help-38

[tribal jetty]

prob this image is easier to understand idk

if you have seen physics

is the same logic

[torpid moat]

No I understand it

the different graphs really helped!

thanks sm!

[tribal jetty]

ok

[torpid moat]

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[torn ether]

how should I solve this. I honestly have no clue where to start

[wanton rune]

seems like a trig-sub integral

[torn ether]

I think so based on my teachers notes but Ive never done one like this

[wanton rune]

just try it using this as a basis

[round mango]

There's a 2/z³ outside, I'd start by letting 1/z²=u

Multiply and divide by -1 first to form du at the top

[wanton rune]

huh

[torn ether]

??

[wanton rune]

ah

[round mango]

1/z²=u

[wanton rune]

i see

[round mango]

=> -2/z³=du/dz

[wanton rune]

it may be better to let u=sqrt(x^2-1)

hmm nah

[torn ether]

im so lost rn

[round mango]

Did you try letting 1/z²=u

Should get you somewhere

[torn ether]

sorry I just don't get this one

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[long iris]

Sorry

My bad

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[stiff briar]

i need help on 52 😭

[stiff briar]

i'm not sure if it's oscillating or not

but it converges to two values

0 and 2

so is that considered oscillating

and therefore diverging

[jagged fiber]

yes ur correct

[stiff briar]

okay thanks

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[jagged fiber]

np

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[ocean lake]

Hello! This is not a math related topic but i was wondering about how the availability bot and channels are made? I would like to do something similar in one of my study servers

[sly pewter]

not the person who made the bot but you can probably just add the bot to your server

[ocean lake]

may i know how?

[sly pewter]

hmm i couldn't find it in the app directory

but you can use a ticket bot

[ocean lake]

and it would do the same function?

[sly pewter]

it would be similar

[austere cedar]

https://github.com/discord-math/bot

[sly pewter]

ok cool 🤩

do that

[austere cedar]

That's the github. Not necessarily how you might actually add it to a server.

I am not certain myself

[ocean lake]

yea that was my next question lol

i'll try out a ticket bot since i cant download this one.. but thanks for your help!

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[tropic sequoia]

How to solve this do I make the denominator the same?

[tropic sequoia]

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[tropic sequoia]

.reopen

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✅

[tranquil widget]

isolate the x’s by splitting the numerators

Or find a common denominator

[tropic sequoia]

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[odd spear]

is the region from 1 to y the same as y to 1 for an integral?

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[thorny whale]

i think so, because area is always be positive right?

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[verbal nebula]

tried changing all to common bases in the expression t3-t2=t2-t1 but it got messy

[gritty harness]

write all of them with the base k using this rule

then try to simplify

[verbal nebula]

i got 2/log_k(m)=1+1/log_k(n)

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[orchid wagon]

explain like im 5, when we say function f: A -> B, what does A and B mean?

[marble wharf]

sets

[orchid wagon]

elaborate?

ive never seen this notation ever in my hs curriculum, yet it seems so prevalent in american hs

[marble wharf]

A, B are sets. A is the domain of the function f and B is the codomain

f takes inputs from the set A and gives outputs from the set B

[orchid wagon]

so A is the domain of the func, and B is the range?

[marble wharf]

no

range and codomain are not the same thing

[fathom ivy]

A: all valid inputs
B: all things that you think has a chance of being the output

for example if i define a function and call it shredder

i might have shredder: {paper, printed letters im not gonna read, some idiot’s laptop} -> {shredded paper, shredded letters, the shredder catches fire, the universe explodes}

and we can assign a mapping:
shredder(paper) = shredded paper
shredder(letter) = shredded letter
shredder(some idiot’s laptop)=shredder catches fire

note that nothing maps to the universe explodes, not everything in the codomain must have something map to it

[orchid wagon]

so codomain is just range, except its not reliable?

[fathom ivy]

it always contains the range, but sometimes the function is so not well understood that it might contain extra stuff

[orchid wagon]

so it must contain the range

[fathom ivy]

a silly example is f: R -> {0, 1}, where

f(x) = 1 for all x if we establish contact with aliens in the next 50 years

f(x)=0 for all x if we don’t

[marble wharf]

and its just generally easier to be able to say f:R->R and not worry about the actual range of f for now

[orchid wagon]

makes sense

[marble wharf]

you are just saying: f takes as input a real number and produces a real number as output

which numbers it can actually output (aka the range) is not important yet at that point

[orchid wagon]

alright i think i get it now

thanks

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[sage harness]

A light house of height 53m stands on horizontal ground level with the base of a 80m cliff. The angle of elevation of the top of the cliff from the top of the light house is 27 degrees. Calculate theangle of depression of the base of the cliff when viewed from the top of the light house, leaving your answers to 1 decimal place.

[sage harness]

i just dont get how it would look like so could somebody draw a diagram for me?

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[sage harness]

<@&286206848099549185>

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[distant abyss]

hello! I need help on number 3

[devout drift]

What have you tried?

