User:Tikai/math

It is fairly easy to see a spectrum produced by white light when it passes through a prism, the bevelled edge of a mirror or a special pane of glass, or through drops of rain to form a rainbow. When samples of single elements are caused to emit light they may emit light at several characteristic frequencies. The frequency profile produced is characteristic of that element. Instead of there being a wide band filled with colors from violet to red, there will be isolated bands of single colors separated by darkness. Such a display is called a line spectrum. Some lines go beyond the visible frequencies and can only be detected by special photographic film or other such devices. Scientists hypothesized that an atom could radiate light the way the string on a fine violin radiates sound – not only with a fundamental frequency (in which the entire string moves the same way at once) but with several higher harmonics (formed when the string divides itself into halves and other divisions that vibrate in coordination with each other as when one half of the string is going one way as the other half of the string is going the opposite way). For a long time nobody could find a mathematical way to relate the frequencies of the line spectrum of any element.

In 1885, Johann Jakob Balmer (1825-1898) figured out how the frequencies of atomic hydrogen are related to each other. The formula is a simple one: $$ \frac{1}{L} = R\left ( \frac{1}{2^2} - \frac{1}{n^2} \right )$$ where L is wavelength, R is the Rydberg constant and n is an integer (n =3, 4,...) This formula can be generalized to apply to atoms that are more complicated than hydrogen, but we will stay with hydrogen for this general exposition. (That is the reason that the denominator in the first fraction is expressed as a square.)

The next development was the discovery of the Zeeman effect, named after Pieter Zeeman (1865-1943). The physical explanation of the Zeeman effect was worked out by Hendrik Antoon Lorentz (1853-1928). Lorentz hypothesized that the light emitted by hydrogen was produced by vibrating electrons. It was possible to get feedback on what goes on within the atom because moving electrons create a magnetic field and so can be influenced by the imposition of an external magnetic field in a manner analogous to the way that one iron magnet will attract or repel another magnet.

The Zeeman effect could be interpreted to mean that light waves are originated by electrons vibrating in their orbits, but classical physics could not explain why electrons should not fall out of their orbits and into the nucleus of their atoms, nor could classical physics explain why their orbits would be such as to produce the series of frequencies derived by Balmer’s formula and displayed in the line spectra. Why did the electrons not produce a continuous spectrum?

fundamental tone
The fundamental tone, often referred to simply as the fundamental and abbreviated fo, is the lowest frequency in a harmonic series.

The fundamental frequency (also called a natural frequency) of a periodic signal is the inverse of the pitch 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 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 sinusoids (for example, fourier series), the fundamental frequency is the lowest frequency sinusoidal in the sum.

To find the fundamental frequency of a sound wave in a tube that has a closed end you will use the equation:

$$F=\frac{V}{4L}$$

To find L you will use:

$$L=\frac{\lambda}{4}$$

To find λ (lambda) you will use:

$$\lambda = \frac{V}{F}$$

To find the fundamental frequency of a sound wave in a tube that has open ends you will use the equation:

$$F=\frac{V}{2L}$$

To find L you will use:

$$L=\frac{\lambda}{2}$$

To find Wavelength which is the distance in the medium between the beginning and end of a cycle and is found using the following equation: - WAVELENGTH = Velocity/Frequency or

$$-\lambda=\frac{V}{F}$$

At 70 °F the speed of sound in air is approximately 1130 ft/s or 340 m/s. This speed is temperature dependent and does increase at a rate of 1.1 ft/s for each degree Fahrenheit increase in temperature.

The velocity of a sound wave at different temperatures:
 * V = 343.7 m/s at 20 °C
 * V = 331.5 m/s at 0 °C

WHERE:

F = fundamental Frequency L = length of the tube V = velocity of the sound wave λ = wavelength

Application to simple gas thermodynamics
In classical statistical mechanics, this average is predicted to hold exactly for homogeneous ideal gases. Monatomic ideal gases possess 3 degrees of freedom per atom, corresponding to the three spatial directions, which means a thermal energy of 1.5kT per atom. As indicated in the article on heat capacity, this corresponds very well with experimental data. The thermal energy can be used to calculate the root mean square speed of the atoms, which is inversely proportional to the square root of the atomic mass. The root mean square speeds found at room temperature accurately reflect this, ranging from 1370 m/s for helium, down to 240 m/s for xenon.

From kinetic theory one can show that for an ideal gas the average pressure P is given by:


 * $$ P = \frac{1}{3}\frac{N}{V} m {\overline{v^2}}$$

Substituting that the average translational kinetic energy is:


 * $$ \frac{1}{2}m \overline{v^2} = \frac{3}{2} k T $$

gives:


 * $$ P = \frac{N}{V} k T $$

and so the ideal gas equation is regained.

The ideal gas equation is also followed quite well for molecular gases; but the form for the heat capacity is more complicated, because the molecules possess new internal degrees of freedom, as well as the three degrees of freedom for movement of the molecule as a whole. Diatomic gases, for example, possess in total approximately 5 degrees of freedom per molecule.

kinetic theory
Consider a continuous mass enlarging at the speed of V in all directions. This is to speculate that any molecule has equal chances of direction when talking about its speed. We then put it into a cuboid to exert pressure on its surface. We also presume that upon hitting the surface of the cuboid, the molecule does not lose energy and instead do a perfect bounce.

the surface density is given by $$\sigma=\frac{m}{4\pi r^2}\,$$

a perfect bounce with an incidence angle of $$\theta$$ leads to$$\Delta V = 2 v \sin\theta\,$$

the frequency is also given by $$\frac{\Delta v}{l_{axis}}=\frac{2 v \sin\theta}{l_{axis}}$$

the force is $$F_{out}=F_x+F_y+F_z=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dF_x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dF_y+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dF_z$$

$$=\sum_{axis=x}^z\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dA \times\sigma \times\Delta v\times frequency$$

$$=\sum_{axis=x}^z\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2\pi r^2\cos\theta d\theta)(\frac{m}{4\pi r^2})(2 v \sin\theta)\times frequency$$

$$=\sum_{axis=x}^z\frac{mv^2}{l_{axis}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta\cos\theta d\theta$$

