precalculus

[grim sand]

Now time for me to try question 8

[viscid thistle]

LOL

[echo wagon]

Didn't you already do it?

[viscid thistle]

We almost did lol

[grim sand]

No I just kept asking questions

We never finished it

[viscid thistle]

It's good to ask questions though

[echo wagon]

We did get to the answers, do you remember what they were?

[grim sand]

So now the vertex is

(3,-3)

[viscid thistle]

Let him redo

[echo wagon]

Yes

[viscid thistle]

Yessir

[grim sand]

Range would be [-3, INF)

[echo wagon]

Are they a sir?

[grim sand]

I am a sir

[echo wagon]

No

[viscid thistle]

Idk i was memeing

[echo wagon]

Range would be [-3, INF)
@grim sand no

[grim sand]

Oh

[echo wagon]

Going up or down?

[grim sand]

It’s going down

So

(INF, -3]?

[viscid thistle]

Sign

[grim sand]

Lmao

[echo wagon]

Why ) and not ] ?

[grim sand]

That one was a mistake

[echo wagon]

Okay

[grim sand]

Is it still wrong

[viscid thistle]

And the negative sign

[grim sand]

Oh

Why isn’t it a negative

[echo wagon]

What's wrong with the negative?

[grim sand]

Yea

[echo wagon]

Al is trolling

[grim sand]

That’s my question

Oh

[viscid thistle]

Wait

[echo wagon]

And by trolling I mean confused

[viscid thistle]

No

He said (inf and not (-inf

[echo wagon]

Oh lol

[grim sand]

Oh

[viscid thistle]

Lol see

[echo wagon]

I'm confused

I'll just leave now

[grim sand]

Cuz it’s going down

It has to be -INF

Yes

😃

[viscid thistle]

I gtg too lol

We've been here too long

[grim sand]

True

But is it right

Is that the reason it’s negative

[viscid thistle]

Yes

[grim sand]

Don’t leave me with curiosity

Ok thx

[echo wagon]

You always have a -inf in the first position, and a +inf in the second position

[grim sand]

Huh I missed a box

[echo wagon]

It's never the other way around

[grim sand]

I got a 6.75/8

[viscid thistle]

Yeah from left to right

[echo wagon]

I got a 6.75/8
@grim sand Al's fault not mine

[grim sand]

😂

[viscid thistle]

BRUH

No u

[echo wagon]

He's been here longer

[viscid thistle]

Nah prolly the prev questions

He did by himself

[grim sand]

Question 7 is free so it’s 7.75/8

[viscid thistle]

Right?

[grim sand]

What’s that .25% I missed

🤔

[viscid thistle]

Are you asking me?

[grim sand]

Anyone in general

[viscid thistle]

What you did with me it's fully correct lol

We can't know

[echo wagon]

Ego much

[grim sand]

@echo wagon what did u teach me wrong

😔

[echo wagon]

It was AL

[grim sand]

😂😂

[viscid thistle]

No u

[grim sand]

It’s just class work so it doesn’t really matter

[viscid thistle]

Bruh but like you did 5 questions alonr

[grim sand]

But I’m wondering what I got wrong it won’t tell me

[viscid thistle]

It was those

100% sure

[grim sand]

No

Someone helped me with 5 of those questions

Lol

[viscid thistle]

But I’m wondering what I got wrong it won’t tell me
@grim sand you can ask your teacher

LOL

[grim sand]

I don’t remember who

[viscid thistle]

He's prolly wrong then anyway

[grim sand]

There was like 10 diff ppl teaching me

so I just kept getting confused

Since everyone teaches differently

[viscid thistle]

See?

[echo wagon]

I bet it was Ann leading you astray /s

[viscid thistle]

That was it not mine

Lol

[grim sand]

Ann gave up tryna help me last night when I called her a sir on accident

[viscid thistle]

Well

[echo wagon]

Well her name is Ann, I wouldn't have guessed sir

[viscid thistle]

I can imagine what happened

[grim sand]

No I was just like

[viscid thistle]

It's logic tho

[grim sand]

“Yessir”

U know the meme

[viscid thistle]

Oh but thats a meme

Yeah

[grim sand]

Yeah

[viscid thistle]

Well bye, it's been a fun 4h

[grim sand]

I don’t assume genders so at first I was like “yes sir/ma’am”

Then I said yessir n she got mad

Well bye, it's been a fun 4h
@viscid thistle yes, thank y’all for teaching me

Much appreciated

[undone pawn]

lol she does that

[grim sand]

@echo wagon u too

[stuck lark]

$\bR=(-\inf,\inf)$ /s

[obsidian monolith]

RokettoJanpu:

[grim sand]

What’s that

[undone pawn]

it's an Ann trap

[grim sand]

I’m always up for learning

[stuck lark]

abusing the macro for infimum to denote R

[grim sand]

@echo wagon

Can u dumb that down for me

😔

[undone pawn]

lmao

[stuck lark]

yes luna teach us infimum

also supremum

[grim sand]

Super mum

A superb mom

[echo wagon]

LOL

[grim sand]

😀👍🎶

[stuck lark]

while you're at it also teach us liminf+limsup

[grim sand]

Liminf

Infinite lemons

Limsup

Like

Wassup lim

[echo wagon]

If you don't understand, just ignore me. But in the interval (1,3) we say that 1 is a lower bound of the set because every number in the set is greater than or equal to 1

But 0 is also a lower bound of the set, because everything in the set is greater than 0

[stuck lark]

oh no you're actually doing it

[echo wagon]

The biggest upper bound of a set is called it's infimum, and usually written inf.

