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# Lines in Polar Coordinates - 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.

We're talking about the equation of a line and polar coordinates. Here is a line, L with equation r equals 5 over cosine of the quantity theta minus pi over 6. It says find the distance to the origin. Here we mean the distance between the line, and the origin. So you have to find the distance between the closest point on the line and the origin.

Of course, this equation includes as its parameters 5, and pi over 6, which are the coordinates of the point closest to the origin. So its distance will be 5. The polar coordinates of n, the point on L closest to the origin, are 5 pi over 6. These are the parameters for our equation. You have to know the coordinates of the point closest to the origin, in order to write an equation like this. Now it says graph the line in part C.

The first thing I want to do is, plot the point 5 pi over 6. So pi over 6 is this direction, so 1, 2, 3, 4, 5. That's one point of the line. Remember that to graph a line, all you need is two points, and then you can just set up your ruler and draw a straight line. We need a second point. I use my equation; cosine of theta minus pi over 6. Let's plug in pi over 2. So we get r equals 5 over cosine of pi over 2 minus pi over 6. Pi over 2 minus pi over 6. Pi over 2 is 3 pi over 6, minus pi over 6 is 2 pi over 6 which is pi over 3.

So this is 5 over cosine pi over 3, and the cosine of pi over 3 is 1/2. So five over 1/2 is 10. So when the angle is pi over 2, the distance is 10. Well pi over 2 is this angle, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. This is a second point of the line, and now I can draw it. Here is our line.

Now find the rectangular equation of the line. Well let me erase this. First, let's start with the equation in polar form 5 over cosine of theta minus pi over 6. Let's multiply both sides by cosine. We get r cosine theta minus pi over 6 equals 5.

On this left side, in order to convert this into xs, and ys, I'm going to need just cosine thetas, and sine thetas. The way to get that is just use the cosine of a difference formula. We get r times cosine theta, cosine pi over 6, plus sine theta, sine pi over 6, equals 5. Now cosine of pi over 6 is root 3 over 2. So this is r cosine theta times root 3 over 2 plus, sine of pi over 6 is 1/2, so I get 1/2 times r sine pi. All that equals 5.

I could further multiply through by 2 and get root 3 r cosine theta, plus r sine theta equals 10. Finally, r cosine theta, remember my goal here is to write a rectangular equation. This is x. Root 3 times x plus, and our sine theta is y, equals 10. So y equals negative root 3x plus 10. That's the equation of our line.

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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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