[Tex/LaTex] Curly brace arrow and labeling

tikz-pgf

I have the following code which generates a plot of $x^2+1$ and then marks a distance by a curly brace on it.

\documentclass{article}
\usepackage{tikz}
\usepgflibrary{arrows}
\usetikzlibrary{arrows.meta,automata, decorations.pathreplacing}
\usepackage{verbatim}
\begin{document}
\begin{tikzpicture}[domain=0:1.8, xscale = 6, yscale = 1.5]

\draw[->]  (-0.2,0)  --  (2,0)  node[right]  {$x$};
\draw[->]  (0,-1.2)  --  (0,4)  node[above]  {$f(x)$};

\draw[color=blue]      plot  (\x,{\x^2+1})        node[right]  {};
\node(1) at (0.5,0) [circle,draw, fill, scale = 0.5 pt]{};
\node(2) at (0.5,0.5^2+1) [circle,draw, fill, scale = 0.5pt]{};
\node(3) at (1.5,0) [circle,draw, fill, scale = 0.5pt]{};
\node(4) at (1.5,1.5^2+1) [circle,draw, fill, scale = 0.5pt]{};
\node(5) at (1.0, 0) [circle,draw, fill, scale = 0.5pt]{};
\node(6) at (1.0, 1.0^2+1) [circle,draw, fill, scale = 0.5pt]{};

\node[below of = 1](leftlabel){$x-dx$};
\node[below of = 5](middlelabel){$x$};
\node[below of = 3](rightlabel){$x+dx$};

\draw [decorate,decoration={brace,amplitude=10pt,mirror}, yshift = -0.2cm]
(1.0,1^2+1) -- (1.5,1^2+1) node (curly_bracket)[black,midway, yshift =- 0.3 cm] 
{};
\node[below right of =  curly_bracket, xshift = 1 cm, yshift = -0.1 cm](test){};
\draw[<-] (curly_bracket) -- (test) node[at end,label=below right:{$dx$}]{};

\draw[dashed] (1.0,1^2+1)--(1.5,1^2+1) node[below =0.3cm, right = 0.7 cm](7){} node[midway](8){};


\end{tikzpicture}
\end{document}

The above results in the following plot:

As you can see the positioning of the arrow and its label to the curly brace are not very good. I would like for the arrow to point to exactly the tip of the brace with a small vertical spacing, and for the label to the opposite side of the arrow to appear a lot closer to the end of the line.

Ideally I am looking to achieve something like this:

How do I bring the arrow from the tip of the curly brace, and label its end with adequate anchor spacing?

EDIT

To respond to @Zarko, and to those skeptical of why I would be embarking on such an endeavor, I would like to show the following plot

As you can see this would be challenging to do without the arrow leaders provided by the accepted answer

Best Answer

\documentclass{article}
\usepackage{tikz}
\usepgflibrary{arrows}
\usetikzlibrary{arrows.meta,automata, decorations.pathreplacing}
\usepackage{verbatim}
\begin{document}
\begin{tikzpicture}[domain=0:1.8, xscale = 6, yscale = 1.5]

\draw[->]  (-0.2,0)  --  (2,0)  node[right]  {$x$};
\draw[->]  (0,-1.2)  --  (0,4)  node[above]  {$f(x)$};

\draw[color=blue]      plot  (\x,{\x^2+1})        node[right]  {};
\node(1) at (0.5,0) [circle,draw, fill, scale = 0.5 pt]{};
\node(2) at (0.5,0.5^2+1) [circle,draw, fill, scale = 0.5pt]{};
\node(3) at (1.5,0) [circle,draw, fill, scale = 0.5pt]{};
\node(4) at (1.5,1.5^2+1) [circle,draw, fill, scale = 0.5pt]{};
\node(5) at (1.0, 0) [circle,draw, fill, scale = 0.5pt]{};
\node(6) at (1.0, 1.0^2+1) [circle,draw, fill, scale = 0.5pt]{};

\node[below of = 1](leftlabel){$x-dx$};
\node[below of = 5](middlelabel){$x$};
\node[below of = 3](rightlabel){$x+dx$};

