help-13

[dire geode]

Yes I saw that

[forest scarab]

we will have:

( 1 + sin x )^2024 + sin^2024(x) = 1

[dire geode]

Use binomial theorem

Oh hmm maybe that doesn't work

Could also try complex exponential

[cedar kiln]

@forest scarab Has your question been resolved?

[gritty viper]

That's not true in general so there's nothing to try

cos(x)-sin(x) = 1 is only true for very specific values of x

it's not too hard to find what those are

[cedar kiln]

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[warped stump]

this is my work in that second picture and here "(5/36)100 =13.88"

[warped stump]

question is how did I get it wrong

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[novel glade]

Hi guys I’m having a brain fart lol I can’t remember how to solve this lol

[remote trail]

well you know that absolute value will always make your value of g(x) positive

so you could write out your boundaries as the 2 equations for g(x)<=20 and g(x)>=-20

[novel glade]

So like -10+3(x+5)<20?

[novel glade]

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[random kelp]

anyone know why in the circled bit, when infinity is inputted into the function, it results to 0?

[random kelp]

f(x) isnt even defined btw, they're just showing how fourier transforms work here

cos(inf) would be undefined i think, f(inf) doesnt really mean anything so I just dont know what to do here

[cedar kiln]

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[strong pilot]

Hello I had a quick physics question I'm not sure if I did this correctly?

[strong pilot]

I'm not sure if I got the final velocity and final momentum correct which is Vf1 Vf2 and Pf1 Pf2

The formula I used is m1v1 + m2v2 iniital = m1v'1 + m2v'2

[cedar kiln]

@strong pilot Has your question been resolved?

[cedar kiln]

@strong pilot Has your question been resolved?

[strong pilot]

<@&286206848099549185>

[void spoke]

Physics woohoo

[strong pilot]

It's quite difficult for me 🥲 but I'm trying

[void spoke]

Where is the coefficient of restitution

All are elastic collisions?

My guy say something

[strong pilot]

It’s 1D

[void spoke]

1D

Ok but still

Coefficient of restitution tells how elastic the collision is

Understand if e =1

You can conserve both momentum and K.E.

[dusk goblet]

yo oppai

[void spoke]

Yo

[strong pilot]

Okay ima send a ss of what I’m looking at

[dusk goblet]

weren’t you grinding for green all those months ago

what happened

[void spoke]

Imma different guy now bro

Got bz

[void spoke]

Aah

I see

Elastic moment

Conserve momentum and energy

[strong pilot]

I’ve been trying to get it down but I’m struggling

[void spoke]

m1u1+m2u2=m1v1+m2v2

Conserve momentum

Conserve kinetic energy too

[strong pilot]

Okay!

[void spoke]

Ok

Imma doing mechanics bro

Want a question?

[strong pilot]

Oof I feel like I would suck at it i wanna see tho

[void spoke]

Fbd of rigid body systems

Ok imma send

later first do this

Your work

[dusk goblet]

@void spoke

where’s the question bruh

[void spoke]

Yo

Imma send it later

Let him do it

His work

[dusk goblet]

ahh man i thought you were gonna send a fun problem

[void spoke]

Its a kind of simple problem.if you make the fbd correct

[strong pilot]

I'm reaching out, hopefully, this is how I start it

Sorry if my handwriting is pretty bad

[strong pilot]

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[serene bison]

find vert asympt
f(x)=sec(pi*x), so 1/cosx, the vert asymptotes are at integers of kpi?

[serene bison]

like this?

(want to make sure my notation is correct as well)

[cedar kiln]

@serene bison Has your question been resolved?

[drifting marlin]

cos is either 1 or -1 at integer multiples of pi

[serene bison]

so is what i did correct

[drifting marlin]

well you're looking for vertical asymptotes right?

[serene bison]

correct

[drifting marlin]

if you get f(x) = 1 or f(x) = -1, do you have a vertical asymptote?

[serene bison]

if i understand your question

yes

[drifting marlin]

no

refresh your understanding of what a vertical asymptote requires

[serene bison]

well a denominator of 0

[drifting marlin]

indeed

[serene bison]

and its non removable

[drifting marlin]

so you're looking for where cos is 0

[serene bison]

all integers?

