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# Lines in Polar Coordinates - 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.

In the study of polar equations we must learn how to write the equation of a polar coordinates line. Students are sometimes asked to use the distance formula for polar coordinates, or to find the equation of the **polar coordinates line** in rectangular form, so students should know how to convert points on the polar coordinates line from polar to rectangular, and from rectangular to polar coordinates.

I want to talk about lines in polar coordinates. Now let's take a look at the graph of one of them. Theta equals pi over 6. It looks like this. Notice that every point has the same second coordinate pi over 6, that's the angle the angle from the horizontal axis, pi over 6. And they do form a straight line. Now equations of this type theta equals k are all lines that pass through the origin and conversely all lines through the origin have equation of this form. So let's use that and find the equations for some lines that pass through the origin like y equals negative root 3x.

Now the first thing I would do is write this in the form y over x equals negative root 3 and then make the observation that y over x is the same as tangent theta. So, what we have to do is figure out what theta equals. So what angle has a tangent of negative root 3? Well, it's negative pi over 3. So theta equals negative pi over 3, is the same as the line y equals negative root 3x.

What about this one? Here we have y over x equals 5 and therefore tangent theta equals 5. Well, unfortunately 5 is is not a real nice number. I could write theta equals inverse tangent of 5 though and that's a perfectly good equation. so this this number is a constant and so we have theta equals inverse tangent of 5. This is going to be the equation of the line y=5x in our polar coordinates. Now if you want something you know, something that's a little more numbery you can use your calculator to get inverse tangent of 5 and it's approximately 1.37. So theta equals approximately 1.37.

Now, what about other lines? What about lines that don't pass through the origin? So draw a picture here of a line that doesn't pass through the origin. I want to find the equation of this line. In order to do that I need to know the coordinates of the point that's closest to the origin. That's what this point n is. And the point that's closest to the origin, if you drew a line from that point to the origin, it will be perpendicular to the given line. So this is perpendicular to this. And we need that because it forms a right triangle here. Point of the origin would be point o and I can use right triangle trigonometry to get me an equation for this line. First of all let me observe that since the coordinates of point p here, this is just any arbitrary point on the line. Our r theta, this distance is r. And this angle is theta. And likewise, since the coordinates of this point are d beta, this length is d and this angle is beta. And so the angle between the two between the 2 segments is going to be theta minus beta. And we're going to use that.

So let's use that right now. Cosine of this angle theta minus beta equals side adjacent over hypotenuse and this is the side adjacent, this is the hypotenuse. So it's going to equal d over r. And I just multiply both sides by r. And I get r cosine theta minus beta equals d and therefore r equals d over cosine of theta minus beta. So here the perameters of the lines are d, the distance between this point and the origin right? The closest point and beta, the angle between the positive axis and this line segment drawn to the closest point.

So you really need to know the polar coordinates of that closest point in order to come up with this equation. But this is the general equation for a line in polar coordinates.

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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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