help-13

[rich blaze]

I need to find all the lines that are tangent to a certain parabola

some Ay=Bx+C

[iron wren]

there are infinite lol

[rich blaze]

family of functions

[rich blaze]

yeah but I can find a general equation with constants right?

[dull oxide]

yes

[rich blaze]

its for a differential equations problem

[iron wren]

you mean with equation x^2?

thats just calculus

[old dirge]

that is not a parabola

[rich blaze]

for this parabola : x^2=4y

[iron wren]

so y=x^2/4

[rich blaze]

I mean, thats the family of lines I have to find

[rich blaze]

yea

[iron wren]

you find the derivative of x^2/4

thats the answer

[rich blaze]

x/2

but thats just 1 line

[iron wren]

oh yeah

so the slope of a tangent line is x/2

and you know the height of one point is x^2/4

so use that to find a general equation of a line

[dire geode]

!original

[cedar kiln]

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.

[rich blaze]

hmm I'm still lost

[dull oxide]

Given $x^2=4y$, and some fixed $x_0\in\bR$, what is $\frac{\dd y}{\dd x}$ at $x=x_0$?

[wraith dagger]

SWR

[rich blaze]

Ok so I'm with 3)e) and it says:
3)Find the differential equation of the next families:
e)Tangent lines to the parabola x^2=4y.

[rich blaze]

x0/2

?

[dull oxide]

Yes. So, that is the slope of the line tangent to $x^2=4y$ at $x=x_0$. Next, how would you find the equation of the tangent. Do you know?

[wraith dagger]

SWR

[rich blaze]

If I have the slope and a point I can find it

slope x0/2
point (x0, y0)

[dull oxide]

Yes, now you just need $y_0$

[wraith dagger]

SWR

[rich blaze]

y0= (x0^2)/4

[dull oxide]

Perfect. So, you put it all together, and you have your family of tangent lines

[rich blaze]

Ok perfectly, I just checked and that works

[dull oxide]

"Ok perfect", not "perfectly"

[rich blaze]

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

oh it doesn't embed well its fine I can finish the exercise now

[rich blaze]

I meant to say perfectly done but ate that word completely

thanks again

.close

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[cedar kiln]

@strong marten Has your question been resolved?

[dull oxide]

Start with what you do know. Any side lengths you can calculate right now?

[wraith dagger]

brandon

brandon

[cedar kiln]

@strong marten Has your question been resolved?

[dull oxide]

@strong marten hint: what shape does PRM1M2 make?

Not quite

No. PM2 is variable

It varies based on P.

Yes, for any p

No. You'll need to consider what the parallelogram gives you

Also, take a look at the triangle PQR. See what that can give you

[dull oxide]

yes]

but not just that. What kind of triangle is PQR?

what can you say about angles PQR and PRQ?

[dull oxide]

No you cannot

It's still always variable

But, you can solve for one angle in terms of another

[dull oxide]

sorry bro, got to head out 🙁

<@&286206848099549185>

sure

[tropic merlin]

Whats the help

He needs

[dull oxide]

Just finding stuff out. Generally, finding what sides and angles you have is always a good idea

[dull oxide]

#help-13 message

[ebon stump]

Oh, you're doing the 3rd problem you mentioned now

Oh, another one?

[tropic merlin]

The only thing im bad at is geometry

But il give it a shot

Oh wait that might be solvable

Since i do have a similar syllabus wait

[dull oxide]

@strong marten are you supposed to use trig to solve this, or just geometry?

[tropic merlin]

Im pretty sure we have to use pure geometry

But wait

I can help with dis

Just need a notebook

Try making two traingles

And using the concept of medians

Il try it

But i thinks thats a good approach

[tropic merlin]

Im unable to

Since i dont live alone

And its 4 AM ;-;

Did you try my approach?

Medians of a traingle?

Well if the ratio of the sides

Are the same as that of the median

That proves they are similar

Am i wrong somewhere?

Pls tell me if i am

I think this can be concluded with thales theorem

Ive certainly used thales theorem to prove this in my exam!

BUT THIS WORKS

Yeah

Its simple too

Yeah

Il help in the morning if its still unsolved

Rn i need to go and workout in a bit

It should be the right track

Im pretty sure

Yeah!

Your welcome if your able to crack it!

Cya

[cedar kiln]

@strong marten Has your question been resolved?

[cedar kiln]

@strong marten Has your question been resolved?

[cedar kiln]

@strong marten Has your question been resolved?

[wise void]

can anyone help me with this maths

[cedar kiln]

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[long mesa]

How do I know if the transformation from 2^x to 4^x is f(2x) or 2f(x)

[proven summit]

f(2x) means a stretch parallel to the x axis by scale factor of 1/2. 2f(x) means a stretch parallel to the y axis by scale factor of 2

[pastel vault]

can you rewrite 4 as a power of 2? then you can use laws of indices on 4^x

that will tell you straight away

[proven summit]

yeah rewrite it as as a power of 2

[long mesa]

oo okay

IK it says describe but I’m trying to learn to write it with notation

Is this correct? Since if you put it inside the brackets it’s 1/2 and I want to multiply it by 2

[nova snow]

mix rigged boy

[cedar kiln]

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[dusky marsh]

yo

[sullen cipher]

What have u tried

[dusky marsh]

ive done like (k-1)x^2 = 9-x^2

then re arrange for x which gave me

3/squareroot 3

sorruy idk how to put math symbols in discord

[carmine bronze]

$\frac{3}{\sqrt{3}}$

[wraith dagger]

Kookiemon

[wicked mantle]

What does this achieve?

[proven summit]

I don't think you need to do that since you can just look at the graph and identify the intersection point

[wicked mantle]

Could you explain the reasoning behind this?

