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jana

  • 5 years ago

how do i test for symmetry? r=16cos3(theta)

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  1. anonymous
    • 5 years ago
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    symmetry around what?

  2. anonymous
    • 5 years ago
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    http://www.most.gov.mm/techuni/media/EM_02011_chap5b.pdf

  3. Jana
    • 5 years ago
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    symmetry with respect to theta=pie/2, the polar axis, and the pole

  4. Jana
    • 5 years ago
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    thanks but how do i actually work it to find the answer?

  5. anonymous
    • 5 years ago
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    This 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, pi-theta), 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)).

  6. anonymous
    • 5 years ago
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    The 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))).

  7. anonymous
    • 5 years ago
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    If the function's symmetric about pi/2, then replacing (r, theta) with either (r,pi-theta) 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 pi-theta, 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,\[(r,\pi+\theta) \rightarrow 16\cos 3 (\pi + \theta)= -16\cos 3 \theta = -r \ne r\]

  8. anonymous
    • 5 years ago
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    So it's only symmetric about the polar axis.

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