Problem 39 - Algebra and Trigonometry

Question

Test for symmetry and then graph each polar equation. $$r=\frac{1}{1-\cos \theta}$$

Step-by-Step Solution

Answer

The polar equation \( r = \frac{1}{1 - \cos \theta} \) is symmetric with respect to the line \( \theta = \pi/2 \) but not with respect to the pole. The graph of polar equation is a rose with an infinite number of petals.

Step 1: Test for Symmetry

To test for symmetry, one can examine if the equation is symmetrical with respect to the line \( \theta = \pi / 2 \) (the vertical axis), and the pole (the origin, r=0). \n 1. Symmetry with respect to the vertical line: Replace \( \theta \) with \( -\theta \), if the equation remains same then it is symmetrical with respect to the vertical line. \n 2. Symmetry with respect to the pole: Replace \( r \) with \( -r \), if the equation remains same, then it is symmetrical with respect to the pole. \n After replacing \( \theta \) with \( -\theta \) in the given equation \( r = \frac{1}{1 - \cos \theta} \), we get \( r = \frac{1}{1 - \cos(-\theta)} = \frac{1}{1 - \cos \theta} \). Therefore, the equation is symmetrical with respect to the vertical line. But, when we replace \( r \) with \( -r \), equation becomes \( -r = \frac{1}{1 - \cos \theta} \). This isn't the same as the original polar equation, so it is not symmetrical about the pole.

Step 2: Graph the Polar Equation

Because the equation is symmetric with respect to the vertical line, only the graph for \( 0 \leq \theta \leq \pi \) needs to be plotted. For other values of \( \theta \), the graph will reflect about the vertical line. In the domain \( 0 \leq \theta \leq \pi \), calculate the value of r for various values of \( \theta \). For better results, select values for \( \theta \) that makes \( \cos \theta \) equal to 0,1, and -1. Plot the corresponding points (r, \( \theta \)) in the polar coordinate system and then draw a smooth curve through these points. Finally, draw a mirror image of the plotted graph about the vertical line to complete the entire graph because the function is symmetric about this line.