help-36

[viral jungle]

how about you show me how you got 9

[livid tangle]

Oh

Im stupid nevermind

There are 8 sorry

So 8×2 is 16

Do I find the surface area of these 16 faces then subtract it from the final answer?

[viral jungle]

yes

[livid tangle]

So like 16×(3)^2?

[viral jungle]

yes

[livid tangle]

But the answers don't match up then

[viral jungle]

also since I have come this far, I've checked your original answer

I don't think it's right

[livid tangle]

Is it not?

Can you tell what I did wrong?

[viral jungle]

how many exposed faces did you find?

don't calculate their total area. just tell me how many exposed faces you found.

[livid tangle]

In the last method or this one?

[viral jungle]

original

[livid tangle]

36 faces

[viral jungle]

cuz from your answer of 540 and a surface area per face of 9, that means you somehow found 60 exposed faces, when there are only 54 TOTAL faces on 9 cubes, not even including the hidden ones

[livid tangle]

So I did it wrong then

What did I do wrong tho I don't get it

The front has 9 faces and so does the the back

[viral jungle]

well ok, I see why now

[livid tangle]

What do you mean?

[viral jungle]

where did this 210 come from?

you only have 5 exposed faces on the top and bottom

should be clear what I mean now, right?

look through your work again

[livid tangle]

Oh is it supposed to be 180?

[viral jungle]

explain your working.

[livid tangle]

Okay so the top has 5 faces right

Oh

I see what I did wrong

[viral jungle]

ok nvm I miscounted the hidden faces
yes, there are 9

I missed one junction at the left

my bad on that one. but either way, the two methods must agree

[livid tangle]

Okay so let me do this again

Since there are 5 faces at the top

The formula should be 5×(3)^2?

[viral jungle]

yes

in fact, just precalculate the surface area per face

[livid tangle]

Oh okay that's what I was doing wrong

[viral jungle]

you know they're all cubes of the same size. they should be constant anyway

[livid tangle]

Mhm

[viral jungle]

the fewer calculations you make, the less the chances of making mistakes

[livid tangle]

Okay can you give me a sec to do the calculations?

[viral jungle]

so might as well just make it 9 instead of always having to write 3^2 over and over again

it's your channel, mate. take as long as you want, so long as the channel doesn't close

[livid tangle]

Oh right that's genius

Ty

Okay so the top and bottom are both 45

The left and right have 4 faces so they are 36

The front and back have 9 faces so they're 81

So do I add these now?

[viral jungle]

do you need to ask me again

[livid tangle]

You're right sorry

[viral jungle]

pretty sure I told you what to do if you're asked to find a total

[livid tangle]

I DID IT

Thank you soo much

You're a life saver genuinely

[viral jungle]

well done

also, normally if you're given this kind of problem

toss away any area shit first

just count faces

[livid tangle]

Okay

[viral jungle]

whether you want to do direct counting (exposed faces) or complementary counting (subtract hidden faces from total), just count faces first

[livid tangle]

Got it

[viral jungle]

you already know the area per face is constant, so once you're done counting faces, mult the number of faces by the area per face and you're done

[livid tangle]

Right

Again thank you so much for helping me out with this I really appreciate it

So do I just close the channel now?

[wet egret]

Nice

[viral jungle]

yes

unless you have other questions

[livid tangle]

No that's all thanks alot

.close

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[warm python]

$dim(Null(T^*))= dim(Range T)^{\perp}$

[soft zealot]

wai

[warm python]

now what

[silver dew]

You want to solve a) or b)?

[warm python]

(a)

I suppose dim(Range(T))^perp=. dim(W)-dim(range(T))

and dim(range(T)= dim (V) = dim(null(T))

is that it?

[silver dew]

View linear map as matrices, then f: W->F will be a one times dim(W) row vector and T is a dim(W) times dim(V) matrix. f•T=0 becomes the usual sense of matrix fT=0, thus dim(kernel(T*)) will be dim(V)-dim(rank(T))=dim(V)-(dim(W)-dim(ker(T)))

[warm python]

Is my reasoning using rank nullity not right here

[silver dew]

Let me see

Explain your orthogonal symbol

[silver dew]

Definition of this ^perp

[warm python]

That's the orthogonal complement

[silver dew]

You wanted to say dim(V)-dim(Null(T)) right? Sure. It is correct

Looks good to me

[warm python]

Yea

thanks

Uh, so

Is it just 1

Like let $T$ be an eigenvector then we have $\mathcal{A}T=T$

[silver dew]

Just came back. Let me see

[soft zealot]

wai

[silver dew]

What is L(V)? Is it short for L(V,V) or something

[warm python]

yes

[silver dew]

Doesn’t seem right, you have picture showing the definition?

[warm python]

Of?

[silver dew]

L(V) and T*

[warm python]

oh, I get my mistake

$\mathcal{A}T =T^*= \lambda T$

[soft zealot]

wai

[warm python]

so I need to find the solutions to $T^*=\lambda T$

[soft zealot]

wai

[silver dew]

This bracket, <>, what does this mean

[warm python]

inner product

[silver dew]

Then i think the correct expression is A in L(L(V), L(V*)). Anyway, its eigenvalues, let me think…

[warm python]

Lambda has to be 1

$\langle Tv,v \rangle = \langle v, T^*v \rangle = \langle v, \lambda T v \rangle = \lambda \langle Tv, v\rangle$

[soft zealot]

wai

[warm python]

so $\lambda =1$

[soft zealot]

wai

[silver dew]

Matrix wise, A maps T to transpose of T

choose a basis (e_i) of T, T(e_i)=(e_i)B, then T* e_j* (e_k)=e_j* (T(e_k))=B_jk. Thus T* corresponds to B^T

So it’s the transpose. Clearly, eigenvalues are 1 and -1

[warm python]

han't my working shown it must just be 1 though

when the field is R

[silver dew]

Oh this space L(V) then is the direct sum of those two eigenspaces, symmetric matrices and anti symmetric matrices (A=(A+A^T)/2+(A-A^T)/2)

A^2=id so 1 and -1

Two eigenspaces, consisting of symmetric matrices and anti symmetric matrices respectively. The direct sum of two is exactly L(V)

[warm python]

sure

but what's the gap in my reasoning

[silver dew]

(v, Tv) can be zero

[warm python]

oh right

[silver dew]

Though, you can modify your thought like this:

(Tv, w)=(v, λTw)=λ(Tw, v)=λ^2(Tv, w)

[warm python]

right

[silver dew]

Find v,w such that (Tv, w) doesn’t equal 0 then λ^2=1

This can be done since T as a eigenvector is nonzero

[warm python]

thanks!

