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 1

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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Let’s graph a rose, r equals 8 sine 3 theta. Now first of all let’s recall that any of the sine roses are going to be symmetric about the y axis. So I want to use that because it takes a lot of time. You have to plot a lot of points to get these graphs. Notice I’ve already plotted many points, but what I did was I entered values for 3 theta because it's easy to evaluate sine of 3 theta and multiply by 8. So what I’m going to do to get theta is just divide all these values by 3 and then I’ll plot those values and we’ll take a look at the graph.

So 0 divide by 3, 0. 30 divide by 3 is 10, 60 divide by 3 is 20, 30, 40, 50, 60, 70, 80 and 90. So I’ve plotted from 0 to 90 in increments of 10. Now, these points I’ve already plotted on the graph but want I want to remind you of is. first of all the symmetry with respect to y axis. And also when you are graphing a polar graph, make sure to draw the curve in the order that the points are plotted. So we start with 0,0 and then it’s 10 degrees 4, or 4,10 degrees and we go in this direction.

This is kind of important because, polar curves can cross over themselves. And it's really important that you now which way the curve goes at a point where intersects itself. At this point, we're back at 60,0 or 0,60. R equals 0 the angle 60 degrees. And then we are going to go to -4, 70.

So pointing at 70 degrees, which is roughly this direction, I need to go -4, and that lands me down here. So I’m going in this direction. So what I have here is half of my graph and if I fill in the rest of the points I’ll get the rest of my graph. So this point is reflected here, this one is reflected here and this one is reflected right there. So I complete this leaf. And then to finish the graph, I’m going to have another leaf here and it's going to be a reflection of this one. So I need to reflect these points.

This point here at 4 goes right here, this one which is at 7 goes 5, 6, 7, right about here. And then the one at 8 and back to 7 again and then back to 4, and then back to 0. So it’s going to come around this way. It will actually plot this way first and then back.

By the time we arrive back at the origin, we are at one complete cycle and that’s our 3 leaf rose. This is r equals 8, times the sine of 3 theta. It’s a 3 leaf rose, symmetric about the y axis. If you want to graph a cosine 3 theta, you get a three leaf rose symmetric about the x axis.

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