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The '''fundamental [[frequency]]''' (also called a '''natural frequency''') of a [[periodic signal|periodic]] [[Signal (information theory)|signal]] is the inverse of the [[Pitch (music)|pitch]] [[period (music)|period]] length. The pitch period is, in turn, the smallest repeating unit of a signal. One pitch period thus describes the periodic signal completely. The significance of defining the pitch period as the ''smallest'' repeating unit can be appreciated by noting that two or more concatenated pitch periods form a repeating pattern in the signal. However, the concatenated signal unit obviously contains redundant information.
A 'fundamental bass' is the [[root (chord)|root note]], or lowest note or pitch in a chord or sonority when that chord is in root position or [[normal form]].
In terms of a superposition of [[sinusoid]]s (for example, [[fourier series]]), the fundamental frequency is the lowest frequency sinusoidal in the sum.
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a perfect bounce with an incidence angle of <math>\theta</math> leads to<math>\Delta V = 2 v \sin\theta\,</math>
the frequency is also given by <math>\frac{\Delta
the force is <math>F_{out}=F_x+F_y+F_z=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dF_x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dF_y+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dF_z</math>
<math>=\sum_{axis=x}^z\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dA \times\sigma \times\Delta v\times frequency</math>
<math>=\sum_{axis=x}^z\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2\pi r^2\cos\theta d\theta)(\frac{m}{
<math>=\sum_{axis=x}^z\frac{mv^2}{l_{axis}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta\cos\theta d\theta</math>
Line 97:
<math>=\frac{1}{3}mv^2\Rightarrow \frac{3}{2}PV=\frac{1}{2}mv^2=\mathrm{K}\mathrm{E}</math>
==trigonometry==
<math>R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C}</math>
<math>H_a = \frac{bc}{2R} = \sin C b = \sin B c</math>
<math>\sin A = \frac{a}{2R} = \sqrt{1-\cos^2}</math>
<math>\cos A = \frac{b^2+c^2-a^2}{2bc}</math>
<math>a^2 = b^2+ c^2-2bc\cos A\,</math>
<math>\Delta = \frac{1}{2} ab\sin C =...=\frac{abc}{4R}=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}= \frac{a+b+c}{2}\times r =2 R^2 \sin A\sin B\sin C</math>
<math>\tan \frac{A}{2}= \frac{2r}{b+c-a}</math>
<math>r=\cfrac{(b+c-a)\tan \frac{A}{2}}{2} = \frac{2\Delta}{a+b+c}</math>
<math>\Delta = \frac{1}{2} ab\sin C = \frac{ab}{2} \sqrt{1-\cos ^2 C} = \frac{ab}{2} \sqrt{\left (1-\frac{a^2+b^2+c^2}{2ab}\right )\left ( 1+\frac{a^2+b^2+c^2}{2ab}\right )}= \frac{1}{4} \sqrt{\left (\left (a+b\right )^2-c^2\right )\left (c^2-\left (a-b\right )^2\right )}= \sqrt{\left (a+b-c\right )\left (a+b+c\right )\left (c+b-a\right )\left (c+a-b\right )}=\sqrt{s\left (s-a\right )\left (s-b\right )\left (s-c\right )}</math>
==calculus==
<math>\frac{d(uv)}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}</math>
<math>Insert formula here</math>
==pi==
*The area of the [[unit disc]]:
::<math>2\int_{-1}^1 \sqrt{1-x^2}\,dx = \pi</math>
*Half the [[circumference]] of the [[unit circle]]:
::<math>\int_{-1}^1\frac{dx}{\sqrt{1-x^2}} = \pi</math>
*[[François Viète]], 1593 ([[Viète formula|proof]]):
::<math>\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots = \frac2\pi</math>
*[[Gottfried Leibniz|Leibniz]]' formula ([[Leibniz formula for pi|proof]]):
::<math>\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \frac{\pi}{4}</math>
*[[John Wallis|Wallis]] product, 1655 ([[Wallis product|proof]]):
::<math> \prod_{n=1}^{\infty} \left ( \frac{n+1}{n} \right )^{(-1)^{n-1}} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2} </math>
*Faster product (see Sondow, 2005 and [http://home.earthlink.net/~jsondow/ Sondow web page])
::<math> \left ( \frac{2}{1} \right )^{1/2} \left (\frac{2^2}{1 \cdot 3} \right )^{1/4} \left (\frac{2^3 \cdot 4}{1 \cdot 3^3} \right )^{1/8} \left (\frac{2^4 \cdot 4^4}{1 \cdot 3^6 \cdot 5} \right )^{1/16} \cdots = \frac{\pi}{2} </math>
:where the ''n''th factor is the 2<sup>''n''</sup>th root of the product
::<math>\prod_{k=0}^n (k+1)^{(-1)^{k+1}{n \choose k}}.</math>
*Symmetric formula (see Sondow, 1997)
::<math> \frac {\displaystyle \prod_{n=1}^{\infty} \left (1 + \frac{1}{4n^2-1} \right )}{\displaystyle\sum_{n=1}^{\infty} \frac {1}{4n^2-1}} = \frac {\displaystyle\left (1 + \frac{1}{3} \right ) \left (1 + \frac{1}{15} \right ) \left (1 + \frac{1}{35} \right ) \cdots} {\displaystyle \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \cdots} = \pi </math>
*[[Bailey-Borwein-Plouffe formula|Bailey-Borwein-Plouffe]] algorithm (See Bailey, 1997 and [http://www.nersc.gov/~dhbailey/ Bailey web page])
::<math>\sum_{k=0}^\infty\frac{1}{16^k}\left(\frac {4}{8k+1} - \frac {2}{8k+4} - \frac {1}{8k+5} - \frac {1}{8k+6}\right) = \pi</math>
* [[Pafnuty Chebyshev|Chebyshev]] series [http://www.jstor.org/journals/08916837.html Y. Luke, Math. Tabl. Aids Comp. 11 (1957) 16]
::<math>\sum_{k=0}^\infty\frac{(-1)^k(\sqrt{2}-1)^{2k+1}}{2k+1} = \frac{\pi}{8}.</math>
::<math>\sum_{k=0}^\infty\frac{(-1)^k(2-\sqrt{3})^{2k+1}}{2k+1}=\frac{\pi}{12}.</math>
*An [[integral]] formula from [[calculus]] (see also [[Error function]] and [[Normal distribution]]):
::<math>\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}</math>
*[[Basel problem]], first solved by [[Leonhard Euler|Euler]] (see also [[Riemann zeta function]]):
::<math>\zeta(2)= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}</math>
::<math>\zeta(4)= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90}</math>
::and generally, <math>\zeta(2n)</math> is a rational multiple of <math>\pi^{2n}</math> for positive integer n
*[[Gamma function]] evaluated at <sup>1</sup>/<sub>2</sub>:
::<math>\Gamma\left({1 \over 2}\right)=\sqrt{\pi}</math>
*[[Stirling's approximation]]:
::<math>n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n</math>
*[[Euler's identity]] (called by [[Richard Feynman]] "the most remarkable formula in mathematics"):
::<math>e^{i \pi} + 1 = 0\;</math>
*A property of [[Euler's totient function]] (see also [[Farey sequence]]):
::<math>\sum_{k=1}^{n} \phi (k) \sim \frac{3n^2}{\pi^2}</math>
*An application of the [[residue theorem]]
::<math>\oint\frac{dz}{z}=2\pi i ,</math>
:where the path of integration is a closed curve around the origin, traversed in the standard anticlockwise direction.
