Derive Cartesian cubic Möbius strip from parametric

Question

The following link: http://mathworld.wolfram.com/MoebiusStrip.html shows the Möbius strip parametrized as \begin{eqnarray} x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\ y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\ z = s \sin \left ( \frac12 t \right ) \end{eqnarray} The symbols for $R$ and $s$ and angle $t$ are explained there.

Then they say that from this parametrization we can derive the cubic.

\begin{equation} -R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z + y z^2 = 0. \end{equation}

Any ideas about how to do this? I have tried with no success.

Thanks.

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This is what I have done so far:

Square the first two equations above and add to find \begin{equation} x^2 + y^2 = \left ( R + s \left ( \cos \frac{t}{2} \right ) \right )^2 \end{equation} Take the square root of this \begin{equation} \sqrt{x^2 + y^2 }= R + s \left ( \cos \frac{t}{2} \right ) \Longrightarrow \sqrt{x^2 + y^2 } - R = s \left ( \cos \frac{t}{2} \right ) \end{equation} Square this and the third equation (for $z$) and add to find \begin{equation} s^2 = \left ( \sqrt{x^2+y^2}- R \right )^2 + z^2 \end{equation}

Now, let us divide the second by the first equation

That is \begin{equation} \frac{y}{x} = \tan t = \frac{2 \tan (t/2)}{1 - \tan^2 (t/2)} \end{equation} multiply numerator and denominator by $\cos^2 (t/2)$ \begin{equation} \frac{y}{x} = \frac{2 \sin(t/2) (\sqrt{1-\sin^2(t/2)} }{\cos^2 (t/2) - \sin^2(t/2)} = \frac{2 \sin(t/2) \sqrt{1 - \sin^2(t/2)}}{1 - 2 \sin^2(t/2)}. \end{equation} Multiply numerator and denominator by $s^2$, then \begin{equation} \frac{y}{x} = \frac{2 z \sqrt{s^2 - z^2}}{s^2 - 2 z^2} \end{equation}

That is \begin{equation} \frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ ( \sqrt{x^2 + y^2} - R)^2 - z^2} \end{equation} or \begin{equation} \frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ (x^2 + y^2) + R^2 - 2 R \sqrt{x^2+y^2} - z^2} \end{equation}

Or \begin{equation} y (x^2 + y^2) + y R^2 - 2 R y \sqrt{x^2+y^2} - z^2 y = 2 z x \sqrt{x^2 + y^2} -2 R x z \end{equation}

Answer 1

I'd just substitute the given coordinates into the equation and show that it's satisfied. The terms can be grouped according to the powers of $R$ and $s$ they contain, with the two exponents always summing to $3$, and the equation has to be satisfied for all four groups separately; that makes the calculation more manageable. For example, for $R^3s^0$ I get

$$-\sin t+\cos^2t\sin t+\sin^3t=0\;,$$

which is indeed satisfied.

Answer 2

Consider the Moebius strip with midline the circle of radius $R$ in the $(x,y)$-plane and having width $2a>0$: $$M:\quad(\phi,s)\mapsto\left\{\eqalign{x&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi \cr y&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi \cr z&=s\sin{\textstyle{\phi\over2}}\ .\cr}\right.\tag{1}$$ The parameter domain is $-\pi<\phi<\pi$, $\ -a<s<a$. For "technical reasons" we have excluded $\phi=\pm\pi$. This guarantees $\cos{\phi\over2}>0$ over the whole parameter domain.

It is claimed that $M$ is part of a certain cubic surface $S\subset{\mathbb R}^3$. This surface results from revoking the condition $-a<s<a$ in $(1)$, so that now the parameter $s$ runs from $-\infty$ to $\infty$. Consider a fixed value $\phi\in\ ]{-\pi},\pi[\ $. Then $$g_\phi:\quad s\mapsto\left(\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi, \ \bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi , \ s\sin{\textstyle{\phi\over2}}\right)\tag{2}$$ describes a straight line lying on $S$. We now replace the parameter $s$ in $(2)$ by the new parameter $u:=R+s\cos{\phi\over2}$. In this way $g_\phi$ appears in the form $$g_\phi:\quad u\mapsto\bigl(u\cos\phi, u\sin\phi, \tan{\textstyle{\phi\over2}}(u-R)\bigr)\qquad(-\infty<u<\infty)\ .\tag{3}$$ When we let $\phi$ vary as well in $(3)$ we obtain another parametrization of our surface $S$. In order to get rid of the trigonometric functions we introduce the new parameter $t:=\tan{\phi\over2}$, which runs from $-\infty$ to $\infty$. In this way we obtain the following rational representation of $S$: $$S:\quad(t,u)\mapsto\left\{\eqalign{x&=u\ {1-t^2\over 1+t^2} \cr y&=u\ {2t\over1+t^2}\cr z&=t\ (u-R)\cr}\right.\qquad\qquad\bigl((t,u)\in{\mathbb R}^2\bigr)\ .$$ One has $2(z+Rt)=2ut=y(1+t^2)$ and $$2tx=y(1-t^2)\ .\tag{4}$$ Adding these two equations leads to $$t={y-z\over R+x}\ .$$ We insert this value of $t$ into $(4)$ and obtain after clearing denominators $$2x(R+x)(y-z)=y\bigl((R+x)^2-(y-z)^2\bigr)\ .$$ This equation is valid for all points $(x,y,z)\in S$, and as well on the line $g_{\pm \pi}$ omitted from consideration. It expands to $$-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z - 2 y^2 z + y z^2=0\ ,$$ as given in the quoted Wikipedia link (in your question you have forgotten one term).