help-36

[exotic rose]

But there is some area, which means that area is being lost in the translation

[pliant shore]

no area is being lost

[exotic rose]

Are we trying to find the curved surface area of the blue vertical slice?

[pliant shore]

the variable of integration in the diagram is dx

[pliant shore]

yes

[pliant shore]

but then if it's east-west, there is no change in x

[exotic rose]

Right. But the curved surface area of the east-west is non zero. The dx is zero so, that translated area is zero.

will you also take the dy projection

and somehow combine them?

[pliant shore]

okay, consider this example

if we take a vertical segment of the line x = 5, so that would be x = 5 and y = t for say, 0 <= t <= 3

and our function is f(x, y) = 10 (or any constant function), such that we have a rectangle as the area

if you go through the calculations, ds = sqrt(dx^2 + dy^2) = sqrt((dx/dt)^2 + (dy/dt)^2) dt = sqrt(0 + 1) dt = dt

so now you have $\int_0^3 10 \ dt$, and indeed that gives $30$

[soft zealot]

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

[pliant shore]

.

yeah, so you don't have to project the slices onto the x-axis

there are some projections which give you zero area

but any other non-zero projection will give you the same area

you don't want to have the case where the area collapses to zero

[final saddle]

@exotic rose Has your question been resolved?

[exotic rose]

How do I take the vertical segment of the line x=5

[final saddle]

@exotic rose Has your question been resolved?

[bronze grove]

huh

path is a circle

take the x y coordinates rcostheta

and r sinotheta?

here in the equation of circle

r = √2

and x + 1 = rcos theta

and then put the values of x and y in surface equation

[final saddle]

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[worldly mesa]

which one

!status

[final saddle]

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

[oak dawn]

Both

I don’t know where to begin

[deep plume]

for a just split them off

[plucky rover]

Are you aware that integral is linear

[oak dawn]

What do you mean?

I’m very new to this

[plucky rover]

God I don't wanna latex this one sec

$\int (a+b) = \int a + \int b$

[soft zealot]

Xavier 🌺

[worldly mesa]

$\int f(x)+g(x)dx = \int f(x)dx + \int g(x) dx$

[soft zealot]

ExpertEsquieESQUIE

[plucky rover]

Are you familiar with this

[oak dawn]

You mean for question 1 right?

1a

[plucky rover]

Both

[deep plume]

it'll work for both

[oak dawn]

Ahhhhh, I’m not too familiar with that. I’ve seen it but I don’t understand how to apply it

[plucky rover]

Okay let's take this one step at a time

Let's simplify 1a

[oak dawn]

Also, is that method still using algebraic manipulation?

No other tricks

[worldly mesa]

yes

[oak dawn]

Okay, I’m ready to learn

can i simplify 1a by factoring out x^2 and reducing it to 5x^4+1 all over x?

[plucky rover]

You can reduce if further

$\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$

[soft zealot]

Xavier 🌺

[oak dawn]

5x^3 + 1/x

right?

[deep plume]

yes

now you can integrate them seperately

[plucky rover]

Those should be standard integrals that you have the answers for

Also note that

[oak dawn]

Yes, I got (5/4)x^4 + ln(x) + c

What does that mean?

[plucky rover]

$\int a f(x) dx = a \int f(x) dx$

[soft zealot]

Xavier 🌺

[plucky rover]

That looks correct yea

[oak dawn]

That’s a question I’ve got, why can you just move the a like that?

[plucky rover]

Intuitively, it's like multiplying one side of a rectangle

Cuz an integral is the sum of areas of small rectangles

If you multiply one of their sides by a, you multiple the overall area by a

[oak dawn]

Okay, that kind of makes sense

For 1b, how can I simplify that? I’ve forgotten some basic rules

[plucky rover]

Well first step is multiply

[deep plume]

you can multiply and distribute e^(3x) and do the same thing by seperating them

[plucky rover]

This time you gotta use another integration rule

[oak dawn]

what does integration rule mean?

i have 1/e^x +e^3x

oh

okay, is it -e^-x +1/3e^3x

e outside of 1/3

[plucky rover]

Yup

[oak dawn]

how to i close this

close

[plucky rover]

.close

[final saddle]

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[worldly mesa]

<@&268886789983436800> spam

[final saddle]

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[brittle vector]

this question is probably super easy but I don’t understand how you get the answer 😭😭 u have to click on it because it’s pretty small

[rare girder]

what have you tried / what tools do you have?

[brittle vector]

i started a level a week ago so i’m pretty new to all of this stuff - so i tried expanding it and using discriminant because i was hopeless😭 but i don’t think that works

and i have a calculator, would it be allowed if i just inputted that into the polynomial part

but i’m just wondering if there is a method or easier way to see the answer

[rare girder]

you're right, discriminants are only valid for quadratics, and the left-hand side is a quartic.

I don't remember what a level is, do you only have algebra and not calculus?

it's probably more rewarding to work out algebraically, but it depends on context.

the easiest method doesn't exist! whatever thoughts you first get, pursue those first. it's healthier to try ideas of your own making, even if they don't work, rather than following someone else's instructions

...it's a quartic

not a cubic

[brittle vector]

uhmm 😭 i’m not sure - i’m 16 so it’s pre university but post gcse,

so by doing it algebraically

i would expand and then

i’d have to put it into my calculator probably

[rare girder]

you might not need computers

[brittle vector]

why is this only 1 mark 😔

it seems like too much work

i got

let me send a photo

ok nvm basically i just got this long quartic

idk what to do

[pastel iron]

try to graph it

[brittle vector]

i only know how to graph it by factorising it

idk how to factorise a quartic 😭 we haven’t learnt this yet

[pastel iron]

differentiate it twice maybe

you'll see something

[brittle vector]

uh what does differentiate mean

[rare girder]

what does differentiate mean

[brittle vector]

oh is that the uhm

[rare girder]

we don't know that

[brittle vector]

the thingy where u do the indices

[pastel iron]

derivative

[brittle vector]

to the number

[pastel iron]

are you not allowed to use calculus

[rare girder]

no it's a calculus thing, sorry

[brittle vector]

i only just started a level idk this stuff

[rare girder]

context matters

[brittle vector]

but yes

u are allowed im pretty sure

idk guys

[pastel iron]

well

you can try to graph it to get a rough idea of the roots but you'll need calculus to be sure

