Families of Polar Curves: Circles, Cardiods, and Limacon - Problem 2

Let’s graph another member of this family of curves; r equals a plus b cosine theta. This time with a equals 5 and b equals 3. That will give us the equation r equals 5 plus 3 equals sine theta.

So the first thing I want to do is find symmetry if there is any. And I can do that by plugging in negative theta, and so I get 5 plus 3 cosine of negative theta. Cosine is an even function, so cosine of negative theta, equals cosine theta. So this is 5 plus 3 cosine theta and that is exactly r. And that means that r negative theta is in the graph of my function, my curve, and this point is the mirror image of our theta across the x axis. And that means this curve is going to have symmetry across the x axis. And I want to remember that because it will save me the trouble of plotting some points.

Let’s get started plotting points now. Theta, cosine theta and I'll have r equals 5 plus 3 cosine theta on here. So these will be the plots I’m plotting, this is a sort of intermediate calculation.

Let me start with 0. Cosine of 0 is 1, so I get 5 plus 3 times 1, 8. And then pi over 6, cosine of pi over 6, root 3 over 2. Now when I’m graphing I use an approximation for root 3 over 2 of about 7/8. So I’ll write 5 plus 3, this is approximately 5 plus 3 times 7/8. And 3 times 7/8 is 21 over 8, so this is 40 over 8 plus 21 over 8, 61 over 8. Now that’s 3/8 short of 8, so 7 and 5/8.

When I plot that, I’ll plot it as roughly 7 and 1/2. And then pi over 3, cosine of pi over 3 is 1/2. So 5 plus 3 times 1/2, 1.5, gives me 6.5. Pi over 2, cosine of pi over 2 is 0, so I get 5 plus 3 times 0, 5. Let me plot these points.

So I have 8,0 right here. At pi over 6, I got that strange number 7 and 5/8. Pi over 6 is this direction, this is 7, this is 8. 7 and 1/2 will be here, 7 and 5/8 just a little further. Pi over 3 I get 6 and 1/2. Pi over 3 is this direction 6 and 1/2, there is 6, 7, 6.5 will be right here. At pi over 2, I get 5; 1, 2, 3, 4, 5. And then let’s remember that this graph is symmetric about the axis, so I can reflect these points down across the axis. So 5 down 1, 2, 3, 4, 5, this point is 6 and 1/2 out, so I go 1, 2, 3, 4, 5, 6 and 1/2. And then this point was 7 and 5/8, so 1, 2, 3, 4, 5, 6, 7, 7 and 1/2 would be here, 7 and 5/8 about here.

And now let me continue the points in around. I’ll plot, let’s see 2 pi over 3, 5 pi over 6 and pi. Cosine of 2 pi over 3 is -1/2. So I get 5 plus 3 times -1/2,, 5 minus 1.5 is 3.5. 5 pi over 6, cosine of that is negative root 3 over 2, again I'll use the approximation root 3 over 2 is about 7/8. So this is approximately 5 plus 3 times 7/8, sorry, minus. That’s 40 over 8, minus 21 over 8 which is 19 over 8. 19 over 8, 16 over 8 will be 2. So this is 2 and 3/8. I’ll come back to that, and then at pi the cosine is -1. So I get 5 plus 3 times -1, 5 minus 3 is 2.

Let me plot these points; 3.5, 2 pi over 3, this is the 2 pi over 3 direction. 3.5 is 1, 2, 3 and 1/2. 5 pi over 6, I get this funny point 2 and 3/8. So this is 5 pi over 6, 1, 2 that would be 2 and 1/2. 2 and 3/8 is a little short and then finally 2 at pi. So this is the pi direction I go to 2. And let me reflect all these points.

So the reflection of this point would go down here, this point would go, that was 3 and 1/2, 2, 3 and 1/2. And there we go. And let me fill this in to a nice curve.

If you’ve already seen the graph of the cardiod, you'll notice that this looks kind of similar. Only it doesn’t quite make it back to the original pole. It kind of puckers in a little bit and then just goes back out again. It looks kind of like an apple or something on its side.

This graph is called a limacon. L-i-m-a-c-o-n. And this is another class of curves. You get this kind of a shape when a and b are different. And specifically this kind of sort of slightly dimpled in shape, when a is bigger than b.