Question

identify the graph of $r = 2 \sin(\theta)$.

Explanation:

Step1: Recall Polar Equation Form

The polar equation \( r = a\sin(\theta) \) (where \( a
eq0 \)) represents a circle. The general form for such a circle: convert to Cartesian coordinates. Multiply both sides by \( r \): \( r^{2}=2r\sin(\theta) \). Since \( r^{2}=x^{2}+y^{2} \) and \( y = r\sin(\theta) \), substitute: \( x^{2}+y^{2}=2y \).

Step2: Rewrite in Standard Circle Form

Rearrange \( x^{2}+y^{2}-2y = 0 \). Complete the square for \( y \): \( x^{2}+(y^{2}-2y + 1)=1 \), so \( x^{2}+(y - 1)^{2}=1 \). This is a circle with radius \( 1 \), centered at \( (0,1) \) (in Cartesian coordinates). Now, analyze the graphs: the first graph is a four - leaf rose (for \( r = a\sin(2\theta) \) or similar with multiple leaves), the third is a three - leaf rose (for \( r = a\sin(3\theta) \) or similar), the second and fourth are circles. The center in polar for \( r = 2\sin(\theta) \): when \( \theta=\frac{\pi}{2} \), \( r = 2 \); when \( \theta = 0 \) or \( \pi \), \( r = 0 \). The circle \( x^{2}+(y - 1)^{2}=1 \) has radius \( 1 \), so diameter \( 2 \). The second graph's scale: x - axis (or r - axis) has marks 1,2,3. The circle here should have radius 1, so the second graph (top right) is a circle with appropriate scale (radius 1, centered above the origin in polar, which translates to center \( (0,1) \) in Cartesian, so in polar, the center is at \( (1,\frac{\pi}{2}) \), and the circle passes through the origin. The fourth graph has a larger scale (marks 1,2,3,4) which would imply a larger radius, but our circle has radius 1. So the correct graph is the top right (second) graph.

Answer:

The top right graph (the circle in the top right panel)