==scribbles==
<math>
\frac{\rho _{x2} -\rho _{x1}}{\rho _N}\times \mathbf{P} _N\times \alpha = \alpha \mathbf{V} \rho _N \times \mathbf{V}
\left (
\frac{\rho_N}{\rho_{x2}}-\frac{\rho_N}{\rho_{x2}}\right )</math>
<math>\rho va\frac{dX_2-dX_1}{dt}</math>
<math>v\left (\frac{\rho_0}{\rho_2}-\frac{\rho_0}{\rho_1}\right )=\rho v^2 a \frac{\rho_1-\rho_2}{\rho_2\rho_1}</math>
<math>\rho_1-\rho_2</math>
<math>\begin{align}
mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \\
& =\frac{1}{2}mv^2+\frac{1}{2}I\frac{v^2}{R^2} \\
& =v^2\left ( \frac{m}{2} + \frac{I}{R^2} \right ) \\
I_ball=\int_0^\pi r^2 dm \\
& =\int_0^\pi\left [ ((R\sin \theta)^2 \times \pi )\times (Rd\theta\times\sin \theta)\times \rho\right ]\times\left [ R^2\sin^2\theta \right ] \\
& (\because \rho = \cfrac{m}{\cfrac{4}{3}mR^3}) \\
& =\frac{3}{4}\int_0^\pi mR^2 \sin^5
\end{align}</math>
<math>\rho\int_0^\pi 4R^3\cos^2\theta\sin\theta d\theta \times (R\sin\theta)^2</math>
<math>\begin{align}
& \int_0^{\frac{\pi}{2}}
\rho \times
\left (
4R^2 \pi \sin \theta \cos \theta \times Rd\theta\cos \theta
\right )
\times (R\sin \theta)^2 \\
& \left( \because \rho = \frac{3m}{4\pi R^3} \right ) \\
& =\frac{3m}{4\pi R^3} \times 4 R^5\pi
\int_0^{\frac{\pi}{2}}
\cos^2 \theta \sin^3 \theta d\theta \\
& =3mR^2
\int_0^{\frac{\pi}{2}}
\cos^2\theta\left (1-\cos^2 \theta \right )\sin\theta d\theta \\
& =3mR^2
\left (
\int_0^{\frac{\pi}{2}} \cos^2\theta\sin\theta d\theta -
\int_0^{\frac{\pi}{2}} \cos^4\theta\sin\theta d\theta
\right ) \\
& =3mR^2
\left (
(-\frac{1}{3} \cos^3 \theta \Bigg|_0^{\frac{\pi}{2}}) -
(-\frac{1}{5} \cos^5 \theta \Bigg|_0^{\frac{\pi}{2}})
\right ) \\
& =3mR^2\left (\frac{1}{3}-\frac{1}{5}\right ) \\
& =\frac{2}{5}mR^2
\end{align}</math>
<math>\begin{align}
mgh=KE+RE \\
& =\frac{1}{2}mv^2 + \frac{1}{2} I\omega^2 \\
& =\frac{1}{2}mv^2 + \frac{1}{2} \times \frac{2}{5}mR^2\frac{v^2}{r^2} \\
& =\frac{1}{2}mv^2 +\frac{1}{5}mv^2 \\
& =\frac{7}{10}mv^2 \\
v = \sqrt{\frac{10}{7}gh}
\end{align}</math>
<math>\frac{q}{m}=\frac{2V}{r^2B^2}\Rightarrow r\varpropto \sqrt{\cfrac{m}{q}}</math>
<math>\vec v</math>
<math>\vec v_{tangent}</math>
<math>\vec F
=\frac{d\vec v_{tangent}}{dt}
=\frac{dB}{dt} \times \frac{d\vec v_{tangent}}{dB}
</math>
<math>=\omega_{B} \times \frac{d(\vec v_{A} \times \cos B)}{dB}
=\frac{v\cos C}{R} \times (-\vec v_{A}\sin B)
=-\frac{v\vec v_A \cos C \sin B}{R}</math>
<math>k = \frac{1 + \sqrt{5}}{2} or \frac{1 - \sqrt{5}}{2}</math>
<math>\Phi</math>
<math>\times\,</math><math>\vec v\times\vec B</math>
<math>\vec a\times\vec b = \vec n|a||b|\sin\theta</math>
<math>\vec a\perp\vec n \land \vec b\perp\vec n</math>
<math>V=n\frac{d\Phi}{dt}</math>
<math>a^4-ka^2-\frac{p^2}{4}</math>
<math>a^2=\frac{k\plusmn\sqrt{k^2+p^2}}{2}=\frac{k\plusmn z}{2}</math>
<math>a=\plusmn\sqrt{\frac{k+z}{2}}</math>
<math>\begin{align}
& a=\plusmn (n+1) \\
& b=\frac{p}{2a}=\frac{2n^2+2n}{\plusmn 2(n+1)}= \plusmn n
\end{align}</math>
==scrib 2==
<math>\triangle</math>
<math>\begin{align}
& \vec {PA}+\vec {PB}+\vec {PC} =\vec {BC} \\
& \vec {QA}+\vec {QB}+\vec {QC} =\vec {CA} \\
& \vec {RA}+\vec {RB}+\vec {RC} =\vec {AB}
\end{align}</math>
<math>\begin{align}
& \vec P-\vec A+\vec P-\vec B+\vec P-\vec C=\vec B-\vec C \\
& 3\vec P-(\vec A+\vec B+\vec C)=\vec B-\vec C \\
& 3\vec P = \vec A +2\vec B
\end{align}</math>
<math>\begin{align}
& 4\vec {OA}+\vec {OB}+\vec {OC} =\vec {OA'} \\
& \vec {OA}+4\vec {OB}+\vec {OC} =\vec {OB'} \\
& \vec {OA}+\vec {OB}+4\vec {OC} =\vec {OC'}
\end{align}</math>
<math>\begin{align}
& \vec O-\vec A'=4\vec O-4\vec A+\vec O-\vec B+\vec O-\vec C \cdots (1)\\
& \vec O-\vec B'=\vec O-\vec A+4\vec O-4\vec B+\vec O-\vec C \cdots (2)\\