\draw [decorate,decoration={brace,amplitude=10pt,mirror}, yshift = -0.2cm]
(1.0,1^2+1) -- (1.5,1^2+1) node (curly_bracket)[black,midway, yshift =- 0.3 cm] 
{};
\node[below right of =  curly_bracket, xshift = 1 cm, yshift = -0.1
cm](test){$dx$};
\draw[latex-] (curly_bracket) to[out=-90,in=90,looseness=1.8] (test);

\draw[dashed] (1.0,1^2+1)--(1.5,1^2+1) node[below =0.3cm, right = 0.7 cm](7){} node[midway](8){};
\end{tikzpicture}
\end{document}

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

The source of the difficulty is that ellipses are constructed in a particular way in TikZ. They are paths that start from the x-axis and proceed counter-clockwise around their centre. The vast majority of the time, the exact parametrisation doesn't matter. You appear to have found the one situation where it does!

In the actual question, you only want to be able to mirror the ellipse, and so draw it starting from the negative x-axis (the title of the question suggests a more flexible approach). That's actually not too hard since we can exploit the symmetry of the ellipse. The key is to provide it with a negative x-radius, since then it will start from the negative x-axis (and proceed clockwise, but we could correct for that by negating the y-radius as well). To do this, we interrupt the call from the node shape to the drawing command and flip the sign of the x-radius. The simplest way to do this is to redefine the \pgfpathellipse macro to do the negation and then call the original macro. The following code does this.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations,shapes,decorations.markings}
\makeatletter
\let\origpgfpathellipse=\pgfpathellipse
\def\revpgfpathellipse#1#2#3{%
  #2%
  \pgf@xa=-\pgf@x
  \origpgfpathellipse{#1}{\pgfqpoint{\pgf@xa}{0pt}}{#3}}
\makeatother

\tikzset{
  reversed ellipse/.style={
    ellipse,
    reverse the ellipse%
  },
  reverse the ellipse/.code={
    \let\pgfpathellipse=\revpgfpathellipse
  }
}
\begin{document}
\begin{tikzpicture}
\node[ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,0) {hello world};

\node[reversed ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,-2) {hello world};
\end{tikzpicture}

\end{document}

Here's the result:

rotated ellipse

(the arrow got clipped, but you can see where it lies)

[Tex/LaTex] Numerical conditional within tikz keys

I am definitely unfamiliar with both beamer and tikz (do not quite get what the \only are supposed to do) but perhaps this could go in the direction you want:

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{chains}
\newcounter{count}

% helper macro:
\long\def\GobToSemiColon #1;{}

\newcommand\myPicture{
\begin{tikzpicture}
  \begin{scope}[start chain = going below]
  \ifnum\value{count}<1 \expandafter\GobToSemiColon\fi
  \ifnum\value{count}>3 \expandafter\GobToSemiColon\fi
  \node[draw, rectangle, on chain] {display only when counter is between
    1 and 3};

  \ifnum\value{count}>-1 \expandafter\GobToSemiColon\fi
  \node[draw, rectangle, on chain] {display only when counter is
    negative}; 

  \ifnum\value{count}<100 \expandafter\GobToSemiColon\fi
  \ifnum\value{count}>200 \expandafter\GobToSemiColon\fi  
  \node[draw, rectangle, on chain] {display only if counter is between
    100 and 200};

  \ifnum\value{count}<3 \expandafter\GobToSemiColon\fi
  \ifnum\value{count}>20 \expandafter\GobToSemiColon\fi
  \node[draw, circle, on chain] {only when counter is in the range 3 to 20};
  \end{scope}
\end{tikzpicture}
}

\begin{document}
\begin{frame}
\only{\setcounter{count}{-3}\myPicture}
\only{\setcounter{count}{105}\myPicture}
\only{\setcounter{count}{39}\myPicture}
\only{\setcounter{count}{2}\myPicture}
\only{\setcounter{count}{5}\myPicture}
\end{frame}
\end{document}

beamer

Best Answer

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

[Tex/LaTex] Numerical conditional within tikz keys

Related Question