Video Transcript
Select the graph of 𝑟 equals three minus two cos 𝜃.
In this question, we’ve been given a graph of a polar equation. This is an equation for 𝑟 as some function of 𝜃. The graph of a polar equation is the set of all points in the plane whose polar coordinates satisfy this equation. And one way we have of sketching the graphs of these polar equations is to plot points. We’re going to treat 𝜃 as the independent variable here. Remember, 𝑟 is the distance from the origin, and the angle 𝜃 is made from the positive 𝑥-axis in a counterclockwise direction.
Three minus two cos 𝜃 is a periodic function with a period of two 𝜋. So, we’re going to choose a number of values of the angle 𝜃 in one revolution. Let’s go up in intervals of 𝜋 by six radians. And of course, we could use a table function on a calculator to help us.
Let’s begin with 𝜃 equals zero. When 𝜃 equals zero, 𝑟 equals three minus two cos of zero. Now, cos of zero is one, so we get three minus two, which is simply one. When 𝜃 is equal to 𝜋 by six, 𝑟 is equal to three minus two cos of 𝜋 by six. And now, this is three minus the square root of three. But that’s pretty tricky to plot on a diagram, so instead we’ll use an approximate value of 𝑟. I’ve rounded it to three decimal places. So, we get 1.268.
When 𝜃 is equal to 𝜋 by three, 𝑟 is three minus two cos of 𝜋 by three. Well, cos of 𝜋 three is 0.5, so we get three take away one, which is two. Next, we have 𝜃 equals 𝜋 by two, and then 𝑟 is equal to three minus two cos of 𝜋 by two. Well, cos of 𝜋 by two is zero, so we get three minus two times zero, which is just three. We repeat this process for the remaining values of 𝜃. And we get when 𝜃 is equal to two 𝜋 by three, 𝑟 is equal to four. When 𝜃 is five 𝜋 by six, 𝑟 is equal to 4.732, and so on. And we go all the way up to 𝜃 equals two 𝜋, where we get 𝑟 is equal to one again.
So, let’s find these coordinates on a graph. Remember, 𝜃 is the angle that we make with the positive horizontal in a counterclockwise direction. So, when 𝜃 is equal to zero, we’re actually just interested in the positive horizontal line. 𝑟 represents the distance from the pole, or the origin. And that’s one. We can, therefore, mark the polar coordinates zero, one on each of our graphs as shown. And we quite clearly see that the only graph that passes through this point is A.
We will double check a few of the remaining points. Let’s look at the point 𝜃 equals 𝜋 by two. 𝜋 by two is a quarter of a turn. It’s equivalent to 90 degrees. So, it’s the line shown. When 𝜃 is equal to 𝜋 by two, 𝑟 is equal to three. And so, we plot the given polar coordinate. Next, we will look at 𝜃 equals 𝜋. Measuring an angle of 𝜋 radians from the positive horizontal gives us the line shown.
This time, 𝑟, the distance from the origin, is five. So, we plot this polar coordinate. Three 𝜋 by two as three-quarters of a full turn. And at this point, 𝑟 is three. And then, we notice that when 𝜃 is equal to two 𝜋, we go back to that first point, 𝑟 is equal to one. And so, the graph of 𝑟 equals three minus two cos 𝜃 is A.