[stuck lark]

[echo wagon]

Of course

[grim sand]

Oh

I understand that

[echo wagon]

So the infimum of (1, 3) is 1 because 1 is the biggest lower bound of the set

If you take 1.5 for example then 1.5 is not a lower bound because 1.4 is in the set and is less than 1.5

[grim sand]

The infimum of (2,5) is 2

[echo wagon]

Yes

[grim sand]

Smert

[echo wagon]

Anyway, so inf is usually written to mean infimum

[grim sand]

Oh

I’ll be sure to remember that next time

[echo wagon]

So really infinity should be written as $\infty$

[obsidian monolith]

Lunasong:

[grim sand]

I don’t know how to use this bot

I’m just tryna learn pre calc

[stuck lark]

but in the context of precalc, NOONE will read inf as infimum, only infty

[echo wagon]

Yeah, so it's fine. But that's why people are hounding me for using inf, lol

[grim sand]

I used it today cuz I didn’t know how to use that infinite sign

[stuck lark]

wonder how you'll tackle liminf+limsup

[echo wagon]

You know what, I don't think I will

[grim sand]

😂

[echo wagon]

Jupiter has learned a lot today

[grim sand]

Yes I have

4 hours worth of confusion and knowledge

[echo wagon]

Tomorrow you can substitute for me and teach him limsup and liminf, k? @stuck lark

[grim sand]

Oh bet I’m down to learn

[echo wagon]

There you go

[grim sand]

Ima learn computer science now

[echo wagon]

It's a date

[stuck lark]

shame you had a chance to post this https://i.kym-cdn.com/photos/images/newsfeed/001/497/040/471.jpg

[echo wagon]

Lol

[grim sand]

What time u free tmr @stuck lark

[echo wagon]

Lol

[stuck lark]

also i'm not teaching, i'm using this gif on you

[grim sand]

😔💔

[echo wagon]

They want to know when you are free, Rokabe, tell them

Don't be like this

[stuck lark]

i'll happily teach intro qm but not liminf

[grim sand]

What’s intro qm

[echo wagon]

LOL

[stuck lark]

quantum mechanics

[echo wagon]

Intro to quantum mechanics

[grim sand]

More learning rn

I’ve always wanted to learn that

My school doesn’t offer it

[echo wagon]

Now is your time to shine, Rokabe

[stuck lark]

it takes plenty of prereq math

no it's not luna

[echo wagon]

You can go over the math too

[stuck lark]

i already shine

[echo wagon]

Lmao

[grim sand]

That emote got me dead

Lmao

But yes plz teach quantum mechanics

Learning is always fun

[stuck lark]

tldr $\hat H\Psi=E\Psi$

[obsidian monolith]

RokettoJanpu:

[echo wagon]

Lol

[grim sand]

I want to be a real estate agent but I love math so 👍🎶😃

What is that

That looks like Poseidon’s trident

[stuck lark]

schrodinger eqn rewritten as eigenvalue problem

[echo wagon]

Greek letter psi

[grim sand]

schrodinger eqn written as eigenvalue problem
@stuck lark I only understand written as problem

[stuck lark]

good start

[echo wagon]

eqn is equation

So now you know more

I did my part

[grim sand]

Yes

[stuck lark]

[grim sand]

And schrodigner is a person

[stuck lark]

he's known for cats ofc

[grim sand]

Schrodinger

I don’t know him I just heard that name before so I took a guess

See I’m learning

😎

[blissful ridge]

Schrondinger is known for cats

[stuck lark]

he's a big phys dude

[blissful ridge]

[grim sand]

Do I have to get closer to cats too to be good at math

[stuck lark]

known for killing cats

[grim sand]

Oh

[stuck lark]

of course

[blissful ridge]

Tell that to my professor who asked us to derive the schrondinger's equation

[echo wagon]

I wanted to be a dick and spell schrodinger with the diaeresis on the o, but I can't do it on my phone

:(

[blissful ridge]

Schrödinger

Like this?

[echo wagon]

Yes

[grim sand]

So u down to teach me quantum mechanics 🥺

[echo wagon]

Rokabe is evil

[stuck lark]

$\Sigma ch\ddot{\mathcal O}d\i\eta g\ep\bR$

[echo wagon]

All the honourable are evil here.

[grim sand]

When I’m older?

[blissful ridge]

Then what?

[obsidian monolith]

RokettoJanpu:

[blissful ridge]

What monstrosity is this?

[grim sand]

I’d that his name

Is

[echo wagon]

Σψηροδινγερ

[blissful ridge]

I don't have greek letters on my phone

[stuck lark]

forget diaeresis, latex $\ddot o$

[obsidian monolith]

RokettoJanpu:

[grim sand]

What language is this

It’s intriguing

Please teach me

[stuck lark]

greek

[echo wagon]

Γρεεκλιση

[grim sand]

So teach me? 😃

[echo wagon]

I'll start

[stuck lark]

1st lesson in greek

[echo wagon]

α is alpha

[stuck lark]

read the help channel names

[echo wagon]

β is beta

[stuck lark]

$\aleph$ is aleph /s

[obsidian monolith]

RokettoJanpu:

[echo wagon]

ω σ δ ρ θ π λ κ are omega, sigma, delta, roe, theta, pi, lambda, kappa in that order

[grim sand]

No I wanna learn quantum mechanics

[blissful ridge]

Believe me, it's not fun

I hate phsyics

[echo wagon]

First you have to learn all the greek letters, be able to say them backwards and forwards, then Rokabe will teach you qm

[stuck lark]

didn't agree to that but ok

[echo wagon]

It's that or liminf

[stuck lark]

YOU do liminf

[grim sand]

What is liminf

[blissful ridge]

What is liminf?

