Error of index exceeds the number of array elements, not sure why. Time sensitive, but not important

I have a project for college class so this function is set up specifically for the problem. We have to do the euler equation without using the integrated euler functions. If there's any questions I can explain them, I'm not too sure what all info to give. Everything works fine but at the end I get this error:

Index exceeds the number of array elements (11).

Error in euler1 (line 15)
        V(j+1)=V(j)+step*func1(T(j+1),V(j));

Here are the inputs

funct1=@(t)3*sin(t)^2;
funct2=@(t,V)3*sin(t)^2-3*(1.35*V^(1/3)-1)^1.5;
tstart=0;
tend=10;
y0=0.9;
yo=1;
s=1.6;

The steps are 1, 0.5, 0.1, 0.05, but I didn't know how to implement that so I just do it one at a time.

This specific error is for a step of 1, but other steps just change the number in the error.

And the function

function euler1(func1,func2,tstart,tend,step,y0,yo,s)
%Function 1 is with output, function 2 is without
steps=(tend-tstart)/step+1;
Y=zeros(1,steps);
V=zeros(1,steps);
T=tstart:step:tend;
V0=(pi*y0^3)/(3*s^2);
Y(1)=y0;
V(1)=V0;
for j=1:steps
    if Y(j)<yo
        V(j+1)=V(j)+step*func2(T(j+1));
        Y(j+1)=((3*s^2*V(j+1))/(pi))^(1/3);
    else
        V(j+1)=V(j)+step*func1(T(j+1),V(j));
        Y(j+1)=((3*s^2*V(j+1))/(pi))^(1/3);
    end
end
end

It used to give me the values at the end after the error anyways, but now I can't get the values out of it and I need to make tables of the outputs.

[Walter Roberson]

Thank you for differentiating between time sensitive and important!! +1!

Answer 1

[Walter Roberson]

You index T(j+1). For that to work the maximum value for j cannot exceed one less than the number of elements in T, so make sure that your for loop stops early enough.

Also because of floating point roundoff, the way you calculate steps is not guaranteed to exactly equal the length of T. It would be safer if you calculate T first and then assign steps based on length(T)

[Max Bailey]

So, what do you think I should change things to to fix it? I'm fairly new to Matlab and everything you said is catching and making sense but I'm not sure how to implement it without messing things up in the process.

[Walter Roberson]

Well first make the change on how you calculate steps. Then change the for loop so it ends one earlier.