Hockey pucks and parameters There is one hockey puck with a diameter of $3$ inches. The puck is spinning around its center at a speed of $3$ counterclockwise rotations per second. At the center, the puck is traveling at a speed of $24$ inches per second at an angle of $45^\circ$ to the positive $x$-axis. 
(a) At time $t=0$, the center of the puck is at the origin. Find the location of the center of the puck at time $t$? (Note that time $t$ is measured in seconds.) 
(b) A point on the outer edge of the puck begins at the point $\left(0,\frac32\right)$. Find its location at time $t$?
I am not sure how to find part (a), but this is what I have for part b). 
The puck has diameter of $3$ inches, a radius of $\frac{3}{2}$. 
I used the formula where that in order to find a point on the circle's circumference, given a circle with center $(a,b)$ and radius $r$, 
$x(t) = r\cos (t) + a$
$y(t) = r \sin (t) + b$
the circle's center is at $(0,0)$ because of part a) statement.
Therefore $x(t) = r \cos (t)$ and $y(t) = r \sin (t)$. I plugged in $x(t) = 0$ and $y(t) = \frac{3}{2}$ because of what was given in part b). 
From there I found that $t = \frac{\pi}{2}$ and that $r = \frac{3}{2}$, 
so now the formula is $x(t) = \frac{3}{2} \cos (t)$ and $y(t) = \frac{3}{2} \sin (t)$. But now I am stuck. What do I do next for part b), and how do I solve part a)? 

(a) The hockey puck is traveling along the line $y=x$, traveling 24 inches every second. Since the puck moves 24 inches when $t=1$ and the puck travels at a constant pace, we can graph:

When $t=1$, $x = y = 12\sqrt{2}$. Therefore, the center of the puck at time $t$ is at:
\begin{align*}
x &= 12\sqrt{2}t \\
y &= 12\sqrt{2}t
\end{align*}
(b) Let's make a parametric equation for the puck using variables:
\begin{align*}
x &= a + r \cos \theta \\
y &= b + r \sin \theta
\end{align*}
We know that the center of the puck at time $t$ is at $12 \sqrt{2} t$. We also know that the radius is $\frac{3}{2}$. Filling in the variables, we get: 
\begin{align*}
x &= 12 \sqrt{2}t + \frac{3}{2} \cos \theta \\
y &= 12 \sqrt{2}t + \frac{3}{2} \sin \theta
\end{align*}
But what is $\theta$? Since the puck spins $6\pi$ radians every second and starts at $\frac{\pi}{2}$, $\theta = 6\pi t + \frac{\pi}{2}$
\begin{align*}
x &= 12 \sqrt{2}t + \frac{3}{2} \cos (6\pi t + \frac{\pi}{2}) \\
y &= 12 \sqrt{2}t + \frac{3}{2} \sin (6\pi t + \frac{\pi}{2})
\end{align*}
Since $\sin(\theta+90) = \cos(\theta)$ and $\cos(\theta + 90) = -\sin \theta$
\begin{align*}
x &= 12 \sqrt{2}t - \frac{3}{2} \sin (6\pi t) \\
y &= 12 \sqrt{2}t + \frac{3}{2} \cos (6\pi t)
\end{align*}
