help i dont know why my code isnt working

a = 0; % lower bound of interval
b = 2*π; % upper bound of interval
tol = 0.001; % tolerance
while (b-a) > tol
    c = (a + b) / 2; % midpoint of interval
    if dy(a)*dy(c) < 0
        b = c;
    else
        a = c;
    end
    min_angle = (a + b) / 2; % angle of crank for minimum y-coordinate
    max_angle = (a + b) / 2 + π; % angle of crank for maximum y-coordinate 
    min_angle = (a + b) / 2; 
    % angle of crank for minimum y-coordinate max_angle = (a + b) / 2 + pi;
    % angle of crank for maximum y-coordinate  
    % Initialize arrays to store angles and y-coordinates angles = a:0.001:b;
    y_coords = zeros(size(angles)); 
    % Calculate y-coordinate 
    for each angle for i = 1:length(angles) theta = angles(i); 
        beta = asin((CD/BC)sin(theta)); 
        phi = pi - beta; 
        psi = acos((DE^2 - BC^2 - CE^2)/(2BCCE)); 
        alpha = phi - psi; 
        y_coords(i) = OAcos(theta) + sqrt(OB^2 - OA^2sin(theta)^2) + CDcos(alpha) - DE*cos(beta+alpha) - d; 
    end 
    % Find the minimum and maximum y-coordinates and their corresponding angles 
    [min_y, min_idx] = min(y_coords); 
    [max_y, max_idx] = max(y_coords); 
    min_angle = angles(min_idx); max_angle = angles(max_idx); 
    % Display the results 
    fprintf('Minimum y-coordinate: %f at angle %f\n', min_y, min_angle);
    fprintf('Maximum y-coordinate: %f at angle %f\n', max_y, max_angle);

Answer 1

[Walter Roberson]

b = 2*π; % upper bound of interval

π is a permitted character in MATLAB only in

• comments
• character vectors
• string objects

Never as the name of a function or variable.

[Walter Roberson]

You had part of a comment about for each angle down on a line where it was executable code.

You have not defined BC CD CDcos DE CE OA OAcos OB

2BCCE is not a valid variable name.

2sin is not a valid variable name.

Please note that MATLAB has absolutely no implied multiplication -- not anywhere . Not even in the hidden language of the symbolic toolbox. Every multiplication must be stated using one of the two multiplication operators. I recommend that you use the .* multiplication operator at all times and the ./ division operator at all times (until you start needing to do linear algebra). .* and ./ are not always strictly necessary, but you will have fewer problems if you treat them as the rule instead of treating .* and ./ as the exceptions.

a = 0; % lower bound of interval
b = 2*pi; % upper bound of interval
tol = 0.001; % tolerance
while (b-a) > tol
    c = (a + b) / 2; % midpoint of interval
    if dy(a)*dy(c) < 0
        b = c;
    else
        a = c;
    end
    min_angle = (a + b) / 2; % angle of crank for minimum y-coordinate
    max_angle = (a + b) / 2 + pi; % angle of crank for maximum y-coordinate 
    min_angle = (a + b) / 2; 
    % angle of crank for minimum y-coordinate max_angle = (a + b) / 2 + pi;
    % angle of crank for maximum y-coordinate  
    % Initialize arrays to store angles and y-coordinates angles = a:0.001:b;
    y_coords = zeros(size(angles)); 
    % Calculate y-coordinate for each angle
    for i = 1:length(angles)
        theta = angles(i); 
        beta = asin((CD/BC)sin(theta)); 

Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.

        phi = pi - beta; 
        psi = acos((DE^2 - BC^2 - CE^2)/(2BCCE)); 
        alpha = phi - psi; 
        y_coords(i) = OAcos(theta) + sqrt(OB^2 - OA^2sin(theta)^2) + CDcos(alpha) - DE*cos(beta+alpha) - d; 
    end 
    % Find the minimum and maximum y-coordinates and their corresponding angles 
    [min_y, min_idx] = min(y_coords); 
    [max_y, max_idx] = max(y_coords); 
    min_angle = angles(min_idx); max_angle = angles(max_idx); 
    % Display the results 
    fprintf('Minimum y-coordinate: %f at angle %f\n', min_y, min_angle);
    fprintf('Maximum y-coordinate: %f at angle %f\n', max_y, max_angle);

[Image Analyst]

To expand on that, use

beta = asin((CD/BC) * sin(theta));

instead of

beta = asin((CD/BC)sin(theta));

If CD is actually the product of C and D instead of one variable name, then use

beta = asin(((C*D)/(B*C)) * sin(theta));

And if you want to work in degrees instead of radians, there are versions of those functions that end in d, like sind and asind

If this Answer solves your original question, then could you please click the "Accept this answer" link to award the answerer with "reputation points" for their efforts in helping you? They'd appreciate it. Thanks in advance. 🙂 Note: you can only accept one answer (so pick the best one) but you can click the "Vote" icon for as many Answers as you want. Voting for an answer will also award reputation points.

[Walter Roberson]

B C D E are not defined so BC cannot be B*C

The notation and formulas hint that BC is intended to be the magnitude of the distance between two points labeled B and C, typically points in two dimensions.