Question

Project rules The course projects offer you the chance to earn a total of up to 5% of your grade. They designed to accompany the course material and develop your programming skills. The project needs to be solved in MATLAB or Octave. We will simulate some real applications and you will have to submit individually. To respect your workload from other courses, the deadlines are generous, but be respectful of them. Late submission will not be accepted the project will be submitted on Blackboard in the form of a Word document. Any handwritten parts need to scanned/photographed and inserted to your word document. The last three digits of your student ID are XYZ. This parameter will be used in our problems. 777 Problem 1 You are working as engineers in an oil field. You know from the seismic measurements that the top of your oil reservoir follows the function f(x) = -3.8(x - XXY)2 - 8.6(x - 10) - 500 - YYX, where x is the distance from your current location and y the depth in the ground. For the point closest to the surface (hint: f'(x)=0), you have to find the distance x from your location and depth from the surface. You can modify the bisection method algorithm from BlackBoard (hint: do not use the chain rule of integration)

Answer 1

f @ (x) -3.8* (x-777).^2 - 8.6.* (x-10) 500 - 777 Ploting function to get glimpse fplot (f, [0,1500]); f(x)' xlabel('x') ylabel('Depth y = grid on From this plot it is clear that closest point lies in range x 500 to 100 §Derivative of function f' (x), say g (x) @(x) -3.8 2* (x-777) - 8.6 We have to find roots of g(x) = 0 using bisection Initial guesses xl 500 xu 1000; xr(xl+xu) /2 err abs (g (xr)) 1st approximation using bisection method defining error for 1st approximation tol 1e-6; Convergence tolerance Subsequent iterations while err > tol While error tolerace, continue doing iterations Guess replacement for next iteration §if g(xl) and g (xr) have opposite signs, if g (xl) g (xr) < 0 then, xu = xr for next iteration Xu Xr else else if they have same signs, xl = xr Xl Xr end Next approximation using bisection method (xl+xu)/2 abs (g (xr)) Updating error err end Now our converged root is in xr fprintf'\nFor the point closest to the surface :-'); fprintf'n\t Distance from your location, x = $.4f',xr); fprintf\n\t Depth from surface, y= $.4f \n', f (xr)

Text :-

f =@(x) -3.8*(x-777).^2 - 8.6.*(x-10) - 500 - 777;

% Ploting function to get glimpse
fplot(f,[0,1500]);  xlabel('x');    ylabel('Depth y = f(x)')
grid on
% From this plot it is clear that closest point lies in range
% x = 500 to 100

% Derivative of function f'(x), say g(x)
g = @(x) -3.8*2*(x-777) - 8.6;

% We have to find roots of g(x) = 0 using bisection
% Initial guesses
xl = 500;       xu = 1000;

xr = (xl+xu)/2;          % 1st approximation using bisection method
err = abs(g(xr));        % defining error for 1st approximation
tol = 1e-6;              % Convergence tolerance

% Subsequent iterations
while err > tol         % While error > tolerace, continue doing iterations
    % Guess replacement for next iteration
    if g(xl)*g(xr) < 0      % if g(xl) and g(xr) have opposite signs,
        xu = xr;            % then, xu = xr for next iteration
    else
        xl = xr;            % else if they have same signs, xl = xr
    end

    xr = (xl+xu)/2;         % Next approximation using bisection method
    err = abs(g(xr));       % Updating error
end

% Now our converged root is in xr
fprintf('\nFor the point closest to the surface :-');
fprintf('\n\t Distance from your location, x = %.4f',xr);
fprintf('\n\t Depth from surface, y = %.4f \n', abs(f(xr)));

result :-

Command Window For the point closest to the surface Distance from your location, x = 775.8684 Depth from surface, y = 7868.3342

106 -0.2 -0.4 -0.6 -0.8 -1 1.2 1.4 1.6 -1.8 -2 -2.2 1000 1500 0 500 X O Depth y f(x)