[Tex/LaTex] Tikz and Secant Line diagram

tikz-arrowstikz-pgf

Hi I am looking for feedback to improve an existing program PLUS advice for a desired diagram in the same direction.

Here is my minimal example:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing}

\begin{document}
\begin{center}
\begin{tikzpicture}[scale=1.75,cap=round]
\tikzset{axes/.style={}}
%\draw[style=help lines,step=1cm, dotted] (-5.25,-5.25) grid (5.25,5.25);
% The graphic
\begin{scope}[style=axes]
\draw[->] (-.5,0) -- (4.5,0) node[below] {$x$};
\draw[->] (0,-.5)-- (0,3) node[left] {$y$};
\foreach \x/\xtext in {1.5/x_{1}, 3/x_{2}}
 \draw[xshift=\x cm] (0pt,2pt) -- (0pt,-2pt) 
node[below,fill=white,font=\normalsize]
  {$\xtext$};    
\foreach \y/\ytext in {1/y_{1}=f(x_{1}), 2.125/y_{1}=f(x_{2})}
  \draw[yshift=\y cm] (2pt,0pt) -- (-2pt,0pt) 
  node[left,fill=white,font=\normalsize]
  {$\ytext$};
  %%%
 \draw[domain=.5:3.25,smooth,variable=\x,red,<->,thick] plot ({\x},{.5*(\x-1.5)*(\x-1.5)+1});
  %%%
\filldraw[black] (1.5,1) circle (1pt) node[above] {\scriptsize $P$};
\filldraw[black] (3,2.125) circle (1pt) node[left] {\scriptsize $Q$};
\draw[thick,blue!50,shorten >=-.5cm,shorten <=-.5cm] (1.5,1)--(3,2.125) 
node[midway,left] {\scriptsize Secant Line};
 %%%
 \draw[blue!50,thick,dashed] (1.5,1)--(3,1)--(3,2.125);
 \draw[blue!50] (3,1.1)--(2.9,1.1)--(2.9,1);
 \draw[decoration={brace,mirror,raise=5pt},decorate,blue!50]
    (1.5,-.250) -- node[below=6pt] {$x_{2}-x_{1}$} (3,-.250);
 \draw[decoration={brace,mirror, raise=5pt},decorate,blue!50]
    (3,1) -- node[right=6pt] {$f(x_{2})-f(x_{1})$} (3,2.215);
 %%%
\filldraw[black] (1.5,1) circle (1pt) node[above] {\scriptsize $P$};
\filldraw[black] (3,2.125) circle (1pt) node[left] {\scriptsize $Q$};
\end{scope}
\end{tikzpicture}
\end{center}
\end{document}

This will Output

I am trying to go here with the picture:

This is a bit beyond my programming skills I think ? PLease all suggestions welcome

Best Answer

With decorations.markings you can mark coordinates along the path, which then allow you to draw tangents. Note that drawing tangents has already been discussed at length in this nice answer, and I am implicitly using the same approach. However, my code is an attempt to have a unified treatment of both of your requests, i.e. tangent and secants, so at first sight it looks quite different.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{decorations.pathreplacing,decorations.markings,calc,arrows.meta,bending}

\begin{document}
\begin{tikzpicture}[scale=2.5,cap=round,mark pos/.style args={#1/#2}{%
postaction={decorate,decoration={markings,%
mark=at position #1 with {
\coordinate (#2);}}}}]
\tikzset{axes/.style={}}
%\draw[style=help lines,step=1cm, dotted] (-5.25,-5.25) grid (5.25,5.25);
% The graphic
\begin{scope}[style=axes]
  %%%
  \pgfmathsetmacro{\posP}{0.38}
 \draw[red,{Latex[bend]}-{Latex[bend]},thick,mark
 pos/.list={\posP-0.005/p-0,\posP/P,\posP+0.005/p-2,0.5/q-4,0.62/q-3,0.74/q-2,0.86/q-1}] plot[domain=.5:3.25,samples=101,variable=\x] ({\x},{.5*(\x-1.5)*(\x-1.5)+1});
 \draw[red] let \p1=($(p-2)-(p-0)$),\n1={(\y1/\x1)*(1cm/1pt)}
 in ($(P)-1*(1,\n1)$) -- ($(P)+2*(1,\n1)$) node[right,anchor=north
 west,font=\scriptsize,text width=1cm]{slope $m$ $=$ instaneous rate \dots};
 \fill (P) circle (1pt) node[above,font=\scriptsize] {$P$};
 \foreach \X in {1,...,4}
 {\fill (q-\X) circle (1pt) node[below right,font=\scriptsize] {$Q_\X$};
 \path (P) -- (q-\X) coordinate[pos=-0.5] (L-\X) coordinate[pos={1.2+\X*0.3}] (R-\X);
 \draw[cyan,dashed] (L-\X) -- (R-\X) node[right,font=\scriptsize] (m\X) {slope $m_\X$}; }
 \draw[line width=2mm,-{Latex[bend]},red!20] ($(m1)+(0.5,0.1)$)
 to[out=-90,in=65] ++ (-0.2,-1.2);
  %%%
 %%%
\end{scope}
\end{tikzpicture}
\end{document}

Related Solutions

[Tex/LaTex] TikZ: Cropping the Bounding Box

The code adds some completely useless invisible (or rather white) stuff. The lines

\clip(0pt,403pt) -- (389.957pt,403pt) -- (389.957pt,99.6166pt) -- (0pt,99.6166pt) -- (0pt,403pt);
\color[rgb]{1,1,1}
\fill(3.76406pt,399.236pt) -- (380.923pt,399.236pt) -- (380.923pt,253.19pt) -- (3.76406pt,253.19pt) -- (3.76406pt,399.236pt);
\fill(53.4497pt,394.719pt) -- (374.901pt,394.719pt) -- (374.901pt,289.325pt) -- (53.4497pt,289.325pt) -- (53.4497pt,394.719pt);

draw a white background that is larger than the actual picture. TikZ sees that and thinks it is part of the picture. Simply removing/uncommenting these lines removes most of the whitespace.

Near the end of the first scope,

\color[rgb]{1,1,1}
\fill(3.76406pt,249.426pt) -- (386.193pt,249.426pt) -- (386.193pt,103.381pt) -- (3.76406pt,103.381pt) -- (3.76406pt,249.426pt);

does the same. Additionally (near the end of the second scope),

\pgftext[center, base, at={\pgfpoint{220.95pt}{106.392pt}}]{\sffamily\fontsize{9}{0}\selectfont{\textbf{ }}}

adds a blank node below the picture, again enlarging the bounding box.

Removing all those lines gives a tight bounding box.

As far as I know, TikZ cannot do the cropping for you, as it can't know whether the white stuff is intentional or not (there might for example be a dark background behind the image so that white is visible).

[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)

Best Answer

Related Solutions

[Tex/LaTex] TikZ: Cropping the Bounding Box

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

Related Question