& \vec O-\vec C'=\vec O-\vec A+\vec O-\vec B+4\vec O-4\vec C \cdots (3)\\
& \\
& (1)-(2)......\vec B'-\vec A'=3\vec B-3\vec A \\
& \vec {B'A'}=3\vec {BA}
\end{align}</math>
<math>v'=\frac{m_1v_1-m_2v_2}{m_1+m_2}</math>
<math>\begin{align} if m_1v_1'+m_2v_2'=0 \\
then m_1(-v_1')+m_2(-v_2')=0
\end{align}</math>
<math>\frac{1}{2}m_1v_1'^2+\frac{1}{2}m_2v_2'^2=\frac{1}{2}m_1(-v_1')^2+\frac{1}{2}m_2(-v_2')^2</math>
<math>\begin{align}
v_1'=v_1-v' \\
v_2'=-v_2-v'
\end{align}</math>
<math>\begin{align}
& y_{c.m.}=\frac{\int_0^H y dm}{m}=\cfrac{\int_0^H y\times \rho (w\times dy)}{\rho \cfrac{WH}{2}} \\
& =\cfrac{\int_0^H y\times \rho (\cfrac{H-y}{H}W\times dy)}{\rho \cfrac{WH}{2}} \\
& =\frac{2}{H}\int_0^H y-\frac{y^2}{H}dy \\
& =\frac{2}{H}\left (\frac{1}{2}y^2\bigg|_0^H-\frac{y^3}{3H}\bigg|_0^H\right ) \\
& =\frac{2}{H}\left (\frac{H^2}{2}-\frac{H^3}{3H}\right ) = \frac{H}{3}
\end{align}</math>
<math>\rho\,</math><math>L\,</math><math>g\,</math> <math>\theta=\arctan\frac{l\rho g}{T}\mbox{ or }\tan\theta=\frac{l\rho g}{T}</math> <math>dl\,</math><math>dl\rho\,</math><math>l\,</math> <math>dl\cos\theta\,</math>
<math>\begin{align}
& \left ( \mbox{let it be that} \arctan \cfrac{0.5l}{\rho g} = \omega\right ) \\
& x=\int_0^{0.5l} \cos (\arctan\frac{0.5l\rho g}{T}) dl \\
& =\int_0^{0.5l} \cos u \frac{dl}{du} du \\
& =\frac{T}{\rho g}\int_0^\omega \cos u \cfrac{dl\times(\rho g)}{\arctan\cfrac{0.5l\rho g}{T}} du\\
& =\frac{T}{\rho g}\int_0^\omega \cos u \frac{1}{\cos^2 u} du \\
& =\frac{T}{\rho g}\int_0^\omega \frac{1}{\cos u} du \\
& =\frac{T}{\rho g} \ln \left (\sec u+\tan u\right )\bigg|_0^\omega=\frac{T}{\rho g} \ln \left (\sec \omega+\tan \omega\right )
\end{align}</math>
<math>{\color{Blue}\vec Q_{BDAF}=\overrightarrow{BD}\times(\frac{m_a+m_c}{m_a+m_b+m_c})}</math>
<math>{\color{Blue}\vec Q_{BDCE}=\overrightarrow{BD}\times(\frac{m_a+m_c}{m_a+m_b+m_c})}</math>
<math>\begin{align}
& \left ( \mbox{let it be that} \arctan \cfrac{0.5l}{\rho g} = \omega\right ) \\
& y=\int_0^{0.5l} \sin (\arctan\frac{0.5l\rho g}{T}) dl \\
& =\int_0^{0.5l} \sin u \frac{dl}{du} du \\
& =\frac{T}{\rho g}\int_0^\omega \sin u \cfrac{dl\times(\rho g)}{\arctan\cfrac{0.5l\rho g}{T}} du\\
& =\frac{T}{\rho g}\int_0^\omega \sin u \frac{1}{\cos^2 u} du \\
& =\frac{T}{\rho g}\int_0^\omega \left (\sec u\tan u \right )du \\
& =\frac{T}{\rho g} \sec u\bigg|_0^\omega = \frac{T}{\rho g}\left(\sec\omega-1\right)
\end{align}</math>
<math>\begin{align}
& \left ( \mbox{let it be that} k = \frac{\rho g}{T} \right) \\
\end{align}</math>
<math>\begin{align}
& \left(\mbox{when }x>0,y>0,\omega>0\right)\\
& kx=\ln(\sec\omega+\tan\omega)\\
& e^{kx}=\sec\omega+\tan\omega=\sec\omega+\sqrt{\sec^2\omega-1}\\
& \left(e^{kx}-\sec\omega\right)^2=\sec^2\omega-1\\
& e^{2kx}-2e^{kx}\sec\omega+\sec^2\omega=\sec^2\omega-1\\
& e^{2kx}+1=2e^{kx}\sec\omega\\
& \sec\omega=\frac{1}{2}\left(e^{kx}+e^{-kx}\right)\\
& \\
& y=\frac{1}{k}\sec\omega-1\\
& =\frac{1}{2}\frac{1}{k}\left(e^{kx}+e^{-kx}-2\right)\\
\end{align}</math>
==scrib3==
<math>\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta</math>
<math>\begin{align}
& \vec a\times\vec b=\vec c \\
& |\vec c|=|a||b|\sin\theta \\
& \vec c\perp\vec b,\vec c\perp\vec a
\end{align}</math>
<math>\begin{align}
\vec A(X_a,Y_a,Z_a) \\
\vec B(X_b,Y_b,Z_b)
\end{align}</math>
<math>\begin{align}
X_a x+Y_a y+Z_a z=0 \\
X_b x+Y_b y+Z_b z=0 \\
\end{align}</math>
<math>a^2+b^2+c^2</math>
<math>\begin{align}
& -X_a k=Y_a y+Z_a z=0 \\
& -X_b k=Y_b y+Z_b z=0 \\
& y=\frac{Z_A X_B-X_A Z_B}{Z_B Y_A-Y_A Z_B}k \\
& z=\frac{X_A Y_B-Y_A X_B}{Z_B Y_A-Y_A Z_B}k \\