[dusky marsh]

i have no clue tbh im lowkey kinda lost

just tryna find limits of it

[wicked mantle]

yeah, but look. There are two variables in the equation. How do you manage to find k with it?

[dusky marsh]

wait could u restate the question im lowkey feeling a lil lost

[sullen cipher]

Intersecting points r the limits,u mean?

[dusky marsh]

yeah

[sullen cipher]

Makes sense ig

When u did rhe sqrt over

U get two values

±

So two intersecting points

[dusky marsh]

aaah

wait i feel like im doing this the hard way

any suggestions on what u guys would do?

[sullen cipher]

It's a shortcut typa way

But going from base u can find the area of the two curves separtely and then deduct it to get the area of the shaded region

[dusky marsh]

how would you start that

[sullen cipher]

Well at first can u identify which equation belongs to which curve

[dusky marsh]

y=9-x^2 is the happy face and y=(k-1)x^2 is sad face

wait

[sullen cipher]

That's nice way to term it
But we went opposite here

[dusky marsh]

ah alr alr mb

[sullen cipher]

Happy face is the y=(k-1)x²

And sad face is the other one

[sullen cipher]

Np but did u figure out how?

[dusky marsh]

my graphic calculator

[sullen cipher]

But if u didnt hv access to that how would u identify

[dusky marsh]

uuh -x^2

cause doesnt it like the y = ax^2 + bx + c isnt when a is below 0 it goes downwards

[sullen cipher]

Well that works when
y=ax²(upward) or y=-ax²(downard)

[dusky marsh]

yes

[sullen cipher]

Yea that's a way to identify

Alr now do we agree that if we subtract the area of happy face from area sad face ,we get the shaded region?

[dusky marsh]

coudl u explain why that is

[sullen cipher]

Sorry i stated opposite

[sullen cipher]

Alr wait but lemme clarify subtracting the sad part from happy part gives u the shaded area

This is the happy area

[dusky marsh]

yes

[sullen cipher]

Sad area

[sullen cipher]

Sorry restating
"The area u get after deducting the area u got after deducting the sad area from happy area from the area of the sad part..."

Damn i miscalculated 🥲

[dusky marsh]

alr alr

[dusky marsh]

uh oh

[sullen cipher]

Did u understand what i meant

[dusky marsh]

kind of

[sullen cipher]

Like after subtracting those two u get the area of the part which isnt shaded but undder sad part

[dusky marsh]

ooh yes ok ok but arent we trying to find shaded?

[sullen cipher]

Yes when u get the unshaded part u can deduct it from the area of the sad part to get shaded part

[dusky marsh]

aaaaah ok ok

[sullen cipher]

Do u know how to find area under 5he curve of each

But ur method is easier,and short

[dusky marsh]

no soz

[sullen cipher]

Do u want to proceed with that one?(easier to explain)

[dusky marsh]

yes please

[sullen cipher]

We will go to that

[sullen cipher]

When u found the value of x

What did u get

[dusky marsh]

x= +- 3/ sqaureroot k

[sullen cipher]

Yes so u got the limits right

[dusky marsh]

aah ok ok

so the limits are -3/sqrt k and 3/sqrt k?

[sullen cipher]

Yes

Now since we r going to use integral here ,lemme show u which part we r considering under the limit

[dusky marsh]

alr alr

[sullen cipher]

This area

Is within the determined limits

And it is also under the sad curve

Do we agree

[dusky marsh]

yes

[sullen cipher]

And this is the part under the happy curve

Do we follow

[dusky marsh]

yes

[sullen cipher]

So if we deduct the blue part from red part we get the shaded region right?

[dusky marsh]

aaaah ok ok

[sullen cipher]

U got it? Or i'll explain again?

[dusky marsh]

wait which variables would i use to deduct them

[sullen cipher]

Ur limits r on the basis of x right,

?*

[dusky marsh]

yes

[sullen cipher]

U found the value of x ,so the limiitts were basing on x
So when u deduct to find integral,u need to use x

[dusky marsh]

so would it be sqrt3/k + sqrt3/k?

[sullen cipher]

Nope

Like u r subtracting the area

Not the limits

Since we r working kn the basis of limits on x axis

Wee need to deduct the x portions

[dusky marsh]

😭

[sullen cipher]

Like 9-x²-(k-1)x²

Integral of that is the area of shaded region

Uk how to do integral,

?*

[dusky marsh]

i do but not confident

[dusky marsh]

do i simplyfy then integral?

[sullen cipher]

U should try

[sullen cipher]

Yes

[dusky marsh]

okay holup

[sullen cipher]

I'll be back after a while

[dusky marsh]

alr alr

hopefully ive done something by then

\int _{-\frac{3}{\sqrt{k}}}^{\frac{3}{\sqrt{k}}}:9x-\frac{x^3}{3}-\frac{\left(k-1\right)x^3}{3}

what do i do next

[sullen cipher]

Once u do the integral ,u should remove the integral sign

And put a 3rd bracket to show ur limits

Also i see tha u didnt simplify, but that works as well

[dusky marsh]

how would i simplyfy

[sullen cipher]

Like taking common from x³ terms and gettibg a single x³ term

But urs is right too

Js not the presentation

[dusky marsh]

so would it be 9x- k/3x^3

??

[sullen cipher]

Yes

Well now u r supposssed to implement the limits

[dusky marsh]

alr alr holup

i feel like im doing this so wrong

[sullen cipher]

Lemme check

Yea the 2nd term

Isnt right

Recheck that one

Since 2nd and 3rd is similar ,thae 3rd one is wrong as well

X³/3=((3/sqrt(k))³)/3

[dusky marsh]

would it be 27/3k^3/2?

