vault backup: 2026-02-09 19:20:04

education/calculus/precalculus/Composing Functions.md

@@ -1,3 +0,0 @@
-- To compose a function is to create a new function from multiple smaller functions.
-- They can be solved from the inside out
-- 

education/math/MATH1220 (calc II)/Integral Review.md

@@ -11,14 +11,15 @@ Define $$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i)\Delta x$$
 	- Or, the width of each interval times the interval index, plus the starting offset.
 
 Then let $f$ be a continous function on $[a, b]$ and let $F$ be the antiderivative of $f$ (i.e $F'(x) = f(x)$).
+
 Then $\int_a^b f(x) dx = F(b) - F(a)$.
 
+## Sum of an infinite series
+The sum of an infinite series can be defined as follows:
+$$ \sum_{i = 1}^n i = \frac{n(n+1)}{2}$$
+
 ## Examples
 1. Find the area under the curve between 0 and 1 of the function $f(x) = x^2$
 $$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
 2. Find the Riemann Sum under the curve between -2 and 2 of the function $2x + 2$. 
 $$ \int_{-2}^2 (2x + 2)dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$
-> Using the fact that $x_i = \Delta x + a$, $
-$$ = lim_{n \to \infty} \sum_{i=1}^n(2x_i + 2)\frac{4}{n}$$
-$$ = \lim_{n \to \infty} \sum_{i = 1}^n (2(-2 +\frac{4i}{n}) + 2)\frac{4}{n}$$
-$$ = \lim_{n \to \infty} \sum_{i = 1}^n(-4 + \frac{8i}{n} + 2)\frac{4}{n} $$

education/math/MATH1220 (calc II)/Integration by Parts.md

@@ -1,6 +1,11 @@
 The integration by parts formula is:
 $$ \int udv = uv - \int vdu $$
-
+Broadly speaking, integration by parts is done by:
+1. Pick a part of integral to be $u$. 
+2. The rest of the integral will be $dv$, 
+3. Compute the derivative of $u$, $du$.
+4. Compute the antiderivative of $dv$
+5. Substitute those values in to the integration by parts formula.
 ## Deriving the Integration by Parts Formula
 $$ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) $$
 1. Integrating both sides, we get:
@@ -15,3 +20,23 @@ $$ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$$
 Now, let $u = f(x)$ and $v = g(x)$, then $dv = g'(x)dx$ and $du = f'(x)dx$. 
 
 # Examples
+>  Evaluate the below antiderivative using integration by parts.
+$$\int xe^{2x}dx$$
+1. Define $u$ to be a value you can take the derivative of easily, in this case $u = x$. The rest of the integral will be set to $dv$, in this case, $dv = e^{2x}dx$. 
+	- $u = x$
+	- $du = \frac{d}{dx}(x)= 1dx$ 
+	- $dv = e^{2x}dx$
+	- $v = \frac{1}{2}e^{2x}$ - The antiderivative of $dv$.
+2. Looking back at the integration by parts formula, we know that:
+	$$ \int udv = uv - \int v du $$
+	$$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)-\int (\frac{1}{2}e^{2x}) (1dx) $$
+3. The remaining integral can be solved with $u$ substitution, but we've already defined $u$, so we use $w$ as a replacement.
+	- $w = 2x$
+	- $dw = 2dx$
+	- $\frac{1}{2}dw=dx$ 
+		1. Substituting $w$ and $dw$ into the integral:
+		$$ \int \frac{1}{2}e^w \frac{1}{2}dw $$
+		2. This gives an integral that can be computed naively
+		$$ \int\frac{1}{2}e^{w}\frac{1}{2}dw = \frac{1}{4}\int e^w dw= \frac{1}{4}e^{2x} $$
+4. Combining everything together, we get:
+	$$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)- (\frac{1}{4}e^2x) + C$$