[distant abyss]

Through the set operation, i used
n[(C n E n J) U (C n J n E^c) U (C n E n J^c) U (E n J n C^c)]

I used venn diagram to come up with that notation

After expanding

[devout drift]

It will be easier to understand with the venn diagram

Can I see the venn diagram

[distant abyss]

I used the venn diagram again to cancel out the mutual exclusives (which is not a formal proof sadly), and after all that I was left with

n (C n E n J) + n (C n J n E^c) + n (C n J^c n E) + n (C^c n J n E)

[devout drift]

I meant with the elements of the sets as well (for example how many eat in caffeine) those you have already found out right?

[distant abyss]

so 42 + (126 - 42) + (112 - 42) = 196

[distant abyss]

Ah yes

[devout drift]

You don't expect me to solve this all over again just for last part when you already have that information

[distant abyss]

But how am i supposed to get number 3 tho from number 1

[devout drift]

I don't know just show me the elements

For now

[distant abyss]

uhh number 1 is n [ C n (E U J)^c] = 112

number 2 is 0

[devout drift]

Nvm

I'll just do it

Wait 10 hours

[distant abyss]

damn.

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@distant abyss Has your question been resolved?

[distant abyss]

hello @devout drift are you still on it and willing to continue?

[distant abyss]

ah i see what u mmean now

C= Caffeine, E = Express, J = Joe

n(C) = 308
n(E) = 238
n(J) = 504
n(C n E) = 112
n(C n J) = 126
n (E n J) = 42
n ( C n E n J) = 42

[devout drift]

Yea

[distant abyss]

mb i think im too late on that now

[devout drift]

You made me find all the elements

[distant abyss]

i am so sorry

im incredibly slow

[devout drift]

Your first part is correct

Last one imma check

[distant abyss]

okay, thank you sm

[devout drift]

It's correct

You did

number of elements in exactly two sets + number of elements in all three sets correct?

[distant abyss]

yeah

[devout drift]

It's correct

[distant abyss]

alright, but how did you do it?

[devout drift]

This would have been fast with all those elements

[distant abyss]

im just not sure since my solution involved using venn diagrams to determine those which are mutually exclusive, and im kinda sure that my professor does not find that rigorous or sumn

[devout drift]

$n(C \cap E) + n(C \cap J) + n(J \cap E) - 3n(C \cap J \cap E) + n(C \cap J \cap E)$

[solid kiln]

ColdTee

[devout drift]

Basically

[devout drift]

I find set operations easier sometimes

For questions like these

[distant abyss]

i did use set operations, but when i expanded, some terms like n[ (C n E n J) n (C n J n E^c) ] came up, and since i dont know any properties i can use to get an answer, i used venn diagram instead, and it showed me that they;re exclusive

[distant abyss]

oh i see

you headed directly for the ones with the arithmetic

based on how my prof taught it, she wants us to denote the thing being asked for in n(X) of something

then use properties to expand, blah blah blah

i think i shudve went with that instead

thanks so much for the time and energy spent @devout drift

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[red jolt]

doing it for 5th time now and still can’t solve it. what am i doing wrong here?

[brazen forge]

arent you overcomplicating this?

[digital bolt]

if you think about it, you should be using the negative root

a-sqrt(y-a)

the one you're integrating is like the right side of the parabola

whereas you want the left

if you think about it, sqrt(y - a) + a will never touch the y-axis

you want a - sqrt(y-a)

to match the graph

but also yes you are overcomplicating this

[digital bolt]

just integrate wrt. x to get S+T and then subtract a^2 to get S

[brazen forge]

you should find a with this info alone

dx in integral ofc

[red jolt]

oh for real

thanks for helping guys

i’m not good at understanding integral

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[quick hound]

if i do trigonmetric sbustitutin with this

[quick hound]

i tried complete the square

and this what i get

what do i even sub with this??

[rough goblet]

x - 4?

[quick hound]

really?

[rough goblet]

idk

i dont have much intuition for ints

gimme a moment

hm

[patent crow]

that leaves you with a sqrt(8u^2+2) integral

[quick hound]

it has to be a trigonemtric thing

[rough goblet]

yea ik

[proven sluice]

Do u know standard integrations of such function?

[rough goblet]

am dum lmao

[quick hound]

wdym by this

[proven sluice]

Int of √x²+a²

[quick hound]

like the other functions

the subs were pretty self epxlenatory

but this one is weird

[quick hound]

ye it works with tan

but its not x^2-a^2 form here

[proven sluice]

That's when it's in denominator

[quick hound]

oh

different in nominator?

[proven sluice]

Yeah

It's much longer

[quick hound]

uh oh

do u know the name?

or still trigonemtric substitution

[proven sluice]

Wait

[proven hound]

,calc 16*8

[solid kiln]

Result:

128

[quick hound]

oh is it -2

[proven hound]

Ig it should be -2

Ye

[quick hound]

still dont know how to tdo the mian part;-;

[heady lava]

take out sqrt(2) maybe

so the inside becomes 4(x-4)^2-1, idk if that actually helps tho

[quick hound]

still the (x-4)^2

i dont know what to do with it afterwards

is the problem

[proven sluice]

Use these

[quick hound]

oo

thanks

whats this called?

[heady lava]

might be stupid, what if
(x-4) sqrt(4(x-4)^2-1) + 4 sqrt(4(x-4)^2-1)

[proven hound]

There's an x outside though

[proven sluice]

Special integrals ig?

