User talk:Nfarley88

$$Q = mc\Delta \theta $$

$$m = 150; c = 4.2 \times 10^{-3}; \Delta \theta = 100 - 19 = 81$$

$$Q = 150\times 10^{-3} \times 4.2 \times 10^3 \times 81$$

And now for the hard stuff:

$$\int e^{4x} cos(x)$$

$$\Re \{ \int e^{4x} e^{ix} \}$$

$$\Re \{ \int e^{(4+i)x} \}$$

$$\Re \{ \frac 1 {4+i} e^{(4+i)x} + C \}$$

$$\Re \{ \frac {4-i} {17} e^{4x} e^{ix} + C \}$$

$$\Re \{ \frac {4-i} {17} e^{4x} {cos(x) + i sin(x)} + C \}$$

$$\Re \{ \frac {1} {17} e^{4x} (4 cos(x) + sin(x)) + \frac {-i} {17} e^{4x} (cos(x) + 4 sin(x)) + C \}$$

$$\frac {1} {17} e^{4x} (4 cos(x) + sin(x)) + C $$

Time for the Lagrangian. Sorry, I can't draw the diagram yet, I'll get to that.

$$ T = \frac 1 2 m_1 v_1^2 + \frac 1 2 m_2 v_2^2$$

$$ = \frac 1 2 [ m_1(\frac 1 2 l_1 \dot{\theta})^2 + m_2(\frac 1 2 l_2 \dot{\psi})^2]$$

$$ = \frac 1 8 [ m_1 l_1^2 \dot{\theta}^2 + m_2 l_2^2 \dot{\psi}^2]$$

$$V = m_1 g h_1 + m_2 g h_2 = \frac 1 2 g [m_1 l_1 c_\theta + m_2(2l_1 c_\theta + l_2 c_\psi)]$$Note the different definition of $$h$$ to the norm. It makes life a lot easier.

$$ L = T - V$$

$$L = \frac 1 8 [ m_1 l_1^2 \dot{\theta}^2 + m_2 l_2^2 \dot{\psi}^2] - \frac 1 2 g [m_1 l_1 c_\theta + m_2(2l_1 c_\theta + l_2 c_\psi)]$$

$$ = m_1 \left[ \frac 1 8 l_1^2 \dot{\theta}^2 - \frac 1 2 g l_1 c_\theta \right] + m_2 \left[ \frac 1 8 l_2^2 \dot{\psi}^2 + 2 l_1 c_\theta + l_2 c_\psi \right]$$

$$ = \frac 1 8 m_1 l_1^2 \dot{\theta}^2 + c_\theta \left(-\frac 1 2 m_1 g l_1 + 2 l_1 m_2 \right) + m_2 \left[ \frac 1 8 l_2^2 \dot{\psi}^2 + l_2 c_\psi \right]$$ After that, $$\partial$$iff should be easy!

It should be noted, I got confused with Wolfram, it uses Newtonian mechanics with (x,y) --> (r,$$\theta$$). How unhelpful.

Magic formula time!

$$ \frac d {dt} \left( \frac {\partial L} {\partial \dot{\theta}} \right) = \frac {\partial L} {\partial \theta}$$

$$\frac {\partial L} {\partial \dot{\theta}} = \frac {\partial} {\partial \dot{\theta}} \left( \frac 1 8 m_1 l_1^2 \dot{\theta}^2 + c_\theta \left(-\frac 1 2 m_1 g l_1 + 2 l_1 m_2 \right) + C \right) $$ That $$C$$ is all variables with $$\psi$$ in.

$$ = \frac 1 4 m_1 l_1^2 \dot{\theta}$$

It is important to note that $$\theta \not\equiv \dot{\theta} \Rightarrow \theta = const \Rightarrow \frac {\partial} {\partial \dot{\theta}} c_\theta = 0$$

$$\frac d {dt} \left( \frac {\partial L} {\partial \dot{\theta}} \right) = \frac 1 4 m_1 l_1^2 \ddot{\theta}$$

For $$\psi$$

$$\frac {\partial L} {\partial \dot{\psi}} = \frac {\partial} {\partial \dot{\psi}} \left( C + m_2 \left[ \frac 1 8 l_2^2 \dot{\psi}^2 + l_2 c_\psi \right] \right)$$ With the $$C$$ as before but vice versa

$$ = \frac 1 4 m_2 l_2^2 \dot{\psi}$$

$$\frac d {dt} \left( \frac {\partial L} {\partial \dot{\psi}} \right) = \frac 1 4 m_1 l_1^2 \ddot{\psi}$$

So what now?

Now, for Electro

$$ B = \frac {\mu_0 I} {2 \pi} \frac 1 r$$

$$ B = \frac {\mu_0} {2\pi} \int\limits_{l+w}^{w} \frac { J dr} r$$