Chemistry

Basics

Given some element $\mathrm{X}$

\begin{equation*}\begin{split} \ce{^{A}_{Z}X} \end{split}\end{equation*}

\begin{equation*}\begin{split} \ce{X-A} \end{split}\end{equation*}

Where

$A = \text{neutrons + protons}$
$Z = \text{protons}$

Cations: Positively （＋） Charged
Anions: Negatively （−） Charged

Mnemonic

Cations are Pawsitive

Conventions on homework

When a question says, determine the energy of 1 ㏖ of photons, the unit will be ᴶ/㏖.

Units

\begin{equation*}\begin{split} 1 \pu{mol}&= {{6.022}} \times 10^{{23}}\end{split}\end{equation*}

\begin{equation*}\begin{split} \small 1 \;\mathrm{amu} &\approx \;\text{protons} + \;\text{neutrons} \end{split}\end{equation*}

SI Prefixes

Value	Prefix	Symbol
$10^{1}$	deca	$\mathrm{da}$
$10^{2}$	hecto	$\mathrm{h}$
$10^{3}$	kilo	$\mathrm{k}$
$10^{6}$	mega	$\mathrm{M}$
$10^{9}$	giga	$\mathrm{G}$
$10^{12}$	tera	$\mathrm{T}$

Value	Prefix	Symbol
$10^{-1}$	deci	$\mathrm{d}$
$10^{-2}$	centi	$\mathrm{c}$
$10^{-3}$	mili	$\mathrm{m}$
$10^{-6}$	micro	$\mathrm{\mu}$
$10^{-9}$	nano	$\mathrm{n}$
$10^{-12}$	pico	$\mathrm{p}$

Classification of Matter

Overview

Mixtures

Heterogeneous mixture

Where the prefix Hetero- means different

The Atom and Nuclear Chemistry

Isotopes and Subatomic Particles

Electrons and Quantum Mechanics

If the average atomic mass of boron is $10.81\;\mathrm{amu}$ , what is the percent abundance of boron-11 （mass of $11.009306\;\mathrm{amu}$ ） if the only other isotope is boron-10 （mass of $10.012937\;\mathrm{amu}$ ）? Given the formula for average atomic mass:

\begin{equation*}\begin{split} \small\text{adv. mass}\normalsize&= \sum \left(\small\text{percent abundance}\normalsize\times \small\text{mass}\normalsize\right)\\ \end{split}\end{equation*}

Therefore in summary, we are given the following known quantities

\begin{equation*}\begin{split} \small\text{average atomic mass}\normalsize&= 10.81\;\mathrm{amu}\\ \small\text{boron-11 mass}\normalsize&= 11.009306\;\mathrm{amu}\\ \small\text{boron-10 mass}\normalsize&= 10.012937\;\mathrm{amu} \end{split}\end{equation*}

With the following unknown quantities

\begin{equation*}\begin{split} \small\text{boron-11 \% abundance}\normalsize&= X_{\small\text{B-11}\normalsize}\\ \small\text{boron-10 \% abundance}\normalsize&= X_{\small\text{B-10}\normalsize} \end{split}\end{equation*}

And asked to find the percent abundance of boron-11 （ $X_{\small\text{B-11}\normalsize}$ ）. Therefore our equation is

\small \begin{gather*} \underbrace{10.81\;\mathrm{amu}}_{\mathclap{\small\text{avg}\normalsize_m}} = X_{\small\text{B-11}\normalsize} \times \underbrace{11.009306\;\mathrm{amu}}_{\mathclap{\small\text{M}\normalsize_1}} + X_{\small\text{B-10}\normalsize} \times \underbrace{10.012937\;\mathrm{amu}}_{\mathclap{\small\text{M}\normalsize_2}}\\ \underbrace{\small\text{avg}\normalsize_m = X_{\small\text{B-11}\normalsize} \times M_1 + X_{\small\text{B-10}\normalsize} \times M_2}_{\small\text{shorthand}\normalsize} \end{gather*}

We have two unknowns, but luckily we can use the following fact/relation and therefore express the percent abundance of boron-10 in terms of the percent abundance of boron-11

\begin{equation*}\begin{split} X_{\small\text{B-11}\normalsize} + X_{\small\text{B-10}\normalsize} = 100\% = 1\\ X_{\small\text{B-10}\normalsize} = 1 - X_{\small\text{B-11}\normalsize} \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} \small \small\text{avg}\normalsize_m &= X_{\small\text{B-11}\normalsize} M_1 + X_{\small\text{B-10}\normalsize} M_2\\ \small\text{avg}\normalsize_m &= X_{\small\text{B-11}\normalsize} M_1 + \left(1 - X_{\small\text{B-11}\normalsize}\right) M_2\\ \small\text{avg}\normalsize_m &= X_{\small\text{B-11}\normalsize} M_1 + M_2 - X_{\small\text{B-11}\normalsize} M_2\\ \small\text{avg}\normalsize_m - M_2 &= X_{\small\text{B-11}\normalsize} \left(M_1 - M_2\right)\\ \frac{\small\text{avg}\normalsize_m - M_2}{M_1 - M_2}&= X_{\small\text{B-11}\normalsize}\\ \therefore X_{\small\text{B-11}\normalsize} &= \frac{\small\text{avg}\normalsize_m - M_2}{M_1 - M_2}\\ &= \frac{10.81\;\mathrm{amu} - 10.012937\;\mathrm{amu}}{11.009306\;\mathrm{amu} - 10.012937\;\mathrm{amu}}\\ &\approx \underbrace{0.7995}_{\mathclap{ \begin{gathered} \small\text{decimal form}\normalsize\\ \small\text{multiply by $100$ to get percentage}\normalsize\end{gathered} }}\\ &\approx 79.95\% \end{split}\end{equation*}

But we aren't done, we have to compute significant figures.

\begin{gather*} \small X_{\small\text{B-11}\normalsize} = \begin{rcases} \frac{ \overbrace{10.81\;\mathrm{amu} - 10.012937\;\mathrm{amu}}^{\text{2 sig decimal places}} }{ \underbrace{11.009306\;\mathrm{amu} - 10.012937\;\mathrm{amu}}_{\text{8 sig decimal places}} }\\ \end{rcases}\therefore\;\small\text{2 sig figs}\normalsize\end{gather*}

Therefore, we round our answer to 2 sig figs, yielding $80.\%$

Molecules and Compounds

Terms

Electronegativity: Can be used to to determine type of bond based on electronegativity difference.
Ionic Bond
Covalent Bond
Polar Covalent Bond: Sharing of electrons.
Hydrogen Bond: Hydrogen is bonded with an electronegative element.

Prefixes

Prefix	Value
mono-	$1$
di-	$2$
tri-	$3$
tetra-	$4$
penta-	$5$

Prefix	Value
hexa-	$6$
hepta-	$7$
octa-	$8$
nona-	$9$
deca-	$10$

Formula Summary

Formal Charge （Option 1）

Given some element $\ce{X}$ which has $v_e$ number of valance electrons, and a lewis diagram of bonded （ $b_e$ ） electrons and non-bonded （ $f_e$ ） electrons, the formal charge for the given element in the lewis structure is

\begin{equation*}\begin{split} \mathrm{FC} &= v_e - f_e - \frac{b_e}{2}\end{split}\end{equation*}

For the given quantities

\begin{equation*}\begin{split} \small\text{FC}\normalsize&= \small\text{Formal Change}\normalsize\\ v_e &= \small\text{Valence electrons (from element)}\normalsize\\ f_e &= \small\text{Non-bonded (free) electrons (from diagram)}\normalsize\\ b_e &= \small\text{Bonded Valence electrons (from diagram)}\normalsize\end{split}\end{equation*}

Formal Charge （Option 2）

Given some element $\ce{X}$ which has $v_e$ number of valance electrons, and a lewis diagram of some number of bonds （ $b$ ） and dots （ $d$ ）, the formal charge for the given element in the lewis structure is

\begin{equation*}\begin{split} \mathrm{FC} &= v_e - (b + d) \end{split}\end{equation*}

For the given quantities

\begin{equation*}\begin{split} \small\text{FC}\normalsize&= \small\text{Formal Change}\normalsize\\ v_e &= \small\text{Valence electrons (from element)}\normalsize\\ b &= \small\text{Number of bonds (from diagram)}\normalsize\\ d &= \small\text{Number of dots (from diagram)}\normalsize\end{split}\end{equation*}

Note: each bond is 2 electrons, but counts as one bond.

Polyatomic Ions

Remembering the number of oxygens

General

Given a set of oxygens in increasing order

\set{\ce{ O_{\small\text{Lowest}\normalsize}, O_{\small\text{Lower}\normalsize}, O_{\small\text{Higher}\normalsize}, O_{\small\text{Highest}\normalsize} }}

Where

\set{ \mathrm{Lowest} < \mathrm{Lower} < \mathrm{Higher} < \mathrm{Highest} }

For example

\set{\ce{ O_1, O_2, O_3, O_4 }}

Given some element or compound $\ce{X}$ ionically bonded with such

\set{\ce{ XO_{\small\text{Lowest}\normalsize}, XO_{\small\text{Lower}\normalsize}, XO_{\small\text{Higher}\normalsize}, XO_{\small\text{Highest}\normalsize} }}

For example

\set{\ce{ ClO_1^{-}, ClO_2^{-}, ClO_3^{-}, ClO_4^{-} }}

The following table denotes the naming conventions therein

Prefix	Prefix Meaning （relative to suffix）	Suffix	Oxygen Order	Example
Prefix	Prefix Meaning （relative to suffix）	Suffix	Oxygen Order	Formula	Name
Per-	More than	-ate	Highest #	$\ce{ClO_4^{-}}$	Perchlorate
		-ate	Higher #	$\ce{ClO_3^{-}}$	Chlorate
		-ite	Lower #	$\ce{ClO_2^{-}}$	Chlorite
Hypo-	less than	-ite	Lowest #	$\ce{ClO^{-}}$	Hypochlorite

Oxygen vs No-Oxygen Comparison

Prefix	Suffix	Meaning	Example	Name
Per-	-ate		$\ce{SO_5^{2-}}$ or $\ce{S_2O_8^{2-}}$	Persulfate
	-ate		$\ce{SO_4^{2-}}$	Sulfate
	-ite		$\ce{SO_3^{2-}}$	Sulfite
Hypo-	-ite		$\ce{SO_2^{2-}}$	Hyposulfite
	-ide	No oxygennon-metal	$\ce{S^{2-}}$	Sulfide

In summary

Per*ate

Greater than

*ate
*ite
Hypo*ite

*ate

Less than

Per*ate

Greater than

*ite
Hypo*ite

*ite

Less than

Per*ate
*ate

Greater than

Hypo*ite

Hypo*ite

Less than

Per*ate
*ate
*ite

Generally

per-ate	Used in the ion with the largest number of oxygen atoms
-ate-ite	Polyatomic ions of oxygen
hypo-ite	Used in the ion with the lowest number of oxygen atoms
-ide	Non-metal, no oxygens

Determining The Charge^†

Warning

Phosphate （ $\ce{PO_4^{3-}}$ ） is the only one that violates this rule!

Example for Nitrate

Example for Cyanide

Example for Oxalate

Example for Hydrogen Carbonate

†

Source

Ionic Lewis Structures

Examples

Lewis structure for sulfate ion

Notes

When naming an ionic compound, the name of the cation is followed by the name of the anion. Monatomic anions are named with the ending -ide.

Periodic Properties of the Elements

Note, electron affinity is not the same as electronegativity!

Terms

Isoelectronic: Atoms with the same number of electrons.
Ionization energy
Predict Metallic Character Based on Periodic Trends

Expanded Octet （Exceptions to the Octet Rule）

All non-metals from period 3 to period 8 of the Periodic Table, can have expanded octets.

Quantum Mechanical Models of the Atom

The Electromagnetic Spectrum

Terms

Pauli Exclusion Principle: No two electrons in an atom can have the same four quantum numbers.; Pauli’s Principle prevents two electrons with the same spin from existing in the same subshell, each subshell will be filled with one spin direction before they are filled with the opposite spin. This is the second of Hund’s Rules.
Aufbau Principle: This pattern of orbital filling is known as the aufbau principle （the German word aufbau means “build up”）.
Hund’s rule: When filling degenerate orbitals, electrons fill them singly first, then with parallel spins.; I.e. start by filling boxes with single 'upward' arrows, and then once all of such boxes are maxed out, then you add double arrows pointing in opposite directions.; Pauli’s Principle prevents two electrons with the same spin from existing in the same subshell, each subshell will be filled with one spin direction before they are filled with the opposite spin. This is the second of Hund’s Rules.
Coulomb’s Law: $\begin{equation*}\begin{split} \mathrm{E} &= \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1\;q_2}{r}\end{split}\end{equation*}$

Aufbau Principle

Hund’s rule

Pauli's Exclusion Principle

Each election has a unique set of four quantum numbers （i.e. see quantum numbers）. They are

$\mathrm{n}$
$\mathrm{l}$
$\mathrm{m}_l$
$\mathrm{m}_s$

Overview

Formulas

\begin{equation*}\begin{split} \mathrm{f} = \nu &= \frac{c}{\lambda}\end{split}\end{equation*}

\begin{equation*}\begin{split} \mathrm{\lambda} = \frac{\small\text{speed}\normalsize}{\small\text{frequency}\normalsize}\end{split}\end{equation*}

\begin{equation*}\begin{split} \mathrm{E} &= \mathrm{h}\times\mathrm{f}\\ &= \mathrm{h}\frac{\mathrm{c}}{\lambda}\end{split}\end{equation*}

\begin{equation*}\begin{split} \mathrm{T} &= \frac{1}{f}\end{split}\end{equation*}

\begin{equation*}\begin{split} \lambda &= \frac{c}{f}\end{split}\end{equation*}

Values

Name	Symbol	Unit	Description	Range
Wavelength	$\mathrm{\lambda}$	Any unit for distance	Distance between two analogous points	Always Positive
Frequency	$f$ or $\nu$ （nu）	$㎐ = \frac{\text{1 cycle}}{\text{second}}$ $\mathrm{s}^{-1}$	Number of cycles	Always Positive
Energy	$\mathrm{E}$	$\mathrm{J}$ （joule）	Amount of energy （ $\mathrm{E}$ ） in a light packet

Constants

Name	Symbol	Unit	Value
Speed of Light	$\mathrm{c}$	$\\frac{\\mathrm{m}}{\\mathrm{s}}$	$\mathrm{c} = {{3.00}} \times 10^{{8}}\\frac{\\mathrm{m}}{\\mathrm{s}}$
Planck's constant	$\mathrm{h}$	Energy multiplied by time joule-seconds $\mathrm{J}\cdot\mathrm{s}$	$\mathrm{h} = {{6.626}} \times 10^{{-34}}\;\mathrm{J}\cdot\mathrm{s}$

Other Formulas

de Broglie Relation

\begin{equation*}\begin{split} \lambda &= \frac{\mathrm{h}}{\mathrm{m}\mathrm{v}}\\ \text{where}&\\ m &= \text{mass}\\ v &= \text{velocity} \neq \nu\;\text{(nu)} \end{split}\end{equation*}

Heisenberg's Uncertainty Principle

\begin{equation*}\begin{split} \Delta{x} \times \mathrm{m} \Delta{v} \geq \frac{\mathrm{h}}{4\pi}\end{split}\end{equation*}

Where

$\Delta{x}$ is the uncertainty in position.
$\Delta{v}$ is the uncertainty in velocity.
$\mathrm{m}$ is the mass of the particle.
$\mathrm{h}$ is the plank's constant.

