help-36

[raw patrol]

Mp?

[late gazelle]

minus plus

sub in our values for a and b, you get 4 complex solutions

simplify the radicals into x+yi form

and we get our answer

[raw patrol]

Oh

Is this working wrong

[late gazelle]

is this still for Q:iii?

[raw patrol]

Yh

[late gazelle]

take our answer and we now write: $\sqrt{16 \pm 30i} = x+yi$

[soft zealot]

KB

[late gazelle]

square both sides and equate Re(z) and Im(z) components

[raw patrol]

I did that for the - root is my working wrong

[late gazelle]

this is pretty much your answer

you solved for y and x

[raw patrol]

Aren’t my signs wrong ?

[late gazelle]

both are pm

$\pm (5 \pm 3i)$

[soft zealot]

KB

[raw patrol]

I got -5+3i so the signs are opposite in my working but they’re the same here

Sorry I’m just a bit confused

[late gazelle]

you get part of your solution by solving one combination of sings, you get all solution by solving for all cases of the signs

you should get 4

[late gazelle]

only way that can happen is if we have ^ ^

[late gazelle]

that is one of the solutions

$5 + 3i \
-5 + 3i \
5 - 3i \
-5 - 3i$

[soft zealot]

KB

[raw patrol]

Hmmm

[late gazelle]

looks messy but thats all 4 solutions

[raw patrol]

Ohhh I see I thought this was only two solutions is it really 4?

[late gazelle]

yea

we started with y^4 + ... =0

right?

by the fundamental theorem of algebra, there is at most deg(f(x)) solutions real or complex

[raw patrol]

Yes

[late gazelle]

we simplified it down to a quadratic thus making it 2 solutions

however, when we subbed back y^2 we found all 4

[raw patrol]

I see

Thanks

[late gazelle]

think of it as $(x^2)^2 = k, x^2 = \pm \sqrt{k} \implies x = \pm \sqrt{\pm k}$

[soft zealot]

KB

[raw patrol]

I see

[late gazelle]

of course this is only for when our degree is even

since taking third root doesnt really help (for solving degree(2n+1) polynomials)

hopefully that all makes sense and hopefully i answered your original question

[raw patrol]

Yeah u have tysm for ur time

[late gazelle]

np

[raw patrol]

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[night fjord]

Suppose that the demand and supply for electricity are represented by the given equations. Demand: 𝑄d = 25 – 1/2 𝑃𝑃 Supply: 𝑄s= -5 + 1/3 𝑃𝑃 where P stands for price, Qd stands for quantity demanded, and Qs stands for quantity supplied.

b. Despite the increasing deployment of renewable energy, fossil fuels still dominate electricity generation. As of 2026, around 60% of Australia's electricity was produced using coal and natural gas, while renewables accounted for approximately 40% of the total energy mix. This reliance on fossil fuels continues to contribute to carbon emissions and environmental challenges. Suppose that generating each kilowatt of electricity results in an external cost of $5 due to CO2 emissions. Analyse if the equilibrium quantity in part (a) reflect the socially optimal level of quantity? Support your answer by examining the presence of deadweight loss (DWL) on your graph.

[night fjord]

have i done this correctly

i havent done this one before

[gritty chasm]

so to clarify, your equations are:\

1. $QD = 25 - \frac12 P$\
2. $QS = -5 + \frac13 P$\

[soft zealot]

Annie Maqionde

[night fjord]

yes

[wicked anvil]

what’s 1+1

[wicked anvil]

when will we use that in real life

[gritty chasm]

<@&268886789983436800> troll

[wicked anvil]

no it’s not

hats 1+1

[night fjord]

2 i believe🧐

[gritty chasm]

Please dont engage in trolling.

[wicked anvil]

thank you

[night fjord]

my bad

[wicked anvil]

not troll bih

[white gazelle]

don't troll in the help channels

[wicked anvil]

sorry

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[night fjord]

<@&286206848099549185>

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@night fjord Has your question been resolved?

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@night fjord Has your question been resolved?

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@night fjord Has your question been resolved?

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@night fjord Has your question been resolved?

[torn trench]

GET OUT

[warped hollow]

What is part a?

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[chilly rover]

I feel stuck and I dont know what to do. Please help me math gods

[chilly rover]

I am confused how to solve 61-66

[compact laurel]

I'd recommend finding the traces

[chilly rover]

i mean like for the equationa?

z = sin(xy)

[royal gust]

It's vibes, really.

You should know how similar functions work in 2D, can you recognize these in 3D?

[compact laurel]

That's why I'm telling him to find the traces. lol

[royal gust]

Like, B doesn't have trig in it. See why?

[chilly rover]

i mean B looks like a plane with some transformation to it

why is there no method to do this omg

[chilly rover]

i know how to find traces for quadric surfaces but not

[compact laurel]

Sorry bro, my messages are lagging in here

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@chilly rover Has your question been resolved?

[chilly rover]

can u explain how to solve

<@&286206848099549185>

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[final saddle]

@night fjord Has your question been resolved?

[night fjord]

<@&286206848099549185>

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@night fjord Has your question been resolved?

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@night fjord Has your question been resolved?

[sage osprey]

The diagram appears to be correct

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@night fjord Has your question been resolved?

[ornate token]

Well, as I can see, you noted the formulas for demand and private supplies are correct.

That addition the external cost of $5 to the private supply curve looks fine too, everything seems fine so far.

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[low flint]

Need help

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@low flint Has your question been resolved?

[muted prairie]

Right matrix is the adjugate of the left one

[late spade]

my answer is d

am i right ?