& x:y:z=(Y_AZ_B-Z_AY_B):(Z_AX_B-X_AZ_B):(X_AY_B-Y_AX_B)
\end{align}</math>
<math>C(
\det
\begin{pmatrix}
Y_A & Y_B \\
Z_A & Z_B
\end{pmatrix}
,
\det
\begin{pmatrix}
Z_A & Z_B \\
X_A & X_B
\end{pmatrix}
,
\det
\begin{pmatrix}
X_A & X_B \\
Y_A & Y_B
\end{pmatrix}
)</math>
<math>\left(\frac{2}{3}^3\right)\left(\frac{1}{3}^0\right)</math>
<math>\left(\frac{2}{3}^2\right)\left(\frac{1}{3}^1\right)</math>
<math>\left(\frac{2}{3}^1\right)\left(\frac{1}{3}^2\right)</math>
<math>\left(\frac{2}{3}^0\right)\left(\frac{1}{3}^3\right)</math>
==4==
<math>\begin{align}
& \vec A \cdot \vec B = |\vec A||\vec B|\cos\theta
=|\vec A||\vec B|\times \frac{a^2+b^2-c^2}{2ab}
=\frac{a^2+b^2-c^2}{2} \\
& =\frac{(x_A^2+y_A^2 +z_A^2)+(x_B^2+ y_B^2+z_B^2)-(\left[x_A- x_B\right]^2 +\left[y_A-y_B\right]^2 + \left[z_A-z_B\right]^2)}{2} \\
& =x_Ax_B+y_Ay_B+z_Az_B
\end{align}</math>
<math>\vec V_1 </math><math>\vec V_2</math>
<math>\vec V_c=\frac{m_1\vec V_1 +m_2\vec V_2}{m_1+m_2}</math>
<math>\vec V_1 -\vec V_c </math><math>\vec V_2-\vec V_c</math>
<math>\vec U\mbox{ and } |\vec U|=1</math>
<math>\mbox{reflect}
(\vec A, \vec U)=-\vec A + 2\left( \vec A \cdot\vec U\right)\vec U</math>
<math>\mbox{reflect}
(
\vec V_1-\vec V_c , \vec U)=-( \vec V_1 -\vec V_c)+2 \left ( \left [ \vec V_1 -\vec V_c \right ]\cdot \vec U \right ) \vec U
</math>
<math>\mbox{reflect}
(
\vec V_2-\vec V_c , \vec U)=-( \vec V_2 -\vec V_c)+2 \left ( \left [ \vec V_2 -\vec V_c \right ]\cdot \vec U \right ) \vec U
</math>
<math>\vec V_1'=\vec V_c +\mbox{reflect}(\vec V_1 -\vec V_c, \vec U)
=2\vec V_c - \vec V_1+
2\left( \left[\vec V_1 -\vec V_c\right ] \cdot\vec U\right)\vec U</math>
<math>\vec V_2'=\vec V_c +\mbox{reflect}(\vec V_2 -\vec V_c, \vec U)
=2\vec V_c - \vec V_2+
2\left( \left[\vec V_2 -\vec V_c \right ]\cdot\vec U\right)\vec U</math>
<math>\vec V_1'
=2\vec V_c - \vec V_1+2\left( \left[\vec V_1 -\vec V_c \right ]\cdot\vec U\right)\vec U
= \vec V_2 -\left( \vec V_2\cdot\vec U\right)\vec U</math>
<math>\vec V_2'
=2\vec V_c - \vec V_2+2\left( \left[\vec V_2 -\vec V_c \right ]\cdot\vec U\right)\vec U
= \vec V_1 -\left( \vec V_1 \cdot\vec U\right)\vec U</math>
<math>\vec V_1 \| \vec V_2 \perp \vec U</math>
<math>\alpha\, </math><math>\beta\,</math>
<math>4\beta-2\alpha\,</math>
<math>\begin{align}
& \frac{d(4\beta-2\alpha)}{d\sin\alpha} \\
& \left( \because n_{water}\mbox{is 1.332986 and we set sin} \alpha \mbox{ to be } u \right) \\
& \left( n_w \sin \beta = \sin \alpha \mbox{ and so } \sin\beta = \frac{u}{1.332986}\right) \\
& =\frac{d(4\arcsin (0.7501954259 u)-2\arcsin u)}{du} \\
& =\frac{4}{\sqrt{1-(0.750195 u)^2}} \times 0.750195 - \frac{2}{\sqrt{1-u^2}} \\
& \mbox{because we want to find the biggest value so}\\
& =0
\end{align}</math>
<math>\begin{align}
& \frac{4}{\sqrt{1-(0.750195 u)^2}} \times 0.750195 = \frac{2}{ \sqrt{1-u^2}} \\
& \frac{\sqrt{1-(0.750195 u)^2}}{3.000781704} = \frac{\sqrt{1-u^2}}{2} \\
& 0.3332465 - 0.1875486435 u^2 = 0.5 -0.5 u^2 \\
& 0.3124513565 u^2 = 0.1667535 \\
& u = \sqrt{0.5336942744238} = 0.730543821 \\
& \alpha = 46.932003877481 \\
\end{align}</math>
<math>\begin{align}
& \frac{d(4\beta-2\alpha)}{d\sin\alpha} \\
& =\frac{d(4\arcsin (\frac{u}{n})-2\arcsin u)}{du} \\
& =\cfrac{ 4 }{ \sqrt{ 1-(\frac{u}{n})^2 } } \frac{1}{n} - \frac{2}{\sqrt{1-u^2}} \\
& =0
\end{align}</math>
<math>\begin{align}
& \frac{3}{4}u^2=1-\frac{n^2}{4} \\
& u = \sqrt{\frac{4}{3}- \frac{n^2}{3}} \\
\end{align}</math>
<math>\sin \alpha = u = \sqrt{\frac{4}{3}- \frac{n^2}{3}}</math>
<math>4\beta -2 \alpha= 4 \arcsin \frac{u}{n}- 2\arcsin u = 42 ^\circ</math>
==5==
<math>
\begin{align}
& F_1\times S_1 -F_2\times S_2 \\
