Basics

distance
1-dimensional $$ d = \vert n_2 - n_1\vert $$
2-dimensional $$ d = \pm \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
slope
$$ M(\frac{x_1 + x_2}{2}, \frac{y_1+y_2}{2}) $$
$$\theta = \alpha_x - \alpha_i = m_2 - \arctan{m_1} [?]$$
Terminal = line that "makes angle"
(α) Inclination $$ [0^\circ, 180^\circ), 0^\circ \le inclination \lt 180^\circ $$
(m) Slope = Measure of slantness on line $$ m = \tan{\alpha} = \frac{y_2 - y_1}{x_2 - x_1} $$
Positive Slope: $$ 0^\circ < \alpha < 90^\circ $$
Negative Slope: $$ 90^\circ < \alpha < 180^\circ $$
Undetermined Slope: $$ \alpha = 90^\circ $$
$$ \alpha = 0 $$
Parallel Slopes (||): $$ m_1 = m_2 $$
Perpendicular Slopes (⊥): $$ m_2 = -\frac{1}{m_1} $$

Method of Least Squares

Example math
We are trying to find the correct fit for our observational data
In other words, we are attempting to find m, b for the equation: $$d = tm + b$$
By multiplying both sides by t $$td = t^2m + tb$$
Then averaging the results of these equations
independent dependent observational equation
$$ t $$ $$ d $$ $$ d = tm + b $$ $$ td = t^2m + tb $$
0 6.00 $$(6.00) = (0)m + b$$ $$-$$
5 6.60 $$(6.60) = (5)m + b$$ $$5(660) = (5)^2m + 5b$$
10 7.00 $$(7.00) = (10)m + b$$ $$10(7.00) = (10)^2m + (10)b $$
15 7.40 $$(7.40) = (15)m + b$$ $$15(7.40) = (15)^2m + (15)b $$
20 7.92 $$(7.92) = (20)m + b$$ $$20(7.92) = (20)^2m + (20)b $$
25 8.60 $$(8.60) = (25)m + b$$ $$25(8.60) = (25)^2m + (25)b $$
30 9.10 $$(9.10) = (30)m + b$$ $$30(9.10) = (30)^2m + (30)b $$
n = 7 Add equations togeter --> $$ (52.62 = 105m + 7b) $$ $$ (860.40 = 2275m + 105b) $$
Now find b from one equation: $$ 52.62 = 105m + 7b $$
$$ b = \frac{52.62 - 105m}{7} $$
And plug it into the other: $$ 860.40 = 2275m + 105b $$
$$ 860.40 = 2275m + 105(\frac{52.62 - 105m}{7}) $$
$$ 860.40 = 2275m + 789.30 - 1575m $$
$$ 700m = 71.1 $$
$$ m = 0.1015... $$
Now put that m back into the first one to find b: $$ b = \frac{52.62 - 105(0.1015...)}{7} $$
$$ b = 5.9935 $$
Putting it all together for the final answer: $$ d = 0.102t + 5.99 $$
Some equations for what we just did
$$ m = \frac{n\sum xy - \sum x \cdot \sum y}{n \sum (x^2) - (\sum x)^2} $$
$$ b = \frac{\sum x ^2 \cdot \sum y - \sum xy \cdot \sum x}{n\sum (x^2) - (\sum x)^2} $$
Pearson $$ r = \frac{n \sum xy - \sum x \sum y}{\sqrt{(n \sum x^2 - (\sum x)^2)(n \sum y^2 - (\sum y)^2)}} $$
Definitions
Interpolation = Determining value within range of values
Extrapolation = Determining value outside range of values
Outlier = Data point outside points of value
Correlation Coefficient (r) = Squared is the percent of accuracy $$ -1 \le r \le 1 $$
-r means -m, reliable
+r means +m, reliable
note: closer to 0 means unreliable
Log equation manipulations
$$ y = ax^t $$
$$ \log y = \log (ax^t) $$
$$ \log y = \log a + \log (x^t) $$
$$ \log y = \log a + t\log x $$
Series equation (don't know why this was part of my notes)
$$ y = a + bx + cx^2 + ... + mx^n $$
Amount of multiply by x: $$n + 1$$

