Graph r 2 = 9 sin 2 θ r^2=9\sin2\theta r 2 = 9 sin 2 θ

In this problem, we're asked to graph the equation r squared equals 9 times the sine of 2 theta. Now this is a graph of lemniscate because I know that when I see an r squared value, that is the equation of a lemniscate. Now starting with our first step here, we wanna find out where that first petal is. Now Now we want to get our value of a, which remember this value at the beginning of your equation is a squared. So we need to take that a squared, in this case 9, and take the square root of it to get 3, which is our a value here. So I know that that first petal is going to land at an r value of 3, but I need to determine where theta is. Now remember that whenever we're dealing with a positive sine function, so I have positive a squared sine of 2 theta, that tells me that theta is going to be a value of pi over 4. So I can go ahead and plot this first petal at 3 pi over 4, which will land me right here in this first quadrant. Now from here, I just want to take this petal and reflect it over the pole. So reflecting this point over the pole, I end up right out here in quadrant 3. Now I can just connect these with a smooth continuous curve knowing that this Lemnis Gate looks similar to an infinity symbol. I'm taking these petals or propellers out and graphing them like this. Now we've graphed r squared equals 9 times the sine of 2 theta a lemniscate. Thanks for watching, and let me know if you have questions.

9. Polar Equations

Graphing Other Common Polar Equations

Trigonometry

9. Polar Equations

Graphing Other Common Polar Equations

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Multiple Choice

Graph $r^2=9\sin2\theta$

Verified step by step guidance

Recognize that the given equation is in polar form: $ r^2 = 9 \sin 2\theta $. This is a type of polar equation that can represent a lemniscate.

Recall that a lemniscate is a figure-eight shaped curve. The general form of a lemniscate in polar coordinates is $ r^2 = a^2 \sin 2\theta $ or $ r^2 = a^2 \cos 2\theta $.

In the given equation, $ a^2 = 9 $, so $ a = 3 $. This means the lemniscate will have loops that extend to a maximum radius of 3 units from the origin.

The equation $ r^2 = 9 \sin 2\theta $ suggests that the lemniscate is oriented along the lines $ \theta = \frac{\pi}{4} $ and $ \theta = \frac{3\pi}{4} $, as these are the angles where $ \sin 2\theta $ reaches its maximum and minimum values.

Compare the given images to identify which one matches the orientation and size of the lemniscate described by the equation. The correct graph will have its loops aligned along the lines $ \theta = \frac{\pi}{4} $ and $ \theta = \frac{3\pi}{4} $, with each loop extending to a radius of 3.