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anonymous
 5 years ago
how do i test for symmetry? r=16cos3(theta)
anonymous
 5 years ago
how do i test for symmetry? r=16cos3(theta)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0symmetry around what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0symmetry with respect to theta=pie/2, the polar axis, and the pole

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks but how do i actually work it to find the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This question will, unfortunately, take more time to answer than I have available. I can give you the following transformations and an example of what to do for one of them, and hopefully you can see what you have to do for the others. A polar function is symmetric about the polar axis if, when you replace (r, theta) with EITHER (r,theta) or (r, pitheta), you get the original function back. Here, if (r, theta) is on your graph, then\[(r, \theta) \rightarrow r=16 \cos 3 \theta \]\[\rightarrow r=16 \cos 3 ( \theta))\]\[\rightarrow (r, \theta)\]is on your graph (because cosine is even (i.e. cos(theta) = cos(theta)).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The function's symmetric about the pole if when you replace\[(r, \theta)\] with (r, theta) OR (r, pi+ theta), you get the original function back. So here,\[(r, \theta) \rightarrow r=16\cos 3 \theta\]\[\rightarrow r=16 \cos 3 (\pi+\theta)\]\[\rightarrow r=16\cos 3 \theta\ne 16\cos3\theta\]So neither (r theta) or (r, pi + theta) lie on the graph (note, when I'm giving you two points to choose from in your test, these points on the graph are identical (you just take a different route getting there (draw it if you can't see it))).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the function's symmetric about pi/2, then replacing (r, theta) with either (r,pitheta) or (r, theta) should get you back to (r, theta). Here,\[(r, \pi\theta) \rightarrow 16 \cos 3(\pi\theta))=16\cos 3 \theta=r \ne r\]i.e. if you replace with pitheta, you don't get back to (r, theta), you get to (r, theta). So this isn't symmetric about pi/2. There is a mistake in the reasoning above. I should have said,\[(r,\pi+\theta) \rightarrow 16\cos 3 (\pi + \theta)= 16\cos 3 \theta = r \ne r\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it's only symmetric about the polar axis.
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