如何在 TiKz 中绘制曲线点处矢量的切向和法向分量的图形

Question

如何在 TikZ 中绘制下图？

编译使用 XeLaTeX。

我已经尝试过如下方法

\begin{tikzpicture}[tangent/.pic={\draw[-stealth] (0,0)--(2,0);}]
  
  \draw[mydarkblue,line width=1.2pt] (0,0) .. controls +(70:6) and +(-120:6) .. (5,2) coordinate[pos=0.87] (P) pic[pos=0.87,sloped,red,thick,name path=T]{tangent} pic[pos=0.87,sloped,rotate=90,blue,thick,name path=N]{tangent};


  \fill[red] (P) circle(1.3pt);

\end{tikzpicture}

提前致谢。

顺便说一句，圣诞快乐！

Answer 1

您可以在原点处绘制矩形并旋转tikzpicture。使用该库可以轻松绘制花括号。例如，decorations曲线可以用几个 s 来完成。arc

像这样：

\documentclass[tikz,border=2mm]{standalone}
\usepackage{amsmath} % for \dfrac
\usetikzlibrary{decorations.pathreplacing} % for the curly brackets

\tikzset
{
  every node/.style={text=black},
  curly/.style={decorate,decoration={brace,amplitude=5pt,raise=3pt}},
  vector/.style={very thick,-latex}
}

\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,rotate=20,scale=2,thick]
% curve and rectangle
\draw[cyan] (-1,-1) arc (180:90:1) (0,0) arc (-90:-20:2);
\draw[gray] (0,0) rectangle (1.2,1.4);
% vectors
\draw[vector,yellow!70!orange] (0,0) -- (1.2,1.4) node[above] {\bfseries a};
\draw[vector,magenta]          (0,0) -- (1,0)     node[above] {\bfseries T};
\draw[vector,green!80!blue]    (0,0) -- (0,1)     node[right] {\bfseries N};
% s, P_0
\fill[cyan] (0,0) circle (1pt);
\begin{scope}[shift={(0,-1)}]
  \draw[vector,cyan!60!black] (150:1) arc (150:120:1) node[below right] {$s$} ;
  \draw (150:1.05) -- (150:0.95) node [right] {$P_0$};
\end{scope}
% curly brackets
\draw[curly] (0,0)   -- (0,1.4) node[midway,xshift=-1.6cm] {$a_\mathrm{N}=\kappa\left(\dfrac{ds}{dt}\right)^2$};
\draw[curly] (1.2,0) -- (0,0)   node[midway,yshift=-0.8cm] {$a_\mathrm{T}=\dfrac{d^2s}{dt^2}$};
\end{tikzpicture}
\end{document}