Thanks in advance!
Currently learning polar coordinates where we use the coordinates of (r,t) so r = radius and t = arc length or angle
We've been shown that if our cartesian coordinates end up in the third or second quadrant then we do +pi to arctan(y/x) and if we end up in the fourth quadrant we do +2pi
I can see how this makes each quadrant have its own set of distinct values so you can tell which quadrant each t value lies in but understand how this new t value works to go back to the cartesian coordinates
How come adding pi doesn't change the cartesian coordinates when we work back from the polar coordinates?
How can I type in polar coordinates (in r,ΞΈ) and functions with r and ΞΈ? I plot a point in the graph which should be (1,pi/6), but it shows (0.866025,0.5). (I intended to calculate the area in polar coordinates like the overlap area of sinΞΈ and cosΞΈ)
title
Can someone explain the graph r=a(theta)?
Why isn't it a Circle?
https://imgur.com/a/7stXihn
What are and how do you find the the limits of questions 2?
Is there a way to find a point (or prove that such point DNE) that has the same rectangular and polar coordinates? That is, the point (a, b) in rectangular is the same as (a, b) polar? I believe this would be a matter of solving:
[;a = a\cos{b};]
[;b = a\sin{b};]
since this is the same as converting (a, b) polar to points in rectangular? But I'm not sure how to start with that system besides graphing, which seems to show that no solution exists.
There's the following restrictions for the area:
- It's under the line y = x
- It's in between x= 0 and x = sqrt(2)
- It's within the circle x^2 + y^2 = 4
I need to change this into integral boundaries in polar coordinates(r, Ο). I drew the figure here. The lines x = 0 and y = 0 intersect in the same point as is the center of the circle: (0,0). I figured this out:
- r >= 0 and r <= 4
- Ο > -1/2 Ο , Ο < 1/4 Ο
But that doesn't include the x = sqrt(2) yet. Since you can replace x for rcos( Ο ), you can work out that r must be less than sqrt(2)/cos( Ο ). But if you use that as the upper boundary you won't have r <= 4 anymore. So how do you make a good upper boundary for r that includes both of these?
Hello, I am new to this topic and have trouble starting this problem:
https://preview.redd.it/ufzysmotfdy71.png?width=1018&format=png&auto=webp&s=00bd3de94ddf4ea904e7a0283e858c10b8d743a4
I have not encountered a problem like this. To be able to evaluate it, I should be converted first, though I do not know how to do this. My Univ. send lecture videos about these topics but they do not do it step-by-step. I tried learning through Youtube vids, but I just can't seem to apply it correctly because most are just simple problem types.
So, if anyone knows how to do it, please could you help me with understanding how to convert.
Thank you very much.
I have seen a ton of posts on that - none has really resolved the issue.
I don't necessarily want to learn how to enter the ^0 for degrees: First of all, I want radians, and I have noticed Geogebra intakes radians by default, even if it later expresses the value in degrees depending on settings. Regardless, the Ctrl + 0 suggested in some posts doesn't do the trick on Windows.
I have managed to create points by entering (3; pi), for instance; however it doesn't draw the arrow from the origin: it simply plots the point at (-1,0).
Ultimately, I'd like to use Translate(...) to actually start the vector at some point other than the origin, but this seems secondary at this point.
It is quite easy to visualise (and is part of most textbooks) the elementary area in Cartesian Coordinates which corresponds to a cube, but we don't see the same in case of Spherical/Polar coordinates. Is it not possible/mathematically senseful or is the procedure to state it in those 2 coordinate systems tedious and or useless?
How do I integrate the second term in the second to the last line? When I enter that in software like Symbolab and Wolfram Alpha, the answer is around 2.10075 but I am unsure as to how that term can be integrated.
Note: The upper bound for x is supposed to be negative so it's -sqrt(2)/2. I encoded it wrongly while making it.
When we have the angle, why do we add a pi or 2 pi at times? My brain makes me think thatβll change the coordinates?
