfrac{\partial y_1}{\partial x_2}\\\cfrac{\partial y_2}{\partial x_1}&\cfrac{\partial y_2}{\partial x_2}\end{bmatrix}$

可微与偏导数连续的关系

（1）偏导数连续 $f_x,f_y$ 均连续，说明 $f$ 可微，但 $f$ 可微不一定偏导数连续

函数可微但偏导数不连续的例子：

f (x, y) = {\begin{matrix} (x^{2} +^{2}) \sin \frac{1}{x^{2} + y^{2}} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{matrix}

$f(x,y)=\left\{\begin{matrix}(x^2+^2)\sin\frac{1}{x^2+y^2}&(x,y)\ne (0,0)\\0&(x,y)=(0,0)\end{matrix}\right.\\$

（2） $f$ 可微，则一阶偏导数均存在
（3） $f$ 在 $(x,y)$ 处可微，则 $f$ 一定在 $(x,y)$ 处连续

例题

例1：已知理想气体状态方程 $PV=RT$ （ $R$ 为常数），求证

\frac{\partial P}{\partial V} \cdot \frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} = - 1

$\dfrac{\partial P}{\partial V}\cdot\dfrac{\partial V}{\partial T}\cdot\dfrac{\partial T}{\partial P}=-1\\$

解：

\begin{aligned} P = \frac{R T}{V} \Rightarrow \frac{\partial P}{\partial V} = - \frac{R T}{V^{2}} \\ V = \frac{R T}{P} \Rightarrow \frac{\partial V}{\partial T} = \frac{R}{P} \\ T = \frac{P V}{R} \Rightarrow \frac{\partial T}{\partial P} = \frac{V}{R} \end{aligned}

$\begin{aligned}&P=\dfrac{RT}{V}\Rightarrow\dfrac{\partial P}{\partial V}=-\dfrac{RT}{V^2}\\&V=\dfrac{RT}{P}\Rightarrow \dfrac{\partial V}{\partial T}=\dfrac{R}{P}\\&T=\dfrac{PV}{R}\Rightarrow\dfrac{\partial T}{\partial P}=\dfrac{V}{R}\end{aligned}\\$

则

\frac{\partial P}{\partial V} \cdot \frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} = - \frac{R T}{V^{2}} \cdot \frac{R}{P} \cdot \frac{V}{R} = - \frac{R T}{P V} = - 1

$\dfrac{\partial P}{\partial V}\cdot\dfrac{\partial V}{\partial T}\cdot\dfrac{\partial T}{\partial P}=-\dfrac{RT}{V^2}\cdot\dfrac{R}{P}\cdot\dfrac{V}{R}=-\dfrac{RT}{PV}=-1\\$

方向导数与梯度

方向导数

对函数 $f:v\to \mathbb{R}$ ，记 $f_{\bm v}:t\to f(\bm x+t\bm v)$ ，若 $f_{\bm v}$ 对 $t$ 的微分在 $t=0$ 处存在，那么可定义 $f$ 在 $\bm x$ 处沿方向 $\bm v$ 的导数为

D_{v} f (x) = \frac{\partial f}{\partial v} = \frac{d f_{v}}{d t} |_{t = 0} = lim_{t \to 0} \frac{f (x + t v) - f (x)}{t}

$\mathrm{D}_{\bm v}f(\bm x)=\dfrac{\partial f}{\partial \bm v}=\frac{\mathrm{d}f_{\bm v}}{\mathrm{d}t}\bigg|_{t=0}=\lim_{t\to 0}\dfrac{f(\bm x+t\bm v)-f(\bm x)}{t}\\$

当 $\boldsymbol{v}$ 是单位向量时， $D_{\boldsymbol{v}}f(x_0, y_0)$ 就是函数 $f(x, y)$ 在 $(x_0, y_0)$ 处的方向导数。