education/math/MATH1220 (calc II)/Integration with Trig Identities.md

@@ -0,0 +1,85 @@
+The below integration makes use of the following trig identities:
+1. The Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$
+2. The derivative of sine: $\frac{d}{dx}sin(x) = cos(x)$
+3. The derivative of cosine: $\dfrac{d}{dx} \cos(x) = -\sin(x)$
+4. Half angle cosine identity: $\cos^2(x) = \frac{1}{2}(1 + \cos(2x))$
+5. Half angle sine identity: $\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$
+6. $tan^2(x) + 1 = sec^2(x)$
+7. $\dfrac{d}{dx}(\tan(x)) = \sec^2(x) \Rightarrow \int \sec^2(x)dx = \tan(x) + C$ 
+8. $\dfrac{d}{dx}(\sec x) = \sec(x)\tan(x) \Rightarrow \int\sec(x)\tan(x) dx = \sec(x) + C$
+# Examples
+> Evaluate the integral $\int\sin^5(x)dx$
+1. With trig identities, it's common to work *backwards* with u-sub. In the above example, we can convert the equation into simpler cosine functions by setting $du$ to $-\sin(x)dx$. This means that $u$ is equal to $cos(x)$. 
+$$ \int\sin^4(x)\sin(x)dx$$
+2. Rewrite $sin^4(x)$ to be $(\sin^2(x))^2$ to take advantage of the trig identity $1 - \cos^2(x) = \sin^2(x)$ 
+$$ \int(\sin^2x)^2 \sin(x)dx$$
+3. Apply the above trig identity and substitute $u$:
+$$ \int(1 - u^2)^2 (-du) $$
+4. Foil out and move negative out of integral:
+$$ -\int(1 - 2u^2 + u^4)du $$
+5. Take advantage of the distributive property of integrals:
+ $$ - (u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + C $$
+ 6. Substituting $\cos(x)$ back in for $u$, we get the evaluated (but not entirely simplified) integral:
+ $$-(\cos(x)- \frac{2}{3}\cos^3x + \frac{1}{5}\cos^5x) $$
+# Trigonometric Substitutions
+Trigonometric substitution is useful for equations containing  $\sqrt{a^2 + x^2}$ or $a^2 + x^2$, where $a$ is any constant. It removes any addition or subtraction.
+
+The general process involves the use of a trig identity and multiplying everything in that identity by a constant.
+
+Consider the identity:
+$$ 1 + \tan^2(\theta) = \sec^2(\theta)$$
+Multiplying both sides of the identity by $a^2$, we get:
+$$a^2 + a^2\tan^2(\theta) = a^2\sec^2(\theta)$$
+This enables us to make use of **substitution** to simplify many integrals.
+- $x = a\tan \theta$
+- $dx = a \sec^2\theta d\theta$
+- for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$
+# Examples
+> Evaluate the integral $\int\frac{3}{4+x^2}dx$ 
+1. Move the constant coefficient out of the integral:
+$$ \int \frac{3}{4+x^2}dx = 3\int \frac{1}{4 + x^2}dx$$
+2. Let $x = 2tan\theta$ and $dx = (2sec^2\theta d\theta)$, substitute accordingly
+$$ = 3\int\frac{1}{4 + 4\tan^2\theta}(2\sec^2\theta)d\theta$$
+3. Factor $4$ out in the denominator
+$$ = 3\int\frac{1}{4(1 + \tan^2\theta)}(2\sec^2\theta)d\theta$$
+4. Considering the identity $1 + \tan^2 \theta = \sec^2 \theta$
+$$ = 3\int\frac{1}{4(\sec^2\theta)}(2\sec^2\theta)d\theta$$
+5. $\sec^2\theta$ is present in the numerator and the denominator, so we can cancel those out. This means that:
+$$ 3\int\frac{2}{4}d\theta = \frac{3}{2} \theta + C$$
+6. At this point, we want to determine what $\theta$ is equal to relative to $x$. 
+	1. Look back to step 2 we defined $x = 2\tan\theta$
+	2. Moving $2$ to the other side, we get $\frac{x}{2} = \tan\theta$ 
+	3. Because we defined bounds for our definition of $\theta$, we can take advantage of $\arctan$, therefore:
+$$ \theta = \arctan(\frac{x}{2}) $$
+7. Rewriting the equation with $\theta$ in terms of x, we get:
+$$ \frac{3}{2}\arctan(\frac{x}{2}) + C$$
+This means that:
+$$ \int\frac{3}{4+x^2}dx = \frac{3}{2}\arctan(\frac{x}{2}) + C $$
+
+# A VERY LARGE LIST OF TRIG IDENTITIES FOR CALCULUS
+
+| Reciprocal           |
+| ------------------------------------ |
+| $\csc(x) = \dfrac{1}{sin(x)}$        |
+| $\sec(x) = \dfrac{1}{\cos(x)}$       |
+| $\cot(x) = \dfrac{1}{\tan(x)}$       |
+
+| Quotient |
+| -- |
+| $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$  |
+| $\cot(x) = \dfrac{\cos(x)}{\sin(x)}$  |
+
+| Pythagorean |
+| -- |
+| $\sin^2(x) + \cos^2(x) = 1$          |
+| $1 + \tan^2(x) = \sec^2(x)$          |
+| $1 + \cot^2(x) = \csc^2(x)$          |
+
+| Sum and Difference                                  |
+| --------------------------------------------------- |
+| $\sin(A \pm B) = \sin(A)\cos(B) \pm \cos(A)\sin(B)$ |
+| $\cos(A \pm B) = \cos(A)\cos(B) \mp \sin(A)\sin(B)$ |
+| $\sin(2x) = 2\sin(x)\cos(x)$                        |
+| $\cos(2x) = \cos^2(x) - \sin^2(x)$                  |
+| $\cos^2(x) = \frac{1}{2}(1 + \cos(2x)$              |
+|                                                     |

