Problem 14 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The polar equation \(r=2 \sin \theta\) is symmetric about the y-axis but not symmetric about the x-axis or the origin. Its graph represents a semicircle with a 2 units radius lying in the top half of the xy-plane.

Step 1: Testing for Symmetry

1. Symmetry in respect to the x-axis: If an equation in polar coordinates remains unchanged when \(\theta\) is replaced by \(-\theta\), then the graph of the equation is symmetrical in respect to the x-axis. However, if we replace \(\theta\) with \(-\theta\) in the equation \(r = 2\sin(\theta)\), we would get \(r = 2\sin(-\theta) = -2\sin(\theta)\), which is not equal to the original equation. Hence, there is no symmetry about the x-axis. 2. Symmetry in respect to the y-axis: If an equation in polar coordinates remains unchanged when \(\theta\) is replaced by \(π-\theta\), then the graph of the equation is symmetrical in respect to the y-axis. If we replace \(\theta\) with \(π-\theta\) in the equation \(r = 2\sin(\theta)\), we would get \(r = 2\sin(π-\theta) = 2\sin(\theta)\), which is equal to the original equation. Hence, there is symmetry about the y-axis. 3. Symmetry in respect to the origin: If an equation in polar coordinates remains unchanged when \(r\) is replaced by \(-r\), then the graph of the equation is symmetrical with respect to the origin. If the equation remains the same when \(\theta\) is replaced by \(\theta + π\), it's also symmetric about the origin. However, if we replace \(\theta\) with \(\theta + π\) in the equation \(r = 2\sin(\theta)\), we would get \(r = 2\sin(\theta + π) = -2sin(\theta)\), which is not equal to the original equation. Therefore, there is no symmetry about the origin.

Step 2: Graphing the Polar Equation

The graph of the polar equation \(r=2 \sin \theta\) is a circle with radius 2 units centered at the origin, that lies solely in the top half of the xy-plane due to the range of the \(\sin\) function (between -1 and 1). The range of \(r\) also lies between -1 and 1 (since \(r = 2sin(\theta)\), \(r\) ranges between -2 and 2). But since the radius cannot be negative, the graph only includes values from 0 to 2, creating a semicircle in polar coordinates.