How to determine X,Y position from point P based on time, velocity and rate of turn? We need to figure out an estimated position along an arc using a set of known variables. The starting position on the arc would be called point P, here are the known/measured variables:

P = initial position of point P
x = x coordinate of P (longitude)
y = y coordinate of P (latitude)
V = velocity of point P (nautical miles / min, or nautical miles / hour)
H = heading / directional angle of point P in degrees (000 = North, 180 = South)
R = rate of turn of P in degrees/minute.

All of those variables are known. We actually have a device that is able to calculate point P's rate of turn. See this link here for the device.
We need to calculate the future X,Y positions of point P along the arc based on the variables above with respect to time. The dynamically changing variable to plug into the formula would be Time (in minutes or hours).
For example: at future time T, what would be the position of P (or Px1, Py1) along the arc. My thinking is that it would require some kind of calculus based derivative calculation. Thanks.
 A: Let consider the coordinate $x$ first.
Set the origin of your coordinate system at the initial position $P$: the positive direction of the $x$ axis points to East, and time $t = 0$
At any point of time $t$, the $x$-coordinate of your ship will be given by 
$$ x(t) = \int_0 ^ t v(s) \cos\, \Big(\frac{\pi}{2} - h (s) \Big) \mathrm{d}s$$ where $h$ is the heading (converted to radians), and $v$ is the velocity. 
What this hopefully does is get the value of the velocity, compute its projection with respect to the $x$-axis (the $\frac{\pi}{2}$ should take care of the different conventions used for angles, as sailors measure counterclockwise from the $y$ axis), and integrate.
Same for the $y$-coordinate 
$$ y(t) = \int_0 ^ t v(s) \sin\, \Big(\frac{\pi}{2} - h (s) \Big) \mathrm{d}s$$
You could also consider your point is moving on the complex plane, and wrap both the integrals in
$$ z(t) = \int_0 ^ t v(s) \exp \Bigg( i\Big( \frac{\pi}{2} - h (s) \Big) \Bigg) \mathrm{d}s$$
and now your $x$- and $y$-coordinates correspond to $\Re (z)$ and $\Im (z)$, the real and imaginary parts of $z$ respectively.
EDIT FOLLOWING FURTHER CLARIFICATION
It seems that my understanding of the proviced information was fallacious. It turns out that velocity is constant, as well as rate of turn.
The ship traces an arc of circumference, the radius $r$ of which satisfies the relationship $$ 2 \pi r = v \frac{2\pi}{R}$$ where $R$ is the rate of turn in radians / unit of time. 
The above equation simply states that the circumference (traced by a ship after a full round) equals the velocity times the time required for a 360 degress rotation. 
The ship position vector $X$ as a function of time could be found by the parametric equations
$$X(t) =
\left\{
 \begin{array}{ll}
    r \cos (tR + H)    \\
   r \sin (tR + H) 
 \end{array}
\right.
$$
to which a easy to figure correction should be added to account for the position $P$ at the beginning of the arc.
