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    (Original post by Star-girl)
    Solution 56

    By the t-substitution t=\tan \dfrac{1}{2} x , the limits get transformed into t_2= \tan \dfrac{1}{2} \cdot 2\pi = 0 ,\ t_1 = \tan \dfrac{1}{2} \cdot 0 = 0 and since the limits are the same, the integral evaluates to  0 .
    I get \dfrac{\pi}{\sqrt{6}}
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    (Original post by Felix Felicis)
    I get \dfrac{\pi}{\sqrt{6}}
    Seems more plausible than 0. My solution is probably dodgy then... but why? :holmes:
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    (Original post by Star-girl)
    Solution 56

    By the t-substitution t=\tan \dfrac{1}{2} x , the limits get transformed into t_2= \tan \dfrac{1}{2} \cdot 2\pi = 0 ,\ t_1 = \tan \dfrac{1}{2} \cdot 0 = 0 and since the limits are the same, the integral evaluates to  0 .

    Seems a bit too quick - maybe this is a little dodgy...
    The integrand is positive over the interval, so it's impossible for the integral to be 0.
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    (Original post by Star-girl)
    Seems more plausible than 0. My solution is probably dodgy then... but why? :holmes:
    Maybe because of the asymptotes of tan between the limits? :holmes:
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    (Original post by und)
    The integrand is positive over the interval, so it's impossible for the integral to be 0.
    (Original post by Felix Felicis)
    Maybe because of the asymptotes of tan between the limits? :holmes:
    Ha! What a noobish error I made... I think it's sleepy-time for me... :facepalm:
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    Solution 56

    Solution 56
    \displaystyle\int_{0}^{2 \pi} \dfrac{dx}{\cos x + 5}  = \displaystyle\int_{0}^{\pi} \dfrac{dx}{\cos x + 5}  + \displaystyle\int_{\pi}^{2 \pi} \dfrac{dx}{\cos x + 5}

    Let t = \tan \frac{x}{2} :

    \displaystyle\int_{0}^{2 \pi} \dfrac{dx}{\cos x + 5}  = \displaystyle\int_{-\infty}^{\infty} \dfrac{dt}{2t^{2} + 3} = \dfrac{1}{\sqrt{6}} \Big[ \arctan t \Big]^{\infty}_{-\infty} = \dfrac{\pi}{\sqrt{6}}
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    (Original post by MsDanderson)
    hey! IF YOU'RE A FEMALE, could you fill this out please? it literally only takes 2mins do it -> http://www.surveymonkey.com/s/B3MVL58

    I need 100 done in the next 24 hours or I'll fail my project Please help! I'm so grateful
    You come looking for females... in a maths thread?
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    (Original post by und)
    You come looking for females... in a maths thread?
    I only just read that it's for females...I already filled it in :facepalm: Ah well
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    (Original post by MsDanderson)
    it's okay! i appreciate it anyway!
    You might want to discard my results anyway...my answers may skew your data for a survey targeted at females to put it lightly
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    (Original post by Star-girl)
    Solution 56

    By the t-substitution t=\tan \dfrac{1}{2} x , the limits get transformed into t_2= \tan \dfrac{1}{2} \cdot 2\pi = 0 ,\ t_1 = \tan \dfrac{1}{2} \cdot 0 = 0 and since the limits are the same, the integral evaluates to  0 .

    Seems a bit too quick - maybe this is a little dodgy...

    EDIT: Ignore this rubbish.
    Can't just sub in t=tan(x/2) as it is not continuous across the range of integration.
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    Problem 59**

    Prove that there are no integers x,y for which x^2+3xy-2y^2=122.
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    (Original post by dknt)
    Well I guess I'll provide another physicsy mathsy problem

    Problem 58 **/***

    A double pendulum consists of a mass m2 suspended by a rod of length l2 from a mass m1, which is itself suspended by a rod of length l1 from a fixed pivot, as shown below.

