help-49

[cedar mason]

!occupied .

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[cedar mason]

also what.

#math-discussion

[dark locust]

.close

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[dark wharf]

I need to find a counting problem that proves this combinatorical identity

[lethal path]

if you want to choose n people from m + n + 1 people

[dark wharf]

That's what I thought but I have no idea how to prove the right hand side is a valid solution

[lethal path]

you can choose 0 people from m people, then there is only one way to choose n + 1 people from n + 1 people

or you could choose 1 person from m + 1 people, then there is also only one way to choose n people from n people

the total number of people is always m + n + 1

[dark wharf]

But then it's (m+1 choose 1) times (n choose n-1), isn't it? Because there are multiple ways to choose the n-1 people out of n

[lethal path]

so you just divide them into two groups

wait yeah there's +1 stuff going on

[dark wharf]

After choosing the k people we are left to choose n-k out of n-m-k+1

[lethal path]

and the big group is split into (m, n + 1)
then (m + 1, n)
then (m + 2, n - 1)
then (m + 3, n - 2) and so on

and the total number of people you choose is 0 + (n + 1) = 1 + (n) + 2 + (n - 1) = ...

ok fixed finally

[dark wharf]

What do you mean the group is split into (m, n+1)?

Like the big group is m and the small one is n+1?

[lethal path]

this

[lethal path]

yes

like one group has m people and the other one has n + 1

then that should imply the LHS is (m + n + 1)C(n + 1) no

[dark wharf]

We want m+n+1 C n, not n+1

[lethal path]

idk

[dark wharf]

I understand your solution and I think it's very close but it needs to add one less person

[lethal path]

yeah I hate combinatorial proofs

hopefully you can spot where I've gone wrong and come up with the correct version

[midnight plank]

@dark wharf Has your question been resolved?

[midnight plank]

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[misty laurel]

Hello, I would like to know if there is a way to create a formula to describe: (1/800)+(1/799)+(1/798)+(1/797)...+(1/n)

[placid spoke]

there is not unfortunately

approximations are the best we can do

there's nothing closed form at least

[misty laurel]

damn

thank you tho

[runic hamlet]

wdym with "formula"

[misty laurel]

like a harmonic summation

[lethal path]

no closed form

[runic hamlet]

there is no "short" formula which you can enter into your calculator to "quickly" compute the value of the whole sum

if thats what you mean

[tired osprey]

It's approximately equal to ln(n) + euler mascharoni constant

asymptotically* as n gets greater

[sharp coral]

well depending on your definition of a closed form, it can be expressed using the euler-mascheroni constant and the digamma function

but ofc neither of those are buttons on a typical calculator

[misty laurel]

yea, I want to summarize it in a way that make calculating the answer easier (meaning I dont have to write individual parts of the summation ex: 1/800+1/799+1/798

[tired osprey]

with approximations or an exact fraction?

[placid spoke]

no way to get an exact answer without a lot of adding

[misty laurel]

aproximation is also okay

[tired osprey]

ln(n) + γ

so 1/800 + 1/799 + ... + 1/1 is ln(800) + 0.57721..

7,2618217277

[misty laurel]

what does 0.57721 represents?

[tired osprey]

the euler mascheroni constant.

https://en.wikipedia.org/wiki/Euler's_constant

[misty laurel]

oh ok

what if I want it to not be until 1/1 but 1/500 for ex

[tired osprey]

first do the 1/800 + 1/799 + ... + 1/1 and then subtract 1/499 + 1/498 + ... + 1/1.

[misty laurel]

ohhh okok

how exact is the aproximate?

[sharp coral]

,w sum from k = 500 to 800 of 1/k

,w ln(800) - ln(500)

[grand pond]

Results provided by WolframAlpha

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Requested by cloudedhorizons

[tired osprey]

lmao

[misty laurel]

Is this another way of calculating ?

[sharp coral]

it's the "exact result"

[grand pond]

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[sharp coral]

so to answer your question it looks like you get 2 decimal places from that

[misty laurel]

oh, so I can either use summation to find the exact no. or the euler's constant to finding an approximation (up to two decimal point)

[tired osprey]

The bigger your n, how closer the approximations

[sharp coral]

you can also add more terms to the approximation if you want more decimal places

[misty laurel]

ok thank you

.close

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[sharp coral]

the formula with more terms is in the bottom of the calculation section https://en.wikipedia.org/wiki/Harmonic_number

[misty laurel]

more terms for the constant right?

[sharp coral]

more terms for the approximation. so a better approximation is γ + ln(n) + 1/(2n)

and a better approximation than that is γ + ln(n) + 1/(2n) - 1/(12n^2) and so on for however many terms you're willing to use

[misty laurel]

where is it ?

[sharp coral]

bottom of "calculation"

[misty laurel]

kkk thank u

.close

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[midnight plank]

@woeful turret Has your question been resolved?

[midnight plank]

@woeful turret Has your question been resolved?

[woeful turret]

someone help

[tidal turret]

is O the centroid

https://physics.stackexchange.com/questions/783951/number-of-images-formed-by-3-plane-mirrors-forming-equilateral-triangle

[midnight plank]

@woeful turret Has your question been resolved?

[midnight plank]

@woeful turret Has your question been resolved?

