2

Can anyone helped me with this? I've looked up questions with similar problems but nome of them seem to work for me. I'm writing a book and I like to use wrapfig for site-notes for recalls or notation etc. Throughout the book, the wrapped-figure reserves space below it for the rest of the page. \clearpage fixes the problem but I don't want to start a new page after each wrapfigure. Help :(

Here's the particular case I'm dealing with: enter image description here

I want the paragraph under the diagram to spread out over the whole page. An MWE of the above can be found below.

    \documentclass{article}
    \usepackage{tikz}
    \usepackage{amsmath}
    \usepackage{wrapfig}
    \usepackage{framed}
    \usepackage{geometry}
    \geometry{a4paper, portrait, margin=1in}
        \begin{document}
    \subsection*{What is Differentiation?}
    Differentiation is a mathematical tool used to find the \textbf{gradient of a tangent} to any general curve $y=f(x)$ at any desired point ($P$).
    \begin{center}
        \begin{tikzpicture}[domain=0:4]
        \draw (0,-0.3) node[left]{$O$};
        \draw[thick, color=gray,->] (-4,0) -- (5,0) node[right] {\textcolor{black}{$x$}};
        \draw[thick, color=gray, ->] (0,-1) -- (0,5) node[above] {\textcolor{black}{$f(x)$}};
        \draw [color=red, semithick](-4,1).. controls(1,1.5).. (3.5,5);
        \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=below right:{$P(x,y)$}] (P) at (1.15,2.15) {};
        \draw (3,4) node[right]{$y=f(x)$};
        \draw (-1,3/5+0.05)--(3,34/10+0.05);
        \draw (-0.3,0.3) node[left]{tangent};
        \end{tikzpicture}
    \end{center}
    In general, the steepness (i.e. gradient) of a curve at any point $P$ is the same as the gradient of the tangent at that point; i.e.
    \[m_{\text{tangent at P}}=m_{f(x)\text{ at P}}\]
    In calculus, $m_{f(x)\text{ at P}}$ is denoted \[\displaystyle\frac{d}{dx}\left(f(x)\right)\text{~~or~~~~} \displaystyle\frac{dy}{dx}\] when the equation is defined in the form $y=f(x)$, and we call this general gradient the \textbf{derivative} of the curve. \paragraph{}
    \begin{wrapfigure}{r}{5.5cm}
        \vspace{-1cm}
        \begin{center}
            \begin{minipage}{5cm}
                \colorlet{shadecolor}{green!15}
                \begin{shaded}
                    \normalsize \textbf{Notation}
                    \Large $$\delta x$$
                    \normalsize The Greek letter $\delta$ (small-case delta) is a \textbf{prefix} to a variable and it represents an infinitesimally small increase in that variable. It is not a distinct value. 
                \end{shaded}
            \end{minipage}
        \end{center}
    \end{wrapfigure}
    Consider now another point on our general curve, the point Q. This point is $\delta x$ away from $P$ horizontally and $\delta y$ away from $P$ vertically:\\
        \begin{tikzpicture}[domain=0:4]
        \draw (0,-0.3) node[left]{$O$};
        \draw[thick, color=gray,->] (-4,0) -- (5,0) node[right] {\textcolor{black}{$x$}};
        \draw[thick, color=gray, ->] (0,-1) -- (0,5) node[above] {\textcolor{black}{$f(x)$}};
        \draw[color=blue] (1.15,2.15)--(3.3,4.7);
        \draw [color=red, semithick](-4,1).. controls(1,1.5).. (3.5,5);
        \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=below right:{$P(x,y)$}] (P) at (1.15,2.15) {};
        \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=right:{$Q(x+\delta x,y+\delta y)$}] (Q) at (3.3,4.7) {};
        \draw (3,4) node[right]{$y=f(x)$};
        \draw (-1,3/5+0.05)--(3,34/10+0.05);
        \draw (-0.3,0.3) node[left]{tangent};
        \draw[dotted] (3.3,4.7)--(3.3,4.3);
        \draw[dotted] (3.3,3.7)--(3.3,0);
        \draw[dotted] (1.12,2.15)--(1.12,0);
        \draw[dotted] (3.3,4.7)--(0,4.7);
        \draw[dotted] (1.12,2.15)--(0,2.15);
        \draw[<->,dashed](-0.3,2.15)--(-0.3,4.7);
        \draw(-0.6, 3.4) node{$\delta y$};
        \draw[<->,dashed](1.15,-0.3)--(3.3,-0.3);
        \draw(2.3, -0.6) node{$\delta x$};
        \end{tikzpicture}\\
        The coordinates of $Q$ are $(x+\delta x,y+\delta y)$, shown above. We notice that if the values of $\delta x$ and $\delta y$ were to get smaller and smaller, the gradient of the chord $PQ$ (in blue) would approach that of the tangent at $P$, the gradient we wish to find ($\frac{dy}{dx}$). We also know that the $y$-values of this graph are dependent on their respective $x$-value, since $y$ is defined as a function of $x$ ($y=f(x)$). So as $\delta x$ decreases, $\delta y$ will consequentially decrease because $(y+\delta y)$ is dependent on $(x+\delta x)$. So we can say that as $\boldsymbol{\delta x\longrightarrow 0}$;  $\boldsymbol{m_{PQ}\longrightarrow\frac{dy}{dx}}$. 
    \end{document}

