Graph r = 2 − 2 cos θ r=2-2\cos\theta r = 2 − 2 cos θ

In this problem, we're asked to graph r equals 2 minus 2 cosine theta. Now I know that this is a graph of a cardioid because I'm dealing with subtraction subtraction and both my a and b values are equal to each other. Now we know the general shape of a cardioid, this sort of heart shape, but it remains to find the orientation and the exact positioning of it. So let's go ahead and dive into our steps here. 1st, starting with our symmetry. Now because this is a cosine function inside of my equation, I know that my graph will be symmetric about the polar axis, which I can keep in mind as we proceed into step 2 and plot our points. Now we want to plot points specifically at our quadrantal angles, so I'm going to start with 0 and plug that into my equation for theta. 2 minus 2 times the cosine of 0. Now I know that the cosine of 0 is just 1, so this is 2 minus 2, which is just 0. Now plotting this on my graph at 0 that's just my pull. So I can proceed to plugging in pi over 2, my next quadrantal angle. 2 minus 2 times the cosine of pi over 2. The cosine of pi over 2 is 0, so this is just 2. So I can go ahead and plot that at pi over 2, 2, which will be right here. Now remember that we have some symmetry going on. So because we're symmetric about this polar axis, I can go ahead and just reflect that point over onto one of my other quadrantal angles. And I know that there exists a point at 3pi over 2, 2 having used that symmetry. Now one final point here plugging in pi, 2 minus 2 times the cosine of pi. The cosine of pi is negative 1, so this ends up being 2+2 or 4. So I plot that point at pi, at 4, pi. So going out here, I end up with my point all the way at the end of this axis, and I have all of those points that I need. Now I can connect this with a smooth and continuous curve. Now remember for a cardioid, the furthest point from our pole will always be the sort of rounded bottom of that cardioid, which in this case is this point over here. Then I have that sort of dimple in the middle that kind of makes our heart shape right here at a pole. So I'm going to go ahead and connect everything using a sort of cardioid heart shape rounding around. And this is the basic shape of my cardioid. We have graphed r equals 2 minus 2 times the cosine of theta. Let me know if you have any questions here. Thanks so much for watching.

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15. Polar Equations

Graphing Other Common Polar Equations

Precalculus

15. Polar Equations

Graphing Other Common Polar Equations

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Multiple Choice

Graph $r=2-2\cos\theta$

Graph of r=2-2cosθ showing a heart-shaped polar curve.

Graph of r=2-2cosθ depicting a heart-shaped polar curve.

Graph of r=2-2cosθ illustrating a heart-shaped polar curve.

Graph of r=2-2cosθ representing a heart-shaped polar curve.

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Verified step by step guidance

Identify the polar equation given: r = 2 - 2cos(θ). This is a limaçon with an inner loop.

Recall that the general form of a limaçon is r = a ± bcos(θ) or r = a ± bsin(θ). Here, a = 2 and b = 2, which means the limaçon will have an inner loop since a = b.

Determine the key points: When θ = 0, r = 2 - 2cos(0) = 0, indicating the inner loop touches the pole. When θ = π, r = 2 - 2cos(π) = 4, indicating the maximum distance from the pole.

Analyze the symmetry: The equation is symmetric about the horizontal axis (polar axis) because it involves cos(θ).

Compare the given graphs with the characteristics of the limaçon with an inner loop. The correct graph should have an inner loop touching the pole and a maximum radius of 4 at θ = π.