help-36

[distant breach]

sigh..

[supple mantle]

Hopefully you'll be the highest performing person :')

Also

Nothing is impossible

[distant breach]

everyone already took their final on monday :') ive been bedridden for the past couple days

so this is rly my only studying.

[supple mantle]

[distant breach]

🙏

[supple mantle]

How many chapters do you have left?

[distant breach]

wait cud u help me out a bit more with understanding stuff

he has some practice problems

[supple mantle]

Yes I'm more than happy

[distant breach]

so im just trying to go thru them

[supple mantle]

Hm

[distant breach]

do i solve for theta?

[supple mantle]

First thing

[distant breach]

or just make them equal eachother

[supple mantle]

Wait

It needs some trigonometric magic you'll remember

Ok so

First step

sin(2θ) = 2sin(θ)cos(θ)

[distant breach]

oh wait yes

[supple mantle]

Now

It needs some thinking

2sin(θ)cos(θ) = 2sin(θ)

What is missing here for them to be exactly the same?

[distant breach]

cos(theta)

[supple mantle]

Exactly

This means that cos(θ) should equal 1

[distant breach]

wait why 1

[supple mantle]

So you can remove it and not change 2sin(θ)

[distant breach]

wait what did u mean by cosθ = 1

[supple mantle]

If we removed cosθ from 2sinθcosθ, wouldn't it be exactly the same as 2sinθ?

[distant breach]

mhm

it would

[supple mantle]

Ok what's the thing that when multiplied by a number keeps the number as it is?

[distant breach]

1

[supple mantle]

So here

We need to remove cosθ and make it 1

This needs some values of θ

[distant breach]

oh

cos60 = 1 right?

[supple mantle]

Nope

[distant breach]

wait

cos30?

[supple mantle]

cos(0) ans cos(180

[distant breach]

hold up thinking of unit circle atm

[supple mantle]

Ayooo

[distant breach]

180? .o.

[supple mantle]

Yep

[distant breach]

isnt that -1 tho

[supple mantle]

Oh yeah

True

So you only have θ = 0°

[distant breach]

ok ok

so

wait are we allowed to plug into theta tho?

[supple mantle]

You don't need to

You took trigonometric inverse functions right?

$\cos^{-1} (x)$

[distant breach]

uhm..

[soft zealot]

VulcanOne

[distant breach]

we didn't learn much about them

[supple mantle]

How

[distant breach]

my prof has a weird way of teaching things ngl 🥲

[supple mantle]

Reminds me of my concrete professor

Makes us memorize stuff

Anyways

You have your calculator right?

[distant breach]

so..we arent allowed to use that on the exam..

[supple mantle]

Wow

[distant breach]

yeahhh...

😭 im bout to fail bro

[supple mantle]

You will win

Dw

Quickly here

[distant breach]

wait is this called arccos?

[supple mantle]

We use trig inverses like how we use logarithms

[supple mantle]

Yep

[distant breach]

the cos^-1

he made us write out

[supple mantle]

Yepp

[distant breach]

arccos.

everytime 😭

so thats why

[supple mantle]

Yes that's the correct way

[distant breach]

ok

perf

then i do know

[supple mantle]

Cos^-1 is bad

[distant breach]

would i do arccos(cosθ)

[supple mantle]

Yep

And arccos (1)

You apply it on both sides

[distant breach]

so i would write it as

[supple mantle]

θ = arccos(1)

[distant breach]

wait now im confused

theta = arccos(1)?

[supple mantle]

Yep

[distant breach]

so id plug in

cos(arccos(1))

[supple mantle]

Should give you 1 back

Yep

[distant breach]

so when i write it

wdym apply it on both sides

im trying to write notes atm so i can study them ltr :')

[supple mantle]

We have cosθ = 1

Ooo

[distant breach]

2sinθ*cosθ = 2sinθ

[supple mantle]

I became blue

[distant breach]

yeee

u got active role :>

[supple mantle]

No wait

[distant breach]

its the original one

[supple mantle]

cosθ is multiplied by 2sinθ

[distant breach]

o

[supple mantle]

sin(2θ) = 2sinθcosθ

[distant breach]

okok

and then we use the arccos?

[supple mantle]

Not now

We have 2sinθcosθ = 2sinθ

[distant breach]

mhm

[supple mantle]

I think we're allowed to divide by 2sinθ on both sides

[distant breach]

o

[supple mantle]

So that we get cosθ=1

Then here we use arccos

[distant breach]

oh wait true

ok so

that makes complete sense

[supple mantle]

:')

[distant breach]

for the arccos do we add it to both sides?

