help-23

[severe pond]

although slayla basically said it already

[thin lion]

I can give it a go but I think your intuition is much nicer

[severe pond]

yea i don't like definitions that use nonexistence over ones with existence

[thin lion]

Fairs

[severe pond]

another instance of this sort of way of defining things comes up with connectedness if you get to it

[thin lion]

I think I’m just having hard time just juggling real analysis mindset rn

[severe pond]

not sure if this is for a real analysis course or if your course even dabbles in metric spaces/topology

[thin lion]

It’s real analysis

[severe pond]

my analysis course covered metric spaces in the second half

not all courses do though

[thin lion]

I mean I’m up to continuity, have done ratio tests, sequences series etc

I’m just revising rn

And it seems to be more difficult

[severe pond]

well i guess you're studying for finals

[thin lion]

Yea end of term exam

[severe pond]

so it wouldn't be in the course

[thin lion]

Yea

[severe pond]

good luck sir

[thin lion]

Ty ❤️

.close

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[thin lion]

.reopen

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✅ Original question: #help-23 message

[thin lion]

@severe pond sorry for taking ur time again, was this the overall logic we had?

Just rewriting it

Wrong way at the end

b>=M

Definition should also say s of E instead of s in E

It’s calm I see it

.close

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[obtuse kettle]

@opaque fern hiii, im back

[obtuse kettle]

(need some help with part b with determining which parts/sections of each curve i need to subtract to find the shaded area)

[safe radish]

@obtuse kettle Has your question been resolved?

[severe pond]

then double it

you should be thinking of polar area as standing at the pole and shining a flashlight at the wall like i’ve drawn here

the wall for the first bit is the circle

then afterwards it changes to the whatever you call it

C_1

[obtuse kettle]

So for the yellow part, im integrating using C2 and for the red part, im integrating C1?

[severe pond]

yes

[obtuse kettle]

thankss!

[severe pond]

then you double it since it’s symmetrical

[obtuse kettle]

Yuppp

[severe pond]

you’re welcome

[obtuse kettle]

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[prime belfry]

Hi, yuchanie

[safe radish]

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[lean otter]

sum1/(x^(m - n) + x^(n - p) + 1)

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@lean otter Has your question been resolved?

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[lean otter]

I was looking at the gradient vector when studying for my multi calc final, and I still dont understand how it is defined as being perpendicular to the tangent plane, because i remember that the gradient vector is the normal vector for said plane, so the reasoning feels circular. Is there something im missing here?

[median vigil]

the gradient vector is the normal vector for said plane, so the reasoning feels circular
is that the definition they gave for the gradient vector? if not then there is no circular reasoning

[timid ridge]

normals should be perpendicular to the tangent plane

so like, it's just saying the same thing

[lean otter]

but like why is it the normal for the tangent plane

[median vigil]

in this presentation, that's the definition of the tangent plane

which is motivated as being the plane containing the tangent vectors to any curve on the level surface

and as they showed, the gradient is perpendicular to all of those tangent vectors

[safe radish]

@lean otter Has your question been resolved?

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[indigo scaffold]

what are the components of v if v = 2t + 3p -s t = (20,49), p(21,12) and s(-13,18)?

[urban ermine]

hi opal

[normal moss]

First find 2t, 3p and -s

[indigo scaffold]

whats up

[urban ermine]

nm jst OVERWHELMED because the exam got postponed again

[indigo scaffold]

actually i finished it but my answer gave none of the right answers i have 2t 3p and -s

[indigo scaffold]

more tim to study

[normal moss]

Well 2t 3p and -3 aren't the solution

[indigo scaffold]

i put it in the equation wait ill do it again

[normal moss]

You need 2t + 3p - s

So just a bit of addition and subtraction

[normal moss]

What equation?

[indigo scaffold]

oh wait i think i put them in norm

[indigo scaffold]

to find v

i found the norm for the components

then i multiplied like norm for t is 2(t)

nvm

[normal moss]

What are components of a vector

[indigo scaffold]

the sides

all i have to do is multiply t by 2 and p by 3

then add them?

after subtract by s

[normal moss]

$\vec{m} = (m_1, m_2)$

$m_1, m_2$ are the components

[flat frigate]

USS-Enterprise

USS-Enterprise

[normal moss]

In your case

Vector v is $\vec{v} = (v_1, v_2)$

[flat frigate]

USS-Enterprise

[normal moss]

And you get vector v with (2t + 3p - s)

And I assume you have learned how to add and subtract vectors & multiply vectors with scalars when they are given in component form

[indigo scaffold]

so add vector t and p?

[indigo scaffold]

yup

[normal moss]

Then that's all you have to do here

2t + 3p - s

You need 2t, 3p, and -3

So just multiplication by scalars

Then add 2t and 3p - addition

And add -s or subtract s from that

[indigo scaffold]

why multiply t and p

[normal moss]

We don't

We add 2t and 3p

[indigo scaffold]

oh alright

[normal moss]

But we need to multiply t with 2

and p with 3

[indigo scaffold]

yeah

[normal moss]

so multiplication with a scalar

So, what is 2t and 3p, and -s

[indigo scaffold]

ill find it

[normal moss]

👍

[indigo scaffold]

(40,98) + (63,36) - (-13,18)

[normal moss]

Yes

[indigo scaffold]

gotcha

[normal moss]

And how do we do that?

