polar coordinates

[Anonymous No.6790479]

>currently in a2 year of sixth form
>studying further maths
>teacher comes in to teach us about this exciting new topic called 'polar coordinates'

he teaches us them and i understand the basics but what i dont understand is why? why is this better than the typical/ordinary coordinate system we're all used to? i'm sure there's a reason so would someone be kind enough to explain it so i'll understand, thanks

[Anonymous No.6790488]

>>6790479
makes some integrals easier to evaluate

[Anonymous No.6790493]

>>6790479
In some calculations you can make use of the radial symmetry of the problem. There are many problems with such symmetries. For example the movement of planets. In cartesian coordinates the movement is pretty complicated, whereas in polar coordinates the angle is changing linearly and the radius is constant (in case it's moving in a perfect circle).
The simplification is pretty easy to understand in integrals. To calculate the area of a circle you have to integrate sqrt(r^2-x^2) dx from 0 to r and multiply it by four, which is pretty complicated, whereas in polar coodinates all you have to do is integrate r dr d phi from 0 to 2pi and from 0 to R, which is trivial.

[Anonymous No.6790497]

>>6790493
>>6790488
thanks, the teacher never actually mentioned why it was better - just that it was :/ but thanks a bunch

[Anonymous No.6790506]

>>6790479
there are some interesting shapes that can be graphed in polar coordinates where the functions that generate them are relatively simple where as in Cartesian coordinates the functions would be ugly. There are also other obvious answers listed by the people above me as well.

[Anonymous No.6790521]

what would we do without latitude and longitude

[Anonymous No.6790532]

>>6790479
Its essential to do any vector mechanics.

[Anonymous No.6790572]

Spherical Coordinates for the win

[Anonymous No.6790583]

>>6790479
>>he teaches us them and i understand the basics but what i dont understand is why? why is this better than the typical/ordinary coordinate system we're all used to? i'm sure there's a reason so would someone be kind enough to explain it so i'll understand, thanks

Complex numbers. a+bi is shit compared to re^(iθ). Multiplication is scaling by r and a rotation of θ, addition is classic vector addition.

[Anonymous No.6790651]

While working with trigonometric functions it is easier to do mathematical operations with a combination of Euler's formula and complex numbers/polar coordinates. Communication systems and feedback systems use a lot of the techniques to simplify arithmetic.

[Anonymous No.6790931]

>>6790479
On one hand, it is just another way to define coordinates.

On the other hand it can give you advantages over the Cartesian Coordinate system depending on what you are doing. Integration is one example. Another example is when using complex numbers. If you try to multiply a bunch of complex numbers in a Cartesian System (a+i*b) then it becomes a bunch FOILing and is messy. In Polar (r*Exp[i*(angle)]) you just multiply by each value of r and add the angles. Polar isn't a better system though, it just has it's own advantages. Adding complex numbers in Cartesian Coordinates, for example, is easier than polar.

[Anonymous No.6790939]

It's easy OP.

First, you do all the calculations in a hyperspherical coordinate system of arbitrary dimension n.

Then you let n=2

[Anonymous No.6790996]

>>6790532
>vector mechanics
why do I come here

[Anonymous No.6791038]

When a pilot needs to know his magnetical bearing toward an airfield or more specifically a runway in use.

[Anonymous No.6791067]

Pretend you took the Cartesian grid and rolled it up into a ball.

[Anonymous No.6791080]

ayy lmao
capcha: 2333

[Anonymous No.6791086]

If you want to study dynamics there are like 5 different coordinate systems that are commonly used. Why? Because each one has a use where it turns 10 pages of math into 1 simple relation. Right place and time for everything.

[Anonymous No.6791090]

>>6790996
vector analysis?

[Anonymous No.6791093]

>not using the perfect comformal coordinate system
stay pleb.

[Anonymous No.6791094]

>>6791086
>polar
>spherical
>cylindrical
>cartesian
What's the last one?

