help-36

[paper zodiac]

Use Soh cah Toa again to find line CB

[unique vapor]

YES

ok

[paper zodiac]

Line CB -Line DB = Line CD

[unique vapor]

righ

t

thank you

[paper zodiac]

I think that's it

[unique vapor]

ok]

we are done them

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[pseudo socket]

If you have a 6 period schedule and 3 elective classes. How many schedule arrangements are possible where you have at least 2 elective classes in a row?

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[faint zealot]

for this i think i need to use

d/dx( f + g ) = d/dx(f) + d/dx(g)

but how would i identify what if f and what is g?
from the sin(t^3)? or from the upper and lower limit?

[strange shore]

fundamental theorem of calculus ?

oh wait you cant

[faint zealot]

sry idek what that means

;-;

never heard my prof mention that

he skips too many steps

and scream at the class when we ask questions

[strange shore]

maybe try to integrate ?

[tulip coyote]

Isn't there a version that deals with cases like this?

[strange shore]

idk

[strange shore]

ah this

Ah I see

basically chain rule

[faint zealot]

the f(b) and f(a) is basically setting the upper and lower limit for one or the other to 0 while keep the other the same?

for example for this

lower limit 0, upper limit x^2

then

lower limit 2x, upper limit 0

and those two different ones would be the f(a) and f(b)?

so basically like this???

or no?

and then i find F'(x) for that

by substituting in the 2x for t in the first one, and the x^2 into the t in the second one?

@tulip coyote umm sry for ping but would those steps be right?

[strange shore]

just do like in the pic he sends, what r u doin

[tulip coyote]

Hmm, well that does work to be fair! And with the first integral you can swap the limits around

[tulip coyote]

Think they’re trying to see why it’s true, unless I’m being slow?

[faint zealot]

im just trying to find F'(x)

its what the problem says

[tulip coyote]

Well, you can literally just use this

[strange shore]

$F'(x) = \sin(x^6) \times 6x^5 - \sin(8x^3) \times 24x^2$

[soft zealot]

Ｈｅｒｅｌｓ

[faint zealot]

whats a(x) and b(x) tho?

like how would i put them in the equation

[strange shore]

a(x) is 2x and b(x) = x^2 in your expression

[faint zealot]

so sin( (x^2)^3 ) * 2x - sin( (2x)^3 ) * 2

??

[strange shore]

?

why *2x ?

[faint zealot]

for the b'(x)

sry if its confusing im new to calc 1

so idk how to write most of the things properly

[faint zealot]

@strange shore

[strange shore]

basically :

$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t) dt= F'(b(x)) - F'(a(x))$

[soft zealot]

Ｈｅｒｅｌｓ

[strange shore]

where $F'(x) = f(x)$

[soft zealot]

Ｈｅｒｅｌｓ

[strange shore]

basically :

$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t) dt= \frac{d}{dx} \left(F(b(x)) - F(a(x)) \right) = F'(b(x)) - F'(a(x))$

[soft zealot]

Ｈｅｒｅｌｓ

[strange shore]

and $F'(b(x)) - F'(a(x)) = b'(x)f(b(x)) - a'(x)f(a(x))$

[soft zealot]

Ｈｅｒｅｌｓ

[faint zealot]

nvm im lost now

[faint zealot]

so this isnt the correct way of doing it?

but it has the [ f( b(x) ) * b'(x) ] - [ f( a(x) ) * a'(x) ]

???

isnt that all i need to find the F'(x)

[strange shore]

no its not

[faint zealot]

k

[strange shore]

thats sin(x^6) you differentiate

with chain rule and shit

well not sin(x^6) but there is a chain rule here

[faint zealot]

ye the chain rule i did and got 2x

after i got that i made the (x^2)^3 into x^6

???

the 2x goes outside in front of sin

[strange shore]

thats x^6 you differentiate, not x^2

[faint zealot]

so 2x * sin(x^6)

ok

[strange shore]

No

[faint zealot]

?

[strange shore]

I said its 6x^5 sin(x^6)

[faint zealot]

wait but....

if its b'(x) y would i apply the ^3 first and then find the b'(x)

wouldnt i need to find the b'(x) from b(x) before doing the step with ^3

nvm im stupid, tysm for the help.

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[worthy dagger]

Anyone that can help me out to start of? I seam to get a grasp on how I am suppose to start :/

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[maiden current]

What would a suitable change of variables be to solve this integral?

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<@&286206848099549185>

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[terse dagger]

Culinder but elliptic

[maiden current]

why not spherical?

[terse dagger]

You can try both i dont know for sure which is best

[maiden current]

understandable

well then I feel a bit more confident im on the right track

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[small inlet]

anyone able to find the derivative of the equation?

