Problem 15 - Algebra and Trigonometry

Question

Test for symmetry and then graph each polar equation. $$r=1-\sin \theta$$

Step-by-Step Solution

Answer

The polar equation \(r = 1 - \sin \theta\) is neither symmetrical about the x-axis, y-axis, nor the origin. The graph curves from (1,0) downward towards the origin and then back up to (1,\(\pi\)), completing a loop.

Step 1: Symmetry Testing

Test for symmetry about the x-axis, y-axis, and origin. In a polar coordinate system, symmetry about the x-axis can be tested by replacing \(\theta\) with \(-\theta\). If we get the same equation, then the function is symmetrical about the x-axis. Symmetry about the y-axis can be tested by replacing \(\theta\) with \(\pi - \theta\), and symmetry about the origin can be tested by replacing \(r\) with \(-r\).\n\nFor the expression \(r = 1 - \sin\theta\), let's proceed with these replacements:\nFor x-axis symmetry, replacing \(\theta\) with \(-\theta\), we get: \(r = 1 - \sin(-\theta) = 1 + \sin\theta\), which is different from the original expression, so the function is not symmetric about the x-axis.\nFor y-axis symmetry, replacing \(\theta\) with \(\pi - \theta\), we get: \(r = 1 - \sin(\pi - \theta) = 1 - \sin\pi + \sin\theta = 1 + \sin\theta\), which is different from the original expression, so the function is not symmetric about the y-axis.\nFor origin symmetry, replacing \(r\) with \(-r\), we get: \(-r = 1 - \sin\theta\), which is different from the original expression, so the function is not symmetric about the origin.

Step 2: Graphing the Polar Equation

After determining that the function lacks symmetry, we move on to draw the polar plot. Some critical points can be identified easily as \(\sin \theta\) ranges from -1 to 1. For \(\theta=0, r=1-0=1\), for \(\theta=\pi/2, r=1-1=0\) and for \(\theta=\pi, r=1-0=1\). This tells us the plot is starting from 1 at \(\theta=0\), going to 0 at \(\theta=\pi/2\), and back to 1 at \(\theta=\pi\). As a result, we get a figure that curves from (1,0) downward towards the origin and then back up to (1,\(\pi\)), completing a loop.