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Graphing Polar Equations - Problem 3
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Let’s graph another polar curve. This time, r equals 1 plus 2 sine theta. First, I want to find out if there’s any kind of symmetry because knowing that there’s symmetry will save me a lot of plotting points.
I want to test this for symmetry across the y axis or x axis by plugging in negative theta. So I get 1 plus 2 sine negative theta, which equals, the sine of negative theta is the same as minus the sine of theta because sine’s an odd function. So this is 1 minus 2 sine theta.
Now unfortunately 1 minus 2 sine theta is neither equal to r, or its opposite, so this doesn’t actually help me. But that doesn’t mean that there’s no symmetry. There’s another test I can use. I can plug in pi minus theta, 1 plus 2 times sine of pi minus theta. Now recall the supplementary angle identity, sine of pi minus theta equals sine theta. Let me just separate these.
This is 1 plus 2 sine theta and that is equal to r. So that means that when I plug in pi minus theta I get r. So the point r pi minus theta is in the graph of our curve. Now this is the reflection of r theta about the y axis. So that means that this curve is going to be symmetric about the y axis, or you can call it in polar, theta equals pi over 2.
That will help us not have to plot too many points. Let’s start plotting points keeping in mind that we’re going to get some negative values for r. This is the graph of, the rectangular version of y equals 1 plus 2 sine theta, so it give us some idea of what the r values are going to end up being. So theta and r. Let me start with zero. Sine of zero is zero, so I’m just going to get 1. Pi over 6; sine of pi over 6 is ½ times 2 is 1, plus 1, 2. Pi over 3; sine of pi over 3 is root 3 over 2 and this is 1 plus 2 root 3 over 2.
Now when I’m graphing I use the approximation root 3 over 2 is about 7/8. So this is about 1 plus 2 times 7/8, that’s 7/4 plus 1 is 11/4. This is going to be 2.75. And then pi over 2; the sine of pi over 2 is 1, so 1 plus 2, 3.
Let’s start with these points, 0, 1. Now this is r equals 1, theta equals zero. Theta equals zero is this direction, r equals 1 is right here. So that’s my first point. And then r equals 2, theta is pi over 6. Pi over 6 is 30 degrees, so this direction, I get 2. With pi over 3 which is another 30 degrees I got 2.75. This is 2, that’s 3, 2½, 2.75 is about here. And pi over 2 I got 3.
Now if I keep going further, I’m going on the other side of the y axis and I know I have symmetry with respect to the y axis. So I can fill in those points right now. This point has a reflection over here. This one has a reflection over here and this one has a reflection right here.
So instead of going forward, I should go backward. Let me try negative pi over 6. So I have negative pi over 6 is minus the sine of pi over 6. So minus ½. So get 1 plus 2 times -1/2, that’s -1 plus 1, zero. So r equals zero; I’m in this direction but r equals zero, doesn’t matter what the direction is and then minus pi over 3. Sine of pi over 3 is root 3 over 2. Sine of negative pi over 3 is negative root 3 over 2. So this is 1 plus 2 times negative root 3 over 2 and again I’ll use the approximation 7/8 for root 3 over 2. So this is 1 plus 2 times negative 7/8. That gives me -7/4 plus 4/4 is -3/4. -.75.
Finally negative pi over 2; sine of negative pi over 2 is going to be -1. This is 1 plus 2 times -1. This is -2, plus 1, so it is actually negative 1. Let’s plot these points. I’ve got, I already plotted zero, negative pi over 6. I need to plot -.75, negative pi over 3. So negative pi over 3 direction is this direction. I want to go -.75 in this direction which means I go backwards, .75. That’s ½, that’s 1, so that’s .75. Actually going backwards here. And then for minus pi over 2, I go backwards, 1. This is the minus pi over 2 direction, so I go backwards 1 to here. Again, use symmetry with the y axis to plot a mirror image of this point over here. I think that’s in the wrong place, this should actually go here.
So let me complete this graph. It starts by going around this way, looks roughly circular but then it comes, going backwards, it goes like this. This is called a limacon with inner loop. And that’s it. We’ll be talking more about this family of curves in a later episode.
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