[serene bison]

but would the notation for the Z work

or should i just write "1 or -1 integer multiples of pi"

[drifting marlin]

you don't understand

x just belonging to the integers is not correct

you're looking for where cos(pi*x) = 0. This does not happen when x is an integer

[serene bison]

not sure i understand

[drifting marlin]

,calc cos(0)

[wraith dagger]

Result:

1

[drifting marlin]

,calc cos(pi)

[wraith dagger]

Result:

-1

[drifting marlin]

,calc cos(2*pi)

[wraith dagger]

Result:

1

[serene bison]

oh wait

then its just multiples of half pi?

[drifting marlin]

that sounds better

[serene bison]

yeah oops

so

it would be xpi/2 x is an integer

like that?

wait

no

xpi/2 + x?

yeah no

[drifting marlin]

perhaps you should more specifically look for odd multiples

[serene bison]

pi/2 + kpi, 3pi/2 + kpi

99% sure thats correct

[drifting marlin]

that works

[serene bison]

is there a more efficient way

[drifting marlin]

i mean pi/2 + kpi is equivalent

[serene bison]

ah okay

i see

thanks for the explanation

[drifting marlin]

[serene bison]

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[fading temple]

can someone help me w this math q

[fading temple]

(this is vectors)

like why does that equal that

[upper ruin]

Well, I think you can just calculate those vectors and check that the equation there is true

[fading temple]

oh ye i can def do that

but conceptually im a bit like

not there

hm

[upper ruin]

You can prove it with geometric argument or also with vectors in a general case

[fading temple]

yessir

[upper ruin]

It seems they only ask you to do that, not to prove it for the general case

[fading temple]

ye but is it oka if i ask u to prove it

please

[upper ruin]

I don't remember it well, but I can suggest you some videos

[fading temple]

oh oka thank you

[upper ruin]

You can search in YouTube this:
"parallelogram diagonals bisect each other proof"

I just looked it up and there are plenty of videos, both with geometry and also vectors

[fading temple]

oh okay

thank you

:D

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[eager hemlock]

I tried kings rule, failed

[upper ruin]

https://youtu.be/hGqSj0FBreo?feature=shared

[eager hemlock]

Lifesaver

Tysm

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[open mauve]

Help with iv)

Thanks

<@&286206848099549185>

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@open mauve Has your question been resolved?

[open mauve]

<@&286206848099549185>

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@open mauve Has your question been resolved?

[open mauve]

<@&286206848099549185>

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[oblique lynx]

is the I in 2IX - AX = -2A necessary? Next step makes step since we factored it out, but the step before it i have just as 2X - AX = -2A

[oblique lynx]

those are matrices

[dire geode]

No

[oblique lynx]

And so is it just a matter of convetion and to see it better to use I?

[dire geode]

Depends on context

[oblique lynx]

okey, thank you

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[vital rain]

I need help in explaining something for me please

[crimson sedge]

What is it?

[vital rain]

hold on let me send

why does the domain of g not have an equality

[odd verge]

If x is 2 what happens to the denominator

[odd verge]

To be clear you are talking about number one or number 2

[vital rain]

question 1

[odd verge]

Yes so
If x is 2 what happens to the denominator

[vital rain]

i dont know

[odd verge]

What is √(2-x)

If x is 2

@vital rain

[vital rain]

im sorry but i still dont get it

[odd verge]

Replace x with 2

[vocal crest]

do you know what can go in a sqrt symbol and what can't go in the denominator?

[odd verge]

Then what do you get if you calculate √(2-2)

@vital rain

[vital rain]

0?

[vocal crest]

and can u divide with 0?

[vital rain]

no i dont think so

[odd verge]

So now what is the domain of a function

The values of x which the function has a 'definite' answer

[vocal crest]

find the range of 2-x then that of x

[odd verge]

Since 2/0 is not defined

X = 2 can't be in the domain of g(x)

@vital rain

[vital rain]

ohhhh

i get it thanks!

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[heavy crypt]

Am I allowed to do this?

It feels wrong haha

[vocal crest]

if it helps sure u may

i use that too

[heavy crypt]

is the last step right?

[cedar kiln]

@heavy crypt Has your question been resolved?

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[earnest crest]

what is a difference quotient i need to find it at f(2.5)

[vocal crest]

difference coefficients is the derivative at a specific point

[earnest crest]

what does that mean

my teacher didn't teach us that im ngl

we only learned limits

[vocal crest]

the difference coefficient of a function at a point is the slope of the tangent of the function at that point

[tulip stream]

You mean at x=2.5?

Also i dont see why there exists another graph of function g here

Is there more context?