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[hidden tulip]

Hi, I'm having a lot of trouble determining the bounds for the z part of the integral, the two planes don't seem to intersect at all in the first octant. Maybe I've completely misinterpreted the way you need to solve triple integrals.

[final saddle]

@hidden tulip Has your question been resolved?

[final saddle]

@hidden tulip Has your question been resolved?

[glossy zephyr]

Well, i got a slight clue of what it could be

This is the two planes restricted to the first octant

If you connect that little triangle looking directly perpendicular to the plane, you get something as follows.

Something like a slab

[pliant shore]

it's weird that slab is unbounded though

[glossy zephyr]

ik

But thats about the best you can do ig. Its just the positive part of the function used as floor.

We assume the bounds are also restricted to the 1st octant

[hidden tulip]

These problems are computer generated by my school (not ai)

[glossy zephyr]

they might have missed something

[hidden tulip]

So there is a non-zero chance this is a bugged question

[drifting talon]

hiii

[hidden tulip]

I'll try working with the bounds of the slab I guess

[hidden tulip]

I've given up and will just make a discussion with the professor

There's no point doing it if it's so crudely defined

.close

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[unborn matrix]

How would you approach an exercise where you're asked to give the classification (up to isomorphism) of covers of the Klein Bottle ?

[unborn matrix]

I know that you can cover the klein bottle with R^2, T^2, mobius strip etc, but how do you formalize it ?

[silver dew]

What do you mean a “cover”?

[unborn matrix]

https://en.wikipedia.org/wiki/Covering_space

[silver dew]

Then two covers how are they defined to be “isomorphic”?

Oh nvm got it

[unborn matrix]

if p1 and p2 is the covering space function of resp the first and second covering, then two covering are isomorphic if there's a function that renders the diagram commutative

[silver dew]

There is a chapter 10 in Rotman, algebraic topology. In it, covering mappings are classified kind of like an analogy of Galois theory

A bit long to put it here…

You need to find the universal covering, then all covering is between this universal covering and the Klein bottle itself. Very similar to Galois theory

[unborn matrix]

ok, should you use a group theory approach to it then ? With only generators etc. ?

[silver dew]

No. It’s just very similar to Galois theory from appearance. But totally within algebraic topology.

[unborn matrix]

hmmm ok, and how would I know when I have all the coverings ?

[silver dew]

It’s proven in the chap 10

The result is
Let G be a subgroup of π(X, x_0), this case X is the Klein bottle. Let X_G be the set of equivalent classes of paths in X starting from x_0. Where the equivalent relation is defined by: two paths f, g. f(0)=g(0)=x_0, are equivalent if f(1)=g(1) and f* g^-1 is in G. Denote the equivalent class of f by <f>. This set has a topology defined by a basis A(U, <f>)={<f * λ>: λ is a path in U with λ(0)=f(1)} for all open U and <f>.

X_G is a covering space , i: X_G->X, mapping <f> to f(1) is a covering map

It has the property that i_{*} π(X_G, <constant path at x_0>) is isomorphic to G. When you take G={e}, X_{e} is the universal covering

So a one to one correspondence between coverings of X and subgroups of π(X, x_0). Very much like Galois theory

Though you need to read that chapter for proofs

[unborn matrix]

perfect, I think I can get throught it now, thanks!

.close

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[bronze grove]

.

.

yo

[bronze grove]

any other way to do this other than example of S3?

[trail mango]

what do you mean by example?

[warm python]

show the prouduct of two n-cycles doesn't commute if n≥3

[trail mango]

that’s overkill and not true in general

[bronze grove]

(1,2) and (2,3) ig

[bronze grove]

but we have to take an example for that right

[trail mango]

sure. problem done

[bronze grove]

i was asking for any other way

since i just don't want to prove for S3

[trail mango]

you have proven it for all the other ones too

[bronze grove]

OH WAIT

FUCK

what am i doing bruh

[trail mango]

those are also elements in S_4

[bronze grove]

mentally going insane

[trail mango]

that don’t commute

[bronze grove]

just realized that omg 😭

[warm python]

just take two 3 cycles that don't commute

[bronze grove]

thanks

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[vivid crystal]

Can I get some help

[cyan kayak]

I can't promise to help you depending on what the question is, but it's faster if you just ask the question outright, rather than waiting for someone to respond.

[vivid crystal]

Sure

I’m taking Calulus 2 about integration and I have a homework I need to solve by tomorrow I need some help with explaining 3 questions

Question 1
Question 2

Question 6

Regarding to question 2
For the integration the start point should be 4 and end point could be 5

Because we need a number that is higher than 3 right for the start point

[willow tiger]

Aha

!show

[final saddle]

Show your work, and if possible, explain where you are stuck.

[willow tiger]

!da2a

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question!

[vivid crystal]

Sure I’ll show my work the formulas and writing are not in English does that effect?

