The graph of curve $|x|+|y|=1 $ is a rhombus, How can I obtain a parameterization in a counterclockwise sense in such a way that it can be expressed as the curve $\alpha:I\subset R\to R^2 $,$\alpha (t)=(x(t),y(t))$ ?
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3You'll need to parametrize the four line segments individually. – Ted Shifrin Apr 08 '17 at 16:34
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Just use a piece-wise function. You can't really do it any other way. – Isaac Browne Apr 08 '17 at 16:46
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Why do you call "rhombus" a square ? – Jean Marie Apr 08 '17 at 23:12
2 Answers
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If one uses the unit step function
\begin{equation} u(t)=\begin{cases}0\text{ for }t<0\\1\text{ for }t\ge0\end{cases} \end{equation}
then
\begin{aligned} x(t) &= 1-t+2(t-2)u(t-2) \\ y(t) &= t+2(1-t)u(t-1)+2(t-3)u(t-3) \end{aligned} for $0 \leq t \leq 4$.
The only advantage with this version is constant speed.
Desmos animation of constant speed rhombus
John Wayland Bales
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Ted's suggestion is probably best for computation, but $$ \left. \begin{aligned} x(t) &= \cos t |\cos t| \\ y(t) &= \sin t |\sin t| \end{aligned} \right\} \qquad 0 \leq t \leq 2\pi, $$ does the job.
Andrew D. Hwang
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1@Ted: On the plus side, one deduces a real-analytic parametrization, for each odd positive integer $p$, of $|x|^{2/p} + |y|^{2/p} = 1$. :) – Andrew D. Hwang Apr 08 '17 at 17:22
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