[supple mantle]

We write it on both sides yes

arccos(cos(θ)) = arccos(1)

[distant breach]

arccos(cosθ) = arccos(1)

[supple mantle]

Yeppp

[distant breach]

ye

[supple mantle]

And I think you can leave it on arccos(1)

[distant breach]

o ok

perf

[supple mantle]

Now on the second question

[distant breach]

mhm

[supple mantle]

$2^{x-1} = 3^{1-x}$

[soft zealot]

VulcanOne

[supple mantle]

First thing is to take natural log

So that we drop the powers

[distant breach]

drop the powers?

[supple mantle]

Yep

$\ln(x^a) = a\ln(x)$

[soft zealot]

VulcanOne

[supple mantle]

That's a property of logarithms

[distant breach]

so the left would be

x-1ln_2?

[supple mantle]

Yep

You also take the log on the right side

[distant breach]

1-xln_3

[supple mantle]

Yepp

Don't forget your parentheses

[distant breach]

mhm

[supple mantle]

I think your prof will deduct marks for simple parentheses

[distant breach]

yeah probs

i have it on paper atm

i just didnt use it comp

[supple mantle]

Alrighty:)

Now we can distribute

(x-1)(ln2) = (1-x)ln(3)

[distant breach]

distribute?

[supple mantle]

xln2 - ln2 = ln3 - xln3

Yep

[distant breach]

o

[distant breach]

and the 2's and 3's are all subscripts still right?

[supple mantle]

Nope those are the arguments

[distant breach]

o wait

[supple mantle]

Our subscript throughout the problem is the same

e

[distant breach]

oh wait by subscript i mean

like

lemme get a pic

[supple mantle]

$\log_a (b)$

[soft zealot]

VulcanOne

[supple mantle]

Actually ln by itself has its own subscipt

[distant breach]

oh

[supple mantle]

$\ln(x) = \log_e (x)$

[soft zealot]

VulcanOne

[supple mantle]

It's so special that it deserves its own notation

[distant breach]

wait so how would we write this

after distributing

[supple mantle]

$x\ln(2) - \ln(2) = \ln(3) - x\ln(3)$

[soft zealot]

VulcanOne

[supple mantle]

No subscripts with natural log

[distant breach]

oh

[supple mantle]

Also sometimes people write log instead of ln to mean natural log

[distant breach]

mhm

[supple mantle]

Anyways

Treat ln(2) and ln(3) as numbers

And solve for x

And remember something that will make your life easy

$\ln(a) + \ln(b) = \ln(a\cdot b)$

[soft zealot]

VulcanOne

[distant breach]

and if its subtraction, its division?

[supple mantle]

Yepp

[distant breach]

uhh

xln2/1ln2

?

for left side?

[supple mantle]

Hmm

I don't think there are any multiplication or division now

[distant breach]

o

[supple mantle]

We add xln(3) to both sides

And then we add ln(2) to both sides

So that we make the x's all in one side

And the numbers in the other

[distant breach]

oh

[summer shadow]

Can I drop a question

[supple mantle]

In another help channel please

[summer shadow]

Why

[supple mantle]

This one's already occupied

Please

[summer shadow]

Ok

[supple mantle]

Take #help-2

Thank you

Anyways

[distant breach]

ok so rn i have

xln2+xln3=ln3+ln2

[supple mantle]

Yepp

Take x as a common factor

[distant breach]

so

x(ln2+ln3)

[supple mantle]

Yep

Last step?

[distant breach]

divide?

[supple mantle]

Yeppp

[distant breach]

x=1

[supple mantle]

Your professor likes having simple numbers as answers

Weird

[distant breach]

these are practice questions :')

nothing like the real exam sadly

[supple mantle]

[distant breach]

his tests are actually hard 😭

im struggling with this, how will i do with the exammm :')

[supple mantle]

Do you have a copy of his past exams?

[distant breach]

nope

[supple mantle]

Damn

[distant breach]

i dont think anyone has it online

its all paper n stuff so no one bothers to post pdfs

and either way

he grades on the last day of the semester

aka today

[supple mantle]

[distant breach]

so no ones rly even there to collect tests back

🥲

[supple mantle]

You will pass easily though

[distant breach]

🙏 i pray i do

[supple mantle]

Be sure to get a nap or something because sleeping will help you store the info you got

[distant breach]

can we do 2 more if u still have time? simple ones tho cus im dead atm

[supple mantle]

Sure

[distant breach]

gonna take a hr nap soon

inequalities are pretty easy i think

[supple mantle]

Yep

Simple algebra

d) is a sneaky one though

[distant breach]

o fr?

[supple mantle]

Yep

That absolute value will have stuff going on

But Imma simplify it for you when we get to it

[distant breach]

okk

so for the first one

cant i just multiply the whole denominator?

to the other side?

