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 one plus two times the sine of theta. Now this is the graph of a limosome 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 to be a limacon. Now let's go ahead and get started with our steps here. Now we first want to take a deeper look at those a and b values. I see that my a value is one, my b value is two, showing me that a is less than b, meaning that this limacon will have an inner loop and also have a zero. So I can go ahead and plot that zero right at the pole and I can proceed to step two 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 two, which is something that we can keep in mind as we proceed to step three. Now for step three, we actually wanna plug in some angles and get some points to plot using specifically our quadrantal angles. So I'm gonna start by plugging in zero, one plus two times the sine of zero. Now the sine of zero is just zero, so this ends up just being one. So I'm gonna plot my first point at one zero, 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 two, giving me another point at one of my other quadrantal angles theta equals pi. So I now have that other point at one pi. Now I want to plug in these other two quadrantal angles starting with pi over two. So one plus two times the sine of pi over two. The sine of pi over two is positive one, so this ends up being one plus two which gives me three. So I can plot another point at three pi over two which ends up being right up here. Then for my final quadrantal angle, I'm going to plug in three pi over two, one plus two times the sine of three pi over two. The sine of three pi over two is negative one, so this will give me one minuteus two, which ends up giving me negative one. So when I come over to my point at negative one, three pi over two, 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 three pi over two and then counting back in the opposite direction one, 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 limacon 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 one plus two times the sine of theta. Thanks for watching, and let me know if you have any questions.

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Calculus

16. Parametric Equations & Polar Coordinates

Polar Coordinates

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Multiple Choice

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

Graph of r=1+2sinθ showing a loop in polar coordinates.

Graph of r=1+2sinθ with a loop in polar coordinates.

Graph of r=1+2sinθ depicting a larger loop in polar coordinates.

Graph of r=1+2sinθ illustrating a symmetrical loop in polar coordinates.

Verified step by step guidance

Step 1: Recognize that the given equation is in polar form, r = 1 + 2sinθ. This represents a polar curve where r is the radial distance and θ is the angle in radians.

Step 2: Analyze the behavior of the equation. The term 2sinθ oscillates between -2 and 2, so r will vary between 1 - 2 = -1 (negative radius) and 1 + 2 = 3. Negative values of r indicate that the curve loops back through the origin.

Step 3: Plot key points by substituting specific values of θ (e.g., θ = 0, π/2, π, 3π/2, 2π) into the equation to calculate corresponding values of r. This helps identify the shape and symmetry of the graph.

Step 4: Observe that the curve exhibits symmetry about the vertical axis (polar axis) because the sine function is symmetric about θ = π/2. This symmetry helps simplify the graphing process.

Step 5: Use the calculated points and symmetry to sketch the graph. The curve will form a loop near the origin due to the negative values of r and extend outward to a maximum radius of 3. The resulting graph is a limacon with an inner loop.