help-27

[remote sun]

yeah k=0 gives error

look into that ig

[magic gust]

ok

I will check this function

thany you so much !

and thanks guys 😉

[devout snow]

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[scarlet aspen]

can someone help me with this problem? i forgot how to do this

[scarlet aspen]

idk bro im lost

i would think theres supposed to be a y variable but there isnt

[misty crest]

A(x)

well you call the other side of the rectangle y

if you’d like

the long side

is that what you mean

[scarlet aspen]

i mean the program doesnt allow you to type y

so i dont know what to type instead

[misty crest]

yea because you can rewrite y in terms of x

we have a constraint

the perimeter is 160

not of the rectangle but of the fence

write an equation relating the perimeter (160) to y and x

[scarlet aspen]

ok so i know the perimeter of the three sides is 160

[misty crest]

yep

write it in terms of y and x

[scarlet aspen]

what does that mean, in terms of y and x

[misty crest]

160 = some function of y + some function of x

do you understand

[scarlet aspen]

not really im gonna be honest

[misty crest]

i mean, how would you write the perimeter of a rectangle with side lengths x and y

P = ..

[scarlet aspen]

i would write 2x+2y i think

[misty crest]

great

but in our scenario one of the sides is missing

so what would it be now

(the y side is missing)

[scarlet aspen]

2x+y

[misty crest]

so 160 = ..

[scarlet aspen]

ok i know thats for the perimeter but i have to find the area

[misty crest]

right but we need this constraint to solve for y in terms of x

160 = 2x + y

[scarlet aspen]

ok

[misty crest]

what does y =

[scarlet aspen]

y=2x-160

[misty crest]

not quite

[scarlet aspen]

or 160-2x

[misty crest]

so A = xy = x(160 - 2x)

[scarlet aspen]

ohhh

ok that makes sense actually

[misty crest]

it should

[scarlet aspen]

for b and c is there a correct way of doing it or can i trial and error it

[misty crest]

all of these optimization problems come down to using your "constraint equation" which in this case is a fixed perimeter or amount of fence to then write the desired quantity (area in this case) in terms of one variable

because that constraint equation fixes a relationship between the two variables allowing you to solve for whichever in terms of the other

[scarlet aspen]

and thats what 160-2x is

[misty crest]

derivative

this is calculus yea?

[scarlet aspen]

this is algebra 2

[misty crest]

oh

[scarlet aspen]

at least my program is for algebra 2

[misty crest]

ok well you have a parabola

vertex

you can use the vertex to get your answers for the next two parts

[scarlet aspen]

ok im going to watch videos for those

[misty crest]

you know how to find the vertex right

[scarlet aspen]

i have to brush up on it

[misty crest]

x(160 - 2x) = 160x - 2x^2

this is a concave down quadratic so it has a maximum at it’s vertex/turning point

[scarlet aspen]

vertex form was always my least favorite form

[misty crest]

just complete the square to put it in vertex form

id factor out -2 first

-2(x^2 - 80x)

[scarlet aspen]

ok ima start writing this down

[misty crest]

completing the square is quite simple if you remember the motivation

you’re using the fact that

$(x \pm a)^2 = x^2 \pm 2ax + a^2$

[woven radish]

knief

[misty crest]

so you’re basically manipulating the standard form to look like a perfect square

[scarlet aspen]

wait doesnt completing the square just find the solutions

[misty crest]

completing the square puts it in vertex form where the vertex is easily identifiable

which we’re clearly interested in for this problem

[scarlet aspen]

ok so im at x^2-80x+160

[misty crest]

🔥

which can be rewritten as

[scarlet aspen]

x(x-80)+160

[misty crest]

well you’re missing a 0 here

40^2 =1,600

[scarlet aspen]

oh yeah

[misty crest]

nono, write it as a square

x^2 - 80x + 1600 = (what)^2

[scarlet aspen]

idk where to put the -80x

nothing squares to that

[misty crest]

this is the square of a binomial

[misty crest]

^

you’re at the right side rn

make it look like the left

[scarlet aspen]

(x-40)^2

[misty crest]

so

[-2(x^2 - 80x + 1600 - 1600) = -2((x - 40)^2 - 1600)]

[woven radish]

knief

[misty crest]

so just distribute the -2

[scarlet aspen]

wait whered the -2 come from

[misty crest]

well we factored that out from the start

[misty crest]

^

[scarlet aspen]

so we are supposed to readd it?

[misty crest]

so we can’t leave it behind

[scarlet aspen]

ok

[misty crest]

it would change our area formula

we just did that for convenience

so what do you get after distributing

[scarlet aspen]

-2(x-40)^2+3200

[misty crest]

yep

which means the vertex occurs where

[scarlet aspen]

thats vertex form ok

so is it (-40, 3200)

that doesnt seem right

[misty crest]

close

not -40

[scarlet aspen]

is it 80 cuz we distribute

[misty crest]

nope

[scarlet aspen]

oh wait its 40?

[misty crest]

the idea is that the max/min occurs when the squared term is 0

[misty crest]

yes

$a(x - h)^2 + k$

[woven radish]

knief

[misty crest]

so since (x - h)^2 >= 0

if a > 0 then we will have a minimum because the left term can be at least 0

but if a < 0 then (x - h)^2 being nonzero means you’re really subtracting from k

since it will always be negative so long as (x - h)^2 ≠ 0

which means if we want a maximum then we want (x - h)^2 = 0

which occurs precisely when x = h

likewise for the minimum

[misty crest]

and plugging in x = h leaves us just with k since a(h - h)^2 = 0

make sense?