[woeful turret]

<@&286206848099549185>

[last slate]

I feel sorry for you

[woeful turret]

😔

[tawny apex]

uhh so first if we take 2 mirrors AB and AC

by the formula, $n=theta/360$

[grand pond]

Aria

[tawny apex]

we can find the number of images formed, which would be 6

and as it is an even number, we will subtract one

giving us 5 images by 2 sides

so similarly, each pair of sides will form 5 images each

giving us 3*5=15 images

but 3 of them will coincide, and thus there would be 12 images

[midnight plank]

@woeful turret Has your question been resolved?

[brave patio]

Yo im gonna cry what is this

[woeful turret]

ohh ok thanks a lot

[woeful turret]

think u mean 360 / theta btw

[midnight plank]

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[tranquil marsh]

Need help with part C. This is my work so far

after thatt work i subbed in everything to that equation and got the wrong answer

[final mural]

<@&286206848099549185> I am stuck aswell

[jagged sandal]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hollow sigil]

Isn’t it like 10:30

pm

Or sum for you

[final mural]

wait sorry

[hollow sigil]

.

[final mural]

yes its 11:30

[jagged sandal]

This isn’t an American-exclusive server

[final mural]

yeah

[hollow sigil]

sleep?

[final mural]

its morning time?

[jagged sandal]

But let’s not talk in this channel

[final mural]

yesplease

if we could

[hollow sigil]

k bye

[final mural]

fattie

[hollow sigil]

bro what

y so mean? And I’m not fat

[hollow sigil]

you sure about that?

[remote tangle]

!topic

[midnight plank]

Please read the channel description before posting, and stay on topic.

[tranquil marsh]

could someone help me with the problem please ?

[hollow sigil]

alr now don’t say crap again and let me leave u be with your math problem

not you tweets

the other dude

Okay bye

My apologies for interfering

[jagged sandal]

@tranquil marsh Content-wise, I’m right behind you (I’m getting to the end of the Series megaunit. I should be at Parametric/Polar in a week or two)

[final mural]

@jagged sandal we're trying to find the slope of just the x values but the question isnt accepting the values when plugging in the values for theta after deriving

[tranquil marsh]

im abt to do my series homework too LOL

[final mural]

wait so do you think you can help?

[jagged sandal]

Please find another help channel. This channel belongs to Tweeks at this moment

[tranquil marsh]

we r doing the same hw its ok

[jagged sandal]

Go to #❓how-to-get-help if you need help with that

[final mural]

what if we're doing the same hw

[jagged sandal]

@tranquil marsh I’m giving ~ ~ the benefit of the doubt by having you confirm; is what ~~ is saying true?

[tranquil marsh]

Im confused

yea we have the same hw

[jagged sandal]

Huh…

[tranquil marsh]

we have the same question essentially

we both need the same help

do u think u can help on the problem even tho if u still havent reached this unit yet ?

[jagged sandal]

I’m going to pull up my textbook

It might take a bit, but I could probably figure it out quickly

[tranquil marsh]

Ok thanks its ok if you cant

[jagged sandal]

I do know a little bit about Polar Coordinates thanks to University Physics I

But I may need to understand it from a Calculus perspective

[tranquil marsh]

ok

if it helps i think youre basically implicitly deriving the polar to rectangular equations

[jagged sandal]

I found the equation

[tranquil marsh]

like the x=rcostheta equation is used i believe

[jagged sandal]

I think

[proper veldt]

Hi

[jagged sandal]

Hello

[tranquil marsh]

Hi did u see the problem i neededd help with ?

[fickle oriole]

Sorry

I figured it out

I was struggling for 5 minutes

Because I forgot to multiply by 5

[tranquil marsh]

ooh

[jagged sandal]

I’m actually going to stay so I can get a head start

[fickle oriole]

🙏

Ok I don’t remember the equation stuff they teach you

[jagged sandal]

But maybe chime in every once in a while if I can

[fickle oriole]

start with x=rcostheta right

[tranquil marsh]

hold on ima resend my work so i can follow u

[fickle oriole]

I think doing it like that is annoying

[tranquil marsh]

yea i starteds with that and dervied with respect to t using the product rule in the first line of work, does that look good ?

[tranquil marsh]

ok

[fickle oriole]

what is x in terms of theta

[jagged sandal]

Wait

[tranquil marsh]

what do you mean ?

[proper veldt]

Wheres the problem I might be able to help

[tranquil marsh]

its pinned, i sent my work just right there

[fickle oriole]

rewrite x

using x=rcostheta

[jagged sandal]

$d\theta\divdt = 5$ right?

[tranquil marsh]

dtheta/dt equals 5, not just dtheta

[fickle oriole]

right

[jagged sandal]

Sorry, I rushed to compile the teXit

I’m new to it

[fickle oriole]

right

[proper veldt]

Is it the physics problem?

[jagged sandal]

But, dtheta/dt = 5, right?

[static plume]

let theta=k cuz can’t find symbol
x(t)=cos^2(k) + cos(k) * k,
so dx/dt=[-2cos(k)sin(k) + cos(k) + -sin(k) * k] dk/dt

[fickle oriole]

hell nah cuh

[static plume]

then plug in dk/dt and k

[tranquil marsh]

oh so you substitued r a step earlierr than did right ?

[fickle oriole]

ye just do product rule now

[tranquil marsh]

ok

[fickle oriole]

with implicit diffirneation

[tranquil marsh]

should we not get the same answer though ?

