help-33

[hazy lion]

im going to take a walk, i think im being dense, can i ping u if i get something worked out?

[glass silo]

Sure thing

[marsh citrus]

@hazy lion Has your question been resolved?

[hazy lion]

Okay here's the thoughts

[elfin berry]

jan Niku

[hazy lion]

We can create a subspace, call it Z, which is the span of $x_0$: $$Z = { \alpha x_0 : \alpha \in \mathbb K }$$ and as $X$ is a vector space then $Z$ is for sure a subspace

[elfin berry]

jan Niku

jan Niku

jan Niku

jan Niku

jan Niku

[hazy lion]

then exists some linear functional g on X such that norm of g is equal to norm of f

but norm of f is norm of x_0

but norm of x_0 isnt 0

then g cant be equivalently 0

then g is missing from {0}

@glass silo how close tried to get it all on one page

[glass silo]

Happy with that

[hazy lion]

okay, thank you appreciate it

.close

[marsh citrus]

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[desert prairie]

for 5

[desert prairie]

can i do it right to left?

or only left to righr

[obtuse sluice]

I think its left to right... but idk they are the same actually

[sour burrow]

You can find the answer from the left side to right

also you can do the other way round which is easier

[desert prairie]

ok ty

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[tranquil elbow]

(non calc paper)

[marsh citrus]

Please don't occupy multiple help channels.

[tranquil elbow]

it becomes square root of 51

lxb

[elfin berry]

faiyrose

[tranquil elbow]

bro

i get it now

oml im dumb asf

i was doing it right i just didnt have to continue

i was finding the square root of x which I had to re square

thx for the help

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[main mica]

a car moves 60km/h. How many km/h does the car need to increase the speed so the car goes 1km half a minute faster

[small berry]

How long does the car take to go 1 km now?

[main mica]

1 min?

[small berry]

yes

So how long must it take to go 1km after it accelerates

[main mica]

1 min

[small berry]

No, I mean after it accelerates

[main mica]

Wdym after it accelerates

[small berry]

Right now it takes 1 min to cover 1km

[main mica]

It just gies 60km/h always

[small berry]

however, it needs to speed up no?

so it can go 1km half a minute faster

[main mica]

Yes

Ok

Soo

[small berry]

so after it speeds up (accelerates), how much time should it cover 1km in?

previously it was 1 min

[main mica]

2

[small berry]

No, the question says "so the car goes 1km half a minute faster"

If it currently goes 1km in 1 minute

[main mica]

Oh

1.5 * 1?

[small berry]

No

half a minute faster means

in half a minute less time

[main mica]

0,3

so x in 1.5mins

So i need x

Is x 2/3?

So 2/3 times faster?

@small berry

So 40km/h faster?

30km/faster

[small berry]

No, half a minute is 0.5 minutes. The car needs to cover 1 km in 0.5 minutes less. If it initially covered 1km in 1 minute, it needs to cover it in 1-0.5 = 0.5 minutes

[main mica]

Yes so it has to incrise the speed by 30km/h

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[rigid vigil]

Cylinder A and Cylinder B are similar right cylinders. The radius of Cylinder A is 4, and its surface area is dπ. The surface area of Cylinder B is tπ. What is the value of t-d ?
Volume of Cylinder A is 480π
Volume of Cylinder B is 12960π

[rigid vigil]

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[mental kayak]

I lost an argument against irrationals but I know he is wrong. He claims that irrationals do not exist, and gave this proof.

My friend said that

Have any “irrational” number a, and let n be a value so that a is a whole number.

We can represent this number as:
a10^n/10^(n+1)

This makes a rational number, since both the numerator and denominator are integers, and integers are rationals, so irrational numbers do not exist.

[mental kayak]

This looks correct and I wasn’t able to think of a way to contradict this proof.

[crystal lintel]

let n be a value so that a is a whole number

?

[mental kayak]

That’s what he said

[crystal lintel]

what the heck does that mean

[broken dome]

Assume this irrational number isn't irrational, therefore it isn't irrational

[mental kayak]

He’s basically trying to say turn an irrational number into an integer

I think

[broken dome]

So we start with assume a is irrational, then assume there exists an n that makes a whole number in the form of a10^n/10^(n+1)

ok, even if this number is whole

it doesn't change a

[mental kayak]

No he just means the numerator be a whole number

[broken dome]

this is a new number

not a

so a is still irrational

[mental kayak]

Wait what

Elaborate

[broken dome]

the number a10^n/10^(n+1) is not necessarily rational, why would you assume it to be?

if a is irrational then this number must also be irrational

therefore this number is completely irrational

this proof is irrational

what are you guys on about?

[mental kayak]

Irrationals being a real thing

I argued that irrationals were a thing and cannot be represented as a fraction

His proof basically is opposite of what I said

[broken dome]

Has he ever heard of the square root of 2 proof?

[mental kayak]

I don’t think so

Middle schoolers btw so o

[broken dome]

You know pythagoras died on that hill

[mental kayak]

what hill

[broken dome]

of there being no irrational numbers

[mental kayak]

Oh

[broken dome]

https://www.youtube.com/watch?v=X1E7I7_r3Cw

[mental kayak]

What the flip 😭

[broken dome]

This video has the proof for irrational numbers in it

it's also funny go watch it

[mental kayak]

Ok

[broken dome]

https://www.mathsisfun.com/numbers/euclid-square-root-2-irrational.html

also this

[broken dome]

also this expression simplifies to a/10 and since a is irrational this new number must be too so it is irrational

I'm done with this lol

[marsh citrus]

@mental kayak Has your question been resolved?

[marsh citrus]

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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

Show your work, and if possible, explain where you are stuck.

[cunning fiber]

I’d start by finding the relationship between the areas of the triangles

Then turn that into the ratio with the quadrilaterals

[cunning fiber]

Do you have an idea of how to do this?