[fickle crater]

sup dr senku

[small inlet]

yo whats up 🤣

[fickle crater]

ngl

that derivative scares me

[small inlet]

yeah

im not sure how to do it

even when i put it on a app it still wasn't correct

[tranquil pine]

nah

just look at i

it

you need to use

the uh

chain rule if

you make a substituion that

2.6 times square x = 0

t*

[small inlet]

i dont understand

[tranquil pine]

you know how the chain rule works right

[small inlet]

yes

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[topaz siren]

How can I show these are equal?

[royal gust]

Can always just compute each side separately and show they simplify to the same thing

[topaz siren]

How can we compute, for example, the first one?

Using cofactor method?

[royal gust]

Sure! Any method that can take down a 3x3 really

I like this one, it's worth knowing.

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[spring violet]

Did I get the right equation for this ?

[tranquil pine]

,w maclaurin of sin(ax)

[soft zealot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ♡LexQa♡#5050

[tranquil pine]

Ding ding!

[spring violet]

Sick!! Thank you haha

How goes it btw ?

[tranquil pine]

Oh what do you mean?

[spring violet]

Ahh you’ve helped me a few times just saying hey

Always appreciate it. Thank you !

[tranquil pine]

Oh yeah ahaha I am just chilling, I am mainly done with calculus, I am kind of stressed out with discrete math stuff instead now ahaha

[strange shore]

what kind of discrete maths

👀

[spring violet]

That’s me next semester! Got 2 discrete math courses gonna be spinning I think haha

[tranquil pine]

Graph theory, a lot of counting, combinatrics

Etc

[spring violet]

Proving shit to death !

[tranquil pine]

Good luck I find them so so much harder than calculus

[strange shore]

ive never had a course on graph theory ;-;

[spring violet]

I had a python lab using a graph that's as deep as i've gone, hopefully it clicks for you !

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[tranquil pine]

Oh happy you

[tranquil pine]

Thank youu

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[slate girder]

What did I do wrong? Answer should be -2i

[fiery bluff]

what is C

[slate girder]

From i to -i

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[full mango]

Is the answer 518.4

[final saddle]

Please don't occupy multiple help channels.

[full mango]

<@&286206848099549185>

[tulip coyote]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[full mango]

I forgot

[tulip coyote]

Don't worry about it

Could you please also close the other channel you have open (20) too?

[full mango]

Sorry my bad

[tulip coyote]

It's all good, thank you

How did you get your [£]518.4, if you don't mind me asking?

[full mango]

I did 1.2 x 432

I think I'm supposed to do 1.1 x 432 then do the answer x 1.2

[tulip coyote]

Hmmm I don't think I agree with what I think you're trying

If I decrease something by, say, 30%, then increase what I get from that by 30%, would I get my original number back?

[full mango]

I didn't add 10% and 20%

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[mystic quartz]

Did I setup the integral below correctly? By drawing the region, it's clear that $ 0 \leq \theta \leq 2\pi$ and letting $x = r\cos(\theta)$, $y = r\sin(\theta)$, $D$ becomes $1 \leq r^2 \leq 4$ which is the same as $1 \leq r \leq 2$, therefore the new integral is $\int^{2\pi}{0}{\int{2}^{1}{r^3\cos(\theta)\ln(r^2) dr}d\theta}$

[soft zealot]

Chippotle Maths

[mystic quartz]

Just wondering if I setted up the integral correctly since I'm unsure if I did some mistake integrating or if I described the region wrong

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[lament eagle]

heyy

[karmic lark]

hey

btw i know a lot about maths

[maiden canopy]

well you never know lot about maths

[karmic lark]

what do u need help with

i know 8th grade math which is decent

harvard level

[maiden canopy]

hmm

[karmic lark]

shes into me

[fossil geyser]

@lament eagle Did you have a question? Because I'm kind of afraid of what this channel is about to turn into

[karmic lark]

nono dont worry be patient

im waiting

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[rare quarry]

is this MML

im getting flashbacks

[random niche]

what that

[rare quarry]

what platform is that question on

[random niche]

ya it says mylab

[rare quarry]

mymathlab and mastering

[random niche]

yup

[rare quarry]

so go ahead and tell me what ur struggling on

[random niche]

what is v and w?

and i and j

[rare quarry]

v and w are the vectors defined there

[random niche]

i tried just multiplying them together like with foil and it said wrong

[rare quarry]

and i and j are just part of the definition of a vector

so first of all, do you know what a vector is?