[earnest crest]

nah thats for part b

this is part a

[tulip stream]

Are you given anything about function f? Or just the graph

[earnest crest]

wait one sec

bruh its not letting me send

jsut said assume f(x) is real numbers

f(x) is interval 0<=x<=5

[tulip stream]

I think you failed to copu

Copy

As it sent the same image

[earnest crest]

and right the instantaneous change of g(x) at x = 2.5

[tulip stream]

First thing to do is to express f as a function

[earnest crest]

its g(x) not f

thats mb

in the question

[tulip stream]

For a) ?

Ah i see

[earnest crest]

is it just the slope?

[tulip stream]

You can easily express g(x) as a linear function on interval 2 < x <=3

[tulip stream]

Slope of the line for the interval i typed, yeah

Wait oops

Interval is wrong

Ok fixed, graph of g(x) on interval 2<x <=4

Then use definition of limit at a point

Or simply take the slope as you did

[earnest crest]

alr think i got it

so it should be -3

[tulip stream]

Yes

[earnest crest]

thanks

close

!close

.close

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[dire thorn]

What have you tried

[magic garden]

I just tried to differentiate the options instead of integrating the question

;-;

[dusk goblet]

good strategy tbh

[dire thorn]

To integrate this you should try u sub

[magic garden]

I just wanna know the correct steps of integration

Like

I can do but it's all messy

[magic garden]

You mean by parts integration?

[dire thorn]

Nah

[magic garden]

Oh

[dusk goblet]

no he said u sub

[magic garden]

Umm oki lemme trt

Try

I know it's wrong

Im struggling in integration

How do I do it?

[dusk goblet]

let x be t?

that’s what you wrote

[magic garden]

Yeah

[dusk goblet]

what difference does that make

the name of the variable is irrelevant

[magic garden]

True

[dusk goblet]

and if x = t why did you then write t(t+2)

did you mean log(x) = t

[magic garden]

Yeah

My bad

[dusk goblet]

you cant integrate t if you still have dx

[magic garden]

Yeah it makes 2 variables

[dusk goblet]

find dt

[magic garden]

Oki

Shall I open the bracket?

[dusk goblet]

what

[magic garden]

Like shall the bracket in the beginning itself?

[dusk goblet]

no clue what you’re saying

[magic garden]

Sorry 😔 I'm confused myself

[dusk goblet]

$dt = \frac{1}{x} dx$

[wraith dagger]

knief

[magic garden]

Yeah

But

1/x Doesn't come anywhere

Before integrating

Dx = xdt

How do we solve this

Is this what you're telling to do

[glacial field]

Derivative of logx is 1/xln10

$\frac{d}{dx} (\log x) = \frac{1}{x \cdot \ln 10}$

[magic garden]

Yep

[dire thorn]

Logx=lnx for respectable people

[wraith dagger]

mari

[dire thorn]

Hmm I think my u sub idea might have been bad

[magic garden]

Yep I know this

[glacial field]

You wrote the derivative of lnx

[magic garden]

Oh

Lnx = 1/x?

[glacial field]

You did not write the derivative of logx

Despite letting t = logx

So dt does not equal 1/x

[magic garden]

Ohhh

[cedar kiln]

@magic garden Has your question been resolved?

[magic garden]

Is this portion correct?

Are we supposed to integrate 2logx again?

I have doubt in just that one portion other then that I understood

@dire thorn

@glacial field

Um guys?

@dusk goblet

.close

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[magic garden]

. reopen

.reopen

[cedar kiln]

✅

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[royal loom]

Is this the proper interpretation of "the union of right cosets of its stabilizer"

I took h in S because if I took h in G then I don't think that would be correct as for all subsets S the stabilizer contains the identity so if h ranged over G, then the uniion would be G, not S

[cedar kiln]

@royal loom Has your question been resolved?

[obtuse mango]

i guess so?

[obtuse mango]

yeah i feel like they could be a little clearly in the question

but yeah that works in proving that the stab. has order p^r

[royal loom]

Thank you

Would you mind looking at the next part also?

[obtuse mango]

sure

[royal loom]

It is sort of lengthy

[obtuse mango]

yeah looks fine

actually this is a new proof of sylow's for me

i know 2 proofs of sylow's thm and this is neither of them

[royal loom]

it is new for me too lol

took me so long

thank you for checking it

[obtuse mango]

nw!