[quick hound]

yeah

i needa know the proof

im gonna google it

[proven sluice]

That's long , sure u can

[heady lava]

I think this is doable, not sure

[proven sluice]

Well yeahh that might be challenging

[heady lava]

im guess you can integration by parts? it’s just super annoying to do

[proven sluice]

By parts maybe?

[quick hound]

it has to involve

some sort of trognmeitc sub

[proven sluice]

Yeah

[quick hound]

i dont know what it is

[proven sluice]

Idk u can try

[quick hound]

how do i remove the -64x

[proven sluice]

I think just take x² as t

[proven hound]

What will be the first function then

[proven sluice]

And it's dt is 2x

So it's solvable ig

[heady lava]

differentiate x and integrate that monstrosity

lol

[proven hound]

Lol

[heady lava]

u=x-4 looks like it works for the first part

[quick hound]

if i do that using x^2 sub is easier

[proven sluice]

Ok nvm

Cuz then x will be root t

Much more difficult

[quick hound]

https://www.youtube.com/watch?v=NeJZKIQhELE

mighve found it

[charred trail]

isnt it x = t + 4

[quick hound]

no

[charred trail]

no like because when completing the square you get something (x-4)^2 or whatever, so let t = x-4, i.e. x = t+4

unless im getting confused

[heady lava]

the integral calculator agrees with this

although it tends to use stupid methods sometimes

the antiderivative is a heck of an expression btw

[quick hound]

yeah

im trying to make it into integration form

[proven sluice]

But again by parts right

[quick hound]

look at this guy

hemade the complete the square

by making the inside of the square

into smth else

[heady lava]

yeah but in this case it's very hard to find a good substitution

[quick hound]

then using the shit outside as like a trigonemtric sub

lemme cook i have an ideda

[charred trail]

but isnt the question asking what substitution to make

[quick hound]

what trig substitution

[heady lava]

judging from the integral calculator result, it would be absolutely insane if there was a trig sub for this

[charred trail]

if the integral does come to the form of some sqrt(8u^2+2) or whatever it is, probs best to use the fact that sec^2(x) = tan^2(x) + 1

[quick hound]

ye

im trying

[heady lava]

oh I see

also it's sqrt(8(x-4)²-2), so it should be ||x=sec(theta)/2 + 4||

[quick hound]

GOT IT

oh

you guys got it too

nice 😄

ok thank god for that yt vid

that was hard to wrap around

[proven sluice]

Ok maybe I got it

[heady lava]

I keep forgetting you can just add random constants to the trig sub lol

[proven sluice]

Ok nice lol

[quick hound]

ok trying the next one

ok so this one i use sin

i think

use 1-sin^2(x)

ok nice got it

thanks for the help guys

how do i integrrate tan^2x

nvm i did it

but i got smth wrong somewhere else

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[viscid pagoda]

Como eu posso práticar mais matemática, tipo eu assisto uma aula sobre frações, e depois de um tempo eu vejo que frações é muito mais aprofundado do que aquilo que eu pensava, o que eu faço?
Experiência? Tentativa e erro? Eu vim aqui para poder perguntar para vocês que obviamente estão a milhares de kilometros a minha frente, por favor alguem poderia me ajudar nisso?

[worthy eagle]

translate it for us

[fathom gull]

How can I practice more math, like I watch a class on fractions, and after a while I see that fractions are much more in-depth than what I thought, what do I do?
Experience? Trial and error? I came here so I could ask you who are obviously thousands of kilometers ahead of me, could someone please help me with this?

google translate

[worthy eagle]

maybe if you can show an example I'd be appreciate it

[viscid pagoda]

Sorry, I understand English, I just asked the question on the wrong server

[worthy eagle]

no worries

[viscid pagoda]

what can I do?

[worthy eagle]

do you have a question that you don't get

that's what I asked

I dont usually get "like I watch a class on fractions, and after a while I see that...depth than what I thought"

[viscid pagoda]

Relax, I managed to think of something here, thank you very much

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[blazing socket]

I have this example problem that my instructor didn't get to finish in class, could someone explain where the csc comes from in the second to last step?

[fierce bay]

yes

its just the coefficient sec/tan factored into tan^2+1 and simplified

[vagrant marsh]

in line 5 you also need to divide the tan to the 1

oh

i see, this is a poorly written solution

[blazing socket]

yea my instructor was running low on time and tried to get it done as quickly as possible 😭

[vagrant marsh]

sec/tan simplifies to csc

when multiplying sec/tan with tan^2, you get sec*tan, and multiplying sec/tan with 1, you still get sec/tan which you can simplify

[blazing socket]

ahh gotcha

i've got it figured out now, thank y'all :)

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[vagrant marsh]

yw

[blazing socket]

.reopen

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✅

[blazing socket]

i'm now attempting to finish the problem myself and i have no idea how to get back to x terms from here 🗿

do i need to get all of these trig functions into the form of tan theta to convert it back?

