help-28

[cursive condor]

“First branch”

Like the most used/common

[clever flicker]

Sorry but what is first branch

In this context, especially

[cursive condor]

The one that makes the most sense

For example

[wide sundial]

think about square roots

what's √(-1)

[clever flicker]

i

[wide sundial]

not quite

[cursive condor]

A “principal branch” of sqrt(i) is 1//sqrt(2)+1/sqrt(2)i

[clever flicker]

-i as well

[wide sundial]

it's i if you take the principle branch of the sqrt function

[cursive condor]

There are infinitely many solutions

[wide sundial]

exactly

[cursive condor]

So you take the one that has the first application

[clever flicker]

ohh ok

[wide sundial]

perhaps you may think of -1 as e^iπ, then sqrt will be e^iπ/2

[clever flicker]

so in this case i should take the positive branch

?

well err

[cursive condor]

i.e., when you have k=0 if it is some form e^(itheta+2i*pik)

[clever flicker]

"positive" idk

[cursive condor]

Just a regular 10th root

[clever flicker]

this is my work so far for the problem:

[wide sundial]

but if you think of -1 as e^i3π then you end up at e^3π/2 which isn't the same as e^iπ/2

[clever flicker]

ohh gotcha

yeah he explained that in class

[wide sundial]

which means the square root function, in order to be a function needs to send every input to only ONE output

we've just seen a case where representing the input differently resulted in a different output

so instead we take the "principle" branch of sqrt so that it actually is a function

[clever flicker]

ok i understand thank u both @wide sundial @cursive condor

is my approach to (c) reasonable?

[wide sundial]

it looks right

[clever flicker]

Was just gonna say "f'(z) is defined for the open ball of radius (blank) centered at 0 containing those points for z on the boundary"

[clever flicker]

ok ok ty

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[quick summit]

can someone double check my answer

[hot herald]

show work

[quick summit]

wait mb other way around he got 1.42 i got 1.73

that 3 on the arctan when I sub it back in is supposed to be a 6 just wrote it down wrong but plugged in my calc correctly

[hot herald]

seems alright

[quick summit]

he did pretty much the same exact work as me but for his theta equation

he had 3/a and 6/10-a

in the arctan

[hot herald]

he had the angles the other way around,
if he's doing it that way,
he'd need to subtract that sum from pi to get theta

as those give these angles

[quick summit]

ohhhhhhhhhhhhh

Thank you so much

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[clever flicker]

need help continuing the argument:

[clever flicker]

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[dreamy spoke]

I dont really understand derivatives
D(4x)=4

D(x^2)=2x

Why is it not x since y grows by x for every x value

[honest hill]

!original

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[dreamy spoke]

There is not an original problem, this is just one of the things i've never understood about derivative

[long sky]

Oh by D() do you mean d/dx()?

[dreamy spoke]

Arent they the same thing

[onyx glen]

different notations for the same thing

[long sky]

I've never heard of anyone using that notation

[onyx glen]

Why is it not x since y grows by x for every x value
what are you trying to say with this?

[onyx glen]

i have seen it.

[long sky]

Oh oke

[slender onyx]

informally, try estimating the tangent slope of x^2 at various points. you should find it to match 2x instead of x
rigorously, go through the limit definition of derivative to get 2x

[dreamy spoke]

I mean with 4x for example, its pretty simple

For every x the y is 4 times bigger

So it would make sense for xx to just be x, since for every x value y value grows by x

But for some reason its 2x

[long sky]

Yea I mean that's not how it works

[dreamy spoke]

I dont get why

[long sky]

Hold on

[slender onyx]

make a plot of x^2 and estimate the tangent slope at various points

[onyx glen]

So it would make sense for xx to just be x, since for every x value y value grows by x
you're trying to treat x as both a constant and a variable at the same time

[long sky]

Yea

[onyx glen]

try actually working out what happens when you increment x by 1 from, say, x=10

when going from x=10 to x=11, does the value of y really increase only by something like 10?

[long sky]

The whole D(Kx)=K is only true if K is a constant

[slender onyx]

ah applying constant rule to powers

[long sky]

I mean it's a honest mistake though

[normal spoke]

df(x)/dx = lim h-> 0 (f(x+h)-f(x))/h

[slender onyx]

yes, its just that their wording suggested a graphical error to me

[dreamy spoke]

Its not easy

[long sky]

So the rule applied here for both is the power rule where:
$$\frac{d}{dx}(x^n)=nx^{n-1}$$
Where n is a constant

[glossy valve]

denzio321

[dreamy spoke]

Yes i know the rule im maybe starting to get this now

[long sky]

Nah this is like day one stuff

You need to read a bit more I figure

But yea good start

[dreamy spoke]

thx guys

[drifting ferry]

I suggest u the 3b1b calculus series

He explains all this very intuitively and geometrically

[dreamy spoke]

On yt?

