help-49

[dreamy lichen]

yeah, but getting any information about ab from that is almost impossible

[solemn silo]

This was my only idea

[dreamy lichen]

Here is a little warmup exercise, which of these are perfect squares:
A) 3^2 * 5^4 * 7^9
B) 3^2 * 5 * 7^2
C) 3^4 * 2^2 * 5^4

[steel crest]

it might help because you can turn it into two compound inequalities neither of which has n^2

[solemn silo]

C

[steel crest]

note that my suggestion is not independent of mathisalways right's suggestion, you would still need their approach

[solemn silo]

I don t see how it helps

[dreamy lichen]

correct, prime factorizations of perfect squares has only even powers

[solemn silo]

Is not what i suggested

[dreamy lichen]

prime factorize a and b

their product will be just product of their prime factorizations

[solemn silo]

Well how i do that

I don t know who a and b are

What form would they ahve

Have

[dreamy lichen]

Yeah, right

we can make up some form to it though

write a = a' * k1^2, b = b' * k2^2, where k1 and k2 are the largest possible squares

[solemn silo]

Idk man the problem looks so easy first time but it s kinda tough lol

I have an headache rn

[dreamy lichen]

so e.g.
3^2 * 5^4 * 7^9
would be written just as
7 * (7^4 * 3 * 5^2)^2

[solemn silo]

Then a' and b' can are from R, you can t expect them to be natural

[dreamy lichen]

the point is that we will only focus on the non-square parts, that is a' and b'

[solemn silo]

And it s even more messed up

[dreamy lichen]

i define them to be natural

k1 and k2 are supposed to be the largest squares, such that a and b are still integers

[dreamy lichen]

here is an example, that might be easier to understand than the wording

it's basically just seperating the square part of the factorization from the non-square part

[solemn silo]

Ok,ok, so the form is that, how do I continue

[dreamy lichen]

well, when we multiply ab, we get a'b' * (k1k2)^2

[solemn silo]

Yes

[dreamy lichen]

if ab was a square, then so would be a'b'

and this already massively reduces the question

because we at least have an idea of how could the factorization of a' and b' look like

it will only have primes with exponent 1

[steel crest]

typo, I meant

[dreamy lichen]

i.e. no exponents at all, no extra powers

[steel crest]

$$n^2 + 1 \le a + 1 \le b \le n^2 + 2n +1$$
$$-n^2 -2n -2 \le -b \le -a-1 \le -n^2 -1$$

[solemn silo]

a' and b' are primes right?

[dreamy lichen]

not necessariyl primes

it could be something like 3 * 7

or 3 * 5 * 7

but it could never be something like 3^3 * 5, because then we could take the 3^2 and put it to the k1^2 part

the point is that it will always be just product of unique primes

[solemn silo]

Ok, but how i continue tho

I mean correct

[dreamy lichen]

what would a' and b' have to be in order for a'b' to be a square?

[grand pond]

gfauxpas

[dreamy lichen]

think

make some example a' and b' if you want

but you'll have to figure this out by yourself

[solemn silo]

Equal?

[dreamy lichen]

Exactly

[solemn silo]

No, that s wrong

4 and 9 ?

[dreamy lichen]

wdym? What could a' and b' be?

[solemn silo]

Or other example

[solemn silo]

4*9=36=6^2

So they are not equal

But their product is p.s.

[dreamy lichen]

Yes, but 4 = 2^2

and 9 = 3^2

we dont allow exponents in a' and b'

that was the whole point of seperating them out

[solemn silo]

So a' and b' can t be perfect squares?

[dreamy lichen]

no, they cant

[solemn silo]

Ahh yeha

Yeah yeah

[dreamy lichen]

and they cant even have exponents such as ^3

[solemn silo]

So they are equal for sure right?

[dreamy lichen]

Yep

because when we have sth like
a' = 2 * 3 * 5, then in order for a'b' to be a square, b' must be divisible by 2. So it must include 2 (and it cnat include exponent of it, since thats disallowed for b')

similarly, it must also include 3 and 5

so b' = 2 * 3 * 5

and it cant include anything extra, because then a' would have to include that as well

[solemn silo]

Ok, but how does a'=b' go to a contradiction

[dreamy lichen]

Good question, can you answer it?

or can you at least make some progress?

[solemn silo]

Wait a sec

So you said that a' and b' are chosen so they could be natural

And to have the other propriety with k1 and k2

And a and b are between n^2 and (n+1)^2

That would mean that a'*k1^2

And a'*k2^2

Are between these two guys

That s the idea?

[dreamy lichen]

yep, continue thinking

try examples if you get stuck

[solemn silo]

k1 and k2 are smaller than n?

[dreamy lichen]

can be

can you prove it?

[solemn silo]

So they can be higher?

[dreamy lichen]

well, a' is certainly at least 1. But it cannot be exactly 1, since then you'd have k1^2 between 2 consecutive squares

so it must be more than 1

[solemn silo]

Yes

[dreamy lichen]

sorry, confused some stuff i think

uh actually i think that k1 and k2 can be both smaller and larger

it shouldnt be necessary for the proof though

[solemn silo]

Yeah you re right

I tried now

So how do I continue

I m stuck af

[dreamy lichen]

I'd try some examples

take 2 squares, maybe 64 and 81

and try to find 2 numbers of form a'*k1^2 and a'*k2^2 between them

[solemn silo]

Well first i d chose some a and b

Let s say 70 and 76

[dreamy lichen]

just try finding 2 a'-multiples of distinct squares, that should be enough to give you a hard time lol

[solemn silo]

Nah man tell me green in front

How I finish this one

Bro don t let me now

[dreamy lichen]

well, a' cant be 1 because you'd have 2 perfect squares between n^2 and (n+1)^2

can it be 2?

can it be 3?

