help-13

[queen sundial]

but i dont wanna completely fry my brain before it yk

[dreamy void]

don'T fry it

you seem you practiced alot

[queen sundial]

its not enough

this teacher pulls out the most devious questions on the real exam

yk our last test original was out of 50

she curved it to 35

thats how bad it was

ppl were complaining to school admin to fire her 💀

[dreamy void]

hahahaha

typical

[dreamy void]

nonoononono

you can do all of them

just be happy and confident

[queen sundial]

okok hopefullyyyy 🤞

[cedar kiln]

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[muted timber]

Obtain a plane, parallel to x+y+z=6, that goes through P.
Obtain a plane, perpendicular to the given line, that goes through P.
Obtain the intersection of those planes. It will be the line asked for.

[cedar kiln]

@unborn plank Has your question been resolved?

[unborn plank]

how would I obtain the intersection of the two planes?

<@&286206848099549185>

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[crimson sedge]

Can someone please give me a nice problem set on indefinite integration for practice? I had an exam on it yesterday, and I completely flunked the integration portion.

[raven shard]

there are also definite integrals on here but w/e

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Okay! I go through them

[raven shard]

more. again there are definite integrals

also if u have a textbook that will be useful...

[crimson sedge]

Thank you so so much!!!

[raven shard]

yw

[crimson sedge]

Can you recommend one

[raven shard]

(this one is sorted by type of integral)

[crimson sedge]

My calc portion in general could use the extra help

[raven shard]

never used

[crimson sedge]

Okay

Thank you so much

I’ll work on those problems

I really appreciate it

Have a good one

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[azure cedar]

Does anyone know how i'm suppose to prove this using definitions and properties to support my answer? I know the statement is true but i'm not sure how to say that other than by plugging in numbers.

[cedar kiln]

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[lament hollow]

Where did I go wrong? The answers are supposed to be 4 and 8 but I got only 8

[raven shard]

no

it is 16

[fallen moat]

yay

[raven shard]

also "4 and 8" is not the correct solution set

[fallen moat]

too nervous haha, so i got wrong

[raven shard]

note this: 2sqrt((x - 8)^2) = 2|x - 8|

hence u r solving x + |x - 8| = 8

for x in the domain of the original left hand expression

[lament hollow]

Doesn't it still result in 8

[raven shard]

to deal with absolute value bars, consider splitting

for x > 8, expression equals [this]. for x < 8, expression equals [this]

etc

[lament hollow]

x >= 8 -> 2x = 16
x <= 8 -> 8=8, which is true for all x <= 8, accounting for the splitting

From what I see but it's wrong

[raven shard]

it's not wrong it is the answer

the correct solution set to this problem is the closed interval [4, 8]

[lament hollow]

Oh

Yeah

I forgot the restriction from the radical

Thanks!

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[raven shard]

yvw 🎉

[cedar kiln]

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[fervent arch]

Considering C as a constant in the following items, we have that y is implicitly defined as a function of t by the corresponding equation. In each of the items, determine a differential equation for which y is a solution.

[fervent arch]

i tried to differenciate both sides

a)

ty` - y´/y = 0

is this correct ?

why ? 😭

[cerulean sail]

Forgot product rule on ty

[fervent arch]

d/dy(ty) = d/dy(t)y + t d/dy(y) right ?

[cerulean sail]

Remember you’re differentiating with respect to t here

Hence y turned into y’, rather than just 1

[fervent arch]

ahhhhh yeah

mb

yeah so i will diferenciate with respect to t , so i get y` in the equation ( which is what they want right ? )

ty

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[terse bramble]

I have the sequence an+1=root(1+2an) where a1=1

[terse bramble]

I need to study its monotony

[vagrant elbow]

You can use induction

[stable hull]

or check recurrence

[terse bramble]

Lets use induction

?

@stable hull

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[dawn gulch]

Hello

[dawn gulch]

Need a hand solving c

With working out if possible

[wicked mantle]

,rotate

[hot crag]

i, ii, or iii?

or a combination?

[dawn gulch]

combo

please

[hot crag]

what?

[dawn gulch]

all three

[hot crag]

no my question is which of i, ii, and ii- oh

ahk

the coordinates of the vertex are just gonna be -b/2a

[dawn gulch]

if possible, thank you

[hot crag]

nw

[dawn gulch]

stressing so frigging much

test tmr and I do not even have a cheat sheet

[hot crag]

it's fine my guy

it isn't that big of a deal

not like it's going to change ur future in one fell swoop

[dawn gulch]

I suppose so

so the (x) is a?

[hot crag]

for part i, the vertex's x-coord is just -b/2a
for part ii, it's a minimum if the parabola points up (if the coeff of x^2 is positive) and vice versa
and for part iii it's slightly more complicated

[hot crag]

?

[dawn gulch]

my bad...