[drifting ferry]

Yes

[dreamy spoke]

ok i'll look into it

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[slate violet]

also if you remember sequences

if you take the sequence 1, 4, 9, 16......
the first differences are 3, 5, 7...... and the first differences increase by 2 each time

you can think of the first differences as an approximation to the derivative, where h = 1

so (f(x + h) - f(x)) / h would just be f(x + 1) - f(x), as expected

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[craggy tapir]

is part (a) wrong?

if i set dx/dt = 0 and dy/dt = 0, for the former i get 3/2, while i get 1 and -1 for the latter

which means no tangents

[slate violet]

you need to find dy/dx = (dy/dt)/(dx/dt) when t^2 = 3 and t^3 - 3t = 0

yeah you are confusing parts a and b together

[craggy tapir]

hm

can you explain this?

the curve doesn't even cross (3, 0), i don't see why it has any tangent in the first place

[honest hill]

mby just a typo? (1,0)?

[craggy tapir]

ok thanks

got y = (sqrt3)x - 3sqrt3 and y = -(sqrt3)x + 3sqrt3

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[lunar orbit]

has anyone created deployable structures by using math and origami. if so, what softwares did you use and how did you do it

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[thick hedge]

so $x \geq 0; y \geq 0;y\leq 3$

[leaden ermine]

you are integrating z = 3-y over a a quarter circle

[glossy valve]

What a wonderful world !

[thick hedge]

yeah, guess as much

hmm

okie

so $\int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} 3-y dy dx$

[leaden ermine]

no

check your upper bound

[glossy valve]

What a wonderful world !

[leaden ermine]

yes

you could convert this into polar

[thick hedge]

yeah,

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r(3-r\sin(\theta)) dr d(\theta)$

[leaden ermine]

jacobi

[thick hedge]

F

Forgot that

tq

[leaden ermine]

still wrong?

you factored wrong

[thick hedge]

yea

[leaden ermine]

you gon let that slide

[glossy valve]

What a wonderful world !

[leaden ermine]

nic

[thick hedge]

tq

[leaden ermine]

np

[thick hedge]

So here I'll have to integrate z over the part of x^2+y=4 in Q1, right

huh

I'm getting $\int_{0}^{2} \int_{0}^{4-x^2} (4-x^2-y) dy dx$

[leaden ermine]

first octant means x,y are non-negative and so there will be are level curve at z = 0

btw your bounds are messed up

0 = 4-x^2-y solve properly for y

[thick hedge]

F

[leaden ermine]

why are you taking the square root?

[thick hedge]

Yeah, my abd

sorry

[glossy valve]

What a wonderful world !

[thick hedge]

Which is going to be a bit of a pain to integrate

😭

[leaden ermine]

sometimes pain is good

makes you grow

[thick hedge]

$\int_{0}^{2} 4(4-x^2) -x^2(4-x^2) - \frac{(4-x^2)^2}{2} dx$

[glossy valve]

What a wonderful world !

[thick hedge]

tq

I don't think I'm going to gain anything useful from this IMO( Such tedious integrals)

[sharp flame]

for a few of these you can simplify them nicely

[glossy valve]

neon

[sharp flame]

Eh

yeah it helps

[thick hedge]

hmm

mhm

thanks

I was thinking $\int_{-\pi /3}^{ \pi/3} \int_{-\sec(x)}^{\sec(x)} 1+y^2 dy dx$

[glossy valve]

What a wonderful world !

[thick hedge]

Is this right

Which is $\int_{-\pi /3}^{\pi/3} 2 \sec^3(x) dx$

[glossy valve]

What a wonderful world !

[thick hedge]

Does this work

<@&286206848099549185>

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[timber blade]

the aim of the exercise is to simplify the expression, which to say the least is very simple, but due to a long period of fever it led me not to remember how to do it at all 😅

[austere breach]

First of all, you're going to want to expand all terms

Square everything that's squared and multiply everything out until you have no parentheses

Then combine like terms

Once you do that, you will likely able to factor out the result

[timber blade]

i feel like something is wrong with this

[austere breach]

It looks like you aren't expanding your squared terms properly

but let me check

[timber blade]

i didn’t finish the third passage

[jade radish]

You can't distribute the exponent into the terms like that

i.e. (a+b)^2 isn't a^2+b^2

[austere breach]

Yeah, if something is squared, you have to multiply every term by every other term

[jade radish]

also for the second term, you have to distribute the 2 to the (1-a) term and then b to the result of whatever is then inside

[austere breach]

so @timber blade, (a + b)^2 is a*a + a*b + b*a + b*b, or it simplifies to a^2 + 2ab + b^2

[jade radish]

It does simplify to a^2+2ab+b^2. The other option you wrote a*a ... etc. is wrong though ( I originally read it wrong )

when you see it like this

[timber blade]

so you are saying that this is a remarkable product

[glossy valve]

Juke | ping me if no response

[jade radish]

You may have heard this being called FOIL

[austere breach]

It's not wrong, it's just further simplified

[jade radish]

you're right, I read it wrong, my bad

[timber blade]

ima write it again and send it again

wait

wait but

[austere breach]

what?

[timber blade]

nope i don’t get it

i mean i don’t understand if i did smth wrong or correct

and if i did the first thing what do i have to do? how do i start

[austere breach]

Can you try to expand (a + b + c)^2 ?