[grim sand]

I don’t hate any subject

[echo wagon]

See what I mean? All the yellow people are evil

[grim sand]

Learning is fun if the teachers make it fun

[blissful ridge]

Genuinely QM is hard

[grim sand]

All the more reason I wanna learn

[stuck lark]

@blissful ridge https://ptb.discordapp.com/channels/268882317391429632/409596215697866773/735685438987305014

[blissful ridge]

What is this?

[echo wagon]

Okay, I'm really going to go now. My tv is waiting for me. Have fun learning liminf from Rokabe. Don't stop tagging them until they teach you.

[stuck lark]

a way to define limsup

[grim sand]

😀

Liam sup

[blissful ridge]

Tell me that in simple words

[echo wagon]

Omg that's too cute

[stuck lark]

you never saw copswing?

[echo wagon]

No

[stuck lark]

that's a crime

you will get to know

[echo wagon]

OMG

Cuteness overload

[blissful ridge]

Download the gif I posted

[grim sand]

So

[blissful ridge]

You can copswing without nitro then

[grim sand]

Quantum mechanics?

[echo wagon]

Pandas are my favourite animal, and all the panda emojis are like 90% of the reason I haven't left the server

[blissful ridge]

I thought you liked maths

😂

[echo wagon]

No

I like pandas

[blissful ridge]

Can't argue with that

[grim sand]

I accidentally stepped into the

Question that looks like a S and 8

[echo wagon]

So many of them, and they're all adorable

[grim sand]

And

[blissful ridge]

Anyway tell me what is liminf in simple words

[grim sand]

I was just confused

So many words

[stuck lark]

do you know sup

[grim sand]

Who doesn’t

Sup dude

U free now?

[stuck lark]

that's right everyone should know sup, even jupiter knows

[blissful ridge]

No, I don't

[grim sand]

It’s proper greetings man

[stuck lark]

in all seriousness luna just went over sup & inf

[grim sand]

Where

He went over

[stuck lark]

https://ptb.discordapp.com/channels/268882317391429632/363224154469826562/746053655668719627

[grim sand]

Infinite mums

That’s what he explain

[stuck lark]

tldr sup=least upper bound, inf=greatest lower bound

[echo wagon]

inf of a sequence is the greatest lower bound of the sequence. liminf of the sequence is the limit of the infimums as you only look at the later terms in the sequence.

[blissful ridge]

Understood

[grim sand]

I keep reading infimum as infinite mums

[echo wagon]

I want to live in a world of superb mums instead of supremums too

Anyway, my tv is really waiting for me. Bye. Thanks for all the pandas.

[stuck lark]

[grim sand]

Lmao

Thank u for teaching me

[stuck lark]

being serious again, qm has a somewhat big list of prereq math. you should at the very least have taken and done well in calculus 1, maybe multi too, linear algebra, and some probability before diving in @grim sand

[blissful ridge]

Despite studying all that, I found it really hard,
Teachers didn't help either

[stuck lark]

i'm not doing it rn but there are fragments of a talk i had in offtopic, it's very scattershot and again, it has math you've not seen https://ptb.discordapp.com/channels/268882317391429632/504761543846658049/739342635633999962

[blissful ridge]

And it was kinda hard to teach myself

[stuck lark]

i wouldn't manage to self study either

[grim sand]

Oh man

I can’t even do good in pre calc

Cuz I’m just not good with graphs

[novel cargo]

I may be wrong but some of advanced topics in precalc are more difficult than beginner calculus stuff

[viscid thistle]

@stuck lark wait you like physics?

High five ✋

[stuck lark]

[novel cargo]

$\frac{x^2}{1} - \frac{y^2}{169/3} = 1$

[obsidian monolith]

az:

[novel cargo]

for y = 18, I get x = 2.598...

my book shows x = 2.403

can someone check?

[novel cargo]

<@&286206848099549185>

did you remove your comment?

[echo wagon]

I also get 2,598, but of course it can be + or -

And yes, because what I said is dumb

[novel cargo]

oh, I'm going to say something dumb now

the unit being in feet doesn't change anything, r?

[echo wagon]

No

Are you sure your equation is right? Did the book agree with it?

[novel cargo]

this is the answer section from the book

probably an error?

[echo wagon]

Think so

[novel cargo]

wow

we found an error

[echo wagon]

Bad book, burn it

[novel cargo]

I don't have experience with this stuff but it doesn't seem that bad

[echo wagon]

Wait

[novel cargo]

Larson's Precalc 10th

I'm gonna do Larson's Calc after this

yeeey

[uncut mulch]

1 sec

the book is right, you're not applying the equation properly

y doesn't represent the height

the base value is at y=-13

[novel cargo]

y/2 then?

[uncut mulch]

no

[novel cargo]

I mean 18/2

[uncut mulch]

a height of 18 would be at y = -13+18 = 5

[echo wagon]

Lol, I also realized there is a mistake because x is not the width, but I was trying to just multiply it by 2 which is still wrong. Thanks for the save.

[novel cargo]

Oh

makes sense

[uncut mulch]

which you then solve to get |x|, the semi-width
and then double it

[novel cargo]

oh man

I feel ashamed

[echo wagon]

Don't be

[proud crypt]

can u guys tell me if my proof of derivatives is correct

sry im not good with texit

ok nvm i need better texit skills first

[stuck lark]

paper works

[gritty dew]

How do I know that

is equal to

without a calculator?

