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I'm learning parametric equations in this section. Although I understand why the following works, I'm having difficulty understanding why the method employed for solving it is the correct one.

I'm instructed to derive a Cartesian equation for the parametric equations given and describe the path the particle traverses. Describing the path amounts to simply graphing the path, i.e. the Cartesian function or parametric equation, and superimpose arrows indicating direction.

So, without further ado, the parametric equations give are:

$$ \begin{array}{rcl} x & = & \cos2t \\ y & = & \sin2t \end{array} $$ Over the interval: $0 \le t \le \pi$

The solution involves observing that $x^2 + y^2 = 1 => \cos^2(2t) + \sin^2(2t) = 1$

Squaring these functions to make them produce something convenient seems like "fixing" the problem to do what I'd like it to do. Why is this a valid approach? I hope I'm asking my question well. Squaring both of these functions seems like turning the problem into something it isn't, but yet, it's the right thing to do. Why?

Thanks, Andy

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In general, we want to use the two equations to "eliminate" $t$. That cannot always be done explicitly, and the strategy is quite equations-dependent. –  André Nicolas Mar 16 '13 at 18:54
    
You are not "fixing" anything "to do what you want" (??) but realizing some relation exists between the parametric functions. The trigonometric Pythagoras Theorem is hardly something out of the blue in this particular case, and in other cases you will have to come up with soem other ideas. –  DonAntonio Mar 16 '13 at 18:54
    
I apologize for the lateness of approving an answer or commenting. I didn't see the application of the Pythagorean Theorem as something out of the blue. Rather, it is simply that $x=cos2t$ and $y=sin2t$ weren't squared. By squaring them, I was considering that as "changing" what was there to something that wasn't. However, I think I see that it's because the parametric equations aren't a single Cartesian relationship but two. I'm trying to make a single relationship from them. Am I getting warm? –  Andrew Falanga Mar 25 '13 at 21:01
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It is the right approach.

Take a candidate cartesian equation, replace x and y withe their expression x(t), y(t) and massage the formula until the parameter t disappears.

Then you are sure that every point of your parametric curve is also on the curve with the candidate cartesian equation.

You may have points of the parametric curve which are not on the cartesian curve. Like the equations $(x,y) = (e^t, e^{-t})$ which gives only one branch of the hyperbola $xy=1$, the one for which $x,y>0$.

A part of that, finding the cartesian equation is some kind of a guess work.

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