$$=mv^2\left(\frac{1}{l_x}+\frac{1}{l_y}+\frac{1}{l_z}\right)\times\frac{2}{3}$$

$$PV=\frac{F}{A}=\left(l_xl_yl_z\right)\times\frac{mv^2\left(\frac{1}{l_x}+\frac{1}{l_y}+\frac{1}{l_z}\right)\times\frac{2}{3}}{2(l_xl_y+l_yl_z+l_zl_x)}$$

$$=\frac{1}{3}mv^2\Rightarrow \frac{3}{2}PV=\frac{1}{2}mv^2=\mathrm{K}\mathrm{E}$$

trigonometry
$$R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C}$$

$$H_a = \frac{bc}{2R} = \sin C b = \sin B c$$

$$\sin A = \frac{a}{2R} = \sqrt{1-\cos^2}$$

$$\cos A = \frac{b^2+c^2-a^2}{2bc}$$

$$a^2 = b^2+ c^2-2bc\cos A\,$$

$$\Delta = \frac{1}{2} ab\sin C =...=\frac{abc}{4R}=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}= \frac{a+b+c}{2}\times r =2 R^2 \sin A\sin B\sin C$$

$$\tan \frac{A}{2}= \frac{2r}{b+c-a}$$

$$r=\cfrac{(b+c-a)\tan \frac{A}{2}}{2} = \frac{2\Delta}{a+b+c}$$

$$\Delta = \frac{1}{2} ab\sin C = \frac{ab}{2} \sqrt{1-\cos ^2 C} = \frac{ab}{2} \sqrt{\left (1-\frac{a^2+b^2+c^2}{2ab}\right )\left ( 1+\frac{a^2+b^2+c^2}{2ab}\right )}= \frac{1}{4} \sqrt{\left (\left (a+b\right )^2-c^2\right )\left (c^2-\left (a-b\right )^2\right )}= \sqrt{\left (a+b-c\right )\left (a+b+c\right )\left (c+b-a\right )\left (c+a-b\right )}=\sqrt{s\left (s-a\right )\left (s-b\right )\left (s-c\right )}$$

calculus
$$\frac{d(uv)}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}$$

$$Insert formula here$$

pi

 * The area of the unit disc:
 * $$2\int_{-1}^1 \sqrt{1-x^2}\,dx = \pi$$


 * Half the circumference of the unit circle:
 * $$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}} = \pi$$


 * François Viète, 1593 (proof):
 * $$\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots = \frac2\pi$$


 * Leibniz' formula (proof):
 * $$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \frac{\pi}{4}$$


 * Wallis product, 1655 (proof):
 * $$ \prod_{n=1}^{\infty} \left ( \frac{n+1}{n} \right )^{(-1)^{n-1}} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2} $$


 * Faster product (see Sondow, 2005 and Sondow web page)
 * $$ \left ( \frac{2}{1} \right )^{1/2} \left (\frac{2^2}{1 \cdot 3} \right )^{1/4} \left (\frac{2^3 \cdot 4}{1 \cdot 3^3} \right )^{1/8} \left (\frac{2^4 \cdot 4^4}{1 \cdot 3^6 \cdot 5} \right )^{1/16}  \cdots = \frac{\pi}{2}  $$
 * where the nth factor is the 2nth root of the product
 * $$\prod_{k=0}^n (k+1)^{(-1)^{k+1}{n \choose k}}.$$


 * Symmetric formula (see Sondow, 1997)
 * $$ \frac {\displaystyle \prod_{n=1}^{\infty} \left (1 + \frac{1}{4n^2-1} \right )}{\displaystyle\sum_{n=1}^{\infty} \frac {1}{4n^2-1}}  =  \frac {\displaystyle\left (1 + \frac{1}{3} \right ) \left (1 + \frac{1}{15} \right ) \left (1 + \frac{1}{35} \right ) \cdots} {\displaystyle \frac{1}{3} +  \frac{1}{15} +  \frac{1}{35} + \cdots}  = \pi $$


 * Bailey-Borwein-Plouffe algorithm (See Bailey, 1997 and Bailey web page)
 * $$\sum_{k=0}^\infty\frac{1}{16^k}\left(\frac {4}{8k+1} - \frac {2}{8k+4} - \frac {1}{8k+5} - \frac {1}{8k+6}\right) = \pi$$


 * Chebyshev series Y. Luke, Math. Tabl. Aids Comp. 11 (1957) 16
 * $$\sum_{k=0}^\infty\frac{(-1)^k(\sqrt{2}-1)^{2k+1}}{2k+1} = \frac{\pi}{8}.$$
 * $$\sum_{k=0}^\infty\frac{(-1)^k(2-\sqrt{3})^{2k+1}}{2k+1}=\frac{\pi}{12}.$$


 * An integral formula from calculus (see also Error function and Normal distribution):
 * $$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$$


 * Basel problem, first solved by Euler (see also Riemann zeta function):
 * $$\zeta(2)= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}$$
 * $$\zeta(4)= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90}$$
 * and generally, $$\zeta(2n)$$ is a rational multiple of $$\pi^{2n}$$ for positive integer n


 * Gamma function evaluated at 1/2:
 * $$\Gamma\left({1 \over 2}\right)=\sqrt{\pi}$$


 * Stirling's approximation:
 * $$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$


 * Euler's identity (called by Richard Feynman "the most remarkable formula in mathematics"):
 * $$e^{i \pi} + 1 = 0\;$$


 * A property of Euler's totient function (see also Farey sequence):
 * $$\sum_{k=1}^{n} \phi (k) \sim \frac{3n^2}{\pi^2}$$


 * An application of the residue theorem
 * $$\oint\frac{dz}{z}=2\pi i ,$$
 * where the path of integration is a closed curve around the origin, traversed in the standard anticlockwise direction.

