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.

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0. Fundamental Concepts of Algebra3h 32m
1. Equations and Inequalities3h 27m
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15. Polar Equations2h 5m
16. Parametric Equations1h 6m
17. Graphing Complex Numbers1h 7m
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20. Sequences, Series & Induction1h 15m
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22. Limits & Continuity1h 49m
23. Intro to Derivatives & Area Under the Curve2h 9m

15. Polar Equations

Graphing Other Common Polar Equations

Precalculus

15. Polar Equations

Graphing Other Common Polar Equations

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Multiple Choice

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

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Verified step by step guidance

Identify the polar equation given: $ r = 1 + 2\sin\theta $. This is a polar equation that represents a limaçon.

Understand the general form of a limaçon: $ r = a + b\sin\theta $ or $ r = a + b\cos\theta $. In this case, $ a = 1 $ and $ b = 2 $.

Determine the type of limaçon based on the values of $ a $ and $ b $. Since $ b > a $, the graph will have an inner loop.

Plot key points by substituting values of $ \theta $ into the equation. 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 $.

Sketch the graph using these points and the understanding that the graph is symmetric about the vertical axis. The graph will have a loop inside, and the outer part will resemble a distorted circle.