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I need to find $l_{2}$ and $\theta$ numerically by solving below equations. How could I do that? At least do i have some iterative way of finding those two unknowns. All others parameters are constants.

\begin{eqnarray} \frac{\mu_{0}\sin(\theta-\theta_{0})}{\sqrt{l_{2}^2 +\mu_{0}l_{2}\cos(\theta-\theta_{0}) +\lambda_{0}}} + \frac{\mu_{1}\sin(\theta-\theta_{1})}{\sqrt{l_{2}^2 +\mu_{1}l_{2}\cos(\theta-\theta_{1}) +\lambda_{1}}} =0 \\ \frac{2l_{2}+\beta_{0}+\mu_{0}\cos(\theta-\theta_{0})}{\sqrt{l_{2}^2 +\mu_{0}l_{2}\cos(\theta-\theta_{0}) +\lambda_{0}}}+ \frac{2l_{2}+\beta_{1}+\mu_{1}\cos(\theta-\theta_{1})}{\sqrt{l_{2}^2 +\mu_{1}l_{2}\cos(\theta-\theta_{1}) +\lambda_{1}}} =\frac{-2}{1+Vr^{l_{2}/D}} \end{eqnarray}

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Your system is like $\frac{\partial f_0}{\partial \theta}+\frac{\partial f_1}{\partial \theta}=\frac{\partial g}{\partial \theta}\land \frac{\partial f_0}{\partial l_2}+\frac{\partial f_1}{\partial l_2}=\frac{\partial g}{\partial l_2}$. Is this how you got it? –  xavierm02 Mar 1 '13 at 10:45
    
actually it looks like $\bigg(\frac{\partial f_{0}}{\partial \theta} +\frac{\partial f_{1}}{\partial \theta}\bigg)= 0$ and $\bigg(\frac{\partial f_{0}}{\partial l_{2}} +\frac{\partial f_{1}}{\partial l_{2}}\bigg)=\frac{-1}{1+Vr^{l_{2}/D}}$ –  Norman Mar 1 '13 at 10:48

1 Answer 1

Try multivariate Newton Raphson method

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Is this method easy? it is hard to find the derivatives. –  Norman Mar 1 '13 at 10:53
    
Ask any computer algebra system, i.e. maxima or WolframAlpha.com. –  vonbrand Mar 1 '13 at 14:50

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