scribbles
$$ \frac{\rho _{x2} -\rho _{x1}}{\rho _N}\times \mathbf{P} _N\times \alpha = \alpha \mathbf{V} \rho _N \times \mathbf{V} \left ( \frac{\rho_N}{\rho_{x2}}-\frac{\rho_N}{\rho_{x2}}\right )$$

$$\rho va\frac{dX_2-dX_1}{dt}$$

$$v\left (\frac{\rho_0}{\rho_2}-\frac{\rho_0}{\rho_1}\right )=\rho v^2 a \frac{\rho_1-\rho_2}{\rho_2\rho_1}$$

$$\rho_1-\rho_2$$

$$\begin{align} mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \\ & =\frac{1}{2}mv^2+\frac{1}{2}I\frac{v^2}{R^2} \\ & =v^2\left (            \frac{m}{2}  +  \frac{I}{R^2}            \right ) \\ I_ball=\int_0^\pi r^2 dm \\ & =\int_0^\pi\left [ ((R\sin \theta)^2 \times \pi )\times (Rd\theta\times\sin \theta)\times \rho\right ]\times\left [ R^2\sin^2\theta \right ] \\ & (\because \rho = \cfrac{m}{\cfrac{4}{3}mR^3}) \\ & =\frac{3}{4}\int_0^\pi mR^2 \sin^5 \end{align}$$

$$\rho\int_0^\pi 4R^3\cos^2\theta\sin\theta d\theta \times (R\sin\theta)^2$$

$$\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}$$

$$\begin{align} mgh=KE+RE \\ & =\frac{1}{2}mv^2 + \frac{1}{2} I\omega^2 \\ & =\frac{1}{2}mv^2 + \frac{1}{2} \times \frac{2}{5}mR^2\frac{v^2}{r^2} \\ & =\frac{1}{2}mv^2 +\frac{1}{5}mv^2 \\ & =\frac{7}{10}mv^2 \\ v = \sqrt{\frac{10}{7}gh} \end{align}$$

$$\frac{q}{m}=\frac{2V}{r^2B^2}\Rightarrow r\varpropto \sqrt{\cfrac{m}{q}}$$ $$\vec v$$ $$\vec v_{tangent}$$

$$\vec F =\frac{d\vec v_{tangent}}{dt} =\frac{dB}{dt}    \times     \frac{d\vec v_{tangent}}{dB}

$$

$$=\omega_{B}   \times    \frac{d(\vec v_{A} \times \cos B)}{dB} =\frac{v\cos C}{R}   \times     (-\vec v_{A}\sin B) =-\frac{v\vec v_A \cos C \sin B}{R}$$

$$k = \frac{1 + \sqrt{5}}{2} or \frac{1 - \sqrt{5}}{2}$$

$$\Phi$$

$$\times\,$$$$\vec v\times\vec B$$

$$\vec a\times\vec b = \vec n|a||b|\sin\theta$$

$$\vec a\perp\vec n \land \vec b\perp\vec n$$

$$V=n\frac{d\Phi}{dt}$$ $$a^4-ka^2-\frac{p^2}{4}$$

$$a^2=\frac{k\plusmn\sqrt{k^2+p^2}}{2}=\frac{k\plusmn z}{2}$$ $$a=\plusmn\sqrt{\frac{k+z}{2}}$$ $$\begin{align} & a=\plusmn (n+1) \\ & b=\frac{p}{2a}=\frac{2n^2+2n}{\plusmn 2(n+1)}= \plusmn n \end{align}$$

scrib 2
$$\triangle$$ $$\begin{align} & \vec {PA}+\vec {PB}+\vec {PC} =\vec {BC} \\ & \vec {QA}+\vec {QB}+\vec {QC} =\vec {CA} \\ & \vec {RA}+\vec {RB}+\vec {RC} =\vec {AB} \end{align}$$

$$\begin{align} & \vec P-\vec A+\vec P-\vec B+\vec P-\vec C=\vec B-\vec C \\ & 3\vec P-(\vec A+\vec B+\vec C)=\vec B-\vec C \\ & 3\vec P = \vec A +2\vec B \end{align}$$

$$\begin{align} & 4\vec {OA}+\vec {OB}+\vec {OC} =\vec {OA'} \\ & \vec {OA}+4\vec {OB}+\vec {OC} =\vec {OB'} \\ & \vec {OA}+\vec {OB}+4\vec {OC} =\vec {OC'} \end{align}$$

$$\begin{align} & \vec O-\vec A'=4\vec O-4\vec A+\vec O-\vec B+\vec O-\vec C \cdots (1)\\ & \vec O-\vec B'=\vec O-\vec A+4\vec O-4\vec B+\vec O-\vec C \cdots (2)\\ & \vec O-\vec C'=\vec O-\vec A+\vec O-\vec B+4\vec O-4\vec C \cdots (3)\\ & \\ & (1)-(2)......\vec B'-\vec A'=3\vec B-3\vec A \\ & \vec {B'A'}=3\vec {BA} \end{align}$$

$$v'=\frac{m_1v_1-m_2v_2}{m_1+m_2}$$

$$\begin{align} if m_1v_1'+m_2v_2'=0 \\ then m_1(-v_1')+m_2(-v_2')=0 \end{align}$$

$$\frac{1}{2}m_1v_1'^2+\frac{1}{2}m_2v_2'^2=\frac{1}{2}m_1(-v_1')^2+\frac{1}{2}m_2(-v_2')^2$$

$$\begin{align} v_1'=v_1-v' \\ v_2'=-v_2-v' \end{align}$$

$$\begin{align} & y_{c.m.}=\frac{\int_0^H y dm}{m}=\cfrac{\int_0^H y\times \rho (w\times dy)}{\rho \cfrac{WH}{2}} \\ & =\cfrac{\int_0^H y\times \rho (\cfrac{H-y}{H}W\times dy)}{\rho \cfrac{WH}{2}} \\ & =\frac{2}{H}\int_0^H y-\frac{y^2}{H}dy \\ & =\frac{2}{H}\left (\frac{1}{2}y^2\bigg|_0^H-\frac{y^3}{3H}\bigg|_0^H\right ) \\ & =\frac{2}{H}\left (\frac{H^2}{2}-\frac{H^3}{3H}\right ) = \frac{H}{3} \end{align}$$ $$\rho\,$$$$L\,$$$$g\,$$ $$\theta=\arctan\frac{l\rho g}{T}\mbox{ or }\tan\theta=\frac{l\rho g}{T}$$  $$dl\,$$$$dl\rho\,$$$$l\,$$ $$dl\cos\theta\,$$

$$\begin{align} & \left ( \mbox{let it be that} \arctan \cfrac{0.5l}{\rho g} = \omega\right ) \\ & x=\int_0^{0.5l} \cos (\arctan\frac{0.5l\rho g}{T}) dl \\ & =\int_0^{0.5l} \cos u \frac{dl}{du} du \\ & =\frac{T}{\rho g}\int_0^\omega \cos u \cfrac{dl\times(\rho g)}{\arctan\cfrac{0.5l\rho g}{T}} du\\ & =\frac{T}{\rho g}\int_0^\omega \cos u \frac{1}{\cos^2 u} du \\ & =\frac{T}{\rho g}\int_0^\omega \frac{1}{\cos u} du \\ & =\frac{T}{\rho g} \ln \left (\sec u+\tan u\right )\bigg|_0^\omega=\frac{T}{\rho g} \ln \left (\sec \omega+\tan \omega\right ) \end{align}$$

