anonymous
  • anonymous
Explain polar coords and give an example.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
With rectangular coords we typically represent locations as the intersection of lines on a horizontal/vertical axes. With polar coords, we take the magnitude of the x and y parts that we are used to on a cartesian coord system. so if we had a point (3,4), to find the magnitude we would just that the square root of (3^2 + 4^2). Just think of it like finding the hypotenuse if you were doing pythagorean theoram. What you now have, in polar, is rhe "r", which is just the length of the ray (line) from the origin (0,0). But, we also need something else called the argument, which is the angle (or theta) from the x-axis. If we had some point on the cartesian plane, like (1,1) then it would have an "r" of sqrt(2), and an argument of pi/4. I would write it like sqrt(2)[cos(pi/4) + sin(pi/4)]. You might also be able to write it like sqrt(2)*e^i*(pi/4).
anonymous
  • anonymous
What is the Cartesian Plane?
anonymous
  • anonymous
just the normal x,y coords that we are used to.

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anonymous
  • anonymous
Okay.
anonymous
  • anonymous
Where did you get the r's from (1,1)?
anonymous
  • anonymous
if I take the sqrt of 1^2 + 1^2, I have sqrt(2). Just think of it like finding the hypotenuse of a right triangle.
anonymous
  • anonymous
So if this (1,1) was (x,y), then it is x^2+y^2?
anonymous
  • anonymous
Or the sqrt(x^2+y^2)?
anonymous
  • anonymous
with x,y coords we look for whee they intersect and that is the point. with polar, we have a ray that extends from the origen. the length of that ray is found just by the same way that you would find the hyp of a right triangle.
anonymous
  • anonymous
sqrt(x^2 + y^2)
anonymous
  • anonymous
Oh! Okay, thanks!
anonymous
  • anonymous
remember, that just gives you the length of the ray from the origin. You still need to include an angle, or else we would have no idea where the endpoint of the ray is.
anonymous
  • anonymous
Wait, so how do I get the angle?
anonymous
  • anonymous
You will likely have to use your trig and the length of the x (cosine) and y (sine) parts to determine angle. If you have an angle that can be represented at y/x, then you also have something that can be represented as tangent theta, right? (because sin/cos). So once you have that, then you take the arctan (y/x) and that will give you the angle.
anonymous
  • anonymous
so for the example I used above, if I have x = 1 and y = 1, then what I really have is y/x = 1/1 = 1. So I take tangent theta = 1, ----> theta = arctan (or tan inverse) (1), and that is pi/4 (or 45 degrees if you are in degree mode)
anonymous
  • anonymous
Can you help me with this problem? I think it might be simpler to explain the steps. Use your grapher to determine which of the graphs matches the polar equation r = 3 sin 2θ.
anonymous
  • anonymous
what is the theta?
anonymous
  • anonymous
IT says 3 sin 2(theta)
anonymous
  • anonymous
Without rectangular coords to convert to polar, or a polar coord to convert back to rectangular, I'm not really sure what to do with that. Where are the graphs the question is asking you to compare it to?
anonymous
  • anonymous
The answer choices?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
ah, ok. hold on, let me plot it.
anonymous
  • anonymous
it is the third option
anonymous
  • anonymous
How did you get that?
anonymous
  • anonymous
do you have a graphing calculator? On mine, I switched the graph-type from rectangular to polar, then I just entered what you gave me: 3sin(2theta). It plotted the 3rd option.
anonymous
  • anonymous
Oh, okay.
anonymous
  • anonymous
Can you do another one? Find the rectangular coordinates of the point with the polar coordinates (8,3/2*pi ).
anonymous
  • anonymous
My calculator isn't cooperating, but lets see if we can reason the answer. We know r=8, and we have a theta of 3/2*pi. So I think then that the point is on the y-axis below zero. We have a problem trying to use tangent right away, because cosine of (3/2*pi) = 0, that means tangent give a divide by zero error. Even so, we know the ray is length 8, so I think the rectangular coords are (0,-8)
anonymous
  • anonymous
Okay,
anonymous
  • anonymous
That's an answer! gj!
