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