$\mathrm{D}_{\bm v}(f+g)=\mathrm{D}_{\bm v}f+\mathrm{D}_{\bm v}g$
$\mathrm{D}_{\bm v}(cf)=c\mathrm{D}_{\bm v}f$
$\mathrm{D}_{\bm v}(fg)=g\mathrm{D}_{\bm v}f+f\mathrm{D}_{\bm v}g$
$h:\mathbb{R}\to\mathbb{R},g:U\to \mathbb{R}$ ，则 $\mathrm{D}_{\bm v}(h\circ g)(\bm x)=h'(g(\bm x))\mathrm{D}_{\bm v}g(\bm x)$
方向导数存在时，偏导数不一定存在；可微是方向导数存在的充分条件，而不是必要条件。例： $f=\sqrt{x^2+y^2}$

梯度

梯度是一个矢量，梯度的方向是方向导数中取到最大值的方向，梯度的值是方向导数的最大值，函数 $f(\bm x)$ 在 $\bm x_0$ 处的梯度记为 $\nabla f(\bm x_0)$ 、 $\cfrac{\partial f(\bm x_0)}{\partial \bm x}$ 或 $\text{grad}f(\bm x_0)$ ，且

\nabla f (x_{0}) = {[\begin{matrix} \frac{\partial f}{\partial x_{1}} & \dots & \frac{\partial f}{\partial x_{n}} \end{matrix}]}^{T}

$\nabla f(\bm x_0)=\begin{bmatrix}\cfrac{\partial f}{\partial x_1}&\cdots&\cfrac{\partial f}{\partial x_n}\end{bmatrix}^T\\$

$\nabla (c_1f+c_2g)=c_1\nabla f+c_2\nabla g$
$\nabla(fg)=f\nabla g+g\nabla f$
$\nabla\left(\cfrac{f}{g}\right)=\cfrac{g\nabla f-f\nabla g}{g^2}$
$h:\mathbb{R}\to\mathbb{R},g:U\to \mathbb{R}$ ，则 $\nabla(h\circ g)(\bm x)=h'(g(\bm x))\nabla g(\bm x)$

方向导数与梯度的关系

梯度的方向是方向导数中取到最大值的方向
梯度的值是方向导数的最大值
$\mathrm{D}_{\bm v}f=\nabla f\cdot \bm v$

例题

例1：设 $\displaystyle f(x,y)=\left\{\begin{matrix}\cfrac{xy^2}{x^2+y^4}&x^2+y^2\ne 0\\0&x^2+y^2=0\end{matrix}\right.$ ，求 $f$ 沿 $\vec{e}=(\cos \theta,\sin \theta)$ 在点 $(0,0)$ 的方向导数
解：
当 $\cos\theta\ne 0$ 时

\begin{aligned} \frac{\partial f}{\partial \vec{e}} |_{(0, 0)} & = lim_{ρ \to 0} \frac{f (ρ \cos θ, ρ \sin θ) - f (0, 0)}{ρ} \\ = lim_{ρ \to 0} \frac{\cos θ \sin^{2} θ}{\cos^{2} θ + ρ^{2} \sin^{4} θ} \\ = \frac{\sin^{2} θ}{\cos θ} \end{aligned}

$\begin{aligned}\frac{\partial f}{\partial \vec{e}}\bigg|_{(0,0)}&=\lim_{\rho\to 0}\frac{f(\rho \cos\theta,\rho\sin\theta)-f(0,0)}{\rho}\\&=\lim_{\rho\to 0}\frac{ \cos\theta\sin^2\theta}{\cos^2\theta+\rho^2\sin^4\theta}\\&=\frac{\sin^2\theta}{\cos\theta}\end{aligned}\\$

当 $\cos\theta =0$ 时， $f(\rho \cos\theta,\rho\sin\theta)=0$

\frac{\partial f}{\partial \vec{e}} |_{(0, 0)} = 0

$\frac{\partial f}{\partial \vec{e}}\bigg|_{(0,0)}=0\\$

例2：求函数 $z=xe^{2y}$ 在点 $P(1,0)$ 处沿着从点 $P(1,0)$ 到点 $Q(2,-1)$ 的方向的方向导数
解：

\vec{v} = \vec{P Q} = (1, - 1)

$\vec{v}=\overrightarrow{PQ}=(1,-1)$

\hat{v} = (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2})

$\hat{v}=\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\\$

\nabla z (1, 0) = (1, 2)

