help-27

[restive river]

Yeah I don’t understand

[deft trout]

so you're trying to get rid of the 4y^2 on the bottom now

you have to multiply the left term (that you're dividing the inside term with) to a certain number to be able to get rid of that y^4

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

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[noble mirage]

Am i losing my mind or is this algebra just wrong?

[wind mason]

context?

[noble mirage]

But im just following thru the solution

[wind mason]

nevermind i cant do that

[noble mirage]

How has he factored that...

[wind mason]

i don't see a problem with this frankly

[noble mirage]

yeah lol me neither. But its just the algebra of how my lecturer has factored the 3rd line

[wind mason]

because if you expand out all the terms and recollect them you get that given result

[noble mirage]

You factor out -1 in 2nd line to get -r^2sin etc but you get A_+... -1

[wind mason]

no it makes complete sense

im not fucking latexing that but it does distribute correctly

[noble mirage]

okay... then ive just lost my mind looking at this for centuries

[wind mason]

question, where in the flying fuck did the $\theta(A_+^2)$ term go?

[woven radish]

;(

[wind mason]

bro just made it magically disappear with a wave of his hand

poof!

[noble mirage]

question says to to first order A+

i hate this module and got exam tmr :|

[wind mason]

no clue what that means so ill assume its relevant!

but yes, i am sanity checking myself and this does factor correctly (note that cos^2+sin^2=1)

[noble mirage]

Its $O(A^2_+) not theta. The terms get smaller so we approximate solution

yeah but i cant see how he goes from the first lot to the 2nd. Since he takes out a -1 from $(1-A_+\cos(wt)$ but ill just kms

[woven radish]

SollyPolly

[wind mason]

it looks like a theta wth

anyways yeah it makes sense since its big O notation blah blah blah

[noble mirage]

ye well lectuere a physist so

[wind mason]

you can just remove it since it will become insignificant

[wind mason]

!

ok fine ill latex it whatever

i feel bad for u man

[noble mirage]

nah its all good

ive got this

only need to get 26/100 on this exam to pass

then i can forget this module

[wind mason]

💀

this statement is sending me in tears

[noble mirage]

we done well on the coursework!

godbless the 40/60 split

[wind mason]

$r^2\cos^2(\theta)+r^2\cos^2(\theta)A_+\cos(wt)+r^2\sin^2(\theta)-r^2\sin^2(\theta)A_+\cos(wt)$

[woven radish]

;(

[wind mason]

so that $r^2\cos^2(\theta)+r^2\sin^2(\theta)$ term simplifies to $r^2$

[woven radish]

;(

[noble mirage]

ye i see it

i was trynna do it all in 1 step idk where i was going wrong

ty

[wind mason]

👍

[noble mirage]

.close

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[serene zephyr]

i thought i had to just subtract the two functions and then integrate?

[misty crest]

how is that area?

you gave an anti derivative

find the area of the region between the curves

[serene zephyr]

i thought i was doing this from my notes

[misty crest]

yes

a and b

that’s a definite integral

[strange arch]

yes that's correct but you need to determine a and b

[misty crest]

you just gave an anti derivative

[strange arch]

since they're not asking for the integral function of their difference, but the area in between

[serene zephyr]

okay so there has to be a definit integral

okay how do i find a and b

when its not given

[misty crest]

intersection of the curves

[serene zephyr]

okay so i need to set each function to 0

[misty crest]

no you’re finding where they intersect (here they intersect at the roots but this is not always the case)

you set the functions equal to each other

[serene zephyr]

okay so generally how would i find where they intersect

[unique vortex]

x=x^(1/3)
And find roots

[misty crest]

^

you didn’t read what i said

[serene zephyr]

haha oops

[unique vortex]

x^3-x=0
x(x^2-1)=0
x=0
x=+-1

[misty crest]

you don’t need to solve it for them

[serene zephyr]

can u explain how to get this? i dont understand how the power 1/3 became 3

[misty crest]

x = x^1/3 just cube both sides

[serene zephyr]

oh that makes sense

okay so now do i have a= 0 b=1?

becaue it cant be negative

[unique vortex]

Here is the solution.

[misty crest]

what

[serene zephyr]

wasn't that to find the bounds for the integral so that i could find the area?

[misty crest]

yes but what can’t be negative?

[serene zephyr]

the bounds? dont they have yo be positive if im finding the area between them?

[unique vortex]

Cubic root of x?

[misty crest]

why?

you’re saying $\int_{-2}^{-1} x^2 \dd{x}$ is invalid?

[woven radish]

knief

[serene zephyr]

i dont know i just thought that was a rule when im getting the area between function?

i thought so

[unique vortex]

Why?

[misty crest]

why would the bounds have to be positive?

do you mean the area?

[unique vortex]

Only area itself can't be negative.

[serene zephyr]

kind of... i thought that since u plug in the bounds when solving for area they couldnt be negative

[misty crest]

the bounds simply define the region we’re considering (what interval/subset of the domain to integrate over)

[serene zephyr]

okay so they can be negative

so how do i know if the bounds r 0, -1, or1?

[unique vortex]

For example you have area under x axis.
If you integrate it you will get negative number.
But area cannot be negative, so you multiply it by 1 or reverse boundaries.