In general it states that the more you know about an electrons position, the less you know about it's velocity.

Energy of an Electron in an Orbital with Quantum Number $\mathrm{n}$ in a Hydrogen Atom

\begin{equation*}\begin{split} E_n &= {{-2.18}} \times 10^{{-18}}\mathrm{J} \left(\frac{1}{n^2}\right) \end{split}\end{equation*}

Energy of an Electron in an Orbital with Quantum Number $\mathrm{n}$ for any atom

\begin{equation*}\begin{split} E_n &= {{-2.18}} \times 10^{{-18}}\mathrm{J} \left(\frac{1}{n^2}\right) \cdot \mathrm{Z}^2 \end{split}\end{equation*}

Where $\mathrm{Z}$ is the atomic number of the given element.

Change in Energy That Occurs in an Atom When It Undergoes a Transition between Levels （Further Details）

$n_{\small\text{initial}}$ and $n_{\small\text{final}}$

\begin{equation*}\begin{split} \Delta{E} &= {{-2.18}} \times 10^{{-18}}\mathrm{J} \left(\frac{1}{n^2_f}- \frac{1}{n^2_i}\right) \end{split}\end{equation*}

If $\Delta{E}$ is negative, energy is being released.
If $\Delta{E}$ is positive, energy is being absorbed.

Ionization Energy

\begin{equation*}\begin{split} \mathrm{IE}_{n,\small\text{Big Jump}\normalsize} -1 &= \small\text{\# of valance electrons}\normalsize\\ &= \small\text{Element row \#}\normalsize\end{split}\end{equation*}

Where

$n$ is the electron number
$\small\text{Big jump}\normalsize$ the point on the table where you see the highest difference/delta.

Atomic Spectroscopy

The Principal Quantum Number （n）（Hydrogen Atom）

For the hydrogen atom, the energy of an electron in an orbital with quantum number $n$ is given by

\begin{equation*}\begin{split} E_n = {{-2.18}} \times 10^{{-18}}\mathrm{J}\;\frac{1}{n^2}\end{split}\end{equation*}

Therefore the difference in energy is given by the following

\begin{equation*}\begin{split} \Delta E &= \Delta E_{\small\text{Final}\normalsize} - \Delta E_{\small\text{Initial}\normalsize}\\ &= \left({{-2.18}} \times 10^{{-18}}\mathrm{J}\;\frac{1}{{n_f}^2}\right) - \left({{-2.18}} \times 10^{{-18}}\mathrm{J}\;\frac{1}{{n_i}^2}\right)\\ &= {{-2.18}} \times 10^{{-18}}\mathrm{J}\;\frac{1}{{n_f}^2}+ {{2.18}} \times 10^{{-18}}\mathrm{J}\;\frac{1}{{n_i}^2}\\ &= {{-2.18}} \times 10^{{-18}}\mathrm{J}\; \left(\frac{1}{{n_f}^2}- \frac{1}{{n_i}^2}\right) \end{split}\end{equation*}

The Principal Quantum Number （n）（Any Atom）

For the hydrogen atom, the energy of an electron in an orbital with quantum number $n$ is given by

TODO

\begin{equation*}\begin{split} E_n &= {{-2.18}} \times 10^{{-18}}\mathrm{J} \left(\frac{1}{n^2}\right) \cdot \mathrm{Z}^2 \end{split}\end{equation*}

Where $\mathrm{Z}$ is the atomic number of the given element.

Electron Configuration

Traditional Chart

Better Method

Examples

Electron configuration for $_{26}\mathrm{Fe}$

\begin{equation*}\begin{split} \ce{\underbrace{1s^2 2s^2 2p^6 3s^2 3p^6}_{\small\text{Equal to Argon}\normalsize} 4s^2 3d^6} &= \ce{[Ar] 4s^2 3d^6} \end{split}\end{equation*}

Since the electron configuration for Argon is

\begin{equation*}\begin{split} \ce{[Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6} \end{split}\end{equation*}

Electron configuration for $_{26}\mathrm{Fe}^{+2}$

Beginning with the electron configuration for $_{26}\mathrm{Fe}$

\begin{equation*}\begin{split} \underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}_{\small\text{Equal to Argon}\normalsize}\; \overbrace{4s^2\; 3d^6}^{\mathclap{ \begin{gathered} \small\text{Which should we remove 2 $\mathrm{e}^{-}$ from?}\normalsize\\ \small\text{($4s^2$ or $3d^6$)?}\normalsize\\ \end{gathered} }} &= \ce{[Ar] \underbrace{4s^2\; 3d^6}_{\mathclap{ \begin{gathered} \small\text{Which should we remove 2 $\mathrm{e}^{-}$ from?}\normalsize\\ \small\text{($4s^2$ or $3d^6$)?}\normalsize\\ \end{gathered} }} } \end{split}\end{equation*}

Remove the electrons from the term with the higher electron state. Warning! Do not just remove the electrons from the rightmost term since the rightmost term may be a lower electron state. For instance given $4s^2\; 3d^6$

$4s^2$ is in a higher electron state
$3d^6$ is in a lower electron state

Alternatively, to compute the lowest energy orbital, add the principle quantum number （ $n$ ） to the The angular momentum quantum number （ $l$ ） to get the orbital with the lowest energy. Therefore

\begin{equation*}\begin{split} \small\text{orbital with the lowest energy}\normalsize&= O_{\small\text{min}\normalsize} = n + l\\ \end{split}\end{equation*}

Given $\ce{4s^2}$ and $\ce{3d^6}$

\begin{equation*}\begin{split} \ce{4s^2}\;\begin{dcases} n &= 4\\ i &= 0\\ \therefore\; O_{\small\text{min}\normalsize} &= 4 \end{dcases}\end{split}\end{equation*}

\begin{equation*}\begin{split} \ce{3d^6}\; \begin{dcases} n &= 3\\ i &= 2\\ \therefore\; O_{\small\text{min}\normalsize} &= 5 \end{dcases}\end{split}\end{equation*}

NVM this is an exception to the rule...

TODO move this somewhere else...

As shown

\begin{equation*}\begin{split} \underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}\; \overbrace{4s^{\left(2 - 2\right)}}^{ \mathclap{ \begin{gathered} \small\text{higher state}\normalsize\\ \small\text{remove $2\mathrm{e^{-}}$}\normalsize\end{gathered} }}\; \underbrace{3d^6}_{ \mathclap{\small\text{lower state}\normalsize}}\; &= [\mathrm{Ar}]\; \overbrace{4s^{\left(2 - 2\right)}}^{ \mathclap{ \begin{gathered} \small\text{higher state}\normalsize\\ \small\text{remove $2\mathrm{e^{-}}$}\normalsize\end{gathered} }}\; \underbrace{3d^6}_{ \mathclap{\small\text{lower state}\normalsize}}\\\\ \underbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}\; \underbrace{3d^6}_{ \mathclap{\small\text{unchanged}\normalsize}}\; &= [\mathrm{Ar}]\; \underbrace{3d^6}_{ \mathclap{\small\text{unchanged}\normalsize}} \end{split}\end{equation*}

Therefore the electron configuration for $_{26}\mathrm{Fe}^{+2}$ is:

\begin{equation*}\begin{split} \ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^6} &= \ce{[Ar] 3d^6} \end{split}\end{equation*}

Electron configuration for $_{24}\mathrm{Cr}$

It would appear that the electron configuration for $_{24}\mathrm{Cr}$ would be

\begin{equation*}\begin{split} \overbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}^{\small\text{Equal to Argon}\normalsize}\; \underbrace{4s^2\; 3d^4}_{ \mathclap{\small\text{this is wrong!}\normalsize}}\; &= [\mathrm{Ar}]\; \underbrace{4s^2\; 3d^4}_{ \mathclap{\small\text{this is wrong!}\normalsize}}\; \end{split}\end{equation*}

But this is wrong! It's actually

\begin{equation*}\begin{split} \overbrace{1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^6}^{\small\text{Equal to Argon}\normalsize}\; \underbrace{4s^1\; 3d^5}_{ \mathclap{\small\text{notice the superscripts}\normalsize}}\; &= [\mathrm{Ar}]\; \underbrace{4s^1\; 3d^5}_{ \mathclap{\small\text{notice the superscripts}\normalsize}}\; \end{split}\end{equation*}

How-tos

What are the valence electrons?

Given

\begin{equation*}\begin{split} \ce{1s^2 2s^2 2p^6 3s^2 3p^4} \end{split}\end{equation*}

The valance electrons will be the ones in the highest energy state. Therefore

\begin{equation*}\begin{split} \overbrace{1s^2\; 2s^2\; 2p^6}^{\mathclap{ \small\text{Core electrons}\normalsize}}\; \underbrace{3s^2\;3p^4}_{\mathclap{ \begin{gathered} \small\text{Highest state}\normalsize\\ \small\text{Therefore these are the valence electrons}\normalsize\end{gathered} }} \end{split}\end{equation*}

Therefore there are $6$ valence electrons.

Given

\begin{equation*}\begin{split} \ce{[Ne] 3s^2 3p^2} \end{split}\end{equation*}

The valance electrons will be the ones in the highest energy state. Therefore

\begin{equation*}\begin{split} [\mathrm{Ne}] \underbrace{3s^2 3p^2}_{\mathclap{ \begin{gathered} \small\text{Highest state}\normalsize\\ \small\text{Therefore these are the valence electrons}\normalsize\end{gathered} }} \end{split}\end{equation*}

Therefore there are $4$ valence electrons.

Quantum Numbers

Overview

Symbol	Description
$\mathrm{n}$	The principle quantum number
$l$	The angular momentum quantum number
$\mathrm{m}_1$	The magnetic quantum number
$\mathrm{m}_s$	The spin quantum number

The Principle Quantum Number （ $\mathrm{n}$ ）

Value of $\mathrm{n}$	Value of $l$	Orbital Sublevel
$\mathrm{n} = 1$	$l = 0$	$1\mathrm{s}$
$\mathrm{n} = 2$	$l = 0$ $l = 1$	$2\mathrm{s}$ $2\mathrm{p}$
$\mathrm{n} = 3$	$l = 0$ $l = 1$ $l = 2$	$3\mathrm{s}$ $3\mathrm{p}$ $3\mathrm{d}$
$\mathrm{n} = 4$	$l = 0$ $l = 1$ $l = 2$ $l = 3$	$4\mathrm{s}$ $4\mathrm{p}$ $4\mathrm{d}$ $4\mathrm{f}$

Angular Momentum Quantum Number

Value	Result
$l = 0$	$\mathrm{s}$
$l = 1$	$\mathrm{p}$
$l = 2$	$\mathrm{d}$
$l = 3$	$\mathrm{f}$

Value of $l$	Value of $\mathrm{m_l}$
$l = 0$	$\mathrm{m_l} = 0$
$l = 1$	$\mathrm{m_l} = -1$ $\mathrm{m_l} = 0$ $\mathrm{m_l} = 1$
$l = 2$	$\mathrm{m_l} = -2$ $\mathrm{m_l} = -1$ $\mathrm{m_l} = 0$ $\mathrm{m_l} = 1$ $\mathrm{m_l} = 2$
$l = 2$	$\mathrm{m_l} = -3$ $\mathrm{m_l} = -2$ $\mathrm{m_l} = -1$ $\mathrm{m_l} = 0$ $\mathrm{m_l} = 1$ $\mathrm{m_l} = 2$ $\mathrm{m_l} = 3$

Summary

\begin{equation*}\begin{split} l \leq \mathrm{n} - 1 \end{split}\end{equation*}

\begin{equation*}\begin{split} -l \leq \mathrm{m_l} \leq l \end{split}\end{equation*}

Useful Formulas

The equation for a maximum number of electrons a given energy level can hold given some value for $n$

\begin{equation*}\begin{split} \text{max}_{\mathrm{e}^{-}} &= 2n^2 \end{split}\end{equation*}

How many orbitals are possible given some value for $n$

\begin{equation*}\begin{split} \text{max}_{\small\text{orbitals}\normalsize} &= n^2 \end{split}\end{equation*}

Examples

Light

Interference and Diffraction

Constructive Interference

If two waves of equal amplitude are in phase when they interact—that is, they align with overlapping crests—a wave with twice the amplitude results. This is called constructive interference.

Destructive Interference

If two waves are completely out of phase when they interact—that is, they align so that the crest from one overlaps with the trough from the other—the waves cancel by destructive interference.

Mathematics

Algebra

Miscellaneous

Functional Utilities & Notation Conveniences

Right to Left Evaluation

\begin{equation*}\begin{split} f \triangleleft x &= f(x) \end{split}\end{equation*}

\begin{equation*}\begin{split} f \circ g \triangleleft x &= f(g(x)) \\ &= f \triangleleft g \triangleleft x \end{split}\end{equation*}

Left to Right Evaluation

\begin{equation*}\begin{split} x \triangleright f &= f(x) \end{split}\end{equation*}

Derivative Shorthand

\begin{equation*}\begin{split} \delta f(x) &= \frac{\mathrm{d}}{\mathrm{d}x} f(x) = f^\prime(x) \end{split}\end{equation*}

For this notation, the derivative with respect to a given variable, is implicit.