[chrome rain]

hey so I also have a different method
suppose ab=k^2, so sqrt(ab)=k
we have that n<sqrt(a)<sqrt(b)<n+1

if a and b are coprime, then sqrt(a) would be a number (try out why) between n and n+1, not possible

similarly if gcd of a and b is d, then sqrt(a) = rsqrt(d), sqrt(b) = s*sqrt(d) for some number r and s (try out why) so that n/sqrt(d) < r < s < (n+1)/sqrt(d), but that is not possible (try out why)

sorry for interrupting lol

[dreamy lichen]

if a and b are coprime, then sqrt(a) would be a number
What does this mean?

[chrome rain]

that a=s^2

[dreamy lichen]

oh so perfect square

[chrome rain]

sqrt(a) will be a positive integer I should have said

[dreamy lichen]

i think that this is actually pretty similar to the method of mine

[chrome rain]

ohh

I thought you were doing prime factorisation

didn't see

[dreamy lichen]

there is a connection between gcd and prime factorization

$\frac{n}{\sqrt{a'}}<k_{1}<k_{2}<\frac{n+1}{\sqrt{a'}}$

[solemn silo]

So we remained here

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

this is the inequality we have basically

[solemn silo]

What a timing :))

Yeah

[dreamy lichen]

in this format, is it easier to see why it cant work?

[solemn silo]

I don t see it

[dreamy lichen]

sqrt(a') is gonna be at least 1 for sure

$n<k_{1}<k_{2}<n+1$

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

so what about now?

[solemn silo]

Oh yeah this can t be

They should be equal

[dreamy lichen]

and can this be?

[dreamy lichen]

Yeah, the only way would be if k1 and k2 are equal to either n or n+1

[solemn silo]

I d say yes but i don t see it

I mean it becomes less restrictive

[dreamy lichen]

the reason why this cant be is that there isnt enough space between n and n+1 for 2 integers

there is just one unit of space, definitely not enough to fit 2 integers

[dreamy lichen]

is the gap here smaller or larger?

[solemn silo]

OH YEAH I M STUPID

It was so easy :))

Yeah man

Yeah

I understand

[dreamy lichen]

perfect

[dreamy lichen]

how do you deal with the final inequality here btw?

n/sqrt(d) < r < (n+1)/sqrt(d)
This can sometimes be true, unless we know more info

[solemn silo]

@dreamy lichen can you help me with one more thing please? I write right now the solution(with english) and can you say if it s ok?

[dreamy lichen]

sure

[solemn silo]

I m sending it in like 6 minutes

I will tag you

Thanks man

[dreamy lichen]

you can check konvux's solution as well, im not sure how to fill in some details but it looks pretty similar to what we did

at least the idea, the wording is drastically different and probably better than mine (since it's always harder to talk about prime factorization than about gcds)

[chrome rain]

oh sorry actually the last one should instead be :
sqrt(a) = rsqrt(d) and sqrt(b) = s*sqrt(d)
so r AND s are in the interval I said before, which is not possible as r =/= s

[dreamy lichen]

Ah okay, so actually it's exactly the same as mine lol

[chrome rain]

oh lol

[dreamy lichen]

they are the same ideas, yours is just worded in the language of gcds which makes it simpler

[chrome rain]

i see, I just thought initially you were doing smthin else so just started typing

[solemn silo]

@dreamy lichen

It s ok?

@dreamy lichen

[midnight plank]

@solemn silo Has your question been resolved?

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[odd lagoon]

hello

[odd lagoon]

First they want me to calculate Nabla x F and then show that something wonky symbols = 0 for all closed curves C on the unitsphere

[nova yoke]

did you do (a)?

[odd lagoon]

i was just working on it but i think i know how to do it

i get a 3x3 matrix

with first column i, j k

and then the partial derivatives with respect to x, y and z in that order for the 2nd column

or is it 3x2?

[hasty delta]

3x3

[odd lagoon]

sry i mean 2x3

ok 3x3

good

yes third and final would be

last column is -2xz, 0, y^2

[hasty delta]

only square matrices have determinants

yeah i think you're right

[odd lagoon]

right

and

then i use Sarrus funny rule

[nova yoke]

yea now find the "determinant" of that

[odd lagoon]

to find the determinant

so i write i, j, k to the right

then to the right of that i write uhh

uuuuuuh

the partial derivatives

again?

it feels very abstract

but okay then i do the Sarrus thingy

i need to do this on paper to arrive at a result

but i have no clue how t odo the next part

0 clue

only little clue for a)

[hasty delta]

nabla is just a vector so all you're doing is finding the cross product between two vectors

[odd lagoon]

i will take photo

once i have worked it out

or tried to

and you can tell me if im wrong or right

(if you want to)

[nova yoke]

$$\det\begin{pmatrix}i & j & k \ d/dx & d/dy & d/dz \ -2xz & 0 & y^2\end{pmatrix}$$

[grand pond]

Bungo

[nova yoke]

just writing for reference what you're calculating

[odd lagoon]

mine looks different

i hav them in columns

instead

[nova yoke]

that's ok

[odd lagoon]

ok

[nova yoke]

determinant of a matrix = determinant of its transpose

so you can do it with either rows or columnes

[odd lagoon]

ah ok

d/dx k is just 0 right?