2xsquared is A and 4 is B

[hot crag]

no...

not quite

[dawn gulch]

keep throwing myself off

[wraith dagger]

ren

[dawn gulch]

cannot remember some simple frigging rules

[hot crag]

don't beat yourself up so much

these distinctions are hard to make at first

[dawn gulch]

so what does the F(X) stand for currently?

[hot crag]

?

what do you mean f(x)

[dawn gulch]

yes

[hot crag]

what is the function equal to?

just the quadratic

[dawn gulch]

okay, got it

[hot crag]

i.e. f(x) = ax^2 + bx + c

[dawn gulch]

f being the question

so where does a and b fall in the equation?

Just need to fully wrap my head around it before tmr

[hot crag]

??

i don't get where you're getting f(x) from

a is the coefficient of x^2

b is the coefficient of x

c is the constant

so for example

in 3x^2 + 5x -7
a is 3
b is 5
c is -7

[dawn gulch]

so what would that be for my equation?

[hot crag]

try it

[dawn gulch]

seeing as there is only 2 numerals it would be, 2 is a and 4 is b?

[hot crag]

no

[dawn gulch]

or is b the x2?

[hot crag]

no no no

if i have a quadratic

the coefficient of x^2 is ALWAYS a

the coefficient of x is ALWAYS b

and the constant is ALWAYS c

but HERE

since you have 2x^2 + 0x + 4

b = 0

and c = 4

a = 2

[dawn gulch]

c being what you deal with at the end, got it

so when it displays itself as X without a numeral we Identify it as a 0?

[hot crag]

...?

i don't understand you

[dawn gulch]

I do not even know what I am frigging saying now

I am so lost

where did the 0x come from?

I think that is what is throwing me off

[hot crag]

because-

0x = 0

[wraith dagger]

ren

[dawn gulch]

so, seeing as there is no "3rd" number in there we replace that as a 0?

[hot crag]

yes...

[dawn gulch]

alrighty

[dawn gulch]

I like how this here is re-arranged, making it easier to understand.

so would it look something like

0/2x2

[cedar kiln]

@dawn gulch Has your question been resolved?

[dawn gulch]

<@&286206848099549185>

[sudden swallow]

yo mb

tell wht u dont understan

d

@dawn gulch

[dawn gulch]

Think you can scroll up have a squiz at the previous conversation?

[sudden swallow]

nah i dont have that much time u just explian i also got exam tmrw

[dawn gulch]

alright

I do not entirely get the result

[sudden swallow]

so u didnt got the values for x?

explain more wht happened

[dawn gulch]

the axis of symmetry

[sudden swallow]

wah

[dawn gulch]

I do not understand that

[sudden swallow]

wht u mean by that

send a pic

[dawn gulch]

Gotta quickly have dinner

[sudden swallow]

just find the value of x

b will be 0

in simple words

its the x co ordinates

[dawn gulch]

so would it look like

[sudden swallow]

or any co ordinates

wait

[dawn gulch]

0/2x2

[sudden swallow]

write it on ur copy i cant understand like this

[sudden swallow]

waaaah how did u get this

use quadratic formula

b^2-4ac

then use the formula for x

[dawn gulch]

can you place the answer below and i will try work back from it

i gotta go quickly

[sudden swallow]

ok

whts the statement

nvm

i cant do the calculation in my head

so

x= 2+4 times root of 2/4

the other value is just negative

[sudden swallow]

solve it u will get the value

oh wait

x= root 2

and -root 2

check if its correct

[dawn gulch]

hmm, okay

[sudden swallow]

so is it correct?

[dawn gulch]

no

I was correct

I appreciate your help

But I am stumped in getting the vertex

[sudden swallow]

ok

[dawn gulch]

I cannot figure out the layout

<@&286206848099549185>

[sudden swallow]

ill tell ;ater u ski[ it

later

skip

[arctic inlet]

huh

[dawn gulch]

just need help fixing up my process

As you can see I have made an issue regarding the equation under x=0 and above =4+

I am well and truely effed for my test tomorrow

you know what I am reffering to?

[arctic inlet]

above as in [4,->〉?

[dawn gulch]

I need the equation to be fixed for me'

"when x=0, f(x)= . . .

I need that fixed

hello?

[sudden swallow]

idk isnt this supposed to be normal quadratic equation question

oh i sea

[sudden swallow]

it is 2x^2+4 put the value of x 0

wait

does ur question say to make a graph?

like this

as i was saying put value of x 0 in the eq.

u get 6

so it should be 6,4 absed on wht u wrote

[dawn gulch]

n

o

[dawn gulch]

I know what the final answer is (0,4) but I am trying to figure out where I went wrong in the equation

[sudden swallow]

no ur going correct way

[dawn gulch]

for the bit in the middle

I cannot correcly lay it out

nvm, I got it

[sudden swallow]

oh ok

[dawn gulch]

Maybe I just get kidnapped

then I can avoid this

[sudden swallow]

lol

[dawn gulch]

thanks for your assistance monkey

[dawn gulch]

.close

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[cedar kiln]

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[faint egret]

Somone can help with this,? solve for all x

[hearty arch]

what have you tried so far

[faint egret]

Take tangens on both sides, geting tha awnser 1. But its not rights.