$${\color{Blue}\vec Q_{BDAF}=\overrightarrow{BD}\times(\frac{m_a+m_c}{m_a+m_b+m_c})}$$

$${\color{Blue}\vec Q_{BDCE}=\overrightarrow{BD}\times(\frac{m_a+m_c}{m_a+m_b+m_c})}$$

$$\begin{align} & \left ( \mbox{let it be that} \arctan \cfrac{0.5l}{\rho g} = \omega\right ) \\ & y=\int_0^{0.5l} \sin (\arctan\frac{0.5l\rho g}{T}) dl \\ & =\int_0^{0.5l} \sin u \frac{dl}{du} du \\ & =\frac{T}{\rho g}\int_0^\omega \sin u \cfrac{dl\times(\rho g)}{\arctan\cfrac{0.5l\rho g}{T}} du\\ & =\frac{T}{\rho g}\int_0^\omega \sin u \frac{1}{\cos^2 u} du \\ & =\frac{T}{\rho g}\int_0^\omega \left (\sec u\tan u \right )du \\ & =\frac{T}{\rho g} \sec u\bigg|_0^\omega = \frac{T}{\rho g}\left(\sec\omega-1\right) \end{align}$$

$$\begin{align} & \left ( \mbox{let it be that} k = \frac{\rho g}{T} \right) \\ \end{align}$$

$$\begin{align} & \left(\mbox{when }x>0,y>0,\omega>0\right)\\ & kx=\ln(\sec\omega+\tan\omega)\\ & e^{kx}=\sec\omega+\tan\omega=\sec\omega+\sqrt{\sec^2\omega-1}\\ & \left(e^{kx}-\sec\omega\right)^2=\sec^2\omega-1\\ & e^{2kx}-2e^{kx}\sec\omega+\sec^2\omega=\sec^2\omega-1\\ & e^{2kx}+1=2e^{kx}\sec\omega\\ & \sec\omega=\frac{1}{2}\left(e^{kx}+e^{-kx}\right)\\ & \\ & y=\frac{1}{k}\sec\omega-1\\ & =\frac{1}{2}\frac{1}{k}\left(e^{kx}+e^{-kx}-2\right)\\ \end{align}$$

scrib3
$$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$$

$$\begin{align} & \vec a\times\vec b=\vec c \\ & |\vec c|=|a||b|\sin\theta \\ & \vec c\perp\vec b,\vec c\perp\vec a \end{align}$$

$$\begin{align} \vec A(X_a,Y_a,Z_a) \\ \vec B(X_b,Y_b,Z_b) \end{align}$$

$$\begin{align} X_a x+Y_a y+Z_a z=0 \\ X_b x+Y_b y+Z_b z=0 \\ \end{align}$$ $$a^2+b^2+c^2$$

$$\begin{align} & -X_a k=Y_a y+Z_a z=0 \\ & -X_b k=Y_b y+Z_b z=0 \\ & y=\frac{Z_A X_B-X_A Z_B}{Z_B Y_A-Y_A Z_B}k \\ & z=\frac{X_A Y_B-Y_A X_B}{Z_B Y_A-Y_A Z_B}k \\ & x:y:z=(Y_AZ_B-Z_AY_B):(Z_AX_B-X_AZ_B):(X_AY_B-Y_AX_B) \end{align}$$

$$C( \det \begin{pmatrix} Y_A & Y_B \\  Z_A & Z_B \end{pmatrix}, \det \begin{pmatrix}  Z_A & Z_B \\  X_A & X_B \end{pmatrix} , \det \begin{pmatrix}  X_A & X_B \\  Y_A & Y_B \end{pmatrix} )$$ $$\left(\frac{2}{3}^3\right)\left(\frac{1}{3}^0\right)$$ $$\left(\frac{2}{3}^2\right)\left(\frac{1}{3}^1\right)$$ $$\left(\frac{2}{3}^1\right)\left(\frac{1}{3}^2\right)$$ $$\left(\frac{2}{3}^0\right)\left(\frac{1}{3}^3\right)$$

4
$$\begin{align} & \vec A \cdot \vec B = |\vec A||\vec B|\cos\theta =|\vec A||\vec B|\times \frac{a^2+b^2-c^2}{2ab} =\frac{a^2+b^2-c^2}{2} \\ & =\frac{(x_A^2+y_A^2 +z_A^2)+(x_B^2+ y_B^2+z_B^2)-(\left[x_A- x_B\right]^2 +\left[y_A-y_B\right]^2 + \left[z_A-z_B\right]^2)}{2} \\ & =x_Ax_B+y_Ay_B+z_Az_B \end{align}$$

$$\vec V_1 $$$$\vec V_2$$

$$\vec V_c=\frac{m_1\vec V_1 +m_2\vec V_2}{m_1+m_2}$$

$$\vec V_1 -\vec V_c $$$$\vec V_2-\vec V_c$$

$$\vec U\mbox{ and } |\vec U|=1$$

$$\mbox{reflect} (\vec A, \vec U)=-\vec A + 2\left( \vec A \cdot\vec U\right)\vec U$$

$$\mbox{reflect} ( \vec V_1-\vec V_c, \vec U)=-( \vec V_1 -\vec V_c)+2 \left ( \left [ \vec V_1 -\vec V_c \right ]\cdot \vec U \right ) \vec U $$

$$\mbox{reflect} ( \vec V_2-\vec V_c, \vec U)=-( \vec V_2 -\vec V_c)+2 \left ( \left [ \vec V_2 -\vec V_c \right ]\cdot \vec U \right ) \vec U $$

$$\vec V_1'=\vec V_c +\mbox{reflect}(\vec V_1 -\vec V_c, \vec U) =2\vec V_c - \vec V_1+ 2\left( \left[\vec V_1 -\vec V_c\right ] \cdot\vec U\right)\vec U$$

$$\vec V_2'=\vec V_c +\mbox{reflect}(\vec V_2 -\vec V_c, \vec U) =2\vec V_c - \vec V_2+ 2\left( \left[\vec V_2 -\vec V_c \right ]\cdot\vec U\right)\vec U$$

