help-36

[storm haven]

Product rule

[sacred night]

then chain rule?

k thx

[storm haven]

Theres gonna be a lot of chain rule

[sacred night]

so which one first

i am so confused

bc if i take $d/dx (t^4 + 8t^2 - 3)^9$

[soft zealot]

calvin

[sacred night]

do i use power rule first or do i do derivative of inside

@storm haven

[storm haven]

chain rule ofc

[sacred night]

?

so inside?

[storm haven]

up to us

d/dx f(g(x)) = f'(g(x)) * g'(x)

omg

[sacred night]

so i wrote $(4t^2-9)9(4t^3+16t)^8$

[soft zealot]

calvin

[sacred night]

would that work

[storm haven]

thats wrong

[sacred night]

why

wait

mb

ill fix it

[storm haven]

lol

[sacred night]

$9(t^4 + 8t^2 -3)^8$

ughhh calculus is confusing

i don't think its rigt

right

[storm haven]

almost

the derivative of the "inside" is wrong

[sacred night]

wdym

i am sooooooooooooooooooooooooooooooooooooooooooooooooooo confused

[storm haven]

derivative of t^4 + 8t^2 -3

[sacred night]

i didn't do the derivative of the inside yet

[storm haven]

what

oh

omg bro

[sacred night]

i just applied power rule to $(t^4 + 8t^2 -3)^9$

[soft zealot]

calvin

[storm haven]

i was confused why there is 4t^2 - 9 term here

[sacred night]

uhm

[storm haven]

just remove 4t^2 - 9 first

[soft zealot]

calvin

[storm haven]

just compute this and forget abt 4t^2 - 9, that will be done seperately

[storm haven]

chain rule

[sacred night]

so derivative of inside now?

$9(t^4 + 8t^2 -3)^8 \cdot (4t^3 + 16t)$

[soft zealot]

calvin

[storm haven]

yes

[sacred night]

now i multiply by $(4t^2 - 9)$

[soft zealot]

calvin

[storm haven]

thats not how productt rule works

[sacred night]

oh wait

[storm haven]

d/dx f(x) * g(x) = f'(x) * g(x) + f(x) * g'(x)

[sacred night]

$9(4t^2 - 9)(t^4 + 8t^2 -3)^8(4t^3 + 16t)$
using product rule

[soft zealot]

calvin

[storm haven]

uhm

[sacred night]

right?

i am actually so lost

wait

[storm haven]

bro

[sacred night]

jjmoiqfnoinfefionfof\

[storm haven]

you are saying d/dx f(x) g(x) = f'(x) g(x)

[sacred night]

😭

[storm haven]

but that is wrong

[sacred night]

can you tell me what i ned to do now

after applying chain rule to the t^4 term

[storm haven]

read this

[sacred night]

ye i know that

OH

SO NOW I JUST SUB THAT BACK IN??

[storm haven]

find the derivative of 4t^2 - 9

[storm haven]

what

[sacred night]

ohhh

ok

thats 8t

[storm haven]

yep

[sacred night]

why isn't my windows key working wth

[storm haven]

now plug in f(x) , g(x), f'(x) , g'(x) into the product rule formula

[sacred night]

wait lemme restart my pc

its breakuing

ok back

is g' the whole thing that we found

using chain rule

[storm haven]

yeah

[sacred night]

so $9(4t^3+16t)(t^4+8t^2-3)^8$

[soft zealot]

calvin

[sacred night]

all of that??

[storm haven]

yeah this is the derivative of g(x)

[sacred night]

k thx

[storm haven]

problem solved?

[sacred night]

ok so i got

$(4t^2-9) \cdot (9)(t^4 + 8t^2 -3)^8(4t^3 + 16t) + (8t)(t^4 + 8t^2 -3)^9$

[soft zealot]

calvin

[storm haven]

looks all correct

!done

[final saddle]

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[sacred night]

.close

[final saddle]

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[sacred night]

alr

[final saddle]

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[storm haven]

can someone explain the things i have underlined?

[storm haven]

Melo alr tried helping me with why P(C chosen) = 1/ 2^(n-k)

but i dont rlly get it

Why i think P(C chosen) is hard to compute :
1/(2^n-1) is the probability that C is chosen on the first day. To find the probability that C is chosen (in total), we need to sum over all days.

ive tried writing out the sum and its just too ugly

Melo said that a cookie can be : chosen or a superset of it is chosen

P( C chosen) + P(C 's superset chosen) = 1

The logic behind 1/2^(n-k) is really faulty in my opinion

[storm haven]

cause this is true over all days. Not just 1 day

you cant just say P(C chosen on first day) + P(C's superset chosen on first day) = 1

[storm haven]

but this seems to be what the book is saying.

and how did they apply binomial theorem?? I dont see how they used it there

Ping me to reply guys

[final saddle]

@storm haven Has your question been resolved?

[floral nova]

It would probably help if you wrote some basic examples to convince yourself the formula makes sense.

[storm haven]

I want to understand the logic tho

[floral nova]

Take n = 2, then the subsets are {1,2,{1,2}}. Then convince yourself that P(Cookie {1,2}) = 1 etc

[storm haven]

Okkk

[floral nova]

Yeah and if you can't convince yourself when n = 2 then you can't convince yourself in the general case

[storm haven]

Well, is there any way other than just looking at examples?

[floral nova]

Math is built on starting with examples to build intuition then use that intuition to develop more complex things. I don't know why you are so reluctant and want to skip ahead.

[storm haven]

Chill I know what math is bro...

Js asking if there's a faster method 🙄

My examples are not exampling

[severe verge]

the sum doesnt include k=0

which is why you have that -(1/2)^n at the end

[storm haven]

I want to know why P(C chosen) is what it is

Why is it 1/2^(n-|C|)?

