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
BOOK ANSWERS
a) $x(t) = -2\cos(t)$, $y(t) = 2\sin(t)$, $0 \le t \le \pi/2$
b) $x(t) = -2\cos(t)$, $y(t) = 3\sin(t)$, $0 \le t \le \pi$
MY ANSWERS
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.
Thanks
Best Answer
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