education/math/MATH1220 (calc II)/Sequences and Series.md

@@ -0,0 +1,159 @@
+A sequence is defined as an ordered list of numbers.
+- Sequences are ordered, meaning two sequences that contain the same values but in a different order are not equal.
+- Sequences can be infinite if a rule is defined, i.e $\{1, 1, 1, 1, ...\}; a_i = 1$ 
+
+# Behavior
+- A sequence is considered **increasing** if $a_n$ is smaller than $a_{n+1}$ for all $n$.
+- A sequence is considered **decreasing** if $a_n$ is greater than or equal to $a_{n+1}$ for all $n$.
+- Sequences exist that do not fall into either category, i.e, $a_n = (-1)^n$
+
+- If the terms of a sequence grow $\{a_n\}$ get arbitrarily close to a single number $L$ as $n$ grows larger, this is noted by writing:
+$$\lim_{n\to\infty} a_n = L$$ OR
+$$ a_n \to L \text{ as } n \to \infty $$
+
+and say that $a_n$ *converges* to $L$. If no $L$ exists, we say $\{a_n\}$ *diverges*.
+
+# Properties of Sequences
+> The below properties assume two sequences are defined, $a_n \to L$ and $b_n \to M$
+1. $a_n + b_n \to L + M$
+2. $C*a_n \to CL$ 
+3. $a_n b_n \to LM$
+4. $\frac{a_n}{b_n} \to \frac{L}{M}$ holds true where all values are defined
+5. If $L = M$ and a sequence $c_n$ exists such that $a_n \le c_n \le b_n$ for all $n$, then $c_n \to L = M$
+6. If $a_n$ and $b_n$ both approach infinity at a similar rate, $\frac{a_n}{b_n}$  will approach an arbitrary value. This value can be found by rewriting $\frac{a_n}{b_n}$ in such a manner that the end behavior of the series is more easily identifiable
+	> For example, given the series $c_n = \frac{n}{2n+1}$, both the numerator and the denominator approach infinity at a similar rate. However, when the numerator and denominator are both multiplied by $\frac{1}{n}$, it becomes $\frac{1}{2+\frac{1}{n}}$, an equivalent sequence that more clearly converges on $1/2$. 
+
+# Sum of an infinite sequence
+- If $f(x)$ is a function and $\{a_n\}$ is a sequence such that $f(n) = a(n)$, then we say $f(x)$ *agrees* with the sequence  $\{a_n\}$ 
+- If $f(x)$ agrees with $\{a_n\}$ then if $\lim_{x \to \infty}f(x) = L$  then $\lim_{n \to \infty}a_n = L$
+- Given the above knowledge, we can apply L'Hospital's rule to sequences that seem to approach $\frac{\infty}{\infty}$.
+
+Remember, L'Hospital's rule states that:
+
+>If you have a limit of the indeterminate form $\dfrac{0}{0}$, the limit can be found by taking the derivative of the numerator, divided by the derivative of the denominator.
+> $$ \lim_{x \to 2} \dfrac{x-2}{x^2-4} = \lim_{x \to 2} \dfrac{1}{2x}$$
+> L'Hospital's Rule can also be used when both the numerator and denominator approach some form of infinity.
+>$$ \lim_{x \to \infty} \dfrac{x^2-2}{3x^2-4} = \lim_{x \to \infty} \dfrac{2x}{6x}$$
+>The above problem can be solved more easily *without* L'Hospital's rule, the leading coefficients are 1/3, so the limit as $x$ approaches $\infty$ is 1/3.
+> L'Hospital's rule **cannot** be used in any other circumstance. 
+
+# Series
+Vocabulary: A **series** is another name for a sum of numbers. 
+
+## The Limit Test
+You can break a series into *partial sums*:
+$$\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + ...$$
+Given the above series, we can define the following:
+- $S_1 = a_1 = \sum_{i=1}^\infty a_i$