    Attachment 208929

    Show that the equations of motion for small displacements can be written as

    M \ddot{\Theta} = -K \Theta where,

    M= \begin{pmatrix} {l_1} ^2(m_1 +m_2) & l_1 l_2 m_2 \\l_1 l_2 m_2 & {l_2} ^2 m_2 \end{pmatrix}

     K = \begin{pmatrix} gl_1 (m_1 +m_2) & 0 \\ 0 & gl_2 m_2  \end{pmatrix}

    and  \Theta = \begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix}
    Spoiler:
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    Would you use the small displacement approximation before or after taking the lagrangian to find the equations of motion? :holmes:
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    (Original post by cpdavis)
    Spoiler:
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    Would you use the small displacement approximation before or after taking the lagrangian to find the equations of motion? :holmes:
    Spoiler:
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    Far easier to do it right at the beginning when finding the potentials and kinetic energies. Of course, you could do it after and you should still get the same result, however it makes applying Lagrange's equation and therefore the equations of motion much harder! Of course for a challenge try finding the EOMs for arbitrary displacements
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    (Original post by dknt)
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    Far easier to do it right at the beginning when finding the potentials and kinetic energies. Of course, you could do it after and you should still get the same result, however it makes applying Lagrange's equation and therefore the equations of motion much harder! Of course for a challenge try finding the EOMs for arbitrary displacements
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    Yeah I realised that it would be easier to do it at an earlier stage I remember coming across a problem like this but we assumes that the lengths and masses were the same. Now working through this solution, good practice for my dynamics course
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    (Original post by bananarama2)
    I just lost the will to live with that question after I completed a load of school work (binomial expansion). I generally go for energy, but just didn't in that instance, nice one Star-girl (I had to think about that name.)
    For some reason I didn't see this yesterday. Ah yes... school work... what I should have been doing this holiday... :ninjagirl:

    Thanks (and haha). :cute:
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    (Original post by und)
    You come looking for females... in a maths thread?


    (Original post by james22)
    Can't just sub in t=tan(x/2) as it is not continuous across the range of integration.
    Yeah - I realised afterwards, hence the edits...
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    Solution 59

    More general:

    Denote f(x,y)=ax^{2}+bxy+cy^{2}. This is called binary quadratic form. Its discriminant is defined as  \Delta= b^2-4ac.
    We say that f represents given number n properly if there are integers x_{0},y_{0} such that f(x_{0},y_{0})=n and \gcd(x_{0},y_{0})=1.

    We have the following proposition:
    Let n \not= 0 and d be given integers. Then there exists a binary quadratic form of discriminant d that represents n properly if and only if x^2 \equiv d \pmod {4|n|} is solvable.

    Let \gcd(x_{0},y_{0}) and f(x_{0},y_{0})=n. Since \gcd(x_{0},y_{0})=1, we can choose integers n_{1},n_{2} such that n_{1}n_{2}=4|n|, \gcd(n_{1},n_{2})=1, \gcd(n_{1},x_{0})=1, and \gcd(n_{2},y_{0})=1. Notice that 4an=(2ax_{0}+by_{0})^{2}-dy_{0}^{2}. Now there are integers k_{1},k_{2} such that k_{1}^{2} \equiv d \pmod {n_{1}} and k_{2}^{2} \equiv d \pmod {n_{2}}. It remains to cite the Chinese remainder theorem.
    Conversely, let b^2-d=4cn. Then the form f(x,y)=nx^{2}+bxy+cy^{2} has discriminant d and represents n properly.