9/k^3/2

[sullen cipher]

Yes

[dusky marsh]

ok ok

[sullen cipher]

U can write k^3/2=ksqrt(k)

To make life easier

[sullen cipher]

Then xan u show me the correct form

[sullen cipher]

U didnt fix the 3rd terms

Those had the same issue

And u cant put a (-) here

This makes it look like u r subtracting 3/k

When u r multiplying

[dusky marsh]

wait which one

[sullen cipher]

This and the other side too

[dusky marsh]

ah ok wait

[sullen cipher]

Now simplify

[dusky marsh]

alr holup

ts is so confusing

so much stuff going on all at once

[sullen cipher]

Lemme try

Just noticed but this is supposed to be plus

[dusky marsh]

oh ok mb

[sullen cipher]

Hopefully this is right

Xan u check as well

[dusky marsh]

ill try its like 1 am so i might be like tipsy

[sullen cipher]

Thas late

[dusky marsh]

im too young to be experiencing this bro 😭

imma take the night off ill just look over this tomorrow morning

tysm for helping me <33

[sullen cipher]

Ish fine

U will nail it

[sullen cipher]

Alr

[sullen cipher]

🥲np gn

[dusky marsh]

gn

[cedar kiln]

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[cedar kiln]

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[heady ledge]

I have pasted a picture of how far I could go to, I am just confused what to do ahead.
Also, P(f) is the probability that the plan is found, and P(l) is that it has a locator. I am just confused what to do next.
<@&286206848099549185>

[cedar kiln]

@heady ledge Has your question been resolved?

[iron wren]

So 42% of the overall aircraft had a locator and were found and 3% had a locator and were not found

[heady ledge]

mm no?

I did attach the picture of the question

[iron wren]

Yes but 60% of 70% is 42%

And 10% of 30% is 3%

[heady ledge]

oh u mean that, then yeah

no, it says not found and did not had the locator

if it weren't found and did not had the locator it would be a different story

[iron wren]

60% of the 70% of planes that were found had a locator

And 90% of the 30% that were not found did not have a locator

[heady ledge]

mm yeah ig

[iron wren]

So you are left with 42 and 3 for the percentages

[heady ledge]

oh wait so that means had a locator and were not found is 0.03

[iron wren]

So chance one is not found is 3 in (42+3)

[heady ledge]

mm yeah I did found that out, p(l) but using de morgan's

P(l:f`) = 0.03

but we need to find P(f':l) and P(f:l')

[heady ledge]

thanks you just got me the idea

[cedar kiln]

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[torn marsh]

why does it give me a diff answer when I dont bracket -1

[torn marsh]

and then when I do

[tropic oxide]

$-1^2 = -(1^2) \neq (-1)^2$.

[wraith dagger]

Ann

[torn marsh]

oh so its (-) * 1 * 1

[tropic oxide]

if you're using desmos anyway i strongly suggest making full use of it and defining a function like f(x) = x^2/2 - x^3/3 - x^2 or whatever it is in your case and then doing f(2) - f(-1)

[tropic oxide]

sure

except that this "lone minus" should rather be a -1

[fossil dawn]

-1(1^2)

oh noted already

[torn marsh]

alr ty

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[tropic oxide]

it's impossible to have 10 consecutive squarefree integers tho.

a multiple of 4 already isn't squarefree, and your 10-integer range is guaranteed to catch either 2 or 3 of those...

oh non squarefree.

[wraith dagger]

brandon

[tropic oxide]

well then you've got yourself a chinese remainder thm problem right

with a mighty ten equations, but still

yeah that's. big.

i think you may not need 10 primes though.

you can make do with less, no?

(ngl though i would prob not be able to find such N without a lot of ink)

i would say that just appealing to CRT is fine.

of course, if you made an explicit thing, it would carry a ton more weight! but appealing to CRT and adding multiples of (\prod_i p_i)^2 if desired should do the trick if you're just in the market for existence

[flat mica]

yeah that doesn't say you need to find the numbers, just determine the bounds

i had an exam question in college, prove there exist 100 consecutive squareful numbers

that's the upper bound for sure

can there be 0? 1?

it asks for the lower bound too

rocky numbers?

never heard that. ive heard "squareful"

but not as commonly as "not squarefree"

got it

[cedar kiln]

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@strong marten Has your question been resolved?

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[bold flint]

I need help with math

[proven summit]

!da2a

[cedar kiln]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[proven summit]

And

!xy

[cedar kiln]

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.

[bold flint]

Ok thanks gng

Sorry

[proven summit]

no worries

just post the original problem you're stuck in

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[vale kettle]

How do i understant ts

[dusk goblet]

pretty straightforward

you’re "completing the root" so to speak

you want the radical to go away in the denominator so you multiply by the number with the same base but who’s exponent adds to it to make the root go away

it’s probably clearer if you get rid of this notation

[vale kettle]

Got it

Thks

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[full apex]

hi i have a question similar to this (https://math.stackexchange.com/questions/2498359/probability-of-meeting) but im not sure how they got the 2 lines,
its that there is 2 people one of them has 30 mins waiting time the other has 20 mins. i got it so the graph is similar to how they have it but im not sure how they got the 2 lines

[dire geode]

!occupied

[cedar kiln]

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

[full apex]

bro 😛

are we so fr

hi i have a question similar to this (https://math.stackexchange.com/questions/2498359/probability-of-meeting) but im not sure how they got the 2 lines,
its that there is 2 people one of them has 30 mins waiting time the other has 20 mins. i got it so the graph is similar to how they have it but im not sure how they got the 2 lines

<@&286206848099549185> sorry to ping but any help is appreciated

[raw mason]

If you mean the lines over and under the y=x line, basically:
The line under it represents how much time is B willing to give A before he goes home: for example, if you check the width on the x axis, where B arrives precisely at 7, you can see it leaves A no more than 10 minutes.
Then the line over y=x represents how much time is A willing to give B. If we, again, check the y axis, we see B has 10 minutes since the arrival of A

They are also parallel to y=x

[full apex]

ohhh i think i get it

tysm

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[clever crow]

does anyone know if/where I can get the solutions for even numbered problems for this book? Linear Algebra and its Applications, 4th Edition Gilbert Strang

[clever crow]

im trying to self study and it gets pretty annoyinh having to just ball without knowing if my answers are correct or not

[split ice]

searched up "strang linear algebra 4th edition exercise solutions," found this

oh sorry paywall >w<

[fossil dawn]

well other methods involve some... "soul-searching", if i may

[split ice]

there are solutions for the 5th edition of the book - are you set on the 4th?

