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Simple question: I've been asked find a parameterization of the circle of radius $2$ starting at $(2,0)$, moving in the counterclockwise direction.

Simple enough I get $(2\cos(t),2\sin(t))$ because $x=r\cdot \cos \theta$ and $y=r\cdot \sin \theta$.

Now the second part of the question asks the same but in the clockwise direction and the answer provided is $(2\sin(t),2\cos(t))$. I don't understand how this is possible.

If we take $t$ as $-t$ because of the opposite direction it still only makes sense to say the parameterized equation is $(2\cos(t),-2\sin(t))$. And I don't see how for the $x$ - coordinate we can have

$$ x= r\cdot \sin (t)$$ and similarly for the $y$ - coordinate
$$ y= r\cdot \cos (t)$$

We never covered parametric equations in Calc II and because I'm in a different school, it's assumed(rightly) that we know parametric equations.

Thanks in advance.

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    $\begingroup$ Shift $t$ by a quarter turn. $\endgroup$ Mar 10 '16 at 23:00
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Note that, in their answer, the circle doesn't start at $(2,0)$ but at $(0,2)$. So it's not just a change of direction but also of starting point.

But this actually gives the resulting answer an easy interpretation: I take the motion of my point in the first case (i.e. starting at $(2,0)$ and going counter-clockwise) and redraw it with the axes swapped. This reverses the motion of the point and changes the starting point from $(0,2)$ to $(2,0)$. More to the point, the parametrization is now $(2\sin t,2\cos t)$ instead of $(2\cos t,2\sin t)$, exactly as the problem suggests.

This also gives a simple way to see how your answer results: If you take your initial circle and flip it along the $x$-axis, then the direction reverses but the initial point stays the same. This amounts to changing the parametrization from $(2\cos t,2\sin t)$ to your result of $(2\cos t,-2\sin t)$.

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