    Next, I claim that there is not binary quadratic form of discriminant d=17 which represents 122 properly. Since 122 is square-free, we cannot have improper representations.
    The congruence x^{2} \equiv 17 \pmod8 is obviously solvable. Consider the congruence x^{2} \equiv 17 \pmod {61}. Now \left(\frac{17}{61} \right)\left(\frac{10}{17} \right)=\left(\frac{17}{61} \right)\left(\frac{2}{17} \right)\left(\frac{5}{17} \right)=\left(\frac{17}{61} \right)\left(\frac{2}{5} \right)=-\left(\frac{17}{61} \right)=1. Hence \left(\frac{17}{61} \right)=-1. Therefore the congruence x^{2} \equiv 17 \pmod {4\times 122} has no solutions and my claim holds true.
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    Solution 58

     x_1 = l_1 \sin \theta_1

     \dot{ x_1 } = l_1 \dot{ \theta_1} \cos \theta_1

     y_1 = l_1 \cos \theta_1

     \dot{ y_1} = -l_1 dot{ \theta_1} \sin \theta_1

     x_2 = l_1 \sin \theta_1 + l_2 \sin \theta_2

     \dot {x_2} = l_1 \dot{ \theta_1 } \cos \theta_1 + l_2 \dot{ \theta_2} \cos \theta_2

     y_2 = l_1 \cos \theta_1 + l_2 \cos \theta_2

     \dot {y_2} = -l_1 \dot{ \theta_1 } \sin \theta_1 - l_2 \dot{ \theta_2} \sin \theta_2


    \displaystyle \mathcal{ L} = \frac{1}{2} m_1 l_1^2 \dot{ \theta_1}^2 + \frac{1}{2} m_2 \left( l_1 \dot{ \theta_1 } \cos \theta_1 + l_2 \dot{ \theta_2} \cos \theta_2  \right)^2 + \frac{1}{2}m_2 \left(-l_1 \dot{ \theta_1 } \sin \theta_1 - l_2 \dot{ \theta_2} \sin \theta_2 \right)^2 + m_1 g l_1 \cos \theta_1 + m_2 g ( l_1 \cos \theta_1 + l_2 \cos \theta_2 )

    \displaystyle  = \frac{1}{2} m_1 l_1^2 \dot{ \theta_1}^2  + \frac{1}{2}m_2 l_1^2 \dot{\theta_1}^2 + \frac{1}{2}m_2l_2^2 \dot{\theta_2}^2 + m_2 l_1 l_2 \cos (\theta_1 - \theta_2 ) \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1 \cos \theta_1 + m_2 g ( l_1 \cos \theta_1 + l_2 \cos \theta_2 )

    \displaystyle \frac{\partial \mathcal{ L}}{\partial \theta_1}= -m_2 l_1 l_2 \sin ( \theta_1 - \theta_2 )  \dot{ \theta_1} \dot{ \theta_2} - m_1 g l_1 \sin \theta_1 -m_2 g l_1 \sin \theta_1

    \displaystyle \approx -m_2 l_1 l_2 ( \theta_1 - \theta_2 )  \dot{ \theta_1} \dot{ \theta_2} - m_1 g l_1  \theta_1 -m_2 g l_1 \theta_1

     \displaystyle \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}} =m_1 l_2^2 \dot{ \theta_1} + m_2 l_1^2 \dot{ \theta_1} +  m_2 l_1 l_2 \cos ( \theta_1 - \theta_2 ) \dot{ \theta_2 }

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}}\right) = m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}- \dot{\theta_2} ) m_2 \sin (\theta_1-\theta_2 ) \dot{\theta_2} + m_2 l_1 l_2 \cos ( \theta_1 - \theta_2 ) \ddot{ \theta_2}

    \displaystyle \approx m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}- \dot{\theta_2} ) m_2 (\theta_1-\theta_2 ) \dot{\theta_2} + m_2 l_1 l_2  \ddot{ \theta_2}

