help-0

[lucid cargo]

Ok OK i see let me try ill update if i run into anything

[sharp juniper]

wait i wanna try solving this too

[lucid cargo]

Ok so the equation for AB is 4/3x - 2/3 so i need to find its point at 0

When its 0***

[sharp juniper]

wait are you supposed to find area of the thing

[lucid cargo]

ye

the whole thing

[sharp juniper]

i think i have it

its 36

smthing

idk what the unit is

am i correct?

[lucid cargo]

ok i did it

yeah

its 36

[mental python]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lucid cargo]

i was getting 37 before when i was using 0.5 more of the breadth

so i was confused bc how can i be 1 unit off

[sharp juniper]

i tried to do it myself

im not helper

[lucid cargo]

But i see now

[sharp juniper]

ok sure

sorry

[mental python]

you are a helper, you are not the owner of the channel

[sharp juniper]

oh

yeah

mb

[lucid cargo]

geometry is so aesthetic

Anyways

[mental python]

its okay, just keep it in mind 🙂

[lucid cargo]

Thanksfor the help

ill close

.close

[lone heart]

Channel closed

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[strange fractal]

How would i do the last question?

[strange fractal]

What does minimum value mean

[fierce cipher]

smallest value

[strange fractal]

Which values are z1 and z2?

[fierce cipher]

uh

it's shown in the problem

[strange fractal]

Ohh

I see

Ok but theres an inequality how would i subtract them? Would i move it to the other side

Like |z| -2=0?

[fierce cipher]

no

[strange fractal]

How would i subtract z1 and z2

[fierce cipher]

gotta remind you that each complex number can be represented as a point/vector on the plane

[strange fractal]

Oh..

[strange fractal]

Wait but then for the first one

The point is just the centre of the circle right?

[fierce cipher]

it says z1 is in S, that means z1 is a point of the circle

it doesnt have to be the centre

because |z1| <= 2

the point that represents that inequality creates a circle with radius of 2

so i can take any point on that circle and i get z1

[strange fractal]

Wtf

So i can choose any point i want

But it wants the minimum value

[fierce cipher]

yeah

so there's only either 1, or some choice of points that can satisfy

[strange fractal]

How do i know which one gives me the minimum

[fierce cipher]

remember what i said, that complex numbers are just points/vectors?

|z1 - z2| = |OS - OT| = |TS|

so you just need to find the smallest distance of TS

[strange fractal]

Im kinda confused

This is how they did it

How did they find pi/4

[fierce cipher]

let's put it in basic language

$z=a+bi\longrightarrow \overrightarrow{OM}=\left( a,b \right)$

[ocean seal]

TargetVN

[fierce cipher]

with M being the representation of z

[strange fractal]

um im really bad at vectors.. is there a diff way

[fierce cipher]

this is the easiest way

[strange fractal]

oh ok..

lets see

[fierce cipher]

but its ok, we only need to use 1 vector formula

$\overrightarrow{OS}-\overrightarrow{OT}=\overrightarrow{TS}$

[ocean seal]

TargetVN

[strange fractal]

guys whats OS and OT

ohhh the points

[fierce cipher]

yeahhh

[strange fractal]

how does this give us the shortest way

[fierce cipher]

Because the modulus of complex number is exactly the modulus of vectors

[strange fractal]

wait how do we know which points to use for the vectors if we're trying to find the minimum..

[fierce cipher]

uh idk

those points are given by the problem

different problems require different methods

So, $\left| z_{1}-z_{2} \right|=\left| \overrightarrow{OS}-\overrightarrow{OT} \right|=\left| \overrightarrow{TS} \right|=TS$

[ocean seal]

TargetVN

[fierce cipher]

is it clear?

[strange fractal]

yes..

[fierce cipher]

so you just need TS reach minimum

if you dont know where S and T should be, probably its best to try some positions first

[strange fractal]

wait thats not the minimum 😡😡

wouldnt the minimum be the edge of the circle and then the closest point on the line

[fierce cipher]

exactly

in fact, O, S, T should be on a line

now, what next?

[strange fractal]

uhh

find OS.. which is 2..

and we find OF

OMG inappropriate

OT

and subtract 2

[fierce cipher]

yep

[strange fractal]

so to find OT.. first of all..

[fierce cipher]

also its not pi/4, the actual answer is at the bottom

[strange fractal]

u can do 2^2 + 2^2 =h^2

because pyhtagorean theorem

wWAIT OMG I GUESSED THAT

guys i think im getting smarter

h = sqrt(8)

minimum = sqrt(8) - 2

AND I CEEBS SIMPLIFYING I KNOW OK

ITS MY BED TIME

MY TEST IS TOMORROW FIRST PERIOD AT 8:50 AM

[fierce cipher]

go sleep

better to conserve energy

[strange fractal]

yess ok thank u this was very helpful

at first i was like omg vector

and now im like

omg easy

ok bai

WISH ME LUCK

[fierce cipher]

k

good luck

[strange fractal]

.close

[lone heart]

Channel closed

Closed by @strange fractal

Use .reopen if this was a mistake.