actually nvm it didnt even matter here

[nova yoke]

you're not differentiating i,j,k, just treat them as constants

if you do it correctly, the only things you're computing d/dx of are 0 and y^2

[odd lagoon]

but a constant is 0

so they would be 0 right?

unless the constant is scaling something

[nova yoke]

yea but you're not differentiating the stuff in the first column, you're differentiating the stuff in the 3rd column

[odd lagoon]

that im not treating as constant

when im using Sarrus im getting that everything is beign differentiated in some way

[nova yoke]

yea using the sarrus thing is probably gonna put things in the wrong order

you'll have to rearrange

or just use the cofactor expansion

[odd lagoon]

yes you are right

i have had to rearrange

for things to make sense

but after rearranging im fine right?

the partial derivative

has to come before the function

aka one of the coordinates in F

[nova yoke]

either way, this is form you want
$$i\left(\frac{d}{dy}y^2 - \frac{d}{dz} 0\right) + \cdots$$

[grand pond]

Bungo

[nova yoke]

with similar terms for j and k

[odd lagoon]

i only got 2yi - 2xj

did i do it wrong

that seems too simple

[nova yoke]

no that's correct

the k component of the curl is 0

[odd lagoon]

that is awesome im happy i got it right

[nova yoke]

yep

thoughts for how to proceed with (b)?

[odd lagoon]

0 thoughts unfortunately! i literally learned all of this today

about 2 hours ago

and me and the tutor didnt get that far

[nova yoke]

do you know stokes' theorem?

[odd lagoon]

right he said

[hasty delta]

Usually in questions with cross products and math questions where there is a lot of writing down to get the answer like with vector products, a lot of the stuff cancels or is equal to 0 to make it faster while still testing your knowledge

[odd lagoon]

i need a bunch of theorems

[nova yoke]

stokes should take care of (b)

if you know it

[odd lagoon]

yeah like 4 or 5 diagonals were equal to 0

i will have to look stokes up

[nova yoke]

you know that's the one that will likely be useful because it's the one that relates curve integrals to integrals involving curl

[odd lagoon]

or wait is this an example

okay we i have the thing inside the double integral

nice!

[nova yoke]

yea that second screenshot is the general statement

[odd lagoon]

i have no idea what dS means

[nova yoke]

the left hand side is the thing you need to show is zero, for part b

[odd lagoon]

i have seen dA before

is S an area?

jacobian*drdtheta

[nova yoke]

dS is a vector in the direction of the surface normal

the surface in this question being the unit sphere

the thing you would hope to show here is that this is always zero:

[odd lagoon]

vectors are wild man

okay

d(some kind of vector)

oh it is the dot product

when is the dot product 0 again? is it when the vectors are completely parallel?

SORRY

i mean

normal

completley normal

[nova yoke]

dS is itself a vector, it's a vector pointing in the direction of the surface normal, and its length is dA

[odd lagoon]

to eachother

[nova yoke]

yea when they are orthogonal or perpendicular

[odd lagoon]

yes perpendicular is the word i was looking for

[nova yoke]

so basically you want to show that the vector you found in part (a), namely (2y, -2x, 0), is always perpendicular to the surface normal of the unit sphere

[odd lagoon]

is it forgivable to have a hard time following what that even means

[nova yoke]

do you know what the surface normal vector is, for the unit sphere?

sure, you're just learning this stuff, it takes time to absorb

[odd lagoon]

okay nice ty, and im just trying to think of the words, i dont know

but im just trying to guess intuitively

the surface normal of a unit sphere?

wouldnt that just be a plane

a circle plane?

[nova yoke]

if you draw a vector from the origin to some point (x,y,z) on the sphere, the normal vector at that point continues in the same direction right?

[odd lagoon]

or just surface i guess would be a plane circle

[nova yoke]

"surface normal" is a vector that is perpendicular to the surface at that point

the surface in this problem being the unit sphere

[odd lagoon]

i dont really understand this

so okay hold on

geometrically

im trying to visualise a sphere

but now that is hard so im viewing it as a 2d circle

im drawing a vector from origo if origo is its centerpoint

and you are saying that a normal vector is in the same direction? wouldnt the normal vector literally be 90 degrees i nsome other direction?

[nova yoke]

damn it's surprisingly hard to find a good picture of this with google images haha

[odd lagoon]

oh wait literally

surface of the sphere

no circle inside

[nova yoke]

if C is the center of the sphere and P is a point on the sphere, the arrow shown is the normal vector at that point

[odd lagoon]

okay this is nto at all what i was imagining

[nova yoke]

the normal vector just continues in the same direction as your vector from C to P

so if P = (x,y,z) is some generic point on the sphere, then the normal vector at that point is proportional to (x,y,z)

[odd lagoon]

sorry that picture is great

but can we talk about it some more

it still hasn't quite landed

[nova yoke]

(times some constant scale factor so that when you do the surface integral, you get the correct 4 pi R^2 for the area of a sphere)

btw you can also do this algebraically if you prefer

[odd lagoon]

is that Nabla x F

?

[odd lagoon]

P?