The way that I’ll use the formula

The formula is

In English

[willow tiger]

I suppose ill try to understand

[vivid crystal]

This is the first question I have wrote a list of formulas for integrations but the first one I’m not quite sure how to start with it I’ve wrote the start point which is always the below and the upper is the end point afterwards as I remember you should just look at denominator

On this question I’m stuck at that part

[willow tiger]

I dont understand whats happening . Which one are you trying to do and whattt

[vivid crystal]

These are the rules i use to solve these types of questions depends on the question

I'm not good at explaining

[vivid crystal]

if you could explain for me the question i'll be thankful

[willow tiger]

I would just simplify all the integrals then split them to parts

Like at the first there is the term which can be turned into x^2+x/(x^2+6x+10)

[twilit topaz]

6th one is the easiest

(\int a^{x}dx) = (a^x)/lna

[soft zealot]

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

[twilit topaz]

wtf

yeah

[willow tiger]

Yep that is

[vivid crystal]

Question 1 is a bit tricky for me

could i ask another question please

it would be the last question

[willow tiger]

Yea

[vivid crystal]

I would like to inquire if my steps are right and the reason why that we are allowed to use a calculator which is much easier for us I’ve been studying online but most of the channels explains the material without a calculator

[willow tiger]

I dont understand the confusion here but i doubt you should use a calculator because it will just give you the answer

[vivid crystal]

I meant to ask am I doing the right steps while solving the question

Apologies for not explaining well

[willow tiger]

You are just typing them in without doing any steps?

So i dont see the steps

[vivid crystal]

For this question as I saw on YouTube that you just directly put it on the calculator

[willow tiger]

Its probably that you just started calculus and your professor/teacher prob asked for you to do some steps to help you understand

I dont know what steps you should do, but you’re prob doing it right

[vivid crystal]

I guess so

I appreciate your time roy for assisting me

[willow tiger]

Np

Any more questions?

[vivid crystal]

I'm good for now tyyyyysm

[willow tiger]

!done

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@vivid crystal Has your question been resolved?

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[sour helm]

guys i need help

[sour helm]

idk the size of angle k

[tired walrus]

did you manage to find any other angles

[sour helm]

no

i only know 82

cuz it says 82

wait

do i have to do 82 + 82

and than take that away from 180

and the answer is k

im gonna try that actually

[tired walrus]

mm\

can you justify why this particular sequence of steps

and not anything else

[sour helm]

thanks for the help guys

i solved it

[tired walrus]

...

i mean ok you do you lol

[sour helm]

im srry

[tired walrus]

if you wanna do math as a "find the right answer" exercise while giving 0 shits about understanding anything then thats on you

[sour helm]

😢

i literally figured it out on my own

[tired walrus]

ok, then you phrased it poorly

[sour helm]

https://tenor.com/view/scream-loud-scream-girl-kid-hispanic-girl-gif-3761151244536557927

[tired walrus]

you must have spotted that this red angle was alternate with the purple angle.

[sour helm]

🤔

yeah i did!

[tired walrus]

and then you used the fact that this triangle in the pic is isosceles, so it had angles 82, 82 and k

[sour helm]

how do i end this ticket

[tired walrus]

which add up to 180 as they do in any triangle

you end the thing with .close

[sour helm]

.close

i dont think it worked

[tired walrus]

that's weird

it should

.close

...no?

bot's down, huh.

wait let's try this again

.close

nope ok i guess she is down

.close

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[tired walrus]

@sour helm there you go, now bot's back up

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[teal glen]

part B

[white jay]

what have you tried?

[teal glen]

umm i constructed CG (point G lying on AB) || DF

and then applied BPT but still i can't prove it

[white jay]

is there work you can show? I wanna see what's stopping you

[final saddle]

@teal glen Has your question been resolved?

[teal glen]

not yet

[white jay]

I'm sorry, I didn't mean to dip, I attempted it myself and couldn't conclude it so I can't help you. <@&286206848099549185> if anyone else could

[wary arrow]

So C = 90.
AEF = CED (angles).
D = A
B = F.

then we use these and... it works out, right?
it's just that the sides opposite equal angles are proportional.😭 sorry if im wrong; just saying

maybe it helps.

[teal glen]

C = 90 is written nowhere in the ques

[wary arrow]

it doesn't make a difference anyway, because to prove they are similar, we only need to prove that two of their angles are equal, and that's already the case here

[timber jungle]

hey try using BPT for the traingle BDF with line CG you constructed

you should be able to figure it out after that

just remember BG+GF=BF nad BC+CD=BD

its a pretty old question i think i remember doing it when i was in 10th too

[final saddle]

@teal glen Has your question been resolved?

[tepid ridge]

do you want me to verify or your done
If so close ticket

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[jovial nimbus]

hi i need help with this part of my working out i know i have to multiply something by something else to get the missing ratios however i forgot the process

[trail mango]

seems like some info is missing here

[jovial nimbus]

its just a part of my working out however if the numbers dont work out, then my process is wrong and i need to try something else

[plucky rover]

We need to know what exactly you're working out to know if it's right or wrong

[trail mango]

@vital crag hi

[jovial nimbus]

i can show the question

moment

[plucky rover]

Yes please

[plucky rover]

Ah cool

How many jars are empty in total

[jovial nimbus]

150 are empty

if 3/8 of 400 is 150

[plucky rover]

Yup

And now there's two ways to do this, one is algebraic and one is just some numerical manipulation

[jovial nimbus]

i went through it in class, but i dont remember how we did it

[jovial nimbus]

it was on my mock paper and we used the sort of ratio thing i had here

[plucky rover]

Well the idea is simple

You have S:M = 3:4

And M:L = 1:2

[jovial nimbus]

yea

[plucky rover]

If you can make the M value equal for both, you can combine them into S:M:L

[jovial nimbus]

yeah i was asking how to go about that

[plucky rover]

Well the first one has M = 4 and the second has M = 1

[jovial nimbus]

yea

[plucky rover]

How do we make them equal

[jovial nimbus]

divide four by four

or multiply 1 by 4

[plucky rover]

Well which one would be "cleaner"

[jovial nimbus]

^

[plucky rover]

So do that

[jovial nimbus]

sooo the top ratio would be 3:4:8?