$\nabla z(1,0)=(1,2)\\$

\frac{\partial z}{\partial \hat{v}} |_{(1, 0)} = \nabla z (1, 0) \cdot \hat{v} = - \frac{\sqrt{2}}{2}

$\frac{\partial z}{\partial \hat{v}}\bigg|_{(1,0)}=\nabla z(1,0)\cdot\hat{v}=-\frac{\sqrt{2}}{2}\\$

例3：设 $u=xyz+z^2+5$ ，求在点 $M(0,1,-1)$ 处方向导数的最大值和最小值
解：

\begin{aligned} ∵ \frac{\partial u}{\partial x} = y z \frac{\partial u}{\partial y} = x z \frac{\partial u}{\partial z} = x y + 2 z \\ ∴ grad u (0, 1, - 1) = (y z, x z, x y + 2 z) |_{(0, 1, - 1)} = (- 1, 0, - 2) \end{aligned}

$\begin{aligned}&\because \cfrac{\partial u}{\partial x}=yz\qquad\cfrac{\partial u}{\partial y}=xz\qquad\cfrac{\partial u}{\partial z}=xy+2z\\&\therefore\text{grad}\;u(0,1,-1)=(yz,xz,xy+2z)|_{(0,1,-1)}=(-1,0,-2)\end{aligned}\\$

因此

max {\frac{\partial u}{\partial v}} = ‖ grad u ‖ = \sqrt{5} min {\frac{\partial u}{\partial v}} = - ‖ grad u ‖ = - \sqrt{5}

$\max\left\{\frac{\partial u}{\partial \bm v}\right\}=\|\text{grad}\;u\|=\sqrt{5}\\\min\left\{\frac{\partial u}{\partial \bm v}\right\}=-\|\text{grad}\;u\|=-\sqrt{5}\\$

多元复合函数求导

推导

(x, y) \mapsto (u, v) \mapsto f

$(x,y)\mapsto(u,v)\mapsto f$

则

d f = \frac{\partial f}{\partial u} d u + \frac{\partial f}{\partial v} d v

$\mathrm{d}f=\frac{\partial f}{\partial u}\mathrm{d}u+\frac{\partial f}{\partial v}\mathrm{d}v$

d u = \frac{\partial u}{\partial x} d x + \frac{\partial u}{\partial y} d y

$\mathrm{d}u=\frac{\partial u}{\partial x}\mathrm{d}x+\frac{\partial u}{\partial y}\mathrm{d}y$

d v = \frac{\partial v}{\partial x} d x + \frac{\partial v}{\partial y} d y

$\mathrm{d}v=\frac{\partial v}{\partial x}\mathrm{d}x+\frac{\partial v}{\partial y}\mathrm{d}y\\$

因此

d f = (\frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x}) d x + (\frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y}) d y

$\mathrm{d}f=\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\right)\mathrm{d}x+\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\right)\mathrm{d}y\\$

例题

例1：若 $g(x,y,z)=e^{x^3+y^2+z}$ ， $f(x,y)=g(x,y,x\sin y)$ ，求 $\cfrac{\partial f}{\partial x}$

\begin{aligned} d f = d g & = \frac{\partial g}{\partial x} d x + \frac{\partial g}{\partial y} d y + \frac{\partial g}{\partial z} d z \\ = (\frac{\partial g}{\partial x} + \frac{\partial g}{\partial z} \frac{\partial z}{\partial x}) d x + (\frac{\partial g}{\partial y} + \frac{\partial g}{\partial z} \frac{\partial z}{\partial y}) d y \end{aligned}

$\begin{aligned}\mathrm{d}f=\mathrm{d}g&=\frac{\partial g}{\partial x}\mathrm{d}x+\frac{\partial g}{\partial y}\mathrm{d}y+\frac{\partial g}{\partial z}\mathrm{d}z\\&=\left(\frac{\partial g}{\partial x}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial x}\right)\mathrm{d}x+\left(\frac{\partial g}{\partial y}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial y}\right)\mathrm{d}y\end{aligned}\\$

