Given the polar equation $r^2=\sin \left(2\theta \right)$, which of the following best describes the shape of its graph on the unit circle?

• A. A circle centered at the origin
• B. A straight line through the origin
• C. A lemniscate (figure-eight shape)
• D. A four-petaled rose curve

Answer: (D)

Verified step by step guidance

1. Start by examining the given polar equation: \(r^{2} = \sin(2\theta)\). Notice that \(r^{2}\) is expressed in terms of a trigonometric function of \$2\theta$, which suggests symmetry and periodicity in the graph.
2. Recall that equations of the form \(r^{2} = \sin(n\theta)\) or \(r^{2} = \cos(n\theta)\) often represent rose curves, where the number of petals depends on the value of \(n\). Specifically, if \(n\) is even, the rose has \$2n\( petals; if \)n\( is odd, it has \)n$ petals.
3. In this problem, \(n = 2\), so the rose curve will have \$2 \times 2 = 4$ petals. This matches the description of a four-petaled rose curve.
4. Understand that the shape is not a circle or a straight line because those would have simpler forms such as \(r = \text{constant}\) for a circle or \(\theta = \text{constant}\) for a line. The presence of \(\sin(2\theta)\) and the squared radius indicates a more complex, petal-like shape.
5. Therefore, by analyzing the form of the equation and the value of \(n\), you conclude that the graph represents a four-petaled rose curve.