Hexagonal

Curves

\text{A} = (x, y)

\text{A} = (x, y)

Cartesian Coordinate System

Polar Coordinate System

\text{A} = (x, y)

\text{A} = (x, y)

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

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

Cartesian Coordinate System

\text{A} = (x, y)

\text{A} = (x, y)

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

\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}

\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)

\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)

\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)

\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)

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

Equations in cartesian and polar coordinate systems

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

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

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

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}

\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}}

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

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

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

Equations in cartesian and polar coordinate systems

0\leq \theta \leq 2\pi

0\leq \theta \leq 2\pi

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

r=\cos(3\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=\dfrac{\theta}{2\pi},\,0\leq \theta \leq 10\pi

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

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

\text{Butterfly curve}

\text{Butterfly curve}

\text{Hexagonal 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