Hexagonal

Curves

\text{A} = (x, y)

Cartesian Coordinate System

Polar Coordinate System

\text{A} = (x, y)

\text{A} = (r; \theta)

Cartesian Coordinate System

\text{A} = (x, y)

\text{A} = (r; \theta)

\begin{array}{rcl} x & = & r\cos(\theta)\\ \\ y & = & r\sin(\theta)\\ \\ r^2 & = & x^2+y^2 \\ \\ \tan\theta^* &=& \dfrac{y}{x} \end{array}

* One must consider what quadrant the point is in when computing \(\theta\) from this identity.

Polar Coordinate System

Cartesian Coordinate System

\text{A} = (1,2)

\(x=1\,\) and \(\,y=2\).

\(r^2=(1)^2+(2)^2\)

\(r^2=5\)

\(r=\sqrt{5}.\)

Then

\(r^2=x^2+y^2\)

Since \(\tan\theta = y/x,\)

\(\tan\theta =\dfrac{2}{1}\)

\(\Rightarrow\theta =\arctan(2).\)

From cartesian to polar

\text{A} = (1,2)

From cartesian to polar

Therefore

\((1,2) = \left(\sqrt{5};\arctan(2)\right)\)

\(r=4\,\) and \(\,\theta=\dfrac{\pi}{4}\).

\(x = r\cos\left(\theta\right)\)

Then

\(x = 4\cos\left(\dfrac{\pi}{4}\right)\)

\(x = 2\sqrt{2},\)

\(y = r\sin\left(\theta\right)\)

and

\(y = 4\sin\left(\dfrac{\pi}{4}\right)\)

\(y = 2\sqrt{2}.\)

From polar to cartesian

\text{A} = \left(4;\dfrac{\pi}{4}\right)

From polar to cartesian

Therefore

\(\left(4;\dfrac{\pi}{4}\right) = \bigg(2\sqrt{2},2\sqrt{2}\bigg)\)

\text{A} = \left(4;\dfrac{\pi}{4}\right)

Equations in cartesian and polar coordinate systems

4x^2+9y^2=6^2

r = \dfrac{6}{\sqrt{5\sin^2(\theta)+4}}

\begin{array}{rcl} x & = & r\cos(\theta)\\ \\ y & = & r\sin(\theta)\\ \\ r^2 & = & x^2+y^2 \\ \\ \tan\theta^* &=& \dfrac{y}{x} \end{array}

r = \dfrac{6}{\sqrt{5\sin^2(\theta)+4}}

4x^2+9y^2=6^2

Equations in cartesian and polar coordinate systems

0\leq \theta \leq 2\pi

r=2,\,0\leq \theta \leq 2\pi

r=1+\cos(\theta),\,0\leq \theta \leq 2\pi

r=\cos(3\theta),\,0\leq \theta \leq 2\pi

r=\dfrac{\theta}{2\pi},\,0\leq \theta \leq 10\pi

r=\sqrt{\cos(2\theta)},\,0\leq \theta \leq 2\pi

r=1-\cos(\theta)\sin(3\theta),\,0\leq \theta \leq 2\pi

r=2\cos\left(\frac{4\theta}{7}\right),\,0\leq \theta \leq 15\pi

\text{Butterfly curve}

\text{Hexagonal curve}

n = 50
k = 2.5
m = 3

GeoGebra code

n = 50
k = 2.5
m = 3
t = Slider(0, 2*pi, 0.01)

GeoGebra code

n = 50
k = 2.5
m = 3
t = Slider(0, 2*pi, 0.01)
c = Curve((1/(cos(m*θ/2)^n + sin(m*θ/2)^n)^(1/(k*n)); θ), t, pi + t)

GeoGebra code

n = 50
k = 2.5
m = 3
t = Slider(0, 2*pi, 0.01)
c = Curve((1/(cos(m*θ/2)^n + sin(m*θ/2)^n)^(1/(k*n)); θ), t, pi + t)

GeoGebra code

n = 50
k = 2.5
m = 3
t = Slider(0, 2*pi, 0.01)
f(x) = cos(m*x/2)^n + sin(m*x/2)^n)^(1/(k*n))

GeoGebra code

n = 50
k = 2.5
m = 3
t = Slider(0, 2*pi, 0.01)
f(x) = cos(m*x/2)^n + sin(m*x/2)^n)^(1/(k*n))
c = Curve((1/f(θ); θ), t, pi + t)

GeoGebra code

n = 50
k = 2.5
m = 3
t = Slider(0, 2*pi, 0.01, 0.3, 200)
f(x) = (cos(m*x/2)^n+sin(m*x/2)^n)^(1/(k*n))
Sequence(Curve((j * 1/f(θ); θ), θ, j*t, pi + j*t), j, 1, 10)

Final GeoGebra code