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Chapter 6: Problem 40

Test for symmetry and then graph each polar equation. $$r=\frac{2}{1-\cos \theta}$$

Short Answer

Expert verified
The graph of the polar equation \(r=\frac{2}{1-\cos \theta}\) is symmetric about the y-axis and the origin but not symmetric about the x-axis.

Step by step solution

01

Symmetry about the x-axis

Test for symmetry about the x-axis by replacing 胃 with 蟺-胃 in the equation. If the resulting equation is the same as the original one, the graph is symmetrical about the x-axis, else it is not. In this case, replacing 胃 with 蟺-胃, we obtain \(r = \frac{2}{1-\cos(蟺 - 胃)}\), which simplifies to \(r=\frac{2}{1+cos \theta }\). As this is not the original equation, the graph is not symmetrical about the x-axis.
02

Symmetry about the y-axis

Test for symmetry about the y-axis by replacing 胃 with -胃 in the equation. If the resulting equation is the same as the original one, the graph is symmetrical about the y-axis. In this case, replacing 胃 with -胃, we obtain \(r = \frac{2}{1-\cos(-胃)}\), which simplifies to \(r=\frac{2}{1-\cos \theta }\). As this is the same as the original equation, the graph is symmetrical about the y-axis.
03

Symmetry about the origin

Test for symmetry about the origin by replacing 胃 with 胃 + 蟺. If r(胃+蟺) = -r(胃) holds, the graph is symmetrical about the origin. In this case, replacing 胃 with 胃 + 蟺 results in \(r = \frac{2}{1-\cos(胃 + 蟺)}\), which simplifies to \(r=-\frac{2}{1-\cos \theta }\). Since -r(胃) equals the original polar equation, the graph is symmetrical about the origin.
04

Graphing the equation

To graph the equation, choose different values for 胃 and calculate the corresponding r. For example, 胃 = 0 gives r = 2 and 胃 = 蟺/2 gives r = 1. Continue this for a set of 胃 values and plot the (r, 胃) points on a polar graph.

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