[clever crow]

I have a translation of the 4th, it wouldnt hurt to see the 5th I guess, lets hope theres overlap in the excercises

[surreal cave]

https://dn720003.ca.archive.org/0/items/linear-algebra-by-strang-4-th-edition/linear algebra by strang 4 th edition.pdf

does the textbook have the solutions?

[clever crow]

yeah textbook has solutions for odd numbered excerscises

[surreal cave]

oh what?

strange...

[clever crow]

seeing Im a completionist though and will probably be solving all of them was looking into getting solutions for even ones too

[fossil dawn]

that's very common

[surreal cave]

also hi Hana!

[split ice]

i did find an english 5th edition + solution set, but it seems a little sketchy to just send here, sorry qWq

[surreal cave]

are the solutions at the end of the book?

[clever crow]

dms are open 🤭 🤭

[surreal cave]

this one is 40 pages longer...

https://students.aiu.edu/submissions/profiles/resources/onlineBook/Y5B7M4_Introduction_to_Linear_Algebra-_Fourth_Edition.pdf

but idk where the solutions are in the book so you might wanna check?

[fossil dawn]

i think OP wants the solutions specifically to his book

[surreal cave]

that is the same book, also nvm

I found them, all evens again

[clever crow]

I keep finding the instructors manual for a diffrent book with the same title, but by diffrent author 😭 , ill end my search thx

.close

[cedar kiln]

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[native elbow]

can someone check my work because if its not right my mum is gonna be the death of me

[fossil dawn]

!filetype

[cedar kiln]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[fossil dawn]

preferably send an image instead please

[native elbow]

alr

[fossil dawn]

for Q1, the theta looks like a 0 on one side and a Q on another, might wanna be careful to not get misinterpreted
but otherwise correct

that being said, i don't see a single line of reasoning here

Q2 is correct but very messy, esp the -90 on both sides (you wrote only -9 on the right)

[native elbow]

ok ill make sure to double check my work

[fossil dawn]

everything looks good but 0 reasoning

[cedar kiln]

@native elbow Has your question been resolved?

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[snow quartz]

i have to express x in terms of a

[snow quartz]

it seems rlly ez but ive been stuck on it for half an hour

[oblique flare]

try separating the sqrt to make it easy to simplify after squaring

[snow quartz]

i tried but im just getting a quartic equation

[dreamy void]

Consider $x^2-a=\sqrt{a-x}$

[snow quartz]

i did that

im getting a degree 4 equation which is even harder to solve

it doesnt factor or anything

[dreamy void]

What a cheesy task

[snow quartz]

it looked so ez but now i so confused idk what else to do

[dreamy void]

In my mind I saw a² and thought you could factor then x²-a²

[snow quartz]

<@&286206848099549185> sorry to ping but i rlly need the answer for this, im going crazy!!!

[dreamy void]

Maybe a trigonometric substitution??

[snow quartz]

i tried tan that got me nowhere

[dreamy void]

What else have you tried that you haven't told us before

[snow quartz]

i tried trig, i tried replacing some variables like a-x or x^2 by t or smth but that just complicates it more, i tried factoring the quartic and that got me nowhere

[dreamy void]

,, (a\cos^2(\theta))^2-a=\sqrt{a-a\cos^2(\theta)}=\sqrt{a}\sin(\theta)

[jovial anchor]

i feel like you're just gonna have to be brave and tackle that quartic

[snow quartz]

it doesnt factor, ive tried all the math ik, even wolfram alpha, it doesnt factor

[jovial anchor]

ok if we naively square both sides, what do we get

$x^4 - 2 a x^2 + a^2 = a - x$

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[snow quartz]

yeah

[jovial anchor]

so that's

$x^4 - 2 a x^2 + x + (a^2 - a) = 0$

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[dreamy void]

A quadratic in a

[jovial anchor]

yes, though i don't think that'll help for us

it's not impossible to factor quartics

it's just, uh

supremely annoying

[dreamy void]

pseudo annoying

[jovial anchor]

what you can start with is assuming a factorisation of the following form:

$(x^2 + \alpha x + \beta)(x^2 - \alpha x + \gamma)$

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[snow quartz]

this question is based on quadratic equations i rlly dont think we have to go into quartics

[jovial anchor]

hm, i see...

the issue is that you've got both $x^2$ and $x^{\frac 12}$ kicking around

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[jovial anchor]

i don't see an obvious way to reduce that to a quadratic

[last apex]

Is this the original question? or do you need this for another question?

[snow quartz]

our proff said its an obviious solution once u see the trick, but idk how

[snow quartz]

this is the original

[jovial anchor]

ok i'm gonna try it on my own in one of the latex channels

maybe that'll help me see the trick

will report back

[snow quartz]

ok

[dreamy void]

rip latex channels

[high coyote]

Sorry I don't want to bother you but are you asked exactly that, isolating x in terms of a?

[snow quartz]

ye, we were given 4 equations to find x in terms of the other variable

[idle crystal]

hm

[high coyote]

Ah ok 👌🏻

[pastel vault]

oh, you can just let $u = \sqrt{a - x}$ to get $(u^2 - a)^2 - u = a$

[wraith dagger]

south

[high coyote]

We'll get it

[pastel vault]

I don't know how exactly WA gets those nice solutions from there

[snow quartz]

i alr tried that, it just gets more complex and another quartic

[high coyote]

Is a-1 a root of the quartic?