[obsidian monolith]

Lunasong:

Multiply by $\frac{\sqrt{6} + \sqrt{2} } {\sqrt{6} + \sqrt {2} } $, which is just 1 so it doesn't affect the value. This is called rationalizing the denominator. @Kerry

[gritty dew]

Oh tyvm

btw what do those reaction icons for TeXit do?

[echo wagon]

The first one deletes it

The second one does something weird, idk

I think one of the others gives you like a report if there is something wrong with your code, not sure

[gritty dew]

aight thanks

[proud crypt]

@patent beacon

first of all pls accept my discord friend request

second of all, whats next for derivatives?

ik how to derive the formula

i can prove

it

ik the general formula

y' (ax^n) = anx^n-1

and what derivative all means

also d/dx

multiply every term

linearity i think its called

like distributing

so whats nexxt in my studies of derivatives?

[patent beacon]

Well, okay if you're actually interested in getting into derivatives, get into limits

[proud crypt]

ok...

oh ik how that works

well nto relaly

can u teach me limits

i have a bit of time rn i think

also some texit knowledge would be nice

[patent beacon]

Sure. Let's take a look at this function:

,w graph sin(x)/x

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[patent beacon]

So natural question, what's the value of sin(x)/x, when x = 0?

[proud crypt]

1

[patent beacon]

You'd think so! But sin(0)/0 = 0/0 which is a division by 0

[proud crypt]

oh righ

btu teh graph doesnt say that

[patent beacon]

So it's undefined there. sin(x)/x has a hole there, which Wolfram can't show

[proud crypt]

bruhhhh

activates hawk vision

ok...

[patent beacon]

Still, you just noticed that we're missing information by just saying "it doesn't exist there"

[proud crypt]

yeah

is that what limits are for

[patent beacon]

One might say that "if it did exist there, it would have to be 1, because that's where the function goes"

That's the point of a limit. A limit takes a look "around" the point, and reports back what the most natural definition for a function would be

In the case above...

[proud crypt]

ooh

like a scout

kinda

more accurate then

"is there smth there"

lemme try

[patent beacon]

$\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$

[obsidian monolith]

Kaynex:

[proud crypt]

i see

[patent beacon]

Is a common limit

[proud crypt]

limit of x as it approahces 0 = 1

but it never actually does = 0

1

because lim is approcahing not at

am i right

so if u paired that with x at 0 is undefined

much better than

just x at 0 is undefined

[patent beacon]

Exactly. The limit ignores what actually happens AT the point and instead only uses what happens NEAR it

[proud crypt]

wish online sites had told me that

aka khan academy

ahh

how near?

.1?

[patent beacon]

One might say infinitely near

[proud crypt]

ooh

wait wouldnt that just be on tho

[patent beacon]

You're getting into analysis territory at that point haha

"Every open set containing the limit point"

[proud crypt]

huhhh

ok wut

[patent beacon]

At least for now, a limit is like a scout. I did like that analogy

[proud crypt]

lol

u: "look around"
me: SCOUT

feel free tro use it btw

for others

ok....

[patent beacon]

Limits end up being exactly what you need to define the tangent line, and the derivative

[proud crypt]

yes

ik the proof

but cant texit it

cna u texit the proof

just first part

[patent beacon]

The derivative is actually a particular kind of limit

[proud crypt]

and i can explain rest

(gtg in 3 mins

just to test if i remeeb

wait nvm i gtg

dinner

lol
ahh yes ik derivative
ik hwo to rpove as well since my discord friend told me how to
But im not godo with texit
i can explain i think
lim - 0 h = ((f(x+h) - f(x))/h
so we get x^2 +2xh + h^2 - x^2/h
so (2xh + h^2)h
so thats 2x + h
since lim h approahces 0
2xd
answer is 2x
which is derivative of x^2
did i do it right?

here

lim - 0

just is value of h as it approaches 0

or u can texit if u are quick lol

[patent beacon]

So EVERY derivative can be found using this formula, which is the definition of the derivative:

$\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$

[proud crypt]

yes

[obsidian monolith]

Kaynex:

[proud crypt]

ok so basiclaly

if u have x^2

wiat a sec

huh

oh

so u square the first par t

[patent beacon]

What this is actually doing is finding the slope between two points, and they are a distance h away

[proud crypt]

f(x+h)

and square second

f(x)

minus

divide by h

2x+h is what u get

replace h with 0

2x

derivative

i think that was incorrect but idk

[patent beacon]

Yeah yeah haha

[proud crypt]

wait did i do it righ t

tbh i dont understand the proof

i didnt come up with it

discord firne did

can u generalize it

[patent beacon]

Be careful with the algebra, and know why
f(x + h) = (x + h)²

[proud crypt]

well ik why

[patent beacon]

Given that f(x) = x²

[proud crypt]

oh right

but can u generalize it

so like f(x) = a^x

and prove

to show that x^2 is lucky case

[patent beacon]

For any polynomial, the proof goes the same way

[proud crypt]

oh ok

f(x) = x^2

so f(x+h) would be (x+h)^2

since the right hand side is whatever is in the parantheses

f(whatever is in here) = whatever is in here ^2

[patent beacon]

Something like f(x) = x³:
f(x + h)
= (x + h)³
= x³ + 3x²h + 3xh² + h³

Then
f(x + h) - f(x) = 3x²h + 3xh² + h³

[proud crypt]

f(whatever is in here + h) = (whatever is in here + h)^2

ahh i get it now thanks like alway smate

what will we discuss when i return from dinner

btw

[patent beacon]

Then divide by h, take the limit, you get 3x²

Gratz it makes sense! See ya

[proud crypt]

ah i see

so it works with any polynomail

when we come back can we dive deeper into derivatives

ill ping u if u dont mind

[patent beacon]

Sure sure

[proud crypt]

oh can u accep my discord firend request

@patent beacon

(so no one else reading this will go looking for it lol)

ok bye we will dive in to derivaties when i come back

I FEAR U NO MORE DERIVATIVES

STILL DOTN SEE WHY U ARE SO POPULAR TO USE IN CALC AT ALL CUZ U AND CALC SEEM UNRELATED BUT U SHALL CONFUSE ME NO MORE

[proud crypt]

@patent beacon

lets do derivatives

whats next?

or lims if i need more learning with that first

[patent beacon]

K sure, so a huge use of derivatives are finding minimums and maximums of functions

[proud crypt]

derivatives are completely fascinating

ok...