[indigo scaffold]

first we add the first 2

then subtract

[normal moss]

Sure

[indigo scaffold]

it gives (116,116)

uh

wait

[normal moss]

I am blind

[indigo scaffold]

no wories

[normal moss]

Yes

Correct

[indigo scaffold]

alright thank you

[normal moss]

no problem!

[indigo scaffold]

.close

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[jaunty bridge]

Im studying for my midterm but got stuck on this question, how would you do it?

[jaunty bridge]

Dont know where to even start

[white gate]

f(x) = the value of f when x=something

when finding f(2), find where x=2 is and find the value of f(2)

[kind seal]

Keep in mind that a white dot means the point is not included and a coloured dot means it is

[jaunty bridge]

I think I can do it with that

Thank you

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[keen thicket]

Why can we use [-π, π] as the bound of the integration here?

[keen thicket]

Since in the definition of fourier series its 0 -> T

[last wren]

you can use whatever you want as your period really. but it's almost always -T to T or 0 to T

[safe radish]

@keen thicket Has your question been resolved?

[keen thicket]

I can see that after multiplying some graph

Thanks a lot

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[fathom plume]

lf for someone to explain $\int{\cot{^{-1}(x^2+x+1)}dx$ nothing making sense in ts

[flat frigate]

Cera
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fathom plume]

so i converted to $\tan{^{-1}\frac{1}{1+x+x^2}}$

[flat frigate]

Cera

[fathom plume]

but now most i can do is complete sq on bottom

which w subs gives a sec^2x

[tardy mango]

You can apply the arctan subtraction formula on $\arctan \left(\frac{1}{x^2+x+1} \right)$. You can also do integration by parts on the original integral directly.

[flat frigate]

Civil Service Pigeon

[fathom plume]

whats the subtraction formulae??

ive never heard

[tardy mango]

[https://math.stackexchange.com/questions/1724348/what-is-arctanx-arctany]

you can derive the subtraction version by replacing y with -y

[fathom plume]

also int by parts gives two giant polynomials in fraction

doesnt look right

[fathom plume]

ooh thanks ima try that

[tardy mango]

it's doable - never said it would be nice tbf lol

but if you don't wanna do fanc(-ier) trig stuff, it works

[safe radish]

@fathom plume Has your question been resolved?

[fathom plume]

i might have fucked it since it looks a bit too nice but if it is how do i deal w $\sqrt{\tan{x+c}}$ term?

[flat frigate]

Cera

[tardy mango]

gonna be honest idk where this came from

,w integrate sqrt(tan x - 3/4)

[flat frigate]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[tardy mango]

also the antiderivative of the square root term is quite gnarly

you're better off being boring and doing ||partial fractions||

[grim plover]

i am interested in this, i only know how to do this using the arctan identity

[tardy mango]

it's literally just pfd bash

at least I don't see a nicer way off the top of my head

[grim plover]

oh fuck no ew

[fathom plume]

i have $\int{x} d\theta$ where $\tan{\theta}=x^2+x+1$

[flat frigate]

Cera

[tardy mango]

lol

[tardy mango]

oh god

I see where you're going with this now

I thought you were scribbling out all of that stuff lol

[tardy mango]

that's not gonna go very well though

[fathom plume]

ye i might back up and do the original tangent diff identity lmfao 😭😭

i thought by parts is gon be easier

[safe radish]

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[astral glacier]

1)If we equate the given to K we get that LHS of what we have to prove is K (P+Q)

Now I have no idea how to ab+ac+ab

in the denominator

,rccw

[astral glacier]

Oh I think i got it

can someone verify that , it can be solved using multiply letters in denominator and numerator and adding

it

[lost jewel]

We could multiply the terms by a/a, b/b and c/c respectively

[astral glacier]

yaaaaaaaaa

Exactly I think so too

Iamma trya

Yea Done

THnek you guys

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[steady yoke]

In order to try and recover a car stuck in a muddy field, two tractors pull on it. The first acts at an angle of 20° left of the forwards direction with a force of 2250 N. The second acts 15° to the right of the forwards direction with a force of 2000 N. Draw a scale diagram of the situation and find the resultant force on the stuck car.

[steady yoke]

I don’t understand what I means by left and right of the forward direction

[faint hornet]

it doesn't actually matter what you set as "forwards", be it north, or east

the important part is only the angle of what you set forwards as

so, it is for you to choose what forwards is and then you take the angle from there

@steady yoke

[steady yoke]

Okay I did this

@faint hornet

[faint hornet]

looks ethereal 😂

but yeah that's fine

[steady yoke]

Now?

[faint hornet]

it was 2250not 2500 though

[steady yoke]

Sorry yes

corrected it

it’s To scale tho

[faint hornet]

right, now you are just resolving forces

do you know how to do that

[steady yoke]

yeah

I can make it into a parralolegram?

[faint hornet]

ummm i think it's easier to just work out the forces in each direction and then make a triangle

[steady yoke]

okay let’s do that @faint hornet

hellooo

[faint hornet]

hi

i thought you were doing it?

[steady yoke]

No

I don’t know what to do

[faint hornet]

oh, okay np

let me draw you a triangle

can you find x and y here

[steady yoke]

okay so this is the first triangle?

[faint hornet]

yes

[steady yoke]

now what

[faint hornet]

can you find x and y

[steady yoke]

Yes

[faint hornet]

okay, can you do a similar thing for the other side?

