User:darkpulsaromega/Sandbox

= Math Sandbox =

Wikipedia: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 \!

= Finding the Conservation of Relative Speed =

== 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} \!