help-26

[loud oasis]

so the thing is that there are always two angles between any two lines, which are supplementary, and also always two angles between any two planes, which are supplementary

so if you find one of the two angles between the two normal vectors then you find one of the two angles between the planes

and the other one is just the supplement of that

[shut obsidian]

Something like this

Imagine tilting both of the arrows at the same time to the left

This is what I was describing

Both are offset by 90° so they will lie flat in the planes at the same time

[heavy timber]

I do understand everything you just said... but i still have a problem with the proof that the angles are equal... heres a demonstration of my concern

[loud oasis]

i didn't claim they're equal, i said they are supplementary

[heavy timber]

oh

like 180 - x = y

[loud oasis]

but the point is that there is a pair of supplementary angles between the two planes, and also a pair of supplementary angles between the normal vectors

[heavy timber]

may i ask for the defination of the word supplementary

[loud oasis]

adds up to 180 degrees

i have taken the liberty to draw the full set of angles here:

[heavy timber]

and the |cos()| of any two supplementary angles are equal

so thats why we use doty product with | | to evaluate the angle

lovely.

[loud oasis]

we would use absolute value if you want to only get the acute angle, yes

[heavy timber]

well thats one part the teacher explained that we always use the smaller one. we can use the other one but we dont.

well yes everything makes sense now

[heavy timber]

and i get this proof too

ty cloud and kepe.

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[proud flare]

Part b)

[proud flare]

How do i integrate 1/(x-1)^2

[mystic grove]

man

[proud flare]

yes

[restive inlet]

(reverse) power rule (with trivial application of chain rule)

[mystic grove]

think about can you make x+1 as a single term?

and integrate it wrt that term?

[proud flare]

How

[mystic grove]

assum x+1 = a

[proud flare]

Its confusing me

[mystic grove]

integrate it for a^-2 da

[proud flare]

Why

[mystic grove]

you could do it with as x+1

its easier you could do it normally to

[restive inlet]

if you're struggling with what you currently have
doing a substitution can get you something you may be more comfortable with

[mystic grove]

doom um no offense but do you have an general idea abt integration? the question in part a looks like a algebra one

are you doing precalc rn?

[proud flare]

yes

[mystic grove]

leave it for later

[restive inlet]

partial fractions is a technique in integration

[mystic grove]

there are a few rules you need to know

the person who made it probably included it by mistake

dw about it

[proud flare]

This is my work

[restive inlet]

the two parts have a direct link

there is no mistake in the question setup

[mystic grove]

mb

sorry

it becomes ln2x+3/x-1

you have to divide it ?

[drifting swift]

you're missing like 3 pairs of brackets in this one

ln( (2x+3)/(x-1) )

none of these are droppable

[mystic grove]

excuse me

im sorry

Im not used to typing yet

I normally do it on paper

[restive inlet]

no real need to combine them yet

but anyway, lets focus on
$$\int_2^a \frac{1}{(x-1)^2}\dd{x}$$

[obtuse salmon]

Dx

[thorny flame]

ραμOmeganato5

[restive inlet]

would you know how to integrate something like
$$\int \frac{1}{x^2} \dd{x}$$

[thorny flame]

ραμOmeganato5

[proud flare]

lnx^2/2x i think

[restive inlet]

no, that'd be incorrect

are you familiar with the power rule for derivatives?

[proud flare]

kinda

[sharp pagoda]

try thinking inversely

what derivated would you get 1/x^2

[proud flare]

-1/3 x^-3

[sharp pagoda]

no

[restive inlet]

no

[sharp pagoda]

oh yea

yea wrong direction

[mystic grove]

doom

[proud flare]

Yes

[mystic grove]

do you know what the derivative of x^n is?

[proud flare]

Yes

nx^n-1

[mystic grove]

in the question what do you diffrentiate to get 1/(x-1)^2?

basically d/da (A) = 1/(x-1)^2 solve for a

hello::

[proud flare]

Is it this

[restive inlet]

yes

[mystic grove]

yes

[sharp pagoda]

huh

how is that a yes

[mystic grove]

its is?

+1 in the power

[sharp pagoda]

i mean yea but it skipped some steps but whatever

[mystic grove]

ok doom since you are given a and 2 as your limits you can subsitute them

[sharp pagoda]

no you cannot

[mystic grove]

why not ?

Im getting the ans?

[sharp pagoda]

you cant just integrate 1/(x-1)^2

[mystic grove]

why not 🙁

[sharp pagoda]

you need u substitution for x-1

1/(x-1)^2 != 1/x^2

[mystic grove]

man

he did it directly

it isnt a big deal

[sharp pagoda]

it is

the bounds arent the same

[mystic grove]

I swear when ever i type I keep missing few words

[mystic grove]

wait

[restive inlet]

if you keep the same bounds there is no issue here

[mystic grove]

you telling me

if I say x-1 = a and integrate a from 2 to 3 I cant use the same bounds????