Radians & Radian Conversion

Constants

\begin{equation*}\begin{split} \tau &= 2\pi = 360^{\circ} \\ \pi &= \frac{1}{2}\tau = 180^{\circ} \end{split}\end{equation*}

Conversion

\begin{equation*}\begin{split} \text{Given}\\ \textcolor{blue}{1}^{\circ} &= \frac{\textcolor{blue}{1}}{360}\tau \; {\displaystyle {\mathrm{rad}}} \\ \textcolor{blue}{1} \; \mathrm{rad} &= \frac{\textcolor{blue}{1}}{\tau}\cdot 360^{\circ} = \textcolor{blue}{1} \cdot \frac{360^{\circ}}{\tau}\\ \\\text{Degrees to Radians}\\ \textcolor{blue}{x^{\circ}} &= \frac{\textcolor{blue}{x}}{360}\tau \; {\displaystyle {\mathrm{rad}}} \\ \\\text{Radians to Degrees}\\ \textcolor{blue}{x} \; \mathrm{rad} &= \frac{\textcolor{blue}{x}}{\tau}\cdot 360^{\circ} \\ &= \textcolor{blue}{x} \cdot \frac{360^{\circ}}{\tau}\\ &= \textcolor{blue}{x} \cdot \frac{180^{\circ}}{\pi}\\ \end{split}\end{equation*}

Constants

ℯ （Euler's number）

\begin{equation*}\begin{split} \mathrm{e} &= \sum_{n = 0}^\infty \frac{1}{n!}\\ &= \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n \\ &= \lim_{t \to 0} \left(1 + t\right)^{\frac{1}{t}} \end{split}\end{equation*}

\begin{equation*}\begin{split} \mathrm{e}^x &= 1 + \frac{x}{1!}+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \cdots \\ &= \sum_{n = 0}^\infty \frac{x^n}{n!}\\ &= \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n \end{split}\end{equation*}

Algebra

Properties

\begin{equation*}\begin{split} a^m \cdot b^n &= a^{m+n}\\ \left(a^m\right)^n &= a^{m\cdot\,n} = \left(a^n\right)^m\\ \left(a\cdot\,b\right)^n &= a^n \cdot b^n\\ \left(\frac{a}{b}\right)^{-n} &= \left(\frac{b}{a}\right)^n\\ \frac{x^n}{y^n}&= \left(\frac{x}{y}\right)^n\\ x^{y^z} &= x^{\left(y ^ z\right)} \neq \left(x^y\right)^z \end{split}\end{equation*}

\begin{equation*}\begin{split} |x| &= \sqrt{x^2} \neq x \end{split}\end{equation*}

\begin{equation*}\begin{split} \log_{\beta}(\alpha) &= \gamma\\ \beta^{\gamma} &= \alpha\\ \beta^{\log_{\beta}(N)} &= N\;\text{for all $N > 0$}\\ \log_{\beta}(\beta^x) &= x \end{split}\end{equation*}

Trigonometry

The Unit Circle & Special Angles In Trig

Warning

Never use Pi （ $\pi$ ）! It makes （thinking in terms of） radians confusing, Tao （ $\tau$ ） is what the enlightened trigonometer uses, and won't screw you over.

To easily memorize the special angles in trig, notice the repeating patterns on the above angles.

For values on the x-axis, anything over $\frac{1}{4}\tau$ and under $\frac{3}{4}\tau$ will be negative
For values on the y-axis, anything over $\frac{1}{2}\tau$ will be negative
Diagonals will be $\pm \frac{\sqrt{2}}{2}$ , For ratios of $\frac{1}{12}\tau$ on the sides, i.e. $\frac{1}{12}\tau$ , $\frac{2}{12}\tau$ , $\frac{4}{12}\tau$ , $\frac{5}{12}\tau$ , $\frac{7}{12}\tau$ , $\frac{8}{12}\tau$ , $\frac{10}{12}\tau$ , and $\frac{11}{12}\tau$ . Draw a circle and dot the point where it occurs （which is pretty easy since the above are simple ratios of a circle when expressed in terms of $\tau$ ）. Then with regards to the $x$ and $y$ axis values:
- The longer size will be $\pm \frac{\sqrt{3}}{2}$
- The shorter side will be $\pm \frac{1}{2}$
See the above examples.

Trigonometric Identities

Pythagorean Identities

\begin{equation*}\begin{split} \cos^2(\theta) + \sin^2(\theta) = 1 \end{split}\end{equation*}

\begin{equation*}\begin{split} \sec^2(\theta) - \tan^2(\theta) &= 1 \\ \sec^2(\theta) &= 1 + \tan^2(\theta) \end{split}\end{equation*}

\begin{equation*}\begin{split} \csc^2(\theta) - \cot^2(\theta) &= 1 \\ \csc^2(\theta) &= 1 + \cot^2(\theta) \end{split}\end{equation*}

Sum and Difference Identities

\begin{equation*} \begin{split} \cos(\alpha - \beta) &= \cos(\alpha) \cdot \cos(\beta) + \sin(\alpha) \cdot \sin(\beta) \\ \cos(\alpha + \beta) &= \cos(\alpha) \cdot \cos(\beta) - \sin(\alpha) \cdot \sin(\beta) \\ &\\ \sin(\alpha - \beta) &= \sin(\alpha) \cdot \cos(\beta) - \cos(\alpha) \cdot \sin(\beta) \\ \sin(\alpha + \beta) &= \sin(\alpha) \cdot \cos(\beta) + \cos(\alpha) \cdot \sin(\beta) \\ &\\ \tan(\alpha + \beta) &= \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \cdot \tan(\beta)}\\ \tan(\alpha - \beta) &= \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \cdot \tan(\beta)}\end{split} \end{equation*}

Cofunction Identities

\begin{equation*}\begin{split} \sin(\theta) = \cos(\frac{1}{4}\tau - \theta) \end{split}\end{equation*}

\begin{equation*}\begin{split} \cos(\theta) = \sin(\frac{1}{4}\tau - \theta) \end{split}\end{equation*}

\begin{equation*}\begin{split} \tan(\theta) = \cot(\frac{1}{4}\tau - \theta) \end{split}\end{equation*}

\begin{equation*}\begin{split} \cot(\theta) = \tan(\frac{1}{4}\tau - \theta) \end{split}\end{equation*}

\begin{equation*}\begin{split} \csc(\theta) = \sec(\frac{1}{4}\tau - \theta) \end{split}\end{equation*}

\begin{equation*}\begin{split} \sec(\theta) = \csc(\frac{1}{4}\tau - \theta) \end{split}\end{equation*}

Ratio Identities

\begin{equation*} \begin{split} \tan(90^\circ - x) & = \frac{\sin(90^\circ - x)}{\cos(90^\circ - x)}= \frac{\cos(x)}{\sin(x)}= \cot(x) \\ \\ \cot(90^\circ - x) & = \frac{\cos(90^\circ - x)}{\sin(90^\circ - x)}= \frac{\sin(x)}{\cos(x)}= \tan(x) \\ \end{split} \end{equation*}

Double-Angle Identities

\begin{equation*} \begin{split} \sin(2\alpha) &= 2\sin(\alpha)\cos(\alpha) \\ \cos(2\alpha) &= \cos^2(\alpha) - \sin^2(\alpha) \\ &= 1 - 2\sin^2(\alpha) \\ &= 2\cos^2(\alpha) - 1 \\ &\\ \tan(2\alpha) &= \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)}\end{split} \end{equation*}

Half-Angle Identities

\begin{equation*} \begin{split} \sin \frac{\alpha}{2}&= \pm \sqrt{\frac{1 - \cos(\alpha)}{2}} \\ \cos \frac{\alpha}{2}&= \pm \sqrt{\frac{1 + \cos(\alpha)}{2}} \\ \tan \frac{\alpha}{2}&= \pm \sqrt{\frac{1 - \cos(\alpha)}{1 + \cos(\alpha)}} \\ &= \frac{sin(\alpha)}{1 + \cos(\alpha)}\\ &= \frac{1 - \cos(\alpha)}{sin(\alpha)}\end{split} \end{equation*}

Power-Reducing Identities

\begin{equation*} \begin{split} \sin^2(\alpha) &= \frac{1 - \cos(2\alpha)}{2}\\ &\\ \cos^2(\alpha) &= \frac{1 + \cos(2\alpha)}{2}\\ &\\ \tan^2(\alpha) &= \frac{1 - \cos(2\alpha)}{1 + \cos(2\alpha)}\end{split} \end{equation*}

\begin{equation*}\begin{split} \sin\alpha\cdot\cos\alpha &= \frac{1}{2}\sin(2\alpha) \end{split}\end{equation*}

Product-to-Sum Identities

\begin{equation*} \begin{split} \sin(\alpha) \cdot \cos(\beta) &= \frac{1}{2}\Big[ \sin(\alpha + \beta) + \sin(\alpha - \beta) \Big] \\ \cos(\alpha) \cdot \sin(\beta) &= \frac{1}{2}\Big[ \sin(\alpha + \beta) - \sin(\alpha - \beta) \Big] \\ \cos(\alpha) \cdot \cos(\beta) &= \frac{1}{2}\Big[ \cos(\alpha + \beta) + \cos(\alpha - \beta) \Big] \\ \sin(\alpha) \cdot \sin(\beta) &= \frac{1}{2}\Big[ \cos(\alpha - \beta) - \cos(\alpha + \beta) \Big] \end{split} \end{equation*}

Sum-to-Product-Identities

\begin{equation*} \begin{split} \sin(x) + \sin(y) &= 2 \cdot \sin\left( \frac{x + y}{2}\right) \cdot \cos\left( \frac{x - y}{2}\right) \\ \cos(x) + \cos(y) &= 2 \cdot \cos\left( \frac{x + y}{2}\right) \cdot \cos\left( \frac{x - y}{2}\right) \\ \sin(x) - \sin(y) &= 2 \cdot \cos\left( \frac{x + y}{2}\right) \cdot \sin\left( \frac{x - y}{2}\right) \\ \cos(x) - \cos(y) &= -2 \sin \cos\left( \frac{x + y}{2}\right) \cdot \sin\left( \frac{x - y}{2}\right) \end{split} \end{equation*}

Trigonometric Equations

Euler's Formula

\begin{equation*}\begin{split} e^{i x} &= \cos(x) + \mathrm{i}\sin(x) \end{split}\end{equation*}

\begin{equation*}\begin{split} \mathrm{e} &= \sum_{n = 0}^\infty \frac{1}{n!}\\ &= \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n \\ &= \lim_{t \to 0} \left(1 + t\right)^{\frac{1}{t}} \end{split}\end{equation*}

\begin{equation*}\begin{split} \mathrm{e}^x &= 1 + \frac{x}{1!}+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \cdots \\ &= \sum_{n = 0}^\infty \frac{x^n}{n!}\\ &= \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n \end{split}\end{equation*}

Coordinate & Number Systems

\begin{equation*}\begin{split} x &= r \cdot \cos\,\theta \\ y &= r \cdot \sin\,\theta \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} z &= x + \mathrm{i}y \\ &= r \left(\cos\, \theta + \mathrm{i}\sin\, \theta\right)\\ & = r\;\mathrm{cis}\; \theta \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} e^{i x} &= \cos\, x + \mathrm{i}\sin\, x \\ e ^{i x} &= r \left(\cos\, \theta + \mathrm{i}\sin\, \theta\right) \\ \left(\cos\, x + \mathrm{i}\sin\, x\right)^n &= \cos(n x) + \mathrm{i}\sin(n x) \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} r &= |z| = |x + \mathrm{i}y| = \sqrt{x^2 + y^2} \\ r^2 &= x^2 + y^2 \\ \tan \theta &= \frac{y}{x}\\ \theta &= \arctan\left(\frac{y}{x}\right) \end{split}\end{equation*}

Polar Coordinate System

Given

\begin{equation*}\begin{split} \tan \theta &= \frac{y}{x}\\ r^2 &= x^2 + y^2 \\ \end{split}\end{equation*}

Then

\begin{equation*}\begin{split} x &= r \cdot \cos\,\theta \\ y &= r \cdot \sin\,\theta \\ \end{split}\end{equation*}

Properties

Given

\begin{equation*}\begin{split} z_1 &= r_1\;\mathrm{cis}\;\theta_1 \\ z_2 &= r_1\;\mathrm{cis}\;\theta_1 \end{split}\end{equation*}

Then

\begin{equation*}\begin{split} z_1 \cdot z_2 &= r_1 \cdot r_2 \;\mathrm{cis}\;\left(\theta_1 + \theta_2\right) \\ \frac{z_1}{z_2}&= \frac{r_1}{r_2}\;\mathrm{cis}\;\left(\theta_1 - \theta_2\right) \\ \end{split}\end{equation*}

De Moivre’s Theorem

\begin{equation*}\begin{split} z^n &= r^n \;\mathrm{cis}\;\;n\theta \end{split}\end{equation*}

De Moivre’s Theorem For Finding Roots

\begin{equation*}\begin{split} \underbrace{w_k = r^{\frac{1}{n}}\;\mathrm{cis}\;\frac{\theta + \tau\cdot{k}}{n}} _{\begin{split}\forall\; k &= 0,1,2\cdots,n-1\\ n &\geq 1\end{split}} \end{split}\end{equation*}

Trigonometric form of a complex number

\begin{equation*}\begin{split} z &= x + \mathrm{i}y \\ &= r \left(\cos\, \theta + \mathrm{i}\sin\, \theta\right)\\ & = r\;\mathrm{cis}\;\theta \\ \end{split}\end{equation*}

Vectors

Quick Facts

Two vectors are equal if they share the same magnitude and direction.
Initial points don't matter.
You can define a vector with two points.
Do not divide vectors!
$\vec{v} \cdot \vec{v} = ||\vec{v}||^2 \implies \text{constant}$
$\vec{v} \cdot \left(\vec{v}\right)^\prime = 0$

Vector Operations

Dot Product

\begin{equation*}\begin{split} \vec{a}\cdot\vec{b} &= \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \cdot \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} = a_1b_1 + a_2b_2 + a_3b_3\\\\ \vec{a}\cdot\vec{b} &= \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} \cdot \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} = a_1b_1 + a_2b_2 \end{split}\end{equation*}

Cross Product

\begin{equation*}\begin{split} \vec{a}\times\vec{b} &= \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}\times\begin{bmatrix} b_1 & b_2 & b_3 \end{bmatrix}\\ &= \mathrm{det} \begin{pmatrix} \mathrm{\hat{i}} & \mathrm{\hat{j}} & \mathrm{\hat{k}}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{pmatrix}\\ &= \mathrm{\hat{i}} \begin{vmatrix} e & f\\ h & i \end{vmatrix} - \mathrm{\hat{j}} \begin{vmatrix} d & f\\ g & i \end{vmatrix} + \mathrm{\hat{k}} \begin{vmatrix} d & e\\ g & h \end{vmatrix}\\ &= \mathrm{\hat{i}}\left(e i - f h\right) - \mathrm{\hat{j}}\left(d i - f g\right) + \mathrm{\hat{k}}\left(d h - e g\right)\\ &= \begin{bmatrix} e i - f h & d i - f g & d h - e g \end{bmatrix}\\ &= \vec{c} \end{split}\end{equation*}

Length of a Vector

\begin{equation*}\begin{split} |a| &= \sqrt{(a_1)^2 + (a_2)^2} \\ |a| &= \sqrt{(a_1)^2 + (a_2)^2 + (a_3)^2} \end{split}\end{equation*}

Definition of Vector Addition

If $\vec{u}$ and $\vec{v}$ are positioned so the initial point of $\vec{v}$ is at the terminal point of $\vec{u}$ , then the sum $\vec{u} + \vec{v}$ is the vector from the initial point of $\vec{u}$ to the terminal point of $\vec{v}$ .