[idle crystal]

solve this as a quadratic in a

[high coyote]

I asked bc I didn't try anything

[idle crystal]

and then invert that

[snow quartz]

oh wait i didnt try that

[high coyote]

Maybe try to find by inspection a root of the quartic in terms of a

[idle crystal]

$a = \frac{2x^2 +1 \pm |2x-1|}{2}$

[wraith dagger]

Copter

[idle crystal]

split these into cases, you get your solution

[jovial anchor]

ok i am reporting back

the quartic does factor

[snow quartz]

howw

[jovial anchor]

$(x^2 +x -a)(x^2 - x -a + 1) = 0$

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[snow quartz]

i tried it on wolfram alpha it gave me nothing

[idle crystal]

$x = \frac{-1 \pm \sqrt{1+4a}}{2}$

[wraith dagger]

Copter

[jovial anchor]

ig that's a skill issue on wolfram alpha's part

[jovial anchor]

yeah that's from the first factor

[idle crystal]

oh wait you guys didnt see this?-

[snow quartz]

nice, thxs a lot guys, u saved me atleast an hour

[idle crystal]

this also works if you dont wanna factor a quartic

[pastel vault]

np!

[jovial anchor]

all i did was expand this out and match coefficients btw

it's always possible to factorise a quartic into two quadratics

[dreamy void]

After all the cheese didn't melt

[jovial anchor]

if you expand out this way

(essentially because the quartic formula exists)

oh, one more thing

[opal hinge]

the fact that you did it in TeX is kinda impressive lol

[jovial anchor]

you should check that your solutions are actually valid!!!

[pastel vault]

and impossible to anticipate how hellish the process might be

[jovial anchor]

the original question had a $\sqrt{a - x}$

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[snow quartz]

ill get the answers tmrw, hopefully its right

[jovial anchor]

so you need to check that $a - x \geq 0$, for one

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[snow quartz]

oh ye

[jovial anchor]

essentially you have to be careful that the solutions you get from the quartic actually correspond to solutions of the original equation

after all, you'd get the same quartic if you had $x^2 - a = - \sqrt{a - x}$

[wraith dagger]

Pseudo (Cat theory #1 Fan)

[snow quartz]

okie, thxsss

[jovial anchor]

[snow quartz]

.close

[cedar kiln]

Channel closed

Closed by @snow quartz

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[jovial anchor]

i'm very used to it~

[jovial anchor]

one thing you've gotta remember is the psyhological aspect of the math Qs

unlike the ones you find in research, these ones are designed to be solved by people at your education level

there are a few times where i've successfully identified and corrected typos in math questions because of this

[snow quartz]

genius fr 🙏

[jovial anchor]

nah i'm not a genius

[snow quartz]

pls stop the cap 🙏

[jovial anchor]

i'm not capping!!!!

[cedar kiln]

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[crude brook]

I need help with these 2 exercises please

[crude brook]

I need to find out of a way so that there are no imaginary numbers in the final solution

[abstract breach]

The way it's written, x and y are real valued function no?

[crude brook]

well they want us to solve it with eigenvalues and eigenvectors

[crude brook]

those are the 2 variables

[cedar kiln]

Channel closed

The owner is missing!

[crude brook]

BRO

WHAT

[abstract breach]

Outstanding move

[crude brook]

what?

[abstract breach]

Try to reopen

[crude brook]

.reopen

I DIDNT DO ANYTHING

smh

.reopen

[abstract breach]

.help

[cedar kiln]

Commands:

• clopen: .close, .reopen
• consensus: .poll
• factoids: .tag
• help: .help
• version: .version

Type .help <command name> for more info on a command.

[crude brook]

.reopen

.help

[cedar kiln]

Commands:

• clopen: .close, .reopen
• consensus: .poll
• factoids: .tag
• help: .help
• version: .version

Type .help <command name> for more info on a command.

[crude brook]

.reopen DARNIT

[abstract breach]

Maybe it didn't close?

[crude brook]

????????????????????????/

aight

so

[dreamy void]

you got kidnapped

[crude brook]

virtually lol

[abstract breach]

So did you find the eigen values/ eigen vectors then?

[crude brook]

eigenvalues 1+i, 1-i

eigenvectors (1,i) (1,-i)

[cedar kiln]

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[crude brook]

CLAIMED

[crude brook]

loll

the hard part is how to eliminate the i s

like

[dreamy void]

I think you should repost

[crude brook]

i got c1e^t(cosx-isinx)(1,i)+c2e^t(cosx+isinx)(1,-i)

[crude brook]

this has i s in it, we dont want that

we dont want any imaginary numbers in fact

[abstract breach]

So you've got x+iy= c1e^(1+i)t and x-iy=c2e^(1-i)t

I'm not very experienced in this kind of de but can't you say x=Re(blabla) and y=Re(blabla), then you have some condition on c1 and c2?

[crude brook]

i can turn e^(ix)=cosx+isinx and e^(-ix)=cosx-isinx

yup but there is the i again

[abstract breach]

Maybe from there you enforce Im x=0 thus giving something like c1=c2

[crude brook]

you know what i'll try to do it without having matrices

[abstract breach]

Maybe not the correct equality but you see what I mean

[crude brook]

idk what that means exactly

[abstract breach]

Actually these are complex conjugates so it's most likely c1=c2

[crude brook]

ah no for the time being we want to leave c1,c2 intact

[abstract breach]

I don't see why (I'm not familiar with coupled DE yet)

[crude brook]

we need partial solutions afterwards

[abstract breach]

As in "degrees of freedom" ?