[sharp pagoda]

the bounds are 1 lower

[restive inlet]

the application of chain rule is simple enough here that you don't need to explicitly go through that whole substitution process

[native geode]

[mystic grove]

I would just use that rule for multipication

[restive inlet]

there was no sub, there is no need to change the bounds

[sharp pagoda]

I might be wrong tho

[proud flare]

Is this good

[mystic grove]

yes it is

[restive inlet]

seems ok, also that ln(1) is simply 0

[mystic grove]

now just simplify into ln d and c

[sharp pagoda]

yea

[proud flare]

How

[sharp pagoda]

ln(a) + ln(b) = ln(ab)

[restive inlet]

combine logs with sum/difference → product/quotient laws

[proud flare]

Oh yeah forgot that rule

[sharp pagoda]

ln(a) - ln(b) = ln(a/b)

[obtuse salmon]

I dont think the last equality holds. You dont need to sub x-1 back in, since you’ve changed the limits already

[sharp pagoda]

oh yeeeaaaaa forgot about that

ok sry then i was dumb

[obtuse salmon]

Np

[proud flare]

Is it good

[restive inlet]

looks good

[mystic grove]

yes

correct

[proud flare]

Wait but in the question it says c + ln d and i have that extra +1

[mystic grove]

it is included in c

[restive inlet]

the whole -1/(a-1) + 1 will be the c

[proud flare]

Okay solved then tysm guys

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[simple canopy]

im stuck

[drifting swift]

i think you have come here with this question before and we concluded you didn't know anything about complex numbers

have you learned anything about those since then?

[simple canopy]

how do u even remember all that 😭

[simple canopy]

yeah

a little

[drifting swift]

dunno, it stood out in my memory. but it was also only 5 days ago.

[simple canopy]

i tried replacing i with +1 and then got 2 in the denominator then 2 got cancelled and i got left with 3i

[drifting swift]

i isn't 1

so no

[simple canopy]

wait

so i = -1 ?

[drifting swift]

no

i is i

[simple canopy]

oh

[drifting swift]

it isn't a real number

[simple canopy]

so what should i do

[drifting swift]

ok so let's maybe go over some basics

do you know how to add or subtract complex numbers

[simple canopy]

yes

[drifting swift]

and multiply?

[simple canopy]

um

not sure

maybe

[drifting swift]

ok let's set this one aside for a bit

cause you do need to know how to multiply complex numbers before learning to divide them

i would like you to try your hand at multiplying these complex numbers:

(3+2i) * (7+i)

[simple canopy]

alr

[drifting swift]

[we will get back to your original question later! this one is simply to check what you do and don't know]

i am also gonna disappear in 20 min due to an appointment so would like to make it quick if we can.

[simple canopy]

alr

im stuck on 21+17i+2i^2

[drifting swift]

so far so good, now remember i^2 = -1.

[simple canopy]

oh yes

u are right

[drifting swift]

complete the last simplification step

[simple canopy]

5 is the ans ?

[drifting swift]

how did that all get down to just 5

[simple canopy]

oh no wai

im getting 2 now

[drifting swift]

just 2??

show me your work.

[simple canopy]

alr

[drifting swift]

i didn't name myself after her, no.

@simple canopy btw when i say show your work i mean show your work

i don't mean go and redo the thing again and hope the answer will stick this time

[simple canopy]

ya i was in laptop now switching to phone to send u

[drifting swift]

i = 19-17

wtf is this

i isn't an unknown to solve for, it is the imaginary unit.

19+17i is your final answer

[neon iron]

Yeah it is undefined

[drifting swift]

no

i isn't undefined

i is i

[neon iron]

√-1?

[simple canopy]

oh

[drifting swift]

sure.

[simple canopy]

so i was doing it right

[drifting swift]

you were but then your last 2 lines were bs

[neon iron]

Yeah, √-1 do = I but instead of that we don't know the value of √-1

[drifting swift]

anyway here. practice MULTIPLYING complex numbers. then watch this vid. https://youtu.be/EfRRpVB62Ko

[simple canopy]

i didnt know that we dont have to further solve it

[drifting swift]

i is its own thing. stop trying to paint it as "unknown". it is its own thing, not equal to any real number, but also not some foggy "unknown".

that view is very unhelpful when working with complex numbers.

[simple canopy]

thanks ann u are a huge help everytime

[hidden jetty]

Genius

[nova grail]

well $i \not = -1$

[thorny flame]

Kaladin.

[nova grail]

btw

[simple canopy]

so i is just i right ?

[nova grail]

yes

[neon iron]

Yeah you are saying true, √-1 has no value as a real unit that's why it is imaginary and that's what I was trying to say, just wording was wrong.

[drifting swift]

your wording was wrong enough that i spoke out against it.

[neon iron]

Thanks for pointing out

[simple canopy]

even after learning to multiply and divide complex numbers how am i gonna do this

[drifting swift]

well the first half of the question just asks you to work out a itself

and name its real part and imaginary part

[simple canopy]

whats a conjugate ?

[drifting swift]

for a complex number x+iy the conjugate is x-iy

i.e. to get the conjugate of a complex number you flip only its imaginary component

[simple canopy]

the opposite ?

[neon iron]

I meant that i is a whole new concept instead of just real numbers and that also means that "we can't put it in a real value " that's what I meant but it's just obvious

[drifting swift]

no.

the conjugate of 4+20i is 4-20i
the opposite of 4+20i is -4-20i

[simple canopy]

yeah right

[drifting swift]

idk if that was meant literally or in a sarcastic way.

[simple canopy]

real way

[drifting swift]

be careful with SPECIFICALLY the phrase "yeah right". like those words in that order.

ppl may misunderstand you.

[neon iron]

Ann am I following right?

[simple canopy]

mhm

[drifting swift]

sure. i think the time for yapping is over though.

ive given op instructions on what to do next on his own

[neon iron]

I did something wrong for what you called an op for?