[Anonymous No.6791096]

>>6791094
normal-tangential

[Anonymous No.6791221]

>>6791094
Polar is really just a simplification of either the spherical or cylindrical system on the x-y plane (2D vs 3D).

[Anonymous No.6791231]

>>6790479

Suppose you're on a ship, and you see another ship. Would you say the other ship you see with your binoculars is some distance starboard and some distance aft, or would you say it is some distance away in a particular direction? Polar coordinates are in a way much more natural than cartesian. In your daily life you tend to think of objects to be at a certain distance in a certain direction.

Besides that, it also makes some problems much easier to calculate

[Anonymous No.6791247]

>>6791096
Thanks.

>>6791221
I am aware of it.
But we do state then separately, so I thought it counted as 1 more of said coordinate systems.

[Anonymous No.6791249]

We actually use spherical coordinates most often in real life, at least on a large scale. How do you describe a location: latitude, longitude, altitude.

[Anonymous No.6791287]

Whilst we're talking about polar vs Cartesian, something that's always bugged me; it's really easy to move from polar to Cartesian, but it's awkward to move from Cartesian to polar, the distance bit odd fine but the angle requires you catch edge cases (e.g. 1,0)

[Anonymous No.6791292]

>>6791287
>catch edge cases
what?

[Anonymous No.6791305]

>>6791292
How do you translate from Cartesian to polar? Can you give a straight forward equation for the angle?

[Anonymous No.6791307]

>>6791305
Just do the inverse.

[Anonymous No.6791313]

>>6791307
Which is...

Give an equation to calculate the angle for all cases.

[Anonymous No.6791314]

>>6791313
>what's arccos

[Anonymous No.6791324]

Is it better? No.

Is it more natural? Maybe. The x-y coordinate system is boxy and artificial.

Using polar coordinates, you can describe where a point is purely based on how far it is from an arbitrary point, and what angle it makes with a straight line that comes out of that point.

You can also convert one to the other.

[Anonymous No.6791325]

>>6791314
arccos has edge cases that need filtering
As does atan2

[Anonymous No.6791330]

>>6791325
What are talking about?
arccos(0) = 1
arccos(1) = 0

[Anonymous No.6791336]

>>6791330
>arccos(0) = 1
Ops.
pi/2

[Anonymous No.6791342]

>>6791330
Let me put it this way:
Polar to Cartesian = x=sin(θ) x r, y=cos(θ) x r

That works for all cases

Write the Cartesian to polar equivalent that works for all cases

[Anonymous No.6791379]

>>6790572
>spherical coordinates are essentially radius, x-y angle, and y-z angle
>mathematics, engineering, physics, and astronomy use different formats and representative symbols for them

This was the bane of my existence at uni.

[Anonymous No.6791387]

>>6791379
squirtle im dying

[Anonymous No.6791398]

>>6791342
θ = arctan(y/x); x,y => 0
θ = arctan(y/x) + pi; x < 0
θ = arctan(y/x) + 2pi; y < 0, x => 0

r = (x^2 + y^2)^1/2

[Anonymous No.6791543]

>>6791247
>I am aware of it.
>But we do state then separately, so I thought it counted as 1 more of said coordinate systems.
I assumed you were. I'm just being nitpicky.

[Laurie No.6791553]

If you want to lob a bomb you want to know how far and in what direction.

[Anonymous No.6791613]

>>6791379

I felt like the coordinate systems you're introduced and utilized to in those classes made the math your bitch. Like others have said, it's a bit more of a natural representation + makes a lot of the grunt work easier

[Medical_Physicist !!ST92PtcX00D No.6791721]

>>6790479
In physics there are countless scenarios where solving a problem in a Cartesian coordinate system is way harder than solving it in a spherical coordinate system (this rotated in and out of screen so you have a 3D space).

[Anonymous No.6791901]

>>6791398
Doesn't work for x=0, y=0

[Anonymous No.6791905]

>>6790479
>Fourier Transforms
You use a similar system to getting polar co-ordinates from cartesian ones to get a signal's amplitude and phase angle from the real and imaginary parts of a transform.