Yk how if you have two similar figures

The ratio of their perimeters is the same as the scale factor

[cunning fiber]

I’m asking if you know this

<@&268886789983436800> troll

Let’s say you have two figures

If you multiply each side by the same thing

[cunning fiber]

Then you multiply the perimeter (the sum of all of the sides) by that same value

The idea is similar for finding the ratio of the areas

Let’s keep it simple: you have a base and a height

Both the base and the height are multiplied by that scale factor

Based on this, what can you guess abt the relationship between the areas of two similar figures?

[cunning fiber]

So …?

[midnight folio]

For similar shapes, the ratios between the "parallel" (i forgot the word, but corresponding or whatever) sides are all equal to a factor

[cunning fiber]

Just focus on what I said

We’ll get there eventually

Once you get that, it’s relatively easy to finish

[midnight folio]

how do you find the area of the triangle

you know the ratios, 1:2:3 right

given to you by the problem

the ratios hold true for every part of the triangles (except the angles)

momentarily ignore all triangles except the larger one

its area would be B x H / 2 right?

you dont need it

you're working with theoretical stuff here

BB x HB / 2 = Areas of Big Triangle (B is for Big, M will be for Medium and S for small)

you know that, the Base of the Big Triangle, and its height, have a ratio

in your case its 3:6 or 1:2

use the ratio and rewrite the areas of each triangle as the area based on the sides of one of them

[marsh citrus]

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[proper kiln]

How do you eval this limit?

[turbid mica]

write it as a summation first

[marsh citrus]

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[lost stratus]

not sure how to begin

[broken dome]

How would you approach this?

Oh wait sorry

Well it looks complicated but in simplified terms

what does this really say?

[lost stratus]

It wants the area of the square

[broken dome]

Rectangle to be precise

[lost stratus]

yeah

[broken dome]

Well let's go step by step, what does the first statement say

This one

[lost stratus]

it says that the function is given by h(x,y)

[still temple]

does any1 know how to read vernier calipers?

[broken dome]

I do, but this help channel is occupied go find your own

So we have this gamma which is basically a set of points right?

[vernal forge]

notice that your functions is linear

that would make the surface area a rectangle

[broken dome]

Gamma gives us points from (0, 0) inclusive to (2, 1) inclusive right?

[lost stratus]

mhm

[lost stratus]

mhm

[broken dome]

Going through all values

Okay so this set of points is the domain for function h right?

[lost stratus]

mhm

[broken dome]

This function is given by 2x - y + 1

[vernal forge]

this is the graph if it helps

[broken dome]

In order to get the surface area of the rectangle that results from this function we really only need to look at the edge cases

that is, the end points of the rectnagle

[lost stratus]

mhm

[broken dome]

the first is given by (0, 0) from the domain

another is given by (2, 1) right?

[lost stratus]

right

[broken dome]

What are the other 2?

[lost stratus]

0,1

2,0

[broken dome]

Good

Let's call those points, in order of writing in this chat, A, B, C and D

Calculate h for those points

[lost stratus]

1,4,0,5

[broken dome]

Good

Now you have points in space

in format of (x, y, h)

these are the points of your rectangle

So just calculate the distances between them and then calculate the surface

[lost stratus]

i got sqrt(2) for both

idk if thats correct

[broken dome]

Let me just check that real quick

Did you calculate all the disctances?

or only for 2 pairs?

[lost stratus]

for 2 pairs

and then multiply

is that wrong?

distance between (0,0) to (0,1)

and distance between (2,1) to (2,0)

[broken dome]

Ah, but you calculated the lengths of 2 opposite lines in the rectangle

You need 2 adjacent ones

[lost stratus]

oh

[broken dome]

Be careful not to hit a diagonal

Maybe sketch it out if it helps

[lost stratus]

it should be (0,0) to (0,1) and (0,0) to (2,0) right?

these two

[broken dome]

Yes, that works

[lost stratus]

then I get 2*sqrt(10)

[broken dome]

For the surface?

[lost stratus]

yea

[broken dome]

Seems about right

Just recheck your calculations

and that's it

[lost stratus]

hmm wel

what about this formula ?

[midnight folio]

i love your pfp

[broken dome]

Where did that come from?

what is that?

Oh, well

If you wanna take the integral approach you can

I don't know why you would but... go ahead...

Unless this is some super math basics course where you have to calculate area by definition and not by formula

[lost stratus]

yeah I kinda want

just not sure how

[broken dome]

Yeah, that's a bit above my level atm, sorry

[outer lodge]

Its called a surface integral

you first parametrize the surface

For your surface, a simple parametrization would be
[ \Omega : [0,2]\times[0,1] \rightarrow \mathbb{R}^3 ]
[ \Omega (u,v) = (u, v, 2u - v + 1) ]

[elfin berry]

shsgd

[lost stratus]

mhm

but this returns 2*sqrt(6)

which is different from before

[outer lodge]

can you send your working

[outer lodge]

what does norm return

,w cross product (1,0,2) (0,1,-1)

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by shsgd

[lost stratus]

norm is like sqrt(x_1^2+x_2^2+...+x_n^2)

crossp is cross product

[outer lodge]

yeah, well 2root6 is your answer then

[lost stratus]

ah okay

its just different from what the other dude siad

thanks

[marsh citrus]

@lost stratus Has your question been resolved?

[marsh citrus]

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[obsidian leaf]

the radius is whatever u want it start from

[fervent rampart]

we would have to choose where we make the linear speed measurements

since, in this scenario, their feet would be moving at a different linear speed than their heads

so the radius we consider is based on the speed measurement we consider

[marsh citrus]

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[delicate pond]

Quick question, what does the composite o thingy mean for matricies? Thanks 🙂

[still creek]

you can think of it as just matrix multiplication

basically you are finding T2 times that vector and then applying T1 to that

which is the same as multiplying T1 and T2 and then applying the product to the vector

[delicate pond]

sweet, so I got the vector (7 38 122) from multiplying T2 and the vector, so I would just multiply this new vector by T1?

[still creek]

yup

[delicate pond]

thanks for the help!

