Help me to sketch this parametric curves

Question

Is there any defined process to sketch parametric curves? Thanks in advance.

$$x = \cos^2 t, \quad y = 1 - \sin t, \quad 0 \leq t \leq 2\pi.$$

score 2 · Accepted Answer · answered Dec 02 '12 at 00:14

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\begin{align} x=\cos^2(t)\\1-x=\sin^2(t)\\\sqrt{1-x}=\sin(t)\\1-\sqrt{1-x}=1-\sin(t)=y\\ y=1-\sqrt{1-x}\\ (1-x)=(1-y)^2 \end{align}

answered Dec 02 '12 at 00:14

Inquest

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is there any defined method to do that? – Diego Pacheco Dec 02 '12 at 00:25
Be careful: $\sqrt{1-x}=\sqrt{\sin^2(t)}=|\sin(t)|$. – Sigur Dec 02 '12 at 00:25

score 2 · Answer 2 · answered Dec 02 '12 at 00:33

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Here is the animated curve. Note that it is periodic. gif

answered Dec 02 '12 at 00:33

Sigur

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I understand Sigur, very illustrative. What software did you use? – Diego Pacheco Dec 02 '12 at 00:43
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GeoGebra www.geogebra.org – Sigur Dec 02 '12 at 00:45

score 1 · Answer 3 · answered Dec 02 '12 at 00:16

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You can use the identity $\cos^2t + \sin^2t = 1$. Then $\sin^2t = 1 - \cos^2t$, so $\sin t = \pm \sqrt{1 - x}$. Then you can plug this into the equation for $y$ and sketch the curve.

answered Dec 02 '12 at 00:16

Noah Halford

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Help me to sketch this parametric curves

3 Answers3