[fickle oriole]

i didnt check ur work

one sec

[tranquil marsh]

ohh but now i wont have to solve for dr/dt

[fickle oriole]

no

[tranquil marsh]

i think what u did is right, i had to get rid of r early

[fickle oriole]

i dont see a world

where u could get the right answer

with that

answer is missing a pi

[tranquil marsh]

LOL ok

so with what u did now i wont have to solve for dr/dt

[fickle oriole]

u dont need dr/dt

like its there

but u dont have to mention it

here ill do the first part of product rule to show

[tranquil marsh]

yea i just substitue the stuff they fave me in the question

i understand i think

[fickle oriole]

oops not 2theta

just 2

[tranquil marsh]

wait what ?

i product ruled and got this LOL

[fickle oriole]

what

[fickle oriole]

yes

[tranquil marsh]

I derived that

and made 2theta and costheta my f(x) and g(x) that i was productt ruling

[fickle oriole]

why did u factor a 2 out

it will cancel out with the sqrt2/2

from pi/3

[tranquil marsh]

i just got rid of it before i did the product rule

oh oops

[fickle oriole]

3pi/4*

[tranquil marsh]

is that math right though ?

[fickle oriole]

ye

i made the 2costhetasino

into sin2theta

[tranquil marsh]

oh ok

am getting -18.73

[fickle oriole]

ye

thats right

[tranquil marsh]

YES

THANK U

its right

[fickle oriole]

https://tenor.com/view/dog-swing-golden-retriever-gif-1023548632554749391

[tranquil marsh]

.close

[midnight plank]

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[midnight plank]

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[kind estuary]

I have a problema at this exercise I tried to make all kind of transformarions and notations , I am stuck

[zenith osprey]

Can you please translate

[kind estuary]

Oh yeah

[lyric charm]

,rccw

[lyric charm]

ok so

you have rewritten the logarithmic equations there as $a^{4/3} = b$ and $c^{5/6} = d$, yes?

[grand pond]

Ann

[kind estuary]

Ok

I have got to this but from now...

[lyric charm]

you can raise both sides ^3 on the first equation and ^6 on the second to get a^3 = b^4 and c^5 = d^6

so you know a has to be a fourth power, and c a sixth power

[zenith osprey]

Tbh this can only be solved by hit and trial

[kind estuary]

Ok

[kind estuary]

://

[zenith osprey]

Like we know b will be in some form of a^3k

And d is c^6k

[kind estuary]

Yeah I ve think about that the numbers are natural would help really much

[zenith osprey]

Jus sub some values and check

[kind estuary]

To know the kind of form they have

[zenith osprey]

Okay so

2^6

Amd 3^3

Are satisfying

[zenith osprey]

There is no other way

[kind estuary]

For witch number?

Oh b and d

Or?

[lyric charm]

btw which ≠ witch

[kind estuary]

Thanks 🙂

[midnight plank]

@kind estuary Has your question been resolved?

[midnight plank]

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[leaden seal]

can someone check my answer please

[zenith osprey]

Yes it is correct

Are calculators allowed in exams tho?

[leaden seal]

yes

[zenith osprey]

alright

[leaden seal]

.close

[midnight plank]

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[molten bay]

24

[shrewd tusk]

do u know the rule?

[zenith osprey]

Lebeniz Newton theorem

[molten bay]

Which rule?

Nope

[zenith osprey]

+L hopital rule

[molten bay]

Any other idea?

I know hospital rule

[stiff panther]

hello

[shrewd tusk]

well thats the easiest way

[molten bay]

Could you explain it?

[shrewd tusk]

ig ull have to integrate it then

[molten bay]

But why?

[shrewd tusk]

oh so leibnitz is for diff of integrals

so basically you get d(Upper limit)f(upperlimit)-d(Lowerlimit)f(lowerlimit)
so it becomes its ||2xsin((x^2)^2))-0||

[zenith osprey]

n should be 4

[molten bay]

I know the answer please explain

[zenith osprey]

No

Shi

Mb

It's 5

[shrewd tusk]

tbh a video would help better

[zenith osprey]

Sun

Are you a jee aspirant?

[echo topaz]

sun

[zenith osprey]

If so yea 12th ka part hai

[echo topaz]

🌞

[zenith osprey]

Sup cuh

[molten bay]

Sunayo

[zenith osprey]

That shi looks like allen ka module bruh

[echo topaz]

if u didnt understand limits

[zenith osprey]

Well Newton lebeniz theorem

[molten bay]

Module kisi ka v ho

Question btao

[echo topaz]

is that 11th or 12 th grade

[zenith osprey]

Is in 12th usually integration ya differential equation mai padhate hain

[shrewd tusk]

integral should be in 12th only

[zenith osprey]

No we learn it in 11th too

[tough shale]

Ya, sin ka expansion use kar sakte ho

[zenith osprey]

Tf?

Nahi

[tough shale]

But it works

[zenith osprey]

Kaafi lamba hoga

[zenith osprey]

Maclaren series' laga de

Agar nahi sikhaya toh

[tough shale]

Yeah, but that's the only alternate I can think of

[tough shale]

What's maclaren series?

[zenith osprey]

Wahi tailor series bhai x/1!-x^3/3!

[tough shale]

Oh, ok

[thin tree]

taylor*

[zenith osprey]

Mb

[molten bay]

Sab Indians at one channel🤣😅

[zenith osprey]

Real

[zenith osprey]

Konse grade mai hai tho?