[steady yoke]

Wdym other side

[faint hornet]

other "triangle"

[steady yoke]

Yes

[faint hornet]

this one, is the forces going to the left

now do the same thing for the forces going to the right

[steady yoke]

yes I can do that

[faint hornet]

okay

what do you do next

[steady yoke]

I’m not sure

[faint hornet]

try to think, it's okay to take time

[steady yoke]

why would I want to work out x and y

[faint hornet]

okay, what is the question asking for

[steady yoke]

the resualtant force

[faint hornet]

right, and what is that

[steady yoke]

we’ve been taught ywo ways to do that for non perpendicular forces

[faint hornet]

i know what it is myself, i am asking you to tell me

[steady yoke]

the total force

and the direction

[faint hornet]

oh i see why you said this, okay
we can do it this way as well i guess

i was going to tell you how to just do it straight up without scale

[steady yoke]

Can I do it like this then

[faint hornet]

like how you mentioned originally? yes

[steady yoke]

Not been taught that yet

[faint hornet]

ah okay, my bad

[steady yoke]

Question also dsays

scale diagram

[faint hornet]

yeah i know, i just thought it would be nice conceptually to do it like this

[steady yoke]

Understood

[faint hornet]

well, i suppose you don't need more help then? or is there something you still want

[safe radish]

@steady yoke Has your question been resolved?

[safe radish]

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[opaque fern]

So I have been reading up on Tannaka duality and I understand that it allows us to reconstruct a group from its category of reps, but how does the fiber functor come into play here exactly? Since the group we recover is defined as the automorphisms of the functor, does this mean the dependence on the functor mean the group isnt actually intrinisic to the category?

[opaque fern]

@keen tulip

this isnt my only thign in my mind but i feel like its better to just post this first

[keen tulip]

bruh

okay so the basic input for the Tannaka duality is like
[
(CCC, otimes, 1, omega).
]

is that right

[opaque fern]

yes

[flat frigate]

クーリー

[opaque fern]

yes

i mean thats the Tannakian category ig

[scarlet plank]

Are you both doing forbidden maths?🧐

[median vigil]

please only use the channel if you have something productive to add.

[keen tulip]

{}I see wait so the fiber functor
[
omega colon CCC to Vect_k
]
is what lets you forget the group action but remember the underlying vector space basically.

So, for $CCC = Rep(G)$, $w(V)$ is the vector space underlying the representation of $V$, adm morphisms are sent to linear maps. That provides a linear realisation of the abstract tensor category.

[flat frigate]

クーリー

[keen tulip]

silly preamble

So once you have $omega$, you can define
[
G coloneq Aut^{otimes}(omega)
]
which is the tensor-preserving natural automorphisms of $omega$.

[opaque fern]

wot does linear relisation mean

[flat frigate]

クーリー

[opaque fern]

also btw what happens if the functor doesnt exist

are u just

unable to recover the group

[opaque fern]

i get that

[keen tulip]

A linearisation of a tensor category C is a faithful, exact, tensor functor which realises every object of C as an honest vector space and every morphism as a linear map, compatibly with the tensor structure.

I think

bruh why is my wifi not working

[opaque fern]

wtf does honest mean again

one sec

[keen tulip]

I think a category is called neutral Tannakian precisely when such a fiber functor to Vect_k exists.

[opaque fern]

oh its not a term u r just using it for emphasis

ok 💀

[opaque fern]

okay

okay

okay

this is awesome

[keen tulip]

I only mean like in contrastf to something that's abstractly behaving like a vector space inside a category

[keen tulip]

indeed. I shall be helpful in no time

I think that's the term they use honest.

bruh where are we now

[opaque fern]

lets go back to my question

[keen tulip]

alrigt

[opaque fern]

all we have said so far is like

the functor is the tool that translates abstract category to vector spaces

[keen tulip]

I think the crucial part is like it fixes a base point

[opaque fern]

wdym

[keen tulip]

and it ensures every object is realised in the same linear universe, so a single automorphism can act coherently on all objects.

[keen tulip]

just that it's a choice of base point

[opaque fern]

ok i get that

oh yeah

is the group gonna be unique

like are any two fiber functors isomorphic

i think thats basically another way to restate my og question

[keen tulip]

okay okay wait

I think I do recall this

If C is neutral Tannakian over an algebraically closed field (or more generally if a fiber functor exists over the base field), then any two fiber functors are isomorphic. This is true because (I think) the collection of fibre functors is a torsor under the group you are trying to reconstruct. This means any two fiber functors are locally isomorphic, but there may be no canonical global isomorphism between them.

[opaque fern]

one second lemme look up something

[keen tulip]

Why is blud even doing Tannakian

[opaque fern]

wait so

if the fiber functors form a torsor

then effectively

wait

im confusing myself

one sec

[keen tulip]

[opaque fern]

like

the gerbe of the category is gonna be the stack of all the fiber functions correct?

so saying a category is "neutral" is just asserting that this stack is trivial or whateveer?

[keen tulip]

Yeah

[opaque fern]

what about if its not neutral

[keen tulip]

i.e. admits a section

[opaque fern]

does the group still exist locally

ok ofc it does

but then

the torsor has no global point ig?