则

\frac{\partial f}{\partial x} = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial z} \frac{\partial z}{\partial x} = (3 x^{2} + \sin y) e^{x^{3} + y^{2} + x \sin y}

$\frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial x}=(3x^2+\sin y)e^{x^3+y^2+x\sin y}\\$

例2：若函数 $f(x+y,x-y)=x^2-y^2$ ，求 $\cfrac{\partial f(x,y)}{\partial x}+\cfrac{\partial f(x,y)}{\partial y}$
解：
令

{\begin{matrix} u = x + y \\ v = x - y \end{matrix} \Rightarrow {\begin{matrix} x = \frac{u + v}{2} \\ y = \frac{u - v}{2} \end{matrix}

$\left\{\begin{matrix}u=x+y\\v=x-y\end{matrix}\right.\Rightarrow \left\{\begin{matrix}x=\cfrac{u+v}{2}\\y=\cfrac{u-v}{2}\end{matrix}\right.$

则

\begin{aligned} \frac{\partial f (u, v)}{\partial u} + \frac{\partial f (u, v)}{\partial v} & = (\frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}) + (\frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}) \\ = \frac{1}{2} (u + v) + \frac{1}{2} (u - v) + \frac{1}{2} (u + v) - \frac{1}{2} (u - v) \\ = u + v \end{aligned}

$\begin{aligned}\frac{\partial f(u,v)}{\partial u}+\frac{\partial f(u,v)}{\partial v}&=\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\right)+\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}\right)\\&=\frac{1}{2}(u+v)+\frac{1}{2}(u-v)+\frac{1}{2}(u+v)-\frac{1}{2}(u-v)\\&=u+v\end{aligned}\\$

则

\frac{\partial f (x, y)}{\partial x} + \frac{\partial f (x, y)}{\partial y} = x + y

$\cfrac{\partial f(x,y)}{\partial x}+\cfrac{\partial f(x,y)}{\partial y}=x+y$

隐函数定理及逆映射定理

隐函数定理

设 $F:U\subset \mathbb{R^{m+n}}\to \mathbb{R^m}$ 是一个 $\mathscr{C}^p$ 映射（P阶光滑），若 $\bm x_0\in \mathbb{R^n},\bm y_0\in \mathbb{R^m},(\bm x_0,\bm y_0)\in U$ 满足

$F(\bm x_0,\bm y_0)=\bm 0$
$\dfrac{\partial F}{\partial \bm y}(\bm x_0,\bm y_0)$ 是 $m$ 阶可逆矩阵

则存在 $(\bm x_0,\bm y_0)$ 的领域 $U_x\times U_y\subset U$ ，以及 $\mathscr{C}^p$ 映射 $f:U_x\to U_y$ ，使得

F (x, y) = 0 \Leftrightarrow y = f (x)

$F(\bm x,\bm y)=0\Leftrightarrow \bm y=f(\bm x)\\$

且

f^{'} (x) = - {(\frac{\partial F}{\partial y})}^{- 1} (\frac{\partial F}{\partial x})

$f'(\bm x)=-\left(\frac{\partial F}{\partial \bm y}\right)^{-1}\left(\frac{\partial F}{\partial \bm x}\right)\\$

其中 $f'(\bm x)$ 表示 $f(\bm x)$ 的 $\text{Jacobi}$ 矩阵

:::danger

$F$ 与 $f$ 的像空间的维数相同，即 $F(\bm x,\bm y)$ 的维数与 $f(\bm x)$ 的维数相同
:::

反函数定理

设 $U\subset\mathbb{R^n}$ ， $f:U\to\mathbb{R^n}$ 是一个 $\mathscr{C}^p$ 的映射，若 $f$ 在 $\bm x_0$ 处的 $\text{Jacobi}$ 矩阵是可逆矩阵，则反函数 $f^{-1}$ 也是 $\mathscr{C}^p$ 的映射，且

