Answer
The particle moves clockwise along an ellipse from (0,5) to (-2, 4). The equation of this ellipse is $\frac{1}{4}x^2 + (y-4)^2 = 1$
Work Step by Step
1. Convert the equation to a Cartesian one, so we can identify the type of curve:
$x = 2sin (t)$
$x^2 = 4sin^2(t)$
$\frac{1}{4}x^2 = sin^2(t)$
$y = 4 + cos (t)$
$y - 4 = cos(t)$
$(y-4)^2 = cos^2(t)$
Adding the equations:
$\frac{1}{4}x^2 + (y-4)^2 = sin^2(t) + cos^2(t)$
$\frac{1}{4}x^2 + (y-4)^2 = 1$
- As we can see by the equation, this curve is an ellipse with a radius of $\sqrt 4 = 2$ parallel to the x-axis, and a radius of $1$ along the y-axis. And with its center at $(0,4)$
2. Find the initial and final point:
Initial point: $t = 0$
$ x = 2 sin(0) = 2 (0) = 0$
$y = 4 + cos(0) = 4 + 1 = 5$
Final point: $t = \frac {3\pi} 2$
$ x = 2 sin(\frac {3\pi} 2) = 2 (-1) = -2$
$y = 4 + cos(\frac {3\pi} 2) = 4 + 0 = 4$
- Since $sin(t)$ determines the $x$ position, and $cos(t)$ determines the $y$ position, the particle will move clockwise along the ellipse.