[simple canopy]

im closing the channel now guys thank u all

[drifting swift]

?? no

OP means original poster

[neon iron]

Sry for disturbance if I did.

[drifting swift]

that's ouchyaar in this case

[simple canopy]

.close

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[real kiln]

I've been given the following f(f(x)-2y) = 2x-3y+f(f(y)-x) for x,y real. I've been able to prove bijectivity of f but nothing more comes to my mind, i don't want the answer, only a path to continue searching. Thank you

[exotic dome]

The question is to prove f is bijective ? Do we know if it’s continuous ?

[real kiln]

no the question is to find all functions f which respect this equation

it's not given continuous but it may be

Also i saw that f(x) = x + b for some constant b is a valid solution

[exotic dome]

How did you prove bijective

@real kiln

[thorny flame]

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

[exotic dome]

Ok

Pas mal

Tu peux t’arranger pour simplifier l’expression en prenant 2x -3y = 0

Tu choisis un y tel que l’égalité au dessus marche

Ca te simplifie le truc

[real kiln]

oui ca va donner un truc que je pourrais simplifier dans l'équation initiale

mais ca donne f(3/2x) = f(x) - 1/2

et avec ça je n'ai pas su avancer

Let $P(x,y)$, l'assertion donnée.
Let $t \in R $ $$P(f(y)-t,y) \equiv f(f(f(y)-t)-2y)=2f(y)-2t-3y+f(t)$$
$$P(2y, f(y)-t) \equiv f(f(2y)-2f(y)+2t)) = 4y-3f(y)+3t+f(f(f(y)-t)-2y)$$
There's an equal term, so we get $$2f(y)-2t-3y+f(t) = f(f(2y)-2f(y)+2t)) - 4y+3f(y)-3t$$
$$\Leftrightarrow f(f(2y)-2f(y)+2t)) = y+f(t)-f(y)+t$$
taking $y=t$ : $$f(f(2y)-2f(y)+2y)) = 2y$$
So, $f$ est surjective.

Suppose $u\neq v$ such that $f(u) = f(v) $ then
$$P(u,y) = P(v,y)$$
$$\Leftrightarrow 2u+f(f(y)-u)=2v+f(f(y)-v)$$
by surjectiviy, there exists $y$ such that $f(y) = u+v$
We get $$2u+f(v)=2v+f(u) \implies u = v$$
So f is bijective

[thorny flame]

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

[real kiln]

<@&286206848099549185>

[ruby cove]

Yes?

[real kiln]

I'm asking for solving in f the functional equation f(f(x)-2y) = 2x-3y+f(f(y)-x) for x,y real which I've proved bijectivity

is it even the good channel to post this ?

[exotic dome]

If f is continuous the solution is easy. But are you sure continuity is not a given ?

[real kiln]

yes i'm sure it's not given

[exotic dome]

Then perhaps we should prove it separately

[real kiln]

and showing continuity is difficult or impossible so there has to be another way

uh showing continuity ?

[exotic dome]

Tu peux screenshot l’énoncé ?

Je suis juste curieux

[outer portal]

\R isnt defined in your preamble, go DM the bot to modify their shortcut command to \R

[real kiln]

Dm the bot ?

[outer portal]

that means to message the texit bot, here

try sending ,help

or ,preamble

dont do it directly on this channel though, fixing the \R isnt that related to the problem

[real kiln]

ok but still, nobody has ideas ?

[outer portal]

I got something else when I tried to get this

instead I got f(3t) - f(2t) = t, or f(3/2 x) = f(x) + x/2

[real kiln]

my mistake what you got is true i flipped the sign

[outer portal]

you can then rewrite it to look like this
so we know that, for powers of 3/2, the slope between the points is always 1

from this you can try out f(x) = x + g(x)

[real kiln]

yes yes it gives $g(\frac{3t}{2})=g(t)$ for all real $t$ but sadly without continuity you go nowhere

[thorny flame]

Kawachi

[outer portal]

I also got g(g(x) - x) = g(g(x))

g(x) does not need to be injective and does not need to be surjective, since g(x) = 0 is a valid solution (since f(x) = x is a valid solution)

[real kiln]

yupp i totally agree and if you try f(x) = x + b is valid

so we're left either to prove g is constant which i think is impossible

otherwise prove f(x) = x + t in another way

[topaz sinew]

@real kiln Has your question been resolved?

[topaz sinew]

@real kiln Has your question been resolved?

[topaz sinew]

@real kiln Has your question been resolved?

[mint crescent]

I think considering ||P(f(x),x)||, ||P(x,0)||, ||P(f(0)-x,0)|| might work

[real kiln]

It may helps but i don't know how

if i state f(0) = a i got

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

[thorny flame]

Kawachi

[real kiln]

2. $$f(f(x)) = 2x+ f(a-x)$$

[thorny flame]

Kawachi

[real kiln]

3.$$f(f(a-x) = 2a-2x + f(x) $$

[thorny flame]

Kawachi

[real kiln]

oh man i think you got this

Oh yeay it was clever

i'm able to solve this now thanks to you

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[limber halo]

Can I have a starting point for this problem?

[mystic grove]

diagram

[limber halo]

Ok

[limber halo]

Any hints

[mystic grove]

can you try showing that H lies on MC and BN?

