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# Families of Polar Curves: Circles, Cardiods, and Limacon - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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