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更新曲率部分与相关题目
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Didnelpsun
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\documentclass[UTF8, 12pt]{ctexart}
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% UTF8编码，ctexart现实中文
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\usepackage{color}
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% 使用颜色
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\usepackage{geometry}
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\setcounter{tocdepth}{4}
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\setcounter{secnumdepth}{4}
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% 设置四级目录与标题
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\geometry{papersize={21cm,29.7cm}}
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% 默认大小为A4
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\geometry{left=3.18cm,right=3.18cm,top=2.54cm,bottom=2.54cm}
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% 默认页边距为1英尺与1.25英尺
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\usepackage{indentfirst}
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\setlength{\parindent}{2.45em}
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% 首行缩进2个中文字符
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\usepackage{setspace}
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\renewcommand{\baselinestretch}{1.5}
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% 1.5倍行距
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\usepackage{amssymb}
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% 因为所以
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\usepackage{amsmath}
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% 数学公式
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\author{Didnelpsun}
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\title{导数与微分}
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\date{}
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\begin{document}
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\maketitle
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\pagestyle{empty}
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\thispagestyle{empty}
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\tableofcontents
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\thispagestyle{empty}
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\newpage
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\pagestyle{plain}
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\setcounter{page}{1}
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\section{一阶导数}
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\subsection{导数存在性}
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导数存在即可导。
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\subsection{导数连续性}
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\subsection{已知导数求极限}
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\section{高阶导数}
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\subsection{导数存在性}
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\section{微分}
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\section{隐函数与参数方程}
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\section{导数应用}
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\subsection{单调性与凹凸性}
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\subsection{极值与最值}
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\subsection{函数图像}
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\subsection{曲率}
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曲率公式：$k=\left\lvert\dfrac{\rm{d}\alpha}{\rm{d}s}\right\rvert=\dfrac{\vert y''\vert}{(1+y'^2)^{\frac{3}{2}}}$。
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\subsubsection{一般计算}
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\textbf{例题：}求$y=\sin xx$在$x=\dfrac{\pi}{4}$对应的曲率
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$y'=\cos x$，$y'(\dfrac{\pi}{4})=\dfrac{\sqrt{2}}{2}$。
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$y''=-\sin x$，$y''(\dfrac{\pi}{4})=-\dfrac{\sqrt{2}}{2}$。
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$\therefore k=\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{3}{2}\cdot\sqrt{\dfrac{3}{2}}}=\dfrac{2\sqrt{3}}{9}$。
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所以$y=\sin x$在$x=\dfrac{\pi}{4}$的点$(\dfrac{\pi}{4},\dfrac{\sqrt{2}}{2})$的曲率为$\dfrac{2\sqrt{3}}{9}$。
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\subsubsection{最值}
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\textbf{例题：}求$y=x^2-4x+11$曲率最大值所在的点。
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简单得$y'=2x-4$，$y''=2$。
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曲率为$\dfrac{2}{[1+(2x-4)^2]^{\frac{3}{2}}}$。
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当$2x-4=0$时即在$(2,7)$时曲率最大为2。
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\end{document}
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@@ -31,6 +31,10 @@
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% 圆圈序号
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\usepackage{mathtools}
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% 有字的长箭头
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\usepackage{yhmath}
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% 弧线标识
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\usetikzlibrary{decorations.pathreplacing}
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% tikz的大括号
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\author{Didnelpsun}
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\title{微分中值定理与导数的应用}
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\date{}
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@@ -453,4 +457,159 @@ $\forall x\in U(x_0,\delta)$恒有$f(x)\leqslant f(x_0)$，则$f(x)$在$x_0$取
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\item 若$\lim\limits_{x\to\infty}\dfrac{f(x)}{x}=a,b=\lim\limits_{x\to\infty}(f(x)-ax)$，那么$y=ax+b$就是斜渐近线。
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\end{itemize}