[limber halo]

Ok I’ll try that thanks for the hint

.close

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[mystic grove]

wait

hold on

Im DUMB

just prove p1=p2 powers of each

then the chord is radical axis

@limber halo

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[eternal sierra]

i got
f(x) = cos(arctanx - arccot x)

which i wrote as sin(pi/2+arctanx-arccotx) = sin(2arctanx)

but i seem to be missing a (-) sign somewhere?

cause my integration is supposed to give -arctanx+c

but its giving arctanx

[tight nexus]

What's the question?

[eternal sierra]

find y(-sqrt(3))

[tight nexus]

Surely you can just write

$y = \frac{1}{2} \arcsin{f(x)} + C$

[thorny flame]

Miyagi

[eternal sierra]

ok sure

but that still dosent help with the - sign

[craggy marsh]

which -ve sign?

[eternal sierra]

^

[tight nexus]

I got $f(x) = \frac{2x}{1 + x^2}$

[thorny flame]

Miyagi

[eternal sierra]

🤔

thats def wrong

[tight nexus]

How

Look

[eternal sierra]

f(x) = -tan^-1 x + c

according to the solution

[tight nexus]

In the book?

[eternal sierra]

oh u mean the original f(x)

[tight nexus]

What do you mean mate

[eternal sierra]

yeah that is correct

now put it in the differential eqn

and tell what u get pls

[tight nexus]

How about you do it yourself

I already have my answer but you're the one doing the question

[eternal sierra]

lol?

then why even message

[eternal sierra]

i told you the only problem i had was with the - sign

[coral wraith]

sure, but since you're the one who posted the question and are going to have to do the working out anyways, you might as well get on with it

[eternal sierra]

but i did the working out

[coral wraith]

Then show it here

[eternal sierra]

and i had an error in one of the steps

[eternal sierra]

bro i wrote it out ...

what more do u want

[tight nexus]

That working is not going to help you

[coral wraith]

Then you should phrase this out better

[eternal sierra]

and why not?

[tight nexus]

Because its gibberish

[eternal sierra]

how exactly should i have phrased it?

[coral wraith]

"Okay, but I did put this into the DE, and still had this problem - [PROBLEM]; any hints?"

[eternal sierra]

its not, it almost got me to the answer except without a - sign

[coral wraith]

Verbs without a subject in English come off as instructions, which in certain settings are rude, to say the least, if not properly used

[tight nexus]

$f(x) = \frac{2x}{1+x^2} \
\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} \sin^{-1} {f(x)}$

[eternal sierra]

but i dont want to put whatever function he got, i wanted help with my work

[tight nexus]

You just have to solve for y mate

[eternal sierra]

yes, im off by a - sign again

[thorny flame]

Miyagi

[tight nexus]

To be honest I don't know where they got that negative sign from

[eternal sierra]

are u serious

why not say that at the start

rather than waste my time..

[coral wraith]

I'd like to point out that this isn't even much of a DE, since you can just integrate this

[eternal sierra]

that was already pointed out before

.close

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[sharp meteor]

why is it specifically x-a, is it because were shifting the graph p(x) by a?

[sharp meteor]

cause of the approximation about 0, it makes sense

but say now we do approximation about p(x-a) what does that change?

[loud oasis]

approximating f(x) around a is the same as approximating f(x-a) around 0

[sharp meteor]

oh so were just shifting the grpah back to the orgin

then carrying it out

Ohh i see

why is it x-a and not x+a ?

[topaz onyx]

For the most part, convention.

This is not your thread

lmao

[sharp meteor]

yk if we were to not shift it back to 0

how would the eq look liff

it would be f(x-a) = ...

but have f(a)

?

for the derivative points

[topaz onyx]

If you don't shift it back to 0 then the approximation doesn't work

[loud oasis]

actually let's redo it. so let's define $g(x) = f(x + a)$. so $g(0) = f(a)$ and so on. then
[ g(x) \approx g(0) + g'(0)(x) + \frac{g''(0)}{2}x^2 ]
then since $f(x) = g(x-a)$ we have
[ f(x) \approx g(0) + g'(0) (x-a) + \frac{g''(0)}{2}(x-a)^2 ]
then replacing $g(0) = f(a)$, $g'(0) = f'(a)$, $g''(0) = f''(a)$:
[ f(x) \approx f(a) + f'(a) (x-a) + \frac{f''(a)}{2}(x-a)^2 ]

[thorny flame]

κλαουντ ☁ (cloud)

[sharp meteor]

oh wait

i see

[topaz onyx]

The idea behind taylor series is that a function $f$ with infinitely many non-zero derivatives (for example, trigonometric functions) can be represented as the sum of an infinite polynomial,

$$f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} \cdot (x - a)^n$$

where $f^{(n)}$ is the nth derivative of the function $f$.

We take the first 3 terms of this series to reach the approximation you're seeing

[sharp meteor]

this makes alot more sense now

ty

yea, what i was confused with was when it wasnt centred at 0

but we want to always try and shift it

so that it is

?

[topaz onyx]

Oh sorry, I forgot to do x - a

let me edit it rq

[sharp meteor]

calm ty guys

[thorny flame]

@topaz onyx

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[pure talon]

how should I approach this integral?

[pure talon]

this integral looks so evil

it looks difficult to u-sub

[stiff knoll]

it doesn’t look awful

have you tried $u = 1/x$ ?