anonymous
  • anonymous
Wait, can I try one, and tell me if it's correct/
anonymous
  • anonymous
Sure
anonymous
  • anonymous
Find the rectangular coordinates of the point with the polar coordinates . This is the question.
anonymous
  • anonymous
(-3, 5pi/8)
anonymous
  • anonymous
Those are the coords.
anonymous
  • anonymous
So r=-3, right?
anonymous
  • anonymous
Agh, I'm confused.
anonymous
  • anonymous
r = -3, but remember that r is sqrt(x^2 + y^2). We get the negative sign because 5/8*pi is in quadrant II, which means the cosine (x) part is a negative number.
anonymous
  • anonymous
So I will need a unit circle, right?
anonymous
  • anonymous
That is probably what I would do, look at tan(5/8pi) first
anonymous
  • anonymous
It's not on the unit circle.
anonymous
  • anonymous
if you divide pi (the top half of the circle) into 8, you can see that 5/8pi is just a little past pi/2
anonymous
  • anonymous
So pi/2?
anonymous
  • anonymous
0,1
anonymous
  • anonymous
what I did to find x was 3cos(5/8*pi), and to find y was 3sin(5/8*pi). I had to round off a bit, but I got x = 1.15 and y = -2.78. Just to doublecheck, I did sqrt((1.15)^2 + (-.2.78)^2) and it did come back as 3.
anonymous
  • anonymous
Since 5/8pi is in quadrant 2, I know the x value must be negative, and the y value will be positive.
anonymous
  • anonymous
Can you tell me the equation you useed>
anonymous
  • anonymous
I only used what you gave me. So, my reasoning went like this: first, locate what quadrant 5/8*pi is located. It is in Q2, so we are dealing with a negative x and a positive y. Then, to find x, I just entered 3cos(5/8*pi) into my calculator and got -1.15 (rounded). To find y, I entered 3sin(5/8*pi) into my calculator, and it gave me 2.78 (rounded). So with that I now have my x and y values to plot.
anonymous
  • anonymous
So for r, you used x(sin or cos) of y, right?
anonymous
  • anonymous
Yes, but I just thought of something. If that r is -3, then I should have used it in my computations for x and y, which would have switched the signs of both x and y. I think we need to change my answer to 1.15 and -2.78, sorry about that.
anonymous
  • anonymous
So x =-2.78?
anonymous
  • anonymous
nope, x = 1.15, and y = -2.78. Cosine gives us the x part, and Sine gives us the y part.
anonymous
  • anonymous
Wait, I thought x was negative.
anonymous
  • anonymous
it was, but look at that r. r = -3, so we have to include that in our calculation. If we have some negative value for x and multiply by -3, the answer will turn positive, right?
anonymous
  • anonymous
Oh, okay.
anonymous
  • anonymous
But I don't have that answer choice.
anonymous
  • anonymous
same for y. It would be positive, but times -3 it turns negative.
anonymous
  • anonymous
what are the options?
anonymous
  • anonymous
I only have (-3/2,3/2) (3sqrt(3)/2),-3/2) (-3/2, 3sqrt(3)/2) and 3/2, -3/2
anonymous
  • anonymous
Hello?
anonymous
  • anonymous
I need to leave in half an hour, with the assignment done and Istill dont understand polar coords. I'm osrry, but I need a brief descriptiion that wil help get my assignment done.
anonymous
  • anonymous
@satellite73 Please help!
anonymous
  • anonymous
It looks like there is probably a much easier/better way to do this problem than how I've been working on it. You might be better off spending that 30 minutes reviewing the textbook or getting help from someone better at this than I am.
anonymous
  • anonymous
Soeey Ajk, maybe someome else can explani it better.
anonymous
  • anonymous
Thanks anyway.
anonymous
  • anonymous
:(
anonymous
  • anonymous
@cwrw238
anonymous
  • anonymous
@dpaInc
anonymous
  • anonymous
I'd be interested to see an explanation, if someone else helps. I'll check back and hopefully learn something too.
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
@dpaInc

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