[cedar kiln]

@faint egret Has your question been resolved?

[reef sage]

are u supposed to solve for x?

[faint egret]

Yes, find all answers for x. Ofcourse x =-1 is not ok.

[reef sage]

and whats the final answer?

what u r supposed to get

[faint egret]

I dont have the answer. But if I plot the graph I see that all values below -1 is ok......

[cedar kiln]

@faint egret Has your question been resolved?

[cedar kiln]

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[obtuse mango]

how did u do the first part?

[harsh skiff]

its a bit of a shoddy solution

i did dy/dx = something

then divide top and bottom by 1/x

then u=y/x

then integrate and resub

but i dont think that is the intended solution

[obtuse mango]

yeah you haven't written it as d/dx (F(x,y) ) = 0

[harsh skiff]

yeah exactly

I'm not sure exactly how

[obtuse mango]

so let's suppose i just gave u d/dx (x^2 + y^2) = 0

how would u expand that?

[dawn junco]

well I suggested a way earlier, but everybody was yapping about y=vx

[obtuse mango]

ahaha everyone is an npc

[harsh skiff]

well it would just be 2x + 2y dy/dx

[obtuse mango]

yes exactly

[dawn junco]

#help-0 message

essentially what ly is going for ig

[obtuse mango]

you got the 2x by differentiating x^2 + y^2 w.r.t. x and 2y by differentiating x^2 + y^2 w.r.t. y

[obtuse mango]

so there are technically 2 ways of doing the first part

you could integrate it formally etc. but i think the easiest way to go about it is

[obtuse mango]

can you think of a function F(x,y) where if i differentiate it w.r.t. x, i get 2x+y

and if i differentiate it w.r.t y, i get x+2y

[harsh skiff]

oh yeah x^2 + xy + y^2

= c

is there a more rigourous way to do it than inspection?

[obtuse mango]

basically this

[obtuse mango]

if you think about what u've actually done, you've integrated 2x+y w.r.t. x

that leaves you with x^2 + xy + g(y)

and then you can solve for g(y)

[obtuse mango]

but often depending on how good ur inspection skills are, you can just inspect it

[obtuse mango]

now can u use this method to solve the circled red part?

[harsh skiff]

I haven't seen this before, is this a standard result?

if you integrate y w.r.t x do you get xy + g(y)

[obtuse mango]

also i read this and it seems like they didn't actually give you a reason why you should try y=vx, whilst that isn't the point of the question it's a standard technique so i feel like i should give a quick explanation

[harsh skiff]

oh ok thanks

[obtuse mango]

basically, if you look, the 'degree' of 2x+y is '1' (it's linear)

and the degree of the thing in front of dy/dx is 1 too

whenever you have a homogeneous linear d.e. where everything is of the same degree, y=vx will often simplify the problem

[obtuse mango]

well basically if you think about non-multivariable calculus

if i integrated 5 i would get 5x+c where c is a constant

[harsh skiff]

oh i see

[obtuse mango]

but now if i have a function of x and y

and i say integrated y w.r.t. x

then i'd get xy + g(y) because essentially, g(y) acts as the constant

[harsh skiff]

oh that actually makes a lot of sense

[obtuse mango]

if you think about it, xy + y^2 and xy + e^cos(ln(y)) etc. all differentiate (w.r.t. x) to give y

[harsh skiff]

yeah that makes sense

ok thank you so much

genuinely appreciate it

[obtuse mango]

nw!

[dreamy void]

I actually applied this now on this problem, thanks

[dawn junco]

exact equations baby

[dreamy void]

reversed product rule

[harsh skiff]

I'm starting uni soon and they gave us some prereading stuff but we haven't learned some topics so I wasn't sure about stuff like this

[dawn junco]

yea

[harsh skiff]

thanks

[dreamy void]

damn that method is actually really faster

[cedar kiln]

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[cerulean wind]

How to do part (iii)

[cedar kiln]

Please don't occupy multiple help channels.

[brave aspen]

Are you in calculus?

[cerulean wind]

Yuh

[brave aspen]

You know how to find velocity right

From position

[cerulean wind]

Like in part i?

[brave aspen]

Yea i suppose

I.e. do you know how to take the derivative to find velocity

[cerulean wind]

Yea I think so

[brave aspen]

Basically if you have a position function

Its derivative is velocity

The reason i ask about whether youre in calculus is because the way you'd figure this out or answer is affected by what tools you have

[cerulean wind]

yea but im bad w it

so im not quite sure

[brave aspen]

Well you got it right actually

I just checked

Once you have velocity

Do you know how to check for increase/decrease

[cerulean wind]

the acceleration?

[brave aspen]

Uhh let me rephrase

Once you got velocity do you know how to check for places where v(t) is positive or negative

[cerulean wind]

no

[brave aspen]

Ok so basically

If you have a nice continuous function

It changes signs from + to - at its zeros

So basically

Find the zeros of velocity

Start there

Well you kinda did already right

[cerulean wind]

oh yea

but i just thought that was where they stopped

[brave aspen]

Yes it's true

But like

If I'm moving forward

And then start moving backward

Along the way i had to stop right?