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[timber blade]

well that would be a good idea

ima go now

i appreciate the help

thank you

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[night musk]

For this we use the Archimedean property when x greater than a to disprove the second part of the inequality leading to a contradiction. but how do we know that it is x greater that causes the contradiction but not that that n being greater than 1. Also how do we know that it doesn't hold for x greater than a when n is less than or equal to 1

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@night musk Has your question been resolved?

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@night musk Has your question been resolved?

[cursive dock]

@night musk you still here?

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[hexed kayak]

3+8 got me stumped (I ran out of fingers to count on)

[gritty rose]

<@&268886789983436800>

funny guy is funny

[slender onyx]

@hexed kayak pls dont abuse help channels

[hexed kayak]

erm

[wide star]

Don't post troll questions here

[slender onyx]

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[cursive condor]

Loosah

Be serious

[hexed kayak]

mb g

.close

[cursive condor]

Lol

Someone forced closed it cause of dumb question

[night musk]

Sorry I'm here now

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[craggy tapir]

does this mean the polar coordinate system is ill-defined, in that it has multiple representations for the same point?

[brittle steeple]

Not really

That's like saying addition is ill-defined because 1+4 is the same thing as 2+3

Usually you just choose the representation with positive radius and smallest angle

[sand escarp]

it's more of saying sin(pi + 2kpi) where k is an integer

it's all 0, just more than 1 way of representing it

[umbral dome]

you can make it unique for every point other than the origin by requiring $r \ge 0$ and $0 \le \theta < 2\pi$, or $-\pi < \theta \le \pi$, etc.

[granite torrent]

it fails to be injective at more than that

[glossy valve]

cloud

[granite torrent]

a good point is that these boundary points usually do not cause issues when you integrate

i.e. powerful theorems like change of variable simply require your parameterizations to be injective on the interior of some compact set, so even despite this failure of "unique representation" nothing bad happens

[craggy tapir]

alright, thanks y'all

.lose

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[heady hearth]

I feel like I need someone to help me walk through this to make sure I don't make any mistakes. Can someone help me with this?

[heady hearth]

I think you multiply the second equation by -3, right?

[eternal halo]

multiplying by -3 or 3 can both get the answer

so yes

what do you get after multiplying by -3?

[heady hearth]

The second equation is now 9y - 30x - 93 = 0.

[eternal halo]

-10x?

10x * -3 = ?

[heady hearth]

Sorry, I meant -30x.

[eternal halo]

all good

[heady hearth]

Now we add them both together.

[eternal halo]

yep

[heady hearth]

So that's -26x - 104 = 0.

Or -26x = 104.

[eternal halo]

mhm

[heady hearth]

So that means that x = -4.

[eternal halo]

yep

[heady hearth]

I got x = -4 and y = -3.

[eternal halo]

yea thats correct

nice work

[heady hearth]

Thank you!

I think my main problem with these is not making any mistakes. The general concept feels simple.

I'll close this for now.

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[desert berry]

I need help. I don’t know how to graph this all. This is college algebra.

[spiral hearth]

Well what are the roots…?

Hold on lemme switch

[hot herald]

its already in factored form, so that makes things easier

[soft dew]

Yeah.

[hot herald]

can you identify roots/zeroes/x-intercepts from that?

[soft dew]

So essentially the factored form provides you with roots

Which is what omega is saying (roots are essentially x intercepts, where the graph intersects the x axis, and when y=0)

So first determine the roots

For instance, an example of a root would be: x+4=0 so it intersects it at x=-4 in this specific example

[hot herald]

ideally you'd also make some sort of table of values

[soft dew]

I’ll let you snipe this one omegan, you have more concise and profound answers

(Well more rigorous)

[soft dew]

So there are essentially two steps to this equation. Determining the highest degree of the equation, and determining the roots

[hot herald]

my responses will be limited if op isn't responding
and i gtg soon

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[fathom turtle]

I'm new to graph theory:
let the graph be H, A={1,2,3,4}. What is |E(A,H\A)|? Does (3,9) and (3,8) count as the same edges?

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[steel nest]

Do you see other solution than x(t) = 1 constant function when x_0 = 1

I tried the usual:

[subtle depot]

https://cdn.discordapp.com/attachments/1344607660829311007/1345254642979835914/IMG_1548.jpg?ex=67c3e184&is=67c29004&hm=4e8e2a7543fe9f3c5486e00ae7fccb6dffd3cfd59b85e441dab7ee68de99195d&

help pls

[steel nest]

#❓how-to-get-help

If x_0 = +-1, the constant function x(t) = x_0 is clearly a solution. Find other solutions (not constant) for x_0 = 1 and x_0 = -1.

that’s my translation

[steel nest]

however arcsin(1) is infinite so separation of variable is sus

c can’t be infinite and so i’m here

oh my bad arcsin of 1 is pi/2 my bad life good

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@torn plinth Has your question been resolved?

[torn gust]

r=r

[torn plinth]

so like x = x?

something like that

[torn gust]

idk of a good comparison

[torn plinth]

r is the spherical coordinate variable

distance from the origin

[royal holly]

to find the field at r>R use guass law whats the problem?