& =(P_1A_1)(V_1 \Delta \mathbf {t}) -(P_2 A_2 )(V_2 \Delta \mathbf{t}) \\
& =\Delta E = \Delta \mbox{KE} + \Delta \mbox{PE}\\
& =\frac{1}{2} A_2 V_2 \Delta \mathbf{t}\rho V_2^2 - \frac{1}{2} A_1 V_1 \Delta \mathbf{t}\rho V_1^2 + A_2 V_2 \Delta \mathbf{t} \rho \mathbf{g} y_2 - A_1 V_1 \Delta \mathbf{t} \rho \mathbf{g} y_1
\end{align}</math>
<math>\Delta m \propto \mbox{t and } \Delta m \propto V</math>
<math>\frac{\Delta m}{t\cdot V}=r\mbox{ so the unit of r is} \frac{\mbox{mol}}{s\cdot L}</math>
<math>\frac{k}{k'}=\frac{A}{A'}e^{-\frac{(E_a-E_a')}{RT}}=Ce^{-\frac{(\Delta H)}{RT}}</math>
<math>\left[ \mbox{A}\right]\left[ \mbox{B}\right]</math>
<math>\frac{a\times D}{20\mbox{ sec}}\times AB/100,00</math>
<math>r \propto \left[ \mbox{A}\right]\left[ \mbox{B}\right]\left[ \mbox{C}\right]...</math>
<math>r = k \left[ \mbox{A}\right]\left[ \mbox{B}\right]\left[ \mbox{C}\right]...</math>
<math>e^{{-E_a}/{RT}}</math>
<math>r = k \left[ \mbox{A}\right]\left[ \mbox{B}\right]\left[ \mbox{C}\right]... =r' = k' \left[ \mbox{A}'\right]\left[ \mbox{B}'\right]\left[ \mbox{C}'\right]...</math>
<math>k=k'\times W</math>
<math>W=\frac{k}{k'}=\frac{A}{A'}e^{-\frac{(E_a-E_a')}{RT}}=Ce^{-\frac{(\Delta H)}{RT}}</math>
<math>\frac{\Delta H}{T}</math>
<math>v=\sqrt{2gH}</math>
<math>\begin{align}
& t=\frac{\Delta V}{a}=\frac{2v\sin\theta}{g} \\
& d=v_x \times t = (v\cos\theta)\times (\frac{2v\sin\theta}{g}) \\
& =\frac{2v^2\cos\theta\sin\theta}{g}=\frac{v^2\sin2\theta}{g}=2H\sin\theta \\
& d=H \\
& \therefore \sin 2\theta=\frac{1}{2} & \theta=15^\circ ,75^\circ
\end{align}</math>
<math>\frac{(-v\cos p)^2}{2g}= \frac{4\cos ^2 P}{20}=0.2\cos^2 P =0.2(1-\sin^2 P)</math>
<math>0.5+0.5sinP+0.2-0.2\sin^2 P = -0.2(\sin P - 1.25)^2 +0.7+\frac{5}{16}=\frac{81}{80}</math>
<math>a_{tide}=a_{near}-a_{center}=\frac{GM}{(d-r)^2}-\frac{GM}{d^2}=GM\frac{d^2-(d-r)^2}{d^2(d-r)^2}=GM\frac{2dr-r^2}{d^2(d-r)^2}</math>
<math>\lim_{d>>r}a_{tidal}=\frac{2GMr}{d^3}</math>
==sec 6==
<math>s^2=x^2-(ct)^2\,</math>
<math>\begin{align}
& s^2=(vt)^2-(ct)^2
\end{align}</math>
<math>s^2=(0)^2-(cT)^2\,</math>
<math>-(cT)^2=(vt)^2-(ct)^2\,</math>
<math>\frac{T}{t}=\sqrt{1-v^2/c^2}</math>
<math>\frac{X}{x}=\sqrt{1-v^2/c^2}</math>
<math>\sqrt{0.75}</math>
<math>\frac{\overline{AB}}{\overline{OA}}=\sin{\beta}=\frac{v}{c}</math>
<math>\overline{OB}=\overline{OA}\cos\beta=\sqrt{1-v^2/c^2}\overline{OA}</math>
<math>
\begin{array}{c|c||c|c||c|c} &this & picture & ..tikai & made & it \\
\hline
&x&945&219&y&\\
4&4y&876&207&{3x-12y}&3 \\
\hline
&x-4y&69&12&-3x+13y& \\
2&-15x+65y&60&9&16x-69y&1 \\
\hline
&16x-69y&9&3&-19x+82y& \\
3&-57x+246y&9&&& \\
\hline
&9&0&
\end{array}
</math>
<math>\begin{align}
& \Sigma \vec F = 0 \\
& \rightarrow \vec F_1 +\vec F_2+\vec F_3 ...=0 \\
& \vec F_{x1}+\vec F_{x2}+\vec F_{x3}...=0 \\
& \vec F_{y1}+\vec F_{y2}+\vec F_{y3}...=0 \\
& \vec F_{z1}+\vec F_{z2}+\vec F_{z3}...=0 \\
\end{align}</math>
<math>\begin{align}
& \Sigma \tau = 0\\
& \Sigma \tau= \Sigma \vec F \times \vec d = 0\\
\end{align}</math>
<math>\begin{align}
& \Sigma \vec F \times \vec d = 0\\
& \Sigma (F_x,F_y,F_z)\times (d_x,d_y,d_z) \\
& =\Sigma \left((F_yd_z-F_zd_y),(F_zd_x-F_xd_z),(F_xd_y-F_yd_x)\right)= 0\\
\end{align}</math>
<math>\begin{align}
& \Sigma \vec F \times \vec d' = 0\\
& \Sigma (F_x ,F_y,F_z)\times ((d_x+r_x),(d_y+r_y),(d_z+r_z)) \\
& ={\color{Red}\Sigma\left((F_yd_z-F_zd_y),(F_zd_x-F_xd_z),(F_xd_y-F_yd_x)\right)} \\
& \mbox{ }+\Sigma
\left(
{\color{Blue} (F_yr_z-F_zr_y)},{\color{Brown}(F_zr_x-F_xr_z)},{\color{OliveGreen}(F_xr_y-F_yr_x)}
\right)= 0\\
& ={\color{Red}0}+{\color{Blue}(r_z\Sigma F_y,0,0)-(r_y\Sigma F_z,0,0)}
\\ & \mbox{ }+{\color{Brown}(0,r_z\Sigma F_x,0)-(0,r_x\Sigma F_z,0)}