[thorny flame]

Médicis

[pure talon]

OHH

i tried e^1/x

[stiff knoll]

you’re welcome haha

[pure talon]

howd u figure out it u should be 1/x?

[stiff knoll]

My first thought was that e^1/x is hard to integrate so you better simplify it to e^u

My second thought was that by u-substitution the 1/x^2 would get cancelled out, so that would give you the neat answer

[pure talon]

i actually didnt think of that

I got an answer yippee

now i just need to check it

[stiff knoll]

well played!

[pure talon]

i'll take it that i got it right and i did my math legally

[stiff knoll]

hum

isn’t the answer A ?

show me your calculations pls

[pure talon]

OH MY DAYS I FORGOT A NEGATIVE

[stiff knoll]

yeah since the function is positive on R+ you couldn’t have gotten a negative result (1-e)

[pure talon]

i forgot a negative and that derailed me to get C

[pure talon]

new answer and fixed calculations

[stiff knoll]

well played

[pure talon]

i'll take this as a good sign and i got it correct

YIPPEE

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[proven eagle]

hi, can some explain to me how the denominator breaks down? i'm not getting it

[mint crescent]

$(a^2 +a^2 \tan^2 \theta)^{3/2}=\left[a^2 (1+\tan^2 \theta) \right]^{3/2}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

$=(a^2)^{3/2} (1+\tan^2 \theta)^{3/2}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

$=a^{2(3/2)} (1+\tan^2 \theta)^{3/2}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

$=a^3 (1+\tan^2 \theta)^{3/2}$

[thorny flame]

Civil Service Pigeon

[mint crescent]

where here we used $(x^a)^b=x^{ab}$

[thorny flame]

Civil Service Pigeon

[proven eagle]

THANK YOU SO MUCH!!! ❤️ 🙏🏻

[mint crescent]

!done

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.close

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[royal sky]

what is the level curve related to exact equations in ODE?

[royal sky]

and is it like really importatnt to understand further concepts in ODE

this is pretty much all the context i have

[normal storm]

did u just click on all of the roles

💔

[topaz sinew]

@royal sky Has your question been resolved?

[pseudo sonnet]

given a constant c, a level curve (corresponding to c) of a two variable function f is the set of inputs (x, y) for which f(x, y) = c.

an ODE of the form M(x, y) + N(x, y)y' = 0 is exact if you can find a potential function f satisfying f_x = M(x, y) and f_y = N(x, y). in that case, you can write the ODE as f_x + f_y y' = 0. considering y as a function of x, this equation is equivalent to the equation d/dx f(x, y(x)) = 0 by the (multivariate) chain rule, so f(x, y(x)) = c for some constant c. but this is precisely a level curve of f!! so in other words, the solutions to an exact ODE are given implicitly as the level curves of the potential function

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[north heath]

can someone explain the Rtu thing

[quartz summit]

have you been helped

[north heath]

?

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[chilly walrus]

R_tu is the region over which you are integrating

it's just its name

[quartz summit]

okay so what is the question?

sorry just wondering if it was answered

[topaz sinew]

@north heath Has your question been resolved?

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[north heath]

.reopen

[topaz sinew]

✅

[north heath]

how did this triangle on a u t graph appear

thats the question

[topaz sinew]

@north heath Has your question been resolved?

[north heath]

anyone?

<@&286206848099549185>

[neat jasper]

draw a line from.u=0 to u=t

and from.t=0 to t=M

[chilly walrus]

t goes from 0 to M
u goes from 0 to t

[north heath]

yes but whats that got to do with a triangle

this is a double integral so its calculating volume

[neat jasper]

that shape is a triangle

[north heath]

the shape is an area

[chilly walrus]

the area is the domain

on which you evaluate the function

[north heath]

and where is the function

[neat jasper]

single integral calculating area integrates over a line segment

[chilly walrus]

it's not shown on the graph

[north heath]

so the function is on the 3d space that is inside the triangle cordinates for any z

[neat jasper]

the graph is only the domain of integration

[north heath]

well if it was

[chilly walrus]

exactly

[north heath]

how would it look like

[chilly walrus]

it would be 3d

idk what f and g are

can you send them?

[neat jasper]

this looks like its from proofwiki

[north heath]

just use any functions you want

on a line segment it would be something like this on the line ab

[chilly walrus]

yes exactly

[north heath]

if we have a function on that triangle how would this space we calculate look like

[chilly walrus]

3d volume

let me plot it

[north heath]

can you make a graph where i can see the S same as here for that

[north heath]

ok do that

[neat jasper]

is this the Laplace transform of a convolution?

[north heath]

it is

[neat jasper]

did i write the page? lol i might have, i used to write for the wiki

but this is probably prime mover

yeah he did it for this

anyway, not relevant

[chilly walrus]

this is the volume

[north heath]

can you show me the function itself

without the volume

[chilly walrus]

the full thing?

it looks a bit funny because of z-fighting errors but here you go

[topaz sinew]

@north heath Has your question been resolved?

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[void rune]

simple math question for a distribution is -F(x-l) *(-x+l) = F(x-l)**2

[timber jetty]

$-F(x-l)\times (-x + l) = F(x-l)^2$?

[thorny flame]

This is sad 😢

[timber jetty]

@void rune Is this your question?

[void rune]

yes

[old crypt]

Yes, assuming F is a value or a variable, and not a function

[void rune]

yes F is a variable thx !