[cerulean wind]

yup

[brave aspen]

That's what you're looking for now

The moments where you stop as you change direction

Once you get those zeros

Put them on like a number line

------------|----------|-----
1 3

Like that kinda

[cerulean wind]

okok

[brave aspen]

See how there are three regions

Less than 1

Between 1 and 3

Greater than 3

[cerulean wind]

yea

[brave aspen]

Test numbers in each region

To check their sign

Just checking positive or negative

For example you could plug in 0 for x < 1

[cerulean wind]

wdym by test the numbers

[brave aspen]

Take 0

Plug it into v(t)

Check what you get

Is it a positive or negative?

[cerulean wind]

9

[brave aspen]

So positive

Then you conclude

[cerulean wind]

oh so its moving forward

[brave aspen]

When x < 1, velocity is positive

Yessss

Exactly

Check the region 1 < x < 3

[cerulean wind]

okokk

-3

[brave aspen]

Then v is negative in that region

[cerulean wind]

moving backwards?

[brave aspen]

Yep

[cerulean wind]

so it turned at t=1?

[brave aspen]

Yea basically

[cerulean wind]

then how does it go to t=3

[brave aspen]

Wdym

Time moves forward

That doesn't mean i physically move forward

When i walk backwards, time is still moving forward

[cerulean wind]

how would it show in the timeline

[brave aspen]

Well the original function is position right

When v turns negative the position function would start going downward

On the graph

[cerulean wind]

position function as in displacement?

[brave aspen]

Yes it's a displacement function in this case

[cerulean wind]

alr alr

[brave aspen]

So then the last check is t > 3

Just check if v is positive or negative

[cerulean wind]

9

[brave aspen]

Ok so then positive

Alright so in this interval

You had some time spent moving in the positive direction and some time spent moving in the negative direction right

[cerulean wind]

right

[brave aspen]

Ok so then consider this

If i walk 10 meter forward

Then 10 meter backwards

How much was my total displacement

[cerulean wind]

0

[brave aspen]

But did i actually move 0 meters?

Or did i travel a certain distance

[cerulean wind]

u moved 20m

[brave aspen]

Yes

So essentially

Displacement does not always equal total distance traveled

[cerulean wind]

facts

[brave aspen]

Part iii answer:

Because velocity takes on both positive and negative values for t in that interval, this means that at some point you traveled in opposite directions

Therefore, total distance traveled will be higher than total displacement

[cerulean wind]

ohh i get it now

so to find total distance then we need to find the total displacement between in each interval?

[brave aspen]

Yess perfect

And take absolute value of them all

Then add them

[cerulean wind]

to find the displacements do I input t as the points where they stopped in the eqn for displacement

[brave aspen]

That's right

Because it's at those times that the particle turns around

So basically you'll find 3 displacements for 3 regions

[cerulean wind]

so i put the values 1 and 3 of t into the eqn?

[brave aspen]

Well it's like

Distance traveled during leg 1 of the journey

Leg 1 is the first positive region

So

T = 0 to t = 1

Find displacement between t = 0 and t = 1

That's the first part

[cerulean wind]

put 1 as t?

[brave aspen]

This well tell you how many meters forward you walked

Do

Final position - initial position

[cerulean wind]

ohh ok

[brave aspen]

Which is to say

P(1) - P(0)

That will be distance traveled between t = 0 and t = 1

[cerulean wind]

for the next displacement do i find P(1) - P(3)

[brave aspen]

Well technically other way around

[cerulean wind]

wait what

why

[brave aspen]

As it should be negative in this leg

[cerulean wind]

ohh

[brave aspen]

It's

Final - Initial

So

P(3) - P(1)

That being said....!

You need the absolute value of this either way for this part

[cerulean wind]

then the last one would be p(4) - P(3)

[brave aspen]

Yessss exactly

Then you'll get 2 positive, 1 negative

Your total distance traveled is the absolute value of all those numbers added together

So that when you walk 10 ft forwards and 10 ft backwards we call it 20 ft rather than 0

That's the difference between distance and displacement

[cerulean wind]

alr alr thankss

[brave aspen]

Yw

[cerulean wind]

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[crimson sedge]

how did they split this

[native heath]

it'd be cool if I could get a zoomed in picture but uh factorisation ig?

[crimson sedge]

does this help

[crimson sedge]

how so

is there some identity like $a^2+b^2+c^2 = (a+b+c)(a-b-c)$

[wraith dagger]

sviyyyy

[crimson sedge]

but that's wrong isn't it

[cerulean sail]

Did they even write that correct I think they meant (x^2 + x + 1)(x^2 - x + 1)

[crimson sedge]

WAIT

YES

omg thank you so much!!!