[marsh citrus]

@delicate pond Has your question been resolved?

[marsh citrus]

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[stark trail]

I really have no idea how to do this, I could use any help

[hushed egret]

correspondence theorem

[stark trail]

I haven't heard of that

[sand fable]

but now you have!!

[stark trail]

I don't think I can use it

[hushed egret]

true

use the ideas though

[stark trail]

What are the ideas

[hushed egret]

lift a normal subgroup of G/N to G

[stark trail]

I have no idea what you mean by lift

[crystal lintel]

i thought you worked out?

[stark trail]

not much group theory in the gym

[hushed egret]

worked out what? the problem?

[crystal lintel]

no i don’t think he worked out the problem yet

[stark trail]

I have not

Pls help

[hushed egret]

ok do the easy direction first then

[stark trail]

Which is what direction?

[hushed egret]

let K be an intermediate normal subgroup

show that G/N isnt simple

[stark trail]

Okay

thanks

.close

[marsh citrus]

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[errant loom]

can someone help me how to factor this its for trig identities,

[dawn spear]

break 8sin^2x into 4sin^2x + 4sin^2x

[summer trench]

you can convert cos2x to 1-sin2x

and then let sinx = t

and solve it like a norma quadratic

[dawn spear]

after this you get 4sin^2x + 4 = 7

sin^2x = 3/4

sin x = root3/2

[errant loom]

aight hold on, lemme try

[errant loom]

how is that possible

[summer trench]

$\cos^2{x}+\sin^2{x}=1$

[elfin berry]

yajat

[summer trench]

$\cos^2{x}=1-\sin^2{x}$

[elfin berry]

yajat

[errant loom]

oh i see, its that rule

kk

[summer trench]

that's just an identity

[errant loom]

yh

[summer trench]

idk how u dont know about this if you're solving such questions

[errant loom]

what happened to the other "4sin^2x and 1-sinx

[dawn spear]

4sin^2 x + 4sin^2 x + 4cos^2 x = 7
4sin^2x + 4(sin^2x + cos^2x) = 7

[errant loom]

then after i get
sin^2x=3/4 do i square rooth it

[marsh citrus]

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[unkempt goblet]

A company produces widgets with an average weight of 50 grams and a standard deviation of 8 grams. A quality control inspector randomly selects a sample of 36 widgets.

Is there an error to the problem

[unkempt goblet]

My calculated z score is - 10.5

Or I did smth wrong?

[marsh citrus]

@unkempt goblet Has your question been resolved?

[bleak thistle]

$$ z = \frac{50 - 50}{\frac{8}{\sqrt{36}}} = \frac{0}{\frac{8}{6}} = 0 $$

[elfin berry]

misha

[marsh citrus]

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[quick hollow]

I don’t have any work for this question because I’m unsure where to start as it’s a written question and I suck at them. The question starts like this: A small radio transmitter broadcasts in a 45-mile radius. If you drive along a straight line from a city 63 miles north of the transmitter to a second city 62 miles east of the transmitter, for how much of the drive in miles will you be able to pick up a signal from the transmitter? Round your answer to 4 decimal places.

[late geode]

start by drawing a pic

[quick hollow]

@late geode

At me if you come back

[late geode]

@quick hollowpoor pic, not enough of the info represented

that's a very precise representation

the lengths and radius should be indicate on your diagram
as well as the path the car

[quick hollow]

It just occurred to me that the car can’t be going negative miles so it has to stay in the first quadrant. I assumed the cars path started at the top of the circle then went to the bottom my bad. Do you want me to draw that representation?

@late geode

[late geode]

the car doesn't start on the circle

the circle represents the range of the broadcast
the car start starts outside of that

[quick hollow]

Ik, the circle only has a 45d radius I’m just saying I thought the line started at 63 on top of the circle and went to 62 under the circle.

But that doesn’t really matter, now that ik what the problem actually looks like, how do I solve it?

[late geode]

find the intersection points

between the line and the circle

[quick hollow]

How

Is there an equation I can use?

@late geode

[late geode]

make equations based on the cooridnate system

[quick hollow]

Treat me like I’m learning this for the first time, whst is the coordinate system? @late geode

[late geode]

coordinates of the xy plane

[quick hollow]

Btw tell me if you want me to stop pinging you I just don’t know if your seeing my msgs

And how do I make the equations, all ik is the radius and where the line starts and ends.

@late geode

[late geode]

stop pining me

do you know the equations of circles?

[quick hollow]

Ik 2 but I’m assuming your talking about this one?

If you are then I think I got it I’ll try the rest by myself.

[late geode]

no

that's the quadratic formula that'll you need later

[quick hollow]

This one?

[late geode]

still no

well its related to the equation for a circle, but not quite what i want from you

[marsh citrus]

@quick hollow Has your question been resolved?

[quick hollow]

Then I don’t know

[late geode]

look up equaiton of a circle

[quick hollow]

?

[late geode]

yes

you have you radius r
and for convenience, have your station located at the origin(0,0)

[quick hollow]

Ok

So (0-h)^2 + (0-k)^2 = 45^2

[late geode]

no

you shuld be replacing (h,k with (0,0)

[quick hollow]

Ok

What do I do with that equation

[late geode]

now you'll need to get your equation for the path of the vehicle

i.e. the equation that passes through (63,0) and (0,62)

[quick hollow]

Is it a different formula than the ones I’ve already showed?

[late geode]

yes

[quick hollow]

Well I have to go. Does this (help 33) stay open till tmrw?

[late geode]

its going to time out, just repost when you come back

if you know you're going for an extended amount of time, just close

[quick hollow]

Alright good bye

[marsh citrus]

@quick hollow Has your question been resolved?

[marsh citrus]

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[cerulean oxide]

why does this give me the wrong answer

[outer lodge]

Better off using DeMoivres theorem I believe

[ z^n = r^n(\cos n\theta + i\sin n\theta ) ]

[elfin berry]

shsgd

[late geode]

use is with how you're using arctan

due to its range

you'd need to consider the location of your complex number and perform a quadrant shift if needed (which it is here)
essentially use what's called arctan2

[valid kestrel]

Does the range have to be from -π to π?