[supple mantle]

Yep

[distant breach]

ok

3x-5 <= 2(x+2)

[supple mantle]

Yep

[distant breach]

then distrubute

[supple mantle]

Yepp

[distant breach]

xs on one side and rest on other

[supple mantle]

Yepp

[distant breach]

x<= 9

[supple mantle]

Correct

[distant breach]

thats all he wants? 💀

[supple mantle]

Yeah

Lol

[distant breach]

that was the easiest thing ive done all night 😭 so glad

[supple mantle]

Yeah I was surprised

Lol

Thought there was some tricks

But nope

Simple algebra

[distant breach]

mhm

hope he gives like at least 2 of those on the test

anyways for d

[supple mantle]

Last you should be simple till you isolate |1-x|

[distant breach]

divide 3?

[supple mantle]

Yepp

[distant breach]

ye

we take the 1 out?

[supple mantle]

Nope

[distant breach]

o

[supple mantle]

Absolute value needs a special trick

Ok so

You know that anything inside the absolute value, negative or positive, will make the absolute value in the end positive right?

[distant breach]

mhm

[supple mantle]

Alrighty

Now

You have |1-x| < 2/3

Right?

[distant breach]

yes

[supple mantle]

Ok so

1 - x on its own should be in between -2/3 and 2/3 to have an absolute value less than 2/3

Makes sense or needs some clarification?

[distant breach]

i get that

that makes sense

[supple mantle]

Alrighty so

-2/3 < 1 - x < 2/3

[distant breach]

mhm

[supple mantle]

Then you solve for x

[distant breach]

ohh

i usually just do +/-2/3 -1 = x

but like move the 1 first then the absolute val

but u probs doing the right way

[supple mantle]

You could try your method

[distant breach]

so for this

how do yk the sign on the left?

the inequality one

[supple mantle]

Because the absolute value is less than 2/3

Less than means that it is between the positive value and negative value

[distant breach]

o wait

u right 💀

1/3 < x < 5/3

[supple mantle]

Yeppp

[distant breach]

woo 👏

[supple mantle]

Now what if it was bigger than 2/3?

[supple mantle]

[distant breach]

then the negative would also be >

[supple mantle]

Nope

[distant breach]

right?

o shoot fr

[supple mantle]

You see the graph

[distant breach]

mhm

[supple mantle]

Ok so when it was less than 2/3

The x was between

[distant breach]

mhm

[supple mantle]

But when it was bigger

The x will be everywhere but between

[distant breach]

wait isnt that wut i said tho

OH wait

nvm

it would be like

< x >

[supple mantle]

1 - x > 2/3 and 1 - x < -2/3

It can't have a between thingy

[distant breach]

o

wait

yeh thats what i meant the first time

[supple mantle]

It's bigger than the positive and smaller than the negative

[supple mantle]

Ooo

[distant breach]

o

goshh i cannot comprehend

but ye i think i get wut u saying

ima nap for a hr so i have some energy left to study more 😭

[supple mantle]

Good luckklk

[distant breach]

6:16 am 🥲

[supple mantle]

You got this

[distant breach]

tysm for helping me btw :)

[supple mantle]

I'll be available whenever you need help later today

[distant breach]

Tyyy 🥹

i probs should close this channel but yess ty again 🥹😅

.close

[final saddle]

Channel closed

Closed by @distant breach

Use .reopen if this was a mistake.

[supple mantle]

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[trim temple]

Did I do this correctly?

[silver hedge]

yes

[analog salmon]

yes

[silver hedge]

x=2+y and x=2y would've been enough

[analog salmon]

the double age info isnt necessary

its just extra information

[silver hedge]

you DO need either 2 statements

[analog salmon]

yeah but not all 3

[trim temple]

thankss

.close

[final saddle]

Channel closed

Closed by @trim temple

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tranquil pine]

[
\sqrt[i]i
]

[soft zealot]

♡LexQa♡

[tranquil pine]

Kind of curious on the reasoning behind this root

[soft zealot]

♡LexQa♡

♡LexQa♡

[fiery frost]

Ok

Since i^i=e^pi/2, 1/i^i = 1/e^-pi/2

[strange shore]

you forgot something

[fiery frost]

Oh

[tranquil pine]

hm so why would i^i = e^(pi/2) in this case?

[strange shore]

$i^i = \exp(-\frac{\pi}{2})$

[soft zealot]

Ｈｅｒｅｌｓ

[strange shore]

$i^{\frac{1}{i}} = i^{-i} = \frac{1}{i^i}$

?