[torn plinth]

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[earnest kite]

Calculate the area of the plane figures bounded by the curves:
𝑓1(x) = xx f2(x)= 2xx f3(x)= 1

[earnest kite]

how do I do this<

I can only solve for two curves

[craggy tapir]

sketch the curves

[earnest kite]

done

[onyx glen]

show us the sketch

[earnest kite]

Still no clue how to form an integral out of this

Like

When I was working with two curves it was quite clear

But here I have no idea

[onyx glen]

you can do it the easy way or the hard way

the easy way is restating all this in the form x = g(y), and then you'll have an integral from 0 to 1 of some bullshit dy

[earnest kite]

Whats the hard way

We havent worked with the notation x= g(y)

[onyx glen]

the hard way is to split the thing vertically at x=1 and make it into 2 integrals

[onyx glen]

you have never expressed an area like this as an integral wrt y?

[earnest kite]

No

[onyx glen]

but you have done the same thing wrt x and a horizontally oriented thing?

[earnest kite]

Yeah

[onyx glen]

it really is not a different concept at all, just that the axes are swapped.

[earnest kite]

So I need to calculate this area

?

[onyx glen]

i think maybe only the right half?

but unless you are at gunpoint, i would not worry about maybe having to multiply by 2

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[jade cloak]

I’m kind of confused in what the question is asking

[nocturne shard]

does it just ask to solve the equation?

[onyx glen]

you simplified f(1/x) incorrectly

[onyx glen]

no it asks to show that f(1/x)+f(x) simplifies all the way down to 0

[nocturne shard]

oh

this doesn't look like it

[onyx glen]

show that

[nocturne shard]

bruh

[wide sundial]

You missed an x

The big fraction on the left should have x² on the bottom

It just disappeared when you made the denominators the same

[jade cloak]

Ohhh

I see it now

Thanks I’ll retry it now

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[orchid glen]

Found on Facebook officially posted by a government organization

I really want to understand the math behind it since the Facebook post was just about corruption and such

[full forum]

@orchid glen Has your question been resolved?

[torn jolt]

Interesting

[orchid glen]

@torn jolt think you can solve it?

[torn jolt]

Yeah

[orchid glen]

Please help, I'm curious how a problem like this is solved

[torn jolt]

Aight so this is a physics problem. first we gotta solve for the maximum bending moment to use the formula of the bending stress

[orchid glen]

Alright, I'm not familiar with formulas for this type of solving so forgive my ignorance

What's I and c?

[torn jolt]

I is moment of Inertia, and c is the distance from the neutral axis

[orchid glen]

Alright

[torn jolt]

let's get the maximum bending moment first, then we work our way all to the moment of inertia

so we use this:

W is weight (W=mg where m is the mass, and g is the acceleration due to gravity which is approximately 9.81 m/s^2)

a is the truck's distance from A and b is the truck's distance from B

as mentioned in the problem if i read it correctly

[orchid glen]

Oh alright

[torn jolt]

L is the total length of the bridge

let's go with weight first

so 102 metric tons is a unit of mass

in kilograms that's about 102,000 kg

so we multiply 102,000kg by the acceleration due to gravity (9.81)

[orchid glen]

A and b is 10 • 55 right?

[torn jolt]

that's right!

[orchid glen]

Yay!

[torn jolt]

what's 102,000kg multiplied by 9.81?

[orchid glen]

Uhhhh

[torn jolt]

you can use a calculator. It's okay

[orchid glen]

1000621

[orchid glen]

I was planning to anyway, no way I'm pulling out pen and paper for a big number💀

[orchid glen]

So it's 1,000,620

So 1,000,620 • 10 • 55

550,341,000

Divided by the total length of the bridge, 65

M = 8466784.61538??

Do engineers round up?

[torn jolt]

kinda

[torn jolt]

Or 8,467,553.85 Nm

without writing the units, teachers may get angry at you. Cause physics

[orchid glen]

Wait how'd you convert that

[torn jolt]

they're usually harsh with those

[orchid glen]

Oh right, I understand

[torn jolt]

Since Physics usually requires absolute accuracy, we use your value

because this is the most accurate approximation

8466784.61538

Nm

aight so now we got M right?

let's now move on to the moment of inertia

[orchid glen]

yes

[orchid glen]

alright

[torn jolt]

so we use this formula for the moment of inertia in a rectangular cross-section:

so b is the position of the truck

so that's 10m from the b point

h is the thickness

so its 1m thick

so since any number as exponent of 1 is always 1

10/12 = 0.8333 m^4

m^4 is a unit dont worry

and finally, c

like what we had said, c is the distance from the neutral axis

calculated by h/2

since its a symmetrical rectangle

so its 1/2 or 0.5m

[torn jolt]

finally, we combine all of it back to the first formula I showed

therefore, the bending stress is 5.08 MPa

[orchid glen]

Alright

[torn jolt]

So will the bridge collapse under this load?

it is asked in the question if the bridge will collapse

[torn jolt]

So since the bending stress is 5.08 MPa

what will be our answer for question 2?

