0
$\begingroup$

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

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

  • 0
    *Hint*: try to compute $y^2$.2012-12-02

3 Answers 3

2

\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}

  • 0
    Be careful: $\sqrt{1-x}=\sqrt{\sin^2(t)}=|\sin(t)|$.2012-12-02
2

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

  • 1
    GeoGebra www.geogebra.org2012-12-02
1

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.