[craggy sage]

u can make a right angle triangle in which one angle is theta and side oppo to it is x and adjacent to it is 1 and hypotenuse is root(1+x2) and then write the ratios

[blazing socket]

ohh gotcha

ty

[knotty locust]

Yes when you did the sub x=tan(theta) remember what that tells you

Using SOHCAHTOA to form the triangle

[blazing socket]

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[blazing socket]

i've got another problem now, and i'm not sure how to choose which trig function to use

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[full dock]

No idea how to integrate this thing

[frozen plover]

right path

Do a + 1 - 1 in the numerator

split the fraction

x^2 + 1 and x^2 - 1

then divide both fractions numerators and deniminators by x^2 and complete the square appropriately

[full dock]

Okay I’ll try that

[frozen plover]

this is gonna get a little nasty

you have been warned

[full dock]

I saw the answer for the integral and I’m already scared

[frozen plover]

haha

[full dock]

Is this what you mean

For the split

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[tribal fractal]

for this question, the derivative does not exist at x = 0

[tribal fractal]

is that enough to say why the formula doesnt work?

it has a discontinuity

[vagrant prism]

yes

[tribal fractal]

thanks

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[marble vault]

How would I find the percentage of the triangle that doesn't contain the circle

[sly pewter]

Where are you at right now

[marble vault]

My house

[sly pewter]

Like where are you at in the problem lol

[marble vault]

Lol i was gonna find the area of the circle and then after that find the heighrt of the triangle so i can find the area of it and then subtract the area of the circle from the triangle and get a percent

But im not sure how to fully find the area of this circle

Im in 9th grade btw so this is prob like a stupid easy question for u lol

[sly pewter]

There must be some geometry rule about circles being inside a triangle

I did geometry in 8th grade and it went by so fast 😭 let me see

[marble vault]

Damn thats impressive bro nice work

[sly pewter]

Not impressive if you can’t maintain

[placid juniper]

radius = 2 x area of triangle / perimeter of triangle

[sly pewter]

Real formula

[marble vault]

how do i find the area of the triangle though

i dont have a height

can i juse use herons?

[sly pewter]

What is the name of the theorem

[placid juniper]

idk i made it up

[marble vault]

🔥🔥🔥

[placid juniper]

💀

no too complicated

did u get that off chat gbt

[marble vault]

no we learned it in class dawg

now how do i find the circle

[placid juniper]

https://youtu.be/PM70wEvKnE8?si=xggZMllbpYfFle64

[sly pewter]

Oh use the tangible line rule

Tangent

Two tangent

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[marble vault]

.reopen

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✅

[marble vault]

@sly pewter how do i use two tangent if it isnt a right isoceles

im a little confused

and @placid juniper in the vid i sent u the way he does it isnt possible without right isoceles

also i had to redraw the triangle got new area as 22.27

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[distant bay]

HELPERS IN 31

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Please don't occupy multiple help channels.

[distant bay]

HELP

[worthy eagle]

!1c

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Please stick to your channel.

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[rain light]

Can someone help me with this?

[gaunt wave]

So the sides are going to be equal

for the upper one, the hypotenuses are going to be equal

and the other sides are going to be equal assuming they are the respective sides

@rain light

[rain light]

Yeah

[gaunt wave]

Does what is above make sense?

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[distant bay]

help

[burnt mulch]

Where’s the question

[distant bay]

nevermind

new question

Pls help

<@&286206848099549185>

[distant bay]

HElp

[candid roost]

what do you need help with?

[distant bay]

This question

[candid roost]

The last one you posted, the image.

[distant bay]

Yes

[candid roost]

okay give me a sec to figure it out.

I think you just find the area of a rectangle then subtract the area of a triangle.

What do you think?

[distant bay]

Idk

i think you add instead

[candid roost]

Find the area of the rectangle first then subtract the triangle.

[distant bay]

I got it wrong

[candid roost]

Give me a sec to run a calculation on it.

[split chasm]

show your work

[candid roost]

Okay so the rectangle's area is A=l*w

A=9*12=108

Can you now figure out the triangle and then subtract the two.

You can look at the image to get the base and height of the triangle.

The base is 6 and height is 5

Area of triangle is A=0.565=15

Thus the total area is rectangle - triangle

You should get 93 as the area so see if you can match my answer.

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[full flax]

i can't seem to find online if this is true or not. Can i use sum rule of asymptotic notation to add like this?

[heady swallow]

gpt?

[full flax]

yes idk where else 😭

[tribal jetty]

wtf

[frail heron]

yk it’s totally bs, right?

[full flax]

understood

.close

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[pure sluice]

how would you solve for x when the denominator and numerator have x in both of them

[pure sluice]

oops ignore the negative sign infront of 2

[edgy willow]

you can just ignore the bottom

[pure sluice]

ohhhhh

okok!!!

thank you

[split chasm]

can't fully ignore though, still need to ensure that the denominator isn't 0 at the values you found when determining whether a solution is valid.

[gusty heart]

but for this problem you can just keep in mind that you have to make sure the denominator isnt 0

[pure sluice]

so as long as (3x-x^2)^2 doesn't equal 0, it's fine?

[gusty heart]

yes

[pure sluice]

ok tahnk you!!

.close

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[quiet turret]

how do i solve d

my equation is $500=4x^3-80x^2+400x$

[solid kiln]

pixel

[quiet turret]

right?

someone please?