+{\color{OliveGreen}(0,0,r_x\Sigma F_y)-(0,0,r_y\Sigma F_x)} \\
& ={\color{Red}0}+{\color{Blue}0-0}+{\color{Brown}0-0}+{\color{OliveGreen}0-0}=0
\end{align}</math>
<math>\begin{align}
& \because \sin k =\frac{v}{c}\\
& \therefore \cos k =\sqrt{1-(\frac{v}{c})^2}\\
\end{align}</math>
<math>\color{OliveGreen}\frac{x - vt}{\sqrt{(1 - v^2/c^2)}} \,</math>
<math>\color{OliveGreen}\frac{t - (v/c^2)x}{\sqrt{(1 - v^2/c^2)}} \,</math>
<math>n_a k_a\mbox{ }n_b k_b\,</math>
<math>\frac{dn_a \rho_a}{dt}=-k_b n_b.....(1)</math>
<math>\frac{dn_b \rho_b}{dt}=-k_a n_a.....(2)</math>
<math>\begin{align}
& \frac{(1)}{(2)}\rightarrow dn_a \rho_a k_a n_a=dn_b \rho_b k_b n_b \\
& \int \rho_a k_a n_a dn_a= \int \rho_b k_b n_b dn_b\\
& \rho_a k_a n_a^2=\rho_b k_b n_b^2\\
\end{align}</math>
<math>R_{t+1}=R_t-gG_t \mbox{ ; } G_{t+1}=G_t-rR_t\,</math>
<math>rR_{t+1}^2-gG_{t+1}^2=(1-rR)(rR_{t}^2-gG_{t}^2)</math>
==13==
<math>\begin{align}
& S_n=a_1+a_2+...a_3=3n^2\\
& a_n=S_n-S_{n-1}=3[n^2-(n-1)^2]=6n-3 \\
& a_2+a_4+a_6...a_{2n}= 6\sum^n_{k=1} 2k-3\sum^n_{k=1}1=6n^2+3\\
& a_1+a_3+a_5...a_{2n-1}=6\sum^n_{k=1}(2k-1)-3\sum^n_{k=1}1=6n^2-3 \\
& \lim_{n\to\infty}[\sqrt{a_2+a_4+a_6...a_{2n}}-\sqrt{a_1+a_3+a_5...a_{2n-1}}] \\
& =\lim_{n\to\infty}[\sqrt{6n^2+3n}-\sqrt{6n^2-3n}][\frac{\sqrt{6n^2+3n}+\sqrt{6n^2-3n}}{\sqrt{6n^2+3n}+\sqrt{6n^2-3n}}] \\
& =\lim_{n\to\infty}\frac{6n}{\sqrt{6n^2+3n}+\sqrt{6n^2-3n}}=\frac{6}{2\sqrt{6}}=\frac{\sqrt{6}}{2}
\end{align}</math>
<math>\begin{align}
& a_c=\omega^2r=\frac{F}{m} \\
& m=a\rho dr\mbox{ , }F=(\Delta P)\times a=(\rho g d h)\times a \\
& a_c=\omega^2r=\frac{F}{m}=g\frac{dh}{dr} \\
& \omega^2 \int r dr= g\int dh \\
& h=\omega^2\frac{r^2}{2g} \mbox{which is a parabola}
\end{align}</math>
<math>\sqrt{2}</math>
<math>x\vec x+y\vec y+z\vec z+t\vec {ct}=x'\vec {x'}+y'\vec {y'}+z'\vec {z'}+t'\vec {ct'}</math>
<math>\begin{bmatrix} c t' \\x' \\y' \\z' \end{bmatrix}
=
\begin{bmatrix} a &b&c&d\\e&f&g&h\\i&j&k&l\\m&n&o&p\\ \end{bmatrix} \begin{bmatrix} c t\\x\\y\\z \end{bmatrix}. </math>
<math>\begin{bmatrix}x\\y\\z\\ct\\0\\0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\x'\\y'\\z'\\ct'\end{bmatrix}</math>
<math>\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=a\vec {x}+b\vec y+c\vec z+d\vec {ct}</math>
<math>\begin{matrix}
t'=at+bx \\x'=ct+dx\end{matrix} \Rightarrow \begin{bmatrix}t'\\x'\end{bmatrix}=
\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}t\\x\end{bmatrix}</math>
<math>1=\frac{x}{ct}=\frac{x'}{ct'}</math>
<math>\begin{bmatrix}cT\\0\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}ct\\vt\end{bmatrix}</math>
<math>\begin{bmatrix}ct\\-vt\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}cT\\0\end{bmatrix}</math>
<math>\begin{bmatrix}ct\\vt\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\cT\\0\end{bmatrix}</math>
<math>\begin{bmatrix}cT\\0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\ct\\-vt\end{bmatrix}</math>
<math>x\vec x+ct\vec {ct}+x'\vec {x'} +ct'\vec {ct'}=\begin{bmatrix}x\\ct\\x'\\ct'\end{bmatrix}</math>
<math>\begin{bmatrix}a\\b\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\c\\d\end{bmatrix}</math>
<math>\begin{bmatrix}\end{bmatrix}</math>
<math>\begin{bmatrix}\gamma \\0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\\-\frac{v}{c}\end{bmatrix}</math>
<math>\begin{bmatrix}1\\\frac{v}{c}\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\\gamma\\0\end{bmatrix}</math>
<math>(\frac{\frac{v}{c}-1}{\gamma})\begin{bmatrix}\gamma \\0\\0\\0\end{bmatrix}+\begin{bmatrix}1\\\frac{v}{c}\\0\\0\end{bmatrix}
= \begin{bmatrix}\frac{v}{c}\\\frac{v}{c}\\0\\0\end{bmatrix}