[scarlet aspen]

i have to try and process what you just said

[misty crest]

i have to go eat

but yea x value is 40

[scarlet aspen]

ok thank you for your help

[misty crest]

area is 3200

[misty crest]

you’re welcome

[scarlet aspen]

thanks 👍

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[celest verge]

a) are the columns of A linearly independent vectors?
b) are the columns of the augmented matrix A|b linearly independent vecors?

[celest verge]

i said yes and yes but my teacher said b is not true

why isn't the augmented matrix linearly independent?

[celest verge]

btw. the identity matrix is supposed to be A

[lunar harbor]

The columns of A are linearly independent

[celest verge]

yes

[lunar harbor]

But b is a linear combination of the columns of A

[celest verge]

yes

true

lol

they are dependent on the matrix

so like if 1 = 2 then 5 = 10

so its dependent

[devout snow]

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

[celest verge]

.close

[devout snow]

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[limber bison]

I was wondering how to do these types of problems? I have somewhat of an idea for this problem in specific I thought it wanted the question in terms of Y so the for the function I made it ln(y) and took the pi * integral from 1 to 7 and I made sure to square R

[junior flax]

Do you know about the washer method or shell method?

[junior flax]

Your method looks correct.

[limber bison]

uh yeah sort of

[limber bison]

this is how I set up the integral

and I put in my answer as 35.384

[junior flax]

Yes,

I’m pretty sure that is the correct answer

[limber bison]

yeah but I got it wrong

which im very confused about

[junior flax]

🤔

[limber bison]

im going to try using a ti 84 calculator, not rlly sure how that would change much tho...

same answer

[junior flax]

🤔

Maybe try doing the integral of (7-e^x)^2 from 0 to ln(7)

[limber bison]

yeah i got that one wrong

[limber bison]

I think this wouldve been correct

[junior flax]

It should give the same answer though

[limber bison]

that was the correct answer

[junior flax]

That’s pi times the answer you got

Perhaps the calculators were wonky

[limber bison]

I did it on desmos and the ti84

bruh

[junior flax]

Yes, the ti-84 did not multiply in the pi for some reason

😔

[limber bison]

wow

well there goes that grade

[junior flax]

There may be a difference between integrating with respect to x and y that we’re missing

[limber bison]

I thought it wanted it in terms of y

i just dunno

[junior flax]

Ah, I think I see now, since where revolving around y=7 integrating with respect to y doesn’t work, since the radius that gets squared is not correct

[limber bison]

ohhh

alr

[junior flax]

It’s been a long time since I did one of these. I think that method is to integrate perpendicular to your axis of rotation when using the washer method.

[limber bison]

ohh what math are you doing right now?

[junior flax]

Multivariable calculus. It’s not much more advanced, just the next class up.

[limber bison]

ah I see

ok I got another

question thats similar

since its rotating around y=1

this is the setup

[junior flax]

not quite

[limber bison]

it was right

lmao

im actually clueless to how that works

[junior flax]

the boundry of integration should've been (2-sqrt(x))^2, but apparently those result in the same answer

as the radius is defined by 1-f(x)

I did it using my method and I also got 8.377

[limber bison]

can you show what you inputted in your calculator?

[limber bison]

oh I see okay

[junior flax]

very strange coincidence

[limber bison]

yeah I agree

alr I got 3 more (2 of which I got wrong even tho I did it the way my teacher taught me)

This would be the washer method right? where you have

[junior flax]

I'm not sure if our methods are somehow algebraically equivalent, but I'd rather you use my method, since I know for sure it works

[limber bison]

yeah I will next time

[limber bison]

wait no for this one they just give us the bounds right 0 and 2

[junior flax]

yes, but since we are integrating revolving around the line y=-2, little r = 0

[limber bison]

oh yeah

so would we want 2x^3 in terms of y

[junior flax]

This is why I was confused by the previous question dealing with sqrt(x)-1

[junior flax]

not quite. I'm not sure exactly what you mean, but we are integrating with respect to x, so imagine choping vertically down towards the x-axis to get you radius

[limber bison]

uh

im still kind of confused

[junior flax]

basically, we're integrating with respect to dx

[limber bison]

yeah

so the radius would just be 2x^3 ?

[junior flax]

not quite, 2x^3 would measure from the function f(x) to 0, but we don't stop at zero

[limber bison]

we want to stop at 2 right

[junior flax]

yes, the radius extends to y = -2

[limber bison]

so -2+2x^3?

[junior flax]

radius = absolute value of (the difference between functions)

[junior flax]

very close

but subtracting 2 would decrease the radius

[limber bison]

we want to add 2

errr

[junior flax]

you accidentally added a -2

infront of your function 2x^3

[limber bison]

yeah

I just noticed

355.449 would be the right answer

if we were rounding to the nearest thousandth

[junior flax]

There's been so many strange coincidences. Having 1/pi of the correct answer when we integrated the completely wrong radius in question 1. Arriving at the same answer despite using two different methods. I feel like I'm in limbo

[limber bison]

this is why I hate delta mathb

its never accurate

i swear

or like it feels random as hell

bounds would be 0 to 1.0986123

[junior flax]

delta math's alright. I think you just got the unlucky (or maybe lucky) draw of questions

[limber bison]

i guess so

[junior flax]

well the bounds would be from 0 to ln(7).

since e^x is 7 when x = ln(7)

that part makes sense, logarithms create some strange looking irrational numbers sometimes

[limber bison]

wait why 7?