[nova yoke]

no, the surface normal is dS

[odd lagoon]

okay

[nova yoke]

you already found Nabla x F in (a)

[odd lagoon]

what is c

im trying to relate it to these formulas and theorems

[nova yoke]

C is any closed curve on the unit sphere in this case

i.e. a curve that lies entirely on the surface of the sphere, and which starts and ends at the same point on the sphere

[odd lagoon]

so is C like a circumferance of

of some part of the sphere

idk what to call it

[nova yoke]

here's an example picture

[odd lagoon]

yes

[nova yoke]

C is some curve on the sphere

[odd lagoon]

that is

what i mean

i know i didnt

put it well but i was imagining exactly that

[nova yoke]

and S is the part of the surface "above" that curve (or below, depending on which direction you travel on C)

[odd lagoon]

and it could theoretically be

the ltieral circumferance

literal*

[nova yoke]

it could be (like the equator of the earth is one example for C) but it doesn't have to be

[odd lagoon]

it just doesnt necessarily have to be the circumferance

[nova yoke]

C doesn't even have to be a circle

[odd lagoon]

yes okay then

i understand what C is

but

where in the formula

is C

or is C like baked into dS, like you need to know what C is to find dS

[nova yoke]

it's part of the integral on the left hand side

[odd lagoon]

ooooh

[nova yoke]

the left hand side is the curve integral along the curve C

and then the S on the right hand side is like the S in the figure i showed above

it's the part of the surface that lies "above" the curve

fortunately you don't even need to know what C and S are here

because if you can show that this is always zero:

then the integral on the right hand side is zero regardless of S

[odd lagoon]

but didnt we show V x F is 2yi-2xj

[nova yoke]

and therefore the integral on the left hand side is zero regardless of C

[nova yoke]

yep

[odd lagoon]

they are linearly independent of one another

[nova yoke]

but you're not only integrating that

you're integrating the dot product of that with dS

and that dot product is gonna be zero

always

[odd lagoon]

but to integrate dS

dont i need to know dS

[nova yoke]

you do

[odd lagoon]

but you said this

[nova yoke]

but to know that, all you need to know is that you're on the unit sphere

[nova yoke]

yea S is not the whole sphere, it's the part that is bounded by the curve

in general you'd need to know C and S

but in this particular problem you won't

[odd lagoon]

ah S and dS are not the same

[nova yoke]

right

[odd lagoon]

but what we want S to be 0 then? or constant?

[nova yoke]

similar names but very different things

[odd lagoon]

actually nvm

we want it to be 0

[nova yoke]

no what you want to show is:

[odd lagoon]

actually i dont understand this i think

[nova yoke]

the vector dS
is perpendicular to (Nabla x F)

and therefore their dot product is zero

so then when you compute this integral you get zero

[odd lagoon]

wasnt dS perpendicular to C

[nova yoke]

no, dS knows nothing about C

dS is perpendicular to the sphere

C is just some random curve on the sphere

[odd lagoon]

but if dS is perpendicular to the sphere, it could have the same direction as my Nabla x F, or where the hell would Nabla x F be on the sphere

[nova yoke]

it could be in general, but for your particular Nabla x F, it won't

[odd lagoon]

i dont know how to see that

something you are seeing im not seeing

i see that in Nabla x F there is no z

[nova yoke]

well you have to find dS in order to show this

[odd lagoon]

and i dont need to find S to find dS

?

how would i find dS is a better question

[nova yoke]

well you know the equation for the sphere

x^2 + y^2 + z^2 = 1

and whenever you have a surface defined in the form f(x,y,z) = constant, then the surface normal is just the gradient of f

so what's the gradient of f(x,y,z) = x^2 + y^2 + z^2?

[odd lagoon]

f'(x, y, z) = 2x + 2y + 2z?

[nova yoke]

well it should be a vector

(2x, 2y, 2z)

or

2xi + 2yj + 2zk

whichever form you prefer

[odd lagoon]

i still dont really understand when i can add or remove i j or k

tbh

but

okay

[nova yoke]

well 2x + 2y + 2z is just a scalar

whereas i,j,k are vectors

so 2xi + 2yj + 2zk is a vector

so now what's the dot product of (2y, -2x, 0) (your answer from part a), and (2x, 2y, 2z)?

[odd lagoon]

sorry i dont think i remmeber how to do the dot product, at least not using matrices

i thought it was |x||y|cos(theta)

[nova yoke]

do you remember this formula:
(a,b,c) . (d,e,f) = ad + be + cf

[odd lagoon]

is that for multiplying matrices?

[nova yoke]

no, dot product of two vectors

[odd lagoon]

no i dont remember it

[nova yoke]

like this

[odd lagoon]

oh that is nicer than i could have imagined

[nova yoke]

you take the two x components and multiply them

do the same for the two y components and the two z components

and then add up the results

that's what you want to do here

[odd lagoon]

so i get (4xyi^2) - (4xyj^2)?

wait

wtf

i forgot

what happens to basis vectors

when they get squared

[nova yoke]

i . i is the dot product of the vector (1,0,0) with itself

because i is just (1,0,0)

so instead of writing i^2 you should have i . i

and what is i . i?

[odd lagoon]

(1, 0, 0)

[nova yoke]

that's just i

[odd lagoon]

oh

ehrm

[nova yoke]

what is its dot product with itself

[odd lagoon]

1

?

[nova yoke]

(1,0,0) . (1,0,0)

[odd lagoon]

it is 1

[nova yoke]

yea just 1

so what's your answer now

[odd lagoon]

(4xy) - (4xy) = 0

[nova yoke]

yep

zero!

[odd lagoon]

and the integral of 0 dS

is just S + c?

[nova yoke]

nope

[odd lagoon]

okay!