Cuz there will be a constant (say c1) associated with x+iy and another one associated (say d1) with x-iy, thus giving 2 constants that we can choose just like in the solution

Wait no

[crude brook]

<@&286206848099549185> help please

[abstract breach]

Wdym by partial solution? I don't know that term

[crude brook]

we have to do that part next

this part will be used for the partial solution

now we do the exercise by ignoring it

[abstract breach]

Like for the particular solution?

[crude brook]

sorry i am not familiar with english terms

[abstract breach]

Maybe you could solve it by treating x and y as complex valued functions, then enforce Im=0 afterwards

[abstract breach]

Me neither dw

[crude brook]

so i am trying to gather all sines and all cosines for x and y separately and trying to get 2 new constants, would this be correct?

[abstract breach]

Wdym

[crude brook]

i have this

its a mess so i am trying to gather all sines and all cosines together, but for x and y separately, not as matrices like above

now i got $x=c1e^{t}(cost-isint)+c2e^{t}(cost+isint)$ and $y=ic1e^t(cost-isint)+(-i)c2e^{t}(cost+isint)$

[wraith dagger]

MichaelRafto

[crude brook]

$x=(c1+c2)e^{t}cost+i(c2-c1)e^{t}sint$

[wraith dagger]

MichaelRafto

[crude brook]

(ig)

[raw mason]

I'm pretty sure you can consider the real and imaginary parts as 2 separate solutions (don't know if someone has already said this)

[abstract breach]

That makes sense

[crude brook]

and $y=i(c1-c2)e^{t}cost+(c1+c2)e^{t}sint$

[wraith dagger]

MichaelRafto

[crude brook]

i think i am finally getting somewhere i got it guys for now

$d1=c1+c2, d2=i(c2-c1)$

[wraith dagger]

MichaelRafto

[raw mason]

either this or enforce that c1 and c2 are conjugates so that the sum is real and the difference is imaginary

[raw mason]

so that you can do this sub precisely and you get both d1 and d2 reals

[crude brook]

i got $x=d1e^{t}cost+d2e^{t}sint$ and $y=-d2e^{t}cost+d1e^{t}sint$

[wraith dagger]

MichaelRafto

[raw mason]

seems good to me

[crude brook]

ok so this is for the homogenous

now it says that for the partial (or particular-idk how it is called in english) we need $C'(t)=S^{-1}(t)*B(t)$

[wraith dagger]

MichaelRafto

[crude brook]

S matrix being the solution we just found

wow this is a mess

ahhhh

$C'(t)=((-e^{2t}cos^{2}t-e^{2t}sin^{2}t)/(-e^{2t}cos^{2}t+e^{2t}sin^{2}t),(e^{2t}costsint+e^{2t}costsint)/(-e^{2t}cos^{2}t+e^{2t}sin^{2}t))$

[wraith dagger]

MichaelRafto

[crude brook]

Idek if my matrix is correct because of the replacements

Like

Does it matter if the change the constants?

[raw mason]

the names of the constants shouldn't matter as long as they are consistent

[crude brook]

If the change the constants

[crude brook]

So as long as it is correct it doesn't matter

[crude brook]

So this is correct?

It's a 1x2 matrix

Idk how to do matrices with latex sorry

[raw mason]

dont think so, the second component simplifies to 0 which seems wrong but i can't really be sure

[crude brook]

Mb it was + not -

[raw mason]

$C'(t)=((-e^{2t}cos^{2}t-e^{2t}sin^{2}t)/(-e^{2t}cos^{2}t+e^{2t}sin^{2}t),(e^{2t}costsint+e^{2t}costsint)/(-e^{2t}cos^{2}t+e^{2t}sin^{2}t))$?

[wraith dagger]

DJ Pope

[crude brook]

$C'(t)=((-e^{2t}cos^{2}t-e^{2t}sin^{2}t)/(-e^{2t}cos^{2}t+e^{2t}sin^{2}t),(e^{2t}costsint+e^{2t}costsint)/(-e^{2t}cos^{2}t+e^{2t}sin^{2}t))$

[wraith dagger]

MichaelRafto

[crude brook]

Ye

If these are correct then I need to integrate them (oh my goodness)

[crude brook]

This correct?

the good thing is now i got the idea of the exercise

[raw mason]

i think so but again im not sure

maybe you can simplify some e^2t actually

[crude brook]

true

and maybe i can turn the first fraction numerator into -1

[raw mason]

yeah

[crude brook]

ahh man i dont think it's correct

[crude brook]

they want the end result to be this (for the partial solutions)

no wait

this:

[raw mason]

i think it may just be a miscalculation at play here

i gotta go, sorry

[crude brook]

ok

np

chatgpt told me what you guys were trying to tell me and it is truly easier and more simple to understand

ahhhh crap

wait

wait no it's correct

ok cool

[dire geode]

!done

[cedar kiln]

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@crude brook Has your question been resolved?

[crude brook]

should i make another channel for iii)?

[cedar kiln]

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[crude brook]

aight

[cedar kiln]

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[crude brook]

can you guys please help me do iii) now?

[crude brook]

how do i integrate $\frac{2e^2t}{t}$?

[wraith dagger]

MichaelRafto

[dreamy void]

Did you mean $\f{2e^{2t}}{t}$?

[crude brook]

yeah how do i integrate that?

[dreamy void]

It has no elementary solution

[crude brook]

i want the particular solutions for this problem

.close

[cedar kiln]

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[dreamy void]

[ \bm{x'} = \underbrace{\begin{pmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{pmatrix}}{\eqqcolon , Y} \bm{x} + \underbrace{\begin{pmatrix} 3 \ -1 \ -1 \end{pmatrix}e^t}{\eqqcolon , \bm{b}} ]
One particular solution may be found by
[ \bm{x}_p = Y \int^x Y^{-1} \cdot \bm{b}(t) , dt ]

[crude brook]

i think i got this, thank you

[cedar kiln]

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[vapid kraken]

1+1=37 < MATH

[wicked mantle]

Don’t troll please

.close

[cedar kiln]

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[vapid kraken]

sorry

Im new

[wicked mantle]

You might get banned by a mod for trolling in help channels

just don’t do it next time.