[misty crest]

they’re all of them

there’s two regions

[serene zephyr]

so its -1 to 0 and 0 to 1?

[wind mason]

or, change the sign of one of the regions if there is multiple, as is the case here

[unique vortex]

x^(1/3) Cannot be negative because of cubic root

[misty crest]

note that we can’t just do $\int_{-1}^1 x - x^{\frac{1}{3}} \dd{x}$ here since this would give us zero (odd function)

[woven radish]

knief

[wind mason]

yep

[wind mason]

that’s...quite the opposite.

x^(1/3) can take negative inputs.

[unique vortex]

Yes, sry.

[misty crest]

what

you literally solved as if they did before

[wind mason]

^ @serene zephyr

[serene zephyr]

i remember that from my notes

[wind mason]

which region do you think is negative?

[misty crest]

this has no meaning

[wind mason]

huh

[unique vortex]

Btw, you need to calculate from -1 to 0 and from 0 to 1 separately, otherwise you will get zero.

[misty crest]

a better question is which function is greater than the other in the two regions

[misty crest]

how can a region be negative

you’re assuming we do x - x^{1/3}

[unique vortex]

x^(1/3)

[misty crest]

what

[misty crest]

sure that’s for that region

[unique vortex]

x^(1/3) is upper function

[misty crest]

it’s the opposite for the other region

[wind mason]

well, it can be, since any function that is below the x-axis on an interval will technically have negative area (the point of the comment was to stimulate OP to realize which area to take the absolute value of)

[unique vortex]

Ah, yes, so x-x^(1/3) for -1 to 0.

[misty crest]

you should be more precise so they understand what we’re really doing

the whole idea is "top - bottom"

[wind mason]

yes, i just covered this unit in class

like, i have a test on it in 3 days

[serene zephyr]

so how do u find which region is on top then?

[misty crest]

which function you mean

in a given region

[wind mason]

it’s better to understand that sometimes, when calculating, it will give negative inputs if you’re not working correctly, so this is to make sure OP understands that you need to take the absolute value of certain intervals in order to get the answer correct

if it has multiple regions

[serene zephyr]

okay i think that i got confused while reading everything

[unique vortex]

Plot functions and see.

[misty crest]

no

don’t build a dependence on graphing tools

[wind mason]

you can just plug in values, since both functions are one-to-one, for each interval

so like, x=-1/2 and x=1/2, respectively

[unique vortex]

But you should be able to plot them by yourself, especially basic ones.

[misty crest]

sure

[wind mason]

what exactly are you confused about?

and i’m sorry if i am bombarding too much information

[serene zephyr]

its okay i just didnt know when we were talking about the functions vs the intervals

but im all good now i think

[misty crest]

are you aware that for x in (0,1) x^{1/3} > x?

[serene zephyr]

back to the question i plug in .5 for x into the origional functions?

to figure out which is on top?

[misty crest]

that works yes but do you know the cube root of 0.5

[wind mason]

yep

[misty crest]

a better value would be something like 1/8 since it’s a perfect cube

[wind mason]

you can easily cube both sides

[misty crest]

^

[misty crest]

taking the nth root for any natural number n > 1 of x in (0,1) will result in a number larger than x and i hope it’s clear why that is

mhm

[serene zephyr]

ya

[misty crest]

so 1/8 < 1/2

hence x^{1/3} > x in that interval

since we’ve already determined the points of intersection

it must be greater throughout the entire interval

now consider the other interval

-1 -> 0

what do you expect to happen?

you don’t even need to compute anything here

[serene zephyr]

um would it switch?

[misty crest]

yes but why

there are many intuitive reasons

use one

[serene zephyr]

because -1/8 is larger than -1/2

[misty crest]

mhm

if you wanted to multiply the inequality you had by -1 it would work and you’d get that but what i was getting at was the absolute value of the nth root (take n to be odd here because we can use symmetry for the negative side) of x will be larger than the absolute value of x for x in (0,1)

hence since x is lower than cube root of x when they’re positive it means |x| < |x^1/3|

but we consider values that are more negative to be smaller

so x^1/3 will be below x in the interval (-1,0)

more negative

[serene zephyr]

okay ya that makes sense

so does that mean from (-1.0) it is x-x^1/3 and from (0,1) its x1/3 -x

[misty crest]

yes exactly

[serene zephyr]

okay awesome

[misty crest]

note that you don’t need to calculate both integrals

because of symmetry about the origin

since it’s odd

[serene zephyr]

but its written like this?

[misty crest]

yes this is correct

remember to write dx

[serene zephyr]

oh ya!

so its 3/4 and -3/4?

no its just positive

[misty crest]

hmm how did you get that

what happened to the x?

[wind mason]

don't think so.

[unique vortex]

Should be 1/4 and 1/4.

[wind mason]

dude

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[serene zephyr]

what did i miss in my work?

[unique vortex]

No, you plug 0 into both functions (x and x^(1/3)) and plug -1 into both and then subtract values.

[misty crest]

(-1)^2 =?