I appreciate your feedback :)

  • I suspect wrapfig does not realize how much vertical space your tikzpicture is taking up, and thus continues to wrap lines of text until it has decided it has wrapped enough. The \clearpage probably just "forces" the issue. – jon May 29 '15 at 17:23
2

You only have to set the number of lines that will be wrapped as an optional argument to the environment:

\begin{wrapfigure}[3]{r}{5.5cm}
 …
\end{wrapfigure}

Alternatively, you might try the InsertBoxR generic macro in the place of the wrapfigure environment: use

......
\input{insboxtex}
\begin{document}
......

\InsertBoxR{0}{%
 \begin{minipage}{5cm}
\colorlet{shadecolor}{green!15}
    \vskip\dimexpr-\FrameSep-0.6ex\relax
\begin{shaded}
\normalsize \textbf{Notation}
\Large $$\delta x$$
    \normalsize The Greek letter $\delta$ (small-case delta) is a \textbf{prefix} to a variable and it represents an infinitesimally small increase in that variable. It is not a distinct value.
\end{shaded}
\end{minipage}}[-2]

enter image description here

2

You don't need wrapfig for this. You also don't need framed and amsbsy (you're abusing \boldsymbol, where \boldmath should be used if you really want to make math bold).

I also removed \left, \right and \displaystyle where unnecessary.

Instead of a non working wrapfig environment, use tabular, since you want to place something that won't allow flowing around the insertion.

\documentclass{article}
\usepackage{geometry}
\usepackage{tikz}
\usepackage{amsmath,bm,tabularx,framed}

\geometry{a4paper, portrait, margin=1in}

\newlength{\normalparindent}
\AtBeginDocument{\setlength{\normalparindent}{\parindent}}
\newcommand{\normalindent}{\hspace*{\normalparindent}}

\begin{document}

\subsection*{What is Differentiation?}

Differentiation is a mathematical tool used to find the \textbf{gradient of a tangent} to 
any general curve $y=f(x)$ at any desired point ($P$).
\begin{center}
\begin{tikzpicture}[domain=0:4]
  \draw (0,-0.3) node[left]{$O$};
  \draw[thick, color=gray,->] (-4,0) -- (5,0) node[right] {\textcolor{black}{$x$}};
  \draw[thick, color=gray, ->] (0,-1) -- (0,5) node[above] {\textcolor{black}{$f(x)$}};
  \draw [color=red, semithick](-4,1).. controls(1,1.5).. (3.5,5);
  \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=below right:{$P(x,y)$}] (P) at (1.15,2.15) {};
  \draw (3,4) node[right]{$y=f(x)$};
  \draw (-1,3/5+0.05)--(3,34/10+0.05);
  \draw (-0.3,0.3) node[left]{tangent};
\end{tikzpicture}
\end{center}
In general, the steepness (i.e. gradient) of a curve at any point $P$ is the same as the 
gradient of the tangent at that point; i.e.
\[
m_{\text{tangent at $P$}}=m_{f(x)\text{ at $P$}}
\]
In calculus, $m_{f(x)\text{ at $P$}}$ is denoted
\[
\frac{d}{dx}(f(x))\qquad\text{or}\qquad\frac{dy}{dx}
\]
when the equation is defined in the form $y=f(x)$, and we call this general gradient the 
\textbf{derivative} of the curve.\strut

