Graph r = 1 + 2 sin θ r=1+2\sin\theta r = 1 + 2 sin θ

In this problem, we're asked to graph the equation r equals 1+2 times the sine of theta. Now this is the graph of a limacon because I see that I have addition happening here and I see that my a and b values are not equal to each other, meaning that this cannot be a cardioid and therefore has to be a limosome. Now let's go ahead and get started with our steps here. Now we first wanna take a deeper look at those a and b values. I see that my a value is 1, my b value is 2, showing me that a is less than b, meaning that this limacon will have an inner loop and also have a 0. So I can go ahead and plot that 0 right at the pole, and I can proceed to step 2 and determine the symmetry of my graph. Now because my equation contains a sine function, that tells me that my graph will have symmetry about the line theta equals pi over 2, which is something that we can keep in mind as we proceed to step 3. Now for step 3, we actually want to plug in some angles and get some points to plot using specifically our quadrantal angles. So I'm going to start by plugging in 0, 1 +2 times the sine of 0. Now the sine of 0 is just 0, so this ends up just being 1. So I'm gonna plot my first point at 10, which is right here on my graph. Now knowing that I have this symmetry, I can also go ahead and reflect this point over my line theta equals pi over 2, giving me another point at one of my other quadrantal angles theta equals pi. So I now have that other point at 1 pi. Now I want to plug in these other 2 quadrantal angles starting with pi over 2. So 1 +2 times the sine of pi over 2. The sine of pi over 2 is positive 1 so this ends up being 1 plus 2, which gives me 3. So I can plot another point at 3pi over 2, which ends up being right up here. Then for my final quadrantal angle, I'm going to plug in 3pi over 2, 1 +2 times the sine of 3pi over 2. The sine of 3pi over 2 is negative 1, so this will give me 1 minuteus 2, which ends up giving me negative 1. So when I come over to my point at negative 1, 3pi over 2, remember that having a negative r value means that we're going to count out and away from our angle rather than towards it like we would with a positive r value. So coming to my angle 3 pi over 2 and then counting back in the opposite direction 1, I end up with a point right there. Now I can proceed to my final step here and connect all of these points with a smooth and continuous curve. Recalling that in step one, we said that this is a lima zone with an inner loop, which is also quite obvious based on the points that we have on our graph. This point closest to the pole here represents that of my inner loop, and then my outer loop, the rest of my lima zone, goes all the way up to this point here. So I'm going to connect the rest of those with as smooth of a curve as I can. And this is the graph of my lima zone for the equation r equals 1 +2 times the sine of theta. Thanks for watching, and let me know if you have any 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 $r=1+2\sin\theta$

Verified step by step guidance

Understand the equation given: $ r = 1 + 2\sin\theta $. This is a polar equation where $ r $ is the radius and $ \theta $ is the angle.

Recognize that the equation $ r = 1 + 2\sin\theta $ represents a limaçon. Limaçons are a family of curves that can have an inner loop, a dimple, or be convex, depending on the coefficients.

Determine the type of limaçon by comparing the coefficients. Here, the coefficient of $ \sin\theta $ is 2, which is greater than the constant term 1. This indicates that the graph will have an inner loop.

Plot key points by substituting values of $ \theta $ into the equation to find corresponding $ r $ values. For example, at $ \theta = 0 $, $ r = 1 $; at $ \theta = \frac{\pi}{2} $, $ r = 3 $; at $ \theta = \pi $, $ r = 1 $; and at $ \theta = \frac{3\pi}{2} $, $ r = -1 $.

Use these points to sketch the graph, noting that the graph will loop inside the origin due to the negative $ r $ value at $ \theta = \frac{3\pi}{2} $. The graph should resemble a limaçon with an inner loop, similar to the fourth image provided.