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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).
What is the Cartesian Plane?
just the normal x,y coords that we are used to.
Where did you get the r's from (1,1)?
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.
So if this (1,1) was (x,y), then it is x^2+y^2?
Or the sqrt(x^2+y^2)?
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.
sqrt(x^2 + y^2)
Oh! Okay, thanks!
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.
Wait, so how do I get the angle?
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.
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)
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.
what is the theta?
IT says 3 sin 2(theta)
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?
The answer choices?
ah, ok. hold on, let me plot it.
it is the third option
How did you get that?
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.
Can you do another one? Find the rectangular coordinates of the point with the polar coordinates (8,3/2*pi ).
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)
That's an answer! gj!
Wait, can I try one, and tell me if it's correct/
Find the rectangular coordinates of the point with the polar coordinates . This is the question.
Those are the coords.
So r=-3, right?
Agh, I'm confused.
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.
So I will need a unit circle, right?
That is probably what I would do, look at tan(5/8pi) first
It's not on the unit circle.
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
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.
Since 5/8pi is in quadrant 2, I know the x value must be negative, and the y value will be positive.
Can you tell me the equation you useed>
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.
So for r, you used x(sin or cos) of y, right?
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.
So x =-2.78?
nope, x = 1.15, and y = -2.78. Cosine gives us the x part, and Sine gives us the y part.
Wait, I thought x was negative.
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?
But I don't have that answer choice.
same for y. It would be positive, but times -3 it turns negative.
what are the options?
I only have (-3/2,3/2) (3sqrt(3)/2),-3/2) (-3/2, 3sqrt(3)/2) and 3/2, -3/2
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.
@satellite73 Please help!
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.
Soeey Ajk, maybe someome else can explani it better.
I'd be interested to see an explanation, if someone else helps. I'll check back and hopefully learn something too.