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[devout veldt]

can someone explain to me why this diverges

[devout veldt]

.solved

[topaz sinew]

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[neon iron]

hi

[neon iron]

i need help understanding this question

[timber jetty]

hi

[neon iron]

yoo

whats good

whats the difference between a quadric surface/ cylinder and a surface of revolution

<@&286206848099549185>

[tacit lintel]

I’m not too experienced in these types of problems but here’s my guess, it may help. A quadratic surface is figure that is graphed by a parabola or something. A surface of revolution is a function that is rotated about a point to create a figure.

[topaz sinew]

@neon iron Has your question been resolved?

[patent owl]

technically, cylinder = solid cylinder, which you cannot really create by revolving a function. (bonus question: why?)
a hollow cylinder without "lids", however, are also surfaces of revolution. some shapes and surfaces, such as an ellipsoid, are also surfaces of revolution.
however, every possible surface cannot be created by revolving a function. (e.g. a hyperbolic paraboloid or "pringle" shape, or an elliptic paraboloid)

so some quadric surfaces are solids of revolution, but not all. and of course, all solids of revolution are not inherently quadric

[neon iron]

so the surfaces of revolutions that are rotated have their equations constructed from a generating curve radius function?

[patent owl]

yes. there's a generating "radius function," as you aptly put it, and that's revolved to create the surface

though this generally doesn't give you an implicit equation for the surface

[neon iron]

but are their general equations formed that way?

[patent owl]

in practice, not really. the main "application" of a solid of revolution is finding its volume by integrating something based on the "radius function"

(in this case, you have an 'inner' and 'outer' radius to give the shape some thickness)

in theory, you could possibly find a parametrization of the shape, especially in spherical/cylindrical coordinates oriented along the axis of rotation. but whether that leads to a clean implicit surface, i'm not sure (i doubt it)

[neon iron]

i see, also

a trace of a surface is where the surface intersects the planes right

what do they mean by the coordinates of the focus

[patent owl]

the idea is, assume that the intersection of this surface and the plane looks like a parabola (because it does). what's the coordinate of the vertex of this parabola?

i.e. focus = vertex

[neon iron]

so whats up with the y = 5 and x = 2

[patent owl]

well, those are the planes you are intersecting with

[patent owl]

so you have 2 equations:
\begin{align}
\frac 12 x^2 + \frac 14 y^2 &= z \
y &= 5
\end{align}
and both must be true at the same time

[thorny flame]

χασιβ ♥

[patent owl]

think like a system of equations, but instead of a point of intersection, you're now getting an equation of intersection

[neon iron]

but theyre giving me certain planes

and when finding intersects i would set 2 variables = 0

[patent owl]

yeah this will be different to traces

but you can eliminate one variable here (you should expect to get 2, since you need 2 to define a parabola)

any guesses as to which one?

[neon iron]

wait i thought traces are when the shape intersects the axes themselves

when the generating curve lies on the axes

is that not?

well

i think when y = 5 the shape is up right

shifted up?

thats what i assume

wso the center is y =7

i mean 5

<@&286206848099549185>

[patent owl]

im guessing you meant "plane passing through an axis," and not always. generally we define 'trace' to be an intersection with a parallel plane to y=0,z=0, or x=0, such as y=5

[patent owl]

not necessarily. the shape could change between 0 and 5

you can only assume that if the shape is constant between 0 and 5, in this case it's a different parabola

[neon iron]

im talking about the generating curve

i thought the trace was this

@patent owl this is what i mean, is this what the question is asking for ?

i dont understand tbh

i know they want this intersection

[patent owl]

its not always a "generating curve," because the solid is not always a solid of revolution

[patent owl]

which is why i wouldn't think of it like a trace at all, really

but if you want to, just set y=5 instead of y=0

[patent owl]

you'll note that this is the same thing as subbing equation 2 (y=5) into equation 1 and "solving" for a parabola

so there's a few ways to think about this

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@neon iron Has your question been resolved?

[topaz sinew]

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[oblique sky]

Exo 5 eq2

[oblique sky]

Im confused

I assume my path is so far correct

But idk how to continue it

[neon iron]

what happened to the -1 on the left

[oblique sky]

I forgot to write it 🤦‍♂️

What about now

x²≥-2

[neon iron]

looks right but

[neon iron]

whats the second part of the question?

"and...."

"solve the following equatiuons and.."

[oblique sky]

inequations

equations and inequations in R

[neon iron]

ah

[neon iron]

also do you need to present a domain for x?

[oblique sky]

its analysis

i think so yeah

So with x≥1/2

[neon iron]

then x is greater than or equal to 1/2

[neon iron]

yea

do you know why

[oblique sky]

Square root thing

[neon iron]

right

unless youre dealing with complex numbers which i doubt

this is real analysis right

not that im experienced in the field but i doubt complex numbers would be involved in a RA class right

[oblique sky]

The module has both real and complex numbers

But in this chapter it's only real numbers

So S=[1/2,+inf[?

@neon iron just to make sure

[neon iron]

yeah mb i got a little distracted

[oblique sky]

its ok

[neon iron]

yeah but for the infinity end point you cant have a square bracket

[ ] means include the endpoints

() means dont include the endpoints

can you ever include infinity?

whats infinity

[oblique sky]

so [1/2,+inf)?

[neon iron]

do you think it can be -inf?

