% script for excercise 1 of case study 1 for EMATH171
% in this script we solve for V using newtons method repetating untill we
% have a tolerence of 0.0001
clear
clc
close all

% Coefficients for the function and its deriviative
m = 1000;
g = 9.81;
theta = 0.25 * pi;
coff_friction = 0.2;
a = 1;
b = 100 * pi;
rgb = 20;
R = 0.5;


% functions
j = @(v) (m .* g .* (sin(theta)) .* v) + (coff_friction .* m .* g .* (cos(theta)) .* v);
% motor shaft rotation speed

k = @(v) (a .* ((b .* ((rgb .* v) / R)) - (((rgb.^2) * (v.^2))/(R.^2))));
% power output

f = @(v) j(v) - k(v);
% origional function

g = @(v) (m .* g .* (sin(theta))) + (coff_friction .* m .* g .* (cos(theta))) - (a .* ((b .* (rgb / R)) - ((2 .* (rgb.^2).* v)/(R.^2))));
% deriviative

h = @(old_x) old_x - ((f(old_x))/(g(old_x)));
% newtons method propper

% Coefficients for newtons method

N = 200;
% maximuim amount of repetitions, unlikley to actualy reach this value

x = 2;
% initial guess

tol = 0.0001;
%tolerence of diffrence between iterations

out_array = [x];
% output array of the results of newtons methods approximations

error_array = [abs(f(x))];
% array of the error

delta_array = [];
% array of change between iterations

counter = 0;
% tracks number of iterations

% Newtons method implementaion

for item = 1:N
    old_x = x; % saving old x
    delta_array(item) = (x);
    x = h(x); % creating new x
    counter = counter + 1;
    out_array(item+1) = x;
    error_array(item+1) = abs(f(x));
    delta_array(item) = abs(delta_array(item)- x);
    if and((abs(f(x))<tol),(abs(x-old_x)<tol))
        counter = counter - 1;
        break
    end
end

% display
figure
hold on
x_actual_array = [0:0.1:5];
plot (x_actual_array, j(x_actual_array))
plot (x_actual_array, k(x_actual_array))
hold off
x
counter
out_array
error_array;
delta_array;