pgfplots：如何绘制起始角为 60° 的指数图？

Question

我想绘制一个简化的米氏动力学（单函数）来将其与线性函数进行比较。

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最小工作示例（MWE）：

\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{amsmath}

\pgfplotsset{compat=1.14,   /pgf/declare function={f1(\x)=ln(x);}}% <- This is the exponential function which needs to be optimized

\begin{document}
    \begin{tikzpicture}
        \begin{axis}[
                    ymin        = 0,
                    xmin        = 0,
                    xmax        = 1,
                    ymax        = 0.9,
                    axis x line = bottom,
                    axis y line = left,
                    ]

%           \addplot[no marks, samples=100, draw=blue] {f1(x)};% This is the exponential graph based on the function
            \addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
            \addplot[no marks, samples=100, draw=black, thick] (0.2020725942,0.35) to [out=60,in=180] (0.8,0.7) to [out=0,in=0] (1,0.7);%
            \draw[draw=black, dashed] (0,0.7) -- node[above] {\(y_{\text{tot}}\)} ++(0.8,0.0);%
            \draw[draw=black, dashed] (0,0.35) -- node[above] {\(\frac{y_{\text{tot}}}{2}\)} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
        \end{axis}
    \end{tikzpicture}
\end{document}

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结果截图：

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问题说明：

我怎样才能用指数图替换当前图？

指数图的起点：

• 起点：x = 0.2020725942，y = 0.35，角度 = 60°，
• 终点：y = ~ 0.7（当然，e 函数的终点是这里）

一旦我用指数函数激活图形，我的整个图表就会扭曲。如何正确实现基于上限值的指数图？

Answer 1

一种方法是通过这个（请注意，这使用与您的不同的函数）。你MWE没有错IMO。但是，由于域不同，您的最终轴变得混乱。

尽管如此，您可以通过两个指数相加来获得所需的解决方案。

\documentclass{amsart}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepackage{tikz}
\begin{document} 
    \begin{tikzpicture}
    \begin{axis}[
    scaled ticks=false,
    xmin=0,
    xmax=1,
    ymin=0,
    ymax=1.2,
    xlabel=x axis label,
    ylabel=y axis label,
    axis x line = bottom,
    axis y line = left,
    ]
    \addplot[domain=0.2:1.2, samples=1000, red, ultra thick,smooth] {(1-e^(-5*x)-exp(-10*x))*0.7};
\addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
\draw[draw=black, dashed] (0,0.7) -- node[above] {\(y_{\text{tot}}\)} ++(1,0.0);%
\draw[draw=black, dashed] (0,0.35) -- node[above] {\(\frac{y_{\text{tot}}}{2}\)} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
    \end{axis}
    \end{tikzpicture}
\end{document}

要得到：

Answer 2

我找不到这样的函数。这里，具有精确起始角度 (60°) 和终止角度 (180°) 的曲线就足够了。

另外，你为什么不直接在 Ti 中使用 tan 函数钾Z? 0.2020725942 ≈ 0.35 × tan(30°)，但如果你输入的话，{.35*tan(30)}它肯定比 更准确0.2020725942。

\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}[scale=8,>=stealth]
\draw[<->] (1,0) -- (0,0) -- (0,.9);
\draw[thick] (0,0) -- ({.35*tan(30)},0.35) coordinate (a);
\draw[thick] (a) to[out=60,in=180] (0.8,0.7) -- (1,0.7);
\foreach \i in {0,0.2,0.4,0.6,0.8} {
    \draw (\i,.01) -- (\i,-.01) node[below] {$\i$};
    \draw (.01,\i) -- (-.01,\i) node[left] {$\i$};
}
\draw (1,.01) -- (1,-.01) node[below] {$1$};
\draw[dashed] (0.8,0.7) -- (0,0.7) node[midway,above] {$y_\mathrm{tot}$};
\draw[dashed] ({.35*tan(30)},0) -- ({.35*tan(30)},0.35);
\draw[dashed] ({.35*tan(30)},0.35) -- (0,0.35) node[midway,above] {$y_\mathrm{tot}/2$};
\end{tikzpicture}
\end{document}