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

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

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Hint: try to compute $y^2$. –  Sigur Dec 2 '12 at 0:15

3 Answers 3

up vote 2 down vote accepted

\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}

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is there any defined method to do that? –  Rakisbro Dec 2 '12 at 0:25
    
Be careful: $\sqrt{1-x}=\sqrt{\sin^2(t)}=|\sin(t)|$. –  Sigur Dec 2 '12 at 0:25

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

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I understand Sigur, very illustrative. What software did you use? –  Rakisbro Dec 2 '12 at 0:43
1  
GeoGebra www.geogebra.org –  Sigur Dec 2 '12 at 0:45

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.

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