Given some vectors $\vec{u}$ and $\vec{v}$ , the vector $\vec{u} - \vec{v}$ is the vector that points from the head of $\vec{v}$ to the head of $\vec{u}$

Standard Basis Vectors

\begin{equation*}\begin{split} \mathrm{\hat{i}} &= \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}\\ \mathrm{\hat{j}} &= \begin{bmatrix} 0 & 1 & 0 \end{bmatrix}\\ \mathrm{\hat{k}} &= \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \end{split}\end{equation*}

Orthogonal

Two vectors are orthogonal if and only if

\begin{equation*}\begin{split} \vec{a}\cdot\vec{b} = 0 \end{split}\end{equation*}

The Unit Vector

\begin{equation*}\begin{split} \hat{u} &= \frac{\vec{a}}{|\vec{a}|}\end{split}\end{equation*}

If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$ , then

\begin{equation*}\begin{split} \vec{a} \cdot \vec{a} = |\vec{a}| \cdot |\vec{a}| \cos\theta \end{split}\end{equation*}

If $\theta$ is the angle between the nonzero vectors $\vec{a}$ and $\vec{b}$ , then

\begin{equation*}\begin{split} \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}| |\vec{b}|}\end{split}\end{equation*}

Two nonzero vectors $\vec{a}$ and $\vec{b}$ are parallel if and only if

\begin{equation*}\begin{split} \vec{a}\times\vec{b} = 0 \end{split}\end{equation*}

Properties of the Dot Product

\begin{equation*}\begin{split} \vec{a} \cdot \vec{a} = |\vec{a}|^2 \end{split}\end{equation*}

\begin{equation*}\begin{split} \vec{a} \cdot \left(\vec{b} + \vec{c}\right) = \vec{a}\vec{b} + \vec{a}\vec{c} \end{split}\end{equation*}

\begin{equation*}\begin{split} \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \end{split}\end{equation*}

\begin{equation*}\begin{split} \left(c \cdot \vec{a}\right) \cdot \vec{b} = c\left(\vec{a} \cdot \vec{b}\right) \end{split}\end{equation*}

Direction Cosines & Direction Angles of a Vector

Where

\begin{equation*}\begin{split} \vec{v} = \begin{bmatrix} v_x & v_y & v_z \end{bmatrix} ||\vec{v}|| = \sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2} \end{split}\end{equation*}

Direction Cosines

\begin{equation*}\begin{split} \cos\alpha &= \frac{v_x}{||\vec{v}||}\\ \cos\beta &= \frac{v_y}{||\vec{v}||}\\ \cos\gamma &= \frac{v_z}{||\vec{v}||}\end{split}\end{equation*}

Direction Angles

\begin{equation*}\begin{split} \alpha &= \arccos \left(\frac{v_x}{||\vec{v}||}\right)\\ \beta &= \arccos \left(\frac{v_y}{||\vec{v}||}\right)\\ \gamma &= \arccos \left(\frac{v_z}{||\vec{v}||}\right) \end{split}\end{equation*}

Theorem

\begin{equation*}\begin{split} \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \end{split}\end{equation*}

Proof

\begin{equation*}\begin{split} \vec{v} &= \hat{i}\cos\alpha + \hat{k}\cos\beta + \hat{k}\cos\gamma\\ ||\vec{v}|| &= || \hat{i}\cos\alpha + \hat{k}\cos\beta + \hat{k}\cos\gamma|| = 1 \end{split}\end{equation*}

Given

\begin{equation*}\begin{split} \sqrt{\cos^2\alpha + \cos^2\beta + \cos^2\gamma} &= 1\\ \sqrt{\cos^2\alpha + \cos^2\beta + \cos^2\gamma}^2 &= 1^2\\ \cos^2\alpha + \cos^2\beta + \cos^2\gamma &= 1 \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\\ \end{split}\end{equation*}

Vector Relations

Parallel Vectors

When two vectors are parallel; they never intersect （duh）.

Given some vectors

\begin{equation*}\begin{split} \vec{a} &= \begin{bmatrix} a_x & a_y & a_z \end{bmatrix}\\ \vec{b} &= \begin{bmatrix} b_x & b_y & b_z \end{bmatrix} \end{split}\end{equation*}

The vectors $\vec{a}$ and $\vec{b}$ are parallel if and only if they are scalar multiples of one another.

\begin{equation*}\begin{split} \vec{a} &= k\;\vec{b} \;\;\;\;\forall k \neq 0 \end{split}\end{equation*}

Alternatively

\begin{equation*}\begin{split} \frac{a_x}{b_x}= \frac{a_y}{b_y}= \frac{a_z}{b_z}\end{split}\end{equation*}

Orthogonal Vectors

When two vectors are orthogonal; they meet at right angles.

Given some vectors

\begin{equation*}\begin{split} \vec{a} &= \begin{bmatrix} a_x & a_y & a_z \end{bmatrix}\\ \vec{b} &= \begin{bmatrix} b_x & b_y & b_z \end{bmatrix} \end{split}\end{equation*}

Two vectors are orthogonal if and only if

\begin{equation*}\begin{split} \vec{a}\times\vec{b} = 0 \end{split}\end{equation*}

Reparameterization of the position vector $\vec{v}(t)$ in terms of length $S(t)$

We can parametrize a curve with respect to arc length; because arc length arises naturally from the shape of the curve and does not depend on any coordinate system.

The Arc Length Function

Given

\begin{equation*}\begin{split} \vec{v} &= \begin{bmatrix} x & y & z \end{bmatrix} = \begin{bmatrix} f(t) & g(t) & h(t) \end{bmatrix} \end{split}\end{equation*}

We can redefine $\vec{v}$ in terms of arc length between two endpoints

\newcommand{\Long}{ \int_a^t \sqrt{ \left(f^\prime(u)\right)^2 \left(g^\prime(u)\right)^2 \left(h^\prime(u)\right)^2 } } \newcommand{\AltLong}{ \int_a^t \sqrt{ \left(\frac{\mathrm{d}x}{\mathrm{d}u}\right)^2 \left(\frac{\mathrm{d}y}{\mathrm{d}u}\right)^2 \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 } } \newcommand{\short}{ \int_a^t ||\left(v^\prime(u)\right)|| } \begin{equation} \begin{split} S(t) &= \Long\\ &= \AltLong\\ &= \short \end{split} \end{equation}

That is, $S(t)$ is the length of the curve （ $C$ ） between $r(a)$ and $r(b)$ .

Furthermore from the adjacent definition; we can simply the above to

\begin{equation*}\begin{split} S(t) = \int_a^t \frac{\mathrm{d}S}{\mathrm{d}t}\end{split}\end{equation*}

The Arc Length Function

\begin{equation*}\begin{split} \frac{\mathrm{d}S}{\mathrm{d}t}\equiv || v^\prime(t) || \end{split}\end{equation*}

That is

\newcommand{\Long}{ \sqrt{ \left(f^\prime(u)\right)^2 \left(g^\prime(u)\right)^2 \left(h^\prime(u)\right)^2 } } \newcommand{\AltLong}{ \sqrt{ \left(\frac{\mathrm{d}x}{\mathrm{d}u}\right)^2 \left(\frac{\mathrm{d}y}{\mathrm{d}u}\right)^2 \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 } } \newcommand{\short}{ ||\left(v^\prime(u)\right)|| } \begin{equation} \begin{split} \frac{\mathrm{d}S}{\mathrm{d}t}&= \Long\\ &= \AltLong\\ &= \short \end{split} \end{equation}

Vectors Derived From Some Curve Defined by $\vec{v}$

The Unit Vector

\begin{equation*}\begin{split} \hat{U} \equiv \frac{\vec{v}}{||\vec{v}||}\end{split}\end{equation*}

The Unit Tangent Vector

\begin{equation*}\begin{split} \vec{T} \equiv \frac{v^\prime(t)}{||v^\prime(t)||}\equiv \frac{\mathrm{d}v}{\mathrm{d}S}\end{split}\end{equation*}

The Unit Normal Vector

\begin{equation*}\begin{split} \vec{N} \equiv \frac{T^\prime}{||T^\prime||}\end{split}\end{equation*}

The Binormal Vector

\begin{equation*}\begin{split} \vec{B} \equiv \vec{T}\times\vec{N} \end{split}\end{equation*}

Therefore, the binormal vector is orthogonal to both the tangent vector and the normal vector.
The plane determined by the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P.
The plane determined by the vectors T and N is called the osculating plane of C at P. The name comes from the Latin osculum, meaning “kiss.” It is the plane that comes closest to containing the part of the curve near P. （For a plane curve, the osculating plane is simply the plane that contains the curve.）

Kappa - Curvature of a Vector

\begin{equation*}\begin{split} \kappa \equiv \left|\frac{\mathrm{d}T}{\mathrm{d}S}\right| \equiv \frac{\left| T^\prime \right|}{\left| r^\prime \right|}\equiv \frac{\left| r^\prime \times r^{\prime\prime} \right|}{\left| r^\prime \right|^3}\end{split}\end{equation*}

Tangential & Normal Components of the Acceleration Vector of the Curve

When we study the motion of a particle, it is often useful to resolve the acceleration into two components, one in the direction of the tangent and the other in the direction of the normal.

\begin{equation*}\begin{split} a_{\vec{T}} &= \frac{\mathrm{d}}{\mathrm{d}t}\left|\vec{v}\right| = \frac{r^\prime \cdot r^{\prime\prime}}{|r^\prime|}\\ a_{\vec{N}} &= \kappa \left|\vec{v}\right|^2 = \frac{\left|r^\prime \times r^{\prime\prime}\right|}{|r^\prime|}\\ \vec{a} &= a_{\vec{T}} \vec{T} + a_{\vec{N}} \vec{N} \end{split}\end{equation*}

Specifically

\begin{equation*}\begin{split} \begin{rcases} a_{\vec{T}}\\ a_{\vec{N}} \end{rcases}\text{Tangential \& Normal Components of $\vec{a}$} \end{split}\end{equation*}

Vector Calculus

The Position Vector $\vec{r}(t)$

（Original Function）

The Velocity Vector $\vec{v}(t)$

（First Derivative）

The velocity vector is also the tangent vector and points in the direction of the tangent line.
The speed of the particle at time t is the magnitude of the velocity vector, that is,
$\begin{equation*}\begin{split} \underbrace{|\vec{v}(t)| = |(\vec{r})^\prime(t)| = \frac{\mathrm{d}s}{\mathrm{d}t}} _{\text{rate of change of distance with respect to time}} \end{split}\end{equation*}$

The Acceleration Vector $\vec{a}(t)$

（Second Derivative）

Matrices

Reference

The Determinant of A Matrix

\begin{equation*}\begin{split} |A| &= \begin{vmatrix} a & b\\ c & d \end{vmatrix} = ad - bc\\\\ |A| &= \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}\\ &= a \begin{vmatrix} e & f\\ h & i \end{vmatrix} - b \begin{vmatrix} d & f\\ g & i \end{vmatrix} + c \begin{vmatrix} d & e\\ g & h \end{vmatrix} \end{split}\end{equation*}

Only works for square matrices.

The Cross Product

\begin{equation*}\begin{split} \vec{a}\times\vec{b} &= \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}\times\begin{bmatrix} b_1 & b_2 & b_3 \end{bmatrix}\\ &= \mathrm{det} \begin{pmatrix} \mathrm{\hat{i}} & \mathrm{\hat{j}} & \mathrm{\hat{k}}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{pmatrix}\\ &= \mathrm{\hat{i}} \begin{vmatrix} e & f\\ h & i \end{vmatrix} - \mathrm{\hat{j}} \begin{vmatrix} d & f\\ g & i \end{vmatrix} + \mathrm{\hat{k}} \begin{vmatrix} d & e\\ g & h \end{vmatrix}\\ &= \mathrm{\hat{i}}\left(e i - f h\right) - \mathrm{\hat{j}}\left(d i - f g\right) + \mathrm{\hat{k}}\left(d h - e g\right)\\ &= \begin{bmatrix} e i - f h & d i - f g & d h - e g \end{bmatrix}\\ &= \vec{c} \end{split}\end{equation*}

Geometry

The Circle

\begin{equation*}\begin{split} \small\text{Area}\;\normalsize &= \pi r^2\\ \small\text{Circumference}\;\normalsize &= 2 \pi r \end{split}\end{equation*}

Definition of a Line

Vector Equation of a Line

Given

\begin{equation*}\begin{split} P_1&= \begin{bmatrix} x_1& y_1& z_1\end{bmatrix}\\ P_2&= \begin{bmatrix} x_2& y_2& z_2\end{bmatrix}\\ \end{split}\end{equation*}

\begin{equation*}\begin{split} P_1&= \begin{bmatrix} x_1& y_1& z_1\end{bmatrix}\\ P_2&= \begin{bmatrix} x_2& y_2& z_2\end{bmatrix}\\ \end{split}\end{equation*}

We can define a vector between $P_1$ and $P_2$

\begin{equation*}\begin{split} \overrightarrow{\Delta\mathsf{v}}&= \begin{bmatrix} x_2\\ y_2\\ z_2\end{bmatrix} - \begin{bmatrix} x_1\\ y_1\\ z_1\end{bmatrix} = \begin{bmatrix} x_2-x_1\\ y_2-y_1\\ z_2-z_1\end{bmatrix} = \begin{bmatrix} \Delta_x\\ \Delta_y\\ \Delta_z\end{bmatrix} \end{split}\end{equation*}

Therefore

The equation of a line in 3D space or $\mathbb{R}^3$ can be defined VIA the following options

\begin{equation*}\begin{split} L &= P_1+ t\cdot\overrightarrow{\Delta\mathsf{v}}\\ \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \begin{bmatrix} x_1\\ y_1\\ z_1\end{bmatrix} + t\begin{bmatrix} \Delta_x\\ \Delta_y\\ \Delta_z\end{bmatrix} = \begin{bmatrix} x_1\\ y_1\\ z_1\end{bmatrix} + \begin{bmatrix} t\;\Delta_x\\ t\;\Delta_y\\ t\;\Delta_z\end{bmatrix} \\ &= \begin{bmatrix} x_1+ t\;\Delta_x\\ y_1+ t\;\Delta_y\\ z_1+ t\;\Delta_z\end{bmatrix}\\ &= \begin{bmatrix} x_1+ t(x_2-x_1) \\ x_2+ t(x_2-x_1) \\ x_3 + t(x_2-x_1) \end{bmatrix} \end{split}\end{equation*}

That is

\begin{equation*}\begin{split} L &= P_1+ t\cdot\overrightarrow{\Delta\mathsf{v}}\\ \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \begin{bmatrix} x_1+ t\;\Delta_x\\ y_1+ t\;\Delta_y\\ z_1+ t\;\Delta_z\end{bmatrix} = \begin{bmatrix} x_1+ t(x_2-x_1) \\ y_1+ t(y_2-y_1) \\ z_1+ t(z_2-z_1) \end{bmatrix} \end{split}\end{equation*}

Parametric Equation of a Line

\begin{equation*}\begin{split} \underbrace{ \begin{split} x &= x_1+ t(x_2-x_1) = x_1+ t\;\Delta_x\\ y &= x_1+ t(y_2-y_1) = y_1+ t\;\Delta_y\\ z &= x_1+ t(z_2-z_1) = z_1+ t\;\Delta_z\end{split}}_{ r \;=\; r_0 \;+\; a \;=\; r_0 \;+\; t\,v } \end{split}\end{equation*}

Essentially

\begin{equation*}\begin{split} L &= P_1+ t\cdot\overrightarrow{\Delta\mathsf{v}}\;\;\forall t\in\mathbb{R} \end{split}\end{equation*}

That is, $t$ is the scaling factor. In a way, it's like it's a function of $t$ , but also similar to the slope （ $m$ ） in $y = mx + b$ , except $m$ （i.e. $t$ ） is parameterized.

