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Viewing Version
3
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'asymptote of Lam\'e's cubic'
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| Title of object: |
asymptote of Lam\'e's cubic |
| Canonical Name: |
AsymptoteOfLamesCubic |
| Type: |
Example |
| Created on: |
2008-03-14 19:17:23 |
| Modified on: |
2008-03-15 09:45:45 |
| Classification: |
msc:26C05, msc:53A04 |
| Defines: |
Lam\'e's cubic |
Revision comment (for changes between this and next version):
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
\usepackage{pstricks}
\usepackage{pst-plot}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
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Content:
We will show that the {\em Lam\'e's cubic}
\begin{align}
x^3+y^3 = a^3,
\end{align}
where $a$ is a positive \PMlinkescapetext{constant}, has the line
$$y = \underbrace{-x}_{g(x)}$$
as its asymptote.\\
Because the equation (1) of the curve is symmetric with respect to $x$ and $y$, the curve is symmetric about the line \,$y = x$.\, From the solved form
\begin{align}
y = \underbrace{\sqrt[3]{a^3-x^3}}_{f(x)}
\end{align}
of (1) we see that every real value of $x$ gives one point of the curve.
\begin{center}
\begin{pspicture}(-4,-4)(4,4)
\psaxes[Dx=9,Dy=9]{->}(0,0)(-3.5,-3.5)(3.5,3.5)
\rput(3.6,-0.2){$x$}
\rput(0.2,3.5){$y$}
\psdot[linecolor=blue](1,0)
\psdot[linecolor=blue](0,1)
\rput(1.13,-0.17){$a$}
\rput(-0.17,1.13){$a$}
\psline[linecolor=cyan](-3.5,3.5)(3.5,-3.5)
\psline[linestyle=dashed](-3.5,-3.5)(3.5,3.5)
\psplot[linecolor=blue]{-3.5}{1}{1 x 3 exp sub 1 3 div exp}
\psplot[linecolor=blue]{1}{3.5}{0 x 3 exp 1 sub 1 3 div exp sub}
\rput(0.5,-4){Lam\'e's cubic\, $y = \sqrt[3]{a^3-x^3}$\, (blue)}
\end{pspicture}
\end{center}
The difference \,$\Delta = f(x)\!-\!g(x)$\, \PMlinkescapetext{represents} the distance of a point \,$(x,\,y)$\, of the curve and the point of the asserted asymptote\, $y = -x$\, with the same abscissa $x$.\, We multiply the numerator and denominator with the expression \,$(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2$ for being able to utilise the polynomial formula
$$(u+v)(u^2-uv+v^2) = u^3+v^3,$$
getting
\begin{align*}
\Delta &= f(x)\!-\!g(x)\\
&= \frac{\sqrt[3]{a^3-x^3}+x}{1}\\
&= \frac{(\sqrt[3]{a^3-x^3})^3+x^3}{(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2}\\
&= \frac{a^3}{(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2}.
\end{align*}
Thus,\, $\displaystyle \Delta \to \frac{a^3}{\infty+\infty+\infty} = 0$\; when\; $ |x| \to \infty$\; (see the improper limits).\, According to the definition of \PMlinkname{asymptote}{Asymptote}, the line\, $y = -x$\, is asymptote of Lam\'e's cubic.
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