[brittle vector]

huh

wait what is calculus

[warm ether]

youll be getting to that soon

[brittle vector]

okay

[rare girder]

not even soon, it's senior year or college and distinct from algebra

[pastel iron]

she said she's 16 so it'll come soon

probably

[brittle vector]

why is ts one mark

[warm ether]

calculus is covered in a level maths

[rare girder]

ah, ok sure

[brittle vector]

okay so

basically

i need to give up until i learn more

but my teacher expected us to do this now

[wary juniper]

Did you ask like your teacher how to do it

[deep condor]

you said you knew how to sketch based on factorizing. are you able to graph x^2(x-1)(x-3) by itself, and 2-x by itself? where they intersect is where x^2(x-1)(x-3) = 2-x

[brittle vector]

yes and she said she didn’t know am i cooked

[wary juniper]

Bruh

[rare girder]

you can definitely do it with a graphing utility and just stare at the two curves

[brittle vector]

ahhh

[deep condor]

arguably still not a one-mark question, but i think this is the trick

[rare girder]

one-mark if they assume computer graphing

*?

[deep condor]

or that you are a computer yourself

[pastel iron]

if they're allowed polynomial calculators then they can just graph

[brittle vector]

i don’t think i will get desmos in my exam

[brittle vector]

i’m not naturally gifted

a level maths will probably kill me off

[rare girder]

I was staring and hoping there'd be some obvious root or two when you got to this step

[pastel iron]

the roots aren't even nice bro 💔

[brittle vector]

no edexcel is evil

[rare girder]

do you know rational root theorem?

[brittle vector]

not by name but maybe i would if u explained it

i’m not sure

[rare girder]

if p/q is a zero of a polynomial, then q divides the coefficient of the highest-degree term and p divides the constant

[brittle vector]

huh

[rare girder]

by divides I mean, they are factors

q is a factor of the leading coefficient, p is a factor of the constant term

[brittle vector]

wait are u just saying factorising a quadratic

[deep condor]

was just making a little joke, the sketching and "root analysis" stuff will come with practice, so it only gets better from here :D

[rare girder]

no

[brittle vector]

you give me hope 😭🙏

[brittle vector]

i’m trying to understand

i don’t think i get it sorry

[rare girder]

post your polynomial you got in the step with the quartic you didn't like

the one that's equal to x^2(x-1)(x-3) + x - 2

[rare girder]

looks believable

[brittle vector]

appreciate it

[rare girder]

a rational number is the ratio of two integers; they are always of the form p/q where p and q are integers

they are descendants of integers and they're easy to work with

so when you're asked to factor a polynomial without heavy machinery, they either have at least integer or rational roots

[brittle vector]

ohh

wait

what do u mean by the ratio of p:q what would p and q be in this example

[rare girder]

what the rational root theorem asserts, in this case, is that if p/q is a solution to your equation, then p is a factor of -2 (the constant) and q is a factor of 1 (the leading coefficient)

the upshot is that there aren't many factors to choose from, so this limits your search greatly

1 has factors 1 and -1

-2 has factors 1, -1, 2, and -2

[brittle vector]

ahhhh

[rare girder]

so your candidates are all possible ratios of these numbers, the latter set divided by the former set

so really just 1, -1, 2, and -2 in this case

you can check if these are roots, and if none of them are, then you know it has no rational roots

[brittle vector]

so how many roots would it have

[rare girder]

good question I was hoping it'd have at least 2 so you could attack a quadratic

but if not then at least you learned something lol

[brittle vector]

😔

[jovial grotto]

sorry for the message guys

wrong channel

😭

[brittle vector]

it’s ok i’m cooked for maths anyways

[rare girder]

lol np

[brittle vector]

lmaoo

[jovial grotto]

https://tenor.com/view/sorry-bboying-dogeza-gif-8802783

ok continue ima leave

[brittle vector]

😭😭😭

so in this case

all i can do

is graph it

and cry

[rare girder]

yeah that's fine

genuinely I think if it's a 1-mark question then the purpose is to ask if you know what a solution to an equation is geometrically

so if you're interested in graphing it then you might know?

the question is what do you want to graph

[brittle vector]

i checked the answer page and it just said 2

😭

ok well when you draw it on paper

unless u want to be calculating to infinity it is very hard to see if it crosses twice

wait

nevermind

i’m an idiot

ok yes when u graph it it’s very obviously 2

NO

i’m not an idiot

i graphed it on paper without computer

and mine crossed like 4 times

i’d have to have a graphing calculator IN the exam to do it

and for 1 mark

[rare girder]

yeah that's why that one guy said you needed calculus, what happens with the bump between 0 and 1

how high does that go

calculus helps

[brittle vector]

i don’t want to see this question ever again

[rare girder]

you'll see other stuff like it, maybe. that's the point of toy examples, they're meant to make harder variants easy by association

or they're meant to teach you something

points of intersection between curves f and g are exactly solutions to their equations f(x) = g(x)

[brittle vector]

yeah i know 😔

[rare girder]

that's probably the whole point

[brittle vector]

thank you anyways

[rare girder]

or to get you thinking

ye

ok have fun with your book-reading

or exercise doing etc

[brittle vector]

thank you have fun being employed

.close

[final saddle]

Channel closed

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[final saddle]

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[stoic sage]

Hi! i was wondering if someone could help me with this combined-exercise i dont know its name but, i can't get the answer (-1/2) i get stuck at the root

[obtuse grail]

can u translate the question to english

[stoic sage]

how

<@&286206848099549185> can someone help me too get the answer pls 💔

[sage meteor]

convert the thing into fractions first

its easier to see what cancels out that way

[stoic sage]

what thing 🥀

[sage meteor]

the given expression

[stoic sage]

https://tenor.com/view/jgmm-monkey-think-monkey-meme-ponder-monkey-idea-gif-6401133862108294696

[sage meteor]

...do you not get what i said?