[junior flax]

we integrate from x = 0 to the x-value at which e^x = 7

since that is when y = f(x) = e^x intersects g(x) = 7

[limber bison]

it says y = 6 tho

wait idk im missing something

[junior flax]

whoops

[limber bison]

lol

[junior flax]

I was still on the first question you sent

my bad

[limber bison]

oop

all good

[junior flax]

yes, it would be to ln(6)

actually wait

it would be ln(3)

[limber bison]

so the decimal I put

was right

alr

[junior flax]

my bad. These question got me tripping

[limber bison]

1.0986123 = ln(3)

[limber bison]

ur good

[limber bison]

6-2e^x is the radius?

[junior flax]

yes

[limber bison]

yeah

awesome

I think I have a better understanding now

thank youu for the help

[junior flax]

you're welcome

[limber bison]

have a good rest of your night

.close

[devout snow]

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[devout snow]

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[trim pewter]

can i get some examples of integrable functions that are not continuous?
and please ping if u respond

[timber pebble]

yea theyre very common

and not too hard to construct

you should think of one as an exercise

I'd go as far to say

start here: can you give a function that is not continuous?

[trim pewter]

1/x

[timber pebble]

okay, now lets reign it in

can you think of one that has no asymptotes

but is still discontinuous

[trim pewter]

$\frac{x^2-9}{x-3}$

[woven radish]

Flatus

[timber pebble]

sure! that's a great example

what is $\int _0 ^5 \frac{ (x+3)(x-3) }{ (x-3) } \dd x$

[woven radish]

jan Niku

[trim pewter]

so even though it's discontinuous

you can still integrate it

because that one point

has 0 length

so it doesn't matter

right?

[timber pebble]

yea, its not especially important that were missing a point here either

you dont need to have holes in the domain to have a discontinuous function

how about $f(x) = \frac{1}{\ceil x}$

[woven radish]

jan Niku

[trim pewter]

ooh

i see

[timber pebble]

but its just fine to integrate this function all over the place

and its only missing a single point

it has tons of discontinuities

you can think of some other esoteric jumpy function

maybe $\frac{x}{|x|}$?

[woven radish]

jan Niku

[timber pebble]

perfectly integrable

[trim pewter]

i've heard multiple different opinions on whether

$\int_{-1}^{2} \frac{1}{(2x-1)^2} \ dx$

[woven radish]

Flatus

[trim pewter]

is integrable

because doesn't it cross an asymptote? but some people say the area converges

and it confuses me a lot

[timber pebble]

well lets see

lets be naive

[trim pewter]

i always assumed if it crosses asymptote, it's not integrable

[timber pebble]

what method should we use

[trim pewter]

RCR or u-sub to integrate

[timber pebble]

lets just assume we can and see if problems come up

i like u sub yea

$\int _{x = -1} ^{x=2} \frac{1}{2u^2} \dd u$

seem good?

[trim pewter]

u^2

[woven radish]

jan Niku

[trim pewter]

yea :D

[timber pebble]

$\frac 1 2 \int _{x = -1} ^{x=2} u^{-2} \dd u$

[woven radish]

jan Niku

[timber pebble]

whats the anti deriv?

[trim pewter]

$\frac{1}{2} [-u^{-1}]_{x=-1}^{x=2}$

[woven radish]

Flatus

[timber pebble]

sure lets go $\frac{-1}{2u}$ from $x=-1$ to $x=2$

[woven radish]

jan Niku

[timber pebble]

i like the u so maybe we go into u world

2x-1 = u so x=-1 gives u = -3

2x-1 = u so x=2 gives u=3

$\frac{-1}{2u} \bigg \vert ^3 _{-3}$

[woven radish]

jan Niku

[timber pebble]

anything jump out at you

[trim pewter]

negative area

[timber pebble]

more so than that

you know about even and odd?

[trim pewter]

ye

[timber pebble]

-1/2u is it odd? is it even? neither? both?

,w odd function

can wolfie do it

[woven radish]

Results provided by WolframAlpha

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

Requested by janniku

[timber pebble]

$f(u) = \frac{-1}{2u}$. does $f(-x) = - f(x)$?

[woven radish]

jan Niku

[trim pewter]

yes :0

[timber pebble]

okay

its a quick proof from here

let me give you the theorem

[woven radish]

jan Niku

jan Niku

[timber pebble]

so is $\frac{1}{(2x-1)^2}$ integrable from -2 to 1?

[woven radish]

jan Niku

[trim pewter]

wait brb

[timber pebble]

@misty crest lol

[misty crest]

right at the end

🤣

[timber pebble]

its okay we really have kind of blazed through this without any concern about discontinuity

or improper integrals and issues there

who cares tho

math is about the big picture

[misty crest]

how’s life

[timber pebble]

bad man

hows yours

[misty crest]

worse

[trim pewter]

what did i come back to

[timber pebble]

https://tenor.com/view/impossible-heres-my-answer-not-happening-gif-12692184

[misty crest]

💀💀

[trim pewter]

i believe it isn’t

[timber pebble]

you know i kinda fucked up that explanation

i actually think youre right

i dont think it is either

[misty crest]

just say it’s not bounded and call it a day

[timber pebble]

BUT

we can create functions like this that are

how about $\int _{-3} ^3 \frac{1}{x^3} \dd x$

[woven radish]

jan Niku

[timber pebble]

abosolutely we can evaluate this

[misty crest]

this is cauchy principal value

[misty crest]

define evaluate

[timber pebble]

put a equals 0 and call it a day

[misty crest]

if we’re being pedantic then unfortunately we can’t say anything

but we can use cauchy principal value like sane people

[timber pebble]

its not important

[misty crest]

depends what class this fellas in

if he’s an engineer

then yea who cares

[timber pebble]

hey nothing wrong with being an engineer

[misty crest]

when did i say there was?