[cerulean sail]

In the latter case, you can difference of two squares that, ((x^2 + 1) + x)((x^2 + 1) - x)

[crimson sedge]

no but wait let me process

[crimson sedge]

but it's x^2-x+1

[cerulean sail]

(x^2 + 1) - x is the same thing

Just written differently to make it more convenient to work with

[crimson sedge]

hmm i understand

but how did they just start off that way? 😭

[cerulean sail]

A good question, for which I don't think I have a helpful answer someone may have one that would be of help

This is one of the examples of a polynomial of degree strictly greater than 3 which factors but doesn't have any (real) roots/zeros, probably the best example of one

[cedar kiln]

@crimson sedge Has your question been resolved?

[robust coyote]

It is easy

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[floral terrace]

I might have a brainfart.
Question about derivative.

[floral terrace]

I have this function of a particle's velocity:

and to find the acceleration of the particle, you find the derivative of the function, correct?

[sonic thicket]

With respect to time yes

[floral terrace]

and when you derive... for example... Bcos(wt), it becomes -wBsin(wt)?

[sonic thicket]

Yes

[floral terrace]

why does the answer sheet say this? 💀

[sonic thicket]

Doesn't seem correct

[floral terrace]

Did they make a mistake? lol

this comes from an exam question HAHa

[sonic thicket]

Yeah they probably made a mistake

[floral terrace]

good to know, thank you:))

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[sharp cradle]

Hello, someone help me with this exercise pls?

[sharp cradle]

The rank of the standard matrix is equal to the dimension of the image set, so I thought about reducing the matrix by gaussian but I don't have all the values of a left.

[dreamy void]

Did you ref this matrix regardless?

[sharp cradle]

yes

[dreamy void]

show

[dreamy void]

your a looks like partial

F_2 - 2a * F_1

[sharp cradle]

what do you mean?

[wraith dagger]

bacc

[sharp cradle]

a

[dreamy void]

Also I don't understand your 2nd matrix

You are multiply the first row by -2a

-2a-2a²-2a(a+1)

0 -2a²+2 -2a²+2

[obtuse mango]

for most values of a, the matrix is probably non-singular

so kernel = 0, image = R^3 for most values of a

i'm expecting there to be like only a few values of a where the matrix is singular

so i'd just analyse those 1 by 1

that's what i think would be the easiest way to approach this problem

[sharp cradle]

I was wrong

[dreamy void]

We are all wrong at least once in life

[sharp cradle]

these are the answers

[sharp cradle]

I realized that it comes out equaling the determinant of the matrix to zero, but I do not understand why xd

[obtuse mango]

if det != 0, then you can inverse the matrix

so Mx = 0 => x = 0

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[somber ibex]

Do you see a mistake?

First box is what is given

It seems like a fairly easy question

[crimson sedge]

98/5 is ?🤨

[somber ibex]

I can’t have decimals

I have to round it

The answer is 19.6

[crimson sedge]

Ah well while finding the value of y and z

You should have x as 19.6

[somber ibex]

Ohhh

Alright Ill give that a try give me a couple of min

Thank. USO much

[lone temple]

https://tenor.com/view/gacha-life-gacha-jjk-toji-bleach-gif-17275865596869682801

[crimson sedge]

Sus

Very sus

Oh you didn't send it

[nimble mountain]

why is there a gif in the middle of the channel

[crimson sedge]

Should have seen the other gif which was deleted

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[ancient quarry]

Intuition for proving that $f(x, y) = \frac{x^3}{x^2 + y^4}$ is continuous at $(0, 0)$?

[wraith dagger]

Bean Man

[ancient quarry]

In class we proved it by bounded it by a continuous function that approached zero.

Oh yea and f(0, 0) is defined to be = 0

With the formal definition we want to say that $$\forall \epsilon, \exists \delta \text{ such that } ||(x_1, y_1) - (x_2, y_2)|| < \delta \implies |f(x_1, y_1) - f(x_2, y_2)| < \epsilon$$.

All we did in class was show that $f(x) \leq |x|$ and since we are taking the limit as $x$ approaches zero, then we've proven it?? I don't understand why thats sufficient.

[wraith dagger]

Bean Man

[cedar kiln]

@ancient quarry Has your question been resolved?

[violet jay]

$0 \leq f(x,y) \leq |x|$
it's the squeeze theorem for multivariate functions

[wraith dagger]

Morrow

[violet jay]

You can convert it to an epsilon-delta argument

Let $\varepsilon > 0$, take $\delta = \varepsilon$ (the same $\delta$ we'd take if we were trying to find the limit of $|x|$ as $x \to 0$) then if $|(x,y)| < \delta$ in particular $|x| < \delta$ and hence $$|f(x,y)| \leq |x| < \delta = \varepsilon.$$

[wraith dagger]

Morrow

[violet jay]

(I'm assuming you separately define f(0,0) = 0)

[ancient quarry]

yea

It makes sense when I think about it, but it doesn't immediately go ding ding ding and I think its because bounding the function isn't giving me and idea of what the function is doing outside of that point. Does this kind of argument work when we don't have f(0, 0) = 0? Is it the case whenever we can bound the function?

