11

How can continue the equation of the rectangle red in next line?

\documentclass[11pt,a4paper,twocolumn]{article}

...

\begin{equation*}
    \left| \dot{r} \right| =
    \sqrt{{- \, {e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, {{e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, {(- \, {e}^{-t})}^{2}}
\end{equation*}

enter image description here

Bernard
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LuisPac
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4 Answers4

7

I suggest using the multlined environment from mathtools to break the radicand . You also can choose not to break it, but to type in medsize (~80 % of \displaystyle) with the \mediummath command from nccmath. I took the opportunity to clean your code:

\documentclass[11pt,a4paper,twocolumn]{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools, nccmath}

\begin{document}

\begin{equation*}
    \bigl| \dot{r} \bigr| =
    \sqrt{\begin{multlined}[b] -\!\bigl(e^{-t} ( \cos t + \sin t )\bigr)^2 + \\ \bigl(e^{-t} (\cos t + \sin t)\bigr)^2 + \bigl(-e^{-t}\bigr)^2 \end{multlined}}
\end{equation*}
\begin{align*}
    \bigl| \dot{r} \bigr| & =
    \sqrt{\medmath{ -\!\bigl(e^{-t} ( \cos t + \sin t )\bigr)^2 + \bigl(e^{-t} (\cos t + \sin t)\bigr)^2 + \bigl(-e^{-t}\bigr)^2}} \\
         & = \sqrt{ -\!\bigl(e^{-t} ( \cos t + \sin t )\bigr)^2 + \bigl(e^{-t} (\cos t + \sin t)\bigr)^2 + \bigl(-e^{-t}\bigr)^2}
\end{align*}

\end{document}

enter image description here

Bernard
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    Exactly what I was suggesting (about the first part of your answer)! – GuM Apr 10 '17 at 18:39
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    Great minds think together! – Bernard Apr 10 '17 at 18:44
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    Are you sure the entire term e^{-t}(\cos t + \sin t) should be squared? The OP's code seems to suggest that only (\cos t + \sin t) should be squared. (Of course, the OP's code might be in error...) – Mico Apr 10 '17 at 18:52
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    @Mico: Not sure at all. That's the way I interpreted all those spurious { … }. Anyway, it's only a demo. – Bernard Apr 10 '17 at 19:13
6

As the equation* environment doesn't permit line breaks, consider using a multline* environment. And, instead of creating a multi-line surd expression with a \sqrt instruction, I suggest you use (...)^{1/2} notation.

Two additional comments about your code: Do try to use \left and \right less frequently, and don't needlessly encase various terms in curly braces.

enter image description here

\documentclass[11pt,a4paper,twocolumn]{article}
\usepackage{amsmath} % for 'multline*' env.
\begin{document}
...
\hrule % just to illustrate column width
\begin{multline*}
    | \dot{r} | = \bigl\{
    - e^{-t} ( \cos t + \sin t )^{2} \\
    + e^{-t} ( \cos t + \sin t )^{2}
    + (-e^{-t})^{2} \bigr\}^{1/2}
\end{multline*}
\end{document}
Mico
  • 506,678
4

If you are interested in a square root expression and not a power notation that Mico's answer provides, have a look at the following code:

\begin{equation*}
    \left| \dot{r} \right| =
    \sqrt{
        \begin{aligned}
        &{- \, {e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, \\ 
        &{{e}^{-t}  \left( \cos t + \sin t \right)}^{2} \, + \, {(- \, {e}^{-t})}^{2}
        \end{aligned}   
        }
\end{equation*}

Ouptut

enter image description here

gnikit
  • 459
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    Perhaps you could enhance readability by adding a \quad at the beginning of the second line. Edit: Also note that the mathtools package provides a multlined environment (maybe you already knew… :-) – GuM Apr 10 '17 at 18:23
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    @GustavoMezzetti - I was not aware of that option since I do not usually encounter problems like this. It was a very helpful comment indeed – gnikit Apr 10 '17 at 19:19
  • Thanks for all your answers, apparently the root squared be cant to break. – LuisPac Apr 11 '17 at 01:51
  • @nikjohn Off-topic, but seems that the surd is not perfectly connected to the line in the "Ouptut." Is there a way to avoid this? – L. F. Mar 04 '19 at 10:39
0

I faced this situation today and the simplest solution is to use {align*} within the equation environment and double back slash \\ command at the location you want to start the next line. Although not automatic, it looks neat.

\begin{equation}
\begin{align*}
            \Pi( \phi,\eta)= \int_0^L \left[ N \cdot \Lambda^T \left( \frac{d \eta_0}{ds} - \theta \times \frac{d \phi_0}{ds} \right)+M \cdot \Lambda^T \frac{d \theta}{ds} \right]ds \\
        -\int_0^L \left( \bar{n} \cdot \eta_0 + \bar{m} \cdot \theta \right) ds = 0
\end{align*}
\end{equation}
Miyase
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