[timber pebble]

it was implied

[misty crest]

nope

[timber pebble]

[misty crest]

engineers are great

i couldn’t do that shit

[timber pebble]

how long should we hang out

[misty crest]

🤣

flatus left again

you fr grew up without cable

[timber pebble]

https://tenor.com/view/essica-monthly-bnuy-25th-gif-15850821094290025976

[misty crest]

did you have indoor plumbing

[timber pebble]

in kindergarten

[misty crest]

rough first few years

[timber pebble]

yea i had it

we just tdidnt have cable or a computer

it was the 90s

[misty crest]

didn’t you have like atari?

[timber pebble]

no

[misty crest]

and nintendo games

[timber pebble]

nah

i didnt have a console until i was an adult like 20

[misty crest]

oh your parents didn’t love you

jokes

[timber pebble]

consoles werent big back then

[misty crest]

interesting

what was

grass?

[timber pebble]

my backyard

[misty crest]

sunlight

[timber pebble]

yea

u go to the local big area of grass wheneve u are bored

and all ur friends are there

e z

[misty crest]

y’all would go to each others houses on bikes and ask if y’all can play

[timber pebble]

i assume kids still do this today

but i am in contact with no children

other than my neice

[timber pebble]

yes

[misty crest]

when i was younger we would just call each others house phones or have our parents arrange shit

[timber pebble]

and ur friends mom would say we locked ourselves out of the house

can you crawl in through the doggie door

since ur a small child

[misty crest]

do those really exist

[timber pebble]

yea

[trim pewter]

im in year 11

[misty crest]

oh i’ve seen one actually

he’s back

[trim pewter]

oops

[timber pebble]

whats up flatus

[misty crest]

british

unfortunate

[trim pewter]

australia

[timber pebble]

i thought it was exclamation lol

[misty crest]

sorry

[timber pebble]

unfortunate!

@trim pewter yea discontinuities are not the hugest deal in the world

[misty crest]

dope accent though

[timber pebble]

is the big point im trying to make

differentiation, pretty big problem

integration, less of an issue

[misty crest]

riemann integrable iff bounded

[timber pebble]

https://tenor.com/view/cat-meme-hiding-gif-15505332

[trim pewter]

when do i close 😭

[timber pebble]

whenev ryou want

[trim pewter]

.close

[devout snow]

Channel closed

Closed by @trim pewter

Use .reopen if this was a mistake.

[misty crest]

https://tenor.com/view/over-here-waving-korg-thor-ragnarok-gif-14586261

new zealand accents are dope

[timber pebble]

australian type gif

[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[lost laurel]

Could someone guide me thorugh thi

[lost laurel]

As I'm doing it by hand, I'll only use the first two decimal places

[devout snow]

Show your work, and if possible, explain where you are stuck.

[lost laurel]

I will

one minute

The table I got

so I have to use this fomrula

$f(x) = 1 +0.22 \binom{x/0.2}{1} + 0.05 \binom{x/0.2}{2}+ 0.01\binom{x/0.2}{3}$

feels sus

[woven radish]

What a wonderful world !

[lost laurel]

@supple trench why teh cross

[supple trench]

why only use 2dp?

you're given a table of fct val w/ more dp

[lost laurel]

yea, but computing by hand will be a pain, and we'll get simpler values in the exam

[supple trench]

you're exposing yourself to the risk of truncation errors

[lost laurel]

What matters is I understand the method and am able to execute it

[supple trench]

such truncation explains this

[lost laurel]

I graphed it, seems to match up

Anyway, I'll now do a problem on backward interpolation

.close

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[unique cloud]

Why it says the dimension of an autospace is the Ker of all that stuff? And why on the next page it follows the same line suit, and why it says he uses the Theorem of Nullity and Rank but doesn't include Image

[stone stump]

{X: AX=\lambda X} = {X: (A-\lambda I)X=0} but the second space is just the kernel of A-\lambda I

and you can rearrange in another way to get the other kernel

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[unique cloud]

No i didnt really understand—why, in the highlighted line of the second picture i sent, although the book uses the theorem of nullity and rank, it does not feature any Im L(A-Iλ)

[unique cloud]

Given that m(λ) should be dimension of Vλ(A) it says its equal to Ker but does not feature the Im

Does Ker, written in that way, embodies the Im as well?

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[prisma elm]

Help please

I need video tutorial for these question please

Trying to study for final exam but forget this part

[supple knot]

,tex .FTC2

[woven radish]

riemann

[supple knot]

use that to find g'(x)

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[mortal pagoda]

I have a question regarding finding the normal vector to a plane, and a line. It's an A level question, but i am starting to think there is some kind of error

[mortal pagoda]

tried using the dot product to prove perpendicularity, and that doesn't work

and used the vector product to find the normal vector, no luck there too

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✅

[mortal pagoda]

<@&286206848099549185>

[sand dove]

can you should exactly what you did and where exactly you think there's an error?

[mortal pagoda]

so it is just for the first part, finding the normal to the plane

I have attempted to find a second directional vector from the point P to the plane

used the vector product with this, and the position vector of the plane given

this doesn't give us -i + 5j + 3k

likewise if i tried just the position vector of that point P

along with the position vector of the plane

using the vector product, i still don't get it

and also the dot product for neither works.