(f^{- 1})^{'} (y) = (f^{'} (x))^{- 1}

$(f^{-1})'(\bm y)=(f'(\bm x))^{-1}\\$

或记作

J_{f^{- 1}} (y) = J_{f} (x)^{- 1}

$J_{f^{-1}}(\bm y)=J_f(\bm x)^{-1}\\$

关于逆映射定理的证明？ - 知乎：https://www.zhihu.com/question/67456647

例题

例1：已知方程 $x^2 + y^2 - 4x - 6y + 12 = 0$ ，求在点 $(1,2)$ 处的切线方程。

解：
令 $F(x,y)=x^2+y^2-4x-6y+12$

\frac{\partial F}{\partial x} = 2 x - 4 \frac{\partial F}{\partial y} = 2 y - 6

$\frac{\partial F}{\partial x}= 2x-4 \qquad \frac{\partial F}{\partial y}= 2y - 6$

在点 $(1,2)$ 处，代入可得：

\frac{\partial F}{\partial x} |_{(1, 2)} = - 2 \frac{\partial F}{\partial y} |_{(1, 2)} = - 2

$\frac{\partial F}{\partial x} \bigg|_{(1,2)} = -2 \qquad \frac{\partial F}{\partial y} \bigg|_{(1,2)} = -2$

斜率为

\frac{d y}{d x} |_{(1, 2)} = - \frac{{\frac{\partial F}{\partial x} |}_{(1, 2)}}{{\frac{\partial F}{\partial y} |}_{(1, 2)}} = - 1

$\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(1,2)}=-\dfrac{\left.\frac{\partial F}{\partial x}\right|_{(1,2)}}{\left.\frac{\partial F}{\partial y}\right|_{(1,2)}}=-1\\$

因此切线方程为：

x + y = 3

$x+y=3$

例2：方程 $x^3+y^3+z^3=x+y+z$ 在点 $(x,y,z)=(1,1,-1)$ 附近确定了一个二阶连续可微的隐函数 $z=z(x,y)$ ，满足 $z(1,1)=-1$ ，则 $\cfrac{\partial ^2z}{\partial x\partial y}(1,1)=\_\_\_\_\_$
解：
方程

x^{3} + y^{3} + z^{3} (x, y) = x + y + z (x, y)

$x^3+y^3+z^3(x,y)=x+y+z(x,y)\\$

对 $x$ 求偏导得：

3 x^{2} + 3 z^{2} (x, y) \frac{\partial z}{\partial x} = 1 + \frac{\partial z}{\partial x}

$3x^2+3z^2(x,y)\cfrac{\partial z}{\partial x}=1+\cfrac{\partial z}{\partial x}\\$

再对 $y$ 求偏导得：

6 z (x, y) \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} + 3 z^{2} (x, y) \frac{\partial^{2} z}{\partial x \partial y} = \frac{\partial^{2} z}{\partial x \partial y}

$6z(x,y)\cfrac{\partial z}{\partial x}\cdot\cfrac{\partial z}{\partial y}+3z^2(x,y)\cfrac{\partial^2 z}{\partial x\partial y}=\cfrac{\partial ^2z}{\partial x\partial y}\\$

则

\frac{\partial^{2} z}{\partial x \partial y} = \frac{6 z}{1 - 3 z^{2}} \cdot \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y}

$\cfrac{\partial^2 z}{\partial x\partial y}=\cfrac{6z}{1-3z^2}\cdot\cfrac{\partial z}{\partial x}\cdot\cfrac{\partial z}{\partial y}\\$

又

\frac{\partial z}{\partial x} = - \frac{1 - 3 x^{2}}{1 - 3 z^{2}} \frac{\partial z}{\partial y} = - \frac{1 - 3 y^{2}}{1 - 3 z^{2}}

$\cfrac{\partial z}{\partial x}=-\cfrac{1-3x^2}{1-3z^2}\qquad\cfrac{\partial z}{\partial y}=-\cfrac{1-3y^2}{1-3z^2}\\$

则

\frac{\partial^{2} z}{\partial x \partial y} = \frac{6 z (1 - 3 x^{2}) (1 - 3 y^{2})}{(1 - 3 z^{2})^{3}}

$\cfrac{\partial^2 z}{\partial x\partial y}=\cfrac{6z(1-3x^2)(1-3y^2)}{(1-3z^2)^3}\\$