[vapid kraken]

Understud

[cedar kiln]

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[soft shuttle]

Does anyone know good resources for Calculus 1 & 2? I'm currently taking Calculus 2 and it looks like ancient language. When I took Cal 1 I got a very low C barely passing the class and to be honest barely learning anything. So in order to not fail Cal 2 would anyone know of resources that goes over the whole course?

[austere hull]

paul's notes should be a good resource

[fossil dawn]

paul's online notes and math libretext's calculus shelf

[soft shuttle]

Thank you

.close

[cedar kiln]

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[cedar kiln]

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[round geode]

I'm not certain how to get an exact y for this

velocity is constant

Theres just no info fr

draw a straight line and see where it hits fr

[upper ruin]

Is it part of an exercise or is it an exercise as is?

Because I read "now consider".. does it mean it was previously different?

[fossil dawn]

there probably was a previous section to the question. OP, do you have it?

[cedar kiln]

@round geode Has your question been resolved?

[cedar kiln]

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[torn marsh]

need this checked thank you

[livid hound]

looks fine

[tropic oxide]

52?

[sullen cipher]

Seems correct

[sullen cipher]

64-12

[rapid mauve]

doesn't the binomial there require a chain rule application?

cmiiw

[livid hound]

technically yes

but its trivial here and i'm giving h the botd that they did

[rapid mauve]

fair enough, sorry for intruding

[tropic oxide]

a division by 1?

[rapid mauve]

i apologize, should have checked it first

[torn marsh]

so no mistakes found in op?

[livid hound]

the 4 in the denom is missing a stroke

[torn marsh]

opps

my handwriting just gets badder as time goes by

[tropic oxide]

badder

[cedar kiln]

@torn marsh Has your question been resolved?

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[crude brook]

Hey guys i need help with laplace. How do i laplace these?

[crude brook]

Like

The answers given are different and i dont fully understand why

[arctic comet]

what is fibonacci numbers

[crude brook]

What?

Nel can you help me please?

Or Ann?

[cedar kiln]

@crude brook Has your question been resolved?

[floral arrow]

I can't decipher that, sorry

You should probably post the original question and answer, though

[crude brook]

im doing ii) now

<@&286206848099549185>

Ok i did ii) now i am going to iii)

<@&286206848099549185>

nvm i did all of them

.close

[cedar kiln]

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[cedar kiln]

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[hasty path]

Part b confusing

Why can only certain values of x be used

[radiant yacht]

did you do part a yet

[hasty path]

Yh

[radiant yacht]

so you got an infinite expansion

that is equal to 1/sqrt(4-x)

now here's a questoin: if you plug in x=2 into that infinite expansion, does it become infinity

[hasty path]

Idk, Wym infinite expansion, my teacher hasn't taught me that

[radiant yacht]

ah

basically, what part A is trying to get at is

when you have a function like 1/sqrt(4-x)

this is equal to an infinite polynomial

a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + ...

for some coefficients a0, a1, a2, a3, a4, ...

does that make sense or is that something you haven't learned yet

[hasty path]

Ill try watching a video ill close this for now and come back. Thanks for letting me know my gap in knowledge

.close

[cedar kiln]

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[hasty path]

This it?

Then check if x falls in that range

.reopen

[cedar kiln]

✅

[radiant yacht]

yes!!

what video did you watch that helped?

/ how did you figure it out?

[hasty path]

https://youtu.be/9jHo7X-_1As?si=nFEvzBUnnipmKKgZ

This video

.close

[cedar kiln]

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[radiant yacht]

nice - curious how you figured out to find that video

(seems like a perfect video to explain what iw as trying to say)

[hasty path]

https://sites.google.com/view/tlmaths/home/a-level-maths?authuser=0

[hasty path]

Went on the binomial expansion section on year 2 within sequences

[cedar kiln]

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[fallow drift]

so my prof is trying to prove the volume formula of a cone using shells and washers method, why is it that for shells he had a 2pi in front of the integral but for the washers he had a pi instead?

[tropic oxide]

this all ties in with how the volume element is shaped

for washers, the volume element is a disk with radius equal to the height of the function (y) and thickness dx

so that gives you pi y^2 dx as the thing to integrate, from the area formula of a circle

on the other hand, with the shell method your volume element is instead a cylindrical shell with radius x and height y (and still thickness dx)

and its area is given by 2pi xy

[fallow drift]

wait so

for shells

we use cylinders, so we look at SA whhich is 2pirh?

and for washers we look at circles so it's pir^2h?

@tropic oxide

[tropic oxide]

yes

[fallow drift]

thxx

and like

is disk and washer the same?

they both use circles?

and the formula pir^2h?

[tropic oxide]

i usually see it called "disk" when there's only one bounding function and the SoR engulfs the x-axis, and "washer" when there's both an inner and an outer bounding function

but the formula's still the same-ish in both cases

[fallow drift]

and also is there a specific axis of rotation that each method uses??

like does disk method always rotate around y axis

and shell rotate around x axis

[fallow drift]

ok that makes sense thx

[tropic oxide]

no

[cedar kiln]

@fallow drift Has your question been resolved?

[fallow drift]

wait is shell not always spun around the axis?

[tropic oxide]

shell is spun around the axis but it doesn't have to be the y-axis

[fallow drift]

sorry i meant x axis

[tropic oxide]

not sure it makes sense to formulate any general rule tbh. kinda cba to do that right now & would recommend you look at a whole bunch of examples of this stuff anyway

[fallow drift]

okay thx

[cedar kiln]

@fallow drift Has your question been resolved?

