From 6a9949dbeca864bfaf982467ace222a4d79bbf7e Mon Sep 17 00:00:00 2001 From: unlockable Date: Sat, 17 Dec 2022 18:37:40 +0800 Subject: [PATCH] =?UTF-8?q?=E5=AE=9A=E7=A7=AF=E5=88=86=E7=9A=84=E5=BA=94?= =?UTF-8?q?=E7=94=A8=E3=80=82?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 07函数的积分.tex | 20 ++--- 08定积分应用.tex | 224 +++++++++++++++++++++++++++++++++++++++++++++++ 09常微分方程.tex | 0 高等微积分.tex | 5 +- 4 files changed, 238 insertions(+), 11 deletions(-) create mode 100644 08定积分应用.tex create mode 100644 09常微分方程.tex diff --git a/07函数的积分.tex b/07函数的积分.tex index 101bd20..e5fe2f8 100644 --- a/07函数的积分.tex +++ b/07函数的积分.tex @@ -37,7 +37,7 @@ \end{corollary} \begin{corollary} - 设$\vert f \vert, f \in R[a,b]$，则 + 设$\abs{f}, f \in R[a,b]$，则 \[\left| \int_a^b f(x) \dif x \right| \leq \int_a^b \vert f(x) \vert \dif x \eqper\] \end{corollary} @@ -121,18 +121,18 @@ 使得 \[\left|\sum_{i=1}^n f(c_i) \Delta x - A\right| < 1, c_i \in [x_{i-1}, x_i]\text{任取}, i = 1, 2, \cdots, n\] 由此导出 - \[\left|\sum_{i=1}^n f(c_i) \Delta x \right| < \left|\sum_{i=1}^n f(c_i) \Delta x - A\right| + \vert A \vert < 1 + \vert A \vert\] + \[\left|\sum_{i=1}^n f(c_i) \Delta x \right| < \left|\sum_{i=1}^n f(c_i) \Delta x - A\right| + \abs{A} < 1 + \abs{A}\] 因此 - \[\left|\sum_{i=1}^n f(c_i) \right| \leq \frac{1 + \vert A \vert}{\Delta x}\] + \[\left|\sum_{i=1}^n f(c_i) \right| \leq \frac{1 + \abs{A}}{\Delta x}\] 进一步有 - \[\vert f(c_i) \vert \leq \left| \sum_{i=1}^n f(c_i) \right| + \left| \sum_{i=2}^n f(c_i)\right| \leq \frac{1 + \vert A \vert}{\Delta x} + \left| \sum_{i=2}^n f(x_i) \right|\] + \[\vert f(c_i) \vert \leq \left| \sum_{i=1}^n f(c_i) \right| + \left| \sum_{i=2}^n f(c_i)\right| \leq \frac{1 + \abs{A}}{\Delta x} + \left| \sum_{i=2}^n f(x_i) \right|\] 即 - \[\vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq 1}^n f(x_i)\right| = M_1, \forall x \in [x_0, x_1]\] + \[\vert f(x) \vert \leq \frac{1 + \abs{A}}{\Delta x} + \left|\sum_{i\neq 1}^n f(x_i)\right| = M_1, \forall x \in [x_0, x_1]\] 类似地可以得到 \begin{align*} - \vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq 2}^n f(x_i)\right| & = M_2, \forall x \in [x_1, x_2]\\ + \vert f(x) \vert \leq \frac{1 + \abs{A}}{\Delta x} + \left|\sum_{i\neq 2}^n f(x_i)\right| & = M_2, \forall x \in [x_1, x_2]\\ \vdots & \\ - \vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq n}^n f(x_i)\right| & = M_n, \forall x \in [x_{n-1}, x_n] + \vert f(x) \vert \leq \frac{1 + \abs{A}}{\Delta x} + \left|\sum_{i\neq n}^n f(x_i)\right| & = M_n, \forall x \in [x_{n-1}, x_n] \end{align*} 将这$n$个式子综合起来就可以得到 \[\vert f(x) \vert \leq \max \{M_1, \cdots, M_n\}, \forall x \in [a,b] \eqper \qedhere\] @@ -521,7 +521,7 @@ C^n [a,b] = \{f \in C[a,b] \mid f^{(n)} \in C[a,b]\}\] \begin{corollary} 容易由Lebesgue定理得到以下结论： \begin{enumerate} - \item 若$f \in R[a,b]$，则$\vert f \vert \in R[a,b]$； + \item 若$f \in R[a,b]$，则$\abs{f} \in R[a,b]$； \item 若$f, g \in R[a,b]$，则$fg \in R[a,b]$； \item 若$f \in R[a,b]$且$\dfrac{1}{f}$有界，则$\dfrac{1}{f} \in R[a,b]$； \item 设$f \in R[a,b], \varphi \in C[\alpha, \beta]$且$f([a,b]) \subset [\alpha, \beta]$，则$\varphi \circ f \in R[a,b]$。 @@ -609,7 +609,7 @@ C^n [a,b] = \{f \in C[a,b] \mid f^{(n)} \in C[a,b]\}\] \end{definition} \begin{lemma}[Riemann-Lebesgue] - $f$在$[a,b]$上可积或广义绝对可积（$f$与$\vert f \vert$均在$[a,b]$上广义可积），则 + $f$在$[a,b]$上可积或广义绝对可积（$f$与$\abs{f}$均在$[a,b]$上广义可积），则 \[\tolim{\lambda}{\infty} \int_a^b f(x) \cos \lambda x \dif x = 0, \tolim{\lambda}{\infty} \int_a^b f(x) \sin \lambda x \dif x = 0\eqper\] \end{lemma} @@ -646,4 +646,4 @@ C^n [a,b] = \{f \in C[a,b] \mid f^{(n)} \in C[a,b]\}\] \left|\int_a^b f(x) \cos \lambda x \dif x \right| & \leq \left|\int_a^{a + \delta} f(x) \cos x \lambda x \dif x \right| + \left|\int_{a + \delta}^b f(x) \cos \lambda x \dif x \right|\\ & < \frac{\varepsilon}{2} = \frac{\varepsilon}{2} = \varepsilon\eqper \qedhere \end{align*} -\end{proof} \ No newline at end of file +\end{proof} diff --git a/08定积分应用.tex b/08定积分应用.tex new file mode 100644 index 0000000..eee2828 --- /dev/null +++ b/08定积分应用.tex @@ -0,0 +1,224 @@ +\chapter{定积分的应用} +\section{几何应用} +\subsection{平面图形的面积} +\subsubsection{直角坐标系下平面图形面积的计算} +考虑由直线$x = a, x = b$及$x$轴和连续曲线$y = f(x)$所围成的曲边梯形的面积$A$。根据定积分的定义和几何意义可知， +\[A = \int_a^b \abs{f(x)} \dif x\eqper\] + +再考虑由曲线$y = f(x), y = g(x)$和直线$x = a, x = b$所围成的面积$A$。对于$g(x) \leq f(x), x \in [a,b]$，有 +\[\dif A = [f(x) - g(x)] \dif x\] +因此 +\[A = \int_a^b [f(x) - g(x)] \dif x\] + +进一步若$f(x), g(x)$大小关系不确定，有 +\[A = \int_a^b \abs{f(x) - g(x)} \dif x\eqper\] + +最后考虑：设连续函数$\varphi(y), \psi(y)$满足 +\[0 \leq \psi(y) \leq \varphi(y), y \in [c,d]\] +求由曲线$x = \varphi(y), x = \psi(y)$和直线$y = c, y = d$围成的面积$A$。 +仿照上面，有 +\[A = \int_c^d[\varphi(y) - \psi(y)] \dif y\eqper\] + +\subsubsection{极坐标系下平面图形面积的计算} +求曲线$\rho = \rho(\theta)$及射线$\theta = \alpha, \theta = \beta$所围成的面积$A$。 + +注意面积微元，即小扇形的面积为 +\[\dif A = \frac{1}{2}\rho^2(\theta \dif \theta)\] +因此面积为 +\[A = \frac{1}{2} \int_\alpha^\beta \rho^2(\theta) \dif \theta\eqper\] + +\subsubsection{利用参数方程求图形面积} +\begin{example} + 求星形线： + \(\begin{cases} + x = a\cos^3 t\\ + y = a\sin^3 t + \end{cases} t \in [0,2\pi]\) + 所围成的面积。 +\end{example} + +\begin{proof}[解] + 利用对称性 + \begin{align*} + A & = 4A_1 = 4\int_0^a y \dif x\\ + & = 4 \int_\frac{\pi}{2}^0 a \sin^3 t \cdot 3a\cos^2 t (-\sin t) \dif t\\ + & = 12\int_0^\frac{\pi}{2}a^2 \sin^4 t(1 - \sin^2 t) \dif t\\ + & = 12a^2 \left(\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} - \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right)\\ + & = \frac{3}{8} \pi a^2 \eqper \qedhere + \end{align*} +\end{proof} + +\subsection{空间体的体积} +\subsubsection{已知平行截面面积的空间体的体积} +若已知平行截面面积关于$x$的函数为$A(x)$，那么体积 +\[V = \int_a^b A(x) \dif x\eqper\] + +\subsubsection{旋转体的体积} +将$f(x), x = a, x = b$与$x$轴所围成的图形绕$x$轴旋转一周，求所得的空间体的体积。注意到每个截面的面积$A(x) = \pi y^2 = \pi f^2(x)$。因此 +\[V = \int_a^b \pi f^2(x) \dif x = \pi \int_a^b f^2(x) \dif x\eqper\] + +\subsection{平面曲线的弧长} +将曲线$AB$取细分的点$A = M_0, M_1, M_2, \dots, M_n = B$，曲线的长度定义为 +\[l = \tolim{\lambda}{0} \sum_{i=1}^n \abs{M_{i-1}M_i}\] +即折线的长度的极限。 + +\begin{enumerate} + \item 假设曲线方程由$y = f(x)$给出，且曲线是光滑的，即$\deriv{f}(x)$在$[a,b]$上连续。那么 + \[\abs{M_{i-1}M_i} = \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}(i = 1, 2, \dots, n)\] + 由Lagrange中值定理可以得到 + \[\Delta y_i = f(x_i) - f(x_{i-1}) = \deriv{f}(\xi_i) \cdot \Delta x_i (x_{i-1} < \xi_i < x_i)\] + 这导出 + \[\abs{M_{i-1}M_i} = \sqrt{1 + [\deriv{f}(\xi_i)]^2} \cdot \Delta x_i (i = 1, 2, \dots, n)\] + 因而 + \[\sum_{i=1}^n \abs{M_{i-1}M_i} = \sum_{i=1}^n \sqrt{1 + [\deriv{f}(\xi_i)]^2}\cdot \Delta x_i\] + + 那么令$\lambda = \max \limits_{1 \leq i \leq n} \abs{M_{i-1}M_i}$，$\mu = \max \limits_{1 \leq i \leq n} \{\Delta x_i\}$。$\Delta x_i \leq \abs{M_{i-1}M_i}$导出$\mu \leq \lambda$。因此$\lambda \to 0$时，有$\mu \to 0$。那么 + \begin{align*} + l & = \tolim{\lambda}{0} \sum_{i=1}^n \abs{M_{i-1}M_i}\\ + & = \tolim{\mu}{0} \sum_{i=1}^n \sqrt{1 + [\deriv{f}(\xi_1)]^2} \cdot \Delta x_1\\ + & = \int_a^b \sqrt{1 + [\deriv{f}(x)]^2}\dif x + \end{align*} + + 因此弧长为 + \[l = \int_a^b \sqrt{1 + [\deriv{f}(x)]^2} \dif x = \int_a^b \sqrt{1 + \left(\deriv{y}\right)^2} \dif x\] + \item 假设曲线由参数方程 + \(\begin{cases} + x = x(t)\\ + y = y(t) + \end{cases} (\alpha \leq t \leq \beta)\) + 给出，$\deriv{x}(t), \deriv{y}(t) \in C[\alpha, \beta]$且不同时为零。$t = \alpha$对应起点$A$，$t = \beta$对应重点$B$，即当$\dif t > 0$时，由$\dif l > 0$。因而弧长公式为 + \[l = \int_\alpha^\beta \sqrt{\left(\deriv{x}\right)^2(t) + \left(\deriv{y}\right)^2(t)}\dif t\eqper\] + \item 假设曲线段由极坐标方程$\rho = \rho(\theta)(\alpha \leq \theta \leq \beta)$给出，且$\deriv{\rho}(\theta)$在$[\alpha, \beta]$上连续。选择$\theta$作为参数，那么 + \[\begin{cases} + x = \rho(\theta) \cos \theta\\ + y = \rho(\theta) \sin \theta + \end{cases} (\alpha \leq \theta \leq \beta)\] + 因此弧长公式为 + \[l = \int_\alpha^\beta \sqrt{\rho^2(\theta) + \left(\deriv{\rho}\right)^2(\theta)}\dif \theta\eqper\] + 对应的弧微分公式为 + \[\dif l = \sqrt{\rho^2(\theta) + [\deriv{\rho}(\theta)]^2} \dif \theta\eqper\] +\end{enumerate} + +\subsection{旋转体的侧面积} +将$y = f(x)$绕$x$轴旋转，求其侧面积。