Question

Sketch a graph of the polar equation. $$ r=5-4 \sin \theta $$

Short answer

This limaçon has an inner loop at \( \theta = \frac{\pi}{2} \) and an outer loop at \( \theta = \pi, 2\pi, \) and \( 0 \). It is symmetric with respect to the origin.

Step-by-step solution

Identify the circle or limaçon

We start by analyzing the format of the equation to determine whether it's describing a circle or a limaçon. In this particular case, \( r = 5 - 4 \sin \theta \), it does not represent a circle as the coefficient of the \( \sin \theta \) term does not equal to the constant. Hence, it represents a limaçon.

Determine specific points

Evaluate the polar function for values of \( \theta \) that result in \( r = 0 \), as well as typical values such as \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \) and \( 2\pi \). A few particular points could be: when \( \theta = 0, r = 5 \), when \( \theta = \frac{\pi}{2}, r = 1 \), when \( \theta = \pi, r = 5 \), when \( \theta = \frac{3\pi}{2}, r = 9 \), and when \( \theta = 2\pi, r = 5 \). These points help us define the outer and inner loop of the limaçon.

Sketch the graph

Using the determined points, we can start sketching the graph. We see that the graph has an inner loop at \( \theta = \frac{\pi}{2} \) and comes out to the outer loop at \( \theta = \pi \), \( \theta = 2\pi \), and \( \theta = 0 \). This limaçon is symmetric with respect to the origin.