[orchid glen]

Sorry I went offline, Signal is being weak here. Solve it while showing your process ple

[orchid glen]

It's overloaded by a lot so yeah

Ohh I get it

I understand the solving but I feel like I don't know enough to be able to solve other problems on a conceptual level

How would you solve question no. 3?

[torn jolt]

oops! wrong

not immediately

but it is unsafe for that truck though

since theres a limit to how many metric tons

so the answer is not immediately. The bridge is still pretty strong

[orchid glen]

wait its not a yes or no question??

[torn jolt]

but the truck is unsafe to go in that bridge

[torn jolt]

Its more of a explanation so yeah

[orchid glen]

i see alright

[torn jolt]

we get the volume so 10 times 1 times 65 so thats 650 cubic meters

so the cost to replace with high-strength concrete is 6,500,000 philippine peso

which is extremely expensive

[orchid glen]

i see

[torn jolt]

Okay let's say we bought 10 metric tons of steel

and one metric ton of that steel is 60 thousand

so of course that is 600k yes?

so now let's add

6,500,000 + 600k

so they will need to pay another additional funds of around 7,100,000

[orchid glen]

i see

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[orchid glen]

thanks for helping me out @torn jolt

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[fresh ginkgo]

For what value of a is f(x)=9x⁴+4ax³-6x²-1 convex for every x?

[fresh ginkgo]

Second derivative is 9x²+2ax-1

9x²+2ax-1>0 ? how does it work

Or derivative should be >0

[onyx glen]

9x^2 + 2ax - 1 needs to be ≥0 for all x. find its discriminant and write the inequality that says it's less than 0

и честита баба марта, между другото :)

[fresh ginkgo]

Честита баба Марта 🙂

Объркал съм знак

9x²+2ax+1

Мерси

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[fresh ginkgo]

.reopen

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✅

[fresh ginkgo]

А защо дискриминантата трябва да е по малка от нула

[onyx glen]

по-малка или равна, извинявай

ако беше положителна, f'' щеше да има два корена и щеше да е отрицателна между тях

което не може да има

[fresh ginkgo]

Ок, разбрах

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[formal ridge]

Can someone give me hint

[torn jolt]

ibp

[formal ridge]

ibp?

[torn jolt]

ye

[formal ridge]

Oh right by parts

I'll try that

I'm getting a huge conceptual doubt in ibp- why is t still there? 😳

It's supposed to be from 0 to 1, do I apply the limits for the first term entirely or just for e^t

I'm so confusedd

[torn jolt]

apply ibp on the second integral

[glacial pasture]

yeah in ibp when you get your uv term you apply the limits to it, we just tend to leave it to the end for convenience

[formal ridge]

So this is correct right

[glacial pasture]

yeah

[cursive condor]

Good boy

[formal ridge]

Alright I'm getting the ans thanks

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[sleek torrent]

im rlly confused with these questions, how exactly do I differentiate them? How do I know when to stop differentiating? Did I do it correctly or is there a more efficient way of solving this

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[sleek torrent]

<@&286206848099549185> if anyone can give me some tips on differentiating expressions inside of expressions that would be very much appreciated, ty

[brazen stirrup]

!15m

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[cursive condor]

Square both sides

$(f(t))^2=\sqrt{t^2-1}-\frac{1}{t^2}$

[glossy valve]

mathisfun

[cursive condor]

Then do implicit differentiation

[sleek torrent]

i can do that?

[sleek torrent]

what’s that

[brazen stirrup]

Wait

[cursive condor]

$\dv{x}y=\frac{dy}{dx}$

[glossy valve]

mathisfun

[cursive condor]

You are differentiating something with respect to another variable

[cursive condor]

So we denote it like this

i.e., we use chain rule

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

[brazen stirrup]

@sleek torrent Are you in Calc I or DfQ?

[glossy valve]

mathisfun

[sleek torrent]

idk, im in Canada so it’s just called calculus

[sleek torrent]

huh?

[cursive condor]

This is chain rule in Leibniz form

[sleek torrent]

okay i get the notation itself, but how would that help us in this question specifically?

[cursive condor]

It is the definition of implicit differentiation

[sleek torrent]

oh i see, im in calc I ig @brazen stirrup

[sleek torrent]

😭

how would i do it according to implicit differentiation?

[cursive condor]

I will demonstrate

$\dv{t}f(t)=f'(t)$

[glossy valve]

mathisfun

[cursive condor]

Do you agree?

If so, then I will proceed

[cursive condor]

This would become $2(f(t))(f'(t))=\frac{t}{\sqrt{t^2-1}}+\frac{2}{t^3}$

[glossy valve]

mathisfun

[cursive condor]

Therefore $f'(t)=\frac{1}{2f(t)}\qty(\frac{t}{\sqrt{t^2-1}}+\frac{2}{t^3})$

[glossy valve]

mathisfun

[sleek torrent]

Yes

[sleek torrent]

Oh wait i kinda understand

So you view this whole expression as 2 separate functions?