😭

[young swift]

are you sure that's the equation?

the equation should be $x^3-40x^2+400x$

[solid kiln]

InfinityMango

[young swift]

I got there by using (20-x)(20-x)x

[quiet turret]

lol

its (20-2x)

not (20-x)

[young swift]

oh im dumb wow

[quiet turret]

https://cdn.discordapp.com/attachments/238956921581862913/1218075563785195601/800520659884572692.png?ex=66065845&is=65f3e345&hm=9e1250961cf559353d155adffb8703e2743715489f20aa455f5fd9d6fddae863&

equation should be this

how do i simplify it further now

[young swift]

anyway, you should just be able to bring the 125 over to the left

[quiet turret]

$x^3-20x^2+100x-125=0$

then?

[solid kiln]

pixel

[young swift]

then you can refactor it

[quiet turret]

wdym by refactor

[young swift]

get it back into the form (x-a)(x-b)(x-c)

you can use polynomial division

oh wait, it might be easier to use rational roots

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@quiet turret Has your question been resolved?

[quiet turret]

whats that

[young swift]

its one way of finding your first factor

https://twinklsecondary.blog/how-to-factorise-a-cubic-a-level-maths-revision/

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[static fern]

Can you check this?

[vernal warren]

it is correct

dont know why teacher wrote "no"

[static fern]

Why did I get 2/10

lol

He gave me 2/10

[vernal warren]

it is correct tho u didnt do any thing wrong and proved

[static fern]

Exactly

Thanks

Are you sure it is correct

?

[wind sonnet]

you did mess up the signs

[static fern]

Where

I removed the minus 1

[split chasm]

think a lot of marks were deducted from the laziness of not having theta

[static fern]

Nope

[split chasm]

as in you don't think that's why you lost marks?

i also don't see this question being worth 10 marks

where does it say how much its worth

[static fern]

It doesn’t but he said in class

[static fern]

Yes it isn’t

[split chasm]

did you ask your teacher why you lost marks?

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[dusky tinsel]

how to do ' - would u derive accel twice to get dispalcement, howevre i di that and when t = 0, i got -12 (not 1m to the right as said)

[dusky tinsel]

this question

[trim jolt]

@dusky tinsel Has your question been resolved?

[hallow crane]

You wouldn't be deriving, but rather integrating

[dusky tinsel]

oh yea

[dusky tinsel]

while i do that, can any1 helpw ith this maybe you kanna

what i tried is setting say, point h on the x acis and b on y
then find area of rectangele (b * H)
it but divided/4 should = the area under the curve from 0 to h
but i stuck when i get bh = h^4 since i cant... prove that

[hallow crane]

so what question are we doing? lol

choose 1

[dusky tinsel]

k this one

but im just doign the working rn

[hallow crane]

I mean if you want we can do this

let the length of the rectangle be l and the "width" of the rectangle is f(l)

[dusky tinsel]

i just thought while i try do the acceleratiion one again maybe you can figure out where i went wrong/how to do ^

[hallow crane]

well we can do the rectangle question at one since u said you were working on this

[hallow crane]

what would the area of the rectangle be then?

can you get the area as a function of l?

[dusky tinsel]

lw

[hallow crane]

what is w?

[dusky tinsel]

width and length

[hallow crane]

w = f(l)

I know

I mean w is just f(l)

[dusky tinsel]

ah k

[hallow crane]

so for some l, we have the width is l^3

no?

[dusky tinsel]

hm

area of rectangle is length * width, so whys w = f(l), since the function doesnt even have a 'l' value so its not rly a function i dont think, confused there

[hallow crane]

area = length * width and you can do nothing about it if you're doing single variable calculus

you want area as a function in a single variable

Anyway, look at the red point

if x = l then the y coordinate is necessarily l^3 lol

[dusky tinsel]

oh didnt know that

is that cuz x^3

and how that works

im sorry i have to go eat dinner right now so wont respond just wanted to let you know

[hallow crane]

i mean if you want to find the corresponding y value (on the curve) for a given x then that's exactly what you'd do ig

I.e. x = 2 would give y = (2)^3 = 8 and this point lies on the curve y = x^3

[hallow crane]

[dusky tinsel]

oh ye that totally makes sense, just flew past my mind thx

[hallow crane]

okay so what's the area in terms of l?

[dusky tinsel]

l^4

i think i got it from here hopes

[hallow crane]

nice

now try to integrate the function to get the area under the curve

so integrate the function from 0 to l and then yeah you should immediately have your "proof"

[dusky tinsel]

ye nice thx

[hallow crane]

Now for this, you first integrate a(t) dt to get the velocity function + some constant

You're supposed to figure out the value of that constant by using the fact that it was "initially at rest"

Then you integrate v(t) to get x(t) + some constant again

which you can determine with the "1m to the right at the start"

[dusky tinsel]

yep i got 3t^2 -2(t+1)^4/12 + 2/3t + 7 = s

works for t = 0 to give 1

at t = 2 its displacement is 41/6

[hallow crane]

Hmm what? That's not what your check should be

[dusky tinsel]

hmm

[hallow crane]

You're supposed to make it work for t = 0 by finding the appropriate constant

[dusky tinsel]

oh yea

i got the equation

i did the c's

[dusky tinsel]

like thats the eqaution i integrated for displacement

or am i... delusional and wrong again

[hallow crane]

did you have $v(t) = 6t - \frac{2}{t+1} + C_1$

[dusky tinsel]

uhh

[solid kiln]

kanna

[hallow crane]

no?