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\section{弧微分与曲率}
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\subsection{弧微分}
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\begin{minipage}{0.5\linewidth}
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\begin{tikzpicture}[scale=3]
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\draw[-latex](-0.1,0) -- (1.25,0) node[below]{$x$};
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\draw[-latex](0,-0.1) -- (0,1.25) node[above]{$y$};
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\filldraw[black] (0,0) node[below]{$O$};
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\draw[black, thick,domain=0.4:1.1] plot (\x, \x*\x);
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\filldraw[black] (0.5,1) node {$y=f(x)$};
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\draw[densely dashed](0.5,0.25) -- (0.5, 0) node[below]{$x$};
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\draw[densely dashed](1,1) -- (1, 0) node[below]{$x+\Delta x$};
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\draw[densely dashed](0.5,0.25) -- (1,0.25);
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\filldraw[black](0.5,0.35) node{$y$};
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\filldraw[black](0.95,1.1) node{$y_0$};
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\filldraw[black](0.75,0.35) node{$\Delta x$};
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\filldraw[black](1.1,0.6) node{$\Delta y$};
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\end{tikzpicture}
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\end{minipage}
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\hfill
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\begin{minipage}{0.4\linewidth}
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$\vert\wideparen{yy_0}\vert=S(x)$
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$\Delta y=f(x+\Delta x)-f(x)$
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$(\Delta s)^2\approx(\Delta x)^2+(\Delta y)^2$
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\end{minipage}\medskip
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当偏移量无穷小时，$y=f(x)$所构成的线段就是一条直线，所以：
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\begin{minipage}{0.5\linewidth}
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\begin{tikzpicture}[scale=3]
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\draw[-latex](-0.1,0) -- (1.25,0) node[below]{$x$};
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\draw[-latex](0,-0.1) -- (0,1.25) node[above]{$y$};
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\filldraw[black] (0,0) node[below]{$O$};
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\draw[black, thick,domain=0.4:1.1] plot (\x, \x);
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\filldraw[black] (0.5,1) node {$y=f(x)$};
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\draw[densely dashed](0.5,0.5) -- (0.5, 0) node[below]{$x$};
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\draw[densely dashed](1,1) -- (1, 0) node[below]{$x+\rm{d} x$};
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\draw[densely dashed](0.5,0.5) -- (1,0.5);
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\filldraw[black](0.5,0.6) node{$y$};
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\filldraw[black](0.95,1.1) node{$y_0$};
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\filldraw[black](0.75,0.35) node{$\rm{d} x$};
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\filldraw[black](1.1,0.6) node{$\rm{d} y$};
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\end{tikzpicture}
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\end{minipage}
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\hfill
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\begin{minipage}{0.4\linewidth}
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$\rm{d} y=f(x+\rm{d} x)-f(x)$
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$(\rm{d} s)^2=(\rm{d} x)^2+(\rm{d} y)^2$
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$\rm{d}s=\sqrt{(\rm{d}x)^2+(\rm{d}y)^2}$（弧微分）
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\end{minipage}
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对于弧微分：
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\begin{itemize}
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\item 若直角坐标系下$y=f(x)$，$\rm{d}s=\sqrt{1+\left(\dfrac{\rm{d}y}{\rm{d}x}\right)^2}\rm{d}x$$=\sqrt{1+f'^2(x)}\rm{d}x$，即$\rm{d}s=$$\sqrt{1+f'^2(x)}\rm{d}x$。
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\item 若参数方程下：$x=\phi(t),y=\psi(t)$，$\rm{d}s=\sqrt{\left(\dfrac{\rm{d}x}{\rm{d}t}\right)^2+\left(\dfrac{\rm{d}y}{\rm{d}t}\right)^2}\rm{d}t$\medskip\\$=\sqrt{\psi'^2(t)+\phi'^2(t)}\rm{d}t$，即$\rm{d}s=\sqrt{\psi'^2(t)+\phi'^2(t)}\rm{d}t$。
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\end{itemize}
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\subsection{曲率与曲率半径}
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曲率\textcolor{violet}{\textbf{定义：}}表明曲线在某一点的弯曲程度的数值，针对曲线上某个点的切线方向角对弧长的转动率，通过微分来定义，表明曲线偏离直线的程度。曲率越大，表示曲线的弯曲程度越大。
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曲率的倒数就是曲率半径。\medskip
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\begin{minipage}{0.5\linewidth}
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两点切线改变角相同时，弯曲程度与两点之间的弧长度成反比。
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两点之间的弧长度相同时，弯曲程度与两点切线改变角成正比。
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\end{minipage}
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\hfill
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\begin{minipage}{0.2\linewidth}
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\begin{tikzpicture}[scale=0.6]
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\draw[black, thick,domain=-2:2] plot (\x, {\x*\x});
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\filldraw[black] (-1,1) circle (2pt) node[left]{$M_1$};
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\filldraw[black] (1,1) circle (2pt) node[right]{$N_1$};
