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Below is my current workspace in MATLAB.

theta = linspace(0,2*pi);

for a = [10:0.1:14]

c = 30-a;

bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));

M = [max(bt)]

end

When I run this file it gives me the maximum value of 41 arrays. The answer shows like this:

M =

13.5983

M =

13.6983

M =

13.7983

M =

13.8983

M =

13.9982

I am wondering is there a way I can make my answer into a single array which I can then plot against a(10:14)?

I wish for it to look something like this:

M =

13.5983 13.6983 13.7983 13.8983 13.9982

OR

M =

13.5983

13.6983

13.7983

13.8983

13.9982

Any help would be great

Clayton Gotberg
on 26 Apr 2021

Edited: Clayton Gotberg
on 26 Apr 2021

This is definitely possible! Since you aren't using integers for indexing, there are a couple of ways to do it:

theta = linspace(0,2*pi);

M = []; % Create empty M

for a = [10:0.1:14]

c = 30-a;

bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));

M = [M max(bt)]; % Add max(bt) to the end of M every loop

end

% Also

theta = linspace(0,2*pi);

count = 1; % Create empty M

for a = [10:0.1:14]

c = 30-a;

bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));

M(count) = max(bt);

count = count+1;

end

% Or

theta = linspace(0,2*pi);

area_of_interest = 10:0.1:14;

count = length(area_of_interest);

for k = 1:count

a = area_of_interest(k)

c = 30-a;

bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));

M(k) = max(bt); % Add max(bt) to the end of M every loop

end

For plotting against a(10:14), I want to check about what you mean.

If you want to plot the maximum against a from 10 to 14, just use plot(a,M). If you want to plot only against the five values a = [10 ,11, 12 ,13, 14], you can just change the area of interest from 10:0.1:14 to 10:14. If you want to plot the 10th through 14th elements of a and M, you can say plot(a(10:14),M(10:14)).

Clayton Gotberg
on 26 Apr 2021

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