[keen tulip]

So the group exists locally but doesn't glue to a global group over k

[opaque fern]

what is k

[keen tulip]

The base field

[opaque fern]

oh the base field

ok

ok i get it now

:k

thanks

[keen tulip]

Blud udnerstands it now

[opaque fern]

pogchamp

[keen tulip]

How am I still not green though

This should've given me 5 Greens

[opaque fern]

u barely help bud

[keen tulip]

alright lil blud

[opaque fern]

go help the middle schoolers here

[keen tulip]

You ain't that guy

bruh

I'd rather remain purple

[opaque fern]

LOL

nah bro

the algorithm like

needs u to help a loooooooooot

[keen tulip]

No way

This is gonna be the end of me

who even decided this

I shall have a word

.close

blud can't even close

[opaque fern]

also

btw dude

like

[keen tulip]

blud has no authroity

[opaque fern]

back when i switched accounts

they gaev me access to helpful lounge

manuaklly

without having helpful

bcuz my old account had helpful

[keen tulip]

bruh no one told me about this

[opaque fern]

it only ever happened to me

💀

[keen tulip]

Is it still possible though

[opaque fern]

idk try ur luck. show them ur olad account's id

[keen tulip]

Maybe I should modmail

[opaque fern]

and dm modmail

yeah

thats what i did

[keen tulip]

alright lil blud

[opaque fern]

tell me how it goes

[keen tulip]

okay

[opaque fern]

.close

[safe radish]

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[whole sleet]

Hello

[whole sleet]

Am I doing this correctly

[rigid inlet]

your answer to 2. doesn't look like it's degree 3 from where i'm sitting

[whole sleet]

Yeah

I was gonna ask what degree 3 is referring to

[glacial cairn]

Kinda hard to know what they want tbh

What is that "at, at"

[normal moss]

Well

[vagrant ice]

(x - 2)(x + 3) has leading term x^2 if you expand it out

[rigid inlet]

https://tenor.com/view/abell46s-reface-meme-gif-18920512

type shit

[normal moss]

Judging by the first sentence

[vagrant ice]

so that has degree 2

[normal moss]

"now we reverse the process"

implies we don't see everything

may we see everything 🥺

[whole sleet]

Yes

[glacial cairn]

I think that's just a transition

[whole sleet]

Finding roots

[left igloo]

what is "degree of function"?

[whole sleet]

That is also my question

[left igloo]

don't they mean degree of polynomial?

[normal moss]

Ah yes, very possibly it seems

[whole sleet]

Guys, how can I make it degree 3

[normal moss]

But then this isn't very specific

[whole sleet]

uss-enterprise, I only ask one question

[normal moss]

You need to square one of the brackets

[whole sleet]

Ok

Thank you angels ……. Or should I say angles ahahaahhaaaaaa

How do I close

[normal moss]

.close

I see they forgot to write a number

at "at, at"

So that's why it doesn't make sense

also should be "function with 3 roots"

(╯°□°)╯︵ ┻━┻

[whole sleet]

I’m glad I’m not stupid

[normal moss]

Yeah

But this example formation is very lazy and disappointing

[glacial cairn]

Not necessarily

Presumably they meant there's a double root

[normal moss]

I doubt it

Since "at, at" is written

meaning very likely a number was meant to be placed in between

My guess is they just copied and pasted the first example

But couldn't care enough to change it

Also

Changing my opinion again

Judging by example number 3.

You need to make up requirements which aren't listed yourself

[thick smelt]

you add another root

so it has 3 roots

HOWEVER those roots can repeat

so you could have (x-2)(x-2)(x+3)

which has 3 roots

but one of them has multiplicity 2, because it repeats

[normal moss]

yes that would be my takeaway

It doesn't matter what the third root is

You need to add it yourself

But I stand by my point that these are badly formulated examples

[safe radish]

@whole sleet Has your question been resolved?

Channel closed

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[deep tangle]

how is people so good at math

legit makes 0 sense

[safe radish]

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[open wedge]

so basically

higher mean or higher median

the definition i have here says that higher mean or higher median

i mean

try it for yourself

did the system say that it's the correct answer?

but i think that calculating the mean is typically the stronger option

[vale vessel]

i personally would use mean here

[open wedge]

after getting 15mined for a lot of time, now (ask for pronouns) knows what to do

i'd say im mid in desmos

ah

[meager night]

Most appropriately it's C. Mean gives an overall gist.

[open wedge]

yaya

basic variables

desmos have regression if im not wrong

[meager night]

Let's take a smaller example:

Team A : 66 67 69
Team B : 66 67 68.

Median is 67 in both. Mean differs. Team A, W

[open wedge]

also don't ping helpers when it's not 15 minutes for the sake of god

[safe radish]

@patent musk Has your question been resolved?

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[tight flint]

The ratio of highlighter to felt tips is 1:6. The ratio of felt tips to pencils is 3:2. GIven that she has ten more felt tips than penicls, how many highlighters does she have?

[tight flint]

I did H:F:P = 3:18:12

Then she has x pencils and 10x felt tips but i dont know what to do with it

[vagrant ice]

nope, she does not have 10x felt-tips

[tight flint]

oh nvm 10+x

[vagrant ice]

great, so now 18 parts - 12 parts = (10 + x) - x

[tight flint]

oh okay

so it will be 6 parts = 10

[vagrant ice]

yep

[tight flint]

oh okay i got the answer it is 5

ty for your help

.close

[safe radish]

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[safe radish]

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[granite shoal]

what did i do wrong here?