@sand dove

[devout snow]

@mortal pagoda Has your question been resolved?

[mortal pagoda]

@shrewd flame

<@&286206848099549185>

[sand dove]

Am I right?

Wait

[sand dove]

Are you talking about P1?

If so what do you mean by "second directional vector"?

Just to be clear, just because P is on the plane

Doesn't mean that the position vector of P is a directional vector of the plane

Not at all

I'm also curious about which points do you use to get that second directional vector

[devout snow]

@mortal pagoda Has your question been resolved?

[mortal pagoda]

I have assumed that the line L, is perpendicular to P which is on the plane

I haven't assumed that line L is on the plane

but if they are perpendicular, you should be able to use the vector product

or at the very least, use the dot product to prove they are perpendicular?

[mortal pagoda]

I used the directional vector for Line L

so the lamba ..... part

[sand dove]

Again

L is perpendicular to the plane

so perpendicular to directional vectors of the plane

P is on the plane

OP, the directional vector of P

is not a directional vector of the plane

[mortal pagoda]

so how would you set you find the normal of the plane with the given information please?

[sand dove]

You're given a line that's perpendicular to the plane

so the directional vector of the line

is normal to the plane

here's the equation of said line

thus we read (-i + 2j + 3k) as directional vector of that line

so normal to the plane

[mortal pagoda]

ahh i see

but the answer given in the question isn't -i +2j +3k

[devout snow]

@mortal pagoda Has your question been resolved?

[sand dove]

?

this is what they got

wait

ah yeah they f-ed up

they wrote 5j instead of 2j

mmmh

to try to be in par with the solution

[sand dove]

suppose that in the question statement, equation of L is "r = 3i - 2k + t(-i+5j+3k)"

and continue from there

you're not to blame here, the solution is (or the question statement)

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[steady creek]

A rectangular pen is to be built with 1200 m of fencing. Each side of the pen has to be at least 200m long. The pen is to be divided into three parts using two parallel partitions.
a) Find the maximum possible area of the pen

[steady creek]

can someone help me with this

i know how to do it without the restriction of it being atleast 200 m

but how do i approach it with the restriction?

[restive river]

Do you have any work yet?

[steady creek]

no

[restive river]

Did you try it on your own yet?

[steady creek]

im confused how to start

i drew a diagram

and i get that 4w + 2l = 1200

[restive river]

Yep

Solve for L, do you want help on how to do that?

[steady creek]

l = 1200-4w/2

[restive river]

Good and what is that?

[steady creek]

length

[restive river]

I know

But its l = 600-2w right?

[steady creek]

yea

so we get the equation A(w) = (500-2w)(w)

[restive river]

A = (600 - 2W)(W)

[steady creek]

oh yea

my bad

[restive river]

No worries

[steady creek]

i find the derivative and set it = 0

?

[restive river]

Yes

[steady creek]

i get 150

which isnt possible cuz it says atleast 200m

[restive river]

Well, we have 600-4w = 0

Add 600 to both sides

We have 4 w = 600

Meaning w = 150

[steady creek]

yeaa

[restive river]

So now we can find L

[steady creek]

but the question says each side has to be atleast 200 m long

[restive river]

L = 600 - 2(150)

L = 300

So 300 • 150

[steady creek]

?

[fossil locust]

okay, so that means that l >= 200 and w >= 200

so 600 - 2w >= 200

[restive river]

I hope i didnt mess up the question did I ?

[steady creek]

right

[fossil locust]

can you now try to solve for w?

[steady creek]

w<=200

[fossil locust]

yep, nicely done

yeah you have to flip the inequality sign

[fossil locust]

so it's actually really

[steady creek]

wait how that even make sense

[fossil locust]

silly

[steady creek]

wait

[fossil locust]

cause we have both $w \le 200$ AND $w \ge 200$

[woven radish]

south

[fossil locust]

there's only one value

[steady creek]

why are we doing 600 - 2w >= 200

isnt it (1200-4w)/2 >= 200

and w>=200

[fossil locust]

you needed brackets!

should be (1200 - 4w)/2 >= 200

[steady creek]

right

so 200<=w<=266.67

?

[fossil locust]

nope

[fossil locust]

just roll with this cause that's what you obtained

[steady creek]

why not ?

[fossil locust]

where does 266.67 come from?

[steady creek]

oh 🤔

[steady creek]

mb yea

so w can only be 200

[fossil locust]

yep!

the <= is really really important cause it lets w = 200

[fossil locust]

so if I can explain what's going on with this weird question

you have 6 sides, and each side is 200 minimum

so the minimum total perimeter must be 1200

[steady creek]

right

200*6 = 1200

but im confused w the answer key my teacher provided

[fossil locust]

that means all the sides have to be 200

so length = 200 and width = 200

[fossil locust]

I think they messed up the wording here frankly

with the partitions

I think your teacher only wanted the pen to be divided into 2 parts

so you'd get 3w + 2l = 1200 instead and you'd have new equations

[steady creek]

hmm i see

[fossil locust]

oh wait I see

no the question is correct but you should change 200 to 180

so it's still 4w + 2l = 1200

but now you have w >= 180 and 600 - 2w >= 180 instead

so that does give you -2w >= -420 or w <= 210

and then the vertex (a maximum) occurs at w = 300, and these w-values are before the maximum

so for the maximum area you just take the maximum value of w, 210

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[steel wedge]

Hi, just need to confirm that the coordinate ring of an affine plane without a point is K[x, y] real quick

[steel wedge]

I calculated it using the sheaf condition as a ring pullback, not sure how to do it with elementary methods (I’m reading Harris’s 1st course)

It essentially reduces to calculating K[x, y, 1/x] \cap K[x, y, 1/y] inside K[x, y, 1/x, 1/y]

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[steel wedge]

<@&286206848099549185>

<@&286206848099549185>

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@steel wedge Has your question been resolved?