[random niche]

not really

my teacher always gives us stuff we dont know yet

[rare quarry]

ok so i would recommend getting that down first

what's a vector?

and then come back to the question

[random niche]

line with arrow

[rare quarry]

do you have a book?

[random niche]

direction?

[rare quarry]

uh yeah so thats the graphical representation of a vector

what class is this

[random niche]

precalc

[rare quarry]

do you have a textbook? or does your teacher assign you videos

[random niche]

assigns us online textbook but its very convoluted and doesnt get to the point in a way i understand

[rare quarry]

oh i see

well in any case, whats a vector?

and how is it related to i and j

[random niche]

a direction and i and j are before and after?

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[gloomy fossil]

Graphing cos and sine fucntions, I have a problem but am stuck on graphic the phase shift

[gloomy fossil]

Should I send the problem in here

[vital crag]

yes

and your attempt

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[gloomy fossil]

I have gotten fsr in the equation but am confused on how the phase shift comes into play here

Aka my D or C value

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[heavy crystal]

I need help with this integral

[heavy crystal]

more specifically how they get to this cos

I understand the other stuff but i cant for the life of me get to the cos

[rain crane]

Weierstrass t formulas

[heavy crystal]

could you elaborate if you dont mind

do you mean the weierstrass theorem?

i dont really see how its gonna help here tho

[heavy crystal]

Holy shit

I legit did not know that existed

I dont remember learning this though

yeah we didnt learn this as a proper formula

but i managed to deduce it

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[jolly spear]

19 pls!

This unit kinda confused me. but so far I have <1,1,1> for one vector and <2,3,-1> for the other, then to my understanding I need to metricize them (?) giving me a final vector of <2,,-1,5>

then I used points from my original point giving me 2(x-3)-1(y+1)+5(z-1)

@ me so I get the notification pls

it was a matrix error

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[dusty harbor]

why is this true?

[sweet summit]

do you ever look at your more advanced mathematics work and think "we've gone too far?"

[dusty harbor]

hahah

i havent gone far enough

still taking introductory classes lmao

[sweet summit]

what class is this for?

[rare cradle]

Never seen that notation

[dusty harbor]

linear algebra

[sweet summit]

wut

[dusty harbor]

its the adjoint

the dagger

[sweet summit]

ah we didn't get to this in our LA course

[dusty harbor]

means conjugate transpose

[rare cradle]

Is L a matrix

[dusty harbor]

yes

[rare cradle]

Or linear map

[dusty harbor]

a linear map is a matrix

L is an operator

so its a linear map from dim n to dim n

nxn matrix

[rare cradle]

Almost but not quite

[dusty harbor]

not?

[rare cradle]

Cuz with a matrix you need to specify a basis

[dusty harbor]

what do you mean exactly

do you mean as in L is not necessarily a matrix but [L]_epsilon is?

[rare cradle]

Yea what does the epsilon mean

And what does G mean

I'd say let's just say that L is a matrix for simplicity

[dusty harbor]

G is the metric tensor

or metric matrix

same thing

[rare cradle]

Damn forgot what that was

Isn't it for specifying the basis for a dot product

Or smth

[dusty harbor]

sort of yeah

[sweet summit]

this scares me

[dusty harbor]

it should

(n't)

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[shell scroll]

Did I draw the reimann sum properly?This is the homework for an introductory lesson on Reimann sums

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@shell scroll Has your question been resolved?

[shell scroll]

<@&286206848099549185>

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@shell scroll Has your question been resolved?

[viral belfry]

looks like it to me

[shell scroll]

Ok, thank you

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[formal flicker]

Is there such a concept of $$\frac{d}{dx}!$$ There is some sort of fractional derivative (many definitions, but none are "right"). Does such a thing exist too?

[soft zealot]

ohNoiAmHere

[vital crag]

Not universally defined

[formal flicker]

is there something like fractional ones that arent universally defined either?

[vital crag]

Fractional what

Oh derivatives

https://www.cantorsparadise.com/fractional-calculus-48192f4e9c9f

[formal flicker]

yea, theres nothing like that for factorials

cause theres many fractional derivatives

and theyre usually defined off integration

is there a sort of thing for the factorial too?

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[noble mist]

are these right

[grim badger]

Definitely not 4

[noble mist]

2x + 8 = 28?

[grim badger]

No

Pretty sure you still need similar shapes concept

Meaning ratios still

[noble mist]

oh

what should I do

[grim badger]

Same process that you did for the other problems

Except this time, you don't have x, it's 2x + 8

[noble mist]

quick question

how do I know which order to put the sides in

to solve for the missing side

[grim badger]

How did you do it for the other problems?