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[copper roost]

hello can someone help righthere im frustrated

also the picture translated poorly

so i can just do +(7x-14) right

[brazen breach]

I

I think so

[copper roost]

hold up but

how do i know where the tangent is

or how do i explain that they have 1 more point together

[brazen breach]

you have a point

I mean you literally have a point

P(3, f(3))

[copper roost]

yes right

whats the y value of that tho

[brazen breach]

you have a function

[copper roost]

can i just plugin 3

[brazen breach]

place x in that function

[copper roost]

into the tangent

or in the function

[brazen breach]

both

[copper roost]

ill be back in 20 minutes

dont leave me

[brazen breach]

also you might wanna algebraicly prove that the value you get from f(3) has only one x solution for 1/3(x-3)^2(x+6)+(7x-14) by expansion

[brazen breach]

oh no worries

leaves

[copper roost]

wdym expansion

hell na

i dont have to do that no way

[brazen breach]

you

may

wanna

do it

[copper roost]

okay so how do i know they have 1 more common point

WAIT

nah nvm

yeah how do i show they have 1 more common point

[brazen breach]

hmm

then there'd be 2 solutions to f(x)=1/3(x-3)^2(x+6)+(7x-14)?

I think

or just draw the line on the graph

[copper roost]

shit

[brazen breach]

graph the line y=7x-14 here- hey wait a minute

[copper roost]

yeahh ill try in a bit

[brazen breach]

nah wait u can just sub twin

[copper roost]

what

bro ill try in a bit

[worldly spruce]

I am sorry, I don't really understand what the objective is

Do you just have to show that $f(x) - (7x - 14) = \frac{1}{3} \cdot (x-3)^2 \cdot (x+6)$

[brazen breach]

it seems to be proving y=7x-14 is tangent to f(x) at 2 points

or something

I don't understand german enough for this

@copper roost I will be eating dinner direct questions to someone else in the meantime

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[rugged spade]

Hi guys, I'm stuck on question 3, will you have a small clue on your side that could be useful to me?

[rugged spade]

It's my job for the moment, it's written in French if you don't understand don't hesitate, question 2 is badly written I'm rewriting it

[quasi rapids]

havent taken a look at the question itself but i really just want to say

your handwriting is gorgeous

[rugged spade]

Haha, thank you so much! I must admit that the year of medical school I did helped me quite a bit with that.

[final saddle]

@rugged spade Has your question been resolved?

[waxen lintel]

Looks like Tanguy Rivoal is one of your professors 🙂 Just a guess

[rugged spade]

you know the goat

respect

[waxen lintel]

I don't know if it helps but you may find this proof in a work of Nesterenko himself. It is in Russian, but nowadays with AI it's not a big problem. Here it is.

[rugged spade]

I'll take a look, but I have to admit the question is really, really difficult for someone my age. He went a bit overboard on the subject, even though he told us he modified it to make the first questions more manageable.

[waxen lintel]

Beukers' proof is the easiest.

[rugged spade]

Yes, but here we're dealing with Apery's theorem; otherwise, there's no point.

[waxen lintel]

both prove irrationality of zeta(3) which is a proclaimed goal of your tasks

[rugged spade]

Apery mentionned

but i'm agree with you

But I'm a bit too lazy to get lynched by my teacher.

xD

[waxen lintel]

Apery was the first to prove irrationality of zeta(3). There are many variations of that proof. The easiest one imo is Beukers' proof.

[rugged spade]

Okay, I'll look into that, even though the last questions are really about Apery, and ultimately what this theorem proves.

The questions are actually there to guide us and put us on Apery's trail, even though from what I understand it starts roughly the same way with other proof

you are french ?

[waxen lintel]

If you need to answer exactly those questions then just do them. If you want to prove only the final result about irrationality I think the easiest is this paper. All these proofs on the whole give the same approximation sequence for zeta(3). Though there are many different ways to get it.

[waxen lintel]

no

[rugged spade]

Idk if it's for me and if i have the level

i'm 18 xd

the final result was to hard

[waxen lintel]

You just need to know integration. And it is a way lower level than those complex variable G functions in your questions.

[rugged spade]

Okay, I'll go check that out.

ok its very easy tf

I don't understand why I'm being made to suffer so much, and why I hadn't thought of this before.

[waxen lintel]

Because the irrationality of zeta(5) is still unproven, and no one knows for sure which approach might succeed in that case. So, you should know all of them to be prepared.

[rugged spade]

I'll be the first to prove it, don't worry. ^^

[waxen lintel]

who knows 🙂

[rugged spade]

I'm an idiot, man. The more I improve, the more I realize I could never actually be good.

[waxen lintel]

that's a normal way of thinking. This is motivation to get better

All things should be assessed in comparison

[rugged spade]

Thank you, you're right, I'm a follower of stoicism.

.close

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[serene sonnet]

im struggling with C, I can't understand the question

[topaz kite]

what does the term "donation rate" mean to you?

(in fact you've been using this value all along in parts a and b I presume.)

[serene sonnet]

1/30 people

[topaz kite]

well I was asking for its meaning. also what does 1/30 people mean in this context?

[serene sonnet]

1/30 people making blood donations in australia in a week but also its probably not relvant as it was for a prior question

[topaz kite]

how did you get the figure of 1/30 people when the population of Australia is not given?
also, a week?

the figures in the question are per year, unless this is a continued question from somewhere else.

(in which case your figures do not apply, and you should look at your answer for part a) instead.)

[serene sonnet]

oh ok

.close

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[rain sentinel]

can someone help me find the third derivative

i dont understand why they've done this

[rain patrol]

Do you know the rule for the second derivative?

The third derivative is merely the second derivative, of the first derivative.