用每一点的切线旋转所得圆台的侧面积近似。侧面积为 +\[\text{圆台侧面积} = \pi [y + (y + \dif y)] \cdot \dif l = 2\pi y \dif l + \pi \dif y \cdot \dif l\] +当$\dif x \to 0$时，$\dif y \cdot \dif l = o(\dif x)$，略去。 + +由此得到侧面积微元 +\[\dif S = 2\pi y \dif l = 2\pi y \sqrt{1 + (\deriv{y})^2} \dif x\] +因此侧面积为 +\[S = 2\pi \int_a^b y \sqrt{1 + (\deriv{y})^2} \dif x\] + +\subsection{曲率与曲率半径} +曲率问题就是研究曲线的弯曲程度的问题。 + +\begin{definition} + 设$M_0, M$之间的弧长为$\Delta l$，$\dfrac{\Delta \alpha}{\Delta l}$为$M_0, M$之间的平均曲率。若 + \[\abs{\tolim{\Delta l}{0} \frac{\Delta \alpha}{\Delta l}}\] + 存在，则称$k = \abs{\tolim{\Delta l}{0} \dfrac{\Delta \alpha}{\Delta l}}$为曲线在$M_0$处的曲率。 +\end{definition} + +设曲线$y = f(x)$二阶可导，那么 +\[k = \abs{\tolim{\Delta l}{0} \frac{\Delta \alpha}{\Delta l}} = \abs{\frac{\dif \alpha}{\dif l}}\] +又因为 +\[\tan \alpha = \deriv{y}, \alpha = \arctan \deriv{y}\] +可以得到 +\[\dif \alpha = \frac{1}{1 + (\deriv{y})^2} y^{\prime \prime} \dif x, \dif l = \sqrt{1 + (\deriv{y})^2}\dif x\] +由此我们能得到 + +\begin{theorem} + 曲线的曲率公式为 + \[k = \frac{\abs{y^{\prime \prime}}}{(1+(\deriv{y})^2)^\frac{3}{2}}\eqper\] +\end{theorem} + +\begin{definition} + $R = \dfrac{1}{k}$称为曲线$y = f(x)$在$M_0$处的曲率半径。 +\end{definition} + +\section{物理应用} +\subsection{引力问题} +\begin{example} + 设有一均匀细杆长为$2l$，质量为$M$。另有一质量为$m$的质点，位于细杆所在的直线上，与杆的近端的距离为$a$。求细杆对质点的引力$F$。 +\end{example} + +\begin{figure}[H] + \centering + \begin{tikzpicture} + \draw[-{Stealth[width=5pt]}] (-3,0)--(3,0) node[below] {$x$}; + \node[below] at (-2.8,0) {$O$}; + \node[above] at (-2.8,0) {$m$}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=4pt] at (-2.8,0) {}; + \node[above] at (-1,0) {$a$}; + \node[above] at (0,0) {$x$}; + \node[above] at (1,0) {$x + \dif x$}; + \node[above] at (2.3,0) {$2l + a$}; + \draw[color=red, line width=2pt] (-1,0)--(2,0); + \draw[color=black!50!white, line width=2pt] (0,0)--(0.6,0); + \end{tikzpicture} +\end{figure} + +\begin{proof}[解] + 取一个小区间$[x, x + \dif x]$，视为质点质量$\dfrac{M}{2l}\dif x$。 + 因此 + \[\dif F = k \frac{m\left(\frac{M}{2l}\dif x\right)}{x^2} = \frac{kmM}{2l}\cdot \frac{1}{x^2} \dif x\] + 因此合力应为 + \[F = \int_a^{a + 2l} \frac{kmM}{2l} \cdot \frac{1}{x^2} \dif x = \frac{kmM}{2l} \cdot \eval{\left(-\frac{1}{x}\right)}_a^{a+2l} = \frac{kmM}{a(a+2l)}\eqper \qedhere\] +\end{proof} + +\subsection{变力做功问题} +问题：求物体从$x = a$移到$x = b$时变力$f(x)$所做的功。注意到功的微元$\dif W = f(x) \dif x$。因此 +\[W = \int_a^b f(x) \dif x \eqper\] + +\subsection{静力矩和质心} +\subsubsection{质点系的质心} +设有若干个点$A_1, A_2, \dots, A_n$，它们的坐标分别为$(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$，质量分别为$m_1, m_2, \dots, m_n$。 + +\begin{figure}[H] + \centering + \begin{tikzpicture} + \draw[-{Stealth[width=5pt]}] (-0.5,0)--(4,0); + \draw[-{Stealth[width=5pt]}] (0,-0.5)--(0,3); + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (0.5,1) {}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (1.2,0.8) {}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (1.5,1.4) {}; + \node at (2.3,1.5) {$A_i(m_i)$}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (0.7,2) {}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (2.5,0.5) {}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (3.5,2.8) {}; + \node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (2,2.3) {}; + \draw[dashed] (0,1.4)--(1.5,1.4)--(1.5,0); + \node[left] at (0,1.4) {$y_i$}; + \node[below] at (1.5,0) {$x_i$}; + \end{tikzpicture} +\end{figure} + +质点$A_i$对$x$轴的静力矩为$m_iy_i$，对$y$轴的静力矩为$m_ix_i$。因此质点系对$x$轴的静力矩 +\[M_x = \sum_{i=1}^n m_iy_i\] +对$y$轴的静力矩 +\[M_y = \sum_{i=1}^n m_ix_i\] +同时质点系的总质量 +\[M = \sum_{i=1}^n m_i\] +我们可设质心的坐标为$(\overline{x}, \overline{y})$，那么 +\[M_x = M\overline{y}, M_y = M\overline{x}\] +即 +\[\overline{x} = \frac{\displaystyle_{i=1}^n m_i x_i}{M}, \overline{y} = \frac{\displaystyle_{i=1}^n m_i y_i}{M}\eqper\] + +\subsubsection{平面曲线的质心} +设线密度$\rho$为常数。将弧长区间$[0,L]$分割。任取一个小区间$[l,l + \dif l]$。将其视为坐标为$(x,y)$的质点。因此$\dif M = \rho \dif l$。那么静力矩微元 +\[\dif M_x = y \rho \dif l, \dif M_y = x \rho \dif l\] +于是有 +\[M_x = \int_0^L y \rho \dif l = \rho \int_0^L y \dif l, \dif M_y = \int_0^L x\rho \dif l = \rho \int_0^L x \dif l\] +同时 +\[M = \int_0^L \rho \dif l = \rho \int_0^L \dif l = \rho L\] +因此质心坐标为 +\[\overline{x} = \frac{\rho \int_0^L x \dif l}{\rho L} = \frac{\int_0^L x \dif l}{L}\] + +\subsubsection{平面薄板的质心} +设面密度$\mu$为常数。 +质量$M = \int_a^b \mu y \dif x$。静力矩 +\[M_x = \frac{1}{2} \mu \int_a^b y^2 \dif x, M_y = \mu \int_a^b xy \dif x\] +因此质心坐标为 +\[\overline{x} = \frac{\int_a^b xy \dif x}{\int_a^b y \dif x}, \overline{y} = \frac{\frac{1}{2} \int_a^b y^2 \dif x}{\int_a^b y \dif x}\] + +对极坐标方程$\rho = \rho(\theta)$，质量微元的质心坐标为 +\[\dif M_x = \frac{2}{3}\rho(\theta)\sin \theta \dif \theta, \dif M_y = \frac{2}{3} \rho(\theta) \cos \theta \dif \theta\eqper\] diff --git a/09常微分方程.tex b/09常微分方程.tex new file mode 100644 index 0000000..e69de29 diff --git a/高等微积分.tex b/高等微积分.tex index d4c1a35..0d5b1d2 100644 --- a/高等微积分.tex +++ b/高等微积分.tex @@ -16,6 +16,7 @@ \usepackage{extarrows} \usepackage{physics} % \usepackage{mathptmx} +\usetikzlibrary{arrows.meta} \geometry{a4paper,scale=0.8} @@ -59,7 +60,7 @@ \date{} % linespread{1.5} -\includeonly{07函数的积分.tex} +% \includeonly{08定积分应用.tex} \begin{document} \maketitle @@ -77,4 +78,6 @@ \include{05插值与逼近初步.tex} \include{06求导的逆运算.tex} \include{07函数的积分.tex} + \include{08定积分应用.tex} + \include{09常微分方程.tex} \end{document} \ No newline at end of file