[royal loom]

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[jolly sail]

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[jolly sail]

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[winged heron]

x3+y3+z3

=k

[hollow trail]

first, find the given limit. then find what physical quantity the limit represents given the explanation of what the function represents

[jolly sail]

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[hollow trail]

it asks for $\lim_{t \to \infty} Y(t)$, though

[wraith dagger]

cloud

[jolly sail]

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[hollow trail]

for limits as t goes to +infinity only care about large values of t, so you can focus on the last portion of the function

[jolly sail]

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[drifting marlin]

have you never computed limits directly before?

[hollow trail]

https://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityII.aspx

[jolly sail]

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[young sail]

I am struggling to figure out where to begin

[cedar kiln]

@young sail Has your question been resolved?

[young sail]

<@&286206848099549185>

[buoyant vigil]

plot points first

then connect them with 1 line

[young sail]

I'm not sure what you mean by plot points first

[buoyant vigil]

plug in x

into the integral

the output will be y

[young sail]

So like for the first point, put it on (2,0)?

I am still learning this stuff, i am a bit behind in this class

[buoyant vigil]

yeah

(2, 0)

then plug 3, 4, 5 etc

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[cedar kiln]

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[raven shard]

l'hopital rule

[violet flume]

maybe its easier to use binomial

[raven shard]

no

[violet flume]

[raven shard]

why wouldn't it?

[wild patio]

Where’s this help for my channel wth

[dire geode]

! occupied

[cedar kiln]

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).

[raven shard]

u can show (1 + sin2t)^(1/t) has a removable discontinuity at 0 hence when x approaches 0 so does the integral

[wild patio]

You must love me

[cedar kiln]

@eager hemlock Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[lethal jungle]

can someone help me

[nimble mountain]

yes someone can

[lethal jungle]

is this occupied?

sorry

[nimble mountain]

by u yes

u occupied it

[lethal jungle]

can someone help me with this question

[bleak viper]

what question

[nimble mountain]

just send the question

[lethal jungle]

fourteen cii

[bleak viper]

,rotate

[nimble mountain]

,rccw

which

[bleak viper]

14 c 2

[bleak viper]

u did 14 c i?

just manipulate that to isolate c

[lethal jungle]

i mean the other one

[bleak viper]

what

[lethal jungle]

the ax+by=c

[bleak viper]

yes

thats whta im saying

[lethal jungle]

i don't understand that bit

oh

[bleak viper]

if u have 14ci

[lethal jungle]

so how would you do it

[bleak viper]

then manipulate that equation

lets say

y=3x+2

then -3x + y = 2

[lethal jungle]

Ohhh

[bleak viper]

wait wait wait

misread question

u have the slope right

[lethal jungle]

i got the gradient yes

i've got the equation it's

y=-5x/3 +5

[bleak viper]

ok so

you know slope = (y-y1)/(x-x1)

[lethal jungle]

oh yeah

i've found everything i need to know

[bleak viper]

u have the slope, and for (x1,y1) just use any of the two intercept points, lets say (0,5)

[lethal jungle]

i js don't know how to plug it in

[bleak viper]

put in that

[lethal jungle]

js that the equation for the slope?

[bleak viper]

wdym

[bleak viper]

after u put these in it

simplify it

you'll have an equation in the form of ax+by=c

[lethal jungle]

OH yeah that's the equation for the gradient sorry

confusing myself

wait i think i see how u do it

i get it nvm

i think i'm okay

u js make y zero to find x and x zero to find why

i think i'm okay

[bleak viper]

what

[lethal jungle]

i js sup it in as if i were finding the points for it

.close

[cedar kiln]

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[cedar kiln]

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[broken carbon]

Hello I'm looking for help in Calc 2 on Disk/Washers. I would like someone to help guide me through the steps with an explanation on why/how this certain step takes place, but kept simple.