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}}\right) - \frac{\partial \mathcal{ L}}{\partial \theta_1} = 0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}- \dot{\theta_2} ) m_2 (\theta_1-\theta_2 )\dot{\theta_2}  + m_2 l_1 l_2  \ddot{ \theta_2} +m_2 l_1 l_2 ( \theta_1 - \theta_2 )  \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}\dot{\theta_2}- \dot{\theta_2}^2 ) m_2 (\theta_1-\theta_2 )  + m_2 l_1 l_2  \ddot{ \theta_2} +m_2 l_1 l_2 \theta_1 \dot{ \theta_1} \dot{ \theta_2} - m_2 l_1 l_2 \theta_2  \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} -  \theta_1 \dot{\theta_1 } \dot{\theta_2} m_2 l_1 l_2 + \dot{\theta_1} \dot{\theta_2}\theta_2 m_2 l_1 l_2 + \dot{\theta_2}^2 \theta_1 m_1 l_1 l_2 - \dot{\theta_2}^2 \theta_2 m_2 l_1 l_2   + m_2 l_1 l_2  \ddot{ \theta_2} +m_2 l_1 l_2 \theta_1 \dot{ \theta_1} \dot{ \theta_2} - m_2 l_1 l_2 \theta_2  \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} + \dot{\theta_2}^2 \theta_1 m_1 l_1 l_2 - \dot{\theta_2}^2 \theta_2 m_2 l_1 l_2   + m_2 l_1 l_2  \ddot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    Ignoring none linear terms.

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} + m_2 l_1 l_2  \ddot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

     \displaystyle \frac{\partial \mathcal{ L}}{\partial \theta_2} = m_2 l_1 l_2 \sin ( \theta_1 - \theta_2 ) \dot{\theta_1} \dot{ \theta_2} - m_2 g l_2 \sin \theta_2

     \displaystyle \approx m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{\theta_1} \dot{ \theta_2} - m_2 g l_2 \theta_2

    \displaystyle \frac{\partial \mathcal{ L}}{\partial \dot{\theta_2}}= m_2 l_2^2 \dot{\theta_2 } + m_2 l_1 l_2 \cos ( \theta_1 - \theta_2 ) \dot{ \theta_1}

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}}\right) = m_2 l_2^2 \ddot{\theta_2} - (\dot{\theta_1} - \dot{\theta_2} )m_2 l_1 l_2 \sin ( \theta_1 - \theta_2 ) \dot{ \theta_1} + m_2 l_1 l_2 \cos (\theta_1 - \theta_2 ) \ddot{ \theta_1}

    \displaystyle \approx m_2 l_2^2 \ddot{\theta_2} - (\dot{\theta_1} - \dot{\theta_2} )m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{ \theta_1} + m_2 l_1 l_2 \ddot{ \theta_1}

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_2}}\right) - \frac{\partial \mathcal{ L}}{\partial \theta_2} = 0

     \displaystyle m_2 l_2^2 \ddot{\theta_2} - (\dot{\theta_1} - \dot{\theta_2} )m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{ \theta_1} + m_2 l_1 l_2 \ddot{ \theta_1} -m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{\theta_1} \dot{ \theta_2} + m_2 g l_2 \theta_2 =0

    Ignoring non linear terms.

     m_2l_2^2 \ddot{\theta_2} + m_2 l_1 l_2 \ddot{theta_1} = -m_2gl_2 \theta_2

    We have

    \displaystyle {l_1} ^2(m_1 +m_2) \ddot{\theta_1} + l_1 l_2 m_2 \ddot{\theta_2} = gl_1 (m_1 +m_2 ) \theta_1

    and

     m_2l_2^2 \ddot{\theta_2} + m_2 l_1 l_2 \ddot{theta_1} = -m_2gl_2 \theta_2

    So;

    M \ddot{\Theta} = -K \Theta where,

    M= \begin{pmatrix} {l_1} ^2(m_1 +m_2) & l_1 l_2 m_2 \\l_1 l_2 m_2 & {l_2} ^2 m_2 \end{pmatrix}