[cedar kiln]

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[sick night]

So my exercise 2 is in french but basically its asking to find the left hand limit and right hand limit of 2 and -2. I had to use an ai but i wanted to know why do we have to proceed in intervals ? I just want to understand why

[chilly spear]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[sick night]

Sorry

[limpid glacier]

,rccw

[sick night]

?

[limpid glacier]

i rotated your original pic lol

[sick night]

Oh okay

[livid hound]

wdym "proceed in intervals"

[sick night]

I meant use intervals i dont understand that either... it was ai who anwmswered but now i will avoid

[sick night]

I just want to know how do we get the left hand and right hand limit of 2 and -2 in the function of number 2

[livid hound]

since you can't do much cancellation with those abs val present,
you want to consider the definition of the absval
and express those without those bars when evaluating the limits

[sick night]

Okay how do i remove the bars

[neat dune]

Is it just me or is there not much cancellation going on regardless

[neat dune]

Think of cases. As you get very close to x=2, what will the value be of x-3? Of x+5?

Similarly for x= -2

Technically you should only be thinking of getting very close from the right for x=2 and very close from the left for x=-2 but it doesn't matter too much for these abs values

[sick night]

X - 3 will be very near to -3 when i get near to 2

[neat dune]

Well, leave the expression in terms of x

[sick night]

Hum

[lapis spruce]

|x-3| is 3-x for x < 3

[sick night]

Oh i see

[lapis spruce]

Use similar logic to express numerator "for values close to 2"

Etc

[sick night]

Thanks

But what about x + 5 ? Do i take x > -5 ?

[neat dune]

Yup

[sick night]

Oh okay thanks

Best server thanks

.close

[cedar kiln]

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[cedar kiln]

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[crude brook]

how do i solve this with separation of variables?

[crude brook]

$\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}-12y=1$

[dreamy void]

That one cannot be separated unfortunately

[wraith dagger]

MichaelRafto

[chrome quail]

You cant

[crude brook]

well uh

without using $y=e^{lt}$ how can i solve this?

[wraith dagger]

MichaelRafto

[chrome quail]

undetermined coefficients

[crude brook]

without lagrange either

[chrome quail]

how is lagrange relevent to undetermined coefs

Solve the homogenous equation $\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}-12y=0$ and sum it with some particular solution (which you can get via vop or uc)

[wraith dagger]

Stitches

[dreamy void]

Well it's an autonomous differential equation, you may try the substitution z=y'

[crude brook]

this is what you mean?

[chrome quail]

Do you know how to solve a 2nd order homogenous ode with constant coefficients?

Generally $ay''+by'+cy=0$

[wraith dagger]

Stitches

[crude brook]

yeah by taking $y=e^{lt}$

[wraith dagger]

MichaelRafto

[crude brook]

$al^2+bl+c=0$

[wraith dagger]

MichaelRafto

[crude brook]

(we use lambda)

[chrome quail]

Yeah, you're not allowed to do that?

[chrome quail]

^

[crude brook]

no they want other methods (assuming we are beginners and we dont know the complex ones yet)

[chrome quail]

Are you sure about that? I don't know of any simpler way to solve a generic 2nd order ode

I mean you could do laplace transform but that's even more complex so it wouldnt make sense

[crude brook]

oh ok

so uhh these are just a hodge podge of methods here ig

[fringe garden]

hello, can i ask for help here?

[chrome quail]

Read #❓how-to-get-help and open a channel

[crude brook]

you can make another channel

[chrome quail]

Yeah this looks like a bunch of different methods. All 2nd order methods should be fair game in a worksheet like this

[dreamy void]

(you would turn it into a differential equation of order 1 and could apply variation of parameters)

[chrome quail]

Like (ii) is uc, (v) is higher order (similar process, just with two indepenent sets), (vi) is a euler equation (it has some special substituion you can google), etc

[crude brook]

ok so i get the roots from each one

ah yes then i just replace

[chrome quail]

which one?

[crude brook]

generally all of the problems

i get the roots with every method i need for each one

+particular roots ig

(right?)

[chrome quail]

Well each problem has a different technique. You "get the roots" when doing a regular 2nd order linear constant coefficient ode or any method that requires the complimentary (homogenous) solution

[chrome quail]

yes for ii

not for vi, for example

[crude brook]

let's say we have:

$y=c_1e^{-2t}$ for i)

[wraith dagger]

MichaelRafto

[crude brook]

i just replace y(0)=2 here and find $c_1$ right?

[wraith dagger]

MichaelRafto

[crude brook]

and maybe derive similar results too if needed, right?

yeah this seems right

[chrome quail]

yeah

[crude brook]

ok thank you guys once again

[chrome quail]

that's (i), (ii) is different and so on

[crude brook]

.close

[cedar kiln]

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[cedar kiln]

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[hexed vortex]

Is A, and B subset X part of the hypothesis?

[void sand]

yes, it's stated there too

[hexed vortex]

but it comes after the then

[floral arrow]

I think it's both A in X and B in X

[surreal cave]

I do not believe so, as it is part of the consequent because it is after the then

[void sand]

ah, I see your point

but A and B are arbitrary here too

[floral arrow]

Read it "then, [for any sets A, B both in X], f..."

[hexed vortex]

but it doesnt really make sense to me if its not in the hyp

[void sand]

the statement reads the exact same if you wrote: "Let A, B \subset X. Prove that if f is 1-1, f(A \cap B) = f(A) \cap f(B)"

[hexed vortex]

ok that makes sense, thats how I understood it

[surreal cave]

because the thing after the then is a statement, for A B in X, f(A intersect B) = f(A) intersect f(B)