\noindent\begin{tabular}{
  @{}
  p{\dimexpr\textwidth-2\tabcolsep-5cm-12pt}
  c
  @{}
}
\normalindent Consider now another point on our general curve, the point Q. This point 
is $\delta x$ away from $P$ horizontally and $\delta y$ away from $P$ vertically:
\begin{center}
\begin{tikzpicture}[domain=0:4,scale=.95]
  \draw (0,-0.3) node[left]{$O$};
  \draw[thick, color=gray,->] (-4,0) -- (5,0) node[right] {\textcolor{black}{$x$}};
  \draw[thick, color=gray, ->] (0,-1) -- (0,5) node[above] {\textcolor{black}{$f(x)$}};
  \draw[color=blue] (1.15,2.15)--(3.3,4.7);
  \draw [color=red, semithick](-4,1).. controls(1,1.5).. (3.5,5);
  \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=below right:{$P(x,y)$}] (P) at (1.15,2.15) {};
  \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=right:{$Q(x+\delta x,y+\delta y)$}] (Q) at (3.3,4.7) {};
  \draw (3,4) node[right]{$y=f(x)$};
  \draw (-1,3/5+0.05)--(3,34/10+0.05);
  \draw (-0.3,0.3) node[left]{tangent};
  \draw[dotted] (3.3,4.7)--(3.3,4.3);
  \draw[dotted] (3.3,3.7)--(3.3,0);
  \draw[dotted] (1.12,2.15)--(1.12,0);
  \draw[dotted] (3.3,4.7)--(0,4.7);
  \draw[dotted] (1.12,2.15)--(0,2.15);
  \draw[<->,dashed](-0.3,2.15)--(-0.3,4.7);
  \draw(-0.6, 3.4) node{$\delta y$};
  \draw[<->,dashed](1.15,-0.3)--(3.3,-0.3);
  \draw(2.3, -0.6) node{$\delta x$};
\end{tikzpicture}
\end{center} &
\setlength{\fboxsep}{6pt}%
\smash{\colorbox{green!15}{%
\begin{minipage}[t]{5cm}
\textbf{Notation}
\begin{center}
\Large $\delta x$
\end{center}
The Greek letter $\delta$ (small-case delta) is a \textbf{prefix} to a variable and it 
represents an infinitesimally small increase in that variable. It is not a distinct value.
\end{minipage}}}
\end{tabular}
The coordinates of $Q$ are $(x+\delta x,y+\delta y)$, shown above. We notice that if the 
values of $\delta x$ and $\delta y$ were to get smaller and smaller, the gradient of the 
chord $PQ$ (in blue) would approach that of the tangent at $P$, the gradient we wish to find 
($\frac{dy}{dx}$). We also know that the $y$-values of this graph are dependent on their 
respective $x$-value, since $y$ is defined as a function of $x$ ($y=f(x)$). So as $\delta x$ 
decreases, $\delta y$ will consequentially decrease because $(y+\delta y)$ is dependent on 
$(x+\delta x)$. So we can say that as {\boldmath$\delta x\longrightarrow 0$; 
$m_{PQ}\longrightarrow\frac{dy}{dx}$}.

\end{document}

It takes some tricks to persuade LaTeX into indenting the paragraph in the tabular, but \normalindent can reveal handy also in other situations when you want to indent paragraph in a \parbox or minipage, where normal indentation is suppressed.

Never use $$; the \Large$$\delta x$$ you had is simply achieved with a center environment.

I also scaled a bit the second tikzpicture or it wouldn't fit in the allotted space.

enter image description here

You can do it also without tabular, with some lower level tricks. The insertion is added at the start, as a zero width box containing another box as wide as the text width, where the insertion is placed at the far right margin.

Then the running text is set with a negative \hangindent, which makes the indentation at the right. The last line (containing the diagram) is centered with a nice trick involving \leftskip and \rightskip. We cover our tracks by issuing \noindent at the start of the text after the diagram.

\documentclass{article}
\usepackage{geometry}
\usepackage{tikz}
\usepackage{amsmath,bm,tabularx,framed}