=(\frac{\frac{v}{c}-1}{\gamma})\begin{bmatrix}0\\0\\1\\-\frac{v}{c}\end{bmatrix}+\begin{bmatrix}0\\0\\\gamma\\0\end{bmatrix}
= \begin{bmatrix}0\\0\\\frac{v}{c}-(\frac{v}{c})^2\\\frac{v}{c}-1+\gamma^2\end{bmatrix}</math>
==sec 14==
<math>\begin{align}
& \frac{\delta([\sin y\cos x][\sin (x-y)])}{\delta x}\\
& =-\sin x \sin y \sin (x-y)+\sin y\cos x\cos(x-y)=0\\
& \Rightarrow \tan (x-y)= \tan y \\
& \frac{\delta([\sin y\cos x][\sin (x-y)]}{\delta y)}\\
& =\cos y \cos x\sin(x-y)+[-1\sin y \cos x\cos(x-y)]=0\\
& \Rightarrow \tan (x-y)= \tan y \\
& x=2y+180n^\circ \mbox{ , } n\in \mathbb{R}
\end{align}</math>
<math>\begin{align}
& |K||L|\sin \theta = \sqrt{(x^2+y^2+z^2)(a^2+b^2+c^2)-(\vec K\cdot \vec L)^2}\\
& = \sqrt{x^2(b^2+c^2)+y^2(c^2+a^2)+z^2(a^2+b^2)-2(xaby+bycz+czxa)}\\
& = \sqrt{(yc-zb)^2+(za-xc)^2+(xb-ay)^2} = |\vec K\times \vec L|
\end{align}</math>
<math>v_{shm}=\sqrt{gr}=\sqrt{GM/r_0^2 \times r_0}=\sqrt{\frac{GM}{r_0}}</math>
<math>\Delta v= v_{shm}[(-\sin\theta,\cos\theta)-(1,0)]\,</math>
<math>\begin{align}
& V(\theta)=V_i+\Delta v=(0,V_0)+\cfrac{g_0r_0}{V_0}[(-\sin\theta,\cos\theta)-(0,1)]\\
& =(-\cfrac{g_0r_0}{V_0}\sin\theta] , [\cfrac{g_0r_0}{V_0}\cos\theta+ V_0 - \cfrac{g_0r_0}{V_0}])\\
& L=r\times v =(r\cos\theta,r\sin\theta) \times (-[\cfrac{g_0r_0}{V_0}\sin\theta] , [\cfrac{g_0r_0}{V_0}\cos\theta+ V_0 - \cfrac{g_0r_0}{V_0}])\\
& = (0,0,r[\cfrac{g_0r_0}{V_0}\cos\theta -V_0\cos\theta-\cfrac{g_0r_0}{V_0}])=r_0V_0 \\
& \Rightarrow r=\frac{r_0V_0}{[\cfrac{g_0r_0}{V_0}\cos\theta -V_0\cos\theta-\cfrac{g_0r_0}{V_0}])} \\
& =\frac{r_0V_0}{ \cos\theta[\cfrac{g_0r_0}{V_0} -V_0]-\cfrac{g_0r_0}{V_0})}
\end{align}</math>
<math>(g_0,\omega_0)\,</math>
<math>(g_0,r_0)\,</math>
<math>g_0=\omega^2r_{shm}\mbox{ and }r_{shm}\omega=V_{shm}=\frac{g_0}{\omega_0}=\cfrac{g_0r_0}{V_0}</math>
==sec15==<math>Insert formula here</math>
<math>Y \overset{\mathrm{def}}{=}\frac{dP}{dr}=\cfrac{dP}{\frac{dl}{l}}=\cfrac{dP}{\frac{dV}{V}}=V\frac{dP}{dV}</math>
<math>v_{rms}\,</math>
<math>F=\frac{m{v_{rms}}^2}{L}</math>
<math>P=\frac{m{v_{rms}}^2}{L^3}=\frac{m{v_{rms}}^2}{V}</math>
<math>KE=3\frac{m{v_{rms}}^2}{2}=\frac{3}{2}PV</math>
<math>\frac{{mv_{rms}^2}}{2}=\frac{PV}{2}</math>
<math>E=\frac{f}{2}PV</math>
<math>dW=-PdV=dE=d\left(\frac{f}{2}PV\right)=\frac{f}{2}PdV+\frac{f}{2}VdP</math>
<math>\begin{align}
& -PdV=\frac{f}{2}PdV+\frac{f}{2}VdP \\
& \left(\frac{f}{2}+1\right)\frac{dV}{V}+\frac{f}{2}\frac{dP}{P}=0 \\
& \int\left[\left(\frac{f}{2}+1\right)\frac{dV}{V}+\frac{f}{2}\frac{dP}{P}\right] \\
& \left(\frac{f}{2}+1\right)\ln V+ \frac{f}{2}\ln P=const \\
& P^\frac{f}{2}V^{\left(\frac{f}{2}+1\right)}=const \\
& PV^{\frac{f+2}{f}}=const\\
& \because \gamma \overset{\underset{\mathrm{def}}{}}{=} \frac{f+2}{f} \\
& \therefore PV^{\gamma}= const \\
\end{align}</math>
<math>PV^{\gamma}= const\,</math>
<math>\frac{d(PV^\gamma)}{dV}=V^\gamma \frac{dP}{dV}+\gamma PV^{\gamma-1}=0 \Rightarrow \gamma P =-V\frac{dP}{dV}=K</math>
<math>K=\gamma P\,</math>
<math>\int</math>
<math>c\times a\times\rho=\frac{dm}{dt}</math>
<math>x_1</math><math>x_2</math><math>\rho_1</math><math>\rho_2</math>
<math>\begin{align}
& dF=dP\times a\\
& =\cfrac{d\left(\cfrac{d\left(\sum dm\times x\right)}{dt}\right)}{dt}\\
& =\frac{d\left(\frac{dm}{dt}\times\Delta x\right)}{dt}=\frac{d\left(\frac{dm}{dt}\times(x_2-x_1)\right)}{dt}\\
& =\frac{dm}{dt}\frac{d(x_2-x_1)}{dt}\\
& =ca\rho[c\frac{\rho}{\rho_2}-c\frac{\rho}{\rho_1}]\\
& =c^2a\frac{\rho^2}{\rho_1\rho_2}[\rho_1-\rho_2]\\
& =c^2a\frac{\rho^2}{\rho_1\rho_2}[d\rho]\\
& \because \rho_1\rho_2\approx\rho^2\\
& \therefore dP\times a =c^2a[d\rho] \\
\end{align}</math>
<math>\begin{align}