[steel wedge]

<@&286206848099549185>

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[tawny pewter]

#groups-rings-fields #advanced-algebra

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[young olive]

You have an equilateral triangle with a circle perfectly inscribed inside it (the incircle), and the circle's radius is 3.

You want to find the area of the triangle.

[young olive]

Stuck af on this

feeling stupid

.close

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[topaz sparrow]

How big a subject calculus is?? I am about to get calculus 1 course. what else will I come across in calculus.? I heard there's is calculus 2 and 3 ... What are difference in these courses?? And are these hard to do??

[topaz sparrow]

And also whats are suggested textbooks should I get to support me doing calculus from beginning to advance?

[ebon coyote]

That'll depend on a variety of factors unique to your situation - what year are you in, what country, what exam board you're taking (if you're in HS)...

[autumn girder]

Calculus is a very big and beautiful branch of mathematics the only downside is you're pretty much always playing with fire. If you're working with non-standard stuff you're basically fucked. It is fun tho.

[supple knot]

#book-recommendations and or #calculus

[topaz sparrow]

So as there are several books written on calculus. I want to know if I try to read any calculus book then am I going to understand it?? I mean, how do I get to know my level of calculus knowledge so that I tend to know which books will I get to understand and which books are a little difficult? And how to start from beginning about calculus? @autumn girder

[topaz sparrow]

What do you mean when you say non- standard stuff?? Plz elaborate?

[autumn girder]

Someone posted a question on this server asking for integration of $x^{\sqrt x}$ and it doesn't have a standard integration. This monstrosity was what I got when I tried to push through anyways

[woven radish]

@autumn girder

[autumn girder]

$$f(x) = x^{\sqrt{x}}\cdot\left(\frac{2x}{\ln\abs{x}}\right) + f(x) \cdot \left(\sum_{i = 1}^{\infty} (-1)^{i}\frac{d^i}{dx^i}\left(\frac{2x}{\ln\abs{x}}\right)\right) + c$$

[woven radish]

@autumn girder

[ebon coyote]

And yet you went ahead with that!?

Honestly that's pretty rad ngl mad props to you

[autumn girder]

For the record this isn't verifiable because this is just never gonna convergr

[autumn girder]

Nah I am just hard headed lol

[autumn girder]

Now TBVF you'll prolly never do this in any standard HS or even Grad math course, this is just me being an idiot. But this is just an example to show you what happens when you don't listen to calculus Gods and try to manufacture something they think shouldn't exist (which I feel is easier to do in Calc than it is in many other fields). They'll fuck you up. Keep in mind tho even as I say this, Calc is still beautiful. It is crazy but that's the beauty of it.

[devout snow]

@topaz sparrow Has your question been resolved?

[autumn girder]

@topaz sparrow wait you're in PLD server? that's rare

especially pre-calc

[topaz sparrow]

Do you mean calculus has no usage in any field? No application?

Kindly answer

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[topaz sparrow]

.reopen

[devout snow]

✅

[autumn girder]

OH FUCK NO I didn't mean to imply that

It is very useful

It's the heart of applied math

[mystic scarab]

#calculus and/or #math-discussion please

[topaz sparrow]

.close

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[odd yacht]

Is the horse right?

[odd yacht]

Is the results of the horse an aproximation, or is it exact?

[acoustic leaf]

,calc cbrt(81)

[woven radish]

Result:

4.3267487109222

[acoustic leaf]

,calc cbrt(5)

[woven radish]

Result:

1.7099759466767

[acoustic leaf]

well it seems like it's correct if you count "rounded to the nearest 11 decimal places" as correct

ofc it can't be fully precise because there are an infinite number of decimals

[odd yacht]

So an aproximation?

Ty

[acoustic leaf]

in general the cube root (or any root) of an integer is either an integer or it's irrational

[odd yacht]

So eather irrational or natural

[acoustic leaf]

so since both of these are irrational there's no repeating decimal and we have to settle for a few decimal places

[acoustic leaf]

well you can have a negative cube root of a negative number

[odd yacht]

Ok

Ty

/close

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[wheat pawn]

.close

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[dark warren]

I got nerd sniped by a silly problem that showed up on a social media feed. I've attached the pic. The goal is to find the radius r. My mathematical background is your basic engineering curriculum (algebra, calculus, differential equations, some linear algebra). Yet these types of problems tend to keep me busy for what seems like way too long, often with me going in circles never actually finishing them.

In my work so far I've denoted the functions as f(x) = x^2 and g(x) = x^2 - 1/2. And I've denoted the points of intersection to the circle as (a, f(a)) = (a, a^2) and (b, g(b)) = (b, b^2 - 1/2). I also denoted the center of the circle as (c, 0).

From there I derived expressions for the radius using right triangles on those points of intersection with a and b. I set them equal to each other to get an equation for c in terms of a and b.

I also derived equations for the lines perpendicular to the points of intersection between the functions and the circle. They are f_p(x) = -x/(2a) + a^2 + 1/2 and g_p(x) = -x/(2b) + b^2. I then set these equal to each other, realizing that they intersect at (c, 0). This yielded a different equation for c in terms of a and b.