[noble mist]

tbh

I could barley do those

cheated on 1

and for 2 and 3

I just got help but they didnt really help tbh

they just kinda told me the equation

[grim badger]

https://www.youtube.com/watch?v=YiFwvAFk-xs&ab_channel=TheOrganicChemistryTutor

Here's a video that should help

[noble mist]

ty

[final saddle]

@noble mist Has your question been resolved?

[noble mist]

@grim badger

I think I learned it

[grim badger]

Looks right

[noble mist]

I think I did this wrong

@grim badger

what u think

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[grim badger]

.reopen

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✅

[noble mist]

hi

[grim badger]

Because 14 * 32 is not 768

[noble mist]

wtf

oops

idk how I even got that

[final saddle]

@noble mist Has your question been resolved?

[noble mist]

@grim badger

looking back on number 2

wouldnt it be this?

it looks wrong but

arnt the angles equal this way

isnt HU = FH

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[noble mist]

.reopen

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✅

[slow tapir]

m8 dont ping people

[slow tapir]

how

they literally gave the measures of HF(84) and HU(45)

84 is not equal to 45

[noble mist]

fam get off my d1ck

he was previously helping me

[slow tapir]

ik but dont ping ppl

[noble mist]

ight

[noble mist]

and yeha I just realized but

im watching this video

idk its confusing me

like

these should be correct then right

[slow tapir]

idu what u r saying

[noble mist]

idk bruv

are these correct

[slow tapir]

yea ig

theyre right

[noble mist]

so then how isnt number 2 right @slow tapir

i pinged

[slow tapir]

look at the triangles

[noble mist]

u myb

[slow tapir]

FGH ~ TUH

[noble mist]

yea

[slow tapir]

so

FH/GH = TH/UH

so 84/108 = x/45

now solve it

[noble mist]

liket his

[final saddle]

@noble mist Has your question been resolved?

[slow tapir]

yeahh

good job

[noble mist]

so all these good

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[noble mist]

.reopen

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✅

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[tranquil pine]

really confused on D

there is no difference right?

The area u gave in part d of the question?

Heh

yeah that was a typo in the question

confuses me even more

wtf do i write for this question

Well the thing is

The integral kind of minuses out the negative part of the graph

So it is not the true "area"

You need to split up the integral and take the absolute value of the negative part and add it up to get the actual "area"

Or u can just multiply by 2 since that's cos which the negative part is symmetrical to the positive one

thank you

I'm just learning about integrals so still a lil confused

but u helped

Yeah so the integral kind of subtracts the thing under the graph so to get the area

You should write:
[
\text{Area = } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}f(x) dx+ \abs{\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x)dx}
]

Painnn

damn

also what ???

[soft zealot]

♡LexQa♡

[tranquil pine]

There you go

thank you!

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[dusk canyon]

Find the measure of angle z.
help with this problem

[vital comet]

Which ones 65

Use triangle sum and equate opposite angles

[oblique portal]

yeah thats what i would do

dunno why they gave u those x y w and v angles you dont need em

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[tranquil pine]

ahm

[tranquil pine]

hello

first of all how can i claim a channel please?

i wanna have my own

[sturdy cypress]

it's yours

[tranquil pine]

is it acc???

[sturdy cypress]

it's even called chubby

[tranquil pine]

omd

[tranquil pine]

ima cry hows that tho

who gave it to me?

[sturdy cypress]

christmas magics

[tranquil pine]

nah acc what’s going on

[tranquil pine]

thats mental

anyways back to the question that i was posing like 50times

is there a quick easy way for me to know when to use tan and when tan inverse?

im solving some polar coordinates problems and getting confused by this point

come on my chubbys help me i got quiz tmrw

[sturdy cypress]

tan works on angles

and doesn't give an angle

and arctan gives an angle and doesn't work on angles

and it's the same idea as with sine and cos, the "basic" function with a shorter name is the one that works on angles

ok i've no idea, maybe there's something called tan inverse and it's not what i think

[tranquil pine]

oh that’s something new

thats the tan inverse im talking about

it got mince one above it

ALSO that question

i never get the same answer as (which is 0.714) in the question even tho im putting the same values in my calculator

i got 0.0124

and tan inverse give me ≈35°

OHHHH wait i think they just use tan inverse in these questions, they got 216° cos they added 180° to 35° which u should do

[vital crag]

arctan and inverse tan are the same thing

[tranquil pine]

oh, yh i thought so but he said that tan inverse may be smth diff

[vital crag]

common awful math notation/jargon

[tranquil pine]

lol gotta be having a meaning behind it but still

[vital crag]

doesn't make it not awful

[tranquil pine]

is the answer a?

