Dated: 30-10-2024

Sketching

Graph of \(r = \sin(\theta)\)

Let us take some samples.
For that, we will increment \(\theta\) by \(\frac \pi 2 30^{\circ}\) and construct a table containing our samples

\(\theta\) (radians)	\(0\)	\(\frac{\pi}{6}\)	\(\frac{\pi}{3}\)	\(\frac{\pi}{2}\)	\(\frac{2\pi}{3}\)	\(\frac{5\pi}{6}\)
\(r = \sin \theta\)	\(0\)	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(1\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)

\(\theta\) (radians)	\(\pi\)	\(\frac{7\pi}{6}\)	\(\frac{4\pi}{3}\)	\(\frac{3\pi}{2}\)	\(\frac{5\pi}{3}\)	\(\frac{11\pi}{6}\)	\(2\pi\)
\(r = \sin \theta\)	\(0\)	\(-\frac{1}{2}\)	\(-\frac{\sqrt{3}}{2}\)	\(-1\)	\(-\frac{\sqrt{3}}{2}\)	\(-\frac{1}{2}\)	\(0\)

To convert the polar coordinates into the cartesian ones, we need to multiply both sides of equation with \(r\).

\[\because r = \sin(\theta)\]

\[r^2 = r \cdot \sin(\theta)\]

\[x^2 + y^2 = y\]

\[x^2 + y^2 - y = 0\]

By completing the square method, we get

\[x^2 + \left(y - \frac 1 2\right)^2 = \frac 1 4\]

Hence, it is a circle with \(radius = \frac 1 2\) centered at \(\left(0, \frac 1 2\right)\).

Following form of equations are of limacons.

\[r = a + b \sin (\theta)\]

\[r = a - b \sin (\theta)\]

\[r = a + b \cos (\theta)\]

\[r = a - b \cos (\theta)\]

Special case for limacons in which \(a = b\), creates a cardioid.

if \(b > a\) or \(\frac a b < 1\) then the limacons have an inner ring.

There is a greek word called lemnicos which means looped ribon.
If \(a > 0\) then following are the equations for leminscate.

\[r^2 = a^2 \cos 2\theta\]

\[r^2 = -a^2 \cos 2\theta\]

\[r^2 = a^2 \sin 2\theta\]

\[r^2 = -a^2 \sin 2\theta\]

The \(a\) determines the length of each petal.

\[r = a \sin(n \cdot \theta)\]

\[r = a \cos(n \cdot \theta)\]

Where \(n \in \mathbb{Z}\).

\[r = a \cdot \theta\]