So far what I know I need to do for this problem:

• im working around the y-axis
• i think its washer's method, although im not sure
• i need to set the terms to y
• have them equal to 0 to find the bounds (this is where i stop because i got confused)

[void sage]

i dont know that much calculas but wont the volume be the same as original

since ur just rotating the solid

idk i could be wrong

[broken carbon]

what

no because i need to get some sort of value in the end

[dire geode]

The shape is defined by the two curves and their intersection points

The volume formula has two endpoints, those are usually the intersection, either x value or y value

[remote trail]

I think it would be better to use shell method from first glance

[void sage]

u can graph it on desmos

[broken carbon]

the practice problem is under a disk/washer pdf

[remote trail]

ah alright

[broken carbon]

ive been at it for 3 hours now with no luck

[remote trail]

then since it's already in x = f(y) I don't think you need to invert the function

[broken carbon]

of understanding it

[remote trail]

since you should be applying it around x = 8 which indicates a rotation left to right

[broken carbon]

im still not sure what to do

[remote trail]

so this is the formula right?

except using washer

so the radius will be x2-x1

[broken carbon]

yes

im not sure how to set it up though

[remote trail]

you can see in your drawing here what your radius should be

since you're rotating around the line y = 8

you can see the radius will be 8-x in this case from that line

to form that disc

[broken carbon]

when i was looking at disks, it told me to find R(x) i needed to get a value in between the bounds to find the proper equation for R(x)

is it different with washer?

[remote trail]

and because you have x=4y as your lower line it should be (8-(y^3-4y) inverting eqs

[broken carbon]

because im not sure how 8 - x happens

[remote trail]

I think

[remote trail]

no it should be the same

in this case your bounds should be from 0 to 8 looking at your drawn areas (assuming they are right)

you can basically think of the washer method as taking the larger disc and subtracting the smaller disc

so you can think of doing your disc for x^1/3 and then subtracting the disc for 4y

which will give you the area of the region

[broken carbon]

its in terms of x?

wait i read that wrong

[remote trail]

forgive me for the bad drawing but

[broken carbon]

i think

[remote trail]

it's in terms of y

[broken carbon]

yeah

[remote trail]

if you look at this you would substitute x = f(y)

[broken carbon]

this is also where im confused since im basing it off of the disk strategy

cuz if my bounds are 0 to 8 and im using y^3 and 4y

if i input 1 then 4y would be bigger

if it were 2 then y^3 would be bigger

[remote trail]

they should be the same function

[broken carbon]

now im more lost

[remote trail]

here

this is the desmos graph

[broken carbon]

yes

[remote trail]

try inputting x into these instead of y maybe?

input one into x = 4y, y=1/4

and then y=1^1/3 = 1

so x=y^3 would be larger

sorry into x=y^3 it is still 1

so basically it shouldn't change whether it's larger

I think you were just inputting the wrong value

[broken carbon]

maybe? im not too sure since im following the

pdf

i think im using the wrong function(?) to input values in

[remote trail]

ok so for x = y^3 and x = 4y

if you try substituting x = 1

you get 1 = y^3, y = 1

[remote trail]

and 1 = 4y, y = 1/4

[broken carbon]

cuz this is what i did for my bounds

but i also have y outside which is 0

so i kinda wanna backtrack a bit so ik the steps

[remote trail]

that's right yeah

you can see in the graph those are the intercepts

actually yeah you would use from 0 to 2 for the integral since you need to rewrite as y

[broken carbon]

ok so it is 0 to 2

cuz y = 0

[remote trail]

yeah

[broken carbon]

and i just found 2

[remote trail]

yeah they intercept at (8,2)

so because we are using Y for integration

it would be y =2

[broken carbon]

yea

so this is from my notes

as an example*

[remote trail]

yes

[broken carbon]

im not entirely sure where 1/2 comes from

im just assuming its a value picked within the bounds

[remote trail]

and what is our two radii atm

for this q

[remote trail]

yeah to see which one is larger over the interval

which we have already established

[broken carbon]

can we go over it again

sorry

cuz i didnt really understand it

[remote trail]

ok

so what step is confusing you

[broken carbon]

so after i found my bounds

im not sure how to go about finding R(x) or R(y) (general scenario)

based on knowing if larger or smaller

[remote trail]

so what's the general formula we will be using here

as a starting point

[broken carbon]

i think

its using one of the functions

and the line given

so in this case x = 8

which gives me

function - 8 or 8 - function

but im not sure what function i should be using

[remote trail]

just the general integral formula

for the washer method

for the axis of integration that we are rotating around

here is the general formulas

[remote trail]

we won't be using this one since we are rotating around the y-axis

instead of the x axis

[broken carbon]

yeah

so instead of x its y

[remote trail]

instead we will be using the second one here yeah

where x2 = r2(y)

[broken carbon]

what is r2(y)

my things are denoted by R and r

[remote trail]

sorry

basically in this case x2 is the larger radius

and x is a function of y

[broken carbon]

oo

would x2 be considered the

top function?

