User:Darkpulsaromega/Sandbox

= Math Sandbox = FORMULA

Calculus

 * $$f'(c) = \lim_{h \to 0}\frac{f(c + h) - f(c)}{h}$$
 * $$\int_{a}^{b}f(x)dx = \lim_{n \to \infty}\sum_{i=1}^n f(x_i^*)\Delta x$$

Physics

 * $$v = \sqrt{\frac{F}{p}}$$
 * $$v = f\lambda\!$$

Math Problem Solution

 * $$F = \frac{ -G m_E m_O }{ r_E^2 } \!$$


 * $$F = m_O g \!$$


 * $$G m_E = r_E^2 g \!$$


 * $$T_M = 2 \pi \sqrt{ \frac{ r }{ a_M } } \!$$


 * $$4 \pi^2 r = T_M^2 a_M \!$$


 * $$4 \pi^2 r = T_M^2 \frac{ G m_E }{ r^2 } \!$$


 * $$4 \pi^2 r^3 = T_M^2 R_E^2 g \!$$


 * $$r = \left ( \frac{ T_M R_E }{ 2 \pi } \right )^{ \frac{ 2 }{ 3 } } g^{ \frac{ 1 }{ 3 } } \!$$


 * $$F = m_M a_M \!$$


 * $$F = \frac{ -G m_E m_M }{ r^2 } \!$$


 * $$a_M = \frac{ G M_E }{ r^2 } \!$$


 * $$\vec F_{s,max} = -\mu_s \cdot |\vec F_N| \cdot \hat F_{ap} \!$$


 * $$\mu_s = \frac{ m_T }{ m_b } \!$$


 * $$\vec F_k = -\mu_k \cdot |\vec F_N| \cdot \hat v \!$$


 * $$\mu_k = \frac{ m_T }{ m_b } - \frac{ a }{ g } \cdot (1 + \frac{ m_T }{ m_b }) \!$$


 * $$\mu_s = \frac{ 1 }{ \sqrt{ (L / h_{max})^2 - 1 } } \!$$


 * $$\mu_k = \frac{ 1 }{ \sqrt{ (L / h_{max})^2 - 1 } } \cdot (1 - \frac{ a }{ g } \frac{ L }{ h }) \!$$


 * $$\mu$$


 * $$\frac{\tan{x}}{\sec{x}-cos{x}}$$


 * $$\frac{\sin{x}}{\cos{x}(\sec{x} - \cos{x})}$$


 * $$\frac{\sin{x}}{1 - \cos^2{x}}$$


 * $$\frac{\sin{x}}{\sin^2{x}}$$


 * $$\frac{1}{\sin{x}} = \csc{x}$$


 * $$V = l w h\!$$


 * $$\Delta V = \sqrt{ (w h \Delta l)^2 + (l h \Delta w)^2 + (l w \Delta h)^2 } \!$$


 * $$V = \pi (\frac{ d }{ 2 })^2 h \!$$


 * $$V = \pi r^2 h\!$$


 * $$\Delta V = \sqrt{ (\frac{ \pi h }{ 2 } d \Delta d)^2 + (\frac{ d }{ 2 }^2 \pi \Delta h)^2 } \!$$


 * $$f = f(x_1, x_2, ..., x_n) \!$$


 * $$\Delta f = \sqrt{ \sum_{i=1}^n \left (\frac{ \partial f }{ \partial x_i } \Delta x_i \right )^2 } \!$$


 * $$t = \sqrt{ \frac{ 2d }{ g } } \!$$


 * $$P = 0.67 \frac{ \sqrt{ \frac{ 1 }{ n-1 } \sum_{i=1}^n (x_i - \overline{ x })^2 } }{ \sqrt{ n } } \!$$


 * $$v(t) = \frac{ x(t) - x(t - \Delta t) }{ \Delta t } \!$$


 * $$rv(t) = \frac{ x(t + \Delta t) - x(t) }{ \Delta t } \!$$


 * $$a(t) = \frac{ rv(t) - v(t) }{ \Delta t } \!$$


 * $$a(t) = \frac{ \frac{ x(t + \Delta t) - x(t) }{ \Delta t } - \frac{ x(t) - x(t - \Delta t) }{ \Delta t } }{ \Delta t } \!$$