[late geode]

not necessarily

some places use 0 to 2pi for the principal argument

the range of arctan being from -pi/2 to pi/2 is insufficient regardless

[valid kestrel]

How do you tell when a quadrant shift is needed?

[late geode]

plot your point on the argand diagram/plane or otherwise to determine which quadrant its in

[cerulean oxide]

here it is i - 1

so it is in the second quadrant

[valid kestrel]

Right

So if it's in the second quadrant, why can't you still use arctan(b/a)

[late geode]

because note that in your work, your result from arctan is -pi/4

which is in Q4

[valid kestrel]

Oh

Ok

Right

So how would you fix that then?

[late geode]

consider the periodicity of tan
identify
tan(what angle in Q2) = -1

[cerulean oxide]

what even is your question jonnyboy

you are in the one i asked

make your own, it is easier

[valid kestrel]

3pi/4

[valid kestrel]

Sorry my bad

[late geode]

didn't realise who's channel this was

[cerulean oxide]

so we are here

[late geode]

yeh

[cerulean oxide]

what formula do i use

[late geode]

consider the periodicity of tan
identify
tan(what angle in Q2) = -1

[cerulean oxide]

3pi/4

i think

[late geode]

yes

[cerulean oxide]

ok, but dont i sitll need you use the formula tan^-1(b/a)

[late geode]

not really

if you want to explicitly use that,
you'd perform a quadrant shift here by adding pi to that result to get something in Q2

[cinder quartz]

Hi im so happy im back

[late geode]

how you do it, whether a shift is needed will depend on what you have

as a basic trig question you're essentially looking for theta in (-pi,pi] where
tan(theta) = -1
sin(theta) > 0
cos(theta) < 0
and use whatever method you want to find that

[cerulean oxide]

but

this still isnt quite right

[late geode]

why -3pi/4

[cerulean oxide]

that is the angle

[late geode]

no?

[cerulean oxide]

how no

[late geode]

how are you getting -3pi/4 for the angle

[cerulean oxide]

the picture i made

[late geode]

yeh and you said 3pi/4 earlier

which i said was correct

you've now changed it to something incorrect

[cerulean oxide]

oh, yeah

mb

yeah i works now

[marsh citrus]

@cerulean oxide Has your question been resolved?

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[tight meteor]

I need help determining the magnetic moment of a spinning charged sphere

[marsh citrus]

@tight meteor Has your question been resolved?

[loud kernel]

you need to explain your question a lot more than that

!status

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tight meteor]

Well uh

I'm given the radius, angular velocity, and charge uniformly distributed acrosss the volume

[tight meteor]

7 I guess?

I have a vague idea but I'm not quite sure

[loud kernel]

well you need to post the actual question haha

[tight meteor]

Alright hold on let me translate it and type it out

Determine the magnetic moment $\mu$ of a solid sphere of radius $r$, and charge $q$ distributed uniformly across its volume, with an angular velocity $\omega$ about its diameter.

[elfin berry]

jewels!

[tight meteor]

Is that better?

[loud kernel]

yes, but i have no idea how to solve this one sorry, i think ill be learning it soon, but now that the question is more visible, people who know will now try to help you

[tight meteor]

It's alright, I appreciate your time!

[loud kernel]

also, if you dont get anyone to help, there is a physics server which will be a better place to put it

[tight meteor]

It's pretty dead 😭

I was helped with electromagnetism last time here by an indivudal in college

[loud kernel]

yea

[tight meteor]

If they show up again I'll be lucky :)

[loud kernel]

im learning electromagnetism rn but im in year 12

i got a test on friday

[tight meteor]

Good luck with that!

[loud kernel]

havent done moments + magnetism rn so im sorry

[tight meteor]

Haha don't apologize its alright

[loud kernel]

maybe try calculating the force from the magnetic field

and do the torque formula

[still temple]

Year 12?

[loud kernel]

"grade 12?" i dont know, im in australia

[still temple]

Is it undergraduate?

[loud kernel]

yes

last year of highschool

[still temple]

I don't think there is electromagnetic momentum in my specification (A level)

[loud kernel]

yeah, im not doing electromagnetic momentum, OP is

[still temple]

Give us some equations from your book

[still temple]

What induce the motion?

[tight meteor]

What equation?

[tight meteor]

Uh nothing, it's just spinning

[still temple]

But is force applied to charge?

[tight meteor]

It's not in a magnetic field so I assume not

[still temple]

Do you know Biot-Savart

should be a direct triple integral to compute

[tight meteor]

yay the cavalry has arrived

[tight meteor]

I do but how would I use it here

[tight meteor]

out of my scope 😭

[still temple]

https://youtu.be/VvoXhQbFBUI?feature=shared

I found it easy

[tight meteor]

What I thought was $dq = \rho 4\pi x^2 \dd{x}$

[elfin berry]

jewels!

[still temple]

Onky thing I don't know is how to derive expression for infinitely small volume dt but who cares

[tight meteor]

doesnt that feel like cheating

I just need some guidance

[still temple]

What kind of cheating? That's just proof you suppose to know, by using what you have been taught.
For me it is not cheating because I haven't new what is magnetic even is

But once you define variables precisely it just trivial

[tight meteor]

I'm supposed to derive it

[tight meteor]

and then $di = \frac{dq}{T} = \frac{\omega dq}{2\pi}$

[still temple]

dt only important part to define infinitely small volume on a sphere

[elfin berry]

jewels!

[tight meteor]

bruh finally

[tight meteor]

yeah im doing that

i know it goes down to elements

[tight meteor]

and I guess then $d\mu = Adi$ but i dont know what A would be

[elfin berry]

jewels!

[still temple]

I don't know maybe you can use left hand rule + definition of angular momentum to derive first equation

[tight meteor]

angular momentum?