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\draw[black](2,3) -- (0,-1);
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\draw[black](-2,3) -- (0,-1);
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\end{tikzpicture}
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\end{minipage}
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\hfill
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\begin{minipage}{0.2\linewidth}
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\begin{tikzpicture}[scale=0.6]
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\draw[black, thick,domain=-1.5:1.5] plot (\x, {2*\x*\x});
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\filldraw[black] (-1/2,1/2) circle (2pt) node[left]{$M_2$};
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\filldraw[black] (1/2,1/2) circle (2pt) node[right]{$N_2$};
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\draw[black](2,3.5) -- (0,-1/2);
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\draw[black](-2,3.5) -- (0,-1/2);
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\end{tikzpicture}
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\end{minipage}
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\begin{minipage}{0.3\linewidth}
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\begin{tikzpicture}[scale=3]
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\draw[-latex](-0.1,0) -- (1.25,0) node[below]{$x$};
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\draw[-latex](0,-0.1) -- (0,1.25) node[above]{$y$};
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\filldraw[black] (0,0) node[below]{$O$};
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\draw[black, thick,domain=-0.1:1.1] plot (\x, \x*\x);
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\draw[densely dashed](0.5,0.25) -- (0.5, 0) node[below]{$x$};
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\draw[densely dashed](1,1) -- (1, 0) node[below]{$x+\Delta x$};
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\filldraw[black](0.5,0.35) node{$y$};
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\filldraw[black](0.95,1.1) node{$y_0$};
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\filldraw[black] (1/2,1/4) circle (0.5pt);
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\filldraw[black] (1,1) circle (0.5pt);
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\draw[black](1,3/4) -- (1/4,0);
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\draw[black](0.6,0.2) -- (9/8,5/4);
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\draw[line width=0.1] (0.85,0.7) arc (50:0:0.1);
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\filldraw[black](1,0.8) node{$\Delta\alpha$};
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\filldraw[black](0.5,0.8) node{$\vert\wideparen{yy_0}\vert=\Delta s$};
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\end{tikzpicture}
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\end{minipage}
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\hfill
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\begin{minipage}{0.6\linewidth}
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$y-y_0$平均曲率：$\hat{k}=\dfrac{\vert\Delta\alpha\vert}{\vert\Delta s\vert}$。\medskip
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$y$曲率：$k=\lim\limits_{\Delta x\to 0}\left\lvert\dfrac{\Delta\alpha}{\Delta s}\right\rvert=\left\lvert\dfrac{\rm{d}\alpha}{\rm{d}s}\right\rvert$（$\alpha$为$y$处切线与$x$轴所成角）。
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\end{minipage}\medskip
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需要对曲率公式进行化简，得到$s$与$\alpha$关于$x$的表示。根据弧微分的定义：$\rm{d}s=$$\sqrt{1+f'^2(x)}\rm{d}x$。
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而对于$\alpha$：$\tan\alpha=y'=f'(x)$。
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两边对$x$求导：$\sec^2\alpha\cdot\dfrac{\rm{d}\alpha}{\rm{d}x}=y''=f''(x)$。
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又$\because\sec^2\alpha=1+\tan^2\alpha=1+y'^2$。
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$\therefore\dfrac{\rm{d}\alpha}{\rm{d}x}=\dfrac{y''}{1+y'^2}\Rightarrow\rm{d}\alpha=\dfrac{y''}{1+y'^2}\rm{d}x$。
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$\therefore k=\left\lvert\dfrac{\rm{d}\alpha}{\rm{d}s}\right\rvert=\dfrac{\vert y''\vert}{(1+y'^2)^{\frac{3}{2}}}$。
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\begin{minipage}{0.5\linewidth}
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$\bigcirc O$为函数$L$在点$X$处的曲率圆，该圆与$L$在$X$处相切，切线为$T$。
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该点的曲率半径为$R$，其中$R=\dfrac{1}{K}$。
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\end{minipage}
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\hfill
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\begin{minipage}{0.4\linewidth}
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\begin{tikzpicture}[scale=2]
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\draw (-1,0) -- (1,0);
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\node (X) at (0,0)[below]{X};
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\node (O) at (0,0.5)[above]{O};
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\draw[densely dashed] (X) -- (O);
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\filldraw[black] (0.75,0.25) node{$L$};
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\draw[decorate,decoration={brace,mirror,raise=2pt}] (X) -- (O);
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\filldraw[black] (0.2,0.25) node{$R$};
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\filldraw[black] (-0.75,0) node{$T$};
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\draw[black, thick,domain=-1:1] plot (\x, \x*\x);
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\draw[black] (0,0.5) circle (0.5);
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\end{tikzpicture}
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\end{minipage}
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\end{document}
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