[granite shoal]

i mean i know its the 5th step but idk why its wrong

[round egret]

I'm not sure why you putted 1.75 at the left top. But when you log both sides in step 4, you get $\log(1.75^t) = \log(\frac{N}{20})$

[flat frigate]

MxRgD

[granite shoal]

i tried to do like the ^ to log

like when you have 2^3=8, 2log(8)=3

[granite shoal]

wym put 1.75? all the way at the top is the question

ohh nvm

[round egret]

Okay i see what you mean, but you don't have to convert them into that form in this case

[granite shoal]

i tried to use this

[round egret]

do you know your log laws?

[granite shoal]

ah okay

[granite shoal]

not all no

i get them on my test

[round egret]

these specfically

[round egret]

ah okay

yeah you need to apply log laws in this case

for example, we can apply the power rule to $\log(1.75^t)$

[flat frigate]

MxRgD

[granite shoal]

yeah thats what they did in the book

[round egret]

yeah, log laws are pretty useful for rearranging to make a specific variable the subject

so in this case, what would log(1.75^t) be equal to then?

[granite shoal]

txlog(1.75)

[round egret]

nice, it would be better to write it as tlog(1.75) as some people might confuse the x as a variable rather than a multiplication

[granite shoal]

i had already seen it in my book though

[granite shoal]

can you tell me what i did here that i can not do?

[round egret]

I can go through and explain the steps if you want me to?

[round egret]

sure

[granite shoal]

no thank you i understand the steps now👍🏼

[round egret]

for example, in step 5 you didn't log both sides

think of log as an operation itself

for example let's say 5 = 5. If you times both sides by 2, you get 10 = 10

[granite shoal]

but is it "illegal" to do my method this way around

[round egret]

yeah

[granite shoal]

because the other way around you can do it right?

[round egret]

it's the same as doing log(5) = log(5)

for both sides

[granite shoal]

ah okay

so you can only do this one way?

[round egret]

yep, it wouldn't make sense if you did multipication on one side and end up with 10 = 5

[granite shoal]

okay

[round egret]

you have to do it to both sides and the same concept applies to logging both sides

[granite shoal]

only thing i dont understand

is why it is allowed here and not the other way around

but maybe i should just let that be idk

[round egret]

sorry if i was gone for a bit

had to do something

[granite shoal]

np

[round egret]

anyways from here that's because it's just using the definition of a logarithim in that being $\log_a(b) = c$ is equivalent to $a^c = b$

[flat frigate]

MxRgD

[round egret]

so they converted between the two

[granite shoal]

but why can i not do it here?

[round egret]

as in logging both sides?

[granite shoal]

no this method of converting

like converting a ^ to a log

[round egret]

oh okay i see what you mean, it's not that you can't do it. You can, but it would not help you rearrange the equation

for example

[granite shoal]

ah okay

i understand

i find it hard to notice when i can and can't do things like that

[round egret]

if you notice, we bring the exponetial down when we do the power rule

[granite shoal]

like how do i know which rule i should use

[round egret]

yeah it's practice

[granite shoal]

okay, i haven't done much of that yet so that is probably the answer then

[round egret]

main thing is when you see something in the form of a^b = c, log both sides to get log(a^b) = log(c), apply power rule to get blog(a) = log(c)

[granite shoal]

okay

i understand everything now

thank you very much

im going to study like until 3:00 probably hahah

[round egret]

hope your studies goes well!

[granite shoal]

thank you

i have the test tomorrow

[round egret]

i'd also like to point something out if you have some time?

[granite shoal]

ofc

[round egret]

there are cases where the power rule won't work take for example log(x^2) = 2log(x)

the problem is that it won't work for x < 0 for the domain of the logarithim

[granite shoal]

okay

I don't think that they will put that in the test tho

[round egret]

yeah this is like

[granite shoal]

but thank you for letting me know

[round egret]

additional understanding

do you know what absoulte values are?

[granite shoal]

no

[round egret]

they look like this |5|

you can think of it like returning the size/magnitude

so |5| = 5

but |-5| = 5

[granite shoal]

hm

[round egret]

so in the case, we can apply this to say that log(x^2) = 2log(|x|)

[granite shoal]

and then you can apply the rule?

[round egret]

yes

[granite shoal]

ah okay

[round egret]

notice how you always get a posistive number?

[granite shoal]

interesting

yes

[round egret]

but yeah that's like a heads up, they can kinda do these weird tricky examples if they wanted to

[granite shoal]

yeah they wont tho

students would get very mad

we only get things that we practice (its high school)

[round egret]

alright, if you have no further questions. You can type .close, to close the channel :p

[granite shoal]

yeah sure

thank you for everything!

.close

[safe radish]

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[patent lotus]

I don't know where to begin

[round egret]

for starters, i'd look at the roots that's being shown

[patent lotus]

What are the roots?

[round egret]

when it crosses the x-axis basically, so y = 0

look at the x-coordinates

[patent lotus]

Oh okay, so that means the roots are x = -3, -2, 1, 2

[round egret]

that's right

[patent lotus]

now we put it into the equation?

y = (x+3)(x+2)(x-1)(x-2)?

[onyx valley]

you have to know the dominant coefficient'

[patent lotus]

no i dont know

[patent lotus]

oh i read the statement wrong i thought u were asking a question

whats the dominant coefficient

[cloud hound]

Essentially, you have the right equation, but you have to figure out what coefficient goes in front: $y = a(x+3)(x+2)(x-1)(x-2)$

[flat frigate]

Coolempire93

[patent lotus]

i think i remember this, we have to use f(0)

[cloud hound]

Exactly!

[patent lotus]

alright let me try to solve the whole problem, when i say the answer could u check it for me?

