多元微积分期中
目录极限与连续
定义
对于函数\(f:U\to V\)
重极限:
连续:
\[\lim_{\bm x\to\bm x_0}f(\bm x)=f(\bm x_0)\\ \]累次极限:
\[形如\lim_{x\to x_0}\lim_{y\to y_0}f(x,y)、\lim_{y\to y_0}\lim_{x\to x_0}f(x,y)\\ \]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:
\[\lim_{x→0,y→0}\frac{\sqrt{xy+1}-1}{xy} \]解:
\[\lim_{x→0,y→0}\frac{\sqrt{xy+1}-1}{xy}=\lim_{x→0,y→0} \dfrac{xy}{xy\sqrt{xy+1}+1}=\lim_{x→0,y→0}\dfrac{1}{\sqrt{xy+1}+1}=\frac{1}{2} \]例2:求极限
\[\lim_{x→∞,y→∞} \dfrac{x+y}{x^2-xy+y^2} \]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
\[\lim_{x\to0,y\to 0}\dfrac{xy^2}{x^2+y^2}=0\\ \]解:
\[\left|\dfrac{xy^2}{x^2+y^2}\right|=|y|\cdot\dfrac{|xy|}{x^2+y^2}\le \frac{1}{2}|y|\to 0\\ \]例4:求极限
\[\lim_{x\to0,y\to0}\dfrac{\sin(x^2y+y^4)}{x^2+y^2}\\ \]解:
由\(|\sin x|\le |x|\),则
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
当 \((x,y)\) 沿着 \(y=0\) 趋近于 \((0,0)\) 时
\[\lim_{x\to 0,y=0}\dfrac{x^2y^2}{x^2y^2+(x-y)^2}=0\\ \]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
\[g(t)=f(x_0+t(x-x_0),y_0+t(y-y_0)) \]则
\[g(1)=f(x,y)\\ \]由泰勒公式,有
\[g(t)=g(0)+g'(0)t+\frac{1}{2}g''(0)t^2+\cfrac{1}{6}g'''(\xi)t^2\quad \xi\in(0,t)\\ \]则
\[g(1)=g(0)+g'(0)+\cfrac{1}{2}g''(0)+\cfrac{1}{6}g'''(\xi)\quad\xi\in(0,1)\\ \]又
\[\begin{aligned}g(0)&=f(x_0,y_0)\\g'(0)&=(x-x_0)\cfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}+(y-y_0)\cfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\\g''(0)&=(x-x_0)^2\cfrac{\partial^2 f}{\partial x^2}\bigg|_{(x_0,y_0)}+2(x-x_0)(y-y_0)\cfrac{\partial^2 f}{\partial x\partial y}\bigg|_{(x_0,y_0)}+(y-y_0)^2\cfrac{\partial ^2f}{\partial y^2}\end{aligned}\\ \]则
\[\begin{aligned}f(x,y)&=f(x_0,y_0)\\&+(x-x_0)\cfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}+(y-y_0)\cfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\\&+\cfrac{1}{2}(x-x_0)^2\cfrac{\partial^2 f}{\partial x^2}\bigg|_{(x_0,y_0)}+(x-x_0)(y-y_0)\cfrac{\partial^2 f}{\partial x\partial y}\bigg|_{(x_0,y_0)}+\cfrac{1}{2}(y-y_0)^2\cfrac{\partial ^2f}{\partial y^2}\\&+o(x^2+y^2)\end{aligned}\\ \]或
\[\begin{aligned}f(x,y)&=f(x_0,y_0)\\&+(x-x_0)\cfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}+(y-y_0)\cfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\\&+\cfrac{1}{2}(x-x_0)^2\cfrac{\partial^2 f}{\partial x^2}\bigg|_{M}+(x-x_0)(y-y_0)\cfrac{\partial ^2f}{\partial x\partial y}\bigg|_{M}+\cfrac{1}{2}(y-y_0)^2\cfrac{\partial^2 f}{\partial y^2}\bigg|_{M}\end{aligned}\\M\in(x_0,x)\times(y_0,y)\\ \]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
\[\lim_{\bm h\to 0}\dfrac{\|\bm f(\bm x_0+\bm h)-\bm f(\bm h)-L(\bm h)\|}{\|\bm h\|}=0\\ \]则称其在\(\bm x_0\)处可微,记微分
\[\mathrm{D}\bm f(\bm h)=L(\bm h)\\ \]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
\[\Delta f = A\Delta x + B\Delta y + o(\rho) \]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
\[\lim_{\begin{matrix}\small\scriptstyle\Delta x\to 0\\\small\scriptstyle\Delta y\to 0\end{matrix}}\frac{\Delta f-A\Delta x-B\Delta y}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=0 \]或
\[\lim_{x\to x_0,y\to y_0}\dfrac{f(x,y)-f(x_0,y_0)-A(x-x_0)-B(y-y_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0\\ \]记微分
