Problem 16 - 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 symmetrical with respect to the line \( \theta = \pi/2 \) and not symmetrical about the polar axis or the pole. Thus, the graph of this equation forms a cardioid shape.

Step 1: Symmetry with respect to the polar axis

Test for symmetry with respect to the polar axis, by replacing \( \theta \) with \( -\theta \) . If the equation remains the same, it's symmetrical along polar axis. Our equation becomes \( r = 1+ \sin (-\theta) = 1 - \sin \theta \). This isn't the same as our original equation, so it's not symmetrical with respect to the polar axis.

Step 2: Symmetry with respect to the pole

Test for symmetry with respect to the pole, by replacing \( r \) with \( -r \) . If the equation remains valid, it's symmetrical. For our equation, \( -r = 1+ \sin \theta \) doesn't make sense as it implies \( r < 0 \) in the original equation, so it's not symmetrical with respect to the pole.

Step 3: Symmetry with respect to the line theta = pi/2

Test for symmetry with respect to the line \( \theta = \pi/2 \) by replacing \( \theta \) with \( \pi - \theta \) . If the equation remains the same, it's symmetrical along that line. For our equation, this becomes \( r = 1+ \sin (\pi - \theta) = 1 + \sin \theta \), which equals to the original equation. Hence, it's symmetrical with respect to the line \( \theta = \pi/2 \).

Step 4: Graph the equation

After determining the symmetry, you can plot the equation on a polar coordinate system. Notice that the radius \( r \) is a function of \( \theta \). For each value of \( \theta \), find the corresponding value of \( r = 1 + \sin \theta \) and plot this on the polar grid. The graph will reveal a cardioid shape as we move from \( \theta = 0 \) to \( \theta = 2\pi \).