I could go on with all the things I've tried. But I've been slogging through algebra, looking for simplifications or substitutions that move the needle in some way. Instead I wind up with large power-4 polynomials of a and b. Maybe there's a key piece I'm missing? I even wondered if this is a valid construction. My rough intuition says yes, but I could be wrong of course.

[undone chasm]

Pythagoras theorem? equate hypotenuse to red and hypotenuse to blue?

[gaunt iron]

not quite a great way

[mortal crypt]

doesn't guarantee perpendicularity?

[dark warren]

Should note that the tangent lines on the parabolas are skew. So yes, two different right triangles can be made with hypotenuse r. But equating those expressions gets messy.

[mortal crypt]

(a,a^2) and then (b,b^2 - 1/2), and then (u,v) for the center of the circle is getting me

$-2au+a^2-2a^2v+a^4=-2bu+b^2-2b^2v+b^4-v+\frac{1}{4}+b^2$

[woven radish]

Yeatte

[mortal crypt]

a bit ew yeah

[dark warren]

v would be zero in this scenario, yeah?

[mortal crypt]

v is just the y coord for the center of the sphere

i mena, for your scenario yes

[dark warren]

I didn't want to try generalizing it too soon since it's hard enough for me in the specific case here. But by all means go ahead if you can figure it out :p

[mortal crypt]

I mean, there is a quartic formula.. but

It's just going to be ugly :\

[dark warren]

$f_p(x)=\frac{-x}{2a}+a^2+\frac{1}{2}$

[woven radish]

Jerry

[dark warren]

$g_p(x)=\frac{-x}{2b}+b^2$

[woven radish]

Jerry

[mortal crypt]

yep

[dark warren]

Why is my latex so huge? Sorry I'm new to this. Anyway those are the lines perpendicular to the intersection points of each function f and g to the circle.

[mortal crypt]

mm almost

$y-a^2 = -\frac{1}{2a}(x-a)$

[woven radish]

Yeatte

[mortal crypt]

I used a for the red one

[dark warren]

Yeah that's the same, no? I moved everything to the rhs

[mortal crypt]

yep

[dark warren]

They intersect at the center of the circle.

But more algebra didn't really get me anywhere. Anyway I'm going to take a walk. I'll check back later.

[mortal crypt]

$U=\frac{a^3-b^3+\frac{a}{2}}{a-b}$

[woven radish]

Yeatte

[mortal crypt]

$V=\frac{2ab^3-2ba^3-ab}{a-b}$

[woven radish]

Yeatte

[devout snow]

@dark warren Has your question been resolved?

[mortal crypt]

,w solve ((a^3-b^3+a/2)/(a-b) - a)^2 + ((2ab^3-2ba^3-ab)/(a-b) -a^2)^2= ((a^3-b^3+a/2)/(a-b) - b)^2 + ((2ab^3-2ba^3-ab)/(a-b) -(b^2 -1/2))^2

[woven radish]

Results provided by WolframAlpha

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

Requested by yeatte137

[dark warren]

Hey those are nice and round

but why negative, I wonder

[mortal crypt]

its supposed to spit out a relationship between a and b, but instead wolfram gave an example of an answer along the line

so its not really what w'ere looking for

[dark warren]

Oh duh

[mortal crypt]

It ends up instead of being a polynomial of degree 4, it ends up being degree 6

which is not guaranteed to have a 'nice' root equation

[dark warren]

Ah ok. I never got far enough to a degree 6 expression. Crazy that such a simple looking problem blows up into this mess.

[mortal crypt]

with the 2ab^3 -2ba^3 its a 4th degree

but with the squaring, and the (a-b) on the bottom we would have a 8th deg on top and a 2nd deg on bottom

multiplying by the bottom to get rid of any a or b on the bottom, we would get 10 degree polynomial

but stuff cancels out so it ends up being back down to a 6th degree

tho if you try to solve for either a or b, it can be thought of as a 5th deg

still most likely not doable

that being said

if you instead took a circle between 2 ellipses

I ended up doing that math, and it ends up quite elegant

So I would recommend trying that instead

[dark warren]

You mean instead of between two parabolas? That would be a different problem right?

[mortal crypt]

yep

I think one of the problems was that the exponent of y and x didnt match each other

a lot more things wouldve cancelled out if they did

just like the U^2 and the V^2 terms are on both and cancelled each other out as well

[topaz crag]

What are you trying to determine precisely ?

[mortal crypt]

oh the radius

lel

hm

[dense lynx]

$r=\frac{1}{54}\sqrt{189-78\sqrt{3}+12\sqrt{342\sqrt{3}-582}}$

[woven radish]

Mqnic_

[mortal crypt]

mm?

dja do the same thing I did or something different?

cuz for that (c,0) point we could find the 2 tangents to the circle, and then cuz of V = 0 there, it reduces to a 4th deg

of a in terms of b or vice versa

and then plug that into one of the circle eqs

[devout snow]

@dark warren Has your question been resolved?

[dark warren]

I uh, think so. @dense lynx can you elaborate on your solution?

This isn’t homework, I’m just trying to understand the solution.

[devout snow]

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[devout snow]

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[scarlet fulcrum]

Hi I'm doing induction problems and I don't see how to go on further

[heavy current]

can you show the original problem?

[wispy oyster]

prove that
$1^3 + 2^3 + \cdots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$

[woven radish]

mia

[heavy current]

my first instinct is to factor out (k + 1)^2 from both terms in the numerator

[scarlet fulcrum]

This part?

[heavy current]

no

the numerator is $k^2 (k + 1)^2 + 4(k + 1)^3$

[woven radish]

higher!

