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I've got some problems solving (numerically) this system of equations.

\begin{array}{l} 40 \cdot \cos (2t) + 105 \cdot \cos ({\theta _3}) - 75 \cdot \cos ({\theta _4}) - 91.924 \cdot \cos ({337.62}) = 0\\ 40 \cdot \sin(2t) + 105 \cdot \sin({\theta _3}) - 75 \cdot \sin({\theta _4}) - 91.924 \cdot \sin({337.62}) = 0 \end{array}

Now t is an array of numbers (variable?) ranging from 0 to 0.785 like this t = 0:0.01:0.785.

I was wondering if it is possible to find $\theta_3$ and $\theta_4$ for every t (like t=0 --> $\theta_3$=something, $\theta_4$=something...t=0.4-->$\theta_3$="something else", $\theta_4$="something else") put those values(solutions) in an array (vector)so that I can plot them.

I've tried to do it symbolically with solve but MATLAB couldn't find any (useful) solution. I tried to solve it numerically but I couldn't make it spit out more than two solutions at a time.

I've done it in Mathcad but I need to solve this with MATLAB (I'm new to MATLAB).

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  • $\begingroup$ Is $\theta_1$ also an unknown? $\endgroup$ – GertVdE Apr 7 '14 at 18:14
  • $\begingroup$ No, theta1 is 337.62 (deg). Sorry I didn't point that out earlier. $\endgroup$ – John Dr.Cox Dorian Apr 7 '14 at 18:17
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I would loop over t and solve the system using the fsolve command. For $t=0$, it shouldn't be too difficult with a reasonable initial guess. And then solve each problem using as initial value the solution of the previous value for $t$...

Define your function as

function [res] = problem(x,t)

t1 = 337.62/180.0*pi;
t3 = x(1);
t4 = x(2);

res = zeros(2,1);
res(1) = 40.0*cos(2*t)+105*cos(t3)-75*cos(t4)-91.924*cos(t1);
res(2) = 40.0*sin(2*t)+105*sin(t3)-75*sin(t4)-91.924*sin(t1);

return;

Define the values of the parameter:

t = [0:0.01:0.785]

Make room for the solutions:

sols = zeros(2,length(t))

Solve for $t=0$ with an initial guess of $\theta_3 = \theta_4 = 1$ (arbitrary)

theta0 = [1,1];
sols(:,1) = fsolve(@(x) problem(x,t(1)),theta0)

And then loop over the remaining values for $t$ using the previous solution as initial guess:

for i=2:length(t),
  sols(:,i) = fsolve(@(x) problem(x,t(i)),sols(:,i-1))
end

and plot

plot(t,sols(1,:),t,sols(2,:))

Caveat: my Matlab is a bit rusty (my switch to Python/Numpy/Scipy is already a couple of years ago...)

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  • $\begingroup$ Thank you for your answer Gert. Even though your method might work I don't think I'd be able to do it since I'm new to MATLAB and this sounds kinda complicated. $\endgroup$ – John Dr.Cox Dorian Apr 7 '14 at 18:22
  • $\begingroup$ Just another thing. Here sols(:,i) = fsolve(@(x) problem(x,t(i)),sols(:,i-1)) with @ you're passing a reference of the function problem(x,t) to fsolve right? And the previous solutions sols(:,i-1) is used as a guess for the next one right? $\endgroup$ – John Dr.Cox Dorian Apr 7 '14 at 19:08
  • $\begingroup$ Yes, that is correct. $\endgroup$ – GertVdE Apr 7 '14 at 19:22
  • $\begingroup$ Thank you GertVdE. I'll just have to look at how exactly the thing with fsolve and functions as parameters works and it'll be done i guess. Thanks again. :) $\endgroup$ – John Dr.Cox Dorian Apr 7 '14 at 19:34

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