[misty crest]

yes

now you know the other region will have the same area

so just double it

or add them together

same thing

[serene zephyr]

1/2

[serene zephyr]

thx

.close

[misty crest]

you’re welcome

[devout snow]

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[finite briar]

Boo

[finite briar]

Uhh so say I have an odd prime p

Is it always true that p is a quadratic non residue mod p^2?

I think this is doable via something called jacobi symbols but idk anything about them

[finite briar]

Found this on wikipedia

Apparently it referrences gauss's arithmetica but idk how to read latin

[tender cobalt]

dont know anything about this, but I tried testing this out and, for 2 < n < 141, "there exists a number k such that k^2 = n mod n^2" only seems to be true for perfect square n

I didnt test higher than that for now

[finite briar]

So meaning it isn't true for odd primes right

[devout snow]

@finite briar Has your question been resolved?

[finite briar]

<@&286206848099549185>

[devout snow]

@finite briar Has your question been resolved?

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[thick schooner]

what kinda justification is required here?

i mean asin(x) is 2pi-periodic

and tan(x) is pi-periodic

so f is 2pi-periodic

do i have to show smth else?

[strange nimbus]

As long as there's a sort of LCM in their periods going on, it's periodic.

You can show it by dividing the periods by pi to get integers and then find the LCM of those integers and then multiply that by pi.

[thick schooner]

i mean isn't that exactly what i claimed

[strange nimbus]

Yes, but it's a bit more detailed about why.

[thick schooner]

yeah i mean

sin(x + 2pik) = sin(x)

tan(x + pi n) = tan(x) = tan(x + 2pin)

but i think to be fair

u need more justification

like why period can't be less than 2pi

[strange nimbus]

Yes, that's fine, but it doesn't show that that's the smallest combined period.

The LCM can do that.

[thick schooner]

some multiple of 2pi and pi has to coincide

and yes it's obvious

the lcm does that

but we still haven't showed

that is the smallest period

do we not have to?

[strange nimbus]

Well, you have the smallest periods of the individual functions.

Then, you divide them by pi, take the LCM, and multiply it by pi, and then that will be the smallest period for the combination.

[thick schooner]

i mean the book shows that it is the smallest period

most of the time

after my-kinda justification

but i thought it was redundant

also brb

[devout snow]

@thick schooner Has your question been resolved?

[slender mirage]

Suppose T is a period smaller than 2pi, then 2pi=kT for some natural k => T = 2pi/k, k \in {2, 3, ...} but it is manageable to check that no T of such form satisfies f(x + T) = f(x)

[thick schooner]

but is this required?

[slender mirage]

not necessarily. you can simply claim sin x has period 2pi, tan x has period pi, so f must have period lcm(1, 2) pi

[thick schooner]

okay makes sense

thanks

[sand dove]

complicated to claim that too

but since they have different periods maybe this is a possible result

otherwise subtracting two functions of the same period can get a function of smaller period

[slender mirage]

f(x) = (tan x)(2cos x - 1)

= (sec x/2)(tan x)(cos 3x/2)```

hmm? should work now :p

[devout snow]

@thick schooner Has your question been resolved?

[thick schooner]

oh

maybe that's why they're checking

ok thx

[devout snow]

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[heavy peak]

idk how to factorise that

[heavy peak]

the 1/2 (156g square - 80gh + 15h square)

[solid osprey]

ok you prolly made some sort of mistake? its unfactorable

[devout snow]

@heavy peak Has your question been resolved?

[latent quarry]

I think you miss the (11x-2y)(x+4y)

[devout snow]

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[solar goblet]

i saw what you deleted

unspeakable horrors

[devout snow]

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[slender mirage]

Send f to the left hand side, take reciprocal of the exponential function and integrate both sides

It's an exact differential

[eternal aspen]

Haven’t learn integrals yet

[ancient sluice]

Solving differential eqns without knowledge of integrals ?

[eternal aspen]

Yeah I am using some kind of theorem

[arctic field]

which theorem?

[ancient sluice]

I think he means applying that exact differential thing!

[eternal aspen]

Constant function theorem

[arctic field]

let the man speak rather than speaking for him smh

and as you see, it is not what your misconception was

[arctic field]

is that like f'(x) = 0 iff f(x) = constant?

[eternal aspen]

I am translating give me a minute

[arctic field]

👍

[eternal aspen]

First one should be saying continuous not differentiable

[arctic field]

okay

so we should probably find ourselves a constant function

we expect that $f(x) = x^2 e^{3x}$, so how about we try
[ g(x) = f(x) e^{-3x} - x^2? ]

[eternal aspen]

Yeah I should assume a g(x) but idk what to assume

Ok I am trying that

[arctic field]

aight

[eternal aspen]

Yeah that propably works as I can see , but how should I know how to reform so it can work like that

[arctic field]

well if $f(x) = x^2 e^{3x}$, then [ f(x) e^{-3x} - x^2 = 0 ]

[arctic field]

so thats one of the things you should try

obviously there are other things to try

your first thought might be [ g(x) = f(x) - x^2 e^{3x} ] but if you try that, you'll see it doesnt work out as nicely

[eternal aspen]

Yeah I mean it’s mostly experience I guess

[eternal aspen]

Yeah that’s what I tried

Anyways thanks for your time !