[plucky rover]

Yup

[jovial nimbus]

thank you

[plucky rover]

Now you wanna find the number of empty small jard

That's just a ratio problem

[jovial nimbus]

i got this now twin

i just needed a refresh

i appreciate it

how to close

[plucky rover]

.close

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[quiet garden]

How I do this

[formal trail]

which part?

[quiet garden]

A

[formal trail]

look at the examples near the end here: https://tutorial.math.lamar.edu/Classes/CalcIII/ConservativeVectorField.aspx
it has a systematic procedure for finding potential functions

[quiet garden]

I only understand the first part for integrating x

I don’t get the stuff afterwards

[formal trail]

so once you integrate with respect to x, you are left with some function
(result of integral) + g(y,z)
the g(y,z) has to go there because it stands for all the functions depending only on y and z which were lost due to the partial derivative

[quiet garden]

But it says g(y,z)=0 so what’s the point

And it says differentiate but I don’t see any chnages

[formal trail]

g(y,z) is only sometimes 0

[quiet garden]

So what am I suppose to do after integrating the x

All they did was add g(y,z) and removed it. And then it wasn’t even part of final solution. Same with z

[formal trail]

let's take a super basic example, $\vb F(x,y,z) = \langle 1,1,1 \rangle$
Since $\grad f(x,y,z) = \vb F(x,y,z)$ we know $\pdv fx = 1$, $\pdv fy = 1$, $\pdv fz = 1.$ therefore we have
[ f(x,y,z) = \int \pdv fx \dd x = \int 1 \dd x = x + g(y,z) ]
Taking derivative wrt $y$,
[ \pdv fy = 0 + \pdv gy = 1 \implies \pdv gy = 1 ]
Integrating wrt y:
[ g(y,z) = \int \pdv gy \dd y = \int 1 \dd y = y + h(z) ]
So so far we have $f(x,y,z) = x + g(y,z) = x + y + h(z)$. Then again differentiating:
[ \pdv fz = 0 + 0 + \odv hz = 1 \implies \odv hz = 1 ]
Therefore
[ h(z) = \int \odv hz \dd z = \int 1 \dd z = z + C ]
So our potential function is
[ f(x,y,z) = x + y + z + C ]
we can choose any value of $C$ we want

[soft zealot]

cloud ☁

[hollow bough]

this preschooler math is so easy

but im too lazy to do it

[hollow bough]

fuck yoiu man

[quiet garden]

How is dg/dy =1

[hollow bough]

dont ask quesitons. it just equals one

it will always equal one if its just dg/dy

im lying

[vital crag]

<@&268886789983436800>

[formal trail]

@hollow bough if all you are going to do is come into here is be rude and unhelpful, you can take some time off

[quiet garden]

Also how is df/dz 0+0 and =1

[formal trail]

we know that $\pdv fy = 1$ and we previously found a formula thay says $f(x,y,z) = x + g(x,y)$. just comparing the two gives
[ \pdv fy = \pdv{}y (x) + \pdv gy = 0 + \pdv gy ]
for the first expression and
[ \pdv fy = 1 ] for the second

[soft zealot]

cloud ☁

[formal trail]

they must be equal, therefore dg/gy = 1

[quiet garden]

So I just put dg/dy same as the y equation from the given vector

?

[formal trail]

you find df/dy from the expression you got from integrating f wrt x and compare that to the df/dy from the vector. that gives you an equation to solve for dg/dy

[quiet garden]

Bruh what

[quiet garden]

I don’t understand anything after this

[formal trail]

ok so if i know f(x,y,z) = x + g(y,z), what is df/dy?

[quiet garden]

So I find derivative of x+g(x,y)?

[formal trail]

g(y,z) sorry

but yes

[quiet garden]

Why not x

And idk how u do that for g(y,z)

[formal trail]

you just write dg/dy, nothing more to it than that

[quiet garden]

0+dg/dy

[formal trail]

it should depend only on the variables we haven't integrated yet

[formal trail]

yes

so we also know from the vector that df/dy = 1, right?

[quiet garden]

I meant do I derivative the x also

[formal trail]

no, there is no need

[quiet garden]

That’s how I got 0+ here tho

[formal trail]

we are doing this in order to figure out what g(y,z) is so differentiating wrt x would be unhelpful

[quiet garden]

I did it with respect to y

[formal trail]

oh you do take the derivative of x yes

i thought you meant with respect to x

[quiet garden]

Ok

Ye

[formal trail]

so both of our expressions for df/dy must be equal

the one from the vector, and the one we got from differentiating

[quiet garden]

Ok

So mine I have x^8 +dg/dy

[formal trail]

yes

so if you set them equal, this gives you an equation to solve for dg/gy

[quiet garden]

dg/dy =-x^8?

[formal trail]

no

what is df/dy from your vector?

[quiet garden]

So I replace dg/dy with the y in the vector

[formal trail]

no

you have df/dy = x^8 + dg/dy from your previous work and df/dy = x^8 - 2y/(y^2+z^2) from your vector

knowing only these two facts, what is dg/dy?

[quiet garden]

-2y/y^2+z^2

[formal trail]

yes

[quiet garden]

Now I derivative with respect to z?

[formal trail]

not yet, now that you have dg/dy you should use it to find what g(y,z) is

[quiet garden]

So I integrate -2y/y^2+z^2

[formal trail]

yes

[quiet garden]

With respect to y?

[formal trail]

yes, you have dg/dy so integrating wrt y is what undoes that

[quiet garden]

Ok I got -ln|y^2+z^2|

[formal trail]

ok that's mostly correct but you must add a "constant of integration"

since you had a function which dependend on y and z, and you integrated dg/dy, this erased any information you have about functions depending only on z

so you must add a function depending only on z now

[quiet garden]

What

So what I do

[formal trail]

add a function depending on z

call it h(z) maybe

[quiet garden]

So I set -ln.. +h(z) = to the z from the vector?

