Solving for theta at different time steps

I have the following equation:

 47.9618*theta + 15.9809*sin(2*theta) = 20.780136*t

Given, at time t = 0, theta = 0 radians. I want to find theta value at t = 1 second, 2 seconds, etc.

Could this be automated on matlab?

Answer 1

[James Tursa]

E.g.,

>> t = 0;
>> f = @(theta) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
>> fzero(f,0)
ans =
     0
>> t = 1;
>> f = @(theta) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
>> fzero(f,0)
ans =
    0.2649
>> t = 2;
>> f = @(theta) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
>> fzero(f,0)
ans =
    0.5651

[Lee Quan]

So for getting results from t = 0 to 10 seconds at once, I could do this:

for t=0:10
f = @(theta)47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
fzero(f,0)
end

Awesome! Thanks James!

[Lee Quan]

Could you point me to some resources where the '@' notation is used so that I can familiarize myself with it? The @ part is something I can't find in matlab documentation online.

[James Tursa]

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html?searchHighlight=anonymous%20function&s_tid=doc_srchtitle

Answer 2

[Walter Roberson]

There is no closed form solution for that. Use fzero or fsolve:

eqn = @(theta,t) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
for t = 1 : 5
  theta0 = randn()
  fzero( @(theta) eqn(theta,t), theta0)
end