[cursive condor]

Yes

[sleek torrent]

oh okay

thank you

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[craggy tartan]

Hello, I need help to check if I done the thing right. So here is the integral (I need to change the order of integration) :

[craggy tartan]

,, \int\limits_{0}^{2}{\mathrm{d}x}\int\limits_{6-x}^{2x}{f(x,y)\mathrm{d}y}

[glossy valve]

Ulerand

[craggy tartan]

And here what I’ve got

[summer fable]

the region is right
but the sign isn't, try evaluating it with f(x, y) = 1

also, that second integral is from 4 to 6 right? not 9 to 6

[craggy tartan]

Yes, 4 to 6

[summer fable]

that's good then, just the sign issue

[craggy tartan]

So basically I need to change + to - ?

[summer fable]

the negative of all of this, yeah

[craggy tartan]

I can only know this by evaluating f(x,y) = 1?

Or I can see it in graph somehow

[summer fable]

you can kinda see it here
notice that in 0≤x≤2, 2x ≤ 6-x

but I only noticed it after I graphed it

[steel nest]

just to be sure is this multivariate calculus, and changing order of integration on a non-rectangular domain?

[craggy tartan]

Yes I guess

I also graphed it, but not sure I can see it is negative

[steel nest]

oh looked at your awnser again looks good

[craggy tartan]

Ah, you need to go from up to the bottom?

[summer fable]

oh on paper,
you can see that the bottom bound is the one on top

[craggy tartan]

Hm, yes

I think I understand it now

Thank you for your help and time

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[clever sequoia]

can anyone explain to me what really is the dot product

[clever sequoia]

is it similar to scalar projection of vector onto another vector?

and that yellow is the projection of the black vector onto the blue right?

i see

ah i see now thanks

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[naive canyon]

ss coz couldn't be btoehred to retyrpe

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[visual smelt]

help

[full forum]

Please don't occupy multiple help channels.

[visual smelt]

.

[hazy lance]

.

[visual smelt]

how do i solve this?

i can only solve homogeneous

[hazy lance]

method of undetermined coefficients

https://tutorial.math.lamar.edu/classes/de/undeterminedcoefficients.aspx
Differential Equations - Undetermined Coefficients

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[hallow walrus]

how does the second last line follow? Am I being dumb

like how did limit t -> 0 go to that

ohh because they are linear transformations

I see

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[tall ibex]

why its not making sense what i m doing wrong

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[near wing]

prove that the functions $f(n)=2n^3+n^2+6n+3$ always produces a number that is divisible by an odd number greater than 1, for any natural number, n

[glossy valve]

haku

[near wing]

so uhm

i looked at it

and i realized i needed to use a proof

then i gave up 💀

[onyx glen]

well it does say to PROVE something..

but ok, let's try poking around to see if anything comes up

[near wing]

i checked the answer sheet and you somehow do it by factoring

but idk why that works

[onyx glen]

can you calculate f(1), f(2) and f(3) for me?

[near wing]

$f(1)= 2(1)^3 + 1^2 + 6(1)+3 = 7$

[glossy valve]

haku

[onyx glen]

2+1+6+3=7?

[near wing]

wait holsd on

that doesn't seem right

oh its $12$

[glossy valve]

haku

[onyx glen]

ok, and f(2)?

[near wing]

35

and 84

[onyx glen]

,calc 2 * 8 + 4 + 6 * 2 + 3

[glossy valve]

Result:

35

[onyx glen]

ok yeah

so notice that

f(1) is divisible by 3
f(2) is divisible by 5
f(3) is divisible by 7

this may let you hazard a guess that MAYBE f(n) is divisible by 2n+1 on the level of polynomials

[near wing]

when you factor it

one of ur factors is 2n+1

but

what does that prove

[onyx glen]

so it does factor as (2n+1)(something) then?

[near wing]

yes

[onyx glen]

2n+1 is always odd and greater than 1, yes?

[near wing]

yep

because of

idk i forgot the proofs name

[onyx glen]

what name

[near wing]

idk i just heard that 2n+1 was always odd when i was researching it

and you can prove it with direct substitution and a few others i believe

[onyx glen]

do you know what an odd number is?

[near wing]

yeah yeah

[onyx glen]

ok then tell me

what is the definition of an odd number

emphasis on the word definition

you have to know the formal definition if you want to be able to prove shit

[near wing]

uhh..

damn how do i not know this

isnt it like

acc

im out

of ideas

is it a number that cant be derived from any even number

[onyx glen]

in fact, an odd number is precisely a number expressible as 2n+1 for an integer n

[near wing]

so is an even number just

n

wait no

2n

[onyx glen]

an even number is 2n yes

[near wing]

okay okay now

how do we prove this

[onyx glen]

2n+1 is always odd and greater than 1, yes?

[near wing]

yes

[onyx glen]

what exactly was your factorization for f(n)?