\[\mathrm{d}f(x,y)=Ax+By \]偏导数的定义如下:
\[f_x(x_0,y_0)=\dfrac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}=\lim_{x\to x_0}\dfrac{f(x,y_0)-f(x_0,y_0)}{x-x_0}\\f_y(x_0,y_0)=\dfrac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}=\lim_{y\to y_0}\dfrac{f(x_0,y)-f(x_0,y_0)}{y-y_0} \]全微分为
\[\mathrm{d}f=\dfrac{\partial f}{\partial x}\mathrm{d}x+\dfrac{\partial f}{\partial y}\mathrm{d}y\\ \]雅各比矩阵
\[\cfrac{\partial (y_1,y_2)}{\partial (x_1,x_2)}=\begin{bmatrix}\cfrac{\partial y_1}{\partial x_1}&\cfrac{\partial y_1}{\partial x_2}\\\cfrac{\partial y_2}{\partial x_1}&\cfrac{\partial y_2}{\partial x_2}\end{bmatrix} \]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:
\[f(x,y)=\left\{\begin{matrix}(x^2+^2)\sin\frac{1}{x^2+y^2}&(x,y)\ne (0,0)\\0&(x,y)=(0,0)\end{matrix}\right.\\ \](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
\[\dfrac{\partial P}{\partial V}\cdot\dfrac{\partial V}{\partial T}\cdot\dfrac{\partial T}{\partial P}=-1\\ \]解:
\[\begin{aligned}&P=\dfrac{RT}{V}\Rightarrow\dfrac{\partial P}{\partial V}=-\dfrac{RT}{V^2}\\&V=\dfrac{RT}{P}\Rightarrow \dfrac{\partial V}{\partial T}=\dfrac{R}{P}\\&T=\dfrac{PV}{R}\Rightarrow\dfrac{\partial T}{\partial P}=\dfrac{V}{R}\end{aligned}\\ \]则
\[\dfrac{\partial P}{\partial V}\cdot\dfrac{\partial V}{\partial T}\cdot\dfrac{\partial T}{\partial P}=-\dfrac{RT}{V^2}\cdot\dfrac{R}{P}\cdot\dfrac{V}{R}=-\dfrac{RT}{PV}=-1\\ \]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
\[\mathrm{D}_{\bm v}f(\bm x)=\dfrac{\partial f}{\partial \bm v}=\frac{\mathrm{d}f_{\bm v}}{\mathrm{d}t}\bigg|_{t=0}=\lim_{t\to 0}\dfrac{f(\bm x+t\bm v)-f(\bm x)}{t}\\ \]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
\[\nabla f(\bm x_0)=\begin{bmatrix}\cfrac{\partial f}{\partial x_1}&\cdots&\cfrac{\partial f}{\partial x_n}\end{bmatrix}^T\\ \]- \(\nabla (c_1f+c_2g)=c_1\nabla f+c_2\nabla g\)
- \(\nabla(fg)=f\nabla g+g\nabla f\)
- \(\nabla\left(\cfrac{f}{g}\right)=\cfrac{g\nabla f-f\nabla g}{g^2}\)
- \(h:\mathbb{R}\to\mathbb{R},g:U\to \mathbb{R}\),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
当 \(\cos\theta =0\) 时,\(f(\rho \cos\theta,\rho\sin\theta)=0\)
\[\frac{\partial f}{\partial \vec{e}}\bigg|_{(0,0)}=0\\ \]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:
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:
因此
\[\max\left\{\frac{\partial u}{\partial \bm v}\right\}=\|\text{grad}\;u\|=\sqrt{5}\\\min\left\{\frac{\partial u}{\partial \bm v}\right\}=-\|\text{grad}\;u\|=-\sqrt{5}\\ \]多元复合函数求导
推导
\[(x,y)\mapsto(u,v)\mapsto f \]则
\[\mathrm{d}f=\frac{\partial f}{\partial u}\mathrm{d}u+\frac{\partial f}{\partial v}\mathrm{d}v \]\[\mathrm{d}u=\frac{\partial u}{\partial x}\mathrm{d}x+\frac{\partial u}{\partial y}\mathrm{d}y \]\[\mathrm{d}v=\frac{\partial v}{\partial x}\mathrm{d}x+\frac{\partial v}{\partial y}\mathrm{d}y\\ \]因此
\[\mathrm{d}f=\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\right)\mathrm{d}x+\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\right)\mathrm{d}y\\ \]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 \(\cfrac{\partial f}{\partial x}\)