[nova yoke]

the integral of 0 dS is just 0

[odd lagoon]

oh wait

[nova yoke]

well actually it's not even 0 dS

[odd lagoon]

actually no i dont get why that is

[nova yoke]

(Nabla x F) . dS is 0

and you're integrating 0

integral of 0 is just 0

[odd lagoon]

i thought we had to add a constant

or is it because

it is definite

[nova yoke]

for indefinite integrals you do

[odd lagoon]

ah

[nova yoke]

but these are definite even though we haven't really specified what C and S are

[odd lagoon]

wait but that means i dont even have to care about C

[nova yoke]

right, because...

yep, the right hand side integral is 0

so the left hand side integral must also be 0

[odd lagoon]

so lefthand is also 0

yes

[nova yoke]

regardless of C

[odd lagoon]

yes

[nova yoke]

and that's what you wanted to show

[odd lagoon]

how quickly did you get an intuition for this

[nova yoke]

i've taken multivar calculus before, haha

[odd lagoon]

like to be able to reason yourself into dS being perpendicular to the sphere

is it less about intuition for you

and more about practice then

?

no wrong answers

im just curious

if that is okay

[nova yoke]

oh you do a bunch of problems like this and that's the first thing you think of

[odd lagoon]

you dont have to answer

okay

[nova yoke]

yea it's not like my mind just naturally knew this stuff, i had to learn it just like you are

but you'll get an intuition for it by doing more problems

[odd lagoon]

Stokes Theorem

seemed very important here

[nova yoke]

yes

[odd lagoon]

are there any other theorems

that appear commonly

or are there too many to name

at once

[nova yoke]

and the problem with most multivar classes is that they teach stokes at the very last minute right before the final exam

and people barely have time to learn it

[odd lagoon]

my final exam is in 6 days and im not passing unfortunately, but i am also behind in other courses, and the exam after multi is single variable which i think i will pass this time aroud

[nova yoke]

the other main ones that are related are: green's theorem (basically a 2d version of stokes' theorem) and the divergence theorem

[odd lagoon]

i have practied ODEs 1st and 2nd order

a lot

and on the exam there always seems to be 1 first order, seperable or linear, and one 2nd order

[nova yoke]

ah well best advice i can give is to do old practice exams, if your instructor provides them

[odd lagoon]

if i get them right i get half-way to passing

[odd lagoon]

yes yes i have been

so for that course

i really do think i will pass

i am saying that not out of arrogance but because i have practiced so much

[nova yoke]

and if they offer exam week help sessions or review sessions definitely attend those

[odd lagoon]

i have been practicing for over a month now

[odd lagoon]

yes i have been doing that too XD

[nova yoke]

nice

[odd lagoon]

i really do think i will pass im so ready for

single variable

[nova yoke]

cool cool

[odd lagoon]

but i wanted to jump in on multivariable as well

it is what im SUPPOSED to be ready for

but im not

and i wont be but

i still wanted to take a peek

i feel like i have improved my thinking a lot by just practicing single variable calc

i can understand things better now even if i still cant understand right away

[nova yoke]

yea you still have to use single variable calc in multivar, so that's useful

[odd lagoon]

i have a confidence that i will understand eventually

i didnt have that confidence to start with

[nova yoke]

like actually computing curve integrals you end up converting them to a single variable integral

[nova yoke]

you'll get it eventually for sure, just takes some time to sink in

[odd lagoon]

can i ask you some more stuff

[nova yoke]

and a lot of practice

[odd lagoon]

about this problem

that we just went over

i understand everyone has limited time

to engage

[nova yoke]

well i have to take off now but if you have conceptual questions about it you could just open a new channel and someone else can help i'm sure

(new channel to bump it to the top of the list)

[odd lagoon]

alright nvm then i am gonna take a break from this and keep practicing single variable calc

thanks for all the help

[nova yoke]

ok best of luck!

[odd lagoon]

.close

[midnight plank]

Channel closed

Closed by @odd lagoon

Use .reopen if this was a mistake.

[midnight plank]

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[odd lagoon]

hi serious question

[odd lagoon]

how to integrate cos^2x dx

and sin^2x dx

i have let these two integrals intimidate me for no reason

for far too long

im supposed to swap cos with something

i forgot what

cos^2x = cos2x + sin^2x

but that doesnt help me

[grim vector]

euler formula

[cedar pawn]

Ehhh

Double angle is cleaner

[grim vector]

yeah double angle

i call it euler

mb

[odd lagoon]

but isnt what i did here double angle? how does this help?

[grim vector]

cos(2x) = 2cos²(x) - 1

isolate cos²(x)

cos²(x) = cos(2x)/2 + 1/2

integrate...

[odd lagoon]

but if i swap sin^2x then i get 1-cos^2x

and then im going in circles

[dawn dagger]

no

[grim vector]

no you don't

[odd lagoon]

no?

[cedar pawn]

I thought you meant turn it into 1/4(e^ix + e^-ix)^2 and go from there

[dawn dagger]

if you bring it to the other side you get a constant

[odd lagoon]

okay so i swap cos^2x again

cos^2x = cos2x + 1 - 1 + sin^2x

i swear i am going in circles

[grim vector]

swap ?

[odd lagoon]

this happens everytime

[cedar pawn]

I'd suggest keeping the cos(2x) on the LHS

[odd lagoon]

but i need to isolate cos^2x

how else am i supposed to sub it in

[cedar pawn]

We'll do that once we've made it nicer

[odd lagoon]

okay

sure

but i still dont see how this helps

now we have turned cos^2x into 2 cos^2x

[dawn dagger]

you can integrate cos(2x) easily

[dawn dagger]

divide by 2

it's a silly constant

[dawn dagger]

.