$$\vec V_1' =2\vec V_c - \vec V_1+2\left( \left[\vec V_1 -\vec V_c \right ]\cdot\vec U\right)\vec U = \vec V_2 -\left( \vec V_2\cdot\vec U\right)\vec U$$

$$\vec V_2' =2\vec V_c - \vec V_2+2\left( \left[\vec V_2 -\vec V_c \right ]\cdot\vec U\right)\vec U = \vec V_1 -\left( \vec V_1 \cdot\vec U\right)\vec U$$

$$\vec V_1 \| \vec V_2 \perp \vec U$$

$$\alpha\, $$$$\beta\,$$

$$4\beta-2\alpha\,$$

$$\begin{align} & \frac{d(4\beta-2\alpha)}{d\sin\alpha} \\ & \left( \because n_{water}\mbox{is 1.332986 and we set sin} \alpha \mbox{ to be } u \right) \\ & \left( n_w \sin \beta = \sin \alpha \mbox{ and so } \sin\beta = \frac{u}{1.332986}\right) \\ & =\frac{d(4\arcsin (0.7501954259 u)-2\arcsin u)}{du} \\ & =\frac{4}{\sqrt{1-(0.750195 u)^2}} \times 0.750195 - \frac{2}{\sqrt{1-u^2}} \\ & \mbox{because we want to find the biggest value so}\\ & =0 \end{align}$$

$$\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}$$

$$\begin{align} & \frac{d(4\beta-2\alpha)}{d\sin\alpha} \\ & =\frac{d(4\arcsin (\frac{u}{n})-2\arcsin u)}{du} \\ & =\cfrac{ 4 }{  \sqrt{    1-(\frac{u}{n})^2    }  } \frac{1}{n} - \frac{2}{\sqrt{1-u^2}} \\ & =0 \end{align}$$

$$\begin{align} & \frac{3}{4}u^2=1-\frac{n^2}{4} \\ & u = \sqrt{\frac{4}{3}- \frac{n^2}{3}} \\ \end{align}$$ $$\sin \alpha = u = \sqrt{\frac{4}{3}- \frac{n^2}{3}}$$

$$4\beta -2 \alpha= 4 \arcsin \frac{u}{n}- 2\arcsin u = 42 ^\circ$$

5
$$ \begin{align} & F_1\times S_1 -F_2\times S_2 \\ & =(P_1A_1)(V_1 \Delta \mathbf {t}) -(P_2 A_2 )(V_2 \Delta \mathbf{t}) \\ & =\Delta E = \Delta \mbox{KE} + \Delta \mbox{PE}\\ & =\frac{1}{2} A_2 V_2 \Delta \mathbf{t}\rho V_2^2 - \frac{1}{2} A_1 V_1 \Delta \mathbf{t}\rho V_1^2 +  A_2 V_2 \Delta \mathbf{t} \rho \mathbf{g} y_2 - A_1 V_1 \Delta \mathbf{t} \rho \mathbf{g} y_1 \end{align}$$

$$\Delta m \propto \mbox{t and } \Delta m \propto V$$

$$\frac{\Delta m}{t\cdot V}=r\mbox{ so the unit of r is} \frac{\mbox{mol}}{s\cdot L}$$

$$\frac{k}{k'}=\frac{A}{A'}e^{-\frac{(E_a-E_a')}{RT}}=Ce^{-\frac{(\Delta H)}{RT}}$$

$$\left[ \mbox{A}\right]\left[ \mbox{B}\right]$$

$$\frac{a\times D}{20\mbox{ sec}}\times AB/100,00$$

$$r \propto \left[ \mbox{A}\right]\left[ \mbox{B}\right]\left[ \mbox{C}\right]...$$

$$r = k \left[ \mbox{A}\right]\left[ \mbox{B}\right]\left[ \mbox{C}\right]...$$ $$e^{{-E_a}/{RT}}$$

$$r = k \left[ \mbox{A}\right]\left[ \mbox{B}\right]\left[ \mbox{C}\right]... =r' = k' \left[ \mbox{A}'\right]\left[ \mbox{B}'\right]\left[ \mbox{C}'\right]...$$

$$k=k'\times W$$

$$W=\frac{k}{k'}=\frac{A}{A'}e^{-\frac{(E_a-E_a')}{RT}}=Ce^{-\frac{(\Delta H)}{RT}}$$

$$\frac{\Delta H}{T}$$

$$v=\sqrt{2gH}$$

$$\begin{align} & t=\frac{\Delta V}{a}=\frac{2v\sin\theta}{g} \\ & d=v_x \times t = (v\cos\theta)\times (\frac{2v\sin\theta}{g}) \\ & =\frac{2v^2\cos\theta\sin\theta}{g}=\frac{v^2\sin2\theta}{g}=2H\sin\theta \\ & d=H \\ & \therefore \sin 2\theta=\frac{1}{2} & \theta=15^\circ ,75^\circ \end{align}$$

$$\frac{(-v\cos p)^2}{2g}= \frac{4\cos ^2 P}{20}=0.2\cos^2 P =0.2(1-\sin^2 P)$$

$$0.5+0.5sinP+0.2-0.2\sin^2 P = -0.2(\sin P - 1.25)^2 +0.7+\frac{5}{16}=\frac{81}{80}$$

$$a_{tide}=a_{near}-a_{center}=\frac{GM}{(d-r)^2}-\frac{GM}{d^2}=GM\frac{d^2-(d-r)^2}{d^2(d-r)^2}=GM\frac{2dr-r^2}{d^2(d-r)^2}$$

$$\lim_{d>>r}a_{tidal}=\frac{2GMr}{d^3}$$

sec 6
$$s^2=x^2-(ct)^2\,$$ $$\begin{align} & s^2=(vt)^2-(ct)^2 \end{align}$$ $$s^2=(0)^2-(cT)^2\,$$

$$-(cT)^2=(vt)^2-(ct)^2\,$$

$$\frac{T}{t}=\sqrt{1-v^2/c^2}$$

$$\frac{X}{x}=\sqrt{1-v^2/c^2}$$

$$\sqrt{0.75}$$

$$\frac{\overline{AB}}{\overline{OA}}=\sin{\beta}=\frac{v}{c}$$

$$\overline{OB}=\overline{OA}\cos\beta=\sqrt{1-v^2/c^2}\overline{OA}$$

$$

\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} $$

$$\begin{align} & \Sigma \vec F = 0 \\ & \rightarrow \vec F_1 +\vec F_2+\vec F_3 ...=0 \\ & \vec F_{x1}+\vec F_{x2}+\vec F_{x3}...=0 \\ & \vec F_{y1}+\vec F_{y2}+\vec F_{y3}...=0 \\ & \vec F_{z1}+\vec F_{z2}+\vec F_{z3}...=0 \\ \end{align}$$

