Why does r=cos(4t) have 8 petals while r=cos(4t)+2 has only 4 petals?
I can see why this happens. Half of the first graph's petals are created by negative radii (radiuses?). The second equation's radius is stretched by 2 units and never goes below 0.
How would you describe this (on a test)?
I think your answer is good. But I would recommend graphing r = cos(4t) + a for different values of a ( t in [0,2*pi]) with your favorite graphing program/calculator. I think this would be very instructive. Here's the graph with a = 1/2:
Originally Posted by johng
I think your answer is good. But I would recommend graphing r = cos(4t) + a for different values of a ( t in [0,2*pi]) with your favorite graphing program/calculator. I think this would be very instructive. Here's the graph with a = 1/2:
I don't know what was the teacher expected.
This is actually for a student I'm tutoring. The question was quoted exactly as on the test (no calculators allowed on test). The answer was how I had described it to her and is a paraphrase of what she said on the test. The teacher took off partial credit.
My student also correctly graphed cos 4t and cos 4t+2 in polar coordinates. She had intermediate steps where she plotted r vs. t on a regular Cartesian coordinate, but those graphs were of sin 4t and sin 4t + 2. The teacher remarked that the final graphs did not match the intermediate steps. But that's not relevant to the grade (it is to me and the student!). I don't think the teacher would've taken off for that.
My goal (in posting) was to see how you all would've answered it and help her understand what was wrong. If you mathelpforum's residents like the answer, I guess the only step left is to ask the teacher what was wrong.
P.S. Nice drawing johng. How do you make them on a PC?
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