\frac{\partial^{2} z}{\partial x \partial y} |_{(1, 1, - 1)} = 3

$\cfrac{\partial ^2z}{\partial x\partial y}\bigg|_{(1,1,-1)}=3\\$

极值

多元函数的无条件极值

当我们在优化一个函数时，Hesse矩阵是一个非常重要的概念。它是一个二阶导数矩阵，表示函数的曲率和凸性。在本篇笔记中，我们将学习Hesse矩阵的定义、性质以及如何计算它。

Hesse矩阵的定义

设 $f(x_1,x_2,...,x_n)$ 是一个具有二阶连续偏导数的函数，则 $f$ 的Hesse矩阵 $H$ 定义为：

H_{i, j} = \frac{\partial^{2} f}{\partial x_{i} \partial x_{j}}

$H_{i,j}=\frac{\partial^2 f}{\partial x_i \partial x_j}$

其中 $1\leq i,j \leq n$ 。

Hesse矩阵的性质

Hesse矩阵是一个对称矩阵，即 $H_{i,j}=H_{j,i}$ 。这是由于二阶偏导数的次序不影响结果。

如果 $f$ 是一个凸函数，则 $H$ 是一个半正定矩阵。

如果 $f$ 是一个严格凸函数，则 $H$ 是一个正定矩阵。

Hesse矩阵的计算

计算Hesse矩阵需要对每个变量求二阶偏导数。以下是一些常见函数的Hesse矩阵：

$f(x,y)=x^2+y^2$ 的Hesse矩阵为：

H = [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}]

$H=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$f(x,y)=xy$ 的Hesse矩阵为：

H = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]

$H=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

$f(x,y)=x^3+y^3-3xy$ 的Hesse矩阵为：

H = [\begin{matrix} 6 x - 6 y & - 6 \\ - 6 & 6 y - 6 x \end{matrix}]

$H=\begin{bmatrix} 6x-6y & -6 \\ -6 & 6y-6x \end{bmatrix}$

我们也可以用LaTeX来表示Hesse矩阵。例如， $H=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ 可以表示为 $H_{i,j}=\frac{\partial^2 f}{\partial x_i \partial x_j}=\begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} \\ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} \end{bmatrix}=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ 。

以上就是Hesse矩阵的学习笔记，希望对你有所帮助！

首先令 $\cfrac{\partial f}{\partial x_i}=0$ ，求出满足条件的点 $\bm x$

多元函数的条件极值

对于一个多元函数

f (x_{1}, x_{2}, \dots, x_{n})

$f(x_1,x_2,\cdots,x_n)$

如果存在一些约束条件

g_{1} (x_{1}, x_{2}, \dots, x_{n}) = 0 g_{2} (x_{1}, x_{2}, \dots, x_{n}) = 0 ⋮ g_{m} (x_{1}, x_{2}, \dots, x_{n}) = 0

$g_1(x_1,x_2,\cdots,x_n)=0\\g_2(x_1,x_2,\cdots,x_n)=0\\\vdots\\g_m(x_1,x_2,\cdots,x_n)=0$

可以通过拉格朗日乘数法来找到其条件极值点。

具体来说，我们需要建立拉格朗日函数：

L (x_{1}, x_{2}, \dots, x_{n}, λ_{1}, λ_{2}, \dots, λ_{m}) = f (x_{1}, x_{2}, \dots, x_{n}) - \sum_{i = 1}^{m} λ_{i} g_{i} (x_{1}, x_{2}, \dots, x_{n})

$L(x_1,x_2,\cdots,x_n,\lambda_1,\lambda_2,\cdots,\lambda_m) = f(x_1,x_2,\cdots,x_n) - \sum_{i=1}^m\lambda_i g_i(x_1,x_2,\cdots,x_n)$

其中 $\lambda_1,\lambda_2,\cdots,\lambda_m$ 是拉格朗日乘数。然后我们需要求出拉格朗日函数的偏导数，并令其等于零，即：

\frac{\partial L}{\partial x_{1}} = 0, \frac{\partial L}{\partial x_{2}} = 0, \dots, \frac{\partial L}{\partial x_{n}} = 0 \frac{\partial L}{\partial λ_{1}} = 0, \frac{\partial L}{\partial λ_{2}} = 0, \dots, \frac{\partial L}{\partial λ_{m}} = 0

