1. **State the problem:** We need parametric equations for a particle starting at (6,0) and tracing the top half of the circle $x^2 + y^2 = 36$ twice. 2. **Recall the circle equation and parametric form:** The circle has radius $r=6$ and center at the origin. The standard parametric equations for a circle are: $$x = r \cos(t), \quad y = r \sin(t)$$ where $t$ usually goes from $0$ to $2\pi$ for one full circle. 3. **Adjust for top half and two times:** The top half corresponds to $y \geq 0$, so $t$ goes from $0$ to $\pi$ for one half-circle. To trace the top half twice, let $t$ go from $0$ to $2\pi$ but speed up the angle so that the particle completes two half-circles in that interval. 4. **Parametric equations:** Use $$x = 6 \cos(2t), \quad y = 6 \sin(2t)$$ with $t$ in $[0, \pi]$ to trace the top half twice. 5. **Check starting point:** At $t=0$, $$x=6 \cos(0)=6, \quad y=6 \sin(0)=0$$ which matches the starting point (6,0). 6. **Parameter interval:** Given the problem states $0 \leq t \leq 2\pi$, to trace the top half twice in this interval, use $$x=6 \cos(t), \quad y=6 \sin(t)$$ with $t$ in $[0, 2\pi]$ but restrict the motion to the top half by considering only $y \geq 0$ twice. To do this, we double the angle inside the trig functions: $$x=6 \cos(2t), \quad y=6 \sin(2t), \quad 0 \leq t \leq 2\pi$$ This traces the top half of the circle twice as $t$ goes from $0$ to $2\pi$. **Final answer:** $$x=6 \cos(2t), \quad y=6 \sin(2t), \quad 0 \leq t \leq 2\pi$$