[stoic sage]

twin i don't get you 🥀

[obtuse grail]

basically you convert the thing into fractions first
its easier to see what cancels out that way

[stoic sage]

like should i put a big 1 below everything or how 💔

[obtuse grail]

did you unlock a valorant pfp decoration since i was here last 😭

[stoic sage]

BYE

[viral pasture]

why did I just walk into

[stoic sage]

guys im slow ok

[sage meteor]

fine, i'll do that myself

$$\frac{\left(\frac{\left(\sqrt{\frac{3}{2}}\left(-\frac{2}{5}\right)\right)^{-3}}{\left(\frac{5}{12}\right)^2}\right)^{-1}}{\frac{1}{30}}$$

gih

[stoic sage]

OHHHH

[obtuse grail]

@sage meteor what do the : mean from the question

[stoic sage]

isnt that the same

can i like

multiply the root of 3/2

for -2/5

https://tenor.com/view/jgmm-monkey-think-monkey-meme-ponder-monkey-idea-gif-6401133862108294696

[obtuse grail]

im still confused on what this question is

whats it asking for

[stoic sage]

i dont know how to go forward the root

im so confused ok

[obtuse grail]

what is the :

[stoic sage]

how do i multiply this

https://tenor.com/view/katseye-megan-katseye-reaction-katseye-reaction-katseye-meme-gif-4087191036126461463

[obtuse grail]

shes fixing it i think

anyways dawg what does : mean

[stoic sage]

d i v i d e d

[worthy wren]

it means ratio.
3:2 = 3/2

dont try to multiply first. can u split the exponents?

[obtuse grail]

bro u shoulda translated earlier 😭

[soft zealot]

parabolicinsanity

[sage meteor]

ts pmo how to write cube roots but

this is the expression just cube root instead of square root

[stoic sage]

twin my first language is not english 🥀

[obtuse grail]

https://tenor.com/view/valorant-valorant-flex-valorant-toy-valorant-bear-flex-valorant-bear-gif-17962922962962054108

[worthy wren]

lock in chat

[stoic sage]

Yeah so

[obtuse grail]

you finna get banned bro ;sob

[stoic sage]

WHY

[worthy wren]

.

[sage meteor]

remove it

[stoic sage]

IM 15 MIN OLD

[stoic sage]

yes thatsit

[obtuse grail]

lol its not but close enough

[sage meteor]

Its a cube root though, i just don't know how to write them

:p

[obtuse grail]

$int$

[sage meteor]

oop

[stoic sage]

🆘

[obtuse grail]

auto emoji ggs

[stoic sage]

back again

[sage meteor]

now apply the laws of indices

you'll be fine

[obtuse grail]

just multilply everything by 30 lowk

[stoic sage]

what are indices

[sage meteor]

https://tenor.com/view/балдёж-класс-имба-весело-лайк-gif-14918847556827868253

[sage meteor]

exponents

[obtuse grail]

what is the law of indices 😭

[stoic sage]

OHH

but

what about the cube root * the another frac

[sage meteor]

example?

[obtuse grail]

1/3 * -3 dawg

-1

free

[sage meteor]

😃

[obtuse grail]

mr gamer thats enough from you

[obtuse grail]

is that wrong 🙏

[stoic sage]

lemme

use your method

and i'll be back

[obtuse grail]

yea my method is peak

[stoic sage]

hii

my friends

how do i

multiply the indices

like in what order

[obtuse grail]

you can do it in any order

but you should do it in the order that makes it easiest

[stoic sage]

can i do

wait

-3/5 * 12/5

and then multiply the exponents

[obtuse grail]

like clearly the cube root is hard to deal with so u wanna distrubute the -3 first

[stoic sage]

https://tenor.com/view/jgmm-monkey-think-monkey-meme-ponder-monkey-idea-gif-6401133862108294696

[stoic sage]

oh so it was

lemme erase

[obtuse grail]

where is -3/5 coming from 🗓️

[stoic sage]

cube root of 3/2 * -2/5...

= cube root of -3/5-

[obtuse grail]

ur cooked dawg

[stoic sage]

twin

help m

[obtuse grail]

https://tenor.com/view/cut-it-out-dude-gif-7309611

[stoic sage]

https://tenor.com/view/katseye-sophia-smile-drop-sophia-laforteza-surprised-face-drop-gif-13315436907339782073

BUT

[obtuse grail]

u cant multiply if they have different exponents

[stoic sage]

HHOW DO I

lemme translate

[obtuse grail]

so if its not cube rooted u cant multiply

yo hablo espanol

[stoic sage]

how do I raise 2 independent fractions to NEGATIVE 3

[stoic sage]

u do not

[obtuse grail]

si

mi espanol es bien

por el resto del conversacion solamente espanol

[stoic sage]

okei

entonces

como elevo 2 fracciones independientes a 3 negativo...

[obtuse grail]

no puedes

necesitas distrubutar

el tres

[stoic sage]

distributar el 3

https://tenor.com/view/jgmm-monkey-think-monkey-meme-ponder-monkey-idea-gif-6401133862108294696

[obtuse grail]

si

[stoic sage]

como hago eso

[obtuse grail]

multiplicar

[stoic sage]

multiplicar que

😭

[obtuse grail]

el tres con el otro exponents

[stoic sage]

estoy cocinada

[obtuse grail]

sabes los derechas de exponenettes

[stoic sage]

deberia multiplicar

-3 * 2

primero

??