Or I derivative with respect to z first

[formal trail]

well have g(y,z) = -ln(...) + h(z), so plug that into your f first

[final saddle]

@quiet garden Has your question been resolved?

[quiet garden]

I plug z from the vector as g(y,z)?

[formal trail]

you previously found that f(x,y,z) = x^8 y + g(y,z)

plug in your new expression for g(y,z) into that

[quiet garden]

X^8-ln..

So that +h(z)?

[formal trail]

remember this is what you got, x^8*y + g(y,z)

so x^8 y - ln|...| + h(z)

[quiet garden]

How do I find h(z) now

I derivative with respect to z? Then integrate it?

[formal trail]

same procedure, you differentiate wrt z and compare to your vector to find out what dh/dz is, then integrate that

[final saddle]

@quiet garden Has your question been resolved?

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[zealous stump]

i am writing a report and i have this integral as one of my functions. how do i explain that this cannot be solved by hand and must be done through a calculator?
$\int_0^\nu\sqrt{1+(R_0+\omega x)^2\sin^2(\varsigma x)-2(R_0\omega+\omega^2x)\varsigma\sin(\varsigma x)\cos(\varsigma x)+\omega^2\cos^2(\varsigma x)}dx\
\omega=\frac{R_H-R_0}{H}$ and $\varsigma=\frac{2\pi}{P}$

[soft zealot]

Amby

[graceful wedge]

tf

[zealous stump]

lol

[graceful wedge]

💀 gl

[final saddle]

@zealous stump Has your question been resolved?

[zealous stump]

<@&286206848099549185>

[gritty chasm]

I mean you can just write the integral as it is. Need not solve it if its for a report and its this complicated

[zealous stump]

i know i don't need to solve it but how do i justify that it cannot be solved or that i shouldn't solve it

[woven ledge]

why dont you ask your supervisor

[zealous stump]

i don't have time its due tomorrow...

[woven ledge]

if theres no simple closed form you can say that

[zealous stump]

so i can just say that the integral cannot be solved using elementary functions because it consists of a square root function with multiple parameters?

[final saddle]

@zealous stump Has your question been resolved?

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[north marsh]

im getting nowhere on this question pls help

[silver dew]

Do you mean $\binom{n}{r-k}$?

[soft zealot]

Cogwheels of the mind

[silver dew]

$\sum_{k=0}^{r}(-1)^{k}(k+1)(k+2)\binom{n}{r-k}=2\binom{n-3}{r}$

[soft zealot]

Cogwheels of the mind

[heady moon]

Yeah

Some write it as C

[final saddle]

@north marsh Has your question been resolved?

[silver dew]

I am trying generating function

Done

I will handwrite this one

[silver dew]

^

Oh second line k is still from 0 to r, I wrote infinity

(Even though generating functions we use series, derivative, antiderivative it is still pure algebraic, elements of the Laurent ring Q[x,x^-1]. So it won’t affect your taste.)

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[safe oyster]

Im trying to find the convolution of this function but don’t know how to proceed

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[frosty sigil]

Perhaps calculating the length of the vertical slice inside that region for a given x could help?

[frosty sigil]

.close

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[unborn pollen]

can someone help me integrate this💔

[rugged merlin]

Do you know u substitution?

[atomic moon]

take out the constants first

[unborn pollen]

yes a bit

[unborn pollen]

you mean take out 2?

[atomic moon]

Not only the 2

[rugged merlin]

Yes

[unborn pollen]

a?

ohhhh

[atomic moon]

Don't guess

Tell me what is constant in integral here

[unborn pollen]

a

[atomic moon]

You know it

[unborn pollen]

because we are integrating x

right

[atomic moon]

So what goes outside the integral?

[unborn pollen]

a

[rose rain]

someone help me get the RHS

[atomic moon]

More precisely ?

[atomic moon]

!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).

[rose rain]

like how we got the RHS

[unborn pollen]

root a

[atomic moon]

Yes

[unborn pollen]

OHH

[atomic moon]

So whats lasting now

[unborn pollen]

root x

waitwait

[atomic moon]

You can do it from now im sure

[unborn pollen]

like this?

[next thorn]

this channel is occupied. #❓how-to-get-help to figure out how to claim yours (it's a very complicated process)

[atomic moon]

Exactly

[unborn pollen]

YAYYY THANK YOUUU

[atomic moon]

You're welcome

[unborn pollen]

.closed

[atomic moon]

no d

[unborn pollen]

how do i uhh close this

ooh okay

.close

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[tulip panther]

The answer is C for this question, but I don't know why.

I thought it was A, because if x = y = z, then surely this just forms a diagonal square, which has all right angles?

[tired walrus]

you want necessary and sufficient, not merely sufficient

[opal plinth]

C is less of a constraint than A

[tired walrus]

you can have SPQ=90° without A

[tulip panther]

ah

so how do we determine whats absolutely necessary?

[tired walrus]

SPQ=90° iff triangle SPQ satisfies converse-Pythagoras

SP^2 = x^2+z^2, and PQ^2 = x^2+y^2

and SQ^2 = y^2+z^2+2yz

[tranquil pine]

as someone who sat this exam previously i would suggest censoring that, the Pythagoras hint is already pretty significant and this exam rewards the struggle

[tired walrus]

i mean op is presumably looking at a past paper

[tranquil pine]

the issue isn't academic honesty it's the fact that there are limited resources for this exam and wasting a question is p significant

[tranquil pine]

but also I recommend reading the logic notes by UAT

and maybe book of proof by hammack for enrichment

[tulip panther]

wait wdym?