     K = \begin{pmatrix} gl_1 (m_1 +m_2) & 0 \\ 0 & gl_2 m_2  \end{pmatrix}

    and  \Theta = \begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix}
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    (Original post by bananarama2)
    Solution 58

     x_1 = l_1 \sin \theta_1

     \dot{ x_1 } = l_1 \dot{ \theta_1} \cos \theta_1

     y_1 = l_1 \cos \theta_1

     \dot{ y_1} = -l_1 dot{ \theta_1} \sin \theta_1

     x_2 = l_1 \sin \theta_1 + l_2 \sin \theta_2

     \dot {x_2} = l_1 \dot{ \theta_1 } \cos \theta_1 + l_2 \dot{ \theta_2} \cos \theta_2

     y_2 = l_1 \cos \theta_1 + l_2 \cos \theta_2

     \dot {y_2} = -l_1 \dot{ \theta_1 } \sin \theta_1 - l_2 \dot{ \theta_2} \sin \theta_2


    \displaystyle \mathcal{ L} = \frac{1}{2} m_1 l_1^2 \dot{ \theta_1}^2 + \frac{1}{2} m_2 \left( l_1 \dot{ \theta_1 } \cos \theta_1 + l_2 \dot{ \theta_2} \cos \theta_2  \right)^2 + \frac{1}{2}m_2 \left(-l_1 \dot{ \theta_1 } \sin \theta_1 - l_2 \dot{ \theta_2} \sin \theta_2 \right)^2 + m_1 g l_1 \cos \theta_1 + m_2 g ( l_1 \cos \theta_1 + l_2 \cos \theta_2 )

    \displaystyle  = \frac{1}{2} m_1 l_1^2 \dot{ \theta_1}^2  + \frac{1}{2}m_2 l_1^2 \dot{\theta_1}^2 + \frac{1}{2}m_2l_2^2 \dot{\theta_2}^2 + m_2 l_1 l_2 \cos (\theta_1 - \theta_2 ) \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1 \cos \theta_1 + m_2 g ( l_1 \cos \theta_1 + l_2 \cos \theta_2 )

    \displaystyle \frac{\partial \mathcal{ L}}{\partial \theta_1}= -m_2 l_1 l_2 \sin ( \theta_1 - \theta_2 )  \dot{ \theta_1} \dot{ \theta_2} - m_1 g l_1 \sin \theta_1 -m_2 g l_1 \sin \theta_1

    \displaystyle \approx -m_2 l_1 l_2 ( \theta_1 - \theta_2 )  \dot{ \theta_1} \dot{ \theta_2} - m_1 g l_1  \theta_1 -m_2 g l_1 \theta_1

     \displaystyle \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}} =m_1 l_2^2 \dot{ \theta_1} + m_2 l_1^2 \dot{ \theta_1} +  m_2 l_1 l_2 \cos ( \theta_1 - \theta_2 ) \dot{ \theta_2 }

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}}\right) = m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}- \dot{\theta_2} ) m_2 \sin (\theta_1-\theta_2 ) \dot{\theta_2} + m_2 l_1 l_2 \cos ( \theta_1 - \theta_2 ) \ddot{ \theta_2}

    \displaystyle \approx m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}- \dot{\theta_2} ) m_2 (\theta_1-\theta_2 ) \dot{\theta_2} + m_2 l_1 l_2  \ddot{ \theta_2}

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}}\right) - \frac{\partial \mathcal{ L}}{\partial \theta_1} = 0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}- \dot{\theta_2} ) m_2 (\theta_1-\theta_2 )\dot{\theta_2}  + m_2 l_1 l_2  \ddot{ \theta_2} +m_2 l_1 l_2 ( \theta_1 - \theta_2 )  \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} - (\dot{\theta_1}\dot{\theta_2}- \dot{\theta_2}^2 ) m_2 (\theta_1-\theta_2 )  + m_2 l_1 l_2  \ddot{ \theta_2} +m_2 l_1 l_2 \theta_1 \dot{ \theta_1} \dot{ \theta_2} - m_2 l_1 l_2 \theta_2  \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} -  \theta_1 \dot{\theta_1 } \dot{\theta_2} m_2 l_1 l_2 + \dot{\theta_1} \dot{\theta_2}\theta_2 m_2 l_1 l_2 + \dot{\theta_2}^2 \theta_1 m_1 l_1 l_2 - \dot{\theta_2}^2 \theta_2 m_2 l_1 l_2   + m_2 l_1 l_2  \ddot{ \theta_2} +m_2 l_1 l_2 \theta_1 \dot{ \theta_1} \dot{ \theta_2} - m_2 l_1 l_2 \theta_2  \dot{ \theta_1} \dot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} + \dot{\theta_2}^2 \theta_1 m_1 l_1 l_2 - \dot{\theta_2}^2 \theta_2 m_2 l_1 l_2   + m_2 l_1 l_2  \ddot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

    Ignoring none linear terms.