[remote trail]

so what is our larger and smaller radius is the next part

[remote trail]

yes

[broken carbon]

so i would be using

x = y^3

[remote trail]

yep

[broken carbon]

or y = x^(1/3)

oh ok

[remote trail]

since we are writing in terms of y

[broken carbon]

y^3

[remote trail]

it will be y^3

and we have already shown bounds are from 0 to 2

[broken carbon]

ohhh

so if i input 1 its just

1 and thats less than 8

[remote trail]

one thing to note though that's not the complete radius because we are rotating around the line x = 8

[broken carbon]

so i would be doing

8 - y^3

[remote trail]

yep!

that's the larger radius

[broken carbon]

ok now r(y)

[remote trail]

and then the smaller radius would be 8-4y

[broken carbon]

yeah

[remote trail]

so then you can substitute both your radii into the formula and solve

[broken carbon]

yes

[remote trail]

that should work from my understanding

[broken carbon]

i will be right back in a minute i am writing down my work

is this right?

i forgot to add dy at the end

@remote trail

[remote trail]

yeah that looks right to me

[broken carbon]

ok i will move onto the integration

did i do it right

@remote trail

[remote trail]

did you forget to divide by 7 with the 1/7y^7

[broken carbon]

oops

wait lemme go back

@remote trail is this right?

it looks more wrong than before..

is this how "big" they are supposed to look

[remote trail]

you could check your integral with wolframalpha @broken carbon

[broken carbon]

how do i use it

[remote trail]

go to the site and put in your integral

and it should solve it

https://www.wolframalpha.com/

[broken carbon]

ooh

i was right

i really hope i did the set up correctly then

thank you so much for helping me @remote trail

i was really struggling

[remote trail]

np

[broken carbon]

i think this wraps up my disks/washers

im moving onto shells next

[remote trail]

good luck

[broken carbon]

is there anything i need to look out for?

my test is on monday so i hope i can blitz through everything i need to know

[remote trail]

it would be useful to remember when to use either shell or disk/washer

[broken carbon]

oh yeah

how would i know?

[remote trail]

basically you can use either on a bunch of questions and sometimes you have to decide which is better

[broken carbon]

how would i know which is better

[remote trail]

I would encourage you to go research about it but it basically depends on whether the axis of rotation is parallel or perpendicular to the axis of integration

[broken carbon]

axis of integration as in

[remote trail]

and also it's always useful to look at the graph and ask yourself "what is the radius"

[broken carbon]

what it wants me to rotate it by?

[remote trail]

whether it's written in x or y

[broken carbon]

ohh ok

[remote trail]

because not all functions can be inverted

and then also which way you are rotating

[broken carbon]

if its asking me to work around the x axis

i would be using

[remote trail]

because the shell method does rotation in y while disk/washer does it in x and vice versa

[broken carbon]

dx or dy cuz it confuses me sometimes

[remote trail]

so it can be useful in those cases

[remote trail]

DX indicates write in X and dy indicates write in y

As a word of advice

[broken carbon]

of axis right

ok thank you

i need to go sleep now

how do i close

i joined server and then ask a question just today

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[magic wraith]

Am I right?

[magic wraith]

,rotate

[brazen dagger]

No

[brazen dagger]

5-(-20)=25

@magic wraith

[magic wraith]

Ohh

So the remainder is 25/2x-5?

Im stuck on #4

I cant figure out -x²/2x²

[brazen dagger]

-1/2

[magic wraith]

Am I on the right track?

Or is my process wrong

[brazen dagger]

Something is wrong, because you got 2x³+3x²+3x... So 3x(2x²+3) will give you 6x³ but you don't have a x³ term in(4x⁵+6x⁴....)

[brazen dagger]

This should be -6x³ I think

[magic wraith]

Ohh

[magic wraith]

The constant is still -1/2?

[brazen dagger]

What?

2x²+3)-x²+7x(-1/2
You will be left with 7x+3/2

I think

[magic wraith]

Yeah I did

.close

[cedar kiln]

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[hexed vortex]

Im a bit confused here

[hexed vortex]

Here is what I have

oh dang it put it in the wrong channel

thats what I have so far

I was going to parametrize and then integrate

and then plug in 2 for k

[topaz geyser]

yes thats what i thought

[hexed vortex]

but Im not sure how to write 1/z^k as u(z) + iv(z)

someone mentioned earlier that the question is just saying let k = 2

is that right?

it becomes easier if so but I don't see why they would put a k there then

[topaz geyser]

pretty sure

would just be 0 no?