[severe verge]

oh did you already get the green arrow?

[storm haven]

No

[severe verge]

I'm explaining that green arrow

[storm haven]

Oh ok👌

[severe verge]

continue from this

you can rewrite the general term as $\frac {nCk \cdot 2^k \cdot 1^(n-k)}{2^n}$

[soft zealot]

DeC∆rbonizeD

[severe verge]

wtf

[storm haven]

What

Why is there a 1•(n-k)

[severe verge]

1^(n-k) I meant

[storm haven]

Why is that there tho

[severe verge]

it is basically just 1

[storm haven]

It's just =1

[severe verge]

if you sum up all the terms then it's just $\frac {(2+1)^n}{2^n}$

[soft zealot]

DeC∆rbonizeD

[storm haven]

I don't see binomial theorem anywhere near

[severe verge]

you see it now?

[storm haven]

Why??

[storm haven]

I don't get why

[severe verge]

try expanding (2+1)^n using binomial theorem

[storm haven]

Wait

Sum nCk 2^(n-k) •1^k

Oh that's shmart

Ok I see it

[severe verge]

the other part... I haven't even read the question lol

[storm haven]

It's so confusing

[storm haven]

Wait what?

Why is there -(1/2) ^n

I don't get it

Oh I get it

Nvm

[severe verge]

uhh I don't even understand the problem

[storm haven]

Wdym

[severe verge]

so you have 2^n-1 cookies

[storm haven]

Yes

[severe verge]

and you have n sets right?

[storm haven]

Wdym

We have 2^n - 1 sets

[thin cloud]

The question is there are 2^n-1 subsets and label them for each cookie in 2^n-1 cookies. When eat a cookie we also eat all the cookies labeled its subset

[severe verge]

ahhh I get it now

[storm haven]

Ahem, why on Earth is P(C chosen) = 1/2^(n-|C|)????!

[severe verge]

I gtg

@thin cloud Cho e thử hiểu bài này

thần đồng đất việt

[thin cloud]

Hoi này

[severe verge]

a biến đây

[thin cloud]

|C| là gì

Kí hiệu lạ chắc t.a viết khác

[severe verge]

C chỉ là số mà bánh quy đó đc đặt

[storm haven]

I don't think so

C is the number of elements of the label of the cookie

Or my Google Translate is dumb

<@&286206848099549185>

Ping me to reply

[severe verge]

@storm haven can you check me on this

[storm haven]

Yuh

[severe verge]

so suppose I choose a cookie with C elements

[storm haven]

Yes

[severe verge]

then I have to also eat all the other cookies that are the subset of that set right?

[storm haven]

Yes

[severe verge]

Which means I have to eat 2^C-1 cookies

[storm haven]

Yep

[severe verge]

so the probability of those cookies being chosen is (2^C-1)/(2^n-1)

which is roughly 1/2^(n-C)

Am I getting sth wrong here

[storm haven]

What if there are less than 2^n -1 cookies?

I.e. some cookies are alr eaten?

That's the main problem....

Tbh I should just open this tmrw

I appreciate ur help bro

But I gtg

.close

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[severe verge]

alrighty

[thin cloud]

I also thought of that but that doesn't make sense

[floral nova]

This is why writing down some examples are helpful. The probability that cookie C is chosen is the probability that C is chosen before any of its super set elements are selected.

There are 2^(n-|C|) super sets, and only 1 cookie C. So the probability is 1 divided by 2^(n-|C|)

[floral nova]

e.g., as I've already shown here. the cookie {2,1} has no super set, so the probability of it being chosen is 1.

[severe verge]

ahh got it

[floral nova]

This has been an hour of going in circles over a reluctance to study the problem and understand it when n = 2.

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[errant basin]

ok another one. for two values x and y, you compute the arithmetic mean and geometric mean, and do that again for the new two values. what will they converge to? what if n points?

[errant basin]

first answer first question will be quite enough

[terse crypt]

Wdym ‘new two values’

[errant basin]

the am and gm

[terse crypt]

Oh

[tired walrus]

this has a special name

[errant basin]

really

[tired walrus]

https://en.wikipedia.org/wiki/Arithmetic–geometric_mean?wprov=sfla1

[errant basin]

oh.

.close

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[tranquil pine]

Integrate 2x+3/5x^2 + 1

[tranquil pine]

With ul 1 and ll 0

While changing dx to du

[uncut flint]

use quotient rule?

[tranquil pine]

Wha is that

[uncut flint]

oh wait

why am i thinking of differentiation

[tranquil pine]

Shudnt we change the limits as well

[terse crypt]

split into 2 fracs

do u sub for 1

arctan for the other

[edgy mauve]

1/(a²+x²)

[uncut flint]

yea 2 frac

[tranquil pine]

Yes then then dx becomes du

[uncut flint]

after substitution yeah

[terse crypt]

this

[final tangle]

do you mean
$$\int_0^1 \frac{2x+3}{5x^2+1} \dd{x}$$

[soft zealot]

ℝαμOmeganato5

[uncut flint]

you find u then you differentiation find dx replace and stuff

[uncut flint]

i think he means that yeah

[tranquil pine]

But if we were to change by substitution, shudnt the limit change as well

[uncut flint]

no..?

i think

[tranquil pine]

Bruh how

[final tangle]

what have you done so far

you shouldn't be immediate subbing for this

[tranquil pine]

I got ans without changing the limits

[final tangle]

depends on personal preference

if you don't update the limits as you go
you'll need to be very careful with your notation

for what's written to be valid

[tranquil pine]

I dont understand

Well i have smth for u

[final tangle]

can you take a pic of what you did

[tranquil pine]

Umm ye i have to write again

Mines always messy and random

,rccw

[tranquil pine]

I dont know why i didnt change limits but the ans is correct

[final tangle]

what did you do after

[tranquil pine]

Theres a whole another exercise where i have to sub u and change limits, i aint getting ans without changing limits in the next exercise

[final tangle]

multiple errors may cancel each other out

[tranquil pine]

After using du/ u then i got 1/5 log u

Then sub u = 5x2+1

Then i got 1/5 log 6

[final tangle]

ok, so this is a notation issue

[tranquil pine]

So i used wrong notations?