& \because V\propto \frac{1}{\rho} \\
& \therefore dr=\lim_{dV\rightarrow 0} -\frac{dV}{V}=\lim_{d\rho\rightarrow 0} \frac{d\rho}{\rho} \\
& d\rho=dr\times\rho
\end{align}</math>
<math>c=\sqrt{\frac{dP}{d\rho}}=\sqrt{\frac{dP}{dr \times \rho}}=\sqrt{\frac{K}{\rho}}</math>
<math>\gamma\,</math> <math>\omega</math>
<math>c=\sqrt{\frac{\gamma P}{\rho}}</math>
<math>c=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{1.4(101,325Pa)}{\frac{1.2kg}{m^3}}=343.820m/s</math>
<math>\begin{align}
& R(\cos \theta,\sin\theta)\Rightarrow R _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])\\
\end{align}</math>
<math>\cos\theta \hat X + \cos\theta \hat Y = \cos [\theta-\omega t] \hat x+\sin [\theta-\omega t])\hat y</math>
<math>R(\cos \theta,\sin\theta)= R _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])</math>
<math>\begin{align} & R=r(\cos \theta,\sin\theta)\Rightarrow r _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])\\ \end{align}</math>
==sec 16==
<math>\hat i=\cos\theta \hat X + \sin\theta \hat Y = \cos [\theta-\omega t] \hat x+\sin [\theta-\omega t])\hat y</math>
<math>\dot{\theta}=\frac{d\theta}{dt} , \ddot{\theta}=\frac{d^2\theta}{dt^2} , \dot \omega = 0 \dot{\omega}=\frac{d\omega}{dt},\dot{r}=\frac{dr}{dt},\ddot{r}=\frac{d^2r}{dt^2}</math>
<math>\hat j=\sin\theta \hat X - \cos\theta \hat Y = \sin [\theta-\omega t] \hat x-\cos [\theta-\omega t])\hat y</math>
<math>\begin{align} & R(\cos \theta,\sin\theta)\Rightarrow R _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])\\ \end{align}</math>
<math>\begin{align}
& V=\frac{d r(\cos\theta,\sin\theta)}{dt}=\dot{r}(\cos\theta,\sin\theta)+r\omega(\sin\theta,-\cos\theta) \\
& =\dot r\hat i+ r\dot \theta \hat j \\
\end{align}</math>
<math>(\mbox{replace }\theta \mbox{ with }\theta-\omega t)\Rightarrow </math>
<math>\begin{align}
& A=\frac{d^2 r(\cos\theta,\sin\theta)}{dt^2}
=\frac{d[\dot{r}(\cos\theta,\sin\theta)+r\dot\theta(\sin\theta,-\cos\theta)]}{dt} \\
& =[\ddot r (\cos\theta,\sin\theta)+\dot r\dot\theta(\sin\theta,-\cos\theta)]+[\dot r \dot\theta(\sin\theta,-\cos\theta) \\
& +r\ddot\theta(\sin\theta,-\cos\theta)+r\dot\theta^2(-\cos\theta,-\sin\theta)]\\
& =(\ddot r-r\dot \theta^2)\hat i+ (2\dot r\dot \theta+r\ddot \theta)\hat j \\
\end{align}</math>
<math>\phi = \theta-\omega t\,</math>
<math>\therefore \dot \theta = \dot \phi+\omega</math>
<math>\theta = \phi+\omega t\,</math>
<math>\ddot\theta =\ddot\phi</math>
<math>(\ddot r-r\dot \phi^2)\hat i+ (2\dot r\dot \phi+r\ddot \phi)\hat j</math>
<math>[\ddot r-r(\dot \phi^2+2\dot\phi \omega+\omega^2)]\hat i+[2\dot r(\dot \phi+\omega)+r\ddot \phi]\hat j</math>
<math>V=\dot r\hat i+ (r\dot \phi+r\omega )\hat j</math>
<math>\begin{align} & [\ddot r-r(\dot \phi^2+2\dot\phi \omega+\omega^2)]\hat i+[2\dot r(\dot \phi+\omega)+r\ddot \phi]\hat j\\
& -[(\ddot r-r\dot \phi^2)\hat i+ (2\dot r\dot \phi+r\ddot \phi)\hat j]
=A_f=(-2r\dot \phi\omega-r\omega^2)\hat i+2\dot r \omega \hat j
\end{align}</math>
<math>\hat i \times \hat j=\hat k,\hat k \times \hat i=\hat j, \hat j \times \hat k=-\hat i</math>
<math>\begin{align}
& V\times \Omega = [\dot r\hat i+ (r\dot \phi+r\omega )\hat j]\times [\omega \hat k]\\
& =(-2r\dot \phi\omega-2r\omega^2)\hat i+2\dot r \omega \hat j
\end{align}</math>
<math>\Omega\times (\Omega\times R)</math>
<math>\begin{align}
& Ways=\frac{N!}{N_1!N_2!...N_{\inf}} \\
& \mbox{to insure energy conservation } a+c=2b \\
\end{align}</math>
==sec 17==
<math>|\vec a|=|\vec a_A-\vec a_B|=\frac{Gm_Am_B}{r^2}\times (\frac{1}{m_A}+\frac{1}{m_B})=\frac{G(m_a+m_b)}{r^2}</math>
<math>a=\frac{G(m_a+m_b)}{r^2}</math>
<math>\theta</math>
<math>\phi</math>
| |||