[violet jay]

Well the bounding function tells you that locally, near (0,0), f(x,y) is small. How small? At least as small as |x|. So naturally, f(x,y) goes to zero as (x,y) goes to (0,0), since |x| does and f(x,y) is even smaller.

If f(0,0) is not 0 then you just end up with a removable discontinuity

Which is not a big deal

[ancient quarry]

Would I be able to use the same kind of strategy to prove that it is continuous at (0, 0)?

[violet jay]

Yeah you just have to adjust the argument slightly.
Instead of saying f(x,y) is small you say that f(x,y) - 1 is small.
So you say |f(x,y) - 1| <= |x|
and it works out the same

[ancient quarry]

Sorry, I'm still a little confused

Sorry it should be |x| + 1 I think

[violet jay]

yeah 1 <= |f(x, y)| <= |x| + 1 would work too

actually no

since that shows that |f(x,y)| -> 1

but we need f(x,y) -> 1

in the 0 case the absolute value doesn't matter

but in the 1 case it does

[ancient quarry]

oh god

[violet jay]

Yeah that's the squeeze theorem

https://en.wikipedia.org/wiki/Squeeze_theorem

[ancient quarry]

OHHH WAIT

I was only confused because I wasn't understanding it generally. The point of the upper bound is that it globally upper bounds our f(x,y)

We don't need the fact that its everywhere, but just locally around that point, but I wasn't thinking about it bounding it for some region, just after some point in my head

for whatever reason I was thinking about |x| as a function on [0, infty)

Alright that was my dumb moment of the day, everything makes sense now, thank you

.close

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[cedar kiln]

@hoary mauve Has your question been resolved?

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@hoary mauve Has your question been resolved?

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@hoary mauve Has your question been resolved?

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@hoary mauve Has your question been resolved?

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[dreamy void]

product and quotient rule

[vapid otter]

yes yes

but how

i dont know the actual functions

[dreamy void]

use the definition

[vapid otter]

uh

[dreamy void]

can you write down the product rule

[vapid otter]

nx^n-1

[dreamy void]

not power rule

product rule

[vapid otter]

what

[dreamy void]

ok you don't know the product rule

[vapid otter]

is there another way to do this

[dreamy void]

This is the product rule

[vapid otter]

so then

[wraith dagger]

bacc

[dreamy void]

So just apply the definition

you dont need to know the actual function because you are given the values that you need

[vapid otter]

$P'(x) = F'(x)G(x) + F(x)G'(x)$

[wraith dagger]

Devil Wears Prada

[muted timber]

with the graph you could actually get them

[vapid otter]

reminds me of double angle formula for trig

[vapid otter]

how

by approximation?

[muted timber]

easily. F is a parabola and a line

on the parabola, you got 3 points

[vapid otter]

its not defined so therefore im not doing it

[muted timber]

(1,4),(2,3) and(3,4)

[dreamy void]

you are not defined

[vapid otter]

the parabola could be 1.00001x^2

[dreamy void]

your job is to count squares and use the average rate formula for the first derivative

[muted timber]

then your value for derivative isnt either

[vapid otter]

its a piecewise so how can i even get the derivative

[muted timber]

no. the job is to obtain the derivative at two points

piecewise as well

[vapid otter]

so how would i go about finding F'(x)

[dreamy void]

you can literally see it

[muted timber]

F can be diff everywhere except 3, G everywhere except 4

[vapid otter]

WHAT

[vapid otter]

yes

and

[dreamy void]

you don't need to consider where it peaks for this task

[vapid otter]

im overstimulated

[muted timber]

anyways, assuming you dont want to obtain the function

[vapid otter]

NO

.close

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[muted timber]

and using this instead

[dreamy void]

😂

[muted timber]

ok

[vapid otter]

😈

[dreamy void]

bro is coping

[vapid otter]

Missionary so I can stare into her blue sea-eyes, piercing through the red sea, while conquering the depths of the dark sea in her heart.

.close

wait

@dreamy void

look at the question

find P'(2)

F'(2) = 0

right

G'(2) = 1/2

[muted timber]

now yes

[vapid otter]

my bad

forgot prime

[dreamy void]

at least open your channel

[vapid otter]

.open

it wont

[muted timber]

.

[vapid otter]

.reopen

[cedar kiln]

✅

[vapid otter]

so then

P'(2) = 0 + 0

bro

is this right?