+- $S_2 = a_1 + a_2 = \sum_{i=1}^2 a_i$ 
+- $S_n = a_1 + a_2 + ... = \sum_{i=1}^n a_i$
+- If the limit of $S_n$ as $S_n$ approaches $\infty$ converges to $L$, then we write:
+  $$\sum_{n=1}^\infty a_n = L$$
+and say that the sum converges to $L$. 
+
+## Examples
+> Prove that $\sum_{n = 1}^\infty \frac{1}{2^n} = 1$
+- $S_1 = \frac{1}{2}$
+- $S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ 
+- $S_3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$ 
+- $S_n = \frac{2^n - 1}{2^n}$
+So:
+$$ \sum_{n=1}^\infty \frac{1}{2^n} = \lim_{n \to \infty}S_n = \lim_{n \to \infty} (1 - \frac{1}{2^n}) = 1$$
+
+> Using the limit test, determine whether the series $\sum_{n = 1}^\infty n$  converges or diverges
+- $S_1 = 1$
+- $S_2 = 1 + 2 = 3$
+- $S_n = \frac{n(n+1)}{2}$ 
+So: 
+$$ \lim_{n \to \infty} \frac{n(n+1)}{2} = \infty $$
+
+Given the above info, the limit is non-zero, so we know that the series diverges.
+
+# Geometric Series
+A geometric series of the form:
+$$ \sum_{n = 1}^\infty ar^{n-1} = \sum_{n=0}^\infty ar^n $$
+Converges to $\dfrac{a}{1-r}$  if $|r| < 1$ or diverges if $|r| >= 1$. 
+
+# Examples:
+
+> Determine if the series $\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1})$ diverges or converges. If it converges, state where.
+1. Rewrite $7^{-n}$ as $(\frac{1}{7})^n$ to be closer to the form $ar^n$
+$$ = \sum_{n=1}^\infty 35*(\frac{1}{7})^n*2^{n-1}$$
+2. Pull a $7$ out of the bottom, making use of the fact that $(\frac{1}{7})^n = \frac{1}{7}(\frac{1}{7})^{n-1}$ 
+$$ = \sum_{n=1}^\infty \frac{35}{7}(\frac{1}{7})^{n-1}* 2^{n-1} = \sum_{n = 1}^\infty \frac{35}{7}(\frac{2}{7})^{n-1} $$
+3. This is now of the form $\sum_{n=1}^\infty ar^{n-1}$, so:
+$$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1-\frac{2}{7}}$$
+
+# Divergence Test
+**If $\lim_{n \to \infty} a_n \ne 0$ then $\sum_{n=1}^\infty a_n$diverges.**
+
+The divergence test only tells us that if the limit does not equal zero, then the series diverges. If the limit is zero, it doesn't necessarily mean the series converges.
+
+# Alternating series test
+Sometimes a series is not continually positive for the entire series, meaning most tests on series do not apply. To get around this, you can split the series into two or more parts. A finite negative number + infinity is still infinity, and a finite negative number + a finite number is still a finite number.
+
+A simple example of an alternating series is:
+$$ \sum_{n=1}^\infty (-1)^{n+1}a_n = a_1 - a_2 + a_3 - a_4 $$
+The above series converges if all three of the following hold true:
+- $a_n > 0$
+- Series decreases: $a_n \ge a_{n+1}$  for all $n$
+- $\lim_{n\to\infty} a_n = 0$ as_
+This test does not provide any guarantees about divergence i.e if if the test fails, the series does not necessarily diverge.
+
+A sequence $a_n$ *converges absolutely* if $\sum |a_n|$ converges
+
+Then if the *series converges* absolutely then the sum converges.
+
+## Examples
+> Does the series $\sum_{n=1}^\infty (\frac{(-1)^n}{n^2}$  conditionally converge, absolutely converge, or diverge?
+1. $\sum_{n=1}^\infty|\frac{(-1)^n}{n^2}| = \sum_{n=1}^\infty \frac{1}{n^2}$ 
+2. We know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges by the p test, so the series absolutely converges.
+
