2

This relates to problem 5.7 from "Riley, Hobson and Bence - Mathematical Methods for Physics and Engineering".

I would like to have a 2D projected (looking down from z) version of the surface shown, keeping the color gradient. I would also like to add a family of parabolic curves (black) for different values of `a', as given in the problem. Thanks much.

\documentclass{memoir}

% __________ AMS ________________________
  \usepackage{amsmath}
% __________ PGF TikZ ___________________
  \usepackage{pgfplots}
% \usepackage{tikz}
  \usepackage{tikz-3dplot}

% __________ Differentials _____________________________________________________
% Single
    \newcommand{\diff}{d}           % If you want an upright `d', change it here
    \newcommand{\p}[1]{\partial#1}  
% ___________________ Derivatives ______________________________________________
% 1st derivative:
    \newcommand{\dod}[2]{\dfrac{\diff{#1}}{\diff{#2}}}  % 'differential over differential'
    \newcommand{\pop}[2]{\dfrac{\p#1}{\p#2}}            % 'partial over partial'
    \newcommand{\lpop}[2]{\p#1/\p#2}            % A 'layed down' version

\setlength{\parindent}{0cm}
\begin{document}

\textbf{Problem 5.7}\par
\centering
\textbf{The Chain Rule and Stationary Points}\\[3mm]
\flushleft
  \fbox{\parbox{4.25in}{
    The function $G(t)$ is defined by
    $$ G(t) = F(x,y) = x^2 + y^2 + 3xy $$
    where $x(t) = at^2$ and $y(t) = 2at$. Use the chain rule to find the values of $(x,y)$ at which $G(t)$ has stationary values as a function of $t$. Do any of them correspond to the stationary points of $F(x,y)$ as a function of $x$ and $y$?
    }}
\flushleft

  \vspace{3mm}
  \noindent\emph{Solution:}\\[1mm]
  (To be terse, the derivation has been omitted.)\\[2mm]
  The Stationary points occur at $\lpop Ft = 0$, in which case
  \[ 2a^2t(2t + 1)(t + 4) = 0 \]
  So,
  \[ t\in \{ -4,\ -1/2,\ 0 \} \]
  This corresponds to the stationary points
  \[ (16a,\ -8a)\ ,\quad (a/4,\ -a)\ ,\quad (0,\ 0) \]
  To answer the second part of the question, we differentiate $F(x,y)$ with respect to $x$, and $y$, and set these to zero;
  \[ \pop Fx = 2x + 3y = 0 \]
  \[ \pop Fy = 3x + 2y = 0 \]
  The only solution is $(0,\ 0)$.

\centering
\begin{figure}
  \tdplotsetmaincoords{0}{0}
  \begin{tikzpicture}[tdplot_main_coords,scale=1.5,rotate=0]
    \begin{axis}[domain=-6:6,y domain=-6:6]
      \addplot3[surf] {x^2 + y^2 + 3*x*y};
    \end{axis}
  \end{tikzpicture}
  \caption{$z = x^2 + y^2 + 3xy$}
\end{figure}
\flushleft

The other two solutions for $t$, are the stationary points for the parabolic sheet (not shown) $(at^2, 2at, t)$ that intersects with the surface shown. There is also the two-dimensional version realized by looking down from $z$ (i.e. the projection onto the $z$-plane). The blue to orange colors could then be interpreted, for example, as a scalar field for temperature.

\end{document}
2
  • 4
    Hi, welcome. Do you mean the output you get if you add view={0}{90} to the axis options? Jan 18 '19 at 21:11
  • 2
    Off-topic: $$...$$ is deprecated LaTeX syntax for almost 25 years by now. Use \[...\] or an environment dedicated to math display
    – user31729
    Jan 18 '19 at 21:32
1

For the question in the title, you can add view={0}{90} to the axis options.

To plot f(x) = 2*sqrt(a*x), do something like \addplot [black] {2*sqrt(2*x)};, if a=2. You'll need a second plot for the negative part of the parabola.

In the example below I used a loop like this:

  \pgfplotsinvokeforeach{0.25,1}{% values of a, represented by #1 in the loop code
      \addplot [black, samples at={0,0.01,...,1,1.6}] {2*sqrt(#1*x)} node[above left] {$a=#1$};
      \addplot [black, samples at={0,0.01,...,6}] {-2*sqrt(#1*x)};
      }

The samples at stuff is to make the plot smoother.