[final tangle]

unless explicitly stated, the bounds of the integral are associated with the differential present

if you want to keep the bounds as 0 and 1, after sub with u
you'd need to explicitly write x=0, x=1 instead of just 0 and 1

[tranquil pine]

I dont study that deep

Go ahead with surface level

[edgy mauve]

This is still surface level-

[tranquil pine]

Cuz ncert is trash w/e

[tranquil pine]

Oh💀😭

[edgy mauve]

What he means is when u substitute u can change the limits to the variable u used to substitute

It reduces steps

[tranquil pine]

I dont get it

Can u solve me this question

[final tangle]

that wasn't really the point i was making

[tranquil pine]

Integrate x/(x^2+1)

Ul 1 and ll 0

[edgy mauve]

Oh?

[final tangle]

while i do find it more efficient to update the bounds as you do the sub,
the point i'm trying to make here was in regards to the notation issue

[tranquil pine]

So pls tell what does a notation issue mean

[final tangle]

if you don't want to update the bounds, then you'd need to write
$$\frac15 \int_{\red{x=}0}^{\red{x=}1} \frac{\dd{u}}{u}$$

[soft zealot]

ℝαμOmeganato5

[tranquil pine]

Ok, so the ans is still right?

[edgy mauve]

Eyy i forgot rhe coefficient

Wait

[tranquil pine]

U gettin ans without changing huh

[final tangle]

i would describe a notation issue as something that resembles something close to what's supposed to happen,
but violates/conflicts with mathematical laws in some way

[edgy mauve]

I did change lemme be a lil clear

[tranquil pine]

So like its kinda right, but technically wrong

[final tangle]

yes

like writing \
$=x+1^2 \
= x^2 + 2x + 1$ \
would be an example of a notation issue

[soft zealot]

ℝαμOmeganato5

[tranquil pine]

Well i got the ans without changing the limits

I get it now

Its because i bring back the u in terms of x, so x limits work when its in x

[final tangle]

its an acceptable method and depending on how strict the teacher is, you may lose marks

if you skimp on notation

yes

[tranquil pine]

I see

So which is ezier

[edgy mauve]

It's a step less

[tranquil pine]

Changing the limits or no.)?

[edgy mauve]

Changing i would say

[final tangle]

up to personal preference

what's easier is whatever you find more comfortable

though i would personally recommend updating bounds as you go

[tranquil pine]

Im lazy as hell

Ig ill choose to update

Thanks a lot (both)

.close

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[gusty sedge]

what is the difference bewteen the jacobian matrix and the gradient?

[lean magnet]

the jacobian of a scalar valued function is the gradient

[ripe jewel]

Jacobian is a square matrix

but same principle

[lean magnet]

gradients are typically only defined for scalar valued functions

[gusty sedge]

OOhh

okay that makes a lot of sense

[ripe jewel]

but yeah the fundamental idea behind it is the same

[lean magnet]

i like to think of gradients and jacobians as analogous just different use cases

[ripe jewel]

did you take linear algebra?

[gusty sedge]

yeah i did

[ripe jewel]

ythere's another way to view the gradient

[gusty sedge]

the steepest ascent?

[ripe jewel]

holding look up which one is vector which is covector to not make a mistake

okay the gradient is the vector associated with the covector df, the total differential

is a different way to view it

less intuitive, but interesting

[gusty sedge]

nice i didnt know that

wow

but i have a question though..

so if i had a function that takes vectors from R^2 and spits out vectors in R^2. in order to differentiate the vector valued function, i will need to take the jacobian. which in this case will be a 2x2 matrix right?

[ripe jewel]

yes

[gusty sedge]

okay perfect

[ripe jewel]

however be careful

very often when people say "Jacobian" they mean the determinant of the Jacobian matrix

so you should say "Jacobian matrix" if it's not clear whethe ryou mean the determinant or the matrix itself

[gusty sedge]

what is teh determinant of the jacobian matrix?

what does it say about the function?

[ripe jewel]

represents how much the area/volume/whatever differential element gets shrunk or grown

which makes sense, because the determinant of any matrix cvan be thought of as the volume of a parallelipiped

[gusty sedge]

yeah i just realized

now this is much more manageable

thank you for your time

[ripe jewel]

np

[gusty sedge]

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[topaz charm]

So its my first day in this math course for community college (online) and after watching 3 short videos on how to do linear inequalities I started the first homework. The first question looks nothing like anything in the videos, theres an "or" in there and there's a "|" which is usually used for absolute value but its just open with no other "|" to close it. I'm pretty bad at math and I'm just taking this class because I need it. Thanks.

[stuck juniper]

| here does not mean

absolute value

Do u know sets?

Or means

It can be this

Or that

And means

It has to be both

[topaz charm]

nah

[tulip coyote]

To actually write something clear, the | means "such that", the set is basically saying "all values x such that x <= 5 or x >= 2"

[topaz charm]

oh alr

so its just saying thats what x is, it could be either of the 2

or no, it is

[stuck juniper]

No it says

[topaz charm]

using those 2 i find what x is

[tulip coyote]

Yep, x can either be less than or equal to -5, or it could be greater than or equal to 2

[stuck juniper]

X is more than equal to 2

[topaz charm]

x <= -5
x >= 2

ohhh

yo when you broke it down for me

i understand it now

thanks man

[final saddle]

@topaz charm Has your question been resolved?