+> Does the series $\sum_{n=1}^\infty (\frac{(-1)^n}{n+5}$  conditionally converge, absolutely converge, or diverge?
+1. Consider $\sum_{n=1}^\infty|\frac{(-1)^n}{n+5}| = \sum_{n=1}^\infty \frac{1}{n+5}$.
+2. The above series can be compared to $\frac{1}{n}$ with the limit comparison test. $\lim_{n \to \infty}\dfrac{\frac{1}{n}}{\frac{1}{n+5}} = 1 \ne 0, \infty$. By the L.C.T, $\frac{1}{n+5}$ diverges, so the series does not absolutely converge.
+3. Applying the Alternating Series Test,
+	- $a_n > 0$ for all $n$ - all values in the series are greater than zero
+	- $a_n \ge a_{n+1}$ - the series decreases
+	- $\lim_{n \to \infty} a_n = 0$ - the series approaches zero
+All three conditions hold true, therefore we know that $\sum_{n=1}^\infty \frac{(-1)^n}{n+5}$ conditionally converges.
+
+## Error
+Let $\sum_{n=1}^\infty (-1)^n a_n$ be a series shown to converge by the alternating series test, and that it converges to a $L$. Then the remainder for a given term $N$ is $R_N = L - S_N$ . So $|R_N| \le a_{N+1}$. 
+
+The error for the first $n$ terms of a sequence is less than or equal to the $(n + 1)$ st term of the sequence. 
+
+# Power Series
+A power series centered at $c$ is a series of the form:
+$$ \sum_{n=0}^\infty a_n(x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots$$
+where $x$ is a variable.
+
+$\sum x^n$ converges when $|x| < 1$  and diverges when $|x| \ge 1$. 
+
+The above series is a power series where $a_n = 1$  and $c = 0$.
+
+## Behavior
+The behavior a given power series falls into one of three cases:
+1. The series converges on an interval with radius $R > 0$ When this happens, each interval endpoint needs to be checked independently, because the ratio test will always be indeterminate at those points.
+2. The series converges for all $x = \mathbb{R}$
+3. The series will converge at a single point, $c$. 
+
+# Examples
+> When does the series $\sum_{n=0}^\infty \frac{x^n}{3^n}$ converge?
+1.  Solving for any power series usually starts with the ratio test.
+		1. Create a ratio with $n + 1$ and $n$. 
+		$$ \frac{x^{n+1}}{3^{n+1}} \cdot \frac{3^n}{x^n} $$
+		2. Cancel stuff out
+		$$ \frac{x}{3}$$
+2. For a geometric series, it converges when the ratio $r$ is less than one. Written as an equality, this gives us:
+$$ |r| < 1 \to |\frac{x}{3}| < 1 \to |x| < 3 $$
+This means that the interval of convergence is $(-3, 3)$, given it diverges at both interval endpoints.
+
+## Integral test for series
+The integral test determines if an infinite series converges or diverges by comparing it to an improper integral. If the integral diverges ($= \infty$), then the series also diverges. If the integral converges, then the series also converges. This is somewhat unique because it proves both convergence and divergence, but has the caveat that the integral must be feasibly solvable.
+
+Given the series $\sum_{n=0}^\infty a_n$, the improper integral would be of the below form:
+$$ \int_0^\infty a_x dx$$
+Effectively, substitute $n$ for $x$, and swap the summation for an integral. The lower bound of the integral would match the lower bound of the series, and the upper bound will be infinity.
+
+The integral test only applies if the series is positive, continuous, and decreasing along the entire interval.

