Problem 17 - Algebra and Trigonometry

Question

Test for symmetry and then graph each polar equation. $$r=2+2 \cos \theta$$

Step-by-Step Solution

Answer

The polar equation \( r = 2 + 2 \cos \theta \) is symmetric about the polar axis. It is not symmetric about the pole or the line \( \theta = \frac{\pi}{2} \). The graph of the equation will form a cardioid shape which is centered at 2 units from the origin on the positive x-axis.

Step 1: Test for Symmetry

First, test for Symmetry. There's three types of symmetry to test: Symmetry about the polar axis, symmetry about the pole and symmetry about the line \( \theta = \frac{\pi}{2} \). For the symmetry with respect to the polar axis, replace \( \theta \) with \( -\theta \) in the equation. If the resulting equation matches the original equation, our curve is symmetrical about the polar axis. Here, \(r=2+2 \cos(-\theta)\) simplifies to \(r=2+2 \cos \theta\), which is the original equation - therefore, the equation is symmetric about the polar axis.

Step 2: Check Symmetry about the Pole

To test for symmetry about the pole (origin), replace \( r \) with \( -r \) in the original equation. If the resulting equation matches the original, then the curve has symmetry with respect to the pole. Here, \(-r = 2 + 2 \cos \theta \) doesn't correspond to our original equation therefore, it's not symmetric about the pole.

Step 3: Check Symmetry about the Line \( \theta = \frac{\pi}{2} \)

To test for symmetry about the line \( \theta = \frac{\pi}{2} \), we replace \( \theta \) by \( \pi - \theta \) in the original equation. If the resulting equation matches the original equation, then the graph will be symmetric about the line \( \theta = \frac{\pi}{2} \). So, substituting \( \theta \) with \( \pi - \theta \), we get \( r = 2 + 2 \cos(\pi - \theta) \), which simplifies to \( r = 2 - 2 \cos(\theta) \). Since this doesn't match the original equation, the graph is not symmetric about the line \( \theta = \frac{\pi}{2}\)

Step 4: Graph the Polar Equation

Finally, graphing a polar equation involves plotting the points for various values of \( \theta \) and linking them in order to form a curve. For this polar equation, the graph starts from the pole when \( \theta = 0 \) ends angle \( \theta = 2 \pi \). As \( r \) is always positive, it will form a cardiod shape centered at 2 units from the origin on the positive x-axis