[arctic field]

one thing you could think about is like

[arctic field]

i didnt like this option because it divides by 0 at x=0

[eternal aspen]

Yeah X€R so wouldn’t work out well

[arctic field]

if you'd just tried $g(x) = f(x)/e^{3x} = f(x) e^{-3x}$, that wouldve gotten you most of the way there though

[eternal aspen]

I am getting a 2x as an answer from g’(x) and not a constant c as I should

Idk if I have made a mistake

Nvm I found the mistake e

[devout snow]

@eternal aspen Has your question been resolved?

[devout snow]

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[mortal aspen]

how else can i slolve this prob??
i tried two methods but its still showing incorrect

[fair juniper]

read the first point again

also what do l and n mean?

[mortal aspen]

l nd n means the direction lines

[slender mirage]

So like (l, m, n) are direction ratios?

[mortal aspen]

yess

[mortal aspen]

direction cosines na

[slender mirage]

Alr got it

[slender mirage]

x coordinate is 5 not -5

[mortal aspen]

ohhhh

.close

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[tulip halo]

3√7 + 8√21

How do you add it?

[sand quarry]

nothing to "add" really, as both sqrt(7) and sqrt(21) are irrational

you could factor it if that's what you intend to do, though.

[strange arch]

what remains unclear? c:

[tulip halo]

How?

@sand quarry

[sand quarry]

prime factorise 21

[strange arch]

sqrt(21) = sqrt(3) * sqrt(7) if that is what you seek

[tulip halo]

Ok.

.close

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[mild crypt]

how are you ever meant to do this without mathway?

[valid silo]

whar is the question?

[mild crypt]

factor

[valid silo]

ok

so you can see all three values have d^2 terms

and z^2 terms

so you factor those out first

right?

[mild crypt]

yes i see that

[faint zinc]

That's it.

[mild crypt]

wdym thats it?

[faint zinc]

You see you can factor out these terms, so you factor them out

And that's the end

[mild crypt]

i see how the z^2 would work after

but idk how to do it with the parentheses and stuff after that

OH wait

i think i maybe see

[faint zinc]

Are you not familiar with the distributive property?

[mild crypt]

so this is just the same

[mild crypt]

whats that

i have highschool diploma but this is some next level factoring

i only got basic factoring

[valid silo]

maybe you are askimg how to factor the bracket?

[mild crypt]

like from parabola

[faint zinc]

a(b+c) = ab+ac

[mild crypt]

i know that

[faint zinc]

Then ab + ac = a(b+c) yeah?

[mild crypt]

well yes but thats way simpler

than the questions

[valid silo]

as in what is the factorisation of (d^3z^2+z^3-1)

is that your qn?

[mild crypt]

nah i just dont know how to do these huge things, i got teached the product sum method

where you have

[faint zinc]

In your problem, a = d^2 z^2, b = z^3, c = d^3z^2 - 1

[mild crypt]

x^2 + 6x + 9

and you just do

3 + 3 is 6

and 3 * 3 is 9

so its

(x+3)(x+3)

[faint zinc]

This isn't that complicated though

[mild crypt]

this is the only way i got it thought

and this is completely different from that

[faint zinc]

You just have a d^2 and z^2 in every term that you can pull out

[mild crypt]

ye i understand that question

[mild crypt]

but idk how i would do this

[faint zinc]

There's an f in every term you can pull out

[mild crypt]

i guess first factor out f?

ye

[faint zinc]

Then it's just a quadratic

Not all cubics and higher are easy though

And quintics and higher can be actually impossible

[mild crypt]

idk how i would factor this more tho

i understand i just need to factor whats inside the brackets

but i have no idea how

i can only do this if the second term is a the sum and the third term is the product of the 2 factors

[mild crypt]

how do i do that tho

idk how to do it if its not whole numbers

[faint zinc]

Hmm? It's the same though

[mild crypt]

no

like i said i only got thought the product sum method

idk if thats what its called in english but

[faint zinc]

1/3+1/3 = 2/3 And 1/3*1/3 = 1/9

[mild crypt]

oh

ok but this doesnt have a term thats common in all

[faint zinc]

It's a little trickier when the leading coefficient isn't 1

So why not make it 1 by multiplying everything by 4?

[mild crypt]

OH WAIT I DID IT

[faint zinc]

Be sure to also pull out the 1/4 to outside of the brackets

[mild crypt]

ye

i got it

i figured it out

this also confuses me

there is literally like

nothing there

i dont get how thats even factorable

[worthy heath]

( z + sqrt 19 ) ( z - sqrt 19 )

[mild crypt]

ohhhh

ye ofc

bro

i thought i understood

then this came

this is apparnetly not it

💀

it needs to be more

ok i got it that was so hard

[devout snow]

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[oblique rover]

how to show that $\forall n \in\mathbb{N}:\frac{-2}{n^2+1}\neq 0$?

[woven radish]

Slowaq

[oblique rover]

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[grizzled wyvern]

Hii

[grizzled wyvern]

I need help in Q.2 a d Q.3

[worthy heath]

for 3
|A-xI|=0

then solve for x which will be the eigenvalues
idk about eigenvectors :/

[grizzled wyvern]

Oh.. ok thanks

[brittle inlet]

Alright

2 and 3, right?

