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a) The portion of the circle $x^2 + y^2 = 4$ traversed clockwise from $(-2,0)$ to $(0,2)$

b) The part of the ellipse $(x^2)/(4) + (y^2)/(9) = 1$ that lies above the line $y = 0$, traversed clockwise.

How do you do them... The back of the book says the answer are

a) $x(t) = -2\cos(t)$, $y(t) = 2\sin(t)$, $0 \le t \le \pi/2$
b) $x(t) = -2\cos(t)$, $y(t) = 3\sin(t)$, $0 \le t \le \pi$

a) $x(t) = 2\cos(t)$, $y(t) = -2\sin(t)$, $-\pi \le t \le -3\pi/2$
b) $x(t) = 2\cos(t)$, $y(t) = -3\sin(t)$, $-\pi \le t \le -2\pi$

WHAT AM I DOING WRONG? or what's the correct way to solve these problems.

Sorry about formatting, it wouldn't let me do it? Just kept giving me errors.

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In your answers, check carefully whether either of the inequalities systems $$-\pi\le t\le -\frac{3\pi}{2}\;\;,\;\;-\pi\le t\le -2\pi$$ make sense... – DonAntonio Sep 8 '13 at 11:33
You're right, they don't make sense. I'm not sure how to get what they got though. @donantonio – braziil Sep 8 '13 at 11:36
@lorde I think you don't quite understand what DonAntonio meant. Please think more carefully. – Tunococ Sep 8 '13 at 11:40
@tunococ I have no idea to be honest. Can you hint me, I'm having a brain dead moment – braziil Sep 8 '13 at 11:45
@lorde $-\pi$ is greater than $-\frac{3\pi}2$. – Tunococ Sep 8 '13 at 11:46
up vote 2 down vote accepted

I propose to attack the problem the following way: we know the usual, standard parametrization of the first circle is

$$(2\cos t\;,\;2\sin t)$$

Yet this usual parametrization assumes we "travel" on the circle in the positive direction, meaning anticlockwise. We in fact want to walk the upper semicircle but the negative direction: clockwise, so when we choose the parameter we must make sure we begin at the left hand of the semicircle's main diameter (the horizontal one), i.e. the point $\;(-2,0)\;$ and we go around until $\,(2,0)\;$ , so we need the corresponding coordinates with their corresponding signs! Two basic (imo) ways to achieve this:

$$\begin{align*}(1)&\;\;(-2\cos t\,,\,2\sin t)\;,\;\;&0\le t\le\pi\\{}\\ (2)&\;\;(-2\cos\pi t\,,\,2\sin\pi t)\;,\;\;&0\le t\le 1 \end{align*}$$

You can also try weirder stuff:

$$(2\cos(\pi-t),2\sin(\pi-t))\;,\;\;0\le t\le\pi$$

Something similar goes with the ellipse

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AH ok, So I look at where the starting point is first. Given where that point is, it dictates whether it's -cos, cos, -sin or sin. If it's (-2,0) in this case, it's a flip in the y axis, so I make the x(t) = -2cos(t) and then treat the negative x axis as the new place to START the angle? @donantonio – braziil Sep 8 '13 at 11:57
Well yes, but it is not only a matter of paying attention where is "the first" point, but also and perhaps mainly to observe in what direction we want to move on. – DonAntonio Sep 8 '13 at 11:59
Yeah, sorry, I just realised that lol. Okay, so if it mentions clockwise, make the x(t) negative and start the angle from the negative axis. If it's anticlockwise, use the positive x axis – braziil Sep 8 '13 at 12:03
Now that looks fine, @lorde. Observe that the sign that "matters" more is the cosine's since from the sine's perspective we remain over the $\,y-$axis all the time, so it makes no much difference. – DonAntonio Sep 8 '13 at 12:05

Maybe this diagram will help for the parametric circle: enter image description here

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