[strange fractal]

.reopen

[lone heart]

✅

[strange fractal]

is |z1 - z2| = |z1| - |z2|

[twin nimbus]

no

[strange fractal]

oh shucks

im so scared for tomorrow i dont even understand inequalities but ima yolo

[twin nimbus]

consider z1 = 1, z2 = -1

[strange fractal]

so this is what ive memorised if its like |z| <2 u have to shade everything in the circle

and if its the other way u shade the oufside

[fierce cipher]

What we did was actually this: $\left| z_{1}-z_{2} \right|\ge \left| \left| z_{1} \right|-\left| z_{2} \right| \right|$

[ocean seal]

TargetVN

[fierce cipher]

you can interpret it as the distance of 2 points AB reach minimum when it equals |OA-OB|

[strange fractal]

and this is -3pi/2 < arg(z+1+i) < 0

i think

i just dont get it like im reading it and its like

-3pi/2 is less than arg(z+1+i)

liek it doesnt make sense

id there a YT vid i can watch on this? i tried finding but i cant

[twin nimbus]

you understand the meaning of arg(z), correct?

[strange fractal]

yess the angle it take to get to the point or smt..

the shortest distance

[twin nimbus]

"or smt"?

[strange fractal]

ok the angle it takes to get to the point and no or smt

[twin nimbus]

this is a very basic operation to do to a complex number, you need to be absolutely sure what it is.

ok

[strange fractal]

): its not basic

its hard

to wrap ur head around

[twin nimbus]

So if you had: -3pi/2 < arg(w) < 0

[strange fractal]

yes

[twin nimbus]

would you understand what this looks like?

[strange fractal]

no

[fierce cipher]

you just learned complex numbers as vectors and instantly find it easy

[strange fractal]

tvh complex numbers are easier thsn vectors

[twin nimbus]

It is possible for something to be both basic and difficult to learn.

[strange fractal]

in vectors my teacher gave us 10 mark questions based on a boat getting moved by a current

it was do hard

[strange fractal]

oh

[twin nimbus]

What I'm getting at, is that this is foundational.

[strange fractal]

ohh ok

[twin nimbus]

this is something you need to know.

[strange fractal]

ohhh ok thx i thoughtu. Were being mean

[twin nimbus]

and if you're unsure about it, then it's a problem.

[strange fractal]

theres a lot of mean people here bcuz im slow 😢😢

[strange fractal]

i would not know what this looks like

is there a video i can watch?

[twin nimbus]

I don't know of one off of the top of my head.

[strange fractal]

ok this is what goes through my brain when i read this

[twin nimbus]

if I asked you what the set of all complex numbers, w, such that -3pi/2 = arg(w) could you tell me?

[strange fractal]

no

idek what that means

[twin nimbus]

alright

do you understand the polar form of a complex number?

[strange fractal]

yes

[twin nimbus]

please explain the polar form of a complex number to me.

[strange fractal]

um... so basically.... its like...

a complex number but it has an argument

[twin nimbus]

explain it to me as if you were trying to teach me.

[strange fractal]

and theres a modulus

uhhhh so basically.. the polar form is given by

Rcis(theta)

[twin nimbus]

ok

what does cis mean?

[strange fractal]

To find theta u need to find the argument which is the smallest angle it takes to reach the point

cis is cos(theta) + isin(theta)

[twin nimbus]

ok cool

and so what you're saying is that arg(R cis(theta)) = what?

[strange fractal]

the argument is the smallest angle it takes to reach the point given by the complex number..

in the range

-pi, pi

[twin nimbus]

so given that you have R cis(theta) what is the argument of this number?

[strange fractal]

theta

[twin nimbus]

ok, cool

Do you know what this looks like graphically?

[strange fractal]

um i guess u just need to expand it

and then solve it so its in the cartesian form

[twin nimbus]

like, if you have a complex number a + bi, can you draw for me where the angle theta is?

[strange fractal]

and then u just plot it

yes

wait but im in my bed rn ill just show u in notes

wait i have a pic of one

[twin nimbus]

ok, if I gave you the number -3 - 3i, can you draw for me and take a picture of it, the angle theta?

ok

cool

perfect.

so let's say that angle there is -3pi/2

(it's super close)

or wait, no, that's not quite right

that's -3pi/4

[strange fractal]

yess

[twin nimbus]

So the thing is, the equation R cis(theta) works for any theta, not just theta in the range (-pi, pi]

[strange fractal]

what i thought we had to force it into the range

by adding 2pi or subtracting 2 pi

[twin nimbus]

You can, however, for your question that needs some special care.

[strange fractal]

thats what they make me do when they make me solve like z^5 = -1

ohh ok

[twin nimbus]

and you don't actually need to as long as you're conscious about what you're dealing with.

[strange fractal]

ohh okkk

[twin nimbus]

so just so we're absolutely clear.

your question is -3pi/2 < arg(1 + i + z) < 0, yes?

[strange fractal]

tes

yes

wait no

its -3pi/4

[twin nimbus]

ok

[strange fractal]

like dis one

[twin nimbus]

yes.