[hexed vortex]

yeah it would

but I'd have to show my work and im not sure how

I could expand 1/(x+iy)^k

but

[topaz geyser]

z(t)=e^(it) t:[0,2pi] or am i going crazy

[hexed vortex]

I believe that also works hmmm

wait let me check my notes

[topaz geyser]

dz=ie^(it)dt

seems nice enough

[humble marsh]

you're working with complex numbers here, right

[hexed vortex]

yeah

[humble marsh]

so really (a(t) = \cos(t) + i \sin(t) = e^{i t})

[wraith dagger]

Invariance

[topaz geyser]

yeah

thats what im saying

[humble marsh]

oh you said that, sorry

[hexed vortex]

wait

but thats only for gamma

oh wait

no I see now

oh ok let me try that

[topaz geyser]

youd get int from 0 to 2pi of (ie^(it))/((e^(it))^2)

dt

dont know how to really use that bot, would make it look much nicer explaining lol

then just integral ie^(-it) from 0 to 2pi

right

periodic over the full period

so 0?

[rare vault]

$\int_{0}^{2\pi} \frac{ie^{it}}{{e^{it}}^2} , dt$

[topaz geyser]

thats it thank you!!

let me try use this thing lol

$\i \int_{0}^{2\pi} \frac{e^{-it}

[topaz geyser]

how would i write i as being multiplied outside the integral sorry?

[rare vault]

$i\int_{0}^{2\pi} \frac{e^{it}}{{e^{it}}^2} , dt$

[topaz geyser]

$i\int_{0}^{2\pi} \frac{e^{-it}} , dt$

[wraith dagger]

al ameen ayodele

[rare vault]

\frac is for fracitons, and it needs a top and a bottom. if you dont want a fraction, remove the whole \frac{}

[topaz geyser]

$i\int_{0}^{2\pi} {e^{-it))

[topaz geyser]

this is quite a cool bot

[rare vault]

$i\int_{0}^{2\pi} e^{-it} , dt$

[rapid pumice]

we can cancel e^it, right

[topaz geyser]

yeah

then you get i over e^(it)

[rapid pumice]

and can get 1/e^it

[topaz geyser]

you get i on top

just keep it outside the integral

[rare vault]

the markup language is called latex, you can look up guides - its quite simple once you get used to it

[rapid pumice]

and then simple constant int

[topaz geyser]

i never used latex in my degree so far lol, nice bot

[rapid pumice]

i am new to it

[topaz geyser]

@hexed vortex after a while of fiddling with the bot... this is what youd get. int of e^(-it) over its full period 2pi is zero since its periodic with period 2pi and its just 0

i believe thats enough

residue confirms since theres no pole inside the unit circle

z=0 is of order 2 so 0 residue

nice

[rare vault]

how do you type math!

[topaz geyser]

such a nightmare

[rare vault]

handwritten notes only? I switched to latex starting at like calc - crazy to me that you made it to (at least) complex analysis

yeah learn it for sure :)

[topaz geyser]

always handwritten

ill definitely learn some latex

seems very useful

love some complex analysis

[hexed vortex]

This is what I got using k

not sure what's next though lol

[topaz geyser]

input k!!

[hexed vortex]

shouldn't I integrate it first though

[topaz geyser]

if k=1 youd get 2pi

I_k = 0 for k>(or equal) 2

[hexed vortex]

yeah and shouldnt integrating with k show that for all k >= 2

[topaz geyser]

yup

youd get 0 on top obviously

you can prove it by noting that e^(ix) is periodic

so e^i(x+2npi) = e^(ix)

set x=2pi(k-1)

e^(-2ipi(k-1)) = e^(2ipi(1-k)) = e^(-2inpi) where n=k-1

since e^(2inpi) = 1 for any integer n, it equals 1, and thus 1-e^0 = 0 so it doesnt matter what its divided by, its still 0

i hope that seemed at all comprehendable by my horrible way of writing math in text

when i learn latex ill be better lol

[cedar kiln]

@hexed vortex Has your question been resolved?