[obtuse grail]

no

-3 y 1/3

[stoic sage]

entonces 😭

[obtuse grail]

pregunta

[stoic sage]

DE DONDE SALIO 1/3

[obtuse grail]

como es mi espanol

uno a dies

diez

[stoic sage]

un siete

[obtuse grail]

muy fuego

[stoic sage]

pero de donde salió 1/3 🥀🥀🥀

[obtuse grail]

uhhh

el rooto cubo

[stoic sage]

https://tenor.com/view/jgmm-monkey-think-monkey-meme-ponder-monkey-idea-gif-6401133862108294696

QUE CUBO 💔

[obtuse grail]

eres cocinado

no hay hope

[stoic sage]

SI HAY

AYUDAME PORAVOR

https://tenor.com/view/katseye-megan-katseye-reaction-katseye-reaction-katseye-meme-gif-4087191036126461463

[obtuse grail]

cubo root es igual con el 1/3 exponento

[stoic sage]

enserió

AHHH

YA ENTENDI

la raiz cubica es lo mismo que 1/3

https://tenor.com/view/ardilla-bailando-gif-27331221

verdad...

[obtuse grail]

sii

[stoic sage]

gemela voy a comer

ahora vuelvo ok

@obtuse grail

hermana

entonces

volviendo al tema

si hago la operacion de 1/3 * -3 y el resultado que da como exponente

luego hago sin problemas 3/2 * -2/5?

y lo elevo al resultado de 1-3 * -3?

@obtuse grail

twin are we alive

i didn't get -1/2

.close

[final saddle]

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[junior turtle]

,rccw

[final saddle]

@junior turtle Has your question been resolved?

[wary juniper]

So we have ECD =5x, ECB=3x, DCB =8x (for some angle x)
If you can find DEC in terms of x then you can use the outside angle theorem to find x

[junior turtle]

how to find DEC?

[wary juniper]

Triangle EDC and DAC should help

[junior turtle]

DEC=34+3x right?

oh no I calculated DEC=34+3X34=136

sorry ok now I calculated DEC=2X34=68

what's should I do next?

[wary juniper]

..how did you find DEC?

[junior turtle]

8x+34+ADC=180 so,ADC=146-8x,then CED is a triangle so ADC+5x+DEC=DEC+146-8x+5x=180,am I right?

[wary juniper]

Yeah so you get DEC=?

Oh wait.... fuck im dumb you still cant find x

Uh maybe try finding all the angles in terms of x or sth and see if you can get an equation for x (many of the angles are congruent so not as much of problem as you might think)

[junior turtle]

I know how to do now sorry name I forgot to say CE is a diameter,so EDC is 90 8x+90+34=180 then 3x=ECB and also ECB=EDB because arc EB so EDB=21

it's my fault

[wary juniper]

Damn

Welp at least its solved now

[final saddle]

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[proven bone]

hello

[proven bone]

how do i find the deriviative of the fraction

so so far

it would be

6+ (dydx{frac

[odd seal]

can you write 1/x as x^(n)?

[proven bone]

let me check

what is n here?

[odd seal]

thats what you need to find

[proven bone]

alright

let me compute

[odd seal]

once you find that n, you can simply use the standard formula for the derivative of x^n for that value of n

[proven bone]

could i get a hint please

[odd seal]

(1/x) * x^1 = 1

look thru your textbooks of past year or smthn to see the laws of exponents

[proven bone]

ok

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[zealous oar]

Can anyone help me here

[zealous oar]

,rotate

[zealous oar]

$\ \lim_{x\to 0} 1/x \times sin(\pi/(1-x) )$

[soft zealot]

AymanRhz

[zealous oar]

Yeaah like that

I couldn't solve it

Can anyone help

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[timber plume]

I need help w this q

[timber plume]

number 4

my answer was x^2/2-ln|2x|+c

[versed crater]

Um

[versed crater]

Try differentiating this

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@timber plume Has your question been resolved?

[rain sentinel]

you havnt integrated correctly

take the constant out

so you would take the 1/2 out and have 1/2 integral 1/x

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[timber plume]

Do u mean like this

[timber plume]

like this?

[tired walrus]

derivative of ln|2x| is not 1/(2x).

and derivative of x^2/2 is not 2x

[timber plume]

derivative of x^2/2 is x

No worries I get it now

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[rain sentinel]

Yes it's that

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[sonic crescent]

How do you solve this?

[sonic crescent]

Do i need to sum up P(X = 190) to P(X = 300)

[bold zenith]

Ignorance University 💀

[bold zenith]

Definitely a job for a calculator

[sonic crescent]

The crazy thing is he won't let us use anything besides a basic arithmetic calculator

So like we cant do summations

[bold zenith]

Are you allowed to do integration or normal distribution?

[sonic crescent]

Yeah we can, by hand

[bold zenith]

Side note: This question is badly worded, as it depends on how many students there are. I would assume there are infinitely many students and we’re just taking a sample of 300 of them, (like if there are exactly 300 students at the university, then there cannot be more than 60%*300=180 nursing majors so the probability would be 0)

[bold zenith]

The note: part of the question suggests we should find the variance I guess

[sonic crescent]

Yeahh my professor is notoriously bad at wording lol

[sonic crescent]

I would assume he meant infinitely many students though

[sonic crescent]

I suppose but how does it help find the probability?

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[bold zenith]

Use it to make a normal distribution approximation and then I guess see how many sigma it is away from the mean?