[strange fractal]

how would u find the inequality just by looking at it

[twin nimbus]

that shaded region there is -3pi/4 < arg(z) < 0

[strange fractal]

so our argument is greater than -3pi/2

how

[twin nimbus]

my bad

typo

[strange fractal]

ohhh yea its ok

i make a lot of them

or argument looks like its equal to -3pi/4

[twin nimbus]

ok so, what happens if you shift z to the right before you take the arg, by one unit?

[strange fractal]

uh... u do arg(z-(1))

arg(z-1)

[twin nimbus]

let's stop and think for a second.

We can find, for instance arg(-3 - 3i) = -3pi/4

[strange fractal]

yes

[twin nimbus]

if we have arg(z + 1 + i) what value of z would result in arg being called with the parameter -3 - 3i?

[strange fractal]

uhh

dis is confusing

wth is a pareneter

[twin nimbus]

a parameter is a name for the thing you put in the function

so if I have arg(1 + i) then the parameter to arg is "1 + i"

[strange fractal]

ohh i think i get it the argument must be less than 3pi/4 as its going backwards and its also less than 0

why cant it be equal to 0?

[twin nimbus]

-3pi/4 < arg(z) < 0 means that arg(z) > -3pi/4

[strange fractal]

wait let me do another example

[twin nimbus]

we're not actually done with this example yet though

[strange fractal]

oh

what did i do it wrong

wait i dont understand why it cant be

[twin nimbus]

well, first it sounds as if you don't fully understand though you think you do, and second you never solved the original problem, just a simpler version

[strange fractal]

-3pi/4 < arg(z) <=0

why is it just

<0

[twin nimbus]

!original

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[strange fractal]

u said it here

[twin nimbus]

I took that from your problem you stated earlier

[twin nimbus]

here

[strange fractal]

ohhh

[twin nimbus]

which we now know had a typo

[strange fractal]

wait can i just do 1 and go to bed cuz its late 4 me

and i have a test tmr on this

would dis be

the arg is less than pi/6

so

wait but it is greater than 0

wut

ok

so

[twin nimbus]

that's less than pi/6 and greater than 0

[strange fractal]

pi/6 < arg(z) > 0

[twin nimbus]

no

[strange fractal]

oh

pi/6 > arg(z) > 0

[twin nimbus]

also I don't understand why you have pi/6 on that problem

[strange fractal]

ok im going to do one more when i wake up

and hopefully i think during my test

[strange fractal]

how would i know if its > or >=

[twin nimbus]

dashed vs solid line

[strange fractal]

if its dashed then its just >?

[wind cloak]

yes

[strange fractal]

but my point isnt dashed

so why dont i do pi/6 >= arg(z) >=0

[twin nimbus]

The difference between (for instance) -3pi/4 < arg(z) and -3pi/4 <= arg(z) is whether or not points like -1 - i are included or not.

we represent this using either a dashed (not included) or solid (included) line.

[strange fractal]

omg thanks that sounds like something my teacher would trick us with

ohh he did the dashed line here

[twin nimbus]

so this image

is this the question?

or?

[strange fractal]

nooo i think im done for today

its sleepy time

[twin nimbus]

good night

[strange fractal]

but since my line here isnt dashed

i include it

-3pi/4 <= arg(z) <= 0

ok gn guys

thxxxx sm im so tieed

.close

[lone heart]

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[lone heart]

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[smoky obsidian]

so i convert to 3x^2 -11x -42 = 0 (standard form)

[mental python]

do you know the quadratic formula

[smoky obsidian]

yes but it's not asking me to do that

[long axle]

it says solve by factoring

[smoky obsidian]

what gets me stuck is having to separate the middle term -11x into a combination that multiplies to -126 (which is 3*-42) and adds up to -11

[mental python]

you find the factors by using the quadratic formula

[smoky obsidian]

i think the problem refers to the factoring method

[long axle]

think of some other factors of 126

[smoky obsidian]

or the factoring formula however you wanna call it

[mental python]

you can do that aswell

[smoky obsidian]

-14 * 9 but that doesn't add up to 11

[long axle]

im p sure factoring here means not QF

[smoky obsidian]

-18*7

[long axle]

ya

[smoky obsidian]

!close

uhhh

[long axle]

use .

[smoky obsidian]

.close

[lone heart]

Channel closed

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[lone heart]

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[full nimbus]

Motion of a Particle

The position of a particle traveling along a plane curve after t seconds is given by the parametric equations x(t) = t³ - 3t^2 and y(t) = t^2 - 8t. When is the particle at rest? What is the position of the particle at this time?

[full nimbus]

do I jus do derivative of the x and y?

and then when it equals to 0?

and then plug in the t into the x and y to get the position?

[naive valley]

yes, that's the right approach

when both x' and y' are zero then the velocity is zero

[lone heart]

@full nimbus Has your question been resolved?

[lone heart]

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[warped salmon]

Can somebody explain me, why does we pick |x+ 4| < 7 and |x-2| < 5 but not 5 and -3 ?