[patent beacon]

Remember how that works? Want to go over it?

[proud crypt]

ik how to do with quadratics but nothing else

-a/2b

i think

then plug that in for x

[patent beacon]

Hey, we can prove that!

[proud crypt]

oh sure

idk how but sure

also could we maybe prove quadratic formula

i think its completeing square on standard form

[patent beacon]

So let's say your quadratic is
y = ax² + bx + c

[proud crypt]

uhhh

yes

wait can we prove using derivatives

anywya contineu 🙂

[patent beacon]

In order to find the maximum/minimum of a function you:

• Take its derivative
• Set that equal to 0

[proud crypt]

ok

so we can use it to rpove vertex of quadratic!!

cool

so if we have ax^2 + bx+ c

[patent beacon]

y' = 2ax + b
Set that equal to 0:
0 = 2ax + b
x = -b/2a

[proud crypt]

woahh

[patent beacon]

But that's "been there done that". We could have gotten that with algebra

[proud crypt]

d/dx(ax^2) + d/dx(bx) + d/dx(c) = 2ax+b

oh how?

[patent beacon]

One way is to complete the square

[proud crypt]

ok...

so lemme see

ax^2 + bx + c = 0

can we prove quadratic formula

using complete the square

[past meadow]

Yes

[proud crypt]

o k

saitama

[past meadow]

Thats me

[proud crypt]

hey its ya boi sonic get ready to get yeeted out of existence

ha killedm eoh well

morphs into subterraens

i never wtached one punch

tho my friend did

and i watched epic battles in itlol

[past meadow]

Do watch it, it’s great

[proud crypt]

no time

i have alg 2 this year in 8th grade and other stuff

GENOS OP

can u help me prove quadratic formula

with compelte the swquare

@past meadow @patent beacon

[past meadow]

I mean when you complete the square, a good first step is always to have your leading coefficient as 1

So start with that

[proud crypt]

ok...

[patent beacon]

That's not calculus but we can

That's 100% algebra

[proud crypt]

ok lets go to prealg algebra

[turbid torrent]

Hey guys, I’ve recently started making math help videos like this one https://youtu.be/y2EsTXtewLU

If you want help with a particular concept, let me know and I’ll try my best to make one for you 🙂

You can also check out the website, where we have heaps of awesome resources. Here’s the corresponding topic for this video https://treena.org/courses/hsc-mathematics-advanced/exponentials-and-logarithms

[proud crypt]

eyeyy lotr fan nice

that used to be my name as well

[tardy ridge]

@proud crypt Learn stuff here: https://courseware.cemc.uwaterloo.ca/
Discord is debatably the worst place to learn.

(Unless you're trying to get help)

[turbid torrent]

There are some pretty bad places to learn 😛

eyeyy lotr fan nice
@proud crypt haha definitely love lotr

[novel cargo]

this is a chapter about hyperbolas

I can deduce the hyerbola equation from the given information

what confuses me is that in a real world scenario, would it always be correct?

also, if it's always correct, to find the explosions location, do we need to travel the hyperbola to find it because there are infinitely many solutions?

[novel cargo]

<@&286206848099549185>

[willow bear]

yes, the given info is only enough to constrain the strike location to a hyperbola with you and your friend's houses as the foci

,calc 4*5280

[obsidian monolith]

Result:

21120

[willow bear]

,calc 1100*18

[obsidian monolith]

Result:

19800

[novel cargo]

is this always true?

if I have the coords of two locations and the difference of sound arrival at each location

I can always limit the coord of the sound's event to all points on a hyperbola?

[willow bear]

i mean that's the defn of a hyperbola

in your case specifically it's one of the arms of a hyperbola

[novel cargo]

it just feels unreal intuitively

probably bc of my lack of experience

[novel cargo]

$\cot^2{x} + \csc{x} - 1$

[obsidian monolith]

az:

[novel cargo]

how would I simplify this?

[viscid thistle]

Start by writing cot and csc in terms of cos and sin

[novel cargo]

don't I need to use the pethagorean theorem?

like cot^2 x + 1 = csc^2x

[viscid thistle]

That's another way of doing it

You can go ahead

[novel cargo]

either way, I don't see how to procedd

if it helps the answer according to my book is:

[viscid thistle]

Sure it helps let me see

[novel cargo]

$(\csc x - 1)(\csc x + 2)$

[obsidian monolith]

az:

[novel cargo]

I see some factoring

ohh

so substituring csc^2 x - 1 for cot^2 x

we get

csc^2 x - 1 + csc x - 1

csc^2 x + csc x - 2

this is

second one is answer

isn't it working the way I was doing?

by using cot^2 x + 1 = csc x

[viscid thistle]

It is, i was not understanding before what you were saying

[novel cargo]

csc^2 x

[viscid thistle]

Yeah it is correct

[novel cargo]

thanks

was really struggling with this

[novel cargo]

https://www.desmos.com/calculator/v6zoy6gyeu

this is what I got so far

the solution for station A and B is:

$\frac{x^2}{2200^2} - \frac{y^2}{2459^2} = 1$

[obsidian monolith]

az:

[novel cargo]

Now when I try to setup an equation with A and C as foci to calculate the exact coordinate of the explosion, my b becomes 0.