$\frac{\partial L}{\partial x_1} = 0,\frac{\partial L}{\partial x_2} = 0,\cdots,\frac{\partial L}{\partial x_n} = 0\\\;\\\frac{\partial L}{\partial \lambda_1} = 0,\frac{\partial L}{\partial \lambda_2} = 0,\cdots,\frac{\partial L}{\partial \lambda_m} = 0$

解这个方程组，就可以得到函数的条件极值点。

例题

求解函数 $f(x,y)=x^2+y^2$ 在约束条件 $g(x,y)=x+y-1=0$ 下的条件极值点。

首先，建立拉格朗日函数：

L (x, y, λ) = f (x, y) - λ g (x, y) = x^{2} + y^{2} - λ (x + y - 1)

$L(x,y,\lambda)=f(x,y)-\lambda g(x,y)=x^2+y^2-\lambda(x+y-1)$

然后求出偏导数：

\frac{\partial L}{\partial x} = 2 x - λ = 0

$\frac{\partial L}{\partial x}=2x-\lambda=0$

\frac{\partial L}{\partial y} = 2 y - λ = 0

$\frac{\partial L}{\partial y}=2y-\lambda=0$

\frac{\partial L}{\partial λ} = x + y - 1 = 0

$\frac{\partial L}{\partial \lambda}=x+y-1=0$

解这个方程组，得到 $x=y=\frac{1}{2}$ ，代入原函数可得条件极值点 $(\frac{1}{2},\frac{1}{2})$ 。

接下来，我们需要判断这个点是否为条件极值点。我们可以通过求二阶偏导数来判断。计算得到：

\frac{\partial^{2} L}{\partial x^{2}} = 2, \frac{\partial^{2} L}{\partial y^{2}} = 2, \frac{\partial^{2} L}{\partial x \partial y} = 0

$\frac{\partial^2 L}{\partial x^2}=2,\frac{\partial^2 L}{\partial y^2}=2,\frac{\partial^2 L}{\partial x \partial y}=0$

几何应用

曲面

对于曲面 $z(x,y)$ 上一点 $(x_0,y_0,z_0)$ ，由于

\begin{matrix} (1) & d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y \end{matrix}

$\tag1\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y$

同时

[\begin{matrix} d x \\ d y \\ d z \end{matrix}] = [\begin{matrix} x - x_{0} \\ y - y_{0} \\ z - z_{0} \end{matrix}]

$\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\\\mathrm{d}z\end{bmatrix}=\begin{bmatrix}x-x_0\\y-y_0\\z-z_0\end{bmatrix}\\$

得到切平面方程

z - z_{0} = \frac{\partial z}{\partial x} (x - x_{0}) + \frac{\partial z}{\partial y} (y - y_{0})

$z-z_0=\frac{\partial z}{\partial x}(x-x_0)+\frac{\partial z}{\partial y}(y-y_0)\\$

将 $(1)$ 变形得到

\frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y - d z = 0

$\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y-\mathrm{d}z=0\\$

即

[\begin{matrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} & - 1 \end{matrix}] [\begin{matrix} d x \\ d y \\ d z \end{matrix}] = 0

$\begin{bmatrix}\cfrac{\partial z}{\partial x}&\cfrac{\partial z}{\partial y}&-1\end{bmatrix}\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\\\mathrm{d}z\end{bmatrix}=0\\$

又 $\begin{bmatrix}\mathrm{d}x&\mathrm{d}y&\mathrm{d}z\end{bmatrix}^T$ 为切平面中的向量，则 $(x_0,y_0,z_0)$ 处的法向量为

\vec{n} = {[\begin{matrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} & - 1 \end{matrix}]}^{T}

$\vec{n}=\begin{bmatrix}\cfrac{\partial z}{\partial x}&\cfrac{\partial z}{\partial y}&-1\end{bmatrix}^T\\$

同理，对于曲面 $f(x,y,z)=0$ 上一点 $(x_0,y_0,z_0)$

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z = 0

$\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z=0\\$

切平面为

\frac{\partial f}{\partial x} (x - x_{0}) + \frac{\partial f}{\partial y} (y - y_{0}) + \frac{\partial f}{\partial z} (z - z_{0}) = 0