[lyric gale]

Please give the full question please?

[warped salmon]

Delta epsilon proof of limits

[exotic canopy]

are you talking about 30 or 31

[warped salmon]

both of them

[exotic canopy]

let's go one by one

[warped salmon]

i just dont understand. why do we pick larger ablosute value in such situation

[exotic canopy]

which line is confusing you

[warped salmon]

4

[exotic canopy]

okay i see

[warped salmon]

|x+ 4| < 7

why is there 7 but not 5 ?

[exotic canopy]

because |x+4| can be larger than 5

we need an upper bound

[exotic canopy]

we started with |x+4||x-2| < epsilon, but this expression as is has too many "moving parts". so if we restrict the delta < 1, we get that |x+4| is DEFINITELY smaller than 7. indeed it is larger than 5 but that does not really matter: we are safest picking the largest bound

we do this to obtain that |x-2| < epsilon/7

and we know that |x-2| < delta

so we pick delta = min{1, epsilon/7}

if we would've picked 5 instead, we would not have have the guarantee that delta = epsilon/5 works

[lone heart]

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[hollow shard]

Can anyone check my work and confirm my answers are correct?

[lone heart]

@hollow shard Has your question been resolved?

[lone heart]

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[short void]

I need help with continuing this

I am also confused b/c woudln't anythin divided by 0 be undefined

[north rover]

Okay. There isn’t really a nice way around this; what you want to do is start from scratch, prove the derivative of e^x, then chain rule, then apply it.

[north rover]

You still have a 0/0 form.

[north rover]

$e^{x^2}\lim_{h\to 0}\frac{e^{2xh+h^2}-1}{h}$

[ocean seal]

;(

[north rover]

That is very stupid.

[north rover]

Hmm.

Let me get my notebook. I remember proving this.

[sterile mirage]

so far i have this

actually made a mistake

with that e^h

cant split it up like that

@north rover it says associated terminology

what if we use the ln x limit definition

which comes when we take the regular derivative of e^x like this

[north rover]

That’s...kinda obvious.

It is a standard technique in proving this.

[sterile mirage]

a

[sterile mirage]

so for this can we then write this as ln(e^2x)

[north rover]

Yes.

[sterile mirage]

les gooooo

[lone heart]

@short void Has your question been resolved?

[short void]

Thank you Everyone!!

[lone heart]

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[alpine sable]

why is this defined only for x >=1

shouldn't it be also for x <= -1?

[tacit arch]

is there more context

it doesn't say "only defiend" for x>=1

[alpine sable]

true

it doesn't

and there is no more context

really

so im assuming they just wanted to restrict its domain for reasons unknown?

they are testing for just the positive numbers anyway so im assuming

thats why they did it

[north rover]

Okay this makes sense a little bit

It would be better for me to plot it side by side

,w plot arccot(x) and arctan(1/x)

[ocean seal]

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[tacit arch]

they are giving a weaker result intentionally to make the proof easier for readers

[north rover]

There are certain discontinuities, etc. that may cause some stuff

At least that is what I learned

[tacit arch]

keep reading to find the full result

[alpine sable]

ahhh i see

alright thank you guys

.close

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[rough path]

I just need help with how to start I dont want the answer please

[swift grail]

Im guessing you are allowed to use a calculator?

[rough path]

Yes

[swift grail]

And im guessing you know trig?

[rough path]

Yeah

I always get confused with triangles for some reason though

[swift grail]

Ok well start by calculating everything you can.

[rough path]

do i use sine rule to calculate the angle D?

[swift grail]

Every angle , every side. Until you eventually can calculate the area

[swift grail]

Yes

[rough path]

x = 43.97

[swift grail]

Idk what is x.

[rough path]

oh sorry D

like the first side

[swift grail]

AD?

[rough path]

yes

then after idk what to do

[swift grail]

I doubt 43.97 is a length of any side.

Is that maybe some angle?

[rough path]

oh yes

its angle ADB

hello?

@swift grail

[swift grail]

Get some other angle

Then get a side

And etc

Formula for area is
$$A = \frac{ab\sin\gamma}{2}$$

[ocean seal]

casework

[rough path]

okay

[lone heart]

@rough path Has your question been resolved?

[lone heart]

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[civic harbor]

Good Afternoon,
I must insert the word "running" into this alphabetically sorted tree.
hockey is the root/center.
It should go to the left node of soccer, right?

My logic is that it's

Higher than hockey.
Lower than tennis.
Lower than soccer

Thank you.

[limpid turret]

[lone heart]

@civic harbor Has your question been resolved?

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[lone heart]

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[red echo]

Answer: 1.5

I found the first step which is to find that CD is 19 through Law of Cosines

But after that I'm lost

You do not get a calculator for this by the way, the solution is purely by hand

[clear jungle]

Out of curiousity, what test is this from?

[swift grail]

You can get ED and AE

You can also get BD

[slim belfry]

Are you using cosine to get ED? That's the only way I see right now.

