Polar Coordinates

Imaginary Numbers
Standard: (x + yi) or (a + bi)
Conversions from cartesian (x, y) to polar (r, θ)
$$ r^2 = x^2 + y^2 $$
$$ \tan{\theta} = \frac{y}{x} $$
$$ x = r\cos{\theta} $$
$$ y = r\sin{\theta} $$
Trig form / Polar form $$ r(\cos{\theta} + i\sin{\theta}) $$
Example: Convert $$2 + 2i$$ to polar
$$ r = \pm\sqrt{a^2 + b^2} $$
$$ r = \pm\sqrt{(2)^2 + (2)^2} $$
$$ r = \pm 2 \sqrt{2} $$
$$ \tan{\theta} = \frac{a}{b} $$
(A)S(T)C $$ \tan{\theta} = \frac{2}{2} = 1$$
$$ \theta = 45^\circ, 225^\circ $$
Polar form: $$ 2\sqrt{2}(\cos45^\circ + i\sin45^\circ) $$
$$ -2\sqrt2(\cos225^\circ + i\sin225^\circ) $$
Some more equations
$$ n(\cos\theta + i\sin\theta) \cdot m(\cos\phi + i\sin\phi) = nm(\cos{(\theta + \phi)} + i\sin{(\theta + \phi)}) $$
$$ \frac{n(\cos\theta + i\sin\theta)}{m(\cos\phi + i\sin\phi)} = \frac{n}{m}(\cos{(\theta - \phi)} + i\sin{(\theta - \phi)})$$
$$ [r(\cos\theta + i\sin\phi)]^n = r^n(\cos{(n \cdot \theta)} + i\sin{(n \cdot \phi)}) $$
The origin is called the "pole"
The radius out is called the "polar axis"
$$ r == \rho? $$
Branched Tangents
1: ρ = 0
2: solve for θ
Symmetry
"Horizontal" $$ \theta = 0^\circ, 180^\circ $$
Pole = $$ \theta = 90^\circ, 270^\circ $$
Q II Q I
$$ (\rho, 180^\circ - \theta), (-\rho, -\theta) $$ $$ (\rho, \theta) $$
$$ (\rho, 180^\circ + \theta), (-\rho, \theta) $$ $$ (\rho, -\theta), (-\rho, 180^\circ - \theta) $$
Q III Q IV
Basic shapes
$$ \rho = a \pm b\sin\theta $$
$$ \rho = a \pm b\cos\theta $$
if a or b = 0: circle
example: $$ \rho = \sin\theta $$
if a ≠ b: limacon
example: $$ \rho = 1 + 2\cos\theta $$
if a = b: cardioid
example: $$ -1 - \sin\theta $$
$$ \rho = r\sin{(n\theta)} $$
$$ \rho = r\cos{(n\theta)} $$
A rose, where r is constant
n - leaved when n is odd
Example: $$ \sin{(3\theta)} $$
2n - leaved when n is even
Example: $$ \cos{(2\theta)} $$
Something cool to solidify concept
I approached a problem where the graphs couldn't handle a negative ρ
The solution? everytime we find a negative ρ, set it positive and add 180° to the respective θ
This way, negative ρ is on the opposite side of the graph, as intended
This possible via the ambiguity of polar graphs

Parametric Coordinates

Parametric equation
Third variable
Example: $$ x = 2t $$
$$ y = t - t^2 $$
t x y
0 0 0
1 2 0
2 4 -2
3 6 -6
Something cool to solidify concept
I wanted my code to be cleaner, and to defer the logic of calculating all normal x, y graphs (including inverses) to parametric
Lets define the original x, y: $$ y = func(x) $$
What you can do is replace the independent variable for t
The parametric equivalent would be: $$ x = t $$
$$ y = func(t) $$
This works because now x and t are interchangeable
When x and y are switched, just pick y for t instead, meaning now I can have both types of graphs deferring to parametric calculations