How can I make linspace work in the case where I want to plot a function

I want to plot the maximum theta for acceleration(af) where acceleration is in a range between 0.4905,49.05.

How can I do this ? This script is only using 49.05 as af, and not as a linear vector. When plotting it gives me a single point.

this is the function

function theta2dot=pendulumcode2(t,theta,g,d,af,kg)
theta2dot=zeros(1,1);
theta2dot=(af*d*cos(theta)-g*d*sin(theta))/(kg^2+d^2);

this is the script

clc;clear
g=9.81; 
kg=0.26; d=0.46;
tspan=[0,6];
theta0=[0;0];

N=300;
for af=linspace(0.4905,49.05,N)
    [t,theta]=ode45(@(t,theta)pendulumcode2(t,theta,g,d,af,kg),tspan,theta0);
end

[maxt,maxtheta]=max(theta);
c2=max(theta);
y2=t(maxtheta);
degree2 = c2(1,1);
degree = degree2*180/pi
time = y2(1,1)

figure;
hold on
plot(degree,af(:,1));
xlabel('theta')
ylabel('af')
hold off

Answer 1

[DGM]

If you want to preserve a vector, don't use it as the loop iterator.

N=300;
af=linspace(0.4905,49.05,N); % make the vector

for nn=1:numel(af) % index into the vector
    [t,theta]=ode45(@(t,theta) pendulumcode2(t,theta,g,d,af(nn),kg),tspan,theta0);
end

% this doesn't do what you think it does
[maxt,maxtheta]=max(theta); % with this syntax, the second output argument is an index
c2=max(theta); % with this syntax, this is a row vector
y2=t(maxtheta); 
degree2 = c2(1,1); % this is a scalar
degree = degree2*180/pi % this is a scalar
time = y2(1,1)

hold on
plot(degree,af(:,1)); % you're literally plotting a scalar versus a scalar.
xlabel('theta')
ylabel('af')
hold off

I don't know what you're expecting to get out of the loop, You might want to save the results instead of overwriting them.

[Bence Mészáros]

Yeah, i am just starting MATLAB, so I am not familiar with these yet. What do you think how can I correct these to get the answer. Which is supposed to be a plot with max theta and af in that range?

I was struggling to get the maximum for theta and the time it takes to reach that theta. What is an easier way to get it?