[soft zealot]

Ｈｅｒｅｌｓ

[wraith crater]

Lexqa are you asking why i^i is exp(-pi/2)

[tranquil pine]

yes

[strange shore]

$i=\exp(i\frac{\pi}{2})$

[soft zealot]

Ｈｅｒｅｌｓ

[tranquil pine]

hmm

[barren drum]

well
i can be written as cis(pi/2)

and since (cis(x))^n = cis(nx)

[wraith crater]

Strictly in principle branch i = exp(ipi/2) raising both sides to the i-th power gives you the result

[grim nebula]

did you see the youtube short as well

[tranquil pine]

LMAO YES

exactly

[grim nebula]

[tranquil pine]

snow read through me

[grim nebula]

[tranquil pine]

talk about imaginary number shenanigans

[supple mantle]

Blackpenredpen?

[grim nebula]

no

some bri guy

[tranquil pine]

yeah

[supple mantle]

Oh

[wraith crater]

[supple mantle]

Well I saw the i^i video by bprp

[grim nebula]

oh wait no

https://youtube.com/shorts/EhuAMPjuEks?feature=share

[supple mantle]

Lol

[wraith crater]

Oh yes

Brithemathsguy

[grim nebula]

[wraith crater]

I remember seeing the integral of x^x^x^x^… video

That was wild

[grim nebula]

[fiery bluff]

complex exponentials are pretty much multivalued

we need to choose a principal branch for Log

then we can define x^y = e^(y Log x)

[wraith crater]

Actually it’s this one https://youtu.be/gjjo2uKfRvs

[fiery bluff]

but there's many ways to choose the principal branch

[supple mantle]

Ok so

We can express any complex number as $re^{i\theta}$

[soft zealot]

VulcanOne

[supple mantle]

And we know that $i^{\frac{1}{i}} = i^{-i}$

[soft zealot]

VulcanOne

[supple mantle]

But we also know that the magnitude of i = 1

So r = 1

Then we do $i^{-i} = e^{i\theta}$

[soft zealot]

VulcanOne

[supple mantle]

ln on both sides

And we're left with $-i\ln(i) = i\theta$

[soft zealot]

VulcanOne

[supple mantle]

Divide both sides by i

And let's look at i on the complex plane

It's 0 + i

So it's 90°

[tranquil pine]

i regret asking a complex analysis question

[supple mantle]

π/2

[grim nebula]

this was wild

[supple mantle]

So

[wraith crater]

[grim nebula]

[supple mantle]

ln(i) = π/2

Don't forget we had a - sign

So

θ = -π/2

Wait

Something doesn't add up

Oh forgor

We need to do the

Ok so in the ln(i) step

Forgot to make it in terms of

$\ln(i) = \ln(e^{i\frac{\pi}{2}} ) = i\frac{\pi}{2}$

[soft zealot]

VulcanOne

[supple mantle]

And the negative sign

[supple mantle]

Yeah that works

Or do it hard mode

[wraith crater]

“Complex analysis question”

[supple mantle]

Lol just some angles

And some i's

Anyways

[wraith crater]

[supple mantle]

Yeah that's why $\sqrt[i]{i} = e^{\frac{\pi}{2}}$

[soft zealot]

VulcanOne

[supple mantle]

Woke moment

[wraith crater]

[supple mantle]

Did lex sleep or what?

[grim nebula]

no lexy is busy learning how to use latex

[supple mantle]

I need to get into latex more tbh

Lots of things missing to use it 100%

[tulip coyote]

At one point I thought I was good at LaTeX, then people introduced me to some packages 😂

[supple mantle]

😂

I wanna write a research paper with LaTeX someday

But I use word

[tulip coyote]

I’m pretty lazy when it comes to it, as long as I get what I’m wanting

[tulip coyote]

I have to say I respect you for that there

Initially I was gonna try doing my final year project in Word, but gave up on that idea

Thought LaTeX would be nuts, but it isn’t too bad once you get into it a little

Don’t think I’ll ever track the tikz diagram level tho

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

Channel closed

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lucid moon]

In a survey of 120 college students living in the dorms, 60 said that they had only a stereo set in their rooms, 40 said they had only a microcomputer in their room, and 15 said they had both a stereo and a microcomputer in their rooms. The remaining 5 students had neither
If a student is randomly chosen from this group, the probability that the student does not have a microcomputer given that the student has a stereo is

[tired walrus]

1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@lucid moon what stage are you on?

[lucid moon]

3

let me type it

A = Student does not have a microcomp
B = student has a stereo

P(A) = 60+5/120
P(B) = 60+15/120

so I was looking for a formula for conditional probability and I found out that
P(A|B) = P( A and B)/P(B)

Now I tried to get P (A and B) but I’m not sure if I did it correctly… so what I did was I just multiplied P(A) * P(B)

[tired walrus]

missing parentheses here.