[cloud hound]

Sure 👍

[patent lotus]

f(x) = 3(x+3)(x+2)(x-1)(x-2)

$f(x) = 3(x+3)(x+2)(x-1)(x-2)$

[flat frigate]

sherman

[patent lotus]

i wrote it in latex just for practice

[cloud hound]

Looks a little large to me 🤔

Plugging in 0 I get $f(0) = 3 \cdot 3 \cdot 2 \cdot -1 \cdot-2 = 36$

[flat frigate]

Coolempire93

[cloud hound]

(I think you meant ||1/3||)

[onyx valley]

oh uh yes sory wrong calculations

[patent lotus]

oh wait I did $4a = 12$ instead of $12a = 4$

[flat frigate]

sherman

[patent lotus]

thank you for teaching me to check my mistakes

[cloud hound]

😆 Always important

[patent lotus]

could u help me with this one as well, i don't know where to begin

im guessing i have to open a new channel, alr see ya guys

.close

[safe radish]

Channel closed

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[cloud hound]

Haha not that I didn't enjoy the last one, I just have no idea how to do this one 😂

[patent lotus]

oh ok. i see

[cloud hound]

But it's best to open a new one, new people will see it and help

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[fathom adder]

Get regex-ed

[quasi timber]

what just happened

[safe radish]

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[graceful gull]

hey,
if we have a function f: R^n --> R^m
how can we determine if f is deffirentiable in R and in a certain point (x_0,y_0), and find the deffirentiability at this point.
I'd be very thankful if there are more than a way.
thanks

[onyx valley]

$f$ is differentiable in $(x_0) in \mathcal R^n$ if there exists a linear map L such as :
$\forall h \in \mathbb R^n$
$f(x_0+h) = f(x_0) + L(h) + o(| |h| |)$

as o tends to 0

[flat frigate]

robins

[graceful gull]

so ill have to find L(h) that satisfy small-o-notation of ||h|| ?

[onyx valley]

yes

but

in fact we know it

this is $\nabla f(x) . h$

[flat frigate]

robins

[graceful gull]

wait is it the same as Jacobian matrix ??

[onyx valley]

yes kind of.. but i made an error

this is for functions $f : \mathbb R^n \rightarrow \mathbb R$

[flat frigate]

robins

[graceful gull]

yeah i got you

[onyx valley]

so for functions $f : \mathbb R^n \rightarrow \mathbb R^m$, you can write $f = (f_1, ..., f_m)$ with every $f_i$ mapping to $\mathbb R$

[flat frigate]

robins

[onyx valley]

actually, for $f : \mathbb R^n \rightarrow \mathbb R$, we have $J_f(x) = \nabla f(x)^T$

[flat frigate]

robins

[graceful gull]

aaah i see that's what i was looking for

i got it thanks robins

.close

[safe radish]

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[safe radish]

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[indigo scaffold]

A motorboat must travel from point A(-1, 4) to point B(5, 7). The units on the plane are in km. There is a 30 km/h wind with an orientation of 25° and a 10 km/h ocean current with an orientation of 110°. The boat's captain wants to complete the journey in 30 minutes. What must be the speed and orientation of the boat's motor?

[indigo scaffold]

why do we not divide 30/2 and 25/2

[noble mango]

Well why do you want to divide them?

[indigo scaffold]

it says in 30 minutes but they're in hour

[noble mango]

Well yeah, so how many hours are there in 30 minutes?

[indigo scaffold]

there isn't

[noble mango]

Well yeah, but we can write it as a decimal

[indigo scaffold]

0.5

[noble mango]

Right, so just use that for your time.

[indigo scaffold]

oh alright

so like i have to divide by 0.5

[noble mango]

Divide what by 0.5?

[indigo scaffold]

wait nvm

[noble mango]

Well, you do have to divide something by 0.5, my question is what :)

[indigo scaffold]

hmm

im not sure

[noble mango]

Okay well before we figure out these details, let's consider the question at hand - what was your initial plan for solving the question?

[indigo scaffold]

well from what i understood

to get to point AB

its influenced by 3 vectors

the motor the current and the wind

so m + c + w = AB

so i was gonna find the components of AB c and w

then i subtract c and w from AB

[noble mango]

Good! I just wanted to interrupt you for a second,

There's a small issue with this approach.

[indigo scaffold]

what is it

[noble mango]

m, c and w, what kind of vectors are they?

I mean,

If you think about the vectors, are they displacement vectors, velocity vectors or are they acceleration vectors?

[indigo scaffold]

acceleration

[noble mango]

Close but nope :)

[indigo scaffold]

ah

[noble mango]

Remember that acceleration is distance/time^2, so if you look at the units of these vectors, they aren't acceleration vectors.

[indigo scaffold]

wait dw about this cause i dont think i learned that

[noble mango]

Wdym

Wait

[indigo scaffold]

like those names

[noble mango]

o

You've never heard of those names before?

[indigo scaffold]

nope

[noble mango]

That's not good, your teacher should be teaching you what those are

Let me quickly go over them

Because it is actually important you know the difference

[indigo scaffold]

my role is wrong im in highschool

[indigo scaffold]

alright

[noble mango]

Yeah this is high school stuff

[noble mango]

Motion is the study of how things move. There are three main concepts that you need to know. The first is distance.

Distance is how far you have 'travelled' so to speak.