$$\begin{align} & \Sigma \tau = 0\\ & \Sigma \tau= \Sigma \vec F \times \vec d = 0\\ \end{align}$$

$$\begin{align} & \Sigma \vec F \times \vec d = 0\\ & \Sigma (F_x,F_y,F_z)\times (d_x,d_y,d_z) \\ & =\Sigma \left((F_yd_z-F_zd_y),(F_zd_x-F_xd_z),(F_xd_y-F_yd_x)\right)= 0\\ \end{align}$$

$$\begin{align} & \Sigma \vec F \times \vec d' = 0\\ & \Sigma (F_x ,F_y,F_z)\times ((d_x+r_x),(d_y+r_y),(d_z+r_z)) \\ & ={\color{Red}\Sigma\left((F_yd_z-F_zd_y),(F_zd_x-F_xd_z),(F_xd_y-F_yd_x)\right)} \\ & \mbox{  }+\Sigma \left( {\color{Blue} (F_yr_z-F_zr_y)},{\color{Brown}(F_zr_x-F_xr_z)},{\color{OliveGreen}(F_xr_y-F_yr_x)} \right)= 0\\ & ={\color{Red}0}+{\color{Blue}(r_z\Sigma F_y,0,0)-(r_y\Sigma F_z,0,0)} \\ & \mbox{  }+{\color{Brown}(0,r_z\Sigma F_x,0)-(0,r_x\Sigma F_z,0)} +{\color{OliveGreen}(0,0,r_x\Sigma F_y)-(0,0,r_y\Sigma F_x)} \\ & ={\color{Red}0}+{\color{Blue}0-0}+{\color{Brown}0-0}+{\color{OliveGreen}0-0}=0 \end{align}$$

$$\begin{align} & \because \sin k =\frac{v}{c}\\ & \therefore \cos k =\sqrt{1-(\frac{v}{c})^2}\\ \end{align}$$

$$\color{OliveGreen}\frac{x - vt}{\sqrt{(1 - v^2/c^2)}} \,$$

$$\color{OliveGreen}\frac{t - (v/c^2)x}{\sqrt{(1 - v^2/c^2)}} \,$$

$$n_a k_a\mbox{ }n_b k_b\,$$

$$\frac{dn_a \rho_a}{dt}=-k_b n_b.....(1)$$

$$\frac{dn_b \rho_b}{dt}=-k_a n_a.....(2)$$

$$\begin{align} & \frac{(1)}{(2)}\rightarrow dn_a \rho_a k_a n_a=dn_b \rho_b k_b n_b \\ & \int \rho_a k_a n_a dn_a= \int \rho_b k_b n_b dn_b\\ & \rho_a k_a n_a^2=\rho_b k_b n_b^2\\ \end{align}$$

$$R_{t+1}=R_t-gG_t \mbox{ ; } G_{t+1}=G_t-rR_t\,$$

$$rR_{t+1}^2-gG_{t+1}^2=(1-rR)(rR_{t}^2-gG_{t}^2)$$

13
$$\begin{align} & S_n=a_1+a_2+...a_3=3n^2\\ & a_n=S_n-S_{n-1}=3[n^2-(n-1)^2]=6n-3 \\ & a_2+a_4+a_6...a_{2n}= 6\sum^n_{k=1} 2k-3\sum^n_{k=1}1=6n^2+3\\ & a_1+a_3+a_5...a_{2n-1}=6\sum^n_{k=1}(2k-1)-3\sum^n_{k=1}1=6n^2-3 \\ & \lim_{n\to\infty}[\sqrt{a_2+a_4+a_6...a_{2n}}-\sqrt{a_1+a_3+a_5...a_{2n-1}}] \\ & =\lim_{n\to\infty}[\sqrt{6n^2+3n}-\sqrt{6n^2-3n}][\frac{\sqrt{6n^2+3n}+\sqrt{6n^2-3n}}{\sqrt{6n^2+3n}+\sqrt{6n^2-3n}}] \\ & =\lim_{n\to\infty}\frac{6n}{\sqrt{6n^2+3n}+\sqrt{6n^2-3n}}=\frac{6}{2\sqrt{6}}=\frac{\sqrt{6}}{2} \end{align}$$

$$\begin{align} & a_c=\omega^2r=\frac{F}{m} \\ & m=a\rho dr\mbox{, }F=(\Delta P)\times a=(\rho g d h)\times a \\ & a_c=\omega^2r=\frac{F}{m}=g\frac{dh}{dr} \\ & \omega^2 \int r dr= g\int dh \\ & h=\omega^2\frac{r^2}{2g} \mbox{which is a parabola} \end{align}$$

$$\sqrt{2}$$

$$x\vec x+y\vec y+z\vec z+t\vec {ct}=x'\vec {x'}+y'\vec {y'}+z'\vec {z'}+t'\vec {ct'}$$

$$\begin{bmatrix} c t' \\x' \\y' \\z' \end{bmatrix} = \begin{bmatrix} a &b&c&d\\e&f&g&h\\i&j&k&l\\m&n&o&p\\ \end{bmatrix} \begin{bmatrix} c t\\x\\y\\z \end{bmatrix}. $$

$$\begin{bmatrix}x\\y\\z\\ct\\0\\0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\x'\\y'\\z'\\ct'\end{bmatrix}$$

$$\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=a\vec {x}+b\vec y+c\vec z+d\vec {ct}$$

$$\begin{matrix} t'=at+bx \\x'=ct+dx\end{matrix} \Rightarrow \begin{bmatrix}t'\\x'\end{bmatrix}= \begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}t\\x\end{bmatrix}$$

$$1=\frac{x}{ct}=\frac{x'}{ct'}$$

$$\begin{bmatrix}cT\\0\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}ct\\vt\end{bmatrix}$$ $$\begin{bmatrix}ct\\-vt\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}cT\\0\end{bmatrix}$$

