Answer
(a)
$x = 2cos(t)$
$y = 1 - 2sin(t)$
$0 \leq t \leq 2 \pi$
(b)
$x = 2cos(t)$
$y = 2sin(t) + 1$
$0 \leq t \leq 6\pi$
(c)
$x = 2cos(t)$
$y = 2sin(t) + 1$
$\frac{\pi}{2} \leq t \leq \frac{3 \pi}{2}$
Work Step by Step
$(x - x_0)^2 + (y-y_0)^2 = (radius)^2$
This equation describes a circle centered at $(x_0,y_0)$.
1. As we can identify by that cartesian equation, that curve is a circle with:
Radius equal to $\sqrt 4 = 2$
Its center at $(0 , 1)$
2. The general parametric equations** for a circle is:
$x = (radius)cos(t) + x_0$
$y = (radius)sin(t) + y_0$
** In this case, the particle will move counterclockwise, and starts at $(radius + x_0, y_0)$.
And the particle goes around the circle completely for every $2\pi$.
For that circle:
$x = 2cos(t)$
$y = 2sin(t) + 1$
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(a)
Since the particle moves once around, $t$ will have the following values:
$0 \leq t \leq 2 \pi$
To make the particle move counterclockwise, we should multiply $sin(t)$ by (-1);
$y = 2sin(t) + 1$
$y = 2(-sin(t)) + 1$
$y = 1 - 2sin(t)$
(b) Since the particle moves three times around the circle, $t$ should go from 0 to $3*(2\pi)$
$0 \leq t \leq 6\pi$.
$x = 2cos(t)$
$y = 2sin(t) + 1$
(c) Find the values for $t$ when the particle is at $(0,3)$.
$0 = 2cos(t)$
$t = \frac{\pi}{2}$ or $t = \frac{3 \pi}{2}$
$3 = 2sin(t) + 1$
$2 = 2sin(t)$
$1 = sin(t)$
$t = \frac{\pi}{2}$
Therefore, $t$ should start at $\frac{\pi}{2}$
Since the particle goes halfway around the circle, it should increase by $\frac{1}{2}(2\pi) = \pi$
$\frac{\pi}{2} + \pi = \frac{3\pi }{2}$
Therefore: $\frac{\pi}{2} \leq t \leq \frac{3 \pi}{2}$