\geometry{a4paper, portrait, margin=1in}

\begin{document}

\subsection*{What is Differentiation?}

Differentiation is a mathematical tool used to find the \textbf{gradient of a tangent} to 
any general curve $y=f(x)$ at any desired point ($P$).
\begin{center}
\begin{tikzpicture}[domain=0:4]
  \draw (0,-0.3) node[left]{$O$};
  \draw[thick, color=gray,->] (-4,0) -- (5,0) node[right] {\textcolor{black}{$x$}};
  \draw[thick, color=gray, ->] (0,-1) -- (0,5) node[above] {\textcolor{black}{$f(x)$}};
  \draw [color=red, semithick](-4,1).. controls(1,1.5).. (3.5,5);
  \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=below right:{$P(x,y)$}] (P) at (1.15,2.15) {};
  \draw (3,4) node[right]{$y=f(x)$};
  \draw (-1,3/5+0.05)--(3,34/10+0.05);
  \draw (-0.3,0.3) node[left]{tangent};
\end{tikzpicture}
\end{center}
In general, the steepness (i.e. gradient) of a curve at any point $P$ is the same as the 
gradient of the tangent at that point; i.e.
\[
m_{\text{tangent at $P$}}=m_{f(x)\text{ at $P$}}
\]
In calculus, $m_{f(x)\text{ at $P$}}$ is denoted
\[
\frac{d}{dx}(f(x))\qquad\text{or}\qquad\frac{dy}{dx}
\]
when the equation is defined in the form $y=f(x)$, and we call this general gradient the 
\textbf{derivative} of the curve.

\noindent
\makebox[0pt][l]{%
  \makebox[\textwidth][r]{%
    \setlength{\fboxsep}{6pt}%
    \smash{\colorbox{green!15}{%
    \begin{minipage}[t]{5cm}
    \textbf{Notation}
    \begin{center}
    \Large $\delta x$
    \end{center}
    The Greek letter $\delta$ (small-case delta) is a \textbf{prefix} to a variable and it 
    represents an infinitesimally small increase in that variable. It is not a distinct value.
    \end{minipage}}}%
  }%
}\indent
\hangindent=-\dimexpr5cm+12pt+6pt\relax\hangafter=0
\begingroup
\leftskip=0pt plus 0.5fil \rightskip=0pt plus -0.5fil
Consider now another point on our general curve, the point Q. This point 
is $\delta x$ away from $P$ horizontally and $\delta y$ away from $P$ vertically:\\
\begin{tikzpicture}[domain=0:4,scale=.95]
  \draw (0,-0.3) node[left]{$O$};
  \draw[thick, color=gray,->] (-4,0) -- (5,0) node[right] {\textcolor{black}{$x$}};
  \draw[thick, color=gray, ->] (0,-1) -- (0,5) node[above] {\textcolor{black}{$f(x)$}};
  \draw[color=blue] (1.15,2.15)--(3.3,4.7);
  \draw [color=red, semithick](-4,1).. controls(1,1.5).. (3.5,5);
  \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=below right:{$P(x,y)$}] (P) at (1.15,2.15) {};
  \node[circle,fill=black,inner sep=0pt,minimum size=3pt,label=right:{$Q(x+\delta x,y+\delta y)$}] (Q) at (3.3,4.7) {};
  \draw (3,4) node[right]{$y=f(x)$};
  \draw (-1,3/5+0.05)--(3,34/10+0.05);
  \draw (-0.3,0.3) node[left]{tangent};
  \draw[dotted] (3.3,4.7)--(3.3,4.3);
  \draw[dotted] (3.3,3.7)--(3.3,0);
  \draw[dotted] (1.12,2.15)--(1.12,0);
  \draw[dotted] (3.3,4.7)--(0,4.7);
  \draw[dotted] (1.12,2.15)--(0,2.15);
  \draw[<->,dashed](-0.3,2.15)--(-0.3,4.7);
  \draw(-0.6, 3.4) node{$\delta y$};
  \draw[<->,dashed](1.15,-0.3)--(3.3,-0.3);
  \draw(2.3, -0.6) node{$\delta x$};
\end{tikzpicture}
\par\endgroup

\noindent
The coordinates of $Q$ are $(x+\delta x,y+\delta y)$, shown above. We notice that if the 
values of $\delta x$ and $\delta y$ were to get smaller and smaller, the gradient of the 
chord $PQ$ (in blue) would approach that of the tangent at $P$, the gradient we wish to find 
($\frac{dy}{dx}$). We also know that the $y$-values of this graph are dependent on their 
respective $x$-value, since $y$ is defined as a function of $x$ ($y=f(x)$). So as $\delta x$ 
decreases, $\delta y$ will consequentially decrease because $(y+\delta y)$ is dependent on 
$(x+\delta x)$. So we can say that as {\boldmath$\delta x\longrightarrow 0$; 
$m_{PQ}\longrightarrow\frac{dy}{dx}$}.

\end{document}
  • @LukeCollins I was going to add a variant without tabular – egreg May 29 '15 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.