[swift grail]

You get $\sin \angle ADC$

[ocean seal]

casework

[slim belfry]

Ah, okay, that's the same process as getting cos(ADC), but much less calculation.

[swift grail]

From that you get that its $2 \sin \angle ADE \cos \angle ADE = 2\frac{ |AE| \cdot |ED| }{|AD|^2}$

[ocean seal]

casework

[swift grail]

And ofc. $|AE|^2 + |ED|^2 = |AD|^2$

[ocean seal]

casework

[swift grail]

Solve for that and hope you get something nice.

[lone heart]

@red echo Has your question been resolved?

[tough ember]

ehm guys

anyone here ?

[tacit arch]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[red echo]

!cloes

!close

[red echo]

.close

[lone heart]

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[lone heart]

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[mental stag]

i need help understanding on how to find the perimeter of triangles in a packet its a packet

[charred flint]

you're trying to find how much angle each circular arc covers, and for the square it's slightly under half of a full 360 turn

[mental stag]

It would be 150 right

These area in my math class is geometry and I took it online so I never learned it

[charred flint]

right, so 2pi*150/360*r times 4 for the perimeter

[mental stag]

Gotcha thanks

That makes more sense

How would I put that into a calculator

Is this right or wrong

If it's right I get how to do it

[marsh hamlet]

How did you derive it?

[lone heart]

@mental stag Has your question been resolved?

[mental stag]

I took the 150 from the missing angle subtracted it from 360 and did 2pi*150/360*r times 4 for the perimeter

[marsh hamlet]

Valid

[mental stag]

ok thx

[lone heart]

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[hearty tree]

Guys does this seem right to you? I am watching organic chemistry tutor tutorial and he used different quadratic method to solve for x but i used quadratic since that is what i am most comfortable with but the answers are bit different, mine are here:

organic chemistry tutors were x=2 or x = -1

[buoyant saddle]

is that not a 9?

9/x

[crimson dawn]

It's a 4 I think

[hearty tree]

its 4

[buoyant saddle]

thats a 9 bro

[crimson dawn]

If you observe as time goes on it takes the shape of a 4

As you follow it down the page it evolves

[dire gazelle]

i can't tell apart your 9's and 4's

[rustic ice]

9x(x+2) should expand to 9x^2+18x not 9x^2+8x

[hearty tree]

its really hard to draw with mouse

its 4

[rustic ice]

unless it's a 4

[marsh hamlet]

Another case of confusing handwriting

[buoyant saddle]

(-4)^2 \neq -16

[marsh hamlet]

What the hell?

[hearty tree]

yes and neg 16 * neg 8 = 112

[buoyant saddle]

🤔

[floral maple]

👍

lol

they are debating ab if its 9 or 4

[hearty tree]

nevermind

i suck at math

[floral maple]

not rly

[floral maple]

ahh

that 128

[buoyant saddle]

he subtracted 16 from it when he should have added 16

also

[hearty tree]

actually nevermind it should be 112

[buoyant saddle]

bet the house on it

how much would you put up bro

[buoyant saddle]

LMAO

bet the house bro

[hearty tree]

nope

[buoyant saddle]

yea

ok

(-4)^2

not -4^2

-4^2 is interpreted as -(4)^2

,av tonzafan

[ocean seal]

tonzafan's Avatar

Click here to view the image.

[hearty tree]

yeah thats me?

are you shaming?

[buoyant saddle]

nah

you look good bro

[hearty tree]

<@&268886789983436800>

[buoyant saddle]

🤔

[marsh hamlet]

||Coping?||

[hearty tree]

What?

man i have lost like 5kg why do you shame

[mental python]

youre doing good bro

good shit

keep going

[marsh hamlet]

You pinged moderators because someone looked at your avatar?

[buoyant saddle]

[hearty tree]

yes

its non related

[crimson dawn]

yea slight error in your quadratic formula, it should be (-4)^2=16 not -4^2=-16

[marsh hamlet]

You seem hypersensitive. Calm down, not everyone has malicious intentions.

[crimson dawn]

Oh nvm y'all pointed it out alr

[buoyant saddle]

the boogie bomb was covering it bro

[crimson dawn]

I think we should just stop talking about this

[hearty tree]

i agree

[crimson dawn]

Uh @hearty tree has your problem been resolved

[hearty tree]

yeah, i got it right now

.close

[lone heart]

Channel closed

Closed by @hearty tree

Use .reopen if this was a mistake.

[hearty tree]

.reopen

[lone heart]

✅

[hearty tree]

this cant be right because it leads to imaginary numbers.

[buoyant saddle]

it’s -8

so it’s + 128

[hearty tree]

damn

.close

[lone heart]

Channel closed

Closed by @hearty tree

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[buoyant saddle]

-(4)(4)(-8)

[lone heart]

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[broken pivot]

Can someone check my vector projection work?

[buoyant saddle]

2 * 4 ≠ 6

[broken pivot]

oop

is it good now

[lone heart]

@broken pivot Has your question been resolved?