[tired walrus]

yeah this is wrong

you assumed that the events "student doesn't have a computer" and "student has a stereo" are independent, which they are not.

(or at least not known to be)

if A and B were independent then P(A|B) would just be equal to P(A) with this logic!

[lucid moon]

Ohh

what should I do to get the P( A and B) then?

or should I still get it?

[tired walrus]

P(student has a stereo but no computer)

it's almost directly given to you in the data

[lucid moon]

60/120

?

[tired walrus]

exactly.

[lucid moon]

Then?

sorr of confused here

[tired walrus]

well you now know P(A&B) and you know P(B)

[tired walrus]

and you have this definition of conditional probability on hand

[lucid moon]

wait I was right in P(B)?

I’ve got 0.8

[tired walrus]

the number of students who have a stereo is 60+15, so yes.

[lucid moon]

Can you verify my answer?

[tired walrus]

,calc 75/120

[soft zealot]

Result:

0.625

[tired walrus]

well, it appears you screwed up the arithmetic.

it's 5/8 and not 4/5 as you said.

[lucid moon]

I mean 0.8 is the answer for the whole question

[lucid moon]

this one is only for event B (student has a stereo)

and I am looking for the probability that the student does not have a comp given that the student has a stereo

so P(B|A) = (60/120)/(75/120)

got 0.8

[lucid moon]

???

[tired walrus]

oh

,calc (60/120)/(75/120)

[soft zealot]

Result:

0.8

[lucid moon]

this is the answer to my question now, right?

[tired walrus]

yes

[lucid moon]

Thanks!

[final saddle]

@lucid moon Has your question been resolved?

Channel closed

Closed by @lucid moon

Use .reopen if this was a mistake.

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[unique fox]

Let 𝑅(𝑚,𝑛) be the predicate "If 𝑚 is a factor of 𝑛^2 then is 𝑚 a factor of 𝑛"
𝑅(25,10)

Why is R(25,10) false?

[tired walrus]

when you say n2 do you mean n^2?

@unique fox

[unique fox]

oops yeah

[tired walrus]

right

can you write out the predicate R(25, 10) in plain english?

[unique fox]

u mean "if 25 is a factor of 100 then 25 is a factor of 10"?

[tired walrus]

yes, exactly.

do you see it now?

[unique fox]

no

because 25 is a factor of 100

or do both sides have to be true

in that case yes because 25 is not a factor of 10

[tired walrus]

for an implication to be true, either both the premise and the conclusion have to be true, or the premise has to be false.

the only way for A -> B to be false is for A to be true and B not

[unique fox]

yeah it makes sense now

.close

[final saddle]

Channel closed

Closed by @unique fox

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[lofty kindle]

why do they take away an i in the second bracket?

[void crest]

i think

its a typo

mis wrote

[lofty kindle]

alright thanks

.close

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[sand dagger]

Can’t do ii and iii

I mean specifically iii

[final saddle]

@sand dagger Has your question been resolved?

[kind glen]

I might be wrong here, but if it’s repeated indefinitely wouldn’t that mean it gets evened out completely?

[sand dagger]

Ya but the ratio wld change again and again

I tried

[kind glen]

Yeah but if they are both the same ratio, wouldn’t it stop changing?

[sand dagger]

Wdym

[kind glen]

I mean

[sand dagger]

Can u show me ur working

I’m rlly confused

[kind glen]

I don’t exactly have any math on paper but think about it

Eventually after mixing a lot the ratio evens out and is the same in both bottles, right?

[sand dagger]

Hm lemme try

[sand dagger]

I tried still can’t figure it out

[kind glen]

Did you discover anything new?

Does it end up with the ratios balancing out?

[sand dagger]

Nop

Just gets way diff

[final saddle]

@sand dagger Has your question been resolved?

[final saddle]

@sand dagger Has your question been resolved?

[thorn dragon]

the total ratio of honey : water is 1 : 2

when you do it infinitely it will tend to 1 : 2

[sand dagger]

How tho

[final saddle]

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[native flame]

Help please, I can't figure this out. I need to do the area of this shape, wrapped around the y-axis. I went through several attempts, including the latest that felt best but still got it wrong somewhere

[final saddle]

@native flame Has your question been resolved?

[final saddle]

@native flame Has your question been resolved?

[native flame]

<@&286206848099549185>

[mint orbit]

wheres your picture?

[native flame]

You can't see it?

[native flame]

This one

[mint orbit]

no i mean

well idk id start by drawing a picture but whatever

,w graph 1/12 * x^2

[soft zealot]

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[mint orbit]

rotated about the y axis, yea?

[native flame]

Yup

[mint orbit]

hmm id be tempted to do wrt x axis and then subtract

,w 9 = 1/12 * x^2

[soft zealot]

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[mint orbit]

,w int from 0 to 6*sqrt(3) of 1/12 * x^2 dx

[soft zealot]

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[mint orbit]

do you know the answer?