[tulip panther]

i see, will do

[tranquil pine]

i assume you are aiming for top UK unis if u are doing TMUA, with only ~16-18 papers every question should be done with struggle

if you're applying for CS/math at least

[tranquil pine]

this was a comment to Ann though, not you so dw

[tired walrus]

ok well what's done is done ig, im not gonna say more

[tranquil pine]

you can get upset if you want but I'd argue helping OP takes priority, @tulip panther, if you'd like extra support feel free to DM or ping me in a help channel

[tulip panther]

im struggling to understand what you've done here

[tulip panther]

i understand this

[tulip panther]

i dont understand where this comes from

and whats converse pythag?

[tulip panther]

i see

[tired walrus]

if a triangle's sides satisfy a^2+b^2=c^2 then the angle opposite c is right

(regular pythag is if the angle opposite c is right then a^2+b^2=c^2)

[tired walrus]

well then look at the diagram. what's SQ

[tulip panther]

ah

[tulip panther]

the hyp of SPQ?

[white jay]

I believe what confused you here is the wording, because when A is sufficient, it is not necessary

[tulip panther]

i do realise that now

[white jay]

it never hurts to go back to basics and plug in your own set of values to double check

[tired walrus]

what's its length

[tulip panther]

i would assume its root of SQ^2

which is root of SP^2 + PQ^2

so root of 2x^2 + y^2 + z^2

i dont understand where 2yz comes from

[final saddle]

@tulip panther Has your question been resolved?

[harsh ice]

$(y+z)^2 = y^2 + 2yz + y^2$

[soft zealot]

maxtsg

[tulip panther]

ohhhhh

i was thinking of it through the two sides

i get it nowwww

thank you

.close

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[carmine plaza]

this is a past final

[carmine plaza]

i still am super shaky on all types of problems relating volumes and applications of integrals

[carmine plaza]

LOL

idk just sketch out this but idk what lnx looks like in the 4th quadrant

[marsh temple]

what does lnx look like in general?

[magic sparrow]

that is unfortunate because it is one of 2 quadrants ln(x) is in

[carmine plaza]

well yeah

[magic sparrow]

how is ln(x) defined

[carmine plaza]

uhm

[magic sparrow]

like what is it

[carmine plaza]

ln of x

[magic sparrow]

what's ln(1)

[carmine plaza]

infinity i think

0

[magic sparrow]

No

[carmine plaza]

nvm

[magic sparrow]

Why?

[carmine plaza]

cuz e to the 0 is 1

[magic sparrow]

Yes

so what is ln(x) defined as

e^(0)=1 => ln(1)=0

[carmine plaza]

yes

[magic sparrow]

soooo

[carmine plaza]

e^y=x idk

and y = lnx

[magic sparrow]

yea exactly

its the inverse of e^x

[carmine plaza]

yes

[magic sparrow]

so to plot it you just need to know how to plot e^x

and that's easier

do you know how to plot e^x

[carmine plaza]

yeah

exponential function sort of like half a parabola

[magic sparrow]

So r we good on this now?

[cursive cliff]

lurking,,,

[carmine plaza]

idk

[magic sparrow]

Well, the fourth quadrant is when x is positive and y is negative

first question is ln(x) increasing

[carmine plaza]

no

[magic sparrow]

?

as x gets bigger, does ln(x) get bigger

[carmine plaza]

yes

[magic sparrow]

so it is increasing

and we saw earlier ln(1)=0

[carmine plaza]

but in the 4th quadrant like its still increasing but like

i just got confused cuz its negative

[magic sparrow]

its increasing but its negative

yah

and then, the fourth quadrant well it starts at x=0

and from what we've just observed

ln(x) will become positive at x=1

so

we only have to figure out how to plot ln(x) for 0<x<1

does that make sense

(because that is what will remain in the fourth quadrant)

[carmine plaza]

well as ln(0) approaches infinity so its just like this curve until it his 1 in the fourth quadrant

[magic sparrow]

yes

that's right

[carmine plaza]

so its the area in that curve

now what do i do with that

[magic sparrow]

well, if it was rotated about the x-axis would you know what to do

[carmine plaza]

no

[magic sparrow]

There is a formula for that

[carmine plaza]

whats it called like cross section washer disc idk

theres cylindrical shells

im like aware of them but idk how to use any

i dont feel like i dont understand the process of how to use them

[magic sparrow]

I can recommend you a good video

it would be better and have visuals than what someone can explain here just chatting

[carmine plaza]

ok

[magic sparrow]

https://youtu.be/ydyXf01WNYA?si=xB9xtKXmw_rwBjRA

this one is quite good

and if you have questions about something in the video, the u can like point out the timestamp and ask here

[carmine plaza]

i watched that one

[magic sparrow]

optional 2nd video

https://youtu.be/M85_r3pZ5YA?si=FzC07egFD34g3jPL

[carmine plaza]

ok ill try the second one

also what about cross sections is that something that i need to worry about or is it more simple

[cursive cliff]

that’s how we derive the formula, you don’t have to directly deal with them in the computation

[carmine plaza]

oh perfect

ok

[cursive cliff]

it’s just we sum up the cross sections to make the volume

infinitely many times -> integration

[carmine plaza]

yup

[cursive cliff]

were u taught any formulas tho, like a standard one to use in ur hw

[carmine plaza]

ok so i draw like a sort of infinite cone going down

[cursive cliff]

yes !!!

[carmine plaza]

so the intregral im guessing is from -infinity to 0

and A(y) would be the area of a circle

if the radius is x then it would x=e^y

[cursive cliff]

YES but would be x²

[carmine plaza]

yes

[cursive cliff]

yesss

[carmine plaza]

ok so using u sub i can cancel one of e^y and left with pi*e^y +C?

wait no

[cursive cliff]

no u cannot

[carmine plaza]

pi/2 (e^y)^2

or idk whats the right approach here

[cursive cliff]

,w \int_{0}^{\inf} \pi e^2y dx

wait

[soft zealot]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by aoba0158_14440

[cursive cliff]

,tex \int_{0}^{\inf} \pi e^2y dx

[soft zealot]

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

[cursive cliff]

i’m so bad at latex

lemme just write w my hand

[magic sparrow]

$\int_{0}^{\infty}\pi e^{2y}dx$

[soft zealot]

Austin

[cursive cliff]

THANK U

wait a minute

[carmine plaza]

Ok it's dx?

