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如何绘制隐式极坐标方程的图形？

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如何绘制隐式极坐标方程的图形？

我需要使用 Latex 绘制以下隐式极坐标方程

在此处输入图片描述

我将非常感激任何帮助。谢谢

答案1

使用方法如下渐近线以及一些借用自渐近线路径的变换/映射。

镜面夏威夷耳环

% file: foo.tex
% to compile: pdflatex --shell-escape foo
%
% For MikTeX users: Asymptote requires a separate program that cannot be installed
% by the package manager. You can get the installation file from
% https://sourceforge.net/projects/asymptote/files/2.35/
% (specifically, the file ending in setup.exe).
\documentclass[margin=10pt]{standalone}
\usepackage{asypictureB}

\begin{asyheader}
struct planeTransformation {
  int nInterpolate = 4;
  pair apply(real, real);
  pair apply(pair uv) { return apply(uv.x, uv.y); }
  transform derivative(real, real);
  transform derivative(pair uv) { return derivative(uv.x, uv.y); }
  transform linearization(real u, real v) {
    return shift(apply(u,v)) * derivative(u,v) * shift(-(u,v));
  }
  transform linearization(pair uv) {
    return linearization(uv.x, uv.y);
  }
  /* Apply to a single Bezier spline. */
  guide _apply(pair p1, pair c1, pair c2, pair p2) {
    return apply(p1) .. controls linearization(p1)*c1 and linearization(p2)*c2 .. apply(p2);
  }
  guide _apply(path g) {
    assert((length(g)) == 1);
    return _apply(point(g,0), postcontrol(g,0), precontrol(g,1), point(g,1));
  }
  path apply(path g, int nInterpolate = nInterpolate) {
    guide toreturn;
    for (int i = 0; i < nInterpolate*length(g); ++i) {
      real currentpos = i / nInterpolate;
      real nextpos = (i+1) / nInterpolate;
      toreturn = toreturn & _apply(subpath(g, currentpos, nextpos));
    }
    if (cyclic(g)) toreturn = toreturn & cycle;
    return toreturn;
  }
}

planeTransformation polar;

polar.apply = new pair(real theta, real r) {
  return r * expi(theta);
};

polar.derivative = new transform(real theta, real r) {
  transform t = (0, 0, -r*sin(theta), cos(theta), r*cos(theta), sin(theta));
  return t;
};
\end{asyheader}

\begin{document}
\begin{asypicture}{name=ImplicitPolarGraph}
settings.outformat = "pdf";
import contour;
size(10cm);

real F(real theta, real r) {
  return r^2*sin(theta) + sin(theta) - r*theta;
}

real rmax = 2;

path[] xygraphcomponents = contour(F, (-6 pi, -rmax), (6 pi, rmax), new real[] {0}, nx=600, ny=100)[0];

// Draw polar "axes"
for (real r = 1/2; r < rmax; r += 1/2) {
  draw(circle(r=r, c=(0,0)), gray + linewidth(0.4),
       L=Label("$"+(string)r+"$", black, position=BeginPoint, align=N, filltype=UnFill));
}
draw(circle(r=2, c=(0,0)), black);

for (int degrees = 0; degrees < 360; degrees += 30) {
  draw((0,0) -- scale(rmax)*dir(degrees), p=gray + linewidth(0.4), L=Label("$"+(string)degrees+"^{\circ}$", black, position=EndPoint, align=dir(degrees)));
}

for (path xycomponent : xygraphcomponents) {
  path rthetacomponent = polar.apply(xycomponent);
  draw(rthetacomponent, blue);
}

\end{asypicture}
\end{document}

答案2

作为@egreg建议，你可以用极坐标图绘制数据。假设你想绘制r = sin(2a)绘制A是角度，r是幅度，

在此处输入图片描述

\documentclass{standalone}
\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{pgfplots.polar}

\begin{document}

\begin{tikzpicture}
    \begin{polaraxis}
    \addplot+[mark=none,domain=0:360,samples=100] 
        {sin(2*x)}; 
    % the expression is the RADIUS
    \end{polaraxis}
\end{tikzpicture}