[native flame]

I have no idea

The formula for the area I have is different from that integral though..

[mint orbit]

yea

[native flame]

it's $2pi * \int (f(x) sqrt(1+(f'(x))^2)$

[mint orbit]

probably easier to do by y i guess

[soft zealot]

Kienai

[mint orbit]

its rotated about the y axis though

so you can just use like

[native flame]

Ok, so what do I do to change that? This is the only formula I have

[mint orbit]

well more or less $\pi r ^2$

[soft zealot]

jan Niku

[mint orbit]

this is area of a circle at any one y value

[native flame]

I thought I had to change the limit from 9 to 6sqrt3 but that was it

[mint orbit]

unless youre afraid of integrating wrt y its easier to do it this way here

[native flame]

wrt is with respect to?

[mint orbit]

yea, like our integral will end in dy

wait

[native flame]

Right, ok

[mint orbit]

well gimme a sec

lemme scribble

sorry downloading game so everything is running slow lol

but first thought is to get x(y), x as a function of y

then treat x(y) as your radius function, the distance away from the y-axis for some given y

so call it $x(y) = R(y) = \sqrt{12y}$

[soft zealot]

jan Niku

[mint orbit]

then, you imagine calculating the volumes of a bunch of little cylinders

theyll be sorta stacked through the y axis, like that old stacking game where you put the donuts on a spike

each one will have some small height dy

and will have radius R(y)

so you just "sum" up all of these little cylinders

$\int_0^9 \pi \qty( \sqrt{12y} )^2 \dd y$

[soft zealot]

jan Niku

[native flame]

That's it?

[mint orbit]

yea

[native flame]

goddamnit

[mint orbit]

well, its unclear to me which region exactly theyre asking for

can you tell which region this integral is giving you?

and do you know how to switch to the other without re-integrating?

[native flame]

It litterally just says the area that appears when it rotates around the y axis

[mint orbit]

why its good to draw a picture 😄 since to me thats ambiguous

[native flame]

Gimme a sec, I'll try it

[mint orbit]

sure

[native flame]

486pi seems like a lot..

[mint orbit]

we can approximate

like thats what i did bc it does seem like a lot

say you just take this region

the box bounded by the blue and green lines

rotate that around the y axis

it has radius 6sqrt3

and height 9

so

,w 9pi(6*sqrt(3))^2

[soft zealot]

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[mint orbit]

assume our parabola is approximately slicing this cylinder in half

,w 972pi/2

[soft zealot]

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[mint orbit]

pretty close

suspiciously close, really

[native flame]

That's not close

That is exactly what I got

[mint orbit]

yea

bizarre

[native flame]

I tried it though, it's wrong

[mint orbit]

[native flame]

Always so angry with me

[mint orbit]

[native flame]

Are you sure there isn't some way to adapt the other formula I have for this y axis shenanigans?

[mint orbit]

tbh your formula looks like arclength

[native flame]

I've literally never seen it before, and only have the one formula

[mint orbit]

but i am dumb at calculus 😌

[native flame]

They called it rotational area

[mint orbit]

wait

area

?

[native flame]

Yes

[mint orbit]

i thought this was a solid of revolution

they want the surface area?

[native flame]

Yeah, the area of the shape wrapped around the y axis

[mint orbit]

oh

well geez i got volume lol

okay your formula makes more sense then

[native flame]

Explains why it's so big

[mint orbit]

so

[hard vapor]

oh so basically first u wanna sketch this

oops too lazy to properly read

smh

[mint orbit]

$\int _0 ^{6\sqrt3} 2 \pi x \sqrt{1 + \qty(\frac x6)^2} \dd x$

[soft zealot]

jan Niku

[mint orbit]

i think

[hard vapor]

@mint orbit no

[mint orbit]

[hard vapor]

you just integrate along y axis

i think

method if disks or something

[still void]

There is also something called the shell method

[glossy ravine]

volume or surface area?

[hard vapor]

oh this is SA?

my bad i thought it was volume

[glossy ravine]

it can be one of the two

I'm asking

@native flame

[native flame]

It's area, yes

[hard vapor]

Volume I believe is $\int_0^9 \sqrt{12y}dy$ which is what jan said

[soft zealot]

Invictus

[hard vapor]

oh i see

[glossy ravine]

no it's not

you forgot the pi and the squared

[native flame]

How did you arrive at this?

[hard vapor]

yeah 🤦‍♂️

[native flame]

I see the derivative in the sqrt there but why is it a lone x where the original function was in my formula?