[tranquil pine]

Mathematician arent the best at naming stuff as we can see

s

[vital crag]

indeed

[tranquil pine]

thats acc really lovely that i have my own channels ima be posing questions all the time no one can say nothing

[vital crag]

with great power, comes great responsibility

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

no it hasn’t

dont take the channel away from me

[vital crag]

no

calculate the magnitude of all the vectors using the equation for R = sqrt(...) above

[tranquil pine]

waaiiittt

[tranquil pine]

IS IT C??

[vital crag]

did you show it by using the equation for R?

[tranquil pine]

yess i square rooted 5

[vital crag]

Why 5

,calc sqrt(5)

[soft zealot]

Result:

2.2360679774998

[tranquil pine]

urm bcs they wanted the magnitude and to find it u have to square root the given number, and since i have one component i dont have to do any operations

[tranquil pine]

wait what, well i think i didn’t really get the question then

[tranquil pine]

thanks mr. bot

[vital crag]

you still have something to plug into the right side

[tranquil pine]

thats stupid it cant be that hard to get

[tranquil pine]

should i be having two components?

bcs on the 3rd option we just got one

which well be z (which they should have added to their role)

[tranquil pine]

oh even the 2nd

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

can u stop asking me that question and let me have my channel on peace?

[vital crag]

$R = \sqrt{R_x^2 + R_y^2 + R_z^2}$

[soft zealot]

riemann

[vital crag]

don't know why your book only gave the formula for two components

[vital crag]

Here,
Avec = [Ax, Ay, Az] = [2, 5, 0]
Bvec = [Bx, By, Bz] = [0, -3, 0]
Cvec = [Cx, Cy, Cz] = [0, 0, 5]

[vital crag]

use this formula to find magnitude of each Avec, Bvec, Cvec

[tranquil pine]

ahh yess thats exactly how it should be, so when the components isn’t mentioned it’ll be 0

[tranquil pine]

it will still be Cvec right? bcs they said which of the following their magnitude vector equals the component itself so itll be C as the answer

[tranquil pine]

i think it depends in their mood bcs sometimes they don’t mention on other formulas as well even tho they should

[vital crag]

you've still yet to show any calculation

the way you describe it sounds like you haven't calculated anything

[tranquil pine]

there u go

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[grim nebula]

<@&268886789983436800>

[tribal condor]

Dead

[stable kraken]

.close

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[cosmic dawn]

If f : [a,b] to R is riemann integrable, is g : [a,b] to R also riemann integrable?

[dry light]

r and g seem independent of each other

[vital crag]

need more information. see ^

[final saddle]

@cosmic dawn Has your question been resolved?

[cosmic dawn]

If f is riemann integrable on some interval, is g riemann integrable on the exact same interval? It is unknown if f=g, but it is possible.

[vital crag]

can you just screenshot or take a picture of the problem

[cosmic dawn]

I'm just saying in general. Are two functions, (let's just say that they are different), RI on the same interval if one of the functions is?

[vital crag]

then no

not every function is integrable

[cosmic dawn]

What does it mean for a function to be riemann integrable?

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[heavy crystal]

Need help with this improper integral

[heavy crystal]

I tried parts but couldn't finish

I legit can't find a way to do this

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@heavy crystal Has your question been resolved?

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@heavy crystal Has your question been resolved?

[fluid sky]

try the substitution u = 2 - arctan(x)

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[pearl wraith]

How did they pick the interval [0,250] in this optimization problem?

[dry light]

It would be really funny if the fence had a length of -100

It's gotta make sense

You can't have negative length

Or rather negative area

[pearl wraith]

why [0,250] instead of (0,250)

[dry light]

Because you have 0 area

And also makes critical points easier

Since you can include endpoints

And so those are also candidates

You won't have to fret whether the best option are the endpoints or not

[pearl wraith]

oh okay

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[stray stump]

im having some trouble solving for v1 and v2

what should i do here?

[sinful folio]

what have you tried?

[stray stump]

solving for v1 in second eqn and subbing it in first eqn

gave me 0=0

[sinful folio]

do you have any other ideas?

[stray stump]

the matrix stuff

the calculator said to first multiply row 1 by (1+i)/4

[sinful folio]

thats one way of doing it yeah

[stray stump]

then divide it by 2-2i

that didnt match up when i did it by hand

[sinful folio]

i dont think you need to divide?

you defo dont need to

[stray stump]

oh

made some dumb algebra mistkae

nvm

ty

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[timid furnace]

(x-h)^3-x^3

[timid furnace]

Is expanding the cubic (x-h)^3 the fastest way to do it

I expanded everything

And then factored everything by h

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[proud sage]

Having trouble with question C, some help would be appreciated

[final saddle]

@proud sage Has your question been resolved?