$\odv[order=3]{y}{x} = \odv[order=2]{}{x} \left( \odv{y}{x} \right)$

[soft zealot]

Cark Marney 🐢

[rain patrol]

Or t in your case

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@rain sentinel Has your question been resolved?

[rain sentinel]

yh ik that rule

[drowsy epoch]

general derivative

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[rain sentinel]

howmdo we rearrange to get Ys

[drowsy epoch]

Factor y

[worthy wren]

factory

[worn trout]

hello guys i need your help. I have a mathmatics exam about limits , integration and other concepts related so I want some videos to solve with calcluater

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

[worn trout]

??!

[gritty chasm]

Please read the message below it.

[worn trout]

ok

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[pure pine]

I have to simplify these vectors, how do i do it? for example on a) im stuck on OB + AD and i dont know what to do next since they are different letters

[ornate token]

I think that is the simplest form.

[brazen breach]

it's just those unless they have values

[pure pine]

ohh

okaay thank you

[ornate token]

!done

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[pure pine]

.close

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[severe verge]

$y=xy'+y'-y' \ln{(y')}$

[soft zealot]

DeC∆rbonizeD

[marsh mountain]

looks like clairaut

[drowsy epoch]

ln(dy)-ln(dx)

[severe verge]

never heard of it

[marsh mountain]

cookin

[marsh mountain]

its equations of the form y = xy' + f(y')

standard way is usually to differentiate wrt x and then you should see something pop out

[severe verge]

so 0 = xy'' - y''ln(y')

?

[drowsy epoch]

yes

[severe verge]

ok thanks

been trying everything for hours and couldnt crack this

.close

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[tired iron]

The AM of two positive integral nos exceeds their positive GM by 2 and their sum is 20 find the numbers

[severe canyon]

! 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

[tired iron]

1

[severe canyon]

Start by giving a name to the two numbers

[tired iron]

Let a and b be two positive real nos

[severe canyon]

Good

Now we can say that x + y = 20, right?

[tired iron]

Yes

[severe canyon]

Awesome

Now, can you write an equation using the first info?

[tired iron]

Nope

[severe canyon]

Do you know what AM and GM mean?b😅

[tired iron]

Oh ye

I mean i dont understand the q

Thats why i came here

How am and gm are related here

[severe canyon]

The question simply asks you to find the two numbers

[severe canyon]

Do you know the meaning of the word "exceed", like the English meaning, not math related?

[tired iron]

Crossing it

Like going forward

By 2

Am plus 2 = Gm

[severe canyon]

That's a possible interpretation yeah

[severe canyon]

The other way round

[tired iron]

How

[severe canyon]

If A exceeds B by some quantity, it means that A is larger than B

[tired iron]

Yes

[severe canyon]

Then that's it 😅

[tired iron]

Am plus 2 greater than gm

[severe canyon]

No ...

AM = GM + 2

Also, it's an equality not an inequality

[tired iron]

Bro but airthmetic mean is exceeding gm

[severe canyon]

Yeah

[tired iron]

Like am is 2 steps away from gm

[severe canyon]

If A exceeds B by some quantity, it means that A is larger than B

[tired iron]

Alr so to make it equal i get it

U subtracted 2

?

From am

[severe canyon]

I don't get you

[tired iron]

Nvm man

I was confusing myself

I got it now

Really sry too much sequences got me

.close

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[quartz vigil]

I need help with 61-66

[quartz vigil]

please for the love of god

I have my exam and i feel suicidal

looking at these graphs

[blissful meadow]

Look at the function on the axes like I suggested last time.

[quartz vigil]

Ur messages were lagging last time

so I thought u went

but

[blissful meadow]

And avoid pinging helpers right away please

[quartz vigil]

Okay can u elaborate

function on the axes

[blissful meadow]

You can rule out graphs based on "easy" places to evaluate your function.
Say for 61, if you plug in y=0, you get z=0. That means that for any point (x,0), the output is 0.

Similarly if you plug in x=0, you get z=0, so any point (0,y) has output 0.

This means you should expect the graph of 61 to be 0 everywhere along the x and y axes.

[quartz vigil]

Omg

Wait

I get that it will be 0 along x and y axis

but only when z = 0 right

[blissful meadow]

Well z is a function of x and y

So along the axes, z should be 0.

A doesn't have that for instance. If you follow along the axes, it doesn't look like z=0.

[quartz vigil]

Oh

But the curly waves

[blissful meadow]

Orange is following the output z along the x axis.
Green is following the output z along the y axis

[quartz vigil]

in C

why are they there

cus of sin?

but that’s an elevation too right

[modest cairn]

oh mai gah

[blissful meadow]

Well if you follow the x and y axes for C., it's not very clear but the function is 0 along the axes.

Usually, setting x=0, y=0 and x=y should give you a lot of information.

[quartz vigil]

Oh so for x and y axes for C i can see it being like a flat

oh

[blissful meadow]

If you set x=y, then you get sin(x^2), so that means that following the line y=x in the xy-plane it should oscillate.

So here green is the output on the y axis, orange is the output on the x axis and pink is the output along the line y=x.

You can try the other graphs to see if they work too, but they all fail one of those requirements, so 61 would be associated to C.

[grave lichen]

Wow as someone who is in high school seeing this im shocked what is that

[quartz vigil]

Oh

yeah this makes sense now

[blissful meadow]

-# When you learn about functions in highschool they're usually functions of one variable. Their graph is like a curve in the plane, like a string in a certain shape.
-# If you add a variable, what you get is a surface in space, like a bedsheet.