[hexed vortex]

I think I almost got it

im just getting -1-1 insteadof 1-1

[topaz geyser]

where did you get upto

can you show me pls

[hexed vortex]

one sec ill show

nvmm I got it

I thought e^(-i2pi) was -1

but its 1

[topaz geyser]

yes yes its just 1

my explanation probably seemed awful in text form lol sorry about that

[hexed vortex]

no your good you helped me solve it

[topaz geyser]

great! glad to help

[hexed vortex]

thanks

.close

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

[topaz geyser]

e^it / e^itk = e^-it(k-1) not e^it(k-1)

dont want you losing a mark over it

[hexed vortex]

oh your right

thanks gotta fix that

[topaz geyser]

still the same answer but didnt want you to drop a mark for something like that

[hexed vortex]

do you know if it will still be 0 when k is less than 1?

what I have says it will

but I wanted to be sure

like is it always 0 except for when k = 1

[topaz geyser]

for k=0, itll just be 2ipi

for 0<k<1 itll be 0

to summarise all cases, you can say I_k ={2ipi if k=1, 0 if k>=2, 2ipi if k=0, 0 if 0<k<1}

[cedar kiln]

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[minor wren]

Hello, I have no idea what I'm doing wrong on this maths question....

[coral jewel]

whats your integral?

[normal cipher]

Integrate

First find the bounds

[minor wren]

Here is my working

ignore the part in the top left, that was from the question before

basically I tried to take the inverse of the two functions put together and integrate it

ingore the bounds on the integral, Ive got this question wrong multiple times so Ive tried different bounds that dont work

initially I had 0 and 2

sorry my working is a bit hard to read

basically I got to ln(2y) + y + yln(y) from 0 to 2 (although I dont even know if the solve integral is correct)

Which ones?

The point of intersection between the two curves is at x=ln2

Well- I’m not sure

Yes

Wait what?

e^0=0?

Yes, so how do you solve e^x = 0

[wraith dagger]

wtr.on

[minor wren]

Ohh

Ok

Yes

But I think the integral I got might be wrong

basically I got to ln(2y) + y + yln(y) from 0 to ln2

As in, this is already integrated

I think I see what you mean

But do I not need to find the inverse becuase it’s the area against the y axis?

[wraith dagger]

wtr.on

[minor wren]

Ok

[fossil sedge]

0.5?

[minor wren]

yeah, me too

it's correct, thanks a lot

I was overcomplicating it

[fossil sedge]

Do u have any other integral questions?

[minor wren]

plenty before my maths test on friday lmao

[fossil sedge]

Ohh are tests hard in there?

[minor wren]

yeah, tests are quite hard

im also preparing for the MAT

[fossil sedge]

Good luck(ʘᴗʘ✿)

[minor wren]

thanks! 😄

bye

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[small cloak]

in this question

i reached this:

so i got 13r = 82

r = 6.3

[haughty osprey]

Distance isn't the sum of radii

Look closely

[small cloak]

what does this mean

oh

distance is 82/2

so 41/13

i still didnt get the answer @haughty osprey

[haughty osprey]

You take to take diameter of the middle 3

Or 2r

But only of the middle 3

Not all

[small cloak]

i do not get it

can you please explain

[haughty osprey]

Radius is the distance from the centre of the circle to one end

And distance from end to end and passing through centre is 2x radius or diameter

[small cloak]

ok i get this

[haughty osprey]

In the image

You have to take the diameter of the middle 3 cuz the line passes end to end

[small cloak]

oh so it will be 1 + 2(2) + 2.5(2)+3(2)+4.5

[haughty osprey]

Yea

[small cloak]

oh so its 4

i get it now

thank you

another question

i am trying to find the diagonal

since its isocleces , i got 40 as the base

and the height is 10

no i do not know what to do

[south lagoon]

feels like pythagorean theorem

[small cloak]

i did 10^2 + 40^2

[haughty osprey]

Yes

[small cloak]

then sqrt

i got 41.5

[south lagoon]

huh

[small cloak]

can u try please

[south lagoon]

sqrt(20^2 + 20^2) is not 40

so it should be sqrt(sqrt(20^2 + 20^2)^2 + 100)

= sqrt(20^2 + 20^2 + 100)

= sqrt(500)

[small cloak]

i applied the 45 45 90 rule

[south lagoon]

which ig is 20sqrt(3)

idk the 45 45 90 rule

it should be 20sqrt(2)

not 40

[small cloak]

ohhhh

i got 30

thanks

[south lagoon]

idk just use pythagorean theorem

[small cloak]

can you help me with this? the wording confuses me alot

[cedar kiln]

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