I can't use B and C as focal points because the transverse axis then becomes not paralel to x or y and I haven't learned rotation yet.

[novel cargo]

<@&286206848099549185>

[grim sand]

It must be nice being that smart 😔💔

[novel cargo]

hey Jupiter, how you doin?

remember your today's level (write it down, document it somehow, human memory is weird)

keep up the grind for 14 days

come back and see how much you have improved

compare yourself to your former self and not to other ppl

@grim sand

[near monolith]

az, are you only using conic sections then?

Also, have you learned translation?

@novel cargo

[grim sand]

I try to make time to learn pre calc

But I just got more work added I barely have time

😔💔

Especially with sports practice too

[novel cargo]

@near monolith yes, I'm only using hyperbolas in this problem (the section is only about hyperbolas). I haven't still learned translation. Rotation is next chapter.

[near monolith]

Hmmm, what about a translated circle?

[novel cargo]

I can imagine a translated hyperbola working but circle I don't know

circle needs radius

we only have difference of distance

which makes hyperbola useful

like you hear the sound of explosion in station A 1 sec before in station B

you know coords of both stations

this will restrict the locations of the explosion to a hyperbola with both stations at the foci

the possible locations

[rapid lance]

just looking at the problem, it seems fairly easy to get they coordinate, then just use that to get the x

[novel cargo]

look at my solution so far

I have an equation for the hyperbola between Station A and B

Now I'm given station C

and need to find exact coord

I need to set up a new hyperbola between C and B or C and A

[rapid lance]

i think there might be more than one point possible, hence KD mentioning a circle

[novel cargo]

and see where this hyperbola intersects with the firtst equation

with three stations?

[rapid lance]

in b i dont see anything aobut station b, am i missing something?

is it still the 4 second one

[novel cargo]

in a you get an equation between station A and B

yes

[rapid lance]

oh

[novel cargo]

it's the same explosion

in b you need to find exact coords

[near monolith]

Well the circle idea probably won't work out. I was thinking about you drawing a circle for where the sound could have been with respect to C upon reaching A, but a hyperbola is what we really want here

[rapid lance]

just use the fact that sound goes 1100 feet per second

and go from there

[novel cargo]

I actually worked out the second hyperbola between C and A

[rapid lance]

maybe try distance formula???

idrk

[near monolith]

Then you should be able to simply find the intersection

[novel cargo]

the problem is that the denominator of its second part becomes zero

normally when this happens we are dealing with a parabola

[near monolith]

What was your second hyperbola?

[novel cargo]

let me fetch it

so, third station is C

center: (3300, 550)

foci: (3300, 1100) Station C and (3300, 0) Station A

detects explosion one second after station A, that's 1100 feet difference.

We have d2 - d1 = 2a = 1100, hence a = 550

c = 550 (we know this from the foci)

c^2 = a^2 + b^2

so, b^2 = c^2 - a^2

here is the problem

c and a are both 550

so be becomes 0

*b

this will remove the y term completely

and make the equation a parabola

now, this is out of my knowledge and exp

I can put together something and get an interseciton

but that's guess work

Makes sense?

[rapid lance]

what was your answer?

[novel cargo]

I didn't further pursue this

[rapid lance]

oh

[novel cargo]

I needed someone with more experience and stuff

[near monolith]

Actually

It becomes a straight line, I think

[rapid lance]

thats what i thought too

[near monolith]

You have the line x=3300

[novel cargo]

sounds promising KD

I had some such thought but was too confused about it

lol

I have to go guys

will check that later

thank you for your feedback

[near monolith]

Basically, the distance between the y coordinates matches the distance that had to be traveled, so there couldn't be any distance x direction.

So the x position must be 3300

Probably should have realized this sooner just by looking at the problem

[rapid lance]

i think you can do similair thing with the y

[near monolith]

The final step is to substitute x=3300 into your hyperbola and find the point closer to A than C

From the drawing it should be pretty obvious

[novel cargo]

wow

thanks KD

makes sense

[hidden willow]

Find the slope-intercept form of the equation of the line that passes through the given point and has the indicated slope m.(-6, 6), m = -2

[novel cargo]

are you asking?

[hidden willow]

yeah

thats on my hw

[novel cargo]

so, what is it you don't know about it?

[hidden willow]

I know the equation

just dont know how to form it with the point given

[novel cargo]

y - y0 = m * (x - x0)

[hidden willow]

oh ok

so

[novel cargo]

is the formula that you need to use to get the equation given the slope and a point

[upper kelp]

Generally speaking, a slope-intercept form is an equation of the form y = mx + b, where m is the slope of the line, and b is the value of the Y-intercept. (meaning, the intercept is (0, b))
In this specific case, you are told that m = -2, and so you get an equation of the form:
y = -2x + b
for some b that is still unknown.
Now, you are also told that the line crosses through (-6, 6).
Do you know how to use that to find b?

[hidden willow]

well I know thats where it crosses

the y intercept

so it would be -6?

[upper kelp]

Not quite, sorry.
Keep in mind that -6 is the X-value of a certain point, but this doesn't mean that the value of the Y-intercept is -6.

[hidden willow]

oh

so

what equation is this

y - y0 = m * (x - x0)

whats it called

[upper kelp]

It's a "formula" to find the equation of a line.
In y0, you substitute the Y value of a certain point.
In x0, you substitute the X value of the same point.
In m, you substitute the slope.