[indigo scaffold]

which is AB

[noble mango]

Now, you also may need to know how fast you are going when you travel this distance. We call this quantity speed. Speed tells you how far you travel per unit of time. In other words,
Speed = Distance/Time

[noble mango]

Ehhh well yes, but also no!

[indigo scaffold]

the current motor and wind

is the speed

[indigo scaffold]

its the distance no?

[noble mango]

Now, the last quantity you need to know is acceleration, and it's basically how fast the speed is changing. In other words, it's the speed of the speed.

So, there are three main concepts involved with motion: Distance, Speed and Acceleration.

[indigo scaffold]

i dont understand the acceleration

that can also be the wind cause the wind can be changed

[noble mango]

The thing is, all of these concepts are missing something. In particular, they are missing a direction.

dw about acceleration for now :)

[noble mango]

For example, we might want to ask for distance in a certain direction.

For example, we go 5 units to the right, or whatever.

I.e in principle, we want to link each of these three concepts to a vector.

Hence, there are three different concepts of motion that are related to vectors:
Displacement (the vector representing distance),
Velocity (the vector representing speed),
Acceleration Vector (the vector representing acceleration).

[noble mango]

So, here's the thing.

The vector AB isn't actually distance, it's displacement.

I can prove it:

The 'distance' from A to B using distance formula is sqrt(36 + 9) = sqrt(45)

So in particular, the vector representing AB is the point from A to B with a distance of sqrt(45).

[noble mango]

Furthermore, these aren't 'speed' quantities, but rather velocities.

[indigo scaffold]

alr listen we shouldnt do this

[noble mango]

Because they are vectors that represent speed.

[indigo scaffold]

this is too smart for me

[noble mango]

My guy this is just definitions 😭

All I'm saying is

If you have a vector

And it tells you the speed

We call that vector velocity.

That's it

So the point is that

[indigo scaffold]

alright but what does that name do

[noble mango]

Well it outlines a key problem

You say that

m + c + w = AB

Right?

[indigo scaffold]

yup

[noble mango]

But the left hand side are velocity vectors

And the right hand side is a displacement vector

So we have a problem, because velocities aren't displacements.

[noble mango]

What you are saying here is that if you add a bunch of speed's up with a certain direction, that will give you a position.

But this is not how speed and distance work.

[indigo scaffold]

oh so we cant do m+ c + w = ab?

[indigo scaffold]

oh alright

[austere goblet]

well I mean

does it make sense to you if someone tells you that 20mph + 20mph = 40 miles in front of you?

[indigo scaffold]

nope

[austere goblet]

that's the problem with adding speeds/velocities and expecting the answers to be distances/displacements - the units are not correct

or rather they are not consistent

[noble mango]

Indeed, we can't - but there is a pretty simple fix. How do we turn AB from a displacement vector into a velocity vector?

Hint: What's the formula for speed?

[indigo scaffold]

displacement vector which is a distance vector

into a speed vector?

[noble mango]

Yep.

Well

They aren't the same obviously,

[indigo scaffold]

distance = time/speed

[noble mango]

But there is something in the situation that allows us to find a corresponding velocity vector.

[noble mango]

mmm this looks like a little strange

If you travel at a speed of 2km/hr for 5 hrs, how far do you travel?

[indigo scaffold]

speed = distance/time

[noble mango]

Good! That's better

So,

[indigo scaffold]

10

[noble mango]

Consider the situation

You know you have a displacement of AB,

How would you figure out the velocity of the boat going through AB?

[indigo scaffold]

i have to do speed(velocity vector) = (6,3)/0.5

[noble mango]

Perfect.

[indigo scaffold]

(6,3) is the components of AB

[noble mango]

Yep.

That's exactly it!

So instead of m + w + c = AB, you'd have AB/0.5.

[noble mango]

And that's why you need all of this :D

[indigo scaffold]

everytime i got vectors that are speed and the AB is a trajectory

i have to turn AB into a velocity vector?

[austere goblet]

velocity* vector

[noble mango]

Yes well, it is largely situationally dependent but yeah

The idea is

You want to look at your equation and see if it makes sense

I.e

[indigo scaffold]

but one time i didnt

[noble mango]

You want sum of velocity = velocity

Not sum of velocity = distance

[indigo scaffold]

ill show you

[austere goblet]

OP, are you interested in a way to check your answers thoroughly

[noble mango]

Yeah I know

[austere goblet]

and perhaps catch this kind of stuff

[noble mango]

They might give you the velocity vector already, so you don't need to do any conversions.

It's all about looking at what the question has given you

Well here you did convert something, you converted the 40km/h into the 360km displacement vector.

[indigo scaffold]

ohhh

alright so even here i did something

[noble mango]

Yep

[indigo scaffold]

alright ty lets go back

[noble mango]

Just as a side remark

There is a small caveat

Oh wait actually

Nvm

I was going to say something but I'll say it later

[noble mango]

Anyway, so you have this now.

[indigo scaffold]

what if i replace by (12,6)

[noble mango]

Sounds like you have a decent idea of how to continue.

[indigo scaffold]

yeah

[noble mango]

Yep, that's fine.

[austere goblet]

\[
\frac{km}{\bcancel{h}} \cdot \bcancel{h} = km
\]

just in case

[indigo scaffold]

alright thank you for making me learn that

[flat frigate]

Hanako(x, y); ∂(fox)/∂x

[noble mango]

As I said, this is a pretty fundamental concept, especially in high school :)

[austere goblet]

that was what you were doing earlier with the multiplication

[indigo scaffold]

yeah now i know what im doing more

[indigo scaffold]

yeah

[noble mango]

If you learn vectord without learning what displacement, velocity and acceleration are, you are missing out.

