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Old 01-27-2009, 10:53 PM    (permalink
JETS5128
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can anyone try and help me in graphing a polar graph? IF someone just PM's me this thread can be locked. This is urgent and any help will receive major +rep
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Old 01-27-2009, 10:55 PM    (permalink
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can anyone try and help me in graphing a polar graph? IF someone just PM's me this thread can be locked. This is urgent and any help will receive major +rep
what are you graphing? what function are you using? what class?
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Old 01-27-2009, 11:00 PM    (permalink
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It's for pre-calculus. I have to graph r/theta=12
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Old 01-27-2009, 11:09 PM    (permalink
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It's for pre-calculus. I have to graph r/theta=12
which is 12theta = r
or the radius = 12 times the degree. i think.

is that the right question for real?
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Old 01-27-2009, 11:14 PM    (permalink
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That's what i did, but isn't 12theta just a diagonal line? How can you graph a diagonal line on polar paper?
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Old 01-27-2009, 11:17 PM    (permalink
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It's not a diagonal line at all. r = theta is a spiral. theta*12 would probably be a pretty zany spiral.
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Old 01-27-2009, 11:17 PM    (permalink
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That's what i did, but isn't 12theta just a diagonal line? How can you graph a diagonal line on polar paper?
it's not. it's a spiral.

you know the spiral of archamedies? function of theta = theta? (the polar equivalent to x = y (fyi))
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Old 01-27-2009, 11:18 PM    (permalink
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Math sucks. You guys might as well be speaking Mandarin right now. So glad I got done with math a couple of years ago.
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Old 01-27-2009, 11:18 PM    (permalink
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It's not a diagonal line at all. r = theta is a spiral. theta*12 would probably be a pretty zany spiral.
zing. zoom. man, i miss the days when math was actually easy.
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Old 01-27-2009, 11:18 PM    (permalink
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your right, and the spiral is ****** HUGE
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Old 01-27-2009, 11:22 PM    (permalink
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CAn anyone help me on how to graph it now?
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Old 01-27-2009, 11:23 PM    (permalink
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it's not. it's a spiral.

you know the spiral of archamedies? function of theta = theta? (the polar equivalent to x = y (fyi))
I think I know what you mean by "polar equivalent," but not to confuse JETS here, r = theta is not the polar equivalent of the xy coordinate line x = y. It's sin(theta) = cos(theta).
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Old 01-27-2009, 11:26 PM    (permalink
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I think I know what you mean by "polar equivalent," but not to confuse JETS here, r = theta is not the polar equivalent of the xy coordinate line x = y. It's sin(theta) = cos(theta).
true, but as far as the most basic function you could acquire in either coordinate system, they're 'equivalent' in concept not in actual form.
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Old 01-27-2009, 11:27 PM    (permalink
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CAn anyone help me on how to graph it now?
change your scale. don't go by 1's, go by 10's or better yet let each tick = 120, so for every 10* you move 1 tick.

is that confusing?
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Old 01-27-2009, 11:28 PM    (permalink
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Can anyone help me on how to graph it now?
Not trying too be annoying, but time is of the essence here
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Old 01-27-2009, 11:30 PM    (permalink
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CAn anyone help me on how to graph it now?
when r = theta, that means the distance a point is from the center (or pole or whatever) is equal to the angle of the point. r = 12*theta means that the distance from the center equals twelve times the angle of the point. So your first point would be (0,0), your second point would be (1, 12), and so on. The problem is that the angles are in terms of pi, with each revolution around the center being 2pi. That means that the angle would be 6/pi, which you would have to estimate.

Basically, your teacher's a prick.
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Old 01-27-2009, 11:31 PM    (permalink
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change your scale. don't go by 1's, go by 10's or better yet let each tick = 120, so for every 10* you move 1 tick.

is that confusing?
I'm supposed to graph each point from 0 to 360, in increments of 15 degrees. But how can i graph the point theta=15, r=180 on polar paper?
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Old 01-27-2009, 11:31 PM    (permalink
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Wait, I'm assuming that you're drawing this. Is that correct?
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Old 01-27-2009, 11:32 PM    (permalink
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Wait, I'm assuming that you're drawing this. Is that correct?
Yes, on polar paper
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Old 01-27-2009, 11:34 PM    (permalink
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I'm supposed to graph each point from 0 to 360, in increments of 15 degrees. But how can i graph the point theta=15, r=180 on polar paper?
to quote basic engineering 101.

engineers choose the coordinate plane and scale. they also do it to specification.

what does your graph paper look like? bullseyes or a regular cartesian coordinate system?
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Old 01-27-2009, 11:35 PM    (permalink
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Old 01-27-2009, 11:40 PM    (permalink
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bitchin.

each one of those circles do NOT equal 1. they equal 60. in distance.

so. for every 5 degrees, you move out one concentric circle (which the scale is 1r=60? i forget). meaning at 15 degrees, you'd be moved out 180, or three concentric circles. which rtheta = 15*12 (or 180)

when you move out to 30 degrees, it'd be 30*12 or 360, which would be six concentric circles.

just make sure you put on the graph that you have changed the scale.
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Old 01-27-2009, 11:42 PM    (permalink
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Originally Posted by awfullyquiet View Post
bitchin.

each one of those circles do NOT equal 1. they equal 60. in distance.

so. for every 5 degrees, you move out one concentric circle (which the scale is 1r=60? i forget). meaning at 15 degrees, you'd be moved out 180, or three concentric circles. which rtheta = 15*12 (or 180)

when you move out to 30 degrees, it'd be 30*12 or 360, which would be six concentric circles.

just make sure you put on the graph that you have changed the scale.
this. my suggestion is way wrong.
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Old 01-27-2009, 11:42 PM    (permalink
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Old 01-27-2009, 11:43 PM    (permalink
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