[hidden willow]

so

y+6=-2(x-6)?

[upper kelp]

Almost. You need to add the y on the left hand side.

[hidden willow]

fixed it

y-6=-2(x+6)?

[upper kelp]

You're close.
Keep in mind that the y0 value is 6 and the x0 value is -6, not the other way around.

[hidden willow]

y-6=-2(x+6)?

[upper kelp]

Exactly! 😃
Do you know how to proceed from there to find the equation?

[hidden willow]

uhhh

no

also just to double check becuase this is graded, these are neither parallel or perpendicular?

i got neither as the answer

but not sure

[upper kelp]

I'll reiterate:
The equation that you got, y - 6 = -2(x + 6), is an equation for the line. We would like to convert this equation to the form y = mx + b by moving terms around. Do you know how to do that?

Also, I'm sorry, but I can't see the image you've just sent.
I can only say that, if you have two lines of the form:
y1 = m1 x + b1
y2 = m2 x + b2
then they're parallel if m1 = m2 and b1 =/= b2, and they're perpendicular if m1·m2 = -1.

[hidden willow]

oh

so what would the answer be to the first question i asked

or how do i get to it

i got

[upper kelp]

For the first question, you can move those terms around to get an equation of the form y = mx + b.
I'll give an example with other numbers to illustrate the point:
5y - 5x + 13 = 99(x + 1)
In this case, we can move terms around to have only y on one side:
5y = 5x - 13 + 99(x + 1)
y = 0.2x - 13 + 19.8(x + 1)
and then open the brackets to get:
y = 0.2x - 13 + 19.8x + 19.8 = 20x + 6.8.

[hidden willow]

y=-2/b x -6/b

[upper kelp]

Where is the b from, if I may ask?

[hidden willow]

oh that was wrong

[upper kelp]

Would you mind if I show the steps?

[hidden willow]

sure

[upper kelp]

We aim to get to an equation that looks like $y = mx + b$, where $m$ and $b$ are both real numbers. We start with the equation $y - 6 = -2(x + 6)$ and move on from there. $\$
First, we can open the brackets to get $-2(x + 6) = -2x - 12$, and we are now left with the equation: $\$
$y - 6 = -2x - 12$ $\$
Now, we need to isolate $y$, and for that, we add 6 to either side and get: $\$
$y = -2x - 6$ $\$
Does this make sense?

[obsidian monolith]

RoiKadmon:

[hidden willow]

yeah it does

[upper kelp]

In this equation, we have: $\$
$y = \underbrace{-2}{=m, \text{ the slope}}x + \underbrace{(-6)}{= b, \text{ the Y-intercept}}$ $\$
where $m, b$ are both real numbers.

[obsidian monolith]

RoiKadmon:

[hidden willow]

ah ok

thank you

and these are not perpendicular, correct

y-3x=5

x+3y=12

[upper kelp]

Let's try to convert them to y = mx + b form and check:
y - 3x = 5 is equivalent to y = 3x + 5
x + 3y = 12 is equivalent to 3y = -x + 12, which is equivalent to y = -1/3 x + 4.
Seeing them that way, do you think they are perpendicular or not?

[hidden willow]

they are

[upper kelp]

Yup! And do you know why that is?

[hidden willow]

the slop is the reciprocal

[upper kelp]

Exactly! 😃 (and of opposite signs)

By the way, may I show another way to solve the first question you've asked?

[hidden willow]

sure

that would be nice

[upper kelp]

We are told that the slope is $m = -2$. This means that we know the line is of the form: $y = -2x + b$. We now need to find the value of $b$. $\$
To do that, we note that we are told that $(-6, 6)$ is on the line. Meaning, it satisfies the equation of a line. In this case, we can substitute the coordinates of the point in the equation of the line, so we get: $\$
$y = -2x + b, (-6, 6): \ \ \ \underbrace{6}{y_0} = -2 \underbrace{(-6)}{x_0} + b \Rightarrow 6 = 12 + b \Rightarrow b = -6$ $\$
This means that $b = -6$, and therefore, the equation of the line is $y = -2x + (-6) = -2x - 6$.

[obsidian monolith]

RoiKadmon:

[hidden willow]

oh

thanks a lot

could i ask one more

i skipped alg 2 so im pretty lost :P

this is supposed to be a review so the teacher breezed past it but it not for me

lol

[upper kelp]

Please feel free to ask. I hope I'll be able to explain it.

[hidden willow]

can you see that?

[upper kelp]

Yup.
Once again, I believe that you can solve this by using the y = mx + b form, except this has the extra step where you have to calculate these forms from the beginning.

Are you familiar with how to calculate the slope of a line, given two points?

[hidden willow]

yes

y=3/4x + 5

for the first line

[upper kelp]

Exactly! 😃

[hidden willow]

y=-4/3x -1

thats for the second one

[upper kelp]

Correct as well. 😁

Now, given those lines, and in particular, the slopes, which are 3/4 and -4/3, can you tell whether the lines are parallel, perpendicular, or neither?

[hidden willow]

perpendicular

[upper kelp]

Yup! 😃
And once again, as you said previously, this is because their product is -1. (or, rather, that they are opposite signs and reciprocals)

[hidden willow]

ok

thanks a lot

im all done

[viscid thistle]

Need help w someone verifying my work

That it’s correctly done

[upper kelp]

Feel free to send it. I hope I'll be able to say something meaningful.

