Problem 70 - Essential Calculus

Question

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

Step-by-Step Solution

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 1: 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.

Step 2: 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.

Step 3: 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.