[broken pivot]

<@&286206848099549185>

[tepid drum]

how did you get the orthogonal vector

[broken pivot]

i just assumed i needed to get from the terminating point of the projection vector to the terminating point of vector u

[tepid drum]

ah i see what you are getting at

did you try to find the dot product between this vector and v?

to check if your assumption is correct

[broken pivot]

i didnt

[broken pivot]

oh am i just completely misunderstanding this topic

[tepid drum]

well the dot product will not give 0

so lets do this together

[broken pivot]

is my projection correct though

$\proj_{\vec v} \vec u = \left<0, \frac{22}{\sqrt{13}}, \frac{33}{\sqrt{13}}\right>$

[tepid drum]

?

[ocean seal]

King Leo

[tepid drum]

let me check

given 2 vectors a and b, what is proj_a b in general

[broken pivot]

$$\frac{\vec a \cdot \vec b}{||\vec b||^2} \vec b$$

[ocean seal]

King Leo

[tepid drum]

alright so here u.v=11 and |v|=5

also v= <0,3,4>

so u.v/|v|^2=11/25 then u.v/|v|^2 v=11/25 <0,3,4> = <0,33/25,44/25> no ?

[tepid drum]

so how did you get this

[broken pivot]

oh shoot

i was looking at the wrong problem when i wrote that

[tepid drum]

oh i see np

[broken pivot]

thats what i got for the actual problem #help-0 message

[tepid drum]

this is almost correct

you divided by |v| instead of |v|^2

it seems that this was just a slip since you just wrote the general formula as it should be

so np

[broken pivot]

edited

[tepid drum]

alright so now the projection is correct

it remains to find an orthogonal vector

idk but i feel that i am misunderstanding part b

[broken pivot]

how

wait who exactly is misunderstanding

[tepid drum]

because it says the vector component of u orthognal to v

[tepid drum]

me

[broken pivot]

i just assumed i needed to get from the terminating point of the projection vector to the terminating point of vector u

[tepid drum]

not just find an orthogonal vector to v

[tepid drum]

yes but as you saw, this didnt give you an orthogonal vector to v

[broken pivot]

,w dot product <0, 33/25, 44/25>, <2, 1 - 33/25, 2 - 44/25>

[ocean seal]

Results provided by WolframAlpha

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Requested by .king.leo.

[broken pivot]

is that bc my previous projection was wrong

[tepid drum]

ah yes indeed

[broken pivot]

so does that mean my work is now correct

[tepid drum]

yes great job

[broken pivot]

thanks!

.close

[lone heart]

Channel closed

Closed by @broken pivot

Use .reopen if this was a mistake.

[tepid drum]

you did the whole work i didnt do anything

[broken pivot]

#help-0 message + my teacher taught the orthogonal part in a different way, so i wanted to see if [my assumption](#help-0 message) also works

[tepid drum]

what did he say ?

[broken pivot]

idr exactly, but it was something about spamming the pythagorean theorem

[tepid drum]

well if you want any orthogonal vector

[broken pivot]

i just use kn (where n is the orthogonal vector and k is some constant)

[tepid drum]

then you can find the vector u that satisfies u.v=0

you can also check if your method in general by doing (u-proj_v u).v

[broken pivot]

interesting, thanks!

[tepid drum]

np

have a great day

[broken pivot]

you too

[lone heart]

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[summer hemlock]

So what I currently know is:

[zealous rose]

Just that info?

[zealous rose]

You know the bottom triangle is a right one so solve for the height

[zealous rose]

Wouldn't that just be the answer then

Then do that and 6 and use the Pythagorean theorem

[summer hemlock]

Yeah I put that as my answer but it says it's wrong

[summer hemlock]

What would I use the Pythagoras theorem for

[zealous rose]

Wait let me draw it out rq

[still prism]

you can try and show that the triangles are similar

[zealous rose]

Is the answer sqrt(42)

[summer hemlock]

Alr

[zealous rose]

Do you have an answer key

[summer hemlock]

It's 4 root 3

[zealous rose]

Hmm

[rustic coral]

?

Triangle ACD is similar to triangle BCA

And hence ||x/8 = 6/x||

[summer hemlock]

I think I solved it I got square root 48

Thanks for the help

.close

[lone heart]

Channel closed

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[paper turtle]

Does anybody know how to do absolute value?

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[cunning fog]

Is my second derivative right

Idk how to get the final answer too plz help

[zealous rose]

For what? Derivatives?

[cunning fog]

I am never getting help danm

[zealous rose]

I barely know what it's asking

[cunning fog]

I need proof that the second derivative +4y =8y^3

[zealous rose]

That's not what it looks like tho

Wait second derivative of

Y=sec2x?

[cunning fog]

8sec(2x)tan(2x)^2+4sec(2x)^3

That what I found but I am not 100% sure

[zealous rose]

Ok I'm working on it

I need to rewrite it though

[cunning fog]

For got the 4

[zealous rose]

I'm getting somewhere

[cunning fog]

How did you get 8 sec^3

[zealous rose]

That's my issue

[cunning fog]

Maby I am wrong

[zealous rose]

Yeah idk how you got this

[cunning fog]

I did the f'.g+g'.f

[zealous rose]

Okay

So did I

Your work is a bit sparse

[cunning fog]

How is there sec.sec .sec?