[oblique sky]

im confused

[neon iron]

no im just asking

like to make sure you understand the bounds

[oblique sky]

in class the prof usually uses [] to include/not include endpoints

[neon iron]

did he ever use ] on infinity?

or she

im not saying you cant use ] ever

[oblique sky]

] on -inf

[ on +inf

[neon iron]

huh...

[oblique sky]

different education systems maybe?

does the symboles change overtime

do highschool students and uni students use the same symboles to include and not include end points?

[neon iron]

well as far as i know its a matter of convinient notation, i dont see inf being includes in the domain

cause think about it

whats infinity

when you "include" infinity by writing ]

what are you including?

you arent reaching any number it just keeps increasing

its a notation convinience thing

and an intuitive thing too now that i think about it

[oblique sky]

ok so theoratically lets say we're not including a real number like 12

[neon iron]

so i dont think its universally correct to use ] in inf

[neon iron]

okay

[oblique sky]

is the interval gonna be [1/2,12[

or is it [1/2,12)

[neon iron]

brother just note that the square brackets go in this order []

[neon iron]

itll be this

youre including 1/2 and values all up until 5 but not 5

so you can include 4.9999

for example

[oblique sky]

so its gonna be [1/2, 5)?

[neon iron]

my bad i came up with my own example in my head of [1/2, 5) and therefore mentioned 5

i just forgot to mention the example

[neon iron]

for my example yes

[oblique sky]

ok

[neon iron]

in your case you arent including 12, so itll be all numbers approaching 12 but not 12

[oblique sky]

i understand what you're saying

i think the symboles differ in each country

i'll go ask the prof tomorrow

maybe he was wrong

maybe he wasnt

but i understand your point in here

@neon iron after some research

it appears that we were both right

[oblique sky]

this is the european way of writing intervals

yours is the standard one

[neon iron]

😂

[neon iron]

interesting

[oblique sky]

Ikr

I never knew that till now

So anyway

How do we close this channel

[topaz sinew]

@oblique sky Has your question been resolved?

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[orchid hatch]

it asks for a+c

isn't that just 62+78

so 140

[exotic dome]

Yaah

[orchid hatch]

ok

and for b it asks for what is b

so is b just 40

180-140?

[orchid hatch]

am i right?

ok

[exotic dome]

https://tenor.com/view/confetti-yas-queen-girl-celebrate-gif-17104362

[orchid hatch]

todays hw is ez

i helped teacher

[exotic dome]

He hates you less (your teacher)

[orchid hatch]

if l and m are parallel what is x

[orchid hatch]

lol

[exotic dome]

This is easy too

[orchid hatch]

yea

is it 62?

wait

[torpid shard]

i think you got the idea right but the math wrong anyways

[orchid hatch]

42

use the triangle?

[torpid shard]

yes

[orchid hatch]

this

In the diagram below, AD is perpendicular to BC and DE is parallel to CA. If ∠BCA = 68◦,
find ∠ADE, in degrees

similar triangles?

[torpid shard]

oh yeah, nevermind

[orchid hatch]

aww

find DAC?

[torpid shard]

didnt understand the idea for a sec

[orchid hatch]

ok

[torpid shard]

Yeah, basically, BE is colinear to AB (in fact, its a part of it) and DE and AC are similar.

[orchid hatch]

ye

[torpid shard]

So they fit the criteria for being similar triangles

[orchid hatch]

is EBD right or smth

no mb

[torpid shard]

Nope, consider the 90 degree you got near it

[orchid hatch]

EBD and ABC are similar

OH

its 180-90-68

[torpid shard]

yep

[orchid hatch]

so 22

In triangle ABC, AB = AC. Let D be a point on side AC such that AD = BD. If ∠DBC = 66◦,
then find ∠BAC, in degrees.

isocelese triangle so

hmm how do i do this

[orchid hatch]

wha

[exotic dome]

Do sum of angles in ABC

[orchid hatch]

ok

is ADB 180-2x

ok

2x+66+y=180

[exotic dome]

What is y

[orchid hatch]

hmm

[exotic dome]

ABC is isocèles

[orchid hatch]

x+66

so 3x+132=180

x=16

so its just 16?

[exotic dome]

Yes I didn’t check the math. But should be yeah

[orchid hatch]

ok

[exotic dome]

Test it

[orchid hatch]

i have a darn pdf this week

let me get online

ok

yea

ima ask him

In the diagram, BC is parallel to DE. If BD = 2, BC = 12, and DE = 21, then find AD.

set up an equation

AB = x

(2+x)/21=x/12

x=24/9?

or 8/3

so in total 14/3

i think

[exotic dome]

The equation is correct

The rest I don’t know

I never do the math

[orchid hatch]

correct

[orchid hatch]

lol

im 1/2 done

In triangle ABC, ∠ABC = 90◦. Points P and Q are on side AB so that AP = PQ = QB, and
CP = 7 and CQ = 5. Find BC

hmm

AC is 9 or smth?

BC is 3???