 * $$a(t) = \frac{ x(t - \Delta t) - 2x(t) + x(t + \Delta t) }{ \Delta t^2 } \!$$


 * $$\vec V_x = \left | \vec V \right | \cos(\vec V_{angle}) \!$$


 * $$\vec V_y = \left | \vec V \right | \sin(\vec V_{angle}) \!$$


 * $$ \sqrt{ \vec V_x^2 + \vec V_y^2 } \!$$ @ $$\arctan{ \left( \frac{ \vec V_y }{ \vec V_x } \right) } \!$$


 * $$ \left| \vec C \right| = \sqrt{ \left| \vec A \right|^2 + \left| \vec B \right|^2 - 2 \left| \vec A \right| \left| \vec B \right| \cos( \vec B_{angle} ) } \!$$


 * $$ \vec C_{angle} = \arccos \left( \frac{ \left| \vec C \right|^2 + \left| \vec A \right|^2 - \left| \vec B \right|^2 }{ 2 \left| \vec C \right| \left| \vec A \right| } \right) + \vec A_{angle} \!$$


 * $$ \vec B_{angle} = 180 - \arccos \left( \frac{ \left| \vec B \right|^2 + \left| \vec A \right|^2 - \left| \vec C \right|^2 }{ 2 \left| \vec B \right| \left| \vec A \right| } \right) + \vec A_{angle} \!$$


 * $$ \beta \gamma \theta \Phi $$

Conservation of Momentum

 * $$ m_A v_{A0} + m_B v_{B0} = m_A v_{AF} + m_B v_{BF} \!$$

Conservation of Kinetic Energy

 * $$ \frac{ 1 }{ 2 } m_A v_{A0}^2 + \frac{ 1 }{ 2 } m_B v_{B0}^2 = \frac{ 1 }{ 2 } m_A v_{AF}^2 + \frac{ 1 }{ 2 } m_B v_{BF}^2 \!$$

Rearranging the Conservation of Momentum

 * $$ m_A v_{A0} - m_A v_{AF} = m_B v_{BF} - m_B v_{B0} \!$$


 * $$ m_A (v_{A0} - v_{AF}) = m_B (v_{BF} - v_{B0}) \!$$

Rearranging the Conservation of Kinetic Energy

 * $$ \frac{ 1 }{ 2 } m_A v_{A0}^2 - \frac{ 1 }{ 2 } m_A v_{AF}^2 = \frac{ 1 }{ 2 } m_B v_{BF}^2 - \frac{ 1 }{ 2 } m_B v_{B0}^2 \!$$


 * $$ m_A (v_{A0}^2 - v_{AF}^2) = m_B (v_{BF}^2 - v_{B0}^2) \!$$

Because:
 * $$ (v_{A0}^2 - v_{AF}^2) = (v_{A0} + v_{AF}) (v_{A0} - v_{AF}) \!$$

And:
 * $$ (v_{BF}^2 - v_{B0}^2) = (v_{BF} + v_{B0}) (v_{BF} - v_{B0}) \!$$

Therefore:
 * $$ m_A (v_{A0} + v_{AF}) (v_{A0} - v_{AF}) = m_B (v_{BF} + v_{B0}) (v_{BF} - v_{B0}) \!$$

Dividing the two equations

 * $$ \frac{ m_A (v_{A0} + v_{AF}) (v_{A0} - v_{AF}) }{ m_A (v_{A0} - v_{AF}) } = \frac{ m_B (v_{BF} + v_{B0}) (v_{BF} - v_{B0}) }{

m_B (v_{BF} - v_{B0}) } \!$$


 * $$ v_{A0} + v_{AF} = v_{BF} + v_{B0} \!$$


 * $$ v_{A0} - v_{B0} = v_{BF} - v_{AF} \!$$