I doubt OP has done fourier transforms cause he's in sixth form, but still, it's a very very very useful thing to know.

But for now, passing exams is fine.

[Anonymous No.6791910]

>>6791379
BANE!?

[Anonymous No.6792161]

>>6791901
It doesn't work, because if r = 0 (x=0 and y=0) then θ can be anything.

[Anonymous No.6792179]

>>6791901
It doesn't work, because if r = 0 (x=0 and y=0) then θ can be anything.

However, the angle of the zero vector with any other vector is defined as pi/2.

[Anonymous No.6792195]

>>6792179
So instead we can define it like this:
θ = arccos(x/r); y => 0
θ = 2pi - arccos(x/r); y < 0

r = (x^2 + y^2)^1/2

[Anonymous No.6792198]

>>6792179
So instead we can define it like this:
r = (x^2 + y^2)^1/2

θ = arccos(x/r); y => 0
θ = 2pi - arccos(x/r); y < 0
θ = pi/2; x,y=0

or
θ = arctan(y/x); x,y => 0
θ = arctan(y/x) + pi; x < 0
θ = arctan(y/x) + 2pi; y < 0, x => 0
θ = pi/2; x,y=0

[noko No.6792217]

>>6792198

Which proves the point. That's a complete mess compared to the straight forward polar to Cartesian conversion.

Yeah, you can do it, if you filter for edge cases and make arbitrary guesses (for the 0,0 case).

[Anonymous No.6792439]

>>6792217
That is pretty straightforward.
You only have two different rules (for y =>0 and y <0) and just use the definition that the 0 vector is perpendicular to every other vector.

[Anonymous No.6792578]

>>6792439

a) It's not as simple as the p2c conversion (p2c has one set of equations, c2p has several equations)
b) It doesn't cover the whole plane (unless you consider guessing for the 0,0 case good practice, meanwhile p2c works for the whole plane)

In short; objectively c2p conversion is more complex and incomplete.

I think you misunderstood the point to begin with and have painted yourself into a corner arguing for something that wasn't in question.

I'd suggest it's the math telling us that he polar system is 'better' (more natural?) than Cartesian, which ties back to the OP question.

[Anonymous No.6792606]

>>6792578
>(unless you consider guessing for the 0,0 case good practice)
It's not a guess. That is the definition of it.

You call a vector perpendicular to another one if <span class="math">\langle v,u \rangle = [/spoiler], which is true if one of the vectors is the zero vector.

[Anonymous No.6792610]

>>6792606
<span class="math">\langle v,u \rangle = 0[/spoiler]

[Anonymous No.6792639]

>>6792606

Isn't it reasonable to want the conversion between polar and Cartesian to be fully reversible?

That is, you'd want:
(r,θ) = p2c(c2p(r,θ)) for all r and θ

It seems odd to be accepting of loss of information like this.

And still the original issue of the c2p equations being more complex than the p2c ones stands. That is undeniable.

[Anonymous No.6792663]

>>6790479
>tfw our sick cunt lecturer asked us to learn it by ourselves in physics 101

I will scan some of my notes for you soon, hold on

[Anonymous No.6792675]

>>6791067
That's a Riemann sphere.

[Anonymous No.6792740]

>>6792639
It is if you define it piecewise.

This happens because sine and cosine are only invertible on a certain interval and not on the whole domain.

You can do something like this (which probably can be simplified), if you want to write it in one line:
<div class="math">\theta = \frac{\pi}{2} - sgn[ |x|+|y| ]\frac{\pi}{2} + sgn[ |x|+|y| ]\cdot\left(\pi - sgn(y)\pi + sgn(y)\arccos\left(\frac{x}{r}\right)\right)</div>

<span class="math">r = \sqrt{x^2 + y^2}[/spoiler]

[Anonymous No.6792754]

>>6792639
It is if you define it piecewise.