education/physics/PHYS2210/Chapter 15.md

@@ -0,0 +1,23 @@
+Hydrostatic equilibrium and fluid motion.
+
+Fluids include gasses and liquids.
+
+- For a gas, the density changes readily, and the molecules are far apart.
+- For liquids, molecules are close together, and the density remains basically constant
+
+A fluid is defined as a matter that flows under the influence of external forces.
+
+Fluids cannot maintain a fixed structure, but fit the shape of the container they're confined to.
+
+# Pressure
+
+Pressure is defined as **force *per* unit area*** exerted by a fluid.
+
+$$ p = \frac{F (N)}{A(m^2)} $$
+The SI unit for pressure is called a pascal ($Pa$), equal to $\frac{N}{m^2}$.
+
+Pressure is exerted on the fluid's container. 
+
+Pressure is the same everywhere at the same depth.
+
+$$ 1atm = 101.3 kPa = 14.7 psi $$

education/physics/PHYS2210/Unit 1.md

@@ -0,0 +1,67 @@
+Motion in a straight line is one dimensional.
+
+Kinematics or the physics of motion has 4 noteworthy parameters: time ($t$), position ($x$), velocity ($v$), and acceleration ($a$). 
+
+Kinematic problems have a start and end of motion.
+
+# Displacement
+
+Displacement is calculated with the formula:
+$$\Delta x = \text{x-value of final position} - \text{x-value of initial posiion}$$
+
+# Velocity
+Average velocity over a time interval $\Delta t$ is defined to be: **the displacement** (net change in position), **divided by** **the time taken**.
+
+$$ \bar{v} = \dfrac{\text{final position-initial position}}{\text{final time - initial time}} = \dfrac{x_2 - x_1}{t_2 - t_1} = \dfrac{\Delta x}{\Delta t}$$
+**Speed (m/s)** is defined to be the total distance traveled divided by the time taken. Speed and velocity are *not the same*. 
+
+
+$$ v_{\text{instant}} = v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$
+- $x(t)$ -> **position** as a function of time
+- $v(t)$ -> **slope** of position-vs-time (derivative of $x(t)$)
+- $a(t)$ -> **slope** of velocity-vs-time (derivative of $v(t)$)
+# Acceleration
+To find the instantaneous acceleration, we can apply the formula:
+
+$$a_{\text{instant}} = a = \frac{dv}{dt} = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}$$
+## Equations of Motion for Constant Acceleration
+1. $v = v_0 + at$  - Use when missing $x$
+2. $x = x_0 + \frac{1}{2}(v_0 + v)t$  - Use when missing $a$
+3. $x = x_0 + v_0 t + \frac{1}{2} a t^2$  - Use when missing $v$
+4. $v^2 = v_0^2 + 2a(x - x_0)$  - Use when missing $t$ 
+
+(Actually useful equations)
+$$ t = \frac{v_f - v_i}{a} $$
+$$ a = \frac{v - v_0}{t}
+$$s = v_it + \frac{1}{2}at^2 $$
+| -------------- | ----- |
+| $t_0$          | $t$   |
+| $v_0$          | $v$   |
+| $x_0$          | $x$   |
+| $a$ (constant) | $a$   |
+## Examples
+
+ > Sally aggressively drives her Alfa Romeo from rest to 50 m/s in 6s. What is her acceleration?
+1. Start:
+	- $v_0 = 0 m/s$
+	- $t_0 = 0s$
+	- $x_0 = 0m$
+2. Final:
+	- $v = 50 m/s$
+	- $t = 6s$
+	- $x = ?$ 
+3. Using the equation $v = v_0 + at \to a = \frac{(v - v_0)}{t}$, $a$ = $(50 m/s - 0 m/s)/ 6s = 8.3m/s^2$ 
+4. Assess: do the units make sense? Is the answer reasonable?
+
+> An object slides until coming to a rest. It traveled 15 meters in 3 seconds. What was the object's acceleration?
+1. The problem is a kinematics problem, refer to the above formula.
+2. Acceleration is a constant
+3. Problem has two unknowns, $v_0$ and $a$, let's solve for $v_0$ first.
+4. $x = x_0 + \frac{1}{2}(v_0 + v) t \to v_0 = 2*(x - x_0)/ t \to 2*(15)/3 = 10$