[grizzled wyvern]

Yess

I have done Q.2 half

[brittle inlet]

Ok

[grizzled wyvern]

I need help further I'm stuck

[brittle inlet]

Well, in Q2, you're given that for some nxn matrix, some row is full of 0s, right?

[grizzled wyvern]

Yepp

[brittle inlet]

Alright, let's forget for now about that

Suppose you're given a 4x4 matrix, what would you do to calculate its determinant?

[grizzled wyvern]

Try to solve it with echelon form

Or diagonal form

[brittle inlet]

As in diagonalize it and calculate the determinant of the diagonal matrix?

Hm, well, that's not what I was expecting

What theorems/propositions have you got to work with?

[grizzled wyvern]

.close

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[errant arch]

How do I find B? I don't understand the circle theorem. Please help

[wind mason]

they contain the same inscribed angle.

also, let me draw something real quick, to get an ease of notation.

[errant arch]

alright

[sonic parrot]

I always see you here

[mellow raft]

Then hes a nice guy

[wind mason]

?

[wind mason]

sorry for the bad handwriting; im doing this on pc.

[sonic parrot]

I see you often in this doscord server you must be a nice guy

[wind mason]

@errant arch does this make sense?

[errant arch]

Not really

It's just totally different from whatever my teacher gave us

[wind mason]

ah.

well, what did your teacher give you?

[errant arch]

just like the theory but no examples on how to find the angles

[wind mason]

ah.

[errant arch]

Someone help 🙏

[devout snow]

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[knotty sage]

just to check the general form of solutions to this ODE
y'-4y=x-2 is y'=-(x/4)+(7/16)+Ce^(4x)?

[fair juniper]

the numbers check out

seems good to me

[knotty sage]

thx

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[brisk bobcat]

im so confused rn. i have a calc test tomorrow but idk anything about this unit everything looks so confusing. ive been sick this past like week

[serene quest]

have you began solving it?

[brisk bobcat]

not yet, im kinda lost

[misty crest]

what’s the area of the rectangle

for a given x and y

[brisk bobcat]

the only thinf they give us are on the problem, i have no idea the dimensions of it

[misty crest]

no

you’re not understanding

what is the formula for the area of a rectangle

[brisk bobcat]

ohh

[misty crest]

that is what we wish to maximize

then we can write it in terms of x and y

[brisk bobcat]

y= (6-x)/2

[misty crest]

and see if we can write it in terms of just one

[brisk bobcat]

wait lemme do that

[latent quarry]

The maximum area of a rectangle inscribed in a triangle is 1/2 area of the triangle, am I right?

Then the area of the triangle is (3*6)/2

Since then then the maximum area of the rectangle is 4.5 ???

[misty crest]

sure but i think this is for a calculus class and seeing as they’re struggling with the process of optimization it’s probably worth the practice anyway

but yes

[devout snow]

@brisk bobcat Has your question been resolved?

[latent quarry]

It also could be proved using calculus right

[devout snow]

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[sand dove]

yeah 8C2 is correct

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[gaunt cliff]

the empty set isnt in every set. the condition for part (b) is that the pair of subsets don't share elements in common

[gaunt cliff]

seems good

what are your answers for 1. then (a) (b) (c) (d) ?

[devout snow]

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[prisma elm]

how to solve for e)?

[devout snow]

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[devout snow]

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[undone comet]

I'm confused with b)

What are the elements I'm adding/multiplying together?

I know with mod m the elements are 1,2,3,4,5,....,m-1

[devout snow]

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[lost laurel]

$\int_{0}^{3} \int_{\sqrt{x/3}}^{1} \frac{dy dx}{y^4+1}$

[devout snow]

Please don't occupy multiple help channels.

[woven radish]

What a wonderful world!

[lost laurel]

so I want to change the bounds of integration gere

sorry

rather the order

so I have $0<x<3$

[woven radish]

What a wonderful world!

[lost laurel]

and $\sqrt{x/3}<y<1$

[woven radish]

What a wonderful world!

[lost laurel]

when I switch the order, I find that the bounds of x remain the same

and 1<\sqrt{x/3{<1

which is sus

[stone stump]

have you made a drawing

[lost laurel]

yes

and shaded the region too

[lost laurel]

yes

[stone stump]

then I dont know how you got 1 < sqrt(x/3) < 1, in particular something independent of y

[lost laurel]

the horizontal line enteres at y =\sqrt{x/3} and exists at y=1

oops

really confused

[stone stump]

if you pick some y, eg y=0.5, then which x are allowed?

graphically, you have the horizontal line y=0.5 and ask yourself what the intersection with your shaded area is

[lost laurel]

Okay

so if we consider horizontal lines intersecting the region

the enter at x=0

and exit at x=3y^2

is that right

[lost laurel]

Now what

so I have $\int_{a}^{b} \int_{0}^{3y^2} \frac{dx dy}{y^4+1}$

[woven radish]

What a wonderful world!

[lost laurel]

$0<x<3, 0<x/3<1, 0<\sqrt{\frac{x}{3}}<1$

[woven radish]

What a wonderful world!