[near wing]

uhh

hold on

(n^2+3)(2n+1)

[onyx glen]

ok

so then you know that for any natural n, f(n) is equal to an integer (n^2+3) times the odd number 2n+1

[near wing]

yes yes

im understanding

[onyx glen]

yeah so

[near wing]

is it because

a factor of a funtion means something that can devide evenly into a the function

so (n^2+3) can divide evenly into f

and so can 2n+1

and since we know 2n+1 is a N thats always odd given our definition of n

we proved that the function is always going to be divided by 2n+1

[onyx glen]

well this is quite wordy but yes you said the same thing as i did

f(n) is divisible by (2n+1)

[near wing]

oh lol

so if i got that question what would i write on the test

cause ive never written a proof

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[torpid stone]

could someone please help me with (iii)? sending what I've done so far in the next msg

[torpid stone]

i don't even know if im going in the right direction

[vestal moat]

You are

Now keep simplifying and match the coefficients for each exponent

[torpid stone]

okay let me try

[vestal moat]

Keep going

[torpid stone]

what would i substitute though?

im not sure

i got b = -1/4

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[ruby ice]

hlw guys

[gritty rose]

!occupied

[full forum]

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).

[fathom osprey]

oh sorry I'm new, how can we open our own channels?

[flint hull]

#❓how-to-get-help

[full forum]

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[ruby ice]

chat GPT ... bahahaha

[full forum]

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[ruby ice]

...
🥲

[gritty rose]

ahh yes clear, concise question

and lots of work shown

[ruby ice]

tell yourself that our math teacher just asked us this equation and he told us you have until Monday 🥲🥲..

[ruby ice]

am french

..

[stiff musk]

there's no actual question stated here

just an equation

what is the question?

[ruby ice]

wait

[glacial pasture]

havent seen the g term before

[ruby ice]

Does it appear to be the Schrödinger equation?

he told us that

the question is :

"How does a quantum wave evolve in a nonlinear interacting system?"

[lunar veldt]

force à oit chef

[ruby ice]

c'est la fin

[hazy lance]

it seems like the answer is already given

[ruby ice]

nah

[ruby ice]

mm chat GPT il galère

[lunar veldt]

y’a une faute d’orthographe ou je rêve

[ruby ice]

ah bon ???

ouuu ??

[lunar veldt]

bah normalement c’est « ou » et pas « où »💀💀

[ruby ice]

mdrrr je vais lui faire la remarque 🤣🤣🤣

[hazy lance]

this equation describes how a quantum wave evolves over time

[ruby ice]

je vais gagner un peu daura ..🤣🤣

[lunar veldt]

résout d’abord l’équation ouais mdr

[hazy lance]

without some initial or boundary conditions theres still no real question here

[ruby ice]

OK, what do I deduce from this?
?

[lunar veldt]

bah que sa question est conne MDR

[lunar veldt]

il te demande ça mais il te décrit directement l’évolution en te donnant l’équation

[ruby ice]

c'est vrm la fin 🥲

force a oim

merci en tout cas

[lunar veldt]

résout la au pire t’auras un état quantique indexé sur des doubles sommes

et une exponentielle

[ruby ice]

ok ok merci mec t'es un bon

[lunar veldt]

mdrr force

[full forum]

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[safe hearth]

can someone expalin how i went horribly wrong?

[safe hearth]

the answer is like…26.6

but i got two different answers, my friend got a different answer and two other people got different answers

[full forum]

@safe hearth Has your question been resolved?

[brittle steeple]

,w 328/10 + 186/10

[glossy valve]

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Requested by a_mog

[brittle steeple]

when did theta go from 2sin^-1(6/10) to sin^-1(6/10)?

[full forum]

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[dark flume]

just need someone to verify if my solution is proper (rt 5, 5) (-rt5, 5)

[cursive condor]

Yeah

Evidenced by desmos

[dark flume]

alr ty

now then

[tulip marlin]

ok

now try doing it using discriminant

[cursive condor]

Yeah

[tulip marlin]

no calculus

[dark flume]

discriminant sounds mad harder

[tulip marlin]

easier

neater

[cursive condor]

Not too much

[topaz valley]

,w (x^2 + 5)/x = 2x

[glossy valve]

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Requested by tearfox

[cursive condor]

You just let y=mx-5 then subsitute

[dark flume]

what do you mean

[cursive condor]

The tangent line will have form y=mx-5

[dark flume]

oh yea it has to go through (0,-5)

[cursive condor]

So sub it into y=x^2, check when discriminant is 0

Those are your values for m

[dark flume]

ahhh

because when 0 then its not intersecting twice with 2 sol

its just touching

and thats the tangent thingy we want

[cursive condor]

Yep

[dark flume]

alr let me write this down hol on

[dark flume]

whoah that is much cleaner actually

got me to the 2rt5 gradients much faster

[topaz valley]

here you literally solve 1 quadratic

and it's like

[tulip marlin]

but calculus is sin

[topaz valley]

x^2 = 5

[tulip marlin]

dont be sinful snow

[dark flume]

sin 😭

[topaz valley]

it's fast and easy

[tulip marlin]

sinful

cosful

tanful

[dark flume]

where does this come from

[dark flume]

secful 🥶

[topaz valley]

whatever i said in the other channel

[tulip marlin]

derivative is 2x

[dark flume]

yea

[dark flume]

wait

[topaz valley]

set slope between (x, x^2) and (0, -5) equal to derivative

[dark flume]

OHHH

change in y / change in x = gradient

[tulip marlin]

now what if the slope was infinite snow

[dark flume]

so theres 3 ways then thats sick

[tulip marlin]

gonna make another question where u cant cheat

[topaz valley]

then the derivative would also be infinite

[dark flume]

which means were fucked right

[topaz valley]

no it means shuri is

[dark flume]

💀

[tulip marlin]

[topaz valley]

i am never

[cursive condor]

In the act?