Some other changes:

  • As Christian mentioned in a comment, don't use $$ ... $$ for display math, better stick to \[ ... \]. (See Why is \[ … \] preferable to $$?.)
  • I added \nonzeroparskip to the preamble, removed all \\[<some length>], and added some paragraph breaks (empty lines).
  • Instead of the various \centering ... \flushleft, I added the title inside a center environment, and moved the one \centering to inside the figure environment.
  • I removed the tikz-3dplot stuff, you're not using it.
  • I removed the scale=1.5 and added a width=... in the axis options.

enter image description here

\documentclass{memoir}

% __________ AMS ________________________
  \usepackage{amsmath}
% __________ PGF TikZ ___________________
  \usepackage{pgfplots}

% __________ Differentials _____________________________________________________
% Single
    \newcommand{\diff}{d}           % If you want an upright `d', change it here
    \newcommand{\p}[1]{\partial#1}  
% ___________________ Derivatives ______________________________________________
% 1st derivative:
    \newcommand{\dod}[2]{\dfrac{\diff{#1}}{\diff{#2}}}  % 'differential over differential'
    \newcommand{\pop}[2]{\dfrac{\p#1}{\p#2}}            % 'partial over partial'
    \newcommand{\lpop}[2]{\p#1/\p#2}            % A 'layed down' version

\setlength{\parindent}{0pt}
\nonzeroparskip % <--- added
\begin{document}

\textbf{Problem 5.7}
\begin{center}
\textbf{The Chain Rule and Stationary Points}
\end{center}
  \fbox{\parbox{4.25in}{
    The function $G(t)$ is defined by
    \[ G(t) = F(x,y) = x^2 + y^2 + 3xy \]
    where $x(t) = at^2$ and $y(t) = 2at$. Use the chain rule to find the values of $(x,y)$ at which $G(t)$ has stationary values as a function of $t$. Do any of them correspond to the stationary points of $F(x,y)$ as a function of $x$ and $y$?
    }}

  \medskip
  \emph{Solution:}

  (To be terse, the derivation has been omitted.)

  The Stationary points occur at $\lpop Ft = 0$, in which case
  \[ 2a^2t(2t + 1)(t + 4) = 0 \]
  So,
  \[ t\in \{ -4,\ -1/2,\ 0 \} \]
  This corresponds to the stationary points
  \[ (16a,\ -8a)\ ,\quad (a/4,\ -a)\ ,\quad (0,\ 0) \]
  To answer the second part of the question, we differentiate $F(x,y)$ with respect to $x$, and $y$, and set these to zero;
  \begin{align*}
   \pop Fx &= 2x + 3y = 0 \\
   \pop Fy &= 3x + 2y = 0
  \end{align*}
  The only solution is $(0,\ 0)$.

\begin{figure}
  \centering
  \begin{tikzpicture}
    \begin{axis}[
        width=0.8\linewidth, % <-- instead of scale
        domain=-6:6,
        y domain=-6:6,
        view={0}{90}
        ]
      \addplot3[surf] {x^2 + y^2 + 3*x*y};

      \pgfplotsinvokeforeach{0.25,1}{% values of a, represented by #1 in the loop code
          \addplot [black, smooth, samples at={0,0.01,...,6}] {2*sqrt(#1*x)} node[above left] {$a=#1$};
          \addplot [black, smooth, samples at={0,0.01,...,6}] {-2*sqrt(#1*x)};
          }

    \end{axis}
  \end{tikzpicture}
  \caption{$z = x^2 + y^2 + 3xy$}
\end{figure}

The other two solutions for $t$, are the stationary points for the parabolic sheet (not shown) $(at^2, 2at, t)$ that intersects with the surface shown. There is also the two-dimensional version realized by looking down from $z$ (i.e. the projection onto the $z$-plane). The blue to orange colors could then be interpreted, for example, as a scalar field for temperature.

\end{document}
2
  • The curves that I would like to add are of the form x = y^2/(4a), so the curve x = y^2/2, and another x = y^2/4 would be nice. If it's better to have y on the left, then y = \sqrt{x}, and y = 2\sqt{x}, would be ok. Thanks. Jan 19 '19 at 21:51
  • @ScotParker How's that? If you don't want the a=.. in the plot, remove node[above left] {$a=#1$} in the loop. Jan 19 '19 at 22:12

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