[molten bay]

grade choro

[shrewd tusk]

or just take x^2 as t and integrate by parts if yk integration

[zenith osprey]

Are fir lebeniz theorem nahi aata kya?

[molten bay]

Integration k andar lagao kya maclaurie

[zenith osprey]

Idt he knows that too

[zenith osprey]

He was saying ki

Sinx^2 ko

x^2/1!-x^6/3!+x^10/5!

[molten bay]

I got n=5

[zenith osprey]

Likhe ke integrate karo

[molten bay]

@zenith osprey

[zenith osprey]

6 aayegi bhai

Is hisab se

Lemme check maine kya galti ki

[zenith osprey]

(x^3/3!) ke limits x^2 to 0 hai na

Toh x^6 hoga

[molten bay]

Yes

(x^2)^3 hmm

x^6/6x^n

[zenith osprey]

Yea

[molten bay]

Can I do all questions?

Like this

Where integration is included

[tough shale]

Honestly, depends

Like, in the 26th question

U can't get the series for (tan inv x)² easily

[zenith osprey]

Lebeniz theorem se bhi 6 hai btw maine substitution mai galti ki thi

[zenith osprey]

Woh toh yaad hi rakhni padti hai

3 terms tak

[tough shale]

Par square hai

[zenith osprey]

Oh shit

[molten bay]

Lebiez wala send karna

[zenith osprey]

Han abhi dekha

[molten bay]

Which rule you applied?

@zenith osprey

[tough shale]

2nd wala

[zenith osprey]

It's not multivariable bhai

2nd wala

[molten bay]

How you put sin(x^4) as x^4?

[zenith osprey]

Substitution karde limits ka

Sin(x^2) tha na

[molten bay]

No

[zenith osprey]

So limits dalne ke baad

[molten bay]

Next step

[zenith osprey]

Oh

Woh toh

[molten bay]

Around 0

??

So

[zenith osprey]

Sin(x)/x=1 se kiya hai

Toh sin(x^4)/x^4=1

And sin(x^4)=x^4

[midnight plank]

@molten bay Has your question been resolved?

[midnight plank]

@molten bay Has your question been resolved?

[midnight plank]

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[twilit field]

I'm trying to parametrize this , would apprciate any inputs.
I start by re-writing $z=2-x^2-y^2$as $z=2-z^2$, whose soutions are $z=-2; z=1$.
\
As the cone opens upwards, $z=1$.
\
As is, the paraboloid is given by$(\sqrt{\cos^2(t)-2}, r \sin(t), 2-r^2); 0≤t≤2π; 0≤r≤1$

[grand pond]

What a wonderful world !

[twilit field]

oops

yea, I'm lost

so uh, let's start again

x= rcos(t), y= rsin(t), z= 2-r^2

[midnight plank]

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[shrewd tusk]

How do i find the height of centre of mass of a hemisphere?

[cedar mason]

integration

[shrewd tusk]

um well how exactly?

like what do i integrate

[cedar mason]

do you know the centre of mass is.

[shrewd tusk]

yeah ik that i can do it for a semicricle

[cedar mason]

can you do it for a circle.

[shrewd tusk]

isnt that r,r by symmetry

[cedar mason]

sure

so

[dusty portal]

Would this not be using the centroid formula?

[cedar mason]

now you can compose a hemisphere out of a bunch of disks.

which are just circles with some infinitesimal height.

[shrewd tusk]

ryt

[cedar mason]

can you proceed from here

it's literally just like how you would find the area of a hemisphere.

well, the integral setup, i mean.

[shrewd tusk]

I'm thinking give me a moment, ill ping if i need help

[sharp coral]

there is also a derivation using the pappus centroid theorem, if you've heard of that

[shrewd tusk]

I haven't, Im still in high school

[sharp coral]

so what class do you need this for? are you expected to derive the formula or just use it?

[shrewd tusk]

i am expected to memorize it, for physics

but i kinda keep forgetting

if i knew how to find it, it will stick in my head

[sharp coral]

i think that the pappus theorem would work for that

so we can imagine a sphere as a surface which is generated by revolving a semicircle around its axis

https://en.m.wikipedia.org/wiki/File:Pappus_centroid_theorem_areas.gif

[dreamy lichen]

(Can you do hemisphere with that btw? I can only think of how to use it for semicircle)

[sharp coral]

well by symmetry the centroid of a hemisphere should be the same, if we imagine it as being a semicircle pivoted about its center

[dreamy lichen]

It isn't

[sharp coral]

well that's annoying

[dreamy lichen]

the mass thats further from the centroid gets more spread out sort of

[shrewd tusk]

its ||3r/8|| if you wer wondering

[dreamy lichen]

you'd need volume of a hypersphere to make the pappus theorem argument

maybe we could just do the integrating

[shrewd tusk]

my problem with this is like isnt the radius diff for each disc?

like how does it vary

[sharp coral]

back to integration i think

if we look at it from the side the hemisphere looks like a semicircle

[shrewd tusk]

I think the disc one would work but i just dont know how to form the integral

[cedar mason]

use the arc

and radin stuff

so the radius is just $r \cos(\theta)$

[grand pond]

Percy

[shrewd tusk]

uh where's theta from?