[bold zenith]

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[steady vale]

I need help with question 2b

[steady vale]

ill show the way i solved it vs the teacher

theres a discrepancy in one of our numbers

instead of the 0, my teacher got a 1 but i don’t understand how

could someone clarify where i might have went wrong?

[bronze grove]

in the seconf one

-1 * -1/2

will be 1/2

not -1/2

so it will be 1/2 + 1/2

[steady vale]

OHHHH

ok thanks so much!!!

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[dire sky]

May anyone help me to understand part a ?

[dire sky]

I understand that parallel vectors are having cross product of zero and we got each component to make it equal to zero

I dont get how we got Bx By Bz

[pliant shore]

there's actually no need for the determinant method at all

if B = A, then of course B is parallel to A

then we can multiply B by a scalar multiple and it will still be parallel to A

[dire sky]

That's how my teacher solved it

[dire sky]

U mean A=kB?

[pliant shore]

I must assume the other condition is that $(B_x, B_y, B_z)$ has unit magnitude

[pliant shore]

indeed

[soft zealot]

south

[pliant shore]

the magnitude is $\sqrt{5^2 + (-2)^2 + 1^2} = \sqrt{30}$

[soft zealot]

south

[dire sky]

Alr stupid question.. why not B=kA?

[pliant shore]

they both represent the same thing

if B = kA, then (1/k) * B = A

k and 1/k are both scalars

[dire sky]

Makes sense

So we got magnitude of A

[pliant shore]

yeah and then if you divide every component by sqrt(30), the magnitude will now be 1

[dire sky]

U mean getting unit vector?

[pliant shore]

yes

normalising the vector

[dire sky]

Why we'd need to get unit vector tho

[pliant shore]

I don't know

you don't need it if the only condition is that B is parallel to A

[dire sky]

I don't remember this tbh

This problem is weird

[pliant shore]

exactly

[dire sky]

I spent 4 days trying to get it

Alr man

Thanks for ur time

[pliant shore]

no worries!

[dire sky]

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[sonic crescent]

I am not sure if you knew of this, but I have now learned that there is a normal approximation to binomial distribution, which can be used for problems with large values like this.

A question of similar size to the question I created came up in my homework, and the only reasonable way for me to solve it is via approximation.

[vital crag]

wut

don't ping me out of nowhere, thanks

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[zinc kayak]

I can't figure out how to deal with this exponent

[green grail]

perhaps u can break it down?

divide sin^3(u) by sin(u) and what does it give u

[zinc kayak]

sin^2(u)

[green grail]

yes so take out a sin(u) and re write it as sin(u)(1- sin^2(u))

[late rose]

factor the denominator and then apply pythagorean identities

afterwards, you can simplify by comparing with the sine double angle formula on the denominator (or use product-to-sum rule)

and it should be smooth sailing from then on

[zinc kayak]

I'm confused on how to factor sin³u would it just be this?

[green grail]

so many sin(u)

[late rose]

factor sin u from the denominator.

that is, sin u - sin^3 u

[zinc kayak]

Pardon the handwriting

Forgot how much I hated factoring. Like this?

[late rose]

you just ended up the same thing...

[zinc kayak]

No I redid it just to make sure I was right so the factored expression would be the second line right?

[late rose]

oh yes the second line is correct

but note that $a - a^3$ can be factored as $a(1-a^2)$

so you can just skip straight to that

[zinc kayak]

So it would be
$sin(u) (1-sin^2(u))$

[late rose]

no

[soft zealot]

Python The Mage

[late rose]

yeah

that's correct

[zinc kayak]

Woops

Ok so would I break that down then like this

[late rose]

no you cant do that

[zinc kayak]

Okay I thought I was wrong putting the second cosine there

Or is that no it

[late rose]

you cant break fractions like that

the numerators will be different

try something else

[late rose]

to clarify, if you have $\frac{a}{bc}$, then you cannot split it as $\frac{a}{b} \cdot \frac{a}{c}$

[zinc kayak]

Right okay I thought that felt weird I just wanted to make sure

[late rose]

yeah

[zinc kayak]

I have this but now I'm having trouble understanding where to go from here

[late rose]

do you remember you pythagorean identities?

[zinc kayak]

I'm still getting used to those we just learned them in the last week

[late rose]

just curious, what level of schooling are you at?

[zinc kayak]

Precalculus 2 as a college class so we're moving pretty fast and it's kind of hard to retain.

[late rose]

pythagorean identities only in college is pretty surprising

[zinc kayak]

I dropped out before I could take it in high School.

[late rose]

oh well you'll need to apply the pythagroean identities here

if you remember them, can you state them here?

[zinc kayak]

sin^2(theta) + cos^2(theta) = 1
1 + cot^2(theta) = csc^2(theta)
tan^2(theta) + 1 = sec^2(theta)

this is what I have written down in my notes

[vivid fog]

so which one would be most helpful here?

[zinc kayak]

(1-sin^2(u)) would cos^2(u) right?

[vivid fog]

yep perfect

[late rose]

👍

[late rose]

yep

i think you can figure out the next step

[zinc kayak]

I'm assuming I divide cos(u) by cos^2(u)

To make the bottom cos just cos(u)

[late rose]

yup

you also have to divide the numerator

[zinc kayak]

Right so is this correct

[late rose]

yup

[zinc kayak]

Thank you for being patient with me. That actually helped a lot cuz it forced me to figure out the Pythagorean identities and how to actually apply them so thank you.

[late rose]

no problem

it's not over yet though

you can simplify it more

[zinc kayak]

I can simplify it further into secant and cosecant

[late rose]

nono there's a better simplification

[zinc kayak]

I realize that when I typed it in but in my online homework took the first one anyway

Oh

[late rose]

do you recall your double angle formula?

for sine

[zinc kayak]

I don't think we've covered that yet.

[late rose]

well do you know your addition formula for sine?