+5. $v = v_0 + at \to a = (v - v_0) /t \to a = (-10 m/s) / 3s = -3.33m/s^2$ 
+6. Asses: Units? Answer make sense? Significant figures? Acceleration speeding up or slowing down?
+# Gravity
+- Is the problem 1 dimensional?
+- Could draw axis pointing upwards, call it y axis
+- If not specified, assume acceleration is $g = 9.8 m/s^2$.
+- Direction is important, $g$ is down towards the earth, so it's often negative. The sign depends on your choice of axis. 

education/physics/PHYS2220/Electric Charge.md

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+# Electric Charge
+- Charges come in two varieties, positive and negative.
+- Net charge is the *algebraic sum* of an object's charges
+- Protons and electrons have the same magnitude of charge (designated $1e$; a unit, **not** Euler's number)
+	- The SI Unit of charge is the *Coulomb* (abbreviated C)
+	- The smallest discrete quantity of charge is $\frac{1}{3}e$.
+- In an isolated system, the net charge will always remain constant.
+
+# Coulomb's Law
+- Two charges will exert a force on each other along the line joining them.
+	- The magnitude of this force is proportional to the *product of the charges* and inversely proportional to the to the $\sqrt{dist}$.
+	- The equation to determine the force between two charges is as follows:
+	$$ \vec{F}_{12} = \vec{r}k\frac{q_1q_2}{r^2} $$
+		- $\vec{r}$ is a unit vector pointing from charge 1 to charge 2
+		- $k$ is Coulomb's constant, or $8.99 * 10^9 \frac{Nm^2}{C^2}$ 
+		- $q_1$ and $q_2$ are the charges
+		- $r$ is the distance between those charges
+		- The resulting force will push away if $q_1q_2$ is *positive*, and attract if $q_1q_2$ is negative. This is where the rule "opposites attract, like repels" comes from
+- Coulomb's law only holds exactly true for *point charges* i.e a proton or electron.
+
+# The Superposition Principle
+The superposition principle states that:
+> The net force acting on a point charge is equal to the sum of all individual forces.
+
+This means that to find the net force acting on a single charge, you add up all of the individual forces acting on that charge.
+
+# The Electric Dipole
+An electric dipole consists of two point charges of equal magnitude but opposite sign. Many molecules behave like dipoles.
+- **Electric dipole moment** ($p$) is defined as the product of the charge $q$ and the separation $d$ between the two charges making up the dipole. $p = qd$
+- The dipole field at large distances decreases as the inverse *cube* of the distance. This is because the dipole has zero *net* charge.
+- In an electric field, a dipole experiences a torque that aligns it with the field.
+# Continuous Charge Distributions
+It's largely impossible to sum the electric field from every particle in a piece of matter, so the approximation is made that the charge is spread continuously over the distribution.
+- The number of dimensions involved changes the unit and terminology used:
+	-  If the charge distribution extends throughout a *3d volume*, we describe it in terms of the **volume charge density** $\rho$, with units of $\frac{C}{m^3}$. 
+	- For charge distributions spread over *surfaces*, we use **surface charge density** $\sigma$ ($\frac{C}{m^2}$).
+	- For charge distributions spread over *lines*, we use **line charge density** $\lambda$ ($\frac{C}{m}$).
+- To find the point charge, we can use this formula:
+$$ \vec{E} = \int d \vec{E} = \int \frac{kdq}{r^2}\hat{r}$$
+- 