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[soft zealot]

vegeta

[fossil flume]

How to approach using de moivres theorem?

[final saddle]

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[tired walrus]

!msgdel

[final saddle]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[outer knot]

hi

i am new !.......

[sullen gale]

Hello, if you need help, check out #❓how-to-get-help , if you just want to chat, you can go to #360643390594875392.

[sullen gale]

Welcome, by the way! But please do not comment in occupied channels.

[outer knot]

so you can expilane algebra

[sullen gale]

For this, you can check out #prealg-and-algebra .

[outer knot]

no i am in std 9th

[sullen gale]

If you need someone to help you with algebra (no matter your grade), you should be going to #prealg-and-algebra . If you have a specific math problem, see how to open your own help channel in #❓how-to-get-help.

[final saddle]

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[rough moss]

A

[rough moss]

I’m confused on how come I’m not subtracting one point from the other and that’s my direction vector? How did they get their answer?

[bleak granite]

likely scaled the direction vector so that all the components are integer values? that's my guess

,calc 3*(5)-(-2)

[soft zealot]

Result:

17

[bleak granite]

yeah that's probably it :), so when the subtraction occurs we have a fraction, so multiply each component by 3

the direction doesn't change, but integers in the components make everything nicer i suppose

[rough moss]

Nvm they js simplified

Yea

Was looking at it again n js game back

Thanks

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[rough moss]

A

[rough moss]

Does it matter which one I subtract from which or are they the same vector js in opposite directions

Like my answers had opposite signs for the t coeff

-17,11,9

[bleak granite]

yeah that'd just a sign difference

cause B-A=-(A-B)

[rough moss]

so it’s still right

js the opp correction

direction

[bleak granite]

[rough moss]

K thanks

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[rough moss]

How come for #19 and #21 I use both vectors as my coeff for t even though one problem says it’s parallel and another says it’s perpendicular. Having a hard time visualizing it and understanding why if someone could explain or like draw it out please and thanks.

@ me if u reply

[final saddle]

@rough moss Has your question been resolved?

[rough moss]

No

[rough moss]

Are the coefficients of any equations the normal vector

And since in 21 I was given the vector its not the normal vector its the direction one

[rough moss]

I also have another question

I need to find the equation of the plane for this

[rough moss]

41 btw

But

So

If im finding the plane that’s perpendicular to n

Yea bro idk what to do ts is confusing the fuck out of me

I need help 😭

<@&286206848099549185>

For both questions

[sage osprey]

For 41, I believe we should start by trying to find all points (x,y,z) such that the vector with tip at (x,y,z) and tail at (3,2,2) is perpendicular to the vector 2i+3j-k

[rough moss]

So the tail of the vector n should be at 3,2,2?

and I need to find the points at the tip?

[sage osprey]

Not vector n, but some new vector that should be perpendicular to vector n

[rough moss]

Could you draw it out for me 😭

I’m so lost

I swear it made sense in class but now it just doesn’t

https://youtu.be/2sZKZHyaQJ8?si=KhtWQrvSPgL2NhZx

Should I watch this

[sage osprey]

Sure

Probably something like this ( I can't draw)

[rough moss]

is the vector like level with the plane?

[sage osprey]

Yes

[rough moss]

And I want to find the vector?

Well first

[sage osprey]

In terms of x,y and z

[rough moss]

And then the plane

[rough moss]

How 😭

I can’t do dot product can I and set it equal to 0

[sage osprey]

Subtract the tail point from the tip point

[rough moss]

So x-3,y-2, etc

[sage osprey]

It's probably like in the video you linked

[rough moss]

Oops

Other way around?

[sage osprey]

This is correct

[rough moss]

Oh

Yea

ok

So

what’s that do though

[sage osprey]

The dot product of that vector and vector n should be 0

[rough moss]

How do I get x y and z tho 😭

[sage osprey]

So we have our vector (x-3,y-2,z-2) and (2,3,-1)

[rough moss]

yes

and if I do the dot product don’t I get (2x-6)+(3y-6)+(-z+2)=0?

[sage osprey]

yes

you basically have the answer at this point

[rough moss]

how 😭

DO I PLUG IN THE POINT!

Or no

im dumb

That’d make everything 0

😭😭😭

[sage osprey]

we already have (2x-6)+(3y-6)+(-z+2)=0

[rough moss]

Yea

Should I move the constant over or nah

[sage osprey]

yes, move the constants onto the right side

[rough moss]

2x+3y-z=10

wait is that it

YO WHAY

HOWD I GET THAT

😭

i don’t need to like actually find what the vector equals

or what it is

[sage osprey]

not really

[rough moss]

Are you like

A genius

[sage osprey]

not really

[rough moss]

Are you sure

[sage osprey]

yes

[rough moss]

Can I friend you and ask you questions later bc I like Just started this course and I’m alr struggling 3 days in 😭

And you’re like Lagrange reborn

[sage osprey]

i might not be available all the time though

[rough moss]

that’s okay

Any help is better than none

😭

[final saddle]

@rough moss Has your question been resolved?