[still temple]

Yeah maybe. I mean look you have current passing through the area of a sphere inducing rotation at particular strip of a sphere this strip define circle since sphere solid there is charges on this strip then Area * charge density over time * volume which have this charge density

[tight meteor]

yes that is what im doing

I dont know what the area would be

[tight meteor]

here

[still temple]

I think first equation do no need prove, and dt component to describe infinitely small volume is also trivial approximation, and maybe you can just quot it?

[tight meteor]

ugh ill come back to this later when my brain has cleared up

[still temple]

Area will be just pir^2
But r is not radius of a sphere

[tight meteor]

right its the area of the enclosed loop

which varies with distance

[still temple]

So you should multiply radius of sphere by angle sinx to find the r. Because you make angle x with r and radius of a sphere upon touching required circle. Just do right angle triangle there for horizontal conponent

[tight meteor]

but for the shell element that I do have it would be 4pir^2 for all the current

[tight meteor]

wha

what sin(x)

[still temple]

Give me a minute

[tight meteor]

$d\mu = 4\pi x^2 \frac{\omega dq}{2\pi}$

[elfin berry]

jewels!

[tight meteor]

$d\mu = 4\pi x^2 \frac{\omega 4\pi \rho x^2 \dd{x}}{2\pi}$

[elfin berry]

jewels!

[tight meteor]

uhh

rho = q/V

[tight meteor]

okay uh I have

[still temple]

Then define current as Q/T
Where T=time period
For w=2pif
f-frequency=1/T
Therefor I= wQ/2pi

[tight meteor]

thats what im doing yes

[tight meteor]

.

$\mu = \frac{6q\omega}{r^3}\int_0^r x^4 dx$

[elfin berry]

jewels!

[tight meteor]

Is this right?

[still temple]

Maybe I can't read this notations

Define derivative of I then Area separately

Then you can combine them

[tight meteor]

this is the simplified expression

im waiting for qylo to approve it or something 😭

[still temple]

Yeah I know.

Have you defined dQ as pdt?

[tight meteor]

pdt?

[still temple]

P=Q/volume of sphere

[tight meteor]

oh

yeah

rho

dt I assume is volume for some reason

[still temple]

Yeah dt is infinitely small volume of a point where rotation occurs

[tight meteor]

idk i used spherical shell elements

[still temple]

Like rotation occurs at the strip of a circle but in reality there is infinitely small vertical component

[still temple]

I mean just compare with video, there is no cheating in video as long as you can remember the proof because it is trivial and not a problem solving

[tight meteor]

😭 man I need to derive it myself

[still temple]

Yeah you do by knowing a couple of identities which are used in the video, otherwise you can't derive it from scratch

[tight meteor]

but I did

??

[still temple]

So if you can't remember required identities just look them up, that's what this video is doing

[tight meteor]

here

[tight meteor]

why is he using 3 integrals

what

[still temple]

I don't know, I am not pr0 either I just can explain steps done on the video, so I can't really do much of a rearangments

[still temple]

Just multiplication :/

You know what makes function integradable?

Terms like dt, dx, ds

If you have them separately just multiply out :/

That's pretty simple actually

I mean it is really nice proof, really clear unlike those maths proofs

[tight meteor]

i havent been taught stuff like that before

i doubt thats the way im meant to do it

[still temple]

You mean that you can distribute integral?

[tight meteor]

anything involving more than one integral

[still temple]

look if you have
dy/dx=y/x
Find y

[tight meteor]

y = mx

[still temple]

Just isolate variables and integrate both sides

It is similar here because you have dt, dx terms

Alright so what is dt
Represent if t is time?

[tight meteor]

uh

ill come back to this later with a fresh mind honestly

my brains hurting

[still temple]

Alright Integral of dt =dt because integration reverse differentiation, so you need to have this dt term under integration

[tight meteor]

@still temple You seemed to have been typing for a while 😅 I don't want any of it to go to waste I'll be back in like an hour save it for then

[still temple]

Is it the only question you have?

[tight meteor]

integral of dt = dt what

[tight meteor]

for now yes

the only one left

[still temple]

Ah sorry integral of dt =t

Typi

Typo

[tight meteor]

I thank you for your time as well

I'll be back later

.close

[marsh citrus]

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[marsh citrus]

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[tight meteor]

@still temple I checked the answer with a friend of mine and I think what I did was right

[tight meteor]

Although I'd still be interested in listening to what you've got to say

[still temple]

what did u do? i forgor

[tight meteor]

I used spherical shell elements

[still temple]

also my explanation was more or less building up the intuition to forming the triple integral in spherical coordinates

[still temple]

yeah

[tight meteor]

I didn't have no angles involved though

I defined $dq = \rho dV$ where $dV = 4\pi x^2 dx$

[elfin berry]

jewels!

[tight meteor]

And then $d\mu = Adi$ where $di = \frac{dq}{T}$, and since $T = \frac{2\pi}{\omega}$ we end up with $d\mu = (4\pi x^2 ) \frac{4\pi\rho x^2\omega dx}{2\pi}$

[elfin berry]

jewels!

[still temple]

right

how did u get dx tho

[tight meteor]

volume element?

surface area times the thickness

[still temple]

what are u using x to denote

the radial component?

[tight meteor]

yeah

[still temple]

ok thats cursed notation

[tight meteor]

x is the distance from the center

[still temple]

use dr instead

[tight meteor]

sorry im used to it 😭

[tight meteor]

could have but the radius of the shell is r so

[still temple]

dr'

[tight meteor]

🙄

[tight meteor]

okay 😔

[still temple]

ok but yeah carry on

[tight meteor]

yeah so thats the volume element

the only part that im still a little doubtful is the value of A

I just chucked in 4pi r^2 because it felt right??

and it is right

[still temple]

i mean

[tight meteor]

:/

[still temple]

the surface area of a spherical shell is [
4\pi r'^2
]

[still temple]

so whats the issue

[tight meteor]

look how ugly r' looks with an exponent

i would never

[still temple]

[tight meteor]

yeah but

the magnetic moment of a loop has the area be that enclosed by the current

if I had a loop, the moment would be iA where A is the area of the loop

in this case though is the current really "enclosing" the surface area of the sphere?