[zealous rose]

Because tangent derives to sec^2 and then you have to derive that because of the chain rule

[cunning fog]

Man I have exam next week I might suffer

[zealous rose]

Do you know the chain rule?

[cunning fog]

Yea

[zealous rose]

Okay good

@bronze oar when you're done can you help him I'm so close to the answer

[cunning fog]

But my question is if tan derives to sec^2 why do I need ues chain rule?

[zealous rose]

That's the only way to get 8

You do chain rule because idk you just have to

[cunning fog]

Bruh

I have answered something like it before

[zealous rose]

You take the derivative of the inside

[cunning fog]

Nah man shouldn't the sec still be ^3

Cuz the n going be -1

So it's back to 1

[zealous rose]

You always do, but if it is just x then the derivative is 1 so you don't even write it

[zealous rose]

Yes but I don't know how to get that

[cunning fog]

You did something wrong

Sec^2 should become sec

And the ^2 gose outside and becomes like this (sec)^2-1

[zealous rose]

Where is the subtraction from

[cunning fog]

I just saw something similar to to it and answered it right the ^2 dosent change on sec after applying chain rules most of the time

[zealous rose]

Hmm interesting

Where does the 8 come from then

[cunning fog]

Chain rule

[zealous rose]

Can you write it out then

[cunning fog]

I think

[zealous rose]

I want to see it

[cunning fog]

Alr I am do it

[zealous rose]

Okayy

[zealous rose]

Oh I see

[cunning fog]

Don't pay attention to the 2sec

I shouldv put it sec2x

And put 2.(sec)

[zealous rose]

Hm okay

You shouldn't have secxtanx both times I don't think

[cunning fog]

But isn't sec'=sectan?

[zealous rose]

Yes but first you derive sec then you derive tan

You're not deriving sec both times like before and after the plus

[cunning fog]

There might be a trick I think

[zealous rose]

Okay 😭

[cunning fog]

But the chain rule says n(f)^n-1.f'

Can I just delete a part of the rule and change it to my liking?

[zealous rose]

I'm saying the uv'+u'v

[cunning fog]

Ho

Then what is this?

[zealous rose]

The first part

sec2x is u and tan2x is v

[cunning fog]

Yes but what I ment is you said chain rule I used it

Wait I think I found something interesting

What if i take common factor

And then make it into solving an equation

[zealous rose]

Yes try that

[cunning fog]

Now it looks way too weird

I might be cooked for the exam bruh

[zealous rose]

Yeah that can't be it 😭😭😭

This is just a long one I'm sure you'll do fine on the rest

Wait for Mello I pinged him he will be back in like an hour and a half and he can help

[cunning fog]

Well the one after this is y= cosx proof that it's equal to d^4y/dx^4=y

[zealous rose]

That's not bad

[cunning fog]

Well yea ever 4 for sin it becomes -

And everything 4 for cos it stays the same

[zealous rose]

It's cos -sin -cos sin cos

[cunning fog]

Yea

[zealous rose]

Yeah

Easy peasy

You'll ace the test

[cunning fog]

I try at less

[zealous rose]

Try at more

[cunning fog]

I am doing what the teacher haven't yet taught us for fun

[zealous rose]

Girl 😭😭 so no wonder you can't do it

Me on the other hand 🥶 I'm just tired

[cunning fog]

Him saying don't spoil class for others to someone gave me motivation

[cunning fog]

I am a boy

[zealous rose]

I know that

I'm not colorblind 🤓

I might not be able to do that derivative but I can identify colors

[cunning fog]

But tbh the we study like a single paper a day

[zealous rose]

So not much?

[cunning fog]

Yea

[zealous rose]

What year are you

[zealous rose]

Weird

[cunning fog]

Last year highschool

[zealous rose]

Okay same

We do like five a day 😭😭😭

[cunning fog]

Well it depend on which one lol

[zealous rose]

On what? The day?

[cunning fog]

But the derivative one has one paper

[zealous rose]

How is that even possible

[cunning fog]

for ever lesson

[zealous rose]

We do like many practice problems during notes

[cunning fog]

He tells us to practice at home

[zealous rose]

I mean we also do that tho

I'm sure kids aren't practicing

[cunning fog]

Well if you count the challenge question that isn't way too hard it's 2 paper

[zealous rose]

Oh well that makes more sense

Our notes are like spread out too so that's why it's so many pages

Are you in regular or advanced or what

[cunning fog]

Wait idk what that means?

[zealous rose]

Also where're you from

Wdym

Which part

[cunning fog]

Regular or advance?