[manic elm]

Partial sum of first 5 terms is simply the sum of the first 5 terms of the sequence. If you've figured out the common ratio r, and are able to find the 6th term, then finding the sum of the first 5 terms should follow

It's called a partial sum because the entire sequence is infinite in terms, and you're only choosing to evaluate the first finite number of terms

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[muted pelican]

If Sin(theta) < 0 and tan(theta) = 5/3, 0<=theta<360, then theta, correct to the nearest whole angle must be

[muted pelican]

<@&286206848099549185>

[muted pelican]

<@&286206848099549185>

[thorny tendon]

Do you have a screenshot of the original problem? θ can’t be < 0 and > 360. You made a typo

[muted pelican]

fixed

@thorny tendon

[thorny tendon]

What have you tried?

[muted pelican]

idk where to begin

@thorny tendon

[thorny tendon]

Do you know the unit circle and what these trig functions mean?

[muted pelican]

yes

k

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[muted pelican]

lolol

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[next pagoda]

f(x)= 1/x

[next pagoda]

Why would this be a odd function ?

[cold gorge]

Because $f(x) \neq f(-x)$

[soft zealot]

What the hell am I doing here?

[tranquil pine]

because f(x)= -f(-x)

[next pagoda]

I'm getting two different answers

[cold gorge]

The answers aren't different when you properly look at them.

[tranquil pine]

not even function does not imply odd function

[cold gorge]

Yes, I wanted them to work the - for themself

Yours is right, i know it.

[next pagoda]

bet

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[slim cove]

Hello, i have a problem with a polynomal division and how to get to the result.

[slim cove]

it solves to = (2x-) * 1/2x + 1/4 + (1/((4(2x-1)))

the asymptote <= 1/2x+1/4

y=

[tranquil pine]

the numerator is xx^2(x-1)

aka u can simplify the fraction by cancelling

[slim cove]

so it would be like x^3(x-1) so i can cancel out to x^3/(x(2x-1)) = x^2/(2x-1) ?

[tranquil pine]

yup!!!

[slim cove]

and this simplified term i need to use for the polynomal division to find the y for the asymptote?

I think i got it now. I did overlook the fact to use the simplified term for the division. Thank you.

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[broken dove]

I need help setting this up

[broken dove]

I know that the radius of the sphere is 2

and that it goes from 0 to 2pi

[tranquil pine]

radians is for 2-D angle measurement

[broken dove]

but what goes in the integrand is confusing me

yeah, isn't that necessary for surface area?

[tranquil pine]

yes

[broken dove]

would I use spherical coordinates as well?

[tranquil pine]

I'm not familiar with this, someone else will help you

[broken dove]

set up like this, right?

So what would go in the integrand?

<@&286206848099549185> give me a hand here

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@broken dove Has your question been resolved?

[robust sedge]

We have that z = f(x, y) = sqrt(4 - x^2 - y^2) = sqrt(4 - r^2) .
We need to calculate sqrt((f_x)^2 + (f_y)^2 + 1) then calculate

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[distant breach]

can someone help me with this:

[distant breach]

im confused on where to start atm

[supple mantle]

You see how the function is a fraction?

[distant breach]

mhm

the domain is based off the bottom part right?

[supple mantle]

Yep

If you divide by 0, the function becomes undefined

So you basically find all the x values that make the function not work basically

[distant breach]

o

ok so basically 1 and -1?

[tranquil pine]

assume that a function's domain is $\bR$ and then chop off the points that dont match that

[soft zealot]

♡LexQa♡

[tranquil pine]

aka when 1/0 for example

[distant breach]

so from what im getting/understanding, domain should be (-infinity, -1) U (-1,1) U (1, infinity)

[tranquil pine]

inbetween -1 and 1?

[distant breach]

o wait

right

[supple mantle]

The x values are correct but the condition needs some tweaking

[tranquil pine]

yeah there u go

golden

[supple mantle]

Yep

[distant breach]

then x = -1 and x = 1 are my vertical asymptotes too

[supple mantle]

Yep

[tranquil pine]

yes

[distant breach]

how would i find intercepts?

[supple mantle]

What about the horizontal ones?

[distant breach]

o

[tranquil pine]

wait

wait

[tranquil pine]

x = 1 isnt a vertical asymptote

[distant breach]

oh

[tranquil pine]

it is removeable

[distant breach]

wdym by removable

[tranquil pine]

because

[supple mantle]

admira you took Limits I'm assuming right?

You know how they work?

[distant breach]

ive learnt about them somewhat but not rly

it was like a class..