$\frac{\partial f}{\partial x}(x-x_0)+\frac{\partial f}{\partial y}(y-y_0)+\frac{\partial f}{\partial z}(z-z_0)=0\\$

法向量为

\vec{n} = {[\begin{matrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{matrix}]}^{T}

$\vec{n}=\begin{bmatrix}\cfrac{\partial f}{\partial x}&\cfrac{\partial f}{\partial y}&\cfrac{\partial f}{\partial z}\end{bmatrix}^T\\$

曲线

参数方程确定的空间曲线

若空间曲线 $\Gamma$ 的参数方程为

{\begin{matrix} x = φ (t) \\ y = ψ (t) & t \in [α, β] \\ z = ω (t) \end{matrix}

$\left\{\begin{matrix}x=\varphi(t)\\y=\psi(t)&t\in[\alpha,\beta]\\z=\omega(t)\end{matrix}\right.\\$

曲线上一点 $M$ 对应 $t=t_0$ ，在曲线上靠近 $M$ 的地方有一点 $M'$ ，则

\vec{M^{'} M} = (Δ x, Δ y, Δ z)

$\overrightarrow{M'M}=(\Delta x,\Delta y,\Delta z)\\$

割线 $M'M$ 的方程为

\frac{x - x_{0}}{Δ x} = \frac{y - y_{0}}{Δ y} = \frac{z - z_{0}}{Δ z} ⇓ \frac{x - x_{0}}{\frac{Δ x}{Δ t}} = \frac{y - y_{0}}{\frac{Δ y}{Δ t}} = \frac{z - z_{0}}{\frac{Δ z}{Δ t}}

$\frac{x-x_0}{\Delta x}=\frac{y-y_0}{\Delta y}=\frac{z-z_0}{\Delta z}\\\Downarrow\\\frac{x-x_0}{\frac{\Delta x}{\Delta t}}=\frac{y-y_0}{\frac{\Delta y}{\Delta t}}=\frac{z-z_0}{\frac{\Delta z}{\Delta t}}$

当 $M'\to M$ 时， $\Delta t\to 0$ ，则切线方程为

\frac{x - x_{0}}{φ^{'} (t)} = \frac{y - y_{0}}{ψ^{'} (t)} = \frac{z - z_{0}}{ω^{'} (t)}

$\frac{x-x_0}{\varphi'(t)}=\frac{y-y_0}{\psi'(t)}=\frac{z-z_0}{\omega'(t)}\\$

则切向量为

(φ^{'} (t), ψ^{'} (t), ω^{'} (t))

$(\varphi'(t),\psi'(t),\omega'(t))$

又由切向量是法平面的法向量，则法平面为

(x - x_{0}) φ^{'} (t) + (y - y_{0}) ψ^{'} (t) + (z - z_{0}) ω^{'} (t) = 0

$(x-x_0)\varphi'(t)+(y-y_0)\psi'(t)+(z-z_0)\omega'(t)=0\\$

空间曲面相交形成的空间曲线

对于空间曲线 $z_1(x,y)\cap z_2(x,y)$ 上一点 $(x_0,y_0,z_0)$ ，其切线为两曲线切平面的交线，即为

{\begin{matrix} z - z_{0} = \frac{\partial z_{1}}{\partial x} (x - x_{0}) + \frac{\partial z_{1}}{\partial y} (y - y_{0}) \\ z - z_{0} = \frac{\partial z_{2}}{\partial x} (x - x_{0}) + \frac{\partial z_{2}}{\partial y} (y - y_{0}) \end{matrix}

$\left\{\begin{matrix}z-z_0=\cfrac{\partial z_1}{\partial x}(x-x_0)+\cfrac{\partial z_1}{\partial y}(y-y_0)\\z-z_0=\cfrac{\partial z_2}{\partial x}(x-x_0)+\cfrac{\partial z_2}{\partial y}(y-y_0)\end{matrix}\right.\\$

切向量、法平面均借助同样的方法求得。