我正在尝试使用 tikz 根据此绘制参数方程邮政。我最终得到的是一条刚性曲线，与期望的轨迹完全不同。
我认为我输入参数方程的方式可能存在问题。我该如何解决这个问题？
\begin{tikzpicture}[scale=1]
\draw[gray,very thin] (-3.9,-3.9) grid (3.9,3.9);
\draw[->] (-4,0) -- (4,0) node[right] {$x(t)$};
\draw[->] (0,-4) -- (0,4) node[above] {$y(t)$};
\foreach \pos in {-3,-2,-1,2,3}
\draw[shift={(\pos,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\pos$};
\foreach \pos in {-3,-2,-1,1,2,3}
\draw[shift={(0,\pos)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\pos$};
\fill (0,0) circle (0.064cm);
\draw[thick,variable=\t,domain=-0.2:7,samples=100]
plot ({e^(-\t)*(-2*cos(2\t)+2.3817218*sin(2\t)) },{e^(-\t)*(2.3817218*cos(2\t)+2*sin(2\t))})
node[right] {$\bigl(x(t),y(t)\bigr)$};
\fill[red] (1,0) circle (2pt)
node [below right,yshift=-4pt] {$(1,0)$};
\end{tikzpicture}
答案1
\documentclass[tikz, border=1cm]{standalone}
\begin{document}
\begin{tikzpicture}
\draw[gray, very thin] (-3.9,-3.9) grid (3.9,3.9);
\draw[->] (-4,0) -- (4,0) node[right] {$x(t)$};
\draw[->] (0,-4) -- (0,4) node[above] {$y(t)$};
\foreach \pos in {-3,-2,-1,2,3}
\draw[shift={(\pos,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\pos$};
\foreach \pos in {-3,-2,-1,1,2,3}
\draw[shift={(0,\pos)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\pos$};
\fill (0,0) circle[radius=2pt];
\draw[thick, variable=\t, domain=-0.2:7, samples=100, trig format=rad, smooth]
plot ( {e^(-\t)*(-2*cos(2*\t)+2.3817218*sin(2*\t))} , {e^(-\t)*(2.3817218*cos(2*\t)+2*sin(2*\t))} )
node[above] {$\bigl(x(t),y(t)\bigr)$};
\fill[red] (1,0) circle[radius=2pt] node[below right] {$(1,0)$};
\end{tikzpicture}
\end{document}