[scarlet fulcrum]

And I factor out the (k+1)^2

Ok let me do

[scarlet fulcrum]

I'm lowkey dyslexic and struggle with moving numbers around when factoring so 🙈

[heavy current]

hm, it does seem like you've made a mistake

namely, the k^2 factor in your first term has vanished

[scarlet fulcrum]

OMG

I SEE IT

I'm lying wait

Give me a sepcnd

[heavy current]

take your time

[scarlet fulcrum]

I thought I factored it out

[heavy current]

you factored (k + 1)^2 out

[scarlet fulcrum]

if I do that

Doesn't that leave me with

[heavy current]

remember that your original expression is k^2 (k + 1)^2 + 4(k + 1)^3

[heavy current]

when you factor (k + 1)^2 out from both terms, you need to divide each term by (k + 1)^2

[heavy current]

hm, first things first, that /4 extends to (k + 1)^3 too

[scarlet fulcrum]

oh yes pretend it does

[heavy current]

secondly, your sum of terms has become a product after factoring (k + 1)^2 out of both terms

and (k + 1)^2 is nowhere to be seen

[scarlet fulcrum]

is it possible you can write it out from your side? I like

Have trouble visualizing these kinds of factoring

😞

[heavy current]

I'll give a different example

[scarlet fulcrum]

Okay tysm

[heavy current]

say we have $k^3 (k + 2)^2 + 7(k + 2)^3$

[woven radish]

higher!

[scarlet fulcrum]

Yes

[heavy current]

then we can factor $(k + 2)^2$ out of both terms to get $(k + 2)^2 (k^3 + 7(k + 2))$

[woven radish]

higher!

[heavy current]

this is just reversing the distrubutive law

[scarlet fulcrum]

I see where (k+2)^2 got factored out of the right term

But the left term it's like

Nothing ever got factored but rearranged?

Does that makes sense

[heavy current]

I'm a bit confused

let me write it out more explicitly

\begin{align*}
k^3 (k + 2)^2 + 7(k + 2)^3 & = (k + 2)^2 \left(\frac{k^3 (k + 2)^2}{(k + 2)^2} + \frac{7(k + 2)^3}{(k + 2^2)}\right) \
& = (k + 2)^2 (k^3 + 7(k + 2))
\end{align*}

[woven radish]

higher!

[scarlet fulcrum]

To me this is what I saw from your display

Like I feel like the k^3 just gets moved a spot I don't see the actual factoring

[heavy current]

here

[heavy current]

the parentheses I used contain all that information

you might've noticed I added an additional pair

that was on purpose

[scarlet fulcrum]

I'm starting to understand let me

Look at it longer

I understand everything, but I don't understand where you pulled

This from

Can anyone explain to me where the extra (k+1)^2 comes from

[heavy current]

it's not really that it comes from anywhere

do you agree that $a(b + c) = ab + ac$?

[woven radish]

higher!

[heavy current]

this is the distributive law

[scarlet fulcrum]

Yes

[heavy current]

then in our case, a = (k + 1)^2, b = k^2, and c = 4(k + 1)

[scarlet fulcrum]

Ohhh

Oh my gosh

I get it

I see it

Wow

And factoring is like simplifying it it's not getting rid of anything

Which is why the (k+1)^2 is still there

Sorry if that doesn't make sense that's just how I try to make sense out of it in my head but that helps a lot tysm

[heavy current]

glad you finally see it!

[scarlet fulcrum]

So by factoring out (k+1)^2 I essentially just took it out of the k^2 and the 4(k+1)^3

Correct?

[heavy current]

yes

[scarlet fulcrum]

I think you just explained factoring to me 🤩

it never made sense to me
Omg

Thank god I had this epiphany before going into AP calc tysm

[heavy current]

glad I could help

I wish you the best of luck with AP calc

[scarlet fulcrum]

thank you

.close

[devout snow]

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[fallen harness]

for question 1, why do the answers say 3cm?

[pseudo basin]

that formula you got for the box volume looks sus

why are you subtracting the area off?

[fallen harness]

bc it’s an open box

so the top bit is removed

[pseudo basin]

yeah but that doesnt affect the volume

[fallen harness]

oh wait

ur right

[pseudo basin]

it affects the SURFACE AREA if you were writing that down

[fallen harness]

yea

[pseudo basin]

but the volume is the same whether it's an open or a closed box

[fallen harness]

oopsie

is the domain for this question 0 <= x < 8?

[pseudo basin]

i mean, why include one endpoint but exclude the other

[fallen harness]

bc the area of the cardboard can’t be zero

oh but x can’t be zero either

for 2a, how the hell am i meant to get 7500

7500/pir^2

$h = \frac {7500}{\pi r^2}$

[woven radish]

beluwuga

[fallen harness]

ooo it worked

[devout snow]

@fallen harness Has your question been resolved?

[fallen harness]

.close

[devout snow]

Channel closed

Closed by @fallen harness

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[raven plaza]

,rotate

[raven plaza]

I get different equations because of lnC and just C

Both of these are valid right?

[autumn girder]

Yes both of them are equivalent

[raven plaza]

@pseudo basin

What do you think master

They're equivalent right

[autumn girder]

!noping

[devout snow]

Please do not ping individual helpers unprompted.

[raven plaza]

We friends chill

[pseudo basin]

we are not friends.

and i do not appreciate being pinged like this.

[raven plaza]

😢

[raven plaza]

Okay sorry

.close

[devout snow]

Channel closed

Closed by @raven plaza

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