[dusky tinsel]

6t - (2(t+1)^3)/3 + c

where c was 2/3

[dusky tinsel]

sorry long reply

did you... derive the t+1^2 shouldt it add a power

[hallow crane]

no it looks like you're deriving the second term

[dusky tinsel]

oh

uh

[hallow crane]

we're integrating

do you know u substitution?

[dusky tinsel]

wait, i thought u.. add a power in integration

[hallow crane]

,, \int x^n \text{ dx} = \frac{x^{n+1}}{n+1} + C

[dusky tinsel]

like t --> t^2

yep

[solid kiln]

kanna

[hallow crane]

for n != -1

[dusky tinsel]

oh wait its ^-2

hence ^-1

not 3

[hallow crane]

yeah and (t+1)^-2 should have a -1 after the addition of exponent lol

[dusky tinsel]

k i got 3t^2 - 2ln(t+1) + 2t + 1 = s

[hallow crane]

okay but what did you have for C_1?

[hallow crane]

here

[dusky tinsel]

2 = c1

[hallow crane]

Looks good

[dusky tinsel]

so i can make t = 2 for the dispalcement equaiton

but that gives displacement

how would i find the coords of particle b

[hallow crane]

,w integrate 6x - 2/(x+1) + 2

[solid kiln]

Results provided by WolframAlpha

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

Requested by kannaaaaaaaaaaaaaaaaaaaaaaa

[hallow crane]

,w -1 + 2 x + 3 x^2 - 2 log(1 + x) + c = 1 when x = 0

[solid kiln]

Results provided by WolframAlpha

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

Requested by kannaaaaaaaaaaaaaaaaaaaaaaa

[hallow crane]

Sure your s(t) is correct

[hallow crane]

yeah you'd plug t = 2

[dusky tinsel]

i got +1 not -1

wierd

[hallow crane]

+1 is correct

[dusky tinsel]

k

[hallow crane]

c = 2 so -1 + 2 = 1

[dusky tinsel]

so is the answer just s(2)

or do you need cords of the particle - which idk how to get

[hallow crane]

I mean that's where the collision presumably occurs lol

well for those particles to collide, they have to be at the same place

disregarding the dimensions of the "particle" itself

[dusky tinsel]

yep

[dusky tinsel]

just that we know dispalcement = 17 -2ln(3) = s(2) but question asks where, so idk if we need coords on a plane as in (x, y)

or are you saying no then to that

[hallow crane]

It's like a 1dimensional problem

[dusky tinsel]

ah lol

[hallow crane]

x = 1 when t = 0

[dusky tinsel]

so basicly dont include physics into it knowing what disaplcement actually is

[hallow crane]

No i mean it's not in 2d where you have the variance of y

It's just along a line

[dusky tinsel]

hmm

[hallow crane]

your function is x(t) = something

there's no y(t)

[dusky tinsel]

hence anwer is just how far away from origin (in a straight line ofc) the particles collide
just when i see colliding particles i imagine a certgain point where they coolide, not a dispalcement from origin

[hallow crane]

Yeah, that's what you'd be dealing with in 1 dimensional physics problems too

It's either along a vertical straight line or a horizontal straight line

[dusky tinsel]

k ty

[hallow crane]

Sure np

Are you done with your questions?

[dusky tinsel]

yep

.close

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[quiet wasp]

hey quick question

[trim jolt]

Please don't occupy multiple help channels.

[quiet wasp]

sorry

[wraith hinge]

say?

[quiet wasp]

can anyone do this

[radiant quartz]

@quiet wasp which part(s) are you stuck on

[quiet wasp]

c and d

[radiant quartz]

okay we'll start with (c)

[quiet wasp]

thx

[radiant quartz]

it tells us the range of h is the same as the range of g

so you should be able to immediately write down γ and δ

[quiet wasp]

so 0 and sqrt(3/2) ?

[radiant quartz]

because its bijective hence surjective

yep

[quiet wasp]

ohhhh thanks so much

what about alpha/beta?

[radiant quartz]

you can work out which value of x does h(x)=0

and which value of x does h(x)=√(3/2)

[quiet wasp]

-pi/6 +2kpi?

[radiant quartz]

yes but you need only one value

because we dont want every interval that maps it, but only one, because its bijective hence injective

[quiet wasp]

ohh right

so which should i use

[radiant quartz]

the one that has 0 in it

right?

it says h(0) = ½

[quiet wasp]

yeah

[radiant quartz]

So the domain of h better contain 0

[quiet wasp]

so alpha = 0?

[radiant quartz]

nuh uh

[quiet wasp]

oh damm

[radiant quartz]

does 0 = any of these?

[proven sluice]

[-π/6,π/2]

[radiant quartz]

Chill aditi im getting there

[proven sluice]

aight

[quiet wasp]

no it doesnt*

oh wait it does

sin(-pi/6) = -1/2

so yes

[radiant quartz]

yep

so we have α = -π/6

[quiet wasp]

ohhhh rightt

[radiant quartz]

and by a similar process you can pick β

[quiet wasp]

so it = 3/2 right?

sqrt(3/2) *

so therefore pi/2?