[quartz vigil]

I feel like it’s also vibe game

I regret taking calculus 3

and i regret signing up for engineering

[blissful meadow]

Yes it's not an exact science you just have to rule out some of the choices

[quartz vigil]

How would u do 66

I think i got all of them except 66

[blissful meadow]

y=0 gives x/(1+x^2)
x=0 gives -y/(1+y^2)
x=y gives 0

[quartz vigil]

I can’t still imagine how we make this contour plot

[blissful meadow]

Those graphs mean that along the x and y axes the function needs to have one bump above the xy-plane and one bump below the xy-plane.

Like this

[quartz vigil]

I m gonna cry

what

[blissful meadow]

Those are lines you expect to be on the graph.

They're what you get when you set x=0, y=0 or y=x in the function

[final saddle]

@quartz vigil Has your question been resolved?

[final saddle]

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[mystic lotus]

Guys what the flip am I supposed to do

[mystic lotus]

I AM DESPERATE

[compact laurel]

<@&268886789983436800>

[ornate token]

Simplify the bracket.

[mystic lotus]

Okay wait can I show you what I did so far and you tell me if its good?

[ornate token]

Sure.

Hint: Turn them into perfect squares.

[mystic lotus]

My bad for my bad writing

And now I dont know what to do

[ornate token]

You can cancel the second fraction.

[mossy inlet]

shared factor in second fraction

notice (x+4) is in the top and bottom...

[ornate token]

Notice they both have (x+4) on the numerator and denominator.

[mystic lotus]

Omg YOU GUYS ARE RIGHT

Wait so I just like cancel it

[ornate token]

Yes.

[obtuse grail]

English please

[mystic lotus]

Like that

[ornate token]

Because they share same term in multiplication.

[mossy inlet]

mmgm

*mmhm

so this would cancel out to...

[ornate token]

So many bots today.

[mystic lotus]

But do I do it for the other one too

Like the 3x-2

[ornate token]

Correct, they do the same method.

[mossy inlet]

Right, both can be simplified in the same manner of canceling out common terms.

[ornate token]

(3x-2)^2 is just (3x-2)(3x-2)

[mossy inlet]

Which I saw you already did with the 3x-2

[mystic lotus]

Like that?

[ornate token]

Yes.

[mossy inlet]

yup!

now back to the original problem (division right..?)

[mystic lotus]

Okayy omg THANK YOU GUYS

[ornate token]

Now, let's clear up the right hand side.

[mystic lotus]

Yess

[mossy inlet]

now you wanna divide that trinomial by 2x-1, right?

[mystic lotus]

Yup

[mossy inlet]

have you learned synthetic division?

[mystic lotus]

Hm i dont think so

[ornate token]

Long division?

[mossy inlet]

you can do long division too

an example of the methods

[ornate token]

Long division is more familiar by the students, but depends.

[mystic lotus]

Ya no I need to do it the long way or my teacher will think I used ai or something

[mossy inlet]

lmao

sounds like a teacher

[ornate token]

Don't think that will be the case.

As long as you understand it and can redo similar problems.

[brazen breach]

synthetic division is just the same thing as long division it's just less complex looking 😭

[mossy inlet]

if you wanna do long division, you can set it up the same way say you woul do 500/12

[brazen breach]

stop the long division hate

[mossy inlet]

tbf u need long division when its not a linear factor

so igs long division still good

[ornate token]

Fair point, or when you got a solution.

[brazen breach]

real people use factor and remainder theorem

[ornate token]

Can you solve a polynomial that can't be used by neither of them?

[mystic lotus]

Guys something is not working

WHY DO I HAVE A 6 LEFT

[mossy inlet]

uh

wait nvm

[ornate token]

-7 -(-1)=-6

[brazen breach]

it looks right

[ornate token]

The second subtracting part.

[mystic lotus]

OMG YOU ARE A GENIUS

[ornate token]

No worries, you are doing fine so far!

[brazen breach]

oh yeah didn't see that

[mossy inlet]

yeah should be -6x+3

[mystic lotus]

OMG TY GUYS I DID IT

[mossy inlet]

yay!

good job

[mystic lotus]

I also didnt understand an other problem but its in french so-

[ornate token]

Ask away!

[mossy inlet]

we can try

[mystic lotus]

Aww ty I couldn't have done it without you guys 🫶

[ornate token]

Just tell us what the questions mean in the scope of your understanding.

[mossy inlet]

i mean i dont know french but um as long as its not a word problem or u can translate or we can use google translate

[mystic lotus]

Okay I will try to translate it

[ornate token]

Perimeter

Ah okay.

You use Pythagorean formula?

To find x.

Then you can substitute back into the result P=10x+30 you just got.

[mossy inlet]

Do you know the pythagorean theorem?

[mystic lotus]

The triangle rectangle PQR has these characteristics

Put your answer in numerical form what is the perimeter of the triangle

[ornate token]

I think she is still translating

[mystic lotus]

Yup

[ornate token]

So can you apply them to the 3 sides?

[mossy inlet]

ah... i lwk just used gemini to translate-

[ornate token]

Since it is a right-triangle.

[ornate token]

Wow. I can read a bit French so I got it straight away.

[mystic lotus]

Ayee period queen

[mossy inlet]

i know no french ... i knew the quetoin asked something about ... a triangle. i only speak english and chinese.

[ornate token]

Anyways, the Pythagorean formula can solve this by setting RQ^2+PQ^2=PR^2

[loud sundial]

.pin

[ornate token]

Haha!

[mossy inlet]

well... same result anyway xd so how would you apply pythag theorem to this?

[ornate token]

Just use google translate....

[mossy inlet]

i didn't wanan type allat...