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[kindred sun]

Hi

[kindred sun]

howdo i get inverse of 3x3 and 4x4 matrices with the inverse next to the matrix

[rough moss]

Wdym the inverse next to the matrix

Oh

U mean like

idk how to code it with the bot I’ll write it down

[kindred sun]

ok

like identity is next to the matrix i nede to invert

[rough moss]

Yea yea

[kindred sun]

yea idk how to apply the changes to the right side properly

[rough moss]

Ignore the top but you have your 3x3 and then you add the 100,010,001 I believe right

Then all u gotta do is make ur original matrix that 100,010,001

[kindred sun]

yea

i just can not seem to get the rightside correct

i cant apply the changes correctly

and i dont even know how to do it on a 4x4

[rough moss]

for a 4x4 I think you just add another row

And have it be

1000,0100,0010,0001

[coarse hull]

its the same process for 4x4

[rough moss]

You want to make the left part/side of the matrix into this

[rough moss]

Or this

[dense badge]

https://homepages.math.uic.edu/~bpower6/math160/math160-2-5.pdf

some worked examples

[rough moss]

once u make the left side match the uh wtv that’s called I forget the right side will be what ur inverse is

[kindred sun]

\textbf B = \left[\begin{array}{rrr}2 & 1 & 2 \ 1 & 2 & 2 \ 2 & 0 & 0\end{array}\right]

[soft zealot]

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

[kindred sun]

yea this is an example oen

like first step can i swap r1 with r2

i would have to do same on right side aswell right

[coarse hull]

yep

[kindred sun]

but then once i do that i need to get 0s so i do -2r1 + r2

[rough moss]

Yea it’s basically all one matrix

[kindred sun]

and then that 2 turns into a 0

but then what do i write onto theother side

[rough moss]

U do the same

[kindred sun]

i switch that 1 for a -2?

[rough moss]

one sec I’ll try to do it out rq

[kindred sun]

ok

[rough moss]

U just add it like normal

And once ur done the 3 right columns will be ur inverse

[dapper hull]

,rccw

[dapper hull]

please write the matrix symbol this is a little unreadable lol

[rough moss]

lmao

mb

[kindred sun]

Ok i think i get it

give me an example question

[rough moss]

Finish the one I started

I’ll do it too

[kindred sun]

Ok

[coarse hull]

matrices without the brackets 🥀

[dense badge]

tbh it's efficient for computing ever so faster

[rough moss]

I was too lazy to draw them lol

[dense badge]

but it can be harder to read

[coarse hull]

also like to draw a line in between the given matrix and identity matrix

[rough moss]

Yea I used to do that

Then I got too lazy to

And my precalc teacher hated me

I drew the brackets at least back then

||0,0,1/2
-1,1,1/2
1,-1/2,-3/4||

watch it be wrong

that’d be horrible

[magic coyote]

My solutions be like

[kindred sun]

ok i got it

lt me send

[remote portal]

i do not understand anything

[remote portal]

how

[magic coyote]

It's not your question lel

Open a new help channel

[kindred sun]

shit man

i got 2 of them wrong

where did i mess up

in mywork

[rough moss]

I got no idea sorry I erased all my work 😭😭

Lemme try to check

[kindred sun]

lol

[rough moss]

When u swapped u js wrote it wrong

lol

R3 and R2

lol

Wait

Nvm

[kindred sun]

\textbf A = \left[\begin{array}{rrr}1 & 0 & 1 \ 1 & 2 & 2 \ 1 & 2 & 1\end{array}\right]

[soft zealot]

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

[kindred sun]

let me try this one

[rough moss]

Yea mb idk lmao

I gtg do my hw tho so

[kindred sun]

np

[rough moss]

Gl

😭😭

[kindred sun]

aight

cya

[rough moss]

Cya

[kindred sun]

i got the answer

let me check

its fucking wrong

im gona tweak

[rough moss]

💀

same

what u know abt vectors

[kindred sun]

i alr did them last unit

bro thats the reason i failed my math test

cuz of a stupid 4 day break

i didnt study shit cuz i wasnt home

and then on test day i droped a 29.8/100

i have no choice butto get 100s on every test to get a B

i have like 3-4 more test

then my max grade can be like an 84 with all 100s

[deep condor]

you didnt do the last row operation to all the entries in that row

so that might be something to work on

[final saddle]

@kindred sun Has your question been resolved?

[final saddle]

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[dim oasis]

I need help with this question like idk how to solve it and AI says to integrate both sides but like by what?? im pretty sure u can't just put an integral sign on both sides when the left is by dx and the right is by dt.

[edgy mauve]

V(x) refers to velocity

Which means it's dx/dt

Where x is the displacement

I think it's allowed to put integrals kn both sides even if the variables r different

[worn knot]

can't you just integrate the velocity function?

you don't need separation of variables

[final saddle]

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[plucky vigil]

Hi

[pliant shore]

!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! See https://dontasktoask.com/

[sullen gale]

Hello, do you have a question? If you don't, kindly type .close so that other people can use this channel. If you have a question, please post it here. If you are looking to chat generally, go to #360643390594875392.

[final saddle]

@plucky vigil Has your question been resolved?

[vital crag]

.close

[final saddle]

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[odd urchin]

I keep getting confused on the chain rule and when to use it. My calc class is going over implicit differentiation and it seems like everything uses the chain rule. I got the product and quotient rule just fine, but now for some reason this section is super confusing.

[mint orbit]

chain rule is for function composition

[mint orbit]

do you know what composition is?

like a function inside a function

$\sin \qty( \sqrt x )$

[soft zealot]

jan Niku

[mint orbit]

here square root is composed with sine

@odd urchin 👀

[odd urchin]

Yes

But what if it was a 2x instead of just an x...would that be chain rule or product rule or both? Or 2xy in the square root?

[mint orbit]

you mean like $\dv x \qty( \sqrt{2xy} )$?

sorry

[odd urchin]

Yeah, something like that for example. As these get more complicated, I lose track of what rule to use, how to tell, and if I have to use multiple, what order to do them in

[soft zealot]

jan Niku

[mint orbit]

they describe chain rule like peeling an onion

[mint orbit]

you can do anything which is helpful

for me, I liked to rename functions that made me confused, because I like the symbolic process

like I would define $S(x) = \sqrt x$ here

[soft zealot]

jan Niku

[mint orbit]

then do $\dv x \qty( S[2xy] )$

[soft zealot]

jan Niku

[mint orbit]

you can abuse notation, too

@odd urchin do you feel like you understand when chain rule, product rule, etc are applied individually?

like just given some situation where only one was required, do you think youd reliably identify it?