[lost laurel]

oh

so $\int_{0}^{1} \int_{0}^{3y^2} \frac{dxdy}{ y^4+1}$

[woven radish]

What a wonderful world!

[lost laurel]

Is this right

,w is is \int{0}^{1} \int{0}^{3y^2} \frac{dxdy}{ y^4+1} = \int{0}^{3} \int{\sqrt{x/3}}^{1} \frac{dy dx}{y^4+1}

[woven radish]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by math_rocks

[lost laurel]

<@&286206848099549185>

[devout snow]

@lost laurel Has your question been resolved?

[devout snow]

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[frail ginkgo]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[frail ginkgo]

anyone can help on how i can like do the graph

like how do i graph this graph

help me please

.close

[devout snow]

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[wind mason]

i can help, if you're still willing.

[frail ginkgo]

yes please

im willing

anything

im desperate

[wind mason]

ok...

you can use the command .reopen to repoen the post.

[frail ginkgo]

.reopen

[devout snow]

✅

[wind mason]

okay.

so, what exactly are you confused about?

[frail ginkgo]

like the graph

like how did they get that particular graph from the data

like ik how to do they asyptotes and why they are there

its just the curve thingies

[wind mason]

you create what fits from the data.

[frail ginkgo]

yeah but like do i just make a curve that connects to the points?

do i like make curves that connects to the intercepts?

[wind mason]

i suppose, but i don't really think it would matter, so long as you just graph one point in between each asymptote.

[frail ginkgo]

oh

ok

thanks

[wind mason]

but do make sure, if you want full marks, to graph the characteristics given

[frail ginkgo]

okya

[wind mason]

because those are the most obvious things to do

[frail ginkgo]

mb bruh im jus stupid

.close

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[knotty orchid]

Question

[knotty orchid]

If I am on a train with a constant velocity

And walk in the same direction as travel with a constant velocity

Would the momentum of the train change?

Compared to if I was sitting stil?

[devout snow]

@knotty orchid Has your question been resolved?

[iron kindle]

veeeery slightly

[knotty orchid]

How would you express that in a formula?

[iron kindle]

any momentum you gain when walking

the train loses

[knotty orchid]

So like

Relative speed?

[iron kindle]

total momentum: m_train * v_train + m_you * v_you = P

so when youre not moving

P=m_train*v_train

when you start moving, P stays the same, so m_train * v_train1 = m_train * v_train2 + m_you * v_you

v_train1 is the speed before you moved

v_train2 is the speed after you moved

[knotty orchid]

Aaaah I see

Cool

Thank you :)

.close

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[knotty orchid]

Wait, but if I sitt stil in this equation, then my mass is canceled

[iron kindle]

yes, because your speed is 0, and anything *0 =0

if you sit still, your momentum is 0

[knotty orchid]

But my mass is still in the train

Like

Let's say the entire train satt still

(Pasangers)

200 people

80 kg each

That's 16 tonns

I find it hard to believe that if they sitt still on the train

That their mass just disapears

Right?

[iron kindle]

thats kinetic energy

$p=mv\E_k=\frac12mv^2$

[woven radish]

Bonk

[iron kindle]

and you are confusing a few things

[knotty orchid]

I don't see why that would change it 😭

[iron kindle]

when finding the total momentum you take the whole mass

[knotty orchid]

Yeah

[iron kindle]

so train and everyoneone on it

[knotty orchid]

Yeah

[iron kindle]

since momentum is conserved, if you start moving, you "take away" momentum from the train and use it for yourself

[knotty orchid]

Yeah

[iron kindle]

do you have an example question?

[knotty orchid]

I'll make one

[iron kindle]

we need to look at our reference point

is our reference moving along with the train or a spectator watching the train go past

if its moving with the train, if everyone is sitting still, momentum is 0

if its watching the train, then the real speed is train speed+walking speed

[knotty orchid]

What effect does the pasangers moving have on the train

[iron kindle]

newtons 3rd law

if you push yourself to walk forward

you push the train backwards

[knotty orchid]

Yeah that was what I was thinking originaly

But does the total stay the same then?

[iron kindle]

momentum is conserved, yes

[knotty orchid]

Okay

Thank you

.closr

.close

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[rich parrot]

$z = arcsin(x-y)\newline x=3t \newline y=4t^2$

I need to use partial differentiation to show that...

$\frac{dz}{dt}=\frac{3}{\sqrt{1-t^2}}$

I tried using this formula:

$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt}+\frac{\partial z}{\partial y} \frac{dz}{dt}$$

But I got...

$$\frac{3-8t}{\sqrt{1-(x-y)^2}}$$

which doesn't seem to prove it

[woven radish]

radian

[feral agate]

You don't really need partial differentiation here, you can just get z as a function of t

Also the required derivative just looks wrong

Yours is correct

[rich parrot]

That's what I was thinking too

Then I ran it through chatgpt to tell me I got stuck and it got stuck too

.close

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[velvet rose]

Am I on the right track?