[cursive cliff]

sry forgot the q by now

we are rotating around x=-1

[carmine plaza]

Isn't it negativity infinity to 0

[cursive cliff]

dy ur right

[carmine plaza]

Yeah if it was dx do we use shells?

[cursive cliff]

u can use either for y or x

but also

$\int_{0}^{\infty}\pi e^{2y}dy$

[soft zealot]

aoba

[cursive cliff]

wish there was some sorta desmos here

what do u have for ur diagram rn?

[tepid narwhal]

im tryna put the graph on this

hold on

wait how to put the shaded region on the graph again on desmos

oh nvm i did it

[cursive cliff]

yayy let’s see

also in terms of formulas, now that u already derived em let’s re paste em

$\int_{a}^{b}\pi y^2 dx$

[soft zealot]

aoba

[cursive cliff]

this is what we are looking at for a revolution around a vertical line

[tepid narwhal]

i think this is the picture for the question

for rotating around x=-1

[carmine plaza]

Right

That's what I have

[cursive cliff]

legendd

[tepid narwhal]

is from int -infinity to 0 right

[carmine plaza]

Arent the bounds -inf to 0

[tepid narwhal]

we have to double it

im tryna do dy here

[cursive cliff]

waitt isn’t this dx

[tepid narwhal]

you can do it both ways

[cursive cliff]

and cone has radius 1

[tepid narwhal]

dx or dy

[cursive cliff]

yeah true

[tepid narwhal]

since is the volume

[cursive cliff]

we can stick w dy

[tepid narwhal]

man is the inf

im only done calc 1

lim when ln x goes to -inf

let me calc that

[cursive cliff]

okay i don’t wannna confuse u ninoo does it make sense u can do dx and dy

[tepid narwhal]

nvm

i done it wrong

lim when ln x goes to -inf does not exist

[carmine plaza]

No it doesnt

Nothing makes sense I just had int -inf to 0 pi (e^y)^2 dy

[tepid narwhal]

oh oops

i forgot to change the

y=ln x

i forgor it was on dx form

[cursive cliff]

okay for now let’s stick with doing dx when rotating vertically since the radius would be on x axis and that dy for rotating horizontally would be dy since radius is on y axis

does this make sense ^

[carmine plaza]

Sort of

[cursive cliff]

if i spin it around vertically, the x axis will be on the flat top of the cone right

[carmine plaza]

I guess yeah

[tepid narwhal]

wait isnt this should be 2pi?

since this is doubled the region

from -infinity to 0

if you times 2 then it should be flipped upwards from 0 to inf, and 2pi times [e^y]^2?

[carmine plaza]

I'm just confused atp

[tepid narwhal]

idk if this one helps

hold on lemme sent this one to you

[cursive cliff]

ykw i don’t like about this question is that it tries to confuse u by rotating around x = -1, when rly the function doesn’t exist beyond zero

i found my pencil so imma draw to not confuse u gimme a sec

[tepid narwhal]

my syllabus question it just tells you

the function

or draws it for you

and rotating 360 degrees vertical or horizontal

to avoid confusion

basically just like that

but to beyond inf?

this is like involving calc 2 (improper integral)

a bit

[tepid narwhal]

i assume this question just ask rotating x=-1 which is horizontally

since the x=-1 is legit on the right side of the graph ln x so yea is gotta be the cone shape towards upside down

[tepid narwhal]

so this is gotta be

[soft zealot]

EpicalNerds
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[tepid narwhal]

nice diagram for 3D, Upvote THIS GUY

[cursive cliff]

tyty

[tepid narwhal]

yea it should be 2pi

[cursive cliff]

i’m ngl the x=-1 is throwing me off

[tepid narwhal]

since is -inf to 0

[cursive cliff]

what do u think epic

[tepid narwhal]

the volume of the revolution is

[cursive cliff]

perhaps we can just f(x+1)

[tepid narwhal]

it should be times 2

[tepid narwhal]

alr what about this

@cursive cliff

a=0, b is 1

[cursive cliff]

i agree w this formula, radius changes right

[tepid narwhal]

radius should be

x - (-1)

it should be x+1 for the shaded region to the centre

[carmine plaza]

im not sure about this after evaluatin the integral using dx if you'll get the same thing

u also changed the dy bounds from -inf to 0 to 0 to inf

[tepid narwhal]

yea then the height is gotta be -lnx

since we looking from upside down

[tepid narwhal]

ans should be 5pi/3

[cursive cliff]

mbmb

[tepid narwhal]

OH YEA BOUNDS

when you change the x functions

the bounds also change

so thats why that integral for dy is so odd to me

[cursive cliff]

i’m ngl tho i don’t rly understand using dy for this so i’ll leave it for someone else to explain cus i’ve confused myself tryna understand that

but i can do this with dx though thats not what u may be looking for

[tepid narwhal]

i mean definite and indefinite integral for it

definite is just that troublesome

[cursive cliff]

sry to confuse u even more nino !!

[tepid narwhal]

i always forgot to change bounds during tests

anyways @carmine plaza

[tepid narwhal]

for that question just look at this

or that vid i just sent you

ans should be that

it should fully detailed explained

[tepid narwhal]

lemme fix it for you the dy bound.
$\int_{a}^{b}\pi y^{2}dx$
where y is= $e^y$, a is $-\infty$, b is 0
so the integral becomes
$\int_{-\infty}^{0}\pi e^{2y}dy$

[carmine plaza]

to learn latex u can use overleaf, desmos also natively writes the equations in latex so you can copy paste from desmos paste in latex

[soft zealot]

EpicalNerds

[final saddle]

@carmine plaza Has your question been resolved?

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Use .reopen if this was a mistake.

