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# Graphing Polar Equations - Concept

###### 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.

The most basic method of graphing polar equations is by plotting points and doing a quick sketch. **Graphing polar equations** is a skill that requires the ability to plot points and sometimes recognize a special case of polar curves, such as cardioids, and roses and conic sections. However, we need to understand the polar coordinate system and how to plot points for graphing polar equations.

Let's graph a polar equation. I have a pretty easy polar equation here. r equals theta over pi for theta greater than or equal to 0. Now the best approach when you're trying to graph something new is to plot some points. And so let's start with theta equals 0. If theta equals 0 r=0 so that's going to be a point. And let's try multiples of pi over 4. So when theta equals pi over 4, I get pi over 4 divided by pi which is a quarter. Pi over 2. Pi over 2 divided by pi is a half.

Let's try 3 pi over 4. 3 pi over 4 divided by pi is three quarters. And you can kind of see the pattern. The number I'm going to get here is basically this number without the pi, right? Divide out the pi. So pi will give me 1 and so on. Let me plot some of these points, and see what kind of a curve I'm getting.

So I have 0 0, right? r is 0, theta could be anything as long as r is 0, that gives me the origin. And then pi over 4, one quarter. Now I've made this so that each of these each of these circles represent a quarter unit. So pi over 4 one quarter is right here. And then pi over 2 this direction one half is right here. 3 pi over 4 gives me three quarters here. Negative pi gives me 4 quarters or 1 and following in this pattern if we wanted to keep going, 5 pi over 4 would give me 5 quarters. 1, 2, 3, 4, 5. 3 pi over 2 is 6 pi over 4. So I go 1, 2, 3, 4, 5, 6 and then 7 pi over 4 would give me 7 quarters. 1, 2, 3, 4, 5, 6, 7. And finally, let's just finish at 2 pi. 2 pi is the same as 8 pi over 4. So I go out to 2, right? 8 quarters.

Alright, let's see if if we can draw this. It looks kind of like a spiral. And just about done. There and the graph will continue forever, right? So it's just spirals around and around. this is the equation, the graph of the equation r equals theta over pi, for theta greater than or equal to 0.

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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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