-- Exercise 4.5
-- George C. Privon
-- 2024-07-20

type R = Double
type Derivative = (R -> R) -> R -> R

type Time = R
type Position = R
type Velocity = R
type Acceleration = R

type PositionFunction = Time -> Position
type VelocityFunction = Time -> Velocity
type AccelerationFunction = Time -> Acceleration
--
-- a function to take the numerical derivative, approximated for some dt
derivative :: R -> Derivative
derivative dt x t = (x (t + dt/2) - x (t - dt/2)) / dt


pos1 :: PositionFunction
pos1 t = if t < 0
         then 0
         else 5 * t**2

vel1Analytic :: VelocityFunction
vel1Analytic t = if t < 0
                 then 0
                 else 10 * t

acc1Analytic :: AccelerationFunction
acc1Analytic t = if t < 0
                 then 0
                 else 10

vel1Numerical :: VelocityFunction
vel1Numerical t = derivative 0.01 pos1  t

acc1Numerical :: AccelerationFunction
acc1Numerical t = derivative 0.01 (derivative 0.01 pos1 ) t

-- Solution
-- The numerical and analytic velocity functions seem to agree closely across
-- the range, except when t~<0.001.
-- The numerical acceleration function differs the most around t=0 but is
-- in agreement across the rest of the range.