$$\begin{bmatrix}ct\\vt\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\cT\\0\end{bmatrix}$$ $$\begin{bmatrix}cT\\0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\ct\\-vt\end{bmatrix}$$

$$x\vec x+ct\vec {ct}+x'\vec {x'} +ct'\vec {ct'}=\begin{bmatrix}x\\ct\\x'\\ct'\end{bmatrix}$$

$$\begin{bmatrix}a\\b\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\c\\d\end{bmatrix}$$

$$\begin{bmatrix}\end{bmatrix}$$

$$\begin{bmatrix}\gamma \\0\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\\-\frac{v}{c}\end{bmatrix}$$

$$\begin{bmatrix}1\\\frac{v}{c}\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0\\\gamma\\0\end{bmatrix}$$

$$(\frac{\frac{v}{c}-1}{\gamma})\begin{bmatrix}\gamma \\0\\0\\0\end{bmatrix}+\begin{bmatrix}1\\\frac{v}{c}\\0\\0\end{bmatrix} = \begin{bmatrix}\frac{v}{c}\\\frac{v}{c}\\0\\0\end{bmatrix} =(\frac{\frac{v}{c}-1}{\gamma})\begin{bmatrix}0\\0\\1\\-\frac{v}{c}\end{bmatrix}+\begin{bmatrix}0\\0\\\gamma\\0\end{bmatrix} = \begin{bmatrix}0\\0\\\frac{v}{c}-(\frac{v}{c})^2\\\frac{v}{c}-1+\gamma^2\end{bmatrix}$$

sec 14
$$\begin{align} & \frac{\delta([\sin y\cos x][\sin (x-y)])}{\delta x}\\ & =-\sin x \sin y \sin (x-y)+\sin y\cos x\cos(x-y)=0\\ & \Rightarrow \tan (x-y)= \tan y \\ & \frac{\delta([\sin y\cos x][\sin (x-y)]}{\delta y)}\\ & =\cos y \cos x\sin(x-y)+[-1\sin y \cos x\cos(x-y)]=0\\ & \Rightarrow \tan (x-y)= \tan y \\ & x=2y+180n^\circ \mbox{  ,    } n\in \mathbb{R} \end{align}$$

$$\begin{align} & |K||L|\sin \theta = \sqrt{(x^2+y^2+z^2)(a^2+b^2+c^2)-(\vec K\cdot \vec L)^2}\\ & = \sqrt{x^2(b^2+c^2)+y^2(c^2+a^2)+z^2(a^2+b^2)-2(xaby+bycz+czxa)}\\ & = \sqrt{(yc-zb)^2+(za-xc)^2+(xb-ay)^2} = |\vec K\times \vec L| \end{align}$$

$$v_{shm}=\sqrt{gr}=\sqrt{GM/r_0^2 \times r_0}=\sqrt{\frac{GM}{r_0}}$$

$$\Delta v= v_{shm}[(-\sin\theta,\cos\theta)-(1,0)]\,$$

$$\begin{align} & V(\theta)=V_i+\Delta v=(0,V_0)+\cfrac{g_0r_0}{V_0}[(-\sin\theta,\cos\theta)-(0,1)]\\ & =(-\cfrac{g_0r_0}{V_0}\sin\theta], [\cfrac{g_0r_0}{V_0}\cos\theta+ V_0 - \cfrac{g_0r_0}{V_0}])\\ & L=r\times v =(r\cos\theta,r\sin\theta) \times (-[\cfrac{g_0r_0}{V_0}\sin\theta], [\cfrac{g_0r_0}{V_0}\cos\theta+ V_0 - \cfrac{g_0r_0}{V_0}])\\ & = (0,0,r[\cfrac{g_0r_0}{V_0}\cos\theta -V_0\cos\theta-\cfrac{g_0r_0}{V_0}])=r_0V_0 \\ & \Rightarrow r=\frac{r_0V_0}{[\cfrac{g_0r_0}{V_0}\cos\theta -V_0\cos\theta-\cfrac{g_0r_0}{V_0}])} \\ & =\frac{r_0V_0}{ \cos\theta[\cfrac{g_0r_0}{V_0} -V_0]-\cfrac{g_0r_0}{V_0})} \end{align}$$

$$(g_0,\omega_0)\,$$

$$(g_0,r_0)\,$$

$$g_0=\omega^2r_{shm}\mbox{ and }r_{shm}\omega=V_{shm}=\frac{g_0}{\omega_0}=\cfrac{g_0r_0}{V_0}$$

sec15
$$Insert formula here$$ $$Y \overset{\mathrm{def}}{=}\frac{dP}{dr}=\cfrac{dP}{\frac{dl}{l}}=\cfrac{dP}{\frac{dV}{V}}=V\frac{dP}{dV}$$

$$v_{rms}\,$$

$$F=\frac{m{v_{rms}}^2}{L}$$

$$P=\frac{m{v_{rms}}^2}{L^3}=\frac{m{v_{rms}}^2}{V}$$

$$KE=3\frac{m{v_{rms}}^2}{2}=\frac{3}{2}PV$$

$$\frac{2}=\frac{PV}{2}$$

$$E=\frac{f}{2}PV$$

$$dW=-PdV=dE=d\left(\frac{f}{2}PV\right)=\frac{f}{2}PdV+\frac{f}{2}VdP$$

$$\begin{align} & -PdV=\frac{f}{2}PdV+\frac{f}{2}VdP \\ & \left(\frac{f}{2}+1\right)\frac{dV}{V}+\frac{f}{2}\frac{dP}{P}=0 \\ & \int\left[\left(\frac{f}{2}+1\right)\frac{dV}{V}+\frac{f}{2}\frac{dP}{P}\right] \\ & \left(\frac{f}{2}+1\right)\ln V+ \frac{f}{2}\ln P=const \\ & P^\frac{f}{2}V^{\left(\frac{f}{2}+1\right)}=const \\ & PV^{\frac{f+2}{f}}=const\\ & \because \gamma \overset{\underset{\mathrm{def}}{}}{=} \frac{f+2}{f} \\ & \therefore PV^{\gamma}= const \\ \end{align}$$

$$PV^{\gamma}= const\,$$

$$\frac{d(PV^\gamma)}{dV}=V^\gamma \frac{dP}{dV}+\gamma PV^{\gamma-1}=0 \Rightarrow \gamma P =-V\frac{dP}{dV}=K$$