[cedar mason]

we're using theta for the integrating variable

not x or y

for example if this was the area of a hemisphere, we'd get the integral as $\int^{\frac{\pi}2}_{-\frac{\pi}2} \pi (r \cos(\theta))^2 r d\theta \sin(\theta)$.

I think.

Aaaand that's prolly spoiling it too much lol

,w \int^{\frac{\pi}2}_{-\frac{\pi}2} (r \cos(\theta) r \sin(\theta)) d\theta

nice

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by percykel

[cedar mason]

nvm

its right dw

[shrewd tusk]

i see give some time to process it lol

[cedar mason]

whoops i forgot what area even is

[shrewd tusk]

isnt it like pi r^2 h

ah yeah

that makes sense

[grand pond]

Percy

[shrewd tusk]

i think u put pi in bracket bymistake?

[cedar mason]

IM NOT SURE THIS IS RIGHT

NVM IGNORE ME

[shrewd tusk]

hmm but it makes sense tho except for the bounds, how d you get -pi/2

[cedar mason]

yeah i was doing a up and down thing

idk whats going on

but the general idea is fine 😭

[shrewd tusk]

i think its 0

for hemisphere

[cedar mason]

the correct integral is $4\int_{0}^{\frac{\pi}{2}}\pi r^{2}\cos^{2}\left(x\right)r\sin\left(x\right)dx$

[grand pond]

Percy

[cedar mason]

but not sure why

[cedar mason]

no i was doing sphere

[shrewd tusk]

ah ok

[cedar mason]

the 4 makes no sense

AH

LMAO

alright i just brain fogged the wrong trig function

[woeful turret]

solid hemisphere?

no it is not 0

[cedar mason]

it's $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\pi r^{2}\cos^{2}\left(x\right)r\cos\left(x\right)dx$

[grand pond]

Percy

[cedar mason]

we're talking about general integral for sphere stuff

[woeful turret]

oh i thought he was jee wala

[cedar mason]

:sparkes

[shrewd tusk]

I see so its correct

[woeful turret]

where are u from

[shrewd tusk]

India lmao

[woeful turret]

so ur preparing for jee?

[last slate]

jee over

today

[shrewd tusk]

there's advanced 💀

[last slate]

oooh

[woeful turret]

u dont need integration of hemisphere just remember the results

[last slate]

idk if ill qualify for that

im on the brink

[woeful turret]

mine next year

[last slate]

only math and physics carrying me

[last slate]

ooh cool

chem i fked up

[shrewd tusk]

well i wanna know anyway

[woeful turret]

which part?

[shrewd tusk]

the integration stuff helps

in other problems

[woeful turret]

ur writing next year?

[last slate]

entire of chemistry

[shrewd tusk]

this year

[woeful turret]

yeah learn for sphere its more useful

[woeful turret]

no i mean like physical inorganic or organic??

all 3?

[woeful turret]

how was mains

[last slate]

all of chem

[woeful turret]

oh but why

[last slate]

i did not study for it

i hate chem

[shrewd tusk]

4s1 💀

[last slate]

its just memorising and shit

and there are hell lot of exceptions

[woeful turret]

oh

yeah fr

[woeful turret]

4s1?

[last slate]

yup yup

[last slate]

4 April shift 1

[shrewd tusk]

well ill get around 170-180 ish

[last slate]

Damn

[woeful turret]

oh did something happen in that shift?

[last slate]

ill get around 100

[last slate]

its harder than others

[shrewd tusk]

it was lengthy as hell

i only did 50 Qs

[woeful turret]

oh no ;/

[last slate]

i did 49

bye anyways

[shrewd tusk]

byee

[woeful turret]

thats pretty good no

99%ile higher u should get

[thorny plinth]

[last slate]

i find advance easier than mains😭

[woeful turret]

.

[last slate]

spookiee

[woeful turret]

naw

[last slate]

yup

[shrewd tusk]

yeah more time to solve helps a lot

[thorny plinth]

hi is there a question

[shrewd tusk]

mains is more memory based which sucks

[last slate]

would you love me if i were a worm

[last slate]

YEAAA IKRRR

[thorny plinth]

i love worms

[last slate]

where in India

[thorny plinth]

nice goodluck then

i feel the same

[shrewd tusk]

oh i was trying to find the centroid of hemisphere

[woeful turret]

bro but some of the questions are too hard to understand what is going on

[woeful turret]

for hollow hemisphere- R/2 from base along y axis
solid hemisphere- 3R/8 from base along y axis

[shrewd tusk]

we can just skip them, cutoff for advanced isnt that high

[thorny plinth]

oh nice did u do it?

[shrewd tusk]

kinda, i got the method, i think

like split into discs

[last slate]

yayy

[last slate]

cool

[shrewd tusk]

and assume rcostheta and rsintheta smtn smtn

[woeful turret]

yeah thats true..2024 cutoff was kinda high tho

[shrewd tusk]

that was exception, the paper was pretty easy

[thorny plinth]

2024 was seriously fked

(experience)

[shrewd tusk]

maths was like mains

[woeful turret]

ohh

[thorny plinth]

||i got 243 in mains and dint even get 99.9||

[woeful turret]

wth

not possible

no shift cutoff was that high

[thorny plinth]

29jan s2 2024

[woeful turret]

except that one

people were saying that paper got leaked

[last slate]

center of mass?

[shrewd tusk]

tf jan attempt???