[mystic lotus]

Holy molly my translation is kinda good yall

[ornate token]

It does!

[mossy inlet]

(I believe after pythag theorem, you would end up with a quadratic that you'd solve for)

[brazen breach]

^

place the x on one side and solve for (usually) positive x values

[ornate token]

You realise you can just react, not to type "^" right.

[brazen breach]

-# I'm stupid and my default is reply and ^

[mossy inlet]

lwk real i do that too

[ornate token]

Just set "^" as your default emote.

Ah, I think that emote to be set by default, you need Nitro.

[mossy inlet]

^ @mystic lotus how are you doing on this so far?

[mystic lotus]

Okay so

I dont know if what i did is right

[mossy inlet]

looks right

[mystic lotus]

Okay but like what do I do now

[mossy inlet]

just keep on solving

try moving everything to one side..

[mystic lotus]

Bring it back to 0?

[mossy inlet]

Make one side 0 by moving the stuff on that side to the other side by performing the appropriate addition/subtraction

id recommend moving the left side to the right side

[ornate token]

Yes.

[ornate token]

Can you cancel out the coefficients?

Or, simplify them.

[mossy inlet]

mmhm.. look for anything to make factoring easier (any gcfs youre noticing?)

[mystic lotus]

Like that?

[final tangle]

did they tell you where the right angle is?
you may need to go through all 3 cases

[ornate token]

It did.

[ornate token]

Here for your information.

[final tangle]

where

[ornate token]

^

[final tangle]

that doesn't say where the right angle is

[ornate token]

Le triangle rectangle means right triangle in French.

Ah, I get what you said, you do not know where the right angle is.

[mossy inlet]

igs since q is in the middle its supposed to be more implied

[ornate token]

In this case we need to implies 3 cases.

[ornate token]

No, that's too random and intuitive.

[mystic lotus]

Wha-

I need to do it 3 times???

[final tangle]

one case flat out fails,
one has ugly values
one has nice values

[ornate token]

Did the question specify where is the right angle?

[mossy inlet]

er.... probably

[mystic lotus]

Hm no

[mossy inlet]

did ur teacher mention anything bout this problem?

[mystic lotus]

Nope

[ornate token]

Hmm, I think in this case, solve for the assumption you are doing then ask your teacher a clarification.

[mystic lotus]

Okay so my final answer is

This?

[ornate token]

Yes, let's cancel out 4.

I do not think this will give us a nice solution.

After this case, let's try if 26 is hypotenuse, that one should provide a nicer answer.

[final saddle]

@mystic lotus Has your question been resolved?

[mystic lotus]

I think i will leave it like that bc I learned that my teacher wont even correct jt

[ornate token]

Alright, all the best!

[mystic lotus]

Thank you so much tho 🫶

[ornate token]

No worries!

If you have no other questions, feel free to type .close to close the channel!

[final saddle]

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[topaz kite]

(this might be a better question for the physics server, but) at what angle is the wind relative to you?

and also, for future helpers, what have you tried and what do you not get?

[jovial sapphire]

we need more info i think

[topaz kite]

do you intend to take over this channel again? if yes, I'll leave this one as well.

[jovial sapphire]

lol sorry

you can have this dw

[topaz kite]

no, it's fine. carry on please.

[ornate token]

-# The profile picture is quite dodgy.

[jovial sapphire]

and the nickname

<@&268886789983436800> ?

[ornate token]

The nickname is not bad, but the pfp does.

[jovial sapphire]

i think the nickname refers to epstein's email

[final saddle]

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as this action is irreversible, and this channel may abruptly lock.

[final saddle]

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[dense garnet]

Basic question perhaps. The author says gamma is a curve in R^3. Why is the 1-form using the coordinates x and y, when the curve could have non-zero z-coordinates?

[dense garnet]

.close

[final saddle]

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[raw patrol]

I don’t really know where to start

I wrote out the equations in Cartesian form

[winter dagger]

the line of intersection is given by cross product of their normal vectors

[raw patrol]

Ohhhh

And then I have to find any point at which they intersect ?

[winter dagger]

Yes

[winter dagger]

Any point works

[final saddle]

@raw patrol Has your question been resolved?

[raw patrol]

Wdym

Can I use 1 2 1

[ornate token]

For the first question (27a), just set a cross product.

Then find a point on l by setting z.

Hope this helps.

[raw patrol]

Can you elaborate on setting z

[ornate token]

Well, suppose we let z=1

Then we can continue from there.

[raw patrol]

Ok

[raw patrol]

Thx

I have a second question

[ornate token]

Please ask straight away!

[ornate token]

Have you finished the previous question by any chance?

[raw patrol]

Yes I have

It makes sense now tysm

[ornate token]

No worries.

[ornate token]

Let's do the next one.

What have you tried?

[raw patrol]

I kinda gave up and looked at the MARKSCHEME and I’m just a bit confused where they got the second line from

[ornate token]

Ah I see.

Let's resolve it together.

The problem states that for any normal to plane A, then components of x and y are normal, right?

[raw patrol]

Yeah

[ornate token]

So, let this normal vector be n = (a a b)

Right?

[raw patrol]

Yep

[ornate token]

Now, what is a general equation for a plane?

[raw patrol]

Ax+by+cz=d

Or R.n=d

[ornate token]

Correct, we will do with the first definition.

[raw patrol]

Oki

[ornate token]

Now, for plane A, it contains point A (3,-2,-1).

Can you substitute it into the equation?

[raw patrol]

3x-2y-z=d

[ornate token]

Substitute in here.

Then it would be something like 3a-2a-b=d_A.