[odd urchin]

Product rule yes. Chain rule no.

[rapid sky]

You look at what's done last. Like in that example, square root is done last, so you handle it first. It's a function, and you have something other than x inside, so you use the chain rule.

[odd urchin]

If only one is required, I might be able to identify chain rule. When it's tougher, probably not

[mint orbit]

Chain rule is just for functions inside functions

if the input to a function is another function, then you need the chain rule

[mint orbit]

like here, can you identify the function composition?

[odd urchin]

like an example of a chain rule question that I'm not sure of what to do is y=cos(a^3 + x^3)

[mint orbit]

sure

[odd urchin]

I know cos (x) is the outside function

[mint orbit]

$\dv x \qty[ \cos (a^3+x^3) ]$

[soft zealot]

jan Niku

[mint orbit]

yea, thats right

[odd urchin]

the inside is a^3 + x^3

[mint orbit]

yup

[odd urchin]

but taking the derivative of a^3 I'd need to use the implicit differentiation, right?

[mint orbit]

is a a function of x?

[odd urchin]

y

[mint orbit]

a is a function of y?

[odd urchin]

I think so, yes

because it's y=

[mint orbit]

oh

[odd urchin]

right?

[mint orbit]

well in the case that you have y=

we usually think of y as a function of x

at least here it's true

[odd urchin]

y= on the left side

[mint orbit]

so when you have something like this

y = (a bunch of stuff with only x, and not y)

we'd call that an explicit function

you want the value of y, you plug in x on the other side

this is compared to an implicit function, say something like $\sqrt{ y(x) } + x^2 = 4$

[soft zealot]

jan Niku

[mint orbit]

when you have an implicit function, you need implicit differentiation

[mint orbit]

this is an explicit function, so normal differentiation is fine

does that help

if its hard to isolate y = ..., you need implicit

[odd urchin]

Okay, that makes sense on the explicit versus implicit...I'm gonna try to use the chain rule on this example I sent earlier and see what I get. Gimme one second and I'll show you what I get

[mint orbit]

on the cosine? alright

[odd urchin]

y=-sin(a^3 + x^3) (3a^2 + 3x^2)

[rapid sky]

Almost. a is considered a constant. What's the derivative of a constant?

[odd urchin]

How do you know a is a constant and x isn't?

Derivative of a constant is 1

[rapid sky]

It's just the normal way of doing things.

[odd urchin]

I mean 0

derivative of x is 1

[rapid sky]

x, y, and z are frequently variables.

m, n can be variables.

But like a, b, c, k are constants.

[odd urchin]

okay, let me retry with that in mind and see what I get

[rapid sky]

OK.

[odd urchin]

y=-sin(3x^2)(3x^2)

[mint orbit]

not quite

but close

[rapid sky]

No, remember that you do -sin(leave the stuff inside the same).

[mint orbit]

the mnemonic is, derivative of outside, times derivative of the inside

[odd urchin]

ohhhhh wait I think I see what I did wrontg

I need to keep the a^3 inside of the sin, right?

it is only 0 on the second set of parenthesis because we are differentiating it

[mint orbit]

yea, since it wouldnt be differentiated there

yup

[odd urchin]

so it is y=-sin(a^3+x^3)(3x^2) ?

[rapid sky]

Also, a small mistake is you have y =, but it should be y' = because now you have the derivative of y.

[odd urchin]

ahh yeah that's right

[mint orbit]

minor mistake

that looks right

[odd urchin]

😮

[mint orbit]

you wanna try an implicit one

[odd urchin]

Yeah, I've got one for that, too

[mint orbit]

sure

[odd urchin]

or wait actually this is a related rates question, i'm not sure if it is implicit or not yet

[mint orbit]

they can be implicit, yea

a lot of times they involve the chain rule

hard to say without seeing the problem though yea

[odd urchin]

#27

[mint orbit]

How do you feel about starting

[odd urchin]

First I think we should list what we know already

Which for me, isn't really anything lol

[mint orbit]

Well, we have the flux, right

[odd urchin]

It's usually something that I draw, but I'm not sure what to draw

we do know the flux formula

[mint orbit]

Do you understand the geometry of the problem?

what they're describing

[odd urchin]

I imagine a blood vessel as a cylindrical shape, no?

[mint orbit]

yea, sure, like a long tube

circular shape

[odd urchin]

does that mean I need the volume of a cylinder formula or something?

[mint orbit]

but were not talking about the whole cylinder

nah

[odd urchin]

then I am lost

[mint orbit]

just because its an application question

you notice the flux formula

they give you two things, one is k, they dont bother defining it

[odd urchin]

and radius

[mint orbit]

the other is R, and its given as the radius of the blood vessel

[mint orbit]

where is radius measured here?

which measurement is the angioplasty increasing

[odd urchin]

the halfway point between the yellow-ish parts of the vessel?

[mint orbit]

yea, its really like, from the center of the blood vessel, to the outside

[odd urchin]

It's an up and down type of measurement as opposed to side to side?

[mint orbit]

you can measure it any way you want, actually

Here's a piece of our blood vessel

You'll notice height is conspicuously missing from your flux formula

it must be talking about only one tiny slice

a circular cross section

make sense?

[odd urchin]

so it's just a 2D circle?

[mint orbit]

in our model, yea

so we're imagining something like this

and we think of the blood as flowing either into our out of the page

[odd urchin]

and we are looking at how the volume changes as the radius increases?