[thorny plinth]

yes theres 2 isnt there

[shrewd tusk]

centroid and centre of mass is same for uniform

[thorny plinth]

no way

[last slate]

oohk

[thorny plinth]

also 27th was like that

[woeful turret]

@thorny plinth any tips for me 🙏

[thorny plinth]

uh sure dms ig we should stop flooding the channel

[last slate]

yup

[woeful turret]

yeah okay xd

[shrewd tusk]

yeah ill close ig

.close

[midnight plank]

Channel closed

Closed by @shrewd tusk

Use .reopen if this was a mistake.

[slow thorn]

3R/8

it just sticks after a while

after you see some problems

[shrewd tusk]

well i dont have time to practice a ton of problems on one formula, so its faster to just learn the derivation, and as a bonus ill also learn a new method

[slow thorn]

if you really want integration

1 min

[shrewd tusk]

.reopen

[midnight plank]

✅

[slow thorn]

maybe this

filled is again another layer of integrating with hollow hemispheres

so it's quite a lengthy proof

[cedar mason]

almostttt there

the height isnt actually r d theta

thats the slant

wait

youre not op

freaking

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[slow thorn]

yes

what

he asked for another method

[cedar mason]

also its wrong

[cedar mason]

aaaah

okay

but its still wrong tho

[slow thorn]

why so

isn't it R/2?

[cedar mason]

.

[slow thorn]

i never said the height was R d theta?

that's the thickness of the strip

[shrewd tusk]

where th did you get the 6 from?

[cedar mason]

[slow thorn]

im so sorry

past me handwriting was horrible

that's a sigma

[shrewd tusk]

oh

[slow thorn]

areal density

[shrewd tusk]

btw i think as we take small enough i think it becomes the same

[cedar mason]

you still got the thing super wrong right

[cedar mason]

no it doesnt

[slow thorn]

hmm

looks safe

[cedar mason]

you need that trig factor there

[slow thorn]

answer also matches

meaning?

[cedar mason]

no?

[shrewd tusk]

oh he took a hollow sphere first

[slow thorn]

that's a solid

[cedar mason]

AH

okay

[shrewd tusk]

then using that hes finding for solid

[cedar mason]

apologies

[slow thorn]

solid is a bit more arduous than this

[shrewd tusk]

hmm well yeah i think Percy's disk method is better if we just want the solid (although im still figuring it out)

[slow thorn]

maybe

[shrewd tusk]

thanks for your help!

.close

[midnight plank]

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[midnight plank]

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[molten bay]

I didn't understand the question properly and option and the =p part?

[cedar mason]

use more words

[molten bay]

If I check the condition random values like i out integer x=2

2-2=0

[2+h]+[2-h]

2-3=-1

Not continuous at integers

Now let me check at non Integers

like 2.5

At x=2.5

[shrewd tusk]

btw percy i got the 3r/8 using the method you suggested, thank you very much

[shrewd tusk]

sorry, continue

[molten bay]

2-3=-1

2.5+h it will be same

And at x=2.5-h same

So non integers yooo

B and D

Am I right?@shrewd tusk

@cedar mason

[cedar mason]

i dont really understnand man sorry

[shrewd tusk]

one sec let me check

yeah your ryt @molten bay

the funcn is 0 whenever its an integer

and -1 otherwise

[molten bay]

Thanks

[molten bay]

But i am saying it is not continuous at integers

@shrewd tusk

[shrewd tusk]

yeah its not continuous cuz at integers it suddenly jumps from -1 to 0 and back to -1

for continuity you should be able to draw the graph without lifting your pen up

clearly you cant do that at the integers

[molten bay]

Thanks

.close

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[wet cairn]

How would I find the perimeter of thisss??

[wet cairn]

Would the red dotted lines be 10cm each?

[cerulean solar]

Not necessarily, but using the pythagorean formula we can confirm that it is the case in this question

I’m assuming those are semi-circles

[wet cairn]

ohh how'd u use the pythagores therom?

oh like a squared plus a squared

[cerulean solar]

a^2+b^2 = c^2

[wet cairn]

is 14.14 squared?

a2+a2=14.14 squared???

is that how u got it?

IM LIKE 13 DUDE 🙏

[cerulean solar]

Actually the other side length technically isn’t 10, I was too hasty and rounded up by accident

[wet cairn]

aren't the two red dotted lineds equal?

[cerulean solar]

No, they are different by just the tiniest fraction

[wet cairn]

💀

this a year 9 problem

surely

like

there is some leisure

and we're supposed to asumme

[fathom atlas]

They are...the question makers rounded it up I think...

[wet cairn]

they're equal rightt?

[cerulean solar]

The other side is about 9.996979

[wet cairn]

ohh okk

[cerulean solar]

Yeah, but I’d rather not

[wet cairn]

fair enough

9.99??

duh

anyways

so if the red dotted lines are 10 each

then

the perimeter would be 20+14.14??

for the traingle at least

[cerulean solar]

The dotted lines are not part of the perimeter

[wet cairn]

rllyy??

oh

im so sorry

ur right that makes sense

[cerulean solar]

You now need to find the perimeters of the semi-circles

[wet cairn]

ok

yeh

so

the perimeter of the semicircle

[cerulean solar]

The formula for the circumference of a circle is 2(pi)(radius)

[wet cairn]

since they are equal i can just double my answer right?

so

pi times diamter here is

10 pi

31.4159265359

as accurate as it gets

and then double that?