[raw patrol]

Ohh ok

[ornate token]

Now, can you do the same to B?

• The point B.

[raw patrol]

13a+10a +9b =d_B

[ornate token]

Now, do you know the distance formula?

[raw patrol]

Not for Planes

[ornate token]

This one.

[raw patrol]

Oh ok

Thanks

[ornate token]

No worries!

[raw patrol]

If I didn’t know that is there another way to get there

[ornate token]

I just googled the formula up, haha.

[raw patrol]

Cuz I only know these + the angle ones which are in my formula
Booklet

[raw patrol]

Ah ok

[ornate token]

Ah..

[raw patrol]

Yeah that’s why I was so confused

[timber leaf]

They should have added that formula into your formula sheet in my opinion.

[raw patrol]

Is there any way to do it without knowing the formula ?

[ornate token]

Maybe you could try using distance between point and plane?

[final saddle]

@raw patrol Has your question been resolved?

[final saddle]

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[faint monolith]

how do i do for b?

[final saddle]

@faint monolith Has your question been resolved?

[final saddle]

@faint monolith Has your question been resolved?

[faint monolith]

.close

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[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

[hollow pollen]

can u offer a translation in english?

[gentle zephyr]

1

[hollow pollen]

thx

[blissful meadow]

You're doing 8?

Given σ(t) can you compute the velocity?

[gentle zephyr]

r'

[blissful meadow]

How do you compute the velocity of a parametrized curve?

[gentle zephyr]

r'

[blissful meadow]

Yes I get that it's denoted r' (well here σ'), but can you compute it?

[gentle zephyr]

component wise computation

[blissful meadow]

Okay which gives?

[gentle zephyr]

r'

[blissful meadow]

Can you write down what σ' is?

[gentle zephyr]

,w differentiate r(t) = (t - sin(t), t - cos(t)

[blissful meadow]

Sure well that's your velocity.

[soft zealot]

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[blissful meadow]

Do you know what speed is?

[gentle zephyr]

r

[gentle zephyr]

ohh

[blissful meadow]

So now that you have σ'(t), what is the speed?

[gentle zephyr]

||r'||

[blissful meadow]

Yes

[gentle zephyr]

,w magnitude (t - sin(t), t - cos(t)

[soft zealot]

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

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[blissful meadow]

This is neither your position function nor your velocity.

[gentle zephyr]

is sped

[blissful meadow]

You need to take the magnitude of the velocity

(t- sin(t), t-cos(t)) is not your velocity.

[gentle zephyr]

what

,w magnitude of (1-cos(t), sin(t) + 1)

[soft zealot]

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[blissful meadow]

Your position function is (t-sin(t), 1-cos(t))

[gentle zephyr]

@blissful meadow

[blissful meadow]

Not (t-sin(t), t-cos(t)).

[gentle zephyr]

,w differentiate r(t) = (t-sin(t), 1-cos(t))

[soft zealot]

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[blissful meadow]

You're making this harder than it should be by not computing it yourself.

[gentle zephyr]

,w magnitude (1-cos(t), sin(t))

[soft zealot]

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[gentle zephyr]

i found the speed

@blissful meadow

[blissful meadow]

Okay you may want to simplify it.

Then you'll need the arclength.

Do you know how to compute the arclength?

[gentle zephyr]

yes

[gentle zephyr]

care to elaborate

[proud jolt]

It's written in your book the arc legth formula

[blissful meadow]

This is not the arclength of a parametrized curve.

It's the arclenght of a function y = f(x).

[blissful meadow]

Then compute that. You've already gotten the speed.

[gentle zephyr]

what is ab

[blissful meadow]

The endpoints for your parameter.

You're told to compute the arclength between the points r(0) and r(2pi), so the endpoints for your parameter are 0 and 2pi.

[gentle zephyr]

,, \int_0^{2\pi} \sqrt{(\sin(t))^2 + (1-\cos(t))^2}

@blissful meadow

[blissful meadow]

You've already computed the speed.

The integrand should be the magnitude of the velocity, i.e. the speed.

[soft zealot]

Renato

[gentle zephyr]

@blissful meadow

[blissful meadow]

Sure, you've already computed that.

[gentle zephyr]

what

[blissful meadow]

You'd already computed the speed there's nothing to check

[gentle zephyr]

the integral of this?

[blissful meadow]

No

Now you can compute the integral.

You may want to expand things under the root.

[gentle zephyr]

is hard

[blissful meadow]

It's not that hard if you expand things and simplify.

[gentle zephyr]

sin^2 + 1^2 -2cos(t) + cos^2

2-2cos(t)

2(1-cos(t))

2sin(t)

@blissful meadow

[blissful meadow]

1-cos(t) is not sin(t).

[gentle zephyr]

how to conclude

[blissful meadow]

Trig identities. $\sin^2(\frac{t}{2}) = \frac{1-\cos(t)}{2}$.

[soft zealot]

Azyrashacorki

[gentle zephyr]

yeah I'm not that advanced

[blissful meadow]

This is not advanced.

It's a common trig identity. If you don't know it then you should get to know it.

[gentle zephyr]

name

[modest sequoia]

this comes from rearranging the double angle identity

cos(2x) = 1-2sin^2 (x) where in this case x = t/2

[gentle zephyr]

is it possible without

[blissful meadow]

Not really

[gentle zephyr]

trig is scary

[severe canyon]

You can't be afraid of trig if you're doing calculus!