This happens because sine and cosine are only invertible on a certain interval and not on the whole domain.

You can do something like this (which probably can be simplified), if you want to write it in one line:
<div class="math">\theta = \frac{\pi}{2} - sgn( |x|+|y| )\frac{\pi}{2} + sgn( |x|+|y| )\cdot\left(\pi - sgn(y)\pi + sgn(y)\arccos\left(\frac{x}{r^{sgn( |x|+|y| )} } \right) \right) </div>

<span class="math">r = \sqrt{x^2 + y^2}[/spoiler]

[Anonymous No.6792779]

>>6792754
Oops, made a mistake.
<div class="math">\theta = \frac{\pi}{2} - sgn( |x|+|y| )\frac{\pi}{2} + sgn( |x|+|y| )\cdot\left(\pi - sgn(y)\pi + sgn(y)\arccos\left(\frac{x}{r +1 - 1^{sgn( |x|+|y| )} } \right) \right) </div>

There you go!.

[Anonymous No.6792784]

ok i'm not reading all of the replies but i can tell you if you have a problem that is rotational symmetric, it is way easier to set up the equations in polar coordinates, because you only consider r, since phi is irrelevant because of symmetry

[Anonymous No.6792787]

>>6792754
Made a mistake.

<div class="math">\theta = \frac{\pi}{2} - sgn( |x|+|y| )\frac{\pi}{2} + sgn( |x|+|y| )\cdot\left(\pi - sgn(y)\pi + sgn(y)\arccos\left(\frac{x}{r + 1 - sgn( |x|+|y| ) } \right) \right) </div>

[Anonymous No.6792821]

>>6792754
It's not about it being a single line, it's the overall complexity of the procedure.

In this case your just using sgn to hide the edge case filtering and producing a monster of an equation (compared to the r*{sin,cos} of the p2c method). Might just as well use atan2, which itself just hides the complexity.

That's the point, conversion one direction is almost effortlessly easy, conversion in the opposite direction is a lot more complex, requiring you to check for awkward situations and accept the loss of information.

It's an odd imbalance. It looks like math is telling us something along the lines that Cartesian systems are more limited than polar.

[Anonymous No.6792835]

>>6792821
>In this case your just using sgn to hide the edge case filtering
If you have problem with piecewise functions, you might as well have a problem with the absolute values.

The conversion "simplicity" simply comes from the fact that you call:

<span class="math">\displaystyle \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i+1}}{(2i+1)!} = \sin(x) [/spoiler]
You could take that equation and call it <span class="math">ang(x,y)[/spoiler] and then it would look just as simple.

[Anonymous No.6792853]

>>6792835
Well yeah, the trig functions hide a lot of complexity of their own, but then you are using arccos, so add that on too, your methods are sill more complex.

I don't know why you're so insistent on trying to prove an obviously complicated procedure is as simple as the p2c ones, you can't, that's obvious. It's not a denial of the usefulness of Cartesian systems, just an observation.

[Anonymous No.6792924]

>>6792835
See that <span class="math">(-1)^i[/spoiler] on the sine:

That's simply another way of writing:
1, if i is even.
-1, if i is odd.

And that's all it does, switch the sign of the terms.

>>6792853
Well, you wanted a formula that worked for all cases.

But yes, converting back from polar to cartesian is harder, here we can do something like that.
On higher dimensions it's too difficult to work with polar, you simply can't visualize the angles.
And converting back to cartesian is a nightmare.

[Anonymous No.6793830]

>>6792787
And just noticed a problem with (1,0), oh well. but now it should work for all cases.

<div class="math">\theta = \frac{\pi}{2} - sgn( |x|+|y| )\frac{\pi}{2} + sgn( |x|+|y| )\cdot\left( sgn[1-sgn(x)+sgn(|y|)](\pi - sgn(y)\pi) + sgn(y)\arccos\left(\frac{x}{r + 1 - sgn( |x|+|y| ) } \right) \right) </div>