Sometimes this will be （confusingly） denoted as

\begin{equation*}\begin{split} \vec{r} &= \vec{r_0} + \vec{a} = \vec{r_0} + t\vec{v}\\ \end{split}\end{equation*}

Symmetric Equation of a Line

\begin{equation*}\begin{split} t &= \frac{x - x_1}{x_2-x_1}= \frac{x - x_1}{\Delta_x}\\ t &= \frac{y - y_1}{y_2-y_1}= \frac{y - y_1}{\Delta_y}\\ t &= \frac{z - z_1}{z_2-z_1}= \frac{z - z_1}{\Delta_z}\end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} \frac{x - x_1}{\Delta_x}&= \frac{y - y_1}{\Delta_y}= \frac{z - z_1}{\Delta_z}\\\\ \frac{x - x_1}{x_2-x_1}&= \frac{y - y_1}{y_2-y_1}= \frac{z - z_1}{z_2-z_1}\end{split}\end{equation*}

Rationale

We rewrite $r = r_0 + a = r_0 + t v$ in terms of $t$ .

That is

\begin{equation*}\begin{split} x &= x_1+ t(x_2-x_1) = x_1+ t\;\Delta_x\\ t\;\Delta_x&= x - x_1= t(x_2-x_1)\\ t &= \frac{x - x_1}{x_2-x_1}= \frac{x - x_1}{\Delta_x}\\\\ y &= y_1+ t(y_2-y_1) = y_1+ t\;\Delta_y\\ t\;\Delta_y&= y - y_1= t(y_2-y_1)\\ t &= \frac{y - y_1}{y_2-y_1}= \frac{y - y_1}{\Delta_y}\\\\ z &= z_1+ t(z_2-z_1) = z_1+ t\;\Delta_z\\ t\;\Delta_z&= z - z_1= t(z_2-z_1) \\ t &= \frac{z - z_1}{z_2-z_1}= \frac{z - z_1}{\Delta_z}\end{split}\end{equation*}

Parameterizations of a curve

Parametrized curve: A curve in the plane is said to be parameterized if the set of coordinates on the curve, （x,y）, are represented as functions of a variable t.; A parametrized Curve is a path in the xy-plane traced out by the point $\left(x(t), y(t)\right)$ as the parameter $t$ ranges over an interval $I$ .; A parametrized Curve is a path in the xyz-plane traced out by the point $\left(x(t), y(t), z(t)\right)$ as the parameter $t$ ranges over an interval $I$ .

Curvature Properties

Length of a Curve

\newcommand{\Long}{ \int_a^b \sqrt{ \left(f^\prime(t)\right)^2 \left(g^\prime(t)\right)^2 \left(h^\prime(t)\right)^2 } } \newcommand{\AltLong}{ \int_a^b \sqrt{ \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2 } } \newcommand{\short}{ \int_a^b ||\left(r^\prime(t)\right)|| } \begin{equation} \begin{split} L &= \Long\\ &= \AltLong\\ &= \short\\ &\implies \text{some constant} \end{split} \end{equation}

The Arc Length Function

Suppose

Given some curve $C$ defined by some vector $\vec{r}$ in $\mathbb{R}^3$
where $r^\prime$ is continuous and $C$ is traversed exactly once as $t$ increases from $a$ to $b$

We can define it's arc length function VIA

\newcommand{\Long}{ \int_a^t \sqrt{ \left(f^\prime(u)\right)^2 \left(g^\prime(u)\right)^2 \left(h^\prime(u)\right)^2 } } \newcommand{\AltLong}{ \int_a^t \sqrt{ \left(\frac{\mathrm{d}x}{\mathrm{d}u}\right)^2 \left(\frac{\mathrm{d}y}{\mathrm{d}u}\right)^2 \left(\frac{\mathrm{d}z}{\mathrm{d}u}\right)^2 } } \newcommand{\short}{ \int_a^t ||\left(r^\prime(u)\right)|| } \begin{equation*} \begin{split} S(t) &= \Long\\ &= \AltLong\\ &= \short \end{split} \end{equation*}

Calculus

Derivative Tables

\begin{equation*}\begin{split} \delta\sin(x) &= \cos(x) \\ \delta\csc(x) &= -\cot(x) \cdot \csc(x) \end{split}\end{equation*}

\begin{equation*}\begin{split} \delta\cos(x) &= -\sin(x) \\ \delta\sec(x) &= \tan(x) \cdot \sec(x) \end{split}\end{equation*}

\begin{equation*}\begin{split} \delta\tan(x) &= \sec^2(x) \\ \delta\cot(x) &= -\csc^2(x) \end{split}\end{equation*}

\begin{equation*}\begin{split} \delta\sin^{-1}(x) &= \frac{1}{\sqrt{1 - x^2}}\end{split}\end{equation*}

\begin{equation*}\begin{split} \delta\cos^{-1}(x) &= -\frac{1}{\sqrt{1 - x^2}}\end{split}\end{equation*}

\begin{equation*}\begin{split} \delta\tan^{-1}(x) &= \frac{1}{1 + x^2}\end{split}\end{equation*}

Integration Tables

\begin{equation*}\begin{split} \int\sin(x)\;\mathrm{d}x &= -\cos x \\ \int\csc^2(x)\;\mathrm{d}x &= -\cot x \\ \int\csc(x)\cdot\cot(x)\;\mathrm{d}x &= -\csc x \\ \int\csc(x)\;\mathrm{d}x &= \ln\left|\csc x - \cot x\right| \\ \int\sinh(x)\;\mathrm{d}x &= \cosh(x) \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} \int\cos(x)\;\mathrm{d}x &= \sin x \\ \int\sec^2(x)\;\mathrm{d}x &= \tan x \\ \int\sec(x)\cdot\tan(x)\;\mathrm{d}x &= \sec x \\ \int\sec(x)\;\mathrm{d}x &= \ln\left|\sec x + \tan x\right| \\ \int\cosh(x)\;\mathrm{d}x &= \sinh(x) \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} \int\tan(x)\;\mathrm{d}x &= \ln| \sec x | \\ \int\cot(x)\;\mathrm{d}x &= \ln| \sin x | \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} \int b^x\;\mathrm{d}x &= \frac{b^x}{\ln(b)}\end{split}\end{equation*}

\begin{equation*}\begin{split} \int\frac{1}{x^2 + a^2}\;\mathrm{d}x &= \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \end{split}\end{equation*}

\begin{equation*}\begin{split} \int\frac{1}{x^2 - a^2}\;\mathrm{d}x &= \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| \end{split}\end{equation*}

\begin{equation*}\begin{split} \int\frac{1}{\sqrt{a^2 - x^2}}\;\mathrm{d}x &= \sin^{-1}\left(\frac{x}{a}\right),\;a>0 \end{split}\end{equation*}

\begin{equation*}\begin{split} \int\frac{1}{\sqrt{x^2 \pm a^2}}\;\mathrm{d}x &= \ln\left|x+\sqrt{x^2\pm a^2}\right| \end{split}\end{equation*}

Riemann Sums

Given

\begin{equation*}\begin{split} A &= \int_{a}^{b} f(x) = \lim_{n\to\infty}\sum_{i=1}^{n} \Delta{x} \cdot f(x) \;\text{where}\; \begin{dcases} \Delta{x} = \frac{b - a}{n}\end{dcases}\end{split}\end{equation*}

Left Riemann Sum

\begin{equation*}\begin{split} A &= \int_{a}^{b} f(x) \;\mathrm{d}x \approx L_n = \sum_{i = 0}^{n-1}\, \Delta{x}\cdot f\left(a + i\cdot\Delta{x}\right) \end{split}\end{equation*}

Right Riemann Sum

\begin{equation*}\begin{split} A &= \int_{a}^{b} f(x) \;\mathrm{d}x \approx R_n = \sum_{i = 1}^{n}\, \Delta{x}\cdot f\left(a + i\cdot\Delta{x}\right) \end{split}\end{equation*}

Midpoint Riemann Sum

\newcommand{\generalFormat}{ \sum_{\small{\cdots}}^{\small{\cdots}}\, \Delta{x}\cdot f\left(a + \text{“avg. of $x_i$ and $x_{i-1}$”} \cdot \Delta{x}\right) } \newcommand{\SigmaExampleOne}{ \sum_{i = 1}^{n}\, \Delta{x}\cdot f\left(a + \frac{x_{i - 1} + x_i}{2}\cdot \Delta{x}\right) } \newcommand{\SigmaExampleTwo}{ \sum_{i = 0}^{n-1}\, \Delta{x}\cdot f\left(a + \frac{x_{i + 1} + x_i}{2}\cdot \Delta{x}\right) } \begin{equation} \begin{split} A = \int_{a}^{b} f(x) \;\mathrm{d}x &\approx \generalFormat\\ &\approx \SigmaExampleOne\\ \text{or alternatively}&\\ A &\approx \SigmaExampleTwo \end{split} \end{equation}

We can also do away with the index notation and simplify things.

\newcommand{\SigmaExampleOne}{ \sum_{i = 1}^{n}\, \Delta{x}\cdot f\left(a + \frac{(i - 1) + i}{2}\cdot \Delta{x}\right) = \sum_{i = 1}^{n}\, \Delta{x}\cdot f\left(a + \frac{2i - 1}{2}\cdot \Delta{x}\right) } \newcommand{\SigmaExampleTwo}{ \sum_{i = 0}^{n-1}\, \Delta{x}\cdot f\left(a + \frac{(i + 1) + i}{2}\cdot \Delta{x}\right) = \sum_{i = 0}^{n-1}\, \Delta{x}\cdot f\left(a + \frac{2i + 1}{2}\cdot \Delta{x}\right) } \begin{equation} \begin{split} A = \int_{a}^{b} f(x) \;\mathrm{d}x &\approx \SigmaExampleOne\\ &\approx \SigmaExampleTwo \end{split} \end{equation}

Trapezoidal Riemann Sum

\dots

Simpson's Rule

\begin{equation*}\begin{split} \dots \end{split}\end{equation*}

Improper Integrals

\begin{equation*}\begin{split} \text{Given}&\;L = \int_{a}^{\infty} f(x) \;\mathrm{d}x = \lim_{t\to\infty} \int_{a}^{\infty} f(x) \;\mathrm{d}x = \lim_{t\to\infty} F(x)\\ \text{If}&\;L\;\text{“exists”}\;\text{then $L$ is}\;\mathbf{\text{convergent}}\\ \text{If}&\;L\;\text{“does not exists”}\;\text{then $L$ is}\;\mathbf{\text{divergent}} \end{split}\end{equation*}

Infinite Sequences

Infinite Sequence

\begin{equation*}\begin{split} \text{Given}&\\ &S_n = \{a_n\}_{n=1}^{\infty}\\ \text{Tests}&\\ &\text{If}\;\lim_{n\to\infty}\,S_n \text{“exists”}\;\text{then}\;\mathbf{\text{$S_n$ is convergent}}\\ &\text{If}\;\lim_{n\to\infty}\,S_n \text{“does not exists”}\;\text{then}\;\mathbf{\text{$S_n$ is Divergent}} \end{split}\end{equation*}

Helpful Theorem

\begin{equation*}\begin{split} \text{If}&\\ \lim_{x\to\infty}\,|a_n| &= 0\\ \text{Then}&\\ \lim_{x\to\infty}\,a_n &= 0\\ \end{split}\end{equation*}

Example

Given

\begin{equation*}\begin{split} S_n &= \{\cos\left(\frac{n\pi}{2}\right)\}\\ L &=\lim_{n\to\infty}\,\cos\left(\frac{n\pi}{2}\right)\\ &= \cos\left(\lim_{n\to\infty}\,\frac{n\pi}{2}\right) \\ &= \cos\left(\infty\right) \\ &= \text{undefined} \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} \therefore\;\text{$S_n$ is Divergent} \end{split}\end{equation*}

Example

Given

\begin{equation*}\begin{split} S_n &= \{\sin\left(\frac{\pi}{n}\right)\}\\ L &= \lim_{n\to\infty}\,\sin\left(\frac{\pi}{n}\right)\\ &= \sin\left(\lim_{n\to\infty}\,\frac{\pi}{n}\right) \\ &= \sin\left(0\right) \\ &= 0 \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} \therefore\;\text{$S_n$ is Convergent} \end{split}\end{equation*}

Infinite Series

\begin{equation*}\begin{split} \text{Given}&\\ &S_n = \sum_{n = 1}^{\infty}\,a_n\\ \text{Tests}&\\ &\text{If}\;\lim_{n\to\infty}\,a_n = 0\;\text{then}\;\text{$S_n$ may be $\mathbf{\text{convergent}}$}\\ &\text{If}\;\lim_{n\to\infty}\,a_n \neq 0\;\text{then}\;\mathbf{\text{$S_n$ is divergent}}\\ &\text{If}\;\lim_{n\to\infty}\,a_n \text{“does not exists”}\;\text{then}\;\mathbf{\text{$S_n$ is divergent}}\\ \end{split}\end{equation*}

Note that the limit of every convergent series is equal to zero. But the inverse isn't always true. If the limit is equal to zero, it may not be convergent.

For example, $\sum_{n=1}^\infty \frac{1}{n}$ does diverge; but it's limit is equal to zero.

If the limit is equal to zero; the test is inconclusive.