$\sin(a\pm b)=\sin a\cos b\pm\sin b\cos a$

[zinc kayak]

Oh yeah I've seen that. I have it somewhere in my notes. I'll do more research into that but I have to work now but thanks a lot

[late rose]

oh alright

so if we let $b=a$ then we have
$$\sin(a+a)=\sin a\cos a +\sin a\cos a$$
which simplifies down to
$$\sin(2a)=2\sin a\cos a$$
this is the double angle formula

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[zinc kayak]

Yes

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[sand patrol]

i dont understand this question, what it means, and how to even begin....

[sand patrol]

i know i just need to prove something is reflexive, symmetic, and transitive lol

[bleak granite]

and what have you done thus far?

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[rocky tusk]

simplest way is to consider the asymptotes

what are the asymptotes for sec and csc

i.e. when is cosx = 0 and when is sinx = 0

sure

[craggy plank]

@slow escarp I’ll close your other channel since knief is here

[rocky tusk]

you didn't answer my question

why -3?

did you plug in 0?

oh is it at -4

looks like -5 if you don't click it

yea

i was just saying if you don't click on the image it looks like -5 instead

seems so

you would just put -secx - 3 {-pi <= x <= pi}

i have to go eat

have a nice night sir

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[olive dirge]

What happened between steps 2 and 3 that allowed the equation at step 2 to turn into the one at step 3?

[royal gust]

Multiply top and bottom by √x and √x + Δx

[rocky tusk]

get a common denominator

[bleak granite]

subtract fractions in numerator

and (a/b)/c=a/(bc)

lol three different answers at once

[olive dirge]

That was my idea as well, but wouldn't that cause the numerator and denominator of the upper fraction to cancel out? I'll visually describe it.

[bleak granite]

[\frac{\frac{1}{\sqrt{x+\Delta x}}-\frac{1}{\sqrt{x}}}{\Delta x}=\frac{\frac{\sqrt{x}-\sqrt{x+\Delta x}}{\sqrt{x+\Delta x}\sqrt{x}}}{\Delta x}]

[soft zealot]

PajamaMamaLlama

[olive dirge]

Ooohh.

[royal gust]

Remember that when you multiply the top by something, both terms get multiplied by that thing

[olive dirge]

Right.

Thank you.

[bleak granite]

in general 1/a-1/b=b/ab-a/ab=(b-a)/ab, you did 1/a-1/b=a/a+b/b (not equal!)

[olive dirge]

I'm gonna do this on paper.

[olive dirge]

What are these... equivalences called again? Don't they have a name?

Like what you showed, with 1/a - 1/b = b/ab - a/ab = (b-a)/ab ?

[bleak granite]

just simplifications

I'm getting a common denominator

[bleak granite]

tho the way you've done it is wrong

[olive dirge]

Whoops.

For these sorts of fractions, if you want to remove the bottom-most thing, don't you reverse it, so Delta-X becomes 1/Delta-X, and then you multiply that, as a separate fraction, with the original?

Wait, I'm stupid.

Never mind!!

Now how do I get the bot to close this channel...

[bleak granite]

.close

[olive dirge]

Thanks.

.close

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[toxic cypress]

Prove that $A_n$ does not have a proper subgroup of index < n for all $n \geq 5$.

[soft zealot]

Zerofall

[toxic cypress]

I am honestly struggling a lot and I have NO clue where to start

Dummit and Foote provides a theorem and a proof for said Thm, it is that A_n is simple for all n \geq 5

The proof starts by induction, is this the same route I should tkae for this proof?

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[tranquil pine]

is this graph right?

[craggy island]

That function is three-dimensional

You input two numbers and obtain a third

[tranquil pine]

what

[craggy island]

You can't graph that in two dimensions

[tranquil pine]

do u know what ur saying

[craggy island]

https://www.desmos.com/3d/jxdroah1bq

That's the function

Unbounded of course

[tranquil pine]

but my teacher asked to sketch the graph

“find the domain and sketch it on the xy plane”

[craggy island]

The domain

Not the function itself

You're correct then, but you should have specified the question

My bad

[tranquil pine]

oh sorry i should’ve clarified

[craggy island]

That's alright mate

[tranquil pine]

i’m having a lot of trouble with this maths

[craggy island]

Well you got this one right

[tranquil pine]

my exam is in 2 days im really scared

[craggy island]

What field of maths are you doing?

[tranquil pine]

multivariable do you know this is?

[final saddle]

@tranquil pine Has your question been resolved?

[craggy island]

Yeah I have an idea

How about you DM me

[tranquil pine]

i can help u

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@tranquil pine Has your question been resolved?

[tranquil pine]

I Dmed

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[rose hamlet]

it says find unit nor,mal to surface z^2=x^2-y^2at 1,0,-1

[rose hamlet]

the book solves for the gradien then normalises it but i thought gradient is supposed to find direction of maximum change

how is it solving for the normal to the surface

idk if i am making sense

[glossy zephyr]

!original

[final saddle]

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.

[rose hamlet]

Find a unit normal to the surface z^2 = x^2-y^2 at the point (1,0,-1)

[glossy zephyr]

With that i meant, a photo of the book / screenshot of the problem, etc...
Anyways, the gradient is the higher dimension version of the derivative.

[rose hamlet]

my question is why is gradient suddenly the normal vector

[glossy zephyr]

If the derivative is tangent to the graph (and therefore the gradient too)
then you can find the normal using it

[rose hamlet]

i thought its the dir of max change

yeah i get it the analogy of derivative and how gradient is with partial derivative but still idk

[plain breach]

yoo

[rose hamlet]

ig ill take the word for it

[plain breach]

my math exam

[rose hamlet]

!occupied

[final saddle]

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

[plain breach]

is hard

asf

[proper dagger]

!redir, please.