I'm stuck

[worldly crystal]

its negative for sure

to goes to negative infinity

cuz

Instead of $\int_{0}^9 \frac{1}{\sqrt{x}-3} dx$

[woven radish]

kronium_

[worldly crystal]

we can interchange the values

that gives us

$\int_0^9 -\frac{1}{3-\sqrt{x}} dx$

[woven radish]

kronium_

kronium_

[velvet rose]

but don't I have to find the anti-derivative to be able to say anything like that?

[velvet rose]

and 3 - sqrt(x) approaches 0 I think?

[finite briar]

This isn't terribly tough to integrate btw

Play around with a couple substitutions (hint- x=u^2)

[worldly crystal]

yes

0

lemme rewrite it

[worldly crystal]

as $x --> 9$, $3 - $\sqrt{x} --> 0$
Therefore $\frac{1}{3-\sqrt{x}} --> \infty$

[woven radish]

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

[velvet rose]

.

nvm solved it

[devout snow]

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[opaque skiff]

need help with this question: f,g,k,h are different isometries if gof=koh then fog=hok

[opaque skiff]

?

[rotund widget]

hello

[opaque skiff]

hi

can u help me?

[rotund widget]

how can a halp

help you

?

[opaque skiff]

help me solve it

disprove or prove

[rotund widget]

prove fs

[opaque skiff]

wdym

[toxic grove]

Have you tried it with some examples in 2d?

[opaque skiff]

yes

from this gof=koh i can tell that f-1og-1=h-1ok-1]

[toxic grove]

Yeah I'm not sure that's gonna get anywhere

What happens if k is the identity?

[opaque skiff]

then koh=hok

and gof=hok

i think its not true

[toxic grove]

So what does that tell you about g and f?

[opaque skiff]

what

[toxic grove]

You've got gof = koh by definition

You've got koh = hok

And you've got hok = gof by assumption (if you want to do it by contradiction)

Sorry hok = fog

[opaque skiff]

but gof is not neccesarly eqale to fog

[toxic grove]

Exactly

That's a contradiction

[opaque skiff]

can u give me am example

to disprove claim

?

[toxic grove]

Do you understand proof by contradiction?

And do you understand why IF you assume the statement you're trying to prove, then gof = fog?

[opaque skiff]

but the claim is false

[toxic grove]

I'll take that as a no then

[opaque skiff]

is that true?

[toxic grove]

One way of proving that a statement is false is by assuming it to be true, then finding a contradiction

So for example, if you wanted to prove that "10000 is the largest number" was false, you could assume it's true, then show that "10001 > 10000"

Which would be a contradiction, showing that the original statement was false

It's a bit contrived, but does that make sense?

[opaque skiff]

yes but u can give an example to the 4 isometris and show that the condition is true buts the claim is false

[toxic grove]

I'm not gonna give you an example but I'll help you find one

[opaque skiff]

i didnt find an example i tried

[toxic grove]

If you assume that the claim is true, then since it applies to all possible g, f, h, k, it means that gof = fog for all f and g

So all you need to do is find f and g such that fog isn't equal to gof

Then that will disprove the statement

Does that make sense?

[opaque skiff]

and h and k

[toxic grove]

Yes, but you can take

Sorry didn't mean to press enter

You can take h=1 and k=gof

[opaque skiff]

what is h=1 its not an isometry

[toxic grove]

The identity map

It's an abbreviation

[opaque skiff]

oh oh ok

then k-1=f-1og-1

[toxic grove]

Yes but that doesn't really get you anywhere I don't think

I could be wrong

Which isometries do you know?

[opaque skiff]

there are only 5 no?

[toxic grove]

There are quite a lot more than 5 lol

Which ones do you know?

[opaque skiff]

i know identity ,rotation ,transformation, mirroring, offset mirroring
sry id its not the exact name its google translate

[toxic grove]

Ah okay I see what you mean

Yeah that sounds right

Those are 5 families of isometries rather than individual transformations but yes I think that's all of them

How about you let f be a "transformation" (in english we call them translations) and let g be a rotation

And see if fog = gof

[opaque skiff]

but transformation in how much and rotation by how much

[toxic grove]

Doesn't matter

Pick some easy numbers

[opaque skiff]

ok

[toxic grove]

Also in case you're wondering, we call "mirroring" reflections

[opaque skiff]

In relation to which straight line the transformation and whether the axis of rotation should be on the line

[toxic grove]

Again just pick something easy

[opaque skiff]

i called the line l

[toxic grove]

Translate 1 unit to the right, and rotate 180 degrees

[opaque skiff]

and o is the axis

o is on l

[toxic grove]

Yes the origin

[opaque skiff]

origin?

[toxic grove]

The middle of the graph

[opaque skiff]

what is the second line for

[toxic grove]

Are you doing isometries in one dimension?

[opaque skiff]

wdym

[toxic grove]

Are you doing isometries on numbers on the number line?

I assumed you were doing it on points in a plane

[opaque skiff]

no

[toxic grove]

Okay good, that would've been a bit weird

I drew two lines to represent points (x,y) on a plane, that you can apply isometries to

Anyway so you've got f and g now

So pick a point and apply fog and gof to that point

[opaque skiff]

fog=!gof

[toxic grove]

Great

So do you understand why that disproves the statement?

