Question

To sketch the curves for the polar equation \(r = 2 + \sin \theta \) and its Cartesian coordinates.

Short answer

It is observed that the curve for the given polar equation is inner looped limacon.

Step-by-step solution

Given data

The given polar equation is \(r = 2 + \sin \theta \).

Concept of Polar curve

Polar curves are defined by points that are a variable distance from the origin (the pole) which depends on the angle measured off the positive\(x\)-axis.

Find the values from the table

Observe the values from given figure as shown below.

Relationship between angle and the polar curve

When \(\theta \) increases from 0 to \(\frac{\pi }{6}\), the polar curve \(r\) increases from \(2\) to \(2.5\), so the polar curve starts from \(2\) end with \(2.5\).

When \(\theta \) increases from \(\frac{\pi }{6}\) to \(\frac{\pi }{3}\), the polar curve \(r\) increases from \(2.5\) to \(2\), so the polar curve starts from \(2.5\) end with \(2\).

When \(\theta \) increases from \(\frac{\pi }{3}\) to \(\frac{\pi }{2}\), the polar curve \(r\) increases from \(2\) to \(0\), so the polar curve starts from \(2\) end with \(0\).

When \(\theta \) increases from \(\frac{\pi }{2}\) to \(\frac{{2\pi }}{3}\), the polar curve \(r\) decreases from \(0\) to \( - 2\), so the polar curve starts from \(0\) end with \( - 2\).

When \(\theta \) increases from \(\frac{{2\pi }}{3}\) to \(\frac{{5\pi }}{6}\), the polar curve \(r\) decreases from \( - 2\) to \( - 2.5\), so the polar curve starts from \( - 2\) end with \( - 2.5\).

When \(\theta \) increases from \(\frac{{5\pi }}{6}\) to \(\pi \), the polar curve \(r\) increases from \( - 2.5\) to \( - 2\), so the polar curve starts from \( - 2.5\) end with \( - 2\).

When \(\theta \) increases from \(\pi \) to \(\frac{{3\pi }}{2}\), the polar curve \(r\) increases from \( - 2\) to \(0\), so the polar curve starts from \( - 2\) end with \(0\).

When \(\theta \) increases from \(\frac{{3\pi }}{2}\) to \(2\pi \), the polar curve \(r\) increases from \(0\) to \(2\), so the polar curve starts from \(0\) end with \(2\).

Thus, the graph is inner looped limacon.

Sketch the polar curve for the Cartesian coordinates

Sketch the polar curve for Cartesian coordinates by above procedure.

From figure, it is observed that the curve for the given polar equation is inner looped limacon.