[still temple]

hmmm

[tight meteor]

I think

every shell can be deconstructed into an infinite number of rings

and the total area enclosed by those rings perhaps?

and magically that is the surface area of the sphere 😭

[still temple]

i mean i am not sure about the theory of it rn but i am pretty sure the two situations between just having a loop and a solid sphere are not completely synonymous in their physical meaning here

[tight meteor]

I actually understand that

Because a bar magnet's magnetic moment is defined analagous to that of a loop

Despite having no "area"

[still temple]

also i can see what u did being purely mathematical in terms of ease. Because basically setting up the triple integrals i mentioned involves creating differential volumes of "loops" in the sphere

(so instead of dV being a spherical shell concentric to ur actual sphere, it is a sort of plate-ish circular volume which u integrate over)

[tight meteor]

I see

[still temple]

like uh

[tight meteor]

It's alright if I'm too dumb for it 😂

[still temple]

hard to draw with a mouse

but ont he right is what u did more or less

[tight meteor]

yeah

[still temple]

on the left is the other method that gets u the loops and the triple integrals

i mean i guess both are equivalent at the end of the day but the first is more reasonable in like

the loop idea u mentioned

[tight meteor]

yeah was gonna say

it maintains the intuition

[still temple]

yes sadly hs EM is really constrained in what it can explain

[tight meteor]

It's alrighttt

I appreciate the effort

If there's nothing left to say, I'll close the channel?

[still temple]

yeah sure

good luck on ur studying adventures

[tight meteor]

you might wanna turn that off btw haha

[tight meteor]

And you!

.close

[marsh citrus]

Channel closed

Closed by @tight meteor

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[still temple]

thats a thing? damn

[tight meteor]

It is for me 😭

[marsh citrus]

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[novel juniper]

if $H(x) = max {f(x),g(x)}$ where $ f C \in [a,b]$ is $H \in C[a,b]$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

wait

butchered the notation

[upbeat sable]

Also, put \s before the { and }

[novel juniper]

if $H(x) = max {f(x),g(x)}$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

where $g,f \in C [a,b]$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

then is $H \in C [a,b]$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

now I feel no

as f(x), g(x) can be continuous

without H being continuous

[upbeat sable]

Can you give an example?

[upbeat sable]

Dont worry about giving exact representations for f and g, just a picture that proves this statement

[novel juniper]

uh

I feel such a function will exist

[upbeat sable]

You will see why H must also be continuous, and then we can formulate a formal argument

[novel juniper]

just a minute

$sin(x), tan(x)-1$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

on $[0 , \frac{\pi}{2}$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[upbeat sable]

And what is the point of discontinuity here?

Also, tan(x)-1 isnt continuous on [0,π/2]

[novel juniper]

oh, because of the asymptote at $\frac{\pi}{2}$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

ok

[novel juniper]

is this supposed to be an easy problem, feel like I'm missing something obvious

[upbeat sable]

You feel that there are continuous f and g such that H is discontinuous, so try drawing the graphs of f and g which give you discontinuous H

You will see why H must be continuous (|| intermediate value theorem ||)

[novel juniper]

I don't get what IVt has to do with this

[upbeat sable]

The only points that can be candidates for the discontinuity of H are when H switches from taking the values of g to taking the values of f (or vice versa). And that switch can happen only at points where f=g, so H must be continuous. If that makes sense

It will become very obvious if you draw a picture(and try to construct a counterexample), which is why ive been advocating for you to do it

[novel juniper]

wait, why must $f(x)=g(x)$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

I mean for f(x)=x^2, g(x)=x^3

this makes sense

[glacial hedge]

did you draw anything yet?

[novel juniper]

yeah

[novel juniper]

teh graph for this

$Max { x^2, x^3}$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[glacial hedge]

why does that one work?

[novel juniper]

it's continuous at x=1

which is where the function change

[upbeat sable]

I hope this picture makes clear what i was trying to say

[novel juniper]

oh

[glacial hedge]

H is a piecewise mix of f and g
it's slightly cumbersome to iron out the details, but the idea is simple

[novel juniper]

right,makes sense

wait, what if g(x) starts above f(x)?

[glacial hedge]

no difference, it's symmetric wrt f and g

[novel juniper]

but then H won't be continuous

[upbeat sable]

Not sure what you mean

[novel juniper]

wait, let me try to sketch a function

something like this

[upbeat sable]

Can you draw a diagram like this? What you sent rn doesnt make much sense

[novel juniper]

how do I do that on my computer

[upbeat sable]

Ms paint

[novel juniper]

I don't have paint

mac

wait,let me try doing it on my phone

[upbeat sable]

F and g must be defined for the whole interval

Here f and g are defined for different intervals

[novel juniper]

ah

that's what I was missing

got ti

*it

seems so trivial now

D

thanks

.close

[marsh citrus]

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[proper kiln]

How should I approach such problems? Is there a trick?

[versed hatch]

draw it out

do some doodling

usually it works

[marsh citrus]

@proper kiln Has your question been resolved?

[marsh citrus]

@proper kiln Has your question been resolved?

[proper kiln]

.close

[marsh citrus]

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[void elm]

does this suffice?

[winter smelt]

the proof looks good. Since we assumed |x| >= |y|, you can drop the outermost modulus after the first equality, but it's fine either way

[void elm]

thanks

.close

[marsh citrus]

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[novel juniper]

quick question, couldn't you just use the result from C here

[void elm]

.reopen

[marsh citrus]

✅

[void elm]

C?