[mint orbit]

how much the flow changes as the radius changes, yea

[odd urchin]

so basically how much more flow we get with a 5% radius increase

[mint orbit]

They ask you a handful of questions

"Show that the relative change in F is about four times the relative change in R"

so, if we change R a little bit, we hope that F changes about 4 times as much

Can you see the function they're talking about

they do tell you $F = kR^2$, but whats a function of what

[odd urchin]

F=kR^4

[soft zealot]

jan Niku

[mint orbit]

we want to differentiate the equation, with respect to what

[odd urchin]

R is a function of F

so respect to F

[mint orbit]

well what parameter are we imagining varying

i should say variable

the idea with a function is you control one thing, and then the function tells you how something else changes

[odd urchin]

well I assume k is a constant, so it cannot be that

[mint orbit]

you dont change the something else, it just reacts to the thing you can change

[odd urchin]

and we are changing both F and R

[mint orbit]

when we do the angioplasty, we're inflating the vessel, growing the radius

we hope this has an impact on the flux, to restore a more normal blood flow

so we control the radius, and the flux changes as a result

[odd urchin]

ohhhhh

I see

[mint orbit]

This means $F(R) = kR^4$

[soft zealot]

jan Niku

[mint orbit]

which maybe makes it more clear what to differentiate with respect to what

well, we have to differentiate both sides

but we usually do it with respect to the variable we can control

[odd urchin]

So since this is with respect to R, we do implicit differentiation on F?

[mint orbit]

is the function implicit or explicit

[odd urchin]

so the derivative of F is equal to 1 (dF/dR)?

[odd urchin]

ah well we could separate them, so it would just be explicit

[mint orbit]

they are separated, yea

maybe we change the variables for a moment

$y(x) = kx^4$

[soft zealot]

jan Niku

[mint orbit]

looks separated

[odd urchin]

yes

[mint orbit]

so, we can just differentiatie both sides

$\dv R F = \dv R \qty( k R^4 )$

[soft zealot]

jan Niku

[mint orbit]

can you apply the derivative to both sides? whats the resulting equation?

(btw k is just some constant, that's another one thats common used to mean "some number")

[odd urchin]

so the derivative of F would just be 1?

[mint orbit]

well F is a function of R

just like if we were doing $\dv x \qty[ y(x) ] = y'(x)$

[soft zealot]

jan Niku

[mint orbit]

hopefully you've seen that notation before

or we can just write $\dv{y}{x}$

[soft zealot]

jan Niku

[odd urchin]

but x = R here?

[mint orbit]

it does, yea

so $\dv R \qty[ F(R) ]$

[odd urchin]

so it would be 1 (R)

[soft zealot]

jan Niku

[mint orbit]

unfortunately, no

[odd urchin]

huhhhhhh

but this is explicit, right?

[mint orbit]

it is, yea

[odd urchin]

So I don't need to do anything aside from take the derivative of F, which is 1

[mint orbit]

the derivative of F isnt 1, though

the derivative of F with respect to R is another function

we usually call it $F'(R)$

[soft zealot]

jan Niku

[mint orbit]

or $\dv{F}{R}$

[soft zealot]

jan Niku

[odd urchin]

I thought you only did the "with respect to" something if it is implicit?

[mint orbit]

well every derivative is with respect to something

[odd urchin]

I have no idea what to do now

[mint orbit]

the special thing about an explicit function is that you don't need to collect a bunch of derivatives afterwards

like if we have $y^2 + x^2 = 1$ then differentiation gives us $2yy'+2x=0$

[soft zealot]

jan Niku

[mint orbit]

its not important at the moment to be able to track that result fully out

but you should notice that the derivative is an implicit function here

just like we started with something that wasnt y = ....

we ended up with something that wasnt y' = ...

if you differentiate an explicit function, you get an explicit derivative

y(x) = something

then y'(x) = something

idk if that helps

[odd urchin]

I'm not sure what this is getting at for F here

[mint orbit]

we started with $F(R)$

[soft zealot]

jan Niku

[mint orbit]

that tells us, for some radius, how much blood is flowing through that circle over some time

if we plug in different R, we get different flux

if you differentiate F with respect to R, you get a derivative

F'(R) tells you how the flow changes as you change the radius

[odd urchin]

I'm with you so far

[mint orbit]

to get F'(R), we just differentiate our explicit function

$F(R) = kR^4$

[soft zealot]

jan Niku

[mint orbit]

so $F'(R) = \dv R \qty( kR^4)$

[soft zealot]

jan Niku

[mint orbit]

can you do the derivative

[odd urchin]

well the derivative uses the product rule on the right side

I still have no idea what the left side would be if not 1

I'm sorry, I'm not quite sure what it is I'm missing here 😢

[mint orbit]

its just a function

you know how if you have like

$y(x) = x^2$ then $y'(x) = 2x$

[soft zealot]

jan Niku

[mint orbit]

y' doesn't evaluate to anything, if you dont plug in an x

[odd urchin]

So I don't need to take the derivative of F at all?

[mint orbit]

no, you do

but its a function

so it goes from F to F'

@rapid sky you still around

my sleeping pills are kicking in

[odd urchin]

But we're trying to see how one side's rate of change effects the other, so we take the derivative of both sides?

[mint orbit]

we do, yea

we started with $F(R) = kR^4$

[soft zealot]

jan Niku

[odd urchin]

so the derivative of the right side is 4kR^3

[mint orbit]

then $\dv R F = \dv R \qty( kR^4)$

[soft zealot]

jan Niku

[mint orbit]

so $\dv{F}{R} = 4kR^3$

[soft zealot]

jan Niku

[odd urchin]

and the derivative of F would just be 1 or it would be 1(dF/dR)

[mint orbit]

the 1 is worrying me

its just dF/dR

[odd urchin]

yeah bc 1 x dF/dR = dF/dR

[mint orbit]

i guess the 1 makes me nervous where its coming from, but sure

[odd urchin]

derivative of F = 1 but since it's respect to R, it would be 1 x dF/dR

[mint orbit]

derivative of F is not 1

its F'

I'm sorry to do this but I have to go to bed

I asked a couple places to see if someoe else can come help

[odd urchin]

thank you!

is anyone here that can help out?