Middle east

[zealous rose]

Okayy

[zealous rose]

Like there's usually the class and then a more advanced option of it

[cunning fog]

Well I took physic and math

[zealous rose]

Hmm

[cunning fog]

I think in my country there ain't advanced option or regular

[zealous rose]

Oh okay

[cunning fog]

Good part the majority of teacher that teach us are good

[zealous rose]

That's good

I like all my teachers this year so that's nice

[cunning fog]

I had English teacher at the start of the semester

Was changed like 3 weeks later

[zealous rose]

Hmm

[cunning fog]

Reading was a little bad and he reads PowerPoint he made

[zealous rose]

That's annoying

[cunning fog]

Well last semester the teacher was the head of the English department

Was very good teacher tbh

I see if I can help someone lol

[zealous rose]

Oh cool

[zealous rose]

Okayy

[lone heart]

If you are done with this channel, please mark your problem as solved by typing .close

@cunning fog Has your question been resolved?

[thin meadow]

@cunning fog ur question not resolved yet?

[cunning fog]

Yea

[thin meadow]

ok could you resend?

[cunning fog]

Now I have this

[thin meadow]

ok

[cunning fog]

Y=sec2x

[thin meadow]

so whats the question?

you want to prove LS = RS?

[cunning fog]

The second derivative +4y=8y^8

Proce it

Prove it

[thin meadow]

ok

do you know the derivative of secx?

[cunning fog]

This

Second derivative is sectan

Tan

[thin meadow]

oh then wheres the original question?

i cant find it

[cunning fog]

,rotate

[cunning fog]

The question is if y=sec2x then prove d^2/dx^2 +4y=8y^3

[thin meadow]

ah k

alr

so

can you substitue sec(2x) into the y spots and write the full equation out

@cunning fog

[cunning fog]

Wait I just understood

[thin meadow]

oh

lmao ok

[cunning fog]

I was stuck at a part where I had 4ytan^2+4y=4y^3

I divide by 4y

[thin meadow]

nah you substitue

then use trig identities

[cunning fog]

Then I have tan^2 +1=y^2

Which is same as tan^2 +1=sec^2

[thin meadow]

oh wait

yea

that works perfectly too

[cunning fog]

Man I am feeling so dumb

[thin meadow]

same bro

same

[cunning fog]

I will make sure to take them into account when my exam start

[thin meadow]

mhm

you should review a list of trig identities

itll help

gl on ur exam man

[cunning fog]

Can you send me a pic

[thin meadow]

?

[cunning fog]

Wai

[thin meadow]

pic of?

[cunning fog]

This works?

[thin meadow]

this has more but im not sure how many of these u learned

[thin meadow]

yea those are fundementals

[cunning fog]

The 3theta I haven't learned

[thin meadow]

also remember the common cos and sin angle values

[cunning fog]

Entire right side I haven't learned anything there

[thin meadow]

like sin0 = 0 or cos0 = 1

[thin meadow]

well it isnt anything new, just applications of what you know

[cunning fog]

True gonna memorize or understand them

[thin meadow]

itll be nice to look through, but if you havent learned it dont worry about them too much

[cunning fog]

So I can ues old one to find the new one

,close

[thin meadow]

yup

.

[cunning fog]

,closs

[thin meadow]

.close

[cunning fog]

.close

[lone heart]

Channel closed

Closed by @cunning fog

Use .reopen if this was a mistake.

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bronze oar]

Uh revise me on the question

[zealous rose]

^

[zealous rose]

^

[lone heart]

@bronze oar Has your question been resolved?

[lavish cave]

,w 2nd derivative of sec 2x

[ocean seal]

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Requested by sooouth.

[bronze oar]

y' = 2sec2x tan2x

Well yeah

That's what I got

And it does satisfy the equation

Idk what else

[lone heart]

Channel closed

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[lavish cave]

yes

so they just need to do the second derivative properly

and then just calculate 8y^3 - 4y

[lone heart]

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[lone heart]

@mild oyster Has your question been resolved?

[vivid anvil]

!status

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vivid anvil]

we need to know what you can take for granted

like can we assume det != 0 implies invertible?

[mild oyster]

If determinant is 0 then it’s invertible

[vivid anvil]

you mean nonzero right

[mild oyster]

Wdym

[vivid anvil]

determinant being nonzero is equivalent for invertibility

[median oar]

If something is invertible finding the inverse would be great

[mild oyster]

Ic

Oh I think I messed up the meaning of vertible and invertible

[median oar]

It’s invertible or singular

Vertible isn’t real

[mild oyster]

Wdym by that

[median oar]

That’s not a word

Invertible comes from the word inverse

[mild oyster]

Ok

[median oar]

We call matrices that aren’t invertible “singular”

[mild oyster]

Oh that’s singular

Okk

But how can I prove it there is no way for me to find the determinant

[median oar]

I don’t think determinant is the way to do this

[mild oyster]

Yes

So is there any other ways

[median oar]

What’s the definition of linear independence

[mild oyster]

ax1+bx2… =0 is linearly independent

[median oar]

What about that

Only a set of vectors can be linearly independent

An equation can’t be linearly independent

[mild oyster]

(x1, x2 …are vectors)

Idk what is the convention of naming a vector

Sorry about that

[median oar]

You should revise what the definition of linear independence is

It talks about the coefficients

It’s hard to do any math when you can’t recall the definitions of words

[mild oyster]

Ok