Derivatives

Limit Examples
Example 1: $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x + 2) \cancel{(x - 2)}}{\cancel{x - 2}} = \lim_{x \to 2} x + 2 = 2 + 2 = 4$$
Example 2: $$\lim_{x \to -3} x + 2 = -3 + 2 = -1 $$
Derivative Long Method
$$m = \lim_{\triangle x \to 0} \frac{f(x_1 + \triangle x) - f(x_1)}{\triangle x} $$
Example: find the tangential slope to the following equation when x is 1: $$ f(x) = x^2 $$
$$ m_{tan} = \lim_{\triangle x \to 0} \frac{f(x_1 + \triangle x) - f(x_1)}{\triangle x} = \lim_{\triangle x \to 0} \frac{f(x_1 + \triangle x)^2 - f(x_1)^2}{\triangle x}$$
$$ m_{tan} = \lim_{\triangle x \to 0} \frac{\cancel{x_1^2} + 2x_1 \cancel{\triangle x} + (\triangle x)^{\cancel{2}} - \cancel{x_1^2}} {\cancel{\triangle x}} $$
$$ m_{tan} = \lim_{\triangle x \to 0} (2x_1) + \triangle x) = 2x_1 + 0 = 2x_1 $$
$$ m_{tan} = 2x_1 $$
$$ m_{x = 1} = 2(1) = 2 $$
Example: find the slope for: $$ g(x) = x^3 $$
$$ g'(x) = m_{tan} = \lim_{\triangle x \to 0} \frac{g(x_1 + \triangle x) - g(x_1)}{\triangle x}$$
$$ = \lim_{\triangle x \to 0} \frac{(x_1 + \triangle x)^3 - x_1^3}{\triangle x} $$
$$ = \lim_{\triangle x \to 0} \frac{\cancel{x_1^3} + 3x_1^2 \cancel{\triangle x} + 3x_1(\triangle x)^{\cancel{2}} + (\triangle x)^{\cancel{3} = 2} - \cancel{x_1^3}} {\triangle x} $$
$$ = \lim_{\triangle x \to 0} 3x_1^2 + 3x_1 \triangle x + (\triangle x)^2 = 3x_1^2 + 3x_1(0) + (0)^2 $$
$$ = 3x_1^2 $$
The better method
$$f(x) = cx^n$$
$$ f'(x) = n \cdot cx^{n - 1} $$
Example: $$ f(x) = 4x^2 + 2x + 1 $$
$$ f'(x) = 2 \cdot 4x^{2 - 1} + 1 \cdot 2x^{1 - 1} + \cancel{0 \cdot 1 x^{0-1}} $$
$$ f'(x) = 8x + 2 $$
Physics application
a(t) is acceleration at time t
v(t) is the instantaneous velocity at time t
d(t) is the total distance moved at time t
$$ a(t) = v'(t) = d''(t) $$
Double derivative visualization
$$ f''(x) < 0 $$
$$ f''(x) > 0 $$
Note
When finding when m = 0: set f'(x) = 0, solve for x, then put that value into f(x) to find y

Antiderivatives

Formulas
If we sum together the areas of the infinite skinny rectangles that make up a curve, we find the area under the curve
$$ \lim_{\triangle x \to 0} \sum^{\infty}_{i = 1} f(x_i) \cdot \triangle x = Area $$
$$ f(x) = cx^n $$
$$ F(x) = \frac{c}{n} x^{n + 1} + C $$
Integration Example
$$ f(x) = 3x^2 $$
$$ F(x) = \int f(x) dx = \int 3x^2 dx = \frac{3}{3} x^{2 + 1} $$
$$ F(x) = x^3 $$
Double check: $$ F'(x) = f(x) = 3x^2 $$
$$ H(x) = x^2 + 3 $$ $$ G(x) = x^2 $$ $$ F(x) = x^2 - 4 $$
$$ H'(x) = h(x) = 2x $$ $$ G'(x) = g(x) = 2x $$ $$ F'(x) = f(x) = 2x $$
$$ H(x) = \frac{2}{2}x^2 + C $$ $$ G(x) = \frac{2}{2}x^2 + C $$ $$ F(x) = \frac{2}{2}x^2 + C $$
Find the Area Examples
Find the area under the curve $$ y = \frac{1}{2} + 2 $$ between x = 2 and x = 5
$$ Area = \int^5_2 f(x) dx $$
$$ = \int^5_2 (\frac{1}{2}x + 2) dx $$
$$ = (\frac{x^2}{4} + 2x + C) |^5_2 $$
$$ = (\frac{(5)^2}{4} + 2(5) + \cancel{C}) - (\frac{(2)^2}{4} + 2(2) + \cancel{C}) $$
$$ 16\frac{1}{4} - 5 = $$
$$ 11\frac{1}{4} (units)^2 $$
Another Example: $$ f(x) = x^2, [1, 2] $$
$$ Area = \int^2_1 f(x)dx = \int^2_1 x^2 dx = \frac{x^3}{3}|^2_1 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{7}{3} = $$
$$ 2 \frac{1}{3} (units)^2 $$
More Integration Examples
$$ f(x) = x^2 - 4x + 2 $$
$$ F(x) = \int f(x)dx = \int x^2 - 4x + 2 dx $$
$$ F(x) = \frac{x^3}{3} - \frac{4x^2}{2} + \frac{2x}{1} + C $$
$$ F(x) = \frac{x^3}{3} - 2x^2 + 2x + C $$
Another Example: $$ g(x) = 6x + 2, G(2) = 9 $$
$$ G(x) = \int g(x) dx = \int 6x + 2 dx $$
$$ G(x) = 3x^2 + 2x + C $$
$$ G(2) = 3(2)^2 + 2(2) + C = 9 $$
$$ C = -7 $$
$$ G(x) = 3x^2 + 2x - 7 $$