Oh any one more thing, about acceleration:

Acceleration is how fast the speed is changing.

Imagine a car travelling at 10km/h, 20km/h then 30km/h, and so on so forth.

You can see the car's speed is increasing.

[indigo scaffold]

yeah

[noble mango]

So we say that the acceleration is how fast the velocity vector is.

Anyway you won't be using acceleration a lot, it is mostly used in physics and mechanics, but it is still good to know.

I don't think I've ever seen much acceleration vector questions

[indigo scaffold]

alright but i dont think its gonna be accelerationin the exam

[noble mango]

Yea, fair enough

[indigo scaffold]

its mostly velocity and the distance

displacement

[noble mango]

Yep, velocity and displacement 👍

[indigo scaffold]

yea

alright thats good to know thank you so much

[noble mango]

No problemo

[indigo scaffold]

have a good day

.close

[safe radish]

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[indigo scaffold]

.reopen

[safe radish]

✅ Original question: #help-23 message

[indigo scaffold]

@noble mango i got a question

[noble mango]

What's up

[indigo scaffold]

could you also just turn those velocity vectors into displacement or nah

[noble mango]

Ah yes that's also wanted to say

So

Let me ask you this

If a car is travelling at a speed of 50km/h,

How far does it travel after x hours?

[indigo scaffold]

it dependso n the hours

[noble mango]

Right.

So how far does it travel after 2 hrs?

[indigo scaffold]

100 km

[noble mango]

How about 10 hrs?

[indigo scaffold]

500 km

[noble mango]

Good, so how far does it travel after x hours?

Hint: Remember that x is a variable

[noble mango]

Consider how you got these answers

[indigo scaffold]

50x = 500

[noble mango]

Eh well

Drop the equal sign

So like,

After x hours, would you agree the car travels 50x km?

[indigo scaffold]

yeah

[noble mango]

Good so

It is possible to turn velocitt vectors into displacement vectors, but there's a small caveat.

Actually there's two:

[noble mango]

The first one is exemplified here

If you don't know the actual time, you will need to have a variable next to it

So in this case, the variable is x

This is pretty useful if you are given questions where the time is unknown.

[indigo scaffold]

here we know

its 30 minutes

[noble mango]

The second one is that you will also need to factor in the starting position

This is not so much a problem with vector questions

But just to illustrate what I mean,

If you have that a plane starts at (0, 100) and is travelling with velocity (10, 0),

Then to write the vector of that plane, you'd need to write it as:

(0, 100) + (10, 0)x

But this I am guessing is a little too advanced

[indigo scaffold]

yeah

[noble mango]

So don't worry about this for now

[indigo scaffold]

alright

i appreciate it but yeah thats hard

[noble mango]

This one is the important one I would say to remember.

[noble mango]

Just as an example,

[indigo scaffold]

what if i were to just multiply those velocity vectors into displacement

[noble mango]

How would you turn the velocity vector of (1, 2) given that some particle is travelling for 3 hrs under this vector?

[indigo scaffold]

by 0.5

[indigo scaffold]

3x1 and 3x2

[noble mango]

This works too! But there problem is a small conceptual note you have to take in mind

Just to clarify,

It is easier to turn displacement into velocity

Than it is to turn velocity into displacement.

[indigo scaffold]

why

[noble mango]

This will be explained a little more once you go to calculus

[indigo scaffold]

alright

[noble mango]

In displacement, you need a starting position, right?

[indigo scaffold]

yeah

oh so if i dont have a starting position i cant

[noble mango]

(Constant) velocity vectors don't need a starting velocity

All of these problems assume constant velocity

[noble mango]

Displacement vectors do need a starting position though.

[indigo scaffold]

yeah

[noble mango]

In calculus, we call this an 'initial condition'.

[indigo scaffold]

nice name

wait

what if instead of the time

which is 0.5 we had the distance

i can do distance x speed

then it gives me displacement aswell?

[noble mango]

Not so unfortunately

Because distance * speed is not displacement

[indigo scaffold]

oh alright

[noble mango]

Displacement is distance with direction

If you want to be technical about it

Its actually

distance * (vector with distance)

For example,

If I travelled 10 units in the direction (1, 2), what is my displacement?

[indigo scaffold]

(10,20)

[noble mango]

Actually did a bad example because the distance of this vector is not 10 😭

Its technically

(10, 20)/sqrt(1 + 4)

[indigo scaffold]

oh nah

[noble mango]

Because you have to turn (1, 2) into a unit vector

Yeah dw about that if you havent seen it lol

[indigo scaffold]

yeah 😭

[noble mango]

Is this like a geometry vector class?

Yr9 or Yr10 stuff?

[indigo scaffold]

its about linear combinations

forces

[noble mango]

Oh forces

[indigo scaffold]

scalar product

[noble mango]

You might learn it soon actually

[indigo scaffold]

like 210 N and one person pulls it

[noble mango]

Sounds like an actual vectors course

[indigo scaffold]

and it tells me to find the vectical force

[noble mango]

Yeaaah

[indigo scaffold]

like physics

[noble mango]

Anyways if your teacher hasnt covered it, then you probably dont need to know it. But you should eventually stumble across it in your math study of vectors.

Vectors are pretty pretty important

[indigo scaffold]

yeah

alright ty again

.close

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