    \displaystyle m_1 l_2^2 \ddot{ \theta_1} + m_2 l_1^2 \ddot{ \theta_1} + m_2 l_1 l_2  \ddot{ \theta_2} + m_1 g l_1  \theta_1 +m_2 g l_1 \theta_1 =0

     \displaystyle \frac{\partial \mathcal{ L}}{\partial \theta_2} = m_2 l_1 l_2 \sin ( \theta_1 - \theta_2 ) \dot{\theta_1} \dot{ \theta_2} - m_2 g l_2 \sin \theta_2

     \displaystyle \approx m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{\theta_1} \dot{ \theta_2} - m_2 g l_2 \theta_2

    \displaystyle \frac{\partial \mathcal{ L}}{\partial \dot{\theta_2}}= m_2 l_2^2 \dot{\theta_2 } + m_2 l_1 l_2 \cos ( \theta_1 - \theta_2 ) \dot{ \theta_1}

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_1}}\right) = m_2 l_2^2 \ddot{\theta_2} - (\dot{\theta_1} - \dot{\theta_2} )m_2 l_1 l_2 \sin ( \theta_1 - \theta_2 ) \dot{ \theta_1} + m_2 l_1 l_2 \cos (\theta_1 - \theta_2 ) \ddot{ \theta_1}

    \displaystyle \approx m_2 l_2^2 \ddot{\theta_2} - (\dot{\theta_1} - \dot{\theta_2} )m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{ \theta_1} + m_2 l_1 l_2 \ddot{ \theta_1}

    \displaystyle \frac{d}{dt} \left( \frac{\partial \mathcal{ L}}{\partial \dot{\theta_2}}\right) - \frac{\partial \mathcal{ L}}{\partial \theta_2} = 0

     \displaystyle m_2 l_2^2 \ddot{\theta_2} - (\dot{\theta_1} - \dot{\theta_2} )m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{ \theta_1} + m_2 l_1 l_2 \ddot{ \theta_1} -m_2 l_1 l_2 ( \theta_1 - \theta_2 ) \dot{\theta_1} \dot{ \theta_2} + m_2 g l_2 \theta_2 =0

    Ignoring non linear terms.

     m_2l_2^2 \ddot{\theta_2} + m_2 l_1 l_2 \ddot{theta_1} = -m_2gl_2 \theta_2

    We have

    \displaystyle {l_1} ^2(m_1 +m_2) \ddot{\theta_1} + l_1 l_2 m_2 \ddot{\theta_2} = gl_1 (m_1 +m_2 ) \theta_1

    and

     m_2l_2^2 \ddot{\theta_2} + m_2 l_1 l_2 \ddot{theta_1} = -m_2gl_2 \theta_2

    So;

    M \ddot{\Theta} = -K \Theta where,

    M= \begin{pmatrix} {l_1} ^2(m_1 +m_2) & l_1 l_2 m_2 \\l_1 l_2 m_2 & {l_2} ^2 m_2 \end{pmatrix}

     K = \begin{pmatrix} gl_1 (m_1 +m_2) & 0 \\ 0 & gl_2 m_2  \end{pmatrix}

    and  \Theta = \begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix}
    Beaten was planning to type this later
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    (Original post by cpdavis)
    Beaten was planning to type this later
    Sorry Although I think I may have done you a favor, it took and hour and a half to type that. :pierre:
 
 
 
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