$$K=\gamma P\,$$

$$\int$$

$$c\times a\times\rho=\frac{dm}{dt}$$

$$x_1$$$$x_2$$$$\rho_1$$$$\rho_2$$

$$\begin{align} & dF=dP\times a\\ & =\cfrac{d\left(\cfrac{d\left(\sum dm\times x\right)}{dt}\right)}{dt}\\ & =\frac{d\left(\frac{dm}{dt}\times\Delta x\right)}{dt}=\frac{d\left(\frac{dm}{dt}\times(x_2-x_1)\right)}{dt}\\ & =\frac{dm}{dt}\frac{d(x_2-x_1)}{dt}\\ & =ca\rho[c\frac{\rho}{\rho_2}-c\frac{\rho}{\rho_1}]\\ & =c^2a\frac{\rho^2}{\rho_1\rho_2}[\rho_1-\rho_2]\\ & =c^2a\frac{\rho^2}{\rho_1\rho_2}[d\rho]\\ & \because \rho_1\rho_2\approx\rho^2\\ & \therefore dP\times a =c^2a[d\rho] \\ \end{align}$$

$$\begin{align} & \because V\propto \frac{1}{\rho} \\ & \therefore dr=\lim_{dV\rightarrow 0} -\frac{dV}{V}=\lim_{d\rho\rightarrow 0} \frac{d\rho}{\rho} \\ & d\rho=dr\times\rho \end{align}$$

$$c=\sqrt{\frac{dP}{d\rho}}=\sqrt{\frac{dP}{dr \times \rho}}=\sqrt{\frac{K}{\rho}}$$

$$\gamma\,$$ $$\omega$$

$$c=\sqrt{\frac{\gamma P}{\rho}}$$

$$c=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{1.4(101,325Pa)}{\frac{1.2kg}{m^3}}=343.820m/s$$

$$\begin{align} & R(\cos \theta,\sin\theta)\Rightarrow R _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])\\ \end{align}$$

$$\cos\theta \hat X + \cos\theta \hat Y = \cos [\theta-\omega t] \hat x+\sin [\theta-\omega t])\hat y$$

$$R(\cos \theta,\sin\theta)= R _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])$$

$$\begin{align} & R=r(\cos \theta,\sin\theta)\Rightarrow r _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])\\ \end{align}$$

sec 16
$$\hat i=\cos\theta \hat X + \sin\theta \hat Y = \cos [\theta-\omega t] \hat x+\sin [\theta-\omega t])\hat y$$

$$\dot{\theta}=\frac{d\theta}{dt}, \ddot{\theta}=\frac{d^2\theta}{dt^2} , \dot \omega = 0 \dot{\omega}=\frac{d\omega}{dt},\dot{r}=\frac{dr}{dt},\ddot{r}=\frac{d^2r}{dt^2}$$

$$\hat j=\sin\theta \hat X - \cos\theta \hat Y = \sin [\theta-\omega t] \hat x-\cos [\theta-\omega t])\hat y$$

$$\begin{align} & R(\cos \theta,\sin\theta)\Rightarrow R _2 (\cos [\theta-\omega t],\sin [\theta-\omega t])\\ \end{align}$$

$$\begin{align} & V=\frac{d r(\cos\theta,\sin\theta)}{dt}=\dot{r}(\cos\theta,\sin\theta)+r\omega(\sin\theta,-\cos\theta) \\ & =\dot r\hat i+ r\dot \theta \hat j \\ \end{align}$$

$$(\mbox{replace }\theta \mbox{ with }\theta-\omega t)\Rightarrow $$

$$\begin{align} & A=\frac{d^2 r(\cos\theta,\sin\theta)}{dt^2} =\frac{d[\dot{r}(\cos\theta,\sin\theta)+r\dot\theta(\sin\theta,-\cos\theta)]}{dt} \\ & =[\ddot r (\cos\theta,\sin\theta)+\dot r\dot\theta(\sin\theta,-\cos\theta)]+[\dot r \dot\theta(\sin\theta,-\cos\theta) \\ & +r\ddot\theta(\sin\theta,-\cos\theta)+r\dot\theta^2(-\cos\theta,-\sin\theta)]\\ & =(\ddot r-r\dot \theta^2)\hat i+ (2\dot r\dot \theta+r\ddot \theta)\hat j \\ \end{align}$$

$$\phi = \theta-\omega t\,$$ $$\therefore \dot \theta = \dot \phi+\omega$$ $$\theta = \phi+\omega t\,$$ $$\ddot\theta =\ddot\phi$$

$$(\ddot r-r\dot \phi^2)\hat i+ (2\dot r\dot \phi+r\ddot \phi)\hat j$$

$$[\ddot r-r(\dot \phi^2+2\dot\phi \omega+\omega^2)]\hat i+[2\dot r(\dot \phi+\omega)+r\ddot \phi]\hat j$$

$$V=\dot r\hat i+ (r\dot \phi+r\omega )\hat j$$

$$\begin{align} & [\ddot r-r(\dot \phi^2+2\dot\phi \omega+\omega^2)]\hat i+[2\dot r(\dot \phi+\omega)+r\ddot \phi]\hat j\\ & -[(\ddot r-r\dot \phi^2)\hat i+ (2\dot r\dot \phi+r\ddot \phi)\hat j] =A_f=(-2r\dot \phi\omega-r\omega^2)\hat i+2\dot r \omega \hat j \end{align}$$

$$\hat i \times \hat j=\hat k,\hat k \times \hat i=\hat j, \hat j \times \hat k=-\hat i$$

$$\begin{align} & V\times \Omega = [\dot r\hat i+ (r\dot \phi+r\omega )\hat j]\times [\omega \hat k]\\ & =(-2r\dot \phi\omega-2r\omega^2)\hat i+2\dot r \omega \hat j \end{align}$$

$$\Omega\times (\Omega\times R)$$

$$\begin{align} & Ways=\frac{N!}{N_1!N_2!...N_{\inf}} \\ & \mbox{to insure energy conservation } a+c=2b \\

\end{align}$$

sec 17
$$|\vec a|=|\vec a_A-\vec a_B|=\frac{Gm_Am_B}{r^2}\times (\frac{1}{m_A}+\frac{1}{m_B})=\frac{G(m_a+m_b)}{r^2}$$ $$a=\frac{G(m_a+m_b)}{r^2}$$

$$\theta$$

$$\phi$$