# TikZ: non-linear tangent curve

I'd like to draw a tangent curve from point (a) to point (b), intersecting point (c) (the center of the square). The curve should be stretched according to the gray lines (being a logarithmic scale). I've played around with controls, but I didn't get very far. Is there a better way than using controls?

My code so far:

\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}


The result: What I want to achieve is the following (although the left half of the curve should decrease more rapidly): • Among your trials, have you come across this \draw (0,-1) .. controls (5,-1.0) and (5,1.0) .. (10,1); Further, you could try (5,-1.1) and (5,1.1) pairs too. – Jesse May 12 '14 at 11:50
• I don't undersand the "logarithmic" nature of your curve. What is is supposed to increase logarithmically? The derivative of the curve? Then, how could it be tangent to (b)? – JLDiaz May 12 '14 at 12:23
• Will this produce what you want? \draw[cyan] (a) to[out=0,in=180] (b); – Gonzalo Medina May 12 '14 at 12:26
• Or what about this: \draw[cyan] (a) .. controls +(4, 0) and +(-.5, -.5) .. (c.center) .. controls +(.5,.5) and +(-2,0) .. (b);  ? – JLDiaz May 12 '14 at 12:30
• Hi, watain, just a thought: Continue searching \draw (0,-1) .. controls (6,-1.0) and (4,1.0) .. (10,1); (7,-1) and (3,1); (8,-1) and (2,1) or any pairs among them, should get one of your expected output. – Jesse May 12 '14 at 12:30

May be this comes some what closer?

\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} • At first I wasn't too keen about using the to[...] syntax, but it seems like it's really the easiest way. It pretty much matches what I needed. Thank you! – watain May 13 '14 at 6:49

This is what I get from your description in one of your comments. Was I close to your idea? If not, perhaps you could provide a hand-drawn mockup. \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}

• Thanks for your answer, too bad I can't accept two correct answers - yours seems to be a good alternative to what Harish Kumar posted. – watain May 13 '14 at 6:50
• Not worry about the acceptance. More important to me is, if I got it right. Was this the kind of slope you wanted? – JLDiaz May 13 '14 at 8:17
• Yes, this is the kind of slope that I wanted. Your answer made me understand how controls really works. Harish Kumar's answer just seemed simpler to me, but both are correct. – watain May 13 '14 at 8:43