| Version current |
Version 3 |
| We will show that the {\em Lam\'e's cubic} |
We will show that the {\em Lam\'e's cubic} |
| \begin{align} |
\begin{align} |
| x^3+y^3 = a^3, |
x^3+y^3 = a^3, |
| \end{align} |
\end{align} |
| where $a$ is a positive \PMlinkescapetext{constant}, has the line |
where $a$ is a positive \PMlinkescapetext{constant}, has the line |
| $$y = \underbrace{-x}_{g(x)}$$ |
$$y = \underbrace{-x}_{g(x)}$$ |
| as its asymptote.\\ |
as its asymptote.\\ |
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| 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 |
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} |
\begin{align} |
| y = \underbrace{\sqrt[3]{a^3-x^3}}_{f(x)} |
y = \underbrace{\sqrt[3]{a^3-x^3}}_{f(x)} |
| \end{align} |
\end{align} |
| of (1) we see that every real value of $x$ gives one point of the curve. |
of (1) we see that every real value of $x$ gives one point of the curve. |
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| \begin{center} |
\begin{center} |
| \begin{pspicture}(-4,-4)(4,4) |
\begin{pspicture}(-4,-4)(4,4) |
| \psaxes[Dx=9,Dy=9]{->}(0,0)(-3.5,-3.5)(3.5,3.5) |
\psaxes[Dx=9,Dy=9]{->}(0,0)(-3.5,-3.5)(3.5,3.5) |
| \rput(3.6,-0.2){$x$} |
\rput(3.6,-0.2){$x$} |
| \rput(0.2,3.5){$y$} |
\rput(0.2,3.5){$y$} |
| \psdot[linecolor=blue](1,0) |
\psdot[linecolor=blue](1,0) |
| \psdot[linecolor=blue](0,1) |
\psdot[linecolor=blue](0,1) |
| \rput(1.13,-0.17){$a$} |
\rput(1.13,-0.17){$a$} |
| \rput(-0.17,1.13){$a$} |
\rput(-0.17,1.13){$a$} |
| \psline[linecolor=cyan](-3.5,3.5)(3.5,-3.5) |
\psline[linecolor=cyan](-3.5,3.5)(3.5,-3.5) |
| \psline[linestyle=dashed](-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]{-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} |
\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)} |
\rput(0.5,-4){Lam\'e's cubic\, $y = \sqrt[3]{a^3-x^3}$\, (blue)} |
| \end{pspicture} |
\end{pspicture} |
| \end{center} |
\end{center} |
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| 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 |
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 |
| $$(u+v)(u^2-uv+v^2) = u^3+v^3,$$ |
$$(u+v)(u^2-uv+v^2) = u^3+v^3,$$ |
| getting |
getting |
| \begin{align*} |
\begin{align*} |
| \Delta &= f(x)\!-\!g(x)\\ |
\Delta &= f(x)\!-\!g(x)\\ |
| &= \frac{\sqrt[3]{a^3-x^3}+x}{1}\\ |
&= \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{(\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}. |
&= \frac{a^3}{(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2}. |
| \end{align*} |
\end{align*} |
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