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.

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Families of Polar Curves: Roses - Problem 2

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.

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One of the things you'll notice when you are graphing roses is the big difference between a rose with an even value of b and a rose with an add value of b. We graphed before r equals 8 sine 3 theta and we got a 3-leaf rose. Let’s see what we will get when we graph this function.

So I’ve plotted a bunch of points 4, 2 theta and then r equals 8 sine 2 theta. All I have to do to get theta is just to divide all these by 2. So I get 0, 15, 30, 45, 60, 75, 90, and then down here, 105 120 135, 150, 165 and 180. And I’ve plotted these points over here.

Now let’s keep in mind that because we have a sine rose, this is going to be symmetric about the y axis. So I’ll flip all these points over in a moment. First of all, roses have a lot of intersection with themselves. And so, I want to show you how to trace through the curve, in the order that the plots are pointed, so that you make sure that you get the right kind of shape.

So we start at 0,0 and then we go out in this direction and around. And you’ll notice that the leaves for this rose are a little bit thicker than the leaves for the 8 sine of 3 theta rose. Just sort of interesting. But you can already see that we’ve got two leaves and if I reflect this, I’m going to get two more. So this is going to end up being 4-leaf rose.

So it’s going in this direction now. So the next point will, be a reflection of this guy here and a reflection of this guy here, and then here, here, here. So that’s the third leaf. And then we need a reflection of this leaf over here. So the first point is going to be this point, then this point, this point, this point and back to this point, so something like this.

So this is r equals 8 sine 2 theta, a four-leaf rose. One thing to remember when you are graphing rose with an even value of b, is you’ll get a number of leaves double this value. Only for even values of b. With the odd values of b you get the number of leaves equal to the b value.

Also notice for all our sine roses, we have symmetry across the y axis. For this one we have symmetry across the x axis as well and the test doesn’t show that. So that’s a good reminder that sometimes you get symmetry but the tests don’t indicate. So, that’s a four leaf rose.

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