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A curve has parametric equations $$\begin{array}{rcl} x & = & \cos(t)\\\ y & = & \frac{1}{2}\sin(2t) \end{array}\qquad \text{ for }0\leq t \leq 2\pi.$$ Show that the Cartesian equation of the curve is $y^2=(x^2)(1-x^2).$

What are the steps for showing this?

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2 Answers 2

up vote 2 down vote accepted

First, evaluate $x^2(1-x^2)$ with $x=\cos(t)$ and verify that you get $y^2$. This will prove that every point in the parametric curve lies on the curve $y^2 = x^2(1-x^2)$.

But this is not enough; this only shows that the parametric curve is contained inside the curve $y^2 = x^2(1-x^2)$. To show that the paramatric curve is the curve $x^2(1-x^2)$, you also need to show that every point on the latter curve is "covered" by the parametric curve.

So show that every point on that curve can be obtained by picking an appropriate value of $t$ and plugging into the parametric equations. (For example, notice that if $(a,b)$ lies in the curve $y^2 = x^2(1-x^2)$, then you must have $-1\leq a\leq 1$ (otherwise, $1-x^2\lt 0$, and $y^2\lt 0$, which is impossible; that means that there exist a $t$ on $[0,2\pi]$ (in fact, two of them usually) for which $\cos(t) = a$. Then check to see if one of those values of $t$ "works" for $y$).

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Hint:

Use the trigonometric identities:

$$\sin (2x) = 2 \sin x \cos x$$

$$ \sin^2 x + \cos^2 x = 1$$

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Thanks very much –  user7761 Mar 3 '11 at 22:48

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