[final saddle]

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[carmine plaza]

volume question again not sure how to draw this one

[storm haven]

You want to draw the graph?

[carmine plaza]

do i have to

[storm haven]

No

[carmine plaza]

ah

[storm haven]

You can just solve for the intersection of y= 0 and y= x^2-x^3 without drawing any graphs

But drawing a graph will definitely make things more intuitive imo

[carmine plaza]

x=0,1

[storm haven]

Can you do the rest of the question?

[carmine plaza]

no im not sure how to do the rotations

[storm haven]

Have you learnt washers method?

[carmine plaza]

yeah i know about it

the general formula

i still dont understand how to use it

[storm haven]

Okay then, if we rotate it about the x axis, what do you think the radius (r) should be?

[carmine plaza]

x? idk

[storm haven]

Then maybe we should graph it first

[storm haven]

That's okay let's try to graph x^2 - x^3 first

As we know, its roots are at x=0, 1 right?

[carmine plaza]

yes

[storm haven]

Okay, so, when x< 0 is x^2-x^3 positive or negative?

[carmine plaza]

negative

[storm haven]

You sure 😭

[carmine plaza]

when x is negative x^2 is positive - a larger x^3 number which is negative

oh but its negative

[storm haven]

So you end up getting
Positive - (negative) = positive + positive..?

[storm haven]

Yeah

[carmine plaza]

yeah

mb

[storm haven]

Okay, so when x<0 the graph is above the x axis

[loud sundial]

hi

[storm haven]

And at exactly x= 0, the graph meets the x axis

How about when 0<x<1?

[carmine plaza]

positive?

[storm haven]

And when x>1?

[carmine plaza]

negative

[storm haven]

You should know what cubic equations look like in general

[carmine plaza]

i do

[storm haven]

So this information is enough to piece together a very sketchy graph

[carmine plaza]

but this one had x^2 as well so i assume it would change something in it

[storm haven]

Actually not

It's still a cubic

Anyways you should go ahead and graph that

And then I can continue explaining what r should be

[carmine plaza]

i mean if i graphed a normal cubic it would sitll be different

ok

what now

[storm haven]

Could you send the picture of your graph

[carmine plaza]

my camera is terrible at best i can send you a desmos picture

[storm haven]

Okay that's alright

I'll send the desmos picture then hold up

[carmine plaza]

ur good its fine

we should only care about this

\

[storm haven]

If we are going to rotate it around the x axis, the radius at 0.5should be the green line

And in general r would be the distance from y=0 to the point f(x) at any point 0<=x<=1

Do you get why?

[carmine plaza]

no

[storm haven]

Wait imma plot it in 3d to help you understand better

[carmine plaza]

like for this problem i understand everything except for how they get r0 and ri why do they add +1

[storm haven]

So you see this is a rough estimate on what the volume looks like

[storm haven]

We want to chop up the area into a bunch of slices, and calculate the radius of each of those slices

[carmine plaza]

Wha

Sry there's too many lines

[storm haven]

Those lines are the graph rotated 0°, 90°, 180°, and 270°

Do you get it now 😭

[carmine plaza]

I mean I understand the image

[storm haven]

That's great

What's wrong then

[carmine plaza]

It's ok 😭 we can keep going

[storm haven]

You understand this right

[carmine plaza]

Yes

[storm haven]

The center of each of those slices would be on the x axis right ?

Because we rotate it around the x axis

radius = distance from any point on the circumference to the center of the circle

[carmine plaza]

Right

[storm haven]

From this, we know that r = x^2-x^3 - 0

y= 0 is the x axis, and that's where 0 comes from, because that's the center of the circle

[carmine plaza]

So the thing you're subtracting is the place where the intersection happens?

Not intersection

[storm haven]

Wdym

[carmine plaza]

The like line it's being rotated by

[carmine plaza]

So here it's rotated by -1 so you subtract -1

That's the one part I find confusing but I think I understand ish

[storm haven]

Because y = -1 is the center

[storm haven]

Do you understand this or do you want me to elaborate more

[carmine plaza]

Right

Ok let me see if I can do it

Yup I understand

If y= something else it would just be minus that right

So integral from 0 to 1
pi (x^2-x^3)^2 dx

[final saddle]

@carmine plaza Has your question been resolved?

[final saddle]

@carmine plaza Has your question been resolved?

Channel closed

Closed by @carmine plaza

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[carmine plaza]

whats the definition of an improper integral?

[carmine plaza]

i see ppl saying like -inf to inf integrals are improper

but arent indefinite integrals proper

im not sure like what it properly means

haha

yeah idk 😔

[uncut aspen]

search on ggole

google

[robust mulch]

thats kinda not the point of asking here

an improper integral is generally one that is unbounded

[steady locust]

its when the integral either has infinite bounds or goes to infinity within the bounds

[lilac oar]

Google is best for small stuff like this

[carmine plaza]

i dont udnerstand google

[robust mulch]

ok

[carmine plaza]

indefinite intergals are unbounded

[robust mulch]

good for you

[carmine plaza]

ok yk ill just psot a problem

[robust mulch]

unbounded as in (in the most basic sense) you cant draw a rectangle around it

[carmine plaza]

so this one seems like it goes to infinity within the bounsd due to the denominator

[steady locust]

yeah, at x=0

[carmine plaza]

but if u have 1/x for that same integral

is it improper

[robust mulch]

an integral without bounds is an indefinite integral though

[carmine plaza]

thats what sorta confuses me

[steady locust]

yeah

[steady locust]

indefinite is completely different to improper

[carmine plaza]

what does it mean to not have bounds

i thought it meant -inf to inf

[steady locust]

when its just $\int$ instead of $\int_a^b$

[soft zealot]

c2b7

[carmine plaza]

what does that mean

like if u decompose it into sum

[steady locust]

like you've done integral problems right?

[supple jolt]

😭

[carmine plaza]

yes

[robust mulch]

my god can people stop interrupting if they have nothing useful to say