jana
  • jana
how do i test for symmetry? r=16cos3(theta)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
symmetry around what?
anonymous
  • anonymous
http://www.most.gov.mm/techuni/media/EM_02011_chap5b.pdf
jana
  • jana
symmetry with respect to theta=pie/2, the polar axis, and the pole

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jana
  • jana
thanks but how do i actually work it to find the answer?
anonymous
  • anonymous
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)).
anonymous
  • anonymous
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))).
anonymous
  • anonymous
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,\[(r,\pi+\theta) \rightarrow 16\cos 3 (\pi + \theta)= -16\cos 3 \theta = -r \ne r\]
anonymous
  • anonymous
So it's only symmetric about the polar axis.

Looking for something else?

Not the answer you are looking for? Search for more explanations.