[cedar pawn]

cos^2(x) = 1/2 cos (2x) + 1/2

[odd lagoon]

i have let that silly constant

[cedar pawn]

Which is easy to integrate

[odd lagoon]

ruin my integrating

for months now

[dawn dagger]

how?

you can pull 1/2 behind the integral

[odd lagoon]

idk i just left it there i didnt know what to do with it

i just assumed i was going in circles

[dawn dagger]

you carry it usually

[odd lagoon]

because i couldnt get cos^2x

to be nice

but you are telling me the reason i couldnt

was because i cant

[odd lagoon]

so i need to extend f(x) with the something to fix the 1/2

so i multiply it by 2?

[dawn dagger]

you can do that too

[odd lagoon]

and only break out 1/2

[cedar pawn]

You can integrate 1/2

[odd lagoon]

im so confused

can we start over

[dawn dagger]

okay

[dawn dagger]

what do you wanna do next?

[odd lagoon]

i dont know apparently im supposed to use that thing i thought was circular

[cedar pawn]

$$ \int (\cos(2x) + 1) dx $$

[dawn dagger]

you can use an identity for example to integrate this

[grand pond]

😈 Greenie The Demon Queenie 😈

[cedar pawn]

Can you do this integral?

[dawn dagger]

getting rid of that square by replacing it withsomething linear

[grim vector]

i think one helper to explain is the best for the moment in regards of avoiding multiples questions or confusions

[cedar pawn]

Oki

[dawn dagger]

okay I will leave it to you guys

[odd lagoon]

yes i think so, i would get sin2x/2 + x + c?

sry

not divided by x

i get that?

i edited it

i was just doing it in my head so i dont know for sure if it is right or not

but i dont get what that has to do with this

i have cos2x + sin^2x

and if i replace sin^2x i get 1 - cos^2x

cos2x + 1 - cos^2x

then i replace cos^2x

srry i saw

cos2x + 1 - 1 + sin^2x

urgh

making many oopsies

then i replace sin^2x

[dawn dagger]

if you swap it again it's pointless of course

[odd lagoon]

well what the hell else do i do with it

[dawn dagger]

the point is to get rid of sine once and kinda establish a relation to cos^2(x) again

[odd lagoon]

okay

so i got this

cos2x + 1 - cos^2x

[dawn dagger]

if you do it like whatever times it will just keep cancelling out

[odd lagoon]

i should stop there

but problem is

i still have the cos^2x

but now it is even more complicated

than it was originally

[dawn dagger]

not really

[odd lagoon]

why not

[odd lagoon]

idk that still looks dreadful to me

[dawn dagger]

it's like a linear equation

in u

[fathom onyx]

Just rearrange to make u the subject

[odd lagoon]

2u = cos2x + 1

[dawn dagger]

yes

[odd lagoon]

this feels really counterintuitive

all of this

[fathom onyx]

Well, your aim is to integrate u

[dawn dagger]

it's just a derivation of the formula

[odd lagoon]

like even if it works

i dont understand how anyone could look at this and understand what to do

[fathom onyx]

That is now equivalent to half the integral of the RHS of this

You should be able to do the integral of cos(2x), for example

[odd lagoon]

sin2x/2 + c

but that is intuitive

[fathom onyx]

And the integral of 1 is...?

[odd lagoon]

just x

but that is intuitive

because of how integrals relate to derivatives

[fathom onyx]

We're using identities we do know in order to rewrite the function whose integral we don't know into something more manageable

[dawn dagger]

Again, you know that cos²(x) = cos(2x)+sin²(x) but thats less helpful to replace with, so that begs the next question how this can be still useful.
Now actually, intuitively you would want to get rid of sin²(x) and replace it with something in terms of cos²(x) and luckily there just is the Pythagoras identity which you proceed to apply.

[odd lagoon]

idk even then i dont find it intuitive

i have rewritten cos^2x, it is in a simple form but hard in this context

and now i still have that term

[fathom onyx]

This isn't really a thing about "intuition"

[odd lagoon]

but negative

and some other stuff

[fathom onyx]

This is about problem-solving

[odd lagoon]

okay i am really bad at solving problems

[fathom onyx]

We're trying to find ways to rewrite the question in a way that makes it easier to approach

[odd lagoon]

yes how can i do this

[fathom onyx]

...we just did?

[odd lagoon]

i dont see how it is easier cos2x + 1 - cos^2x is easier

it is

by all means

harder

[fathom onyx]

$\cos^2{x} = \frac{\cos(2x) + 1}{2}$

[grand pond]

Waes (Wires)

[odd lagoon]

than what we started with

[fathom onyx]

(acc imma check this0

[dawn dagger]

well, idk else, maybe re read it and then it clicks? or tells us what's intuitive to you and whats not here?

[odd lagoon]

we added two terms and made the original term negative

[odd lagoon]

wait this is the first time im seeing this

we never divided the 2

because we replaced it with u

[dawn dagger]

we did

[odd lagoon]

no

we jsut left it as

[fathom onyx]

You've got 2u here

[odd lagoon]

2u = cos2x + 1

[dawn dagger]

since here

[fathom onyx]

So to get u on its own?