Geometric Series

Given

\begin{equation*}\begin{split} S_n = \sum_{n = 1}^{\infty}\,a_n = \sum_{n = 1}^{\infty}\,a\cdot r^{n - 1}\;\text{where}\;\begin{dcases}a &= a_1\\ r &= \frac{S_2}{S_1}\end{dcases}\end{split}\end{equation*}

Alternatively

\begin{equation*}\begin{split} S_n = \sum_{n=0}^{\infty}\,a_n = \sum_{n=0}^{\infty}\,a\cdot r^{n} \end{split}\end{equation*}

Tests

\begin{equation*}\begin{split} \text{If}\;|r|\geq{1}\;\text{then}\;\mathbf{\text{$S_n$ is divergent}}\\ \text{If}\;|r|<1\;\text{then}\;\mathbf{\text{$S_n$ is convergent}}\\ \end{split}\end{equation*}

Furthermore

\begin{equation*}\begin{split} \sum_{n = 1}^{\infty}\,a_n = \sum_{n = 1}^{\infty}\,a\cdot r^{n - 1} = \frac{a}{1 - r}\;\text{for all}\;|r|<1 \end{split}\end{equation*}

The Integral Test

\begin{equation*}\begin{split} &\text{Given}\\ &\;\;\;\;a_n = f(n)\;\text{$\forall$n on}\;[1,n)\\ &\;\;\;\;S_n = \sum_{n = 1}^{\infty}\,a_n\\ &\;\;\;\;F(x)= \int_{1}^{\infty}f(x)\;\mathrm{d}x\\ &\text{Where}\\ &\;\;\;\;f(x)> 0, \forall\,x\in\,[1, \infty)\\ &\;\;\;\;f^\prime(x)< 0, \forall\,x\in\,[1, \infty)\\ &\text{Tests}\\ &\;\;\;\;\text{If $S_n$ convergent; then $F(x)$ is $\mathbf{convergent}$}\\ &\;\;\;\;\text{If $S_n$ divergent; then $F(x)$ is $\mathbf{divergent}$}\end{split}\end{equation*}

Constraints on $[1,n)$

Continuous
Positive
Decreasing （i.e. use derivative test）

P-Series -or- Harmonic Series

\begin{equation*}\begin{split} &\text{Given}\\ &\;\;\;\;S_n=\sum_{n=1}^{\infty}\frac{1}{n^p}\\ &\text{Tests}\\ &\;\;\;\;\text{If $p>1$ then $S_n$ is $\mathbf{convergent}$}\\ &\;\;\;\;\text{If $0 < p \leq{1}$ then $S_n$ is $\mathbf{divergent}$} \end{split}\end{equation*}

Note: the Harmonic series is the special case where $p=1$

Comparison Test

\begin{equation*}\begin{split} &\text{Given}\\ &\;\;\;\;A_n = \sum_{n=1}^\infty\,a_n\\ &\;\;\;\;B_n = \sum_{n=1}^\infty\,b_n\\ &\text{Where}\\ &\;\;\;\;a_n, b_n \geq 0\\ &\;\;\;\;a_n \leq b_n\\ &\text{Tests}\\ &\;\;\;\;\text{If $B_n$ converges $\implies A_n$ converges}\\ &\;\;\;\;\text{If $A_n$ diverges $\implies B_n$ diverges}\\ \end{split}\end{equation*}

Limit Comparison Test

\begin{equation*}\begin{split} &\text{Given}\\ &\;\;\;\;A_n = \sum_{n=1}^\infty\,a_n\\ &\;\;\;\;B_n = \sum_{n=1}^\infty\,b_n\\ &\;\;\;\; L = \lim_{n\to\infty}\frac{a_n}{b_n}\\ &\text{Where}\\ &\;\;\;\;L > 0,\;L \neq \pm \infty\\ &\text{Therefore either both converge or diverge}\\ &\;\;\;\;\text{$A_n$ converges $\Longleftrightarrow B_n$ converges}\\ &\;\;\;\;\text{$A_n$ diverges $\Longleftrightarrow B_n$ diverges} \end{split}\end{equation*}

Warning

If $L > 0$ , this only means that the limit comparison test can be used. You still need to determine if either $A_n$ or $B_b$ converges or diverges.
Therefore, this does not apply to any arbitrary rational function.

Notes

For many series, we find a suitable comparison, $B_n$ , by keeping only the highest powers in the numerator and denominator of $A_n$ .

Estimating Infinite Series

\begin{equation*}\begin{split} \cdots \end{split}\end{equation*}

Differential Equations

Separable Differential Equations

\begin{equation*}\begin{split} \text{Given}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&= g(x) \cdot f(y)\\ &= \frac{g(x)}{\frac{1}{f(y)}}\\ &= \frac{g(x)}{h(y)}\;\text{where}\;h(x) = \frac{1}{f(y)}\\ \\\text{Therefore (restated)}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&= \frac{g(x)}{h(y)}\\ \\\text{Multiply reciprocals}\\ h(y)\;\mathrm{d}y &= g(x)\;\mathrm{d}x\\ \\\text{Integrate}\\ \int h(y)\;\mathrm{d}y &= \int g(x)\;\mathrm{d}x\\ \\\text{Differentiate}\\ \frac{d}{dx}\left(\int h(y)\;\mathrm{d}y\right) &= \frac{d}{dx}\left(\int g(x)\;\mathrm{d}x\right)\\ \\\text{Given}\\ \frac{\mathrm{d}}{dx}&= \frac{\mathrm{d}}{dx}\cdot \frac{\mathrm{d}y}{\mathrm{d}y}= \textcolor{blue}{\frac{\mathrm{d}}{dy}} \cdot \textcolor{JungleGreen}{\frac{\mathrm{d}y}{dx}}\\ \\\text{Therefore the LHS is equal to}\\ \frac{\mathrm{d}}{dx}\left(\int h(y) \mathrm{d}y\right) &= \textcolor{blue}{\frac{\mathrm{d}}{dy}} \left(\int h(y) \mathrm{d}y\right) \textcolor{JungleGreen}{\frac{\mathrm{d}y}{dx}}\\ &= h(y) \textcolor{JungleGreen}{\frac{\mathrm{d}y}{dx}} \\\text{Therefore (in conclusion)}\\ \therefore \; h(y)\;\frac{\mathrm{d}y}{dx}&= g(x) \end{split}\end{equation*}

Growth and Decay Models

\begin{equation*}\begin{split} \text{Given}\\ \frac{\mathrm{d}y}{\mathrm{d}t}&= \textcolor{Periwinkle}{k} y\\ \text{Proof}\\ \frac{1}{y}\;\mathrm{d}y &= \textcolor{Periwinkle}{k}\;\mathrm{d}t \\ \int \frac{1}{y}\;\mathrm{d}y &= \int \textcolor{Periwinkle}{k}\;\mathrm{d}t \\ \ln(y) &= \textcolor{Periwinkle}{k} t + \textcolor{SeaGreen}{C} \\ e^{\ln(y)} &= e^ {\textcolor{Periwinkle}{k} t + \textcolor{SeaGreen}{C}} \\ y &= e^{\textcolor{Periwinkle}{k} t} \cdot e^{\textcolor{SeaGreen}{C}} \\ y &= e^{\textcolor{Periwinkle}{k} t} \cdot \textcolor{SeaGreen}{C} \\ \text{Therefore}\\ y &= \textcolor{SeaGreen}{C} e^{\textcolor{Periwinkle}{k} t} \end{split}\end{equation*}

The above states that all solutions for $y^\prime = k y$ are of the form $y = C e^{k t}$ .

Where

\begin{equation*}\begin{split} \textcolor{SeaGreen}{C} &= \textcolor{SeaGreen}{\text{Initial value of $y$}}\\ \textcolor{Periwinkle}{k} &= \textcolor{Periwinkle}{\text{Proportionality constant}}\\ \end{split}\end{equation*}

Exponential growth occurs when $\textcolor{Periwinkle}{k > 0}$ , and exponential decay occurs when $\textcolor{Periwinkle}{k < 0}$ .

The Law of Natural Growth:

\begin{equation*}\begin{split} \frac{\mathrm{d}\textcolor{DarkOrchid}{P}}{\mathrm{d}t}= \textcolor{Periwinkle}{k} \textcolor{DarkOrchid}P \end{split}\end{equation*}

The Logistic Model of Population Growth:

\begin{equation*}\begin{split} \frac{\mathrm{d}\textcolor{DarkOrchid}P}{\mathrm{d}t}= \textcolor{Periwinkle}{k} \textcolor{DarkOrchid}P \left(1 - \frac{\textcolor{DarkOrchid}{P}}{\textcolor{Aquamarine}{L}}\right) \end{split}\end{equation*}

Where

\begin{equation*}\begin{split} \textcolor{Periwinkle}{k} &= \textcolor{Periwinkle}{\text{Constant of proportionality}}\\ \textcolor{DarkOrchid}{P} &= \textcolor{DarkOrchid}{\text{Population at time $t$}}\\ \textcolor{Aquamarine}{L} &= \textcolor{Aquamarine}{\text{Max size of population}}\\ \end{split}\end{equation*}

Solving the Logistic Equation

\begin{equation*}\begin{split} \frac{\mathrm{d}P}{\mathrm{d}t}&= kP \left({1 - \frac{P}{L}}\right)\\ \mathrm{d}P &= kP \left({1 - \frac{P}{L}}\right)\mathrm{d}t \\ \frac{1}{P \left({1 - \frac{P}{L}}\right)}\mathrm{d}P &= k \mathrm{d}t \end{split}\end{equation*}

Via partial fraction decomposition

\begin{equation*}\begin{split} \frac{1}{P \left({1 - \frac{P}{L}}\right)}&= \frac{A}{P}+ \frac{B}{\left({1 - \frac{P}{L}}\right)}\\ 0P + 1 &= A\left({1 - \frac{P}{L}}\right)+ BP\\ 0P + 1 &= A - \frac{P A}{L}+ BP\\ 0P + 1 &= P\left({B - \frac{A}{L}}\right)+ A \\ \text{Where}\\ \begin{rcases} B - \frac{A}{L}&= 0\\ A &= 1 \end{rcases}\;\begin{array}{ll} A &= 1\\ B &= \frac{1}{L}\end{array}\\ \text{Therefore}\\ \frac{1}{P \left({1 - \frac{P}{L}}\right)}&= \frac{1}{P}+ \frac{\frac{1}{L}}{\left({1 - \frac{P}{L}}\right)}\end{split}\end{equation*}

Rewriting the differential equation

\begin{equation*}\begin{split} \frac{1}{P \left({1 - \frac{P}{L}}\right)}\mathrm{d}P &= \frac{1}{P}+ \frac{\frac{1}{L}}{\left({1 - \frac{P}{L}}\right)}\mathrm{d}P\\ &= k \mathrm{d}t \\ \int \frac{1}{P}+ \frac{\frac{1}{L}}{\left({1 - \frac{P}{L}}\right)}&= \int k \mathrm{d}t \\ \ln\; P - \frac{1}{L}\ln\left({1 - \frac{P}{L}}\right)&= k t + C\\ \ln\left({\frac{P}{L \cdot \left({1 - \frac{P}{L}}\right)}}\right)\end{split}\end{equation*}

Second Order Homogeneous Linear Differential Equations with Constant Coefficients

Properties

If $f(x)$ and $g(x)$ are solutions; then $f + g$ is also a solution. Therefore, the most general solution to some second order homogeneous linear differential equations with constant coefficients would be $y = C_1 f(x) + C_2 g(x)$ .
This is why, when you find two solutions to the characteristic equation $r_1$ and $r_2$ respectively, we write it like so.

$r_1 = r_2$

Given some:

\begin{equation*}\begin{split} a y^{\prime\prime} + b y^\prime + c y = 0\;\text{where $a \neq 0$}\\ \end{split}\end{equation*}

We can presume that $y$ is of the form $e^{r t}$ , and therefore:

\begin{equation*}\begin{split} \text{if}\\ y &= e^{r t}\\ \text{then}\\ y^{\prime} &= r e^{r t}\\ y^{\prime\prime} &= r^2 e^{r t} \end{split}\end{equation*}

Substituting this back into the original equation, we have:

\begin{equation*}\begin{split} 0 &= a r^2 e^{r t} + b r e^{r t} + c e^{r t}\\ &= e^{r t} \left(a r^2 + b r + c\right) \end{split}\end{equation*}

Where:

\begin{equation*}\begin{split} &= \underbrace{e^{r t}}_\text{never $0$} \underbrace{\left(a r^2 + b r + c\right)}_\text{?} \end{split}\end{equation*}

So therefore:

\begin{equation*}\begin{split} &\underbrace{a r^2 + b r + c = 0} _\text{characteristic equation}\\ &r = \frac{-b \pm \sqrt{b^2 - 4ac}} \implies r_1, r_2 \end{split}\end{equation*}

Where the general solution is of the form:

\begin{equation*}\begin{split} \begin{rcases} y &=\; &C_1\; e^{r_1 t} &+\; C_2\; e^{r_2 t}\\ y^\prime &=\; &C_1\; r_1\; e^{r_1 t} &+\; C_2\; r_2\; e^{r_2 t}\\ y^{\prime\prime} &=\; &C_1\; \left(r_1\right)^2\; e^{r_1 t} &+\; C_2\; \left(r_2\right)^2\; e^{r_2 t} \end{rcases}\text{$\forall r_1 \; r_2$ where $r_1 \neq r_2$} \end{split}\end{equation*}

Parametric Equations

First Derivative Formula

To find the derivative of a given function defined parametrically by the equations $x = u(t)$ and $y = v(t)$ .

\begin{equation*}\begin{split} \text{Given}\\ x &= u(t)\\ y &= v(t)\\ \text{Therefore}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&= \frac{\frac{\mathrm{d}y}{\mathrm{d}t}} = \frac{v^\prime(t)}{u^\prime(t)}\end{split}\end{equation*}

Second Derivative Formula

To find the second derivative of a given function defined parametrically by the equations $x = u(t)$ and $y = v(t)$ .