[rough furnace]

Where did you get that

[orchid hatch]

i got QB as root(8)

[orchid hatch]

1st thing that came to my mind

[rough furnace]

Yeah that's right

[orchid hatch]

ok

[rough furnace]

This's right

[orchid hatch]

do BC is

[orchid hatch]

ye lol AC is NOT 9

BC is sqrt(17)

[rough furnace]

Seems true

[orchid hatch]

ok

then

AC is

root(89)

[rough furnace]

Yeah

[orchid hatch]

ok

In triangle ABC, M is the midpoint of AB, N is the midpoint of AC, and G is the
intersection of BN and CM. The area of triangle GMN is 6. Find the area of triangle GBC

the two triangles are similar

[rough furnace]

Yeah

[orchid hatch]

ok

what ratio though

i think 12

stupid guess

[rough furnace]

I don't think it's 12

[orchid hatch]

ye

its 9 or 12

[rough furnace]

NM should be half BC

That's obvious

[orchid hatch]

oh

[exotic dome]

24

[orchid hatch]

dam

[rough furnace]

Mannnn

[exotic dome]

Not 12

[rough furnace]

nosolsssssss

[orchid hatch]

.nosols

[rough furnace]

[orchid hatch]

!nosols

[topaz sinew]

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.

[exotic dome]

Ok ok

[orchid hatch]

YES I GOT IT IM HIM

[orchid hatch]

but how

i might be randomly picked to go over this problem

[exotic dome]

No sols ! Think !

You got half the answer already

[orchid hatch]

?

its .2 and then another .2

cuz base ratio is 1/2

[rough furnace]

Think about the height of each triangle

[orchid hatch]

and height ratio is 1/2

[rough furnace]

You already have the base of each

Yeahhhh

[orchid hatch]

yea

so 6.2.2

so 24 ye

10 is just area of ABC

[rough furnace]

What are you talking about

[exotic dome]

GMN and GBC are similar

[orchid hatch]

number 10

[exotic dome]

With a ratio of 1/2

[rough furnace]

Ohhh

[orchid hatch]

yea

and are wise that is .4

[exotic dome]

Meaning the ratio between their areas is 1/4

[orchid hatch]

oh shoot

i realized there is a second page ☠️☠️⚰️⚰️😭😭💀💀✌️✌️

quesitons look ez

like that asks for degree ABC

Let A, B, C, and D be four consecutive vertices of a regular nonagon. Two of the sides are
extended to meet at P, as shown below. Find the degree measure of ∠ABC

isn't that just 140

[rough furnace]

Huhhhhh

ABC???

[orchid hatch]

yea

[rough furnace]

So what's P even used for

[orchid hatch]

next question

Using the same diagram as in Problem 14, find the degree measure of ∠APD.

[exotic dome]

Peepee

[orchid hatch]

https://tenor.com/view/luffy-wapol-one-piece-anime-punch-gif-16694500

[exotic dome]

https://tenor.com/view/luffy-gif-24920153

[orchid hatch]

lol

how do you find APD

its a pentagon

so angles add to 540

BAD and CAP are 40 each

Bad boy likes to cap

long DCB and long CBA are both 220

so APD is 20

[orchid hatch]

https://tenor.com/view/sanji-luffy-gif-8363120696706789420

[exotic dome]

What ! 😂

[orchid hatch]

kinda weird lol

[exotic dome]

Me or the 20 ?

[orchid hatch]

no

[exotic dome]

I was thinking of a more complicated way so if you found 20
Easily

Stick with it

[orchid hatch]

oh

so it is 20

i need to find ABC

[exotic dome]

The area ?

[orchid hatch]

yea

[exotic dome]

I’m not sure .. I never do the math

Remind me of what we know

[orchid hatch]

oh no

this diff

its not 20

[orchid hatch]

oh

for that

GBC is 24

and GMN is 6

fot this DPA is 20

[rough furnace]

[exotic dome]

Are you asking me to check ?

[orchid hatch]

WAT

[orchid hatch]

yea

pls

is it correct?

[orchid hatch]

wah

[exotic dome]

I never do the math let’s try it another way

Look at triangle OND

NOD is 90

OND is 140/2 =70

[orchid hatch]

no

its a pentagon
so angles add to 540
BAD and CAP are 40 each
Bad boy likes to cap
long DCB and long CBA are both 220
so APD is 20

[exotic dome]

That leaves ODN = 180 -70 -90 =20

So I agree

Meow

[orchid hatch]

ok

[exotic dome]

https://tenor.com/view/bored-gif-7996817

[orchid hatch]

we have to find ABC

[exotic dome]

I need to get a beer

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wha

[exotic dome]

Ok can we call MNG = X

[orchid hatch]

ok

[exotic dome]

X is the area of MNG ok ?

[orchid hatch]

ye so its 6

[exotic dome]

Do you agree that MNB = 2 MBC ?

[orchid hatch]

mhmm

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Same height , basis x2

[orchid hatch]

yea

[exotic dome]

So 6 + MGB = 2(24+ MGB)

[orchid hatch]

mhmm

wiat

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I just split the areas in this equation

[exotic dome]

I just split the areas here

[orchid hatch]

ok

-42 = MGB/

[exotic dome]

So find MGB

Huh ?

Wait

[orchid hatch]

6+MGB = 2(24+MGB)

6+x=48+2x

[exotic dome]

Wait

[exotic dome]

The 2 here should be on the other side !

It’s MBC = 2 MNB

[orchid hatch]

oh lol

[exotic dome]

Now split each of them to find MGB

[orchid hatch]

i was confused for a sec

2(6+MGB) = 24+MGB

12+2MGB = 24 + MBG

MBG =12

wiat

[exotic dome]

Ok now do you agree that MNB = MNC

[orchid hatch]

ye

mnc?

oh ye

nvm

[exotic dome]

So that means that : MGB = NGC

[orchid hatch]

yea

[exotic dome]

Cause they share MGN