[odd lagoon]

yes i see it now

but i didnt before

that does make it easier

i agree

because now there are no cos^2 stuff going on

[cedar pawn]

At this point I wanna do it by parts lol

[odd lagoon]

that's usually what i do

[odd lagoon]

im serious

i thought as an anti-algebraist you would understand

[dawn dagger]

well doing by parts is where you actually may end up going in circles ♾️

[odd lagoon]

i learned that the hard way

for xsinx

[dawn dagger]

nah i dont like group and ring theory that much

[odd lagoon]

i picked the wrong u

and wrong dv

[odd lagoon]

ok that is way beyond what i know

[fathom onyx]

$$
\begin{aligned}
\cos{2x} &= 2\cos^2{x} - 1 \
2\cos^2{x} &= \cos{2x} + 1 \
\cos^2{x} &= \frac{\cos{2x} + 1}{2} \
\ &= \frac{1}{2} \cos{2x} + \frac{1}{2}
\end{aligned}
$$

[cedar pawn]

Throw it into Mount Doom

[odd lagoon]

okay

thanks for helping, so cos^2x = (cos2x + 1)/2?

what about sin^2x

[dawn dagger]

do that as an exercise

[odd lagoon]

i usually end up going in circles there as well

[grand pond]

Waes (Wires)

[dawn dagger]

you know that cos²(x) = cos(2x)+sin²(x)

[odd lagoon]

sin^2x = 2sinxcosx

[cedar pawn]

No you use the cos identity again

[dawn dagger]

(you can even apply the same swap trick as before but for cos²(x))

[cedar pawn]

The sine identity just sits in the corner with no friends

[odd lagoon]

i dont know the cos identity for that 😭

[dawn dagger]

pythagoras

[fathom onyx]

for $\sin^2{x}$ use the other part of the $\cos{2x}$ identity - namely, that
$$\cos{2x} = 1 - 2\sin^2{x}$$

[cedar pawn]

The same one we just used

[odd lagoon]

but the pythagaros one is squared!!

this one isnt

😦

oh wiat

you mean

[grand pond]

Waes (Wires)

[odd lagoon]

1-cos^2x

[dawn dagger]

do you mean sin(2x)?

[odd lagoon]

ooooooh

no i meant sin^2x

[dawn dagger]

yes

[odd lagoon]

and also i realise

i used the double angle

for some reason

idk why id id that

no good reason

i cant really explain why i did that

[dawn dagger]

nobody asked anyway so we good

[odd lagoon]

it looks similar to the cos substition

i think that's why i did it

anyway 1 - sin^2x

okay

but yeah i do end up going i ncircles here

as well

because i replace sin^2x

[fathom onyx]

$$\cos{2x} = \cos^2{x} - \sin^2{x} = 2\cos^2{x} - 1 = 1 - 2\sin^2{x}$$
These three ways of expressing cos(2x) you need to commit to memory

[grand pond]

Waes (Wires)

[odd lagoon]

with 1 - cos^2x

and then cos^2x

with 1 - sin^2x

[cedar pawn]

cos(2x) = cos^2(x) - sin^2(x)
= (1 - sin^2(x)) - sin^2(x)
= 1 - 2 sin^2(x)

2 sin^2(x) = 1 - cos(2x)
sin^2(x) = 1/2 - 1/2 cos(2x)

[odd lagoon]

i have a serious problem

i need help

[dawn dagger]

actually a straight path

[cedar pawn]

The first one is sufficient tbh

[odd lagoon]

the way im doing it

im just changing the squared to 1 - the other one squared

over and over

[fathom onyx]

(well, now it is because we have an integral for cos^2, but I'd still recommend learning these - even if the last two can be achieved by using the Pythagorean identity)

[cedar pawn]

You don't need that integral

Indeed

It's a 1 step derivation

[fathom onyx]

(my point is I'd still advise learning them)

[cedar pawn]

I prefer to err on the side of minimal memorisation

[odd lagoon]

im trying to get away

from memorisation

[dawn dagger]

I just prefer to err

[odd lagoon]

it is how i got through high school and the bits of uni i have passed so far

but that changes in 6 days when i pass my calc 2 exam

there i have more understanding

[dawn dagger]

well you gotta know something to start somewhere

[odd lagoon]

but i dont know anything

i just know gibberish

and i never understood any of it

it just let me get funneled further along the system

but anyway so

how do i stop myself from

going in circles again

[dawn dagger]

then empty your current glass and refill it with something new

[odd lagoon]

sin^2x = 1 - cos^2x

[odd lagoon]

it is what i have been doing

for the last 4 months

[dawn dagger]

with cola?

[odd lagoon]

yaes

i actually have 11 cans of cola

right next to me

i bought them today

[dawn dagger]

nice

[odd lagoon]

im putting them in the fridge when i get home

anyway

HELP HELP

😦

sin^2x = 1 - cos^2x

i just see circles here

im trying to remembr what we did for cos^2x

cos^2x = cos2x + sin^2x

oh man

[dawn dagger]

now it depends on what you want (an identity for sin^2x or cos^2x)

regardless

[odd lagoon]

we literally just talked

about cos^2x = cos2x + sin^2x

[dawn dagger]

you apply the pythagoras identity

[odd lagoon]

i already forgot how to prevent it from going circular

i think i tried these integral and identities and got it wrong so many times that now i only know how to do them wrong

[fathom onyx]

From this line, integrating the left is now the same as integrating the right

[dawn dagger]

just try it

[fathom onyx]

We can integrate 1 easily

[dawn dagger]

you learn by doing

[fathom onyx]

And we've just integrated cos^2 (x)

[odd lagoon]

no no

not true

i have been doing

i swear to god i have been

i just keep getting them wrong

[dawn dagger]

then do it again

[odd lagoon]

i just see circles

no

i go in circles

[dawn dagger]

well show it

[odd lagoon]

i did earlier

i just keep

subbing in the wrong thing

im trying to remember

what you did

without looking up in chat

to stop that sin^2x

from spiraling into an infinite series of 1 - 1 + 1 - 1...