[muted timber]

no

[solid juniper]

:bending_skull:

[vapid otter]

what

[muted timber]

you need also F(2) and G(2)

[vapid otter]

i did

(0)(2) + (3)(1/2)

oh

[muted timber]

p' = f' g + f g'

[vapid otter]

7/2

is that it

[muted timber]

bruh

[muted timber]

look at this again

[vapid otter]

7/4

oh

[muted timber]

...

lemme rewrite your line

p' = 0*2 + 3*(1/2)

[bleak viper]

how

$0\cdot2 + 3\cdot\frac{1}{2}$

[wraith dagger]

Astar777

[bleak viper]

0 times 2 is zero

why are u taking lcm

[vapid otter]

yes 7/4

[bleak viper]

[vapid otter]

yo im actually dying

3/2

[bleak viper]

yes

[dreamy void]

other helpful gave up

[vapid otter]

fire him

[solid juniper]

:bending_skull:

[muted timber]

i mean, i would not expect him to make such elementary mistakes

like not distinguishing a sum from a product

[dreamy void]

there are worse

someone did (2)(3)(6) = 2(3) + 2(6)

[raven shard]

the skull that bends

[muted timber]

i know

[dreamy void]

and they were calculating derivatives

and surprisingly they understood derivatives

but struggled with arithmetic

[muted timber]

not enough training

thats what happens when calculators are given to kids

[solid juniper]

:bending_skull:

[dreamy void]

you are arriving every-time layla is somewhere

[fallen moat]

is this done? 👍

[vapid otter]

im actaully not

@muted timber im reporting you to mathmetics corporate

.close

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[vapid otter]

^^for excessive hate

[solid juniper]

:bending_skull:

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[lofty light]

if y=3u^2-cosx

[lofty light]

is y'=3+sinx

just wanna confirm

[grand raft]

whats u

[lofty light]

mb

u=root x

[grand raft]

so y = 3x - cosx?

[lofty light]

yeah

but our teacher made a mistake i think

so is it 3+sinx

[grand raft]

The derivative of 3x - cosx is 3+sinx

but

[lofty light]

or something else

[grand raft]

if u = sqrt(x)

[lofty light]

btw its dy/dx

alr so im right yeah

[grand raft]

yea but it's only defined for x >= 0

[lofty light]

alright

!close

how to close

[grand raft]

.close

[frail root]

.close

[lofty light]

.close

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[cold timber]

the first digit can be any from 1-9 ,the second can be any from 0-9 except the first digit , and the third can be any from 0-9 except the first two digits. Thus, the total possible numbers are 9* 9* 8

[cedar kiln]

@compact gulch Has your question been resolved?

[compact gulch]

okay thanks., what about part (ii)

[cold timber]

For (ii), there are 5 odd numbers from 0 to 9. For the 700s and 900s, the last digit has 4 options (excluding 7 or(excluding or) 9), and the second digit has 8 options (excluding first digit and the last digit). For the 800s, the last digit has 5 options and the second digit still has 8 options. Therefore, the total possible numbers is 8 * 4 + 8 * 4 + 8 * 5 = 104

[compact gulch]

okay tysm!

.close

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[half vault]

I have the following four premises and would like to know whether they are enough to entail the conclusion (and if now what other premise is required). The 'grounding relationship' is a relationship between two propositions which is transitive (see P2.) and factive (see P4.) (it's also asymmetric, but that doesn't matter for now, I think). p grounds q is symbolised as "<p ← q.>''. 'Q', 'T' and 'S' are names of particular propositions:

[P1.] ◻(Q → <Q ← T.>)
[P2.] ∀p, q, r ◻((<p ← q.> & <q ← r.>) → <p ← r.>)]
[P3.] ◻(T → <T ← S.>)
[P4.] ∀p, q ◻( (p ← q.) → (p & q))
[C.] ◻(Q → <Q ← S.>)

[cedar kiln]

@half vault Has your question been resolved?

[floral arrow]

Looks ok to me

• Q and P1 and P4 give you T
• Q and P1 give you <Q ← T.>
• T and P3 give you <T ← S.>
• <Q ← T.> and <T ← S.> and P2 give you <Q ← S.>
• so Q and P{1,2,3,4} give you <Q ← S.>, which is just C

[half vault]

thansk!

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[half vault]

solved, I think:
[P1.] ◻(Q → <Q ← T.>)
[P2.] ∀p, q, r ◻((<p ← q.> & <q ← r.>) → <p ← r.>)]
[P3.] ◻(T → <T ← S.>)
[P4.] ∀p, q ◻( (p ← q.) → (p & q))
[C1.] ◻(Q → (<Q ← T.> & T) ) {P1. and P4.}
[C2.] ◻(Q → T) {C1. and P0.}
[C3.] ◻(Q → <T ← S.>) {C2. and P3. and the transitivity of strict implication}

[C.] ◻(Q → <Q ← S.>) {P1., C3. and P2.}

[floral arrow]

Do you have another question?

[cedar kiln]

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[half vault]

no tahnsk

:)

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[cloud warren]

how did they get 24/pi on the outside

[cloud warren]

dont you just pull 4 out

[void sand]

u sub

[carmine bronze]

They multiplied 4 by a whole fraction. (6/pi)/(6/pi).