Willing or not

[modest sequoia]

trig is a crucial part of calculus and it is extremely important you know your way around it for calc

[blissful meadow]

-# especially when you're at the point of doing multivariable calculus

[gentle zephyr]

so 4sin^2(t/2) -> 2|sin(t/2)|

then wat

[severe canyon]

Then you solve the integral

That's a single variable integral which you should know how to solve

[gentle zephyr]

,, \int_0^{2\pi} 2\cdot |\sin(t/2)| dt

[soft zealot]

Renato

[gentle zephyr]

how

[severe canyon]

How what?

[gentle zephyr]

abs is giving trouble

[severe canyon]

That shouldn't...

If it does, revise the definition of absolute value

As a piecewise function

[gentle zephyr]

abs(x) = x or -x

x >= 0 or x < 0

[severe canyon]

Yeah

That's it

[gentle zephyr]

what about it

[severe canyon]

Maybe.... use it???!!!

[gentle zephyr]

can I just ignore the abs

[blissful meadow]

Look into the sign of sin(t/2) in the interval [0,2pi]

[blissful meadow]

No, you can't.

[gentle zephyr]

• 1<= sin <= 1

[blissful meadow]

Yes that's in general.

[gentle zephyr]

0 <= abs sin <= 1

[gentle zephyr]

trig scary

[blissful meadow]

But for t in [0,2pi], what can you say about the sign of sin(t/2)?

[gentle zephyr]

sin0 is 0

,calc sin(pi)

[soft zealot]

Result:

1.2246467991474e-16

[severe canyon]

STOP SAYING THIS

[blissful meadow]

Also sin(0) and sin(pi) are things you should not need a calculator to check

[gentle zephyr]

we would need to find the critical points and analyze the func

[severe canyon]

Also because this might deceive you and think that it some non zero number

[blissful meadow]

No. If you remember the unit circle you should be able to tell instantly what the sign of sin(t/2) is for t in [0,2pi].

[gentle zephyr]

(x,y) = (cos, sin)

[blissful meadow]

You may think of it as studying the sign of sin(x) for x in [0,pi].

What is the sign of the sine of an angle if that angle is between 0 and pi?

[gentle zephyr]

posi

[blissful meadow]

So sin(t/2) is positive on [0,2pi].

[gentle zephyr]

how?

[blissful meadow]

Wdym how

t ranges from 0 to 2pi, so t/2 ranges from 0 to pi.
You agreed that the sine of an angle between 0 and pi is positive.

[gentle zephyr]

this argument doesnt uses unit ci3cle

@blissful meadow

[blissful meadow]

You literally drew the unit circle

[gentle zephyr]

also not necessarily if thta is in third or 4th qudrant we have negative teta

is just above 180

ohh

[blissful meadow]

Yes, hence why sin(t/2) is positive between 0 and 2pi.

[gentle zephyr]

but if we are above pi then sin is y and y is negative

wait wat

[blissful meadow]

t ranges from 0 to 2pi.
t/2 ranges from 0 to pi.

So t/2 is below pi.

So sin(t/2) is positive.

[gentle zephyr]

how

[gentle zephyr]

how

[blissful meadow]

The sine of an angle between 0 and pi is positive.

You drew it

[gentle zephyr]

sin(x) > 0 forall x in [0, pi]

[blissful meadow]

t/2 is an angle

Between 0 and pi

[gentle zephyr]

between pi and 2pi what happens

@blissful meadow

[blissful meadow]

Then it's negative but that's not in your interval

[gentle zephyr]

how is negativo

@blissful meadow

[blissful meadow]

Wdym by how?

It's like that.

same argument as your unit circle

But the angle is between pi and 2pi

So it's below the x axis.

So the y-coordinate is negative.

[gentle zephyr]

i see

yeah

[blissful meadow]

In any case, sin(t/2) is positive between 0 and 2pi.

[gentle zephyr]

and y coordinate is sin

[blissful meadow]

What can you say about |sin(t/2)| between 0 and 2pi?

[gentle zephyr]

= sin(t/2

[blissful meadow]

Yes

[gentle zephyr]

so i can ignoee abs

[blissful meadow]

So then you don't really need to deal with the absolute value

In this specific case.

[gentle zephyr]

,, 2\cdot \int_0^{2\pi} \sin(t/2) dt

[soft zealot]

Renato

[gentle zephyr]

2[-2cos(t/2)]_(0)^(2pi)

@blissful meadow

u = t/2
du = 1/2 dt
dt = 2du

[modest sequoia]

this is right

now evalute from 0 to 2pi

[gentle zephyr]

,w cospi

[soft zealot]

Results provided by WolframAlpha

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

Requested by renatoxbz

[modest sequoia]

you should know the value of cos(pi)...

[gentle zephyr]

trig is hard

[modest sequoia]

if cos(pi) is hard you should definitely go relearn and fully understand the unit circle

[gentle zephyr]

2[-2 + 2] = 0

[modest sequoia]

no

check your signs again

[gentle zephyr]

2[2] - 2[-2]

is 8

[blissful meadow]

-# (Note that a good way to see something went wrong with the signs is that it makes no sense that your curve would have 0 length between 0 and 2pi.)

[gentle zephyr]

why

[blissful meadow]

Because the only way your curve would have no length is if it's just a point, which it clearly isn't.

[gentle zephyr]

why

[drowsy epoch]

ragebait

[gentle zephyr]

why do you say that the arclength would be 0 iff its a point

is that very intuitive to everyone?

[drowsy epoch]

What's the dimension of a point

[gentle zephyr]

say for example P is a point and O is the origin

PO = ||P||

[drowsy epoch]

What

OP is also a location vector not a point

[drowsy epoch]

It's sloppy but P and Q are vectors

[blissful meadow]

The arclength is the lenght of the curve. It's always nonnegative. The only way it can be 0 is if your curve is just a point, otherwise it would have a length

In other words you would need ||r'(t)||= 0 for all t. And this describes a parametrized curve which is constant, i.e. a point.

