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# Introduction to Polar Coordinates - 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.

Let's plot some points with polar coordinates. Here are three points. I've got a which is 10, pi over 6. Remember with polar coordinates, the distance from 0 comes first, and the angle pi over 6, comes second. So we'll plot this one first.

10 units away from the origin with an angle of pi over 16. Now I'm going to make each of these two. So; 2, 4, 6, 8, 10. This will be 10 units away and an angle of pi over 6. Well I've got a lot of lines here. I have to know that pi over 6 is 1/3 of pi over 2. So 2, 4, 6, 8, 10. This would be my point A. I'm going to get in the habit, at least in the beginning of drawing a line from my point to the pole, just so that it's easier to see.

Now let's find its rectangular coordinates. Now for that we need some conversion formulas. I have r and theta. So my first conversion formula is going to be x equals r cosine theta. My second one is y equals r sine theta. So the x coordinate of this point is going to be r cosine theta, which is 10 cosine pi over 6. The cosine of pi over 6 is root 3 over 2. So 10 times root 3 over 2, and that's going to be 5 root 3.

The y coordinate is going to be 10 sine pi over 6. Now the sine of pi over 6 is 1/2. So this is 10 times 1/2 or 5. That means the coordinates of this point are going to be 5 root 3,5. These are the rectangular coordinates. The polar coordinates are 10 pi over 6.

What about point B? Point B is 10 away from the origin at an angle of negative pi over 6. Now I think it's easiest if we actually did the angle first. Negative pi over 6, remember the negative direction is clockwise in Mathematics. So negative pi over 6 is going to be this direction, so a point along this line. Imagine yourself standing at the pole, you turn pi over 6 clockwise, and you walk forward 10 units 2, 4, 6, 8, 10. This is going to be my point B. As before I'm going to draw a line to the origin.

Notice what the effect of changing the angle has done, is actually reflected the point around the horizontal axis. Let's find the coordinates. X equals 10 cosine of negative pi over 6. Well, the cosine of negative pi over 6 is the same as cosine of pi over 6, because cosine is an even function. That's going to be root 3 over 2 again. 10 root 3 over 2; 5 root 3.

Then the y coordinate will be 10 sine of negative pi over 6. Now sine is an odd function, so the sine of negative pi over 6, will be minus the sine of pi over 6 which was 1/2. So this will be 10 times minus 1/2, -5. My coordinates for B are going to be 5 root 3, -5. Now these are the rectangular coordinates.

Finally, point C. -10, negative pi over 6. So to locate that point, imagine you're standing on the pole, you're facing in the direction of the polar axis, which is the theta equals zero direction. You turn in the direction negative pi over 6. Negative pi over 6 is the direction I want to face. But then you walk backwards 10 units because of the -10. So -2, -4, -6, -8, -10, and you end up here. This is my point C.

So notice that reversing both the angle, and the r value, gets me a point that is a reflection of the original point around the vertical axis. So this point, point C is a reflection of this point.

X coordinate; -10, the r value times the cosine of negative pi over 6. We already calculated that was root 3 over 2. So -10 times root 3 over 2, -5 root 3. The y value; -10, sine of negative pi over 6. We calculated that before as well, it was -1/2. So -10 times -1/2, 5. So my coordinates are to be -5 root 3, 5. These are the rectangular coordinates for point C.

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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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