\[\begin{aligned}\mathrm{d}f=\mathrm{d}g&=\frac{\partial g}{\partial x}\mathrm{d}x+\frac{\partial g}{\partial y}\mathrm{d}y+\frac{\partial g}{\partial z}\mathrm{d}z\\&=\left(\frac{\partial g}{\partial x}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial x}\right)\mathrm{d}x+\left(\frac{\partial g}{\partial y}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial y}\right)\mathrm{d}y\end{aligned}\\ \]则
\[\frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial x}=(3x^2+\sin y)e^{x^3+y^2+x\sin y}\\ \]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
则
\[\begin{aligned}\frac{\partial f(u,v)}{\partial u}+\frac{\partial f(u,v)}{\partial v}&=\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\right)+\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}\right)\\&=\frac{1}{2}(u+v)+\frac{1}{2}(u-v)+\frac{1}{2}(u+v)-\frac{1}{2}(u-v)\\&=u+v\end{aligned}\\ \]则
\[\cfrac{\partial f(x,y)}{\partial x}+\cfrac{\partial f(x,y)}{\partial y}=x+y \]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
\[F(\bm x,\bm y)=0\Leftrightarrow \bm y=f(\bm x)\\ \]且
\[f'(\bm x)=-\left(\frac{\partial F}{\partial \bm y}\right)^{-1}\left(\frac{\partial F}{\partial \bm x}\right)\\ \]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
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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
\[(f^{-1})'(\bm y)=(f'(\bm x))^{-1}\\ \]或记作
\[J_{f^{-1}}(\bm y)=J_f(\bm x)^{-1}\\ \]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\(F(x,y)=x^2+y^2-4x-6y+12\)
在点 \((1,2)\) 处,代入可得:
\[\frac{\partial F}{\partial x} \bigg|_{(1,2)} = -2 \qquad \frac{\partial F}{\partial y} \bigg|_{(1,2)} = -2 \]斜率为
\[\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(1,2)}=-\dfrac{\left.\frac{\partial F}{\partial x}\right|_{(1,2)}}{\left.\frac{\partial F}{\partial y}\right|_{(1,2)}}=-1\\ \]因此切线方程为:
\[x+y=3 \]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
对 \(x\) 求偏导得:
\[3x^2+3z^2(x,y)\cfrac{\partial z}{\partial x}=1+\cfrac{\partial z}{\partial x}\\ \]再对 \(y\) 求偏导得:
\[6z(x,y)\cfrac{\partial z}{\partial x}\cdot\cfrac{\partial z}{\partial y}+3z^2(x,y)\cfrac{\partial^2 z}{\partial x\partial y}=\cfrac{\partial ^2z}{\partial x\partial y}\\ \]则
\[\cfrac{\partial^2 z}{\partial x\partial y}=\cfrac{6z}{1-3z^2}\cdot\cfrac{\partial z}{\partial x}\cdot\cfrac{\partial z}{\partial y}\\ \]又
\[\cfrac{\partial z}{\partial x}=-\cfrac{1-3x^2}{1-3z^2}\qquad\cfrac{\partial z}{\partial y}=-\cfrac{1-3y^2}{1-3z^2}\\ \]则
\[\cfrac{\partial^2 z}{\partial x\partial y}=\cfrac{6z(1-3x^2)(1-3y^2)}{(1-3z^2)^3}\\ \]\[\cfrac{\partial ^2z}{\partial x\partial y}\bigg|_{(1,1,-1)}=3\\ \]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:
\[H_{i,j}=\frac{\partial^2 f}{\partial x_i \partial x_j} \]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:
- \(f(x,y)=xy\)的Hesse矩阵为:
- \(f(x,y)=x^3+y^3-3xy\)的Hesse矩阵为:
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
\[f(x_1,x_2,\cdots,x_n) \]如果存在一些约束条件
\[g_1(x_1,x_2,\cdots,x_n)=0\\g_2(x_1,x_2,\cdots,x_n)=0\\\vdots\\g_m(x_1,x_2,\cdots,x_n)=0 \]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:
\[L(x_1,x_2,\cdots,x_n,\lambda_1,\lambda_2,\cdots,\lambda_m) = f(x_1,x_2,\cdots,x_n) - \sum_{i=1}^m\lambda_i g_i(x_1,x_2,\cdots,x_n) \]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:
\[\frac{\partial L}{\partial x_1} = 0,\frac{\partial L}{\partial x_2} = 0,\cdots,\frac{\partial L}{\partial x_n} = 0\\\;\\\frac{\partial L}{\partial \lambda_1} = 0,\frac{\partial L}{\partial \lambda_2} = 0,\cdots,\frac{\partial L}{\partial \lambda_m} = 0 \]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:
\[L(x,y,\lambda)=f(x,y)-\lambda g(x,y)=x^2+y^2-\lambda(x+y-1) \]然后求出偏导数:
\[\frac{\partial L}{\partial x}=2x-\lambda=0 \]\[\frac{\partial L}{\partial y}=2y-\lambda=0 \]\[\frac{\partial L}{\partial \lambda}=x+y-1=0 \]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:
\[\frac{\partial^2 L}{\partial x^2}=2,\frac{\partial^2 L}{\partial y^2}=2,\frac{\partial^2 L}{\partial x \partial y}=0 \]几何应用
曲面
对于曲面 \(z(x,y)\)上一点\((x_0,y_0,z_0)\),由于
\[\tag1\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y \]同时
\[\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\\\mathrm{d}z\end{bmatrix}=\begin{bmatrix}x-x_0\\y-y_0\\z-z_0\end{bmatrix}\\ \]得到切平面方程
\[z-z_0=\frac{\partial z}{\partial x}(x-x_0)+\frac{\partial z}{\partial y}(y-y_0)\\ \]将\((1)\)变形得到
\[\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y-\mathrm{d}z=0\\ \]即
\[\begin{bmatrix}\cfrac{\partial z}{\partial x}&\cfrac{\partial z}{\partial y}&-1\end{bmatrix}\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\\\mathrm{d}z\end{bmatrix}=0\\ \]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
\[\vec{n}=\begin{bmatrix}\cfrac{\partial z}{\partial x}&\cfrac{\partial z}{\partial y}&-1\end{bmatrix}^T\\ \]同理,对于曲面\(f(x,y,z)=0\)上一点\((x_0,y_0,z_0)\)
\[\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z=0\\ \]切平面为
\[\frac{\partial f}{\partial x}(x-x_0)+\frac{\partial f}{\partial y}(y-y_0)+\frac{\partial f}{\partial z}(z-z_0)=0\\ \]法向量为
\[\vec{n}=\begin{bmatrix}\cfrac{\partial f}{\partial x}&\cfrac{\partial f}{\partial y}&\cfrac{\partial f}{\partial z}\end{bmatrix}^T\\ \]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
\[\left\{\begin{matrix}x=\varphi(t)\\y=\psi(t)&t\in[\alpha,\beta]\\z=\omega(t)\end{matrix}\right.\\ \]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
\[\overrightarrow{M'M}=(\Delta x,\Delta y,\Delta z)\\ \]割线 \(M'M\) 的方程为
\[\frac{x-x_0}{\Delta x}=\frac{y-y_0}{\Delta y}=\frac{z-z_0}{\Delta z}\\\Downarrow\\\frac{x-x_0}{\frac{\Delta x}{\Delta t}}=\frac{y-y_0}{\frac{\Delta y}{\Delta t}}=\frac{z-z_0}{\frac{\Delta z}{\Delta t}} \]当\(M'\to M\)时,\(\Delta t\to 0\),则切线方程为
\[\frac{x-x_0}{\varphi'(t)}=\frac{y-y_0}{\psi'(t)}=\frac{z-z_0}{\omega'(t)}\\ \]则切向量为
\[(\varphi'(t),\psi'(t),\omega'(t)) \]又由切向量是法平面的法向量,则法平面为
\[(x-x_0)\varphi'(t)+(y-y_0)\psi'(t)+(z-z_0)\omega'(t)=0\\ \]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
\[\left\{\begin{matrix}z-z_0=\cfrac{\partial z_1}{\partial x}(x-x_0)+\cfrac{\partial z_1}{\partial y}(y-y_0)\\z-z_0=\cfrac{\partial z_2}{\partial x}(x-x_0)+\cfrac{\partial z_2}{\partial y}(y-y_0)\end{matrix}\right.\\ \]切向量、法平面均借助同样的方法求得。
标签:frac,复习,bm,cfrac,期中,dfrac,partial,复盘,mathrm From: https://www.cnblogs.com/lzxzy-blog/p/17319639.html