[dusk goblet]

notice how in the integrand there is a pi/6

[cloud warren]

right usub is 4 sec u * 6/pi

[dusk goblet]

well that’s not how i’d integrate secant

[cloud warren]

you pull the 4 and 6/pi out and get 24/pi yes but that just leaves you with 24/pi integral sec u

[cloud warren]

oh mb i forgor

ln|secu + tanu|

[dusk goblet]

no +C necessary

it’s a definite integral

[cloud warren]

yup

[dusk goblet]

but yea

your u would just be the angle

[cloud warren]

so 24/pi * integral from [0,2] (ln|sec pi/6 + tan pi/6)?

[dusk goblet]

no

you already integrated

why the integral again

you’re evaluating the anti derivative from 0 to 2 sure

but it’s not an integral

[cloud warren]

ok i got it right

ty

.close

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[cinder walrus]

no idea how to approach this problem

[cinder walrus]

<@&286206848099549185>

[cedar kiln]

@cinder walrus Has your question been resolved?

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[ionic wolf]

idk

can u do infinity minus infinity

wat if it was infinity + infinity

idk

[slate lintel]

hmm

what class are you in?

[ionic wolf]

calculus

[ionic wolf]

calculus

[slate lintel]

hmm ok

[ionic wolf]

ok i think it's false

[slate lintel]

do you think this is true or false? try some okok

[ionic wolf]

cuz there is not enough info

[slate lintel]

why's that

[ionic wolf]

like

f(x) could be like something like x^2 and g could be like x

idk

[slate lintel]

yep that's good

what you've produced is called a counterexample

and it's enough to show something is false

[ionic wolf]

fr

[slate lintel]

yep

good job

[ionic wolf]

ty

.close

[cedar kiln]

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[tidal snow]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tidal snow]

I want integration questions

[obtuse mango]

https://math.mit.edu/~yyao1/integrationbee.html

[cedar kiln]

@tidal snow Has your question been resolved?

[raven shard]

school level? or..

[cedar kiln]

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[solar edge]

I recently was introduced to trignometry and OMFG that shit is hard af, I have no clue what I am proving nor what I am doing, I feel like I am aimless, How to be better at dis shiii

[brave aspen]

Well the most important thing imo is understanding the first quadrant

Because all the "important" angles and numbers basically repeat from there

So you can just learn the values for a few angles

Instead of for idk how many in the whole circle

[solar edge]

2cos theta= x+1/x then the value of 2 cos 3theta?

Weird shit like this is asked

I am a rookie so dont have much of an idea

[brave aspen]

They ask you what is 2cos(3x) ?

[solar edge]

nah

[silk moss]

Trigonometry may seem tough at first, but with some practice and patience, you'll start to grasp the concepts. Try focusing on small steps, learning the basic definitions, and connecting trigonometric functions to the geometry of triangles. Regular practice will help, and don’t get discouraged—progress comes gradually!

[solar edge]

3*theta

[brave aspen]

Yea sorry let x = theta

[solar edge]

But

[brave aspen]

Do you have a pic of the problems

[solar edge]

its already given as x+1/x right?

Yea gimme a sec

Thats the one I am dealing currently with

[brave aspen]

Yea it's a hard one i tried working it out

[cedar kiln]

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[raw hinge]

i need help with this proof

[raw hinge]

so far i made the ratios x/y, y/z, and z/x and used componendo dividendo to make them in the correct format
For Example:

but when i tried with the other ratios and then plugged it all back into the question it didn't cancel out for some reason

algebraic trigonometry

JEE

lmao

its a JEE question thats why

i was just doing for fun because its known as the hardest exam in the world

JEE is 10-12th grade math

but i dont live in india so i cant take it anyways just for fun

same

but like its not that hard

im just getting the negatives wrong

did you take trig yet?

or calculus

because often there are no numbers

ah i see

these are trig identities

learned later in the year

im practicing these because i want to try for the IMO USA team

kind of

but the identities are not like differences of squares

they are more like proofs about the properties of basic sin cos tan

for example sin^2(theta)+cos^2(theta)=1

it works for all angles subbed in for theta

or radians if you know what those are

yeah

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[cerulean root]

Could somebody explain part a(i) to me please I don't get it

How do they know this ?

I thought if it was an asymptote that would mean that it always get's close to the line but never actually touches it

So how are they saying that y=0 ?

[hollow trail]

it's saying the vertical asymptote is found where the denominator is 0

[cerulean root]

Oh ok I get that and I guess they sub in x=2 because that's when the denominator is 0 right ?

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[serene bane]

Is there a shorthand for $X_1 \times X_2 \times ... \times X_n$ where $X_i$ is a set?

[wraith dagger]

Improv

[crimson delta]

$\prod_{i=1}^n X_i$ is sometimes used

[wraith dagger]

Denascite

[opaque root]

$\times X_i$

[wraith dagger]

Samuel

[serene bane]

cheers

.close

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[keen nova]

i have a general question: i'm working with a 3d shape assignment and thus with linear algebra and geometry. I am applying a translation so that the barycenter of the shape coincides with the origin of the coordinate system. Thereafter I am scaling it so that "it fits inside the unit cube"