\end{document}

现在回答你的问题，根据 Matlab，解决方案是

在此处输入图片描述

如果角度等于零或 pi，则上述函数未定义。

\documentclass{standalone}
\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{pgfplots.polar}

\begin{document}

\begin{tikzpicture}
    \begin{polaraxis}
    \addplot+[mark=none,domain=30:80,samples=100] 
        { (x + sqrt( (x - 2*sin(x) )*(x + 2*sin(x)) ))  /(2*sin(x) };
    \end{polaraxis}
\end{tikzpicture}

\end{document}

结果是

在此处输入图片描述

答案3

这是函数的图形，将根的表达式简化为

r=\frac{1 \pm \sqrt{1-4\sinc^2\theta}}{2\sinc\theta}

在哪里sinc是正弦基数。它定义为 sinc θ <= 1/2，θ 不同于 kπ。恰好pst-math定义了一个SINC函数，这简化了代码：

\documentclass[11pt, a2paper, pdf, svgnames]{standalone}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage{ pst-plot, pst-math}
\usepackage{auto-pst-pdf}
\let\TwoPi\psPiTwo
\let\Pi\psPi

\def\Fone{(1 + sqrt(1-(2*SINC(x))^2))/(2*SINC(x))}
\def\Ftwo{(1 - sqrt(1-(2*SINC(x))^2))/(2*SINC(x))}

\begin{document}

\psset{arrowinset=0.15, ticksize=2.5pt -2.5pt, labelFontSize=\footnotesize, tickwidth =0.6pt}
\begin{pspicture*}(-11,-10)(11,9)
\psaxes[arrows=->, linecolor=LightSteelBlue, tickcolor=LightSteelBlue, Dx=2,ytrigLabels ,dy = \psPi] (0,0)(-11,-9)(11,9)[$x$,-110][$y$,-140]
\psset{linewidth=1.2pt, plotpoints=200, plotstyle=curve, polarplot, algebraic, labelsep=0.5em}
%First series
\psset{linecolor=IndianRed}
\psplot[arrows=*-]{1.896}{3.14159}{\Fone}%
\psplot{6.28}{9.42}{\Fone}
\psplot{12.57}{15.71}{\Fone}%
\psplot[arrows=*-]{-1.896}{-3.14}{\Fone}%
\psplot{-6.28}{-9.42}{\Fone}%
\psplot{-12.57}{-15.71}{\Fone}
\psset{linecolor=Coral}
\psplot{3.15160}{6.28}{\Ftwo}%
\psplot{9.42}{12.57}{\Ftwo}%
\psplot{-3.15}{-6.28}{\Ftwo}%
\psplot{-9.42}{-12.57}{\Ftwo}%
\psline[linecolor=LightSteelBlue](-11, \TwoPi)(11,\TwoPi)
\psline[linecolor=LightSteelBlue](-11, \Pi)(11,\Pi)
\psline[linecolor=LightSteelBlue](-11, -\Pi)(11,-\Pi)%
\psline[linecolor=LightSteelBlue](-11, -\TwoPi)(11,-\TwoPi)

%Second series
\psset{linecolor =OliveDrab }
\psplot{1.896}{3.14160}{\Ftwo}
\psplot{-1.896}{-3.14160}{\Ftwo}
\psplot{6.28}{9.42}{\Ftwo}%
\psplot{-6.28}{-9.42}{\Ftwo}%
\psset{linecolor =DarkSeaGreen}
\psplot{3.14160}{6.28}{\Fone}%
\psplot{9.42}{12.57}{\Fone}%
\psplot{-3.15}{-6.28}{\Fone}%
\psplot{-9.42}{-12.57}{\Fone}%%%
\end{pspicture*}

\end{document}

在此处输入图片描述

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