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Families of Polar Curves: Circles, Cardiods, and Limacon - Problem 1
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Let’s graph another member of the family of curves r equals a plus b cosine theta, when a equals b equals 4. So that means I will have r equals 4 plus 4 cosine theta. So I’m specifically interested in when a and b are the same value.

First thing I want to observe is, does this curve have any kind of symmetry? So I’m going to plug negative theta in to the formula and I get 4 plus 4 cosine of negative theta. Now cosine is an even function, so the cosine of negative theta is the same as cosine theta. So this is 4 plus 4 cosine theta, and that’s exactly r.

And that means that the point r negative theta is on the graph. Now r negative theta is a reflection of r theta across the x axis. So this is going to give me the symmetry across the x axis. And I want to remember that because it might save me saving me plotting a few points.

So let’s actually do that, let’s plot some points. Let me start with, I’ll put theta up here. I have a table here that has three rows. So I’ll put my cosine theta here, and my r equals 4 plus 4 cosine theta in here.

Now first of all, theta equals 0. Start easy, cosine of 0 is 1, so I get 4 plus 4 times 1 is 8. So I’ll get 8,0 as one point. Then let me try pi over 6, cosine of pi over 6 is root 3 over 2, and 4 plus 4 times root 3 over 2. Here I use the approximation 7/8 for root 3 over 2. So this is approximately 4 plus 4 times 7/8 is 7/2 which is 3.5, so 7.5. Pi over 3, cosine of pi over 3 is 1/2. So I get 4 plus 4 times 1/2, 4 plus 2, 6, 1/2 and 6. And then pi over 2, cosine of pi over 2 is 0, so I get 4 plus 4 times 0, and that’s 4.

Let me plot the points I have so far. I have 8,0 and that point goes here, I’ve got 7.5 pi over 6. Now pi over 6 is this direction, 30 degrees above the horizontal axis. I have 7.5, 7 is here, 7.5 is here so that’s not the point. And then for pi over 3, I’ve got 6, 6 is here and for pi over 2, I have 4, 4 is here. So let me keep going, keeping in mind. Actually you know I do have the symmetry, I can make use of that now.

This graph should be symmetric about the x axis, so I could just plot my mirror images now. This point has a mirror image down here, this point has a mirror image. Let’s see that’s right and this point has a mirror image right here. Now let me plot some points past pi over 2 and see what happens. So let’s try 2 pi over 3. Cosine of 2 pi over 3 is -1/2, so I have 4 plus 4 times -1/2, 4 plus -2 is 2. 5 pi over 6, negative root 3 over 2. So I get 4 plus 4 times negative root 3 over 2 and this is approximately 4 plus, negative root 3 over 2 is approximately -7/8. So 4 times that is approximately -7/2. So this is actually 4 minus 3.5, that’s about .5.

And then for pi, the cosine is -1. And so I get 4 plus 4 times -1, 4 minus 4 is 0. Let me plot these points; 2, 2pi over 3. This is the 2 pi over 3 direction, this is 2. Then I’ve got 5 pi over 6, 1/2. So this is the pi over 6 direction, 1/2 is right here, and then in the pi direction I get 0. And then I’m going to plot my reflections.

So this is reflected down here, this point is reflected down here, and then I fill in. Started here, trying to get a smooth curve and it comes in kind of like so. This curve has kind of a hard shape. And thus it’s called a cardioid. C-a-r-d-i-o-i-d. So anytime that a and b have the same value, in this family of curves, we get a cardioid like this. It comes in to a point. If you have a cardioid with a sine function instead of a cosine it would be symmetric above the y axis, that’s the only difference.

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