function [t, v] = skydiver_params(f,dt,tol,varargin)
% Function to calculate skydiver falling speed
% for specified parameter values and time increment
% using an iterative, numerical solution with
% a while loop. For this problem, the goal is to show
% how varargin can be used to pass input parameters to
% the function passed into this function.
%
% Inputs:  Function handle to first derivative function
%          time increment (s)
%          convergence tolerance tol (m/s)
%          varargin
%          which should be
%          mass m (kg)
%          gravity g (m/s^2)
%          drag coefficient c (kg/m)
% if we pass in derivparams as the function to be called.
%
% All inputs are passed into the function.
%
% Outputs: falling speed vector v (m/s)

% Iterate the solution until the final velocity stops changing
tolnew = 1; % Why do we need this statement?
tnew = 0;
t(1,1) = 0;
vnew = 0;
v(1,1) = vnew; % Can/should I pre-allocate memory for v?
i = 1; % Why is this line needed?

while tolnew > tol
    % Increment counter
    i = i+1;
    
    % Update v
    vold = vnew;
    dvdt = f(vold,varargin{:});
    vnew = vold + dvdt*dt;
    
    % Update t
    told = tnew;
    tnew = told + dt;

    % Update tol
    tolnew = abs(vold-vnew); % Why not abs((vold-vnew)/vnew)?
    
    % Store new solution
    t(i,1) = tnew;
    v(i,1) = vnew;
end

% Compare with analytic solution to see how good our final value
% of v is. Note that we cannot normally do this, but we will do it
% here to develop an understanding of what factors influence the
% accuracy of the numerical solution.
% vtrue = sqrt(m*g/c)*tanh(sqrt(g*c/m)*t(i,1));
% Note that we can no longer do the line above since m, g, and c
% are no longer known to the program.

fprintf('The numerical final velocity is %7.4f m/s.\n', v(i,1))
% fprintf('and the analytical final velocity is %7.4f m/s.\n', vtrue)
fprintf('The final time is %7.4f sec.\n', t(i,1))

% Question: How accurately can we find the time required to reach
% terminal velocity? Only to within dt of the true value.