[final saddle]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[glossy zephyr]

Ill try to explain it for a regular f(x)

[plain breach]

redir?

mb

[glossy zephyr]

This is essentially the derivative of f(x) expressed in vector form

If the tangent has an angle 0 to the point of tangency
and the normal has angle 90

Then the tangent vector and the normal are perpendicular to eachother

[rose hamlet]

ik what a derivative is trust

[glossy zephyr]

in f(x)

[finite burrow]

e

[rose hamlet]

i have done a lot of single variable calculus this multivaiable stuff is new so it is kind of confusing for now

[glossy zephyr]

that makes -1/f'(x) the slope of the normal vector

[finite burrow]

? ☠️

[rose hamlet]

i know these things

[finite burrow]

someone explain the mean table

🥺

[rose hamlet]

m1 . m2 = -1 for normal and f'x is slope of tangent so normal is -1/f'x i know

[glossy zephyr]

Visual representation of it

Well, we can generalize this same process

for these vectors i did
$F(y) = f(x) \implies g(x,y) = f(x) - f(y) = 0$, and did gradient to g(x,y)

We can have an equation
$F(z) = f(x,y) \implies g(x,y,z) = f(x,y) - F(z) = 0$

[soft zealot]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[glossy zephyr]

calculate the gradient of $x^2 - y^2 - z^2$, and evaluate at the point you want, and voala, thats the normal

[soft zealot]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[final saddle]

@rose hamlet Has your question been resolved?

[rose hamlet]

i still dont get it if gradient is analogosus to derivative then how is it normal not tangent

[glossy zephyr]

By extending the function one dimension higher and equating it to = 0, you basically get an analogous process to doing -1/f'(x)

consider: we started from a function z = f(x,y)
and created a g(x,y,z)

So g(x,y,z) in reality is 4d, but we just simply take a single slice of g(x,y,z) = w

You have to consider that the tangent and normal vectors are really related

[rose hamlet]

yeah this is going over my head for now ill accept this fact

[glossy zephyr]

You remember cross product?

From two vectors tangent to a plane, you can find the normal of said plane.
the tangents and normals are quite similar, you can usually find one from the other

[rose hamlet]

yeah i get that but idk its hard relating the two here

ig i gotta give it time

[final saddle]

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[jagged flare]

find all functions $f:\bR\to\bR$ such that
$$f(f(x)+xy)+f(y^2)=2025x+yf(x+y)$$

[soft zealot]

ihave<skissue>

[jagged flare]

gimme a sec

[warm python]

well, I suppose f(0)=0 here plays a role ( but that's all I can see)

[raven marsh]

How yummy an fe

[plucky rover]

And I'm thinking the function needs to be odd

[raven marsh]

So quite clearly we see f(x) = 45x works

[warm python]

I think it is , yea

[raven marsh]

maybe bijectivity

[plucky rover]

So does -45x

[raven marsh]

Indeed

[plucky rover]

I feel confident in saying those are the only ones

[warm python]

think you can first show ( perhaps) that all such functions are continuous

[plucky rover]

I think it would be easier to prove linearity

[jagged flare]

P(0,0) implies f(f(0))=-f(0)
P(0,y) implies f(f(0))+f(y^2)=yf(y) <=> f(y^2)=yf(y)+f(0)
when y=1 we get f(1)=f(1)+f(0), thus f(0)=0, and f(y^2)=yf(y) [property *]
from property [*], we get that yf(y)=f(y^2)=f((-y)^2)=-yf(-y) <=> -f(y)=f(-y) thus f is odd
P(x,0) implies f(f(x))=2025x, subbing x=f(x) we get f(f(f(x)))=2025f(x), and putting both sides into the function gets f(f(f(x)))=f(2025x), thus f(2025x)=2025f(x)

thats where im stuck

[plucky rover]

Can you show that f is bijective

Injective should be easy, surjective is where I'm a bit dodgy on the details

[raven marsh]

Fix y = 0 and let x vary

[jagged flare]

thhis idea might work? but i dont understand it at all :p

[plucky rover]

Oh no surjectivity is easy too

[raven marsh]

Then RHS = 2025 x

[plucky rover]

Yeah get an expression for f(f(x))

[raven marsh]

OTIS FE part how civilised

[raven marsh]

this works btw @jagged flare

[plucky rover]

That gives bijectivity

[jagged flare]

i mean i didnt get the fe from here but yeah

[raven marsh]

ah ok

so we know is bijective

[jagged flare]

wait what how?

how does f(f(x))=2025x imply bijectivity

[raven marsh]

firstly its surjective right

because 2025x can take all real values

[plucky rover]

Injectivity is clear right

[jagged flare]

f(x) is surjective iff f(f(x)) is surjective?

[raven marsh]

then its injective since if f(a) = f(b) then f(f(a)) = f(f(b)) = 2025a = 2025b so a = b

[raven marsh]

nono

[plucky rover]

You can't get a surjective function by composing a non-surjective function to itsrlf

[raven marsh]

surjective means f can output all real values

[jagged flare]

ohhh

[plucky rover]

That should be easy to see cuz if you're missing smth the first time, you will miss it the next time too

[raven marsh]

Since f(smth) = all real values we are done

[whole halo]

where was this from?

[jagged flare]

national oly simulation

[jagged flare]

ohhh

[whole halo]

I swear Ive seen something like this before

[plucky rover]

Do you see injectivity

[raven marsh]

.