education/physics/PHYS2220/Gauss's Law.md

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+# Flux
+Flux refers to a flow of matter or energy. Examples include water through a pipe, blood through veins, or air over an airplane's wing.
+
+While there's nothing directly *flowing* in an electric field, the term flux is used to describe the total strength of a field.
+
+In the simplest case with a uniform field of magnitude $E$ perpendicular to an area $A$, the flux is described as follows:
+$$ \Phi = EA$$
+- $E$ refers to the amplitude
+- $A$ refers to the area
+
+If the area is tilted relative to the field, then the strength of the field is reduced by a factor of $\cos \theta$, where $\theta$ is the angle between the electric field $\vec{E}$ and a vector $\vec{A}$ that's normal to the surface. This generalizes our flux equation to $\Phi = EA\cos\theta$ .
+
+---
+I don't understand what the below section means, but copying it for posterity:
+The electric flux through any closed surface is proportional to the net charge enclosed by that surface. This would be written mathematically as:
+$$ \Phi = \oint \vec{E} \cdot d\vec{A} \propto q_{enclosed} $$
+> interjection: $\propto$ means "is proportional to", and $\oint dA$ can *possibly* be treated as the area of the surface.
+
+$$ \Phi = \oint \vec{E} \cdot d\vec{A} = \oint EdA \cos\theta$$
+For a closed sphere, the equation becomes:
+$$ \Phi = E(4\pi r^2) $$
+
+Before proceeding, we introduce the so-called permittivity constant $\in_0$, defined as $\in_0 = 1/4 \pi k$, where $k$ is the Coulomb constant. 
+
+---

education/physics/PHYS2220/Magnetism.md

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+- Magnetic fields are represented with the symbol $\vec{B}$
+- Magnetism is based in *electric charge*, specifically the motion of that charge.
+- The magnetic force is always at right angles to both the velocity $\vec{v}$, and the magnetic field $\vec{B}$
+- The force is greatest when the charge is moving at right angles and is zero for motion parallel to the field. THe force is generally proportional to $\sin(\theta)$, where $\theta$ is the angle between the velocity $\vec{v}$, and the field $\vec{B}$.
+The formula that describes magnetic force compactly is:
+$$ \vec(F_B) = q\vec(v) \times \vec{B} $$
+	- $F_B$ is the magnetic force
+	- $q$ is the charge the force is acting on
+	- $v$ is the velocity of the charge
+	- $B$ is the magnetic field
+For the magnitude of a magnetic force:
+	$$ |\vec(F_B)| = |q|vB\sin(\theta) $$
+For the radius of a particle's circular path:
+$$ r = \frac{mv}{qB} $$For the period of a particle's circular orbit in a uniform magnetic field:
+$$ T = \frac{2\pi r}{v} = \frac{2\pi}{v}\frac{mv}{qB} = \frac{2\pi m}{qB} $$
+- The period is independent of the particle's speed and orbital radius.