[tulip marlin]

xx + yy = 1
do the same question

oh wait no no

[dark flume]

x^2 + y^2 = 1 so unit circle

[topaz valley]

it's an exercise for the reader

[tulip marlin]

xx + yy = 1
do it for the point (1, 5)

i think that works to foil snows plans

[topaz valley]

does it though

[dark flume]

one is a straight line vertical 💀

[cursive condor]

[topaz valley]

(y - 5)/(x - 1) = -x/y
y^2 - 5y = -x^2 + x
1 - 5y = x

[dark flume]

what

ah

x^2 + y^2 = 1

wait

now its gone to hell fr

[tulip marlin]

snow should think of a question snow cant solve

[topaz valley]

1 - 10y + 25y^2 = x^2
1 - 10y + 25y^2 = 1 - y^2
26y^2 - 10y = 0
y = 0 or 5/13

i think your plan has been foiled shuri

[analog shale]

but this doesn't work if y=0

[dark flume]

this definitely doesnt work when x = 1

[analog shale]

ig u just separate the case

[tulip marlin]

hes cheating, so to speak

[topaz valley]

no no you see even if i write the fraction, i actually work in CP^1

[dark flume]

🔥

whats CP^1

[topaz valley]

complex projective space

[dark flume]

bro is trying to justify it 😭

[analog shale]

high school question❌
ah, alg geo✅

[topaz valley]

slopes actually live in CP^1, i see nothing wrong with it

[tulip marlin]

[analog shale]

someone should tell eisenbud to make a new book: pre-calc 12 with a view towards algebraic geometry

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[torpid stone]

need some urgent help with (iii) in this question please 🙏

[full forum]

Please don't occupy multiple help channels.

[torpid stone]

here’s what i’ve done so far

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[hoary ember]

my first idea was to just use trig identities to break it down, but i would need sinA, sinB,, cosB, ect right? so im not so sure about that idea yet

im not sure how to utilize thst 3a^2+3b^2=5c^2 either

[simple ridge]

i am not sure if this is helpful , but the given equation tells you that the triangle with sides a\sqrt(3), b\sqrt(3) and c\sqrt(5) is a right triangle with hypotenuse of length c\sqrt(5) no ?

[bright bronze]

dumb way that I can think of is, simplify the cot expression, so it becomes sin^2C/(sinAsinBcosC), and then use cos rule to find cosC, from in find sinC, and sine rule to find the other sines

[hoary ember]

sinC/sinA=c/a
sinC/sinB=c/b
cosC=(a^2+b^2-c^2)/2ab

[bright bronze]

You can rewrite cosC as cosC=3(a^2+b^2-c^2)/6ab = c^2/3ab

[hoary ember]

$\frac{c^2}{ab}\cdot\frac{2ab}{a^2+b^2-c^2}=\frac{2c^2}{\frac{2}{3}c^2}=3$ right

[glossy valve]

skissue.in.a.teacup

[bright bronze]

yea

[hoary ember]

oh ty

.close

[full forum]

Channel closed

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Use .reopen if this was a mistake.

[full forum]

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[spring vigil]

wouldnt the bottom be

2 -2x /
2(x^2 +1 )

[fallen loom]

no

[spring vigil]

why not

[fallen loom]

hmm

[fallen loom]

why do you think it would be this

explain me your logic

[spring vigil]

because theres 2
1 + x^2

[fallen loom]

hmm i see

what about 1/3+1/3

how would you do this?

[spring vigil]

2/3

[fallen loom]

are you adding the denominators?

[amber patrol]

there is 2 of (1/1 + x^2), that 2 is in numerator

[spring vigil]

i think im mixing it up with the root ones

if i see two sqrt 1 - x^2 in the denom

wouldnt that be

2 sqrt 1- x^2

[fallen loom]

hmm can you explain further

i dont think i understand

[limber rapids]

anyone need help with school work? DM me. Civil engineering student have passed all courses

[signal lance]

does anyone know the different ways u can sketch a linear graph

[amber patrol]

occupy a different channel

this one is in use

[spring vigil]

the bottom example here

[amber patrol]

im on laptop and i cant see a thing unfortunately

[spring vigil]

i t hink i want to skip this problem for now

[spring vigil]

i want to do this

[signal lance]

oh

sorry 😭

[amber patrol]

its fine

[amber patrol]

yeah?

[spring vigil]

i tried to solve its inverse

[amber patrol]

differentiate it using chain rule

[spring vigil]

x^3 - 1 = y

[lean tangle]

do you know anything geometrically about inverses

and its function

[amber patrol]

ohhh

yeah mb

USE A DIFFERENT CHANNEL

this is occubpied

[lean tangle]

i think u put this in the wrong channel chief

[spring vigil]

is this a good direction