[cerulean solar]

Yes

[wet cairn]

62.8318530718

but then

we gotta half right?

since its a semicircle

so go back

to 31.4159265359

correct?

[cerulean solar]

However, technically one of the sides isn’t a diameter of 10,

[cerulean solar]

This is correct

[wet cairn]

for the perimeter of BOTH semicircles?

[wet cairn]

(lets just say!)

cool

so

[cerulean solar]

Do it my way just to rage bait your teacher

[wet cairn]

31.4159265359 added to 14.14

is the perimeter of the entire composite shape?

45.5559265359

which

rounded to 2

is 45.56

yess thank u

wbt area?

rq

how would i find the area of the triangle?

[wet cairn]

its reivison for a test shes just gonna laugh her ass off

[cerulean solar]

Math teachers will probably like the pendanticness

[wet cairn]

aight aight

is the area of the triangle

50cm sqaured??

10 times 10 divided by 2

i got it thnks (;

.closed

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[midnight plank]

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[hazy seal]

isnt the answer 130 because inscribed angles are half of the arc?

[lyric charm]

mm careful though

there is an arc here that measures 130° but it isn't MW

[hazy seal]

🤔

[lyric charm]

look carefully at what arc the 65° angle (AHW) actually subtends

[hazy seal]

AW?

[lyric charm]

i would call it AMW just to be safe

but yes

so that arc measures 130°

[hazy seal]

yes and then you take away 60

[lyric charm]

exactly

[hazy seal]

3 letter arcs are for the ones over 180 degrees though

[lyric charm]

was that a strict rule in your class?

[hazy seal]

i dont know I just follow the course i thought it was proper notation

[lyric charm]

cause in my experience naming an arc by 3 letters is just a more general disambiguation for when there's any uncertainty which way to go

2 letters is fine for arcs that are visually small

with no hard boundary for where such usage is acceptable

[hazy seal]

ok thanks

.close

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[rigid turret]

Is the answer .0625 or 1/12
I have 2 answers and don't remember which was verified

[rigid turret]

Screenshot if u prefer

[small jasper]

1/16 = 0.0625

Idk why this is 2/3

it should be 4/3

If it's 4/3, then you also get 1/16

[midnight plank]

@rigid turret Has your question been resolved?

[rigid turret]

@small jasper is 3/2 = 4/b false

[small jasper]

At the instant when the height is 2? No.

But you want h in terms of b in general

so

[rigid turret]

yea

yep im good

[small jasper]

[rigid turret]

db/dt = 4/3 dh/dt

[midnight plank]

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[rigid turret]

both my method were good

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[lavish venture]

Show that the map

[d_A \colon X \to \mathbb{R}^{\geq 0}, \quad d_A(x) = \inf{d(a, x) \colon a \in A}]

is continuous

[grand pond]

knief

[lavish venture]

so i've taken an open set $U \subset \mathbb{R}^{\geq 0}$ and then let $x \in d_A^{-1}(U)$ or equivalently, $d_A(x) \in U$ and then since $U$ is open i know there exists $\varepsilon > 0$ so that the ball of radius epsilon (open interval) is contained in $U$

[grand pond]

knief

[lavish venture]

but i am struggling to see how i can extract a delta from this to show there is a ball of radius delta for some delta that is contained in the inverse image of U

like if i have some epsilon neighborhood around d_A(x) contained in U then this epsilon certainly has to be important

is it true that the distances between my points in the space X will also be less than epsilon if their corresponding distances to A are less than epsilon?

it has to be related somehow

surely

ping me if you have an idea

[midnight plank]

@lavish venture Has your question been resolved?

[lavish venture]

.close

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[slender lagoon]

Just straight up idk what to do here

[midnight plank]

@slender lagoon Has your question been resolved?

[slender lagoon]

No

[gaunt plume]

See if you can work out what $\cos\theta$ is

[grand pond]

depression

[midnight plank]

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[manic bison]

in such an expression is TV a constant

(C is a constant here)

[polar star]

T and V can vary such that they output the same value C

[lyric charm]

zero context...

[manic bison]

its a derivation in thermodyanimcs

[manic bison]

what about T^Cv/R . V

sorry if im being dumb

[polar star]

do you mean VT^(C_v/R)

[manic bison]

yes

[polar star]

and you are keeping C constant?

[manic bison]

yes

[polar star]

like we saw in the log form, there exists some T and V such that VT^(C_v/R) = C where C is constant but T≠V

u can visualize a range of values by looking at VT^(C_v/R) = 5 for example

[manic bison]

how

[polar star]

with defined heat capacity and 8.31

,w plot x*y^(4/8.31) = 5

[grand pond]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by _mia_34

[manic bison]

so if sum of logs is constant their arguments rpoudct is also constant?

[manic bison]

i understand this

[polar star]

what exactly are you asking

[polar star]

what all things do u wanna keep constant

from those variables

[manic bison]

this

[polar star]

are you asking if log(x)+log(y) is constant then xy is constant?