[viscid thistle]

C is my own problem

Set builder notation

[upper kelp]

Seems good!
By the way, it's a bit specific in this case, but generally, you can use the notation [n] which denotes all the integers from 1 to n, so, for example: [10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
In that case, for example, you can write, for B:
B = {2x | x ∈ [100]}

[viscid thistle]

Huh

There was a logarithms question too, gotta find the image

[upper kelp]

Is it finding the image of the "regular" log (x) function, or something else?

[upper kelp]

Ooh, I see.
Are you familiar with logarithm rules?

[viscid thistle]

I saw a photo that explained all of it a while ago

Completely forgot it

[upper kelp]

May I reiterate the rules?

[viscid thistle]

Go ahead

[upper kelp]

Sum of logs: $\$
$\log_{a} (b) + \log_{a} (c) = \log_{a} (bc)$ $\$
Multiplying log: $\$
$a \cdot \log_{b} (c) = \log_{b} (c^{a})$ $\$
In particular, for example, if $a = -1$, then: $\$
$- \log_{b} (c) = \log_{b} (c^{-1}) = \log_{b} (\frac{1}{c})$. $\$
Can you think of which law could be used here?

[obsidian monolith]

RoiKadmon:

[viscid thistle]

Thanks

[prisma sedge]

Can I send a picture in here?

For help?

[past meadow]

ofc

[old nimbus]

hello

Can I get some help?

[jade heron]

hi

no

fuck the tories

sry

[old nimbus]

how would I integrate (y-2)/3

[uncut mulch]

can you post the exact question

[old nimbus]

sure

Can I send a photo?

[jade heron]

yeap =)

[old nimbus]

So Im doing finding the area as the y axis as your boundary

[jade heron]

O.K. so i would start with a sketch

[old nimbus]

yeah Ive used desmos to visualise my graph

[jade heron]

O.k. so what are you thinking of doing next

[old nimbus]

My brain just i9snt working atm and I have forgotten how to integrate (y-2)/3

[jade heron]

O.K. so here we should recall

[uncut mulch]

you'd be integrating wrt y, so just basic power laws here.

[obsidian monolith]

L'Âne 🍐:

[jade heron]

yeap i agree with ramonov

[old nimbus]

so would I be right in saying that the integral (y^2-2y)/6

[uncut mulch]

be sure to also calculate the lower bound

[jade heron]

not quite because

integral of y is actually y^2/2

Not just y^2

[old nimbus]

but it starts out as y-2/3

so you would raise the power of the y

times that by the bottom?

[uncut mulch]

the second part is wrong

[jade heron]

ok ramonov i will let u continue so we dont have two people talking if you'd like

[old nimbus]

which part?

[uncut mulch]

antiderivative wrt y of 2/3 isn't (2/6) y

[old nimbus]

oh

then how do I integrate it

im really confused at the moment

[uncut mulch]

don't overthink it

its just (2/3) * y

[old nimbus]

so would the top half of the fraction be y^2 -2y

[jade heron]

u should of got this

$\frac{\frac{y^{2}}{2}-2y}{3}$

[obsidian monolith]

L'Âne 🍐:

[uncut mulch]

i'd just split it into two fractions

[jade heron]

if u integrated y-2 correctly

[old nimbus]

is that not just the same as (y^2-2y)/6

[jade heron]

Nop

[old nimbus]

hmmmm

[jade heron]

||multiply top and bottom by 2||

[old nimbus]

ramanov

relating back to what u said earlier by splitting it into 2 fractions

would I do that before or after I have integrated

[jade heron]

doesn't rly matter but i think the idea was it'll be easier if u split it first

[uncut mulch]

before, makes things relatively simpler

[old nimbus]

so y/3 -2/3 ?

[uncut mulch]

and minimises risks of mistakes

integrate that yes

[jade heron]

yea like that then u integrate each one seperately

ramonov

what r ur pronouns my friend

[old nimbus]

so you would get (y^2/2)/3? for the first?

[jade heron]

yeap

[old nimbus]

and 2y/3?

[jade heron]

minus sign but yea

[old nimbus]

I then have to sub both 8 and 2 into those and find out the area

[jade heron]

yep!

[uncut mulch]

he or equivalent (but i don't really care)

[jade heron]

ok thank u =)

[old nimbus]

Thanks

My college teaches me different so thats why I was confused

[jade heron]

how does ur college teach u

[old nimbus]

I dont really remember with fractions

lemme check

no nevermind

Ive been taught the splitting into fractions

Thnaks for the help guys appreciate it

[jade heron]

no problem mister prime minister

[old nimbus]

haha

I have a question

[jade heron]

hi

[old nimbus]

In my graphic calculator I have used my x value as 2

[jade heron]

yep?

[old nimbus]

I have subbed 2 both into y^2/6 and (y^2/2)/3

and have both got 2/3

[jade heron]

they r the same thing

[old nimbus]

thats what I was trying to explain before

[jade heron]

\frac{\frac{y^2}2}3=\frac{y^2}6

yea that wasnt the part that was wrong, the -2y part was wrong ||should have been -4y||

[old nimbus]

ohh

cheers

Finally got the answer thanks heaps

[jade heron]

no problem friend

[old nimbus]

Is it ok if I keep asking questions? or no?

[uncut mulch]

yes

[old nimbus]

so Im doing the same kind of question

ive drawn out the graph

its y=x^2

and y=4

they enclose

but do I calculate the area just of the right side of the y axis orboth sides?

[jade heron]

depends what the question is asking

[old nimbus]

"Calculate the area enclosed by"

[jade heron]

enclosed by what

will u send the qn

[uncut mulch]

if they only mention enclosed between those two equations, then both sides

[uncut mulch]

though due to symmetry you could calculate the right side and double it