TikZ：非线性切线曲线

Question

我想从点到点绘制一条切线曲线(a)，(b)交点(c)（正方形的中心）。曲线应根据灰线拉伸（对数刻度）。我尝试过controls，但进展不大。有没有比使用更好的方法controls？

我迄今为止的代码：

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[every node/.style = {draw, circle, fill=white, inner sep=.01cm}]

\draw (0, 1) -- (10, 1) -- (10, -1) -- (0, -1) -- cycle;
\node (a) at (0, -1) {a};   % south-west corner
\node (b) at (10, 1) {b};   % north-east corner
\node (c) at (5, 0) {c};    % center
\foreach \n in {1,...,50} \draw[gray, very thin]
        ({10/pow(2, \n/12)}, 1)
    --  ({10/pow(2, \n/12)}, -1);

\draw[cyan] (a) -- (b);

\end{tikzpicture}
\end{document}

结果：

我想要实现以下目标（尽管曲线的左半部分应该下降得更快）：

Answer 1

这也许更接近一点吧？

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[every node/.style = {draw, circle, fill=white, inner sep=.01cm}]

\draw (0, 1) -- (10, 1) -- (10, -1) -- (0, -1) -- cycle;
\node (a) at (0, -1) {a};   % south-west corner
\node (b) at (10, 1) {b};   % north-east corner
\node (c) at (5, 0) {c};    % center
\foreach \n in {1,...,50} \draw[gray, very thin]
        ({10/pow(2, \n/12)}, 1)
    --  ({10/pow(2, \n/12)}, -1);

\draw[thick,cyan] (a) to[out=0,in=270,looseness=0.55] (c.center) to[out=65,in=180,looseness=0.5] (b);

\end{tikzpicture}
\end{document}

Answer 2

这是我从你的评论描述中得到的。我的想法接近你的想法了吗？如果不接近，也许你可以提供一个手绘模型。

\usetikzlibrary{calc}
\begin{tikzpicture}[every node/.style = {draw, circle, fill=white, inner sep=.01cm}]

\draw (0, 1) -- (10, 1) -- (10, -1) -- (0, -1) -- cycle;
\node (a) at (0, -1) {a};   % south-west corner
\node (b) at (10, 1) {b};   % north-east corner
\node (c) at (5, 0) {c};    % center
\foreach \n in {1,...,50} \draw[gray, very thin]
        ({10/pow(2, \n/12)}, 1)
    --  ({10/pow(2, \n/12)}, -1);

\draw[cyan] (a) .. controls +(5, 0) and +(-.5, -.5) .. (c.center) .. controls +(.5,.5) and +(-1,0) .. (b);
\end{tikzpicture}