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# Lines in Polar Coordinates - Problem 3

###### 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've got this circle x² plus y² equals 4, and a tangent line drawn to it at the point negative root 3,1. Let's answer the following questions. We want to find first our polar coordinates of point P.

Well, one thing I can see about P right away is that, it's on this circle. That means that means that its distance from the origin had to be 2, since 2 is the radius of the circle. So r is 2. Then I need to find its angle, theta. I can do that by observing that the cosine of theta is going to equal this x coordinate, negative root 3 over r. The sine of the angle equals the y coordinate, 1, divided by r.

So what angle has cosine of negative root 3 over 2, and sine of 1/2? 5 pi over 6. So that's our angle. The polar coordinates would be 2, 5 pi over 6. Now we're supposed to find the polar equation of the tangent line.

Now here, we need to use the fact that, this point is actually the closest to the origin. If this line, if this segment is perpendicular to the line, that means this point is closest to the origin. Remember, the equation for a line in polar coordinates requires that we know the coordinates of the point closest to the origin. The equation looks like this; r equals d over cosine of theta, minus beta; where d is the distance between that closest point and the origin, and beta is its angle. So I just plug in; r equals 2 over cosine of theta minus the angle 5 pi over 6. That's our equation.

Part C asks us to find the x and the y intercepts of the line. That's a question you're usually asked when you have a rectangular equation, but you can answer it in polar coordinates. Think about what the x and the y intercepts would be. This x intercept we would get when theta is pi. The y intercept which is not shown, we would get when theta equals pi over 2. So let's plug those numbers in here.

When theta equals pi over 2, we get 2 over the cosine of pi over 2 minus 5 pi over 6. Pi over 2 minus 5 pi over 6. Now pi over 2 is 3 pi over 6, 3 pi over 6 minus 5 pi over 6 is -2 pi over 6, or negative pi over 3. So we have the cosine of negative pi over 3.

Now the cosine of pi over 3 is the same as the cosine of over 3, because cosine is an even function. The cosine of pi over 3 is 1/2, so 2 over 1/2 is 4. That means that my y-intercept is 4 units up the axis. That means that my y-intercept in polar coordinates is 4, pi over 2 and in rectangular, it would be 0, 4.

How about the x intercept? Well, we got to plug in theta equals pi. So we get r equals 2 over cosine of pi minus 5 pi over 6. Now pi is the same as 6 pi over 6, minus 5 pi over 6 is pi over 6. The cosine of pi over 6 is root 3 over 2. So this is 2 over root 3 over 2, or 2 times 2 over root 3. That's 4 over root 3.

Now we need to rationalize that. I multiply top and bottom by root 3, and I get 4 root 3 over 3. That's the distance to the origin of this x intercept; 4 root 3 over 3. So the x intercept is 4 root 3 over 3, pi, or in rectangular coordinates, -4 root 3 over 3, 0.

Now let's find a rectangular equation for the line. We'll start with our polar equation; r equals 2 over cosine of theta minus 5 pi over 6. I multiply both sides by the cosine, and I get r cosine theta minus 5 pi over 6 equals 2. I may have to use the cosine of a difference formula. I get r times the quantity cosine theta cosine 5 pi over 6, plus sine theta, sine 5 pi over 6 equals 2.

Now the cosine of 5 pi over 6 is negative root 3 over 2. So this is negative root 3 over 2 times r cosine theta plus, the sine of 5 pi over 6 is 1/2. So 1/2r sine theta equals 2. I multiply through by 2 to get rid of the fractions. I'll get minus root 3 r cosine theta plus r sine theta equals 4.

Now r cosine theta is x, and r sine theta is y. So this whole equation becomes minus root 3x plus y equals 4. I just add root 3x to both sides. Y equals root 3x plus 4. That's the rectangular equation.

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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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