[novel juniper]

Compelx numbers

[void elm]

but this is R^n

[novel juniper]

oh

right

sorry

[void elm]

np problem haha

.close

[marsh citrus]

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[hallow edge]

can u show me different ways u can type this equation

[hallow edge]

help

[devout mauve]

wdym

[hallow edge]

@devout mauve

[devout mauve]

ok?

you can divide by b to get $\overline a = \frac{\sqrt3}{b}$

[elfin berry]

Denascite

[devout mauve]

is that what you are looking for?

[left wadi]

what the dash above a mean?

[marsh citrus]

@hallow edge Has your question been resolved?

[hallow edge]

sqrt of 3

[marsh citrus]

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[foggy pier]

Which part are you stuck on?

[orchid zealot]

a)

why is a) 4 not -4?

4+p=0
p=-4

[loud stratus]

you don't need to solve an equation there

[orchid zealot]

i didn't solve?

i just sub in (4,2)

[loud stratus]

the function is already in his canonical form

[orchid zealot]

(4+p)=0
p=-4

[loud stratus]

where (p, q) are the vertex coordinates

and here vertex is (4, 2)

[loud stratus]

where does = 0 come from

[orchid zealot]

[loud stratus]

they're giving you the function in its vertex form

where p is the x coordinate of the vertex

and q the y coordinate of the vertex

and since you already know p and q, you only need to state their values

[orchid zealot]

.close

[marsh citrus]

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[marsh citrus]

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[drowsy bane]

can anyone help me with this one

[drowsy bane]

why is this wrong solution

[storm dagger]

im so confused

ah nvm

you forgot about the differential after substituting

m = z^3 => dm = 3z^2 dz

are you sure they asked u to solve the integral? this is pretty hard

[drowsy bane]

i dont remember the problem description exactly but i think so

[drowsy bane]

wait so

it would be (tanm)^-1 x 3z^2 ?

on the second line

[storm dagger]

no

you cant integrate m with respect to z. You need the entire integral to be in terms of one var iable in order to integrate

[drowsy bane]

so how would i write the differential in terms of m

[storm dagger]

$m = z^3 => z = m^{\frac{1}{3}} => dz = \frac{1}{3m^{\frac{2}{3}}}dm$

[elfin berry]

Obotron

[storm dagger]

as i said earlier this integral is hard. unfortunately, substitution is not the way to go about it. it''s more likely the original problem involves differentiating the integral with respect to x

[drowsy bane]

okay thanks a lot though

.close

[marsh citrus]

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[marsh citrus]

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[jagged zealot]

Hello, can somebody teach me how to solve this? Thank you!

[jagged zealot]

The answer key says the answer is 86.60 but I just don't understand how they got there.. I need the solution

[tight meteor]

What function relates the opposite and hypotenuse?

[jagged zealot]

sine

[rough trellis]

Oh hey im seeing this type of questions often today that is nice

Math 9 innit?

[jagged zealot]

yeah

[rough trellis]

i recently got this problem, if u need help somewhere ask

[jagged zealot]

I wanna know how they got to 86.60 m

[rough trellis]

simple

A kite string is 100 meters long, and it makes an angle of 60°

[jagged zealot]

is it sin 60° = 100/AC?

[rough trellis]

its to find height

ur looking for AB

[jagged zealot]

oh

[rough trellis]

From my point, Hypotenuse and Adjacents could be remembered as lengths or distance of an object

And opposites are just Heights

u got it?

[jagged zealot]

opposite

OHHHHHHHHHH

II GET IT NOW

i just misunderstood the question this whole time

sin 60° = h/100
100 sin 60° = h
h = 86.60

thank u

so much

.close

[marsh citrus]

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[jagged zealot]

ur amazing

[marsh citrus]

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[gaunt gale]

LOGIC AND PROBABILITY PROOF
Is this proof valid or does it contain some mistake? see #help-35 message for context.

1. P(A|B&C) ≈ 1 [Premise 1]

2. P(A|B&C) = P(A&B&C)/P(B&C) [Definition of conditional probability]

3. P(A&B&C)/P(B&C) ≈ 1

4. P(A&B&C) ≈ P(B&C) [Result 1]

5. P( ~A | ~B) ≈ P( ~A & C | ~B) ≈ 1 [Premise 2]

6. P( ~A | ~B) = P( ~A & ~B)/P(~B) [Definition of conditional probability]

7. P( ~A & ~B)/P(~B) ≈ 1

8. P( ~A & ~B) ≈ P(~B)

9. P( ~A & C | ~B) = P( ~A & C & ~B)/P(~B) [Definition of conditional probability]

10. P( ~A & C & ~B)/P(~B) ≈ 1

11. P( ~A & C & ~B) ≈ P(~B)

12. P(C|A&B) ≈ 0 [Premise 3]

13. P(C|A&B) = P(A&B&C)/P(A&B) [Definition of conditional probability]

14. P(A&B&C)/P(A&B) ≈ 0

15. P(A&B&C) ≈ 0

16. P(A&B&C) ≈ P(B&C) from Result 1

17. P(A&B&C) ≈ 0 from Result 15

18. P(B&C) ≈ 0

19. P(~B) = P(A&B&C) + P( ~A & C & ~B) + P( ~A & ~B) [Law of total probability]

20. P(A&B&C) ≈ 0 from Result 15

21. P( ~A & C & ~B) ≈ P(~B) from Result 11

22. P( ~A & ~B) ≈ P(~B) from Result 8

23. P(~B) ≈ 0

24. P(B|~A&C) = P( ~A & B & C)/P( ~A & C) [Definition of conditional probability]

25. P( ~A & B & C) ≈ 0 [Given P(~B) ≈ 0, and sub-events of ~B must be ≈ 0]

26. P(B|~A&C) ≈ 0

Therefore, we have proved ¬(P(B|~A&C) ≈ 1) using the given premises in the style of a logical proof.

[marsh citrus]

@gaunt gale Has your question been resolved?