[supple mantle]

Damn

[tranquil pine]

[
\lim_{x \to -1} (\frac{x^2 -x}{x^2 -1}) = \text{DNE}
]
[
\lim_{x \to 1} (\frac{x^2 -x}{x^2 -1}) = \frac{1}{2}
]

[supple mantle]

Ok so

[soft zealot]

♡LexQa♡

[distant breach]

so if i plug it in?

[supple mantle]

Limits basically is an operation to find how the function behaves at certain points

[distant breach]

oh ok

so like increasing decreasing?

ish

or

[supple mantle]

If they approach a certain value or not

[distant breach]

oh ok

[supple mantle]

Sometimes they go to infinity

[tranquil pine]

its a removeable discontinuity

[supple mantle]

Sometimes they don't approach anything

[tranquil pine]

anyways i will go back to the Fourier shit

[supple mantle]

Limits make this possible

[supple mantle]

[distant breach]

so how do ik when ones a vert asymptote or horizontal asymptote?

is it based off limits?

[supple mantle]

Yep

You take the limit

[distant breach]

ok so

[distant breach]

how do i do this without calculator and stuff..

do i plug in -1 and 1 to x?

[supple mantle]

You can do some factoring

x(x-1)

And on the denominator (x-1)(x+1)

[tranquil pine]

[f(x) = 0 \text{ (tells you x intercept) }]
[x = 0 \text{ (tells you y intercept)}]
[\lim_{x \to C} f(x) \text{ (tells you vertical asymptotes for any}]
[\text{non removable infinite discontinuty C) }]
[\lim_{x \to \pm \infty} f(x) \text{ (tells you horizontal asymptotes) }]

[supple mantle]

[soft zealot]

♡LexQa♡

[distant breach]

then after factoring?

[supple mantle]

Yep now works

[supple mantle]

If you have a value, then the limit exists

[distant breach]

o

[distant breach]

wdym by value..

[supple mantle]

Like a number

If you got -5 for example

[distant breach]

yeah

[supple mantle]

Then the limit approaches -5

And this means no asymptotes

[distant breach]

im kind of lost still..

[supple mantle]

Let's try the limit when x approaches-1

[distant breach]

ok

[supple mantle]

$$\lim_{x\rightarrow -1} \frac{x^2 - x}{x^2 - 1}$$

[soft zealot]

VulcanOne

[supple mantle]

We factored it right?

[distant breach]

yes

[supple mantle]

$$\lim_{x\rightarrow -1} \frac{x(x-1)}{(x-1)(x+1)}$$

[soft zealot]

VulcanOne

[distant breach]

mhm

[tranquil pine]

there are two types of removable discontinuities

[distant breach]

o

[supple mantle]

We can cancel x-1 right?

[distant breach]

yes we can

[supple mantle]

Now we are left with this

[distant breach]

x/x+1

[supple mantle]

$$\lim_{x\rightarrow -1} \frac{x}{x+1}$$

[soft zealot]

VulcanOne

[supple mantle]

Here we stop

[distant breach]

and then what do we do

[tranquil pine]

There are two types of removable discontinuities I believe, one where the limit is finite but the function is undefined, and second when the limit is finite but the value of that function at that point is not the same as the limit's.

You are having the first case with x = 1

[supple mantle]

We need to see if the value from the right equals the value from the left

[distant breach]

o

[distant breach]

how do we do that..?

[supple mantle]

If we plug in -1, we get -1/0 right?

0 by nature doesn't have any signs

[distant breach]

yeah

[supple mantle]

So we need to see if we added a sign to 0

Will the function behave the same or not

[distant breach]

wdym by a sign

[supple mantle]

-0 and +0

[distant breach]

oh ok

[supple mantle]

If we made 0 negative, the function goes to + infinity

And if we made the 0 positive

The function goes to - infinity

Now here does the limit from the left equal the limit on the right?

[distant breach]

im sorry about this again..

but how do yk it'll go to positive

infinity

and negative infinity

[supple mantle]

We add signs to the zero

[distant breach]

yeah

[supple mantle]

$$\lim_{x \rightarrow -1} \frac{x}{x+1} = \frac{-1}{-1+1}$$

[soft zealot]

VulcanOne

[distant breach]

oh

[supple mantle]

$$\frac{-1}{0}$$

[soft zealot]

VulcanOne

[supple mantle]

We're gonna be unsure about the behavior of the function if we kept the zero with a sign

So we take the limit from the left

[distant breach]

ok..

