# This worksheet contains the Maple commands from Chapter 8 of
# Introduction to Scientific Programming by Joseph L. Zachary.
# 
# (8.1)  We record the mass, maximum power output, and drag coefficient
# for the destroyer.
# 
> mass := 4.5e6;
> maxPower := 5e6;
> drag := 5.5e3;
# 
# (8.2)  We define a symbolic expression that gives the position of the
# destroyer in terms of  t.  We will be ignoring the effects of drag in
# the first part of this worksheet.
# 
> pos := .032 * t^2;
# 
# (8.3)  We plot the position of the destroyer when it is near the
# origin.
# 
> plot(pos, t=-100..100,
>       labels=[`time (seconds)`, `position (meters)`],
>       title=`Position of the destroyer`);
# 
# (8.4)  We plot the position of the destroyer for positive times.
# 
> plot(pos, t=0..200,
>       labels=[`time (seconds)`, `position (meters)`],
>       title=`Position of the destroyer`);
# 
# (8.6)  The velocity of the destroyer is the first derivative of
# position.
# 
> vel := diff(pos, t);
# 
# (8.7)  The acceleration of the destroyer is the first derivative of
# velocity.
# 
> acc := diff(vel, t);
# 
# (8.8)  The force required to move the destroyer along its trajectory
# is the product of mass and acceleration.
# 
> force := mass * acc;
# 
# (8.9)  The power required to move the destroyer along its trajectory
# is the product of force and velocity.
# 
> pow := force * vel;
# 
# (8.10)  We plot the power required to move the destroyer when it is
# close to the origin.
# 
> plot(pow/1e6, t=-100..100,
>       labels=[`time (seconds)`,  `power (megawatts)`],
>       title=`Power required for trajectory .032*t^2, no drag`);
# 
# (8.11)  We plot the absolute value of the power required to move the
# destroyer when it is close to the origin.
# 
> plot(abs(pow)/1e6, t=-100..100,
>       labels=[`time(seconds)`,  `power (megawatts)`],
>       title=`Magnitude of power required for trajectory .032*t^2, no
> drag`);
# 
# (8.12)  We use solve to find the two times at which the maximum power
# capacity of the destroyer is reached.
# 
> threshold := solve(abs(pow) = maxPower, t);
# 
# (8.13)  The solution to (8.12) is a sequence of two values.
# 
> threshold;
# 
# (8.14)  We calculate the speed of the destroyer when it is at the
# first power threshold.  We extract the first value of the sequence by
# subscripting with 1.
# 
> .064 * threshold[1];
# 
# (8.15)  We calculte the speed of the destroyer when it is at the
# second power threshold.  We extract the second value of the sequence
# by subscripting with 2.
# 
> .064 * threshold[2];
# 
# (8.16)  We use subs to substitute a value for t into the velocity
# expression.
# 
> subs(t = threshold[1], vel);
# 
# (8.17)  We create a function makePower to calculate the power
# required, in the absence of drag, at any point of time along an
# arbitrary trajectory.  Here, pos is the trajectory expression (which
# must be a symbolic expression in t) and mass is the mass of the
# destroyer.  In the body of the function, we use the diff function to
# determine velocity and acceleration functions.
# 
> makePower := (pos, mass) -> mass * diff(diff(pos,t),t) * diff(pos,t);
# 
# (8.18)  We use makePower to calculate the power required to follow
# trajectory pos.  As expected, we get the same result as in (8.9).
# 
> pow := makePower(pos, mass);
# 
# (8.19)  We calculate power expressions corresponding to two new
# trajectories.
# 
> pow1 := makePower(.016 * t^2, mass);
> pow2 := makePower(.002 * t^2, mass);
# 
# (8.20)  We find the times at which power threshold is reached for the
# first of the new trajectories.
# 
> threshold1 := solve(abs(pow1) = maxPower, t);
# 
# (8.21)  We find the velocity corresponding to the first power
# threshold for the first of the new trajectories. 
# 
> subs(t = threshold1[1], diff(.016*t^2, t));
# 
# (8.22)  We find the times at which power threshold is reached for the
# second of the new trajectories.
# 
> solve(abs(pow2) = maxPower, t);
# 
# (8.23)  We find the power required to follow a constant-velocity
# trajectory (in the absence of drag).
# 
> makePower(v*t, mass);
# 
# (8.24)  This function computes the power expression for a destroyer of
# the given mass and drag coefficient that is following the trajectory
# pos (where pos must be a symbolic expression in t).  For the remainder
# of this worksheet, we will consider the effects of drag on power
# requirements.
# 
> makePowerDrag := (pos, mass, drag) -> (mass * diff(diff(pos,t),t) +
> drag * diff(pos,t)) * diff(pos,t);
# 
# (8.25)  We calculate a power expression for a destroyer following our
# original trajectory of .032*t^2. 
# 
> pow := makePowerDrag(pos, mass, drag);
# 
# (8.26)  We plot the power consumption calculated in (8.25).
# 
> plot(pow/1e6, t=-100..100,
>       labels=[`time (seconds)`, `power (megawatts)`],
>       title=`Power required for trajectory .032*t^2, with drag`);
# 
# (8.27)  We find the power in watts required at times 100 and -100
# seconds when following the tajectory .032*t^2.
# 
> abs(subs(t = 100, pow));
> abs(subs(t = -100, pow));
# 
# (8.28) We plot the power consumption calculated in (8.25) over a
# different time range than we used in (8.26).
# 
> plot(pow/1e6, t=-1000..200,
>       labels=[`time (seconds)`, `power(megawatts)`],
>       title=`Power required for trajectory .032*t^2, with drag`);
# 
# (8.29)  We determine the times at which the power threshold is reached
# when following the trajectory from (8.25).
# 
> solve(pow = maxPower, t);
# 
# (8.30)  We determine the power required to follow a constant-velocity
# trajectory (in the presence of drag).
# 
> constantPower := makePowerDrag(v*t, mass, drag);
# 
# (8.31)  We find the maximum sustainable velocities (in m/sec) for the
# destroyer in the presence of drag.
# 
> solve(constantPower = maxPower, v);
# 
# (8.33)  We define a piecewise trajectory in which the destroyer begins
# at a rest, moves 10,000 meters, and ends at rest.
# 
> move := piecewise(t < -500, -5000,
>                   t < 500, 5000*sin(Pi*t/1000),
>                   5000);
# 
# (8.34)  We plot the piecewise trajectory from (8.33).
# 
> plot(move, t=-600..600,
>       labels=[`time (seconds)`, `position (meters)`],
>       title=`A Piecewise Trajectory`);
# 
# (8.35)  We plot the power required to follow the piecewise trajectory
# from (8.34).
# 
> plot(makePowerDrag(move, mass, drag)/1e6,
>       t=-600..600,
>       labels=[`time (seconds)`, `power (megawatts)`],
>       title=`Power Required to Follow Piecewise Trajectory`);
>