(1,1) is a parametrized curve which is the point (1,1) for all t.

[gentle zephyr]

ok this is better explanation

[blissful meadow]

Well it shouldn't be unintuitive that a curve having no length means it's a point

[gentle zephyr]

thr thing is rhat

the tangent line of the curve is

L: k.r'(t) + r(t)

[blissful meadow]

and?

[gentle zephyr]

i thought i had something

[gentle zephyr]

And this describes a parametrized curve which is constant

wdym?

[blissful meadow]

If r'(t) = 0 then that means r(t) = (a,b) where a and b are constant.

Simply by integrating.

But a and b don't depend on t

So r(t) is just the point (a,b)

[gentle zephyr]

isee

[blissful meadow]

In this case (and whenever the velocity is 0 for that matter), you can't define a tangent line.

[gentle zephyr]

yeah

can you help with 10

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[burnt geyser]

eh the 11th question

i just didnt understand how that proves abcd is a square

if anyone can help and give me a tool to practice such problems would be much appreciated

oh and help with this problem too

tysm 🙏

question 11

[void valley]

Do you know the definition of a square?

[burnt geyser]

yes

[void valley]

it is a quadrilateral with all sides equal and all angles equal (90 degrees), they shown these features hold and hence it's a square

What part\line is confusing you there?

[burnt geyser]

so basically

they are showing triangle pas and pbo are congruent

[void valley]

rather that SP = PQ and that angle SPQ is a right angle (because it is equal to 180 - 2 * 45), but yes - these triangles are congruent

[burnt geyser]

um

im kinda lost

would you mind explaining slowly 🙏

the problem

[void valley]

Okay, slowly, let's start from the very beginning

[burnt geyser]

yes ty

[void valley]

Do you understand why AS = BQ?

[burnt geyser]

yeah

ad+bc

sorry equal to

because they are opposite sides

[void valley]

Ok, do you understand how does that imply that AS = 1/2AD and BQ = 1/2BC?

[burnt geyser]

yeah s is midpoint

so is q

[void valley]

Okay, so the first line is clear

[burnt geyser]

yeah

[void valley]

same thing happened in the 2nd line, where AP = BP = 1/2AB (midpoint)

[burnt geyser]

yeah

[void valley]

angle SAP = QBP = 90 deg because ABCD is a square, right

[burnt geyser]

yeah so sas congruence?

[void valley]

Yeah, exactly

[burnt geyser]

okay got that

[void valley]

so ASP and BQP are congruent

[burnt geyser]

yeah

[void valley]

this means that SP = QP (because if SAS holds then also SSS holds)

[burnt geyser]

wait i didnt get that

1 sec let me chek

oh yeah i got it

[void valley]

you could repeat this pattern 3 times for each pair of adjacent triangles (corners) and you would get that all sides of the quadrilateral PQRS are equal

Do you see that?

[burnt geyser]

yeah

[void valley]

Now notice that these corners are isosceles right triangles, so their acute angles are 45 degrees, right?

[burnt geyser]

oh yeah

[void valley]

e.g. in SAP you have SAP = 90 deg, so ASP = APS = (180 - 90)/2 = 45

[burnt geyser]

uh huh

i got it

[void valley]

one last thing - angles SPA, SPQ and QPB are adjacent, so their sum is 180 (the same for the rest of the triangles)

and hence SPQ = 180 - 45 - 45 = 90

this proves that all sides and all angles of the quadrilateral PQRS are equal, so (by definition) it is a square

[burnt geyser]

oh

this was really help ful

i rllly appreciate it

do you reccomend any place to get practice problems like this one

[void valley]

isn't there more of this in the book/textbook?

[burnt geyser]

nope

its kinda only problems

little to no explantatioin

[void valley]

If you have the answers (as above), you can practice, understanding comes with time, do 10 such tasks and you'll get the hang of it

[burnt geyser]

uh huh

i want the ten problems or tasks like this one tho lol

[void valley]

the first picture I found on the internet (the quality is terrible, I know, but it's readable)

[burnt geyser]

i will try to solve it

Thank you so much

[void valley]

just search "basic geometric proofs exercises" or smth like that

[burnt geyser]

i will

goodnight]

.close

[final saddle]

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[fallen valve]

How do we get that the supremum exists on this open set from the fact that g is continuous?

[drowsy epoch]

isn't that extreme value theorem

[jovial sapphire]

hello!

do you still need help?

[jovial sapphire]

it is

since $r < 1$, the closed ball $\bar{B}(0, r) = {z : |z| \le r}$ is a compact set contained within the domain where $g$ is continuous (assuming $g$ is defined on the unit disk)

[soft zealot]

W*

[jovial sapphire]

so by EVT, $|g|$ has to achieve a maximum on that compact set, so it's bounded

[soft zealot]

W*

[jovial sapphire]

the sup on the open ball is then just bounded by that max value lol

[final saddle]

@fallen valve Has your question been resolved?

[final saddle]

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[manic leaf]

(k) $|{f : \mathbb{R} \to \mathbb{R} : f$ is continuous}|

[soft zealot]

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

[manic leaf]

alright ill try this problem now

(k) ${f : \mathbb{R} \to \mathbb{R} : f$ is continuous}

[soft zealot]

toast

[manic leaf]

I am trying to find the cardinality of this set

what does the conitnuity imply here?

[tiny gorge]

one thing that it implies is that if you specify f(x) for all x in Q, then that uniquely determines f

[manic leaf]

hmm wdym by that