[supple mantle]

And the limit from the right

Lemme try showing this on a graph

[distant breach]

wait

ok so thats the thing

how do i know without the graph

[supple mantle]

Seeing a zero on the denominator should be a red flag to you

[distant breach]

o like that

[supple mantle]

Yep

You should expect 2 things

It either becomes (infinity or negative infinity)

Or the limit doesn't exist

[distant breach]

ok ok..

[supple mantle]

The limit does not exist when

You have 2 different values from the left and from the right

[distant breach]

ok..

yeach could you show me on a graph..its confusing me a bit

[supple mantle]

Alrighty

This is a graph of 1/x

[distant breach]

o

[supple mantle]

You see how around x = 0

The function goes into 2 places?

[distant breach]

mhm..

[supple mantle]

From the left, it goes down

[distant breach]

yeah

[supple mantle]

From the right, it goes up

This means that the limit doesn't exist at x=0

[distant breach]

so what does that tell us

that its discontinuity?

[supple mantle]

Yep

[distant breach]

but how would ik if its horizontal or vertical?

[supple mantle]

You take the limit of x at a finite number

That's vertical

But when you take the limit for x at infinity

That's horizontal

[distant breach]

is there another way to figure it out..

i really dont remember doing this in class

[supple mantle]

You need to solve more limits to understand it

Because 1 class isn't enough

Lemme give you an example

[distant breach]

the thing is that our prof never really even went over limits

im in precalc atm

[supple mantle]

$\lim_{x\rightarrow \infty} \frac{x^2 -x}{x^2 - 1}$

[soft zealot]

VulcanOne

[supple mantle]

What will this find you?

Vertical or horizontal?

[distant breach]

since its infinity

horizontal

[supple mantle]

Yep

[distant breach]

can i send u an example that my prof did?

[supple mantle]

Sure

Here's a graph of your function

[supple mantle]

The dotted line is the horizontal asymptote

[distant breach]

this is what he showed us for notes

[supple mantle]

That's

That's sad

[distant breach]

🥲

[supple mantle]

He didn't explain why this works

[distant breach]

not rly

[supple mantle]

All of that works through limits

[distant breach]

oh

[supple mantle]

Ok don't worry about it

You can take the limit by going super close to the value

Like

$$\lim_{x\rightarrow -1} \frac{x^2 - x}{x^2 - 1} = \frac{(-1.001)^2 - (-1.001)}{(-1.001)^2 - 1}$$

[distant breach]

o

[soft zealot]

VulcanOne

[supple mantle]

If this worked

Then you try from the other side

$$\lim_{x\rightarrow -1} \frac{x^2 - x}{x^2 - 1} = \frac{(-0.999)^2 - (-0.999)}{(-0.999)^2 - 1}$$

[soft zealot]

VulcanOne

[distant breach]

whats the way he was trying to show then?

the n = m things

[supple mantle]

The highest powers of the numerator and denominator

[distant breach]

and the reason it works is because of limits?

[supple mantle]

Yep

When you have this for example

$$\lim_{x\rightarrow \infty} \frac{x^2 - x}{x^2 - 1}$$

[soft zealot]

VulcanOne

[supple mantle]

You take x and replace it with a big number

Like 1000000

[distant breach]

o

[supple mantle]

Now 1 compared 1000000^2 is like comparing an ant to the sky

In size

Same thing for 1000000 compared to 1000000^2

So you take only x with the highest powers from the numerator and denominator

And the coefficients do the job for you

1 minute brb

[distant breach]

ok

[supple mantle]

Oki I'm back

That's the reason why taking the highest powers works

There is also something else relating to how fast functions approach infinity

Which is called "the list"

That's something else you'll take in the future

Anyways

[distant breach]

o

[distant breach]

wb

i think i kinda get the limits thingy

[supple mantle]

[distant breach]

i have my final tmr :')

well today.

[supple mantle]

Damn

[distant breach]

been up all night studying loll

its 5 am atm

[supple mantle]

What

[distant breach]

yeppp..

[supple mantle]

Damn

How much left do you have?

[distant breach]

uh

[supple mantle]

For studies and time?

[distant breach]

• 1 hr of driving to campus

5 hours

[supple mantle]

Wow

[distant breach]

:') yep

[supple mantle]

Are you sure you can pull it off?

[distant breach]

nope

ive been calculating my grade

[supple mantle]

Wait don't

[distant breach]

i need a 62% to pass the class on this final.

[supple mantle]

You're gonna hopefully pass

[distant breach]

too late, been doing it from week 1 😭

[distant breach]

honestly im scared

[cobalt lodge]

there's always scaling

[distant breach]

wuts that

[supple mantle]

You'll do good I'm sure

He means taking the grade of the highest performing person and scaling all the others based off on that score

[distant breach]

my prof would never do that