这与“Riley、Hobson 和 Bence - 物理和工程的数学方法”中的 5.7 问题有关。
我想要显示一个 2D 投影(从 z 向下看)版本的表面,并保持颜色渐变。我还想为不同的“a”值添加一组抛物线(黑色),如问题中所示。非常感谢。
\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}
答案1
对于标题中的问题,大家可以补充view={0}{90}
一下axis
选项。
要绘制 f(x) = 2*sqrt(a*x),请执行类似 的操作\addplot [black] {2*sqrt(2*x)};
,如果 a=2。您需要为抛物线的负部分绘制第二个图。
在下面的例子中我使用了这样的循环:
\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)};
}
这些samples at
东西是为了让剧情更加流畅。
其他一些变化:
- 正如 Christian 在评论中提到的那样,不要使用
$$ ... $$
它来显示数学,最好坚持使用\[ ... \]
。(参见为什么\[ … \]
优于$$
?。 - 我添加了
\nonzeroparskip
序言,删除了所有内容\\[<some length>]
,并添加了一些段落分隔符(空行)。 - 我没有使用各种
\centering ... \flushleft
,而是在环境中添加了标题center
,然后将其移动\centering
到里面环境figure
。 - 我把这些
tikz-3dplot
东西删掉了,你不用它。 - 我删除了并在选项中
scale=1.5
添加了。width=...
axis
\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}