[glossy ravine]

ignore what invictus said

it's wrong

[native flame]

I was talking to Jan

[mint orbit]

sorry, i called in the supports

im going by the formula from pauls notes

[native flame]

: )

[mint orbit]

but im bad at calculus

you should trust these other folks

[glossy ravine]

basically what you want to do is write x as a function f(y) of y
and then use the formula you have but replacing x with y

[native flame]

ok, sure

so sqrt12y then...

[glossy ravine]

yep

and plug that into the formula

simplify a bit and it comes out fairly nice

[native flame]

Yeah the simplifying is not my strong suit, hang on

I normally substitute the roots when it's like this

I think otherwise I have to do integration by parts, right?

[glossy ravine]

for this one just use √a√b = √(ab)

[native flame]

🤦‍♂️

Yeah... always forgetting my basics when doing this stuff..

[deft sandal]

Who doesnt

[native flame]

Yeah I still can't get it right..

I integrated the formula and got $2pi * \frac{( 2y+6 )^{3/2}}{3}$ between 0 and 9

[glossy ravine]

yeah and then plug the values in

[native flame]

I can't get the bot right but it's supposed to be ^(3/2)

[glossy ravine]

$2\pi * \frac{( 2y+6 )^{3/2}}{3}$

[native flame]

I did, I got 2pi(24^(3/2))/3 and it's wrong apparently..

[soft zealot]

lambda cube (alison)

[glossy ravine]

don't put it in normal brackets

[native flame]

ok

[soft zealot]

Kienai

[glossy ravine]

also \pi instead of pi

is better

[native flame]

for the latex? right

[glossy ravine]

regardless

just plug values in

[native flame]

But if you got the answer I got, what the hell

[glossy ravine]

your answer looks like it's probably correct

it's in a form i expected at least

[native flame]

oh wait, hjang on

I was getting to used to ignoring the 0 because all my previous forms had no value there

Still wrong, I get $2\pi((8^{3/2} -2^{3/2}))$

[soft zealot]

Kienai

[native flame]

I don't know what I'm doing wrong... I take the upper end where y=9 from $2\pi * \frac{( 2y+6 )^{3/2}}{3}$ and subtract where y=0

[soft zealot]

Kienai

[native flame]

It was 168 pi.. I tried it with one pi and it worked.. somehow

No idea how or why but that's it for me. Thanks for trying to help me understand!

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[native solstice]

Not really sure where to start with this one.

[final saddle]

@native solstice Has your question been resolved?

[native solstice]

<@&286206848099549185>

[vital crag]

Probably should try #numerical-analysis

[final saddle]

@native solstice Has your question been resolved?

[native solstice]

Oh I'm stupid, T has the same eigenvalues as the resulting diagonal matrix

So det(T) = det(diagonal)

so just find det(T) which is trivial

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[rugged fern]

I do not understand how this is wrong?

[rugged fern]

<@&286206848099549185>

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@rugged fern Has your question been resolved?

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@rugged fern Has your question been resolved?

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[rugged fern]

.close

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[wanton sentinel]

why in the first one its in brackets

And the second is modulus?

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@wanton sentinel Has your question been resolved?

[vital crag]

The first one is always positive so taking the absolute value is unnecessary

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[amber belfry]

These two cycles do be the same yeah

[rugged idol]

,rotate

[strange shore]

yes they are

[amber belfry]

Thanks

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[vernal mica]

Question is to derivate the function f(x)
The answer should be 3cosx^2x+3sin^2x

[fluid sky]

u call it "differentiate" not "derivate"

🙂

[vernal mica]

oh

[fluid sky]

but that seems wrong

[ashen shadow]

Ans would be 3cos²x+3sin²x yes but it's exactly 3

[fluid sky]

yeah

[tranquil pine]

aint u ask this earlier? i thought it was solved

[fluid sky]

i mean $f(x)=3x\sin^2x+3x\cos^2x=3x\left(\sin^2x+\cos^2x\right)=3x$

[vernal mica]

i was afk sorry

[soft zealot]

SilverSoldier

[ashen shadow]

Yupp and just solve d/dx(3x) that's it

Which is 3

Remember sin²x+cos²x=1

[vernal mica]

how are these the same tho

this is an example of how my material solves these kind of problems

[fluid sky]

yeah right

[fluid sky]

this is not the derivative

this is the function

i simplified the function

so $f(x)=3x$

[soft zealot]

SilverSoldier

[fluid sky]

to find the derivative u differentiate 3x

[vernal mica]

uhh, do you know a thing called geogebra?

because that gave me another answer

[fluid sky]

yeaha

same thing

do u know $\sin^2x+\cos^2x=1$

[soft zealot]

SilverSoldier

[fluid sky]

from trigonometry

[vernal mica]

yes, i thought that has no use when finding derivatives for trigonometric functions

[fluid sky]

lol nothing has no use

trigonometric identities are very useful especially for integration

even in differentiation

[vernal mica]

same thing as what

the answer i had?