[gaunt gale]

**LOGIC AND APPROXIMATE PROBABILITIES (slightly advanced) with additional premise (in p2.) **
*the problem: *
I want to prove ¬( P(B | ¬A & C) ≈ 1 )
from the premises (probably not all parts of all premises are required)

• P(A | B & C) ≈ 1 ≈ P(A | B) [premise 1 (hereafter, p1.)]
• P(~A | ~B & C) ≈ 1 ≈ P(~A | ~B) ≈ P( ~A & C | ~B) [p2.]
• P(C & A | B) ≈ 0 ≈ P(~C & ~A | ~B) [p3.]
• P(C | A & B) ≈ P(C | A & ~B) ≈ 0 ≈ P(C | A) ≈ P(~C | ~A) [p4.]
• P(C | ~A & B) ≈ P(C | ~A & B) ≈ 1 ≈ P(C | ~A) ≈ P( ~C | A) [p5.]
  we don't have precise numbers, just approximate one's.
  can this be done? If not, what other premise would be required to prove this? Is this set of premises consistent?

[marsh citrus]

@gaunt gale Has your question been resolved?

[marsh citrus]

@gaunt gale Has your question been resolved?

[gaunt gale]

.close

[marsh citrus]

Channel closed

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Use .reopen if this was a mistake.

[gaunt gale]

.reopen

[marsh citrus]

✅

[gaunt gale]

LOGIC/PROBABILITY PROOF
there are 15 premises (you probably only need 4 of them tho)

1. P(A | B & C) ≈ 1
2. P(A | B) ≈ 1
3. P(~A | ~B & C) ≈ 1
4. P(~A | ~B) ≈ 1
5. P( ~A & C | ~B) ≈ 1
6. P(C & A | B) ≈ 0
7. P(~C & ~A | ~B) ≈ 0
8. P(C | A & ~B) ≈ 0
9. P(C | ~A & B) ≈ 1
10. P(C | ~A & ~B) ≈ 1
11. P( ~C | A) ≈ 1

we want to conclude that ¬(P(B | ¬A & C) ≈ 1) it's very important that these are approximate premises. This for example, means that you cannot use P(C | ~A) ≈ 1 to mean if ~A then C, that makes an assumption that you would need to support, i.e. P(C | ~A) ≈ 1 ≠ P(C | ~A) ≈ 1. Curious to hear what you come up with. Ping me if you want some preliminary results

<@&286206848099549185> anyone here who knows a bit about logic/probability theory? I really need some help, im super stuck....

[gaunt gale]

1. P(ABC) ≈ P(BC)
2. P(AB) ≈ P(B)
3. P(~A~BC) ≈ P(~BC)
4. P(~A~B) ≈ P(~B)
5. P(~A~BC) ≈ P(~B)
6. P(ABC) << P(B)
7. P(~A~B~C) << P(~B)
8. P(A~BC) << P(A~B)
9. P(~ABC) ≈ P(~AB)
10. P(~A~BC) ≈ P(~A~B)
11. P(A~C) ≈ P(A)
    this is in a different format

Along with the proposition to prove:
P(~ABC) << P(~AC)
(note that I just removed the & signs for brevity, ~AB means ~A & B, not ~(A & B))

@knotty trellis are you still interested in this problem? I got a more advanced and probably provable version

[knotty trellis]

I'm interested, but my free time is quite limited in present days :/

[marsh citrus]

@gaunt gale Has your question been resolved?

[marsh citrus]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[onyx sentinel]

Can someone walk me through step by step on how to do question 11 im so very lost 😭😭🙏🙏🙏🙏

[stark trail]

(a+b)/c = a/c + b/c

[onyx sentinel]

ok now i have 1 - 1^n/n

where do i go from there>

?

[stark trail]

what do you think

[onyx sentinel]

im not really sure because i dont really understand any of the steps im supposed to take for this problem

like i know i need to find the limit as n approaches infinity

but i dont know why

or like what doing that does

[stark trail]

you don't need to do that

[onyx sentinel]

oh

[stark trail]

you need to determine if the sequence is bounded, monotonic, or both

do you know what those mean

[onyx sentinel]

bounded means like it cant go above or below a certain value?

monotonic means its either increasing or decreasing

[stark trail]

yes but to be more specific

the sequence {a_n} is bounded if |a_n| <= C for a constant C, for all n

agreed?

[onyx sentinel]

why is it only less than or equal to?

[stark trail]

if it's less than or equal to C

it's strictly less than C+1

or C+ (anything positive)

so it doesn't matter

[onyx sentinel]

ok

alright agreed

[stark trail]

Okay so try determining if the sequence is bounded then

[onyx sentinel]

oh

is it

[stark trail]

i.e determine the maximum value of |a_n|

[onyx sentinel]

it has to be greater than 0?

and less than or equal to1

[stark trail]

this sequence is not less than or equal to 1, as n=2 is 1 + 1/2 = 3/2

[onyx sentinel]

oh

[onyx sentinel]

would it be less than 2?

[stark trail]

Why do you think that?

[onyx sentinel]

wait sorry one sec isnt this supposed to be 1 - 1/2?

[stark trail]

and also be specific what you mean by "it"

[stark trail]

(-1)^(2) = 1

[onyx sentinel]

ohhh

[onyx sentinel]

would it be less than 3.2

3/2

sorry idk the proper way to say it

like the sequence is bounded by 3/2?

[stark trail]

I gave the proper way to say it here

and if you think it's bounded by 3/2, you should try proving why

you'd have to prove
|a_n| <= 3/2 for all n

[onyx sentinel]

is there a formula or something i can use to prove that?

[stark trail]

I just told you

|a_n| <= 3/2 for all n is what you must prove

start by replacing a_n = 1+ (-1)^n/n

then what

[onyx sentinel]

ok now i have (-1)^n <= 1/2n

[stark trail]

?

[onyx sentinel]

like i subtracted both sides by 1

[stark trail]

(-1)^2 = 2 which is not less than 1/2(2) = 1/4

You need to pay more attention to what I'm saying

[stark trail]

reread from here

[onyx sentinel]

Ok

ok so you tell me to replace a_n