[opaque skiff]

but what with koh

f,g,k,h are different isometries if gof=koh then fog=hok

[toxic grove]

It's like I was saying back here

If that statement is true, then

Let k be the identity, so hok = koh

Then gof = koh = hok = fog

So gof = fog

[opaque skiff]

but why

[toxic grove]

Which part doesn't make sense?

[opaque skiff]

from what i understood fog=!gof an bc gof=koh and k is identity gof=hok so thats wht fog=!hok bc fog=!gof

[toxic grove]

Exactly

So that's a contradiction isn't it?

[opaque skiff]

so it doesnt meter what h is?

yes

[toxic grove]

Not really no

There's a false statement in there whatever h is

So you assumed a statement and concluded a contradiction

[opaque skiff]

thats what i said

[toxic grove]

So the statement has to be false

[toxic grove]

Yeah sorry I was agreeing with you

[soft wyvern]

Hi

Do I get help here?

[opaque skiff]

no

[toxic grove]

Go to one of the "available" help channels and ask there

#help-20 for instnace

[opaque skiff]

but gof suposed to be eqale to koh

[toxic grove]

If k is the identity, and h = gof, then that is true

But fog is not equal to hok

[opaque skiff]

ye tytytytytytytyty u saved me

[toxic grove]

Np

[opaque skiff]

gof is offset mirroring

[toxic grove]

Make sure you really understand it though, it shouldn't seem like I pulled it from nowhere

[toxic grove]

It's actually another rotation I'm pretty sure

Just a different one

[toxic grove]

gof and fog are both rotations but with different centers

Yes but that's only in one dimension

In two dimensions or more they're different

[opaque skiff]

its offset mirroring relation to l and offest in 1

ithing

isnt that true

[toxic grove]

If you do the rotation first then the translation, then it's actually equivalent to a rotation by 180 degrees around the point 0.5 to the right of 0

It doesn't look like it, but it is actually a rotation

It's definitely not a reflection

The point (0.5, 0) goes to itself, which you never get with offset reflections

[opaque skiff]

But it's g first

[toxic grove]

Yes, g is a rotation isn't it?

[opaque skiff]

Yea

[toxic grove]

And f is a translation

So fog (so g first) is a 180 degree rotation around (0.5, 0)

And gof is a 180 degree rotation around (-0.5, 0)

[opaque skiff]

But I didn't learned it

(-0.5,0)

[toxic grove]

Those are the points I mean

[opaque skiff]

How

[toxic grove]

You can try it with some examples

Work out fog(0.5, 0) and you'll get 0.5, 0

[opaque skiff]

Idk how to prove it

[toxic grove]

I'm sure you'll get to it quite soon

Dw about it for now

It doesn't really matter for the question you were stuck on

[opaque skiff]

Didn't learn to write it like this

[toxic grove]

I was just correcting you because you said something wrong

How are you writing it? Sorry

[opaque skiff]

0.5,0

[toxic grove]

Okay

[opaque skiff]

What is h

[toxic grove]

So if you work out fog on that point, you'll just get that same point

Wdym?

h is just gof

h doesn't really matter

[opaque skiff]

Ok

Wait a sec

[toxic grove]

It doesn't matter what fog and gof are, the only thing that matters for that proof is that they aren't the same

[opaque skiff]

How can I prove that gof=!fog

[toxic grove]

You managed to prove it back here

So however you did it then

Otherwise you just have to find any point x such that gof(x) != fog(x)

[opaque skiff]

I didn't really prove it I found a point A that (fog)(A)=!(gof)(A) is that enough?

[toxic grove]

Yes that proves they aren't the same

[opaque skiff]

I Should I just show it on a graph?

?

[toxic grove]

Any way you like

A graph works

[devout snow]

@opaque skiff Has your question been resolved?

[devout snow]

@opaque skiff Has your question been resolved?

[opaque skiff]

Is that OK?

?

?

@toxic grove

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[opaque skiff]

.reopen

[devout snow]

✅

[toxic grove]

Yeah looks good

[opaque skiff]

ok ty

where did u learned this?

[toxic grove]

I looked at isometries in lots of detail in my 2nd and 4th years of university

[opaque skiff]

in what university

[toxic grove]

I can't recommend any books about it that are suitable for your level, but the course textbook for my 2nd year course was the Miles Reid textbook "geometry and topology" and it has some great chapters about that type of thing

But it might be a tough read

I'd rather not say, I quite like my anonymity

[opaque skiff]

?

[toxic grove]

I'm not telling you which university I went to

I like being anonymous

[opaque skiff]

oh oh

[toxic grove]

Nothing personal

[opaque skiff]

its fine

[toxic grove]

Most universities will have a course that covers that though

[opaque skiff]

now

its tough

[toxic grove]

Oh you're in first year?

[opaque skiff]

ye

[toxic grove]

How are you at school and university at the same time?

[opaque skiff]

i got in

[toxic grove]

Interesting

I've not heard of that before

But you're not a full time student presumably?

[opaque skiff]

but i can

[devout snow]

@opaque skiff Has your question been resolved?

Channel closed

Closed by @opaque skiff

Use .reopen if this was a mistake.

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[finite gale]

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