[manic bison]

yes

oh wait

you conbine log

sorry

i was being dumb

.close

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[twilit field]

Consider the seqeunce given by $b_n=.n -\sqrt{n^2+2n}$ Taking $(1/n} \to 0$ as given , and using both the algebric limit theorm and the fact that if $x_n \to 0$, then $\sqrt{x_n} \to 0$. Show lim $(b_n)$ exists and find the limit

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

so to start , I re-write this as $b_n= \frac{ -2n}{ n+ \sqrt{n^2+2n}}$
\
$\frac{-2n}{ n( 1+ \sqrt{1+ \frac{2}{n}}$
\
dividing the numerator and denominator by $n$
\
$\frac{-2}{ \sqrt{ 1+ \frac{ 1 + \frac{2}}{n}}$
\
as $\frac{1}{n}$ goes to 0 and $1+ \frac{2}{n} \to 0$
\
$\sqrt{ 1+ \frac{2}{n}}\to 0$
\
We then have $1 + \sqrt{ 1+ \frac{2}{n}} \to 2$
\
We then have by quotient rule, the limit is $-1$

[grand pond]

What a wonderful world !

so to start , I re-write this as $b_n= \frac{ -2n}{ n+ \sqrt{n^2+2n}}$
\\
$\frac{-2n}{ n( 1+ \sqrt{1+ \frac{2}{n}}$
\\
dividing the numerator and denominator by $n$
\\
$\frac{-2}{ \sqrt{ 1+ \frac{ 1 + \frac{2}}{n}}$
\\
as $\frac{1}{n}$ goes to 0 and $1+ \frac{2}{n} \to 0$
\\
$\sqrt{ 1+ \frac{2}{n}}\to 0$
\\
We then have $1 + \sqrt{ 1+ \frac{2}{n}} \to 2$
\\
We then have by quotient rule, the limit is $-1$
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                \par 
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to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[normal prawn]

do we need to find the limit of n-√(n²-2n) when n approaches infinity?

[twilit field]

yes

it's -1

I know that

[normal prawn]

Limit is -1?

how

[twilit field]

isn't it

[normal prawn]

idk I'm not good when it comes to limits lol

[twilit field]

,w lim n -sqrt{n^2+2n}

okay, 1

my bad

[grand pond]

Results provided by WolframAlpha

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Requested by math_rocks

[twilit field]

wait

no

[normal prawn]

1 how?

0?

[twilit field]

it's -1

[normal prawn]

Howw

[twilit field]

-1

[normal prawn]

howow

[twilit field]

this should explain it , if I can render it

so to start , I re-write this as $b_n= \frac{ -2n}{ n+ \sqrt{n^2+2n}}$
\
$\frac{-2n}{ n( 1 + \sqrt{1+ \frac{1}{n}}}$
\
dividing the numerator and denominator by $n$
\
$\frac{-2}{\sqrt{1+ \frac{1}{n}}}$
\
as $\frac{1}{n}$ goes to 0 and $1+ \frac{2}{n} \to 0$
\
$\sqrt{ 1+ \frac{2}{n}}\to 0$
\
We then have $1 + \sqrt{ 1+ \frac{2}{n}} \to 2$
\
We then have by quotient rule, the limit is $-1$

[grand pond]

What a wonderful world !

[twilit field]

finally

.close

[midnight plank]

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[midnight plank]

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[twilit field]

Prove that there's no such seqeunce or give a counter example
\
An unbounded seqeunce (a_n) and a convergent seqeunce (b_n) with (a_n-b_n) bounded

[placid spoke]

what have you tried

[twilit field]

I was thinking that such a seqeunce would exits

[placid spoke]

do you have any reason why you think that

[twilit field]

idk, just a feeling

[placid spoke]

well what do you know about convergent sequences

can those be unbounded

[twilit field]

A convergent seqeunce is always bounded

right

[placid spoke]

yes

so an unbounded sequence minus something bounded

[twilit field]

An unbounded seqeunce - a convegrent sequence is always not convergent

same proof as previosu problem

[placid spoke]

well an unbounded sequence - a bounded sequence cannot be bounded

it's like infinity - some number

and if a sequence isn't bounded, you know it can't converge

so it's very similar to the previous one yeah

[grim vector]

the question is wierd

[west sapphire]

Hello can I submit a theroy of mine?

[grim vector]

you write a_n = a_n - b_n + b_n :)

if you have a_n - b_n bounded

and b_n convergent (so bounded)

a_n should be bounded

(sum of bounded sequences is bounded sequence)

[west sapphire]

It's possible how we got four dimensions or more and time

[grim vector]

interesting question but open a new channel

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

@twilit field Has your question been resolved?

[night axle]

Or, just good old triangle inequality that says ||a_n| - |b_n|| < |a_n + b_n|

Which is ultimately the same but a different way of seeing the same thing

[grim vector]

sure sir

[twilit field]

Got it thanks

.close

[midnight plank]

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[violet dune]

would i use integration by parts for this? $\int(2^{x+1}*3^{x+4})dx$

[grand pond]

Krish

[runic hamlet]

combine them into a single exponential

[violet dune]

how do i do that if they dont have the same base or exponent?

[fallow scarab]

$a^b = e^{b \log(a)}$

[grand pond]

riemann

[violet dune]

what if i make it $2^x23^x*3^4$?

[grand pond]

Krish

[violet dune]

then would it be $162\int(2^x3^x)dx$?

[grand pond]

Krish

[fallow scarab]

that's a good step yes

[violet dune]

and does the inside simplify to 6^x or does it not work like that?

because then integral of a^x = a^x/lna right?