[jagged flare]

f(a)=f(b) => f(f(a))=f(f(b)) <=> 2025a=2025b <=> a=b this is what you mean right

so at the end f(a)=f(b) => a=b

[plucky rover]

Yes

[jagged flare]

ooh thats rly smart

yeah so f is bijective

[plucky rover]

Next step, linearity

[raven marsh]

Yes

[raven marsh]

Indeed

[whole halo]

I swear I mustve seen this before, this is very familiar

final step being linearity

[jagged flare]

how do you prove something is linear

i checked online and it said it needed additivity and homogeneity but for this problem its abit iffy, so maybe theres some other way to prove linearity/

[plucky rover]

Let's revisit the initial expression

[plucky rover]

My first thought is using bijectivity to plug in smth

Sorry in class, intermittent responses

[final saddle]

@jagged flare Has your question been resolved?

[whole halo]

I cant find it anywhere, Ive hallucinated an entire functional problem to solve

what in the goddamn

[jagged flare]

ive been staring at this for a while and i couldnt come up with any progress, could i get more hints?

[whole halo]

as a hint from the problem I never solved, I can show you a method I never used

[whole halo]

what is a something a line has everywhere that other functions cant?

[jagged flare]

a constant slope??

[slender basin]

lack of continuous breaks everything, are you sure its not continuous in original problem?

[whole halo]

if you can prove that the slope between two points is always constant, the function can only ever be a line

keep in mind the expression inside of the points doesnt have to be f(x) and f(y)

it could be f(something) and f(something else), as long as both of those span R (maybe even less)

[jagged flare]

does f(x+y)-f(x)=ky imply linearity here?

"plugging in" screams to plug in like t=f(x) but thats kind of wierd here as you can only put it for one thing and it aint even clean

[whole halo]

did you manage to find that here?

[jagged flare]

im thinking either like t=f(x)+xy or t=x+y

[jagged flare]

no not yet, like for some reason reading what you said here it pointed to it implying linearity, but i remember there was a diffrent problem and it said that it was bogus

[whole halo]

that wouuld imply linearity, yes, as long as k remains constant

this is for any choice of x or y, so the usual arguments against it dont hold up

for one, (f(x+y) - f(x)) / y = k

[jagged flare]

ooh ok i think i remember why it was bogus cause it fixed y i think?

[whole halo]

if you say thats the case, that would be a case where this doesnt work

the problem I didnt solve didnt have this problem I never saw

[jagged flare]

im for the first one cause that nest is kinda annoying to see
f(t)=2025x+yf(x+y)-yf(y)=2025x+y(f(x+y)-f(y))
this is kinda nice but i havent used bijectivity here yet so im kinda worrying?

[timber pilot]

Assume that f(x) is a linear function

[whole halo]

if this is done, it restricts f(x) to be 45x or -45x
other than that, you still need to prove that f can only be linear

hm

I cant even see the mistake in this

oh right y f(y) does not span R

[final saddle]

@jagged flare Has your question been resolved?

[whole halo]

cant see any use out of this, this is from (-x,-y)

any real a, b

[whole halo]

if f(x) = -g(x), then g(x) also works

so if f(x) works, then so does -f(x)

using from earlier that f(2025x) = 2025 f(x) and f(x f(x)) = 2025x^2,
f(x) = 2025 implies x = ±45

so (±45, ±2025) or (±45, -±2025)
so f(45) = ±2025

[raven marsh]

😭 china tst ahh

[jagged flare]

45×2025=45f(45)=f(45×45)=f(2025)
f(2025)=91125=f(2025×1)=2025×f(1)
f(1)=45

with the ± brah

P(1,-45) seems promising

[whole halo]

2025 f(1) = f(2025) = f(45 * 45) = 45 f(45) = 45 * ±2025
so f(45) = ±2025 means that f(1) = ±45

[jagged flare]

f(0)+f(2025)=2025×1-45f(-44)
-f(-44)=f(44)=1980

[whole halo]

(1, -±45) shows that f(45 - ±1) = ±45(45 - ±1)

[final saddle]

@jagged flare Has your question been resolved?

[final saddle]

@jagged flare Has your question been resolved?

[viscid stirrup]

The same FE (with only change being 2016 instead of 2025) can be found on AOPS btw

ofc it makes it much nicer knowing that 2025 is a square for this problem

[jagged flare]

i cant find it on aops

prolly cus i suck at searching

[viscid stirrup]

Have you heard of approach zero (a math search engine)?

[jagged flare]

no

DAMN

[viscid stirrup]

you can use it to search for problems that usually come from math stack exchange and aops

[viscid stirrup]

that’s my reaction when I first found out about it

[jagged flare]

ok so tldr the trick is to sum P(x,y) and P(-x,y) to prove that its equivalent to cauchy over reals

actually maybe im wrong but dont you need something extra to prove that something thats cauchy is linear

nvm

what did they even do here after the "Let x=1"

ok guys thank you!

.solved

[final saddle]

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[final saddle]

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[tired walrus]

rewrite the system in matrix form and set det=0 perhaps

that should give you something to work with

[vague anchor]

There comes Ann

[vague anchor]

I found determinant
I was trying to solve it via determinant

Determinant=0 then it would have non-zero solutions

[tired walrus]

what did you actually get for the determinant

[vague anchor]

Ugly

[tired walrus]

how ugly are we talking

also we know that at least one of a, b, c is rational & between 0 and 1

this will have to somehow come into play

[vague anchor]

c^2b + a^2b + 2ac +b^2

[tired walrus]

aight wait, can you share your matrix too

[vague anchor]

Proper fraction can't be less than 0
I didn't know that fr
Most probably i would have to use AM GM

[vague anchor]

Lemme write it with good writing

[tired walrus]

ok

im gonna be busy for the next half hour or so, so slow responses

[vague anchor]

Ohk

I see my mistake

[vague anchor]

Now it's 1 + abc + a^2 + b^2 + c^2

[candid pulsar]

i think its 2abc

and -1

[vague anchor]

Yea 💀

I just typed it wrong

[vague anchor]

Any idea to do it further??

[tired walrus]

missing ^ (3 times)?

aight im off to do irl stuff, sorry