[radiant quartz]

yeag, h(β) should = rt(3/2)

Yep

[quiet wasp]

so beta = pi/2

[radiant quartz]

correct

[proven sluice]

If that's the case then beta diff ryt?

[radiant quartz]

and we r done with (c)

[quiet wasp]

thanks so much appreciiate it fr

[radiant quartz]

on to (d)?

[proven sluice]

root3/2 is for π/3?

[radiant quartz]

Or u get it now

[quiet wasp]

yes sirr

[radiant quartz]

you're getting confused. h(π/2) = rt(3/2). Youre thinking of sin(π/3) = rt(3)/2

Note the change in parenthesis

and the different functions

[proven sluice]

Oh alrightt mb thanks

[radiant quartz]

okay so h is bijective which makes finding h^-1 s domain and codomain a lot easier

theyre just going to be the opposite to h

[quiet wasp]

but how do we know where to restrict the domain to?

[radiant quartz]

the domain for h^-1 will be the codomain of h

and we already worked out the codomain of h

[quiet wasp]

[0. sqrt(3/2) is new domain?

[radiant quartz]

correct

forgot to close your square bracket but yes

[quiet wasp]

and the new range is [-pi/6, 7pi/6]

[radiant quartz]

Codomain

but yes

[quiet wasp]

yeah sorry

ok that makes sense, but how would i explain part c

mathematically and backed up with reason

are u able to write it out and send, if not its fine

[radiant quartz]

we already talked through it

[quiet wasp]

sorry to sound annoying but r we able to clarify c one more time?

[radiant quartz]

just explain what reasoning you used

Okay sure

[quiet wasp]

thx

cause i get confused with bijective surjective and injective and what it all allows us to do

[radiant quartz]

the range of h is the range of g
so the range of h is [0, rt(3/2)]
h is bijective so it is surjective
h is surjective so its range is its entire codomain
so its codomain is [0, rt(3/2)]

[quiet wasp]

ohhh this makes total sense now

how can we now relate this to find alpha, beta, gamma, delta?

[radiant quartz]

each of the intervals that make up g's domain individually map the entire range twice
h is bijective so it is injective
h is injective so each value should only get mapped to once
each value gets mapped to once so h's domain contains only one half of one of the intervals from g's domain
h(0) exists so 0 is inside h's domain
only one of the halves of the intervals from dom(g) contains 0
this interval is [-π/6, 7π/6]
the relevant half of the interval is [-π/6, π/2] because it contains 0
this is the domain of h

so now we have α, β, γ, δ

[quiet wasp]

it makes sense but to clarify, what does this mean "ach value gets mapped to once so h's domain contains only one half of one of the intervals from g's domain"

like the half of one interval part

[radiant quartz]

okay so if you see how g maps each interval

at the lower end of the interval g is at a minimum

then g increases towards a maximum

and symmetrically decreases back to a minimum at the end of the interval

[quiet wasp]

right yeah

[radiant quartz]

so each value that g maps to, gets mapped twice by each interval

once in the lower half and once in the upper half

[quiet wasp]

OHHH

so we restrict it so we only map each interval once

[radiant quartz]

yep

[quiet wasp]

so the domain of h = [-pi/6, pi/2] ?

[radiant quartz]

yep

[quiet wasp]

so that means alpha and beta = -pi/6 and pi/2 respectively

and, gamma and delta = h(-pi/6) and h(pi/2) respectively?

[radiant quartz]

yep

[quiet wasp]

so does gamma = 1/sqrt(2) and delta = sqrt(3/2)

[radiant quartz]

well gamma = 0 but delta yes

[quiet wasp]

oh how come gamma wrong?

[radiant quartz]

you just evaluated h wrong

,calc sqrt( 1/2 + sin(-pi/6) )

[solid kiln]

Result:

7.4505805969238e-9

[radiant quartz]

0

(calculator has rounding error

[quiet wasp]

i thought it was sqrt(1/2)

[radiant quartz]

well you can just plug it in and find out its not

[quiet wasp]

oh shit its 0 haahhaha

my bad

i was plugging in wrong sorryy

[radiant quartz]

i did tell you so

[radiant quartz]

🤷‍♀️

[quiet wasp]

hahahahah fair enough thanks very much

u literally saved me

this topic was garbage, now its less garbage

alr cya thanks alot appreciate the help!

[radiant quartz]

o7

happy to help 👍

[quiet wasp]

bye!

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[crimson void]

least value of n such that 25^n + 16^n leaves a remainder of 1 upon being divided by 121

[fathom gull]

!status

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[crimson void]

2

i tried putting n = 3k+1,3k+2,3k and taking mod 121

but couldnt continue from there

the reason for taking n = 3k,3k+1,3k+2 was that 5^3 and 16^3 ,both leave remainder 4 upon being divided by 121

i thought maybe that could lead somewhere

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<@&286206848099549185>

[void anchor]

So you're saying you're going to find it manually?

step by step?

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[crimson void]

<@&286206848099549185>

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[river yoke]

Where is 1 represented here?

Shouldn't the beginning of the series begin with 1 and k? But the beginning of the sequence they wrong begins with k(k-1).

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[proven palm]

I'd like to know how this is not a counterexample to the inverse function theorem: let f(x) = x^2, let a = 0. Then f' is invertible and continuous everywhere, but f is not invertible in any neighborhood of a.