Given

\begin{equation*}\begin{split} x &= u(t)\\ y &= v(t)\\ \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} \frac{\mathrm{d}{^2} y}{\mathrm{d}x^2}&= \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\\ &= \frac{\frac{d}{\mathrm{d}t}\left( \frac{\mathrm{d}y}{\mathrm{d}x}\right)}{\frac{\mathrm{d}x}{\mathrm{d}t}}\\ &= \frac{\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{v^\prime(t)}{u^\prime(t)}\right)}{u^\prime(t)}\\ &= \underbrace{ \frac{\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{v^\prime(t)}{u^\prime(t)}\right)}{\frac{\mathrm{d}}{\mathrm{d}t}u(t)}= \frac{\frac{\mathrm{d}}{\mathrm{d}t}}{\frac{\mathrm{d}}{\mathrm{d}t}}\frac{\left(\frac{v^\prime(t)}{u^\prime(t)}\right)}{u(t)}}_{ \text{notice the common $\frac{\mathrm{d}}{\mathrm{d}t}$} } \end{split}\end{equation*}

The above shows different ways of representing $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^2}$ . （I.e. it doesn't correspond to some final solution.）

Arc Length

Formula for the arc length of a parametric curve over the interval $[a, b]$ .

\begin{equation*}\begin{split} \int_a^b \sqrt{ \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 } \mathrm{d}t \end{split}\end{equation*}

Physics

Kinematic Equations in 1D

Conventions

\begin{equation*}\begin{split} \bar{v} &= \text{Average velocity}\\ \bar{a} &= \text{Average acceleration}\\ \text{Time} = \Delta{t} &= t_2 - t_1\\ \text{Displacement} = \Delta{x} &= x_2 - x_1\\ \Delta{v} &= v_2 - v_1 \end{split}\end{equation*}

Graphical Representation

Standard Equations

\begin{equation*}\begin{split} v_2 &= v_1 + a\Delta{t} \end{split}\end{equation*}

\begin{equation*}\begin{split} (v_2)^2 &= (v_1)^2 + 2a\Delta{x} \end{split}\end{equation*}

\begin{equation*}\begin{split} \Delta{x} &= v_1 \Delta{t} + \frac{1}{2}a\Delta{t^2}\\ &= v_2\Delta{t} - \frac{1}{2}a\Delta{t^2}\\ &= \left(\frac{v_2 + v_1}{2}\right)\Delta{t} \end{split}\end{equation*}

Summary

Formula	Missing	Quantities Present
$v_2 = v_1 + a\Delta{t}$	$\Delta{x}$	$v_1$ $v_2$ $a$ $\Delta{t}$
$\Delta{x} = \left(\frac{v_2 + v_1}{2}\right)\Delta{t}$	$a$	$\Delta{x}$ $v_1$ $\Delta{t}$ $v_2$
$\Delta{x} = v_1 \Delta{t} + \frac{1}{2}a\Delta{t^2}$	$v_2$	$\Delta{x}$ $v_1$ $\Delta{t}$ $a$
$\Delta{x} = v_2\Delta{t} - \frac{1}{2}a\Delta{t^2}$	$v_1$	$\Delta{x}$ $v_2$ $\Delta{t}$ $a$
$(v_2)^2 = (v_1)^2 + 2a\Delta{x}$	$\Delta{t}$	$\Delta{x}$ $v_1$ $v_2$ $a$

Basics

Constant Velocity

\begin{equation*}\begin{split} \bar{v} &= \frac{\Delta{x}}{\Delta{t}}\\ \Delta{x} &= \bar{v}\Delta{t} \end{split}\end{equation*}

Uniform acceleration

\begin{equation*}\begin{split} \bar{a} &= \frac{\Delta{v}}{\Delta{t}}\\ v_2 &= v_1 + a\Delta{t}\\ \Delta{x} &= v_0\Delta{t}+\frac{1}{2}a\Delta{t^2}\\ (v_2)^2 &= (v_1)^2 - 2\bar{a}\Delta{x} \end{split}\end{equation*}

Miscellaneous

\begin{equation*}\begin{split} \bar{v} &= \frac{\Delta{x}}{\Delta{t}}= \frac{v_2 + v_1}{2}\\ \bar{a} &= \frac{\Delta{v}}{\Delta{t}}\end{split}\end{equation*}

Deriving Displacement Formulas

Displacement when object moves with constant velocity

Deriving $\Delta{x} = \bar{v}\Delta{t}$

\begin{equation*}\begin{split} \Delta{x} &= \bar{v}\Delta{t} \end{split}\end{equation*}

Displacement when object accelerates from rest

Deriving $\Delta{x} = \frac{1}{2}\bar{a}\Delta{t^2}$

\begin{equation*}\begin{split} \Delta{x} &= \frac{1}{2}\Delta{v}\Delta{t}\\ &= \frac{1}{2}\bar{a}\Delta{t^2} \end{split}\end{equation*}

Displacement when object accelerates with initial velocity

Deriving $\Delta{x} = v_1\Delta{t} + \frac{1}{2}\bar{a}\Delta{t^2}$

\begin{equation*}\begin{split} \Delta{x} &= v_1\Delta{t} + \frac{1}{2}\Delta{v}\Delta{t} \\ &= v_1\Delta{t} + \frac{1}{2}\bar{a}\Delta{t^2} \end{split}\end{equation*}

Deriving The Other Kinematic Formulas

Deriving $v_2 = v_1 + \bar{a}\Delta{t}$

Given

\begin{equation*}\begin{split} \bar{a} &= \frac{\Delta{v}}{\Delta{t}}= \frac{v_2 - v_1}{\Delta{t}}\;(1) \end{split}\end{equation*}

We can rearrange $v_2$ from equation （1） like so

\begin{equation*}\begin{split} \bar{a} &= \frac{v_2 - v_1}{\Delta{t}}\\ \bar{a}\Delta{t} &= v_2 - v_1\\ v_1 + \bar{a}\Delta{t} &= v_2\\ \therefore v_2 &= v_1 + \bar{a}\Delta{t} \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} v_2 &= v_1 + \bar{a}\Delta{t} \end{split}\end{equation*}

Deriving $v_2^2 = v_1^2 + 2\bar{a}\Delta{x}$

Given

\begin{equation*}\begin{split} \bar{a} &= \frac{\Delta{v}}{\Delta{t}}\;(1)\\ \bar{v} &= \frac{\Delta{x}}{\Delta{t}}\;(2)\\ &= \frac{v_2 + v_1}{2}\;(3) \end{split}\end{equation*}

$\Delta{t}$ from equation （1） can be rearranged as

\begin{equation*}\begin{split} \Delta{t} &= \frac{\Delta{v}}{\bar{a}}= \frac{v_2 - v_1}{\bar{a}}\;(4) \end{split}\end{equation*}

$\Delta{x}$ from equation （2） can be rearranged like so

\begin{equation*}\begin{split} \bar{v} &= \frac{\Delta{x}}{\Delta{t}}\\ \bar{v}\Delta{t} &= \Delta{x}\\ \therefore \Delta{x} &= \bar{v}\cdot\Delta{t} \end{split}\end{equation*}

Using the following equations from above

$\bar{v} = \frac{v_2 + v_1}{2}$ from equation （3）
$\Delta{t} = \frac{v_2 - v_1}{\bar{a}}$ from equation （4）

\begin{equation*}\begin{split} \Delta{x} &= \bar{v}\cdot\Delta{t}\\ &= \frac{v_2 + v_1}{2}\cdot\frac{v_2 - v_1}{\bar{a}}\\ &= \frac{(v_2)^2 - (v_1)^2}{2\bar{a}}\;(5) \end{split}\end{equation*}

Rearranging equation （5）

\begin{equation*}\begin{split} 2\bar{a}\Delta{x} &= (v_2)^2 - (v_1)^2 \end{split}\end{equation*}

Rearrange again to obtain the more common form

\begin{equation*}\begin{split} (v_2)^2 &= (v_1)^2 + 2\bar{a}\Delta{x} \end{split}\end{equation*}

TODO

Two-dimensional Projectile Motion

Conventions

Summary

It's easy to see in the above visualization that $t$ and $x$ increase linearly, while $y$ is non-linear.

Formulas

Displacement & Projectile Position

Generalized

In general （without respect to any $x$ or $y$ axis values）

\begin{equation*}\begin{split} \text{Displacement} &= \Delta_{\text{general}}\\ &= \bar{v}\cdot\Delta{t} \end{split}\end{equation*}

Where the distance traveled or displaced is

\begin{equation*}\begin{split} \text{Displacement} &= \Delta_{\text{general}}\\ &= V_1\cdot\Delta{t} + \frac{1}{2}a\Delta{t^2} \end{split}\end{equation*}

In terms of $x$ and $y$ axis values

\begin{equation*}\begin{split} \Delta{x} &= x_2 - x_1\\ \Delta{y} &= y_2 - y_1 \end{split}\end{equation*}

With respect to the $y$ axis

The displacement of a given projectile in terms of the $y$ axis is

\begin{equation*}\begin{split} \Delta{y} = V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2} \end{split}\end{equation*}

Since $\Delta{y} = y_2 - y_1$

\begin{equation*}\begin{split} \Delta{y} &= V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2}\\ y_2 - y_1 &= V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2}\\ y_2 &= y_1 + V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2} \end{split}\end{equation*}

Which can be read as （in terms of the $y$ axis）

\begin{equation*}\begin{split} \small{\text{the final position}}\; &= \small{\text{the initial position}}\; + V_{y1}\cdot\Delta{t} + \frac{1}{2}a_y\Delta{t^2} \end{split}\end{equation*}

With respect to the $x$ axis

The displacement of a given projectile in terms of the $x$ axis is

\begin{equation*}\begin{split} \Delta{x} = V_{x}\cdot\Delta{t} + \frac{1}{2}a_x\Delta{t^2} (1) \end{split}\end{equation*}

Note that $a_x = 0$ （because there is no force acting on the projectile in the horizontal direction）, and therefore the initial and final velocities are the same. I.e. it's constant throughout. Therefore in summary

$a_x = 0$
$v_{x1} = v_{x2}$ and therefore we will simple refer to the velocity vector as as $v_x$ .

Therefore we can simplify equation （1） considerably

\begin{equation*}\begin{split} \Delta{x} &= V_{x}\cdot\Delta{t} + 0\\ &= V_{x}\cdot\Delta{t} \end{split}\end{equation*}

Solving Projectile Motion Problems

Projectile Motion

In terms of the $x$ axis

TODO

\begin{equation*}\begin{split} \Delta{x} &= x_2 - x_1 = V_{0,x} \dot t\\ v_{x} &= v_{0,x} + a_x \end{split}\end{equation*}

TODO

In terms of the $y$ axis

TODO

\begin{equation*}\begin{split} \Delta{y} &= y_2 - y_1 = V_{0,x} \dot t\\ v_{y} &= v_{0,y} + a_y \end{split}\end{equation*}

TODO

In Summary

Initial Quantities

\begin{equation*}\begin{split} v_x &= v_o\cdot\cos\theta\\ v_y &= v_o\cdot\sin\theta \end{split}\end{equation*}

\begin{equation*}\begin{split} a_y &= -g\\ a_x &= 0 \end{split}\end{equation*}

Derived expressions

\begin{equation*}\begin{split} \Delta{x} &= v_{0,x} \cdot \Delta{t}\\ v_x &= v_{0,x} \end{split}\end{equation*}

\begin{equation*}\begin{split} \Delta{y} &= v_{0,y} \cdot \Delta{t} - \frac{1}{2} g \Delta{t^2}\\ v_y &= v_{0,y} - g \Delta{t} \end{split}\end{equation*}

Solutions

\begin{equation*}\begin{split} t_{\text{top}} &= \frac{v_0\cdot\sin\theta}{g}\\ \Delta{y_{\text{max}}} &= \frac{v_0^2 + sin^2\theta}{2g}\\ \text{Range} &= \frac{2 \cdot v_0^2 \cdot \sin\theta \cdot \cos\theta}{g}\end{split}\end{equation*}

Projectile Motion from an initial height, with given initial velocity and angle

Given

A projectile angle $\theta$
The initial height $y_0$
The initial velocity $v_0$

We can therefore derive the the initial velocities for $x$ and $y$ in terms of the given angle and initial velocity.

\begin{equation*}\begin{split} v_{0x} &= v_0 \cdot \cos\theta\\ v_{0y} &= v_0 \cdot \sin\theta \end{split}\end{equation*}

Given the general formulas for displacement and velocity

\begin{equation*}\begin{split} \small \text{displacement} &= \text{initial displacement} + \text{initial velocity} \cdot \Delta{t} + \frac{1}{2}a\Delta{t^2}\\ \text{velocity} &= \text{initial velocity} + a\cdot\Delta{t} \end{split}\end{equation*}

Which this information, we will derive specific equations in terms of the $x$ and $y$ axes governing the projectile.

In terms of the $x$ axis

Deriving displacement as a function of time

Using the general formula from above in terms of $x$ as a function of time.

\begin{equation*}\begin{split} x(t) &= x_0 + v_{0x} t + \frac{1}{2}a_x t^2 \end{split}\end{equation*}

Which we can simplify using the following facts

From the given depiction of the problem, we know that $x(0) = 0$ .
There is no acceleration along the $x$ axis, so $a_x = 0$ .
$v_{0x} = v_0\cdot\cos\theta$ as shown above.

Therefore

\begin{equation*}\begin{split} x(t) &= 0 + v_{0x}\cdot t + \frac{1}{2}0 t^2\\ &= v_{0x} t\\ &= v_0\cos\theta\cdot t \end{split}\end{equation*}

Deriving velocity

\begin{equation*}\begin{split} v_x &= v_{0x} + a_x\cdot t\\ &= v_0\cdot\cos\theta + 0\cdot t\\ &= v_0\cdot\cos\theta \end{split}\end{equation*}

In terms of the $y$ axis

Deriving displacement as a function of time

Using the general formula from above in terms of $y$ as a function of time.

\begin{equation*}\begin{split} y(t) &= y_0 + v_{0y} t + \frac{1}{2}a_y t^2 \end{split}\end{equation*}

Which we can simplify using the following facts

Initial height is given to us which we will represent as $y_0$ , for the sake of generality.
Acceleration along the $y$ axis is the constant for gravity, so $a_y = -9.8\frac{\mathrm{m}}{\mathrm{s^2}}$ .
$v_{0y} = v_0\cdot\sin\theta$ as shown above.

