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# Graphing Polar Equations - Problem 2

###### 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 polar curve. I want to graph r equals 2 cosine theta plus 2. Now first let’s check it for symmetry. The way we do that is we plug in negative theta. So 2 cosine of –theta plus 2. Now cosine’s an even function so the cosine of minus theta is the same as the cosine of theta. And that means 2 cosine theta, plus 2, but that’s exactly r.

Now what this means is that the point r, negative theta is in the graph of this curve. Now r negative theta is the reflection of r theta across the x axis. So this graph is going to be symmetric about the x axis. Let me just make a note of that. Symmetric about the x axis. Sometimes knowing the symmetry ahead of time can save us graphing some points.

The other thing I need to be aware of, is that this is the rectangular graph, if 2 cosine theta plus 2l. I need to be aware that the r value is going to go as high as 4 and as low as zero. So this graph will come in handy when I want to figure out qualitative behavior. Let me plot some points though. Starting with zero. When theta equals zero, cosine theta is 1 so I get 2 times 1 plus 2, 4 and then pi over 6. Cosine of pi over 6 is root 3 over 2, so it’s 2 times root 3 over 2 plus 2.

Now when I’m graphing I use the approximation 7/8, for root 3 over 2. So this is approximately 2 times 7/8 plus 2 so that’s 7/4 plus 2, 15/4. And that’s ¼ short of 4, so 3 ¾. Cosine of pi over 3 is ½ so this is 2 times ½ plus 2, 3. Pi over 2, cosine of pi over 2, is zero. So this is 2 times zero, plus 2, 2.

Let me plot these points and see how far I’ve gotten. I have 4, 0. That’s this point. Then I’ve got approximately 15/4, pi over 6. So pi over 6 is this direction, 15/4 is ¼ short of 4, this is 4, so if tha5t’s 3 ½, 3 ¾ is right here. 3 pi over 3, this is a distance of 3 and this is pi over 3, so here’s my point. And then 2 pi over 2, so that’s 1, 2, here. Looks like we’ve got some kind of curve doing this.

Let’s see what happens in this part, in the second quadrant here. I’m going to plot 2 pi over 3, 5 pi over 6 and pi. 2 pi over 3; the cosine of 2 pi over 3 is -1/2. So that’s 2 times -1/2 plus 2, -1 plus 2, 1. 5 pi over 6, the cosine’s negative root 3 over 2. So 2 times negative root 3 over 2 plus 2, let me lower this guy. I will have to do some calculations. This is approximately negative 7/8. So this is about -7/4 plus 8/4, ¼ finally when I plug in pi, cosine of pi is -1. So I get 2 times -1 plus 2, 0.

Let me plot these points. 2 pi over 3, 1 2 pi over 3, ¼, 5 pi over 6, and zero pi. 2 pi over 3 I get 1. ¼ 5 pi over 6, this is the 5 pi over 6 direction. Pi over 6 short of pi. ¼ is right here. And then I get 0, pi which is right here. Let’s not forget that our graph will be symmetrical across the x axis. So I can duplicate these points. So this point duplicated down here, this point is duplicated down here. This point’s duplicated down here. This point will see where, pi over 3 and where, 1, 2, 3, 4, 5, 6, so 1, 2, 3, 4, 5, 6, out. And then this point is duplicated down here.

That should be enough to get me a graph. So first the left side, and then it comes around, a sort of interesting shape. And the bottom, remember; use symmetry this should come out exactly the same shape as the top. There you go. This shape is called a cardioid. The equation once again r equals 2 cosine theta plus 2. This is a cardioid and it's part of a family of functions that we’re going to study a little bit later.

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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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