[fluid sky]

same thing as what we said

[fluid sky]

ur answer is wrong

[fluid sky]

what havea u done here

first of all u dont have to use the product rule or chain rule or anything, u can factor out the 3x in the beginning

and then u find that $f(x)=3x$

[soft zealot]

SilverSoldier

[green birch]

Bro forgot abt my existence

[fluid sky]

if u still want to use the chain rule and all that

well what u have done is wrong

[fluid sky]

😝

[vernal mica]

I tried to derivate these two seperatively as two combined functions

[ashen shadow]

,w d/dx(sin²x)

[soft zealot]

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[ashen shadow]

Bro don't make it that complicated

Just simplify and then differentiate

[fluid sky]

but maybe we must show him how to differentiate sin^2x correctly

[ashen shadow]

Yes

[green birch]

Yep

💀

[ashen shadow]

Like use the power rule

[fluid sky]

yeh

[ashen shadow]

For sin²x

[green birch]

Ye

[ashen shadow]

And also for Trigonometric functions you have to differentiate the inside

By using chain rule

d/dx(sinx)²

[fluid sky]

@vernal mica do u know what the chain rule and product rule are

[ashen shadow]

He's Offline 🥲

[vernal mica]

i dont think i do

[ashen shadow]

Oh online

Good

[vernal mica]

but i know what the product rule is

[ashen shadow]

Can you understand chain rule by this or you didn't understand my handwriting

[vernal mica]

do you think its possible that this rule isn't being teached in everywhere

i have my last exam of the course next monday

[ashen shadow]

Well you couldn't solve these without using this rules

Well if you want then don't learn it but then you wouldn't understand differentiation properly

[vernal mica]

oh wait

if u meant that did i know the chain rule, yeah i do

it's just looks different the way its presented in my school

but thanks

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[latent fog]

What am i doing wrong?

[sharp forge]

not sure why you removed the real parts?

[latent fog]

Our teacher removed real parts as we found them in the past

Is it not acceptable?

[sharp forge]

maybe not "remove"

you can keep them on the side, like $\frac{3}{2\sqrt{2}}\cdot\frac{1-i}{1+i}$

[soft zealot]

pramana

[latent fog]

Yes, but outside the Re(---)

[sharp forge]

yeah, you can do that

Once you find Re(--) though, you have to multiply those things you took out and the $\sqrt{2}\cdot4$

[soft zealot]

pramana

[latent fog]

Yes i know, but the correct answer is 0

so the fact that i am left with 2 -2i is wrong

[sharp forge]

oh i see

(1-i)(1-i)

-i times -i is what?

[latent fog]

-i^2

[mellow cedar]

-i * -i = i²

[sharp forge]

but negative times negative is positive no?

[latent fog]

ohhhh, so i would have 1-1 which is 0

thank you a lot

[sharp forge]

ofc 😃

[latent fog]

Thank you once again 🙂

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[mellow cedar]

i thought you are the one who ask in the channel 😆

[sharp forge]

oops T_T

sorry for speaking in rhetorical questions

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[quick bronze]

Ok so like high order derivatives

[dusty harbor]

yeah

123rd derivative

[quick bronze]

Like I know I should probably start with the first four derivatives and a pattern is gonna show up

[dusty harbor]

exactly

[fallow wren]

just do that

[quick bronze]

Yeah I just need to like understand which one the 123rd derivative will fall into

Cus like there's a faster way than just counting and wondering which one it falls into

[dusty harbor]

see what the period is

[grim badger]

Divide by 4, find remainder

[quick bronze]

It's probably really easy but like I zoned out in class for 5 minutes 💀

[quick bronze]

Wait that's literally it

Just divide by 4 and see what happens

Ig

[dusty harbor]

dont just give the answers 🤓

[grim badger]

That wasn't an answer

[quick bronze]

It wasn't

Now I know what I'm doing wowie

[grim badger]

It was a really helpful hint

[quick bronze]

That's all I needed 💀 tyy

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[tranquil pine]

If $X$ is a separable Banach space of infinite dimension, is it true that every linear bounded operator $T: X \to X$ has a closed image?

[soft zealot]

♡LexQa♡

[desert mantle]

I doubt it but I don't have a counterexample rn

[tranquil pine]

Same i am unsure either because it doesn't sound like it should

wait

i dont think it is? @desert mantle

if we consider the identity operator

which is a linear bounded operator

then the image of the identity operator is the entire space X, right? so it is not a closed subset of X

[desert mantle]

how is X not a closed subset of X