Therefore

\begin{equation*}\begin{split} y(t) &= y_0 + v_{0y} t + \frac{1}{2}a_y t^2\\ &= y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}\left(-9.8\right) t^2\\ &= y_0 + v_0\cdot\sin\theta\cdot t - 4.9 t^2 \end{split}\end{equation*}

Deriving velocity

\begin{equation*}\begin{split} v_y &= v_{0y} + a_y\cdot t = v_{0y} + g \cdot t\\ &= v_0\cdot\sin\theta - 9.8\frac{\mathrm{m}}{\mathrm{s^2}}\end{split}\end{equation*}

In summary

\begin{equation*}\begin{split} x(t) &= v_0\cos\theta\cdot t\;(1)\\ v_x &= v_0\cdot\cos\theta\;(2) \end{split}\end{equation*}

\begin{equation*}\begin{split} y(t) &= y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}g t^2\;(3)\\ &= y_0 + v_0\cdot\sin\theta\cdot t - 4.9 t^2\\ v_y &= v_0\cdot\sin\theta + g \cdot t\;(4)\\ &= v_0\cdot\sin\theta - 9.8\frac{\mathrm{m}}{\mathrm{s^2}}\cdot t \end{split}\end{equation*}

To find the range

We know that at the moment of impact $y = 0$ , therefore we can use equation $(3)$

\begin{equation*}\begin{split} y = y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}g t^2 \end{split}\end{equation*}

Rearranging a bit and setting $y = 0$ , we can see that solving for $t$ will yield the time at which $y = 0$ .

\begin{equation*}\begin{split} 0 = \underbrace{\frac{1}{2}g}_{\text{a}}\;t^2 + \underbrace{v_0\cdot\sin\theta}_{\text{b}}\; t + \underbrace{y_0}_{\text{c}} \end{split}\end{equation*}

Therefore

\begin{equation*}\begin{split} t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ &= \frac{-v_0\cdot\sin\theta\pm\sqrt{\left(v_0\cdot\sin\theta\right)^2 - 4\left(\frac{1}{2}g\right)y_0}}{2\frac{1}{2}g}\\ &= \frac{-v_0\cdot\sin\theta\pm\sqrt{\left(v_0\cdot\sin\theta\right)^2 - 2 g y_0}}{g}\end{split}\end{equation*}

Plugging the solution for $t$ （and ignoring the negative or non-real solutions for $t$ ） into $x(t)$ will yield the horizontal displacement （range） at the time $y = 0$ . Therefore:

\begin{equation*}\begin{split} \text{range}\;=x(t)\;\small\text{where $t$ is the point at which $y=0$} \end{split}\end{equation*}

To find the maximum vertical displacement （i.e. peak height）

We begin with equation $\text{(4)}$

\begin{equation*}\begin{split} v_y &= v_0\cdot\sin\theta + g t \end{split}\end{equation*}

We know that at the moment our projectile crests its trajectory, the vertical component of our projectile will be zero. Therefore $v_y = 0$ . To find the time, we simply solve for $t$ .

\begin{equation*}\begin{split} 0 &= v_0\cdot\sin\theta + g \cdot t\\ -v_0\cdot\sin\theta &= g \cdot t\\ \frac{-v_0\cdot\sin\theta}{g}&= t\\ \therefore\;t &= \frac{-v_0\cdot\sin\theta}{g}\end{split}\end{equation*}

Therefore, knowing the time at which our projectile crests its trajectory, we simply plugin our solution for $t$ into the function given in equation $(4)$ . I.e:

\begin{equation*}\begin{split} y_{\text{max}} &= y_0 + v_0\cdot\sin\theta\cdot t + \frac{1}{2}g t^2\\ &= \small y_0 + v_0\cdot\sin\theta\cdot\left(\frac{-v_0\cdot\sin\theta}{g}\right) + \frac{1}{2}g \left(\frac{-v_0\cdot\sin\theta}{g}\right)^2 \end{split}\end{equation*}

To find the velocity at a given moment of time

Given some time which we will denote as $t_n$ , to find the velocity we simply plug in our given values for $\theta$ and $v_0$ into equations $(2)$ and $(4)$ . I.e.

\begin{equation*}\begin{split} v_x &= v_0\cdot\cos\theta\\ v_y &= v_0\cdot\sin\theta + g \cdot t \end{split}\end{equation*}

With the given value for $t_n$ , yielding the vector at time $t = t_n$ , which we will denote as $\vec{v_n}$

\begin{equation*}\begin{split} A &= v_y(t_n)\\ B &= v_x(t_n)\\ \end{split}\end{equation*}

To define the vector in terms of engineering notation, （i.e. $v_x\hat{i} + v_y\hat{j}$ ）

\begin{equation*}\begin{split} \vec{v_n} &= B\hat{\textbf{i}} + A\hat{\textbf{j}} \end{split}\end{equation*}

To define the vector in terms of magnitude （which we will denote as $v_n$ ） and direction （which we will denote as $\theta_n$ ）

\begin{equation*}\begin{split} v_n &= \sqrt{A^2 + B^2}\\ \theta_n &= \tan^{-1}\left(\frac{A}{B}\right) \end{split}\end{equation*}

Range

The distance a projectile travels is called its range.

\begin{equation*}\begin{split} \underbrace {\text{range}\;=\frac{v^2 \cdot sin\left(2\theta\right)}{g}} _{\small{\text{start/end elevation must be the same}}} \end{split}\end{equation*}

Only applies in situations where the projectile lands at the same elevation from which it was fired.

Reasoning About Projectile Motion

Notes

An object is in free fall when the only force acting on it is the force of gravity.

Question

Based on the figure, for which trajectory was the object in the air for the greatest amount of time?

Answer

Trajectory A

Explanation

All that matters is the vertical height of the trajectory, which is based on the component of the initial velocity in the vertical direction （ $v_0\sin\theta$ ）. The higher the trajectory, the more time the object will be in the air, regardless of the object's range or horizontal velocity.

Problems

The function in this graph represents an object that is speeding up, or accelerating at a constant rate.

When you throw a ball directly upward, what is true about its acceleration after the ball has left your hand?

Answer: The ball’s acceleration is always directed downward.

Wrong: The ball’s acceleration is always directed downward, except at the top of the motion, where the acceleration is zero.

Question

As an object moves in the x-y plane, which statement is true about the object’s instantaneous velocity at a given moment?

Answer

The instantaneous velocity is tangent to the object’s path

Wrong

The instantaneous velocity is perpendicular to the object’s path.
The instantaneous velocity can point in any direction, independent of the object’s path.

Explanation

As an object moves in the x-y plane the instantaneous velocity is tangent to the object‘s path at a given moment. This is because the displacement vector during an infinitesimally small time interval is always directed along the object’s path and the velocity vector always has the same direction as the displacement vector.

Relative Motion

Galilean transformation of velocity

The velocity $\vec{v}$ of some object P as seen from a stationary frame must be the sum of $\vec{w}$ and $\vec{v_F}$

\begin{equation*}\begin{split} \vec{v} &= \vec{w} + \vec{v_F} \end{split}\end{equation*}

Where

Symbol	Description
$\vec{v}$	Velocity as measured in a stationary frame
$\vec{w}$	Velocity of an object measured in the moving frame relative to the moving frame
$\vec{v_F}$	velocity of the moving frame - with respect to the stationary frame

Galilean transformation of velocity （alternate notation）

Given two reference frames \text｛ A ｝ and $B$ and some object $O$ . The velocity of the object can be defined in terms of $A$ or $B$ as shown

Symbol	Description
$\vec{v_{O,A}}$	The velocity of $O$ relative to $A$
$\vec{v_{O,B}}$	The velocity of $O$ relative to $B$
$\vec{v_{A,B}}$	The velocity of $A$ relative to $B$
$\vec{v_{B,A}}$	The velocity of $B$ relative to $A$ . It locates the origin of $A$ relative to the origin of $B$ .

Therefore

\begin{equation*}\begin{split} \vec{v_{O,B}} &= \vec{v_{O,A}} + \vec{v_{A,B}}\\ \vec{v_{O,A}} &= \vec{v_{O,B}} + \vec{v_{B,A}} \end{split}\end{equation*}

Rotational Motion & Kinematics

Basics

\begin{equation*}\begin{split} \small\text{Angular velocity}\normalsize= \omega &= \tau f\\ &= \frac{\tau}{T}\\ \end{split}\end{equation*}

\begin{equation*}\begin{split} \small\text{Centripetal acceleration}\normalsize= a_C &= \frac{v^2}{r}\\ &= \frac{\omega^2 r^2}{r}\\ &= \omega^2 r \end{split}\end{equation*}

\begin{equation*}\begin{split} \text{Period} = T &= \frac{1}{f}= \frac{\tau}{\omega}\end{split}\end{equation*}

\begin{equation*}\begin{split} \vec{v} &\perp \vec{r}\\ \vec{a} &\perp \vec{v}\\ \end{split}\end{equation*}

\begin{equation*}\begin{split} \begin{rcases} \vec{a} &\parallel \vec{r}\\ \vec{a} &\propto \vec{r} \end{rcases}\;\small\text{They are anti-parallel}\normalsize\end{split}\end{equation*}

Auxiliary Formula Reference

\begin{equation*}\begin{split} \text{Period} = T &= \frac{1}{f}= \frac{\tau}{\omega}\end{split}\end{equation*}

\begin{equation*}\begin{split} \theta &= \omega \cdot t\\ &= \omega_1 \cdot t + \frac{1}{2}\alpha t^2\\ \omega_2 &= \omega_1 + \alpha \cdot t\\ \omega_2^2 &= \omega_1^2 + 2\cdot\alpha\cdot\theta\\ v &= \small{\frac{1 \text{circumference}}{1 \text{period}}}\normalsize\\ &= \frac{2 \pi r}{T}\end{split}\end{equation*}

Formula	Missing	Quantities Present
$\omega_2 = \omega_1 + \alpha\Delta{t}$	$\Delta\theta$	$\omega_1$ $\omega_2$ $\alpha$ $\Delta{t}$
$\Delta\theta = \left(\frac{\omega_2 + \omega_1}{2}\right)\Delta{t}$	$a$	$\Delta\theta$ $\omega_1$ $\Delta{t}$ $\omega_2$
$\Delta\theta = \omega_1 \Delta{t} + \frac{1}{2}\alpha\Delta{t^2}$	$\omega_2$	$\Delta\theta$ $\omega_1$ $\Delta{t}$ $\alpha$
$\Delta\theta = \omega_2\Delta{t} - \frac{1}{2}\alpha\Delta{t^2}$	$\omega_1$	$\Delta\theta$ $\omega_2$ $\Delta{t}$ $\alpha$
$(\omega_2)^2 = (\omega_1)^2 + 2\alpha\Delta\theta$	$\Delta{t}$	$\Delta\theta$ $\omega_1$ $\omega_2$ $\alpha$

\begin{equation*}\begin{split} \underbrace{\small\text{Arc Length}\normalsize}_{S} &= \underbrace{\small\text{radius}\normalsize}_{r} \cdot \underbrace{\small\text{Central angle}\normalsize}_{\theta}\\ S &= r\cdot\theta\\ \end{split}\end{equation*}

\begin{equation*}\begin{split} \underbrace{\small\text{Linear displacement}\normalsize}_{\Delta{x}} &= \underbrace{\small\text{Angular displacement}\normalsize}_{\theta} \cdot \underbrace{\small\text{radius}\normalsize}_{r}\\ \Delta{x} &= \theta \cdot r\\ \underbrace{\small\text{Linear velocity}\normalsize}_{v} &= \underbrace{\small\text{Angular Velocity}\normalsize}_{\omega} \cdot \underbrace{\small\text{radius}\normalsize}_{r}\\ v &= \omega \cdot r\\ \underbrace{\small\text{Linear acceleration}\normalsize}_{a} &= \underbrace{\small\text{Angular acceleration}\normalsize}_{\alpha} \cdot \underbrace{\small\text{radius}\normalsize}_{r}\\ a &= \alpha \cdot r \end{split}\end{equation*}

\begin{equation*}\begin{split} \underbrace{\small\text{Angular displacement}\normalsize}_{\theta} &= \underbrace{\small\text{Angular speed}\normalsize}_{\omega} \cdot \underbrace{\small\text{time}\normalsize}_{t}\\ \theta &= \omega \cdot t \end{split}\end{equation*}

A particle moves with uniform circular motion if and only if its angular velocity V is constant and unchanging.

Uniform Circular Motion

Uniform means content speed

Forces and Newton's laws of motion

Newton's laws of motion

Normal force and contact force

Balanced and unbalanced forces

Inclined planes and friction

Tension

Test Page

Hello world

Math

\begin{equation*}\begin{split} \vec{v} &= \begin{bmatrix} v_x & v_y & v_z \end{bmatrix}\\ \;\;\;\; ||\vec{v}|| &= \sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2} \end{split}\end{equation*}

\begin{dcases} a & x\\ a & y\\ a & z \end{dcases}

\begin{rcases} a & x\\ a & y\\ a & z \end{rcases}

\begin{align*} 2x - 5y &= 8 \\ 3x + 9y &= -12 \end{align*}

\begin{align} 2x - 5y &= 8 \\ 3x + 9y &= -12 \end{align}

\begin{align}\tag{A} 2x - 5y &= 8 \\ 3x + 9y &= -12 \end{align}

\begin{gather*} 2x - 5y = 8 \\ 3x^2 + 9y = 3a + c \end{gather*}

\begin{gather} 2x - 5y = 8 \\ 3x^2 + 9y = 3a + c \end{gather}

\begin{gather}\tag{B} 2x - 5y = 8 \\ 3x^2 + 9y = 3a + c \end{gather}

Drawings

Top-Level Title

Top-level Title

Subsection Title

Some Drawing Entry

Some Other Drawing Entry

Hello world

Quisque eget posuere odio, mollis consectetur purus. Morbi iaculis porta diam, quis rhoncus orci dictum tempor. Pellentesque viverra quam erat, sagittis luctus nulla condimentum in. Morbi dui ligula, laoreet eget urna sed, accumsan semper metus. Orci varius natoque penatibus et magnis dis parturient montes, nascetur ridiculus mus. Mauris ex enim, dapibus vitae sollicitudin vitae, placerat et lorem. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Etiam vel viverra ante. Ut ac lacus erat. Nunc eleifend elementum ex a posuere.

Hello world

\begin{equation}\tag{A}\begin{split} A = \int_{a}^{b} f(x) \;\mathrm{d}x &\approx \sum_{i = 1}^{n}\, \Delta{x}\cdot f\left(a + \frac{(i - 1) + i}{2}\cdot \Delta{x}\right) = \sum_{i = 1}^{n}\, \Delta{x}\cdot f\left(a + \frac{2i - 1}{2}\cdot \Delta{x}\right)\\ &\approx \sum_{i = 0}^{n-1}\, \Delta{x}\cdot f\left(a + \frac{(i + 1) + i}{2}\cdot \Delta{x}\right) = \sum_{i = 0}^{n-1}\, \Delta{x}\cdot f\left(a + \frac{2i + 1}{2}\cdot \Delta{x}\right) \end{split}\end{equation}

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Hello world

Pellentesque viverra quam erat, sagittis luctus nulla condimentum in. Morbi dui ligula, laoreet eget urna sed, accumsan semper metus. Orci varius natoque penatibus et magnis dis parturient montes, nascetur ridiculus mus. Mauris ex enim, dapibus vitae sollicitudin vitae, placerat et lorem. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Etiam vel viverra ante. Ut ac lacus erat. Nunc eleifend elementum ex a posuere.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa. Aenean a purus aliquam justo varius feugiat. Quisque eget posuere odio, mollis consectetur purus. Morbi iaculis porta diam, quis rhoncus orci dictum tempor. Pellentesque viverra quam erat, sagittis luctus nulla condimentum in. Morbi dui ligula, laoreet eget urna sed, accumsan semper metus. Orci varius natoque penatibus et magnis dis parturient montes, nascetur ridiculus mus. Mauris ex enim, dapibus vitae sollicitudin vitae, placerat et lorem. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Etiam vel viverra ante. Ut ac lacus erat. Nunc eleifend elementum ex a posuere.

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa.
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa.
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa.
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa.
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa.
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Etiam a nisi eu sapien bibendum scelerisque. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia curae; Fusce id blandit lacus, at interdum massa.

Morbi iaculis porta diam, quis rhoncus orci dictum tempor. Pellentesque viverra quam erat, sagittis luctus nulla condimentum in. Morbi dui ligula, laoreet eget urna sed, accumsan semper metus. Orci varius natoque penatibus et magnis dis parturient montes, nascetur ridiculus mus. Mauris ex enim, dapibus vitae sollicitudin vitae, placerat et lorem. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Etiam vel viverra ante. Ut ac lacus erat. Nunc eleifend elementum ex a posuere.