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# Graphing Polar Equations - 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.

I want to graph a polar equation. I’ve got r equals 12 sine theta. One of the first things I like to do is check for symmetry. The reason is, if the graph does posses some symmetry, it can save me plotting a lot of points.

Let’s try testing for symmetry by plugging in negative theta. So 2 sine negative theta. Remember that sine is an odd function so the sine of negative theta is minus the sine of theta. So this is minus 2, sine theta. Which is exactly the opposite of r. now what that tells me is that negative r, negative theta is in the graph. Negative r negative theta is a reflection of the point r theta around the y axis. So this graph is going to be symmetric about the y axis. I should put a little note here, symmetric about the y axis. That will help me when I’m graphing.

Now I’ve graphed the equation y equals sine theta, this is the rectangular graph, just to give me some heads up on what’s going to happen with the magnitudes of r. I notice that sine graph goes between 2 and -2, so the r values will be a maximum of 2, a minimum of -2 and that will equal zero at 0, pi, 2pi, etcetera. So that’s good to know. Let me plot some points starting with theta equals zero. Now remember the sine of zero is zero so 2 times that is still zero.

Let’s try pi over 6. I’m going to get 2 times the sine of pi over 6. The sine of pi over 6 is ½ so 2 times ½, 1. Pi over 3; the sine of pi over 3 is root 3 over 2. Twice that, now I’m going to need to come up with a value that I can actually plot. I need to get some idea for how big root 3 over 2. When I’m graphing I use the approximation 7/8 for root 3 over 2. So this is approximately, 2 times 7/8. That equals 7 quarters. 7/4 is 1.75 so that’s what I’m going to use as my approximation and it works pretty well for graphs. Pi over 2; sine of pi over 2 is 1. 2 times that is 2.

Let’s start with those points. So (0, 0) that’s here and we’ve got pi over 6, 1. Pi over 6 is this direction and I want to go 1 unit. I’ve made two of these ticks one unit so this will be my second point. And I’ve got pi over 3, 7/4. This is the direction pi over 3. Pi over 3 is 2 times pi over 6. So that’s pi over 6, and that’s another one. So I have 1, 1.5, 2, 1.75 is right here. And then pi over 2, 2. Or rather 2 pi over 2. So these are 4 points that I can use to get started.

Now let’s make the observation that, for the rest of these angles we use the fact that sine of pi minus theta equals sine of theta. In other words, if I take, for example 2 pi over 3, that’s the supplement of this angle, and they’re going to have the same sine value. And therefore, I’m going to get the exact same y value here. I’ll get 2 times root 3 over 2, which is approximately 7/4. And then 5 pi over 6 is the supplement of this point. And so I’m going to get the same r value down here.

Finally pi, I’m going to get the same r value that I got here. Supplementary angles have the same sine value. Let me plot these points. 2 pi over 3, 7/4. 2 pi over 3 is pi over 6 past pi over 2 so 7/4. This is 1, 1.5, 1 ¾. And then 5 pi over 6, 1; this is 5 pi over 6, 1 unit is right there. And then at pi I get zero, so I’m back here again. So far this is what we’ve got.

Now what’s going to happen afterwards? Well if I go to 7 pi over 6, I get sine is -1/2. So 2 times -1/2 is -1. 7 pi over 6, -1, I’m pointing in this direction but I’m going backwards 1 unit. I’m back to point again. So, this graph just keeps repeating itself every pi and we’re not going to get any more. This is the final graph of r equals 2 sine theta. It’s a circle, centered at the point 1, pi over 2 and has a radius of 1.

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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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