[dawn dagger]

dont try to remember what i did, try to think about what you have to do

use your logic

[odd lagoon]

cos^2x = cos2x + sin^2x

[dawn dagger]

now ask yourself what you want and how can you get there

[odd lagoon]

okay unfortunately

i remembered

what we did

but honestly

i still dont understand it

i mean i do in the sense that

it is literally just arithmetic

after subbing

and dividing by 2

but i dont understand it

i dont think i have good logic

[dawn dagger]

you need to empty the glass, you keep drinking from the old one

[odd lagoon]

^

very good way to put it

i have used my memory as a crutch my entire life

i dont think i have developed very good logic

[dawn dagger]

you do that by trying and doing until it clicks

[odd lagoon]

can i tell you

the worst part now

so you understand why im having a serious problem

so i when you told me to

use my logic

and not drink from the old glass

[odd lagoon]

and you told me this

try to think what to do

i remember, what you said, you wanted to do with it

so i remember your logic

im not using my own

you wanted cos alone

for some reason

cos^2x

im butchering

your logic

because i dont understand it

but

for some reason

you wanted to put everything in terms of cos

no sin

and then you moved the cos^2x

over to the LHS

[dawn dagger]

are there any objections why you wouldnt?

[odd lagoon]

so you got two of them

and divided by 2

[odd lagoon]

well see i dont understand

why that is important

except

for the meta reasons

like i remember

what it led to

is what i mean by meta reason

i dont understand why it led to that

[dawn dagger]

if you dont get cos²(x) then you won't derive a relation that will be helpful

[odd lagoon]

campus is closing and i need to pack my things and leave unfortunately

[odd lagoon]

i know what the function cos does

[dawn dagger]

cos(2x) is nice because you keep in mind it's easy to integrate

[odd lagoon]

and the sides of a right angled triangle it cares about

[dawn dagger]

unlike sin^2(x)

[odd lagoon]

yes sin^2x is not nice

[dawn dagger]

so you focus on this

[odd lagoon]

anyway ty for trying to help i will ask my tutor about this tomorrow

[dawn dagger]

next question you ask yourself is, if there exists a nicer form for sin^2x

[odd lagoon]

i get that this is super easy for most people, 99% of people at this level of maths

[dawn dagger]

and that can be any (pythagoras just happened to work here)

you need to ask yourself small questions and keep track of what you do

and what you want

else you get lost in the sauce

it's like a puzzle in the end

[odd lagoon]

i never asked any questions regarding maths throughout my entire schooling except

how to solve something

like a specific problem

[dawn dagger]

i meant in your mind

[odd lagoon]

then i learned how to

by just memorising

i didnt understand what i was doing and never asked any small question

in my mind

[dawn dagger]

i bet you do that all the time except in maths

when you wanna buy ice cream you know what you want and intuitively you already know how to get it

[odd lagoon]

in math i just get stuck until i know how to solve something

[dawn dagger]

because the questions on how to get it are already answered by your mind

you just need to apply and work that skill on math

[odd lagoon]

then when someone tells me, i might not remember the first or 2nd or third

but fourth or fifth i might have a memory of it

but i never understand any of it

when i buy ice cream

i dont understand how the cash register works

i just know i lose money and get ice cream

[dawn dagger]

maybe you dont need to

[odd lagoon]

same with card transactions

i want to

and it seems more fun

the few things i feel like i understand somewhat have been very rewarding

but yeah sry i really need to go now

ty again for the help

[dawn dagger]

you only care about the easiest way to get ice cream and thats also in math, to find the best and most efficient solution

it's natural

[odd lagoon]

okay im glad

im not alone in that then

but i feel like im on the extreme end

to be honest

[dawn dagger]

you take the bus and dont walk 20km to school

it's more efficient

[odd lagoon]

again ty!!!!

.close

[midnight plank]

Channel closed

Closed by @odd lagoon

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[last slate]

I didn’t know how to graph that and I’m so lost

[night hawk]

You have indicated the midpoint of segment BA. That's good, but make sure you label it more specifically as point O.

[last slate]

Is the others right

🙁

[night hawk]

You also put point C on the perpendicular bisector. I recommend adding a right-angle symbol to show that the lines intersect each other at a right angle.

[night hawk]

Can you show the previous page of the question paper?

[last slate]

OKG THANK U SO MUCH UR SAVING MY LUFE

I’m crying 😢

[night hawk]

1 - 7 are all correct.

[last slate]

IKAG

[latent kite]

what the heckin flip

[last slate]

STOP I WAS LIKE SO CONFUSED CECAUSE I DIDNT STUDY IT AT ALL

IM DOING ONLINE SCCUOOL

SCHOOL

AND I DIDNT DO ANYTHINF

😭 😭 😭

[latent kite]

u played urself g

[last slate]

AND I DIDNT KNOW THERE WAS A TEAT

HELP

DEAD

[latent kite]

r u taking a remedial algebra course

[last slate]

No smt like geometry or smt

[latent kite]

oh sry playa ive never worked with shapes in math before

[last slate]

FUCKKK