[naive crest]

you forgot that the derivative being non zero at a is a condition

[proven palm]

huh it doesn't say that in my textbook

wait unless being a C'-mapping implies nonvanishing derivative...

[trail ingot]

send pic of statement in book

[proven palm]

being a C' mapping is just continuous differentiability so the condition is not implied from there...

[marble wharf]

f'(a)=0 is not invertible

[proven palm]

so the texbook just has an error then?

[marble wharf]

in your example f'(a)=0

which is not invertible

[proven palm]

oh wait

but why would that imply non-invertibility? In my example f'(x) = 2x which should be invertible everywhere by the inverse function g(x) = (1/2)x ... ?

Or am I totally misunderstanding inverses here...?

wait....ohh

OHHH

i see now

because f'(x) is the linear transformation, not just a value

thanks so much!

.close

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[languid coyote]

Hi! Wanted to confirm some intuition regarding linear maps, null-spaces, and matrix representations.

[languid coyote]

From Axler, we know that:

1. Let $\dim \text{null} : T = k < n$. If this is the case, $\exists$ a matrix representation where $k$ of the columns have all $0$, is that correct?

2. For any matrix representation of $T$ however, this is not necessarily true, but with some matrix manipulation, we should be able to get to the matrix representation where $k$ columns have all $0$, is that correct?

3. Could you help demonstrate how to go from the matrix representation using the standard basis, to the matrix representation with a column of $0$s for the following linear map: $Te_1 = e_1 + e_2$, $Te_2 = e_1 + e_2$?

Just those three question, in order from top to bottom!

[solid kiln]

txptr

txptr

txptr

[neon dirge]

@languid coyote is dim null = dim ker?

in your notation

[languid coyote]

yes, sorry just using Axler notation

[neon dirge]

kk np

yes first is correct

bc you can bring it into Rref form via Gauß algorithm

for columns and get k zero-columns

@languid coyote 2nd is true as well, slightly oddly worded but yes

[languid coyote]

okay, i guess what im failing to connect then is

seeing it in action

so wanted to see if you could walk through one with me

[neon dirge]

sure

[languid coyote]

using the linear map above I stated

where $e_i$ are the standard basis vectors

[solid kiln]

txptr

[neon dirge]

ok so we have the following matrix:

now via Gauß-Algorithm, we can subtract rows and columns from each other

if we subtract the first column from the second we receive

which gives us the form with a 0-column

[languid coyote]

oh wait hmmm

i thought gaussian elimination you can only use rows

also, in this case, how have the basis vectors changed

for both $V$ and $W$

[solid kiln]

txptr

[neon dirge]

oh in case you didn't have that yet, you can transpose a matrix, which preserves its core properties, but changes the way everything is arranged

which also allows you to perform manipulations both on rows and columns

[languid coyote]

ah i see hmmm i think transpose comes much later in the axler book

[neon dirge]

kk, transposing just means flipping the matrix diagonally

so in the above case

[languid coyote]

so im just tryna understand in terms of whatever basic properties have been explained so far

[neon dirge]

before our basis was (e1, e2) right

[languid coyote]

yep

[neon dirge]

for V

and now after the column manipulation

notice that we subtracted the first from the second

[languid coyote]

mhm

[neon dirge]

so now our basis is (e1, e2-e1)

which makes it so convenient to have done it column-wise

bc now we can simply subtract/add the basis vectors too

[languid coyote]

ahhh i see

and T(e2 - e1) = 0

[neon dirge]

you can perform this manipulation as you like!

[languid coyote]

so that's why that column is 0

[neon dirge]

definitely experiment a little with manipulating a matrix

[languid coyote]

so i guess the basis is (e2-e1, e2)

[neon dirge]

in case you have some visualisation tool

that can help as well

[languid coyote]

do you have a recommendation for this?

[neon dirge]

hm maybe something like Desmos for starting

[languid coyote]

ye for sure; ive taken linear algebra before computationally, but im self-studying axler now to really understand it inside and out from the ground up

so i wanted to make sure i really understand why everything is true and then connect it to all the computations

so i can have a strong intuition

okay this was super helpful, apologies if it was too simple of a question haha

thank you!

[neon dirge]

nono, the essence of linear algebra and analysis are crucial

and some glance over them leading to uncertainties later on

[languid coyote]

for sure...haha

[neon dirge]

there's also a few neat channels such as 3Blue1Brown which elegantly visualise quite a bit

[languid coyote]

oh yeah i've watched some of his linear algebra stuff, pretty helpful

i guess while we're at it one more quick question

so for any linear transformation, we know that there exists a matrix representation

[neon dirge]

ys

[languid coyote]

in particular, one where we map the basis vectors of V to the basis vectors of W

*to some linear combination of basis vectors of W

so it is fair to say that

if dim ker T and dim range T are k, n-k

then there should exist a matrix representation where

i have n-k non-zero columns and k zero columns?

[neon dirge]

and?

[languid coyote]

yes

[neon dirge]

hm 2nd yes, currently thinking about first

[languid coyote]

like let's say the dimension of V is n, dim ker T is k, dim range T is n-k

[neon dirge]

dym the same linear map

has two matrix representations

one with k zero columns

and one with n-k zero-clumns?