😦

[ornate knot]

whats ur question again?

[odd urchin]

27

[ornate knot]

and where are u at ?

like have u started it and gotten stuck or smthing

cuz it seems u guys started the question

[odd urchin]

I got confused as to when to implicity differentiate versus explicitly, but I used chat GPT to explain some of it to me. Why don't we start with a new question from scratch instead?

[ornate knot]

ok sure

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[ornate knot]

btw

[odd urchin]

mm I see

[ornate knot]

well , first id just rewrite the word problem into equations

and maybe draw a diagram if u want , but not necessary anyway

[odd urchin]

so it is a right triangle

and we know the bottom side of the triangle is 3m

I'm not too sure how the velocities help for this

[ornate knot]

not really , its just the distance at that instant , but yeah we will consider its 3m

[ornate knot]

they are not really velocities , top of the ladder slides down a vertical wall at a rate of .15 can be rewritten as dy/dt = 0.15

cuz our vetical distance (which is usually called y-axis) is changing with respect to time , and thats what derivatives are about

[odd urchin]

ohhh and dx/dt would be 0.2

so the bottom side of the triangle is x, 3 is just a point for x

and x changes at a rate of 0.2 m/s

[ornate knot]

no no , thats not what i meant

the bottom of triangle is 3 at the instant when dx/dt = 0.2 , its changing

like after a second , our bottom would be 3+0.2 = 3.2 , and so on

but for now we will consider x = 3 m

and we need to find length of the ladder (which is just the hypotenuse of the triangle basically) , any equation / theorem comes to find that relates all these info?

[odd urchin]

so we can just call that side h for simplicity

pythagorean theorem

so h^2 = y^2 + x^2

[odd urchin]

how would you write this in a formula?

dx/dt = 3 + 0.2(t)?

[ornate knot]

i was just explaining that , x = 3 and dx/dt=0.2 , nothing fancy

cuz we are calculating at the instant x=3 , so we are just neglecting change basically u can say

[ornate knot]

u can differentaite both sides with t , cuz doing smthing to both sides dont change the equation (most of the time)

have u not done any questions like this in class/lecture??

[odd urchin]

we did a singular question in class together

and now we have homework on it

quite literally one question we did together

[ornate knot]

oof

the main thing in these kinda questions is basically , trying to figure out what known equation to use and then u differentiate that

plug in the values and try to find out the unknown

[final saddle]

@odd urchin Has your question been resolved?

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@odd urchin Has your question been resolved?

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@odd urchin Has your question been resolved?

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[blissful zenith]

Help

[blissful zenith]

Let $A$ be index set. Suppose $A$ is countable, the product $\sigma$-Algebra $\bigotimes_{\alpha\in A}\mathcal{M}$ is generated by $\mathcal{F}{1}:=\big{\prod{\alpha \in A} E_\alpha:E_\alpha\in \mathcal{M}_{\alpha}\big}$.

However I still find it very struggling understanding this part. Let $\pi:\prod_{a\in A}X_\alpha\to X_\alpha$ be coordinate map, suppose $A$ is uncountable then the product $\sigma$-Algebra $\bigotimes_{\alpha\in A} \mathcal{M}_\alpha$ is generated by $\mathcal{F}_2:=\{\pi_{\alpha}^{-1}(E_\alpha):E_\alpha\in \mathcal{E}_\alpha, \alpha\in A\}$ where $\mathcal{M}_\alpha$ is generated by $\mathcal{E}_\alpha$

I found it really hard understanding, why if the index set $A$ is uncountable, we have to define the product $\sigma$-Algebra with an inverse coordinate map to define this generator $\mathcal{F}_2$, and why it only coincides with $\mathcal{F}_1$ whenever the index set $A$ is countable. 

The inverse map $\pi_\alpha^{-1}(E_\alpha)$ essentially will yields $\pi_\alpha^{-1}(E_\alpha)=\prod_{\alpha\in A} E_\alpha$ which is the same hypercube as the usual defined generator for countable case $\big\{\prod_{\alpha\in A} E_\alpha: E_\alpha\in \mathcal{M}_\alpha\big\}$ 


Maybe, due to lack of example, it made me really confused with this additional layer in definition, making me question why it is needed and why we can't define it without this additional layer of abstraction using coordinate maps. What would happen if I define generator of generating $\sigma$-Algebra to be $\mathcal{F}:=\big\{\prod_{\alpha\in A} E_\alpha: E_\alpha\in \mathcal{M}_\alpha\big\}$ and is it ill-defined? if so why?

Or is there any good examples illustrating why this way of defining product $\sigma$-Algebra is necessary or good examples related to this in general. I am studying Folland's for once more however, reading it the second time, I found the details only more abstract and nuanced or simply hard. Thanks so much in advanced.

[soft zealot]

Emmaaaaa

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[blissful zenith]

<@&286206848099549185>

[blissful zenith]

<@&286206848099549185> please help me 😭😭 I am too confused for this and also too weak for tackling such confusion myself

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[arctic fiber]

What is value of sinA * tanA while 5*cosA-√5 = 0?

[terse crypt]

ok

find cosA first

then find the possible sinA and tanA using it

cosA can only be in two quadrants (but the same of sinA and tanA are the same so any is fine)

[arctic fiber]

There is no exact triangle

[tired walrus]

no need for a triangle

you just need to know cos(A)

and some trig identity magic

[terse crypt]

wdym there isnt

[terse crypt]

or this

[ 5\cos A = \sqrt{5} \implies \cos A = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}]

pytahgorean

[tired walrus]

\cos(A)